Expand your learning experience with the
Tools for Enriching Calculus CD-ROM The Tools for Enriching Calculus CD-ROM is the ideal complement to Essential Calculus. This innovative learning tool uses a discovery and exploratory approach to help you explore calculus in new ways. Visuals and Modules on the CD-ROM provide geometric visualizations and graphical applications to enrich your understanding of major concepts. Exercises and examples, built from the content in the applets, take a discovery approach, allowing you to explore open-ended questions about the way certain mathematical objects behave. The CD-ROM’s simulation modules include audio explanations of the concept, along with exercises, examples, and instructions. Tools for Enriching Calculus also contains Homework Hints for representative exercises from the text (indicated in blue in the text).
Hours of interactive video instruction!
Interactive Video Skillbuilder CD-ROM The Interactive Video Skillbuilder CD-ROM contains more than eight hours of video instruction.The problems worked during each video lesson are shown first so that you can try working them before watching the solution.To help you evaluate your progress, each section of the text contains a ten-question web quiz (the results of which can be e-mailed to your instructor), and each chapter contains a chapter test, with answers to every problem. Icons in the text direct you to examples that are worked out on the CD-ROM. If you would like to purchase these resources, visit www.cengage.com/highered
REFERENCE PAGES
Cut here and keep for reference
ALGEBRA
G E O M E T RY
ARITHMETIC OPERATIONS
GEOMETRIC FORMULAS
a c ad bc b d bd a b a d ad c b c bc d
ab c ab ac a c ac b b b
Formulas for area A, circumference C, and volume V: Triangle
Circle
Sector of Circle
A 12 bh 12 ab sin
A r 2 C 2 r
A 12 r 2 s r in radians
a
EXPONENTS AND RADICALS xm x mn xn 1 xn n x
x m x n x mn x x m n
mn
n
x y
xyn x n y n
n
n n n xy s xs y s
r
s
¨
b
r
xn yn
Sphere V 43 r 3 A 4 r 2
n n x mn s x m (s x )m
n x 1n s x
r
h
¨
n x x s n y sy
Cylinder V r 2h
Cone V 13 r 2h
r r
h
h
FACTORING SPECIAL POLYNOMIALS
r
x 2 y 2 x yx y x 3 y 3 x yx 2 xy y 2 x 3 y 3 x yx 2 xy y 2
DISTANCE AND MIDPOINT FORMULAS BINOMIAL THEOREM
Distance between P1x1, y1 and P2x 2, y2:
x y2 x 2 2xy y 2
x y2 x 2 2xy y 2
d sx 2 x12 y2 y12
x y3 x 3 3x 2 y 3xy 2 y 3 x y3 x 3 3x 2 y 3xy 2 y 3 x yn x n nx n1y
nn 1 n2 2 x y 2
Midpoint of P1 P2 :
n nk k x y nxy n1 y n k
x1 x 2 y1 y2 , 2 2
LINES
n nn 1 n k 1 where k 1 2 3 k
Slope of line through P1x1, y1 and P2x 2, y2: m
QUADRATIC FORMULA If ax 2 bx c 0, then x
b sb 2 4ac . 2a
y2 y1 x 2 x1
Point-slope equation of line through P1x1, y1 with slope m: y y1 mx x1
INEQUALITIES AND ABSOLUTE VALUE If a b and b c, then a c.
Slope-intercept equation of line with slope m and y-intercept b:
If a b, then a c b c.
y mx b
If a b and c 0, then ca cb. If a b and c 0, then ca cb. If a 0, then
x a x a x a
means
x a or
CIRCLES
x a
Equation of the circle with center h, k and radius r:
means a x a means
x a or
x h2 y k2 r 2
x a 1
REFERENCE PAGES T R I G O N O M E T RY ANGLE MEASUREMENT
FUNDAMENTAL IDENTITIES
radians 180 1
rad 180
180
1 rad
s
r r
in radians
RIGHT ANGLE TRIGONOMETRY
cos tan
hyp csc opp
adj hyp
sec
opp adj
cot
1 sin
sec
1 cos
tan
sin cos
cot
cos sin
cot
1 tan
sin 2 cos 2 1
¨
s r
opp sin hyp
csc
hyp
hyp adj
opp
¨ adj
1 tan 2 sec 2
1 cot 2 csc 2
sin sin
cos cos
tan tan
sin
cos 2
tan
cot 2
adj opp
cos
sin 2
TRIGONOMETRIC FUNCTIONS sin
y r
csc
cos
x r
sec
r x
tan
y x
cot
x y
y
sin A sin B sin C a b c
(x, y)
C c
¨
THE LAW OF COSINES x
y y=sin x
y
b
a 2 b 2 c 2 2bc cos A b 2 a 2 c 2 2ac cos B c 2 a 2 b 2 2ab cos C
y=tan x
A
y=cos x
1
1 π
a
r
GRAPHS OF THE TRIGONOMETRIC FUNCTIONS y
B
THE LAW OF SINES
r y
2π
ADDITION AND SUBTRACTION FORMULAS
2π x
_1
π
2π x
sinx y sin x cos y cos x sin y
x
π
sinx y sin x cos y cos x sin y
_1
cosx y cos x cos y sin x sin y y
y=csc x
y
y=sec x
y
cosx y cos x cos y sin x sin y
y=cot x
1
1 π
2π x
π
π
2π x
2π x
tanx y
tan x tan y 1 tan x tan y
tanx y
tan x tan y 1 tan x tan y
_1
_1
DOUBLE-ANGLE FORMULAS sin 2x 2 sin x cos x
TRIGONOMETRIC FUNCTIONS OF IMPORTANT ANGLES
radians
sin
cos
tan
0 30 45 60 90
0 6 4 3 2
0 12 s22 s32 1
1 s32 s22 12 0
0 s33 1 s3 —
cos 2x cos 2x sin 2x 2 cos 2x 1 1 2 sin 2x tan 2x
2 tan x 1 tan2x
HALF-ANGLE FORMULAS sin 2x
2
1 cos 2x 2
cos 2x
1 cos 2x 2
ESSENTIAL CALCULUS EARLY TRANSCENDENTALS JAMES STEWART McMaster University and University of Toronto
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Essential Calculus: Early Transcendentals James Stewart Publisher: Bob Pirtle Assistant Editor: Stacy Green Editorial Assistant: Magnolia Molcan Technology Project Manager: Earl Perry Senior Marketing Manager: Karin Sandberg Marketing Communications Manager: Darlene Amidon-Brent Project Manager, Editorial Production: Cheryll Linthicum Creative Director: Rob Hugel Senior Art Director: Vernon T. Boes
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Library of Congress Control Number: 2009939223
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ISBN-13: 978-0-538-49739-8
Text Designer: Stephanie Kuhns, Kathi Townes
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Printed in Canada 1 2 3 4 5 6 7 8 9 1 0 1 3 1 2 11 1 0 0 9
K08T09
Trademarks Derive is a registered trademark of Soft Warehouse, Inc. Maple is a registered trademark of Waterloo Maple, Inc. Mathematica is a registered trademark of Wolfram Research, Inc. Tools for Enriching is a trademark used herein under license.
CONTENTS
1
FUNCTIONS AND LIMITS 1.1 1.2 1.3 1.4 1.5 1.6
2
DERIVATIVES 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
3
1
Functions and Their Representations 1 A Catalog of Essential Functions 10 The Limit of a Function 24 Calculating Limits 35 Continuity 45 Limits Involving Infinity 56 Review 69
73 Derivatives and Rates of Change 73 The Derivative as a Function 83 Basic Differentiation Formulas 94 The Product and Quotient Rules 106 The Chain Rule 113 Implicit Differentiation 121 Related Rates 127 Linear Approximations and Differentials Review 138
133
INVERSE FUNCTIONS: Exponential, Logarithmic, and Inverse Trigonometric Functions 142 3.1 3.2 3.3 3.4 3.5 3.6 3.7
Exponential Functions 142 Inverse Functions and Logarithms 148 Derivatives of Logarithmic and Exponential Functions Exponential Growth and Decay 167 Inverse Trigonometric Functions 175 Hyperbolic Functions 181 Indeterminate Forms and L’Hospital’s Rule 187 Review 195
160
iii
iv
■
CONTENTS
4
APPLICATIONS OF DIFFERENTIATION 4.1 4.2 4.3 4.4 4.5 4.6 4.7
5
INTEGRALS 5.1 5.2 5.3 5.4 5.5
6
7
251 Areas and Distances 251 The Definite Integral 262 Evaluating Definite Integrals 274 The Fundamental Theorem of Calculus The Substitution Rule 293 Review 300
SERIES
304
357
Areas between Curves 357 Volumes 362 Volumes by Cylindrical Shells 373 Arc Length 378 Applications to Physics and Engineering Differential Equations 397 Review 407
410 8.1 8.2
284
Integration by Parts 304 Trigonometric Integrals and Substitutions 310 Partial Fractions 320 Integration with Tables and Computer Algebra Systems Approximate Integration 333 Improper Integrals 345 Review 354
APPLICATIONS OF INTEGRATION 7.1 7.2 7.3 7.4 7.5 7.6
8
Maximum and Minimum Values 198 The Mean Value Theorem 205 Derivatives and the Shapes of Graphs 211 Curve Sketching 220 Optimization Problems 226 Newton’s Method 236 Antiderivatives 241 Review 247
TECHNIQUES OF INTEGRATION 6.1 6.2 6.3 6.4 6.5 6.6
198
Sequences 410 Series 420
384
328
CONTENTS
8.3 8.4 8.5 8.6 8.7 8.8
9
PARAMETRIC EQUATIONS AND POLAR COORDINATES 9.1 9.2
9.3 9.4 9.5
10
The Integral and Comparison Tests 429 Other Convergence Tests 437 Power Series 447 Representing Functions as Power Series 452 Taylor and Maclaurin Series 458 Applications of Taylor Polynomials 471 Review 479
Parametric Curves 482 Calculus with Parametric Curves 488 Polar Coordinates 496 Areas and Lengths in Polar Coordinates 504 Conic Sections in Polar Coordinates 509 Review 515
VECTORS AND THE GEOMETRY OF SPACE
517
10.1 Three-Dimensional Coordinate Systems 10.2 Vectors
517
522
10.3 The Dot Product 10.4 10.5 10.6 10.7 10.8 10.9
11
530 The Cross Product 537 Equations of Lines and Planes 545 Cylinders and Quadric Surfaces 553 Vector Functions and Space Curves 559 Arc Length and Curvature 570 Motion in Space: Velocity and Acceleration Review 587
PARTIAL DERIVATIVES
591
11.1 Functions of Several Variables 11.2 Limits and Continuity
11.5 11.6 11.7 11.8
591
601 609 Tangent Planes and Linear Approximations 617 The Chain Rule 625 Directional Derivatives and the Gradient Vector 633 Maximum and Minimum Values 644 Lagrange Multipliers 652 Review 659
11.3 Partial Derivatives 11.4
578
482
■
v
vi
■
CONTENTS
12
MULTIPLE INTEGRALS 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8
13
Double Integrals over Rectangles 663 Double Integrals over General Regions 674 Double Integrals in Polar Coordinates 682 Applications of Double Integrals 688 Triple Integrals 693 Triple Integrals in Cylindrical Coordinates 703 Triple Integrals in Spherical Coordinates 707 Change of Variables in Multiple Integrals 713 Review 722
VECTOR CALCULUS 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9
APPENDIXES A B C D E
INDEX
A83
663
725
Vector Fields 725 Line Integrals 731 The Fundamental Theorem for Line Integrals Green’s Theorem 751 Curl and Divergence 757 Parametric Surfaces and Their Areas 765 Surface Integrals 775 Stokes’ Theorem 786 The Divergence Theorem 791 Review 797
A1 Trigonometry A1 Proofs A10 Sigma Notation A26 The Logarithm Defined as an Integral Answers to Odd-Numbered Exercises
A31 A39
742
PREFACE
This book is a response to those instructors who feel that calculus textbooks are too big. In writing the book I asked myself: What is essential for a three-semester calculus course for scientists and engineers? The book is about two-thirds the size of my other calculus books (Calculus, Fifth Edition and Calculus, Early Transcendentals, Fifth Edition) and yet it contains almost all of the same topics. I have achieved relative brevity mainly by condensing the exposition and by putting some of the features on the website www.stewartcalculus.com. Here, in more detail are some of the ways I have reduced the bulk: ■ I have organized topics in an efficient way and rewritten some sections with briefer exposition. ■ The design saves space. In particular, chapter opening spreads and photographs have been eliminated. ■ The number of examples is slightly reduced. Additional examples are provided online. ■ The number of exercises is somewhat reduced, though most instructors will find that there are plenty. In addition, instructors have access to the archived problems on the website. ■ Although I think projects can be a very valuable experience for students, I have removed them from the book and placed them on the website. ■ A discussion of the principles of problem solving and a collection of challenging problems for each chapter have been moved to the web. Despite the reduced size of the book, there is still a modern flavor: Conceptual understanding and technology are not neglected, though they are not as prominent as in my other books.
CONTENT This book treats the exponential, logarithmic, and inverse trigonometric functions early, in Chapter 3. Those who wish to cover such functions later, with the logarithm defined as an integral, should look at my book titled simply Essential Calculus. CHAPTER 1 FUNCTIONS AND LIMITS After a brief review of the basic functions, limits and continuity are introduced, including limits of trigonometric functions, limits involving infinity, and precise definitions. ■
DERIVATIVES The material on derivatives is covered in two sections in order to give students time to get used to the idea of a derivative as a function. The CHAPTER 2
■
vii
viii
■
PREFACE
formulas for the derivatives of the sine and cosine functions are derived in the section on basic differentiation formulas. Exercises explore the meanings of derivatives in various contexts. CHAPTER 3
■
INVERSE FUNCTIONS: EXPONENTIAL, LOGARITHMIC, AND INVERSE TRIGONOMETRIC FUNCTIONS
Exponential functions are defined first and the number e is defined as a limit. Logarithms are then defined as inverse functions. Applications to exponential growth and decay follow. Inverse trigonometric functions and hyperbolic functions are also covered here. L’Hospital’s Rule is included in this chapter because limits of transcendental functions so often require it. APPLICATIONS OF DIFFERENTIATION The basic facts concerning extreme values and shapes of curves are deduced from the Mean Value Theorem. The section on curve sketching includes a brief treatment of graphing with technology. The section on optimization problems contains a brief discussion of applications to business and economics.
CHAPTER 4
■
CHAPTER 5 INTEGRALS The area problem and the distance problem serve to motivate the definite integral, with sigma notation introduced as needed. (Full coverage of sigma notation is provided in Appendix C.) A quite general definition of the definite integral (with unequal subintervals) is given initially before regular partitions are employed. Emphasis is placed on explaining the meanings of integrals in various contexts and on estimating their values from graphs and tables. ■
TECHNIQUES OF INTEGRATION All the standard methods are covered, as well as computer algebra systems, numerical methods, and improper integrals. CHAPTER 6
■
APPLICATIONS OF INTEGRATION General methods are emphasized. The goal is for students to be able to divide a quantity into small pieces, estimate with Riemann sums, and recognize the limit as an integral. The chapter concludes with an introduction to differential equations, including separable equations and direction fields.
CHAPTER 7
■
SERIES The convergence tests have intuitive justifications as well as formal proofs. The emphasis is on Taylor series and polynomials and their applications to physics. Error estimates include those based on Taylor’s Formula (with Lagrange’s form of the remainder term) and those from graphing devices.
CHAPTER 8
■
CHAPTER 9 PARAMETRIC EQUATIONS AND POLAR COORDINATES This chapter introduces parametric and polar curves and applies the methods of calculus to them. A brief treatment of conic sections in polar coordinates prepares the way for Kepler’s Laws in Chapter 10. ■
VECTORS AND THE GEOMETRY OF SPACE In addition to the material on vectors, dot and cross products, lines, planes, and surfaces, this chapter covers vectorvalued functions, length and curvature of space curves, and velocity and acceleration along space curves, culminating in Kepler’s laws. CHAPTER 10
■
CHAPTER 11 PARTIAL DERIVATIVES In view of the fact that many students have difficulty forming mental pictures of the concepts of this chapter, I’ve placed a special emphasis on graphics to elucidate such ideas as graphs, contour maps, directional derivatives, gradients, and Lagrange multipliers. ■
CHAPTER 12 MULTIPLE INTEGRALS Cylindrical and spherical coordinates are introduced in the context of evaluating triple integrals. ■
PREFACE
■
ix
VECTOR CALCULUS The similarities among the Fundamental Theorem for line integrals, Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem are emphasized. CHAPTER 13
■
WEBSITE The website www.stewartcalculus.com includes the following. ■
Review of Algebra, Analytic Geometry, and Conic Sections
■
Additional Examples
■
Projects
■
Archived Problems (drill exercises that have appeared in previous editions of my other books), together with their solutions
■
Challenge Problems
■
Complex Numbers
■
Graphing Calculators and Computers
■
Lies My Calculator and Computer Told Me
■
Additional Topics (complete with exercise sets): Principles of Problem Solving, Strategy for Integration, Strategy for Testing Series, Fourier Series, Area of a Surface of Revolution, Linear Differential Equations, SecondOrder Linear Differential Equations, Nonhomogeneous Linear Equations, Applications of Second-Order Differential Equations, Using Series to Solve Differential Equations, Complex Numbers, Rotation of Axes
■
Links, for particular topics, to outside web resources
■
History of Mathematics, with links to the better historical websites
ACKNOWLEDGMENTS I thank the following reviewers for their thoughtful comments. Ulrich Albrecht, Auburn University Christopher Butler, Case Western Reserve University Joe Fisher, University of Cincinnati John Goulet, Worchester Polytechnic Institute Irvin Hentzel, Iowa State University Joel Irish, University of Southern Maine Mary Nelson, University of Colorado, Boulder Ed Slaminka, Auburn University Li (Jason) Zhongshan, Georgia State University I also thank Marv Riedesel for accuracy in proofreading and Dan Clegg for detailed discussions on how to achieve brevity. In addition, I thank Kathi Townes, Stephanie
x
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PREFACE
Kuhns, Jenny Turney, and Brian Betsill of TECHarts for their production services and the following Brooks/Cole staff: Cheryll Linthicum, editorial production project manager; Vernon Boes, art director; Karin Sandberg and Darlene Amidon-Brent, marketing team; Earl Perry, technology project manager; Stacy Green, assistant editor; Magnolia Molcan, editorial assistant; Bob Kauser, permissions editor; Rebecca Cross, print/media buyer; and William Stanton, cover designer. They have all done an outstanding job. The idea for this book came from my editor, Bob Pirtle, who had been hearing of the desire for a much shorter calculus text from numerous instructors. I thank him for encouraging me to pursue this idea and for his advice and assistance whenever I needed it. JA M E S S T E WA RT
PREFACE
ANCILLARIES FOR INSTRUCTORS
ANCILLARIES FOR STUDENTS
COMPLETE SOLUTIONS MANUAL ISBN 0495014303
STUDENT SOLUTIONS MANUAL ISBN 049501429X
The Complete Solutions Manual provides worked-out solutions to all of the problems in the text.
The Student Solutions Manual provides completely worked-out solutions to all odd-numbered exercises within the text, giving students a way to check their answers and ensure that they took the correct steps to arrive at an answer.
SOLUTIONS BUILDER CD ISBN 0495106925
This CD is an electronic version of the complete solutions manual. It provides instructors with an efficient method for creating solution sets to homework or exams. Instructors can easily view, select, and save solution sets that can then be printed or posted. TOOLS FOR ENRICHING CALCULUS ISBN 0495107638
TEC contains Visuals and Modules for use as classroom demonstrations. Exercises for each Module allow instructors to make assignments based on the classroom demonstration. TEC also includes Homework Hints for representative exercises. Students can benefit from this additional help when instructors assign these exercises.
■
INTERACTIVE VIDEO SKILLBUILDER CD ISBN 0495113719
The Interactive Video Skillbuilder CD-ROM contains more than eight hours of instruction. To help students evaluate their progress, each section contains a tenquestion web quiz (the results of which can be e-mailed to the instructor) and each chapter contains a chapter test, with the answer to each problem on each test. TOOLS FOR ENRICHING CALCULUS ISBN 0495107638
TEC provides a laboratory environment in which students can enrich their understanding by revisiting and exploring selected topics. TEC also includes Homework Hints for representative exercises.
Ancillaries for students are available for purchase at www.cengage.com
xi
TO THE STUDENT
Reading a calculus textbook is different from reading a newspaper or a novel, or even a physics book. Don’t be discouraged if you have to read a passage more than once in order to understand it. You should have pencil and paper and calculator at hand to sketch a diagram or make a calculation. Some students start by trying their homework problems and read the text only if they get stuck on an exercise. I suggest that a far better plan is to read and understand a section of the text before attempting the exercises. In particular, you should look at the definitions to see the exact meanings of the terms. And before you read each example, I suggest that you cover up the solution and try solving the problem yourself. You’ll get a lot more from looking at the solution if you do so. Part of the aim of this course is to train you to think logically. Learn to write the solutions of the exercises in a connected, step-by-step fashion with explanatory sentences—not just a string of disconnected equations or formulas. The answers to the odd-numbered exercises appear at the back of the book, in Appendix E. Some exercises ask for a verbal explanation or interpretation or description. In such cases there is no single correct way of expressing the answer, so don’t worry that you haven’t found the definitive answer. In addition, there are often several different forms in which to express a numerical or algebraic answer, so if your answer differs from mine, don’t immediately assume you’re wrong. For example, if the answer given in the back of the book is s2 1 and you obtain 1(1 s2 ), then you’re right and rationalizing the denominator will show that the answers are equivalent. The icon ; indicates an exercise that definitely requires the use of either a graphing calculator or a computer with graphing software. But that doesn’t mean that graphing devices can’t be used to check your work on the other exercises as well. The symbol CAS is reserved for problems in which the full resources of a computer algebra system (like Derive, Maple, Mathematica, or the TI-89/92) are required. You will also encounter the symbol | , which warns you against committing an error. I have placed this symbol in the margin in situations where I have observed that a large proportion of my students tend to make the same mistake. xii
The CD-ROM Tools for Enriching™ Calculus is referred to by means of the symbol . It directs you to Visuals and Modules in which you can explore aspects of calculus for which the computer is particularly useful. TEC also provides Homework Hints for representative exercises that are indicated by printing the exercise number in blue: 43. . These homework hints ask you questions that allow you to make progress toward a solution without actually giving you the answer. You need to pursue each hint in an active manner with pencil and paper to work out the details. If a particular hint doesn’t enable you to solve the problem, you can click to reveal the next hint. (See the front endsheet for information on how to purchase this and other useful tools.) The Interactive Video Skillbuilder CD-ROM contains videos of instructors explaining two or three of the examples in every section of the text. (The symbol V has been placed beside these examples in the text.) Also on the CD is a video in which I offer advice on how to succeed in your calculus course. I also want to draw your attention to the website www.stewartcalculus.com. There you will find an Algebra Review (in case your precalculus skills are weak) as well as Additional Examples, Challenging Problems, Projects, Lies My Calculator and Computer Told Me (explaining why calculators sometimes give the wrong answer), History of Mathematics, Additional Topics, chapter quizzes, and links to outside resources. I recommend that you keep this book for reference purposes after you finish the course. Because you will likely forget some of the specific details of calculus, the book will serve as a useful reminder when you need to use calculus in subsequent courses. And, because this book contains more material than can be covered in any one course, it can also serve as a valuable resource for a working scientist or engineer. Calculus is an exciting subject, justly considered to be one of the greatest achievements of the human intellect. I hope you will discover that it is not only useful but also intrinsically beautiful. JA M E S S T E WA RT
1
FUNCTIONS AND LIMITS Calculus is fundamentally different from the mathematics that you have studied previously. Calculus is less static and more dynamic. It is concerned with change and motion; it deals with quantities that approach other quantities. So in this first chapter we begin our study of calculus by investigating how the values of functions change and approach limits.
1.1
Year
Population (millions)
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
1650 1750 1860 2070 2300 2560 3040 3710 4450 5280 6080
FUNCTIONS AND THEIR REPRESENTATONS Functions arise whenever one quantity depends on another. Consider the following four situations. A. The area A of a circle depends on the radius r of the circle. The rule that connects r and A is given by the equation A r 2. With each positive number r there is associated one value of A, and we say that A is a function of r. B. The human population of the world P depends on the time t. The table gives estimates of the world population Pt at time t, for certain years. For instance, P1950 2,560,000,000 But for each value of the time t there is a corresponding value of P, and we say that P is a function of t. C. The cost C of mailing a first-class letter depends on the weight w of the letter. Although there is no simple formula that connects w and C, the post office has a rule for determining C when w is known. D. The vertical acceleration a of the ground as measured by a seismograph during an earthquake is a function of the elapsed time t. Figure 1 shows a graph generated by seismic activity during the Northridge earthquake that shook Los Angeles in 1994. For a given value of t, the graph provides a corresponding value of a. a {cm/s@} 100
50
5
FIGURE 1
Vertical ground acceleration during the Northridge earthquake
10
15
20
25
30
t (seconds)
_50 Calif. Dept. of Mines and Geology
Each of these examples describes a rule whereby, given a number (r, t, w, or t), another number ( A, P, C, or a) is assigned. In each case we say that the second number is a function of the first number. 1
2
■
CHAPTER 1
FUNCTIONS AND LIMITS
A function f is a rule that assigns to each element x in a set A exactly one element, called f x, in a set B.
x (input)
f
ƒ (output)
FIGURE 2
Machine diagram for a function ƒ
x
ƒ a
f(a)
f
A
We usually consider functions for which the sets A and B are sets of real numbers. The set A is called the domain of the function. The number f x is the value of f at x and is read “ f of x.” The range of f is the set of all possible values of f x as x varies throughout the domain. A symbol that represents an arbitrary number in the domain of a function f is called an independent variable. A symbol that represents a number in the range of f is called a dependent variable. In Example A, for instance, r is the independent variable and A is the dependent variable. It’s helpful to think of a function as a machine (see Figure 2). If x is in the domain of the function f, then when x enters the machine, it’s accepted as an input and the machine produces an output f x according to the rule of the function. Thus we can think of the domain as the set of all possible inputs and the range as the set of all possible outputs. Another way to picture a function is by an arrow diagram as in Figure 3. Each arrow connects an element of A to an element of B. The arrow indicates that f x is associated with x, f a is associated with a, and so on. The most common method for visualizing a function is its graph. If f is a function with domain A, then its graph is the set of ordered pairs
x, f x
B
x A
(Notice that these are input-output pairs.) In other words, the graph of f consists of all points x, y in the coordinate plane such that y f x and x is in the domain of f . The graph of a function f gives us a useful picture of the behavior or “life history” of a function. Since the y-coordinate of any point x, y on the graph is y f x, we can read the value of f x from the graph as being the height of the graph above the point x. (See Figure 4.) The graph of f also allows us to picture the domain of f on the x-axis and its range on the y-axis as in Figure 5.
FIGURE 3
Arrow diagram for ƒ
y
y
{ x, ƒ}
y ⫽ ƒ(x)
range
ƒ f (2) f(1) 0
1
2
x
x
x
0
domain FIGURE 4
FIGURE 5
y
EXAMPLE 1 The graph of a function f is shown in Figure 6.
(a) Find the values of f 1 and f 5. (b) What are the domain and range of f ? 1 0
FIGURE 6
SOLUTION 1
x
(a) We see from Figure 6 that the point 1, 3 lies on the graph of f , so the value of f at 1 is f 1 3. (In other words, the point on the graph that lies above x 1 is 3 units above the x-axis.) When x 5, the graph lies about 0.7 unit below the x-axis, so we estimate that f 5 0.7.
SECTION 1.1
■ The notation for intervals is given on Reference Page 3. The Reference Pages are located at the front and back of the book.
FUNCTIONS AND THEIR REPRESENTATIONS
■
3
(b) We see that f x is defined when 0 x 7, so the domain of f is the closed interval 0, 7 . Notice that f takes on all values from 2 to 4, so the range of f is
y
2 y 4 2, 4
■
REPRESENTATIONS OF FUNCTIONS
There are four possible ways to represent a function: ■ ■
verbally (by a description in words) numerically (by a table of values)
■ ■
visually (by a graph) algebraically (by an explicit formula)
If a single function can be represented in all four ways, it is often useful to go from one representation to another to gain additional insight into the function. But certain functions are described more naturally by one method than by another. With this in mind, let’s reexamine the four situations that we considered at the beginning of this section. A. The most useful representation of the area of a circle as a function of its radius
is probably the algebraic formula Ar r 2, though it is possible to compile a table of values or to sketch a graph (half a parabola). Because a circle has to have a positive radius, the domain is r r 0 0, , and the range is also 0, .
Year
Population (millions)
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
1650 1750 1860 2070 2300 2560 3040 3710 4450 5280 6080
B. We are given a description of the function in words: Pt is the human population
of the world at time t. The table of values of world population provides a convenient representation of this function. If we plot these values, we get the graph (called a scatter plot) in Figure 7. It too is a useful representation; the graph allows us to absorb all the data at once. What about a formula? Of course, it’s impossible to devise an explicit formula that gives the exact human population Pt at any time t. But it is possible to find an expression for a function that approximates Pt. In fact, we could use a graphing calculator with exponential regression capabilities to obtain the approximation Pt f t 0.008079266 1.013731t and Figure 8 shows that it is a reasonably good “fit.” The function f is called a mathematical model for population growth. In other words, it is a function with an explicit formula that approximates the behavior of our given function. We will see, however, that the ideas of calculus can be applied to a table of values; an explicit formula is not necessary.
P
P
6x10'
6x10'
1900
1920
1940
1960
1980
2000 t
FIGURE 7 Scatter plot of data points for population growth
1900
1920
1940
1960
1980
2000 t
FIGURE 8 Graph of a mathematical model for population growth
4
■
CHAPTER 1
FUNCTIONS AND LIMITS
■ A function defined by a table of values is called a tabular function.
w (ounces)
Cw (dollars)
0w 1 1w 2 2w 3 3w 4 4w 5
0.39 0.63 0.87 1.11 1.35
12 w 13
3.27
The function P is typical of the functions that arise whenever we attempt to apply calculus to the real world. We start with a verbal description of a function. Then we may be able to construct a table of values of the function, perhaps from instrument readings in a scientific experiment. Even though we don’t have complete knowledge of the values of the function, we will see throughout the book that it is still possible to perform the operations of calculus on such a function. C. Again the function is described in words: Cw is the cost of mailing a first-class letter with weight w. The rule that the US Postal Service used as of 2006 is as follows: The cost is 39 cents for up to one ounce, plus 24 cents for each successive ounce up to 13 ounces. The table of values shown in the margin is the most convenient representation for this function, though it is possible to sketch a graph (see Example 6). D. The graph shown in Figure 1 is the most natural representation of the vertical acceleration function at. It’s true that a table of values could be compiled, and it is even possible to devise an approximate formula. But everything a geologist needs to know—amplitudes and patterns—can be seen easily from the graph. (The same is true for the patterns seen in electrocardiograms of heart patients and polygraphs for lie-detection.) In the next example we sketch the graph of a function that is defined verbally. EXAMPLE 2 When you turn on a hot-water faucet, the temperature T of the water
depends on how long the water has been running. Draw a rough graph of T as a function of the time t that has elapsed since the faucet was turned on.
T
0
FIGURE 9
t
SOLUTION The initial temperature of the running water is close to room temperature because the water has been sitting in the pipes. When the water from the hotwater tank starts flowing from the faucet, T increases quickly. In the next phase, T is constant at the temperature of the heated water in the tank. When the tank is drained, T decreases to the temperature of the water supply. This enables us to make the rough sketch of T as a function of t in Figure 9. ■
EXAMPLE 3 Find the domain of each function.
(a) f x sx 2
(b) tx
1 x x 2
SOLUTION If a function is given by a formula and the domain is not stated explicitly, the convention is that the domain is the set of all numbers for which the formula makes sense and defines a real number. ■
(a) Because the square root of a negative number is not defined (as a real number), the domain of f consists of all values of x such that x 2 0. This is equivalent to x 2, so the domain is the interval 2, . (b) Since 1 1 tx 2 x x xx 1 and division by 0 is not allowed, we see that tx is not defined when x 0 or x 1. Thus the domain of t is x x 0, x 1 , which could also be written in interval notation as , 0 0, 1 1, . ■
The graph of a function is a curve in the xy-plane. But the question arises: Which curves in the xy-plane are graphs of functions? This is answered by the following test. THE VERTICAL LINE TEST A curve in the xy-plane is the graph of a function of
x if and only if no vertical line intersects the curve more than once.
SECTION 1.1
■
FUNCTIONS AND THEIR REPRESENTATIONS
5
The reason for the truth of the Vertical Line Test can be seen in Figure 10. If each vertical line x a intersects a curve only once, at a, b, then exactly one functional value is defined by f a b. But if a line x a intersects the curve twice, at a, b and a, c, then the curve can’t represent a function because a function can’t assign two different values to a. y
y
x=a
(a, c)
x=a
(a, b) (a, b) x
a
0
a
0
x
FIGURE 10
PIECEWISE DEFINED FUNCTIONS
The functions in the following three examples are defined by different formulas in different parts of their domains. V EXAMPLE 4
A function f is defined by f x
1 x if x 1 x2 if x 1
Evaluate f 0, f 1, and f 2 and sketch the graph. SOLUTION Remember that a function is a rule. For this particular function the rule is the following: First look at the value of the input x. If it happens that x 1, then the value of f x is 1 x. On the other hand, if x 1, then the value of f x is x 2.
Since 0 1, we have f 0 1 0 1. Since 1 1, we have f 1 1 1 0.
y
Since 2 1, we have f 2 2 2 4.
1
1
x
FIGURE 11
How do we draw the graph of f ? We observe that if x 1, then f x 1 x, so the part of the graph of f that lies to the left of the vertical line x 1 must coincide with the line y 1 x, which has slope 1 and y-intercept 1. If x 1, then f x x 2, so the part of the graph of f that lies to the right of the line x 1 must coincide with the graph of y x 2, which is a parabola. This enables us to sketch the graph in Figure l1. The solid dot indicates that the point 1, 0 is included on the graph; the open dot indicates that the point 1, 1 is excluded from the graph. ■ The next example of a piecewise defined function is the absolute value function. Recall that the absolute value of a number a, denoted by a , is the distance from a to 0 on the real number line. Distances are always positive or 0, so we have
www.stewartcalculus.com For a more extensive review of absolute values, click on Review of Algebra. ■
For example,
3 3
3 3
a 0 0 0
for every number a
s2 1 s2 1
3 3
6
■
CHAPTER 1
FUNCTIONS AND LIMITS
In general, we have
a a a a
if a 0 if a 0
(Remember that if a is negative, then a is positive.)
EXAMPLE 5 Sketch the graph of the absolute value function f x x .
y
SOLUTION From the preceding discussion we know that
y=| x |
x 0
x
x x
if x 0 if x 0
Using the same method as in Example 4, we see that the graph of f coincides with the line y x to the right of the y-axis and coincides with the line y x to the left of the y-axis (see Figure 12). ■
FIGURE 12
EXAMPLE 6 In Example C at the beginning of this section we considered the cost
Cw of mailing a first-class letter with weight w. In effect, this is a piecewise defined function because, from the table of values, we have
C 1
0
1
FIGURE 13
2
3
4
5
0.39 0.63 Cw 0.87 1.11
w
if if if if
0w 1w 2w 3w
1 2 3 4
The graph is shown in Figure 13. You can see why functions similar to this one are called step functions—they jump from one value to the next. ■ SYMMETRY
If a function f satisfies f x f x for every number x in its domain, then f is called an even function. For instance, the function f x x 2 is even because f x x2 x 2 f x The geometric significance of an even function is that its graph is symmetric with respect to the y-axis (see Figure 14). This means that if we have plotted the graph of y
y
f(_x)
ƒ _x
0
x
FIGURE 14 An even function
_x x
ƒ
0 x
FIGURE 15 An odd function
x
SECTION 1.1
f
_1
■
7
f for x 0, we obtain the entire graph simply by reflecting this portion about the y-axis. If f satisfies f x f x for every number x in its domain, then f is called an odd function. For example, the function f x x 3 is odd because
y 1
FUNCTIONS AND THEIR REPRESENTATIONS
x
1
f x x3 x 3 f x
_1
The graph of an odd function is symmetric about the origin (see Figure 15 on page 6). If we already have the graph of f for x 0, we can obtain the entire graph by rotating this portion through 180 about the origin.
(a) y 1
V EXAMPLE 7 Determine whether each of the following functions is even, odd, or neither even nor odd. (a) f x x 5 x (b) tx 1 x 4 (c) hx 2x x 2
g 1
SOLUTION
x
f x x5 x 15x 5 x
(a)
x 5 x x 5 x f x
(b)
Therefore, f is an odd function.
y
tx 1 x4 1 x 4 tx
(b)
h
1
So t is even. x
1
(c)
hx 2x x2 2x x 2
Since hx hx and hx hx, we conclude that h is neither even nor odd.
■
(c)
The graphs of the functions in Example 7 are shown in Figure 16. Notice that the graph of h is symmetric neither about the y-axis nor about the origin.
FIGURE 16
INCREASING AND DECREASING FUNCTIONS y
B
The graph shown in Figure 17 rises from A to B, falls from B to C, and rises again from C to D. The function f is said to be increasing on the interval a, b , decreasing on b, c , and increasing again on c, d . Notice that if x 1 and x 2 are any two numbers between a and b with x 1 x 2, then f x 1 f x 2 . We use this as the defining property of an increasing function.
D
y=ƒ C f(x™) A
f(x¡)
A function f is called increasing on an interval I if 0 a x¡
FIGURE 17
x™
b
c
d
x
f x 1 f x 2
whenever x 1 x 2 in I
It is called decreasing on I if f x 1 f x 2
whenever x 1 x 2 in I
8
■
CHAPTER 1
FUNCTIONS AND LIMITS
In the definition of an increasing function it is important to realize that the inequality f x 1 f x 2 must be satisfied for every pair of numbers x 1 and x 2 in I with x 1 x 2. You can see from Figure 18 that the function f x x 2 is decreasing on the interval , 0 and increasing on the interval 0, . y
y=≈
1.1
EXERCISES
1. The graph of a function f is given.
(a) (b) (c) (d) (e) (f )
x
0
FIGURE 18
■ Determine whether the curve is the graph of a function of x. If it is, state the domain and range of the function.
3–6
State the value of f 1. Estimate the value of f 2. For what values of x is f x 2? Estimate the values of x such that f x 0. State the domain and range of f . On what interval is f increasing?
y
3.
y
4.
1
1 0
0
x
1
1
x
1
x
y
1 0
y
5.
y
6.
x
1
1
1 0
1
0
x
2. The graphs of f and t are given.
(a) (b) (c) (d) (e) (f )
State the values of f 4 and t3. For what values of x is f x tx? Estimate the solution of the equation f x 1. On what interval is f decreasing? State the domain and range of f. State the domain and range of t.
■
■
■
■
■
■
■
■
■
■
7. The graph shown gives the weight of a certain person as a
function of age. Describe in words how this person’s weight varies over time. What do you think happened when this person was 30 years old?
y 200
g f
Weight (pounds)
2
0
2
x
■
150 100 50 0
10
20 30 40
50
60 70
Age (years)
SECTION 1.1
8. The graph shown gives a salesman’s distance from his home
as a function of time on a certain day. Describe in words what the graph indicates about his travels on this day.
8 AM
■ Evaluate the difference quotient for the given function. Simplify your answer.
f 3 h f 3 h
f a h f a h
20. f x x 3,
10
NOON
2
4
6 PM
Time (hours)
water, and then let the glass sit on a table. Describe how the temperature of the water changes as time passes. Then sketch a rough graph of the temperature of the water as a function of the elapsed time. 10. Sketch a rough graph of the number of hours of daylight as
a function of the time of year. 11. Sketch a rough graph of the outdoor temperature as a func-
tion of time during a typical spring day. 12. Sketch a rough graph of the market value of a new car as a
function of time for a period of 20 years. Assume the car is well maintained. 13. Sketch the graph of the amount of a particular brand of cof-
fee sold by a store as a function of the price of the coffee.
1 , x
22. f x
x3 , x1
15. A homeowner mows the lawn every Wednesday afternoon.
Sketch a rough graph of the height of the grass as a function of time over the course of a four-week period. 16. A jet takes off from an airport and lands an hour later at
another airport, 400 miles away. If t represents the time in minutes since the plane has left the terminal, let xt be the horizontal distance traveled and yt be the altitude of the plane. (a) Sketch a possible graph of xt. (b) Sketch a possible graph of yt. (c) Sketch a possible graph of the ground speed. (d) Sketch a possible graph of the vertical velocity. 17. If f x 3x 2 x 2, find f 2, f 2, f a, f a,
■
23–27
■
■
Vr 43 r 3 . Find a function that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r 1 inches.
■
■
■
■
■
■
■
■
■
■
■
■
Find the domain of the function. x 3x 1
24. f x
5x 4 x 2 3x 2
3 t 25. f t st s
26. tu su s4 u 27. hx ■
■
1 4 x 2 5x s ■
■
■
■
■
28. Find the domain and range and sketch the graph of the
function hx s4 x 2 . 29– 40
■
Find the domain and sketch the graph of the function.
29. f x 5
30. Fx 2 x 3
31. f t t 2 6t
32. Ht
33. tx sx 5
34. Fx 2x 1
35. Gx 37. f x 38. f x 39. f x
f a 1, 2 f a, f 2a, f a 2 , [ f a] 2, and f a h.
18. A spherical balloon with radius r inches has volume
f x f 1 x1
23. f x
14. You place a frozen pie in an oven and bake it for an
hour. Then you take it out and let it cool before eating it. Describe how the temperature of the pie changes as time passes. Then sketch a rough graph of the temperature of the pie as a function of time.
f x f a xa
21. f x
■
9. You put some ice cubes in a glass, fill the glass with cold
9
19–22
19. f x 4 3x x 2 ,
Distance from home (miles)
■
FUNCTIONS AND THEIR REPRESENTATIONS
40. f x
■
■
1
3x x x
x2 1x 3 12 x 2x 5
36.
4 t2 2t
x tx
x2
if x 0 if x 0 if x 2 if x 2
x 2 if x 1 x2 if x 1
1 if x 1 3x 2 if x 1 7 2x if x 1
■
■
■
■
■
■
■
■
■
10
■
CHAPTER 1
FUNCTIONS AND LIMITS
■ Find an expression for the function whose graph is the given curve.
stairs. Give two other examples of step functions that arise in everyday life.
41– 44
41. The line segment joining the points 2, 1 and 4, 6 42. The line segment joining the points 3, 2 and 6, 3
53–54 ■ Graphs of f and t are shown. Decide whether each function is even, odd, or neither. Explain your reasoning.
43. The bottom half of the parabola x y 12 0
53.
54.
y
y
g
44. The top half of the circle x 1 y 1 2
2
f
f ■
■
45– 49
■
■
■
■
■
■
■
■
■
■
Find a formula for the described function and state its
x
x g
domain. 45. A rectangle has perimeter 20 m. Express the area of the
rectangle as a function of the length of one of its sides. ■
■
■
■
■
■
■
■
■
■
■
46. A rectangle has area 16 m2. Express the perimeter of the
rectangle as a function of the length of one of its sides.
55. (a) If the point 5, 3 is on the graph of an even function,
what other point must also be on the graph? (b) If the point 5, 3 is on the graph of an odd function, what other point must also be on the graph?
47. Express the area of an equilateral triangle as a function of
the length of a side. 48. Express the surface area of a cube as a function of its
56. A function f has domain 5, 5 and a portion of its graph
volume.
is shown. (a) Complete the graph of f if it is known that f is even. (b) Complete the graph of f if it is known that f is odd.
3
49. An open rectangular box with volume 2 m has a square
base. Express the surface area of the box as a function of the length of a side of the base. ■
■
■
■
■
■
■
■
■
■
y
■
50. A taxi company charges two dollars for the first mile (or
part of a mile) and 20 cents for each succeeding tenth of a mile (or part). Express the cost C (in dollars) of a ride as a function of the distance x traveled (in miles) for 0 x 2, and sketch the graph of this function.
0
_5
x
5
51. In a certain country, income tax is assessed as follows.
There is no tax on income up to $10,000. Any income over $10,000 is taxed at a rate of 10%, up to an income of $20,000. Any income over $20,000 is taxed at 15%. (a) Sketch the graph of the tax rate R as a function of the income I. (b) How much tax is assessed on an income of $14,000? On $26,000? (c) Sketch the graph of the total assessed tax T as a function of the income I. 52. The functions in Example 6 and Exercises 50 and 51(a)
are called step functions because their graphs look like
1.2
57–62 ■ Determine whether f is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually.
x2 x 1
57. f x
x x 1
58. f x
59. f x
x x1
60. f x x x
2
61. f x 1 3x 2 x 4 ■
■
■
4
■
■
62. f x 1 3x 3 x 5 ■
■
■
■
■
■
A CATALOG OF ESSENTIAL FUNCTIONS In solving calculus problems you will find that it is helpful to be familiar with the graphs of some commonly occurring functions. These same basic functions are often used to model real-world phenomena, so we begin with a discussion of mathematical modeling. We also review briefly how to transform these functions by shifting, stretching, and reflecting their graphs as well as how to combine pairs of functions by the standard arithmetic operations and by composition.
SECTION 1.2
A CATALOG OF ESSENTIAL FUNCTIONS
■
11
MATHEMATICAL MODELING
A mathematical model is a mathematical description (often by means of a function or an equation) of a real-world phenomenon such as the size of a population, the demand for a product, the speed of a falling object, the concentration of a product in a chemical reaction, the life expectancy of a person at birth, or the cost of emission reductions. The purpose of the model is to understand the phenomenon and perhaps to make predictions about future behavior. Figure 1 illustrates the process of mathematical modeling. Given a real-world problem, our first task is to formulate a mathematical model by identifying and naming the independent and dependent variables and making assumptions that simplify the phenomenon enough to make it mathematically tractable. We use our knowledge of the physical situation and our mathematical skills to obtain equations that relate the variables. In situations where there is no physical law to guide us, we may need to collect data (either from a library or the Internet or by conducting our own experiments) and examine the data in the form of a table in order to discern patterns. From this numerical representation of a function we may wish to obtain a graphical representation by plotting the data. The graph might even suggest a suitable algebraic formula in some cases. Real-world problem
Formulate
Mathematical model
Solve
Mathematical conclusions
Interpret
Real-world predictions
Test FIGURE 1 The modeling process
The second stage is to apply the mathematics that we know (such as the calculus that will be developed throughout this book) to the mathematical model that we have formulated in order to derive mathematical conclusions. Then, in the third stage, we take those mathematical conclusions and interpret them as information about the original real-world phenomenon by way of offering explanations or making predictions. The final step is to test our predictions by checking against new real data. If the predictions don’t compare well with reality, we need to refine our model or to formulate a new model and start the cycle again. A mathematical model is never a completely accurate representation of a physical situation—it is an idealization. A good model simplifies reality enough to permit mathematical calculations but is accurate enough to provide valuable conclusions. It is important to realize the limitations of the model. In the end, Mother Nature has the final say. There are many different types of functions that can be used to model relationships observed in the real world. In what follows, we discuss the behavior and graphs of these functions and give examples of situations appropriately modeled by such functions. ■
www.stewartcalculus.com To review the coordinate geometry of lines, click on Review of Analytic Geometry. ■
LINEAR MODELS
When we say that y is a linear function of x, we mean that the graph of the function is a line, so we can use the slope-intercept form of the equation of a line to write a formula for the function as y f x mx b where m is the slope of the line and b is the y-intercept.
12
■
CHAPTER 1
FUNCTIONS AND LIMITS
A characteristic feature of linear functions is that they grow at a constant rate. For instance, Figure 2 shows a graph of the linear function f x 3x 2 and a table of sample values. Notice that whenever x increases by 0.1, the value of f x increases by 0.3. So f x increases three times as fast as x. Thus the slope of the graph y 3x 2, namely 3, can be interpreted as the rate of change of y with respect to x. y
y=3x-2
0
x
_2
x
f x 3x 2
1.0 1.1 1.2 1.3 1.4 1.5
1.0 1.3 1.6 1.9 2.2 2.5
FIGURE 2
V EXAMPLE 1
(a) As dry air moves upward, it expands and cools. If the ground temperature is 20 C and the temperature at a height of 1 km is 10 C, express the temperature T (in °C) as a function of the height h (in kilometers), assuming that a linear model is appropriate. (b) Draw the graph of the function in part (a). What does the slope represent? (c) What is the temperature at a height of 2.5 km? SOLUTION
(a) Because we are assuming that T is a linear function of h, we can write T mh b We are given that T 20 when h 0, so 20 m 0 b b In other words, the y-intercept is b 20. We are also given that T 10 when h 1, so 10 m 1 20
T
The slope of the line is therefore m 10 20 10 and the required linear function is T 10h 20
20
T=_10h+20 10
0
1
FIGURE 3
3
h
(b) The graph is sketched in Figure 3. The slope is m 10 Ckm, and this represents the rate of change of temperature with respect to height. (c) At a height of h 2.5 km, the temperature is T 102.5 20 5 C
■
SECTION 1.2
■
A CATALOG OF ESSENTIAL FUNCTIONS
■
13
POLYNOMIALS
A function P is called a polynomial if Px a n x n a n1 x n1 a 2 x 2 a 1 x a 0 where n is a nonnegative integer and the numbers a 0 , a 1, a 2 , . . . , a n are constants called the coefficients of the polynomial. The domain of any polynomial is ⺢ , . If the leading coefficient a n 0, then the degree of the polynomial is n. For example, the function Px 2x 6 x 4 25 x 3 s2 is a polynomial of degree 6. A polynomial of degree 1 is of the form Px mx b and so it is a linear function. A polynomial of degree 2 is of the form Px ax 2 bx c and is called a quadratic function. Its graph is always a parabola obtained by shifting the parabola y ax 2. The parabola opens upward if a 0 and downward if a 0. (See Figure 4.) y
y
2 2
x
1 0
FIGURE 4
The graphs of quadratic functions are parabolas.
1
x
(b) y=_2≈+3x+1
(a) y=≈+x+1
A polynomial of degree 3 is of the form Px ax 3 bx 2 cx d
a0
and is called a cubic function. Figure 5 shows the graph of a cubic function in part (a) and graphs of polynomials of degrees 4 and 5 in parts (b) and (c). We will see later why the graphs have these shapes. y
y
1
2
0
FIGURE 5
y 20 1
1
(a) y=˛-x+1
x
x
(b) y=x$-3≈+x
1
x
(c) y=3x%-25˛+60x
Polynomials are commonly used to model various quantities that occur in the natural and social sciences. For instance, in Chapter 2 we will explain why economists often use a polynomial Px to represent the cost of producing x units of a commodity.
14
■
CHAPTER 1
FUNCTIONS AND LIMITS
■
POWER FUNCTIONS
A function of the form f x x a, where a is a constant, is called a power function. We consider several cases. (i) a n, where n is a positive integer
The graphs of f x x n for n 1, 2, 3, 4, and 5 are shown in Figure 6. (These are polynomials with only one term.) You are familiar with the shape of the graphs of y x (a line through the origin with slope 1) and y x 2 (a parabola). y
y=x
0
1
x
0
y=x #
y
1
1
FIGURE 6
y=≈
y
x
0
1
x
y=x%
y
1
1
1
y=x$
y
1
0
1
x
0
x
1
Graphs of ƒ=x n for n=1, 2, 3, 4, 5
The general shape of the graph of f x x n depends on whether n is even or odd. If n is even, then f x x n is an even function and its graph is similar to the parabola y x 2. If n is odd, then f x x n is an odd function and its graph is similar to that of y x 3. Notice from Figure 7, however, that as n increases, the graph of y x n becomes flatter near 0 and steeper when x 1. (If x is small, then x 2 is smaller, x 3 is even smaller, x 4 is smaller still, and so on.)
y
y
y=x $ y=x ^
y=x # y=≈
(_1, 1)
FIGURE 7
Families of power functions
(1, 1) y=x %
(1, 1)
x
0
(_1, _1) x
0
(ii) a 1n, where n is a positive integer n The function f x x 1n s x is a root function. For n 2 it is the square root function f x sx , whose domain is 0, and whose graph is the upper half of the parabola x y 2. [See Figure 8(a).] For other even values of n, the graph of n ys x is similar to that of y sx . For n 3 we have the cube root function
y
y
(1, 1) 0
(1, 1) x
0
FIGURE 8
Graphs of root functions
x (a) ƒ=œ„
x (b) ƒ=Œ„
x
SECTION 1.2
A CATALOG OF ESSENTIAL FUNCTIONS
15
■
3 f x s x whose domain is ⺢ (recall that every real number has a cube root) and n whose graph is shown in Figure 8(b). The graph of y s x for n odd n 3 is 3 similar to that of y sx .
(iii) a 1
y
The graph of the reciprocal function f x x 1 1x is shown in Figure 9. Its graph has the equation y 1x, or xy 1, and is a hyperbola with the coordinate axes as its asymptotes. This function arises in physics and chemistry in connection with Boyle’s Law, which says that, when the temperature is constant, the volume V of a gas is inversely proportional to the pressure P:
y=Δ 1 0
x
1
V FIGURE 9
C P
where C is a constant. Thus the graph of V as a function of P has the same general shape as the right half of Figure 9.
The reciprocal function
■
y
RATIONAL FUNCTIONS
A rational function f is a ratio of two polynomials: 20
f x
0
x
2
Px Qx
where P and Q are polynomials. The domain consists of all values of x such that Qx 0. A simple example of a rational function is the function f x 1x, whose domain is x x 0 ; this is the reciprocal function graphed in Figure 9. The function
f x
FIGURE 10
ƒ=
2x$-≈+1 ≈-4
is a rational function with domain x ■
2x 4 x 2 1 x2 4
x 2 . Its graph is shown in Figure 10.
TRIGONOMETRIC FUNCTIONS
Trigonometry and the trigonometric functions are reviewed on Reference Page 2 and also in Appendix A. In calculus the convention is that radian measure is always used (except when otherwise indicated). For example, when we use the function f x sin x, it is understood that sin x means the sine of the angle whose radian measure is x. Thus the graphs of the sine and cosine functions are as shown in Figure 11. y _ _π
π 2
y 3π 2
1 0 _1
π 2
π
_π 2π
5π 2
3π
_
π 2
π 0
x _1
(a) ƒ=sin x FIGURE 11
1 π 2
3π 3π 2
2π
5π 2
x
(b) ©=cos x
Notice that for both the sine and cosine functions the domain is , and the range is the closed interval 1, 1 . Thus, for all values of x, we have 1 sin x 1
1 cos x 1
16
■
CHAPTER 1
FUNCTIONS AND LIMITS
or, in terms of absolute values,
sin x 1
cos x 1
Also, the zeros of the sine function occur at the integer multiples of ; that is, sin x 0
when
x n
n an integer
An important property of the sine and cosine functions is that they are periodic functions and have period 2. This means that, for all values of x, sinx 2 sin x
y
The periodic nature of these functions makes them suitable for modeling repetitive phenomena such as tides, vibrating springs, and sound waves. The tangent function is related to the sine and cosine functions by the equation
1 _
cosx 2 cos x
0
3π _π π _ 2 2
π 2
3π 2
π
x
tan x
sin x cos x
and its graph is shown in Figure 12. It is undefined whenever cos x 0, that is, when x 2, 32, . . . . Its range is , . Notice that the tangent function has period :
FIGURE 12
y=tan x
tanx tan x y
0
The remaining three trigonometric functions (cosecant, secant, and cotangent) are the reciprocals of the sine, cosine, and tangent functions. Their graphs are shown in Appendix A.
y
1
1 0
x
1
(a) y=2®
■
x
1
(b) y=(0.5)®
FIGURE 13 y
y=log™ x y=log£ x
1
0
1
for all x
x
EXPONENTIAL FUNCTIONS AND LOGARITHMS
The exponential functions are the functions of the form f x a x, where the base a is a positive constant. The graphs of y 2 x and y 0.5 x are shown in Figure 13. In both cases the domain is , and the range is 0, . Exponential functions will be studied in detail in Section 3.1, and we will see that they are useful for modeling many natural phenomena, such as population growth (if a 1) and radioactive decay (if a 1. The logarithmic functions f x log a x, where the base a is a positive constant, are the inverse functions of the exponential functions. They will be studied in Section 3.2. Figure 14 shows the graphs of four logarithmic functions with various bases. In each case the domain is 0, , the range is , , and the function increases slowly when x 1.
y=log∞ x y=log¡¸ x TRANSFORMATIONS OF FUNCTIONS
FIGURE 14
By applying certain transformations to the graph of a given function we can obtain the graphs of certain related functions. This will give us the ability to sketch the graphs of many functions quickly by hand. It will also enable us to write equations for given
SECTION 1.2
■
A CATALOG OF ESSENTIAL FUNCTIONS
17
graphs. Let’s first consider translations. If c is a positive number, then the graph of y f x c is just the graph of y f x shifted upward a distance of c units (because each y-coordinate is increased by the same number c). Likewise, if tx f x c, where c 0, then the value of t at x is the same as the value of f at x c (c units to the left of x). Therefore, the graph of y f x c is just the graph of y f x shifted c units to the right.
■ Figure 15 illustrates these shifts by showing how the graph of y x 3 2 1 is obtained from the graph of the parabola y x 2 : Shift 3 units to the left and 1 unit upward.
y y=≈
VERTICAL AND HORIZONTAL SHIFTS Suppose c 0. To obtain the graph of (_3, 1) _3
1 0
x
FIGURE 15
y f x c, shift the graph of y f x a distance c units upward y f x c, shift the graph of y f x a distance c units downward y f x c, shift the graph of y f x a distance c units to the right y f x c, shift the graph of y f x a distance c units to the left
y=(x+3)@+1
Now let’s consider the stretching and reflecting transformations. If c 1, then the graph of y cf x is the graph of y f x stretched by a factor of c in the vertical direction (because each y-coordinate is multiplied by the same number c). The graph of y f x is the graph of y f x reflected about the x-axis because the point x, y is replaced by the point x, y. The following chart also incorporates the results of other stretching, compressing, and reflecting transformations. VERTICAL AND HORIZONTAL STRETCHING AND REFLECTING
Suppose c 1. To obtain the graph of y cf x, stretch the graph of y f x vertically by a factor of c y 1cf x, compress the graph of y f x vertically by a factor of c y f cx, compress the graph of y f x horizontally by a factor of c y f xc, stretch the graph of y f x horizontally by a factor of c y f x, reflect the graph of y f x about the x-axis y f x, reflect the graph of y f x about the y-axis
Figure 16 illustrates these stretching transformations when applied to the cosine function with c 2. For instance, in order to get the graph of y 2 cos x we multiply the y-coordinate of each point on the graph of y cos x by 2. This means that the graph of y cos x gets stretched vertically by a factor of 2. y
y=2 cos x
y
2
y=cos x
2
1 2
1
1 0
y= cos x 1
x
y=cos 1 x 2
0
x
y=cos x FIGURE 16
y=cos 2x
18
■
CHAPTER 1
FUNCTIONS AND LIMITS
V EXAMPLE 2 Given the graph of y sx , use transformations to graph y sx 2, y sx 2 , y sx , y 2sx , and y sx .
SOLUTION The graph of the square root function y sx , obtained from Figure 8(a), is shown in Figure 17(a). In the other parts of the figure we sketch y sx 2 by shifting 2 units downward, y sx 2 by shifting 2 units to the right, y sx by reflecting about the x-axis, y 2sx by stretching vertically by a factor of 2, and y sx by reflecting about the y-axis. y
y
y
y
y
y
1 0
1
x
x
0
0
x
2
x
0
x
0
0
x
_2
(a) y=œ„x
(b) y=œ„-2 x
(c) y=œ„„„„ x-2
(d) y=_ œ„x
(f ) y=œ„„ _x
(e) y=2 œ„x
■
FIGURE 17
EXAMPLE 3 Sketch the graph of the function y 1 sin x. SOLUTION To obtain the graph of y 1 sin x, we start with y sin x. We reflect about the x -axis to get the graph y sin x and then we shift 1 unit upward to get y 1 sin x. (See Figure 18.) y
y 2
y=sin x 1
y=1-sin x
1 0
π 2
π
x
0
π 2
π
3π 2
2π
x
■
FIGURE 18
COMBINATIONS OF FUNCTIONS
Two functions f and t can be combined to form new functions f t, f t, ft, and ft in a manner similar to the way we add, subtract, multiply, and divide real numbers. The sum and difference functions are defined by f tx f x tx
f tx f x tx
If the domain of f is A and the domain of t is B, then the domain of f t is the intersection A B because both f x and tx have to be defined. For example, the domain of f x sx is A 0, and the domain of tx s2 x is B , 2 , so the domain of f tx sx s2 x is A B 0, 2 . Similarly, the product and quotient functions are defined by ftx f xtx
f f x x t tx
SECTION 1.2
A CATALOG OF ESSENTIAL FUNCTIONS
■
19
The domain of ft is A B, but we can’t divide by 0 and so the domain of ft is x A B tx 0 . For instance, if f x x 2 and tx x 1, then the domain of the rational function ftx x 2x 1 is x x 1 , or , 1 1, . There is another way of combining two functions to get a new function. For example, suppose that y f u su and u tx x 2 1. Since y is a function of u and u is, in turn, a function of x, it follows that y is ultimately a function of x. We compute this by substitution:
y f u f tx f x 2 1 sx 2 1 The procedure is called composition because the new function is composed of the two given functions f and t. In general, given any two functions f and t, we start with a number x in the domain of t and find its image tx. If this number tx is in the domain of f , then we can calculate the value of f tx. The result is a new function hx f tx obtained by substituting t into f . It is called the composition (or composite) of f and t and is denoted by f ⴰ t (“f circle t”). DEFINITION Given two functions f and t, the composite function f ⴰ t (also called the composition of f and t) is defined by
f ⴰ tx f tx The domain of f ⴰ t is the set of all x in the domain of t such that tx is in the domain of f . In other words, f ⴰ tx is defined whenever both tx and f tx are defined. Figure 19 shows how to picture f ⴰ t in terms of machines.
FIGURE 19
g
x (input)
The f • g machine is composed of the g machine (first) and then the f machine.
g(x)
f
f { ©} (output)
f•g
EXAMPLE 4 If f x x 2 and tx x 3, find the composite functions f ⴰ t
and t ⴰ f .
SOLUTION We have
f ⴰ tx f tx f x 3 x 32 t ⴰ f x t f x tx 2 x 2 3
|
■
NOTE You can see from Example 4 that, in general, f ⴰ t t ⴰ f . Remember, the notation f ⴰ t means that the function t is applied first and then f is applied second. In Example 4, f ⴰ t is the function that first subtracts 3 and then squares; t ⴰ f is the function that first squares and then subtracts 3.
20
■
CHAPTER 1
FUNCTIONS AND LIMITS
V EXAMPLE 5
domain. (a) f ⴰ t
If f x sx and tx s2 x , find each function and its (b) t ⴰ f
(c) f ⴰ f
(d) t ⴰ t
SOLUTION
(a)
4 f ⴰ tx f tx f (s2 x ) ss2 x s 2x
The domain of f ⴰ t is x
x
x 2 , 2 .
t ⴰ f x t f x t(sx ) s2 sx
(b) If 0 a b, then a 2 b 2.
2 x 0
For sx to be defined we must have x 0. For s2 sx to be defined we must have 2 sx 0, that is, sx 2, or x 4. Thus we have 0 x 4, so the domain of t ⴰ f is the closed interval 0, 4 . 4 f ⴰ f x f f x f (sx ) ssx s x
(c)
The domain of f ⴰ f is 0, . t ⴰ tx ttx t(s2 x ) s2 s2 x
(d)
This expression is defined when both 2 x 0 and 2 s2 x 0. The first inequality means x 2, and the second is equivalent to s2 x 2, or 2 x 4, or x 2. Thus, 2 x 2, so the domain of t ⴰ t is the closed interval 2, 2 .
■
It is possible to take the composition of three or more functions. For instance, the composite function f ⴰ t ⴰ h is found by first applying h, then t, and then f as follows: f ⴰ t ⴰ hx f thx So far we have used composition to build complicated functions from simpler ones. But in calculus it is often useful to be able to decompose a complicated function into simpler ones, as in the following example. EXAMPLE 6 Given Fx cos2x 9, find functions f , t, and h such that
F f ⴰ t ⴰ h.
SOLUTION Since Fx cosx 9 2, the formula for F says: First add 9, then
take the cosine of the result, and finally square. So we let hx x 9
tx cos x
f x x 2
Then f ⴰ t ⴰ hx f thx f tx 9 f cosx 9 cosx 9 2 Fx
■
SECTION 1.2
1.2
slope 2 and sketch several members of the family. (b) Find an equation for the family of linear functions such that f 2 1 and sketch several members of the family. (c) Which function belongs to both families? 2. What do all members of the family of linear functions
f x 1 mx 3 have in common? Sketch several members of the family. 3. What do all members of the family of linear functions
f x c x have in common? Sketch several members of the family. 4. Find expressions for the quadratic functions whose graphs
are shown. y
y (_2, 2)
f
(0, 1) (4, 2)
0
x
g 3
■
21
EXERCISES
1. (a) Find an equation for the family of linear functions with
0
A CATALOG OF ESSENTIAL FUNCTIONS
x
(1, _2.5)
5. Find an expression for a cubic function f if f 1 6 and
f 1 f 0 f 2 0.
6. Some scientists believe that the average surface temperature
of the world has been rising steadily. They have modeled the temperature by the linear function T 0.02t 8.50, where T is temperature in C and t represents years since 1900. (a) What do the slope and T -intercept represent? (b) Use the equation to predict the average global surface temperature in 2100. 7. If the recommended adult dosage for a drug is D (in mg),
then to determine the appropriate dosage c for a child of age a, pharmacists use the equation c 0.0417Da 1. Suppose the dosage for an adult is 200 mg. (a) Find the slope of the graph of c. What does it represent? (b) What is the dosage for a newborn? 8. The manager of a weekend flea market knows from past
experience that if he charges x dollars for a rental space at the flea market, then the number y of spaces he can rent is given by the equation y 200 4x. (a) Sketch a graph of this linear function. (Remember that the rental charge per space and the number of spaces rented can’t be negative quantities.) (b) What do the slope, the y-intercept, and the x-intercept of the graph represent? 9. The relationship between the Fahrenheit F and Celsius
C temperature scales is given by the linear function F 95 C 32. (a) Sketch a graph of this function.
(b) What is the slope of the graph and what does it represent? What is the F-intercept and what does it represent? 10. Jason leaves Detroit at 2:00 PM and drives at a constant
speed west along I-96. He passes Ann Arbor, 40 mi from Detroit, at 2:50 PM. (a) Express the distance traveled in terms of the time elapsed. (b) Draw the graph of the equation in part (a). (c) What is the slope of this line? What does it represent? 11. Biologists have noticed that the chirping rate of crickets of
a certain species is related to temperature, and the relationship appears to be very nearly linear. A cricket produces 113 chirps per minute at 70 F and 173 chirps per minute at 80 F. (a) Find a linear equation that models the temperature T as a function of the number of chirps per minute N. (b) What is the slope of the graph? What does it represent? (c) If the crickets are chirping at 150 chirps per minute, estimate the temperature. 12. The manager of a furniture factory finds that it costs $2200
to manufacture 100 chairs in one day and $4800 to produce 300 chairs in one day. (a) Express the cost as a function of the number of chairs produced, assuming that it is linear. Then sketch the graph. (b) What is the slope of the graph and what does it represent? (c) What is the y-intercept of the graph and what does it represent? 13. At the surface of the ocean, the water pressure is the same
as the air pressure above the water, 15 lbin2. Below the surface, the water pressure increases by 4.34 lbin2 for every 10 ft of descent. (a) Express the water pressure as a function of the depth below the ocean surface. (b) At what depth is the pressure 100 lbin2 ? 14. The monthly cost of driving a car depends on the number of
miles driven. Lynn found that in May it cost her $380 to drive 480 mi and in June it cost her $460 to drive 800 mi. (a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model. (b) Use part (a) to predict the cost of driving 1500 miles per month. (c) Draw the graph of the linear function. What does the slope represent? (d) What does the y-intercept represent? (e) Why does a linear function give a suitable model in this situation?
22
■
CHAPTER 1
FUNCTIONS AND LIMITS
15. Suppose the graph of f is given. Write equations for the
19. The graph of f is given. Use it to graph the following
graphs that are obtained from the graph of f as follows. (a) Shift 3 units upward. (b) Shift 3 units downward. (c) Shift 3 units to the right. (d) Shift 3 units to the left. (e) Reflect about the x-axis. (f ) Reflect about the y-axis. (g) Stretch vertically by a factor of 3. (h) Shrink vertically by a factor of 3. 16. Explain how the following graphs are obtained from the
graph of y f x. (a) y 5 f x (b) y f x 5 (c) y f x (d) y 5 f x (e) y f 5x (f ) y 5 f x 3
its graph and give reasons for your choices. (a) y f x 4 (b) y f x 3 (c) y 13 f x (d) y f x 4 (e) y 2 f x 6 y 6
3
!
_3
%
y 1 0
x
1
20. (a) How is the graph of y 2 sin x related to the graph of
y sin x ? Use your answer and Figure 18(a) to sketch the graph of y 2 sin x. (b) How is the graph of y 1 sx related to the graph of y sx ? Use your answer and Figure 17(a) to sketch the graph of y 1 sx .
■ Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions and then applying the appropriate transformations.
f
0
21. y x 3
22. y 1 x 2
23. y x 12
24. y x 2 4x 3
25. y 1 2 cos x
26. y 4 sin 3x
27. y sin x2
28. y
29. y sx 3
30. y x 24 3
31. y 2 x 2 8x
3 x1 32. y 1 s
1
33. y
#
$ _6
(b) y f ( 12 x) (d) y f x
21–34
17. The graph of y f x is given. Match each equation with
@
functions. (a) y f 2x (c) y f x
■
3
6
x
■
35–36
2 x1
34. y
■
■
■
■
35. f x x 3 2x 2,
■
18. The graph of f is given. Draw the graphs of the following
■
■
■
■
■
(c) y 2 f x
(d) y f x 3
38. f x 1 x ,
tx 1x
3
40. f x 1 3x, 41. f x x
■
■
■
■
■
1 , x
tx 5x 2 3x 2 tx
42. f x s2x 3 , x
■
tx 1 sx
39. f x sin x,
1
■
■ Find the functions (a) f ⴰ t, (b) t ⴰ f , (c) f ⴰ f , and (d) t ⴰ t and their domains.
tx 2x 1
0
■
37– 42
37. f x x 2 1,
1
■
tx s1 x
(b) y f x 4
y
■
tx 3x 2 1
functions. (a) y f x 4
1 2
1 tan x 4 4
Find f t, f t, f t, and ft and state their domains.
36. f x s1 x ,
_3
■
1 x4
■
■
■
■
x1 x2
tx x 2 1 ■
■
■
■
■
■
■
SECTION 1.2
43– 44
■
Find f ⴰ t ⴰ h.
44. f x ■
■
2 , x1 ■
tx cos x,
■
■
23
55. A stone is dropped into a lake, creating a circular ripple that
tx x 2 2,
43. f x sx 1,
A CATALOG OF ESSENTIAL FUNCTIONS
■
■
hx x 3
hx sx 3 ■
■
■
■
■
travels outward at a speed of 60 cms. (a) Express the radius r of this circle as a function of the time t (in seconds). (b) If A is the area of this circle as a function of the radius, find A ⴰ r and interpret it. 56. An airplane is flying at a speed of 350 mih at an altitude
45– 48
■
Express the function in the form f ⴰ t.
45. Fx x 1
46. Fx sin( sx )
47. ut scos t
48. ut
2
■
■
49–51
10
■
■
■
■
■
■
tan t 1 tan t
■
■
■
■
57. The Heaviside function H is defined by
Express the function in the form f ⴰ t ⴰ h.
49. Hx 1 3
50. Hx s2 x
x2
8
Ht
51. Hx sec4 (sx ) ■
■
■
■
■
■
■
■
■
■
■
52. Use the table to evaluate each expression.
(a) f t1 (d) t t1
(b) t f 1 (e) t ⴰ f 3
(c) f f 1 (f ) f ⴰ t6
x
1
2
3
4
5
6
f x
3
1
4
2
2
5
tx
6
3
2
1
2
3
53. Use the given graphs of f and t to evaluate each expression,
or explain why it is undefined. (a) f t2 (b) t f 0 (d) t ⴰ f 6 (e) t ⴰ t2
(c) f ⴰ t0 (f ) f ⴰ f 4
y
g
f
2
0
2
of one mile and passes directly over a radar station at time t 0. (a) Express the horizontal distance d (in miles) that the plane has flown as a function of t. (b) Express the distance s between the plane and the radar station as a function of d. (c) Use composition to express s as a function of t.
x
0 1
if t 0 if t 0
It is used in the study of electric circuits to represent the sudden surge of electric current, or voltage, when a switch is instantaneously turned on. (a) Sketch the graph of the Heaviside function. (b) Sketch the graph of the voltage Vt in a circuit if the switch is turned on at time t 0 and 120 volts are applied instantaneously to the circuit. Write a formula for Vt in terms of Ht. (c) Sketch the graph of the voltage Vt in a circuit if the switch is turned on at time t 5 seconds and 240 volts are applied instantaneously to the circuit. Write a formula for Vt in terms of Ht. (Note that starting at t 5 corresponds to a translation.) 58. The Heaviside function defined in Exercise 57 can also be
used to define the ramp function y ctHt, which represents a gradual increase in voltage or current in a circuit. (a) Sketch the graph of the ramp function y tHt. (b) Sketch the graph of the voltage Vt in a circuit if the switch is turned on at time t 0 and the voltage is gradually increased to 120 volts over a 60-second time interval. Write a formula for Vt in terms of Ht for t 60. (c) Sketch the graph of the voltage Vt in a circuit if the switch is turned on at time t 7 seconds and the voltage is gradually increased to 100 volts over a period of 25 seconds. Write a formula for Vt in terms of Ht for t 32. 59. Let f and t be linear functions with equations
54. A spherical balloon is being inflated and the radius of the
balloon is increasing at a rate of 2 cms. (a) Express the radius r of the balloon as a function of the time t (in seconds). (b) If V is the volume of the balloon as a function of the radius, find V ⴰ r and interpret it.
f x m1 x b1 and tx m 2 x b 2. Is f ⴰ t also a linear function? If so, what is the slope of its graph?
60. If you invest x dollars at 4% interest compounded annually,
then the amount Ax of the investment after one year is Ax 1.04x. Find A ⴰ A, A ⴰ A ⴰ A, and A ⴰ A ⴰ A ⴰ A. What do these compositions represent? Find a formula for the composition of n copies of A.
24
■
CHAPTER 1
FUNCTIONS AND LIMITS
63. (a) Suppose f and t are even functions. What can you say
61. (a) If tx 2x 1 and hx 4x 2 4x 7, find a
function f such that f ⴰ t h. (Think about what operations you would have to perform on the formula for t to end up with the formula for h.) (b) If f x 3x 5 and hx 3x 2 3x 2, find a function t such that f ⴰ t h.
62. If f x x 4 and hx 4x 1, find a function t such
that t ⴰ f h.
about f t and f t ? (b) What if f and t are both odd?
64. Suppose f is even and t is odd. What can you say about f t ? 65. Suppose t is an even function and let h f ⴰ t. Is h always
an even function? 66. Suppose t is an odd function and let h f ⴰ t. Is h always
an odd function? What if f is odd? What if f is even?
1.3
THE LIMIT OF A FUNCTION Our aim in this section is to explore the meaning of the limit of a function. We begin by showing how the idea of a limit arises when we try to find the velocity of a falling ball. V EXAMPLE 1 Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds.
SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by st and measured in meters, then Galileo’s law is expressed by the equation
st 4.9t 2 The difficulty in finding the velocity after 5 s is that we are dealing with a single instant of time t 5, so no time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the brief time interval of a tenth of a second from t 5 to t 5.1: average velocity
Time interval
Average velocity (ms)
5 t 6 5 t 5.1 5 t 5.05 5 t 5.01 5 t 5.001
53.9 49.49 49.245 49.049 49.0049
change in position time elapsed
s5.1 s5 0.1
4.95.12 4.952 49.49 ms 0.1
The table shows the results of similar calculations of the average velocity over successively smaller time periods. It appears that as we shorten the time period, the average velocity is becoming closer to 49 ms. The instantaneous velocity when t 5 is defined to be the limiting value of these average velocities over shorter and shorter time periods that start at t 5. Thus the (instantaneous) velocity after 5 s is v 49 ms
■
SECTION 1.3
THE LIMIT OF A FUNCTION
■
25
INTUITIVE DEFINITION OF A LIMIT
Let’s investigate the behavior of the function f defined by f x x 2 x 2 for values of x near 2. The following table gives values of f x for values of x close to 2, but not equal to 2. y
ƒ approaches 4.
y=≈-x+2
4
0
2
As x approaches 2, FIGURE 1
x
f x
x
f x
1.0 1.5 1.8 1.9 1.95 1.99 1.995 1.999
2.000000 2.750000 3.440000 3.710000 3.852500 3.970100 3.985025 3.997001
3.0 2.5 2.2 2.1 2.05 2.01 2.005 2.001
8.000000 5.750000 4.640000 4.310000 4.152500 4.030100 4.015025 4.003001
x
From the table and the graph of f (a parabola) shown in Figure 1 we see that when x is close to 2 (on either side of 2), f x is close to 4. In fact, it appears that we can make the values of f x as close as we like to 4 by taking x sufficiently close to 2. We express this by saying “the limit of the function f x x 2 x 2 as x approaches 2 is equal to 4.” The notation for this is lim x 2 x 2 4 x l2
In general, we use the following notation. 1 DEFINITION
We write lim f x L
xla
and say
“the limit of f x, as x approaches a, equals L”
if we can make the values of f x arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a. Roughly speaking, this says that the values of f x tend to get closer and closer to the number L as x gets closer and closer to the number a (from either side of a) but x a. An alternative notation for lim f x L
xla
is
f x l L
as
xla
which is usually read “ f x approaches L as x approaches a.” Notice the phrase “but x a” in the definition of limit. This means that in finding the limit of f x as x approaches a, we never consider x a. In fact, f x need not even be defined when x a. The only thing that matters is how f is defined near a.
26
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CHAPTER 1
FUNCTIONS AND LIMITS
Figure 2 shows the graphs of three functions. Note that in part (c), f a is not defined and in part (b), f a L. But in each case, regardless of what happens at a, it is true that lim x l a f x L. y
y
y
L
L
L
0
a
0
x
a
(a)
0
x
(b)
x
a
(c)
FIGURE 2 lim ƒ=L in all three cases x a
EXAMPLE 2 Guess the value of lim x l1
x1
f x
0.5 0.9 0.99 0.999 0.9999
0.666667 0.526316 0.502513 0.500250 0.500025
x1 . x2 1
SOLUTION Notice that the function f x x 1x 2 1 is not defined when
x 1, but that doesn’t matter because the definition of lim x l a f x says that we consider values of x that are close to a but not equal to a. The tables at the left give values of f x (correct to six decimal places) for values of x that approach 1 (but are not equal to 1). On the basis of the values in the tables, we make the guess that lim x l1
x1
f x
1.5 1.1 1.01 1.001 1.0001
0.400000 0.476190 0.497512 0.499750 0.499975
x1 0.5 x2 1
■
Example 2 is illustrated by the graph of f in Figure 3. Now let’s change f slightly by giving it the value 2 when x 1 and calling the resulting function t :
t(x)
x1 x2 1
if x 1
2
if x 1
This new function t still has the same limit as x approaches 1. (See Figure 4.) y
y 2
y=
x-1 ≈-1
y=©
0.5
0
FIGURE 3
0.5
1
x
0
FIGURE 4
1
x
SECTION 1.3
EXAMPLE 3 Estimate the value of lim tl0
THE LIMIT OF A FUNCTION
■
27
st 2 9 3 . t2
SOLUTION The table lists values of the function for several values of t near 0.
t
st 2 9 3 t2
1.0 0.5 0.1 0.05 0.01
0.16228 0.16553 0.16662 0.16666 0.16667
As t approaches 0, the values of the function seem to approach 0.1666666 . . . and so we guess that lim
st 2 9 3 t2
t
tl0
1 st 2 9 3 2 t 6
■
In Example 3 what would have happened if we had taken even smaller values of t? The table in the margin shows the results from one calculator; you can see that something strange seems to be happening. If you try these calculations on your own calculator you might get different values, but eventually you will get the value 0 if you make t sufficiently small. Does this mean 1 that the answer is really 0 instead of 6? No, the value of the limit is 16 , as we will show | in the next section. The problem is that the calculator gave false values because st 2 9 is very close to 3 when t is small. (In fact, when t is sufficiently small, a cal■ www.stewartcalculus.com culator’s value for st 2 9 is 3.000. . . to as many digits as the calculator is capable For a further explanation of why calof carrying.) culators sometimes give false values, Something similar happens when we try to graph the function 0.0005 0.0001 0.00005 0.00001
0.16800 0.20000 0.00000 0.00000
click on Lies My Calculator and Computer Told Me. In particular, see the section called The Perils of Subtraction.
f t
st 2 9 3 t2
of Example 3 on a graphing calculator or computer. Parts (a) and (b) of Figure 5 show quite accurate graphs of f , and when we use the trace mode (if available) we can estimate easily that the limit is about 16. But if we zoom in too much, as in parts (c) and (d), then we get inaccurate graphs, again because of problems with subtraction.
0.2
0.2
0.1
0.1
(a) _5, 5 by _0.1, 0.3 FIGURE 5
(b) _0.1, 0.1 by _0.1, 0.3
(c) _10–^, 10–^ by _0.1, 0.3
(d) _10–&, 10–& by _ 0.1, 0.3
28
■
CHAPTER 1
FUNCTIONS AND LIMITS
sin x . x SOLUTION The function f x sin xx is not defined when x 0. Using a calculator (and remembering that, if x ⺢, sin x means the sine of the angle whose radian measure is x), we construct the table of values correct to eight decimal places. From the table at the left and the graph in Figure 6 we guess that V EXAMPLE 4
x
sin x x
1.0 0.5 0.4 0.3 0.2 0.1 0.05 0.01 0.005 0.001
0.84147098 0.95885108 0.97354586 0.98506736 0.99334665 0.99833417 0.99958339 0.99998333 0.99999583 0.99999983
Guess the value of lim
xl0
lim
xl0
sin x 1 x
This guess is in fact correct, as will be proved in the next section using a geometric argument. y
_1
FIGURE 6
V EXAMPLE 5 COMPUTER ALGEBRA SYSTEMS Computer algebra systems (CAS) have commands that compute limits. In order to avoid the types of pitfalls demonstrated in Examples 3 and 5, they don’t find limits by numerical experimentation. Instead, they use more sophisticated techniques such as computing infinite series. If you have access to a CAS, use the limit command to compute the limits in the examples of this section and to check your answers in the exercises of this chapter. ■
Investigate lim sin xl0
1
y=
0
1
sin x x
x
■
. x
SOLUTION Again the function f x sinx is undefined at 0. Evaluating the
function for some small values of x, we get f 1 sin 0
f ( 12 ) sin 2 0
f ( 13) sin 3 0
f ( 14 ) sin 4 0
f 0.1 sin 10 0
f 0.01 sin 100 0
Similarly, f 0.001 f 0.0001 0. On the basis of this information we might be tempted to guess that lim sin 0 xl0 x
| but this time our guess is wrong. Note that although f 1n sin n 0 for any integer n, it is also true that f x 1 for infinitely many values of x that approach 0. The graph of f is given in Figure 7. y
y=sin(π/x)
1
_1 1
_1
FIGURE 7
x
SECTION 1.3
THE LIMIT OF A FUNCTION
■
29
The dashed lines near the y-axis indicate that the values of sinx oscillate between 1 and 1 infinitely often as x approaches 0. (Use a graphing device to graph f and zoom in toward the origin several times. What do you observe?) Since the values of f x do not approach a fixed number as x approaches 0, lim sin
xl0
|
x
The Heaviside function H is defined by
Ht y
0
FIGURE 8
t
■
Examples 3 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is easy to guess the wrong value if we use inappropriate values of x, but it is difficult to know when to stop calculating values. And, as the discussion after Example 3 shows, sometimes calculators and computers give the wrong values. In the next section, however, we will develop foolproof methods for calculating limits. V EXAMPLE 6
1
does not exist
0 1
if t 0 if t 0
[This function is named after the electrical engineer Oliver Heaviside (1850–1925) and can be used to describe an electric current that is switched on at time t 0.] Its graph is shown in Figure 8. As t approaches 0 from the left, Ht approaches 0. As t approaches 0 from the right, Ht approaches 1. There is no single number that Ht approaches as t approaches 0. Therefore, lim t l 0 Ht does not exist. ■
ONE-SIDED LIMITS
We noticed in Example 6 that Ht approaches 0 as t approaches 0 from the left and Ht approaches 1 as t approaches 0 from the right. We indicate this situation symbolically by writing lim Ht 0
t l0
and
lim Ht 1
t l0
The symbol “t l 0 ” indicates that we consider only values of t that are less than 0. Likewise, “t l 0 ” indicates that we consider only values of t that are greater than 0.
2 DEFINITION
We write lim f x L
x la
and say the left-hand limit of f x as x approaches a [or the limit of f x as x approaches a from the left] is equal to L if we can make the values of f x arbitrarily close to L by taking x to be sufficiently close to a and x less than a.
30
■
CHAPTER 1
FUNCTIONS AND LIMITS
Notice that Definition 2 differs from Definition 1 only in that we require x to be less than a. Similarly, if we require that x be greater than a, we get “the right-hand limit of f x as x approaches a is equal to L” and we write lim f x L
x l a
Thus, the symbol “x l a” means that we consider only x a. These definitions are illustrated in Figure 9. y
y
L
ƒ 0
x
a
0
x
a
(a) lim ƒ=L
FIGURE 9
ƒ
L
x
x
(b) lim ƒ=L
x a_
x a+
By comparing Definition l with the definitions of one-sided limits, we see that the following is true.
3
3
y=©
lim f x L
x la
(a) lim tx
(b) lim tx
(c) lim tx
(d) lim tx
(e) lim tx
(f ) lim tx
xl2
xl5
1
FIGURE 10
if and only if
and
lim f x L
x la
V EXAMPLE 7 The graph of a function t is shown in Figure 10. Use it to state the values (if they exist) of the following:
y 4
0
lim f x L
xla
1
2
3
4
5
x
xl2
xl2
xl5
xl5
SOLUTION From the graph we see that the values of tx approach 3 as x
approaches 2 from the left, but they approach 1 as x approaches 2 from the right. Therefore (a) lim tx 3
and
xl2
(b) lim tx 1 xl2
(c) Since the left and right limits are different, we conclude from (3) that lim x l 2 tx does not exist. The graph also shows that (d) lim tx 2
and
xl5
(e) lim tx 2 xl5
(f ) This time the left and right limits are the same and so, by (3), we have lim tx 2
xl5
Despite this fact, notice that t5 2.
■
SECTION 1.3
EXAMPLE 8 Find lim
xl0
THE LIMIT OF A FUNCTION
■
31
1 if it exists. x2
SOLUTION As x becomes close to 0, x 2 also becomes close to 0, and 1x 2 becomes
very large. (See the following table.) In fact, it appears from the graph of the function f x 1x 2 shown in Figure 11 that the values of f x can be made arbitrarily large by taking x close enough to 0. Thus the values of f x do not approach a number, so lim x l 0 1x 2 does not exist.
x
1 x2
1 0.5 0.2 0.1 0.05 0.01 0.001
1 4 25 100 400 10,000 1,000,000
y
y=
1 ≈
x
0
■
FIGURE 11
PRECISE DEFINITION OF A LIMIT
Definition 1 is appropriate for an intuitive understanding of limits, but for deeper understanding and rigorous proofs we need to be more precise. We want to express, in a quantitative manner, that f x can be made arbitrarily close to L by taking x to be sufficiently close to a (but x a. This means that f x can be made to lie within any preassigned distance from L (traditionally denoted by , the Greek letter epsilon) by requiring that x be within a specified distance (the Greek letter delta) from a . That is, f x L when x a and x a. Notice that we can stipulate that x a by writing 0 x a . The resulting precise definition of a limit is as follows.
4 DEFINITION Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then we say that the limit of f x as x approaches a is L, and we write
lim f x L
xla
if for every number 0 there is a corresponding number 0 such that if
In Module 1.3/1.6 you can explore the precise definition of a limit both graphically and numerically.
0 xa
then
f x L
Definition 4 is illustrated in Figures 12 –14. If a number 0 is given, then we draw the horizontal lines y L and y L and the graph of f . (See Figure 12.) If limx l a f x L, then we can find a number 0 such that if we restrict x to lie in the interval a , a and take x a, then the curve y f x lies
32
■
CHAPTER 1
FUNCTIONS AND LIMITS
between the lines y L and y L . (See Figure 13.) You can see that if such a has been found, then any smaller will also work. y
y
y
y=ƒ
L+∑ y=L+∑
ƒ is in here
∑ ∑
L
y=L+∑
y=L-∑
y=L+∑
∑ ∑
L
y=L-∑
y=L-∑ L-∑
0
0
x
a
a-∂
0
x
a
a+∂
x
a
a-∂
a+∂
when x is in here (x≠ a) FIGURE 12
FIGURE 13
FIGURE 14
It’s important to realize that the process illustrated in Figures 12 and 13 must work for every positive number , no matter how small it is chosen. Figure 14 shows that if a smaller is chosen, then a smaller may be required. In proving limit statements it may be helpful to think of the definition of limit as a challenge. First it challenges you with a number . Then you must be able to produce a suitable . You have to be able to do this for every 0, not just a particular . Prove that lim 4x 5 7.
V EXAMPLE 9
Figure 15 shows the geometry behind Example 9. ■
y
and L 7, we need to find a number such that
then 4x 5 7 But 4x 5 7 4x 12 4x 3 4 x 3 . Therefore, we want: if 0 x 3 then 4 x 3 0 x3
if
y=4x-5
7+∑
x l3
SOLUTION Let be a given positive number. According to Definition 4 with a 3
7
7-∑
We can choose to be 4 because
0
4
then
4 x3 4
x
3
3-∂
0 x3
if
4
Therefore, by the definition of a limit,
3+∂
lim 4x 5 7
FIGURE 15
■
x l3
For a left-hand limit we restrict x so that x a , so in Definition 4 we replace 0 x a by a x a . Similarly, for a right-hand limit we use a x a .
y
y=œ„ x
V EXAMPLE 10
y=∑
∑
Prove that lim sx 0. x l0
SOLUTION Let be a given positive number. We want to find a number such that
if 0
∂=∑@
FIGURE 16
x
0x
then
sx 0
that is
sx
But sx &? x 2 . So if we choose 2 and 0 x 2, then sx . (See Figure 16.) This shows that sx l 0 as x l 0.
■
SECTION 1.3
1.3
(g) t2
height in feet t seconds later is given by y 40t 16t 2. (a) Find the average velocity for the time period beginning when t 2 and lasting (i) 0.5 second (ii) 0.1 second (iii) 0.05 second (iv) 0.01 second (b) Estimate the instantaneous velocity when t 2.
(h) lim tt tl4
y 4 2
2. If an arrow is shot upward on the moon with a velocity of 2
58 ms, its height in meters t seconds later is given by h 58t 0.83t 2. (a) Find the average velocity over the given time intervals: (i) [1, 2] (ii) [1, 1.5] (iii) [1, 1.1] (iv) [1, 1.01] (v) [1, 1.001] (b) Estimate the instantaneous velocity when t 1.
determine the values of a for which lim x l a f x exists:
2x f x x x 12
if it exists. If it does not exist, explain why. (a) lim f x (b) lim f x (c) lim f x xl1
xl1
xl1
(e) f 5
■ Sketch the graph of an example of a function f that satisfies all of the given conditions.
7. lim f x 2,
x l 1
lim f x 2,
8. lim f x 1,
x l 0
xl1
4
lim f x 1,
2
4
9. lim f x 4,
f 3 3,
each quantity, if it exists. If it does not exist, explain why. (a) lim f x (b) lim f x (c) lim f x xl3
xl3
xl3
(e) f 3
■
■
lim f x 3,
x l 4
f 4 1 ■
■
■
■
■
x 2 2x x l2 x x 2 x 2.5, 2.1, 2.05, 2.01, 2.005, 2.001, 1.9, 1.95, 1.99, 1.995, 1.999
11. lim
2
0
2
4
x
each quantity, if it exists. If it does not exist, explain why. (a) lim tt (b) lim tt (c) lim tt tl0
(e) lim tt tl2
tl0
( f ) lim tt tl2
2
x 2 2x xl 1 x x 2 x 0, 0.5, 0.9, 0.95, 0.99, 0.999, 2, 1.5, 1.1, 1.01, 1.001
12. lim
5. For the function t whose graph is given, state the value of
tl2
lim f x 3,
x l 4
■
■
■ Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).
4
(d) lim tt
f 2 1
xl1
■
lim f x 2,
x l 2
11–14
y
tl0
lim f x 2,
10. lim f x 3,
f 1 1,
lim f x 0,
x l 2
f 0 is undefined
x l 3
xl3
x
4. For the function f whose graph is given, state the value of xl0
f 2 1,
x l 2
2
f 1 2
lim f x 1,
xl0
(d) lim f x
if x 1 if 1 x 1 if x 1
7–10
y
0
t
4
6. Sketch the graph of the following function and use it to
3. Use the given graph of f to state the value of each quantity,
xl5
33
EXERCISES
1. If a ball is thrown into the air with a velocity of 40 fts, its
(d) lim f x
■
THE LIMIT OF A FUNCTION
2
sin x x tan x x 1, 0.5, 0.2, 0.1, 0.05, 0.01
13. lim
xl0
■
34
■
CHAPTER 1
FUNCTIONS AND LIMITS
sx 4 , x 17, 16.5, 16.1, 16.05, 16.01, x 16 15, 15.5, 15.9, 15.95, 15.99
23. Use the given graph of f x 1x to find a number such
14. lim
x l 16
that
x 2
if ■
■
■
■
■
■
■
■
■
■
■
■
■
■
x 1 x10 1 ■
9 5 x x
18. lim
xl0
■
■
■
■
■
0.5 0.3 x
0 ■
■
ing in toward the point where the graph crosses the y-axis, estimate the value of lim x l 0 f x. (b) Check your answer in part (a) by evaluating f x for values of x that approach 0.
that
x 1
if
x
then
2
y
6x 2x x
xl0
y=≈
1 0.5
by graphing the function y 6 2 x. State your answer correct to two decimal places. (b) Check your answer in part (a) by evaluating f x for values of x that approach 0. x
x
21. (a) Evaluate the function f x x 2 2 x1000 for x 1,
0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of 2x lim x xl0 1000 2
(b) Evaluate f x for x 0.04, 0.02, 0.01, 0.005, 0.003, and 0.001. Guess again. 22. (a) Evaluate hx tan x xx for x 1, 0.5, 0.1, 0.05, 3
0.01, and 0.005.
tan x x . x3 (c) Evaluate hx for successively smaller values of x until you finally reach 0 values for hx. Are you still confident that your guess in part (b) is correct? Explain why you eventually obtained 0 values. (In Section 3.7 a method for evaluating the limit will be explained.) (d) Graph the function h in the viewing rectangle 1, 1 by 0, 1 . Then zoom in toward the point where the graph crosses the y-axis to estimate the limit of hx as x approaches 0. Continue to zoom in until you observe distortions in the graph of h. Compare with the results of part (c). (b) Guess the value of lim
xl0
1 21
1.5
lim
x
10 3
24. Use the given graph of f x x 2 to find a number such
; 20. (a) Estimate the value of
2
10 7
■
; 19. (a) By graphing the function f x tan 4xx and zoom-
;
1 y= x
1
tan 3x 16. lim x l 0 tan 5x
6
xl1
1 0.5 0.2 x
0.7
sx 4 2 15. lim xl0 x 17. lim
y
Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically.
15–18
then
0
?
1
x
?
; 25. Use a graph to find a number such that
x 2
if
then
s4x 1 3 0.5
; 26. Use a graph to find a number such that if
x
6
then
| sin x | 0.1 1 2
27. A machinist is required to manufacture a circular metal disk
with area 1000 cm2. (a) What radius produces such a disk? (b) If the machinist is allowed an error tolerance of 5 cm2 in the area of the disk, how close to the ideal radius in part (a) must the machinist control the radius? (c) In terms of the , definition of limx l a f x L , what is x ? What is f x ? What is a? What is L ? What value of is given? What is the corresponding value of ?
; 28. A crystal growth furnace is used in research to determine
how best to manufacture crystals used in electronic components for the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input power. Suppose the relationship is given by Tw 0.1w 2 2.155w 20
SECTION 1.4
39. lim x 2 0
where T is the temperature in degrees Celsius and w is the power input in watts. (a) How much power is needed to maintain the temperature at 200 C ? (b) If the temperature is allowed to vary from 200 C by up to 1 C , what range of wattage is allowed for the input power? (c) In terms of the , definition of limx l a f x L, what is x ? What is f x ? What is a? What is L ? What value of is given? What is the corresponding value of ?
31. lim 1 4x 13
32. lim 7 3x 5
xl1
x l3
■
■
33– 44
■
■
33. lim x l3
35. lim
■
■
■
■
■
CAS
■
34. lim
3x 4 5
xl6
7
37. lim x a
■
■
■
■
■
■
■
■
■
45. (a) For the limit lim x l 1 x 3 x 1 3, use a graph to
find a value of that corresponds to 0.4. (b) By using a computer algebra system to solve the cubic equation x 3 x 1 3 , find the largest possible value of that works for any given 0. (c) Put 0.4 in your answer to part (b) and compare with your answer to part (a).
■
x 9 3 4 2
46. If H is the Heaviside function defined in Example 6, prove,
x 2 x 12 7 36. lim x l3 x3
using Definition 4, that lim t l 0 Ht does not exist. [Hint: Use an indirect proof as follows. Suppose that the limit is L. Take 21 in the definition of a limit and try to arrive at a contradiction.]
xla
1.4
■
38. lim c c
xla
[Hint: If x 3 1 , what can you say about x 4 ?]
Prove the statement using the , definition of limit.
x 3 5 5
x l5
■
[Hint: Write x 2 9 x 3 x 3 .
x l3
xl4
■
xl9
44. lim x 2 x 4 8
( 12 x 3) 2 x l2
30. lim
4 9x 0 42. lim s
Show that if x 3 1 , then x 3 7 . If you let be the smaller of the numbers 1 and 7 , show that this works.]
■
29. lim 2x 3 5
43. lim x 2 9
Prove the statement using the , definition of limit and illustrate with a diagram like Figure 15. 29–32
xl0
41. lim x 0
xl3
35
40. lim x 3 0
xl0
xl0
■
CALCULATING LIMITS
CALCULATING LIMITS In Section 1.3 we used calculators and graphs to guess the values of limits, but we saw that such methods don’t always lead to the correct answer. In this section we use the following properties of limits, called the Limit Laws, to calculate limits. LIMIT LAWS Suppose that c is a constant and the limits
lim f x
xla
and
exist. Then 1. lim f x tx lim f x lim tx xla
xla
xla
2. lim f x tx lim f x lim tx xla
xla
xla
3. lim cf x c lim f x xla
xla
4. lim f xtx lim f x lim tx xla
xla
5. lim
lim f x f x xla tx lim tx
xla
xla
xla
if lim tx 0 xla
lim tx
xla
36
■
CHAPTER 1
FUNCTIONS AND LIMITS
Sum Law Difference Law Constant Multiple Law
These five laws can be stated verbally as follows: 1. The limit of a sum is the sum of the limits. 2. The limit of a difference is the difference of the limits. 3. The limit of a constant times a function is the constant times the limit of the function.
Product Law
4. The limit of a product is the product of the limits.
Quotient Law
5. The limit of a quotient is the quotient of the limits (provided that the limit of
the denominator is not 0). It is easy to believe that these properties are true. For instance, if f x is close to L and tx is close to M, it is reasonable to conclude that f x tx is close to L M. This gives us an intuitive basis for believing that Law 1 is true. All of these laws can be proved using the precise definition of a limit. (See Appendix B.) If we use the Product Law repeatedly with tx f x, we obtain the following law. Power Law
6. lim f x n lim f x x la
[
x la
n
]
where n is a positive integer
In applying these six limit laws, we need to use two special limits: 7. lim c c
8. lim x a
xla
xla
These limits are obvious from an intuitive point of view (state them in words or draw graphs of y c and y x), but they can be proved from the precise definition. (See Exercises 37 and 38 in Section 1.3.) If we now put f x x in Law 6 and use Law 8, we get another useful special limit. 9. lim x n a n
where n is a positive integer
xla
A similar limit holds for roots as follows. n n 10. lim s x s a
where n is a positive integer
xla
(If n is even, we assume that a 0.)
More generally, we have the following law. Root Law
n 11. lim s f x)
x la
f x) s lim x la n
where n is a positive integer
[If n is even, we assume that lim f x 0.] x la
SECTION 1.4
NEWTON AND LIMITS Isaac Newton was born on Christmas Day in 1642, the year of Galileo’s death. When he entered Cambridge University in 1661 Newton didn’t know much mathematics, but he learned quickly by reading Euclid and Descartes and by attending the lectures of Isaac Barrow. Cambridge was closed because of the plague in 1665 and 1666, and Newton returned home to reflect on what he had learned. Those two years were amazingly productive for at that time he made four of his major discoveries: (1) his representation of functions as sums of infinite series, including the binomial theorem; (2) his work on differential and integral calculus; (3) his laws of motion and law of universal gravitation; and (4) his prism experiments on the nature of light and color. Because of a fear of controversy and criticism, he was reluctant to publish his discoveries and it wasn’t until 1687, at the urging of the astronomer Halley, that Newton published Principia Mathematica. In this work, the greatest scientific treatise ever written, Newton set forth his version of calculus and used it to investigate mechanics, fluid dynamics, and wave motion, and to explain the motion of planets and comets. The beginnings of calculus are found in the calculations of areas and volumes by ancient Greek scholars such as Eudoxus and Archimedes. Although aspects of the idea of a limit are implicit in their “method of exhaustion,” Eudoxus and Archimedes never explicitly formulated the concept of a limit. Likewise, mathematicians such as Cavalieri, Fermat, and Barrow, the immediate precursors of Newton in the development of calculus, did not actually use limits. It was Isaac Newton who was the first to talk explicitly about limits. He explained that the main idea behind limits is that quantities “approach nearer than by any given difference.” Newton stated that the limit was the basic concept in calculus, but it was left to later mathematicians like Cauchy to clarify his ideas about limits. ■
CALCULATING LIMITS
■
37
EXAMPLE 1 Evaluate the following limits and justify each step.
(a) lim 2x 2 3x 4
(b) lim
x l5
x l 2
x 3 2x 2 1 5 3x
SOLUTION
lim 2x 2 3x 4 lim 2x 2 lim 3x lim 4
(a)
x l5
x l5
x l5
x l5
(by Laws 2 and 1)
2 lim x 2 3 lim x lim 4
(by 3)
25 2 35 4
(by 9, 8, and 7)
x l5
x l5
x l5
39 (b) We start by using Law 5, but its use is fully justified only at the final stage when we see that the limits of the numerator and denominator exist and the limit of the denominator is not 0. lim
x l2
lim x 3 2x 2 1 x 3 2x 2 1 x l2 5 3x lim 5 3x
(by Law 5)
x l2
lim x 3 2 lim x 2 lim 1
x l2
x l2
x l2
x l2
lim 5 3 lim x
23 222 1 5 32
(by 1, 2, and 3)
x l2
(by 9, 8, and 7)
1 11
■
NOTE If we let f x 2x 2 3x 4, then f 5 39. In other words, we would
have gotten the correct answer in Example 1(a) by substituting 5 for x. Similarly, direct substitution provides the correct answer in part (b). The functions in Example 1 are a polynomial and a rational function, respectively, and similar use of the Limit Laws proves that direct substitution always works for such functions (see Exercises 49 and 50). We state this fact as follows. DIRECT SUBSTITUTION PROPERTY If f is a polynomial or a rational function
and a is in the domain of f , then lim f x f a x la
y
P(cos ¨, sin ¨) 1 ¨ 0
FIGURE 1
(1, 0)
x
The trigonometric functions also enjoy the Direct Substitution Property. We know from the definitions of sin and cos that the coordinates of the point P in Figure 1 are cos , sin . As l 0, we see that P approaches the point 1, 0 and so cos l 1 and sin l 0. Thus 1
lim cos 1
l0
lim sin 0
l0
Since cos 0 1 and sin 0 0, the equations in (1) assert that the cosine and sine
38
■
CHAPTER 1
FUNCTIONS AND LIMITS
■ Another way to establish the limits in (1) is to use the inequality sin (for 0 ), which is proved on page 42.
functions satisfy the Direct Substitution Property at 0. The addition formulas for cosine and sine can then be used to deduce that these functions satisfy the Direct Substitution Property everywhere (see Exercises 51 and 52). In other words, for any real number a, lim sin sin a lim cos cos a la
la
This enables us to evaluate certain limits quite simply. For example, lim x cos x lim x lim cos x cos
(
xl
)(
xl
xl
)
Functions with the Direct Substitution Property are called continuous at a and will be studied in Section 1.5. However, not all limits can be evaluated by direct substitution, as the following examples show. EXAMPLE 2 Find lim
xl1
x2 1 . x1
SOLUTION Let f x x 2 1x 1. We can’t find the limit by substituting
x 1 because f 1 isn’t defined. Nor can we apply the Quotient Law, because the limit of the denominator is 0. Instead, we need to do some preliminary algebra. We factor the numerator as a difference of squares: x2 1 x 1x 1 x1 x1 The numerator and denominator have a common factor of x 1. When we take the limit as x approaches 1, we have x 1 and so x 1 0. Therefore, we can cancel the common factor and compute the limit as follows: lim
xl1
x2 1 x 1x 1 lim x l 1 x1 x1 lim x 1 xl1
112
■
NOTE In Example 2 we were able to compute the limit by replacing the given function f x x 2 1x 1 by a simpler function, tx x 1, with the same limit. This is valid because f x tx except when x 1, and in computing a limit as x approaches 1 we don’t consider what happens when x is actually equal to 1. In general, we have the following useful fact.
If f x tx when x a, then lim f x lim tx, provided the limits exist. xla
xla
EXAMPLE 3 Find lim tx where x l1
tx
x 1 if x 1 if x 1
SECTION 1.4
y
y=ƒ
3 2
xl1
1
2
3
x
y=©
V EXAMPLE 4
Evaluate lim
hl0
SOLUTION If we define
Fh
1 1
2
3
x
■
xl1
3 h2 9 . h
2
0
39
Note that the values of the functions in Examples 2 and 3 are identical except when x 1 (see Figure 2) and so they have the same limit as x approaches 1.
y 3
■
SOLUTION Here t is defined at x 1 and t1 , but the value of a limit as x approaches 1 does not depend on the value of the function at 1. Since tx x 1 for x 1, we have lim tx lim x 1 2
1 0
CALCULATING LIMITS
3 h2 9 h
then, as in Example 2, we can’t compute lim h l 0 Fh by letting h 0 since F0 is undefined. But if we simplify Fh algebraically, we find that
FIGURE 2
The graphs of the functions f (from Example 2) and g (from Example 3)
Fh
9 6h h 2 9 6h h 2 6h h h
(Recall that we consider only h 0 when letting h approach 0.) Thus lim
hl0
EXAMPLE 5 Find lim tl0
3 h2 9 lim 6 h 6 hl0 h
■
st 2 9 3 . t2
SOLUTION We can’t apply the Quotient Law immediately, since the limit of the denominator is 0. Here the preliminary algebra consists of rationalizing the numerator:
lim tl0
st 2 9 3 st 2 9 3 st 2 9 3 lim tl0 t2 t2 st 2 9 3 lim
t 2 9 9 t2 lim 2 2 2 t l 0 t (st 9 3) t (st 9 3)
lim
1 st 9 3
tl0
tl0
2
2
1 1 1 s lim t 9 3 3 3 6 2
tl0
This calculation confirms the guess that we made in Example 3 in Section 1.3.
■
Some limits are best calculated by first finding the left- and right-hand limits. The following theorem is a reminder of what we discovered in Section 1.3. It says that a two-sided limit exists if and only if both of the one-sided limits exist and are equal.
2
THEOREM
lim f x L
xla
if and only if
lim f x L lim f x
xla
x la
40
■
CHAPTER 1
FUNCTIONS AND LIMITS
When computing one-sided limits, we use the fact that the Limit Laws also hold for one-sided limits.
EXAMPLE 6 Show that lim x 0. xl0
SOLUTION Recall that
x
if x 0 if x 0
x x
Since x x for x 0, we have
The result of Example 6 looks plausible from Figure 3. ■
lim x lim x 0
x l 0
y
y=| x |
xl0
For x 0 we have x x and so
lim x lim x 0
x l 0
0
x
Therefore, by Theorem 2,
lim x 0
FIGURE 3
xl0
V EXAMPLE 7
SOLUTION
y | x|
y= x
xl0
lim
x
lim
x
xl0
x
x
xl0
1 0
x does not exist.
Prove that lim
x l 0
■
x
x
lim
x lim 1 1 xl0 x
lim
x lim 1 1 xl0 x
x l 0
xl0
_1
Since the right- and left-hand limits are different, it follows from Theorem 2 that lim x l 0 x x does not exist. The graph of the function f x x x is shown in Figure 4 and supports the one-sided limits that we found. ■
FIGURE 4 ■ Other notations for x are x and ⎣ x⎦. The greatest integer function is sometimes called the floor function.
EXAMPLE 8 The greatest integer function is defined by x the largest integer
y
that is less than or equal to x. (For instance, 4 4, 4.8 4, 3, s2 1, 12 1.) Show that lim x l3 x does not exist.
4
SOLUTION The graph of the greatest integer function is shown in Figure 5. Since
3
x 3 for 3 x 4, we have y=[ x]
2
lim x lim 3 3
x l3
1 0
1
2
3
4
5
x
x l3
Since x 2 for 2 x 3, we have lim x lim 2 2
x l3
FIGURE 5
Greatest integer function
x l3
Because these one-sided limits are not equal, lim x l3 x does not exist by Theorem 2.
■
SECTION 1.4
CALCULATING LIMITS
■
41
The next two theorems give two additional properties of limits. Their proofs can be found in Appendix B. 3 THEOREM If f x tx when x is near a (except possibly at a) and the limits of f and t both exist as x approaches a, then
lim f x lim tx
xla
4
xla
THE SQUEEZE THEOREM If f x tx hx when x is near a (except
possibly at a) and lim f x lim hx L
y
xla
h g
xla
lim tx L
then
xla
L
f 0
x
a
FIGURE 6
The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the Pinching Theorem, is illustrated by Figure 6. It says that if tx is squeezed between f x and hx near a, and if f and h have the same limit L at a, then t is forced to have the same limit L at a. 1 0. x SOLUTION First note that we cannot use V EXAMPLE 9
Show that lim x 2 sin xl0
lim x 2 sin
xl0
1 1 lim x 2 lim sin xl0 xl0 x x
because lim x l 0 sin1x does not exist (see Example 5 in Section 1.3). However, since 1 sin y
1
1 x
we have, as illustrated by Figure 7,
y=≈
x 2 x 2 sin
1
x2 x
x
0
We know that y=_≈ FIGURE 7
y=≈ sin(1/x)
lim x 2 0
xl0
and
lim x 2 0
xl0
Taking f x x 2, tx x 2 sin1x, and hx x 2 in the Squeeze Theorem, we obtain 1 lim x 2 sin 0 xl0 ■ x
42
■
CHAPTER 1
FUNCTIONS AND LIMITS
In Example 4 in Section 1.3 we made the guess, on the basis of numerical and graphical evidence, that
lim
5
D
l0
We can prove Equation 5 with help from the Squeeze Theorem. Assume first that lies between 0 and 2. Figure 8(a) shows a sector of a circle with center O, central angle , and radius 1. BC is drawn perpendicular to OA. By the definition of radian measure, we have arc AB . Also, BC OB sin sin . From the diagram we see that
B
¨
sin
Therefore 1
C
A
(a)
sin 1
so
Let the tangent lines at A and B intersect at E. You can see from Figure 8(b) that the circumference of a circle is smaller than the length of a circumscribed polygon, and so arc AB AE EB . Thus
B E A
O
BC AB arc AB
E
O
sin 1
arc AB AE EB AE ED
AD OA tan tan (b) FIGURE 8
(In Appendix B the inequality tan is proved directly from the definition of the length of an arc without resorting to geometric intuition as we did here.) Therefore, we have
sin cos
and so
cos
sin 1
We know that lim l 0 1 1 and lim l 0 cos 1, so by the Squeeze Theorem, we have lim
l 0
sin 1
But the function sin is an even function, so its right and left limits must be equal. Hence, we have lim
l0
so we have proved Equation 5.
EXAMPLE 10 Find lim
xl0
sin 7x . 4x
sin 1
SECTION 1.4
CALCULATING LIMITS
■
43
SOLUTION In order to apply Equation 5, we first rewrite the function by multiplying and dividing by 7:
sin 7x 7 4x 4
Note that sin 7x 7 sin x.
sin 7x 7x
Notice that as x l 0, we have 7x l 0, and so, by Equation 5 with 7x, sin 7x sin7x lim 1 7x l 0 7x 7x
lim
xl0
Thus
lim
xl0
sin 7x 7 lim xl0 4 4x
EXAMPLE 11 Evaluate lim
l0
sin 7x 7x
7 sin 7x 7 7 lim 1 4 x l 0 7x 4 4
■
cos 1 .
SOLUTION ■ We multiply numerator and denominator by cos 1 in order to put the function in a form in which we can use the limits we know.
lim
l0
cos 1 lim l0 lim
l0
sin 2 lim l0 cos 1
lim
l0
1
1.4
x la
lim tx 0
lim hx 8
x la
(b) lim f x 2
3 hx (c) lim s
(d) lim
1 f x
(f ) lim
tx f x
(e) lim x la
(g) lim x la
f x hx f x tx
sin sin cos 1
sin sin lim l 0 cos 1
0 11
0
(by Equation 5)
■
limit, if it exists. If the limit does not exist, explain why. y
x la
(a) lim f x hx
xla
l0
cos2 1 cos 1
2. The graphs of f and t are given. Use them to evaluate each
find the limits that exist. If the limit does not exist, explain why. x la
lim
EXERCISES
1. Given that
lim f x 3
cos 1 cos 1 cos 1
y=ƒ
x la
(h) lim x la
2 f x hx f x
y=©
1 1
x la
x la
y
x
1
0
1
(a) lim f x tx
(b) lim f x tx
(c) lim f xtx
(d) lim
(e) lim x 3f x
(f ) lim s3 f x
x l2
x l0
x l2
x l1
x l 1
x l1
f x tx
x
44
■
CHAPTER 1
FUNCTIONS AND LIMITS
■ Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).
3–9
3. lim 3x 4 2x 2 x 1
4. lim t 2 13t 35
3 5. lim (1 s x )2 6x 2 x 3
6. lim su 4 3u 6
x l 2
xl8
7. lim x l1
9. ■
1 3x 1 4x 2 3x 4
f x
t l 1
3
8. lim
xl0
cos 4 x 5 2x 3
; 27. Use the Squeeze Theorem to show that
lim sin ■
■
■
■
■
■
■
■
■
lim x l 0 x 2 cos 20 x 0. Illustrate by graphing the functions f x x 2, tx x 2 cos 20 x, and hx x 2 on the same screen.
■
10. (a) What is wrong with the following equation?
; 28. Use the Squeeze Theorem to show that
x2 x 6 x3 x2
lim sx 3 x 2 sin x l0
lim x l2
x2 x 6 lim x 3 x l2 x2
29. If 4x 9 f x x 2 4x 7 for x 0, find
lim x l 4 f x.
is correct.
11. lim x l2
30. If 2x tx x 4 x 2 2 for all x, evaluate lim x l 1 tx.
Evaluate the limit, if it exists.
■
x2 x 6 x2
31. Prove that lim x 4 cos
12. lim
x l4
x2 x 6 13. lim x l2 x2 t2 9 2t 2 7t 3
16. lim
17. lim
4 h 16 h
18. lim
x l1
2
hl0
19. lim
x l2
h l0
x2 x3 8
20. lim
■
24. lim tl0
■
x l0
■ Find the limit, if it exists. If the limit does not exist, explain why.
33–36
x 2 4x x 2 3x 4
33. lim (2x x 3 xl3
35. lim x l0
x 2 2x 1 x4 1
■
■
■
■
1 1 2 t t t ■
■
■
■
x l0
)
x l6
■
■
36. lim x l0
■
■
■
x tx 1 x 2 x1
■
34. lim
1 1 x x
37. Let
■
2x 12 x6
1 1 x x ■
■
■
if x 1 if 1 x 1 if x 1
(a) Evaluate each of the following limits, if it exists. ■
; 25. (a) Estimate the value of lim
2 0. x
32. Prove that lim sx 1 sin2 2x 0 .
3 h1 3 1 22. lim hl0 h
1 1 4 x 23. lim x l 4 4 x ■
x 2 5x 4 x 2 3x 4
s1 h 1 h
x l1
sx 2 3 21. lim x l7 x7
■
x l0
x 2 4x 14. lim 2 x l 4 x 3x 4
15. lim
t l3
0 x
Illustrate by graphing the functions f, t, and h (in the notation of the Squeeze Theorem) on the same screen.
(b) In view of part (a), explain why the equation
11–24
s3 x s3 x
to estimate the value of lim x l 0 f x to two decimal places. (b) Use a table of values of f x to estimate the limit to four decimal places. (c) Use the Limit Laws to find the exact value of the limit.
u l2
l2 ■
; 26. (a) Use a graph of
x s1 3x 1
by graphing the function f x x(s1 3x 1). (b) Make a table of values of f x for x close to 0 and guess the value of the limit. (c) Use the Limit Laws to prove that your guess is correct.
(i) lim tx x l1
(ii) lim tx
(iii) lim tx
(v) lim tx
(vi) lim tx
x l1
(iv) lim tx x l1
x l 1
(b) Sketch the graph of t. 38. Let Fx
(a) Find
x2 1 . x1
(i) lim Fx x l1
(ii) lim Fx x l1
x l0
x l 1
■
SECTION 1.5
we need to show that lim x l a sin x sin a for every real number a. If we let h x a, then x a h and x l a &? h l 0 . So an equivalent statement is that
39. (a) If the symbol denotes the greatest integer function
defined in Example 8, evaluate (i) lim x (ii) lim x
(iii) lim x
x l 2
lim sina h sin a
x l 2.4
h l0
(b) If n is an integer, evaluate (i) lim x (ii) lim x x ln
Use (1) to show that this is true.
xln
(c) For what values of a does lim x l a x exist?
52. Prove that cosine has the Direct Substitution Property.
40. Let f x x x.
53. Show by means of an example that lim x l a f x tx
may exist even though neither limx l a f x nor limx l a tx exists.
(a) Sketch the graph of f. (b) If n is an integer, evaluate (i) lim f x (ii) lim f x x ln
54. Show by means of an example that limx l a f xtx may
x ln
(c) For what values of a does lim x l a f x exist?
exist even though neither lim x l a f x nor limx l a tx exists.
41. If f x x x , show that lim x l 2 f x exists but is
55. Is there a number a such that
not equal to f 2.
42. In the theory of relativity, the Lorentz contraction formula
lim
x l2
L L 0 s1 v 2c 2
43. lim
xl0
45. lim tl0
47. lim
l0
■
■
■
56. The figure shows a fixed circle C1 with equation
x 12 y 2 1 and a shrinking circle C2 with radius r and center the origin. P is the point 0, r, Q is the upper point of intersection of the two circles, and R is the point of intersection of the line PQ and the x-axis. What happens to R as C2 shrinks, that is, as r l 0 ?
Find the limit. sin 3x x
44. lim
xl0
sin 4x sin 6x
y
2
tan 6t sin 2t
46. lim tl0
■
sin 3t t2
P
Q
C™
sin tan ■
3x 2 ax a 3 x2 x 2
exists? If so, find the value of a and the value of the limit.
expresses the length L of an object as a function of its velocity v with respect to an observer, where L 0 is the length of the object at rest and c is the speed of light. Find lim v l c L and interpret the result. Why is a left-hand limit necessary? 43– 48
45
■
51. To prove that sine has the Direct Substitution Property
(b) Does lim x l 1 Fx exist? (c) Sketch the graph of F.
x l2
CONTINUITY
48. lim x cot x xl0
■
■
■
■
■
■
■
■
49. If p is a polynomial, show that lim xl a px pa.
■
0
R
x
C¡
50. If r is a rational function, use Exercise 49 to show that
lim x l a rx ra for every number a in the domain of r.
1.5
CONTINUITY We noticed in Section 1.4 that the limit of a function as x approaches a can often be found simply by calculating the value of the function at a. Functions with this property are called continuous at a. We will see that the mathematical definition of continuity corresponds closely with the meaning of the word continuity in everyday language. (A continuous process is one that takes place gradually, without interruption or abrupt change.)
46
■
CHAPTER 1
FUNCTIONS AND LIMITS
1 DEFINITION
A function f is continuous at a number a if lim f x f a x la
■ As illustrated in Figure 1, if f is continuous, then the points x, f x on the graph of f approach the point a, f a on the graph. So there is no gap in the curve.
y
ƒ approaches f(a).
f(a)
x
a
As x approaches a, FIGURE 1
y
0
1. f a is defined (that is, a is in the domain of f ) 2. lim f x exists x la
3. lim f x f a x la
y=ƒ
0
Notice that Definition l implicitly requires three things if f is continuous at a:
The definition says that f is continuous at a if f x approaches f a as x approaches a. Thus a continuous function f has the property that a small change in x produces only a small change in f x. In fact, the change in f x can be kept as small as we please by keeping the change in x sufficiently small. If f is defined near a (in other words, f is defined on an open interval containing a, except perhaps at a), we say that f is discontinuous at a (or f has a discontinuity at a) if f is not continuous at a. Physical phenomena are usually continuous. For instance, the displacement or velocity of a vehicle varies continuously with time, as does a person’s height. But discontinuities do occur in such situations as electric currents. [See Example 6 in Section 1.3, where the Heaviside function is discontinuous at 0 because lim t l 0 Ht does not exist.] Geometrically, you can think of a function that is continuous at every number in an interval as a function whose graph has no break in it. The graph can be drawn without removing your pen from the paper. EXAMPLE 1 Figure 2 shows the graph of a function f. At which numbers is f discontinuous? Why?
1
2
3
4
5
x
SOLUTION It looks as if there is a discontinuity when a 1 because the graph has a break there. The official reason that f is discontinuous at 1 is that f 1 is not defined. The graph also has a break when a 3, but the reason for the discontinuity is different. Here, f 3 is defined, but lim x l3 f x does not exist (because the left and right limits are different). So f is discontinuous at 3. What about a 5? Here, f 5 is defined and lim x l5 f x exists (because the left and right limits are the same). But
FIGURE 2
lim f x f 5
xl5
■
So f is discontinuous at 5.
Now let’s see how to detect discontinuities when a function is defined by a formula. V EXAMPLE 2
x2 x 2 (a) f x x2 (c) f x
Where are each of the following functions discontinuous?
x2 x 2 x2 1
(b) f x if x 2 if x 2
1 x2 1
(d) f x x
if x 0 if x 0
SECTION 1.5
■
CONTINUITY
47
SOLUTION
(a) Notice that f 2 is not defined, so f is discontinuous at 2. Later we’ll see why f is continuous at all other numbers. (b) Here f 0 1 is defined but lim f x lim
xl0
xl0
1 x2
does not exist. (See Example 8 in Section 1.3.) So f is discontinuous at 0. (c) Here f 2 1 is defined and lim f x lim x l2
x l2
x2 x 2 x 2x 1 lim lim x 1 3 x l2 x l2 x2 x2
exists. But lim f x f 2 x l2
so f is not continuous at 2. (d) The greatest integer function f x x has discontinuities at all of the integers because lim x ln x does not exist if n is an integer. (See Example 8 and Exercise 39 in Section 1.4.) ■ Figure 3 shows the graphs of the functions in Example 2. In each case the graph can’t be drawn without lifting the pen from the paper because a hole or break or jump occurs in the graph. The kind of discontinuity illustrated in parts (a) and (c) is called removable because we could remove the discontinuity by redefining f at just the single number 2. [The function tx x 1 is continuous.] The discontinuity in part (b) is called an infinite discontinuity. The discontinuities in part (d) are called jump discontinuities because the function “jumps” from one value to another. y
y
y
y
1
1
1
1
0
(a) ƒ=
1
2
≈-x-2 x-2
x
0
1 if x≠0 (b) ƒ= ≈ 1 if x=0
0
x
(c) ƒ=
1
2
x
≈-x-2 if x≠2 x-2 1 if x=2
0
1
2
3
(d) ƒ=[ x ]
FIGURE 3 Graphs of the functions in Example 2
2 DEFINITION
A function f is continuous from the right at a number a if lim f x f a
x la
and f is continuous from the left at a if lim f x f a
x la
x
48
■
CHAPTER 1
FUNCTIONS AND LIMITS
EXAMPLE 3 At each integer n, the function f x x [see Figure 3(d)] is continu-
ous from the right but discontinuous from the left because lim f x lim x n f n
x ln
x ln
lim f x lim x n 1 f n
but
x ln
x ln
■
3 DEFINITION A function f is continuous on an interval if it is continuous at every number in the interval. (If f is defined only on one side of an endpoint of the interval, we understand continuous at the endpoint to mean continuous from the right or continuous from the left.)
EXAMPLE 4 Show that the function f x 1 s1 x 2 is continuous on the
interval 1, 1 .
SOLUTION If 1 a 1, then using the Limit Laws, we have
lim f x lim (1 s1 x 2 )
xla
xla
1 lim s1 x 2
(by Laws 2 and 7)
xla
1 s lim 1 x 2
(by 11)
xla
1 s1 a 2
(by 2, 7, and 9)
f a Thus, by Definition l, f is continuous at a if 1 a 1. Similar calculations show that
y
ƒ=1-œ„„„„„ 1-≈
lim f x 1 f 1
1
-1
FIGURE 4
0
x l1
1
x
and
lim f x 1 f 1
x l1
so f is continuous from the right at 1 and continuous from the left at 1. Therefore, according to Definition 3, f is continuous on 1, 1 . The graph of f is sketched in Figure 4. It is the lower half of the circle x 2 y 12 1
■
Instead of always using Definitions 1, 2, and 3 to verify the continuity of a function as we did in Example 4, it is often convenient to use the next theorem, which shows how to build up complicated continuous functions from simple ones. 4 THEOREM If f and t are continuous at a and c is a constant, then the following functions are also continuous at a : 1. f t 2. f t 3. cf f 4. ft 5. if ta 0 t
SECTION 1.5
CONTINUITY
■
49
PROOF Each of the five parts of this theorem follows from the corresponding Limit Law in Section 1.4. For instance, we give the proof of part 1. Since f and t are continuous at a, we have
lim f x f a
lim tx ta
and
xla
xla
Therefore lim f tx lim f x tx
xla
xla
lim f x lim tx xla
xla
(by Law 1)
f a ta f ta This shows that f t is continuous at a.
■
It follows from Theorem 4 and Definition 3 that if f and t are continuous on an interval, then so are the functions f t, f t, cf, ft, and (if t is never 0) ft. The following theorem was stated in Section 1.4 as the Direct Substitution Property. 5 THEOREM
(a) Any polynomial is continuous everywhere; that is, it is continuous on ⺢ , . (b) Any rational function is continuous wherever it is defined; that is, it is continuous on its domain.
PROOF
(a) A polynomial is a function of the form Px cn x n cn1 x n1 c1 x c0 where c0 , c1, . . . , cn are constants. We know that lim c0 c0
xla
and
lim x m a m
xla
(by Law 7)
m 1, 2, . . . , n
(by 9)
This equation is precisely the statement that the function f x x m is a continuous function. Thus, by part 3 of Theorem 4, the function tx cx m is continuous. Since P is a sum of functions of this form and a constant function, it follows from part 1 of Theorem 4 that P is continuous. (b) A rational function is a function of the form f x
Px Qx
where P and Q are polynomials. The domain of f is D x ⺢ Qx 0 . We know from part (a) that P and Q are continuous everywhere. Thus, by part 5 of ■ Theorem 4, f is continuous at every number in D.
50
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CHAPTER 1
FUNCTIONS AND LIMITS
As an illustration of Theorem 5, observe that the volume of a sphere varies continuously with its radius because the formula Vr 43 r 3 shows that V is a polynomial function of r. Likewise, if a ball is thrown vertically into the air with a velocity of 50 fts, then the height of the ball in feet t seconds later is given by the formula h 50t 16t 2. Again this is a polynomial function, so the height is a continuous function of the elapsed time. Knowledge of which functions are continuous enables us to evaluate some limits very quickly, as the following example shows. Compare it with Example 1(b) in Section 1.4. EXAMPLE 5 Find lim
x l 2
x 3 2x 2 1 . 5 3x
SOLUTION The function
f x
x 3 2x 2 1 5 3x
is rational, so by Theorem 5 it is continuous on its domain, which is {x Therefore lim
x l2
23 222 1 1 5 32 11
lim sin sin a
la
1 _
π 2
0
π 2
π
3π 2
x
y=tan x
lim cos cos a
la
In other words, the sine and cosine functions are continuous everywhere. It follows from part 5 of Theorem 4 that tan x
FIGURE 5
■
It turns out that most of the familiar functions are continuous at every number in their domains. For instance, Limit Law 10 (page 36) is exactly the statement that root functions are continuous. From the appearance of the graphs of the sine and cosine functions (Figure 11 in Section 1.2), we would certainly guess that they are continuous. And in Section 1.4 we showed that
y
3π _π
5 3
x 3 2x 2 1 lim f x f 2 x l2 5 3x
_ 2
x }.
sin x cos x
is continuous except where cos x 0. This happens when x is an odd integer multiple of 2, so y tan x has infinite discontinuities when x 2, 32, 52, and so on (see Figure 5).
6 THEOREM The following types of functions are continuous at every number in their domains: polynomials, rational functions, root functions, trigonometric functions
SECTION 1.5
CONTINUITY
■
51
EXAMPLE 6 On what intervals is each function continuous?
(a) f x x 100 2x 37 75 (c) hx sx
(b) tx
x1 x1 2 x1 x 1
x 2 2x 17 x2 1
SOLUTION
(a) f is a polynomial, so it is continuous on , by Theorem 5(a). (b) t is a rational function, so by Theorem 5(b) it is continuous on its domain, which is D x x 2 1 0 x x 1 . Thus, t is continuous on the intervals , 1, 1, 1, and 1, . (c) We can write hx Fx Gx Hx, where
Fx sx
Gx
x1 x1
Hx
x1 x2 1
F is continuous on 0, by Theorem 6. G is a rational function, so it is continuous everywhere except when x 1 0, that is, x 1. H is also a rational function, but its denominator is never 0, so H is continuous everywhere. Thus, by parts 1 and 2 of Theorem 4, h is continuous on the intervals 0, 1 and 1, . ■ Another way of combining continuous functions f and t to get a new continuous function is to form the composite function f ⴰ t. This fact is a consequence of the following theorem. ■ This theorem says that a limit symbol can be moved through a function symbol if the function is continuous and the limit exists. In other words, the order of these two symbols can be reversed.
7 THEOREM
If f is continuous at b and lim tx b, then lim f tx f b. x la
In other words,
lim f tx f lim tx
(
xla
xla
x la
)
Intuitively, Theorem 7 is reasonable because if x is close to a, then tx is close to b, and since f is continuous at b, if tx is close to b, then f tx is close to f b. A proof of Theorem 7 is given in Appendix B. 8 THEOREM If t is continuous at a and f is continuous at ta, then the composite function f ⴰ t given by f ⴰ tx f tx is continuous at a.
This theorem is often expressed informally by saying “a continuous function of a continuous function is a continuous function.” PROOF Since t is continuous at a, we have
lim tx ta
xla
Since f is continuous at b ta, we can apply Theorem 7 to obtain lim f tx f ta
xla
which is precisely the statement that the function hx f tx is continuous at a; that is, f ⴰ t is continuous at a. ■
52
■
CHAPTER 1
FUNCTIONS AND LIMITS
Where are the following functions continuous? 1 (a) hx sinx 2 (b) Fx sx 2 7 4 V EXAMPLE 7
SOLUTION
(a) We have hx f tx, where tx x 2
and
f x sin x
Now t is continuous on ⺢ since it is a polynomial, and f is also continuous everywhere by Theorem 6. Thus, h f ⴰ t is continuous on ⺢ by Theorem 8. (b) Notice that F can be broken up as the composition of four continuous functions: Ffⴰtⴰhⴰk where
f x
1 x
or
tx x 4
Fx f thkx hx sx
kx x 2 7
We know that each of these functions is continuous on its domain (by Theorems 5 and 6), so by Theorem 8, F is continuous on its domain, which is
{ x ⺢ sx 2 7
4} x
x 3 , 3 3, 3 3,
■
An important property of continuous functions is expressed by the following theorem, whose proof is found in more advanced books on calculus. Suppose that f is continuous on the closed interval a, b and let N be any number between f a and f b, where f a f b. Then there exists a number c in a, b such that f c N . 9 THE INTERMEDIATE VALUE THEOREM
The Intermediate Value Theorem states that a continuous function takes on every intermediate value between the function values f a and f b. It is illustrated by Figure 6. Note that the value N can be taken on once [as in part (a)] or more than once [as in part (b)]. y
y
f(b)
f(b)
y=ƒ
N N
y=ƒ
f(a) 0
FIGURE 6
a
f(a)
c
(a)
b
x
0
a c¡
c™
c£
b
x
(b)
If we think of a continuous function as a function whose graph has no hole or break, then it is easy to believe that the Intermediate Value Theorem is true. In geometric terms it says that if any horizontal line y N is given between y f a and
SECTION 1.5
y=ƒ y=N
N f(b) 0
a
b
■
53
y f b as in Figure 7, then the graph of f can’t jump over the line. It must intersect y N somewhere. It is important that the function f in Theorem 9 be continuous. The Intermediate Value Theorem is not true in general for discontinuous functions (see Exercise 34). One use of the Intermediate Value Theorem is in locating roots of equations as in the following example.
y f(a)
CONTINUITY
x
V EXAMPLE 8
Show that there is a root of the equation
FIGURE 7
4x 3 6x 2 3x 2 0 between 1 and 2. SOLUTION Let f x 4x 3 6x 2 3x 2. We are looking for a solution of the
given equation, that is, a number c between 1 and 2 such that f c 0. Therefore, we take a 1, b 2, and N 0 in Theorem 9. We have f 1 4 6 3 2 1 0 f 2 32 24 6 2 12 0
and
Thus f 1 0 f 2; that is, N 0 is a number between f 1 and f 2. Now f is continuous since it is a polynomial, so the Intermediate Value Theorem says there is a number c between 1 and 2 such that f c 0. In other words, the equation 4x 3 6x 2 3x 2 0 has at least one root c in the interval 1, 2. In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f 1.2 0.128 0
f 1.3 0.548 0
and
a root must lie between 1.2 and 1.3. A calculator gives, by trial and error, f 1.22 0.007008 0
f 1.23 0.056068 0
and
so a root lies in the interval 1.22, 1.23.
■
We can use a graphing calculator or computer to illustrate the use of the Intermediate Value Theorem in Example 8. Figure 8 shows the graph of f in the viewing rectangle 1, 3 by 3, 3 and you can see that the graph crosses the x-axis between 1 and 2. Figure 9 shows the result of zooming in to the viewing rectangle 1.2, 1.3 by 0.2, 0.2 . 3
0.2
3
_1
_3
FIGURE 8
1.2
_0.2
FIGURE 9
1.3
54
■
CHAPTER 1
FUNCTIONS AND LIMITS
In fact, the Intermediate Value Theorem plays a role in the very way these graphing devices work. A computer calculates a finite number of points on the graph and turns on the pixels that contain these calculated points. It assumes that the function is continuous and takes on all the intermediate values between two consecutive points. The computer therefore connects the pixels by turning on the intermediate pixels.
1.5
EXERCISES
1. Write an equation that expresses the fact that a function f
(b) Discuss the discontinuities of this function and their significance to someone who parks in the lot.
is continuous at the number 4. 2. If f is continuous on , , what can you say about its
8. Explain why each function is continuous or discontinuous.
(a) The temperature at a specific location as a function of time (b) The temperature at a specific time as a function of the distance due west from New York City (c) The altitude above sea level as a function of the distance due west from New York City (d) The cost of a taxi ride as a function of the distance traveled (e) The current in the circuit for the lights in a room as a function of time
graph? 3. (a) From the graph of f , state the numbers at which f is
discontinuous and explain why. (b) For each of the numbers stated in part (a), determine whether f is continuous from the right, or from the left, or neither. y
9. If f and t are continuous functions with f 3 5 and
lim x l 3 2 f x tx 4, find t3.
_4
0
_2
2
4
x
6
■ Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
10 –11
10. f x x 2 s7 x , 11. f x x 2x ,
4. From the graph of t, state the intervals on which t is
■
continuous.
■
■
■
a4
a 1
3 4
■
■
■
■
■
■
■
■
12. Use the definition of continuity and the properties of
y
limits to show that the function f x x s16 x 2 is continuous on the interval 4, 4 . ■ Explain why the function is discontinuous at a 1. Sketch the graph of the function.
13–16 _4
_2
2
4
6
8
x
13. f x
5. Sketch the graph of a function that is continuous every-
where except at x 3 and is continuous from the left at 3.
14. f x
6. Sketch the graph of a function that has a jump discontinuity
at x 2 and a removable discontinuity at x 4, but is continuous elsewhere. 7. A parking lot charges $3 for the first hour (or part of an
hour) and $2 for each succeeding hour (or part), up to a daily maximum of $10. (a) Sketch a graph of the cost of parking at this lot as a function of the time parked there.
15. f x
1 x 12
1 x1 2
if x 1
1 x2 1x
if x 1 if x 1
x2 x 16. f x x 2 1 1 ■
■
■
■
if x 1
if x 1 if x 1 ■
■
■
■
■
■
■
■
SECTION 1.5
3 18. Gx s x 1 x 3
19. Rx x 2 s2 x 1
20. hx
21. Fx sx sin x ■
■
■
■
tx
■
■
■
■
■
■
■
■ Locate the discontinuities of the function and illustrate by graphing.
■
■
■
25–26
■
25. lim x l4
■
28. f x ■
■
■
■
■
■
26. lim sinx sin x ■
■
■
■
■
■
■
if x 4 if x 4
sin x cos x
■
■
■
■
■
■
■
■
is discontinuous. At which of these points is f continuous from the right, from the left, or neither? Sketch the graph of f . 30. The gravitational force exerted by the Earth on a unit mass
at a distance r from the center of the planet is if r R if r R
where M is the mass of the Earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r ? 31. For what value of the constant c is the function f continu-
ous on , ? f x
3 sx 9x
a9
that f c 1000.
■
x 2 if x 0 if 0 x 1 f x 2x 2 2 x if x 1
Fr
(d) f x
35. If f x x 2 10 sin x, show that there is a number c such ■
36. Use the Intermediate Value Theorem to prove that there is a
positive number c such that c 2 2. (This proves the existence of the number s2 .)
29. Find the numbers at which the function
GMr R3 GM r2
a 4
0.25 and that f 0 1 and f 1 3. Let N 2. Sketch two possible graphs of f , one showing that f might not satisfy the conclusion of the Intermediate Value Theorem and one showing that f might still satisfy the conclusion of the Intermediate Value Theorem (even though it doesn’t satisfy the hypothesis).
■
if x 1 if x 1
x2 sx
x 64 x4
34. Suppose that a function f is continuous on [0, 1] except at
x l
■
(c) f x ■
Show that f is continuous on , .
27. f x
■
■
5 sx s5 x ■
■
■
if x 4 if x 4
3
Use continuity to evaluate the limit.
■
27–28
24. y tan sx
■
x2 c2 cx 20
tinuity at a ? If the discontinuity is removable, find a function t that agrees with f for x a and is continuous at a. x 2 2x 8 (a) f x a 2 x2 x7 (b) f x a7 x7
; 23–24
1 23. y 1 sin x
33. Which of the following functions f has a removable discon-
sin x x1
22. Fx sincossin x ■
55
32. Find the constant c that makes t continuous on , .
■ Explain, using Theorems 4, 5, 6, and 8, why the function is continuous at every number in its domain. State the domain.
17–22
x 17. Fx 2 x 5x 6
■
CONTINUITY
cx 2 2x if x 2 x 3 cx if x 2
■ Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
37– 40
37. x 4 x 3 0, 39. cos x x, ■
■
■
1, 2
3 38. s x 1 x,
0, 1 ■
40. tan x 2x, ■
■
■
■
■
0, 1 0, 1.4
■
■
■
41– 42 ■ (a) Prove that the equation has at least one real root. (b) Use your calculator to find an interval of length 0.01 that contains a root. 41. cos x x 3 ■
■
■
42. x 5 x 2 2x 3 0 ■
■
■
■
■
■
■
■
■
; 43– 44
■ (a) Prove that the equation has at least one real root. (b) Use your graphing device to find the root correct to three decimal places.
43. x 5 x 2 4 0 ■
■
■
■
44. sx 5 ■
■
■
■
■
1 x3 ■
■
45. Is there a number that is exactly 1 more than its cube?
■
56
■
CHAPTER 1
FUNCTIONS AND LIMITS
46. (a) Show that the absolute value function Fx x is
47. A Tibetan monk leaves the monastery at 7:00 AM and
takes his usual path to the top of the mountain, arriving at 7:00 PM. The following morning, he starts at 7:00 AM at the top and takes the same path back, arriving at the monastery at 7:00 PM. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.
continuous everywhere. (b) Prove that if f is a continuous function on an interval, then so is f . (c) Is the converse of the statement in part (b) also true? In other words, if f is continuous, does it follow that f is continuous? If so, prove it. If not, find a counterexample.
1.6
LIMITS INVOLVING INFINITY In this section we investigate the global behavior of functions and, in particular, whether their graphs approach asymptotes, vertical or horizontal. INFINITE LIMITS
1 0.5 0.2 0.1 0.05 0.01 0.001
In Example 8 in Section 1.3 we concluded that
1 x2
x
lim x l0
1 4 25 100 400 10,000 1,000,000
1 x2
by observing, from the table of values and the graph of y 1x 2 in Figure 1, that the values of 1x 2 can be made arbitrarily large by taking x close enough to 0. Thus the values of f x do not approach a number, so lim x l 0 1x 2 does not exist. To indicate this kind of behavior we use the notation lim x l0
y
does not exist
1 x2
| This does not mean that we are regarding as a number. Nor does it mean that the y=
limit exists. It simply expresses the particular way in which the limit does not exist: 1x 2 can be made as large as we like by taking x close enough to 0. In general, we write symbolically
1 ≈
lim f x x la
0
x
FIGURE 1
to indicate that the values of f x become larger and larger (or “increase without bound”) as x approaches a. 1 DEFINITION
■ A more precise version of Definition 1 is given at the end of this section.
The notation lim f x x la
means that the values of f x can be made arbitrarily large (as large as we please) by taking x sufficiently close to a (on either side of a) but not equal to a. Another notation for lim x l a f x is f x l
as
xla
SECTION 1.6
y=ƒ
a
x
x=a
57
or
“ f x becomes infinite as x approaches a”
or
“ f x increases without bound as x approaches a ”
This definition is illustrated graphically in Figure 2. Similarly, as shown in Figure 3,
FIGURE 2
lim f x
lim ƒ=`
x la
x a
■ When we say that a number is “large negative,” we mean that it is negative but its magnitude (absolute value) is large.
y
means that the values of f x are as large negative as we like for all values of x that are sufficiently close to a, but not equal to a. The symbol lim x l a f x can be read as “the limit of f x, as x approaches a, is negative infinity” or “ f x decreases without bound as x approaches a.” As an example we have
x=a
lim x l0
0
a
x
1 x2
Similar definitions can be given for the one-sided infinite limits
y=ƒ
lim f x
lim f x
x la
x la
lim f x
lim f x
x la
FIGURE 3 x a
y
y
a
0
(a) lim ƒ=` a_
x la
remembering that “x l a” means that we consider only values of x that are less than a, and similarly “x l a” means that we consider only x a. Illustrations of these four cases are given in Figure 4.
lim ƒ=_`
x
■
Again, the symbol is not a number, but the expression lim x l a f x is often read as “the limit of f x, as x approaches a, is infinity”
y
0
LIMITS INVOLVING INFINITY
x
y
a
0
x
(b) lim ƒ=` x
a+
y
a
0
(c) lim ƒ=_` x
a
0
x
x
(d) lim ƒ=_`
a_
x
a+
FIGURE 4 2 DEFINITION The line x a is called a vertical asymptote of the curve y f x if at least one of the following statements is true:
lim f x x la
lim f x x la
lim f x
x la
lim f x
x la
lim f x
x la
lim f x
x la
For instance, the y-axis is a vertical asymptote of the curve y 1x 2 because lim x l 0 1x 2 . In Figure 4 the line x a is a vertical asymptote in each of the four cases shown.
58
■
CHAPTER 1
FUNCTIONS AND LIMITS
EXAMPLE 1 Find lim x l3
2x 2x and lim . x l3 x 3 x3
SOLUTION If x is close to 3 but larger than 3, then the denominator x 3 is a small positive number and 2x is close to 6. So the quotient 2xx 3 is a large positive number. Thus, intuitively, we see that
y
y=
2x x-3
lim
x l3
5
Likewise, if x is close to 3 but smaller than 3, then x 3 is a small negative number but 2x is still a positive number (close to 6). So 2xx 3 is a numerically large negative number. Thus
x
0
2x x3
x=3
lim
x l3
2x x3
The graph of the curve y 2xx 3 is given in Figure 5. The line x 3 is a vertical asymptote.
FIGURE 5
■
EXAMPLE 2 Find the vertical asymptotes of f x tan x. SOLUTION Because
y
tan x 1 3π _π
_ 2
_
π 2
0
π 2
π
3π 2
x
sin x cos x
there are potential vertical asymptotes where cos x 0. In fact, since cos x l 0 as x l 2 and cos x l 0 as x l 2, whereas sin x is positive (and not near 0) when x is near 2, we have lim tan x
x l2
lim tan x
and
x l2
This shows that the line x 2 is a vertical asymptote. Similar reasoning shows that the lines x 2n 12, where n is an integer, are all vertical asymptotes of f x tan x. The graph in Figure 6 confirms this. ■
FIGURE 6
y=tan x
LIMITS AT INFINITY x
f x
0 1 2 3 4 5 10 50 100 1000
1 0 0.600000 0.800000 0.882353 0.923077 0.980198 0.999200 0.999800 0.999998
In computing infinite limits, we let x approach a number and the result was that the values of y became arbitrarily large (positive or negative). Here we let x become arbitrarily large (positive or negative) and see what happens to y. Let’s begin by investigating the behavior of the function f defined by f x
as x becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of f has been drawn by a computer in Figure 7. y
y=1
0
FIGURE 7
x2 1 x2 1
1
y=
≈-1 ≈+1
x
SECTION 1.6
LIMITS INVOLVING INFINITY
59
■
As x grows larger and larger you can see that the values of f x get closer and closer to 1. In fact, it seems that we can make the values of f x as close as we like to 1 by taking x sufficiently large. This situation is expressed symbolically by writing x2 1 1 x2 1
lim
x l
In general, we use the notation lim f x L
x l
to indicate that the values of f x approach L as x becomes larger and larger. 3 DEFINITION
Let f be a function defined on some interval a, . Then lim f x L
x l
means that the values of f x can be made as close to L as we like by taking x sufficiently large. Another notation for lim x l f x L is f x l L
as
xl
The symbol does not represent a number. Nonetheless, the expression lim f x L x l is often read as “the limit of f x, as x approaches infinity, is L” or
“the limit of f x, as x becomes infinite, is L”
or
“the limit of f x, as x increases without bound, is L”
The meaning of such phrases is given by Definition 3. A more precise definition, similar to the , definition of Section 1.3, is given at the end of this section. Geometric illustrations of Definition 3 are shown in Figure 8. Notice that there are many ways for the graph of f to approach the line y L (which is called a horizontal asymptote) as we look to the far right of each graph. y
y
y=L
y
y=L
y=ƒ
y=ƒ y=ƒ
y=L 0
x
FIGURE 8
Examples illustrating lim ƒ=L x `
0
0
x
x
Referring back to Figure 7, we see that for numerically large negative values of x, the values of f x are close to 1. By letting x decrease through negative values without bound, we can make f x as close to 1 as we like. This is expressed by writing lim
x l
x2 1 1 x2 1
60
■
CHAPTER 1
FUNCTIONS AND LIMITS
y
In general, as shown in Figure 9, the notation y=ƒ
lim f x L
x l
means that the values of f x can be made arbitrarily close to L by taking x sufficiently large negative. Again, the symbol does not represent a number, but the expression lim x l f x L is often read as
y=L 0
x y
“the limit of f x, as x approaches negative infinity, is L”
y=ƒ
4 DEFINITION The line y L is called a horizontal asymptote of the curve y f x if either
y=L
0
lim f x L
x
or
x l
lim f x L
x l
FIGURE 9
Examples illustrating lim ƒ=L
For instance, the curve illustrated in Figure 7 has the line y 1 as a horizontal asymptote because
x _`
lim
x l
x2 1 1 x2 1
The curve y f x sketched in Figure 10 has both y 1 and y 2 as horizontal asymptotes because lim f x 1
xl
and
lim f x 2
x l
y 2
y=2
0
y=_1
y=ƒ x
_1
FIGURE 10
EXAMPLE 3 Find the infinite limits, limits at infinity, and asymptotes for the funcy
tion f whose graph is shown in Figure 11. SOLUTION We see that the values of f x become large as x l 1 from both sides,
so lim f x
2
x l1
0
2
x
Notice that f x becomes large negative as x approaches 2 from the left, but large positive as x approaches 2 from the right. So lim f x
x l2
FIGURE 11
and
lim f x
x l2
Thus, both of the lines x 1 and x 2 are vertical asymptotes.
SECTION 1.6
LIMITS INVOLVING INFINITY
■
61
As x becomes large, it appears that f x approaches 4. But as x decreases through negative values, f x approaches 2. So lim f x 4
lim f x 2
and
x l
x l
This means that both y 4 and y 2 are horizontal asymptotes. EXAMPLE 4 Find lim
x l
■
1 1 and lim . x l x x
SOLUTION Observe that when x is large, 1x is small. For instance,
1 0.01 100
1 0.0001 10,000
In fact, by taking x large enough, we can make 1x as close to 0 as we please. Therefore, according to Definition 3, we have
y
y=Δ
lim
x l
0
x
lim
lim x `
1 1 =0, lim =0 x x _` x
1 0 x
Similar reasoning shows that when x is large negative, 1x is small negative, so we also have
x l
FIGURE 12
1 0.000001 1,000,000
1 0 x
It follows that the line y 0 (the x-axis) is a horizontal asymptote of the curve y 1x. (This is an equilateral hyperbola; see Figure 12.)
■
Most of the Limit Laws that were given in Section 1.4 also hold for limits at infinity. It can be proved that the Limit Laws listed in Section 1.4 (with the exception of Laws 9 and 10) are also valid if “x l a” is replaced by “x l ” or “ x l .” In particular, if we combine Law 6 with the results of Example 4 we obtain the following important rule for calculating limits.
5
If n is a positive integer, then lim
x l
V EXAMPLE 5
1 0 xn
lim
x l
1 0 xn
Evaluate lim
x l
3x 2 x 2 5x 2 4x 1
SOLUTION As x becomes large, both numerator and denominator become large, so it isn’t obvious what happens to their ratio. We need to do some preliminary algebra. To evaluate the limit at infinity of any rational function, we first divide both the numerator and denominator by the highest power of x that occurs in the denomi-
62
■
CHAPTER 1
FUNCTIONS AND LIMITS
nator. (We may assume that x 0, since we are interested only in large values of x.) In this case the highest power of x is x 2, and so, using the Limit Laws, we have 3x 2 x 2 1 2 3 2 3x x 2 x2 x x lim lim lim x l 5x 2 4x 1 x l 5x 2 4x 1 x l 4 1 5 2 x2 x x 2
■ Figure 13 illustrates Example 5 by showing how the graph of the given rational function approaches the horizontal asymptote y 53 .
y
y=0.6
1 2 2 x x
lim 5
4 1 2 x x
x l
0
1
x
lim 3
x l
1 2 lim x l x 1 lim 5 4 lim lim x l x l x x l lim 3 lim
x l
FIGURE 13
y=
3≈-x-2 5≈+4x+1
x l
300 500
3 5
1 x2 1 x2
[by (5)]
A similar calculation shows that the limit as x l is also 35.
■
EXAMPLE 6 Compute lim (sx 2 1 x). x l
SOLUTION Because both sx 2 1 and x are large when x is large, it’s difficult to We can think of the given function as having a denominator of 1. ■
see what happens to their difference, so we use algebra to rewrite the function. We first multiply numerator and denominator by the conjugate radical: lim (sx 2 1 x) lim (sx 2 1 x)
x l
x l
lim
x l
x 2 1 x 2 1 lim 2 1 x 2 1 x x l sx sx
The Squeeze Theorem could be used to show that this limit is 0. But an easier method is to divide numerator and denominator by x. Doing this and remembering that x sx 2 for x 0, we obtain 1 1 x lim (sx 2 1 x) lim lim x l x l sx 2 1 x x l sx 2 1 x x
y
y=œ„„„„„-x ≈+1 1
lim 0
FIGURE 14
1
sx 2 1 x sx 2 1 x
x l
x
Figure 14 illustrates this result.
1
1 x 1 1 x2
0 0 s1 0 1 ■
SECTION 1.6
EXAMPLE 7 Evaluate lim sin xl
LIMITS INVOLVING INFINITY
■
63
1 . x
SOLUTION If we let t 1x, then t l 0 as x l . Therefore
1 lim sin t 0 tl0 x
lim sin
xl
■
(See Exercise 55.) EXAMPLE 8 Evaluate lim sin x. x l
SOLUTION As x increases, the values of sin x oscillate between 1 and 1 infinitely often. Thus lim x l sin x does not exist. ■ INFINITE LIMITS AT INFINITY
lim f x
The notation
x l
is used to indicate that the values of f x become large as x becomes large. Similar meanings are attached to the following symbols: lim f x
lim f x
x l
x l
lim f x
x l
EXAMPLE 9 Find lim x 3 and lim x 3. xl
y
x l
SOLUTION When x becomes large, x 3 also becomes large. For instance,
10 3 1000
y=˛
0
x
100 3 1,000,000
1000 3 1,000,000,000
In fact, we can make x 3 as big as we like by taking x large enough. Therefore, we can write lim x 3 xl
Similarly, when x is large negative, so is x 3. Thus lim x 3
x l
FIGURE 15
lim x#=`, lim x#=_` x `
x _`
These limit statements can also be seen from the graph of y x 3 in Figure 15.
■
EXAMPLE 10 Find lim x 2 x. x l
| SOLUTION It would be wrong to write lim x 2 x lim x 2 lim x
x l
x l
x l
The Limit Laws can’t be applied to infinite limits because is not a number ( can’t be defined). However, we can write lim x 2 x lim xx 1
x l
x l
because both x and x 1 become arbitrarily large.
■
64
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CHAPTER 1
FUNCTIONS AND LIMITS
EXAMPLE 11 Find lim
x l
x2 x . 3x
SOLUTION We divide numerator and denominator by x (the highest power of x that
occurs in the denominator): lim
x l
x2 x x1 lim x l 3 3x 1 x
because x 1 l and 3x 1 l 1 as x l .
■
PRECISE DEFINITIONS
The following is a precise version of Definition 1. 6 DEFINITION Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then
lim f x
xla
means that for every positive number M there is a positive number such that
y
y=M
M
0
x
a
a-∂ FIGURE 16
a+∂
0 xa
if
then
f x M
This says that the values of f x can be made arbitrarily large (larger than any given number M ) by taking x close enough to a (within a distance , where depends on M , but with x a). A geometric illustration is shown in Figure 16. Given any horizontal line y M , we can find a number 0 such that if we restrict x to lie in the interval a , a but x a, then the curve y f x lies above the line y M . You can see that if a larger M is chosen, then a smaller may be required. 1 . x2 SOLUTION Let M be a given positive number. According to Definition 6, we need to find a number such that V EXAMPLE 12
Use Definition 6 to prove that lim
xl0
0 x
if
But x 2 1M &? if
1 M x2
then
that is
x 1sM . We can choose 1sM
0 x
1 s
then
x2
1 M
because
1 1 2 M x2
Therefore, by Definition 6, lim
xl0
1 x2
■
SECTION 1.6
LIMITS INVOLVING INFINITY
■
65
Similarly, limx l a f x means that for every negative number N there is a positive number such that if 0 x a , then f x N. Definition 3 can be stated precisely as follows.
7 DEFINITION
Let f be a function defined on some interval a, . Then lim f x L
xl
means that for every 0 there is a corresponding number N such that xN
if
Module 1.3/1.6 illustrates Definition 7 graphically and numerically.
f x L
then
In words, this says that the values of f x can be made arbitrarily close to L (within a distance , where is any positive number) by taking x sufficiently large (larger than N , where N depends on ). Graphically it says that by choosing x large enough (larger than some number N ) we can make the graph of f lie between the given horizontal lines y L and y L as in Figure 17. This must be true no matter how small we choose . Figure 18 shows that if a smaller value of is chosen, then a larger value of N may be required. y
y=ƒ
y=L +∑ ∑ L ∑ y=L -∑
ƒ is in here
0
x
N
FIGURE 17
lim ƒ=L
when x is in here
x `
y
y=ƒ y=L+∑
L
y=L-∑ 0
N
FIGURE 18
x
lim ƒ=L x `
Similarly, limx l f x L means that for every 0 there is a corresponding number N such that if x N, then f x L .
1 0. x
EXAMPLE 13 Use Definition 7 to prove that lim
xl
SOLUTION Given 0, we want to find N such that
if
xN
then
1 0 x
66
■
CHAPTER 1
FUNCTIONS AND LIMITS
In computing the limit we may assume that x 0. Then 1x &? x 1. Let’s choose N 1. So xN
if
1
then
1 1 0 x x
Therefore, by Definition 7, lim
xl
1 0 x
Figure 19 illustrates the proof by showing some values of and the corresponding values of N . y
y
y
∑=1 ∑=0.2 0
x
N=1
∑=0.1
0
N=5
x
0
N=10
x
FIGURE 19
Finally we note that an infinite limit at infinity can be defined as follows. The geometric illustration is given in Figure 20.
y
y=M
8 DEFINITION
M
Let f be a function defined on some interval a, . Then lim f x
xl
0
x
N
FIGURE 20
means that for every positive number M there is a corresponding positive number N such that if xN then f x M
lim ƒ=` x `
Similar definitions apply when the symbol is replaced by .
1.6
EXERCISES
1. For the function f whose graph is given, state the following.
(a) lim f x
(b)
(c) lim f x
(d) lim f x
x l2
x l1
y
lim f x
x l1
x l
(e) lim f x x l
(f ) The equations of the asymptotes
1 1
x
SECTION 1.6
2. For the function t whose graph is given, state the following.
(a) lim tx
(b) lim tx
(c) lim tx
(d) lim tx
(e) lim tx
(f ) The equations of the asymptotes
x l
asymptotes of the curve y
x l0
x l2
x
2
13. 3– 8 ■ Sketch the graph of an example of a function f that satisfies all of the given conditions.
lim f x 0,
x l0
lim f x ,
x l
lim f x ,
x l2
lim f x ,
x l
lim f x 0,
x l
16. lim cot x x l
19. lim
x 3 5x 2x x 2 4
20. lim
21. lim
4u 4 5 u 22u 2 1
22. lim
ul
lim f x
x l0
xl5
18. lim
x l
5. lim f x , x l0
lim
6 x5
14. lim
2x x 12
x l2
x l
lim f x 1
x2 x3
sec x
lim f x 1,
x l0
lim
x l3
x l1
17. 4. lim f x ,
x
Find the limit.
■
15. lim
f is odd
x l
2 x
to estimate the value of lim x l f x correct to two decimal places. (b) Use a table of values of f x to estimate the limit to four decimal places. 13–31
f 1 1,
f x 1
1
3. f 0 0,
x3 x 2x 1 3
; 12. (a) Use a graph of
y
0
67
; 11. Use a graph to estimate all the vertical and horizontal
x l
x l3
■
LIMITS INVOLVING INFINITY
x l
3
3x 5 x4
t l
2
x l
t2 2 t t2 1 3
x2 s9x 2 1
23. lim (s9x 2 x 3x) x l
6. lim f x ,
lim f x 3,
lim f x 3
24. lim (sx 2 ax sx 2 bx
)
25. lim cos x
26. lim
sin 2x x2
lim f x 3
27. lim ( x sx )
28. lim
x 3 2x 3 5 2x 2
8. lim f x ,
29. lim x x
30. lim x 2 x 4
x l2
7. f 0 3,
x l
lim f x 4,
x l0
lim f x ,
x l
lim f x 2,
x l0
lim f x ,
x l 4
x l
x l
x l
lim f x ,
x l 4
xl
x l
lim f x 2,
xl3
■
■
■
x l
■
■
■
■
f 0 0, ■
■
■
■
x2 lim x x l 2 by evaluating the function f x x 22 x for x 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 50, and 100. Then use a graph of f to support your guess. 1 1 and lim 3 x l1 x 1 x 1 (a) by evaluating f x 1x 3 1 for values of x that approach 1 from the left and from the right, (b) by reasoning as in Example 1, and (c) from a graph of f .
10. Determine lim x l1
4
f is even
; 9. Guess the value of the limit
;
x l
xl
5
xl
x l
■
xx x 1 x2 x4 3
31. lim
xl
■
■
■
5
■
■
■
■
■
■
■
■
; 32. (a) Graph the function f x
s2x 2 1 3x 5
How many horizontal and vertical asymptotes do you observe? Use the graph to estimate the values of the limits
3
lim
x l
s2x 2 1 3x 5
and
lim
x l
s2x 2 1 3x 5
(b) By calculating values of f x, give numerical estimates of the limits in part (a).
■
68
■
CHAPTER 1
FUNCTIONS AND LIMITS
(b) Two functions are said to have the same end behavior if their ratio approaches 1 as x l . Show that P and Q have the same end behavior.
(c) Calculate the exact values of the limits in part (a). Did you get the same value or different values for these two limits? [In view of your answer to part (a), you might have to check your calculation for the second limit.]
; 33–34
41. Let P and Q be polynomials. Find
Find the horizontal and vertical asymptotes of each curve. Check your work by graphing the curve and estimating the asymptotes. ■
33. y ■
■
2x 2 x 1 x2 x 2 ■
■
34. Fx ■
■
■
■
lim
xl
x9 s4x 2 3x 2 ■
■
■
Px Qx
if the degree of P is (a) less than the degree of Q and (b) greater than the degree of Q. ■
42. Make a rough sketch of the curve y x n (n an integer)
; 35. (a) Estimate the value of
for the following five cases: (i) n 0 (ii) n 0, n odd (iii) n 0, n even (iv) n 0, n odd (v) n 0, n even Then use these sketches to find the following limits. (a) lim x n (b) lim x n
lim (sx 2 x 1 x)
x l
by graphing the function f x sx 2 x 1 x. (b) Use a table of values of f x to guess the value of the limit. (c) Prove that your guess is correct.
x l0
x l0
(c) lim x n
(d) lim x n
x l
; 36. (a) Use a graph of
x l
43. Find lim x l f x if, for all x 5,
f x s3x 2 8x 6 s3x 2 3x 1
4x 1 4x 2 3x f x x x2
to estimate the value of lim x l f x to one decimal place. (b) Use a table of values of f x to estimate the limit to four decimal places. (c) Find the exact value of the limit.
44. In the theory of relativity, the mass of a particle with velocity v is
m
; 37. Estimate the horizontal asymptote of the function f x
3x 3 500x 2 x 500x 2 100x 2000
where m 0 is the mass of the particle at rest and c is the speed of light. What happens as v l c?
3
by graphing f for 10 x 10. Then calculate the equation of the asymptote by evaluating the limit. How do you explain the discrepancy?
45. (a) A tank contains 5000 L of pure water. Brine that con-
tains 30 g of salt per liter of water is pumped into the tank at a rate of 25 Lmin. Show that the concentration of salt t minutes later (in grams per liter) is
38. Find a formula for a function that has vertical asymptotes
x 1 and x 3 and horizontal asymptote y 1.
Ct
39. Find a formula for a function f that satisfies the following
conditions: lim f x 0, x l
lim f x ,
x l3
lim f x , x l0
30t 200 t
(b) What happens to the concentration as t l ?
f 2 0,
lim f x
4x 2 5x 2. x l 2x 2 1 (b) By graphing the function in part (a) and the line y 1.9 on a common screen, find a number N such that
46. (a) Show that lim
x l3
; 40. By the end behavior of a function we mean the behavior of
its values as x l and as x l . (a) Describe and compare the end behavior of the functions Px 3x 5 5x 3 2x
m0 s1 v 2c 2
Qx 3x 5
by graphing both functions in the viewing rectangles 2, 2 by 2, 2 and 10, 10 by 10,000, 10,000 .
;
4x 2 5x 1.9 2x 2 1
when
What if 1.9 is replaced by 1.99?
xN
CHAPTER 1
47. How close to 3 do we have to take x so that
x l3
49. Prove that lim x l 1
lim
xl
1 . x 34
(b) Taking n 2 in (5), we have the statement lim
xl
s4x 2 1 2 x1
54. Prove, using Definition 8, that lim x 3 . xl
55. Prove that
lim f x lim f 1t
; 51. Use a graph to find a number N such that
6x 2 5x 3 3 0.2 2x2 1
1
xl
xN
REVIEW
2. Discuss four ways of representing a function. Illustrate your
3. (a) What is an even function? How can you tell if a func-
tion is even by looking at its graph? (b) What is an odd function? How can you tell if a function is odd by looking at its graph? 4. What is a mathematical model? 5. Give an example of each type of function.
(b) Power function (d) Quadratic function (f ) Rational function
6. Sketch by hand, on the same axes, the graphs of the follow-
(b) tx x 2 (d) jx x 4
7. Draw, by hand, a rough sketch of the graph of each function.
8. Suppose that f has domain A and t has domain B.
(a) What is the domain of f t ? (b) What is the domain of f t ? (c) What is the domain of ft ?
9. How is the composite function f ⴰ t defined? What is its
domain?
discussion with examples.
(a) y sin x (c) y 2 x (e) y x
tl0
CONCEPT CHECK
(b) What is the graph of a function? (c) How can you tell whether a given curve is the graph of a function?
ing functions. (a) f x x (c) hx x 3
x l
if these limits exist.
1. (a) What is a function? What are its domain and range?
(a) Linear function (c) Exponential function (e) Polynomial of degree 5
tl0
lim f x lim f 1t
and whenever
1 0 x2
Prove this directly using Definition 7.
illustrate Definition 7 by finding values of N that correspond to 0.5 and 0.1.
2x 1 sx 1
53. (a) How large do we have to take x so that 1x 2 0.0001?
; 50. For the limit lim
69
illustrate Definition 8 by finding a value of N that corresponds to M 100.
5 . x 1 3
xl
■
; 52. For the limit
1 10,000 x 34 48. Prove, using Definition 6, that lim
REVIEW
(b) y tan x (d) y 1x (f ) y sx
10. Suppose the graph of f is given. Write an equation for
each of the graphs that are obtained from the graph of f as follows. (a) Shift 2 units upward. (b) Shift 2 units downward. (c) Shift 2 units to the right. (d) Shift 2 units to the left. (e) Reflect about the x-axis. (f ) Reflect about the y-axis. (g) Stretch vertically by a factor of 2. (h) Shrink vertically by a factor of 2. (i) Stretch horizontally by a factor of 2. ( j) Shrink horizontally by a factor of 2. 11. Explain what each of the following means and illustrate
with a sketch. (a) lim f x L
(b) lim f x L
(c) lim f x L
(d) lim f x
x la
x la
(e) lim f x L x l
x la
x la
70
■
CHAPTER 1
FUNCTIONS AND LIMITS
12. Describe several ways in which a limit can fail to exist.
(b) What does it mean for f to be continuous on the interval , ? What can you say about the graph of such a function?
Illustrate with sketches. 13. State the following Limit Laws.
(a) (c) (e) (g)
Sum Law Constant Multiple Law Quotient Law Root Law
(b) Difference Law (d) Product Law (f ) Power Law
16. What does the Intermediate Value Theorem say? 17. (a) What does it mean to say that the line x a is a vertical
asymptote of the curve y f x? Draw curves to illustrate the various possibilities. (b) What does it mean to say that the line y L is a horizontal asymptote of the curve y f x? Draw curves to illustrate the various possibilities.
14. What does the Squeeze Theorem say? 15. (a) What does it mean for f to be continuous at a?
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
11. If p is a polynomial, then lim x l b px pb. 12. If lim x l 0 f x and lim x l 0 tx , then
lim x l 0 f x tx 0.
1. If f is a function, then f s t f s f t. 2. If f s f t, then s t.
13. A function can have two different horizontal asymptotes.
3. A vertical line intersects the graph of a function at most
14. If f has domain 0, and has no horizontal asymptote,
then lim x l f x or lim x l f x .
once. 4. If f and t are functions, then f ⴰ t t ⴰ f .
2x 8 5. lim x l4 x4 x4
x l1
lim x 2 6x 7
x 6x 7 x l1 x 2 5x 6 lim x 2 5x 6 2
6. lim
2x 8 lim lim x l4 x 4 x l4 x 4
x l1
15. If the line x 1 is a vertical asymptote of y f x, then f
is not defined at 1. 16. If f 1 0 and f 3 0, then there exists a number c
between 1 and 3 such that f c 0.
17. If f is continuous at 5 and f 5 2 and f 4 3, then
lim x l 2 f 4x 2 11 2.
lim x 3
7. lim x l1
x3 x l1 x 2 2x 4 lim x 2 2x 4
18. If f is continuous on 1, 1 and f 1 4 and f 1 3,
then there exists a number r such that r 1 and f r .
x l1
8. If lim x l 5 f x 2 and lim x l 5 tx 0, then
limx l 5 f xtx does not exist.
19. Let f be a function such that lim x l 0 f x 6. Then
9. If lim x l5 f x 0 and lim x l 5 tx 0, then
lim x l 5 f xtx does not exist.
10. If lim x l 6 f xtx exists, then the limit must be f 6t6.
there exists a number such that if 0 x , then f x 6 1.
20. If f x 1 for all x and lim x l 0 f x exists, then
lim x l 0 f x 1.
EXERCISES 1. Let f be the function whose graph is given.
(a) (b) (c) (d) (e) (f )
Estimate the value of f 2. Estimate the values of x such that f x 3. State the domain of f. State the range of f. On what interval is f increasing? Is f even, odd, or neither even nor odd? Explain.
y
f 1 1
x
CHAPTER 1
2. Determine whether each curve is the graph of a function
■
71
16. Find an expression for the function whose graph consists of
the line segment from the point 2, 2 to the point 1, 0 together with the top half of the circle with center the origin and radius 1.
of x. If it is, state the domain and range of the function. y y (a) (b) 2
REVIEW
2
17. If f x sx and tx sin x, find the functions (a) f ⴰ t,
(b) t ⴰ f , (c) f ⴰ f , (d) t ⴰ t, and their domains.
0
x
1
0
x
1
18. Express the function Fx 1sx sx as a composition
of three functions.
; 19. Use graphs to discover what members of the family of func3–6
Find the domain and range of the function.
■
3. f x s4 3x 2
4. tx 1x 1
5. y 1 sin x
6. y tan 2x
■
tions f x sin n x have in common, where n is a positive integer. How do they differ? What happens to the graphs as n becomes large?
■
■
■
■
■
■
■
■
20. A small-appliance manufacturer finds that it costs $9000 to ■
■
■
7. Suppose that the graph of f is given. Describe how the
graphs of the following functions can be obtained from the graph of f. (a) y f x 8 (b) y f x 8 (c) y 1 2 f x (d) y f x 2 2 (e) y f x (f ) y 3 f x 8. The graph of f is given. Draw the graphs of the following
functions. (a) y f x 8 (c) y 2 f x
produce 1000 toaster ovens a week and $12,000 to produce 1500 toaster ovens a week. (a) Express the cost as a function of the number of toaster ovens produced, assuming that it is linear. Then sketch the graph. (b) What is the slope of the graph and what does it represent? (c) What is the y-intercept of the graph and what does it represent? 21. The graph of f is given.
(a) Find each limit, or explain why it does not exist. (i) lim f x (ii) lim f x
(b) y f x (d) y 12 f x 1
x l2
x l3
(iii) lim f x
(iv) lim f x
(v) lim f x
(vi) lim f x
(vii) lim f x
(viii) lim f x
x l3
y
x l4
x l0
x l2
x l
1 0
1
x l
(b) State the equations of the horizontal asymptotes. (c) State the equations of the vertical asymptotes. (d) At what numbers is f discontinuous? Explain.
x
y
9–14
■
Use transformations to sketch the graph of the function.
9. y sin 2 x
10. y x 2 2
11. y 1 x 3
12. y 2 sx
1 2
1
1 13. f x x2 14. f x ■
■
0
1x 1 x2
■
■
if x 0 if x 0 ■
■
■
■
■
■
■
15. Determine whether f is even, odd, or neither even nor odd.
(a) (b) (c) (d)
f x 2x 3x 2 f x x 3 x 7 f x cosx 2 f x 1 sin x 5
x
1
2
■
22. Sketch the graph of an example of a function f that satisfies
all of the following conditions: lim f x 2, lim f x 0, x l
lim f x ,
x l3
xl
lim f x 2,
x l3
f is continuous from the right at 3
lim f x ,
x l3
72
■
23–38
CHAPTER 1
FUNCTIONS AND LIMITS
41. If 2x 1 f x x 2 for 0 x 3, find lim x l1 f x.
Find the limit.
■
x 9 x 2 2x 3 2
23. lim cosx sin x
24. lim
x2 9 25. lim 2 x l3 x 2x 3
x2 9 26. lim 2 x l1 x 2x 3
xl0
x l3
h 1 1 h 3
27. lim
h l0
r l9
43– 46
28. lim t l2
vl4
45. lim
xl
4v
4 v
■
1 2x x 2 33. lim x l 1 x 2x 2
1 2x 2 x 4 34. lim x l 5 x 3x 4
v 2 2v 8
xl1
cot 2x csc x
37. lim
xl0
■
■
; 39– 40
1 1 2 x1 x 3x 2
■
■
■
■
■
■
■
■
■
if x 0 if 0 x 3 if x 3
(v) lim f x x l3
■
t3 tan3 2t
(iii) lim f x x l0
(vi) lim f x x l3
■
48. Show that each function is continuous on its domain. State ■
■
the domain. sx 2 9 (a) tx 2 x 2
■
■
■
■
■
4 (b) hx s x x 3 cos x
■ Use the Intermediate Value Theorem to show that there is a root of the equation in the given interval.
49–50
50. 2 sin x 3 2x, ■
■
x l0
49. 2x 3 x 2 2 0,
■
■
x l3
cos2 x x2
■
■
2 sx 4
(b) Where is f discontinuous? (c) Sketch the graph of f .
40. y sx 2 x 1 sx 2 x ■
■
(a) Evaluate each limit, if it exists. (i) lim f x (ii) lim f x
Use graphs to discover the asymptotes of the curve. Then prove what you have discovered.
■
■
sx f x 3 x x 32
■
39. y
x l4
(iv) lim f x
tl0
■
46. lim
x l0
38. lim ■
■
v 4 16
35. lim (sx 2 4x 1 x) 36. lim
■
xl0
1 0 x4
47. Let
32. lim
xl
3 x 0 44. lim s
x l5
4 ss 31. lim s l16 s 16
v l2
Prove the statement using the precise definition of
43. lim 7x 27 8
t 4 t3 8
30. lim
■
a limit.
2
sr r 94
29. lim
42. Prove that lim x l 0 x 2 cos1x 2 0.
■
■
■
■
■
2, 1 0, 1
■
■
■
■
■
■
■
■
DERIVATIVES
2
In this chapter we study a special type of limit, called a derivative, that occurs when we want to find the slope of a tangent line, or a velocity, or any instantaneous rate of change.
2.1
DERIVATIVES AND RATES OF CHANGE The problem of finding the tangent line to a curve and the problem of finding the velocity of an object involve finding the same type of limit, which we call a derivative. THE TANGENT PROBLEM
The word tangent is derived from the Latin word tangens, which means “touching.” Thus a tangent to a curve is a line that touches the curve. In other words, a tangent line should have the same direction as the curve at the point of contact. How can this idea be made precise? For a circle we could simply follow Euclid and say that a tangent is a line that intersects the circle once and only once as in Figure 1(a). For more complicated curves this definition is inadequate. Figure l(b) shows two lines L and T passing through a point P on a curve C. The line L intersects C only once, but it certainly does not look like what we think of as a tangent. The line T, on the other hand, looks like a tangent but it intersects C twice.
T P T
C
L FIGURE 1
(a)
To be specific, let’s look at the problem of trying to find a tangent line T to the parabola y x 2 in the following example.
y
Q { x, ≈}
T V EXAMPLE 1
y=≈
(b)
Find an equation of the tangent line to the parabola y x 2 at the
point P1, 1.
P (1, 1)
SOLUTION We will be able to find an equation of the tangent line T as soon as we 0
FIGURE 2
x
know its slope m. The difficulty is that we know only one point, P, on T, whereas we need two points to compute the slope. But observe that we can compute an approximation to m by choosing a nearby point Qx, x 2 on the parabola (as in Figure 2) and computing the slope mPQ of the secant line PQ. 73
74
■
CHAPTER 2
DERIVATIVES
We choose x 1 so that Q P. Then mPQ
x2 1 x1
What happens as x approaches 1? From Figure 3 we see that Q approaches P along the parabola and the secant lines PQ rotate about P and approach the tangent line T. y
y
y
Q T
T
T
Q Q P
P
0
x
P
0
0
x
x
Q approaches P from the right y
y
y
T
Q
T
P
T
P
P
Q 0
x
Q 0
0
x
x
Q approaches P from the left FIGURE 3
It appears that the slope m of the tangent line is the limit of the slopes of the secant lines as x approaches 1: In Visual 2.1A you can see how the process in Figure 3 works for additional functions.
m lim x l1
x2 1 x 1x 1 lim xl1 x1 x1
lim x 1 1 1 2 x l1
■ Point-slope form for a line through the point x1 , y1 with slope m : y y1 mx x 1
Using the point-slope form of the equation of a line, we find that an equation of the tangent line at 1, 1 is y 1 2x 1
or
y 2x 1
■
We sometimes refer to the slope of the tangent line to a curve at a point as the slope of the curve at the point. The idea is that if we zoom in far enough toward the point, the curve looks almost like a straight line. Figure 4 illustrates this procedure for the curve y x 2 in Example 1. The more we zoom in, the more the parabola looks like a line. In other words, the curve becomes almost indistinguishable from its tangent line.
SECTION 2.1
2
DERIVATIVES AND RATES OF CHANGE
1.5
(1, 1)
2
75
1.1
(1, 1)
0
■
(1, 1)
1.5
0.5
1.1
0.9
FIGURE 4 Zooming in toward the point (1, 1) on the parabola y=≈
Visual 2.1B shows an animation of Figure 4.
In general, if a curve C has equation y f x and we want to find the tangent line to C at the point Pa, f a, then we consider a nearby point Qx, f x, where x a, and compute the slope of the secant line PQ : mPQ
f x f a xa
Then we let Q approach P along the curve C by letting x approach a. If mPQ approaches a number m, then we define the tangent T to be the line through P with slope m. (This amounts to saying that the tangent line is the limiting position of the secant line PQ as Q approaches P. See Figure 5.) y
y
t Q
Q{ x, ƒ }
Q
ƒ-f(a) P
P { a, f(a)}
Q
x-a
0
a
x
x
x
0
FIGURE 5 1 DEFINITION The tangent line to the curve y f x at the point Pa, f a is the line through P with slope f x f a m lim xla xa
provided that this limit exists. Q { a+h, f(a+h)} y
t
There is another expression for the slope of a tangent line that is sometimes easier to use. If h x a, then x a h and so the slope of the secant line PQ is
P { a, f(a)} f(a+h)-f(a)
h 0
FIGURE 6
a
a+h
x
mPQ
f a h f a h
(See Figure 6 where the case h 0 is illustrated and Q is to the right of P. If it happened that h 0, however, Q would be to the left of P.) Notice that as x approaches a, h approaches 0 (because h x a) and so the expression for the slope of the
76
■
CHAPTER 2
DERIVATIVES
tangent line in Definition 1 becomes
2
m lim
hl0
f a h f a h
EXAMPLE 2 Find an equation of the tangent line to the hyperbola y 3x at the
point 3, 1.
SOLUTION Let f x 3x. Then the slope of the tangent at 3, 1 is
m lim
hl0
f 3 h f 3 h
3 3 3 h 1 3h 3h lim lim hl0 hl0 h h y
lim
3 y= x
x+3y-6=0
hl0
h 1 1 lim h l 0 h3 h 3h 3
Therefore, an equation of the tangent at the point 3, 1 is
(3, 1)
y 1 13 x 3
x
0
x 3y 6 0
which simplifies to
The hyperbola and its tangent are shown in Figure 7.
FIGURE 7
■
THE VELOCITY PROBLEM position at time t=a
position at time t=a+h s
0
f(a+h)-f(a)
f(a) f(a+h) FIGURE 8 s
Q { a+h, f(a+h)} P { a, f(a)} h
0
mPQ=
a
a+h
f(a+h)-f(a) average = h velocity
FIGURE 9
t
In Section 1.3 we investigated the motion of a ball dropped from the CN Tower and defined its velocity to be the limiting value of average velocities over shorter and shorter time periods. In general, suppose an object moves along a straight line according to an equation of motion s f t, where s is the displacement (directed distance) of the object from the origin at time t. The function f that describes the motion is called the position function of the object. In the time interval from t a to t a h the change in position is f a h f a. (See Figure 8.) The average velocity over this time interval is displacement f a h f a average velocity time h which is the same as the slope of the secant line PQ in Figure 9. Now suppose we compute the average velocities over shorter and shorter time intervals a, a h . In other words, we let h approach 0. As in the example of the falling ball, we define the velocity (or instantaneous velocity) va at time t a to be the limit of these average velocities:
3
va lim
hl0
f a h f a h
SECTION 2.1
DERIVATIVES AND RATES OF CHANGE
■
77
This means that the velocity at time t a is equal to the slope of the tangent line at P (compare Equations 2 and 3). Now that we know how to compute limits, let’s reconsider the problem of the falling ball. V EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. (a) What is the velocity of the ball after 5 seconds? (b) How fast is the ball traveling when it hits the ground?
Recall from Section 1.3: The distance (in meters) fallen after t seconds is 4.9t 2. ■
SOLUTION We first use the equation of motion s f t 4.9t 2 to find the velocity va after a seconds: va lim
hl0
lim
hl0
f a h f a 4.9a h2 4.9a 2 lim hl0 h h 4.9a 2 2ah h 2 a 2 4.92ah h 2 lim hl0 h h
lim 4.92a h 9.8a hl0
(a) The velocity after 5 s is v5 9.85 49 ms. (b) Since the observation deck is 450 m above the ground, the ball will hit the ground at the time t1 when st1 450, that is, 4.9t12 450 This gives t12
450 4.9
t1
and
450 9.6 s 4.9
The velocity of the ball as it hits the ground is therefore
vt1 9.8t1 9.8
450 94 ms 4.9
■
DERIVATIVES
We have seen that the same type of limit arises in finding the slope of a tangent line (Equation 2) or the velocity of an object (Equation 3). In fact, limits of the form lim
h l0
f a h f a h
arise whenever we calculate a rate of change in any of the sciences or engineering, such as a rate of reaction in chemistry or a marginal cost in economics. Since this type of limit occurs so widely, it is given a special name and notation. 4 DEFINITION
The derivative of a function f at a number a, denoted by
f a, is ■
f a lim
f a is read “ f prime of a .”
h l0
if this limit exists.
f a h f a h
78
■
CHAPTER 2
DERIVATIVES
If we write x a h, then h x a and h approaches 0 if and only if x approaches a. Therefore, an equivalent way of stating the definition of the derivative, as we saw in finding tangent lines, is f a lim
5
xla
V EXAMPLE 4
f x f a xa
Find the derivative of the function f x x 2 8x 9 at the
number a. SOLUTION From Definition 4 we have
f a h f a h
f a lim
h l0
lim
a h2 8a h 9 a 2 8a 9 h
lim
a 2 2ah h 2 8a 8h 9 a 2 8a 9 h
lim
2ah h 2 8h lim 2a h 8 h l0 h
h l0
h l0
h l0
2a 8
■
We defined the tangent line to the curve y f x at the point Pa, f a to be the line that passes through P and has slope m given by Equation 1 or 2. Since, by Definition 4, this is the same as the derivative f a, we can now say the following. The tangent line to y f x at a, f a is the line through a, f a whose slope is equal to f a, the derivative of f at a. If we use the point-slope form of the equation of a line, we can write an equation of the tangent line to the curve y f x at the point a, f a:
y
y=≈-8x+9
x
0 (3, _6)
y=_2x FIGURE 10
y f a f ax a V EXAMPLE 5 Find an equation of the tangent line to the parabola y x 2 8x 9 at the point 3, 6.
SOLUTION From Example 4 we know that the derivative of f x x 2 8x 9 at
the number a is f a 2a 8. Therefore, the slope of the tangent line at 3, 6 is f 3 23 8 2. Thus, an equation of the tangent line, shown in Figure 10, is y 6 2x 3
or
y 2x
■
RATES OF CHANGE
Suppose y is a quantity that depends on another quantity x. Thus y is a function of x and we write y f x. If x changes from x 1 to x 2 , then the change in x (also called the increment of x) is x x 2 x 1
SECTION 2.1
Q { ¤, ‡}
y
DERIVATIVES AND RATES OF CHANGE
■
79
and the corresponding change in y is y f x 2 f x 1
P {⁄, fl}
Îy
The difference quotient y f x 2 f x 1 x x2 x1
Îx ⁄
0
¤
x
average rate of change ⫽ mPQ instantaneous rate of change ⫽ slope of tangent at P FIGURE 11
is called the average rate of change of y with respect to x over the interval x 1, x 2 and can be interpreted as the slope of the secant line PQ in Figure 11. By analogy with velocity, we consider the average rate of change over smaller and smaller intervals by letting x 2 approach x 1 and therefore letting x approach 0. The limit of these average rates of change is called the (instantaneous) rate of change of y with respect to x at x x 1 , which is interpreted as the slope of the tangent to the curve y f x at Px 1, f x 1:
6
instantaneous rate of change lim
x l 0
y f x2 f x1 lim x l x x x2 x1 2 1
We recognize this limit as being the derivative f x 1. We know that one interpretation of the derivative f a is as the slope of the tangent line to the curve y f x when x a . We now have a second interpretation: y
The derivative f a is the instantaneous rate of change of y f x with respect to x when x a.
Q
P
x
FIGURE 12
The y-values are changing rapidly at P and slowly at Q.
t
Dt
1980 1985 1990 1995 2000
930.2 1945.9 3233.3 4974.0 5674.2
The connection with the first interpretation is that if we sketch the curve y f x, then the instantaneous rate of change is the slope of the tangent to this curve at the point where x a. This means that when the derivative is large (and therefore the curve is steep, as at the point P in Figure 12), the y-values change rapidly. When the derivative is small, the curve is relatively flat and the y-values change slowly. In particular, if s f t is the position function of a particle that moves along a straight line, then f a is the rate of change of the displacement s with respect to the time t. In other words, f a is the velocity of the particle at time t a. The speed of the particle is the absolute value of the velocity, that is, f a . In the following example we estimate the rate of change of the national debt with respect to time. Here the function is defined not by a formula but by a table of values.
V EXAMPLE 6 Let Dt be the US national debt at time t. The table in the margin gives approximate values of this function by providing end of year estimates, in billions of dollars, from 1980 to 2000. Interpret and estimate the value of D1990.
SOLUTION The derivative D1990 means the rate of change of D with respect to t when t 1990, that is, the rate of increase of the national debt in 1990. According to Equation 5,
D1990 lim
t l1990
Dt D1990 t 1990
80
■
CHAPTER 2
DERIVATIVES
So we compute and tabulate values of the difference quotient (the average rates of change) as follows.
A NOTE ON UNITS The units for the average rate of change Dt are the units for D divided by the units for t , namely, billions of dollars per year. The instantaneous rate of change is the limit of the average rates of change, so it is measured in the same units: billions of dollars per year. ■
t
Dt D1990 t 1990
1980 1985 1995 2000
230.31 257.48 348.14 244.09
From this table we see that D1990 lies somewhere between 257.48 and 348.14 billion dollars per year. [Here we are making the reasonable assumption that the debt didn’t fluctuate wildly between 1980 and 2000.] We estimate that the rate of increase of the national debt of the United States in 1990 was the average of these two numbers, namely D1990 303 billion dollars per year Another method would be to plot the debt function and estimate the slope of the ■ tangent line when t 1990. The rate of change of the debt with respect to time in Example 6 is just one example of a rate of change. Here are a few of the many others: The velocity of a particle is the rate of change of displacement with respect to time. Physicists are interested in other rates of change as well—for instance, the rate of change of work with respect to time (which is called power). Chemists who study a chemical reaction are interested in the rate of change in the concentration of a reactant with respect to time (called the rate of reaction). A steel manufacturer is interested in the rate of change of the cost of producing x tons of steel per day with respect to x (called the marginal cost). A biologist is interested in the rate of change of the population of a colony of bacteria with respect to time. In fact, the computation of rates of change is important in all of the natural sciences, in engineering, and even in the social sciences. All these rates of change can be interpreted as slopes of tangents. This gives added significance to the solution of the tangent problem. Whenever we solve a problem involving tangent lines, we are not just solving a problem in geometry. We are also implicitly solving a great variety of problems involving rates of change in science and engineering.
2.1
EXERCISES 2. (a) Find the slope of the tangent line to the curve y x 3 at
1. (a) Find the slope of the tangent line to the parabola
y x 2x at the point 3, 3 (i) using Definition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). (c) Graph the parabola and the tangent line. As a check on your work, zoom in toward the point (3, 3) until the parabola and the tangent line are indistinguishable. 2
;
;
the point 1, 1 (i) using Definition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). (c) Graph the curve and the tangent line in successively smaller viewing rectangles centered at (1, 1) until the curve and the line appear to coincide.
SECTION 2.1
4. y 2x 5x, 3
5. y sx , ■
■
■
3, 2
(1, 1 ■
0, 0 ■
■
■
■
■
■
■
■
7. (a) Find the slope of the tangent to the curve
;
y 3 4x 2 2x 3 at the point where x a. (b) Find equations of the tangent lines at the points 1, 5 and 2, 3. (c) Graph the curve and both tangents on a common screen. 8. (a) Find the slope of the tangent to the curve y 1sx at
;
the point where x a. (b) Find equations of the tangent lines at the points 1, 1 and (4, 12 ). (c) Graph the curve and both tangents on a common screen.
9. The graph shows the position function of a car. Use the
shape of the graph to explain your answers to the following questions. (a) What was the initial velocity of the car? (b) Was the car going faster at B or at C ? (c) Was the car slowing down or speeding up at A, B, and C ? (d) What happened between D and E ?
13. The displacement (in meters) of a particle moving in a
straight line is given by the equation of motion s 1t 2, where t is measured in seconds. Find the velocity of the particle at times t a, t 1, t 2, and t 3. 14. The displacement (in meters) of a particle moving in a
straight line is given by s t 2 8t 18, where t is measured in seconds. (a) Find the average velocity over each time interval: (i) 3, 4 (ii) 3.5, 4 (iii) 4, 5 (iv) 4, 4.5 (b) Find the instantaneous velocity when t 4. (c) Draw the graph of s as a function of t and draw the secant lines whose slopes are the average velocities in part (a) and the tangent line whose slope is the instantaneous velocity in part (b). 15. For the function t whose graph is given, arrange the follow-
ing numbers in increasing order and explain your reasoning: 0
s
81
58 ms, its height (in meters) after t seconds is given by H 58t 0.83t 2. (a) Find the velocity of the arrow after one second. (b) Find the velocity of the arrow when t a. (c) When will the arrow hit the moon? (d) With what velocity will the arrow hit the moon?
1, 3
6. y 2x x 1 2,
■
12. If an arrow is shot upward on the moon with a velocity of
■ Find an equation of the tangent line to the curve at the given point.
3–6
3. y x 1x 2,
DERIVATIVES AND RATES OF CHANGE
t2
t0
t2
t4
E
D
y
C
y=©
B A 0
t
_1
0
1
2
3
4
x
10. Shown are graphs of the position functions of two runners,
A and B, who run a 100-m race and finish in a tie. 16. (a) Find an equation of the tangent line to the graph of
s (meters) 80
y tx at x 5 if t5 3 and t5 4. (b) If the tangent line to y f x at (4, 3) passes through the point (0, 2), find f 4 and f 4.
A
40
17. Sketch the graph of a function f for which f 0 0,
B 0
4
8
f 0 3, f 1 0, and f 2 1.
12
t (seconds)
(a) Describe and compare how the runners run the race. (b) At what time is the distance between the runners the greatest? (c) At what time do they have the same velocity? 11. If a ball is thrown into the air with a velocity of 40 fts, its
height (in feet) after t seconds is given by y 40t 16t 2. Find the velocity when t 2.
18. Sketch the graph of a function t for which t0 t0 0,
t1 1, t1 3, and t2 1.
19. If f x 3x 2 5x, find f 2 and use it to find an equa-
tion of the tangent line to the parabola y 3x 2 5x at the point 2, 2.
20. If tx 1 x 3, find t0 and use it to find an equation of
the tangent line to the curve y 1 x 3 at the point 0, 1.
82
■
CHAPTER 2
DERIVATIVES
21. (a) If Fx 5x1 x 2 , find F2 and use it to find
37. The table shows the estimated percentage P of the popula-
an equation of the tangent line to the curve y 5x1 x 2 at the point 2, 2. (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.
;
tion of Europe that use cell phones. (Midyear estimates are given.)
22. (a) If Gx 4x x , find Ga and use it to find equa2
3
tions of the tangent lines to the curve y 4x 2 x 3 at the points 2, 8 and 3, 9. (b) Illustrate part (a) by graphing the curve and the tangent lines on the same screen.
;
23–28
■
25. f t
■
26. f x
■
■
x2 1 x2
28. f x s3x 1 ■
■
■
■
■
■
■
29. lim
1 h10 1 h
30. lim
31. lim
2 x 32 x5
32. lim
33. lim
cos h 1 h
34. lim
x l5
h l0
■
■
■
2002
2003
P
28
39
55
68
77
83
h l0
■
t l1
■
■
■
tan x 1 x 4
t4 t 2 t1 ■
■
■
Year
1998
1999
2000
2001
2002
N
1886
2135
3501
4709
5886
(a) Find the average rate of growth (i) from 2000 to 2002 (ii) from 2000 to 2001 (iii) from 1999 to 2000 In each case, include the units. (b) Estimate the instantaneous rate of growth in 2000 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 2000 by measuring the slope of a tangent.
4 16 h 2 s h
x l 4
■
2001
■
■
h l0
2000
is given in the table. (The numbers of locations as of June 30 are given.)
Each limit represents the derivative of some function f at some number a. State such an f and a in each case. 29–34
1999
38. The number N of locations of a popular coffeehouse chain
1 sx 2
27. f x ■
24. f t t 4 5t
2t 1 t3
1998
(a) Find the average rate of cell phone growth (i) from 2000 to 2002 (ii) from 2000 to 2001 (iii) from 1999 to 2000 In each case, include the units. (b) Estimate the instantaneous rate of growth in 2000 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 2000 by measuring the slope of a tangent.
Find f a.
23. f x 3 2x 4x 2
Year
■
39. The cost (in dollars) of producing x units of a certain 35. A warm can of soda is placed in a cold refrigerator. Sketch
the graph of the temperature of the soda as a function of time. Is the initial rate of change of temperature greater or less than the rate of change after an hour? 36. A roast turkey is taken from an oven when its temperature
has reached 185°F and is placed on a table in a room where the temperature is 75°F. The graph shows how the temperature of the turkey decreases and eventually approaches room temperature. By measuring the slope of the tangent, estimate the rate of change of the temperature after an hour.
commodity is Cx 5000 10x 0.05x 2. (a) Find the average rate of change of C with respect to x when the production level is changed (i) from x 100 to x 105 (ii) from x 100 to x 101 (b) Find the instantaneous rate of change of C with respect to x when x 100. (This is called the marginal cost. Its significance will be explained in Section 2.3.) 40. If a cylindrical tank holds 100,000 gallons of water, which
can be drained from the bottom of the tank in an hour, then Torricelli’s Law gives the volume V of water remaining in the tank after t minutes as
T (°F)
Vt 100,000 (1
200
P 100
0
30
60
90
120 150
t (min)
1 60
t)
2
0 t 60
Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t ) as a function of t. What are its units? For times t 0, 10, 20, 30, 40, 50, and 60 min, find the flow rate and the amount of water remaining in the tank. Summarize your findings in a sentence or two. At what time is the flow rate the greatest? The least?
SECTION 2.2
mine is C f x dollars. (a) What is the meaning of the derivative f x? What are its units? (b) What does the statement f 800 17 mean? (c) Do you think the values of f x will increase or decrease in the short term? What about the long term? Explain.
S (mg / L) 16 12 8 4
42. The number of bacteria after t hours in a controlled labora-
0
2
4
6
8
10
12
14
T
73
73
70
69
72
81
88
91
45. The quantity of oxygen that can dissolve in water depends
on the temperature of the water. (So thermal pollution influences the oxygen content of water.) The graph shows how oxygen solubility S varies as a function of the water temperature T. (a) What is the meaning of the derivative ST ? What are its units?
2.2
24
32
40
T (°C)
S (cm/s) 20
0
44. The quantity (in pounds) of a gourmet ground coffee that is
sold by a coffee company at a price of p dollars per pound is Q f p. (a) What is the meaning of the derivative f 8? What are its units? (b) Is f 8 positive or negative? Explain.
16
maximum sustainable swimming speed S of Coho salmon. (a) What is the meaning of the derivative ST ? What are its units? (b) Estimate the values of S15 and S25 and interpret them.
midnight on June 2, 2001. The table shows values of this function recorded every two hours. What is the meaning of T 10? Estimate its value. 0
8
46. The graph shows the influence of the temperature T on the
43. Let Tt be the temperature (in F ) in Dallas t hours after
t
83
(b) Estimate the value of S16 and interpret it.
41. The cost of producing x ounces of gold from a new gold
tory experiment is n f t. (a) What is the meaning of the derivative f 5? What are its units? (b) Suppose there is an unlimited amount of space and nutrients for the bacteria. Which do you think is larger, f 5 or f 10? If the supply of nutrients is limited, would that affect your conclusion? Explain.
■
THE DERIVATIVE AS A FUNCTION
47– 48
■
Determine whether f 0 exists.
47. f x
48. f x ■
■
T (°C)
20
10
x sin
1 x
if x 0 if x 0
0
x 2 sin
1 x
if x 0 if x 0
0
■
■
■
■
■
■
■
■
■
■
THE DERIVATIVE AS A FUNCTION In Section 2.1 we considered the derivative of a function f at a fixed number a: 1
f a lim
hl0
f a h f a h
Here we change our point of view and let the number a vary. If we replace a in Equation 1 by a variable x, we obtain
2
f x lim
hl0
f x h f x h
84
■
CHAPTER 2
DERIVATIVES
Given any number x for which this limit exists, we assign to x the number f x. So we can regard f as a new function, called the derivative of f and defined by Equation 2. We know that the value of f at x, f x, can be interpreted geometrically as the slope of the tangent line to the graph of f at the point x, f x. The function f is called the derivative of f because it has been “derived” from f by the limiting operation in Equation 2. The domain of f is the set x f x exists and may be smaller than the domain of f .
V EXAMPLE 1 The graph of a function f is given in Figure 1. Use it to sketch the graph of the derivative f .
y y=ƒ
SOLUTION We can estimate the value of the derivative at any value of x by drawing
1 0
1
x
FIGURE 1
the tangent at the point x, f x and estimating its slope. For instance, for x 5 we draw the tangent at P in Figure 2(a) and estimate its slope to be about 32 , so f 5 1.5. This allows us to plot the point P5, 1.5 on the graph of f directly beneath P. Repeating this procedure at several points, we get the graph shown in Figure 2(b). Notice that the tangents at A, B, and C are horizontal, so the derivative is 0 there and the graph of f crosses the x-axis at the points A, B, and C, directly y
B m=0 1
m=0
y=ƒ
P mÅ3 2
A
0
1
5
x
m=0
C
Visual 2.2 shows an animation of Figure 2 for several functions.
(a) y
Pª (5, 1.5) y=fª(x)
1
Bª 0
FIGURE 2
Aª
Cª
1
(b)
5
x
SECTION 2.2
THE DERIVATIVE AS A FUNCTION
■
85
beneath A, B, and C. Between A and B the tangents have positive slope, so f x is positive there. But between B and C the tangents have negative slope, so f x is negative there. ■ V EXAMPLE 2
(a) If f x x 3 x, find a formula for f x. (b) Illustrate by comparing the graphs of f and f . SOLUTION
2
(a) When using Equation 2 to compute a derivative, we must remember that the variable is h and that x is temporarily regarded as a constant during the calculation of the limit.
f _2
2
f x lim
hl0
_2
lim
x 3 3x 2h 3xh 2 h 3 x h x 3 x h
lim
3x 2h 3xh 2 h 3 h h
hl0
2
fª
hl0
_2
2
f x h f x x h3 x h x 3 x lim hl0 h h
lim 3x 2 3xh h 2 1 3x 2 1 hl0
_2
FIGURE 3
(b) We use a graphing device to graph f and f in Figure 3. Notice that f x 0 when f has horizontal tangents and f x is positive when the tangents have positive slope. So these graphs serve as a check on our work in part (a). ■ EXAMPLE 3 If f x sx , find the derivative of f . State the domain of f . SOLUTION
f x lim
h l0
lim
h l0
Here we rationalize the numerator.
f x h f x h sx h sx h
lim
lim
x h x 1 lim h l0 sx h sx h (sx h sx )
h l0
h l0
sx h sx sx h sx h sx h sx
1 1 2sx sx sx
We see that f x exists if x 0, so the domain of f is 0, . This is smaller than the domain of f , which is 0, . ■ Let’s check to see that the result of Example 3 is reasonable by looking at the graphs of f and f in Figure 4. When x is close to 0, sx is also close to 0, so f x 1(2sx ) is very large and this corresponds to the steep tangent lines near 0, 0 in Figure 4(a) and the large values of f x just to the right of 0 in Figure 4(b).
86
■
CHAPTER 2
DERIVATIVES
When x is large, f x is very small and this corresponds to the flatter tangent lines at the far right of the graph of f and the horizontal asymptote of the graph of f . y
y
1
1
0
x
1
(a) ƒ=œ„ x
FIGURE 4
EXAMPLE 4 Find f if f x
SOLUTION a c b d ad bc 1 e bd e
0
x
1
(b) f ª (x)=
1 2œ„ x
1x . 2x
1 x h 1x f x h f x 2 x h 2x f x lim lim hl0 hl0 h h lim
1 x h2 x 1 x2 x h h2 x h2 x
lim
2 x 2h x 2 xh 2 x h x 2 xh h2 x h2 x
lim
3h h2 x h2 x
lim
3 3 2 x h2 x 2 x2
hl0
hl0
hl0
hl0
■
OTHER NOTATIONS
If we use the traditional notation y f x to indicate that the independent variable is x and the dependent variable is y, then some common alternative notations for the derivative are as follows: f x y
dy df d f x Df x Dx f x dx dx dx
The symbols D and ddx are called differentiation operators because they indicate the operation of differentiation, which is the process of calculating a derivative. The symbol dydx, which was introduced by Leibniz, should not be regarded as a ratio (for the time being); it is simply a synonym for f x. Nonetheless, it is a very useful and suggestive notation, especially when used in conjunction with increment notation. Referring to Equation 2.1.6, we can rewrite the definition of derivative in Leibniz notation in the form dy y lim x l 0 x dx
SECTION 2.2
Gottfried Wilhelm Leibniz was born in Leipzig in 1646 and studied law, theology, philosophy, and mathematics at the university there, graduating with a bachelor’s degree at age 17. After earning his doctorate in law at age 20, Leibniz entered the diplomatic service and spent most of his life traveling to the capitals of Europe on political missions. In particular, he worked to avert a French military threat against Germany and attempted to reconcile the Catholic and Protestant churches. His serious study of mathematics did not begin until 1672 while he was on a diplomatic mission in Paris. There he built a calculating machine and met scientists, like Huygens, who directed his attention to the latest developments in mathematics and science. Leibniz sought to develop a symbolic logic and system of notation that would simplify logical reasoning. In particular, the version of calculus that he published in 1684 established the notation and the rules for finding derivatives that we use today. Unfortunately, a dreadful priority dispute arose in the 1690s between the followers of Newton and those of Leibniz as to who had invented calculus first. Leibniz was even accused of plagiarism by members of the Royal Society in England. The truth is that each man invented calculus independently. Newton arrived at his version of calculus first but, because of his fear of controversy, did not publish it immediately. So Leibniz’s 1684 account of calculus was the first to be published. ■
THE DERIVATIVE AS A FUNCTION
■
87
If we want to indicate the value of a derivative dydx in Leibniz notation at a specific number a, we use the notation dy dx
dy dx
or xa
xa
which is a synonym for f a. DIFFERENTIABLE FUNCTIONS
A function f is differentiable at a if f a exists. It is differentiable on an open interval a, b [or a, or , a or , ] if it is differentiable at every number in the interval. 3 DEFINITION
Where is the function f x x differentiable?
V EXAMPLE 5
SOLUTION If x 0, then x x and we can choose h small enough that x h 0 and hence x h x h. Therefore, for x 0 we have
f x lim
hl0
lim
hl0
x h x h
x h x h lim lim 1 1 hl0 h hl0 h
and so f is differentiable for any x 0. Similarly, for x 0 we have x x and h can be chosen small enough that x h 0 and so x h x h. Therefore, for x 0,
f x lim
hl0
lim
hl0
x h x h
x h x h lim lim 1 1 hl0 h hl0 h
and so f is differentiable for any x 0. For x 0 we have to investigate f 0 lim
hl0
lim
hl0
f 0 h f 0 h
0 h 0
if it exists
h
Let’s compute the left and right limits separately: lim
h l 0
and
lim
hl0
0 h 0 h
0 h 0 h
lim
h l 0
lim
hl0
h h
h h
lim
h l 0
lim
hl0
h lim 1 1 hl0 h
h lim 1 1 hl0 h
Since these limits are different, f 0 does not exist. Thus f is differentiable at all x except 0.
88
■
CHAPTER 2
DERIVATIVES
A formula for f is given by
y
f x
if x 0 if x 0
1 1
and its graph is shown in Figure 5(b). The fact that f 0 does not exist is reflected geometrically in the fact that the curve y x does not have a tangent line at 0, 0. [See Figure 5(a).] ■
x
0
(a) y=ƒ=| x |
Both continuity and differentiability are desirable properties for a function to have. The following theorem shows how these properties are related.
y 1
4 THEOREM
If f is differentiable at a, then f is continuous at a.
x
0
PROOF To prove that f is continuous at a, we have to show that lim x l a f x f a. We do this by showing that the difference f x f a approaches 0. The given information is that f is differentiable at a, that is,
_1
(b) y=fª(x)
f a lim
FIGURE 5
xla
f x f a xa
exists (see Equation 2.1.5). To connect the given and the unknown, we divide and multiply f x f a by x a (which we can do when x a): f x f a
f x f a x a xa
Thus, using the Product Law and (2.1.5), we can write lim f x f a lim
xla
xla
lim
xla
f x f a x a xa f x f a lim x a xla xa
f a 0 0 To use what we have just proved, we start with f x and add and subtract f a: lim f x lim f a f x f a
xla
xla
lim f a lim f x f a xla
xla
f a 0 f a ■
Therefore, f is continuous at a. |
NOTE The converse of Theorem 4 is false; that is, there are functions that are continuous but not differentiable. For instance, the function f x x is continuous at 0 because lim f x lim x 0 f 0
xl0
xl0
(See Example 6 in Section 1.4.) But in Example 5 we showed that f is not differentiable at 0.
SECTION 2.2
THE DERIVATIVE AS A FUNCTION
■
89
HOW CAN A FUNCTION FAIL TO BE DIFFERENTIABLE?
We saw that the function y x in Example 5 is not differentiable at 0 and Figure 5(a) shows that its graph changes direction abruptly when x 0. In general, if the graph of a function f has a “corner” or “kink” in it, then the graph of f has no tangent at this point and f is not differentiable there. [In trying to compute f a, we find that the left and right limits are different.] Theorem 4 gives another way for a function not to have a derivative. It says that if f is not continuous at a, then f is not differentiable at a. So at any discontinuity (for instance, a jump discontinuity) f fails to be differentiable. A third possibility is that the curve has a vertical tangent line when x a; that is, f is continuous at a and
y
vertical tangent line
lim f x
0
a
xla
x
This means that the tangent lines become steeper and steeper as x l a. Figure 6 shows one way that this can happen; Figure 7(c) shows another. Figure 7 illustrates the three possibilities that we have discussed.
FIGURE 6
y
y
0
0
x
a
y
x
a
0
a
x
FIGURE 7
Three ways for ƒ not to be differentiable at a
(a) A corner
(b) A discontinuity
(c) A vertical tangent
A graphing calculator or computer provides another way of looking at differentiability. If f is differentiable at a, then when we zoom in toward the point a, f a the graph straightens out and appears more and more like a line. (See Figure 8. We saw a specific example of this in Figure 4 in Section 2.1.) But no matter how much we zoom in toward a point like the ones in Figures 6 and 7(a), we can’t eliminate the sharp point or corner (see Figure 9). y
0
y
a
x
0
a
FIGURE 8
FIGURE 9
ƒ is differentiable at a.
ƒ is not differentiable at a.
x
90
■
CHAPTER 2
DERIVATIVES
HIGHER DERIVATIVES
If f is a differentiable function, then its derivative f is also a function, so f may have a derivative of its own, denoted by f f . This new function f is called the second derivative of f because it is the derivative of the derivative of f . Using Leibniz notation, we write the second derivative of y f x as d dx
dy dx
d 2y dx 2
EXAMPLE 6 If f x x 3 x, find and interpret f x. SOLUTION In Example 2 we found that the first derivative is f x 3x 2 1. So
the second derivative is f x lim
2 f·
_1.5
h l0
fª
f
lim
3x h2 1 3x 2 1 h
lim
3x 2 6xh 3h 2 1 3x 2 1 h
h l0
1.5
f x h f x h
h l0
_2
lim 6x 3h 6x
FIGURE 10
In Module 2.2 you can see how changing the coefficients of a polynomial f affects the appearance of the graphs of f , f , and f .
h l0
The graphs of f , f , f are shown in Figure 10. We can interpret f x as the slope of the curve y f x at the point x, f x. In other words, it is the rate of change of the slope of the original curve y f x. Notice from Figure 10 that f x is negative when y f x has negative slope and positive when y f x has positive slope. So the graphs serve as a check on our calculations. ■ In general, we can interpret a second derivative as a rate of change of a rate of change. The most familiar example of this is acceleration, which we define as follows. If s st is the position function of an object that moves in a straight line, we know that its first derivative represents the velocity v t of the object as a function of time: v t st
ds dt
The instantaneous rate of change of velocity with respect to time is called the acceleration at of the object. Thus the acceleration function is the derivative of the velocity function and is therefore the second derivative of the position function: at vt st or, in Leibniz notation, a
dv d 2s 2 dt dt
SECTION 2.2
■
THE DERIVATIVE AS A FUNCTION
91
The third derivative f is the derivative of the second derivative: f f . So f x can be interpreted as the slope of the curve y f x or as the rate of change of f x. If y f x, then alternative notations for the third derivative are y f x
d dx
d2y dx 2
d 3y dx 3
The process can be continued. The fourth derivative f is usually denoted by f 4. In general, the nth derivative of f is denoted by f n and is obtained from f by differentiating n times. If y f x, we write dny dx n
y n f nx
EXAMPLE 7 If f x x 3 x, find f x and f 4x. SOLUTION In Example 6 we found that f x 6x . The graph of the second derivative has equation y 6x and so it is a straight line with slope 6. Since the derivative f x is the slope of f x, we have
f x 6 for all values of x. So f is a constant function and its graph is a horizontal line. Therefore, for all values of x, f 4x 0
■
We have seen that one application of second derivatives occurs in analyzing the motion of objects using acceleration. We will investigate another application of second derivatives in Section 4.3, where we show how knowledge of f gives us information about the shape of the graph of f . In Section 8.7 we will see how second and higher derivatives enable us to represent functions as sums of infinite series.
2.2
EXERCISES 2. (a) f 0
■ Use the given graph to estimate the value of each derivative. Then sketch the graph of f .
1–2
1. (a) f 3
(d) f 0 (g) f 3
(b) f 2 (e) f 1
(b) f 1 (d) f 3 (f ) f 5
(c) f 2 (e) f 4
(c) f 1 (f ) f 2
y y
y=f(x)
y=f(x) 1 1 0
1
x 0 ■
■
■
■
■
x
1 ■
■
■
■
■
■
■
92
■
CHAPTER 2
DERIVATIVES
3. Match the graph of each function in (a)–(d) with the graph
of its derivative in I–IV. Give reasons for your choices. y
(a)
0
x
0
x ■
y
(c)
x
y
I
y
(d)
0
y
11.
y
(b)
0
y
10.
0
■
■
■
■
■
■
■
■
x
■
■
■
12. Shown is the graph of the population function Pt for yeast
cells in a laboratory culture. Use the method of Example 1 to graph the derivative Pt. What does the graph of P tell us about the yeast population?
x
P (yeast cells)
y
II
0
x
500 0
x
y
III
0
0
y
IV
0
x
x
0
5
10
15
t (hours)
13. The graph shows how the average age of first marriage of
x
Japanese men has varied in the last half of the 20th century. Sketch the graph of the derivative function Mt. During which years was the derivative negative?
■ Trace or copy the graph of the given function f . (Assume that the axes have equal scales.) Then use the method of Example 1 to sketch the graph of f below it.
4 –11
4.
y
5.
M
y 27
25 0
0
x
x 1960
6.
y
7.
1970
1980
1990
2000 t
y
14. Make a careful sketch of the graph of the sine function and 0 0
8.
x
y
0
x
below it sketch the graph of its derivative in the same manner as in Exercises 4–11. Can you guess what the derivative of the sine function is from its graph?
9.
x
2 ; 15. Let f x x .
y
0
x
(a) Estimate the values of f 0, f ( 12 ), f 1, and f 2 by using a graphing device to zoom in on the graph of f. (b) Use symmetry to deduce the values of f ( 12 ), f 1, and f 2. (c) Use the results from parts (a) and (b) to guess a formula for f x.
SECTION 2.2
(d) Use the definition of a derivative to prove that your guess in part (c) is correct.
29.
_2
30.
17. f x 2 x 1
18. f x 1.5x 2 x 3.7
1 3
19. f x x 3 3x 5
20. f x x sx
21. tx s1 2x
22. f x
■
■
■
■
■
■
■
3x 1 3x
■
4 x
■
■
0
■
x
2
■
■
■
■
■
■
■
■
; 31. Graph the function f x x s x . Zoom in repeatedly,
first toward the point (1, 0) and then toward the origin. What is different about the behavior of f in the vicinity of these two points? What do you conclude about the differentiability of f ?
; 32. Zoom in toward the points (1, 0), (0, 1), and (1, 0) on
the graph of the function tx x 2 123. What do you notice? Account for what you see in terms of the differentiability of t.
4t 23. Gt t1 ■
■
0
y
_2
17–23 ■ Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.
93
y
3 ; 16. Let f x x .
(a) Estimate the values of f 0, f ( 12 ), f 1, f 2, and f 3 by using a graphing device to zoom in on the graph of f. (b) Use symmetry to deduce the values of f ( 12 ), f 1, f 2, and f 3. (c) Use the values from parts (a) and (b) to graph f . (d) Guess a formula for f x. (e) Use the definition of a derivative to prove that your guess in part (d) is correct.
■
THE DERIVATIVE AS A FUNCTION
■
■
■
33. The figure shows the graphs of f , f , and f . Identify each
curve, and explain your choices. 24. (a) Sketch the graph of f x s6 x by starting with the
graph of y sx and using the transformations of Section 1.2. (b) Use the graph from part (a) to sketch the graph of f . (c) Use the definition of a derivative to find f x. What are the domains of f and f ? (d) Use a graphing device to graph f and compare with your sketch in part (b).
;
y
a b x
c
25. (a) If f x x 4 2x, find f x.
;
;
(b) Check to see that your answer to part (a) is reasonable by comparing the graphs of f and f . 34. The figure shows graphs of f, f , f , and f . Identify each
26. (a) If f t t 2 st , find f t.
curve, and explain your choices.
(b) Check to see that your answer to part (a) is reasonable by comparing the graphs of f and f .
y
a b c d
27–30 ■ The graph of f is given. State, with reasons, the numbers at which f is not differentiable. 27.
28.
y
y x 0
_2
0
2
x
2
4
x
94
■
CHAPTER 2
DERIVATIVES 3 x has a vertical tangent line at 0, 0. (c) Show that y s (Recall the shape of the graph of f . See Figure 8 in Section 1.2.)
35. The figure shows the graphs of three functions. One is the
position function of a car, one is the velocity of the car, and one is its acceleration. Identify each curve, and explain your choices. y
40. (a) If tx x 23, show that t0 does not exist.
a b
;
c
(b) If a 0, find ta. (c) Show that y x 23 has a vertical tangent line at 0, 0. (d) Illustrate part (c) by graphing y x 23.
41. Show that the function f x x 6 is not differentiable
at 6. Find a formula for f and sketch its graph.
t
0
42. Where is the greatest integer function f x x not differ-
entiable? Find a formula for f and sketch its graph.
43. Recall that a function f is called even if f x f x for
all x in its domain and odd if f x f x for all such x. Prove each of the following. (a) The derivative of an even function is an odd function. (b) The derivative of an odd function is an even function.
Use the definition of a derivative to find f x and f x. Then graph f , f , and f on a common screen and check to see if your answers are reasonable.
; 36 –37
■
36. f x 1x ■
■
■
44. When you turn on a hot-water faucet, the temperature T of
37. f x 1 4x x 2 ■
■
■
■
■
■
■
■
2 3 4 ; 38. If f x 2x x , find f x, f x, f x, and f x.
■
Graph f , f , f , and f on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives?
3 39. Let f x s x.
(a) If a 0, use Equation 2.1.5 to find f a. (b) Show that f 0 does not exist.
2.3
y=c slope=0
0
FIGURE 1
The graph of ƒ=c is the line y=c, so fª(x)=0.
45. Let ᐍ be the tangent line to the parabola y x 2 at the point
1, 1. The angle of inclination of ᐍ is the angle that ᐍ makes with the positive direction of the x-axis. Calculate correct to the nearest degree.
BASIC DIFFERENTIATION FORMULAS
y c
the water depends on how long the water has been running. (a) Sketch a possible graph of T as a function of the time t that has elapsed since the faucet was turned on. (b) Describe how the rate of change of T with respect to t varies as t increases. (c) Sketch a graph of the derivative of T.
x
If it were always necessary to compute derivatives directly from the definition, as we did in the preceding section, such computations would be tedious and the evaluation of some limits would require ingenuity. Fortunately, several rules have been developed for finding derivatives without having to use the definition directly. These formulas greatly simplify the task of differentiation. In this section we learn how to differentiate constant functions, power functions, polynomials, and the sine and cosine functions. Then we use this knowledge to compute rates of change. Let’s start with the simplest of all functions, the constant function f x c. The graph of this function is the horizontal line y c, which has slope 0, so we must have f x 0. (See Figure 1.) A formal proof, from the definition of a derivative, is also easy: f x h f x cc f x lim lim hl0 hl0 h h lim 0 0 hl0
SECTION 2.3
BASIC DIFFERENTIATION FORMULAS
■
95
In Leibniz notation, we write this rule as follows. DERIVATIVE OF A CONSTANT FUNCTION
d c 0 dx
POWER FUNCTIONS
We next look at the functions f x x n, where n is a positive integer. If n 1, the graph of f x x is the line y x, which has slope 1. (See Figure 2.) So
y
y=x slope=1
d x 1 dx
1 0 x
FIGURE 2
The graph of ƒ=x is the line y=x, so fª(x)=1.
(You can also verify Equation 1 from the definition of a derivative.) We have already investigated the cases n 2 and n 3. In fact, in Section 2.2 (Exercises 15 and 16) we found that 2
d x 2 2x dx
d x 3 3x 2 dx
For n 4 we find the derivative of f x x 4 as follows: f x lim
f x h f x x h4 x 4 lim hl0 h h
lim
x 4 4x 3h 6x 2h 2 4xh 3 h 4 x 4 h
lim
4x 3h 6x 2h 2 4xh 3 h 4 h
hl0
hl0
hl0
lim 4x 3 6x 2h 4xh 2 h 3 4x 3 hl0
Thus 3
d x 4 4x 3 dx
Comparing the equations in (1), (2), and (3), we see a pattern emerging. It seems to be a reasonable guess that, when n is a positive integer, ddxx n nx n1. This turns out to be true. THE POWER RULE If n is a positive integer, then
d x n nx n1 dx
96
■
CHAPTER 2
DERIVATIVES
PROOF If f x x n, then
f x lim
hl0
The Binomial Theorem is given on Reference Page 1. ■
f x h f x x hn x n lim hl0 h h
In finding the derivative of x 4 we had to expand x h4. Here we need to expand x hn and we use the Binomial Theorem to do so:
x n nx n1h
f x lim
hl0
nx n1h lim
hl0
lim nx n1 hl0
nn 1 n2 2 x h nxh n1 h n x n 2 h
nn 1 n2 2 x h nxh n1 h n 2 h
nn 1 n2 x h nxh n2 h n1 2
nx n1 because every term except the first has h as a factor and therefore approaches 0.
■
We illustrate the Power Rule using various notations in Example 1. EXAMPLE 1
(a) If f x x 6, then f x 6x 5. (c) If y t 4, then
(b) If y x 1000, then y 1000x 999.
dy 4t 3. dt
(d)
d 3 r 3r 2 dr
■
What about power functions with negative integer exponents? In Exercise 55 we ask you to verify from the definition of a derivative that d dx
1 x
1 x2
We can rewrite this equation as d x 1 1x 2 dx and so the Power Rule is true when n 1. In fact, we will show in the next section [Exercise 55(c)] that it holds for all negative integers. What if the exponent is a fraction? In Example 3 in Section 2.2 we found that d 1 sx dx 2sx which can be written as d 12 x 12 x12 dx
SECTION 2.3
BASIC DIFFERENTIATION FORMULAS
■
97
This shows that the Power Rule is true even when n 12 . In fact, we will show in Section 3.3 that it is true for all real numbers n.
THE POWER RULE (GENERAL VERSION) If n is any real number, then
d x n nx n1 dx
EXAMPLE 2 Differentiate:
(a) f x ■ Figure 3 shows the function y in Example 2(b) and its derivative y. Notice that y is not differentiable at 0 (y is not defined there). Observe that y is positive when y increases and is negative when y decreases.
1 x2
SOLUTION In each case we rewrite the function as a power of x.
(a) Since f x x2, we use the Power Rule with n 2: f x
2
(b)
y yª _3
3
_2
FIGURE 3
y=#œ≈ „
3 (b) y s x2
d 2 x 2 2x 21 2x 3 3 dx x
dy d 3 2 d ( x 23 23 x 231 23 x13 sx ) dx dx dx
■
The Power Rule enables us to find tangent lines without having to resort to the definition of a derivative. It also enables us to find normal lines. The normal line to a curve C at a point P is the line through P that is perpendicular to the tangent line at P. (In the study of optics, one needs to consider the angle between a light ray and the normal line to a lens.) Find equations of the tangent line and normal line to the curve y xsx at the point 1, 1. Illustrate by graphing the curve and these lines. V EXAMPLE 3
SOLUTION The derivative of f x xsx xx 12 x 32 is
f x 32 x 321 32 x 12 32 sx
3
So the slope of the tangent line at (1, 1) is f 1 32 . Therefore, an equation of the tangent line is
tangent
y 1 32 x 1
or
y 32 x 12
normal _1
3
_1
FIGURE 4
The normal line is perpendicular to the tangent line, so its slope is the negative 3 reciprocal of 2, that is, 23. Thus an equation of the normal line is y 1 23 x 1
or
y 23 x 53
We graph the curve and its tangent line and normal line in Figure 4.
■
NEW DERIVATIVES FROM OLD
When new functions are formed from old functions by addition, subtraction, or multiplication by a constant, their derivatives can be calculated in terms of derivatives of
98
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CHAPTER 2
DERIVATIVES
the old functions. In particular, the following formula says that the derivative of a constant times a function is the constant times the derivative of the function. THE CONSTANT MULTIPLE RULE If c is a constant and f is a differentiable
function, then d d cf x c f x dx dx
GEOMETRIC INTERPRETATION OF THE CONSTANT MULTIPLE RULE ■
PROOF Let tx cf x. Then
tx lim
y
hl0
tx h tx cf x h cf x lim hl0 h h
y=2ƒ
lim c hl0
y=ƒ 0
c lim
hl0
x
Multiplying by c 2 stretches the graph vertically by a factor of 2. All the rises have been doubled but the runs stay the same. So the slopes are doubled, too.
f x h f x h
f x h f x h
(by Law 3 of limits)
cf x
■
EXAMPLE 4
d d 3x 4 3 x 4 34x 3 12x 3 dx dx d d d (b) x 1x 1 x 11 1 dx dx dx (a)
■
The next rule tells us that the derivative of a sum of functions is the sum of the derivatives. THE SUM RULE If f and t are both differentiable, then Using prime notation, we can write the Sum Rule as f t f t ■
d d d f x tx f x tx dx dx dx PROOF Let Fx f x tx. Then
Fx lim
hl0
lim
hl0
lim
hl0
lim
hl0
Fx h Fx h f x h tx h f x tx h
f x h f x tx h tx h h
f x h f x tx h tx lim hl0 h h
f x tx
(by Law 1)
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SECTION 2.3
BASIC DIFFERENTIATION FORMULAS
■
99
The Sum Rule can be extended to the sum of any number of functions. For instance, using this theorem twice, we get f t h f t h f t h f t h By writing f t as f 1t and applying the Sum Rule and the Constant Multiple Rule, we get the following formula. THE DIFFERENCE RULE If f and t are both differentiable, then
d d d f x tx f x tx dx dx dx
The Constant Multiple Rule, the Sum Rule, and the Difference Rule can be combined with the Power Rule to differentiate any polynomial, as the following examples demonstrate. EXAMPLE 5
d x 8 12x 5 4x 4 10x 3 6x 5 dx d d d d d d x 8 12 x 5 4 x 4 10 x 3 6 x 5 dx dx dx dx dx dx 8x 7 125x 4 44x 3 103x 2 61 0 8x 7 60x 4 16x 3 30x 2 6
■
Find the points on the curve y x 4 6x 2 4 where the tangent line is horizontal. V EXAMPLE 6
SOLUTION Horizontal tangents occur where the derivative is zero. We have
dy d d d x 4 6 x 2 4 dx dx dx dx 4x 3 12x 0 4xx 2 3 Thus dydx 0 if x 0 or x 2 3 0, that is, x s3 . So the given curve has horizontal tangents when x 0, s3 , and s3 . The corresponding points are 0, 4, (s3 , 5), and (s3 , 5). (See Figure 5.) y (0, 4)
0
x
FIGURE 5
The curve y=x$-6x@+4 and its horizontal tangents
{_ œ„ 3, _5}
3, _5} {œ„
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100
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CHAPTER 2
DERIVATIVES
THE SINE AND COSINE FUNCTIONS
If we sketch the graph of the function f x sin x and use the interpretation of f x as the slope of the tangent to the sine curve in order to sketch the graph of f (see Exercise 14 in Section 2.2), then it looks as if the graph of f may be the same as the cosine curve (see Figure 6). ƒ=sin x
0
π 2
π
2π
x
π 2
π
2π
x
Visual 2.3 shows an animation of Figure 6. fª(x)
0
FIGURE 6
To prove that this is true we need to use two limits from Section 1.4 (see Equation 5 and Example 11 in that section): lim
l0
sin 1
lim
l0
cos 1 0
d sin x cos x dx
4
PROOF If f x sin x, then
f x lim
hl0
We have used the addition formula for sine. See Appendix A. ■
lim
hl0
lim
hl0
f x h f x sinx h sin x lim hl0 h h sin x cos h cos x sin h sin x h
lim sin x hl0
Note that we regard x as a constant when computing a limit as h l 0 , so sin x and cos x are also constants. ■
cos h 1 h
lim sin x lim hl0
sin x cos h sin x cos x sin h h h
hl0
cos x
sin h h
cos h 1 sin h lim cos x lim hl0 hl0 h h
sin x 0 cos x 1 cos x
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SECTION 2.3
BASIC DIFFERENTIATION FORMULAS
■
101
Using the same methods as in the proof of Formula 4, one can prove (see Exercise 56) that d cos x sin x dx
5
EXAMPLE 7 Differentiate y 3 sin 4 cos . SOLUTION
dy d d 3 sin 4 cos 3 cos 4 sin d d d
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EXAMPLE 8 Find the 27th derivative of cos x. SOLUTION The first few derivatives of f x cos x are as follows:
f x sin x f x cos x f x sin x f 4x cos x f 5x sin x Looking for a pattern, we see that the successive derivatives occur in a cycle of length 4 and, in particular, f nx cos x whenever n is a multiple of 4. Therefore f 24x cos x and, differentiating three more times, we have f 27x sin x
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APPLICATIONS TO RATES OF CHANGE
We discussed velocity and other rates of change in Section 2.1, but now that we know some differentiation formulas we can solve problems involving rates of change more easily. V EXAMPLE 9
The position of a particle is given by the equation s f t t 3 6t 2 9t
where t is measured in seconds and s in meters. (a) Find the velocity at time t. (b) What is the velocity after 2 s? After 4 s? (c) When is the particle at rest? (d) When is the particle moving forward (that is, in the positive direction)? (e) Draw a diagram to represent the motion of the particle. (f ) Find the total distance traveled by the particle during the first five seconds. (g) Find the acceleration at time t and after 4 s.
102
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CHAPTER 2
DERIVATIVES
(h) Graph the position, velocity, and acceleration functions for 0 t 5. (i) When is the particle speeding up? When is it slowing down? SOLUTION
(a) The velocity function is the derivative of the position function. s f t t 3 6t 2 9t vt
ds 3t 2 12t 9 dt
(b) The velocity after 2 s means the instantaneous velocity when t 2, that is, v2
ds dt
t2
322 122 9 3 ms
The velocity after 4 s is v4 342 124 9 9 ms
(c) The particle is at rest when vt 0, that is, 3t 2 12t 9 3t 2 4t 3 3t 1t 3 0 and this is true when t 1 or t 3. Thus the particle is at rest after 1 s and after 3 s. (d) The particle moves in the positive direction when vt 0, that is, 3t 2 12t 9 3t 1t 3 0
t=3 s=0
t=0 s=0 FIGURE 7
t=1 s=4
s
This inequality is true when both factors are positive t 3 or when both factors are negative t 1. Thus the particle moves in the positive direction in the time intervals t 1 and t 3. It moves backward (in the negative direction) when 1 t 3. (e) Using the information from part (d) we make a schematic sketch in Figure 7 of the motion of the particle back and forth along a line (the s-axis). (f ) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately. The distance traveled in the first second is
f 1 f 0 4 0 4 m From t 1 to t 3 the distance traveled is
f 3 f 1 0 4 4 m From t 3 to t 5 the distance traveled is
f 5 f 3 20 0 20 m The total distance is 4 4 20 28 m. (g) The acceleration is the derivative of the velocity function: at
d 2s dv 6t 12 dt 2 dt
a4 64 12 12 ms 2
SECTION 2.3
25
√
a s
0
5
-12
BASIC DIFFERENTIATION FORMULAS
■
103
(h) Figure 8 shows the graphs of s, v, and a. (i) The particle speeds up when the velocity is positive and increasing (v and a are both positive) and also when the velocity is negative and decreasing (v and a are both negative). In other words, the particle speeds up when the velocity and acceleration have the same sign. (The particle is pushed in the same direction it is moving.) From Figure 8 we see that this happens when 1 t 2 and when t 3. The particle slows down when v and a have opposite signs, that is, when 0 t 1 and when 2 t 3. Figure 9 summarizes the motion of the particle.
FIGURE 8
a
√
In Module 2.3 you can see an animation of Figure 9 with an expression for s that you can choose yourself.
s
5 0 _5
forward
FIGURE 9
t
1
slows down
backward speeds up
slows down
forward speeds up
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V EXAMPLE 10 Suppose Cx is the total cost that a company incurs in producing x units of a certain commodity. The function C is called a cost function. If the number of items produced is increased from x 1 to x 2 , then the additional cost is C Cx 2 Cx 1 , and the average rate of change of the cost is
C Cx 2 Cx 1 Cx 1 x Cx 1 x x2 x1 x The limit of this quantity as x l 0, that is, the instantaneous rate of change of cost with respect to the number of items produced, is called the marginal cost by economists: marginal cost lim
x l 0
C dC x dx
[Since x often takes on only integer values, it may not make literal sense to let x approach 0, but we can always replace Cx by a smooth approximating function.] Taking x 1 and n large (so that x is small compared to n), we have Cn Cn 1 Cn Thus the marginal cost of producing n units is approximately equal to the cost of producing one more unit [the n 1st unit]. It is often appropriate to represent a total cost function by a polynomial Cx a bx cx 2 dx 3
104
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CHAPTER 2
DERIVATIVES
where a represents the overhead cost (rent, heat, maintenance) and the other terms represent the cost of raw materials, labor, and so on. (The cost of raw materials may be proportional to x, but labor costs might depend partly on higher powers of x because of overtime costs and inefficiencies involved in large-scale operations.) For instance, suppose a company has estimated that the cost (in dollars) of producing x items is Cx 10,000 5x 0.01x 2 Then the marginal cost function is Cx 5 0.02x The marginal cost at the production level of 500 items is C500 5 0.02500 $15item This gives the rate at which costs are increasing with respect to the production level when x 500 and predicts the cost of the 501st item. The actual cost of producing the 501st item is C501 C500 10,000 5501 0.015012
10,000 5500 0.015002
$15.01 Notice that C500 C501 C500.
2.3 1–24
■
EXERCISES
Differentiate the function. 2. f x s30
3. f x 5x 1
4. Fx 4x 10
5. f x x 3 4x 6
6. f t 2 t 6 3t 4 t
7. f x x 3 sin x
8. y sin t cos t
25
( 12 x) 5
17. y 4
x 4x 3 19. y sx 21. v t 2 23. z ■
1 4 3 t s
22. y
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28. y 3x 2 x 3,
1, 2
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33. Find
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■
1, 2
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■
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3 30. G r sr s r
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32. ht st 5 sin t ■
■
■
■
■
■
d 99 sin x. dx 99
34. Find the nth derivative of each function by calculating the ■
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Find the first and second derivatives of the function.
31. tt 2 cos t 3 sin t
sin c 2
■
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29. f x x 4 3x 3 16x
3 2 24. u s t 2 st 3 ■
■
29–32
x 2 2 sx 20. y x
A B cos y y 10 ■
■
18. tu s2 u s3u
2
■
27. y x sx ,
16. y sx x 1
2
■
1, 9
■ Find an equation of the tangent line to the curve at the given point. Illustrate by graphing the curve and the tangent line on the same screen.
s10 14. Rx x7
13. Vr r 3
■
26. y 1 2x2,
; 27–28
12. Rt 5t 35
4 3
15. F x
■
10. hx x 22x 3
1
3, 3
25. y 6 cos x ,
1
9. f t 4 t 4 8
■
■ Find equations of the tangent line and normal line to the curve at the given point.
25–26
1. f x 186.5
11. y x
■
first few derivatives and observing the pattern that occurs. (a) f x x n (b) f x 1x
■
SECTION 2.3
35. For what values of x does the graph of f x x 2 sin x
BASIC DIFFERENTIATION FORMULAS
■
105
47. If a stone is thrown vertically upward from the surface of
the moon with a velocity of 10 ms, its height (in meters) after t seconds is h 10t 0.83t 2. (a) What is the velocity of the stone after 3 s? (b) What is the velocity of the stone after it has risen 25 m?
have a horizontal tangent? 36. For what values of x does the graph of
f x x 3 3x 2 x 3 have a horizontal tangent? 37. Show that the curve y 6x 3 5x 3 has no tangent line
48. If a ball is thrown vertically upward with a velocity of
with slope 4.
80 fts, then its height after t seconds is s 80t 16t 2. (a) What is the maximum height reached by the ball? (b) What is the velocity of the ball when it is 96 ft above the ground on its way up? On its way down?
38. Find an equation of the tangent line to the curve y x sx
that is parallel to the line y 1 3x.
39. Find an equation of the normal line to the parabola
y x 2 5x 4 that is parallel to the line x 3y 5.
49. Suppose that the cost (in dollars) for a company to produce
x pairs of a new line of jeans is
40. Where does the normal line to the parabola y x x 2
Cx 2000 3x 0.01x 2 0.0002x 3
at the point (1, 0) intersect the parabola a second time? Illustrate with a sketch. 41. The equation of motion of a particle is s t 3 3t, where s
is in meters and t is in seconds. Find (a) the velocity and acceleration as functions of t, (b) the acceleration after 2 s, and (c) the acceleration when the velocity is 0.
(a) Find the marginal cost function. (b) Find C100 and explain its meaning. What does it predict? (c) Compare C100 with the cost of manufacturing the 101st pair of jeans. 50. The cost function for a certain commodity is
42. The equation of motion of a particle is
s 2t 3 7t 2 4t 1, where s is in meters and t is in seconds. (a) Find the velocity and acceleration as functions of t. (b) Find the acceleration after 1 s. (c) Graph the position, velocity, and acceleration functions on the same screen.
;
Cx 84 0.16x 0.0006x 2 0.000003x 3 (a) Find and interpret C100. (b) Compare C100 with the cost of producing the 101st item. 51. A spherical balloon is being inflated. Find the rate of
A particle moves according to a law of motion s f t, t 0, where t is measured in seconds and s in feet. (a) Find the velocity at time t. (b) What is the velocity after 3 s? (c) When is the particle at rest? (d) When is the particle moving in the positive direction? (e) Find the total distance traveled during the first 8 s. (f ) Draw a diagram like Figure 7 to illustrate the motion of the particle. 43– 44
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43. f t t 3 12t 2 36t 44. f t t 3 9t 2 15t 10 ■
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45. The position function of a particle is given by
s t 3 4.5t 2 7t, t 0. (a) When does the particle reach a velocity of 5 ms? (b) When is the acceleration 0? What is the significance of this value of t ? 46. If a ball is given a push so that it has an initial velocity of
5 ms down a certain inclined plane, then the distance it has rolled after t seconds is s 5t 3t 2. (a) Find the velocity after 2 s. (b) How long does it take for the velocity to reach 35 ms?
increase of the surface area S 4 r 2 with respect to the radius r when r is (a) 1 ft, (b) 2 ft, and (c) 3 ft. What conclusion can you make? 52. If a tank holds 5000 gallons of water, which drains from the
bottom of the tank in 40 minutes, then Torricelli’s Law gives the volume V of water remaining in the tank after t minutes as V 5000 (1 401 t )
2
0 t 40
Find the rate at which water is draining from the tank after (a) 5 min, (b) 10 min, (c) 20 min, and (d) 40 min. At what time is the water flowing out the fastest? The slowest? Summarize your findings. 53. Boyle’s Law states that when a sample of gas is compressed
at a constant temperature, the product of the pressure and the volume remains constant: PV C. (a) Find the rate of change of volume with respect to pressure. (b) A sample of gas is in a container at low pressure and is steadily compressed at constant temperature for 10 minutes. Is the volume decreasing more rapidly at the beginning or the end of the 10 minutes? Explain.
106
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CHAPTER 2
DERIVATIVES
54. Newton’s Law of Gravitation says that the magnitude F of
the force exerted by a body of mass m on a body of mass M is GmM F r2 where G is the gravitational constant and r is the distance between the bodies. (a) Find dFdr and explain its meaning. What does the minus sign indicate? (b) Suppose it is known that the Earth attracts an object with a force that decreases at the rate of 2 Nkm when r 20,000 km. How fast does this force change when r 10,000 km?
60. (a) Find equations of both lines through the point 2, 3
that are tangent to the parabola y x 2 x. (b) Show that there is no line through the point 2, 7 that is tangent to the parabola. Then draw a diagram to see why.
61. For what values of a and b is the line 2x y b tangent to
the parabola y ax 2 when x 2?
62. Find a parabola with equation y ax 2 bx c that has
slope 4 at x 1, slope 8 at x 1, and passes through the point 2, 15.
63. Find a cubic function y ax 3 bx 2 cx d whose graph
has horizontal tangents at the points 2, 6 and 2, 0.
55. Use the definition of a derivative to show that if f x 1x,
64. A tangent line is drawn to the hyperbola xy c at a point P.
56. Prove, using the definition of derivative, that if f x cos x,
(a) Show that the midpoint of the line segment cut from this tangent line by the coordinate axes is P. (b) Show that the triangle formed by the tangent line and the coordinate axes always has the same area, no matter where P is located on the hyperbola.
then f x 1x 2. (This proves the Power Rule for the case n 1.) then f x sin x.
57. The equation y y 2y sin x is called a differential
equation because it involves an unknown function y and its derivatives y and y. Find constants A and B such that the function y A sin x B cos x satisfies this equation. (Differential equations will be studied in detail in Section 7.6.) 58. Find constants A, B, and C such that the function
y Ax 2 Bx C satisfies the differential equation y y 2y x 2.
xl1
x 1000 1 . x1
66. Draw a diagram showing two perpendicular lines that inter-
sect on the y-axis and are both tangent to the parabola y x 2. Where do these lines intersect? 67. If c 2 , how many lines through the point 0, c are normal 1
59. Draw a diagram to show that there are two tangent lines to
the parabola y x 2 that pass through the point 0, 4. Find the coordinates of the points where these tangent lines intersect the parabola.
2.4
65. Evaluate lim
lines to the parabola y x 2 ? What if c 12 ?
68. Sketch the parabolas y x 2 and y x 2 2x 2. Do you
think there is a line that is tangent to both curves? If so, find its equation. If not, why not?
THE PRODUCT AND QUOTIENT RULES The formulas of this section enable us to differentiate new functions formed from old functions by multiplication or division. THE PRODUCT RULE
| By analogy with the Sum and Difference Rules, one might be tempted to guess, as Leibniz did three centuries ago, that the derivative of a product is the product of the derivatives. We can see, however, that this guess is wrong by looking at a particular example. Let f x x and tx x 2. Then the Power Rule gives f x 1 and tx 2x. But ftx x 3, so ftx 3x 2. Thus ft f t. The correct formula was discovered by Leibniz (soon after his false start) and is called the Product Rule.
We can write the Product Rule in prime notation as ft ft t f ■
THE PRODUCT RULE If f and t are both differentiable, then
d d d f xtx f x tx tx f x dx dx dx
SECTION 2.4
THE PRODUCT AND QUOTIENT RULES
■
107
PROOF Let Fx f xtx. Then
Fx lim
hl0
lim
hl0
Fx h Fx h f x htx h f xtx h
In order to evaluate this limit, we would like to separate the functions f and t as in the proof of the Sum Rule. We can achieve this separation by subtracting and adding the term f x htx in the numerator: Fx lim
hl0
f x htx h f x htx f x htx f xtx h
lim f x h hl0
tx h tx f x h f x tx h h
lim f x h lim hl0
hl0
tx h tx f x h f x lim tx lim hl0 hl0 h h
f xtx txf x Note that lim h l 0 tx tx because tx is a constant with respect to the variable h. Also, since f is differentiable at x, it is continuous at x by Theorem 2.2.4, and so lim h l 0 f x h f x. ■ ■ Figure 1 shows the graphs of the function of Example 1 and its derivative. Notice that y 0 whenever y has a horizontal tangent.
5 yª _4
y
In words, the Product Rule says that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. V EXAMPLE 1
4
_5
FIGURE 1
Differentiate y x 2 sin x.
SOLUTION Using the Product Rule, we have
dy d d x2 sin x sin x x 2 dx dx dx x 2 cos x 2x sin x EXAMPLE 2 Differentiate the function f t st a bt.
■ In Example 2, a and b are constants. It is customary in mathematics to use letters near the beginning of the alphabet to represent constants and letters near the end of the alphabet to represent variables.
SOLUTION 1 Using the Product Rule, we have
f t st
d d a bt a bt (st ) dt dt
st b a bt 12 t 12 bst
a bt a 3bt 2st 2st
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108
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CHAPTER 2
DERIVATIVES
SOLUTION 2 If we first use the laws of exponents to rewrite f t, then we can proceed directly without using the Product Rule.
f t ast btst at 12 bt 32 f t 12 at12 32 bt 12 which is equivalent to the answer given in Solution 1.
■
Example 2 shows that it is sometimes easier to simplify a product of functions than to use the Product Rule. In Example 1, however, the Product Rule is the only possible method. EXAMPLE 3 If hx xtx and it is known that t3 5 and t3 2, find h3. SOLUTION Applying the Product Rule, we get
hx
d d d xtx x tx tx x dx dx dx
xtx tx h3 3t3 t3 3 2 5 11
Therefore
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THE QUOTIENT RULE
The following rule enables us to differentiate the quotient of two differentiable functions. THE QUOTIENT RULE If f and t are differentiable, then In prime notation we can write the Quotient Rule as t f ft f t2 t ■
d dx
f x tx
tx
d d f x f x tx dx dx tx 2
PROOF Let Fx f xtx. Then
f x h f x Fx h Fx tx h tx Fx lim lim hl0 hl0 h h lim
hl0
f x htx f xtx h htx htx
We can separate f and t in this expression by subtracting and adding the term f xtx in the numerator: Fx lim
hl0
f x htx f xtx f xtx f xtx h htx htx tx
lim
hl0
f x h f x tx h tx f x h h tx htx
SECTION 2.4
lim tx lim
hl0
hl0
THE PRODUCT AND QUOTIENT RULES
109
f x h f x tx h tx lim f x lim hl0 hl0 h h lim tx h lim tx hl0
■
hl0
txf x f xtx tx 2
Again t is continuous by Theorem 2.2.4, so lim h l 0 tx h tx.
■
In words, the Quotient Rule says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. The Quotient Rule and the other differentiation formulas enable us to compute the derivative of any rational function, as the next example illustrates. We can use a graphing device to check that the answer to Example 4 is plausible. Figure 2 shows the graphs of the function of Example 4 and its derivative. Notice that when y grows rapidly (near 2 ), y is large. And when y grows slowly, y is near 0 . ■
V EXAMPLE 4
Let y
Then x 3 6 y
1.5 yª _4
d d x 2 x 2 x 2 x 2 x 3 6 dx dx x 3 62
x 3 62x 1 x 2 x 23x 2 x 3 62
2x 4 x 3 12x 6 3x 4 3x 3 6x 2 x 3 62
x 4 2x 3 6x 2 12x 6 x 3 62
4 y _1.5
x2 x 2 . x3 6
FIGURE 2
EXAMPLE 5 Find an equation of the tangent line to the curve y sx1 x 2 at
the point (1, 12 ).
SOLUTION According to the Quotient Rule, we have
dy dx
1 x 2
d d (sx ) sx dx 1 x 2 dx 1 x 2 2
1 sx 2x 2sx 1 x 2 2
1 x 2
1 x 2 4x 2 1 3x 2 2sx 1 x 2 2 2sx 1 x 2 2
So the slope of the tangent line at (1, 12 ) is dy dx
x1
1 3 12 1 2 2 2s11 1 4
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110
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CHAPTER 2
DERIVATIVES
We use the point-slope form to write an equation of the tangent line at (1, 12 ):
1
y 12 14 x 1
”1, 21 ’ y=
0
FIGURE 3
œ„ x 1+≈
or
y 14 x 34
The curve and its tangent line are graphed in Figure 3. 4
■
NOTE Don’t use the Quotient Rule every time you see a quotient. Sometimes it’s easier to rewrite a quotient first to put it in a form that is simpler for the purpose of differentiation. For instance, although it is possible to differentiate the function
Fx
3x 2 2sx x
using the Quotient Rule, it is much easier to perform the division first and write the function as Fx 3x 2x 12 before differentiating.
TRIGONOMETRIC FUNCTIONS
Knowing the derivatives of the sine and cosine functions, we can use the Quotient Rule to find the derivative of the tangent function: d d tan x dx dx
cos x
sin x cos x
d d sin x sin x cos x dx dx cos2x
cos x cos x sin x sin x cos2x
cos2x sin2x cos2x
1 sec2x cos2x
d tan x sec2x dx
The derivatives of the remaining trigonometric functions, csc, sec, and cot , can also be found easily using the Quotient Rule (see Exercises 37–39). We collect all the differentiation formulas for trigonometric functions in the following table. Remember that they are valid only when x is measured in radians.
SECTION 2.4
THE PRODUCT AND QUOTIENT RULES
■
111
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
d sin x cos x dx d cos x sin x dx d tan x sec2x dx
When you memorize this table, it is helpful to notice that the minus signs go with the derivatives of the “cofunctions,” that is, cosine, cosecant, and cotangent. ■
EXAMPLE 6 Differentiate f x
have a horizontal tangent?
d csc x csc x cot x dx d sec x sec x tan x dx d cot x csc 2x dx
sec x . For what values of x does the graph of f 1 tan x
SOLUTION The Quotient Rule gives
1 tan x f x
3
_3
1 tan x sec x tan x sec x sec2x 1 tan x2
sec x tan x tan2x sec2x 1 tan x2
sec x tan x 1 1 tan x2
5
_3
FIGURE 4
The horizontal tangents in Example 6
2.4
In simplifying the answer we have used the identity tan2x 1 sec2x. Since sec x is never 0, we see that f x 0 when tan x 1, and this occurs when x n 4, where n is an integer (see Figure 4).
by using the Product Rule and by performing the multiplication first. Do your answers agree? 2. Find the derivative of the function
Fx
x 3x sx sx
in two ways: by using the Quotient Rule and by simplifying first. Show that your answers are equivalent. Which method do you prefer? ■
Differentiate.
3. tt t 3 cos t 5. F y
■
EXERCISES
1. Find the derivative of y x 2 1x 3 1 in two ways:
3–26
d d sec x sec x 1 tan x dx dx 1 tan x2
1 3 y 5y 3 2 y y4
4. f x sx sin x
6. Yu u2 u3 u 5 2u 2 7. f x sin x 2 cot x 1
9. h csc cot 11. tx 13. y 15. y 17. y
3x 1 2x 1
t2 3t 2t 1 2
v 3 2v sv v
r2 1 sr
8. y 2 csc x 5 cos x 10. y ua cos u b cot u 12. f t
2t 4 t2
14. y
t3 t t4 2
16. y
sx 1 sx 1
18. y
cx 1 cx
112
■
19. y
CHAPTER 2
x cos x sec 1 sec
21. f 23. y
DERIVATIVES
sin x x2 x
25. f x
x
20. y
1 sin x x cos x
22. y
1 sec x tan x
24. y
u 6 2u 3 5 u2
26. f x
c x
42. If f 3 4, t3 2, f 3 6, and t3 5, find the
following numbers. (a) f t3
(b) ft3
43. If f and t are the functions whose graphs are shown, let ux f xtx and vx f xtx.
(b) Find v5.
(a) Find u1. y
ax b cx d
f g ■
1
27–30 ■ Find an equation of the tangent line to the curve at the given point.
0
■
■
27. y
■
■
2x , x1
■
4, 1
■
■
■
■
■
■
■
28. y
1, 1
29. y tan x, ■
■
■
■
■
44. Let Px FxGx and Qx FxGx, where F and
sx , 4, 0.4 x1
30. y 1 x cos x, ■
■
■
■
0, 1 ■
x
1
G are the functions whose graphs are shown. (a) Find P2. (b) Find Q7. y
■
F
31. (a) The curve y 11 x 2 is called a witch of Maria
;
Agnesi. Find an equation of the tangent line to this curve at the point (1, 12 ). (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.
0
an equation of the tangent line to this curve at the point 3, 0.3. (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen. 33. If f x x 21 x, find f 1.
G
1
32. (a) The curve y x1 x 2 is called a serpentine. Find
;
(c) ft3
x
1
45. If t is a differentiable function, find an expression for the
derivative of each of the following functions. x tx (a) y xtx (b) y (c) y tx x 46. If f is a differentiable function, find an expression for the
34. If f x sec x, find f 4.
derivative of each of the following functions.
35. If H sin , find H and H .
(a) y x 2 f x
36. Find
d 35 x sin x. dx 35
37. Prove that
d csc x csc x cot x. dx
38. Prove that
d sec x sec x tan x. dx
39. Prove that
d cot x csc 2x. dx
(c) y
x2 f x
(b) y
f x x2
(d) y
1 x f x sx
47. A mass on a spring vibrates horizontally on a smooth level
surface (see the figure). Its equation of motion is xt 8 sin t, where t is in seconds and x in centimeters. (a) Find the velocity and acceleration at time t. (b) Find the position, velocity, and acceleration of the mass at time t 23 . In what direction is it moving at that time? Is it speeding up or slowing down?
40. Suppose f 3 4 and f 3 2, and let
tx f x sin x and hx cos xf x. Find (a) t3 (b) h3
equilibrium position
41. Suppose that f 5 1, f 5 6, t5 3, and
t5 2. Find the following values. (a) ft5 (b) ft5
(c) tf 5
0
x
x
SECTION 2.5
48. An object with weight W is dragged along a horizontal
plane by a force acting along a rope attached to the object. If the rope makes an angle with the plane, then the magnitude of the force is
W F sin cos
;
where is a constant called the coefficient of friction. (a) Find the rate of change of F with respect to . (b) When is this rate of change equal to 0? (c) If W 50 lb and 0.6, draw the graph of F as a function of and use it to locate the value of for which dFd 0. Is the value consistent with your answer to part (b)? 49. The gas law for an ideal gas at absolute temperature T
(in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV nRT, where n is the number of moles of the gas and R 0.0821 is the gas constant. Suppose that, at a certain instant, P 8.0 atm and is increasing at a rate of 0.10 atmmin and V 10 L and is decreasing at a rate of 0.15 Lmin. Find the rate of change of T with respect to time at that instant if n 10 mol. 50. If R denotes the reaction of the body to some stimulus of
strength x, the sensitivity S is defined to be the rate of change of the reaction with respect to x. A particular example is that when the brightness x of a light source is increased, the eye reacts by decreasing the area R of the pupil. The experimental formula 40 24x 0.4 R 1 4x 0.4
2.5
;
THE CHAIN RULE
■
113
has been used to model the dependence of R on x when R is measured in square millimeters and x is measured in appropriate units of brightness. (a) Find the sensitivity. (b) Illustrate part (a) by graphing both R and S as functions of x. Comment on the values of R and S at low levels of brightness. Is this what you would expect? 51. How many tangent lines to the curve y xx 1) pass
through the point 1, 2? At which points do these tangent lines touch the curve?
52. Find the points on the curve y cos x2 sin x at
which the tangent is horizontal. 53. (a) Use the Product Rule twice to prove that if f , t, and h
are differentiable, then fth f th fth fth. (b) Use part (a) to differentiate y x sin x cos x .
54. (a) If Fx f xtx, where f and t have derivatives
of all orders, show that F f t 2 f t f t . (b) Find similar formulas for F and F 4. (c) Guess a formula for F n.
55. (a) If t is differentiable, the Reciprocal Rule says that
d dx
1 tx
tx tx 2
Use the Quotient Rule to prove the Reciprocal Rule. (b) Use the Reciprocal Rule to differentiate the function y 1x 4 x 2 1. (c) Use the Reciprocal Rule to verify that the Power Rule is valid for negative integers, that is, d x n nxn1 dx for all positive integers n.
THE CHAIN RULE Suppose you are asked to differentiate the function Fx sx 2 1
■ See Section 1.2 for a review of composite functions.
The differentiation formulas you learned in the previous sections of this chapter do not enable you to calculate Fx. Observe that F is a composite function. In fact, if we let y f u su and let u tx x 2 1, then we can write y Fx f tx, that is, F f ⴰ t. We know how to differentiate both f and t, so it would be useful to have a rule that tells us how to find the derivative of F f ⴰ t in terms of the derivatives of f and t. It turns out that the derivative of the composite function f ⴰ t is the product of the derivatives of f and t. This fact is one of the most important of the differentiation rules and is called the Chain Rule. It seems plausible if we interpret derivatives as rates of change. Regard dudx as the rate of change of u with respect to x, dydu as the rate
114
■
CHAPTER 2
DERIVATIVES
of change of y with respect to u, and dydx as the rate of change of y with respect to x. If u changes twice as fast as x and y changes three times as fast as u, then it seems reasonable that y changes six times as fast as x, and so we expect that dy dy du dx du dx THE CHAIN RULE If f and t are both differentiable and F f ⴰ t is the com-
posite function defined by Fx f tx, then F is differentiable and F is given by the product Fx f tx tx
In Leibniz notation, if y f u and u tx are both differentiable functions, then dy dy du dx du dx
COMMENTS ON THE PROOF OF THE CHAIN RULE Let u be the change in u corre-
sponding to a change of x in x, that is,
u tx x tx Then the corresponding change in y is y f u u f u It is tempting to write dy y lim x l 0 x dx 1
lim
y u u x
lim
y u lim u x l 0 x
lim
y u lim u x l 0 x
x l 0
x l 0
u l 0
(Note that u l 0 as x l 0 since t is continuous.)
dy du du dx
The only flaw in this reasoning is that in (1) it might happen that u 0 (even when x 0) and, of course, we can’t divide by 0. Nonetheless, this reasoning does at least suggest that the Chain Rule is true. A full proof of the Chain Rule is given at the end of this section. ■
SECTION 2.5
THE CHAIN RULE
■
115
The Chain Rule can be written either in the prime notation f ⴰ tx f tx tx
2
or, if y f u and u tx, in Leibniz notation: dy dy du dx du dx
3
Equation 3 is easy to remember because if dydu and dudx were quotients, then we could cancel du. Remember, however, that du has not been defined and dudx should not be thought of as an actual quotient. EXAMPLE 1 Find Fx if Fx sx 2 1. SOLUTION 1 (using Equation 2): At the beginning of this section we expressed F as Fx f ⴰ tx f tx where f u su and tx x 2 1. Since
f u 12 u12
1 2su
tx 2x
and
Fx f tx tx
we have
1 x 2x 2 2 2sx 1 sx 1
SOLUTION 2 (using Equation 3): If we let u x 2 1 and y su , then
Fx
dy du 1 2x du dx 2su 1 x 2x 2sx 2 1 sx 2 1
■
When using Formula 3 we should bear in mind that dydx refers to the derivative of y when y is considered as a function of x (called the derivative of y with respect to x), whereas dydu refers to the derivative of y when considered as a function of u (the derivative of y with respect to u). For instance, in Example 1, y can be considered as a function of x ( y sx 2 1 ) and also as a function of u ( y su ). Note that dy x Fx dx sx 2 1
whereas
dy 1 f u du 2su
NOTE In using the Chain Rule we work from the outside to the inside. Formula 2 says that we differentiate the outer function f [at the inner function tx] and then we multiply by the derivative of the inner function.
d dx
f
tx
outer function
evaluated at inner function
f
tx
derivative of outer function
evaluated at inner function
tx derivative of inner function
116
■
CHAPTER 2
DERIVATIVES
V EXAMPLE 2
Differentiate (a) y sinx 2 and (b) y sin2x.
SOLUTION
(a) If y sinx 2 , then the outer function is the sine function and the inner function is the squaring function, so the Chain Rule gives dy d dx dx
sin
x 2
outer function
evaluated at inner function
2x cosx 2
cos
x 2
derivative of outer function
evaluated at inner function
2x derivative of inner function
(b) Note that sin2x sin x2. Here the outer function is the squaring function and the inner function is the sine function. So dy d sin x2 dx dx inner function
■
See Reference Page 2 or Appendix A.
2
derivative of outer function
sin x
evaluated at inner function
cos x derivative of inner function
The answer can be left as 2 sin x cos x or written as sin 2x (by a trigonometric identity known as the double-angle formula).
■
In Example 2(a) we combined the Chain Rule with the rule for differentiating the sine function. In general, if y sin u, where u is a differentiable function of x, then, by the Chain Rule, dy dy du du cos u dx du dx dx Thus
d du sin u cos u dx dx
In a similar fashion, all of the formulas for differentiating trigonometric functions can be combined with the Chain Rule. Let’s make explicit the special case of the Chain Rule where the outer function f is a power function. If y tx n, then we can write y f u u n where u tx. By using the Chain Rule and then the Power Rule, we get dy dy du du nu n1 n tx n1tx dx du dx dx 4 THE POWER RULE COMBINED WITH THE CHAIN RULE
If n is any real
number and u tx is differentiable, then d du u n nu n1 dx dx Alternatively,
d tx n n tx n1 tx dx
Notice that the derivative in Example 1 could be calculated by taking n 12 in Rule 4.
SECTION 2.5
THE CHAIN RULE
■
117
EXAMPLE 3 Differentiate y x 3 1100. SOLUTION Taking u tx x 3 1 and n 100 in (4), we have
dy d d x 3 1100 100x 3 199 x 3 1 dx dx dx 100x 3 199 3x 2 300x 2x 3 199 V EXAMPLE 4
Find f x if f x
SOLUTION First rewrite f :
■
1 . 3 x2 x 1 s
f x x 2 x 113. Thus
f x 13 x 2 x 143
d x 2 x 1 dx
13 x 2 x 1432x 1
■
EXAMPLE 5 Find the derivative of the function
tt
t2 2t 1
9
SOLUTION Combining the Power Rule, Chain Rule, and Quotient Rule, we get
t2 2t 1
8
tt 9
d dt
t2 2t 1
t2 2t 1
8
9
2t 1 1 2t 2 45t 28 2 2t 1 2t 110
■
EXAMPLE 6 Differentiate y 2x 15x 3 x 14. SOLUTION In this example we must use the Product Rule before using the Chain
Rule: ■ The graphs of the functions y and y in Example 6 are shown in Figure 1. Notice that y is large when y increases rapidly and y 0 when y has a horizontal tangent. So our answer appears to be reasonable. 10
yª _2
dy d d 2x 15 x 3 x 14 x 3 x 14 2x 15 dx dx dx d 2x 15 4x 3 x 13 x 3 x 1 dx d x 3 x 14 52x 14 2x 1 dx 42x 15x 3 x 133x 2 1 5x 3 x 142x 14 2
1
Noticing that each term has the common factor 22x 14x 3 x 13, we could factor it out and write the answer as
y _10
FIGURE 1
dy 22x 14x 3 x 1317x 3 6x 2 9x 3 dx
■
The reason for the name “Chain Rule” becomes clear when we make a longer chain by adding another link. Suppose that y f u, u tx, and x ht, where f , t, and
118
■
CHAPTER 2
DERIVATIVES
h are differentiable functions. Then, to compute the derivative of y with respect to t, we use the Chain Rule twice: dy dy dx dy du dx dt dx dt du dx dt V EXAMPLE 7
If f x sincostan x, then f x coscostan x
d costan x dx
coscostan x sintan x
d tan x dx
coscostan x sintan x sec2x ■
Notice that we used the Chain Rule twice. EXAMPLE 8 Differentiate y ssec x 3 . SOLUTION Here the outer function is the square root function, the middle function is the secant function, and the inner function is the cubing function. So we have
dy 1 d sec x 3 dx 2ssec x 3 dx
1 d sec x 3 tan x 3 x 3 2ssec x 3 dx
3x 2 sec x 3 tan x 3 2ssec x 3
■
HOW TO PROVE THE CHAIN RULE
Recall that if y f x and x changes from a to a x, we defined the increment of y as y f a x f a According to the definition of a derivative, we have lim
x l 0
y f a x
So if we denote by the difference between the difference quotient and the derivative, we obtain lim lim
x l 0
But
x l 0
y f a x
y f a f a f a 0 x ?
y f a x x
If we define to be 0 when x 0, then becomes a continuous function of x.
SECTION 2.5
■
THE CHAIN RULE
119
Thus, for a differentiable function f, we can write y f a x x
5
where l 0 as x l 0
and is a continuous function of x. This property of differentiable functions is what enables us to prove the Chain Rule. PROOF OF THE CHAIN RULE Suppose u tx is differentiable at a and y f u
is differentiable at b ta. If x is an increment in x and u and y are the corresponding increments in u and y, then we can use Equation 5 to write u ta x 1 x ta 1 x
6
where 1 l 0 as x l 0. Similarly y f b u 2 u f b 2 u
7
where 2 l 0 as u l 0. If we now substitute the expression for u from Equation 6 into Equation 7, we get y f b 2 ta 1 x y f b 2 ta 1 x
so
As x l 0, Equation 6 shows that u l 0. So both 1 l 0 and 2 l 0 as x l 0. Therefore dy y lim lim f b 2 ta 1 x l 0 x x l 0 dx f bta f tata ■
This proves the Chain Rule.
2.5
EXERCISES
■ Write the composite function in the form f tx. [Identify the inner function u tx and the outer function y f u.] Then find the derivative dydx .
15. tx 1 4x53 x x 2 8
1–6
16. ht t 4 13t 3 14
1. y sin 4x
2. y s4 3x
17. y 2x 548x 2 53
3 18. y x 2 1 s x2 2
3. y 1 x 2 10
4. y tansin x
19. y x 3 cos nx
20. y x sin sx
5. y ssin x
6. y sin sx
21. y sinx cos x
22. f x
■
■
7–38
■
■
■
■
■
■
■
■
■
■
Find the derivative of the function.
4 7. Fx s 1 2x x 3
1 9. tt 4 t 13
■
23. Fz
8. Fx x 2 x 13 10. f t s1 tan t 3
11. y cosa 3 x 3
12. y a 3 cos3x
13. y cotx2
14. y 4 sec 5x
25. y
z1 z1
r sr 1 2
24. G y 26. y
x s 7 3x
y2 y1
sin2x cos x
27. y tancos x
28. y tan 23
29. y sins1 x 2
30. y x sin
1 x
5
120
■
CHAPTER 2
DERIVATIVES
31. y 1 cos 2x6
32. y cotx 2 cot 2 x
33. y sec 2x tan2x
34. y sinsinsin x
35. y cot 2sin
36. y
37. y sin(tan ssin x )
38. y scossin 2 x
■
■
■
■
■
■
■
each derivative, if it exists. If it does not exist, explain why. (a) u1 (b) v1 (c) w1 y
sx sx sx
■
■
■
■
f ■
g 1
Find an equation of the tangent line to the curve at the given point.
39– 40
■
39. y 1 2 x10 ,
■
■
■
x
1
0, 1
40. y sin x sin 2 x , ■
0
52. If f is the function whose graph is shown, let
0, 0 ■
■
■
■
■
■
■
■
hx f f x and tx f x 2 . Use the graph of f to estimate the value of each derivative. (a) h2 (b) t2
41. (a) Find an equation of the tangent line to the curve
y tan x 24 at the point 1, 1. (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.
;
y
y=ƒ
42. (a) The curve y x s2 x 2 is called a bullet-nose
1
curve. Find an equation of the tangent line to this curve at the point 1, 1. (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.
;
43– 46
■
44. hx sx 2 1
45. y x 1
46. Ht tan 3t
3
■
23
■
■
■
■
■
■
■
x
1
53. Suppose f is differentiable on ⺢. Let Fx f cos x and
Gx cos f x. Find expressions for (a) Fx and (b) Gx.
Find the first and second derivatives of the function.
43. Ft 1 7t6
■
0
54. Suppose f is differentiable on ⺢ and is a real number.
Let Fx f x and Gx f x . Find expressions for (a) Fx and (b) Gx.
■
■
47. If Fx f tx, where f 2 8, f 2 4,
f 5 3, t5 2, and t5 6, find F5.
48. If hx s4 3f x , where f 1 7 and f 1 4,
find h1.
■
55. Let rx f thx, where h1 2, t2 3, h1 4,
t2 5, and f 3 6. Find r1.
56. If t is a twice differentiable function and f x xtx 2 ,
find f in terms of t, t, and t .
57. Find all points on the graph of the function
49. A table of values for f , t, f , and t is given. x
f x
tx
f x
tx
1 2 3
3 1 7
2 8 2
4 5 7
6 7 9
f x 2 sin x sin 2x at which the tangent line is horizontal. 58. Find the 50th derivative of y cos 2 x . 59. The displacement of a particle on a vibrating string is given
by the equation st 10 4 sin10 t 1
(a) If hx f tx, find h1. (b) If Hx t f x, find H1. 50. Let f and t be the functions in Exercise 49.
(a) If Fx f f x, find F2. (b) If Gx ttx, find G3.
51. If f and t are the functions whose graphs are shown, let ux f tx, vx t f x, and w x t tx. Find
where s is measured in centimeters and t in seconds. Find the velocity of the particle after t seconds. 60. If the equation of motion of a particle is given by
s A cos t , the particle is said to undergo simple harmonic motion. (a) Find the velocity of the particle at time t. (b) When is the velocity 0?
SECTION 2.6
nately increases and decreases. The most easily visible such star is Delta Cephei, for which the interval between times of maximum brightness is 5.4 days. The average brightness of this star is 4.0 and its brightness changes by 0.35. In view of these data, the brightness of Delta Cephei at time t, where t is measured in days, has been modeled by the function
(a) The derivative of an even function is an odd function. (b) The derivative of an odd function is an even function. 67. Use the Chain Rule to show that if is measured in
degrees, then
d sin cos d 180
(a) Find the rate of change of the brightness after t days. (b) Find, correct to two decimal places, the rate of increase after one day.
(This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: The differentiation formulas would not be as simple if we used degree measure.)
62. A model for the length of daylight (in hours) in Philadel-
phia on the t th day of the year is given by the function
2 t 80 365
68. Suppose y f x is a curve that always lies above the
x-axis and never has a horizontal tangent, where f is differentiable everywhere. For what value of y is the rate of change of y 5 with respect to x eighty times the rate of change of y with respect to x ?
Use this model to compare how the number of hours of daylight is increasing in Philadelphia on March 21 and May 21. 63. A particle moves along a straight line with displacement st, velocity vt, and acceleration at. Show that
at vt
69. If y f u and u tx, where f and t are twice differen-
dv ds
tiable functions, show that d 2y dy d 2u d 2y 2 2 dx du dx du 2
Explain the difference between the meanings of the derivatives dvdt and dvds.
du dx
2
70. (a) Write x sx 2 and use the Chain Rule to show that
64. Air is being pumped into a spherical weather balloon. At
any time t, the volume of the balloon is Vt and its radius is rt. (a) What do the derivatives dVdr and dVdt represent? (b) Express dVdt in terms of drdt.
d x dx
x
x
(b) If f x sin x , find f x and sketch the graphs of f and f . Where is f not differentiable? (c) If tx sin x , find tx and sketch the graphs of t and t. Where is t not differentiable?
65. (a) If n is a positive integer, prove that
d sinn x cos nx n sinn1x cosn 1x dx
2.6
121
66. Use the Chain Rule to prove the following.
Bt 4.0 0.35 sin2 t5.4
■
(b) Find a formula for the derivative of y cosnx cos nx that is similar to the one in part (a).
61. A Cepheid variable star is a star whose brightness alter-
Lt 12 2.8 sin
IMPLICIT DIFFERENTIATION
IMPLICIT DIFFERENTIATION The functions that we have met so far can be described by expressing one variable explicitly in terms of another variable—for example, y sx 3 1
or
y x sin x
or, in general, y f x. Some functions, however, are defined implicitly by a relation between x and y such as 1
x 2 y 2 25
2
x 3 y 3 6xy
or
122
■
CHAPTER 2
DERIVATIVES
In some cases it is possible to solve such an equation for y as an explicit function (or several functions) of x. For instance, if we solve Equation 1 for y, we obtain y s25 x 2 , so two of the functions determined by the implicit Equation l are f x s25 x 2 and tx s25 x 2 . The graphs of f and t are the upper and lower semicircles of the circle x 2 y 2 25. (See Figure 1.) y
y
0
FIGURE 1
x
(a) ≈+¥=25
y
0
0
x
25-≈ (b) ƒ=œ„„„„„„
x
25-≈ (c) ©=_ œ„„„„„„
It’s not easy to solve Equation 2 for y explicitly as a function of x by hand. (A computer algebra system has no trouble, but the expressions it obtains are very complicated.) Nonetheless, (2) is the equation of a curve called the folium of Descartes shown in Figure 2 and it implicitly defines y as several functions of x. The graphs of three such functions are shown in Figure 3. When we say that f is a function defined implicitly by Equation 2, we mean that the equation x 3 f x 3 6x f x is true for all values of x in the domain of f . y
y
y
y
˛+Á=6xy
0
x
FIGURE 2 The folium of Descartes
0
x
0
x
0
x
FIGURE 3 Graphs of three functions defined by the folium of Descartes
Fortunately, we don’t need to solve an equation for y in terms of x in order to find the derivative of y. Instead we can use the method of implicit differentiation: This consists of differentiating both sides of the equation with respect to x and then solving the resulting equation for y. In the examples and exercises of this section it is always assumed that the given equation determines y implicitly as a differentiable function of x so that the method of implicit differentiation can be applied. V EXAMPLE 1
dy . dx (b) Find an equation of the tangent to the circle x 2 y 2 25 at the point 3, 4. (a) If x 2 y 2 25, find
SECTION 2.6
IMPLICIT DIFFERENTIATION
■
123
SOLUTION 1
(a) Differentiate both sides of the equation x 2 y 2 25: d d x 2 y 2 25 dx dx d d x 2 y 2 0 dx dx Remembering that y is a function of x and using the Chain Rule, we have d dy dy d y 2 y 2 2y dx dy dx dx 2x 2y
Thus
dy 0 dx
Now we solve this equation for dydx : x dy dx y (b) At the point 3, 4 we have x 3 and y 4, so dy 3 dx 4 An equation of the tangent to the circle at 3, 4 is therefore y 4 34 x 3
or
3x 4y 25
SOLUTION 2
(b) Solving the equation x 2 y 2 25, we get y s25 x 2 . The point 3, 4 lies on the upper semicircle y s25 x 2 and so we consider the function f x s25 x 2 . Differentiating f using the Chain Rule, we have f x 12 25 x 2 12
d 25 x 2 dx
12 25 x 2 122x
Example 1 illustrates that even when it is possible to solve an equation explicitly for y in terms of x , it may be easier to use implicit differentiation. ■
So
f 3
x s25 x 2
3 3 4 s25 3 2
and, as in Solution 1, an equation of the tangent is 3x 4y 25. V EXAMPLE 2
(a) Find y if x 3 y 3 6xy. (b) Find the tangent to the folium of Descartes x 3 y 3 6xy at the point 3, 3. (c) At what point in the first quadrant is the tangent line horizontal?
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CHAPTER 2
DERIVATIVES
SOLUTION
(a) Differentiating both sides of x 3 y 3 6xy with respect to x, regarding y as a function of x, and using the Chain Rule on the y 3 term and the Product Rule on the 6xy term, we get 3x 2 3y 2 y 6y 6xy x 2 y 2 y 2y 2xy
or We now solve for y :
y 2 y 2xy 2y x 2 y 2 2xy 2y x 2
y
y
(3, 3)
0
2y x 2 y 2 2x
(b) When x y 3,
x
y
2 3 32 1 32 2 3
and a glance at Figure 4 confirms that this is a reasonable value for the slope at 3, 3. So an equation of the tangent to the folium at 3, 3 is
FIGURE 4
y 3 1x 3
4
or
xy6
(c) The tangent line is horizontal if y 0. Using the expression for y from part (a), we see that y 0 when 2y x 2 0 (provided that y 2 2x 0). Substituting y 12 x 2 in the equation of the curve, we get x 3 ( 12 x 2)3 6x ( 12 x 2) which simplifies to x 6 16x 3. Since x 0 in the first quadrant, we have x 3 16. If x 16 13 2 43, then y 12 2 83 2 53. Thus the tangent is horizontal at 2 43, 2 53 , which is approximately (2.5198, 3.1748). Looking at Figure 5, we see that our answer is reasonable. ■
4
0
FIGURE 5
EXAMPLE 3 Find y if sinx y y 2 cos x. SOLUTION Differentiating implicitly with respect to x and remembering that y is a function of x, we get
cosx y 1 y 2yy cos x y 2sin x 2
(Note that we have used the Chain Rule on the left side and the Product Rule and Chain Rule on the right side.) If we collect the terms that involve y, we get cosx y y 2 sin x 2y cos xy cosx y y
_2
2
So
_2
FIGURE 6
y
y 2 sin x cosx y 2y cos x cosx y
Figure 6, drawn with the implicit-plotting command of a computer algebra system, shows part of the curve sinx y y 2 cos x. As a check on our calculation, notice that y 1 when x y 0 and it appears from the graph that the slope is ■ approximately 1 at the origin.
SECTION 2.6
IMPLICIT DIFFERENTIATION
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125
EXAMPLE 4 Find y if x 4 y 4 16. SOLUTION Differentiating the equation implicitly with respect to x, we get
4x 3 4y 3 y 0 Solving for y gives y
3
To find y we differentiate this expression for y using the Quotient Rule and remembering that y is a function of x :
Figure 7 shows the graph of the curve x y 4 16 of Example 4. Notice that it’s a stretched and flattened version of the circle x 2 y 2 4 . For this reason it’s sometimes called a fat circle. It starts out very steep on the left but quickly becomes very flat. This can be seen from the expression x3 x 3 y 3 y y ■
4
d dx
y
y
x3 y3
x3 y3
y 3 ddxx 3 x 3 ddxy 3 y 3 2
y 3 3x 2 x 33y 2 y y6
If we now substitute Equation 3 into this expression, we get
x $+y$=16
3x 2 y 3 3x 3 y 2
2
y
2 x
0
x3 y3
y6 3x 2 y 4 x 6 3x 2 y 4 x 4 7 y y7
But the values of x and y must satisfy the original equation x 4 y 4 16. So the answer simplifies to y
FIGURE 7
2.6 1–2
9. 4 cos x sin y 1
10. y sinx 2 x sin y 2
11. tanxy x y
12. sx y 1 x 2 y 2
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1. xy 2x 3x 2 4 ■
3–14
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EXERCISES
(a) Find y by implicit differentiation. (b) Solve the equation explicitly for y and differentiate to get y in terms of x. (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a). ■
3x 216 x2 48 y7 y7
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2. 4x 2 9y 2 36 ■
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Find dydx by implicit differentiation.
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13. sxy 1 x 2 y ■
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14. sin x cos y sin x cos y ■
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15. If f x x 2 f x 3 10 and f 1 2, find f 1. 16. If tx x sin tx x 2, find t0.
3. x x y 4y 6
4. x 2xy y c
17–22 ■ Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
5. x y xy 3x
6. y 5 x 2 y 3 1 x 4 y
17. x 2 xy y 2 3,
7. x 2 y 2 x sin y 4
8. 1 x sinxy 2
18. x 2 2xy y 2 x 2,
3 2
2
2
2
2
3
1, 1 (ellipse) 1, 2 (hyperbola)
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CHAPTER 2
DERIVATIVES
(c) Find the exact x-coordinates of the points in part (a). (d) Create even more fanciful curves by modifying the equation in part (a).
19. x 2 y 2 2x 2 2y 2 x2 20. x 23 y 23 4
(0, )
(3 s3 , 1)
(cardioid)
(astroid)
1 2
CAS
y
y
30. (a) The curve with equation
2y 3 y 2 y 5 x 4 2x 3 x 2 x
0
has been likened to a bouncing wagon. Use a computer algebra system to graph this curve and discover why. (b) At how many points does this curve have horizontal tangent lines? Find the x-coordinates of these points.
x
8
31. Find the points on the lemniscate in Exercise 21 where the 21. 2x 2 y 2 2 25x 2 y 2
22. y 2 y 2 4 x 2x 2 5
tangent is horizontal.
(0, 2) (devil’s curve)
(3, 1) (lemniscate) y
32. Show by implicit differentiation that the tangent to the
ellipse y2 x2 1 2 a b2
y
at the point x 0 , y 0 is
x
0
x
y0 y x0 x 2 1 a2 b ■
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23–26
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23. 9x y 2 9
24. sx sy 1
25. x y 1
26. x 4 y 4 a 4
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■ Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Show that the given families of curves are orthogonal trajectories of each other, that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes.
33–36
Find y by implicit differentiation.
2
3
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27. (a) The curve with equation y 2 5x 4 x 2 is called a
;
kampyle of Eudoxus. Find an equation of the tangent line to this curve at the point 1, 2. (b) Illustrate part (a) by graphing the curve and the tangent line on a common screen. (If your graphing device will graph implicitly defined curves, then use that capability. If not, you can still graph this curve by graphing its upper and lower halves separately.)
; CAS
3
34. x 2 y 2 ax,
x 2 y 2 by
35. y cx 2,
x 2 2y 2 k
36. y ax 3,
x 2 3y 2 b
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37. Show, using implicit differentiation, that any tangent line at
a point P to a circle with center O is perpendicular to the radius OP.
2
38. Show that the sum of the x- and y-intercepts of any tangent
Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point 1, 2. (b) At what points does this curve have a horizontal tangent? (c) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen.
line to the curve sx sy sc is equal to c. 39. The equation x 2 xy y 2 3 represents a “rotated
ellipse,” that is, an ellipse whose axes are not parallel to the coordinate axes. Find the points at which this ellipse crosses the x-axis and show that the tangent lines at these points are parallel.
29. Fanciful shapes can be created by using the implicit plotting
capabilities of computer algebra systems. (a) Graph the curve with equation
40. (a) Where does the normal line to the ellipse
y y 2 1 y 2 xx 1x 2 At how many points does this curve have horizontal tangents? Estimate the x-coordinates of these points. (b) Find equations of the tangent lines at the points (0, 1) and (0, 2).
ax by 0
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28. (a) The curve with equation y x 3x is called the 2
33. x 2 y 2 r 2,
;
x 2 xy y 2 3 at the point 1, 1 intersect the ellipse a second time? (b) Illustrate part (a) by graphing the ellipse and the normal line. 41. Find all points on the curve x 2 y 2 xy 2 where the slope
of the tangent line is 1.
SECTION 2.7
RELATED RATES
127
x 2 4y 2 5. If the point 5, 0 is on the edge of the shadow, how far above the x-axis is the lamp located?
42. Find equations of both the tangent lines to the ellipse
x 2 4y 2 36 that pass through the point 12, 3.
y
43. The Bessel function of order 0, y Jx, satisfies the dif-
ferential equation xy y xy 0 for all values of x and its value at 0 is J0 1 . (a) Find J0. (b) Use implicit differentiation to find J 0.
? 0
_5
44. The figure shows a lamp located three units to the right of
the y-axis and a shadow created by the elliptical region
2.7
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3
x
≈+4¥=5
RELATED RATES If we are pumping air into a balloon, both the volume and the radius of the balloon are increasing and their rates of increase are related to each other. But it is much easier to measure directly the rate of increase of the volume than the rate of increase of the radius. In a related rates problem the idea is to compute the rate of change of one quantity in terms of the rate of change of another quantity (which may be more easily measured). The procedure is to find an equation that relates the two quantities and then use the Chain Rule to differentiate both sides with respect to time. Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3s. How fast is the radius of the balloon increasing when the diameter is 50 cm? V EXAMPLE 1
SOLUTION We start by identifying two things:
the given information: the rate of increase of the volume of air is 100 cm3s and the unknown: the rate of increase of the radius when the diameter is 50 cm In order to express these quantities mathematically, we introduce some suggestive notation: Let V be the volume of the balloon and let r be its radius. The key thing to remember is that rates of change are derivatives. In this problem, the volume and the radius are both functions of the time t. The rate of increase of the volume with respect to time is the derivative dVdt, and the rate of increase of the radius is drdt . We can therefore restate the given and the unknown as follows: Given:
dV 100 cm3s dt
Unknown:
dr dt
when r 25 cm
128
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CHAPTER 2
DERIVATIVES
In order to connect dVdt and drdt , we first relate V and r by the formula for the volume of a sphere: V 43 r 3 In order to use the given information, we differentiate each side of this equation with respect to t. To differentiate the right side, we need to use the Chain Rule: dV dV dr dr 4 r 2 dt dr dt dt Now we solve for the unknown quantity: ■ Notice that, although dVdt is constant, drdt is not constant.
dr 1 dV dt 4r 2 dt If we put r 25 and dVdt 100 in this equation, we obtain dr 1 1 2 100 dt 4 25 25 The radius of the balloon is increasing at the rate of 125 cms.
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EXAMPLE 2 A ladder 10 ft long rests against a vertical wall. If the bottom of the
ladder slides away from the wall at a rate of 1 fts, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall? SOLUTION We first draw a diagram and label it as in Figure 1. Let x feet be the
distance from the bottom of the ladder to the wall and y feet the distance from the top of the ladder to the ground. Note that x and y are both functions of t (time, measured in seconds). We are given that dxdt 1 fts and we are asked to find dydt when x 6 ft. (See Figure 2.) In this problem, the relationship between x and y is given by the Pythagorean Theorem:
Wall
10
y
x 2 y 2 100 Differentiating each side with respect to t using the Chain Rule, we have
x
Ground
FIGURE 1
2x
dx dy 2y 0 dt dt
and solving this equation for the desired rate, we obtain
dy dt
dy x dx dt y dt
=?
When x 6, the Pythagorean Theorem gives y 8 and so, substituting these values and dxdt 1, we have
y
dy 6 3 1 fts dt 8 4
x dx dt
FIGURE 2
=1
The fact that dydt is negative means that the distance from the top of the ladder to the ground is decreasing at a rate of 34 fts. In other words, the top of the ladder is sliding down the wall at a rate of 34 fts. ■
SECTION 2.7
RELATED RATES
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129
EXAMPLE 3 A water tank has the shape of an inverted circular cone with base
radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m3min, find the rate at which the water level is rising when the water is 3 m deep. SOLUTION We first sketch the cone and label it as in Figure 3. Let V , r, and h be
2
r 4
the volume of the water, the radius of the surface, and the height at time t, where t is measured in minutes. We are given that dVdt 2 m3min and we are asked to find dhdt when h is 3 m. The quantities V and h are related by the equation V 13 r 2h
h
but it is very useful to express V as a function of h alone. In order to eliminate r, we use the similar triangles in Figure 3 to write FIGURE 3
r 2 h 4
r
h 2
and the expression for V becomes V
1 h 3 2
2
h
3 h 12
Now we can differentiate each side with respect to t : dV 2 dh h dt 4 dt dh 4 dV dt h 2 dt
so
Substituting h 3 m and dVdt 2 m3min, we have dh 4 8 2 2 dt 3 9 The water level is rising at a rate of 89 0.28 mmin.
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STRATEGY Examples 1–3 suggest the following steps in solving related rates problems: 1. Read the problem carefully. 2. Draw a diagram if possible. 3. Introduce notation. Assign symbols to all quantities that are functions of time.
|
WARNING A common error is to substitute the given numerical information (for quantities that vary with time) too early. This should be done only after the differentiation. (Step 7 follows Step 6.) For instance, in Example 3 we dealt with general values of h until we finally substituted h 3 at the last stage. (If we had put h 3 earlier, we would have gotten dVdt 0 , which is clearly wrong.)
4. Express the given information and the required rate in terms of derivatives. 5. Write an equation that relates the various quantities of the problem. If necessary,
use the geometry of the situation to eliminate one of the variables by substitution (as in Example 3). 6. Use the Chain Rule to differentiate both sides of the equation with respect to t. 7. Substitute the given information into the resulting equation and solve for the unknown rate. The following examples are further illustrations of the strategy.
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DERIVATIVES
V EXAMPLE 4 Car A is traveling west at 50 mih and car B is traveling north at 60 mih. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?
C
x
y
z
B
A
SOLUTION We draw Figure 4, where C is the intersection of the roads. At a given
time t, let x be the distance from car A to C , let y be the distance from car B to C, and let z be the distance between the cars, where x, y, and z are measured in miles. We are given that dxdt 50 mih and dydt 60 mih. (The derivatives are negative because x and y are decreasing.) We are asked to find dzdt . The equation that relates x, y, and z is given by the Pythagorean Theorem: z2 x 2 y 2
FIGURE 4
Differentiating each side with respect to t, we have 2z
dz dx dy 2x 2y dt dt dt dz 1 dt z
x
dx dy y dt dt
When x 0.3 mi and y 0.4 mi, the Pythagorean Theorem gives z 0.5 mi, so dz 1 0.350 0.460 dt 0.5 78 mih The cars are approaching each other at a rate of 78 mih.
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V EXAMPLE 5 A man walks along a straight path at a speed of 4 fts. A searchlight is located on the ground 20 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight?
SOLUTION We draw Figure 5 and let x be the distance from the man to the point on
x
the path closest to the searchlight. We let be the angle between the beam of the searchlight and the perpendicular to the path. We are given that dxdt 4 fts and are asked to find ddt when x 15. The equation that relates x and can be written from Figure 5: x tan 20
20 ¨
x 20 tan
Differentiating each side with respect to t, we get FIGURE 5
dx d 20 sec2 dt dt so
d dx 201 cos2 201 cos2 4 15 cos2 dt dt
SECTION 2.7
RELATED RATES
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131
When x 15 ft, the length of the beam is 25 ft, so cos 45 and d 1 dt 5
4 5
2
16 0.128 125 ■
The searchlight is rotating at a rate of 0.128 rads.
2.7
EXERCISES
1. If V is the volume of a cube with edge length x and the
12. A street light is mounted at the top of a 15-ft-tall pole.
cube expands as time passes, find dVdt in terms of dxdt.
A man 6 ft tall walks away from the pole with a speed of 5 fts along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?
2. (a) If A is the area of a circle with radius r and the circle
expands as time passes, find dAdt in terms of drdt. (b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 ms, how fast is the area of the spill increasing when the radius is 30 m? 3. Each side of a square is increasing at a rate of 6 cms. At
what rate is the area of the square increasing when the area of the square is 16 cm2 ? 4. The length of a rectangle is increasing at a rate of 8 cms
and its width is increasing at a rate of 3 cms. When the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing? 5. If y x 3 2x and dxdt 5, find dydt when x 2. 6. If x 2 y 2 25 and dydt 6, find dxdt when y 4. 7. If z 2 x 2 y 2, dxdt 2, and dydt 3, find dzdt when
x 5 and y 12.
8. A particle moves along the curve y s1 x 3 . As it
reaches the point 2, 3, the y-coordinate is increasing at a rate of 4 cms. How fast is the x-coordinate of the point changing at that instant?
9–12
(a) (b) (c) (d) (e)
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13. Two cars start moving from the same point. One travels
south at 60 mih and the other travels west at 25 mih. At what rate is the distance between the cars increasing two hours later? 14. A spotlight on the ground shines on a wall 12 m away. If a
man 2 m tall walks from the spotlight toward the building at a speed of 1.6 ms, how fast is the length of his shadow on the building decreasing when he is 4 m from the building? 15. A man starts walking north at 4 fts from a point P. Five
minutes later a woman starts walking south at 5 fts from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking? 16. A baseball diamond is a square with side 90 ft. A batter hits
the ball and runs toward first base with a speed of 24 fts. (a) At what rate is his distance from second base decreasing when he is halfway to first base? (b) At what rate is his distance from third base increasing at the same moment?
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What quantities are given in the problem? What is the unknown? Draw a picture of the situation for any time t. Write an equation that relates the quantities. Finish solving the problem.
9. If a snowball melts so that its surface area decreases at a
90 ft
rate of 1 cm min, find the rate at which the diameter decreases when the diameter is 10 cm. 2
10. At noon, ship A is 150 km west of ship B. Ship A is sailing
east at 35 kmh and ship B is sailing north at 25 kmh. How fast is the distance between the ships changing at 4:00 PM ? 11. A plane flying horizontally at an altitude of 1 mi and a
speed of 500 mih passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.
17. The altitude of a triangle is increasing at a rate of 1 cmmin
while the area of the triangle is increasing at a rate of 2 cm2min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm2 ? 18. A boat is pulled into a dock by a rope attached to the bow
of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in
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CHAPTER 2
DERIVATIVES
at a rate of 1 ms, how fast is the boat approaching the dock when it is 8 m from the dock?
always equal. How fast is the height of the pile increasing when the pile is 10 ft high?
19. At noon, ship A is 100 km west of ship B. Ship A is sailing
south at 35 kmh and ship B is sailing north at 25 kmh. How fast is the distance between the ships changing at 4:00 PM ? 20. A particle is moving along the curve y sx . As the par-
ticle passes through the point 4, 2, its x-coordinate increases at a rate of 3 cms. How fast is the distance from the particle to the origin changing at this instant?
21. Two carts, A and B, are connected by a rope 39 ft long that
passes over a pulley P. The point Q is on the floor 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 2 fts. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q ? P
27. Two sides of a triangle are 4 m and 5 m in length and the
angle between them is increasing at a rate of 0.06 rads. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is 3. angle between them is increasing at a rate of 2 min. How fast is the length of the third side increasing when the angle between the sides of fixed length is 60 ?
A
B
29. Boyle’s Law states that when a sample of gas is compressed
Q 22. Water is leaking out of an inverted conical tank at a rate
of 10,000 cm3min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cmmin when the height of the water is 2 m, find the rate at which water is being pumped into the tank. 23. A trough is 10 ft long and its ends have the shape of isos-
celes triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 12 ft3min, how fast is the water level rising when the water is 6 inches deep? 24. A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the
shallow end, and 9 ft deep at its deepest point. A crosssection is shown in the figure. If the pool is being filled at a rate of 0.8 ft 3min, how fast is the water level rising when the depth at the deepest point is 5 ft? 3 6 12
speed of 8 fts. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?
28. Two sides of a triangle have lengths 12 m and 15 m. The
12 f t
6
26. A kite 100 ft above the ground moves horizontally at a
16
at a constant temperature, the pressure P and volume V satisfy the equation PV C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPamin. At what rate is the volume decreasing at this instant? 30. When air expands adiabatically (without gaining or losing
heat), its pressure P and volume V are related by the equation PV 1.4 C, where C is a constant. Suppose that at a certain instant the volume is 400 cm3 and the pressure is 80 kPa and is decreasing at a rate of 10 kPamin. At what rate is the volume increasing at this instant? 31. If two resistors with resistances R1 and R2 are connected in
parallel, as in the figure, then the total resistance R, measured in ohms (), is given by 1 1 1 R R1 R2 If R1 and R2 are increasing at rates of 0.3 s and 0.2 s, respectively, how fast is R changing when R1 80 and R2 100 ?
6
25. Gravel is being dumped from a conveyor belt at a rate of
30 ft 3min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are
R¡
R™
SECTION 2.8
32. Brain weight B as a function of body weight W in fish has
been modeled by the power function B 0.007W 23, where B and W are measured in grams. A model for body weight as a function of body length L (measured in centimeters) is W 0.12L2.53. If, over 10 million years, the average length of a certain species of fish evolved from 15 cm to 20 cm at a constant rate, how fast was this species’ brain growing when the average length was 18 cm? 33. A television camera is positioned 4000 ft from the base of a
rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let’s assume the rocket rises vertically and its speed is 600 fts when it has risen 3000 ft. (a) How fast is the distance from the television camera to the rocket changing at that moment? (b) If the television camera is always kept aimed at the rocket, how fast is the camera’s angle of elevation changing at that same moment?
2.8
y=ƒ
0
FIGURE 1
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133
34. A lighthouse is located on a small island 3 km away from
the nearest point P on a straight shoreline and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P ? 35. A plane flying with a constant speed of 300 kmh passes
over a ground radar station at an altitude of 1 km and climbs at an angle of 30 . At what rate is the distance from the plane to the radar station increasing a minute later? 36. Two people start from the same point. One walks east at
3 mih and the other walks northeast at 2 mih. How fast is the distance between the people changing after 15 minutes? 37. A runner sprints around a circular track of radius 100 m at
a constant speed of 7 ms. The runner’s friend is standing at a distance 200 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 m? 38. The minute hand on a watch is 8 mm long and the hour
hand is 4 mm long. How fast is the distance between the tips of the hands changing at one o’clock?
LINEAR APPROXIMATIONS AND DIFFERENTIALS
y
{a, f(a)}
LINEAR APPROXIMATIONS AND DIFFERENTIALS
y=L(x)
x
We have seen that a curve lies very close to its tangent line near the point of tangency. In fact, by zooming in toward a point on the graph of a differentiable function, we noticed that the graph looks more and more like its tangent line. (See Figure 4 in Section 2.1.) This observation is the basis for a method of finding approximate values of functions. The idea is that it might be easy to calculate a value f a of a function, but difficult (or even impossible) to compute nearby values of f. So we settle for the easily computed values of the linear function L whose graph is the tangent line of f at a, f a. (See Figure 1.) In other words, we use the tangent line at a, f a as an approximation to the curve y f x when x is near a. An equation of this tangent line is y f a f ax a and the approximation 1
f x f a f ax a
is called the linear approximation or tangent line approximation of f at a. The linear function whose graph is this tangent line, that is, 2
Lx f a f ax a
is called the linearization of f at a. Find the linearization of the function f x sx 3 at a 1 and use it to approximate the numbers s3.98 and s4.05 . Are these approximations overestimates or underestimates? V EXAMPLE 1
SOLUTION The derivative of f x x 312 is
f x 12 x 312
1 2sx 3
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CHAPTER 2
DERIVATIVES
and so we have f 1 2 and f 1 14 . Putting these values into Equation 2, we see that the linearization is 7 x Lx f 1 f 1x 1 2 14 x 1 4 4 The corresponding linear approximation (1) is sx 3
7 x 4 4
(when x is near 1)
In particular, we have y 7
7 0.98 s3.98 4 4 1.995
x
y= 4 + 4 (1, 2) _3
FIGURE 2
0
1
y= x+3 œ„„„„ x
7 1.05 s4.05 4 4 2.0125
and
The linear approximation is illustrated in Figure 2. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near l. We also see that our approximations are overestimates because the tangent line lies above the curve. Of course, a calculator could give us approximations for s3.98 and s4.05 , but the linear approximation gives an approximation over an entire interval. ■ In the following table we compare the estimates from the linear approximation in Example 1 with the true values. Notice from this table, and also from Figure 2, that the tangent line approximation gives good estimates when x is close to 1 but the accuracy of the approximation deteriorates when x is farther away from 1.
s3.9 s3.98 s4 s4.05 s4.1 s5 s6
x
From Lx
Actual value
0.9 0.98 1 1.05 1.1 2 3
1.975 1.995 2 2.0125 2.025 2.25 2.5
1.97484176 . . . 1.99499373 . . . 2.00000000 . . . 2.01246117 . . . 2.02484567 . . . 2.23606797 . . . 2.44948974 . . .
How good is the approximation that we obtained in Example 1? The next example shows that by using a graphing calculator or computer we can determine an interval throughout which a linear approximation provides a specified accuracy. EXAMPLE 2 For what values of x is the linear approximation
sx 3
7 x 4 4
accurate to within 0.5? What about accuracy to within 0.1? SOLUTION Accuracy to within 0.5 means that the functions should differ by less
than 0.5:
sx 3
7 x 4 4
0.5
Equivalently, we could write sx 3 0.5
7 x sx 3 0.5 4 4
SECTION 2.8
4.3 Q y= œ„„„„ x+3+0.5
L (x)
P
y= œ„„„„ x+3-0.5
_4
10
LINEAR APPROXIMATIONS AND DIFFERENTIALS
135
This says that the linear approximation should lie between the curves obtained by shifting the curve y sx 3 upward and downward by an amount 0.5. Figure 3 shows the tangent line y 7 x4 intersecting the upper curve y sx 3 0.5 at P and Q. Zooming in and using the cursor, we estimate that the x-coordinate of P is about 2.66 and the x-coordinate of Q is about 8.66. Thus we see from the graph that the approximation sx 3
_1
FIGURE 3
■
7 x 4 4
is accurate to within 0.5 when 2.6 x 8.6. (We have rounded to be safe.) Similarly, from Figure 4 we see that the approximation is accurate to within 0.1 when 1.1 x 3.9. ■
3 Q y= œ„„„„ x+3+0.1
APPLICATIONS TO PHYSICS y= œ„„„„ x+3-0.1
P _2
5
1
FIGURE 4
Linear approximations are often used in physics. In analyzing the consequences of an equation, a physicist sometimes needs to simplify a function by replacing it with its linear approximation. For instance, in deriving a formula for the period of a pendulum, physics textbooks obtain the expression a T t sin for tangential acceleration and then replace sin by with the remark that sin is very close to if is not too large. [See, for example, Physics: Calculus, 2d ed., by Eugene Hecht (Pacific Grove, CA: Brooks/Cole, 2000), p. 431.] You can verify that the linearization of the function f x sin x at a 0 is Lx x and so the linear approximation at 0 is sin x x (see Exercise 26). So, in effect, the derivation of the formula for the period of a pendulum uses the tangent line approximation for the sine function. Another example occurs in the theory of optics, where light rays that arrive at shallow angles relative to the optical axis are called paraxial rays. In paraxial (or Gaussian) optics, both sin and cos are replaced by their linearizations. In other words, the linear approximations sin
and
cos 1
are used because is close to 0. The results of calculations made with these approximations became the basic theoretical tool used to design lenses. [See Optics, 4th ed., by Eugene Hecht (San Francisco: Addison Wesley, 2002), p. 154.] In Section 8.8 we will present several other applications of the idea of linear approximations to physics. DIFFERENTIALS
■ If dx 0, we can divide both sides of Equation 3 by dx to obtain
dy f x dx We have seen similar equations before, but now the left side can genuinely be interpreted as a ratio of differentials.
The ideas behind linear approximations are sometimes formulated in the terminology and notation of differentials. If y f x, where f is a differentiable function, then the differential dx is an independent variable; that is, dx can be given the value of any real number. The differential dy is then defined in terms of dx by the equation 3
dy f x dx
So dy is a dependent variable; it depends on the values of x and dx. If dx is given a specific value and x is taken to be some specific number in the domain of f , then the numerical value of dy is determined.
136
■
CHAPTER 2
DERIVATIVES
The geometric meaning of differentials is shown in Figure 5. Let Px, f x and Qx x, f x x be points on the graph of f and let dx x. The corresponding change in y is
y
Q
R
Îy
P dx=Îx
0
x
y=ƒ FIGURE 5
dy
y f x x f x
S
x+Î x
x
The slope of the tangent line PR is the derivative f x. Thus the directed distance from S to R is f x dx dy. Therefore, dy represents the amount that the tangent line rises or falls (the change in the linearization), whereas y represents the amount that the curve y f x rises or falls when x changes by an amount dx. Notice from Figure 5 that the approximation y dy becomes better as x becomes smaller. If we let dx x a, then x a dx and we can rewrite the linear approximation (1) in the notation of differentials: f a dx f a dy For instance, for the function f x sx 3 in Example 1, we have dy f x dx
dx 2sx 3
If a 1 and dx x 0.05, then dy and
0.05 0.0125 2s1 3
s4.05 f 1.05 f 1 dy 2.0125
just as we found in Example 1. Our final example illustrates the use of differentials in estimating the errors that occur because of approximate measurements. V EXAMPLE 3 The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?
SOLUTION If the radius of the sphere is r, then its volume is V 3 r 3. If the error 4
in the measured value of r is denoted by dr r, then the corresponding error in the calculated value of V is V, which can be approximated by the differential dV 4 r 2 dr When r 21 and dr 0.05, this becomes dV 4 212 0.05 277 The maximum error in the calculated volume is about 277 cm3.
■
NOTE Although the possible error in Example 3 may appear to be rather large, a better picture of the error is given by the relative error, which is computed by dividing the error by the total volume:
V dV 4r 2 dr dr 4 3 3 V V r 3 r
SECTION 2.8
LINEAR APPROXIMATIONS AND DIFFERENTIALS
■
137
Therefore, the relative error in the volume is approximately three times the relative error in the radius. In Example 3 the relative error in the radius is approximately drr 0.0521 0.0024 and it produces a relative error of about 0.007 in the volume. The errors could also be expressed as percentage errors of 0.24% in the radius and 0.7% in the volume.
2.8 1– 4
■
EXERCISES
1. f x x 3x , 4
3. f x cos x, ■
■
a0
a 2
■
(a) Find the differential dy. (b) Evaluate dy and y if x 1 and dx x 1. (c) Sketch a diagram like Figure 5 showing the line segments with lengths dx, dy, and y.
a 1
2
2. f x 1s2 x ,
■
20. Let y sx .
Find the linearization Lx of the function at a.
■
■
4. f x x 34, ■
■
■
a 16 ■
■
■
sible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing (a) the volume of the cube and (b) the surface area of the cube.
; 5. Find the linear approximation of the function
f x s1 x at a 0 and use it to approximate the numbers s0.9 and s0.99 . Illustrate by graphing f and the tangent line.
; 6. Find the linear approximation of the function
22. The radius of a circular disk is given as 24 cm with a maxi-
3 tx s 1 x at a 0 and use it to approximate the 3 3 numbers s 0.95 and s 1.1 . Illustrate by graphing t and the tangent line.
mum error in measurement of 0.2 cm. (a) Use differentials to estimate the maximum error in the calculated area of the disk. (b) What is the relative error? What is the percentage error?
■ Verify the given linear approximation at a 0. Then determine the values of x for which the linear approximation is accurate to within 0.1.
; 7–10
3 7. s 1 x 1 3x
9. 11 2x4 1 8x ■
■
■
■
23. The circumference of a sphere was measured to be 84 cm
8. tan x x
1
■
10. 1s4 x 2 1
■
■
■
■
■
1 16
x
■
21. The edge of a cube was found to be 30 cm with a pos-
■
■ Use a linear approximation (or differentials) to estimate the given number.
with a possible error of 0.5 cm. (a) Use differentials to estimate the maximum error in the calculated surface area. What is the relative error? (b) Use differentials to estimate the maximum error in the calculated volume. What is the relative error?
11–14
11. 2.001
13. 8.06 23 ■
■
■
24. Use differentials to estimate the amount of paint needed to
apply a coat of paint 0.05 cm thick to a hemispherical dome with diameter 50 m.
12. s99.8
5
14. 11002 ■
■
■
■
■
■
■
■
■
Explain, in terms of linear approximations or differentials, why the approximation is reasonable.
15–16
■
15. sec 0.08 1 ■
■
17–18
■ ■
16. 1.016 1.06 ■
■
■
■
■
■
■
(b) y s4 5x
18. (a) y s1 2s
(b) y 1x 1
■
■
■
■
Find the differential of each function.
17. (a) y x 2 sin 2x
■
■
■
■
■
■
■
■
■
■
25. When blood flows along a blood vessel, the flux F (the vol-
ume of blood per unit time that flows past a given point) is proportional to the fourth power of the radius R of the blood vessel: F kR 4 (This is known as Poiseuille’s Law.) A partially clogged artery can be expanded by an operation called angioplasty, in which a balloon-tipped catheter is inflated inside the artery in order to widen it and restore the normal blood flow. Show that the relative change in F is about four times the relative change in R. How will a 5% increase in the radius affect the flow of blood?
19. Let y tan x.
(a) Find the differential dy. (b) Evaluate dy and y if x 4 and dx 0.1.
26. On page 431 of Physics: Calculus, 2d ed., by Eugene Hecht
(Pacific Grove, CA: Brooks/Cole, 2000), in the course of
138
■
CHAPTER 2
DERIVATIVES
deriving the formula T 2 sLt for the period of a pendulum of length L, the author obtains the equation a T t sin for the tangential acceleration of the bob of the pendulum. He then says, “for small angles, the value of in radians is very nearly the value of sin ; they differ by less than 2% out to about 20°.” (a) Verify the linear approximation at 0 for the sine function:
(b) Are your estimates in part (a) too large or too small? Explain. y
y=fª(x) 1
sin x x
;
0
(b) Use a graphing device to determine the values of x for which sin x and x differ by less than 2%. Then verify Hecht’s statement by converting from radians to degrees. 27. Suppose that the only information we have about a function
f is that f 1 5 and the graph of its derivative is as shown. (a) Use a linear approximation to estimate f 0.9 and f 1.1.
2
REVIEW
x
1
28. Suppose that we don’t have a formula for tx but we know
that t2 4 and tx sx 2 5 for all x. (a) Use a linear approximation to estimate t1.95 and t2.05. (b) Are your estimates in part (a) too large or too small? Explain.
CONCEPT CHECK
1. Write an expression for the slope of the tangent line to the
curve y f x at the point a, f a. 2. Suppose an object moves along a straight line with position
f t at time t. Write an expression for the instantaneous velocity of the object at time t a. How can you interpret this velocity in terms of the graph of f ? 3. Define the derivative f a. Discuss two ways of interpreting
this number. 4. If y f x and x changes from x 1 to x 2 , write expressions
for the following. (a) The average rate of change of y with respect to x over the interval x 1, x 2 . (b) The instantaneous rate of change of y with respect to x at x x 1. 5. Define the second derivative of f . If f t is the position
(c) Sketch the graph of a function that is continuous but not differentiable at a 2. 7. Describe several ways in which a function can fail to be
differentiable. Illustrate with sketches. 8. State each differentiation rule both in symbols and in words.
(a) (c) (e) (g)
The Power Rule The Sum Rule The Product Rule The Chain Rule
(b) The Constant Multiple Rule (d) The Difference Rule (f ) The Quotient Rule
9. State the derivative of each function.
(a) y x n (d) y tan x (g) y cot x
(b) y sin x (e) y csc x
(c) y cos x (f ) y sec x
10. Explain how implicit differentiation works.
function of a particle, how can you interpret the second derivative?
11. (a) Write an expression for the linearization of f at a.
6. (a) What does it mean for f to be differentiable at a?
(b) What is the relation between the differentiability and continuity of a function?
(b) If y f x, write an expression for the differential dy. (c) If dx x, draw a picture showing the geometric meanings of y and dy.
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. If f is continuous at a, then f is differentiable at a. 2. If f and t are differentiable, then
d f x tx f x tx dx
3. If f and t are differentiable, then
d f xtx f xtx dx 4. If f and t are differentiable, then
d f tx f txtx dx
CHAPTER 2
d f x sf x dx 2 sf x
lim
xl2
6. If f is differentiable, then 10.
d f x f (sx ) dx 2 sx d x 2 x 2x 1 dx
139
9. If tx x 5, then
5. If f is differentiable, then
7.
■
REVIEW
d2y dx 2
dy dx
tx t2 80 x2
2
11. An equation of the tangent line to the parabola y x 2 at
2, 4 is y 4 2xx 2.
12.
8. If f r exists, then lim x l r f x f r.
d d tan2x sec 2x dx dx
EXERCISES 7. The figure shows the graphs of f , f , and f . Identify each
1. For the function f whose graph is shown, arrange the
curve, and explain your choices.
following numbers in increasing order: 0
f 2
1
f 3
f 5
y
f 5
a
y
b x
0
c 1 0
x
1
8. The total fertility rate at time t, denoted by Ft, is an esti2. Find a function f and a number a such that
lim
h l0
2 h6 64 f a h
3. The total cost of repaying a student loan at an interest rate
of r% per year is C f r. (a) What is the meaning of the derivative f r? What are its units? (b) What does the statement f 10 1200 mean? (c) Is f r always positive or does it change sign? ■ Trace or copy the graph of the function. Then sketch a graph of its derivative directly beneath.
4 –6 4.
5.
y 3.5
baby boom
3.0 2.5
6. y
y
mate of the average number of children born to each woman (assuming that current birth rates remain constant). The graph of the total fertility rate in the United States shows the fluctuations from 1940 to 1990. (a) Estimate the values of F1950, F1965, and F1987. (b) What are the meanings of these derivatives? (c) Can you suggest reasons for the values of these derivatives?
baby bust baby boomlet
y=F(t)
y 2.0
0
x
x
■
■
■
■
■
■
1.5
x
0 ■
■
■
■
■
■
1940
1950
1960
1970
1980
1990
t
140
■
CHAPTER 2
DERIVATIVES
39. If f t s4t 1, find f 2.
9. Let Ct be the total value of US currency (coins and
banknotes) in circulation at time t. The table gives values of this function from 1980 to 2000, as of September 30, in billions of dollars. Interpret and estimate the value of C1990. t
1980
1985
1990
1995
2000
Ct
129.9
187.3
271.9
409.3
568.6
40. If t sin , find t 6. 41. Find y if x 6 y 6 1. 42. Find f nx if f x 12 x.
Find an equation of the tangent to the curve at the given point.
43– 46
■
6, 1
43. y 4 sin2 x,
Find f x from first principles, that is, directly from the definition of a derivative. 10 –11
■
10. f x ■
■
4x 3x ■
■
44. y
45. y s1 4 sin x ,
11. f x x 3 5x 4 ■
■
■
■
■
■
■
■
find f x. (b) Find the domains of f and f . (c) Graph f and f on a common screen. Compare the graphs to see whether your answer to part (a) is reasonable.
13–38
■
14. y costan x
1 15. y sx 3 4 sx
3x 2 16. y s2x 1
17. y 2xsx 2 1
18. y
19. y
t 1 t2
1 x2
2, 1
■
■
■
■
sec 2 1 tan 2
3 x sx 28. y 1s
29. sinxy x 2 y
30. y ssin sx
31. y cot3x 5
x 4 32. y 4 x 4
Px f xtx, Qx f xtx, and Cx f tx. Find (a) P2, (b) Q2, and (c) C2. y
g f
1 0
51–58
35. y tan2sin
36. x tan y y 1
5 x tan x 37. y s
38. y ■
■
■
Find f in terms of t. 52. f x tx 2
53. f x tx 2
54. f x x atx b
55. f x t tx
56. f x sin tx 58. f x t(tan sx )
57. f x tsin x ■
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■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
Find h in terms of f and t.
59. hx
x 1x 4 x 2x 3 ■
■
x
1
51. f x x 2tx
59–60
34. y
■
■
49. Suppose that hx f xtx and Fx f tx, where
sin mx x
33. y sin(tan s1 x 3 )
■
■
gent line has slope 1.
1 sinx sin x
27. y 1 x 1 1
■
■
is the tangent line horizontal?
26. x 2 cos y sin 2y xy
2
■
48. Find the points on the ellipse x 2 2y 2 1 where the tan-
20. y sincos x
24. y sec1 x 2
■
■
s7
23. xy 4 x 2 y x 3y
■
■
50. If f and t are the functions whose graphs are shown, let
22. y
■
■
47. At what points on the curve y sin x cos x, 0 x 2,
x
2
f 2 3, t2 5, t2 4, f 2 2, and f 5 11. Find (a) h2 and (b) F2.
21. y tan s1 x
25. y
■
Calculate y.
13. y x 4 3x 2 53
0, 1
46. x 4xy y 13, 2
12. (a) If f x s3 5x , use the definition of a derivative to
;
x 1 , 0, 1 x2 1 2
f xtx f x tx
60. hx f tsin 4x ■
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CHAPTER 2
61. The graph of f is shown. State, with reasons, the numbers
REVIEW
■
141
68. A waterskier skis over the ramp shown in the figure at a
at which f is not differentiable.
speed of 30 fts. How fast is she rising as she leaves the ramp?
y
4 ft 15 ft _1 0
2
4
6
x
69. The angle of elevation of the Sun is decreasing at a rate of
0.25 radh. How fast is the shadow cast by a 400-ft-tall building increasing when the angle of elevation of the Sun is 6?
62. The volume of a right circular cone is V r 2h3, where r
is the radius of the base and h is the height. (a) Find the rate of change of the volume with respect to the height if the radius is constant. (b) Find the rate of change of the volume with respect to the radius if the height is constant.
; 70. (a) Find the linear approximation to f x s25 x 2
near 3. (b) Illustrate part (a) by graphing f and the linear approximation. (c) For what values of x is the linear approximation accurate to within 0.1?
63. A particle moves on a vertical line so that its coordinate at
time t is y t 3 12t 3, t 0. (a) Find the velocity and acceleration functions. (b) When is the particle moving upward and when is it moving downward? (c) Find the distance that the particle travels in the time interval 0 t 3. 64. The cost, in dollars, of producing x units of a certain com-
modity is
3 71. (a) Find the linearization of f x s 1 3x at a 0.
State the corresponding linear approximation and use it 3 to give an approximate value for s 1.03 . (b) Determine the values of x for which the linear approximation given in part (a) is accurate to within 0.1.
;
72. Evaluate dy if y x 3 2x 2 1, x 2, and dx 0.2. 73. A window has the shape of a square surmounted by a semi-
Cx 920 2x 0.02x 2 0.00007x 3 (a) Find the marginal cost function. (b) Find C100 and explain its meaning. (c) Compare C100 with the cost of producing the 101st item. 65. The volume of a cube is increasing at a rate of 10 cm min.
circle. The base of the window is measured as having width 60 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum error possible in computing the area of the window. 74 –76
■
3
How fast is the surface area increasing when the length of an edge is 30 cm? 66. A paper cup has the shape of a cone with height 10 cm and
radius 3 cm (at the top). If water is poured into the cup at a rate of 2 cm3s, how fast is the water level rising when the water is 5 cm deep? 67. A balloon is rising at a constant speed of 5 fts. A boy is
cycling along a straight road at a speed of 15 fts. When he passes under the balloon, it is 45 ft above him. How fast is the distance between the boy and the balloon increasing 3 s later?
Express the limit as a derivative and evaluate.
x 1 x1 17
74. lim x l1
76. lim
l 3
■
■
75. lim
hl0
4 16 h 2 s h
cos 0.5 3 ■
■
77. Evaluate lim
xl0
■
■
■
■
■
■
■
■
s1 tan x s1 sin x . x3
78. Show that the length of the portion of any tangent line to
the astroid x 23 y 23 a 23 cut off by the coordinate axes is constant.
3
INVERSE FUNCTIONS EXPONENTIAL, LOGARITHMIC, AND INVERSE TRIGONOMETRIC FUNCTIONS The common theme that links the functions of this chapter is that they occur as pairs of inverse functions. In particular, two of the most important functions that occur in mathematics and its applications are the exponential function f x a x and its inverse function, the logarithmic function tx log a x . Here we investigate their properties, compute their derivatives, and use them to describe exponential growth and decay in biology, physics, chemistry, and other sciences.We also study the inverses of the trigonometric and hyperbolic functions. Finally we look at a method (l’Hospital’s Rule) for computing limits of such functions.
3.1
EXPONENTIAL FUNCTIONS The function f x 2 x is called an exponential function because the variable, x, is the exponent. It should not be confused with the power function tx x 2, in which the variable is the base. In general, an exponential function is a function of the form f x a x where a is a positive constant. Let’s recall what this means. If x n, a positive integer, then an a a a n factors
If x 0, then a 0 1, and if x n, where n is a positive integer, then a n
1 an
If x is a rational number, x pq, where p and q are integers and q 0, then q p q a x a pq sa (sa )
y
1 0
1
x
But what is the meaning of a x if x is an irrational number? For instance, what is meant by 2 s3 or 5 ? To help us answer this question we first look at the graph of the function y 2 x, where x is rational. A representation of this graph is shown in Figure 1. We want to enlarge the domain of y 2 x to include both rational and irrational numbers. There are holes in the graph in Figure 1 corresponding to irrational values of x. We want to fill in the holes by defining f x 2 x, where x ⺢, so that f is an increasing continuous function. In particular, since the irrational number s3 satisfies
FIGURE 1
Representation of y=2®, x rational 142
p
1.7 s3 1.8
SECTION 3.1
EXPONENTIAL FUNCTIONS
■
143
we must have 2 1.7 2 s3 2 1.8 and we know what 21.7 and 21.8 mean because 1.7 and 1.8 are rational numbers. Similarly, if we use better approximations for s3 , we obtain better approximations for 2 s3:
■ A proof of this fact is given in J. Marsden and A. Weinstein, Calculus Unlimited (Menlo Park, CA: Benjamin/ Cummings, 1981). For an online version, see
1.73 s3 1.74
?
2 1.73 2 s3 2 1.74
1.732 s3 1.733
?
2 1.732 2 s3 2 1.733
1.7320 s3 1.7321
?
2 1.7320 2 s3 2 1.7321
1.73205 s3 1.73206 . . . . . .
?
2 1.73205 2 s3 2 1.73206 . . . . . .
It can be shown that there is exactly one number that is greater than all of the numbers 2 1.7,
2 1.73,
2 1.732,
2 1.7320,
2 1.73205,
...
2 1.733,
2 1.7321,
2 1.73206,
...
and less than all of the numbers
www.cds.caltech.edu/~marsden/ volume/cu/CU.pdf
2 1.8,
2 1.74,
We define 2 s3 to be this number. Using the preceding approximation process we can compute it correct to six decimal places:
y
2 s3 3.321997
1 0
FIGURE 2
1
x
Similarly, we can define 2 x (or a x, if a 0) where x is any irrational number. Figure 2 shows how all the holes in Figure 1 have been filled to complete the graph of the function f x 2 x, x ⺢. In general, if a is any positive number, we define a x lim a r
1
r lx
y=2®, x real
r rational
This definition makes sense because any irrational number can be approximated as closely as we like by a rational number. For instance, because s3 has the decimal representation s3 1.7320508 . . . , Definition 1 says that 2 s3 is the limit of the sequence of numbers 21.7,
21.73,
21.732,
21.7320,
21.73205,
21.732050,
21.7320508,
...
53.1415926,
...
Similarly, 5 is the limit of the sequence of numbers 53.1,
53.14,
53.141,
53.1415,
53.14159,
53.141592,
It can be shown that Definition 1 uniquely specifies a x and makes the function f x a x continuous. The graphs of members of the family of functions y a x are shown in Figure 3 for various values of the base a. Notice that all of these graphs pass through the same
144
■
CHAPTER 3
INVERSE FUNCTIONS
point 0, 1 because a 0 1 for a 0. Notice also that as the base a gets larger, the exponential function grows more rapidly (for x 0). 1 ® ” ’ 2
1 ® ” ’ 4
y
10®
4®
y
2®
y
1.5®
y=2®
y=2® 200
y=≈
1®
100
y=≈ 10
0
0
x
1
FIGURE 3 Members of the family of exponential functions
0
x
4
2
FIGURE 4
2
6
4
x
FIGURE 5
Figure 4 shows how the exponential function y 2 x compares with the power function y x 2. The graphs intersect three times, but ultimately the exponential curve y 2 x grows far more rapidly than the parabola y x 2. (See also Figure 5.) You can see from Figure 3 that there are basically three kinds of exponential functions y a x. If 0 a 1, the exponential function decreases; if a 1, it is a constant; and if a 1, it increases. These three cases are illustrated in Figure 6. Since 1a x 1a x a x, the graph of y 1a x is just the reflection of the graph of y a x about the y-axis. y
y
y
1
(0, 1)
(0, 1) 0
FIGURE 6
0
x
(a) y=a®, 0
x
(b) y=1®
0
x
(c) y=a®, a>1
The properties of the exponential function are summarized in the following theorem. If a 0 and a 1, then f x a x is a continuous function with domain ⺢ and range 0, . In particular, a x 0 for all x. If a, b 0 and x, y ⺢, then 2 THEOREM
1. a xy a xa y
■ In Appendix D we present a definition of the exponential function that enables us to give an easy proof of the Laws of Exponents.
2. a xy
ax ay
3. a x y a xy
4. ab x a xb x
The reason for the importance of the exponential function lies in properties 1–4, which are called the Laws of Exponents. If x and y are rational numbers, then these laws are well known from elementary algebra. For arbitrary real numbers x and y these laws can be deduced from the special case where the exponents are rational by using Equation 1.
SECTION 3.1
EXPONENTIAL FUNCTIONS
■
145
The following limits can be read from the graphs shown in Figure 6 or proved from the definition of a limit at infinity. (See Exercise 77 in Section 3.2.)
3
If a 1, then
lim a x
and
lim a x 0
and
x l
If 0 a 1, then
x l
lim a x 0
x l
lim a x
x l
In particular, if a 1, then the x-axis is a horizontal asymptote of the graph of the exponential function y a x. EXAMPLE 1
(a) Find lim x l 2x 1. (b) Sketch the graph of the function y 2x 1. SOLUTION y
lim 2x 1 lim [( 12 ) x 1]
(a)
y=2–®-1
x l
x l
01
[by (3) with a 12 1]
1 0
x
(b) We write y ( 12 ) x 1 as in part (a). The graph of y ( 12 ) x is shown in Figure 3, so we shift it down one unit to obtain the graph of y ( 12 ) x 1 shown in Figure 7. (For a review of shifting graphs, see Section 1.2.) Part (a) shows that the line y 1 is a horizontal asymptote. ■
y=_1
FIGURE 7
THE NUMBER y
Of all possible bases for an exponential function, there is one that is most convenient for the purposes of calculus. We will see in Section 3.3 that the differentiation formula for an exponential function is simplest when the base is chosen to be the number e , which is defined as follows:
3 2
e AND THE NATURAL EXPONENTIAL FUNCTION
y=(1+x)!?®
1 4 0
x
FIGURE 8
x
(1 x)1/x
0.1 0.01 0.001 0.0001 0.00001 0.000001 0.0000001 0.00000001
2.59374246 2.70481383 2.71692393 2.71814593 2.71826824 2.71828047 2.71828169 2.71828181
e lim 1 x1x xl0
The graph of the function y 1 x 1x is shown in Figure 8. It is not defined when x 0, but its behavior when x is near 0 is indicated by the table of values correct to eight decimal places. These values suggest (but don’t prove) that the limit in Definition 4 exists and that e 2.71828. The existence of the limit is proved in Appendix B. The approximate value to 20 decimal places is e 2.71828182845904523536 The decimal expansion of e is nonrepeating because e is an irrational number. (See Exercise 30 in Section 8.8.) The notation e for this number was chosen by the Swiss mathematician Leonhard Euler in 1727, probably because it’s the first letter of the word exponential.
146
■
CHAPTER 3
y
INVERSE FUNCTIONS
y=3® y=2® y=e®
1
The exponential function y e x with base e is called the natural exponential function. Because e lies between 2 and 3, the graph of y e x lies between the graphs of y 2 x and y 3 x, as shown in Figure 9. We will see in Section 3.3 that the natural exponential function is the exponential function whose graph crosses the y -axis with a slope of 1. (See Figure 10.) In fact, we can see why this might be true if we look at the limit in Definition 4. The slope of the tangent line to the graph of f x e x at the point 0, 1 is f 0 lim
x
0
hl0
f 0 h f 0 eh 1 lim hl0 h h
Replacing x by h in Definition 4, we see that, for small values of h ,
FIGURE 9
e 1 h1h
y
y=´
eh 1 h
so
and
eh 1 h
Thus if h is near 0, we have m=1
eh 1 1 h
1
0
x
FIGURE 10
The natural exponential function crosses the y-axis with a slope of 1.
and so it seems plausible that f 0 1. The exponential function f x e x is one of the most frequently occurring functions in calculus and its applications, so it is important to be familiar with its graph (Figure 10) and properties. We summarize these properties as follows, using the fact that this function is just a special case of the exponential functions considered before but with base a e 1. 5 PROPERTIES OF THE NATURAL EXPONENTIAL FUNCTION The exponential function f x e x is a continuous function with domain ⺢ and range 0, . Thus e x 0 for all x. Also
lim e x 0
x l
lim e x
xl
So the x-axis is a horizontal asymptote of f x e x. Evaluate lim e 1x.
V EXAMPLE 2
x l0
SOLUTION If we let t 1x, we know from Section 1.6 that t l as x l 0.
Therefore, by (5), lim e 1x lim e t 0
x l0
EXAMPLE 3 Find lim
x l
t l
■
e 2x . e 1 2x
SOLUTION We divide numerator and denominator by e 2x :
lim
x l
e 2x 1 1 1 lim 1 2x 2x 2x x l e 1 1e 1 lim e 10 x l
We have used the fact that t 2x l as x l and so lim e2x lim e t 0
x l
t l
■
SECTION 3.1
3.1
with base a 0. (b) What is the domain of this function? (c) If a 1, what is the range of this function? (d) Sketch the general shape of the graph of the exponential function for each of the following cases. (i) a 1 (ii) a 1 (iii) 0 a 1
Find the exponential function f x Ca x whose graph is given. 17–18
■
17.
y
18.
2
(1, 6)
(b) What is an approximate value for e? (c) What is the natural exponential function? ■ Graph the given functions on a common screen. How are these graphs related?
y e x,
4. y e x,
y e x,
y 8 x,
5. y 3 x,
y 10 x,
y ( 13 ) ,
■
■
y 5 x,
y 0.6 , x
■
■
■
y (101 )
y 0.3 ,
■
■
■
■
■
■
■
■
■
■
■
■
■
23–30 ■
■
xl
■
■
■
xl
■
■
■
■
2
2 10 x 3 10 x
28. lim e 32x x l2
29. lim e
■
cos x ■
lim e tan x
30. ■
■
■
x l2 ■
■
■
■
■
; 31. If you graph the function 1 e 1x 1 e 1x
you’ll see that f appears to be an odd function. Prove it.
; 32. Graph several members of the family of functions
1 1 ex
f x
(b) tt s1 2 t ■
xl
f x
(b) f x
■
26. lim
32x
2x
Find the domain of each function.
■
■
xl
x l2
graph that results from (a) reflecting about the line y 4 (b) reflecting about the line x 2
16. (a) tt sinet
24. lim ex
e 3x e3x e 3x e3x
27. lim e
14. Starting with the graph of y e , find the equation of the
1 1 ex
■
Find the limit.
23. lim 1.001 x 25. lim
x
15. (a) f x
■
xl
graph that results from (a) shifting 2 units downward (b) shifting 2 units to the right (c) reflecting about the x-axis (d) reflecting about the y-axis (e) reflecting about the x-axis and then about the y-axis
■
■
x
e x 1,000,000,000.
13. Starting with the graph of y e x, write the equation of the
15–16
■
; 22. Use a graph to estimate the values of x such that
12. y 21 e x
1
■
ing both f and t in several viewing rectangles. When does the graph of t finally surpass the graph of f ?
10. y 1 2e x
11. y 1 2 ex
■
10 x ; 21. Compare the functions f x x and tx e by graph-
8. y 4 x3
9. y 2 x
■
tx 5 x by graphing both functions in several viewing rectangles. Find all points of intersection of the graphs correct to one decimal place.
■
■
7. y 4 x 3
■
5 ; 20. Compare the rates of growth of the functions f x x and
Make a rough sketch of the graph of the function. Do not use a calculator. Just use the graphs given in Figures 3 and 9 and, if necessary, the transformations of Section 1.2. 7–12
■
19. Suppose the graphs of f x x and tx 2 are drawn
x
y 0.1 ■
■
on a coordinate grid where the unit of measurement is 1 inch. Show that, at a distance 2 ft to the right of the origin, the height of the graph of f is 48 ft but the height of the graph of t is about 265 mi. x
■
■
x
0
x
2
y 8 x
x
■
■
y 20 x
x
2
”2, 9 ’
0
; 3–6
3. y 2 x,
y
(3, 24)
2. (a) How is the number e defined?
6. y 0.9 ,
147
■
EXERCISES
1. (a) Write an equation that defines the exponential function
x
EXPONENTIAL FUNCTIONS
■
■
■
■
1 1 ae bx
where a 0. How does the graph change when b changes? How does it change when a changes?
148
■
CHAPTER 3
3.2
INVERSE FUNCTIONS
INVERSE FUNCTIONS AND LOGARITHMS Table 1 gives data from an experiment in which a bacteria culture started with 100 bacteria in a limited nutrient medium; the size of the bacteria population was recorded at hourly intervals. The number of bacteria N is a function of the time t : N f t. Suppose, however, that the biologist changes her point of view and becomes interested in the time required for the population to reach various levels. In other words, she is thinking of t as a function of N. This function is called the inverse function of f, denoted by f 1, and read “ f inverse.” Thus t f 1N is the time required for the population level to reach N. The values of f 1 can be found by reading Table 1 from right to left or by consulting Table 2. For instance, f 1550 6 because f 6 550. TABLE 1 N as a function of t
4
10
3
7
2
4
1
2 f
A
t (hours)
N f t population at time t
N
t f 1N time to reach N bacteria
0 1 2 3 4 5 6 7 8
100 168 259 358 445 509 550 573 586
100 168 259 358 445 509 550 573 586
0 1 2 3 4 5 6 7 8
B
4
TABLE 2 t as a function of N
10
3
Not all functions possess inverses. Let’s compare the functions f and t whose arrow diagrams are shown in Figure 1. Note that f never takes on the same value twice (any two inputs in A have different outputs), whereas t does take on the same value twice (both 2 and 3 have the same output, 4). In symbols,
4
2 1
2 g
A
B
t2 t3
FIGURE 1 f is one-to-one; g is not
but
f x 1 f x 2
whenever x 1 x 2
Functions that share this property with f are called one-to-one functions. In the language of inputs and outputs, Definition 1 says that f is one-to-one if each output corresponds to only one input. ■
y
1 DEFINITION A function f is called a one-to-one function if it never takes on the same value twice; that is,
f x 1 f x 2
whenever x 1 x 2
y=ƒ fl
0
⁄
‡
¤
FIGURE 2
This function is not one-to-one because f(⁄)=f(¤).
x
If a horizontal line intersects the graph of f in more than one point, then we see from Figure 2 that there are numbers x 1 and x 2 such that f x 1 f x 2 . This means that f is not one-to-one. Therefore, we have the following geometric method for determining whether a function is one-to-one. HORIZONTAL LINE TEST A function is one-to-one if and only if no horizontal line intersects its graph more than once.
SECTION 3.2
149
SOLUTION 1 If x 1 x 2 , then x 13 x 23 (two different numbers can’t have the same
y=˛
cube). Therefore, by Definition 1, f x x 3 is one-to-one.
SOLUTION 2 From Figure 3 we see that no horizontal line intersects the graph of
x
0
■
Is the function f x x 3 one-to-one?
V EXAMPLE 1
y
INVERSE FUNCTIONS AND LOGARITHMS
f x x 3 more than once. Therefore, by the Horizontal Line Test, f is one-to-one.
■
Is the function tx x 2 one-to-one?
FIGURE 3
V EXAMPLE 2
ƒ=˛ is one-to-one.
SOLUTION 1 This function is not one-to-one because, for instance,
y
t1 1 t1
y=≈
and so 1 and 1 have the same output.
0
SOLUTION 2 From Figure 4 we see that there are horizontal lines that intersect the graph of t more than once. Therefore, by the Horizontal Line Test, t is not one-toone. ■
x
FIGURE 4
One-to-one functions are important because they are precisely the functions that possess inverse functions according to the following definition.
©=≈ is not one-to-one.
2 DEFINITION Let f be a one-to-one function with domain A and range B. Then its inverse function f 1 has domain B and range A and is defined by
f 1y x
&?
f x y
for any y in B. This definition says that if f maps x into y, then f 1 maps y back into x. (If f were not one-to-one, then f 1 would not be uniquely defined.) The arrow diagram in Figure 5 indicates that f 1 reverses the effect of f . Note that
x
A f B
f –! y
domain of f 1 range of f
FIGURE 5
range of f 1 domain of f For example, the inverse function of f x x 3 is f 1x x 13 because if y x 3, then f 1y f 1x 3 x 3 13 x
|
CAUTION Do not mistake the 1 in f 1 for an exponent. Thus
f 1x does not mean
1 f x
The reciprocal 1f x could, however, be written as f x 1.
150
■
CHAPTER 3
A
INVERSE FUNCTIONS
B
V EXAMPLE 3
f 1
5
3
7
8
_10
1
10.
If f 1 5, f 3 7, and f 8 10, find f 17, f 15, and
SOLUTION From the definition of f 1 we have
f A
B
1
5
3
7
8
_10
f 17 3
because
f 3 7
f 15 1
because
f 1 5
f 110 8
because
f 8 10
The diagram in Figure 6 makes it clear how f 1 reverses the effect of f in this case. ■
The letter x is traditionally used as the independent variable, so when we concentrate on f 1 rather than on f , we usually reverse the roles of x and y in Definition 2 and write
f –! FIGURE 6
The inverse function reverses inputs and outputs.
f 1x y &?
3
f y x
By substituting for y in Definition 2 and substituting for x in (3), we get the following cancellation equations:
4
f 1 f x x
for every x in A
f f 1x x
for every x in B
The first cancellation equation says that if we start with x, apply f , and then apply f 1, we arrive back at x, where we started (see the machine diagram in Figure 7). Thus f 1 undoes what f does. The second equation says that f undoes what f 1 does.
x
f
ƒ
f –!
x
FIGURE 7
For example, if f x x 3, then f 1x x 13 and so the cancellation equations become f 1 f x x 3 13 x f f 1x x 13 3 x These equations simply say that the cube function and the cube root function cancel each other when applied in succession. Now let’s see how to compute inverse functions. If we have a function y f x and are able to solve this equation for x in terms of y, then according to Definition 2 we must have x f 1y. If we want to call the independent variable x, we then interchange x and y and arrive at the equation y f 1x.
SECTION 3.2
INVERSE FUNCTIONS AND LOGARITHMS
■
151
5 HOW TO FIND THE INVERSE FUNCTION OF A ONE-TO-ONE FUNCTION f
STEP 1
Write y f x.
STEP 2
Solve this equation for x in terms of y (if possible).
STEP 3 To express f 1 as a function of x, interchange x and y.
The resulting equation is y f 1x.
V EXAMPLE 4
Find the inverse function of f x x 3 2.
SOLUTION According to (5) we first write
y x3 2 Then we solve this equation for x : x3 y 2 3 xs y2
Finally, we interchange x and y : In Example 4, notice how f 1 reverses the effect of f . The function f is the rule “Cube, then add 2”; f 1 is the rule “Subtract 2, then take the cube root.”
3 ys x2
■
3 Therefore, the inverse function is f 1x s x 2.
■
The principle of interchanging x and y to find the inverse function also gives us the method for obtaining the graph of f 1 from the graph of f . Since f a b if and only if f 1b a, the point a, b is on the graph of f if and only if the point b, a is on the graph of f 1. But we get the point b, a from a, b by reflecting about the line y x. (See Figure 8.) y
y
(b, a)
f –! (a, b) 0
0 x
x
y=x
y=x
FIGURE 8
FIGURE 9
f
Therefore, as illustrated by Figure 9: The graph of f 1 is obtained by reflecting the graph of f about the line y x.
152
■
CHAPTER 3
INVERSE FUNCTIONS
EXAMPLE 5 Sketch the graphs of f x s1 x and its inverse function using the same coordinate axes. SOLUTION First we sketch the curve y s1 x (the top half of the parabola
y 2 1 x, or x y 2 1) and then we reflect about the line y x to get the graph of f 1. (See Figure 10.) As a check on our graph, notice that the expression for f 1 is f 1x x 2 1, x 0. So the graph of f 1 is the right half of the parabola y x 2 1 and this seems reasonable from Figure 10. y
y=ƒ y=x 0 (_1, 0)
x
(0, _1)
y=f –!(x) FIGURE 10
■ THE CALCULUS OF INVERSE FUNCTIONS
Now let’s look at inverse functions from the point of view of calculus. Suppose that f is both one-to-one and continuous. We think of a continuous function as one whose graph has no break in it. (It consists of just one piece.) Since the graph of f 1 is obtained from the graph of f by reflecting about the line y x, the graph of f 1 has no break in it either (see Figure 9). Thus we might expect that f 1 is also a continuous function. This geometrical argument does not prove the following theorem but at least it makes the theorem plausible. A proof can be found in Appendix B. 6 THEOREM If f is a one-to-one continuous function defined on an interval, then its inverse function f 1 is also continuous.
y
¨ (a, b)
y=x ˙ ¨
graph of f –!
(b, a) 0
x
graph of f
Now suppose that f is a one-to-one differentiable function. Geometrically we can think of a differentiable function as one whose graph has no corner or kink in it. We get the graph of f 1 by reflecting the graph of f about the line y x, so the graph of f 1 has no corner or kink in it either. We therefore expect that f 1 is also differentiable (except where its tangents are vertical). In fact, we can predict the value of the derivative of f 1 at a given point by a geometric argument. In Figure 11 the graphs of f and its inverse f 1 are shown. If f b a, then f 1a b and f 1a is the slope of the tangent to the graph of f 1 at a, b, which is tan . Likewise, f b tan . From Figure 11 we see that 2, so
f 1a tan tan
FIGURE 11
that is,
1 1 cot 2 tan f b
f 1a
1 f f 1a
SECTION 3.2
INVERSE FUNCTIONS AND LOGARITHMS
■
153
7 THEOREM If f is a one-to-one differentiable function with inverse function f 1 and f f 1a 0, then the inverse function is differentiable at a and 1 f 1a f f 1a
PROOF Write the definition of derivative as in Equation 2.1.5:
f 1a lim
xla
f 1x f 1a xa
If f b a , then f 1a b . And if we let y f 1x, then f y x . Since f is differentiable, it is continuous, so f 1 is continuous by Theorem 6. Thus if x l a, then f 1x l f 1a, that is, y l b. Therefore ■ Note that x a ? f y f b because f is one-to-one.
f 1a lim
xla
lim y lb
f 1x f 1a yb lim y lb f y f b xa 1 1 f y f b f y f b lim y lb yb yb
1 1 f b f f 1a
■
NOTE 1 Replacing a by the general number x in the formula of Theorem 7, we get 8
f 1x
1 f f 1x
If we write y f 1x, then f y x, so Equation 8, when expressed in Leibniz notation, becomes dy 1 dx dx dy NOTE 2 If it is known in advance that f 1 is differentiable, then its derivative can
be computed more easily than in the proof of Theorem 7 by using implicit differentiation. If y f 1x, then f y x. Differentiating the equation f y x implicitly with respect to x, remembering that y is a function of x, and using the Chain Rule, we get f y Therefore
dy 1 dx dy 1 1 dx f y dx dy
154
■
CHAPTER 3
INVERSE FUNCTIONS
V EXAMPLE 6
If f x 2x cos x, find f 1 1.
SOLUTION Notice that f is differentiable and one-to-one. (Its graph is shown in Figure 12.) To use Theorem 7 we need to know f 11 and we can find it by inspection:
f 11 0
f 0 1 ? Therefore
f 1 1
1 1 1 1 f f 11 f 0 2 sin 0 2 y
10
0
5
x
■
FIGURE 12
LOGARITHMIC FUNCTIONS
If a 0 and a 1, the exponential function f x a x is either increasing or decreasing and so it is one-to-one by the Horizontal Line Test. It therefore has an inverse function f 1, which is called the logarithmic function with base a and is denoted by log a . If we use the formulation of an inverse function given by (3),
y
y=x
f 1x y &?
f y x
then we have y=a®, a>1
9 0
x
log a x y &?
ay x
Thus if x 0, then log a x is the exponent to which the base a must be raised to give x. For example, log10 0.001 3 because 103 0.001. The cancellation equations (4), when applied to the functions f x a x and 1 f x log a x, become
y=log a x, a>1
FIGURE 13 y
10
y=log™ x y=log£ x
log aa x x
for every x ⺢
a log a x x
for every x 0
1
0
1
x
y=log∞ x y=log¡¸ x
FIGURE 14
The logarithmic function log a has domain 0, and range ⺢ and is continuous since it is the inverse of a continuous function, namely, the exponential function. Its graph is the reflection of the graph of y a x about the line y x. Figure 13 shows the case where a 1. (The most important logarithmic functions have base a 1.) The fact that y a x is a very rapidly increasing function for x 0 is reflected in the fact that y log a x is a very slowly increasing function for x 1. Figure 14 shows the graphs of y log a x with various values of the base a 1. Since log a 1 0, the graphs of all logarithmic functions pass through the point 1, 0.
SECTION 3.2
INVERSE FUNCTIONS AND LOGARITHMS
■
155
The following properties of logarithmic functions follow from the corresponding properties of exponential functions given in Section 3.1. LAWS OF LOGARITHMS If x and y are positive numbers, then 1.
log axy log a x log a y
2.
log a
3.
log ax r r log a x
x y
log a x log a y (where r is any real number)
EXAMPLE 7 Use the laws of logarithms to evaluate log 2 80 log 2 5. SOLUTION Using Law 2, we have
log 2 80 log 2 5 log 2
80 5
log 2 16 4
because 2 4 16.
■
The limits of exponential functions given in Section 3.1 are reflected in the following limits of logarithmic functions. (Compare with Figure 13.) 11
If a 1, then lim log a x
xl
lim log a x
and
x l 0
In particular, the y-axis is a vertical asymptote of the curve y log a x. EXAMPLE 8 Find lim log10 tan2x. xl0
SOLUTION As x l 0, we know that t tan2x l tan2 0 0 and the values of t are
positive. So by (11) with a 10 1, we have
lim log10 tan2x lim log10 t
xl0
tl0
■
NATURAL LOGARITHMS NOTATION FOR LOGARITHMS Most textbooks in calculus and the sciences, as well as calculators, use the notation ln x for the natural logarithm and log x for the “common logarithm,” log 10 x . In the more advanced mathematical and scientific literature and in computer languages, however, the notation log x usually denotes the natural logarithm. ■
Of all possible bases a for logarithms, we will see in the next section that the most convenient choice of a base is the number e, which was defined in Section 3.1. The logarithm with base e is called the natural logarithm and has a special notation: log e x ln x If we put a e and replace log e with “ln” in (9) and (10), then the defining properties of the natural logarithm function become 12
ln x y &? e y x
156
■
CHAPTER 3
INVERSE FUNCTIONS
13
lne x x
x⺢
e ln x x
x0
In particular, if we set x 1, we get ln e 1 EXAMPLE 9 Find x if ln x 5. SOLUTION 1 From (12) we see that
ln x 5
means
e5 x
Therefore, x e 5. (If you have trouble working with the “ln” notation, just replace it by log e . Then the equation becomes log e x 5; so, by the definition of logarithm, e 5 x.) SOLUTION 2 Start with the equation
ln x 5 and apply the exponential function to both sides of the equation: e ln x e 5 But the second cancellation equation in (13) says that e ln x x. Therefore, x e 5. V EXAMPLE 10
■
Solve the equation e 53x 10.
SOLUTION We take natural logarithms of both sides of the equation and use (13):
lne 53x ln 10 5 3x ln 10 3x 5 ln 10 x 13 5 ln 10 Since the natural logarithm is found on scientific calculators, we can approximate the solution to four decimal places: x 0.8991. ■ V EXAMPLE 11
Express ln a 12 ln b as a single logarithm.
SOLUTION Using Laws 3 and 1 of logarithms, we have
ln a 12 ln b ln a ln b 12 ln a ln sb ln(asb )
■
The following formula shows that logarithms with any base can be expressed in terms of the natural logarithm.
SECTION 3.2
14 CHANGE OF BASE FORMULA
INVERSE FUNCTIONS AND LOGARITHMS
■
157
For any positive number a a 1, we have log a x
ln x ln a
PROOF Let y log a x . Then, from (9), we have a y x. Taking natural logarithms
of both sides of this equation, we get y ln a ln x. Therefore y
ln x ln a
■
Scientific calculators have a key for natural logarithms, so Formula 14 enables us to use a calculator to compute a logarithm with any base (as shown in the following example). Similarly, Formula 14 allows us to graph any logarithmic function on a graphing calculator or computer (see Exercises 55 and 56). EXAMPLE 12 Evaluate log 8 5 correct to six decimal places. SOLUTION Formula 14 gives
log 8 5
y
y=´ y=x
1
y=ln x
0 1
x
ln 5 0.773976 ln 8
■
The graphs of the exponential function y e x and its inverse function, the natural logarithm function, are shown in Figure 15. Because the curve y e x crosses the y-axis with a slope of 1, it follows that the reflected curve y ln x crosses the x-axis with a slope of 1. In common with all other logarithmic functions with base greater than 1, the natural logarithm is a continuous, increasing function defined on 0, and the y-axis is a vertical asymptote. If we put a e in (11), then we have the following limits:
15
FIGURE 15
V EXAMPLE 13
lim ln x
xl
lim ln x
x l0
Sketch the graph of the function y lnx 2 1.
SOLUTION We start with the graph of y ln x as given in Figure 15. Using the
transformations of Section 1.2, we shift it 2 units to the right to get the graph of y lnx 2 and then we shift it 1 unit downward to get the graph of y lnx 2 1. (See Figure 16 on page 158.) Notice that the line x 2 is a vertical asymptote since lim lnx 2 1
x l2
158
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CHAPTER 3
INVERSE FUNCTIONS
y
y
y
x=2
y=ln x 0
x=2 y=ln(x-2)-1
y=ln(x-2) 0
x
(1, 0)
2
x
(3, 0)
2
0
x (3, _1)
■
FIGURE 16
3.2
EXERCISES 11. tx 1x
1. (a) What is a one-to-one function?
(b) How can you tell from the graph of a function whether it is one-to-one? 2. (a) Suppose f is a one-to-one function with domain A and
range B. How is the inverse function f 1 defined? What is the domain of f 1? What is the range of f 1? (b) If you are given a formula for f , how do you find a formula for f 1? (c) If you are given the graph of f , how do you find the graph of f 1?
■ A function is given by a table of values, a graph, a formula, or a verbal description. Determine whether it is one-to-one.
3–14
3.
4.
5.
x
1
2
3
4
5
6
f x
1.5
2.0
3.6
5.3
2.8
2.0
x
1
2
3
4
5
6
f x
1
2
4
8
16
32
6.
y
12. tx cos x
13. f t is the height of a football t seconds after kickoff. 14. f t is your height at age t. ■
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15. If f is a one-to-one function such that f 2 9, what
is f 19?
16. If f x x cos x, find f 11. 17. If tx 3 x e x, find t14. 18. The graph of f is given.
(a) (b) (c) (d)
Why is f one-to-one? What are the domain and range of f 1? What is the value of f 12? Estimate the value of f 10. y
1 0
1
x
y
19. The formula C 9 F 32, where F 459.67, 5
x
x
7.
8.
y
y
9. f x x 2 2x
20. In the theory of relativity, the mass of a particle with speed v is x
x
10. f x 10 3x
expresses the Celsius temperature C as a function of the Fahrenheit temperature F. Find a formula for the inverse function and interpret it. What is the domain of the inverse function?
m f v
m0 s1 v 2c 2
where m 0 is the rest mass of the particle and c is the speed of light in a vacuum. Find the inverse function of f and explain its meaning.
■
SECTION 3.2
21–26
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23. f x e x
24. y 2 x 3 3
25. y lnx 3 ■
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26. y
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1 ex 1 ex ■
42. (a) What is the natural logarithm? ■
■
(b) What is the common logarithm? (c) Sketch the graphs of the natural logarithm function and the natural exponential function with a common set of axes.
■
Find an explicit formula for f 1 and use it to graph f , f , and the line y x on the same screen. To check your work, see whether the graphs of f and f 1 are reflections about the line.
; 27–28
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1
27. f x x 4 1, ■
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29–30
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29.
x0
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28. f x 2 e x ■
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Use the given graph of f to sketch the graph of f 30.
y
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1
.
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31–34
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47–50
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33. f x 9 x , ■
35–38
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0 x 3,
a8
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Use the properties of logarithms to expand the
x 3y z2
48. ln sab 2 c 2 50. ln
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3x 2 x 15
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Express the given quantity as a single logarithm. 52. ln x a ln y b ln z 1 2
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decimal places. (a) log12 10
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(b) log 2 8.4
; 55–56
■ Use Formula 14 to graph the given functions on a common screen. How are these graphs related?
a1
55. y log 1.5 x ,
a2
37. f x 3 x 2 tan x2, 1 x 1, 38. f x sx 3 x 2 x 1,
a2
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54. Use Formula 14 to evaluate each logarithm correct to six
x 1, a 2
3
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51. 2 ln 4 ln 2
a2
36. f x x x 2x,
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53. ln1 x ln x ln sin x
35. f x x 3 x 1,
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51–53
Find f 1 a. 5
■
2
34. f x 1x 1, ■
■
47. log 2
■
a8 2
■
49. lnuv10
Show that f is one-to-one. Use Theorem 7 to find f 1a. Calculate f 1x and state the domain and range of f 1. Calculate f 1a from the formula in part (c) and check that it agrees with the result of part (b). (e) Sketch the graphs of f and f 1 on the same axes.
32. f x sx 2 ,
■
(b) e 3 ln 2
quantity.
(a) (b) (c) (d)
31. f x x ,
(b) ln e s2
■
x
2
■
3
44. (a) log 8 2
46. (a) 2log 2 3 log 2 5
x
■
(b) log 6 36
(b) log 5 10 log 5 20 3 log 5 2
0 1
1
43. (a) log 2 64
45. (a) log 10 1.25 log 10 80
1
0
■ Find the exact value of each expression (without a calculator).
43– 46
y
1
159
(b) What is the domain of this function? (c) What is the range of this function? (d) Sketch the general shape of the graph of the function y log a x if a 1.
4x 1 22. f x 2x 3
3
■
41. (a) How is the logarithmic function y log a x defined?
Find a formula for the inverse of the function.
21. f x s10 3x
INVERSE FUNCTIONS AND LOGARITHMS
■
■
56. y ln x,
a3
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■
39. Suppose f 1 is the inverse function of a differentiable func-
tion f and f 4 5, f 4 23. Find f 15. 40. Suppose f 1 is the inverse function of a differentiable func-
tion f and let Gx 1f find G2.
1
x. If f 3 2 and f 3 , 1 9
■
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y ln x, y log 10 x ,
y log 10 x , ■
■
ye , x
■
■
y log 50 x
y 10 x ■
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57. Suppose that the graph of y log 2 x is drawn on a coordi-
nate grid where the unit of measurement is an inch. How many miles to the right of the origin do we have to move before the height of the curve reaches 3 ft? 0.1 ; 58. Compare the functions f x x and tx ln x by
graphing both f and t in several viewing rectangles. When does the graph of f finally surpass the graph of t ?
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160
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CHAPTER 3
INVERSE FUNCTIONS
73. lim ln1 x 2 ln1 x
■ Make a rough sketch of the graph of each function. Do not use a calculator. Just use the graphs given in Figures 14 and 15 and, if necessary, the transformations of Section 1.2.
59–60
59. (a) y log 10x 5 ■
61–64
■
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(b) y ln x
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CAS
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62. (a) e 2x3 7 0
(b) ln5 2 x 3
x5
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(b) ln x 1
66. (a) 2 ln x 9
(b) e 23x 4
67–68
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75. Graph the function f x sx 3 x 2 x 1 and explain
why it is one-to-one. Then use a computer algebra system to find an explicit expression for f 1x. (Your CAS will produce three possible expressions. Explain why two of them are irrelevant in this context.) begin to recharge the flash’s capacitor, which stores electric charge given by Qt Q 0 1 e ta (The maximum charge capacity is Q 0 and t is measured in seconds.) (a) Find the inverse of this function and explain its meaning. (b) How long does it take to recharge the capacitor to 90% of capacity if a 2 ? 77. Let a 1. Prove, using precise definitions, that
(a) lim a x 0
68. f x ln2 ln x
2x
■
■
Find (a) the domain of f and (b) f 1 and its domain.
67. f x s3 e ■
■
Solve each inequality for x.
65. (a) e 10
■
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76. When a camera flash goes off, the batteries immediately
(b) e ax Ce bx, where a b
x
■
■
(b) ln x lnx 1 1
3
64. (a) lnln x 1
65–66
■
Solve each equation for x. (b) ex 5
■
xl
■
61. (a) 2 ln x 1
63. (a) 2
74. lim ln2 x ln1 x
(b) y ln x
60. (a) y lnx ■
xl
x l
■
(b) lim a x xl
■
78. (a) If we shift a curve to the left, what happens to its reflec69–74
■
Find the limit.
69. lim ln2 x
70. lim log10x 5x 6
71. lim lncos x
72. lim lnsin x
xl2
xl0
2
xl3
xl0
3.3
tion about the line y x ? In view of this geometric principle, find an expression for the inverse of tx f x c, where f is a one-to-one function. (b) Find an expression for the inverse of hx f cx, where c 0.
DERIVATIVES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS In this section we find formulas for the derivatives of logarithmic functions and then use them to calculate the derivatives of exponential functions. DERIVATIVES OF LOGARITHMIC FUNCTIONS
In using the definition of a derivative to differentiate the function f x log a x , we use the fact that it is continuous, together with some of the laws of logarithms. We also need to recall the definition of e from Section 3.1: e lim 1 x1x xl0
1 THEOREM
The function f x log a x is differentiable and f x
1 log a e x
SECTION 3.3
DERIVATIVES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS
■
161
PROOF
f x lim
hl0
f x h f x log a x h log a x lim h l 0 h h
xh x h
log a lim
hl0
lim
hl0
lim
hl0
1 h log a 1 h x
1 x h log a 1 x h x
1 x h lim log a 1 x hl0 h x
(by Limit Law 3)
xh
1 h lim log a 1 x hl0 x
1 h log a lim 1 h l 0 x x
1 h log a lim 1 hl0 x x
(by Law 3 of Logarithms)
xh
(since log a is continuous)
1hx
1 log a e x
The final step may be seen more clearly by making the change of variable t hx . As h l 0, we also have t l 0, so
lim 1
hl0
h x
1hx
lim 1 t1t e tl0
by the definition of e . Thus f x
1 log a e x
■
NOTE We know from the Change of Base Formula (3.2.14) that
log a e
ln e 1 ln a ln a
and so the formula in Theorem 1 can be rewritten as follows: d 1 log a x dx x ln a
2
EXAMPLE 1 Differentiate f x log 102 sin x. SOLUTION Using Formula 2 with a 10, together with the Chain Rule, we have
f x
d 1 d log 102 sin x 2 sin x dx 2 sin x ln 10 dx cos x 2 sin x ln 10
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162
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CHAPTER 3
INVERSE FUNCTIONS
If we put a e in Formula 2, then the factor ln a on the right side becomes ln e 1 and we get the formula for the derivative of the natural logarithmic function log e x ln x : 3 DERIVATIVE OF THE NATURAL LOGARITHMIC FUNCTION
d 1 ln x dx x By comparing Formulas 2 and 3, we see one of the main reasons that natural logarithms (logarithms with base e) are used in calculus: The differentiation formula is simplest when a e because ln e 1. V EXAMPLE 2
Differentiate y lnx 3 1.
SOLUTION To use the Chain Rule, we let u x 3 1. Then y ln u, so
dy dy du 1 du 1 3x 2 3 3x 2 3 dx du dx u dx x 1 x 1
■
In general, if we combine Formula 3 with the Chain Rule as in Example 2, we get d 1 du ln u dx u dx
4
V EXAMPLE 3
Find
or
d tx ln tx dx tx
d lnsin x. dx
SOLUTION Using (4), we have
d 1 d 1 lnsin x sin x cos x cot x dx sin x dx sin x
■
EXAMPLE 4 Differentiate f x sln x . SOLUTION This time the logarithm is the inner function, so the Chain Rule gives Figure 1 shows the graph of the function f of Example 5 together with the graph of its derivative. It gives a visual check on our calculation. Notice that f x is large negative when f is rapidly decreasing. ■
f x 12 ln x12 EXAMPLE 5 Find
y
d 1 1 1 ln x dx 2sln x x 2x sln x
d x1 ln . dx sx 2
SOLUTION 1 f
d x1 ln dx sx 2
1 0
x
fª
FIGURE 1
1 d x1 x 1 dx sx 2 sx 2
1 sx 2 sx 2 1 x 1( 2 )x 212 x1 x2
x 2 12 x 1 x5 x 1x 2 2x 1x 2
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SECTION 3.3
DERIVATIVES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS
■
163
SOLUTION 2 If we first simplify the given function using the laws of logarithms, then the differentiation becomes easier:
d x1 d ln [lnx 1 12 lnx 2] dx dx sx 2
1 1 x1 2
1 x2
(This answer can be left as written, but if we used a common denominator we would see that it gives the same answer as in Solution 1.) ■ ■ Figure 2 shows the graph of the function f x ln x in Example 6 and its derivative f x 1x . Notice that when x is small, the graph of y ln x is steep and so f x is large (positive or negative).
3
Find f x if f x ln x .
V EXAMPLE 6
SOLUTION Since
f x
ln x if x 0 lnx if x 0
it follows that fª
f _3
f x
3
1 x 1 1 1 x x
if x 0 if x 0
Thus f x 1x for all x 0.
■
_3
FIGURE 2
The result of Example 6 is worth remembering: d 1 ln x dx x
5
LOGARITHMIC DIFFERENTIATION
The calculation of derivatives of complicated functions involving products, quotients, or powers can often be simplified by taking logarithms. The method used in the following example is called logarithmic differentiation. V EXAMPLE 7
Differentiate y
x 34 sx 2 1 . 3x 25
SOLUTION We take logarithms of both sides of the equation and use the Laws of
Logarithms to simplify: ln y 34 ln x 12 lnx 2 1 5 ln3x 2 Differentiating implicitly with respect to x gives 1 dy 3 1 1 2x 3 2 5 y dx 4 x 2 x 1 3x 2
164
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CHAPTER 3
INVERSE FUNCTIONS
Solving for dydx, we get
3 x 15 dy y 2 dx 4x x 1 3x 2 ■ If we hadn’t used logarithmic differentiation in Example 7, we would have had to use both the Quotient Rule and the Product Rule. The resulting calculation would have been horrendous.
Because we have an explicit expression for y, we can substitute and write dy x 34 sx 2 1 dx 3x 25
3 x 15 2 4x x 1 3x 2
■
STEPS IN LOGARITHMIC DIFFERENTIATION 1. Take natural logarithms of both sides of an equation y f x and use the
Laws of Logarithms to simplify. 2. Differentiate implicitly with respect to x. 3. Solve the resulting equation for y.
If f x 0 for some values of x, then ln f x is not defined, but we can write y f x and use Equation 5. We illustrate this procedure by proving the general version of the Power Rule, as promised in Section 2.3.
THE POWER RULE If n is any real number and f x x n, then
f x nx n1 PROOF Let y x n and use logarithmic differentiation: ■ If x 0 , we can show that f 0 0 for n 1 directly from the definition of a derivative.
ln y ln x
n ln x
x0
y n y x
Therefore
Hence
n
y n
y xn n nx n1 x x
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DERIVATIVES OF EXPONENTIAL FUNCTIONS
To compute the derivative of the exponential function y a x we use the fact that exponential and logarithmic functions are inverse functions. 6 THEOREM
The exponential function f x a x, a 0, is differentiable
and d a x a x ln a dx PROOF We know that the logarithmic function y log a x is differentiable (and its derivative is nonzero) by Theorem 1. So its inverse function y a x is differentiable by Theorem 3.2.7.
SECTION 3.3
DERIVATIVES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS
■
165
If y a x , then log a y x . Differentiating this equation implicitly with respect to x , we get
Another method for proving Theorem 6 is to use logarithmic differentiation. ■
1 dy 1 y ln a dx dy y ln a a x ln a dx
Thus
■
EXAMPLE 8 Combining Formula 6 with the Chain Rule, we have
d d ( 10 x ) 10 x ln 10 x 2 2 ln 10x10 x dx dx 2
2
2
■
If we put a e in Theorem 6, the differentiation formula for exponential functions takes on a particularly simple form: 7 DERIVATIVE OF THE NATURAL EXPONENTIAL FUNCTION
Visual 3.3 uses the slope-a-scope to illustrate this formula.
d e x e x dx This equation says that the exponential function f x e x is its own derivative. Comparing Equations 6 and 7, we see that the simplest differentiation formula for an exponential function occurs when a e . This is the reason that the natural exponential function is most often used in calculus. The geometric significance of Equation 7 is that the slope of a tangent to the curve y e x at any point is equal to the y-coordinate of the point. In particular, if f x e x , then f 0 e 0 1. This means that of all the possible exponential functions y a x , y e x is the one that crosses the y -axis with a slope of 1. (See Figure 3.)
y {x, e ® } slope=e®
1
slope=1
y=e® 0
FIGURE 3
x
EXAMPLE 9 Differentiate the function y e tan x. SOLUTION To use the Chain Rule, we let u tan x. Then we have y e u, so
dy dy du du eu e tan x sec2x dx du dx dx
■
In general if we combine Formula 7 with the Chain Rule, as in Example 9, we get
8
d du e u e u dx dx
EXAMPLE 10 Find y if y e4x sin 5x. SOLUTION Using Formula 8 and the Product Rule, we have
y e4xcos 5x5 sin 5xe4x4 e4x5 cos 5x 4 sin 5x
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166
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CHAPTER 3
INVERSE FUNCTIONS
To differentiate a function of the form y f x tx, where both the base and the exponent are functions, logarithmic differentiation can be used as in the following example. V EXAMPLE 11
Differentiate y x sx .
SOLUTION 1 Using logarithmic differentiation, we have
ln y ln x sx sx ln x ■ Figure 4 illustrates Example 11 by showing the graphs of f x x sx and its derivative.
y 1 1 sx ln x y x 2sx
y
y y
f fª
2 ln x 2sx
d sx d sx ln x d ( x ) dx (e ) e sx ln x dx (sx ln x) dx
x
1
x sx
FIGURE 4
3.3
x sx
SOLUTION 2 Another method is to write x sx e ln x sx :
1 0
1 ln x 2sx sx
2 ln x 2sx
(as in Solution 1)
■
EXERCISES
1. f x log 21 3x
2. f x lnx 2 10
3. f lncos
4. f x cosln x
29. y 2 sin x
5 5. f x s ln x
5 6. f x ln s x
31. f u e 1u
1 xe x x ex u e e u 30. y u e e u 32. y e k tan sx
7. f x sin x ln5x
8. f x log 5 xe x
33. y lnex xex
34. y ln1 e x 2
35. Ft e t sin 2t
36. y 23
1–36
■
Differentiate the function.
9. tx ln
ax ax
10. f t
1 ln t 1 ln t
2t 1 3 11. Ft ln 3t 1 4
12. f x log10
ln u 13. f u 1 ln2u
14. y lnx sin x
15. y ln 2 x 5x 17. f x x 2e x 19. y
2
16. Gu ln
20. y
21. y xex
2
25. ht t 3 3
3u 2 3u 2
ex 1x
22. y e5x cos 3x 24. y 10
x cos x
t
■
ae x b ce x d
■
37– 40
■
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28. f x
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■
37. y e x sin x
38. y
39. y x ln x ■
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■
■
ln x x2
40. y lnsec x tan x ■
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41– 42 ■ Find an equation of the tangent line to the curve at the given point. 41. y ln ln x, ■
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43– 44
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43. f x ■
■
42. y e xx,
e, 0 ■
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1, e
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Differentiate f and find the domain of f .
1x 2
1 26. y s ke s
x2
Find y and y .
2
18. tx sx e x
ex x2
23. y e
4
x x1
27. y
x 1 lnx 1 ■
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44. f x ln ln ln x ■
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SECTION 3.4
45–54 ■ Use logarithmic differentiation or an alternative method to find the derivative of the function. x2
45. y 2x 1 x 3 5
47. y
4
sin2x tan4x x 2 12
48. y
4
51. y cos x x
52. y sx
53. y tan x 1x
54. y sin x ln x
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10
pt
x
■
; ■
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2
55. Find y if e x y x y.
xe ye 1 at the point 0, 1. x
differential equation y 2y y 0. equation y 5y 6y 0?
2
58. Find y if x y y x.
63. If f x e 2x, find a formula for f nx.
; 59. The motion of a spring that is subject to a frictional force or
a damping force (such as a shock absorber in a car) is often modeled by the product of an exponential function and a sine or cosine function. Suppose the equation of motion of a point on such a spring is st 2e1.5t sin 2 t where s is measured in centimeters and t in seconds. Find the velocity after t seconds and graph both the position and velocity functions for 0 t 2.
3.4
where pt is the proportion of the population that knows the rumor at time t and a and k are positive constants. (a) Find lim t l pt. (b) Find the rate of spread of the rumor. (c) Graph p for the case a 10, k 0.5 with t measured in hours. Use the graph to estimate how long it will take for 80% of the population to hear the rumor.
62. For what values of r does the function y e rx satisfy the
57. Find y if y lnx y . 2
1 1 ae k t
61. Show that the function y Aex Bxex satisfies the
56. Find an equation of the tangent line to the curve y
167
the equation
2
x2 1 x2 1
50. y x cos x
■
60. Under certain circumstances a rumor spreads according to
46. y sx e x 1
6
49. y x x
■
EXPONENTIAL GROWTH AND DECAY
64. Find the thousandth derivative of f x xex. 65. Find a formula for f nx if f x lnx 1. 66. Find
d9 x 8 ln x. dx 9
67. If f x 3 x e x, find f 14. 68. Evaluate lim
xl
e sin x 1 . x
EXPONENTIAL GROWTH AND DECAY In many natural phenomena, quantities grow or decay at a rate proportional to their size. For instance, if y f t is the number of individuals in a population of animals or bacteria at time t, then it seems reasonable to expect that the rate of growth f t is proportional to the population f t; that is, f t kf t for some constant k . Indeed, under ideal conditions (unlimited environment, adequate nutrition, immunity to disease) the mathematical model given by the equation f t kf t predicts what actually happens fairly accurately. Another example occurs in nuclear physics where the mass of a radioactive substance decays at a rate proportional to the mass. In chemistry, the rate of a unimolecular first-order reaction is proportional to the concentration of the substance. In finance, the value of a savings account with continuously compounded interest increases at a rate proportional to that value. In general, if yt is the value of a quantity y at time t and if the rate of change of y with respect to t is proportional to its size yt at any time, then
1
dy ky dt
where k is a constant. Equation 1 is sometimes called the law of natural growth
168
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CHAPTER 3
INVERSE FUNCTIONS
(if k 0) or the law of natural decay (if k 0). It is called a differential equation because it involves an unknown function y and its derivative dydt . It’s not hard to think of a solution of Equation 1. This equation asks us to find a function whose derivative is a constant multiple of itself. We have met such functions in this chapter. Any exponential function of the form yt Ce kt , where C is a constant, satisfies yt Cke kt kCe kt kyt We will see in Section 7.6 that any function that satsifies dydt ky must be of the form y Ce kt . To see the significance of the constant C , we observe that y0 Ce k0 C Therefore C is the initial value of the function. 2 THEOREM The only solutions of the differential equation dydt ky are the exponential functions
yt y0e kt
POPULATION GROWTH
What is the significance of the proportionality constant k? In the context of population growth, where Pt is the size of a population at time t , we can write 3
dP kP dt
or
1 dP k P dt
The quantity 1 dP P dt is the growth rate divided by the population size; it is called the relative growth rate. According to (3), instead of saying “the growth rate is proportional to population size” we could say “the relative growth rate is constant.” Then (2) says that a population with constant relative growth rate must grow exponentially. Notice that the relative growth rate k appears as the coefficient of t in the exponential function Ce kt . For instance, if dP 0.02P dt and t is measured in years, then the relative growth rate is k 0.02 and the population grows at a relative rate of 2% per year. If the population at time 0 is P0 , then the expression for the population is Pt P0 e 0.02t V EXAMPLE 1 Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 to model the population of the world in the second half of the 20th century. (Assume that the growth rate is proportional to the population size.)
SECTION 3.4
EXPONENTIAL GROWTH AND DECAY
■
169
What is the relative growth rate? Use the model to estimate the world population in 1993 and to predict the population in the year 2020. SOLUTION We measure the time t in years and let t 0 in the year 1950. We mea-
sure the population Pt in millions of people. Then P0 2560 and P10) 3040. Since we are assuming that dPdt kP , Theorem 2 gives Pt P0e kt 2560e kt P10 2560e 10k 3040 k
1 3040 ln 0.017185 10 2560
The relative growth rate is about 1.7% per year and the model is Pt 2560e 0.017185t We estimate that the world population in 1993 was P43 2560e 0.01718543 5360 million The model predicts that the population in 2020 will be P70 2560e 0.01718570 8524 million The graph in Figure 1 shows that the model is fairly accurate to date (the dots represent the actual population), so the estimate for 1993 is quite reliable. But the prediction for 2020 is riskier. P 6000
P=2560e 0.017185t
Population (in millions)
FIGURE 1 20
A model for world population growth in the second half of the 20th century
Years since 1950
40
t
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RADIOACTIVE DECAY
Radioactive substances decay by spontaneously emitting radiation. If mt is the mass remaining from an initial mass m0 of the substance after time t, then the relative decay rate 1 dm m dt has been found experimentally to be constant. (Since dmdt is negative, the relative decay rate is positive.) It follows that dm km dt
170
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CHAPTER 3
INVERSE FUNCTIONS
where k is a negative constant. In other words, radioactive substances decay at a rate proportional to the remaining mass. This means that we can use (2) to show that the mass decays exponentially: mt m0 e kt Physicists express the rate of decay in terms of half-life, the time required for half of any given quantity to decay. The half-life of radium-226 ( .226 88 Ra) is 1590 years. (a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of .226 88 Ra that remains after t years. (b) Find the mass after 1000 years correct to the nearest milligram. (c) When will the mass be reduced to 30 mg? V EXAMPLE 2
SOLUTION
(a) Let mt be the mass of radium-226 (in milligrams) that remains after t years. Then dmdt km and y0 100, so (2) gives mt m0e kt 100e kt In order to determine the value of k, we use the fact that y1590 12 100. Thus 100e 1590k 50
so
e 1590k 12
1590k ln 12 ln 2
and
k
ln 2 1590
mt 100eln 2t1590
Therefore
We could use the fact that e ln 2 2 to write the expression for mt in the alternative form mt 100 2 t1590 (b) The mass after 1000 years is m1000 100eln 210001590 65 mg (c) We want to find the value of t such that mt 30, that is, 100eln 2t1590 30
or
eln 2t1590 0.3
We solve this equation for t by taking the natural logarithm of both sides: 150
ln 2 t ln 0.3 1590
m=100e_(ln 2)t/1590
Thus m=30 0
FIGURE 2
4000
t 1590
ln 0.3 2762 years ln 2
■
As a check on our work in Example 2, we use a graphing device to draw the graph of mt in Figure 2 together with the horizontal line m 30. These curves intersect when t 2800, and this agrees with the answer to part (c).
SECTION 3.4
EXPONENTIAL GROWTH AND DECAY
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171
NEWTON’S LAW OF COOLING
Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. (This law also applies to warming.) If we let Tt be the temperature of the object at time t and Ts be the temperature of the surroundings, then we can formulate Newton’s Law of Cooling as a differential equation: dT kT Ts dt where k is a constant. This equation is not quite the same as Equation 1, so we make the change of variable yt Tt Ts . Because Ts is constant, we have yt Tt and so the equation becomes dy ky dt We can then use (2) to find an expression for y, from which we can find T . EXAMPLE 3 A bottle of soda pop at room temperature (72 F) is placed in a refrig-
erator where the temperature is 44 F. After half an hour the soda pop has cooled to 61 F. (a) What is the temperature of the soda pop after another half hour? (b) How long does it take for the soda pop to cool to 50 F? SOLUTION
(a) Let Tt be the temperature of the soda after t minutes. The surrounding temperature is Ts 44 F, so Newton’s Law of Cooling states that dT kT 44) dt If we let y T 44, then y0 T0 44 72 44 28, so y satisfies dy ky dt
y0 28
and by (2) we have yt y0e kt 28e kt We are given that T30 61, so y30 61 44 17 and 28e 30k 17
17 e 30k 28
Taking logarithms, we have k
ln ( 17 28 ) 0.01663 30
172
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CHAPTER 3
INVERSE FUNCTIONS
Thus yt 28e 0.01663t Tt 44 28e 0.01663t T60 44 28e 0.0166360 54.3 So after another half hour the pop has cooled to about 54 F. (b) We have Tt 50 when 44 28e 0.01663t 50 e 0.01663t 286 T 72
t
44
ln ( 286 ) 92.6 0.01663
The pop cools to 50 F after about 1 hour 33 minutes.
■
Notice that in Example 3, we have 0
FIGURE 3
30
60
90
t
lim Tt lim 44 28e 0.01663t 44 28 0 44
tl
tl
which is to be expected. The graph of the temperature function is shown in Figure 3. CONTINUOUSLY COMPOUNDED INTEREST
EXAMPLE 4 If $1000 is invested at 6% interest, compounded annually, then after
1 year the investment is worth $10001.06 $1060, after 2 years it’s worth $ 10001.06 1.06 $1123.60, and after t years it’s worth $10001.06t. In general, if an amount A0 is invested at an interest rate r r 0.06 in this example), then after t years it’s worth A0 1 r t. Usually, however, interest is compounded more frequently, say, n times a year. Then in each compounding period the interest rate is rn and there are nt compounding periods in t years, so the value of the investment is
A0 1
r n
nt
For instance, after 3 years at 6% interest a $1000 investment will be worth
$10001.063 $1191.02
with annual compounding
$10001.036 $1194.05
with semiannual compounding
$10001.01512 $1195.62
with quarterly compounding
$10001.00536 $1196.68
with monthly compounding
$1000 1
0.06 365
365 3
$1197.20 with daily compounding
SECTION 3.4
EXPONENTIAL GROWTH AND DECAY
■
173
You can see that the interest paid increases as the number of compounding periods n increases. If we let n l , then we will be compounding the interest continuously and the value of the investment will be
At lim A0 1 nl
xl0
If we put n 1x , then n l as x l 0 and so an alternative expression for e is
e lim 1 nl
1 n
nl
1
r n
A0 lim
1
1 m
ml
Recall: e lim 1 x 1x
lim A0
A0 lim
nl
■
nt
r n
nr
rt
m
rt
1
r n
nr
rt
(where m nr)
But the limit in this expression is equal to the number e. So with continuous compounding of interest at interest rate r, the amount after t years is At A0 e rt
n
If we differentiate this equation, we get dA rA0 e rt rAt dt which says that, with continuous compounding of interest, the rate of increase of an investment is proportional to its size. Returning to the example of $1000 invested for 3 years at 6% interest, we see that with continuous compounding of interest the value of the investment will be A3 $1000e 0.063 $1197.22 Notice how close this is to the amount we calculated for daily compounding, $1197.20. But the amount is easier to compute if we use continuous compounding.
3.4
■
EXERCISES
1. A population of protozoa develops with a constant relative
growth rate of 0.7944 per member per day. On day zero the population consists of two members. Find the population size after six days. 2. A common inhabitant of human intestines is the bacterium
Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 60 cells. (a) Find the relative growth rate. (b) Find an expression for the number of cells after t hours. (c) Find the number of cells after 8 hours. (d) Find the rate of growth after 8 hours. (e) When will the population reach 20,000 cells?
3. A bacteria culture initially contains 100 cells and grows at a
rate proportional to its size. After an hour the population has increased to 420. (a) Find an expression for the number of bacteria after t hours. (b) Find the number of bacteria after 3 hours. (c) Find the rate of growth after 3 hours. (d) When will the population reach 10,000? 4. A bacteria culture grows with constant relative growth rate.
After 2 hours there are 600 bacteria and after 8 hours the count is 75,000. (a) Find the initial population. (b) Find an expression for the population after t hours.
174
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CHAPTER 3
INVERSE FUNCTIONS
(c) Find the number of cells after 5 hours. (d) Find the rate of growth after 5 hours. (e) When will the population reach 200,000?
(b) How long will the reaction take to reduce the concentration of N2O5 to 90% of its original value? 8. Bismuth-210 has a half-life of 5.0 days.
5. The table gives estimates of the world population, in
millions, from 1750 to 2000: Year
Population
Year
Population
1750 1800 1850
790 980 1260
1900 1950 2000
1650 2560 6080
(a) Use the exponential model and the population figures for 1750 and 1800 to predict the world population in 1900 and 1950. Compare with the actual figures. (b) Use the exponential model and the population figures for 1850 and 1900 to predict the world population in 1950. Compare with the actual population. (c) Use the exponential model and the population figures for 1900 and 1950 to predict the world population in 2000. Compare with the actual population and try to explain the discrepancy. 6. The table gives the population of the United States, from
census figures in millions, for the years 1900–2000.
;
Year
Population
Year
Population
1900 1910 1920 1930 1940 1950
76 92 106 123 131 150
1960 1970 1980 1990 2000
179 203 227 250 275
(a) Use the exponential model and the census figures for 1900 and 1910 to predict the population in 2000. Compare with the actual figure and try to explain the discrepancy. (b) Use the exponential model and the census figures for 1980 and 1990 to predict the population in 2000. Compare with the actual population. Then use this model to predict the population in the years 2010 and 2020. (c) Graph both of the exponential functions in parts (a) and (b) together with a plot of the actual population. Are these models reasonable ones? 7. Experiments show that if the chemical reaction 1 N2O5 l 2NO 2 2 O 2
takes place at 45 C, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows:
d N2O5 0.0005 N2O5 dt
(a) Find an expression for the concentration N2O5 after t seconds if the initial concentration is C.
(a) A sample originally has a mass of 800 mg. Find a formula for the mass remaining after t days. (b) Find the mass remaining after 30 days. (c) When is the mass reduced to 1 mg? (d) Sketch the graph of the mass function. 9. The half-life of cesium-137 is 30 years. Suppose we have a
100-mg sample. (a) Find the mass that remains after t years. (b) How much of the sample remains after 100 years? (c) After how long will only 1 mg remain? 10. A sample of tritium-3 decayed to 94.5% of its original
amount after a year. (a) What is the half-life of tritium-3? (b) How long would it take the sample to decay to 20% of its original amount? 11. Scientists can determine the age of ancient objects by a
method called radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14 C, with a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14 C through food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14 C begins to decrease through radioactive decay. Therefore, the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 74% as much 14 C radioactivity as does plant material on Earth today. Estimate the age of the parchment. 12. A curve passes through the point 0, 5 and has the property
that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of the curve? 13. A roast turkey is taken from an oven when its temperature
has reached 185 F and is placed on a table in a room where the temperature is 75 F. (a) If the temperature of the turkey is 150 F after half an hour, what is the temperature after 45 minutes? (b) When will the turkey have cooled to 100 F? 14. A thermometer is taken from a room where the temperature
is 20 C to the outdoors, where the temperature is 5 C. After one minute the thermometer reads 12 C. (a) What will the reading on the thermometer be after one more minute? (b) When will the thermometer read 6 C? 15. When a cold drink is taken from a refrigerator, its tempera-
ture is 5 C. After 25 minutes in a 20 C room its temperature has increased to 10 C. (a) What is the temperature of the drink after 50 minutes? (b) When will its temperature be 15 C?
SECTION 3.5
16. A freshly brewed cup of coffee has temperature 95 C in a
20 C room. When its temperature is 70 C, it is cooling at a rate of 1 C per minute. When does this occur?
;
■
175
(b) Suppose $500 is borrowed and the interest is compounded continuously. If At is the amount due after t years, where 0 t 2, graph At for each of the interest rates 14%, 10%, and 6% on a common screen.
17. The rate of change of atmospheric pressure P with respect
to altitude h is proportional to P, provided that the temperature is constant. At 15 C the pressure is 101.3 kPa at sea level and 87.14 kPa at h 1000 m. (a) What is the pressure at an altitude of 3000 m? (b) What is the pressure at the top of Mount McKinley, at an altitude of 6187 m?
INVERSE TRIGONOMETRIC FUNCTIONS
19. If $3000 is invested at 5% interest, find the value of
the investment at the end of 5 years if the interest is compounded (a) annually (b) semiannually (c) monthly (d) weekly (e) daily (f ) continuously
18. (a) If $500 is borrowed at 14% interest, find the amounts
due at the end of 2 years if the interest is compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) daily, (v) hourly, and (vi) continuously.
3.5
20. (a) How long will it take an investment to double in value if
the interest rate is 6% compounded continuously? (b) What is the equivalent annual interest rate?
INVERSE TRIGONOMETRIC FUNCTIONS In this section we apply the ideas of Section 3.2 to find the derivatives of the so-called inverse trigonometric functions. We have a slight difficulty in this task: Because the trigonometric functions are not one-to-one, they do not have inverse functions. The difficulty is overcome by restricting the domains of these functions so that they become one-to-one. You can see from Figure 1 that the sine function y sin x is not one-to-one (use the Horizontal Line Test). But the function f x sin x, 2 x 2 (see Figure 2), is one-to-one. The inverse function of this restricted sine function f exists and is denoted by sin 1 or arcsin. It is called the inverse sine function or the arcsine function. y
y
y=sin x _ π2 0
_π
π 2
0
x
π
π 2
π
x
π
FIGURE 2 y=sin x, _ 2 ¯x¯ 2
FIGURE 1
Since the definition of an inverse function says that f 1x y &?
f y x
we have
1
| sin 1x
1 sin x
sin1x y
&? sin y x
and
y 2 2
Thus if 1 x 1, sin 1x is the number between 2 and 2 whose sine is x.
176
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CHAPTER 3
INVERSE FUNCTIONS
EXAMPLE 1 Evaluate (a) sin1( 2) and (b) tan(arcsin 3 ). 1
1
SOLUTION
(a) We have sin1( 2) 1
3 1 ¨
6
because sin6 12 and 6 lies between 2 and 2. (b) Let arcsin 13 , so sin 13 . Then we can draw a right triangle with angle as in Figure 3 and deduce from the Pythagorean Theorem that the third side has length s9 1 2s2 . This enables us to read from the triangle that
2 œ„ 2
tan(arcsin 13 ) tan
FIGURE 3
1 2s2
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The cancellation equations for inverse functions become, in this case,
2
y π 2
_1
0
1
x
_ π2
FIGURE 4 y=sin–! x=arcsin x
x 2 2
sin1sin x x
for
sinsin1x x
for 1 x 1
The inverse sine function, sin1, has domain 1, 1 and range 2, 2 , and its graph, shown in Figure 4, is obtained from that of the restricted sine function (Figure 2) by reflection about the line y x. We know that the sine function f is continuous, so the inverse sine function is also continuous. We also know from Section 2.3 that the sine function is differentiable, so the inverse sine function is also differentiable. We could calculate the derivative of sin 1 by the formula in Theorem 3.2.7, but since we know that sin 1 is differentiable, we can just as easily calculate it by implicit differentiation as follows. Let y sin1x. Then sin y x and 2 y 2. Differentiating sin y x implicitly with respect to x, we obtain cos y
dy 1 dx dy 1 dx cos y
and
Now cos y 0 since 2 y 2, so cos y s1 sin 2 y s1 x 2 Therefore
3
dy 1 1 dx cos y s1 x 2 d 1 sin1x dx s1 x 2
1 x 1
SECTION 3.5
INVERSE TRIGONOMETRIC FUNCTIONS
■
177
If f x sin 1x 2 1, find (a) the domain of f , (b) f x, and (c) the domain of f .
4
V EXAMPLE 2
fª
SOLUTION _2
2 f
(a) Since the domain of the inverse sine function is 1, 1 , the domain of f is
x
1 x
2
0 x 2 {x x s2 } [s2 , s2 ]
1 1 x
_4
2
(b) Combining Formula 3 with the Chain Rule, we have
FIGURE 5 ■ The graphs of the function f of Example 2 and its derivative are shown in Figure 5. Notice that f is not differentiable at 0 and this is consistent with the fact that the graph of f makes a sudden jump at x 0 .
f x
1 d x 2 1 2 2 s1 x 1 dx 1 2x 2x 2 s1 x 2x 1 s2x 2 x 4 4
(c) The domain of f is
x
y
1 x
2
0 x 2 {x 0 x s2 } (s2 , 0) (0, s2 )
1 1 x
1 0
π 2
π
x
FIGURE 6 y=cos x, 0¯x¯π y
2
■
The inverse cosine function is handled similarly. The restricted cosine function f x cos x, 0 x , is one-to-one (see Figure 6) and so it has an inverse function denoted by cos 1 or arccos. 4
cos1x y
&? cos y x
and 0 y
The cancellation equations are
π
cos 1cos x x
5 π 2
coscos1x x
0
_1
x
1
for 0 x for 1 x 1
The inverse cosine function, cos1, has domain 1, 1 and range 0, and is a continuous function whose graph is shown in Figure 7. Its derivative is given by
FIGURE 7 y=cos–! x=arccos x y
_ π2
6
0
π 2
FIGURE 8
y=tan x,
π π _ 2 <x< 2
x
d 1 cos1x dx s1 x 2
1 x 1
Formula 6 can be proved by the same method as for Formula 3 and is left as Exercise 11. The tangent function can be made one-to-one by restricting it to the interval 2, 2. Thus the inverse tangent function is defined as the inverse of the function f x tan x, 2 x 2. (See Figure 8.) It is denoted by tan1 or arctan.
7
tan1x y
&? tan y x
and
y 2 2
178
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CHAPTER 3
INVERSE FUNCTIONS
EXAMPLE 3 Simplify the expression costan1x. SOLUTION 1 Let y tan1x. Then tan y x and 2 y 2. We want to
find cos y but, since tan y is known, it’s easier to find sec y first: sec2 y 1 tan2 y 1 x 2 sec y s1 x 2
costan1x cos y
Thus
œ„„„„„ 1+≈ x y
since sec y 0 for 2 y 2
1 1 sec y s1 x 2
SOLUTION 2 Instead of using trigonometric identities as in Solution 1, it is perhaps easier to use a diagram. If y tan1x, then tan y x, and we can read from Figure 9 (which illustrates the case y 0) that
1
costan1x cos y
FIGURE 9 π 2
lim tan x
0
y=tan–! x=arctan x
and
x l2
x
FIGURE 10
■
The inverse tangent function, tan1 arctan, has domain ⺢ and its range is 2, 2. Its graph is shown in Figure 10. We know that
y
_ π2
1 s1 x 2
lim
x l2
tan x
and so the lines x 2 are vertical asymptotes of the graph of tan. Since the graph of tan1 is obtained by reflecting the graph of the restricted tangent function about the line y x, it follows that the lines y 2 and y 2 are horizontal asymptotes of the graph of tan 1. This fact is expressed by the following limits:
8
lim tan1x
xl
EXAMPLE 4 Evaluate lim arctan x l2
2
lim tan1x
x l
2
1 . x2
SOLUTION Since
1 l x2 the first equation in (8) gives lim arctan
x l2
as x l 2
1 x2
2
■
Since tan is differentiable, tan1 is also differentiable. To find its derivative, let y tan1x. Then tan y x. Differentiating this last equation implicitly with respect to x, we have dy sec2 y 1 dx
SECTION 3.5
INVERSE TRIGONOMETRIC FUNCTIONS
■
179
dy 1 1 1 2 2 dx sec y 1 tan y 1 x2
and so
d 1 tan1x dx 1 x2
9
The remaining inverse trigonometric functions are not used as frequently and are summarized here. 10
x 1
y csc1x x 1
&? csc y x
and
y 0, 2 , 32
y sec1x
&? sec y x
and
y 0, 2 , 32
&? cot y x
and
y 0,
y cot1x x ⺢ y
_1
0
FIGURE 11
π
2π
x
The choice of intervals for y in the definitions of csc1 and sec1 is not universally agreed upon. For instance, some authors use y 0, 2 2, in the definition of sec1. [You can see from the graph of the secant function in Figure 11 that both this choice and the one in (10) will work.] The reason for the choice in (10) is that the differentiation formulas are simpler (see Exercise 41). We collect in Table 11 the differentiation formulas for all of the inverse trigonometric functions. The proofs of the formulas for the derivatives of csc 1, sec 1, and cot 1 are left as Exercises 13–15.
y=sec x 11 TABLE OF DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
d 1 sin1x dx s1 x 2
d 1 csc1x dx x sx 2 1
d 1 cos1x dx s1 x 2
d 1 sec1x dx xsx 2 1
d 1 tan1x dx 1 x2
d 1 cot1x dx 1 x2
Each of these formulas can be combined with the Chain Rule. For instance, if u is a differentiable function of x, then d 1 du sin1u dx s1 u 2 dx
and
d 1 du tan1u dx 1 u 2 dx
EXAMPLE 5 Differentiate f x x tan1sx . SOLUTION
f x x
1 1 12 sx x tan1 sx tan1sx 1 (sx )2 2 21 x
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180
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CHAPTER 3
INVERSE FUNCTIONS
3.5 1–6
EXERCISES
Find the exact value of each expression.
■
1. (a) sin1(s32)
(b) cos11
2. (a) arctan1
(b) csc1 2
3. (a) tan 1s3
(b) arcsin(1s2 )
4. (a) sec1s2
(b) arcsin 1
5. (a) sinsin 1 0.7
(b) tan1 tan
6. (a) secarctan 2
(b) cos(2 sin1 (135 ))
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4 3
■
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9. sintan x
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31. tx cos13 2x ■
■
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■
■
■
■
■
y 3 arccosx2 at the point 1, .
10. cscarctan 2x
■
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■
■
■
■
■
■
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method as for Formula 3. 12. (a) Prove that sin1x cos1x 2.
35–38 35.
14. Prove that
d 1 sec1x . dx x sx 2 1
15. Prove that
d 1 csc1x . dx x sx 2 1
Find the derivative of the function. Simplify where
possible. 16. y stan1 x 17. y tan1 sx
18. hx s1 x 2 arcsin x
1
19. y sin 2x 1 21. Hx 1 x arctan x 2
1
22. ht e sec
t
23. y cos1e 2x 1
24. y x cos x s1 x 2 25. y arctancos 26. y tan1( x s1 x 2 ) 27. ht cot1t cot11t
20. f x x lnarctan x
Find the limit.
lim sin1x
xl
x l 1
■
■
36. lim arccos
■
1 x2 1 2x 2
38. lim tan1ln x
xl
■
d 1 13. Prove that cot1x . dx 1 x2
■
37. lim arctane x
(b) Use part (a) to prove Formula 6.
■
■
33. If tx x sin 1x4 s16 x 2 , find t2.
11. Prove Formula 6 for the derivative of cos1 by the same
16 –29
■
0 x , a b 0
34. Find an equation of the tangent line to the curve
1
■
■
b a cos x , a b cos x
32. Find y if tan1 xy 1 x 2 y.
8. tansin1x
■
xa xa
30. f x arcsine x
Simplify the expression.
■
29. y arccos
ln
30 –31 ■ Find the derivative of the function. Find the domains of the function and its derivative.
7. Prove that cossin1x s1 x 2 . 8 –10
x a
■
■
28. y tan1
xl0
■
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39. A ladder 10 ft long leans against a vertical wall. If the bot-
tom of the ladder slides away from the base of the wall at a speed of 2 fts, how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 ft from the base of the wall? 40. A lighthouse is located on a small island, 3 km away from
the nearest point P on a straight shoreline, and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P ? 41. Some authors define y sec1x &? sec y x and
y 0, 2 2, . Show that with this definition, we have (instead of the formula given in Exercise 14) d 1 sec1x dx x sx 2 1
x 1
42. (a) Sketch the graph of the function f x sinsin1x.
(b) Sketch the graph of the function tx sin1sin x, x ⺢. cos x (c) Show that tx . cos x (d) Sketch the graph of hx cos1sin x, x ⺢, and find its derivative.
SECTION 3.6
3.6
HYPERBOLIC FUNCTIONS
■
181
HYPERBOLIC FUNCTIONS Certain combinations of the exponential functions e x and ex arise so frequently in mathematics and its applications that they deserve to be given special names. In many ways they are analogous to the trigonometric functions, and they have the same relationship to the hyperbola that the trigonometric functions have to the circle. For this reason they are collectively called hyperbolic functions and individually called hyperbolic sine, hyperbolic cosine, and so on. DEFINITION OF THE HYPERBOLIC FUNCTIONS
sinh x
e x ex 2
csch x
1 sinh x
cosh x
e x ex 2
sech x
1 cosh x
tanh x
sinh x cosh x
coth x
cosh x sinh x
The graphs of hyperbolic sine and cosine can be sketched using graphical addition as in Figures 1 and 2. y
y
y=cosh x 1
y= 2 ´
y
y=1
y=sinh x 0
x
1
y=tanh x
1
1
y= 2 ´
y= e–® 2
1 y=_ 2 e–®
0
x
y=_1 0
FIGURE 1
FIGURE 2
y=sinh x= 21 ´- 21 e–®
y=cosh x= 21 ´+ 21 e–®
y
0
FIGURE 4
A catenary y=c+a cosh(x/a)
x
x
FIGURE 3
Note that sinh has domain ⺢ and range ⺢, while cosh has domain ⺢ and range 1, . The graph of tanh is shown in Figure 3. It has the horizontal asymptotes y 1. (See Exercise 19.) Applications of hyperbolic functions to science and engineering occur whenever an entity such as light, velocity, electricity, or radioactivity is gradually absorbed or extinguished, for the decay can be represented by hyperbolic functions. The most famous application is the use of hyperbolic cosine to describe the shape of a hanging wire. It can be proved that if a heavy flexible cable (such as a telephone or power line) is suspended between two points at the same height, then it takes the shape of a curve with equation y c a coshxa called a catenary (see Figure 4). (The Latin word catena means “chain.”)
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The hyperbolic functions satisfy a number of identities that are similar to wellknown trigonometric identities. We list some of them here and leave most of the proofs to the exercises. HYPERBOLIC IDENTITIES
sinhx sinh x
coshx cosh x
cosh2x sinh2x 1
1 tanh2x sech2x
sinhx y sinh x cosh y cosh x sinh y coshx y cosh x cosh y sinh x sinh y
V EXAMPLE 1
Prove (a) cosh2x sinh2x 1 and (b) 1 tanh2x sech2x.
SOLUTION
(a)
cosh2x sinh2x
e x ex 2
2
e x ex 2
2
e 2x 2 e2x e 2x 2 e2x 4 4
44 1 (b) We start with the identity proved in part (a):
y P(cos t, sin t)
cosh2x sinh2x 1 If we divide both sides by cosh2x, we get
O
Q
x
1 ≈ +¥ =1
or
FIGURE 5 y P(cosh t, sinh t)
0
≈-¥=1 FIGURE 6
x
sinh2x 1 cosh2x cosh2x
1 tanh2x sech2x
■
The identity proved in Example 1(a) gives a clue to the reason for the name “hyperbolic” functions: If t is any real number, the point Pcos t, sin t lies on the unit circle x 2 y 2 1 because cos2t sin2t 1. In fact, t can be interpreted as the radian measure of ⬔POQ in Figure 5. For this reason the trigonometric functions are sometimes called circular functions. Likewise, if t is any real number, then the point Pcosh t, sinh t lies on the right branch of the hyperbola x 2 y 2 1 because cosh2t sinh2t 1 and cosh t 1. This time, t does not represent the measure of an angle. However, it turns out that t represents twice the area of the shaded hyperbolic sector in Figure 6, just as in the trigonometric case t represents twice the area of the shaded circular sector in Figure 5. The derivatives of the hyperbolic functions are easily computed. For example, d d sinh x dx dx
e x ex 2
e x ex cosh x 2
SECTION 3.6
■
HYPERBOLIC FUNCTIONS
183
We list the differentiation formulas for the hyperbolic functions as Table 1. The remaining proofs are left as exercises. Note the analogy with the differentiation formulas for trigonometric functions, but beware that the signs are different in some cases. 1 DERIVATIVES OF HYPERBOLIC FUNCTIONS
d sinh x cosh x dx
d csch x csch x coth x dx
d cosh x sinh x dx
d sech x sech x tanh x dx
d tanh x sech2 x dx
d coth x csch2 x dx
V EXAMPLE 2 Any of these differentiation rules can be combined with the Chain Rule. For instance,
d d sinh sx (cosh sx ) sinh sx dx sx dx 2sx
■
INVERSE HYPERBOLIC FUNCTIONS
You can see from Figures 1 and 3 that sinh and tanh are one-to-one functions and so they have inverse functions denoted by sinh1 and tanh1. Figure 2 shows that cosh is not one-to-one, but when restricted to the domain 0, it becomes one-to-one. The inverse hyperbolic cosine function is defined as the inverse of this restricted function. 2
y sinh1x
&?
y cosh1x
&? cosh y x
y tanh1x
&? tanh y x
sinh y x and
y0
The remaining inverse hyperbolic functions are defined similarly (see Exercise 24). We can sketch the graphs of sinh1, cosh1, and tanh1 in Figures 7, 8, and 9 by using Figures 1, 2, and 3. y y
y
0 0
_1
x 0
1
FIGURE 7
FIGURE 8
FIGURE 9
y=sinh–! x domain=R
y=cosh–! x domain=[1, `} range=[0, `}
y=tanh–! x domain=(_1, 1)
range=R
1
x
x
range=R
184
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INVERSE FUNCTIONS
Since the hyperbolic functions are defined in terms of exponential functions, it’s not surprising to learn that the inverse hyperbolic functions can be expressed in terms of logarithms. In particular, we have:
■ Formula 3 is proved in Example 3. The proofs of Formulas 4 and 5 are requested in Exercises 22 and 23.
3
sinh1x ln( x sx 2 1 )
x⺢
4
cosh1x ln( x sx 2 1 )
x1
5
tanh1x 12 ln
1x 1x
1 x 1
EXAMPLE 3 Show that sinh1x ln( x sx 2 1 ). SOLUTION Let y sinh1x. Then
x sinh y
e y ey 2
e y 2x ey 0
so or, multiplying by e y,
e 2y 2xe y 1 0 This is really a quadratic equation in e y: e y 2 2xe y 1 0 Solving by the quadratic formula, we get ey
2x s4x 2 4 x sx 2 1 2
Note that e y 0, but x sx 2 1 0 (because x sx 2 1 ). Thus the minus sign is inadmissible and we have e y x sx 2 1 Therefore
y lne y ln( x sx 2 1 ) ■
(See Exercise 21 for another method.) 6 DERIVATIVES OF INVERSE HYPERBOLIC FUNCTIONS
Notice that the formulas for the derivatives of tanh1x and coth1x appear to be identical. But the domains of these functions have no numbers in common: tanh1x is defined for x 1 , whereas coth1x is defined for x 1. ■
d 1 sinh1x dx s1 x 2
d 1 csch1x dx x sx 2 1
d 1 cosh1x 2 dx sx 1
d 1 sech1x dx xs1 x 2
d 1 tanh1x dx 1 x2
d 1 coth1x dx 1 x2
SECTION 3.6
HYPERBOLIC FUNCTIONS
■
185
The inverse hyperbolic functions are all differentiable because the hyperbolic functions are differentiable. The formulas in Table 6 can be proved either by the method for inverse functions or by differentiating Formulas 3, 4, and 5. V EXAMPLE 4
Prove that
d 1 sinh1x . dx s1 x 2
SOLUTION Let y sinh1x. Then sinh y x. If we differentiate this equation
implicitly with respect to x, we get cosh y
dy 1 dx
Since cosh2 y sinh2 y 1 and cosh y 0, we have cosh y s1 sinh2 y , so
■ Another method for solving Example 4 is to differentiate Formula 3.
dy 1 1 1 2y dx cosh y sinh s1 s1 x 2 V EXAMPLE 5
Find
■
d tanh1sin x . dx
SOLUTION Using Table 6 and the Chain Rule, we have
d 1 d tanh1sin x sin x 2 dx 1 sin x dx
3.6 1–6
11. sinhx y sinh x cosh y cosh x sinh y
1. (a) sinh 0
(b) cosh 0
12. coshx y cosh x cosh y sinh x sinh y
2. (a) tanh 0
(b) tanh 1
13. sinh 2x 2 sinh x cosh x
3. (a) sinhln 2
(b) sinh 2
4. (a) cosh 3
(b) coshln 3
14.
5. (a) sech 0
(b) cosh1 1
15. cosh x sinh xn cosh nx sinh nx
1
6. (a) sinh 1 ■
■
7–15
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■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
16. If sinh x 4 , find the values of the other hyperbolic func3
Prove the identity.
tions at x.
7. sinhx sinh x
17. If tanh x 5 , find the values of the other hyperbolic func4
(This shows that sinh is an odd function.)
tions at x.
8. coshx cosh x
(This shows that cosh is an even function.) 9. cosh x sinh x e
1 tanh x e 2x 1 tanh x (n any real number)
(b) sinh 1 ■
■
EXERCISES
Find the numerical value of each expression.
■
1 cos x sec x 2 cos x 1 sin x cos2x
18. (a) Use the graphs of sinh, cosh, and tanh in Figures 1–3 to
x
10. cosh x sinh x ex
;
draw the graphs of csch, sech, and coth. (b) Check the graphs that you sketched in part (a) by using a graphing device to produce them.
186
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INVERSE FUNCTIONS
19. Use the definitions of the hyperbolic functions to find each
of the following limits. (a) lim tanh x
(b) lim tanh x
(c) lim sinh x
(d) lim sinh x
(e) lim sech x
(f ) lim coth x
(g) lim coth x
(h) lim coth x
xl
43. A telephone line hangs between two poles 14 m apart in the
shape of the catenary y 20 coshx20 15, where x and y are measured in meters. (a) Find the slope of this curve where it meets the right pole. (b) Find the angle between the line and the pole.
x l
xl
x l
xl
xl
x l0
y
x l0
¨
(i) lim csch x
5
x l
20. Prove the formulas given in Table 1 for the derivatives of
the functions (a) cosh, (b) tanh, (c) csch, (d) sech, and (e) coth.
_7
7 x
0
21. Give an alternative solution to Example 3 by letting
y sinh1x and then using Exercise 9 and Example 1(a) with x replaced by y.
44. Using principles from physics it can be shown that when a
cable is hung between two poles, it takes the shape of a curve y f x that satisfies the differential equation
22. Prove Equation 4.
d2y t dx 2 T
23. Prove Formula 5 using (a) the method of Example 3 and
(b) Exercise 14 with x replaced by y. 24. For each of the following functions (i) give a definition like
those in (2), (ii) sketch the graph, and (iii) find a formula similar to Formula 3. (a) csch 1 (b) sech1 (c) coth1 25. Prove the formulas given in Table 6 for the derivatives of
the following functions. (a) cosh1 (b) tanh1 26 – 41
■
y f x
Find the derivative. 27. f x x cosh x
28. Fx sinh x tanh x
29. hx sinhx 2
30. f t e sech t
31. ht coth s1 t 2
32. f t lnsinh t
33. Ht tanhe t
34. y sinhcosh x
35. y e cosh 3x
36. y x 2 sinh12x
37. y tanh1sx
satisfies the differential equation y m 2 y. (b) Find y yx such that y 9y, y0 4, and y0 6. 46. Evaluate lim
xl
■
■
■
sinh x . ex
47. At what point of the curve y cosh x does the tangent have
x0
slope 1?
41. y coth1sx 2 1 ■
T tx cosh t T
y A sinh mx B cosh mx
39. y x sinh1x3 s9 x 2
■
2
45. (a) Show that any function of the form
38. y x tanh1x ln s1 x 2
■
dy dx
is a solution of this differential equation.
26. tx sinh2x
40. y sech1s1 x 2 ,
1
where is the linear density of the cable, t is the acceleration due to gravity, and T is the tension in the cable at its lowest point, and the coordinate system is chosen appropriately. Verify that the function
(c) sech1
t
■
■
■
■
■
; 42. A flexible cable always hangs in the shape of a catenary
y c a coshxa, where c and a are constants and a 0 (see Figure 4 and Exercise 44). Graph several members of the family of functions y a coshxa. How does the graph change as a varies?
■
48. If x lnsec tan , show that sec cosh x . 49. Show that if a 0 and b 0, then there exist numbers
and such that ae x bex equals either sinhx or coshx . In other words, almost every function of the form f x ae x bex is a shifted and stretched hyperbolic sine or cosine function.
SECTION 3.7
3.7
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
■
187
INDETERMINATE FORMS AND L’HOSPITAL’S RULE Suppose we are trying to analyze the behavior of the function Fx
ln x x1
Although F is not defined when x 1, we need to know how F behaves near 1. In particular, we would like to know the value of the limit lim
1
x l1
ln x x1
In computing this limit we can’t apply Law 5 of limits (the limit of a quotient is the quotient of the limits, see Section 1.4) because the limit of the denominator is 0. In fact, although the limit in (1) exists, its value is not obvious because both numerator and denominator approach 0 and 00 is not defined. In general, if we have a limit of the form lim
xla
f x tx
where both f x l 0 and tx l 0 as x l a, then this limit may or may not exist and is called an indeterminate form of type 00 . We met some limits of this type in Chapter 1. For rational functions, we can cancel common factors: lim x l1
x2 x xx 1 x 1 lim lim x l1 x 1x 1 x l1 x 1 x2 1 2
We used a geometric argument to show that lim
xl0
sin x 1 x
But these methods do not work for limits such as (1), so in this section we introduce a systematic method, known as l’Hospital’s Rule, for the evaluation of indeterminate forms. Another situation in which a limit is not obvious occurs when we look for a horizontal asymptote of F and need to evaluate the limit 2
lim
xl
ln x x1
It isn’t obvious how to evaluate this limit because both numerator and denominator become large as x l . There is a struggle between numerator and denominator. If the numerator wins, the limit will be ; if the denominator wins, the answer will be 0. Or there may be some compromise, in which case the answer may be some finite positive number. In general, if we have a limit of the form lim
xla
f x tx
where both f x l (or ) and tx l (or ), then the limit may or may not
188
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CHAPTER 3
INVERSE FUNCTIONS
exist and is called an indeterminate form of type . We saw in Section 1.6 that this type of limit can be evaluated for certain functions, including rational functions, by dividing numerator and denominator by the highest power of x that occurs in the denominator. For instance, 1 1 x 1 x2 10 lim lim 2 x l 2x 1 xl 20 2 1 2 2 x 1
2
This method does not work for limits such as (2), but l’Hospital’s Rule also applies to this type of indeterminate form. L’Hospital’s Rule is named after a French nobleman, the Marquis de l’Hospital (1661–1704), but was discovered by a Swiss mathematician, John Bernoulli (1667–1748). See Exercise 45 for the example that the Marquis used to illustrate his rule. ■
L’HOSPITAL’S RULE Suppose f and t are differentiable and tx 0 near a
(except possibly at a). Suppose that
or that y
f
lim f x 0
and
lim f x
and
xla
xla
lim tx 0
xla
lim tx
xla
(In other words, we have an indeterminate form of type 00 or .) Then g
lim
xla
0
a
f x f x lim x l a tx tx
x
if the limit on the right side exists (or is or ). y
y=m¡(x-a)
NOTE 1 L’Hospital’s Rule says that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives, provided that the given conditions are satisfied. It is especially important to verify the conditions regarding the limits of f and t before using l’Hospital’s Rule.
y=m™(x-a) 0
a
x
FIGURE 1 ■ Figure 1 suggests visually why l’Hospital’s Rule might be true. The first graph shows two differentiable functions f and t, each of which approaches 0 as x l a . If we were to zoom in toward the point a, 0, the graphs would start to look almost linear. But if the functions actually were linear, as in the second graph, then their ratio would be m1x a m1 m2x a m2
which is the ratio of their derivatives. This suggests that lim
xla
f x f x lim x l a tx tx
NOTE 2 L’Hospital’s Rule is also valid for one-sided limits and for limits at infinity or negative infinity; that is, “ x l a” can be replaced by any of the symbols x l a, x l a, x l , or x l . NOTE 3 For the special case in which f a ta 0, f and t are continuous, and ta 0, it is easy to see why l’Hospital’s Rule is true. In fact, using the alternative form of the definition of a derivative, we have
f x f a f x f a f x f a xa xa lim lim x l a tx x l a tx ta ta tx ta lim xla xa xa lim
xla
lim xla
f x f a f x lim x l a tx tx ta
The general version of l’Hospital’s Rule is more difficult; its proof can be found in Appendix B.
SECTION 3.7
V EXAMPLE 1
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
■
189
ln x . x1
Find lim x l1
SOLUTION Since
lim ln x ln 1 0 x l1
and
lim x 1 0 x l1
we can apply l’Hospital’s Rule: d ln x ln x dx 1x 1 lim lim lim lim 1 x l1 x 1 x l1 d x l1 1 x l1 x x 1 dx
| Notice that when using l’Hospital’s Rule we differentiate the numerator and denominator separately. We do not use the Quotient Rule.
V EXAMPLE 2 ■ The graph of the function of Example 2 is shown in Figure 2. We have noticed previously that exponential functions grow far more rapidly than power functions, so the result of Example 2 is not unexpected. See also Exercise 39.
Calculate lim
xl
■
ex . x2
SOLUTION We have lim x l e x and lim x l x 2 , so l’Hospital’s Rule gives
d e x e dx ex lim 2 lim lim xl x xl d x l 2x x 2 dx x
20
Since e x l and 2x l as x l , the limit on the right side is also indeterminate, but a second application of l’Hospital’s Rule gives y= ´ ≈
lim
xl
10
0
ex ex ex lim lim x l 2x xl 2 x2
FIGURE 2 V EXAMPLE 3 ■ The graph of the function of Example 3 is shown in Figure 3. We have discussed previously the slow growth of logarithms, so it isn’t surprising that this ratio approaches 0 as x l . See also Exercise 40.
xl
ln x . 3 x s
3 SOLUTION Since ln x l and s x l as x l , l’Hospital’s Rule applies:
lim
xl
ln x 1x lim 1 23 3 xl 3 x x s
Notice that the limit on the right side is now an indeterminate of type 00 . But instead of applying l’Hospital’s Rule a second time as we did in Example 2, we simplify the expression and see that a second application is unnecessary:
2 y= ln x Œ„ x 0
Calculate lim
■
10,000
lim
xl
_1
EXAMPLE 4 Find lim
xl0
FIGURE 3
ln x 1x 3 lim 1 23 lim 3 0 3 xl 3 x x l sx x s
tan x x . [See Exercise 22 in Section 1.3] x3
SOLUTION Noting that both tan x x l 0 and x 3 l 0 as x l 0, we use l’Hospi-
tal’s Rule: lim
xl0
tan x x sec2x 1 lim 3 xl0 x 3x 2
■
190
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CHAPTER 3
INVERSE FUNCTIONS
■ The graph in Figure 4 gives visual confirmation of the result of Example 4. If we were to zoom in too far, however, we would get an inaccurate graph because tan x is close to x when x is small. See Exercise 22(d) in Section 1.3.
1
Since the limit on the right side is still indeterminate of type 00 , we apply l’Hospital’s Rule again: sec2x 1 2 sec2x tan x lim lim 2 xl0 xl0 3x 6x Because lim x l 0 sec2 x 1, we simplify the calculation by writing 2 sec2x tan x 1 tan x 1 tan x lim sec2 x lim lim xl0 6x 3 xl0 x 3 xl0 x
lim
xl0
We can evaluate this last limit either by using l’Hospital’s Rule a third time or by writing tan x as sin xcos x and making use of our knowledge of trigonometric limits. Putting together all the steps, we get
tan x- x y= ˛ _1
1 0
lim
xl0
FIGURE 4
tan x x sec 2 x 1 2 sec 2 x tan x lim lim 3 2 xl0 xl0 x 3x 6x
V EXAMPLE 5
Find lim xl
1 tan x 1 sec 2 x 1 lim lim 3 xl0 x 3 xl0 1 3
■
sin x . 1 cos x
SOLUTION If we blindly attempted to use l’Hospital’s Rule, we would get
|
lim
xl
sin x cos x lim xl 1 cos x sin x
This is wrong! Although the numerator sin x l 0 as x l , notice that the denominator 1 cos x does not approach 0, so l’Hospital’s Rule can’t be applied here. The required limit is, in fact, easy to find because the function is continuous at and the denominator is nonzero there: lim
xl
sin x sin 0 0 1 cos x 1 cos 1 1
■
Example 5 shows what can go wrong if you use l’Hospital’s Rule without thinking. Other limits can be found using l’Hospital’s Rule but are more easily found by other methods. (See Examples 2 and 4 in Section 1.4, Example 5 in Section 1.6, and the discussion at the beginning of this section.) So when evaluating any limit, you should consider other methods before using l’Hospital’s Rule. INDETERMINATE PRODUCTS
If lim x l a f x 0 and lim x l a tx (or ), then it isn’t clear what the value of lim x l a f xtx, if any, will be. There is a struggle between f and t. If f wins, the answer will be 0; if t wins, the answer will be (or ). Or there may be a compromise where the answer is a finite nonzero number. This kind of limit is called an indeterminate form of type 0 . We can deal with it by writing the product ft as a quotient: f t ft or ft 1t 1f
SECTION 3.7
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
■
191
This converts the given limit into an indeterminate form of type 00 or so that we can use l’Hospital’s Rule. ■ Figure 5 shows the graph of the function in Example 6. Notice that the function is undefined at x 0 ; the graph approaches the origin but never quite reaches it.
y
y=x ln x
V EXAMPLE 6
Evaluate lim x ln x . x l0
SOLUTION The given limit is indeterminate because, as x l 0 , the first factor x
approaches 0 while the second factor ln x approaches . Writing x 11x, we have 1x l as x l 0 , so l’Hospital’s Rule gives lim x ln x lim
x l 0
xl0
ln x 1x lim lim x 0 x l 0 xl0 1x 1x 2
■
NOTE In solving Example 6 another possible option would have been to write 0
1
x
lim x ln x lim
x l 0
FIGURE 5
xl0
x 1ln x
This gives an indeterminate form of the type 00, but if we apply l’Hospital’s Rule we get a more complicated expression than the one we started with. In general, when we rewrite an indeterminate product, we try to choose the option that leads to the simpler limit. INDETERMINATE DIFFERENCES
If lim x l a f x and lim x l a tx , then the limit lim f x tx
xla
is called an indeterminate form of type . Again there is a contest between f and t. Will the answer be ( f wins) or will it be ( t wins) or will they compromise on a finite number? To find out, we try to convert the difference into a quotient (for instance, by using a common denominator, or rationalization, or factoring out a common factor) so that we have an indeterminate form of type 00 or . EXAMPLE 7 Compute
lim sec x tan x.
x l2
SOLUTION First notice that sec x l and tan x l as x l 2, so the limit
is indeterminate. Here we use a common denominator: lim sec x tan x
x l2
lim
x l2
lim
x l2
1 sin x cos x cos x
1 sin x cos x lim 0 x l2 cos x sin x
Note that the use of l’Hospital’s Rule is justified because 1 sin x l 0 and cos x l 0 as x l 2.
■
192
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CHAPTER 3
INVERSE FUNCTIONS
INDETERMINATE POWERS
Several indeterminate forms arise from the limit lim f x tx
xla
1. lim f x 0 xla
and
2. lim f x and xla
3. lim f x 1 xla
and
lim tx 0
type 0 0
lim tx 0
type 0
lim tx
type 1
xla xla xla
Each of these three cases can be treated either by taking the natural logarithm: y f x tx,
let
then ln y tx ln f x
or by writing the function as an exponential: f x tx e tx ln f x (Recall that both of these methods were used in differentiating such functions.) In either method we are led to the indeterminate product tx ln f x, which is of type 0 . EXAMPLE 8 Calculate lim 1 sin 4xcot x. xl0
SOLUTION First notice that as x l 0 , we have 1 sin 4x l 1 and cot x l , so
the given limit is indeterminate. Let y 1 sin 4xcot x ln y ln 1 sin 4xcot x cot x ln1 sin 4x
Then
so l’Hospital’s Rule gives 4 cos 4x ln1 sin 4x 1 sin 4x lim ln y lim lim 4 x l 0 xl0 xl0 tan x sec2x The graph of the function y x x , x 0 , is shown in Figure 6. Notice that although 0 0 is not defined, the values of the function approach 1 as x l 0. This confirms the result of Example 9. ■
So far we have computed the limit of ln y, but what we want is the limit of y. To find this we use the fact that y e ln y : lim 1 sin 4xcot x lim y lim e ln y e 4
x l 0
2
V EXAMPLE 9
xl0
xl0
Find lim x x. xl0
SOLUTION Notice that this limit is indeterminate since 0 x 0 for any x 0 but
_1
FIGURE 6
0
2
■
x 0 1 for any x 0. We could proceed as in Example 8 or by writing the function as an exponential: x x e ln x x e x ln x
SECTION 3.7
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
■
193
In Example 6 we used l’Hospital’s Rule to show that lim x ln x 0
x l 0
lim x x lim e x ln x e 0 1
Therefore
3.7
x l 0
EXERCISES
Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.
1–36
■
1. lim
x l1
3.
x2 1 x1 cos x 1 sin x
lim
x l2
tl0
e 1 t3
35. lim cos x1x
2
36. lim x ln 21 ln x
xl0
■
■
■
■
xl
■
■
x tan x sin x
37. lim x lnx 5 ln x
4. lim
e 1 t
39. Prove that
xl0
tl0
■
■
■
■
■
xl0
tan px tan qx
9. lim xl0
8.
ln x x
lim
l 2
10. lim
xl
lim
1 sin csc
ln ln x x
lim
ex 1 x 13. lim xl0 x2
ex 14. lim 3 xl x
x 15. lim x l ln1 2e x
cos mx cos nx 16. lim xl0 x2
xl
18. lim
x tan14x
19. lim
x a ax a 1 x 12
20. lim
1 e2x sec x
21. lim sx ln x
22. lim x 2e x
23. lim cot 2x sin 6x
24. lim sin x ln x
xl0
xl0
3 x
25. lim x e
2
27. lim xe xl
x
29. lim x ln x xl
31. lim x x
2
xl0
33. lim 1 2 x1x xl0
34. lim
xl
1 1 ln x x1
1
r n
nt
account, is
32. lim tan 2 x x xl0
ln x 0 xp
42. If an object with mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into
28. lim csc x cot x
xl1
■
A A0 e rt
xl0
30. lim
■
rate r compounded n times a year, the value of the investment after t years is
x l
xl0
■
If we let n l , we refer to the continuous compounding of interest. Use l’Hospital’s Rule to show that if interest is compounded continuously, then the amount after t years is
xl
1x
■
ex xn
A A0 1
26. lim x tan1x
xl
■
41. If an initial amount A0 of money is invested at an interest
1 x ln x 1 cos x
xl0
■
for any number p 0. This shows that the logarithmic function approaches more slowly than any power of x.
17. lim
xl1
■
40. Prove that
ln x 12. lim x l 1 sin x
xl0
■
for any positive integer n. This shows that the exponential function approaches infinity faster than any power of x.
5t 3t 11. lim tl0 t
xl1
■
x l 4
■
xl
7. lim
■
38. lim tan xtan 2x
xl
3t
6. lim
■
■ Use a graph to estimate the value of the limit. Then use l’Hospital’s Rule to find the exact value.
xa 1 xb 1
xl1
■
; 37–38
2. lim
t
5. lim
■
xl0
a x
bx
v
mt 1 e ctm c
where t is the acceleration due to gravity and c is a positive constant. (a) Calculate lim t l v. What is the meaning of this limit? (b) For fixed t, use l’Hospital’s Rule to calculate lim c l 0 v. What can you conclude about the velocity of a falling object in a vacuum?
194
■
CHAPTER 3
INVERSE FUNCTIONS
43. If an electrostatic field E acts on a liquid or a gaseous polar
dielectric, the net dipole moment P per unit volume is PE
e E e E 1 e E e E E
Show that lim E l 0 PE 0.
47. If f is continuous, f 2 0, and f 2 7 , evaluate
lim
xl0
48. For what values of a and b is the following equation true?
44. A metal cable has radius r and is covered by insulation, so
that the distance from the center of the cable to the exterior of the insulation is R. The velocity v of an electrical impulse in the cable is
lim
xl0
r R
2
ln
r R
where c is a positive constant. Find the following limits and interpret your answers. (a) lim v (b) lim v R lr
r l0
lim
hl0
y
3 aax s2a 3x x 4 a s 4 3 a sax
as x approaches a, where a 0. (At that time it was common to write aa instead of a 2.) Solve this problem. 46. The figure shows a sector of a circle with central angle .
Let A be the area of the segment between the chord PR and the arc PR. Let B be the area of the triangle PQR. Find lim l 0 .
sin 2x b a 2 x3 x
0
f x h f x h f x 2h
Explain the meaning of this equation with the aid of a diagram. 50. If f is continuous, show that
45. The first appearance in print of l’Hospital’s Rule was in
the book Analyse des Infiniment Petits published by the Marquis de l’Hospital in 1696. This was the first calculus textbook ever published and the example that the Marquis used in that book to illustrate his rule was to find the limit of the function
49. If f is continuous, use l’Hospital’s Rule to show that
v c
f 2 3x f 2 5x x
lim
hl0
f x h 2 f x f x h f x h2
51. Let
f x
e1x 0
2
if x 0 if x 0
(a) Use the definition of derivative to compute f 0. (b) Show that f has derivatives of all orders that are defined on ⺢. [Hint: First show by induction that there is a polynomial pnx and a nonnegative integer k n such that f nx pnxf xx k n for x 0.]
; 52. Let f x
x 1
x
if x 0 if x 0
P A(¨) B(¨) ¨ O
Q
R
(a) Show that f is continuous at 0. (b) Investigate graphically whether f is differentiable at 0 by zooming in several times toward the point 0, 1 on the graph of f . (c) Show that f is not differentiable at 0. How can you reconcile this fact with the appearance of the graphs in part (b)?
CHAPTER 3
3
REVIEW
REVIEW
function is one-to-one by looking at its graph? (b) If f is a one-to-one function, how is its inverse function f 1 defined? How do you obtain the graph of f 1 from the graph of f ? (c) Suppose f is a one-to-one function. If f f 1a 0 , write a formula for f 1a.
(c) How is the inverse tangent function f x tan1x defined? What are its domain and range? Sketch its graph. 5. Write the definitions of the hyperbolic functions sinh x,
cosh x, and tanh x. 6. State the derivative of each function.
2. (a) Express e as a limit.
(b) What is the value of e correct to five decimal places? (c) Why is the natural exponential function y e x used more often in calculus than the other exponential functions y a x ? (d) Why is the natural logarithmic function y ln x used more often in calculus than the other logarithmic functions y log a x ?
(a) (d) (g) ( j) (m)
tial function f x e x ? (b) What are the domain and range of the natural logarithmic function f x ln x ? (c) How are the graphs of these functions related? Sketch these graphs, by hand, using the same axes. (d) If a is a positive number, a 1, write an equation that expresses log a x in terms of ln x.
y ex y log a x y tan1x y tanh x y tanh1x
(b) (e) (h) (k)
y ax y sin1x y sinh x y sinh1x
(c) (f ) (i) (l)
y ln x y cos1x y cosh x y cosh1x
7. (a) Write a differential equation that expresses the law of
natural growth. What does it say in terms of relative growth rate? (b) Under what circumstances is this an appropriate model for population growth? (c) What are the solutions of this equation?
3. (a) What are the domain and range of the natural exponen-
8. (a) What does l’Hospital’s Rule say?
(b) How can you use l’Hospital’s Rule if you have a product f xtx where f x l 0 and tx l as x l a ? (c) How can you use l’Hospital’s Rule if you have a difference f x tx where f x l and tx l as x l a? (d) How can you use l’Hospital’s Rule if you have a power f x tx where f x l 0 and tx l 0 as x l a ?
4. (a) How is the inverse sine function f x sin1x defined?
What are its domain and range? (b) How is the inverse cosine function f x cos1x defined? What are its domain and range?
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
2. If f is one-to-one and differentiable, with domain ⺢,
then f 16 1f 6.
3. The function f x cos x, 2 x 2, is
one-to-one.
10.
d 10 x x10 x1 dx
11.
d 1 ln 10 dx 10
12. The inverse function of y e 3x is y 3 ln x. 1
13. cos1x
1 cos x
14. tan1x
sin1x cos1x
4. tan11 3 4 5. If 0 a b, then ln a ln b. 6. s5 e s5 ln 7. You can always divide by e x. 8. If a 0 and b 0, then lna b ln a ln b. 9. If x 0, then ln x 6 ln x. 6
195
CONCEPT CHECK
1. (a) What is a one-to-one function? How can you tell if a
1. If f is one-to-one, with domain ⺢, then f 1 f 6 6 .
■
15. cosh x 1
for all x
tan x sec2x lim x l 1 cos x x l sin x
16. lim
196
■
CHAPTER 3
INVERSE FUNCTIONS
EXERCISES 19. tan1x 1
1. The graph of f is shown. Is f one-to-one? Explain. y
■
■
21– 47
0
x
20. sin x 0.3
■
■
■
■
■
y
g
23. h e tan 2
24. hu 10 su
29. y lnsec2x
30. y lnx 2e x
31. y xe1x
32. y x re sx
33. y 2t
34. y e cos x cose x
2
35. Hv v tan1v
36. Fz log101 z 2
37. y x sinhx 2
38. y cos x x
3. Suppose f is one-to-one, f 7 3, and f 7 8 . Find
(a) f 13 and (b) f 1 3.
40. y arctan(arcsin sx )
46. y x tanh1sx
■
■
7. y ln x
8. y lnx 1
d dx 49–52
■
■
■
■
■
■
■
■
■
■
10. Let a 1. For large values of x, which of the functions
11–12
■
■
■
■
1 2
tan1 x 14 ln
x 1 2 x2 1
■
■
■
■
■
■
1 1 x1 x 2
Find f in terms of t.
50. f x te x
■
■
52. f x tln x ■
■
■
■
■
■
■
■
■
■
■
■
x
53–54
11. (a) e 2 ln 3
(b) log 10 25 log 10 4
12. (a) ln e
(b) tan(arcsin 12 )
■
■
■
■
■
■
■
■
Find f nx.
53. f x 2 x
Find the exact value of each expression.
■
■
51. f x ln tx ■
y x , y a , and y log a x has the largest values and which has the smallest values? a
1
■
49. f x e tx
9. y 2 arctan x
■
■
2
48. Show that
6. y ex
x 2 1 4 2x 1 33x 1 5
45. y cosh1sinh x 47. f x e sin lnx
5. y 5 x 1
■
42. xe y y 1 44. y
■
■
1 ln x
43. y lncosh 3x
Sketch a rough graph of the function without using a calculator.
5–9
1 x
3
x1 4. Find the inverse function of f x . 2x 1
■
26. y ett 2 2t 2 28. y sin1e x
41. y ln
x
■
27. y e cxc sin x cos x
1
0 1
■
et 1 et
22. tt
39. y ln sin x 2 sin2x
1
■
21. f t t 2 ln t
Why is t one-to-one? Estimate the value of t12. Estimate the domain of t1. Sketch the graph of t1.
(a) (b) (c) (d)
■
Differentiate.
25. y ln sec 5x tan 5x 2. The graph of t is given.
■
■
■
■
■
■
54. f x ln2x ■
■
■
■
■
55. Use mathematical induction to show that if f x xe x,
then f nx x ne x.
■
56. Find y if y x arctan y. 13–20
■
Solve the equation for x.
13. ln x
1 3
x
14. e x
1 3
15. e e 17
16. ln1 ex 3
17. ln x 1 ln x 1 1
18. log 5 c x d
Find an equation of the tangent to the curve at the given point. 57–58
■
57. y 2 xex, ■
■
■
■
0, 2 ■
58. y x ln x, ■
■
■
■
e, e ■
■
■
CHAPTER 3
59. At what point on the curve y lnx 4 2 is the tangent
66. A cup of hot chocolate has temperature 80 C in a room kept
screen and comment.
at 20 C. After half an hour the hot chocolate cools to 60 C. (a) What is the temperature of the chocolate after another half hour? (b) When will the chocolate have cooled to 40 C ?
61. (a) Find an equation of the tangent to the curve y e x that
is parallel to the line x 4y 1. (b) Find an equation of the tangent to the curve y e x that passes through the origin.
62. The function Ct Keat ebt , where a, b, and K are
positive constants and b a, is used to model the concentration at time t of a drug injected into the bloodstream. (a) Show that lim t l Ct 0. (b) Find Ct, the rate at which the drug is cleared from circulation. (c) When is this rate equal to 0?
63. A bacteria culture contains 200 cells initially and grows at a
rate proportional to its size. After half an hour the population has increased to 360 cells. (a) Find the number of bacteria after t hours. (b) Find the number of bacteria after 4 hours. (c) Find the rate of growth after 4 hours. (d) When will the population reach 10,000? 64. Cobalt-60 has a half-life of 5.24 years.
(a) Find the mass that remains from a 100-mg sample after 20 years. (b) How long would it take for the mass to decay to 1 mg? 65. Let Ct be the concentration of a drug in the bloodstream.
As the body eliminates the drug, Ct decreases at a rate that is proportional to the amount of the drug that is present at the time. Thus Ct kCt, where k is a positive number called the elimination constant of the drug. (a) If C0 is the concentration at time t 0, find the concentration at time t.
197
(b) If the body eliminates half the drug in 30 hours, how long does it take to eliminate 90% of the drug?
horizontal? sin x ; 60. If f x xe , find f x. Graph f and f on the same
■
REVIEW
67– 82
Evaluate the limit.
■
3x
68. lim ln100 x 2
67. lim e x l
69. lim e
x l10
70. lim arctanx 3 x
2x3
xl
xl3
72. lim ex sin x
71. lim lnsinh x
xl
xl0
73. lim
1 2x 1 2x
74. lim
75. lim
tan x ln1 x
76. lim
1 cos x x2 x
77. lim
e 4x 1 4x x2
78. lim
e 4x 1 4x x2
xl
xl0
xl0
xl
xl0
xl
79. lim x 3ex xl
81. lim xl1
■
■
x 1 x1 ln x ■
■
■
1
4 x
x
80. lim x 2 ln x xl0
lim tan xcos x
82. ■
■
x l2 ■
■
■
83. If f x ln x tan1 x , find f 14. 84. Show that
cos arctan sinarccot x
x2 1 x2 2
■
■
APPLICATIONS OF DIFFERENTIATION
4
We have already investigated some of the applications of derivatives, but now that we know the differentiation rules we are in a better position to pursue the applications of differentiation in greater depth. Here we learn how derivatives affect the shape of a graph of a function and, in particular, how they help us locate maximum and minimum values of functions. Many practical problems require us to minimize a cost or maximize an area or somehow find the best possible outcome of a situation. In particular, we will be able to investigate the optimal shape of a can and to explain the shape of cells in beehives.
4.1
MAXIMUM AND MINIMUM VALUES Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. Here are examples of such problems that we will solve in this chapter: ■
What is the shape of a can that minimizes manufacturing costs?
■
What is the maximum acceleration of a space shuttle? (This is an important question to the astronauts who have to withstand the effects of acceleration.)
■
What is the radius of a contracted windpipe that expels air most rapidly during a cough?
■
How far should you stand from a painting in an art gallery to get the best view of the painting?
These problems can be reduced to finding the maximum or minimum values of a function. Let’s first explain exactly what we mean by maximum and minimum values. 1 DEFINITION A function f has an absolute maximum (or global maximum) at c if f c f x for all x in D, where D is the domain of f . The number f c is called the maximum value of f on D. Similarly, f has an absolute minimum at c if f c f x for all x in D and the number f c is called the minimum value of f on D. The maximum and minimum values of f are called the extreme values of f .
y
f(d) f(a) a
0
b
c
FIGURE 1
Minimum value f(a), maximum value f(d)
198
d
e
x
Figure 1 shows the graph of a function f with absolute maximum at d and absolute minimum at a. Note that d, f d is the highest point on the graph and a, f a is the lowest point. In Figure 1, if we consider only values of x near b [for instance, if we restrict our attention to the interval a, c], then f b is the largest of those values of f x and is called a local maximum value of f . Likewise, f c is called a local minimum value of f because f c f x for x near c [in the interval b, d, for instance]. The function f also has a local minimum at e. In general, we have the following definition.
SECTION 4.1
MAXIMUM AND MINIMUM VALUES
■
199
2 DEFINITION A function f has a local maximum (or relative maximum) at c if f c f x when x is near c. [This means that f c f x for all x in some open interval containing c.] Similarly, f has a local minimum at c if f c f x when x is near c.
EXAMPLE 1 The function f x cos x takes on its (local and absolute) maximum value of 1 infinitely many times, since cos 2n 1 for any integer n and 1 cos x 1 for all x. Likewise, cos2n 1 1 is its minimum value, where n is any integer. ■
y
y=≈
EXAMPLE 2 If f x x 2, then f x f 0 because x 2 0 for all x. Therefore,
0
x
FIGURE 2
Minimum value 0, no maximum
f 0 0 is the absolute (and local) minimum value of f . This corresponds to the fact that the origin is the lowest point on the parabola y x 2. (See Figure 2.) However, there is no highest point on the parabola and so this function has no maximum value. ■ EXAMPLE 3 From the graph of the function f x x 3, shown in Figure 3, we see
y
that this function has neither an absolute maximum value nor an absolute minimum value. In fact, it has no local extreme values either. ■
y=˛
0
x
V EXAMPLE 4
The graph of the function f x 3x 4 16x 3 18x 2
is shown in Figure 4. You can see that f 1 5 is a local maximum, whereas the absolute maximum is f 1 37. (This absolute maximum is not a local maximum because it occurs at an endpoint.) Also, f 0 0 is a local minimum and f 3 27 is both a local and an absolute minimum. Note that f has neither a local nor an absolute maximum at x 4. ■
FIGURE 3
No minimum, no maximum
y (_1, 37)
y=3x$-16˛+18≈
We have seen that some functions have extreme values, whereas others do not. The following theorem gives conditions under which a function is guaranteed to possess extreme values.
(1, 5) _1
1 x 4
1
2
3
4
5
x
3 THE EXTREME VALUE THEOREM If f is continuous on a closed interval a, b , then f attains an absolute maximum value f c and an absolute minimum value f d at some numbers c and d in a, b .
(3, _27)
The Extreme Value Theorem is illustrated in Figure 5. Note that an extreme value can be taken on more than once. Although the Extreme Value Theorem is intuitively very plausible, it is difficult to prove and so we omit the proof.
FIGURE 4
y
FIGURE 5
0
y
y
a
c
d b
x
0
a
c
d=b
x
0
a c¡
d
c™ b
x
200
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
Figures 6 and 7 show that a function need not possess extreme values if either hypothesis (continuity or closed interval) is omitted from the Extreme Value Theorem. y
y
3
g 1
0
y {c, f(c)}
{d, f(d)} 0
c
d
x
FIGURE 8
Fermat’s Theorem is named after Pierre Fermat (1601–1665), a French lawyer who took up mathematics as a hobby. Despite his amateur status, Fermat was one of the two inventors of analytic geometry (Descartes was the other). His methods for finding tangents to curves and maximum and minimum values (before the invention of limits and derivatives) made him a forerunner of Newton in the creation of differential calculus. ■
f
1
2
x
0
2
x
FIGURE 6
FIGURE 7
This function has minimum value f(2)=0, but no maximum value.
This continuous function g has no maximum or minimum.
The function f whose graph is shown in Figure 6 is defined on the closed interval [0, 2] but has no maximum value. [Notice that the range of f is [0, 3). The function takes on values arbitrarily close to 3, but never actually attains the value 3.] This does not contradict the Extreme Value Theorem because f is not continuous. [Nonetheless, a discontinuous function could have maximum and minimum values. See Exercise 13(b).] The function t shown in Figure 7 is continuous on the open interval (0, 2) but has neither a maximum nor a minimum value. [The range of t is 1, . The function takes on arbitrarily large values.] This does not contradict the Extreme Value Theorem because the interval (0, 2) is not closed. The Extreme Value Theorem says that a continuous function on a closed interval has a maximum value and a minimum value, but it does not tell us how to find these extreme values. We start by looking for local extreme values. Figure 8 shows the graph of a function f with a local maximum at c and a local minimum at d . It appears that at the maximum and minimum points the tangent lines are horizontal and therefore each has slope 0. We know that the derivative is the slope of the tangent line, so it appears that f c 0 and f d 0. The following theorem says that this is always true for differentiable functions. 4 FERMAT’S THEOREM If f has a local maximum or minimum at c, and if f c exists, then f c 0.
PROOF Suppose, for the sake of definiteness, that f has a local maximum at c. Then, according to Definition 2, f c f x if x is sufficiently close to c. This implies that if h is sufficiently close to 0, with h being positive or negative, then
f c f c h and therefore 5
f c h f c 0
SECTION 4.1
MAXIMUM AND MINIMUM VALUES
■
201
We can divide both sides of an inequality by a positive number. Thus if h 0 and h is sufficiently small, we have f c h f c
0 h Taking the right-hand limit of both sides of this inequality (using Theorem 1.4.3), we get lim
h l0
f c h f c
lim 0 0 h l0 h
But since f c exists, we have f c lim
hl0
f c h f c f c h f c lim h l0 h h
and so we have shown that f c 0. If h 0, then the direction of the inequality (5) is reversed when we divide by h : f c h f c 0 h
h0
So, taking the left-hand limit, we have f c lim
hl0
y
We have shown that f c 0 and also that f c 0. Since both of these inequalities must be true, the only possibility is that f c 0. We have proved Fermat’s Theorem for the case of a local maximum. The case of a local minimum can be proved in a similar manner, or we could use Exercise 62 to deduce it from the case we have just proved (see Exercise 63). ■
y=˛
0
x
FIGURE 9
If ƒ=˛, then fª(0)=0 but ƒ has no maximum or minimum.
|
y
y=| x| 0
x
FIGURE 10
If ƒ=| x |, then f(0)=0 is a minimum value, but fª(0) does not exist.
f c h f c f c h f c lim 0 h l0 h h
Although Fermat’s Theorem is very useful, we have to guard against reading too much into it. If f x x 3, then f x 3x 2, so f 0 0. But f has no maximum or minimum at 0, as you can see from its graph in Figure 9. The fact that f 0 0 simply means that the curve y x 3 has a horizontal tangent at 0, 0. Instead of having a maximum or minimum at 0, 0, the curve crosses its horizontal tangent there. Thus, when f c 0, f doesn’t necessarily have a maximum or minimum at c. (In other words, the converse of Fermat’s Theorem is false in general.) We should bear in mind that there may be an extreme value where f c does not exist. For instance, the function f x x has its (local and absolute) minimum value at 0 (see Figure 10), but that value cannot be found by setting f x 0 because, as was shown in Example 5 in Section 2.2, f 0 does not exist. Fermat’s Theorem does suggest that we should at least start looking for extreme values of f at the numbers c where f c 0 or where f c does not exist. Such numbers are given a special name.
202
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
6 DEFINITION A critical number of a function f is a number c in the domain of f such that either f c 0 or f c does not exist.
Figure 11 shows a graph of the function f in Example 5. It supports our answer because there is a horizontal tangent when x 1.5 and a vertical tangent when x 0 . ■
V EXAMPLE 5
Find the critical numbers of f x x 354 x.
SOLUTION The Product Rule gives
f x x 351 35 x254 x x 35
3.5
_0.5
5
_2
FIGURE 11
34 x 5x 2 5
5x 34 x 12 8x 5x 25 5x 25
[The same result could be obtained by first writing f x 4x 35 x 85.] Therefore, f x 0 if 12 8x 0, that is, x 32 , and f x does not exist when x 0. Thus the critical numbers are 32 and 0. ■ In terms of critical numbers, Fermat’s Theorem can be rephrased as follows (compare Definition 6 with Theorem 4): 7
If f has a local maximum or minimum at c, then c is a critical number
of f. To find an absolute maximum or minimum of a continuous function on a closed interval, we note that either it is local [in which case it occurs at a critical number by (7)] or it occurs at an endpoint of the interval. Thus the following three-step procedure always works. THE CLOSED INTERVAL METHOD To find the absolute maximum and mini-
mum values of a continuous function f on a closed interval a, b : 1. Find the values of f at the critical numbers of f in a, b. 2. Find the values of f at the endpoints of the interval. 3. The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.
V EXAMPLE 6
Find the absolute maximum and minimum values of the function f x x 3 3x 2 1
12 x 4
SOLUTION Since f is continuous on [2 , 4], we can use the Closed Interval 1
Method: f x x 3 3x 2 1 f x 3x 2 6x 3xx 2 Since f x exists for all x, the only critical numbers of f occur when f x 0, that is, x 0 or x 2. Notice that each of these critical numbers lies in the interval
SECTION 4.1
y
(12 , 4). The values of
MAXIMUM AND MINIMUM VALUES
■
203
f at these critical numbers are
y=˛-3≈+1
20
f 0 1
(4, 17)
f 2 3
15
The values of f at the endpoints of the interval are
10
f (12 ) 18
5 1 _1 0 _5
3
Comparing these four numbers, we see that the absolute maximum value is f 4 17 and the absolute minimum value is f 2 3. Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in Figure 12. ■
x
4
(2, _3)
FIGURE 12
4.1
EXERCISES
1. Explain the difference between an absolute minimum and a
■ Use the graph to state the absolute and local maximum and minimum values of the function.
5–6
local minimum.
5.
2. Suppose f is a continuous function defined on a closed
y
interval a, b . (a) What theorem guarantees the existence of an absolute maximum value and an absolute minimum value for f ?
y=ƒ
1
(b) What steps would you take to find those maximum and minimum values?
0
6.
For each of the numbers a, b, c, d, e, r, s, and t, state whether the function whose graph is shown has an absolute maximum or minimum, a local maximum or minimum, or neither a maximum nor a minimum.
3– 4
3.
f 4 17
2
■
x
1
y
y=ƒ
1
y
0 ■
■
x
1 ■
■
■
■
■
■
■
■
■
■
■ Sketch the graph of a function f that is continuous on [1, 5] and has the given properties.
7–10
7. Absolute minimum at 2, absolute maximum at 3, 0
a
b
c
d
e
r
s
local minimum at 4
x
t
8. Absolute minimum at 1, absolute maximum at 5,
local maximum at 2, local minimum at 4 4.
y
9. Absolute maximum at 5, absolute minimum at 2,
local maximum at 3, local minima at 2 and 4 10. f has no local maximum or minimum, but 2 and 4 are
critical numbers ■
■
■
a
b ■
c d ■
■
e ■
■
r ■
s ■
x ■
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11. (a) Sketch the graph of a function that has a local maximum
t 0
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■
at 2 and is differentiable at 2. (b) Sketch the graph of a function that has a local maximum at 2 and is continuous but not differentiable at 2.
204
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(c) Sketch the graph of a function that has a local maximum at 2 and is not continuous at 2. 12. (a) Sketch the graph of a function on [1, 2] that has an
1, 2
43. f t t s4 t ,
1, 2
x 44. f x 2 , x 4
47. f x xe
Sketch the graph of f by hand and use your sketch to find the absolute and local maximum and minimum values of f . (Use the graphs and transformations of Section 1.2.)
16. f x 3 2x,
x 5
17. f x x 2, 18. f x e
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■
0x2
20. f tan ,
4 2
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of f x x a1 x b , 0 x 1.
; 50. Use a graph to estimate the critical numbers of
f x x 3 3x 2 2 correct to one decimal place. ■
52. f x e x x,
1 x 1
1 x 0
53. f x x sx x 2 54. f x x 2 cos x, ■
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2 x 0 ■
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23. f x 5x 4x
24. f x x x x
meters) of 1 kg of water at a temperature T is given approximately by the formula
25. f x x 3x 24x
26. f x x 3 x 2 x
V 999.87 0.06426T 0.0085043T 2 0.0000679T 3
27. st 3t 4 4t 3 6t 2
28. tt 3t 4
3
3
29. t y
2
31. Fx x
x 4
45
2
y1 y2 y 1
30. h p
p1 p2 4 3
34. t 4 tan
35. f x x ln x
36. f x xe 2x
■
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■
56. An object with weight W is dragged along a horizontal
plane by a force acting along a rope attached to the object. If the rope makes an angle with the plane, then the magnitude of the force is
2
33. f 2 cos sin2
■
Find the temperature at which water has its maximum density.
32. Gx sx x
2
■
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■
F ■
■
Find the absolute maximum and absolute minimum values of f on the given interval.
37– 48
■
37. f x 3x 2 12x 5, 38. f x x 3 3x 1,
0, 3
0, 3
■
55. Between 0 C and 30 C, the volume V (in cubic centi-
Find the critical numbers of the function. 2
■
49. If a and b are positive numbers, find the maximum value
3
if 2 x 0 if 0 x 2 ■
■
51. f x x 5 x 3 2,
21. f x 1 sx
4 x2 22. f x 2x 1
■
1 2
(a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. (b) Use calculus to find the exact maximum and minimum values.
x
2 2
, 1, 4
[ , 2]
; 51–54
19. f sin ,
x 28
48. f x x ln x,
■
x1
0, 3 ,
46. f x x 2 cos x,
maxima, one local minimum, and no absolute minimum. (b) Sketch the graph of a function that has three local minima, two local maxima, and seven critical numbers.
15. f x 8 3x,
0, 3
45. f x sin x cos x,
14. (a) Sketch the graph of a function that has two local
23–36
42. f x x 1 , 3
2
absolute maximum but no absolute minimum. (b) Sketch the graph of a function on [1, 2] that is discontinuous but has both an absolute maximum and an absolute minimum.
2, 3
2
2
2, 3
1, 4
2
41. f x x 2x 3,
13. (a) Sketch the graph of a function on [1, 2] that has an
■
40. f x x 6x 9x 2, 3 4
absolute maximum but no local maximum. (b) Sketch the graph of a function on [1, 2] that has a local maximum but no absolute maximum.
15–22
39. f x 2x 3 3x 2 12x 1,
■
W sin cos
where is a positive constant called the coefficient of friction and where 0 2. Show that F is minimized when tan . 57. A model for the food-price index (the price of a representa-
tive “basket” of foods) between 1984 and 1994 is given by
SECTION 4.2
the function It 0.00009045t 5 0.001438t 4 0.06561t 3 0.4598t 2 0.6270t 99.33 where t is measured in years since midyear 1984, so 0 t 10, and It is measured in 1987 dollars and scaled such that I3 100. Estimate the times when food was cheapest and most expensive during the period 1984 –1994. 58. The Hubble Space Telescope was deployed April 24, 1990,
by the space shuttle Discovery. A model for the velocity of the shuttle during this mission, from liftoff at t 0 until the solid rocket boosters were jettisoned at t 126 s , is given by v t 0.001302t 3 0.09029t 2 23.61t 3.083
(in feet per second). Using this model, estimate the absolute maximum and minimum values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters. 59. When a foreign object lodged in the trachea (windpipe)
forces a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. This is accompanied by a contraction of the trachea, making a narrower channel for the expelled air to flow through. For a given amount of air to escape in a fixed time, it must move faster through the narrower channel than the wider one. The greater the velocity of the airstream, the greater the force on the foreign object. X rays show that the radius of the circular tracheal tube contracts to about two-thirds of its normal radius during a cough. According to a mathematical model
4.2
THE MEAN VALUE THEOREM
■
205
of coughing, the velocity v of the airstream is related to the radius r of the trachea by the equation vr kr0 rr 2
r r r0
1 2 0
where k is a constant and r0 is the normal radius of the trachea. The restriction on r is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than 1 2 r0 is prevented (otherwise the person would suffocate). (a) Determine the value of r in the interval 12 r0 , r0 at which v has an absolute maximum. How does this compare with experimental evidence? (b) What is the absolute maximum value of v on the interval? (c) Sketch the graph of v on the interval 0, r0 .
[
]
60. Show that 5 is a critical number of the function
tx 2 x 5 3 but t does not have a local extreme value at 5. 61. Prove that the function f x x 101 x 51 x 1 has
neither a local maximum nor a local minimum. 62. If f has a local minimum value at c, show that the function
tx f x has a local maximum value at c. 63. Prove Fermat’s Theorem for the case in which f has a local
minimum at c. 64. A cubic function is a polynomial of degree 3; that is, it has
the form f x ax 3 bx 2 cx d, where a 0. (a) Show that a cubic function can have two, one, or no critical number(s). Give examples and sketches to illustrate the three possibilities. (b) How many local extreme values can a cubic function have?
THE MEAN VALUE THEOREM We will see that many of the results of this chapter depend on one central fact, which is called the Mean Value Theorem. But to arrive at the Mean Value Theorem we first need the following result.
Rolle’s Theorem was first published in 1691 by the French mathematician Michel Rolle (1652–1719) in a book entitled Méthode pour résoudre les égalitéz. Later, however, he became a vocal critic of the methods of his day and attacked calculus as being a “collection of ingenious fallacies.” ■
ROLLE’S THEOREM Let f be a function that satisfies the following three
hypotheses: 1. f is continuous on the closed interval a, b . 2. f is differentiable on the open interval a, b. 3. f a f b
Then there is a number c in a, b such that f c 0.
Before giving the proof let’s take a look at the graphs of some typical functions that satisfy the three hypotheses. Figure 1 shows the graphs of four such functions. In each
206
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
case it appears that there is at least one point c, f c on the graph where the tangent is horizontal and therefore f c 0. Thus Rolle’s Theorem is plausible. y
0
y
a
c¡
c™ b
0
x
(a)
y
y
a
c
b
x
(b)
0
a
c¡
(c)
c™
b
x
0
a
c
b
x
(d)
FIGURE 1
PROOF There are three cases: CASE I f ( x) k, a constant Then f x 0, so the number c can be taken to be any number in a, b. CASE II f ( x) > f (a) for some x in (a, b) [as in Figure 1(b) or (c)] By the Extreme Value Theorem (which we can apply by hypothesis 1), f has a maximum value somewhere in a, b . Since f a f b, it must attain this maximum value at a number c in the open interval a, b. Then f has a local maximum at c and, by hypothesis 2, f is differentiable at c. Therefore, f c 0 by Fermat’s Theorem. CASE III f ( x) < f (a) for some x in (a, b) [as in Figure 1(c) or (d)] By the Extreme Value Theorem, f has a minimum value in a, b and, since f a f b, it attains this minimum value at a number c in a, b. Again f c 0 by Fermat’s Theorem. ■
EXAMPLE 1 Let’s apply Rolle’s Theorem to the position function s f t of a
moving object. If the object is in the same place at two different instants t a and t b, then f a f b. Rolle’s Theorem says that there is some instant of time t c between a and b when f c 0; that is, the velocity is 0. (In particular, you can see that this is true when a ball is thrown directly upward.) ■
EXAMPLE 2 Prove that the equation x 3 x 1 0 has exactly one real root. ■ Figure 2 shows a graph of the function f x x 3 x 1 discussed in Example 2. Rolle’s Theorem shows that, no matter how much we enlarge the viewing rectangle, we can never find a second x-intercept.
3
_2
2
SOLUTION First we use the Intermediate Value Theorem (1.5.9) to show that a root exists. Let f x x 3 x 1. Then f 0 1 0 and f 1 1 0. Since f is a polynomial, it is continuous, so the Intermediate Value Theorem states that there is a number c between 0 and 1 such that f c 0. Thus the given equation has a root. To show that the equation has no other real root, we use Rolle’s Theorem and argue by contradiction. Suppose that it had two roots a and b. Then f a 0 f b and, since f is a polynomial, it is differentiable on a, b and continuous on a, b . Thus by Rolle’s Theorem there is a number c between a and b such that f c 0. But f x 3x 2 1 1 for all x
(since x 2 0 ) so f x can never be 0. This gives a contradiction. Therefore, the equation can’t have two real roots. _3
FIGURE 2
■
Our main use of Rolle’s Theorem is in proving the following important theorem, which was first stated by another French mathematician, Joseph-Louis Lagrange.
SECTION 4.2
■
THE MEAN VALUE THEOREM
207
THE MEAN VALUE THEOREM Let f be a function that satisfies the following
hypotheses: The Mean Value Theorem is an example of what is called an existence theorem. Like the Intermediate Value Theorem, the Extreme Value Theorem, and Rolle’s Theorem, it guarantees that there exists a number with a certain property, but it doesn’t tell us how to find the number. ■
1. f is continuous on the closed interval a, b . 2. f is differentiable on the open interval a, b.
Then there is a number c in a, b such that f c
1
f b f a ba
or, equivalently, f b f a f cb a
2
Before proving this theorem, we can see that it is reasonable by interpreting it geometrically. Figures 3 and 4 show the points Aa, f a and Bb, f b on the graphs of two differentiable functions. The slope of the secant line AB is mAB
3
f b f a ba
which is the same expression as on the right side of Equation 1. Since f c is the slope of the tangent line at the point c, f c, the Mean Value Theorem, in the form given by Equation 1, says that there is at least one point Pc, f c on the graph where the slope of the tangent line is the same as the slope of the secant line AB. In other words, there is a point P where the tangent line is parallel to the secant line AB. y
y
P¡
P { c, f(c)}
B
P™
A
A{ a, f(a)} B { b, f(b)} 0
a
c
b
x
FIGURE 3 y
0
a
c¡
c™
b
x
FIGURE 4
PROOF We apply Rolle’s Theorem to a new function h defined as the difference y=ƒ h (x)
A
between f and the function whose graph is the secant line AB. Using Equation 3, we see that the equation of the line AB can be written as
ƒ B 0
x f(b)-f(a) f(a)+ (x-a) b-a
FIGURE 5
y f a
f b f a x a ba
y f a
f b f a x a ba
x
or as So, as shown in Figure 5,
208
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
hx f x f a
4
f b f a x a ba
First we must verify that h satisfies the three hypotheses of Rolle’s Theorem. 1. The function h is continuous on a, b because it is the sum of f and a first-
The Mean Value Theorem was first formulated by Joseph-Louis Lagrange (1736–1813), born in Italy of a French father and an Italian mother. He was a child prodigy and became a professor in Turin at the tender age of 19. Lagrange made great contributions to number theory, theory of functions, theory of equations, and analytical and celestial mechanics. In particular, he applied calculus to the analysis of the stability of the solar system. At the invitation of Frederick the Great, he succeeded Euler at the Berlin Academy and, when Frederick died, Lagrange accepted King Louis XVI’s invitation to Paris, where he was given apartments in the Louvre. Despite all the trappings of luxury and fame, he was a kind and quiet man, living only for science. ■
degree polynomial, both of which are continuous. 2. The function h is differentiable on a, b because both f and the first-degree polynomial are differentiable. In fact, we can compute h directly from Equation 4: f b f a hx f x ba (Note that f a and f b f a b a are constants.) 3.
ha f a f a
f b f a a a 0 ba
hb f b f a
f b f a b a ba
f b f a f b f a 0 Therefore, ha hb. Since h satisfies the hypotheses of Rolle’s Theorem, that theorem says there is a number c in a, b such that hc 0. Therefore 0 hc f c and so
f c
f b f a ba
f b f a ba
■
To illustrate the Mean Value Theorem with a specific function, let’s consider f x x 3 x, a 0, b 2. Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on 0, 2 and differentiable on 0, 2. Therefore, by the Mean Value Theorem, there is a number c in 0, 2 such that V EXAMPLE 3
y
y=˛- x B
f 2 f 0 f c2 0 Now f 2 6, f 0 0, and f x 3x 2 1, so this equation becomes 6 3c 2 12 6c 2 2
O c
FIGURE 6
2
x
which gives c 2 43 , that is, c 2s3 . But c must lie in 0, 2, so c 2s3 . Figure 6 illustrates this calculation: The tangent line at this value of c is parallel to the secant line OB. ■ V EXAMPLE 4 If an object moves in a straight line with position function s f t, then the average velocity between t a and t b is
f b f a ba
SECTION 4.2
THE MEAN VALUE THEOREM
■
209
and the velocity at t c is f c. Thus the Mean Value Theorem (in the form of Equation 1) tells us that at some time t c between a and b the instantaneous velocity f c is equal to that average velocity. For instance, if a car traveled 180 km in 2 hours, then the speedometer must have read 90 kmh at least once. In general, the Mean Value Theorem can be interpreted as saying that there is a number at which the instantaneous rate of change is equal to the average rate of change over an interval. ■ The main significance of the Mean Value Theorem is that it enables us to obtain information about a function from information about its derivative. The next example provides an instance of this principle. V EXAMPLE 5 Suppose that f 0 3 and f x 5 for all values of x. How large can f 2 possibly be?
SOLUTION We are given that f is differentiable (and therefore continuous) every-
where. In particular, we can apply the Mean Value Theorem on the interval 0, 2 . There exists a number c such that f 2 f 0 f c2 0 f 2 f 0 2f c 3 2f c
so
We are given that f x 5 for all x, so in particular we know that f c 5. Multiplying both sides of this inequality by 2, we have 2f c 10, so f 2 3 2f c 3 10 7 The largest possible value for f 2 is 7.
■
The Mean Value Theorem can be used to establish some of the basic facts of differential calculus. One of these basic facts is the following theorem. Others will be found in the following sections. 5 THEOREM
on a, b.
If f x 0 for all x in an interval a, b, then f is constant
PROOF Let x 1 and x 2 be any two numbers in a, b with x 1 x 2 . Since f is differentiable on a, b, it must be differentiable on x 1, x 2 and continuous on x 1, x 2 . By applying the Mean Value Theorem to f on the interval x 1, x 2 , we get a number c such that x 1 c x 2 and 6
f x 2 f x 1 f cx 2 x 1
Since f x 0 for all x, we have f c 0, and so Equation 6 becomes f x 2 f x 1 0
or
f x 2 f x 1
Therefore, f has the same value at any two numbers x 1 and x 2 in a, b. This means ■ that f is constant on a, b.
210
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
7 COROLLARY If f x tx for all x in an interval a, b, then f t is constant on a, b; that is, f x tx c where c is a constant.
PROOF Let Fx f x tx. Then
Fx f x tx 0 for all x in a, b. Thus, by Theorem 5, F is constant; that is, f t is constant.
■
NOTE Care must be taken in applying Theorem 5. Let
f x
x 1 x 1
if x 0 if x 0
The domain of f is D x x 0 and f x 0 for all x in D. But f is obviously not a constant function. This does not contradict Theorem 5 because D is not an interval. Notice that f is constant on the interval 0, and also on the interval , 0. We will make extensive use of Theorem 5 and Corollary 7 when we study antiderivatives in Section 4.7.
4.2
EXERCISES
Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem.
1– 4
1. f x x 2 4x 1, 2
3. f x sin 2 x, ■
■
5. Let f x 1 x
; 9. (a) Graph the function f x x 4x in the viewing rect 0, 2
1, 1
4. f x x sx 6 , ■
values of c that satisfy the conclusion of the Mean Value Theorem for the interval 1, 7 .
0, 4
2. f x x 3x 2x 5, 3
■
8. Use the graph of f given in Exercise 7 to estimate the
■
6, 0 ■
■
■
■
■
■
■
■
. Show that f 1 f 1 but there is no number c in 1, 1 such that f c 0. Why does this not contradict Rolle’s Theorem? 23
6. Let f x x 12. Show that f 0 f 2 but there is no
number c in 0, 2 such that f c 0. Why does this not contradict Rolle’s Theorem?
7. Use the graph of f to estimate the values of c that satisfy
the conclusion of the Mean Value Theorem for the interval 0, 8 . y
angle 0, 10 by 0, 10 . (b) Graph the secant line that passes through the points 1, 5 and 8, 8.5 on the same screen with f . (c) Find the number c that satisfies the conclusion of the Mean Value Theorem for this function f and the interval 1, 8 . Then graph the tangent line at the point c, f c and notice that it is parallel to the secant line.
; 10. (a) In the viewing rectangle 3, 3 by 5, 5 , graph the
function f x x 3 2 x and its secant line through the points 2, 4 and 2, 4. Use the graph to estimate the x-coordinates of the points where the tangent line is parallel to the secant line. (b) Find the exact values of the numbers c that satisfy the conclusion of the Mean Value Theorem for the interval 2, 2 and compare with your answers to part (a).
■ Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers c that satisfy the conclusion of the Mean Value Theorem.
11–14 y =ƒ
11. f x 3x 2 2x 5, 12. f x x 3 x 1,
1 0
1
x
13. f x e2x,
0, 3
1, 1
0, 2
SECTION 4.3
14. f x ■
■
x , x2 ■
211
1
■
■
■
■
■
■
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15. Let f x x 1 . Show that there is no value of c such
that f 3 f 0 f c3 0. Why does this not contradict the Mean Value Theorem?
16. Let f x x 1x 1. Show that there is no value of
c such that f 2 f 0 f c2 0. Why does this not contradict the Mean Value Theorem?
17. Show that the equation 1 2x x 4x 0 has exactly 3
5
28. Suppose f is an odd function and is differentiable every-
where. Prove that for every positive number b, there exists a number c in b, b such that f c f bb. 29. Use the Mean Value Theorem to prove the inequality
sin a sin b a b
for all a and b
30. If f x c (c a constant) for all x, use Corollary 7 to show
that f x cx d for some constant d.
31. Let f x 1x and
one real root. 18. Show that the equation 2x 1 sin x 0 has exactly one
real root. 19. Show that the equation x 15x c 0 has at most one 3
tx
20. Show that the equation x 4 4x c 0 has at most two
real roots. 21. (a) Show that a polynomial of degree 3 has at most three
real roots. (b) Show that a polynomial of degree n has at most n real roots. 22. (a) Suppose that f is differentiable on ⺢ and has two roots.
Show that f has at least one root. (b) Suppose f is twice differentiable on ⺢ and has three roots. Show that f has at least one real root. (c) Can you generalize parts (a) and (b)?
23. If f 1 10 and f x 2 for 1 x 4, how small can
f 4 possibly be?
24. Suppose that 3 f x 5 for all values of x. Show that
18 f 8 f 2 30.
1 x 1
root in the interval 2, 2 .
if x 0 1 x
if x 0
Show that f x tx for all x in their domains. Can we conclude from Corollary 7 that f t is constant? 32. Use Theorem 5 to prove the identity
2 sin1x cos11 2x 2
x0
33. Prove the identity
arcsin
x1 2 arctan sx x1 2
34. At 2:00 PM a car’s speedometer reads 30 mih. At 2:10 PM
it reads 50 mih. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mih2. 35. Two runners start a race at the same time and finish in a tie.
25. Does there exist a function f such that f 0 1,
f 2 4, and f x 2 for all x ?
26. Suppose that f and t are continuous on a, b and differ-
entiable on a, b. Suppose also that f a ta and f x tx for a x b. Prove that f b tb. [Hint: Apply the Mean Value Theorem to the function h f t.]
4.3
■
27. Show that s1 x 1 2 x if x 0.
1, 4
■
DERIVATIVES AND THE SHAPES OF GRAPHS
Prove that at some time during the race they have the same speed. [Hint: Consider f t tt ht, where t and h are the position functions of the two runners.] 36. A number a is called a fixed point of a function f if
f a a. Prove that if f x 1 for all real numbers x, then f has at most one fixed point.
DERIVATIVES AND THE SHAPES OF GRAPHS Many of the applications of calculus depend on our ability to deduce facts about a function f from information concerning its derivatives. Because f x represents the slope of the curve y f x at the point x, f x, it tells us the direction in which the curve proceeds at each point. So it is reasonable to expect that information about f x will provide us with information about f x.
212
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
y
WHAT DOES f SAY ABOUT f ?
D B
C
A
x
0
To see how the derivative of f can tell us where a function is increasing or decreasing, look at Figure 1. (Increasing functions and decreasing functions were defined in Section 1.1.) Between A and B and between C and D, the tangent lines have positive slope and so f x 0. Between B and C, the tangent lines have negative slope and so f x 0. Thus it appears that f increases when f x is positive and decreases when f x is negative. To prove that this is always the case, we use the Mean Value Theorem. INCREASING/ DECREASING TEST
FIGURE 1 Let’s abbreviate the name of this test to the I/D Test. ■
(a) If f x 0 on an interval, then f is increasing on that interval. (b) If f x 0 on an interval, then f is decreasing on that interval. PROOF
(a) Let x 1 and x 2 be any two numbers in the interval with x1 x2 . According to the definition of an increasing function (page 7) we have to show that f x1 f x2 . Because we are given that f x 0, we know that f is differentiable on x1, x2 . So, by the Mean Value Theorem there is a number c between x1 and x2 such that f x 2 f x 1 f cx 2 x 1
1
Now f c 0 by assumption and x 2 x 1 0 because x 1 x 2 . Thus the right side of Equation 1 is positive, and so f x 2 f x 1 0
or
f x 1 f x 2
This shows that f is increasing. Part (b) is proved similarly.
■
Find where the function f x 3x 4 4x 3 12x 2 5 is increasing and where it is decreasing. V EXAMPLE 1
SOLUTION
To use the ID Test we have to know where f x 0 and where f x 0. This depends on the signs of the three factors of f x, namely, 12x, x 2, and x 1. We divide the real line into intervals whose endpoints are the critical numbers 1, 0, and 2 and arrange our work in a chart. A plus sign indicates that the given expression is positive, and a minus sign indicates that it is negative. The last column of the chart gives the conclusion based on the ID Test. For instance, f x 0 for 0 x 2, so f is decreasing on (0, 2). (It would also be true to say that f is decreasing on the closed interval 0, 2 .)
20
_2
3
_30
FIGURE 2
f x 12x 3 12x 2 24x 12xx 2x 1
Interval
12x
x2
x1
f x
x 1 1 x 0 0x2 x2
f decreasing on ( , 1) increasing on (1, 0) decreasing on (0, 2) increasing on (2, )
The graph of f shown in Figure 2 confirms the information in the chart.
■
SECTION 4.3
DERIVATIVES AND THE SHAPES OF GRAPHS
■
213
Recall from Section 4.1 that if f has a local maximum or minimum at c, then c must be a critical number of f (by Fermat’s Theorem), but not every critical number gives rise to a maximum or a minimum. We therefore need a test that will tell us whether or not f has a local maximum or minimum at a critical number. You can see from Figure 2 that f 0 5 is a local maximum value of f because f increases on 1, 0 and decreases on 0, 2. Or, in terms of derivatives, f x 0 for 1 x 0 and f x 0 for 0 x 2. In other words, the sign of f x changes from positive to negative at 0. This observation is the basis of the following test. THE FIRST DERIVATIVE TEST Suppose that c is a critical number of a continu-
ous function f . (a) If f changes from positive to negative at c, then f has a local maximum at c. (b) If f changes from negative to positive at c, then f has a local minimum at c. (c) If f does not change sign at c (that is, f is positive on both sides of c or negative on both sides), then f has no local maximum or minimum at c. The First Derivative Test is a consequence of the ID Test. In part (a), for instance, since the sign of f x changes from positive to negative at c, f is increasing to the left of c and decreasing to the right of c. It follows that f has a local maximum at c. It is easy to remember the First Derivative Test by visualizing diagrams such as those in Figure 3. y
y
y
y
fª(x)<0 fª(x)>0
fª(x)<0
fª(x)>0 fª(x)<0
0
c
(a) Local maximum
x
0
fª(x)<0
fª(x)>0
c
(b) Local minimum
fª(x)>0 x
0
c
x
(c) No maximum or minimum
0
c
x
(d) No maximum or minimum
FIGURE 3 V EXAMPLE 2
Find the local minimum and maximum values of the function f in
Example 1. SOLUTION From the chart in the solution to Example 1 we see that f x changes from negative to positive at 1, so f 1 0 is a local minimum value by the First Derivative Test. Similarly, f changes from negative to positive at 2, so f 2 27 is also a local minimum value. As previously noted, f 0 5 is a local maximum value because f x changes from positive to negative at 0. ■
EXAMPLE 3 Find the local maximum and minimum values of the function
tx x 2 sin x
0 x 2
SOLUTION To find the critical numbers of t, we differentiate:
tx 1 2 cos x
214
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
So tx 0 when cos x 12 . The solutions of this equation are 23 and 43. Because t is differentiable everywhere, the only critical numbers are 23 and 43 and so we analyze t in the following table.
The + signs in the table come from the fact that tx 0 when cos x 21 . From the graph of y cos x , this is true in the indicated intervals.
Interval
tx 1 2 cos x
t
0 x 23 23 x 43 43 x 2
increasing on (0, 23) decreasing on (23, 43) increasing on (43, 2)
■
Because tx changes from positive to negative at 23, the First Derivative Test tells us that there is a local maximum at 23 and the local maximum value is 6
t23
2 2 2 s3 2 sin 2 3 3 3 2
2 s3 3.83 3
Likewise, tx changes from negative to positive at 43 and so 2π
0
t43
4 4 4 s3 2 sin 2 3 3 3 2
4 s3 2.46 3
FIGURE 4
is a local minimum value. The graph of t in Figure 4 supports our conclusion.
y=x+2 sin x
■
WHAT DOES f SAY ABOUT f ?
Figure 5 shows the graphs of two increasing functions on a, b. Both graphs join point A to point B but they look different because they bend in different directions. How can we distinguish between these two types of behavior? In Figure 6 tangents to these curves have been drawn at several points. In (a) the curve lies above the tangents and f is called concave upward on a, b. In (b) the curve lies below the tangents and t is called concave downward on a, b. y
y
B
a
b
(a) FIGURE 5
x
0
g A
A a
b
(b)
B
f
A
A
y
B
g
f
0
y
B
x
0
x
(a) Concave upward
0
x
(b) Concave downward
FIGURE 6
DEFINITION If the graph of f lies above all of its tangents on an interval I , then it is called concave upward on I . If the graph of f lies below all of its tangents on I, it is called concave downward on I .
SECTION 4.3
■
DERIVATIVES AND THE SHAPES OF GRAPHS
215
Figure 7 shows the graph of a function that is concave upward (abbreviated CU) on the intervals b, c, d, e, and e, p and concave downward (CD) on the intervals a, b, c, d, and p, q. y
D B
0 a
b
FIGURE 7
CD
P
C
c
CU
d
CD
e
CU
p
CU
q
x
CD
DEFINITION A point P on a curve y f x is called an inflection point if f is continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at P.
For instance, in Figure 7, B, C, D, and P are the points of inflection. Notice that if a curve has a tangent at a point of inflection, then the curve crosses its tangent there. Let’s see how the second derivative helps determine the intervals of concavity. Looking at Figure 6(a), you can see that, going from left to right, the slope of the tangent increases. This means that the derivative f is an increasing function and therefore its derivative f is positive. Likewise, in Figure 6(b) the slope of the tangent decreases from left to right, so f decreases and therefore f is negative. This reasoning can be reversed and suggests that the following theorem is true. A proof is given in Appendix B with the help of the Mean Value Theorem.
CONCAVITY TEST
(a) If f x 0 for all x in I , then the graph of f is concave upward on I . (b) If f x 0 for all x in I , then the graph of f is concave downward on I . In view of the Concavity Test, there is a point of inflection at any point where the second derivative changes sign. V EXAMPLE 4
Sketch a possible graph of a function f that satisfies the following
conditions: i f x 0 on , 1, f x 0 on 1, ii f x 0 on , 2 and 2, , f x 0 on 2, 2 iii lim f x 2, lim f x 0 x l
xl
SOLUTION Condition (i) tells us that f is increasing on , 1 and decreasing on
1, . Condition (ii) says that f is concave upward on , 2 and 2, , and concave downward on 2, 2. From condition (iii) we know that the graph of f has two horizontal asymptotes: y 2 and y 0.
216
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
y
0
-2
x
2
1
y=_2
We first draw the horizontal asymptote y 2 as a dashed line (see Figure 8). We then draw the graph of f approaching this asymptote at the far left, increasing to its maximum point at x 1 and decreasing toward the x-axis at the far right. We also make sure that the graph has inflection points when x 2 and 2. Notice that we made the curve bend upward for x 2 and x 2, and bend downward when x is between 2 and 2. ■ Another application of the second derivative is the following test for maximum and minimum values. It is a consequence of the Concavity Test.
FIGURE 8 y
THE SECOND DERIVATIVE TEST Suppose f is continuous near c.
f
(a) If f c 0 and f c 0, then f has a local minimum at c. (b) If f c 0 and f c 0, then f has a local maximum at c.
P ƒ
fª(c)=0
f(c) c
0
x
For instance, part (a) is true because f x 0 near c and so f is concave upward near c. This means that the graph of f lies above its horizontal tangent at c and so f has a local minimum at c. (See Figure 9.)
x
FIGURE 9 f·(c)>0, f is concave upward
Discuss the curve y x 4 4x 3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. V EXAMPLE 5
SOLUTION If f x x 4 4x 3, then
f x 4x 3 12x 2 4x 2x 3 f x 12x 2 24x 12xx 2 To find the critical numbers we set f x 0 and obtain x 0 and x 3. To use the Second Derivative Test we evaluate f at these critical numbers: f 0 0
Since f 3 0 and f 3 0, f 3 27 is a local minimum. Since f 0 0, the Second Derivative Test gives no information about the critical number 0. But since f x 0 for x 0 and also for 0 x 3, the First Derivative Test tells us that f does not have a local maximum or minimum at 0. [In fact, the expression for f x shows that f decreases to the left of 3 and increases to the right of 3.] Since f x 0 when x 0 or 2, we divide the real line into intervals with these numbers as endpoints and complete the following chart.
y
y=x$-4˛ (0, 0)
inflection points
2
3
(2, _16)
(3, _27)
FIGURE 10
f 3 36 0
x
Interval
f x 12xx 2
Concavity
( , 0) (0, 2) (2, )
upward downward upward
The point 0, 0 is an inflection point since the curve changes from concave upward to concave downward there. Also 2, 16 is an inflection point since the curve changes from concave downward to concave upward there. Using the local minimum, the intervals of concavity, and the inflection points, we ■ sketch the curve in Figure 10.
SECTION 4.3
DERIVATIVES AND THE SHAPES OF GRAPHS
■
217
NOTE The Second Derivative Test is inconclusive when f c 0. In other words, at such a point there might be a maximum, there might be a minimum, or there might be neither (as in Example 5). This test also fails when f c does not exist. In such cases the First Derivative Test must be used. In fact, even when both tests apply, the First Derivative Test is often the easier one to use.
EXAMPLE 6 Sketch the graph of the function f x x 236 x13. SOLUTION You can use the differentiation rules to check that the first two deriva-
tives are f x ■ Try reproducing the graph in Figure 11 with a graphing calculator or computer. Some machines produce the complete graph, some produce only the portion to the right of the y -axis, and some produce only the portion between x 0 and x 6 . An equivalent expression that gives the correct graph is
y x 2 13
4x x 136 x23
f x
8 x 436 x53
Since f x 0 when x 4 and f x does not exist when x 0 or x 6, the critical numbers are 0, 4, and 6.
6x 6 x 13 6 x
Interval
4x
x 13
6 x23
f x
f
x0 0x4 4x6 x6
decreasing on ( , 0) increasing on (0, 4) decreasing on (4, 6) decreasing on (6, )
y 4
To find the local extreme values we use the First Derivative Test. Since f changes from negative to positive at 0, f 0 0 is a local minimum. Since f changes from positive to negative at 4, f 4 2 53 is a local maximum. The sign of f does not change at 6, so there is no minimum or maximum there. (The Second Derivative Test could be used at 4 but not at 0 or 6 since f does not exist at either of these numbers.) Looking at the expression for f x and noting that x 43 0 for all x, we have f x 0 for x 0 and for 0 x 6 and f x 0 for x 6. So f is concave downward on , 0 and 0, 6 and concave upward on 6, , and the only inflection point is 6, 0. The graph is sketched in Figure 11. Note that the curve has vertical tangents at 0, 0 and 6, 0 because f x l as x l 0 and as x l 6. ■
(4, 2%?#)
3 2
0
1
2
3
4
7 x
5
y=x@?#(6-x)!?#
FIGURE 11
4.3 1– 8
EXERCISES 9–10 ■ Find the local maximum and minimum values of f using both the First and Second Derivative Tests. Which method do you prefer? x 9. f x x s1 x 10. f x 2 x 4
■
(a) Find the intervals on which f is increasing or decreasing. (b) Find the local maximum and minimum values of f . (c) Find the intervals of concavity and the inflection points. 1. f x x 3 12x 1
■
2. f x x 4 4x 1 3. f x x 2 sin x, 4. f x
■
7. f x ln xsx
8. f x x ln x ■
■
■
■
■
■
■
■
■
■
■
■
12. (a) Find the critical numbers of f x x 4x 13.
6. f x x 2e x ■
■
(a) If f 2 0 and f 2 5, what can you say about f ? (b) If f 6 0 and f 6 0, what can you say about f ?
x2 2 x 3
■
■
11. Suppose f is continuous on , .
0 x 3
5. f x xe x ■
■
■
■
■
■
(b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers? (c) What does the First Derivative Test tell you?
218
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION ■ The graph of the derivative f of a continuous function f is shown. (a) On what intervals is f increasing or decreasing? (b) At what values of x does f have a local maximum or minimum? (c) On what intervals is f concave upward or downward? (d) State the x-coordinate(s) of the point(s) of inflection. (e) Assuming that f 0 0, sketch a graph of f.
13. In each part state the x-coordinates of the inflection points
21–22
of f . Give reasons for your answers. (a) The curve is the graph of f . (b) The curve is the graph of f . (c) The curve is the graph of f . y
y
21. 0
2
4
6
8
x
y=fª(x) 2
14. The graph of the first derivative f of a function f is shown.
(a) On what intervals is f increasing? Explain. (b) At what values of x does f have a local maximum or minimum? Explain. (c) On what intervals is f concave upward or concave downward? Explain. (d) What are the x-coordinates of the inflection points of f ? Why?
0
6
8 x
6
8 x
4
2
_2
22.
y
y=fª(x) 2
y
y=fª(x) 0 0
1
3
5
7
■
■
(a) (b) (c) (d)
15. f x and f x are always negative 16. f x 0 for all x 1,
vertical asymptote x 1,
f x 0 if x 1 or x 3,
f x 0 if 1 x 3
f x 0 if x 0 or 2 x 4,
f x 0 if x 1,
f 2 0,
20. f x 0 if x 2,
f 2 0,
lim f x 1,
f x 0 if 0 x 3, ■
■
■
■
■
f x 0 if x 2
■
■
■
■
■
24. f x 2 3x x 3
■
■
■
27. hx 3x 5 5x 3 3
28. hx x 2 13
29. Ax x sx 3
30. Bx 3x 23 x
■
■
■
0 2
34. f t t cos t,
2 t 2
■
■
■
35– 42
f x f x, f x 0 if x 3
■
■
Find the intervals of increase or decrease. Find the local maximum and minimum values. Find the intervals of concavity and the inflection points. Use the information from parts (a)–(c) to sketch the graph. Check your work with a graphing device if you have one.
33. f 2 cos cos 2,
f x 0 if x 2,
xl
■
32. f x lnx 4 27
f x 0 if x 2,
lim f x ,
xl2
■
31. Cx x 13x 4
f x 1 if x 2,
f x 0 if 2 x 0, inflection point 0, 1 19. f x 0 if x 2,
■
26. tx 200 8x 3 x 4
f x 0 if x 1 or x 3
f x 0 if 1 x 2,
■
25. f x x 4 6x 2
f x 0 if 0 x 2 or x 4, 18. f 1 f 1 0,
■
23. f x 2x 3 3x 2 12x
17. f 0 f 2 f 4 0,
f x 0 if 1 x 3,
■
23–34
Sketch the graph of a function that satisfies all of the given conditions.
15–20
4
_2
x
9
2
■
■
■
■
■
■
■
(a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values.
■
■
■
SECTION 4.3
x2 x 1
36. f x
x2 x 22
2
f x ex
37. f x sx 2 1 x
;
2 x 2
38. f x x tan x,
40. f x
41. f x e 1x1
42. f x lntan2x
■
■
■
■
■
■
■
■
■
■
■
f x axe bx
51. Show that tan x x for 0 x 2. [Hint: Show that
f x tan x x is increasing on 0, 2.]
y x 3 3a 2x 2a 3, where a is a positive constant. What do the members of this family of curves have in common? How do they differ from each other?
52. (a) Show that e x 1 x for x 0.
(b) Deduce that e x 1 x 12 x 2 for x 0. (c) Use mathematical induction to prove that for x 0 and any positive integer n,
■
(a) Use a graph of f to estimate the maximum and minimum values. Then find the exact values. (b) Estimate the value of x at which f increases most rapidly. Then find the exact value.
■
■
ex 1 x
x2 xn 2! n!
53. Show that a cubic function (a third-degree polynomial)
x1 sx 2 1 ■
2
have the maximum value f 2 1?
44. Use the methods of this section to sketch the curve
■
(a) Find the asymptote, maximum value, and inflection points of f . (b) What role does ! play in the shape of the curve? (c) Illustrate by graphing four members of this family on the same screen.
50. For what values of the numbers a and b does the function ■
f x x 12 x 35 x 6 4 . On what interval is f increasing?
45. f x
2! 2
2
a local maximum value of 3 at 2 and a local minimum value of 0 at 1.
43. Suppose the derivative of a function f is
; 45– 46
219
49. Find a cubic function f x ax 3 bx 2 cx d that has
ex 1 ex
39. f x ln1 ln x
■
and the positive constant ! is called the standard deviation. For simplicity, let’s scale the function so as to remove the factor 1(! s2 ) and let’s analyze the special case where 0. So we study the function
(d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)–(d) to sketch the graph of f . 35. f x
DERIVATIVES AND THE SHAPES OF GRAPHS
46. f x x 2ex ■
■
■
■
■
■
■
■
; 47. For the period from 1980 to 2000, the percentage of households in the United States with at least one VCR has been modeled by the function Vt
85 1 53e0.5t
where the time t is measured in years since midyear 1980, so 0 t 20. Use a graph to estimate the time at which the number of VCRs was increasing most rapidly. Then use derivatives to give a more accurate estimate. 48. The family of bell-shaped curves
1 2 2 y ex 2! ! s2 occurs in probability and statistics, where it is called the normal density function. The constant is called the mean
always has exactly one point of inflection. If its graph has three x-intercepts x 1, x 2, and x 3, show that the x-coordinate of the inflection point is x 1 x 2 x 3 3.
; 54. For what values of c does the polynomial
Px x 4 cx 3 x 2 have two inflection points? One inflection point? None? Illustrate by graphing P for several values of c. How does the graph change as c decreases?
55. Prove that if c, f c is a point of inflection of the graph
of f and f exists in an open interval that contains c, then f c 0. [Hint: Apply the First Derivative Test and Fermat’s Theorem to the function t f .]
56. Show that if f x x 4, then f 0 0, but 0, 0 is not an
inflection point of the graph of f .
57. Show that the function tx x x has an inflection point
at 0, 0 but t0 does not exist.
58. Suppose that f is continuous and f c f c 0, but
f c 0. Does f have a local maximum or minimum at c ? Does f have a point of inflection at c ?
220
■
CHAPTER 4
4.4
APPLICATIONS OF DIFFERENTIATION
CURVE SKETCHING So far we have been concerned with some particular aspects of curve sketching: domain, range, symmetry, limits, continuity, and asymptotes in Chapter 1; derivatives and tangents in Chapters 2 and 3; l’Hospital’s Rule in Chapter 3; and extreme values, intervals of increase and decrease, concavity, and points of inflection in this chapter. It’s now time to put all of this information together to sketch graphs that reveal the important features of functions. You may ask: Why don’t we just use a graphing calculator or computer to graph a curve? Why do we need to use calculus? It’s true that modern technology is capable of producing very accurate graphs. But even the best graphing devices have to be used intelligently. The use of calculus enables us to discover the most interesting aspects of curves and to detect behavior that we might otherwise overlook. We will see in Example 4 how calculus helps us to avoid the pitfalls of technology. GUIDELINES FOR SKETCHING A CURVE
The following checklist is intended as a guide to sketching a curve y f x by hand. Not every item is relevant to every function. (For instance, a given curve might not have an asymptote or possess symmetry.) But the guidelines provide all the information you need to make a sketch that displays the most important aspects of the function. A. Domain It’s often useful to start by determining the domain D of f , that is, the set of values of x for which f x is defined. B. Intercepts The y-intercept is f 0 and this tells us where the curve intersects the y-axis. To find the x-intercepts, we set y 0 and solve for x. (You can omit this step if the equation is difficult to solve.)
y
C. Symmetry
0
x
(a) Even function: reflectional symmetry y
x
0
(b) Odd function: rotational symmetry FIGURE 1
(i) If f x f x for all x in D, that is, the equation of the curve is unchanged when x is replaced by x, then f is an even function and the curve is symmetric about the y-axis. This means that our work is cut in half. If we know what the curve looks like for x 0, then we need only reflect about the y-axis to obtain the complete curve [see Figure 1(a)]. Here are some examples: y x 2, y x 4, y x , and y cos x. (ii) If f x f x for all x in D, then f is an odd function and the curve is symmetric about the origin. Again we can obtain the complete curve if we know what it looks like for x 0. [Rotate 180° about the origin; see Figure 1(b).] Some simple examples of odd functions are y x, y x 3, y x 5, and y sin x. (iii) If f x p f x for all x in D, where p is a positive constant, then f is called a periodic function and the smallest such number p is called the period. For instance, y sin x has period 2 and y tan x has period . If we know what the graph looks like in an interval of length p, then we can use translation to sketch the entire graph (see Figure 2).
y
FIGURE 2
Periodic function: translational symmetry
a-p
0
a
a+p
a+2p
x
SECTION 4.4
CURVE SKETCHING
■
221
D. Asymptotes
(i) Horizontal Asymptotes. Recall from Section 1.6 that if lim x l f x L or lim x l f x L , then the line y L is a horizontal asymptote of the curve y f x. If it turns out that lim x l f x (or ), then we do not have an asymptote to the right, but that is still useful information for sketching the curve. (ii) Vertical Asymptotes. Recall from Section 1.6 that the line x a is a vertical asymptote if at least one of the following statements is true: lim f x
1
x la
lim f x
x la
E.
F.
G.
H.
In Module 4.4 you can practice using information about f and f to determine the shape of the graph of f .
lim f x
x la
lim f x
x la
(For rational functions you can locate the vertical asymptotes by equating the denominator to 0 after canceling any common factors. But for other functions this method does not apply.) Furthermore, in sketching the curve it is very useful to know exactly which of the statements in (1) is true. If f a is not defined but a is an endpoint of the domain of f , then you should compute lim x l a f x or lim x l a f x, whether or not this limit is infinite. Intervals of Increase or Decrease Use the I / D Test. Compute f x and find the intervals on which f x is positive ( f is increasing) and the intervals on which f x is negative ( f is decreasing). Local Maximum and Minimum Values Find the critical numbers of f [the numbers c where f c 0 or f c does not exist]. Then use the First Derivative Test. If f changes from positive to negative at a critical number c, then f c is a local maximum. If f changes from negative to positive at c, then f c is a local minimum. Although it is usually preferable to use the First Derivative Test, you can use the Second Derivative Test if f c 0 and f c 0. Then f c 0 implies that f c is a local minimum, whereas f c 0 implies that f c is a local maximum. Concavity and Points of Inflection Compute f x and use the Concavity Test. The curve is concave upward where f x 0 and concave downward where f x 0. Inflection points occur where the direction of concavity changes. Sketch the Curve Using the information in items A–G, draw the graph. Sketch the asymptotes as dashed lines. Plot the intercepts, maximum and minimum points, and inflection points. Then make the curve pass through these points, rising and falling according to E, with concavity according to G, and approaching the asymptotes. If additional accuracy is desired near any point, you can compute the value of the derivative there. The tangent indicates the direction in which the curve proceeds.
V EXAMPLE 1
Use the guidelines to sketch the curve y
2x 2 . x2 1
A. The domain is
x
x
2
1 0 x
x 1 , 1 1, 1 1,
B. The x- and y-intercepts are both 0. C. Since f x f x, the function f is even. The curve is symmetric about the
y-axis.
222
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
lim
D.
x l
Therefore, the line y 2 is a horizontal asymptote. Since the denominator is 0 when x 1, we compute the following limits:
y
y=2
lim
2x 2 x 1
lim
2x 2 x2 1
x l1
0
x
x l1
x=_1
2x 2 2 lim 2 2 x l x 1 1 1x 2
x=1
2
lim
2x 2 x 1
lim
2x 2 x2 1
x l1
x l1
2
Therefore, the lines x 1 and x 1 are vertical asymptotes. This information about limits and asymptotes enables us to draw the preliminary sketch in Figure 3, showing the parts of the curve near the asymptotes.
FIGURE 3
Preliminary sketch
f x
E. ■ We have shown the curve approaching its horizontal asymptote from above in Figure 3. This is confirmed by the intervals of increase and decrease.
4xx 2 1 2x 2 2x 4x 2 2 2 x 1 x 12
Since f x 0 when x 0 x 1 and f x 0 when x 0 x 1, f is increasing on , 1 and 1, 0 and decreasing on 0, 1 and 1, . F. The only critical number is x 0. Since f changes from positive to negative at 0, f 0 0 is a local maximum by the First Derivative Test.
y
G. y=2
f x
4x 2 12 4x 2x 2 12x 12x 2 4 x 2 14 x 2 13
Since 12x 2 4 0 for all x, we have 0
f x 0 &?
x
x=_1
x=1
FIGURE 4
Finished sketch of y=
2≈ ≈-1
x2 1 0
&?
x 1
and f x 0 &? x 1. Thus the curve is concave upward on the intervals , 1 and 1, and concave downward on 1, 1. It has no point of inflection since 1 and 1 are not in the domain of f . H. Using the information in E–G, we finish the sketch in Figure 4. ■ Sketch the graph of f x xe x. The domain is ⺢. The x- and y-intercepts are both 0. Symmetry: None Because both x and e x become large as x l , we have lim x l xe x . As x l , however, e x l 0 and so we have an indeterminate product that requires the use of l’Hospital’s Rule: x 1 lim xe x lim x lim lim e x 0 x l x l e x l ex x l
V EXAMPLE 2
A. B. C. D.
Thus the x-axis is a horizontal asymptote. E.
f x xe x e x x 1e x
Since e x is always positive, we see that f x 0 when x 1 0, and f x 0 when x 1 0. So f is increasing on 1, and decreasing on , 1. F. Because f 1 0 and f changes from negative to positive at x 1, f 1 e1 is a local (and absolute) minimum.
SECTION 4.4
y
y=x´
■
223
f x x 1e x e x x 2e x
G.
Since f x 0 if x 2 and f x 0 if x 2, f is concave upward on 2, and concave downward on , 2. The inflection point is 2, 2e2 . H. We use this information to sketch the curve in Figure 5. ■
1 _2
CURVE SKETCHING
_1 x
EXAMPLE 3 Sketch the graph of f x
(_1, _1/e)
cos x . 2 sin x
A. The domain is ⺢. 1 B. The y -intercept is f 0 2. The x -intercepts occur when cos x 0, that is,
FIGURE 5
x 2n 12, where n is an integer.
C. f is neither even nor odd, but f x 2 f x for all x and so f is periodic
and has period 2. Thus in what follows we need to consider only 0 x 2 and then extend the curve by translation in part H. D. Asymptotes: None E.
f x
2 sin xsin x cos x cos x 2 sin x 1 2 sin x 2 2 sin x 2
Thus f x 0 when 2 sin x 1 0 &? sin x 12 &? 76 x 116. So f is increasing on 76, 116 and decreasing on 0, 76 and 116, 2. F. From part E and the First Derivative Test, we see that the local minimum value is f 76 1s3 and the local maximum value is f 116 1s3 . G. If we use the Quotient Rule again and simplify, we get f x
2 cos x 1 sin x 2 sin x 3
Because 2 sin x 3 0 and 1 sin x 0 for all x , we know that f x 0 when cos x 0, that is, 2 x 32. So f is concave upward on 2, 32 and concave downward on 0, 2 and 32, 2. The inflection points are 2, 0 and 32, 0. H. The graph of the function restricted to 0 x 2 is shown in Figure 6. Then we extend it, using periodicity, to the complete graph in Figure 7. y 1 2
π 2
”
11π 1 6 , œ„ 3’
π
3π 2
y 1 2
2π x
_π
π
2π
3π
x
1 - ’ ” 7π 6 , œ„ 3
FIGURE 6
FIGURE 7
■
GRAPHING WITH TECHNOLOGY
When we use technology to graph a curve, our strategy is different from that in Examples 1–3. Here we start with a graph produced by a graphing calculator or computer and then we refine it. We use calculus to make sure that we reveal all the important features of the curve. And with the use of graphing devices we can tackle curves that would be far too complicated to consider without technology.
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EXAMPLE 4 Graph the polynomial f x 2x 6 3x 5 3x 3 2x 2. Use the graphs
41,000
of f and f to estimate all maximum and minimum points and intervals of concavity. y=ƒ
_5
5 _1000
FIGURE 8 100 y=ƒ
_3
2
SOLUTION If we specify a domain but not a range, many graphing devices will deduce a suitable range from the values computed. Figure 8 shows the plot from one such device if we specify that 5 x 5. Although this viewing rectangle is useful for showing that the asymptotic behavior (or end behavior) is the same as for y 2x 6, it is obviously hiding some finer detail. So we change to the viewing rectangle 3, 2 by 50, 100 shown in Figure 9. From this graph it appears that there is an absolute minimum value of about 15.33 when x 1.62 (by using the cursor) and f is decreasing on , 1.62 and increasing on 1.62, . Also there appears to be a horizontal tangent at the origin and inflection points when x 0 and when x is somewhere between 2 and 1. Now let’s try to confirm these impressions using calculus. We differentiate and get
f x 12x 5 15x 4 9x 2 4x f x 60x 4 60x 3 18x 4
_50
FIGURE 9
When we graph f in Figure 10 we see that f x changes from negative to positive when x 1.62; this confirms (by the First Derivative Test) the minimum value that we found earlier. But, perhaps to our surprise, we also notice that f x changes from positive to negative when x 0 and from negative to positive when x 0.35. This means that f has a local maximum at 0 and a local minimum when x 0.35, but these were hidden in Figure 9. Indeed, if we now zoom in toward the origin in Figure 11, we see what we missed before: a local maximum value of 0 when x 0 and a local minimum value of about 0.1 when x 0.35. 20
1 y=ƒ
y=fª(x) _1 _3
2 _5
FIGURE 10 10 _3
2 y=f·(x)
_30
FIGURE 12
1
_1
FIGURE 11
What about concavity and inflection points? From Figures 9 and 11 there appear to be inflection points when x is a little to the left of 1 and when x is a little to the right of 0. But it’s difficult to determine inflection points from the graph of f , so we graph the second derivative f in Figure 12. We see that f changes from positive to negative when x 1.23 and from negative to positive when x 0.19. So, correct to two decimal places, f is concave upward on , 1.23 and 0.19, and concave downward on 1.23, 0.19. The inflection points are 1.23, 10.18 and 0.19, 0.05. We have discovered that no single graph reveals all the important features of this polynomial. But Figures 9 and 11, when taken together, do provide an accurate picture. ■
SECTION 4.4
4.4 1– 44
■
Use the guidelines of this section to sketch the curve.
1. y x x 3. y 2 15x 9x x
2
4. y 8x 2 x 4
3
5. y x 4 4x 3
6. y xx 23
7. y 2x 5 5x 2 1
8. y 20x 3 3x 5
x x1 1 11. y 2 x 9 x 13. y 2 x 9 x1 15. y x2
x x 1 2 x 12. y 2 x 9 x2 14. y 2 x 9 x3 1 16. y 3 x 1
17. y xs5 x
18. y 2sx x
9. y
y
20. y
s1 x 21. y x 23. y x 3x 13 2
0
L
46. Coulomb’s Law states that the force of attraction between
two charged particles is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The figure shows particles with charge 1 located at positions 0 and 2 on a coordinate line and a particle with charge 1 at a position x between them. It follows from Coulomb’s Law that the net force acting on the middle particle is
x x5
22. y xs2 x
Fx
2
3 26. y s x 2 12
27. y 3 sin x sin3x
28. y sin x tan x
30. y 2x tan x,
2 x 2
31. y 2 x sin x,
0 x 3
1
+1
_1
+1
0
x
2
x
■ The line y mx b is called a slant asymptote if f x mx b l 0 as x l or x l because the vertical distance between the curve y f x and the line y mx b approaches 0 as x becomes large. Find an equation of the slant asymptote of the function and use it to help sketch the graph. [For rational functions, a slant asymptote occurs when the degree of the numerator is one more than the degree of the denominator. To find it, use long division to write f x mx b RxQx.]
47–50
32. y cos2x 2 sin x
sin x 1 cos x
34. y sin x x
35. y 11 e x
36. y e 2 x e x
37. y x ln x
38. y e xx
39. y xex
40. y xln x2
41. y lnsin x
42. y e x 3e x 4x
x 2
■
0x2
2 x 2
29. y x tan x,
■
k k x2 x 22
where k is a positive constant. Sketch the graph of the net force function. What does the graph say about the force?
24. y x 53 5x 23
25. y x s x
■
W
10. y
x sx 2 1
43. y xe
225
where E and I are positive constants. (E is Young’s modulus of elasticity and I is the moment of inertia of a cross-section of the beam.) Sketch the graph of the deflection curve.
2. y x 6x 9x 3
2
33. y
■
EXERCISES
3
19. y
CURVE SKETCHING
1
44. y tan ■
■
■
■
■
47. y
■
■
45. The figure shows a beam of length L embedded in concrete
walls. If a constant load W is distributed evenly along its length, the beam takes the shape of the deflection curve WL 3 WL2 2 W x4 x x y 24EI 12EI 24EI
48. y
■
■
■
■
x 2 12 x2
50. y e x x
49. xy x 2 4
x1 x1 ■
2x 2 5x 1 2x 1
■
■
■
■
■
■
■
■
51. Show that the curve y x tan1x has two slant asymp-
■
totes: y x 2 and y x 2. Use this fact to help sketch the curve.
52. Show that the curve y sx 2 4x has two slant
asymptotes: y x 2 and y x 2. Use this fact to help sketch the curve.
226
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
; 53–56
; 59–63
■ Describe how the graph of f varies as c varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when c changes. You should also identify any transitional values of c at which the basic shape of the curve changes.
■ Produce graphs of f that reveal all the important aspects of the curve. In particular, you should use graphs of f and f to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points.
53. f x 4x 4 32 x 3 89x 2 95x 29 54. f x x 6 15x 5 75x 4 125x 3 x 55. f x x 4x 7 cos x, 2
59. f x x 4 cx 2
4 x 4
61. f x ecx
56. f x tan x 5 cos x ■
■
■
■
■
■
■
■
■
■
■
■
; 57–58
■ Produce graphs of f that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavity, and use calculus to find these intervals exactly.
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4.5
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■
■
■
■
62. f x lnx 2 c
■
■
■
■
■
■
■
■
■
■
■
■
; 64. Investigate the family of curves given by the equation
1 2 10 8 x8 x4 ■
2
63. f x cx sin x
1 8 1 57. f x 1 2 3 x x x 58. f x
60. f x x 3 cx
■
f x x 4 cx 2 x. Start by determining the transitional value of c at which the number of inflection points changes. Then graph several members of the family to see what shapes are possible. There is another transitional value of c at which the number of critical numbers changes. Try to discover it graphically. Then prove what you have discovered.
OPTIMIZATION PROBLEMS The methods we have learned in this chapter for finding extreme values have practical applications in many areas of life. A businessperson wants to minimize costs and maximize profits. A traveler wants to minimize transportation time. Fermat’s Principle in optics states that light follows the path that takes the least time. In this section and the next we solve such problems as maximizing areas, volumes, and profits and minimizing distances, times, and costs. In solving such practical problems the greatest challenge is often to convert the word problem into a mathematical optimization problem by setting up the function that is to be maximized or minimized. The following steps may be useful.
STEPS IN SOLVING OPTIMIZATION PROBLEMS
1. Understand the Problem The first step is to read the problem carefully until it is
2. 3.
4. 5.
clearly understood. Ask yourself: What is the unknown? What are the given quantities? What are the given conditions? Draw a Diagram In most problems it is useful to draw a diagram and identify the given and required quantities on the diagram. Introduce Notation Assign a symbol to the quantity that is to be maximized or minimized (let’s call it Q for now). Also select symbols a, b, c, . . . , x, y for other unknown quantities and label the diagram with these symbols. It may help to use initials as suggestive symbols—for example, A for area, h for height, t for time. Express Q in terms of some of the other symbols from Step 3. If Q has been expressed as a function of more than one variable in Step 4, use the given information to find relationships (in the form of equations) among these
SECTION 4.5
OPTIMIZATION PROBLEMS
■
227
variables. Then use these equations to eliminate all but one of the variables in the expression for Q. Thus Q will be expressed as a function of one variable x, say, Q f x. Write the domain of this function. 6. Use the methods of Sections 4.1 and 4.3 to find the absolute maximum or minimum value of f . In particular, if the domain of f is a closed interval, then the Closed Interval Method in Section 4.1 can be used. EXAMPLE 1 A farmer has 2400 ft of fencing and wants to fence off a rectangular
field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area? SOLUTION In order to get a feeling for what is happening in this problem, let’s experiment with some special cases. Figure 1 (not to scale) shows three possible ways of laying out the 2400 ft of fencing. We see that when we try shallow, wide fields or deep, narrow fields, we get relatively small areas. It seems plausible that there is some intermediate configuration that produces the largest area.
400
1000 2200
700
100
1000
700
1000
100
Area=100 · 2200=220,000 ft@
Area=700 · 1000=700,000 ft@
Area=1000 · 400=400,000 ft@
FIGURE 1
Figure 2 illustrates the general case. We wish to maximize the area A of the rectangle. Let x and y be the depth and width of the rectangle (in feet). Then we express A in terms of x and y:
y
A xy x
A
x
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the fencing is 2400 ft. Thus FIGURE 2
2x y 2400 From this equation we have y 2400 2x, which gives A x2400 2x 2400x 2x 2 Note that x 0 and x 1200 (otherwise A 0). So the function that we wish to maximize is Ax 2400x 2x 2
0 x 1200
The derivative is Ax 2400 4x, so to find the critical numbers we solve the equation 2400 4x 0
228
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
which gives x 600. The maximum value of A must occur either at this critical number or at an endpoint of the interval. Since A0 0, A600 720,000, and A1200 0, the Closed Interval Method gives the maximum value as A600 720,000. [Alternatively, we could have observed that Ax 4 0 for all x, so A is always concave downward and the local maximum at x 600 must be an absolute maximum.] Thus the rectangular field should be 600 ft deep and 1200 ft wide. ■ V EXAMPLE 2 A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.
SOLUTION Draw the diagram as in Figure 3, where r is the radius and h the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). From Figure 4 we see that the sides are made from a rectangular sheet with dimensions 2 r and h. So the surface area is A 2 r 2 2 rh
h
To eliminate h we use the fact that the volume is given as 1 L, which we take to be 1000 cm3. Thus r 2h 1000
r FIGURE 3 2πr
which gives h 1000 r 2 . Substitution of this into the expression for A gives
r
A 2 r 2 2 r h
1000 r 2
Area (2πr)h
FIGURE 4
y=A(r)
h
FIGURE 5
10
r0
2000 4 r 3 500 2 r r2
3 Then Ar 0 when r 3 500, so the only critical number is r s 500 . Since the domain of A is 0, , we can’t use the argument of Example 1 con3 cerning endpoints. But we can observe that Ar 0 for r s 500 and 3 Ar 0 for r s500 , so A is decreasing for all r to the left of the critical 3 number and increasing for all r to the right. Thus r s 500 must give rise to an absolute minimum. [Alternatively, we could argue that Ar l as r l 0 and Ar l as r l , so there must be a minimum value of Ar, which must occur at the critical number. See Figure 5.] 3 The value of h corresponding to r s 500 is
y
0
2000 r
To find the critical numbers, we differentiate: Ar 4 r
1000
2000 r
Therefore, the function that we want to minimize is Ar 2 r 2
Area 2{πr@}
2 r 2
r
1000 1000 2 2 r 50023
3
500 2r
3 Thus to minimize the cost of the can, the radius should be s 500 cm and the height should be equal to twice the radius, namely, the diameter.
■
SECTION 4.5
OPTIMIZATION PROBLEMS
■
229
NOTE 1 The argument used in Example 2 to justify the absolute minimum is a variant of the First Derivative Test (which applies only to local maximum or minimum values) and is stated here for future reference. Module 4.5 takes you through six additional optimization problems, including animations of the physical situations.
FIRST DERIVATIVE TEST FOR ABSOLUTE EXTREME VALUES Suppose that c is a critical number of a continuous function f defined on an interval. (a) If f x 0 for all x c and f x 0 for all x c, then f c is the absolute maximum value of f . (b) If f x 0 for all x c and f x 0 for all x c, then f c is the absolute minimum value of f . NOTE 2 An alternative method for solving optimization problems is to use implicit differentiation. Let’s look at Example 2 again to illustrate the method. We work with the same equations A 2 r 2 2 rh r 2h 100
but instead of eliminating h, we differentiate both equations implicitly with respect to r : A 4 r 2 rh 2 h r 2h 2 rh 0 The minimum occurs at a critical number, so we set A 0, simplify, and arrive at the equations 2r h rh 0 2h rh 0 and subtraction gives 2r h 0, or h 2r. V EXAMPLE 3
point 1, 4.
Find the point on the parabola y 2 2x that is closest to the
SOLUTION The distance between the point 1, 4 and the point x, y is y (1, 4)
(See Figure 6.) But if x, y lies on the parabola, then x y 22, so the expression for d becomes
(x, y)
1 0
d sx 12 y 42
¥=2x
1 2 3 4
x
d s( 12 y 2 1 ) 2 y 42 (Alternatively, we could have substituted y s2x to get d in terms of x alone.) Instead of minimizing d, we minimize its square:
FIGURE 6
d 2 f y ( 12 y 2 1 ) 2 y 42 (You should convince yourself that the minimum of d occurs at the same point as the minimum of d 2, but d 2 is easier to work with.) Differentiating, we obtain f y 2( 12 y 2 1) y 2 y 4 y 3 8 so f y 0 when y 2. Observe that f y 0 when y 2 and f y 0 when y 2, so by the First Derivative Test for Absolute Extreme Values, the absolute minimum occurs when y 2. (Or we could simply say that because of the geometric nature of the problem, it’s obvious that there is a closest point but not a
230
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
farthest point.) The corresponding value of x is x y 22 2. Thus the point on y 2 2x closest to 1, 4 is 2, 2.
■
EXAMPLE 4 A man launches his boat from point A on a bank of a straight river,
3 km wide, and wants to reach point B, 8 km downstream on the opposite bank, as quickly as possible (see Figure 7). He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between C and B and then run to B. If he can row 6 kmh and run 8 kmh, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared with the speed at which the man rows.)
3 km A
C
D
SOLUTION If we let x be the distance from C to D, then the running distance is DB 8 x and the Pythagorean Theorem gives the rowing distance as AD sx 2 9 . We use the equation
8 km
time B
distance rate
Then the rowing time is sx 2 96 and the running time is 8 x8, so the total time T as a function of x is
FIGURE 7
Tx
8x sx 2 9 6 8
The domain of this function T is 0, 8 . Notice that if x 0 he rows to C and if x 8 he rows directly to B. The derivative of T is Tx
x 1 6sx 2 9 8
Thus, using the fact that x 0, we have Tx 0 &?
y=T(x)
T0 1.5
1
FIGURE 8
&? 4x 3sx 2 9
&?
16x 2 9x 2 9 &? 7x 2 81
&?
x
9 s7
The only critical number is x 9s7 . To see whether the minimum occurs at this critical number or at an endpoint of the domain 0, 8 , we evaluate T at all three points:
T
0
x 1 6sx 2 9 8
2
4
6
x
T
9 s7
1
s7 1.33 8
T8
s73 1.42 6
Since the smallest of these values of T occurs when x 9s7 , the absolute minimum value of T must occur there. Figure 8 illustrates this calculation by showing the graph of T. Thus the man should land the boat at a point 9s7 km ( 3.4 km) downstream from his starting point. ■
SECTION 4.5
OPTIMIZATION PROBLEMS
■
231
V EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of radius r.
SOLUTION 1 Let’s take the semicircle to be the upper half of the circle x 2 y 2 r 2
y
(x, y)
2x _r
y r x
0
with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the x-axis as shown in Figure 9. Let x, y be the vertex that lies in the first quadrant. Then the rectangle has sides of lengths 2x and y, so its area is A 2xy To eliminate y we use the fact that x, y lies on the circle x 2 y 2 r 2 and so y sr 2 x 2 . Thus A 2x sr 2 x 2
FIGURE 9
The domain of this function is 0 x r. Its derivative is A
2x 2 2r 2 2x 2 2 x2 2sr sr 2 x 2 sr 2 x 2
which is 0 when 2x 2 r 2, that is, x rs2 (since x 0). This value of x gives a maximum value of A since A0 0 and Ar 0. Therefore, the area of the largest inscribed rectangle is
A
r s2
2
r s2
r2
r2 r2 2
SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let be the angle shown in Figure 10. Then the area of the rectangle is r ¨ r cos ¨ FIGURE 10
r sin ¨
A 2r cos r sin r 22 sin cos r 2 sin 2 We know that sin 2 has a maximum value of 1 and it occurs when 2 2. So A has a maximum value of r 2 and it occurs when 4. Notice that this trigonometric solution doesn’t involve differentiation. In fact, we didn’t need to use calculus at all. ■ APPLICATIONS TO BUSINESS AND ECONOMICS
In Example 10 in Section 2.3 we introduced the idea of marginal cost. Recall that if Cx, the cost function, is the cost of producing x units of a certain product, then the marginal cost is the rate of change of C with respect to x. In other words, the marginal cost function is the derivative, Cx, of the cost function. Now let’s consider marketing. Let px be the price per unit that the company can charge if it sells x units. Then p is called the demand function (or price function) and we would expect it to be a decreasing function of x. If x units are sold and the price per unit is px, then the total revenue is Rx xpx and R is called the revenue function. The derivative R of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
If x units are sold, then the total profit is Px Rx Cx and P is called the profit function. The marginal profit function is P, the derivative of the profit function. In Exercises 35– 40 you are asked to use the marginal cost, revenue, and profit functions to minimize costs and maximize revenues and profits. V EXAMPLE 6 A store has been selling 200 DVD burners a week at $350 each. A market survey indicates that for each $10 rebate offered to buyers, the number of units sold will increase by 20 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize its revenue?
SOLUTION If x is the number of DVD burners sold per week, then the weekly increase in sales is x 200. For each increase of 20 units sold, the price is decreased by $10. So for each additional unit sold, the decrease in price will be 201 10 and the demand function is 1 px 350 10 20 x 200 450 2 x
The revenue function is Rx xpx 450x 12 x 2 Since Rx 450 x, we see that Rx 0 when x 450. This value of x gives an absolute maximum by the First Derivative Test (or simply by observing that the graph of R is a parabola that opens downward). The corresponding price is p450 450 12 450 225 and the rebate is 350 225 125. Therefore, to maximize revenue the store should offer a rebate of $125.
4.5
■
EXERCISES
1. Consider the following problem: Find two numbers whose
sum is 23 and whose product is a maximum. (a) Make a table of values, like the following one, so that the sum of the numbers in the first two columns is always 23. On the basis of the evidence in your table, estimate the answer to the problem.
3. Find two positive numbers whose product is 100 and whose
sum is a minimum. 4. Find a positive number such that the sum of the number and
its reciprocal is as small as possible. 5. Find the dimensions of a rectangle with perimeter 100 m
whose area is as large as possible. First number
Second number
Product
1 2 3 . . .
22 21 20 . . .
22 42 60 . . .
(b) Use calculus to solve the problem and compare with your answer to part (a). 2. Find two numbers whose difference is 100 and whose prod-
uct is a minimum.
6. Find the dimensions of a rectangle with area 1000 m2 whose
perimeter is as small as possible. 7. Consider the following problem: A farmer with 750 ft of
fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? (a) Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these configurations. Does it appear that there is a maximum area? If so, estimate it.
SECTION 4.5
(b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols. (c) Write an expression for the total area. (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the total area as a function of one variable. (f ) Finish solving the problem and compare the answer with your estimate in part (a). 8. Consider the following problem: A box with an open top is
to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have. (a) Draw several diagrams to illustrate the situation, some short boxes with large bases and some tall boxes with small bases. Find the volumes of several such boxes. Does it appear that there is a maximum volume? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols. (c) Write an expression for the volume. (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the volume as a function of one variable. (f ) Finish solving the problem and compare the answer with your estimate in part (a). 2
9. If 1200 cm of material is available to make a box with a
square base and an open top, find the largest possible volume of the box.
OPTIMIZATION PROBLEMS
233
16. Find the dimensions of the rectangle of largest area that has
its base on the x-axis and its other two vertices above the x-axis and lying on the parabola y 8 x 2. 17. A right circular cylinder is inscribed in a sphere of radius r.
Find the largest possible volume of such a cylinder. 18. Find the area of the largest rectangle that can be inscribed in
the ellipse x 2a 2 y 2b 2 1. 19. A Norman window has the shape of a rectangle surmounted
by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 30 ft, find the dimensions of the window so that the greatest possible amount of light is admitted. 20. A right circular cylinder is inscribed in a cone with height h
and base radius r. Find the largest possible volume of such a cylinder. 21. A piece of wire 10 m long is cut into two pieces. One piece
is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) A minimum? 22. A fence 8 ft tall runs parallel to a tall building at a distance
of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? 23. A cone-shaped drinking cup is made from a circular piece
of paper of radius R by cutting out a sector and joining the edges CA and CB. Find the maximum capacity of such a cup. A
10. A box with a square base and open top must have a volume
of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used.
■
R
B
C
11. (a) Show that of all the rectangles with a given area, the one
with smallest perimeter is a square. (b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square. 12. A rectangular storage container with an open top is to have
a volume of 10 m3. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container. 13. Find the points on the ellipse 4x 2 y 2 4 that are farthest
away from the point 1, 0.
; 14. Find, correct to two decimal places, the coordinates of the
point on the curve y tan x that is closest to the point 1, 1.
15. Find the dimensions of the rectangle of largest area that can
be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle.
24. A cone-shaped paper drinking cup is to be made to hold
27 cm3 of water. Find the height and radius of the cup that will use the smallest amount of paper. 25. A cone with height h is inscribed in a larger cone with
height H so that its vertex is at the center of the base of the larger cone. Show that the inner cone has maximum volume when h 13 H . 26. The graph (on page 234) shows the fuel consumption c of
a car (measured in gallons per hour) as a function of the speed v of the car. At very low speeds the engine runs inefficiently, so initially c decreases as the speed increases. But at high speeds the fuel consumption increases. You can see that cv is minimized for this car when v 30 mih. However, for fuel efficiency, what must be minimized is not the consumption in gallons per hour but rather the fuel
234
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APPLICATIONS OF DIFFERENTIATION
consumption in gallons per mile. Let’s call this consumption G. Using the graph, estimate the speed at which G has its minimum value.
Note: Actual measurements of the angle in beehives have been made, and the measures of these angles seldom differ from the calculated value by more than 2 .
c
trihedral angle ¨
rear of cell
0
20
40
60
h
√
27. If a resistor of R ohms is connected across a battery of
b
E volts with internal resistance r ohms, then the power (in watts) in the external resistor is P
E 2R R r 2
If E and r are fixed but R varies, what is the maximum value of the power? 28. For a fish swimming at a speed v relative to the water, the energy expenditure per unit time is proportional to v 3. It is
believed that migrating fish try to minimize the total energy required to swim a fixed distance. If the fish are swimming against a current u u v, then the time required to swim a distance L is Lv u and the total energy E required to swim the distance is given by Ev av 3
L vu
where a is the proportionality constant. (a) Determine the value of v that minimizes E. (b) Sketch the graph of E.
30. A boat leaves a dock at 2:00 PM and travels due south at a
speed of 20 kmh. Another boat has been heading due east at 15 kmh and reaches the same dock at 3:00 PM. At what time were the two boats closest together? 31. The illumination of an object by a light source is directly
proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 ft apart, where should an object be placed on the line between the sources so as to receive the least illumination? 32. A woman at a point A on the shore of a circular lake with
radius 2 mi wants to arrive at the point C diametrically opposite A on the other side of the lake in the shortest possible time. She can walk at the rate of 4 mih and row a boat at 2 mih. How should she proceed? B
Note: This result has been verified experimentally; migrating fish swim against a current at a speed 50% greater than the current speed. A 29. In a beehive, each cell is a regular hexagonal prism, open
at one end with a trihedral angle at the other end as in the figure. It is believed that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of wax in cell construction. Examination of these cells has shown that the measure of the apex angle is amazingly consistent. Based on the geometry of the cell, it can be shown that the surface area S is given by 3 S 6sh 2 s 2 cot (3s 2s32) csc
where s, the length of the sides of the hexagon, and h, the height, are constants. (a) Calculate dSd. (b) What angle should the bees prefer? (c) Determine the minimum surface area of the cell (in terms of s and h).
front of cell
s
¨ 2
2
C
33. Find an equation of the line through the point 3, 5 that
cuts off the least area from the first quadrant. 34. At which points on the curve y 1 40x 3 3x 5 does the
tangent line have the largest slope? 35. (a) If Cx is the cost of producing x units of a commodity,
then the average cost per unit is cx Cxx . Show that if the average cost is a minimum, then the marginal cost equals the average cost. (b) If Cx 16,000 200x 4 x 32 , in dollars, find (i) the cost, average cost, and marginal cost at a productional level of 1000 units; (ii) the production
SECTION 4.5
level that will minimize the average cost; and (iii) the minimum average cost. 36. (a) Show that if the profit Px is a maximum, then the mar-
ginal revenue equals the marginal cost. (b) If Cx 16,000 500x 1.6x 2 0.004 x 3 is the cost function and px 1700 7x is the demand function, find the production level that will maximize profit. 37. A baseball team plays in a stadium that holds 55,000 specta-
tors. With ticket prices at $10, the average attendance had been 27,000. When ticket prices were lowered to $8, the average attendance rose to 33,000. (a) Find the demand function, assuming that it is linear. (b) How should ticket prices be set to maximize revenue?
OPTIMIZATION PROBLEMS
■
235
43. Let v1 be the velocity of light in air and v2 the velocity of
light in water. According to Fermat’s Principle, a ray of light will travel from a point A in the air to a point B in the water by a path ACB that minimizes the time taken. Show that sin 1 v1 sin 2 v2 where 1 (the angle of incidence) and 2 (the angle of refraction) are as shown. This equation is known as Snell’s Law. A ¨¡ C
38. During the summer months Terry makes and sells necklaces
on the beach. Last summer he sold the necklaces for $10 each and his sales averaged 20 per day. When he increased the price by $1, he found that he lost two sales per day. (a) Find the demand function, assuming that it is linear. (b) If the material for each necklace costs Terry $6, what should the selling price be to maximize his profit? 39. A manufacturer has been selling 1000 television sets a week
at $450 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week. (a) Find the demand function. (b) How large a rebate should the company offer the buyer in order to maximize its revenue? (c) If its weekly cost function is Cx 68,000 150x, how should the manufacturer set the size of the rebate in order to maximize its profit?
¨™ B 44. Two vertical poles PQ and ST are secured by a rope PRS
going from the top of the first pole to a point R on the ground between the poles and then to the top of the second pole as in the figure. Show that the shortest length of such a rope occurs when 1 2. P S
experience that all units will be occupied if the rent is $800 per month. A market survey suggests that, on average, one additional unit will remain vacant for each $10 increase in rent. What rent should the manager charge to maximize revenue? 41. Let a and b be positive numbers. Find the length of the
shortest line segment that is cut off by the first quadrant and passes through the point a, b.
¨™
¨¡
40. The manager of a 100-unit apartment complex knows from
Q
R
T
45. The upper right-hand corner of a piece of paper, 12 in. by
8 in., as in the figure, is folded over to the bottom edge. How would you fold it so as to minimize the length of the fold? In other words, how would you choose x to minimize y ? 12
CAS
42. The frame for a kite is to be made from six pieces of wood.
The four exterior pieces have been cut with the lengths indicated in the figure. To maximize the area of the kite, how long should the diagonal pieces be? a
y
x
8
b 46. A steel pipe is being carried down a hallway 9 ft wide.
a
b
At the end of the hall there is a right-angled turn into a narrower hallway 6 ft wide. What is the length of the longest
236
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
pipe that can be carried horizontally around the corner?
49. Where should the point P be chosen on the line segment AB
so as to maximize the angle ? B
2
6 ¨
P
¨
3 9 5
A 47. Find the maximum area of a rectangle that can be circum-
scribed about a given rectangle with length L and width W. [Hint: Express the area as a function of an angle .] 48. A rain gutter is to be constructed from a metal sheet of
width 30 cm by bending up one-third of the sheet on each side through an angle . How should be chosen so that the gutter will carry the maximum amount of water?
50. A painting in an art gallery has height h and is hung so that
its lower edge is a distance d above the eye of an observer (as in the figure). How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle subtended at his eye by the painting?) h
¨ 10 cm
4.6
¨ 10 cm
¨
d
10 cm
NEWTON’S METHOD Suppose that a car dealer offers to sell you a car for $18,000 or for payments of $375 per month for five years. You would like to know what monthly interest rate the dealer is, in effect, charging you. To find the answer, you have to solve the equation 1
0.15
0
0.012
_0.05
FIGURE 1 ■ Try to solve Equation 1 using the numerical rootfinder on your calculator or computer. Some machines are not able to solve it. Others are successful but require you to specify a starting point for the search.
48x1 x60 1 x60 1 0
(The details are explained in Exercise 29.) How would you solve such an equation? For a quadratic equation ax 2 bx c 0 there is a well-known formula for the roots. For third- and fourth-degree equations there are also formulas for the roots, but they are extremely complicated. If f is a polynomial of degree 5 or higher, there is no such formula. Likewise, there is no formula that will enable us to find the exact roots of a transcendental equation such as cos x x. We can find an approximate solution to Equation 1 by plotting the left side of the equation. Using a graphing device, and after experimenting with viewing rectangles, we produce the graph in Figure 1. We see that in addition to the solution x 0, which doesn’t interest us, there is a solution between 0.007 and 0.008. Zooming in shows that the root is approximately 0.0076. If we need more accuracy we could zoom in repeatedly, but that becomes tiresome. A faster alternative is to use a numerical rootfinder on a calculator or computer algebra system. If we do so, we find that the root, correct to nine decimal places, is 0.007628603. How do those numerical rootfinders work? They use a variety of methods, but most of them make some use of Newton’s method, which is also called the Newton-
SECTION 4.6
y {x ¡, f(x¡)}
y=ƒ L x™ x ¡
r
0
x
FIGURE 2
NEWTON’S METHOD
■
237
Raphson method. We will explain how this method works, partly to show what happens inside a calculator or computer, and partly as an application of the idea of linear approximation. The geometry behind Newton’s method is shown in Figure 2, where the root that we are trying to find is labeled r. We start with a first approximation x 1, which is obtained by guessing, or from a rough sketch of the graph of f , or from a computergenerated graph of f. Consider the tangent line L to the curve y f x at the point x 1, f x 1 and look at the x-intercept of L, labeled x 2. The idea behind Newton’s method is that the tangent line is close to the curve and so its x-intercept, x2 , is close to the x-intercept of the curve (namely, the root r that we are seeking). Because the tangent is a line, we can easily find its x-intercept. To find a formula for x2 in terms of x1 we use the fact that the slope of L is f x1 , so its equation is y f x 1 f x 1 x x 1 Since the x-intercept of L is x 2 , we set y 0 and obtain 0 f x 1 f x 1 x 2 x 1 If f x 1 0, we can solve this equation for x 2 : x2 x1
f x 1 f x 1
We use x2 as a second approximation to r. Next we repeat this procedure with x 1 replaced by x 2 , using the tangent line at x 2 , f x 2 . This gives a third approximation:
y {x¡, f(x¡)}
x3 x2
If we keep repeating this process, we obtain a sequence of approximations x 1 , x 2 , x 3 , x 4 , . . . as shown in Figure 3. In general, if the nth approximation is x n and f x n 0, then the next approximation is given by
{x™, f(x™)}
r x£
0
f x 2 f x 2
x™ x ¡
x
x¢ 2
FIGURE 3 ■ The convergence of infinite sequences is discussed in detail in Section 8.1.
x n1 x n
f x n f x n
If the numbers x n become closer and closer to r as n becomes large, then we say that the sequence converges to r and we write lim x n r
y
nl
| Although the sequence of successive approximations converges to the desired root for x™
0
x£
FIGURE 4
x¡
r
x
functions of the type illustrated in Figure 3, in certain circumstances the sequence may not converge. For example, consider the situation shown in Figure 4. You can see that x 2 is a worse approximation than x 1. This is likely to be the case when f x 1 is close to 0. It might even happen that an approximation (such as x 3 in Figure 4) falls outside the domain of f . Then Newton’s method fails and a better initial approximation x 1 should be chosen. See Exercises 23–25 for specific examples in which Newton’s method works very slowly or does not work at all.
238
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
V EXAMPLE 1 Starting with x 1 2, find the third approximation x 3 to the root of the equation x 3 2x 5 0.
SOLUTION We apply Newton’s method with
f x x 3 2x 5 In Module 4.6 you can investigate how Newton’s method works for several functions and what happens when you change x1 .
and
f x 3x 2 2
Newton himself used this equation to illustrate his method and he chose x 1 2 after some experimentation because f 1 6, f 2 1, and f 3 16. Equation 2 becomes x n1 x n
x n3 2x n 5 3x n2 2
With n 1 we have x2 x1
Figure 5 shows the geometry behind the first step in Newton’s method in Example 1. Since f 2 10 , the tangent line to y x 3 2x 5 at 2, 1 has equation y 10x 21 and so its x -intercept is x 2 2.1. ■
2
1.8
x3 x2
2.2 x™
x 23 2x 2 5 3x 22 2
2.1 y=10x-21
FIGURE 5
2 3 22 5 2.1 322 2
Then with n 2 we obtain
1
_2
x13 2x 1 5 3x12 2
2.13 22.1 5 2.0946 32.12 2
It turns out that this third approximation x 3 2.0946 is accurate to four decimal places.
■
Suppose that we want to achieve a given accuracy, say to eight decimal places, using Newton’s method. How do we know when to stop? The rule of thumb that is generally used is that we can stop when successive approximations x n and x n1 agree to eight decimal places. (A precise statement concerning accuracy in Newton’s method will be given in Exercise 29 in Section 8.8.) Notice that the procedure in going from n to n 1 is the same for all values of n. (It is called an iterative process.) This means that Newton’s method is particularly convenient for use with a programmable calculator or a computer. V EXAMPLE 2
6 Use Newton’s method to find s 2 correct to eight decimal places.
6 SOLUTION First we observe that finding s 2 is equivalent to finding the positive
root of the equation x6 2 0 so we take f x x 6 2. Then f x 6x 5 and Formula 2 (Newton’s method) becomes x n1 x n
x n6 2 6x n5
SECTION 4.6
NEWTON’S METHOD
■
239
If we choose x 1 1 as the initial approximation, then we obtain x 2 1.16666667 x 3 1.12644368 x 4 1.12249707 x 5 1.12246205 x 6 1.12246205 Since x 5 and x 6 agree to eight decimal places, we conclude that 6 2 1.12246205 s
■
to eight decimal places. V EXAMPLE 3
Find, correct to six decimal places, the root of the equation
cos x x. SOLUTION We first rewrite the equation in standard form:
cos x x 0 Therefore, we let f x cos x x. Then f x sin x 1, so Formula 2 becomes cos x n x n cos x n x n x n1 x n xn sin x n 1 sin x n 1 y
y=x
y=cos x 1
π 2
x
π
In order to guess a suitable value for x 1 we sketch the graphs of y cos x and y x in Figure 6. It appears that they intersect at a point whose x-coordinate is somewhat less than 1, so let’s take x 1 1 as a convenient first approximation. Then, remembering to put our calculator in radian mode, we get x 2 0.75036387 x 3 0.73911289
FIGURE 6
x 4 0.73908513 x 5 0.73908513 Since x 4 and x 5 agree to six decimal places (eight, in fact), we conclude that the root of the equation, correct to six decimal places, is 0.739085. ■
1
Instead of using the rough sketch in Figure 6 to get a starting approximation for Newton’s method in Example 3, we could have used the more accurate graph that a calculator or computer provides. Figure 7 suggests that we use x1 0.75 as the initial approximation. Then Newton’s method gives
y=cos x
y=x
0
FIGURE 7
1
x 2 0.73911114
x 3 0.73908513
x 4 0.73908513
and so we obtain the same answer as before, but with one fewer step.
240
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
4.6
EXERCISES
1. The figure shows the graph of a function f . Suppose that
9–10 ■ Use Newton’s method to approximate the given number correct to eight decimal places.
Newton’s method is used to approximate the root r of the equation f x 0 with initial approximation x 1 1. (a) Draw the tangent lines that are used to find x 2 and x 3, and estimate the numerical values of x 2 and x 3. (b) Would x 1 5 be a better first approximation? Explain.
3 9. s 30 ■
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7 10. s 1000 ■
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■ Use Newton’s method to approximate the indicated root of the equation correct to six decimal places.
11–12
y
11. The positive root of sin x x 2 12. The positive root of 2 cos x x 4 ■
1 0
r
1
s
■
; 13–20
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y f x when x 3. If Newton’s method is used to locate a root of the equation f x 0 and the initial approximation is x1 3, find the second approximation x 2.
4. For each initial approximation, determine graphically what
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y
4 x2 1
15. ex 2 x
16. ln4 x 2 x
17. x 2 s2 x x 2 1
18. 3 sinx 2 2x
19. tan1x 1 x
20. tan x s9 x 2
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happens if Newton’s method is used for the function whose graph is shown. (a) x1 0 (b) x1 1 (c) x1 3 (d) x1 4 (e) x1 5
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21. (a) Apply Newton’s method to the equation x 2 a 0 to
derive the following square-root algorithm (used by the ancient Babylonians to compute sa ) : x n1
1 a xn 2 xn
(b) Use part (a) to compute s1000 correct to six decimal places.
x
5
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14. x 2 4 x 2
3. Suppose the line y 5x 4 is tangent to the curve
3
■
13. x 5 x 4 5x 3 x 2 4x 3 0
the starting approximation for finding the root s.
1
■
Use Newton’s method to find all the roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.
x
2. Follow the instructions for Exercise 1(a) but use x 1 9 as
0
■
22. (a) Apply Newton’s method to the equation 1x a 0 to
derive the following reciprocal algorithm: Use Newton’s method with the specified initial approximation x 1 to find x 3 , the third approximation to the root of the given equation. (Give your answer to four decimal places.)
5–6
■
5. x 3 2x 4 0, 6. x 2 0, ■
■
(This algorithm enables a computer to find reciprocals without actually dividing.) (b) Use part (a) to compute 11.6984 correct to six decimal places.
x1 1
x1 1
5
■
x n1 2x n ax n2
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; 7. Use Newton’s method with initial approximation x1 1
to find x 2 , the second approximation to the root of the equation x 3 x 3 0. Explain how the method works by first graphing the function and its tangent line at 1, 1.
; 8. Use Newton’s method with initial approximation x1 1
to find x 2 , the second approximation to the root of the equation x 4 x 1 0 . Explain how the method works by first graphing the function and its tangent line at 1, 1.
23. Explain why Newton’s method doesn’t work for finding the
root of the equation x 3 3x 6 0 if the initial approximation is chosen to be x 1 1. 24. (a) Use Newton’s method with x 1 1 to find the root of the
equation x 3 x 1 correct to six decimal places. (b) Solve the equation in part (a) using x 1 0.6 as the initial approximation. (c) Solve the equation in part (a) using x 1 0.57. (You definitely need a programmable calculator for this part.)
SECTION 4.7
;
(d) Graph f x x 3 x 1 and its tangent lines at x1 1, 0.6, and 0.57 to explain why Newton’s method is so sensitive to the value of the initial approximation. 25. Explain why Newton’s method fails when applied to the 3 equation s x 0 with any initial approximation x 1 0. Illustrate your explanation with a sketch.
26. Use Newton’s method to find the absolute minimum value
of the function f x x 2 sin x correct to six decimal places. 27. Use Newton’s method to find the coordinates of the inflec-
tion point of the curve y e cos x, 0 x , correct to six decimal places. 28. Of the infinitely many lines that are tangent to the curve
y sin x and pass through the origin, there is one that has the largest slope. Use Newton’s method to find the slope of that line correct to six decimal places.
A
■
241
30. The figure shows the Sun located at the origin and the Earth
at the point 1, 0. (The unit here is the distance between the centers of the Earth and the Sun, called an astronomical unit: 1 AU 1.496 10 8 km.) There are five locations L 1 , L 2 , L 3 , L 4 , and L 5 in this plane of rotation of the Earth about the Sun where a satellite remains motionless with respect to the Earth because the forces acting on the satellite (including the gravitational attractions of the Earth and the Sun) balance each other. These locations are called libration points. (A solar research satellite has been placed at one of these libration points.) If m1 is the mass of the Sun, m 2 is the mass of the Earth, and r m 2m1 m 2 , it turns out that the x-coordinate of L 1 is the unique root of the fifthdegree equation px x 5 2 rx 4 1 2rx 3 1 rx 2 21 rx r 1 0 and the x-coordinate of L 2 is the root of the equation px 2rx 2 0
29. A car dealer sells a new car for $18,000. He also offers to
sell the same car for payments of $375 per month for five years. What monthly interest rate is this dealer charging? To solve this problem you will need to use the formula for the present value A of an annuity consisting of n equal payments of size R with interest rate i per time period:
ANTIDERIVATIVES
Using the value r 3.04042 10 6, find the locations of the libration points (a) L 1 and (b) L 2. y L¢
R 1 1 i n i
Earth Sun L∞
Replacing i by x, show that 48x1 x60 1 x60 1 0
L¡
L™
x
L£
Use Newton’s method to solve this equation.
4.7
ANTIDERIVATIVES A physicist who knows the velocity of a particle might wish to know its position at a given time. An engineer who can measure the variable rate at which water is leaking from a tank wants to know the amount leaked over a certain time period. A biologist who knows the rate at which a bacteria population is increasing might want to deduce what the size of the population will be at some future time. In each case, the problem is to find a function F whose derivative is a known function f. If such a function F exists, it is called an antiderivative of f. DEFINITION A function F is called an antiderivative of f on an interval I if Fx f x for all x in I .
For instance, let f x x 2. It isn’t difficult to discover an antiderivative of f if we keep the Power Rule in mind. In fact, if Fx 13 x 3, then Fx x 2 f x. But the function Gx 13 x 3 100 also satisfies Gx x 2. Therefore, both F and G are antiderivatives of f . Indeed, any function of the form Hx 13 x 3 C, where C is a constant, is an antiderivative of f . The question arises: Are there any others?
242
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
To answer this question, recall that in Section 4.2 we used the Mean Value Theorem to prove that if two functions have identical derivatives on an interval, then they must differ by a constant (Corollary 4.2.7). Thus, if F and G are any two antiderivatives of f , then y
˛
Fx f x Gx
˛
so Gx Fx C, where C is a constant. We can write this as Gx Fx C, so we have the following result.
y= 3 +3 y= 3 +2 ˛
y= 3 +1 y= ˛ 0
x
1 THEOREM If F is an antiderivative of f on an interval I , then the most general antiderivative of f on I is
3
˛
y= 3 -1
Fx C
˛
y= 3 -2
FIGURE 1
Members of the family of antiderivatives of ƒ=≈
where C is an arbitrary constant. Going back to the function f x x 2, we see that the general antiderivative of f is x C. By assigning specific values to the constant C, we obtain a family of functions whose graphs are vertical translates of one another (see Figure 1). This makes sense because each curve must have the same slope at any given value of x.
1 3
3
EXAMPLE 1 Find the most general antiderivative of each of the following functions.
(a) f x sin x
(b) f x 1x
(c) f x x n,
n 1
SOLUTION
(a) If Fx cos x, then Fx sin x, so an antiderivative of sin x is cos x. By Theorem 1, the most general antiderivative is Gx cos x C. (b) Recall from Section 3.3 that d 1 ln x dx x So on the interval 0, the general antiderivative of 1x is ln x C. We also learned that d 1 ln x dx x
for all x 0. Theorem 1 then tells us that the general antiderivative of f x 1x is ln x C on any interval that doesn’t contain 0. In particular, this is true on each of the intervals , 0 and 0, . So the general antiderivative of f is
Fx
ln x C1 lnx C2
if x 0 if x 0
(c) We use the Power Rule to discover an antiderivative of x n. In fact, if n 1, then d dx
x n1 n1
n 1x n xn n1
Thus the general antiderivative of f x x n is Fx
x n1 C n1
SECTION 4.7
ANTIDERIVATIVES
■
243
This is valid for n 0 since then f x x n is defined on an interval. If n is negative (but n 1), it is valid on any interval that doesn’t contain 0. ■ As in Example 1, every differentiation formula, when read from right to left, gives rise to an antidifferentiation formula. In Table 2 we list some particular antiderivatives. Each formula in the table is true because the derivative of the function in the right column appears in the left column. In particular, the first formula says that the antiderivative of a constant times a function is the constant times the antiderivative of the function. The second formula says that the antiderivative of a sum is the sum of the antiderivatives. (We use the notation F f , G t.) 2 TABLE OF ANTIDIFFERENTIATION FORMULAS
Function
■ To obtain the most general antiderivative from the particular ones in Table 2, we have to add a constant (or constants), as in Example 1.
Particular antiderivative
Function
Particular antiderivative
c f x
cFx
sin x
cos x
f x tx
Fx Gx
sec2x
tan x
x n1 n1
sec x tan x
sec x
1 s1 x 2
sin1x
1 1 x2
tan1x
xn
n 1
1x
ln x
ex
ex
cos x
sin x
EXAMPLE 2 Find all functions t such that
tx 4 sin x
2x 5 sx x
SOLUTION We first rewrite the given function as follows:
tx 4 sin x
2x 5 1 sx 4 sin x 2x 4 x x sx
Thus we want to find an antiderivative of tx 4 sin x 2x 4 x12 Using the formulas in Table 2 together with Theorem 1, we obtain tx 4cos x 2
x5 x12 1 C 5 2
4 cos x 25 x 5 2sx C
■
In applications of calculus it is very common to have a situation as in Example 2, where it is required to find a function, given knowledge about its derivatives. An equation that involves the derivatives of a function is called a differential equation. These will be studied in some detail in Section 7.6, but for the present we can solve some elementary differential equations. The general solution of a differential equation involves an arbitrary constant (or constants) as in Example 2. However, there may be some extra conditions given that will determine the constants and therefore uniquely specify the solution.
244
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
■ Figure 2 shows the graphs of the function f in Example 3 and its antiderivative f . Notice that f x 0 , so f is always increasing. Also notice that when f has a maximum or minimum, f appears to have an inflection point. So the graph serves as a check on our calculation.
40
EXAMPLE 3 Find f if f x e x 201 x 2 1 and f 0 2. SOLUTION The general antiderivative of
f x e x
20 1 x2
f x e x 20 tan1 x C
is
To determine C we use the fact that f 0 2: f 0 e 0 20 tan1 0 C 2
fª _2
3
Thus we have C 2 1 3, so the particular solution is
f
f x e x 20 tan1 x 3
■
_25
FIGURE 2
V EXAMPLE 4
Find f if f x 12x 2 6x 4, f 0 4, and f 1 1.
SOLUTION The general antiderivative of f x 12x 2 6x 4 is
f x 12
x3 x2 6 4x C 4x 3 3x 2 4x C 3 2
Using the antidifferentiation rules once more, we find that f x 4
x4 x3 x2 3 4 Cx D x 4 x 3 2x 2 Cx D 4 3 2
To determine C and D we use the given conditions that f 0 4 and f 1 1. Since f 0 0 D 4, we have D 4. Since f 1 1 1 2 C 4 1 we have C 3. Therefore, the required function is f x x 4 x 3 2x 2 3x 4
■
RECTILINEAR MOTION
Antidifferentiation is particularly useful in analyzing the motion of an object moving in a straight line. Recall that if the object has position function s f t, then the velocity function is v t st. This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration function is at vt, so the velocity function is an antiderivative of the acceleration. If the acceleration and the initial values s0 and v0 are known, then the position function can be found by antidifferentiating twice. V EXAMPLE 5 A particle moves in a straight line and has acceleration given by at 6t 4 . Its initial velocity is v0 6 cms and its initial displacement is s0 9 cm. Find its position function st.
SOLUTION Since vt at 6t 4, antidifferentiation gives vt 6
t2 4t C 3t 2 4t C 2
SECTION 4.7
ANTIDERIVATIVES
■
245
Note that v 0 C. But we are given that v 0 6, so C 6 and v t 3t 2 4t 6
Since v t st, s is the antiderivative of v : st 3
t3 t2 4 6t D t 3 2t 2 6t D 3 2
This gives s0 D. We are given that s0 9, so D 9 and the required position function is st t 3 2t 2 6t 9 ■ An object near the surface of the Earth is subject to a gravitational force that produces a downward acceleration denoted by t. For motion close to the ground we may assume that t is constant, its value being about 9.8 ms2 (or 32 fts2 ). EXAMPLE 6 A ball is thrown upward with a speed of 48 fts from the edge of a cliff 432 ft above the ground. Find its height above the ground t seconds later. When does it reach its maximum height? When does it hit the ground? SOLUTION The motion is vertical and we choose the positive direction to be upward. At time t the distance above the ground is st and the velocity v t is decreasing. Therefore, the acceleration must be negative and we have
at
dv 32 dt
Taking antiderivatives, we have v t 32t C
To determine C we use the given information that v 0 48. This gives 48 0 C, so v t 32t 48 Figure 3 shows the position function of the ball in Example 6. The graph corroborates the conclusions we reached: The ball reaches its maximum height after 1.5 s and hits the ground after 6.9 s. ■
The maximum height is reached when v t 0, that is, after 1.5 s. Since st v t, we antidifferentiate again and obtain st 16t 2 48t D Using the fact that s0 432, we have 432 0 D and so
500
st 16t 2 48t 432 The expression for st is valid until the ball hits the ground. This happens when st 0, that is, when 0
FIGURE 3
16t 2 48t 432 0
8
or, equivalently,
t 2 3t 27 0
246
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
Using the quadratic formula to solve this equation, we get t
3 3s13 2
We reject the solution with the minus sign since it gives a negative value for t. Therefore, the ball hits the ground after 3(1 s13 )2 6.9 s.
4.7
EXERCISES 27. f x x 2,
■ Find the most general antiderivative of the function. (Check your answer by differentiation.)
1–12
1. f x 6x 2 8x 3
2. f x 1 x 3 12 x 5
3. f x 5x 14 7x 34
4. f x 2x 3x 1.7
3
5 5. f x sx 6 x
6. f x sx sx
u 4 3su u2
8. tx
7. f u
4
9. t cos 5 sin
3
3
x 0,
f 1 0,
f 2 0
28. f t 2e 3 sin t,
f 0 0,
f 0
■
■
t
■
; 13–14
■
■
a ■
■
■
■
f
b
a
■
x
x
b
■
■
■
■
■
c c
F0 4
14. f x 4 31 x 2 1, ■
■
F1 0
■
■
■
■
■
■
■
■
■
15. f x 6 x 12 x 2
16. f x 2 x 3 x 6
17. f x 1 x 45
18. f x cos x
19. f x sx 6 5x, 20. f x 2x 3x , 4
2 t 2,
21. f t 2 cos t sec 2 t, 22. f x 4s1 x , 2
f(
1 2
24. f x 4 6x 40x , 25. f sin cos ,
f 3 4
)1
23. f x 24x 2 2 x 10 , 3
f 1 3
f 1 5, f 0 2,
f 0 3,
f 4 20,
f 1 3 f 0 1
f 0 4
f 4 7
■
■
■
■
■
■
A particle is moving with the given data. Find the position of the particle.
34. vt 1.5 st ,
s0 0
s4 10
35. at 10 sin t 3 cos t,
f 1 10 x 0,
■
■
33. vt sin t cos t,
Find f .
26. f t 3st ,
■
33–36
■
■
y
f ■
■
32.
y
■
■
31–32 ■ The graph of a function f is shown. Which graph is an antiderivative of f and why? 31.
■
■
x y 0 is tangent to the graph of f .
10. f x 3e x 7 sec2x
■
13. f x 5x 4 2x 5,
15–28
■
30. Find a function f such that f x x 3 and the line
5 4x 3 2x 6 x6
Find the antiderivative F of f that satisfies the given condition. Check your answer by comparing the graphs of f and F.
■
■
and that the slope of its tangent line at x, f x is 2x 1, find f 2.
x x1 x ■
■
29. Given that the graph of f passes through the point 1, 6
2
■
■
4
11. f x 2 x 51 x 2 12 12. f x
■
s0 0,
s2 12
36. at 10 3t 3t 2,
s0 0, s2 10
■
■
■
■
■
■
■
■
■
■
■
■
37. A stone is dropped from the upper observation deck (the
Space Deck) of the CN Tower, 450 m above the ground. (a) Find the distance of the stone above ground level at time t. (b) How long does it take the stone to reach the ground? (c) With what velocity does it strike the ground? (d) If the stone is thrown downward with a speed of 5 ms, how long does it take to reach the ground?
CHAPTER 4
38. Show that for motion in a straight line with constant acceleration a, initial velocity v 0 , and initial displacement s 0 , the
displacement after time t is
■
247
If the raindrop is initially 500 m above the ground, how long does it take to fall? 44. A car is traveling at 50 mih when the brakes are fully
applied, producing a constant deceleration of 22 fts2. What is the distance traveled before the car comes to a stop?
s at v 0 t s 0 1 2
REVIEW
2
39. An object is projected upward with initial velocity v 0 meters
per second from a point s0 meters above the ground. Show that
45. What constant acceleration is required to increase the speed
of a car from 30 mih to 50 mih in 5 s? 46. A car braked with a constant deceleration of 16 fts2, pro-
vt 2 v02 19.6 st s0 40. Two balls are thrown upward from the edge of the cliff in
Example 6. The first is thrown with a speed of 48 fts and the other is thrown a second later with a speed of 24 fts. Do the balls ever pass each other? 41. A stone was dropped off a cliff and hit the ground with a
speed of 120 fts. What is the height of the cliff? 42. If a diver of mass m stands at the end of a diving board with
length L and linear density , then the board takes on the shape of a curve y f x, where 1 EI y mtL x 2 tL x2
E and I are positive constants that depend on the material of the board and t 0 is the acceleration due to gravity. (a) Find an expression for the shape of the curve. (b) Use f L to estimate the distance below the horizontal at the end of the board. y
ducing skid marks measuring 200 ft before coming to a stop. How fast was the car traveling when the brakes were first applied? 47. A car is traveling at 100 kmh when the driver sees an acci-
dent 80 m ahead and slams on the brakes. What constant deceleration is required to stop the car in time to avoid a pileup? 48. A model rocket is fired vertically upward from rest. Its
acceleration for the first three seconds is at 60t, at which time the fuel is exhausted and it becomes a freely “falling” body. Fourteen seconds later, the rocket’s parachute opens, and the (downward) velocity slows linearly to 18 fts in 5 s. The rocket then “floats” to the ground at that rate. (a) Determine the position function s and the velocity function v (for all times t). Sketch the graphs of s and v. (b) At what time does the rocket reach its maximum height, and what is that height? (c) At what time does the rocket land? 49. A high-speed bullet train accelerates and decelerates at the
0
x
43. Since raindrops grow as they fall, their surface area
increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 ms and its downward acceleration is a
4
9 0.9t 0
if 0 t 10 if t 10
REVIEW
CONCEPT CHECK
1. Explain the difference between an absolute maximum and a
local maximum. Illustrate with a sketch. 2. (a) What does the Extreme Value Theorem say?
(b) Explain how the Closed Interval Method works. 3. (a) State Fermat’s Theorem.
(b) Define a critical number of f .
rate of 4 fts2. Its maximum cruising speed is 90 mih. (a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes? (b) Suppose that the train starts from rest and must come to a complete stop in 15 minutes. What is the maximum distance it can travel under these conditions? (c) Find the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart. (d) The trip from one station to the next takes 37.5 minutes. How far apart are the stations?
4. (a) State Rolle’s Theorem.
(b) State the Mean Value Theorem and give a geometric interpretation. 5. (a) State the Increasing/ Decreasing Test. (b) What does it mean to say that f is concave upward on an interval I ? (c) State the Concavity Test. (d) What are inflection points? How do you find them?
248
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(b) Write an expression for x 2 in terms of x 1, f x 1 , and f x 1 . (c) Write an expression for x n1 in terms of x n , f x n , and f x n . (d) Under what circumstances is Newton’s method likely to fail or to work very slowly?
6. (a) State the First Derivative Test.
(b) State the Second Derivative Test. (c) What are the relative advantages and disadvantages of these tests? 7. If you have a graphing calculator or computer, why do you
need calculus to graph a function? 8. (a) Given an initial approximation x 1 to a root of the equa-
tion f x 0, explain geometrically, with a diagram, how the second approximation x 2 in Newton’s method is obtained.
9. (a) What is an antiderivative of a function f ?
(b) Suppose F1 and F2 are both antiderivatives of f on an interval I . How are F1 and F2 related?
T R U E - FA L S E Q U I Z 10. There exists a function f such that f x 0, f x 0,
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
and f x 0 for all x.
1. If f c 0, then f has a local maximum or minimum at c. 2. If f has an absolute minimum value at c, then f c 0. 3. If f is continuous on a, b, then f attains an absolute maxi-
mum value f c and an absolute minimum value f d at some numbers c and d in a, b.
4. If f is differentiable and f 1 f 1, then there is a num-
ber c such that c 1 and f c 0.
11. If f and t are increasing on an interval I , then f t is
increasing on I . 12. If f and t are increasing on an interval I , then f t is
increasing on I . 13. If f and t are increasing on an interval I , then ft is
increasing on I . 14. If f and t are positive increasing functions on an interval I ,
5. If f x 0 for 1 x 6, then f is decreasing on (1, 6).
then ft is increasing on I .
6. If f 2 0, then 2, f 2 is an inflection point of the
15. If f is increasing and f x 0 on I , then tx 1f x is
curve y f x.
decreasing on I .
7. If f x tx for 0 x 1, then f x tx for
0 x 1.
16. The most general antiderivative of f x x 2 is
8. There exists a function f such that f 1 2, f 3 0,
and f x 1 for all x.
Fx
9. There exists a function f such that f x 0, f x 0,
1 C x
17. If f x exists and is nonzero for all x, then f 1 f 0.
and f x 0 for all x.
EXERCISES ■ Find the local and absolute extreme values of the function on the given interval.
Sketch the graph of a function that satisfies the given conditions.
1– 4
1. f x 10 27x x 3, 2. f x x sx , 3. f x
■
x , x2 x 1
■
■
0, 4
f 2 f 1 f 9 0, lim x l f x 0, lim x l 6 f x , f x 0 on , 2, 1, 6, and 9, , f x 0 on 2, 1 and 6, 9, f x 0 on , 0 and 12, , f x 0 on 0, 6 and 6, 12
2, 0
6. f 0 0,
1, 3 ■
■
■
5. f 0 0,
0, 4
4. f x ln xx 2, ■
5–7
■
■
■
■
■
■
f is continuous and even, f x 2x if 0 x 1, f x 1 if 1 x 3, f x 1 if x 3
CHAPTER 4
f x 0 for 0 x 2, f x 0 for x 2, f x 0 for 0 x 3, f x 0 for x 3, lim x l f x 2 ■
■
■
■
■
■
■
■
26. f x sin x cos2x, ■
■
■
CAS
_2 3
4
5
6
CAS
x
7
(a) (b) (c) (d) (e)
Find the vertical and horizontal asymptotes, if any. Find the intervals of increase or decrease. Find the local maximum and minimum values. Find the intervals of concavity and the inflection points. Use the information from parts (a)–(d) to sketch the graph of f . Check your work with a graphing device. 10. f x
13. y e x e3x
14. y lnx 2 1
■
15–22
■
■
■
■
■
■
■
■
1 xx 32
■
■
28. (a) Graph the function f x 11 e 1x .
29. If f x arctancos3 arcsin x, use the graphs of f , f ,
30. If f x ln2x x sin x, use the graphs of f , f , and f
happens to the maximum and minimum points and the inflection points as c changes? Illustrate your conclusions by graphing several members of the family.
■
■
33. Show that the equation x 101 x 51 x 1 0 has exactly
one real root. 34. Suppose that f is continuous on 0, 4 , f 0 1 , and
2 f x 5 for all x in 0, 4. Show that 9 f 4 21.
35. By applying the Mean Value Theorem to the function
1 1 x x1
f x x 15 on the interval 32, 33 , show that
3 20. y sx s x
19. y x s2 x
■
2
16. y x 3 6x 2 15x 4 18. y
■
cx ; 32. Investigate the family of functions f x cxe . What
Use the guidelines of Section 4.4 to sketch the curve.
15. y x 4 3x 3 3x 2 x 17. y
■
■
What features do the members of this family have in common? How do they differ? For which values of C is f continuous on , ? For which values of C does f have no graph at all? What happens as C l ?
2
12. y e 2xx
■
1x 2
; 31. Investigate the family of functions f x lnsin x C .
1 1 x2
11. y sin2x 2 cos x
■
■
0 x 2
to estimate the intervals of increase and the inflection points of f on the interval 0, 15 .
■
9. f x 2 2 x x 3
■
and f to estimate the x-coordinates of the maximum and minimum points and inflection points of f .
CAS
9–14
■
(b) Explain the shape of the graph by computing the limits of f x as x approaches , , 0, and 0. (c) Use the graph of f to estimate the coordinates of the inflection points. (d) Use your CAS to compute and graph f . (e) Use the graph in part (d) to estimate the inflection points more accurately.
y=fª(x)
2
■
in a viewing rectangle that shows all the main aspects of this function. Estimate the inflection points. Then use calculus to find them exactly.
y
1
■
; 27. Graph f x e
function f . (a) On what intervals is f increasing or decreasing? (b) For what values of x does f have a local maximum or minimum? (c) Sketch the graph of f . (d) Sketch a possible graph of f .
0
249
■
8. The figure shows the graph of the derivative f of a
_1
■
25. f x 3x 6 5x 5 x 4 5x 3 2x 2 2
7. f is odd,
■
REVIEW
5 33 2.0125 2s
1
21. y sin 1x 22. y 4x tan x,
2 x 2
■
■
■
; 23–26
■
■
■
■
■
36. For what values of the constants a and b is 1, 6 a point of
inflection of the curve y x 3 ax 2 bx 1 ?
■
■
■
Produce graphs of f that reveal all the important aspects of the curve. Use graphs of f and f to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points. In Exercise 23 use calculus to find these quantities exactly. ■
x 1 x3 2
23. f x
sx 1x 3
24. f x
■
37. Find two positive integers such that the sum of the first
number and four times the second number is 1000 and the product of the numbers is as large as possible. 38. Find the point on the hyperbola x y 8 that is closest to the
point 3, 0.
39. Find the smallest possible area of an isosceles triangle that
is circumscribed about a circle of radius r.
250
■
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
40. Find the volume of the largest circular cone that can be
51–54
inscribed in a sphere of radius r.
■
Find f x.
51. f x 21 x 2 ,
BD 4 cm, and CD⬜ AB. Where should a point P be chosen on CD so that the sum PA PB PC is a minimum? What if CD 2 cm?
41. In ABC, D lies on AB, CD 5 cm, AD 4 cm,
52. f u
u 2 su , u
f 0 1 f 1 3
53. f x 1 6x 48x 2,
42. An observer stands at a point P, one unit away from a track.
Two runners start at the point S in the figure and run along the track. One runner runs three times as fast as the other. Find the maximum value of the observer’s angle of sight between the runners. [Hint: Maximize tan .]
f 0 1,
54. f x 2x 3 3x 2 4x 5, ■
■
■
■
■
■
f 0 2
f 0 2,
■
■
■
f 1 0 ■
■
■
55–56 ■ A particle is moving with the given data. Find the position of the particle.
P
55. at t 2, ¨
s0 1, v0 3
56. at cos t sin t,
1
■
■
■
■
■
s0 0, ■
■
v0 5 ■
■
■
■
■
57. A canister is dropped from a helicopter 500 m above the
ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100 ms. Will it burst?
S 43. The velocity of a wave of length L in deep water is
vK
; 58. Investigate the family of curves given by
L C C L
f x x 4 x 3 cx 2
where K and C are known positive constants. What is the length of the wave that gives the minimum velocity? 44. A metal storage tank with volume V is to be constructed in
the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal?
In particular you should determine the transitional value of c at which the number of critical numbers changes and the transitional value at which the number of inflection points changes. Illustrate the various possible shapes with graphs. 59. A rectangular beam will be cut from a cylindrical log of
radius 10 inches. (a) Show that the beam of maximal cross-sectional area is a square. (b) Four rectangular planks will be cut from the four sections of the log that remain after cutting the square beam. Determine the dimensions of the planks that will have maximal cross-sectional area. (c) Suppose that the strength of a rectangular beam is proportional to the product of its width and the square of its depth. Find the dimensions of the strongest beam that can be cut from the cylindrical log.
45. A hockey team plays in an arena with a seating capacity of
15,000 spectators. With the ticket price set at $12, average attendance at a game has been 11,000. A market survey indicates that for each dollar the ticket price is lowered, average attendance will increase by 1000. How should the owners of the team set the ticket price to maximize their revenue from ticket sales? 46. Use Newton’s method to find all roots of the equation
sin x x 2 3x 1 correct to six decimal places. 47. Use Newton’s method to find the absolute maximum value
of the function f t cos t t t 2 correct to eight decimal places. 48. Use the guidelines in Section 4.4 to sketch the curve
y x sin x, 0 x 2. Use Newton’s method when necessary. 49–50
■
■
10
Find the most general antiderivative of the function.
49. f x e x (2sx ) ■
depth
■
■
■
50. tt 1 tst ■
■
■
■
■
■
■
width
5
INTEGRALS
5.1
AREAS AND DISTANCES
In Chapter 2 we used the tangent and velocity problems to introduce the derivative, which is the central idea in differential calculus. In much the same way, this chapter starts with the area and distance problems and uses them to formulate the idea of a definite integral, which is the basic concept of integral calculus.We will see in Chapter 7 how to use the integral to solve problems concerning volumes, lengths of curves, work, forces on a dam, and centers of mass, among many others. There is a connection between integral calculus and differential calculus.The Fundamental Theorem of Calculus relates the integral to the derivative, and we will see in this chapter that it greatly simplifies the solution of many problems.
In this section we discover that in trying to find the area under a curve or the distance traveled by a car, we end up with the same special type of limit. THE AREA PROBLEM y
y=ƒ x=a S
x=b
a
0
x
b
FIGURE 1
S=s(x, y) | a¯x¯b, 0¯y¯ƒd
We begin by attempting to solve the area problem: Find the area of the region S that lies under the curve y f x from a to b. This means that S, illustrated in Figure 1, is bounded by the graph of a continuous function f [where f x 0], the vertical lines x a and x b, and the x-axis. In trying to solve the area problem we have to ask ourselves: What is the meaning of the word area ? This question is easy to answer for regions with straight sides. For a rectangle, the area is defined as the product of the length and the width. The area of a triangle is half the base times the height. The area of a polygon is found by dividing it into triangles (as in Figure 2) and adding the areas of the triangles.
A™ w
h l
FIGURE 2
A=lw
A¡
A£ A¢
b A= 21 bh
A=A¡+A™+A£+A¢
However, it isn’t so easy to find the area of a region with curved sides. We all have an intuitive idea of what the area of a region is. But part of the area problem is to make this intuitive idea precise by giving an exact definition of area. Recall that in defining a tangent we first approximated the slope of the tangent line by slopes of secant lines and then we took the limit of these approximations. We pursue a similar idea for areas. We first approximate the region S by rectangles and then we take the limit of the areas of these rectangles as we increase the number of rectangles. The following example illustrates the procedure.
y (1, 1)
y=≈
S
Use rectangles to estimate the area under the parabola y x 2 from 0 to 1 (the parabolic region S illustrated in Figure 3). SOLUTION We first notice that the area of S must be somewhere between 0 and 1 because S is contained in a square with side length 1, but we can certainly do better V EXAMPLE 1
0
FIGURE 3
1
x
251
252
■
CHAPTER 5
INTEGRALS
than that. Suppose we divide S into four strips S1, S2 , S3, and S4 by drawing the vertical lines x 14 , x 12 , and x 34 as in Figure 4(a). y
y
(1, 1)
(1, 1)
y=≈
S¢ S™
S£
S¡ 0
1 4
1 2
3 4
x
1
0
1 4
(a)
FIGURE 4
1 2
3 4
x
1
(b)
We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the strip [see Figure 4(b)]. In other words, the heights of these rectangles are the values of the function f x x 2 at the right endpoints of the subintervals [0, 14 ], [ 14 , 12 ], [ 12 , 34 ], and [ 34 , 1]. Each rectangle has width 41 and the heights are ( 14 )2, ( 12 )2, ( 34 )2, and 12. If we let R 4 be the sum of the areas of these approximating rectangles, we get R 4 14 ( 14 )2 14 ( 12 )2 14 ( 34 )2 14 12 15 32 0.46875 From Figure 4(b) we see that the area A of S is less than R 4 , so A 0.46875 y
Instead of using the rectangles in Figure 4(b) we could use the smaller rectangles in Figure 5 whose heights are the values of f at the left endpoints of the subintervals. (The leftmost rectangle has collapsed because its height is 0.) The sum of the areas of these approximating rectangles is
(1, 1)
y=≈
L 4 14 0 2 14 ( 14 )2 14 ( 12 )2 14 ( 34 )2 327 0.21875
0
1 4
1 2
3 4
1
x
FIGURE 5
We see that the area of S is larger than L 4 , so we have lower and upper estimates for A: 0.21875 A 0.46875 We can repeat this procedure with a larger number of strips. Figure 6 shows what happens when we divide the region S into eight strips of equal width. y
y (1, 1)
(1, 1)
y=≈
0
FIGURE 6
Approximating S with eight rectangles
1 8
1
(a) Using left endpoints
x
0
1 8
1
(b) Using right endpoints
x
SECTION 5.1
AREAS AND DISTANCES
■
253
By computing the sum of the areas of the smaller rectangles L 8 and the sum of the areas of the larger rectangles R 8 , we obtain better lower and upper estimates for A: 0.2734375 A 0.3984375
n
Ln
Rn
10 20 30 50 100 1000
0.2850000 0.3087500 0.3168519 0.3234000 0.3283500 0.3328335
0.3850000 0.3587500 0.3501852 0.3434000 0.3383500 0.3338335
So one possible answer to the question is to say that the true area of S lies somewhere between 0.2734375 and 0.3984375. We could obtain better estimates by increasing the number of strips. The table at the left shows the results of similar calculations (with a computer) using n rectangles whose heights are found with left endpoints L n or right endpoints R n . In particular, we see by using 50 strips that the area lies between 0.3234 and 0.3434. With 1000 strips we narrow it down even more: A lies between 0.3328335 and 0.3338335. A good estimate is obtained by averaging these numbers: A 0.3333335. ■ From the values in the table in Example 1, it looks as if R n is approaching 13 as n increases. We confirm this in the next example. V EXAMPLE 2 For the region S in Example 1, show that the sum of the areas of 1 the upper approximating rectangles approaches 3 , that is,
lim R n 13
nl
y
SOLUTION R n is the sum of the areas of the n rectangles in Figure 7. Each rectangle
has width 1n and the heights are the values of the function f x x 2 at the points 1n, 2n, 3n, . . . , nn; that is, the heights are 1n2, 2n2, 3n2, . . . , nn2. Thus
(1, 1)
y=≈
Rn 0
1
x
1 n
2
1 n
2 n
2
1 n
3 n
2
1 1 2 1 2 2 3 2 n 2 n n2
1 2 1 2 2 3 2 n 2 n3
1 n
FIGURE 7
1 n
1 n
n n
2
Here we need the formula for the sum of the squares of the first n positive integers:
1
12 2 2 3 2 n 2
nn 12n 1 6
Perhaps you have seen this formula before. It is proved in Example 5 in Appendix C. Putting Formula 1 into our expression for R n , we get Rn
1 nn 12n 1 n 12n 1 3 n 6 6n 2
254
■
CHAPTER 5
INTEGRALS
Thus we have
■ Here we are computing the limit of the sequence R n . Sequences and their limits will be studied in detail in Section 8.1. The idea is very similar to a limit at infinity (Section 1.6) except that in writing lim n l we restrict n to be a positive integer. In particular, we know that 1 lim 0 nl n
lim R n lim
nl
nl
lim
1 6
lim
1 6
nl
nl
When we write lim n l R n 31 we mean that we can make R n as close to 31 as we like by taking n sufficiently large.
n 12n 1 6n 2
n1 n
1
2n 1 n
1 n
2
1 n
16 1 2 13
■ 1
It can be shown that the lower approximating sums also approach 3 , that is, lim L n 13
nl
In Visual 5.1 you can create pictures like those in Figures 8 and 9 for other values of n.
From Figures 8 and 9 it appears that, as n increases, both L n and R n become better and better approximations to the area of S. Therefore, we define the area A to be the limit of the sums of the areas of the approximating rectangles, that is, A lim R n lim L n 13 nl
y
nl
y
n=10 R¡¸=0.385
0
y
n=50 R∞¸=0.3434
n=30 R£¸Å0.3502
1
x
0
1
x
0
1
x
1
x
FIGURE 8
y
y
n=10 L¡¸=0.285
0
y
n=50 L∞¸=0.3234
n=30 L£¸Å0.3169
1
x
0
1
x
0
FIGURE 9 The area is the number that is smaller than all upper sums and larger than all lower sums
SECTION 5.1
AREAS AND DISTANCES
■
255
Let’s apply the idea of Examples 1 and 2 to the more general region S of Figure 1. We start by subdividing S into n strips S1, S2 , . . . , Sn of equal width as in Figure 10. y
y=ƒ
S¡
0
a
S™
⁄
S£
¤
Si
‹
. . . xi-1
Sn
. . . xn-1
xi
b
x
FIGURE 10
The width of the interval a, b is b a, so the width of each of the n strips is x
ba n
These strips divide the interval [a, b] into n subintervals x 0 , x 1 ,
x 1, x 2 ,
x 2 , x 3 ,
...,
x n1, x n
where x 0 a and x n b. The right endpoints of the subintervals are x 1 a x, x 2 a 2 x, x 3 a 3 x, Let’s approximate the ith strip Si by a rectangle with width x and height f x i , which is the value of f at the right endpoint (see Figure 11). Then the area of the ith rectangle is f x i x . What we think of intuitively as the area of S is approximated by the sum of the areas of these rectangles, which is R n f x 1 x f x 2 x f x n x y
Îx
f(xi)
0
FIGURE 11
a
⁄
¤
‹
xi-1
xi
b
x
256
■
CHAPTER 5
INTEGRALS
Figure 12 shows this approximation for n 2, 4, 8, and 12. Notice that this approximation appears to become better and better as the number of strips increases, that is, as n l . Therefore, we define the area A of the region S in the following way.
y
0
a
b x
⁄
(a) n=2
2 DEFINITION The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:
y
A lim R n lim f x 1 x f x 2 x f x n x nl
0
a
⁄
¤
‹
b
x
nl
It can be proved that the limit in Definition 2 always exists, since we are assuming that f is continuous. It can also be shown that we get the same value if we use left endpoints:
(b) n=4 y
A lim L n lim f x 0 x f x 1 x f x n1 x
3
0
b
a
x
(c) n=8 y
nl
nl
In fact, instead of using left endpoints or right endpoints, we could take the height of the ith rectangle to be the value of f at any number x*i in the ith subinterval x i1, x i . We call the numbers x1*, x2*, . . . , x n* the sample points. Figure 13 shows approximating rectangles when the sample points are not chosen to be endpoints. So a more general expression for the area of S is A lim f x1* x f x2* x f x*n x
4
nl
y 0
b
a
x
Îx
(d) n=12 FIGURE 12 f(x *) i
0
a x*¡
⁄
¤
‹
x™*
xi-1
x£*
xi
b
xn-1
x *i
x
x n*
FIGURE 13 This tells us to end with i=n. This tells us to add. This tells us to start with i=m.
n
μ f(xi) Îx i=m
We often use sigma notation to write sums with many terms more compactly. For instance, n
f x x f x x f x x f x x i
i1
1
2
n
SECTION 5.1
AREAS AND DISTANCES
■
257
So the expressions for area in Equations 2, 3, and 4 can be written as follows: n
A lim
If you need practice with sigma notation, look at the examples and try some of the exercises in Appendix C. ■
f x x i
n l i1 n
A lim
f x
A lim
f x* x
n l i1
i1
x
n
i
n l i1
We can also rewrite Formula 1 in the following way: n
i
2
i1
nn 12n 1 6
EXAMPLE 3 Let A be the area of the region that lies under the graph of f x ex
between x 0 and x 2. (a) Using right endpoints, find an expression for A as a limit. Do not evaluate the limit. (b) Estimate the area by taking the sample points to be midpoints and using four subintervals and then ten subintervals. SOLUTION
(a) Since a 0 and b 2, the width of a subinterval is x
20 2 n n
So x 1 2n, x 2 4n, x 3 6n, x i 2in, and x n 2nn. The sum of the areas of the approximating rectangles is Rn f x 1 x f x 2 x f x n x ex1 x ex 2 x exn x e2n
2 n
e4n
2 n
e2nn
2 n
According to Definition 2, the area is A lim Rn lim nl
nl
2 2n e e4n e6n e2nn n
Using sigma notation we could write A lim
nl
2 n
n
e
2in
i1
It is difficult to evaluate this limit directly by hand, but with the aid of a computer algebra system it isn’t hard (see Exercise 18). In Section 5.3 we will be able to find A more easily using a different method. (b) With n 4 the subintervals of equal width x 0.5 are 0, 0.5 , 0.5, 1 , 1, 1.5 , and 1.5, 2 . The midpoints of these subintervals are x1* 0.25, x2* 0.75, x3* 1.25, and x4* 1.75, and the sum of the areas of the four approximating
258
■
CHAPTER 5
INTEGRALS
y 1
rectangles (see Figure 14) is 4
y=e–®
M4
f x* x i
i1
f 0.25 x f 0.75 x f 1.25 x f 1.75 x 0
1
2
x
e0.250.5 e0.750.5 e1.250.5 e1.750.5 12 e0.25 e0.75 e1.25 e1.75 0.8557
FIGURE 14
So an estimate for the area is A 0.8557 y 1
With n 10 the subintervals are 0, 0.2 , 0.2, 0.4 , . . . , 1.8, 2 and the midpoints * 1.9. Thus are x1* 0.1, x2* 0.3, x3* 0.5, . . . , x10
y=e–®
A M10 f 0.1 x f 0.3 x f 0.5 x f 1.9 x 0.2e0.1 e0.3 e0.5 e1.9 0.8632 0
1
2
x
From Figure 15 it appears that this estimate is better than the estimate with n 4.
FIGURE 15
■
THE DISTANCE PROBLEM
Now let’s consider the distance problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times. (In a sense this is the inverse problem of the velocity problem that we discussed in Section 2.1.) If the velocity remains constant, then the distance problem is easy to solve by means of the formula distance velocity time But if the velocity varies, it’s not so easy to find the distance traveled. We investigate the problem in the following example. V EXAMPLE 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30-second time interval. We take speedometer readings every five seconds and record them in the following table:
Time (s) Velocity (mih)
0
5
10
15
20
25
30
17
21
24
29
32
31
28
In order to have the time and the velocity in consistent units, let’s convert the velocity readings to feet per second (1 mih 52803600 fts): Time (s) Velocity (fts)
0
5
10
15
20
25
30
25
31
35
43
47
46
41
During the first five seconds the velocity doesn’t change very much, so we can estimate the distance traveled during that time by assuming that the velocity is constant. If we take the velocity during that time interval to be the initial velocity (25 fts),
SECTION 5.1
AREAS AND DISTANCES
■
259
then we obtain the approximate distance traveled during the first five seconds: 25 fts 5 s 125 ft Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t 5 s. So our estimate for the distance traveled from t 5 s to t 10 s is 31 fts 5 s 155 ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: 25 5 31 5 35 5 43 5 47 5 46 5 1135 ft We could just as well have used the velocity at the end of each time period instead of the velocity at the beginning as our assumed constant velocity. Then our estimate becomes 31 5 35 5 43 5 47 5 46 5 41 5 1215 ft If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second. ■ √ 40
20
0
10
FIGURE 16
20
30
t
Perhaps the calculations in Example 4 remind you of the sums we used earlier to estimate areas. The similarity is explained when we sketch a graph of the velocity function of the car in Figure 16 and draw rectangles whose heights are the initial velocities for each time interval. The area of the first rectangle is 25 5 125, which is also our estimate for the distance traveled in the first five seconds. In fact, the area of each rectangle can be interpreted as a distance because the height represents velocity and the width represents time. The sum of the areas of the rectangles in Figure 16 is L 6 1135, which is our initial estimate for the total distance traveled. In general, suppose an object moves with velocity v f t, where a t b and f t 0 (so the object always moves in the positive direction). We take velocity readings at times t0 a, t1, t2 , . . . , tn b so that the velocity is approximately constant on each subinterval. If these times are equally spaced, then the time between consecutive readings is t b an. During the first time interval the velocity is approximately f t0 and so the distance traveled is approximately f t0 t. Similarly, the distance traveled during the second time interval is about f t1 t and the total distance traveled during the time interval a, b is approximately n
f t0 t f t1 t f tn1 t
f t
i1
t
i1
If we use the velocity at right endpoints instead of left endpoints, our estimate for the total distance becomes n
f t1 t f t2 t f tn t
f t t i
i1
The more frequently we measure the velocity, the more accurate our estimates become, so it seems plausible that the exact distance d traveled is the limit of such expressions: n
5
d lim
f t
n l i1
i1
n
t lim
f t t
n l i1
We will see in Section 5.3 that this is indeed true.
i
260
■
CHAPTER 5
INTEGRALS
Because Equation 5 has the same form as our expressions for area in Equations 2 and 3, it follows that the distance traveled is equal to the area under the graph of the velocity function. In Chapter 7 we will see that other quantities of interest in the natural and social sciences—such as the work done by a variable force—can also be interpreted as the area under a curve. So when we compute areas in this chapter, bear in mind that they can be interpreted in a variety of practical ways.
5.1
EXERCISES
1. (a) By reading values from the given graph of f , use five
rectangles to find a lower estimate and an upper estimate for the area under the given graph of f from x 0 to x 10. In each case sketch the rectangles that you use. (b) Find new estimates using ten rectangles in each case. y
4. (a) Estimate the area under the graph of f x 25 x 2
from x 0 to x 5 using five approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? (b) Repeat part (a) using left endpoints.
5. (a) Estimate the area under the graph of f x 1 x 2
5
y=ƒ
0
10 x
5
from x 1 to x 2 using three rectangles and right endpoints. Then improve your estimate by using six rectangles. Sketch the curve and the approximating rectangles. (b) Repeat part (a) using left endpoints. (c) Repeat part (a) using midpoints. (d) From your sketches in parts (a)–(c), which appears to be the best estimate? 2
2. (a) Use six rectangles to find estimates of each type for the
area under the given graph of f from x 0 to x 12. (i) L 6 (sample points are left endpoints) (ii) R 6 (sample points are right endpoints) (iii) M6 (sample points are midpoints) (b) Is L 6 an underestimate or overestimate of the true area? (c) Is R 6 an underestimate or overestimate of the true area? (d) Which of the numbers L 6, R 6, or M6 gives the best estimate? Explain. y
x ; 6. (a) Graph the function f x e , 2 x 2.
(b) Estimate the area under the graph of f using four approximating rectangles and taking the sample points to be (i) right endpoints and (ii) midpoints. In each case sketch the curve and the rectangles. (c) Improve your estimates in part (b) by using 8 rectangles.
7. The speed of a runner increased steadily during the first
three seconds of a race. Her speed at half-second intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds.
8
y=ƒ
t (s)
0
0.5
1.0
1.5
2.0
2.5
3.0
v (fts)
0
6.2
10.8
14.9
18.1
19.4
20.2
4
8. Speedometer readings for a motorcycle at 12-second inter-
vals are given in the table. 0
4
8
12 x
3. (a) Estimate the area under the graph of f x 1x from
x 1 to x 5 using four approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? (b) Repeat part (a) using left endpoints.
t (s)
0
12
24
36
48
60
v (fts)
30
28
25
22
24
27
(a) Estimate the distance traveled by the motorcycle during this time period using the velocities at the beginning of the time intervals.
SECTION 5.1
AREAS AND DISTANCES
■
261
12. The velocity graph of a car accelerating from rest to a speed
(b) Give another estimate using the velocities at the end of the time periods. (c) Are your estimates in parts (a) and (b) upper and lower estimates? Explain.
of 120 kmh over a period of 30 seconds is shown. Estimate the distance traveled during this period. √ (km / h)
9. Oil leaked from a tank at a rate of rt liters per hour. The
rate decreased as time passed and values of the rate at twohour time intervals are shown in the table. Find lower and upper estimates for the total amount of oil that leaked out.
t h rt (Lh)
0
2
4
6
8
10
8.7
7.6
6.8
6.2
5.7
5.3
80 40 0
■ Use Definition 2 to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.
13–14
4 13. f x s x,
1 x 16 ln x 14. f x , 3 x 10 x
10. When we estimate distances from velocity data, it is some-
times necessary to use times t0 , t1, t2 , t3 , . . . that are not equally spaced. We can still estimate distances using the time periods ti ti ti1. For example, on May 7, 1992, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table, provided by NASA, gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Use these data to estimate the height above the Earth’s surface of the space shuttle Endeavour, 62 seconds after liftoff.
Event Launch Begin roll maneuver End roll maneuver Throttle to 89% Throttle to 67% Throttle to 104% Maximum dynamic pressure Solid rocket booster separation
Time (s)
Velocity (fts)
0 10 15 20 32 59 62 125
0 185 319 447 742 1325 1445 4151
■
■
■
■
■
n
lim
n l i1
■
■
■
■
■
■
i tan 4n 4n
Do not evaluate the limit. 16. (a) Use Definition 2 to find an expression for the area under
the curve y x 3 from 0 to 1 as a limit. (b) The following formula for the sum of the cubes of the first n integers is proved in Appendix C. Use it to evaluate the limit in part (a). 13 2 3 3 3 n 3 CAS
nn 1 2
2
17. (a) Express the area under the curve y x 5 from 0 to 2 as
a limit. (b) Use a computer algebra system to find the sum in your expression from part (a). (c) Evaluate the limit in part (a). 18. Find the exact area of the region under the graph of y ex
from 0 to 2 by using a computer algebra system to evaluate the sum and then the limit in Example 3(a). Compare your answer with the estimate obtained in Example 3(b).
11. The velocity graph of a braking car is shown. Use it to estiCAS
19. Find the exact area under the cosine curve y cos x from
x 0 to x b, where 0 b 2. (Use a computer algebra system both to evaluate the sum and to compute the limit.) In particular, what is the area if b 2?
√ (ft /s) 60
20. (a) Let A n be the area of a polygon with n equal sides
40 20 0
■
15. Determine a region whose area is equal to
CAS
mate the distance traveled by the car while the brakes are applied.
30 t (seconds)
20
10
2
4
6 t (seconds)
inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2n, show that A n 12 nr 2 sin2n. (b) Show that lim n l A n r 2. [Hint: Use Equation 1.4.5 on page 42.]
262
■
CHAPTER 5
5.2
INTEGRALS
THE DEFINITE INTEGRAL We saw in Section 5.1 that a limit of the form n
1
lim
f x* x lim f x * x f x * x f x * x i
n l i1
1
nl
n
2
arises when we compute an area. We also saw that it arises when we try to find the distance traveled by an object. It turns out that this same type of limit occurs in a wide variety of situations even when f is not necessarily a positive function. Here we consider limits similar to (1) but in which f need not be positive or continuous and the subintervals don’t necessarily have the same length. In general we start with any function f defined on a, b and we divide a, b into n smaller subintervals by choosing partition points x 0, x1, x 2, . . . , x n so that a x 0 x 1 x 2 x n1 x n b The resulting collection of subintervals x 0, x 1 ,
x 1, x 2 ,
x 2, x 3 ,
...,
x n1, x n
is called a partition P of a, b . We use the notation xi for the length of the ith subinterval xi1, xi . Thus xi xi xi1 Then we choose sample points x1*, x2*, . . . , x *n in the subintervals with x*i in the ith subinterval x i1, x i . These sample points could be left endpoints or right endpoints or any numbers between the endpoints. Figure 1 shows an example of a partition and sample points. Î⁄
Τ
΋
Îxi
Îxn
FIGURE 1
A partition of [a, b] with sample points x*i
■ The Riemann sum is named after the German mathematician Bernhard Riemann (1826–1866). See the biographical note on page 263.
a=x¸
⁄
¤
⁄*
¤*
‹
‹*
. . . xi-1
xn=b
xi . . . xn-1
x*i
x
xn*
A Riemann sum associated with a partition P and a function f is constructed by evaluating f at the sample points, multiplying by the lengths of the corresponding subintervals, and adding: n
f x* x i
i1
i
f x1* x1 f x2* x 2 f x * n xn
The geometric interpretation of a Riemann sum is shown in Figure 2. Notice that if f x* i is negative, then f x* i xi is negative and so we have to subtract the area of the corresponding rectangle.
SECTION 5.2
THE DEFINITE INTEGRAL
■
263
y
y=ƒ
A¡
A™
FIGURE 2
A Riemann sum is the sum of the areas of the rectangles above the x-axis and the negatives of the areas of the rectangles below the x-axis.
0
⁄*
a
A∞
x¢*
‹*
¤*
x∞* b
A¢
A£
x
5
μ f{x*i } Îxi=A¡+A™-A£-A¢+A∞
i=1
If we imagine all possible partitions of a, b and all possible choices of sample points, we can think of taking the limit of all possible Riemann sums as n becomes large by analogy with the definition of area. But because we are now allowing subintervals with different lengths, we need to ensure that all of these lengths xi approach 0. We can do that by insisting that the largest of these lengths, which we denote by max xi , approaches 0. The result is called the definite integral of f from a to b. Bernhard Riemann received his Ph.D. under the direction of the legendary Gauss at the University of Göttingen and remained there to teach. Gauss, who was not in the habit of praising other mathematicians, spoke of Riemann’s “creative, active, truly mathematical mind and gloriously fertile originality.” The definition (2) of an integral that we use is due to Riemann. He also made major contributions to the theory of functions of a complex variable, mathematical physics, number theory, and the foundations of geometry. Riemann’s broad concept of space and geometry turned out to be the right setting, 50 years later, for Einstein’s general relativity theory. Riemann’s health was poor throughout his life, and he died of tuberculosis at the age of 39.
2 DEFINITION OF A DEFINITE INTEGRAL If f is a function defined on a, b , the definite integral of f from a to b is the number
■
y
b
a
n
f x dx
lim
f x* x i
max xi l 0 i1
i
provided that this limit exists. If it does exist, we say that f is integrable on a, b . The precise meaning of the limit that defines the integral in Definition 2 is as follows:
x
b a
f x dx I means that for every 0 there is a corresponding number 0 such that
n
I
f x* x i
i1
i
for all partitions P of a, b with max x i and for all possible choices of x*i in x i1, x i .
This means that a definite integral can be approximated to within any desired degree of accuracy by a Riemann sum. NOTE 1 The symbol x was introduced by Leibniz and is called an integral sign. It is an elongated S and was chosen because an integral is a limit of sums. In the notation xab f x dx, f x is called the integrand and a and b are called the limits of integration; a is the lower limit and b is the upper limit. The symbol dx has no official meaning by itself; xab f x dx is all one symbol. The procedure of calculating an integral is called integration. NOTE 2 The definite integral
xab f x dx is a number; it does not depend on x. In
fact, we could use any letter in place of x without changing the value of the integral:
y
b
a
f x dx y f t dt y f r dr b
a
b
a
264
■
CHAPTER 5
INTEGRALS
We have defined the definite integral for an integrable function, but not all functions are integrable. The following theorem shows that the most commonly occurring functions are in fact integrable. The theorem is proved in more advanced courses. 3 THEOREM If f is continuous on a, b , or if f has only a finite number of jump discontinuities, then f is integrable on a, b ; that is, the definite integral xab f x dx exists.
If f is integrable on a, b , then the Riemann sums in Definition 2 must approach
xab f x dx as max x i l 0 no matter how the partitions and sample points are chosen. So in calculating the value of an integral we are free to choose partitions P and sample points x*i to simplify the calculation. It’s often convenient to take P to be a regular partition; that is, all the subintervals have the same length x . Then ba n
x x 1 x 2 x n x0 a, x 1 a x,
and
x 2 a 2 x,
...,
x i a i x
If we choose x*i to be the right endpoint of the ith subinterval, then x*i xi a i x a i
ba n
In this case, max x i x b an l 0 as n l , so Definition 2 gives
y
b
a
4 THEOREM
n
f x dx lim
n
x l 0 i1
f xi x lim
f x x
n l i1
i
If f is integrable on a, b , then
y
b
a
where
x
n
f x dx lim
f x x
n l i1
ba n
and
i
x i a i x
In computing the value of an integral, Theorem 4 is much simpler to use than Definition 2. EXAMPLE 1 Express n
lim
x
n l i1
3 i
x i sin x i x
as an integral on the interval 0, . SOLUTION Comparing the given limit with the limit in Theorem 4, we see that they will be identical if we choose f x x 3 x sin x. We are given that a 0 and
SECTION 5.2
■
THE DEFINITE INTEGRAL
265
b . Therefore, by Theorem 4, we have n
lim
x
n l i1
3 i
x i sin x i x y x 3 x sin x dx
■
0
Later, when we apply the definite integral to physical situations, it will be important to recognize limits of sums as integrals, as we did in Example 1. When Leibniz chose the notation for an integral, he chose the ingredients as reminders of the limiting process. In general, when we write n
lim
f x* x y i
n l i1
b
a
f x dx
we replace lim by x, x*i by x, and x by dx. NOTE 3 If f happens to be positive, then the Riemann sum can be interpreted as a sum of areas of approximating rectangles (see Figure 3). By comparing Theorem 4 with the definition of area in Section 5.1, we see that the definite integral xab f x dx can be interpreted as the area under the curve y f x from a to b. (See Figure 4.) y
y
Îx
0
y=ƒ
x *i
a
x
b
0
a
b
x
FIGURE 3
FIGURE 4
If ƒ˘0, the Riemann sum μ f(xi*) Îx is the sum of areas of rectangles.
If ƒ˘0, the integral ja ƒ dx is the area under the curve y=ƒ from a to b.
b
If f takes on both positive and negative values, as in Figure 5, then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis and the negatives of the areas of the rectangles that lie below the x-axis (the areas of the dark blue rectangles minus the areas of the light blue rectangles). When we take the limit of such Riemann sums, we get the situation illustrated in Figure 6. A definite integral can be interpreted as a net area, that is, a difference of areas:
y
b
a
f x dx A 1 A 2
where A 1 is the area of the region above the x-axis and below the graph of f , and A 2 is the area of the region below the x-axis and above the graph of f . y
y
y=ƒ
0 a
y=ƒ b
0 a
x
FIGURE 5
FIGURE 6
μ f(x*i ) Î x is an approximation to the net area
j
b
a
ƒ dx is the net area
b x
266
■
CHAPTER 5
INTEGRALS
EVALUATING INTEGRALS
When we use the definition or Theorem 4 to evaluate a definite integral, we need to know how to work with sums. The following three equations give formulas for sums of powers of positive integers. Equation 5 may be familiar to you from a course in algebra. Equations 6 and 7 were discussed in Section 5.1 and are proved in Appendix C. nn 1 2
n
i
5
i1
2
nn 12n 1 6
3
n
i
6
i1 n
i
7
i1
nn 1 2
2
The remaining formulas are simple rules for working with sigma notation: n
■ Formulas 8–11 are proved by writing out each side in expanded form. The left side of Equation 9 is
ca 1 ca 2 ca n The right side is ca 1 a 2 a n
c nc
8
i1 n
ca
9
n
i
c
i1 n
a
10
These are equal by the distributive property. The other formulas are discussed in Appendix C.
bi
a
n
i
i1
n
a
i
i1 n
i
i1
11
a
bi
i1
a
i
i1
n
i
b n
i
i1
b
i
i1
EXAMPLE 2
(a) Evaluate the Riemann sum for f x x 3 6x taking the sample points to be right endpoints and a 0, b 3, and n 6. (b) Evaluate y x 3 6x dx. 3
0
SOLUTION
(a) With n 6 the interval width is x
ba 30 1 n 6 2
and the right endpoints are x 1 0.5, x 2 1.0, x 3 1.5, x 4 2.0, x 5 2.5, and x 6 3.0. So the Riemann sum is 6
R6
f x x i
i1
f 0.5 x f 1.0 x f 1.5 x f 2.0 x f 2.5 x f 3.0 x 12 2.875 5 5.625 4 0.625 9 3.9375
SECTION 5.2
y
5
0
x
3
x
y
3
0
n
x 3 6x dx lim
nl
lim
nl
lim
nl
lim
y
nl
y=˛-6x
lim
A¡ 0
A™
nl
3
x
FIGURE 8 3
ba 3 n n
(˛-6x) dx=A¡-A™=_6.75
n
3i n
f x x lim f i
n l i1
lim
In the sum, n is a constant (unlike i ), so we can move 3n in front of the sign. ■
0
267
Thus x 0 0, x 1 3n, x 2 6n, x 3 9n, and, in general, x i 3in. Since we are using right endpoints, we can use Theorem 4:
FIGURE 7
j
■
Notice that f is not a positive function and so the Riemann sum does not represent a sum of areas of rectangles. But it does represent the sum of the areas of the dark blue rectangles (above the x-axis) minus the sum of the areas of the light blue rectangles (below the x-axis) in Figure 7. (b) With n subintervals we have
y=˛-6x
5
THE DEFINITE INTEGRAL
3 n 3 n
n l i1
n
i1 n
3
3i n
6
(Equation 9 with c 3n)
27 3 18 i i n3 n
i1
81 n4
3i n
3 n
n
i3
i1
54 n2
81 n4
nn 1 2
81 4
1
1 n
n
i
(Equations 11 and 9)
i1 2
54 nn 1 n2 2
2
27 1
(Equations 7 and 5)
1 n
81 27 27 6.75 4 4
This integral can’t be interpreted as an area because f takes on both positive and negative values. But it can be interpreted as the difference of areas A 1 A 2 , where A 1 and A 2 are shown in Figure 8. Figure 9 illustrates the calculation by showing the positive and negative terms in the right Riemann sum R n for n 40. The values in the table show the Riemann sums approaching the exact value of the integral, 6.75, as n l . y
5
0
y=˛-6x
3
x
n
Rn
40 100 500 1000 5000
6.3998 6.6130 6.7229 6.7365 6.7473
FIGURE 9
R¢¸Å_6.3998
■
268
■
CHAPTER 5
INTEGRALS
A much simpler method for evaluating the integral in Example 2 will be given in Section 5.3 after we have proved the Evaluation Theorem. Evaluate the following integrals by interpreting each in terms of
V EXAMPLE 3
areas. (a)
y
1
0
s1 x 2 dx
(b)
y
3
0
x 1 dx
SOLUTION
(a) Since f x s1 x 2 0, we can interpret this integral as the area under the curve y s1 x 2 from 0 to 1. But, since y 2 1 x 2, we get x 2 y 2 1, which shows that the graph of f is the quarter-circle with radius 1 in Figure 10. Therefore
y
y= œ„„„„„ 1-≈ or ≈+¥=1
1
y s1 x 1
2
0
0
1
x
FIGURE 10
dx 14 12
4
(In Section 6.2 we will be able to prove that the area of a circle of radius r is r 2.) (b) The graph of y x 1 is the line with slope 1 shown in Figure 11. We compute the integral as the difference of the areas of the two triangles:
y
3
0
x 1 dx A 1 A 2 12 2 2 12 1 1 1.5 y (3, 2)
y=x-1 A¡ 0 A™
1
3
x
_1
FIGURE 11
■
THE MIDPOINT RULE
We often choose the sample point x*i to be the right endpoint of the i th subinterval because it is convenient for computing the limit. But if the purpose is to find an approximation to an integral, it is usually better to choose x*i to be the midpoint of the interval, which we denote by x i . Any Riemann sum is an approximation to an integral, but if we use midpoints and a regular partition we get the following approximation.
Module 5.2/6.5 shows how the Midpoint Rule estimates improve as n increases.
MIDPOINT RULE
y
b
a
where and
n
f x dx
f x x x f x f x i
1
i1
x
ba n
x i 12 x i1 x i midpoint of x i1, x i
n
SECTION 5.2
y
1 y= x
y
2
1
1
■
269
1 dx. x SOLUTION The endpoints of the five subintervals are 1, 1.2, 1.4, 1.6, 1.8, and 2.0, so the midpoints are 1.1, 1.3, 1.5, 1.7, and 1.9. The width of the subintervals is x 2 15 15 , so the Midpoint Rule gives V EXAMPLE 4
0
THE DEFINITE INTEGRAL
2
x
FIGURE 12
Use the Midpoint Rule with n 5 to approximate y
2
1
1 dx x f 1.1 f 1.3 f 1.5 f 1.7 f 1.9 x 1 1 1 1 1 1 0.691908 5 1.1 1.3 1.5 1.7 1.9
Since f x 1x 0 for 1 x 2, the integral represents an area, and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles shown in Figure 12. ■ At the moment we don’t know how accurate the approximation in Example 4 is, but in Section 6.5 we will learn a method for estimating the error involved in using the Midpoint Rule. At that time we will discuss other methods for approximating definite integrals. If we apply the Midpoint Rule to the integral in Example 2, we get the picture in Figure 13. The approximation M40 6.7563 is much closer to the true value 6.75 than the right endpoint approximation, R 40 6.3998, shown in Figure 9. y
In Visual 5.2 you can compare left, right, and midpoint approximations to the integral in Example 2 for different values of n.
5
y=˛-6x
0
3
x
FIGURE 13
M¢¸Å_6.7563
PROPERTIES OF THE DEFINITE INTEGRAL
We now develop some basic properties of integrals that will help us to evaluate integrals in a simple manner. We assume that f and t are integrable functions. When we defined the definite integral xab f x dx, we implicitly assumed that a b. But the definition as a limit of Riemann sums makes sense even if a b. Notice that if we reverse a and b in Theorem 4, then x changes from b an to a bn. Therefore
y
a
b
f x dx y f x dx b
a
If a b, then x 0 and so
y
a
a
f x dx 0
270
■
CHAPTER 5
INTEGRALS
PROPERTIES OF THE INTEGRAL Suppose all the following integrals exist.
y
area=c(b-a) 0
a
b
x
FIGURE 14 b
a
y
b
2.
y
b
3.
y
b
4.
y
b
a
a
a
a
c dx cb a,
where c is any constant
f x tx dx y f x dx y tx dx b
b
a
a
cf x dx c y f x dx, b
a
where c is any constant
f x tx dx y f x dx y tx dx b
b
a
a
y=c
c
j
1.
c dx=c(b-a)
y
Property 1 says that the integral of a constant function f x c is the constant times the length of the interval. If c 0 and a b, this is to be expected because cb a is the area of the shaded rectangle in Figure 14. Property 2 says that the integral of a sum is the sum of the integrals. For positive functions it says that the area under f t is the area under f plus the area under t. Figure 15 helps us understand why this is true: In view of how graphical addition works, the corresponding vertical line segments have equal height. In general, Property 2 follows from Theorem 4 and the fact that the limit of a sum is the sum of the limits:
y
f+g
b
a
n
f x tx dx lim
f x tx x i
n l i1
i
n
g
lim
nl
f
i1
tx x i
i1
n
lim 0
n
f x x lim tx x
n l i1
b x
a
FIGURE 15
j
a
i
n l i1
i
y f x dx y tx dx b
b
n
f x i x
b
a
[ƒ+©] dx=
j
b
a
b
ƒ dx+j © dx a
■ Property 3 seems intuitively reasonable because we know that multiplying a function by a positive number c stretches or shrinks its graph vertically by a factor of c. So it stretches or shrinks each approximating rectangle by a factor c and therefore it has the effect of multiplying the area by c.
a
Property 3 can be proved in a similar manner and says that the integral of a constant times a function is the constant times the integral of the function. In other words, a constant (but only a constant) can be taken in front of an integral sign. Property 4 is proved by writing f t f t and using Properties 2 and 3 with c 1. EXAMPLE 5 Use the properties of integrals to evaluate y 4 3x 2 dx . 1
0
SOLUTION Using Properties 2 and 3 of integrals, we have
y
1
0
4 3x 2 dx y 4 dx y 3x 2 dx y 4 dx 3 y x 2 dx 1
0
1
0
1
0
We know from Property 1 that
y
1
0
4 dx 41 0 4
1
0
SECTION 5.2
THE DEFINITE INTEGRAL
■
271
and we found in Example 2 in Section 5.1 that y x 2 dx 13 . So 1
0
y
1
0
4 3x 2 dx y 4 dx 3 y x 2 dx 1
1
0
0
4 3 13 5
■
The next property tells us how to combine integrals of the same function over adjacent intervals: y
y
5. y=ƒ
0
a
FIGURE 16
c
c
a
b
x
f x dx y f x dx y f x dx b
b
c
a
Property 5 is more complicated and is proved in Appendix B, but for the case where f x 0 and a c b it can be seen from the geometric interpretation in Figure 16: The area under y f x from a to c plus the area from c to b is equal to the total area from a to b. If it is known that x010 f x dx 17 and x08 f x dx 12,
V EXAMPLE 6
find x f x dx . 10 8
SOLUTION By Property 5, we have
y
8
0
so
y
10
8
f x dx y f x dx y f x dx 10
10
8
0
f x dx y f x dx y f x dx 17 12 5 10
8
0
0
■
Notice that Properties 1–5 are true whether a b, a b, or a b. The following properties, in which we compare sizes of functions and sizes of integrals, are true only if a b. COMPARISON PROPERTIES OF THE INTEGRAL 6. If f x 0 for a x b, then
y
b
a
7. If f x tx for a x b, then
f x dx 0.
y
b
a
f x dx y tx dx. b
a
8. If m f x M for a x b, then
mb a y f x dx Mb a b
a
If f x 0, then xab f x dx represents the area under the graph of f , so the geometric interpretation of Property 6 is simply that areas are positive. (It also follows directly from the definition because all the quantities involved are positive.). Property 7 says that a bigger function has a bigger integral. It follows from Properties 6 and 4 because f t 0.
272
■
CHAPTER 5
INTEGRALS
Property 8 is illustrated by Figure 17 for the case where f x 0. If f is continuous we could take m and M to be the absolute minimum and maximum values of f on the interval a, b . In this case Property 8 says that the area under the graph of f is greater than the area of the rectangle with height m and less than the area of the rectangle with height M. In general, since m f x M, Property 7 gives
y M
y=ƒ m 0
a
b
x
y
b
a
FIGURE 17
m dx y f x dx y M dx b
b
a
a
Using Property 1 to evaluate the integrals on the left- and right-hand sides, we obtain mb a y f x dx Mb a b
a
Property 8 is useful when all we want is a rough estimate of the size of an integral without going to the bother of using the Midpoint Rule. EXAMPLE 7 Use Property 8 to estimate y ex dx. 1
2
0
2
SOLUTION Because f x ex is a decreasing function on 0, 1 , its absolute max-
imum value is M f 0 1 and its absolute minimum value is m f 1 e1. Thus, by Property 8,
y
y=1
1
e11 0 y ex dx 11 0 1
2
0
y=e–x
2
e1 y ex dx 1 1
or y=1/e
2
0
Since e1 0.3679, we can write 0.367 y ex dx 1 1
2
0
0
1
■
x
The result of Example 7 is illustrated in Figure 18. The integral is greater than the area of the lower rectangle and less than the area of the square.
FIGURE 18
5.2
EXERCISES
1. Evaluate the Riemann sum for f x 2 x 2, 0 x 2,
with four equal subintervals, taking the sample points to be right endpoints. Explain, with the aid of a diagram, what the Riemann sum represents. 2. If f x ln x 1, 1 x 4, evaluate the Riemann sum
for a regular partition with n 6, taking the sample points to be left endpoints. (Give your answer correct to six decimal places.) What does the Riemann sum represent? Illustrate with a diagram.
3. If f x sx 2, 1 x 6 , find the Riemann sum for a
regular partition with n 5 correct to six decimal places, taking the sample points to be midpoints. What does the Riemann sum represent? Illustrate with a diagram.
4. (a) Find the Riemann sum for f x x 2 sin 2x,
0 x 3, with a regular partition and six terms, taking the sample points to be right endpoints. (Give your answer correct to six decimal places.) Explain what the Riemann sum represents with the aid of a sketch. (b) Repeat part (a) with midpoints as the sample points.
SECTION 5.2
■
THE DEFINITE INTEGRAL
273
■ Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places.
5. Find the Riemann sum for f x x 3 , 1 x 1 , if the
11–14
6. Find the Riemann sum for f x x x 2 , 2 x 0 , if
11.
y
10
13.
y
1
partition points are 1, 0.5, 0, 0.5, 1 and the sample points are 1, 0.4, 0.2, 1.
the partition points are 2, 1.5, 1, 0.7, 0.4, 0 and the sample points are left endpoints. What is max x i ?
7. The graph of a function f is given. Estimate x f x dx 8 0
0
■
using four subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints.
sx 3 1 dx, n 4
12.
y
sinx 2 dx,
14.
y
5
2
■
■
15–18
n5 ■
■
■
secx3 dx,
0
1
■
x 2ex dx,
■
n6
n4
■
■
■
■
Express the limit as a definite integral on the given
■
interval.
y
n
f
15. lim
x
16. lim
1 0
sin x i x,
0,
e xi x, 1 xi
1, 5
n l i1 n
n l i1
x
1
i
n
17.
lim
s2 x* x*
lim
4 3x *
i
max x i l 0 i1
i
2
x i ,
1, 8]
n
3 8. The graph of t is shown. Estimate x3 tx dx with six sub-
18.
intervals using (a) right endpoints, (b) left endpoints, and (c) midpoints.
i
max x i l 0 i1
■
■
■
■
■
2
6x *i 5 x i , ■
■
■
0, 2 ■
■
■
■
■ Use the form of the definition of the integral given in Theorem 4 to evaluate the integral.
19–23
y
g 19.
y
5
21.
y
2
23.
y
2
1
0
x
1
1
0
1
■
1 3x dx
2 x 2 dx
20.
y
4
22.
y
5
1
0
x 2 2x 5 dx 1 2x 3 dx
x 3 dx
■
■
■
■
■
■
■
■
■
■
■
24. (a) Find an approximation to the integral x x 3x dx 4 0
using a Riemann sum with right endpoints and n 8. (b) Draw a diagram like Figure 2 to illustrate the approximation in part (a). (c) Use Theorem 4 to evaluate x04 x 2 3x dx. (d) Interpret the integral in part (c) as a difference of areas and illustrate with a diagram like Figure 6.
9. A table of values of an increasing function f is shown. Use
the table to find lower and upper estimates for x025 f x dx. x f x
0
5
10
15
20
25
42
37
25
6
15
36
■ Express the integral as a limit of Riemann sums. Do not evaluate the limit.
25–26
10. The table gives the values of a function obtained from an
experiment. Use them to estimate x06 f x dx using three equal subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints. If the function is known to be a decreasing function, can you say whether your estimates are less than or greater than the exact value of the integral? x
f x
0 9.3
1 9.0
2 8.3
3 6.5
4
5
6
2.3
7.6
10.5
2
25.
y
6
2
■
CAS
■
x dx 1 x5 ■
26. ■
■
■
y
10
1
■
■
x 4 ln x dx ■
■
■
■
27–28 ■ Express the integral as a limit of sums. Then evaluate, using a computer algebra system to find both the sum and the limit. 27. ■
y
0 ■
sin 5x dx ■
28. ■
■
■
■
y
10
2 ■
x 6 dx ■
■
■
■
274
■
CHAPTER 5
INTEGRALS
37. Given that y sx dx 9
29. The graph of f is shown. Evaluate each integral by inter-
preting it in terms of areas. (a)
y
2
(c)
y
7
0
5
f x dx
(b)
y
5
f x dx
(d)
y
9
0
0
, what is y st dt ? 4
9
38. Evaluate y x 2 cos x dx. 1
f x dx
1
39. Write as a single integral in the form xab f x dx :
f x dx
y
y
2
2
y
f x dx
5
2
f x dx y
1
2
f x dx
40. If x15 f x dx 12 and x45 f x dx 3.6, find x14 f x dx.
y=ƒ
2
38 3
4
41. If x09 f x dx 37 and x09 tx dx 16, find 0
2
4
6
x09 2 f x 3tx dx.
x
8
42. Find x05 f x dx if
3 x
f x 30. The graph of t consists of two straight lines and a semi-
43. In Example 2 in Section 5.1 we showed that x01 x 2 dx 3 . 1
circle. Use it to evaluate each integral.
y
(a)
2
0
tx dx
(b)
y
6
2
tx dx
(c)
y
7
0
for x 3 for x 3
Use this fact and the properties of integrals to evaluate x01 5 6x 2 dx.
tx dx
44. Suppose f has absolute minimum value m and absolute y 4
maximum value M. Between what two values must x02 f x dx lie? Which property of integrals allows you to make your conclusion?
y=©
2
45. Use the properties of integrals to verify that
0 y ln x dx 2 ln 3 3
0
7 x
4
1
without evaluating the integral. 46 –50 31–36
■
Evaluate the integral by interpreting it in terms of
46.
y
2
s4 x 2 dx
48.
y
2
3 2x dx
50.
y
areas. 31.
y(
33.
y (1 s9 x ) dx
35.
y x dx
3 1 2
0
x 1 dx
0
2
3
32.
y
2
34.
y
3
36.
y x 5 dx
2
1
■
■
2
1 ■
■
■
5.3
■
■
■
10
0
0
Use Property 8 to estimate the value of the integral.
■
sx 3 1 dx
47.
y
x 3 3x 3 dx
49.
y
34 4
■
■
1 dx x
3 4
tan x dx
sin2x dx
■
■
■
■
■
■
■
■
■
■
■
51. Express the following limit as a definite integral:
0
■
2
1
n
■
■
lim
n l i1
i4 n5
EVALUATING DEFINITE INTEGRALS In Section 5.2 we computed integrals from the definition as a limit of Riemann sums and we saw that this procedure is sometimes long and difficult. Sir Isaac Newton discovered a much simpler method for evaluating integrals and a few years later Leibniz made the same discovery. They realized that they could calculate xab f x dx if they happened to know an antiderivative F of f . Their discovery, called the Evaluation Theorem, is part of the Fundamental Theorem of Calculus, which is discussed in the next section.
SECTION 5.3
EVALUATING DEFINITE INTEGRALS
■
275
EVALUATION THEOREM If f is continuous on the interval a, b , then
y
b
f x dx Fb Fa
a
where F is any antiderivative of f , that is, F f . This theorem states that if we know an antiderivative F of f , then we can evaluate
xab f x dx simply by subtracting the values of F at the endpoints of the interval a, b . It is very surprising that xab f x dx, which was defined by a complicated procedure
involving all of the values of f x for a x b, can be found by knowing the values of Fx at only two points, a and b. For instance, we know from Section 4.7 that an antiderivative of f x x 2 is Fx 13 x 3, so the Evaluation Theorem tells us that
y
1
0
x 2 dx F1 F0 13 13 13 0 3 13
Comparing this method with the calculation in Example 2 in Section 5.1, where we found the area under the parabola y x 2 from 0 to 1 by computing a limit of sums, we see that the Evaluation Theorem provides us with a simple and powerful method. Although the Evaluation Theorem may be surprising at first glance, it becomes plausible if we interpret it in physical terms. If vt is the velocity of an object and st is its position at time t, then vt st, so s is an antiderivative of v. In Section 5.1 we considered an object that always moves in the positive direction and made the guess that the area under the velocity curve is equal to the distance traveled. In symbols:
y
b
a
vt dt sb sa
That is exactly what the Evaluation Theorem says in this context. PROOF OF THE EVALUATION THEOREM We divide the interval a, b into n subinter-
vals with endpoints x 0 a, x1, x 2 , … , xn b and with length x b an. Let F be any antiderivative of f . By subtracting and adding like terms, we can express the total difference in the F values as the sum of the differences over the subintervals: Fb Fa Fxn Fx 0
Fx n Fxn1 F xn1 Fxn2 F x 2 F x1 Fx1 Fx0 n
Fx Fx i
i1
i1
■ See Section 4.2 for The Mean Value Theorem.
Now F is continuous (because it’s differentiable) and so we can apply the Mean Value Theorem to F on each subinterval x i1, x i . Thus, there exists a number x*i between x i1 and x i such that Fx i Fx i1 Fx*i x i x i1 f x*i x
276
■
CHAPTER 5
INTEGRALS n
Fb Fa
Therefore
f x* x i
i1
Now we take the limit of each side of this equation as n l . The left side is a constant and the right side is a Riemann sum for the function f , so n
f x* x y
Fb Fa lim
i
n l i1
b
a
f x dx
■
When applying the Evaluation Theorem we use the notation b
]
Fx a Fb Fa and so we can write
y
b
a
b
]
f x dx Fx
where
a
F f
Other common notations are Fx ba and Fx ba . V EXAMPLE 1
In applying the Evaluation Theorem we use a particular antiderivative F of f . It is not necessary to use the most general antiderivative e x C. ■
Evaluate
y
3
1
e x dx.
SOLUTION An antiderivative of f x e x is Fx e x, so we use the Evaluation
Theorem as follows:
y
3
1
3
e x dx e x]1 e 3 e
■
y 1
EXAMPLE 2 Find the area under the cosine curve from 0 to b, where 0 b 2.
y=cos x
SOLUTION Since an antiderivative of f x cos x is Fx sin x, we have area=1 0
FIGURE 1
π 2
x
A y cos x dx sin x 0 sin b sin 0 sin b b
0
]
b
In particular, taking b 2, we have proved that the area under the cosine curve from 0 to 2 is sin2 1. (See Figure 1.)
■
When the French mathematician Gilles de Roberval first found the area under the sine and cosine curves in 1635, this was a very challenging problem that required a great deal of ingenuity. If we didn’t have the benefit of the Evaluation Theorem, we would have to compute a difficult limit of sums using obscure trigonometric identities (or a computer algebra system as in Exercise 19 in Section 5.1). It was even more difficult for Roberval because the apparatus of limits had not been invented in 1635. But in the 1660s and 1670s, when the Evaluation Theorem was discovered by Newton and Leibniz, such problems became very easy, as you can see from Example 2. INDEFINITE INTEGRALS
We need a convenient notation for antiderivatives that makes them easy to work with. Because of the relation given by the Evaluation Theorem between antiderivatives and integrals, the notation x f x dx is traditionally used for an antiderivative of f and is called an indefinite integral. Thus
SECTION 5.3
y f x dx Fx |
EVALUATING DEFINITE INTEGRALS
■
277
Fx f x
means
You should distinguish carefully between definite and indefinite integrals. A definite integral xab f x dx is a number, whereas an indefinite integral x f x dx is a function (or family of functions). The connection between them is given by the Evaluation Theorem: If f is continuous on a, b , then
y
b
a
f x dx y f x dx
b
a
Recall from Section 4.7 that if F is an antiderivative of f on an interval I , then the most general antiderivative of f on I is Fx C, where C is an arbitrary constant. For instance, the formula 1 y x dx ln x C
is valid (on any interval that doesn’t contain 0) because ddx ln x 1x. So an indefinite integral x f x dx can represent either a particular antiderivative of f or an entire family of antiderivatives (one for each value of the constant C ). The effectiveness of the Evaluation Theorem depends on having a supply of antiderivatives of functions. We therefore restate the Table of Antidifferentiation Formulas from Section 4.7, together with a few others, in the notation of indefinite integrals. Any formula can be verified by differentiating the function on the right side and obtaining the integrand. For instance,
y sec x dx tan x C 2
because
d tan x C sec2x dx
1 TABLE OF INDEFINITE INTEGRALS
y f x tx dx y f x dx y tx dx
■ We adopt the convention that when a formula for a general indefinite integral is given, it is valid only on an interval.
x n1 C n1
yx
n
dx
ye
x
dx e x C
n 1
y cf x dx c y f x dx y
1 dx ln x C x
ya
x
dx
ax C ln a
y sin x dx cos x C
y cos x dx sin x C
y sec x dx tan x C
y csc x dx cot x C
y sec x tan x dx sec x C
y csc x cot x dx csc x C
2
yx
2
1 dx tan1x C 1
2
y
1 dx sin1x C s1 x 2
278
■
CHAPTER 5
INTEGRALS
EXAMPLE 3 Find the general indefinite integral ■ The indefinite integral in Example 3 is graphed in Figure 2 for several values of C. Here the value of C is the y -intercept.
y 10x
2 sec 2x dx
4
SOLUTION Using our convention and Table 1, we have 4
y 10x _1.5
2 sec2x dx 10 y x 4 dx 2 y sec2x dx
4
1.5
10
x5 2 tan x C 5
2x 5 2 tan x C
_4
■
You should check this answer by differentiating it.
FIGURE 2
EXAMPLE 4 Evaluate y x 3 6x dx. 3
0
SOLUTION Using the Evaluation Theorem and Table 1, we have
y
3
0
x 3 6x dx
x4 x2 6 4 2
3
0
( 3 3 3 2 ) ( 14 0 4 3 0 2 ) 1 4
4
814 27 0 0 6.75 ■
Compare this calculation with Example 2(b) in Section 5.2.
V EXAMPLE 5
Find
y
2
0
of areas. ■ Figure 3 shows the graph of the integrand in Example 5. We know from Section 5.2 that the value of the integral can be interpreted as the sum of the areas labeled with a plus sign minus the area labeled with a minus sign.
2x 3 6x
3 x2 1
dx and interpret the result in terms
SOLUTION The Evaluation Theorem gives
y
2
0
3 2x 6x 2 x 1 3
x4 x2 dx 2 6 3 tan1x 4 2
2
0
2
]
12 x 4 3x 2 3 tan1x
y
0
12 2 4 32 2 3 tan1 2 0 3
4 3 tan1 2 0
FIGURE 3
2 x
This is the exact value of the integral. If a decimal approximation is desired, we can use a calculator to approximate tan1 2. Doing so, we get
y
2
0
2x 3 6x
3 x 1 2
dx 0.67855
■
SECTION 5.3
EXAMPLE 6 Evaluate y
9
1
EVALUATING DEFINITE INTEGRALS
■
279
2t 2 t 2 st 1 dt. t2
SOLUTION First we need to write the integrand in a simpler form by carrying out the division:
y
9
1
2t 2 t 2 st 1 9 dt y 2 t 12 t2 dt 2 1 t 2t
[
t 32 3 2
t1 1
2 9 9 2 3
32
9
2t 23 t 32
1
1 9
] (2 1
2 3
1 t
1
32
9
1
11 )
18 18 19 2 23 1 32 49
■
APPLICATIONS
The Evaluation Theorem says that if f is continuous on a, b , then
y
b
a
f x dx Fb Fa
where F is any antiderivative of f . This means that F f , so the equation can be rewritten as b y Fx dx Fb Fa a
We know that Fx represents the rate of change of y Fx with respect to x and Fb Fa is the change in y when x changes from a to b. [Note that y could, for instance, increase, then decrease, then increase again. Although y might change in both directions, Fb Fa represents the net change in y.] So we can reformulate the Evaluation Theorem in words as follows. NET CHANGE THEOREM The integral of a rate of change is the net change:
y
b
a
Fx dx Fb Fa
This principle can be applied to all of the rates of change in the natural and social sciences. Here are a few instances of this idea: ■
If Vt is the volume of water in a reservoir at time t, then its derivative Vt is the rate at which water flows into the reservoir at time t. So
y
t2
t1
■
Vt dt Vt2 Vt1
is the change in the amount of water in the reservoir between time t1 and time t2 . If C t is the concentration of the product of a chemical reaction at time t, then the rate of reaction is the derivative d C dt. So
y
t2
t1
d C dt C t2 C t1 dt
is the change in the concentration of C from time t1 to time t2 .
280
■
CHAPTER 5
INTEGRALS
■
If the rate of growth of a population is dndt, then
y
t2
t1
dn dt nt 2 nt1 dt
is the net change in population during the time period from t1 to t2 . (The population increases when births happen and decreases when deaths occur. The net change takes into account both births and deaths.) ■
If an object moves along a straight line with position function st, then its velocity is vt st, so
y
2
t2
t1
■
vt dt st2 st1
is the net change of position, or displacement, of the particle during the time period from t1 to t2 . In Section 5.1 we guessed that this was true for the case where the object moves in the positive direction, but now we have proved that it is always true. If we want to calculate the distance traveled during the time interval, we have to consider the intervals when vt 0 (the particle moves to the right) and also the intervals when vt 0 (the particle moves to the left). In both cases the distance is computed by integrating vt , the speed. Therefore
√
√(t)
y vt dt total distance traveled t2
3
t1
A¡ A£ 0
t¡
A™
t™
Figure 4 shows how both displacement and distance traveled can be interpreted in terms of areas under a velocity curve.
t
t™
displacement=j √(t) dt=A¡-A™+A£
■
The acceleration of the object is at vt, so
t¡
y
t™
distance=j | √ (t)| dt=A¡+A™+A£
t1
t¡
FIGURE 4
t2
at dt vt2 vt1
is the change in velocity from time t1 to time t2 . V EXAMPLE 7
A particle moves along a line so that its velocity at time t is
vt t 2 t 6 (measured in meters per second).
(a) Find the displacement of the particle during the time period 1 t 4. (b) Find the distance traveled during this time period. SOLUTION
(a) By Equation 2, the displacement is s4 s1
y
4
1
vt dt
y
4
1
t 2 t 6 dt
t3 t2 6t 3 2
4
1
9 2
This means that the particle’s position at time t 4 is 4.5 m to the left of its position at the start of the time period.
SECTION 5.3
EVALUATING DEFINITE INTEGRALS
■
281
(b) Note that vt t 2 t 6 t 3t 2 and so vt 0 on the interval 1, 3 and vt 0 on 3, 4 . Thus, from Equation 3, the distance traveled is
y vt dt y 4
To integrate the absolute value of vt, we use Property 5 of integrals from Section 5.2 to split the integral into two parts, one where vt 0 and one where vt 0 . ■
1
3
1
vt dt y vt dt 4
3
y t 2 t 6 dt y t 2 t 6 dt 3
4
1
3
t3 t2 6t 3 2
3
1
t3 t2 6t 3 2
4
3
61 10.17 m 6
■
Figure 5 shows the power consumption in the city of San Francisco for a day in September (P is measured in megawatts; t is measured in hours starting at midnight). Estimate the energy used on that day. V EXAMPLE 8
P 800 600 400 200
0
3
FIGURE 5
6
9
12
15
18
21
t
Pacific Gas & Electric
SOLUTION Power is the rate of change of energy: Pt Et. So, by the Net Change Theorem,
y
24
0
Pt dt y Et dt E24 E0 24
0
is the total amount of energy used on that day. We approximate the value of the integral using the Midpoint Rule with 12 subintervals and t 2:
y
24
0
Pt dt P1 P3 P5 P21 P23 t 440 400 420 620 790 840 850 840 810 690 670 5502 15,840
The energy used was approximately 15,840 megawatt-hours. ■
A note on units
■
How did we know what units to use for energy in Example 8? The integral
x024 Pt dt is defined as the limit of sums of terms of the form Pti* t. Now Pti* is measured in megawatts and t is measured in hours, so their product is measured in megawatt-hours. The same is true of the limit. In general, the unit of measurement for xab f x dx is the product of the unit for f x and the unit for x.
282
■
CHAPTER 5
INTEGRALS
5.3 1–28
Evaluate the integral.
■
1.
y
3
3.
y
2
5.
y
1
7.
y
0
9.
y
2
0
; 33. Use a graph to estimate the x-intercepts of the curve 2.
y
3
6x 2 4x 5 dx
4.
y
0
x 45 dx
6.
y
8 3
8.
y
10.
y
x 5 dx
1
0
EXERCISES
2x e x dx
1
1
y x x 2 x 4. Then use this information to estimate the area of the region that lies under the curve and above the x-axis.
1 2x 4x 3 dx
2
u 5 u 3 u 2 du
4 6 ; 34. Repeat Exercise 33 for the curve y 2x 3x 2x . ■ Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch.
35–36
sx dx
1
2
35.
cos d
y
2
1
■
3u 1 du 2
2
4
0
11.
y
2 dy y3
12.
y
2
13.
y x (sx sx ) dx
14.
y
9
1
4y 3
2 1
3
4
0
1
1
2v 53v 1 dv
■
15.
y
17.
y
9
19.
y
s32
21.
y
64
23.
y
1
y
4
2
sec t dt
0
1
1 dx 2x 6 dt s1 t 2
12
25.
27.
y
1
■
29. 30. ■
y
3
y
1
0 ■
y
1
20.
y
1
0
y x cos x dx x sin x cos x C
2e 4 cos x dx
3
y
32
■
■
■
■
■
■
39.
3
1
]
sec2x dx tan x ■
■
0
3 4
sec tan d
1 x 2 3 dx
0
0
sin sin tan2 d sec2
sin x dx
■
■
■
■
■
4 3
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
Find the general indefinite integral.
41.
y 1 t2 t
43.
y 1 sin x dx
■
2
dt
sin x
2
■
■
■
■
■
42.
y x1 2x
44.
y
■
4
dx
sin 2x dx sin x ■
■
■
45. The area of the region that lies to the right of the y-axis and
to the left of the parabola x 2y y 2 (the shaded region in the figure) is given by the integral x02 2y y 2 dy. (Turn your head clockwise and think of the region as lying below the curve x 2y y 2 from y 0 to y 2.) Find the area of the region.
x=2y-¥ ■
■
■
y cos x 2 sin x dx
2
■
■
■
■
■
32. y sec2x, 0 x 3 ■
■
40.
0
31. y sin x, 0 x ■
■
y
■ Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area.
■
■
■
41–44
; 31–32
■
x dx sx 2 1 C 1
2
y x sx dx
■
What is wrong with the equation?
1 x1 2 dx x 1
■
■ Find the general indefinite integral. Illustrate by graphing several members of the family on the same screen.
4 dt t 1
y
28.
■
; 39– 40
10 x dx
1
x x1 dx x
■
x
y
2
■
■
Verify by differentiation that the formula is correct.
■
38.
24.
26.
■
3x 2 dx sx
e u1 du 1 cos2 d cos2
■
sin x dx
2
0
0
■
y sx
y
■
29–30
18.
0
■
4
37.
22.
0 e
y
■
52
y
y 5y dy y3
3 1s x dx sx
1
1
16.
5
36.
■
37–38
7
■
4
x 3 dx
■
■
■
■
0 1
x
46. The boundaries of the shaded region are the y-axis, the line ■
4 y 1, and the curve y s x . Find the area of this region
SECTION 5.3
10-second intervals and recorded in the table. Use the Midpoint Rule to estimate the distance traveled by the car.
y=1
1
y=$œ„ x
0
x
1
charge: It Qt. What does xab It dt represent?
49. If oil leaks from a tank at a rate of rt gallons per minute at
time t, what does x0120 rt dt represent?
50. A honeybee population starts with 100 bees and increases
at a rate of nt bees per week. What does 100 x nt dt represent? 15 0
Rx as the derivative of the revenue function Rx, where 5000 x is the number of units sold. What does x1000 Rx dx represent? the start of the trail, what does x f x dx represent? 5 3
53. If x is measured in meters and f x is measured in newtons,
what are the units for x0100 f x dx ?
2
3
4
5
6
rt
2
10
24
36
46
54
60
(a) Give upper and lower estimates for the total quantity Q6 of erupted materials after 6 seconds. (b) Use the Midpoint Rule to estimate Q6. 61. Water flows from the bottom of a storage tank at a rate of
rate of change rt of the volume of water in the tank, in liters per day, is shown. If the amount of water in the tank at time t 0 is 25,000 L, use the Midpoint Rule to estimate the amount of water four days later.
1000
■
■
■
■
58. at 2t 3, ■
■
■
■
■
2
3
4 t
_1000
■
■
■
■
64. The area labeled B is three times the area labeled A.
Express b in terms of a. y
y
y=´
y=´
0 t 10
v 0 4, ■
1
h1 3, h2 6, h2 5, h2 13, and h is continuous everywhere. Evaluate x12 hu du.
1 t 6
v 0 5,
0
63. Suppose h is a function such that h1 2, h1 2,
0 t 3
2
57. at t 4,
56 53 50 47 45
r 2000
The acceleration function (in ms ) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time t and (b) the distance traveled during the given time interval. 57–58
60 70 80 90 100
1
55–56 ■ The velocity function (in meters per second) is given for a particle moving along a line. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval.
■
0 38 52 58 55 51
0
per foot, what are the units for dadx ? What units does x28 ax dx have?
■
0 10 20 30 40 50
t
54. If the units for x are feet and the units for ax are pounds
■
v (mih)
62. Water flows into and out of a storage tank. A graph of the
52. If f x is the slope of a trail at a distance of x miles from
■
t (s)
rt 200 4t liters per minute, where 0 t 50. Find the amount of water that flows from the tank during the first 10 minutes.
51. In Section 4.5 we defined the marginal revenue function
■
v (mih)
rt at which solid materials are spewed into the atmosphere are given in the table. The time t is measured in seconds and the units for rt are tonnes (metric tons) per second.
48. The current in a wire is defined as the derivative of the
56. vt t 2 2t 8,
t (s)
60. Suppose that a volcano is erupting and readings of the rate
47. If wt is the rate of growth of a child in pounds per year, what does x510 wt dt represent?
55. vt 3t 5,
283
■
59. The velocity of a car was read from its speedometer at
by writing x as a function of y and integrating with respect to y (as in Exercise 45). y
EVALUATING DEFINITE INTEGRALS
■
■
B
A
0 t 3 ■
■
■
■
0
a
x
0
b
x
284
■
CHAPTER 5
INTEGRALS
5.4
THE FUNDAMENTAL THEOREM OF CALCULUS The Fundamental Theorem of Calculus is appropriately named because it establishes a connection between the two branches of calculus: differential calculus and integral calculus. Differential calculus arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, the area problem. Newton’s teacher at Cambridge, Isaac Barrow (1630–1677), discovered that these two problems are actually closely related. In fact, he realized that differentiation and integration are inverse processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the derivative and the integral. It was Newton and Leibniz who exploited this relationship and used it to develop calculus into a systematic mathematical method. The first part of the Fundamental Theorem deals with functions defined by an equation of the form tx y f t dt x
1
a
y
y=f(t) area=©
0
a
x
b
t
FIGURE 1
where f is a continuous function on a, b and x varies between a and b. Observe that t depends only on x, which appears as the variable upper limit in the integral. If x is a fixed number, then the integral xax f t dt is a definite number. If we then let x vary, the number xax f t dt also varies and defines a function of x denoted by tx. If f happens to be a positive function, then tx can be interpreted as the area under the graph of f from a to x, where x can vary from a to b. (Think of t as the “area so far” function; see Figure 1.) V EXAMPLE 1 If f is the function whose graph is shown in Figure 2 and tx x0x f t dt, find the values of t0, t1, t2, t3, t4, and t5. Then sketch a rough graph of t.
y 2
y=f(t) 1
SOLUTION First we notice that t0 0
1
2
t
4
x00 f t dt 0. From Figure 3 we see that t1
is the area of a triangle: t1 y f t dt 12 1 2 1 1
0
To find t2 we add to t1 the area of a rectangle:
FIGURE 2
t2
y
2
0
f t dt y f t dt y f t dt 1 1 2 3 1
2
0
1
y 2
y 2
y 2
y 2
y 2
1
1
1
1
1
0
1
g(1)=1
FIGURE 3
t
0
1
2
g(2)=3
t
0
1
2
3
t
0
1
2
4
t
0
1
2
g(3)Å4.3
g(4)Å3
g(5)Å1.7
4
t
SECTION 5.4
THE FUNDAMENTAL THEOREM OF CALCULUS
■
285
We estimate that the area under f from 2 to 3 is about 1.3, so
y 4
t3 t2 y f t dt 3 1.3 4.3 3
g
2
3
For t 3, f t is negative and so we start subtracting areas:
2
t4 t3 y f t dt 4.3 1.3 3.0 4
1
3
0
1
2
4
3
t5 t4 y f t dt 3 1.3 1.7 5
5 x
4
FIGURE 4
We use these values to sketch the graph of t in Figure 4. Notice that, because f t is positive for t 3, we keep adding area for t 3 and so t is increasing up to x 3, where it attains a maximum value. For x 3, t decreases because f t is negative. ■
x
©=j f(t) dt a
EXAMPLE 2 If tx
and calculate tx.
xax f t dt, where a 1 and
f t t 2, find a formula for tx
SOLUTION In this case we can compute tx explicitly using the Evaluation Theorem:
tx y t 2 dt x
1
Then
h ƒ a
FIGURE 5
x
d dx
x
1
x3 1 3
(13 x 3 13) x 2
■
For the function in Example 2 notice that tx x 2, that is t f . In other words, if t is defined as the integral of f by Equation 1, then t turns out to be an antiderivative of f , at least in this case. And if we sketch the derivative of the function t shown in Figure 4 by estimating slopes of tangents, we get a graph like that of f in Figure 2. So we suspect that t f in Example 1 too. To see why this might be generally true we consider any continuous function f with f x 0. Then tx xax f t dt can be interpreted as the area under the graph of f from a to x, as in Figure 1. In order to compute tx from the definition of derivative we first observe that, for h 0, tx h tx is obtained by subtracting areas, so it is the area under the graph of f from x to x h (the shaded area in Figure 5). For small h you can see from the figure that this area is approximately equal to the area of the rectangle with height f x and width h :
y
0
tx
t3 3
x+h
b
tx h tx hf x
t
so
tx h tx f x h
Intuitively, we therefore expect that tx lim
hl0
tx h tx f x h
The fact that this is true, even when f is not necessarily positive, is the first part of the Fundamental Theorem of Calculus.
286
■
CHAPTER 5
INTEGRALS
THE FUNDAMENTAL THEOREM OF CALCULUS, PART 1 If f is continuous on We abbreviate the name of this theorem as FTC1. In words, it says that the derivative of a definite integral with respect to its upper limit is the integrand evaluated at the upper limit. ■
a, b , then the function t defined by tx y f t dt x
a x b
a
is an antiderivative of f , that is, tx f x for a x b. PROOF If x and x h are in the open interval a, b, then
tx h tx y
xh
a
Module 5.4 provides visual evidence for FTC1.
y
x
f t dt y
xh
x
xh
x
x
a
a
y
f t dt y f t dt
f t dt y f t dt x
(by Property 5)
a
f t dt
and so, for h 0, y
tx h tx 1 h h
2
y=ƒ
M
x u
√=x+h
xh
x
f t dt
For now let’s assume that h 0. Since f is continuous on x, x h , the Extreme Value Theorem says that there are numbers u and v in x, x h such that f u m and f v M, where m and M are the absolute minimum and maximum values of f on x, x h . (See Figure 6.) By Property 8 of integrals, we have
m
0
y
y
xh
f uh y
xh
mh
x
FIGURE 6
that is,
x
x
f t dt Mh f t dt f vh
Since h 0, we can divide this inequality by h : f u
1 h
y
xh
x
f t dt f v
Now we use Equation 2 to replace the middle part of this inequality: f u
3
tx h tx
f v h
Inequality 3 can be proved in a similar manner for the case h 0. Now we let h l 0. Then u l x and v l x, since u and v lie between x and x h. Thus lim f u lim f u f x
hl0
ulx
and
lim f v lim f v f x
hl0
vlx
SECTION 5.4
THE FUNDAMENTAL THEOREM OF CALCULUS
■
287
because f is continuous at x. We conclude, from (3) and the Squeeze Theorem, that tx lim
4
hl0
tx h tx f x h
If x a or b, then Equation 4 can be interpreted as a one-sided limit. Then Theorem 2.2.4 (modified for one-sided limits) shows that t is continuous on a, b . ■ Using Leibniz notation for derivatives, we can write FTC1 as d dx
y
x
a
f t dt f x
when f is continuous. Roughly speaking, this equation says that if we first integrate f and then differentiate the result, we get back to the original function f . V EXAMPLE 3
Find the derivative of the function tx y s1 t 2 dt . x
0
SOLUTION Since f t s1 t 2 is continuous, Part 1 of the Fundamental Theo-
rem of Calculus gives y
tx s1 x 2
1
f S 0
EXAMPLE 4 Although a formula of the form tx
xax f t dt may seem like a
strange way of defining a function, books on physics, chemistry, and statistics are full of such functions. For instance, the Fresnel function
x
1
■
Sx y sin t 22 dt x
0
FIGURE 7
is named after the French physicist Augustin Fresnel (1788–1827), who is famous for his works in optics. This function first appeared in Fresnel’s theory of the diffraction of light waves, but more recently it has been applied to the design of highways. Part 1 of the Fundamental Theorem tells us how to differentiate the Fresnel function: Sx sin x 22
ƒ=sin(π≈/2) x
S(x)= j sin(πt@/2) dt 0
y 0.5
1
FIGURE 8
The Fresnel function x
S(x)= j sin(πt@/2) dt 0
x
This means that we can apply all the methods of differential calculus to analyze S (see Exercise 29). Figure 7 shows the graphs of f x sin x 22 and the Fresnel function Sx x0x f t dt. A computer was used to graph S by computing the value of this integral for many values of x. It does indeed look as if Sx is the area under the graph of f from 0 to x [until x 1.4 , when Sx becomes a difference of areas]. Figure 8 shows a larger part of the graph of S. If we now start with the graph of S in Figure 7 and think about what its derivative should look like, it seems reasonable that Sx f x. [For instance, S is increasing when f x 0 and decreasing when f x 0.] So this gives a visual confirmation of Part 1 of the Fundamental Theorem of Calculus. ■
288
■
CHAPTER 5
INTEGRALS
EXAMPLE 5 Find
d dx
y
x4
1
sec t dt.
SOLUTION Here we have to be careful to use the Chain Rule in conjunction with Part 1 of the Fundamental Theorem. Let u x 4. Then
d dx
y
x4
1
sec t dt
d dx
d du
y
u
1
sec t dt
y
u
1
sec t dt
du dx
du dx
sec u
(by the Chain Rule)
(by FTC1)
secx 4 4x 3
■
DIFFERENTIATION AND INTEGRATION AS INVERSE PROCESSES
We now bring together the two parts of the Fundamental Theorem. We regard Part 1 as fundamental because it relates integration and differentiation. But the Evaluation Theorem from Section 5.3 also relates integrals and derivatives, so we rename it as Part 2 of the Fundamental Theorem.
THE FUNDAMENTAL THEOREM OF CALCULUS Suppose f is continuous
on a, b . 1. If tx 2.
xax f t dt, then tx f x.
xab f x dx Fb Fa, where F is any antiderivative of
f , that is, F f.
We noted that Part 1 can be rewritten as d dx
y
x
a
f t dt f x
which says that if f is integrated and the result is then differentiated, we arrive back at the original function f . In Section 5.3 we reformulated Part 2 as the Net Change Theorem:
y
b
a
Fx dx Fb Fa
This version says that if we take a function F, first differentiate it, and then integrate the result, we arrive back at the original function F, but in the form Fb Fa. Taken together, the two parts of the Fundamental Theorem of Calculus say that differentiation and integration are inverse processes. Each undoes what the other does. The Fundamental Theorem of Calculus is unquestionably the most important theorem in calculus and, indeed, it ranks as one of the great accomplishments of the human
SECTION 5.4
THE FUNDAMENTAL THEOREM OF CALCULUS
■
289
mind. Before it was discovered, from the time of Eudoxus and Archimedes to the time of Galileo and Fermat, problems of finding areas, volumes, and lengths of curves were so difficult that only a genius could meet the challenge. But now, armed with the systematic method that Newton and Leibniz fashioned out of the Fundamental Theorem, we will see in the chapters to come that these challenging problems are accessible to all of us. AVERAGE VALUE OF A FUNCTION T
It’s easy to calculate the average value of finitely many numbers y1 , y2 , . . . , yn :
15
yave
10 5
Tave
6 0
12
18
24
FIGURE 9
t
y1 y2 yn n
But how do we compute the average temperature during a day if infinitely many temperature readings are possible? Figure 9 shows the graph of a temperature function Tt, where t is measured in hours and T in C, and a guess at the average temperature, Tave. In general, let’s try to compute the average value of a function y f x, a x b. We start by dividing the interval a, b into n equal subintervals, each with length x b an. Then we choose points x1*, . . . , x n* in successive subintervals and calculate the average of the numbers f x1*, . . . , f x n*: f x1* f x *n n (For example, if f represents a temperature function and n 24, this means that we take temperature readings every hour and then average them.) Since x b an, we can write n b ax and the average value becomes f x 1* f x n* 1 f x1* x f x n* x ba ba x n 1 f x i* x b a i1 If we let n increase, we would be computing the average value of a large number of closely spaced values. (For example, we would be averaging temperature readings taken every minute or even every second.) The limiting value is lim
nl
1 ba
n
1
f x * x b a y i
b
a
i1
f x dx
by the definition of a definite integral. Therefore, we define the average value of f on the interval a, b as ■ For a positive function, we can think of this definition as saying area average height width
fave
1 ba
y
b
a
f x dx
290
■
CHAPTER 5
INTEGRALS
V EXAMPLE 6
interval 1, 2 .
Find the average value of the function f x 1 x 2 on the
SOLUTION With a 1 and b 2 we have
1 fave ba
y
b
a
1 f x dx 2 1
1 y1 1 x dx 3 2
2
x3 x 3
2
2
■
1
If Tt is the temperature at time t, we might wonder if there is a specific time when the temperature is the same as the average temperature. For the temperature function graphed in Figure 9, we see that there are two such times––just before noon and just before midnight. In general, is there a number c at which the value of a function f is exactly equal to the average value of the function, that is, f c fave ? The following theorem says that this is true for continuous functions. THE MEAN VALUE THEOREM FOR INTEGRALS If f is continuous on a, b , then
there exists a number c in a, b such that f c fave
y
that is,
PROOF Let Fx
b
a
1 ba
y
b
a
f x dx
f x dx f cb a
xax f t dt
for a x b. By the Mean Value Theorem for derivatives, there is a number c between a and b such that
y
y=ƒ
Fb Fa Fcb a But Fx f x by FTC1. Therefore f(c)=fave
y
b
a
0 a
c
b
■ You can always chop off the top of a (two-dimensional) mountain at a certain height and use it to fill in the valleys so that the mountaintop becomes completely flat.
The geometric interpretation of the Mean Value Theorem for Integrals is that, for positive functions f , there is a number c such that the rectangle with base a, b and height f c has the same area as the region under the graph of f from a to b. (See Figure 10 and the more picturesque interpretation in the margin note.) Since f x 1 x 2 is continuous on the interval 1, 2 , the Mean Value Theorem for Integrals says there is a number c in 1, 2 such that V EXAMPLE 7
y
1 x 2 dx f c 2 1
In this particular case we can find c explicitly. From Example 6 we know that fave 2, so the value of c satisfies f c fave 2
(_1, 2)
fave=2 _1
2
1
(2, 5) y=1+≈
■
x
FIGURE 10
y
f t dt 0 f cb a
0
FIGURE 11
1
2
x
Therefore
1 c2 2
so
c2 1
Thus in this case there happen to be two numbers c 1 in the interval 1, 2 that work in the Mean Value Theorem for Integrals. ■ Examples 6 and 7 are illustrated by Figure 11.
SECTION 5.4
5.4 1. Let tx
THE FUNDAMENTAL THEOREM OF CALCULUS
291
■
EXERCISES
x0x f t dt, where
f is the function whose graph
9. hx
is shown. (a) Evaluate t0, t1, t2, t3, and t6. (b) On what interval is t increasing? (c) Where does t have a maximum value? (d) Sketch a rough graph of t.
11. y
f
0
t
5
■
10. hx
arctan t dt
cos t dt t
sx
y
3x
2x
12. y
y
0
ex
y
x2
0
s1 r 3 dr
sin3t dt
u2 1 du u2 1
Hint: y f u du y f u du y f u du 3x
0
2x
14. y
1
1x
2
3
13. tx
y
1
y
y
y
cos x
sin x
■
15–18
0
1 v 210 dv
■
■
3x
2x
■
■
■
■
■
■
■
■
■
Find the average value of the function on the given
interval. 2. Let tx
x0x f t dt, where
15. f x x 2,
f is the function whose graph
18. f sec tan , ■
19–20
■
■
4
1
6
0
4. tx
1 t 2 dt
■
■
y
7. t y
y
8. Fx
y
x
0 y
2
■
■
■
■
■
s1 2t dt
■
■
6. tx
t 2 sin t dt
x
tan d
Hint: y tan d y tan d 10
x
x
10
■
■
■
2, 5
0, 4 ■
■
■
■
■
■
■
■
■
21. The table gives values of a continuous function. Use the
Midpoint Rule to estimate the average value of f on 20, 50 .
y0 (1 st ) dt
x
20
25
30
35
40
45
50
f x
42
38
31
29
35
48
60
■
■
■
y
x
1
ln t dt
22. The velocity graph of an accelerating car is shown. ■
(a) Estimate the average velocity of the car during the first 12 seconds. (b) At what time was the instantaneous velocity equal to the average velocity? √ (km/h) 60 40
10
■
x
■
5. tx
■
■
20. f x sx ,
t
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
5–14
■
19. f x x 32,
Sketch the area represented by tx. Then find tx in two ways: (a) by using Part 1 of the Fundamental Theorem and (b) by evaluating the integral using Part 2 and then differentiating.
y
0, 4
■
1
■
x
■
y
■
3. tx
■
(a) Find the average value of f on the given interval. (b) Find c such that fave f c. (c) Sketch the graph of f and a rectangle whose area is the same as the area under the graph of f .
0
3– 4
■
1, 4
0, 2
17. tx cos x,
is shown. (a) Evaluate tx for x 0, 1, 2, 3, 4, 5, and 6. (b) Estimate t7. (c) Where does t have a maximum value? Where does it have a minimum value? (d) Sketch a rough graph of t.
16. f x 1x,
1, 1
20 0
4
8
12 t (seconds)
292
■
CHAPTER 5
23. If Fx
y
x
1
INTEGRALS
f t dt, where f t y
t2
1
s1 u 4 du, u
CAS
find F 2.
(b) On what intervals is the function concave upward? (c) Use a graph to solve the following equation correct to two decimal places:
24. Find the interval on which the curve
yy
y
1 dt 1 t t2
x
0
CAS
Six y
■ Let tx x f t dt , where f is the function whose graph is shown. (a) At what values of x do the local maximum and minimum values of t occur? (b) Where does t attain its absolute maximum value? (c) On what intervals is t concave downward? (d) Sketch the graph of t.
f
1 0 _1
2
4
6
sin t dt t
is important in electrical engineering. [The integrand f t sin tt is not defined when t 0, but we know that its limit is 1 when t l 0. So we define f 0 1 and this makes f a continuous function everywhere.] (a) Draw the graph of Si. (b) At what values of x does this function have local maximum values? (c) Find the coordinates of the first inflection point to the right of the origin. (d) Does this function have horizontal asymptotes? (e) Solve the following equation correct to one decimal place:
y 3 2
x
0
x 0
25.
sin t 22 dt 0.2
30. The sine integral function
is concave upward. 25–26
x
0
t
8
y
sin t dt 1 t
x
0
_2
31. Find a function f and a number a such that 26.
y
f
6y
0.4 0.2
1
3
5
7
32. A high-tech company purchases a new computing system
t
9
_0.2
■
f t dt 2 sx t2
for all x 0.
0
■
x
a
■
■
■
■
■
■
■
■
■
■
27. If f 1 12, f is continuous, and x14 f x dx 17, what is
whose initial value is V. The system will depreciate at the rate f f t and will accumulate maintenance costs at the rate t tt, where t is the time measured in months. The company wants to determine the optimal time to replace the system. (a) Let
the value of f 4?
Ct
28. The error function
erfx
2 s
y
x
0
2
et dt
is used in probability, statistics, and engineering. 2 (a) Show that xab et dt 12 s erfb erfa . 2 (b) Show that the function y e x erfx satisfies the differential equation y 2xy 2s .
y
t
0
f s ts ds
Show that the critical numbers of C occur at the numbers t where Ct f t tt. (b) Suppose that
f t
29. The Fresnel function S was defined in Example 4 and
graphed in Figures 7 and 8. (a) At what values of x does this function have local maximum values?
1 t
and
V V t 15 450 0 tt
if 0 t 30 if t 30
Vt 2 12,900
t0
SECTION 5.5
■
293
(a) Explain why x0t f s ds represents the loss in value of the machine over the period of time t since the last overhaul. (b) Let C Ct be given by
Determine the length of time T for the total depreciation Dt x0t f s ds to equal the initial value V. (c) Determine the absolute minimum of C on 0, T . (d) Sketch the graphs of C and f t in the same coordinate system, and verify the result in part (a) in this case.
Ct
33. A manufacturing company owns a major piece of equip-
ment that depreciates at the (continuous) rate f f t, where t is the time measured in months since its last overhaul. Because a fixed cost A is incurred each time the machine is overhauled, the company wants to determine the optimal time T (in months) between overhauls.
5.5
THE SUBSTITUTION RULE
1 t
A y f s ds t
0
What does C represent and why would the company want to minimize C? (c) Show that C has a minimum value at the numbers t T where CT f T .
THE SUBSTITUTION RULE Because of the Fundamental Theorem, it’s important to be able to find antiderivatives. But our antidifferentiation formulas don’t tell us how to evaluate integrals such as
y 2xs1 x
1
■ Differentials were defined in Section 2.8. If u f x, then
du f x dx
2
dx
To evaluate this integral our strategy is to simplify the integral by changing from the variable x to a new variable u. Suppose that we let u be the quantity under the root sign in (1), u 1 x 2. Then the differential of u is du 2x dx. Notice that if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2x dx would occur in (1) and, so, formally, without justifying our calculation, we could write
y 2xs1 x
2
2
dx y s1 x 2 2x dx y su du 23 u 32 C 23 x 2 132 C
But now we can check that we have the correct answer by using the Chain Rule to differentiate the final function of Equation 2: d dx
[ 23 x 2 132 C] 23 32 x 2 112 2x 2xsx 2 1
In general, this method works whenever we have an integral that we can write in the form x f txtx dx. Observe that if F f , then 3
y Ftxtx dx F tx C
because, by the Chain Rule, d Ftx Ftxtx dx
294
■
CHAPTER 5
INTEGRALS
If we make the “change of variable” or “substitution” u tx, then from Equation 3 we have
y Ftxtx dx Ftx C Fu C y Fu du or, writing F f , we get
y f txtx dx y f u du Thus we have proved the following rule. 4 THE SUBSTITUTION RULE If u tx is a differentiable function whose range is an interval I and f is continuous on I , then
y f txtx dx y f u du Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation. Notice also that if u tx, then du tx dx, so a way to remember the Substitution Rule is to think of dx and du in (4) as differentials. Thus the Substitution Rule says: It is permissible to operate with dx and du after integral signs as if they were differentials. EXAMPLE 1 Find y x 3 cosx 4 2 dx. SOLUTION We make the substitution u x 4 2 because its differential is
du 4x 3 dx, which, apart from the constant factor 4, occurs in the integral. Thus, using x 3 dx 14 du and the Substitution Rule, we have
yx
3
cosx 4 2 dx y cos u 14 du 14 y cos u du 14 sin u C
■
Check the answer by differentiating it.
14 sinx 4 2 C Notice that at the final stage we had to return to the original variable x.
■
The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral. This is accomplished by changing from the original variable x to a new variable u that is a function of x. Thus in Example 1 we replaced the integral x x 3 cosx 4 2 dx by the simpler integral 14 x cos u du. The main challenge in using the Substitution Rule is to think of an appropriate substitution. You should try to choose u to be some function in the integrand whose differential also occurs (except for a constant factor). This was the case in Example 1. If that is not possible, try choosing u to be some complicated part of the integrand (perhaps the inner function in a composite function). Finding the right substitution is a bit of an art. It’s not unusual to guess wrong; if your first guess doesn’t work, try another substitution.
SECTION 5.5
THE SUBSTITUTION RULE
■
295
EXAMPLE 2 Evaluate y s2x 1 dx. SOLUTION 1 Let u 2x 1. Then du 2 dx, so dx 2 du. Thus the Substitution 1
Rule gives
y s2x 1 dx y su
12 du 12 y u 12 du
1 u 32 C 13 u 32 C 2 32
13 2x 132 C SOLUTION 2 Another possible substitution is u s2x 1 . Then
du
dx s2x 1
so
dx s2x 1 du u du
(Or observe that u 2 2x 1, so 2u du 2 dx.) Therefore
y s2x 1 dx y u u du y u
V EXAMPLE 3
Find y
2
du
u3 C 13 2x 132 C 3
■
x dx . s1 4x 2
SOLUTION Let u 1 4x 2. Then du 8x dx, so x dx 8 du and 1
1
y
f _1
x 1 dx 18 y du 18 y u 12 du 2 s1 4x su
1
18 (2su ) C 14 s1 4x 2 C
©= ƒ dx _1
FIGURE 1
ƒ=
x 1-4≈ œ„„„„„„
1-4≈ ©=j ƒ dx=_ 41 œ„„„„„„
■
The answer to Example 3 could be checked by differentiation, but instead let’s check it with a graph. In Figure 1 we have used a computer to graph both the integrand f x xs1 4x 2 and its indefinite integral tx 14 s1 4x 2 (we take the case C 0). Notice that tx decreases when f x is negative, increases when f x is positive, and has its minimum value when f x 0. So it seems reasonable, from the graphical evidence, that t is an antiderivative of f . EXAMPLE 4 Calculate y e 5x dx. SOLUTION If we let u 5x, then du 5 dx, so dx 5 du. Therefore 1
ye
5x
dx 15 y e u du 15 e u C 15 e 5x C
■
296
■
CHAPTER 5
INTEGRALS
V EXAMPLE 5
Calculate y tan x dx.
SOLUTION First we write tangent in terms of sine and cosine:
y tan x dx y
sin x dx cos x
This suggests that we should substitute u cos x, since then du sin x dx and so sin x dx du: sin x 1 dx y du cos x u
y tan x dx y
ln u C ln cos x C
Since ln cos x ln cos x Example 5 can also be written as
1
■
ln1 cos x ln sec x , the result of
y tan x dx ln sec x C
5
DEFINITE INTEGRALS
When evaluating a definite integral by substitution, two methods are possible. One method is to evaluate the indefinite integral first and then use the Evaluation Theorem. For instance, using the result of Example 2, we have
y
4
0
s2x 1 dx y s2x 1 dx
4
]
0
4
]
13 2x 132
0
13 932 13 132 13 27 1 263 Another method, which is usually preferable, is to change the limits of integration when the variable is changed. ■ This rule says that when using a substitution in a definite integral, we must put everything in terms of the new variable u, not only x and dx but also the limits of integration. The new limits of integration are the values of u that correspond to x a and x b.
6 THE SUBSTITUTION RULE FOR DEFINITE INTEGRALS
a, b and f is continuous on the range of u tx, then
y
b
a
f txtx dx y
tb
ta
If t is continuous on
f u du
PROOF Let F be an antiderivative of f . Then, by (3), F tx is an antiderivative of
f txtx and so, by the Evaluation Theorem, we have
y
b
a
]
b
f txtx dx Ftx a Ftb F ta
But, applying the Evaluation Theorem a second time, we also have
y
tb
ta
]
f u du Fu
tb ta
Ftb Fta
■
SECTION 5.5
■ The integral given in Example 6 is an abbreviation for
y
2
1
1 dx 3 5x2
EXAMPLE 6 Evaluate y
1
limits of integration we note that
y
2
1
dx 1 3 5x2 5 1 5u
Since the function f x ln xx in Example 7 is positive for x 1 , the integral represents the area of the shaded region in Figure 2. ■
Calculate y
e
1
y
7
2
7
2
du 1 u2 5 1 5
7
1 u
2
1 1 7 2
1 14
ln x dx. x
SOLUTION We let u ln x because its differential du dxx occurs in the integral.
ln x x
When x 1, u ln 1 0; when x e, u ln e 1. Thus
y
e
1
1
when x 2, u 3 52 7
and
Observe that when using (6) we do not return to the variable x after integrating. We simply evaluate the expression in u between the appropriate values of u. ■ V EXAMPLE 7
y
0
297
SOLUTION Let u 3 5x. Then du 5 dx, so dx 5 du . To find the new
Therefore
y=
■
dx . 3 5x2
2
1
when x 1, u 3 51 2
0.5
THE SUBSTITUTION RULE
e
x
ln x 1 u2 dx y u du 0 x 2
1
0
1 2
■
SYMMETRY FIGURE 2
The next theorem uses the Substitution Rule for Definite Integrals (6) to simplify the calculation of integrals of functions that possess symmetry properties.
7 INTEGRALS OF SYMMETRIC FUNCTIONS
Suppose f is continuous on a, a .
a (a) If f is even f x f x , then xa f x dx 2 x0a f x dx . a (b) If f is odd f x f x , then xa f x dx 0.
PROOF We split the integral in two: 8
y
a
a
f x dx y f x dx y f x dx y 0
a
a
0
a
0
f x dx y f x dx a
0
In the first integral on the far right side we make the substitution u x. Then du dx and when x a, u a. Therefore y
a
0
f x dx y f u du y f u du a
0
a
0
298
■
CHAPTER 5
INTEGRALS
and so Equation 8 becomes
y
9
a
a
f x dx y f u du y f x dx a
a
0
0
y
(a) If f is even, then f u f u so Equation 9 gives
_a
0
a
a
y
a
x
f x dx y f u du a
a
0
y
a
0
f x dx 2 y f x dx a
0
a
(a) ƒ even, j ƒ dx=2 j ƒ dx _a
(b) If f is odd, then f u f u and so Equation 9 gives
0
y
y
a
a
_a
f x dx y f u du a
0
y
a
0
f x dx 0
■
0 a a
(b) ƒ odd, j ƒ dx=0 _a
FIGURE 3
x
Theorem 7 is illustrated by Figure 3. For the case where f is positive and even, part (a) says that the area under y f x from a to a is twice the area from 0 to a because of symmetry. Recall that an integral xab f x dx can be expressed as the area above the x-axis and below y f x minus the area below the axis and above the curve. Thus part (b) says the integral is 0 because the areas cancel. V EXAMPLE 8
Since f x x 6 1 satisfies f x f x, it is even and so
y
2
2
x 6 1 dx 2 y x 6 1 dx 2
0
[
2
]
128 284 2 x 7 x 0 2 ( 7 2) 7 1 7
■
EXAMPLE 9 Since f x tan x1 x 2 x 4 satisfies f x f x, it is odd
and so
y
1
1
5.5 1–6
■
y cos 3x dx,
2.
y x4 x
3.
yx
4.
y
5.
y 1 2x
2
u 3x
dx,
u 4 x2
sx 3 1 dx,
u x3 1
2 10
sin sx dx , sx 4
3
dx,
■
EXERCISES
Evaluate the integral by making the given substitution.
1.
tan x dx 0 1 x2 x4
u sx u 1 2x
6.
ye
■
sin
■
7–34
■
cos d, u sin ■
■
y 2xx
9.
y 3x 2
2
34 dx 20
y
■
■
■
■
■
Evaluate the indefinite integral.
7.
11.
■
ln x2 dx x
dx
y x x
3
10.
y xe
dx
12.
y 2 x
8.
2
x2
5 9 dx
6
dx
■
■
SECTION 5.5
13.
15.
dx
y 5 3x
14.
a bx 2 dx s3ax bx 3
y
16.
y x
2
x dx 12
49. ■
1 dt 5t 42.7
y
y
e4
e
dx x sln x
■
51–54
THE SUBSTITUTION RULE
50.
■
■
■
■
■
y
12
0
■
■
■
299
■
■
sin1 x dx s1 x 2 ■
■
Find the average value of the function on the given
interval. 17.
y sin t dt
18.
yy
3
s2y 4 1 dy
2
51. f t tet ,
0, 5
52. tx x s1 x 3 ,
0, 2
53. hx cos4x sin x,
0,
2
19.
21.
ye
x
s1 e dx x
y cos
20.
sin6 d
23.
y scot x csc x dx
25.
y s1 x
22.
2
dx 2
sin1 x
y sec 2 y
tan 2 d
x dx x2 1
54. hr 31 r ,
tan1 x
2 55. Evaluate x2 x 3s4 x 2 dx by writing it as a sum of
24.
y 1x
26.
y
2
■
dx
cosx dx x2
y sin t sec cos t dt
29.
y
ex 1 dx ex
30.
ye
y
sin 2x dx 1 cos2x
32.
y
1x dx 1 x2
■
■
■
35–50
■
35.
y
2
37.
y
1
39.
y
41.
y
4
43.
y
2
45.
y
1
47.
y
0
0
0
0
■
■
■
y
■
y y
x dx 1 x4
■
■
■
7
x 21 2x 3 5 dx
38.
y
s
sec 2t4 dt
40.
y
12
2
0
0
■
■
s4 3x dx
0
16
y
■
■
■
■
y
y=e sin x sin 2x
y=eœ„x
Evaluate the definite integral.
x cosx 2 dx csc t cot t dt
■
1x
0
0
1x
1
πx 2
58. A bacteria population starts with 400 bacteria and grows at
a rate of rt 450.268e1.12567t bacteria per hour. How many bacteria will there be after three hours? 59. Breathing is cyclic and a full respiratory cycle from the
beginning of inhalation to the end of exhalation takes about 5 s. The maximum rate of air flow into the lungs is about 0.5 Ls. This explains, in part, why the function f t 12 sin2 t5 has often been used to model the rate of air flow into the lungs. Use this model to find the volume of inhaled air in the lungs at time t. 60. Alabama Instruments Company has set up a production line
e sx dx sx
42.
y
x sx 1 dx
44.
y
ez 1 dz ez z
46.
y
4
48.
y
a
tan3 d
■
y=2x´
sin x dx 1 cos2x
y
6
■
57. Which of the following areas are equal? Why?
ex dx 1
x
36.
6
■
2
x 125 dx
1
1
34.
■
interpreting the resulting integral in terms of an area.
28.
33.
■
56. Evaluate x01 x s1 x 4 dx by making a substitution and
y sec x tan x dx
31.
■
two integrals and interpreting one of those integrals in terms of an area.
27.
3
1, 6
2
0
cos x sinsin x dx
2
2
0
0
x 2 sin x dx 1 x6
x dx s1 2x x sa 2 x 2 dx
to manufacture a new calculator. The rate of production of these calculators after t weeks is
dx 100 5000 1 dt t 102
calculatorsweek
(Notice that production approaches 5000 per week as time goes on, but the initial production is lower because of the workers’ unfamiliarity with the new techniques.) Find the number of calculators produced from the beginning of the third week to the end of the fourth week.
300
■
CHAPTER 5
INTEGRALS
64. If f is continuous on ⺢, prove that
61. If f is continuous and y f x dx 10, find y f 2x dx. 4
2
0
0
y
62. If f is continuous and y f x dx 4, find y x f x 2 dx. 9
3
0
y
a
y
f x dx
a
bc
ac
f x dx
For the case where f x 0, draw a diagram to interpret this equation geometrically as an equality of areas.
b
f x dx
65. If a and b are positive numbers, show that
For the case where f x 0 and 0 a b, draw a diagram to interpret this equation geometrically as an equality of areas.
5
y
f x c dx
0
63. If f is continuous on ⺢, prove that b
b
a
REVIEW
y
1
0
x a1 xb dx
y
1
0
x b1 xa dx
CONCEPT CHECK
1. (a) Write an expression for a Riemann sum of a function f .
Explain the meaning of the notation that you use. (b) If f x 0, what is the geometric interpretation of a Riemann sum? Illustrate with a diagram. (c) If f x takes on both positive and negative values, what is the geometric interpretation of a Riemann sum? Illustrate with a diagram. 2. (a) Write the definition of the definite integral of f from a
to b. How does the definition simplify if you know that f is continuous and you use equal subintervals? (b) What is the geometric interpretation of xab f x dx if f x 0? (c) What is the geometric interpretation of xab f x dx if f x takes on both positive and negative values? Illustrate with a diagram. 3. State the Midpoint Rule.
5. (a) Explain the meaning of the indefinite integral x f x dx.
(b) What is the connection between the definite integral xab f x dx and the indefinite integral x f x dx ? 6. State both parts of the Fundamental Theorem of Calculus. 7. Suppose a particle moves back and forth along a straight line with velocity vt, measured in feet per second, and
acceleration at. (a) What is the meaning of x60120 vt dt ?
(b) What is the meaning of x60120 vt dt ? (c) What is the meaning of x
120 60
at dt ?
8. (a) What is the average value of a function f on an
interval a, b ? (b) What does the Mean Value Theorem for Integrals say? What is its geometric interpretation? 9. Explain exactly what is meant by the statement that
4. (a) State the Evaluation Theorem.
(b) State the Net Change Theorem. (c) If rt is the rate at which water flows into a reservoir, what does xtt rt dt represent? 2
1
“differentiation and integration are inverse processes.” 10. State the Substitution Rule. In practice, how do you
use it?
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
y
a
y
b
a
1. If f and t are continuous on a, b , then b
2. If f and t are continuous on a, b , then
f xtx dx
y
b
a
f x dx
y
b
a
3. If f is continuous on a, b , then
f x tx dx y f x dx y tx dx b
b
a
a
y
b
a
5f x dx 5 y f x dx b
a
tx dx
CHAPTER 5
4. If f is continuous on a, b , then
y
b
a
REVIEW
then f x tx for a x b.
x f x dx x y f x dx b
a
sin x 1 x 4 2
y
1
10.
y
5
y
1
6. If f is continuous on 1, 3 , then y f v dv f 3 f 1.
11.
7. If f and t are continuous and f x tx for a x b,
12.
x02 x x 3 dx represents the area under the curve
5. If f is continuous on a, b and f x 0, then
y
a
sf x dx
301
8. If f and t are differentiable and f x tx for a x b,
9.
b
■
y
b
a
f x dx 3
1
5
2
x 5 6x 9
dx 0
ax 2 bx c dx 2 y ax 2 c dx 5
0
1 3 dx 4 x 8
1
y x x 3 from 0 to 2.
then
y
b
a
f x dx
y
b
a
tx dx
13. All continuous functions have antiderivatives.
EXERCISES 1. Use the given graph of f to find the Riemann sum with six
3. Evaluate
subintervals. Take the sample points to be (a) left endpoints and (b) midpoints. In each case draw a diagram and explain what the Riemann sum represents.
y ( x s1 x ) dx 1
2
0
by interpreting it in terms of areas.
y
4. Express n
lim
y=ƒ
2
0
2
sin x
n l i1
6
x
i
x
as a definite integral on the interval 0, and then evaluate the integral. 5. The following figure shows the graphs of f, f , and
x0x f t dt. Identify each graph, and explain your choices. y
b
2. (a) Evaluate the Riemann sum for
f x x 2 x
0 x 2
c
with four subintervals, taking the sample points to be right endpoints. Explain, with the aid of a diagram, what the Riemann sum represents. (b) Use the definition of a definite integral (with right endpoints) to calculate the value of the integral
y
2
0
x 2 x dx
(c) Use the Fundamental Theorem to check your answer to part (b). (d) Draw a diagram to explain the geometric meaning of the integral in part (b).
x
a
6. Evaluate:
(a)
y
(c)
d dx
1
0
d arctan x e dx dx
y
x
0
e arctan t dt
(b)
d dx
y
1
0
e arctan x dx
302
■
7–32
■
y
2
9.
y
1
11.
y
9
13.
y
1
15.
y
1
17.
y
1
19.
y
4
21.
7.
CHAPTER 5
INTEGRALS
Evaluate the integral, if it exists.
8x 3 3x 2 dx
1
8.
y
T
1
39. Use Property 8 of integrals to estimate the value of
x 4 8x 7 dx
0
10.
y
su 2u 2 du u
12.
y (su
y y 2 15 dy
14.
y
2
v 2 cos v 3 dv
16.
y
1
e t dt
18.
y
1
1 x x2 dx x2
20.
y
y sx
x2 dx 2 4x
22.
y csc
23.
y sin t cos t dt
24.
y sin x coscos x dx
25.
y
26.
y
cosln x dx x
28.
y
x dx s1 x 4
30.
y sinh1 4x dx
32.
y
1
0
0
0
2
e sx dx sx
27.
y tan x lncos x dx
29.
y 1x
31.
y
■
x3
dx
4
sec tan d 1 sec ■
■
■
0
1 x9 dx
1
4
0
0
0
■
■
40. Use the properties of integrals to verify that
0 y x 4 cos x dx 0.2 1
1 2 du
0
41. Use the Midpoint Rule with n 5 to approximate
sin3 t dt
42. A particle moves along a line with velocity function vt t 2 t, where v is measured in meters per second.
x01 s1 x 3
1 dx 2 3x
2
1
2
4
43. Let rt be the rate at which the world’s oil is consumed,
where t is measured in years starting at t 0 on January 1, 2000, and rt is measured in barrels per year. What does x03 rt dt represent?
3t dt
44. A radar gun was used to record the speed of a runner at the
times given in the table. Use the Midpoint Rule to estimate the distance the runner covered during those 5 seconds.
1 tan t3 sec2t dt
■
■
■
■
guess the value of the integral x02 f x dx. Then evaluate the integral to confirm your guess. Find the derivative of the function.
35. Fx 37. y ■
■
y
y
1
et dt t
x
sx ■
■
38. y ■
■
■
■
y
t (s)
v (ms)
0 0.5 1.0 1.5 2.0 2.5
0 4.67 7.34 8.86 9.73 10.22
3.0 3.5 4.0 4.5 5.0
10.51 10.67 10.76 10.81 10.81
bees per week, where the graph of r is as shown. Use the Midpoint Rule with six subintervals to estimate the increase in the bee population during the first 24 weeks. r 12000 8000
36. tx
s1 t dt 4
v (ms)
45. A population of honeybees increased at a rate of rt
2 3 ; 34. Graph the function f x cos x sin x and use the graph to
x
t (s)
■
region that lies under the curve y x sx , 0 x 4 . Then find the exact area.
■
dx .
Find (a) the displacement and (b) the distance traveled by the particle during the time interval 0, 5 .
sin x dx 1 x2
1
; 33. Use a graph to give a rough estimate of the area of the
35–38
sx 2 3 dx
y 2s1 y 3 dy
0
■
3
1
1 x 9 dx
0
y
y
cos x 3
1
3x1
2x
s1 t dt 2
4000
sint 4 dt 0
■
■
■
■
4
8
12
16
20
24
t (weeks)
CHAPTER 5
46. Find the average value of the function f x x 2 s1 x 3
on the interval 0, 2 .
initial hot spot concentrated at the origin, we need to compute lim Tx, t
47. If f is a continuous function, what is the limit as h l 0 of
the average value of f on the interval x, x h ?
48. Let
x 1 f x s1 x 2
if 3 x 0 if 0 x 1
1 f x dx by interpreting the integral as a differEvaluate x3 ence of areas.
; 49. Estimate the value of the number c such that the area under
al0
Use l’Hospital’s Rule to find this limit. 51. If f is a continuous function such that
y
x
0
along the x-axis is initially C2a if x a and 0 if x a. It can be shown that if the heat diffusivity of the rod is k, then the temperature of the rod at the point x at time t is C a 2 Tx, t y e xu 4kt du a s4 kt 0
To find the temperature distribution that results from an
f t dt xe 2x y e t f t dt x
0
for all x, find an explicit formula for f x. 52. Find a function f and a value of the constant a such that
the curve y sinh cx between x 0 and x 1 is equal to 1.
50. Suppose that the temperature in a long, thin rod placed
■
REVIEW
2 y f t dt 2 sin x 1 x
a
53. If f is continuous on a, b , show that
2 y f x f x dx f b 2 f a 2 b
a
54. Evaluate
lim
nl
1 n
1 n
9
2 n
9
3 n
9
n n
9
303
6
TECHNIQUES OF INTEGRATION Because of the Fundamental Theorem of Calculus, we can integrate a function if we know an antiderivative, that is, an indefinite integral.We summarize here the most important integrals that we have learned so far. x n1 C n1
yx
n
dx
ye
x
dx e x C
n 1
y
1 dx ln x C x
ya
x
dx
ax C ln a
y sin x dx cos x C
y cos x dx sin x C
y sec x dx tan x C
y csc x dx cot x C
y sec x tan x dx sec x C
y csc x cot x dx csc x C
y sinh x dx cosh x C
y cosh x dx sinh x C
y tan x dx ln sec x C
y cot x dx ln sin x C
2
yx
2
2
1 1 x dx tan1 a2 a a
C
y sa
2
1 x dx sin1 x2 a
C,
a0
In this chapter we develop techniques for using these basic integration formulas to obtain indefinite integrals of more complicated functions.We learned the most important method of integration, the Substitution Rule, in Section 5.5.The other general technique, integration by parts, is presented in Section 6.1.Then we learn methods that are special to particular classes of functions such as trigonometric functions and rational functions.
6.1
INTEGRATION BY PARTS Every differentiation rule has a corresponding integration rule. For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation. The rule that corresponds to the Product Rule for differentiation is called the rule for integration by parts. The Product Rule states that if f and t are differentiable functions, then d f xtx f xtx txf x dx In the notation for indefinite integrals this equation becomes
y f xtx txf x dx f xtx or 304
y f xtx dx y txf x dx f xtx
SECTION 6.1
INTEGRATION BY PARTS
■
305
We can rearrange this equation as
1
y f xtx dx f xtx y txf x dx
Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u f x and v tx. Then the differentials are du f x dx and dv tx dx, so, by the Substitution Rule, the formula for integration by parts becomes
y u dv uv y v du
2
EXAMPLE 1 Find y x sin x dx. SOLUTION USING FORMULA 1 Suppose we choose f x x and tx sin x.
Then f x 1 and tx cos x. (For t we can choose any antiderivative of t.) Thus, using Formula 1, we have
y x sin x dx f xtx y txf x dx xcos x y cos x dx x cos x y cos x dx x cos x sin x C It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as expected. SOLUTION USING FORMULA 2 Let ■
It is helpful to use the pattern: u䊐 dv 䊐 du 䊐 v䊐
Then and so
ux
dv sin x dx
du dx
v cos x
u
y x sin x dx y x
d√
u
√
√
du
sin x dx x cos x y cos x dx
x cos x y cos x dx x cos x sin x C
■
NOTE Our aim in using integration by parts is to obtain a simpler integral than the one we started with. Thus in Example 1 we started with x x sin x dx and expressed it in terms of the simpler integral x cos x dx. If we had chosen u sin x and dv x dx,
306
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
then du cos x dx and v x 22, so integration by parts gives
y x sin x dx sin x
x2 1 2 2
yx
2
cos x dx
Although this is true, x x 2 cos x dx is a more difficult integral than the one we started with. In general, when deciding on a choice for u and dv, we usually try to choose u f x to be a function that becomes simpler when differentiated (or at least not more complicated) as long as dv tx dx can be readily integrated to give v. V EXAMPLE 2
Evaluate y ln x dx.
SOLUTION Here we don’t have much choice for u and dv. Let
u ln x du
Then
dv dx
1 dx x
vx
Integrating by parts, we get
y ln x dx x ln x y x
dx x
■
It’s customary to write x 1 dx as x dx .
x ln x y dx
■
Check the answer by differentiating it.
x ln x x C Integration by parts is effective in this example because the derivative of the function f x ln x is simpler than f . V EXAMPLE 3
■
Find y t 2 e t dt.
SOLUTION Notice that t 2 becomes simpler when differentiated (whereas e t is
unchanged when differentiated or integrated), so we choose u t2 Then
dv e t dt
du 2t dt
v et
Integration by parts gives 3
y t e dt t e 2 t
2 t
2 y te t dt
The integral that we obtained, x te t dt, is simpler than the original integral but is still not obvious. Therefore, we use integration by parts a second time, this time with u t and dv e t dt. Then du dt, v e t, and
y te dt te t
t
y e t dt
te t e t C
SECTION 6.1
INTEGRATION BY PARTS
■
307
Putting this in Equation 3, we get
yt
e dt t 2 e t 2 y te t dt
2 t
t 2 e t 2te t e t C t 2 e t 2te t 2e t C1 V EXAMPLE 4
where C1 2C
■
Evaluate y e x sin x dx.
SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try choosing u e x and dv sin x dx anyway. Then du e x dx and v cos x, so
www.stewartcalculus.com An easier method, using complex numbers, is given under Additional Topics. Click on Complex Numbers and see Exercise 50. ■
integration by parts gives
ye
4
x
sin x dx e x cos x y e x cos x dx
The integral that we have obtained, x e x cos x dx, is no simpler than the original one, but at least it’s no more difficult. Having had success in the preceding example integrating by parts twice, we persevere and integrate by parts again. This time we use u e x and dv cos x dx. Then du e x dx, v sin x, and
ye
5
Figure 1 illustrates Example 4 by showing the graphs of f x e x sin x and Fx 12 e x sin x cos x. As a visual check on our work, notice that f x 0 when F has a maximum or minimum. ■
12
cos x dx e x sin x y e x sin x dx
At first glance, it appears as if we have accomplished nothing because we have arrived at x e x sin x dx, which is where we started. However, if we put the expression for x e x cos x dx from Equation 5 into Equation 4 we get
ye
x
sin x dx e x cos x e x sin x y e x sin x dx
This can be regarded as an equation to be solved for the unknown integral. Adding
x e x sin x dx to both sides, we obtain
F f _3
x
2 y e x sin x dx e x cos x e x sin x
6
Dividing by 2 and adding the constant of integration, we get _4
ye
FIGURE 1
x
sin x dx 12 e x sin x cos x C
■
If we combine the formula for integration by parts with the Evaluation Theorem, we can evaluate definite integrals by parts. Evaluating both sides of Formula 1 between a and b, assuming f and t are continuous, and using the Evaluation Theorem, we obtain
6
y
b
a
f xtx dx f xtx a y txf x dx b
]
b
a
308
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
EXAMPLE 5 Calculate y tan1x dx. 1
0
SOLUTION Let
u tan1x dx 1 x2
du
Then
dv dx vx
So Formula 6 gives
y
1
0
tan1x dx x tan1x 0 y 1
]
x dx 1 x2
1
0
1 tan1 1 0 tan1 0 y
1
0
1 ■ Since tan x 0 for x 0 , the integral in Example 5 can be interpreted as the area of the region shown in Figure 2.
1 x y dx 0 4 1 x2
To evaluate this integral we use the substitution t 1 x 2 (since u has another meaning in this example). Then dt 2x dx, so x dx 12 dt. When x 0, t 1; when x 1, t 2; so
y
y=tan–!x
0 1
x dx 1 x2
y
x
1
0
x 2 dt dx 12 y 12 ln t 1 t 1 x2
]
2 1
12 ln 2 ln 1 12 ln 2
FIGURE 2
y
Therefore
1
0
tan1x dx
1 x ln 2 y 2 dx 0 1 x 4 4 2
EXAMPLE 6 Prove the reduction formula ■ Equation 7 is called a reduction formula because the exponent n has been reduced to n 1 and n 2 .
1
y sin x dx n cos x sin n
7
x
n1
n1 n
y sin
n2
x dx
where n 2 is an integer. SOLUTION Let
u sin n1x
dv sin x dx
du n 1 sin n2x cos x dx
Then
v cos x
so integration by parts gives
y sin x dx cos x sin n
x n 1 y sin n2x cos 2x dx
n1
Since cos 2x 1 sin 2x, we have
y sin x dx cos x sin n
x n 1 y sin n2x dx n 1 y sin n x dx
n1
■
SECTION 6.1
■
INTEGRATION BY PARTS
309
As in Example 4, we solve this equation for the desired integral by taking the last term on the right side to the left side. Thus we have n y sin n x dx cos x sin n1x n 1 y sin n2x dx 1
y sin x dx n cos x sin n
or
x
n1
n1 n
y sin
■
n2
x dx
The reduction formula (7) is useful because by using it repeatedly we could eventually express x sin n x dx in terms of x sin x dx (if n is odd) or x sin x0 dx x dx (if n is even).
6.1
EXERCISES
■ Evaluate the integral using integration by parts with the indicated choices of u and dv.
1–2
■
1.
y x ln x dx ;
2.
y sec d ;
u ln x, dv x dx
2
■
■
3–24
■
■
■
3.
y x cos 5x dx
■
5.
y re
4.
2
1
ln x2 dx
■
y
24.
■
■
■
■
t
e s sint s ds
0
■
■
■
■
■
■
■
■
y xe
x
dx
■
■
25.
y sin sx dx
27.
y
■
s
s 2
3 cos 2 d
■
■
■
■
■
26.
yx
28.
y
4
1
■
5
cosx 3 dx
e sx dx
■
■
■
■
29. (a) Use the reduction formula in Example 6 to show that
7.
yx
9.
r2
dr
6.
y t sin 2t dt
8.
yx
2
y ln2x 1 dx
10.
yp
5
11.
y arctan 4t dt
12.
y t e dt
13.
y e 2 sin 3 d
14.
y e cos 2 d
15.
y
16.
y
0
2
t sin 3t dt
20.
y
s3
22.
y
1
1
21.
y
12
sin1x dx
1
1
0
x sin 2x C 2 4
(b) Use part (a) and the reduction formula to evaluate x sin 4x dx. 30. (a) Prove the reduction formula
y cos x dx n
1 n1 cos n1x sin x n n
y cos
n2
x dx
(b) Use part (a) to evaluate x cos 2x dx. (c) Use parts (a) and (b) to evaluate x cos 4x dx.
y
2
0
y dy e 2y
y
0
0
2
31. (a) Use the reduction formula in Example 6 to show that
4
19.
ln p dp
y sin x dx
x 2 1ex dx
y
2
0
1
18.
y
cos mx dx
3 t
ln x dx x2
17.
1
sin x dx
■
■ First make a substitution and then use integration by parts to evaluate the integral.
u , dv sec 2 d
■
y
25–28
Evaluate the integral.
■
23.
st ln t dt arctan1x dx
r3 dr s4 r 2
sin n x dx
n1 n
y
2
0
sin n2x dx
where n 2 is an integer. (b) Use part (a) to evaluate x02 sin 3x dx and x02 sin 5x dx. (c) Use part (a) to show that, for odd powers of sine,
y
2
0
sin 2n1x dx
2 4 6 2n 3 5 7 2n 1
■
310
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
32. Prove that, for even powers of sine,
y
2
0
33–36
constant velocity ve (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation
1 3 5 2n 1 2 4 6 2n 2
sin 2nx dx
vt tt ve ln
where t is the acceleration due to gravity and t is not too large. If t 9.8 ms 2, m 30,000 kg, r 160 kgs, and ve 3000 ms, find the height of the rocket one minute after liftoff.
Use integration by parts to prove the reduction
■
formula. 33.
y ln x dx x ln x n
n
n y ln xn1 dx
34.
yx e
dx x ne x n y x n1e x dx
35.
y x
a dx
n x
it travel during the first t seconds?
2 n
2na 2 x x 2 a 2 n 2n 1 2n 1
y y x
2
a 2 n1 dx
tan x sec x n2 36. y sec x dx n1 n1 n 1 n
■
■
■
■
a
0
n2
■
41. A particle that moves along a straight line has velocity vt t 2et meters per second after t seconds. How far will 42. If f 0 t0 0 and f and t are continuous, show that
2
■
m rt m
■
■
y sec
(n 12 )
a
0
43. Suppose that f 1 2, f 4 7, f 1 5, f 4 3,
and f is continuous. Find the value of x14 x f x dx.
n2
x dx
■
■
44. (a) Use integration by parts to show that ■
37. Use Exercise 33 to find x ln x3 dx.
y f x dx x f x y x f x dx
■
(b) If f and t are inverse functions and f is continuous, prove that
38. Use Exercise 34 to find x x 4e x dx.
y
39. Find the average value of f x x 2 ln x on the
interval 1, 3 .
40. A rocket accelerates by burning its onboard fuel, so its
mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel is consumed at rate r, and the exhaust gases are ejected with
6.2
f xt x dx f ata f ata y f xtx dx
b
a
f x dx bf b af a y
f b
f a
t y dy
[Hint: Use part (a) and make the substitution y f x.] (c) In the case where f and t are positive functions and b a 0, draw a diagram to give a geometric interpretation of part (b). (d) Use part (b) to evaluate x1e ln x dx.
TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS In this section we look at integrals involving trigonometric functions and integrals that can be transformed into trigonometric integrals by substitution. TRIGONOMETRIC INTEGRALS
Here we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE 1 Evaluate y cos 3x dx. SOLUTION Simply substituting u cos x isn’t helpful, since then du sin x dx.
In order to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of sine would require an extra cos x factor. Thus here we can separate one cosine factor and convert the remaining cos2x factor to an expression involving sine
SECTION 6.2
TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS
■
311
using the identity sin 2x cos 2x 1: cos 3x cos 2x cos x 1 sin 2x cos x We can then evaluate the integral by substituting u sin x, so du cos x dx and
y cos x dx y cos x cos x dx y 1 sin x cos x dx 3
2
2
y 1 u 2 du u 13 u 3 C sin x 13 sin 3x C
■
In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor (and the remainder of the expression in terms of cosine) or only one cosine factor (and the remainder of the expression in terms of sine). The identity sin 2x cos 2x 1 enables us to convert back and forth between even powers of sine and cosine. 5 2 Find y sin x cos x dx
V EXAMPLE 2
SOLUTION We could convert cos 2x to 1 sin 2x, but we would be left with an
expression in terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and rewrite the remaining sin 4x factor in terms of cos x : ■ Figure 1 shows the graphs of the integrand sin 5x cos 2x in Example 2 and its indefinite integral (with C 0 ). Which is which?
sin 5x cos 2x sin2x2 cos 2x sin x 1 cos 2x2 cos 2x sin x Substituting u cos x, we have du sin x dx and so
y sin x cos x dx y sin x
0.2
5
2
2
_π
y 1 u
π
2
cos 2x sin x dx y 1 cos 2x2 cos 2x sin x dx u 2 du y u 2 2u 4 u 6 du
2 2
u3 u5 u7 2 3 5 7
FIGURE 1
Example 3 shows that the area of the region shown in Figure 2 is 2. ■
and
cos 2x 12 1 cos 2x
Evaluate y sin 2x dx.
V EXAMPLE 3
0
SOLUTION If we write sin 2x 1 cos 2x, the integral is no simpler to evaluate.
1.5
Using the half-angle formula for sin 2x, however, we have
y=sin@ x
y
0
FIGURE 2
■
In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. If the integrand contains even powers of both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix A): sin 2x 12 1 cos 2x
_0.5
C
13 cos 3x 25 cos 5x 17 cos 7x C
_0.2
0
π
sin 2x dx 12 y 1 cos 2x dx 0
[ (x 1 2
1 2
0
]
sin 2x)
12 ( 12 sin 2) 12 (0 12 sin 0) 12 Notice that we mentally made the substitution u 2x when integrating cos 2x. Another method for evaluating this integral was given in Exercise 29 in Section 6.1. ■
312
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CHAPTER 6
TECHNIQUES OF INTEGRATION
EXAMPLE 4 Find y sin 4x dx. SOLUTION We could evaluate this integral using the reduction formula for
x sin n x dx
(Equation 6.1.7) together with Example 3 (as in Exercise 29 in Section 6.1), but a better method is to write sin 4x sin 2x2 and use a half-angle formula:
y sin x dx y sin x dx 4
2
How to Integrate Powers of sin x and cos x From Examples 1– 4 we see that the following strategy works: ■
(i) If the power of cos x is odd, save one cosine factor and use cos2x 1 sin2x to express the remaining factors in terms of sin x . Then substitute u sin x.
sin2x 2 1 cos 2x 1
cos2x 12 1 cos 2x
It is sometimes helpful to use the identity sin x cos x 12 sin 2x
1 cos 2x 2
2
dx
14 y 1 2 cos 2x cos 2 2x dx Since cos 2 2x occurs, we must use another half-angle formula cos 2 2x 12 1 cos 4x This gives
y sin x dx y 1 2 cos 2x 1 4
4
(ii) If the power of sin x is odd, save one sine factor and use sin2x 1 cos2x to express the remaining factors in terms of cos x . Then substitute u cos x . (iii) If the powers of both sine and cosine are even, use the half-angle identities:
y
2
14 y
1 2
1 cos 4x dx
( 32 2 cos 2x 12 cos 4x) dx
14 ( 32 x sin 2x 18 sin 4x) C
■
We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx . Since ddx tan x sec 2x , we can separate a sec 2x factor and convert the remaining (even) power of secant to an expression involving tangent using the identity sec 2x 1 tan 2x. Or, since ddx sec x sec x tan x , we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant. V EXAMPLE 5
Evaluate y tan 6x sec 4x dx .
SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x
factor in terms of tangent using the identity sec 2x 1 tan 2x. We can then evaluate the integral by substituting u tan x with du sec 2x dx :
y tan x sec x dx y tan x sec x sec x dx 6
4
6
2
2
y tan 6x 1 tan 2x sec 2x dx y u 61 u 2 du y u 6 u 8 du
u9 u7 C 7 9
17 tan 7x 19 tan 9x C
■
SECTION 6.2
TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS
■
313
EXAMPLE 6 Find y tan 5 sec 7 d. SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left
with a sec 5 factor, which isn’t easily converted to tangent. However, if we separate a sec tan factor, we can convert the remaining power of tangent to an expression involving only secant using the identity tan 2 sec 2 1. We can then evaluate the integral by substituting u sec , so du sec tan d :
y tan 5
How to Integrate Powers of tan x and sec x From Examples 5 and 6 we have a strategy for two cases: ■
sec 7 d y tan 4 sec 6 sec tan d
(i) If the power of sec x is even, save a factor of sec2 x and use sec2 x 1 tan 2 x to express the remaining factors in terms of tan x . Then substitute u tan x.
y sec 1 sec 2
2
6
sec tan d
y u 2 12 u 6 du y u 10 2u 8 u 6 du
(ii) If the power of tan x is odd, save a factor of sec x tan x and use tan 2x sec 2x 1 to express the remaining factors in terms of sec x . Then substitute u sec x.
u 11 u9 u7 2 C 11 9 7
111 sec 11 29 sec 9 17 sec 7 C
■
For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to integrate tan x by using the formula established in (5.5.5):
y tan x dx ln sec x C We will also need the indefinite integral of secant:
1
y sec x dx ln sec x tan x C
We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x : sec x tan x
y sec x dx y sec x sec x tan x dx y
sec 2x sec x tan x dx sec x tan x
If we substitute u sec x tan x, then du sec x tan x sec 2x dx , so the integral becomes x 1u du ln u C. Thus we have
y sec x dx ln sec x tan x C
314
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CHAPTER 6
TECHNIQUES OF INTEGRATION
EXAMPLE 7 Find y tan 3x dx. SOLUTION Here only tan x occurs, so we use tan 2x sec 2x 1 to rewrite a tan 2x
factor in terms of sec 2x :
y tan x dx y tan x tan x dx y tan x sec x 1 dx 3
2
2
y tan x sec 2x dx y tan x dx
tan 2x ln sec x C 2
In the first integral we mentally substituted u tan x so that du sec 2x dx.
■
If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of sec x. Powers of sec x may require integration by parts, as shown in the following example. EXAMPLE 8 Find y sec 3x dx. SOLUTION Here we integrate by parts with
u sec x du sec x tan x dx Then
dv sec 2x dx v tan x
y sec x dx sec x tan x y sec x tan x dx 3
2
sec x tan x y sec x sec 2x 1 dx sec x tan x y sec 3x dx y sec x dx Using Formula 1 and solving for the required integral, we get
y sec x dx (sec x tan x ln sec x tan x ) C 3
1 2
■
Integrals such as the one in Example 8 may seem very special but they occur frequently in applications of integration, as we will see in Chapter 7. Integrals of the form x cot m x csc n x dx can be found by similar methods because of the identity 1 cot 2x csc 2x. TRIGONOMETRIC SUBSTITUTIONS
In finding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 dx arises, where a 0. If it were x xsa 2 x 2 dx , the substitution u a 2 x 2 would be effective but, as it stands, x sa 2 x 2 dx is more difficult. If we change the variable from x to by the substitution x a sin , then the identity 1 sin 2 cos 2 allows us to get rid of the root sign because
sa 2 x 2 sa 2 a 2 sin 2 sa 21 sin 2 sa 2 cos 2 a cos
Notice the difference between the substitution u a 2 x 2 (in which the new vari-
SECTION 6.2
TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS
■
315
able is a function of the old one) and the substitution x a sin (the old variable is a function of the new one). In general we can make a substitution of the form x tt by using the Substitution Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is, t is one-to-one. In this case, if we replace u by x and x by t in the Substitution Rule (Equation 5.5.4), we obtain
y f x dx y f tttt dt This kind of substitution is called inverse substitution. We can make the inverse substitution x a sin provided that it defines a one-toone function. We accomplish this by restricting to lie in the interval 2, 2 . In the following table we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities. In each case the restriction on is imposed to ensure that the function that defines the substitution is one-to-one. (These are the same intervals used in Section 3.5 in defining the inverse functions.) TABLE OF TRIGONOMETRIC SUBSTITUTIONS Expression
Substitution
Identity
sa 2 x 2
x a sin ,
2 2
1 sin 2 cos 2
sa 2 x 2
x a tan ,
2 2
1 tan 2 sec 2
sx 2 a 2
x a sec ,
0
V EXAMPLE 9
Evaluate y
3 or 2 2
sec 2 1 tan 2
s9 x 2 dx . x2
SOLUTION Let x 3 sin , where 2 2. Then dx 3 cos d and
s9 x 2 s9 9 sin 2 s9 cos 2 3 cos 3 cos (Note that cos 0 because 2 2.) Thus, using inverse substitution, we get
y 3 x ¨ œ„„„„„ 9-≈ FIGURE 3
sin ¨=
x 3
3 cos cos 2 s9 x 2 dx 3 cos d d y y x2 9 sin 2 sin 2 2 y cot 2 d y csc 1 d cot C
Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot in terms of sin x3 or by drawing a diagram, as in Figure 3, where is interpreted as an angle of a right triangle. Since sin x3, we label the opposite side and the hypotenuse as having lengths x and 3. Then the Pythagorean Theorem gives the length of
316
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
the adjacent side as s9 x 2 , so we can simply read the value of cot from the figure: cot
s9 x 2 x
(Although 0 in the diagram, this expression for cot is valid even when 0.) Since sin x3, we have sin1x3 and so
y V EXAMPLE 10
x s9 x 2 s9 x 2 dx sin1 2 x x 3
C
■
Find the area enclosed by the ellipse x2 y2 1 a2 b2
SOLUTION Solving the equation of the ellipse for y, we get
y
y2 x2 a2 x2 1 b2 a2 a2
(0, b)
y
or
b sa 2 x 2 a
(a, 0) 0
x
Because the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant (see Figure 4). The part of the ellipse in the first quadrant is given by the function b y sa 2 x 2 0 x a a
FIGURE 4
¥ ≈ + =1 b@ a@
1 4
and so
Ay
a
0
b sa 2 x 2 dx a
To evaluate this integral we substitute x a sin . Then dx a cos d. To change the limits of integration we note that when x 0, sin 0, so 0; when x a, sin 1, so 2. Also
sa 2 x 2 sa 2 a 2 sin 2 sa 2 cos 2 a cos a cos since 0 2. Therefore A4
b a
y
a
0
4ab y
2
0
[
sa 2 x 2 dx 4
b a
cos 2 d 4ab y
2ab 12 sin 2
2 0
2
0
2 1 2
0
]
y
2ab
a cos a cos d
1 cos 2 d
0 0 ab 2
We have shown that the area of an ellipse with semiaxes a and b is ab. In particular, taking a b r, we have proved the famous formula that the area of a circle with radius r is r 2. ■ NOTE Since the integral in Example 10 was a definite integral, we changed the limits of integration and did not have to convert back to the original variable x.
SECTION 6.2
V EXAMPLE 11
TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS
■
317
1 dx . x sx 2 4
Find y
2
SOLUTION Let x 2 tan , 2 2. Then dx 2 sec 2 d and
sx 2 4 s4tan 2 1 s4 sec 2 2 sec 2 sec Thus we have dx
y x sx 2
2
4
y
2 sec 2 d 1 2 4 tan 2 sec 4
y
sec d tan 2
To evaluate this trigonometric integral we put everything in terms of sin and cos : sec 1 cos 2 cos 2 2 tan cos sin sin 2 Therefore, making the substitution u sin , we have dx
y x sx 2
2
4
œ„„„„„ ≈+4
2
1 4
cos 1 d sin 2 4
1 u
y
C
du u2 1 C 4 sin
csc C 4
dx sx 2 4 C 2 x sx 4 4x
y
x 2
EXAMPLE 12 Find y
■ Example 12 illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution. You should look for a simpler method first.
y
We use Figure 5 to determine that csc sx 2 4 x and so
FIGURE 5
tan ¨=
1 4
x ¨
■
2
x dx . sx 4 2
SOLUTION It would be possible to use the trigonometric substitution x 2 tan here (as in Example 11). But the direct substitution u x 2 4 is simpler, because then du 2x dx and
y sx
2
x 1 dx 4 2
EXAMPLE 13 Evaluate y
du
y su
su C sx 2 4 C
dx , where a 0. sx a 2 2
SOLUTION We let x a sec , where 0 2 or 32. Then
dx a sec tan d and
sx 2 a 2 sa 2sec 2 1 sa 2 tan 2 a tan a tan
■
318
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
Therefore x
œ„„„„„ ≈-a@
y
¨
dx a sec tan y d 2 a tan sx a 2
a
FIGURE 6
sec ¨=
x a
y sec d ln sec tan C
The triangle in Figure 6 gives tan sx 2 a 2 a , so we have
y
dx x sx 2 a 2 ln 2 a a sx a 2
C
ln x sx 2 a 2 ln a C Writing C1 C ln a, we have
y EXAMPLE 14 Find y
dx ln x sx 2 a 2 C1 sx a 2
3 s32
0
2
■
x3 dx. 4x 932 2
SOLUTION First we note that 4x 2 932 s4x 2 9 )3 so trigonometric substi-
tution is appropriate. Although s4x 2 9 is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution u 2x. When we combine this with the tangent substitution, we have x 32 tan , which gives dx 32 sec 2 d and s4x 2 9 s9 tan 2 9 3 sec When x 0, tan 0, so 0; when x 3s32, tan s3 , so 3.
y
3 s32
0
27 3 x3 3 8 tan dx y 2 32 0 4x 9 27 sec3
163 y
3
163 y
3
3 2
sec 2 d
3 tan 3 3 sin d 163 y d 0 sec cos2
0
1 cos 2 sin d cos 2
0
Now we substitute u cos so that du sin d. When 0, u 1; when 3, u 12. Therefore
y
3 s32
0
2 x3 12 1 u 12 3 dx du 163 y 1 u 2 du 16 y 2 32 2 1 1 4x 9 u
163 u
1 u
12
1
163 [( 12 2) 1 1] 323
■
SECTION 6.2
6.2 1–34 1.
y sin x cos x dx
3.
y
5.
3
34 2
y
2
2
sin 5x cos 3x dx cos2 d
y
9.
y 1 cos
sin 43t dt
0
2.
y sin x cos x dx
4.
y
6.
7.
2
4
d
11.
y
13.
y cos x tan x dx
15.
y
17.
0
sin 4x cos 2x dx
6
2
show that
3
sin A cos B 12 sinA B sinA B (b) Use part (a) to evaluate x sin 3x cos x dx.
cos 5x dx
■ Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle.
37–39
y sin mx dx 3
37.
yx
2
10.
y
cos6 d
38.
yx
12.
y x cos x dx
39.
y sx
0
0
sin 2 2 d
2
2
y
1 sin x dx cos x
16.
y cos x sin 2x dx
y sec x tan x dx
18.
y
19.
y tan x dx
20.
y tan x dx
21.
y sec t dt
22.
y
23.
y
24.
y tan 2x sec 2x dx
2
2
6
3
0
tan 5 x sec 4 x dx
0
2
sin x cos x dx
2
sec 4t2 dt 4
4
0
sec 4 tan 4 d 3
3
5
y tan x sec x dx
26.
y
27.
y tan x dx
28.
y tan ay dy
29.
y
30.
y
5
2 6
cot 2x dx
0
tan 5x sec6x dx 6
2 4
cot 3x dx
31.
y cot csc d
32.
y csc
33.
y csc x dx
34.
y
■
3
■
3
■
■
■
■
■
■
4
x cot 6 x dx
1 tan 2x dx sec 2x ■
■
■
35. (a) Use the formulas for cosA B and cosA B to
show that 1 sin A sin B 2 cosA B cosA B
(b) Use part (a) to evaluate x sin 5x sin 2x dx.
s9 x 2 dx ; x 3 sin x3 dx ; 9
x 3 tan
2
■
40 –58
2
0
3
■
■
■
■
■
■
■
■
■
■
2
25.
3
1 dx ; x 3 sec sx 2 9
y
14.
3
2
8.
■
2
319
36. (a) Use the formulas for sinA B and sinA B to
0
0
■
EXERCISES
Evaluate the integral.
■
TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS
■
Evaluate the integral.
■
40.
y
2 s3
41.
y
2
0
s2
x3 dx s16 x 2
1 dt t 3 st 2 1
yx
45.
y sx
47.
y s1 4x
49.
y
51.
y a
53.
y sx
55.
y
57.
y x s1 x
y
2
x 3 sx 2 4 dx
0
1 dx s25 x 2
43.
■
42.
sx 2 a 2 dx x4
44.
y
46.
y st
dx
48.
y
sx 2 9 dx x3
50.
y u s5 u
x2 dx x 2 32
52.
yx
x dx 7
54.
y
s1 x 2 dx x
56.
y s25 t
58.
y
■
2
2
dx 2 16
2
■
2
4
dx ■
■
■
■
1
0
t5 dt 2 2
x sx 2 4 dx du
1
0
2
2
dx s16x 2 9
sx 2 1 dx t
2
0
■
2
dt
cos t dt s1 sin 2 t ■
■
■
59. Evaluate the integral
y s9x
2
1 dx 6x 8
by first completing the square and using the substitution u 3x 1.
■
320
■
60 –62
CHAPTER 6
Evaluate the integral by first completing the square.
■
60.
y st
61.
y x
■
■
TECHNIQUES OF INTEGRATION
2
2
67. Prove the formula A 2 r 2 for the area of a sector of 1
a circle with radius r and central angle . [Hint: Assume 0 2 and place the center of the circle at the origin so it has the equation x 2 y 2 r 2. Then A is the sum of the area of the triangle POQ and the area of the region PQR in the figure.]
dt 6t 13 dx 2x 22 ■
■
62. ■
■
■
x2
y s4x x ■
■
2
dx
■
■
■
y
P
63. A particle moves on a straight line with velocity function vt sin t cos 2 t. Find its position function s f t
if f 0 0.
¨
64. Household electricity is supplied in the form of alternating O
current that varies from 155 V to 155 V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation
Q
R
x
68. A charged rod of length L produces an electric field at point
Pa, b given by
Et 155 sin120 t where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of Et 2 over one cycle. (a) Calculate the RMS voltage of household current. (b) Many electric stoves require an RMS voltage of 220 V. Find the corresponding amplitude A needed for the voltage Et A sin120 t.
EP
y
La
a
b dx 4 0 x 2 b 2 32
where is the charge density per unit length on the rod and 0 is the free space permittivity (see the figure). Evaluate the integral to determine an expression for the electric field EP. y
P (a, b)
65. Find the average value of f x sx 2 1x , 1 x 7.
0
L
x
66. Find the area of the region bounded by the hyperbola
9x 2 4y 2 36 and the line x 3.
6.3
PARTIAL FRACTIONS In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions 2x 1 and 1x 2 to a common denominator we obtain 2 1 2x 2 x 1 x5 2 x1 x2 x 1x 2 x x2 If we now reverse the procedure, we see how to integrate the function on the right side of this equation:
yx
2
x5 dx x2
y
2 1 x1 x2
dx
2 ln x 1 ln x 2 C To see how the method of partial fractions works in general, let’s consider a rational function Px f x Qx
SECTION 6.3
PARTIAL FRACTIONS
■
321
where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper. Recall that if Px a n x n a n1 x n1 a 1 x a 0 where a n 0, then the degree of P is n and we write degP n. If f is improper, that is, degP degQ, then we must take the preliminary step of dividing Q into P (by long division) until a remainder Rx is obtained such that degR degQ. The division statement is f x
1
Px Rx Sx Qx Qx
where S and R are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required. x3 x dx. x1 SOLUTION Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write V EXAMPLE 1
≈+x +2 x-1 ) ˛ +x ˛-≈ ≈+x ≈-x 2x 2x-2 2
Find y
y
x3 x dx x1
y
x2 x 2
2 x1
dx
x3 x2 2x 2 ln x 1 C 3 2
■
The next step is to factor the denominator Qx as far as possible. It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax b and irreducible quadratic factors (of the form ax 2 bx c , where b 2 4ac 0). For instance, if Qx x 4 16, we could factor it as Qx x 2 4x 2 4 x 2x 2x 2 4 The third step is to express the proper rational function RxQx (from Equation 1) as a sum of partial fractions of the form A ax b i
or
Ax B ax 2 bx c j
A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur. CASE I The denominator Qx is a product of distinct linear factors.
This means that we can write Qx a 1 x b1 a 2 x b 2 a k x bk where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants A1, A2 , . . . , Ak such
322
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
that Rx A1 A2 Ak Qx a 1 x b1 a2 x b2 a k x bk
2
These constants can be determined as in the following example. V EXAMPLE 2
Evaluate y
x 2 2x 1 dx . 2x 3 3x 2 2x
SOLUTION Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide. We factor the denominator as
2x 3 3x 2 2x x2x 2 3x 2 x2x 1x 2 Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form x 2 2x 1 A B C x2x 1x 2 x 2x 1 x2
3
■ Another method for finding A, B, and C is given in the note after this example.
To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators, x2x 1x 2, obtaining 4
Figure 1 shows the graphs of the integrand in Example 2 and its indefinite integral (with K 0 ). Which is which? ■
2
3
_3
x 2 2x 1 A2x 1x 2 Bx x 2 Cx2x 1
Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get x 2 2x 1 2A B 2Cx 2 3A 2B C x 2A
5
The polynomials in Equation 5 are identical, so their coefficients must be equal. The coefficient of x 2 on the right side, 2A B 2C, must equal the coefficient of x 2 on the left side—namely, 1. Likewise, the coefficients of x are equal and the constant terms are equal. This gives the following system of equations for A, B, and C: 2A B 2C 1 3A 2B C 2
_2
2A
FIGURE 1
1
Solving, we get A 12 , B 15 , and C 101 , and so We could check our work by taking the terms to a common denominator and adding them. ■
x 2 2x 1 dx 3 3x 2 2x
y 2x
y
1 1 1 1 1 1 2 x 5 2x 1 10 x 2
dx
12 ln x 101 ln 2x 1 101 ln x 2 K In integrating the middle term we have made the mental substitution u 2x 1, which gives du 2 dx and dx du2. ■ NOTE We can use an alternative method to find the coefficients A , B , and C in Example 2. Equation 4 is an identity; it is true for every value of x . Let’s choose val-
SECTION 6.3
PARTIAL FRACTIONS
■
323
ues of x that simplify the equation. If we put x 0 in Equation 4, then the second and third terms on the right side vanish and the equation then becomes 2A 1, or A 12 . Likewise, x 12 gives 5B4 14 and x 2 gives 10C 1, so B 15 and C 101 . (You may object that Equation 3 is not valid for x 0, 12 , or 2, so why should Equation 4 be valid for those values? In fact, Equation 4 is true for all values 1 of x , even x 0, 2 , and 2. See Exercise 45 for the reason.) CASE II Qx is a product of linear factors, some of which are repeated.
Suppose the first linear factor a 1 x b1 is repeated r times; that is, a 1 x b1 r occurs in the factorization of Qx. Then instead of the single term A1a 1 x b1 in Equation 2, we would use 6
A1 A2 Ar 2 a 1 x b1 a 1 x b1 a 1 x b1 r
By way of illustration, we could write x3 x 1 A B C D E 2 2 3 2 x x 1 x x x1 x 1 x 13 but we prefer to work out in detail a simpler example. EXAMPLE 3 Find y
x 4 2x 2 4x 1 dx. x3 x2 x 1
SOLUTION The first step is to divide. The result of long division is
x 4 2x 2 4x 1 4x x1 3 3 2 2 x x x1 x x x1 The second step is to factor the denominator Qx x 3 x 2 x 1. Since Q1 0, we know that x 1 is a factor and we obtain x 3 x 2 x 1 x 1x 2 1 x 1x 1x 1 x 12x 1 Since the linear factor x 1 occurs twice, the partial fraction decomposition is 4x A B C x 12x 1 x1 x 12 x1 Multiplying by the least common denominator, x 12x 1, we get 7
4x Ax 1x 1 Bx 1 Cx 12 A Cx 2 B 2Cx A B C
■ Another method for finding the coefficients: Put x 1 in (7): B 2 . Put x 1: C 1. Put x 0 : A B C 1.
Now we equate coefficients: A
C0 B 2C 4
A B C 0
324
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
Solving, we obtain A 1, B 2, and C 1, so
y
■
Here we use ln
x 4 2x 2 4x 1 dx x3 x2 x 1
a ln a ln b. b
y
x1
1 2 1 x1 x 12 x1
dx
x2 2 x ln x 1 ln x 1 K 2 x1
x2 2 x1 x ln K 2 x1 x1
■
CASE III Qx contains irreducible quadratic factors, none of which is repeated.
If Qx has the factor ax 2 bx c, where b 2 4ac 0, then, in addition to the partial fractions in Equations 2 and 6, the expression for RxQx will have a term of the form Ax B ax bx c
8
2
where A and B are constants to be determined. For instance, the function given by f x x x 2x 2 1x 2 4 has a partial fraction decomposition of the form x A Bx C Dx E 2 2 x 2x 2 1x 2 4 x2 x 1 x 4 The term in (8) can be integrated by completing the square and using the formula
y
9
V EXAMPLE 4
Evaluate y
dx 1 x tan1 x2 a2 a a
C
2x 2 x 4 dx. x 3 4x
SOLUTION Since x 3 4x xx 2 4 can’t be factored further, we write
2x 2 x 4 A Bx C 2 xx 2 4 x x 4 Multiplying by xx 2 4, we have 2x 2 x 4 Ax 2 4 Bx C x A Bx 2 Cx 4A Equating coefficients, we obtain AB2
C 1
4A 4
Thus A 1, B 1, and C 1 and so
y
2x 2 x 4 dx x 3 4x
y
x1 1 2 x x 4
dx
SECTION 6.3
PARTIAL FRACTIONS
■
325
In order to integrate the second term we split it into two parts:
y
x1 x 1 dx y 2 dx y 2 dx x2 4 x 4 x 4
We make the substitution u x 2 4 in the first of these integrals so that du 2x dx. We evaluate the second integral by means of Formula 9 with a 2:
y
2x 2 x 4 1 x 1 dx y dx y 2 dx y 2 dx 2 xx 4 x x 4 x 4
ln x 12 lnx 2 4 12 tan1x2 K EXAMPLE 5 Evaluate y
■
4x 2 3x 2 dx. 4x 2 4x 3
SOLUTION Since the degree of the numerator is not less than the degree of the
denominator, we first divide and obtain 4x 2 3x 2 x1 1 4x 2 4x 3 4x 2 4x 3 Notice that the quadratic 4x 2 4x 3 is irreducible because its discriminant is b 2 4ac 32 0. This means it can’t be factored, so we don’t need to use the partial fraction technique. To integrate the given function we complete the square in the denominator: 4x 2 4x 3 2x 12 2 This suggests that we make the substitution u 2x 1. Then, du 2 dx and x u 12, so 4x 2 3x 2 dx 2 4x 3
y 4x
y
1
x 12 y x 14 y
x1 4x 2 4x 3 1 2
dx
u 1 1 u1 du x 14 y 2 du 2 u 2 u 2
u 1 du 14 y 2 du u 2 u 2 2
x 18 lnu 2 2
1 1 u tan1 4 s2 s2
x 18 ln4x 2 4x 3
C
1 2x 1 tan1 4s2 s2
C ■
NOTE Example 5 illustrates the general procedure for integrating a partial fraction of the form
Ax B ax 2 bx c
where b 2 4ac 0
326
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
We complete the square in the denominator and then make a substitution that brings the integral into the form
y
Cu D u 1 du C y 2 du D y 2 du u2 a2 u a2 u a2
Then the first integral is a logarithm and the second is expressed in terms of tan1. CASE IV Qx contains a repeated irreducible quadratic factor.
If Qx has the factor ax 2 bx c r , where b 2 4ac 0, then instead of the single partial fraction (8), the sum A1 x B1 A2 x B2 Ar x Br 2 2 2 2 ax bx c ax bx c ax bx c r
10
occurs in the partial fraction decomposition of RxQx. Each of the terms in (10) can be integrated by first completing the square. ■ It would be extremely tedious to work out by hand the numerical values of the coefficients in Example 6. Most computer algebra systems, however, can find the numerical values very quickly. For instance, the Maple command
convertf, parfrac, x or the Mathematica command Apart[f] gives the following values: A 1, E
15 8
1 B 8 , C D 1,
, F 18 , G H 34 ,
EXAMPLE 6 Write out the form of the partial fraction decomposition of the function
x3 x2 1 xx 1x 2 x 1x 2 13 SOLUTION
x3 x2 1 xx 1x 2 x 1x 2 13
A B Cx D Ex F Gx H Ix J 2 2 2 2 x x1 x x1 x 1 x 1 x 2 13
I 12 , J 21
EXAMPLE 7 Evaluate y
■
1 x 2x 2 x 3 dx. xx 2 12
SOLUTION The form of the partial fraction decomposition is
1 x 2x 2 x 3 A Bx C Dx E 2 2 xx 2 12 x x 1 x 12 Multiplying by xx 2 12, we have x 3 2x 2 x 1 Ax 2 12 Bx Cxx 2 1 Dx Ex Ax 4 2x 2 1 Bx 4 x 2 Cx 3 x Dx 2 Ex A Bx 4 Cx 3 2A B Dx 2 C Ex A If we equate coefficients, we get the system AB0
C 1
2A B D 2
C E 1
A1
SECTION 6.3
■
PARTIAL FRACTIONS
327
which has the solution A 1, B 1, C 1, D 1, and E 0. Thus
y
1 x 2x 2 x 3 dx x x 2 12
y
2x x 33x 1
(b)
x1 2. (a) 3 x x2
1 x 3 2x 2 x
x1 (b) 3 x x
1
y x 5 x 1 dx
20.
y x 3x 2
21.
y
5x 2 3x 2 dx x 3 2x 2
22.
y
23.
y x 1x
24.
y x 1 x
26.
y
28.
y
1 dx 3 1
30.
yx
dx x2
32.
y
34.
y x x
2
10
(b)
x2 x 1x 2 x 1
25.
y
4. (a)
x3 x 2 4x 3
(b)
2x 1 x 1 3x 2 4 2
27.
yx
5. (a)
x4 4 x 1
(b)
t4 t2 1 2 t 1t 2 42
29.
yx
6. (a)
x4 x 3 xx 2 x 3
(b)
1 6 x x3
31.
yx
33.
y x
7–34 7.
9.
11.
13.
15. 17.
■
■
x9
y x 5x 2 dx y
3
2
1 dx x2 1
yx y
1
0
y
2
1
■
■
■
■
■
■
Evaluate the integral.
x dx x6
y
■
2
ax dx bx
2x 3 dx x 12 4y 7y 12 dy y y 2 y 3
8.
10.
12.
14.
16.
2
18.
r2 dr r4
y
1
y t 4t 1 dt y
1
0
x1 dx x 2 3x 2
■
■
y
1
0
y
x 4x 10 dx x2 x 6 x 2x 1 dx x3 x
2
4
x4 dx 2x 5
2
■
x3 dx 2x 42 ■
■
■
■
dx
x2 x 6 dx x 3 3x x 2 2x 1 dx 2 2 1
x 3 2x 2 x 1 dx x 4 5x 2 4 x dx x 2 4x 13
1
0
x3 dx 3 1
2x 3 5x dx x 5x 2 4
1
4
0
x4 1 dx 2 12
■
■
■
■
■
■
■ Make a substitution to express the integrand as a rational function and then evaluate the integral.
sx dx x4
35.
y
16
36.
y
1
37.
y sx
x3 dx 2 1
38.
y
39.
ye
e 2x dx 3e x 2
40.
y
3
2
9
dx
2
35– 40
1
y x ax b dx
2
x 3 x 2 2x 1 dx x 2 1x 2 2
2 x 2 3x 4
■
x2
19.
3. (a)
■
2
EXERCISES
■ Write out the form of the partial fraction decomposition of the function (as in Example 6). Do not determine the numerical values of the coefficients.
■
1 K 2x 1
1–6
■
dx
ln x 12 lnx 2 1 tan1x
In the second and fourth terms we made the mental substitution u x 2 1. ■
1. (a)
dx x dx x dx y 2 dx y 2 y 2 x x 1 x 1 x 12
y
6.3
1 x1 x 2 2 x x 1 x 12
■
9
0
1 dx 3 1s x 3
■
(Let u sx .)
2x
■
■
(Let u sx .) 3
■
■
■
3
13
sx dx x2 x
cos x dx sin 2x sin x ■
■
■
■
■
328
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
the integral to give an equation relating the female population to time. (Note that the resulting equation can’t be solved explicitly for P.)
41– 42 ■ Use integration by parts, together with the techniques of this section, to evaluate the integral. 41. 42. ■
y lnx
2
x 2 dx 44. Factor x 4 1 as a difference of squares by first adding and
y x tan
subtracting the same quantity. Use this factorization to evaluate x 1x 4 1 dx.
1
■
x dx
■
■
■
■
■
■
■
■
■
■
45. Suppose that F, G, and Q are polynomials and 43. One method of slowing the growth of an insect population
without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. If P represents the number of female insects in a population, S the number of sterile males introduced each generation, and r the population’s natural growth rate, then the female population is related to time t by PS ty dP P r 1P S
Gx Fx Qx Qx for all x except when Qx 0. Prove that Fx Gx for all x. [Hint: Use continuity.] 46. If f is a quadratic function such that f 0 1 and
Suppose an insect population with 10,000 females grows at a rate of r 0.10 and 900 sterile males are added. Evaluate
6.4
y
f x dx x 2x 13
is a rational function, find the value of f 0.
INTEGRATION WITH TABLES AND COMPUTER ALGEBRA SYSTEMS In this section we describe how to evaluate integrals using tables and computer algebra systems. TABLES OF INTEGRALS
Tables of indefinite integrals are very useful when we are confronted by an integral that is difficult to evaluate by hand and we don’t have access to a computer algebra system. A relatively brief table of 120 integrals, categorized by form, is provided on the Reference Pages at the back of the book. More extensive tables are available in CRC Standard Mathematical Tables and Formulae, 31st ed, by Daniel Zwillinger (Boca Raton, FL: CRC Press, 2002) (709 entries) or in Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products, 6e, edited by A. Jefferey and D. Zwillinger (San Diego: Academic Press, 2000), which contains hundreds of pages of integrals. It should be remembered, however, that integrals do not often occur in exactly the form listed in a table. Usually we need to use the Substitution Rule or algebraic manipulation to transform a given integral into one of the forms in the table.
The Table of Integrals appears on Reference Pages 6–10 at the back of the book. ■
V EXAMPLE 1
Use the Table of Integrals to find y
x2 dx . s5 4x 2
SOLUTION If we look at the section of the table entitled Forms involving sa 2 u 2 ,
we see that the closest entry is number 34:
y sa
u2 u a2 u 2 u2 du sin1 sa 2 u2 2 2 a
C
SECTION 6.4
■
INTEGRATION WITH TABLES AND COMPUTER ALGEBRA SYSTEMS
329
This is not exactly what we have, but we will be able to use it if we first make the substitution u 2x : x2
y s5 4x
2
dx y
u22 du 1 8 s5 u 2 2
u2
y s5 u
2
du
Then we use Formula 34 with a 2 5 (so a s5 ): x2
y s5 4x
dx
2
1 8
u2
y s5 u
2
du
1 8
u 5 u s5 u 2 sin1 2 2 s5
x 5 2x sin1 s5 4x 2 8 16 s5
C
C
■
EXAMPLE 2 Use the Table of Integrals to find y x 3 sin x dx. SOLUTION If we look in the section called Trigonometric Forms, we see that none
of the entries explicitly includes a u 3 factor. However, we can use the reduction formula in entry 84 with n 3:
yx 85.
yu
n
cos u du
u n sin u n y u n1 sin u du
3
sin x dx x 3 cos x 3 y x 2 cos x dx
We now need to evaluate x x 2 cos x dx . We can use the reduction formula in entry 85 with n 2, followed by entry 82:
yx
2
cos x dx x 2 sin x 2 y x sin x dx x 2 sin x 2sin x x cos x K
Combining these calculations, we get
yx
3
sin x dx x 3 cos x 3x 2 sin x 6x cos x 6 sin x C
where C 3K. V EXAMPLE 3
■
Use the Table of Integrals to find y x sx 2 2x 4 dx.
SOLUTION Since the table gives forms involving sa 2 x 2 , sa 2 x 2 , and
sx 2 a 2 , but not sax 2 bx c , we first complete the square: x 2 2x 4 x 12 3
If we make the substitution u x 1 (so x u 1), the integrand will involve the pattern sa 2 u 2 :
y xsx
2
2x 4 dx y u 1 su 2 3 du
y usu
2
3 du y su 2 3 du
330
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
The first integral is evaluated using the substitution t u 2 3:
y usu 21.
y sa
2
u 2 du
u sa 2 u 2 2
2
3 du 12 y st dt 12 23 t 32 13 u 2 332
For the second integral we use Formula 21 with a s3 :
y su
a2 ln (u sa 2 u 2 ) C 2
2
3 du
u 3 su 2 3 2 ln(u su 2 3 ) 2
Thus
y xsx
2
2x 4 dx
13x 2 2x 432
x1 3 sx 2 2x 4 2 ln( x 1 sx 2 2x 4 ) C 2 ■
COMPUTER ALGEBRA SYSTEMS
We have seen that the use of tables involves matching the form of the given integrand with the forms of the integrands in the tables. Computers are particularly good at matching patterns. And just as we used substitutions in conjunction with tables, a CAS can perform substitutions that transform a given integral into one that occurs in its stored formulas. So it isn’t surprising that computer algebra systems excel at integration. That doesn’t mean that integration by hand is an obsolete skill. We will see that a hand computation sometimes produces an indefinite integral in a form that is more convenient than a machine answer. To begin, let’s see what happens when we ask a machine to integrate the relatively simple function y 13x 2. Using the substitution u 3x 2, an easy calculation by hand gives 1 y 3x 2 dx 13 ln 3x 2 C
whereas Derive, Mathematica, and Maple all return the answer 1 3
ln3x 2
The first thing to notice is that computer algebra systems omit the constant of integration. In other words, they produce a particular antiderivative, not the most general one. Therefore, when making use of a machine integration, we might have to add a constant. Second, the absolute value signs are omitted in the machine answer. That is fine if our problem is concerned only with values of x greater than 23 . But if we are interested in other values of x, then we need to insert the absolute value symbol. In the next example we reconsider the integral of Example 3, but this time we ask a machine for the answer. EXAMPLE 4 Use a computer algebra system to find y xsx 2 2x 4 dx . SOLUTION Maple responds with the answer 1 3
x 2 2x 432 14 2x 2 sx 2 2x 4
3 s3 arcsinh 1 x 2 3
SECTION 6.4
INTEGRATION WITH TABLES AND COMPUTER ALGEBRA SYSTEMS
■
331
This looks different from the answer we found in Example 3, but it is equivalent because the third term can be rewritten using the identity ■
arcsinh x ln( x sx 2 1 )
This is Equation 3.6.3.
Thus arcsinh
s3 s3 1 x ln 1 x s 13 1 x2 1 3 3 ln
1 1 x s1 x2 3 s3
ln
1 ln( x 1 sx 2 2x 4 ) s3
[
]
The resulting extra term 32 ln(1s3 ) can be absorbed into the constant of integration. Mathematica gives the answer
5 x x2 6 6 3
sx 2 2x 4
3 1x arcsinh 2 s3
Mathematica combined the first two terms of Example 3 (and the Maple result) into a single term by factoring. Derive gives the answer 1 6
3 sx 2 2x 4 2x 2 x 5 2 ln (sx 2 2x 4 x 1)
The first term is like the first term in the Mathematica answer, and the second term is identical to the last term in Example 3. ■ EXAMPLE 5 Use a CAS to evaluate y x x 2 58 dx. SOLUTION Maple and Mathematica give the same answer: 1 18
12 x 18 52 x 16 50x 14 1750 4375x 10 21875x 8 218750 x 6 156250x 4 390625 x2 3 x 3 2
It’s clear that both systems must have expanded x 2 58 by the Binomial Theorem and then integrated each term. If we integrate by hand instead, using the substitution u x 2 5, we get
y xx
■ Derive and the TI-89 and TI-92 also give this answer.
2
58 dx 181 x 2 59 C
For most purposes, this is a more convenient form of the answer. EXAMPLE 6 Use a CAS to find y sin 5x cos 2x dx. SOLUTION In Example 2 in Section 6.2 we found that 1
y sin x cos x dx 5
2
1 3
cos3x 5 cos5x 7 cos7x C 2
1
■
332
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
Derive and Maple report the answer 8 17 sin 4x cos 3x 354 sin 2x cos 3x 105 cos 3x
whereas Mathematica produces 1 3 1 645 cos x 192 cos 3x 320 cos 5x 448 cos 7x
We suspect that there are trigonometric identities which show these three answers are equivalent. Indeed, if we ask Derive, Maple, and Mathematica to simplify their expressions using trigonometric identities, they ultimately produce the same form of the answer as in Equation 1. ■ CAN WE INTEGRATE ALL CONTINUOUS FUNCTIONS ?
The question arises: Will our basic integration formulas, together with the Substitution Rule, integration by parts, tables of integrals, and computer algebra systems, enable us to find the integral of every continuous function? In particular, can we use it to eval2 uate x e x dx ? The answer is No, at least not in terms of the functions that we are familiar with. Most of the functions that we have been dealing with in this book are what are called elementary functions. These are the polynomials, rational functions, power functions x a , exponential functions a x , logarithmic functions, trigonometric and inverse trigonometric functions, and all functions that can be obtained from these by the five operations of addition, subtraction, multiplication, division, and composition. For instance, the function f x
x2 1 lncos x xe sin 2x x 3 2x 1
is an elementary function. If f is an elementary function, then f is an elementary function but x f x dx need 2 not be an elementary function. Consider f x e x . Since f is continuous, its integral exists, and if we define the function F by Fx y e t dt x
2
0
then we know from Part 1 of the Fundamental Theorem of Calculus that Fx e x
2
2
Thus f x e x has an antiderivative F, but it can be proved that F is not an elementary function. This means that no matter how hard we try, we will never succeed in 2 evaluating x e x dx in terms of the functions we know. (In Chapter 8, however, we will 2 see how to express x e x dx as an infinite series.) The same can be said of the following integrals:
y
ex dx x
y sx 3 1 dx
y sinx 2 dx 1
y ln x dx
y cose x dx y
sin x dx x
In fact, the majority of elementary functions don’t have elementary antiderivatives.
SECTION 6.5
6.4
■ Use the Table of Integrals on Reference Pages 6 –10 to evaluate the integral.
y
3.
y sec 3 x dx
5. 7. 9.
1
2x cos1x dx
0
yx y
dx s4x 2 9
y e 2 sin 3 d
4.
y
8.
tan 1z dz z2 3
11.
y y s6 4y 4y
13.
y sin x cos x lnsin x dx
15.
y 3e
17.
y sx
19.
y
21.
y se
■
2.
6.
x 3 sin x dx
0
y
2
2
dy
2
ex
2x
dx
x 4 dx 10 2
s4 ln x 2 dx x
■
2x
1 dx ■
■
■
■
1 dx x 2 s4x 2 7
3
2
s2y 2 3 dy y2
y
e 2x
y s2 e
x
333
y sin
12.
y x sinx
14.
y
16.
y
2
18.
y
1
20.
y s9 tan
22.
ye
1
0
25.
y x s5 x
27.
y sin x
29.
31. ■
CAS
cos3x 2 dx
dx
26.
y x 21 x 3 4 dx
cos 2x dx
28.
y tan x
y x s1 2x dx
30.
y sin 4x dx
y tan5x dx
32.
y x 5sx
2
3
■
■
2
■
■
y2
■
d
■
1 dx
■
■
■
■
s4 x 1 dx
2x
1 dx
instead. Why do you think it was successful with this form of the integrand?
sin t 3 dt ■
x
y 2 x s2 2
■
2
If it doesn’t return an answer, ask it to try
x 4ex dx sec 2 tan 2
■
sec 4x dx
from human beings. Ask your CAS to evaluate
x 3 s4x 2 x 4 dx
t
■
2
33. Computer algebra systems sometimes need a helping hand
cos43 d
0
0
sx dx 2
25–32 ■ Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent.
dx
10.
■
CAS
CAS ■
34. Try to evaluate
■
y 1 ln x s1 x ln x
2
23. Verify Formula 53 in the Table of Integrals (a) by differenti-
ation and (b) by using the substitution t a bu.
dx
with a computer algebra system. If it doesn’t return an answer, make a substitution that changes the integral into one that the CAS can evaluate.
24. Verify Formula 31 (a) by differentiation and (b) by substi-
tuting u a sin .
6.5
■
EXERCISES
1–22
1.
APPROXIMATE INTEGRATION
APPROXIMATE INTEGRATION There are two situations in which it is impossible to find the exact value of a definite integral. The first situation arises from the fact that in order to find xab f x dx using the Evaluation Theorem we need to know an antiderivative of f . Sometimes, however, it is difficult, or even impossible, to find an antiderivative (see Section 6.4). For example, it is impossible to evaluate the following integrals exactly:
y
1
0
2
e x dx
y
1
1
s1 x 3 dx
334
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
The second situation arises when the function is determined from a scientific experiment through instrument readings or collected data. There may be no formula for the function (see Example 5). In both cases we need to find approximate values of definite integrals. We already know one such method. Recall that the definite integral is defined as a limit of Riemann sums, so any Riemann sum could be used as an approximation to the integral: If we divide a, b into n subintervals of equal length x b an, then we have
y
y
b
a
0
x¸
⁄
¤
‹
x¢
x
(a) Left endpoint approximation
n
f x dx
f x* x i
i1
where x*i is any point in the ith subinterval x i1, x i . If x*i is chosen to be the left endpoint of the interval, then x*i x i1 and we have
y
y
1
n
b
f x dx L n
a
f x
i1
x
i1
If f x 0, then the integral represents an area and (1) represents an approximation of this area by the rectangles shown in Figure 1(a). If we choose x*i to be the right endpoint, then x*i x i and we have 0
x¸
⁄
¤
‹
x¢
x
(b) Right endpoint approximation y
y
2
b
a
n
f x x
f x dx Rn
i
i1
[See Figure 1(b).] The approximations L n and Rn defined by Equations 1 and 2 are called the left endpoint approximation and right endpoint approximation, respectively. In Section 5.2 we also considered the case where x*i is chosen to be the midpoint xi of the subinterval x i1, x i . Figure 1(c) shows the midpoint approximation Mn , which appears to be better than either L n or Rn. MIDPOINT RULE
0
⁄ –
¤ –
– ‹
–x¢
y
x
b
a
(c) Midpoint approximation FIGURE 1
f x dx Mn x f x1 f x 2 f xn x
where
ba n
xi 12 x i1 x i midpoint of x i1, x i
and
Another approximation, called the Trapezoidal Rule, results from averaging the approximations in Equations 1 and 2:
y
b
a
f x dx
1 2
n
n
f x i1 x
i1
f x x i
i1
x 2
( n
f x i1 f x i )
i1
x 2
x f x 0 2 f x 1 2 f x 2 2 f x n1 f x n 2
[( f x f x ) ( f x f x ) ( f x 0
1
1
2
n1
f x n )
]
SECTION 6.5
y
APPROXIMATE INTEGRATION
■
335
TRAPEZOIDAL RULE
y
b
a
f x dx Tn
x f x0 2 f x1 2 f x2 2 f xn1 f x n 2
where x b an and xi a i x.
0
x¸
⁄
¤
‹
x¢
x
The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates the case f x 0. The area of the trapezoid that lies above the ith subinterval is
FIGURE 2
x
Trapezoidal approximation
y=
f x i1 f x i 2
x f x i1 f x i 2
and if we add the areas of all these trapezoids, we get the right side of the Trapezoidal Rule.
1 x
EXAMPLE 1 Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n 5 to
approximate the integral x12 1x dx.
SOLUTION
(a) With n 5, a 1, and b 2, we have x 2 15 0.2, and so the Trapezoidal Rule gives
1
2
y
2
1
1 0.2 dx T5 f 1 2 f 1.2 2 f 1.4 2 f 1.6 2 f 1.8 f 2 x 2
0.1
FIGURE 3
1 2 2 2 2 1 1 1.2 1.4 1.6 1.8 2
0.695635 y=
This approximation is illustrated in Figure 3. (b) The midpoints of the five subintervals are 1.1, 1.3, 1.5, 1.7, and 1.9, so the Midpoint Rule gives
1 x
y
2
1
1 dx x f 1.1 f 1.3 f 1.5 f 1.7 f 1.9 x 1 5
1
2
FIGURE 4
1 1 1 1 1 1.1 1.3 1.5 1.7 1.9
0.691908 This approximation is illustrated in Figure 4.
■
In Example 1 we deliberately chose an integral whose value can be computed explicitly so that we can see how accurate the Trapezoidal and Midpoint Rules are. By the Fundamental Theorem of Calculus,
y
2
1
y
b
a
f x dx approximation error
1 2 dx ln x]1 ln 2 0.693147 . . . x
The error in using an approximation is defined to be the amount that needs to be added to the approximation to make it exact. From the values in Example 1 we see that the
336
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
errors in the Trapezoidal and Midpoint Rule approximations for n 5 are ET 0.002488
and
EM 0.001239
In general, we have ET y f x dx Tn b
a
Module 5.2/6.5 allows you to compare approximation methods. Approximations to y
2
1
Corresponding errors
3. 4.
A
D x i-1
x–i
xi
C R P B Q A FIGURE 5
D
a
n
Ln
Rn
Tn
Mn
5 10 20
0.745635 0.718771 0.705803
0.645635 0.668771 0.680803
0.695635 0.693771 0.693303
0.691908 0.692835 0.693069
n
EL
ER
ET
EM
5 10 20
0.052488 0.025624 0.012656
0.047512 0.024376 0.012344
0.002488 0.000624 0.000156
0.001239 0.000312 0.000078
1. In all of the methods we get more accurate approximations when we increase
2.
B
b
We can make several observations from these tables:
C P
EM y f x dx Mn
The following tables show the results of calculations similar to those in Example 1, but for n 5, 10, and 20 and for the left and right endpoint approximations as well as the Trapezoidal and Midpoint Rules.
1 dx x
■ It turns out that these observations are true in most cases.
and
5.
the value of n. (But very large values of n result in so many arithmetic operations that we have to beware of accumulated round-off error.) The errors in the left and right endpoint approximations are opposite in sign and appear to decrease by a factor of about 2 when we double the value of n. The Trapezoidal and Midpoint Rules are much more accurate than the endpoint approximations. The errors in the Trapezoidal and Midpoint Rules are opposite in sign and appear to decrease by a factor of about 4 when we double the value of n. The size of the error in the Midpoint Rule is about half the size of the error in the Trapezoidal Rule.
Figure 5 shows why we can usually expect the Midpoint Rule to be more accurate than the Trapezoidal Rule. The area of a typical rectangle in the Midpoint Rule is the same as the trapezoid ABCD whose upper side is tangent to the graph at P. The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid AQRD used in the Trapezoidal Rule. [The midpoint error (shaded gray) is smaller than the trapezoidal error (shaded blue).] These observations are corroborated in the following error estimates, which are proved in books on numerical analysis. Notice that Observation 4 corresponds to the n 2 in each denominator because 2n2 4n 2. The fact that the estimates depend on the size of the second derivative is not surprising if you look at Figure 5, because f x measures how much the graph is curved. [Recall that f x measures how fast the slope of y f x changes.]
SECTION 6.5
APPROXIMATE INTEGRATION
■
337
3 ERROR BOUNDS Suppose f x K for a x b. If ET and EM are the errors in the Trapezoidal and Midpoint Rules, then
E T
Kb a3 12n 2
and
E M
Kb a3 24n 2
Let’s apply this error estimate to the Trapezoidal Rule approximation in Example 1. If f x 1x, then f x 1x 2 and f x 2x 3. Since 1 x 2, we have 1x 1, so
f x ■ K can be any number larger than all the values of f x , but smaller values of K give better error bounds.
2 2 2 3 x 13
Therefore, taking K 2, a 1, b 2, and n 5 in the error estimate (3), we see that
ET
22 13 1 0.006667 2 125 150
Comparing this error estimate of 0.006667 with the actual error of about 0.002488, we see that it can happen that the actual error is substantially less than the upper bound for the error given by (3). V EXAMPLE 2 How large should we take n in order to guarantee that the Trapezoidal and Midpoint Rule approximations for x12 1x dx are accurate to within 0.0001?
SOLUTION We saw in the preceding calculation that f x 2 for 1 x 2, so
we can take K 2, a 1, and b 2 in (3). Accuracy to within 0.0001 means that the size of the error should be less than 0.0001. Therefore, we choose n so that 213 0.0001 12n 2 Solving the inequality for n, we get n2
or ■ It’s quite possible that a lower value for n would suffice, but 41 is the smallest value for which the error bound formula can guarantee us accuracy to within 0.0001.
n
2 120.0001 1 40.8 s0.0006
Thus n 41 will ensure the desired accuracy. For the same accuracy with the Midpoint Rule we choose n so that 213 0.0001 24n 2 which gives
n
1 29 s0.0012
■
338
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
y
V EXAMPLE 3
(a) Use the Midpoint Rule with n 10 to approximate the integral x01 e x dx. (b) Give an upper bound for the error involved in this approximation. 2
y=e x
SOLUTION
2
(a) Since a 0, b 1, and n 10, the Midpoint Rule gives
y
1
0
2
e x dx x f 0.05 f 0.15 f 0.85 f 0.95 0.1 e 0.0025 e 0.0225 e 0.0625 e 0.1225 e 0.2025 e 0.3025 e 0.4225 e 0.5625 e 0.7225 e 0.9025
0
1
x
1.460393
FIGURE 6
Figure 6 illustrates this approximation. 2 2 2 (b) Since f x e x , we have f x 2xe x and f x 2 4x 2 e x . Also, since 0 x 1, we have x 2 1 and so 2
0 f x 2 4x 2 e x 6e ■ Error estimates are upper bounds for the error. They give theoretical, worstcase scenarios. The actual error in this case turns out to be about 0.0023.
Taking K 6e, a 0, b 1, and n 10 in the error estimate (3), we see that an upper bound for the error is 6e13 e 0.007 24102 400
■
SIMPSON’S RULE
Another rule for approximate integration results from using parabolas instead of straight line segments to approximate a curve. As before, we divide a, b into n subintervals of equal length h x b an, but this time we assume that n is an even number. Then on each consecutive pair of intervals we approximate the curve y f x 0 by a parabola as shown in Figure 7. If yi f x i , then Pi x i , yi is the point on the curve lying above x i . A typical parabola passes through three consecutive points Pi , Pi1 , and Pi2 . y
y
P¸
P¡
P∞
y=ƒ
P¸(_h, y¸)
Pß
P¡ (0, › )
P™ P£
0
a=x¸
FIGURE 7
⁄
x™
x£
P™ (h, fi)
P¢
x¢
x∞
xß=b
x
_h
0
h
x
FIGURE 8
To simplify our calculations, we first consider the case where x 0 h, x 1 0, and x 2 h. (See Figure 8.) We know that the equation of the parabola through P0 , P1 , and P2 is of the form y Ax 2 Bx C and so the area under the parabola from x h
SECTION 6.5
APPROXIMATE INTEGRATION
■
339
to x h is
y
Here we have used Theorem 5.5.7. Notice that Ax 2 C is even and Bx is odd. ■
h
h
Ax 2 Bx C dx 2 y Ax 2 C dx h
0
2 A 2 A
x3 Cx 3 3
h
0
h h Ch 2Ah 2 6C 3 3
But, since the parabola passes through P0 h, y0 , P10, y1 , and P2h, y2 , we have y0 Ah2 Bh C Ah 2 Bh C y1 C y2 Ah 2 Bh C and therefore
y0 4y1 y2 2Ah 2 6C
Thus we can rewrite the area under the parabola as h y0 4y1 y2 3 By shifting this parabola horizontally we do not change the area under it. This means that the area under the parabola through P0 , P1 , and P2 from x x 0 to x x 2 in Figure 7 is still h y0 4y1 y2 3 Similarly, the area under the parabola through P2 , P3 , and P4 from x x 2 to x x 4 is h y2 4y3 y4 3 If we compute the areas under all the parabolas in this manner and add the results, we get
y
b
a
f x dx
h h h y0 4y1 y2 y2 4y3 y4 yn2 4yn1 yn 3 3 3 h y0 4y1 2y2 4y3 2y4 2yn2 4yn1 yn 3
Although we have derived this approximation for the case in which f x 0, it is a reasonable approximation for any continuous function f and is called Simpson’s Rule after the English mathematician Thomas Simpson (1710–1761). Note the pattern of coefficients: 1, 4, 2, 4, 2, 4, 2, . . . , 4, 2, 4, 1.
340
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
SIMPSON’S RULE
Thomas Simpson was a weaver who taught himself mathematics and went on to become one of the best English mathematicians of the 18th century. What we call Simpson’s Rule was actually known to Cavalieri and Gregory in the 17th century, but Simpson popularized it in his best-selling calculus textbook, entitled A New Treatise of Fluxions. ■
y
b
a
f x dx Sn
x f x 0 4 f x 1 2 f x 2 4 f x 3 3 2 f xn2 4 f xn1 f xn
where n is even and x b an. EXAMPLE 4 Use Simpson’s Rule with n 10 to approximate x12 1x dx. SOLUTION Putting f x 1x, n 10, and x 0.1 in Simpson’s Rule, we obtain
y
2
1
1 dx S10 x x f 1 4 f 1.1 2 f 1.2 4 f 1.3 2 f 1.8 4 f 1.9 f 2 3
0.1 3
1 4 2 4 2 4 2 4 2 4 1 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
0.693150
■
Notice that, in Example 4, Simpson’s Rule gives us a much better approximation S10 0.693150 to the true value of the integral ln 2 0.693147. . . than does the Trapezoidal Rule T10 0.693771 or the Midpoint Rule M10 0.692835. It turns out (see Exercise 40) that the approximations in Simpson’s Rule are weighted averages of those in the Trapezoidal and Midpoint Rules: S2n 13 Tn 23 Mn
(Recall that ET and EM usually have opposite signs and EM is about half the size of ET .) In many applications of calculus we need to evaluate an integral even if no explicit formula is known for y as a function of x. A function may be given graphically or as a table of values of collected data. If there is evidence that the values are not changing rapidly, then the Trapezoidal Rule or Simpson’s Rule can still be used to find an approximate value for xab y dx, the integral of y with respect to x.
Figure 9 shows data traffic on the link from the United States to SWITCH, the Swiss education and research network, on February 10, 1998. Dt is the data throughput, measured in megabits per second Mbs. Use Simpson’s Rule to estimate the total amount of data transmitted on the link up to noon on that day. V EXAMPLE 5
D 8 6 4 2
FIGURE 9
0
3
6
9
12
15
18
21
24 t (hours)
SECTION 6.5
APPROXIMATE INTEGRATION
■
341
SOLUTION Because we want the units to be consistent and Dt is measured in megabits per second, we convert the units for t from hours to seconds. If we let At be the amount of data (in megabits) transmitted by time t, where t is measured in seconds, then At Dt. So, by the Net Change Theorem (see Section 5.3), the total amount of data transmitted by noon (when t 12 60 2 43,200) is
A43,200 y
43,200
0
Dt dt
We estimate the values of Dt at hourly intervals from the graph and compile them in the table. t hours
t seconds
Dt
t hours
t seconds
Dt
0 1 2 3 4 5 6
0 3,600 7,200 10,800 14,400 18,000 21,600
3.2 2.7 1.9 1.7 1.3 1.0 1.1
7 8 9 10 11 12
25,200 28,800 32,400 36,000 39,600 43,200
1.3 2.8 5.7 7.1 7.7 7.9
Then we use Simpson’s Rule with n 12 and t 3600 to estimate the integral:
y
43,200
0
At dt
t D0 4D3600 2D7200 4D39,600 D43,200 3 3600 3.2 42.7 21.9 41.7 21.3 41.0 3 21.1 41.3 22.8 45.7 27.1 47.7 7.9
143,880 Thus the total amount of data transmitted up to noon is about 144,000 megabits, or 144 gigabits. ■ In Exercise 24 you are asked to demonstrate, in a particular case, that the error in Simpson’s Rule decreases by a factor of about 16 when n is doubled. That is consistent with the appearance of n 4 in the denominator of the following error estimate for Simpson’s Rule. It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules, but it uses the fourth derivative of f .
Suppose that f 4x K for a x b. If ES is the error involved in using Simpson’s Rule, then 4 ERROR BOUND FOR SIMPSON’S RULE
ES
Kb a5 180n 4
342
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
EXAMPLE 6 How large should we take n in order to guarantee that the Simpson’s
Rule approximation for x12 1x dx is accurate to within 0.0001?
SOLUTION If f x 1x, then f 4x 24x 5. Since x 1, we have 1x 1 and so
f ■ Many calculators and computer algebra systems have a built-in algorithm that computes an approximation of a definite integral. Some of these machines use Simpson’s Rule; others use more sophisticated techniques such as adaptive numerical integration. This means that if a function fluctuates much more on a certain part of the interval than it does elsewhere, then that part gets divided into more subintervals. This strategy reduces the number of calculations required to achieve a prescribed accuracy.
4
x
24
24 x5
Therefore, we can take K 24 in (4). Thus for an error less than 0.0001 we should choose n so that 2415 0.0001 180n 4 24 1800.0001
n4
This gives
1 6.04 s0.00075
n
or
4
Therefore, n 8 (n must be even) gives the desired accuracy. (Compare this with Example 2, where we obtained n 41 for the Trapezoidal Rule and n 29 for the Midpoint Rule.) ■ EXAMPLE 7
(a) Use Simpson’s Rule with n 10 to approximate the integral x01 e x dx. (b) Estimate the error involved in this approximation. 2
SOLUTION
(a) If n 10, then x 0.1 and Simpson’s Rule gives Figure 10 illustrates the calculation in Example 7. Notice that the parabolic 2 arcs are so close to the graph of y e x that they are practically indistinguishable from it. ■
y
y
1
0
2
e x dx
x f 0 4 f 0.1 2 f 0.2 2 f 0.8 4 f 0.9 f 1 3 0.1 0 e 4e 0.01 2e 0.04 4e 0.09 2e 0.16 4e 0.25 2e 0.36 3 4e 0.49 2e 0.64 4e 0.81 e 1
1.462681 2
y=e
(b) The fourth derivative of f x e x is
x2
f 4x 12 48x 2 16x 4 e x
2
and so, since 0 x 1, we have 0 f 4x 12 48 16e 1 76e
0
FIGURE 10
1
x
Therefore, putting K 76e, a 0, b 1, and n 10 in (4), we see that the error is at most 76e15 0.000115 180104 [Compare this with Example 3(b).] Thus, correct to three decimal places, we have
y
1
0
2
e x dx 1.463
■
SECTION 6.5
6.5 1. Let I
y 3
5.
y
0
■
■
■
2
7.
y
2 4
1
9.
y
2
11.
y
4
13.
y
5
15.
y
3
2
3
■
■
6. ■
y
1
0
■
esx dx, ■
■
■
n6
■
■
■ Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.)
f
1
n8
x 2 sin x dx,
7–16
4 x
2. The left, right, Trapezoidal, and Midpoint Rule approxima-
tions were used to estimate x f x dx, where f is the function whose graph is shown. The estimates were 0.7811, 0.8675, 0.8632, and 0.9540, and the same number of subintervals were used in each case. (a) Which rule produced which estimate? (b) Between which two approximations does the true value of x02 f x dx lie? 2 0
■
s1 x 2 dx ,
0
1
0
1
0 ■
n8
8.
y
12
0
ln x dx, 1x
n 10
10.
y
3
e st sin t dt,
n8
12.
y
4
cos x dx, n 8 x
14.
y
6
1 dy, 1 y5
16.
y
1
■
■
n6 ■
■
■
0
0
4
0
sinx 2 dx,
dt , 1 t2 t4
n4 n6
s1 sx dx , n 8 lnx 3 2 dx,
n 10
sz ez dz, n 10 ■
■
■
■
■
17. (a) Find the approximations T10 and M10 for the integral
x02 ex
2
dx. (b) Estimate the errors in the approximations of part (a). (c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to within 0.00001?
y 1
18. (a) Find the approximations T8 and M8 for x01 cosx 2 dx.
y=ƒ
2
(b) Estimate the errors involved in the approximations of part (a). (c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to within 0.00001?
x
19. (a) Find the approximations T10 and S10 for x01 e x dx and the
1 2 ; 3. Estimate x0 cosx dx using (a) the Trapezoidal Rule and
corresponding errors ET and ES . (b) Compare the actual errors in part (a) with the error estimates given by (3) and (4). (c) How large do we have to choose n so that the approximations Tn , Mn , and Sn to the integral in part (a) are accurate to within 0.00001?
(b) the Midpoint Rule, each with n 4. From a graph of the integrand, decide whether your answers are underestimates or overestimates. What can you conclude about the true value of the integral?
2 ; 4. Draw the graph of f x sinx 2 in the viewing rect-
angle 0, 1 by 0, 0.5 and let I x01 f x dx. (a) Use the graph to decide whether L 2 , R2 , M2, and T2 underestimate or overestimate I . (b) For any value of n, list the numbers L n , Rn , Mn , Tn , and I in increasing order. (c) Compute L 5 , R5 , M5, and T5. From the graph, which do you think gives the best estimate of I ?
■ Use (a) the Midpoint Rule and (b) Simpson’s Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) Compare your
5–6
343
results to the actual value to determine the error in each approximation.
f is the function whose graph is shown. (a) Use the graph to find L 2 , R2, and M2 . (b) Are these underestimates or overestimates of I ? (c) Use the graph to find T2 . How does it compare with I ? (d) For any value of n, list the numbers L n , Rn , Mn , Tn , and I in increasing order.
0
■
EXERCISES
x04 f x dx, where
0
APPROXIMATE INTEGRATION
20. How large should n be to guarantee that the Simpson’s Rule
approximation to x01 e x dx is accurate to within 0.00001? 2
CAS
21. The trouble with the error estimates is that it is often very
difficult to compute four derivatives and obtain a good upper bound K for f 4x by hand. But computer algebra systems have no problem computing f 4 and graphing it, so we can easily find a value for K from a machine graph. This exercise deals with approximations to the integral I x02 f x dx, where f x e cos x. (a) Use a graph to get a good upper bound for f x .
344
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
(b) Use M10 to approximate I . (c) Use part (a) to estimate the error in part (b). (d) Use the built-in numerical integration capability of your CAS to approximate I . (e) How does the actual error compare with the error estimate in part (c)? (f ) Use a graph to get a good upper bound for f 4x . (g) Use S10 to approximate I . (h) Use part (f ) to estimate the error in part (g). (i) How does the actual error compare with the error estimate in part (h)? ( j) How large should n be to guarantee that the size of the error in using Sn is less than 0.0001?
CAS
27. The graph of the acceleration at of a car measured in fts2
is shown. Use Simpson’s Rule to estimate the increase in the velocity of the car during the 6-second time interval. a 12
22. Repeat Exercise 21 for the integral y s4 x 3 dx . 1
1
8 4 0
where the graph of r is as shown. Use Simpson’s Rule to estimate the total amount of water that leaked out during the first six hours. r 4
x01 x 3 dx for n 4, 8, and 16. Then compute the corre-
sponding errors EL , ER, ET , and EM. (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when n is doubled? 2 x1 xe x dx for n 6 and 12. Then compute the corre-
sponding errors ET , EM , and ES . (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when n is doubled?
2
0
6 t
4
the power consumption in megawatts in San Diego County from midnight to 6:00 AM on a day in December. Use Simpson’s Rule to estimate the energy used during that time period. (Use the fact that power is the derivative of energy.)
(a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule, each with n 4. y 4
0
2
29. The table (supplied by San Diego Gas and Electric) gives
25. Estimate the area under the graph in the figure by using
2
6 t
4
28. Water leaked from a tank at a rate of rt liters per hour,
23. Find the approximations L n , Rn , Tn , and Mn to the integral
24. Find the approximations Tn , Mn , and Sn to the integral
2
t
P
t
P
0:00 0:30 1:00 1:30 2:00 2:30 3:00
1814 1735 1686 1646 1637 1609 1604
3:30 4:00 4:30 5:00 5:30 6:00
1611 1621 1666 1745 1886 2052
30. Shown is the graph of traffic on an Internet service pro1
2
3
4 x
26. A radar gun was used to record the speed of a runner during
the first 5 seconds of a race (see the table). Use Simpson’s Rule to estimate the distance the runner covered during those 5 seconds. t (s)
v (ms)
t (s)
v (ms)
0 0.5 1.0 1.5 2.0 2.5
0 4.67 7.34 8.86 9.73 10.22
3.0 3.5 4.0 4.5 5.0
10.51 10.67 10.76 10.81 10.81
vider’s T1 data line from midnight to 8:00 AM. D is the data throughput, measured in megabits per second. Use Simpson’s Rule to estimate the total amount of data transmitted during that time period. D 0.8
0.4
0
2
4
6
8 t (hours)
SECTION 6.6
31. (a) Use the Midpoint Rule and the given data to estimate
the value of the integral x03.2 f x dx. x
f x
x
f x
0.0 0.4 0.8 1.2 1.6
6.8 6.5 6.3 6.4 6.9
2.0 2.4 2.8 3.2
7.6 8.4 8.8 9.0
32. The figure shows a pendulum with length L that makes a
maximum angle 0 with the vertical. Using Newton’s Second Law it can be shown that the period T (the time for one complete swing) is given by
T4
L t
y
2
0
■
345
33. The intensity of light with wavelength traveling through
a diffraction grating with N slits at an angle is given by I N 2 sin 2kk 2, where k Nd sin and d is the distance between adjacent slits. A helium-neon laser with wavelength 632.8 109 m is emitting a narrow band of light, given by 106 106, through a grating with 10,000 slits spaced 104 m apart. Use the Midpoint Rule with n 10 to estimate the total light intensity 10 x10 I d emerging from the grating. 6
6
(b) If it is known that 4 f x 1 for all x, estimate the error involved in the approximation in part (a). CAS
IMPROPER INTEGRALS
dx s1 k 2 sin 2x
where k sin( 12 0 ) and t is the acceleration due to gravity. If L 1 m and 0 42 , use Simpson’s Rule with n 10 to find the period.
34. Sketch the graph of a continuous function on 0, 2 for
which the right endpoint approximation with n 2 is more accurate than Simpson’s Rule.
35. Sketch the graph of a continuous function on 0, 2 for
which the Trapezoidal Rule with n 2 is more accurate than the Midpoint Rule.
36. Use the Trapezoidal Rule with n 10 to approximate
x020 cos x dx. Compare your result to the actual value. Can you explain the discrepancy?
37. If f is a positive function and f x 0 for a x b,
show that Tn y f x dx Mn b
a
38. Show that if f is a polynomial of degree 3 or lower, then
¨¸
Simpson’s Rule gives the exact value of xab f x dx.
39. Show that 2 Tn Mn T2n . 1
40. Show that 3 Tn 3 Mn S2n . 1
6.6
2
IMPROPER INTEGRALS In defining a definite integral xab f x dx we dealt with a function f defined on a finite interval a, b . In this section we extend the concept of a definite integral to the case where the interval is infinite and also to the case where f has an infinite discontinuity in a, b . In either case the integral is called an improper integral. TYPE 1: INFINITE INTERVALS
Consider the infinite region S that lies under the curve y 1x 2, above the x-axis, and to the right of the line x 1. You might think that, since S is infinite in extent, its area must be infinite, but let’s take a closer look. The area of the part of S that lies to the left of the line x t (shaded in Figure 1) is
y
y=
1 ≈ area=1=1
x=1 0
FIGURE 1
1
t
1 t x
At y
t
1
1 1 dx x2 x
t
1
1
Notice that At 1 no matter how large t is chosen.
1 t
346
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
y
We also observe that
0
1
2
tl
y
2
3
1 t
1
1 t 1 dx 1 y 2 dx tlim l 1 x x2
1
area= 3 1
tl
The area of the shaded region approaches 1 as t l (see Figure 2), so we say that the area of the infinite region S is equal to 1 and we write
x
y
0
lim At lim 1
1
area= 2
Using this example as a guide, we define the integral of f (not necessarily a positive function) over an infinite interval as the limit of integrals over finite intervals.
x
y 1 DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1
(a) If xat f x dx exists for every number t a, then
4 area= 5
0 y
1
a
f x dx lim y f x dx t
tl
a
provided this limit exists (as a finite number). (b) If xtb f x dx exists for every number t b, then
area=1 0
y
5 x
1
y
b
x
f x dx lim
t l
y
t
b
f x dx
provided this limit exists (as a finite number).
FIGURE 2
b The improper integrals xa f x dx and x f x dx are called convergent if the corresponding limit exists and divergent if the limit does not exist. a (c) If both xa f x dx and x f x dx are convergent, then we define
y
f x dx y
a
f x dx y f x dx a
In part (c) any real number a can be used (see Exercise 52). Any of the improper integrals in Definition 1 can be interpreted as an area provided that f is a positive function. For instance, in case (a) if f x 0 and the integral xa f x dx is convergent, then we define the area of the region
S x, y x a, 0 y f x in Figure 3 to be
AS y f x dx a
y
y=ƒ S
FIGURE 3
0
a
x
SECTION 6.6
IMPROPER INTEGRALS
■
347
This is appropriate because xa f x dx is the limit as t l of the area under the graph of f from a to t. Determine whether the integral x1 1x dx is convergent or divergent.
V EXAMPLE 1
SOLUTION According to part (a) of Definition 1, we have
y
1
1 t 1 dx lim y dx lim ln x tl 1 x tl x
]
t
1
lim ln t ln 1 lim ln t tl
The limit does not exist as a finite number and so the improper integral x1 1x dx is divergent. ■
y
y=
1 ≈
Let’s compare the result of Example 1 with the example given at the beginning of this section: 1 1 y1 x 2 dx converges y1 x dx diverges
finite area 0
x
1
FIGURE 4
y
y=
1 x
Geometrically, this says that although the curves y 1x 2 and y 1x look very similar for x 0, the region under y 1x 2 to the right of x 1 (the shaded region in Figure 4) has finite area whereas the corresponding region under y 1x (in Figure 5) has infinite area. Note that both 1x 2 and 1x approach 0 as x l but 1x 2 approaches 0 faster than 1x. The values of 1x don’t decrease fast enough for its integral to have a finite value. EXAMPLE 2 Evaluate y
0
xe x dx.
SOLUTION Using part (b) of Definition 1, we have
infinite area
y 0
1
tl
x
xe x dx lim
t l
y
t
0
xe x dx
We integrate by parts with u x, dv e x dx , so that du dx, v e x :
FIGURE 5
y
0
t
In Module 6.6 you can investigate visually and numerically whether several improper integrals are convergent or divergent.
0
xe x dx xe x t y e x dx te t 1 e t
]
0
0
t
We know that e l 0 as t l , and by l’Hospital’s Rule we have t
lim te t lim
t l
t l
t 1 lim t l et et
lim e t 0 t l
Therefore
y
0
xe x dx lim te t 1 e t t l
0 1 0 1 EXAMPLE 3 Evaluate y
1 dx. 1 x2
SOLUTION It’s convenient to choose a 0 in Definition 1(c):
y
1 1 1 0 2 dx y 2 dx y 2 dx 1 x 0 1 x 1x
■
348
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CHAPTER 6
TECHNIQUES OF INTEGRATION
We must now evaluate the integrals on the right side separately:
y
0
1 t dx dx lim y lim tan1x tl 0 1 x2 tl 1 x2
]
t
0
lim tan 1 t tan1 0 lim tan1 t tl
y
0
tl
1 0 dx lim tan1x y 2 dx t lim l t 1 x 2 t l 1x
2
0
]
t
lim tan 1 0 tan 1 t t l
0
2
2
Since both of these integrals are convergent, the given integral is convergent and y=
1 1+≈
y
y
area=π 0
FIGURE 6
x
1 2 dx 1x 2 2
Since 11 x 2 0, the given improper integral can be interpreted as the area of the infinite region that lies under the curve y 11 x 2 and above the x-axis (see Figure 6). ■ EXAMPLE 4 For what values of p is the following integral convergent?
y
1
1 dx xp
SOLUTION We know from Example 1 that if p 1, then the integral is divergent,
so let’s assume that p 1. Then
y
1
1 t p xp1 x dx lim y p dx tlim l 1 t l p 1 x lim
tl
xt
x1
1 1 1 1 p t p1
If p 1, then p 1 0, so as t l , t p1 l and 1t p1 l 0. Therefore
y
1
1 1 p dx x p1
if p 1
and so the integral converges. But if p 1, then p 1 0 and so 1 t 1p l t p1 and the integral diverges.
as t l ■
SECTION 6.6
IMPROPER INTEGRALS
■
349
We summarize the result of Example 4 for future reference:
2
y
1
1 dx xp
is convergent if p 1 and divergent if p 1.
TYPE 2: DISCONTINUOUS INTEGRANDS y
y=ƒ
0
a
x=b
Suppose that f is a positive continuous function defined on a finite interval a, b but has a vertical asymptote at b. Let S be the unbounded region under the graph of f and above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a horizontal direction. Here the region is infinite in a vertical direction.) The area of the part of S between a and t (the shaded region in Figure 7) is
x
t b
FIGURE 7
At
y
t
a
f x dx
If it happens that At approaches a definite number A as t l b, then we say that the area of the region S is A and we write
y
b
a
f x dx lim ya f x dx t
tlb
We use this equation to define an improper integral of Type 2 even when f is not a positive function, no matter what type of discontinuity f has at b. 3 DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2
Parts (b) and (c) of Definition 3 are illustrated in Figures 8 and 9 for the case where f x 0 and f has vertical asymptotes at a and c, respectively. ■
(a) If f is continuous on a, b and is discontinuous at b, then
y
b
a
f x dx lim ya f x dx t
tlb
if this limit exists (as a finite number). (b) If f is continuous on a, b and is discontinuous at a, then
y
y
b
a
f x dx lim y f x dx b
tla
t
if this limit exists (as a finite number). 0
a t
b
x
The improper integral xab f x dx is called convergent if the corresponding limit exists and divergent if the limit does not exist. (c) If f has a discontinuity at c, where a c b, and both xac f x dx and xcb f x dx are convergent, then we define
FIGURE 8 y
y
b
a
EXAMPLE 5 Find y
5
2
0
a
FIGURE 9
c
b x
f x dx y f x dx y f x dx c
a
b
c
1 dx . sx 2
SOLUTION We note first that the given integral is improper because
f x 1sx 2 has the vertical asymptote x 2. Since the infinite discontinuity
350
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
occurs at the left endpoint of 2, 5 , we use part (b) of Definition 3:
y
y=
1 œ„„„„ x-2
y
dx 5 dx lim y lim 2sx 2 tl2 sx 2 tl2 t sx 2
5
2
1
2
3
4
5 t
lim 2(s3 st 2 ) 2s3 tl2
area=2œ„ 3 0
]
5
Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure 10.
x
FIGURE 10
V EXAMPLE 6
Determine whether y
/2
0
■
sec x dx converges or diverges.
SOLUTION Note that the given integral is improper because lim x l /2 sec x .
Using part (a) of Definition 3 and Formula 14 from the Table of Integrals, we have
y
/2
0
sec x dx lim y sec x dx t
t l2
0
lim ln sec x tan x t l2
]
t
0
lim lnsec t tan t ln 1 t l2
because sec t l and tan t l as t l 2. Thus the given improper integral is divergent. ■ EXAMPLE 7 Evaluate y
3
0
dx if possible. x1
SOLUTION Observe that the line x 1 is a vertical asymptote of the integrand. Since it occurs in the middle of the interval 0, 3 , we must use part (c) of Definition 3 with c 1: 3 dx 1 dx 3 dx y0 x 1 y0 x 1 y1 x 1
where
y
1
0
dx t dx lim y lim ln x 1 tl1 tl1 0 x 1 x1
lim (ln t 1 ln 1 tl1
]
t
0
)
lim ln1 t tl1
because 1 t l 0 as t l 1. Thus x01 dxx 1 is divergent. This implies that x03 dxx 1 is divergent. [We do not need to evaluate x13 dxx 1.] ■ |
WARNING If we had not noticed the asymptote x 1 in Example 7 and had instead confused the integral with an ordinary integral, then we might have made the following erroneous calculation:
y
3
0
dx ln x 1 x1
]
3 0
ln 2 ln 1 ln 2
This is wrong because the integral is improper and must be calculated in terms of limits.
SECTION 6.6
IMPROPER INTEGRALS
■
351
From now on, whenever you meet the symbol xab f x dx you must decide, by looking at the function f on a, b , whether it is an ordinary definite integral or an improper integral. A COMPARISON TEST FOR IMPROPER INTEGRALS
Sometimes it is impossible to find the exact value of an improper integral and yet it is important to know whether it is convergent or divergent. In such cases the following theorem is useful. Although we state it for Type 1 integrals, a similar theorem is true for Type 2 integrals. COMPARISON THEOREM Suppose that f and t are continuous functions with
f x tx 0 for x a.
(a) If xa f x dx is convergent, then xa tx dx is convergent. (b) If xa tx dx is divergent, then xa f x dx is divergent.
y
f g
0
x
a
FIGURE 11
We omit the proof of the Comparison Theorem, but Figure 11 makes it seem plausible. If the area under the top curve y f x is finite, then so is the area under the bottom curve y tx. And if the area under y tx is infinite, then so is the area under y f x. [Note that the reverse is not necessarily true: If xa tx dx is convergent, xa f x dx may or may not be convergent, and if xa f x dx is divergent, xa tx dx may or may not be divergent.] V EXAMPLE 8
Show that y ex dx is convergent. 2
0
SOLUTION We can’t evaluate the integral directly because the antiderivative of ex
2
is not an elementary function (as explained in Section 6.4). We write y
y=e _x
y
2
y=e _x
0
0
x
1
1
2
2
0
2
1
and observe that the first integral on the right-hand side is just an ordinary definite integral. In the second integral we use the fact that for x 1 we have x 2 x, so 2 x 2 x and therefore ex ex . (See Figure 12.) The integral of ex is easy to evaluate:
y
FIGURE 12
ex dx y ex dx y ex dx
1
ex dx lim y ex dx lim e1 et e1 t
tl
tl
1
2
Thus, taking f x ex and tx ex in the Comparison Theorem, we see that 2 2 x1 ex dx is convergent. It follows that x0 ex dx is convergent. ■
TABLE 1
t 1 2 3 4 5 6
x0t ex
2
dx
0.7468241328 0.8820813908 0.8862073483 0.8862269118 0.8862269255 0.8862269255
In Example 8 we showed that x0 ex dx is convergent without computing its value. In Exercise 58 we indicate how to show that its value is approximately 0.8862. In probability theory it is important to know the exact value of this improper integral; using the methods of multivariable calculus it can be shown that the exact value is of an improper integral by showing how the s 2. Table 1 illustrates the definition 2 (computer-generated) values of x0t ex dx approach s 2 as t becomes large. In fact, 2 these values converge quite quickly because ex l 0 very rapidly as x l . 2
352
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CHAPTER 6
TECHNIQUES OF INTEGRATION
EXAMPLE 9 The integral y
TABLE 2
t
x1t 1 ex x dx
2 5 10 100 1000 10000
0.8636306042 1.8276735512 2.5219648704 4.8245541204 7.1271392134 9.4297243064
6.6
because
(a)
y
(c)
y
2
1
0
4 x 4
x e
dx
x dx x 2 5x 6
and x1 1x dx is divergent by Example 1 [or by (2) with p 1].
1 dx 2x 1 sin x dx (c) y 2 1 x
y
2
1
EXERCISES
(b)
y
2
(d)
y
0
0
(b) (d)
y
1
y
2
0
1
15.
y
17.
y
19.
y
21.
y
23.
y
1
25.
y
14
27.
y
33
29.
y
1
31.
y
2
1 dx x2 5
1 dx 2x 1 lnx 1 dx
and evaluate it for t 10, 100, and 1000. Then find the total area under this curve for x 1.
1.1 0.9 ; 4. (a) Graph the functions f x 1x and tx 1x in
the viewing rectangles 0, 10 by 0, 1 and 0, 100 by 0, 1 . (b) Find the areas under the graphs of f and t from x 1 to x t and evaluate for t 10, 100, 10 4, 10 6, 10 10, and 10 20. (c) Find the total area under each curve for x 1, if it exists.
5–32 ■ Determine whether each integral is convergent or divergent. Evaluate those that are convergent. 5.
y
7.
y
1
y
9. 11. 13.
1
1 dx 3x 12
6.
y
0
1 dw s2 w
8.
y
0
e y2 dy
10.
y
1
y sin d
12.
y
y
4
2
y
2
xex dx
14.
0
se 5s ds
16.
y
ln x dx x
18.
y
6
ln x dx x2
20.
y
22.
y
24.
y
3
dx 4 x2 s
26.
y
8
x 1 15 dx
28.
y
1
ex dx e 1
30.
y
1
32.
y
1
cos t dt
sec x dx
3. Find the area under the curve y 1x 3 from x 1 to x t
■
Table 2 illustrates the divergence of the integral in Example 9. It appears that the values are not approaching any fixed number.
2. Which of the following integrals are improper? Why?
(a)
1
1 ex dx is divergent by the Comparison Theorem x 1 ex 1 x x
1. Explain why each of the following integrals is improper.
1 dx 2x 5 x dx 2 x 2 2 e2t dt 2 v 4 dv 3
x 2ex dx
■
1
1
3 dx x5
0
2
0
1
0
x2 dx 9 x6
x
z 2 ln z dz
■
33–38
■
■
■
■
■
■
; 35. ; 36. ; 37. ; 38. ■
1
0
2
6
0
0
0 ■
re r3 dr ln x dx x3 ex dx e 2x 3
1 dx s3 x 4 dx x 63 1 dy 4y 1 dx s1 x 2 ln x dx sx ■
■
■
■
Sketch the region and find its area (if the area is finite).
x 1, 0 y e S x, y x 2, 0 y e S x, y 0 y 2x 9 S x, y x 0, 0 y xx 9 S x, y 0 x 2, 0 y sec x S {x, y 2 x 0, 0 y 1sx 2 }
33. S x, y 34.
x
x/2
2
2
2
■
■
■
■
■
■
■
■
■
■
■
SECTION 6.6
2 2 ; 39. (a) If tx sin xx , use your calculator or computer to
make a table of approximate values of x tx dx for t 2, 5, 10, 100, 1000, and 10,000. Does it appear that x1 tx dx is convergent? (b) Use the Comparison Theorem with f x 1x 2 to show that x1 tx dx is convergent. (c) Illustrate part (b) by graphing f and t on the same screen for 1 x 10. Use your graph to explain intuitively why x1 tx dx is convergent. t 1
; 40. (a) If tx 1(sx 1), use your calculator or computer
to make a table of approximate values of x2t tx dx for t 5, 10, 100, 1000, and 10,000. Does it appear that x2 tx dx is convergent or divergent? (b) Use the Comparison Theorem with f x 1sx to show that x2 tx dx is divergent. (c) Illustrate part (b) by graphing f and t on the same screen for 2 x 20. Use your graph to explain intuitively why x2 tx dx is divergent.
41– 46 ■ Use the Comparison Theorem to determine whether the integral is convergent or divergent. 41.
y
43.
y
1
1
y
45.
cos 2x dx 1 x2
42.
y
dx x e 2x
44.
y
dx x sin x
/2
0
■
■
■
46. ■
■
■
2 e x dx x
1
y
1
0
■
e dx sx ■
■
■
■
47. The integral
y
0
1 dx sx 1 x
0
1 1 1 1 dx y dx y dx 0 sx 1 x 1 sx 1 x sx 1 x
48. ■
y
e ■
■
1 dx x ln x p ■
■
49. ■
■
■
y
1
0 ■
lim y x dx 0 t
tl
t
This shows that we can’t define
y
f x dx lim y f x dx t
tl
1 dx xp ■
t
52. If x f x dx is convergent and a and b are real numbers,
show that
y
a
f x dx
y
a
f x dx y
b
f x dx y f x dx b
53. A manufacturer of lightbulbs wants to produce bulbs that
last about 700 hours but, of course, some bulbs burn out faster than others. Let Ft be the fraction of the company’s bulbs that burn out before t hours, so Ft always lies between 0 and 1. (a) Make a rough sketch of what you think the graph of F might look like. (b) What is the meaning of the derivative rt Ft? (c) What is the value of x0 rt dt ? Why?
4 s
M 2RT
32
y
0
2
v 3eMv 2RT dv
where M is the molecular weight of the gas, R is the gas constant, T is the gas temperature, and v is the molecular speed. Show that
8RT M
55. As we saw in Section 3.4, a radioactive substance decays
exponentially: The mass at time t is mt m0e kt, where m0 is the initial mass and k is a negative constant. The mean life M of an atom in the substance is
M k y te kt dt 0
For the radioactive carbon isotope, 14 C, used in radiocarbon dating, the value of k is 0.000121. Find the mean life of a 14 C atom.
Find the values of p for which the integral converges and evaluate the integral for those values of p. 48 – 49
(b) Show that
v
is improper for two reasons: The interval 0, is infinite and the integrand has an infinite discontinuity at 0. Evaluate it by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows:
y
353
51. (a) Show that x x dx is divergent.
v
x
■
■
54. The average speed of molecules in an ideal gas is
x dx s1 x 6
1
IMPROPER INTEGRALS
56. Astronomers use a technique called stellar stereography to ■
■
50. (a) Evaluate the integral x0 x nex dx for n 0, 1, 2, and 3.
(b) Guess the value of x0 x nex dx when n is an arbitrary positive integer. (c) Prove your guess using mathematical induction.
■
determine the density of stars in a star cluster from the observed (two-dimensional) density that can be analyzed from a photograph. Suppose that in a spherical cluster of radius R the density of stars depends only on the distance r from the center of the cluster. If the perceived star density is given by ys, where s is the observed planar distance from
354
■
CHAPTER 6
TECHNIQUES OF INTEGRATION
the center of the cluster, and x r is the actual density, it can be shown that
59. Show that x0 x 2ex dx 2
60. Show that x0 ex dx 2
ys
y
R
s
2r x r dr sr 2 s 2
y
0
1 dx 0.001 x2 1
the sum of x04 ex dx and x4 ex dx. Approximate the first integral by using Simpson’s Rule with n 8 and show that the second integral is smaller than x4 e4x dx, which is less than 0.0000001.
y
2
REVIEW
dy by interpreting the
1 C 4 x 2 sx 2
dx
62. Find the value of the constant C for which the integral
2
6
x01 sln y
dx.
converges. Evaluate the integral for this value of C.
58. Estimate the numerical value of x0 ex dx by writing it as 2
2
61. Find the value of the constant C for which the integral
57. Determine how large the number a has to be so that
a
x0 ex
integrals as areas.
If the actual density of stars in a cluster is x r 12 R r2, find the perceived density ys.
y
1 2
0
x C x2 1 3x 1
dx
converges. Evaluate the integral for this value of C.
CONCEPT CHECK
1. State the rule for integration by parts. In practice, how do
5. State the rules for approximating the definite integral
xab f x dx with the Midpoint Rule, the Trapezoidal Rule, and
you use it?
Simpson’s Rule. Which would you expect to give the best estimate? How do you approximate the error for each rule?
2. How do you evaluate x sin mx cos nx dx if m is odd? What if
n is odd? What if m and n are both even? 3. If the expression sa 2 x 2 occurs in an integral, what sub-
stitution might you try? What if sa 2 x 2 occurs? What if sx 2 a 2 occurs?
4. What is the form of the partial fraction expansion of a
rational function PxQx if the degree of P is less than the degree of Q and Qx has only distinct linear factors? What if a linear factor is repeated? What if Qx has an irreducible quadratic factor (not repeated)? What if the quadratic factor is repeated?
6. Define the following improper integrals.
(a)
y
a
f x dx
(b)
y
b
f x dx
(c)
y
f x dx
7. Define the improper integral xab f x dx for each of the fol-
lowing cases. (a) f has an infinite discontinuity at a. (b) f has an infinite discontinuity at b. (c) f has an infinite discontinuity at c, where a c b. 8. State the Comparison Theorem for improper integrals.
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1.
x x 2 4 B A can be put in the form . x2 4 x2 x2
A x2 4 B C 2. can be put in the form . x x 2 4 x x2 x2 x 4 A B can be put in the form 2 . x 2x 4 x x4 2
3.
4.
A x2 4 B can be put in the form 2 . x x 2 4 x x 4
5.
y
4
6.
y
0
1
x dx 12 ln 15 x2 1 1 dx is convergent. x s2
t 7. If f is continuous, then x f x dx lim t l xt f x dx.
8. The Midpoint Rule is always more accurate than the Trape-
zoidal Rule.
CHAPTER 6
9. (a) Every elementary function has an elementary derivative.
(b) Every elementary function has an elementary antiderivative. 10. If f is continuous on 0, and x1 f x dx is convergent,
lim x l f x 0 , then x f x dx is convergent. 1
■
355
12. If xa f x dx and xa tx dx are both convergent, then
xa f x tx dx is convergent.
13. If xa f x dx and xa tx dx are both divergent, then
xa f x tx dx is divergent.
then x0 f x dx is convergent.
11. If f is a continuous, decreasing function on 1, and
REVIEW
14. If f x tx and x0 tx dx diverges, then x0 f x dx also
diverges.
EXERCISES 1– 40 1.
3.
y
5
0
y
Evaluate the integral.
■
x dx x 10
2
0
cos d 1 sin
2.
4.
29.
y
1
31.
y
ln 10
y
1 dy y 2 4y 12
33.
y 4 x
dx x 2s1 x 2
35.
y sx x
37.
y cos x sin x
39.
y
y
5
0
y
1
y tan x sec x dx
6.
7.
y
sinln t dt t
8.
y
9.
y
4
x 32 ln x dx
10.
y
1
11.
y
2
12.
y
1
13.
15.
17.
1
1
3
sx 2 1 dx x
yx
3
dx x
y sin 2
cos 5 d
y x sec x tan x dx
14.
16.
18.
dt 2t 1 3
4
5.
7
ye0.6y dy
sarctan x dx 1 x2
0
1
y
x 2 dx x2
y y
45.
y
4
y
dt sin 2t cos 2t x3 dx x 110
47.
y
3
49.
y
y csc
24.
ye
0
cos 3x sin 2x dx
■
41–50
23.
2
■
y
y
y
■
43.
y
27.
0
x 2 8x 3 dx x 3 3x 2
22.
y
dx
cos 2x dx
xe 2x dx 1 2x 2
12
sec d tan 2
dx sx 2 4x
25.
32
2
21.
3x 3 x 2 6x 4 dx x 2 1x 2 2
dx
2 32
1
y
y
4x dx
x2
41.
6
20.
4
0
e xse x 1 dx ex 8
■
■
■
dx
30.
y e s1 e
32.
y
34.
y arcsin x dx
36.
y 1 tan d
38.
y x tan
40.
y
■
x
2x
4
x sin x dx cos 3 x
0
2
1 tan
3 4
■
1
x2 dx
stan d sin 2 ■
■
■
■
2
x1 dx 9x 2 6x 5
19.
sin x dx 1 x2
x 5 sec x dx
1
x
cos x dx
■
dx 1 ex
26.
y
28.
y sx 1 dx
3 x 1 s 3
■
1 dx 2x 13
42.
y
1
dx x ln x
44.
y
6
ln x dx sx
46.
y
1
dx x2 x 2
48.
y
1
50.
y
1
2
0
0
Evaluate the integral or show that it is divergent.
dx 4x 4x 5 2
■
■
■
■
■
■
0
2
0
1
1
■
t2 1 dt t2 1 y dy sy 2 1 dx 2 3x x1 dx 3 x4 s tan1x dx x2 ■
■
■
51–54 ■ Use the Table of Integrals on the Reference Pages to evaluate the integral. 51.
ye
x
s1 e 2x dx
52.
y csc t dt 5
■
356
53.
■
CHAPTER 6
y sx
■
■
2
TECHNIQUES OF INTEGRATION
x 1 dx ■
■
54. ■
■
■
cot x
y s1 2 sin x ■
■
■
62. A population of honeybees increased at a rate of rt bees
dx ■
per week, where the graph of r is as shown. Use Simpson’s Rule with six subintervals to estimate the increase in the bee population during the first 24 weeks.
■
55. Is it possible to find a number n such that x x dx is 0
n
convergent?
r
56. For what values of a is x e 0
12000
ax
cos x dx convergent? Evaluate the integral for those values of a.
8000
Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule with n 10 to approximate the given integral. Round your answers to six decimal places. 57–58
57. ■
y
1
0 ■
■
s1 x 4 dx ■
■
58. ■
■
■
y
2
0
■
4000
ssin x dx ■
■
0 ■
4
8
12
16
20
24
t (weeks)
■
59. Estimate the errors involved in Exercise 57, parts (a) and
(b). How large should n be in each case to guarantee an error of less than 0.00001? 60. Use Simpson’s Rule with n 6 to estimate the area under
the curve y e xx from x 1 to x 4.
CAS
63. (a) If f x sinsin x, use a graph to find an upper bound
for f 4x . (b) Use Simpson’s Rule with n 10 to approximate x0 f x dx and use part (a) to estimate the error. (c) How large should n be to guarantee that the size of the error in using Sn is less than 0.00001?
61. The speedometer reading (v) on a car was observed at
1-minute intervals and recorded in the chart. Use Simpson’s Rule to estimate the distance traveled by the car. t (min)
v (mih)
t (min)
v (mih)
0 1 2 3 4 5
40 42 45 49 52 54
6 7 8 9 10
56 57 57 55 56
64. Use the Comparison Theorem to determine whether the
integral
y
1
x3 dx x 2 5
is convergent or divergent. 65. If f is continuous on 0, and lim x l f x 0, show
that
y
0
f x dx f 0
7
APPLICATIONS OF INTEGRATION In this chapter we explore some of the applications of the definite integral by using it to compute areas between curves, volumes of solids, lengths of curves, the work done by a varying force, the center of gravity of a plate, and the force on a dam.The common theme in most of these applications is the following general method, which is similar to the one we used to find areas under curves:We break up a quantity Q into a large number of small parts.We next approximate each small part by a quantity of the form f x*i x and thus approximate Q by a Riemann sum.Then we take the limit and express Q as an integral. Finally we evaluate the integral by using the Evaluation Theorem, or Simpson’s Rule, or technology. In the final section we look at what is perhaps the most important of all the applications of integration: differential equations.When a scientist uses calculus, more often than not it is to solve a differential equation that has arisen in the description of some physical process.
7.1
AREAS BETWEEN CURVES
y
y=ƒ
S 0
a
b
x
y=©
In Chapter 5 we defined and calculated areas of regions that lie under the graphs of functions. Here we use integrals to find areas of regions that lie between the graphs of two functions. Consider the region S that lies between two curves y f x and y tx and between the vertical lines x a and x b, where f and t are continuous functions and f x tx for all x in a, b . (See Figure 1.) Just as we did for areas under curves in Section 5.1, we divide S into n strips of equal width and then we approximate the ith strip by a rectangle with base x and height f x*i tx*i . (See Figure 2. If we like, we could take all of the sample points to be right endpoints, in which case x*i x i .) The Riemann sum
FIGURE 1
n
f x* tx* x
S=s(x, y) | a¯x¯b, ©¯y¯ƒd
i
i
i1
is therefore an approximation to what we intuitively think of as the area of S. y
y
f (x *i )
0
a
f (x *i )-g(x *i )
b
_g(x *i ) Îx
FIGURE 2
x
0
a
b
x
x *i
(a) Typical rectangle
(b) Approximating rectangles
357
358
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
This approximation appears to become better and better as n l . Therefore, we define the area A of S as the limiting value of the sum of the areas of these approximating rectangles.
n
f x* tx* x
A lim
1
n l i1
i
i
We recognize the limit in (1) as the definite integral of f t. Therefore, we have the following formula for area. 2 The area A of the region bounded by the curves y f x, y tx, and the lines x a, x b, where f and t are continuous and f x tx for all x in a, b , is
A y f x tx dx b
y
a
y=ƒ S y=© 0
a
b
x
FIGURE 3 b
b
A=j ƒ dx-j © dx a
Notice that in the special case where tx 0, S is the region under the graph of f and our general definition of area (1) reduces to our previous definition (Definition 5.1.2). In the case where both f and t are positive, you can see from Figure 3 why (2) is true: A area under y f x area under y tx
a
y f x dx y tx dx y f x tx dx b
b
a
b
a
a
y
EXAMPLE 1 Find the area of the region bounded above by y e x, bounded below
by y x, and bounded on the sides by x 0 and x 1.
y=´
x=1
SOLUTION The region is shown in Figure 4. The upper boundary curve is y e x
and the lower boundary curve is y x. So we use the area formula (2) with f x e x, tx x, a 0, and b 1:
1
y=x Îx 0
1
x
A y e x x dx e x 12 x 2] 0 e 12 1 e 1.5 1
1
0
■
FIGURE 4 y
yT yT-yB yB 0
a
FIGURE 5
In Figure 4 we drew a typical approximating rectangle with width x as a reminder of the procedure by which the area is defined in (1). In general, when we set up an integral for an area, it’s helpful to sketch the region to identify the top curve yT , the bottom curve yB , and a typical approximating rectangle as in Figure 5. Then the area of a typical rectangle is yT yB x and the equation n
A lim
Îx
y
n l i1
b
T
yB x y yT yB dx b
a
x
summarizes the procedure of adding (in a limiting sense) the areas of all the typical rectangles.
SECTION 7.1
AREAS BETWEEN CURVES
359
■
Notice that in Figure 5 the left-hand boundary reduces to a point, whereas in Figure 3 the right-hand boundary reduces to a point. In the next example both of the side boundaries reduce to a point, so the first step is to find a and b. Find the area of the region enclosed by the parabolas y x 2 and
V EXAMPLE 2
y 2x x . 2
SOLUTION We first find the points of intersection of the parabolas by solving their
equations simultaneously. This gives x 2 2x x 2, or 2x 2 2x 0. Therefore 2xx 1 0, so x 0 or 1. The points of intersection are 0, 0 and 1, 1. We see from Figure 6 that the top and bottom boundaries are
yT=2x-≈ y
yT 2x x 2
(1, 1)
yB x 2
and
The area of a typical rectangle is Îx
yB=≈
yT yB x 2x x 2 x 2 x 2x 2x 2 x x
(0, 0)
and the region lies between x 0 and x 1. So the total area is A y 2x 2x 2 dx 2 y x x 2 dx 1
FIGURE 6
1
0
0
x2 x3 2 2 3 √ (mi/ h)
30
B
20 10 2
4
0
1 3
■
side and move along the same road. What does the area between the curves represent? Use Simpson’s Rule to estimate it.
A
40
0
1 1 2 3
2
EXAMPLE 3 Figure 7 shows velocity curves for two cars, A and B, that start side by
60 50
1
6
8 10 12 14 16 t (seconds)
SOLUTION We know from Section 5.3 that the area under the velocity curve A represents the distance traveled by car A during the first 16 seconds. Similarly, the area under curve B is the distance traveled by car B during that time period. So the area between these curves, which is the difference of the areas under the curves, is the distance between the cars after 16 seconds. We read the velocities from the graph and convert them to feet per second 1 mih 5280 3600 fts.
FIGURE 7
t
0
2
4
6
8
10
12
14
16
vA
0
34
54
67
76
84
89
92
95
vB
0
21
34
44
51
56
60
63
65
vA v B
0
13
20
23
25
28
29
29
30
Using Simpson’s Rule with n 8 intervals, so that t 2, we estimate the distance between the cars after 16 seconds:
y
16
0
vA v B dt 23 0 413 220 423 225 428 229 429 30 367 ft
■
360
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
Some regions are best treated by regarding x as a function of y. If a region is bounded by curves with equations x f y, x ty, y c, and y d, where f and t are continuous and f y ty for c y d (see Figure 8), then its area is A y f y ty dy d
c
y
y
x=g(y) y=d
d
d
xR
xL Îy
Îy
x=f(y) c
xR -x L y=c
c
0
0
x
x
FIGURE 9
FIGURE 8
If we write x R for the right boundary and x L for the left boundary, then, as Figure 9 illustrates, we have A y x R x L dy d
c
Here a typical approximating rectangle has dimensions x R x L and y. y
(5, 4)
V EXAMPLE 4
4
y 2 2x 6.
1 x L=2 ¥-3
Find the area enclosed by the line y x 1 and the parabola
SOLUTION By solving the two equations we find that the points of intersection are
Îy
1, 2 and 5, 4. We solve the equation of the parabola for x and notice from Figure 10 that the left and right boundary curves are
xR=y+1 x
0
x L 12 y 2 3
_2
(_1, _2)
xR y 1
and
We must integrate between the appropriate y-values, y 2 and y 4. Thus FIGURE 10
A y x R x L dy 4
2
y
y
2
y= œ„„„„„ 2x+6
(5, 4)
A™ y=x-1 ⫺3
A¡
4
0
x
y [ y 1 ( 4
2
1 2
(12 y 2 y 4) dy
1 2
y3 3
y2 4y 2
4
2
64 8 16 ( 2 8) 18 1 6
]
y 2 3) dy
4 3
■
(_1, _2)
y=_ œ„„„„„ 2x+6 FIGURE 11
We could have found the area in Example 4 by integrating with respect to x instead of y, but the calculation is much more involved. It would have meant splitting the region in two and computing the areas labeled A1 and A2 in Figure 11. The method we used in Example 4 is much easier.
SECTION 7.1
7.1 1– 4 1.
y
y sin 2x and between x 0 and x 2. Notice that the region consists of two separate parts. Find the area of this region.
y
x+2 y=œ„„„„ (4, 4)
2 3 2 ; 22. Graph the curves y x x and y x 4x 3x on a
x=2
y=x y= x
common screen and observe that the region between them consists of two parts. Find the area of this region.
x
1 x+1
4.
y
23. Racing cars driven by Chris and Kelly are side by side at
the start of a race. The table shows the velocities of each car (in miles per hour) during the first ten seconds of the race. Use Simpson’s Rule to estimate how much farther Kelly travels than Chris does during the first ten seconds.
y
x=¥-4y x=¥-2
y=1
(_3, 3) x
x=e y
x
y=_1 x=2y-¥ ■
■
■
■
■
■
■
■
■
■
■
■
5–16 ■ Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region. 5. y x 1, 6. y sin x,
y9x ,
y e x, x 0, x 2
8. y 1 sx , 9. y 12 x 2,
y 1 13 x
y 4x x 2
11. x 2y 2,
xy1
12. 4x y 12, 2
14. y sin x, 15. y 1x,
16. y x , ■
■
■
y 14 x,
■
■
■
■
y x 4,
■
0 22 37 52 61 71
6 7 8 9 10
69 75 81 86 90
80 86 93 98 102
1
2
t (min)
25. The widths (in meters) of a kidney-shaped swimming pool ■
■
■
■
■
■
■
were measured at 2-meter intervals as indicated in the figure. Use Simpson’s Rule to estimate the area of the pool.
x0 6.2
y xex 2 ■
0 20 32 46 54 62
x0
y 2 x2
19. y x 2,
0 1 2 3 4 5
0
■ Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves.
18. y e x,
vK
B
x0
; 17–20
17. y x sinx 2 ,
vC
A
y x2 2 ■
t
2
y 2x, y x,
vK
24. Two cars, A and B, start side by side and accelerate from
xy
x4y
13. x 2y , 2
vC
√
y x2 6
10. y x 2,
t
rest. The figure shows the graphs of their velocity functions. (a) Which car is ahead after one minute? Explain. (b) What is the meaning of the area of the shaded region? (c) Which car is ahead after two minutes? Explain. (d) Estimate the time at which the cars are again side by side.
x 1, x 2
2
y x2
7. y x,
361
21. Sketch the region that lies between the curves y cos x and
2.
y=5x-≈
3.
■
EXERCISES
Find the area of the shaded region.
■
AREAS BETWEEN CURVES
20. y x cos x, ■
■
■
■
y x 10 ■
■
■
7.2
6.8
5.6 5.0 4.8
4.8
362
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
26. A cross-section of an airplane wing is shown. Measurements
of the height of the wing, in centimeters, at 20-centimeter intervals are 5.8, 20.3, 26.7, 29.0, 27.6, 27.3, 23.8, 20.5, 15.1, 8.7, and 2.8. Use Simpson’s Rule to estimate the area of the wing’s cross-section.
30. Sketch the region in the xy-plane defined by the inequalities
x 2y 2 0, 1 x y 0 and find its area. 31. Find the values of c such that the area of the region
bounded by the parabolas y x 2 c 2 and y c 2 x 2 is 576. 32. Find the area of the region bounded by the parabola y x 2,
the tangent line to this parabola at 1, 1, and the x-axis.
200 cm 27. If the birth rate of a population is bt 2200e 0.024t people
per year and the death rate is dt 1460e people per year, find the area between these curves for 0 t 10. What does this area represent? 0.018t
28. A water storage tank has the shape of a cylinder with diam-
eter 10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage of the total capacity is being used? 29. Find the area of the crescent-shaped region (called a lune)
bounded by arcs of circles with radii r and R (see the figure).
33. Find the number b such that the line y b divides the
region bounded by the curves y x 2 and y 4 into two regions with equal area.
34. (a) Find the number a such that the line x a bisects the
area under the curve y 1x 2, 1 x 4. (b) Find the number b such that the line y b bisects the area in part (a).
35. Find a positive continuous function f such that the area
under the graph of f from 0 to t is At t 3 for all t 0. 36. Suppose that 0 c 2. For what value of c is the
area of the region enclosed by the curves y cos x, y cosx c, and x 0 equal to the area of the region enclosed by the curves y cosx c, x , and y 0?
r R
37. For what values of m do the line y mx and the curve
y xx 2 1 enclose a region? Find the area of the region.
7.2
VOLUMES In trying to find the volume of a solid we face the same type of problem as in finding areas. We have an intuitive idea of what volume means, but we must make this idea precise by using calculus to give an exact definition of volume. We start with a simple type of solid called a cylinder (or, more precisely, a right cylinder). As illustrated in Figure 1(a), a cylinder is bounded by a plane region B1, called the base, and a congruent region B2 in a parallel plane. The cylinder consists of all points on line segments that are perpendicular to the base and join B1 to B2 . If the area of the base is A and the height of the cylinder (the distance from B1 to B2 ) is h, then the volume V of the cylinder is defined as V Ah In particular, if the base is a circle with radius r, then the cylinder is a circular cylinder with volume V r 2h [see Figure 1(b)], and if the base is a rectangle with length l and width w, then the cylinder is a rectangular box (also called a rectangular parallelepiped ) with volume V lwh [see Figure 1(c)]. For a solid S that isn’t a cylinder we first “cut” S into pieces and approximate each piece by a cylinder. We estimate the volume of S by adding the volumes of the cylin-
SECTION 7.2
VOLUMES
■
363
B™ h h
h w
r
B¡
l FIGURE 1
(a) Cylinder V=Ah
(b) Circular cylinder V=πr@h
(c) Rectangular box V=lwh
ders. We arrive at the exact volume of S through a limiting process in which the number of pieces becomes large. We start by intersecting S with a plane and obtaining a plane region that is called a cross-section of S. Let Ax be the area of the cross-section of S in a plane Px perpendicular to the x-axis and passing through the point x, where a x b. (See Figure 2. Think of slicing S with a knife through x and computing the area of this slice.) The cross-sectional area Ax will vary as x increases from a to b. y
Px
A A(b)
0
x
a
FIGURE 2
b
x
We consider a partition of the interval a, b into n subintervals with partition points x0 , x1, x 2, . . . , x n. We divide S into n “slabs” of width x i x i x i1 by using the planes Px1 , Px 2 , . . . to slice the solid. (Think of slicing a loaf of bread.) If we choose sample points x*i in x i1, x i , we can approximate the ith slab Si (the part of S that lies between the planes Px i1 and Px i ) by a cylinder with base area Ax*i and “height” x i . (See Figure 3.) y
y
Î xi
S
0
FIGURE 3
a
xi-1 x*i xi
b
x
0
a=x¸
⁄
¤
‹
x¢
x∞
xß
x¶=b
x
364
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
The volume of this cylinder is Ax*i x i , so an approximation to our intuitive conception of the volume of the ith slab Si is VSi Ax*i x i Adding the volumes of these slabs, we get an approximation to the total volume (that is, what we think of intuitively as the volume): n
V
Ax* x i
i
i1
This approximation appears to become better and better as the slices become thinner and thinner. So we define the volume as the limit of these sums as max x i l 0. But we recognize the limit of Riemann sums as a definite integral and so we have the following definition.
It can be proved that this definition is independent of how S is situated with respect to the x -axis. In other words, no matter how we slice S with parallel planes, we always get the same answer for V. ■
DEFINITION OF VOLUME Let S be a solid that lies between x a and x b. If the cross-sectional area of S in the plane Px , through x and perpendicular to the x-axis, is Ax, where A is an integrable function, then the volume of S is n
V
Ax* x
lim
max x i l 0 i1
i
y Ax dx b
i
a
When we use the volume formula V xab Ax dx it is important to remember that Ax is the area of a moving cross-section obtained by slicing through x perpendicular to the x-axis. Notice that, for a cylinder, the cross-sectional area is constant: Ax A for all x. So our definition of volume gives V xab A dx Ab a; this agrees with the formula V Ah. EXAMPLE 1 Show that the volume of a sphere of radius r is V 3 r 3. 4
SOLUTION If we place the sphere so that its center is at the origin (see Figure 4), then the plane Px intersects the sphere in a circle whose radius (from the Pythagorean Theorem) is y sr 2 x 2 . So the cross-sectional area is
y
Ax y 2 r 2 x 2 _r
r
x
Using the definition of volume with a r and b r, we have V y Ax dx y r 2 x 2 dx r
r
r
r
2 y r 2 x 2 dx r
0
FIGURE 4
x3 2 r x 3 43 r 3
2
r
2 r 3
0
(The integrand is even.)
r3 3
■
Figure 5 illustrates the definition of volume when the solid is a sphere with radius r 1. From the result of Example 1, we know that the volume of the sphere is 4 3 4.18879. Here the slabs are circular cylinders, or disks, and the three parts of
SECTION 7.2
VOLUMES
■
365
Figure 5 show the geometric interpretations of the Riemann sums n
n
Ax x 1
2
i
i1
Visual 7.2A shows an animation of Figure 5.
(a) Using 5 disks, VÅ4.2726
x i2 x
i1
when n 5, 10, and 20 if we use regular partitions and choose the sample points x*i to be the midpoints xi . Notice that as we increase the number of approximating cylinders, the corresponding Riemann sums become closer to the true volume.
(b) Using 10 disks, VÅ4.2097
(c) Using 20 disks, VÅ4.1940
FIGURE 5
Approximating the volume of a sphere with radius 1
V EXAMPLE 2 Find the volume of the solid obtained by rotating about the x-axis the region under the curve y sx from 0 to 1. Illustrate the definition of volume by sketching a typical approximating cylinder.
SOLUTION The region is shown in Figure 6(a). If we rotate about the x-axis, we get
the solid shown in Figure 6(b). When we slice through the point x, we get a disk with radius sx . The area of this cross-section is Did we get a reasonable answer in Example 2? As a check on our work, let’s replace the given region by a square with base 0, 1 and height 1. If we rotate this square, we get a cylinder with radius 1, height 1, and volume 12 1 . We computed that the given solid has half this volume. That seems about right.
Ax (sx ) 2 x
■
and the volume of the approximating cylinder (a disk with thickness x) is Ax x x x The solid lies between x 0 and x 1, so its volume is V y Ax dx y x dx 1
1
0
0
y
x2 2
1
0
2
y
y=œ„
œ„
Visual 7.2B shows how the solids of revolution in Examples 2–6 are formed.
0
1
x
x
0
1
x
Îx
FIGURE 6
(a)
(b)
■
366
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
V EXAMPLE 3 Find the volume of the solid obtained by rotating the region bounded by y x 3, y 8, and x 0 about the y-axis.
SOLUTION The region is shown in Figure 7(a) and the resulting solid is shown in Figure 7(b). Because the region is rotated about the y-axis, it makes sense to slice the solid perpendicular to the y-axis and therefore to integrate with respect to y. If we 3 slice at height y, we get a circular disk with radius x, where x s y. So the area of a cross-section through y is 3 Ay x 2 (s y ) 2 y 23
and the volume of the approximating cylinder pictured in Figure 7(b) is Ay y y 23 y Since the solid lies between y 0 and y 8, its volume is V y Ay dy y y 23 dy 8
8
0
0
[
3 5
y 53
y
]
8 0
96 5
y
y=8
8
(x, y)
Îy x=0 y=˛ or 3 x=œ„ y 0
0
x
(a)
FIGURE 7
x
(b)
■
EXAMPLE 4 The region enclosed by the curves y x and y x 2 is rotated
about the x-axis. Find the volume of the resulting solid. SOLUTION The curves y x and y x 2 intersect at the points 0, 0 and 1, 1. The
region between them, the solid of rotation, and a cross-section perpendicular to the x-axis are shown in Figure 8. A cross-section in the plane Px has the shape of a washer (an annular ring) with inner radius x 2 and outer radius x, so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle: Ax x 2 x 2 2 x 2 x 4 Therefore, we have V y Ax dx y x 2 x 4 dx 1
0
x3 x5 3 5
1
0
1
0
2 15
SECTION 7.2
y
VOLUMES
367
■
y (1, 1)
A(x)
y=x y=≈ ≈
x
(0, 0)
(a)
FIGURE 8
x
x
0
( b)
(c)
EXAMPLE 5 Find the volume of the solid obtained by rotating the region in
Example 4 about the line y 2. SOLUTION The solid and a cross-section are shown in Figure 9. Again the crosssection is a washer, but this time the inner radius is 2 x and the outer radius is 2 x 2. The cross-sectional area is
Ax 2 x 2 2 2 x2 and so the volume of S is V y Ax dx y 2 x 2 2 2 x2 dx 1
1
0
0
y x 4 5x 2 4x dx 1
0
x5 x3 x2 5 4 5 3 2
1
0
8 15
y 4
y=2
y=2
2-x 2-≈ y=≈
y=x 0
FIGURE 9
x
≈
x
1
x
x
x
■
The solids in Examples 1–5 are all called solids of revolution because they are obtained by revolving a region about a line. In general, we calculate the volume of a
368
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
solid of revolution by using the basic defining formula V y Ax dx b
V y Ay dy d
or
a
c
and we find the cross-sectional area Ax or Ay in one of the following ways: ■
If the cross-section is a disk (as in Examples 1–3), we find the radius of the disk (in terms of x or y) and use A radius2
■
If the cross-section is a washer (as in Examples 4 and 5), we find the inner radius r in and outer radius rout from a sketch (as in Figures 9 and 10) and compute the area of the washer by subtracting the area of the inner disk from the area of the outer disk: A outer radius2 inner radius2
rin rout
FIGURE 10
The next example gives a further illustration of the procedure. EXAMPLE 6 Find the volume of the solid obtained by rotating the region in
Example 4 about the line x 1. SOLUTION Figure 11 shows a horizontal cross-section. It is a washer with inner radius 1 y and outer radius 1 sy , so the cross-sectional area is
Ay outer radius 2 inner radius 2 (1 sy ) 2 1 y 2 The volume is V y Ay dy y 1
0
1
0
[(1 sy )
2
y (2sy y y 2 ) dy 1
0
4y 32 y2 y3 3 2 3
1
0
]
1 y 2 dy
2
SECTION 7.2
VOLUMES
■
369
y
1+œ„ y 1+y 1 x=œ„ y y x=y x
0
x=_1
FIGURE 11
Visual 7.2C shows how the solid in Figure 12 is generated.
■
We now find the volumes of two solids that are not solids of revolution.
EXAMPLE 7 Figure 12 shows a solid with a circular base of radius 1. Parallel cross-
sections perpendicular to the base are equilateral triangles. Find the volume of the solid. y
SOLUTION Let’s take the circle to be x 2 y 2 1. The solid, its base, and a typical
cross-section at a distance x from the origin are shown in Figure 13. x
y
y
FIGURE 12
≈ y=œ„„„„„„ „
C
B(x, y)
Computer-generated picture of the solid in Example 7
C
y B _1
0
0
x
x
œ 3y œ„
x
1
A
A
A FIGURE 13
(a) The solid
(b) Its base
60° y
60° y
B
(c) A cross-section
Since B lies on the circle, we have y s1 x 2 and so the base of the triangle ABC is AB 2s1 x 2 . Since the triangle is equilateral, we see from Figure 13(c) that its height is s3 y s3 s1 x 2 . The cross-sectional area is therefore
Ax 12 2s1 x 2 s3 s1 x 2 s3 1 x 2 and the volume of the solid is V y Ax dx y s3 1 x 2 dx 1
1
1
1
2 y s3 1 x 2 dx 2s3 x 1
0
x3 3
1
0
4s3 3
■
370
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
V EXAMPLE 8 Find the volume of a pyramid whose base is a square with side L and whose height is h.
SOLUTION We place the origin O at the vertex of the pyramid and the x-axis along
its central axis as in Figure 14. Any plane Px that passes through x and is perpendicular to the x-axis intersects the pyramid in a square with side of length s, say. We can express s in terms of x by observing from the similar triangles in Figure 15 that x s2 s h L2 L and so s Lxh. [Another method is to observe that the line OP has slope L2h and so its equation is y Lx2h.] Thus the cross-sectional area is Ax s 2
L2 2 x h2
y
y
P
x
h
s
O
O
x
L
x
x
h
FIGURE 14
FIGURE 15
The pyramid lies between x 0 and x h, so its volume is y
V y Ax dx y h
h
0
0
L2 2 L2 x 3 2 x dx h h2 3
h
0
L2 h 3
■
NOTE We didn’t need to place the vertex of the pyramid at the origin in Example 8. We did so merely to make the equations simple. If, instead, we had placed the center of the base at the origin and the vertex on the positive y-axis, as in Figure 16, you can verify that we would have obtained the integral
y
0
x
Vy
FIGURE 16
h
0
7.2
L2 L2 h 2 h y dy h2 3
EXERCISES
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. 1–12
h
■
1. y 1x, x 1, x 2, y 0 ;
about the x-axis
2. y 1 x 2, y 0 ;
about the x-axis
3. x 2sy , x 0, y 9;
about the y-axis
4. y ln x, y 1, y 2, x 0 ;
about the y-axis
SECTION 7.2
5. y x 3, y x, x 0; 6. y 4 x 2, y 5 x 2; 1
7. y 2 x, x 2y ;
about the y-axis
about y 1
10. y 1x, y 0, x 1, x 3; about y 1 11. y x 2, x y 2;
about x 2
■
■
■
■
24. A log 10 m long is cut at 1-meter intervals and its cross-
sectional areas A (at a distance x from the end of the log) are listed in the table. Use the Midpoint Rule with n 5 to estimate the volume of the log.
about x 1
12. y x, y sx ; ■
■
371
of a human organ that provide information about the organ otherwise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas, in square centimeters, are 0, 18, 58, 79, 94, 106, 117, 128, 63, 39, and 0. Use the Midpoint Rule to estimate the volume of the liver.
about the x-axis
about the y-axis
9. y x, y sx ;
■
23. A CAT scan produces equally spaced cross-sectional views
about the x-axis
8. y x 23, x 1, y 0;
VOLUMES
■
■
■
■
■
■
13. The region enclosed by the curves y x and y sx is 3
rotated about the line x 1. Find the volume of the resulting solid.
14. Find the volume of the solid obtained by rotating the region
in Exercise 13 about the line y 1. ■ Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
x (m)
A (m2 )
x (m)
A (m2 )
0 1 2 3 4 5
0.68 0.65 0.64 0.61 0.58 0.59
6 7 8 9 10
0.53 0.55 0.52 0.50 0.48
15–16
15. y tan 3 x, y 1, x 0;
about y 1
16. y x 2 4, 8x y 16; ■
■
; 17–18
■
■
■
■
25–37
■
Find the volume of the described solid S.
25. A right circular cone with height h and base radius r
about x 10 ■
■
■
26. A frustum of a right circular cone with height h, lower base ■
■
■
radius R, and top radius r r
Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the volume of the solid obtained by rotating about the x-axis the region bounded by these curves. ■
17. y x 2,
CAS
■
R
y lnx 1 y e x2 e2x
18. y 3 sinx 2 , ■
h
■
■
■
■
27. A cap of a sphere with radius r and height h
■
■
■
■
■
■
h
■ Use a computer algebra system to find the exact volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
19–20
19. y sin2 x, y 0, 0 x ; 20. y x, y xe
;
■
■
1x2
■
■
■
about y 1
about y 3 ■
r
■
28. A frustum of a pyramid with square base of side b, square ■
■
■
■
■
top of side a, and height h a
Each integral represents the volume of a solid. Describe the solid. 21–22
■
21. (a) y
2
0
(b) y y 4 y 8 dy 1
cos2x dx
0
22. (a) y y dy
(b) y
5
2
■
■
■
2
0
■
■
■
■
1 cos x2 12 dx ■
■
■
■
b ■
What happens if a b ? What happens if a 0 ?
372
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
29. A pyramid with height h and rectangular base with dimen-
(b) Show that the volume enclosed by the barrel is
sions b and 2b V 13 h (2R 2 r 2 25 d 2 )
30. A pyramid with height h and base an equilateral triangle
with side a (a tetrahedron)
CAS
40. (a) A model for the shape of a bird’s egg is obtained by
rotating about the x-axis the region under the graph of f x ax 3 bx 2 cx ds1 x 2 Use a CAS to find the volume of such an egg. (b) For a Red-throated Loon, a 0.06, b 0.04, c 0.1, and d 0.54. Graph f and find the volume of an egg of this bird.
a a
a
41. (a) Set up an integral for the volume of a solid torus (the
31. A tetrahedron with three mutually perpendicular faces and
three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5 cm 32. The base of S is a circular disk with radius r. Parallel cross-
donut-shaped solid shown in the figure) with radii r and R. (b) By interpreting the integral as an area, find the volume of the torus.
sections perpendicular to the base are squares. 33. The base of S is an elliptical region with boundary curve R
9x 2 4y 2 36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.
r
34. The base of S is the parabolic region x, y x 2 y 1 .
Cross-sections perpendicular to the y-axis are equilateral triangles.
42. A wedge is cut out of a circular cylinder of radius 4 by two
planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle of 30 along a diameter of the cylinder. Find the volume of the wedge.
35. S has the same base as in Exercise 34, but cross-sections
perpendicular to the y-axis are squares. 36. The base of S is the triangular region with vertices 0, 0,
3, 0, and 0, 2. Cross-sections perpendicular to the y-axis are semicircles.
37. S has the same base as in Exercise 36, but cross-sections
perpendicular to the y-axis are isosceles triangles with height equal to the base. ■
■
■
■
■
■
■
■
■
■
■
43. (a) Cavalieri’s Principle states that if a family of parallel
planes gives equal cross-sectional areas for two solids S1 and S2 , then the volumes of S1 and S2 are equal. Prove this principle. (b) Use Cavalieri’s Principle to find the volume of the oblique cylinder shown in the figure.
■
38. The base of S is a circular disk with radius r. Parallel cross-
sections perpendicular to the base are isosceles triangles with height h and unequal side in the base. (a) Set up an integral for the volume of S. (b) By interpreting the integral as an area, find the volume of S.
h r
44. Find the volume common to two circular cylinders, each 39. Some of the pioneers of calculus, such as Kepler and New-
ton, were inspired by the problem of finding the volumes of wine barrels. (In fact Kepler published a book Stereometria doliorum in 1715 devoted to methods for finding the volumes of barrels.) They often approximated the shape of the sides by parabolas. (a) A barrel with height h and maximum radius R is constructed by rotating about the x-axis the parabola y R cx 2, h2 x h2, where c is a positive constant. Show that the radius of each end of the barrel is r R d, where d ch 24.
with radius r, if the axes of the cylinders intersect at right angles.
SECTION 7.3
46. A bowl is shaped like a hemisphere with diameter 30 cm. A
48. A hole of radius r is bored through the center of a sphere of
radius R r. Find the volume of the remaining portion of the sphere.
ball with diameter 10 cm is placed in the bowl and water is poured into the bowl to a depth of h centimeters. Find the volume of water in the bowl.
VOLUMES BY CYLINDRICAL SHELLS
y
y=2≈-˛ 1
xR=?
0
2
x
Some volume problems are very difficult to handle by the methods of the preceding section. For instance, let’s consider the problem of finding the volume of the solid obtained by rotating about the y-axis the region bounded by y 2x 2 x 3 and y 0. (See Figure 1.) If we slice perpendicular to the y-axis, we get a washer. But to compute the inner radius and the outer radius of the washer, we would have to solve the cubic equation y 2x 2 x 3 for x in terms of y; that’s not easy. Fortunately, there is a method, called the method of cylindrical shells, that is easier to use in such a case. Figure 2 shows a cylindrical shell with inner radius r1, outer radius r2, and height h. Its volume V is calculated by subtracting the volume V1 of the inner cylinder from the volume V2 of the outer cylinder:
FIGURE 1
r
373
R r at right angles to the axis of the cylinder. Set up, but do not evaluate, an integral for the volume cut out.
if the center of each sphere lies on the surface of the other sphere.
xL=?
■
47. A hole of radius r is bored through a cylinder of radius
45. Find the volume common to two spheres, each with radius r,
7.3
VOLUMES BY CYLINDRICAL SHELLS
V V2 V1 r22 h r12 h r22 r12 h
Îr
r¡
r2 r1 r2 r1 h
r™
2 h
r2 r1 hr2 r1 2
If we let r r2 r1 (the thickness of the shell) and r 12 r2 r1 (the average radius of the shell), then this formula for the volume of a cylindrical shell becomes V 2 rh r
1
FIGURE 2
and it can be remembered as V [circumference][height][thickness] Now let S be the solid obtained by rotating about the y-axis the region bounded by y f x [where f is continuous and f x 0], y 0, x a, and x b, where b a 0. (See Figure 3.) y
y
y=ƒ
y=ƒ
0
FIGURE 3
a
b
x
0
a
b
x
374
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
We divide the interval a, b into n subintervals x i1, x i of equal width x and let xi be the midpoint of the ith subinterval. If the rectangle with base x i1, x i and height f xi is rotated about the y-axis, then the result is a cylindrical shell with average radius xi , height f xi , and thickness x (see Figure 4), so by Formula 1 its volume is
y
y=ƒ
0
a x i-1 x–i
Vi 2 xi f xi x
x
b
Therefore, an approximation to the volume V of S is given by the sum of the volumes of these shells:
xi
n
y
V
y=ƒ
n
V 2 x f x x i
i
i1
i
i1
This approximation appears to become better as n l . But, from the definition of an integral, we know that b
n
x
lim
2 x f x x y i
n l i1
i
b
a
2 xf x dx
Thus the following appears plausible; a proof is outlined in Exercise 43.
FIGURE 4
2 The volume of the solid in Figure 3, obtained by rotating about the y-axis the region under the curve y f x from a to b, is
V y 2 x f x dx b
a
where 0 a b
The best way to remember Formula 2 is to think of a typical shell, cut and flattened as in Figure 5, with radius x, circumference 2 x, height f x, and thickness x or dx :
y
b
a
2 x circumference
f x dx height
y
ƒ
ƒ x
x
2πx
Îx
FIGURE 5
This type of reasoning will be helpful in other situations, such as when we rotate about lines other than the y-axis. EXAMPLE 1 Find the volume of the solid obtained by rotating about the y-axis the
region bounded by y 2x 2 x 3 and y 0. SOLUTION From the sketch in Figure 6 we see that a typical shell has radius x, circumference 2 x, and height f x 2x 2 x 3. So, by the shell method, the
SECTION 7.3
y
VOLUMES BY CYLINDRICAL SHELLS
■
375
volume is V y 2 x2x 2 x 3 dx 2 y 2x 3 x 4 dx 2
2
0
0
2
2≈-˛
1 2
x 4 15 x 5
]
2 0
2 (8 325 ) 165 ■
It can be verified that the shell method gives the same answer as slicing.
2 x
x
[
y
FIGURE 6
Visual 7.3 shows how the solid and shells in Example 1 are formed. ■ Figure 7 shows a computer-generated picture of the solid whose volume we computed in Example 1.
x
FIGURE 7
NOTE Comparing the solution of Example 1 with the remarks at the beginning of this section, we see that the method of cylindrical shells is much easier than the washer method for this problem. We did not have to find the coordinates of the local maximum and we did not have to solve the equation of the curve for x in terms of y. However, in other examples the methods of the preceding section may be easier. V EXAMPLE 2 Find the volume of the solid obtained by rotating about the y-axis the region between y x and y x 2.
SOLUTION The region and a typical shell are shown in Figure 8. We see that the shell has radius x, circumference 2 x, and height x x 2. So the volume is
y
y=x y=≈
V y 2 xx x 2 dx 2 y x 2 x 3 dx 1
0
shell height=x-≈ 0
x
3
4
x x 3 4
0
1
0
6
■
As the following example shows, the shell method works just as well if we rotate about the x-axis. We simply have to draw a diagram to identify the radius and height of a shell.
FIGURE 8 y
2
x
1
shell height=1-¥
1
V EXAMPLE 3 Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve y sx from 0 to 1.
y
x=¥ ¥
0
x=1
shell radius=y 1
x
SOLUTION This problem was solved using disks in Example 2 in Section 7.2. To use shells we relabel the curve y sx (in the figure in that example) as x y 2 in Figure 9. For rotation about the x-axis we see that a typical shell has radius y, circumference 2 y, and height 1 y 2. So the volume is
V y 2 y1 y 2 dy 2 y y y 3 dy 2 1
FIGURE 9
0
1
0
In this problem the disk method was simpler.
y2 y4 2 4
1
0
2 ■
376
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
V EXAMPLE 4 Find the volume of the solid obtained by rotating the region bounded by y x x 2 and y 0 about the line x 2.
SOLUTION Figure 10 shows the region and a cylindrical shell formed by rotation about the line x 2. It has radius 2 x, circumference 2 2 x, and height x x 2. y
y
x=2
y=x-≈
0
0
x
1
1
x
FIGURE 10
2
3
4
x
2-x
The volume of the given solid is V y 2 2 xx x 2 dx 2 y x 3 3x 2 2x dx 1
0
2
7.3
1
0
1
x4 x3 x2 4
0
4. y x 2,
the figure about the y-axis. Explain why it is awkward to use slicing to find the volume V of S. Sketch a typical approximating shell. What are its circumference and height? Use shells to find V. y
y 0,
x 2
5. y e
y 0, x 0,
,
6. y 3 2x x , 7. y 4x 2 , ■
■
x1
xy3
y x 2 4x 7
2
■
■
■
■
■
■
■
■
■
8. Let V be the volume of the solid obtained by rotating about
y=x(x-1)@
0
1
the y-axis the region bounded by y sx and y x 2. Find V both by slicing and by cylindrical shells. In both cases draw a diagram to explain your method.
x
2. Let S be the solid obtained by rotating the region shown
in the figure about the y-axis. Sketch a typical cylindrical shell and find its circumference and height. Use shells to find the volume of S. Do you think this method is preferable to slicing? Explain.
■ Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. Sketch the region and a typical shell.
9–14
9. x 1 y 2, 10. x sy ,
y
11. y x 3, y=sin{≈}
x 0,
x 0, y 8,
12. x 4y 2 y 3, 13. y 4x 2,
0
π œ„
x
■ Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis. Sketch the region and a typical shell.
3–7
x 1,
x1 2
■
y 0,
■
EXERCISES
1. Let S be the solid obtained by rotating the region shown in
3. y 1x,
2
x2
14. x y 3, ■
■
■
y 1,
y2
y1 x0
x0
2x y 6 x 4 y 12 ■
■
■
■
■
■
■
■
■
15–20 ■ Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. Sketch the region and a typical shell. 15. y x 2, y 0, x 1, x 2;
about x 1
SECTION 7.3
16. y x 2, y 0, x 2, x 1;
about x 4
18. y 4x x , y 8x 2x ;
about x 2
2
2
19. y sx 1, y 0, x 5; 20. y x , x y ; 2
■
■
35. y 5, y x 4x;
about y 3
about y 1
2
■
34. y x 2 3x 2, y 0;
about the y-axis
17. y x , y 0, x 1, x 2; 2
■
■
■
■
■
■
■
■
■
37. x 2 y 12 1;
about the y-axis
38. x 2 y 12 1;
about the x-axis
Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
■
■
■
■
■
■
■
■
■
■
21. y ln x, y 0, x 2; 22. y x, y 4x x 2;
■
■
■
41. A right circular cone with height h and base radius r ■
about x 2
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
27. Use the Midpoint Rule with n 4 to estimate the volume
obtained by rotating about the y-axis the region under the curve y tan x, 0 x 4 . 28. (a) If the region shown in the figure is rotated about the
ferent diameters through two wooden balls (which also have different diameters). You discover that both napkin rings have the same height h, as shown in the figure. (a) Guess which ring has more wood in it. (b) Check your guess: Use cylindrical shells to compute the volume of a napkin ring created by drilling a hole with radius r through the center of a sphere of radius R and express the answer in terms of h.
y-axis to form a solid, use Simpson’s Rule with n 8 to estimate the volume of the solid. (b) Estimate the volume if the region is rotated about the x-axis.
h
y 4 2
43. Use the following steps to prove Formula 2 for the case 0
2
4
6
where f is one-to-one and therefore has an inverse function f 1 : Use the figure to show that
10 x
8
V b 2d a 2c y f 1 y 2 dy
29–32 ■ Each integral represents the volume of a solid. Describe the solid. 29.
y
3
31.
y
1
y
4
32. ■
0
0
30. 2 y
2 x 5 dx
2
0
d
c
y dy 1 y2
Make the substitution y f x and then use integration by parts on the resulting integral to prove that
2 3 y1 y 2 dy
V y 2 x f x dx b
a
0 ■
2 xcos x sin x dx ■
■
■
■
■
y ■
■
■
■
■ The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
■
x=f –! (y)
y=ƒ
d
33–38
33. y x 2 x 2, y 0;
about the x-axis
■
42. Suppose you make napkin rings by drilling holes with dif-
about y 4
about y 5
■
Use cylindrical shells to find the volume of the solid.
40. The solid torus of Exercise 41 in Section 7.2
about x 1
25. x ssin y , 0 y , x 0;
■
39. A sphere of radius r
about x 7
24. y 11 x 2 , y 0, x 0, x 2;
26. x 2 y 2 7, x 4;
39– 41
about the y-axis
23. y x 4, y sin x2;
■
■
377
about x 1
about x 2
■
■
about the y-axis
36. x 1 y 4, x 0;
■
21–26
VOLUMES BY CYLINDRICAL SHELLS
x=b
c
x=a 0
a
b
x
378
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
7.4
ARC LENGTH What do we mean by the length of a curve? We might think of fitting a piece of string to the curve in Figure 1 and then measuring the string against a ruler. But that might be difficult to do with much accuracy if we have a complicated curve. We need a precise definition for the length of an arc of a curve, in the same spirit as the definitions we developed for the concepts of area and volume. If the curve is a polygon, we can easily find its length; we just add the lengths of the line segments that form the polygon. (We can use the distance formula to find the distance between the endpoints of each segment.) We are going to define the length of a general curve by first approximating it by a polygon and then taking a limit as the number of segments of the polygon is increased. This process is familiar for the case of a circle, where the circumference is the limit of lengths of inscribed polygons (see Figure 2).
FIGURE 1
Visual 7.4 shows an animation of Figure 2.
FIGURE 2
y
P™ P¡
y=ƒ Pi-1 Pi
Pn
P¸
0
a
x¡ ¤
x i-1 x i
b x
Now suppose that a curve C is defined by the equation y f x, where f is continuous and a x b. We obtain a polygonal approximation to C by dividing the interval a, b into n subintervals with endpoints x 0 , x 1, . . . , x n and equal width x. If yi f x i , then the point Pi x i , yi lies on C and the polygon with vertices P0 , P1 , . . . , Pn , illustrated in Figure 3, is an approximation to C. The length L of C is approximately the length of this polygon and the approximation gets better as we let n increase. (See Figure 4, where the arc of the curve between Pi1 and Pi has been magnified and approximations with successively smaller values of x are shown.) Therefore, we define the length L of the curve C with equation y f x, a x b, as the limit of the lengths of these inscribed polygons (if the limit exists):
FIGURE 3
Pi
n
L lim
1
P
n l i1
i1
Pi
Pi Pi-1
Pi-1
Pi-1
Pi-1 FIGURE 4
Pi
Pi
Notice that the procedure for defining arc length is very similar to the procedure we used for defining area and volume: We divided the curve into a large number of small parts. We then found the approximate lengths of the small parts and added them. Finally, we took the limit as n l . The definition of arc length given by Equation 1 is not very convenient for computational purposes, but we can derive an integral formula for L in the case where f has a continuous derivative. [Such a function f is called smooth because a small change in x produces a small change in f x.] If we let yi yi yi1 , then
P
i1
Pi sx i x i1 2 yi yi1 2 sx2 yi 2
By applying the Mean Value Theorem to f on the interval x i1, x i , we find that there
SECTION 7.4
ARC LENGTH
■
379
is a number xi* between x i1 and x i such that f x i f x i1 f xi*x i x i1 yi f xi* x
that is, Thus we have
P
i1
Pi sx2 yi 2 sx2 f xi* x 2 s1 [ f xi* 2 sx2 s1 f xi* 2 x
(since x 0 )
Therefore, by Definition 1, n
L lim
n l i1
n
Pi1 Pi lim
s1 f x* i
n l i1
2
x
We recognize this expression as being equal to
y
b
a
s1 f x 2 dx
by the definition of a definite integral. This integral exists because the function tx s1 f x 2 is continuous. Thus we have proved the following theorem: 2 THE ARC LENGTH FORMULA If f is continuous on a, b , then the length of the curve y f x, a x b, is
L y s1 f x 2 dx b
a
If we use Leibniz notation for derivatives, we can write the arc length formula as follows: L
3
y
b
a
1
dy dx
2
dx
EXAMPLE 1 Find the length of the arc of the semicubical parabola y 2 x 3 between
y
the points 1, 1 and 4, 8. (See Figure 5.)
(4, 8)
SOLUTION For the top half of the curve we have ¥=˛
(1, 1) 0
FIGURE 5
dy 32 x 12 dx
y x 32
x
and so the arc length formula gives L
y
4
1
1
dy dx
2
dx y s1 94 x dx 4
1
380
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
As a check on our answer to Example 1, notice from Figure 5 that it ought to be slightly larger than the distance from 1, 1 to 4, 8, which is ■
If we substitute u 1 94 x , then du 94 dx . When x 1, u 134 ; when x 4, u 10. Therefore L 49 y
s58 7.615773 According to our calculation in Example 1, we have L
1 27
(80 s10
13 s13 ) 7.633705
Sure enough, this is a bit greater than the length of the line segment.
10
]
4 2 su du 9 3 u 32
134
[
27 10 32 ( 4 ) 8
13 32
10 134
] (80s10 13s13 ) 1 27
■
If a curve has the equation x t y, c y d, and t y is continuous, then by interchanging the roles of x and y in Formula 2 or Equation 3, we obtain the following formula for its length: L y s1 t y 2 dy d
4
c
V EXAMPLE 2
y
d
1
c
dx dy
2
dy
Find the length of the arc of the parabola y 2 x from 0, 0 to 1, 1.
SOLUTION Since x y 2, we have dxdy 2y, and Formula 4 gives
L
y 1
dx dy
1
0
2
dy y s1 4y 2 dy 1
0
We make the trigonometric substitution y 12 tan , which gives dy 12 sec 2 d and s1 4y 2 s1 tan 2 sec . When y 0, tan 0, so 0; when y 1, tan 2, so tan1 2 , say. Thus
L y sec 12 sec 2 d 12 y sec 3 d 0
0
[
12 12 sec tan ln sec tan
1 4
(sec
tan ln sec tan
)
]
0
(from Example 8 in Section 6.2)
(We could have used Formula 21 in the Table of Integrals.) Since tan 2, we have sec 2 1 tan 2 5, so sec s5 and L Figure 6 shows the arc of the parabola whose length is computed in Example 2, together with polygonal approximations having n 1 and n 2 line segments, respectively. For n 1 the approximate length is L 1 s2 , the diagonal of a square. The table shows the approximations L n that we get by dividing 0, 1 into n equal subintervals. Notice that each time we double the number of sides of the polygon, we get closer to the exact length, which is ■
L
ln(s5 2) s5 1.478943 2 4
ln(s5 2) s5 2 4
■
y 1
x=¥
0
FIGURE 6
1
x
n
Ln
1 2 4 8 16 32 64
1.414 1.445 1.464 1.472 1.476 1.478 1.479
SECTION 7.4
ARC LENGTH
■
381
Because of the presence of the square root sign in Formulas 2 and 4, the calculation of an arc length often leads to an integral that is very difficult or even impossible to evaluate explicitly. Thus we sometimes have to be content with finding an approximation to the length of a curve as in the following example. V EXAMPLE 3
(a) Set up an integral for the length of the arc of the hyperbola xy 1 from the point 1, 1 to the point (2, 12 ). (b) Use Simpson’s Rule with n 10 to estimate the arc length. SOLUTION
(a) We have y
1 x
dy 1 2 dx x
and so the arc length is
L
y
2
1
1
dy dx
2
dx
y
2
1
1
1 dx x4
y
2
1
sx 4 1 dx x2
(b) Using Simpson’s Rule (see Section 6.5) with a 1, b 2, n 10, x 0.1, and f x s1 1x 4 , we have L Checking the value of the definite integral with a more accurate approximation produced by a computer algebra system, we see that the approximation using Simpson’s Rule is accurate to four decimal places.
y
2
1
■
1
1 dx x4
x f 1 4 f 1.1 2 f 1.2 4 f 1.3 2 f 1.8 4 f 1.9 f 2 3
1.1321
■
THE ARC LENGTH FUNCTION
We will find it useful to have a function that measures the arc length of a curve from a particular starting point to any other point on the curve. Thus if a smooth curve C has the equation y f x, a x b, let sx be the distance along C from the initial point P0a, f a to the point Qx, f x. Then s is a function, called the arc length function, and, by Formula 2, sx y s1 f t 2 dt x
5
a
(We have replaced the variable of integration by t so that x does not have two meanings.) We can use Part 1 of the Fundamental Theorem of Calculus to differentiate Equation 5 (since the integrand is continuous): 6
ds s1 f x 2 dx
1
dy dx
2
382
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
Equation 6 shows that the rate of change of s with respect to x is always at least 1 and is equal to 1 when f x, the slope of the curve, is 0. The differential of arc length is
ds
7
dy dx
1
2
dx
and this equation is sometimes written in the symmetric form ds2 dx2 dy2
8 y
ds
The geometric interpretation of Equation 8 is shown in Figure 7. It can be used as a mnemonic device for remembering both of the Formulas 3 and 4. If we write L x ds, then from Equation 8 either we can solve to get (7), which gives (3), or we can solve to get
dy Îs
Îy
dx
0
FIGURE 7
ds
x
dx dy
1
2
dy
which gives (4). Find the arc length function for the curve y x 2 18 ln x taking P01, 1 as the starting point. V EXAMPLE 4
SOLUTION If f x x 2 8 ln x, then 1
1 8x
f x 2x
1 f x 2 1 2x 4x 2
1 8x
2
1 4x 2
1 1 1 2x 2 64x 2 8x
1 1 2 64x 2
2
1 8x
s1 f x 2 2x
Thus the arc length function is given by sx y s1 f t 2 dt x
1
y
x
1
2t
1 8t
]
dt t 2 18 ln t
x
1
x 2 18 ln x 1 For instance, the arc length along the curve from 1, 1 to 3, f 3 is s3 32 18 ln 3 1 8
ln 3 8.1373 8
■
SECTION 7.4
ARC LENGTH
■
383
y
y
1
s(x) ■ Figure 8 shows the interpretation of the arc length function in Example 4. Figure 9 shows the graph of this arc length function. Why is sx negative when x is less than 1?
1
P¸
0
0
1
s(x)=≈+18 ln x-1
x
x
FIGURE 8
7.4
FIGURE 9
EXERCISES 19–22 ■ Use Simpson’s Rule with n 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produced by your calculator.
1. Use the arc length formula (3) to find the length of the
curve y 2 3x, 2 x 1. Check your answer by noting that the curve is a line segment and calculating its length by the distance formula.
19. y xex,
2. Use the arc length formula to find the length of the curve
■
Find the length of the curve.
3. y 1 6x 32, 0 x 1 4. y 2 4x 43,
0 x 2,
5. y
x5 1 , 6 10x 3
6. y
x2 ln x , 2 4
y0
CAS
2 x 4
CAS
8. y lncos x, 0 x 3 9. y lnsec x, 11. y cosh x, 12. y 2 4x, 13. y e x,
0 x 4
1 x s3
10. y ln x,
0 y 2
ex 1 , ex 1
■
■
■
■
■
■
■
■
■
■
■
0 x 2
16. y 2 ,
■
■
■
1 y 4 ■
■
■
■
■
■
■
■
■
■
■
23. Use either a computer algebra system or a table of
24. Use either a computer algebra system or a table of integrals
■
18. ■
1 3 ; 28. (a) Graph the curve y 3 x 14x, x 0.
2
x y 2 1 a2 b ■
■
■
■
(b) Use Formulas 3 and 4 to set up two integrals for the arc length from 0, 0 to 1, 1. Observe that one of these is an improper integral and evaluate both of them. (c) Find the length of the arc of this curve from 1, 1 to 8, 4. starting point P0 1, 2.
0 x 3
x
2
17. x y y 3,
■
■
27. Find the arc length function for the curve y 2x 32 with
15–18 ■ Set up, but do not evaluate, an integral for the length of the curve. 15. y cos x,
1 x 3
■
26. (a) Sketch the curve y 3 x 2.
a x b, a 0 ■
22. y x ln x,
symmetry to find its length.
0 x 1
0 x 3
25. Sketch the curve with equation x 23 y 23 1 and use
0 x 1
14. y ln
21. y sec x,
to find the exact length of the arc of the curve y x 43 that lies between the points 0, 0 and 1, 1. If your CAS has trouble evaluating the integral, make a substitution that changes the integral into one that the CAS can evaluate.
1 y 9
1
1 y 2
integrals to find the exact length of the arc of the curve x ln1 y 2 that lies between the points 0, 0 and (ln 34 , 12 ).
1 x 2
7. x 3 sy y 3,
0 x 5
20. x y sy ,
y s4 x 2, 0 x 2. Check your answer by noting that the curve is a quarter-circle. 3–14
x
1
y=≈- 18 ln x
■
(b) Find the arc length function for this curve with starting point P0 (1, 127 ). (c) Graph the arc length function.
384
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
processing flat sheets of metal as shown in the figure. The profile of the roofing takes the shape of a sine wave. Verify that the sine curve has equation y sin x 7 and find the width w of a flat metal sheet that is needed to make a 28-inch panel. (Use your calculator to evaluate the integral correct to four significant digits.)
29. A hawk flying at 15 ms at an altitude of 180 m accidentally
drops its prey. The parabolic trajectory of the falling prey is described by the equation y 180
x2 45
until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter.
2 in
w
28 in
30. A steady wind blows a kite due west. The kite’s height
above ground from horizontal position x 0 to x 80 ft is given by y 150 401 x 502. Find the distance traveled by the kite.
n n ; 32. The curves with equations x y 1, n 4, 6, 8, . . . ,
are called fat circles. Graph the curves with n 2, 4, 6, 8, and 10 to see why. Set up an integral for the length L 2k of the fat circle with n 2k. Without attempting to evaluate this integral, state the value of lim k l L 2k .
31. A manufacturer of corrugated metal roofing wants to
produce panels that are 28 in. wide and 2 in. thick by
7.5
APPLICATIONS TO PHYSICS AND ENGINEERING
As a consequence of a calculation of work, you will be able to compute the velocity needed for a rocket to escape the Earth’s gravitational field. (See Exercise 22.) ■
Among the many applications of integral calculus to physics and engineering, we consider three: work, force due to water pressure, and centers of mass. As with our previous applications to geometry (areas, volumes, and lengths), our strategy is to break up the physical quantity into a large number of small parts, approximate each small part, add the results, take the limit, and evaluate the resulting integral. WORK
The term work is used in everyday language to mean the total amount of effort required to perform a task. In physics it has a technical meaning that depends on the idea of a force. Intuitively, you can think of a force as describing a push or pull on an object—for example, a horizontal push of a book across a table or the downward pull of the Earth’s gravity on a ball. In general, if an object moves along a straight line with position function st, then the force F on the object (in the same direction) is defined by Newton’s Second Law of Motion as the product of its mass m and its acceleration: Fm
1
d 2s dt 2
In the SI metric system, the mass is measured in kilograms (kg), the displacement in meters (m), the time in seconds (s), and the force in newtons ( N kg ms2 ). Thus a force of 1 N acting on a mass of 1 kg produces an acceleration of 1 ms2. In the US Customary system the fundamental unit is chosen to be the unit of force, which is the pound. In the case of constant acceleration, the force F is also constant and the work done is defined to be the product of the force F and the distance d that the object moves: 2
W Fd
work force distance
If F is measured in newtons and d in meters, then the unit for W is a newton-meter,
SECTION 7.5
APPLICATIONS TO PHYSICS AND ENGINEERING
■
385
which is called a joule (J). If F is measured in pounds and d in feet, then the unit for W is a foot-pound (ft-lb), which is about 1.36 J. For instance, suppose you lift a 1.2-kg book off the floor to put it on a desk that is 0.7 m high. The force you exert is equal and opposite to that exerted by gravity, so Equation 1 gives F mt 1.29.8 11.76 N and then Equation 2 gives the work done as W Fd 11.760.7 8.2 J But if a 20-lb weight is lifted 6 ft off the ground, then the force is given as F 20 lb, so the work done is W Fd 20 6 120 ft-lb Here we didn’t multiply by t because we were given the weight (a force) and not the mass. Equation 2 defines work as long as the force is constant, but what happens if the force is variable? Let’s suppose that the object moves along the x-axis in the positive direction, from x a to x b, and at each point x between a and b a force f x acts on the object, where f is a continuous function. We divide the interval a, b into n subintervals with endpoints x 0 , x 1, . . . , x n and equal width x. We choose a sample point x*i in the i th subinterval x i1, x i . Then the force at that point is f x*i . If n is large, then x is small, and since f is continuous, the values of f don’t change very much over the interval x i1, x i . In other words, f is almost constant on the interval and so the work Wi that is done in moving the particle from x i1 to x i is approximately given by Equation 2: Wi f x*i x Thus we can approximate the total work by n
W
3
f x* x i
i1
It seems that this approximation becomes better as we make n larger. Therefore, we define the work done in moving the object from a to b as the limit of this quantity as n l . Since the right side of (3) is a Riemann sum, we recognize its limit as being a definite integral and so n
W lim
4
f x* x y
n l i1
i
b
a
f x dx
EXAMPLE 1 When a particle is located a distance x feet from the origin, a force
of x 2 2x pounds acts on it. How much work is done in moving it from x 1 to x 3? SOLUTION
W y x 2 2x dx 3
1
2 3
The work done is 16 ft-lb.
x3 x2 3
3
1
50 3 ■
386
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
In the next example we use a law from physics: Hooke’s Law states that the force required to maintain a spring stretched x units beyond its natural length is proportional to x : f x kx frictionless surface
x
0
where k is a positive constant (called the spring constant). Hooke’s Law holds provided that x is not too large (see Figure 1).
(a) Natural position of spring ƒ=kx
0
x
x
(b) Stretched position of spring
V EXAMPLE 2 A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How much work is done in stretching the spring from 15 cm to 18 cm?
SOLUTION According to Hooke’s Law, the force required to hold the spring stretched x meters beyond its natural length is f x kx. When the spring is stretched from 10 cm to 15 cm, the amount stretched is 5 cm 0.05 m. This means that f 0.05 40, so 40 0.05k 40 k 0.05 800
Thus f x 800x and the work done in stretching the spring from 15 cm to 18 cm is
FIGURE 1
Hooke’s Law
Wy
0.08
0.05
800x dx 800
x2 2
0.08
0.05
400 0.08 0.05 1.56 J 2
■
2
V EXAMPLE 3 A 200-lb cable is 100 ft long and hangs vertically from the top of a tall building. How much work is required to lift the cable to the top of the building?
0
x*i
Îx
100
SOLUTION Here we don’t have a formula for the force function, but we can use an argument similar to the one that led to Definition 4. Let’s place the origin at the top of the building and the x -axis pointing downward as in Figure 2. We divide the cable into small parts with length x . If x*i is a point in the ith such interval, then all points in the interval are lifted by approximately the same amount, namely x*i . The cable weighs 2 pounds per foot, so the weight of the ith part is 2x . Thus the work done on the ith part, in foot-pounds, is
x
2x x*i
FIGURE 2
force
■ If we had placed the origin at the bottom of the cable and the x-axis upward, we would have gotten
100
0
2100 x dx
distance
We get the total work done by adding all these approximations and letting the number of parts become large (so x l 0 ): n
Wy
2x*i x
W lim
2x* x y
n l i1
i
100
0
]
2x dx x 2
100 0
10,000 ft-lb
■
which gives the same answer.
EXAMPLE 4 A tank has the shape of an inverted circular cone with height 10 m and
base radius 4 m. It is filled with water to a height of 8 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kgm3.) SOLUTION Let’s measure depths from the top of the tank by introducing a vertical coordinate line as in Figure 3. The water extends from a depth of 2 m to a depth of 10 m and so we divide the interval 2, 10 into n subintervals with endpoints x 0 , x 1, . . . , x n and choose x*i in the i th subinterval. This divides the water into n
SECTION 7.5
■
APPLICATIONS TO PHYSICS AND ENGINEERING
387
layers. The ith layer is approximated by a circular cylinder with radius ri and height x. We can compute ri from similar triangles, using Figure 4, as follows:
4m 0
2m
ri 4 10 x*i 10
x i* 10 m
Îx
ri 25 10 x*i
Thus an approximation to the volume of the ith layer of water is
ri
Vi ri2 x
4 10 x*i 2 x 25
and so its mass is
x
mi density volume
FIGURE 3
1000
4
4 10 x*i 2 x 160 10 x*i 2 x 25
The force required to raise this layer must overcome the force of gravity and so ri
10 10-xi*
FIGURE 4
Fi mi t 9.8160 10 x*i 2 x 1568 10 x*i 2 x Each particle in the layer must travel a distance of approximately x*i . The work Wi done to raise this layer to the top is approximately the product of the force Fi and the distance x*i : Wi Fi x*i 1568 x*i 10 x*i 2 x To find the total work done in emptying the entire tank, we add the contributions of each of the n layers and then take the limit as n l : n
W lim
1568 x*10 x* i
n l i1
2
i
x y 1568 x10 x2 dx 10
2
1568 y 100x 20x 2 x 3 dx 1568 50x 2 10
2
1568 (
2048 3
) 3.4 10
6
20x 3 x4 3 4
10
2
■
J
HYDROSTATIC PRESSURE AND FORCE
surface of fluid
Deep-sea divers realize that water pressure increases as they dive deeper. This is because the weight of the water above them increases. In general, suppose that a thin horizontal plate with area A square meters is submerged in a fluid of density kilograms per cubic meter at a depth d meters below the surface of the fluid as in Figure 5. The fluid directly above the plate has volume V Ad, so its mass is m V Ad. The force exerted by the fluid on the plate is therefore F mt tAd
FIGURE 5
where t is the acceleration due to gravity. The pressure P on the plate is defined to be the force per unit area: P
F td A
388
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
■ When using US Customary units, we write P td d, where t is the weight density (as opposed to , which is the mass density). For instance, the weight density of water is 62.5 lbft 3.
The SI unit for measuring pressure is newtons per square meter, which is called a pascal (abbreviation: 1 Nm2 1 Pa). Since this is a small unit, the kilopascal (kPa) is often used. For instance, because the density of water is 1000 kgm3, the pressure at the bottom of a swimming pool 2 m deep is P td 1000 kgm 3 9.8 ms 2 2 m 19,600 Pa 19.6 kPa An important principle of fluid pressure is the experimentally verified fact that at any point in a liquid the pressure is the same in all directions. (A diver feels the same pressure on nose and both ears.) Thus the pressure in any direction at a depth d in a fluid with mass density is given by P td d
5
This helps us determine the hydrostatic force against a vertical plate or wall or dam in a fluid. This is not a straightforward problem because the pressure is not constant but increases as the depth increases. 50 m
20 m
V EXAMPLE 5 A dam has the shape of the trapezoid shown in Figure 6. The height is 20 m and the width is 50 m at the top and 30 m at the bottom. Find the force on the dam due to hydrostatic pressure if the water level is 4 m from the top of the dam.
SOLUTION We choose a vertical x-axis with origin at the surface of the water as in
30 m
Figure 7(a). The depth of the water is 16 m, so we divide the interval 0, 16 into subintervals with endpoints x i and we choose xi* x i1, x i . The ith horizontal strip of the dam is approximated by a rectangle with height x and width wi , where, from similar triangles in Figure 7(b),
FIGURE 6
_4 0
15
10
a 10 * 16 xi 20
Îx
and so 15 x
(a)
a
16 xi* xi* 8 2 2
1 wi 215 a 2(15 8 2 xi*) 46 xi*
If Ai is the area of the ith strip, then
If x is small, then the pressure Pi on the ith strip is almost constant and we can use Equation 5 to write Pi 1000txi*
16-x i* (b) FIGURE 7
a
Ai wi x 46 xi* x 10
20
or
The hydrostatic force Fi acting on the ith strip is the product of the pressure and the area: Fi Pi Ai 1000txi*46 xi* x Adding these forces and taking the limit as n l , we obtain the total hydrostatic
SECTION 7.5
APPLICATIONS TO PHYSICS AND ENGINEERING
■
389
force on the dam: n
F lim
1000tx*46 x* x y i
n l i1
10009.8 y
16
0
i
16
1000tx46 x dx
0
x3 46x x dx 9800 23x 3 2
2
16
0
4.43 10 7 N
■
EXAMPLE 6 Find the hydrostatic force on one end of a cylindrical drum with radius
3 ft if the drum is submerged in water 10 ft deep. SOLUTION In this example it is convenient to choose the axes as in Figure 8 so that the origin is placed at the center of the drum. Then the circle has a simple equation, x 2 y 2 9. As in Example 5 we divide the circular region into horizontal strips of equal width. From the equation of the circle, we see that the length of the ith strip is 2s9 yi*2 and so its area is
y
di
œ„„„„„„„ œ 9-(yi*)@
7 10
Îy
Ai 2s9 yi*2 y
y i* 0
x
The pressure on this strip is approximately ≈+¥=9
di 62.57 yi* FIGURE 8
and so the force on the strip is approximately
di Ai 62.57 yi*2s9 yi*2 y The total force is obtained by adding the forces on all the strips and taking the limit: n
F lim
62.57 y*2s9 y* i
n l i1
i
2
y
125 y 7 y s9 y 2 dy 3
3
125 7 y s9 y 2 dy 125 y y s9 y 2 dy 3
3
3
3
The second integral is 0 because the integrand is an odd function (see Theorem 5.5.7). The first integral can be evaluated using the trigonometric substitution y 3 sin , but it’s simpler to observe that it is the area of a semicircular disk with radius 3. Thus F 875 y s9 y 2 dy 875 12 32 3
3
7875 12,370 lb 2
■
390
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
MOMENTS AND CENTERS OF MASS
P
Our main objective here is to find the point P on which a thin plate of any given shape balances horizontally as in Figure 9. This point is called the center of mass (or center of gravity) of the plate. We first consider the simpler situation illustrated in Figure 10, where two masses m1 and m2 are attached to a rod of negligible mass on opposite sides of a fulcrum and at distances d1 and d2 from the fulcrum. The rod will balance if
FIGURE 9
m1 d1 m2 d2
6
d¡
d™
m¡
m™
This is an experimental fact discovered by Archimedes and called the Law of the Lever. (Think of a lighter person balancing a heavier one on a seesaw by sitting farther away from the center.) Now suppose that the rod lies along the x-axis with m1 at x 1 and m2 at x 2 and the center of mass at x. If we compare Figures 10 and 11, we see that d1 x x 1 and d2 x 2 x and so Equation 6 gives
fulcrum
m1 x x 1 m2x 2 x
FIGURE 10
m1 x m2 x m1 x 1 m2 x 2 m1 x 1 m2 x 2 m1 m2
x
7
The numbers m1 x 1 and m2 x 2 are called the moments of the masses m1 and m2 (with respect to the origin), and Equation 7 says that the center of mass x is obtained by adding the moments of the masses and dividing by the total mass m m1 m2 . x–
⁄ 0
m¡
¤
x–-⁄
m™
¤-x–
x
FIGURE 11
In general, if we have a system of n particles with masses m1, m2, . . . , mn located at the points x 1, x 2, . . . , x n on the x-axis, it can be shown similarly that the center of mass of the system is located at n
n
mx
i i
8
x
i1 n
m
mx
i i
i1
m
i
i1
where m mi is the total mass of the system, and the sum of the individual moments n
M
mx
i i
i1
is called the moment of the system about the origin. Then Equation 8 could be rewritten as mx M, which says that if the total mass were considered as being concentrated at the center of mass x, then its moment would be the same as the moment of the system.
SECTION 7.5
y m£
⁄
‹
› 0
391
x
fi ¤
■
Now we consider a system of n particles with masses m1, m2, . . . , mn located at the points x 1, y1 , x 2 , y2 , . . . , x n , yn in the xy-plane as shown in Figure 12. By analogy with the one-dimensional case, we define the moment of the system about the y-axis to be
m¡
y£
APPLICATIONS TO PHYSICS AND ENGINEERING
n
m™
My
9
mx i
i
i1
and the moment of the system about the x-axis as
FIGURE 12
n
Mx
10
my
i i
i1
Then My measures the tendency of the system to rotate about the y-axis and Mx measures the tendency to rotate about the x-axis. As in the one-dimensional case, the coordinates x, y of the center of mass are given in terms of the moments by the formulas x
11
My m
y
Mx m
where m mi is the total mass. Since mx My and my Mx , the center of mass x, y is the point where a single particle of mass m would have the same moments as the system. V EXAMPLE 7 Find the moments and center of mass of the system of objects that have masses 3, 4, and 8 at the points 1, 1, 2, 1, and 3, 2.
SOLUTION We use Equations 9 and 10 to compute the moments:
My 31 42 83 29 y
center of mass 8
Mx 31 41 82 15
3
Since m 3 4 8 15, we use Equations 11 to obtain 0
4
x
x FIGURE 13
My m
29 15
y
Mx 15 1 m 15
Thus the center of mass is (1 14 15 , 1). (See Figure 13.)
■
Next we consider a flat plate (called a lamina) with uniform density that occupies a region of the plane. We wish to locate the center of mass of the plate, which is called the centroid of . In doing so we use the following physical principles: The symmetry principle says that if is symmetric about a line l, then the centroid of lies on l. (If is reflected about l, then remains the same so its centroid remains fixed. But the only fixed points lie on l.) Thus the centroid of a rectangle is its center. Moments should be defined so that if the entire mass of a region is concentrated at the center of mass, then its moments remain unchanged. Also, the moment of the union of two nonoverlapping regions should be the sum of the moments of the individual regions.
392
■
CHAPTER 7
y
APPLICATIONS OF INTEGRATION
y=ƒ
0
a
b
x
Suppose that the region is of the type shown in Figure 14(a); that is, lies between the lines x a and x b, above the x-axis, and beneath the graph of f , where f is a continuous function. We divide the interval a, b into n subintervals with endpoints x 0 , x 1, . . . , x n and equal width x. We choose the sample point xi* to be the midpoint xi of the ith subinterval, that is, xi xi1 xi 2. This determines the polygonal approximation to shown in Figure 14(b). The centroid of the ith approx1 imating rectangle Ri is its center Ci (xi , 2 f xi ). Its area is f xi x, so its mass is
f xi x
(a) y
The moment of Ri about the y-axis is the product of its mass and the distance from Ci to the y-axis, which is xi . Thus
{ xi , f(xi)} 1 Ci ”xi , 2 f(xi)’
My Ri f xi x xi xi f xi x Adding these moments, we obtain the moment of the polygonal approximation to , and then by taking the limit as n l we obtain the moment of itself about the y-axis: 0
a
R¡ R™
xi_ 1 R£
xi
xi
b
x n
My lim
x f x x y i
n l i1
(b)
i
b
x f x dx
a
FIGURE 14
In a similar fashion we compute the moment of Ri about the x-axis as the product of its mass and the distance from Ci to the x-axis: Mx Ri f xi x 12 f xi 12 f xi 2 x Again we add these moments and take the limit to obtain the moment of about the x-axis: n
Mx lim
n l i1
1 2
f xi 2 x y
b 1 2 a
f x 2 dx
Just as for systems of particles, the center of mass of the plate is defined so that mx My and my Mx . But the mass of the plate is the product of its density and its area: m A y f x dx b
a
and so
ya xf x dx My x b m y f x dx b
a
y
b
a
y
ya 21 f x 2 dx Mx y b m y f x dx b
a
xf x dx
b
a
f x dx
y
b 1 2
a
y
f x 2 dx
b
a
f x dx
Notice the cancellation of the ’s. The location of the center of mass is independent of the density.
SECTION 7.5
APPLICATIONS TO PHYSICS AND ENGINEERING
■
393
In summary, the center of mass of the plate (or the centroid of ) is located at the point x, y, where 1 A
x
12
y
b
xf x dx
a
y
1 A
y
b 1 2
a
f x 2 dx
EXAMPLE 8 Find the center of mass of a semicircular plate of radius r. SOLUTION In order to use (12) we place the semicircle as in Figure 15 so that
y
f x sr 2 x 2 and a r, b r. Here there is no need to use the formula to calculate x because, by the symmetry principle, the center of mass must lie on the 1 y-axis, so x 0. The area of the semicircle is A 2 r 2, so
y=œ„„„„„ r@-≈ 4r
” 0, 3π ’ 0
_r
x
r
y
FIGURE 15
1 A
y
r 1 2 r
2 r 2
y
r
0
1 1 r 2 f x 2 dx 1 r 2 2 yr (sr 2 x 2 ) dx 2 r 2 x 2 dx
2 x3 2 2 r x r 3
r
0
3
2 2r 4r r 2 3 3
The center of mass is located at the point 0, 4r3. y
■
If the region lies between two curves y f x and y tx, where f x tx, as illustrated in Figure 16, then the same sort of argument that led to Formulas 12 can be used to show that the centroid of is x, y, where
C i ” xi , 21 f(xi )+g(xi ) ’ y=ƒ
13
y=© 0
a
b
xi
x
1 A
y
b
a
x f x tx dx
1 A
y
y
b 1 2
a
f x 2 tx 2 dx
x
(See Exercise 51.)
FIGURE 16
EXAMPLE 9 Find the centroid of the region bounded by the line y x and the
parabola y x 2.
SOLUTION The region is sketched in Figure 17. We take f x x, tx x 2, a 0,
y
y=x
and b 1 in Formulas 13. First we note that the area of the region is
(1, 1)
” 21 , 25 ’
Ay
0
y=≈ 0
FIGURE 17
1
x
x2 x3 x x dx 2 3 2
1
0
1 6
Therefore x
1 A
y
1
0
x f x tx dx
6 y x 2 x 3 dx 6 1
0
1 1 6
y
1
0
x3 x4 3 4
x x x 2 dx
1
0
1 2
394
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CHAPTER 7
APPLICATIONS OF INTEGRATION
y
1 A
y
1 1 2
0
3
f x 2 tx 2 dx
x3 x5 3 5
1
0
1 1 6
y
1 1 2
0
x 2 x 4 dx
2 5
The centroid is ( 12 , 25 ).
■
We end this section by showing a surprising connection between centroids and volumes of revolution. THEOREM OF PAPPUS Let be a plane region that lies entirely on one side of a line l in the plane. If is rotated about l, then the volume of the resulting solid is the product of the area A of and the distance d traveled by the centroid of .
This theorem is named after the Greek mathematician Pappus of Alexandria, who lived in the fourth century A.D. ■
PROOF We give the proof for the special case in which the region lies between
y f x and y tx as in Figure 16 and the line l is the y-axis. Using the method of cylindrical shells (see Section 7.3), we have V y 2 x f x tx dx 2 y x f x tx dx b
b
a
a
2 xA
(by Formulas 13)
2 xA Ad where d 2 x is the distance traveled by the centroid during one rotation about the y-axis. ■ V EXAMPLE 10 A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R r from the center of the circle. Find the volume of the torus.
SOLUTION The circle has area A r 2. By the symmetry principle, its centroid is
its center and so the distance traveled by the centroid during a rotation is d 2 R . Therefore, by the Theorem of Pappus, the volume of the torus is V Ad r 2 2 R 2 2r 2R
■
The method of Example 10 should be compared with the method of Exercise 41 in Section 7.2.
7.5
EXERCISES
1. A particle is moved along the x-axis by a force that
measures 101 x2 pounds at a point x feet from the origin. Find the work done in moving the particle from the origin to a distance of 9 ft. 2. When a particle is located a distance x meters from the ori-
gin, a force of cos x3 newtons acts on it. How much
work is done in moving the particle from x 1 to x 2? Interpret your answer by considering the work done from x 1 to x 1.5 and from x 1.5 to x 2. 3. Shown is the graph of a force function (in newtons) that
increases to its maximum value and then remains constant.
SECTION 7.5
How much work is done by the force in moving an object a distance of 8 m?
■
395
12. A bucket that weighs 4 lb and a rope of negligible weight
are used to draw water from a well that is 80 ft deep. The bucket is filled with 40 lb of water and is pulled up at a rate of 2 fts, but water leaks out of a hole in the bucket at a rate of 0.2 lbs. Find the work done in pulling the bucket to the top of the well.
F (N) 30 20 10 0
APPLICATIONS TO PHYSICS AND ENGINEERING
13. A leaky 10-kg bucket is lifted from the ground to a height 1
2 3 4 5 6 7 8
of 12 m at a constant speed with a rope that weighs 0.8 kgm. Initially the bucket contains 36 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 12 m level. How much work is done?
x (m)
4. The table shows values of a force function f x where x is
measured in meters and f x in newtons. Use Simpson’s Rule to estimate the work done by the force in moving an object a distance of 18 m. x
0
3
6
9
12
15
18
f x
9.8
9.1
8.5
8.0
7.7
7.5
7.4
14. A 10-ft chain weighs 25 lb and hangs from a ceiling. Find
the work done in lifting the lower end of the chain to the ceiling so that it’s level with the upper end. 15. An aquarium 2 m long, 1 m wide, and 1 m deep is full of
water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kgm3.)
5. A force of 10 lb is required to hold a spring stretched 4 in.
16. A circular swimming pool has a diameter of 24 ft, the sides
beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?
are 5 ft high, and the depth of the water is 4 ft. How much work is required to pump all of the water out over the side? (Use the fact that water weighs 62.5 lbft 3.) ■
6. A spring has a natural length of 20 cm. If a 25-N force is
required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?
the edge of a building 120 ft high. (a) How much work is done in pulling the rope to the top of the building? (b) How much work is done in pulling half the rope to the top of the building?
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■
3m 2m 3m 8m
8. If 6 J of work is needed to stretch a spring from 10 cm to
9. A heavy rope, 50 ft long, weighs 0.5 lbft and hangs over
■
(a) Find the work required to pump the water out of the spout.
its natural length of 30 cm to a length of 42 cm. (a) How much work is needed to stretch it from 35 cm to 40 cm? (b) How far beyond its natural length will a force of 30 N keep the spring stretched?
9–16 ■ Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it.
■
17. The tank shown is full of water.
7. Suppose that 2 J of work is needed to stretch a spring from
12 cm and another 10 J is needed to stretch it from 12 cm to 14 cm, what is the natural length of the spring?
■
;
(b) Suppose that the pump breaks down after 4.7 10 5 J of work has been done. What is the depth of the water remaining in the tank? 18. The hemispherical tank shown is full of water. Given that
water weighs 62.5 lbft3, find the work required to pump the water out of the tank. 5 ft
10. A chain lying on the ground is 10 m long and its mass is
80 kg. How much work is required to raise one end of the chain to a height of 6 m? 11. A cable that weighs 2 lbft is used to lift 800 lb of coal up a
mine shaft 500 ft deep. Find the work done.
19. When gas expands in a cylinder with radius r, the pressure
at any given time is a function of the volume: P PV . The force exerted by the gas on the piston (see the figure) is the product of the pressure and the area: F r 2P. Show
396
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CHAPTER 7
APPLICATIONS OF INTEGRATION
that the work done by the gas when the volume expands from volume V1 to volume V2 is
25.
26.
12 ft
W y P dV V2
V1
■
x
piston head
20. In a steam engine the pressure P and volume V of steam
satisfy the equation PV 1.4 k, where k is a constant. (This is true for adiabatic expansion, that is, expansion in which there is no heat transfer between the cylinder and its surroundings.) Use Exercise 19 to calculate the work done by the engine during a cycle when the steam starts at a pressure of 160 lbin2 and a volume of 100 in3 and expands to a volume of 800 in3. 21. (a) Newton’s Law of Gravitation states that two bodies with
masses m1 and m2 attract each other with a force FG
m1 m2 r2
where r is the distance between the bodies and G is the gravitational constant. If one of the bodies is fixed, find the work needed to move the other from r a to r b. (b) Compute the work required to launch a 1000-kg satellite vertically to an orbit 1000 km high. You may assume that the Earth’s mass is 5.98 10 24 kg and is concentrated at its center. Take the radius of the Earth to be 6.37 10 6 m and G 6.67 10 11 N m2kg 2.
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27. A trough is filled with a liquid of density 840 kgm3 . The
ends of the trough are equilateral triangles with sides 8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. 28. A large tank is designed with ends in the shape of the
region between the curves y 12 x 2 and y 12, measured in feet. Find the hydrostatic force on one end of the tank if it is filled to a depth of 8 ft with gasoline. (Assume the gasoline’s density is 42.0 lbft3.) 29. A swimming pool is 20 ft wide and 40 ft long and its bot-
tom is an inclined plane, the shallow end having a depth of 3 ft and the deep end, 9 ft. If the pool is full of water, find the hydrostatic force on (a) the shallow end, (b) the deep end, (c) one of the sides, and (d) the bottom of the pool. 30. A vertical dam has a semicircular gate as shown in the fig-
ure. Find the hydrostatic force against the gate.
2m 12 m
22. (a) Use an improper integral and information from Exer-
cise 21 to find the work needed to propel a 1000-kg satellite out of the Earth’s gravitational field. (b) Find the escape velocity v0 that is needed to propel a rocket of mass m out of the gravitational field of a planet with mass M and radius R. (Use the fact that the initial kinetic energy of 12 mv 20 supplies the needed work.) ■ A vertical plate is submerged in water and has the indicated shape. Explain how to approximate the hydrostatic force against one side of the plate by a Riemann sum. Then express the force as an integral and evaluate it.
23–26
23.
20 m
4m 31. A vertical, irregularly shaped plate is submerged in water.
The table shows measurements of its width, taken at the indicated depths. Use Simpson’s Rule to estimate the force of the water against the plate. Depth (m)
2.0
2.5
3.0
3.5
4.0
4.5
5.0
Plate width (m)
0
0.8
1.7
2.4
2.9
3.3
3.6
24. 32. Point-masses m i are located on the x-axis as shown. Find
the moment M of the system about the origin and the center of mass x. m¡=25 _2
0
m™=20
m£=10
3
7
x
SECTION 7.6 ■ The masses m are located at the points P . Find the i i moments Mx and My and the center of mass of the system.
of intersection of the medians. [Hints: Place the axes so that the vertices are a, 0, 0, b, and c, 0. Recall that a median is a line segment from a vertex to the midpoint of the opposite side. Recall also that the medians intersect at a point two-thirds of the way from each vertex (along the median) to the opposite side.]
33. m1 6, m2 5, m3 10;
P11, 5, P23, 2, P32, 1 34. m1 6, m2 5, m3 1, m4 4;
P11, 2, P23, 4, P33, 7, P46, 1 ■
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■
■
■
■
■
■
■
■
Sketch the region bounded by the curves, and visually estimate the location of the centroid. Then find the exact coordinates of the centroid.
35–38
■
35. y 4 x 2,
y0
36. 3x 2y 6, 37. y e x,
y 0,
38. y 1x, ■
■
y 0,
y 0,
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■
46.
■
x2 ■
■
■
■
■
_2
■
47.
y
x1
x 1,
■
46 – 47 ■ Find the centroid of the region shown, not by integration, but by locating the centroids of the rectangles and triangles (from Exercise 45) and using additivity of moments.
x0
x 0,
397
■
45. Prove that the centroid of any triangle is located at the point
33–34
■
DIFFERENTIAL EQUATIONS
y
3
2
2 1
1
0
_2
3 x
1
0
_1
1
2
x
_1
39– 42
■
Find the centroid of the region bounded by the given
■
curves. 39. y sx ,
yx
40. y x 2,
y x2
41. y sin x,
y cos x,
42. y x, ■
■
y 0, ■
■
■
■
■
x 0, ■
48. A sphere of radius r
x 4
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43. 1
51. Prove Formulas 13.
44. 2 y
y 2
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■
r
0
x
■
7.6
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■
■
■
r ■
(Use Example 8.)
■
x ■
■
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52. Let be the region that lies between the curves y x m
quarter-circle
1
■
2, 3, 2, 5, and 5, 4 about the x-axis ■
0
■
50. The solid obtained by rotating the triangle with vertices
■
_1
■
49. A cone with height h and base radius r
x2
Calculate the moments Mx and My and the center of mass of a lamina with the given density and shape.
43– 44
■
48 –50 ■ Use the Theorem of Pappus to find the volume of the given solid.
y 1x,
■
■
■
and y x n, 0 x 1, where m and n are integers with 0 n m. (a) Sketch the region . (b) Find the coordinates of the centroid of . (c) Try to find values of m and n such that the centroid lies outside .
DIFFERENTIAL EQUATIONS A differential equation is an equation that contains an unknown function and one or more of its derivatives. Here are some examples: 1
y xy
2
y 2y y 0
3
d 3y d 2y dy 2y ex 3 x dx dx 2 dx
398
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CHAPTER 7
APPLICATIONS OF INTEGRATION
In each of these differential equations y is an unknown function x . The importance of differential equations lies in the fact that when a scientist or engineer formulates a physical law in mathematical terms, it frequently turns out to be a differential equation. The order of a differential equation is the order of the highest derivative that occurs in the equation. Thus Equations 1, 2, and 3 are of order 1, 2, and 3, respectively. A function f is called a solution of a differential equation if the equation is satisfied when y f x and its derivatives are substituted into the equation. Thus f is a solution of Equation 1 if f x xf x for all values of x in some interval. When we are asked to solve a differential equation we are expected to find all possible solutions of the equation. We have already solved some particularly simple differential equations, namely, those of the form y f x. For instance, we know that the general solution of the differential equation y x 3 is given by y 14 x 4 C, where C is an arbitrary constant. But, in general, solving a differential equation is not an easy matter. There is no systematic technique that enables us to solve all differential equations. In this section we learn how to solve a certain type of differential equation called a separable equation. At the end of the section, however, we will see how to sketch a rough graph of a solution of a first-order differential equation, even when it is impossible to find a formula for the solution. SEPARABLE EQUATIONS
A separable equation is a first-order differential equation that can be written in the form dy txf y dx The name separable comes from the fact that the expression on the right side can be “separated” into a function of x and a function of y. Equivalently, if f y 0, we could write 4
The technique for solving separable differential equations was first used by James Bernoulli (in 1690) in solving a problem about pendulums and by Leibniz (in a letter to Huygens in 1691). John Bernoulli explained the general method in a paper published in 1694. ■
dy tx dx hy
where hy 1f y. To solve this equation we rewrite it in the differential form hy dy tx dx so that all y’s are on one side of the equation and all x’s are on the other side. Then we integrate both sides of the equation: 5
y hy dy y tx dx
Equation 5 defines y implicitly as a function of x. In some cases we may be able to solve for y in terms of x.
SECTION 7.6
DIFFERENTIAL EQUATIONS
■
399
We use the Chain Rule to justify this procedure: If h and t satisfy (5), then d dx
so
d dy
y
y
y
hy dy
tx dx
dy tx dx
h y dy
and
d dx
h y
dy tx dx
Thus Equation 4 is satisfied. When applying differential equations, we are usually not as interested in finding a family of solutions (the general solution) as we are in finding a solution that satisfies some additional requirement. In many physical problems we need to find the particular solution that satisfies a condition of the form yx 0 y0 . This is called an initial condition, and the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem. EXAMPLE 1
dy x2 2. dx y (b) Find the solution of this equation that satisfies the initial condition y0 2. (a) Solve the differential equation
SOLUTION
(a) We write the equation in terms of differentials and integrate both sides: y 2 dy x 2 dx
yy
Figure 1 shows graphs of several members of the family of solutions of the differential equation in Example 1. The solution of the initial-value problem in part (b) is shown in blue. ■
3
2
dy y x 2 dx
1 3
y 3 13 x 3 C
where C is an arbitrary constant. (We could have used a constant C1 on the left side and another constant C 2 on the right side. But then we could combine these constants by writing C C 2 C1.) Solving for y, we get 3 ys x 3 3C
We could leave the solution like this or we could write it in the form 3 ys x3 K
_3
3
where K 3C. (Since C is an arbitrary constant, so is K .) 3 K . To sat(b) If we put x 0 in the general solution in part (a), we get y 0 s 3 isfy the initial condition y0 2, we must have sK 2 and so K 8. Thus the solution of the initial-value problem is
_3
FIGURE 1
3 ys x3 8
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400
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CHAPTER 7
APPLICATIONS OF INTEGRATION
Some computer algebra systems can plot curves defined by implicit equations. Figure 2 shows the graphs of several members of the family of solutions of the differential equation in Example 2. As we look at the curves from left to right, the values of C are 3, 2, 1, 0, 1, 2, and 3. ■
V EXAMPLE 2
SOLUTION Writing the equation in differential form and integrating both sides, we
have 2y cos ydy 6x 2 dx
y 2y cos ydy y 6x
4
2
dx
y 2 sin y 2x 3 C
6 _2
dy 6x 2 . dx 2y cos y
Solve the differential equation
2
where C is a constant. Equation 6 gives the general solution implicitly. In this case it’s impossible to solve the equation to express y explicitly as a function of x. ■ _4
EXAMPLE 3 Solve the equation y x 2 y.
FIGURE 2
SOLUTION First we rewrite the equation using Leibniz notation:
dy x2y dx ■ If a solution y is a function that satisfies yx 0 for some x, it follows from a uniqueness theorem for solutions of differential equations that yx 0 for all x.
If y 0, we can rewrite it in differential notation and integrate: dy x 2 dx y
y
dy y x 2 dx y
ln y
Several solutions of the differential equation in Example 3 are graphed in Figure 3. The values of A are the same as the y -intercepts. 6 ■
2
x3 C 3
This equation defines y implicitly as a function of x. But in this case we can solve explicitly for y as follows:
y e so
_2
y0
e x 3C e Ce x 3 3
ln y
3
3
y e Ce x 3
We can easily verify that the function y 0 is also a solution of the given differential equation. So we can write the general solution in the form 3
y Ae x 3
_6
FIGURE 3
where A is an arbitrary constant ( A e C, or A e C, or A 0). EXAMPLE 4 Solve the equation
■
dy ky . dt
SOLUTION This differential equation was studied in Section 3.4, where it was called the law of natural growth (or decay). Since it is a separable equation, we can solve it
SECTION 7.6
DIFFERENTIAL EQUATIONS
■
401
by the methods of this section as follows: dy y k dt y
y
y e
y0
ln y kt C ktC
e Ce kt
y Ae kt where A e C or 0 is an arbitrary constant.
■
LOGISTIC GROWTH
The differential equation of Example 4 is appropriate for modeling population growth y ky says that the rate of growth is proportional to the size of the population) under conditions of unlimited environment and food supply. However, in a restricted environment and with limited food supply, the population cannot exceed a maximal size M (called the carrying capacity) at which it consumes its entire food supply. If we make the assumption that the rate of growth of population is jointly proportional to the size of the population y and the amount by which y falls short of the maximal size M y, then we have the equation dy kyM y dt
7
where k is a constant. Equation 7 is called the logistic differential equation and was used by the Dutch mathematical biologist Pierre-François Verhulst in the 1840s to model world population growth. The logistic equation is separable, so we write it in the form dy
y yM y Using partial fractions, we have 1 1 yM y M 1 M
and so
y
dy dy y y My
1 (ln y ln M y M
y k dt
1 1 y My
y k dt kt C
) kt C
But y y and M y M y , since 0 y M, so we have ln
y Mkt C My y Ae kMt My
A e MC
402
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CHAPTER 7
APPLICATIONS OF INTEGRATION
If the population at time t 0 is y0 y0 , then A y0 M y0, so y y0 e kMt My M y0 If we solve this equation for y , we get
8
y
y0 Me kMt y0 M kMt M y0 y0 e y0 M y0ekMt
Using the latter expression for y , we see that lim yt M
tl
FIGURE 4
Logistic growth function
which is to be expected. The graph of the logistic growth function is shown in Figure 4. At first the graph is concave upward and the growth curve appears to be almost exponential, but then it is concave downward and approaches the limiting population M. MIXING PROBLEMS
A typical mixing problem involves a tank of fixed capacity filled with a thoroughly mixed solution of some substance, such as salt. A solution of a given concentration enters the tank at a fixed rate and the mixture, thoroughly stirred, leaves at a fixed rate, which may differ from the entering rate. If yt denotes the amount of substance in the tank at time t, then yt is the rate at which the substance is being added minus the rate at which it is being removed. The mathematical description of this situation often leads to a first-order separable differential equation. We can use the same type of reasoning to model a variety of phenomena: chemical reactions, discharge of pollutants into a lake, injection of a drug into the bloodstream. EXAMPLE 5 A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that
contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 Lmin. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt remains in the tank after half an hour? SOLUTION Let yt be the amount of salt (in kilograms) after t minutes. We are given that y0 20 and we want to find y30. We do this by finding a differential equation satisfied by yt. Note that dydt is the rate of change of the amount of salt, so 9
dy rate in rate out dt
where (rate in) is the rate at which salt enters the tank and (rate out) is the rate at which salt leaves the tank. We have
rate in 0.03
kg L
25
L min
0.75
kg min
SECTION 7.6
DIFFERENTIAL EQUATIONS
■
403
The tank always contains 5000 L of liquid, so the concentration at time t is yt5000 (measured in kilograms per liter). Since the brine flows out at a rate of 25 Lmin, we have rate out
yt kg 5000 L
25
L min
yt kg 200 min
Thus from Equation 9 we get dy yt 150 yt 0.75 dt 200 200 Solving this separable differential equation, we obtain dy
y 150 y
y
ln 150 y
ln 150 y
y
t ln 130 200
150 y 130e
t200
Therefore 150
Since yt is continuous and y0 20 and the right side is never 0, we deduce that 150 yt is always positive. Thus 150 y 150 y and so
100 50
FIGURE 5
t C 200
Since y0 20, we have ln 130 C, so
Figure 5 shows the graph of the function yt of Example 5. Notice that, as time goes by, the amount of salt approaches 150 kg. ■
0
dt 200
yt 150 130et200 200
400
t
The amount of salt after 30 min is y30 150 130e30200 38.1 kg
■
DIRECTION FIELDS
Suppose we are given a first-order differential equation of the form y Fx, y where Fx, y is some expression in x and y. [Recall that a separable equation is the special case in which Fx, y can be factored as a function of x times a function of y .] Even if it is impossible to find a formula for the solution, we can still visualize the solution curves by means of a direction field. If a solution curve passes through a point x, y, then its slope at that point is y, which is equal to Fx, y. If we draw short line segments with slope Fx, y at several points x, y, the result is called a direction field (or slope field). These line segments indicate the direction in which a solution curve is heading, so the direction field helps us visualize the general shape of these curves.
404
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CHAPTER 7
APPLICATIONS OF INTEGRATION
y
V EXAMPLE 6
2
(a) Sketch the direction field for the differential equation y x 2 y 2 1. (b) Use part (a) to sketch the solution curve that passes through the origin.
1
SOLUTION _2
0
_1
1
2
(a) We start by computing the slope at several points in the following chart:
x
-1
x
2
1
0
1
2
2
1
0
1
2
...
_2
y
0
0
0
0
0
1
1
1
1
1
...
y x 2 y 2 1
3
0
1
0
3
4
1
0
1
4
...
FIGURE 6
Now we draw short line segments with these slopes at these points. The result is the direction field shown in Figure 6. (b) We start at the origin and move to the right in the direction of the line segment (which has slope 1 ). We continue to draw the solution curve so that it moves parallel to the nearby line segments. The resulting solution curve is shown in Figure 7. Returning to the origin, we draw the solution curve to the left as well. ■
y 2
1
_2
0
_1
1
2
x
The more line segments we draw in a direction field, the clearer the picture becomes. Of course, it’s tedious to compute slopes and draw line segments for a huge number of points by hand, but computers are well suited for this task. Figure 8 shows a more detailed, computer-drawn direction field for the differential equation in Example 6. It enables us to draw, with reasonable accuracy, the solution curves shown in Figure 9 with y-intercepts 2, 1, 0, 1, and 2.
-1
_2
FIGURE 7
3
Module 7.6 shows direction fields and solution curves for a variety of differential equations.
3
_3
3
_3
3
_3
_3
FIGURE 9
FIGURE 8
7.6 1– 8 1.
■
EXERCISES
Solve the differential equation.
dy y dx x
2.
dy sx y dx e
9.
4. y y 2 sin x
3. x 2 1y xy 5. 1 tan yy x 1
e y sin2 dy 6. d y sec
du 2 2u t tu 7. dt
dz e tz 0 8. dt
2
■
Find the solution of the differential equation that satisfies the given initial condition.
9–14
■
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■
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■
■
10.
■
du 2t sec 2t , u0 5 dt 2u dy y cos x , dx 1 y2
y0 1
11. x cos x 2y e 3y y, 12. ■
■
dP sPt , dt
P1 2
y0 0
SECTION 7.6
13. y tan x a y, y3 a, 14.
25–26
0 x 2
■
■
■
■
■
■
405
Refer to the direction fields in Exercises 21–24.
25. Use field II to sketch the graphs of the solutions that satisfy
dL kL 2 ln t, L1 1 dt
■
■
DIFFERENTIAL EQUATIONS
■
■
■
■
■
the given initial conditions. (a) y0 1 (b) y0 2
■
(c) y0 1
26. Use field IV to sketch the graphs of the solutions that
15. Find an equation of the curve that satisfies dydx 4x 3 y
satisfy the given initial conditions. (a) y0 1 (b) y0 0
and whose y-intercept is 7.
(c) y0 1
16. Find an equation of the curve that passes through the point
1, 1 and whose slope at x, y is y 2x 3.
Sketch a direction field for the differential equation. Then use it to sketch three solution curves. 27–28
17. (a) Solve the differential equation y 2x s1 y . 2
27. y 1 y
(b) Solve the initial-value problem y 2x s1 y 2 , y0 0, and graph the solution. (c) Does the initial-value problem y 2x s1 y 2 , y0 2, have a solution? Explain.
;
■
; 18. Solve the equation e y cos x 0 and graph several
members of the family of solutions. How does the solution curve change as the constant C varies?
19. Solve the initial-value problem y sin xsin y,
y0 2, and graph the solution (if your CAS does implicit plots).
CAS
members of the family of solutions (if your CAS does implicit plots). How does the solution curve change as the constant C varies?
23. y x y 1
24. y sin x sin y
y
I
II
y
■
■
■
■
■
■
1, 0
31. y y x y,
0, 1
■
■
■
■
30. y 1 x y,
0, 0
32. y x x y, ■
■
■
■
1, 0 ■
■
0
_2
x
2
33. Psychologists interested in learning theory study learning
k a positive constant
inside a concentric sphere with radius 2 m and temperature 25 C. The temperature T r at a distance r from the common center of the spheres satisfies the differential equation
_2
0
2
x
y
IV
2 dT d 2T 0 dr 2 r dr
y
4
If we let S dTdr, then S satisfies a first-order differential equation. Solve it to find an expression for the temperature T r between the spheres.
2
2
0
_2
x
2
35. A glucose solution is administered intravenously into the _2
■
0 ■
■
■
34. A sphere with radius 1 m has temperature 15 C. It lies
2
_2
■
is a reasonable model for learning. (b) Solve the differential equation in part (a) to find an expression for Pt. What is the limit of this expression?
2
III
■
dP kM P dt
4
_2
■
curves. A learning curve is the graph of a function Pt, the performance of someone learning a skill as a function of the training time t. The derivative dPdt represents the rate at which performance improves. (a) If M is the maximum level of performance of which the learner is capable, explain why the differential equation
21–24 ■ Match the differential equation with its direction field (labeled I–IV). Give reasons for your answer. 22. y x2 y
■
29. y y 2x,
■
20. Solve the equation y x sx 2 1 ye y and graph several
21. y 2 y
■
28. y x 2 y 2
29–32 ■ Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point.
y
CAS
■
2 ■
■
x ■
■
■
■
■
■
■
bloodstream at a constant rate r. As the glucose is added, it is converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at that time. Thus a model for the concentration
406
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
C Ct of the glucose solution in the bloodstream is dC r kC dt where k is a positive constant. (a) Suppose that the concentration at time t 0 is C0. Determine the concentration at any time t by solving the differential equation. (b) Assuming that C0 rk, find lim t l Ct and interpret your answer. 36. A certain small country has $10 billion in paper currency in
circulation, and each day $50 million comes into the country’s banks. The government decides to introduce new currency by having the banks replace old bills with new ones whenever old currency comes into the banks. Let x x t denote the amount of new currency in circulation at time t, with x 0 0. (a) Formulate a mathematical model in the form of an initial-value problem that represents the “flow” of the new currency into circulation. (b) Solve the initial-value problem found in part (a). (c) How long will it take for the new bills to account for 90% of the currency in circulation? 37. Write the solution of the logistic initial-value problem
dP 0.00008P1000 P dt
P0 100
and use it to find the population sizes P40 and P80. At what time does the population reach 900? 38. The Pacific halibut fishery has been modeled by the differ-
40. Biologists stocked a lake with 400 fish of one species and
estimated the species’ carrying capacity in the lake to be 10,000. The number of fish tripled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years. (b) How long will it take for the population to increase to 5000? 41. (a) Show that if y satisfies the logistic equation (7), then
d 2y k 2 yM yM 2y dt 2 (b) Deduce that a population grows fastest when it reaches half its carrying capacity.
; 42. For a fixed value of M (say M 10 ), the family of logistic functions given by Equation 8 depends on the initial value y0 and the proportionality constant k. Graph several members of this family. How does the graph change when y0 varies? How does it change when k varies?
43. A tank contains 1000 L of brine with 15 kg of dissolved
salt. Pure water enters the tank at a rate of 10 Lmin. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank (a) after t minutes and (b) after 20 minutes? 44. The air in a room with volume 180 m 3 contains 0.15% car-
bon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2 m 3min and the mixed air flows out at the same rate. Find the percentage of carbon dioxide in the room as a function of time. What happens in the long run?
ential equation 45. A vat with 500 gallons of beer contains 4% alcohol (by
dy kyM y dt where yt is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be M 8 10 7 kg , and k 8.875 109 per year. (a) If y0 2 10 7 kg, find the biomass a year later. (b) How long will it take the biomass to reach 4 10 7 kg? 39. One model for the spread of a rumor is that the rate of
spread is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by y. (b) Solve the differential equation. (c) A small town has 1000 inhabitants. At 8 AM, 80 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor?
volume). Beer with 6% alcohol is pumped into the vat at a rate of 5 galmin and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour? 46. A tank contains 1000 L of pure water. Brine that contains
0.05 kg of salt per liter of water enters the tank at a rate of 5 Lmin. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 Lmin. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 Lmin. How much salt is in the tank (a) after t minutes and (b) after one hour? 47. When a raindrop falls, it increases in size and so its mass at
time t is a function of t, mt. The rate of growth of the mass is kmt for some positive constant k. When we apply Newton’s Law of Motion to the raindrop, we get mv tm, where v is the velocity of the raindrop (directed downward) and t is the acceleration due to gravity. The terminal velocity of the raindrop is lim t l vt. Find an expression for the terminal velocity in terms of t and k.
CHAPTER 7
48. An object of mass m is moving horizontally through a
CAS
medium which resists the motion with a force that is a function of the velocity; that is, m
d 2s dv m f v dt 2 dt
407
(b) Solve the differential equation to find an expression for At. Use a computer algebra system to perform the integration.
gravitational force on an object of mass m that has been projected vertically upward from the Earth’s surface is F
the final area of the tissue when growth is complete. Most cell divisions occur on the periphery of the tissue and the number of cells on the periphery is proportional to sAt. So a reasonable model for the growth of tissue is obtained by assuming that the rate of growth of the area is jointly proportional to sAt and M At. (a) Formulate a differential equation and use it to show that the tissue grows fastest when At 13 M .
mtR 2 x R2
where x xt is the object’s distance above the surface at time t, R is the Earth’s radius, and t is the acceleration due to gravity. Also, by Newton’s Second Law, F ma m dvdt and so m
dv mtR 2 dt x R2
(a) Suppose a rocket is fired vertically upward with an initial velocity v 0. Let h be the maximum height above the surface reached by the object. Show that
49. Let At be the area of a tissue culture at time t and let M be
REVIEW
■
50. According to Newton’s Law of Universal Gravitation, the
where v vt and s st represent the velocity and position of the object at time t, respectively. For example, think of a boat moving through the water. (a) Suppose that the resisting force is proportional to the velocity, that is, f v k v, k a positive constant. (This model is appropriate for small values of v.) Let v0 v0 and s0 s0 be the initial values of v and s. Determine v and s at any time t. What is the total distance that the object travels from time t 0? (b) For larger values of v a better model is obtained by supposing that the resisting force is proportional to the square of the velocity, that is, f v k v 2, k 0. (This model was first proposed by Newton.) Let v0 and s0 be the initial values of v and s. Determine v and s at any time t. What is the total distance that the object travels in this case?
7
REVIEW
v0
2tRh Rh
[Hint: By the Chain Rule, m dvdt mv dvdx.] (b) Calculate ve lim h l v 0 . This limit is called the escape velocity for the Earth. (c) Use R 3960 mi and t 32 fts2 to calculate ve in feet per second and in miles per second.
CONCEPT CHECK
1. (a) Draw two typical curves y f x and y tx, where
f x tx for a x b. Show how to approximate the area between these curves by a Riemann sum and sketch the corresponding approximating rectangles. Then write an expression for the exact area. (b) Explain how the situation changes if the curves have equations x f y and x t y, where f y t y for c y d.
(b) If S is a solid of revolution, how do you find the crosssectional areas? 4. (a) What is the volume of a cylindrical shell?
(b) Explain how to use cylindrical shells to find the volume of a solid of revolution. (c) Why might you want to use the shell method instead of slicing? 5. (a) How is the length of a curve defined?
2. Suppose that Sue runs faster than Kathy throughout a
1500-meter race. What is the physical meaning of the area between their velocity curves for the first minute of the race? 3. (a) Suppose S is a solid with known cross-sectional areas.
Explain how to approximate the volume of S by a Riemann sum. Then write an expression for the exact volume.
(b) Write an expression for the length of a smooth curve given by y f x, a x b. (c) What if x is given as a function of y? 6. Suppose that you push a book across a 6-meter-long table
by exerting a force f x at each point from x 0 to x 6. What does x06 f x dx represent? If f x is measured in newtons, what are the units for the integral?
408
■
CHAPTER 7
APPLICATIONS OF INTEGRATION
7. Describe how we can find the hydrostatic force against a
10. (a) What is a differential equation?
vertical wall submersed in a fluid.
(b) What is the order of a differential equation? (c) What is an initial condition?
8. (a) What is the physical significance of the center of mass
of a thin plate? (b) If the plate lies between y f x and y 0, where a x b, write expressions for the coordinates of the center of mass.
11. What is a direction field for the differential equation
y Fx, y? 12. What is a separable differential equation? How do you
solve it?
9. What does the Theorem of Pappus say?
EXERCISES 1– 4
■
Find the area of the region bounded by the given curves.
1. y x 2 x 6, 2. y 20 x 2,
y x 2 12 y x 2 x,
4. x y 0,
x y 3y
■
■
x 1, and y 0. Use the Midpoint Rule with n 4 to estimate the following quantities. (a) The area of (b) The volume obtained by rotating about the x-axis
y0
3. y e x 1,
■
15. Let be the region bounded by the curves y tanx 2 ,
x1
2 ; 16. Let be the region bounded by the curves y 1 x and
2
■
■
■
■
■
■
■
■
y x 6 x 1. Estimate the following quantities. (a) The x-coordinates of the points of intersection of the curves (b) The area of (c) The volume generated when is rotated about the x-axis (d) The volume generated when is rotated about the y-axis
■
■ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
5–9
5. y 2x, y x 2;
about the x-axis
6. x 1 y , y x 3; 2
7. x 0, x 9 y ; 2
about the y-axis
about x 1
8. y x 1, y 9 x 2; 2
17–20 ■ Each integral represents the volume of a solid. Describe the solid.
about y 1
9. x 2 y 2 a 2, x a h (where a 0, h 0);
about the y-axis ■
■
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■
■
■
■
■
■
■
■
■
10. y cos x, y 0, x 32, x 52;
■
■
■
19.
y
2
20.
y [2 x
■
about x 2 ■
■
0
2 x cos x dx
18.
y
2
0
2 cos2x dx
0
2 y4 y 2 dy
1
0
■
■
(2 sx )2 ] dx
2 2
■
■
■
■
■
■
■
■
■
21. The base of a solid is a circular disk with radius 3. Find the
about y 1
12. y x 3, y 8, x 0; ■
2
■
about the y-axis 11. y x 3, y x 2;
y
■
Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
10 –12
17.
■
■
■
■
13. Find the volumes of the solids obtained by rotating the
region bounded by the curves y x and y x about the following lines. (a) The x-axis (b) The y-axis (c) y 2 2
14. Let be the region in the first quadrant bounded by the
curves y x 3 and y 2x x 2. Calculate the following quantities. (a) The area of (b) The volume obtained by rotating about the x-axis (c) The volume obtained by rotating about the y-axis
■
volume of the solid if parallel cross-sections perpendicular to the base are isosceles right triangles with hypotenuse lying along the base. 22. The base of a solid is the region bounded by the parabolas
y x 2 and y 2 x 2. Find the volume of the solid if the cross-sections perpendicular to the x-axis are squares with one side lying along the base. 23. The height of a monument is 20 m. A horizontal cross-
section at a distance x meters from the top is an equilateral triangle with side 14 x meters. Find the volume of the monument. 24. (a) The base of a solid is a square with vertices located at
1, 0, 0, 1, 1, 0, and 0, 1. Each cross-section
CHAPTER 7
perpendicular to the x-axis is a semicircle. Find the volume of the solid. (b) Show that by cutting the solid of part (a), we can rearrange it to form a cone. Thus compute its volume more simply. 25–26
■
0 x 3
26. y 2 ln (sin 2 x) ,
3 x
1
1
■
409
just covers the gate. Find the hydrostatic force on one side of the gate. 34. Find the centroid of the region shown. y
(3, 2)
■
■
■
■
■
x
the curve y e
2
■
■
■
■
■
, 0 x 3.
35–36
■
Find the centroid of the region bounded by the given
curves. 35. y 4 x 2,
28. Find the length of the curve
y y sst 1 dt x
1 x 16
1
x
0
27. Use Simpson’s Rule with n 6 to estimate the length of
29. A force of 30 N is required to maintain a spring stretched
from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm?
36. y sin x, ■
weighs 10 lbft. How much work is required to raise the elevator from the basement to the third floor, a distance of 30 ft? 31. A tank full of water has the shape of a paraboloid of revolu-
tion as shown in the figure; that is, its shape is obtained by rotating a parabola about a vertical axis. (a) If its height is 4 ft and the radius at the top is 4 ft, find the work required to pump the water out of the tank. (b) After 4000 ft-lb of work has been done, what is the depth of the water remaining in the tank?
■
■
yx2 x 4,
y 0, ■
■
■
■
32. A trough is filled with water and its vertical ends have the
shape of the parabolic region in the figure. Find the hydrostatic force on one end of the trough. 8 ft
4 ft
33. A gate in an irrigation canal is constructed in the form of a
trapezoid 3 ft wide at the bottom, 5 ft wide at the top, and 2 ft high. It is placed vertically in the canal so that the water
■
■
■
■
38. Use the Theorem of Pappus and the fact that the volume of
a sphere of radius r is 43 r 3 to find the centroid of the semicircular region bounded by the curve y sr 2 x 2 and the x-axis. 39– 40
■
Solve the differential equation.
39. 3y 2 2yy x cos x ■
■
41– 42 41.
■
4 ft
■
center 1, 0 is rotated about the y-axis.
■
■
■
■
40. ■
■
dr 2tr r, dt
■
dx 1 t x tx dt ■
■
■
■
■
■
■
■
Solve the initial-value problem. r0 5
42. 1 cos xy 1 ey sin x , 4 ft
x 34
37. Find the volume obtained when the circle of radius 1 with
30. A 1600-lb elevator is suspended by a 200-ft cable that
;
■
Find the length of the curve.
25. y 6 x 2 4 32,
■
REVIEW
■
■
■
■
■
y0 0 ■
■
43. (a) Sketch a direction field for the differential equation
y xy. Then use it to sketch the four solutions that satisfy the initial conditions y0 1, y0 1, y2 1, and y2 1. (b) Check your work in part (a) by solving the differential equation explicitly. What type of curve is each solution curve? 44. Let 1 be the region bounded by y x 2, y 0, and x b,
where b 0. Let 2 be the region bounded by y x 2, x 0, and y b 2. (a) Is there a value of b such that 1 and 2 have the same area? (b) Is there a value of b such that 1 sweeps out the same volume when rotated about the x-axis and the y-axis? (c) Is there a value of b such that 1 and 2 sweep out the same volume when rotated about the x-axis? (d) Is there a value of b such that 1 and 2 sweep out the same volume when rotated about the y-axis?
8
SERIES Infinite series are sums of infinitely many terms. (One of our aims in this chapter is to define exactly what is meant by an infinite sum.) Their importance in calculus stems from Newton’s idea of representing functions as sums of infinite series. For instance, in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series. We will pursue his idea in Section 8.7 in order to integrate such functions as ex . (Recall that we have previously been unable to do this.) Many of the functions that arise in mathematical physics and chemistry, such as Bessel functions, are defined as sums of series, so it is important to be familiar with the basic concepts of convergence of infinite sequences and series. Physicists also use series in another way, as we will see in Section 8.8. In studying fields as diverse as optics, special relativity, and electromagnetism, they analyze phenomena by replacing a function with the first few terms in the series that represents it. 2
8.1
SEQUENCES A sequence can be thought of as a list of numbers written in a definite order: a1 , a2 , a3 , a4 , . . . , an , . . . The number a 1 is called the first term, a 2 is the second term, and in general a n is the nth term. We will deal exclusively with infinite sequences and so each term a n will have a successor a n1 . Notice that for every positive integer n there is a corresponding number a n and so a sequence can be defined as a function whose domain is the set of positive integers. But we usually write a n instead of the function notation f n for the value of the function at the number n. NOTATION The sequence {a 1 , a 2 , a 3 , . . .} is also denoted by
a n
or
a n n1
EXAMPLE 1 Some sequences can be defined by giving a formula for the nth term.
In the following examples we give three descriptions of the sequence: one by using the preceding notation, another by using the defining formula, and a third by writing out the terms of the sequence. Notice that n doesn’t have to start at 1. (a)
(b) (c) (d) 410
n n1
an
n n1
an
1nn 1 3n
n1
1nn 1 3n
{sn 3 } n3
a n sn 3 , n 3
a n cos
n cos 6
n0
n , n0 6
1 2 3 4 n , , , ,..., ,... 2 3 4 5 n1
2 3 4 5 1nn 1 , , , ,..., ,... 3 9 27 81 3n
{0, 1, s2 , s3 , . . . , sn 3 , . . .}
1,
n s3 1 , , 0, . . . , cos ,... 2 2 6
■
SECTION 8.1
V EXAMPLE 2
SEQUENCES
■
411
Find a formula for the general term a n of the sequence
3 4 5 6 7 , , , , ,... 5 25 125 625 3125
assuming that the pattern of the first few terms continues. SOLUTION We are given that
a1
3 5
a2
4 25
a3
5 125
a4
6 625
a5
7 3125
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator n 2. The denominators are the powers of 5, so a n has denominator 5 n. The signs of the terms are alternately positive and negative, so we need to multiply by a power of 1. In Example 1(b) the factor 1 n meant we started with a negative term. Here we want to start with a positive term and so we use 1 n1 or 1 n1. Therefore, a n 1 n1
n2 5n
■
EXAMPLE 3 Here are some sequences that don’t have a simple defining equation.
(a) The sequence pn , where pn is the population of the world as of January 1 in the year n. (b) If we let a n be the digit in the nth decimal place of the number e, then a n is a well-defined sequence whose first few terms are
7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . . (c) The Fibonacci sequence fn is defined recursively by the conditions f1 1
f2 1
fn fn1 fn2
n3
Each term is the sum of the two preceding terms. The first few terms are a¡
a™ a£
1 2
0
1, 1, 2, 3, 5, 8, 13, 21, . . .
a¢
This sequence arose when the 13th-century Italian mathematician known as Fibonacci solved a problem concerning the breeding of rabbits (see Exercise 41). ■
1
A sequence such as the one in Example 1(a), a n nn 1, can be pictured either by plotting its terms on a number line as in Figure 1 or by plotting its graph as in Figure 2. Note that, since a sequence is a function whose domain is the set of positive integers, its graph consists of isolated points with coordinates
FIGURE 1 an
1
1, a1
2, a2
3, a3
...
n, a n
...
7
a¶= 8 0
1 2 3 4 5 6 7
FIGURE 2
n
From Figure 1 or 2 it appears that the terms of the sequence a n nn 1 are approaching 1 as n becomes large. In fact, the difference 1
n 1 n1 n1
412
■
CHAPTER 8
SERIES
can be made as small as we like by taking n sufficiently large. We indicate this by writing n lim 1 nl n 1 In general, the notation lim a n L
nl
means that the terms of the sequence a n approach L as n becomes large. Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 1.6. 1 DEFINITION
A sequence a n has the limit L and we write lim a n L
a n l L as n l
or
nl
if we can make the terms a n as close to L as we like by taking n sufficiently large. If lim n l a n exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent). Figure 3 illustrates Definition 1 by showing the graphs of two sequences that have the limit L. an
an
L
L
FIGURE 3
Graphs of two sequences with lim an= L
0
0
n
n
n `
A more precise version of Definition 1 is as follows. 2 DEFINITION
A sequence an has the limit L and we write lim an L
■ Compare this definition with Definition 1.6.7.
a n l L as n l
or
nl
if for every 0 there is a corresponding integer N such that if
nN
a
then
n
L
Definition 2 is illustrated by Figure 4, in which the terms a 1 , a 2 , a 3 , . . . are plotted on a number line. No matter how small an interval L , L is chosen, there exists an N such that all terms of the sequence from a N1 onward must lie in that interval. a¡ FIGURE 4
0
a£
a™
aˆ
aN+1 aN+2 L-∑
L
L+∑
a˜
aß
a∞
a¢
a¶
SECTION 8.1
SEQUENCES
413
■
Another illustration of Definition 2 is given in Figure 5. The points on the graph of
an must lie between the horizontal lines y L and y L if n N . This picture must be valid no matter how small is chosen, but usually a smaller requires a larger N. y
y=L+∑ L y=L-∑ 0
1 2 3 4
FIGURE 5
n
N
If you compare Definition 2 with Definition 1.6.7, you will see that the only difference between lim n l a n L and lim x l f x L is that n is required to be an integer. Thus we have the following theorem, which is illustrated by Figure 6. 3 THEOREM If lim x l f x L and f n a n when n is an integer, then lim n l a n L.
y
y=ƒ
L
0
FIGURE 6
x
1 2 3 4
In particular, since we know that limx l 1x r 0 when r 0, we have lim
4
nl
1 0 nr
if r 0
If a n becomes large as n becomes large, we use the notation lim n l a n . The following precise definition is similar to Definition 1.6.8.
5 DEFINITION lim n l a n means that for every positive number M there is an integer N such that
if
nN
then
an M
If lim n l a n , then the sequence a n is divergent but in a special way. We say that a n diverges to . The Limit Laws given in Section 1.4 also hold for the limits of sequences and their proofs are similar.
414
■
CHAPTER 8
SERIES
If a n and bn are convergent sequences and c is a constant, then
Limit Laws for Sequences
lim a n bn lim a n lim bn
nl
nl
nl
lim a n bn lim a n lim bn
nl
nl
nl
lim ca n c lim a n
nl
lim c c
nl
nl
lim a n bn lim a n lim bn
nl
nl
lim
lim a n an nl bn lim bn
nl
nl
if lim bn 0 nl
nl
[
lim a np lim a n
nl
nl
]
p
if p 0 and a n 0
The Squeeze Theorem can also be adapted for sequences as follows (see Figure 7).
If a n bn cn for n n 0 and lim a n lim cn L, then lim bn L.
Squeeze Theorem for Sequences
nl
cn
nl
nl
Another useful fact about limits of sequences is given by the following theorem, whose proof is left as Exercise 45.
bn an 0
If lim a n 0, then lim a n 0.
6 THEOREM
nl
nl
n
FIGURE 7
The sequence bn is squeezed between the sequences a n and cn .
EXAMPLE 4 Find lim
nl
n . n1
SOLUTION The method is similar to the one we used in Section 1.6: Divide numerator and denominator by the highest power of n that occurs in the denominator and then use the Limit Laws.
lim
nl
n lim nl n1
This shows that the guess we made earlier from Figures 1 and 2 was correct. ■
nl
ln n . n
1
1 n
lim 1
nl
lim 1 lim
nl
nl
1 n
1 1 10
Here we used Equation 4 with r 1. EXAMPLE 5 Calculate lim
1
■
SECTION 8.1
SEQUENCES
■
415
SOLUTION Notice that both numerator and denominator approach infinity as n l . We can’t apply l’Hospital’s Rule directly because it applies not to sequences but to functions of a real variable. However, we can apply l’Hospital’s Rule to the related function f x ln xx and obtain
ln x 1x lim 0 xl 1 x
lim
xl
Therefore, by Theorem 3 we have lim
nl
an
ln n 0 n
■
EXAMPLE 6 Determine whether the sequence a n 1 n is convergent or divergent.
1
SOLUTION If we write out the terms of the sequence, we obtain
0
1
2
3
4
1, 1, 1, 1, 1, 1, 1, . . .
n
_1
The graph of this sequence is shown in Figure 8. Since the terms oscillate between 1 and 1 infinitely often, a n does not approach any number. Thus, lim n l 1 n does not exist; that is, the sequence 1 n is divergent. ■
FIGURE 8
EXAMPLE 7 Evaluate lim
The graph of the sequence in Example 7 is shown in Figure 9 and supports the answer. ■
nl
1 n if it exists. n
SOLUTION
an 1
lim
nl
1 n n
lim
nl
1 0 n
Therefore, by Theorem 6, 0
1
lim
n
nl
1 n 0 n
■
Discuss the convergence of the sequence a n n!n n, where n! 1 2 3 n. V EXAMPLE 8
_1
FIGURE 9
SOLUTION Both numerator and denominator approach infinity as n l but here we have no corresponding function for use with l’Hospital’s Rule (x! is not defined when x is not an integer). Let’s write out a few terms to get a feeling for what happens to a n as n gets large:
a1 1
7
a2 an
12 22
a3
123 333
1 2 3 n n n n n
It appears from these expressions and the graph in Figure 10 that the terms are decreasing and perhaps approach 0. To confirm this, observe from Equation 7 that an
1 n
2 3 n n n n
416
■
CHAPTER 8
SERIES
CREATING GRAPHS OF SEQUENCES Some computer algebra systems have special commands that enable us to create sequences and graph them directly. With most graphing calculators, however, sequences can be graphed by using parametric equations. For instance, the sequence in Example 8 can be graphed by entering the parametric equations ■
xt
y t!t t
Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator. So 0 an
We know that 1n l 0 as n l . Therefore, a n l 0 as n l by the Squeeze Theorem. V EXAMPLE 9
and graphing in dot mode starting with t 1, setting the t -step equal to 1. The result is shown in Figure 10. 1
1 n ■
For what values of r is the sequence r n convergent?
SOLUTION We know from Section 1.6 and the graphs of the exponential functions
in Section 3.1 that lim x l a x for a 1 and lim x l a x 0 for 0 a 1. Therefore, putting a r and using Theorem 3, we have lim r n
nl
0
if r 1 if 0 r 1
For the cases r 1 and r 0 we have 0
10
lim 1n lim 1 1
nl
lim 0 n lim 0 0
and
nl
nl
nl
FIGURE 10
If 1 r 0, then 0 r 1, so
lim r n lim r
nl
nl
n
0
and therefore lim n l r n 0 by Theorem 6. If r 1, then r n diverges as in Example 6. Figure 11 shows the graphs for various values of r. (The case r 1 is shown in Figure 8.) an
an
r>1 1
1
_1
0
r=1
n
1 0
FIGURE 11
1
0
n
r<_1
The sequence an=r n
■
The results of Example 9 are summarized for future use as follows.
The sequence r n is convergent if 1 r 1 and divergent for all other values of r. 8
lim r n
nl
0 1
if 1 r 1 if r 1
SECTION 8.1
SEQUENCES
■
417
9 DEFINITION A sequence a n is called increasing if a n a n1 for all n 1, that is, a1 a2 a3 . It is called decreasing if a n a n1 for all n 1. A sequence is monotonic if it is either increasing or decreasing.
EXAMPLE 10 The sequence
3 n5
is decreasing because
3 3 3 n5 n 1 5 n6
■ The right side is smaller because it has a larger denominator.
and so a n a n1 for all n 1.
■
EXAMPLE 11 Show that the sequence a n
n is decreasing. n2 1
SOLUTION We must show that a n1 a n , that is,
n1 n 2 n 1 2 1 n 1 ■ Another way to do Example 11 is to show that the function
f x
x x2 1
x1
is decreasing because f x 0 for x 1.
This inequality is equivalent to the one we get by cross-multiplication: n1 n 2 n 1 2 1 n 1
&? n 1n 2 1 n n 12 1 &?
n 3 n 2 n 1 n 3 2n 2 2n
&?
1 n2 n
Since n 1, we know that the inequality n 2 n 1 is true. Therefore, a n1 a n and so a n is decreasing. ■
10 DEFINITION
A sequence a n is bounded above if there is a number M
such that an M
for all n 1
It is bounded below if there is a number m such that m an
for all n 1
If it is bounded above and below, then a n is a bounded sequence. For instance, the sequence a n n is bounded below a n 0 but not above. The sequence a n nn 1 is bounded because 0 a n 1 for all n. We know that not every bounded sequence is convergent [for instance, the sequence a n 1n satisfies 1 a n 1 but is divergent from Example 6] and not every monotonic sequence is convergent a n n l . But if a sequence is
418
■
CHAPTER 8
SERIES
an
M L
0 1 23
n
FIGURE 12
both bounded and monotonic, then it must be convergent. This fact is proved as Theorem 11, but intuitively you can understand why it is true by looking at Figure 12. If a n is increasing and a n M for all n, then the terms are forced to crowd together and approach some number L. The proof of Theorem 11 is based on the Completeness Axiom for the set ⺢ of real numbers, which says that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S ), then S has a least upper bound b. (This means that b is an upper bound for S, but if M is any other upper bound, then b M .) The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line. 11 MONOTONIC SEQUENCE THEOREM
Every bounded, monotonic sequence is
convergent. PROOF Suppose a n is an increasing sequence. Since a n is bounded, the set
S a n n 1 has an upper bound. By the Completeness Axiom it has a least upper bound L . Given 0, L is not an upper bound for S (since L is the least upper bound). Therefore aN L
for some integer N
But the sequence is increasing so a n a N for every n N. Thus if n N we have an L 0 L an
so since a n L . Thus
L a n
whenever n N
so lim n l a n L . A similar proof (using the greatest lower bound) works if a n is decreasing.
■
The proof of Theorem 11 shows that a sequence that is increasing and bounded above is convergent. (Likewise, a decreasing sequence that is bounded below is convergent.) This fact is used many times in dealing with infinite series in Sections 8.2 and 8.3. Another use of Theorem 11 is indicated in Exercises 38– 40.
8.1
EXERCISES
1. (a) What is a sequence?
(b) What does it mean to say that lim n l a n 8? (c) What does it mean to say that lim n l a n ? 2. (a) What is a convergent sequence? Give two examples.
(b) What is a divergent sequence? Give two examples.
3. List the first six terms of the sequence defined by
an
n 2n 1
Does the sequence appear to have a limit? If so, find it.
SECTION 8.1
5– 8 ■ Find a formula for the general term a n of the sequence, assuming that the pattern of the first few terms continues. 6.
7. 2, 7, 12, 17, . . . ■
■
■
■
32. (a) If a n is convergent, show that
lim a n1 lim a n
{ 14 , 29 , 163 , 254 , . . .}
nl
■
■
■
■
■
■
■
■ Determine whether the sequence converges or diverges. If it converges, find the limit.
9–28
■ Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?
33–36
3 5n 2 n n2
10. a n
n1 3n 1
33. a n
11. a n
2n 3 n1
12. a n
sn 1 sn
35. a n cosn2
13. a n
n 2! n!
14. a n
n 1 sn
15. a n
1 n n2 1
16. a n
1 n n 3 2n 2 1
9. a n
n1
17.
e n e n e 2n 1
24. a n
sin 2n 1 sn
26. a n
ln n 2 n
27. a n ln2n 2 1 lnn 2 1 28. a n
3n n!
23. a n
1
2 n
25. 0, 1, 0, 0, 1, 0, 0, 0, 1, . . .
■
■
■
■
■
■
■
■
■
■
an 3a n 1
■
■
■
■
■
■
■
a n1 3 1a n is increasing and a n 3 for all n. Deduce that a n is convergent and find its limit.
40. Show that the sequence defined by
a1 2
a n1
1 3 an
satisfies 0 a n 2 and is decreasing. Deduce that the sequence is convergent and find its limit. ■
30. Find the first 40 terms of the sequence defined by 1 2
■
1 n
39. Use induction to show that the sequence defined by a 1 1,
then after n years the investment is worth a n 10001.06n dollars. (a) Find the first five terms of the sequence a n . (b) Is the sequence convergent or divergent? Explain.
36. a n n
(a) By induction or otherwise, show that a n is increasing and bounded above by 3. Apply the Monotonic Sequence Theorem to show that lim n l a n exists. (b) Find lim n l a n .
29. If $1000 is invested at 6% interest, compounded annually,
a n1
■
2n 3 3n 4
38. A sequence a n is given by a 1 s2 , a n1 s2 a n .
22. n cos n n
■
34. a n
{s2 , s2s2 , s2s2s2 , . . .}
20. arctan 2n
cos 2n 2n
■
1 2n 3
37. Find the limit of the sequence 3
18. a n cos2n
19. n 2e n 21. a n
■
n
nl
(b) A sequence a n is defined by a 1 1 and a n1 11 a n for n 1. Assuming that a n is convergent, find its limit.
8. 5, 1, 5, 1, 5, 1, . . . ■
419
all its terms lie between the numbers 5 and 8. Explain why the sequence has a limit. What can you say about the value of the limit?
this sequence appear to have a limit? If so, find it. If not, explain why.
{1, 23 , 49 , 278 , . . .}
■
31. Suppose you know that a n is a decreasing sequence and
4. List the first nine terms of the sequence cosn3 . Does
5.
SEQUENCES
if a n is an even number if a n is an odd number
and a1 11. Do the same if a1 25. Make a conjecture about this type of sequence.
■
41. (a) Fibonacci posed the following problem: Suppose that
rabbits live forever and that every month each pair produces a new pair which becomes productive at age 2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the nth month? Show that the answer is fn , where fn is the Fibonacci sequence defined in Example 3(c). (b) Let a n fn1fn and show that a n1 1 1a n2. Assuming that a n is convergent, find its limit. 42. (a) Let a 1 a, a 2 f a, a 3 f a 2 f f a, . . . ,
a n1 f a n , where f is a continuous function. If lim n l a n L, show that f L L. (b) Illustrate part (a) by taking f x cos x, a 1, and estimating the value of L to five decimal places.
420
■
CHAPTER 8
SERIES
43. We know that lim n l 0.8 n 0 [from (8) with r 0.8].
47. The size of an undisturbed fish population has been
modeled by the formula
Use logarithms to determine how large n has to be so that 0.8 n 0.000001. 44. Use Definition 2 directly to prove that lim n l r n 0
pn1
when r 1.
45. Prove Theorem 6.
[Hint: Use either Definition 2 or the Squeeze Theorem.] 46. (a) Show that if lim n l a 2n L and lim n l a2n1 L ,
then a n is convergent and lim n l a n L . (b) If a 1 1 and 1 an1 1 1 an
find the first eight terms of the sequence a n . Then use part (a) to show that lim n l a n s2 . This gives the continued fraction expansion 1
s2 1 2
8.2
bpn a pn
where pn is the fish population after n years and a and b are positive constants that depend on the species and its environment. Suppose that the population in year 0 is p 0 0. (a) Show that if pn is convergent, then the only possible values for its limit are 0 and b a. (b) Show that pn1 ba pn . (c) Use part (b) to show that if a b, then lim n l pn 0 ; in other words, the population dies out. (d) Now assume that a b. Show that if p 0 b a, then
pn is increasing and 0 pn b a. Show also that if p 0 b a, then pn is decreasing and pn b a. Deduce that if a b, then lim n l pn b a.
1 2
SERIES If we try to add the terms of an infinite sequence a n n1 we get an expression of the form a1 a2 a3 an
1
which is called an infinite series (or just a series) and is denoted, for short, by the symbol
a
n1
n
or
a
n
But does it make sense to talk about the sum of infinitely many terms? It would be impossible to find a finite sum for the series 1 2 3 4 5 n
n
Sum of first n terms
1 2 3 4 5 6 7 10 15 20 25
0.50000000 0.75000000 0.87500000 0.93750000 0.96875000 0.98437500 0.99218750 0.99902344 0.99996948 0.99999905 0.99999997
because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15, 21, . . . and, after the nth term, we get nn 12, which becomes very large as n increases. However, if we start to add the terms of the series 1 1 1 1 1 1 1 n 2 4 8 16 32 64 2 we get 2 , 4 , 8 , 16 , 32 , 64 , . . . , 1 12 n, . . . . The table shows that as we add more and more terms, these partial sums become closer and closer to 1. In fact, by adding sufficiently many terms of the series we can make the partial sums as close as we like to 1. So it seems reasonable to say that the sum of this infinite series is 1 and to write 1
3
7
15 31 63
n1
1 1 1 1 1 1 n 1 2n 2 4 8 16 2
SECTION 8.2
SERIES
■
421
We use a similar idea to determine whether or not a general series (1) has a sum. We consider the partial sums s1 a 1 s2 a 1 a 2 s3 a 1 a 2 a 3 s4 a 1 a 2 a 3 a 4 and, in general, n
sn a 1 a 2 a 3 a n
a
i
i1
These partial sums form a new sequence sn , which may or may not have a limit. If lim n l sn s exists (as a finite number), then, as in the preceding example, we call it the sum of the infinite series a n . Given a series n1 a n a 1 a 2 a 3 , let sn denote its nth partial sum: 2 DEFINITION
n
sn
a
i
a1 a2 an
i1
If the sequence sn is convergent and lim n l sn s exists as a real number, then the series a n is called convergent and we write
a1 a2 an s
a
or
n
s
n1
The number s is called the sum of the series. If the sequence sn is divergent, then the series is called divergent. ■
Compare with the improper integral
y
1
f x dx lim tl
y
t
1
f x dx
Thus the sum of a series is the limit of the sequence of partial sums. So when we write n1 an s we mean that by adding sufficiently many terms of the series we can get as close as we like to the number s. Notice that
To find this integral we integrate from 1 to t and then let t l . For a series, we sum from 1 to n and then let n l .
n
a
n
n1
lim
a
n l i1
i
EXAMPLE 1 An important example of an infinite series is the geometric series
a ar ar 2 ar 3 ar n1
ar
n1
a0
n1
Each term is obtained from the preceding one by multiplying it by the common ratio r. (We have already considered the special case where a 12 and r 12 on page 420.) If r 1, then sn a a a na l . Since lim n l sn doesn’t exist, the geometric series diverges in this case. If r 1, we have sn a ar ar 2 ar n1 and
rsn
ar ar 2 ar n1 ar n
422
■
CHAPTER 8
SERIES
■ Figure 1 provides a geometric demonstration of the result in Example 1. If the triangles are constructed as shown and s is the sum of the series, then, by similar triangles,
s a a a ar
a s 1r
so
Subtracting these equations, we get sn rsn a ar n sn
3
a1 r n 1r
If 1 r 1, we know from (8.1.8) that r n l 0 as n l , so ar# ar@
lim sn lim
nl
ar@
ar a-ar
nl
a1 r n a a a lim r n 1r 1r 1 r nl 1r
ar s
Thus when r 1 the geometric series is convergent and its sum is a1 r. If r 1 or r 1, the sequence r n is divergent by (8.1.8) and so, by Equation 3, lim n l sn does not exist. Therefore, the geometric series diverges in those cases. ■ We summarize the results of Example 1 as follows.
a
a
The geometric series
4
ar
a
n1
a ar ar 2
n1
is convergent if r 1 and its sum is
FIGURE 1
ar
In words: The sum of a convergent geometric series is ■
first term 1 common ratio
n1
a 1r
n1
r 1
If r 1, the geometric series is divergent. Find the sum of the geometric series
V EXAMPLE 2
5 103 209 40 27 SOLUTION The first term is a 5 and the common ratio is r 3 . Since 2
r
2 3
1, the series is convergent by (4) and its sum is 5
■ What do we really mean when we say that the sum of the series in Example 2 is 3? Of course, we can’t literally add an infinite number of terms, one by one. But, according to Definition 2, the total sum is the limit of the sequence of partial sums. So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3. The table shows the first ten partial sums sn and the graph in Figure 2 shows how the sequence of partial sums approaches 3.
n
sn
1 2 3 4 5 6 7 8 9 10
5.000000 1.666667 3.888889 2.407407 3.395062 2.736626 3.175583 2.882945 3.078037 2.947975
10 20 40 5 5 5 3 2 3 9 27 1 ( 3 ) 3
■
sn
3
0
FIGURE 2
20 n
SECTION 8.2
SERIES
■
423
EXAMPLE 3 Is the series
2
2n
3 1n convergent or divergent?
n1
SOLUTION Let’s rewrite the nth term of the series in the form ar n1: Another way to identify a and r is to write out the first few terms: ■
4
16 3
2
649
2n
3 1n
n1
2 3 2 n
n1
n1
n1
4n n1 4 ( 43 ) 3 n1 n1
We recognize this series as a geometric series with a 4 and r 43 . Since r 1, the series diverges by (4). ■ V EXAMPLE 4
Write the number 2.317 2.3171717. . . as a ratio of integers.
SOLUTION
2.3171717. . . 2.3
Module 8.2 explores a series that depends on an angle in a triangle and enables you to see how rapidly the series converges when varies.
17 17 17 3 5 10 10 10 7
After the first term we have a geometric series with a 1710 3 and r 110 2. Therefore 17 17 3 10 1000 2.317 2.3 2.3 1 99 1 2 10 100 23 17 1147 10 990 495
■
EXAMPLE 5 Find the sum of the series
x , where x 1. n
n0
SOLUTION Notice that this series starts with n 0 and so the first term is x 0 1.
(With series, we adopt the convention that x 0 1 even when x 0.) Thus
x
n
1 x x2 x3 x4
n0
This is a geometric series with a 1 and r x. Since r x 1, it converges and (4) gives
x
5
n0
EXAMPLE 6 Show that the series
n1
n
1 1x
1 is convergent, and find its sum. nn 1
SOLUTION This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums. n
sn
i1
1 1 1 1 1 ii 1 12 23 34 nn 1
We can simplify this expression if we use the partial fraction decomposition 1 1 1 ii 1 i i1
■
424
■
CHAPTER 8
SERIES
(see Section 6.3). Thus we have n
sn
Notice that the terms cancel in pairs. This is an example of a telescoping sum: Because of all the cancellations, the sum collapses (like a pirate’s collapsing telescope) into just two terms. ■
i1
n 1 ii 1 i1
1 1 i i1
1 2
1 1 2 3
1 n1 lim sn lim
and so
nl
nl
1 1 3 4
1
1 n1
1 1 n n1
101
Therefore, the given series is convergent and
n1
1
1 1
■ Figure 3 illustrates Example 6 by showing the graphs of the sequence of terms a n 1[nn 1] and the sequence sn of partial sums. Notice that a n l 0 and sn l 1. See Exercises 36 and 37 for two geometric interpretations of Example 6.
1 1 nn 1
■
Show that the harmonic series
V EXAMPLE 7
sn
n1
1 1 1 1 1 n 2 3 4
is divergent. SOLUTION For this particular series it’s convenient to consider the partial sums s2 , s4 , s8 , s16 , s32 , . . . and show that they become large.
a n 0
n
s2 1 12 FIGURE 3
s4 1 12 ( 13 14 ) 1 12 ( 14 14 ) 1 22 s8 1 12 ( 13 14 ) ( 15 16 17 18 ) 1 12 ( 14 14 ) ( 18 18 18 18 ) 1 12 12 12 1 32 s16 1 12 ( 13 14 ) ( 15 18 ) ( 19 161 ) 1 12 ( 14 14 ) ( 18 18 ) ( 161 161 ) 1 12 12 12 12 1 42 Similarly, s32 1 52 , s64 1 62 , and in general s2 n 1
The method used in Example 7 for showing that the harmonic series diverges is due to the French scholar Nicole Oresme (1323–1382). ■
n 2
This shows that s2 n l as n l and so sn is divergent. Therefore, the harmonic series diverges. ■
6 THEOREM
If the series
a
n1
n
is convergent, then lim an 0. nl
SECTION 8.2
■
SERIES
425
PROOF Let sn a 1 a 2 a n . Then a n sn sn1. Since a n is convergent, the sequence sn is convergent. Let lim n l sn s. Since n 1 l as n l , we also have lim n l sn1 s. Therefore
lim a n lim sn sn1 lim sn lim sn1 s s 0
nl
nl
nl
■
nl
NOTE 1 With any series a n we associate two sequences: the sequence sn of its partial sums and the sequence a n of its terms. If a n is convergent, then the limit of the sequence sn is s (the sum of the series) and, as Theorem 6 asserts, the limit of the sequence a n is 0.
|
NOTE 2 The converse of Theorem 6 is not true in general. If lim n l a n 0, we cannot conclude that a n is convergent. Observe that for the harmonic series 1n we have a n 1n l 0 as n l , but we showed in Example 7 that 1n is divergent.
If lim a n does not exist or if lim a n 0,
7 THE TEST FOR DIVERGENCE
nl
then the series
a
n
nl
is divergent.
n1
The Test for Divergence follows from Theorem 6 because, if the series is not divergent, then it is convergent, and so lim n l a n 0.
n2 diverges. 5n 2 4
EXAMPLE 8 Show that the series
n1
SOLUTION
lim a n lim
nl
nl
n2 1 1 lim 0 n l 5 4n 2 5n 4 5 2
■
So the series diverges by the Test for Divergence.
NOTE 3 If we find that lim n l a n 0, we know that a n is divergent. If we find that lim n l a n 0, we know nothing about the convergence or divergence of a n . Remember the warning in Note 2: If lim n l a n 0, the series a n might converge or it might diverge. 8 THEOREM If a n and bn are convergent series, then so are the series ca n (where c is a constant), a n bn , and a n bn , and
(i)
ca n c
n1
a
n1
an
(ii)
n1
(iii)
n
bn
n1
a
n1
n
a n bn
n1
an
b
n
n1
b
n
n1
These properties of convergent series follow from the corresponding Limit Laws for Sequences in Section 8.1. For instance, here is how part (ii) of Theorem 8 is proved:
426
■
CHAPTER 8
SERIES
Let
n
sn
a
a
s
i
i1
n
b
tn
n
t
i
n1
i1
b
n
n1
The nth partial sum for the series a n bn is n
a
un
bi
i
i1
and, using Equation 5.2.10, we have n
lim u n lim
nl
i
n l i1
bi lim
nl
n
lim
n
ai
i1
b
i
i1
n
a
i
n l i1
n
a
b
lim
n l i1
i
lim sn lim tn s t nl
nl
Therefore, a n bn is convergent and its sum is
a
n
a
bn s t
n1
n1
EXAMPLE 9 Find the sum of the series
n1
SOLUTION The series
n
b
n
■
n1
3 1 n . nn 1 2
12 n is a geometric series with a 12 and r 12 , so
n1
1 1 2 1 2n 1 12
In Example 6 we found that
n1
1 1 nn 1
So, by Theorem 8, the given series is convergent and
n1
3 1 n nn 1 2
3
1 1 n 3114 nn 1 n1 2
n1
■
NOTE 4 A finite number of terms doesn’t affect the convergence or divergence of a series. For instance, suppose that we were able to show that the series
n4
n n3 1
is convergent. Since
n1
n 1 2 3 n 3 n 1 2 9 28 n4 n 1 3
it follows that the entire series n1 nn 3 1 is convergent. Similarly, if it is known that the series nN1 a n converges, then the full series
a
n1
is also convergent.
N
n
a
n1
n
nN1
an
SECTION 8.2
8.2
(b) What is a convergent series? What is a divergent series?
27.
n1
3– 8 ■ Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.
29.
n0
209 40 27
■
4. 1 0.4 0.16 0.064
5.
2 n1 3
6.
n1
7.
n0
3 n1
■
■
■
n1
8. ■
cos n x 2n ■
■
■
n1
1
(s2 )
n
■
■
10.
k2 2 k 1
12.
12 3n
11.
k2
13.
1 2n
n1
n1
n
14.
■
n
16.
■
■
■
■
■
13 2n
n
n1
cos 1
0.3 n
k
■
n2
21.
n1 ■
20.
3 nn 3
22.
■
23–26
2 n 1 2
■
■
n1
■
■
■
2 n 2 4n 3
n
ln n 1
n1 ■
■
■
■
■
■
■
■
34. A certain ball has the property that each time it falls from a
■
height h onto a hard, level surface, it rebounds to a height rh, where 0 r 1. Suppose that the ball is dropped from an initial height of H meters. (a) Assuming that the ball continues to bounce indefinitely, find the total distance that it travels. (b) Calculate the total time that the ball travels. (Use the fact that the ball falls 12 tt 2 meters in t seconds.) (c) Suppose that each time the ball strikes the surface with velocity v it rebounds with velocity k v, where
Express the number as a ratio of integers.
24. 0.73 0.73737373 . . . 25. 3.417 3.417417417 . . . 26. 6.254 6.2545454 . . . ■
■
■
■
■
■
■
■
■
■
receive the money also spend some of it. The people receiving some of the twice-spent money will spend some of that, and so on. Economists call this chain reaction the multiplier effect. In a hypothetical isolated community, the local government begins the process by spending D dollars. Suppose that each recipient of spent money spends 100c% and saves 100s% of the money that he or she receives. The values c and s are called the marginal propensity to consume and the marginal propensity to save and, of course, c s 1. (a) Let Sn be the total spending that has been generated after n transactions. Find an equation for Sn . (b) Show that lim n l Sn kD, where k 1s. The number k is called the multiplier. What is the multiplier if the marginal propensity to consume is 80%? Note: The federal government uses this principle to justify deficit spending. Banks use this principle to justify lending a large percentage of the money that they receive in deposits.
■
23. 0.2 0.2222 . . .
■
n1 n1
find a n and n1 a n .
■ Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in Example 6). If it is convergent, find its sum.
■
1 n
33. When money is spent on goods and services, those that
0.8
■
1
kk 2 k 32
19–22
19.
■
32. If the nth partial sum of a series n1 a n is sn 3 n 2 n,
k1 ■
■
31. If the nth partial sum of a series n1 a n is
find a n and n1 a n .
18.
n1
■
■
sn
n1
arctan n
■
n1 2n 3
s2
■
n1
■
k1
n1
17.
n1
15.
■
is another series with this property.
■
■ Determine whether the series is convergent or divergent. If it is convergent, find its sum.
9.
■
9–18
2 nx 1 n
n0
ln
n0 ■
28.
whose terms approach 0. Show that
6 5 n1
n1
n
■
■
xn 3n
30. We have seen that the harmonic series is a divergent series
5( )
427
27–29 ■ Find the values of x for which the series converges. Find the sum of the series for those values of x.
2. Explain what it means to say that n1 a n 5.
10 3
■
EXERCISES
1. (a) What is the difference between a sequence and a series?
3. 5
SERIES
428
■
CHAPTER 8
SERIES
0 k 1. How long will it take for the ball to come to rest?
35. What is the value of c if
39. What is wrong with the following calculation?
0 0 0 0
1 cn 2?
1 1 1 1 1 1
n2
1 1 1 1 1 1
n ; 36. Graph the curves y x , 0 x 1, for n 0, 1, 2, 3,
4, . . . on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that
n1
1 1 1 1 1 1 1 1 0 0 0 1 (Guido Ubaldus thought that this proved the existence of God because “something has been created out of nothing.”)
1 1 nn 1
40. Suppose that n1 a n a n 0 is known to be a convergent 37. The figure shows two circles C and D of radius 1 that touch
at P. T is a common tangent line; C1 is the circle that touches C, D, and T ; C2 is the circle that touches C, D, and C1; C3 is the circle that touches C, D, and C2. This procedure can be continued indefinitely and produces an infinite sequence of circles Cn . Find an expression for the diameter of Cn and thus provide another geometric demonstration of Example 6.
series. Prove that n1 1a n is a divergent series.
41. Prove part (i) of Theorem 8. 42. If a n is divergent and c 0, show that ca n is divergent. 43. If a n is convergent and bn is divergent, show that
the series a n bn is divergent. [Hint: Argue by contradiction.]
44. If a n and bn are both divergent, is a n bn neces-
sarily divergent? 45. Suppose that a series a n has positive terms and its partial
P
sums sn satisfy the inequality sn 1000 for all n. Explain why a n must be convergent.
C£ C™
1
1 46. The Fibonacci sequence was defined in Section 8.1 by the
C
D
C¡
equations T
38. A right triangle ABC is given with ⬔A and AC b.
CD is drawn perpendicular to AB, DE is drawn perpendicular to BC, EF ⬜ AB, and this process is continued indefinitely as shown in the figure. Find the total length of all the perpendiculars
f1 1,
(a)
1 1 1 fn1 fn1 fn1 fn fn fn1
(b)
1 1 fn1 fn1
fn 2 fn1 fn1
n2
D
¨
F H
B
n2
(c)
A
b
G
E
C
fn fn1 fn2
n3
Show that each of the following statements is true.
CD DE EF FG in terms of b and .
f2 1,
47. The Cantor set, named after the German mathematician
Georg Cantor (1845–1918), is constructed as follows. We start with the closed interval [0, 1] and remove the open interval ( 13 , 23 ). That leaves the two intervals [0, 13 ] and [ 23, 1] and we remove the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of
SECTION 8.3
the numbers that remain in [0, 1] after all those intervals have been removed. (a) Show that the total length of all the intervals that are removed is 1. Despite that, the Cantor set contains infinitely many numbers. Give examples of some numbers in the Cantor set. (b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center one-ninth of a square of side 1, then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0.
THE INTEGRAL AND COMPARISON TESTS
■
429
49. Consider the series
n1
n n 1!
(a) Find the partial sums s1, s2, s3, and s4. Do you recognize the denominators? Use the pattern to guess a formula for sn . (b) Use mathematical induction to prove your guess. (c) Show that the given infinite series is convergent, and find its sum. 50. In the figure there are infinitely many circles approaching
the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles.
48. (a) A sequence a n is defined recursively by the equation
a n 12 a n1 a n2 for n 3, where a 1 and a 2 can be any real numbers. Experiment with various values of a 1 and a 2 and use your calculator to guess the limit of the sequence. (b) Find lim n l a n in terms of a 1 and a 2 by expressing a n1 a n in terms of a 2 a 1 and summing a series.
8.3
THE INTEGRAL AND COMPARISON TESTS In general, it is difficult to find the exact sum of a series. We were able to accomplish this for geometric series and the series 1 nn 1 because in each of those cases we could find a simple formula for the nth partial sum sn . But usually it isn’t easy to compute lim n l sn . Therefore, in this section and the next we develop tests that enable us to determine whether a series is convergent or divergent without explicitly finding its sum. In this section we deal only with series with positive terms, so the partial sums are increasing. In view of the Monotonic Sequence Theorem, to decide whether a series is convergent or divergent, we need to determine whether the partial sums are bounded or not.
n
n
sn
i1
5 10 50 100 500 1000 5000
1 i2
1.4636 1.5498 1.6251 1.6350 1.6429 1.6439 1.6447
TESTING WITH AN INTEGRAL
Let’s investigate the series whose terms are the reciprocals of the squares of the positive integers: 1 1 1 1 1 1 2 2 2 2 2 2 n 1 2 3 4 5 n1 There’s no simple formula for the sum sn of the first n terms, but the computergenerated table of values given in the margin suggests that the partial sums are approaching a number near 1.64 as n l and so it looks as if the series is convergent.
430
■
CHAPTER 8
SERIES
We can confirm this impression with a geometric argument. Figure 1 shows the curve y 1x 2 and rectangles that lie below the curve. The base of each rectangle is an interval of length 1; the height is equal to the value of the function y 1x 2 at the right endpoint of the interval. So the sum of the areas of the rectangles is 1 1 1 1 1 1 2 2 2 2 2 2 1 2 3 4 5 n n1
y
y=
1 ≈
area= 1 1@ 0
1
2
area= 1 2@
FIGURE 1
3
area= 1 3@
4
area= 1 4@
x
5
area= 1 5@
If we exclude the first rectangle, the total area of the remaining rectangles is smaller than the area under the curve y 1x 2 for x 1, which is the value of the integral x1 1x 2 dx. In Section 6.6 we discovered that this improper integral is convergent and has value 1. So the picture shows that all the partial sums are less than 1 1 y 2 dx 2 1 x 12 Thus the partial sums are bounded and the series converges. The sum of the series (the limit of the partial sums) is also less than 2:
1 1 1 1 1 2 2 2 2 2 n 1 2 3 42
n1
[The exact sum of this series was found by the Swiss mathematician Leonhard Euler (1707–1783) to be 26, but the proof of this fact is beyond the scope of this book.] Now let’s look at the series
n1
n
n
sn
i1
5 10 50 100 500 1000 5000
1 1 1 1 1 1 sn s1 s2 s3 s4 s5
The table of values of sn suggests that the partial sums aren’t approaching a finite number, so we suspect that the given series may be divergent. Again we use a picture for confirmation. Figure 2 shows the curve y 1sx , but this time we use rectangles whose tops lie above the curve.
1 si
3.2317 5.0210 12.7524 18.5896 43.2834 61.8010 139.9681
FIGURE 2
y
y= 1 x œ„
0
1
2
area= 1 1 œ„
3
area= 1 2 œ„
4
area= 1 3 œ„
5
area= 1 4 œ„
x
SECTION 8.3
THE INTEGRAL AND COMPARISON TESTS
■
431
The base of each rectangle is an interval of length 1. The height is equal to the value of the function y 1sx at the left endpoint of the interval. So the sum of the areas of all the rectangles is 1 1 1 1 1 1 s1 s2 s3 s4 s5 n1 sn
This total area is greater than the area under the curve y 1sx for x 1, which is equal to the integral x1 (1sx ) dx . But we know from Section 6.6 that this improper integral is divergent. In other words, the area under the curve is infinite. So the sum of the series must be infinite, that is, the series is divergent. The same sort of geometric reasoning that we used for these two series can be used to prove the following test. (The proof is given at the end of this section.) THE INTEGRAL TEST Suppose f is a continuous, positive, decreasing function on 1, and let a n f n. Then the series n1 a n is convergent if and only if the improper integral x1 f x dx is convergent. In other words:
(a) If y f x dx is convergent, then 1
(b) If y f x dx is divergent, then 1
a
n
is convergent.
n1
a
n
is divergent.
n1
NOTE When we use the Integral Test it is not necessary to start the series or the integral at n 1. For instance, in testing the series
n4
1 n 32
y
we use
4
1 dx x 32
Also, it is not necessary that f be always decreasing. What is important is that f be ultimately decreasing, that is, decreasing for x larger than some number N. Then nN a n is convergent, so n1 a n is convergent by Note 4 of Section 8.2.
V EXAMPLE 1
Determine whether the series
n1
ln n converges or diverges. n
SOLUTION The function f x ln xx is positive and continuous for x 1
because the logarithm function is positive and continuous there. But it is not obvious whether or not f is decreasing, so we compute its derivative: f x ■ In order to use the Integral Test we need to be able to evaluate x1 f x dx and therefore we have to be able to find an antiderivative of f . Frequently this is difficult or impossible, so we need other tests for convergence too.
x1x ln x 1 ln x x2 x2
Thus f x 0 when ln x 1, that is, x e. It follows that f is decreasing when x e and so we can apply the Integral Test:
y
1
ln x t ln x ln x2 dx lim y dx lim tl 1 tl x x 2
t
1
ln t 2 2
lim
tl
Since this improper integral is divergent, the series ln nn is also divergent by ■ the Integral Test.
432
■
CHAPTER 8
SERIES
V EXAMPLE 2
For what values of p is the series
n1
1 convergent? np
SOLUTION If p 0, then lim n l 1n . If p 0, then lim n l 1n p 1. p
In either case lim n l 1n p 0, so the given series diverges by the Test for Divergence [see (8.2.7)]. If p 0, then the function f x 1x p is clearly continuous, positive, and decreasing on 1, . We found in Chapter 6 [see (6.6.2)] that
■ Exercises 29–34 show how to estimate the sum of a series that is convergent by the Integral Test.
y
1
1 dx xp
converges if p 1 and diverges if p 1
It follows from the Integral Test that the series 1n p converges if p 1 and diverges if 0 p 1. (For p 1, this series is the harmonic series discussed in Example 7 in Section 8.2.) ■ The series in Example 2 is called the p-series. It is important in the rest of this chapter, so we summarize the results of Example 2 for future reference as follows.
1
The p-series
n1
1 is convergent if p 1 and divergent if p 1. np
For instance, the series
n1
1 1 1 1 1 3 3 3 3 n 1 2 3 43
is convergent because it is a p-series with p 3 1. But the series
n1
1 1 1 1 1 1 3 3 3 13 3 n s2 s3 s4 n1 sn
is divergent because it is a p-series with p 13 1. TESTING BY COMPARING
The series
2
n1
1 2n 1
reminds us of the series n1 12 n, which is a geometric series with a 12 and r 12 and is therefore convergent. Because the series (2) is so similar to a convergent series, we have the feeling that it too must be convergent. Indeed, it is. The inequality 1 1 n 2n 1 2 shows that our given series (2) has smaller terms than those of the geometric series and therefore all its partial sums are also smaller than 1 (the sum of the geometric series). This means that its partial sums form a bounded increasing sequence, which is convergent. It also follows that the sum of the series is less than the sum of the geometric series: 1 1 n 2 1 n1
SECTION 8.3
THE INTEGRAL AND COMPARISON TESTS
■
433
Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive. The first part says that if we have a series whose terms are smaller than those of a known convergent series, then our series is also convergent. The second part says that if we start with a series whose terms are larger than those of a known divergent series, then it too is divergent. THE COMPARISON TEST Suppose that a n and bn are series with positive terms. (a) If bn is convergent and a n bn for all n, then a n is also convergent. (b) If bn is divergent and a n bn for all n, then a n is also divergent.
■ It is important to keep in mind the distinction between a sequence and a series. A sequence is a list of numbers, whereas a series is a sum. With every series a n there are associated two sequences: the sequence a n of terms and the sequence sn of partial sums.
Standard Series for Use with the Comparison Test
PROOF
(i) Let n
sn
a
n
tn
i
i1
b
t
i
i1
b
n
n1
Since both series have positive terms, the sequences sn and tn are increasing sn1 sn a n1 sn . Also tn l t, so tn t for all n. Since a i bi , we have sn tn . Thus sn t for all n. This means that sn is increasing and bounded above and therefore converges by the Monotonic Sequence Theorem. Thus a n converges. (ii) If bn is divergent, then tn l (since tn is increasing). But a i bi so sn tn . Thus sn l . Therefore, a n diverges. ■ In using the Comparison Test we must, of course, have some known series bn for the purpose of comparison. Most of the time we use one of these series: ■ ■
A p -series [ 1n p converges if p 1 and diverges if p 1; see (1)] A geometric series [ ar n1 converges if r 1 and diverges if r 1; see (8.2.4)]
V EXAMPLE 3
Determine whether the series
n1
diverges.
5 converges or 2n 4n 3 2
SOLUTION For large n the dominant term in the denominator is 2n 2, so we compare
the given series with the series 52n 2 . Observe that 5 5 2n 2 4n 3 2n 2
because the left side has a bigger denominator. (In the notation of the Comparison Test, a n is the left side and bn is the right side.) We know that
n1
5 5 2 2n 2
n1
1 n2
is convergent ( p-series with p 2 1). Therefore
n1
5 2n 2 4n 3
is convergent by part (a) of the Comparison Test.
■
434
■
CHAPTER 8
SERIES
Although the condition a n bn or a n bn in the Comparison Test is given for all n, we need verify only that it holds for n N, where N is some fixed integer, because the convergence of a series is not affected by a finite number of terms. This is illustrated in the next example.
V EXAMPLE 4
Test the series
n1
ln n for convergence or divergence. n
SOLUTION We used the Integral Test to test this series in Example 1, but we can
also test it by comparing it with the harmonic series. Observe that ln n 1 for n 3 and so ln n 1 n3 n n We know that 1n is divergent ( p-series with p 1). Thus the given series is divergent by the Comparison Test.
■
NOTE The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesn’t apply. Consider, for instance, the series
n1
1 2 1 n
The inequality 1 1 n 2 1 2 n
is useless as far as the Comparison Test is concerned because bn ( 12 ) is convergent and a n bn . Nonetheless, we have the feeling that 12 n 1 ought to be n convergent because it is very similar to the convergent geometric series ( 12 ) . In such cases the following test can be used. n
THE LIMIT COMPARISON TEST Suppose that a n and bn are series with
positive terms. If ■ Exercises 38 and 39 deal with the cases c 0 and c .
lim
nl
an c bn
where c is a finite number and c 0, then either both series converge or both diverge. PROOF Let m and M be positive numbers such that m c M. Because a n bn is close to c for large n, there is an integer N such that
m and so
an M bn
mbn a n Mbn
when n N when n N
If bn converges, so does Mbn . Thus a n converges by part (i) of the Comparison Test. If bn diverges, so does mbn and part (ii) of the Comparison Test shows ■ that a n diverges.
SECTION 8.3
EXAMPLE 5 Test the series
n1
THE INTEGRAL AND COMPARISON TESTS
■
435
1 for convergence or divergence. 2n 1
SOLUTION We use the Limit Comparison Test with
an
1 2n 1
bn
1 2n
and obtain lim
nl
an 12 n 1 2n 1 lim lim n lim 10 n nl nl 2 1 n l 1 12 n bn 12
Since this limit exists and 12 n is a convergent geometric series, the given series converges by the Limit Comparison Test. ■ PROOF OF THE INTEGRAL TEST y
We have already seen the basic idea behind the proof of the Integral Test in Figures 1 and 2 for the series 1n 2 and 1sn . For the general series a n look at Figures 3 and 4. The area of the first shaded rectangle in Figure 3 is the value of f at the right endpoint of 1, 2 , that is, f 2 a 2. So, comparing the areas of the shaded rectangles with the area under y f x from 1 to n, we see that
y=ƒ
a™ a£ a¢ a∞ 0
1
2
3
4
an 5 ...
n x
a 2 a 3 a n y f x dx n
3
1
FIGURE 3 y
(Notice that this inequality depends on the fact that f is decreasing.) Likewise, Figure 4 shows that y=ƒ
y
4
n
1
an-1
f x dx a 1 a 2 a n1
(i) If y f x dx is convergent, then (3) gives 1
a¡ a™ a£ a¢ 0
1
2
3
4
n
5 ...
a
n x
i2
FIGURE 4
y f x dx y f x dx n
i
1
1
since f x 0. Therefore n
sn a 1
a
i2
i
a 1 y f x dx M 1
where M is a constant. Since sn M for all n, the sequence sn is bounded above. Also sn1 sn a n1 sn since a n1 f n 1 0. Thus sn is an increasing bounded sequence and so it is convergent by the Monotonic Sequence Theorem (8.1.11). This means that a n is convergent. (ii) If x1 f x dx is divergent, then x1n f x dx l as n l because f x 0. But (4) gives
y
n
1
n1
f x dx
a
i
sn1
i1
and so sn1 l . This implies that sn l and so a n diverges.
■
436
■
CHAPTER 8
SERIES
8.3
EXERCISES
1. Draw a picture to show that 13.
n2
15.
17.
for x 1 and an f n. By drawing a picture, rank the following three quantities in increasing order:
y
1
5
6
ai
i1
19.
ai
5. It is important to distinguish between
b
n1 ■
■
7.
n1 ■
■
■
■
8.
n1 ■
n1 ■
10.
■
■
■
n2
■
■
■
26.
■
■
■
■
n2
1 n n1 2
1 n
n1
■
n1
1 1 1 1 8 27 64 125 5 4 n4 nsn
n2 1 3n 4 1 4 3n 2n 1 sn 3 1
■
■
■
n1
■
■
n0
n1 ■
■
■
■
1 2n 3 1 sin n 10 n n5 3 n 7 n2 s
■
■
■
■
1 nln n p ■
28.
ln n np
n1 ■
■
■
■
■
■
■
■
R n s sn a n1 a n2 a n3
■
Thus R n is the error made when sn , the sum of the first n terms, is used as an approximation to the total sum s. (a) By comparing areas in a diagram like Figures 3 and 4 (but with x n ), show that
y
n1
f x dx R n y f x dx n
(b) Deduce from part (a) that ■
■
■
■
Find the values of p for which the series is convergent.
convergent by the Integral Test and let f x be the function in that test. The remainder after n terms is
sn n1 ■
■
29. Let s be the sum of a series a n that has been shown to be
1 2 n 1
■ Determine whether the series is convergent or divergent.
11–26
12.
n1
sn y
n1
11. 1
n1
sin
9–10 ■ Use the Comparison Test to determine whether the series is convergent or divergent. 9.
24.
n1
27.
■ Use the Integral Test to determine whether the series is convergent or divergent.
6.
25.
n
6–8
1 n4
2 1 n n sn
27–28
n1
22.
23.
What name is given to the first series? To the second? For what values of b does the first series converge? For what values of b does the second series converge?
1 4 sn
1 sn 2 1
(a) If a n bn for all n, what can you say about an ? Why? (b) If a n bn for all n, what can you say about an ? Why?
20.
n1
bn is known to be divergent.
and
n1 n4n
n1
4. Suppose a n and bn are series with positive terms and
b
21.
(a) If a n bn for all n, what can you say about a n? Why? (b) If a n bn for all n, what can you say about a n? Why?
n
18.
n2 n 1 3
i2
bn is known to be convergent.
n1
cos 2 n n2 1
n1
3. Suppose a n and bn are series with positive terms and
n1
n1
16.
2. Suppose f is a continuous positive decreasing function
f x dx
14.
1 n ln n
n2
What can you conclude about the series?
6
n
n1
1 1 y 1.3 dx 1 x n 1.3
ne
f x dx s sn y f x dx n
30. (a) Find the partial sum s10 of the series n1 1n 4. Use
Exercise 29(a) to estimate the error in using s10 as an approximation to the sum of the series. (b) Use Exercise 29(b) with n 10 to give an improved estimate of the sum. (c) Find a value of n so that s n is within 0.00001 of the sum.
■
SECTION 8.4
31. (a) Use the sum of the first 10 terms and Exercise 29(a) to
estimate the sum of the series n1 1n 2. How good is this estimate? (b) Improve this estimate using Exercise 29(b) with n 10. (c) Find a value of n that will ensure that the error in the approximation s sn is less than 0.001.
38. (a) Suppose that an and bn are series with positive terms
and bn is convergent. Prove that if lim
nl
tial sum of the harmonic series, then
39. (a) Suppose that an and bn are series with positive terms
and bn is divergent. Prove that if lim
nl
n1 n1.001 so that the error is less than 5 in the ninth decimal place, then we need to add more than 10 11,301 terms!
40. Give an example of a pair of series a n and bn with posi-
tive terms where lim n l a nbn 0 and bn diverges, but a n converges. [Compare with Exercise 38.]
41. Prove that if a n 0 and a n converges, then a n2 also
Show that this series always converges. 36.
converges.
Show that if a n 0 and a n is convergent, then ln1 a n is convergent.
8.4
an bn
then an is also divergent. (b) Use part (a) to show that the series diverges. 1 ln n (ii) (i) ln n n n2 n1
34. Show that if we want to approximate the sum of the series
0.d1 d2 d3 . . . (where the digit d i is one of the numbers 0, 1, 2, . . . , 9) is that d1 d2 d3 d4 0.d1 d2 d3 d4 . . . 10 10 2 10 3 10 4
an 0 bn
then an is also convergent. (b) Use part (a) to show that the series converges. ln n ln n (ii) (i) 3 n n1 n n1 sn e
33. (a) Use a graph of y 1x to show that if sn is the nth par-
35. The meaning of the decimal representation of a number
437
that sina n is also convergent?
places.
(b) The harmonic series diverges, but very slowly. Use part (a) to show that the sum of the first million terms is less than 15 and the sum of the first billion terms is less than 22.
■
37. If a n is a convergent series with positive terms, is it true
32. Find the sum of the series n1 1n 5 correct to three decimal
sn 1 ln n
OTHER CONVERGENCE TESTS
42. Find all positive values of b for which the series n1 b ln n
converges.
OTHER CONVERGENCE TESTS The convergence tests that we have looked at so far apply only to series with positive terms. In this section we learn how to deal with series whose terms are not necessarily positive. ALTERNATING SERIES
An alternating series is a series whose terms are alternately positive and negative. Here are two examples: 1
1 1 1 1 1 1 1n1 2 3 4 5 6 n n1
1 2 3 4 5 6 n 1n 2 3 4 5 6 7 n1 n1
We see from these examples that the nth term of an alternating series is of the form a n 1n1bn
a n 1nbn
or
where bn is a positive number. (In fact, bn a n .)
438
■
CHAPTER 8
SERIES
The following test says that if the terms of an alternating series decrease to 0 in absolute value, then the series converges. THE ALTERNATING SERIES TEST If the alternating series
1
bn b1 b2 b3 b4 b5 b6
n1
bn 0
n1
satisfies (i) bn1 bn (ii)
for all n
lim bn 0
nl
then the series is convergent. Before giving the proof let’s look at Figure 1, which gives a picture of the idea behind the proof. We first plot s1 b1 on a number line. To find s2 we subtract b2 , so s2 is to the left of s1 . Then to find s3 we add b3 , so s3 is to the right of s2 . But, since b3 b2 , s3 is to the left of s1 . Continuing in this manner, we see that the partial sums oscillate back and forth. Since bn l 0, the successive steps are becoming smaller and smaller. The even partial sums s2 , s4 , s6 , . . . are increasing and the odd partial sums s1 , s3 , s5 , . . . are decreasing. Thus it seems plausible that both are converging to some number s, which is the sum of the series. Therefore, in the following proof we consider the even and odd partial sums separately. b¡ -b™ +b£ -b¢ +b∞ -bß FIGURE 1
s™
0
s¢
sß
s
s∞
s£
s¡
PROOF OF THE ALTERNATING SERIES TEST We first consider the even partial sums:
In general
s2 b1 b2 0
since b2 b1
s4 s2 b3 b4 s2
since b4 b 3
s2n s2n2 b2n1 b2n s2n2
since b2n b2n1
0 s2 s4 s6 s2n
Thus But we can also write
s2n b1 b2 b3 b4 b5 b2n2 b2n1 b2n Every term in brackets is positive, so s2n b1 for all n. Therefore, the sequence s2n of even partial sums is increasing and bounded above. It is therefore convergent by the Monotonic Sequence Theorem. Let’s call its limit s, that is, lim s2n s
nl
SECTION 8.4
OTHER CONVERGENCE TESTS
■
439
Now we compute the limit of the odd partial sums: lim s2n1 lim s2n b2n1
nl
nl
lim s2n lim b2n1 nl
nl
s0
[by condition (ii)]
s Since both the even and odd partial sums converge to s, we have lim n l sn s (see Exercise 46 in Section 8.1) and so the series is convergent. ■ V EXAMPLE 1
■ Figure 2 illustrates Example 1 by showing the graphs of the terms a n 1 n1n and the partial sums sn . Notice how the values of sn zigzag across the limiting value, which appears to be about 0.7. In fact, it can be proved that the exact sum of the series is ln 2 0.693.
The alternating harmonic series 1
1 1 1 1n1 2 3 4 n n1
satisfies
1
sn
(i) bn1 bn
because
(ii) lim bn lim
1 0 n
nl
nl
1 1 n1 n
■
so the series is convergent by the Alternating Series Test.
V EXAMPLE 2
a n 0
The series
1n 3n is alternating, but 4n 1
n1
n
lim bn lim
nl
FIGURE 2
nl
3n lim nl 4n 1
3 4
1 n
3 4
so condition (ii) is not satisfied. Instead, we look at the limit of the nth term of the series: lim a n lim
nl
nl
1n 3n 4n 1
This limit does not exist, so the series diverges by the Test for Divergence.
EXAMPLE 3 Test the series
1n1
n1
n2 for convergence or divergence. n3 1
SOLUTION The given series is alternating so we try to verify conditions (i) and (ii) of the Alternating Series Test. Unlike the situation in Example 1, it is not obvious that the sequence given by bn n 2n 3 1 is decreasing. However, if we consider the related function f x x 2x 3 1, we find that
f x
x2 x 3 x 3 12
■
440
■
CHAPTER 8
SERIES
Instead of verifying condition (i) of the Alternating Series Test by computing a derivative, we could verify that bn1 bn directly by using the technique of Example 11 in Section 8.1. ■
Since we are considering only positive x, we see that f x 0 if 2 x 3 0, 3 3 that is, x s 2 . Thus f is decreasing on the interval (s 2 , ). This means that f n 1 f n and therefore bn1 bn when n 2. (The inequality b2 b1 can be verified directly but all that really matters is that the sequence bn is eventually decreasing.) Condition (ii) is readily verified: n2 lim bn lim 3 lim nl nl n 1 nl
1 n 1
1 n3
0
Thus the given series is convergent by the Alternating Series Test.
■
A partial sum sn of any convergent series can be used as an approximation to the total sum s, but this is not of much use unless we can estimate the accuracy of the approximation. The error involved in using s sn is the remainder Rn s sn . The next theorem says that for series that satisfy the conditions of the Alternating Series Test, the size of the error is smaller than bn1 , which is the absolute value of the first neglected term.
You can see geometrically why the Alternating Series Estimation Theorem is true by looking at Figure 1 (on page 438). Notice that s s4 b5 , s s5 b6 , and so on. Notice also that s lies between any two consecutive partial sums. ■
ALTERNATING SERIES ESTIMATION THEOREM If s
1n1bn is the sum of
an alternating series that satisfies (i) 0 bn1 bn
(ii) lim bn 0
and
nl
R s s b
then
n
n
n1
PROOF We know from the proof of the Alternating Series Test that s lies between any two consecutive partial sums sn and sn1 . It follows that
s s s n
n1
sn bn1
Find the sum of the series n0 places. (By definition, 0! 1.) V EXAMPLE 4
1n correct to three decimal n!
SOLUTION We first observe that the series is convergent by the Alternating Series
Test because (i) bn1 (ii) 0
1 1 1 bn n 1! n!n 1 n!
1 1 l0 n! n
so bn
1 l 0 as n l n!
To get a feel for how many terms we need to use in our approximation, let’s write out the first few terms of the series: s
1 1 1 1 1 1 1 1 0! 1! 2! 3! 4! 5! 6! 7!
1 1 1 1 1 12 16 241 120 720 5040
■
SECTION 8.4
OTHER CONVERGENCE TESTS
■
441
1 1 b 7 5040 5000 0.0002
Notice that
1 1 s6 1 1 12 16 241 120 720 0.368056
and
By the Alternating Series Estimation Theorem we know that
s s b 6
7
0.0002
This error of less than 0.0002 does not affect the third decimal place, so we have
In Section 8.7 we will prove that e x n0 x nn! for all x, so what we have obtained in Example 4 is actually an approximation to the number e 1 . ■
s 0.368 ■
correct to three decimal places. |
NOTE The rule that the error (in using sn to approximate s) is smaller than the first neglected term is, in general, valid only for alternating series that satisfy the conditions of the Alternating Series Estimation Theorem. The rule does not apply to other types of series. ABSOLUTE CONVERGENCE
Given any series a n , we can consider the corresponding series
a a a a n
1
2
3
n1
whose terms are the absolute values of the terms of the original series. ■ We have convergence tests for series with positive terms and for alternating series. But what if the signs of the terms switch back and forth irregularly? We will see in Example 7 that the idea of absolute convergence sometimes helps in such cases.
DEFINITION A series a n is called absolutely convergent if the series of
absolute values a n is convergent.
Notice that if a n is a series with positive terms, then a n a n and so absolute convergence is the same as convergence.
EXAMPLE 5 The series
n1
1n1 1 1 1 1 2 2 2 n2 2 3 4
is absolutely convergent because
n1
1n1 1 1 1 1 2 2 1 2 2 n 2 3 42 n1 n
is a convergent p-series ( p 2). EXAMPLE 6 We know that the alternating harmonic series
n1
1n1 1 1 1 1 n 2 3 4
■
442
■
CHAPTER 8
SERIES
is convergent (see Example 1), but it is not absolutely convergent because the corresponding series of absolute values is
n1
1n1 1 1 1 1 1 n 2 3 4 n1 n
which is the harmonic series ( p-series with p 1) and is therefore divergent.
■
DEFINITION A series a n is called conditionally convergent if it is convergent but not absolutely convergent. ■ It can be proved that if the terms of an absolutely convergent series are rearranged in a different order, then the sum is unchanged. But if a conditionally convergent series is rearranged, the sum could be different.
Example 6 shows that the alternating harmonic series is conditionally convergent. Thus it is possible for a series to be convergent but not absolutely convergent. However, the next theorem shows that absolute convergence implies convergence. 1 THEOREM
If a series a n is absolutely convergent, then it is convergent.
PROOF Observe that the inequality
0 an an 2 an
is true because a n is either a n or a n . If a n is absolutely convergent, then a n is convergent, so 2 a n is convergent. Therefore, by the Comparison Test, (a n a n ) is convergent. Then
a
n
) a
(a n a n
n
■
is the difference of two convergent series and is therefore convergent. ■ Figure 3 shows the graphs of the terms a n and partial sums sn of the series in Example 7. Notice that the series is not alternating but has positive and negative terms.
V EXAMPLE 7
Determine whether the series
n1
cos n cos 1 cos 2 cos 3 n2 12 22 32
is convergent or divergent. 0.5
SOLUTION This series has both positive and negative terms, but it is not alternating. (The first term is positive, the next three are negative, and the following three are positive. The signs change irregularly.) We can apply the Comparison Test to the series of absolute values
sn
a n 0
FIGURE 3
n
n1
cos n cos n 2 n n2 n1
Since cos n 1 for all n, we have
cos n n2
1 n2
We know that 1n 2 is convergent ( p-series with p 2) and therefore cos n n 2 is convergent by the Comparison Test. Thus the given series cos nn 2 is ■ absolutely convergent and therefore convergent by Theorem 1.
SECTION 8.4
OTHER CONVERGENCE TESTS
■
443
THE RATIO TEST
The following test is very useful in determining whether a given series is absolutely convergent. THE RATIO TEST
a n1 L 1, then the series a n is absolutely convergent nl an n1 (and therefore convergent).
(i) If lim
a n1 a n1 L 1 or lim , then the series a n nl nl an an n1 is divergent.
(ii) If lim
a n1 1, the Ratio Test is inconclusive; that is, no conclusion an can be drawn about the convergence or divergence of a n .
(iii) If lim
nl
PROOF
(i) The idea is to compare the given series with a convergent geometric series. Since L 1, we can choose a number r such that L r 1. Since lim
nl
a n1 L an
Lr
and
the ratio a n1a n will eventually be less than r ; that is, there exists an integer N such that
a n1 r an
whenever n N
or, equivalently,
a a r
2
n1
whenever n N
n
Putting n successively equal to N , N 1, N 2, . . . in (2), we obtain
a a r a a r a r a a r a r N1
N
N2
N1
N
N3
N2
N
2
3
and, in general,
a a r
3
Nk
N
k
for all k 1
Now the series
a r N
k1
k
aN r aN r 2 aN r 3
444
■
CHAPTER 8
SERIES
is convergent because it is a geometric series with 0 r 1. So the inequality (3), together with the Comparison Test, shows that the series
a a a a
an
nN1
Nk
N1
N2
N3
k1
is also convergent. It follows that the series n1 a n is convergent. (Recall that a finite number of terms doesn’t affect convergence.) Therefore, a n is absolutely convergent. (ii) If a n1a n l L 1 or a n1a n l , then the ratio a n1a n will eventually be greater than 1; that is, there exists an integer N such that
a n1 1 an
whenever n N
This means that a n1 a n whenever n N and so lim a n 0
nl
Therefore, a n diverges by the Test for Divergence.
■
NOTE Part (iii) of the Ratio Test says that if lim n l a n1a n 1, the test gives no information. For instance, for the convergent series 1n 2 we have
an1 an
1 n 12 n2 1 n 12 n2
1
l1
2
1 1 n
as n l
whereas for the divergent series 1n we have
a n1 an
1 n1 n 1 l1 1 n1 1 1 n n
as n l
Therefore, if lim n l a n1a n 1, the series a n might converge or it might diverge. In this case the Ratio Test fails and we must use some other test.
EXAMPLE 8 Test the series
1n
n1
Series that involve factorials or other products (including a constant raised to the nth power) are often conveniently tested using the Ratio Test. ■
n3 for absolute convergence. 3n
SOLUTION We use the Ratio Test with a n 1nn 33 n:
an1 an
1n1n 13 3 n1 n 13 3 n 3 n 3 1 n 3 n1 n n 3
|
1 3
|
n1 n
3
1 3
1
1 n
3
l
1 1 3
Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent.
■
SECTION 8.4
OTHER CONVERGENCE TESTS
Test the convergence of the series
V EXAMPLE 9
n1
■
445
nn . n!
SOLUTION Since the terms a n n n! are positive, we don’t need the absolute n
value signs.
■
a n1 n 1n1 n! n 1n 1n n! n n an n 1! n n 1n! n
We know that lim 1 x 1x e
xl0
by the definition of e. If we let n 1x, then n l as x l 0 and so lim 1 1n n e
nl
n1 n
n
1 n
1
n
le
as n l
Since e 1, the given series is divergent by the Ratio Test.
■
The following test is convenient to apply when nth powers occur. Its proof is similar to the proof of the Ratio Test and is left as Exercise 43.
THE ROOT TEST
n (i) If lim s a n L 1, then the series
nl
a
n
is absolutely convergent
n1
(and therefore convergent).
n n (ii) If lim s a n L 1 or lim s a n , then the series
nl
nl
a
n
is
n1
divergent.
n (iii) If lim s a n 1, the Root Test is inconclusive.
nl
n If lim n l s a n 1, then part (iii) of the Root Test says that the test gives no information. The series a n could converge or diverge. (If L 1 in the Ratio Test, don’t try the Root Test because L will again be 1.)
V EXAMPLE 10
Test the convergence of the series
n1
SOLUTION www.stewartcalculus.com We now have several tests for convergence of series. So, given a series, how do you know which test to use? For advice, click on Additional Topics and then on Strategy for Testing Series. ■
an
n an s
2n 3 3n 2
2n 3 3n 2
n
.
n
3 2n 3 n 2 l 1 3n 2 2 3 3 n 2
Thus the given series converges by the Root Test.
■
446
■
CHAPTER 8
SERIES
8.4
EXERCISES
1. (a) What is an alternating series?
18. For what values of p is the following series convergent?
(b) Under what conditions does an alternating series converge? (c) If these conditions are satisfied, what can you say about the remainder after n terms? 2. What can you say about the series a n in each of the
following cases? (a) lim
nl
(c) lim
nl
3– 8 3.
a n1 8 an
(b) lim
nl
■ Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
a n1 0.8 an
19.
a n1 1 an
21.
1
n1
7.
2
3
23.
1n
■
■
25.
sn 6. 1 1 2sn n1 n
■
8.
1 n1
n1 ■
■
■
■
■
27.
29. ■
■
Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?
9–12
■
9.
n1
10.
n1
11.
n1
31.
n1
n1
15.
n1 ■
■
1n1 4 n s
24.
10 n n 142n1
26.
cosn3 n!
28.
1 arctan n n2
30.
n 3
( error 0.0001)
35. 1
1 n1 n6
( error 0.00005)
n1
ne
n
■
■
( error 0.01) ■
■
■
■
■
■
■
1 n1 n5
14.
1 n1 n 2 10 n
16.
■
n1
■
n1 ■
■
■
■
n n2 1
1
n1
n1
n1
n1
n2
32.
13n
n 1 2n 2 1 2
n2
n
34.
n1
2n n4
sin 4n 4n 1 n1 5 n1 n 1 2 4 n2 1n ln nn 1n n ln n 1n arctan nn
135 1357 13 3! 5! 7! 1 3 5 2n 1 1 n1 2n 1!
37.
2 4 6 2n n!
1
n
n1 ■
■
2 n n! 5 8 11 3n 2
■
■
■
■
■
■
■
■
■
■
39. For which of the following series is the Ratio Test inconclu-
sive (that is, it fails to give a definite answer)? ■
■
■
(a)
n1
17. Is the 50th partial sum s50 of the alternating series
n1 1 n1n
n
n1
n
1 n 3 n n! ■
1
2 26 2 6 10 2 6 10 14 5 58 5 8 11 5 8 11 14
38.
1 n n 8n
n2 2n
36.
■
n1
n
1 n n5 n
n1
13–16 ■ Approximate the sum of the series correct to four decimal places.
22.
n1
13.
10 n n!
( error 0.01)
1 ■
n1
20.
33.
n1 ■
3 n n3
2n n!
12.
n1
ln n n ■
n1
3n 1 2n 1
n1
5
1 n1 sn
n1 ■
4
n0
48 49 104 114
4 7
n1
4. 3 4 5 6 7 5.
n1
1 n1 np
19–38
Test the series for convergence or divergence.
■
an overestimate or an underestimate of the total sum? Explain.
(c)
n1
1 n3
(b)
n1
3 sn
n1
(d)
n1
n 2n sn 1 n2
SECTION 8.5
40. For which positive integers k is the following series
convergent?
n1
n! kn!
(b) Deduce that lim n l x nn! 0 for all x.
42. Around 1910, the Indian mathematician Srinivasa
8.5
n0
447
43. Prove the Root Test.
[Hint for part (i): Take any number r such that L r 1 and use the fact that there is an integer n N such that s a n r whenever n N .]
Ramanujan discovered the formula
■
William Gosper used this series in 1985 to compute the first 17 million digits of . (a) Verify that the series is convergent. (b) How many correct decimal places of do you get if you use just the first term of the series? What if you use two terms?
2
41. (a) Show that n0 x nn! converges for all x.
1 2 s2 9801
POWER SERIES
4n!1103 26390n n! 4 396 4n
POWER SERIES A power series is a series of the form
cx
1
n
n
c0 c1 x c2 x 2 c3 x 3
n0
where x is a variable and the cn’s are constants called the coefficients of the series. For each fixed x, the series (1) is a series of constants that we can test for convergence or divergence. A power series may converge for some values of x and diverge for other values of x. The sum of the series is a function f x c0 c1 x c2 x 2 cn x n
Trigonometric series A power series is a series in which each term is a power function. A trigonometric series ■
x
n
1 x x2 xn
n0
a
whose domain is the set of all x for which the series converges. Notice that f resembles a polynomial. The only difference is that f has infinitely many terms. For instance, if we take cn 1 for all n, the power series becomes the geometric series
n
cos nx bn sin nx
n0
is a series whose terms are trigonometric functions.This type of series is discussed on the website
which converges when 1 x 1 and diverges when x 1 (see Equation 8.2.5). More generally, a series of the form
c x a
2
n
www.stewartcalculus.com
Click on Additional Topics and then on Fourier Series.
n
c0 c1x a c2x a2
n0
is called a power series in x a or a power series centered at a or a power series about a. Notice that in writing out the term corresponding to n 0 in Equations 1 and 2 we have adopted the convention that x a0 1 even when x a. Notice also that when x a all of the terms are 0 for n 1 and so the power series (2) always converges when x a.
V EXAMPLE 1
For what values of x is the series
n!x
n
convergent?
n0
SOLUTION We use the Ratio Test. If we let a n , as usual, denote the nth term of the
series, then a n n!x n. If x 0, we have Notice that n 1! n 1nn 1 . . . 3 2 1
■
n 1n!
lim
nl
a n1 n 1! x n1 lim lim n 1 x nl nl an n! x n
448
■
CHAPTER 8
SERIES
By the Ratio Test, the series diverges when x 0. Thus the given series converges only when x 0. ■
V EXAMPLE 2
For what values of x does the series
n1
SOLUTION Let a n x 3 n. Then
x 3n converge? n
n
a n1 x 3 n1 n an n1 x 3 n
1 1 n
1
x 3
l x3
as n l
By the Ratio Test, the given series is absolutely convergent, and therefore convergent, when x 3 1 and divergent when x 3 1. Now
x 3 1
&? 1 x 3 1 &? 2 x 4
so the series converges when 2 x 4 and diverges when x 2 or x 4. The Ratio Test gives no information when x 3 1 so we must consider x 2 and x 4 separately. If we put x 4 in the series, it becomes 1n, the harmonic series, which is divergent. If x 2, the series is 1 nn , which converges by the Alternating Series Test. Thus the given power series converges for 2 x 4. ■
National Film Board of Canada
We will see that the main use of a power series is that it provides a way to represent some of the most important functions that arise in mathematics, physics, and chemistry. In particular, the sum of the power series in the next example is called a Bessel function, after the German astronomer Friedrich Bessel (1784 –1846), and the function given in Exercise 23 is another example of a Bessel function. In fact, these functions first arose when Bessel solved Kepler’s equation for describing planetary motion. Since that time, these functions have been applied in many different physical situations, including the temperature distribution in a circular plate and the shape of a vibrating drumhead. EXAMPLE 3 Find the domain of the Bessel function of order 0 defined by
J0x
n0
1 n x 2n 2 2nn!2
SOLUTION Let a n 1 n x 2n 2 2nn!2 . Then Notice how closely the computergenerated model (which involves Bessel functions and cosine functions) matches the photograph of a vibrating rubber membrane.
a n1 1 n1x 2n1 2 2nn!2 2n1 2 an 2 n 1! 1 nx 2n
2
x 2n2 2 2nn!2 2 2 n 1 n! x 2n
2n2
x2 l 01 4n 12
for all x
Thus, by the Ratio Test, the given series converges for all values of x. In other words, the domain of the Bessel function J0 is , ⺢.
■
SECTION 8.5
y
s¸
n
J0x lim snx x
1
s¡ s£
s0x 1
FIGURE 1
y 1
y=J¸(x)
_10
10 0
449
x
i0
1ix 2i 2 2ii!2
The first few partial sums are
J¸
Partial sums of the Bessel function J¸
snx
where
nl
s¢ 0
■
Recall that the sum of a series is equal to the limit of the sequence of partial sums. So when we define the Bessel function in Example 3 as the sum of a series we mean that, for every real number x,
s™ 1
POWER SERIES
s3x 1
s1x 1
x4 x6 x2 4 64 2304
x2 4
s2x 1
s4x 1
x2 x4 4 64
x4 x6 x8 x2 4 64 2304 147,456
Figure 1 shows the graphs of these partial sums, which are polynomials. They are all approximations to the function J0 , but notice that the approximations become better when more terms are included. Figure 2 shows a more complete graph of the Bessel function. For the power series that we have looked at so far, the set of values of x for which the series is convergent has always turned out to be an interval [a finite interval for the geometric series and the series in Example 2, the infinite interval , in Example 3, and a collapsed interval 0, 0 0 in Example 1]. The following theorem, proved in Appendix B, says that this is true in general.
FIGURE 2
For a given power series
c x a
n
there are only three possibilities: (i) The series converges only when x a. (ii) The series converges for all x. (iii) There is a positive number R such that the series converges if x a R and diverges if x a R. 3 THEOREM
n
n0
The number R in case (iii) is called the radius of convergence of the power series. By convention, the radius of convergence is R 0 in case (i) and R in case (ii). The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In case (i) the interval consists of just a single point a. In case (ii) the interval is , . In case (iii) note that the inequality x a R can be rewritten as a R x a R. When x is an endpoint of the interval, that is, x a R, anything can happen—the series might converge at one or both endpoints or it might diverge at both endpoints. Thus in case (iii) there are four possibilities for the interval of convergence:
a R, a R
a R, a R
a R, a R
The situation is illustrated in Figure 3. convergence for |x-a|
FIGURE 3
a
divergence for |x-a|>R
a+R
a R, a R
450
■
CHAPTER 8
SERIES
We summarize here the radius and interval of convergence for each of the examples already considered in this section. Series
Radius of convergence
Interval of convergence
R1
1, 1
n
R0
0
x 3n n
R1
2, 4
1 nx 2n 2 2nn!2
R
,
Geometric series
x
n
n0
Example 1
n! x
n0
Example 2
n1
Example 3
n0
The Ratio Test (or sometimes the Root Test) should be used to determine the radius of convergence R in most cases. The Ratio and Root Tests always fail when x is an endpoint of the interval of convergence, so the endpoints must be checked with some other test. EXAMPLE 4 Find the radius of convergence and interval of convergence of the series
n0
3 n x n sn 1
SOLUTION Let a n 3 n x nsn 1. Then
a n1 3 n1x n1 sn 1 3x an 3 nx n sn 2
3
1 1n x l 3 x 1 2n
n1 n2
as n l
By the Ratio Test, the given series converges if 3 x 1 and diverges if 3 x 1. Thus it converges if x 13 and diverges if x 13 . This means that the radius of convergence is R 13 . We know the series converges in the interval ( 13 , 13 ), but we must now test for convergence at the endpoints of this interval. If x 13 , the series becomes
n0
n
3 n (13 ) 1 1 1 1 1 1 1 sn s1 s2 s3 s4 n0 sn
which diverges. (Use the Integral Test or simply observe that it is a p-series with p 12 1.) If x 13 , the series is
n0
n
3 n ( 13 ) 1 n sn 1 n0 sn 1
which converges by the Alternating Series Test. Therefore, the given power series ■ converges when 13 x 13 , so the interval of convergence is (13 , 13 ].
SECTION 8.5
V EXAMPLE 5
POWER SERIES
■
451
Find the radius of convergence and interval of convergence of the
series
n0
nx 2 n 3 n1
SOLUTION If a n nx 2 n3 n1, then
a n1 n 1x 2 n1 3 n1 n2 an 3 nx 2 n 1 n
1
x2 3
x 2
l
as n l
3
Using the Ratio Test, we see that the series converges if x 2 3 1 and it diverges if x 2 3 1. So it converges if x 2 3 and diverges if x 2 3. Thus the radius of convergence is R 3. The inequality x 2 3 can be written as 5 x 1, so we test the series at the endpoints 5 and 1. When x 5, the series is
n3 n 13 1 n n n1 3 n0
n0
which diverges by the Test for Divergence [1nn doesn’t converge to 0]. When x 1, the series is
n0
n3 n 1 n1 3 n 3 n0
which also diverges by the Test for Divergence. Thus the series converges only when 5 x 1, so the interval of convergence is 5, 1.
8.5
EXERCISES
1. What is a power series? 2. (a) What is the radius of convergence of a power series?
How do you find it? (b) What is the interval of convergence of a power series? How do you find it? Find the radius of convergence and interval of convergence of the series.
3–18
11.
3.
n1
5.
n1
7.
n0
9.
n1
4.
1 n1 x n n3
6.
xn n!
8.
n0
n1
2 x 4 n s
n
10.
15.
n1
n
n!2x 1
n1
n1
17. ■
■
n
18.
n1
x 2n 2n!
2 n x 3 n sn nx 4 n n3 1 n2xn 2 4 6 2n
■
■
■
■
■
■
■
■
■
19. If n0 cn 4 n is convergent, does it follow that the following
series are convergent?
n
5 n
16.
n
n0
n1
xn n3 n x
n x a n , b 0 bn
n
1
n
n1
14.
n
n1
sn x
n
1 nx n n1
x 2 n n2 n
1
n1
xn sn
12.
n
13.
xn 4 ln n
1
n2
■
■
5
(a)
c 2 n
n0
n
(b)
c 4 n
n0
n
■
452
■
CHAPTER 8
SERIES
20. Suppose that n0 cn x n converges when x 4 and
24. The function A defined by
diverges when x 6. What can be said about the convergence or divergence of the following series?
(a)
c
(b)
n
n0
c 3 n
c8 n
n
n0
(c)
n
(d)
n0
1 c 9 n
n
n
;
n0
21. If k is a positive integer, find the radius of convergence of
the series
n0
n! k n x kn!
n0 x n, together with the sum function
f x 11 x, on a common screen. On what interval do these partial sums appear to be converging to f x?
n0
CAS
f x 1 2x x 2 2x 3 x 4 that is, its coefficients are c2n 1 and c2n1 2 for all n 0. Find the interval of convergence of the series and find an explicit formula for f x. 26. If f x
n0 cn x n, where cn4 cn for all n 0, find the
interval of convergence of the series and a formula for f x.
radius of convergence of the power series cn x n is R 1c.
1 n x 2n1 n!n 1! 2 2n1
28. Suppose that the power series cn x a n satisfies c n 0
for all n. Show that if lim n l cn cn1 exists, then it is equal to the radius of convergence of the power series.
is called the Bessel function of order 1. (a) Find its domain. (b) Graph the first several partial sums on a common screen. (c) If your CAS has built-in Bessel functions, graph J1 on the same screen as the partial sums in part (b) and observe how the partial sums approximate J1.
8.6
is called the Airy function after the English mathematician and astronomer Sir George Airy (1801–1892). (a) Find the domain of the Airy function. (b) Graph the first several partial sums snx on a common screen. (c) If your CAS has built-in Airy functions, graph A on the same screen as the partial sums in part (b) and observe how the partial sums approximate A.
n 27. Show that if lim n l s cn c , where c 0, then the
23. The function J1 defined by
J1x
CAS
x6 x9 x3 2 3 2 3 5 6 2 3 5 6 8 9
25. A function f is defined by
; 22. Graph the first several partial sums snx of the series
;
Ax 1
29. Suppose the series cn x n has radius of convergence 2 and
the series dn x n has radius of convergence 3. What is the radius of convergence of the series cn dnx n ?
30. Suppose that the radius of convergence of the power series
cn x n is R. What is the radius of convergence of the power series cn x 2n ?
REPRESENTING FUNCTIONS AS POWER SERIES In this section we learn how to represent certain types of functions as sums of power series by manipulating geometric series or by differentiating or integrating such a series. You might wonder why we would ever want to express a known function as a sum of infinitely many terms. This strategy is useful for integrating functions that don’t have elementary antiderivatives, for solving differential equations, and for approximating functions by polynomials. (Scientists do this to simplify the expressions they deal with; computer scientists do this to represent functions on calculators and computers.) We start with an equation that we have seen before: 1
1 1 x x2 x3 xn 1x n0
x 1
We first encountered this equation in Example 5 in Section 8.2, where we obtained it by observing that the series is a geometric series with a 1 and r x. But here our
SECTION 8.6
■ A geometric illustration of Equation 1 is shown in Figure 1. Because the sum of a series is the limit of the sequence of partial sums, we have
REPRESENTING FUNCTIONS AS POWER SERIES
■
453
point of view is different. We now regard Equation 1 as expressing the function f x 11 x as a sum of a power series. s¡¡
y
1 lim snx nl 1x
sˆ s∞ f
where snx 1 x x 2 x n is the nth partial sum. Notice that as n increases, snx becomes a better approximation to f x for 1 x 1.
s™
0
_1
FIGURE 1
x
1
Express 11 x 2 as the sum of a power series and find the interval of convergence. V EXAMPLE 1
SOLUTION Replacing x by x 2 in Equation 1, we have 1 1 x 2 n 1 x2 1 x 2 n0
1 x
n 2n
1 x2 x4 x6 x8
n0
Because this is a geometric series, it converges when x 2 1, that is, x 2 1, or x 1. Therefore, the interval of convergence is 1, 1. (Of course, we could have determined the radius of convergence by applying the Ratio Test, but that much work is unnecessary here.) ■
EXAMPLE 2 Find a power series representation for 1x 2. SOLUTION In order to put this function in the form of the left side of Equation 1 we first factor a 2 from the denominator:
1 2x
1
1
2 1 1 2
x 2
n0
2 1
x 2
n
n0
x 2
1n n x 2 n1
This series converges when x2 1, that is, x 2. So the interval of convergence is 2, 2. ■ EXAMPLE 3 Find a power series representation of x 3x 2. SOLUTION Since this function is just x 3 times the function in Example 2, all we
have to do is to multiply that series by x 3: It’s legitimate to move x 3 across the sigma sign because it doesn’t depend on n. [Use Theorem 8.2.8(i) with c x 3.] ■
x3 1 1 n 1 n x3 x 3 n1 x n n1 x n3 x2 x2 n0 2 n0 2
12 x 3 14 x 4 18 x 5 161 x 6
454
■
CHAPTER 8
SERIES
Another way of writing this series is as follows: x3 1 n1 n x x2 2 n2 n3
As in Example 2, the interval of convergence is 2, 2.
■
DIFFERENTIATION AND INTEGRATION OF POWER SERIES
The sum of a power series is a function f x n0 cnx a n whose domain is the interval of convergence of the series. We would like to be able to differentiate and integrate such functions, and the following theorem (which we won’t prove) says that we can do so by differentiating or integrating each individual term in the series, just as we would for a polynomial. This is called term-by-term differentiation and integration. If the power series cnx a n has radius of convergence R 0, then the function f defined by 2 THEOREM
c x a
f x c0 c1x a c2x a2
n
n
n0
is differentiable (and therefore continuous) on the interval a R, a R and (i) f x c1 2c2x a 3c3x a2
nc x a n
n1
n1
In part (ii), x c0 dx c0 x C1 is written as c0 x a C, where C C1 ac0 , so all the terms of the series have the same form. ■
(ii)
y f x dx C c x a c 0
C
1
c
n
n0
x a x a3 c2 2 3 2
x a n1 n1
The radii of convergence of the power series in Equations (i) and (ii) are both R. NOTE 1 Equations (i) and (ii) in Theorem 2 can be rewritten in the form
(iii)
y d dx
n0
www.stewartcalculus.com The idea of differentiating a power series term by term is the basis for a powerful method for solving differential equations. Click on Additional Topics and then on Using Series to Solve Differential Equations. ■
(iv)
n0
cnx a n
n0
cnx a n dx
d cnx a n dx
y c x a n
n
dx
n0
We know that, for finite sums, the derivative of a sum is the sum of the derivatives and the integral of a sum is the sum of the integrals. Equations (iii) and (iv) assert that the same is true for infinite sums, provided we are dealing with power series. (For other types of series of functions the situation is not as simple; see Exercise 36.) NOTE 2 Although Theorem 2 says that the radius of convergence remains the same when a power series is differentiated or integrated, this does not mean that the interval of convergence remains the same. It may happen that the original series converges at an endpoint, whereas the differentiated series diverges there. (See Exercise 37.)
SECTION 8.6
REPRESENTING FUNCTIONS AS POWER SERIES
■
455
EXAMPLE 4 In Example 3 in Section 8.5 we saw that the Bessel function
J0x
n0
1 n x 2n 2 2nn!2
is defined for all x. Thus by Theorem 2, J0 is differentiable for all x and its derivative is found by term-by-term differentiation as follows: J0x
d 1 nx 2n 1 n 2nx 2n1 2n 2 dx 2 n! 2 2nn!2 n1
n0
■
Express 11 x2 as a power series by differentiating Equation 1. What is the radius of convergence? V EXAMPLE 5
SOLUTION Differentiating each side of the equation 1 1 x x2 x3 xn 1x n0 1 2 1 2x 3x nx n1 1 x2 n1
we get
If we wish, we can replace n by n 1 and write the answer as 1 n 1x n 1 x2 n0
According to Theorem 2, the radius of convergence of the differentiated series is the same as the radius of convergence of the original series, namely, R 1. ■ EXAMPLE 6 Find a power series representation for ln1 x and its radius of
convergence. SOLUTION We notice that, except for a factor of 1, the derivative of this function
is 11 x. So we integrate both sides of Equation 1: ln1 x y
1 dx y 1 x x 2 dx 1x
x
n1
x2 x3 x n1 C C 2 3 n0 n 1
xn C n
x 1
To determine the value of C we put x 0 in this equation and obtain ln1 0 C. Thus C 0 and ln1 x x
x2 x3 xn 2 3 n1 n
x 1
The radius of convergence is the same as for the original series: R 1.
■
Notice what happens if we put x 12 in the result of Example 6. Since ln 12 ln 2, we see that 1 1 1 1 1 ln 2 n 2 8 24 64 n2 n1
456
■
CHAPTER 8
SERIES
V EXAMPLE 7
Find a power series representation for f x tan1x.
SOLUTION We observe that f x 11 x 2 and find the required series by
integrating the power series for 11 x 2 found in Example 1. tan1x y
1 dx y 1 x 2 x 4 x 6 dx 1 x2
Cx 1 ■ The power series for tan x obtained in Example 7 is called Gregory’s series after the Scottish mathematician James Gregory (1638–1675), who had anticipated some of Newton’s discoveries. We have shown that Gregory’s series is valid when 1 x 1, but it turns out (although it isn’t easy to prove) that it is also valid when x 1. Notice that when x 1 the series becomes
1 1 1 1 4 3 5 7 This beautiful result is known as the Leibniz formula for .
x3 x5 x7 3 5 7
To find C we put x 0 and obtain C tan1 0 0. Therefore tan1x x
x3 x5 x7 x 2n1 1 n 3 5 7 2n 1 n0
Since the radius of convergence of the series for 11 x 2 is 1, the radius of convergence of this series for tan1x is also 1. ■ EXAMPLE 8 Evaluate x 11 x 7 dx as a power series. SOLUTION The first step is to express the integrand, 11 x 7 , as the sum of a
power series. As in Example 1, we start with Equation 1 and replace x by x 7: 1 1 7 n 7 7 x 1x 1 x n0
1 x
n 7n
1 x 7 x 14
n0 ■ This example demonstrates one way in which power series representations are useful. Integrating 11 x 7 by hand is incredibly difficult. Different computer algebra systems return different forms of the answer, but they are all extremely complicated. (If you have a CAS, try it yourself.) The infinite series answer that we obtain in Example 8 is actually much easier to deal with than the finite answer provided by a CAS.
8.6
Now we integrate term by term:
y
1 x 7n1 n 7n n 7 dx y 1 x dx C 1 1x 7n 1 n0 n0
Cx
x8 x 15 x 22 8 15 22
This series converges for x 7 1, that is, for x 1.
■
EXERCISES
1. If the radius of convergence of the power series n0 cn x n
is 10, what is the radius of convergence of the series n1 ncn x n1 ? Why? 2. Suppose you know that the series n0 bn x n converges for
3–10 ■ Find a power series representation for the function and determine the interval of convergence. 3. f x
1 1x
4. f x
3 1 x4
5. f x
1 1 x3
6. f x
1 1 9x 2
x 2. What can you say about the following series? Why?
n0
bn x n1 n 1
SECTION 8.6
■
REPRESENTING FUNCTIONS AS POWER SERIES
7. f x
1 x5
8. f x
x 4x 1
23–26
9. f x
x 9 x2
10. f x
x2 a x3
23.
y 1t
25.
y
■
■
■
■
■
■
■
■
■
■
Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence.
11–12
457
■ Evaluate the indefinite integral as a power series. What is the radius of convergence?
3
■
■
t
8
dt
x tan 1 x dx x3
■
■
■
■
■
■
■
ln1 t dt t
24.
y
26.
y tan
■
1
■
x 2 dx
■
■
■
■
11. f x
3 x2 x 2
27–30 ■ Use a power series to approximate the definite integral to six decimal places.
12. f x
7x 1 3x 2x 1
27.
y
0.2
29.
y
0.1
■
■
2
■
■
■
■
■
■
■
■
■
for
■
1 f x 1 x2
1 1 x3
y
0.3
0 ■
■
■
■
■
■
■
■
■
■
■
■
■
■ Find a power series representation for f , and graph f and several partial sums snx on the same screen. What happens as n increases?
20. f x
1x 21. f x ln 1x ■
■
■
x 2 J0x x J0x x 2 J0x 0 (b) Evaluate x01 J0x dx correct to three decimal places. 34. The Bessel function of order 1 is defined by
■
■
■
J1x
1 n x 2n1 n!n 1!2 2n1
(b) Show that J0x J1x. 35. (a) Show that the function
f x
xn n!
is a solution of the differential equation
22. f x tan 2x ■
■
1 n x 2n 2n!
n0
1 x 25 2
1
■
x 2J1x x J1x x 2 1J1x 0
; 19–22
19. f x ln3 x
■
(a) Show that J1 satisfies the differential equation
18. f x arctanx3 ■
■
Example 4) satisfies the differential equation
n0
x2 1 2x2
3
■
■
33. (a) Show that J0 (the Bessel function of order 0 given in
16. f x
x x 22
0
x2 dx 1 x4
f x f x 0
■
17. f x
■
ln1 x 4 dx
is a solution of the differential equation
x2 1 x3
15. f x ln5 x
■
30.
n0
Find a power series representation for the function and determine the radius of convergence.
■
x arctan3x dx
f x
What is the radius of convergence? (b) Use part (a) to find a power series for f x x ln1 x. (c) Use part (a) to find a power series for f x lnx 2 1.
■
0.4
0
32. Show that the function
14. (a) Find a power series representation for f x ln1 x.
■
y
five decimal places.
(c) Use part (b) to find a power series for
15–18
28.
31. Use the result of Example 6 to compute ln 1.1 correct to
What is the radius of convergence? (b) Use part (a) to find a power series for
f x
1 dx 1 x5
■
13. (a) Use differentiation to find a power series representation
f x
0
■
f x f x ■
(b) Show that f x e x.
■
458
■
CHAPTER 8
SERIES
36. Let fnx sin nxn 2. Show that the series fnx
converges for all values of x but the series of derivatives fnx diverges when x 2n, n an integer. For what values of x does the series f nx converge?
(b) Find the sum of each of the following series. n (i) nx n, x 1 (ii) n n1 n1 2
(c) Find the sum of each of the following series.
37. Let
f x
n1
(i)
n
x n2
38. (a) Starting with the geometric series n0 x n, find the sum
of the series
n1
8.7
n1
n
n2
(ii)
Find the intervals of convergence for f , f , and f .
nx
nn 1x , x 1
n2
n2 n 2n
(iii)
n1
39. Use the power series for tan
n2 2n
1
x to prove the following expression for as the sum of an infinite series:
x 1
2s3
n0
1 n 2n 1 3 n
TAYLOR AND MACLAURIN SERIES In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions have power series representations? How can we find such representations? We start by supposing that f is any function that can be represented by a power series: 1
f x c0 c1x a c2x a2 c3x a3 c4x a4
x a R
Let’s try to determine what the coefficients cn must be in terms of f . To begin, notice that if we put x a in Equation 1, then all terms after the first one are 0 and we get f a c0 By Theorem 8.6.2, we can differentiate the series in Equation 1 term by term: 2
f x c1 2c2x a 3c3x a2 4c4x a3
x a R
and substitution of x a in Equation 2 gives f a c1 Now we differentiate both sides of Equation 2 and obtain 3
f x 2c2 2 3c3x a 3 4c4x a2
x a R
Again we put x a in Equation 3. The result is f a 2c2 Let’s apply the procedure one more time. Differentiation of the series in Equation 3 gives 4
f x 2 3c3 2 3 4c4x a 3 4 5c5x a2
x a R
SECTION 8.7
TAYLOR AND MACLAURIN SERIES
■
459
and substitution of x a in Equation 4 gives f a 2 3c3 3!c3 By now you can see the pattern. If we continue to differentiate and substitute x a, we obtain f na 2 3 4 ncn n!cn Solving this equation for the nth coefficient cn , we get cn
f na n!
This formula remains valid even for n 0 if we adopt the conventions that 0! 1 and f 0 f . Thus we have proved the following theorem. If f has a power series representation (expansion) at a, that
5 THEOREM
is, if f x
c x a
n
n
n0
x a R
then its coefficients are given by the formula cn
f na n!
Substituting this formula for cn back into the series, we see that if f has a power series expansion at a, then it must be of the following form.
6
f x
n0
The Taylor series is named after the English mathematician Brook Taylor (1685–1731) and the Maclaurin series is named in honor of the Scottish mathematician Colin Maclaurin (1698–1746) despite the fact that the Maclaurin series is really just a special case of the Taylor series. But the idea of representing particular functions as sums of power series goes back to Newton, and the general Taylor series was known to the Scottish mathematician James Gregory in 1668 and to the Swiss mathematician John Bernoulli in the 1690s. Taylor was apparently unaware of the work of Gregory and Bernoulli when he published his discoveries on series in 1715 in his book Methodus incrementorum directa et inversa. Maclaurin series are named after Colin Maclaurin because he popularized them in his calculus textbook Treatise of Fluxions published in 1742.
f na x an n!
f a
■
f a f a f a x a x a2 x a3 1! 2! 3!
The series in Equation 6 is called the Taylor series of the function f at a (or about a or centered at a). For the special case a 0 the Taylor series becomes
7
f x
n0
f n0 n f 0 f 0 2 x f 0 x x n! 1! 2!
This case arises frequently enough that it is given the special name Maclaurin series. NOTE We have shown that if f can be represented as a power series about a, then f is equal to the sum of its Taylor series. But there exist functions that are not equal to the sum of their Taylor series. An example of such a function is given in Exercise 70.
460
■
CHAPTER 8
SERIES
V EXAMPLE 1
Find the Maclaurin series of the function f x e x and its radius
of convergence. SOLUTION If f x e x, then f nx e x, so f n0 e 0 1 for all n. Therefore,
the Taylor series for f at 0 (that is, the Maclaurin series) is
n0
f n0 n xn x x2 x3 x 1 n! 1! 2! 3! n0 n!
To find the radius of convergence we let a n x nn!. Then
a n1 x n1 x n! l 01 an n 1! x n n1
so, by the Ratio Test, the series converges for all x and the radius of convergence is R . ■ The conclusion we can draw from Theorem 5 and Example 1 is that if e x has a power series expansion at 0, then ex
n0
xn n!
So how can we determine whether e x does have a power series representation? Let’s investigate the more general question: Under what circumstances is a function equal to the sum of its Taylor series? In other words, if f has derivatives of all orders, when is it true that f na f x x a n n! n0 As with any convergent series, this means that f x is the limit of the sequence of partial sums. In the case of the Taylor series, the partial sums are n
Tnx
i0
y
f a
y=´ y=T£(x)
y=T™(x)
y=T¡(x)
0
f a f a f na x a x a2 x a n 1! 2! n!
Notice that Tn is a polynomial of degree n called the nth-degree Taylor polynomial of f at a. For instance, for the exponential function f x e x, the result of Example 1 shows that the Taylor polynomials at 0 (or Maclaurin polynomials) with n 1, 2, and 3 are
y=T™(x) (0, 1)
f ia x a i i!
x
T1x 1 x
T2x 1 x
x2 2!
T3x 1 x
x2 x3 2! 3!
y=T£(x) FIGURE 1 ■ As n increases, Tnx appears to approach e x in Figure 1. This suggests that e x is equal to the sum of its Taylor series.
The graphs of the exponential function and these three Taylor polynomials are drawn in Figure 1. In general, f x is the sum of its Taylor series if f x lim Tnx nl
SECTION 8.7
TAYLOR AND MACLAURIN SERIES
■
461
If we let Rnx f x Tnx
so that
f x Tnx Rnx
then Rnx is called the remainder of the Taylor series. If we can somehow show that lim n l Rnx 0, then it follows that lim Tnx lim f x Rnx f x lim Rnx f x
nl
nl
nl
We have therefore proved the following. 8 THEOREM If f x Tnx Rnx, where Tn is the nth-degree Taylor polynomial of f at a and lim Rnx 0 nl
for x a R, then f is equal to the sum of its Taylor series on the interval x a R.
In trying to show that lim n l Rnx 0 for a specific function f , we usually use the expression in the next theorem. 9 TAYLOR’S FORMULA If f has n 1 derivatives in an interval I that contains the number a , then for x in I there is a number z strictly between x and a such that the remainder term in the Taylor series can be expressed as
R nx
f n1z x a n1 n 1!
NOTE 1 For the special case n 0, if we put x b and z c in Taylor’s Formula, we get f b f a f cb a, which is the Mean Value Theorem. In fact, Theorem 9 can be proved by a method similar to the proof of the Mean Value Theorem. The proof is given at the end of this section. NOTE 2 Notice that the remainder term
R nx
10
f n1z x a n1 n 1!
is very similar to the terms in the Taylor series except that f n1 is evaluated at z instead of at a . All we can say about the number z is that it lies somewhere between x and a . The expression for Rnx in Equation 10 is known as Lagrange’s form of the remainder term. NOTE 3 In Section 8.8 we will explore the use of Taylor’s Formula in approximating functions. Our immediate use of it is in conjunction with Theorem 8. In applying Theorems 8 and 9 it is often helpful to make use of the following fact.
11
lim
nl
xn 0 n!
for every real number x
462
■
CHAPTER 8
SERIES
This is true because we know from Example 1 that the series x nn! converges for all x and so its nth term approaches 0. EXAMPLE 2 Prove that e x is equal to the sum of its Taylor series. SOLUTION If f x e x, then f n1x e x, so the remainder term in Taylor’s
Formula is Rnx
ez x n1 n 1!
where z lies between 0 and x . (Note, however, that z depends on n .) If x 0, then 0 z x , so e z e x . Therefore 0 Rnx
ez x n1 x n1 e x l0 n 1! n 1!
by Equation 11, so Rnx l 0 as n l by the Squeeze Theorem. If x 0, then x z 0, so e z e 0 1 and x R x n 1! n1
n
l0
Again Rnx l 0. Thus, by Theorem 8, e x is equal to the sum of its Taylor series, that is, ex
12
n0
In 1748 Leonard Euler used Equation 13 to find the value of e correct to 23 digits. In 2003 Shigeru Kondo, again using the series in (13), computed e to more than fifty billion decimal places! The special techniques employed to speed up the computation are explained on the web page ■
numbers.computation.free.fr
xn n!
for all x
■
In particular, if we put x 1 in Equation 12, we obtain the following expression for the number e as a sum of an infinite series:
13
e
n0
1 1 1 1 1 n! 1! 2! 3!
EXAMPLE 3 Find the Taylor series for f x e x at a 2. SOLUTION We have f n2 e 2 and so, putting a 2 in the definition of a Taylor
series (6), we get
n0
f n2 e2 x 2 n x 2 n n! n0 n!
Again it can be verified, as in Example 1, that the radius of convergence is R . As in Example 2 we can verify that lim n l Rnx 0, so 14
ex
n0
e2 x 2 n n!
for all x
■
We have two power series expansions for e x, the Maclaurin series in Equation 12 and the Taylor series in Equation 14. The first is better if we are interested in values of x near 0 and the second is better if x is near 2.
SECTION 8.7
TAYLOR AND MACLAURIN SERIES
■
463
EXAMPLE 4 Find the Maclaurin series for sin x and prove that it represents sin x for
all x. SOLUTION We arrange our computation in two columns as follows:
x3 3!
T5x x
x3 x5 3! 5!
f 0 0
f x cos x
f 0 1
f x sin x
f 0 0
f x cos x
f 0 1
4
Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows: f 0 2 f 0 f 0 3 x x x f 0 1! 2! 3! x
Notice that, as n increases, Tnx becomes a better approximation to sin x.
x3 x5 x7 3! 5! 7!
T¡
Rnx
1
T∞
f n1z n1 x n 1!
f z x n 1! x n 1!
n1
0 Rnx
15
n1
n1
By Equation 11 the right side of this inequality approaches 0 as n l , so Rnx l 0 by the Squeeze Theorem. It follows that Rnx l 0 as n l , so sin x is equal to the sum of its Maclaurin series by Theorem 8. ■
FIGURE 2
n0
x 2n1 2n 1!
x
T£
1 n
where f x sin x and z lies between 0 and x. But f n1z is sin z or cos z. In any case, f n1z 1 and so
y=sin x 1
Using the remainder term (10) with a 0, we have
y
0
f 40 0
f x sin x
■ Figure 2 shows the graph of sin x together with its Taylor (or Maclaurin) polynomials T1x x
T3x x
f x sin x
We state the result of Example 4 for future reference.
sin x x
16
x3 x5 x7 3! 5! 7!
1
x 2n1 2n 1!
n
n0
for all x
EXAMPLE 5 Find the Maclaurin series for cos x. SOLUTION We could proceed directly as in Example 4 but it’s easier to differentiate
the Maclaurin series for sin x given by Equation 16: cos x
d d sin x dx dx
1
x
x3 x5 x7 3! 5! 7!
3x 2 5x 4 7x 6 x2 x4 x6 1 3! 5! 7! 2! 4! 6!
464
■
CHAPTER 8
SERIES
x ■ The Maclaurin series for e , sin x , and cos x that we found in Examples 2, 4, and 5 were discovered, using different methods, by Newton. These equations are remarkable because they say we know everything about each of these functions if we know all its derivatives at the single number 0.
Since the Maclaurin series for sin x converges for all x, Theorem 8.6.2 tells us that the differentiated series for cos x also converges for all x. Thus cos x 1
17
x2 x4 x6 2! 4! 6!
1
n
n0
x 2n 2n!
■
for all x
EXAMPLE 6 Find the Maclaurin series for the function f x x cos x. SOLUTION Instead of computing derivatives and substituting in Equation 7, it’s easier to multiply the series for cos x (Equation 17) by x :
x cos x x
1
n
n0
x 2n x 2n1 1 n 2n! 2n! n0
■
The power series that we obtained by indirect methods in Examples 5 and 6 and in Section 8.6 are indeed the Taylor or Maclaurin series of the given functions because Theorem 5 asserts that, no matter how we obtain a power series representation f x cnx an, it is always true that cn f nan!. In other words, the coefficients are uniquely determined. EXAMPLE 7 Find the Maclaurin series for f x 1 x k , where k is any real
number. SOLUTION Arranging our work in columns, we have
f x 1 xk
f 0 1
f x k1 xk1
f 0 k
f x kk 11 xk2
f 0 kk 1
f x kk 1k 21 xk3 . . . n f x kk 1 k n 11 xkn
f 0 kk 1k 2 . . . n f 0 kk 1 k n 1
Therefore, the Maclaurin series of f x 1 xk is
n0
f n0 n kk 1 k n 1 n x x n! n! n0
This series is called the binomial series. If its nth term is a n , then
a n1 kk 1 k n 1k nx n1 n! an n 1! kk 1 k n 1x n
kn n1
k n x 1 1 n
1
x l x
as n l
■
465
Thus by the Ratio Test the binomial series converges if x 1 and diverges if x 1.
■
SECTION 8.7
TAYLOR AND MACLAURIN SERIES
The traditional notation for the coefficients in the binomial series is
k n
kk 1k 2 k n 1 n!
and these numbers are called the binomial coefficients. The following theorem states that 1 xk is equal to the sum of its Maclaurin series. It is possible to prove this by showing that the remainder term Rnx approaches 0, but that turns out to be quite difficult. The proof outlined in Exercise 69 is much easier. 18 THE BINOMIAL SERIES
1 x k
n0
If k is any real number and x 1, then
k n kk 1 2 kk 1k 2 3 x 1 kx x x n 2! 3!
Although the binomial series always converges when x 1, the question of whether or not it converges at the endpoints, 1, depends on the value of k. It turns out that the series converges at 1 if 1 k 0 and at both endpoints if k 0. Notice that if k is a positive integer and n k, then the expression for ( nk ) contains a factor k k, so ( nk ) " 0 for n k. This means that the series terminates and reduces to the ordinary Binomial Theorem when k is a positive integer. (See Reference Page 1.) 1 Find the Maclaurin series for the function f x and its s4 x radius of convergence. V EXAMPLE 8
SOLUTION We write f x in a form where we can use the binomial series:
1 s4 x
1
4 1
x 4
2
1
1
x 4
1 2
1
x 4
12
Using the binomial series with k 12 and with x replaced by x4, we have 1 1 2 s4 x
1 2
( )( ) 1
x 4
1
12
1 2
1 2
1
1 2
12 n
n0
x 4
x 4
12 32 2!
n
x 4
2
)
( 12)( 32)( 52)
( 12)( 32)( 52) ( 12 n 1 n!
3!
x 4
x 4
3
n
1 135 3 1 3 5 2n 1 n 13 2 x x x x 8 2!8 2 3!8 3 n!8 n
466
■
CHAPTER 8
SERIES
We know from (18) that this series converges when x4 1, that is, x 4, so the radius of convergence is R 4. ■ We collect in the following table, for future reference, some important Maclaurin series that we have derived in this section and the preceding one. Important Maclaurin series and their radii of convergence
1 xn 1 x x2 x3 1x n0
ex
xn x x2 x3 1 n! 1! 2! 3!
n0
1
sin x
n
n0
1
R
x 2n x2 x4 x6 1 2n! 2! 4! 6!
R
n
n0
tan1x
1
n0
1 x k
n0
R
x 2n1 x3 x5 x7 x 2n 1! 3! 5! 7!
cos x
Module 8.7/8.8 enables you to see how successive Taylor polynomials approach the original function.
R1
n
x 2n1 x3 x5 x7 x 2n 1 3 5 7
R1
k n kk 1 2 kk 1k 2 3 x 1 kx x x n 2! 3!
R1
One reason that Taylor series are important is that they enable us to integrate functions that we couldn’t previously handle. In fact, in the introduction to this chapter we mentioned that Newton often integrated functions by first expressing them as power 2 series and then integrating the series term by term. The function f x ex can’t be integrated by techniques discussed so far because its antiderivative is not an elementary function (see Section 6.4). In the following example we use Newton’s idea to integrate this function. V EXAMPLE 9
(a) Evaluate x ex dx as an infinite series. 2 (b) Evaluate x01 ex dx correct to within an error of 0.001. 2
SOLUTION 2
(a) First we find the Maclaurin series for f x ex . Although it’s possible to use the direct method, let’s find it simply by replacing x with x 2 in the series for e x given in the table of Maclaurin series. Thus, for all values of x, 2
ex
n0
x 2 n x 2n x2 x4 x6 1 n 1 n! n! 1! 2! 3! n0
Now we integrate term by term:
y ex dx y 1 2
x2 x4 x6 x 2n 1 n dx 1! 2! 3! n!
Cx
x3 x5 x7 x 2n1 1 n 3 1! 5 2! 7 3! 2n 1n! 2
This series converges for all x because the original series for ex converges for all x.
SECTION 8.7
TAYLOR AND MACLAURIN SERIES
■
467
(b) The Evaluation Theorem gives
y
1
0
2
ex dx x
x3 x5 x7 x9 3 1! 5 2! 7 3! 9 4!
1
0
1 1 13 101 421 216
We can take C 0 in the antiderivative in part (a). ■
1 1 13 101 421 216 0.7475
The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 1 1 0.001 11 5! 1320
■
Another use of Taylor series is illustrated in the next example. The limit could be found with l’Hospital’s Rule, but instead we use a series. EXAMPLE 10 Evaluate lim
xl0
ex 1 x . x2
SOLUTION Using the Maclaurin series for e x, we have
e 1x lim xl0 x2 x
lim
xl0
1
x x2 x3 1 x 1! 2! 3! x2
x2 x3 x4 2! 3! 4! lim xl0 x2
Some computer algebra systems compute limits in this way. ■
lim
xl0
1 x x2 x3 1 2 3! 4! 5! 2 ■
because power series are continuous functions. MULTIPLICATION AND DIVISION OF POWER SERIES
If power series are added or subtracted, they behave like polynomials (Theorem 8.2.8 shows this). In fact, as the following example illustrates, they can also be multiplied and divided like polynomials. We find only the first few terms because the calculations for the later terms become tedious and the initial terms are the most important ones. EXAMPLE 11 Find the first three nonzero terms in the Maclaurin series for
(a) e x sin x and (b) tan x. SOLUTION
(a) Using the Maclaurin series for e x and sin x in the table on page 466, we have
e x sin x 1
x x2 x3 1! 2! 3!
x
x3 3!
468
■
CHAPTER 8
SERIES
We multiply these expressions, collecting like terms just as for polynomials: 1 x 12 x 2 16 x 3
x 16 x 3
x x 2 12 x 3 16 x 4 16 x 3 16 x 4 x x 2 13 x 3 e x sin x x x 2 13 x 3
Thus
(b) Using the Maclaurin series in the table, we have x5 x3 3! 5! sin x tan x cos x x2 x4 1 2! 4! x
We use a procedure like long division: x 13 x 3
2 15
x5
1 12 x 2 241 x 4 ) x 16 x 3 x 12 x 3
1 120 1 24
x5 x 5
x3 x3
1 30 1 6
x5 x5
2 15
x5
1 3 1 3
tan x x 13 x 3 152 x 5
Thus
■
Although we have not attempted to justify the formal manipulations used in Example 11, they are legitimate. There is a theorem which states that if both f x cn x n and tx bn x n converge for x R and the series are multiplied as if they were polynomials, then the resulting series also converges for x R and represents f xtx. For division we require b0 0; the resulting series converges for sufficiently small x .
PROOF OF TAYLOR’S FORMULA
We conclude this section by giving the promised proof of Theorem 9. Let Rnx f x Tnx, where Tn is the nth-degree Taylor polynomial of f at a . The idea for the proof is the same as that for the Mean Value Theorem: We apply Rolle’s Theorem to a specially constructed function. We think of x as a constant, x a, and we define a function t on I by tt f x f t f tx t
f t x t 2 2!
f nt x t n1 x t n Rnx n! x a n1
SECTION 8.7
TAYLOR AND MACLAURIN SERIES
■
469
Then tx f x f x 0 0 0 ta f x Tnx Rnx f x f x 0 Thus, by Rolle’s Theorem (applied to t on the interval from a to x ), there is a number z between x and a such that tz 0. If we differentiate the expression for t, then most terms cancel. We leave it to you to verify that the expression for tt simplifies to tt
f n1t x t n x t n n 1Rnx n! x a n1
Thus we have tz
f n1z x z n x z n n 1Rnx 0 n! x a n1 R nx
and so
8.7 1. If f x
■
EXERCISES
n0 bnx 5 n for all x, write a formula for b 8.
2. The graph of f is shown.
y
5–10 ■ Find the Maclaurin series for f x using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that Rn x l 0 .] Also find the associated radius of convergence.
f 1 0
f n1z x a n1 n 1!
5. f x cos x
6. f x sin 2x
7. f x e
8. f x xe x
5x
9. f x sinh x 1
x
■
■
■
■
10. f x cosh x ■
■
■
■
■
■
■
■
■ Find the Taylor series for f x centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn x l 0 .]
11–18
(a) Explain why the series 1.6 0.8x 1 0.4x 12 0.1x 13 is not the Taylor series of f centered at 1. (b) Explain why the series 2.8 0.5x 2 1.5x 22 0.1x 23 is not the Taylor series of f centered at 2. 3. If f n0 n 1! for n 0, 1, 2, . . . , find the Maclaurin
series for f and its radius of convergence. 4. Find the Taylor series for f centered at 4 if
1 n n! f 4 n 3 n 1 n
What is the radius of convergence of the Taylor series?
11. f x 1 x x 2, 12. f x x ,
a 1
13. f x e ,
a3
3 x
15. f x cos x, 17. f x 1sx , ■
■
■
■
a2 14. f x ln x,
a
a 2
16. f x sin x, 2
18. f x x ,
a9 ■
a2
■
■
■
■
a1 ■
■
■
19. Prove that the series obtained in Exercise 5 represents cos x
for all x. 20. Prove that the series obtained in Exercise 16 represents
sin x for all x. 21. Prove that the series obtained in Exercise 9 represents
sinh x for all x.
470
■
CHAPTER 8
SERIES
22. Prove that the series obtained in Exercise 10 represents
47–50 ■ Use series to approximate the definite integral to within the indicated accuracy.
cosh x for all x. Use the binomial series to expand the function as a power series. State the radius of convergence.
23–26
■
23. s1 x 25. ■
24.
1 2 x3 ■
1 1 x4
26. 1 x23
■
■
■
■
■
■
■
■
■
1
48.
y
0.2
49.
y
0.1
50.
y
0.5
■
27. f x cos x
28. f x e
29. f x x tan1x
30. f x sinx 4
2 x
31. f x x e
35. f x sin x
36. f x ■
34. f x
x sin x x3
1 6 ■
x2 s2 x
■
dx s1 x 3
0
0
( error 0.001)
2
x 2ex dx
■
■
■
■
■
■
■
53. lim
sin x x 16 x 3 x5
■
■
■
52. lim x l0
■
■
lim
■
■
■
■
■
xl0
■
38. f x ex cos x ■
■
■
■
■
■
■
■
■
1 cos x 1 x ex
■
■
54. Use the series in Example 11(b) to evaluate
2
■
■
Use series to evaluate the limit.
■
if x 0
■
(five decimal places)
( error 108)
x tan1x x3
■
if x 0
37. f x cosx 2 ■
tan 1 x 3 sinx 3 dx
x l0
■ Find the Maclaurin series of f (by any method) and its radius of convergence. Graph f and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and f ?
■
0
(three decimal places)
51. lim
; 37–38
■
x cos x 3 dx
x l0
1 2
2
■
0
51–53
[Hint: Use sin x 1 cos 2x.]
2
■
x2
32. f x x cos 2x
x s4 x 2
33. f x
■
y
■
Use a Maclaurin series derived in this section to obtain the Maclaurin series for the given function. 27–36
47.
■
■
■
x
39. Use the Maclaurin series for e to calculate e
■
0.2
■
■
We found this limit in Example 4 in Section 3.7 using l’Hospital’s Rule three times. Which method do you prefer? 55–58 ■ Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the function. 2
55. y ex cos x
correct to
five decimal places.
57. y
40. Use the Maclaurin series for sin x to compute sin 3 correct
tan x x x3
■
■
56. y sec x
x sin x
58. y e x ln1 x
■
■
■
■
■
■
■
■
■
■
■
■
to five decimal places. 41. (a) Use the binomial series to expand 1s1 x 2 .
59–64 59.
4 (b) Use part (a) to estimate 1s 1.1 correct to three decimal places.
61.
n0
43.
y x cosx
45.
y sx
■
■
60.
1 n 2n1 4 2n 1!
62.
Evaluate the indefinite integral as an infinite series.
3
3
dx
44.
1 dx ■
■
46. ■
■
■
63. 3
y
sin x dx x
y
ex 1 dx x
■
■
■
n
■
■
■
n0
2n1
n0
1 n 2n 6 2n2n! 3n 5n n!
27 81 9 2! 3! 4!
64. 1 ln 2 ■
x 4n n!
1
n0
4 42. (a) Expand 1s 1 x as a power series.
43– 46
Find the sum of the series.
1
(b) Use part (a) to find the Maclaurin series for sin x.
■
■
■
ln 22 ln 23 2! 3! ■
■
■
■
■
■
■
SECTION 8.8
n1
n 2n
66. (a) Expand f x x x 2 1 x3 as a power series.
(b) Let hx 1 xktx and show that hx 0. (c) Deduce that tx 1 xk.
(b) Use part (a) to find the sum of the series
n1
n2 2n
70. (a) Show that the function defined by
67. (a) Use the binomial series to find the Maclaurin series
f x
of f x s1 x 2 . (b) Use part (a) to evaluate f 100. 68. (a) Use the binomial series to find the Maclaurin series
of f x 1s1 x 3 . (b) Use part (a) to evaluate f 90.
8.8
471
(a) Let tx n0 ( nk ) x n. Differentiate this series to show that ktx tx 1 x 1 1x
(b) Use part (a) to find the sum of the series
■
69. Use the following steps to prove (18).
65. (a) Expand f x x1 x2 as a power series.
APPLICATIONS OF TAYLOR POLYNOMIALS
;
e1x 0
2
if x 0 if x 0
is not equal to its Maclaurin series. (b) Graph the function in part (a) and comment on its behavior near the origin.
APPLICATIONS OF TAYLOR POLYNOMIALS In this section we explore two types of applications of Taylor polynomials. First we look at how they are used to approximate functions––computer scientists like them because polynomials are the simplest of functions. Then we investigate how physicists and engineers use them in such fields as relativity, electric dipoles, the velocity of water waves, and building highways across a desert. APPROXIMATING FUNCTIONS BY POLYNOMIALS
Suppose that f x is equal to the sum of its Taylor series at a: f x
n0
f na x a n n!
In Section 8.7 we introduced the notation Tnx for the nth partial sum of this series and called it the n th-degree Taylor polynomial of f at a. Thus n
Tnx
i0
f ia x ai i!
f a
f a f a f na x a x a2 x an 1! 2! n!
Since f is the sum of its Taylor series, we know that Tnx l f x as n l and so Tn can be used as an approximation to f : f x Tnx. Notice that the first-degree Taylor polynomial T1x f a f ax a is the same as the linearization of f at a that we discussed in Section 2.8. Notice also that T1 and its derivative have the same values at a that f and f have. In general, it can be shown that the derivatives of Tn at a agree with those of f up to and including derivatives of order n.
472
■
CHAPTER 8
SERIES
y
y=´ y=T£(x)
y=T™(x)
y=T™(x) (0, 1)
y=T¡(x)
0
x
y=T£(x) FIGURE 1
x
x 0.2
x 3.0
T2x T4x T6x T8x T10x
1.220000 1.221400 1.221403 1.221403 1.221403
8.500000 16.375000 19.412500 20.009152 20.079665
ex
1.221403
20.085537
To illustrate these ideas let’s take another look at the graphs of y e x and its first few Taylor polynomials, as shown in Figure 1. The graph of T1 is the tangent line to y e x at 0, 1; this tangent line is the best linear approximation to e x near 0, 1. The graph of T2 is the parabola y 1 x x 22, and the graph of T3 is the cubic curve y 1 x x 22 x 36, which is a closer fit to the exponential curve y e x than T2. The next Taylor polynomial T4 would be an even better approximation, and so on. The values in the table give a numerical demonstration of the convergence of the Taylor polynomials Tnx to the function y e x. We see that when x 0.2 the convergence is very rapid, but when x 3 it is somewhat slower. In fact, the farther x is from 0, the more slowly Tnx converges to e x. When using a Taylor polynomial Tn to approximate a function f , we have to ask the questions: How good an approximation is it? How large should we take n to be in order to achieve a desired accuracy? To answer these questions we need to look at the absolute value of the remainder:
R x f x T x n
n
There are three possible methods for estimating the size of the error:
1. If a graphing device is available, we can use it to graph Rnx and thereby
estimate the error. 2. If the series happens to be an alternating series, we can use the Alternating Series Estimation Theorem. 3. In all cases we can use Taylor’s Formula (8.7.9), which says that R nx
f n1z x a n1 n 1!
where z is a number that lies between x and a . V EXAMPLE 1 3 (a) Approximate the function f x s x by a Taylor polynomial of degree 2 at a 8. (b) How accurate is this approximation when 7 x 9?
SOLUTION
(a)
3 f x s x x 13
f x 13 x23 f x 29 x53
f 8 2 f 8 121 1 f 8 144
83 f x 10 27 x
Thus the second-degree Taylor polynomial is T2x f 8
f 8 f 8 x 8 x 82 1! 2!
1 2 121 x 8 288 x 82
The desired approximation is 1 3 x T2x 2 121 x 8 288 x 82 s
SECTION 8.8
APPLICATIONS OF TAYLOR POLYNOMIALS
■
473
(b) The Taylor series is not alternating when x 8, so we can’t use the Alternating Series Estimation Theorem in this example. But using Taylor’s Formula we can write R2x
3 f z 5x 8 3 83 x 8 x 8 3 10 27 z 3! 3! 81z 83
where z lies between 8 and x . In order to estimate the error we note that if 7 x 9, then 1 x 8 1, so x 8 1 and therefore x 8 3 1. Also, since z 7, we have
z 83 7 83 179 and so
5 x8 R x 81z
3
2
83
51 0.0004 81 179
Thus if 7 x 9, the approximation in part (a) is accurate to within 0.0004.
Let’s use a graphing device to check the calculation in Example 1. Figure 2 shows 3 that the graphs of y s x and y T2x are very close to each other when x is near 8. Figure 3 shows the graph of R2x computed from the expression
2.5
T™
R x sx T x 3
2
#x „ y= œ 15
0
■
2
We see from this graph that
R x 0.0003 2
FIGURE 2
when 7 x 9. Thus the error estimate from graphical methods is slightly better than the error estimate from Taylor’s Formula in this case.
0.0003
V EXAMPLE 2
y=|R™(x)|
7
(a) What is the maximum error possible in using the approximation sin x x
9 0
FIGURE 3
x3 x5 3! 5!
when 0.3 x 0.3? Use this approximation to find sin 12 correct to six decimal places. (b) For what values of x is this approximation accurate to within 0.00005? SOLUTION
(a) Notice that the Maclaurin series sin x x
x3 x5 x7 3! 5! 7!
is alternating for all nonzero values of x, and the successive terms decrease in size because x 1, so we can use the Alternating Series Estimation Theorem. The error in approximating sin x by the first three terms of its Maclaurin series is at most
x7 x 7 7! 5040
474
■
CHAPTER 8
SERIES
If 0.3 x 0.3, then x 0.3, so the error is smaller than 0.37 4.3 108 5040 To find sin 12 we first convert to radian measure.
12 180
sin 12 sin
15
15
sin
15
3
1 3!
15
5
1 5!
0.20791169 Thus, correct to six decimal places, sin 12 0.207912. (b) The error will be smaller than 0.00005 if
x
7
5040
0.00005
Solving this inequality for x, we get
x
7
0.252
or
x 0.252
17
0.821
So the given approximation is accurate to within 0.00005 when x 0.82. Module 8.7/8.8 graphically shows the remainders in Taylor polynomial approximations.
■
What if we had used Taylor’s Formula to solve Example 2? The remainder term is R6x
f 7z 7 x7 x cos z 7! 7!
(Note that T5 T6.) But cos z 1, so R6x x 77! and we get the same estimates as with the Alternating Series Estimation Theorem. What about graphical methods? Figure 4 shows the graph of
4.3 10–*
R x sin x ( x x x ) and we see from it that R x 4.3 10 when x 0.3. This is the same estimate that we obtained in Example 2. For part (b) we want R x 0.00005, so we graph both y R x and y 0.00005 in Figure 5. By placing the cursor on the right intersection point we find that the inequality is satisfied when x 0.82. Again 1 6
6
y=| Rß(x)|
6
3
1 120
5
8
6
_0.3
0.3 0
6
this is the same estimate that we obtained in the solution to Example 2. If we had been asked to approximate sin 72 instead of sin 12 in Example 2, it would have been wise to use the Taylor polynomials at a 3 (instead of a 0) because they are better approximations to sin x for values of x close to 3. Notice that 72 is close to 60 (or 3 radians) and the derivatives of sin x are easy to compute at 3. Figure 6 shows the graphs of the Maclaurin polynomial approximations
FIGURE 4 0.00006 y=0.00005
y=| Rß(x)|
T1x x _1
x3 3!
T7x x
x3 x5 x7 3! 5! 7!
1 0
FIGURE 5
T3x x
T5x x
x3 x5 3! 5!
SECTION 8.8
APPLICATIONS OF TAYLOR POLYNOMIALS
■
475
to the sine curve. You can see that as n increases, Tnx is a good approximation to sin x on a larger and larger interval. y
T¡
T∞
x
0
y=sin x T£
FIGURE 6
T¶
One use of the type of calculation done in Examples 1 and 2 occurs in calculators and computers. For instance, when you press the sin or e x key on your calculator, or when a computer programmer uses a subroutine for a trigonometric or exponential or Bessel function, in many machines a polynomial approximation is calculated. The polynomial is often a Taylor polynomial that has been modified so that the error is spread more evenly throughout an interval. APPLICATIONS TO PHYSICS
Taylor polynomials are also used frequently in physics. In order to gain insight into an equation, a physicist often simplifies a function by considering only the first two or three terms in its Taylor series. In other words, the physicist uses a Taylor polynomial as an approximation to the function. Taylor’s Formula can then be used to gauge the accuracy of the approximation. The following example shows one way in which this idea is used in special relativity. V EXAMPLE 3 In Einstein’s theory of special relativity the mass of an object moving with velocity v is m0 m s1 v 2c 2
where m 0 is the mass of the object when at rest and c is the speed of light. The kinetic energy of the object is the difference between its total energy and its energy at rest: K mc 2 m 0 c 2 (a) Show that when v is very small compared with c, this expression for K agrees with classical Newtonian physics: K 12 m 0 v 2. (b) Use Taylor’s Formula to estimate the difference in these expressions for K when v 100 ms.
SOLUTION
(a) Using the expressions given for K and m, we get K mc 2 m 0 c 2
m0 c 2 m0 c 2 s1 v 2c 2
m0 c 2
1
v2
c
12
2
1
With x v 2c 2, the Maclaurin series for 1 x12 is most easily computed as a binomial series with k 12 . (Notice that x 1 because v c.) Therefore, we
476
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CHAPTER 8
SERIES
■ The upper curve in Figure 7 is the graph of the expression for the kinetic energy K of an object with velocity v in special relativity. The lower curve shows the function used for K in classical Newtonian physics. When v is much smaller than the speed of light, the curves are practically identical.
have 1 x12 1 12 x
( 12 )( 32 ) x 2 ( 12 )( 32 )( 52) x 3 2!
1 x x 1 2
3 8
K m0 c 2
and
K
m0 c 2 K=mc@-m¸c@
1
2
3!
5 16
x 3
1 v2 3 v4 5 v6 1 2 4 2 c 8 c 16 c 6
1 v2 3 v4 5 v6 2 c2 8 c4 16 c 6
If v is much smaller than c, then all terms after the first are very small when compared with the first term. If we omit them, we get
0
c
1 v2 2 c2
K m0 c 2
K = 21 m ¸ √ @ √
12 m 0 v 2
(b) By Taylor’s Formula we can write the remainder term as
FIGURE 7
f z 2 x 2!
R1x
where f x m0 c2 1 x12 1 and x v 2c 2. Since f x 34 m0 c 2 1 x52, we get R1x
3m0 c 2 v4 81 z52 c 4
where z lies between 0 and v 2c 2. We have c 3 10 8 ms and v 100 ms, so R1x
3 8
m09 10 16100c 4 4.17 1010 m0 1 100 2c 2 52
Thus when v 100 ms, the magnitude of the error in using the Newtonian expression for kinetic energy is at most 4.2 1010 m 0 .
8.8
EXERCISES
; 1. (a) Find the Taylor polynomials up to degree 6 for
f x cos x centered at a 0. Graph f and these polynomials on a common screen. (b) Evaluate f and these polynomials at x 4, 2, and . (c) Comment on how the Taylor polynomials converge to f x.
; 2. (a) Find the Taylor polynomials up to degree 3 for
f x 1x centered at a 1. Graph f and these polynomials on a common screen. (b) Evaluate f and these polynomials at x 0.9 and 1.3.
(c) Comment on how the Taylor polynomials converge to f x. ■ Find the Taylor polynomial T x for the function f at n the number a. Graph f and Tn on the same screen.
; 3– 8
3. f x sin x, 4. f x e x,
a 6,
a 2,
5. f x arcsin x, 6. f x
ln x , x
n3
n3
a 0, n 3
a 1,
n3
■
SECTION 8.8
7. f x xe 2x,
a 0,
8. f x s3 x 2 , ■
■
9–16
■
■
n3
a 1, ■
■
■
■
■
■
■
10. f x 1x ,
a 0,
a 1,
n 1,
n 3,
0 x 6
a 3, n 4, 0 x 23
14. f x cosh x , 15. f x x
n 3,
a 0,
0 x 0.1
x 1
n 5,
,
a 16 ,
n 3,
15 x 17
16. f x ln x ,
a 4,
n 3,
3 x 5
■
■
34
■
■
■
■
t 20 e t20
0.8 x 1.2
12. f x cos x,
a 0,
the conductivity and is measured in units of ohm-meters ( -m). The resistivity of a given metal depends on the temperature according to the equation
a 0 , n 3,
13. f x e x ,
24. The resistivity of a conducting wire is the reciprocal of
0 x 0.1
11. f x tan x ,
2
at a given instant. Using a second-degree Taylor polynomial, estimate how far the car moves in the next second. Would it be reasonable to use this polynomial to estimate the distance traveled during the next minute?
■
(a) Approximate f by a Taylor polynomial with degree n at the number a. (b) Use Taylor’s Formula to estimate the accuracy of the approximation f x Tnx when x lies in the given interval. (c) Check your result in part (b) by graphing Rnx . ; 9. f x s1 x ,
477
23. A car is moving with speed 20 ms and acceleration 2 ms2
n2
■
■
APPLICATIONS OF TAYLOR POLYNOMIALS
■
■
■
■
; ■
■
;
17. Use the information from Exercise 3 to estimate sin 35
correct to five decimal places.
where t is the temperature in C. There are tables that list the values of (called the temperature coefficient) and 20 (the resistivity at 20 C) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for t by its first- or second-degree Taylor polynomial at t 20. (a) Find expressions for these linear and quadratic approximations. (b) For copper, the tables give 0.0039 C and 20 1.7 10 8 -m. Graph the resistivity of copper and the linear and quadratic approximations for 250 C t 1000 C. (c) For what values of t does the linear approximation agree with the exponential expression to within one percent?
18. Use the information from Exercise 12 to estimate cos 69 25. An electric dipole consists of two electric charges of equal
correct to five decimal places.
magnitude and opposite sign. If the charges are q and q and are located at a distance d from each other, then the electric field E at the point P in the figure is
19. Use Taylor’s Formula to determine the number of terms of
the Maclaurin series for e x that should be used to estimate e 0.1 to within 0.00001.
E
20. Suppose you know that
f n4
1 n n! 3 n n 1
and the Taylor series of f centered at 4 converges to f x for all x in the interval of convergence. Show that the fifthdegree Taylor polynomial approximates f 5 with error less than 0.0002. ■ Use the Alternating Series Estimation Theorem or Taylor’s Formula to estimate the range of values of x for which the given approximation is accurate to within the stated error. Check your answer graphically.
x3 6
22. cos x 1 ■
■
■
4
x x 2 24 ■
■
■
■
■
■
■
_q d
D
v2
( error 0.005) ■
q
P
a body of water with depth d, as in the figure, then
( error 0.01)
2
By expanding this expression for E as a series in powers of dD, show that E is approximately proportional to 1D 3 when P is far away from the dipole.
26. If a water wave with length L moves with velocity v across
; 21–22
21. sin x x
q q 2 D D d2
■
tL 2 d tanh 2 L
(a) If the water is deep, show that v stL2 . (b) If the water is shallow, use the Maclaurin series for tanh to show that v std . (Thus in shallow water the velocity of a wave tends to be independent of the length of the wave.)
478
■
CHAPTER 8
SERIES
(c) Use the Alternating Series Estimation Theorem to show that if L 10d, then the estimate v 2 td is accurate to within 0.014tL.
two terms: T 2
d
2 27. If a surveyor measures differences in elevation when mak-
ing plans for a highway across a desert, corrections must be made for the curvature of the Earth. (a) If R is the radius of the Earth and L is the length of the highway, show that the correction is C R secLR R L2 5L 4 2R 24R 3
28. The period of a pendulum with length L that makes a maxi-
mum angle 0 with the vertical is
y
2
0
T 2
12 12 3 2 12 3 25 2 1 2 k2 2 2 k4 2 2 2 k6 2 24 246
L 4 3k 2 t 4 4k 2
x n1 x n
f x n f x n
Use Taylor’s Formula with n 1, a x n , and x r to show that if f x exists on an interval I containing r, x n , and x n1, and f x M , f x K for all x I , then
n1
r
M xn r 2K
2
[This means that if x n is accurate to d decimal places, then x n1 is accurate to about 2d decimal places. More precisely, if the error at stage n is at most 10m, then the error at stage n 1 is at most M2K 102m.] 30. Use the following outline to prove that e is an irrational
dx s1 k 2 sin 2x
where k sin ( 12 0 ) and t is the acceleration due to gravity. (In Exercise 32 in Section 6.5 we approximated this integral using Simpson’s Rule.) (a) Expand the integrand as a binomial series and use the result of Exercise 32 in Section 6.1 to show that L t
(c) Use the inequalities in part (b) to estimate the period of a pendulum with L 1 meter and 0 10 . How does it compare with the estimate T 2 sLt ? What if 0 42 ?
x
R
L t
(1 14 k 2 ) T 2
C
R
L t
mating a root r of the equation f x 0, and from an initial approximation x 1 we obtained successive approximations x 2 , x 3 , . . . , where
(c) Compare the corrections given by the formulas in parts (a) and (b) for a highway that is 100 km long. (Take the radius of the Earth to be 6370 km.)
T4
29. In Section 4.6 we considered Newton’s method for approxi-
(b) Use a Taylor polynomial to show that
L
L (1 14 k 2 ) t
(b) Notice that all the terms in the series after the first one have coefficients that are at most 41. Use this fact to compare this series with a geometric series and show that
L
C
If 0 is not too large, the approximation T 2 sLt , obtained by using only the first term in the series, is often used. A better approximation is obtained by using
number. (a) If e were rational, then it would be of the form e pq, where p and q are positive integers and q 2. Use Taylor’s Formula to write p 1 1 1 ez e1 q 1! 2! q! q 1! sq
ez q 1!
where 0 z 1 . (b) Show that q!e sq is an integer. (c) Show that q!e sq 1. (d) Use parts (b) and (c) to deduce that e is irrational.
CHAPTER 8
8
REVIEW
■
479
CONCEPT CHECK
1. (a) What is a convergent sequence?
7. If a series is convergent by the Alternating Series Test, how
(b) What is a convergent series? (c) What does lim n l an 3 mean? (d) What does n1 an 3 mean?
do you estimate its sum? 8. (a) Write the general form of a power series.
(b) What is the radius of convergence of a power series? (c) What is the interval of convergence of a power series?
2. (a) What is a bounded sequence?
(b) What is a monotonic sequence? (c) What can you say about a bounded monotonic sequence?
9. Suppose f x is the sum of a power series with radius of
convergence R. (a) How do you differentiate f ? What is the radius of convergence of the series for f ? (b) How do you integrate f ? What is the radius of convergence of the series for x f x dx ?
3. (a) What is a geometric series? Under what circumstances
is it convergent? What is its sum? (b) What is a p-series? Under what circumstances is it convergent? 4. Suppose a n 3 and s n is the nth partial sum of the series.
10. (a) Write an expression for the nth-degree Taylor polyno-
mial of f centered at a. (b) Write an expression for the Taylor series of f centered at a. (c) Write an expression for the Maclaurin series of f . (d) How do you show that f x is equal to the sum of its Taylor series? (e) State Taylor’s Formula.
What is lim n l a n ? What is lim n l s n ? 5. State the following.
(a) (b) (c) (d) (e) (f ) (g)
REVIEW
The Test for Divergence The Integral Test The Comparison Test The Limit Comparison Test The Alternating Series Test The Ratio Test The Root Test
11. Write the Maclaurin series and the interval of convergence
for each of the following functions. (a) 11 x (b) e x (d) cos x (e) tan1x
6. (a) What is an absolutely convergent series?
(c) sin x
12. Write the binomial series expansion of 1 x k. What is the
(b) What can you say about such a series? (c) What is a conditionally convergent series?
radius of convergence of this series?
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
10.
n0
1 n 1 n! e
1. If lim n l a n 0, then a n is convergent.
11. If 1 1, then lim n l n 0.
2. The series n1 n sin 1 is convergent.
12. If a n is divergent, then a n is divergent.
3. If lim n l a n L, then lim n l a 2n1 L.
13. If f x 2x x x converges for all x,
4. If cn 6 n is convergent, then cn2n is convergent.
2
then f 0 2.
1 3
3
5. If cn 6 n is convergent, then cn6n is convergent.
14. If a n and bn are divergent, then a n bn is divergent.
6. If cn x diverges when x 6, then it diverges when x 10.
15. If a n and bn are divergent, then a n bn is divergent.
n
7. The Ratio Test can be used to determine whether 1n 3
converges. 8. The Ratio Test can be used to determine whether 1n!
converges. 9. If 0 a n bn and bn diverges, then a n diverges.
16. If a n is decreasing and a n 0 for all n, then a n is
convergent. 17. If a n 0 and a n converges, then 1 n a n converges. 18. If a n 0 and lim n l a n1a n 1, then lim n l a n 0.
480
■
CHAPTER 8
SERIES
EXERCISES
1– 8 ■ Determine whether the sequence is convergent or divergent. If it is convergent, find its limit.
2 n3 1 2n 3
1. a n
■
■
■
■
■
■
■
■
9.
n1
11.
n1
13.
n2
15.
n1
n n3 1
10.
n1
3
n 5n
12.
1 nsln n
14.
n1
n1
■
■
cos 3n 1 1.2 n
16.
n1
■
■
■
■
n1
18.
n1
n1
decimal places.
1 sn 1 n 3n 1
20. ■
n1
the series
n 2n 1 2n 2 n
36.
sn n1
■
1
n1
38.
■
n1 ■ ■
■
■
■
22.
■
■
■
n1 ■
■
25–29
25.
n1
■
■
2
1
n
2 x 2 n 2! n
■
■
■
■
n2 ■
■
■
n 3
■
n1
37.
n1
n
■
■
1 nn 3
39.
n0 ■
■
■
■
x 2 n n 4n 2 n x 3 n sn 3 ■
■
■
■
40. Find the radius of convergence of the series
2n! n x n!2
41. Find the Taylor series of f x sin x at a 6. 42. Find the Taylor series of f x cos x at a 3.
1nsn ln n
Find the sum of the series. 26.
xn n2 5n
n1
24.
2n1
5n
n1 an n
n1
1nn 13 n 2 2n1
nn is convergent. 2n!
n1
n 13
n1
n1
23.
1 n1 correct to four n5
is also absolutely convergent.
■
1
■
35. Prove that if the series n1 an is absolutely convergent, then
21–24 ■ Determine whether the series is conditionally convergent, absolutely convergent, or divergent. 21.
■
nn 0. (b) Deduce that lim n l 2n!
sn 1 sn 1 n
■
■
■ Find the radius of convergence and interval of convergence of the series.
n1
■
36 –39
1
n1
34. (a) Show that the series
n
5 2n n 2 9n
n1
33. Find the sum of the series
1 3 5 2n 1 5 n n!
19.
■
31. Show that cosh x 1 2 x 2 for all x.
n1
17.
■
32. For what values of x does the series n1 ln x n converge?
n 1 n3 1
■
e2 e3 e4 2! 3! 4!
1
■
2
ln
1 n n 3 2n 2n!
fraction.
■
n 1 tan1n
30. Express the repeating decimal 4.17326326326 . . . as a
Determine whether the series is convergent or divergent.
9–20
1
29. 1 e
8. 10 nn! ■
n0
ln n sn
6. a n
7. 1 3n4n ■
28.
4. a n cosn2
n sin n n2 1
5. a n
tan
n1
9 n1 10 n
2. a n
n3 3. a n 1 n2
27.
■ Find the Maclaurin series for f and its radius of convergence. You may use either the direct method (definition of a Maclaurin series) or known series such as geometric series, binomial series, or the Maclaurin series for e x, sin x, and tan1x.
43–50 ■
■
43. f x
x2 1x
44. f x tan1x 2
CHAPTER 8
46. f x xe 2x
47. f x sinx 4
48. f x 10 x
4 49. f x 1s 16 x
50. f x 1 3x5
■
■
51. Evaluate y
■
■
■
■
■
■
■
■
■
e dx as an infinite series. x
;
decimal places. ■
(a) Approximate f by a Taylor polynomial with degree n at the number a. ; (b) Graph f and Tn on a common screen. (c) Use Taylor’s Formula to estimate the accuracy of the approximation f x Tnx when x lies in the given interval. ; (d) Check your result in part (c) by graphing Rn x . 53. f x sx ,
a 1,
54. f x sec x, ■
■
■
F
x
■
n 3,
■
■
0 x 6
■
■
■
■
57. Suppose that f x
lim
sin x x x3
n0 cn x n for all x.
(a) If f is an odd function, show that c0 c2 c4 0 (b) If f is an even function, show that
■
■
c1 c3 c5 0
55. Use series to evaluate the following limit. xl0
mtR 2 R h2
where R is the radius of the Earth and t is the acceleration due to gravity. (a) Express F as a series in powers of hR. (b) Observe that if we approximate F by the first term in the series, we get the expression F mt that is usually used when h is much smaller than R. Use the Alternating Series Estimation Theorem to estimate the range of values of h for which the approximation F mt is accurate to within one percent. (Use R 6400 km.)
0.9 x 1.1
a 0, n 2,
481
height h above the surface of the Earth is
52. Use series to approximate x01 s1 x 4 dx correct to two 53–54
■
56. The force due to gravity on an object with mass m at a
45. f x ln1 x
■
REVIEW
2
58. If f x e x , show that f 2n0
2n! . n!
9
PARAMETRIC EQUATIONS AND POLAR COORDINATES So far we have described plane curves by giving y as a function of x y f x or x as a function of y x ty or by giving a relation between x and y that defines y implicitly as a function of x f x, y 0 . In this chapter we discuss two new methods for describing curves. Some curves, such as the cycloid, are best handled when both x and y are given in terms of a third variable t called a parameter x f t, y tt . Other curves, such as the cardioid, have their most convenient description when we use a new coordinate system, called the polar coordinate system.
9.1 y
PARAMETRIC CURVES
C (x, y)={ f(t), g(t)}
0
x
Imagine that a particle moves along the curve C shown in Figure 1. It is impossible to describe C by an equation of the form y f x because C fails the Vertical Line Test. But the x- and y-coordinates of the particle are functions of time and so we can write x f t and y tt. Such a pair of equations is often a convenient way of describing a curve and gives rise to the following definition. Suppose that x and y are both given as functions of a third variable t (called a parameter) by the equations x f t
FIGURE 1
Module 9.1A gives an animation of the relationship between motion along a parametric curve x f t, y tt and motion along the graphs of f and t as functions of t .
y tt
(called parametric equations). Each value of t determines a point x, y, which we can plot in a coordinate plane. As t varies, the point x, y f t, tt varies and traces out a curve C, which we call a parametric curve. The parameter t does not necessarily represent time and, in fact, we could use a letter other than t for the parameter. But in many applications of parametric curves, t does denote time and therefore we can interpret x, y f t, tt as the position of a particle at time t. EXAMPLE 1 Sketch and identify the curve defined by the parametric equations
x t 2 2t
yt1
SOLUTION Each value of t gives a point on the curve, as shown in the table. For instance, if t 0, then x 0, y 1 and so the corresponding point is 0, 1. In Figure 2 we plot the points x, y determined by several values of the parameter t and we join them to produce a curve. y
t 2 1 0 1 2 3 4
x 8 3 0 1 0 3 8
t=4
y 1 0 1 2 3 4 5
t=3 t=2 t=1
(0, 1) 0
x
t=_1 t=_2
FIGURE 2 482
8
t=0
SECTION 9.1
■ This equation in x and y describes where the particle has been, but it doesn’t tell us when the particle was at a particular point. The parametric equations have an advantage––they tell us when the particle was at a point. They also indicate the direction of the motion.
(8, 5)
(0, 1) x
and so the curve represented by the given parametric equations is the parabola x y 2 4y 3.
■
yt1
0 t 4
shown in Figure 3 is the part of the parabola in Example 1 that starts at the point 0, 1 and ends at the point 8, 5. The arrowhead indicates the direction in which the curve is traced as t increases from 0 to 4. In general, the curve with parametric equations x f t
FIGURE 3
y tt
a t b
has initial point f a, ta and terminal point f b, tb.
y
V EXAMPLE 2
(cos t, sin t)
t 0
(1, 0)
What curve is represented by the following parametric equations? x cos t
t=0
t=π
x
t=2π t=
483
A particle whose position is given by the parametric equations moves along the curve in the direction of the arrows as t increases. Notice that the consecutive points marked on the curve appear at equal time intervals but not at equal distances. That is because the particle slows down and then speeds up as t increases. It appears from Figure 2 that the curve traced out by the particle may be a parabola. This can be confirmed by eliminating the parameter t as follows. We obtain t y 1 from the second equation and substitute into the first equation. This gives x t 2 2t y 12 2y 1 y 2 4y 3
x t 2 2t
π t= 2
■
No restriction was placed on the parameter t in Example 1, so we assumed that t could be any real number. But sometimes we restrict t to lie in a finite interval. For instance, the parametric curve
y
0
PARAMETRIC CURVES
y sin t
0 t 2
SOLUTION If we plot points, it appears that the curve is a circle. We can confirm this impression by eliminating t. Observe that
x 2 y 2 cos 2t sin 2t 1
3π 2
Thus the point x, y moves on the unit circle x 2 y 2 1. Notice that in this example the parameter t can be interpreted as the angle (in radians) shown in Figure 4. As t increases from 0 to 2, the point x, y cos t, sin t moves once around the circle in the counterclockwise direction starting from the point 1, 0. ■
FIGURE 4
y
t=0, π, 2π
EXAMPLE 3 What curve is represented by the given parametric equations? (0, 1)
0
x sin 2t x
y cos 2t
0 t 2
SOLUTION Again we have
x 2 y 2 sin 2 2t cos 2 2t 1
FIGURE 5
so the parametric equations again represent the unit circle x 2 y 2 1. But as t increases from 0 to 2, the point x, y sin 2t, cos 2t starts at 0, 1 and moves ■ twice around the circle in the clockwise direction as indicated in Figure 5.
484
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
Examples 2 and 3 show that different sets of parametric equations can represent the same curve. Thus we distinguish between a curve, which is a set of points, and a parametric curve, in which the points are traced in a particular way. y
EXAMPLE 4 Find parametric equations for the circle with center h, k and radius r.
r (h, k)
0
SOLUTION If we take the equations of the unit circle in Example 2 and multiply the expressions for x and y by r, we get x r cos t, y r sin t. You can verify that these equations represent a circle with radius r and center the origin traced counterclockwise. We now shift h units in the x-direction and k units in the y-direction and obtain parametric equations of the circle (Figure 6) with center h, k and radius r :
x
FIGURE 6 x=h+r cos t, y=k+r sin t
(_1, 1)
y
x h r cos t V EXAMPLE 5
(1, 1)
y k r sin t
0 t 2
■
Sketch the curve with parametric equations x sin t, y sin 2t.
SOLUTION Observe that y sin t 2 x 2 and so the point x, y moves on the
0
parabola y x 2. But note also that, since 1 sin t 1, we have 1 x 1, so the parametric equations represent only the part of the parabola for which 1 x 1. Since sin t is periodic, the point x, y sin t, sin 2t moves back and forth infinitely often along the parabola from 1, 1 to 1, 1. (See Figure 7.) ■
x
FIGURE 7
GRAPHING DEVICES
Most graphing calculators and computer graphing programs can be used to graph curves defined by parametric equations. In fact, it’s instructive to watch a parametric curve being drawn by a graphing calculator because the points are plotted in order as the corresponding parameter values increase. EXAMPLE 6 Use a graphing device to graph the curve x y 4 3y 2.
3
SOLUTION If we let the parameter be t y, then we have the equations _3
3
_3
FIGURE 8
x t 4 3t 2
yt
Using these parametric equations to graph the curve, we obtain Figure 8. It would be possible to solve the given equation x y 4 3y 2 for y as four functions of x and graph them individually, but the parametric equations provide a much easier method. ■ In general, if we need to graph an equation of the form x ty, we can use the parametric equations x tt
yt
Notice also that curves with equations y f x (the ones we are most familiar with— graphs of functions) can also be regarded as curves with parametric equations xt
y f t
SECTION 9.1
PARAMETRIC CURVES
■
485
Graphing devices are particularly useful for sketching complicated curves. For instance, the curves shown in Figures 9, 10, and 11 would be virtually impossible to produce by hand. 8
2.5
_6.5
6.5
1
2.5
_2.5
_8
1
_1
_2.5
_1
FIGURE 9
FIGURE 10
FIGURE 11
x=t+2 sin 2t y=t+2 cos 5t
x=1.5 cos t-cos 30t y=1.5 sin t-sin 30t
x=sin(t+cos 100t) y=cos(t+sin 100t)
THE CYCLOID
EXAMPLE 7 The curve traced out by a point P on the circumference of a circle as
An animation in Module 9.1B shows how the cycloid is formed as the circle moves.
the circle rolls along a straight line is called a cycloid (see Figure 12). If the circle has radius r and rolls along the x-axis and if one position of P is the origin, find parametric equations for the cycloid. P P
FIGURE 12
P
SOLUTION We choose as parameter the angle of rotation of the circle 0
y
when P is at the origin). Suppose the circle has rotated through radians. Because the circle has been in contact with the line, we see from Figure 13 that the distance it has rolled from the origin is
r P
OT arc PT r
C (r¨, r )
¨
Q
Therefore, the center of the circle is Cr , r. Let the coordinates of P be x, y. Then from Figure 13 we see that
y x T
O
x
y TC QC r r cos r1 cos
x OT PQ r r sin r sin
r¨ FIGURE 13
Therefore, parametric equations of the cycloid are 1
x r sin
y r 1 cos
⺢
One arch of the cycloid comes from one rotation of the circle and so is described by 0 2. Although Equations 1 were derived from Figure 13, which illustrates the case 0 2, it can be seen that these equations are still valid for other values of (see Exercise 33).
486
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
A
Although it is possible to eliminate the parameter from Equations 1, the resulting Cartesian equation in x and y is very complicated and not as convenient to work with as the parametric equations. ■ One of the first people to study the cycloid was Galileo, who proposed that bridges be built in the shape of cycloids and who tried to find the area under one arch of a cycloid. Later this curve arose in connection with the brachistochrone problem: Find the curve along which a particle will slide in the shortest time (under the influence of gravity) from a point A to a lower point B not directly beneath A. The Swiss mathematician John Bernoulli, who posed this problem in 1696, showed that among all possible curves that join A to B, as in Figure 14, the particle will take the least time sliding from A to B if the curve is part of an inverted arch of a cycloid. The Dutch physicist Huygens had already shown that the cycloid is also the solution to the tautochrone problem; that is, no matter where a particle P is placed on an inverted cycloid, it takes the same time to slide to the bottom (see Figure 15). Huygens proposed that pendulum clocks (which he invented) should swing in cycloidal arcs because then the pendulum takes the same time to make a complete oscillation whether it swings through a wide or a small arc.
cycloid B FIGURE 14
P
P P
P P
FIGURE 15
9.1
EXERCISES
■ Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases.
1– 4
y t 2 4 t,
1. x 1 st , 2. x 2 cos t,
y t cos t,
3. x 5 sin t,
y t 2,
4. x et t, ■
■
5– 8
■
■
13. x e , 2t
■
■
0 t 2
■
7. x st , ■
■
9–14
y1t
■
■
■
■
6. x 1 3t, 8. x t 2,
■
■
■
■
■
■
11. x sin t,
y csc t,
■
2 2
0 t 2
■
■
■
■
■
y 2 t2
■
■
■
y cos t, 2
■
2 t 32
y 1 2 sin t,
y 2 cos t,
■
y t3
0
y 5 sin ,
■
17. x 5 sin t,
■
■
0 t 32
t 5
2 t 2 ■
■
■
■
■
■
19–21 ■ Use the graphs of x f t and y tt to sketch the parametric curve x f t, y tt. Indicate with arrows the direction in which the curve is traced as t increases. 19.
x
y 1
■
1
10. x 4 cos ,
■
y 4 cos t,
■
y cos ,
■
16. x 2 sin t, 18. x sin t,
(a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. 9. x sin ,
y 2 cos 1
■
15. x 3 2 cos t, ■
■
y 2t 1
■
15–18 Describe the motion of a particle with position x, y as t varies in the given interval.
(a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases. (b) Eliminate the parameter to find a Cartesian equation of the curve. 5. x 3t 5 ,
yt1
■
2 t 2 ■
y tan , 2 2
14. x 1 cos ,
0 t 5
t
y e t t, ■
12. x sec ,
t
1
t
1
t
_1
20.
x
y
1
1 1
t
■
SECTION 9.1
21. x
1
■
■
■
■
; 29–30
■
■
■
■
■
■
■
■
29. y
30.
22. Match the parametric equations with the graphs labeled
y
I–VI. Give reasons for your choices. (Do not use a graphing device.) (a) x t 3 2t, y t 2 t (b) x t 3 1, y 2 t 2 (c) x sin 3t, y sin 4t (d) x t sin 2t, y t sin 3t (e) x sint sin t, y cost cos t (f ) x cos t, y sint sin 5t y
I
II
487
Use a graphing calculator or computer to reproduce the picture.
1 t
t
■
■
(b) Use these parametric equations to graph the ellipse when a 3 and b 1, 2, 4, and 8. (c) How does the shape of the ellipse change as b varies?
y 1
1
PARAMETRIC CURVES
y
III
y
2
4 2
0 ■
■
x
2 ■
■
■
0 ■
■
3
■
x
8
■
■
■
■
31–32 ■ Compare the curves represented by the parametric equations. How do they differ? 31. (a) x t 3,
y t2 (b) x t , y t 4 (c) x e3t, y e2t 6
0
x
x
0
0
x
y t 2 (b) x cos t, y sec2 t (c) x e t, y e2t
32. (a) x t, IV y
V
y
VI
y ■
■
■
■
■
■
■
■
■
■
■
33. Derive Equations 1 for the case 2 .
0
34. Let P be a point at a distance d from the center of a circle
x
0
0
x
x
3 5 ; 23. Graph the curve x y 3y y . 5 2 ; 24. Graph the curves y x and x y y 1 and find their
points of intersection correct to one decimal place.
25. (a) Show that the parametric equations
x x 1 x 2 x 1 t
y y1 y 2 y1 t
where 0 t 1, describe the line segment that joins the points P1x 1, y1 and P2x 2 , y 2 . (b) Find parametric equations to represent the line segment from 2, 7 to 3, 1.
of radius r. The curve traced out by P as the circle rolls along a straight line is called a trochoid. (Think of the motion of a point on a spoke of a bicycle wheel.) The cycloid is the special case of a trochoid with d r. Using the same parameter as for the cycloid and assuming the line is the x-axis and 0 when P is at one of its lowest points, show that parametric equations of the trochoid are x r d sin
y r d cos
Sketch the trochoid for the cases d r and d r. 35. If a and b are fixed numbers, find parametric equations for
the curve that consists of all possible positions of the point P in the figure, using the angle as the parameter. Then eliminate the parameter and identify the curve.
; 26. Use a graphing device and the result of Exercise 25(a) to
y
draw the triangle with vertices A 1, 1, B 4, 2, and C 1, 5.
27. Find parametric equations for the path of a particle that
moves along the circle x 2 y 12 4 in the manner described. (a) Once around clockwise, starting at 2, 1 (b) Three times around counterclockwise, starting at 2, 1 (c) Halfway around counterclockwise, starting at 0, 3
; 28. (a) Find parametric equations for the ellipse
x 2a 2 y 2b 2 1. [Hint: Modify the equations of the circle in Example 2.]
a
b ¨ O
P x
■
488
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
36. A curve, called a witch of Maria Agnesi, consists of all
tance is assumed to be negligible, then its position after t seconds is given by the parametric equations
possible positions of the point P in the figure. Show that parametric equations for this curve can be written as x 2a cot
x v 0 cos t
y 2a sin 2
Sketch the curve. y
C
y=2a
A
; P
a
¨ O
x
; 37. Suppose that the position of one particle at time t is given by
x 1 3 sin t
0 t 2
y1 2 cos t
and the position of a second particle is given by x 2 3 cos t
y 2 1 sin t
0 t 2
(a) Graph the paths of both particles. How many points of intersection are there? (b) Are any of these points of intersection collision points? In other words, are the particles ever at the same place at the same time? If so, find the collision points. (c) Describe what happens if the path of the second particle is given by x 2 3 cos t
y 2 1 sin t
0 t 2
38. If a projectile is fired with an initial velocity of v 0 meters
per second at an angle above the horizontal and air resis-
9.2
1 y v 0 sin t 2 tt 2
where t is the acceleration due to gravity (9.8 ms2 ). (a) If a gun is fired with 30 and v 0 500 ms, when will the bullet hit the ground? How far from the gun will it hit the ground? What is the maximum height reached by the bullet? (b) Use a graphing device to check your answers to part (a). Then graph the path of the projectile for several other values of the angle to see where it hits the ground. Summarize your findings. (c) Show that the path is parabolic by eliminating the parameter.
; 39. Investigate the family of curves defined by the parametric
equations x t 2, y t 3 ct. How does the shape change as c increases? Illustrate by graphing several members of the family.
; 40. The swallowtail catastrophe curves are defined by the
parametric equations x 2ct 4t 3, y ct 2 3t 4. Graph several of these curves. What features do the curves have in common? How do they change when c increases?
; 41. The curves with equations x a sin nt, y b cos t are
called Lissajous figures. Investigate how these curves vary when a, b, and n vary. (Take n to be a positive integer.)
; 42. Investigate the family of curves defined by the parametric equations
x sin t c sin t
y cos t c sin t
How does the shape change as c changes? In particular, you should identify the transitional values of c for which the basic shape of the curve changes.
CALCULUS WITH PARAMETRIC CURVES Having seen how to represent curves by parametric equations, we now apply the methods of calculus to these parametric curves. In particular, we solve problems involving tangents, areas, and arc length. TANGENTS
Suppose f and t are differentiable functions and we want to find the tangent line at a point on the parametric curve x f t, y tt where y is also a differentiable function of x. Then the Chain Rule gives dy dy dx dt dx dt
SECTION 9.2
CALCULUS WITH PARAMETRIC CURVES
■
489
If dxdt 0, we can solve for dydx : ■ If we think of the curve as being traced out by a moving particle, then dydt and dxdt are the vertical and horizontal velocities of the particle and Formula 1 says that the slope of the tangent is the ratio of these velocities.
dy dt dy dx dx dt
1
dx 0 dt
if
Equation 1 (which you can remember by thinking of canceling the dt’s) enables us to find the slope dydx of the tangent to a parametric curve without having to eliminate the parameter t. We see from (1) that the curve has a horizontal tangent when dydt 0 (provided that dxdt 0) and it has a vertical tangent when dxdt 0 (provided that dydt 0). As we know from Chapter 4, it is also useful to consider d 2 ydx 2. This can be found by replacing y by dydx in Equation 1:
|
d 2y d y dt 2 Note that 2 . dx 2 d x dt 2
2
d y d dx 2 dx
2
dy dx
d dt
dy dx dx dt
EXAMPLE 1 A curve C is defined by the parametric equations x t 2, y t 3 3t.
(a) (b) (c) (d)
Show that C has two tangents at the point (3, 0) and find their equations. Find the points on C where the tangent is horizontal or vertical. Determine where the curve is concave upward or downward. Sketch the curve.
SOLUTION
(a) Notice that y t 3 3t tt 2 3 0 when t 0 or t s3 . Therefore, the point 3, 0 on C arises from two values of the parameter, t s3 and t s3 . This indicates that C crosses itself at 3, 0. Since dy dydt 3t 2 3 3 dx dxdt 2t 2
t
1 t
the slope of the tangent when t s3 is dydx 6(2s3 ) s3 , so the equations of the tangents at 3, 0 are y s3 x 3
y s3 x 3
and
(b) C has a horizontal tangent when dydx 0, that is, when dydt 0 and dxdt 0. Since dydt 3t 2 3, this happens when t 2 1, that is, t 1. The corresponding points on C are 1, 2 and (1, 2). C has a vertical tangent when dxdt 2t 0, that is, t 0. (Note that dydt 0 there.) The corresponding point on C is (0, 0). (c) To determine concavity we calculate the second derivative: 2
d y dx 2
d dt
dy dx dx dt
3 2
1 2t
1 t2
3t 2 1 4t 3
490
■
CHAPTER 9
y
PARAMETRIC EQUATIONS AND POLAR COORDINATES
y=œ„ 3 (x-3) t=_1
Thus the curve is concave upward when t 0 and concave downward when t 0. (d) Using the information from parts (b) and (c), we sketch C in Figure 1. ■
(1, 2) V EXAMPLE 2 (3, 0) 0
x
(a) Find the tangent to the cycloid x r sin , y r1 cos at the point where 3. (See Example 7 in Section 9.1.) (b) At what points is the tangent horizontal? When is it vertical? SOLUTION
t=1
(a) The slope of the tangent line is
(1, _2)
y=_ œ„ 3 (x-3)
dy dyd r sin sin dx dxd r1 cos 1 cos
FIGURE 1
When 3, we have xr
sin 3 3
r
s3 3 2
y r 1 cos
3
r 2
dy sin3 s32 s3 dx 1 cos3 1 12
and
Therefore, the slope of the tangent is s3 and its equation is y
r s3 2
x
r r s3 3 2
or
s3 x y r
2 s3
The tangent is sketched in Figure 2. (_πr, 2r)
y
(πr, 2r)
(3πr, 2r)
(5πr, 2r)
π
¨= 3 0
FIGURE 2
2πr
4πr
x
(b) The tangent is horizontal when dydx 0, which occurs when sin 0 and 1 cos 0, that is, 2n 1, n an integer. The corresponding point on the cycloid is (2n 1 r, 2r). When 2n, both dxd and dyd are 0. It appears from the graph that there are vertical tangents at those points. We can verify this by using l’Hospital’s Rule as follows: lim
l 2n
dy sin cos lim lim l 2n 1 cos l 2n dx sin
A similar computation shows that dydx l as l 2n , so indeed there are ■ vertical tangents when 2n, that is, when x 2n r.
SECTION 9.2
CALCULUS WITH PARAMETRIC CURVES
■
491
AREAS
We know that the area under a curve y Fx from a to b is A xab Fx dx, where Fx 0. If the curve is given by parametric equations x f t, y tt and is traversed once as t increases from to , then we can adapt the earlier formula by using the Substitution Rule for Definite Integrals as follows:
A y y dx y ttf t dt b
a
y ttf t dt
or
V EXAMPLE 3 Find the area under one arch of the cycloid x r sin , y r 1 cos . (See Figure 3.)
y
0
if ( f , t ) is the leftmost endpoint
2πr
x
SOLUTION One arch of the cycloid is given by 0 2. Using the Substitution
Rule with y r 1 cos and dx r1 cos d, we have
FIGURE 3
Ay
2r
y dx y
0
0
The result of Example 3 says that the area under one arch of the cycloid is three times the area of the rolling circle that generates the cycloid (see Example 7 in Section 9.1). Galileo guessed this result but it was first proved by the French mathematician Roberval and the Italian mathematician Torricelli.
2
r 1 cos r1 cos d
■
r2 y
2
r2 y
2
0
0
1 cos 2 d r 2 y
2
0
[1 2 cos
[
1 2
1 2 cos cos 2 d
]
1 cos 2 d
r 2 32 2 sin 14 sin 2
2 0
]
r 2 ( 32 2) 3 r 2
■
ARC LENGTH
We already know how to find the length L of a curve C given in the form y Fx, a x b. Formula 7.4.3 says that if F is continuous, then
y
L
2
b
dy dx
1
a
2
dx
Suppose that C can also be described by the parametric equations x f t, y tt, t , where dxdt f t 0. This means that C is traversed once, from left to right, as t increases from to and f a, f b. Putting Formula 1 into Formula 2 and using the Substitution Rule, we obtain L
y
b
a
1
2
dy dx
Since dxdt 0, we have 3
L
y
dx
y
1
dydt dxdt
dx dt
2
dy dt
2
dx dt dt
2
dt
Even if C can’t be expressed in the form y Fx, Formula 3 is still valid but we obtain it by polygonal approximations. We divide the parameter interval , into n subintervals of equal width t. If t0 , t1 , t2 , . . . , tn are the endpoints of these subinter-
492
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
y
C
Pi-1
P™
Pi
vals, then xi f ti and yi tti are the coordinates of points Pi xi , yi that lie on C and the polygon with vertices P0 , P1 , . . . , Pn approximates C (see Figure 4). As in Section 7.4, we define the length L of C to be the limit of the lengths of these approximating polygons as n l :
P¡
n
L lim
Pn
i1
n l i1
P¸ 0
P
x
Pi
The Mean Value Theorem, when applied to f on the interval ti1, ti , gives a number ti* in ti1, ti such that
FIGURE 4
f ti f ti1 f ti*ti ti1 If we let x i x i x i1 and yi yi yi1 , this equation becomes x i f ti* t Similarly, when applied to t, the Mean Value Theorem gives a number ti** in ti1, ti such that yi tti** t Therefore
P
i1
Pi sx i 2 yi 2 s f ti*t 2 tti**t 2 s f ti* 2 tti** 2 t
and so n
4
L lim
s f t*
n l i1
i
2
tti** 2 t
The sum in (4) resembles a Riemann sum for the function s f t 2 tt 2 but it is not exactly a Riemann sum because ti* ti** in general. Nevertheless, if f and t are continuous, it can be shown that the limit in (4) is the same as if ti* and ti** were equal, namely,
L y s f t 2 tt 2 dt
Thus, using Leibniz notation, we have the following result, which has the same form as (3). 5 THEOREM If a curve C is described by the parametric equations x f t, y tt, t , where f and t are continuous on , and C is traversed exactly once as t increases from to , then the length of C is
L
y
dx dt
2
dy dt
2
dt
Notice that the formula in Theorem 5 is consistent with the general formulas L x ds and ds2 dx2 dy2 of Section 7.4.
SECTION 9.2
CALCULUS WITH PARAMETRIC CURVES
■
493
EXAMPLE 4 If we use the representation of the unit circle given in Example 2 in
Section 9.1, x cos t
0 t 2
y sin t
then dxdt sin t and dydt cos t, so Theorem 5 gives L
y
2
0
dx dt
2
dy dt
2
2
2
dt y ssin 2 t cos 2 t dt y dt 2 0
0
as expected. If, on the other hand, we use the representation given in Example 3 in Section 9.1, x sin 2t
0 t 2
y cos 2t
then dxdt 2 cos 2t, dydt 2 sin 2t, and the integral in Theorem 5 gives
y
2
0
dx dt
2
dy dt
2
dt y
2
0
s4 cos 2 2t 4 sin 2 2t dt y
2
0
2 dt 4
| Notice that the integral gives twice the arc length of the circle because as t increases from 0 to 2, the point sin 2t, cos 2t traverses the circle twice. In general, when finding the length of a curve C from a parametric representation, we have to be careful to ensure that C is traversed only once as t increases from to . ■ V EXAMPLE 5 Find the length of one arch of the cycloid x r sin , y r 1 cos .
SOLUTION From Example 3 we see that one arch is described by the parameter interval 0 2. Since
dx r1 cos d
dy r sin d
and
we have The result of Example 5 says that the length of one arch of a cycloid is eight times the radius of the generating circle (see Figure 5). This was first proved in 1658 by Sir Christopher Wren, who later became the architect of St. Paul’s Cathedral in London. ■
y
L
y
2
0
y
2
0
dx d
2
dy d
2
d y
2
0
sr 2 1 cos 2 r 2 sin 2 d
sr 21 2 cos cos 2 sin 2 d r y
2
0
s21 cos d
To evaluate this integral we use the identity sin 2x 12 1 cos 2x with 2x, which gives 1 cos 2 sin 22. Since 0 2, we have 0 2 and so sin2 0. Therefore
L=8r r
s21 cos s4 sin 2 2 2 sin2 2 sin2 0
FIGURE 5
2πr
x
and so
L 2r y
2
0
]
sin2 d 2r 2 cos2
2r 2 2 8r
2 0
■
494
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
9.2 1–2
20. x t 4 4t 3 8t 2,
Find dydx .
■
1. x t t 3, ■
EXERCISES
■
y 2 5t
■
■
■
2. x te t,
■
■
■
■
■
y t et ■
■
■
■
y t 3 t;
3. x t 4 1, 4. x 2t 2 1 , 5. x e st ,
6. x cos sin 2, ■
■
■
■
■
■
■
■
■
■
7. Find an equation of the tangent to the curve x e t,
y t 1 2 at the point 1, 1 by two methods: (a) without eliminating the parameter and (b) by first eliminating the parameter.
; 8. Find equations of the tangents to the curve x sin t,
y sint sin t at the origin. Then graph the curve and the tangents.
■
■
■
21. Show that the curve x cos t, y sin t cos t has two tan-
x r d sin , y r d cos in terms of . (See Exercise 34 in Section 9.1.) (b) Show that if d r, then the trochoid does not have a vertical tangent. 24. (a) Find the slope of the tangent to the astroid x a cos 3,
y a sin 3 in terms of . (b) At what points is the tangent horizontal or vertical? (c) At what points does the tangent have slope 1 or 1? tangent parallel to the line with equations x 7t, y 12t 5? y 2t 3 1 that pass through the point 4, 3.
27. Use the parametric equations of an ellipse, x a cos ,
y t2 1
y b sin , 0 2, to find the area that it encloses.
y t e t
11. x t e t, 12. x t ln t, ■
26. Find equations of the tangents to the curve x 3t 2 1,
y t2 t3
10. x t 3 12t,
■
28. Find the area bounded by the curve x t 1t,
y t ln t ■
■
■
y t 1t and the line y 2.5.
■
■
■
■
■
■
Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. 13–16
■
y t 12t
13. x 10 t , 2
3
14. x 2t 3 3t 2 12t, 15. x 2 cos , ■
■
■
■
30. Find the area of the region enclosed by the astroid
x a cos 3, y a sin 3.
32. Let be the region enclosed by the loop of the curve in ■
■
■
■
■
■
; 17. Use a graph to estimate the coordinates of the leftmost point on the curve x t 4 t 2, y t ln t. Then use calculus to find the exact coordinates.
; 18. Try to estimate the coordinates of the highest point and the leftmost point on the curve x te t, y tet. Then find the exact coordinates. What are the asymptotes of this curve?
; 19–20
0 t 2, and the lines y 1 and x 0.
in Section 9.1 for the case d r.
y 2 sin ■
29. Find the area bounded by the curve x cos t, y e t,
31. Find the area under one arch of the trochoid of Exercise 34
y 2t 3 3t 2 1
y sin 2
16. x cos 3, ■
■
■
9. x 4 t 2,
■
■
25. At what points on the curve x t 3 4t, y 6t 2 is the
Find dydx and d 2 ydx 2. For which values of t is the curve concave upward?
9–12
■
23. (a) Find the slope of the tangent line to the trochoid
t1 ■
■
y tan t1 2 cos 2t cross itself? Find the equations of both tangents at that point.
t 1
y sin cos 2 ; 0
■
■
22. At what point does the curve x 1 2 cos 2t,
y 13 t 3 t ; t 3
y t ln t 2 ;
■
gents at 0, 0 and find their equations. Sketch the curve.
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
3–6
■
y 2t 2 t
Graph the curve in a viewing rectangle that displays all the important aspects of the curve.
Example 1. (a) Find the area of . (b) If is rotated about the x-axis, find the volume of the resulting solid. (c) Find the centroid of . ■ Set up, but do not evaluate, an integral that represents the length of the curve.
33–36
33. x t t 2,
y 43 t 32,
34. x 1 e t,
y t 2,
1 t 2
■
19. x t 4 2t 3 2t 2,
y t3 t
35. x t cos t,
3 t 3
y t sin t,
0 t 2
■
SECTION 9.2
36. x ln t, ■
■
37– 40
y st 1,
■
■
■
■
1 t 5 ■
37. x 1 3t ,
■
■
■
■
■
y 11 sin t 4 sin11t2 What parameter interval gives the complete curve? (b) Use your CAS to find the approximate length of this curve.
equations
0 t 3 ■
■
■
■
■
■
y e t sin t, 0 t
43. x e t t,
y sin t,
■
4 t 34
y 4e t2 , 8 t 3 ■
■
■
■
■
■
■
■
■
44. Find the length of the loop of the curve x 3t t 3,
x Ct
y
y St
y
the curve x t e t, y t e t, 6 t 6.
46. In Exercise 36 in Section 9.1 you were asked to derive the
parametric equations x 2a cot , y 2a sin 2 for the curve called the witch of Maria Agnesi. Use Simpson’s Rule with n 4 to estimate the length of the arc of this curve given by 4 2.
48. x cos 2t,
y cos t, 0 t 4
■
■
■
■
■
■
■
■
■
2
■
s1 e 2 sin 2 d
where e is the eccentricity of the ellipse (e ca, where c sa 2 b 2 ) . 50. Find the total length of the astroid x a cos 3, y a sin 3,
where a 0.
x r cos sin
y r sin cos
y
T
O
y b cos , a b 0, is
0
where C and S are the Fresnel functions that were introduced in Chapter 5. (a) Graph this curve. What happens as t l and as t l ? (b) Find the length of Cornu’s spiral from the origin to the point with parameter value t.
¨
49. Show that the total length of the ellipse x a sin ,
L 4a y
sin u 22 du
r
47– 48 ■ Find the distance traveled by a particle with position x, y as t varies in the given time interval. Compare with the length of the curve.
y cos 2 t, 0 t 3
t
0
cos u 22 du
being held taut. The curve traced by the point P at the end of the string is called the involute of the circle. If the circle has radius r and center O and the initial position of P is r, 0, and if the parameter is chosen as in the figure, show that parametric equations of the involute are
45. Use Simpson’s Rule with n 6 to estimate the length of
47. x sin 2 t,
t
0
53. A string is wound around a circle and then unwound while
y 3t 2.
■
52. A curve called Cornu’s spiral is defined by the parametric
Graph the curve and find its length.
1
■
x 11 cos t 4 cos11t2
y asin cos ,
y 5 2t,
42. x cos t ln(tan 2 t),
■
495
51. (a) Graph the epitrochoid with equations
■
y ln1 t, 0 t 2
41. x e t cos t,
■
■
CAS
t , 1t
40. x e t et,
; 41– 43
■
y 4 2t , 0 t 1
0
■
■
■
3
38. x acos sin ,
■
CAS
■
Find the length of the curve. 2
39. x
■
CALCULUS WITH PARAMETRIC CURVES
P x
54. A cow is tied to a silo with radius r by a rope just long
enough to reach the opposite side of the silo. Find the area available for grazing by the cow.
496
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CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
9.3
POLAR COORDINATES
P (r, ¨ )
r
¨
O
polar axis
x
FIGURE 1
(r, ¨ )
¨+π
¨ O
A coordinate system represents a point in the plane by an ordered pair of numbers called coordinates. Usually we use Cartesian coordinates, which are directed distances from two perpendicular axes. Here we describe a coordinate system introduced by Newton, called the polar coordinate system, which is more convenient for many purposes. We choose a point in the plane that is called the pole (or origin) and is labeled O. Then we draw a ray (half-line) starting at O called the polar axis. This axis is usually drawn horizontally to the right and corresponds to the positive x-axis in Cartesian coordinates. If P is any other point in the plane, let r be the distance from O to P and let be the angle (usually measured in radians) between the polar axis and the line OP as in Figure 1. Then the point P is represented by the ordered pair r, and r, are called polar coordinates of P. We use the convention that an angle is positive if measured in the counterclockwise direction from the polar axis and negative in the clockwise direction. If P O, then r 0 and we agree that 0, represents the pole for any value of . We extend the meaning of polar coordinates r, to the case in which r is negative by agreeing that, as in Figure 2, the points r, and r, lie on the same line through O and at the same distance r from O, but on opposite sides of O. If r 0, the point r, lies in the same quadrant as ; if r 0, it lies in the quadrant on the opposite side of the pole. Notice that r, represents the same point as r, .
EXAMPLE 1 Plot the points whose polar coordinates are given. (_r, ¨)
(a) 1, 54
FIGURE 2
SOLUTION The points are plotted in Figure 3. In part (d) the point 3, 34 is located three units from the pole in the fourth quadrant because the angle 34 is in the second quadrant and r 3 is negative.
5π 4
(b) 2, 3
3π O
(2, 3π)
(c) 2, 23
(d) 3, 34
3π 4
O O
O
_ 2π 3
5π ”1, ’ 4 2π ”2, _ ’ 3
”_3, 3π ’ 4
FIGURE 3
■
In the Cartesian coordinate system every point has only one representation, but in the polar coordinate system each point has many representations. For instance, the point 1, 54 in Example 1(a) could be written as 1, 34 or 1, 134 or 1, 4. (See Figure 4.)
5π 4
13π 4
O O
”1, 5π ’ 4
FIGURE 4
_ 3π 4 ”1, _ 3π ’ 4
”1, 13π ’ 4
π 4
O
O π
”_1, ’ 4
SECTION 9.3
POLAR COORDINATES
■
497
In fact, since a complete counterclockwise rotation is given by an angle 2, the point represented by polar coordinates r, is also represented by r, 2n y P (r, ¨ ) =P (x, y)
r
y
cos
FIGURE 5
x
x
(r, 2n 1)
where n is any integer. The connection between polar and Cartesian coordinates can be seen from Figure 5, in which the pole corresponds to the origin and the polar axis coincides with the positive x-axis. If the point P has Cartesian coordinates x, y and polar coordinates r, , then, from the figure, we have
¨ O
and
x r
sin
y r
and so 1
x r cos
y r sin
Although Equations 1 were deduced from Figure 5, which illustrates the case where r 0 and 0 2, these equations are valid for all values of r and . (See the general definition of sin and cos in Appendix A.) Equations 1 allow us to find the Cartesian coordinates of a point when the polar coordinates are known. To find r and when x and y are known, we use the equations
2
r2 x2 y2
tan
y x
which can be deduced from Equations 1 or simply read from Figure 5. EXAMPLE 2 Convert the point 2, 3 from polar to Cartesian coordinates. SOLUTION Since r 2 and 3, Equations 1 give
x r cos 2 cos y r sin 2 sin
1 2 1 3 2 s3 2 s3 3 2
Therefore, the point is (1, s3 ) in Cartesian coordinates. EXAMPLE 3 Represent the point with Cartesian coordinates 1, 1 in terms of
polar coordinates. SOLUTION If we choose r to be positive, then Equations 2 give
r sx 2 y 2 s1 2 1 2 s2 tan
y 1 x
■
498
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
Since the point 1, 1 lies in the fourth quadrant, we can choose 4 or 74. Thus one possible answer is (s2 , 4); another is s2 , 74.
■
NOTE Equations 2 do not uniquely determine when x and y are given because, as increases through the interval 0 2, each value of tan occurs twice. Therefore, in converting from Cartesian to polar coordinates, it’s not good enough just to find r and that satisfy Equations 2. As in Example 3, we must choose so that the point r, lies in the correct quadrant.
POLAR CURVES
The graph of a polar equation r f , or more generally Fr, 0, consists of all points P that have at least one polar representation r, whose coordinates satisfy the equation. V EXAMPLE 4
1
r= 2
r=4 r=2 r=1
What curve is represented by the polar equation r 2?
SOLUTION The curve consists of all points r, with r 2. Since r represents the distance from the point to the pole, the curve r 2 represents the circle with center O and radius 2. In general, the equation r a represents a circle with center O and radius a . (See Figure 6.) ■
EXAMPLE 5 Sketch the polar curve 1. SOLUTION This curve consists of all points r, such that the polar angle is 1 radian. It is the straight line that passes through O and makes an angle of 1 radian with the polar axis (see Figure 7). Notice that the points r, 1 on the line with r 0 are in the first quadrant, whereas those with r 0 are in the third quadrant.
FIGURE 6
(3, 1) (2, 1)
¨=1 (1, 1) O
1
(_1, 1) (_2, 1)
FIGURE 7
■
EXAMPLE 6
(a) Sketch the curve with polar equation r 2 cos . (b) Find a Cartesian equation for this curve. SOLUTION
(a) In Figure 8 we find the values of r for some convenient values of and plot the corresponding points r, . Then we join these points to sketch the curve, which appears to be a circle. We have used only values of between 0 and , since if we let increase beyond , we obtain the same points again.
SECTION 9.3
The curve in Example 6 is symmetric about the polar axis because cos cos . ■
r 2 cos
0 6 4 3 2 23 34 56
FIGURE 8
Table of values and graph of r=2 cos ¨
” œ„, ’ 2 π4
π ”1, ’ 3
2 s3 s2 1 0 1 s2 s3 2
■
499
” œ„, ’ 3 π6
(2, 0) π ”0, ’ 2
2π ”_1, ’ 3
”_ œ„, ’ 2 3π 4
3 5π ’ ”_ œ„, 6
(b) To convert the given equation into a Cartesian equation we use Equations 1 and 2. From x r cos we have cos xr, so the equation r 2 cos becomes r 2xr, which gives
■ Figure 9 shows a geometrical illustration that the circle in Example 6 has the equation r 2 cos . The angle OPQ is a right angle (Why?) and so r2 cos .
2x r 2 x 2 y 2
y
P
x 2 y 2 2x 0
or
Completing the square, we obtain
r ¨
O
POLAR COORDINATES
2
Q
x 12 y 2 1
x
which is an equation of a circle with center 1, 0 and radius 1. V EXAMPLE 7
FIGURE 9
■
Sketch the curve r 1 sin .
SOLUTION Instead of plotting points as in Example 6, we first sketch the graph of
r 1 sin in Cartesian coordinates in Figure 10 by shifting the sine curve up one unit. This enables us to read at a glance the values of r that correspond to increasing values of . For instance, we see that as increases from 0 to 2, r (the distance from O ) increases from 1 to 2, so we sketch the corresponding part of the polar curve in Figure 11(a). As increases from 2 to , Figure 10 shows that r decreases from 2 to 1, so we sketch the next part of the curve as in Figure 11(b). As increases from to 32, r decreases from 1 to 0 as shown in part (c). Finally, as increases from 32 to 2, r increases from 0 to 1 as shown in part (d). If we let increase beyond 2 or decrease beyond 0, we would simply retrace our path. Putting together the parts of the curve from Figure 11(a)–(d), we sketch the complete curve in part (e). It is called a cardioid because it’s shaped like a heart.
r 2 1 0
π 2
π
3π 2
2π ¨
FIGURE 10
r=1+sin ¨ in Cartesian coordinates, 0¯¨¯2π π
π
¨= 2
¨= 2
2 O O
(a)
1
¨=0
¨=π
O
¨=π
(b)
FIGURE 11 Stages in sketching the cardioid r=1+sin ¨
O
O ¨=2π
3π
3π
¨= 2
¨= 2
(c)
(d)
(e)
500
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
EXAMPLE 8 Sketch the curve r cos 2.
Module 9.3 helps you see how polar curves are traced out by showing animations similar to Figures 10–13.
SOLUTION As in Example 7, we first sketch r cos 2, 0 2, in Cartesian coordinates in Figure 12. As increases from 0 to 4, Figure 12 shows that r decreases from 1 to 0 and so we draw the corresponding portion of the polar curve in Figure 13 (indicated by !). As increases from 4 to 2, r goes from 0 to 1. This means that the distance from O increases from 0 to 1, but instead of being in the first quadrant this portion of the polar curve (indicated by @) lies on the opposite side of the pole in the third quadrant. The remainder of the curve is drawn in a similar fashion, with the arrows and numbers indicating the order in which the portions are traced out. The resulting curve has four loops and is called a four-leaved rose.
r
π
¨= 2
1
¨=
!
$
π 4
@
π 2
3π 4
%
π
*
5π 4
#
3π 2
^
7π 4
2π
¨
π
3π 4
&
¨= 4
^
$
!
%
⑧
¨=π
&
¨=0
@
#
FIGURE 12
FIGURE 13
r=cos 2¨ in Cartesian coordinates
Four-leaved rose r=cos 2¨
■
TANGENTS TO POLAR CURVES
To find a tangent line to a polar curve r f we regard as a parameter and write its parametric equations as x r cos f cos
y r sin f sin
Then, using the method for finding slopes of parametric curves (Equation 9.2.1) and the Product Rule, we have
3
dy dr sin r cos dy d d dx dx dr cos r sin d d
We locate horizontal tangents by finding the points where dyd 0 (provided that dxd 0 ). Likewise, we locate vertical tangents at the points where dxd 0 (provided that dyd 0). Notice that if we are looking for tangent lines at the pole, then r 0 and Equation 3 simplifies to dy dr tan if 0 dx d For instance, in Example 8 we found that r cos 2 0 when 4 or 34. This means that the lines 4 and 34 (or y x and y x) are tangent lines to r cos 2 at the origin.
SECTION 9.3
POLAR COORDINATES
■
501
EXAMPLE 9
(a) For the cardioid r 1 sin of Example 7, find the slope of the tangent line when 3. (b) Find the points on the cardioid where the tangent line is horizontal or vertical. SOLUTION Using Equation 3 with r 1 sin , we have
dr sin r cos dy d cos sin 1 sin cos dx dr cos cos 1 sin sin cos r sin d
cos 1 2 sin cos 1 2 sin 2 1 2 sin sin 1 sin 1 2 sin
(a) The slope of the tangent at the point where 3 is dy dx
3
1 cos31 2 sin3 2 (1 s3 ) 1 sin31 2 sin3 (1 s32)(1 s3 )
1 s3 1 s3 (2 s3 )(1 s3 ) 1 s3 1
(b) Observe that
π
”2, ’ 2 3 π ”1+ œ„ , ’ 2 3
m=_1
dy cos 1 2 sin 0 d
when
3 7 11 , , , 2 2 6 6
dx 1 sin 1 2 sin 0 d
when
3 5 , , 2 6 6
Therefore, there are horizontal tangents at the points 2, 2, ( 12 , 76), ( 12 , 116) and vertical tangents at ( 32 , 6) and ( 32 , 56). When 32, both dyd and dxd are 0, so we must be careful. Using l’Hospital’s Rule, we have lim
l 32
” 32 , π6 ’
3 5π ” , ’ 2 6
By symmetry,
lim
l 32
(0, 0) 1 7π 1 11π ” , ’ ” , ’ 2 6 2 6
dy dx
1 3
lim
1 2 sin 1 2 sin
l 32
lim
l 32
cos 1 1 sin 3
lim
l 32
cos 1 sin lim
l 32
sin cos
dy dx
FIGURE 14
Tangent lines for r=1+sin ¨
Thus there is a vertical tangent line at the pole (see Figure 14).
■
NOTE Instead of having to remember Equation 3, we could employ the method used to derive it. For instance, in Example 9 we could have written
x r cos 1 sin cos cos 12 sin 2 y r sin 1 sin sin sin sin 2
502
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
Then we have dy dyd cos 2 sin cos cos sin 2 dx dxd sin cos 2 sin cos 2 which is equivalent to our previous expression. GRAPHING POLAR CURVES WITH GRAPHING DEVICES
Although it’s useful to be able to sketch simple polar curves by hand, we need to use a graphing calculator or computer when we are faced with a curve as complicated as the one shown in Figure 15. Some graphing devices have commands that enable us to graph polar curves directly. With other machines we need to convert to parametric equations first. In this case we take the polar equation r f and write its parametric equations as
2
_1.4
1.4
x r cos f cos
_0.3
FIGURE 15
y r sin f sin
Some machines require that the parameter be called t rather than .
r=sin ¨+sin#(5¨/2)
EXAMPLE 10 Graph the curve r sin85. SOLUTION Let’s assume that our graphing device doesn’t have a built-in polar graphing command. In this case we need to work with the corresponding parametric equations, which are
x r cos sin85 cos
In any case we need to determine the domain for . So we ask ourselves: How many complete rotations are required until the curve starts to repeat itself? If the answer is n, then 8 2n 8 16n 8 sin sin sin 5 5 5 5
1
_1
y r sin sin85 sin
1
and so we require that 16n5 be an even multiple of . This will first occur when n 5. Therefore, we will graph the entire curve if we specify that 0 10. Switching from to t, we have the equations _1
x sin8t5 cos t
FIGURE 16
9.3
EXERCISES
■ Plot the point whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with r 0 and one with r 0.
■ Plot the point whose polar coordinates are given. Then find the Cartesian coordinates of the point.
1–2
1. (a) 1, 2
(b) 2, 4
■
■
■
■
■
3– 4
(c) 3, 2
(b) 2, 7
2. (a) 3, 0 ■
■
and Figure 16 shows the resulting curve. Notice that this rose has 16 loops.
r=sin(8¨/5)
■
0 t 10
y sin8t5 sin t
(c) 1, 2 ■
■
■
■
■
■
3. (a) 3, 2
(b) (2 s2 , 34)
(c) 1, 3
4. (a) 2, 23
(b) 4, 3
(c) 2, 56
■
■
■
■
■
■
■
■
■
■
■
SECTION 9.3 ■ The Cartesian coordinates of a point are given. (i) Find polar coordinates r, of the point, where r 0 and 0 2. (ii) Find polar coordinates r, of the point, where r 0 and 0 2.
5–6
6. (a) (1, s3 ) ■
■
■
■
■
■
■
■
■
■
■
Sketch the region in the plane consisting of points whose polar coordinates satisfy the given conditions. 7–12
34. r sin 5
35. r 2 4 cos 2
36. r 2 sin 2
37. r 2 cos32
38. r 2 1
■
(b) 2, 3 ■
33. r 2 cos 4
39. r 1 2 cos 2
(b) (2 s3 , 2)
5. (a) 1, 1
■
41.
9. 0 r 4,
2 6
10. 2 r 5,
34 54
11. 2 r 3,
53 73
■
■
■
■
■
40. r 1 2 cos2 ■
■
■
■
r
42. r
2
2
0
■
■
■
■
0
2π ¨
π
2π ¨
π
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
conchoid) has the line x 2 as a vertical asymptote by showing that lim r l x 2. Use this fact to help sketch the conchoid.
■
44. Sketch the curve x 2 y 2 3 4x 2 y 2.
13. r 3 sin
14. r 2 sin 2 cos
45. Show that the curve r sin tan (called a cissoid of
15. r csc
16. r tan sec
■
■
■
■
■
■
■
■
■
■
■
■
46. Match the polar equations with the graphs labeled I–VI.
Give reasons for your choices. (Don’t use a graphing device.) (a) r sin2 (b) r sin4 (c) r sec3 (d) r sin (e) r 1 4 cos 5 (f ) r 1s
18. x y 9
19. x 2 y 2 2cx ■
Diocles) has the line x 1 as a vertical asymptote. Show also that the curve lies entirely within the vertical strip 0 x 1. Use these facts to help sketch the cissoid.
■
17–20 ■ Find a polar equation for the curve represented by the given Cartesian equation. 17. x y 2 ■
20. x 2 y 2 1 ■
■
■
■
43. Show that the polar curve r 4 2 sec (called a
13–16 ■ Identify the curve by finding a Cartesian equation for the curve.
■
■
_2
4 34 ■
■
1
3 23
12. 1 r 1,
■
503
41– 42 ■ The figure shows the graph of r as a function of in Cartesian coordinates. Use it to sketch the corresponding polar curve.
7. 1 r 2 8. r 0,
■
■
POLAR COORDINATES
■
■
■
■
■
■ For each of the described curves, decide if the curve would be more easily given by a polar equation or a Cartesian equation. Then write an equation for the curve.
21–22
I
II
III
IV
V
VI
21. (a) A line through the origin that makes an angle of 6
with the positive x-axis (b) A vertical line through the point 3, 3 22. (a) A circle with radius 5 and center 2, 3
(b) A circle centered at the origin with radius 4 ■
■
23– 40
■
■
■
■
■
■
■
■
■
■
■
Sketch the curve with the given polar equation.
23. 6
24. r 2 3r 2 0
25. r sin
26. r 3 cos
27. r 21 sin , 0
28. r 1 3 cos
29. r , 0
30. r ln , 1
47–50 ■ Find the slope of the tangent line to the given polar curve at the point specified by the value of .
31. r sin 2
32. r 2 cos 3
47. r 2 sin ,
6
48. r 2 sin ,
3
504
■
CHAPTER 9
49. r 1, ■
■
PARAMETRIC EQUATIONS AND POLAR COORDINATES
■
■
50. r sin 3, ■
■
■
■
■
6 ■
■
; 65. A family of curves has polar equations ■
r
Find the points on the given curve where the tangent line is horizontal or vertical. 51–54
■
51. r 3 cos
52. r e
53. r 1 cos
54. r 2 sin 2
■
■
■
■
■
■
■
■
■
■
■
Investigate how the graph changes as the number a changes. In particular, you should identify the transitional values of a for which the basic shape of the curve changes. ■
; 66. The astronomer Giovanni Cassini (1625–1712) studied the family of curves with polar equations
55. Show that the polar equation r a sin b cos , where
r 4 2c 2 r 2 cos 2 c 4 a 4 0
ab 0, represents a circle, and find its center and radius.
where a and c are positive real numbers. These curves are called the ovals of Cassini even though they are oval shaped only for certain values of a and c. (Cassini thought that these curves might represent planetary orbits better than Kepler’s ellipses.) Investigate the variety of shapes that these curves may have. In particular, how are a and c related to each other when the curve splits into two parts?
56. Show that the curves r a sin and r a cos intersect
at right angles.
; 57–60
■ Use a graphing device to graph the polar curve. Choose the parameter interval to make sure that you produce the entire curve.
57. r e sin 2 cos4
(butterfly curve)
67. Let P be any point (except the origin) on the curve r f .
58. r sin 4 cos4 2
59. r 2 5 sin6 ■
■
■
■
■
1 a cos 1 a cos
If # is the angle between the tangent line at P and the radial line OP, show that r tan # drd
60. r cos 2 cos 3 ■
■
■
■
■
■
■
; 61. How are the graphs of r 1 sin 6 and
[Hint: Observe that # in the figure.]
r 1 sin 3 related to the graph of r 1 sin ? In general, how is the graph of r f related to the graph of r f ?
r=f(¨ ) ÿ
; 62. Use a graph to estimate the y-coordinate of the highest
P
points on the curve r sin 2. Then use calculus to find the exact value. ¨
; 63. (a) Investigate the family of curves defined by the polar
equations r sin n, where n is a positive integer. How is the number of loops related to n ? (b) What happens if the equation in part (a) is replaced by r sin n ?
; 64. A family of curves is given by the equations
r 1 c sin n, where c is a real number and n is a positive integer. How does the graph change as n increases? How does it change as c changes? Illustrate by graphing enough members of the family to support your conclusions.
9.4
r ¨ FIGURE 1
˙
O
68. (a) Use Exercise 67 to show that the angle between the tan-
;
gent line and the radial line is # 4 at every point on the curve r e . (b) Illustrate part (a) by graphing the curve and the tangent lines at the points where 0 and 2. (c) Prove that any polar curve r f with the property that the angle # between the radial line and the tangent line is a constant must be of the form r Ce k, where C and k are constants.
AREAS AND LENGTHS IN POLAR COORDINATES In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the area of a sector of a circle 1
A 12 r 2
where, as in Figure 1, r is the radius and is the radian measure of the central angle.
SECTION 9.4
b
O
■
505
Formula 1 follows from the fact that the area of a sector is proportional to its central angle: A 2 r 2 12 r 2. (See also Exercise 67 in Section 6.2.) Let be the region, illustrated in Figure 2, bounded by the polar curve r f and by the rays a and b, where f is a positive continuous function and where 0 b a 2. We divide the interval a, b into subintervals with endpoints 0 , 1 , 2 , . . . , n and equal width . The rays i then divide into n smaller regions with central angle i i1 . If we choose i* in the ith subinterval i1, i , then the area Ai of the ith region is approximated by the area of the sector of a circle with central angle and radius f i*. (See Figure 3.) Thus from Formula 1 we have
r=f(¨) ¨=b
AREAS AND LENGTHS IN POLAR COORDINATES
¨=a a
FIGURE 2
Ai 12 f i* 2 ¨=¨ i
f(¨ i*)
¨=¨ i-1
and so an approximation to the total area A of is n
A
2
¨=b
1 2
f i* 2
i1
Ψ
It appears from Figure 3 that the approximation in (2) improves as n l . But the sums in (2) are Riemann sums for the function t 12 f 2, so
¨=a O
n
FIGURE 3
lim
n l i1
1 2
f i* 2 y
b 1 2
a
f 2 d
It therefore appears plausible (and can in fact be proved) that the formula for the area A of the polar region is Ay
3
b 1 2
a
f 2 d
Formula 3 is often written as
Ay
4
b 1 2
a
r 2 d
with the understanding that r f . Note the similarity between Formulas 1 and 4. When we apply Formula 3 or 4 it is helpful to think of the area as being swept out by a rotating ray through O that starts with angle a and ends with angle b. r=cos 2¨
π
¨= 4
V EXAMPLE 1
Find the area enclosed by one loop of the four-leaved rose
r cos 2. SOLUTION The curve r cos 2 was sketched in Example 8 in Section 9.3. Notice from Figure 4 that the region enclosed by the right loop is swept out by a ray that rotates from 4 to 4 . Therefore, Formula 4 gives
π ¨=_ 4
FIGURE 4
4 1 2 4
Ay y
4 1 2
0
r 2 d 12 y
4
4
cos 2 2 d y
4
0
[
1 cos 4 d 12 14 sin 4
cos 2 2 d 4 0
]
8
■
506
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
Find the area of the region that lies inside the circle r 3 sin and outside the cardioid r 1 sin . V EXAMPLE 2
SOLUTION The cardioid (see Example 7 in Section 9.3) and the circle are sketched in Figure 5 and the desired region is shaded. The values of a and b in Formula 4 are determined by finding the points of intersection of the two curves. They intersect when 3 sin 1 sin , which gives sin 12 , so 6, 56. The desired area can be found by subtracting the area inside the cardioid between 6 and 56 from the area inside the circle from 6 to 56. Thus
r=3 sin ¨
π
5π
¨= 6
¨= 6
O
A 12 y
r=1+sin ¨
56
6
3 sin 2 d 12 y
1 sin 2 d
Since the region is symmetric about the vertical axis 2, we can write
FIGURE 5
y
A2
1 2
y
2
y
2
6
6
2
6
9 sin 2 d 12 y
2
6
1 2 sin sin 2 d
8 sin 2 1 2 sin d 3 4 cos 2 2 sin d
3 2 sin 2 2 cos
2 6
]
[because sin 2 12 1 cos 2 ]
■
Example 2 illustrates the procedure for finding the area of the region bounded by two polar curves. In general, let be a region, as illustrated in Figure 6, that is bounded by curves with polar equations r f , r t , a, and b, where f t 0 and 0 b a 2. The area A of is found by subtracting the area inside r t from the area inside r f , so using Formula 3 we have
r=f(¨) ¨=b
56
6
r=g(¨) ¨=a
O
Ay
b 1 2
a
FIGURE 6
|
f 2 d y
b 1 2
a
t 2 d 12 y ( f 2 t 2 ) d b
a
CAUTION The fact that a single point has many representations in polar coordinates sometimes makes it difficult to find all the points of intersection of two polar curves. For instance, it is obvious from Figure 5 that the circle and the cardioid have three points of intersection; however, in Example 2 we solved the equations r 3 sin and r 1 sin and found only two such points, ( 32, 6) and ( 32, 56). The origin is also a point of intersection, but we can’t find it by solving the equations of the curves because the origin has no single representation in polar coordinates that satisfies both equations. Notice that, when represented as 0, 0 or 0, , the origin satisfies r 3 sin and so it lies on the circle; when represented as 0, 32, it satisfies r 1 sin and so it lies on the cardioid. Think of two points moving along the curves as the parameter value increases from 0 to 2. On one curve the origin is reached at 0 and ; on the other curve it is reached at 32. The points don’t collide at the origin because they reach the origin at different times, but the curves intersect there nonetheless. Thus, to find all points of intersection of two polar curves, it is recommended that you draw the graphs of both curves. It is especially convenient to use a graphing calculator or computer to help with this task.
SECTION 9.4
■
AREAS AND LENGTHS IN POLAR COORDINATES
507
EXAMPLE 3 Find all points of intersection of the curves r cos 2 and r 2 . 1
1 π ” , 3 2 ’ 1 π ” 2 , ’ 6
r=21
r=cos 2¨
SOLUTION If we solve the equations r cos 2 and r 2 , we get cos 2 2 and, 1
1
therefore, 2 3, 53, 73, 113. Thus the values of between 0 and 2 that satisfy both equations are 6, 56, 76, 116. We have found four points of intersection: ( 12, 6), ( 12, 56), ( 12, 76), and ( 12, 116). However, you can see from Figure 7 that the curves have four other points of intersection—namely, ( 12, 3), ( 12, 23), ( 12, 43), and ( 12, 53). These can be found using symmetry or by noticing that another equation of the circle is r 12 and then solving the equations r cos 2 and r 12 . ■ ARC LENGTH
FIGURE 7
To find the length of a polar curve r f , a b, we regard as a parameter and write the parametric equations of the curve as x r cos f cos
y r sin f sin
Using the Product Rule and differentiating with respect to , we obtain dx dr cos r sin d d
dy dr sin r cos d d
so, using cos 2 sin 2 1, we have
dx d
2
dy d
2
2
dr d
2
dr d
cos 2 2r
dr cos sin r 2 sin 2 d
sin 2 2r
dr sin cos r 2 cos 2 d
2
dr d
r2
Assuming that f is continuous, we can use Formula 9.2.5 to write the arc length as L
y
b
a
dx d
2
2
dy d
d
Therefore, the length of a curve with polar equation r f , a b, is
5
L
y b
a
r2
dr d
2
d
O V EXAMPLE 4
FIGURE 8
r=1+sin ¨
Find the length of the cardioid r 1 sin .
SOLUTION The cardioid is shown in Figure 8. (We sketched it in Example 7 in Section 9.3.) Its full length is given by the parameter interval 0 2, so
508
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
Formula 5 gives L
y
2
0
y
2
0
r2
2
dr d
d y
2
0
s1 sin 2 cos 2 d
s2 2 sin d
We could evaluate this integral by multiplying and dividing the integrand by s2 2 sin , or we could use a computer algebra system. In any event, we find that the length of the cardioid is L 8.
9.4
EXERCISES
■ Find the area of the region that is bounded by the given curve and lies in the specified sector.
15–18 ■ Find the area of the region enclosed by one loop of the curve.
1– 4
1. r s , 2. r e
0 4
, 2
■
5– 8
■
■
■
0 ■
■
16. r 4 sin 3
17. r 1 2 sin (inner loop)
3 23
4. r ssin , ■
15. r sin 2
2
3. r sin ,
■
■
■
■
■
■
■
■
■
■
18. r 2 cos sec ■
■
■
■
■
r2
20. r 1 sin , 21. r 3 cos ,
■
■
23–26 r=¨
■
■
25. r sin 2,
8.
■
■
r 3 sin
■
■
■
■
■
■
■
■
■
r cos
24. r sin 2,
r cos 2 ■
■
r sin
26. r 2 2 sin 2, ■
■
■
■
■
r1 ■
27. Find the area inside the larger loop and outside the smaller
loop of the limaçon r 12 cos . 28. When recording live performances, sound engineers often r=sin 4¨
r=4+3 sin ¨ ■
9–12
■
■
■
■
■
■
■
■
■
■
■
■
use a microphone with a cardioid pickup pattern because it suppresses noise from the audience. Suppose the microphone is placed 4 m from the front of the stage (as in the figure) and the boundary of the optimal pickup region is stage
12. r 2 cos 2 ■
■
■
■
■
■
■
■
12 m
■
Graph the curve and find the area that it encloses.
13. r 1 2 sin 6 ■
■
10. r 31 cos
11. r 2 cos 3
; 13–14
■
Sketch the curve and find the area that it encloses.
9. r 2 4 cos 2
■
■
■
■
4m
14. r 2 sin 3 sin 9 ■
■
■
■
■
■
■
microphone ■
■
Find the area of the region that lies inside both curves.
23. r sin ,
r=1+sin ¨
r1
r 1 cos
22. r 2 sin ,
■
■
■ Find the area of the region that lies inside the first curve and outside the second curve.
19. r 4 sin ,
6.
7.
■
19–22 ■
Find the area of the shaded region.
5.
■
audience
■
SECTION 9.5
given by the cardioid r 8 8 sin , where r is measured in meters and the microphone is at the pole. The musicians want to know the area they will have on stage within the optimal pickup range of the microphone. Answer their question. 29–32
■
Find all points of intersection of the given curves.
29. r cos , 31. r sin ,
■
■
■
■
9.5
35. r 2,
0 2
■
■
0 2 ■
■
■
■
■
■
■
■
■ Use a calculator to find the length of the curve correct to four decimal places.
r 2 cos 2 ■
0 2
37–38
r sin 2
■
509
0 3
34. r e 2,
■
■
Find the exact length of the polar curve.
33. r 3 sin ,
■
r sin 3
32. r 2 sin 2,
■
36. r ,
r 1 cos
30. r cos 3,
33–36
CONIC SECTIONS IN POLAR COORDINATES
37. r 3 sin 2 ■
■
■
■
■
■
■
■
■
38. r 4 sin 3 ■
■
■
■
■
■
■
■
■
CONIC SECTIONS IN POLAR COORDINATES In your previous study of conic sections, parabolas were defined in terms of a focus and directrix whereas ellipses and hyperbolas were defined in terms of two foci. After reviewing those definitions and equations, we present a more unified treatment of all three types of conic sections in terms of a focus and directrix. Furthermore, if we place the focus at the origin, then a conic section has a simple polar equation. In Chapter 10 we will use the polar equation of an ellipse to derive Kepler’s laws of planetary motion. CONICS IN CARTESIAN COORDINATES
parabola
axis focus
F
directrix
vertex FIGURE 1
y
Here we provide a brief reminder of what you need to know about conic sections. A more thorough review can be found on the website www.stewartcalculus.com. Recall that a parabola is the set of points in a plane that are equidistant from a fixed point F (called the focus) and a fixed line (called the directrix). This definition is illustrated by Figure 1. Notice that the point halfway between the focus and the directrix lies on the parabola; it is called the vertex. The line through the focus perpendicular to the directrix is called the axis of the parabola. A parabola has a very simple equation if its vertex is placed at the origin and its directrix is parallel to the x -axis or y -axis. If the focus is on the y -axis at the point 0, p, then the directrix has the equation y p and an equation of the parabola is x 2 4py . [See parts (a) and (b) of Figure 2.] If the focus is on the x -axis at p, 0, then the directrix is x p and an equation is y 2 4px as in parts (c) and (d). y
y
y=_p
(0, p) x
y=_p
(a) ≈=4py, p>0 FIGURE 2
( p, 0)
( p, 0)
0 0
y
0
x
(0, p)
(b) ≈=4py, p<0
x=_p
(c) ¥=4px, p>0
x
0
x
x=_p
(d) ¥=4px, p<0
510
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
y
P(x, y)
F¡(_c, 0)
0
x
F™(c, 0)
FIGURE 3
An ellipse is the set of points in a plane the sum of whose distances from two fixed points F1 and F2 is a constant. These two fixed points are called the foci (plural of focus). An ellipse has a simple equation if we place the foci on the x-axis at the points c, 0 and c, 0 as in Figure 3 so that the origin is halfway between the foci. If the sum of the distances from a point on the ellipse to the foci is 2a, then the points a, 0 and a, 0 where the ellipse meets the x -axis are called the vertices. The y -intercepts are b , where b 2 a 2 c 2. (See Figure 4.) 1
y
The ellipse
(0, b)
(_a, 0)
a
b (_c, 0)
c
0
(a, 0)
FIGURE 4
≈ ¥ + =1 a@ b@
b
y
b
y= a x
(_a, 0) (_c, 0)
(a, 0) (c, 0)
0
ab0
has foci c, 0, where c 2 a 2 b 2, and vertices a, 0.
x
(c, 0)
(0, _b)
y=_ a x
x2 y2 1 a2 b2
x
If the foci of an ellipse are located on the y-axis at 0, c, then we can find its equation by interchanging x and y in (1). A hyperbola is the set of all points in a plane the difference of whose distances from two fixed points F1 and F2 (the foci) is a constant. Notice that the definition of a hyperbola is similar to that of an ellipse; the only change is that the sum of distances has become a difference of distances. If the foci are on the x-axis at c, 0 and the difference of distances is 2a , then the equation of the hyperbola is x 2a 2 y 2b 2 1, where b 2 c 2 a 2. The x-intercepts are a and the points a, 0 and a, 0 are the vertices of the hyperbola. There is no y-intercept and the hyperbola consists of two parts, called its branches. (See Figure 5.) When we draw a hyperbola it is useful to first draw its asymptotes, which are the dashed lines y bax and y bax shown in Figure 5. Both branches of the hyperbola approach the asymptotes; that is, they come arbitrarily close to the asymptotes. 2
The hyperbola x2 y2 1 2 a b2
FIGURE 5 ≈ ¥ - =1 a@ b@
has foci c, 0, where c 2 a 2 b 2, vertices a, 0, and asymptotes y bax.
y
(0, c) a y=_ b x
a
y= b x (0, a) (0, _a)
0
(0, _c) FIGURE 6 ¥ ≈ - =1 a@ b@
x
If the foci of a hyperbola are on the y-axis, then by reversing the roles of x and y we get the graph shown in Figure 6. We have given the standard equations of the conic sections, but any of them can be shifted by replacing x by x h and y by y k. For instance, an ellipse with center h, k has an equation of the form x h2 y k2 1 a2 b2 CONICS IN POLAR COORDINATES
In the following theorem we show how all three types of conic sections can be characterized in terms of a focus and directrix.
SECTION 9.5
CONIC SECTIONS IN POLAR COORDINATES
■
511
3 THEOREM Let F be a fixed point (called the focus) and l be a fixed line (called the directrix) in a plane. Let e be a fixed positive number (called the eccentricity). The set of all points P in the plane such that
PF e Pl (that is, the ratio of the distance from F to the distance from l is the constant e) is a conic section. The conic is (a) an ellipse if e 1 (b) a parabola if e 1 (c) a hyperbola if e 1
y
l (directrix) P r
x=d ¨
F
x
r cos ¨
PROOF Notice that if the eccentricity is e 1, then PF Pl and so the given condition simply becomes the definition of a parabola. Let us place the focus F at the origin and the directrix parallel to the y-axis and d units to the right. Thus the directrix has equation x d and is perpendicular to the polar axis. If the point P has polar coordinates r, , we see from Figure 7 that
PF r Pl d r cos Thus the condition PF Pl e, or PF e Pl , becomes r ed r cos
4
d C
FIGURE 7
If we square both sides of this polar equation and convert to rectangular coordinates, we get x 2 y 2 e 2d x2 e 2d 2 2dx x 2 1 e 2 x 2 2de 2x y 2 e 2d 2
or
After completing the square, we have 5
x
e 2d 1 e2
2
y2 e 2d 2 2 1e 1 e 2 2
If e 1, we recognize Equation 5 as the equation of an ellipse. In fact, it is of the form x h2 y2 1 a2 b2 where 6
h
e 2d 1 e2
a2
e 2d 2 1 e 2 2
b2
e 2d 2 1 e2
We know that the foci of an ellipse are at a distance c from the center, where 7
c2 a2 b2
e 4d 2 1 e 2 2
512
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
e 2d h 1 e2
c
This shows that
and confirms that the focus as defined in Theorem 3 means the same as the focus defined earlier. It also follows from Equations 6 and 7 that the eccentricity is given by c e a y
x=d directrix
If e 1, then 1 e 2 0 and we see that Equation 5 represents a hyperbola. Just as we did before, we could rewrite Equation 5 in the form x h2 y2 1 a2 b2
x
F
and see that e
ed (a) r= 1+e cos ¨
c a
where c 2 a 2 b 2
By solving Equation 4 for r, we see that the polar equation of the conic shown in Figure 7 can be written as
y
r
x=_ d directrix F
(b) r=
■
x
ed 1 e cos
If the directrix is chosen to be to the left of the focus as x d, or if the directrix is chosen to be parallel to the polar axis as y d, then the polar equation of the conic is given by the following theorem, which is illustrated by Figure 8. (See Exercises 19–21.)
ed 1-e cos ¨ 8 THEOREM
A polar equation of the form
y
y=d
directrix
x
F
(c) r=
r
r
ed 1 e sin
represents a conic section with eccentricity e. The conic is an ellipse if e 1, a parabola if e 1, or a hyperbola if e 1.
SOLUTION Using Theorem 8 with e 1 and d 6, and using part (d) of Figure 8, we see that the equation of the parabola is
r
x
F
(d) r=
or
V EXAMPLE 1 Find a polar equation for a parabola that has its focus at the origin and whose directrix is the line y 6.
ed 1+e sin ¨
y
y=_d
ed 1 e cos
directrix ed 1-e sin ¨
V EXAMPLE 2
6 1 sin
A conic is given by the polar equation r
10 3 2 cos
FIGURE 8
Polar equations of conics
Find the eccentricity, identify the conic, locate the directrix, and sketch the conic.
■
SECTION 9.5
CONIC SECTIONS IN POLAR COORDINATES
■
513
SOLUTION Dividing numerator and denominator by 3, we write the equation as
r y
d
focus 0
(10, 0)
x
10 3
e
10 3 2 3
5
so the directrix has Cartesian equation x 5. When 0, r 10; when , r 2. So the vertices have polar coordinates 10, 0 and 2, . The ellipse is sketched in Figure 9. ■
(2, π)
FIGURE 9
EXAMPLE 3 Sketch the conic r
π
r
”6, ’ 2 π
”2, ’ 2
12 . 2 4 sin
SOLUTION Writing the equation in the form
y
6 1 2 sin
y=3 (directrix) we see that the eccentricity is e 2 and the equation therefore represents a hyper(6, π) 0
(6, 0)
x
focus FIGURE 10
r=
1 cos
From Theorem 8 we see that this represents an ellipse with e 23 . Since ed 103 , we have
10
r= 3-2 cos ¨
x=_5 (directrix)
10 3 2 3
12 2+4 sin ¨
bola. Since ed 6, d 3 and the directrix has equation y 3. The vertices occur when 2 and 32, so they are 2, 2 and 6, 32, which can be written as 6, 2. It is also useful to plot the x-intercepts. These occur when 0, ; in both cases r 6. For additional accuracy we could draw the asymptotes. Note that r l when 2 4 sin l 0 or 0 and 2 4 sin 0 when sin 12 . Thus the asymptotes are parallel to the rays 76 and 116. The hyperbola is sketched in Figure 10. ■ In Figure 11 we use a computer to sketch a number of conics to demonstrate the effect of varying the eccentricity e. Notice that when e is close to 0 the ellipse is nearly circular, whereas it becomes more elongated as e l 1. When e 1, of course, the conic is a parabola.
e=0.1
e=1 FIGURE 11
e=0.5
e=0.68
e=1.1
e=0.86
e=1.4
e=0.96
e=4
514
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
9.5
EXERCISES 22. Show that the parabolas r c1 cos and
Write a polar equation of a conic with the focus at the origin and the given data. 1– 8
■
1. Hyperbola,
x d can be written in the form
3 eccentricity 4, directrix x 5
3. Ellipse,
4. Hyperbola,
eccentricity 2, directrix y 2
6. Ellipse,
eccentricity 0.8, vertex 1, 2
7. Ellipse,
eccentricity 12, directrix r 4 sec
8. Hyperbola, ■
r
24. (a) The planets move around the Sun in elliptical orbits
eccentricity 3, directrix r 6 csc ■
■
■
■
■
■
■
■
■
9–16 ■ (a) Find the eccentricity, (b) identify the conic, (c) give an equation of the directrix, and (d) sketch the conic. 9. r
1 1 sin
10. r
6 3 2 sin
12 11. r 4 sin
4 12. r 2 3 cos
9 13. r 6 2 cos
5 14. r 2 2 sin
15. r ■
■
3 4 8 cos ■
■
16. r ■
■
■
■
■
r aphelion
■
; 17. Graph the conics r e1 e cos with e 0.4, 0.6,
0.8, and 1.0 on a common screen. How does the value of e affect the shape of the curve?
; 18. (a) Graph the conics r ed1 e sin for e 1 and
various values of d. How does the value of d affect the shape of the conic? (b) Graph these conics for d 1 and various values of e. How does the value of e affect the shape of the conic?
19. Show that a conic with focus at the origin, eccentricity e,
and directrix x d has polar equation r
ed 1 e cos
20. Show that a conic with focus at the origin, eccentricity e,
and directrix y d has polar equation r
ed 1 e sin
21. Show that a conic with focus at the origin, eccentricity e,
and directrix y d has polar equation ed r 1 e sin
with the Sun at one focus. The positions of a planet that are closest to and farthest from the Sun are called its perihelion and aphelion, respectively. Use Exercise 23(a) to show that the perihelion distance from a planet to the Sun is a1 e and the aphelion distance is a1 e. planet
4 2 cos ■
a1 e 2 1 e cos
(b) Find an approximate polar equation for the elliptical orbit of the Earth around the Sun (at one focus) given that the eccentricity is about 0.017 and the length of the major axis is about 2.99 10 8 km.
vertex 4, 32
5. Parabola,
■
23. (a) Show that the polar equation of an ellipse with directrix
directrix x 4
2. Parabola,
■
r d1 cos intersect at right angles.
eccentricity 74, directrix y 6
■
¨ Sun
perihelion
(b) Use the data of Exercise 23(b) to find the distances from the Earth to the Sun at perihelion and at aphelion. 25. The orbit of Halley’s comet, last seen in 1986 and due to
return in 2062, is an ellipse with eccentricity 0.97 and one focus at the Sun. The length of its major axis is 36.18 AU. [An astronomical unit (AU) is the mean distance between the Earth and the Sun, about 93 million miles.] Find a polar equation for the orbit of Halley’s comet. What is the maximum distance from the comet to the Sun? 26. The Hale-Bopp comet, discovered in 1995, has an elliptical
orbit with eccentricity 0.9951 and the length of the major axis is 356.5 AU. Find a polar equation for the orbit of this comet. How close to the Sun does it come? 27. The planet Mercury travels in an elliptical orbit with eccen-
tricity 0.206. Its minimum distance from the Sun is 4.6 10 7 km. Use the results of Exercise 24(a) to find its maximum distance from the Sun. 28. The distance from the planet Pluto to the Sun is
4.43 10 9 km at perihelion and 7.37 10 9 km at aphelion. Use Exercise 24 to find the eccentricity of Pluto’s orbit. 29. Using the data from Exercise 27, find the distance traveled
by the planet Mercury during one complete orbit around the Sun. (If your calculator or computer algebra system evaluates definite integrals, use it. Otherwise, use Simpson’s Rule.)
CHAPTER 9
9
REVIEW
■
REVIEW
515
CONCEPT CHECK
1. (a) What is a parametric curve?
(c) What equations would you use to find the polar coordinates of a point if you knew the Cartesian coordinates?
(b) How do you sketch a parametric curve?
5. (a) How do you find the slope of a tangent line to a polar
2. (a) How do you find the slope of a tangent to a parametric
curve? (b) How do you find the area of a region bounded by a polar curve? (c) How do you find the length of a polar curve?
curve? (b) How do you find the area under a parametric curve? 3. Write an expression for the length of a parametric
6. (a) What is the eccentricity of a conic section?
curve.
(b) What can you say about the eccentricity if the conic section is an ellipse? A hyperbola? A parabola? (c) Write a polar equation for a conic section with eccentricity e and directrix x d. What if the directrix is x d ? y d ? y d ?
4. (a) Use a diagram to explain the meaning of the polar coor-
dinates r, of a point. (b) Write equations that express the Cartesian coordinates x, y of a point in terms of the polar coordinates.
T R U E - FA L S E Q U I Z 4. If a point is represented by x, y in Cartesian coordinates
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
(where x 0) and r, in polar coordinates, then tan 1 yx.
1. If the parametric curve x f t, y tt satisfies
5. The polar curves r 1 sin 2 and r sin 2 1 have
t1 0, then it has a horizontal tangent when t 1.
the same graph.
2. If x f t and y tt are twice differentiable, then 2
2
6. The equations r 2, x 2 y 2 4, and x 2 sin 3t,
y 2 cos 3t 0 t 2 all have the same graph.
2
d ydt d y 2 dx 2 d xdt 2
7. The parametric equations x t 2, y t 4 have the same
graph as x t 3, y t 6.
3. The length of the curve x f t, y tt, a t b,
is x s f t tt dt. b a
2
2
8. A hyperbola never intersects its directrix.
EXERCISES ■ Sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve.
1– 4
1. x t 4t,
y 2 t, 4 t 1
2. x 1 e 2t,
y et
2
6. Use the graphs of x f t and y tt to sketch the para-
metric curve x f t, y tt. Indicate with arrows the direction in which the curve is traced as t increases. x
3. x tan ,
■
■
1
y cot
4. x 2 cos , ■
1
y 1 sin ■
■
■
t
1
_1 ■
■
■
■
■
5. Write three different sets of parametric equations for the
curve y sx .
y
■
7–14
■
Sketch the polar curve.
7. r 1 cos
8. r sin 4
t
516
■
CHAPTER 9
PARAMETRIC EQUATIONS AND POLAR COORDINATES
9. r 1 cos 2 11. r 2 sec 2
12. r 2 cos2
1 13. r 1 cos
8 14. r 4 3 sin
■
■
■
■
31. Find the points of intersection of the curves r 2 and
10. r 3 cos 3
■
■
■
■
■
■
r 4 cos .
32. Find the points of intersection of the curves r cot and
r 2 cos .
■
■
Find a polar equation for the curve represented by the given Cartesian equation.
15–16
■
15. x y 2 ■
■
■
■
■
■
■
■
r 2 sin and r sin cos . 34. Find the area of the region that lies inside the curve
r 2 cos 2 but outside the curve r 2 sin .
16. x y 2 2
33. Find the area of the region that lies inside both of the circles
2
■
■
■
■
35–38
cochleoid. Use a graph of r as a function of in Cartesian coordinates to sketch the cochleoid by hand. Then graph it with a machine to check your sketch.
37. r 1,
Find the slope of the tangent line to the given curve at the point corresponding to the specified value of the parameter.
■
0 t 1
2 0
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19. x ln t, y 1 t ; 2
20. x t 6t 1 ,
21. r e ;
t1 2
22. r 3 cos 3 ;
2
■
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■
■
■
■
y t sin t
24. x 1 t 2,
y t t3
■
■
■
t2 c t2 1
y
tt 2 c t2 1
are called strophoids (from a Greek word meaning “to turn or twist”). Investigate how these curves vary as c varies. ■
■
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a ; 40. A family of curves has polar equations r sin 2 where
a is a positive number. Investigate how the curves change as a changes.
Find dydx and d 2 ydx 2 .
23. x t cos t,
; 39. The curves defined by the parametric equations x
y 2t t ; t 1
3
■
■
0 t 2
y cosh 3t,
38. r sin 33, ■
■
y 2t 3,
36. x 2 3t,
; 18. Graph the ellipse r 24 3 cos and its directrix.
23–24
Find the length of the curve.
35. x 3t 2,
; 17. The curve with polar equation r sin is called a
19–22
■
41. Find a polar equation for the ellipse with focus at the origin,
■
eccentricity 13 , and directrix with equation r 4 sec . ■
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; 25. Use a graph to estimate the coordinates of the lowest point
on the curve x t 3 3t, y t 2 t 1. Then use calculus to find the exact coordinates.
26. Find the area enclosed by the loop of the curve in
Exercise 25. 27. At what points does the curve
x 2a cos t a cos 2t
42. Show that the angles between the polar axis and the
asymptotes of the hyperbola r ed1 e cos , e 1, are given by cos11e. 43. In the figure the circle of radius a is stationary, and for
every , the point P is the midpoint of the segment QR. The curve traced out by P for 0 is called the longbow curve. Find parametric equations for this curve. y
y 2a sin t a sin 2t
R
2a
y=2 a
have vertical or horizontal tangents? Use this information to help sketch the curve. 28. Find the area enclosed by the curve in Exercise 27.
P a
Q
29. Find the area enclosed by the curve r 2 9 cos 5. 30. Find the area enclosed by the inner loop of the curve
r 1 3 sin .
¨ 0
x
10
VECTORS AND THE GEOMETRY OF SPACE In this chapter we introduce vectors and coordinate systems for three-dimensional space.This will be the setting for the study of functions of two variables in Chapter 11 because the graph of such a function is a surface in space. In this chapter we will see that vectors provide particularly simple descriptions of lines, planes, and curves.We will also use vector-valued functions to describe the motion of objects through space. In particular, we will use them to derive Kepler’s laws of planetary motion.
10.1
THREE-DIMENSIONAL COORDINATE SYSTEMS
z
O y x
FIGURE 1
Coordinate axes z
y x
To locate a point in a plane, two numbers are necessary. We know that any point in the plane can be represented as an ordered pair a, b of real numbers, where a is the x-coordinate and b is the y-coordinate. For this reason, a plane is called twodimensional. To locate a point in space, three numbers are required. We represent any point in space by an ordered triple a, b, c of real numbers. In order to represent points in space, we first choose a fixed point O (the origin) and three directed lines through O that are perpendicular to each other, called the coordinate axes and labeled the x-axis, y-axis, and z-axis. Usually we think of the x- and y-axes as being horizontal and the z-axis as being vertical, and we draw the orientation of the axes as in Figure 1. The direction of the z-axis is determined by the right-hand rule as illustrated in Figure 2: If you curl the fingers of your right hand around the z-axis in the direction of a 90 counterclockwise rotation from the positive x-axis to the positive y-axis, then your thumb points in the positive direction of the z-axis. The three coordinate axes determine the three coordinate planes illustrated in Figure 3(a). The xy-plane is the plane that contains the x- and y-axes; the yz-plane contains the y- and z-axes; the xz-plane contains the x- and z-axes. These three coordinate planes divide space into eight parts, called octants. The first octant, in the foreground, is determined by the positive axes.
FIGURE 2
z
Right-hand rule
z
y z-plan
ne
la xz-p
x
FIGURE 3
e
left
O
xy-plane (a) Coordinate planes
y
x
right w all
l wal O
floor
y
(b)
Because many people have some difficulty visualizing diagrams of three-dimensional figures, you may find it helpful to do the following [see Figure 3(b)]. Look at any bottom corner of a room and call the corner the origin. The wall on your left is in 517
518
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
the xz-plane, the wall on your right is in the yz-plane, and the floor is in the xy-plane. The x-axis runs along the intersection of the floor and the left wall. The y-axis runs along the intersection of the floor and the right wall. The z-axis runs up from the floor toward the ceiling along the intersection of the two walls. You are situated in the first octant, and you can now imagine seven other rooms situated in the other seven octants (three on the same floor and four on the floor below), all connected by the common corner point O. Now if P is any point in space, let a be the (directed) distance from the yz-plane to P, let b be the distance from the xz-plane to P, and let c be the distance from the xy-plane to P. We represent the point P by the ordered triple a, b, c of real numbers and we call a, b, and c the coordinates of P; a is the x-coordinate, b is the y-coordinate, and c is the z-coordinate. Thus to locate the point a, b, c we can start at the origin O and move a units along the x-axis, then b units parallel to the y-axis, and then c units parallel to the z-axis as in Figure 4. The point Pa, b, c determines a rectangular box as in Figure 5. If we drop a perpendicular from P to the xy-plane, we get a point Q with coordinates a, b, 0 called the projection of P on the xy-plane. Similarly, R0, b, c and Sa, 0, c are the projections of P on the yz-plane and xz-plane, respectively. As numerical illustrations, the points 4, 3, 5 and 3, 2, 6 are plotted in Figure 6.
z P(a, b, c) O
a
c y
x
b
FIGURE 4
z
z
z
3
(0, 0, c) S(a, 0, c)
0
_4
R(0, b, c) 0
P(a, b, c)
_2
_5
FIGURE 5
x
(_4, 3, _5)
(0, b, 0) x
y
y
x 0 (a, 0, 0)
3
_6
y (3, _2, _6)
Q(a, b, 0)
FIGURE 6
The Cartesian product ⺢ ⺢ ⺢ x, y, z x, y, z ⺢ is the set of all ordered triples of real numbers and is denoted by ⺢ 3. We have given a one-to-one correspondence between points P in space and ordered triples a, b, c in ⺢ 3. It is called a three-dimensional rectangular coordinate system. Notice that, in terms of coordinates, the first octant can be described as the set of points whose coordinates are all positive. In two-dimensional analytic geometry, the graph of an equation involving x and y is a curve in ⺢ 2. In three-dimensional analytic geometry, an equation in x, y, and z represents a surface in ⺢ 3. V EXAMPLE 1
(a) z 3 SOLUTION
What surfaces in ⺢ 3 are represented by the following equations? (b) y 5
(a) The equation z 3 represents the set x, y, z z 3 , which is the set of all points in ⺢ 3 whose z-coordinate is 3. This is the horizontal plane that is parallel to the xy-plane and three units above it as in Figure 7(a).
SECTION 10.1
z
■
THREE-DIMENSIONAL COORDINATE SYSTEMS
z
519
y 5
3 0 0
x
(a) z=3, a plane in R#
FIGURE 7
0
5
x
y
x
y
(b) y=5, a plane in R#
(c) y=5, a line in R@
(b) The equation y 5 represents the set of all points in ⺢ 3 whose y-coordinate is 5. This is the vertical plane that is parallel to the xz-plane and five units to the right of it as in Figure 7(b).
■
NOTE When an equation is given, we must understand from the context whether it represents a curve in ⺢ 2 or a surface in ⺢ 3. In Example 1, y 5 represents a plane in ⺢ 3, but of course y 5 can also represent a line in ⺢ 2 if we are dealing with twodimensional analytic geometry. See Figure 7, parts (b) and (c). In general, if k is a constant, then x k represents a plane parallel to the yz-plane, y k is a plane parallel to the xz-plane, and z k is a plane parallel to the xy-plane. In Figure 5, the faces of the rectangular box are formed by the three coordinate planes x 0 (the yz-plane), y 0 (the xz-plane), and z 0 (the xy-plane), and the planes x a, y b, and z c.
z
y
V EXAMPLE 2
0
equation y x.
Describe and sketch the surface in ⺢ 3 represented by the
SOLUTION The equation represents the set of all points in ⺢ 3 whose x- and y-coor-
x
dinates are equal, that is, x, x, z x ⺢, z ⺢ . This is a vertical plane that intersects the xy-plane in the line y x, z 0. The portion of this plane that lies in the first octant is sketched in Figure 8. ■
FIGURE 8
The plane y=x
The familiar formula for the distance between two points in a plane is easily extended to the following three-dimensional formula.
DISTANCE FORMULA IN THREE DIMENSIONS The distance P1 P2 between the
z P¡(⁄, ›, z¡)
points P1x 1, y1, z1 and P2x 2 , y2 , z2 is
P™(¤, fi, z™)
P P sx 1
0 x
B(¤, fi, z¡) A(¤, ›, z¡) y
FIGURE 9
2
2
x 1 2 y2 y1 2 z2 z1 2
To see why this formula is true, we construct a rectangular box as in Figure 9, where P1 and P2 are opposite vertices and the faces of the box are parallel to the coordinate planes. If Ax 2 , y1, z1 and Bx 2 , y2 , z1 are the vertices of the box indicated in the figure, then
P A x 1
2
x1
AB y
2
y1
BP z 2
2
z1
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
Because triangles P1 BP2 and P1 AB are both right-angled, two applications of the Pythagorean Theorem give
P P P B 1
and
2
2
2
1
P A P1 B
2
2
1
AB BP2
2
2
Combining these equations, we get
P P 1
2
2
AB BP x x y y z P1 A
2
2
2
1
2
2
2
1
2
2
2
z1
2
x 2 x 1 2 y2 y1 2 z2 z1 2 Therefore
P P sx 1
2
2
x 1 2 y2 y1 2 z2 z1 2
EXAMPLE 3 The distance from the point P2, 1, 7 to the point Q1, 3, 5 is
PQ s1 2
2
V EXAMPLE 4
■
Find an equation of a sphere with radius r and center Ch, k, l .
SOLUTION By definition, a sphere is the set of all points Px, y, z whose distance from C is r. (See Figure 10.) Thus P is on the sphere if and only if PC r. Squaring both sides, we have PC 2 r 2 or
z
P(x, y, z) r
x h2 y k2 z l 2 r 2
C(h, k, l)
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The result of Example 4 is worth remembering.
0
EQUATION OF A SPHERE An equation of a sphere with center Ch, k, l and
radius r is
x
x h2 y k2 z l 2 r 2
y
FIGURE 10
3 12 5 72 s1 4 4 3
In particular, if the center is the origin O, then an equation of the sphere is x 2 y 2 z2 r 2
EXAMPLE 5 Show that x 2 y 2 z 2 4x 6y 2z 6 0 is the equation of a
sphere, and find its center and radius. SOLUTION We can rewrite the given equation in the form of an equation of a sphere
if we complete squares: x 2 4x 4 y 2 6y 9 z 2 2z 1 6 4 9 1 x 22 y 32 z 12 8 Comparing this equation with the standard form, we see that it is the equation of a sphere with center 2, 3, 1 and radius s8 2s2 . ■
SECTION 10.1
THREE-DIMENSIONAL COORDINATE SYSTEMS
■
521
EXAMPLE 6 What region in ⺢ 3 is represented by the following inequalities?
1 x 2 y 2 z2 4
z 0
z
SOLUTION The inequalities
1 x 2 y 2 z2 4
2 x
y
FIGURE 11
10.1
1 sx 2 y 2 z 2 2
can be rewritten as
0 1
so they represent the points x, y, z whose distance from the origin is at least 1 and at most 2. But we are also given that z 0, so the points lie on or below the xy-plane. Thus the given inequalities represent the region that lies between (or on) the spheres x 2 y 2 z 2 1 and x 2 y 2 z 2 4 and beneath (or on) the xy-plane. It is sketched in Figure 11. ■
EXERCISES 8. Find the distance from 3, 7, 5 to each of the following.
1. Suppose you start at the origin, move along the x-axis a dis-
tance of 4 units in the positive direction, and then move downward a distance of 3 units. What are the coordinates of your position?
(a) The xy-plane (c) The xz-plane (e) The y-axis
9. Determine whether the points lie on straight line.
2. Sketch the points 0, 5, 2, 4, 0, 1, 2, 4, 6, and
(a) A2, 4, 2, B3, 7, 2, C1, 3, 3 (b) D0, 5, 5, E1, 2, 4, F3, 4, 2
1, 1, 2 on a single set of coordinate axes.
3. Which of the points P6, 2, 3, Q5, 1, 4, and
R0, 3, 8 is closest to the xz-plane? Which point lies in the yz-plane? 4. What are the projections of the point (2, 3, 5) on the xy-, yz-,
and xz-planes? Draw a rectangular box with the origin and 2, 3, 5 as opposite vertices and with its faces parallel to the coordinate planes. Label all vertices of the box. Find the length of the diagonal of the box. 5. Describe and sketch the surface in ⺢3 represented by the
equation x y 2.
10. Find an equation of the sphere with center 2, 6, 4 and
radius 5. Describe its intersection with each of the coordinate planes. 11. Find an equation of the sphere that passes through the point
4, 3, 1 and has center 3, 8, 1. 12. Find an equation of the sphere that passes through the ori-
gin and whose center is 1, 2, 3. 13–16 ■ Show that the equation represents a sphere, and find its center and radius. 13. x 2 y 2 z 2 6x 4y 2z 11
6. (a) What does the equation x 4 represent in ⺢ ? What 2
does it represent in ⺢3 ? Illustrate with sketches. (b) What does the equation y 3 represent in ⺢3 ? What does z 5 represent? What does the pair of equations y 3, z 5 represent? In other words, describe the set of points x, y, z such that y 3 and z 5. Illustrate with a sketch.
7. Find the lengths of the sides of the triangle PQR. Is it a
right triangle? Is it an isosceles triangle? (a) P3, 2, 3, Q7, 0, 1, R1, 2, 1 (b) P2, 1, 0, Q4, 1, 1, R4, 5, 4
(b) The yz-plane (d) The x-axis (f ) The z-axis
14. x 2 y 2 z 2 4x 2y 15. x 2 y 2 z 2 x y z 16. 4x 2 4y 2 4z 2 8x 16y 1 ■
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17. (a) Prove that the midpoint of the line segment from
P1x 1, y1, z1 to P2x 2 , y2 , z2 is
x 1 x 2 y1 y2 z1 z2 , , 2 2 2
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
(b) Find the lengths of the medians of the triangle with vertices A1, 2, 3, B2, 0, 5, and C4, 1, 5.
31–34
Write inequalities to describe the region.
31. The half-space consisting of all points to the left of the
xz-plane
18. Find an equation of a sphere if one of its diameters has end-
points 2, 1, 4 and 4, 3, 10.
■
32. The solid rectangular box in the first octant bounded by the
planes x 1, y 2, and z 3
19. Find equations of the spheres with center 2, 3, 6 that
touch (a) the xy-plane, (b) the yz-plane, (c) the xz-plane.
33. The region consisting of all points between (but not on)
the spheres of radius r and R centered at the origin, where r R
20. Find an equation of the largest sphere with center (5, 4, 9)
that is contained in the first octant.
34. The solid upper hemisphere of the sphere of radius 2
centered at the origin
21–30 Describe in words the region of ⺢ represented by the equation or inequality. 3
■
21. y 4
22. x 10
23. x 3
24. y 0
25. 0 z 6
26. z 2 1
27. x 2 y 2 z 2 3
28. x z
29. x 2 z 2 9
30. x 2 y 2 z 2 2z
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10.2
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35. Find an equation of the set of all points equidistant from the
points A1, 5, 3 and B6, 2, 2. Describe the set. 36. Find the volume of the solid that lies inside both of the
spheres
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■
x 2 y 2 z 2 4x 2y 4z 5 0 ■
and
x 2 y 2 z2 4
VECTORS D
B
■
■
u
v C A
FIGURE 1
Equivalent vectors
The term vector is used by scientists to indicate a quantity (such as displacement or velocity or force) that has both magnitude and direction. A vector is often represented by an arrow or a directed line segment. The length of the arrow represents the magnitude of the vector and the arrow points in the direction of the vector. We denote a vector by printing a letter in boldface v or by putting an arrow above the letter vl. For instance, suppose a particle moves along a line segment from point A to point B. The corresponding displacement vector v, shown in Figure 1, has initial point A l (the tail) and terminal point B (the tip) and we indicate this by writing v AB. l Notice that the vector u CD has the same length and the same direction as v even though it is in a different position. We say that u and v are equivalent (or equal) and we write u v. The zero vector, denoted by 0, has length 0. It is the only vector with no specific direction. COMBINING VECTORS
C B
A FIGURE 2
l Suppose a particle moves from A to B, so its displacement vector is AB. Then the parl ticle changes direction and moves from B to C, with displacement vector BC as in Figure 2. The combined effect of these displacements is that the particle has moved l l l from A to C. The resulting displacement vector AC is called the sum of AB and BC and we write l l l AC AB BC In general, if we start with vectors u and v, we first move v so that its tail coincides with the tip of u and define the sum of u and v as follows.
SECTION 10.2
VECTORS
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523
DEFINITION OF VECTOR ADDITION If u and v are vectors positioned so the
initial point of v is at the terminal point of u, then the sum u v is the vector from the initial point of u to the terminal point of v. The definition of vector addition is illustrated in Figure 3. You can see why this definition is sometimes called the Triangle Law. u u+v
v
v+
u
v
v u+
v
u
u FIGURE 3 The Triangle Law
FIGURE 4 The Parallelogram Law
In Figure 4 we start with the same vectors u and v as in Figure 3 and draw another copy of v with the same initial point as u. Completing the parallelogram, we see that u v v u. This also gives another way to construct the sum: If we place u and v so they start at the same point, then u v lies along the diagonal of the parallelogram with u and v as sides. (This is called the Parallelogram Law.) Draw the sum of the vectors a and b shown in Figure 5.
V EXAMPLE 1
a
b
SOLUTION First we translate b and place its tail at the tip of a, being careful to draw a copy of b that has the same length and direction. Then we draw the vector a b [see Figure 6(a)] starting at the initial point of a and ending at the terminal point of the copy of b. Alternatively, we could place b so it starts where a starts and construct a b by the Parallelogram Law as in Figure 6(b).
FIGURE 5
a
Visual 10.2 shows how the Triangle and Parallelogram Laws work for various vectors u and v.
FIGURE 6
a
b a+b
a+b b
(a)
■
(b)
It is possible to multiply a vector by a real number c. (In this context we call the real number c a scalar to distinguish it from a vector.) For instance, we want 2v to be the same vector as v v, which has the same direction as v but is twice as long. In general, we multiply a vector by a scalar as follows. DEFINITION OF SCALAR MULTIPLICATION If c is a scalar and v is a vector, then the scalar multiple cv is the vector whose length is c times the length of v and whose direction is the same as v if c 0 and is opposite to v if c 0. If c 0 or v 0, then cv 0.
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1 2v
2v
v
_v
This definition is illustrated in Figure 7. We see that real numbers work like scaling factors here; that’s why we call them scalars. Notice that two nonzero vectors are parallel if they are scalar multiples of one another. In particular, the vector v 1v has the same length as v but points in the opposite direction. We call it the negative of v. By the difference u v of two vectors we mean u v u v
_1.5v
So we can construct u v by first drawing the negative of v, v, and then adding it to u by the Parallelogram Law as in Figure 8(a). Alternatively, since v u v u, the vector u v, when added to v, gives u. So we could construct u v as in Figure 8(b) by means of the Triangle Law.
FIGURE 7
Scalar multiples of v
v
u u-v
u-v
_v
v u
FIGURE 8
Drawing u-v
(a)
(b)
EXAMPLE 2 If a and b are the vectors shown in Figure 9, draw a 2b. SOLUTION We first draw the vector 2b pointing in the direction opposite to b and
twice as long. We place it with its tail at the tip of a and then use the Triangle Law to draw a 2b as in Figure 10. a
_2b a b
a-2b
FIGURE 9 y
For some purposes it’s best to introduce a coordinate system and treat vectors algebraically. If we place the initial point of a vector a at the origin of a rectangular coordinate system, then the terminal point of a has coordinates of the form a1, a2 or a1, a2, a3, depending on whether our coordinate system is two- or three-dimensional (see Figure 11). These coordinates are called the components of a and we write
a x
a=ka¡, a™l
a a 1, a 2
z (a¡, a™, a£)
a O y
x
a=ka¡, a™, a£l FIGURE 11
■
COMPONENTS
(a¡, a™)
O
FIGURE 10
or
a a 1, a 2 , a 3
We use the notation a1, a2 for the ordered pair that refers to a vector so as not to confuse it with the ordered pair a1, a2 that refers to a point in the plane. For instance, the vectors shown in Figure 12 are all equivalent to the vector l OP 3, 2 whose terminal point is P3, 2. What they have in common is that the terminal point is reached from the initial point by a displacement of three units to the right and two upward. We can think of all these geometric vectors as representations l of the algebraic vector a 3, 2 . The particular representation OP from the origin to the point P3, 2 is called the position vector of the point P.
SECTION 10.2
y
525
■
z
(4, 5) (1, 3)
VECTORS
position vector of P
P(3, 2)
P(a¡, a™, a£) O
x
O
A(x, y, z)
x
FIGURE 12
B(x+a¡, y+a™, z+a£) y
FIGURE 13 Representations of a=ka¡, a™, a£l
Representations of the vector a=k3, 2l
l In three dimensions, the vector a OP a1, a2, a3 is the position vector of the l point Pa1, a2, a3. (See Figure 13.) Let’s consider any other representation AB of a, where the initial point is Ax 1, y1, z1 and the terminal point is Bx 2 , y2 , z2 . Then we must have x 1 a 1 x 2, y1 a 2 y2, and z1 a 3 z2 and so a 1 x 2 x 1, a 2 y2 y1, and a 3 z2 z1. Thus we have the following result. Given the points Ax 1, y1, z1 and Bx 2 , y2 , z2 , the vector a with represenl tation AB is a x 2 x 1, y2 y1, z2 z1 1
V EXAMPLE 3 Find the vector represented by the directed line segment with initial point A2, 3, 4) and terminal point B2, 1, 1. l SOLUTION By (1), the vector corresponding to AB is y
(a¡+b¡, a™+b™)
a+b
■
The magnitude or length of the vector v is the length of any of its representations and is denoted by the symbol v or v . By using the distance formula to compute the length of a segment OP, we obtain the following formulas.
b™
b
a 2 2, 1 3, 1 4 4, 4, 3
b¡ a 0
a™
a¡
The length of the two-dimensional vector a a 1, a 2 is
a™ x
b¡
a sa
2 1
The length of the three-dimensional vector a a 1, a 2 , a 3 is
FIGURE 14
a sa
2 1
ca a
ca™
a™ a¡
FIGURE 15
a 22
ca¡
a 22 a 32
How do we add vectors algebraically? Figure 14 shows that if a a 1, a 2 and b b 1, b 2 , then the sum is a b a1 b1, a2 b2 , at least for the case where the components are positive. In other words, to add algebraic vectors we add their components. Similarly, to subtract vectors we subtract components. From the similar triangles in Figure 15 we see that the components of ca are ca1 and ca2. So to multiply a vector by a scalar we multiply each component by that scalar.
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CHAPTER 10
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If a a 1, a 2 and b b1, b2 , then a b a 1 b1, a 2 b2
a b a 1 b1, a 2 b2
ca ca1, ca2 Similarly, for three-dimensional vectors, a 1, a 2 , a 3 b1, b2 , b3 a 1 b1, a 2 b2 , a 3 b3 a 1, a 2 , a 3 b1, b2 , b3 a 1 b1, a 2 b2 , a 3 b3 c a 1, a 2 , a 3 ca1, ca2 , ca3
V EXAMPLE 4 If a 4, 0, 3 and b 2, 1, 5, find a and the vectors a b, a b, 3b, and 2a 5b.
a s4
SOLUTION
2
0 2 32 s25 5
a b 4, 0, 3 2, 1, 5 4 2, 0 1, 3 5 2, 1, 8 a b 4, 0, 3 2, 1, 5 4 2, 0 1, 3 5 6, 1, 2 3b 32, 1, 5 32, 31, 35 6, 3, 15 2a 5b 24, 0, 3 52, 1, 5 8, 0, 6 10, 5, 25 2, 5, 31 ■ Vectors in n dimensions are used to list various quantities in an organized way. For instance, the components of a six-dimensional vector
■
We denote by V2 the set of all two-dimensional vectors and by V3 the set of all three-dimensional vectors. More generally, we will later need to consider the set Vn of all n-dimensional vectors. An n-dimensional vector is an ordered n-tuple:
p p1 , p2 , p3 , p4 , p5 , p6
a a1, a 2, . . . , a n
might represent the prices of six different ingredients required to make a particular product. Four-dimensional vectors x, y, z, t are used in relativity theory, where the first three components specify a position in space and the fourth represents time.
where a1, a 2, . . . , a n are real numbers that are called the components of a. Addition and scalar multiplication are defined in terms of components just as for the cases n 2 and n 3. PROPERTIES OF VECTORS If a, b, and c are vectors in Vn and c and d are
scalars, then 1. a b b a
2. a b c a b c
3. a 0 a
4. a a 0
5. ca b ca cb
6. c da ca da
7. cd a cda
8. 1a a
These eight properties of vectors can be readily verified either geometrically or algebraically. For instance, Property 1 can be seen from Figure 4 (it’s equivalent to the
SECTION 10.2
■
VECTORS
527
Parallelogram Law) or as follows for the case n 2: a b a 1, a 2 b1, b2 a 1 b1, a 2 b2 b1 a 1, b2 a 2 b1, b2 a 1, a 2 ba We can see why Property 2 (the associative law) is true by looking at Figure 16 and l applying the Triangle Law several times: The vector PQ is obtained either by first constructing a b and then adding c or by adding a to the vector b c. Three vectors in V3 play a special role. Let
Q c (a+b)+c =a+(b+c)
b
a+b b+c
i 1, 0, 0
j 0, 1, 0
k 0, 0, 1
a
P
These vectors i , j, and k are called the standard basis vectors. They have length 1 and point in the directions of the positive x-, y-, and z-axes. Similarly, in two dimensions we define i 1, 0 and j 0, 1 . (See Figure 17.)
FIGURE 16
y
z
j
k
(0, 1)
0
x
i
j
i
(1, 0)
FIGURE 17
y
x
Standard basis vectors in V™ and V£
(a)
(b)
If a a 1, a 2 , a 3 , then we can write
y (a¡, a™)
a
a¡i
0
a a 1, a 2 , a 3 a 1, 0, 0 0, a 2 , 0 0, 0, a 3
a™ j
a 1 1, 0, 0 a 2 0, 1, 0 a 3 0, 0, 1 x
2
(a) a=a¡i+a™ j
Thus any vector in V3 can be expressed in terms of i , j, and k. For instance, 1, 2, 6 i 2j 6k
z
Similarly, in two dimensions, we can write
(a¡, a™, a£)
a 3
a£k
a¡i
y
x
a™ j (b) a=a¡i+a™j+a£k
FIGURE 18
a a1 i a2 j a3 k
a a1, a2 a1 i a2 j
See Figure 18 for the geometric interpretation of Equations 3 and 2 and compare with Figure 17. EXAMPLE 5 If a i 2j 3k and b 4i 7 k, express the vector 2a 3b in terms of i , j, and k. SOLUTION Using Properties 1, 2, 5, 6, and 7 of vectors, we have
2a 3b 2i 2j 3k 34i 7k 2i 4j 6k 12i 21k 14i 4j 15k
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
A unit vector is a vector whose length is 1. For instance, i , j, and k are all unit vectors. In general, if a 0, then the unit vector that has the same direction as a is 1 a a a a
u
4
In order to verify this we let c 1 a . Then u ca and c is a positive scalar, so u has the same direction as a. Also
1
u ca c a a a 1 EXAMPLE 6 Find the unit vector in the direction of the vector 2i j 2k. SOLUTION The given vector has length
2i j 2k s2
2
12 22 s9 3
so, by Equation 4, the unit vector with the same direction is 1 3
2i j 2k 23 i 13 j 23 k
■
APPLICATIONS
Vectors are useful in many aspects of physics and engineering. In Section 10.9 we will see how they describe the velocity and acceleration of objects moving in space. Here we look at forces. A force is represented by a vector because it has both a magnitude (measured in pounds or newtons) and a direction. If several forces are acting on an object, the resultant force experienced by the object is the vector sum of these forces. 50°
32°
T¡
T™
EXAMPLE 7 A 100-lb weight hangs from two wires as shown in Figure 19. Find the
tensions (forces) T1 and T2 in both wires and their magnitudes. SOLUTION We first express T1 and T2 in terms of their horizontal and vertical com-
ponents. From Figure 20 we see that 100 FIGURE 19
50° T¡
T™
50°
32°
32°
w
T cos 32 i T sin 32 j
5
T1 T1 cos 50 i T1 sin 50 j
6
T2
2
2
The resultant T1 T2 of the tensions counterbalances the weight w and so we must have T1 T2 w 100 j Thus
( T1 cos 50 T2 cos 32 ) i ( T1 sin 50 T2 sin 32 ) j 100 j Equating components, we get
FIGURE 20
T sin 50 T sin 32 100 Solving the first of these equations for T and substituting into the second, we get T cos 50 sin 32 100 T sin 50 T1 cos 50 T2 cos 32 0 1
2
2
1
1
cos 32
SECTION 10.2
VECTORS
■
529
So the magnitudes of the tensions are 100 85.64 lb sin 50 tan 32 cos 50
T 1
T cos 50 T cos 32 64.91 lb 1
and
2
Substituting these values in (5) and (6), we obtain the tension vectors T1 55.05 i 65.60 j
10.2
T2 55.05 i 34.40 j
EXERCISES
1. Name all the equal vectors in the parallelogram shown. A
B
Find a vector a with representation given by the directed l l line segment AB. Draw AB and the equivalent representation starting at the origin. 5– 8
■
5. A2, 3,
E
D
C
B2, 3, 1
■
■
■
l l (b) RP PS l l l (d) RS SP PQ
Q
9. 3, 1 ,
2, 4
11. 0, 1, 2 , ■
13–16
■
■
■
(b) u v (d) w v u w
4. Copy the vectors in the figure and use them to draw the
a
(b) a b (d) 12 b (f ) b 3a
b
■
B4, 2, 1 ■
■
■
■
■
0, 4, 0 ■
■
■
■
■
■
b i 2j b 2 i j 5 k
16. a 2 i 4 j 4 k, ■
■
5, 7
b 3, 6
15. a i 2 j 3 k,
■
12. 1, 0, 2 , ■
14. a 4 i j,
3. Copy the vectors in the figure and use them to draw the
following vectors. (a) a b (c) 2a (e) 2a b
■
Find a b, 2a 3b, a , and a b .
13. a 5, 12 , S
v
■
10. 2, 1 ,
0, 0, 3
■
P
u
8. A4, 0, 2, ■
■ Find the sum of the given vectors and illustrate geometrically.
■
R
B5, 3
9–12
2. Write each combination of vectors as a single vector.
l l (a) PQ QR l l (c) QS PS
■
6. A2, 2,
B2, 1
7. A0, 3, 1, ■
following vectors. (a) u v (c) v w
■
■
■
■
b 2j k ■
■
■
■
17. Find a unit vector with the same direction as 8 i j 4 k. 18. Find a vector that has the same direction as 2, 4, 2 but
has length 6. 19. If v lies in the first quadrant and makes an angle 3 with
the positive x-axis and v 4, find v in component form.
20. If a child pulls a sled through the snow with a force of 50 N
exerted at an angle of 38 above the horizontal, find the horizontal and vertical components of the force. 21. Two forces F1 and F2 with magnitudes 10 lb and 12 lb act
on an object at a point P as shown in the figure. Find the resultant force F acting at P as well as its magnitude and its
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
direction. (Indicate the direction by finding the angle shown in the figure.)
a mass of 0.8 kg is hung at the middle of the line, the midpoint is pulled down 8 cm. Find the tension in each half of the clothesline.
F F¡
F™
26. The tension T at each end of the chain has magnitude 25 N.
What is the weight of the chain?
¨ 45°
30° P
37°
37°
22. Velocities have both direction and magnitude and thus are
vectors. The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45 W at a speed of 50 kmh. (This means that the direction from which the wind blows is 45 west of the northerly direction.) A pilot is steering a plane in the direction N60 E at an airspeed (speed in still air) of 250 kmh. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. 23. A woman walks due west on the deck of a ship at 3 mih.
The ship is moving north at a speed of 22 mih. Find the speed and direction of the woman relative to the surface of the water. 24. Ropes 3 m and 5 m in length are fastened to a holiday deco-
ration that is suspended over a town square. The decoration has a mass of 5 kg. The ropes, fastened at different heights, make angles of 52 and 40 with the horizontal. Find the tension in each wire and the magnitude of each tension.
52° 3 m
40°
27. (a) Draw the vectors a 3, 2 , b 2, 1 , and
c 7, 1 . (b) Show, by means of a sketch, that there are scalars s and t such that c sa t b. (c) Use the sketch to estimate the values of s and t. (d) Find the exact values of s and t.
28. Suppose that a and b are nonzero vectors that are not paral-
lel and c is any vector in the plane determined by a and b. Give a geometric argument to show that c can be written as c sa t b for suitable scalars s and t. Then give an argument using components. 29. If r x, y, z and r0 x 0 , y0 , z0 , describe the set of all
points x, y, z such that r r0 1.
30. If r x, y , r1 x 1, y1 , and r2 x 2 , y2 , describe the
set of all points x, y such that r r1 r r2 k, where k r1 r2 .
31. Figure 16 gives a geometric demonstration of Property 2 of
vectors. Use components to give an algebraic proof of this fact for the case n 2.
5 m
32. Prove Property 5 of vectors algebraically for the case n 3.
Then use similar triangles to give a geometric proof. 33. Use vectors to prove that the line joining the midpoints of 25. A clothesline is tied between two poles, 8 m apart. The line
is quite taut and has negligible sag. When a wet shirt with
10.3
two sides of a triangle is parallel to the third side and half its length.
THE DOT PRODUCT So far we have added two vectors and multiplied a vector by a scalar. The question arises: Is it possible to multiply two vectors so that their product is a useful quantity? One such product is the dot product, whose definition follows. Another is the cross product, which is discussed in the next section.
SECTION 10.3
THE DOT PRODUCT
■
531
1 DEFINITION If a a 1, a 2 , a 3 and b b1, b2 , b3 , then the dot product of a and b is the number a b given by
a b a 1 b1 a 2 b2 a 3 b3 Thus to find the dot product of a and b we multiply corresponding components and add. The result is not a vector. It is a real number, that is, a scalar. For this reason, the dot product is sometimes called the scalar product (or inner product). Although Definition 1 is given for three-dimensional vectors, the dot product of two-dimensional vectors is defined in a similar fashion: a 1, a 2 b1, b2 a 1 b1 a 2 b2 V EXAMPLE 1
2, 4 3, 1 23 41 2 1, 7, 4 6, 2, 12 16 72 4( 12 ) 6 i 2 j 3k 2 j k 10 22 31 7
■
The dot product obeys many of the laws that hold for ordinary products of real numbers. These are stated in the following theorem. 2 PROPERTIES OF THE DOT PRODUCT
is a scalar, then 1. a a a 2 3. a b c a b a c 5. 0 a 0
If a, b, and c are vectors in V3 and c 2. a b b a 4. ca b ca b a cb
These properties are easily proved using Definition 1. For instance, here are the proofs of Properties 1 and 3:
1. a a a12 a 22 a 32 a
2
3. a b c a1, a2, a3 b1 c1, b2 c2 , b3 c3
a 1b1 c1 a 2b2 c2 a 3b3 c3 a 1 b1 a 1 c1 a 2 b2 a 2 c2 a 3 b3 a 3 c3 a 1 b1 a 2 b2 a 3 b3 a 1 c1 a 2 c2 a 3 c3 abac
z
The proofs of the remaining properties are left as exercises. B a-b
b 0 ¨ x
FIGURE 1
a
A
y
■
The dot product a b can be given a geometric interpretation in terms of the angle between a and b, which is defined to be the angle between the representations of a and b that start at the origin, where 0 . In other words, is the angle between l l the line segments OA and OB in Figure 1. Note that if a and b are parallel vectors, then 0 or . The formula in the following theorem is used by physicists as the definition of the dot product.
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
If is the angle between the vectors a and b, then
3 THEOREM
b cos
ab a
PROOF If we apply the Law of Cosines to triangle OAB in Figure 1, we get
AB
2
4
OA
2
OB
2
2 OA
OB cos
(Observe that the Law of Cosines still applies in the limiting cases when 0 or , or a 0 or b 0.) But OA a , OB b , and AB a b , so Equation 4 becomes
a b
2
5
a
2
b
2
b cos
2 a
Using Properties 1, 2, and 3 of the dot product, we can rewrite the left side of this equation as follows:
a b
2
a b a b a a a b b a b b
a
2
2a b b
2
Therefore, Equation 5 gives
a
2
2 a b cos 2a b 2 a b cos a b a b cos
2a b b
Thus or
2
a
2
b
2
■
EXAMPLE 2 If the vectors a and b have lengths 4 and 6, and the angle between them is 3, find a b. SOLUTION Using Theorem 3, we have
b cos3 4 6
ab a
1 2
12
■
The formula in Theorem 3 also enables us to find the angle between two vectors. 6 COROLLARY
If is the angle between the nonzero vectors a and b, then cos
V EXAMPLE 3
b 5, 3, 2 .
ab a b
Find the angle between the vectors a 2, 2, 1 and
SOLUTION Since
a s2 and since
2
2 2 12 3
and
b s5
2
32 2 2 s38
a b 25 23 12 2
we have, from Corollary 6, cos
ab 2 a b 3s38
SECTION 10.3
THE DOT PRODUCT
■
533
So the angle between a and b is
cos1
2 3s38
1.46 or 84
■
Two nonzero vectors a and b are called perpendicular or orthogonal if the angle between them is 2. Then Theorem 3 gives
b cos2 0
ab a
and conversely if a b 0, then cos 0, so 2. The zero vector 0 is considered to be perpendicular to all vectors. Therefore, we have the following method for determining whether two vectors are orthogonal.
7
Two vectors a and b are orthogonal if and only if a b 0.
EXAMPLE 4 Show that 2i 2j k is perpendicular to 5i 4j 2k. SOLUTION Since
2i 2j k 5i 4j 2k 25 24 12 0 a
¨
a
b
b
b
¨
a · b>0
a · b= 0
a · b<0
a FIGURE 2
■
these vectors are perpendicular by (7).
Because cos 0 if 0 2 and cos 0 if 2 , we see that a b is positive for 2 and negative for 2. We can think of a b as measuring the extent to which a and b point in the same direction. The dot product a b is positive if a and b point in the same general direction, 0 if they are perpendicular, and negative if they point in generally opposite directions (see Figure 2). In the extreme case where a and b point in exactly the same direction, we have 0, so cos 1 and ab a b
Visual 10.3A shows an animation of Figure 2.
If a and b point in exactly opposite directions, then and so cos 1 and ab a b .
PROJECTIONS
l l Figure 3 shows representations PQ and PR of two vectors a and b with the same inil tial point P. If S is the foot of the perpendicular from R to the line containing PQ, then l the vector with representation PS is called the vector projection of b onto a and is Visual 10.3B shows how Figure 3 changes when we vary a and b.
R
R b
b
a
a FIGURE 3
Vector projections
P
S proja b
Q S
P
proja b
Q
534
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
denoted by proja b. (You can think of it as a shadow of b). The scalar projection of b onto a (also called the component of b along a) is defined to be numerically the length of the vector projection, which is the number b cos , where is the angle between a and b. (See Figure 4.) This is denoted by compa b. Observe that it is negative if 2 . The equation a b a b cos a ( b cos )
R
b a
¨
P
Q S 兩 b兩 cos ¨ = compa b
FIGURE 4
shows that the dot product of a and b can be interpreted as the length of a times the scalar projection of b onto a. Since
Scalar projection
ab
b cos a
a b a
the component of b along a can be computed by taking the dot product of b with the unit vector in the direction of a. To summarize:
Scalar projection of b onto a:
compa b
Vector projection of b onto a:
proja b
ab a
ab a
a ab a a a 2
Notice that the vector projection is the scalar projection times the unit vector in the direction of a. V EXAMPLE 5 Find the scalar projection and vector projection of b 1, 1, 2 onto a 2, 3, 1 .
SOLUTION Since a s22 3 2 12 s14 , the scalar projection of b onto
a is
compa b
ab 21 31 12 3 a s14 s14
The vector projection is this scalar projection times the unit vector in the direction of a: proja b
R F S
¨ P
Q D
FIGURE 5
3 s14
a 3 3 9 3 a , , a 14 7 14 14
■
One use of projections occurs in physics in calculating work. In Section 7.5 we defined the work done by a constant force F in moving an object through a distance d as W Fd, but this applies only when the force is directed along the line of motion l of the object. Suppose, however, that the constant force is a vector F PR pointing in some other direction as in Figure 5. If the force moves the object from P to Q, then l the displacement vector is D PQ. The work done by this force is defined to be the product of the component of the force along D and the distance moved:
W ( F cos ) D
SECTION 10.3
THE DOT PRODUCT
■
535
But then, from Theorem 3, we have
D cos F D
W F
8
Thus the work done by a constant force F is the dot product F D, where D is the displacement vector. EXAMPLE 6 A crate is hauled 8 m up a ramp under a constant force of 200 N
applied at an angle of 25 to the ramp. Find the work done. F
25°
SOLUTION If F and D are the force and displacement vectors, as pictured in Figure 6, then the work done is
D
D cos 25
WFD F
2008 cos 25 1450 N m 1450 J
FIGURE 6
■
EXAMPLE 7 A force is given by a vector F 3i 4j 5k and moves a particle from the point P2, 1, 0 to the point Q4, 6, 2. Find the work done. l SOLUTION The displacement vector is D PQ 2, 5, 2 , so by Equation 8, the work done is
W F D 3, 4, 5 2, 5, 2 6 20 10 36 If the unit of length is meters and the magnitude of the force is measured in newtons, then the work done is 36 J. ■
10.3
EXERCISES
1. Which of the following expressions are meaningful? Which
are meaningless? Explain. (a) a b c (c) a b c (e) a b c
9–10
■
If u is a unit vector, find u v and u w. 10.
9.
(b) a bc (d) a b c (f ) a b c
u
v
v
2. Find the dot product of two vectors if their lengths are 6
w
and 13 and the angle between them is 4. 3– 8 3.
■
w
Find a b.
a 6, b 5 ,
b 0.7, 1.2
5. a 4, 1,
b 6, 3, 8
,
6. a s, 2s, 3s ,
8. a 4 j 3 k, ■
■
■
■
■
■
■
■
■
■
■
■
■
■
12. A street vendor sells a hamburgers, b hot dogs, and c soft
b 5i 9k ■
■
(b) Show that i i j j k k 1.
b 2i 4 j 6k ■
■
11. (a) Show that i j j k k i 0.
b t, t, 5t
7. a i 2 j 3 k , ■
the angle between a and b is 23
4. a 2, 3 , 1 4
u
■
■
■
■
■
drinks on a given day. He charges $2 for a hamburger, $1.50 for a hot dog, and $1 for a soft drink. If A a, b, c and P 2, 1.5, 1 , what is the meaning of the dot product A P ?
536
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
13–15 ■ Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) 13. a 8, 6 ,
b s7 , 3
14. a 4, 0, 2 ,
b 2, 1, 0
15. a j k, ■
■
■
32. Find the work done by a force of 20 lb acting in the direc-
tion N50 W in moving an object 4 ft due west.
b i 2 j 3k ■
■
■
■
from the point 2, 3, 0 to the point 4, 9, 15. Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
■
■
■
■
■
33. A woman exerts a horizontal force of 25 lb on a crate as she
pushes it up a ramp that is 10 ft long and inclined at an angle of 20 above the horizontal. Find the work done on the box.
16. Find, correct to the nearest degree, the three angles of the
triangle with vertices D0, 1, 1, E2, 4, 3, and F1, 2, 1.
34. A wagon is pulled a distance of 100 m along a horizontal
Determine whether the given vectors are orthogonal, parallel, or neither. 17–18
■
path by a constant force of 50 N. The handle of the wagon is held at an angle of 30 above the horizontal. How much work is done?
17. (a) a 5, 3, 7 ,
b 6, 8, 2 (b) a 4, 6 , b 3, 2 (c) a i 2 j 5 k, b 3 i 4 j k (d) a 2 i 6 j 4 k, b 3 i 9 j 6 k
35. Use a scalar projection to show that the distance from a
point P1x 1, y1 to the line ax by c 0 is
ax
1 by1 c sa 2 b 2
18. (a) u 3, 9, 6 ,
v 4, 12, 8 (b) u i j 2 k, v 2 i j k (c) u a, b, c , v b, a, 0
■
■
■
■
■
■
■
■
■
■
■
■
P1, 3, 2, Q2, 0, 4, and R6, 2, 5 is rightangled.
show that the vector equation r a r b 0 represents a sphere, and find its center and radius.
20. For what values of b are the vectors 6, b, 2 and
37. Find the angle between a diagonal of a cube and one of its
b, b 2, b orthogonal?
edges.
21. Find a unit vector that is orthogonal to both i j and i k. 22. Find two unit vectors that make an angle of 60 with
24. a 1, 2 ,
b 5, 0 b 4, 1
25. a 3, 6, 2 ,
b 1, 2, 3
26. a i j k,
bijk
■
■
■
■
■
■
■
■
■
■
of one of its faces. hydrogen atoms at the vertices of a regular tetrahedron and the carbon atom at the centroid. The bond angle is the angle formed by the H— C—H combination; it is the angle between the lines that join the carbon atom to two of the hydrogen atoms. Show that the bond angle is about 109.5 . [Hint: Take the vertices of the tetrahedron to be the points 1, 0, 0, 0, 1, 0 , 0, 0, 1, and 1, 1, 1 as shown in the figure. Then the centroid is ( 12 , 12 , 12 ).]
Find the scalar and vector projections of b onto a.
23. a 3, 4 ,
38. Find the angle between a diagonal of a cube and a diagonal 39. A molecule of methane, CH 4 , is structured with the four
v 3, 4 . ■
Use this formula to find the distance from the point 2, 3 to the line 3x 4y 5 0. 36. If r x, y, z , a a 1, a 2 , a 3 , and b b1, b2 , b3 ,
19. Use vectors to decide whether the triangle with vertices
23–26
■
z
■
H
27. Show that the vector orth a b b proj a b is orthogonal
to a. (It is called an orthogonal projection of b.) 28. For the vectors in Exercise 24, find orth a b and illustrate by
C
drawing the vectors a, b, proj a b, and orth a b.
H H y
29. If a 3, 0, 1 , find a vector b such that comp a b 2. 30. Suppose that a and b are nonzero vectors.
(a) Under what circumstances is comp a b comp b a? (b) Under what circumstances is proj a b proj b a? 31. A constant force with vector representation
F 10 i 18 j 6 k moves an object along a straight line
x
H
40. If c a b b a, where a, b, and c are all nonzero
vectors, show that c bisects the angle between a and b. 41. Prove Properties 2, 4, and 5 of the dot product (Theorem 2).
SECTION 10.4
42. Suppose that all sides of a quadrilateral are equal in length
■
537
(b) Use the Cauchy-Schwarz Inequality from Exercise 43 to prove the Triangle Inequality. [Hint: Use the fact that a b 2 a b a b and use Property 3 of the dot product.]
and opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular.
43. Use Theorem 3 to prove the Cauchy-Schwarz Inequality:
45. The Parallelogram Law states that
a b ab
a b
2
44. The Triangle Inequality for vectors is
ab
2
2 a
2
2 b
2
(a) Give a geometric interpretation of the Parallelogram Law. (b) Prove the Parallelogram Law. (See the hint in Exercise 44.)
a b a b (a) Give a geometric interpretation of the Triangle Inequality.
10.4
THE CROSS PRODUCT
THE CROSS PRODUCT The cross product a b of two vectors a and b, unlike the dot product, is a vector. For this reason it is also called the vector product. Note that a b is defined only when a and b are three-dimensional vectors. 1 DEFINITION If a a 1, a 2 , a 3 and b b1, b2 , b3 , then the cross product of a and b is the vector
a b a 2 b3 a 3 b2 , a 3 b1 a 1 b3 , a 1 b2 a 2 b1 This may seem like a strange way of defining a product. The reason for the particular form of Definition 1 is that the cross product defined in this way has many useful properties, as we will soon see. In particular, we will show that the vector a b is perpendicular to both a and b. In order to make Definition 1 easier to remember, we use the notation of determinants. A determinant of order 2 is defined by
a c
2 6
For example,
b ad bc d
1 24 16 14 4
A determinant of order 3 can be defined in terms of second-order determinants as follows:
2
a1 b1 c1
a2 b2 c2
a3 b2 b3 a1 c2 c3
b3 b1 a2 c3 c1
b3 b1 a3 c3 c1
b2 c2
Observe that each term on the right side of Equation 2 involves a number a i in the first row of the determinant, and a i is multiplied by the second-order determinant obtained from the left side by deleting the row and column in which a i appears. Notice also the
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
minus sign in the second term. For example,
1 3 5
2 1 0 0 1 1 4 4 2
1 3 2 2 5
1 3 1 2 5
0 4
10 4 26 5 112 0 38
If we now rewrite Definition 1 using second-order determinants and the standard basis vectors i , j, and k, we see that the cross product of a a 1 i a 2 j a 3 k and b b 1 i b 2 j b 3 k is a b
3
a2 b2
a3 a1 i b3 b1
a3 a1 j b3 b1
a2 k b2
In view of the similarity between Equations 2 and 3, we often write
i j a b a1 a2 b1 b2
4
k a3 b3
Although the first row of the symbolic determinant in Equation 4 consists of vectors, if we expand it as if it were an ordinary determinant using the rule in Equation 2, we obtain Equation 3. The symbolic formula in Equation 4 is probably the easiest way of remembering and computing cross products. V EXAMPLE 1
If a 1, 3, 4 and b 2, 7, 5 , then
i j k 3 a b 1 3 4 7 2 7 5
4 1 i 5 2
4 1 j 5 2
3 k 7
15 28 i 5 8 j 7 6 k 43i 13j k V EXAMPLE 2
■
Show that a a 0 for any vector a in V3.
SOLUTION If a a 1, a 2 , a 3 , then
i j a a a1 a2 a1 a2
k a3 a3
a 2 a 3 a 3 a 2 i a 1 a 3 a 3 a 1 j a 1 a 2 a 2 a 1 k 0i 0j 0k 0
■
One of the most important properties of the cross product is given by the following theorem.
SECTION 10.4
5 THEOREM
THE CROSS PRODUCT
■
539
The vector a b is orthogonal to both a and b.
PROOF In order to show that a b is orthogonal to a, we compute their dot product as follows:
a b a
a2 b2
a3 a1 a1 b3 b1
a3 a1 a2 b3 b1
a2 a3 b2
a 1a 2 b3 a 3 b2 a 2a 1 b3 a 3 b1 a 3a 1 b2 a 2 b1 a 1 a 2 b3 a 1 b2 a 3 a 1 a 2 b3 b1 a 2 a 3 a 1 b2 a 3 b1 a 2 a 3 0 A similar computation shows that a b b 0. Therefore, a b is orthogonal to both a and b. ■
axb
a
¨
b
FIGURE 1
The right-hand rule gives the direction of axb.
Visual 10.4 shows how a b changes as b changes.
If a and b are represented by directed line segments with the same initial point (as in Figure 1), then Theorem 5 says that the cross product a b points in a direction perpendicular to the plane through a and b. It turns out that the direction of a b is given by the right-hand rule: If the fingers of your right hand curl in the direction of a rotation (through an angle less than 180 ) from a to b, then your thumb points in the direction of a b. Now that we know the direction of the vector a b, the remaining thing we need to complete its geometric description is its length a b . This is given by the following theorem.
6 THEOREM
If is the angle between a and b (so 0 ), then
a b a b sin PROOF From the definitions of the cross product and length of a vector, we have
a b
2
a 2 b3 a 3 b22 a 3 b1 a 1 b32 a 1 b2 a 2 b12 a 22b 32 2a 2 a 3 b2 b3 a 32 b 22 a 32b12 2a 1 a 3 b1 b3 a12 b 23 a12 b 22 2a 1 a 2 b1 b2 a 22b12 a12 a 22 a 32 b 12 b 22 b 32 a 1 b1 a 2 b2 a 3 b3 2
b a b a b a b cos a b 1 cos a b sin
a
2
2
2
2
2
2
2
2
2
2
2
2
(by Theorem 10.3.3)
2
2
Taking square roots and observing that ssin 2 sin because sin 0 when 0 , we have a b a b sin
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540
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
Geometric characterization of a b
Since a vector is completely determined by its magnitude and direction, we can now say that a b is the vector that is perpendicular to both a and b, whose orientation is determined by the right-hand rule, and whose length is a b sin . In fact, that is exactly how physicists define a b.
Two nonzero vectors a and b are parallel if and only if
7 COROLLARY
a b0 PROOF Two nonzero vectors a and b are parallel if and only if 0 or . In either case sin 0, so a b 0 and therefore a b 0. ■
b
兩 b兩 sin ¨
¨ FIGURE 2
The geometric interpretation of Theorem 6 can be seen by looking at Figure 2. If a and b are represented by directed line segments with the same initial point, then they determine a parallelogram with base a , altitude b sin , and area
( b sin ) a b
A a
a
Thus we have the following way of interpreting the magnitude of a cross product. The length of the cross product a b is equal to the area of the parallelogram determined by a and b. EXAMPLE 3 Find a vector perpendicular to the plane that passes through the points
P1, 4, 6, Q2, 5, 1, and R1, 1, 1. l l l l SOLUTION The vector PQ PR is perpendicular to both PQ and PR and is therefore perpendicular to the plane through P, Q, and R. We know from (10.2.1) that l PQ 2 1 i 5 4 j 1 6 k 3i j 7k l PR 1 1 i 1 4 j 1 6 k 5 j 5k We compute the cross product of these vectors:
i l l PQ PR 3 0
j 1 5
k 7 5
5 35 i 15 0 j 15 0 k 40 i 15 j 15k So the vector 40, 15, 15 is perpendicular to the given plane. Any nonzero scalar multiple of this vector, such as 8, 3, 3 , is also perpendicular to the plane. EXAMPLE 4 Find the area of the triangle with vertices P1, 4, 6, Q2, 5, 1,
and R1, 1, 1.
l
l
SOLUTION In Example 3 we computed that PQ PR 40, 15, 15 . The
area of the parallelogram with adjacent sides PQ and PR is the length of this cross
■
SECTION 10.4
product:
l l
PR s40 PQ
2
THE CROSS PRODUCT
■
541
152 15 2 5s82
The area A of the triangle PQR is half the area of this parallelogram, that is, 52 s82 .
■
If we apply Theorems 5 and 6 to the standard basis vectors i , j, and k using 2, we obtain i jk
j ki
k ij
j i k
k j i
i k j
Observe that |
i jj i Thus the cross product is not commutative. Also i i j i k j whereas i i j 0 j 0 So the associative law for multiplication does not usually hold; that is, in general,
|
a b c a b c However, some of the usual laws of algebra do hold for cross products. The following theorem summarizes the properties of vector products. If a, b, and c are vectors and c is a scalar, then a b b a (ca) b c(a b) a (cb) a (b c) a b a c (a b) c a c b c a b c a b c a b c a cb a bc
8 THEOREM
1. 2. 3. 4. 5. 6.
These properties can be proved by writing the vectors in terms of their components and using the definition of a cross product. We give the proof of Property 5 and leave the remaining proofs as exercises. PROOF OF PROPERTY 5 If a a 1, a 2 , a 3 , b b1, b2 , b3 , and c c1, c2 , c3 ,
then 9
a b c a 1b2 c3 b3 c2 a 2b3 c1 b1 c3 a 3b1 c2 b2 c1 a 1 b2 c3 a 1 b3 c2 a 2 b3 c1 a 2 b1 c3 a 3 b1 c2 a 3 b2 c1 a 2 b3 a 3 b2 c1 a 3 b1 a 1 b3 c2 a 1 b2 a 2 b1 c3 a b c
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
TRIPLE PRODUCTS
The product a b c that occurs in Property 5 is called the scalar triple product of the vectors a, b, and c. Notice from Equation 9 that we can write the scalar triple product as a determinant:
a1 a b c b1 c1
10
¨ a
c b FIGURE 3
a3 b3 c3
The geometric significance of the scalar triple product can be seen by considering the parallelepiped determined by the vectors a, b, and c. (See Figure 3.) The area of the base parallelogram is A b c . If is the angle between a and b c, then the height h of the parallelepiped is h a cos . (We must use cos instead of cos in case 2.) Therefore, the volume of the parallelepiped is
bxc h
a2 b2 c2
V Ah b c
a cos a b c
Thus we have proved the following formula. 11 The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product:
V a b c
If we use the formula in (11) and discover that the volume of the parallelepiped determined by a, b, and c is 0, then the vectors must lie in the same plane; that is, they are coplanar. V EXAMPLE 5 Use the scalar triple product to show that the vectors a 1, 4, 7, b 2, 1, 4, and c 0, 9, 18 are coplanar.
SOLUTION We use Equation 10 to compute their scalar triple product:
1 a b c 2 0 1
4 1 9
1 9
7 4 18
4 2 4 18 0
4 2 7 18 0
1 9
118 436 718 0 Therefore, by (11) the volume of the parallelepiped determined by a, b, and c is 0. This means that a, b, and c are coplanar. ■ The product a b c that occurs in Property 6 is called the vector triple product of a, b, and c. Property 6 will be used to derive Kepler’s First Law of planetary motion in Section 10.9. Its proof is left as Exercise 42.
SECTION 10.4
THE CROSS PRODUCT
■
543
TORQUE
The idea of a cross product occurs often in physics. In particular, we consider a force F acting on a rigid body at a point given by a position vector r. (For instance, if we tighten a bolt by applying a force to a wrench as in Figure 4, we produce a turning effect.) The torque (relative to the origin) is defined to be the cross product of the position and force vectors r F
r ¨ F
and measures the tendency of the body to rotate about the origin. The direction of the torque vector indicates the axis of rotation. According to Theorem 6, the magnitude of the torque vector is
FIGURE 4
r F r F sin where is the angle between the position and force vectors. Observe that the only component of F that can cause a rotation is the one perpendicular to r, that is, F sin . The magnitude of the torque is equal to the area of the parallelogram determined by r and F.
EXAMPLE 6 A bolt is tightened by applying a 40-N force to a 0.25-m wrench as
shown in Figure 5. Find the magnitude of the torque about the center of the bolt. SOLUTION The magnitude of the torque vector is
75° 0.25 m
r F r F sin 75 0.2540 sin 75
40 N
10 sin 75 9.66 N m If the bolt is right-threaded, then the torque vector itself is
n 9.66 n
10.4
EXERCISES
■ Find the cross product a b and verify that it is orthogonal to both a and b.
1–7
1. a 1, 2, 0 ,
b 0, 3, 1
2. a 5, 1, 4 ,
b 1, 0, 2
3. a 2 i j k, 4. a i j k,
b j 2k
t
6. a i e j e k, t
7. a t, t 2, t 3 , ■
■
■
9. State whether each expression is meaningful. If not, explain
why. If so, state whether it is a vector or a scalar. (a) a b c (b) a b c (c) a b c (d) a b c (e) a b c d (f ) a b c d
b i 2 j 3k
10.
b 2 i e t j et k ■
■
■
■
11. | u|=6
b 1, 2t, 3t 2 ■
10 –11 ■ Find u v and determine whether u v is directed into the page or out of the page.
bijk
5. a 3 i 2 j 4 k,
■
■
where n is a unit vector directed down into the page.
FIGURE 5
■
■
| u|=5
■
|v|=8
60°
150°
| v|=10
8. If a i 2 k and b j k, find a b. Sketch a, b, and
a b as vectors starting at the origin.
■
■
■
■
■
■
■
■
■
■
■
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544
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
12. The figure shows a vector a in the xy-plane and a vector b
in the direction of k. Their lengths are a 3 and b 2. (a) Find a b . (b) Use the right-hand rule to decide whether the components of a b are positive, negative, or 0.
31–32 ■ Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS.
31. P2, 0, 1, 32. P3, 0, 1, ■
z
■
■
Q4, 1, 0, Q1, 2, 5, ■
■
R3, 1, 1,
S2, 2, 2
R5, 1, 1,
S0, 4, 2
■
■
■
■
■
■
33. Use the scalar triple product to verify that the vectors
u i 5 j 2 k, v 3 i j, and w 5 i 9 j 4 k are coplanar.
b
34. Use the scalar triple product to determine whether the a
points A1, 3, 2, B3, 1, 6, C5, 2, 0, and D3, 6, 4 lie in the same plane.
y
x
13. If a 1, 2, 1 and b 0, 1, 3 , find a b and b a. 14. If a 3, 1, 2 , b 1, 1, 0 , and c 0, 0, 4 ,
35. A bicycle pedal is pushed by a foot with a 60-N force as
shown. The shaft of the pedal is 18 cm long. Find the magnitude of the torque about P.
show that a b c a b c.
15. Find two unit vectors orthogonal to both 2, 0, 3
and 1, 4, 2 .
60 N
70°
16. Find two unit vectors orthogonal to both i j k
and 2 i k.
10°
P
17. Show that 0 a 0 a 0 for any vector a in V3 . 18. Show that a b b 0 for all vectors a and b in V3 .
36. Find the magnitude of the torque about P if a 36-lb force is
19. Prove Property 1 of Theorem 8.
applied as shown.
20. Prove Property 2 of Theorem 8.
4 ft P
21. Prove Property 3 of Theorem 8. 22. Prove Property 4 of Theorem 8.
4 ft
23. Find the area of the parallelogram with vertices A2, 1,
B0, 4, C4, 2, and D2, 1. 30° 36 lb
24. Find the area of the parallelogram with vertices K1, 2, 3,
L1, 3, 6, M3, 8, 6, and N3, 7, 3. ■ (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R, and (b) find the area of triangle PQR.
25–28
25. P1, 0, 0,
Q0, 2, 0,
26. P2, 1, 5,
Q1, 3, 4,
27. P0, 2, 0,
■
■
■
■
38. Let v 5 j and let u be a vector with length 3 that starts at
R3, 0, 6
Q3, 1, 0, ■
grips a bolt at the origin. A force is applied in the direction 0, 3, 4 at the end of the wrench. Find the magnitude of the force needed to supply 100 N m of torque to the bolt.
R0, 0, 3
Q4, 1, 2,
28. P2, 0, 3,
37. A wrench 30 cm long lies along the positive y-axis and
the origin and rotates in the xy -plane. Find the maximum and minimum values of the length of the vector u v. In what direction does u v point?
R5, 3, 1 R5, 2, 2
■
■
■
■
■
■
■
the points Q and R. Show that the distance d from the point P to the line L is
Find the volume of the parallelepiped determined by the vectors a, b, and c.
29–30
■
29. a 6, 3, 1 ,
b 0, 1, 2 ,
30. a i j k,
b i j k, c i j k
■
■
■
■
■
■
■
c 4, 2, 5
■
■
■
39. (a) Let P be a point not on the line L that passes through
d ■
■
a b a
l l where a QR and b QP.
■
SECTION 10.5
(b) Use the formula in part (a) to find the distance from the point P1, 1, 1 to the line through Q0, 6, 8 and R1, 4, 7. 40. (a) Let P be a point not on the plane that passes through the
points Q, R, and S. Show that the distance d from P to the plane is a b c d a b
EQUATIONS OF LINES AND PLANES
(a) If a b a c, does it follow that b c ? (b) If a b a c, does it follow that b c ? (c) If a b a c and a b a c, does it follow that b c ?
46. If v1, v2, and v3 are noncoplanar vectors, let
41. Prove that a b a b 2a b.
k1
v2 v3 v1 v2 v3
k2
v3 v1 v1 v2 v3
k3
v1 v2 v1 v2 v3
42. Prove Property 6 of Theorem 8, that is,
43. Use Exercise 42 to prove that
a b c b c a c a b 0 44. Prove that
a b c d
10.5
P¸(x¸, y¸, z¸) a P(x, y, z) r¸ O
ac ad
bc bd
(These vectors occur in the study of crystallography. Vectors of the form n1 v1 n 2 v2 n3 v3 , where each n i is an integer, form a lattice for a crystal. Vectors written similarly in terms of k1, k 2 , and k 3 form the reciprocal lattice.) (a) Show that k i is perpendicular to vj if i j. (b) Show that k i vi 1 for i 1, 2, 3. 1 (c) Show that k1 k 2 k 3 . v1 v2 v3
EQUATIONS OF LINES AND PLANES
z
L
545
45. Suppose that a 0.
l l l where a QR, b QS, and c QP. (b) Use the formula in part (a) to find the distance from the point P2, 1, 4 to the plane through the points Q1, 0, 0, R0, 2, 0, and S0, 0, 3.
a b c a cb a bc
■
r
v
x y
A line in the xy-plane is determined when a point on the line and the direction of the line (its slope or angle of inclination) are given. The equation of the line can then be written using the point-slope form. Likewise, a line L in three-dimensional space is determined when we know a point P0x 0 , y0 , z0 on L and the direction of L. In three dimensions the direction of a line is conveniently described by a vector, so we let v be a vector parallel to L. Let Px, y, z be an arbitrary point on L and let r0 and r be the position vectors of P0 and P (that is, they have representations OP A0 and OP A). If a is the vector with representation P A, 0P as in Figure 1, then the Triangle Law for vector addition gives r r0 a. But, since a and v are parallel vectors, there is a scalar t such that a tv. Thus
FIGURE 1 1
z
t=0
t>0 L
t<0 r¸
x
FIGURE 2
r r0 t v
y
which is a vector equation of L. Each value of the parameter t gives the position vector r of a point on L. In other words, as t varies, the line is traced out by the tip of the vector r. As Figure 2 indicates, positive values of t correspond to points on L that lie on one side of P0, whereas negative values of t correspond to points that lie on the other side of P0 . If the vector v that gives the direction of the line L is written in component form as v a, b, c , then we have tv ta, tb, tc . We can also write r x, y, z and
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
r0 x 0 , y0 , z0 , so the vector equation (1) becomes x, y, z x 0 ta, y0 tb, z0 tc Two vectors are equal if and only if corresponding components are equal. Therefore, we have the three scalar equations:
2
x x 0 at
y y0 bt
z z0 ct
where t ⺢. These equations are called parametric equations of the line L through the point P0x 0 , y0 , z0 and parallel to the vector v a, b, c . Each value of the parameter t gives a point x, y, z on L. EXAMPLE 1
(a) Find a vector equation and parametric equations for the line that passes through the point 5, 1, 3 and is parallel to the vector i 4 j 2k. (b) Find two other points on the line.
■ Figure 3 shows the line L in Example 1 and its relation to the given point and to the vector that gives its direction.
SOLUTION
(a) Here r0 5, 1, 3 5i j 3k and v i 4 j 2k, so the vector equation (1) becomes
z
r 5i j 3k ti 4 j 2k
L (5, 1, 3)
r¸
or v=i+4j-2k
x
FIGURE 3
y
r 5 t i 1 4t j 3 2t k
Parametric equations are x5t
y 1 4t
z 3 2t
(b) Choosing the parameter value t 1 gives x 6, y 5, and z 1, so 6, 5, 1 is a point on the line. Similarly, t 1 gives the point 4, 3, 5. ■ The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. For instance, if, instead of 5, 1, 3, we choose the point 6, 5, 1 in Example 1, then the parametric equations of the line become x6t
y 5 4t
z 1 2t
Or, if we stay with the point 5, 1, 3 but choose the parallel vector 2i 8j 4k, we arrive at the equations x 5 2t
y 1 8t
z 3 4t
In general, if a vector v a, b, c is used to describe the direction of a line L, then the numbers a, b, and c are called direction numbers of L. Since any vector parallel to v could also be used, we see that any three numbers proportional to a, b, and c could also be used as a set of direction numbers for L. Another way of describing a line L is to eliminate the parameter t from Equations 2. If none of a, b, or c is 0, we can solve each of these equations for t, equate the results,
SECTION 10.5
EQUATIONS OF LINES AND PLANES
■
547
and obtain
3
x x0 y y0 z z0 a b c
These equations are called symmetric equations of L . Notice that the numbers a, b, and c that appear in the denominators of Equations 3 are direction numbers of L, that is, components of a vector parallel to L. If one of a, b, or c is 0, we can still eliminate t. For instance, if a 0, we could write the equations of L as x x0
y y0 z z0 b c
This means that L lies in the vertical plane x x 0. EXAMPLE 2
■ Figure 4 shows the line L in Example 2 and the point P where it intersects the xy-plane.
B x
z
(a) Find parametric equations and symmetric equations of the line that passes through the points A2, 4, 3 and B3, 1, 1. (b) At what point does this line intersect the xy-plane?
1
SOLUTION 2
1
P
_1
y
v 3 2, 1 4, 1 3 1, 5, 4
L
A FIGURE 4
(a) We are not explicitly given a vector parallel to the line, but observe that the l vector v with representation AB is parallel to the line and
4
Thus direction numbers are a 1, b 5, and c 4. Taking the point 2, 4, 3 as P0, we see that parametric equations (2) are x2t
y 4 5t
z 3 4t
and symmetric equations (3) are x2 y4 z3 1 5 4 (b) The line intersects the xy-plane when z 0, so we put z 0 in the symmetric equations and obtain x2 y4 3 1 5 4 This gives x 114 and y 14 , so the line intersects the xy-plane at the point ( 114 , 14 , 0).
■
In general, the procedure of Example 2 shows that direction numbers of the line L through the points P0x 0 , y0 , z0 and P1x 1, y1, z1 are x 1 x 0 , y1 y0 , and z1 z0 and so symmetric equations of L are x x0 y y0 z z0 x1 x0 y1 y0 z1 z0
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CHAPTER 10
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Often, we need a description, not of an entire line, but of just a line segment. How, for instance, could we describe the line segment AB in Example 2? If we put t 0 in the parametric equations in Example 2(a), we get the point 2, 4, 3 and if we put t 1 we get 3, 1, 1. So the line segment AB is described by the parametric equations x2t
y 4 5t
z 3 4t
0 t 1
or by the corresponding vector equation rt 2 t, 4 5t, 3 4t
0 t 1
In general, we know from Equation 1 that the vector equation of a line through the (tip of the) vector r 0 in the direction of a vector v is r r 0 tv. If the line also passes through (the tip of ) r1, then we can take v r1 r 0 and so its vector equation is r r 0 tr1 r 0 1 tr 0 tr1 The line segment from r 0 to r1 is given by the parameter interval 0 t 1. 4
The line segment from r 0 to r1 is given by the vector equation rt 1 tr 0 t r1
■ The lines L 1 and L 2 in Example 3, shown in Figure 5, are skew lines.
V EXAMPLE 3
z
L¡
5
L™
10
x
Show that the lines L 1 and L 2 with parametric equations x1t
y 2 3t
z4t
x 2s
y3s
z 3 4s
are skew lines; that is, they do not intersect and are not parallel (and therefore do not lie in the same plane). 5
5
0 t 1
y
SOLUTION The lines are not parallel because the corresponding vectors 1, 3, 1 and 2, 1, 4 are not parallel. (Their components are not proportional.) If L 1 and L 2 had a point of intersection, there would be values of t and s such that
1 t 2s
_5
2 3t 3 s 4 t 3 4s
FIGURE 5
But if we solve the first two equations, we get t 115 and s 85 , and these values don’t satisfy the third equation. Therefore, there are no values of t and s that satisfy the three equations, so L 1 and L 2 do not intersect. Thus L 1 and L 2 are skew lines. ■ PLANES
Although a line in space is determined by a point and a direction, a plane in space is more difficult to describe. A single vector parallel to a plane is not enough to convey the “direction” of the plane, but a vector perpendicular to the plane does completely specify its direction. Thus a plane in space is determined by a point P0x 0 , y0 , z0 in the plane and a vector n that is orthogonal to the plane. This orthogonal vector n is
SECTION 10.5
■
549
called a normal vector. Let Px, y, z be an arbitrary point in the plane, and let r0 and r be the position vectors of P0 and P. Then the vector r r0 is represented by P A. 0P (See Figure 6.) The normal vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal to r r0 and so we have
z
n P(x, y, z)
r
EQUATIONS OF LINES AND PLANES
r-r¸
n r r0 0
5
r¸
0
P¸(x¸, y¸, z¸) x
which can be rewritten as
y
FIGURE 6
6
n r n r0
Either Equation 5 or Equation 6 is called a vector equation of the plane. To obtain a scalar equation for the plane, we write n a, b, c , r x, y, z , and r0 x 0 , y0 , z0 . Then the vector equation (5) becomes a, b, c x x 0 , y y0 , z z0 0 or 7
ax x 0 by y0 cz z0 0
Equation 7 is the scalar equation of the plane through P0x 0 , y0 , z0 with normal vector n a, b, c . V EXAMPLE 4 Find an equation of the plane through the point 2, 4, 1 with normal vector n 2, 3, 4 . Find the intercepts and sketch the plane.
SOLUTION Putting a 2, b 3, c 4, x 0 2, y0 4, and z0 1 in Equation 7, we see that an equation of the plane is
z (0, 0, 3)
2x 2 3y 4 4z 1 0 (0, 4, 0) (6, 0, 0) x
FIGURE 7
2x 3y 4z 12
or y
To find the x-intercept we set y z 0 in this equation and obtain x 6. Similarly, the y-intercept is 4 and the z-intercept is 3. This enables us to sketch the portion of the plane that lies in the first octant (see Figure 7). ■ By collecting terms in Equation 7 as we did in Example 4, we can rewrite the equation of a plane as 8
ax by cz d 0
where d ax 0 by0 cz0 . Equation 8 is called a linear equation in x, y, and z. Conversely, it can be shown that if a, b, and c are not all 0, then the linear equation (8) represents a plane with normal vector a, b, c . (See Exercise 55.)
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CHAPTER 10
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■ Figure 8 shows the portion of the plane in Example 5 that is enclosed by triangle PQR.
EXAMPLE 5 Find an equation of the plane that passes through the points P1, 3, 2,
Q3, 1, 6, and R5, 2, 0.
l
l
SOLUTION The vectors a and b corresponding to PQ and PR are
z
a 2, 4, 4
Q(3, _1, 6)
b 4, 1, 2
Since both a and b lie in the plane, their cross product a b is orthogonal to the plane and can be taken as the normal vector. Thus
P(1, 3, 2)
i na b 2 4
y x
R(5, 2, 0) FIGURE 8
j 4 1
k 4 12 i 20 j 14 k 2
With the point P1, 3, 2 and the normal vector n, an equation of the plane is 12x 1 20y 3 14z 2 0 6x 10y 7z 50
or n™ ¨ n¡
■
Two planes are parallel if their normal vectors are parallel. For instance, the planes x 2y 3z 4 and 2x 4y 6z 3 are parallel because their normal vectors are n1 1, 2, 3 and n 2 2, 4, 6 and n 2 2n1 . If two planes are not parallel, then they intersect in a straight line and the angle between the two planes is defined as the acute angle between their normal vectors (see angle in Figure 9).
¨ FIGURE 9
V EXAMPLE 6
(a) Find the angle between the planes x y z 1 and x 2y 3z 1. (b) Find symmetric equations for the line of intersection L of these two planes. ■ Figure 10 shows the planes in Example 6 and their line of intersection L .
SOLUTION
(a) The normal vectors of these planes are
x-2y+3z=1
x+y+z=1
6 4 2 z 0 _2 _4
L
n1 1, 1, 1 and so, if is the angle between the planes, cos
_2
0 y
2
2
0 x
_2
FIGURE 10
■ Another way to find the line of intersection is to solve the equations of the planes for two of the variables in terms of the third, which can be taken as the parameter.
n 2 1, 2, 3
n1 n 2 11 12 13 2 n1 n 2 s1 1 1 s1 4 9 s42
cos1
2 s42
72
(b) We first need to find a point on L. For instance, we can find the point where the line intersects the xy-plane by setting z 0 in the equations of both planes. This gives the equations x y 1 and x 2y 1, whose solution is x 1, y 0. So the point 1, 0, 0 lies on L. Now we observe that, since L lies in both planes, it is perpendicular to both of the normal vectors. Thus a vector v parallel to L is given by the cross product
i j v n1 n 2 1 1 1 2
k 1 5i 2 j 3 k 3
SECTION 10.5
■
EQUATIONS OF LINES AND PLANES
551
and so the symmetric equations of L can be written as x1 y z 5 2 3
■
EXAMPLE 7 Find a formula for the distance D from a point P1x 1, y1, z1 to the plane ax by cz d 0. SOLUTION Let P0x 0 , y0 , z0 be any point in the given plane and let b be the vector
corresponding to PA. 0 P1 Then b x 1 x 0 , y1 y0 , z1 z0 From Figure 11 you can see that the distance D from P1 to the plane is equal to the absolute value of the scalar projection of b onto the normal vector n a, b, c . (See Section 10.3.) Thus
P¡ ¨ b
D
P¸
FIGURE 11
nb ax n
1
D compn b
n
ax
x0 by1 y0 cz1 z0 sa 2 b 2 c 2
by1 cz1 ax0 by0 cz0 sa 2 b 2 c 2
1
Since P0 lies in the plane, its coordinates satisfy the equation of the plane and so we have ax 0 by0 cz0 d 0. Thus the formula for D can be written as
D
9
ax
by1 cz1 d sa 2 b 2 c 2
1
■
EXAMPLE 8 Find the distance between the parallel planes 10x 2y 2z 5
and 5x y z 1.
SOLUTION First we note that the planes are parallel because their normal vectors
10, 2, 2 and 5, 1, 1 are parallel. To find the distance D between the planes, we choose any point on one plane and calculate its distance to the other plane. In particular, if we put y z 0 in the equation of the first plane, we get 10x 5 and so ( 12 , 0, 0) is a point in this plane. By Formula 9, the distance between ( 12 , 0, 0) and the plane 5x y z 1 0 is D
5( ) 10 10 1 1 2
s5 1 1 2
2
2
3 2
3s3
s3 6
So the distance between the planes is s36.
10.5
EXERCISES
1. Determine whether each statement is true or false.
(a) (b) (c) (d) (e)
Two lines parallel to a third line are parallel. Two lines perpendicular to a third line are parallel. Two planes parallel to a third plane are parallel. Two planes perpendicular to a third plane are parallel. Two lines parallel to a plane are parallel.
(f ) (g) (h) (i) ( j) (k)
Two lines perpendicular to a plane are parallel. Two planes parallel to a line are parallel. Two planes perpendicular to a line are parallel. Two planes either intersect or are parallel. Two lines either intersect or are parallel. A plane and a line either intersect or are parallel.
■
552
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
18. L 1: x 1 2t,
■ Find a vector equation and parametric equations for the line.
2–5
L 2: x 1 s,
2. The line through the point 1, 0, 3 and parallel to the 3. The line through the point 2, 4, 10 and parallel to the
x3 y2 z1 4 3 2
20. L 1:
x1 y3 z2 2 2 1
L 2:
x2 y6 z2 1 1 3
4. The line through the origin and parallel to the line x 2t , 5. The line through the point (1, 0, 6) and perpendicular to the
plane x 3y z 5 ■
■
■
■
■
■
■
■
■
■
■ ■
6 –10
Find parametric equations and symmetric equations for
■
the line.
■
■
■
■
Find an equation of the plane.
z 8 t and is parallel to the plane 2 x 4y 8z 17
10. The line of intersection of the planes x y z 1
25. The plane through the points 0, 1, 1, 1, 0, 1, and 1, 1, 0
and x z 0
■
■
■
■
■
■
11. Is the line through 4, 6, 1 and 2, 0 3 parallel to
the line through 10, 18, 4 and 5, 3, 14 ?
12. Is the line through 4, 1, 1 and 2, 5, 3 perpendicular to
the line through 3, 2, 0 and 5, 1, 4 ?
13. (a) Find symmetric equations for the line that passes
through the point 0, 2, 1 and is parallel to the line with parametric equations x 1 2t, y 3t, z 5 7t. (b) Find the points in which the required line in part (a) intersects the coordinate planes. 14. (a) Find parametric equations for the line through 5, 1, 0
that is perpendicular to the plane 2x y z 1. (b) In what points does this line intersect the coordinate planes?
15. Find a vector equation for the line segment from 2, 1, 4
■
26. The plane through the origin and the points 2, 4, 6
and 5, 1, 3
27. The plane that passes through the point 6, 0, 2 and con-
tains the line x 4 2t, y 3 5t, z 7 4 t
28. The plane that passes through the point 1, 1, 1 and
contains the line with symmetric equations x 2y 3z
29. The plane that passes through the point 1, 2, 1 and con-
tains the line of intersection of the planes x y z 2 and 2 x y 3z 1
30. The plane that passes through the line of intersection of the
planes x z 1 and y 2z 3 and is perpendicular to the plane x y 2z 1 ■
16. Find parametric equations for the line segment from
10, 3, 1 to 5, 6, 3. Determine whether the lines L 1 and L 2 are parallel, skew, or intersecting. If they intersect, find the point of intersection. ■
y 1 9t,
z 3t
y 4 3s,
zs
■
■
■
■
■
■
■
■
■
■
■
31. Find the point at which the line x 3 t, y 2 t,
z 5t intersects the plane x y 2z 9.
32. Where does the line through 1, 0, 1 and 4, 2, 2
intersect the plane x y z 6 ?
to 4, 6, 1.
L 2: x 1 2s,
■
24. The plane that contains the line x 3 2t, y t,
1 x 2 2y z 3
17. L 1: x 6t,
■
2 x y 3z 1
9. The line through 1, 1, 1 and parallel to the line
17–20
■
23. The plane through the origin and parallel to the plane
and j k
■
■
vector j 2 k
8. The line through 2, 1, 0 and perpendicular to both i j
■
■
22. The plane through the point 4, 0, 3 and with normal
7. The line through the points (0, , 1) and 2, 1, 3
■
■ ■
the vector 2, 1, 5
1 2
■
■
21–30
z 1 3s
21. The plane through the point 6, 3, 2 and perpendicular to
6. The line through the points 6, 1, 3 and 2, 4, 5
■
y 4 s,
L 2:
vector 3, 1, 8
y 1 t, z 4 3t
z2t
x y1 z2 1 2 3
19. L 1:
vector 2 i 4 j 5 k
■
y 3t,
■ Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them.
33–36
33. x y z 1,
xyz1
34. 2 x 3y 4z 5 , 35. x 4y 2z,
8y 1 2 x 4z
36. x 2y 2z 1, ■
■
■
■
x 6y 4z 3 2x y 2z 1
■
■
■
■
■
■
■
■
SECTION 10.6
37. (a) Find symmetric equations for the line of intersection of
47– 48
the planes x y z 2 and 3x 4y 5z 6. (b) Find the angle between these planes.
■
are equidistant from the points 4, 2, 1 and 2, 4, 3. 39. Find an equation of the plane with x-intercept a, y-intercept
b, and z-intercept c.
4x 6y z 5
■
■
49–50
■
■
■
■
r 2, 0, 2 s 1, 1, 0
■
■
■
D
P1 : 4x 2y 6z 3
P2 : 4x 2y 2z 6
P3 : 6x 3y 9z 5
P4 : z 2 x y 3
y 4 t, z 1 t
Use the formula in Exercise 39 in Section 10.4 to find the distance from the point to the given line. ■
■
■
■
10.6
■
■
■
■
■
d
1 d2 b2 c2 sa 2
a x
d a
b y 0 cz 0 0
56. Give a geometric description of each family of planes.
z 5t
z 1 2t ■
■
ax by cz d 0 represents a plane and a, b, c is a normal vector to the plane. Hint: Suppose a 0 and rewrite the equation in the form
L 4 : r 2, 1, 3 t 2, 2, 10
■
■
55. If a, b, and c are not all 0, show that the equation
z 2 5t
L2: x 1 y 2 1 z
y 3t,
■
metric equations x 1 t, y 1 6t, z 2t, and x 1 2s, y 5 15s, z 2 6s.
them identical?
x 5 t,
■
54. Find the distance between the skew lines with para-
44. Which of the following four lines are parallel? Are any of
46. 1, 0, 1;
■
and x 1 y2 z3 are skew, and find the distance between these lines. [Hint: The skew lines lie in parallel planes.]
them identical?
y 2 3t,
■
53. Show that the lines with symmetric equations x y z
43. Which of the following four planes are parallel? Are any of
x 2 t,
■
x 2y 2z 1 and two units away from it.
0, 1, 2 that is perpendicular to the line x 1 t, y 1 t, z 2t and intersects this line.
45. 1, 2, 3;
■
52. Find equations of the planes that are parallel to the plane
42. Find parametric equations for the line through the point
45– 46
■
ax by cz d1 0 and ax by cz d2 0 is
0, 1, 2 that is parallel to the plane x y z 2 and perpendicular to the line x 1 t, y 1 t, z 2t.
L 3 : x 1 t,
■
x 2y 3z 1
■
41. Find parametric equations for the line through the point
y t,
■
51. Show that the distance between the parallel planes
(b) Find an equation of the plane that contains these lines.
L 1 : x 1 t,
■
3x 6y 3z 4
50. 3x 6y 9z 4,
r 1, 1, 0 t 1, 1, 2
■
Find the distance between the given parallel planes.
49. z x 2y 1,
40. (a) Find the point at which the given lines intersect:
553
x 2y 2z 1
48. 3, 2, 7, ■
■
Find the distance from the point to the given plane.
47. 2, 8, 5,
38. Find an equation for the plane consisting of all points that
CYLINDERS AND QUADRIC SURFACES
■
■
■
(a) x y z c (b) x y cz 1 (c) y cos z sin 1
CYLINDERS AND QUADRIC SURFACES We have already looked at two special types of surfaces—planes (in Section 10.5) and spheres (in Section 10.1). Here we investigate two other types of surfaces—cylinders and quadric surfaces. In order to sketch the graph of a surface, it is useful to determine the curves of intersection of the surface with planes parallel to the coordinate planes. These curves are called traces (or cross-sections) of the surface.
554
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
CYLINDERS
A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass through a given plane curve. z
V EXAMPLE 1
Sketch the graph of the surface z x 2.
SOLUTION Notice that the equation of the graph, z x 2, doesn’t involve y. This
means that any vertical plane with equation y k (parallel to the xz-plane) intersects the graph in a curve with equation z x 2. So these vertical traces are parabolas. Figure 1 shows how the graph is formed by taking the parabola z x 2 in the xz-plane and moving it in the direction of the y-axis. The graph is a surface, called a parabolic cylinder, made up of infinitely many shifted copies of the same parabola. Here the rulings of the cylinder are parallel to the y-axis. ■
0 x
y
We noticed that the variable y is missing from the equation of the cylinder in Example 1. This is typical of a surface whose rulings are parallel to one of the coordinate axes. If one of the variables x, y, or z is missing from the equation of a surface, then the surface is a cylinder.
FIGURE 1
The surface z=≈ is a parabolic cylinder.
EXAMPLE 2 Identify and sketch the surfaces.
(a) x 2 y 2 1
(b) y 2 z 2 1
SOLUTION
(a) Since z is missing and the equations x 2 y 2 1, z k represent a circle with radius 1 in the plane z k, the surface x 2 y 2 1 is a circular cylinder whose axis is the z-axis (see Figure 2). Here the rulings are vertical lines. (b) In this case x is missing and the surface is a circular cylinder whose axis is the x-axis (see Figure 3). It is obtained by taking the circle y 2 z 2 1, x 0 in the yz-plane and moving it parallel to the x-axis. z
z
y 0
x y
x
FIGURE 2 ≈+¥=1
|
FIGURE 3 ¥+z@=1
■
NOTE When you are dealing with surfaces, it is important to recognize that an equation like x 2 y 2 1 represents a cylinder and not a circle. The trace of the cylinder x 2 y 2 1 in the xy-plane is the circle with equations x 2 y 2 1, z 0. QUADRIC SURFACES
A quadric surface is the graph of a second-degree equation in three variables x, y, and z. The most general such equation is Ax 2 By 2 Cz 2 Dxy Eyz Fxz Gx Hy Iz J 0
SECTION 10.6
CYLINDERS AND QUADRIC SURFACES
■
555
where A, B, C, . . . , J are constants, but by translation and rotation it can be brought into one of the two standard forms Ax 2 By 2 Cz 2 J 0
or
Ax 2 By 2 Iz 0
Quadric surfaces are the counterparts in three dimensions of the conic sections in the plane. (See Section 9.5 for a review of conic sections.) EXAMPLE 3 Use traces to sketch the quadric surface with equation
x2
y2 z2 1 9 4
SOLUTION By substituting z 0, we find that the trace in the xy-plane is
x 2 y 29 1, which we recognize as an equation of an ellipse. In general, the horizontal trace in the plane z k is x2
(0, 0, 2)
0
zk
which is an ellipse, provided that k 2 4, that is, 2 k 2. Similarly, the vertical traces are also ellipses:
z
(1, 0, 0)
y2 k2 1 9 4
(0, 3, 0) y
y2 z2 1 k2 9 4
xk
if 1 k 1
z2 k2 1 4 9
yk
if 3 k 3
x2
x
FIGURE 4
The ellipsoid ≈+
z@ y@ + =1 4 9
Figure 4 shows how drawing some traces indicates the shape of the surface. It’s called an ellipsoid because all of its traces are ellipses. Notice that it is symmetric with respect to each coordinate plane; this is a reflection of the fact that its equation involves only even powers of x, y, and z. ■ EXAMPLE 4 Use traces to sketch the surface z 4x 2 y 2. SOLUTION If we put x 0, we get z y 2, so the yz-plane intersects the surface in a
parabola. If we put x k (a constant), we get z y 2 4k 2. This means that if we slice the graph with any plane parallel to the yz-plane, we obtain a parabola that opens upward. Similarly, if y k, the trace is z 4x 2 k 2, which is again a parabola that opens upward. If we put z k, we get the horizontal traces 4x 2 y 2 k, which we recognize as a family of ellipses. Knowing the shapes of the traces, we can sketch the graph in Figure 5. Because of the elliptical and parabolic traces, the quadric surface z 4x 2 y 2 is called an elliptic paraboloid. z
FIGURE 5 The surface z=4≈+¥ is an elliptic paraboloid. Horizontal traces are ellipses; vertical traces are parabolas.
0 x
y
■
556
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
V EXAMPLE 5
Sketch the surface z y 2 x 2.
SOLUTION The traces in the vertical planes x k are the parabolas z y 2 k 2,
which open upward. The traces in y k are the parabolas z x 2 k 2, which open downward. The horizontal traces are y 2 x 2 k, a family of hyperbolas. We draw the families of traces in Figure 6, and we show how the traces appear when placed in their correct planes in Figure 7. z
z
y
1
2 0
_1
1
_1 0
y
1
FIGURE 6
x
x
0
Vertical traces are parabolas; horizontal traces are hyperbolas. 2 All traces are labeled with the Traces in x=k are z=¥-k@ value of k.
1 Traces in y=k are z=_≈+k@
z
Traces in z=k are ¥-≈=k
z
z
1
0 x
_1
x
0
FIGURE 7
x
_1
_1
0
1
Traces moved to their correct planes
Traces in x=k
In Module 10.6A you can investigate how traces determine the shape of a surface.
y
y
y
1
Traces in y=k
Traces in z=k
In Figure 8 we fit together the traces from Figure 7 to form the surface z y 2 x 2, a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles that of a saddle. This surface will be investigated further in Section 12.7 when we discuss saddle points. z
0 y x
FIGURE 8
The surface z=¥-≈ is a hyperbolic paraboloid.
■
EXAMPLE 6 Sketch the surface
x2 z2 y2 1. 4 4
SOLUTION The trace in any horizontal plane z k is the ellipse
x2 k2 y2 1 4 4
zk
SECTION 10.6
z
CYLINDERS AND QUADRIC SURFACES
■
557
but the traces in the xz- and yz-planes are the hyperbolas x2 z2 1 4 4
y0
y2
and
z2 1 4
x0
(0, 1, 0)
(2, 0, 0)
y
x
This surface is called a hyperboloid of one sheet and is sketched in Figure 9.
FIGURE 9
In Module 10.6B you can see how changing a, b, and c in Table 1 affects the shape of the quadric surface.
The idea of using traces to draw a surface is employed in three-dimensional graphing software for computers. In most such software, traces in the vertical planes x k and y k are drawn for equally spaced values of k, and parts of the graph are eliminated using hidden line removal. Table 1 shows computer-drawn graphs of the six basic types of quadric surfaces in standard form. All surfaces are symmetric with respect to the z-axis. If a quadric surface is symmetric about a different axis, its equation changes accordingly. TABLE 1
Surface
x2 y2 z2 1 2 2 a b c2 All traces are ellipses. If a b c, the ellipsoid is a sphere.
z
y
x
z x2 y2 2 2 c a b Horizontal traces are ellipses. Vertical traces are parabolas. The variable raised to the first power indicates the axis of the paraboloid.
z
x
Surface
Equation
Cone z
x
Elliptic Paraboloid
y
Hyperboloid of One Sheet z
x
y
z2 x2 y2 2 2 c a b2 Horizontal traces are ellipses. Vertical traces in the planes x k and y k are hyperbolas if k 0 but are pairs of lines if k 0.
x2 y2 z2 1 2 2 a b c2 Horizontal traces are ellipses. Vertical traces are hyperbolas. The axis of symmetry corresponds to the variable whose coefficient is negative.
y
Hyperbolic Paraboloid z
y x
Graphs of Quadric Surfaces
Equation
Ellipsoid
■
z x2 y2 2 2 c a b Horizontal traces are hyperbolas. Vertical traces are parabolas. The case where c 0 is illustrated.
Hyperboloid of Two Sheets z
x
y
x2 y2 z2 1 2 2 a b c2 Horizontal traces in z k are ellipses if k c or k c. Vertical traces are hyperbolas. The two minus signs indicate two sheets.
558
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
z
EXAMPLE 7 Classify the quadric surface x 2 2z 2 6x y 10 0. SOLUTION By completing the square we rewrite the equation as
0
y 1 x 32 2z 2 y
Comparing this equation with Table 1, we see that it represents an elliptic paraboloid. Here, however, the axis of the paraboloid is parallel to the y-axis, and it has been shifted so that its vertex is the point 3, 1, 0. The traces in the plane y k k 1 are the ellipses x 32 2z 2 k 1 yk
(3, 1, 0)
x
The trace in the xy-plane is the parabola with equation y 1 x 32, z 0. The paraboloid is sketched in Figure 10. ■
FIGURE 10
≈+2z@-6x-y+10=0
10.6
EXERCISES
1. (a) What does the equation y x 2 represent as a curve
17. x 2 4z 2 y 0
in ⺢ 2 ? (b) What does it represent as a surface in ⺢ 3 ? (c) What does the equation z y 2 represent?
19. y z 2 x 2 ■
2. (a) Sketch the graph of y e x as a curve in ⺢ 2.
3– 8
3. y 4z 4
4. z 4 x
5. x y 0
6. yz 4
7. z cos x
8. x 2 y 2 1
2
2
■
3
Describe and sketch the surface.
2
■
■
■
■
■
■
■
2
■
■
■
2
10. (a) Find and identify the traces of the quadric surface
x 2 y 2 z 2 1 and explain why the graph looks like the graph of the hyperboloid of two sheets in Table 1. (b) If the equation in part (a) is changed to x 2 y 2 z 2 1, what happens to the graph? Sketch the new graph. Find the traces of the given surface in the planes x k, y k, z k. Then identify the surface and sketch it.
11–20
■
11. 4x 2 9y 2 36z 2 36
12. 4y x 2 z 2
13. y 2 x 2 z 2
14. z x 2 y 2
16. 25y z 100 4x 2
■
■
■
■
■
■
■
21. z 2 4x 2 9y 2 36
22. x 2 2y 2 3z 2
23. x 2y 2 3z 2
24. 4x y 2 4z 2 0
■
2
■
27. x 2 y 2 z 2 4x 2y 2z 4 0 28. x 2 y 2 z 2 2x 2y 4z 2 0 ■
■
■
■
■
■
■
■
■
■
■
■
29. Sketch the region bounded by the surfaces z sx 2 y 2
and x 2 y 2 1 for 1 z 2.
30. Sketch the region bounded by the paraboloids z x 2 y 2
and z 2 x 2 y 2.
31. Find an equation for the surface consisting of all points that
are equidistant from the point 1, 0, 0 and the plane x 1. Identify the surface. 32. Find an equation for the surface consisting of all points P
for which the distance from P to the x-axis is twice the distance from P to the yz-plane. Identify the surface. 2 2 2 ; 33. Graph the surfaces z x y and z 1 y on a com-
mon screen using the domain x 1.2, y 1.2 and observe the curve of intersection of these surfaces. Show that the projection of this curve onto the xy-plane is an ellipse.
34. Show that the curve of intersection of the surfaces
15. x 2 4y 2 z 2 4 2
20. 16x 2 y 2 4z 2 ■
26. 4y 2 z 2 x 16y 4z 20 0
x y z 1 and explain why the graph looks like the graph of the hyperboloid of one sheet in Table 1. (b) If we change the equation in part (a) to x 2 y 2 z 2 1, how is the graph affected? (c) What if we change the equation in part (a) to x 2 y 2 2y z 2 0? 2
■
25. 4x 2 y 2 4 z 2 4y 24z 36 0
9. (a) Find and identify the traces of the quadric surface 2
■
21–28 ■ Reduce the equation to one of the standard forms, classify the surface, and sketch it.
(b) Sketch the graph of y e as a surface in ⺢ . (c) Describe and sketch the surface z e y. x
■
18. x 2 4y 2 z 2 4
x 2 2y 2 z 2 3x 1 and 2x 2 4y 2 2z 2 5y 0 lies in a plane.
SECTION 10.7
10.7
VECTOR FUNCTIONS AND SPACE CURVES
■
559
VECTOR FUNCTIONS AND SPACE CURVES In general, a function is a rule that assigns to each element in the domain an element in the range. A vector-valued function, or vector function, is simply a function whose domain is a set of real numbers and whose range is a set of vectors. We are most interested in vector functions r whose values are three-dimensional vectors. This means that for every number t in the domain of r there is a unique vector in V3 denoted by rt. If f t, tt, and ht are the components of the vector rt, then f , t, and h are real-valued functions called the component functions of r and we can write rt f t, tt, ht f t i tt j ht k We use the letter t to denote the independent variable because it represents time in most applications of vector functions. EXAMPLE 1 If
rt t 3, ln3 t, st then the component functions are f t t 3
tt ln3 t
ht st
By our usual convention, the domain of r consists of all values of t for which the expression for rt is defined. The expressions t 3, ln3 t, and st are all defined when 3 t 0 and t 0. Therefore, the domain of r is the interval 0, 3. ■ The limit of a vector function r is defined by taking the limits of its component functions as follows.
1
If rt f t, tt, ht , then
If lim t l a rt L, this definition is equivalent to saying that the length and direction of the vector rt approach the length and direction of the vector L. ■
lim rt lim f t, lim tt, lim ht tla
tla
tla
tla
provided the limits of the component functions exist. Equivalently, we could have used an - definition (see Exercise 68). Limits of vector functions obey the same rules as limits of real-valued functions (see Exercise 67).
EXAMPLE 2 Find lim rt, where rt 1 t 3 i tet j tl0
sin t k. t
SOLUTION According to Definition 1, the limit of r is the vector whose components
are the limits of the component functions of r:
lim rt lim 1 t 3 i lim tet j lim tl0
tl0
ik
tl0
(by Equation 1.4.5)
tl0
sin t k t ■
560
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
A vector function r is continuous at a if lim rt ra tla
z
P { f(t), g(t), h(t)} C
In view of Definition 1, we see that r is continuous at a if and only if its component functions f , t, and h are continuous at a. There is a close connection between continuous vector functions and space curves. Suppose that f , t, and h are continuous real-valued functions on an interval I. Then the set C of all points x, y, z in space, where 2
0
r(t)=kf(t), g(t), h(t)l
x
y
FIGURE 1
C is traced out by the tip of a moving position vector r(t).
x f t
y tt
z ht
and t varies throughout the interval I, is called a space curve. The equations in (2) are called parametric equations of C and t is called a parameter. We can think of C as being traced out by a moving particle whose position at time t is ( f t, tt, ht). If we now consider the vector function rt f t, tt, ht , then rt is the position vector of the point P ( f t, tt, ht) on C. Thus any continuous vector function r defines a space curve C that is traced out by the tip of the moving vector rt, as shown in Figure 1. V EXAMPLE 3
Describe the curve defined by the vector function rt 1 t, 2 5t, 1 6t
SOLUTION The corresponding parametric equations are
Visual 10.7A shows several curves being traced out by position vectors, including those in Figures 1 and 2.
x1t
y 2 5t
z 1 6t
which we recognize from Equations 10.5.2 as parametric equations of a line passing through the point 1, 2, 1 and parallel to the vector 1, 5, 6 . Alternatively, we could observe that the function can be written as r r0 tv, where r0 1, 2, 1 and v 1, 5, 6 , and this is the vector equation of a line as given by Equation 10.5.1.
■
Plane curves can also be represented in vector notation. For instance, the curve given by the parametric equations x t 2 2t and y t 1 (see Example 1 in Section 9.1) could also be described by the vector equation rt t 2 2t, t 1 t 2 2t i t 1 j where i 1, 0 and j 0, 1 . z V EXAMPLE 4
Sketch the curve whose vector equation is rt cos t i sin t j t k
SOLUTION The parametric equations for this curve are
x cos t π
”0, 1, 2 ’
x
FIGURE 2
(1, 0, 0)
y
y sin t
zt
Since x 2 y 2 cos 2t sin 2t 1, the curve must lie on the circular cylinder x 2 y 2 1. The point x, y, z lies directly above the point x, y, 0, which moves counterclockwise around the circle x 2 y 2 1 in the xy-plane. (See Example 2 in Section 9.1.) Since z t, the curve spirals upward around the cylinder as t ■ increases. The curve, shown in Figure 2, is called a helix.
SECTION 10.7
VECTOR FUNCTIONS AND SPACE CURVES
■
561
The corkscrew shape of the helix in Example 4 is familiar from its occurrence in coiled springs. It also occurs in the model of DNA (deoxyribonucleic acid, the genetic material of living cells). In 1953 James Watson and Francis Crick showed that the structure of the DNA molecule is that of two linked, parallel helixes that are intertwined as in Figure 3. In Examples 3 and 4 we were given vector equations of curves and asked for a geometric description or sketch. In the next two examples we are given a geometric description of a curve and are asked to find parametric equations for the curve. EXAMPLE 5 Find a vector equation and parametric equations for the line segment
FIGURE 3
that joins the point P1, 3, 2 to the point Q2, 1, 3.
Figure 4 shows the line segment PQ in Example 5. ■
SOLUTION In Section 10.5 we found a vector equation for the line segment that joins the tip of the vector r 0 to the tip of the vector r1:
rt 1 t r 0 t r1
z
Q(2, _1, 3)
0 t 1
(See Equation 10.5.4.) Here we take r 0 1, 3, 2 and r1 2, 1, 3 to obtain a vector equation of the line segment from P to Q :
y
x
or
rt 1 t1, 3, 2 t2, 1, 3
0 t 1
rt 1 t, 3 4t, 2 5t
0 t 1
The corresponding parametric equations are P(1, 3, _2) FIGURE 4
x1t
y 3 4t
z 2 5t
0 t 1
■
Find a vector function that represents the curve of intersection of the cylinder x 2 y 2 1 and the plane y z 2. V EXAMPLE 6
SOLUTION Figure 5 shows how the plane and the cylinder intersect, and Figure 6 shows the curve of intersection C, which is an ellipse. z
z
y+z=2
(0, _1, 3)
(_1, 0, 2)
C (1, 0, 2)
(0, 1, 1)
≈+¥=1 0 x
FIGURE 5
y
x
FIGURE 6
y
562
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
The projection of C onto the xy-plane is the circle x 2 y 2 1, z 0. So we know from Example 2 in Section 9.1 that we can write x cos t
y sin t
0 t 2
From the equation of the plane, we have z 2 y 2 sin t So we can write parametric equations for C as x cos t
y sin t
0 t 2
z 2 sin t
The corresponding vector equation is rt cos t i sin t j 2 sin t k
0 t 2
This equation is called a parametrization of the curve C. The arrows in Figure 6 indicate the direction in which C is traced as the parameter t increases.
■
USING COMPUTERS TO DRAW SPACE CURVES
Space curves are inherently more difficult to draw by hand than plane curves; for an accurate representation we need to use technology. For instance, Figure 7 shows a computer-generated graph of the curve with parametric equations x 4 sin 20t cos t
y 4 sin 20t sin t
z cos 20t
It’s called a toroidal spiral because it lies on a torus. Another interesting curve, the trefoil knot, with equations x 2 cos 1.5t cos t
y 2 cos 1.5t sin t
z sin 1.5t
is graphed in Figure 8. It wouldn’t be easy to plot either of these curves by hand. z
z
y
x
y
x
FIGURE 7 A toroidal spiral
FIGURE 8 A trefoil knot
Even when a computer is used to draw a space curve, optical illusions make it difficult to get a good impression of what the curve really looks like. (This is especially true in Figure 8.) The next example shows how to cope with this problem. EXAMPLE 7 Use a computer to draw the curve with vector equation rt t, t 2, t 3 .
This curve is called a twisted cubic. SOLUTION We start by using the computer to plot the curve with parametric equa-
tions x t, y t 2, z t 3 for 2 t 2. The result is shown in Figure 9(a), but it’s hard to see the true nature of the curve from that graph alone. Most three-
SECTION 10.7
In Visual 10.7B you can rotate the box in Figure 9 to see the curve from any viewpoint.
z 6
VECTOR FUNCTIONS AND SPACE CURVES
_2
x
_6
6
6
4
z 0
_6 0
2
563
dimensional computer graphing programs allow the user to enclose a curve or surface in a box instead of displaying the coordinate axes. When we look at the same curve in a box in Figure 9(b), we have a much clearer picture of the curve. We can see that it climbs from a lower corner of the box to the upper corner nearest us, and it twists as it climbs. We get an even better idea of the curve when we view it from different vantage points. Part (c) shows the result of rotating the box to give another viewpoint.
z 0 2
■
_2
y
y2 4
(a)
_6
0 x
_2
0
2
y2
4
(b)
0 x
2
(c)
■
FIGURE 9
Views of the twisted cubic DERIVATIVES
The derivative r of a vector function r is defined in much the same way as for realvalued functions: dr rt h rt rt lim hl0 dt h
3
if this limit exists. The geometric significance of this definition is shown in Figure 10. l If the points P and Q have position vectors rt and rt h, then PQ represents the vector rt h rt, which can therefore be regarded as a secant vector. If h 0, the scalar multiple 1hrt h rt has the same direction as rt h rt. As h l 0, it appears that this vector approaches a vector that lies on the tangent line. For this reason, the vector rt is called the tangent vector to the curve defined by r at the point P, provided that rt exists and rt 0. The tangent line to C at P is defined to be the line through P parallel to the tangent vector rt. We will also have occasion to consider the unit tangent vector, which is Tt z
Visual 10.7C shows an animation of Figure 10.
rt rt
z
r(t+h)-r(t)
rª(t)
Q
P
P r(t)
r(t) r(t+h)
Q r(t+h)
C
C 0
0 y
x
FIGURE 10
r(t+h)-r(t) h
(a) The secant vector
y
x
(b) The tangent vector
564
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
The following theorem gives us a convenient method for computing the derivative of a vector function r : just differentiate each component of r. 4 THEOREM If rt f t, tt, ht f t i tt j ht k, where f , t, and h are differentiable functions, then
rt f t, tt, ht f t i tt j ht k
PROOF
rt lim
t l 0
lim
t l 0
lim
t l 0
1 rt t rt t 1 f t t, tt t, ht t f t, tt, ht t
lim
t l 0
f t t f t tt t tt ht t ht , , t t t
f t t f t tt t tt ht t ht , lim , lim t l 0 t l 0 t t t
f t, tt, ht
■
V EXAMPLE 8
(a) Find the derivative of rt 1 t 3 i tet j sin 2t k. (b) Find the unit tangent vector at the point where t 0. SOLUTION
(a) According to Theorem 4, we differentiate each component of r: rt 3t 2 i 1 tet j 2 cos 2t k (b) Since r0 i and r0 j 2k, the unit tangent vector at the point 1, 0, 0 is T0
r0 j 2k 1 2 j k r0 s1 4 s5 s5
■
EXAMPLE 9 For the curve rt st i 2 t j, find rt and sketch the position y
vector r1 and the tangent vector r1.
2
SOLUTION We have (1, 1)
r(1) 0
FIGURE 11
rt
rª(1) 1
x
1 ij 2st
and
r1
1 ij 2
The curve is a plane curve and elimination of the parameter from the equations x st , y 2 t gives y 2 x 2, x 0. In Figure 11 we draw the position vector r1 i j starting at the origin and the tangent vector r1 starting at the corre■ sponding point 1, 1.
SECTION 10.7
VECTOR FUNCTIONS AND SPACE CURVES
■
565
V EXAMPLE 10 Find parametric equations for the tangent line to the helix with parametric equations
x 2 cos t
y sin t
zt
at the point 0, 1, 2. SOLUTION The vector equation of the helix is rt 2 cos t, sin t, t , so
rt 2 sin t, cos t, 1 The parameter value corresponding to the point 0, 1, 2 is t 2, so the tangent vector there is r2 2, 0, 1 . The tangent line is the line through 0, 1, 2 parallel to the vector 2, 0, 1 , so by Equations 10.5.2 its parametric equations are x 2t
y1
z
t 2
■
12 8
■ The helix and the tangent line in Example 10 are shown in Figure 12.
z 4 0 _1
FIGURE 12
■ In Section 10.9 we will see how rt and rt can be interpreted as the velocity and acceleration vectors of a particle moving through space with position vector rt at time t.
_0.5
y 0
0.5
1
2
_2 0 x
Just as for real-valued functions, the second derivative of a vector function r is the derivative of r, that is, r r. For instance, the second derivative of the function in Example 10 is rt 2 cos t, sin t, 0 A curve given by a vector function rt on an interval I is called smooth if r is continuous and rt 0 (except possibly at any endpoints of I ). For instance, the helix in Example 10 is smooth because rt is never 0. EXAMPLE 11 Determine whether the semicubical parabola rt 1 t 3, t 2 is
y
smooth. cusp 0
1
FIGURE 13
The curve r(t)=k1+t # , t@l is not smooth.
SOLUTION Since x
rt 3t 2, 2t we have r0 0, 0 0 and, therefore, the curve is not smooth. The point that corresponds to t 0 is (1, 0), and we see from the graph in Figure 13 that there is a sharp corner, called a cusp, at (1, 0). Any curve with this type of behavior—an abrupt change in direction—is not smooth. ■ A curve, such as the semicubical parabola, that is made up of a finite number of smooth pieces is called piecewise smooth.
566
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
DIFFERENTIATION RULES
The next theorem shows that the differentiation formulas for real-valued functions have their counterparts for vector-valued functions. 5 THEOREM Suppose u and v are differentiable vector functions, c is a scalar, and f is a real-valued function. Then d 1. ut vt ut vt dt d 2. cut cut dt d 3. f tut f tut f tut dt d 4. ut vt ut vt ut vt dt d 5. ut vt ut vt ut vt dt d 6. u f t f tu f t (Chain Rule) dt
This theorem can be proved either directly from Definition 3 or by using Theorem 4 and the corresponding differentiation formulas for real-valued functions. The proof of Formula 4 follows; the remaining proofs are left as exercises. PROOF OF FORMULA 4 Let
ut f1t, f2t, f3t
vt t1t, t2t, t3t 3
Then
ut vt f1tt1t f2tt2t f3tt3t
f tt t i
i
i1
so the Product Rule for scalar functions gives d d ut vt dt dt
3
3
f tt t i
i
i1
i1
d fi tti t dt
3
f tt t f ttt i
i
i
i
i1 3
3
f tt t f ttt i
i
i
i1
i
i1
ut vt ut vt V EXAMPLE 12
for all t . SOLUTION Since
■
Show that if rt c (a constant), then rt is orthogonal to rt
rt rt rt
2
c2
SECTION 10.7
VECTOR FUNCTIONS AND SPACE CURVES
■
567
and c 2 is a constant, Formula 4 of Theorem 5 gives 0
d rt rt rt rt rt rt 2rt rt dt
Thus rt rt 0, which says that rt is orthogonal to rt. Geometrically, this result says that if a curve lies on a sphere with center the origin, then the tangent vector rt is always perpendicular to the position vector rt. ■ INTEGRALS
The definite integral of a continuous vector function rt can be defined in much the same way as for real-valued functions except that the integral is a vector. But then we can express the integral of r in terms of the integrals of its component functions f, t, and h as follows. (We use the notation of Chapter 5.)
y
b
a
n
rt dt lim
rt * t i
n l i1
n
lim
nl
n
f ti* t i
i1
n
tti* t j
i1
hti* t k
i1
and so
y
b
a
rt dt
y y y b
a
f t dt i
b
a
tt dt j
b
a
ht dt k
This means that we can evaluate an integral of a vector function by integrating each component function. We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows:
y
b
a
rt dt Rt]ba Rb Ra
where R is an antiderivative of r, that is, Rt rt. We use the notation x rt dt for indefinite integrals (antiderivatives). EXAMPLE 13 If rt 2 cos t i sin t j 2t k, then
y y
y rt dt y 2 cos t dt
i
sin t dt j
2t dt k
2 sin t i cos t j t 2 k C where C is a vector constant of integration, and
y
2
0
[
rt dt 2 sin t i cos t j t 2 k
]
2 0
2i j
2 k 4
■
568
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
10.7 1–2
EXERCISES
Find the domain of the vector function.
■
z
III
z
IV
1. rt t 2, st 1, s5 t 2. rt ■
■
3– 4
t2 i sin t j ln9 t 2 k t2 ■
■
■
■
■
Find the limit.
■
3. lim cos t, sin t, t ln t
4. lim
tl0
■
■
■
■
■
■
tl
■
y x
■
■
■
■
arctan t, e2t, ■
■
■
ln t t ■
■
x
x
■
6. rt t , t
7. rt t, cos 2t, sin 2t
8. rt 1 t, 3t, t
9. rt 1, cos t, 2 sin t
10. rt t 2 i t j 2 k
4
■
■
■
■
■
Q1, 2, 3
14. P1, 0, 1,
Q2, 3, 1
■
■
■
■
■
■
■
■
■
■
sect the paraboloid z x 2 y 2 ?
; 26. Graph the curve with parametric equations x s1 0.25 cos 2 10t cos t
16. P2, 4, 0,
Q6, 1, 2
y s1 0.25 cos 2 10t sin t
■
■
■
■
■
■
■
■
■
z 0.5 cos 10t
■
Explain the appearance of the graph by showing that it lies on a sphere.
17–22 ■ Match the parametric equations with the graphs (labeled I–VI). Give reasons for your choices. 17. x cos 4t,
y t,
27. Show that the curve with parametric equations x t 2,
z sin 4t
y 1 3t, z 1 t 3 passes through the points (1, 4, 0) and (9, 8, 28) but not through the point (4, 7, 6).
z et
18. x t,
y t 2,
19. x t,
y 11 t 2 ,
20. x et cos 10t,
z t2
y et sin 10t ,
28 –30 ■ Find a vector function that represents the curve of intersection of the two surfaces.
z et
21. x cos t,
y sin t,
z sin 5t
28. The cylinder x 2 y 2 4 and the surface z xy
22. x cos t,
y sin t,
z ln t
29. The cone z sx 2 y 2 and the plane z 1 y
I
z
■
25. At what points does the curve rt t i 2t t 2 k inter-
Q4, 1, 7
■
■
y cos t, z sin 2t is the curve of intersection of the surfaces z x 2 and x 2 y 2 1. Use this fact to help sketch the curve.
■
15. P1, 1, 2, ■
■
24. Show that the curve with parametric equations x sin t,
■
13–16 ■ Find a vector equation and parametric equations for the line segment that joins P to Q. 13. P0, 0, 0,
■
y t sin t, z t lies on the cone z 2 x 2 y 2, and use this fact to help sketch the curve.
6
■
■
23. Show that the curve with parametric equations x t cos t,
12. rt cos t i cos t j sin t k ■
y
2
11. rt t i t j t k 2
y x
5. rt sin t, t
3
z
VI
■
Sketch the curve with the given vector equation. Indicate with an arrow the direction in which t increases.
5–12
z
V
y
II
30. The paraboloid z 4x 2 y 2 and the parabolic cylinder
z
y x2
■
■
■
■
■
■
■
■
■
■
■
■
; 31. Try to sketch by hand the curve of intersection of the circux
y
x
y
lar cylinder x 2 y 2 4 and the parabolic cylinder z x 2. Then find parametric equations for this curve and use these equations and a computer to graph the curve.
SECTION 10.7
56. At what point do the curves r1t t, 1 t, 3 t 2 and
r2s 3 s, s 2, s 2 intersect? Find their angle of intersection correct to the nearest degree.
■
(a) Sketch the plane curve with the given vector equation. (b) Find rt. (c) Sketch the position vector rt and the tangent vector rt for the given value of t. 34. rt 1 t, st , 36. rt e i e t
t
t0
3t
t 6
38. rt 1 cos t i 2 sin t j, ■
■
39– 44
■
■
■
■
■
■
■
■
■
■
1
58.
y
1
59.
y
2
60.
y (t
61.
y e
62.
y cos t i sin t j t k dt
■
Find the derivative of the vector function.
39. rt t , 1 t, st
40. rt cos 3t, t, sin 3t
2
■
t2
41. rt e i j ln1 3t k
0
0
16t 3 i 9t 2 j 25t 4 k dt
3 sin 2 t cos t i 3 sin t cos 2 t j 2 sin t cos t k dt
0 2
4 2t j k dt 1 t2 1 t2
2
1
t
i tst 1 j t sin t k) dt
i 2t j ln t k dt
■
■
■
■
■
■
■
■
■
■
■
63. Find rt if rt 2t i 3t j st k and r1 i j. 2
42. rt at cos 3t i b sin t j c cos t k
64. Find rt if rt t i e t j te t k and r0 i j k.
43. rt a t b t 2 c
65. If two objects travel through space along two different
3
3
44. rt t a b t c ■
■
■
■
■
■
■
■
■
■
■
■
■ Find the unit tangent vector Tt at the point with the given value of the parameter t.
45– 46
45. rt cos t i 3t j 2 sin 2t k,
t0
■
■
■
■
■
■
■
■
■
■
■
■
47. If rt t, t , t , find rt, T1, rt, and rt rt. 2
curves, it’s often important to know whether they will collide. (Will a missile hit its moving target? Will two aircraft collide?) The curves might intersect, but we need to know whether the objects are in the same position at the same time. Suppose the trajectories of two particles are given by the vector functions r1 t t 2, 7t 12, t 2
t 4
46. rt 2 sin t i 2 cos t j tan t k, 3
r2 t 4t 3, t 2, 5t 6
for t 0. Do the particles collide? 66. Two particles travel along the space curves
48. If rt e 2t, e2t, te 2t , find T0, r0, and rt rt.
r1 t t, t 2, t 3
49–52 Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
Do the particles collide? Do their paths intersect?
■
49. x t 5, y t 4, z t 3; 51. x et cos t, 52. x ln t, ■
■
■
y et sin t,
y 2st , ■
1, 1, 1
z et;
1, 0, 1
■
■
tla
■
■
■
■
(a) rt t , t , t (c) rt cos 3t, sin 3t 4
5
tla
tla
(c) lim ut vt lim ut lim vt tla
53. Determine whether the curve is smooth. 3
tla
(b) lim cut c lim ut
2
■
67. Suppose u and v are vector functions that possess limits as
tla
z t ; 0, 2, 1
■
r2 t 1 2t, 1 6t, 1 14t
t l a and let c be a constant. Prove the following properties of limits. (a) lim ut vt lim ut lim vt
(1, 1, 1)
50. x t 2 1, y t 2 1, z t 1;
(b) rt t t, t , t 3
4
5
54. (a) Find the point of intersection of the tangent lines to the
;
y
t1 t 4
Evaluate the integral.
■
57.
j, t 0
37. rt e i e j, t
57–62
t 1
35. rt sin t i 2 cos t j,
569
intersect at the origin. Find their angle of intersection correct to the nearest degree.
parabolic cylinder y x and the top half of the ellipsoid x 2 4y 2 4z 2 16. Then find parametric equations for this curve and use these equations and a computer to graph the curve. 2
33. rt t 2, t 2 1 ,
■
55. The curves r1t t, t 2, t 3 and r2t sin t, sin 2t, t
; 32. Try to sketch by hand the curve of intersection of the
33–38
VECTOR FUNCTIONS AND SPACE CURVES
curve rt sin t, 2 sin t, cos t at the points where t 0 and t 0.5. (b) Illustrate by graphing the curve and both tangent lines.
tla
tla
(d) lim ut vt lim ut lim vt tla
tla
tla
68. Show that lim t l a rt b if and only if for every 0
there is a number 0 such that rt b whenever 0 t a .
69. Prove Formula 1 of Theorem 5.
570
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
70. Prove Formula 3 of Theorem 5.
76. Find an expression for
71. Prove Formula 5 of Theorem 5. 72. Prove Formula 6 of Theorem 5.
77. If rt 0, show that
73. If ut i 2t j 3t k and 2
3
[Hint: rt
vt t i cos t j sin t k, find ddt ut vt .
2
74. If u and v are the vector functions in Exercise 73, find
d 1 rt rt rt. dt rt
rt rt]
78. If a curve has the property that the position vector rt is
ddt ut vt .
always perpendicular to the tangent vector rt, show that the curve lies on a sphere with center the origin.
75. Show that if r is a vector function such that r exists, then
79. If ut rt rt rt , show that
d rt rt rt rt dt
10.8
d ut vt wt . dt
ut rt rt rt
ARC LENGTH AND CURVATURE In Section 9.2 we defined the length of a plane curve with parametric equations x f t, y tt, a t b, as the limit of lengths of inscribed polygons and, for the case where f and t are continuous, we arrived at the formula L y s f t 2 tt 2 dt b
1
a
y
b
a
2
dx dt
dy dt
2
dt
The length of a space curve is defined in exactly the same way (see Figure 1). Suppose that the curve has the vector equation rt f t, tt, ht , a t b, or, equivalently, the parametric equations x f t, y tt, z ht, where f , t, and h are continuous. If the curve is traversed exactly once as t increases from a to b, then it can be shown that its length is
z
0
L y s f t 2 tt 2 ht 2 dt b
y
2
a
x
FIGURE 1
y
b
a
The length of a space curve is the limit of lengths of inscribed polygons.
dx dt
2
dy dt
2
dz dt
2
dt
Notice that both of the arc length formulas (1) and (2) can be put into the more compact form 3
Ly
b
a
rt dt
because, for plane curves rt f t i tt j,
rt f t i tt j s f t
2
tt 2
2
tt 2 ht 2
whereas, for space curves rt f t i tt j ht k,
rt f t i tt j ht k s f t
SECTION 10.8
ARC LENGTH AND CURVATURE
■
571
V EXAMPLE 1 Find the length of the arc of the circular helix with vector equation rt cos t i sin t j t k from the point 1, 0, 0 to the point 1, 0, 2.
■ Figure 2 shows the arc of the helix whose length is computed in Example 1.
SOLUTION Since rt sin t i cos t j k, we have
z
rt ssin t
cos 2 t 1 s2
2
The arc from 1, 0, 0 to 1, 0, 2 is described by the parameter interval 0 t 2 and so, from Formula 3, we have
(1, 0, 2π)
L
(1, 0, 0) x
2
y rt dt y 0
2
s2 dt 2s2
0
■
y
A single curve C can be represented by more than one vector function. For instance, the twisted cubic
FIGURE 2
r1t t, t 2, t 3
4
1 t 2
could also be represented by the function r2u e u, e 2u, e 3u
5
where the connection between the parameters t and u is given by t e u. We say that Equations 4 and 5 are parametrizations of the curve C. If we were to use Equation 3 to compute the length of C using Equations 4 and 5, we would get the same answer. In general, it can be shown that when Equation 3 is used to compute the length of any piecewise-smooth curve, the arc length is independent of the parametrization that is used. Now we suppose that C is a piecewise-smooth curve given by a vector function rt f ti ttj htk, a t b, and C is traversed exactly once as t increases from a to b. We define its arc length function s by
■ Piecewise-smooth curves were introduced on page 565.
z
s(t) C r(t)
6
r(a) 0 x
FIGURE 3
0 u ln 2
y
st
y ru du y t
t
a
a
dx du
2
dy du
2
dz du
2
du
Thus st is the length of the part of C between ra and rt. (See Figure 3.) If we differentiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus, we obtain ds 7 rt dt
It is often useful to parametrize a curve with respect to arc length because arc length arises naturally from the shape of the curve and does not depend on a particular coordinate system. If a curve rt is already given in terms of a parameter t and st is the arc length function given by Equation 6, then we may be able to solve for t as a function of s: t ts. Then the curve can be reparametrized in terms of s by substituting for t : r rts. Thus if s 3 for instance, rt3 is the position vector of the point 3 units of length along the curve from its starting point. EXAMPLE 2 Reparametrize the helix rt cos t i sin t j t k with respect to
arc length measured from 1, 0, 0 in the direction of increasing t.
SOLUTION The initial point 1, 0, 0 corresponds to the parameter value t 0.
572
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
From Example 1 we have ds rt s2 dt
s st y ru du y s2 du s2 t t
and so
0
t
0
Therefore, t ss2 and the required reparametrization is obtained by substituting for t : ■ rts cos(ss2 ) i sin(ss2 ) j (ss2 ) k CURVATURE
If C is a smooth curve defined by the vector function r, then rt 0. Recall that the unit tangent vector Tt is given by
z
Tt 0 x
C
y
FIGURE 4
Unit tangent vectors at equally spaced points on C
Visual 10.8A shows animated unit tangent vectors, like those in Figure 4, for a variety of plane curves and space curves.
rt rt
and indicates the direction of the curve. From Figure 4 you can see that Tt changes direction very slowly when C is fairly straight, but it changes direction more quickly when C bends or twists more sharply. The curvature of C at a given point is a measure of how quickly the curve changes direction at that point. Specifically, we define it to be the magnitude of the rate of change of the unit tangent vector with respect to arc length. (We use arc length so that the curvature will be independent of the parametrization.) The curvature of a curve is
8 DEFINITION
$
dT ds
where T is the unit tangent vector. The curvature is easier to compute if it is expressed in terms of the parameter t instead of s, so we use the Chain Rule (Theorem 10.7.5, Formula 6) to write dT dT ds dt ds dt
and
$
dT dTdt ds dsdt
But dsdt rt from Equation 7, so
9
V EXAMPLE 3
$t
Tt rt
Show that the curvature of a circle of radius a is 1a.
SOLUTION We can take the circle to have center the origin, and then a parametriza-
tion is rt a cos t i a sin t j
SECTION 10.8
ARC LENGTH AND CURVATURE
rt a sin t i a cos t j
Therefore
Tt
so
■
573
rt a
and
rt sin t i cos t j rt
Tt cos t i sin t j
and
This gives Tt 1, so using Equation 9, we have
Tt 1 rt a
$t
■
The result of Example 3 shows that small circles have large curvature and large circles have small curvature, in accordance with our intuition. We can see directly from the definition of curvature that the curvature of a straight line is always 0 because the tangent vector is constant. Although Formula 9 can be used in all cases to compute the curvature, the formula given by the following theorem is often more convenient to apply. 10 THEOREM
The curvature of the curve given by the vector function r is
rt rt rt
$t
3
PROOF Since T r r and r dsdt, we have
ds T dt
r r T
so the Product Rule (Theorem 10.7.5, Formula 3) gives r
d 2s ds T T dt 2 dt
Using the fact that T T 0 (see Example 2 in Section 10.4), we have r r
ds dt
2
T T
Now Tt 1 for all t, so T and T are orthogonal by Example 12 in Section 10.7. Therefore, by Theorem 10.4.6,
Thus
and
r r
ds dt
2
T T
ds dt
2
T
T
r r r r T dsdt r T r r $ r r 2
ds dt
2
T
2
3
■
574
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
EXAMPLE 4 Find the curvature of the twisted cubic rt t, t 2, t 3 at a general
point and at 0, 0, 0.
SOLUTION We first compute the required ingredients:
rt 1, 2t, 3t 2
rt s1 4t
2
rt 0, 2, 6t
9t 4
i rt rt 1 0
j 2t 2
rt rt s36t
4
k 3t 2 6t 2 i 6t j 2 k 6t
36t 2 4 2s9t 4 9t 2 1
Theorem 10 then gives
$t
rt rt 2s1 9t 9t 1 4t 9t rt 2
3
2
4
4 32
At the origin, where t 0, the curvature is $0 2.
■
For the special case of a plane curve with equation y f x, we choose x as the parameter and write rx x i f x j. Then rx i f x j and rx f x j. Since i j k and j j 0, we have rx rx f x k. We also have rx s1 f x 2 and so, by Theorem 10,
$x
11
f x 1 f x2 32
y 2
EXAMPLE 5 Find the curvature of the parabola y x 2 at the points 0, 0, 1, 1,
y=≈
and 2, 4.
SOLUTION Since y 2x and y 2, Formula 11 gives
$x
y=k(x) 0
FIGURE 5
The parabola y=≈ and its curvature function y=k(x)
1
y
1 y
2 32
2 1 4x 2 32
x
The curvature at 0, 0 is $0 2. At 1, 1 it is $1 25 32 0.18. At 2, 4 it is $2 217 32 0.03. Observe from the expression for $x or the graph of $ in Figure 5 that $x l 0 as x l . This corresponds to the fact that the parabola appears to become flatter as x l . ■ THE NORMAL AND BINORMAL VECTORS
At a given point on a smooth space curve rt, there are many vectors that are orthogonal to the unit tangent vector Tt. We single out one by observing that, because Tt 1 for all t, we have Tt Tt 0 by Example 12 in Section 10.7, so Tt
SECTION 10.8
■ We can think of the normal vector as indicating the direction in which the curve is turning at each point.
B(t)
■
575
is orthogonal to Tt. Note that Tt is itself not a unit vector. But if r is also smooth, we can define the principal unit normal vector Nt (or simply unit normal) as Nt
T(t)
ARC LENGTH AND CURVATURE
Tt Tt
The vector Bt Tt Nt is called the binormal vector. It is perpendicular to both T and N and is also a unit vector. (See Figure 6.)
N(t)
EXAMPLE 6 Find the unit normal and binormal vectors for the circular helix
rt cos t i sin t j t k FIGURE 6
SOLUTION We first compute the ingredients needed for the unit normal vector:
Figure 7 illustrates Example 6 by showing the vectors T, N, and B at two locations on the helix. In general, the vectors T, N, and B, starting at the various points on a curve, form a set of orthogonal vectors, called the TNB frame, that moves along the curve as t varies. This TNB frame plays an important role in the branch of mathematics known as differential geometry and in its applications to the motion of spacecraft. ■
rt sin t i cos t j k
rt s2
Tt
1 cos t i sin t j s2
Nt
z
rt 1 sin t i cos t j k rt s2
Tt
1
Tt s2
Tt cos t i sin t j cos t, sin t, 0 Tt
T B N
This shows that the normal vector at a point on the helix is horizontal and points toward the z-axis. The binormal vector is
T
B
Bt Tt Nt
N y x
FIGURE 7
Visual 10.8B shows how the TNB frame moves along several curves.
1 s2
i sin t cos t
1 sin t, cos t, 1 s2
j cos t sin t
k 1 0
■
The plane determined by the normal and binormal vectors N and B at a point P on a curve C is called the normal plane of C at P. It consists of all lines that are orthogonal to the tangent vector T. The plane determined by the vectors T and N is called the osculating plane of C at P. The name comes from the Latin osculum, meaning “kiss.” It is the plane that comes closest to containing the part of the curve near P. (For a plane curve, the osculating plane is simply the plane that contains the curve.) The circle that lies in the osculating plane of C at P, has the same tangent as C at P, lies on the concave side of C (toward which N points), and has radius 1$ (the
576
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
reciprocal of the curvature) is called the osculating circle (or the circle of curvature) of C at P. It is the circle that best describes how C behaves near P; it shares the same tangent, normal, and curvature at P. (See Figure 8 and Web Examples A and B.) We summarize here the formulas for unit tangent, unit normal and binormal vectors, and curvature.
y
y=≈
osculating circle
1 2
0
Tt
x
1
rt rt
Nt
FIGURE 8
$
Visual 10.8C shows how the osculating circle changes as a point moves along a curve.
10.8 1– 4
■
1. rt 2 sin t, 5t, 2 cos t ,
11–14
3. rt i t j t k, 3
0 t
■
■
■
■
■
■
■
■
■
■
14. rt t, t , t ■
y sin 3t, z sin t. Find the total length of this curve correct to four decimal places.
Reparametrize the curve with respect to arc length measured from the point where t 0 in the direction of increasing t. ■
■
■
■
■
■
■
10. Reparametrize the curve
2 2t 1 i 2 j t 1 t 1 2
■
■
■
■
■
■
■
■
■
■
Use Theorem 10 to find the curvature.
15. rt t 2 i t k 16. rt t i t j 1 t 2 k 17. rt 3t i 4 sin t j 4 cos t k ■
■
■
■
■
■
■
■
■
■
t
; 20. Graph the curve with parametric equations xt
along the curve x 3 sin t, y 4t, z 3 cos t in the positive direction. Where are you now?
■
19. Find the curvature of rt t, t 2, t 3 at the point (1, 1, 1). ■
9. Suppose you start at the point 0, 0, 3 and move 5 units
rt
point (1, 0, 0).
2t
■
■
■
t0
18. Find the curvature of rt e cos t, e sin t, t at the
8. rt e cos 2t i 2 j e sin 2t k ■
■
2
t
t
7. rt 2t i 1 3t j 5 4t k
■
■
15–17
; 6. Graph the curve with parametric equations x cos t,
■
13. rt s2 t, e , e t
1 2 2
the arc of the twisted cubic x t, y t 2, z t 3 from the origin to the point 2, 4, 8.
2t
■
12. rt t 2, sin t t cos t, cos t t sin t ,
0 t 1
5. Use Simpson’s Rule with n 10 to estimate the length of
7– 8
3
11. rt 2 sin t, 5t, 2 cos t
0 t 1
4. rt 12t i 8t 32 j 3t 2 k, ■
Tt rt rt rt rt
(a) Find the unit tangent and unit normal vectors Tt and Nt. (b) Use Formula 9 to find the curvature.
10 t 10
2. rt t 2, sin t t cos t, cos t t sin t ,
■
dT ds
Bt Tt Nt
EXERCISES
Find the length of the curve.
2
Tt Tt
with respect to arc length measured from the point (1, 0) in the direction of increasing t. Express the reparametrization in its simplest form. What can you conclude about the curve?
y 4t 32
z t 2
and find the curvature at the point 1, 4, 1. 21–23
■
Use Formula 11 to find the curvature.
21. y xe x ■
■
■
23. y 4x 52
22. y cos x ■
■
■
■
■
■
■
■
■
■ At what point does the curve have maximum curvature? What happens to the curvature as x l ?
24 –25
24. y ln x ■
■
■
25. y e x ■
■
■
■
■
■
■
■
■
SECTION 10.8
26. Find an equation of a parabola that has curvature 4 at the
577
■ Find equations of the normal plane and osculating plane of the curve at the given point.
37. x 2 sin 3t , y t, z 2 cos 3t ;
27. (a) Is the curvature of the curve C shown in the figure
greater at P or at Q ? Explain. (b) Estimate the curvature at P and at Q by sketching the osculating circles at those points.
38. x t, y t 2, z t 3;
1, 1, 1
■
■
■
■
■
■
■
0, , 2 ■
■
■
■
■
; 39. Find equations of the osculating circles of the ellipse
P
9x 2 4y 2 36 at the points 2, 0 and 0, 3. Use a graphing calculator or computer to graph the ellipse and both osculating circles on the same screen.
C
1
of the parabola ; 40. Find 1equations of the osculating circles 1
y 2 x 2 at the points 0, 0 and (1, 2 ). Graph both osculating circles and the parabola on the same screen.
Q 0
■
37–38
origin.
y
ARC LENGTH AND CURVATURE
x
1
41. At what point on the curve x t 3, y 3t , z t 4 is the
normal plane parallel to the plane 6x 6y 8z 1?
; 28 –29
Use a graphing calculator or computer to graph both the curve and its curvature function $x on the same screen. Is the graph of $ what you would expect? ■
■
■
■
■
■
■
■
■
■
■
■
■
where is the angle between T and i ; that is, is the angle of inclination of the tangent line.
a
a b
b
45. (a) Show that d Bds is perpendicular to B. x
x ■
■
■
■
■
■
■
■
■
■
■
32. Use Theorem 10 to show that the curvature of a plane para-
metric curve x f t, y tt is
$
■
are of fundamental importance in differential geometry: 1. dTds $ N 2. dNds $ T % B 3. dBds % N (Formula 1 comes from Exercise 43 and Formula 3 comes from Exercise 45.) Use the fact that N B T to deduce Formula 2 from Formulas 1 and 3.
x y yx
x 2 y 2 32
Use the formula in Exercise 32 to find the curvature.
33. x e t cos t,
y e t sin t
■
■
35–36
■
■
47. Use the Frenet-Serret formulas to prove each of the follow-
y t t2
34. x 1 t 3, ■
■
■
■
■
2
■
■
■
(1, 23 , 1)
36. rt e t, e t sin t, e t cos t ,
1, 0, 1
■
■
■
■
Find the vectors T, N, and B at the given point.
35. rt t 2, 3 t 3, t ,
■
■
■
■
(b) Show that d Bds is perpendicular to T. (c) Deduce from parts (a) and (b) that d Bds % sN for some number % s called the torsion of the curve. (The torsion measures the degree of twisting of a curve.) (d) Show that for a plane curve the torsion is % s 0. 46. The following formulas, called the Frenet-Serret formulas,
where the dots indicate derivatives with respect to t. 33–34
44. Show that the curvature of a plane curve is $ dds ,
y
y
■
dT $N ds
31.
30.
43. Show that the curvature $ is related to the tangent and nor-
mal vectors by the equation
Two graphs, a and b, are shown. One is a curve y f x and the other is the graph of its curvature function y $x. Identify each curve and explain your choices.
30 –31
42. Is there a point on the curve in Exercise 41 where the
osculating plane is parallel to the plane x y z 1? [Note: You will need a CAS for differentiating, for simplifying, and for computing a cross product.]
29. y x 2
28. y x 4 2x 2 ■
CAS
■
ing. (Primes denote derivatives with respect to t. Start as in the proof of Theorem 10.) (a) r sT $ s2 N (b) r r $ s3 B 2 3 (c) r s $ s T 3 $ss $s2 N $% s3 B (d) %
■
■
■
■
r r r r r 2
578
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
48. Show that the circular helix
Existing track along the negative x-axis is to be joined smoothly to a track along the line y 1 for x 1. (a) Find a polynomial P Px of degree 5 such that the function F defined by
rt a cos t, a sin t, bt where a and b are positive constants, has constant curvature and constant torsion. [Use the result of Exercise 47(d).]
0 if x 0 Fx Px if 0 x 1 1 if x 1
49. The DNA molecule has the shape of a double helix (see
Figure 3 on page 561). The radius of each helix is about 10 angstroms (1 Å 108 cm). Each helix rises about 34 Å during each complete turn, and there are about 2.9 10 8 complete turns. Estimate the length of each helix. 50. Let’s consider the problem of designing a railroad track to
make a smooth transition between sections of straight track.
10.9
MOTION IN SPACE: VELOCITY AND ACCELERATION
r(t+h)-r(t) h rª(t) Q
z
;
is continuous and has continuous slope and continuous curvature. (b) Use a graphing calculator or computer to draw the graph of F .
In this section we show how the ideas of tangent and normal vectors and curvature can be used in physics to study the motion of an object, including its velocity and acceleration, along a space curve. In particular, we follow in the footsteps of Newton by using these methods to derive Kepler’s First Law of planetary motion. Suppose a particle moves through space so that its position vector at time t is rt. Notice from Figure 1 that, for small values of h, the vector rt h rt h
P 1
r(t) r(t+h) C O x
y
approximates the direction of the particle moving along the curve rt. Its magnitude measures the size of the displacement vector per unit time. The vector (1) gives the average velocity over a time interval of length h and its limit is the velocity vector vt at time t :
FIGURE 1
vt lim
2
hl0
rt h rt rt h
Thus the velocity vector is also the tangent vector and points in the direction of the tangent line. The speed of the particle at time t is the magnitude of the velocity vector, that is, vt . This is appropriate because, from (2) and from Equation 10.8.7, we have
vt rt
ds rate of change of distance with respect to time dt
As in the case of one-dimensional motion, the acceleration of the particle is defined as the derivative of the velocity: at vt rt
SECTION 10.9
MOTION IN SPACE: VELOCITY AND ACCELERATION
■
579
EXAMPLE 1 The position vector of an object moving in a plane is given by
rt t 3 i t 2 j. Find its velocity, speed, and acceleration when t 1 and illustrate geometrically. SOLUTION The velocity and acceleration at time t are
vt rt 3t 2 i 2t j
y
at rt 6t i 2 j
v(1)
and the speed is
a(1) (1, 1)
vt s3t
x
0
2t2 s9t 4 4t 2
2 2
When t 1, we have
FIGURE 2
Visual 10.9 shows animated velocity and acceleration vectors for objects moving along various curves. ■ Figure 3 shows the path of the particle in Example 2 with the velocity and acceleration vectors when t 1.
v1 3 i 2 j
a1 6 i 2 j
These velocity and acceleration vectors are shown in Figure 2.
a(1)
EXAMPLE 2 Find the velocity, acceleration, and speed of a particle with position
vector rt t 2, e t, te t . vt rt 2t, e t, 1 te t
v(1)
at vt 2, e t, 2 te t
vt s4t 1 y
FIGURE 3
■
SOLUTION
z
x
v1 s13
2
e 2t 1 t2 e 2t
■
The vector integrals that were introduced in Section 10.7 can be used to find position vectors when velocity or acceleration vectors are known, as in the next example. V EXAMPLE 3 A moving particle starts at an initial position r0 1, 0, 0 with initial velocity v0 i j k. Its acceleration is at 4t i 6t j k. Find its velocity and position at time t .
SOLUTION Since at vt, we have
vt y at dt y 4t i 6t j k dt 2t 2 i 3t 2 j t k C To determine the value of the constant vector C, we use the fact that v0 i j k. The preceding equation gives v0 C, so C i j k and vt 2t 2 i 3t 2 j t k i j k 2t 2 1 i 3t 2 1 j t 1 k
580
■
CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
■ The expression for rt that we obtained in Example 3 was used to plot the path of the particle in Figure 4 for 0 t 3.
Since vt rt, we have rt y vt dt y 2t 2 1 i 3t 2 1 j t 1 k dt
6
( 23 t 3 t) i t 3 t j ( 12 t 2 t) k D
z 4 2
(1, 0, 0)
0 0
5
10 y
0 15
20
20
Putting t 0, we find that D r0 i, so
x
rt
FIGURE 4
( 23 t 3 t 1) i t 3 t j ( 12 t 2 t) k
■
In general, vector integrals allow us to recover velocity when acceleration is known and position when velocity is known: vt vt0 y au du t
t0
rt rt0 y vu du t
t0
If the force that acts on a particle is known, then the acceleration can be found from Newton’s Second Law of Motion. The vector version of this law states that if, at any time t, a force Ft acts on an object of mass m producing an acceleration at, then Ft mat ■ The angular speed of the object moving with position P is ddt, where is the angle shown in Figure 5.
y
EXAMPLE 4 An object with mass m that moves in a circular path with constant
angular speed has position vector rt a cos t i a sin t j. Find the force acting on the object and show that it is directed toward the origin. vt rt a sin t i a cos t j
SOLUTION
P
at vt a 2 cos t i a 2 sin t j ¨ 0
Therefore, Newton’s Second Law gives the force as
x
Ft mat m 2a cos t i a sin t j Notice that Ft m 2 rt. This shows that the force acts in the direction opposite to the radius vector rt and therefore points toward the origin (see Figure 5). Such a force is called a centripetal (center-seeking) force. ■
FIGURE 5
V EXAMPLE 5 A projectile is fired with angle of elevation and initial velocity v0. (See Figure 6.) Assuming that air resistance is negligible and the only external force is due to gravity, find the position function rt of the projectile. What value of maximizes the range (the horizontal distance traveled)?
y
v¸
a
SOLUTION We set up the axes so that the projectile starts at the origin. Since the
0
x
force due to gravity acts downward, we have
d
F ma mt j FIGURE 6
where t a 9.8 ms2 . Thus a t j
SECTION 10.9
MOTION IN SPACE: VELOCITY AND ACCELERATION
■
581
Since vt a, we have vt tt j C where C v0 v0 . Therefore rt vt tt j v0 Integrating again, we obtain rt 2 tt 2 j t v0 D 1
But D r0 0, so the position vector of the projectile is given by rt 12 tt 2 j t v0
3
If we write v0 v0 (the initial speed of the projectile), then v0 v0 cos i v0 sin j and Equation 3 becomes rt v0 cos t i [v0 sin t 12 tt 2 ] j The parametric equations of the trajectory are therefore ■ If you eliminate t from Equations 4, you will see that y is a quadratic function of x. So the path of the projectile is part of a parabola.
4
x v0 cos t
y v0 sin t 12 tt 2
The horizontal distance d is the value of x when y 0. Setting y 0, we obtain t 0 or t 2v0 sin t. The latter value of t then gives d x v0 cos
2v0 sin v02 2 sin cos v02 sin 2 t t t
Clearly, d has its maximum value when sin 2 1, that is, 4.
■
V EXAMPLE 6 A projectile is fired with muzzle speed 150 ms and angle of elevation 45 from a position 10 m above ground level. Where does the projectile hit the ground, and with what speed?
SOLUTION If we place the origin at ground level, then the initial position of the projectile is (0, 10) and so we need to adjust Equations 4 by adding 10 to the expression for y. With v 0 150 ms, 45 , and t 9.8 ms2, we have
x 150 cos4t 75s2 t y 10 150 sin4t 12 9.8t 2 10 75s2 t 4.9t 2 Impact occurs when y 0, that is, 4.9t 2 75s2t 10 0. Solving this quadratic equation (and using only the positive value of t), we get t
75s2 s11,250 196 21.74 9.8
582
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
Then x 75s2 21.74 2306 , so the projectile hits the ground about 2306 m away. The velocity of the projectile is vt rt 75s2 i (75s2 9.8t) j So its speed at impact is
v21.74 s(75s2 )
2
(75s2 9.8 21.74)2 151 ms
■
TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION
When we study the motion of a particle, it is often useful to resolve the acceleration into two components, one in the direction of the tangent and the other in the direction of the normal. If we write v v for the speed of the particle, then
Tt
rt vt v rt vt v
v vT
and so
If we differentiate both sides of this equation with respect to t, we get a v vT v T
5
If we use the expression for the curvature given by Equation 10.8.9, then we have 6
$
T T v r
T $ v
so
The unit normal vector was defined in the preceding section as N T T , so (6) gives
T T N $ v N and Equation 5 becomes 7
a vT $ v 2 N
aT
Writing a T and a N for the tangential and normal components of acceleration, we have
T
a aT T aN N
a N
aN
FIGURE 7
where 8
a T v
and
aN $v2
This resolution is illustrated in Figure 7. Let’s look at what Formula 7 says. The first thing to notice is that the binormal vector B is absent. No matter how an object moves through space, its acceleration always lies in the plane of T and N (the osculating plane). (Recall that T gives the direction of motion and N points in the direction the curve is turning.) Next we notice that the
SECTION 10.9
MOTION IN SPACE: VELOCITY AND ACCELERATION
■
583
tangential component of acceleration is v, the rate of change of speed, and the normal component of acceleration is $ v 2, the curvature times the square of the speed. This makes sense if we think of a passenger in a car—a sharp turn in a road means a large value of the curvature $, so the component of the acceleration perpendicular to the motion is large and the passenger is thrown against a car door. High speed around the turn has the same effect; in fact, if you double your speed, aN is increased by a factor of 4. Although we have expressions for the tangential and normal components of acceleration in Equations 8, it’s desirable to have expressions that depend only on r, r, and r. To this end we take the dot product of v v T with a as given by Equation 7: v a v T vT $ v 2 N vvT T $ v 3 T N vv
(since T T 1 and T N 0)
Therefore 9
a T v
va v
rt rt rt
Using the formula for curvature given by Theorem 10.8.10, we have
10
aN $v2
rt rt rt rt
2
3
rt rt rt
EXAMPLE 7 A particle moves with position function rt t 2, t 2, t 3 . Find the
tangential and normal components of acceleration. rt t 2 i t 2 j t 3 k
SOLUTION
rt 2t i 2t j 3t 2 k rt 2 i 2 j 6t k
rt s8t
2
9t 4
Therefore, Equation 9 gives the tangential component as aT
Since
rt rt 8t 18t 3 rt s8t 2 9t 4
i rt rt 2t 2
j k 2t 3t 2 6t 2 i 6t 2 j 2 6t
Equation 10 gives the normal component as
rt rt 6s2 t s8t 9t rt 2
aN
2
4
■
584
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
KEPLER’S LAWS OF PLANETARY MOTION
We now describe one of the great accomplishments of calculus by showing how the material of this chapter can be used to prove Kepler’s laws of planetary motion. After 20 years of studying the astronomical observations of the Danish astronomer Tycho Brahe, the German mathematician and astronomer Johannes Kepler (1571–1630) formulated the following three laws. KEPLER’S LAWS 1. A planet revolves around the Sun in an elliptical orbit with the Sun at one
focus. 2. The line joining the Sun to a planet sweeps out equal areas in equal times. 3. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit. In his book Principia Mathematica of 1687, Sir Isaac Newton was able to show that these three laws are consequences of two of his own laws, the Second Law of Motion and the Law of Universal Gravitation. In what follows we prove Kepler’s First Law. The remaining laws are proved as a Web Project (with hints). Since the gravitational force of the Sun on a planet is so much larger than the forces exerted by other celestial bodies, we can safely ignore all bodies in the universe except the Sun and one planet revolving about it. We use a coordinate system with the Sun at the origin and we let r rt be the position vector of the planet. (Equally well, r could be the position vector of the Moon or a satellite moving around the Earth or a comet moving around a star.) The velocity vector is v r and the acceleration vector is a r. We use the following laws of Newton: Second Law of Motion: F ma Law of Gravitation:
F
GMm GMm r 2 u r3 r
where F is the gravitational force on the planet, m and M are the masses of the planet and the Sun, G is the gravitational constant, r r , and u 1rr is the unit vector in the direction of r. We first show that the planet moves in one plane. By equating the expressions for F in Newton’s two laws, we find that
a
GM r r3
and so a is parallel to r. It follows that r a 0. We use Formula 5 in Theorem 10.7.5 to write d r v r v r v dt v vr a000 Therefore
r vh
SECTION 10.9
MOTION IN SPACE: VELOCITY AND ACCELERATION
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585
where h is a constant vector. (We may assume that h 0 ; that is, r and v are not parallel.) This means that the vector r rt is perpendicular to h for all values of t, so the planet always lies in the plane through the origin perpendicular to h. Thus the orbit of the planet is a plane curve. To prove Kepler’s First Law we rewrite the vector h as follows: h r v r r r u r u r u r u ru r 2 u u rru u r 2 u u Then a h
GM u r 2 u u GM u u u r2
GM u uu u uu
(by Theorem 10.4.8, Property 6)
But u u u 2 1 and, since ut 1, it follows from Example 12 in Section 10.7 that u u 0. Therefore a h GM u v h v h a h GM u
and so z
Integrating both sides of this equation, we get
c
¨
y
r x
FIGURE 8
v h GM u c
11
h
v u
where c is a constant vector. At this point it is convenient to choose the coordinate axes so that the standard basis vector k points in the direction of the vector h. Then the planet moves in the xy-plane. Since both v h and u are perpendicular to h, Equation 11 shows that c lies in the xy-plane. This means that we can choose the x- and y-axes so that the vector i lies in the direction of c, as shown in Figure 8. If is the angle between c and r, then r, are polar coordinates of the planet. From Equation 11 we have r v h r GM u c GM r u r c
c cos GMr rc cos
GMr u u r
where c c . Then r
r v h 1 r v h GM c cos GM 1 e cos
where e cGM. But
r v h r v h h h h
2
h2
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
where h h . So r
h 2GM eh 2c 1 e cos 1 e cos
Writing d h 2c, we obtain the equation r
12
Comparing with Theorem 9.5.8, we see that Equation 12 is the polar equation of a conic section with focus at the origin and eccentricity e. We know that the orbit of a planet is a closed curve and so the conic must be an ellipse. This completes the derivation of Kepler’s First Law. A Web Project will guide you through the derivation of the Second and Third Laws. The proofs of these three laws show that the methods of this chapter provide a powerful tool for describing some of the laws of nature.
www.stewartcalculus.com Click on Projects and select Applied Project : Kepler’s Laws. ■
10.9
EXERCISES
■ Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of t.
1–6
1. rt 2 t 2, t ,
2. rt 2 t, 4st , 4. rt e t i e 2t j ,
t1
14. at t i e t j e t k,
t 3
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■ Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.
11–12
11. at i 2 j,
v0 k, r0 i
12. at 2 i 6t j 12t k, 2
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17. A force with magnitude 20 N acts directly upward from
19. A projectile is fired with an initial speed of 500 ms and
10. rt t sin t i t cos t j t 2 k ■
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velocity and acceleration vectors are orthogonal.
9. rt s2 t i e t j et k
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v0 k, r0 j k
18. Show that if a particle moves with constant speed, then the
8. rt 2 cos t, 3t, 2 sin t
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the xy-plane on an object with mass 4 kg. The object starts at the origin with initial velocity v0 i j. Find its position function and its speed at time t.
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position function rt t 3 i t 2 j t 3 k?
7. rt t 2 1, t 3, t 2 1
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16. What force is required so that a particle of mass m has the
t0
Find the velocity, acceleration, and speed of a particle with the given position function. 7–10
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r0 j
rt t 2, 5t, t 2 16t . When is the speed a minimum?
t1
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v0 i,
15. The position function of a particle is given by
6. rt t i 2 cos t j sin t k , ■
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t0
5. rt t i t 2 j 2 k ,
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(a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. ; (b) Use a computer to graph the path of the particle. 13. at 2t i sin t j cos 2t k,
3. rt 3 cos t i 2 sin t j ,
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13–14
t2
1
ed 1 e cos
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20. Rework Exercise 19 if the projectile is fired from a position
200 m above the ground. 21. A ball is thrown at an angle of 45 to the ground. If the ball
lands 90 m away, what was the initial speed of the ball?
v0 i, r0 j k ■
angle of elevation 30 . Find (a) the range of the projectile, (b) the maximum height reached, and (c) the speed at impact.
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22. A gun is fired with angle of elevation 30 . What is the ■
muzzle speed if the maximum height of the shell is 500 m?
CHAPTER 10
tion that can be used to hit a target 800 m away.
25. A medieval city has the shape of a square and is protected
by walls with length 500 m and height 15 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 ms ). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall.) 26. A ball with mass 0.8 kg is thrown southward into the air
with a speed of 30 ms at an angle of 30 to the ground. A west wind applies a steady force of 4 N to the ball in an easterly direction. Where does the ball land and with what speed?
; 27. Water traveling along a straight portion of a river normally
flows fastest in the middle, and the speed slows to almost zero at the banks. Consider a long straight stretch of river flowing north, with parallel banks 40 m apart. If the maximum water speed is 3 ms, we can use a quadratic function as a basic model for the rate of water flow x units from the 3 west bank: f x 400 x40 x. (a) A boat proceeds at a constant speed of 5 ms from a point A on the west bank while maintaining a heading perpendicular to the bank. How far down the river on the opposite bank will the boat touch shore? Graph the path of the boat. (b) Suppose we would like to pilot the boat to land at the point B on the east bank directly opposite A. If we maintain a constant speed of 5 ms and a constant heading, find the angle at which the boat should head. Then
REVIEW
■ Find the tangential and normal components of the acceleration vector.
28. rt 1 t i t 2 2t j 29. rt cos t i sin t j t k 30. rt t i t 2 j 3t k 31. rt e t i s2 t j et k ■
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2. How do you add two vectors geometrically? How do you
add them algebraically? 3. If a is a vector and c is a scalar, how is ca related to a
geometrically? How do you find ca algebraically?
vector rt, then its angular momentum is defined as Lt mrt vt and its torque as t mrt at. Show that Lt t. Deduce that if t 0 for all t, then Lt is constant. (This is the law of conservation of angular momentum.) 33. The position function of a spaceship is
rt 3 t i 2 ln t j 7
4 t 1 2
k
and the coordinates of a space station are 6, 4, 9. The captain wants the spaceship to coast into the space station. When should the engines be turned off? 34. A rocket burning its onboard fuel while moving through
space has velocity vt and mass mt at time t. If the exhaust gases escape with velocity ve relative to the rocket, it can be deduced from Newton’s Second Law of Motion that m
dv dm ve dt dt
m0 ve . mt (b) For the rocket to accelerate in a straight line from rest to twice the speed of its own exhaust gases, what fraction of its initial mass would the rocket have to burn as fuel? (a) Show that vt v0 ln
7. Write expressions for the scalar and vector projections of b
onto a. Illustrate with diagrams. 8. How do you find the cross product a b of two vectors if
you know their lengths and the angle between them? What if you know their components? 9. How are cross products useful?
4. How do you find the vector from one point to another? 5. How do you find the dot product a b of two vectors if you
know their lengths and the angle between them? What if you know their components?
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32. If a particle with mass m moves with position
CONCEPT CHECK
1. What is the difference between a vector and a scalar?
6. How are dot products useful?
587
28 –31
center field fence, which is 10 ft high and 400 ft from home plate. The ball leaves the bat with speed 115 fts at an angle 50 above the horizontal. Is it a home run? (In other words, does the ball clear the fence?)
10
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graph the actual path the boat follows. Does the path seem realistic?
23. A gun has muzzle speed 150 ms. Find two angles of eleva24. A batter hits a baseball 3 ft above the ground toward the
REVIEW
10. (a) How do you find the area of the parallelogram deter-
mined by a and b? (b) How do you find the volume of the parallelepiped determined by a, b, and c? 11. How do you find a vector perpendicular to a plane?
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
12. How do you find the angle between two intersecting planes? 13. Write a vector equation, parametric equations, and sym-
metric equations for a line. 14. Write a vector equation and a scalar equation for a plane. 15. (a) How do you tell if two vectors are parallel?
(b) How do you tell if two vectors are perpendicular? (c) How do you tell if two planes are parallel? 16. (a) Describe a method for determining whether three points
P, Q, and R lie on the same line. (b) Describe a method for determining whether four points P, Q, R, and S lie in the same plane. 17. (a) How do you find the distance from a point to a line?
(b) How do you find the distance from a point to a plane? 18. What are the traces of a surface? How do you find them? 19. Write equations in standard form of the six types of quadric
surfaces.
(b) How do you find the tangent vector to a smooth curve at a point? How do you find the tangent line? The unit tangent vector? 23. If u and v are differentiable vector functions, c is a scalar,
and f is a real-valued function, write the rules for differentiating the following vector functions. (a) ut vt (b) cut (c) f tut (d) ut vt (e) ut vt (f ) u f t 24. How do you find the length of a space curve given by a
vector function rt? 25. (a) What is the definition of curvature?
(b) Write a formula for curvature in terms of rt and Tt. (c) Write a formula for curvature in terms of rt and rt. (d) Write a formula for the curvature of a plane curve with equation y f x. 26. Write formulas for the unit normal and binormal vectors of
a smooth space curve rt.
20. What is a vector function? How do you find its derivative
and its integral? 21. What is the connection between vector functions and space
curves? 22. (a) What is a smooth curve?
27. (a) How do you find the velocity, speed, and acceleration of
a particle that moves along a space curve? (b) Write the acceleration in terms of its tangential and normal components. 28. State Kepler’s Laws.
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
13. The set of points {x, y, z
x
2
y 2 1} is a circle.
14. If u u1, u2 and v v1, v2 , then u v u1v1, u2 v2 .
1. For any vectors u and v in V3 , u v v u.
15. If u v 0 , then u 0 or v 0.
2. For any vectors u and v in V3 , u v v u.
16. If u and v are in V3 , then u v u
3. For any vectors u and v in V3 , u v v u . 4. For any vectors u and v in V3 and any scalar k,
ku v k u v. 5. For any vectors u and v in V3 and any scalar k,
ku v k u v. 6. For any vectors u, v, and w in V3,
u v w u w v w. 7. For any vectors u, v, and w in V3,
u v w u v w. 8. For any vectors u, v, and w in V3 ,
u v w u v w. 9. For any vectors u and v in V3 , u v u 0.
v .
17. The curve with vector equation rt t 3 i 2t 3 j 3t 3 k is
a line. 18. The curve with vector equation rt t, t 3, t 5 is smooth. 19. The curve with vector equation rt cos t, t 2, t 4 is
smooth. 20. The derivative of a vector function is obtained by differen-
tiating each component function. 21. If ut and vt are differentiable vector functions, then
d ut vt ut vt dt 22. If rt is a differentiable vector function, then
10. For any vectors u and v in V3 , u v v u v.
d rt rt dt
11. The cross product of two unit vectors is a unit vector.
23. If Tt is the unit tangent vector of a smooth curve, then the
12. A linear equation Ax By Cz D 0 represents a
line in space.
curvature is $ dTdt . 24. The binormal vector is Bt Nt Tt.
CHAPTER 10
REVIEW
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589
EXERCISES 1. (a) Find an equation of the sphere that passes through the
point 6, 2, 3 and has center 1, 2, 1. (b) Find the curve in which this sphere intersects the yz-plane. (c) Find the center and radius of the sphere
11. (a) Find a vector perpendicular to the plane through the
points A1, 0, 0, B2, 0, 1, and C1, 4, 3. (b) Find the area of triangle ABC. 12. A constant force F 3 i 5 j 10 k moves an object
along the line segment from 1, 0, 2 to 5, 3, 8. Find the work done if the distance is measured in meters and the force in newtons.
x 2 y 2 z 2 8x 2y 6z 1 0 2. Copy the vectors in the figure and use them to draw each of
the following vectors. (a) a b (b) a b
13. A boat is pulled onto shore using two ropes, as shown in the
(c) 12 a
(d) 2 a b
diagram. If a force of 255 N is needed, find the magnitude of the force in each rope.
a b
20° 255 N 30°
3. If u and v are the vectors shown in the figure, find u v and
u v . Is u v directed into the page or out of it?
14. Find the magnitude of the torque about P if a 50-N force is
applied as shown. | v|=3 50 N 30°
45°
| u|=2 40 cm
4. Calculate the given quantity if
a i j 2k (a) (c) (e) (g) (i) (k)
b 3i 2j k
c j 5k
2a 3b (b) b ab (d) a b b c ( f ) a b c (h) a b c c c ( j) proj a b comp a b The angle between a and b (correct to the nearest degree)
5. Find the values of x such that the vectors 3, 2, x and
2x, 4, x are orthogonal.
6. Find two unit vectors that are orthogonal to both j 2 k
and i 2 j 3 k.
15–17
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Find parametric equations for the line.
15. The line through 4, 1, 2 and 1, 1, 5 16. The line through 1, 0, 1 and parallel to the line 1 3
x 4 12 y z 2
17. The line through 2, 2, 4 and perpendicular to the
plane 2x y 5z 12
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Find an equation of the plane.
18. The plane through 2, 1, 0 and parallel to x 4y 3z 1
7. Suppose that u v w 2. Find
(a) u v w (c) v u w
P
(b) u w v (d) u v v
8. Show that if a, b, and c are in V3 , then
a b b c c a a b c 2 9. Find the acute angle between two diagonals of a cube. 10. Given the points A1, 0, 1, B2, 3, 0, C1, 1, 4, and
D0, 3, 2, find the volume of the parallelepiped with adjacent edges AB, AC, and AD.
19. The plane through 3, 1, 1, 4, 0, 2, and 6, 3, 1 20. The plane through 1, 2, 2 that contains the line x 2t,
y 3 t, z 1 3t
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21. Find the point in which the line with parametric equations
x 2 t, y 1 3t, z 4t intersects the plane 2 x y z 2. 22. Find the distance from the origin to the line x 1 t,
y 2 t, z 1 2t.
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CHAPTER 10
VECTORS AND THE GEOMETRY OF SPACE
41. If rt t 2 i t cos t j sin t k, evaluate x01 rt dt.
23. Determine whether the lines given by the symmetric
equations
42. Let C be the curve with equations x 2 t 3, y 2t 1,
x1 y2 z3 2 3 4
z ln t. Find (a) the point where C intersects the xz-plane, (b) parametric equations of the tangent line at 1, 1, 0, and (c) an equation of the normal plane to C at 1, 1, 0.
y3 z5 x1 6 1 2
and
43. Use Simpson’s Rule with n 6 to estimate the length of
are parallel, skew, or intersecting. 24. (a) Show that the planes x y z 1 and
2x 3y 4z 5 are neither parallel nor perpendicular. (b) Find, correct to the nearest degree, the angle between these planes.
Identify and sketch the graph of each surface.
26. x 3
27. x z
28. y z 2
29. x 2 y 2 4z 2
46. Reparametrize the curve rt e t i e t sin t j e t cos t k
with respect to arc length measured from the point 1, 0, 1 in the direction of increasing t .
30. 4x y 2z 4
47. For the curve given by rt
32. y 2 z 2 1 x 2
48. Find the curvature of the ellipse x 3 cos t, y 4 sin t at
33. 4x 2 4y 2 8y z 2 0
the points 3, 0 and 0, 4.
34. x y 2 z 2 2y 4z 5 ■
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35. An ellipsoid is created by rotating the ellipse
4x y 16 about the x-axis. Find an equation of the ellipsoid. 2
13 t 3, 12 t 2, t , find (a) the unit
tangent vector, (b) the unit normal vector, and (c) the curvature.
31. 4x 2 y 2 4z 2 4
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0 t 1.
r2t 1 t i t 2 j t 3 k at the point 1, 0, 0. Find the angle of intersection of these curves.
and 3x y 4z 24. ■
44. Find the length of the curve rt 2t 32, cos 2t, sin 2t , 45. The helix r1t cos t i sin t j t k intersects the curve
25. Find the distance between the planes 3x y 4z 2 26 –34
the arc of the curve with equations x t 2, y t 3, z t 4, 0 t 3.
2
36. A surface consists of all points P such that the distance
from P to the plane y 1 is twice the distance from P to the point 0, 1, 0. Find an equation for this surface and identify it.
; 50. Find an equation of the osculating circle of the curve
y x 4 x 2 at the origin. Graph both the curve and its osculating circle.
51. A particle moves with position function
rt t ln t i t j et k. Find the velocity, speed, and acceleration of the particle. 52. A particle starts at the origin with initial velocity
i j 3 k. Its acceleration is at 6t i 12t 2 j 6t k. Find its position function.
37. (a) Sketch the curve with vector function
rt t i cos t j sin t k
49. Find the curvature of the curve y x 4 at the point 1, 1.
t0
(b) Find rt and rt. 38. Let rt s2 t , e t 1t, lnt 1 .
(a) Find the domain of r. (b) Find lim t l 0 rt. (c) Find rt. 39. Find a vector function that represents the curve of intersec-
53. An athlete throws a shot at an angle of 45 to the horizontal
at an initial speed of 43 fts. It leaves his hand 7 ft above the ground. (a) Where is the shot 2 seconds later? (b) How high does the shot go? (c) Where does the shot land? 54. Find the tangential and normal components of the accelera-
tion vector of a particle with position function
tion of the cylinder x 2 y 2 16 and the plane x z 5.
; 40. Find parametric equations for the tangent line to the curve
x 2 sin t, y 2 sin 2t , z 2 sin 3t at the point (1, s3, 2). Graph the curve and the tangent line on a common screen.
rt t i 2t j t 2 k 55. Find the curvature of the curve with parametric equations
x y sin ( 12 2) d t
0
y
y
t
0
cos ( 12 2) d
11
11.1
PARTIAL DERIVATIVES So far we have dealt with the calculus of functions of a single variable. But, in the real world, physical quantities often depend on two or more variables, so in this chapter we turn our attention to functions of several variables and extend the basic ideas of differential calculus to such functions.
FUNCTIONS OF SEVERAL VARIABLES The temperature T at a point on the surface of the Earth at any given time depends on the longitude x and latitude y of the point. We can think of T as being a function of the two variables x and y, or as a function of the pair x, y. We indicate this functional dependence by writing T f x, y. The volume V of a circular cylinder depends on its radius r and its height h. In fact, we know that V r 2h. We say that V is a function of r and h, and we write Vr, h r 2h. DEFINITION A function f of two variables is a rule that assigns to each
ordered pair of real numbers x, y in a set D a unique real number denoted by f x, y. The set D is the domain of f and its range is the set of values that f takes on, that is, f x, y x, y D .
We often write z f x, y to make explicit the value taken on by f at the general point x, y. The variables x and y are independent variables and z is the dependent variable. [Compare this with the notation y f x for functions of a single variable.] A function of two variables is just a function whose domain is a subset of ⺢2 and whose range is a subset of ⺢. One way of visualizing such a function is by means of an arrow diagram (see Figure 1), where the domain D is represented as a subset of the xy-plane. y (x, y)
f 0
x
f (a, b) 0
f (x, y)
z
D (a, b)
FIGURE 1
If a function f is given by a formula and no domain is specified, then the domain of f is understood to be the set of all pairs x, y for which the given expression is a well-defined real number. 591
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CHAPTER 11
PARTIAL DERIVATIVES
EXAMPLE 1 Find the domains of the following functions and evaluate f 3, 2.
x+y+1=0 y
(a) f x, y
x=1
sx y 1 x1
(b) f x, y x ln y 2 x
SOLUTION _1
0
x
f 3, 2
(a)
_1
The expression for f makes sense if the denominator is not 0 and the quantity under the square root sign is nonnegative. So the domain of f is
FIGURE 2
Domain of f(x, y)=
œ„„„„„„„ x+y+1 x-1
D x, y
y
x y 1 0,
x 1
The inequality x y 1 0, or y x 1, describes the points that lie on or above the line y x 1, while x 1 means that the points on the line x 1 must be excluded from the domain. (See Figure 2.)
x=¥ 0
s3 2 1 s6 31 2
x
f 3, 2 3 ln2 2 3 3 ln 1 0
(b)
Since ln y 2 x is defined only when y 2 x 0, that is, x y 2, the domain of f is D x, y x y 2 . This is the set of points to the left of the parabola x y 2. (See Figure 3.) ■
FIGURE 3
Domain of f(x, y)=x ln(¥-x)
EXAMPLE 2 Find the domain and range of tx, y s9 x 2 y 2 . SOLUTION The domain of t is
D x, y y
9x
2
y 2 0 x, y
x
2
y 2 9
which is the disk with center 0, 0 and radius 3 (see Figure 4). The range of t is
≈+¥=9
z z s9 x 2 y 2 , x, y D _3
3
x
Since z is a positive square root, z 0. Also 9 x2 y2 9
?
s9 x 2 y 2 3
So the range is FIGURE 4
z
9-≈-¥ Domain of g(x, y)=œ„„„„„„„„„
0 z 3 0, 3
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GRAPHS
Another way of visualizing the behavior of a function of two variables is to consider its graph. DEFINITION If f is a function of two variables with domain D, then the graph of f is the set of all points x, y, z in ⺢3 such that z f x, y and x, y is in D.
SECTION 11.1
z { x, y, f(x, y)}
S
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593
Just as the graph of a function f of one variable is a curve C with equation y f x, so the graph of a function f of two variables is a surface S with equation z f x, y. We can visualize the graph S of f as lying directly above or below its domain D in the xy-plane (see Figure 5). EXAMPLE 3 Sketch the graph of the function f x, y 6 3x 2y.
f(x, y) 0
D
y
(x, y, 0)
x
FUNCTIONS OF SEVERAL VARIABLES
FIGURE 5 z
SOLUTION The graph of f has the equation z 6 3x 2y, or 3x 2y z 6, which represents a plane. To graph the plane we first find the intercepts. Putting y z 0 in the equation, we get x 2 as the x-intercept. Similarly, the y-intercept is 3 and the z-intercept is 6. This helps us sketch the portion of the graph that lies in the first octant (Figure 6). ■
The function in Example 3 is a special case of the function
(0, 0, 6)
f x, y ax by c
(0, 3, 0) (2, 0, 0)
y
x
which is called a linear function. The graph of such a function has the equation z ax by c, or ax by z c 0, so it is a plane. In much the same way that linear functions of one variable are important in single-variable calculus, we will see that linear functions of two variables play a central role in multivariable calculus. V EXAMPLE 4
FIGURE 6
Sketch the graph of tx, y s9 x 2 y 2 .
SOLUTION The graph has equation z s9 x 2 y 2 . We square both sides of this
equation to obtain z 2 9 x 2 y 2, or x 2 y 2 z 2 9, which we recognize as an equation of the sphere with center the origin and radius 3. But, since z 0, the graph of t is just the top half of this sphere (see Figure 7). z (0, 0, 3)
0
FIGURE 7
Graph of g(x, y)=œ„„„„„„„„„ 9-≈-¥
(3, 0, 0)
(0, 3, 0) y
x
■
V EXAMPLE 5 Find the domain and range and sketch the graph of hx, y 4x 2 y 2.
z
SOLUTION Notice that hx, y is defined for all possible ordered pairs of real numbers x, y, so the domain is ⺢2, the entire xy-plane. The range of h is the set 0, of all nonnegative real numbers. [Notice that x 2 0 and y 2 0, so hx, y 0 for all x and y.] The graph of h has the equation z 4x 2 y 2, which is the elliptic paraboloid that we sketched in Example 4 in Section 10.6. Horizontal traces are ellipses and vertical traces are parabolas (see Figure 8). ■ x
y
FIGURE 8
Graph of h(x, y)=4≈+¥
Computer programs are readily available for graphing functions of two variables. In most such programs, traces in the vertical planes x k and y k are drawn for equally spaced values of k and parts of the graph are eliminated using hidden line removal.
594
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CHAPTER 11
PARTIAL DERIVATIVES
Figure 9 shows computer-generated graphs of several functions. Notice that we get an especially good picture of a function when rotation is used to give views from different vantage points. In parts (a) and (b) the graph of f is very flat and close to the 2 2 xy-plane except near the origin; this is because ex y is very small when x or y is large. z
z
x y
x
(b) f(x, y)=(≈+3¥)e _≈_¥
(a) f(x, y)=(≈+3¥)e _≈_¥ z
z
x
y
x
(c) f(x, y)=sin x+sin y
y
(d) f(x, y)=
sin x sin y xy
FIGURE 9
LEVEL CURVES
So far we have two methods for visualizing functions: arrow diagrams and graphs. A third method, borrowed from mapmakers, is a contour map on which points of constant elevation are joined to form contour curves, or level curves. DEFINITION The level curves of a function f of two variables are the curves with equations f x, y k, where k is a constant (in the range of f ).
A level curve f x, y k is the set of all points in the domain of f at which f takes on a given value k. In other words, it shows where the graph of f has height k. You can see from Figure 10 the relation between level curves and horizontal traces. The level curves f x, y k are just the traces of the graph of f in the horizontal plane z k projected down to the xy-plane. So if you draw the level curves of a function
SECTION 11.1
FUNCTIONS OF SEVERAL VARIABLES
595
■
z 40 00
45
45 00 50
00
LONESOME MTN.
0
A 55 00
B y 50
x
00
k=45 450
FIGURE 10
0
f(x, y)=20
k=40 k=35 k=30 k=25 k=20
Lon
eso
me
Cree
k
FIGURE 11
Visual 11.1A animates Figure 10 by showing level curves being lifted up to graphs of functions.
FIGURE 12
World mean sea-level temperatures in January in degrees Celsius Tarbuck, Atmosphere: Introduction to Meteorology, 4th Edition, © 1989. Reprinted by permission of Pearson Education, Inc., Upper Saddle River, NJ.
and visualize them being lifted up to the surface at the indicated height, then you can mentally piece together a picture of the graph. The surface is steep where the level curves are close together. It is somewhat flatter where they are farther apart. One common example of level curves occurs in topographic maps of mountainous regions, such as the map in Figure 11. The level curves are curves of constant elevation above sea level. If you walk along one of these contour lines you neither ascend nor descend. Another common example is the temperature function introduced in the opening paragraph of this section. Here the level curves are called isothermals and join locations with the same temperature. Figure 12 shows a weather map of the world indicating the average January temperatures. The isothermals are the curves that separate the shaded bands.
596
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CHAPTER 11
y
PARTIAL DERIVATIVES
EXAMPLE 6 A contour map for a function f is shown in Figure 13. Use it to esti-
50
5
mate the values of f 1, 3 and f 4, 5. SOLUTION The point (1, 3) lies partway between the level curves with z-values 70 and 80. We estimate that
4 3 2
80 70 60
1 0
1
2
50
3
f 1, 3 73 80 70 60 4
Similarly, we estimate that f 4, 5 56 x
5
■
EXAMPLE 7 Sketch the level curves of the function f x, y 6 3x 2y for the
values k 6, 0, 6, 12.
FIGURE 13
SOLUTION The level curves are y
6 3x 2y k
0
x
3x 2y k 6 0
or
_6
k=
0
k=
6
k=
12
k=
This is a family of lines with slope 32 . The four particular level curves with k 6, 0, 6, and 12 are 3x 2y 12 0, 3x 2y 6 0, 3x 2y 0, and 3x 2y 6 0. They are sketched in Figure 14. The level curves are equally spaced parallel lines because the graph of f is a plane (see Figure 6). ■ V EXAMPLE 8
FIGURE 14
Contour map of f(x, y)=6-3x-2y
Sketch the level curves of the function tx, y s9 x 2 y 2
k 0, 1, 2, 3
for
SOLUTION The level curves are
s9 x 2 y 2 k
or
x2 y2 9 k2
This is a family of concentric circles with center 0, 0 and radius s9 k 2 . The cases k 0, 1, 2, 3 are shown in Figure 15. Try to visualize these level curves lifted up to form a surface and compare with the graph of t (a hemisphere) in Figure 7. (See TEC Visual 11.1A.) y
k=3 k=2 k=1 k=0
0
3
x
FIGURE 15
■
Contour map of g(x, y)=œ„„„„„„„„„ 9-≈-¥
EXAMPLE 9 Sketch some level curves of the function hx, y 4x 2 y 2. SOLUTION The level curves are
4x 2 y 2 k
or
y2 x2 1 k4 k
which, for k 0, describes a family of ellipses with semiaxes sk 2 and sk . Figure 16(a) shows a contour map of h drawn by a computer with level curves corre-
SECTION 11.1
FUNCTIONS OF SEVERAL VARIABLES
■
597
sponding to k 0.25, 0.5, 0.75, . . . , 4. Figure 16(b) shows these level curves lifted up to the graph of h (an elliptic paraboloid) where they become horizontal traces. We see from Figure 16 how the graph of h is put together from the level curves. y z
x
Visual 11.1B demonstrates the connection between surfaces and their contour maps.
x
FIGURE 16
y
The graph of h(x, y)=4≈+¥ is formed by lifting the level curves.
(a) Contour map
(b) Horizontal traces are raised level curves
■
Figure 17 shows some computer-generated level curves together with the corresponding computer-generated graphs. Notice that the level curves in part (c) crowd together near the origin. That corresponds to the fact that the graph in part (d) is very steep near the origin. z
y
z
x x
y
(a) Level curves of f(x, y)=_xye_≈_¥
(b) Two views of f(x, y)=_xye_≈_¥
z
y
x
y x
FIGURE 17
(c) Level curves of f(x, y)=
_3y ≈+¥+1
(d) f(x, y)=
_3y ≈+¥+1
598
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CHAPTER 11
PARTIAL DERIVATIVES
FUNCTIONS OF THREE OR MORE VARIABLES
A function of three variables, f , is a rule that assigns to each ordered triple x, y, z in a domain D ⺢ 3 a unique real number denoted by f x, y, z. For instance, the temperature T at a point on the surface of the Earth depends on the longitude x and latitude y of the point and on the time t, so we could write T f x, y, t. EXAMPLE 10 Find the domain of f if f x, y, z lnz y xy sin z. SOLUTION The expression for f x, y, z is defined as long as z y 0, so the domain of f is D x, y, z ⺢ 3 z y
This is a half-space consisting of all points that lie above the plane z y.
■
It’s very difficult to visualize a function f of three variables by its graph, since that would lie in a four-dimensional space. However, we do gain some insight into f by examining its level surfaces, which are the surfaces with equations f x, y, z k, where k is a constant. If the point x, y, z moves along a level surface, the value of f x, y, z remains fixed. z
≈+¥+z@=9
EXAMPLE 11 Find the level surfaces of the function f x, y, z x 2 y 2 z 2.
≈+¥+z@=4 SOLUTION The level surfaces are x 2 y 2 z 2 k, where k 0. These form a
family of concentric spheres with radius sk . (See Figure 18.) Thus, as x, y, z varies over any sphere with center O, the value of f x, y, z remains fixed.
y x
≈+¥+z@=1 FIGURE 18
■
Functions of any number of variables can be considered. A function of n variables is a rule that assigns a number z f x 1, x 2 , . . . , x n to an n-tuple x 1, x 2 , . . . , x n of real numbers. We denote by ⺢ n the set of all such n-tuples. For example, if a company uses n different ingredients in making a food product, ci is the cost per unit of the ith ingredient, and x i units of the ith ingredient are used, then the total cost C of the ingredients is a function of the n variables x 1, x 2 , . . . , x n : 1
C f x 1, x 2 , . . . , x n c1 x 1 c2 x 2 cn x n
The function f is a real-valued function whose domain is a subset of ⺢ n. Sometimes we will use vector notation in order to write such functions more compactly: If x x 1, x 2 , . . . , x n , we often write f x in place of f x 1, x 2 , . . . , x n . With this notation we can rewrite the function defined in Equation 1 as f x c x where c c1, c2 , . . . , cn and c x denotes the dot product of the vectors c and x in Vn . In view of the one-to-one correspondence between points x 1, x 2 , . . . , x n in ⺢ n and their position vectors x x 1, x 2 , . . . , x n in Vn , we have three ways of looking at a function f defined on a subset of ⺢ n : 1. As a function of n real variables x 1, x 2 , . . . , x n 2. As a function of a single point variable x 1, x 2 , . . . , x n 3. As a function of a single vector variable x x 1, x 2 , . . . , x n
We will see that all three points of view are useful.
SECTION 11.1
11.1
599
■
FUNCTIONS OF SEVERAL VARIABLES
EXERCISES
1. Let f x, y x 2e3xy.
22. Two contour maps are shown. One is for a function f
(a) Evaluate f 2, 0. (c) Find the range of f .
whose graph is a cone. The other is for a function t whose graph is a paraboloid. Which is which, and why?
(b) Find the domain of f .
2. Let f x, y lnx y 1.
y
I
(a) Evaluate f 1, 1. (b) Evaluate f e, 1. (c) Find and sketch the domain of f . (d) Find the range of f .
3. Let f x, y, z eszx y . 2
y
II
2
(a) Evaluate f 2, 1, 6. (c) Find the range of f .
x
(b) Find the domain of f .
x
4. Let tx, y, z ln25 x 2 y 2 z 2 .
(a) Evaluate t2, 2, 4. (c) Find the range of t.
5–12
(b) Find the domain of t. 23. Locate the points A and B in the map of Lonesome Moun-
Find and sketch the domain of the function.
■
5. f x, y sx y
tain (Figure 11). How would you describe the terrain near A? Near B?
6. f x, y sxy
7. f x, y ln9 x 9y 2
2
24. Make a rough sketch of a contour map for the function
8. f x, y sy x ln y x 9. f x, y
whose graph is shown.
sy x 1 x2
2
z
10. f x, y sx 2 y 2 1 ln4 x 2 y 2 11. f x, y, z s1 x 2 y 2 z 2 12. f x, y, z ln16 4x 2 4y 2 z 2 ■
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13–20
■ ■
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Sketch the graph of the function.
13. f x, y 6 3x 2y
14. f x, y y
15. f x, y y 1
16. f x, y cos x
17. f x, y 4 x 2 y 2 1
18. f x, y 3 x 2 y 2
2
19. f x, y sx 2 y 2
25–32 ■ Draw a contour map of the function showing several level curves.
20. f x, y s16 x 2 16y 2 ■
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21. A contour map for a function f is shown. Use it to esti-
mate the values of f 3, 3 and f 3, 2. What can you say about the shape of the graph? y
■
25. f x, y y 2x2
26. f x, y x 3 y
27. f x, y y ln x
28. f x, y e yx
29. f x, y ye x
30. f x, y y sec x
31. f x, y sy 2 x 2
32. f x, y yx 2 y 2
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■
0
70 60 50 40 1
30 20 10
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Sketch both a contour map and a graph of the function and compare them.
33–34 1
y
x
x
■
33. f x, y x 2 9y 2 34. f x, y s36 9x 2 4y 2 ■
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600
■
CHAPTER 11
PARTIAL DERIVATIVES
Graphs and Contour Maps for Exercises 41–46
A
B
z
C
z
z
y y
x
y
x
x
z
D
z
E
z
F
x
I
II
y
x
III
y
x
V
y
x
y
x
IV
y
y
x
y
x
VI
y
x
y
x
SECTION 11.2
35. A thin metal plate, located in the xy-plane, has temperature
Tx, y at the point x, y. The level curves of T are called isothermals because at all points on an isothermal the temperature is the same. Sketch some isothermals if the temperature function is given by
■
■
601
■
■
Describe the level surfaces of the function.
47. f x, y, z x 3y 5z 48. f x, y, z x 2 3y 2 5z 2 49. f x, y, z x 2 y 2 z 2
Tx, y 1001 x 2 2y 2
50. f x, y, z x 2 y 2
36. If Vx, y is the electric potential at a point x, y in the
xy-plane, then the level curves of V are called equipotential curves because at all points on such a curve the electric potential is the same. Sketch some equipotential curves if Vx, y csr 2 x 2 y 2 , where c is a positive constant.
; 37– 40
47–50
LIMITS AND CONTINUITY
Use a computer to graph the function using various domains and viewpoints. Get a printout of one that, in your opinion, gives a good view. If your software also produces level curves, then plot some contour lines of the same function and compare with the graph.
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51–52 ■ Describe how the graph of t is obtained from the graph of f . 51. (a) tx, y f x, y 2
(b) tx, y 2 f x, y (d) tx, y 2 f x, y
52. (a) tx, y f x 2, y
(b) tx, y f x, y 2
(c) tx, y f x, y
■
37. f x, y e cos y
■
(c) tx, y f x 3, y 4
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■
■
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x
2
38. f x, y 1 3x 2 y 2 e1x y 39. f x, y xy 2 x 3 40. f x, y xy yx 3
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; 53. Use a computer to investigate the family of functions 2
(monkey saddle) 3
■
; 54. Graph the functions
(dog saddle) ■
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■
Match the function (a) with its graph (labeled A–F on page 600) and (b) with its contour map (labeled I–VI). Give reasons for your choices. 41– 46
■
41. z sinxy
42. z e x cos y
43. z sinx y
44. z sin x sin y
45. z 1 x 2 1 y 2
46. z
■
■
■
■
11.2
2
f x, y e cx y . How does the shape of the graph depend on c?
2
■
■
■
■
■
f x, y e sx y
f x, y lnsx 2 y 2
f x, y sin(sx 2 y 2 )
f x, y
and
2
2
1 sx 2 y 2
In general, if t is a function of one variable, how is the graph of
xy 1 x2 y2 ■
f x, y sx 2 y 2
f x, y t (sx 2 y 2 ) ■
■
obtained from the graph of t?
LIMITS AND CONTINUITY The limit of a function of two or more variables is similar to the limit of a function of a single variable. We use the notation lim
x, y l a, b
f x, y L
to indicate that the values of f x, y approach the number L as the point x, y approaches the point a, b along any path that stays within the domain of f . In other words, we can make the values of f x, y as close to L as we like by taking the point x, y sufficiently close to the point a, b, but not equal to a, b. A more precise definition follows.
602
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CHAPTER 11
PARTIAL DERIVATIVES
1 DEFINITION Let f be a function of two variables whose domain D includes points arbitrarily close to a, b. Then we say that the limit of f x, y as x, y approaches a, b is L and we write
lim
x, y l a, b
f x, y L
if for every number 0 there is a corresponding number 0 such that if x, y D and 0 sx a2 y b2 then
f x, y L
Other notations for the limit in Definition 1 are lim f x, y L
f x, y l L as x, y l a, b
and
xla ylb
Notice that f x, y L is the distance between the numbers f x, y and L, and sx a 2 y b 2 is the distance between the point x, y and the point a, b. Thus Definition 1 says that the distance between f x, y and L can be made arbitrarily small by making the distance from x, y to a, b sufficiently small (but not 0). Figure 1 illustrates Definition 1 by means of an arrow diagram. If any small interval L , L is given around L , then we can find a disk D with center a, b and radius 0 such that f maps all the points in D [except possibly a, b] into the interval L , L . z
y
L+∑ L L-∑
(x, y)
∂
D
f
(a, b) 0
x
(
0
)
L-∑ L L+∑
S
z 0 x
y
b
FIGURE 3
D∂
y
FIGURE 2
FIGURE 1
0
(a, b)
a
x
Another illustration of Definition 1 is given in Figure 2 where the surface S is the graph of f . If 0 is given, we can find 0 such that if x, y is restricted to lie in the disk D and x, y a, b, then the corresponding part of S lies between the horizontal planes z L and z L . For functions of a single variable, when we let x approach a, there are only two possible directions of approach, from the left or from the right. We recall from Chapter 1 that if lim x l a f x lim x l a f x, then lim x l a f x does not exist. For functions of two variables the situation is not as simple because we can let x, y approach a, b from an infinite number of directions in any manner whatsoever (see Figure 3) as long as x, y stays within the domain of f . Definition 1 says that the distance between f x, y and L can be made arbitrarily small by making the distance from x, y to a, b sufficiently small (but not 0). The
SECTION 11.2
LIMITS AND CONTINUITY
■
603
definition refers only to the distance between x, y and a, b. It does not refer to the direction of approach. Therefore, if the limit exists, then f x, y must approach the same limit no matter how x, y approaches a, b. Thus if we can find two different paths of approach along which the function f x, y has different limits, then it follows that lim x, y l a, b f x, y does not exist. If f x, y l L 1 as x, y l a, b along a path C1 and f x, y l L 2 as x, y l a, b along a path C2 , where L 1 L 2 , then lim x, y l a, b f x, y does not exist.
V EXAMPLE 1
Show that
lim
x, y l 0, 0
x2 y2 does not exist. x2 y2
SOLUTION Let f x, y x 2 y 2 x 2 y 2 . First let’s approach 0, 0 along the
x-axis. Then y 0 gives f x, 0 x 2x 2 1 for all x 0, so
y
f x, y l 1
f=_1
f=1
x
as
x, y l 0, 0 along the x-axis
y 2 We now approach along the y-axis by putting x 0. Then f 0, y 2 1 for y all y 0, so f x, y l 1
as
x, y l 0, 0 along the y-axis
(See Figure 4.) Since f has two different limits along two different lines, the given limit does not exist. ■
FIGURE 4
EXAMPLE 2 If f x, y xyx 2 y 2 , does
lim
x, y l 0, 0
f x, y exist?
SOLUTION If y 0, then f x, 0 0x 2 0. Therefore
f x, y l 0
as
x, y l 0, 0 along the x-axis
If x 0, then f 0, y 0y 2 0, so f x, y l 0 y
y=x f=0
1
f= 2 f=0
x
x, y l 0, 0 along the y-axis
Although we have obtained identical limits along the axes, that does not show that the given limit is 0. Let’s now approach 0, 0 along another line, say y x. For all x 0, x2 1 f x, x 2 x x2 2 Therefore
FIGURE 5
as
f x, y l 12
as
x, y l 0, 0 along y x
(See Figure 5.) Since we have obtained different limits along different paths, the given limit does not exist.
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604
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CHAPTER 11
PARTIAL DERIVATIVES
Figure 6 sheds some light on Example 2. The ridge that occurs above the line y x 1 corresponds to the fact that f x, y 2 for all points x, y on that line except the origin. z
In Visual 11.2 a rotating line on the surface in Figure 6 shows different limits at the origin from different directions.
y
x
FIGURE 6
f(x, y)=
xy ≈+¥
V EXAMPLE 3
If f x, y
xy 2 , does lim f x, y exist? x, y l 0, 0 x2 y4
SOLUTION With the solution of Example 2 in mind, let’s try to save time by letting
x, y l 0, 0 along any nonvertical line through the origin. Then y mx, where m is the slope, and f x, y f x, mx ■ Figure 7 shows the graph of the function in Example 3. Notice the ridge above the parabola x y 2.
0.5
f x, y l 0
So
2
FIGURE 7
0 x
2 _2
x, y l 0, 0 along y mx
as
Thus f has the same limiting value along every nonvertical line through the origin. But that does not show that the given limit is 0, for if we now let x, y l 0, 0 along the parabola x y 2, we have
z 0 _0.5
xmx2 m 2x 3 m 2x x 2 mx4 x 2 m 4x 4 1 m 4x 2
y2 y2 y4 1 2 2 4 4 y y 2y 2
f x, y f y 2, y
_2 0 y
f x, y l 12
so
as
x, y l 0, 0 along x y 2
Since different paths lead to different limiting values, the given limit does not exist. ■
Now let’s look at limits that do exist. Just as for functions of one variable, the calculation of limits for functions of two variables can be greatly simplified by the use of properties of limits. The Limit Laws listed in Section 1.4 can be extended to functions of two variables: The limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on. In particular, the following equations are true. 2
lim
x, y l a, b
xa
lim
x, y l a, b
yb
lim
x, y l a, b
cc
The Squeeze Theorem also holds. EXAMPLE 4 Find
lim
x, y l 0, 0
3x 2y if it exists. x y2 2
SOLUTION As in Example 3, we could show that the limit along any line through the origin is 0. This doesn’t prove that the given limit is 0, but the limits along the
SECTION 11.2
LIMITS AND CONTINUITY
■
605
parabolas y x 2 and x y 2 also turn out to be 0, so we begin to suspect that the limit does exist and is equal to 0. Let 0. We want to find 0 such that 0 sx 2 y 2
if
that is,
then
0 sx 2 y 2
if
then
3x 2 y 0 x2 y2
3x 2 y x2 y2
But x 2 x 2 y 2 since y 2 0, so x 2x 2 y 2 1 and therefore
3x 2 y
3 y 3sy 2 3sx 2 y 2 x2 y2
3
Thus if we choose 3 and let 0 sx 2 y 2 , then
Another way to do Example 4 is to use the Squeeze Theorem instead of Definition 1. From (2) it follows that ■
lim
x, y l 0, 0
3 y 0
3x 2 y 0 3sx 2 y 2 3 3 x y2 3 2
Hence, by Definition 1,
and so the first inequality in (3) shows that the given limit is 0.
lim
x, y l 0, 0
3x 2y 0 x2 y2
■
CONTINUITY
Recall that evaluating limits of continuous functions of a single variable is easy. It can be accomplished by direct substitution because the defining property of a continuous function is limx l a f x f a. Continuous functions of two variables are also defined by the direct substitution property.
4 DEFINITION
A function f of two variables is called continuous at a, b if lim
x, y l a, b
f x, y f a, b
We say f is continuous on D if f is continuous at every point a, b in D. The intuitive meaning of continuity is that if the point x, y changes by a small amount, then the value of f x, y changes by a small amount. This means that a surface that is the graph of a continuous function has no hole or break. Using the properties of limits, you can see that sums, differences, products, and quotients of continuous functions are continuous on their domains. Let’s use this fact to give examples of continuous functions. A polynomial function of two variables (or polynomial, for short) is a sum of terms of the form cx my n, where c is a constant and m and n are nonnegative integers. A rational function is a ratio of polynomials. For instance, f x, y x 4 5x 3 y 2 6xy 4 7y 6
606
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CHAPTER 11
PARTIAL DERIVATIVES
is a polynomial, whereas tx, y
2xy 1 x2 y2
is a rational function. The limits in (2) show that the functions f x, y x, tx, y y, and hx, y c are continuous. Since any polynomial can be built up out of the simple functions f , t, and h by multiplication and addition, it follows that all polynomials are continuous on ⺢ 2. Likewise, any rational function is continuous on its domain because it is a quotient of continuous functions. Evaluate
V EXAMPLE 5
lim
x, y l 1, 2
x 2y 3 x 3y 2 3x 2y.
SOLUTION Since f x, y x 2 y 3 x 3 y 2 3x 2y is a polynomial, it is continu-
ous everywhere, so we can find the limit by direct substitution: lim
x, y l 1, 2
x 2y 3 x 3y 2 3x 2y 1 2 2 3 1 3 2 2 3 1 2 2 11
EXAMPLE 6 Where is the function f x, y
■
x2 y2 continuous? x2 y2
SOLUTION The function f is discontinuous at 0, 0 because it is not defined there. Since f is a rational function, it is continuous on its domain, which is the set D x, y x, y 0, 0 . ■
EXAMPLE 7 Let
x2 y2 tx, y x 2 y 2 0
if x, y 0, 0 if x, y 0, 0
Here t is defined at 0, 0 but t is still discontinuous there because lim x, y l 0, 0 tx, y does not exist (see Example 1). ■ ■ Figure 8 shows the graph of the continuous function in Example 8.
EXAMPLE 8 Let
3x 2y f x, y x 2 y 2 0
z
y
if x, y 0, 0 if x, y 0, 0
We know f is continuous for x, y 0, 0 since it is equal to a rational function there. Also, from Example 4, we have
x
lim
x, y l 0, 0
FIGURE 8
f x, y
lim
x, y l 0, 0
3x 2y 0 f 0, 0 x y2 2
Therefore, f is continuous at 0, 0, and so it is continuous on ⺢ 2.
■
Just as for functions of one variable, composition is another way of combining two continuous functions to get a third. In fact, it can be shown that if f is a continuous function of two variables and t is a continuous function of a single variable that is defined on the range of f , then the composite function h t ⴰ f defined by hx, y t f x, y is also a continuous function.
SECTION 11.2
LIMITS AND CONTINUITY
■
607
EXAMPLE 9 Where is the function hx, y arctanyx continuous?
z 0
SOLUTION The function f x, y yx is a rational function and therefore continuous except on the line x 0. The function tt arctan t is continuous everywhere. So the composite function
_2
t f x, y arctan yx hx, y
2
_2
_2 y 0
0 2
2
x
FIGURE 9
The function h(x, y)=arctan(y/x) is discontinuous where x=0.
is continuous except where x 0. The graph in Figure 9 shows the break in the graph of h above the y-axis.
■
FUNCTIONS OF THREE OR MORE VARIABLES
Everything that we have done in this section can be extended to functions of three or more variables. The notation lim
x, y, z l a, b, c
f x, y, z L
means that the values of f x, y, z approach the number L as the point x, y, z approaches the point a, b, c along any path in the domain of f. Because the distance between two points x, y, z and a, b, c is sx a 2 y b 2 z c 2 , we can write the precise definition as follows: For every number 0 there is a corresponding number 0 such that if x, y, z is in the domain of f then
and 0 sx a 2 y b 2 z c 2
f x, y, z L
The function f is continuous at a, b, c if lim
x, y, z l a, b, c
f x, y, z f a, b, c
For instance, the function f x, y, z
1 x y z2 1 2
2
is a rational function of three variables and so is continuous at every point in ⺢ 3 except where x 2 y 2 z 2 1. In other words, it is discontinuous on the sphere with center the origin and radius 1. If we use the vector notation introduced at the end of Section 11.1, then we can write the definitions of a limit for functions of two or three variables in a single compact form as follows. If f is defined on a subset D of ⺢ n, then lim x l a f x L means that for every number 0 there is a corresponding number 0 such that 5
if x D
and 0 x a
then
f x L
Notice that if n 1, then x x and a a, and (5) is just the definition of a limit for functions of a single variable. For the case n 2, we have x x, y , a a, b , and x a sx a 2 y b 2 , so (5) becomes Definition 1. If n 3, then x x, y, z , a a, b, c , and (5) becomes the definition of a limit of a function of three variables. In each case the definition of continuity can be written as
lim f x f a
xla
608
■
CHAPTER 11
11.2
PARTIAL DERIVATIVES
EXERCISES
1. Suppose that lim x, y l 3, 1 f x, y 6. What can you say
19–20
2. Explain why each function is continuous or discontinuous.
19. tt t 2 st ,
■ Find hx, y t f x, y and the set on which h is continuous.
about the value of f 3, 1? What if f is continuous?
(a) The outdoor temperature as a function of longitude, latitude, and time (b) Elevation (height above sea level) as a function of longitude, latitude, and time (c) The cost of a taxi ride as a function of distance traveled and time 3–16 ■ Find the limit, if it exists, or show that the limit does not exist. 3. 4.
5.
7.
9.
11.
13.
15.
16. ■
lim
x, y l 5, 2
lim
y4 x 4 3y 4
6.
lim
x y cos y 3x 2 y 2
8.
x, y l 0, 0
lim
2x 2 y x y2
12.
lim
x2 y2 sx y 2 1 1
14.
4
2
lim
x y yz 2 xz 2 x2 y2 z4
lim
x 2 2y 2 3z 2 x 2 y 2 z2
x, y, z l 0, 0, 0
x, y, z l 0, 0, 0 ■
; 17–18
■
■
■
lim
x 2 sin 2 y 2x 2 y 2
lim
6x 3 y 2x 4 y 4
x, y l 0, 0
x, y l 0, 0
lim
■
lim
x, y l 0, 0 ■
■
■
■
22. Fx, y
■
■
■
xy 1 x2 y2
■
sy x2 y2 z2
lim
lim
xy4 x y8
28. f x, y
lim
x4 y4 x2 y2
x, y l 0, 0
x, y l 0, 0
x, y l 0, 0
2
■
■
■
if x, y 0, 0 if x, y 0, 0
xy x2 xy y2 0 ■
■
if x, y 0, 0 if x, y 0, 0
■
■
■
■
■
■
■
■ Use polar coordinates to find the limit. [If r, are polar coordinates of the point x, y with r 0, note that r l 0 as x, y l 0, 0.]
29–30
29. ■
■
■
■
■
lim
x3 y3 x2 y2
lim
x 2 y 2 lnx 2 y 2
x, y l0, 0
■
30. ■
2x 2 3x y 4y 2 3x 2 5y 2
x, y l0, 0 ■
■
■
■
■
■
■
■
■
■
■
31. Show that the function f given by f x x is continuous
on ⺢ n.
xy x 2 y6 ■
sinx y e x y2
x sin y x 2 2y 2
2
3
18.
■
26. f x, y, z sx y z
■
x, y l 0, 0
■
21–28 ■ Determine the set of points at which the function is continuous.
25. f x, y, z
Use a computer graph of the function to explain why the limit does not exist. 17.
■
x2y3 27. f x, y 2 x 2 y 2 1
2
10.
x, y l 0, 0
■
2
lim
x, y l 0, 0
■
■
24. Fx, y e x y sx y 2
xy sx 2 y 2
x, y l 0, 0
■
f x, y x 2 y
23. Gx, y lnx 2 y 2 4
x y cosx 2y
x, y l 0, 0
st 1 , st 1
21. Fx, y
x 5 4x 3y 5x y 2
lim
x, y l 6, 3
20. tt
f x, y 2 x 3y 6
[Hint: Consider x a
2
x a x a.]
32. If c Vn , show that the function f given by f x c x is ■
■
■
■
■
■
■
■
continuous on ⺢ n.
SECTION 11.3
11.3
PARTIAL DERIVATIVES
■
609
PARTIAL DERIVATIVES If f is a function of two variables x and y, suppose we let only x vary while keeping y fixed, say y b, where b is a constant. Then we are really considering a function of a single variable x, namely, tx f x, b. If t has a derivative at a, then we call it the partial derivative of f with respect to x at a, b and denote it by fx a, b. Thus
1
fx a, b ta
where
tx f x, b
By the definition of a derivative, we have ta h ta h
ta lim
hl0
and so Equation 1 becomes
2
fx a, b lim
hl0
f a h, b f a, b h
Similarly, the partial derivative of f with respect to y at a, b, denoted by fy a, b, is obtained by keeping x fixed x a and finding the ordinary derivative at b of the function Gy f a, y:
3
fy a, b lim
hl0
f a, b h f a, b h
If we now let the point a, b vary in Equations 2 and 3, fx and fy become functions of two variables. 4 If f is a function of two variables, its partial derivatives are the functions fx and fy defined by
fx x, y lim
f x h, y f x, y h
fy x, y lim
f x, y h f x, y h
hl0
hl0
There are many alternative notations for partial derivatives. For instance, instead of fx we can write f1 or D1 f (to indicate differentiation with respect to the first variable) or &f&x. But here &f&x can’t be interpreted as a ratio of differentials.
610
■
CHAPTER 11
PARTIAL DERIVATIVES
NOTATIONS FOR PARTIAL DERIVATIVES If z f x, y, we write
fx x, y fx
&f & &z f x, y f1 D1 f Dx f &x &x &x
fy x, y fy
&f & &z f x, y f2 D2 f Dy f &y &y &y
To compute partial derivatives, all we have to do is remember from Equation 1 that the partial derivative with respect to x is just the ordinary derivative of the function t of a single variable that we get by keeping y fixed. Thus we have the following rule. RULE FOR FINDING PARTIAL DERIVATIVES OF z f x, y 1. To find fx , regard y as a constant and differentiate f x, y with respect to x. 2. To find fy , regard x as a constant and differentiate f x, y with respect to y.
EXAMPLE 1 If f x, y x 3 x 2 y 3 2y 2, find fx 2, 1 and fy 2, 1. SOLUTION Holding y constant and differentiating with respect to x, we get
fx x, y 3x 2 2xy 3 and so
fx 2, 1 3 2 2 2 2 13 16
Holding x constant and differentiating with respect to y, we get fy x, y 3x 2 y 2 4y fy 2, 1 3 2 2 12 4 1 8
■
INTERPRETATIONS OF PARTIAL DERIVATIVES z
T¡ S
C¡ T™ P(a, b, c)
C™
0 y
x (a, b, 0)
FIGURE 1
The partial derivatives of f at (a, b) are the slopes of the tangents to C¡ and C™.
To give a geometric interpretation of partial derivatives, we recall that the equation z f x, y represents a surface S (the graph of f ). If f a, b c, then the point Pa, b, c lies on S. By fixing y b, we are restricting our attention to the curve C1 in which the vertical plane y b intersects S. (In other words, C1 is the trace of S in the plane y b.) Likewise, the vertical plane x a intersects S in a curve C2 . Both of the curves C1 and C2 pass through the point P. (See Figure 1.) Notice that the curve C1 is the graph of the function tx f x, b, so the slope of its tangent T1 at P is ta fx a, b. The curve C2 is the graph of the function Gy f a, y, so the slope of its tangent T2 at P is Gb fy a, b. Thus the partial derivatives fx a, b and fy a, b can be interpreted geometrically as the slopes of the tangent lines at Pa, b, c to the traces C1 and C2 of S in the planes y b and x a. Partial derivatives can also be interpreted as rates of change. If z f x, y, then &z&x represents the rate of change of z with respect to x when y is fixed. Similarly, &z&y represents the rate of change of z with respect to y when x is fixed.
SECTION 11.3
PARTIAL DERIVATIVES
■
611
EXAMPLE 2 If f x, y 4 x 2 2y 2, find fx 1, 1 and fy 1, 1 and interpret these
z
z=4-≈-2¥
numbers as slopes. SOLUTION We have
C¡
fx x, y 2x
fy x, y 4y
y=1
fx 1, 1 2
fy 1, 1 4
(1, 1, 1)
(1, 1)
y
2 x
FIGURE 2
The graph of f is the paraboloid z 4 x 2 2y 2 and the vertical plane y 1 intersects it in the parabola z 2 x 2, y 1. (As in the preceding discussion, we label it C1 in Figure 2.) The slope of the tangent line to this parabola at the point 1, 1, 1 is fx 1, 1 2. Similarly, the curve C2 in which the plane x 1 intersects the paraboloid is the parabola z 3 2y 2, x 1, and the slope of the tangent line at 1, 1, 1 is fy 1, 1 4. (See Figure 3.) ■
z
z=4-≈-2¥
C™ x=1 (1, 1, 1) y 2 x
(1, 1)
FIGURE 3
x &f &f , calculate and . 1y &x &y SOLUTION Using the Chain Rule for functions of one variable, we have V EXAMPLE 3
If f x, y sin
&f x cos &x 1y
& &x
x 1y
cos
&f x cos &y 1y
& &y
x 1y
cos
x 1y
x 1y
1 1y
x 1 y2
■
V EXAMPLE 4 Find &z&x and &z&y if z is defined implicitly as a function of x and y by the equation
x 3 y 3 z 3 6xyz 1 Some computer algebra systems can plot surfaces defined by implicit equations in three variables. Figure 4 shows such a plot of the surface defined by the equation in Example 4. ■
SOLUTION To find &z&x, we differentiate implicitly with respect to x, being careful
to treat y as a constant: 3x 2 3z 2
&z &z 6yz 6xy 0 &x &x
Solving this equation for &z&x, we obtain &z x 2 2yz 2 &x z 2xy Similarly, implicit differentiation with respect to y gives
FIGURE 4
&z y 2 2xz 2 &y z 2xy
■
612
■
CHAPTER 11
PARTIAL DERIVATIVES
FUNCTIONS OF MORE THAN TWO VARIABLES
Partial derivatives can also be defined for functions of three or more variables. For example, if f is a function of three variables x, y, and z, then its partial derivative with respect to x is defined as f x h, y, z f x, y, z h
fx x, y, z lim
hl0
and it is found by regarding y and z as constants and differentiating f x, y, z with respect to x. If w f x, y, z, then fx &w&x can be interpreted as the rate of change of w with respect to x when y and z are held fixed. But we can’t interpret it geometrically because the graph of f lies in four-dimensional space. In general, if u is a function of n variables, u f x 1, x 2 , . . . , x n , its partial derivative with respect to the ith variable x i is &u f x1 , . . . , xi1 , xi h, xi1 , . . . , xn f x1 , . . . , xi , . . . , xn lim hl0 &xi h and we also write &u &f fx i f i Di f &x i &x i EXAMPLE 5 Find fx , fy , and fz if f x, y, z e x y ln z. SOLUTION Holding y and z constant and differentiating with respect to x, we have
fx ye x y ln z Similarly,
fy xe x y ln z
fz
and
e xy z
■
HIGHER DERIVATIVES
If f is a function of two variables, then its partial derivatives fx and fy are also functions of two variables, so we can consider their partial derivatives fx x , fx y , fy x , and fy y , which are called the second partial derivatives of f . If z f x, y, we use the following notation: fx x fxx f11
& &x
fx y fxy f12
& &y
fy x fyx f21
& &x
fy y fyy f22
& &y
&f &x &f &x &f &y &f &y
&2 f &2 z 2 &x &x 2
&2 f &2 z &y &x &y &x
&2 f &2 z &x &y &x &y
&2 f &2 z 2 &y &y 2
Thus the notation fx y (or &2 f&y &x) means that we first differentiate with respect to x and then with respect to y, whereas in computing fyx the order is reversed.
SECTION 11.3
PARTIAL DERIVATIVES
■
613
EXAMPLE 6 Find the second partial derivatives of
f x, y x 3 x 2 y 3 2y 2 SOLUTION In Example 1 we found that
fx x, y 3x 2 2xy 3
fy x, y 3x 2 y 2 4y
Therefore fxx
& 3x 2 2xy 3 6x 2y 3 &x
fxy
& 3x 2 2xy 3 6xy 2 &y
fyx
& 3x 2 y 2 4y 6xy 2 &x
fyy
& 3x 2 y 2 4y 6x 2 y 4 ■ &y
Notice that fx y fyx in Example 6. This is not just a coincidence. It turns out that the mixed partial derivatives fx y and fyx are equal for most functions that one meets in practice. The following theorem, which was discovered by the French mathematician Alexis Clairaut (1713–1765), gives conditions under which we can assert that fx y fyx . The proof is given in Appendix B. Alexis Clairaut was a child prodigy in mathematics, having read l’Hospital’s textbook on calculus when he was ten and presented a paper on geometry to the French Academy of Sciences when he was 13. At the age of 18, Clairaut published Recherches sur les courbes à double courbure, which was the first systematic treatise on three-dimensional analytic geometry and included the calculus of space curves. ■
CLAIRAUT’S THEOREM Suppose f is defined on a disk D that contains the point a, b. If the functions fx y and fyx are both continuous on D, then
fx y a, b fyx a, b Partial derivatives of order 3 or higher can also be defined. For instance, fx yy fx y y
& &y
&2 f &y &x
&3 f &y 2 &x
and using Clairaut’s Theorem it can be shown that fx yy fyx y fyyx if these functions are continuous. V EXAMPLE 7
SOLUTION
Calculate fxx yz if f x, y, z sin3x yz. fx 3 cos3x yz fxx 9 sin3x yz fxx y 9z cos3x yz fxx yz 9 cos3x y z 9yz sin3x yz
■
PARTIAL DIFFERENTIAL EQUATIONS
Partial derivatives occur in partial differential equations that express certain physical laws. For instance, the partial differential equation &2u &2u 0 &x 2 &y 2
614
■
CHAPTER 11
PARTIAL DERIVATIVES
is called Laplace’s equation after Pierre Laplace (1749–1827). Solutions of this equation are called harmonic functions and play a role in problems of heat conduction, fluid flow, and electric potential. EXAMPLE 8 Show that the function ux, y e x sin y is a solution of Laplace’s
equation. SOLUTION
ux e x sin y
u y e x cos y
u xx e x sin y
u yy e x sin y
u xx u yy e x sin y e x sin y 0 ■
Therefore, u satisfies Laplace’s equation. The wave equation 2 &2u 2 & u a &t 2 &x 2
u(x, t) x FIGURE 5
describes the motion of a waveform, which could be an ocean wave, a sound wave, a light wave, or a wave traveling along a vibrating string. For instance, if ux, t represents the displacement of a vibrating violin string at time t and at a distance x from one end of the string (as in Figure 5), then ux, t satisfies the wave equation. Here the constant a depends on the density of the string and on the tension in the string. EXAMPLE 9 Verify that the function ux, t sinx at satisfies the wave
equation. SOLUTION
ux cosx at
uxx sinx at
ut a cosx at
utt a 2 sinx at a 2uxx ■
So u satisfies the wave equation.
11.3
EXERCISES
1. The temperature T at a location in the Northern Hemisphere
depends on the longitude x, latitude y, and time t, so we can write T f x, y, t. Let’s measure time in hours from the beginning of January. (a) What are the meanings of the partial derivatives &T&x, &T&y, and &T&t ? (b) Honolulu has longitude 158 W and latitude 21 N . Suppose that at 9:00 AM on January 1 the wind is blowing hot air to the northeast, so the air to the west and south is warm and the air to the north and east is cooler. Would you expect fx 158, 21, 9, fy 158, 21, 9, and ft 158, 21, 9 to be positive or negative? Explain.
2. A contour map is given for a function f . Use it to estimate
fx 2, 1 and fy 2, 1. y 3 0
_4
6
_2
8 10 12 14 16
4 2 1
3
18
x
SECTION 11.3
32. f u, v, w w tanu v;
■ Determine the signs of the partial derivatives for the function f whose graph is shown.
3– 4
■
z
■
■
■
■
1
■
■
2
■
■
■
■
■
■
■
■
■
2
interpret these numbers as slopes. Illustrate with either hand-drawn sketches or computer plots.
12. f x, y x y
13. w sin cos
14. f s, t st 2s 2 t 2
15. f r, s r lnr 2 s 2
16. f x, t arctan ( xst ) 18. f x, y
y
x
y
22. w sr 2 s 2 t 2
23. u xet sin
24. u x yz
■
29–32
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
fy 6, 4
31. f x, y, z x y z;
fz 3, 2, 1
■
■
■
■
■
■
■
■
■
■
■
■
■
44. f x, y ln3x 5y
45. z xx y
46. z y tan 2x
■
48. v sx y 2
sin t
■
■
■
■
■
■
■
■
■
■
■
■
■
■
50. u x 4 y 2 2xy 5 ■
■
■
■
51. f x, y 3x y 4 x 3 y 2; 52. f x, t x 2ect;
■
55. u e r sin ;
■
■
■
■
fttt ,
fxxy ,
■
■
■
fyyy
ftxx
■
frss ,
fxy z ,
fyzz
frst
& 3u &r 2 & &6u &x &y 2 &z 3
56. u x a y bz c; ■
■
Find the indicated partial derivative.
54. f r, s, t r lnrs t ; ■
■
Find all the second partial derivatives.
fx 3, 4
30. f x, y sin2x 3y;
y 2
43. f x, y x 4 3x 2 y 3
51–56
■
■
(b) z f x y
2 3
■
■
2
Find &z&x and &z&y.
53. f x, y, z cos4x 3y 2z;
Find the indicated partial derivatives.
29. f x, y sx 2 y 2 ;
■
42. (a) z f xt y
■
2 n
■
■
■ Verify that the conclusion of Clairaut’s Theorem holds, that is, u x y u yx .
xy2 26. f x, y, z, t t 2z
■
■
49–50
28. u sinx 1 2x 2 nx n ■
■
49. u x sinx 2y
21. w lnx 2y 3z
27. u sx x x
■
36. f x, y xex
(b) z f x y
■
cost dt
20. f x, y, z x 2e yz
2 2
■
Use implicit differentiation to find &z&x and &z&y.
47. u e
2
19. f x, y, z xy 2z 3 3yz
2 1
■
s
25. f x, y, z, t x yz tan yt
■
(c) z f xy
10. z y ln x
2
■
41. (a) z f x t y
43– 48
17. u te
■
■
8. f x, y x 5 3x 3y 2 3xy 4
wt
■
40. sinx yz x 2y 3z
■
xy xy
■
x x y2
39. x z arctan yz
7. f x, y 3x 2y 4
11. f x, y
■
38. yz lnx z
Find the first partial derivatives of the function.
9. z xe 3y
■
37. x y 2 z 2 3x yz
41– 42
interpret these numbers as slopes. Illustrate with either hand-drawn sketches or computer plots. 7–28
■
2
■
6. If f x, y s4 x 2 4y 2 , find fx 1, 0 and fy 1, 0 and
■
■
■
37– 40
5. If f x, y 16 4x y , find fx 1, 2 and fy 1, 2 and 2
■
34. f x, y
35. f x, y x 2 y 2 x 2 y
(b) fy 1, 2 (d) fyy 1, 2
■
■
■ Find f and f and graph f , f , and f with domains and x y x y viewpoints that enable you to see the relationships between them.
4. (a) fx 1, 2 ■
■
; 35–36
(b) fy 1, 2
■
■
■ Use the definition of partial derivatives as limits (4) to find fx x, y and fy x, y.
3. (a) fx 1, 2
■
615
33–34
y
(c) fxx 1, 2
■
fv 2, 0, 3 ■
33. f x, y xy 2 x 3y x
PARTIAL DERIVATIVES
■
■
■
■
■
■
616
■
CHAPTER 11
PARTIAL DERIVATIVES
57. Verify that the function u e
67. For the ideal gas of Exercise 66, show that
2 2
k t
sin kx is a solution of the heat conduction equation u t u xx . 2
T
58. Determine whether each of the following functions is a
solution of Laplace’s equation u xx u yy 0 . (a) u x 2 y 2 (b) u x 2 y 2 (c) u x 3 3xy 2 (d) u ln sx 2 y 2 (e) u sin x cosh y cos x sinh y (f ) u ex cos y ey cos x
68. The wind-chill index is a measure of how cold it feels in
windy weather. It is modeled by the function W 13.12 0.6215T 11.37v 0.16 0.3965T v 0.16 where T is the temperature C and v is the wind speed kmh. When T 15 C and v 30 kmh, by how much would you expect the apparent temperature to drop if the actual temperature decreases by 1 C ? What if the wind speed increases by 1 kmh ?
59. Verify that the function u 1sx 2 y 2 z 2 is a
solution of the three-dimensional Laplace equation u xx u yy u zz 0 . 60. Show that each of the following functions is a solution of
69. The kinetic energy of a body with mass m and velocity v is K 12 mv 2. Show that
the wave equation u t t a 2u xx . (a) u sink x sinak t (b) u ta 2t 2 x 2 (c) u x at6 x at6 (d) u sinx at lnx at
&K &2K K &m &v 2 70. If a, b, c are the sides of a triangle and A, B, C are the
61. If f and t are twice differentiable functions of a single vari-
opposite angles, find &A&a, &A&b, &A&c by implicit differentiation of the Law of Cosines.
able, show that the function ux, t f x at tx at
71. You are told that there is a function f whose partial deriva-
tives are fx x, y x 4y and fy x, y 3x y. Should you believe it?
is a solution of the wave equation given in Exercise 60. 62. If u e a1 x1a2 x2 an x n, where a 21 a 22 a 2n 1,
show that
2 2 ; 72. The paraboloid z 6 x x 2y intersects the plane
x 1 in a parabola. Find parametric equations for the tangent line to this parabola at the point 1, 2, 4. Use a computer to graph the paraboloid, the parabola, and the tangent line on the same screen.
&2u &2u &2u u 2 2 &x 1 &x 2 &x n2 63. Show that the function z xe y ye x is a solution of the
73. The ellipsoid 4x 2 2y 2 z 2 16 intersects the plane
equation
y 2 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point 1, 2, 2.
&z &z &z &z x y 2 &x 3 &y 3 &x &y 2 &x &y 3
3
3
3
64. The temperature at a point x, y on a flat metal plate is
74. In a study of frost penetration it was found that the tempera-
given by Tx, y 601 x 2 y 2 , where T is measured in C and x, y in meters. Find the rate of change of temperature with respect to distance at the point 2, 1 in (a) the x-direction and (b) the y-direction.
ture T at time t (measured in days) at a depth x (measured in feet) can be modeled by the function Tx, t T0 T1 e x sin t x
65. The total resistance R produced by three conductors with
resistances R1 , R2 , R3 connected in a parallel electrical circuit is given by the formula 1 1 1 1 R R1 R2 R3 Find &R&R1. 66. The gas law for a fixed mass m of an ideal gas at absolute
temperature T, pressure P, and volume V is PV mRT, where R is the gas constant. Show that &P &V &T 1 &V &T &P
&P &V mR &T &T
;
where 2365 and is a positive constant. (a) Find &T&x. What is its physical significance? (b) Find &T&t. What is its physical significance? (c) Show that T satisfies the heat equation Tt kTxx for a certain constant k. (d) If 0.2, T0 0, and T1 10, use a computer to graph Tx, t. (e) What is the physical significance of the term x in the expression sin t x? 75. Use Clairaut’s Theorem to show that if the third-order par-
tial derivatives of f are continuous, then fx yy fyx y fyyx
SECTION 11.4
76. (a) How many nth-order partial derivatives does a function
79. Let
x 3y xy 3 x2 y2 f x, y 0
of two variables have? (b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of three variables.
;
2
77. If f x, y xx 2 y 2 32e sinx y, find fx 1, 0.
[Hint: Instead of finding fx x, y first, note that it’s easier to use Equation 1 or Equation 2.]
CAS
3 x 3 y 3 , find fx 0, 0. 78. If f x, y s
11.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
(a) (b) (c) (d) (e)
■
617
if x, y 0, 0 if x, y 0, 0
Use a computer to graph f . Find fx x, y and fy x, y when x, y 0, 0. Find fx 0, 0 and fy 0, 0 using Equations 2 and 3. Show that fxy 0, 0 1 and fyx 0, 0 1. Does the result of part (d) contradict Clairaut’s Theorem? Use graphs of fxy and fyx to illustrate your answer.
TANGENT PLANES AND LINEAR APPROXIMATIONS One of the most important ideas in single-variable calculus is that as we zoom in toward a point on the graph of a differentiable function, the graph becomes indistinguishable from its tangent line and we can approximate the function by a linear function. (See Section 2.8.) Here we develop similar ideas in three dimensions. As we zoom in toward a point on a surface that is the graph of a differentiable function of two variables, the surface looks more and more like a plane (its tangent plane) and we can approximate the function by a linear function of two variables. We also extend the idea of a differential to functions of two or more variables. TANGENT PLANES
z
y x
FIGURE 1
The tangent plane contains the tangent lines T¡ T and T™ T.
Suppose a surface S has equation z f x, y, where f has continuous first partial derivatives, and let Px 0 , y0 , z0 be a point on S. As in the preceding section, let C1 and C2 be the curves obtained by intersecting the vertical planes y y0 and x x 0 with the surface S. Then the point P lies on both C1 and C2. Let T1 and T2 be the tangent lines to the curves C1 and C2 at the point P. Then the tangent plane to the surface S at the point P is defined to be the plane that contains both tangent lines T1 and T2 . (See Figure 1.) We will see in Section 11.6 that if C is any other curve that lies on the surface S and passes through P, then its tangent line at P also lies in the tangent plane. Therefore, you can think of the tangent plane to S at P as consisting of all possible tangent lines at P to curves that lie on S and pass through P. The tangent plane at P is the plane that most closely approximates the surface S near the point P. We know from Equation 10.5.7 that any plane passing through the point Px 0 , y0 , z0 has an equation of the form Ax x 0 B y y0 Cz z0 0 By dividing this equation by C and letting a AC and b BC, we can write it in the form 1
z z0 ax x 0 by y0
If Equation 1 represents the tangent plane at P, then its intersection with the plane
618
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CHAPTER 11
PARTIAL DERIVATIVES
y y0 must be the tangent line T1. Setting y y0 in Equation 1 gives z z0 ax x 0
y y0
and we recognize these as the equations (in point-slope form) of a line with slope a. But from Section 11.3 we know that the slope of the tangent T1 is fx x 0 , y0 . Therefore, a fx x 0 , y0 . Similarly, putting x x 0 in Equation 1, we get z z0 b y y0 , which must represent the tangent line T2 , so b fy x 0 , y0 . 2 Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface z f x, y at the point Px 0 , y0 , z0 is
■ Note the similarity between the equation of a tangent plane and the equation of a tangent line: y y0 f x 0 x x 0
z z0 fx x 0 , y0 x x 0 fy x 0 , y0 y y0 Find the tangent plane to the elliptic paraboloid z 2x 2 y 2 at the point 1, 1, 3. V EXAMPLE 1
SOLUTION Let f x, y 2x 2 y 2. Then
fx x, y 4x
fy x, y 2y
fx 1, 1 4
fy 1, 1 2
Then (2) gives the equation of the tangent plane at 1, 1, 3 as z 3 4x 1 2 y 1 z 4x 2y 3
or
Visual 11.4 shows an animation of Figure 2.
Figure 2(a) shows the elliptic paraboloid and its tangent plane at (1, 1, 3) that we found in Example 1. In parts (b) and (c) we zoom in toward the point (1, 1, 3) by restricting the domain of the function f x, y 2x 2 y 2. Notice that the more we zoom in, the flatter the graph appears and the more it resembles its tangent plane.
40
40
20
20
20
0
z 0
z 0
_20
_20
40
z
■
_20 _4 _2 y
_2
0 2 4 4
(a)
2
_4
_2
_2
0 x
y
0
0 2
2
(b)
x
0
0 y
1
1 2
2
(c)
FIGURE 2 The elliptic paraboloid z=2≈+¥ appears to coincide with its tangent plane as we zoom in toward (1, 1, 3).
x
SECTION 11.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
■
619
In Figure 3 we corroborate this impression by zooming in toward the point (1, 1) on a contour map of the function f x, y 2x 2 y 2. Notice that the more we zoom in, the more the level curves look like equally spaced parallel lines, which is characteristic of a plane. 1.5
1.2
1.05
FIGURE 3
Zooming in toward (1, 1) on a contour map of f(x, y)=2≈+¥
1.5
0.5
1.2
0.8
0.95
1.05
LINEAR APPROXIMATIONS
In Example 1 we found that an equation of the tangent plane to the graph of the function f x, y 2x 2 y 2 at the point (1, 1, 3) is z 4x 2y 3. Therefore, in view of the visual evidence in Figures 2 and 3, the linear function of two variables Lx, y 4x 2y 3 is a good approximation to f x, y when x, y is near (1, 1). The function L is called the linearization of f at (1, 1) and the approximation f x, y 4x 2y 3 is called the linear approximation or tangent plane approximation of f at (1, 1). For instance, at the point (1.1, 0.95) the linear approximation gives f 1.1, 0.95 41.1 20.95 3 3.3 which is quite close to the true value of f 1.1, 0.95 21.12 0.952 3.3225. But if we take a point farther away from (1, 1), such as (2, 3), we no longer get a good approximation. In fact, L2, 3 11 whereas f 2, 3 17. In general, we know from (2) that an equation of the tangent plane to the graph of a function f of two variables at the point a, b, f a, b is z f a, b fx a, bx a fy a, by b if fx and fy are continuous. The linear function whose graph is this tangent plane, namely 3
Lx, y f a, b fx a, bx a fy a, by b
is called the linearization of f at a, b and the approximation 4
f x, y f a, b fx a, bx a fy a, by b
is called the linear approximation or the tangent plane approximation of f at a, b.
620
■
CHAPTER 11
PARTIAL DERIVATIVES
z
We have defined tangent planes for surfaces z f x, y, where f has continuous first partial derivatives. What happens if fx and fy are not continuous? Figure 4 pictures such a function; its equation is
y
f x, y x
FIGURE 4
xy if (x, y)≠(0, 0), ≈+¥ f(0, 0)=0 f(x, y)=
xy x2 y2 0
if x, y 0, 0 if x, y 0, 0
You can verify (see Exercise 36) that its partial derivatives exist at the origin and, in fact, fx 0, 0 0 and fy 0, 0 0, but fx and fy are not continuous. The linear approximation would be f x, y 0, but f x, y 12 at all points on the line y x. So a function of two variables can behave badly even though both of its partial derivatives exist. To rule out such behavior, we formulate the idea of a differentiable function of two variables. Recall that for a function of one variable, y f x, if x changes from a to a x, we defined the increment of y as y f a x f a In Chapter 2 we showed that if f is differentiable at a, then
■
This is Equation 2.5.5.
5
y f a x x
where l 0 as x l 0
Now consider a function of two variables, z f x, y, and suppose x changes from a to a x and y changes from b to b y. Then the corresponding increment of z is 6
z f a x, b y f a, b
Thus the increment z represents the change in the value of f when x, y changes from a, b to a x, b y. By analogy with (5) we define the differentiability of a function of two variables as follows. 7 DEFINITION If z f x, y, then f is differentiable at a, b if z can be expressed in the form
z fx a, b x fy a, b y 1 x 2 y where 1 and 2 l 0 as x, y l 0, 0. Definition 7 says that a differentiable function is one for which the linear approximation (4) is a good approximation when x, y is near a, b. In other words, the tangent plane approximates the graph of f well near the point of tangency. It’s sometimes hard to use Definition 7 directly to check the differentiability of a function, but the following theorem provides a convenient sufficient condition for differentiability. ■
Theorem 8 is proved in Appendix B.
8 THEOREM If the partial derivatives fx and fy exist near a, b and are continuous at a, b, then f is differentiable at a, b.
SECTION 11.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
■
621
Show that f x, y xe xy is differentiable at (1, 0) and find its linearization there. Then use it to approximate f 1.1, 0.1. V EXAMPLE 2
SOLUTION The partial derivatives are
■ Figure 5 shows the graphs of the function f and its linearization L in Example 2.
fx x, y e xy xye xy
fy x, y x 2e xy
fx 1, 0 1
fy 1, 0 1
Both fx and fy are continuous functions, so f is differentiable by Theorem 8. The linearization is Lx, y f 1, 0 fx 1, 0x 1 fy 1, 0y 0
6 4
1 1x 1 1 y x y
z 2
The corresponding linear approximation is 0
xe xy x y 1 x
_1
0y
0 1
f 1.1, 0.1 1.1 0.1 1
so
Compare this with the actual value of f 1.1, 0.1 1.1e 0.11 0.98542.
FIGURE 5
■
DIFFERENTIALS
For a differentiable function of one variable, y f x, we define the differential dx to be an independent variable; that is, dx can be given the value of any real number. The differential of y is then defined as
y
y=ƒ
Îy dx=Îx 0
a
a+Îx
tangent line y=f(a)+fª(a)(x-a) FIGURE 6
9
dy f x dx
dy
x
(See Section 2.8.) Figure 6 shows the relationship between the increment y and the differential dy : y represents the change in height of the curve y f x and dy represents the change in height of the tangent line when x changes by an amount dx x. For a differentiable function of two variables, z f x, y, we define the differentials dx and dy to be independent variables; that is, they can be given any values. Then the differential dz, also called the total differential, is defined by
10
dz fx x, y dx fy x, y dy
&z &z dx dy &x &y
(Compare with Equation 9.) Sometimes the notation d f is used in place of dz. If we take dx x x a and dy y y b in Equation 10, then the differential of z is dz fx a, bx a fy a, by b So, in the notation of differentials, the linear approximation (4) can be written as f x, y f a, b dz
622
■
CHAPTER 11
PARTIAL DERIVATIVES
Figure 7 is the three-dimensional counterpart of Figure 6 and shows the geometric interpretation of the differential dz and the increment z : dz represents the change in height of the tangent plane, whereas z represents the change in height of the surface z f x, y when x, y changes from a, b to a x, b y. z
{ a+Îx, b+Îy, f (a+Îx, b+Îy)}
surface z=f(x, y) Îz dz {a, b, f(a, b)}
f(a, b)
0
f(a, b)
dx
y
= Îx
(a+Îx, b+Îy, 0)
x (a, b, 0)
Îy=dy
tangent plane z-f(a, b)=ffx (a, b)(x-a)+ff y (a, b)(y-b)
FIGURE 7 V EXAMPLE 3
(a) If z f x, y x 2 3xy y 2, find the differential dz. (b) If x changes from 2 to 2.05 and y changes from 3 to 2.96, compare the values of z and dz. SOLUTION
(a) Definition 10 gives dz ■ In Example 3, dz is close to z because the tangent plane is a good approximation to the surface z x 2 3xy y 2 near 2, 3, 13. (See Figure 8.)
&z &z dx dy 2x 3y dx 3x 2y dy &x &y
(b) Putting x 2, dx x 0.05, y 3, and dy y 0.04, we get dz 22 33 0.05 32 23 0.04 0.65
60
The increment of z is
40
z f 2.05, 2.96 f 2, 3
z 20 0
2.052 32.052.96 2.962 2 2 323 3 2
_20 5
4
FIGURE 8
3
x
2
1
0
0 4 2y
0.6449 Notice that z dz but dz is easier to compute. EXAMPLE 4 The base radius and height of a right circular cone are measured as
10 cm and 25 cm, respectively, with a possible error in measurement of as much as 0.1 cm in each. Use differentials to estimate the maximum error in the calculated volume of the cone.
■
SECTION 11.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
■
623
SOLUTION The volume V of a cone with base radius r and height h is V r 2h3.
So the differential of V is dV
&V &V 2rh r 2 dr dh dr dh &r &h 3 3
Since each error is at most 0.1 cm, we have r 0.1, h 0.1. To find the largest error in the volume we take the largest error in the measurement of r and of h. Therefore, we take dr 0.1 and dh 0.1 along with r 10, h 25. This gives dV
500 100 0.1 0.1 20 3 3
Thus the maximum error in the calculated volume is about 20 cm3 63 cm3.
■
FUNCTIONS OF THREE OR MORE VARIABLES
Linear approximations, differentiability, and differentials can be defined in a similar manner for functions of more than two variables. A differentiable function is defined by an expression similar to the one in Definition 7. For such functions the linear approximation is f x, y, z f a, b, c fx a, b, cx a fy a, b, cy b fza, b, cz c and the linearization Lx, y, z is the right side of this expression. If w f x, y, z, then the increment of w is w f x x, y y, z z f x, y, z The differential dw is defined in terms of the differentials dx, dy, and dz of the independent variables by &w &w &w dw dx dy dz &x &y &z EXAMPLE 5 The dimensions of a rectangular box are measured to be 75 cm, 60 cm,
and 40 cm, and each measurement is correct to within 0.2 cm. Use differentials to estimate the largest possible error when the volume of the box is calculated from these measurements. SOLUTION If the dimensions of the box are x, y, and z, its volume is V xyz and so
dV
&V &V &V dx dy dz yz dx xz dy xy dz &x &y &z
We are given that x 0.2, y 0.2, and z 0.2. To find the largest error in the volume, we therefore use dx 0.2, dy 0.2, and dz 0.2 together with x 75, y 60, and z 40: V dV 60400.2 75400.2 75600.2 1980 Thus an error of only 0.2 cm in measuring each dimension could lead to an error of as much as 1980 cm3 in the calculated volume! This may seem like a large error, but it’s only about 1% of the volume of the box. ■
624
■
CHAPTER 11
PARTIAL DERIVATIVES
11.4
EXERCISES 18 –22
■ Find an equation of the tangent plane to the given surface at the specified point.
1–6
1. z 4x 2 y 2 2y, 3. z s4 x 2 2y 2 , 4. z y ln x, 6. z e x y , 2
■
■
; 7– 8
2
■
21. R 2 cos '
1, 1, 1
■
■
■
■
■
■
■
■
■
Graph the surface and the tangent plane at the given point. (Choose the domain and viewpoint so that you get a good view of both the surface and the tangent plane.) Then zoom in until the surface and the tangent plane become indistinguishable.
8. z arctanxy 2 , ■
CAS
■
■
■
■
■
■
■
■
■
1, 1, 0
10. f x, y exy10 (sx sy sxy ), ■
■
■
■
■
■
■
■
1, 1, 3e0.1 ■
■
■
■
■
11. f x, y x sy , 12. f x, y xy ,
1, 4
13. f x, y tan1x 2y, ■
■
■
■
■
■
■
■
■
■
24. If z x 2 xy 3y 2 and x, y changes from 3, 1 to
2.96, 0.95, compare the values of z and dz.
80 cm, 60 cm, and 50 cm, respectively, with a possible error of 0.2 cm in each dimension. Use differentials to estimate the maximum error in calculating the surface area of the box. 27. Use differentials to estimate the amount of tin in a closed
tin can with diameter 8 cm and height 12 cm if the tin is 0.04 cm thick. 28. Use differentials to estimate the amount of metal in a closed
cylindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1 cm thick and the metal in the sides is 0.05 cm thick. 29. A model for the surface area of a human body is given by S 0.1091w 0.425 h 0.725, where w is the weight (in pounds),
30. The pressure, volume, and temperature of a mole of an ideal
6, 3
14. f x, y sx e 4y ,
■
h is the height (in inches), and S is measured in square feet. If the errors in measurement of w and h are at most 2%, use differentials to estimate the maximum percentage error in the calculated surface area.
Explain why the function is differentiable at the given point. Then find the linearization Lx, y of the function at that point.
11–14
■
26. The dimensions of a closed rectangular box are measured as
9–10 ■ Draw the graph of f and its tangent plane at the given point. (Use your computer algebra system both to compute the partial derivatives and to graph the surface and its tangent plane.) Then zoom in until the surface and the tangent plane become indistinguishable.
xy sinx y 9. f x, y , 1 x2 y2
22. w xye xz ■
and 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.
1, 1, 4 ■
■
25. The length and width of a rectangle are measured as 30 cm
1, 1, 5
■
■
compare the values of z and dz.
■
7. z x 2 xy 3y 2,
■
23. If z 5x 2 y 2 and x, y changes from 1, 2 to 1.05, 2.1,
2, 2, 2
1, 1, 1 ■
20. u et sins 2t
19. z x 3 ln y 2
1, 2, 18
1, 4, 0
5. z y cosx y,
Find the differential of the function.
18. v y cos xy
1, 2, 4
2. z 9x 2 y 2 6x 3y 5,
■
1, 0
3, 0 ■
■
■
■
■
■
15. Find the linear approximation of the function
f x, y s20 x 2 7y 2 at 2, 1 and use it to approximate f 1.95, 1.08.
; 16. Find the linear approximation of the function
f x, y lnx 3y at 7, 2 and use it to approximate f 6.9, 2.06. Illustrate by graphing f and the tangent plane.
17. Find the linear approximation of the function
f x, y, z sx 2 y 2 z 2 at 3, 2, 6 and use it to approximate the number s3.02 2 1.97 2 5.99 2 .
■
gas are related by the equation PV 8.31T , where P is measured in kilopascals, V in liters, and T in kelvins. Use differentials to find the approximate change in the pressure if the volume increases from 12 L to 12.3 L and the temperature decreases from 310 K to 305 K. 31. If R is the total resistance of three resistors, connected in
parallel, with resistances R1 , R2 , R3 , then 1 1 1 1 R R1 R2 R3 If the resistances are measured in ohms as R1 25 , R2 40 , and R3 50 , with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of R.
SECTION 11.5
32. Suppose you need to know an equation of the tangent plane
lim
x, y l 0, 0
r2u 1 u 2, 2u 3 1, 2u 1
36. (a) The function
both lie on S. Find an equation of the tangent plane at P. f x, y
Show that the function is differentiable by finding values of 1 and 2 that satisfy Definition 7. ■
33. f x, y x 2 y 2
■
■
11.5
■
■
■
■
■
■
■
f a x, b y f a, b
xy x2 y2 0
if x, y 0, 0 if x, y 0, 0
was graphed in Figure 4. Show that fx 0, 0 and fy 0, 0 both exist but f is not differentiable at 0, 0. [Hint: Use the result of Exercise 35.] (b) Explain why fx and fy are not continuous at 0, 0.
34. f x, y xy 5y 2 ■
625
tiable at a, b, then f is continuous at a, b. Hint: Show that
r1t 2 3t, 1 t 2, 3 4t t 2
■
■
35. Prove that if f is a function of two variables that is differen-
to a surface S at the point P2, 1, 3. You don’t have an equation for S but you know that the curves
33–34
THE CHAIN RULE
■
THE CHAIN RULE Recall that the Chain Rule for functions of a single variable gives the rule for differentiating a composite function: If y f x and x tt, where f and t are differentiable functions, then y is indirectly a differentiable function of t and dy dy dx dt dx dt
1
For functions of more than one variable, the Chain Rule has several versions, each of them giving a rule for differentiating a composite function. The first version (Theorem 2) deals with the case where z f x, y and each of the variables x and y is, in turn, a function of a variable t. This means that z is indirectly a function of t, z f tt, ht, and the Chain Rule gives a formula for differentiating z as a function of t. We assume that f is differentiable (Definition 11.4.7). Recall that this is the case when fx and fy are continuous (Theorem 11.4.8). 2 THE CHAIN RULE (CASE 1) Suppose that z f x, y is a differentiable function of x and y, where x tt and y ht are both differentiable functions of t. Then z is a differentiable function of t and
dz &f dx &f dy dt &x dt &y dt
PROOF A change of t in t produces changes of x in x and y in y. These, in turn, produce a change of z in z, and from Definition 11.4.7 we have
z
&f &f x y 1 x 2 y &x &y
where 1 l 0 and 2 l 0 as x, y l 0, 0. [If the functions 1 and 2 are not defined at 0, 0, we can define them to be 0 there.] Dividing both sides of this
626
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CHAPTER 11
PARTIAL DERIVATIVES
equation by t, we have &f x &f y z x y 1 2 t &x t &y t t t If we now let t l 0, then x tt t tt l 0 because t is differentiable and therefore continuous. Similarly, y l 0. This, in turn, means that 1 l 0 and 2 l 0, so z dz lim t l 0 t dt
x y x y &f &f lim lim lim 1 lim lim 2 lim t l 0 t l 0 t t l 0 t l 0 t &x t l 0 t &y t l 0 t
&f dx &f dy dx dy 0 0 &x dt &y dt dt dt
&f dx &f dy &x dt &y dt
■
Since we often write &z&x in place of &f&x, we can rewrite the Chain Rule in the form ■ Notice the similarity to the definition of the differential:
dz &z dx &z dy dt &x dt &y dt
&z &z dz dx dy &x &y
EXAMPLE 1 If z x 2 y 3xy 4, where x sin 2t and y cos t, find dzdt when
t 0.
SOLUTION The Chain Rule gives
dz &z dx &z dy dt &x dt &y dt 2xy 3y 4 2 cos 2t x 2 12xy 3 sin t It’s not necessary to substitute the expressions for x and y in terms of t. We simply observe that when t 0 we have x sin 0 0 and y cos 0 1. Therefore,
y (0, 1)
dz dt x
FIGURE 1
The curve x=sin 2t, y=cos t
t0
0 32 cos 0 0 0sin 0 6
■
The derivative in Example 1 can be interpreted as the rate of change of z with respect to t as the point x, y moves along the curve C with parametric equations x sin 2t, y cos t. (See Figure 1.) In particular, when t 0, the point x, y is 0, 1 and dzdt 6 is the rate of increase as we move along the curve C through 0, 1. If, for instance, z Tx, y x 2 y 3xy 4 represents the temperature at the point x, y, then the composite function z Tsin 2t, cos t represents the temperature at points on C and the derivative dzdt represents the rate at which the temperature changes along C. V EXAMPLE 2 The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by the equation PV 8.31T.
SECTION 11.5
THE CHAIN RULE
■
627
Find the rate at which the pressure is changing when the temperature is 300 K and increasing at a rate of 0.1 Ks and the volume is 100 L and increasing at a rate of 0.2 Ls. SOLUTION If t represents the time elapsed in seconds, then at the given instant we have T 300, dTdt 0.1, V 100, dVdt 0.2. Since
P 8.31
T V
the Chain Rule gives dP &P dT &P dV 8.31 dT 8.31T dV dt &T dt &V dt V dt V 2 dt
8.31 8.31300 0.1 0.2 0.04155 100 100 2
The pressure is decreasing at a rate of about 0.042 kPas.
■
We now consider the situation where z f x, y but each of x and y is a function of two variables s and t : x ts, t, y hs, t. Then z is indirectly a function of s and t and we wish to find &z&s and &z&t. Recall that in computing &z&t we hold s fixed and compute the ordinary derivative of z with respect to t. Therefore, we can apply Theorem 2 to obtain &z &z &x &z &y &t &x &t &y &t A similar argument holds for &z&s and so we have proved the following version of the Chain Rule. 3 THE CHAIN RULE (CASE 2) Suppose that z f x, y is a differentiable function of x and y, where x ts, t and y hs, t are differentiable functions of s and t. Then
&z &z &x &z &y &s &x &s &y &s
&z &z &x &z &y &t &x &t &y &t
EXAMPLE 3 If z e x sin y, where x st 2 and y s 2t, find &z&s and &z&t. SOLUTION Applying Case 2 of the Chain Rule, we get
&z &z &x &z &y e x sin yt 2 e x cos y2st &s &x &s &y &s 2
2
t 2e st sins 2t 2ste st coss 2t &z &x &z &y &z e x sin y2st e x cos ys 2 &t &x &t &y &t 2
2
2ste st sins 2t s 2e st coss 2t
■
628
■
CHAPTER 11
PARTIAL DERIVATIVES
Case 2 of the Chain Rule contains three types of variables: s and t are independent variables, x and y are called intermediate variables, and z is the dependent variable. Notice that Theorem 3 has one term for each intermediate variable and each of these terms resembles the one-dimensional Chain Rule in Equation 1. To remember the Chain Rule it’s helpful to draw the tree diagram in Figure 2. We draw branches from the dependent variable z to the intermediate variables x and y to indicate that z is a function of x and y. Then we draw branches from x and y to the independent variables s and t. On each branch we write the corresponding partial derivative. To find &z&s we find the product of the partial derivatives along each path from z to s and then add these products:
z z x
z y
x
x s
x t
s
y
y s
t
y t
s
t
&z &z &x &z &y &s &x &s &y &s
FIGURE 2
Similarly, we find &z&t by using the paths from z to t. Now we consider the general situation in which a dependent variable u is a function of n intermediate variables x 1 , . . . , x n , each of which is, in turn, a function of m independent variables t1 , . . . , tm . Notice that there are n terms, one for each intermediate variable. The proof is similar to that of Case 1. 4 THE CHAIN RULE (GENERAL VERSION) Suppose that u is a differentiable function of the n variables x 1 , x 2 , . . . , x n and each x j is a differentiable function of the m variables t1 , t2 , . . . , tm . Then u is a function of t1 , t2 , . . . , tm and
&u &u &x 1 &u &x 2 &u &x n &ti &x 1 &ti &x 2 &ti &x n &ti for each i 1, 2, . . . , m. V EXAMPLE 4 Write out the Chain Rule for the case where w f x, y, z, t and x xu, v, y yu, v, z zu, v, and t tu, v.
SOLUTION We apply Theorem 4 with n 4 and m 2. Figure 3 shows the tree
w x v
u
z
y u
v
u
diagram. Although we haven’t written the derivatives on the branches, it’s understood that if a branch leads from y to u, then the partial derivative for that branch is &y&u. With the aid of the tree diagram we can now write the required expressions:
t v
u
&w &w &x &w &y &w &z &w &t &u &x &u &y &u &z &u &t &u
v
FIGURE 3
&w &w &x &w &y &w &z &w &t &v &x &v &y &v &z &v &t &v If u x 4 y y 2 z 3, where x rse t, y rs 2e t, and z r 2s sin t, find the value of &u&s when r 2, s 1, t 0. V EXAMPLE 5
u x r
s
y t
FIGURE 4
r
s
SOLUTION With the help of the tree diagram in Figure 4, we have
z t
r
s
t
&u &u &x &u &y &u &z &s &x &s &y &s &z &s 4x 3 yre t x 4 2yz 3 2rset 3y 2z 2 r 2 sin t
■
SECTION 11.5
THE CHAIN RULE
■
629
When r 2, s 1, and t 0, we have x 2, y 2, and z 0, so &u 642 164 00 192 &s
■
EXAMPLE 6 If ts, t f s 2 t 2, t 2 s 2 and f is differentiable, show that t
satisfies the equation t
&t &t s 0 &s &t
SOLUTION Let x s 2 t 2 and y t 2 s 2. Then ts, t f x, y and the Chain
Rule gives &t &f &x &f &y &f &f 2s 2s &s &x &s &y &s &x &y &t &f &x &f &y &f &f 2t 2t &t &x &t &y &t &x &y Therefore t
&t &t &f &f s 2st 2st &s &t &x &y
2st
&f &f 2st &x &y
0
■
EXAMPLE 7 If z f x, y has continuous second-order partial derivatives and
x r 2 s 2 and y 2rs, find (a) &z&r and (b) &2z&r 2.
SOLUTION
(a) The Chain Rule gives &z &z &x &z &y &z &z 2r 2s &r &x &r &y &r &x &y (b) Applying the Product Rule to the expression in part (a), we get
5
2
r
& &r
y
s r
FIGURE 5
&z & 2r &x &r
&z &x
2s
& &r
&z &y
But, using the Chain Rule again (see Figure 5), we have
z x
x
&2z & &z &z 2r 2s &r 2 &r &x &y
s
& &r
&z &x
& &x
&z &x
&x & &r &y
&z &y
& &x
&z &y
&x & &r &y
&z &x
&y &2z &2z 2 2r 2s &r &x &y &x
&z &y
&y &2z &2z 2r 2 2s &r &x &y &y
Putting these expressions into Equation 5 and using the equality of the mixed secondorder derivatives, we obtain
&2z &z &2z &2z 2r 2r 2 2s 2 2 &r &x &x &y &x 2
2s 2r
&z &2z &2z &2z 4r 2 2 8rs 4s 2 2 &x &x &x &y &y
&2z &2z 2s 2 &x &y &y
■
630
■
CHAPTER 11
PARTIAL DERIVATIVES
IMPLICIT DIFFERENTIATION
The Chain Rule can be used to give a more complete description of the process of implicit differentiation that was introduced in Sections 2.6 and 11.3. We suppose that an equation of the form Fx, y 0 defines y implicitly as a differentiable function of x, that is, y f x, where Fx, f x 0 for all x in the domain of f . If F is differentiable, we can apply Case 1 of the Chain Rule to differentiate both sides of the equation Fx, y 0 with respect to x. Since both x and y are functions of x, we obtain &F dx &F dy 0 &x dx &y dx But dxdx 1, so if &F&y 0 we solve for dydx and obtain
6
&F dy &x Fx dx &F Fy &y
To derive this equation we assumed that Fx, y 0 defines y implicitly as a function of x. The Implicit Function Theorem, proved in advanced calculus, gives conditions under which this assumption is valid. It states that if F is defined on a disk containing a, b, where Fa, b 0, Fy a, b 0, and Fx and Fy are continuous on the disk, then the equation Fx, y 0 defines y as a function of x near the point a, b and the derivative of this function is given by Equation 6. EXAMPLE 8 Find y if x 3 y 3 6xy. SOLUTION The given equation can be written as
Fx, y x 3 y 3 6xy 0 so Equation 6 gives ■ The solution to Example 8 should be compared to the one in Example 2 in Section 2.6.
dy Fx 3x 2 6y x 2 2y 2 2 dx Fy 3y 6x y 2x
■
Now we suppose that z is given implicitly as a function z f x, y by an equation of the form Fx, y, z 0. This means that Fx, y, f x, y 0 for all x, y in the domain of f . If F and f are differentiable, then we can use the Chain Rule to differentiate the equation Fx, y, z 0 as follows: &F &y &F &z &F &x 0 &x &x &y &x &z &x
But
& x 1 &x
and
& y 0 &x
SECTION 11.5
THE CHAIN RULE
■
631
so this equation becomes &F &F &z 0 &x &z &x If &F&z 0, we solve for &z&x and obtain the first formula in Equations 7. The formula for &z&y is obtained in a similar manner. &F &z &x &x &F &z
7
&F &z &y &y &F &z
Again, a version of the Implicit Function Theorem gives conditions under which our assumption is valid. If F is defined within a sphere containing a, b, c, where Fa, b, c 0, Fza, b, c 0, and Fx , Fy , and Fz are continuous inside the sphere, then the equation Fx, y, z 0 defines z as a function of x and y near the point a, b, c and this function is differentiable, with partial derivatives given by (7). EXAMPLE 9 Find
&z &z and if x 3 y 3 z 3 6xyz 1. &x &y
SOLUTION Let Fx, y, z x 3 y 3 z 3 6xyz 1. Then, from Equations 7, we
have &z Fx 3x 2 6yz x 2 2yz 2 2 &x Fz 3z 6xy z 2xy
■ The solution to Example 9 should be compared to the one in Example 4 in Section 11.3.
11.5 1– 4
■
x t,
2. z x lnx 2y, 3. w xe ,
4. w x y yz ,
xe,
2
■
5– 8
■
■
■
■
6. z xy ,
■
x se ,
7. z e cos ,
r st,
8. z sin tan , ■
x s t,
t
r
■
■
z e cos t t
■
■
■
■
■
y st
to calculate tu0, 0 and tv0, 0.
t
y 1 se
ss 2 t 2
3s t, s t ■
differentiable, u1, 0 2, us1, 0 2, u t 1, 0 6, v1, 0 3, vs1, 0 5, vt 1, 0 4, Fu2, 3 1, and Fv2, 3 10. Find Ws 1, 0 and Wt 1, 0.
11. Suppose f is a differentiable function of x and y, and tu, v f e u sin v, e u cos v. Use the table of values
Use the Chain Rule to find &z&s and &z&t.
5. z x 2 x y y 2,
■
t
■
10. Let Ws, t Fus, t, vs, t, where F, u, and v are
z 1 2t
y e sin t,
t
t3 2, t3 5, h3 7, h3 4, fx 2, 7 6, and fy 2, 7 8, find dzdt when t 3.
y cos t
y 1 t,
2
9. If z f x, y, where f is differentiable, x tt, y ht,
y st
x sin t,
xt ,
yz
■
EXERCISES
Use the Chain Rule to find dzdt or dwdt.
1. z sin x cos y,
■
&z Fy 3y 2 6xz y 2 2xz 2 2 &y Fz 3z 6xy z 2xy
■
■
■
■
■
■
■
f
t
fx
fy
0, 0
3
6
4
8
1, 2
6
3
2
5
632
■
CHAPTER 11
PARTIAL DERIVATIVES
12. Suppose f is a differentiable function of x and y, and
estimate that the average temperature is rising at a rate of 0.15°Cyear and rainfall is decreasing at a rate of 0.1 cmyear. They also estimate that, at current production levels, &W&T 2 and &W&R 8. (a) What is the significance of the signs of these partial derivatives? (b) Estimate the current rate of change of wheat production, dWdt.
tr, s f 2r s, s 2 4r. Use the table of values in Exercise 11 to calculate tr 1, 2 and ts 1, 2. 13–16 ■ Use a tree diagram to write out the Chain Rule for the given case. Assume all functions are differentiable. 13. u f x, y,
where x xr, s, t, y yr, s, t where x xt, u, y yt, u, z zt, u
14. w f x, y, z,
31. The speed of sound traveling through ocean water with
15. v f p, q, r,
salinity 35 parts per thousand has been modeled by the equation
where p px, y, z, q qx, y, z, r rx, y, z
16. u f s, t,
where s sw, x, y, z, t tw, x, y, z
■
■
■
■
■
■
■
■
■
■
■
C 1449.2 4.6T 0.055T 2 0.00029T 3 0.016D
■
where C is the speed of sound (in meters per second), T is the temperature (in degrees Celsius), and D is the depth below the ocean surface (in meters). A scuba diver began a leisurely dive into the ocean water; the diver’s depth and surrounding water temperature over time are recorded in the following graphs. Estimate the rate of change (with respect to time) of the speed of sound through the ocean water experienced by the diver 20 minutes into the dive. What are the units?
Use the Chain Rule to find the indicated partial derivatives. 17–21
■
17. z x 2 xy 3,
x u v 2 w 3,
&z &z &z , , &u &v &w
when u 2, v 1, w 0
18. u sr 2 s 2 ,
r y x cos t,
&u &u &u , , &x &y &t
19. R lnu v w , u x 2y, v 2x y, 2
20. M xe yz ,
x 2u v,
&u &u &u , , &p &r & 22–24
■
■
y u v,
x pr cos ,
21. u x yz,
■
w 2xy;
z u v;
■
■
y pr sin ,
■
2
■
■
■
■
■
■
25–28
■
■
■
■
23. sxy 1 x 2 y ■
2
■
■
■
■
■
■
■
■
28. yz lnx z ■
■
12
10
10
5
8 20
30
40
t (min)
10
20
30
40 t (min)
33. The length 艎, width w, and height h of a box change with
time. At a certain instant the dimensions are 艎 1 m and
■
■
while h is decreasing at a rate of 3 ms. At that instant find the rates at which the following quantities are changing. (a) The volume (b) The surface area (c) The length of a diagonal 34. The voltage V in a simple electrical circuit is slowly
26. x yz cosx y z
2
27. x z arctan yz ■
■
Use Equations 7 to find &z&x and &z&y.
25. x y z 3x yz 2
■
14
15
w h 2 m, and 艎 and w are increasing at a rate of 2 ms
24. sin x cos y sin x cos y ■
20
1.8 ins while its height is decreasing at a rate of 2.5 ins. At what rate is the volume of the cone changing when the radius is 120 in. and the height is 140 in.?
z p r;
Use Equation 6 to find dydx.
22. y 5 x 2 y 3 1 ye x
T 16
32. The radius of a right circular cone is increasing at a rate of
when p 2, r 3, 0 ■
D
10
when u 3, v 1
2
■
2
when x y 1 2
&M &M , &u &v
s x y sin t ;
when x 1, y 2, t 0
2
&R &R , &x &y
y u ve w;
■
■
29. The temperature at a point x, y is Tx, y, measured in
degrees Celsius. A bug crawls so that its position after t seconds is given by x s1 t , y 2 13 t , where x and y are measured in centimeters. The temperature function satisfies Tx 2, 3 4 and Ty 2, 3 3. How fast is the temperature rising on the bug’s path after 3 seconds? 30. Wheat production in a given year, W, depends on the
average temperature T and the annual rainfall R. Scientists
■
decreasing as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohm’s Law, V IR, to find how the current I is changing at the moment when R 400 , I 0.08 A, dVdt 0.01 Vs, and dRdt 0.03 s. 35. The pressure of 1 mole of an ideal gas is increasing at a rate
of 0.05 kPas and the temperature is increasing at a rate of 0.15 Ks. Use the equation in Example 2 to find the rate of change of the volume when the pressure is 20 kPa and the temperature is 320 K. 36. If a sound with frequency fs is produced by a source traveling along a line with speed vs and an observer is traveling
SECTION 11.6
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
with speed vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is
c vo c vs
fo
is a solution of the wave equation &2z &2z a2 2 2 &t &x
fs
[Hint: Let u x at, v x at.]
where c is the speed of sound, about 332 ms. (This is the Doppler effect.) Suppose that, at a particular moment, you are in a train traveling at 34 ms and accelerating at 1.2 ms 2. A train is approaching you from the opposite direction on the other track at 40 ms, accelerating at 1.4 ms 2, and sounds its whistle, which has a frequency of 460 Hz. At that instant, what is the perceived frequency that you hear and how fast is it changing? 37– 40
■
&z &x
&z &y
&z &r
&z &
1 r2
&u &x
2
&u &y
2
2s
e
39. If z f x y, show that
2
(Compare with Example 7.)
&2z &2z 1 &2z 1 &z &2z 2 2 2 &x &y &r r 2 & 2 r &r 46. Suppose z f x, y, where x ts, t and y hs, t.
&u &s
&u &t
(a) Show that
2
&2z &2z 2 &t &x 2
&z &z 0. &x &y
■
■
■
■
&z &y
2
■
■
2
■
■
■
■
■
Assume that all the given functions have continuous second-order partial derivatives. 41– 46
■
■
■
■
&y &t
2
■
■
■
■
■
■
■
■
■
47. Suppose that the equation Fx, y, z 0 implicitly defines
each of the three variables x, y, and z as functions of the other two: z f x, y, y tx, z, x h y, z. If F is differentiable and Fx , Fy , and Fz are all nonzero, show that
41. Show that any function of the form
&z &x &y 1 &x &y &z
z f x at tx at
11.6
(b) Find a similar formula for &2z&s &t.
&z &z &s &t
■
&x 2 &2z &x &y &2z 2 &t &x &y &t &t &y 2 2 2 &z & x &z & y &x &t 2 &y &t 2
40. If z f x, y, where x s t and y s t, show that
&z &x
43. If z f x, y, where x r 2 s 2, y 2rs, find &2z&r &s.
2
38. If u f x, y, where x e s cos t and y e s sin t, show
that
&2u &2u &2u &2u e2s 2 2 2 2 &x &y &s &t
45. If z f x, y, where x r cos , y r sin , show that
&z&r and &z& and (b) show that
2
that
(a) &z&r, (b) &z&, and (c) &2z&r &.
37. If z f x, y, where x r cos and y r sin , (a) find 2
42. If u f x, y, where x e s cos t and y e s sin t, show
44. If z f x, y, where x r cos , y r sin , find
Assume that all the given functions are differentiable.
2
633
■
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR Recall that if z f x, y, then the partial derivatives fx and fy are defined as fx x0 , y0 lim
f x0 h, y0 f x0 , y0 h
fy x0 , y0 lim
f x0 , y0 h f x0 , y0 h
hl0
1 hl0
and represent the rates of change of z in the x- and y-directions, that is, in the directions of the unit vectors i and j.
634
■
CHAPTER 11
PARTIAL DERIVATIVES
Suppose that we now wish to find the rate of change of z at x 0 , y0 in the direction of an arbitrary unit vector u a, b. (See Figure 1.) To do this we consider the surface S with equation z f x, y (the graph of f ) and we let z0 f x 0 , y0 . Then the point Px 0 , y0 , z0 lies on S. The vertical plane that passes through P in the direction of u intersects S in a curve C. (See Figure 2.) The slope of the tangent line T to C at the point P is the rate of change of z in the direction of u.
y
u
sin ¨
¨ (x¸, y¸)
cos ¨ z
0
x
FIGURE 1
T
P(x¸, y¸, z¸)
A unit vector u=ka, bl=kcos ¨, sin ¨l Q(x, y, z)
Visual 11.6A animates Figure 2 by rotating u and therefore T.
y
FIGURE 2
x
If Qx, y, z is another point on C and P, Q are the projections of P, Q on the B Q is parallel to u and so xy-plane, then the vector P B P Q hu ha, hb for some scalar h. Therefore, x x 0 ha, y y0 hb, so x x 0 ha, y y0 hb, and z z z0 f x 0 ha, y0 hb f x 0 , y0 h h h If we take the limit as h l 0, we obtain the rate of change of z (with respect to distance) in the direction of u, which is called the directional derivative of f in the direction of u. 2 DEFINITION The directional derivative of f at x 0 , y0 in the direction of a unit vector u a, b is
Du f x 0 , y0 lim
hl0
if this limit exists.
f x 0 ha, y0 hb f x 0 , y0 h
SECTION 11.6
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
■
635
By comparing Definition 2 with Equations (1), we see that if u i 1, 0 , then Di f fx and if u j 0, 1 , then Dj f fy . In other words, the partial derivatives of f with respect to x and y are just special cases of the directional derivative. When we compute the directional derivative of a function defined by a formula, we generally use the following theorem. 3 THEOREM If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector u a, b and
Du f x, y fx x, ya fy x, yb PROOF If we define a function t of the single variable h by
th f x 0 ha, y0 hb then by the definition of a derivative we have 4
t0 lim
hl0
th t0 f x 0 ha, y0 hb f x 0 , y0 lim hl0 h h
Du f x 0 , y0 On the other hand, we can write th f x, y, where x x 0 ha, y y0 hb, so the Chain Rule (Theorem 11.5.2) gives th
&f dx &f dy fx x, ya fy x, yb &x dh &y dh
If we now put h 0, then x x 0 , y y0 , and t0 fx x 0 , y0 a fy x 0 , y0 b
5
Comparing Equations 4 and 5, we see that Du f x 0 , y0 fx x 0 , y0 a fy x 0 , y0 b
■ The directional derivative Du f 1, 2 in Example 1 represents the rate of change of z in the direction of u. This is the slope of the tangent line to the curve of intersection of the surface z x 3 3xy 4y 2 and the vertical plane through 1, 2, 0 in the direction of u shown in Figure 3.
■
If the unit vector u makes an angle with the positive x-axis (as in Figure 1), then we can write u cos , sin and the formula in Theorem 3 becomes Du f x, y fx x, y cos fy x, y sin
6
EXAMPLE 1 Find the directional derivative Du f x, y if f x, y x 3 3xy 4y 2
z
and u is the unit vector given by angle 6. What is Du f 1, 2?
SOLUTION Formula 6 gives
Du f x, y fx x, y cos 0 x
FIGURE 3
y
(1, 2, 0) π 6
u
1 2
[3 s3 x
2
fy x, y sin 6 6
3x 2 3y
s3 3x 8y 12 2
]
3x (8 3s3 )y
Therefore
[
]
Du f 1, 2 12 3s3 12 31 (8 3s3 )2
13 3s3 2
■
636
■
CHAPTER 11
PARTIAL DERIVATIVES
THE GRADIENT VECTOR
Notice from Theorem 3 that the directional derivative can be written as the dot product of two vectors: 7
Du f x, y fx x, ya fy x, yb fx x, y, fy x, y a, b fx x, y, fy x, y u
The first vector in this dot product occurs not only in computing directional derivatives but in many other contexts as well. So we give it a special name (the gradient of f ) and a special notation (grad f or (f , which is read “del f ”).
If f is a function of two variables x and y, then the gradient of f is the vector function (f defined by 8 DEFINITION
( f x, y fx x, y, fy x, y
&f &f i j &x &y
EXAMPLE 2 If f x, y sin x e x y, then
(f x, y fx , fy cos x ye x y, xe x y (f 0, 1 2, 0
and
■
With this notation for the gradient vector, we can rewrite the expression (7) for the directional derivative as 9
Du f x, y (f x, y u
This expresses the directional derivative in the direction of u as the scalar projection of the gradient vector onto u. Find the directional derivative of the function f x, y x 2 y 3 4y at the point 2, 1 in the direction of the vector v 2 i 5j. V EXAMPLE 3
SOLUTION We first compute the gradient vector at 2, 1:
(f x, y 2xy 3 i 3x 2 y 2 4j (f 2, 1 4 i 8 j
Note that v is not a unit vector, but since v s29 , the unit vector in the direction of v is u
v 2 5 i j v s29 s29
SECTION 11.6 ■ The gradient vector (f 2, 1 in Example 3 is shown in Figure 4 with initial point 2, 1. Also shown is the vector v that gives the direction of the directional derivative. Both of these vectors are superimposed on a contour plot of the graph of f .
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
■
637
Therefore, by Equation 9, we have Du f 2, 1 (f 2, 1 u 4 i 8 j
y
2 5 i j s29 s29
4 2 8 5 32 s29 s29
■
FUNCTIONS OF THREE VARIABLES ±f(2, _1)
v x
For functions of three variables we can define directional derivatives in a similar manner. Again Du f x, y, z can be interpreted as the rate of change of the function in the direction of a unit vector u.
(2, _1) 10 DEFINITION The directional derivative of f at x 0 , y0 , z0 in the direction of a unit vector u a, b, c is
Du f x 0 , y0 , z0 lim
FIGURE 4
hl0
f x 0 ha, y0 hb, z0 hc f x 0 , y0 , z0 h
if this limit exists. If we use vector notation, then we can write both definitions (2 and 10) of the directional derivative in the compact form
11
Du f x 0 lim
hl0
f x 0 hu f x 0 h
where x 0 x 0 , y0 if n 2 and x 0 x 0 , y0 , z0 if n 3. This is reasonable since the vector equation of the line through x 0 in the direction of the vector u is given by x x 0 tu (Equation 10.5.1) and so f x 0 hu represents the value of f at a point on this line. If f x, y, z is differentiable and u a, b, c , then the same method that was used to prove Theorem 3 can be used to show that 12
Du f x, y, z fx x, y, za fy x, y, zb fzx, y, zc
For a function f of three variables, the gradient vector, denoted by (f or grad f , is (f x, y, z fx x, y, z, fy x, y, z, fzx, y, z or, for short,
13
(f fx , fy , fz
&f &f &f i j k &x &y &z
638
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CHAPTER 11
PARTIAL DERIVATIVES
Then, just as with functions of two variables, Formula 12 for the directional derivative can be rewritten as Du f x, y, z (f x, y, z u
14
V EXAMPLE 4 If f x, y, z x sin yz, (a) find the gradient of f and (b) find the directional derivative of f at 1, 3, 0 in the direction of v i 2 j k.
SOLUTION
(a) The gradient of f is (f x, y, z fx x, y, z, fy x, y, z, fzx, y, z sin yz, xz cos yz, xy cos yz (b) At 1, 3, 0 we have (f 1, 3, 0 0, 0, 3 . The unit vector in the direction of v i 2 j k is 1 2 1 u i j k s6 s6 s6 Therefore, Equation 14 gives Du f 1, 3, 0 ( f 1, 3, 0 u 3k
3
1 2 1 i j k s6 s6 s6
1 s6
3 2
■
MAXIMIZING THE DIRECTIONAL DERIVATIVE
Suppose we have a function f of two or three variables and we consider all possible directional derivatives of f at a given point. These give the rates of change of f in all possible directions. We can then ask the questions: In which of these directions does f change fastest and what is the maximum rate of change? The answers are provided by the following theorem.
Visual 11.6B provides visual confirmation of Theorem 15.
15 THEOREM Suppose f is a differentiable function of two or three variables. The maximum value of the directional derivative Du f x is (f x and it occurs when u has the same direction as the gradient vector (f x.
PROOF From Equation 9 or 14 we have
u cos (f cos
Du f (f u (f
where is the angle between (f and u. The maximum value of cos is 1 and this occurs when 0. Therefore, the maximum value of Du f is (f and it occurs when 0, that is, when u has the same direction as (f . ■
SECTION 11.6
y
■
639
EXAMPLE 5
Q
(a) If f x, y xe y, find the rate of change of f at the point P2, 0 in the direction from P to Q( 12, 2). (b) In what direction does f have the maximum rate of change? What is this maximum rate of change?
2
1
±f(2, 0) 0
(a) We first compute the gradient vector:
3 x
P
1
SOLUTION
(f x, y fx , fy e y, xe y
FIGURE 5
(f 2, 0 1, 2 l
The unit vector in the direction of PQ 1.5, 2 is u 35 , 45 , so the rate of change of f in the direction from P to Q is
■ At 2, 0 the function in Example 5 increases fastest in the direction of the gradient vector ( f 2, 0 1, 2. Notice from Figure 5 that this vector appears to be perpendicular to the level curve through 2, 0. Figure 6 shows the graph of f and the gradient vector.
Du f 2, 0 (f 2, 0 u 1, 2 35 , 45 1( 5 ) 2( 5 ) 1 3
4
(b) According to Theorem 15, f increases fastest in the direction of the gradient vector (f 2, 0 1, 2 . The maximum rate of change is
20
(f 2, 0 1, 2 s5
15 z 10
■
EXAMPLE 6 Suppose that the temperature at a point x, y, z in space is given by
5 0
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
0
1 x
2
3 0
1 y
2
FIGURE 6
Tx, y, z 801 x 2 2y 2 3z 2 , where T is measured in degrees Celsius and x, y, z in meters. In which direction does the temperature increase fastest at the point 1, 1, 2? What is the maximum rate of increase? SOLUTION The gradient of T is
(T
&T &T &T i j k &x &y &z
160x 320y 480z i j k 1 x 2 2y 2 3z 2 2 1 x 2 2y 2 3z 2 2 1 x 2 2y 2 3z 2 2
160 x i 2y j 3z k 1 x 2 2y 2 3z 2 2 At the point 1, 1, 2 the gradient vector is 5 (T1, 1, 2 160 256 i 2 j 6 k 8 i 2 j 6 k
By Theorem 15 the temperature increases fastest in the direction of the gradient vector (T 1, 1, 2 58 i 2 j 6 k or, equivalently, in the direction of i 2 j 6 k or the unit vector i 2 j 6 ks41. The maximum rate of increase is the length of the gradient vector:
(T1, 1, 2 i 2 j 6 k 5 8
5s41 8
Therefore, the maximum rate of increase of temperature is 5s418 4 Cm.
■
640
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CHAPTER 11
PARTIAL DERIVATIVES
TANGENT PLANES TO LEVEL SURFACES
Suppose S is a surface with equation Fx, y, z k, that is, it is a level surface of a function F of three variables, and let Px 0 , y0 , z0 be a point on S. Let C be any curve that lies on the surface S and passes through the point P. Recall from Section 10.7 that the curve C is described by a continuous vector function rt xt, yt, zt . Let t0 be the parameter value corresponding to P ; that is, rt0 x 0 , y0 , z0 . Since C lies on S, any point xt, yt, zt must satisfy the equation of S, that is, 16
F ( xt, yt, zt) k
If x, y, and z are differentiable functions of t and F is also differentiable, then we can use the Chain Rule to differentiate both sides of Equation 16 as follows: 17
&F dx &F dy &F dz 0 &x dt &y dt &z dt
But, since (F Fx , Fy , Fz and rt xt, yt, zt , Equation 17 can be written in terms of a dot product as (F rt 0 In particular, when t t0 we have rt0 x 0 , y0 , z0 , so 18 z
± (x¸, y¸, z¸) ±F tangent plane
Equation 18 says that the gradient vector at P, (Fx0 , y0 , z0 , is perpendicular to the tangent vector rt0 to any curve C on S that passes through P. (See Figure 7.) If (Fx0 , y0 , z0 0, it is therefore natural to define the tangent plane to the level surface Fx, y, z k at Px 0 , y0 , z0 as the plane that passes through P and has normal vector (Fx0 , y0 , z0 . Using the standard equation of a plane (Equation 10.5.7), we can write the equation of this tangent plane as
y x
(Fx0, y0, z0 rt0 0
19
Fx x 0 , y0 , z0 x x 0 Fy x 0 , y0 , z0 y y0 Fzx 0 , y0 , z0 z z0 0
FIGURE 7
The normal line to S at P is the line passing through P and perpendicular to the tangent plane. The direction of the normal line is therefore given by the gradient vector (Fx0 , y0 , z0 and so, by Equation 10.5.3, its symmetric equations are 20
x x0 y y0 z z0 Fx x0 , y0 , z0 Fy x0 , y0 , z0 Fzx0 , y0 , z0
In the special case in which the equation of a surface S is of the form z f x, y (that is, S is the graph of a function f of two variables), we can rewrite the equation as Fx, y, z f x, y z 0
SECTION 11.6
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
■
641
and regard S as a level surface (with k 0) of F. Then Fx x 0 , y0 , z0 fx x 0 , y0 Fy x 0 , y0 , z0 fy x 0 , y0 Fzx 0 , y0 , z0 1 so Equation 19 becomes fx x 0 , y0 x x 0 fy x 0 , y0 y y0 z z0 0 which is equivalent to Equation 11.4.2. Thus our new, more general, definition of a tangent plane is consistent with the definition that was given for the special case of Section 11.4. V EXAMPLE 7 Find the equations of the tangent plane and normal line at the point 2, 1, 3 to the ellipsoid
x2 z2 y2 3 4 9 SOLUTION The ellipsoid is the level surface (with k 3) of the function
Fx, y, z ■ Figure 8 shows the ellipsoid, tangent plane, and normal line in Example 7.
x2 z2 y2 4 9
Therefore, we have Fx x, y, z
x 2
Fy x, y, z 2y
Fzx, y, z
2z 9
4
Fx 2, 1, 3 1
2 0
Fy 2, 1, 3 2
Fz2, 1, 3 23
Then Equation 19 gives the equation of the tangent plane at 2, 1, 3 as
z 2
1x 2 2 y 1 23 z 3 0
4
which simplifies to 3x 6y 2z 18 0. By Equation 20, symmetric equations of the normal line are
6
y
0 2
FIGURE 8
2
0 2 x
x2 y1 z3 1 2 23
■
SIGNIFICANCE OF THE GRADIENT VECTOR
We now summarize the ways in which the gradient vector is significant. We first consider a function f of three variables and a point Px 0 , y0 , z0 in its domain. On the one hand, we know from Theorem 15 that the gradient vector (f x0 , y0, z0 gives the direction of fastest increase of f . On the other hand, we know that (f x0 , y0 , z0 is orthog-
642
■
CHAPTER 11
PARTIAL DERIVATIVES
onal to the level surface S of f through P. (Refer to Figure 7.) These two properties are quite compatible intuitively because as we move away from P on the level surface S, the value of f does not change at all. So it seems reasonable that if we move in the perpendicular direction, we get the maximum increase. In like manner we consider a function f of two variables and a point Px 0 , y0 in its domain. Again the gradient vector (f x0 , y0 gives the direction of fastest increase of f . Also, by considerations similar to our discussion of tangent planes, it can be shown that ( f x0 , y0 is perpendicular to the level curve f x, y k that passes through P. Again this is intuitively plausible because the values of f remain constant as we move along the curve. (See Figure 9.) y
±f(x¸, y¸)
P(x¸, y¸)
level curve f(x, y)=k 0
300 200
curve of steepest ascent
x
FIGURE 9
100
FIGURE 10
If we consider a topographical map of a hill and let f x, y represent the height above sea level at a point with coordinates x, y, then a curve of steepest ascent can be drawn as in Figure 10 by making it perpendicular to all of the contour lines. This phenomenon can also be noticed in Figure 11 in Section 11.1, where Lonesome Creek follows a curve of steepest descent.
11.6
EXERCISES
Find the directional derivative of f at the given point in the direction indicated by the angle .
1–2
1. f x, y s5x 4y , 2. f x, y x sinxy, ■
■
3–6
■ Find the directional derivative of the function at the given point in the direction of the vector v.
■
■
■
6
4, 1, 2, 0,
■
■
7–11
3 ■
■
■
■
■
■
8. f x, y lnx y ,
2, 1,
9. ts, t s e ,
vij
2
2, 0,
2 t
10. f x, y, z x y z,
■
3. f x, y 5xy 4x y, 2
4. f x, y y ln x,
3
■
u ,
P1, 3, u ,
■
■
■
■
■
■
■
■
y
■
■
1, 1, 2, v 2 j k ■
■
■
12. Use the figure to estimate Du f 2, 2. (2, 2)
u ±f (2, 2)
P1, 3, 1, u 27 , 37 , 67 ■
v 1, 2
4, 1, 1, v 1, 2, 3
11. tx, y, z x 2y 3z 32,
P3, 0, 2, u 23 , 23 , 13
6. f x, y, z sx yz , ■
P1, 2,
5 12 13 13
4 3 5 5
5. f x, y, z xe 2 yz,
■
3, 4, v 4, 3
2
(a) Find the gradient of f . (b) Evaluate the gradient at the point P. (c) Find the rate of change of f at P in the direction of the vector u.
■
7. f x, y 1 2x sy ,
■
0 ■
■
x
■
■
■
SECTION 11.6
13. Find the directional derivative of f x, y sxy at P2, 8
26. Suppose you are climbing a hill whose shape is given by the
14. Find the directional derivative of f x, y, z x 2 y 2 z 2
at P2, 1, 3 in the direction of the origin. 15–18 ■ Find the maximum rate of change of f at the given point and the direction in which it occurs.
16. f p, q qe
p
2, 4 pe q,
17. f x, y, z lnxy z , 2 3
0, 0 1, 2, 3
18. f x, y, z tanx 2y 3z, ■
■
■
■
■
■
5, 1, 1
■
■
■
■
■
643
(b) In which direction does V change most rapidly at P ? (c) What is the maximum rate of change at P ?
in the direction of Q5, 4.
15. f x, y y 2x,
■
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
■
19. (a) Show that a differentiable function f decreases most
rapidly at x in the direction opposite to the gradient vector, that is, in the direction of ( f x. (b) Use the result of part (a) to find the direction in which the function f x, y x 4 y x 2 y 3 decreases fastest at the point 2, 3. 20. Find the directions in which the directional derivative of
f x, y x 2 sin xy at the point (1, 0) has the value 1.
equation z 1000 0.005x 2 0.01y 2, where x, y, and z are measured in meters, and you are standing at a point with coordinates 60, 40, 966. The positive x-axis points east and the positive y-axis points north. (a) If you walk due south, will you start to ascend or descend? At what rate? (b) If you walk northwest, will you start to ascend or descend? At what rate? (c) In which direction is the slope largest? What is the rate of ascent in that direction? At what angle above the horizontal does the path in that direction begin? 27. Let f be a function of two variables that has continuous
partial derivatives and consider the points A1, 3, B3, 3, derivative of f at A in C1, 7, and D6, 15. The directional l the direction of the vector AB is 3 and the directional derivl ative at A in the direction of AC is 26. Find the directional l derivative of f at A in the direction of the vector AD . 28. For the given contour map draw the curves of steepest
ascent starting at P and at Q. Q
21. Find all points at which the direction of fastest change of
the function f x, y x 2 y 2 2 x 4y is i j. 60
22. Near a buoy, the depth of a lake at the point with coordi-
nates x, y is z 200 0.02x 2 0.001y 3, where x, y, and z are measured in meters. A fisherman in a small boat starts at the point 80, 60 and moves toward the buoy, which is located at 0, 0. Is the water under the boat getting deeper or shallower when he departs? Explain. 23. The temperature T in a metal ball is inversely proportional
to the distance from the center of the ball, which we take to be the origin. The temperature at the point 1, 2, 2 is 120 . (a) Find the rate of change of T at 1, 2, 2 in the direction toward the point 2, 1, 3. (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin. 24. The temperature at a point x, y, z is given by
Tx, y, z 200ex
2
3y 29z 2
where T is measured in C and x, y, z in meters. (a) Find the rate of change of temperature at the point P2, 1, 2 in the direction toward the point 3, 3, 3. (b) In which direction does the temperature increase fastest at P ? (c) Find the maximum rate of increase at P. 25. Suppose that over a certain region of space the electrical
potential V is given by Vx, y, z 5x 2 3xy xyz. (a) Find the rate of change of the potential at P3, 4, 5 in the direction of the vector v i j k.
20
50
30
40
P 29. Show that the operation of taking the gradient of a function has the given property. Assume that u and v are differen-
tiable functions of x and y and a, b are constants. (a) (au b v a (u b (v (b) (u v u (v v (u
(c) (
u v
v (u u (v
(d) (u n nu n1 (u
v2
30. Sketch the gradient vector ( f 4, 6 for the function f
whose level curves are shown. Explain how you chose the direction and length of this vector. y _5 6
(4, 6)
_3 _1
4
0 1
3
5
2
0
2
4
6
x
644
■
CHAPTER 11
PARTIAL DERIVATIVES
43. Show that the sum of the x-, y-, and z -intercepts of any
31–34 ■ Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point. 31. x 2y z yz 2, 2
2
2
32. x z 4 arctan yz, 33. z 1 xe y cos z, 34. yz lnx z, ■
■
; 35–36
■
2, 1, 1
44. Show that every normal line to the sphere
1 , 1, 1
x 2 y 2 z 2 r 2 passes through the center of the sphere.
1, 0, 0
45. Find parametric equations for the tangent line to the curve
0, 0, 1
■
■
■
■
■
■
■
■
46. (a) The plane y z 3 intersects the cylinder
■
35. x y yz zx 3, 36. x yz 6, ■
of intersection of the paraboloid z x 2 y 2 and the ellipsoid 4x 2 y 2 z 2 9 at the point 1, 1, 2.
■
Use a computer to graph the surface, the tangent plane, and the normal line on the same screen. Choose the domain carefully so that you avoid extraneous vertical planes. Choose the viewpoint so that you get a good view of all three objects.
■
tangent plane to the surface sx sy sz sc is a constant.
■
1, 1, 1
;
47. (a) Two surfaces are called orthogonal at a point of inter-
1, 2, 3 ■
■
x 2 y 2 5 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point 1, 2, 1. (b) Graph the cylinder, the plane, and the tangent line on the same screen.
■
■
■
■
■
■
section if their normal lines are perpendicular at that point. Show that surfaces with equations Fx, y, z 0 and Gx, y, z 0 are orthogonal at a point P where (F 0 and (G 0 if and only if
■
37. If f x, y x 2 4y 2, find the gradient vector ( f 2, 1 and
use it to find the tangent line to the level curve f x, y 8 at the point 2, 1. Sketch the level curve, the tangent line, and the gradient vector.
Fx Gx Fy Gy Fz Gz 0 at P (b) Use part (a) to show that the surfaces z 2 x 2 y 2 and x 2 y 2 z 2 r 2 are orthogonal at every point of intersection. Can you see why this is true without using calculus?
38. If tx, y x y 2, find the gradient vector (t3, 1 and
use it to find the tangent line to the level curve tx, y 2 at the point 3, 1. Sketch the level curve, the tangent line, and the gradient vector.
3 x y is continuous and 48. (a) Show that the function f x, y s
39. Show that the equation of the tangent plane to the ellipsoid
x 2a 2 y 2b 2 z 2c 2 1 at the point x 0 , y0 , z0 can be written as yy0 zz0 xx 0 1 2 2 a b c2 40. Find the points on the ellipsoid x 2 2y 2 3z 2 1 where
the tangent plane is parallel to the plane 3x y 3z 1.
41. Find the points on the hyperboloid x 2 y 2 2z 2 1
where the normal line is parallel to the line that joins the points 3, 1, 0 and 5, 3, 6.
;
the partial derivatives fx and fy exist at the origin but the directional derivatives in all other directions do not exist. (b) Graph f near the origin and comment on how the graph confirms part (a). 49. Suppose that the directional derivatives of f x, y are known
at a given point in two nonparallel directions given by unit vectors u and v. Is it possible to find ( f at this point? If so, how would you do it? 50. Show that if z f x, y is differentiable at x 0 x 0 , y0 ,
then
42. Show that the ellipsoid 3x 2 2y 2 z 2 9 and the sphere
x 2 y 2 z 2 8x 6y 8z 24 0 are tangent to each other at the point 1, 1, 2. (This means that they have a common tangent plane at the point.)
11.7
lim
x l x0
f x f x 0 ( f x 0 x x 0 0 x x0
[Hint: Use Definition 11.4.7 directly.]
MAXIMUM AND MINIMUM VALUES As we saw in Chapter 4, one of the main uses of ordinary derivatives is in finding maximum and minimum values. In this section we see how to use partial derivatives to locate maxima and minima of functions of two variables. In particular, in Example 5 we will see how to maximize the volume of a box without a lid if we have a fixed amount of cardboard to work with.
SECTION 11.7
z
absolute maximum
local maximum
MAXIMUM AND MINIMUM VALUES
■
645
Look at the hills and valleys in the graph of f shown in Figure 1. There are two points a, b where f has a local maximum, that is, where f a, b is larger than nearby values of f x, y. The larger of these two values is the absolute maximum. Likewise, f has two local minima, where f a, b is smaller than nearby values. The smaller of these two values is the absolute minimum.
y
x
absolute minimum
1 DEFINITION A function of two variables has a local maximum at a, b if f x, y f a, b when x, y is near a, b. [This means that f x, y f a, b for all points x, y in some disk with center a, b.] The number f a, b is called a local maximum value. If f x, y f a, b when x, y is near a, b, then f a, b is a local minimum value.
local minimum
FIGURE 1
If the inequalities in Definition 1 hold for all points x, y in the domain of f , then f has an absolute maximum (or absolute minimum) at a, b. ■ Notice that the conclusion of Theorem 2 can be stated in the notation of gradient vectors as (f a, b 0.
2 THEOREM If f has a local maximum or minimum at a, b and the firstorder partial derivatives of f exist there, then fx a, b 0 and fy a, b 0.
PROOF Let tx f x, b. If f has a local maximum (or minimum) at a, b, then
t has a local maximum (or minimum) at a, so ta 0 by Fermat’s Theorem (see Theorem 4.1.4). But ta fx a, b (see Equation 11.3.1) and so fx a, b 0. Similarly, by applying Fermat’s Theorem to the function G y f a, y, we obtain fy a, b 0. ■ If we put fx a, b 0 and fy a, b 0 in the equation of a tangent plane (Equation 11.4.2), we get z z0 . Thus the geometric interpretation of Theorem 2 is that if the graph of f has a tangent plane at a local maximum or minimum, then the tangent plane must be horizontal. A point a, b is called a critical point (or stationary point) of f if fx a, b 0 and fya, b 0, or if one of these partial derivatives does not exist. Theorem 2 says that if f has a local maximum or minimum at a, b, then a, b is a critical point of f . However, as in single-variable calculus, not all critical points give rise to maxima or minima. At a critical point, a function could have a local maximum or a local minimum or neither. z
EXAMPLE 1 Let f x, y x 2 y 2 2x 6y 14. Then
fx x, y 2x 2
These partial derivatives are equal to 0 when x 1 and y 3, so the only critical point is 1, 3. By completing the square, we find that
(1, 3, 4)
f x, y 4 x 12 y 32
0 x
FIGURE 2
z=≈+¥-2x-6y+14
fy x, y 2y 6
y
Since x 12 0 and y 32 0, we have f x, y 4 for all values of x and y. Therefore, f 1, 3 4 is a local minimum, and in fact it is the absolute minimum of f . This can be confirmed geometrically from the graph of f, which is the elliptic ■ paraboloid with vertex 1, 3, 4 shown in Figure 2.
646
■
CHAPTER 11
PARTIAL DERIVATIVES
EXAMPLE 2 Find the extreme values of f x, y y 2 x 2. SOLUTION Since fx 2x and fy 2y, the only critical point is 0, 0. Notice that for points on the x-axis we have y 0, so f x, y x 2 0 (if x 0). However, for points on the y-axis we have x 0, so f x, y y 2 0 (if y 0). Thus every disk with center 0, 0 contains points where f takes positive values as well as points where f takes negative values. Therefore, f 0, 0 0 can’t be an extreme value for f , so f has no extreme value. ■ z
x
FIGURE 3
z=¥-≈
y
Example 2 illustrates the fact that a function need not have a maximum or minimum value at a critical point. Figure 3 shows how this is possible. The graph of f is the hyperbolic paraboloid z y 2 x 2, which has a horizontal tangent plane (z 0) at the origin. You can see that f 0, 0 0 is a maximum in the direction of the x-axis but a minimum in the direction of the y-axis. Near the origin the graph has the shape of a saddle and so 0, 0 is called a saddle point of f . We need to be able to determine whether or not a function has an extreme value at a critical point. The following test, which is proved in Appendix B, is analogous to the Second Derivative Test for functions of one variable. 3 SECOND DERIVATIVES TEST Suppose the second partial derivatives of f are continuous on a disk with center a, b, and suppose that fx a, b 0 and fy a, b 0 [that is, a, b is a critical point of f ]. Let
D Da, b fxx a, b fyy a, b fx y a, b 2 (a) If D 0 and fxx a, b 0, then f a, b is a local minimum. (b) If D 0 and fxx a, b 0, then f a, b is a local maximum. (c) If D 0, then f a, b is not a local maximum or minimum. NOTE 1 In case (c) the point a, b is called a saddle point of f and the graph of f crosses its tangent plane at a, b. NOTE 2 If D 0, the test gives no information: f could have a local maximum or local minimum at a, b, or a, b could be a saddle point of f . NOTE 3 To remember the formula for D it’s helpful to write it as a determinant:
D
fxx fx y fxx fyy fx y 2 fyx fyy
Find the local maximum and minimum values and saddle points of f x, y x 4 y 4 4xy 1.
V EXAMPLE 3
SOLUTION We first locate the critical points:
fx 4x 3 4y
fy 4y 3 4x
Setting these partial derivatives equal to 0, we obtain the equations x3 y 0
and
y3 x 0
To solve these equations we substitute y x 3 from the first equation into the second one. This gives 0 x 9 x xx 8 1 xx 4 1x 4 1 xx 2 1x 2 1x 4 1
SECTION 11.7
MAXIMUM AND MINIMUM VALUES
■
647
so there are three real roots: x 0, 1, 1. The three critical points are 0, 0, 1, 1, and 1, 1. Next we calculate the second partial derivatives and Dx, y:
z
fxx 12x 2
fx y 4
fyy 12y 2
Dx, y fxx fyy fx y 2 144x 2 y 2 16
y x
FIGURE 4
z=x$+y$-4xy+1
Since D0, 0 16 0, it follows from case (c) of the Second Derivatives Test that the origin is a saddle point; that is, f has no local maximum or minimum at 0, 0. Since D1, 1 128 0 and fxx 1, 1 12 0, we see from case (a) of the test that f 1, 1 1 is a local minimum. Similarly, we have D1, 1 128 0 and fxx 1, 1 12 0, so f 1, 1 1 is also a local minimum. The graph of f is shown in Figure 4. ■ y
A contour map of the function f in Example 3 is shown in Figure 5. The level curves near 1, 1 and 1, 1 are oval in shape and indicate that as we move away from 1, 1 or 1, 1 in any direction the values of f are increasing. The level curves near 0, 0, on the other hand, resemble hyperbolas. They reveal that as we move away from the origin (where the value of f is 1), the values of f decrease in some directions but increase in other directions. Thus, the contour map suggests the presence of the minima and saddle point that we found in Example 3. ■
_0.5 0 0.5 0.9 1 1.1 1.5 2
x
3
FIGURE 5 V EXAMPLE 4 Find the shortest distance from the point 1, 0, 2 to the plane x 2y z 4.
In Module 11.7 you can use contour maps to estimate the locations of critical points.
SOLUTION The distance from any point x, y, z to the point 1, 0, 2 is
d sx 12 y 2 z 22 but if x, y, z lies on the plane x 2y z 4, then z 4 x 2y and so we have d sx 1 2 y 2 6 x 2y 2 . We can minimize d by minimizing the simpler expression d 2 f x, y x 12 y 2 6 x 2y2 By solving the equations fx 2x 1 26 x 2y 4x 4y 14 0 fy 2y 46 x 2y 4x 10y 24 0
Example 4 could also be solved using vectors. Compare with the methods of Section 10.5. ■
we find that the only critical point is ( 116, 53 ). Since fxx 4, fx y 4, and fyy 10, we have Dx, y fxx fy y fx y2 24 0 and fxx 0, so by the Second Derivatives
648
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CHAPTER 11
PARTIAL DERIVATIVES
Test f has a local minimum at ( 116, 53 ). Intuitively, we can see that this local minimum is actually an absolute minimum because there must be a point on the given plane that is closest to 1, 0, 2. If x 116 and y 53 , then d sx 12 y 2 6 x 2y2 s(56)2 (53)2 (56)2
5s6 6
The shortest distance from 1, 0, 2 to the plane x 2y z 4 is 5s6 6.
■
A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box. V EXAMPLE 5
SOLUTION Let the length, width, and height of the box (in meters) be x, y, and z, as shown in Figure 6. Then the volume of the box is
V xyz z x y FIGURE 6
We can express V as a function of just two variables x and y by using the fact that the area of the four sides and the bottom of the box is 2xz 2yz xy 12 Solving this equation for z, we get z 12 xy 2x y , so the expression for V becomes V xy
12 xy 12xy x 2 y 2 2x y 2x y
We compute the partial derivatives: &V y 212 2xy x 2 &x 2x y2
&V x 212 2xy y 2 &y 2x y2
If V is a maximum, then &V&x &V&y 0, but x 0 or y 0 gives V 0, so we must solve the equations 12 2xy x 2 0
12 2xy y 2 0
These imply that x 2 y 2 and so x y. (Note that x and y must both be positive in this problem.) If we put x y in either equation we get 12 3x 2 0, which gives x 2, y 2, and z 12 2 2 22 2 1. We could use the Second Derivatives Test to show that this gives a local maximum of V , or we could simply argue from the physical nature of this problem that there must be an absolute maximum volume, which has to occur at a critical point of V, so it must occur when x 2, y 2, z 1. Then V 2 2 1 4, so the maximum volume of the box is 4 m3 . ■ ABSOLUTE MAXIMUM AND MINIMUM VALUES
For a function f of one variable the Extreme Value Theorem says that if f is continuous on a closed interval a, b , then f has an absolute minimum value and an absolute maximum value. According to the Closed Interval Method in Section 4.1, we found these by evaluating f not only at the critical numbers but also at the endpoints a and b.
SECTION 11.7
MAXIMUM AND MINIMUM VALUES
■
649
There is a similar situation for functions of two variables. Just as a closed interval contains its endpoints, a closed set in ⺢ 2 is one that contains all its boundary points. [A boundary point of D is a point a, b such that every disk with center a, b contains points in D and also points not in D.] For instance, the disk (a) Closed sets
D x, y
x
2
y 2 1
which consists of all points on and inside the circle x 2 y 2 1, is a closed set because it contains all of its boundary points (which are the points on the circle x 2 y 2 1). But if even one point on the boundary curve were omitted, the set would not be closed. (See Figure 7.) A bounded set in ⺢ 2 is one that is contained within some disk. In other words, it is finite in extent. Then, in terms of closed and bounded sets, we can state the following counterpart of the Extreme Value Theorem in two dimensions.
(b) Sets that are not closed FIGURE 7
4 EXTREME VALUE THEOREM FOR FUNCTIONS OF TWO VARIABLES If f is continuous on a closed, bounded set D in ⺢ 2, then f attains an absolute maximum value f x 1, y1 and an absolute minimum value f x 2 , y2 at some points x 1, y1 and x 2 , y2 in D.
To find the extreme values guaranteed by Theorem 4, we note that, by Theorem 2, if f has an extreme value at x 1, y1, then x 1, y1 is either a critical point of f or a boundary point of D. Thus we have the following extension of the Closed Interval Method. 5 To find the absolute maximum and minimum values of a continuous function f on a closed, bounded set D :
1. Find the values of f at the critical points of f in D. 2. Find the extreme values of f on the boundary of D. 3. The largest of the values from steps 1 and 2 is the absolute maximum
value; the smallest of these values is the absolute minimum value.
EXAMPLE 6 Find the absolute maximum and minimum values of the function
f x, y x 2 2xy 2y on the rectangle D x, y
SOLUTION Since f is a polynomial, it is continuous on the closed, bounded rectangle D, so Theorem 4 tells us there is both an absolute maximum and an absolute minimum. According to step 1 in (5), we first find the critical points. These occur when fx 2x 2y 0 fy 2x 2 0
y (0, 2)
L£
(2, 2)
L¢ (0, 0)
FIGURE 8
(3, 2)
L™
L¡
(3, 0)
0 x 3, 0 y 2 .
x
so the only critical point is 1, 1, and the value of f there is f 1, 1 1. In step 2 we look at the values of f on the boundary of D, which consists of the four line segments L 1 , L 2 , L 3 , L 4 shown in Figure 8. On L 1 we have y 0 and f x, 0 x 2
0 x 3
650
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CHAPTER 11
PARTIAL DERIVATIVES
This is an increasing function of x, so its minimum value is f 0, 0 0 and its maximum value is f 3, 0 9. On L 2 we have x 3 and f 3, y 9 4y
0 y 2
This is a decreasing function of y, so its maximum value is f 3, 0 9 and its minimum value is f 3, 2 1. On L 3 we have y 2 and f x, 2 x 2 4x 4
9
0 x 3
By the methods of Chapter 4, or simply by observing that f x, 2 x 22, we see that the minimum value of this function is f 2, 2 0 and the maximum value is f 0, 2 4. Finally, on L 4 we have x 0 and f 0, y 2y
0
D
with maximum value f 0, 2 4 and minimum value f 0, 0 0. Thus, on the boundary, the minimum value of f is 0 and the maximum is 9. In step 3 we compare these values with the value f 1, 1 1 at the critical point and conclude that the absolute maximum value of f on D is f 3, 0 9 and the absolute minimum value is f 0, 0 f 2, 2 0. Figure 9 shows the graph of f . ■
L¡ 2
30
L™
FIGURE 9 f(x, y)=≈-2xy+2y
11.7
EXERCISES
1. Suppose (1, 1) is a critical point of a function f with contin-
uous second derivatives. In each case, what can you say about f ? (a) fxx 1, 1 4, fx y 1, 1 1, fyy 1, 1 2 (b) fxx 1, 1 4,
fx y 1, 1 3,
graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. 3. f x, y 9 2x 4y x 2 4y 2 4. f x, y x 3y 12x 2 8y
fyy 1, 1 2
5. f x, y x 4 y 4 4xy 2
2. Use the level curves in the figure to predict the location of
2
6. f x, y e 4yx y
the critical points of f x, y 3x x 3 2y 2 y 4 and whether f has a saddle point or a local maximum or minimum at each of those points. Explain your reasoning. Then use the Second Derivatives Test to confirm your predictions.
8. f x, y 2x 3 x y 2 5x 2 y 2 9. f x, y e x cos y 10. f x, y x 2 y 2
1.5
0 0.5 1 _2 _1 . _1 5
1 x2y2
11. f x, y x sin y
1
_1
2
7. f x, y 1 x yx y
y
_2.9 _2.7 _2.5
0 y 2
12. f x, y 2x x 2 2y y 2
1.5 1.7 1.9
1
2
x
13. f x, y x 2 y 2 e y x 2
14. f x, y x 2 ye x y ■
_1
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2
2
■
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■
; 15–18
■
■
■
■ Use a graph and/or level curves to estimate the local maximum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisely.
Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional
3–14
■
15. f x, y 3x 2 y y 3 3x 2 3y 2 2 2
16. f x, y x ye x y
2
■
SECTION 11.7
MAXIMUM AND MINIMUM VALUES
■
651
17. f x, y sin x sin y sinx y,
computer to produce a graph with a carefully chosen domain and viewpoint to see how this is possible.
18. f x, y sin x sin y cosx y,
31. Find the shortest distance from the point 2, 1, 1 to the
0 x 2, 0 y 2
plane x y z 1.
0 x 4, 0 y 4
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; 19–22
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33. Find the points on the cone z 2 x 2 y 2 that are closest to
the point 4, 2, 0.
34. Find the points on the surface y 2 9 xz that are closest
19. f x, y x 5x y 3x 2 4
2
2
to the origin.
20. f x, y 5 10xy 4x 3y y 2
4
35. Find three positive numbers whose sum is 100 and whose
product is a maximum.
21. f x, y 2x 4x 2 y 2 2xy 2 x 4 y 4
36. Find three positive numbers x, y, and z whose sum is 100
such that x a y bz c is a maximum.
22. f x, y e x y 4 x 3 4 cos y ■
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■
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■
Find the absolute maximum and minimum values of f on the set D.
23–28
■
23. f x, y 1 4 x 5y,
D is the closed triangular region with vertices 0, 0, 2, 0, and 0, 3
octant with three faces in the coordinate planes and one vertex in the plane x 2y 3z 6.
x 1, y 1
26. f x, y 4x 6y x 2 y 2,
40. Find the dimensions of the rectangular box with largest
0 x 4, 0 y 5
volume if the total surface area is given as 64 cm2 .
27. f x, y x y 4xy 2, 4
0 x 3, 0 y 2 f x, y xy , D x, y x 0, y 0, x
41. Find the dimensions of a rectangular box of maximum vol-
D x, y
28. ■
2
■
■
■
■
■
9x 2 36y 2 4z 2 36
39. Find the volume of the largest rectangular box in the first
25. f x, y x 2 y 2 x 2 y 4,
4
parallel to the axes that can be inscribed in the ellipsoid
y2 z2 x2 1 a2 b2 c2
D is the closed triangular region with vertices 1, 0, 5, 0, and 1, 4
D x, y
37. Find the volume of the largest rectangular box with edges
38. Solve the problem in Exercise 37 for a general ellipsoid
24. f x, y 3 xy x 2y,
D x, y
32. Find the point on the plane x y z 4 that is closest to
the point 1, 2, 3.
Use a graphing device (or Newton’s method or a rootfinder) to find the critical points of f correct to three decimal places. Then classify the critical points and find the highest or lowest points on the graph. ■
■
■
■
2
ume such that the sum of the lengths of its 12 edges is a constant c.
y 3 2
■
■
■
; 29. For functions of one variable it is impossible for a con-
tinuous function to have two local maxima and no local minimum. But for functions of two variables such functions exist. Show that the function f x, y x 2 12 x 2 y x 12 has only two critical points, but has local maxima at both of them. Then use a computer to produce a graph with a carefully chosen domain and viewpoint to see how this is possible.
; 30. If a function of one variable is continuous on an interval
and has only one critical number, then a local maximum has to be an absolute maximum. But this is not true for functions of two variables. Show that the function f x, y 3xe y x 3 e 3y has exactly one critical point, and that f has a local maximum there that is not an absolute maximum. Then use a
42. The base of an aquarium with given volume V is made of
slate and the sides are made of glass. If slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials. 43. A cardboard box without a lid is to have a volume of
32,000 cm3. Find the dimensions that minimize the amount of cardboard used. 44. A rectangular building is being designed to minimize
heat loss. The east and west walls lose heat at a rate of 10 unitsm2 per day, the north and south walls at a rate of 8 unitsm2 per day, the floor at a rate of 1 unitm2 per day, and the roof at a rate of 5 unitsm2 per day. Each wall must be at least 30 m long, the height must be at least 4 m, and the volume must be exactly 4000 m3. (a) Find and sketch the domain of the heat loss as a function of the lengths of the sides. (b) Find the dimensions that minimize heat loss. (Check both the critical points and the points on the boundary of the domain.)
652
■
CHAPTER 11
PARTIAL DERIVATIVES
(c) Could you design a building with even less heat loss if the restrictions on the lengths of the walls were removed?
y (x i, yi )
di
45. If the length of the diagonal of a rectangular box must be L ,
(⁄, ›)
what is the largest possible volume?
mx i+b
46. Three alleles (alternative versions of a gene) A, B, and O
determine the four blood types A (AA or AO), B (BB or BO), O (OO), and AB. The Hardy-Weinberg Law states that the proportion of individuals in a population who carry two different alleles is P 2pq 2pr 2rq where p, q, and r are the proportions of A, B, and O in the population. Use the fact that p q r 1 to show that P is at most 23.
0
x
Let di yi mx i b be the vertical deviation of the point x i , yi from the line. The method of least squares determines m and b so as to minimize ni1 di2 , the sum of the squares of these deviations. Show that, according to this method, the line of best fit is obtained when n
m
47. Suppose that a scientist has reason to believe that two quan-
tities x and y are related linearly, that is, y mx b, at least approximately, for some values of m and b. The scientist performs an experiment and collects data in the form of points x 1, y1, x 2 , y2 , . . . , x n , yn , and then plots these points. The points don’t lie exactly on a straight line, so the scientist wants to find constants m and b so that the line y mx b “fits” the points as well as possible. (See the figure.)
11.8
0
n
i
bn
i1 n
m
x
i1
b
i
i1 n
2 i
y
x
i1
n
i
xy
i i
i1
Thus the line is found by solving these two equations in the two unknowns m and b. 48. Find an equation of the plane that passes through the point
1, 2, 3 and cuts off the smallest volume in the first octant.
LAGRANGE MULTIPLIERS
y
g(x, y)=k
x
f(x, y)=11 f(x, y)=10 f(x, y)=9 f(x, y)=8 f(x, y)=7 x
FIGURE 1
Visual 11.8 animates Figure 1 for both level curves and level surfaces.
In Example 5 in Section 11.7 we maximized a volume function V xyz subject to the constraint 2xz 2yz xy 12, which expressed the side condition that the surface area was 12 m2. In this section we present Lagrange’s method for maximizing or minimizing a general function f x, y, z subject to a constraint (or side condition) of the form tx, y, z k. It’s easier to explain the geometric basis of Lagrange’s method for functions of two variables. So we start by trying to find the extreme values of f x, y subject to a constraint of the form tx, y k. In other words, we seek the extreme values of f x, y when the point x, y is restricted to lie on the level curve tx, y k. Figure 1 shows this curve together with several level curves of f . These have the equations f x, y c, where c 7, 8, 9, 10, 11. To maximize f x, y subject to tx, y k is to find the largest value of c such that the level curve f x, y c intersects tx, y k. It appears from Figure 1 that this happens when these curves just touch each other, that is, when they have a common tangent line. (Otherwise, the value of c could be increased further.) This means that the normal lines at the point x 0 , y0 where they touch are identical. So the gradient vectors are parallel; that is, (f x 0 , y0 (tx 0 , y0 for some scalar . This kind of argument also applies to the problem of finding the extreme values of f x, y, z subject to the constraint tx, y, z k. Thus the point x, y, z is restricted to lie on the level surface S with equation tx, y, z k. Instead of the level curves in Figure 1, we consider the level surfaces f x, y, z c and argue that if the maximum value of f is f x 0 , y0 , z0 c, then the level surface f x, y, z c is tangent to the level surface tx, y, z k and so the corresponding gradient vectors are parallel.
SECTION 11.8
LAGRANGE MULTIPLIERS
■
653
This intuitive argument can be made precise as follows. Suppose that a function f has an extreme value at a point Px 0 , y0 , z0 on the surface S and let C be a curve with vector equation rt xt, yt, zt that lies on S and passes through P. If t0 is the parameter value corresponding to the point P, then rt0 x 0 , y0 , z0 . The composite function ht f xt, yt, zt represents the values that f takes on the curve C. Since f has an extreme value at x 0 , y0 , z0 , it follows that h has an extreme value at t0 , so ht0 0. But if f is differentiable, we can use the Chain Rule to write 0 ht0 fx x 0 , y0 , z0 xt0 fy x 0 , y0 , z0 yt0 fzx 0 , y0 , z0 zt0 ( f x0 , y0 , z0 rt0 This shows that the gradient vector (f x 0 , y0 , z0 is orthogonal to the tangent vector rt0 to every such curve C. But we already know from Section 11.6 that the gradient vector of t, (tx 0 , y0 , z0 , is also orthogonal to rt0 . (See Equation 11.6.18.) This means that the gradient vectors (f x 0 , y0 , z0 and (tx 0 , y0 , z0 must be parallel. Therefore, if (tx 0 , y0 , z0 0, there is a number such that Lagrange multipliers are named after the French-Italian mathematician Joseph-Louis Lagrange (1736–1813). See page 208 for a biographical sketch of Lagrange. ■
(f x 0 , y0 , z0 (tx 0 , y0 , z0
1
The number in Equation 1 is called a Lagrange multiplier. The procedure based on Equation 1 is as follows. METHOD OF LAGRANGE MULTIPLIERS To find the maximum and minimum
■ In deriving Lagrange’s method we assumed that (t 0. In each of our examples you can check that (t 0 at all points where tx, y, z k .
values of f x, y, z subject to the constraint tx, y, z k [assuming that these extreme values exist and (t 0 on the surface tx, y, z k]: (a) Find all values of x, y, z, and such that (f x, y, z (tx, y, z tx, y, z k
and
(b) Evaluate f at all the points x, y, z that result from step (a). The largest of these values is the maximum value of f ; the smallest is the minimum value of f . If we write the vector equation (f (t in terms of its components, then the equations in step (a) become fx tx
fy ty
fz tz
tx, y, z k
This is a system of four equations in the four unknowns x, y, z, and , but it is not necessary to find explicit values for . For functions of two variables the method of Lagrange multipliers is similar to the method just described. To find the extreme values of f x, y subject to the constraint tx, y k, we look for values of x, y, and such that (f x, y (tx, y
and
tx, y k
654
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CHAPTER 11
PARTIAL DERIVATIVES
This amounts to solving three equations in three unknowns: fx tx
fy ty
tx, y k
Our first illustration of Lagrange’s method is to reconsider the problem given in Example 5 in Section 11.7. A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box. V EXAMPLE 1
SOLUTION As in Example 5 in Section 11.7 we let x, y, and z be the length, width, and height, respectively, of the box in meters. Then we wish to maximize
V xyz subject to the constraint tx, y, z 2xz 2yz xy 12 Using the method of Lagrange multipliers, we look for values of x, y, z, and such that (V (t and tx, y, z 12. This gives the equations Vx tx
Vy ty
Vz tz
2xz 2yz xy 12
which become 2
yz 2z y
3
xz 2z x
4
xy 2x 2y
5
2xz 2yz xy 12
There are no general rules for solving systems of equations. Sometimes some ingenuity is required. In the present example you might notice that if we multiply (2) by x, (3) by y, and (4) by z, then the left sides of these equations will be identical. Doing this, we have ■ Another method for solving the system of equations (2–5) is to solve each of Equations 2, 3, and 4 for and then to equate the resulting expressions.
6
xyz 2xz xy
7
xyz 2yz xy
8
xyz 2xz 2yz
We observe that 0 because 0 would imply yz xz xy 0 from (2), (3), and (4) and this would contradict (5). Therefore, from (6) and (7) we have 2xz xy 2yz xy which gives xz yz. But z 0 (since z 0 would give V 0 ), so x y. From (7) and (8) we have 2yz xy 2xz 2yz which gives 2xz xy and so (since x 0 ) y 2z. If we now put x y 2z in (5), we get 4z 2 4z 2 4z 2 12 Since x, y, and z are all positive, we therefore have z 1, x 2, and y 2 as before.
■
SECTION 11.8
LAGRANGE MULTIPLIERS
■
655
Find the extreme values of the function f x, y x 2 2y 2 on the circle x y 1.
■ In geometric terms, Example 2 asks for the highest and lowest points on the curve C in Figure 2 that lies on the paraboloid z x 2 2y 2 and directly above the constraint circle x 2 y 2 1.
V EXAMPLE 2 2
2
SOLUTION We are asked for the extreme values of f subject to the constraint
tx, y x 2 y 2 1. Using Lagrange multipliers, we solve the equations (f (t, tx, y 1, which can be written as
z
z=≈+2¥
fx tx
fy ty
tx, y 1
or as
C
≈+¥=1
FIGURE 2
4y 2y
11
x2 y2 1
f 0, 1 2
f 1, 0 1
EXAMPLE 3 Find the extreme values of f x, y x 2 2y 2 on the disk x 2 y 2 1.
x
SOLUTION According to the procedure in (11.7.5), we compare the values of f at the critical points with values at the points on the boundary. Since fx 2x and fy 4y, the only critical point is 0, 0. We compare the value of f at that point with the extreme values on the boundary from Example 2:
f 0, 0 0 ≈+2¥=1
f 1, 0 1
Therefore, the maximum value of f on the circle x 2 y 2 1 is f 0, 1 2 and the minimum value is f 1, 0 1. Checking with Figure 2, we see that these values look reasonable. ■
≈+2¥=2
0
10
f 0, 1 2
■ The geometry behind the use of Lagrange multipliers in Example 2 is shown in Figure 3. The extreme values of f x, y x 2 2y 2 correspond to the level curves that touch the circle x 2 y 2 1.
y
2x 2x
From (9) we have x 0 or 1. If x 0, then (11) gives y 1. If 1, then y 0 from (10), so then (11) gives x 1. Therefore, f has possible extreme values at the points 0, 1, 0, 1, 1, 0, and 1, 0. Evaluating f at these four points, we find that
y x
9
f 1, 0 1
f 0, 1 2
Therefore, the maximum value of f on the disk x 2 y 2 1 is f 0, 1 2 and the minimum value is f 0, 0 0. ■
FIGURE 3
EXAMPLE 4 Find the points on the sphere x 2 y 2 z 2 4 that are closest to and
farthest from the point 3, 1, 1.
SOLUTION The distance from a point x, y, z to the point 3, 1, 1 is
d sx 3 2 y 1 2 z 1 2 but the algebra is simpler if we instead maximize and minimize the square of the distance: d 2 f x, y, z x 32 y 12 z 12 The constraint is that the point x, y, z lies on the sphere, that is, tx, y, z x 2 y 2 z 2 4
656
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CHAPTER 11
PARTIAL DERIVATIVES
According to the method of Lagrange multipliers, we solve (f (t, t 4. This gives 12
2x 3 2x
13
2y 1 2y
14
2z 1 2z
15
x 2 y 2 z2 4
The simplest way to solve these equations is to solve for x, y, and z in terms of from (12), (13), and (14), and then substitute these values into (15). From (12) we have 3 x 3 x or x1 3 or x 1 [Note that 1 0 because 1 is impossible from (12).] Similarly, (13) and (14) give 1 1 y z 1 1 ■ Figure 4 shows the sphere and the nearest point P in Example 4. Can you see how to find the coordinates of P without using calculus?
Therefore, from (15) we have 32 12 12 4 2 2 1 1 1 2
z
which gives 1 2 114 , 1 s112, so
1
s11 2
These values of then give the corresponding points x, y, z: x
P y (3, 1, _1)
6 2 2 , , s11 s11 s11
and
6 2 2 , , s11 s11 s11
It’s easy to see that f has a smaller value at the first of these points, so the closest point is (6s11, 2s11, 2s11 ) and the farthest is (6s11, 2s11, 2s11 ).
FIGURE 4
■
TWO CONSTRAINTS
h=c
±gg ±f C
g=k FIGURE 5
P
±h
Suppose now that we want to find the maximum and minimum values of a function f x, y, z subject to two constraints (side conditions) of the form tx, y, z k and hx, y, z c. Geometrically, this means that we are looking for the extreme values of f when x, y, z is restricted to lie on the curve of intersection C of the level surfaces tx, y, z k and hx, y, z c. (See Figure 5.) Suppose f has such an extreme value at a point Px0 , y0 , z0. We know from the beginning of this section that (f is orthogonal to C there. But we also know that (t is orthogonal to tx, y, z k and (h is orthogonal to hx, y, z c, so (t and (h are both orthogonal to C. This means that the gradient vector (f x 0 , y0 , z0 is in the plane determined by (tx 0 , y0 , z0 and (hx 0 , y0 , z0 . (We assume that these gradient vectors are not zero and not parallel.)
SECTION 11.8
LAGRANGE MULTIPLIERS
■
657
So there are numbers and (called Lagrange multipliers) such that (f x0 , y0 , z0 (tx0 , y0 , z0 (hx0 , y0 , z0
16
In this case Lagrange’s method is to look for extreme values by solving five equations in the five unknowns x, y, z , , and . These equations are obtained by writing Equation 16 in terms of its components and using the constraint equations: fx t x h x fy t y h y fz t z h z tx, y, z k hx, y, z c V EXAMPLE 5 Find the maximum value of the function f x, y, z x 2y 3z on the curve of intersection of the plane x y z 1 and the cylinder x 2 y 2 1.
SOLUTION We maximize the function f x, y, z x 2y 3z subject to the con-
straints tx, y, z x y z 1 and hx, y, z x 2 y 2 1. The Lagrange condition is (f (t (h, so we solve the equations
The cylinder x 2 y 2 1 intersects the plane x y z 1 in an ellipse (Figure 6). Example 5 asks for the maximum value of f when x, y, z is restricted to lie on the ellipse.
17
1 2x
18
2 2y
19
3
20
xyz1
21
x2 y2 1
■
Putting 3 [from (19)] in (17), we get 2x 2, so x 1. Similarly, (18) gives y 52. Substitution in (21) then gives
4 3
1 25 1 2 4 2
2 z 1
and so 2 294 , s292. Then x )2s29 , y 5s29 , and, from (20), z 1 x y 1 7s29 . The corresponding values of f are
0 _1
)
_2 _1
FIGURE 6
0 y
1
5 2 2 s29 s29
3 1
7 s29
3 s29
Therefore, the maximum value of f on the given curve is 3 s29 .
■
658
■
CHAPTER 11
PARTIAL DERIVATIVES
11.8
EXERCISES of f subject to the constraint that tx, y 8. Explain your reasoning.
1–15 ■ Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint(s). 1. f x, y x 2 y 2 ;
xy 1
2. f x, y 4x 6y ;
x 2 y 2 13
3. f x, y x 2 y ;
x 2 2y 2 6
4. f x, y e xy ;
x 3 y 3 16
5. f x, y, z 2x 6y 10z ; 6. f x, y, z 8x 4z ; 7. f x, y, z x yz ;
y g(x, y)=8 40 50 70
x 2 y 2 z 2 35
0
x 2 10y 2 z 2 5
CAS
x 4 y 4 z4 1
10. f x, y, z x 4 y 4 z 4 ;
x 2 y 2 z2 1
11. f x, y, z, t x y z t ;
x 2 y 2 z2 t 2 1
x 12 x 22 x n2 1 x y z 1,
21. The total production P of a certain product depends on the
y 2 z2 4
14. f x, y, z 3x y 3z;
x 2 2z 2 1
15. f x, y, z yz x y ; ■
■
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■
x y 1, ■
■
y 2 z2 1 ■
■
■
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■
Find the extreme values of f on the region described by the inequality.
16 –17
■
16. f x, y 2x 2 3y 2 4x 5, 17. f x, y e xy, ■
■
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■
■
■
■
amount L of labor used and the amount K of capital investment. The Cobb-Douglas model for the production function is P bLK 1, where b and are positive constants and 1. If the cost of a unit of labor is m and the cost of a unit of capital is n, and the company can spend only p dollars as its total budget, then maximizing the production P is subject to the constraint mL nK p. Show that the maximum production occurs when L
x 2 y 2 16
x 2 4y 2 1 ■
20. (a) If your computer algebra system plots implicitly defined
curves, use it to estimate the minimum and maximum values of f x, y x 3 y 3 3xy subject to the constraint x 32 y 32 9 by graphical methods. (b) Solve the problem in part (a) with the aid of Lagrange multipliers. Use your CAS to solve the equations numerically. Compare your answers with those in part (a).
12. f x 1, x 2 , . . . , x n x 1 x 2 x n ;
13. f x, y, z x 2y ;
x
20 10
x 2 y 2 z2 1
9. f x, y, z x 2 y 2 z 2 ;
■
30
x 2 2y 2 3z 2 6
8. f x, y, z x 2 y 2z 2 ;
x y z 0,
60
■
; 18. (a) Use a graphing calculator or computer to graph the
circle x 2 y 2 1. On the same screen, graph several curves of the form x 2 y c until you find two that just touch the circle. What is the significance of the values of c for these two curves? (b) Use Lagrange multipliers to find the extreme values of f x, y x 2 y subject to the constraint x 2 y 2 1. Compare your answers with those in part (a).
19. Pictured are a contour map of f and a curve with equation
tx, y 8. Estimate the maximum and minimum values
■
p m
and
K
1 p n
22. Referring to Exercise 21, we now suppose that the pro-
duction is fixed at bLK 1 Q, where Q is a constant. What values of L and K minimize the cost function CL, K mL nK ? 23. Use Lagrange multipliers to prove that the rectangle with
maximum area that has a given perimeter p is a square. 24. Use Lagrange multipliers to prove that the triangle with
maximum area that has a given perimeter p is equilateral. [Hint: Use Heron’s formula for the area: A sss xs ys z, where s p2 and x, y, z are the lengths of the sides.]
CHAPTER 11
42. f x, y, z x y z;
■ Use Lagrange multipliers to give an alternate solution to the indicated exercise in Section 11.7.
25–37
25. Exercise 31
26. Exercise 32
27. Exercise 33
28. Exercise 34
29. Exercise 35
30. Exercise 36
31. Exercise 37
32. Exercise 38
33. Exercise 39
34. Exercise 40
35. Exercise 41
36. Exercise 42
■
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REVIEW
■
659
■
■
x 2 y 2 z, x 2 z 2 4 ■
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43. (a) Find the maximum value of n f x1 , x2 , . . . , xn s x1 x2 xn
given that x1 , x2 , . . . , xn are positive numbers and x1 x2 xn c, where c is a constant. (b) Deduce from part (a) that if x1 , x2 , . . . , xn are positive numbers, then
37. Exercise 45 ■
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n x1 x2 xn s
38. Find the maximum and minimum volumes of a rectangular
box whose surface area is 1500 cm2 and whose total edge length is 200 cm.
This inequality says that the geometric mean of n numbers is no larger than the arithmetic mean of the numbers. Under what circumstances are these two means equal?
39. The plane x y 2z 2 intersects the paraboloid
z x 2 y 2 in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.
40. The plane 4x 3y 8z 5 intersects the cone
;
CAS
z 2 x 2 y 2 in an ellipse. (a) Graph the cone, the plane, and the ellipse. (b) Use Lagrange multipliers to find the highest and lowest points on the ellipse.
44. (a) Maximize ni1 x i yi subject to the constraints ni1 x i2 1
and ni1 y i2 1. (b) Put xi
REVIEW
and
ab i
yi
bi
s b j2
i
s a j2 s b j2
for any numbers a1, . . . , an, b1, . . . , bn. This inequality is known as the Cauchy-Schwarz Inequality.
9x 2 4y 2 36z 2 36, x y yz 1
11
ai s a j2
to show that
41– 42 ■ Find the maximum and minimum values of f subject to the given constraints. Use a computer algebra system to solve the system of equations that arises in using Lagrange multipliers. (If your CAS finds only one solution, you may need to use additional commands.) 41. f x, y, z ye xz ;
x1 x2 xn n
CONCEPT CHECK
1. (a) What is a function of two variables?
(b) Describe three methods for visualizing a function of two variables. 2. What is a function of three variables? How can you visual-
ize such a function?
(b) How do you interpret fx a, b and fy a, b geometrically? How do you interpret them as rates of change? (c) If f x, y is given by a formula, how do you calculate fx and fy ? 6. What does Clairaut’s Theorem say?
3. What does
lim
x, y l a, b
f x, y L
mean? How can you show that such a limit does not exist? 4. (a) What does it mean to say that f is continuous at a, b?
(b) If f is continuous on ⺢2, what can you say about its graph?
5. (a) Write expressions for the partial derivatives fx a, b and
fy a, b as limits.
7. How do you find a tangent plane to each of the following
types of surfaces? (a) A graph of a function of two variables, z f x, y (b) A level surface of a function of three variables, Fx, y, z k 8. Define the linearization of f at a, b. What is the corre-
sponding linear approximation? What is the geometric interpretation of the linear approximation?
660
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CHAPTER 11
PARTIAL DERIVATIVES
9. (a) What does it mean to say that f is differentiable
15. What do the following statements mean?
at a, b? (b) How do you usually verify that f is differentiable?
(a) (b) (c) (d) (e)
10. If z f x, y, what are the differentials dx, dy, and dz ? 11. State the Chain Rule for the case where z f x, y and x
and y are functions of one variable. What if x and y are functions of two variables?
f f f f f
has a local maximum at a, b. has an absolute maximum at a, b. has a local minimum at a, b. has an absolute minimum at a, b. has a saddle point at a, b.
16. (a) If f has a local maximum at a, b, what can you say
about its partial derivatives at a, b? (b) What is a critical point of f ?
12. If z is defined implicitly as a function of x and y by an
equation of the form Fx, y, z 0, how do you find &z&x and &z&y ?
17. State the Second Derivatives Test.
13. (a) Write an expression as a limit for the directional deriva-
tive of f at x 0 , y0 in the direction of a unit vector u a, b . How do you interpret it as a rate? How do you interpret it geometrically? (b) If f is differentiable, write an expression for Du f x 0 , y0 in terms of fx and fy . 14. (a) Define the gradient vector ( f for a function f of two or
18. (a) What is a closed set in ⺢ 2 ? What is a bounded set?
(b) State the Extreme Value Theorem for functions of two variables. (c) How do you find the values that the Extreme Value Theorem guarantees? 19. Explain how the method of Lagrange multipliers works
in finding the extreme values of f x, y, z subject to the constraint tx, y, z k. What if there is a second constraint hx, y, z c ?
three variables. (b) Express Du f in terms of ( f . (c) Explain the geometric significance of the gradient.
T R U E - FA L S E Q U I Z 7. If f has a local minimum at a, b and f is differentiable at
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. fy a, b lim
ylb
a, b, then ( f a, b 0.
f a, y f a, b yb
8. If f is a function, then
f x, y f 2, 5.
9. If f x, y ln y, then ( f x, y 1y.
2. There exists a function f with continuous second-order
partial derivatives such that fx x, y x y 2 and fy x, y x y 2. 3. fxy
lim
x, y l 2, 5
10. If 2, 1 is a critical point of f and
&2 f &x &y
fxx 2, 1 fyy 2, 1 fx y 2, 1 2
4. Dk f x, y, z fzx, y, z
then f has a saddle point at 2, 1.
5. If f x, y l L as x, y l a, b along every straight line
11. If f x, y sin x sin y, then s2 Du f x, y s2.
through a, b, then lim x, y l a, b f x, y L.
12. If f x, y has two local maxima, then f must have a local
6. If fx a, b and fy a, b both exist, then f is differentiable
at a, b.
minimum.
EXERCISES 1–2
■
Find and sketch the domain of the function.
3– 4
1. f x, y lnx y 1 ■
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■
Sketch the graph of the function.
3. f x, y 1 y 2
2. f x, y s4 x 2 y 2 s1 x 2 ■
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4. f x, y x 2 y 22 ■
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CHAPTER 11
5–6
Sketch several level curves of the function.
■
5. f x, y s4x 2 y 2 ■
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17–20
6. f x, y e y ■
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7. Make a rough sketch of a contour map for the function
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■
20. v r coss 2t ■
■
■
whose graph is shown. z
661
18. z xe2y
19. f x, y, z x k y l z m ■
■
Find all second partial derivatives of f .
17. f x, y 4x 3 xy 2
x
REVIEW
21. If z xy xe yx , show that x
■
■
■
&z &z y xy z. &x &y
22. If z sinx sin t, show that x
2
2
&z & 2z &z & 2z &x &x &t &t &x 2
y
■ Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point.
23–27
8. A contour map of a function f is shown. Use it to make a
rough sketch of the graph of f .
23. z 3x 2 y 2 2x,
y
24. z e cos y,
1
25. x 2y 3z 3, 2
4
9. ■
lim
■
11–15
■
■
2xy x 2 2y 2 ■
10. ■
■
■
lim
x, y l 0, 0 ■
2xy x 2 2y 2
■
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■
■
■
■
■
■
■
■
■
2, 1, 0
■
■
■
■
■
■
■
4
where the tangent plane is parallel to the plane 2x 2y z 5 .
f x, y, z x 3sy 2 z 2 at the point (2, 3, 4) and use it to estimate the number 1.983s3.01 2 3.97 2 . 32. The two legs of a right triangle are measured as 5 m and
■
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■
16. The speed of sound traveling through ocean water is a
function of temperature, salinity, and pressure. It has been modeled by the function C 1449.2 4.6T 0.055T 0.00029T 2
■
31. Find the linear approximation of the function
15. T p, q, r p lnq e r ■
■
30. Find du if u ln1 se 2t .
x 14. w yz
13. tu, v u tan v
■
29. Find the points on the hyperboloid x 2 4y 2 z 2 4
12. u er sin 2
1
■
tangent plane and normal line at 1, 1, 2 on the same screen. Choose the domain and viewpoint so that you get a good view of all three objects.
Find the first partial derivatives.
11. f x, y s2x y 2
1, 1, 1
; 28. Use a computer to graph the surface z x y and its
Evaluate the limit or show that it does not exist.
x, y l 1, 1
2, 1, 1
2
x ■
2
27. sinx yz x 2y 3z, ■
9–10
2
26. x y yz zx 3,
1.5 2
1, 2, 1
0, 0, 1
x
3
1.34 0.01T S 35 0.016D where C is the speed of sound (in meters per second), T is the temperature (in degrees Celsius), S is the salinity (the concentration of salts in parts per thousand, which means the number of grams of dissolved solids per 1000 g of water), and D is the depth below the ocean surface (in meters). Compute &C&T , &C&S, and &C&D when T 10 C, S 35 parts per thousand, and D 100 m. Explain the physical significance of these partial derivatives.
12 m with a possible error in measurement of at most 0.2 cm in each. Use differentials to estimate the maximum error in the calculated value of (a) the area of the triangle and (b) the length of the hypotenuse. 33. If u x 2 y 3 z 4, where x p 3p 2, y pe p, and
z p sin p, use the Chain Rule to find dudp.
34. If v x 2 sin y ye xy, where x s 2t and y st, use the Chain Rule to find &v&s and &v&t when s 0
and t 1.
35. Suppose z f x, y, where x ts, t, y hs, t,
t1, 2 3, ts 1, 2 1, tt 1, 2 4, h1, 2 6, hs 1, 2 5, h t 1, 2 10, fx 3, 6 7, and fy 3, 6 8. Find &z&s and &z&t when s 1 and t 2.
36. Use a tree diagram to write out the Chain Rule for the case where w f t, u, v, t t p, q, r, s, u u p, q, r, s, and v v p, q, r, s are all differentiable functions.
■
CHAPTER 11
PARTIAL DERIVATIVES
37. If z y f x 2 y 2 , where f is differentiable, show that
y
2
&z &z x x &x &y
■
3 ins, the length y of another side is decreasing at a rate of 2 ins, and the contained angle is increasing at a rate of 0.05 radians. How fast is the area of the triangle changing when x 40 in, y 50 in, and 6?
&2z &2z &2z &z 2 2v 2 y 2 4u v &x &y &u &v &v &z &z and . &x &y
56. f x, y
x y z 1,
■
Find the directional derivative of f at the given point in the indicated direction. 43. f x, y 2 sx y ,
1, 5, in the direction toward the point 4, 1
■
■
■
1 1 2 1 x2 y x2 y2 z2 3
■
■
■
x y 2z 2 ■
■
■
■
■
■
■
■
60. A package in the shape of a rectangular box can be mailed
1, 2, 3, in the direction of v 2 i j 2 k ■
■
the origin.
44. f x, y, z x 2 y x s1 z , ■
■
59. Find the points on the surface xy 2z 3 2 that are closest to
2
■
■
58. f x, y, z x 2 2y 2 3z 2;
■
■
■
x2 y2 1
1 1 ; x y
57. f x, y, z xyz;
(b) When is it a minimum? (c) When is it 0? (d) When is it half of its maximum value?
■
■
or a computer algebra system) to find the critical points of f x, y 12 10y 2x 2 8xy y 4 correct to three decimal places. Then classify the critical points and find the highest point on the graph.
55. f x, y x 2 y ;
42. (a) When is the directional derivative of f a maximum?
■
■
55–58 ■ Use Lagrange multipliers to find the maximum and minimum values of f subject to the given constraint(s).
41. Find the gradient of the function f x, y, z z 2e x sy.
■
■
; 54. Use a graphing calculator or computer (or Newton’s method
second partial derivatives, show that
43– 44
■
maximum and minimum values and saddle points of f x, y x 3 3x y 4 2y 2. Then use calculus to find these values precisely.
39. If z f u, v, where u xy, v yx, and f has continuous
40. If yz 4 x 2z 3 e xyz, find
■
D is the disk x 2 y 2 4
; 53. Use a graph and or level curves to estimate the local
38. The length x of a side of a triangle is increasing at a rate of
x2
2
52. f x, y ex y x 2 2y 2 ;
■
■
■
■
45. Find the maximum rate of change of f x, y x y sy 2
at the point 2, 1. In which direction does it occur?
by the US Postal Service if the sum of its length and girth (the perimeter of a cross-section perpendicular to the length) is at most 108 in. Find the dimensions of the package with largest volume that can be mailed. 61. A pentagon is formed by placing an isosceles triangle on a
rectangle, as shown in the figure. If the pentagon has fixed perimeter P, find the lengths of the sides of the pentagon that maximize the area of the pentagon.
46. Find parametric equations of the tangent line at the point
2, 2, 4 to the curve of intersection of the surface z 2x 2 y 2 and the plane z 4.
=
47–50 ■ Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.
=
662
¨
47. f x, y x 2 xy y 2 9x 6y 10 48. f x, y x 3 6xy 8y 3 62. A particle of mass m moves on the surface z f x, y. Let
49. f x, y 3xy x 2 y xy 2 50. f x, y x 2 ye y2 ■
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51–52 ■ Find the absolute maximum and minimum values of f on the set D. 51. f x, y 4xy 2 x 2 y 2 xy 3;
D is the closed triangular region in the xy-plane with vertices 0, 0, 0, 6, and 6, 0
■
x xt, y yt be the x- and y-coordinates of the particle at time t. (a) Find the velocity vector v and the kinetic energy 1 K 2 m v 2 of the particle. (b) Determine the acceleration vector a. (c) Let z x 2 y 2 and xt t cos t, yt t sin t. Find the velocity vector, the kinetic energy, and the acceleration vector.
MULTIPLE INTEGRALS
12
In this chapter we extend the idea of a definite integral to double and triple integrals of functions of two or three variables.These ideas are then used to compute volumes, surface areas, masses, and centroids of more general regions than we were able to consider in Chapter 7. We will see that polar coordinates are useful in computing double integrals over some types of regions. In a similar way, we will introduce two new coordinate systems in three-dimensional space––cylindrical coordinates and spherical coordinates––that greatly simplify the computation of triple integrals over certain commonly occurring solid regions.
12.1
DOUBLE INTEGRALS OVER RECTANGLES In much the same way that our attempt to solve the area problem led to the definition of a definite integral, we now seek to find the volume of a solid and in the process we arrive at the definition of a double integral. REVIEW OF THE DEFINITE INTEGRAL
y
First let’s recall the basic facts concerning definite integrals of functions of a single variable. If f x is defined for a x b, we start by dividing the interval a, b into n subintervals x i1, x i with length x i x i x i1 and we choose sample points x*i in these subintervals. Then we form the Riemann sum n
f x* x
1 0
a
x i _¡
xi
i
x
b
x i*
i
i1
and take the limit of such sums as the largest of the lengths approaches 0 to obtain the definite integral of f from a to b :
FIGURE 1
2
y
b
a
n
f x dx
lim
f x* x
max x i l 0 i1
i
i
In the special case where f x 0, the Riemann sum can be interpreted as the sum of the areas of the approximating rectangles in Figure 1, and xab f x dx represents the area under the curve y f x from a to b. VOLUMES AND DOUBLE INTEGRALS z
In a similar manner we consider a function f of two variables defined on a closed rectangle R a, b c, d x, y ⺢ 2 a x b, c y d
z=f(x, y)
0 a
c
d y
x
b
FIGURE 2
R
and we first suppose that f x, y 0. The graph of f is a surface with equation z f x, y. Let S be the solid that lies above R and under the graph of f, that is,
S x, y, z ⺢ 3 0 z f x, y, x, y R (See Figure 2.) Our goal is to find the volume of S.
663
664
■
CHAPTER 12
MULTIPLE INTEGRALS
The first step is to take a partition P of R into subrectangles. This is accomplished by dividing the intervals a, b and c, d as follows: a x 0 x 1 x i1 x i x m b c y 0 y 1 yj1 yj yn d By drawing lines parallel to the coordinate axes through these partition points as in Figure 3, we form the subrectangles
Rij x i1, x i yj1, yj x, y x i1 x x i , yj1 y yj for i 1, . . . , m and j 1, . . . , n . There are mn of these subrectangles, and they cover R . If we let x i x i x i1 and yj yj yj1 then the area of Rij is Aij x i yj y
R ij
d
Îyj
(xi , yj)
(x *ij , y*ij )
yj y j_1 › c (x *£™, y*£™)
FIGURE 3
0
a
⁄
¤
xi_1 xi
Partition of a rectangle
b
x
Î xi
If we choose a sample point x ij*, y ij* in each Rij , then we can approximate the part of S that lies above each Rij by a thin rectangular box (or “column”) with base Rij and height f x ij*, yij* as shown in Figure 4. (Compare with Figure 1.) The volume of this box is the height of the box times the area of the base rectangle: f x ij*, yij* Aij If we follow this procedure for all the rectangles and add the volumes of the corresponding boxes, we get an approximation to the total volume of S: m
3
V
n
f x *, y * A ij
ij
ij
i1 j1
(See Figure 5.) This double sum means that for each subrectangle we evaluate f at the chosen point and multiply by the area of the subrectangle, and then we add the results.
SECTION 12.1
DOUBLE INTEGRALS OVER RECTANGLES
■
665
z
z
0
c
a
0
f(x *ij , y *ij) d
y
y b x
x
R ij FIGURE 4
■ The meaning of the double limit in Equation 4 is that we can make the double sum as close as we like to the number V [for any choice of x ij*, yij* in Rij ] by making the subrectangles sufficiently small.
FIGURE 5
Our intuition tells us that the approximation given in (3) becomes better as the subrectangles become smaller. So if we denote by max x i , yj the largest of the lengths of all the subintervals, we would expect that m
n
f x *, y * A
lim
ij
max x i , yj l 0 i1 j1
ij
ij
We use the expression in Equation 4 to define the volume of the solid S that lies under the graph of f and above the rectangle R. (It can be shown that this definition is consistent with our formula for volume in Section 7.2.) Limits of the type that appear in Equation 4 occur frequently, not just in finding volumes but in a variety of other situations as well—as we will see in Section 12.4 – even when f is not a positive function. So we make the following definition. 5 DEFINITION
■ Notice the similarity between Definition 5 and the definition of a single integral in Equation 2.
V
4
The double integral of f over the rectangle R is
yy f x, y dA R
m
n
f x *, y * A
lim
max x i , yj l 0 i1 j1
ij
ij
ij
if this limit exists. The precise meaning of the limit in Definition 5 is that for every number 0 there is a corresponding number such that
yy R
m
f x, y dA
n
f x *, y * A ij
i1 j1
ij
ij
for all partitions P of R whose subinterval lengths are less than , and for any choice of sample points x ij*, yij* in Rij. A function f is called integrable if the limit in Definition 5 exists. It is shown in courses on advanced calculus that all continuous functions are integrable. In fact, the double integral of f exists provided that f is “not too discontinuous.” In particular,
666
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CHAPTER 12
MULTIPLE INTEGRALS
if f is bounded [that is, there is a constant M such that f x, y M for all x, y in R ], and f is continuous there, except on a finite number of smooth curves, then f is integrable over R. If we know that f is integrable, we can choose the partitions P to be regular, that is, all the subrectangles Rij have the same dimensions and therefore the same area: A x y . In this case we can simply let m l and n l . In addition, the sample point x ij*, yij* can be chosen to be any point in the subrectangle Rij , but if we choose it to be the upper right-hand corner of Rij [namely x i , yj , see Figure 3], then the expression for the double integral looks simpler:
yy f x, y dA
6
R
m
lim
n
f x , y A i
m, n l i1 j1
j
By comparing Definitions 4 and 5, we see that a volume can be written as a double integral: If f x, y 0, then the volume V of the solid that lies above the rectangle R and below the surface z f x, y is V yy f x, y dA R
y (1, 2) (2, 2)
2
R¡™ 1
R™™
m
R¡¡
ij
ij
is called a double Riemann sum and is used as an approximation to the value of the double integral. [Notice how similar it is to the Riemann sum in (1) for a function of a single variable.] If f happens to be a positive function, then the double Riemann sum represents the sum of volumes of columns, as in Figure 5, and is an approximation to the volume under the graph of f .
x
2
ij
i1 j1
R™¡ 1
n
f x *, y * A
(2, 1)
(1, 1)
0
The sum in Definition 5,
FIGURE 6 z 16
V EXAMPLE 1 Estimate the volume of the solid that lies above the square R 0, 2 0, 2 and below the elliptic paraboloid z 16 x 2 2y 2. Divide R into four equal squares and choose the sample point to be the upper right corner of each square Rij . Sketch the solid and the approximating rectangular boxes.
z=16-≈-2¥
SOLUTION The squares are shown in Figure 6. The paraboloid is the graph of
f x, y 16 x 2 2y 2 and the area of each square is 1. Approximating the volume by the Riemann sum with m n 2, we have 2
V
x
FIGURE 7
i
j
i1 j1
2 2
2
f x , y A
y
f 1, 1 A f 1, 2 A f 2, 1 A f 2, 2 A 131 71 101 41 34 This is the volume of the approximating rectangular boxes shown in Figure 7.
■
SECTION 12.1
DOUBLE INTEGRALS OVER RECTANGLES
■
667
We get better approximations to the volume in Example 1 if we increase the number of squares. Figure 8 shows how the columns start to look more like the actual solid and the corresponding approximations become more accurate when we use 16, 64, and 256 squares. In Example 7 we will be able to show that the exact volume is 48.
FIGURE 8
The Riemann sum approximations to the volume under z=16-≈-2¥ become more accurate as m and n increase.
(a) m=n=4, VÅ41.5
V EXAMPLE 2
(b) m=n=8, VÅ44.875
If R x, y 1 x 1, 2 y 2 , evaluate the integral
yy s1 x (0, 0, 1)
S
FIGURE 9
(1, 0, 0)
2
dA
R
z
x
(c) m=n=16, VÅ46.46875
(0, 2, 0)
y
SOLUTION It would be very difficult to evaluate this integral directly from Definition 5 but, because s1 x 2 0, we can compute the integral by interpreting it as a volume. If z s1 x 2 , then x 2 z 2 1 and z 0, so the given double integral represents the volume of the solid S that lies below the circular cylinder x 2 z 2 1 and above the rectangle R. (See Figure 9.) The volume of S is the area of a semicircle with radius 1 times the length of the cylinder. Thus
yy s1 x
2
dA 12 12 4 2
■
R
THE MIDPOINT RULE
The methods that we used for approximating single integrals (the Midpoint Rule, the Trapezoidal Rule, Simpson’s Rule) all have counterparts for double integrals. Here we consider only the Midpoint Rule for double integrals. This means that we use a double Riemann sum with a regular partition to approximate the double integral, where all the subrectangles have area A and the sample point x ij*, yij* in Rij is chosen to be the center xi , yj of Rij . In other words, xi is the midpoint of x i1, x i and yj is the midpoint of yj1, yj . MIDPOINT RULE FOR DOUBLE INTEGRALS m
n
yy f x, y dA f x , y A i
R
j
i1 j1
where xi is the midpoint of x i1, x i and yj is the midpoint of yj1, yj .
668
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CHAPTER 12
MULTIPLE INTEGRALS
V EXAMPLE 3 Use the Midpoint Rule with m n 2 to estimate the value of the integral xxR x 3y 2 dA, where R x, y 0 x 2, 1 y 2 .
SOLUTION In using the Midpoint Rule with m n 2, we evaluate
f x, y x 3y 2 at the centers of the four subrectangles shown in Figure 10. So x1 12 , x2 32 , y1 54 , and y2 74 . The area of each subrectangle is A 12 . Thus
y (2, 2)
2 3 2
yy x 3y
R¡™
R™™
R¡¡
R™¡
R
2
2
dA
2
f x , y A i
j
i1 j1
f x1, y1 A f x1, y2 A f x2 , y1 A f x2 , y2 A f ( 12 , 54 ) A f ( 12 , 74 ) A f ( 32 , 54 ) A f ( 32 , 74 ) A
1
1 139 1 51 1 123 1 ( 67 16 ) 2 ( 16 ) 2 ( 16 ) 2 ( 16 ) 2
0
1
2
958 11.875
x
yy x 3y
Thus we have
FIGURE 10
2
dA 11.875
■
R
Number of subrectangles
Midpoint Rule approximations
1 4 16 64 256 1024
11.5000 11.8750 11.9687 11.9922 11.9980 11.9995
NOTE In Example 5 we will see that the exact value of the double integral in Example 3 is 12. (Remember that the interpretation of a double integral as a volume is valid only when the integrand f is a positive function. The integrand in Example 3 is not a positive function, so its integral is not a volume. In Examples 4 and 5 we will discuss how to interpret integrals of functions that are not always positive in terms of volumes.) If we keep dividing each subrectangle in Figure 10 into four smaller ones with similar shape, we get the Midpoint Rule approximations displayed in the chart in the margin. Notice how these approximations approach the exact value of the double integral, 12. ITERATED INTEGRALS
Recall that it is usually difficult to evaluate single integrals directly from the definition of an integral, but the Evaluation Theorem (Part 2 of the Fundamental Theorem of Calculus) provides a much easier method. The evaluation of double integrals from first principles is even more difficult, but here we see how to express a double integral as an iterated integral, which can then be evaluated by calculating two single integrals. Suppose that f is a function of two variables that is continuous on the rectangle R a, b c, d . We use the notation xcd f x, y dy to mean that x is held fixed and f x, y is integrated with respect to y from y c to y d. This procedure is called partial integration with respect to y. (Notice its similarity to partial differentiation.) Now xcd f x, y dy is a number that depends on the value of x, so it defines a function of x : Ax y f x, y dy d
c
If we now integrate the function A with respect to x from x a to x b, we get 7
y
b
a
Ax dx y
b
a
y
d
c
f x, y dy dx
The integral on the right side of Equation 7 is called an iterated integral. Usually the
SECTION 12.1
DOUBLE INTEGRALS OVER RECTANGLES
■
669
brackets are omitted. Thus
y y b
8
a
d
c
f x, y dy dx y
y
b
a
d
c
f x, y dy dx
means that we first integrate with respect to y from c to d and then with respect to x from a to b. Similarly, the iterated integral
y y d
9
c
b
a
f x, y dx dy y
y
d
c
b
a
f x, y dx dy
means that we first integrate with respect to x (holding y fixed) from x a to x b and then we integrate the resulting function of y with respect to y from y c to y d. Notice that in both Equations 8 and 9 we work from the inside out. EXAMPLE 4 Evaluate the iterated integrals.
(a)
yy 3
0
2
1
x 2y dy dx
yy 2
(b)
1
3
0
x 2 y dx dy
SOLUTION
(a) Regarding x as a constant, we obtain
y
2
x 2 y dy x 2
1
y2 2
y2
22 2
x2
y1
12 2
x2
32 x 2
Thus the function A in the preceding discussion is given by Ax 32 x 2 in this example. We now integrate this function of x from 0 to 3:
yy 3
0
2
1
x 2 y dy dx y
3
0
y
y
3 3 2 0
2
1
x 2 y dy dx
x 2 dx
3
x3 2
0
27 2
(b) Here we first integrate with respect to x :
yy 2
1
3
0
x y dx dy y 2
2
1
y
3
0
y 9y dy 9 2
1
x y dx dy 2
y2 2
y
2
1
2
1
27 2
x3 y 3
x3
dy
x0
■
Notice that in Example 4 we obtained the same answer whether we integrated with respect to y or x first. In general, it turns out (see Theorem 10) that the two iterated integrals in Equations 8 and 9 are always equal; that is, the order of integration does not matter. (This is similar to Clairaut’s Theorem on the equality of the mixed partial derivatives.) The following theorem gives a practical method for evaluating a double integral by expressing it as an iterated integral (in either order).
670
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CHAPTER 12
MULTIPLE INTEGRALS
Theorem 10 is named after the Italian mathematician Guido Fubini (1879–1943), who proved a very general version of this theorem in 1907. But the version for continuous functions was known to the French mathematician Augustin-Louis Cauchy almost a century earlier. ■
z
0 x x
A(x)
a y
b
10 FUBINI’S THEOREM If f is continuous on the rectangle R x, y a x b , c y d , then
yy f x, y dA y y b
a
d
c
f x, y dy dx y
y
d
c
b
f x, y dx dy
a
R
More generally, this is true if we assume that f is bounded on R, f is discontinuous only on a finite number of smooth curves, and the iterated integrals exist. The proof of Fubini’s Theorem is too difficult to include in this book, but we can at least give an intuitive indication of why it is true for the case where f x, y 0. Recall that if f is positive, then we can interpret the double integral xxR f x, y dA as the volume V of the solid S that lies above R and under the surface z f x, y. But we have another formula that we used for volume in Chapter 7, namely, V y Ax dx b
FIGURE 11
a
Visual 12.1 illustrates Fubini’s Theorem by showing an animation of Figures 11 and 12.
where Ax is the area of a cross-section of S in the plane through x perpendicular to the x-axis. From Figure 11 you can see that Ax is the area under the curve C whose equation is z f x, y, where x is held constant and c y d. Therefore
z
Ax y f x, y dy d
c
and we have 0
c
y
yy f x, y dA V y
d y
b
a
Ax dx y
b
a
y
d
f x, y dy dx
c
R
A similar argument, using cross-sections perpendicular to the y-axis as in Figure 12, shows that d b yy f x, y dA y y f x, y dx dy
x
FIGURE 12
c
a
R ■ Notice the negative answer in Example 5; nothing is wrong with that. The function f in that example is not a positive function, so its integral doesn’t represent a volume. From Figure 13 we see that f is always negative on R, so the value of the integral is the negative of the volume that lies above the graph of f and below R.
Evaluate the double integral xxR x 3y 2 dA, where 0 x 2, 1 y 2 . (Compare with Example 3)
V EXAMPLE 5
R x, y
SOLUTION 1 Fubini’s Theorem gives
yy x 3y
2
dA y
2
0
y
2
1
x 3y 2 dy dx y xy y 3 2
R
y x 7 dx 2
R
0
0
_4
_12
z=x-3¥ 0
0.5
1 y
1.5
2 2
1 x
0
yy x 3y
2
dA y
2
1
R
y2 y1
dx
2
x2 7x 2
12
0
y
2
0
x 3y dx dy 2
y
2
1
1
2
x2 3xy 2 2
y 2 6y 2 dy 2y 2y 3]1 12 2
FIGURE 13
]
SOLUTION 2 Again applying Fubini’s Theorem, but this time integrating with respect to x first, we have
z _8
[
0
x2
dy
x0
■
SECTION 12.1
V EXAMPLE 6
■ For a function f that takes on both positive and negative values, xxR f x, y dA is a difference of volumes: V1 V2 , where V1 is the volume above R and below the graph of f and V2 is the volume below R and above the graph. The fact that the integral in Example 6 is 0 means that these two volumes V1 and V2 are equal. (See Figure 14.)
DOUBLE INTEGRALS OVER RECTANGLES
■
671
Evaluate xxR y sinxy dA, where R 1, 2 0, .
SOLUTION If we first integrate with respect to x, we get
yy y sinxy dA y y 0
2
y sinxy dx dy
1
y [cosxy] 0
x2 x1
dy
R
1 y cos 2y cos y dy 2 sin 2y sin y
0
]
0
0
■
NOTE If we first integrate with respect to y in Example 6, we get
yy y sinxy dA y y 2
1 z 0 _1
1
0
y sinxy dy dx
R
z=y sin(xy)
0
1
y
1 x
but this order of integration is much more difficult than the method given in the example because it involves integration by parts twice. Therefore, when we evaluate double integrals it is wise to choose the order of integration that gives simpler integrals.
3 2
2
FIGURE 14
V EXAMPLE 7 Find the volume of the solid S that is bounded by the elliptic paraboloid x 2 2y 2 z 16, the planes x 2 and y 2, and the three coordinate planes.
SOLUTION We first observe that S is the solid that lies under the surface
z 16 x 2 2y 2 and above the square R 0, 2 0, 2 . (See Figure 15.) This solid was considered in Example 1, but we are now in a position to evaluate the double integral using Fubini’s Theorem. Therefore
16
z 8
0
0
1 y
2 2
1
0
V yy 16 x 2 2y 2 dA y
y
2
0
2
0
16 x 2 2y 2 dx dy
R
x
FIGURE 15
y [16x 2
0
1 3
]
x 3 2y 2x
x2 x0
dy y
2
0
( 883 4y 2 ) dy [ 883 y 43 y 3 ]20 48
■
In the special case where f x, y can be factored as the product of a function of x only and a function of y only, the double integral of f can be written in a particularly simple form. To be specific, suppose that f x, y txh y and R a, b c, d . Then Fubini’s Theorem gives
yy f x, y dA y y d
c
b
a
txhy dx dy y
d
c
R
y
b
a
txhy dx dy
In the inner integral y is a constant, so h y is a constant and we can write
y y d
c
b
a
txhy dx dy y
d
c
y hy
b
a
tx dx
dy y tx dx y h y dy b
a
d
c
since xab tx dx is a constant. Therefore, in this case, the double integral of f can be written as the product of two single integrals:
11
yy tx hy dA y
b
a
R
tx dx y h y dy d
c
where R a, b c, d
672
■
CHAPTER 12
MULTIPLE INTEGRALS
■ The function f x, y sin x cos y in Example 8 is positive on R , so the integral represents the volume of the solid that lies above R and below the graph of f shown in Figure 16.
EXAMPLE 8 If R 0, 2 0, 2 , then, by Equation 11,
yy sin x cos y dA y
2
0
sin x dx y
2
0
cos y dy
R
[
2 0
2 0
] [sin y]
cos x
z
111
■
PROPERTIES OF DOUBLE INTEGRALS 0 y x
We list here three properties of double integrals that can be proved in the same manner as in Section 5.2. We assume that all of the integrals exist. Properties 12 and 13 are referred to as the linearity of the integral.
FIGURE 16 12
yy f x, y tx, y dA yy f x, y dA yy tx, y dA R
Double integrals behave this way because the double sums that define them behave this way. ■
13
R
R
yy c f x, y dA c yy f x, y dA R
where c is a constant
R
If f x, y tx, y for all x, y in R, then 14
yy
f x, y dA yy tx, y dA
R
12.1
R
EXERCISES
1. (a) Estimate the volume of the solid that lies below
the surface z xy and above the rectangle R x, y 0 x 6 , 0 y 4 . Use a Riemann sum with m 3, n 2, and a regular partition, and take the sample point to be the upper right corner of each square. (b) Use the Midpoint Rule to estimate the volume of the solid in part (a).
2. If R 1, 3 0, 2 , use a Riemann sum with m 4,
n 2 to estimate the value of xxR y 2 2x 2 dA. Take the sample points to be the upper left corners of the squares.
m n 2 and choose the sample points to be lower right corners. (b) Use the Midpoint Rule to estimate the volume in part (a). (c) Evaluate the double integral and compare your answer with the estimates in parts (a) and (b). 5. A contour map is shown for a function f on the square
R 0, 4 0, 4 . Use the Midpoint Rule with m n 2 to estimate the value of xxR f x, y dA. y 4
3. (a) Use a Riemann sum with m n 2 to estimate the
value of xxR sinx y dA, where R 0, 0, . Take the sample points to be lower left corners. (b) Use the Midpoint Rule to estimate the integral in part (a). (c) Evaluate the double integral and compare your answer with the estimates in parts (a) and (b).
10
0
10 20 30
0
2
10 20 30
4. (a) Estimate the volume of the solid that lies below the
surface z x 2y 2 and above the rectangle R 0, 2 0, 4 . Use a Riemann sum with
0
2
4
x
SECTION 12.1
6. A 20-ft-by-30-ft swimming pool is filled with water. The
26.
depth is measured at 5-ft intervals, starting at one corner of the pool, and the values are recorded in the table. Estimate the volume of water in the pool.
yy R
■
DOUBLE INTEGRALS OVER RECTANGLES
x dA, 1 xy
■
■
■
■
673
■
■
R 0, 1 0, 1 ■
■
■
■
■
■
27. Find the volume of the solid that lies under the plane 0
5
0
2
3
5
2
3
10
2
4
15
2
20
2
10
15
20
25
4
6
7
8
8
4
7
8
10
8
6
8
10
12
10
3
4
5
6
8
7
2
2
2
3
4
4
3x 2y z 12 and above the rectangle R x, y 0 x 1, 2 y 3 .
30
28. Find the volume of the solid that lies under the hyperbolic
paraboloid z 4 x 2 y 2 and above the square R 1, 1 0, 2 . 29. Find the volume of the solid lying under the elliptic
paraboloid x 24 y 29 z 1 and above the rectangle R 1, 1 2, 2 .
Evaluate the double integral by first identifying it as the volume of a solid.
7–9 7. 8. 9.
■
xxR 3 dA, R x, y 2 x 2, 1 y 6 xxR 5 x dA, R x, y 0 x 5, 0 y 3 xxR 4 2y dA, R 0, 1 0, 1
■
■
■
■
■
■
■
■
■
■
■
30. Find the volume of the solid enclosed by the surface
z 1 e x sin y and the planes x 1, y 0, y , and z 0. 31. Find the volume of the solid bounded by the surface
■
z x sx 2 y and the planes x 0, x 1, y 0, y 1, and z 0.
10. The integral xxR s9 y 2 dA , where R 0, 4 0, 2 ,
represents the volume of a solid. Sketch the solid. 11–20
■
11.
yy 3
1
1
0
13.
yy
2
2
yy
1
0
0
17.
4
yy
2
1
1
19.
y y
15.
2
0
ln 2
■
■
yy R
22.
loid z 1 x 12 4y 2, the planes x 3 and y 2, and the coordinate planes.
14.
yy 4
1
2
1
2x y dx dy 8
x y y x
ln 5
dy dx
e 2xy dx dy ■
■
■
x 2 y 2 dy dx
33. Find the volume of the solid in the first octant bounded by
the cylinder z 9 y 2 and the plane x 2.
y y (x sy ) dx dy 4
1
■
■
12.
x sin y dy dx
0
21–26 21.
1 4xy dx dy
0
0
32. Find the volume of the solid bounded by the elliptic parabo-
Calculate the iterated integral.
2
16.
yy
2
18.
yy
1
20.
yy
1
1
1
0
2
1
■
34. (a) Find the volume of the solid bounded by the surface
0
0
1
0
0
xe x dy dx y
z 6 xy and the planes x 2, x 2, y 0, y 3, and z 0. (b) Use a computer to draw the solid.
; CAS
35. Use a computer algebra system to find the exact value of the
CAS
36. Graph the solid that lies between the surfaces
integral xxR x 5y 3e x y dA, where R 0, 1 0, 1 . Then use the CAS to draw the solid whose volume is given by the integral.
x y2 dx dy xysx 2 y 2 dy dx
■
■
■
■
2
Calculate the double integral.
xy 2 dA, x2 1
R x, y
0 x 1,
3 y 3
37–38 ■ The average value of a function f x, y over a rectangle R is defined to be
yy cosx 2y dA,
fave
R
R x, y 23.
yy
0 x ,
x sinx y dA,
R
24.
yy R
25.
yy R
1 x2 dA, 1 y2 2
xye x y dA,
1 AR
0 y 2
R 0, 6 0, 3
R x, y
0 x 1,
R 0, 1 0, 2
z ex cosx 2 y 2 and z 2 x 2 y 2 for x 1, y 1. Use a computer algebra system to approximate the volume of this solid correct to four decimal places.
■
yy f x, y dA R
(Compare with the definition for functions of one variable in Section 5.4.) Find the average value of f over the given rectangle. 37. f x, y x 2 y,
R has vertices 1, 0, 1, 5, 1, 5, 1, 0
0 y 1
38. f x, y e ysx e y ,
R 0, 4 0, 1
■
■
■
■
■
■
■
■
■
■
■
■
674
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CHAPTER 12
MULTIPLE INTEGRALS
39. If f is a constant function, f x, y k, and
Do the answers contradict Fubini’s Theorem? Explain what is happening.
R a, b c, d , show that xxR k dA kb ad c.
xxR sinx y dA 1.
40. If R 0, 1 0, 1 , show that 0 CAS
42. (a) In what way are the theorems of Fubini and Clairaut
similar? (b) If f x, y is continuous on a, b c, d and
41. Use your CAS to compute the iterated integrals
yy 1
0
1
0
xy dy dx x y3
12.2
and
1
0
0
xy dx dy x y3
x
a
y
y
c
f s, t dt ds
for a x b, c y d, show that txy tyx f x, y.
DOUBLE INTEGRALS OVER GENERAL REGIONS For single integrals, the region over which we integrate is always an interval. But for double integrals, we want to be able to integrate a function f not just over rectangles but also over regions D of more general shape, such as the one illustrated in Figure 1. We suppose that D is a bounded region, which means that D can be enclosed in a rectangular region R as in Figure 2. Then we define a new function F with domain R by
y
D
0
yy 1
tx, y y
x
Fx, y
1
f x, y 0
if x, y is in D if x, y is in R but not in D
FIGURE 1
If the double integral of F exists over R, then we define the double integral of f over D by
y
R 2
D
yy f x, y dA yy Fx, y dA D
0
FIGURE 2
x
where F is given by Equation 1
R
Definition 2 makes sense because R is a rectangle and so xxR Fx, y dA has been previously defined in Section 12.1. The procedure that we have used is reasonable because the values of Fx, y are 0 when x, y lies outside D and so they contribute nothing to the integral. This means that it doesn’t matter what rectangle R we use as long as it contains D. In the case where f x, y 0 we can still interpret xxD f x, y dA as the volume of the solid that lies above D and under the surface z f x, y (the graph of f ). You can see that this is reasonable by comparing the graphs of f and F in Figures 3 and 4 and remembering that xxR Fx, y dA is the volume under the graph of F. z
z
graph of F
graph of f 0
0
y
y
D
D x
x
FIGURE 3
FIGURE 4
SECTION 12.2
■
DOUBLE INTEGRALS OVER GENERAL REGIONS
675
Figure 4 also shows that F is likely to have discontinuities at the boundary points of D. Nonetheless, if f is continuous on D and the boundary curve of D is “well behaved” (in a sense outside the scope of this book), then it can be shown that xxR Fx, y dA exists and therefore xxD f x, y dA exists. In particular, this is the case for the following types of regions. A plane region D is said to be of type I if it lies between the graphs of two continuous functions of x, that is,
D x, y a x b, t1x y t 2x where t1 and t 2 are continuous on a, b . Some examples of type I regions are shown in Figure 5. y
y
y=g™(x)
y
y=g™(x)
y=g™(x)
D
D
D y=g¡(x)
y=g¡(x) 0
y=g¡(x)
a
x
b
0
a
x
b
0
a
b
x
FIGURE 5 Some type I regions
y
In order to evaluate xxD f x, y dA when D is a region of type I, we choose a rectangle R a, b c, d that contains D, as in Figure 6, and we let F be the function given by Equation 1; that is, F agrees with f on D and F is 0 outside D. Then, by Fubini’s Theorem,
y=g™(x)
d
yy f x, y dA yy Fx, y dA y y b
D
a
D
d
c
Fx, y dy dx
R
c
y=g¡(x) 0
a
x
b
x
Observe that Fx, y 0 if y t1x or y t 2x because x, y then lies outside D. Therefore
y
FIGURE 6
d
c
Fx, y dy y
t 2x
t1x
Fx, y dy y
t 2x
t1x
f x, y dy
because Fx, y f x, y when t1x y t 2x. Thus we have the following formula that enables us to evaluate the double integral as an iterated integral. 3
If f is continuous on a type I region D such that
D x, y a x b, t1x y t 2x then
yy f x, y dA y y b
a
D
t 2x
t1x
f x, y dy dx
The integral on the right side of (3) is an iterated integral that is similar to the ones we considered in the preceding section, except that in the inner integral we regard x as being constant not only in f x, y but also in the limits of integration, t1x and t 2x.
676
■
CHAPTER 12
MULTIPLE INTEGRALS
y
We also consider plane regions of type II, which can be expressed as
d
x=h¡(y)
D x, y c y d, h1y x h2 y
4
x=h™(y)
D
where h1 and h2 are continuous. Two such regions are illustrated in Figure 7. Using the same methods that were used in establishing (3), we can show that
c 0
x
y d
D
x=h¡(y)
x=h™(y)
D
0 c
h 2 y
h1 y
f x, y dx dy
where D is a type II region given by Equation 4. x
FIGURE 7
d
c
yy f x, y dA y y
5
V EXAMPLE 1 Evaluate xxD x 2y dA, where D is the region bounded by the parabolas y 2x 2 and y 1 x 2.
SOLUTION The parabolas intersect when 2x 2 1 x 2, that is, x 2 1, so x 1.
Some type II regions
We note that the region D, sketched in Figure 8, is a type I region but not a type II region and we can write
D x, y 1 x 1, 2x 2 y 1 x 2 Since the lower boundary is y 2x 2 and the upper boundary is y 1 x 2, Equation 3 gives
y
y=1+≈
(_1, 2)
(1, 2)
yy x 2y dA y y 1
1
D
1x 2
2x 2
x 2y dy dx
y [ xy y ] 1
2 2 y1x y2x 2
1
dx
y x1 x 2 1 x 2 2 x2x 2 2x 2 2 dx 1
D
1
y=2≈ x
1
_1
FIGURE 8
y
1
1
3
3x 4 x 3 2x 2 x 1 dx
x5 x4 x3 x2 2 x 5 4 3 2
1
1
32 15
■
NOTE When we set up a double integral as in Example 1, it is essential to draw a diagram. Often it is helpful to draw a vertical arrow as in Figure 8. Then the limits of integration for the inner integral can be read from the diagram as follows: The arrow starts at the lower boundary y t1x, which gives the lower limit in the integral, and the arrow ends at the upper boundary y t 2x, which gives the upper limit of integration. For a type II region the arrow is drawn horizontally from the left boundary to the right boundary.
y (2, 4)
y=2x y=≈
EXAMPLE 2 Find the volume of the solid that lies under the paraboloid z x 2 y 2
and above the region D in the xy-plane bounded by the line y 2x and the parabola y x 2.
D 0
1
2
FIGURE 9
D as a type I region
x
SOLUTION 1 From Figure 9 we see that D is a type I region and
D x, y 0 x 2, x 2 y 2x
SECTION 12.2
■
DOUBLE INTEGRALS OVER GENERAL REGIONS
677
Therefore, the volume under z x 2 y 2 and above D is y 4
V yy x 2 y 2 dA y
(2, 4)
2
0
D
x= 12 y
x=œ„ y
y y 2
0
D
x
0
2
0
FIGURE 10
y3 x y 3 2
y
2x
x 2 y 2 dy dx
x2
y2x
dx
y 2
x 22x
0
yx 2
x6 14x 3 x4 3 3
2x3 x 2 3 x 2x 2 3 3
x7 x5 7x 4 dx 21 5 6
2
0
dx
216 35
SOLUTION 2 From Figure 10 we see that D can also be written as a type II region:
D as a type II region
D x, y 0 y 4, 12 y x sy
Figure 11 shows the solid whose volume is calculated in Example 2. It lies above the xy -plane, below the paraboloid z x 2 y 2 , and between the plane y 2x and the parabolic cylinder y x 2. z ■
Therefore, another expression for V is V yy x 2 y 2 dA y
0
D
y=≈ z=≈+¥
y 4
0
x3 y 2x 3
y=2x
y
sy 1 2
y
x 2 y 2 dx dy
xsy
dy
x 12 y
y 4
0
4 152 y 52 27 y 72 13 96 y
x
y
4
]
4 0
y 32 y3 y3 y 52 3 24 2
dy
216 35
■
Evaluate xxD xy dA, where D is the region bounded by the line y x 1 and the parabola y 2 2x 6. V EXAMPLE 3
FIGURE 11
SOLUTION The region D is shown in Figure 12. Again D is both type I and type II, but the description of D as a type I region is more complicated because the lower boundary consists of two parts. Therefore, we prefer to express D as a type II region:
D (x, y) 2 y 4, 12 y 2 3 x y 1
y
y (5, 4)
(5, 4)
¥
y=œ„„„„„ 2x+6
x= -3 2 y=x-1 x
0
_3
(_1, _2)
y=_œ„„„„„ 2x+6
FIGURE 12
x=y+1
(a) D as a type I region
x
0 (_1, _2)
_2
(b) D as a type II region
678
■
CHAPTER 12
MULTIPLE INTEGRALS
Then (5) gives
yy xy dA y y 4
2
D
1 2
y1 y 23
xy dx dy
y
4
2
xy1
x2 y 2
dy
x12 y 23
12 y y y 1 2 ( 12 y 2 3) 2 dy 4
[
2
12
]
y 4
y5 4y 3 2y 2 8y dy 4
2
4
1 y6 y3 y4 2 4y 2 2 24 3
36
2
If we had expressed D as a type I region using Figure 12(a), then we would have obtained 1
yy xy dA y y D
xy dy dx y
s2x6
s2x6
3
5
1
y
s2x6
x1
xy dy dx ■
but this would have involved more work than the other method.
z (0, 0, 2)
EXAMPLE 4 Find the volume of the tetrahedron bounded by the planes
x 2y z 2, x 2y, x 0, and z 0. SOLUTION In a question such as this, it’s wise to draw two diagrams: one of the three-dimensional solid and another of the plane region D over which it lies. Figure 13 shows the tetrahedron T bounded by the coordinate planes x 0, z 0, the vertical plane x 2y, and the plane x 2y z 2. Since the plane x 2y z 2 intersects the xy-plane (whose equation is z 0) in the line x 2y 2, we see that T lies above the triangular region D in the xy-plane bounded by the lines x 2y, x 2y 2, and x 0. (See Figure 14.) The plane x 2y z 2 can be written as z 2 x 2y, so the required volume lies under the graph of the function z 2 x 2y and above
x+2y+z=2
x=2y T
y (0, 1, 0)
0
1
”1, 2 , 0’ x
FIGURE 13 y 1
D x, y
0 x 1,
x2 y 1 x 2
Therefore
x+2y=2 x ”or y=1- ’ 2
V yy 2 x 2y dA y
1
0
y
1x2
x2
2 x 2y dy dx
D
y 2y xy y 2 1
”1, 21 ’
D
0
y= 2x 0
1
x
0
y1x2
]
y 1
yx2
0
V EXAMPLE 5
dx
2xx 1
x 2
y x 2 2x 1 dx 1
FIGURE 14
[
1
x 2
x3 x2 x 3
2
x
1
0
x2 x2 2 4
dx
1 3
Evaluate the iterated integral x01 xx1 siny 2 dy dx.
SOLUTION If we try to evaluate the integral as it stands, we are faced with the task of first evaluating x siny 2 dy. But it’s impossible to do so in finite terms since
■
SECTION 12.2
DOUBLE INTEGRALS OVER GENERAL REGIONS
■
679
x siny 2 dy is not an elementary function. (See the end of Section 6.4.) So we must
y
y=1
change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. Using (3) backward, we have
y y
D y=x
1
1
0
x
siny 2 dy dx yy siny 2 dA D
D x, y 0 x 1, x y 1
where 0
x
1
We sketch this region D in Figure 15. Then from Figure 16 we see that an alternative description of D is
FIGURE 15
D x, y 0 y 1, 0 x y
D as a type I region
This enables us to use (5) to express the double integral as an iterated integral in the reverse order:
y 1
yy 1
0
x=0
1
x
siny 2 dy dx yy siny 2 dA D
D
y
x=y
1
0
0
y
y
0
siny 2 dx dy y [ x siny 2 ]x0 dy 1
xy
0
y y siny 2 dy 12 cos y 2 ]0 1
x
1
0
12 1 cos 1
FIGURE 16
■
D as a type II region PROPERTIES OF DOUBLE INTEGRALS
We assume that all of the following integrals exist. The first three properties of double integrals over a region D follow immediately from Definition 2 and Properties 12, 13, and 14 in Section 12.1. 6
yy f x, y tx, y dA yy f x, y dA yy tx, y dA D
D
D
yy c f x, y dA c yy f x, y dA
7
D
If f x, y tx, y for all x, y in D, then
y
D
yy f x, y dA yy tx, y dA
8
D¡
0
FIGURE 17
D
D™
D
x
D
The next property of double integrals is similar to the property of single integrals given by the equation xab f x dx xac f x dx xcb f x dx. If D D1 D2 , where D1 and D2 don’t overlap except perhaps on their boundaries (see Figure 17), then
9
yy f x, y dA yy f x, y dA yy f x, y dA D
D1
D2
680
■
CHAPTER 12
MULTIPLE INTEGRALS
y
Property 9 can be used to evaluate double integrals over regions D that are neither type I nor type II but can be expressed as a union of regions of type I or type II. Figure 18 illustrates this procedure. (See Exercises 43 and 44.) The next property of integrals says that if we integrate the constant function f x, y 1 over a region D, we get the area of D :
D
0
yy 1 dA AD
10
x
D
(a) D is neither type I nor type II. y
Figure 19 illustrates why Equation 10 is true: A solid cylinder whose base is D and whose height is 1 has volume AD 1 AD, but we know that we can also write its volume as xxD 1 dA. Finally, we can combine Properties 7, 8, and 10 to prove the following property. (See Exercise 47.)
D™ D¡
0
11
x
If m f x, y M for all x, y in D, then mAD yy f x, y dA MAD
(b) D=D¡ D™, D¡ is type I, D™ is type II.
D
FIGURE 18
EXAMPLE 6 Use Property 11 to estimate the integral xxD e sin x cos y dA, where D is the
z
disk with center the origin and radius 2. z=1
SOLUTION Since 1 sin x 1 and 1 cos y 1, we have
1 sin x cos y 1 and therefore
0
e1 e sin x cos y e 1 e y
Thus, using m e 1 1e, M e, and AD 22 in Property 11, we obtain
x
FIGURE 19
4
e
Cylinder with base D and height 1
12.2 1–6 1.
■
x2
yy
ey
1
0
0
3.
5.
1
x 2y dy dx
2
y y
cos
0
0
7–16 2.
sx dx dy
y
0
4.
yy
2
yy
2x
2
1
y
1
0
7.
xy dx dy
■
■
■
■
e sin dr d
6.
x
yy 1
0
v
0
■
■
■
■
Evaluate the double integral.
yy x y
3 2
x 2 y dy dx
8.
yy D
■
dA 4e
D
dA,
D x, y
D
s1 v 2 du dv 9.
■
sin x cos y
EXERCISES
Evaluate the iterated integral.
yy
yy e
■
■
■
■
yy D
0 x 2,
x y x
4y dA, x 2
D x, y
1 x 2,
0 y 2x
2y dA, x2 1
D {x, y
0 x 1,
0 y sx
3
}
SECTION 12.2
10.
yy e
y2
dA,
D x, y
D
11.
yy x cos y dA,
0 y 1,
0 x y
CAS
yy x y dA,
13.
yy y
Use a computer algebra system to find the exact volume of the solid.
29–30
D is bounded by y sx and y x 2
■
yy xy
2
yy
sx
33.
yy
s9y 2
yy 2x y dA, 35.
D
D is bounded by the circle with center the origin and radius 2
yy 2xy dA,
17–26
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
3
■
yy 2
1
ln x
4
18. Under the surface z 2x y 2 and above the region
■
■
37.
yy
3
y y
9
39.
yy
1
40.
41.
yy
2
42.
yy
2
1
0
■
yy
34.
yy
s9y
yy
4
36.
■
1
0
■
4
4x
3
1
0
■
f x, y dy dx
0
0
19. Under the surface z xy and above the triangle with
vertices 1, 1, 4, 1, and 1, 2
20. Enclosed by the paraboloid z x 2 3y 2 and the planes
x 0, y 1, y x, z 0
21. Bounded by the planes x 0, y 0, z 0, and
xyz1
arctan x
■
f x, y dx dy f x, y dy dx
■
■
■
■
1
0
22. Bounded by the planes z x, y x, x y 2, and z 0 23. Enclosed by the cylinders z x , y x and the planes 2
8
1
1
sy
0
sx 3 1 dx dy
cos x s1 cos 2 x dx dy 4
3 sy
e x dx dy
■
■
■
■
■
■
■
■
■
■
■
Express D as a union of regions of type I or type II and evaluate the integral. 43– 44
z 0, y 4
24. Bounded by the cylinder y 2 z 2 4 and the planes x 2y,
arcsin y
0
yy
x 3 sin y 3 dy dx
x2
0
38.
y cosx 2 dx dy
y2
0
■
2
e x dx dy
3y
1
bounded by x y 2 and x y 3
43.
■
yy x
2
dA
44.
D
x 0, z 0 in the first octant
yy xy dA D
y
y
25. Bounded by the cylinder x y 1 and the planes y z, 2
f x, y dx dy
f x, y dy dx
0
3
17. Under the plane x 2y z 0 and above the region
2
s9y 2
0
32.
■ Evaluate the integral by reversing the order of integration.
Find the volume of the given solid.
bounded by y x and y x
f x, y dy dx
0
37– 42
1, 2, and 0, 3 ■
4
0
■
D is the triangular region with vertices 0, 0,
D
■
■
■
31.
dA, D is enclosed by x 0 and x s1 y 2
D
16.
■
Sketch the region of integration and change the order of integration.
dA,
D
15.
■
31–36
D is the triangular region with vertices (0, 2), (1, 1), 3, 2 14.
■
30. Enclosed by z x 2 y 2 and z 2y
D 3
681
29. Enclosed by z 1 x 2 y 2 and z 0
D is bounded by y 0, y x 2, x 1
D
12.
■
DOUBLE INTEGRALS OVER GENERAL REGIONS
2
y=1+≈
1
x 0, z 0 in the first octant
(1, 1)
x=1
D
26. Bounded by the cylinders x 2 y 2 r 2 and y 2 z 2 r 2 ■
■
27–28
■
■
■
■
■
■
■
■
■
■
_1
■
0
Find the volume of the solid by subtracting two
1
x
x=_1
27. The solid enclosed by the parabolic cylinders y 1 x 2, ■
■
28. The solid enclosed by the parabolic cylinder y x 2 and the
45– 46
■
45.
planes z 3y, z 2 y ■
■
■
■
■
■
■
■
x
y=_1
y x 2 1 and the planes x y z 2, 2x 2y z 10 0
■
x=¥ 0
_1
volumes.
■
D
■
■
■
■
■
■
■
■
■
■
Use Property 11 to estimate the value of the integral.
yy sx D
■
3
y 3 dA ,
D 0, 1 0, 1
■
682
46.
■
yy e
CHAPTER 12
x 2y 2
MULTIPLE INTEGRALS
49. Evaluate xxD x 2 tan x y 3 4 dA, where
dA ,
D x, y x 2 y 2 2 . [Hint: Exploit the fact that D is symmetric with respect to both axes.]
D
D is the disk with center the origin and radius ■
■
■
■
■
■
■
■
■
1 2 ■
■
50. Use symmetry to evaluate xxD 2 3x 4y dA, where D
■
is the region bounded by the square with vertices 5, 0 and 0, 5.
47. Prove Property 11.
51. Compute xxD s1 x 2 y 2 dA , where D is the disk
48. In evaluating a double integral over a region D, a sum of
x 2 y 2 1, by first identifying the integral as the volume of a solid.
iterated integrals was obtained as follows:
yy f x, y dA y y 1
0
2y
0
f x, y dx dy
y y 3
1
3y
0
f x, y dx dy
D
Sketch the region D and express the double integral as an iterated integral with reversed order of integration.
12.3
CAS
52. Graph the solid bounded by the plane x y z 1 and
the paraboloid z 4 x 2 y 2 and find its exact volume. (Use your CAS to do the graphing, to find the equations of the boundary curves of the region of integration, and to evaluate the double integral.)
DOUBLE INTEGRALS IN POLAR COORDINATES
■ Polar coordinates were introduced in Section 9.3.
Suppose that we want to evaluate a double integral xxR f x, y dA, where R is one of the regions shown in Figure 1. In either case the description of R in terms of rectangular coordinates is rather complicated but R is easily described using polar coordinates. y
y
≈+¥=4
≈+¥=1 R
R 0
x 0
FIGURE 1
y
P (r, ¨ ) =P (x, y) r
(a) R=s(r, ¨) | 0¯r¯1, 0¯¨¯2πd
x
(b) R=s(r, ¨ ) | 1¯r¯2, 0¯¨¯πd
Recall from Figure 2 that the polar coordinates r, of a point are related to the rectangular coordinates x, y by the equations x r cos
r2 x2 y2
y
≈+¥=1
y r sin
¨ O
FIGURE 2
x
x
The regions in Figure 1 are special cases of a polar rectangle
R r, a r b, which is shown in Figure 3. In order to compute the double integral xxR f x, y dA, where R is a polar rectangle, we divide the interval a, b into m subintervals ri1, ri with lengths ri ri ri1 and we divide the interval , into n subintervals j1, j with lengths j j j1. Then the circles r ri and the rays j divide the polar rectangle R into the small polar rectangles shown in Figure 4.
SECTION 12.3
DOUBLE INTEGRALS IN POLAR COORDINATES
■
683
¨=¨ j ¨=¨ j_ ¡ r=b
R ij
¨=∫
(r i*, ¨ j*)
R
Ψ j r=a
¨=å
r=r i r=r i _ ¡
∫ å O
O
FIGURE 4 Dividing R into polar subrectangles
FIGURE 3 Polar rectangle
The “center” of the polar subrectangle Rij r,
r
i1
r ri , j1 j
has polar coordinates
j* 12 j1 j
ri* 12 ri1 ri
We compute the area of Rij using the fact that the area of a sector of a circle with radius 1 r and central angle is 2 r 2. Subtracting the areas of two such sectors, each of which has central angle j , we find that the area of Rij is 2 2 Aij 12 ri2 j 12 ri1 j 12 ri2 ri1 j
12 ri ri1 ri ri1 j ri* ri j Although we have defined the double integral xxR f x, y dA in terms of ordinary rectangles, it can be shown that, for continuous functions f , we always obtain the same answer using polar rectangles. The rectangular coordinates of the center of Rij are ri* cos j*, ri* sin j* , so a typical Riemann sum is m
1
n
m
f r* cos *, r* sin * A i
j
i
j
ij
i1 j1
n
f r* cos *, r* sin * r* r
i
j
i
j
i
i
j
i1 j1
If we write tr, r f r cos , r sin , then the Riemann sum in Equation 1 can be written as m
n
tr*, * r i
j
i
i1 j1
which is a Riemann sum for the double integral
y y
b
a
tr, dr d
j
684
■
CHAPTER 12
MULTIPLE INTEGRALS
Therefore, we have m
yy f x, y dA
n
lim
f r* cos *, r* sin * A
lim
tr*, * r
i
max ri , j l 0 i1 j1
R
m
y
b
i
j
n
max ri , j l 0 i1 j1
y
j
i
j
i
ij
y
j
y
b
a
tr, dr d
f r cos , r sin r dr d
a
2 CHANGE TO POLAR COORDINATES IN A DOUBLE INTEGRAL If f is continuous on a polar rectangle R given by 0 a r b, , where 0 2, then
yy f x, y dA y y
b
a
f r cos , r sin r dr d
R
dA d¨ dr r
The formula in (2) says that we convert from rectangular to polar coordinates in a double integral by writing x r cos and y r sin , using the appropriate limits of | integration for r and , and replacing dA by r dr d. Be careful not to forget the additional factor r on the right side of Formula 2. A classical method for remembering this is shown in Figure 5, where the “infinitesimal” polar rectangle can be thought of as an ordinary rectangle with dimensions r d and dr and therefore has “area” dA r dr d.
r d¨
EXAMPLE 1 Evaluate xxR 3x 4y 2 dA, where R is the region in the upper half-
plane bounded by the circles x 2 y 2 1 and x 2 y 2 4.
O
SOLUTION The region R can be described as FIGURE 5
R x, y y 0, 1 x 2 y 2 4 It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by 1 r 2, 0 . Therefore, by Formula 2,
yy 3x 4y
2
dA y
y
y
0
y
2
y
2
1
3r cos 4r 2 sin 2 r dr d
R
0
0
■
1
[r
3r 2 cos 4r 3 sin 2 dr d
3
cos r 4 sin 2
]
r2 r1
d y 7 cos 15 sin 2 d 0
Here we use the trigonometric identity sin2 12 1 cos 2
y [7 cos 0
15 2
]
1 cos 2 d
as discussed in Section 6.2.
15 15 7 sin sin 2 2 4
0
15 2
■
SECTION 12.3
DOUBLE INTEGRALS IN POLAR COORDINATES
■
685
V EXAMPLE 2 Find the volume of the solid bounded by the plane z 0 and the paraboloid z 1 x 2 y 2.
SOLUTION If we put z 0 in the equation of the paraboloid, we get x 2 y 2 1.
z
This means that the plane intersects the paraboloid in the circle x 2 y 2 1, so the solid lies under the paraboloid and above the circular disk D given by x 2 y 2 1 [see Figures 6 and 1(a)]. In polar coordinates D is given by 0 r 1, 0 2. Since 1 x 2 y 2 1 r 2, the volume is
(0, 0, 1)
V yy 1 x 2 y 2 dA y
y
x
2
0
D
FIGURE 6
y
2
0
d y r r 3 dr 2 1
0
y
1
0
1 r 2 r dr d
r2 r4 2 4
1
0
2
If we had used rectangular coordinates instead of polar coordinates, then we would have obtained V yy 1 x 2 y 2 dA y
1
1
D
y
s1x 2
s1x 2
1 x 2 y 2 dy dx
which is not easy to evaluate because it involves finding x 1 x 2 32 dx .
r=h™(¨)
¨=∫ D
What we have done so far can be extended to the more complicated type of region shown in Figure 7. It’s similar to the type II rectangular regions considered in Section 12.2. In fact, by combining Formula 2 in this section with Formula 12.2.5, we obtain the following formula.
3
å O
r=h¡(¨)
If f is continuous on a polar region of the form D r,
¨=å
∫
■
then
,
yy f x, y dA y y
h 2
h1
D
h1 r h2
f r cos , r sin r dr d
FIGURE 7 D=s(r, ¨) | 寨¯∫, h¡(¨)¯r¯h™(¨)d
In particular, taking f x, y 1, h1 0, and h2 h in this formula, we see that the area of the region D bounded by , , and r h is AD
yy 1 dA y y D
h
0
y
and this agrees with Formula 9.4.3.
r2 2
h 0
r dr d
d y
1 2
h 2 d
686
■
CHAPTER 12
MULTIPLE INTEGRALS
y
V EXAMPLE 3 Find the volume of the solid that lies under the paraboloid z x 2 y 2, above the xy-plane, and inside the cylinder x 2 y 2 2x.
(x-1)@+¥=1 (or r=2 cos ¨)
SOLUTION The solid lies above the disk D whose boundary circle has equation
x 2 y 2 2x or, after completing the square,
D 0
1
x 12 y 2 1
x
2
(See Figures 8 and 9.) In polar coordinates we have x 2 y 2 r 2 and x r cos , so the boundary circle becomes r 2 2r cos , or r 2 cos . Thus the disk D is given by FIGURE 8
D r, 2 2, 0 r 2 cos
z
and, by Formula 3, we have V yy x y dA y 2
2
4y
2
2
2y
2
0
y
cos 4 d 8 y
2
0
r r dr d
cos 4 d 8 y
0
5
5.
1 cos 2 2
2
d
2
3 2
3 2
■
■
R
7. 0
x
4.
y
R x
■
■
■
r dr d
■
■
■
■
0
■
■
■
xxD xy dA,
xxR cosx 2 y 2 dA, where R is the region that lies above the x-axis within the circle x 2 y 2 9
where R is the region that lies to the left of the y-axis between the circles x 2 y 2 1 and x 2 y 2 4
xxR s4 x 2 y 2
where R x, y
1
■
■
9.
11.
3 x
■
■
■
dA, x 2 y 2 4, x 0
xxR arctan yx dA, where R x, y
12. ■
■
xxR x y dA,
R
1
■
■
4 cos
0
0
8.
10.
2
■
2
y y
6.
where D is the disk with center the origin and radius 3
x
2
y 3
2
0
r dr d
Evaluate the given integral by changing to polar coordinates.
5
0
7
4
y
2
■
2
y y
7–12
2
■
d 0
■ Sketch the region whose area is given by the integral and evaluate the integral.
2
3.
5–6
■
2.
y
R
2
2
2 cos
r4 4
EXERCISES
A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write xxR f x, y dA as an iterated integral, where f is an arbitrary continuous function on R. 1.
y
2
1 2 cos 2 12 1 cos 4 d
■
■
0
2
2
12.3
■
2 cos
2[ 32 sin 2 18 sin 4]0 2
FIGURE 9
1– 4
y
2
D
x
2
xxR ye x dA,
1 x
2
y 2 4, 0 y x
where R is the region in the first quadrant enclosed by the circle x 2 y 2 25 ■
■
■
■
■
■
■
■
■
■
■
SECTION 12.3
13–19
Use polar coordinates to find the volume of the given
■
x2 y2 4
14. Below the paraboloid z 18 2x 2 2y 2 and above the
xy-plane 15. A sphere of radius a 16. Inside the sphere x 2 y 2 z 2 16 and outside the
29. Use polar coordinates to combine the sum
cylinder x 2 y 2 4
y y
17. Above the cone z sx 2 y 2 and below the sphere
1
x y z 1 2
18. Bounded by the paraboloid z 1 2x 2 2y 2 and the
■
■
■
■
■
■
■
■
2
al
■
■
■
■
■
■
■
23.
y y
24.
s9x 2
3
0
yy
0
0
sa 2 y 2
25.
yy 1
s2y 2
0
y
26.
2
yy
s2xx 2
0
0
■
xy dy dx
■
y
2
2
ex y dy dx
dA
2
2
ex y dA
al
⺢2
■
yy e
x 2y 2
dA lim
dA
Sa
where Sa is the square with vertices a, a. Use this to show that
y
ex dx y 2
2
2
ey dy
(c) Deduce that
a
■
s4x 2
0
Da
x 2y 2
sinx y dy dx 2
x 2y 2
yy e
■
3
yy e
y y
Evaluate the iterated integral by converting to polar coordinates.
23–26
2
s2
where Da is the disk with radius a and center the origin. Show that
22. The region enclosed by the curve r 4 3 cos ■
y y
(b) An equivalent definition of the improper integral in part (a) is
Use a double integral to find the area of the region.
■
xy dy dx
lim
21. One loop of the rose r cos 3 ■
2
⺢2
■
through the center of a sphere of radius r 2 . Find the volume of the ring-shaped solid that remains. (b) Express the volume in part (a) in terms of the height h of the ring. Notice that the volume depends only on h, not on r 1 or r 2 . ■
x
0
I yy ex y dA y
20. (a) A cylindrical drill with radius r 1 is used to bore a hole
21–22
s2
1
2
4x 2 4y 2 z 2 64 ■
y y
30. (a) We define the improper integral (over the entire plane ⺢ 2
19. Inside both the cylinder x y 4 and the ellipsoid ■
xy dy dx
into one double integral. Then evaluate the double integral.
plane z 7 in the first octant
■
x
s1x 2
1s2
2
687
tern of radius 100 ft. It supplies water to a depth of er feet per hour at a distance of r feet from the sprinkler. (a) What is the total amount of water supplied per hour to the region inside the circle of radius R centered at the sprinkler? (b) Determine an expression for the average amount of water per hour per square foot supplied to the region inside the circle of radius R.
13. Under the cone z sx 2 y 2 and above the disk
2
■
28. An agricultural sprinkler distributes water in a circular pat-
solid.
2
DOUBLE INTEGRALS IN POLAR COORDINATES
■
y
x 2 y dx dy
y
■
sx 2 y 2 dy dx ■
2
ex dx s
(d) By making the change of variable t s2 x, show that
x y dx dy
■
■
■
■
■
■
27. A swimming pool is circular with a 40-ft diameter. The
depth is constant along east-west lines and increases linearly from 2 ft at the south end to 7 ft at the north end. Find the volume of water in the pool.
2
ex 2 dx s2
(This is a fundamental result for probability and statistics.)
■
31. Use the result of Exercise 30 part (c) to evaluate the follow-
ing integrals. (a)
y
0
2
x 2ex dx
(b)
y
0
sx ex dx
688
■
CHAPTER 12
12.4
MULTIPLE INTEGRALS
APPLICATIONS OF DOUBLE INTEGRALS We have already seen one application of double integrals: computing volumes. Another geometric application is finding areas of surfaces and this will be done in the next chapter. In this section we explore physical applications such as computing mass, electric charge, center of mass, and moment of inertia. DENSITY AND MASS
y
In Chapter 7 we were able to use single integrals to compute moments and the center of mass of a thin plate or lamina with constant density. But now, equipped with the double integral, we can consider a lamina with variable density. Suppose the lamina occupies a region D of the xy-plane and its density (in units of mass per unit area) at a point x, y in D is given by x, y, where is a continuous function on D. This means that
(x, y)
D
0
x
x, y lim
m A
FIGURE 1
y
(x i*j , yi*j )
where m and A are the mass and area of a small rectangle that contains x, y and the limit is taken as the dimensions of the rectangle approach 0. (See Figure 1.) To find the total mass m of the lamina we divide a rectangle R containing D into subrectangles Rij (as in Figure 2) and consider x, y to be 0 outside D. If we choose a point x ij*, yij* in Rij , then the mass of the part of the lamina that occupies Rij is approximately x ij*, yij* Aij , where Aij is the area of Rij . If we add all such masses, we get an approximation to the total mass:
Rij
k
0
m
x
k
1
y
y=1
(1, 1)
2
ij
m
lim
l
x*, y* A ij
max x i , yj l 0 i1 j1
ij
ij
yy x, y dA D
Q yy ! x, y dA D
y=1-x
FIGURE 3
ij
Physicists also consider other types of density that can be treated in the same manner. For example, if an electric charge is distributed over a region D and the charge density (in units of charge per unit area) is given by ! x, y at a point x, y in D, then the total charge Q is given by
D
0
ij
i1 j1
If we now take finer partitions by using smaller subrectangles, we obtain the total mass m of the lamina as the limiting value of the approximations:
FIGURE 2
1
l
x*, y* A
x
EXAMPLE 1 Charge is distributed over the triangular region D in Figure 3 so that
the charge density at x, y is ! x, y xy, measured in coulombs per square meter (Cm2 ). Find the total charge.
SECTION 12.4
APPLICATIONS OF DOUBLE INTEGRALS
■
689
SOLUTION From Equation 2 and Figure 3 we have
Q yy ! x, y dA y
D
y
1
x
0
1 2
y
1
1x
y1x
dx y
5 24
C.
2
1
0
0
1
xy dy dx
y1
y2 2
1 2x x dx 2
y
Thus the total charge is
1
0
3
x 2 1 1 x2 dx 2
2x 3 x4 3 4
1
0
5 24 ■
MOMENTS AND CENTERS OF MASS
In Section 7.5 we found the center of mass of a lamina with constant density; here we consider a lamina with variable density. Suppose the lamina occupies a region D and has density function x, y. Recall from Chapter 7 that we defined the moment of a particle about an axis as the product of its mass and its directed distance from the axis. We divide D into small rectangles as in Figure 2. Then the mass of Rij is approximately x *ij , y*ij Aij , so we can approximate the moment of Rij with respect to the x-axis by x *ij , y*ij Aij y*ij If we now add these quantities and take the limit as the subrectangles become smaller, we obtain the moment of the entire lamina about the x-axis: m
3
Mx
n
y* x*, y* A
lim
max x i , yj l 0 i1 j1
ij
ij
ij
ij
yy y x, y dA D
Similarly, the moment about the y-axis is m
4
(x, y)
FIGURE 4
D
My
n
x* x*, y* A
lim
max x i , yj l 0 i1 j1
ij
ij
ij
ij
yy x x, y dA D
As before, we define the center of mass x, y so that mx My and my Mx . The physical significance is that the lamina behaves as if its entire mass is concentrated at its center of mass. Thus the lamina balances horizontally when supported at its center of mass (see Figure 4). 5 The coordinates x, y of the center of mass of a lamina occupying the region D and having density function x, y are
x
My 1 m m
yy x x, y dA
y
D
Mx 1 m m
where the mass m is given by m
yy x, y dA D
yy y x, y dA D
690
■
CHAPTER 12
MULTIPLE INTEGRALS
V EXAMPLE 2 Find the mass and center of mass of a triangular lamina with vertices 0, 0, 1, 0, and 0, 2 if the density function is x, y 1 3x y.
SOLUTION The triangle is shown in Figure 5. (Note that the equation of the upper boundary is y 2 2x.) The mass of the lamina is
y (0, 2)
y=2-2x
m yy x, y dA y
3 11
D
” , ’ 8 16
D
y
1
0
0
(1, 0)
x
1
0
y 3xy
y
22x
0
y2 2
1 3x y dy dx
y22x
dx 4 y 1 x 2 dx 4 x 1
0
y0
x3 3
1
0
8 3
Then the formulas in (5) give FIGURE 5
x
1 m
3 8
y
3 2
yy x x, y dA y y 3 8
D
y
1
0
1
0
y2 xy 3x y x 2 2
x2 x4 2 4
1
0
3 8
y
1 4
0
1
0
y2 y2 y3 3x 2 2 3
7x 9
3 2
y
1
0
x x 3 dx
3 8 3 8
dx
y0
yy y x, y dA y y 1
x 3x 2 xy dy dx
y22x
1 m
D
22x
0
22x
y 3xy y 2 dy dx
0
y22x
dx 14 y 7 9x 3x 2 5x 3 dx 1
0
y0
x2 x4 x3 5 2 4
1
0
11 16
The center of mass is at the point ( 83 , 11 16 ).
■
V EXAMPLE 3 The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. Find the center of mass of the lamina.
SOLUTION Let’s place the lamina as the upper half of the circle x 2 y 2 a 2. (See
Figure 6.) Then the distance from a point x, y to the center of the circle (the origin) is sx 2 y 2 . Therefore, the density function is
y a
D
_a
FIGURE 6
≈+¥=a@
x, y Ksx 2 y 2
3a
”0, ’ 2π 0
a
x
where K is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then sx 2 y 2 r and the region D is given by 0 r a, 0 . Thus the mass of the lamina is m yy x, y dA yy Ksx 2 y 2 dA D
D
K y d y 0
a
0
r3 r dr K 3 2
a
0
y y 0
a
0
Kr r dr d
K a 3 3
Both the lamina and the density function are symmetric with respect to the y-axis, so
SECTION 12.4
APPLICATIONS OF DOUBLE INTEGRALS
■
691
the center of mass must lie on the y-axis, that is, x 0. The y-coordinate is given by y
Compare the location of the center of mass in Example 3 with Example 8 in Section 7.5 where we found that the center of mass of a lamina with the same shape but uniform density is located at the point 0, 4a3.
1 m
3
3
0
a
0
r sin *r r dr d
D
3 a3
■
yy y x, y dA K a y y y
0
sin d
y
a
0
3 r dr cos a3
[
3
]
0
r4 4
a
0
3 2a 4 3a 3 a 4 2 Therefore, the center of mass is located at the point 0, 3a2.
■
MOMENT OF INERTIA
The moment of inertia (also called the second moment) of a particle of mass m about an axis is defined to be mr 2, where r is the distance from the particle to the axis. We extend this concept to a lamina with density function x, y and occupying a region D by proceeding as we did for ordinary moments. We divide D into small rectangles, approximate the moment of inertia of each subrectangle about the x-axis, and take the limit of the sum as the subrectangles become smaller. The result is the moment of inertia of the lamina about the x-axis: m
6
Ix
n
y * x *, y * A
lim
max x i , yj l 0 i1 j1
ij
2
ij
ij
ij
yy y 2 x, y dA D
Similarly, the moment of inertia about the y-axis is m
7
Iy
n
x * x *, y * A
lim
max x i , yj l 0 i1 j1
ij
2
ij
ij
ij
yy x 2 x, y dA D
It is also of interest to consider the moment of inertia about the origin, also called the polar moment of inertia:
m
8
I0
lim
n
[x*
max x i , yj l 0 i1 j1
ij
2
]
y*ij 2 x *ij , y*ij Aij
yy x 2 y 2 x, y dA D
Note that I 0 I x I y . V EXAMPLE 4 Find the moments of inertia I x , I y , and I 0 of a homogeneous disk D with density x, y , center the origin, and radius a.
SOLUTION The boundary of D is the circle x 2 y 2 a 2 and in polar coordinates
692
■
CHAPTER 12
MULTIPLE INTEGRALS
D is described by 0 2, 0 r a. Let’s compute I 0 first: I 0 yy x 2 y 2 dA y
2
0
y
0
d y
a
0
a
0
r 2 r dr d
D 2
y
r4 r dr 2 4 3
a
0
a 4 2
Instead of computing I x and I y directly, we use the facts that I x I y I 0 and I x I y (from the symmetry of the problem). Thus Ix Iy
I0 a 4 2 4
■
In Example 4 notice that the mass of the disk is m density area a 2 so the moment of inertia of the disk about the origin (like a wheel about its axle) can be written as I0
a 4 12 a 2 a 2 12 ma 2 2
Thus if we increase the mass or the radius of the disk, we thereby increase the moment of inertia. In general, the moment of inertia plays much the same role in rotational motion that mass plays in linear motion. The moment of inertia of a wheel is what makes it difficult to start or stop the rotation of the wheel, just as the mass of a car is what makes it difficult to start or stop the motion of the car. The radius of gyration of a lamina about an axis is the number R such that mR 2 I
9
where m is the mass of the lamina and I is the moment of inertia about the given axis. Equation 9 says that if the mass of the lamina were concentrated at a distance R from the axis, then the moment of inertia of this “point mass” would be the same as the moment of inertia of the lamina. In particular, the radius of gyration y with respect to the x-axis and the radius of gyration x with respect to the y-axis are given by the equations 10
my 2 I x
mx 2 I y
Thus x, y is the point at which the mass of the lamina can be concentrated without changing the moments of inertia with respect to the coordinate axes. (Note the analogy with the center of mass.) V EXAMPLE 5
Find the radius of gyration about the x-axis of the disk in Example 4.
SOLUTION As noted, the mass of the disk is m a 2, so from Equations 10 we
have y2
1 Ix a 4 a2 4 2 m a 4
Therefore, the radius of gyration about the x-axis is y 12 a , half the radius of the disk. ■
SECTION 12.5
12.4
isosceles right triangle with equal sides of length a if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse. 14. A lamina occupies the region inside the circle x 2 y 2 2y
2. Electric charge is distributed over the disk
but outside the circle x 2 y 2 1. Find the center of mass if the density at any point is inversely proportional to its distance from the origin.
x y 4 so that the charge density at x, y is ! x, y x y x 2 y 2 (measured in coulombs per square meter). Find the total charge on the disk. 2
15. Find the moments of inertia I x , I y , I 0 for the lamina of
Find the mass and center of mass of the lamina that occupies the region D and has the given density function . ■
0 x 2, 1 y 1 ; x, y xy D x, y 0 x a, 0 y b ; x, y cxy
3. D x, y 4.
Exercise 7. 16. Find the moments of inertia I x , I y , I 0 for the lamina of
2
Exercise 12. 17. Find the moments of inertia I x , I y , I 0 for the lamina of
Exercise 9.
5. D is the triangular region with vertices 0, 0, 2, 1, 0, 3;
x, y x y
18. Consider a square fan blade with sides of length 2 and the
lower left corner placed at the origin. If the density of the blade is x, y 1 0.1x, is it more difficult to rotate the blade about the x-axis or the y-axis?
6. D is the triangular region with vertices 0, 0, 1, 1, 4, 0;
x, y x
7. D is bounded by y e x, y 0, x 0, and x 1;
CAS
x, y y
8. D is bounded by y sx , y 0, and x 1;
x, y x
10. D x, y ■
■
■
x, y 3
0 y cos x, ■
■
■
0 y sin x,
0 x ; x, y xy
20. D is enclosed by the cardioid r 1 cos ;
x, y sx 2 y 2
0 x 2 ; x, y x ■
■
■
■
■
■
11. A lamina occupies the part of the disk x 2 y 2 1 in the
first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis. 12. Find the center of mass of the lamina in Exercise 11 if the
density at any point is proportional to the square of its distance from the origin.
12.5
■ Use a computer algebra system to find the mass, center of mass, and moments of inertia of the lamina that occupies the region D and has the given density function.
19–20
19. D x, y
9. D is bounded by the parabola x y 2 and the line
y x 2;
693
13. Find the center of mass of a lamina in the shape of an
1 x 3, 0 y 2 so that the charge density at x, y is ! x, y 2x y y 2 (measured in coulombs per square meter). Find the total charge on the rectangle.
3–10
■
EXERCISES
1. Electric charge is distributed over the rectangle
2
TRIPLE INTEGRALS
■
■
■
■
■
■
■
■
■
■
■
■
21. A lamina with constant density x, y occupies a
square with vertices 0, 0, a, 0, a, a, and 0, a. Find the moments of inertia I x and I y and the radii of gyration x and y.
22. A lamina with constant density x, y occupies the
region under the curve y sin x from x 0 to x . Find the moments of inertia I x and I y and the radii of gyration x and y.
TRIPLE INTEGRALS Just as we defined single integrals for functions of one variable and double integrals for functions of two variables, so we can define triple integrals for functions of three variables. Let’s first deal with the simplest case where f is defined on a rectangular box: 1
B x, y, z a x b, c y d, r z s
The first step is to divide B into sub-boxes. We do this by dividing the interval a, b into l subintervals x i1, x i with lengths x i x i x i1, dividing c, d into m subintervals with lengths yj yj yj1, and dividing r, s into n subintervals with lengths zk zk zk1. The planes through the endpoints of these subintervals paral-
694
■
CHAPTER 12
MULTIPLE INTEGRALS
lel to the coordinate planes divide the box B into lmn sub-boxes
z
Bi jk x i1, x i yj1, yj zk1, zk
B
which are shown in Figure 1. The sub-box Bi jk has volume Vijk x i yj zk . Then we form the triple Riemann sum x
y
l
m
n
f x * , y * , z * V
2
ij k
ijk
ijk
ijk
i1 j1 k1
Bijk
where the sample point xi*jk , yi*jk , zi*jk is in Bi jk . By analogy with the definition of a double integral (12.1.5), we define the triple integral as the limit of the triple Riemann sums in (2) as the sub-boxes shrink.
Îz k Îy j
Îx i 3 DEFINITION
z
The triple integral of f over the box B is
yyy f x, y, z dV B
l
m
n
f x * , y * , z * V
lim
i jk
max x i , yj , zk l 0 i1 j1 k1
i jk
i jk
ijk
if this limit exists. x
FIGURE 1
y
Again, the triple integral always exists if f is continuous. We can choose the sample point to be any point in the sub-box, but if we choose it to be the point x i, yj, zk , and if we choose sub-boxes with the same dimensions. so that Vijk V, we get a simplerlooking expression for the triple integral:
yyy f x, y, z dV B
l
m
n
f x , y , z V
lim
l, m, n l i1 j1 k1
i
j
k
Just as for double integrals, the practical method for evaluating triple integrals is to express them as iterated integrals as follows. 4 FUBINI’S THEOREM FOR TRIPLE INTEGRALS If f is continuous on the rectangular box B a, b c, d r, s , then
yyy f x, y, z dV y y y s
r
d
c
b
a
f x, y, z dx dy dz
B
The iterated integral on the right side of Fubini’s Theorem means that we integrate first with respect to x (keeping y and z fixed), then we integrate with respect to y (keeping z fixed), and finally we integrate with respect to z. There are five other possible orders in which we can integrate, all of which give the same value. For instance, if we integrate with respect to y, then z, and then x, we have
yyy f x, y, z dV y y y b
a
B
s
r
d
c
f x, y, z dy dz dx
SECTION 12.5
TRIPLE INTEGRALS
■
695
Evaluate the triple integral xxxB xyz 2 dV, where B is the rectangular
V EXAMPLE 1
box given by
B x, y, z 0 x 1, 1 y 2, 0 z 3 SOLUTION We could use any of the six possible orders of integration. If we choose
to integrate with respect to x, then y, and then z, we obtain
yyy
xyz 2 dV y
3
0
B
y y 2
1
y y 3
2
0
1
1
0
xyz 2 dx dy dz
yz 2 dy dz 2
y
3
0
yy 3
2
0
1
y 2z 2 4
x 2 yz 2 2
x1
dy dz
x0
y2
dz
y1
y
3
0
3z 2 z3 dz 4 4
3
0
27 4
■
Now we define the triple integral over a general bounded region E in threedimensional space (a solid) by much the same procedure that we used for double integrals (12.2.2). We enclose E in a box B of the type given by Equation 1. Then we define a function F so that it agrees with f on E but is 0 for points in B that are outside E. By definition,
yyy f x, y, z dV yyy Fx, y, z dV E
z
z=u™ (x, y) E z=u¡ (x, y)
This integral exists if f is continuous and the boundary of E is “reasonably smooth.” The triple integral has essentially the same properties as the double integral (Properties 6 –9 in Section 12.2). We restrict our attention to continuous functions f and to certain simple types of regions. A solid region E is said to be of type 1 if it lies between the graphs of two continuous functions of x and y, that is,
0 y
D
x
B
5
E x, y, z x, y D, u 1x, y z u 2x, y
where D is the projection of E onto the xy-plane as shown in Figure 2. Notice that the upper boundary of the solid E is the surface with equation z u 2x, y, while the lower boundary is the surface z u1x, y. By the same sort of argument that led to (12.2.3), it can be shown that if E is a type 1 region given by Equation 5, then
FIGURE 2
A type 1 solid region
z
z=u™(x, y)
6
yyy f x, y, z dV yy y E
D
u 2x, y
u1x, y
f x, y, z dz dA
E z=u¡(x, y) a x
b
0
y=g¡(x)
D
FIGURE 3
A type 1 solid region
y
y=g™(x)
The meaning of the inner integral on the right side of Equation 6 is that x and y are held fixed, and therefore u1x, y and u 2x, y are regarded as constants, while f x, y, z is integrated with respect to z. In particular, if the projection D of E onto the xy-plane is a type I plane region (as in Figure 3), then
E x, y, z a x b, t1x y t2x, u1x, y z u 2x, y
696
■
CHAPTER 12
z
MULTIPLE INTEGRALS
and Equation 6 becomes
z=u™(x, y) E
z=u¡(x, y)
t2x
yyy f x, y, z dV y y y b
7
t1x
a
E
u 2x, y
f x, y, z dz dy dx
u1x, y
x=h¡(y) 0
c
d y
x
If, on the other hand, D is a type II plane region (as in Figure 4), then
D
E x, y, z c y d, h1y x h2y, u1x, y z u 2x, y
x=h™(y) FIGURE 4
and Equation 6 becomes
Another type 1 solid region h2 y
yyy f x, y, z dV y y d
8
h1 y
c
E
z
y
u 2x, y
f x, y, z dz dx dy
u1x, y
EXAMPLE 2 Evaluate xxxE z dV, where E is the solid tetrahedron bounded by the
(0, 0, 1)
four planes x 0, y 0, z 0, and x y z 1.
z=1-x-y
SOLUTION When we set up a triple integral it’s wise to draw two diagrams: one of E
the solid region E (see Figure 5) and one of its projection D on the xy-plane (see Figure 6). The lower boundary of the tetrahedron is the plane z 0 and the upper boundary is the plane x y z 1 (or z 1 x y), so we use u1x, y 0 and u 2x, y 1 x y in Formula 7. Notice that the planes x y z 1 and z 0 intersect in the line x y 1 (or y 1 x) in the xy-plane. So the projection of E is the triangular region shown in Figure 6, and we have
(0, 1, 0)
0
y
(1, 0, 0) x
z=0
FIGURE 5 y
E x, y, z 0 x 1, 0 y 1 x, 0 z 1 x y
9
1
This description of E as a type 1 region enables us to evaluate the integral as follows:
y=1-x D
yyy z dV y y y 1
1x
0
0
1xy
0
z dz dy dx
0
E
0
y=0
1
z
x
E
D
1 2
1 6
yy 1
0
y
1
0
1x
0
1 x y dy dx 2
1 2
1 1 x dx 6 3
1 x4 4
dy dx
z0
1 x y3 3
1
0
1
0
1 24
y1x
dx
y0
■
E x, y, z y, z D, u1y, z x u 2y, z
y
where, this time, D is the projection of E onto the yz-plane (see Figure 7). The back surface is x u1y, z, the front surface is x u 2y, z, and we have
x=u™(y, z)
A type 2 region
0
z1xy
A solid region E is of type 2 if it is of the form
x=u¡(y, z)
FIGURE 7
y z2 2
1x
x
FIGURE 6
0
yy 1
10
yyy f x, y, z dV yy y E
D
u 2 y, z
u1 y, z
f x, y, z dx dA
SECTION 12.5
z
TRIPLE INTEGRALS
■
697
Finally, a type 3 region is of the form
E x, y, z x, z D, u1x, z y u 2x, z
y=u™(x, z)
where D is the projection of E onto the xz-plane, y u1x, z is the left surface, and y u 2x, z is the right surface (see Figure 8). For this type of region we have
D E 0
y=u¡(x, z)
y
x
E
FIGURE 8
yyy f x, y, z dV yy y
11
D
u 2x, z
u1x, z
f x, y, z dy dA
In each of Equations 10 and 11 there may be two possible expressions for the integral depending on whether D is a type I or type II plane region (and corresponding to Equations 7 and 8).
A type 3 region
Evaluate xxxE sx 2 z 2 dV, where E is the region bounded by the paraboloid y x 2 z 2 and the plane y 4. V EXAMPLE 3
SOLUTION The solid E is shown in Figure 9. If we regard it as a type 1 region, then we
need to consider its projection D1 onto the xy-plane, which is the parabolic region in Figure 10. (The trace of y x 2 z 2 in the plane z 0 is the parabola y x 2.) y
z
y=≈+z@
Visual 12.5 illustrates how solid regions (including the one in Figure 9) project onto coordinate planes.
y=4 D¡
E
y=≈
0 4
y 0
x
FIGURE 9
FIGURE 10
Region of integration
Projection on xy-plane
x
From y x 2 z 2 we obtain z sy x 2 , so the lower boundary surface of E is z sy x 2 and the upper surface is z sy x 2 . Therefore, the description of E as a type 1 region is
E x, y, z 2 x 2, x 2 y 4, sy x 2 z sy x 2 z
and so we obtain
≈+z@=4 D£ _2
0
FIGURE 11
Projection on xz-plane
2
x
yyy sx E
2
z 2 dV y
2
2
y y 4
syx 2
2
syx 2
x
sx 2 z 2 dz dy dx
Although this expression is correct, it is extremely difficult to evaluate. So let’s instead consider E as a type 3 region. As such, its projection D3 onto the xz-plane is the disk x 2 z 2 4 shown in Figure 11. Then the left boundary of E is the paraboloid y x 2 z 2 and the right boundary is the plane y 4, so taking u1x, z x 2 z 2 and u 2x, z 4 in Equation 11, we
698
■
CHAPTER 12
MULTIPLE INTEGRALS
have
yyy sx
z 2 dV yy
2
E
D3
y
4
x 2z2
sx 2 z 2 dy dA
yy 4 x 2 z 2 sx 2 z 2 dA D3
Although this integral could be written as
y y | The most difficult step in evaluating a triple integral is setting up an expression for the region of integration (such as Equation 9 in Example 2). Remember that the limits of integration in the inner integral contain at most two variables, the limits of integration in the middle integral contain at most one variable, and the limits of integration in the outer integral must be constants.
2
s4x 2
2
s4x 2
4 x 2 z 2 sx 2 z 2 dz dx
it’s easier to convert to polar coordinates in the xz-plane: x r cos , z r sin . This gives
yyy sx
2
z 2 dV yy 4 x 2 z 2 sx 2 z 2 dA
E
2
y y 0
2
0
4 r 2 r r dr d
D3
y
2
0
d y
2
0
4r 3 r5 4r r dr 2 3 5 2
4
2
0
128 15
■
APPLICATIONS OF TRIPLE INTEGRALS
Recall that if f x 0, then the single integral xab f x dx represents the area under the curve y f x from a to b, and if f x, y 0, then the double integral xxD f x, y dA represents the volume under the surface z f x, y and above D. The corresponding interpretation of a triple integral xxxE f x, y, z dV , where f x, y, z 0, is not very useful because it would be the “hypervolume” of a four-dimensional object and, of course, that is very difficult to visualize. (Remember that E is just the domain of the function f ; the graph of f lies in four-dimensional space.) Nonetheless, the triple integral xxxE f x, y, z dV can be interpreted in different ways in different physical situations, depending on the physical interpretations of x, y, z and f x, y, z. Let’s begin with the special case where f x, y, z 1 for all points in E. Then the triple integral does represent the volume of E: VE yyy dV
12
E
For example, you can see this in the case of a type 1 region by putting f x, y, z 1 in Formula 6:
yyy 1 dV yy y E
D
u 2x, y
u1x, y
dz dA yy u 2x, y u1x, y dA D
and from Section 12.2 we know this represents the volume that lies between the surfaces z u1x, y and z u 2x, y. EXAMPLE 4 Use a triple integral to find the volume of the tetrahedron T bounded
by the planes x 2y z 2, x 2y, x 0, and z 0.
SECTION 12.5
z
VT
x+2y+z=2
x=2y T
0
1x2
x2
y
2x2y
0
dz dy dx
1
y
1x2
x2
2 x 2y dy dx 13
by the same calculation as in Example 4 in Section 12.2. (Notice that it is not necessary to use triple integrals to compute volumes. They simply give an alternative method for setting up the calculation.) ■
1
x
FIGURE 12
All the applications of double integrals in Section 12.4 can be immediately extended to triple integrals. For example, if the density function of a solid object that occupies the region E is x, y, z, in units of mass per unit volume, at any given point x, y, z, then its mass is
y
x+2y=2 x ”or y=1- ’ 2
m yyy x, y, z dV
13
”1, 21 ’
E
x y= 2
1
1
0
(0, 1, 0)
D
yyy dV y y
y
”1, 2 , 0’
FIGURE 13
699
T
y
0
0
■
SOLUTION The tetrahedron T and its projection D on the xy-plane are shown in Figures 12 and 13. The lower boundary of T is the plane z 0 and the upper boundary is the plane x 2y z 2, that is, z 2 x 2y. Therefore, we have
(0, 0, 2)
1
TRIPLE INTEGRALS
x
and its moments about the three coordinate planes are 14
Myz
yyy x x, y, z dV
Mxz
yyy y x, y, z dV
E
E
Mx y yyy z x, y, z dV E
The center of mass is located at the point x, y, z, where x
15
Myz m
y
Mxz m
z
Mxy m
If the density is constant, the center of mass of the solid is called the centroid of E. The moments of inertia about the three coordinate axes are 16
Ix yyy y 2 z 2 x, y, z dV E
Iy yyy x 2 z 2 x, y, z dV E
Iz yyy x 2 y 2 x, y, z dV E
As in Section 12.4, the total electric charge on a solid object occupying a region E and having charge density ! x, y, z is Q yyy ! x, y, z dV E
700
■
CHAPTER 12
MULTIPLE INTEGRALS
z
V EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x y 2 and the planes x z, z 0, and x 1.
z=x E
SOLUTION The solid E and its projection onto the xy-plane are shown in Figure 14. The lower and upper surfaces of E are the planes z 0 and z x, so we describe E as a type 1 region:
0 y
1
x
E x, y, z 1 y 1, y 2 x 1, 0 z x Then, if the density is x, y, z , the mass is
y
m x=¥
yyy dV y y y E
D
x=1
x
0
y
1
1
1
y2
1
dz dx dy y
1
1
x1
x2 2
1
x
0
dy
xy 2
2
y
1
1
1
0
1
x dx dy
y2
1 y 4 dy
y 1 y 4 dy y
y
1
y5 5
0
4 5
Because of the symmetry of E and about the xz-plane, we can immediately say that Mxz 0 and, therefore, y 0. The other moments are
FIGURE 14
Myz yyy x dV y E
y
1
1
x3 3
x1
xy 2
Mxy yyy z dV y
2
1
1
E
1
y y 1
1
1
y2
y y 1
1
y2
y y 1
y2
x
0
x 2 dx dy 3
y
1
1
y2
x 2 dx dy
y y
2 1 y dy 3 6
1
1
1
y2
1 y 6 dy
0
y
Myz Mxz Mxy , , m m m
1
y7 y 7
z2 2
0
4 7
zx
dx dy
z0
2 7
( 57 , 0, 145 )
■
EXERCISES
1. Evaluate the integral in Example 1, integrating first with
3–6
respect to z , then x, and then y. 2. Evaluate the integral xxxE xz y 3 dV , where
E x, y, z
y
1
z dz dx dy
0
x, y, z
1
1
2 dy 3
Therefore, the center of mass is
12.5
x dz dx dy y
x
0
1 x 1, 0 y 2, 0 z 1
using three different orders of integration.
■
z
3.
yyy
5.
yyy
1
0
0
3
0
■
Evaluate the iterated integral.
1
0
■
xz
0
6xz dy dx dz
s1z 2
0 ■
ze y dx dz dy ■
■
4.
yy y
6.
yyy
1
0
1
0
■
■
2x
x
■
y
0
z
y
0
0
2xyz dz dy dx 2
zey dx dy dz ■
■
■
■
SECTION 12.5
7–16 7.
Evaluate the triple integral.
■
9.
xxxE yz cosx
11.
xxxE xy dV,
14.
where E is bounded by the planes x 0, y 0, z 0, and 2x 2y z 4
■ Use the Midpoint Rule for triple integrals (Exercise 22) to estimate the value of the integral. Divide B into eight sub-boxes of equal size.
23.
B x, y, z
xxxE xz dV,
where E is the solid tetrahedron with vertices 0, 0, 0, 0, 1, 0, 1, 1, 0, and 0, 1, 1
■
■
xxxE x 2y dV, where E is bounded by the parabolic
yy y
xxxE x dV, where E is bounded by the paraboloid
26.
yy y
■
xxxE z dV, where E is bounded by the cylinder y z 9 and the planes x 0, y 3x, and z 0 in the first octant 2
■
17–20
■
■
■
■
■
■
■
■
■
2
■
■
Use a triple integral to find the volume of the given
plane 2x y z 4 18. The solid bounded by the cylinder y x and the planes 2
1
0
1x
0
2
0
2y
4y 2
0
0
■
22z
0
■
■
■
dx dz dy
■
■
■
■
y 0,
■
■
■
■
■
■
y6
28. z 0,
x 0,
y 2,
z y 2x
29. z 0,
z y,
x 1y 2
■
■
■
■
1
0
■
■
dy dz dx
■
y y y
plane x 16
■
■
■
■
■
■
■
■
31. The figure shows the region of integration for the integral
20. The solid enclosed by the paraboloid x y 2 z 2 and the ■
■
2
planes y z 5 and z 1
■
■
27–30 ■ Express the integral xxxE f x, y, z dV as an iterated integral in six different ways, where E is the solid bounded by the given surfaces.
■
19. The solid enclosed by the cylinder x y 9 and the
■
0 y 2, 0 z 1
■
30. 9x 2 4y 2 z 2 1
z 0, z 4, and y 9
■
■
27. x 2 z 2 4,
17. The tetrahedron enclosed by the coordinate planes and the
2
0 x 4,
Sketch the solid whose volume is given by the iterated
25.
solid.
■
■
■
1
sx
1y
0
f x, y, z dz dy dx
■
21. (a) Express the volume of the wedge in the first octant that
CAS
■
integral.
e dV, where E is bounded by the parabolic cylinder z 1 y 2 and the planes z 0, x 1, and x 1
■
■
■
25–26
2 y
x 4y 2 4z 2 and the plane x 4
16.
xxxB sinxy 2z 3 dV, where
■
cylinder y x 2 and the planes x z, x y, and z 0
15.
1 dV, where ln1 x y z B x, y, z 0 x 4, 0 y 8, 0 z 4
xxxB
24.
where E is the solid tetrahedron with vertices 0, 0, 0, 1, 0, 0, 0, 2, 0, and 0, 0, 3
xxxE x
701
23–24
xxxE 6xy dV, where E lies under the plane z 1 x y and above the region in the xy-plane bounded by the curves y sx , y 0, and x 1 xxxE y dV,
13.
0 x s4 y 2 , 0 z y}
dV, where E x, y, z 0 x 1, 0 y x, x z 2x
10.
12.
0 y 2,
5
■
(b) Use a computer algebra system to approximate the integral in part (a) correct to the nearest integer. Compare with the answer to part (a).
xxxE 2x dV, where E {x, y, z
8.
CAS
TRIPLE INTEGRALS
is cut from the cylinder y 2 z 2 1 by the planes y x and x 1 as a triple integral. (b) Use either the Table of Integrals (on Reference Pages 6 –10) or a computer algebra system to find the exact value of the triple integral in part (a).
Rewrite this integral as an equivalent iterated integral in the five other orders. z 1
z=1-y
22. (a) In the Midpoint Rule for triple integrals we use a
triple Riemann sum to approximate a triple integral over a box B, where f x, y, z is evaluated at the center xi , yj , zk of the box Bijk . Use the Midpoint Rule to estimate xxxB sx 2 y 2 z 2 dV , where B is the cube defined by 0 x 4, 0 y 4, 0 z 4. Divide B into eight cubes of equal size.
0 1
y=œ„ x x
y
702
■
CHAPTER 12
MULTIPLE INTEGRALS
32. The figure shows the region of integration for the integral
yy 1
1x 2
0
0
y
1x
CAS
41. Let E be the solid in the first octant bounded by the cylinder
x 2 y 2 1 and the planes y z, x 0, and z 0 with the density function x, y, z 1 x y z. Use a computer algebra system to find the exact values of the following quantities for E. (a) The mass (b) The center of mass (c) The moment of inertia about the z-axis
f x, y, z dy dz dx
0
Rewrite this integral as an equivalent iterated integral in the five other orders. z 1
CAS
z=1-≈
42. If E is the solid of Exercise 16 with density function
x, y, z x 2 y 2, find the following quantities, correct to three decimal places. (a) The mass (b) The center of mass (c) The moment of inertia about the z-axis
0 1 1
x
y
y=1-x
43. Find the moments of inertia for a cube of constant density k
and side length L if one vertex is located at the origin and three edges lie along the coordinate axes.
Write five other iterated integrals that are equal to the given iterated integral.
33–34
■
33.
yyy
34.
yy y
■
1
0
1
y
1
x2
0
0
■
y
44. Find the moments of inertia for a rectangular brick with
dimensions a, b, and c, mass M, and constant density if the center of the brick is situated at the origin and the edges are parallel to the coordinate axes.
f x, y, z dz dx dy
0 y
0 ■
f x, y, z dz dy dx ■
■
The average value of a function f x, y, z over a solid region E is defined to be 45– 46
■
■
■
■
■
■
■
■
■ Find the mass and center of mass of the solid E with the given density function .
35–38
35. E is the solid of Exercise 9;
fave
x, y, z 2
planes x z 1, x 0, and z 0; x, y, z 4
37. E is the cube given by 0 x a, 0 y a, 0 z a ;
x, y, z x y z 2
x, y, z y
z 0, x y z 1; ■
■
■
■
■
■
■
the cube with side length L that lies in the first octant with one vertex at the origin and edges parallel to the coordinate axes. 46. Find the average value of the function
■
■
■
■
39. The solid of Exercise 19;
x, y, z sx 2 y 2
x, y, z sx 2 y 2 z 2 ■
■
■
■
■
■
■
■
■
■
E ■
■
■
■
■
■
■
■
■
■
47. Find the region E for which the triple integral
yyy 1 x
40. The hemisphere x 2 y 2 z 2 1, z 0; ■
f x, y, z x 2 z y 2 z over the region enclosed by the paraboloid z 1 x 2 y 2 and the plane z 0.
■
Set up, but do not evaluate, integral expressions for (a) the mass, (b) the center of mass, and (c) the moment of inertia about the z -axis.
39– 40
E
45. Find the average value of the function f x, y, z xyz over
2
38. E is the tetrahedron bounded by the planes x 0, y 0, ■
yyy f x, y, z dV
where VE is the volume of E. For instance, if is a density function, then ave is the average density of E.
36. E is bounded by the parabolic cylinder z 1 y 2 and the
2
1 VE
is a maximum.
2
2y 2 3z 2 dV
■
■
■
SECTION 12.6
12.6
TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
703
TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
y P(r, ¨ )=P (x, y)
r
In plane geometry the polar coordinate system is used to give a convenient description of certain curves and regions. (See Section 9.3.) Figure 1 enables us to recall the connection between polar and Cartesian coordinates. If the point P has Cartesian coordinates x, y and polar coordinates r, , then, from the figure,
y
x r cos
¨ O
■
r2 x2 y2
x
x
FIGURE 1
y r sin y tan x
In three dimensions there is a coordinate system, called cylindrical coordinates, that is similar to polar coordinates and gives convenient descriptions of some commonly occurring surfaces and solids. As we will see, some triple integrals are much easier to evaluate in cylindrical coordinates. CYLINDRICAL COORDINATES
In the cylindrical coordinate system, a point P in three-dimensional space is represented by the ordered triple r, , z, where r and are polar coordinates of the projection of P onto the xy-plane and z is the directed distance from the xy-plane to P (see Figure 2). To convert from cylindrical to rectangular coordinates, we use the equations
z P (r, ¨, z)
z
O
r
¨ x
y
1
x r cos
y r sin
zz
(r, ¨, 0)
whereas to convert from rectangular to cylindrical coordinates, we use
FIGURE 2
The cylindrical coordinates of a point 2
r2 x2 y2
tan
y x
zz
EXAMPLE 1
(a) Plot the point with cylindrical coordinates 2, 23, 1 and find its rectangular coordinates. (b) Find cylindrical coordinates of the point with rectangular coordinates 3, 3, 7. SOLUTION
z
(a) The point with cylindrical coordinates 2, 23, 1 is plotted in Figure 3. From Equations 1, its rectangular coordinates are
2π
”2, , 1’ 3 1
2 1 2 3 2
1
y 2 sin
2 s3 2 3 2
s3
2 0 2π 3
x
FIGURE 3
y
x 2 cos
z1 Thus the point is (1, s3 , 1) in rectangular coordinates.
704
■
CHAPTER 12
MULTIPLE INTEGRALS
(b) From Equations 2 we have r s3 2 32 3s2 z
tan
3 1 3
so
7 2n 4
z 7 0
(0, c, 0) y
(c, 0, 0)
Therefore, one set of cylindrical coordinates is (3s2 , 74, 7). Another is (3s2 , 4, 7). As with polar coordinates, there are infinitely many choices.
■
Cylindrical coordinates are useful in problems that involve symmetry about an axis, and the z-axis is chosen to coincide with this axis of symmetry. For instance, the axis of the circular cylinder with Cartesian equation x 2 y 2 c 2 is the z-axis. In cylindrical coordinates this cylinder has the very simple equation r c. (See Figure 4.) This is the reason for the name “cylindrical” coordinates.
x
FIGURE 4
r=c, a cylinder
V EXAMPLE 2
Describe the surface whose equation in cylindrical coordinates
is z r.
z
SOLUTION The equation says that the z-value, or height, of each point on the surface is the same as r, the distance from the point to the z-axis. Because doesn’t appear, it can vary. So any horizontal trace in the plane z k k 0 is a circle of radius k. These traces suggest that the surface is a cone. This prediction can be confirmed by converting the equation into rectangular coordinates. From the first equation in (2) we have
0
y x
z2 r 2 x 2 y 2 We recognize the equation z 2 x 2 y 2 (by comparison with Table 1 in Section 10.6) as being a circular cone whose axis is the z-axis (see Figure 5). ■
FIGURE 5
z=r, a cone
EVALUATING TRIPLE INTEGRALS WITH CYLINDRICAL COORDINATES z
z=u™(x, y)
Suppose that E is a type 1 region whose projection D on the xy-plane is conveniently described in polar coordinates (see Figure 6). In particular, suppose that f is continuous and
E x, y, z x, y D, u1x, y z u 2x, y r=h¡(¨ ) 0
¨=b x
¨=a
where D is given in polar coordinates by
z=u¡(x, y)
D r, , h1 r h 2
y
D r=h™(¨)
We know from Equation 12.5.6 that
FIGURE 6 3
yyy f x, y, z dV yy y E
D
u 2x, y
u1x, y
f x, y, z dz dA
But we also know how to evaluate double integrals in polar coordinates. In fact, com-
SECTION 12.6
TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
■
705
bining Equation 3 with Equation 12.3.3, we obtain
4
yyy f x, y, z dV y y E
dz
r r d¨
y
u 2r cos , r sin
u1r cos , r sin
f r cos , r sin , z r dz dr d
Formula 4 is the formula for triple integration in cylindrical coordinates. It says that we convert a triple integral from rectangular to cylindrical coordinates by writing x r cos , y r sin , leaving z as it is, using the appropriate limits of integration for z, r, and , and replacing dV by r dz dr d. (Figure 7 shows how to remember this.) It is worthwhile to use this formula when E is a solid region easily described in cylindrical coordinates, and especially when the function f x, y, z involves the expression x 2 y 2.
z
d¨
h2
h1
dr
A solid E lies within the cylinder x 2 y 2 1, below the plane z 4, and above the paraboloid z 1 x 2 y 2. (See Figure 8.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E. V EXAMPLE 3
FIGURE 7
Volume element in cylindrical coordinates: dV=r dz dr d¨ z
SOLUTION In cylindrical coordinates the cylinder is r 1 and the paraboloid is
z=4
z 1 r 2, so we can write
(0, 0, 4)
E r, , z 0 2, 0 r 1, 1 r 2 z 4 Since the density at x, y, z is proportional to the distance from the z-axis, the density function is f x, y, z Ksx 2 y 2 Kr
(0, 0, 1)
z=1-r@ 0
y
(1, 0, 0)
where K is the proportionality constant. Therefore, from Formula 12.5.13, the mass of E is m yyy Ksx 2 y 2 dV y
2
0
x
yy 1
0
4
1r 2
Kr r dz dr d
E
y
FIGURE 8
2
0
y
1
0
Kr 2 4 1 r 2 dr d K y
2K r 3
EXAMPLE 4 Evaluate y
z
2
2
r5 5
y
1
s4x 2
s4x 2
d y 3r 2 r 4 dr 1
0
12K 5
0
z=2
2
0
y
2
sx 2 y 2
■
x 2 y 2 dz dy dx .
SOLUTION This iterated integral is a triple integral over the solid region 2
z=œ„„„„„ ≈+¥
2 x
2
FIGURE 9
y
E x, y, z 2 x 2, s4 x 2 y s4 x 2 , sx 2 y 2 z 2 and the projection of E onto the xy-plane is the disk x 2 y 2 4. The lower surface of E is the cone z sx 2 y 2 and its upper surface is the plane z 2. (See Figure 9.) This region has a much simpler description in cylindrical coordinates:
E r, , z 0 2, 0 r 2, r z 2
706
■
CHAPTER 12
MULTIPLE INTEGRALS
Therefore, we have
y y 2
s4x 2
2
s4x 2
y
2
sx y 2 2
x 2 y 2 dz dy dx yyy x 2 y 2 dV y
2
y y
2
d y r 32 r dr 2
[
0
2
0
r 2 r dz dr d
r
E
y
2
0
2
0
1 2
r 4 15 r 5
]
2 0
165 ■
12.6
EXERCISES
■ Plot the point whose cylindrical coordinates are given. Then find the rectangular coordinates of the point.
1–2
1. (a) 2, 4, 1
(b) 4, 3, 5
2. (a) 1, , e
(b) 1, 32, 2
■
■
3– 4
■
■
■
■
■
■
■
15. ■
3. (a) 1, 1, 4
(b) (1, s3 , 2)
4. (a) 3, 3, 2
(b) 3, 4, 5
■
5–6
■
■
■
■
■
■
■
■
■
7. z r 2 ■
9–10
■
■
■
■
■
■
■
■
■
■
■
■
■
11–12
■
■
■
■
■
■
■
■
■
■
■
■
■
■
2
0
■
9r 2
0 ■
r dz dr d ■
■
■
Use cylindrical coordinates.
and z x y 5 and by the cylinders x 2 y 2 4 and x 2 y 2 9.
21. Evaluate xxxE x 2 dV, where E is the solid that lies within
the cylinder x 2 y 2 1, above the plane z 0, and below the cone z 2 4x 2 4y 2.
■
■
■
22. Find the volume of the solid that lies within both the cylin-
der x 2 y 2 1 and the sphere x 2 y 2 z 2 4. 23. (a) Find the volume of the region E bounded by the parabo-
11. r 2 z 2 r 2
■
■
0
z 1 x 2 y 2, the cylinder x 2 y 2 5, and the xy-plane.
Sketch the solid described by the given inequalities.
12. 0 2,
■
2
y yy
20. Evaluate xxxE x dV, where E is enclosed by the planes z 0 ■
■
■
■
16.
19. Evaluate xxxE e z dV, where E is enclosed by the paraboloid
(b) z x 2 y 2
■
■
r dz d dr
octant that lies beneath the paraboloid z 1 x 2 y 2.
(b) x 2 y 2 2y
■
■
4
r
18. Evaluate xxxE x 3 xy 2 dV, where E is the solid in the first
Write the equations in cylindrical coordinates.
10. (a) x 2 y 2 z 2 2 ■
■
0
inside the cylinder x 2 y 2 16 and between the planes z 5 and z 4.
8. r 2 2z 2 4
9. (a) z x 2 y 2 ■
■
4
0
17. Evaluate xxxE sx 2 y 2 dV, where E is the region that lies
Identify the surface whose equation is given.
■
■
■
2
yy y
17–26
6. 3
5. r 3 7– 8
■
■
Describe in words the surface whose equation is given.
■
■
■
Change from rectangular to cylindrical coordinates.
■
■
■ Sketch the solid whose volume is given by the integral and evaluate the integral.
15–16
loids z x 2 y 2 and z 36 3x 2 3y 2. (b) Find the centroid of E (the center of mass in the case where the density is constant).
r z 2
■
■
■
■
■
■
■
■
■
24. (a) Find the volume of the solid that the cylinder
13. A cylindrical shell is 20 cm long, with inner radius 6 cm
and outer radius 7 cm. Write inequalities that describe the shell in an appropriate coordinate system. Explain how you have positioned the coordinate system with respect to the shell.
; 14. Use a graphing device to draw the solid enclosed by the paraboloids z x y and z 5 x y . 2
2
2
2
;
r a cos cuts out of the sphere of radius a centered at the origin. (b) Illustrate the solid of part (a) by graphing the sphere and the cylinder on the same screen. 25. Find the mass and center of mass of the solid S bounded by
the paraboloid z 4x 2 4y 2 and the plane z a a 0 if S has constant density K .
SECTION 12.7
26. Find the mass of a ball B given by x 2 y 2 z 2 a 2 if the
■
■
■
■
■
■
■
■
■
■
■
■
■
27–28 ■ Evaluate the integral by changing to cylindrical coordinates. 27.
y y
s4y 2
28.
y y
s9x 2
■
2
2 3
3
y
2
s4y 2 sx 2y 2
0
y
9x 2y 2
0
■
■
■
707
weight density of the material in the vicinity of a point P is tP and the height is hP. (a) Find a definite integral that represents the total work done in forming the mountain. (b) Assume that Mount Fuji in Japan is in the shape of a right circular cone with radius 62,000 ft, height 12,400 ft, and density a constant 200 lbft3 . How much work was done in forming Mount Fuji if the land was initially at sea level?
density at any point is proportional to its distance from the z -axis. ■
TRIPLE INTEGRALS IN SPHERICAL COORDINATES
xz dz dx dy sx 2 y 2 dz dy dx
■
■
■
■
■
■
■
P 29. When studying the formation of mountain ranges, geolo-
gists estimate the amount of work required to lift a mountain from sea level. Consider a mountain that is essentially in the shape of a right circular cone. Suppose that the
12.7
TRIPLE INTEGRALS IN SPHERICAL COORDINATES Another useful coordinate system in three dimensions is the spherical coordinate system. It simplifies the evaluation of triple integrals over regions bounded by spheres or cones. SPHERICAL COORDINATES
z
The spherical coordinates , , of a point P in space are shown in Figure 1, where OP is the distance from the origin to P, is the same angle as in cylindrical coordinates, and is the angle between the positive z-axis and the line segment OP. Note that
P (∏, ¨, ˙)
∏ ˙
0
O
¨
The spherical coordinate system is especially useful in problems where there is symmetry about a point, and the origin is placed at this point. For example, the sphere with center the origin and radius c has the simple equation c (see Figure 2); this is the reason for the name “spherical” coordinates. The graph of the equation c is a vertical half-plane (see Figure 3), and the equation c represents a half-cone with the z-axis as its axis (see Figure 4).
y
x
0
FIGURE 1
The spherical coordinates of a point
z
z
z
z
c 0
0
0
c
y x
0 y
x
x
y
y x
0
FIGURE 3 ¨=c, a half-plane
c
FIGURE 4 ˙=c, a half-cone
π/2
708
■
CHAPTER 12
MULTIPLE INTEGRALS
z
The relationship between rectangular and spherical coordinates can be seen from Figure 5. From triangles OPQ and OPP we have
Q
z cos
P(x, y, z) P(∏, ¨, ˙)
z
∏
But x r cos and y r sin , so to convert from spherical to rectangular coordinates, we use the equations
˙
˙
r sin
O
x
r
¨
x
y
y P ª(x, y, 0)
1
x sin cos
y sin sin
z cos
Also, the distance formula shows that
FIGURE 5
2 x 2 y 2 z2
2
We use this equation in converting from rectangular to spherical coordinates. V EXAMPLE 1 The point 2, 4, 3 is given in spherical coordinates. Plot the point and find its rectangular coordinates.
SOLUTION We plot the point in Figure 6. From Equations 1 we have
z (2, π/4, π/3) π 3
s3 cos 2 3 4 2
y sin sin 2 sin
s3 sin 2 3 4 2
2
O
x
π 4
x sin cos 2 sin
y
FIGURE 6
z cos 2 cos
1 s2
1 s2
3 2
3 2
2( 12 ) 1 3
Thus the point 2, 4, 3 is (s32 , s32 , 1) in rectangular coordinates.
■
V EXAMPLE 2 The point (0, 2s3 , 2) is given in rectangular coordinates. Find spherical coordinates for this point. WARNING: There is not universal agreement on the notation for spherical coordinates. Most books on physics reverse the meanings of and and use r in place of .
In Module 12.7 you can investigate families of surfaces in cylindrical and spherical coordinates.
SOLUTION From Equation 2 we have
sx 2 y 2 z 2 s0 12 4 4 and so Equations 1 give cos
z 2 1 4 2
2 3
cos
x 0 sin
2
(Note that 32 because y 2s3 the given point are 4, 2, 23.
0.) Therefore, spherical coordinates of
■
SECTION 12.7
TRIPLE INTEGRALS IN SPHERICAL COORDINATES
■
709
EVALUATING TRIPLE INTEGRALS WITH SPHERICAL COORDINATES
In the spherical coordinate system the counterpart of a rectangular box is a spherical wedge E , , a b, , c d
z
where a 0 and 2. Although we defined triple integrals by dividing solids into small boxes, it can be shown that dividing a solid into small spherical wedges always gives the same result. So we divide E into smaller spherical wedges Eijk by means of spheres i , half-planes j , and half-cones k . Figure 7 shows that Eijk is approximately a rectangular box with dimensions i , i k (arc of a circle with radius i , angle k ), and i sin k j (arc of a circle with radius i sin k, angle j ). So an approximation to the volume of Eijk is given by
∏i sin ˙k Ψj Î∏i
˙k Î˙k ∏i Î˙k
0
x
y
Ψj
ri=∏ i sin ˙ k
ri Ψj=∏i sin ˙k Ψj
Vijk i i k i sin k j 2i sin k i j k In fact, it can be shown, with the aid of the Mean Value Theorem (Exercise 41), that the volume of Eijk is given exactly by
FIGURE 7
Vijk 2i sin k i j k
* , y ijk * , z ijk * be the rectangular coordiwhere i , j , k is some point in Eijk . Let x ijk nates of this point. Then
yyy f x, y, z dV E
l
n
f x * , y * , z * V
lim
f
ijk
max i , j, k l 0 i1 j1 k1 l
m
lim
m
n
max i , j, k l 0 i1 j1 k1
ijk
ijk
ijk
sin k cos j, i sin k sin j , i cos k i2 sin k i j k
i
But this sum is a Riemann sum for the function F , , 2 sin f sin cos , sin sin , cos Consequently, we have arrived at the following formula for triple integration in spherical coordinates.
3
yyy f x, y, z dV E
y
z
d
c
∏ sin ˙ d¨
˙
d∏
y y
b
a
f sin cos , sin sin , cos 2 sin d d d
where E is a spherical wedge given by E , ,
∏
a b,
, c d
d˙ ∏ d˙
0
d¨ x
FIGURE 8
Volume element in spherical coordinates: dV=∏@ sin ˙ d∏ d¨ d˙
y
Formula 3 says that we convert a triple integral from rectangular coordinates to spherical coordinates by writing x sin cos
y sin sin
z cos
using the appropriate limits of integration, and replacing dV by 2 sin d d d. This is illustrated in Figure 8.
710
■
CHAPTER 12
MULTIPLE INTEGRALS
This formula can be extended to include more general spherical regions such as
E , , , c d, t1, t 2, In this case the formula is the same as in (3) except that the limits of integration for are t1, and t 2, . Usually, spherical coordinates are used in triple integrals when surfaces such as cones and spheres form the boundary of the region of integration. V EXAMPLE 3
Evaluate xxxB e x y z 2
2
2 32
dV, where B is the unit ball:
B x, y, z x 2 y 2 z 2 1 SOLUTION Since the boundary of B is a sphere, we use spherical coordinates:
B , , 0 1, 0 2, 0 In addition, spherical coordinates are appropriate because x 2 y 2 z2 2 Thus (3) gives
yyy e
x 2y 2z 2 32
dV y
0
2
y y 0
1
e
0
2 sin d d d
2 32
B
y sin d 0
y
2
0
[
]
cos 0 2
d
y
1
0
[e ]
1 3 1 3 0
2e d 3
43 e 1
■
NOTE It would have been extremely awkward to evaluate the integral in Example 3 without spherical coordinates. In rectangular coordinates the iterated integral would have been
y y 1
1
s1x 2 s1x 2
y
s1x 2y 2
s1x 2y 2
e x y z 2
2
2 32
dz dy dx
V EXAMPLE 4 Use spherical coordinates to find the volume of the solid that lies above the cone z sx 2 y 2 and below the sphere x 2 y 2 z 2 z. (See Figure 9.) z (0, 0, 1)
≈+¥+z@=z
π 4
z=œ„„„„„ ≈+¥ y
FIGURE 9
x
SOLUTION Notice that the sphere passes through the origin and has center (0, 0, 2 ). 1
We write the equation of the sphere in spherical coordinates as
2 cos
or
cos
SECTION 12.7
■ Figure 10 gives another look (this time drawn by Maple) at the solid of Example 4.
TRIPLE INTEGRALS IN SPHERICAL COORDINATES
■
711
The equation of the cone can be written as
cos s 2 sin 2 cos 2 2 sin 2 sin 2 sin This gives sin cos , or 4. Therefore, the description of the solid E in spherical coordinates is
E , , 0 2, 0 4, 0 cos Figure 11 shows how E is swept out if we integrate first with respect to , then , and then . The volume of E is VE yyy dV y
FIGURE 10
2
0
4
y y
Visual 12.7 shows an animation of Figure 11.
y
0
2 3
d
y
y
0
4
0
4
∏ varies from 0 to cos ˙ while ˙ and ¨ are constant.
5–6
2. (a) 5, , 2
(b) 4, 34, 3
7– 8
■
■
■
■
■
■
■
■
■
0
8
x
y
¨ varies from 0 to 2π.
■
■
■
■
■
■
■
6. 3 ■
■
■
■
■
■
9–10
■
■
■
■
■
■
■
■
■
■
■
■
■
■
8. 2 cos ■
■
■
■
■
■
■
Write the equation in spherical coordinates. (b) x 2 y 2 2y
10. (a) x 2 y 2 z 2 2 ■
■
Identify the surface whose equation is given.
■
9. (a) z x 2 y 2
(b) (1, 1, s6 ) ■
Describe in words the surface whose equation is given.
■
7. sin 2
■
(b) 0, 1, 1
4. (a) (0, s3 , 1) ■
■
Change from rectangular to spherical coordinates.
3. (a) (1, s3 , 2s3 )
■
4
z
5. 3 ■
3– 4
cos 4 4
EXERCISES
(b) 2, 3, 4
■
y
1. (a) 1, 0, 0
■
2 3
˙ varies from 0 to π/4 while ¨ is constant.
■ Plot the point whose spherical coordinates are given. Then find the rectangular coordinates of the point.
■
d
0
x
y
1–2
■
cos
z
x
12.7
3 sin 3
sin cos 3 d
z
FIGURE 11
2 sin d d d
E 2
cos
0
0
■
■
■
(b) z x 2 y 2 ■
■
■
■
■
712
■
11–14
CHAPTER 12
■
Sketch the solid described by the given inequalities.
11. 2,
0 2,
12. 2 3,
14. 0 3, ■
■
26. Find the volume of the solid that lies within the sphere
x 2 y 2 z 2 4, above the xy-plane, and below the cone z sx 2 y 2 .
0 2
2
13. 2 2,
■
MULTIPLE INTEGRALS
27. (a) Find the volume of the solid that lies above the cone
0 6,
3 and below the sphere 4 cos . (b) Find the centroid of the solid in part (a).
0 sec
2
■
■
■
■
■
■
■
■
28. Let H be a solid hemisphere of radius a whose density at
■
any point is proportional to its distance from the center of the base. (a) Find the mass of H . (b) Find the center of mass of H . (c) Find the moment of inertia of H about its axis.
15. A solid lies above the cone z sx 2 y 2 and below the
sphere x 2 y 2 z 2 z. Write a description of the solid in terms of inequalities involving spherical coordinates.
16. (a) Find inequalities that describe a hollow ball with diame29. (a) Find the centroid of a solid homogeneous hemisphere of
ter 30 cm and thickness 0.5 cm. Explain how you have positioned the coordinate system that you have chosen. (b) Suppose the ball is cut in half. Write inequalities that describe one of the halves.
radius a. (b) Find the moment of inertia of the solid in part (a) about a diameter of its base. 30. Find the mass and center of mass of a solid hemisphere of
Sketch the solid whose volume is given by the integral and evaluate the integral. 17–18
■
6
2
17.
y y y
18.
y y y
■
0
0
2
2
0
■
3
0
2
1
■
radius a if the density at any point is proportional to its distance from the base.
2 sin d d d
■
■
■
■
■
■
■
■
■
19.
■
■
■
■
■
■
■
32. Find the volume of the smaller wedge cut from a sphere of
radius a by two planes that intersect along a diameter at an angle of 6.
z
20.
3
CAS
33. Evaluate xxxE z dV, where E lies above the paraboloid
z x 2 y 2 and below the plane z 2y. Use either the Table of Integrals (on Reference Pages 6 –10) or a computer algebra system to evaluate the integral.
2 y
x ■
■
above the cone z sx 2 y 2 and below the sphere x 2 y 2 z 2 1.
■
z
■
31. Find the volume and centroid of the solid E that lies
Set up the triple integral of an arbitrary continuous function f x, y, z in cylindrical or spherical coordinates over the solid shown.
19–20
■
31–34 ■ Use cylindrical or spherical coordinates, whichever seems more appropriate.
2 sin d d d ■
■
1
■
■
■
■
■
■
■
■
34. (a) Find the volume enclosed by the torus sin .
y
2
x
■
■
■
;
(b) Use a computer to draw the torus. ■
21–30
■
Use spherical coordinates.
21. Evaluate xxxB x y z dV, where B is the unit ball 2
2
2
x 2 y 2 z 2 1.
22. Evaluate xxxH x y dV, where H is the hemispherical 2
23. Evaluate xxxE z dV, where E lies between the spheres
x 2 y 2 z 2 1 and x 2 y 2 z 2 4 in the first octant.
24. Evaluate xxxE e sx y z dV, where E is enclosed by the 2
2
2
sphere x 2 y 2 z 2 9 in the first octant.
25. Evaluate xxxE x 2 dV, where E is bounded by the x z-plane
and the hemispheres y s9 x 2 z 2 and y s16 x 2 z 2 .
■
■
■
■
■
■
■
■
■
■
■
■
■ Evaluate the integral by changing to spherical coordinates.
35–36 35.
yy
36.
y y
2
region that lies above the xy-plane and below the sphere x 2 y 2 z 2 1.
■
■
1
s1x 2
0
0
y
s2x 2y 2
a
sa 2y 2
a
sa 2y 2
■
■
sx 2y 2
y
xy dz dy dx
sa 2x 2y 2
sa 2x 2y 2 ■
■
x 2z y 2z z 3 dz dx dy ■
■
■
■
■
; 37. Use a graphing device to draw a silo consisting of a
cylinder with radius 3 and height 10 surmounted by a hemisphere.
38. The latitude and longitude of a point P in the Northern
Hemisphere are related to spherical coordinates , , as follows. We take the origin to be the center of the Earth and the positive z -axis to pass through the North Pole. The posi-
SECTION 12.8
tive x-axis passes through the point where the prime meridian (the meridian through Greenwich, England) intersects the equator. Then the latitude of P is 90 and the longitude is 360 . Find the great-circle distance from Los Angeles (lat. 34.06 N, long. 118.25 W) to Montréal (lat. 45.50 N, long. 73.60 W). Take the radius of the Earth to be 3960 mi. (A great circle is the circle of intersection of a sphere and a plane through the center of the sphere.) CAS
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
■
713
40. Show that
y y y
2
sx 2 y 2 z 2 ex y
2
z 2
dx dy dz 2
(The improper triple integral is defined as the limit of a triple integral over a solid sphere as the radius of the sphere increases indefinitely.) 41. (a) Use cylindrical coordinates to show that the volume of
the solid bounded above by the sphere r 2 z 2 a 2 and below by the cone z r cot 0 (or 0 ), where 0 0 2, is
39. The surfaces 1 5 sin m sin n have been used as 1
models for tumors. The “bumpy sphere” with m 6 and n 5 is shown. Use a computer algebra system to find the volume it encloses.
V
2a 3 1 cos 0 3
(b) Deduce that the volume of the spherical wedge given by 1 2 , 1 2 , 1 2 is V
23 13 cos 1 cos 2 2 1 3
(c) Use the Mean Value Theorem to show that the volume in part (b) can be written as
V 2 sin
where lies between 1 and 2 , lies between 1 and 2 , 2 1 , 2 1 , and 2 1 .
12.8
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS In one-dimensional calculus we often use a change of variable (a substitution) to simplify an integral. By reversing the roles of x and u, we can write the Substitution Rule (5.5.6) as
y
1
f x dx y f tutu du
b
d
a
c
where x tu and a tc, b td. Another way of writing Formula 1 is as follows:
y
2
b
a
f x dx y f xu d
c
dx du du
A change of variables can also be useful in double integrals. We have already seen one example of this: conversion to polar coordinates. The new variables r and are related to the old variables x and y by the equations x r cos
y r sin
and the change of variables formula (12.3.2) can be written as
yy f x, y dA yy f r cos , r sin r dr d R
S
where S is the region in the r -plane that corresponds to the region R in the xy-plane.
714
■
CHAPTER 12
MULTIPLE INTEGRALS
More generally, we consider a change of variables that is given by a transformation T from the uv-plane to the xy-plane: Tu, v x, y where x and y are related to u and v by the equations 3
x tu, v
y hu, v
x xu, v
y yu, v
or, as we sometimes write, We usually assume that T is a C 1 transformation, which means that t and h have continuous first-order partial derivatives. A transformation T is really just a function whose domain and range are both subsets of ⺢ 2. If Tu1, v1 x 1, y1, then the point x 1, y1 is called the image of the point u1, v1. If no two points have the same image, T is called one-to-one. Figure 1 shows the effect of a transformation T on a region S in the uv-plane. T transforms S into a region R in the xy-plane called the image of S, consisting of the images of all points in S. √
y
T
S
R
(u¡, √¡)
T –!
u
0
(x¡, y¡)
0
x
FIGURE 1
If T is a one-to-one transformation, then it has an inverse transformation T 1 from the xy-plane to the uv-plane and it may be possible to solve Equations 3 for u and v in terms of x and y : u Gx, y v Hx, y
√
S£ (0, 1)
(1, 1)
S¢
S
0
S™
S¡ (1, 0)
V EXAMPLE 1
u
A transformation is defined by the equations x u 2 v2
Find the image of the square S u, v 0 u 1, 0 v 1 .
T
SOLUTION The transformation maps the boundary of S into the boundary of the image. So we begin by finding the images of the sides of S. The first side, S1 , is given by v 0 0 u 1. (See Figure 2.) From the given equations we have x u 2, y 0, and so 0 x 1. Thus S1 is mapped into the line segment from 0, 0 to 1, 0 in the xy-plane. The second side, S 2, is u 1 0 v 1 and, putting u 1 in the given equations, we get
y (0, 2) ¥ x= -1 4
¥
x=1- 4 R
(_1, 0)
FIGURE 2
x 1 v2 0
y 2uv
(1, 0)
x
y 2v
Eliminating v, we obtain 4
x1
y2 4
0 x 1
SECTION 12.8
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
■
715
which is part of a parabola. Similarly, S 3 is given by v 1 0 u 1, whose image is the parabolic arc x
5
y2 1 4
1 x 0
Finally, S4 is given by u 0 0 v 1 whose image is x v 2, y 0, that is, 1 x 0. (Notice that as we move around the square in the counterclockwise direction, we also move around the parabolic region in the counterclockwise direction.) The image of S is the region R (shown in Figure 2) bounded by the x-axis and the parabolas given by Equations 4 and 5. ■ Now let’s see how a change of variables affects a double integral. We start with a small rectangle S in the uv-plane whose lower left corner is the point u0 , v0 and whose dimensions are u and v. (See Figure 3.) y
√
u=u ¸ r (u ¸, √) Î√
S
(u¸, √ ¸)
Îu
T (x¸, y¸)
√=√ ¸ 0
R
r (u, √ ¸) u
0
x
FIGURE 3
The image of S is a region R in the xy-plane, one of whose boundary points is x 0 , y0 Tu0 , v0 . The vector ru, v tu, v i hu, v j is the position vector of the image of the point u, v. The equation of the lower side of S is v v0 , whose image curve is given by the vector function ru, v0. The tangent vector at x 0 , y0 to this image curve is ru tuu0 , v0 i huu0 , v0 j
&x &y i j &u &u
Similarly, the tangent vector at x 0 , y0 to the image curve of the left side of S (namely, u u0 ) is rv tvu0 , v0 i hvu0 , v0 j r (u¸, √¸+Î√)
&x &y i j &v &v
We can approximate the image region R T S by a parallelogram determined by the secant vectors r (u ¸, √¸)
a ru0 u, v0 ru0 , v0
b ru0 , v0 v ru0 , v0
a
shown in Figure 4. But r (u¸+Î u, √¸) FIGURE 4
ru lim
u l 0
ru0 u, v0 ru0 , v0 u
716
■
CHAPTER 12
MULTIPLE INTEGRALS
and so
ru0 u, v0 ru0 , v0 u ru
Similarly
ru0 , v0 v ru0 , v0 v rv
Î √ r√ r (u ¸, √ ¸)
This means that we can approximate R by a parallelogram determined by the vectors u ru and v rv . (See Figure 5.) Therefore, we can approximate the area of R by the area of this parallelogram, which, from Section 10.4, is FIGURE 5
u r v r r
6
u
u
v
rv u v
Computing the cross product, we obtain i &x ru rv &u &x &v
j &y &u &y &v
k
&x 0 &u &x 0 &v
&y &x &u &u k &y &y &v &u
&x &v k &y &v
The determinant that arises in this calculation is called the Jacobian of the transformation and is given a special notation.
The Jacobian is named after the German mathematician Carl Gustav Jacob Jacobi (1804–1851). Although the French mathematician Cauchy first used these special determinants involving partial derivatives, Jacobi developed them into a method for evaluating multiple integrals. ■
7 DEFINITION The Jacobian of the transformation T given by x tu, v and y hu, v is
&x &x, y &u &u, v &y &u
&x &v &x &y &x &y &y &u &v &v &u &v
With this notation we can use Equation 6 to give an approximation to the area A of R: A
8
&x, y u v &u, v
where the Jacobian is evaluated at u0 , v0 . Next we divide a region S in the uv-plane into rectangles Sij and call their images in the xy-plane Rij . (See Figure 6.) √
y
Sij R ij S
Î√
R
Îu
T
(u i , √ j ) 0
FIGURE 6
(x i , yj ) u
0
x
SECTION 12.8
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
■
717
Applying the approximation (8) to each Rij , we approximate the double integral of f over R as follows: m
n
yy f x, y dA f x , y A i
j
i1 j1
R
m
n
f tu , v , hu , v i
j
i
j
i1 j1
&x, y u v &u, v
where the Jacobian is evaluated at ui , vj . Notice that this double sum is a Riemann sum for the integral
yy f tu, v, hu, v S
&x, y du dv &u, v
The foregoing argument suggests that the following theorem is true. (A full proof is given in books on advanced calculus.) Suppose that T is a C 1 transformation whose Jacobian is nonzero and that maps a region S in the uvplane onto a region R in the xy-plane. Suppose that f is continuous on R and that R and S are type I or type II plane regions. Suppose also that T is one-toone, except perhaps on the boundary of S. Then 9 CHANGE OF VARIABLES IN A DOUBLE INTEGRAL
yy f x, y dA yy f xu, v, yu, v R
¨
r=a
v by expressing x and y in terms of u and v and writing r=b
S
dA
å
¨=å
0
a
r
b
y
r=b
&x, y du dv &u, v
Notice the similarity between Theorem 9 and the one-dimensional formula in Equation 2. Instead of the derivative dxdu, we have the absolute value of the Jacobian, that is, &x, y&u, v . As a first illustration of Theorem 9, we show that the formula for integration in polar coordinates is just a special case. Here the transformation T from the r -plane to the xy-plane is given by
T
¨=∫
&x, y du dv &u, v
Theorem 9 says that we change from an integral in x and y to an integral in u and
¨=∫
∫
S
x tr, r cos
y hr, r sin
R r=a ∫ 0
and the geometry of the transformation is shown in Figure 7. T maps an ordinary rectangle in the r -plane to a polar rectangle in the xy-plane. The Jacobian of T is
¨=å
å x
FIGURE 7
The polar coordinate transformation
&x &x, y &r &r, &y &r
&x & cos &y sin &
r sin r cos2 r sin2 r 0 r cos
718
■
CHAPTER 12
MULTIPLE INTEGRALS
Thus Theorem 9 gives
yy f x, y dx dy yy f r cos , r sin R
S
y
y
b
a
&x, y dr d &r,
f r cos , r sin r dr d
which is the same as Formula 12.3.2. Use the change of variables x u 2 v 2, y 2uv to evaluate the integral xxR y dA, where R is the region bounded by the x-axis and the parabolas y 2 4 4x and y 2 4 4x, y 0. V EXAMPLE 2
SOLUTION The region R is pictured in Figure 2 (on page 714). In Example 1 we discovered that TS R, where S is the square 0, 1 0, 1 . Indeed, the reason for making the change of variables to evaluate the integral is that S is a much simpler region than R. First we need to compute the Jacobian:
&x &x, y &u &u, v &y &u Therefore, by Theorem 9,
yy y dA yy 2uv R
S
8y
1
0
y
1
0
&x &v 2u &y 2v &v
2v 4u 2 4v 2 0 2u
&x, y 1 1 dA y y 2uv4u2 v 2 du dv 0 0 &u, v u3v uv 3 du dv 8 y
1
0
y 2v 4v 3 dv v 2 v 4 1
[
0
]
1 0
[
1 4 4 v
u
12 u2v 3
u1
]
u0
dv
2
■
NOTE Example 2 was not a very difficult problem to solve because we were given a suitable change of variables. If we are not supplied with a transformation, then the first step is to think of an appropriate change of variables. If f x, y is difficult to integrate, then the form of f x, y may suggest a transformation. If the region of integration R is awkward, then the transformation should be chosen so that the corresponding region S in the uv-plane has a convenient description.
EXAMPLE 3 Evaluate the integral xxR e xyxy dA, where R is the trapezoidal
region with vertices 1, 0, 2, 0, 0, 2, and 0, 1.
SOLUTION Since it isn’t easy to integrate e xyxy, we make a change of variables
suggested by the form of this function: 10
uxy
vxy
These equations define a transformation T 1 from the xy-plane to the uv-plane. Theorem 9 talks about a transformation T from the uv-plane to the xy-plane. It is
SECTION 12.8
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
■
719
obtained by solving Equations 10 for x and y : x 12 u v
11
y 12 u v
The Jacobian of T is &x &x, y &u &u, v &y &u
&x &v &y &v
1 2 1 2
1 2 1 2
12
To find the region S in the uv-plane corresponding to R, we note that the sides of R lie on the lines √
√=2
(_2, 2)
S
u=_√
y0
(2, 2)
(1, 1)
uv
√=1 0
T
u
v2
u v
v1
S u, v 1 v 2, v u v
y
Theorem 9 gives
x-y=1 2
0 _1
xy1
Thus the region S is the trapezoidal region with vertices 1, 1, 2, 2, 2, 2, and 1, 1 shown in Figure 8. Since
T –!
1
x0
and, from either Equations 10 or Equations 11, the image lines in the uv-plane are
u=√
(_1, 1)
xy2
yy e
x
R
xyxy
dA yy e uv
R
x-y=2
_2
S
y
2
y
1
FIGURE 8
1 2
y
v
v
2
1
&x, y du dv &u, v
e uv ( 12 ) du dv 12 y [ve uv ]uv dv 2
uv
1
e e1 v dv 34 e e1
■
TRIPLE INTEGRALS
There is a similar change of variables formula for triple integrals. Let T be a transformation that maps a region S in u vw-space onto a region R in xyz-space by means of the equations x tu, v, w y hu, v, w z ku, v, w The Jacobian of T is the following 3 3 determinant:
12
&x &u &x, y, z &y &u, v, w &u &z &u
&x &v &y &v &z &v
&x &w &y &w &z &w
720
■
CHAPTER 12
MULTIPLE INTEGRALS
Under hypotheses similar to those in Theorem 9, we have the following formula for triple integrals:
13
yyy f x, y, z dV R
yyy f xu, v, w, yu, v, w, zu, v, w S
&x, y, z du dv dw &u, v, w
V EXAMPLE 4 Use Formula 13 to derive the formula for triple integration in spherical coordinates.
SOLUTION Here the change of variables is given by
x sin cos
y sin sin
We compute the Jacobian as follows:
sin cos &x, y, z sin sin & , , cos cos
sin sin cos cos sin cos cos sin 0 sin
z cos
sin sin cos cos sin cos sin sin sin sin cos cos sin sin cos sin sin
cos 2 sin cos sin2 2 sin cos cos2 sin sin2 cos2 sin2 sin2 2 sin cos2 2 sin sin2 2 sin Since 0 , we have sin 0. Therefore
&x, y, z 2 sin 2 sin & , ,
and Formula 13 gives
yyy f x, y, z dV yyy f sin cos , sin sin , cos R
2
sin d d d
S
which is equivalent to Formula 12.7.3.
■
SECTION 12.8
12.8 1–6
■
1. x u 4 v,
y 3u 2v
2. x u 2 v 2,
y u 2 v2
u , uv
y
4. x sin , 5. x u v,
■
■
■
xy 1, xy 2, xy 2 1, xy 2 2; u xy, v xy 2. Illustrate by using a graphing calculator or computer to draw R.
v
■
■
z e uvw ■
■
■
■
■
■
■
■ Find the image of the set S under the given transformation.
7–10
7. S u, v
0 u 3,
x 2u 3v, y u v
10. S is the disk given by u 2 v 2 1;
11–16 11.
■
■
■
19.
■
x au, y bv ■
■
■
■
xxR x 3y dA,
20. ■
where R is the triangular region with vertices 0, 0, 2, 1, and 1, 2; x 2u v, y u 2v
xxR 4 x 8y dA,
13.
xxR x 2 dA,
15.
■
■
■
■
■
■
■
■
■ Evaluate the integral by making an appropriate change of variables.
Use the given transformation to evaluate the integral.
12.
14.
■
■
19–23
9. S is the triangular region with vertices 0, 0, 1, 1, 0, 1; x u2, y v
■
■
18. Evaluate xxxE x 2 y dV, where E is the solid of Exercise 17(a).
0 v 2 ;
8. S is the square bounded by the lines u 0, u 1, v 0, v 1; x v, y u1 v 2
■
■
ellipsoid x 2a 2 y 2b 2 z 2c 2 1. Use the transformation x au, y bv, z cw. (b) The Earth is not a perfect sphere; rotation has resulted in flattening at the poles. So the shape can be approximated by an ellipsoid with a b 6378 km and c 6356 km. Use part (a) to estimate the volume of the Earth.
z uw
y e uv,
■
17. (a) Evaluate xxxE dV, where E is the solid enclosed by the
y cos
■
721
2 ; 16. xxR y dA, where R is the region bounded by the curves
uv
y vw,
6. x e uv,
■
EXERCISES
Find the Jacobian of the transformation.
3. x
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
where R is the parallelogram with vertices 1, 3, 1, 3, 3, 1, and 1, 5; x 14 u v, y 14 v 3u where R is the region bounded by the ellipse 9x 2 4y 2 36; x 2u, y 3v
xxR x 2 xy y 2 dA,
where R is the region bounded by the ellipse x 2 xy y 2 2; x s2 u s23 v, y s2 u s23 v
xxR xy dA, where R is the region in the first quadrant bounded by the lines y x and y 3x and the hyperbolas xy 1, xy 3; x uv, y v
21.
x 2y dA, where R is the parallelogram enclosed by 3x y R the lines x 2y 0, x 2y 4, 3x y 1, and 3x y 8
yy
xxR x ye x y 2
2
dA , where R is the rectangle enclosed by the lines x y 0, x y 2, x y 0, and x y 3
yx dA, where R is the trapezoidal region yx R with vertices 1, 0, 2, 0, 0, 2, and 0, 1
yy cos
22.
xxR sin9x 2 4y 2 dA,
23.
xxR e xy dA,
■
where R is the region in the first quadrant bounded by the ellipse 9x 2 4y 2 1 where R is given by the inequality x y 1
■
■
■
■
■
■
■
■
■
■
24. Let f be continuous on 0, 1 and let R be the triangular
region with vertices 0, 0, 1, 0, and 0, 1. Show that
yy f x y dA y
1
0
R
u f u du
■
722
■
CHAPTER 12
MULTIPLE INTEGRALS
12
REVIEW
CONCEPT CHECK
1. Suppose f is a continuous function defined on a rectangle
R a, b c, d . (a) Write an expression for a double Riemann sum of f . If f x, y 0, what does the sum represent? (b) Write the definition of xxR f x, y dA as a limit. (c) What is the geometric interpretation of xxR f x, y dA if f x, y 0? What if f takes on both positive and negative values? (d) How do you evaluate xxR f x, y dA? (e) What does the Midpoint Rule for double integrals say?
(d) What is a type 1 solid region? How do you evaluate xxxE f x, y, z dV if E is such a region? (e) What is a type 2 solid region? How do you evaluate xxxE f x, y, z dV if E is such a region? (f ) What is a type 3 solid region? How do you evaluate xxxE f x, y, z dV if E is such a region? 6. Suppose a solid object occupies the region E and has den-
sity function x, y, z. Write expressions for each of the following. (a) The mass (b) The moments about the coordinate planes (c) The coordinates of the center of mass (d) The moments of inertia about the axes
2. (a) How do you define xxD f x, y dA if D is a bounded
region that is not a rectangle? (b) What is a type I region? How do you evaluate xxD f x, y dA if D is a type I region? (c) What is a type II region? How do you evaluate xxD f x, y dA if D is a type II region? (d) What properties do double integrals have?
7. (a) Write the equations for converting from cylindrical to
rectangular coordinates. In what situation would you use cylindrical coordinates? (b) Write the equations for converting from spherical to rectangular coordinates. In what situation would you use spherical coordinates?
3. How do you change from rectangular coordinates to polar
coordinates in a double integral? Why would you want to make the change? 4. If a lamina occupies a plane region D and has density func-
tion x, y, write expressions for each of the following in terms of double integrals. (a) The mass (b) The moments about the axes (c) The center of mass (d) The moments of inertia about the axes and the origin 5. (a) Write the definition of the triple integral of f over a
rectangular box B. (b) How do you evaluate xxxB f x, y, z dV ? (c) How do you define xxxE f x, y, z dV if E is a bounded solid region that is not a box?
8. (a) How do you change from rectangular coordinates to
cylindrical coordinates in a triple integral? (b) How do you change from rectangular coordinates to spherical coordinates in a triple integral? (c) In what situations would you change to cylindrical or spherical coordinates? 9. (a) If a transformation T is given by x tu, v, y hu, v, what is the Jacobian of T ?
(b) How do you change variables in a double integral? (c) How do you change variables in a triple integral?
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1.
2.
y y 2
1
yy 1
0
6
0
x 2 sinx y dx dy y
sx y 2 dy dx y
x
0
6
0
x
0
y
1
0
y
2
1
5. If D is the disk given by x 2 y 2 4, then
yy s4 x 6.
y y (x 4
1
1
0
2
sy ) sinx 2 y 2 dx dy 9
7. The integral
2
y yy 0
3.
yy
4.
y y
2
1
1
1
4
x 2e y dy dx
3
1
0
e
x 2y 2
y
2
1
x 2 dx y e y dy
sin y dx dy 0
4
3
y 2 dA 163
D
x 2 sinx y dy dx
sx y 2 dx dy
2
2
0
2
r
dz dr d
represents the volume enclosed by the cone z sx 2 y 2 and the plane z 2. 8. The integral xxxE kr 3 dz dr d represents the moment of
inertia about the z -axis of a solid E with constant density k.
CHAPTER 12
REVIEW
■
723
EXERCISES 1. A contour map is shown for a function f on the square
■ Calculate the iterated integral by first reversing the order of integration.
13–14
R 0, 3 0, 3 . Use a Riemann sum with nine terms to estimate the value of xxR f x, y dA. Take the sample points to be the upper right corners of the squares. y 3
2
13. ■
9 7 5
1
1
3 x
2
y 2xe dx dy
4.
yy
cosx 2 dy dx
6.
yy
ex
8.
yyy
yy
5.
yy
x
7.
y yy
1
s1y 2
0
0
3.
1
0
1
0
0
0
■
y
■
■
y sin x dz dy dx ■
■
■
■
1
0
1
0
■
0
x
1
y
0
0
1
■
y 4
■
_4 ■
_2 ■
■
2 ■
4 x
0
_4
■
■
■
■
■
2
sin 2
0
0
■
r dr d
12. Describe the solid whose volume is given by the integral 2
2
y y y 0
0
2
1
and evaluate the integral.
2 sin d d d
0 y 1,
0 y 3
y x y 2 2
xxD y dA, where D is the region in the first quadrant that lies above the hyperbola xy 1 and the line y x and below the line y 2 xxD x 2 y 2 32 dA,
where D is the region in the first quadrant bounded by the lines y 0 and y s3 x and the circle x2 y2 9
xxD x dA, where D is the region in the first quadrant that lies
xxxE y 2z 2 dV,
26.
xxxE z dV,
27.
xxxE yz dV,
■
28.
xxxH z 3sx 2 y 2 z 2
11. Describe the region whose area is given by the integral
y y
0 x 2,
1 dA, where D is the triangular region with 1 x2 D vertices 0, 0, 1, 1, and 0, 1
25.
4 x ■
■
y dA, 1 x2 D where D is bounded by y sx , y 0, x 1
xxxT xy dV,
R
0
where D x, y
24.
4
R 2
where R x, y
xxxE xy dV,
■
y
10.
■
Calculate the value of the multiple integral.
23.
■
9.
■
between the circles x 2 y 2 1 and x 2 y 2 2
Write xxR f x, y dA as an iterated integral, where R is the region shown and f is an arbitrary continuous function on R.
9–10
■
xxD y dA, where D is the region in the first quadrant bounded by the parabolas x y 2 and x 8 y 2
22.
6xyz dz dx dy
■
■
19.
ye dx dy
x
sy
0
yy
xy
3xy 2 dy dx
■
ye x dx dy x3
18.
21. 1
2
2
■
1
yy
Calculate the iterated integral.
■
■
1
17.
Exercise 1. 3– 8
■
■
yy
xxD xy dA,
20.
2. Use the Midpoint Rule to estimate the integral in
■
14.
16.
6
3
0
x
■
cos y 2 dy dx
xxR ye xy dA,
2
1
0
1
15.
8
4
1
15–28
10
2
yy
■
where E x, y, z 0 x 3 , 0 y x, 0 z x y
where T is the solid tetrahedron with vertices 0, 0, 0, ( 13 , 0, 0), 0, 1, 0, and 0, 0, 1 where E is bounded by the paraboloid x 1 y 2 z 2 and the plane x 0
where E is bounded by the planes y 0, z 0, x y 2 and the cylinder y 2 z 2 1 in the first octant
where E lies above the plane z 0, below the plane z y, and inside the cylinder x 2 y 2 4
dV, where H is the solid hemisphere that lies above the xy-plane and has center the origin and radius 1 ■
29–34
■
■
■
■
■
■
■
■
■
■
■
Find the volume of the given solid.
29. Under the paraboloid z x 2 4y 2 and above the rectangle
R 0, 2 1, 4
30. Under the surface z x 2 y and above the triangle in the
xy-plane with vertices 1, 0, 2, 1, and 4, 0
724
■
CHAPTER 12
MULTIPLE INTEGRALS
31. The solid tetrahedron with vertices 0, 0, 0, 0, 0, 1,
44. Sketch the solid consisting of all points with spherical coor-
0, 2, 0, and 2, 2, 0
32. Bounded by the cylinder x 2 y 2 4 and the planes z 0
and y z 3
dinates , , such that 0 2, 0 6, and 0 2 cos . 45. Use polar coordinates to evaluate
33. One of the wedges cut from the cylinder x 9y a by 2
2
2
yy 3
the planes z 0 and z mx
34. Above the paraboloid z x 2 y 2 and below the half-cone
y y 2
■
■
■
■
■
■
■
■
■
■
x 3 xy 2 dy dx
46. Use spherical coordinates to evaluate
z sx 2 y 2 ■
s9x 2
s9x 2
0
■
2
s4y 2
0
y
s4x 2y 2
y 2sx 2 y 2 z 2 dz dx dy
s4x 2y 2
35. Consider a lamina that occupies the region D bounded by
the parabola x 1 y 2 and the coordinate axes in the first quadrant with density function x, y y. (a) Find the mass of the lamina. (b) Find the center of mass. (c) Find the moments of inertia and radii of gyration about the x- and y-axes. 36. A lamina occupies the part of the disk x 2 y 2 a 2 that
lies in the first quadrant. (a) Find the centroid of the lamina. (b) Find the center of mass of the lamina if the density function is x, y xy 2. 37. (a) Find the centroid of a right circular cone with height h
and base radius a. (Place the cone so that its base is in the xy-plane with center the origin and its axis along the positive z -axis.) (b) Find the moment of inertia of the cone about its axis (the z -axis). CAS
38. Find the center of mass of the solid tetrahedron with
vertices 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 3 and density function x, y, z x 2 y 2 z 2. 39. The cylindrical coordinates of a point are (2s3 , 3, 2).
Find the rectangular and spherical coordinates of the point. 40. The rectangular coordinates of a point are 2, 2, 1. Find
the cylindrical and spherical coordinates of the point. 41. The spherical coordinates of a point are 8, 4, 6. Find
the rectangular and cylindrical coordinates of the point. 42. Identify the surfaces whose equations are given.
(a) 4
(b) 4
43. Write the equation x y z 2 4 in cylindrical coordi2
2
nates and in spherical coordinates.
47. Rewrite the integral
y y y 1
1
1
x2
1y
0
f x, y, z dz dy dx
as an iterated integral in the order dx dy dz. 48. Give five other iterated integrals that are equal to
yy y 2
0
y3
0
y2
0
f x, y, z dz dx dy
49. Use the transformation u x y, v x y to evaluate
xxR x yx y dA, where R is the square with vertices 0, 2, 1, 1, 2, 2, and 1, 3.
50. Use the transformation x u 2, y v 2, z w 2 to
find the volume of the region bounded by the surface sx sy sz 1 and the coordinate planes. 51. Use the change of variables formula and an appropriate
transformation to evaluate xxR xy dA, where R is the square with vertices 0, 0, 1, 1, 2, 0, and 1, 1. 1 dA, where n is an integer and x y 2 n2 D D is the region bounded by the circles with center the origin and radii r and R, 0 r R. (b) For what values of n does the integral in part (a) have a limit as r l 0 ? 1 (c) Find yyy 2 dV, where E is the region 2 x y z 2 n2 E bounded by the spheres with center the origin and radii r and R, 0 r R. (d) For what values of n does the integral in part (c) have a limit as r l 0 ?
52. (a) Evaluate yy
2
13
13.1
VECTOR CALCULUS In this chapter we study the calculus of vector fields. (These are functions that assign vectors to points in space.) In particular we define line integrals (which can be used to find the work done by a force field in moving an object along a curve).Then we define surface integrals (which can be used to find the rate of fluid flow across a surface).The connections between these new types of integrals and the single, double, and triple integrals that we have already met are given by the higher-dimensional versions of the Fundamental Theorem of Calculus: Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem.
VECTOR FIELDS The vectors in Figure 1(a) are air velocity vectors that indicate the wind speed and direction at points 10 m above the surface elevation in the San Francisco Bay area at noon on June 11, 2002. We see at a glance from the largest arrows that the greatest wind speeds at that time occurred as the winds entered the bay across the Golden Gate Bridge. Associated with every point in the air we can imagine a wind velocity vector. This is an example of a velocity vector field. Another example of a velocity vector field is illustrated in Figure 1(b).
Nova Scotia
(a) San Francisco Bay wind patterns FIGURE 1 Velocity vector fields
Another type of vector field, called a force field, associates a force vector with each point in a region. An example is the gravitational force field that we will look at in Example 4. In general, a vector field is a function whose domain is a set of points in ⺢ 2 (or ⺢ 3 ) and whose range is a set of vectors in V2 (or V3 ).
y F(x, y) (x, y) 0
FIGURE 2
Vector field on R@
(b) Ocean currents off the coast of Nova Scotia
Let D be a set in ⺢ 2 (a plane region). A vector field on ⺢ 2 is a function F that assigns to each point x, y in D a two-dimensional vector Fx, y. 1 DEFINITION
x
The best way to picture a vector field is to draw the arrow representing the vector Fx, y starting at the point x, y. Of course, it’s impossible to do this for all points x, y, but we can gain a reasonable impression of F by doing it for a few representative points in D as in Figure 2. Since Fx, y is a two-dimensional vector, we can write 725
726
■
CHAPTER 13
VECTOR CALCULUS
it in terms of its component functions P and Q as follows: Fx, y Px, y i Qx, y j Px, y, Qx, y FPiQj
or, for short,
Notice that P and Q are scalar functions of two variables and are sometimes called scalar fields to distinguish them from vector fields. Let E be a subset of ⺢ 3. A vector field on ⺢ 3 is a function F that assigns to each point x, y, z in E a three-dimensional vector Fx, y, z. 2 DEFINITION
A vector field F on ⺢ 3 is pictured in Figure 3. We can express it in terms of its component functions P, Q, and R as
z
F (x, y, z)
0
Fx, y, z Px, y, z i Qx, y, z j Rx, y, z k
(x, y, z)
As with the vector functions in Section 10.7, we can define continuity of vector fields and show that F is continuous if and only if its component functions P, Q, and R are continuous. We sometimes identify a point x, y, z with its position vector x x, y, z and write Fx instead of Fx, y, z. Then F becomes a function that assigns a vector Fx to a vector x.
y x
FIGURE 3
Vector field on R#
A vector field on ⺢ 2 is defined by Fx, y y i x j. Describe F by sketching some of the vectors Fx, y as in Figure 2. V EXAMPLE 1
SOLUTION Since F1, 0 j, we draw the vector j 0, 1 starting at the point
y
1, 0 in Figure 4. Since F0, 1 i, we draw the vector 1, 0 with starting point 0, 1. Continuing in this way, we calculate several other representative values of Fx, y in the table and draw the corresponding vectors to represent the vector field in Figure 4.
F (2, 2)
F (0, 3)
F (1, 0) 0
FIGURE 4
x
x, y
Fx, y
x, y
Fx, y
1, 0 2, 2 3, 0 0, 1 2, 2 0, 3
0, 1 2, 2 0, 3 1, 0 2, 2 3, 0
1, 0 2, 2 3, 0 0, 1 2, 2 0, 3
0, 1 2, 2 0, 3 1, 0 2, 2 3, 0
F(x, y)=_y i+x j
It appears from Figure 4 that each arrow is tangent to a circle with center the origin. To confirm this, we take the dot product of the position vector x x i y j with the vector Fx Fx, y: x Fx x i y j y i x j xy yx 0 This shows that Fx, y is perpendicular to the position vector x, y and is therefore tangent to a circle with center the origin and radius x sx 2 y 2 . Notice also that
Fx, y sy
2
x sx 2 y 2 x 2
so the magnitude of the vector Fx, y is equal to the radius of the circle.
■
SECTION 13.1
■
VECTOR FIELDS
727
Some computer algebra systems are capable of plotting vector fields in two or three dimensions. They give a better impression of the vector field than is possible by hand because the computer can plot a large number of representative vectors. Figure 5 shows a computer plot of the vector field in Example 1; Figures 6 and 7 show two other vector fields. Notice that the computer scales the lengths of the vectors so they are not too long and yet are proportional to their true lengths. 5
6
_5
5
_6
5
_5
6
_5
5
_6
_5
FIGURE 5
FIGURE 6
FIGURE 7
F(x, y)=k_y, xl
F(x, y)=ky, sin xl
F(x, y)=k ln(1+¥), ln(1+≈)l
z V EXAMPLE 2
SOLUTION The sketch is shown in Figure 8. Notice that all vectors are vertical and point upward above the xy-plane or downward below it. The magnitude increases with the distance from the xy-plane. ■
0 y x
FIGURE 8
F(x, y, z)=z k In Visual 13.1 you can rotate the vector fields in Figures 9–11 as well as additional fields.
We were able to draw the vector field in Example 2 by hand because of its particularly simple formula. Most three-dimensional vector fields, however, are virtually impossible to sketch by hand and so we need to resort to a computer. Examples are shown in Figures 9, 10, and 11. Notice that the vector fields in Figures 9 and 10 have similar formulas, but all the vectors in Figure 10 point in the general direction of the negative y-axis because their y-components are all 2. If the vector field in Figure 11 represents a velocity field, then a particle would be swept upward and would spiral around the z-axis in the clockwise direction as viewed from above.
1 z
Sketch the vector field on ⺢ 3 given by Fx, y, z z k.
0
z
_1
1
5
0
z3
_1 1 _1
0 y
1
FIGURE 9 F(x, y, z)=y i+z j+x k
_1 0 1 x
_1
0 y
1
FIGURE 10 F(x, y, z)=y i-2 j+x k
1
0
_1 x
_1 _1
y0
0 1
FIGURE 11 y x z F(x, y, z)= i- j+ k z z 4
1
x
728
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CHAPTER 13
VECTOR CALCULUS
EXAMPLE 3 Imagine a fluid flowing steadily along a pipe and let Vx, y, z be the
z
velocity vector at a point x, y, z. Then V assigns a vector to each point x, y, z in a certain domain E (the interior of the pipe) and so V is a vector field on ⺢ 3 called a velocity field. A possible velocity field is illustrated in Figure 12. The speed at any given point is indicated by the length of the arrow. Velocity fields also occur in other areas of physics. For instance, the vector field in Example 1 could be used as the velocity field describing the counterclockwise rotation of a wheel. We have seen other examples of velocity fields in Figure 1. ■
0 y x
FIGURE 12
Velocity field in fluid flow
EXAMPLE 4 Newton’s Law of Gravitation states that the magnitude of the gravita-
tional force between two objects with masses m and M is mMG r2
F
where r is the distance between the objects and G is the gravitational constant. (This is an example of an inverse square law.) Let’s assume that the object with mass M is located at the origin in ⺢ 3. (For instance, M could be the mass of the Earth and the origin would be at its center.) Let the position vector of the object with mass m be x x, y, z . Then r x , so r 2 x 2. The gravitational force exerted on this second object acts toward the origin, and the unit vector in this direction is
x x
Therefore, the gravitational force acting on the object at x x, y, z is
z
x
Fx
3
y
mMG x x 3
[Physicists often use the notation r instead of x for the position vector, so you may see Formula 3 written in the form F mMGr 3 r.] The function given by Equation 3 is an example of a vector field, called the gravitational field, because it associates a vector [the force Fx] with every point x in space. Formula 3 is a compact way of writing the gravitational field, but we can also write it in terms of its component functions by using the facts that x x i y j z k and x sx 2 y 2 z 2 :
Fx, y, z
mMGx mMGy mMGz k 2 2 32 i 2 2 2 32 j 2 x y z x y z x y 2 z 2 32 2
FIGURE 13
Gravitational force field
The gravitational field F is pictured in Figure 13.
■
EXAMPLE 5 Suppose an electric charge Q is located at the origin. According to
Coulomb’s Law, the electric force Fx exerted by this charge on a charge q located at a point x, y, z with position vector x x, y, z is 4
Fx
qQ x x 3
where is a constant (that depends on the units used). For like charges, we have qQ 0 and the force is repulsive; for unlike charges, we have qQ 0 and the force is attractive. Notice the similarity between Formulas 3 and 4. Both vector fields are examples of force fields.
SECTION 13.1
VECTOR FIELDS
■
729
Instead of considering the electric force F, physicists often consider the force per unit charge: 1 Q Ex Fx x q x 3
Then E is a vector field on ⺢ 3 called the electric field of Q.
■
GRADIENT FIELDS
If f is a scalar function of two variables, recall from Section 11.6 that its gradient ∇f (or grad f ) is defined by (f x, y fx x, y i fy x, y j Therefore, ∇ f is really a vector field on ⺢ 2 and is called a gradient vector field. Likewise, if f is a scalar function of three variables, its gradient is a vector field on ⺢ 3 given by (f x, y, z fx x, y, z i fy x, y, z j fz x, y, z k Find the gradient vector field of f x, y x 2 y y 3. Plot the gradient vector field together with a contour map of f. How are they related? V EXAMPLE 6
4
SOLUTION The gradient vector field is given by
(f x, y _4
4
_4
FIGURE 14
&f &f i j 2xy i x 2 3y 2 j &x &y
Figure 14 shows a contour map of f with the gradient vector field. Notice that the gradient vectors are perpendicular to the level curves, as we would expect from Section 11.6. Notice also that the gradient vectors are long where the level curves are close to each other and short where they are farther apart. That’s because the length of the gradient vector is the value of the directional derivative of f and closely spaced level curves indicate a steep graph. ■ A vector field F is called a conservative vector field if it is the gradient of some scalar function, that is, if there exists a function f such that F ∇f . In this situation f is called a potential function for F. Not all vector fields are conservative, but such fields do arise frequently in physics. For example, the gravitational field F in Example 4 is conservative because if we define mMG f x, y, z 2 y2 z2 sx then ( f x, y, z
&f &f &f i j k &x &y &z mMGx mMGy mMGz k 2 2 32 i 2 2 2 32 j 2 x y z x y z x y 2 z 2 32 2
Fx, y, z In Sections 13.3 and 13.5 we will learn how to tell whether or not a given vector field is conservative.
730
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CHAPTER 13
13.1
VECTOR CALCULUS
EXERCISES
1–10 ■ Sketch the vector field F by drawing a diagram like Figure 4 or Figure 8.
I
1. Fx, y 2 i j
2. Fx, y i x j
3. Fx, y y i j
4. Fx, y x y i x j
yixj 5. Fx, y sx 2 y 2
yixj 6. Fx, y sx 2 y 2
1
1 2
7. Fx, y, z k
■
■
■
1
1
z 0
z 0
_1
_1
_1
8. Fx, y, z y k
9. Fx, y, z x k ■
10. Fx, y, z j i ■
■
■
■
■
■
y
0
_1 x
1
z 0
z 0
_1
_1
12. Fx, y 1, sin y _1 0 1 y
13. Fx, y x 2, x 1 ■
14. Fx, y y, 1x CAS
5
II
0
IV
1
11. Fx, y y, x
3
1
■
■
I
_1 0 1 y
_1 1 0x
1
III ■
Match the vector fields F with the plots labeled I–IV. Give reasons for your choices.
11–14
II
■
■
0
1 ■
_1 x
■
_1 y ■
■
■
0
_1 1 0x
1 ■
■
■
■
19. If you have a CAS that plots vector fields (the command
is fieldplot in Maple and PlotVectorField in Mathematica), use it to plot Fx, y y 2 2 x y i 3x y 6 x 2 j
_3
3
_5
Explain the appearance by finding the set of points x, y such that Fx, y 0.
5 CAS
a CAS to plot this vector field in various domains until you can see what is happening. Describe the appearance of the plot and explain it by finding the points where Fx 0.
_5
_3 3
III
20. Let Fx r 2 2rx, where x x, y and r x . Use
5
IV
21–24
_3
3
_5
5
22. f x, y x e x
23. f x, y, z sx 2 y 2 z 2
24. f x, y, z x cos yz
■
25–26
■
■
■
■
■
■
■
■
■
■
Match the vector fields F on ⺢3 with the plots labeled I–IV. Give reasons for your choices. 15–18
■
15. Fx, y, z i 2 j 3 k 16. Fx, y, z i 2 j z k 17. Fx, y, z x i y j 3 k 18. Fx, y, z x i y j z k
■
■
■
■
■
■
■
■
CAS
■
■
■
■
■
■
Find the gradient vector field ∇ f of f and sketch it.
25. f x, y x y 2x
_5 ■
Find the gradient vector field of f .
21. f x, y lnx 2y
■
_3
■
■
■
26. f x, y 4 x y2 1
■
■
■
■
■
■
■
■
27–28 ■ Plot the gradient vector field of f together with a contour map of f . Explain how they are related to each other. 27. f x, y sin x sin y ■
■
■
■
■
28. f x, y sinx y ■
■
■
■
■
■
■
29. A particle moves in a velocity field Vx, y x 2, x y 2 .
If it is at position 2, 1 at time t 3, estimate its location at time t 3.01 .
SECTION 13.2
■
731
(b) If parametric equations of a flow line are x xt, y yt, explain why these functions satisfy the differential equations dxdt x and dydt y. Then solve the differential equations to find an equation of the flow line that passes through the point (1, 1).
30. At time t 1 , a particle is located at position 1, 3. If it
moves in a velocity field Fx, y xy 2, y 2 10 find its approximate location at time t 1.05 .
32. (a) Sketch the vector field Fx, y i x j and then
31. The flow lines (or streamlines) of a vector field are the
sketch some flow lines. What shape do these flow lines appear to have? (b) If parametric equations of the flow lines are x xt, y yt, what differential equations do these functions satisfy? Deduce that dydx x. (c) If a particle starts at the origin in the velocity field given by F, find an equation of the path it follows.
paths followed by a particle whose velocity field is the given vector field. Thus the vectors in a vector field are tangent to the flow lines. (a) Use a sketch of the vector field Fx, y x i y j to draw some flow lines. From your sketches, can you guess the equations of the flow lines?
13.2
LINE INTEGRALS
LINE INTEGRALS In this section we define an integral that is similar to a single integral except that instead of integrating over an interval a, b , we integrate over a curve C. Such integrals are called line integrals, although “curve integrals” would be better terminology. They were invented in the early 19th century to solve problems involving fluid flow, forces, electricity, and magnetism. We start with a plane curve C given by the parametric equations x xt
1
y
P *i (x *i , y *i )
Pi-1
Pi
C
Pn
P™ P¡ P¸
x
0
y yt
or, equivalently, by the vector equation rt xt i yt j, and we assume that C is a smooth curve. [This means that r is continuous and rt 0. See Section 10.7.] If we divide the parameter interval a, b into n subintervals ti1, ti and we let x i xti and yi yti , then the corresponding points Pi x i , yi divide C into n subarcs with lengths s1, s2 , . . . , sn . (See Figure 1.) We choose any point Pi*x i*, yi* in the i th subarc. (This corresponds to a point t*i in ti1, ti .) Now if f is any function of two variables whose domain includes the curve C, we evaluate f at the point x i*, yi*, multiply by the length si of the subarc, and form the sum n
f x *, y* s
t *i a FIGURE 1
t i-1
a t b
i
ti
b t
i
i
i1
which is similar to a Riemann sum. Then we take the limit of these sums and make the following definition by analogy with a single integral. 2 DEFINITION If f is defined on a smooth curve C given by Equations 1, then the line integral of f along C is
y
C
n
f x, y ds
f x *, y* s
lim
max s i l 0 i1
i
i
if this limit exists. In Section 9.2 we found that the length of C is L
y
b
a
dx dt
2
dy dt
2
dt
i
732
■
CHAPTER 13
VECTOR CALCULUS
A similar type of argument can be used to show that if f is a continuous function, then the limit in Definition 2 always exists and the following formula can be used to evaluate the line integral:
y
3
C
b
a
2
dx dt
f x, y ds y f xt, yt
dy dt
2
dt
The value of the line integral does not depend on the parametrization of the curve, provided that the curve is traversed exactly once as t increases from a to b. If st is the length of C between ra and rt, then ds dt
The arc length function s is discussed in Section 10.8. ■
2
dx dt
2
dy dt
So the way to remember Formula 3 is to express everything in terms of the parameter t: Use the parametric equations to express x and y in terms of t and write ds as ds z
dx dt
2
dy dt
2
dt
In the special case where C is the line segment that joins a, 0 to b, 0, using x as the parameter, we can write the parametric equations of C as follows: x x, y 0, a x b. Formula 3 then becomes
0
y
y
C
C
f(x, y) (x, y)
f x, y ds y f x, 0 dx b
a
and so the line integral reduces to an ordinary single integral in this case. Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. In fact, if f x, y 0, xC f x, y ds represents the area of one side of the “fence” or “curtain” in Figure 2, whose base is C and whose height above the point x, y is f x, y.
x
FIGURE 2
EXAMPLE 1 Evaluate xC 2 x 2 y ds, where C is the upper half of the unit circle
x 2 y 2 1.
SOLUTION In order to use Formula 3 we first need parametric equations to represent C. Recall that the unit circle can be parametrized by means of the equations
x cos t y
and the upper half of the circle is described by the parameter interval 0 t . (See Figure 3.) Therefore, Formula 3 gives
≈+¥=1 (y˘0)
y
C
_1
FIGURE 3
0
1
y sin t
x
2 x 2 y ds y 2 cos 2 t sin t 0
dx dt
2
dy dt
2
dt
y 2 cos 2 t sin t ssin 2 t cos 2 t dt 0
y 2 cos 2 t sin t dt 2t 0
2
2 3
cos 3t 3
0
■
SECTION 13.2
y
■
733
Suppose now that C is a piecewise-smooth curve; that is, C is a union of a finite number of smooth curves C1, C2, . . . , Cn , where, as illustrated in Figure 4, the initial point of Ci1 is the terminal point of Ci . Then we define the integral of f along C as the sum of the integrals of f along each of the smooth pieces of C:
C¢
C∞ C£
C™
LINE INTEGRALS
y
C
C¡
f x, y ds y f x, y ds y f x, y ds y f x, y ds C C C 1
2
n
x
0
EXAMPLE 2 Evaluate xC 2x ds, where C consists of the arc C1 of the parabola FIGURE 4
A piecewise-smooth curve
y x 2 from 0, 0 to 1, 1 followed by the vertical line segment C2 from 1, 1 to 1, 2.
SOLUTION The curve C is shown in Figure 5. C1 is the graph of a function of x, so we can choose x as the parameter and the equations for C1 become
y (1, 2)
xx
C™ (1, 1)
Therefore
C¡ (0, 0)
y
x
C1
y x2
0 x 1
dx dx
2x ds y 2x 1
0
2
dy dx
]
5s5 1 6
14 23 1 4x 2 32 10
FIGURE 5
2
C=C¡ C™
dx y 2xs1 4x 2 dx 1
0
On C2 we choose y as the parameter, so the equations of C2 are x1
y
and
C2
Thus
yy
2
1
C
2
dx dy
2x ds y 21
y
1 y 2
2
dy dy
2x ds y 2x ds y 2x ds C1
C2
dy y 2 dy 2 2
1
5s5 1 2 6
■
Any physical interpretation of a line integral xC f x, y ds depends on the physical interpretation of the function f . Suppose that x, y represents the linear density at a point x, y of a thin wire shaped like a curve C. Then the mass of the part of the wire from Pi1 to Pi in Figure 1 is approximately x*i , yi* si and so the total mass of the wire is approximately x*i , yi* si . By taking more and more points on the curve, we obtain the mass m of the wire as the limiting value of these approximations: n
m
lim
x*, y* s
max s i l 0 i1
i
i
i
y x, y ds C
[For example, if f x, y 2 x 2 y represents the density of a semicircular wire, then the integral in Example 1 would represent the mass of the wire.] The center of mass of the wire with density function is located at the point x, y, where 4
x
1 m
y
C
x x, y ds
y
1 m
y
C
y x, y ds
Other physical interpretations of line integrals will be discussed later in this chapter.
734
■
CHAPTER 13
VECTOR CALCULUS
A wire takes the shape of the semicircle x 2 y 2 1, y 0, and is thicker near its base than near the top. Find the center of mass of the wire if the linear density at any point is proportional to its distance from the line y 1. V EXAMPLE 3
SOLUTION As in Example 1 we use the parametrization x cos t, y sin t,
0 t , and find that ds dt. The linear density is
x, y k1 y where k is a constant, and so the mass of the wire is
m y k1 y ds y k1 sin t dt k t cos t C
[
0
0
]
k 2
From Equations 4 we have y
y 1
center of mass
1 m
y
C
1 k 2
y x, y ds
y
C
yk1 y ds
1 2
y
0
sin t sin 2 t dt
1 4 cos t 12 t 14 sin 2t 0 2 2 2
[
]
By symmetry we see that x 0, so the center of mass is _1
FIGURE 6
0
1
x
0,
See Figure 6.
4 2 2
0, 0.38 ■
Two other line integrals are obtained by replacing si by either x i x i x i1 or yi yi yi1 in Definition 2. They are called the line integrals of f along C with respect to x and y: 5
n
y
f x, y dx
y
f x, y dy
C
6
lim
f x*, y* x
lim
f x*, y* y
max x i l 0 i1
i
i
i
n
C
max y i l 0 i1
i
i
i
When we want to distinguish the original line integral xC f x, y ds from those in Equations 5 and 6, we call it the line integral with respect to arc length. The following formulas say that line integrals with respect to x and y can also be evaluated by expressing everything in terms of t : x xt, y yt, dx xt dt, dy yt dt. 7
y
f x, y dx y f xt, yt xt dt
y
f x, y dy y f xt, yt yt dt
C
C
b
a
b
a
It frequently happens that line integrals with respect to x and y occur together. When this happens, it’s customary to abbreviate by writing
y
C
Px, y dx y Qx, y dy y Px, y dx Qx, y dy C
C
SECTION 13.2
LINE INTEGRALS
735
■
When we are setting up a line integral, sometimes the most difficult thing is to think of a parametric representation for a curve whose geometric description is given. In particular, we often need to parametrize a line segment, so it’s useful to remember that a vector representation of the line segment that starts at r0 and ends at r1 is given by rt 1 tr0 t r1
8
0 t 1
(See Equation 10.5.4.) Evaluate xC y 2 dx x dy, where (a) C C1 is the line segment from 5, 3 to 0, 2 and (b) C C2 is the arc of the parabola x 4 y 2 from 5, 3 to 0, 2. (See Figure 7.) V EXAMPLE 4
y (0, 2)
C™
C¡ 0
SOLUTION 4
x
(a) A parametric representation for the line segment is x 5t 5
x=4-¥ (_5, _3)
FIGURE 7
y 5t 3
0 t 1
(Use Equation 8 with r0 5, 3 and r1 0, 2 .) Then dx 5 dt, dy 5 dt, and Formulas 7 give
y
y 2 dx x dy y 5t 325 dt 5t 55 dt 1
C1
0
5 y 25t 2 25t 4 dt 5 1
0
25t 3 25t 2 4t 3 2
1
0
5 6
(b) Since the parabola is given as a function of y, let’s take y as the parameter and write C2 as x 4 y2 yy 3 y 2 Then dx 2y dy and by Formulas 7 we have
y
C2
y 2 dx x dy y y 22y dy 4 y 2 dy y 2y 3 y 2 4 dy 2
2
3
3
y4 y3 4y 2 3
2
40 56
■
3
Notice that we got different answers in parts (a) and (b) of Example 4 even though the two curves had the same endpoints. Thus, in general, the value of a line integral depends not just on the endpoints of the curve but also on the path. (But see Section 13.3 for conditions under which the integral is independent of the path.) Notice also that the answers in Example 4 depend on the direction, or orientation, of the curve. If C1 denotes the line segment from 0, 2 to 5, 3, then using the parametrization x 5t you can verify that
y 2 5t
y
C1
0 t 1
y 2 dx x dy 6 5
736
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CHAPTER 13
VECTOR CALCULUS
In general, a given parametrization x xt, y yt, a t b, determines an orientation of a curve C, with the positive direction corresponding to increasing values of the parameter t. (See Figure 8, where the initial point A corresponds to the parameter value a and the terminal point B corresponds to t b.) If C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point B to terminal point A in Figure 8), then we have
B C A
a
b
t
y
B
A
_C
C
f x, y dx y f x, y dx
y
f x, y dy y f x, y dy
C
C
C
But if we integrate with respect to arc length, the value of the line integral does not change when we reverse the orientation of the curve:
y
FIGURE 8
f x, y ds y f x, y ds
C
C
This is because si is always positive, whereas x i and yi change sign when we reverse the orientation of C. LINE INTEGRALS IN SPACE
We now suppose that C is a smooth space curve given by the parametric equations x xt
y yt
z zt
a t b
or by a vector equation rt xt i yt j zt k. If f is a function of three variables that is continuous on some region containing C, then we define the line integral of f along C (with respect to arc length) in a manner similar to that for plane curves:
y
C
n
f x, y, z ds
f x*, y*, z* s
lim
i
max s i l 0 i1
i
i
i
We evaluate it using a formula similar to Formula 3:
y
9
C
dx dt
f x, y, z ds y f xt, yt, zt b
a
2
dy dt
2
dz dt
2
dt
Observe that the integrals in both Formulas 3 and 9 can be written in the more compact vector notation
y
b
a
f rt rt dt
For the special case f x, y, z 1, we get
y
C
ds y
b
a
rt dt L
where L is the length of the curve C (see Formula 10.8.3). Line integrals along C with respect to x, y, and z can also be defined. For example,
y
C
n
f x, y, z dz
lim
f x*, y*, z* z
max z i l 0 i1
i
i
i
i
y f xt, yt, zt zt dt b
a
SECTION 13.2
LINE INTEGRALS
■
737
Therefore, as with line integrals in the plane, we evaluate integrals of the form
y
10
C
Px, y, z dx Qx, y, z dy Rx, y, z dz
6
by expressing everything x, y, z, dx, dy, dz in terms of the parameter t.
4
V EXAMPLE 5 Evaluate xC y sin z ds, where C is the circular helix given by the equations x cos t, y sin t, z t, 0 t 2. (See Figure 9.)
2
SOLUTION Formula 9 gives
z
C 0 _1
y
C
_1 0
y sin z ds y
2
y
2
0
y
x
0
0
sin t sin t
dx dt
2
dy dt
2
sin 2 tssin 2 t cos 2 t 1 dt s2
2
dz dt
y
2 1 2
0
dt
1 cos 2t dt
1 1
s2 t 12 sin 2t 2
[
FIGURE 9
]
2 0
s2
■
EXAMPLE 6 Evaluate xC y dx z dy x dz, where C consists of the line segment
C1 from 2, 0, 0 to 3, 4, 5 followed by the vertical line segment C2 from 3, 4, 5 to 3, 4, 0. z
SOLUTION The curve C is shown in Figure 10. Using Equation 8, we write C1 as
rt 1 t2, 0, 0 t 3, 4, 5 2 t, 4t, 5t
(3, 4, 5)
or, in parametric form, as C¡
C™ 0
x
x2t
y 4t
z 5t
0 t 1
y
(2, 0, 0)
Thus (3, 4, 0)
y
C1
FIGURE 10
y dx z dy x dz y 4t dt 5t4 dt 2 t5 dt 1
0
y
1
0
t2 10 29t dt 10t 29 2
1
24.5
0
Likewise, C2 can be written in the form rt 1 t3, 4, 5 t 3, 4, 0 3, 4, 5 5t or
x3
y4
z 5 5t
0 t 1
Then dx 0 dy, so
y
y dx z dy x dz y 35 dt 15 1
C2
0
Adding the values of these integrals, we obtain
y
C
y dx z dy x dz 24.5 15 9.5
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738
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CHAPTER 13
VECTOR CALCULUS
LINE INTEGRALS OF VECTOR FIELDS
Recall from Section 7.5 that the work done by a variable force f x in moving a particle from a to b along the x-axis is W xab f x dx. Then in Section 10.3 we found that the work done by a constant force F in moving an object from a point P to another l point Q in space is W F D, where D PQ is the displacement vector. Now suppose that F P i Q j R k is a continuous force field on ⺢ 3, such as the gravitational field of Example 4 in Section 13.1 or the electric force field of Example 5 in Section 13.1. (A force field on ⺢ 2 could be regarded as a special case where R 0 and P and Q depend only on x and y.) We wish to compute the work done by this force in moving a particle along a smooth curve C. We divide C into subarcs Pi1Pi with lengths si by dividing the parameter interval a, b into subintervals. (See Figure 1 for the two-dimensional case or Figure 11 for the three-dimensional case.) Choose a point Pi*x*i , yi*, zi* on the i th subarc corresponding to the parameter value t i*. If si is small, then as the particle moves from Pi1 to Pi along the curve, it proceeds approximately in the direction of Tt i*, the unit tangent vector at Pi*. Thus the work done by the force F in moving the particle from Pi1 to Pi is approximately
z
F(x *i , y*i , z *i ) T(t *i ) Pi-1 0
Pi P i*(x *i , y*i , z *i )
Pn y
x
P¸
F x*i , yi*, zi* si Tt i* Fx*i , yi*, zi* Tt i* si and the total work done in moving the particle along C is approximately n
FIGURE 11
Fx*, y*, z* Tx*, y*, z* s
11
i
i
i
i
i
i
i
i1
where Tx, y, z is the unit tangent vector at the point x, y, z on C. Intuitively, we see that these approximations ought to become better as the subarcs become shorter. Therefore, we define the work W done by the force field F as the limit of the Riemann sums in (11), namely, W y Fx, y, z Tx, y, z ds y F T ds
12
C
C
Equation 12 says that work is the line integral with respect to arc length of the tangential component of the force. If the curve C is given by the vector equation rt xt i yt j zt k, then Tt rt rt , so using Equation 9 we can rewrite Equation 12 in the form
W
y
b
a
Frt
rt rt
rt dt y Frt rt dt b
a
This integral is often abbreviated as xC F dr and occurs in other areas of physics as well. Therefore, we make the following definition for the line integral of any continuous vector field. 13 DEFINITION Let F be a continuous vector field defined on a smooth curve C given by a vector function rt, a t b. Then the line integral of F along C is
y
C
F dr y Frt rt dt y F T ds b
a
C
SECTION 13.2
LINE INTEGRALS
■
739
When using Definition 13, remember that Frt is just an abbreviation for Fxt, yt, zt, so we evaluate Frt simply by putting x xt, y yt, and z zt in the expression for Fx, y, z. Notice also that we can formally write dr rt dt. ■ Figure 12 shows the force field and the curve in Example 7. The work done is negative because the field impedes movement along the curve.
EXAMPLE 7 Find the work done by the force field Fx, y x 2 i xy j in moving
a particle along the quarter-circle rt cos t i sin t j, 0 t 2.
SOLUTION Since x cos t and y sin t, we have
y
Frt cos 2t i cos t sin t j
1
rt sin t i cos t j
and Therefore, the work done is
y
C
0
1
F dr y
2
0
Frt rt dt y
cos 3t 2 3
x
FIGURE 12
2
0
2
0
2 cos 2t sin t dt
2 3
■
NOTE Even though xC F dr xC F T ds and integrals with respect to arc length are unchanged when orientation is reversed, it is still true that
y
F dr y F dr
C
C
because the unit tangent vector T is replaced by its negative when C is replaced by C. EXAMPLE 8 Evaluate xC F dr, where Fx, y, z xy i yz j zx k and C is the twisted cubic given by
xt ■ Figure 13 shows the twisted cubic C in Example 8 and some typical vectors acting at three points on C.
z t3
0 t 1
SOLUTION We have
rt t i t 2 j t 3 k
2
rt i 2t j 3t 2 k
1.5
F { r(1)}
z 1 0.5
y t2
Frt t 3 i t 5 j t 4 k (1, 1, 1)
F { r(3/4)}
0 0 y1 2 2
FIGURE 13
Thus
C
F { r(1/2)}
y
C
F dr y Frt rt dt 1
0
y
1
0
1 x
0
t4 5t 7 t 5t dt 4 7 3
6
1
0
27 28
■
Finally, we note the connection between line integrals of vector fields and line integrals of scalar fields. Suppose the vector field F on ⺢ 3 is given in component form by
740
■
CHAPTER 13
VECTOR CALCULUS
the equation F P i Q j R k. We use Definition 13 to compute its line integral along C :
y
C
F dr y Frt rt dt b
a
y
b
a
P i Q j R k xt i yt j zt k dt
y Pxt, yt, zt xt Qxt, yt, zt yt Rxt, yt, zt zt dt b
a
But this last integral is precisely the line integral in (10). Therefore, we have
y
C
F dr y P dx Q dy R dz
where F P i Q j R k
C
For example, the integral as xC F dr where
xC y dx z dy x dz in Example 6 could be expressed Fx, y, z y i z j x k
13.2 1–14
■
Evaluate the line integral, where C is the given curve.
1.
xC y ds,
2.
xC yx ds,
3.
xC x y
4.
5.
6.
7.
EXERCISES 13.
xC x yz dx 2x d y x yz dz,
14.
xC x 2 dx y 2 d y z 2 dz,
C: x t , y t, 0 t 2 2
C: x t 4, y t 3,
1 2
t 1
4
ds, C is the right half of the circle x 2 y 2 16
xC xe dx, C is the arc of the curve x e y from (1, 0) to e, 1
■
C consists of line segments from 1, 0, 1 to 2, 3, 1 and from 2, 3, 1 to 2, 5, 2
C consists of line segments from 0, 0, 0 to 1, 2, 1 and from 1, 2, 1 to 3, 2, 0 ■
■
■
■
■
■
■
■
■
■
■
y
xC x y dx x y dy,
C consists of line segments from 0, 0 to 2, 0 and from 2, 0 to 3, 2
xC sin x dx cos y dy, C consists of the top half of the circle x 2 y 2 1 from 1, 0 to 1, 0 and the line segment from 1, 0 to 2, 3
15. Let F be the vector field shown in the figure.
(a) If C1 is the vertical line segment from 3, 3 to 3, 3, determine whether xC F dr is positive, negative, or zero. (b) If C2 is the counterclockwise-oriented circle with radius 3 and center the origin, determine whether xC F dr is positive, negative, or zero. 1
2
xC x y 3 ds,
y 3
C: x 4 sin t, y 4 cos t, z 3t, 0 t 2 8.
xC x 2z ds,
C is the line segment from (0, 6, 1) to (4, 1, 5)
9.
xC xe yz ds,
C is the line segment from (0, 0, 0) to (1, 2, 3)
10.
xC 2x 9z ds,
11.
xC x 2 y sz
12.
xC z dx x d y y dz,
dz,
C: x t, y t 2, z t 3, 0 t 1
C : x t 3, y t , z t 2 , 0 t 1
C : x t 2, y t 3 , z t 2 , 0 t 1
2 1 _3
_2
_1 0 _1 _2 _3
1
2
3x
SECTION 13.2
16. The figure shows a vector field F and two curves C1 and C2.
CAS
LINE INTEGRALS
■
741
25. Find the exact value of xC x 3 y 5 ds, where C is the part of the
astroid x cos 3t, y sin 3t in the first quadrant.
Are the line integrals of F over C1 and C2 positive, negative, or zero? Explain.
26. (a) Find the work done by the force field y
C¡
CAS
C™ x
Fx, y x 2 i x y j on a particle that moves once around the circle x 2 y 2 4 oriented in the counterclockwise direction. (b) Use a computer algebra system to graph the force field and circle on the same screen. Use the graph to explain your answer to part (a). 27. A thin wire is bent into the shape of a semicircle
x 2 y 2 4, x 0. If the linear density is a constant k, find the mass and center of mass of the wire. 28. Find the mass and center of mass of a thin wire in the shape
of a quarter-circle x 2 y 2 r 2, x 0, y 0, if the density function is x, y x y.
17–20 ■ Evaluate the line integral xC F dr, where C is given by the vector function rt.
29. (a) Write the formulas similar to Equations 4 for the center
of mass x, y, z of a thin wire with density function x, y, z in the shape of a space curve C. (b) Find the center of mass of a wire in the shape of the helix x 2 sin t, y 2 cos t, z 3t, 0 t 2, if the density is a constant k.
17. Fx, y x 2 y 3 i y sx j,
rt t 2 i t 3 j,
0 t 1
18. Fx, y, z yz i xz j x y k,
rt t i t 2 j t 3 k,
0 t 2
19. Fx, y, z sin x i cos y j xz k,
rt t 3 i t 2 j t k,
20. Fx, y, z z i y j x k,
rt t i sin t j cos t k,
■
CAS
■
■
■
■
30. Find the mass and center of mass of a wire in the shape of
the helix x t, y cos t, z sin t, 0 t 2, if the density at any point is equal to the square of the distance from the origin.
0 t 1
■
0 t ■
■
■
■
■
■
Use a graph of the vector field F and the curve C to guess whether the line integral of F over C is positive, negative, or zero. Then evaluate the line integral. 21–22
■
21. Fx, y x y i x y j,
■
■
■
■
■
■
■
;
C
C
■
■
Fx, y e i x y j and C is given by rt t 2 i t 3 j, 0 t 1. (b) Illustrate part (a) by using a graphing calculator or computer to graph C and the vectors from the vector field corresponding to t 0, 1s2 , and 1 (as in Figure 13). x1
24. (a) Evaluate the line integral xC F dr, where
I y y x 2 x, y ds
I x y y 2 z 2 x, y, z ds
23. (a) Evaluate the line integral xC F dr, where
;
C
curve C, its moments of inertia about the x-, y-, and z -axes are defined as
y x i j, sx 2 y 2 sx 2 y 2 C is the parabola y 1 x 2 from 1, 2 to (1, 2) ■
I x y y 2 x, y ds
32. If a wire with linear density x, y, z lies along a space
22. Fx, y
■
C, its moments of inertia about the x- and y-axes are defined as
Find the moments of inertia for the wire in Example 3.
C is the arc of the circle x 2 y 2 4 traversed counterclockwise from (2, 0) to 0, 2
■
31. If a wire with linear density x, y lies along a plane curve
Fx, y, z x i z j y k and C is given by rt 2t i 3t j t 2 k, 1 t 1. (b) Illustrate part (a) by using a computer to graph C and the vectors from the vector field corresponding to t 1 and 12 (as in Figure 13).
I y y x 2 z 2 x, y, z ds C
I z y x 2 y 2 x, y, z ds C
Find the moments of inertia for the wire in Exercise 29. 33. Find the work done by the force field
Fx, y x i y 2 j in moving an object along an arch of the cycloid rt t sin t i 1 cos t j, 0 t 2. 34. Find the work done by the force field
Fx, y x sin y i y j on a particle that moves along the parabola y x 2 from 1, 1 to 2, 4.
742
■
CHAPTER 13
VECTOR CALCULUS
35. Find the work done by the force field
lies in the plane perpendicular to the wire and whose center is the axis of the wire (as in the figure). Ampère’s Law relates the electric current to its magnetic effects and states that
Fx, y, z y z, x z, x y on a particle that moves along the line segment from 1, 0, 0 to 3, 4, 2. 36. The force exerted by an electric charge at the origin on a
y
charged particle at a point x, y, z with position vector r x, y, z is Fr Kr r 3 where K is a constant. (See Example 5 in Section 13.1.) Find the work done as the particle moves along a straight line from 2, 0, 0 to 2, 1, 5.
C
B dr 0 I
where I is the net current that passes through any surface bounded by a closed curve C and 0 is a constant called the permeability of free space. By taking C to be a circle with radius r, show that the magnitude B B of the magnetic field at a distance r from the center of the wire is
37. A 160-lb man carries a 25-lb can of paint up a helical stair-
case that encircles a silo with a radius of 20 ft. If the silo is 90 ft high and the man makes exactly three complete revolutions, how much work is done by the man against gravity in climbing to the top?
B
0 I 2 r
38. Suppose there is a hole in the can of paint in Exercise 37
I
and 9 lb of paint leaks steadily out of the can during the man’s ascent. How much work is done? 39. (a) Show that a constant force field does zero work on a
particle that moves once uniformly around the circle x 2 y 2 1. (b) Is this also true for a force field Fx k x, where k is a constant and x x, y ?
B
40. Experiments show that a steady current I in a long wire pro-
duces a magnetic field B that is tangent to any circle that
13.3
THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS Recall from Section 5.4 that Part 2 of the Fundamental Theorem of Calculus can be written as
y
1
b
a
Fx dx Fb Fa
where F is continuous on a, b . We also called Equation 1 the Net Change Theorem: The integral of a rate of change is the net change. If we think of the gradient vector ∇f of a function f of two or three variables as a sort of derivative of f , then the following theorem can be regarded as a version of the Fundamental Theorem for line integrals.
2 THEOREM Let C be a smooth curve given by the vector function rt, a t b. Let f be a differentiable function of two or three variables whose gradient vector ∇f is continuous on C. Then
y
C
( f dr f rb f ra
NOTE Theorem 2 says that we can evaluate the line integral of a conservative vector field (the gradient vector field of the potential function f ) simply by knowing the
SECTION 13.3
B(x™, y™)
0
C
■
743
value of f at the endpoints of C. In fact, Theorem 2 says that the line integral of ∇f is the net change in f. If f is a function of two variables and C is a plane curve with initial point Ax 1, y1 and terminal point Bx 2 , y2 , as in Figure 1, then Theorem 2 becomes
y
A(x¡, y¡)
THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
y
x
C
(f dr f x 2 , y2 f x 1, y1
If f is a function of three variables and C is a space curve joining the point Ax 1, y1, z1 to the point Bx 2 , y2 , z2 , then we have z
y
A(x¡, y¡, z¡) B(x™, y™, z™) 0 x
(f dr f x 2 , y2 , z2 f x 1, y1, z1
C
C
Let’s prove Theorem 2 for this case. PROOF OF THEOREM 2 Using Definition 13.2.13, we have
y
y
C
(f dr y (f rt rt dt b
a
FIGURE 1
y
b
y
b
a
a
&f dx &f dy &f dz &x dt &y dt &z dt
d f rt dt dt
dt
(by the Chain Rule)
f rb f ra The last step follows from the Fundamental Theorem of Calculus (Equation 1).
■
Although we have proved Theorem 2 for smooth curves, it is also true for piecewisesmooth curves. This can be seen by subdividing C into a finite number of smooth curves and adding the resulting integrals. EXAMPLE 1 Find the work done by the gravitational field
Fx
mMG x x 3
in moving a particle with mass m from the point 3, 4, 12 to the point 2, 2, 0 along a piecewise-smooth curve C. (See Example 4 in Section 13.1.) SOLUTION From Section 13.1 we know that F is a conservative vector field and, in fact, F ∇f , where mMG f x, y, z sx 2 y 2 z 2
Therefore, by Theorem 2, the work done is W y F dr y (f dr C
C
f 2, 2, 0 f 3, 4, 12
mMG mMG 1 1 mMG 2 22 2 4 2 12 2 2s2 13 s2 s3
■
744
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CHAPTER 13
VECTOR CALCULUS
INDEPENDENCE OF PATH
Suppose C1 and C2 are two piecewise-smooth curves (which are called paths) that have the same initial point A and terminal point B. We know from Example 4 in Section 13.2 that, in general, xC F dr xC F dr. But one implication of Theorem 2 is that 1
2
y
C1
(f dr y (f dr C2
whenever ∇ f is continuous. In other words, the line integral of a conservative vector field depends only on the initial point and terminal point of a curve. In general, if F is a continuous vector field with domain D, we say that the line integral xC F dr is independent of path if xC F dr xC F dr for any two paths C1 and C2 in D that have the same initial and terminal points. With this terminology we can say that line integrals of conservative vector fields are independent of path. A curve is called closed if its terminal point coincides with its initial point, that is, rb ra. (See Figure 2.) If xC F dr is independent of path in D and C is any closed path in D, we can choose any two points A and B on C and regard C as being composed of the path C1 from A to B followed by the path C2 from B to A. (See Figure 3.) Then 1
C FIGURE 2
A closed curve C™ B
y
A
C
C¡ FIGURE 3
2
F dr y F dr y F dr y F dr y C1
C2
C2
C1
F dr 0
since C1 and C2 have the same initial and terminal points. Conversely, if it is true that xC F dr 0 whenever C is a closed path in D, then we demonstrate independence of path as follows. Take any two paths C1 and C2 from A to B in D and define C to be the curve consisting of C1 followed by C2. Then 0 y F dr y F dr y C
C1
C2
F dr y F dr y F dr C1
C2
and so xC F dr xC F dr. Thus we have proved the following theorem. 1
2
3 THEOREM xC F dr is independent of path in D if and only if xC F dr 0 for every closed path C in D.
Since we know that the line integral of any conservative vector field F is independent of path, it follows that xC F dr 0 for any closed path. The physical interpretation is that the work done by a conservative force field (such as the gravitational or electric field in Section 13.1) as it moves an object around a closed path is 0. The following theorem says that the only vector fields that are independent of path are conservative. It is stated and proved for plane curves, but there is a similar version for space curves. We assume that D is open, which means that for every point P in D there is a disk with center P that lies entirely in D. (So D doesn’t contain any of its boundary points.) In addition, we assume that D is connected. This means that any two points in D can be joined by a path that lies in D. Suppose F is a vector field that is continuous on an open connected region D. If xC F dr is independent of path in D, then F is a conservative vector field on D ; that is, there exists a function f such that ∇f F. 4 THEOREM
SECTION 13.3
y (x¡, y)
THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
745
PROOF Let Aa, b be a fixed point in D. We construct the desired potential function f by defining
C™
f x, y y
(x, y)
C¡
■
D (a, b) x
0
x, y
a, b
FIGURE 4
F dr
for any point x, y in D. Since xC F dr is independent of path, it does not matter which path C from a, b to x, y is used to evaluate f x, y. Since D is open, there exists a disk contained in D with center x, y. Choose any point x 1, y in the disk with x 1 x and let C consist of any path C1 from a, b to x 1, y followed by the horizontal line segment C2 from x 1, y to x, y. (See Figure 4.) Then f x, y y F dr y F dr y C1
x1, y
F dr y F dr
a, b
C2
C2
Notice that the first of these integrals does not depend on x, so & & f x, y 0 &x &x
y
C2
F dr
If we write F P i Q j, then
y
C2
F dr y P dx Q dy C2
On C2 , y is constant, so dy 0. Using t as the parameter, where x 1 t x, we have & & f x, y &x &x
y
C2
P dx Q dy
& &x
y
x
x1
Pt, y dt Px, y
y (x, y)
by Part 1 of the Fundamental Theorem of Calculus (see Section 5.4). A similar argument, using a vertical line segment (see Figure 5), shows that
C™ C¡
(x, y¡)
& & f x, y &y &y
D (a, b) 0
FIGURE 5
x
Thus
y
C2
P dx Q dy
FPiQj
& &y
y
y
y1
Qx, t dt Qx, y
&f &f i j ∇f &x &y ■
which says that F is conservative.
The question remains: How is it possible to determine whether or not a vector field F is conservative? Suppose it is known that F P i Q j is conservative, where P and Q have continuous first-order partial derivatives. Then there is a function f such that F ∇f , that is, P
&f &x
and
Q
&f &y
Therefore, by Clairaut’s Theorem, &P &2 f &2 f &Q &y &y &x &x &y &x
746
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CHAPTER 13
VECTOR CALCULUS
5 THEOREM If Fx, y Px, y i Qx, y j is a conservative vector field, where P and Q have continuous first-order partial derivatives on a domain D, then throughout D we have
simple, not closed
&Q &P &y &x
not simple, not closed
not simple, closed
simple, closed FIGURE 6
Types of curves
simply-connected region
regions that are not simply-connected
The converse of Theorem 5 is true only for a special type of region. To explain this, we first need the concept of a simple curve, which is a curve that doesn’t intersect itself anywhere between its endpoints. [See Figure 6; ra rb for a simple closed curve, but rt1 rt2 when a t1 t2 b.] In Theorem 4 we needed an open connected region. For the next theorem we need a stronger condition. A simply-connected region in the plane is a connected region D such that every simple closed curve in D encloses only points that are in D. Notice from Figure 7 that, intuitively speaking, a simply-connected region contains no hole and can’t consist of two separate pieces. In terms of simply-connected regions we can now state a partial converse to Theorem 5 that gives a convenient method for verifying that a vector field on ⺢ 2 is conservative. The proof will be sketched in the next section as a consequence of Green’s Theorem. 6 THEOREM Let F P i Q j be a vector field on an open simply-connected region D. Suppose that P and Q have continuous first-order derivatives and
&P &Q &y &x
FIGURE 7
throughout D
Then F is conservative.
V EXAMPLE 2 ■ Figure 8 shows the vector field in Example 2. The vectors that start on the closed curve C all appear to point in roughly the same direction as C . So it looks as if xC F dr 0 and therefore F is not conservative. The calculation in Example 2 confirms this impression.
10
Fx, y x y i x 2 j is conservative. SOLUTION Let Px, y x y and Qx, y x 2. Then
&P 1 &y
&Q 1 &x
Since &P&y &Q&x, F is not conservative by Theorem 5. V EXAMPLE 3
_10
Determine whether or not the vector field
Determine whether or not the vector field Fx, y 3 2xy i x 2 3y 2 j
10
is conservative. SOLUTION Let Px, y 3 2xy and Qx, y x 2 3y 2. Then
C _10
FIGURE 8
&P &Q 2x &y &x
■
SECTION 13.3
Figure 9 shows the vector field in Example 3. Some of the vectors near the curves C1 and C2 point in approximately the same direction as the curves, whereas others point in the opposite direction. So it appears plausible that line integrals around all closed paths are 0. Example 3 shows that F is indeed conservative. ■
C™
_2
■
747
Also, the domain of F is the entire plane D ⺢ 2 , which is open and simplyconnected. Therefore, we can apply Theorem 6 and conclude that F is conservative. ■
In Example 3, Theorem 6 told us that F is conservative, but it did not tell us how to find the (potential) function f such that F ∇ f . The proof of Theorem 4 gives us a clue as to how to find f . We use “partial integration” as in the following example. EXAMPLE 4
(a) If Fx, y 3 2xy i x 2 3y 2 j, find a function f such that F ∇f . (b) Evaluate the line integral xC F dr, where C is the curve given by rt e t sin t i e t cos t j , 0 t .
2
C¡
THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
2
SOLUTION
(a) From Example 3 we know that F is conservative and so there exists a function f with ∇f F, that is, _2
FIGURE 9
7
fx x, y 3 2xy
8
fy x, y x 2 3y 2
Integrating (7) with respect to x, we obtain f x, y 3x x 2 y t y
9
Notice that the constant of integration is a constant with respect to x, that is, a function of y, which we have called ty. Next we differentiate both sides of (9) with respect to y : fy x, y x 2 ty
10
Comparing (8) and (10), we see that ty 3y 2 Integrating with respect to y, we have ty y 3 K where K is a constant. Putting this in (9), we have f x, y 3x x 2 y y 3 K as the desired potential function. (b) To use Theorem 2 all we have to know are the initial and terminal points of C, namely, r0 0, 1 and r 0, e . In the expression for f x, y in part (a), any value of the constant K will do, so let’s choose K 0. Then we have
y
C
F dr y (f dr f 0, e f 0, 1 e 3 1 e 3 1 C
This method is much shorter than the straightforward method for evaluating line integrals that we learned in Section 13.2.
■
A criterion for determining whether or not a vector field F on ⺢ 3 is conservative is given in Section 13.5. Meanwhile, the next example shows that the technique for finding the potential function is much the same as for vector fields on ⺢ 2.
748
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CHAPTER 13
VECTOR CALCULUS
If Fx, y, z y 2 i 2xy e 3z j 3ye 3z k, find a function f such that ∇f F. V EXAMPLE 5
SOLUTION If there is such a function f , then 11
fx x, y, z y 2
12
fy x, y, z 2xy e 3z
13
fz x, y, z 3ye 3z
Integrating (11) with respect to x, we get f x, y, z xy 2 ty, z
14
where ty, z is a constant with respect to x. Then differentiating (14) with respect to y, we have fy x, y, z 2xy t y y, z and comparison with (12) gives t y y, z e 3z Thus ty, z ye 3z hz and we rewrite (14) as f x, y, z xy 2 ye 3z hz Finally, differentiating with respect to z and comparing with (13), we obtain hz 0 and, therefore, hz K , a constant. The desired function is f x, y, z xy 2 ye 3z K It is easily verified that ∇f F.
■
CONSERVATION OF ENERGY
Let’s apply the ideas of this chapter to a continuous force field F that moves an object along a path C given by rt, a t b, where ra A is the initial point and rb B is the terminal point of C. According to Newton’s Second Law of Motion (see Section 10.9), the force Frt at a point on C is related to the acceleration at rt by the equation Frt mrt So the work done by the force on the object is W y F dr y Frt rt dt b
C
a
m 2
y
b
m 2
y
b
m 2
( rb 2 ra 2 )
a
a
y
d rt rt dt dt d m rt 2 dt dt 2
b
a
mrt rt dt (Theorem 10.7.5, Formula 4)
[ rt ]
2 b a
(Fundamental Theorem of Calculus)
SECTION 13.3
THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
■
749
Therefore
W 12 m vb 2 12 m va
15
2
where v r is the velocity. The quantity 12 m vt 2, that is, half the mass times the square of the speed, is called the kinetic energy of the object. Therefore, we can rewrite Equation 15 as
W KB KA
16
which says that the work done by the force field along C is equal to the change in kinetic energy at the endpoints of C. Now let’s further assume that F is a conservative force field; that is, we can write F ∇f . In physics, the potential energy of an object at the point x, y, z is defined as Px, y, z f x, y, z, so we have F ∇P. Then by Theorem 2 we have W y F dr y (P dr Prb Pra PA PB C
C
Comparing this equation with Equation 16, we see that PA KA PB KB which says that if an object moves from one point A to another point B under the influence of a conservative force field, then the sum of its potential energy and its kinetic energy remains constant. This is called the Law of Conservation of Energy and it is the reason the vector field is called conservative.
13.3
EXERCISES
1. The figure shows a curve C and a contour map of a function
f whose gradient is continuous. Find xC ( f dr.
3–10 ■ Determine whether or not F is a conservative vector field. If it is, find a function f such that F ( f .
y
3. Fx, y 6x 5y i 5x 4y j 40
C
50
4. Fx, y x 3 4 x y i 4 x y y 3 j
60
5. Fx, y xe y i ye x j
30 20
6. Fx, y e y i xe y j
10
7. Fx, y 2 x cos y y cos x i x 2 sin y sin x j
0
8. Fx, y 1 2 x y ln x i x 2 j
x
9. Fx, y ye x sin y i e x x cos y j 2. A table of values of a function f with continuous gradient is
given. Find xC ( f dr, where C has parametric equations xt 1 2
y
yt t 3
0 t 1
0
1
2
0
1
6
4
1
3
5
7
2
8
2
9
x
10. Fx, y x y cos x y sin x y i x 2 cos x y j ■
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■
■
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■
11–16 ■ (a) Find a function f such that F ∇ f and (b) use part (a) to evaluate xC F dr along the given curve C. 11. Fx, y x 3 y 4 i x 4 y 3 j,
C: rt st i 1 t 3 j, 0 t 1 y2 i 2y arctan x j, 1 x2 C: rt t 2 i 2t j, 0 t 1
12. Fx, y
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750
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CHAPTER 13
VECTOR CALCULUS
13. Fx, y, z yz i xz j x y 2z k,
25. Show that if the vector field F P i Q j R k is
C is the line segment from 1, 0, 2 to 4, 6, 3
conservative and P, Q, R have continuous first-order partial derivatives, then
14. Fx, y, z 2 xz y 2 i 2 x y j x 2 3z 2 k,
C: x t 2, y t 1, z 2t 1, 0 t 1
&P &Q &y &x
15. Fx, y, z y cos z i 2 x y cos z j x y sin z k, 2
2
C: rt t 2 i sin t j t k, 0 t
■
■
■
■
■
xC y dx x dy xyz dz is not independent of path.
0 t 1
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■
■
■
■
17.
18.
■
■
29.
xC tan y dx x sec y dy, C is any path from 1, 0 to 2, 4 2
xC 1 ye
x
30. ■
x
dx e dy, C is any path from 0, 1 to 1, 2 ■
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■
■
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■
■
■
■
■ Find the work done by the force field F in moving an object from P to Q.
19. Fx, y 2y 32 i 3x sy j ; 20. Fx, y ey i xey j ; ■
21–22
■
■
■
P1, 1, Q2, 4
■
■
■
■
y
CAS
■
■
■
■
■
x
■
■
■
■
■
23. If Fx, y sin y i 1 x cos y j, use a plot to guess
whether F is conservative. Then determine whether your guess is correct. 24. Let F ( f , where f x, y sinx 2y. Find curves C1
and C2 that are not closed and satisfy the equation. (a)
y
C1
F dr 0
(b)
2
■
■
■
y 2 9
2
■
■
■
■
■
■
y i x j . x2 y2 (a) Show that &P&y &Q&x . (b) Show that xC F dr is not independent of path. [Hint: Compute xC F dr and xC F dr, where C1 and C2 are the upper and lower halves of the circle x 2 y 2 1 from 1, 0 to 1, 0.] Does this contradict Theorem 6?
Fr
y
22.
x
■
■
2
y
C2
2
■
Explain. 21.
2
■
x 0
32. (a) Suppose that F is an inverse square force field, that is,
■
Is the vector field shown in the figure conservative?
■
2
28. x, y
1
P0, 1, Q2, 0 ■
x 0, y 0
x, y 1 x y 4
x, y x y 1 or 4 x
31. Let Fx, y
19–20
■
27–30 ■ Determine whether or not the given set is (a) open, (b) connected, and (c) simply-connected. 27. x, y
Show that the line integral is independent of path and evaluate the integral. 17–18
&Q &R &z &y
26. Use Exercise 25 to show that the line integral
16. Fx, y, z e y i xe y j z 1e z k,
C: rt t i t 2 j t 3 k,
&P &R &z &x
F dr 1
■
cr
r
3
for some constant c, where r x i y j z k. Find the work done by F in moving an object from a point P1 along a path to a point P2 in terms of the distances d1 and d2 from these points to the origin. (b) An example of an inverse square field is the gravitational field F mMG r r 3 discussed in Example 4 in Section 13.1. Use part (a) to find the work done by the gravitational field when the Earth moves from aphelion (at a maximum distance of 1.52 10 8 km from the Sun) to perihelion (at a minimum distance of 1.47 10 8 km). (Use the values m 5.97 10 24 kg, M 1.99 10 30 kg, and G 6.67 10 11 N m 2kg2. (c) Another example of an inverse square field is the electric force field F qQr r 3 discussed in Example 5 in Section 13.1. Suppose that an electron with a charge of 1.6 10 19 C is located at the origin. A positive unit charge is positioned a distance 10 12 m from the electron and moves to a position half that distance from the electron. Use part (a) to find the work done by the electric field. (Use the value 8.985 10 9.)
SECTION 13.4
13.4
■
751
GREEN’S THEOREM
y
D C 0
GREEN’S THEOREM
x
Green’s Theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. (See Figure 1. We assume that D consists of all points inside C as well as all points on C.) In stating Green’s Theorem we use the convention that the positive orientation of a simple closed curve C refers to a single counterclockwise traversal of C. Thus if C is given by the vector function rt, a t b, then the region D is always on the left as the point rt traverses C. (See Figure 2.) y
FIGURE 1
y
C D
D C 0
x
0
(a) Positive orientation
FIGURE 2
x
(b) Negative orientation
GREEN’S THEOREM Let C be a positively oriented, piecewise-smooth, simple
closed curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then ■ Recall that the left side of this equation is another way of writing xC F dr, where F P i Q j.
y
C
P dx Q dy
yy D
&Q &P &x &y
dA
NOTE The notation
y
䊊 C
P dx Q dy
or
gC P dx Q dy
is sometimes used to indicate that the line integral is calculated using the positive orientation of the closed curve C. Another notation for the positively oriented boundary curve of D is &D, so the equation in Green’s Theorem can be written as 1
yy D
&Q &P &x &y
dA y P dx Q dy &D
Green’s Theorem should be regarded as the counterpart of the Fundamental Theorem of Calculus for double integrals. Compare Equation 1 with the statement of the Fundamental Theorem of Calculus, Part 2, in the following equation:
y
b
a
Fx dx Fb Fa
In both cases there is an integral involving derivatives (F, &Q&x, and &P&y) on the left side of the equation. And in both cases the right side involves the values of the original functions (F , Q, and P ) only on the boundary of the domain. (In the one-
752
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CHAPTER 13
VECTOR CALCULUS
Green’s Theorem is named after the self-taught English scientist George Green (1793–1841). He worked full-time in his father’s bakery from the age of nine and taught himself mathematics from library books. In 1828 he published privately An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism, but only 100 copies were printed and most of those went to his friends. This pamphlet contained a theorem that is equivalent to what we know as Green’s Theorem, but it didn’t become widely known at that time. Finally, at age 40, Green entered Cambridge University as an undergraduate but died four years after graduation. In 1846 William Thomson (Lord Kelvin) located a copy of Green’s essay, realized its significance, and had it reprinted. Green was the first person to try to formulate a mathematical theory of electricity and magnetism. His work was the basis for the subsequent electromagnetic theories of Thomson, Stokes, Rayleigh, and Maxwell. ■
dimensional case, the domain is an interval a, b whose boundary consists of just two points, a and b.) Green’s Theorem is not easy to prove in general, but we can give a proof for the special case where the region is both of type I and of type II (see Section 12.2). Let’s call such regions simple regions. PROOF OF GREEN’S THEOREM FOR THE CASE IN WHICH D IS A SIMPLE REGION
Notice that Green’s Theorem will be proved if we can show that
y
2
C
D
y
3
C
D x, y a x b, t1x y t 2x where t1 and t 2 are continuous functions. This enables us to compute the double integral on the right side of Equation 2 as follows:
yy
&P b t x &P b dA y y x, y dy dx y Px, t 2x Px, t1x dx a t x &y a &y 2
1
where the last step follows from the Fundamental Theorem of Calculus. Now we compute the left side of Equation 2 by breaking up C as the union of the four curves C1 , C2 , C3 , and C4 shown in Figure 3. On C1 we take x as the parameter and write the parametric equations as x x, y t1x, a x b. Thus
y=g™(x)
y
D
C™
a
Observe that C3 goes from right to left but C3 goes from left to right, so we can write the parametric equations of C3 as x x, y t 2x, a x b. Therefore
C¡ y=g¡(x) b
Px, y dx y Px, t1x dx b
C1
a
&Q dA &x
We prove Equation 2 by expressing D as a type I region:
C£
0
Q dy yy D
D
C¢
&P dA &y
and
4
y
P dx yy
y
x
C3
Px, y dx y
Px, y dx y Px, t 2x dx b
C3
a
FIGURE 3
On C2 or C4 (either of which might reduce to just a single point), x is constant, so dx 0 and
y
C2
Px, y dx 0 y Px, y dx C4
Hence
y
C
Px, y dx y Px, y dx y Px, y dx y Px, y dx y Px, y dx C1
C2
C3
y Px, t1x dx y Px, t 2x dx b
a
b
a
C4
SECTION 13.4
GREEN’S THEOREM
■
753
Comparing this expression with the one in Equation 4, we see that
y
C
&P dA &y
Px, y dx yy D
Equation 3 can be proved in much the same way by expressing D as a type II region (see Exercise 28). Then, by adding Equations 2 and 3, we obtain Green’s Theorem. ■ EXAMPLE 1 Evaluate xC x 4 dx xy dy, where C is the triangular curve consisting
of the line segments from 0, 0 to 1, 0, from 1, 0 to 0, 1, and from 0, 1 to 0, 0.
SOLUTION Although the given line integral could be evaluated as usual by the methods of Section 13.2, that would involve setting up three separate integrals along the three sides of the triangle, so let’s use Green’s Theorem instead. Notice that the region D enclosed by C is simple and C has positive orientation (see Figure 4). If we let Px, y x 4 and Qx, y xy, then we have
y
y=1-x
(0, 1)
C D (0, 0)
(1, 0)
x
y
C
x 4 dx xy dy
yy D
y
FIGURE 4
1
0
&Q &P &x &y
dA y
1
0
y
1x
0
y 0 dy dx
dx 12 y 1 x2 dx 16 1 x3 1
[y] 1 2
2 y1x y0
0
]
1 0
16
■
Evaluate 䊊xC 3y e sin x dx (7x sy 4 1 ) dy, where C is the circle x y 9. V EXAMPLE 2 2
2
SOLUTION The region D bounded by C is the disk x 2 y 2 9, so let’s change to
polar coordinates after applying Green’s Theorem:
y
䊊 C
3y e sin x dx (7x sy 4 1 ) dy
■ Instead of using polar coordinates, we could simply use the fact that D is a disk of radius 3 and write
yy 4 dA 4 3
2
yy D
y
36
2
0
D
y
& & (7x sy 4 1 ) &y 3y e sin x dA &x 3
0
7 3 r dr d 4 y
2
0
d
y
3
0
r dr 36
■
In Examples 1 and 2 we found that the double integral was easier to evaluate than the line integral. (Try setting up the line integral in Example 2 and you’ll soon be convinced!) But sometimes it’s easier to evaluate the line integral, and Green’s Theorem is used in the reverse direction. For instance, if it is known that Px, y Qx, y 0 on the curve C, then Green’s Theorem gives
yy D
&Q &P &x &y
dA y P dx Q dy 0 C
no matter what values P and Q assume in the region D. Another application of the reverse direction of Green’s Theorem is in computing areas. Since the area of D is xxD 1 dA, we wish to choose P and Q so that &Q &P 1 &x &y
754
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CHAPTER 13
VECTOR CALCULUS
There are several possibilities: Px, y 0
Px, y y
Px, y 12 y
Qx, y x
Qx, y 0
Qx, y 12 x
Then Green’s Theorem gives the following formulas for the area of D : 1 䊊 x dy y dx 䊊 x dy y 䊊 y dx 2 y Ay
5
C
C
C
EXAMPLE 3 Find the area enclosed by the ellipse
x2 y2 1. a2 b2
SOLUTION The ellipse has parametric equations x a cos t and y b sin t, where
0 t 2. Using the third formula in Equation 5, we have A 12 y x dy y dx C
C¡ D¡
D™ C£
C™
_C£
ab 2
y
2
0
1 2
y
2
0
a cos tb cos t dt b sin ta sin t dt
dt ab
■
Although we have proved Green’s Theorem only for the case where D is simple, we can now extend it to the case where D is a finite union of simple regions. For example, if D is the region shown in Figure 5, then we can write D D1 D2 , where D1 and D2 are both simple. The boundary of D1 is C1 C3 and the boundary of D2 is C2 C3 so, applying Green’s Theorem to D1 and D2 separately, we get
y
FIGURE 5
C1C3
P dx Q dy
yy yy D1
C
y
C2C3
P dx Q dy
D2
&Q &P &x &y &Q &P &x &y
dA
dA
If we add these two equations, the line integrals along C3 and C3 cancel, so we get
y
C1C2
P dx Q dy
yy D
FIGURE 6
≈+¥=4
FIGURE 7
≈+¥=1
dA
Evaluate 䊊xC y 2 dx 3xy dy, where C is the boundary of the semiannular region D in the upper half-plane between the circles x 2 y 2 1 and x 2 y 2 4. V EXAMPLE 4
C
0
&Q &P &x &y
which is Green’s Theorem for D D1 D2 , since its boundary is C C1 C2 . The same sort of argument allows us to establish Green’s Theorem for any finite union of nonoverlapping simple regions (see Figure 6).
y
D
x
SOLUTION Notice that although D is not simple, the y-axis divides it into two simple regions (see Figure 7). In polar coordinates we can write
D r, 1 r 2, 0
SECTION 13.4
Therefore, Green’s Theorem gives
y
䊊 C
y 2 dx 3xy dy
yy D
y
0
GREEN’S THEOREM
■
755
& & 2 3xy y 2 dA yy y dA y y r sin r dr d 0 1 &x &y D
y
sin d
2
1
[
r 2 dr cos
1 0 3
][r]
3 2 1
14 3
■
Green’s Theorem can be extended to apply to regions with holes, that is, regions that are not simply-connected. Observe that the boundary C of the region D in Figure 8 consists of two simple closed curves C1 and C2 . We assume that these boundary curves are oriented so that the region D is always on the left as the curve C is traversed. Thus the positive direction is counterclockwise for the outer curve C1 but clockwise for the inner curve C2 . If we divide D into two regions D and D by means of the lines shown in Figure 9 and then apply Green’s Theorem to each of D and D, we get
C™ D C¡ FIGURE 8
yy
Dª
D
&Q &P &x &y
dA
yy D
y
&D
Dªª
&Q &P &x &y
dA
yy D
P dx Q dy y
&D
&Q &P &x &y
dA
P dx Q dy
Since the line integrals along the common boundary lines are in opposite directions, they cancel and we get
FIGURE 9
yy D
&Q &P &x &y
dA y P dx Q dy y P dx Q dy y P dx Q dy C1
C
C2
which is Green’s Theorem for the region D. If Fx, y y i x jx 2 y 2 , show that xC F dr 2 for every positively oriented simple closed path that encloses the origin. V EXAMPLE 5
y
C Cª D
FIGURE 10
x
SOLUTION Since C is an arbitrary closed path that encloses the origin, it’s difficult to compute the given integral directly. So let’s consider a counterclockwise-oriented circle C with center the origin and radius a, where a is chosen to be small enough that C lies inside C. (See Figure 10.) Let D be the region bounded by C and C. Then its positively oriented boundary is C C and so the general version of Green’s Theorem gives
y
C
P dx Q dy y
C
P dx Q dy
yy yy D
D
&Q &P &x &y
y2 x2 y2 x2 2 2 2 x y x 2 y 2 2
0 Therefore
that is,
y
C
P dx Q dy y P dx Q dy C
y
C
F dr y F dr C
dA
dA
756
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CHAPTER 13
VECTOR CALCULUS
We now easily compute this last integral using the parametrization given by rt a cos t i a sin t j , 0 t 2. Thus
y
C
F dr y F dr y C
y
2
0
Frt rt dt
a sin ta sin t a cos ta cos t 2 dt y dt 2 2 2 2 2 0 a cos t a sin t
2
0
■
We end this section by using Green’s Theorem to discuss a result that was stated in the preceding section. SKETCH OF PROOF OF THEOREM 13.3.6 We’re assuming that F P i Q j is a
vector field on an open simply-connected region D, that P and Q have continuous first-order partial derivatives, and that &P &Q &y &x
throughout D
If C is any simple closed path in D and R is the region that C encloses, then Green’s Theorem gives
y
䊊 C
䊊 P dx Q dy F dr y C
yy R
&Q &P &x &y
dA yy 0 dA 0 R
A curve that is not simple crosses itself at one or more points and can be broken up into a number of simple curves. We have shown that the line integrals of F around these simple curves are all 0 and, adding these integrals, we see that xC F dr 0 for any closed curve C. Therefore, xC F dr is independent of path in D by Theorem 13.3.3. It follows that F is a conservative vector field. ■
13.4
EXERCISES 6. Px, y y 2 sin x,
Qx, y x 2 sin y, C consists of the arc of the parabola y x 2 from (0, 0) to (1, 1) followed by the line segment from (1, 1) to (0, 0)
■ Evaluate the line integral by two methods: (a) directly and (b) using Green’s Theorem.
1– 4 1. 2. 3. 4.
■
CAS
x
x y 2 dx x 3 dy, C is the rectangle with vertices (0, 0), (2, 0), (2, 3), and (0, 3) 䊊 C
x y dx x dy, C is the circle with center the origin and radius 1 䊊 C
x
x y dx x 2 y 3 dy, C is the triangle with vertices (0, 0), (1, 0), and (1, 2)
7.
x x dx y dy , C consists of the line segments from 0, 1 to 0, 0 and from 0, 0 to 1, 0 and the parabola y 1 x 2 from 1, 0 to 0, 1 ■
■
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■
■
■
■
■
8.
■
9.
5. Px, y x y ,
Qx, y x y , C is the circle x 2 y 2 1 5
7
6
■
■
■
■
■
■
■
xC e y dx 2 xe y dy,
xC x 2 y 2 dx 4 x y 3 dy, xC ( y e sx ) dx 2x cos y 2 dy, C is the boundary of the region enclosed by the parabolas y x 2 and x y 2
■
4
■
C is the triangle with vertices (0, 0), (1, 3), and (0, 3)
Verify Green’s Theorem by using a computer algebra system to evaluate both the line integral and the double integral.
5–6
■
C is the square with sides x 0, x 1, y 0, and y 1
䊊 C
■
■
■
7–12 ■ Use Green’s Theorem to evaluate the line integral along the given positively oriented curve.
䊊 C
■
■
10.
xC xe2x dx x 4 2x 2 y 2 dy, C is the boundary of the region between the circles x 2 y 2 1 and x 2 y 2 4
SECTION 13.5
11. 12. ■
xC y 3 dx x 3 dy, C is the circle x 2 y 2 4 xC sin y dx x cos y dy, C is the ellipse x 2 xy y 2 1 ■
■
■
■
■
■
■
■
■
■
■
■
13. Fx, y sx y 3, x 2 sy ,
■
■
■
■
■
17. Use Green’s Theorem to find the work done by the force
Fx, y xx y i x y 2 j in moving a particle from the origin along the x-axis to 1, 0, then along the line segment to 0, 1, and then back to the origin along the y-axis. 18. A particle starts at the point 2, 0, moves along the x-axis
to 2, 0, and then along the semicircle y s4 x 2 to the starting point. Use Green’s Theorem to find the work done on this particle by the force field Fx, y x, x 3 3x y 2 .
19. Use one of the formulas in (5) to find the area under one
arch of the cycloid x t sin t, y 1 cos t.
; 20. If a circle C with radius 1 rolls along the outside of the
circle x y 16, a fixed point P on C traces out a curve called an epicycloid, with parametric equations x 5 cos t cos 5t, y 5 sin t sin 5t. Graph the epicycloid and use (5) to find the area it encloses. 2
2
21. (a) If C is the line segment connecting the point x 1, y1 to
the point x 2 , y2, show that
y
C
x dy y dx x 1 y 2 x 2 y1
A 12 x 1 y 2 x 2 y1 x 2 y 3 x 3 y 2 x n1 yn x n yn1 x n y1 x 1 yn
13.5
y
䊊 C
y 2 dx
region in the x y-plane bounded by a simple closed path C. Show that its moments of inertia about the axes are Ix
3
y
䊊 C
Iy
y 3 dx
3
y
䊊 C
x 3 dy
26. Use Exercise 25 to find the moment of inertia of a circular
disk of radius a with constant density about a diameter. (Compare with Example 4 in Section 12.4.) 27. If F is the vector field of Example 5, show that
xC F dr 0 for every simple closed path that does not pass through or enclose the origin.
28. Complete the proof of the special case of Green’s Theorem
by proving Equation 3. 29. Use Green’s Theorem to prove the change of variables
formula for a double integral (Formula 12.8.9) for the case where f x, y 1:
yy dx dy yy R
(b) If the vertices of a polygon, in counterclockwise order, are x 1, y1 , x 2 , y 2 , . . . , x n , yn , show that the area of the polygon is A
1 2A
25. A plane lamina with constant density x, y occupies a
x 22 y 32 1 oriented counterclockwise ■
y
x 2 dy
of radius a.
16. Fx, y y lnx 2 y 2 , 2 tan1yx, C is the circle ■
䊊 C
24. Use Exercise 22 to find the centroid of a semicircular region
C is the circle x 2 y 2 25 oriented clockwise
■
y
vertices 0, 0, 1, 0, and 0, 1.
15. Fx, y e x x 2 y, e y xy 2 ,
■
1 2A
23. Use Exercise 22 to find the centroid of the triangle with
C is the triangle from 0, 0 to 2, 6 to 2, 0 to 0, 0
■
22. Let D be a region bounded by a simple closed path C in the
where A is the area of D.
14. Fx, y y 2 cos x, x 2 2y sin x,
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757
(c) Find the area of the pentagon with vertices 0, 0, 2, 1, 1, 3, 0, 2, and 1, 1.
x
C consists of the arc of the curve y sin x from 0, 0 to , 0 and the line segment from , 0 to 0, 0
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■
xy-plane. Use Green’s Theorem to prove that the coordinates of the centroid x, y of D are
Use Green’s Theorem to evaluate xC F d r. (Check the orientation of the curve before applying the theorem.) 13–16
CURL AND DIVERGENCE
S
&x, y du dv &u, v
Here R is the region in the xy-plane that corresponds to the region S in the uv-plane under the transformation given by x tu, v, y hu, v. [Hint: Note that the left side is AR and apply the first part of Equation 5. Convert the line integral over &R to a line integral over &S and apply Green’s Theorem in the uv-plane.]
CURL AND DIVERGENCE In this section we define two operations that can be performed on vector fields and that play a basic role in the applications of vector calculus to fluid flow and electricity and magnetism. Each operation resembles differentiation, but one produces a vector field whereas the other produces a scalar field.
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CURL
If F P i Q j R k is a vector field on ⺢ 3 and the partial derivatives of P, Q, and R all exist, then the curl of F is the vector field on ⺢ 3 defined by
1
curl F
&R &Q &y &z
i
&P &R &z &x
j
&Q &P &x &y
k
As an aid to our memory, let’s rewrite Equation 1 using operator notation. We introduce the vector differential operator ∇ (“del”) as ∇ i
& & & j k &x &y &z
It has meaning when it operates on a scalar function to produce the gradient of f : ∇f i
&f &f &f &f &f &f j k i j k &x &y &z &x &y &z
If we think of ∇ as a vector with components &&x, &&y, and &&z, we can also consider the formal cross product of ∇ with the vector field F as follows:
i & ( F &x P curl F
j & &y Q
k & &z R
&R &Q &y &z
i
&P &R &z &x
j
&Q &P &x &y
k
Thus the easiest way to remember Definition 1 is by means of the symbolic expression curl F ∇ F
2
EXAMPLE 1 If Fx, y, z xz i xyz j y 2 k, find curl F. SOLUTION Using Equation 2, we have
i & curl F ∇ F &x xz ■ Most computer algebra systems have commands that compute the curl and divergence of vector fields. If you have access to a CAS, use these commands to check the answers to the examples and exercises in this section.
j k & & &y &z xyz y 2
& & y 2 xyz i &y &z
& & y 2 xz j &x &z
& & xyz xz k &x &y
2y xy i 0 x j yz 0 k y2 x i x j yz k
■
Recall that the gradient of a function f of three variables is a vector field on ⺢ 3 and so we can compute its curl. The following theorem says that the curl of a gradient vector field is 0.
SECTION 13.5
CURL AND DIVERGENCE
■
759
3 THEOREM If f is a function of three variables that has continuous secondorder partial derivatives, then
curl(f 0 PROOF We have
■ Notice the similarity to what we know from Section 10.4: a a 0 for every three-dimensional vector a.
i & curl(f ( (f &x &f &x
&2 f &2 f &y &z &z &y
j & &y &f &y
i
k & &z &f &z
&2 f &2 f &z &x &x &z
j
&2 f &2 f &x &y &y &x
k
0i0j0k0 ■
by Clairaut’s Theorem.
■ Compare this with Exercise 25 in Section 13.3.
Since a conservative vector field is one for which F ∇f , Theorem 3 can be rephrased as follows: If F is conservative, then curl F 0. This gives us a way of verifying that a vector field is not conservative. V EXAMPLE 2
Show that the vector field Fx, y, z xz i xyz j y 2 k is not
conservative. SOLUTION In Example 1 we showed that
curl F y2 x i x j yz k This shows that curl F 0 and so, by Theorem 3, F is not conservative.
■
The converse of Theorem 3 is not true in general, but the following theorem says the converse is true if F is defined everywhere. (More generally it is true if the domain is simply-connected, that is, “has no hole.”) Theorem 4 is the three-dimensional version of Theorem 13.3.6. Its proof requires Stokes’ Theorem and is sketched at the end of Section 13.8. If F is a vector field defined on all of ⺢ 3 whose component functions have continuous partial derivatives and curl F 0, then F is a conservative vector field. 4 THEOREM
V EXAMPLE 3
(a) Show that Fx, y, z y 2 z 3 i 2xyz 3 j 3xy 2 z 2 k is a conservative vector field. (b) Find a function f such that F (f .
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SOLUTION
(a) We compute the curl of F:
i j k & & & curl F ( F &x &y &z y 2 z 3 2xyz 3 3xy 2 z 2
6xyz 2 6xyz 2 i 3y 2 z 2 3y 2 z 2 j 2yz 3 2yz 3 k 0 Since curl F 0 and the domain of F is ⺢ 3, F is a conservative vector field by Theorem 4. (b) The technique for finding f was given in Section 13.3. We have 5
fx x, y, z y 2 z 3
6
fy x, y, z 2xyz 3
7
fz x, y, z 3xy 2 z 2
Integrating (5) with respect to x, we obtain 8
f x, y, z xy 2 z 3 t y, z
Differentiating (8) with respect to y, we get fy x, y, z 2xyz 3 ty y, z, so comparison with (6) gives ty y, z 0. Thus ty, z hz and fz x, y, z 3xy 2 z 2 hz Then (7) gives hz 0. Therefore f x, y, z xy 2 z 3 K
curl F(x, y, z)
(x, y, z)
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The reason for the name curl is that the curl vector is associated with rotations. One connection is explained in Exercise 35. Another occurs when F represents the velocity field in fluid flow (see Example 3 in Section 13.1). Particles near (x, y, z) in the fluid tend to rotate about the axis that points in the direction of curl Fx, y, z and the length of this curl vector is a measure of how quickly the particles move around the axis (see Figure 1). If curl F 0 at a point P, then the fluid is free from rotations at P and F is called irrotational at P. In other words, there is no whirlpool or eddy at P. If curl F 0, then a tiny paddle wheel moves with the fluid but doesn’t rotate about its axis. If curl F 0, the paddle wheel rotates about its axis. We give a more detailed explanation in Section 13.8 as a consequence of Stokes’ Theorem.
FIGURE 1
DIVERGENCE
If F P i Q j R k is a vector field on ⺢ 3 and &P&x, &Q&y, and &R&z exist, then the divergence of F is the function of three variables defined by
9
div F
&P &Q &R &x &y &z
Observe that curl F is a vector field but div F is a scalar field. In terms of the gradient operator ( &&x i &&y j &&z k, the divergence of F can be written sym-
SECTION 13.5
CURL AND DIVERGENCE
■
761
bolically as the dot product of ( and F : div F ( F
10
EXAMPLE 4 If Fx, y, z xz i xyz j y 2 k, find div F. SOLUTION By the definition of divergence (Equation 9 or 10) we have
div F ( F
& & & xz xyz y 2 z xz &x &y &z
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If F is a vector field on ⺢ 3, then curl F is also a vector field on ⺢ 3. As such, we can compute its divergence. The next theorem shows that the result is 0. If F P i Q j R k is a vector field on ⺢ 3 and P, Q, and R have continuous second-order partial derivatives, then 11 THEOREM
div curl F 0 PROOF Using the definitions of divergence and curl, we have ■ Note the analogy with the scalar triple product: a a b 0.
div curl F ( ( F
&R &Q &y &z
& &x
&2R &2Q &2P &2R &2Q &2P &x &y &x &z &y &z &y &x &z &x &z &y
& &y
&P &R &z &x
& &z
&Q &P &x &y
0 because the terms cancel in pairs by Clairaut’s Theorem.
■
Show that the vector field Fx, y, z xz i xyz j y 2 k can’t be written as the curl of another vector field, that is, F curl G. V EXAMPLE 5
SOLUTION In Example 4 we showed that
div F z xz and therefore div F 0. If it were true that F curl G, then Theorem 11 would give div F div curl G 0 which contradicts div F 0. Therefore, F is not the curl of another vector field. ■ The reason for this interpretation of div F will be explained at the end of Section 13.9 as a consequence of the Divergence Theorem.
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Again, the reason for the name divergence can be understood in the context of fluid flow. If Fx, y, z is the velocity of a fluid (or gas), then div Fx, y, z represents the net rate of change (with respect to time) of the mass of fluid (or gas) flowing from the point x, y, z per unit volume. In other words, div Fx, y, z measures the tendency of the fluid to diverge from the point x, y, z. If div F 0, then F is said to be incompressible.
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Another differential operator occurs when we compute the divergence of a gradient vector field (f . If f is a function of three variables, we have div(f ( (f
&2 f &2 f &2 f 2 2 &x &y &z 2
and this expression occurs so often that we abbreviate it as ( 2 f . The operator (2 ( ( is called the Laplace operator because of its relation to Laplace’s equation (2 f
&2 f &2 f &2 f 0 2 2 &x &y &z 2
We can also apply the Laplace operator ( 2 to a vector field FPiQjRk in terms of its components: ( 2 F ( 2P i ( 2Q j ( 2R k
VECTOR FORMS OF GREEN’S THEOREM
The curl and divergence operators allow us to rewrite Green’s Theorem in versions that will be useful in our later work. We suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypotheses of Green’s Theorem. Then we consider the vector field F P i Q j. Its line integral is
y
䊊 C
䊊 P dx Q dy F dr y C
and, regarding F as a vector field on ⺢3 with third component 0, we have
i j & & curl F &x &y Px, y Qx, y Therefore
curl F k
&Q &P &x &y
k & &z 0
&Q &P &x &y
kk
k
&Q &P &x &y
and we can now rewrite the equation in Green’s Theorem in the vector form
12
y
䊊 C
F dr yy curl F k dA D
Equation 12 expresses the line integral of the tangential component of F along C as the double integral of the vertical component of curl F over the region D enclosed by C. We now derive a similar formula involving the normal component of F.
SECTION 13.5
CURL AND DIVERGENCE
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If C is given by the vector equation rt xt i yt j
a t b
then the unit tangent vector (see Section 10.7) is y
Tt
T(t) r(t) D
n(t)
yt j rt
You can verify that the outward unit normal vector to C is given by
C 0
xt i rt
nt
x
yt xt i j rt rt
(See Figure 2.) Then, from Equation 13.2.3, we have
FIGURE 2
y
䊊 C
F n ds y F nt rt dt b
a
y
b
a
y
b
a
Pxt, yt yt Qxt, yt xt rt rt
rt dt
Pxt, yt yt dt Qxt, yt xt dt
y P dy Q dx C
yy D
&P &Q &x &y
dA
by Green’s Theorem. But the integrand in this double integral is just the divergence of F. So we have a second vector form of Green’s Theorem.
y
䊊 C
13
F n ds yy div Fx, y dA D
This version says that the line integral of the normal component of F along C is equal to the double integral of the divergence of F over the region D enclosed by C.
13.5 1–7
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EXERCISES 8 –9 ■ The vector field F is shown in the xy-plane and looks the same in all other horizontal planes. (In other words, F is independent of z and its z-component is 0.) (a) Is div F positive, negative, or zero? Explain. (b) Determine whether curl F 0. If not, in which direction does curl F point?
Find (a) the curl and (b) the divergence of the vector
field. 1. Fx, y, z x yz i x 2 y k 2. Fx, y, z x 2 yz i x y 2 z j x yz 2 k 3. Fx, y, z i x yz j ( x y sz ) k
8.
9.
y
y
4. Fx, y, z cos xz j sin x y k 5. Fx, y, z e x sin y i e x cos y j z k 6. Fx, y, z
x y z i 2 j 2 k x2 y2 z2 x y2 z2 x y2 z2
7. Fx, y, z ln x, lnx y, lnx yz ■
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x
0 ■
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x
0 ■
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26. div( f ( t 0
10. Let f be a scalar field and F a vector field. State whether
each expression is meaningful. If not, explain why. If so, state whether it is a scalar field or a vector field. (a) curl f (b) grad f (c) div F (d) curlgrad f (e) grad F (f ) graddiv F (g) divgrad f (h) graddiv f (i) curlcurl F ( j) divdiv F (k) grad f div F (l) divcurlgrad f
27. curlcurl F graddiv F ( 2 F ■
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28 –30
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Let r x i y j z k and r r .
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28. Verify each identity.
(a) ( r 3 (c) ( 2 r 3 12r
(b) ( r r 4r
29. Verify each identity.
(a) (r rr (c) (1r rr 3
Determine whether or not the vector field is conservative. If it is conservative, find a function f such that F ∇ f . 11–16
■
11. Fx, y, z yz i xz j x y k
(b) ( r 0 (d) ( ln r rr 2
30. If F rr p, find div F. Is there a value of p for which
div F 0 ?
12. Fx, y, z 3z i cos y j 2 xz k 2
■
13. Fx, y, z 2 x y i x 2 2yz j y 2 k
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yy f ( t dA y 2
16. Fx, y, z y cos x y i x cos x y j sin z k
䊊 C
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17. Is there a vector field G on ⺢ 3 such that
curl G x y 2 i yz 2 j z x 2 k? Explain.
18. Is there a vector field G on ⺢ 3 such that
curl G yz i x yz j x y k? Explain.
19. Show that any vector field of the form
Fx, y, z f x i t y j hz k where f , t, h are differentiable functions, is irrotational. 20. Show that any vector field of the form
Fx, y, z f y, z i tx, z j hx, y k is incompressible. Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If f is a scalar field and F, G are vector fields, then f F, F G, and F G are defined by f Fx, y, z f x, y, zFx, y, z 21–27
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f (t n ds yy ( f (t dA
D ■
■
Green’s first identity:
15. Fx, y, z yex i ex j 2z k
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31. Use Green’s Theorem in the form of Equation 13 to prove
14. Fx, y, z e z i j xe z k
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F Gx, y, z Fx, y, z Gx, y, z F Gx, y, z Fx, y, z Gx, y, z 21. divF G div F div G 22. curlF G curl F curl G 23. div f F f div F F ( f 24. curl f F f curl F ( f F 25. divF G G curl F F curl G
D
■
where D and C satisfy the hypotheses of Green’s Theorem and the appropriate partial derivatives of f and t exist and are continuous. (The quantity (t n Dn t occurs in the line integral. This is the directional derivative in the direction of the normal vector n and is called the normal derivative of t.) 32. Use Green’s first identity (Exercise 31) to prove Green’s
second identity:
yy f ( t t( f dA y 2
2
䊊 C
f (t t( f n ds
D
where D and C satisfy the hypotheses of Green’s Theorem and the appropriate partial derivatives of f and t exist and are continuous. 33. Recall from Section 11.3 that a function t is called
harmonic on D if it satisfies Laplace’s equation, that is, ( 2t 0 on D . Use Green’s first identity (with the same hypotheses as in Exercise 31) to show that if t is harmonic on D, then x䊊C Dn t ds 0 . Here Dn t is the normal derivative of t defined in Exercise 31. 34. Use Green’s first identity to show that if f is harmonic
on D, and if f x, y 0 on the boundary curve C, then xx D (f 2 dA 0 . (Assume the same hypotheses as in Exercise 31.)
35. This exercise demonstrates a connection between the curl
vector and rotations. Let B be a rigid body rotating about the z-axis. The rotation can be described by the vector w k, where is the angular speed of B, that is, the tangential speed of any point P in B divided by the distance d
SECTION 13.6
from the axis of rotation. Let r x, y, z be the position vector of P. (a) By considering the angle in the figure, show that the velocity field of B is given by v w r. (b) Show that v y i x j. (c) Show that curl v 2w.
PARAMETRIC SURFACES AND THEIR AREAS
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765
36. Maxwell’s equations relating the electric field E and mag-
netic field H as they vary with time in a region containing no charge and no current can be stated as follows: div E 0 curl E
z
div H 0 1 &H c &t
curl H
1 &E c &t
w
where c is the speed of light. Use these equations to prove the following: d
B
v P
(a) ( ( E
1 &2 E c 2 &t 2
(b) ( ( H
1 &2 H c 2 &t 2
¨ 0 y x
13.6
(c) ( 2 E
1 &2 E c 2 &t 2
(d) ( 2 H
1 &2 H c 2 &t 2
[Hint: Use Exercise 27.]
PARAMETRIC SURFACES AND THEIR AREAS We have considered surfaces that are graphs of functions of two variables. Here we use vector functions to discuss more general surfaces, called parametric surfaces, and compute their areas.
√
D
PARAMETRIC SURFACES
(u, √)
0
u
r
In much the same way that we describe a space curve by a vector function rt of a single parameter t, we can describe a surface by a vector function ru, v of two parameters u and v. We suppose that 1
z
ru, v xu, v i yu, v j zu, v k
is a vector-valued function defined on a region D in the uv-plane. So x, y, and z, the component functions of r, are functions of the two variables u and v with domain D. The set of all points x, y, z in ⺢ 3 such that
S
2
r(u, √)
x xu, v
y yu, v
z zu, v
0 x
FIGURE 1
A parametric surface
y
and u, v varies throughout D, is called a parametric surface S and Equations 2 are called parametric equations of S. Each choice of u and v gives a point on S; by making all choices, we get all of S. In other words, the surface S is traced out by the tip of the position vector ru, v as u, v moves throughout the region D. (See Figure 1.)
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EXAMPLE 1 Identify and sketch the surface with vector equation
ru, v 2 cos u i v j 2 sin u k
z (0, 0, 2)
SOLUTION The parametric equations for this surface are
x 2 cos u
0 x
z 2 sin u
So for any point x, y, z on the surface, we have x 2 z 2 4 cos 2u 4 sin 2u 4
y (2, 0, 0)
This means that vertical cross-sections parallel to the xz-plane (that is, with y constant) are all circles with radius 2. Since y v and no restriction is placed on v, the surface is a circular cylinder with radius 2 whose axis is the y-axis (see Figure 2). ■
FIGURE 2
In Example 1 we placed no restrictions on the parameters u and v and so we got the entire cylinder. If, for instance, we restrict u and v by writing the parameter domain as
z
0 u 2
(0, 3, 2) 0 x
yv
y
FIGURE 3
0 v 3
then x 0, z 0, 0 y 3, and we get the quarter-cylinder with length 3 illustrated in Figure 3. If a parametric surface S is given by a vector function ru, v, then there are two useful families of curves that lie on S, one family with u constant and the other with v constant. These families correspond to vertical and horizontal lines in the uv-plane. If we keep u constant by putting u u 0 , then ru 0 , v becomes a vector function of the single parameter v and defines a curve C1 lying on S. (See Figure 4.) z
√
(u¸, √¸) √=√¸
Visual 13.6 shows animated versions of Figures 4 and 5, with moving grid curves, for several parametric surfaces.
D 0
r
C¡
u=u¸ 0
u
C™ y
FIGURE 4
x
Similarly, if we keep v constant by putting v v0 , we get a curve C2 given by ru, v0 that lies on S. We call these curves grid curves. (In Example 1, for instance, the grid curves obtained by letting u be constant are horizontal lines whereas the grid curves with v constant are circles.) In fact, when a computer graphs a parametric surface, it usually depicts the surface by plotting these grid curves, as we see in the following example. EXAMPLE 2 Use a computer algebra system to graph the surface
ru, v 2 sin v cos u, 2 sin v sin u, u cos v Which grid curves have u constant? Which have v constant? SOLUTION We graph the portion of the surface with parameter domain 0 u 4, 0 v 2 in Figure 5. It has the appearance of a spiral tube. To
SECTION 13.6
z
PARAMETRIC SURFACES AND THEIR AREAS
■
767
identify the grid curves, we write the corresponding parametric equations: x 2 sin v cos u
y 2 sin v sin u
z u cos v
If v is constant, then sin v and cos v are constant, so the parametric equations resemble those of the helix in Example 4 in Section 10.7. So the grid curves with v constant are the spiral curves in Figure 5. We deduce that the grid curves with u constant must be the curves that look like circles in the figure. Further evidence for this assertion is that if u is kept constant, u u 0 , then the equation z u 0 cos v shows that the z-values vary from u 0 1 to u 0 1. ■
√ constant
u constant
In Examples 1 and 2 we were given a vector equation and asked to graph the corresponding parametric surface. In the following examples, however, we are given the more challenging problem of finding a vector function to represent a given surface. In the rest of this chapter we will often need to do exactly that. EXAMPLE 3 Find a vector function that represents the plane that passes through the point P0 with position vector r0 and that contains two nonparallel vectors a and b.
y
x
SOLUTION If P is any point in the plane, we can get from P0 to P by moving a
FIGURE 5 P
certain distance in the direction of a and another distance in the direction of b. So there are scalars u and v such that P A 0 P ua v b. (Figure 6 illustrates how this works, by means of the Parallelogram Law, for the case where u and v are positive. See also Exercise 28 in Section 10.2.) If r is the position vector of P, then
√b
r OP A0 P A 0 P r 0 ua v b
b
P¸
So the vector equation of the plane can be written as
a
FIGURE 6
ru, v r0 ua v b
ua
where u and v are real numbers. If we write r x, y, z , r0 x0 , y0 , z0 , a a1 , a2 , a3 , and b b1 , b2 , b3 , then we can write the parametric equations of the plane through the point x0 , y0 , z0 as follows: x x0 ua1 v b1 V EXAMPLE 4
y y0 ua2 v b2
z z0 ua3 v b3
■
Find a parametric representation of the sphere x 2 y 2 z 2 a 2.
SOLUTION The sphere has a simple representation a in spherical coordinates, so let’s choose the angles and in spherical coordinates as the parameters (see Section 12.7). Then, putting a in the equations for conversion from spherical to rectangular coordinates (Equations 12.7.1), we obtain
x a sin cos
y a sin sin
z a cos
as the parametric equations of the sphere. The corresponding vector equation is r, a sin cos i a sin sin j a cos k We have 0 and 0 2, so the parameter domain is the rectangle D 0, 0, 2 . The grid curves with constant are the circles of constant latitude (including the equator). The grid curves with constant are the meridians ■ (semicircles), which connect the north and south poles.
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■ One of the uses of parametric surfaces is in computer graphics. Figure 7 shows the result of trying to graph the sphere x 2 y 2 z 2 1 by solving the equation for z and graphing the top and bottom hemispheres separately. Part of the sphere appears to be missing because of the rectangular grid system used by the computer. The much better picture in Figure 8 was produced by a computer using the parametric equations found in Example 4.
FIGURE 8
FIGURE 7
EXAMPLE 5 Find a parametric representation for the cylinder
x2 y2 4
0 z 1
SOLUTION The cylinder has a simple representation r 2 in cylindrical coordinates, so we choose as parameters and z in cylindrical coordinates. Then the parametric equations of the cylinder are
x 2 cos
y 2 sin
zz
where 0 2 and 0 z 1. In Module 13.6 you can investigate several families of parametric surfaces.
■
Parametric representations (also called parametrizations) of surfaces are not unique. The next example shows two ways to parametrize a cone. EXAMPLE 6 Find a parametric representation for the surface z 2sx 2 y 2 , that
is, the top half of the cone z 2 4x 2 4y 2.
SOLUTION 1 If we regard x and y as parameters, then the parametric equations are
simply xx
yy
z 2sx 2 y 2
and the vector equation is For some purposes the parametric representations in Solutions 1 and 2 are equally good, but Solution 2 might be preferable in certain situations. If we are interested only in the part of the cone that lies below the plane z 1 , for instance, all we have to do in Solution 2 is change the parameter domain to ■
0 r
1 2
0 2
rx, y x i y j 2sx 2 y 2 k SOLUTION 2 Another representation results from choosing as parameters the polar coordinates r and . A point x, y, z on the cone satisfies x r cos , y r sin , and z 2sx 2 y 2 2r . So a vector equation for the cone is
rr, r cos i r sin j 2r k where r 0 and 0 2.
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As in the first solution of Example 6, a general surface given as the graph of a function of x and y, that is, with an equation of the form z f x, y, can always be regarded as a parametric surface by taking x and y as parameters and writing the parametric equations as xx yy z f x, y SURFACES OF REVOLUTION
Surfaces of revolution can be represented parametrically and thus graphed using a computer. For instance, let’s consider the surface S obtained by rotating the curve y f x, a x b, about the x-axis, where f x 0. Let be the angle of rotation
SECTION 13.6
3
0 y
y=ƒ
x
■
769
as shown in Figure 9. If x, y, z is a point on S, then
z
ƒ
PARAMETRIC SURFACES AND THEIR AREAS
xx
y f x cos
z f x sin
Therefore, we take x and as parameters and regard Equations 3 as parametric equations of S. The parameter domain is given by a x b, 0 2.
(x, y, z)
EXAMPLE 7 Find parametric equations for the surface generated by rotating the curve y sin x, 0 x 2, about the x-axis. Use these equations to graph the surface of revolution.
z
¨ ƒ
x
SOLUTION From Equations 3, the parametric equations are
xx
FIGURE 9 z
y sin x cos
z sin x sin
and the parameter domain is 0 x 2, 0 2. Using a computer to plot these equations and rotate the image, we obtain the graph in Figure 10. ■
y
x
We can adapt Equations 3 to represent a surface obtained through revolution about the y- or z-axis. (See Exercise 26.) TANGENT PLANES
FIGURE 10
We now find the tangent plane to a parametric surface S traced out by a vector function √
ru, v xu, v i yu, v j zu, v k at a point P0 with position vector ru0 , v0 . If we keep u constant by putting u u0 , then ru0 , v becomes a vector function of the single parameter v and defines a grid curve C1 lying on S. (See Figure 11.) The tangent vector to C1 at P0 is obtained by taking the partial derivative of r with respect to v :
(u¸, √¸) √=√¸ u=u¸
D 0
u
4
5 z
ru C¡
x
FIGURE 11
ru
&x &y &z u0 , v0 i u0 , v0 j u0 , v0 k &u &u &u
If ru rv is not 0, then the surface S is called smooth (it has no “corners”). For a smooth surface, the tangent plane is the plane that contains the tangent vectors ru and rv , and the vector ru rv is a normal vector to the tangent plane.
P¸
0
&x &y &z u0 , v0 i u0 , v0 j u0 , v0 k &v &v &v
Similarly, if we keep v constant by putting v v0 , we get a grid curve C2 given by ru, v0 that lies on S, and its tangent vector at P0 is
r
r√
rv
V EXAMPLE 8 Find the tangent plane to the surface with parametric equations x u 2, y v 2, z u 2v at the point 1, 1, 3.
C™ y
SOLUTION We first compute the tangent vectors:
ru
&x &y &z i j k 2u i k &u &u &u
rv
&x &y &z i j k 2v j 2 k &v &v &v
770
■
CHAPTER 13
VECTOR CALCULUS
■ Figure 12 shows the self-intersecting surface in Example 8 and its tangent plane at 1, 1, 3.
Thus a normal vector to the tangent plane is i ru rv 2u 0
z
y x
j k 0 1 2v i 4u j 4uv k 2v 2
Notice that the point 1, 1, 3 corresponds to the parameter values u 1 and v 1, so the normal vector there is 2 i 4 j 4 k Therefore, an equation of the tangent plane at 1, 1, 3 is 2x 1 4y 1 4z 3 0 x 2y 2z 3 0
or
FIGURE 12
■
SURFACE AREA
Now we define the surface area of a general parametric surface given by Equation 1. For simplicity we start by considering a surface whose parameter domain D is a rectangle, and we divide it into subrectangles Rij . Let’s choose u i*, vj* to be the lower left corner of Rij. (See Figure 13.) z
√
R ij r
Î √j
Sij
Pij
Î ui
FIGURE 13
The image of the subrectangle Rij is the patch Sij .
(u *i , √ *j )
0
u
0
y
x
The part Sij of the surface S that corresponds to Rij is called a patch and has the point Pij with position vector ru i*, vj* as one of its corners. Let ru* ruu i*, vj*
Pij (a)
rv* rvu i*, vj*
and
be the tangent vectors at Pij as given by Equations 5 and 4. Figure 14(a) shows how the two edges of the patch that meet at Pij can be approximated by vectors. These vectors, in turn, can be approximated by the vectors u i r*u and vj r*v because partial derivatives can be approximated by difference quotients. So we approximate Sij by the parallelogram determined by the vectors u i r*u and vj r*v . This parallelogram is shown in Figure 14(b) and lies in the tangent plane to S at Pij . The area of this parallelogram is
u r* v r* r* r* u v i
Î√√j r *√
u
j
u
v
v
i
j
and so an approximation to the area of S is m
n
r* r* u u
v
i
vj
i1 j1
(b) FIGURE 14
Approximating a patch by a parallelogram
Our intuition tells us that this approximation gets better as we increase the number of subrectangles and their dimensions decrease, and we recognize the double sum as a Riemann sum for the double integral xxD ru rv du dv. This motivates the following definition.
SECTION 13.6
PARAMETRIC SURFACES AND THEIR AREAS
771
■
If a smooth parametric surface S is given by the equation
6 DEFINITION
u, v D
ru, v xu, v i yu, v j zu, v k
and S is covered just once as u, v ranges throughout the parameter domain D, then the surface area of S is
AS yy ru rv dA D
&x &y &z i j k &u &u &u
ru
where
rv
&x &y &z i j k &v &v &v
EXAMPLE 9 Find the surface area of a sphere of radius a. SOLUTION In Example 4 we found the parametric representation
x a sin cos
y a sin sin
z a cos
where the parameter domain is
D , 0 , 0 2 We first compute the cross product of the tangent vectors: i &x r r & &x &
j &y & &y &
k &z i & a cos cos &z a sin sin &
j a cos sin a sin cos
k a sin 0
a 2 sin 2 cos i a 2 sin2 sin j a 2 sin cos k
Thus
r
r sa 4 sin 4 cos 2 a 4 sin 4 sin 2 a 4 sin 2 cos 2 sa 4 sin 4 a 4 sin 2 cos 2 a 2 ssin 2 a 2 sin
since sin 0 for 0 . Therefore, by Definition 6, the area of the sphere is
A yy r r dA y D
a2 y
2
0
2
0
y
0
a 2 sin d d
d y sin d a 222 4a 2 0
■
SURFACE AREA OF THE GRAPH OF A FUNCTION
For the special case of a surface S with equation z f x, y, where x, y lies in D and f has continuous partial derivatives, we take x and y as parameters. The parametric equations are xx
yy
z f x, y
772
■
CHAPTER 13
VECTOR CALCULUS
rx i
so
&f &x
&f &y
ry j
k
k
and 7
i
j
rx ry 1
0
0
1
k &f &f &f i jk &x &x &y &f &y
Thus we have
8
r
x
ry
&f &x
&f &y
2
2
1
&z &x
1
2
&z &y
2
and the surface area formula in Definition 6 becomes ■ Notice the similarity between the surface area formula in Equation 9 and the arc length formula
L
y
b
a
1
AS
9
D
1
&z &x
&z &y
2
2
dA
2
dy dx
from Section 7.4.
dx
Find the area of the part of the paraboloid z x 2 y 2 that lies under the plane z 9. V EXAMPLE 10
z
SOLUTION The plane intersects the paraboloid in the circle x 2 y 2 9, z 9.
9
Therefore, the given surface lies above the disk D with center the origin and radius 3. (See Figure 15.) Using Formula 9, we have A
yy D
3
&z &x
1
&z &y
2
2
dA yy s1 2x 2 2y 2 dA D
yy s1 4x 2 y 2 dA
D x
yy
D
y
Converting to polar coordinates, we obtain Ay
FIGURE 15
2
0
y
3
0
s1 4r 2 r dr d y
2
0
2 ( 18 ) 23 1 4r 2 32
]
3 0
d y r s1 4r 2 dr 3
0
(37s37 1) 6
■
We can use the definition of surface area to find the area of a surface of revolution S obtained by rotating the curve y f x, a x b, about the x-axis, where f x 0 and f is continuous. In Exercise 55 you are asked to use a parametric representation of S (3) and Definition 6 to prove the following formula for the area of a surface of revolution: 10
A 2
y
b
a
f xs1 f x 2 dx
SECTION 13.6
13.6 1– 4
■
773
EXERCISES
Identify the surface with the given vector equation.
■
PARAMETRIC SURFACES AND THEIR AREAS
z
III
z
IV
1. ru, v u v i 3 v j 1 4u 5v k 2. ru, v 2 sin u i 3 cos u j v k,
0 v 2 x
3. rs, t s, t, t 2 s 2
y
y
x
4. rs, t s sin 2t, s , s cos 2t 2
■
■
■
■
■
■
■
■
■
■
■
■ ■
; 5–10
Use a computer to graph the parametric surface. Get a printout and indicate on it which grid curves have u constant and which have v constant.
■
■
■
■
■
■
■
■
■
■
■
■
15–22
6. ru, v u v, u , v , 2
Find a parametric representation for the surface.
15. The plane that passes through the point 1, 2, 3 and con-
5. ru, v u 2 1, v 3 1, u v , 1 u 1, 1 v 1 2
■
tains the vectors i j k and i j k
16. The lower half of the ellipsoid 2 x 2 4y 2 z 2 1
1 u 1, 1 v 1
17. The part of the hyperboloid x 2 y 2 z 2 1 that lies to
the right of the xz-plane
7. ru, v cos 3u cos 3v, sin 3u cos 3v, sin 3v , 0 u , 0 v 2
18. The part of the elliptic paraboloid x y 2 2z 2 4 that
8. ru, v cos u sin v, sin u sin v, cos v ln tanv2 , 0 u 2, 0.1 v 6.2
19. The part of the sphere x 2 y 2 z 2 4 that lies above the
9. x cos u sin 2v,
20. The part of the sphere x 2 y 2 z 2 16 that lies between
y sin u sin 2v,
10. x u sin u cos v, ■
■
■
■
■
■
cone z sx 2 y 2
z sin v
y u cos u cos v, ■
lies in front of the plane x 0
■
the planes z 2 and z 2
z u sin v ■
■
■
21. The part of the cylinder y 2 z 2 16 that lies between the
■
planes x 0 and x 5
22. The part of the plane z x 3 that lies inside the cylinder
Match the equations with the graphs labeled I–IV and give reasons for your answers. Determine which families of grid curves have u constant and which have v constant.
11–14
■
11. ru, v u cos v i u sin v j v k 12. x u , 3
y u sin v,
y 1 cos u sin v,
z
■
■
■
■
■
■
■
■
■
■
■ Use a computer algebra system to produce a graph that looks like the given one.
24.
3
z
z 0
0
z
II
_3 _3 ■
0 y ■
_1
x
0 5 ■
_1 _1
■
y ■
■
■
0
0 1 1 ■
■
x
■
; 25. Find parametric equations for the surface obtained by x
■
23–24 23.
zu
14. x 1 u3 cos v cos 4 u, y 1 u3 cos v sin 4 u, z 3u 1 u sin v I
■
CAS
z u cos v
13. x u sin u cos v,
x2 y2 1
y
y x
■
rotating the curve y e x, 0 x 3, about the x-axis and use them to graph the surface.
■
774
■
CHAPTER 13
VECTOR CALCULUS
; 26. Find parametric equations for the surface obtained by
42. The helicoid (or spiral ramp) with vector equation ru, v u cos v i u sin v j v k, 0 u 1, 0 v
rotating the curve x 4y 2 y 4, 2 y 2, about the y-axis and use them to graph the surface.
43. The surface with parametric equations x u v, y u v, z u v, u 2 v 2 1
; 27. (a) What happens to the spiral tube in Example 2 (see Figure 5) if we replace cos u by sin u and sin u by cos u ? (b) What happens if we replace cos u by cos 2u and sin u by sin 2u?
■
2
■
CAS
yv ,
z u v;
z uv ;
33– 43
■
■
■
■
■
■
■
2
CAS
u 1, v 1
■
■
u 0, v ■
■
■
■
CAS
Find the area of the surface.
33. The part of the plane 3x 2y z 6 that lies in the CAS
2
36. The part of the plane with vector equation ru, v 1 v, u 2v, 3 5u v that is given by 0 u 1, 0 v 1
38. The part of the surface z 1 3x 2y 2 that lies above
; CAS
the triangle with vertices 0, 0, 0, 1, and 2, 1
39. The part of the hyperbolic paraboloid z y 2 x 2 that lies
cylinder y 2 z 2 9
41. The part of the surface y 4x z 2 that lies between the
planes x 0, x 1, z 0, and z 1
■
■
■
■
■
■
■
46. Find, to four decimal places, the area of the part of the
tion 12.1) with four squares to estimate the surface area of the portion of the paraboloid z x 2 y 2 that lies above the square 0, 1 0, 1 . (b) Use a computer algebra system to approximate the surface area in part (a) to four decimal places. Compare with the answer to part (a). 48. Find the area of the surface with vector equation ru, v cos 3u cos 3v, sin 3u cos 3v, sin 3v , 0 u , 0 v 2. State your answer correct to four decimal 49. Find the exact area of the surface z 1 2x 3y 4y 2,
(b) Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area. (c) Use the parametric equations in part (a) with a 2 and b 3 to graph the surface. (d) For the case a 2, b 3, use a computer algebra system to find the surface area correct to four decimal places. 51. (a) Show that the parametric equations x a sin u cos v, y b sin u sin v, z c cos u, 0 u , 0 v 2,
between the cylinders x 2 y 2 1 and x 2 y 2 4
40. The part of the paraboloid x y 2 z 2 that lies inside the
■
50. (a) Set up, but do not evaluate, a double integral for the area of the surface with parametric equations x au cos v, y bu sin v, z u 2, 0 u 2, 0 v 2.
35. The surface z 3 x 32 y 32 , 0 x 1, 0 y 1
x2 y2 1
■
1 x 4, 0 y 1.
cylinder x 2 y 2 9
37. The part of the surface z xy that lies within the cylinder
■
places.
first octant 34. The part of the plane 2x 5y z 10 that lies inside the
■
2
47. (a) Use the Midpoint Rule for double integrals (see Sec-
2, 3, 0
32. ru, v u v i u sin v j v cos u k; ■
■
u 1, v 1
31. ru, v u i lnuv j v k;
■
■
surface z 1 x 2 1 y 2 that lies above the square x y 1. Illustrate by graphing this part of the surface.
Find an equation of the tangent plane to the given parametric surface at the specified point. If you have software that graphs parametric surfaces, use a computer to graph the surface and the tangent plane. ■
30. x u ,
■
x y 4 2
1 2
2
■
45. The part of the surface z ex y that lies above the disk
where r and 0 2, is called a Möbius strip. Graph this surface with several viewpoints. What is unusual about it?
2
■
cylinder x 2 y 2 1
z r sin2
y 3u 2,
■
44. The part of the surface z cosx 2 y 2 that lies inside the
y 2 sin r cos2
29. x u v,
■
■ Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your calculator to estimate the integral.
x 2 cos r cos2
29–32
■
44 – 45
; 28. The surface with parametric equations
1 2
■
;
represent an ellipsoid. (b) Use the parametric equations in part (a) to graph the ellipsoid for the case a 1, b 2, c 3. (c) Set up, but do not evaluate, a double integral for the surface area of the ellipsoid in part (b).
SECTION 13.7
52. (a) Show that the parametric equations x a cosh u cos v, y b cosh u sin v, z c sinh u, represent a hyper-
53. Find the area of the part of the sphere x 2 y 2 z 2 4z
that lies inside the paraboloid z x 2 y 2.
■
775
■ Use Formula 10 to find the area of the surface obtained by rotating the given curve about the x-axis.
56 –57
boloid of one sheet. (b) Use the parametric equations in part (a) to graph the hyperboloid for the case a 1, b 2, c 3. (c) Set up, but do not evaluate, a double integral for the surface area of the part of the hyperboloid in part (b) that lies between the planes z 3 and z 3.
;
SURFACE INTEGRALS
56. y x 3, ■
■
0 x 2 ■
■
57. y sx ,
■
■
■
■
4 x 9
■
■
■
■
58. (a) Find a parametric representation for the torus obtained
;
54. The figure shows the surface created when the cylinder
y 2 z 2 1 intersects the cylinder x 2 z 2 1. Find the area of this surface.
by rotating about the z -axis the circle in the xz-plane with center b, 0, 0 and radius a b. [Hint: Take as parameters the angles and shown in the figure.] (b) Use the parametric equations found in part (a) to graph the torus for several values of a and b. (c) Use the parametric representation from part (a) to find the surface area of the torus. z
z (x, y, z)
0
x
y
å
¨
y
55. Use Definition 6 and the parametric equations for a surface
of revolution (3) to derive Formula 10.
13.7 √
(b, 0, 0)
SURFACE INTEGRALS The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length. Suppose f is a function of three variables whose domain includes a surface S. We will define the surface integral of f over S in such a way that, in the case where f x, y, z 1, the value of the surface integral is equal to the surface area of S. We start with parametric surfaces and then deal with the special case where S is the graph of a function of two variables.
R ij Î√j
D
x
Îui
0
u
PARAMETRIC SURFACES r
Suppose that a surface S has a vector equation z
S
ru, v xu, v i yu, v j zu, v k
P *ij Sij
We first assume that the parameter domain D is a rectangle and we divide it into subrectangles Rij with dimensions ui and vj . Then the surface S is divided into corresponding patches Sij as in Figure 1. We evaluate f at a point Pij* in each patch, multiply by the area Sij of the patch, and form the Riemann sum
0 y x
u, v D
m
n
f P* S ij
ij
i1 j1
FIGURE 1
Then we take the limit as the number of patches increases (and their dimensions
776
■
CHAPTER 13
VECTOR CALCULUS
decrease) and define the surface integral of f over the surface S as
yy f x, y, z dS
1
S
m
n
f P* S
lim
ij
max u i , v j l 0 i1 j1
ij
Notice the analogy with the definition of a line integral (13.2.2) and also the analogy with the definition of a double integral (12.1.5). To evaluate the surface integral in Equation 1 we approximate the patch area Sij by the area of an approximating parallelogram in the tangent plane. In our discussion of surface area in Section 13.6 we made the approximation
Sij ru rv u i vj where
ru
&x &y &z i j k &u &u &u
&x &y &z i j k &v &v &v
rv
are the tangent vectors at a corner of Sij . If the components are continuous and ru and rv are nonzero and nonparallel in the interior of D, it can be shown from Definition 1, even when D is not a rectangle, that ■ We assume that the surface is covered only once as u, v ranges throughout D . The value of the surface integral does not depend on the parametrization that is used.
yy f x, y, z dS yy f ru, v r
2
u
S
rv dA
D
This should be compared with the formula for a line integral:
y
C
f x, y, z ds y f rt rt dt b
a
Observe also that
yy 1 dS yy r
u
S
rv dA AS
D
Formula 2 allows us to compute a surface integral by converting it into a double integral over the parameter domain D. When using this formula, remember that f ru, v is evaluated by writing x xu, v, y yu, v, and z zu, v in the formula for f x, y, z. EXAMPLE 1 Compute the surface integral xxS x 2 dS, where S is the unit sphere
x 2 y 2 z 2 1.
SOLUTION As in Example 4 in Section 13.6, we use the parametric representation
x sin cos that is,
y sin sin
z cos
0
r, sin cos i sin sin j cos k
As in Example 9 in Section 13.6, we can compute that
r
r sin
0 2
SECTION 13.7
SURFACE INTEGRALS
■
777
Therefore, by Formula 2,
yy x
2
dS yy sin cos 2 r r dA y
S
D
y
2
2 1 2 0
cos 2 d y sin 3 d y
0
0
[
12 12 sin 2
2 0
] [cos
2
y
0
1 3
0
sin 2 cos 2 sin d d
1 cos 2 d y sin sin cos 2 d 0
]
cos 3 0
4 3
■
Surface integrals have applications similar to those for the integrals we have previously considered. For example, if a thin sheet (say, of aluminum foil) has the shape of a surface S and the density (mass per unit area) at the point x, y, z is x, y, z, then the total mass of the sheet is m yy x, y, z dS S
and the center of mass is x, y, z, where x
1 m
yy x x, y, z dS
y
S
1 m
yy y x, y, z dS
z
S
1 m
yy z x, y, z dS S
Moments of inertia can also be defined as before (see Exercise 35). GRAPHS
Any surface S with equation z tx, y can be regarded as a parametric surface with parametric equations xx and so we have
rx i
&t &x
yy
z tx, y ry j
k
&t &y
k
Thus rx ry
3
r
and
x
ry
&t &t i jk &x &y
&z &x
2
&z &y
2
1
Therefore, in this case, Formula 2 becomes
4
yy
f x, y, z dS yy f x, y, tx, y
S
D
&z &x
2
&z &y
2
1 dA
Similar formulas apply when it is more convenient to project S onto the yz-plane or xz-plane. For instance, if S is a surface with equation y hx, z and D is its pro-
778
■
CHAPTER 13
VECTOR CALCULUS
jection on the xz-plane, then
z
yy
f x, y, z dS yy f x, hx, z, z
S
D
&y &x
&y &z
2
2
1 dA
EXAMPLE 2 Evaluate xxS y dS, where S is the surface z x y 2, 0 x 1,
0 y 2. (See Figure 2.)
&z 1 &x
SOLUTION Since
y
Formula 4 gives
x
yy y dS yy y
FIGURE 2
S
D
y
&z 2y &y
and
1
0
y
2
0
1
1
&z &y
2
2
dA
ys1 1 4y 2 dy dx
y dx s2 0
&z &x
y
2
0
ys1 2y 2 dy
]
2
s2 ( 14 ) 23 1 2y 2 32 0
13s2 3
■
If S is a piecewise-smooth surface, that is, a finite union of smooth surfaces S1 , S2, . . . , Sn that intersect only along their boundaries, then the surface integral of f over S is defined by
yy f x, y, z dS yy f x, y, z dS yy f x, y, z dS S
S1
Sn
V EXAMPLE 3 Evaluate xxS z dS, where S is the surface whose sides S1 are given by the cylinder x 2 y 2 1, whose bottom S2 is the disk x 2 y 2 1 in the plane z 0, and whose top S3 is the part of the plane z 1 x that lies above S2 .
SOLUTION The surface S is shown in Figure 3. (We have changed the usual position of the axes to get a better look at S.) For S1 we use and z as parameters (see Example 5 in Section 13.6) and write its parametric equations as
z
S£ (z=1+x)
x cos
y
S¡ (≈+¥=1)
zz
where 0 2
x 0
y sin
and
Therefore
S™
i r rz sin 0
FIGURE 3
and
r
0 z 1 x 1 cos
j cos 0
k 0 cos i sin j 1
rz scos 2 sin 2 1
SECTION 13.7
SURFACE INTEGRALS
■
779
3 2
Thus the surface integral over S1 is 2
yy z dS yy z r r dA y y z
S1
0
1cos
0
z dz d y
2 1 2
0
1 cos 2 d
D
12 y
2
0
[
1 2 cos 12 1 cos 2 d 12 32 2 sin 14 sin 2
2 0
]
Since S2 lies in the plane z 0, we have
yy z dS yy 0 dS 0 S2
S2
The top surface S3 lies above the unit disk D and is part of the plane z 1 x. So, taking tx, y 1 x in Formula 4 and converting to polar coordinates, we have
yy z dS yy 1 x S3
D
y
2
0
s2 s2 Therefore
y
1
0
1
&z &x
2
&z &y
2
dA
1 r cos s1 1 0 r dr d
2
y y 0
2
y ( 0
1
0 1 2
r r 2 cos dr d
sin cos ) d s2 2 3 1 3
2
s2
0
yy z dS yy z dS yy z dS yy z dS S
S1
S2
S3
3 0 s2 ( 32 s2 ) 2
■
ORIENTED SURFACES P
To define surface integrals of vector fields, we need to rule out nonorientable surfaces such as the Möbius strip shown in Figure 4. [It is named after the German geometer August Möbius (1790–1868).] You can construct one for yourself by taking a long rectangular strip of paper, giving it a half-twist, and taping the short edges together as in Figure 5. If an ant were to crawl along the Möbius strip starting at a point P, it would end up on the “other side” of the strip (that is, with its upper side pointing in the opposite direction). Then, if the ant continued to crawl in the same direction, it would end up back at the same point P without ever having crossed an edge. (If you have constructed a Möbius strip, try drawing a pencil line down the middle.) Therefore, a Möbius strip really has only one side. You can graph the Möbius strip using the parametric equations in Exercise 28 in Section 13.6.
FIGURE 4
A Möbius strip Visual 13.7 shows a Möbius strip with a normal vector that can be moved along the surface. B
C
A
D
FIGURE 5
Constructing a Möbius strip
B
D
A
C
780
■
CHAPTER 13
VECTOR CALCULUS
z
From now on we consider only orientable (two-sided) surfaces. We start with a surface S that has a tangent plane at every point x, y, z on S (except at any boundary point). There are two unit normal vectors n1 and n 2 n1 at x, y, z. (See Figure 6.) If it is possible to choose a unit normal vector n at every such point x, y, z so that n varies continuously over S, then S is called an oriented surface and the given choice of n provides S with an orientation. There are two possible orientations for any orientable surface (see Figure 7).
n¡
n™ 0
n
n
n
n
n y
x
n
FIGURE 7
FIGURE 6
n
n n
The two orientations of an orientable surface
n
For a surface z tx, y given as the graph of t, we use Equation 3 to associate with the surface a natural orientation given by the unit normal vector
n
5
&t &t i jk &x &y
&t &x
1
2
&t &y
2
Since the k-component is positive, this gives the upward orientation of the surface. If S is a smooth orientable surface given in parametric form by a vector function ru, v, then it is automatically supplied with the orientation of the unit normal vector n
6
ru rv ru rv
and the opposite orientation is given by n. For instance, in Example 4 in Section 13.6 we found the parametric representation
z
r, a sin cos i a sin sin j a cos k for the sphere x 2 y 2 z 2 a 2. Then in Example 9 in Section 13.6 we found that
0
r r a 2 sin 2 cos i a 2 sin 2 sin j a 2 sin cos k
y x
r
and
FIGURE 8
So the orientation induced by r, is defined by the unit normal vector
Positive orientation z
n
y x
FIGURE 9
Negative orientation
r a 2 sin
r r 1 sin cos i sin sin j cos k r, r r a
Observe that n points in the same direction as the position vector, that is, outward from the sphere (see Figure 8). The opposite (inward) orientation would have been obtained (see Figure 9) if we had reversed the order of the parameters because r r r r . For a closed surface, that is, a surface that is the boundary of a solid region E, the convention is that the positive orientation is the one for which the normal vectors point outward from E, and inward-pointing normals give the negative orientation (see Figures 8 and 9).
SECTION 13.7
■
SURFACE INTEGRALS
781
SURFACE INTEGRALS OF VECTOR FIELDS z
F=∏v
n Sij S 0
Suppose that S is an oriented surface with unit normal vector n, and imagine a fluid with density x, y, z and velocity field vx, y, z flowing through S. (Think of S as an imaginary surface that doesn’t impede the fluid flow, like a fishing net across a stream.) Then the rate of flow (mass per unit time) per unit area is v. If we divide S into small patches Sij , as in Figure 10 (compare with Figure 1), then Sij is nearly planar and so we can approximate the mass of fluid crossing Sij in the direction of the normal n per unit time by the quantity v nASij
y
x
where , v, and n are evaluated at some point on Sij . (Recall that the component of the vector v in the direction of the unit vector n is v n.) By summing these quantities and taking the limit we get, according to Definition 1, the surface integral of the function v n over S :
FIGURE 10
7
yy v n dS yy x, y, zvx, y, z nx, y, z dS S
S
and this is interpreted physically as the rate of flow through S. If we write F v, then F is also a vector field on ⺢ 3 and the integral in Equation 7 becomes
yy F n dS S
A surface integral of this form occurs frequently in physics, even when F is not v, and is called the surface integral (or flux integral ) of F over S. 8 DEFINITION If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is
yy F dS yy F n dS S
S
This integral is also called the flux of F across S. In words, Definition 8 says that the surface integral of a vector field over S is equal to the surface integral of its normal component over S (as previously defined). If S is given by a vector function ru, v, then n is given by Equation 6, and from Definition 8 and Equation 2 we have
yy F dS yy F S
S
ru rv dS ru rv
yy D
Fru, v
ru rv ru rv
r
u
rv dA
where D is the parameter domain. Thus we have ■ Compare Equation 9 to the similar expression for evaluating line integrals of vector fields in Definition 13.2.13:
y
C
F dr y Frt rt dt b
9
yy F dS yy F r
u
S
rv dA
D
a
EXAMPLE 4 Find the flux of the vector field Fx, y, z z i y j x k across the unit sphere x 2 y 2 z 2 1.
782
■
CHAPTER 13
VECTOR CALCULUS
■ Figure 11 shows the vector field F in Example 4 at points on the unit sphere.
SOLUTION Using the parametric representation
r, sin cos i sin sin j cos k
z
0
0 2
Fr, cos i sin sin j sin cos k
we have
and, from Example 9 in Section 13.6, r r sin 2 cos i sin 2 sin j sin cos k y
Therefore Fr, r r cos sin 2 cos sin 3 sin 2 sin 2 cos cos
x
and, by Formula 9, the flux is FIGURE 11 2
yy F dS yy F r r dA y y 0
S
0
2 sin 2 cos cos sin 3 sin 2 d d
D
2 y sin2 cos d
y
0
0 y sin 3 d 0
y
2
0
2
0
cos d y sin3 d 0
sin 2 d
y
2
sin2 d
0
2
since y cos d 0 0
4 3 ■
by the same calculation as in Example 1.
If, for instance, the vector field in Example 4 is a velocity field describing the flow of a fluid with density 1, then the answer, 43, represents the rate of flow through the unit sphere in units of mass per unit time. In the case of a surface S given by a graph z tx, y, we can think of x and y as parameters and use Equation 3 to write
F rx ry P i Q j R k
&t &t i jk &x &y
Thus Formula 9 becomes
10
yy F dS yy S
&t &t Q R dA &x &y
V EXAMPLE 5 Evaluate xxS F dS, where Fx, y, z y i x j z k and S is the boundary of the solid region E enclosed by the paraboloid z 1 x 2 y 2 and the plane z 0.
S¡ S™ y
FIGURE 12
P
This formula assumes the upward orientation of S; for a downward orientation we multiply by 1. Similar formulas can be worked out if S is given by y hx, z or x ky, z. (See Exercises 31 and 32.)
z
x
D
SOLUTION S consists of a parabolic top surface S1 and a circular bottom surface S2. (See Figure 12.) Since S is a closed surface, we use the convention of positive (outward) orientation. This means that S1 is oriented upward and we can use
SECTION 13.7
SURFACE INTEGRALS
■
783
Equation 10 with D being the projection of S1 on the xy-plane, namely, the disk x 2 y 2 1. Since Px, y, z y
Qx, y, z x &t 2x &x
on S1 and we have
yy F dS yy S1
D
P
Rx, y, z z 1 x 2 y 2 &t 2y &y
&t &t Q R dA yy y2x x2y 1 x 2 y 2 dA &x &y D
yy 1 4xy x 2 y 2 dA y
2
0
y
1
0
1 4r 2 cos sin r 2 r dr d
D
y
2
0
y
1
0
r r 3 4r 3 cos sin dr d y
2
0
( 14 cos sin ) d 14 2 0
2
The disk S2 is oriented downward, so its unit normal vector is n k and we have
yy F dS yy F k dS yy z dA yy 0 dA 0 S2
S2
D
D
since z 0 on S2 . Finally, we compute, by definition, xxS F dS as the sum of the surface integrals of F over the pieces S1 and S2 :
yy F dS yy F dS yy F dS S
S1
S2
0 2 2
■
Although we motivated the surface integral of a vector field using the example of fluid flow, this concept also arises in other physical situations. For instance, if E is an electric field (see Example 5 in Section 13.1), then the surface integral
yy E dS S
is called the electric flux of E through the surface S. One of the important laws of electrostatics is Gauss’s Law, which says that the net charge enclosed by a closed surface S is 11
Q 0 yy E dS S
where 0 is a constant (called the permittivity of free space) that depends on the units used. (In the SI system, 0 8.8542 10 12 C 2N m 2.) Therefore, if the vector field F in Example 4 represents an electric field, we can conclude that the charge enclosed by S is Q 4 03. Another application of surface integrals occurs in the study of heat flow. Suppose the temperature at a point x, y, z in a body is ux, y, z. Then the heat flow is defined as the vector field F K ∇u where K is an experimentally determined constant called the conductivity of the sub-
784
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CHAPTER 13
VECTOR CALCULUS
stance. The rate of heat flow across the surface S in the body is then given by the surface integral
yy F dS K yy ∇u dS S
S
V EXAMPLE 6 The temperature u in a metal ball is proportional to the square of the distance from the center of the ball. Find the rate of heat flow across a sphere S of radius a with center at the center of the ball.
SOLUTION Taking the center of the ball to be at the origin, we have
ux, y, z Cx 2 y 2 z 2 where C is the proportionality constant. Then the heat flow is Fx, y, z K ∇u KC2x i 2y j 2z k where K is the conductivity of the metal. Instead of using the usual parametrization of the sphere as in Example 4, we observe that the outward unit normal to the sphere x 2 y 2 z 2 a 2 at the point x, y, z is n
1 x i y j z k a
Fn
and so
2KC 2 x y 2 z 2 a
But on S we have x 2 y 2 z 2 a 2, so F n 2aKC. Therefore, the rate of heat flow across S is
yy F dS yy F n dS 2aKC yy dS S
S
S
2aKCAS 2aKC4 a 2 8KC a 3
13.7
EXERCISES
1. Let S be the cube with vertices 1, 1, 1. Approximate
By dividing H into four patches, estimate the value of xxH f x, y, z dS.
xxS sx 2 2y 2 3z 2
dS by using a Riemann sum as in Definition 1, taking the patches Sij to be the squares that are the faces of the cube and the points Pij* to be the centers of the squares.
4. Suppose that f x, y, z t(sx 2 y 2 z 2 ), where t is a
function of one variable such that t2 5. Evaluate xxS f x, y, z dS, where S is the sphere x 2 y 2 z 2 4.
2. A surface S consists of the cylinder x 2 y 2 1,
1 z 1, together with its top and bottom disks. Suppose you know that f is a continuous function with f 1, 0, 0 2, f 0, 1, 0 3, and f 0, 0, 1 4. Estimate the value of xxS f x, y, z dS by using a Riemann sum, taking the patches Sij to be four quarter-cylinders and the top and bottom disks.
3. Let H be the hemisphere x 2 y 2 z 2 50, z 0, and
suppose f is a continuous function with f 3, 4, 5 7, f 3, 4, 5 8, f 3, 4, 5 9, and f 3, 4, 5 12.
5–18 5.
■
Evaluate the surface integral.
xxS yz dS,
S is the surface with parametric equations x u 2, y u sin v, z u cos v, 0 u 1, 0 v 2 6.
xxS s1 x 2 y 2 dS, S is the helicoid with vector equation ru, v u cos v i u sin v j v k, 0 u 1, 0 v
■
SECTION 13.7
7.
xxS x 2 yz dS,
22. Fx, y, z x i y j z 4 k,
xxS x y dS,
23. Fx, y, z x i z j y k,
24. Fx, y, z xz i x j y k,
xxS yz dS,
S is the hemisphere x 2 y 2 z 2 25, y 0, oriented in the direction of the positive y-axis
S is the part of the plane x y z 1 that lies in the first octant
10.
25. Fx, y, z y j z k,
xxS y dS,
S is the surface z x 11.
2 3
32
y
32
S consists of the paraboloid y x 2 z 2, 0 y 1, and the disk x 2 z 2 1, y 1
, 0 x 1, 0 y 1
xxS x 2 z 2 dS,
26. Fx, y, z y i z y j x k,
S is the part of the cone z 2 x 2 y 2 that lies between the planes z 1 and z 3 12.
S is the surface of the tetrahedron with vertices 0, 0, 0, 1, 0, 0, 0, 1, 0, and 0, 0, 1
xxS z dS,
27. Fx, y, z x 2 i y 2 j z 2 k,
S is the boundary of the solid half-cylinder 0 z s1 y 2 , 0 x 2
S is the surface x y 2z 2, 0 y 1, 0 z 1 13.
xxS y dS,
■
S is the part of the paraboloid y x 2 z 2 that lies inside the cylinder x 2 z 2 4 14.
xxS x 2 y z 2 dS,
xxS x 2 y 2 z 2 dS, ■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
(a) Evaluate xxS xyz dS correct to four decimal places. (b) Find the exact value of xxS x 2 yz dS.
19. Fx, y, z x y i yz j zx k,
S is the part of the paraboloid z 4 x 2 y 2 that lies above the square 0 x 1, 0 y 1, and has upward orientation
20. Fx, y, z y i x j z 2 k,
S is the helicoid of Exercise 6 with upward orientation S is the part of the plane x y z 1 in the first octant and has downward orientation
30. Find the flux of Fx, y, z sinx yz i x 2 y j z 2e x5 k
across the part of the cylinder 4y 2 z 2 4 that lies above the xy-plane and between the planes x 2 and x 2 with upward orientation. Illustrate by using a computer algebra system to draw the cylinder and the vector field on the same screen.
31. Find a formula for xxS F dS similar to Formula 10 for the
case where S is given by y hx, z and n is the unit normal that points toward the left.
■
■ Evaluate the surface integral xx F dS for the given S vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
19–27
21. Fx, y, z xze y i xze y j z k,
■
29. Find the value of xxS x 2 y 2z 2 dS correct to four decimal
CAS
xxS x yz dS,
S consists of the cylinder in Exercise 17 together with its top and bottom disks ■
■
places, where S is the part of the paraboloid z 3 2x 2 y 2 that lies above the x y-plane.
S is the part of the cylinder x 2 y 2 9 between the planes z 0 and z 2
18.
■
CAS
xxS x 2 z y 2 z dS,
S is the part of the sphere x 2 y 2 z 2 1 that lies above the cone z sx 2 y 2 17.
■
28. Let S be the surface z xy, 0 x 1, 0 y 1.
S is the hemisphere x 2 y 2 z 2 4, z 0
16.
■
CAS
xxS x y dS, S is the boundary of the region enclosed by the cylinder x 2 z 2 1 and the planes y 0 and x y 2
15.
785
S is the part of the sphere x 2 y 2 z 2 4 in the first octant, with orientation toward the origin
S is the triangular region with vertices (1, 0, 0), (0, 2, 0), and (0, 0, 2) 9.
■
S is the part of the cone z sx 2 y 2 beneath the plane z 1 with downward orientation
S is the part of the plane z 1 2x 3y that lies above the rectangle 0, 3 0, 2
8.
SURFACE INTEGRALS
32. Find a formula for xxS F dS similar to Formula 10 for
the case where S is given by x k y, z and n is the unit normal that points forward (that is, toward the viewer when the axes are drawn in the usual way).
33. Find the center of mass of the hemisphere
x 2 y 2 z 2 a 2, z 0, if it has constant density. 34. Find the mass of a thin funnel in the shape of a cone
z sx 2 y 2 , 1 z 4, if its density function is x, y, z 10 z. 35. (a) Give an integral expression for the moment of inertia I z
about the z -axis of a thin sheet in the shape of a surface S if the density function is . (b) Find the moment of inertia about the z -axis of the funnel in Exercise 34.
786
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CHAPTER 13
VECTOR CALCULUS
36. The conical surface z 2 x 2 y 2, 0 z a, has constant
40. Use Gauss’s Law to find the charge enclosed by the cube
with vertices 1, 1, 1 if the electric field is Ex, y, z x i y j z k.
density k. Find (a) the center of mass and (b) the moment of inertia about the z -axis. 37. A fluid has density 870 kgm3 and flows with velocity
41. The temperature at the point x, y, z in a substance with
v z i y 2 j x 2 k , where x, y, and z are measured in meters and the components of v in meters per second. Find the rate of flow outward through the cylinder x 2 y 2 4 , 0 z 1.
conductivity K 6.5 is ux, y, z 2y 2 2z 2. Find the rate of heat flow inward across the cylindrical surface y 2 z 2 6, 0 x 4.
42. The temperature at a point in a ball with conductivity K is
38. Seawater has density 1025 kgm3 and flows in a velocity
field v y i x j, where x, y, and z are measured in meters and the components of v in meters per second. Find the rate of flow outward through the hemisphere x 2 y 2 z 2 9 , z 0.
inversely proportional to the distance from the center of the ball. Find the rate of heat flow across a sphere S of radius a with center at the center of the ball.
43. Let F be an inverse square field, that is, Fr cr r
for some constant c, where r x i y j z k. Show that the flux of F across a sphere S with center the origin is independent of the radius S.
39. Use Gauss’s Law to find the charge contained in the solid
hemisphere x 2 y 2 z 2 a 2, z 0, if the electric field is Ex, y, z x i y j 2z k.
13.8
STOKES’ THEOREM Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s Theorem. Whereas Green’s Theorem relates a double integral over a plane region D to a line integral around its plane boundary curve, Stokes’ Theorem relates a surface integral over a surface S to a line integral around the boundary curve of S (which is a space curve). Figure 1 shows an oriented surface with unit normal vector n. The orientation of S induces the positive orientation of the boundary curve C shown in the figure. This means that if you walk in the positive direction around C with your head pointing in the direction of n, then the surface will always be on your left.
z
n n
3
S
C 0 x
FIGURE 1
y
STOKES’ THEOREM Let S be an oriented piecewise-smooth surface that is
bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region in ⺢ 3 that contains S. Then
y
C
F dr yy curl F dS S
Stokes’ Theorem is named after the Irish mathematical physicist Sir George Stokes (1819–1903). Stokes was a professor at Cambridge University (in fact he held the same position as Newton, Lucasian Professor of Mathematics) and was especially noted for his studies of fluid flow and light. What we call Stokes’ Theorem was actually discovered by the Scottish physicist Sir William Thomson (1824–1907, known as Lord Kelvin). Stokes learned of this theorem in a letter from Thomson in 1850 and asked students to prove it on an examination at Cambridge University in 1854. We don’t know if any of those students was able to do so. ■
Since
y
C
F dr y F T ds C
and
yy curl F dS yy curl F n dS S
S
Stokes’ Theorem says that the line integral around the boundary curve of S of the tangential component of F is equal to the surface integral of the normal component of the curl of F. The positively oriented boundary curve of the oriented surface S is often written as &S, so Stokes’ Theorem can be expressed as 1
yy curl F dS y S
&S
F dr
SECTION 13.8
787
■
STOKES’ THEOREM
There is an analogy among Stokes’ Theorem, Green’s Theorem, and the Fundamental Theorem of Calculus. As before, there is an integral involving derivatives on the left side of Equation 1 (recall that curl F is a sort of derivative of F ) and the right side involves the values of F only on the boundary of S. In fact, in the special case where the surface S is flat and lies in the xy-plane with upward orientation, the unit normal is k, the surface integral becomes a double integral, and Stokes’ Theorem becomes
y
C
F dr yy curl F dS yy curl F k dA S
S
This is precisely the vector form of Green’s Theorem given in Equation 13.5.12. Thus, we see that Green’s Theorem is really a special case of Stokes’ Theorem. Although Stokes’ Theorem is too difficult for us to prove in its full generality, we can give a proof when S is a graph and F, S, and C are well behaved. z
PROOF OF A SPECIAL CASE OF STOKES’ THEOREM We assume that the equation of S
n
is z tx, y, x, y D, where t has continuous second-order partial derivatives and D is a simple plane region whose boundary curve C1 corresponds to C. If the orientation of S is upward, then the positive orientation of C corresponds to the positive orientation of C1. (See Figure 2.) We are also given that F P i Q j R k, where the partial derivatives of P, Q, and R are continuous. Since S is a graph of a function, we can apply Formula 13.7.10 with F replaced by curl F. The result is
z=g(x, y) S 0 x
C
y
D C¡
2
FIGURE 2
yy curl F dS S
yy D
&R &Q &y &z
&z &x
&P &R &z &x
&z &y
&Q &P &x &y
dA
where the partial derivatives of P, Q, and R are evaluated at x, y, tx, y. If x xt
y yt
a t b
is a parametric representation of C1, then a parametric representation of C is x xt
y yt
z txt, yt
a t b
This allows us, with the aid of the Chain Rule, to evaluate the line integral as follows:
y
C
F dr
y y yy y
b
a
b
P
dx dy dz Q R dt dt dt
PR
a
D
& &x
&z &x
QR
dt
b
P
a
dx &z QR dt &y
&z &y
& &y
PR
dy dt
&z &x
dx dy &z dx &z dy Q R dt dt &x dt &y dt dt
y
C1
dA
PR
&z &x
dt
dx Q R
&z &y
dy
788
■
CHAPTER 13
VECTOR CALCULUS
where we have used Green’s Theorem in the last step. Then, using the Chain Rule again and remembering that P, Q, and R are functions of x, y, and z and that z is itself a function of x and y, we get
y
C
F dr
yy D
&Q &Q &z &R &z &R &z &z &2z R &x &z &x &x &y &z &x &y &x &y
&P &P &z &R &z &R &z &z &2z R &y &z &y &y &x &z &y &x &y &x
dA
Four of the terms in this double integral cancel and the remaining six terms can be arranged to coincide with the right side of Equation 2. Therefore
y
C
F dr yy curl F dS
■
S
Evaluate xC F dr, where Fx, y, z y 2 i x j z 2 k and C is the curve of intersection of the plane y z 2 and the cylinder x 2 y 2 1. (Orient C to be counterclockwise when viewed from above.) V EXAMPLE 1
SOLUTION The curve C (an ellipse) is shown in Figure 3. Although xC F dr could be evaluated directly, it’s easier to use Stokes’ Theorem. We first compute
z
S
C
i & curl F &x y 2
y+z=2
j & &y x
k & 1 2y k &z z2
Although there are many surfaces with boundary C, the most convenient choice is the elliptical region S in the plane y z 2 that is bounded by C. If we orient S upward, then C has the induced positive orientation. The projection D of S on the xy-plane is the disk x 2 y 2 1 and so using Equation 13.7.10 with z tx, y 2 y, we have
D 0 y x
FIGURE 3
y
C
F dr yy curl F dS yy 1 2y dA y
2
0
S
y
2
0
D
r2 r3 2 sin 2 3
1
0
d y
2
0
y
1
0
1 2r sin r dr d
( 12 23 sin ) d 12 2 0
V EXAMPLE 2 Use Stokes’ Theorem to compute the integral xxS curl F dS, where Fx, y, z xz i yz j xy k and S is the part of the sphere x 2 y 2 z 2 4 that lies inside the cylinder x 2 y 2 1 and above the xy-plane. (See Figure 4.)
z
≈+¥+z@=4
SOLUTION To find the boundary curve C we solve the equations x 2 y 2 z 2 4
and x 2 y 2 1. Subtracting, we get z 2 3 and so z s3 (since z 0). Thus C is the circle given by the equations x 2 y 2 1, z s3 . A vector equation of C is
S C
rt cos t i sin t j s3 k
0 t 2
0
y x
FIGURE 4
■
≈+¥=1
so
rt sin t i cos t j
Also, we have Frt s3 cos t i s3 sin t j cos t sin t k
SECTION 13.8
STOKES’ THEOREM
■
789
Therefore, by Stokes’ Theorem,
yy curl F dS y
C
F dr y
2
0
Frt rt dt
S
y
2
0
(s3 cos t sin t s3 sin t cos t) dt
s3
y
2
0
0 dt 0 ■
Note that in Example 2 we computed a surface integral simply by knowing the values of F on the boundary curve C. This means that if we have another oriented surface with the same boundary curve C, then we get exactly the same value for the surface integral! In general, if S1 and S2 are oriented surfaces with the same oriented boundary curve C and both satisfy the hypotheses of Stokes’ Theorem, then
yy curl F dS y
3
C
F dr yy curl F dS
S1
T v
C (a) jC v dr>0, positive circulation
S2
This fact is useful when it is difficult to integrate over one surface but easy to integrate over the other. We now use Stokes’ Theorem to throw some light on the meaning of the curl vector. Suppose that C is an oriented closed curve and v represents the velocity field in fluid flow. Consider the line integral
y
C
T
C
v (b) jC v dr<0, negative circulation FIGURE 5
v dr y v T ds C
and recall that v T is the component of v in the direction of the unit tangent vector T. This means that the closer the direction of v is to the direction of T, the larger the value of v T. Thus xC v dr is a measure of the tendency of the fluid to move around C and is called the circulation of v around C. (See Figure 5.) Now let P0x 0 , y0 , z0 be a point in the fluid and let Sa be a small disk with radius a and center P0. Then (curl FP curl FP0 for all points P on Sa because curl F is continuous. Thus, by Stokes’ Theorem, we get the following approximation to the circulation around the boundary circle Ca :
y
Ca
v dr yy curl v dS yy curl v n dS Sa
Sa
yy curl vP0 nP0 dS curl vP0 nP0 a 2 ■ Imagine a tiny paddle wheel placed in the fluid at a point P , as in Figure 6; the paddle wheel rotates fastest when its axis is parallel to curl v.
Sa
This approximation becomes better as a l 0 and we have 4
curl vP0 nP0 lim
al0
1 a 2
y
Ca
v dr
curl v
FIGURE 6
Equation 4 gives the relationship between the curl and the circulation. It shows that curl v n is a measure of the rotating effect of the fluid about the axis n. The curling effect is greatest about the axis parallel to curl v. Finally, we mention that Stokes’ Theorem can be used to prove Theorem 13.5.4 (which states that if curl F 0 on all of ⺢ 3, then F is conservative). From our previous work (Theorems 13.3.3 and 13.3.4), we know that F is conservative if xC F dr 0 for every closed path C. Given C, suppose we can find an orientable surface S whose boundary is C. (This can be done, but the proof requires advanced tech-
790
■
CHAPTER 13
VECTOR CALCULUS
niques.) Then Stokes’ Theorem gives
y
C
F dr yy curl F dS yy 0 dS 0 S
S
A curve that is not simple can be broken into a number of simple curves, and the integrals around these simple curves are all 0. Adding these integrals, we obtain xC F dr 0 for any closed curve C.
13.8 1– 4
■
EXERCISES
Use Stokes’ Theorem to evaluate xxS curl F dS.
10. (a) Use Stokes’ Theorem to evaluate xC F dr, where
Fx, y, z x 2 y i 13 x 3 j x y k and C is the curve of intersection of the hyperbolic paraboloid z y 2 x 2 and the cylinder x 2 y 2 1 oriented counterclockwise as viewed from above. (b) Graph both the hyperbolic paraboloid and the cylinder with domains chosen so that you can see the curve C and the surface that you used in part (a). (c) Find parametric equations for C and use them to graph C.
1. Fx, y, z x e i y e j z e 2 yz
2 xz
2 xy
k, S is the hemisphere x 2 y 2 z 2 4, z 0, oriented upward
2. Fx, y, z yz i xz j x y k,
;
3. Fx, y, z x yz i x y j x 2 yz k,
;
S is the part of the paraboloid z 9 x 2 y 2 that lies above the plane z 5, oriented upward S consists of the top and the four sides (but not the bottom) of the cube with vertices 1, 1, 1, oriented outward [Hint: Use Equation 3.]
■ Verify that Stokes’ Theorem is true for the given vector field F and surface S.
11–12
11. Fx, y, z y 2 i x j z 2 k,
4. Fx, y, z e xy cos z i x 2 z j x y k,
S is the part of the paraboloid z x 2 y 2 that lies below the plane z 1, oriented upward
S is the hemisphere x s1 y 2 z 2 , oriented in the direction of the positive x-axis [Hint: Use Equation 3.]
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S is the part of the plane 2x y z 2 that lies in the first octant, oriented upward
Use Stokes’ Theorem to evaluate xC F dr. In each case C is oriented counterclockwise as viewed from above. 5– 8
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5. Fx, y, z x y 2 i y z 2 j z x 2 k,
C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1)
7. Fx, y, z yz i 2 xz j e xy k, 8. Fx, y, z xy i 2z j 3y k,
C is the curve of intersection of the plane x z 5 and the cylinder x 2 y 2 9 ■
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9. (a) Use Stokes’ Theorem to evaluate xC F dr, where
Fx, y, z x z i x y j z k 2
; ;
2
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14. Evaluate xC y sin x dx z 2 cos y dy x 3 dz,
C is the circle x 2 y 2 16, z 5
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the points 1, 0, 0, 1, 2, 1, 0, 2, 1, and back to the origin under the influence of the force field Fx, y, z z 2 i 2xy j 4y 2 k. Find the work done.
C is the boundary of the part of the plane 2x y 2z 2 in the first octant
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13. A particle moves along line segments from the origin to
6. Fx, y, z ex i e x j e z k,
■
12. Fx, y, z x i y j x yz k,
2
and C is the curve of intersection of the plane x y z 1 and the cylinder x 2 y 2 9 oriented counterclockwise as viewed from above. (b) Graph both the plane and the cylinder with domains chosen so that you can see the curve C and the surface that you used in part (a). (c) Find parametric equations for C and use them to graph C.
where C is the curve rt sin t, cos t, sin 2t , 0 t 2. [Hint: Observe that C lies on the surface z 2 x y.]
15. If S is a sphere and F satisfies the hypotheses of Stokes’
Theorem, show that xxS curl F dS 0.
16. Suppose S and C satisfy the hypotheses of Stokes’ Theorem
and f , t have continuous second-order partial derivatives. Use Exercises 22 and 24 in Section 13.5 to show the following. (a) xC f (t dr xxS ( f (t dS (b) (c)
xC f ( f dr 0 xC f (t t( f dr 0
SECTION 13.9
13.9
THE DIVERGENCE THEOREM
■
791
THE DIVERGENCE THEOREM In Section 13.5 we rewrote Green’s Theorem in a vector version as
y
C
F n ds yy div Fx, y dA D
where C is the positively oriented boundary curve of the plane region D. If we were seeking to extend this theorem to vector fields on ⺢ 3, we might make the guess that
yy F n dS yyy div Fx, y, z dV
1
S
E
where S is the boundary surface of the solid region E. It turns out that Equation 1 is true, under appropriate hypotheses, and is called the Divergence Theorem. Notice its similarity to Green’s Theorem and Stokes’ Theorem in that it relates the integral of a derivative of a function (div F in this case) over a region to the integral of the original function F over the boundary of the region. At this stage you may wish to review the various types of regions over which we were able to evaluate triple integrals in Section 12.5. We state and prove the Divergence Theorem for regions E that are simultaneously of types 1, 2, and 3 and we call such regions simple solid regions. (For instance, regions bounded by ellipsoids or rectangular boxes are simple solid regions.) The boundary of E is a closed surface, and we use the convention, introduced in Section 13.7, that the positive orientation is outward; that is, the unit normal vector n is directed outward from E.
THE DIVERGENCE THEOREM Let E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then
The Divergence Theorem is sometimes called Gauss’s Theorem after the great German mathematician Karl Friedrich Gauss (1777–1855), who discovered this theorem during his investigation of electrostatics. In Eastern Europe the Divergence Theorem is known as Ostrogradsky’s Theorem after the Russian mathematician Mikhail Ostrogradsky (1801–1862), who published this result in 1826. ■
yy F dS yyy div F dV S
E
Thus the Divergence Theorem states that, under the given conditions, the flux of F across the boundary surface of E is equal to the triple integral of the divergence of F over E. PROOF Let F P i Q j R k. Then
div F
so
yyy div F dV yyy E
E
&P &Q &R &x &y &z
&P &Q &R dV yyy dV yyy dV &x &y &z E E
If n is the unit outward normal of S, then the surface integral on the left side of the
792
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CHAPTER 13
VECTOR CALCULUS
Divergence Theorem is
yy F dS yy F n dS yy P i Q j R k n dS S
S
S
yy P i n dS yy Q j n dS yy R k n dS S
S
S
Therefore, to prove the Divergence Theorem, it suffices to prove the following three equations: &P 2 yy P i n dS yyy &x dV S E
yy Q j n dS yyy
3
S
E
yy R k n dS yyy
4
S
E
&Q dV &y &R dV &z
To prove Equation 4 we use the fact that E is a type 1 region:
E x, y, z x, y D, u1x, y z u 2x, y where D is the projection of E onto the xy-plane. By Equation 12.5.6, we have
yyy E
&R dV yy &z D
y
u 2 x, y
u1 x, y
&R x, y, z dz dA &z
and, therefore, by the Fundamental Theorem of Calculus, 5
yyy E
z
S™ { z=u™(x, y)}
S£
E
The boundary surface S consists of three pieces: the bottom surface S1 , the top surface S2 , and possibly a vertical surface S3 , which lies above the boundary curve of D. (See Figure 1. It might happen that S3 doesn’t appear, as in the case of a sphere.) Notice that on S3 we have k n 0, because k is vertical and n is horizontal, and so
yy R k n dS yy 0 dS 0
0 x
S¡ (z=u¡(x, y)) D
y
S3
S3
Thus, regardless of whether there is a vertical surface, we can write 6
FIGURE 1
&R dV yy Rx, y, u 2 x, y Rx, y, u1 x, y dA &z D
yy R k n dS yy R k n dS yy R k n dS S
S1
S2
The equation of S2 is z u 2x, y, x, y D, and the outward normal n points upward, so from Equation 13.7.10 (with F replaced by R k) we have
yy R k n dS yy Rx, y, u x, y dA 2
S2
D
On S1 we have z u1x, y, but here the outward normal n points downward, so
SECTION 13.9
THE DIVERGENCE THEOREM
■
793
we multiply by 1:
yy R k n dS yy Rx, y, u x, y dA 1
S1
D
Therefore, Equation 6 gives
yy R k n dS yy Rx, y, u x, y Rx, y, u x, y dA 2
S
1
D
Comparison with Equation 5 shows that
yy R k n dS yyy S ■ Notice that the method of proof of the Divergence Theorem is very similar to that of Green’s Theorem.
E
&R dV &z
Equations 2 and 3 are proved in a similar manner using the expressions for E as a type 2 or type 3 region, respectively. ■ V EXAMPLE 1 Find the flux of the vector field Fx, y, z z i y j x k over the unit sphere x 2 y 2 z 2 1.
SOLUTION First we compute the divergence of F :
div F
& & & z y x 1 &x &y &z
The unit sphere S is the boundary of the unit ball B given by x 2 y 2 z 2 1. Thus the Divergence Theorem gives the flux as ■ The solution in Example 1 should be compared with the solution in Example 4 in Section 13.7.
z
yy F dS yyy div F dV yyy 1 dV VB S
V EXAMPLE 2
B
4 3
13
B
4 3
■
Evaluate yy F dS, where S
(0, 0, 1)
y=2-z
2
and S is the surface of the region E bounded by the parabolic cylinder z 1 x 2 and the planes z 0, y 0, and y z 2. (See Figure 2.)
0 (1, 0, 0)
(0, 2, 0) y
x
z=1-≈ FIGURE 2
Fx, y, z xy i ( y 2 e xz ) j sinxy k
SOLUTION It would be extremely difficult to evaluate the given surface integral directly. (We would have to evaluate four surface integrals corresponding to the four pieces of S.) Furthermore, the divergence of F is much less complicated than F itself:
div F
& & & 2 xy ( y 2 e xz ) sin xy y 2y 3y &x &y &z
Therefore, we use the Divergence Theorem to transform the given surface integral into a triple integral. The easiest way to evaluate the triple integral is to express E as a type 3 region:
E x, y, z 1 x 1, 0 z 1 x 2, 0 y 2 z
794
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CHAPTER 13
VECTOR CALCULUS
Then we have
yy F dS yyy div F dV yyy 3y dV S
E
3y
1
E
3y
1
1
y
1x
0
2
2 z2 3 dz dx 2 2
y
y
1
1
1
1x
2
0
y
2z
0
y dy dz dx
2 z3 3
1x 2
dx
0
12 y x 2 13 8 dx y x 6 3x 4 3x 2 7 dx 184 35 1
1
1
n™
0
■
Although we have proved the Divergence Theorem only for simple solid regions, it can be proved for regions that are finite unions of simple solid regions. (The procedure is similar to the one we used in Section 13.4 to extend Green’s Theorem.) For example, let’s consider the region E that lies between the closed surfaces S1 and S2 , where S1 lies inside S2. Let n1 and n 2 be outward normals of S1 and S2 . Then the boundary surface of E is S S1 S2 and its normal n is given by n n1 on S1 and n n 2 on S2. (See Figure 3.) Applying the Divergence Theorem to S, we get
yyy div F dV yy F dS yy F n dS
7
E
S
S
yy F n1 dS yy F n 2 dS
FIGURE 3
S1
S2
yy F dS yy F dS S1
S2
Let’s apply this to the electric field (see Example 5 in Section 13.1): Ex
Q x x 3
where S1 is a small sphere with radius a and center the origin. You can verify that div E 0. (See Exercise 23.) Therefore, Equation 7 gives
yy E dS yy E dS yyy div E dV yy E dS yy E n dS S2
S1
E
S1
S1
The point of this calculation is that we can compute the surface integral over S1 because S1 is a sphere. The normal vector at x is x x . Therefore En since the equation of S1
Q x x 3
S1
x x
Q Q Q 4 x x 2 x x a2
is x a. Thus we have
yy E dS yy E n dS S2
Q a2
yy dS S1
Q Q AS1 2 4 a 2 4 Q a2 a
This shows that the electric flux of E is 4 Q through any closed surface S2 that contains the origin. [This is a special case of Gauss’s Law (Equation 13.7.11) for a single charge. The relationship between and 0 is 14 0 .] Another application of the Divergence Theorem occurs in fluid flow. Let vx, y, z be the velocity field of a fluid with constant density . Then F v is the rate of flow per unit area. If P0x 0 , y0 , z0 is a point in the fluid and Ba is a ball with center P0 and very small radius a, then div FP div FP0 for all points in Ba since div F is con-
SECTION 13.9
THE DIVERGENCE THEOREM
■
795
tinuous. We approximate the flux over the boundary sphere Sa as follows:
yy F dS yyy div F dV Sa
yyy div FP0 dV div FP0 VBa
Ba
Ba
This approximation becomes better as a l 0 and suggests that div FP0 lim
8
y
P¡
x
P™
FIGURE 4
The vector field F=≈ i+¥ j
al0
1 VBa
yy F dS Sa
Equation 8 says that div FP0 is the net rate of outward flux per unit volume at P0. (This is the reason for the name divergence.) If div FP 0, the net flow is outward near P and P is called a source. If div FP 0, the net flow is inward near P and P is called a sink. For the vector field in Figure 4, it appears that the vectors that end near P1 are shorter than the vectors that start near P1. Thus the net flow is outward near P1, so div FP1 0 and P1 is a source. Near P2 , on the other hand, the incoming arrows are longer than the outgoing arrows. Here the net flow is inward, so div FP2 0 and P2 is a sink. We can use the formula for F to confirm this impression. Since F x 2 i y 2 j, we have div F 2x 2y, which is positive when y x. So the points above the line y x are sources and those below are sinks. SUMMARY
The main results of this chapter are all higher-dimensional versions of the Fundamental Theorem of Calculus. To help you remember them, we collect them together here (without hypotheses) so that you can see more easily their essential similarity. Notice that in each case we have an integral of a “derivative” over a region on the left side, and the right side involves the values of the original function only on the boundary of the region. Fundamental Theorem of Calculus
y
b
a
Fx dx Fb Fa
a
b r(b)
Fundamental Theorem for Line Integrals
Green’s Theorem
y
(f dr f rb f ra
yy
C
D
C
r(a)
&Q &P &x &y
C
dA y P dx Q dy
D
C
n
Stokes’ Theorem
yy curl F dS y
C
F dr
S
S
C n S
Divergence Theorem
yyy div F dV yy F dS E
S
E
n
796
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CHAPTER 13
13.9
VECTOR CALCULUS
EXERCISES and below the surface z 2 x 4 y 4, 1 x 1, 1 y 1
Verify that the Divergence Theorem is true for the vector field F on the region E.
1– 4
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1. Fx, y, z 3x i x y j 2 xz k,
E is the cube bounded by the planes x 0, x 1, y 0, y 1, z 0, and z 1
CAS
2
4. Fx, y, z x i y j z k,
E is the unit ball x 2 y 2 z 2 1 ■
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18. Let Fx, y, z z tan1 y 2 i z 3 lnx 2 1 j z k.
Use the Divergence Theorem to calculate the surface integral xxS F dS; that is, calculate the flux of F across S. 5–15
■
1 Fx, y, z z 2 x i ( 3 y 3 tan z) j x 2z y 2 k and S is the top half of the sphere x 2 y 2 z 2 1. [Hint: Note that S is not a closed surface. First compute integrals over S1 and S2, where S1 is the disk x 2 y 2 1, oriented downward, and S2 S S1.]
E is the solid cylinder x y 1, 0 z 1
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17. Use the Divergence Theorem to evaluate xxS F dS, where
3. Fx, y, z x y i yz j zx k,
■
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Fx, y, z sin x cos 2 y i sin 3 y cos 4z j sin 5z cos 6x k in the cube cut from the first octant by the planes x 2, y 2, and z 2. Then compute the flux across the surface of the cube.
E is the solid bounded by the paraboloid z 4 x 2 y 2 and the xy-plane
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16. Use a computer algebra system to plot the vector field
2. Fx, y, z x 2 i x y j z k,
2
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Find the flux of F across the part of the paraboloid x 2 y 2 z 2 that lies above the plane z 1 and is oriented upward.
5. Fx, y, z e x sin y i e x cos y j yz 2 k,
S is the surface of the box bounded by the planes x 0, x 1, y 0, y 1, z 0, and z 2
19. A vector field F is shown. Use the interpretation of
divergence derived in this section to determine whether div F is positive or negative at P1 and at P2.
6. Fx, y, z x 2z 3 i 2 x yz 3 j xz 4 k,
S is the surface of the box with vertices 1, 2, 3
2
7. Fx, y, z 3x y 2 i xe z j z 3 k,
S is the surface of the solid bounded by the cylinder y 2 z 2 1 and the planes x 1 and x 2
P¡ _2
8. Fx, y, z x 3 y i x 2 y 2 j x 2 yz k,
2 P™
S is the surface of the solid bounded by the hyperboloid x 2 y 2 z 2 1 and the planes z 2 and z 2 9. Fx, y, z x y sin z i cosxz j y cos z k,
_2
S is the ellipsoid x 2a 2 y 2b 2 z 2c 2 1
20. (a) Are the points P1 and P2 sources or sinks for the vector
10. Fx, y, z x 2 y i x y 2 j 2 x yz k,
field F shown in the figure? Give an explanation based solely on the picture. (b) Given that Fx, y x, y 2 , use the definition of divergence to verify your answer to part (a).
S is the surface of the tetrahedron bounded by the planes x 0, y 0, z 0, and x 2y z 2 11. Fx, y, z cos z x y 2 i xez j sin y x 2 z k,
S is the surface of the solid bounded by the paraboloid z x 2 y 2 and the plane z 4
2 P¡
12. Fx, y, z x 4 i x 3z 2 j 4 x y 2z k,
S is the surface of the solid bounded by the cylinder x 2 y 2 1 and the planes z x 2 and z 0
_2
2 P™
13. Fx, y, z 4 x 3z i 4 y 3z j 3z 4 k,
S is the sphere with radius R and center the origin _2
14. Fx, y, z x 3 y sin z i y 3 z sin x j 3z k,
S is the surface of the solid bounded by the hemispheres z s4 x 2 y 2 , z s1 x 2 y 2 and the plane z 0 CAS
15. Fx, y, z e y tan z i y s3 x 2 j x sin y k,
S is the surface of the solid that lies above the xy-plane
CAS
21–22 ■ Plot the vector field and guess where div F 0 and where div F 0 . Then calculate div F to check your guess. 21. Fx, y xy, x y 2 ■
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22. Fx, y x 2, y 2 ■
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CHAPTER 13
23. Verify that div E 0 for the electric field Ex
Q x. x 3
24. Use the Divergence Theorem to evaluate
yy 2x 2y z
2
27.
28.
where Fx, y, z x i y j z k
yy curl F dS 0 yy D
n
f dS yyy ( 2 f dV
S
Prove each identity, assuming that S and E satisfy the conditions of the Divergence Theorem and the scalar functions and components of the vector fields have continuous secondorder partial derivatives. ■
yy a n dS 0,
yy F dS,
797
S
where S is the sphere x 2 y 2 z 2 1.
25.
1 3
■
S
dS
S
25–30
26. VE
REVIEW
29.
E
yy f (t n dS yyy f ( t ( f (t dV 2
S
30.
E
yy f (t t( f n dS yyy f ( t t( 2
S
where a is a constant vector
■
2
f dV
E ■
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S
13
REVIEW
CONCEPT CHECK
1. What is a vector field? Give three examples that have physi-
cal meaning. 2. (a) What is a conservative vector field?
(b) What is a potential function? 3. (a) Write the definition of the line integral of a scalar func-
(b) (c)
(d) (e)
tion f along a smooth curve C with respect to arc length. How do you evaluate such a line integral? Write expressions for the mass and center of mass of a thin wire shaped like a curve C if the wire has linear density function x, y. Write the definitions of the line integrals along C of a scalar function f with respect to x, y, and z. How do you evaluate these line integrals?
4. (a) Define the line integral of a vector field F along a
smooth curve C given by a vector function rt. (b) If F is a force field, what does this line integral represent? (c) If F P, Q, R , what is the connection between the line integral of F and the line integrals of the component functions P, Q, and R ? 5. State the Fundamental Theorem for Line Integrals. 6. (a) What does it mean to say that xC F dr is independent
of path? (b) If you know that xC F dr is independent of path, what can you say about F ? 7. State Green’s Theorem.
9. Suppose F is a vector field on ⺢3.
(a) Define curl F. (b) Define div F. (c) If F is a velocity field in fluid flow, what are the physical interpretations of curl F and div F ? 10. If F P i Q j, how do you test to determine whether F
is conservative? What if F is a vector field on ⺢3 ?
11. (a) What is a parametric surface? What are its grid curves?
(b) Write an expression for the area of a parametric surface. (c) What is the area of a surface given by an equation z tx, y? 12. (a) Write the definition of the surface integral of a scalar
function f over a surface S. (b) How do you evaluate such an integral if S is a parametric surface given by a vector function ru, v? (c) What if S is given by an equation z tx, y? (d) If a thin sheet has the shape of a surface S, and the density at x, y, z is x, y, z, write expressions for the mass and center of mass of the sheet. 13. (a) What is an oriented surface? Give an example of a non-
orientable surface. (b) Define the surface integral (or flux) of a vector field F over an oriented surface S with unit normal vector n. (c) How do you evaluate such an integral if S is a parametric surface given by a vector function ru, v? (d) What if S is given by an equation z tx, y? 14. State Stokes’ Theorem. 15. State the Divergence Theorem. 16. In what ways are the Fundamental Theorem for Line
8. Write expressions for the area enclosed by a curve C in
terms of line integrals around C.
Integrals, Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem similar?
798
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CHAPTER 13
VECTOR CALCULUS
T R U E - FA L S E Q U I Z 5. If F P i Q j and Py Q x in an open region D, then F
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
is conservative. 6.
1. If F is a vector field, then div F is a vector field. 2. If F is a vector field, then curl F is a vector field.
xC
f x, y ds xC f x, y ds
7. If S is a sphere and F is a constant vector field, then
xxS F dS 0.
3. If f has continuous partial derivatives of all orders on ⺢ , 3
then divcurl ( f 0.
8. There is a vector field F such that
4. If f has continuous partial derivatives on ⺢ 3 and C is any
circle, then xC ( f dr 0.
curl F x i y j z k
EXERCISES 1. A vector field F, a curve C, and a point P are shown.
10. Find the work done by the force field
(a) Is xC F dr positive, negative, or zero? Explain. (b) Is div FP positive, negative, or zero? Explain.
Fx, y, z z i x j y k in moving a particle from the point 3, 0, 0 to the point 0, 2, 3 along (a) A straight line (b) The helix x 3 cos t, y t, z 3 sin t
y
■ Show that F is a conservative vector field. Then find a function f such that F ∇ f .
11–12 C
11. Fx, y 1 x ye xy i e y x 2e xy j x
12. Fx, y, z sin y i x cos y j sin z k ■
P
■
■
■
■
■
■
■
■
■
■
■
■ Show that F is conservative and use this fact to evaluate xC F dr along the given curve.
13–14
13. Fx, y 4 x 3 y 2 2 x y 3 i 2 x 4 y 3 x 2 y 2 4y 3 j, 2–9 2.
■
C: rt t sin t i 2t cos t j, 0 t 1
Evaluate the line integral.
xC x ds,
14. Fx, y, z e y i xe y e z j ye z k,
C is the arc of the parabola y x 2 from (0, 0) to (1, 1) 3.
C is the line segment from 0, 2, 0 to 4, 0, 3
xC yz cos x ds ,
■
C: x t, y 3 cos t, z 3 sin t, 0 t
4.
xC y dx x y 2 dy,
5.
xC y dx x dy , C is the arc of the parabola x 1 y from 0, 1 to 0, 1
6.
xC sxy dx e
7.
xC x y dx y 2 dy yz dz,
2
2
■
9.
xC F dr,
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■
■
■
■
■
■
■
■
■
■
■
■
y x 2 from 1, 1 to 1, 1 and the line segment from 1, 1 to 1, 1.
17. Use Green’s Theorem to evaluate xC x 2 y dx x y 2 dy,
where C is the circle x 2 y 2 4 with counterclockwise orientation.
18. Find curl F and div F if
Fx, y, z ex sin y i ey sin z j ez sin x k
where Fx, y, z e z i xz j x y k and C is given by rt t 2 i t 3 j t k, 0 t 1 ■
■
where C is the triangle with vertices 0, 0, 1, 0, and 1, 3.
where Fx, y x y i x 2 j and C is given by rt sin t i 1 t j, 0 t
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16. Use Green’s Theorem to evaluate xC s1 x 3 dx 2 xy dy
y
C is the line segment from 1, 0, 1, to 3, 4, 2
xC F dr,
■
xC xy 2 dx x 2 y dy, where C consists of the parabola
dy xz dz, C is given by rt t 4 i t 2 j t 3 k, 0 t 1
8.
■
15. Verify that Green’s Theorem is true for the line integral
C is the ellipse 4x 2 9y 2 36 with counterclockwise orientation 3
■
■
19. Show that there is no vector field G such that ■
curl G 2 x i 3yz j xz 2 k.
CHAPTER 13
20. Show that, under conditions to be stated on the vector fields
F and G, curlF G F div G G div F G ( F F ( G 21. If C is any piecewise-smooth simple closed plane curve
32. Use Stokes’ Theorem to evaluate xxS curl F dS, where
Fx, y, z x 2 yz i yz 2 j z 3e xy k, S is the part of the sphere x 2 y 2 z 2 5 that lies above the plane z 1, and S is oriented upward.
34. Use the Divergence Theorem to calculate the surface
( 2 ft f ( 2t t( 2 f 2( f (t
integral xxS F dS, where Fx, y, z x 3 i y 3 j z 3 k and S is the surface of the solid bounded by the cylinder x 2 y 2 1 and the planes z 0 and z 2.
23. If f is a harmonic function, that is, ( f 0, show that the 2
line integral x fy dx fx dy is independent of path in any simple region D.
35. Verify that the Divergence Theorem is true for the vector
field Fx, y, z x i y j z k, where E is the unit ball x 2 y 2 z 2 1.
24. (a) Sketch the curve C with parametric equations
0 t 2
z sin t
(b) Find xC 2 xe 2y dx 2 x 2e 2y 2y cot z dy y 2 csc 2z dz.
36. Compute the outward flux of
Fx, y, z
25. Find the area of the part of the surface z x 2 2y that lies
above the triangle with vertices 0, 0, 1, 0, and 1, 2.
37. Find xxS F n dS, where Fx, y, z x i y j z k and
4, 2, 1 to the parametric surface S given by
;
0 u 3 , 3 v 3
(b) Use a computer to graph the surface S and the tangent plane found in part (a). (c) Set up, but do not evaluate, an integral for the surface area of S. (d) If
CAS
Fx, y, z
S is the outwardly oriented surface shown in the figure (the boundary surface of a cube with a unit corner cube removed). z
(0, 2, 2) (2, 0, 2)
x2 y2 z2 k 2 i 2 j 1x 1y 1 z2
1
find xxS F dS correct to four decimal places. 27–30 27.
28.
29.
30.
■
■
xiyjzk x 2 y 2 z 2 32
through the ellipsoid 4 x 2 9y 2 6z 2 36.
26. (a) Find an equation of the tangent plane at the point
ru, v v 2 i u v j u 2 k
1
1
y
Evaluate the surface integral. x
xxS z dS,
where S is the part of the paraboloid z x 2 y 2 that lies under the plane z 4
S
(2, 2, 0)
38. Let
xxS x 2 z y 2 z dS,
where S is the part of the plane z 4 x y that lies inside the cylinder x 2 y 2 4
Fx, y
xxS F dS, where Fx, y, z x z i 2y j 3x k and S is the sphere x 2 y 2 z 2 4 with outward orientation
Evaluate x䊊C F dr, where C is shown in the figure.
xxS F dS, where Fx, y, z x 2 i x y j z k and S is the part of the paraboloid z x 2 y 2 below the plane z 1 with upward orientation ■
■
■
■
■
■
■
■
■
■
31. Verify that Stokes’ Theorem is true for the vector field
Fx, y, z x i y j z k where S is the part of the paraboloid z 1 x 2 y 2 that lies above the xy-plane and S has upward orientation. 2
799
Fx, y, z x y i yz j z x k and C is the triangle with vertices 1, 0, 0, 0, 1, 0, and 0, 0, 1, oriented counterclockwise as viewed from above.
22. If f and t are twice differentiable functions, show that
y sin t
■
33. Use Stokes’ Theorem to evaluate xC F dr, where
and f and t are differentiable functions, show that xC f x dx t y dy 0 .
x cos t
REVIEW
2
2
2 x 3 2 x y 2 2y i 2y 3 2 x 2 y 2 x j x2 y2
y
C
■
0
x
APPENDIX A
A
TRIGONOMETRY
■
A1
TRIGONOMETRY ANGLES
Angles can be measured in degrees or in radians (abbreviated as rad). The angle given by a complete revolution contains 360 , which is the same as 2 rad. Therefore
rad 180
1
and 1 rad
2
180 57.3
1
rad 0.017 rad 180
EXAMPLE 1
(a) Find the radian measure of 60 .
(b) Express 54 rad in degrees.
SOLUTION
(a) From Equation 1 or 2 we see that to convert from degrees to radians we multiply by 180. Therefore 60 60 rad 180 3
(b) To convert from radians to degrees we multiply by 180. Thus
5 5 180 rad 4 4
225
■
In calculus we use radians to measure angles except when otherwise indicated. The following table gives the correspondence between degree and radian measures of some common angles. Degrees
0°
30°
45°
60°
90°
120°
135°
150°
180°
270°
360°
Radians
0
6
4
3
2
2 3
3 4
5 6
3 2
2
a r ¨ r
Figure 1 shows a sector of a circle with central angle and radius r subtending an arc with length a. Since the length of the arc is proportional to the size of the angle, and since the entire circle has circumference 2 r and central angle 2, we have a 2 2r Solving this equation for and for a, we obtain
FIGURE 1
r
3
a r
a r
r 1 rad
r
FIGURE 2
Remember that Equations 3 are valid only when is measured in radians. In particular, putting a r in Equation 3, we see that an angle of 1 rad is the angle subtended at the center of a circle by an arc equal in length to the radius of the circle (see Figure 2).
A2
■
APPENDIX A
TRIGONOMETRY
EXAMPLE 2
(a) If the radius of a circle is 5 cm, what angle is subtended by an arc of 6 cm? (b) If a circle has radius 3 cm, what is the length of an arc subtended by a central angle of 38 rad? SOLUTION
(a) Using Equation 3 with a 6 and r 5, we see that the angle is
65 1.2 rad (b) With r 3 cm and 38 rad, the arc length is
y
3 8
a r 3
terminal side
¨
initial side x
0
FIGURE 3 ¨ ˘0 y
initial side 0
x
¨
9 cm 8
■
The standard position of an angle occurs when we place its vertex at the origin of a coordinate system and its initial side on the positive x-axis as in Figure 3. A positive angle is obtained by rotating the initial side counterclockwise until it coincides with the terminal side. Likewise, negative angles are obtained by clockwise rotation as in Figure 4. Figure 5 shows several examples of angles in standard position. Notice that different angles can have the same terminal side. For instance, the angles 34 , 54 , and 114 have the same initial and terminal sides because
terminal side
3 5 2 4 4
3 11 2 4 4
and 2 rad represents a complete revolution. FIGURE 4 ¨<0 y
y
FIGURE 5
x
0
x
¨=_
Angles in standard position
0
y
¨=11π 4
3π ¨= 4
¨=1 0
y
y
0 x
x
π 2
¨=_
0
x
5π 4
THE TRIGONOMETRIC FUNCTIONS
hypotenuse
opposite
For an acute angle the six trigonometric functions are defined as ratios of lengths of sides of a right triangle as follows (see Figure 6).
¨ adjacent
4
sin
opp hyp
csc
hyp opp
cos
adj hyp
sec
hyp adj
tan
opp adj
cot
adj opp
FIGURE 6
This definition doesn’t apply to obtuse or negative angles, so for a general angle in standard position we let Px, y be any point on the terminal side of and we let r
APPENDIX A
y
TRIGONOMETRY
■
A3
be the distance OP as in Figure 7. Then we define
P (x, y)
sin
y r
csc
r y
cos
x r
sec
r x
tan
y x
cot
x y
5
r
¨ O
x
FIGURE 7
Since division by 0 is not defined, tan and sec are undefined when x 0 and csc and cot are undefined when y 0. Notice that the definitions in (4) and (5) are consistent when is an acute angle. If is a number, the convention is that sin means the sine of the angle whose radian measure is . For example, the expression sin 3 implies that we are dealing with an angle of 3 rad. When finding a calculator approximation to this number we must remember to set our calculator in radian mode, and then we obtain sin 3 0.14112 π 4
œ„ 2
π 3
2 1
π 4
1
π 6
1
If we want to know the sine of the angle 3 we would write sin 3 and, with our calculator in degree mode, we find that sin 3 0.05234
œ„ 3
The exact trigonometric ratios for certain angles can be read from the triangles in Figure 8. For instance,
FIGURE 8
y
sin ¨>0
all ratios>0
0
sin
1 4 s2
sin
1 6 2
sin
s3 3 2
cos
1 4 s2
cos
s3 6 2
cos
1 3 2
tan
1 4
tan
1 6 s3
tan
s3 3
x
tan ¨>0
cos ¨>0
FIGURE 9
The signs of the trigonometric functions for angles in each of the four quadrants can be remembered by means of the rule “All Students Take Calculus” shown in Figure 9. EXAMPLE 3 Find the exact trigonometric ratios for 23. SOLUTION From Figure 10 we see that a point on the terminal line for 23 is
P (1, s3 ). Therefore, taking
y P {_1, œ„ 3}
x 1
2π 3
π 3
1 FIGURE 10
r2
in the definitions of the trigonometric ratios, we have
2
3 œ„
y s3
0
sin
2 s3 3 2
cos
2 1 3 2
tan
2 s3 3
csc
2 2 3 s3
sec
2 2 3
cot
2 1 3 s3
x
■
A4
■
APPENDIX A
TRIGONOMETRY
The following table gives some values of sin and cos found by the method of Example 3.
0
6
4
3
2
2 3
3 4
5 6
3 2
2
sin
0
1 2
1 s2
s3 2
1
s3 2
1 s2
1 2
0
1
0
cos
1
s3 2
1 s2
1 2
0
1
0
1
1 2
1 s2
s3 2
EXAMPLE 4 If cos 5 and 0 2, find the other five trigonometric func2
tions of .
SOLUTION Since cos 5 , we can label the hypotenuse as having length 5 and the 2
5
x=œ„„ 21
adjacent side as having length 2 in Figure 11. If the opposite side has length x, then the Pythagorean Theorem gives x 2 4 25 and so x 2 21, or x s21. We can now use the diagram to write the other five trigonometric functions: s21 5
sin
¨ 2
csc
FIGURE 11
5 s21
sec
tan
s21 2
5 2
cot
2 s21
■
16
EXAMPLE 5 Use a calculator to approximate the value of x in Figure 12. SOLUTION From the diagram we see that x
tan 40
40°
FIGURE 12
x
Therefore
16 x
16 19.07 tan 40
■
TRIGONOMETRIC IDENTITIES
A trigonometric identity is a relationship among the trigonometric functions. The most elementary are the following, which are immediate consequences of the definitions of the trigonometric functions.
6
csc
1 sin tan
sec sin cos
1 cos cot
cot
1 tan
cos sin
For the next identity we refer back to Figure 7. The distance formula (or, equivalently, the Pythagorean Theorem) tells us that x 2 y 2 r 2. Therefore sin 2 cos 2
y2 x2 x2 y2 r2 2 1 2 2 2 r r r r
APPENDIX A
TRIGONOMETRY
■
A5
We have therefore proved one of the most useful of all trigonometric identities: 7
sin 2 cos 2 1
If we now divide both sides of Equation 7 by cos 2 and use Equations 6, we get 8
tan 2 1 sec 2
Similarly, if we divide both sides of Equation 7 by sin 2, we get 9
1 cot 2 csc 2
The identities
■ Odd functions and even functions are discussed in Section 1.1.
10a
sin sin
10b
cos cos
show that sin is an odd function and cos is an even function. They are easily proved by drawing a diagram showing and in standard position (see Exercise 39). Since the angles and 2 have the same terminal side, we have 11
sin 2 sin
cos 2 cos
These identities show that the sine and cosine functions are periodic with period 2. The remaining trigonometric identities are all consequences of two basic identities called the addition formulas: 12a
sinx y sin x cos y cos x sin y
12b
cosx y cos x cos y sin x sin y
The proofs of these addition formulas are outlined in Exercises 85, 86, and 87. By substituting y for y in Equations 12a and 12b and using Equations 10a and 10b, we obtain the following subtraction formulas: 13a
sinx y sin x cos y cos x sin y
13b
cosx y cos x cos y sin x sin y
Then, by dividing the formulas in Equations 12 or Equations 13, we obtain the corresponding formulas for tanx y:
14a
tanx y
tan x tan y 1 tan x tan y
14b
tanx y
tan x tan y 1 tan x tan y
A6
■
APPENDIX A
TRIGONOMETRY
If we put y x in the addition formulas (12), we get the double-angle formulas: 15a
sin 2x 2 sin x cos x
15b
cos 2x cos 2 x sin 2 x
Then, by using the identity sin 2x cos 2x 1, we obtain the following alternate forms of the double-angle formulas for cos 2x : 16a
cos 2x 2 cos 2x 1
16b
cos 2x 1 2 sin 2 x
If we now solve these equations for cos 2x and sin 2x, we get the following half-angle formulas, which are useful in integral calculus:
17a
cos 2x
1 cos 2x 2
17b
sin 2x
1 cos 2x 2
Finally, we state the product formulas, which can be deduced from Equations 12 and 13: 18a
sin x cos y 12 sinx y sinx y
18b
cos x cos y 12 cosx y cosx y
18c
sin x sin y 12 cosx y cosx y
There are many other trigonometric identities, but those we have stated are the ones used most often in calculus. If you forget any of them, remember that they can all be deduced from Equations 12a and 12b. EXAMPLE 6 Find all values of x in the interval 0, 2 such that sin x sin 2x. SOLUTION Using the double-angle formula (15a), we rewrite the given equation as
sin x 2 sin x cos x
or
sin x 1 2 cos x 0
Therefore, there are two possibilities: sin x 0 x 0, , 2
or
1 2 cos x 0 cos x 12 x
5 , 3 3
The given equation has five solutions: 0, 3, , 53, and 2.
■
APPENDIX A
TRIGONOMETRY
■
A7
GRAPHS OF TRIGONOMETRIC FUNCTIONS
The graph of the function f x sin x, shown in Figure 13(a), is obtained by plotting points for 0 x 2 and then using the periodic nature of the function (from Equation 11) to complete the graph. Notice that the zeros of the sine function occur at the integer multiples of , that is, whenever x n,
sin x 0 y _
π 2
y
0 _1
1
3π 2
1
_π
n an integer
π 2
_π
π
2π
3π
5π 2
x
π 0
π _ 2 _1
π 2
3π 3π 2
x
5π 2
2π
(b) ©=cos x
(a) ƒ=sin x FIGURE 13
Because of the identity
cos x sin x
2
(which can be verified using Equation 12a), the graph of cosine is obtained by shifting the graph of sine by an amount 2 to the left [see Figure 13(b)]. Note that for both the sine and cosine functions the domain is , and the range is the closed interval 1, 1 . Thus, for all values of x, we have 1 sin x 1
1 cos x 1
The graphs of the remaining four trigonometric functions are shown in Figure 14 and their domains are indicated there. Notice that tangent and cotangent have range , , whereas cosecant and secant have range , 1 1, . All four functions are periodic: tangent and cotangent have period , whereas cosecant and secant have period 2. y
y
y
y
y=sin x 1 0
_π _
π 2
_
_1
π 2
(a) y=tan x FIGURE 14
π 3π 2
x
_π
_
π 2
0
π 2
(b) y=cot x
π
3π x 2
π 2
3π 2
1 0 π
π
_1 2
(c) y=csc x
y=cos x π _π _ 2
1
3π 2
0
x _1
π 2
(d) y=sec x
π
x
A8
■
APPENDIX A
TRIGONOMETRY
A 1–6
EXERCISES
Convert from degrees to radians.
■
1. 210
2. 300
3. 9
4. 315
5. 900
6. 36
■
■
7–12
■
■
■
■
■
■
■
■
■
35. ■
7. 4
8 10. 3
3 11. 8
■
■
■
■
■
■
36.
■
10 cm
25 cm 35°
5 9. 12
37.
38. 22 cm
12. 5 ■
■
■
■
2π 5
12 rad if the radius of the circle is 36 cm.
8 cm
14. If a circle has radius 10 cm, find the length of the arc sub■
tended by a central angle of 72 . center of the circle by an arc 1 m long? 16. Find the radius of a circular sector with angle 34 and arc
length 6 cm. Draw, in standard position, the angle whose measure
20. ■
■
■
■
23.
■
■
1 32. cos x , 3 33. cot 3, 34. csc ■
■
4 3
■
25.
■
■
9 2
■
■
■
■
■
30. tan 2,
■
■
■
■
50. 2 csc 2t sec t csc t
2 tan 1 tan 2 1 1 52. 2 sec 2 1 sin 1 sin
2
54. sin 2x sin 2 y sinx y sinx y 55.
3 2 2
sin csc cot 1 cos
56. tan x tan y ■
■
■
■
■
■
■
■
■
■
46. sin x cos x2 1 sin 2x
53. sin x sin 2x cos x cos 2x cos x
■
■
44. sin x sin x
51. tan 2
0
■
Prove the identity.
49. cot 2 sec 2 tan 2 csc 2
3 x 2
■
■
48. tan 2 sin 2 tan 2 sin 2
2
■
■
■
47. sec y cos y tan y sin y
2 4 , 3
■
45. sin cot cos
11 28. 4
■
3 , 0 5 2
31. sec 1.5,
■
x cos x 2
■
Find the remaining trigonometric ratios.
29. sin
■
43. sin
5 27. 6 ■
■
x sin x 2
24.
■
29–34
(b) Equation 18b
■
26. 5 ■
41. (a) Equation 18a
42. cos
■
■
(b) Equation 14b
22. 3 rad ■
■
40. (a) Equation 14a
21. 2 rad ■
■
(b) Equation 10b
42–58
■
■
Prove each equation.
18. 150
■
3 4
■
3 19. rad 4
Find the exact trigonometric ratios for the angle whose radian measure is given.
23–28
■
■
39. (a) Equation 10a
■
7 rad 3
■
(c) Equation 18c
is given. 17. 315
■
39– 41
15. A circle has radius 1.5 m. What angle is subtended at the
■
x 3π 8
x
■
13. Find the length of a circular arc subtended by an angle of
17–22
x 40°
x
Convert from radians to degrees. 7 8. 2
■
■ Find, correct to five decimal places, the length of the side labeled x.
35–38
sinx y cos x cos y
57. sin 3 sin 2 sin 2 cos
APPENDIX A
■
■
■
■
■
■
■
■
■
■
point C is located as in the figure and the following measurements were recorded:
■
59. sinx y
60. cosx y
61. cosx y
62. sinx y
63. sin 2y ■
A9
■
If sin x 13 and sec y 54, where x and y lie between 0 and 2, evaluate the expression.
59–64
■
84. In order to find the distance AB across a small inlet, a
58. cos 3 4 cos 3 3 cos ■
TRIGONOMETRY
■
■
■
■
■
■
■
BC 910 m
Use the Law of Cosines from Exercise 83 to find the required distance.
64. cos 2y ■
AC 820 m
⬔C 103
■
■
■
A
Find all values of x in the interval 0, 2 that satisfy the equation.
65–72
■
65. 2 cos x 1 0
66. 3 cot 2x 1
67. 2 sin2x 1
68.
69. sin 2x cos x
70. 2 cos x sin 2x 0
71. sin x tan x
72. 2 cos 2x 3 cos x
■
■
■
■
■
■
■
tan x 1 ■
■
■
■
C B ■
85. Use the figure to prove the subtraction formula
Find all values of x in the interval 0, 2 that satisfy the inequality.
73–76
■
73. sin x
75. 1 tan x 1 ■
■
cos cos cos sin sin
74. 2 cos x 1 0
1 2
■
■
76. sin x cos x ■
■
■
■
■
■
■
■
77– 82 ■ Graph the function by starting with the graphs in Figures 13 and 14 and applying the transformations of Section 1.3 where appropriate.
77. y cos x 3
78. y tan 2x
1 79. y tan x 3 2
80. y 1 sec x
81. y sin x ■
■
■
y
A (cos å, sin å) c B (cos ∫, sin ∫ )
1 1 å
82. y 2 sin x ■
[Hint: Compute c 2 in two ways (using the Law of Cosines from Exercise 83 and also using the distance formula) and compare the two expressions.]
■
■
■
■
■
■
■
4
■
83. Prove the Law of Cosines: If a triangle has sides with
lengths a, b, and c, and is the angle between the sides with lengths a and b, then c 2 a 2 b 2 2ab cos y
∫
0
x
86. Use the formula in Exercise 85 to prove the addition
formula for cosine (12b). 87. Use the addition formula for cosine and the identities
cos
sin 2
sin
cos 2
P (x, y)
to prove the subtraction formula for the sine function. b
88. Show that the area of a triangle with sides of lengths a and
c
b and with included angle is
¨ 0
(a, 0)
A 12 ab sin
x
89. Find the area of triangle ABC, correct to five decimal
[Hint: Introduce a coordinate system so that is in standard position as in the figure. Express x and y in terms of and then use the distance formula to compute c.]
places, if
AB 10 cm
BC 3 cm
⬔ABC 107
A10
■
APPENDIX B
B
PROOFS
PROOFS In this appendix we present proofs of several theorems that are stated in the main body of the text. The sections in which they occur are indicated in the margin.
SECTION 1.4
We start by proving the Triangle Inequality, which is an important property of absolute value. THE TRIANGLE INEQUALITY If a and b are any real numbers, then
a b a b Observe that if the numbers a and b are both positive or both negative, then the two sides in the Triangle Inequality are actually equal. But if a and b have opposite signs, the left side involves a subtraction and the right side does not. This makes the Triangle Inequality seem reasonable, but we can prove it as follows. PROOF Notice that
a a a
is always true because a equals either a or a . The corresponding statement for b is b b b
Adding these inequalities, we get
) a b a b If we now use the fact that x a &? a x a (with x replaced by a b and a by a b ), we obtain a b a b ( a b
■
which is what we wanted to show. LIMIT LAWS Suppose that c is a constant and the limits
lim f x L
xla
and
lim tx M
xla
exist. Then 1. lim f x tx L M
2. lim f x tx L M
3. lim cf x cL
4. lim f xtx LM
xla
xla
xla
5. lim
xla
f x L tx M
xla
if M 0
PROOF OF LAW 1 Let 0 be given. We must find 0 such that
if
0 xa
then
f x tx L M
APPENDIX B
PROOFS
■
A11
Using the Triangle Inequality we can write
f x tx L M f x L tx M
f x L tx M
1
We will make f x tx L M less than by making each of the terms f x L and tx M less than 2. Since 2 0 and lim x l a f x L , there exists a number 1 0 such that
0 x a 1
if
f x L 2
then
Similarly, since lim x l a tx M, there exists a number 2 0 such that
0 x a 2
if
tx M 2
then
Let min 1, 2 , the smaller of the numbers 1 and 2 . Notice that
0 xa
if
then
f x L 2
and so
0 x a 1
and
and
0 x a 2
tx M 2
Therefore, by (1),
f x tx L M f x L tx M 2 2 To summarize,
0 xa
if
f x tx L M
then
Thus, by the definition of a limit, lim f x tx L M
■
xla
PROOF OF LAW 4 Let 0 be given. We want to find 0 such that
f xtx LM In order to get terms that contain f x L and tx M , we add and subtract if
0 xa
then
Ltx as follows:
f xtx LM f xtx Ltx Ltx LM f x L tx L tx M
f x L tx L tx M f x L tx L tx M We want to make each of these terms less than 2.
(Triangle Inequality)
A12
■
APPENDIX B
PROOFS
Since lim x l a tx M , there is a number 1 0 such that
0 x a 1
if
tx M 2(1 L )
then
Also, there is a number 2 0 such that if 0 x a 2 , then
tx M 1
and therefore
tx tx M M tx M M 1 M Since lim x l a f x L, there is a number 3 0 such that
0 x a 3
if
f x L 2(1 M )
then
Let min 1, 2 , 3 . If 0 x a , then we have 0 x a 1, 0 x a 2 , and 0 x a 3 , so we can combine the inequalities to obtain
f xtx LM f x L tx L tx M
2(1 M
2 2
(1 M ) L 2(1 L ) )
This shows that lim x l a f xtx LM .
■
PROOF OF LAW 3 If we take tx c in Law 4, we get
lim cf x lim txf x lim tx lim f x
xla
xla
xla
xla
lim c lim f x xla
xla
c lim f x xla
■
(by Law 7)
PROOF OF LAW 2 Using Law 1 and Law 3 with c 1, we have
lim f x tx lim f x 1tx lim f x lim 1tx
xla
xla
xla
xla
lim f x 1 lim tx lim f x lim tx xla
xla
xla
xla
PROOF OF LAW 5 First let us show that
lim
xla
1 1 tx M
To do this we must show that, given 0, there exists 0 such that if
0 xa
then
1 1 tx M
■
APPENDIX B
Observe that
PROOFS
■
A13
M tx Mtx
1 1 tx M
We know that we can make the numerator small. But we also need to know that the denominator is not small when x is near a. Since lim x l a tx M , there is a number 1 0 such that, whenever 0 x a 1 , we have
M tx M 2 and therefore
M M tx tx M tx tx 2 tx M
This shows that
0 x a 1
if
tx 2 M
then
and so, for these values of x,
1 Mtx
1 M tx
1 M
2 2 2 M M
Also, there exists 2 0 such that if
0 x a 2
then
tx M
M tx 2 M Mtx
M2 2
Let min 1, 2 . Then, for 0 x a , we have
1 1 tx M
2
M2 2
It follows that lim x l a 1tx 1M . Finally, using Law 4, we obtain lim
xla
f x 1 lim f x xla tx tx
lim f x lim xla
xla
1 1 L L tx M M
If f x tx for all x in an open interval that contains a (except possibly at a) and 3 THEOREM
lim f x L
xla
and
lim tx M
xla
then L M . PROOF We use the method of proof by contradiction. Suppose, if possible, that
L M . Law 2 of limits says that lim tx f x M L
xla
Therefore, for any 0, there exists 0 such that if
0 xa
then
tx f x M L
■
A14
■
APPENDIX B
PROOFS
In particular, taking L M (noting that L M 0 by hypothesis), we have a number 0 such that
0 xa
if
tx f x M L L M
then
Since a a for any number a, we have
0 xa
if
tx f x M L L M
then
which simplifies to
0 xa
if
tx f x
then
But this contradicts f x tx. Thus the inequality L M must be false. Therefore, L M .
■
4 THE SQUEEZE THEOREM If f x tx hx for all x in an open interval that contains a (except possibly at a) and
lim f x lim hx L
xla
xla
lim tx L
then
xla
PROOF Let 0 be given. Since lim x l a f x L, there is a number 1 0 such
that
0 x a 1
if that is,
0 x a 1
if
f x L
then then
L f x L
Since lim x l a hx L, there is a number 2 0 such that
0 x a 2
if that is,
0 x a 2
if
hx L
then
then
L hx L
Let min 1, 2 . If 0 x a , then 0 x a 1 and 0 x a 2, so
L f x tx hx L L tx L
In particular,
and so tx L . Therefore, lim x l a tx L.
■
The proof of the following result was promised when we proved that lim
l0
THEOREM If 0 2, then tan .
sin 1.
APPENDIX B
D
PROOFS
■
A15
PROOF Figure 1 shows a sector of a circle with center O, central angle , and radius 1. Then
AD OA tan tan B S T
We approximate the arc AB by an inscribed polygon consisting of n equal line segments and we look at a typical segment PQ. We extend the lines OP and OQ to meet AD in the points R and S. Then we draw RT PQ as in Figure 1. Observe that
Q ° °
⬔RTO ⬔PQO 90 °° R
and so ⬔RTS 90 . Therefore, we have
P
O
1
PQ RT RS A
If we add n such inequalities, we get
L n AD tan
FIGURE 1
where L n is the length of the inscribed polygon. Thus, by Theorem 1.4.3, we have lim L n tan
nl
But the arc length is defined in Equation 7.4.1 as the limit of the lengths of inscribed polygons, so
lim L n tan
■
nl
SECTION 1.5
7 THEOREM
If f is continuous at b and lim x l a tx b, then lim f tx f b
xla
PROOF Let 0 be given. We want to find a number 0 such that
0 xa
if
then
f tx f b
Since f is continuous at b, we have lim f y f b
ylb
and so there exists 1 0 such that
0 y b 1
if
then
f y f b
Since lim x l a tx b, there exists 0 such that
tx b Combining these two statements, we see that whenever 0 x a we have tx b , which implies that f tx f b . Therefore, we have if
0 xa
then
1
1
proved that limx l a f tx f b.
■
A16
■
APPENDIX B
SECTION 3.1
PROOFS
We are going to use the Monotonic Sequence Theorem from Section 8.1 to prove the existence of the limit limx l 0 1 x1x that we used to define the number e. We will need the following result. LEMMA If 0 a b and n is a positive integer, then
b n n 1a nb a n1 PROOF We begin by factoring b n1 a n1. Since a b , we have
b n1 a n1 b ab n b n1a b n2a 2 ba n1 a n b ab n b n1b b n2b 2 bb n1 b n b ab n b n b n b n b n b an 1b n
We have shown that b n1 a n1 b an 1b n b n1 n 1b n b a a n1
so
Factoring b n from the left side of this inequality and simplifying, we get b n n 1a nb a n1
■
THEOREM The limit lim 1 x1x exists. xl0
PROOF Let
1 an 1 n
n
To show that a n is an increasing sequence, we put a 1 1n 1 and b 1 1n in the lemma:
1
1 n
n
n 1 1
1 n1
n 1
1 n
1
1 n1
Simplifying the left side of this inequality, we get
1
n
1 n
1
1 n1
n1
This says that a n a n1, that is, a n is an increasing sequence. Next we show that a n is a bounded sequence. If we let a 1 and b 1 12n in the lemma, we get
1
1 2n
n
n1n
1 2
1
n1
APPENDIX B
1 1 2n
so
n
2
1 1 2n
and
PROOFS
■
A17
2n
4
This says that a 2n 4 . Since a n is increasing, we have a n a 2n and so 0 an 4
for all n
Thus a n is a bounded sequence. It follows froom the Monotonic Sequence Theorem that a n is convergent. We denote its limit by e . Now if x is any real number that satisfies 1 1 x n1 n then
1 n1
1
1
n
1 x1x 1
1 n
n1
For instance, the right side of (1) is true because 1 x 1 1n and x
1 n1
But lim
nl
1 n1 x
?
n1
1 n
1
lim
nl
lim
nl
Similarly lim
nl
?
1 x1x 1
1
n
1 n
1
lim
nl
1 1 n1
n1
1 n
n
1 1 n
1 n
1
1 n
e1e
n
e
It follows from the pair of inequalities in (1) that lim 1 x1x exists
x l0
Finally, if x 0, let t x . Then
1 x1x 1 t1t
t 1 1t
1 1t
1t
t 1 1t
As t l 0, we have t1 t l 0 and lim 1 x
1x
x l0
lim t l0
1t
t 1 1t
1tt
1
1tt
1
t 1t
t 1t
e1e
We have therefore shown that lim 1 x1x exists
xl0
■
A18
■
APPENDIX B
PROOFS
6 THEOREM If f is a one-to-one continuous function defined on an interval a, b, then its inverse function f 1 is also continuous.
SECTION 3.2
PROOF First we show that if f is both one-to-one and continuous on a, b, then it must be either increasing or decreasing on a, b. If it were neither increasing nor decreasing, then there would exist numbers x 1 , x 2 , and x 3 in a, b with x 1 x 2 x 3 such that f x 2 does not lie between f x 1 and f x 3 . There are two possibilities: either (1) f x 3 lies between f x 1 and f x 2 or (2) f x 1 lies between f x 2 and f x 3 . (Draw a picture.) In case (1) we apply the Intermediate Value Theorem to the continuous function f to get a number c between x 1 and x 2 such that f c f x 3 . In case (2) the Intermediate Value Theorem gives a number c between x 2 and x 3 such that f c f x 1 . In either case we have contradicted the fact that f is one-to-one. Let us assume, for the sake of definiteness, that f is increasing on a, b. We take any number y0 in the domain of f 1 and we let f 1y0 x 0 ; that is, x 0 is the number in a, b such that f x 0 y0. To show that f 1 is continuous at y0 we take any 0 such that the interval x 0 , x 0 is contained in the interval a, b. Since f is increasing, it maps the numbers in the interval x 0 , x 0 onto the numbers in the interval f x 0 , f x 0 and f 1 reverses the correspondence. If we let denote the smaller of the numbers 1 y0 f x 0 and 2 f x 0 y0, then the interval y0 , y0 is contained in the interval f x 0 , f x 0 and so is mapped into the interval x 0 , x 0 by f 1. (See the arrow diagram in Figure 2.) We have therefore found a number 0 such that if y y0 then f 1y f 1y0
f(x¸-∑)
f(x¸+∑)
y¸
{
}
∂¡ f
FIGURE 2
f
f –!
{
{
a
x¸-∑
y
∂™
x¸
}
}
x¸+∑
b
x
This shows that lim y l y f 1y f 1y0 and so f 1 is continuous at any number y0 in its domain. ■ 0
SECTION 3.7
In order to give the promised proof of l’Hospital’s Rule we first need a generalization of the Mean Value Theorem. The following theorem is named after the French mathematician, Augustin-Louis Cauchy (1789–1857). CAUCHY’S MEAN VALUE THEOREM Suppose that the functions f and t are
continuous on a, b and differentiable on a, b, and tx 0 for all x in a, b. Then there is a number c in a, b such that f c f b f a tc tb ta
Notice that if we take the special case in which tx x, then tc 1 and Cauchy’s Mean Value Theorem is just the ordinary Mean Value Theorem. Furthermore,
APPENDIX B
PROOFS
■
A19
it can be proved in a similar manner. You can verify that all we have to do is change the function h given by Equation 4.2.4 to the function hx f x f a
f b f a tx ta tb ta
and apply Rolle’s Theorem as before. L’HOSPITAL’S RULE Suppose f and t are differentiable and tx 0 on an
open interval I that contains a (except possibly at a). Suppose that lim f x 0
and
lim f x
and
xla
or that
xla
lim tx 0
xla
lim tx
xla
(In other words, we have an indeterminate form of type 00 or .) Then lim
xla
f x f x lim x l a tx tx
if the limit on the right side exists (or is or ). PROOF OF L’HOSPITAL’S RULE We are assuming that lim x l a f x 0 and
lim x l a tx 0. Let
L lim
xla
f x tx
We must show that lim x l a f xtx L. Define Fx
f x if x a 0 if x a
Gx
tx 0
if x a if x a
Then F is continuous on I since f is continuous on x I x a and lim Fx lim f x 0 Fa
xla
xla
Likewise, G is continuous on I . Let x I and x a. Then F and G are continuous on a, x and differentiable on a, x and G 0 there (since F f and G t ). Therefore, by Cauchy’s Mean Value Theorem there is a number y such that a y x and Fy Fx Fa Fx Gy Gx Ga Gx Here we have used the fact that, by definition, Fa 0 and Ga 0. Now, if we let x l a, then y l a (since a y x), so lim
x la
f x Fx Fy f y lim lim lim L x la y la y la tx Gx Gy ty
A20
■
APPENDIX B
PROOFS
A similar argument shows that the left-hand limit is also L. Therefore lim
xla
f x L tx
This proves l’Hospital’s Rule for the case where a is finite. If a is infinite, we let t 1x. Then t l 0 as x l , so we have lim
xl
f x f 1t lim t l0 t1t tx lim
f 1t1t 2 t1t1t 2
lim
f 1t f x lim x l tx t1t
t l0
t l0
(by l’Hospital’s Rule for finite a)
■
CONCAVITY TEST
SECTION 4.3
(a) If f x 0 for all x in I , then the graph of f is concave upward on I . (b) If f x 0 for all x in I , then the graph of f is concave downward on I .
y
PROOF OF (a) Let a be any number in I . We must show that the curve y f x lies
y=ƒ
above the tangent line at the point a, f a. The equation of this tangent is y f a f ax a
ƒ
0
FIGURE 3
a
f(a)+f ª(a)(x-a)
x
So we must show that f x f a f ax a
x
whenever x I x a. (See Figure 3.) First let us take the case where x a. Applying the Mean Value Theorem to f on the interval a, x , we get a number c, with a c x, such that 1
f x f a f cx a
Since f 0 on I we know from the Increasing/ Decreasing Test that f is increasing on I . Thus, since a c, we have f a f c and so, multiplying this inequality by the positive number x a, we get 2
f ax a f cx a
Now we add f a to both sides of this inequality: f a f ax a f a f cx a
APPENDIX B
PROOFS
■
A21
But from Equation 1 we have f x f a f cx a. So this inequality becomes f x f a f ax a
3
which is what we wanted to prove. For the case where x a we have f c f a, but multiplication by the negative number x a reverses the inequality, so we get (2) and (3) as before. ■ PROPERTY 5 OF INTEGRALS
SECTION 5.2
y
b
a
f x dx y f x dx y f x dx c
b
a
c
if all of these integrals exist. PROOF We first assume that a c b . Since we are assuming that xab f x dx
exists, we can compute it as a limit of Riemann sums using only partitions P that include c as one of the partition points. If P is such a partition, let P1 be the corresponding partition of a, c determined by those partition points of P that lie in a, c . Similarly, P2 will denote the corresponding partition of c, b . We introduce the notation P for the length of the longest subinterval in P, that is,
P max x max x , x , . . . , x Note that P P and P P . Thus, if P l 0, it follows that P l 0 and P l 0. If x 1 i n is the set of partition points for P and n k m, where k is the number of subintervals in a, c and m is the number of subintervals in c, b , then x 1 i k is the set of partition points for P . If we write t x for the partition points to the right of c , then t 1 j m is the set of i
1
1
2
n
2
2
1
i
i
j
1
kj
j
partition points for P2. Thus we have
a x 0 x 1 x k x k1 x n b c t1 tm b Choosing x*i x i and letting t j t j t j1, we compute xab f x dx as follows:
y
b
a
n
f x dx lim f x i x i lim P l 0 i1 P l 0 lim P l 0
lim P l 0
f x x
k
i1
m
i
i1
i1
n
f x i x i
f t t
f x i x i
j
j1
k
1
k
lim P l 0 2
f t t j
j1
y f x dx y f t dt c
j
a
Now suppose that c a b. By what we have already proved, we have
y
b
c
f x dx y f x dx y f x dx a
c
f x i x i
ik1
j
m
i
b
a
b
c
A22
■
APPENDIX B
PROOFS
Therefore
y
b
a
f x dx y f x dx y f x dx y f x dx y f x dx a
b
c
c
c
b
a
c
■
The proofs are similar for the remaining four orderings of a , b, and c . SECTION 8.5
In order to prove Theorem 8.5.3 we first need the following results.
THEOREM 1. If a power series
c n x n converges when x b (where b 0), then it
converges whenever x b . 2. If a power series c n x n diverges when x d (where d 0 ), then it diverges whenever x d .
PROOF OF 1 Suppose that
c n b n converges. Then, by Theorem 8.2.6, we have
lim n l c n b 0. According to Definition 8.1.2 with 1, there is a positive integer N such that cn b n 1 whenever n N . Thus for n N we have n
c x n
n
cn b nx n x cn b n bn b
n
x b
n
If x b , then xb 1, so xb n is a convergent geometric series. Therefore, by the Comparison Test, the series nN c n x n is convergent. Thus the series c n x n is absolutely convergent and therefore convergent. ■
c n d n diverges. If x is any number such that x d , then c n x cannot converge because, by part 1, the convergence of c n x n would imply the convergence of c n d n. Therefore, c n x n diverges whenever x d . PROOF OF 2 Suppose that n
■
THEOREM For a power series
c n x n there are only three possibilities:
1. The series converges only when x 0. 2. The series converges for all x.
3. There is a positive number R such that the series converges if x R and
diverges if x R.
PROOF Suppose that neither case 1 nor case 2 is true. Then there are nonzero numbers b and d such that c n x n converges for x b and diverges for x d. Therefore, the set S x c n x n converges is not empty. By the preceding theorem, the series diverges if x d , so x d for all x S. This says that d is an upper bound for the set S. Thus, by the Completeness Axiom (see Section 8.1), S has a least upper bound R. If x R, then x S, so c n x n diverges. If x R, then x is not an upper bound for S and so there exists b S such that b x . Since ■ b S, c n b n converges, so by the preceding theorem c n x n converges.
APPENDIX B
PROOFS
■
A23
For a power series cnx an there are only three possibilities: 1. The series converges only when x a. 3 THEOREM
2. The series converges for all x. 3. There is a positive number R such that the series converges if
and diverges if x a R.
x a R
PROOF If we make the change of variable u x a, then the power series becomes c n u n and we can apply the preceding theorem to this series. In case 3 we have convergence for u R and divergence for u R. Thus we have convergence for x a R and divergence for x a R. ■
SECTION 11.3
CLAIRAUT’S THEOREM Suppose f is defined on a disk D that contains the point a, b. If the functions fxy and fyx are both continuous on D, then fxya, b fyxa, b. PROOF For small values of h, h 0, consider the difference
h f a h, b h f a h, b f a, b h f a, b Notice that if we let tx f x, b h f x, b, then h ta h ta By the Mean Value Theorem, there is a number c between a and a h such that ta h ta tch h fxc, b h fxc, b Applying the Mean Value Theorem again, this time to fx , we get a number d between b and b h such that fxc, b h fxc, b fxyc, dh Combining these equations, we obtain h h 2 fxyc, d If h l 0, then c, d l a, b, so the continuity of fxy at a, b gives lim
hl0
h lim fxyc, d fxya, b c, d l a, b h2
Similarly, by writing h f a h, b h f a, b h f a h, b f a, b and using the Mean Value Theorem twice and the continuity of fyx at a, b, we obtain lim
hl0
It follows that fxya, b fyxa, b.
h fyxa, b h2 ■
A24
■
APPENDIX B
PROOFS
8 THEOREM If the partial derivatives fx and fy exist near a, b and are continuous at a, b, then f is differentiable at a, b.
SECTION 11.4
PROOF Let
z f a x, b y f a, b According to (11.4.7), to prove that f is differentiable at a, b we have to show that we can write z in the form z fxa, b x fya, b y 1 x 2 y
y (a+Îx, b+Îy)
where 1 and 2 l 0 as x, y l 0, 0. Referring to Figure 4, we write
(u, b+Îy) (a, b+Îy)
(a, √)
1
R
(a, b) 0
FIGURE 4
z f a x, b y f a, b y f a, b y f a, b
Observe that the function of a single variable x
tx f x, b y is defined on the interval a, a x and tx fxx, b y. If we apply the Mean Value Theorem to t, we get ta x ta tu x where u is some number between a and a x. In terms of f , this equation becomes f a x, b y f a, b y fxu, b y x This gives us an expression for the first part of the right side of Equation 1. For the second part we let hy f a, y. Then h is a function of a single variable defined on the interval b, b y and hy fya, y. A second application of the Mean Value Theorem then gives hb y hb hv y where v is some number between b and b y. In terms of f , this becomes f a, b y f a, b fya, v y We now substitute these expressions into Equation 1 and obtain z fxu, b y x fya, v y fxa, b x fxu, b y fxa, b x fya, b y fya, v fya, b y fxa, b x fya, b y 1 x 2 y
APPENDIX B
where
PROOFS
■
A25
1 fxu, b y fxa, b 2 fya, v fya, b
Since u, b y l a, b and a, v l a, b as x, y l 0, 0 and since fx and fy are continuous at a, b, we see that 1 l 0 and 2 l 0 as x, y l 0, 0. Therefore, f is differentiable at a, b. ■ SECTION 11.7
SECOND DERIVATIVES TEST Suppose the second partial derivatives of f are
continuous on a disk with center a, b, and suppose that fx a, b 0 and fy a, b 0 [that is, a, b is a critical point of f ]. Let D Da, b fxx a, bfyy a, b fxy a, b 2 (a) If D 0 and fxxa, b 0, then f a, b is a local minimum. (b) If D 0 and fxxa, b 0, then f a, b is a local maximum. (c) If D 0, then f a, b is not a local maximum or minimum. PROOF OF PART (a) We compute the second-order directional derivative of f in the
direction of u h, k . The first-order derivative is given by Theorem 11.6.3: Du f fx h fy k Applying this theorem a second time, we have Du2 f DuDu f
& & Du f h Du f k &x &y
fxx h fyx kh fxy h fyy kk fxx h2 2 fxy hk fyy k 2
(by Clairaut’s Theorem)
If we complete the square in this expression, we obtain
1
Du2 f fxx h
fxy k fxx
2
k2 fxx fyy f xy2 fxx
We are given that fxx a, b 0 and Da, b 0. But fxx and D fxx fyy fxy2 are continuous functions, so there is a disk B with center a, b and radius 0 such that fxx x, y 0 and Dx, y 0 whenever x, y is in B. Therefore, by looking at Equation 1, we see that Du2 f x, y 0 whenever x, y is in B. This means that if C is the curve obtained by intersecting the graph of f with the vertical plane through Pa, b, f a, b in the direction of u, then C is concave upward on an interval of length 2. This is true in the direction of every vector u, so if we restrict x, y to lie in B, the graph of f lies above its horizontal tangent plane at P. Thus, f x, y f a, b whenever x, y is in B. This shows that f a, b is a local minimum.
■
A26
■
APPENDIX C
C
SIGMA NOTATION
SIGMA NOTATION A convenient way of writing sums uses the Greek letter (capital sigma, corresponding to our letter S) and is called sigma notation.
This tells us to end with i=n. This tells us to add.
1 DEFINITION If a m , a m1, . . . , a n are real numbers and m and n are integers such that m n, then
n
μ ai i"m
n
a
This tells us to start with i=m.
i
a m a m1 a m2 a n1 a n
im
With function notation, Definition 1 can be written as n
f i f m f m 1 f m 2 f n 1 f n
im
Thus the symbol nim indicates a summation in which the letter i (called the index of summation) takes on consecutive integer values beginning with m and ending with n, that is, m, m 1, . . . , n. Other letters can also be used as the index of summation. EXAMPLE 1 4
(a)
i
2
12 2 2 3 2 42 30
i1 n
(b)
i 3 4 5 n 1 n
i3 5
(c)
2
j
2 0 2 1 2 2 2 3 2 4 2 5 63
j0 n
(d)
k1 3
(e)
i1
1 1 1 1 1 k 2 3 n i1 11 21 31 1 1 13 2 2 2 0 2 i 3 1 3 2 3 3 3 7 6 42
4
(f)
222228
■
i1
EXAMPLE 2 Write the sum 2 3 3 3 n 3 in sigma notation. SOLUTION There is no unique way of writing a sum in sigma notation. We could
write n
23 33 n 3
i
3
i2
n1
or
23 33 n 3
j 1
3
j1
n2
or
23 33 n 3
k 2
3
k0
■
APPENDIX C
SIGMA NOTATION
■
A27
The following theorem gives three simple rules for working with sigma notation. If c is any constant (that is, it does not depend on i ), then
2 THEOREM n
(a)
ca
n
i
c
im
a
im
n
(c)
a
n
bi
im
n
i
bi
im n
i
a
(b)
i
a
a
n
i
im
b
i
im
n
i
im
b
i
im
PROOF To see why these rules are true, all we have to do is write both sides in expanded form. Rule (a) is just the distributive property of real numbers:
ca m ca m1 ca n ca m a m1 a n Rule (b) follows from the associative and commutative properties: a m bm a m1 bm1 a n bn am am1 an bm bm1 bn ■
Rule (c) is proved similarly. n
EXAMPLE 3 Find
1.
i1
n
1 1 1 1 n
SOLUTION
■
i1
n terms
EXAMPLE 4 Prove the formula for the sum of the first n positive integers: n
i 1 2 3 n
i1
nn 1 2
SOLUTION This formula can be proved by mathematical induction (see page A28) or by the following method used by the German mathematician Karl Friedrich Gauss (1777–1855) when he was ten years old. Write the sum S twice, once in the usual order and once in reverse order:
S1
2
3
n 1 n
S n n 1 n 2
2
1
Adding all columns vertically, we get 2S n 1 n 1 n 1 n 1 n 1 On the right side there are n terms, each of which is n 1, so 2S nn 1
or
S
nn 1 2
■
A28
■
APPENDIX C
SIGMA NOTATION
EXAMPLE 5 Prove the formula for the sum of the squares of the first n positive
integers: n
i
12 2 2 3 2 n 2
2
i1
nn 12n 1 6
SOLUTION 1 Let S be the desired sum. We start with the telescoping sum (or col-
lapsing sum): n
Most terms cancel in pairs.
1 i
3
i 3 2 3 13 3 3 2 3 4 3 3 3 n 13 n 3
i1
n 13 13 n 3 3n 2 3n On the other hand, using Theorem 2 and Examples 3 and 4, we have n
n
1 i
3
i 3
i1
3i
n
2
3i 1 3
i1
3S 3
i
2
3
i1
n
n
i1
i1
i1
nn 1 n 3S 32 n 2 52 n 2
Thus we have n 3 3n 2 3n 3S 32 n 2 52 n Solving this equation for S, we obtain 3S n 3 32 n 2 12 n S
or PRINCIPLE OF MATHEMATICAL INDUCTION Let Sn be a statement involving the positive integer n. Suppose that 1. S1 is true. 2. If Sk is true, then Sk1 is true. Then Sn is true for all positive integers n. ■
2n 3 3n 2 n nn 12n 1 6 6
SOLUTION 2 Let Sn be the given formula.
11 12 1 1 6 2. Assume that Sk is true; that is, 1. S1 is true because
12
12 2 2 3 2 k 2
kk 12k 1 6
Then 12 2 2 3 2 k 12 12 2 2 3 2 k 2 k 12
kk 12k 1 k2k 1 6k 1 k 12 k 1 6 6
k 1
2k 2 7k 6 k 1k 22k 3 6 6
k 1 k 1 1 2k 1 1 6
So Sk1 is true. By the Principle of Mathematical Induction, Sn is true for all n.
■
APPENDIX C
SIGMA NOTATION
■
A29
We list the results of Examples 3, 4, and 5 together with a similar result for cubes (see Exercises 37–40) as Theorem 3. These formulas are needed for finding areas and evaluating integrals in Chapter 5. Let c be a constant and n a positive integer. Then
3 THEOREM n
(a)
n
1n
c nc
(b)
i1 n
(c)
i
i1 n
(e)
i
3
i1
i1
nn 1 2
nn 12n 1 6
n
i
(d)
nn 1 2
2
i1
2
n
EXAMPLE 6 Evaluate
i4i
2
3.
i1
SOLUTION Using Theorems 2 and 3, we have n
i4i
n
2
3
i1
4i
n
3i 4
3
i1
4
nn 1 2
i
n
3
3
i1
2
3
i
i1
nn 1 2
nn 1 2nn 1 3 2 nn 12n 2 2n 3 2
n
The type of calculation in Example 7 arises in Chapter 5 when we compute areas. ■
EXAMPLE 7 Find lim
n l i1
SOLUTION n
lim
n l i1
3 n
3 n
i n
■
i n
2
1 .
2
n
1 lim
n l i1
lim
nl
lim
nl
lim
nl
lim
nl
3 2 3 3 i n n
3 n3
n
i2
i1
3 n
n
1
i1
3 nn 12n 1 3 n n3 6 n 1 n 2 n
n1 n
1 1 1 1 2 n
12 1 1 2 3 4
2n 1 n
2
1 n
3
3
■
A30
■
APPENDIX C
C 1–10
SIGMA NOTATION
EXERCISES
5
1.
2. 4.
i4
5.
k0
i
8.
■
■
■
j
■
10. ■
■
■
1 2
23 34 45
14.
3 7
48 59 106 23 27
f x x i
■
■
■
■
■
■
■
■
■
n
■
■
■
■
j1
4
ii 2 cos k
26.
n1
4
i1
n
(a)
i
i1 4
4
i
i 2
28.
i0
2i
30.
2 5i
n
B
32.
i1
100 4
i 14
1 1 i i1
(b)
3 2i 2
5
i
5 i1
i
a i1
i1
n
(d)
a
i1
42. Prove the generalized triangle inequality:
n
n
i 2 3i 4
99
(c)
i3
i1
n
i1
2 3i
n
i1
31.
i2
n
29.
. . .
5
41. Evaluate each telescoping sum.
100
1 n
2
G¢
3 G£ 2 G™ 1 A1 2 3 4
k0
20
27.
G∞
5
8
24.
. . .
■
i3
6
Gn
.. . ■
6
22.
i4
j1
C
D
Find the value of the sum.
3i 2
i 78.
published by Abu Bekr Mohammed ibn Alhusain Alkarchi in about AD 1010. The figure shows a square ABCD in which sides AB and AD have been divided into segments of lengths 1, 2, 3, . . . , n. Thus the side of the square has length nn 12 so the area is nn 12 2. But the area is also the sum of the areas of the n “gnomons” G1 , G2 , . . . , Gn shown in the figure. Show that the area of Gi is i 3 and conclude that formula (e) is true.
19 20
8
25.
■
40. Prove formula (e) of Theorem 3 using the following method
14 19 161 251 361
3
■
that of Example 5, Solution 1 [start with 1 i 4 i 4 .
20. 1 x x 2 x 3 1n x n
23.
■
39. Prove formula (e) of Theorem 3 using a method similar to
19. x x 2 x 3 x n
21.
■
induction. ■
17. 1 2 4 8 16 32
■
■
38. Prove formula (e) of Theorem 3 using mathematical
i
16. 1 3 5 7 2n 1
21–35
■
37. Prove formula (b) of Theorem 3.
15. 2 4 6 8 2n
■
■
i1
i1
13.
■
■
36. Find the number n such that
12. s3 s4 s5 s6 s7
1 1
■
2
11. 1 2 3 4 10
18.
■
n
Write the sum in sigma notation.
■
i 2
n
j
j0
11–20
k
jn
1
3
i1
n3 10
i1
■
x
i
k5
n1
9.
n
8
6.
ii 1i 2
i1
3
35.
n
7.
i
i4
2k 1 2k 1
4
n
34.
6
i
i 1i 2
i1
1
i1
i1
6
3
33.
6
si
i1
3.
n
Write the sum in expanded form.
■
ai
i1
n
a i
i1
APPENDIX D
43– 46
■
43. lim
n l i1 n
44. lim
n l i1 n
45. lim
n l i1 n
46. lim
n l i1
■
■
1 n 1 n 2 n 3 n
■
i n
n
ar
3
a ar ar 2 ar n1
i1
i1
1
3
2i n
1
3i n
■
3
2 1
■
D
n
■
■
i1 n
2i 2 . i
49. Evaluate
i1
3i n
■
ar n 1 r1
3 . 2 i1
48. Evaluate
2i n
5
A31
with first term a and common ratio r 1:
2
i n
■
47. Prove the formula for the sum of a finite geometric series
Find the limit. n
THE LOGARITHM DEFINED AS AN INTEGRAL
m
■
■
■
n
i j
50. Evaluate
i1
■
j1
.
THE LOGARITHM DEFINED AS AN INTEGRAL The treatment of exponential and logarithmic functions presented in Chapter 3 relied on our intuition, which is based on numerical and visual evidence. Here we use the Fundamental Theorem of Calculus to give an alternative treatment that provides a surer footing for these functions. Instead of starting with a x and defining log a x as its inverse, this time we start by defining ln x as an integral and then define the exponential function as its inverse. In this section you should bear in mind that we do not use any of our previous definitions and results concerning exponential and logarithmic functions. THE NATURAL LOGARITHM
y
We first define ln x as an integral. y= 1t 1 DEFINITION
The natural logarithmic function is the function defined by
area=ln x
ln x y
x
1
0
1
x
1 dt t
t
FIGURE 1
The existence of this function depends on the fact that the integral of a continuous function always exists. If x 1, then ln x can be interpreted geometrically as the area under the hyperbola y 1t from t 1 to t x. (See Figure 1.) For x 1, we have
y
area=_ ln x
ln 1 y
x
FIGURE 2
1
1
1
y= 1t 0
x0
t
For 0 x 1 ,
ln x y
x
1
1 dt 0 t
1 1 1 dt y dt 0 x t t
and so ln x is the negative of the area shown in Figure 2.
A32
■
APPENDIX D
THE LOGARITHM DEFINED AS AN INTEGRAL
V EXAMPLE 1
(a) By comparing areas, show that 12 ln 2 34 . (b) Use the Midpoint Rule with n 10 to estimate the value of ln 2. SOLUTION
(a) We can interpret ln 2 as the area under the curve y 1t from 1 to 2. From Figure 3 we see that this area is larger than the area of rectangle BCDE and smaller than the area of trapezoid ABCD. Thus we have
y
y= 1t
1 2
1 ln 2 1 12 (1 12 )
A
0
1 2
D
E B
(b) If we use the Midpoint Rule with f t 1t, n 10, and t 0.1, we get
C 1
2
3 ln 2 4
t
ln 2 y
FIGURE 3
2
1
1 dt 0.1 f 1.05 f 1.15 f 1.95 t
0.1
1 1 1 1.05 1.15 1.95
0.693
■
Notice that the integral that defines ln x is exactly the type of integral discussed in Part 1 of the Fundamental Theorem of Calculus (see Section 5.4). In fact, using that theorem, we have d dx
y
x
1
1 1 dt t x
and so d 1 ln x dx x
2
We now use this differentiation rule to prove the following properties of the logarithm function. 3 LAWS OF LOGARITHMS
If x and y are positive numbers and r is a rational
number, then 1. lnxy ln x ln y
2. ln
x y
ln x ln y
3. lnx r r ln x
PROOF 1. Let f x lnax, where a is a positive constant. Then, using Equation 2 and
the Chain Rule, we have f x
1 d 1 1 ax a ax dx ax x
Therefore, f x and ln x have the same derivative and so they must differ by a constant: lnax ln x C
APPENDIX D
THE LOGARITHM DEFINED AS AN INTEGRAL
■
A33
Putting x 1 in this equation, we get ln a ln 1 C 0 C C. Thus lnax ln x ln a If we now replace the constant a by any number y, we have lnxy ln x ln y 2. Using Law 1 with x 1y, we have
ln
1 1 ln y ln y ln 1 0 y y
and so
ln
1 ln y y
Using Law 1 again, we have
ln
x y
ln x
1 y
ln x ln
1 ln x ln y y ■
The proof of Law 3 is left as an exercise. In order to graph y ln x, we first determine its limits: 4
(a) lim ln x
(b) lim ln x
xl
xl0
PROOF
(a) Using Law 3 with x 2 and r n (where n is any positive integer), we have ln2 n n ln 2. Now ln 2 0, so this shows that ln2 n l as n l . But ln x is an increasing function since its derivative 1x 0. Therefore, ln x l as x l . (b) If we let t 1x, then t l as x l 0 . Thus, using (a), we have y
0
x l 0
x
1
tl
1 t
lim ln t
and
d2y 1 2 2 0 dx x
tl
■
If y ln x, x 0, then dy 1 0 dx x
which shows that ln x is increasing and concave downward on 0, . Putting this information together with (4), we draw the graph of y ln x in Figure 4. Since ln 1 0 and ln x is an increasing continuous function that takes on arbitrarily large values, the Intermediate Value Theorem shows that there is a number where ln x takes on the value 1 (see Figure 5). This important number is denoted by e.
FIGURE 4 y 1
0
lim ln x lim ln
y=ln x
1
e
x
5 DEFINITION
e is the number such that ln e 1.
y=ln x
FIGURE 5
We will show (in Theorem 19) that this definition is consistent with our previous definition of e.
A34
■
APPENDIX D
THE LOGARITHM DEFINED AS AN INTEGRAL
THE NATURAL EXPONENTIAL FUNCTION
Since ln is an increasing function, it is one-to-one and therefore has an inverse function, which we denote by exp. Thus, according to the definition of an inverse function, f 1x y
&?
f y x
6
expx y
&?
ln y x
and the cancellation equations are f 1 f x x f f 1x x
7
expln x x
and
lnexp x x
In particular, we have exp0 1 since ln 1 0 exp1 e y
y=x
lne r r ln e r
y=ln x 0
ln e 1
We obtain the graph of y exp x by reflecting the graph of y ln x about the line y x. (See Figure 6.) The domain of exp is the range of ln, that is, , ; the range of exp is the domain of ln, that is, 0, . If r is any rational number, then the third law of logarithms gives
y=exp x
1
since
x
1
Therefore, by (6),
expr e r
Thus expx e x whenever x is a rational number. This leads us to define e x, even for irrational values of x, by the equation FIGURE 6
e x expx In other words, for the reasons given, we define e x to be the inverse of the function ln x. In this notation (6) becomes 8
&?
ln y x
and the cancellation equations (7) become
y
y=´
1 0
ex y
1
FIGURE 7
The natural exponential function
x
9
e ln x x
10
lne x x
x0
for all x
The natural exponential function f x e x is one of the most frequently occurring functions in calculus and its applications, so it is important to be familiar with its graph (Figure 7) and its properties (which follow from the fact that it is the inverse of the natural logarithmic function).
APPENDIX D
THE LOGARITHM DEFINED AS AN INTEGRAL
■
A35
PROPERTIES OF THE EXPONENTIAL FUNCTION The exponential function
f x e x is an increasing continuous function with domain ⺢ and range 0, . Thus e x 0 for all x. Also lim e x 0
lim e x
x l
xl
So the x-axis is a horizontal asymptote of f x e x. We now verify that f has the other properties expected of an exponential function. 11 LAWS OF EXPONENTS
1. e xy e xe y
If x and y are real numbers and r is rational, then ex 2. e xy y 3. e x r e rx e
PROOF OF LAW 1 Using the first law of logarithms and Equation 10, we have
lne xe y lne x lne y x y lne xy Since ln is a one-to-one function, it follows that e xe y e xy. Laws 2 and 3 are proved similarly (see Exercises 6 and 7). As we will soon see, Law 3 actually holds when r is any real number. ■ We now prove the differentiation formula for e x.
12
d x e e x dx
PROOF The function y e x is differentiable because it is the inverse function of
y ln x, which we know is differentiable with nonzero derivative. To find its derivative, we use the inverse function method. Let y e x. Then ln y x and, differentiating this latter equation implicitly with respect to x, we get 1 dy 1 y dx dy y ex dx GENERAL EXPONENTIAL FUNCTIONS
If a 0 and r is any rational number, then by (9) and (11), a r e ln a r e r ln a Therefore, even for irrational numbers x, we define 13
a x e x ln a
■
A36
■
APPENDIX D
THE LOGARITHM DEFINED AS AN INTEGRAL
Thus, for instance, 2s3 e s3 ln 2 e1.20 3.32 The function f x a x is called the exponential function with base a. Notice that a x is positive for all x because e x is positive for all x. Definition 13 allows us to extend one of the laws of logarithms. We already know that lna r r ln a when r is rational. But if we now let r be any real number we have, from Definition 13, ln a r lne r ln a r ln a Thus ln a r r ln a
14
for any real number r
The general laws of exponents follow from Definition 13 together with the laws of exponents for e x.
15 LAWS OF EXPONENTS
1. a
y
xy
aa
x y
2. a
xy
If x and y are real numbers and a, b 0, then a xa y
3. a x y a xy
4. abx a xb x
PROOF 1. Using Definition 13 and the laws of exponents for e x, we have
a xy e xy ln a e x ln a y ln a e x ln ae y ln a a xa y
1
0
x
3. Using Equation 14 we obtain x
x
a x y e y lna e yx ln a e xy ln a a xy
lim a®=0, lim a®=` _`
x
`
The remaining proofs are left as exercises.
FIGURE 8
y=a®, a>1
■
The differentiation formula for exponential functions is also a consequence of Definition 13:
y
16
d a x a x ln a dx
1
0 x
x
PROOF
d d d a x e x ln a e x ln a x ln a a x ln a dx dx dx
■
lim a®=`, lim a®=0 _`
FIGURE 9
y=a®, 0
x `
If a 1, then ln a 0, so ddx a x a x ln a 0, which shows that y a x is increasing (see Figure 8). If 0 a 1, then ln a 0 and so y a x is decreasing (see Figure 9).
APPENDIX D
THE LOGARITHM DEFINED AS AN INTEGRAL
■
A37
GENERAL LOGARITHMIC FUNCTIONS
If a 0 and a 1, then f x a x is a one-to-one function. Its inverse function is called the logarithmic function with base a and is denoted by log a . Thus log a x y
17
&?
ay x
In particular, we see that log e x ln x The laws of logarithms are similar to those for the natural logarithm and can be deduced from the laws of exponents (see Exercise 10). To differentiate y log a x, we write the equation as a y x. From Equation 14 we have y ln a ln x, so log a x y
ln x ln a
Since ln a is a constant, we can differentiate as follows: d d ln x 1 d 1 log a x ln x dx dx ln a ln a dx x ln a d 1 log a x dx x ln a
18
THE NUMBER e EXPRESSED AS A LIMIT
In this section we defined e as the number such that ln e 1. The next theorem shows that this is the same as the number e defined in Section 3.1. e lim 1 x1x
19
xl0
PROOF Let f x ln x. Then f x 1x, so f 1 1. But, by the definition of derivative,
f 1 lim
hl0
lim
xl0
f 1 h f 1 f 1 x f 1 lim xl0 h x ln1 x ln 1 1 lim ln1 x xl0 x x
lim ln1 x1x xl0
A38
■
APPENDIX D
THE LOGARITHM DEFINED AS AN INTEGRAL
Because f 1 1, we have lim ln1 x1x 1
xl0
Then, by Theorem 1.5.7 and the continuity of the exponential function, we have e e1 e lim x l 0 ln1x lim e ln1x lim 1 x1x 1x
1x
xl0
D
xl0
EXERCISES
1. (a) By comparing areas, show that 1 3
ln 1.5
5 12
(b) Use the Midpoint Rule with n 10 to estimate ln 1.5. 2. Refer to Example 1.
(a) Find the equation of the tangent line to the curve y 1t that is parallel to the secant line AD. (b) Use part (a) to show that ln 2 0.66. 3. By comparing areas, show that
1 1 1 1 1 1 ln n 1 2 3 n 2 3 n1 4. (a) By comparing areas, show that ln 2 1 ln 3.
(b) Deduce that 2 e 3.
5. Prove the third law of logarithms. [Hint: Start by showing
that both sides of the equation have the same derivative.] 6. Prove the second law of exponents for e x [see (11)]. 7. Prove the third law of exponents for e x [see (11)]. 8. Prove the second law of exponents [see (15)]. 9. Prove the fourth law of exponents [see (15)]. 10. Deduce the following laws of logarithms from (15):
(a) log axy log a x log a y (b) log axy log a x log a y (c) log ax y y log a x
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
E
33. 5,
35. , 0 0, y
y ■
page 8
4
1. (a) 2
(b) 2.8 (c) 3, 1 (d) 2.5, 0.3 (e) 3, 3 , 2, 3 (f ) 1, 3 3. No 5. Yes, 3, 2 , 3, 2 1, 3 7. Diet, exercise, or illness 9.
A39
ANSWERS TO ODD-NUMBERED EXERCISES
CHAPTER 1 Exercises 1.1
■
T
2 0
0
x
x
5
37. ,
39. ,
y
y
(0, 2) (0, 1) t
0
11.
_2
0
1 x
1
1
x
0
T
41. f x 6 x 3 , 2 x 4 7
4
43. f x 1 sx
45. AL 10L L , 0 L 10 2
MIDNIGHT
13.
t
NOON
47. Ax s3 x 24, x 0 51. (a) R (%)
Amount
49. Sx x 2 8x, x 0
15 10
0 0
15.
Price
10,000
20,000
I (in dollars)
(b) $400, $1900 (c) T (in dollars)
Height of grass
2500 1000
Wed.
Wed.
Wed.
Wed.
Wed. t
0
17. 12, 16, 3a 2 a 2, 3a 2 a 2, 3a 2 5a 4,
6a 2a 4, 12a 2a 2, 3a a 2, 9a 4 6a 3 13a 2 4a 4, 3a 2 6ah 3h 2 a h 2 19. 3 h 21. 1ax 1 1 1 23. {x x 3 } ( , 3) ( 3 , ) 25. 0, 27. , 0 5, 29. , 31. , 2
2
4
2
53. f is odd, t is even 55. (a) 5, 3 (b) 5, 3 57. Odd 59. Neither 61. Even Exercises 1.2
■
page 21
1. (a) y 2x b,
where b is the y-intercept
y b=3 b=0
b=_1
y
y 5
0
10,000 20,000 30,000 I (in dollars)
y=2x+b
x
0 _9
6
t
x
A40
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
(b) y mx 1 2m, where m is the slope. See graph at right. (c) y 2x 3
y
25.
m=1
27.
y
m=_1
1
π
m=0
(2, 1)
2π x
0
x
0
y=2 cos x+1
x
y-1=m(x-2)
3. Their graphs have slope 1.
y
y=sin(x/2)
3
29.
y
31.
y=œ„„„„ x+3
y
y
c=_1 c=_2
_3
_4
x
0
0
x
_8 x
0
c=2 c=1
1
33.
y=_(≈+8x) 2
y
c=0
2 y= x+1
2
x=_1
5. f x 3xx 1x 2 7. (a) 8.34, change in mg for every 1 year change
(b) 8.34 mg 9. (a)
(b) 95 , change in F for every 1 C change; 32, Fahrenheit temperature corresponding to 0 C
F (100, 212)
F= 95 C+32 32
11. (a) T 6 N
(b) 16 , change in F for every chirp per minute change (c) 76 F 13. (a) P 0.434d 15 (b) 196 ft 15. (a) y f x 3 (b) y f x 3 (c) y f x 3 (d) y f x 3 (e) y f x (f) y f x (g) y 3f x (h) y 13 f x 17. (a) 3 (b) 1 (c) 4 (d) 5 (e) 2 19. (a) (b) y y 1
307 6
x
0 x
0
(c)
0
21.
y
y
x
0
23.
x
x
y
y=_x# 0
f tx x 3 x 2 1, , ftx 3x 5 6x 4 x 3 2x 2, , ftx x 3 2x 2 3x 2 1, {x x 1s3 } 37. (a) f ⴰ tx 4x 2 4x, , (b) t ⴰ f x 2x 2 1, , (c) f ⴰ f x x 4 2x 2, , (d) t ⴰ tx 4x 3, , 39. (a) f ⴰ tx sin(1 sx ), 0, (b) t ⴰ f x 1 ssin x,
x x 2n, 2n , n an integer (c) f ⴰ f x sinsin x, , (d) t ⴰ tx 1 s1 sx, 0, 1 41. (a) f ⴰ tx 2x 2 6x 5 x 2x 1 ,
x x 2, 1 (b) t ⴰ f x x 2 x 1x 12, {x x 1, 0 (c) f ⴰ f x x 4 3x 2 1 xx 2 1 , {x x 0 5 (d) t ⴰ tx 2x 33x 5 , {x x 2, 3 }
(d)
y
35. f tx x 3 5x 2 1, ,
C
(_40, _40)
x
0
1
_1
y=(x+1)@
0
x
f ⴰ t ⴰ hx sx 2 6x 10 47. tt cos t, f t st tx x 2 1, f x x 10 hx x 2, tx 3 x, f x 1 x hx sx, tx sec x, f x x 4 (a) 4 (b) 3 (c) 0 (d) Does not exist; f 6 6 is not in the domain of t. (e) 4 (f) 2 55. (a) rt 60t (b) A ⴰ rt 3600 t 2; the area of the circle as a function of time 57. (a) (b) H V 43. 45. 49. 51. 53.
1
0
120
t
0
Vt 120Ht
t
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
(c)
Exercises 1.5
V
Vt 240Ht 5
59. 61. 63. 65.
xla
does not exist (b) 4, neither; 2, left; 2, right; 4, right
t
5
5.
Yes; m1 m 2 (a) f (x x 2 6 (b) tx x 2 x 1 (a) Even; even (b) Odd; even Yes
Exercises 1.3
page 54
■
1. lim x l 4 f x f 4 3. (a) f 4 is not defined and lim f x [for a 2 , 2 , and 4 ]
240
0
y
1 0
x
2
page 33
■
1. (a) (i) 32 fts (ii) 25.6 fts (iii) 24.8 fts (iv) 24.16 fts (b) 24 fts 3. (a) 2 (b) 3 (c) Does not exist (d) 4 (e) Does not exist 5. (a) 1 (b) 2 (c) Does not exist (d) 2 (e) 0 (f) Does not exist (g) 1 (h) 3 y 7. . 9. y
7. (a)
Cost (in dollars)
1 0 1
x
0
x
1
Time (in hours)
(b) Discontinuous at t 1, 2, 3, 4 9. 6 13. f 1 is not defined 15.
lim f x does not exist
xl1
y 2
1
1
y
3
11. 3 13. 2 15. 4 17. 5 19. (a) 4 21. (a) 0.998000, 0.638259, 0.358484, 0.158680, 0.038851,
0.008928, 0.001465; 0 (b) 0.000572, 0.000614, 0.000907, 0.000978, 0.000993, 0.001000; 0.001 4 23. 7 (or any smaller positive number) 25. 0.6875 (or any smaller positive number) 27. (a) s1000 cm (b) Within approximately 0.0445 cm (c) Radius; area; s1000 ; 1000; 5; 0.0445 45. (a) 0.093 (b) B 23 126B 13 1, where B 216 108 12 s336 324 81 2 Exercises 1.4
A41
1 x
1
_1
0 y=1-≈
17. x x 3, 2
19.
[
1 2
, )
21. 0,
23. x 2 2n, n an integer 10
page 43
■
(b) 9 (c) 2 (d) 13 (e) 38 6 (g) Does not exist (h) 11 1 3. 59 5. 390 7. 8 9. 2 11. 5 1 6 13. Does not exist 15. 5 17. 8 19. 12 1 2 23. 16 25. (a), (b) 3 29. 7 33. 6 35. Does not exist 37. (a) (i) 0 (ii) 0 (iii) 1 (iv) 1 (v) 0 (vi) Does not exist y (b) 1. (a) 5
(f) 0 10
21.
1 6
10 1
25.
7 3
29. 0, right; 1, left y (0, 2)
(1, 2) (1, 1)
(0, 0)
x
1
_1
39. (a) (i) 2
0
1
x
(ii) Does not exist (iii) 3 (b) (i) n 1 (ii) n (c) a is not an integer. 1 43. 3 45. 3 47. 2 55. 15; 1
y=1/x
0
2
31. 3 33. (a) tx x 4
(c) tx x 2 4x 16
(d) tx 1(3 sx ) 41. (b) 0.86, 0.87 43. (b) 1.434 45. Yes
1
x
A42
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 1.6
11.
page 66
■
1. (a)
(b) (c) (d) 1 (f) x 1, x 2, y 1, y 2 3.
5.
y
13.
y
y= 12 ˛+1
(e) 2 y
1
y= x+2 1 2
1 0
Œ„„ _2
x=2
y
x=_2
0
x
x
1
15. (a) Neither
0 0
x
1
x
(b) Odd (c) Even (d) Neither 17. (a) f ⴰ tx ssin x,
x x 2n, 2n , n an integer 4 (b) t ⴰ f x sin sx, 0, (c) f ⴰ f x s x, 0, (d) t ⴰ tx sinsin x, ⺢ (e) t ⴰ tx x 2 92 9, , 19. All have domain ⺢. The range is 1, 1 for n odd and 0, 1 for n even. The spikes become sharper as n l . 21. (a) (i) 3 (ii) 0 (iii) Does not exist (iv) 2 (v) (vi) (vii) 4 (viii) 1 (b) y 4, y 1 (c) x 0, x 2 (d) 3, 0, 2, 4 3 1 1 23. 1 25. 2 27. 3 29. 31. 8 33. 2 1 37. 2 39. x 0, y 0 41. 1 35. 2 47. (a) (i) 3 (ii) 0 (iii) Does not exist (iv) 0 (v) 0 y (vi) 0 (b) At 0 and 3 (c)
7.
y
y=3 0
x
x=4
9. 0 11. x 1.62, x 0.62, x 1; y 1 1 13. 15. 17. 19. 2 21. 2 1 23. 6 25. Does not exist 27. 29. 1 31. 33. y 2; x 2, x 1 35. (a), (b) 2
2x 37. y 3 39. f x 2 x x 3 41. (a) 0 (b) 43. 4 45. (b) It approaches the concentration of the brine being pumped into the tank. 47. Within 0.1 51. N 13 53. (a) x 100 Chapter 1 Review
■
page 70
True-False Quiz 1. False 11. True
3
0
x
CHAPTER 2 Exercises 2.1
■
1. (a) 4
(b) y 4x 9
page 80
(c)
3. True 13. True
3
5
5. False 7. True 9. False 15. False 17. True 19.
y=≈+2x
(_3, 3)
True _4
Exercises 1. (a) 2.7 (b) 2.3, 5.6 (c) 6, 6 (d) 4, 4 (e) 4, 4 (f) Odd; its graph is symmetric about the origin.
[
]
3. 3 s3, 3 s3 , 0, 2 2
2
5. ⺢, 0, 2
7. (a) Shift the graph 8 units upward.
(b) Shift the graph 8 units to the left. (c) Stretch the graph vertically by a factor of 2, then shift it 1 unit upward. (d) Shift the graph 2 units to the right and 2 units downward. (e) Reflect the graph about the x-axis. (f) Reflect the graph about the line x-axis, then shift it 3 units upward. 9.
y=_sin 2x 0
_2
3. y x 5 7. (a) 8a 6a 2
(c)
π
x
5. y 2 x 1
1 2
(b) y 2x 3, y 8x 19
10
_2
4 _3
9. (a) 0
y
2
(b) C (c) Speeding up, slowing down, neither (d) The car did not move. 11. 24 fts 1 2 13. 2a 3 ms ; 2 ms; 4 ms; 27 ms 15. t0, 0, t4, t2, t2
17.
19. 7; y 7x 12
y
y
5.
y
7. fª
1 0
fª x
0
x
1
0
21. (a) 5 ; y 5 x 3
3
16 5
(b)
y
11.
y
fª _1
fª
6 x
0 _2
x
0
2 8a 25. 5a 32 27. 1 2a 232 f x x 10, a 1 or f x 1 x10 , a 0 f x 2 x, a 5 f x cos x, a or f x cos x, a 0 Temperature Greater (in magnitude) 72
x
4
9.
23. 29. 31. 33. 35.
A43
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
13.
y
y=Mª(t)
0.1
The derivative is negative for the years 1963 to 1971.
0.05 t
_0.03
(in °F)
1950 1960 1970 1980 1990
0
Time 2 (in hours)
1
(a) 0, 1, 2, 4 (b) 1, 2, 4 (c) f x 2x f x 12, ⺢, ⺢ 19. f x 3x 2 3, ⺢, ⺢ tx 1s1 2x, [ 12 , ), ( 12 , ) Gt 4t 12, , 1 1, , , 1 1, 25. (a) f x 4x 3 2
15. 17. 21. 23.
38
37. (a) (i) 11 percentyear
(ii) 13 percentyear (iii) 16 percentyear (b) 14.5 percentyear (c) 15 percentyear 39. (a) (i) $20.25unit (ii) $20.05unit (b) $20unit 41. (a) The rate at which the cost is changing per ounce of gold produced; dollars per ounce (b) When the 800th ounce of gold is produced, the cost of production is $17oz. (c) Decrease in the short term; increase in the long term 43. The rate at which the temperature is changing at 10:00 AM; 4 Fh 45. (a) The rate at which the oxygen solubility changes with respect to the water temperature; mgL C (b) S16 0.25; as the temperature increases past 16 C, the oxygen solubility is decreasing at a rate of 0.25 mgL C. 47. Does not exist
27. 4 corner; 0 discontinuity 29. 1 vertical tangent; 4 corner 31. 2
Differentiable at 1; not differentiable at 0
_2
1
_1
33. a f, b f , c f 35. a acceleration, b velocity, c position 37. 6 fª
f
6
10
f x 4 2x, f x 2
f·
Exercises 2.2 1. (a) 1.5
(c) 0 (e) 0 (g) 1.5
■
page 91
(b) 1 (d) 4 (f) 1
y
10
fª 0
39. (a) a23 1 3
x
41. f x
or f x 3. (a) II
(b) IV
(c) I
(d) III
45. 63
1 1
x6
if x 6 if x 6
x 6
y
fª 1 0 _1
6
x
A44
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 2.3
■
1. f x 0 3. f x 5 5. f x 3x 4 7. f (x) 1 3 cos x 9. f t t 3 2 11. y 5 x75 13. Vr 4r 2 15. Fx 2
5 32
x4
19. y sx (2sx ) 3(2x sx )
17. y 0
3 2
21. v 2t
3 4 3 4t s t
7. f x cos x 2 csc2x 9. h( ) csc csc cot csc 2 11. tx 52x 12 13. y 2t1 t3t 2 2t 12 15. y 2v 1sv cos x x sin x r (4 3sr ) 17. y 19. y cos2x 2(1 sr ) 2 1
page 104
23. z 10 Ay 11 B sin y
21. f
x 25. y 3 s3 x 3 s3 , y 3 3 s3 9 s3 3 1 27. y 2 x 2 3
23. y x cos x 2 sin xx 3
25. f x 2cxx 2 c2 27. y 2 x 2 29. y 2x 1 2 1 31. (a) y 2 x 1 (b) 1.5 1
(1, 2)
_2
sec tan 1 sec 2 1
1
3
(_1, 0.5) 4
4
_2
0.5
29. f x 4x 3 9x 2 16, f x 12x 2 18x 31. tt 2 sin t 3 cos t , tt 2 cos t 3 sin t 1 33. cos x 35. 2n 1 3 , n an integer 1 1 39. y 3 x 3 41. (a) vt 3t 2 3, at 6t (b) 12 ms2
(c) a1 6 ms2 43. (a) 3t 2 24t 36 (b) 9 fts (c) t 2, 6 (d) 0 t 2, t 6 (e) 96 ft (f) See graph at right.
t " 8, s " 32
t " 6, s"0
t " 2, s " 32
t " 0, s"0
s
0
45. (a) t 4 s
(b) t 1.5 s; the velocity has a minimum.
47. (a) 5.02 ms (b) s17 ms 49. (a) Cx 3 0.02x 0.0006x 2
(b) $11pair, the rate at which the cost is changing as the 100th pair of jeans is being produced; the cost of the 101st pair (c) $11.07 51. (a) 8 ft 2ft (b) 16 ft 2ft (c) 24 ft 2ft The rate increases as the radius increases. 53. (a) dVdP CP 2 (b) At the beginning 3 1 57. A 10, B 10 y 59. 2, 4 {a, a@}
y=≈ x
0 (0, _4)
61. a 2 , b 2 63. y 65. 1000 67. 3; 1 1
3 16
33. 35. cos sin ; 2 cos sin 20 2 41. (a) 16 (b) 9 (c) 20 43. (a) 0 (b) 3 45. (a) y xtx tx (b) y tx xtx tx 2 1 4
(c) y xtx tx x 2 47. (a) vt 8 cos t, at 8 sin t (b) 4 s3, 4, 4s3; to the left; speeding up 1 49. 0.2436 Kmin 51. Two, (2 s3, 2 (1 ) s3 )) 2 2 53. (b) y sin x cos x x cos x x sin x 55. (b) y 2 x(2 x 2 1)(x 4 x 2 1)2 Exercises 2.5
■
page 119
cos x 2 ssin x 3 2 3x 2 12t 7. Fx 9. tt 4 41 2x x 334 t 14 1 11. y 3x 2 sina 3 x 3 13. y 2 csc 2x2 15. tx 41 4x43 x x 2 717 9x 21x 2 17. y 82x 538x 2 544x 2 30x 5 19. y 3x 2 cos nx nx 3 sin nx 21. y cos x x sin x cosx cos x 23. Fz 1 z 112z 132 25. y r 2 132 27. y sin x sec 2cos x 29. y ( x coss1 x 2 )s1 x 2 31. y 12 cos x sin x 1 cos 2x5 33. y 4 sec 2x tan x 35. y 2 cos cotsin csc2sin 1 cos x 37. y cos(tan ssin x )(sec2ssin x ) 2 ssin x 39. y 20x 1 41. (a) y x 1 (b) 3 1. 4 cos 4 x
3. 20x1 x 2 9
x 3 94 x 3
5.
(1, 1) 0
Exercises 2.4
■
page 111
1. y 5x 4 3x 2 2x 3. t(t) 3t 2 cos t t 3 sin t 5. F y 5 14y 2 9y 4
2
43. Ft 421 7t , F t 14701 7t4 5
1.4
2x 2 4x 2x 4 13 , y 3 13 3 x 1 x 1 x 143 47. 24 49. (a) 30 (b) 36 3 51. (a) 4 (b) Does not exist (c) 2 53. (a) Fx sin x f cos x (b) Gx sin f x f x 55. 120 57. 2 2n, 3, 32 2n, 1, n an integer 5 59. vt 2 cos10 t cms 45. y
Exercises 2.7
page 131
■
7. 13 1. dVdt 3x dxdt 3. 48 cm 2s 5. 70 9. (a) The rate of decrease of the surface area is 1 cm2min. 46
2
3
(b) The rate of decrease of the diameter when the diameter is 10 cm (c) (d) S x 2 (e) 120 cmmin x
r
(b) 0.16 61. (a) dBdt 54 cos2 t5.4 63. dvdt is the rate of change of velocity with respect to time; dvds is the rate of change of velocity with respect to 7
displacement 65. (b) n cos n1x sin n 1x Exercises 2.6
■
11. (a) The plane’s altitude is 1 mi and its speed is 500 mih.
(b) The rate at which the distance from the plane to the station is increasing when the plane is 2 mi from the station x (c) (d) y 2 x 2 1 (e) 250 s3 mih 1
page 125
1. (a) y y 2 6xx
y
(b) y 4x 2 3x, y 4x 2 3 3. y x 3x 2yx 2 8y 5. y 3 2xy y 2 x 2 2xy 7. y 2xy 2 sin y2x 2 y x cos y 9. y tan x tan y y sec 2xy y 2 4xy sxy y 11. y 2 13. y y x sec 2xy x 2x 2sxy 16 1 15. 13 17. y x 2 19. y x 2 21. y 13 x 9
40 13 5 2
27. (a) y x 9 2
A45
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
23. 81y 3
(b)
15. 837s8674 8.99 fts
13. 65 mih
17. 1.6 cmmin
21. 10s133 0.87 fts
55.4 kmh 4 23. 5 ftmin
25. 65 0.38 ftmin
27. 0.3 m 2s
31.
107 810
19.
0.132 s
720 13
33. (a) 360 fts
35. 1650s31 296 kmh
25. 2xy 5
Exercises 2.8
5
■
3. Lx x 2
5. s1 x 1 x ; 1 2
3
y=1- 21 x
s0.9 0.95, s0.99 0.995
2
_2
(b) 0.096 rads
37. 7 s154 6.78 ms
page 137
1. Lx 10x 6 (1, 2)
29. 80 cm 3min
(0, 1)
y=œ„„„„ 1-x _2
29. (a)
_3
(1, 0)
4
3
_1
Eight; x 0.42, 1.58 _2
7. 1.204 x 0.706 9. 0.045 x 0.055 11. 32.08 13. 4.02 17. (a) dy 2xx cos 2x sin 2x dx
5
(b) dy 5(2s4 5x ) dx 19. (a) dy sec 2x dx (b) dy 0.2, y 0.18237 21. (a) 270 cm 3, 0.01, 1% (b) 36 cm 2, 0.006, 0.6% 1 23. (a) 84 27 cm 2; 84 0.012 (b) 1764 2 179 cm 3; 561 0.018 25. A 5% increase in the radius corresponds to a 20% increase in blood flow. 27. (a) 4.8, 5.2 (b) Too large
_3
(b) y x 1, y 13 x 2 5 5 31. ( 4 s3, 4 ) 33.
(c) 1 ) s33
35.
y
y
x
Chapter 2 Review
x
■
page 138
True-False Quiz 39. (s3, 0)
41. 1, 1, 1, 1
43. (a) 0
(b) 2
1
1. False 9. True
3. False 11. False
5. True
7. False
A46
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises
3.
5 y=20 ® y=5® y=´
1. f 5, 0, f 5, f 2, 1, f 3 3. (a) The rate at which the cost changes with respect to the
interest rate; dollars(percent per year) (b) As the interest rate increases past 10%, the cost is increasing at a rate of $1200(percent per year). (c) Always positive y 5. 7. a f, c f , b f fª
0
y=2®
_1
5.
2
0
The functions with base greater than 1 are increasing and those with base less than 1 are decreasing. The latter are reflections of the former about the y-axis.
1 ® y=” 13 ’® y=” 10 ’ 5 y=10® y=3®
x
_2
37. 5 x tan x45tan x x sec 2x 1
39. 27 4
5x 4y 11 43. y 2s3x 1 s33 y 2x 1 47. (4, s2 ), (54, s2 ) (a) 2 (b) 44 51. f x 2xtx x 2tx 55. f x t tx tx f x 2tx tx f x tsin x cos x f x tx 2 tx f x 2 59. hx f x tx 2 61. 4 (discontinuity), 1 (corner), 2 (discontinuity), 5 (vertical tangent) 63. (a) vt 3t 2 12, at 6t (b) Upward when t 2, downward when 0 t 2 (c) 23 4 65. 3 cm2min 67. 13 fts 69. 400 fth 3 3 71. (a) Lx 1 x; s 1 3x 1 x ; s 1.03 1.01 (b) 0.23 x 0.40 1 1 71. 12 32 16.7 cm 2 75. 32 77. 4 41. 45. 49. 53. 57.
2
0
9. The rate at which the total value of US currency in
circulation is changing in billions of dollars per year; $22.2 billionyear 11. f x 3x 2 5 13. 6xx 4 3x 2 522x 2 3 1 4 22x 2 1 t 2 1 3 7 15. 17. 19. 2 sx 3sx 1 t 2 2 sx 2 1 sec 2s1 x 1 y 4 2xy 21. 23. 2 s1 x 4xy 3 x 2 3 2 sec 2 tan 2 1 25. 27. x 12 1 tan 22 2x y cosxy 29. 31. 6x csc 23x 2 5 x cosxy 1 3x 2 33. cos(tan s1 x 3 )(sec 2s1 x 3 ) 2 s1 x 3 35. 2 cos tansin sec2sin
y
7.
All approach 0 as x l , all pass through 0, 1, and all are increasing. The larger the base, the faster the rate of increase.
9.
0
x
y x
0 _1
y=_2–®
_2
y=4®-3 y=_3
11.
y
y=1 1
”0, 2 ’ 0
x 1
y=1- 2 e–®
13. (a) y e x 2
(d) 15. 17. 23.
(b) y e x2 (c) y e x (e) y ex ye (a) , (b) , 0 0, 21. At x 35.8 f x 3 2 x 25. 1 27. 0 29. 0 x
Exercises 3.2
■
page 158
1. (a) See Definition 1.
(b) It must pass the Horizontal Line Test. 3. No 5. Yes 7. No 9. No 11. Yes 13. No 15. 2 17. 0 9 19. F 5 C 32; the Fahrenheit temperature as a function of the Celsius temperature; 273.15, 1 10 3 21. f 1x 3 x 2 3 , x 0 23. f 1x s ln x x 25. y e 3 4 x1 27. f 1x s 29. 6
y f
f –!
CHAPTER 3 f
Exercises 3.1
■
page 147
0 f –!
1. (a) f x a x, a 0
(b) ⺢ (c) 0, (d) See Figures 6(c), 6(b), and 6(a), respectively. 0
6
x
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
31. (b)
33. (b) 2
1 12
1
1
(c) f x sx , domain ⺢ range y (e)
(c) f x s9 x , domain 0, 9 , range 0, 3 (e) y
ƒ
1
Exercises 3.3
1
3
ƒ
3
f –!
0
2
21. y ex 1 2x 2
35. 1 37. 2 39. 41. (a) It’s defined as the inverse of the exponential function 3 2
with base a, that is, log a x y &? a y x. (b) 0, (c) ⺢ (d) See Figure 13. 43. (a) 6 (b) 2 45. (a) 2 (b) 2 47. 3 log 2 x log 2 y 2 log 2 z 49. 10 ln u 10 ln v 51. ln 8 55.
53. ln
sin x y=log 1.5 x y=ln x y=log 10 x
0
4
y=log 50 x
5
57. About 1,084,588 mi 59. (a)
(b)
y
_5
y
x
61. 63. 65. 67. 69. 75.
x
1
35. Ft e t sin 2t 2t cos 2t sin 2t
2
4
10 24x 3 4 2x 1 x 3
sec2 x ln tan x x tan x x2
4
_2 _1
s4 3 (sD 27x 2 20 s3 D 27x 2 20 s3 2 ) , 6 where D 3 s3 s27x 4 40x 2 16; two of the expressions are complex. 3
f 1x
1 2xye 2x 57. y 2 x 2e x y 1 x y 2 2y 59. v t 2e1.5t2 cos 2 t 1.5 sin 2 t x2y
2
15
2
√
s 0
2
0
63. f (x) 2 e (n)
67.
2
7
1
The graph passes the Horizontal Line Test.
2
4 sec x 4x sin x tan x 2 cot x 2 x 2 12 tan x x 1 49. y x x1 ln x 51. y cos x xx tan x ln cos x 47. y
[
5
45. y 2x 15x 4 36
55. y
(a) se (b) ln 5 (a) 5 log 2 3 or 5 ln 3ln 2 (b) 12 (1 s1 4e ) (a) x ln 10 (b) x 1e 1 1 (a) ( , 2 ln 3 (b) f 1x 2 ln3 x 2 , 0, s3 ) 71. 0 73.
]
31. f u 1u 2 e 1u
y e x 2 2 sin x 2 cos x 39. y 1 ln x, y 1x 41. x ey e 2x 1 x 1 lnx 1 ; 43. f x x 1 1 lnx 1 2 1, 1 e 1 e,
53. y tan x1x 0
ad bce x ce x d 2
37. y e x cos x sin x;
y=-ln x
0
_4
27. y
25. ht 3t 2 3 t ln 3
33. y x1 x
All graphs approach as x l 0, all pass through 1, 0, and all are increasing. The larger the base, the slower the rate of increase.
y=log 10 (x+5)
23. y cos x x sin xe x cos x
29. y 2sin x ln 2 cos x
(1 x 2) sx
3
19. y x 2e xx 3
17. f x x x 2e x
9 x
3
page 166
■
3 3. f tan 3x 1 ln 2 1 sin x 5. f x 7. f x cos x ln5x 5 5xs ln x4 x 2a 6 12 9. tx 2 11. Ft a x2 2t 1 3t 1 10x 1 1 ln 2 13. f u 15. y u 1 ln2u 2 5x 2 x 2
x
0 1
A47
1. f x
9
f –!
■
n 2x
n
65. f x 1n1n 1!x 1n
1 2
Exercises 3.4
■
page 173
1. About 235 3. (a) 1004.2 t
(b) 7409 (c) 10,632 bacteriah (d) ln 100ln 4.2 3.2 h 5. (a) 1508 million, 1871 million (b) 2161 million (c) 3972 million; wars in the first half of century, increased life expectancy in second half
A48
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
7. (a) Ce0.0005t (b) 2000 ln 0.9 211 s 9. (a) 100 2t30 mg (b) 9.92 mg
Exercises 1. No
(c) 199.3 years 11. 2500 years 13. (a) 137 F (b) 116 min 15. (a) 13.3 C (b) 67.74 min 17. (a) 64.5 kPa (b) 39.9 kPa 19. (a) $3828.84 (b) $3840.25 (c) $3850.08 (d) $3851.61 (e) $3852.01 (f) $3852.08
3. (a) 7
5.
(b)
1 8
7.
y
y
y=-ln x y=5x-1 0
0
x
1
x
y=_1
Exercises 3.5
■
page 180
(b) 3. (a) 3 (b) 4 5. (a) 0.7 (b) 3 9. xs1 x 2 1 1 17. y 19. y 2sx 1 x sx 2 x 21. Hx 1 2x arctan x 23. y 2e 2xs1 e 4x sin 25. y 27. ht 0 1 cos 2 29. y sa 2 b 2 a b cos x 2 ; [1, 2], (1, 2) 31. tx s1 3 2x 2 1 33. 6 35. 2 37. 2 39. 4 rads 1. (a) 3
Exercises 3.6
■
page 185
(b) 1 3. (a) 4 (b) 2 e 2 e2 3.62686 1. (a) 0 5. (a) 1 (b) 0 5 3 5 4 3 17. coth x 4 , sech x 5 , cosh x 3 , sinh x 3 , csch x 4 19. (a) 1 (b) 1 (c) (d) (e) 0 (f) 1 3
(g)
(h)
1
(i) 0
27. f x x sinh x cosh x
29. hx 2x coshx 2
t csch 2 s1 t 2 31. ht s1 t 2
33. Ht e sech e t
2
Exercises 3.7 3. 1. 2 1 13. 2 15. 23. 3 25. 35. 1se
■
37. y
y=π y=2 arctan x x
y=_π
11. (a) 9 17. 23. 27. 31.
(b) 2
13. e 13
15. ln ln 17
19. tan 1 21. f t t 2t ln t s1 e h 2 sec 22 e tan 2 25. y 5 sec 5x y 1 c 2 e cx sin x 29. y 2 tan x 2 y e1x1 1x 33. y 2 ln 2 t2t v
tan1v 1 v2 37. y 2x 2 coshx 2 sinhx 2 35. Hv
41. y
43. y 3 tanh 3x
cosh x ssinh 2x 1
45. y
47. f x
1 1 1 x ln x2
39. y cot x sin x cos x
6x sin 2lnx 2 1 coslnx 2 1 e sin lnx 1 x2 1 51. f x txtx f x txe tx 57. y x 2 59. 3, 0 2 xln 2n (a) y 14 x 14 ln 4 1 (b) y ex (a) 2003.24t (b) 22,040 (d) ln 50ln 3.24 3.33 h 25,910 bacteriah (a) C0 ekt (b) 100 h 67. 0 69. 0 71. 1 2 75. 77. 8 1 79. 0 81. 2 83. 3 3
49. 53. 61. 63.
(c) 65. 73.
2
page 193
5 5. 7. pq 9. 11. ln 3 1 2 1 17. 1 19. 2 aa 1 21. 0 0 27. 1 29. 31. 1 33. e2 16 37. 5 45. 9 a 47. 56 51. (a) 0
Chapter 3 Review
■
page 195
True-False Quiz 1. True 11. False
y
0
t
1 2 sx 1 x 1 39. y sinh1x3 41. y x sx 2 1 43. (a) 0.3572 (b) 70.34° 45. (b) y 2 sinh 3x 4 cosh 3x 47. (ln(1 s2 ), s2 ) 35. y 3e cosh 3x sinh 3x
9.
3. False 5. True 7. True 13. False 15. True
9. False
CHAPTER 4 Exercises 4.1
■
page 203
Abbreviations: abs., absolute; loc., local; max., maximum; min., minimum 1. Absolute minimum: smallest function value on the entire domain of the function; local minimum at c: smallest function value when x is near c 3. Abs. max. at b, loc. max. at b and e, abs. min. at d, loc. min. at d and s 5. Abs. max. f 4 4; abs. min. f 7 0; loc. max. f 4 4 and f 6 3; loc. min. f 2 1 and f 5 2
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
7.
9.
y
Exercises 4.2
y
■
page 210
3. , 4 1 4
3
3
3
1. 2
2
2
5. f is not differentiable on (1, 1)
1
1
0
1
2
3
4
5
x
y
11. (a)
(b)
2
3
4
5
x
10
10
y 1
1 0
1
2
3
0
x
1
2
3
x
_1
_1
(c)
1
7. 0.8, 3.2, 4.4, 6.1
(c) 2 s2
9. (a), (b)
0
A49
10
0
[
]
11. 0 13. 2 ln 6 1 e6 15. f is not differentiable at 1 25. No 31. No 1
y 2
10
0
1
23. 16
1
Exercises 4.3 0
1
2
3
x
(b)
y
1. (a) Inc. on , 2, 2, ; dec. on 2, 2
y
_1 0
2
x
0
15. Abs. max. f 1 5 17. None 19. Abs. and loc. max. f 2 f 32 1;
x
abs. and loc. min. f 32 f 2 1 2 21. Abs. max. f 0 1 23. 5 25. 4, 2 1 27. 0, 2 (1 s5 ) 29. 0, 2 8 31. 0, 7, 4 33. n n an integer 35. 1e 39. f 1 8, f 2 19 37. f 0 5, f 2 7 41. f 3 66, f 1 2 43. f (s2 ) 2, f 1 s3 45. f 4 s2, f 0 1 8 47. f 2 2se , f 1 1s e a b a a b 49. f ab a b ab
(b) 256 s 35 2 , 256 s35 2 53. (a) 0.32, 0.00 (b) 163 s3 , 0 55. 3.9665 C 57. Cheapest, t 10; most expensive, t 5.1309 2 59. (a) r 3 r0 (b) v 274 kr 30
(b) Loc. max. f 2 17; loc. min. f 2 15 (c) CU on 0, ; CD on , 0; IP 0, 1 3. (a) Inc. on 3, 53, 73, 3; dec. on 0, 3, 53, 73 (b) Loc. max. f 53 53 s3 ; loc. min. f 3 3 s3 , f 73 73 s3 (c) CU on 0, , 2, 3; CD on , 2; IP , , 2, 2 5. (a) Inc. on 1, ; dec. on , 1 (b) Loc. min. f 1 1e (c) CU on 2, ; CD on , 2; IP 2, 2e2 7. (a) Inc. on 0, e 2 ; dec. on e 2, (b) Loc. max. f e 2 2e (c) CU on e 83, ; CD on 0, e 83 ; IP (e 83, 83 e43 ) 3 5 9. Loc. max. f ( 4 ) 4 11. (a) f has a local maximum at 2. (b) f has a horizontal tangent at 6. 13. (a) 3, 5 (b) 2, 4, 6 (c) 1, 7 15.
17.
y
0
51. (a) 2.19, 1.81
(c)
page 217
Abbreviations: inc., increasing; dec., decreasing; CD, concave downward; CU, concave upward; HA, horizontal asymptote; VA, vertical asymptote; IP, inflection point
_1
13. (a)
■
y
x 0
19.
y
1
2
3
4
y
√ 4 27 kr#¸
_2
0
2 3
r¸
r¸ r
0
x
x=2
_2
0
2
x
x
A50
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
21. (a) Inc. on (0, 2), (4, 6), 8, ; dec. on (2, 4), (6, 8) (b) Loc. max. at x 2, 6; loc. min. at x 4, 8 (c) CU on (3, 6), 6, ; CD on (0, 3) (d) 3 (e) See graph at right. 23. (a) Inc. on , 1, 2, ; dec. on 1, 2 (b) Loc. max. f 1 7; loc. min. f 2 20 1 1 (c) CU on ( 2 , ); CD on ( , 2 ); 1 13 IP ( 2 , 2 ) (d) See graph at right. 25. (a) Inc. on (s3, 0), (s3, ); dec. on ( , s3 ), (0, s3 ) (b) Loc. min. f (s3 ) 9; loc. max. f 0 0 (c) CU on , 1, 1, ; CD on 1, 1; IP 1, 5 (d) See graph at right. 27. (a) Inc. on , 1, 1, ;
dec. on 1, 1 (b) Loc. max. h1 5; loc. min. h1 1 (c) CU on (1s2, 0), (1s2, ); CD on ( , 1s2 ), (0, 1s2 ); 7 IP 0, 3, (1s2, 3 ) 8 s2 ) (d) See graph at right.
0
2
4
6
8
(_1, 7) y
0 1 ” 2,
x
13 _ 2’
(2, _20)
π 3
5π 3
” 3 , 2’
” 3 , 2’
IP 0
IP π _ 2
π
3π 2
IP
35. (a) HA y 1, VA x 1, x 1 (b) Inc. on , 1, 1, 0; dec. on 0, 1, 1, (c) Loc. max. f 0 0 (d) CU on , 1, 1, ; CD on 1, 1 (e) See graph at right.
2π ¨
IP
y
y=1 0 x
x=_1
37. (a) HA y 0
y 0 1
2
(b) (c) (d) (e)
x
(_1, _5) (1, _5)
5
x=1
y
Dec. on , None CU on , See graph at right.
1 x
0
(_œ„ 3, _9) 10
(_1, 5)
(œ„ 3, _9)
39. (a) VA x 0, x e
(b) Dec. on 0, e (c) None (d) CU on (0, 1); CD on 1, e; IP (1, 0) (e) See graph at right.
y
y
x=0
x=e
1
(1, 0) x
0
(1, 1)
0
x
y 2 _3 x
_2 _2
41. (a) HA y 1, VA x 1 (b) Inc. on , 1, 1, (c) None 1 (d) CU on , 1, (1, 2 ); 1 1 2 CD on ( 2 , ); IP ( 2 , 1e ) (e) See graph at right.
y=1
x=_1
x
0
1 4
(3 s17 ) 0.28
47. When t 7.94
y {2, 6 Œ„2 }
49. f x 9 2x 3 3x 2 12x 7 1
Exercises 4.4 _4
y
43. 3, 45. (a) Loc. and abs. max. f 1 s2, no min.
(b)
31. (a) Inc. on 1, ;
dec. on , 1 (b) Loc. min. C1 3 (c) CU on , 0, 2, ; CD on 0, 2; 3 IP 0, 0, (2, 6 s 2) (d) See graph at right.
x
y
(π, _3)
29. (a) Inc. on 2, ;
dec. on 3, 2 (b) Loc. min. A2 2 (c) CU on 3, (d) See graph at right.
CU on 0, , , 2 , 2 , 2; CD on , , 2 , 2 ; IP , y 1, , y 2, 2 , y2, 2 , y1 (d) See graph at right.
y
0 (_1, _3)
33. (a) Inc. on (0, 3), (, 53); dec. on (3, ), (53, 2) (b) Loc. max. f 3) f 53) 32; loc. min. f 3 (c) Let cos1( 18 (1 s33 )), cos1( 18 (1 s33 )), y1 163 (1 s33 ), and y2 163 (1 s33 ).
x
■
page 225
Abbreviation: SA, slant asymptote y
1. A. ⺢
B. y-int. 0; x-int. 0 C. About 0, 0 D. None E. Inc. on , F. None G. CU on 0, ; CD on , 0; IP (0, 0) H. See graph at right.
1 _1
1 _1
x
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
13. A. ⺢
B. y-int. 2; x-int. 2, 12 (7 3 s5 ) y C. None D. None E. Inc. on (1, 5); dec. on , 1, 5, F. Loc. min. f 1 5; loc. max. f 5 27 G. CU on , 3; 0 (1, _5) CD on 3, ; IP 3, 11 H. See graph at right. 3. A. ⺢
x
5. A. ⺢
B. y-int. 0; x-int. 4, 0 C. None D. None E. Inc. on 3, ; dec. on , 3 F. Loc. min. f 3 27 G. CU on , 2, 0, ; CD on 2, 0 ; IP (0, 0), 2, 16 H. See graph at right.
x
1
”_3, _ 6 ’
15. A. , 0 0,
0
x
(_3, _27)
B. y-int. 1 C. None D. None E. Inc. on , 0, 1, ; dec. on 0, 1 F. Loc. max. f 0 1; loc. min. f 1 2 3 G. CU on (1s 4, ); CD on 3 ( , 1s4 ); 3 3 IP (1s 4, 1 9(2 s 16 )) H. See graph at right.
0
x
IP 1, _2
B. y-int. 0; x-int. 0 C. About the origin D. HAs y 1 E. Inc. on , F. None G. CU on , 0; CD on 0, ; IP 0, 0 H. See graph at right.
y"1 x
0
x"1
y
x
x " 3
x"3
2 9
0
x
1
10 10œ„„ 15 ’ 9
” 3 ,
y
x
19. A. ⺢
y
11. A. x x 3
1 4
”2, ’ ”3, ’
B. y-int. 0; x-int. 0, 5 C. None D. None E. Inc. on ( , 103 ); dec. on ( 103 , 5) F. Loc. max. f (103) 109 s15 G. CD on , 5 H. See graph at right.
0, 1
y
B. x-int. 1 C. None D. HA y 0; VA x 0 E. Inc. on 0, 2; dec. on , 0, 2, F. Loc. max. f 2 14 G. CU on 3, ; CD on , 0, 0, 3; IP (3, 29 ) H. See graph at right 17. A. , 5
y
9. A. x x 1
B. y-int. 19 C. About y-axis D. VAs x 3, HA y 0 E. Inc. on , 3, 3, 0; dec. on (0, 3), 3, F. Loc. max. f 0 19 G. CU on , 3, 3, ; CD on 3, 3 H. See graph at right.
1
”3, 6 ’
y
7. A. ⺢
B. y-int. 0; x-int. 0 C. None D. VA x 1, HA y 1 E. Dec. on , 1, 1, F. None G. CU on 1, ; CD on , 1 H. See graph at right.
y
B. y-int. 0; x-int. 0 C. About (0, 0) D. HA y 0 E. Inc. on 3, 3; dec. on , 3, 3, F. Loc. min. f 3 16 ; loc. max. f 3 16 ; G. CU on (3 s3, 0), (3 s3, ); CD on ( , 3 s3 ), (0, 3 s3 ); 1 IPs (0, 0), (3 s3, 12 s3 ) H. See graph at right.
(5, 27)
A51
y
y=1
(0, 0)
x
y=_1
21. A. { x x 1, x 0} 1, 0 0, 1 C. About (0, 0) D. VA x 0 E. Dec. on 1, 0, 0, 1 F. None G. CU on (1, s23 ), (0, s23 ); 1 CD on (s23, 0), (s23, 1); IPs (s23, 1s2 ) H. See graph at right.
B. x-int. 1 y
0
1 x
23. A. ⺢ B. y-int. 0; x-int. 0, 3s3 C. About the origin D. None E. Inc. on , 1, 1, ; dec. on 1, 1 y F. Loc. max. f 1 2; _1, 2 3œ„ 3, 0 loc. min. f 1 2 0, 0 G. CU on 0, ; CD on , 0; 0 x IP 0, 0 _3œ„ 3, 0 1, _2 H. See graph at right.
A52
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
25. A. ⺢ B. y-int. 0; x-int. 1, 0 C. None D. None E. Inc. on ( , 14 ), 0, ; dec. on ( 14 , 0) F. Loc. max. f ( 14 ) 14 ; loc. min. f 0 0 G. CD on , 0, 0, H. See graph at right.
35. A. ⺢
y 1 1 ”_ 4 , 4 ’
1
x
0
1
B. y-int. 2 C. None D. HAs y 0, y 1 E. Inc. on ⺢ F. None G. CU on , 0; CD on 0, ; 1 IP (0, 2 ) H. See graph at right.
29. A. 2, 2 B. y-int. 0; x-int. 0 C. About y-axis D. VA x 2 π x" 2 E. Inc. on 0, 2; dec. on 2, 0 F. Loc. min. f 0 0 G. CU on 2, 2 H. See graph at right. 31. A. 0, 3 B. No y -int. C. None E. Inc. on 3, 53, 73, 3; dec. on 0, 3, 53, 73 1 F. Loc. min. f 3 6 2 s3 , 1 y f 73 76 2 s3 ; loc. max. f 53 56 12 s3 G. CU on 0, , 2, 3; CD on , 2; IP , 2, 2, π 3 H. See graph at right. 0
37. A. 0, B. x-int. 1 C. None D. None E. Inc. on 1e, ; dec. on 0, 1e F. Loc. min. f 1e 1e G. CU on 0, H. See graph at right.
x
y
” π, 2’ 2
0
2π x _1 _2
y
π x" 2
x
0
39. A. ⺢ B. y-int. 0; x-int. 0 C. None D. HA y 0 E. Inc. on , 1; dec. on 1, F. Loc. max. f 1 1e G. CU on 2, ; CD on , 2; IP 2, 2e 2 H. See graph at right.
y
”1, e1 ’ 0
1
_4π _3π _2π _π
π
2π
3π
7π 3
2π
3π x
4π x
0
5π 3
2
41. A. All x in 2n, 2n 1 (n an integer) B. x-int. 2 2n C. Period 2 D. VAs x n E. Inc. on 2n, 2 2n; dec. on 2 2n, 2n 1 F. Loc. max. f 2 2n 0 G. CD on 2n, 2n 1 y H.
D. None
π
x
(1/e, _1/e)
33. A. All reals except 2n 1 (n an integer) B. y-int. 0; x-int. 2n C. About the origin, period 2 D. VAs x 2n 1 E. Inc. on 2n 1, 2n 1 F. None G. CU on 2n, 2n 1; CD on 2n 1, 2n; IPs 2n, 0 x=_π y x=π x=3π H. x=_3π
43. A. ⺢ B. y-int. 0; x-int. 0 C. About (0, 0) D. HA y 0 E. Inc. on (1s2, 1s2 ); dec. on ( , 1s2 ), (1s2, ) F. Loc. min. f (1s2 ) 1s2e; loc. max. f (1s2 ) 1s2e G. CU on (s32, 0), (s32, ); CD on ( , s32 ), (0, s32 ); IPs (s32, s32 e32 ), 0, 0 y H. 1 , 1 ” œ„2
œ„„ 2e
’
x
0 0
y"1
0
27. A. ⺢ B. y-int. 0; x-int. n (n an integer) C. About the origin, period 2 D. None E. Inc. on 2n 2, 2n 2; dec. on 2n 2, 2n 32 y F. Loc. max. f 2n 2 2; 2 1 loc. min. f 2n 32 2 G. CU on 2n 1, 2n; 0 _2π CD on 2n, 2n 1; IPs n, 0 ”_ π, _2’ 2 H. See graph at right.
_2π
y
2π
x
45.
y
0
L/2
L
x
x
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
47. A. ( , 2 ) 1
1 4
( 12 , ) (5 s17 )
49. A. x x 0
0
x
1
75
1
x=_ 2
f f
y=_x+2
B. None
_1
y
C. About (0, 0) D. VA x 0; SA y x E. Inc. on , 2, 2, ; dec. on 2, 0, 0, 2 F. Loc. max. f 2 4; loc. min. f 2 4 G. CU on 0, ; CD on , 0 H. See graph at right. 51.
A53
57. Inc. on (8 s61, 8 s61 ); dec. on ( , 8 s61 ), 8 s61, 0, 0, ; CU on (12 s138, 12 s138 ), 0, ; CD on ( , 12 s138 ), (12 s138, 0)
y
B. y-int. 1; x-int. C. None D. VA x 12; SA y x 2 E. Dec. on ( , 12 ), ( 12, ) F. None G. CU on ( 12, ); CD on ( , 12 ) H. See graph at right.
■
_100
(2, 4)
_10
59. For c 0, there is no IP and only one extreme point, the origin. For c 0, there is a maximum point at the origin, two minimum points, and two IPs, which move downward and away from the origin as c l .
y=x x
0
1
_1 0.95
(_2, _4)
4 1 _1
4
_2 _3
y
_2.1
2.1
π
y"x2
x
0
y"x
_2.3
π 2
61. There is no maximum or minimum, regardless of the value of c. For c 0, there is a vertical asymptote at x 0, lim x l 0 fx , and lim x l f x 1. c 0 is a transitional value at which f x 1 for x 0. For c 0, lim x l 0 fx 0, lim x l f x 1, and there are two IPs, which move away from the y-axis as c l .
53. Inc. on 0.92, 2.5, 2.58, ; dec. on , 0.92, 2.5, 2.58; loc. max. f 2.5 4 ; loc. min. f 0.92 5.12, f 2.58 3.998; CU on , 1.46, 2.54, ; CD on 1.46, 2.54; IPs 1.46, 1.40, 2.54, 3.999 10
2
c=_2 c=_1
c=0.5
4.04
ƒ
c=2
c=1
4
_4
0
4
c=_0.5
_4
ƒ
4
_1
2.4 _6
2.7 3.96
55. Inc. on 1.49, 1.07, 2.89, 4; dec. on 4, 1.49,
1.07, 2.89; loc. max. f 1.07 8.79 ; loc. min. f 1.49 8.75 , f 2.89 9.99 ; CU on 4, 1.28, 1.28, 4; CD on 1.28, 1.28; IPs 1.28, 8.77, 1.28, 1.48 30
_1
63. For c 1, the graph has local maximum and minimum
values; for c 1 it does not. The function increases for c 1 and decreases for c 1 . As c changes, the IPs move vertically but not horizontally. 10
c=3
c=1 c=0.5
10
c=0
ƒ
ƒ
_15
_4
15
4
c=_0.5
_2.5 _10
0 6
4
_10 c=_3
c=_1
A54
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 4.5
■
page 232
(b) 11.5, 11.5 1. (a) 11, 12 5. 25 m by 25 m 7. (a)
3. 10, 10
12,500 ft@
50 250
12,500 ft@
100
9000 ft@
120
75
(c) (d) (e) (f )
x
A xy 5x 2y 750 Ax 375x 52 x 2 14,062.5 ft 2
y
9. 4000 cm 3 13. (3 , 3 s2 ) 15. L2, s3 L4 17. 4r 3(3 s3 ) 19. Width 604 ft; rectangle height 304 ft 21. (a) Use all of the wire for the square 1
Exercises 4.6
Chapter 4 Review
■
page 240 4
■
■
page 248
True-False Quiz 1. False 11. True
3. False 13. False
5. True 15. True
7. False 9. True 17. True
Exercises 1. Abs. min. f 0 10; abs. and loc. max. f 3 64 3. Abs. max. f 0 0; abs. and loc. min. f 1 1 y 5. x"6
2 0
]
(b) No 3. 5 5. 1.1797 1. (a) x2 2.3, x3 3 7. 1.25 9. 3.10723251 11. 0.876726 13. 1.39194691, 1.07739428, 2.71987822 15. 0.44285440 17. 1.97806681, 0.82646233 19. 0.52026899 21. (b) 31.622777 27. (0.904557, 1.855277) 29. 0.76286% Exercises 4.7
(b) s4504.9 9.58 s (c) 9.8s4504.9 93.9 ms (d) About 9.09 s 130 41. 225 ft 43. 11 11.8 s 88 45. 15 5.87 fts 2 47. 62,500 kmh2 4.82 ms2 (b) 21.675 mi (c) 30 min 33 s 49. (a) 22.9125 mi (d) 55.425 mi
4
(b) 40 s3(9 4 s3 ) m for the square 23. V 2 R 3(9 s3 ) 27. E 24r 3 2 (b) cos1(1s3 ) 55 29. (a) 2 s csc csc s3 cot (c) 6s h s(2 s2 ) 3 3 31. 10 s 3(1 s 3 ) ft from the stronger source 5 33. y 3 x 10 35. (b) (i) $342,491; $342unit; $390unit (ii) 400 (iii) $320unit 1 37. (a) px 19 3000 x (b) $9.50 1 39. (a) px 550 10 x (b) $175 (c) $100 41. a 23 b 23 32 45. x 6 in. 1 47. 2 L W 2 49. At a distance 5 2 s5 from A
[
2x 4 13 x 3 5x 2 22x 593 sin cos 5 4 29. 10 31. b ln x ln 2x ln 2 st 1 cos t sin t st 10 sin t 3 cos t 6t 3
37. (a) st 450 4.9t 2
125
(b)
23. 25. 27. 33. 35.
page 246
1. F x 2x 3 4x 2 3x C 3. F x 4x 54 4x 74 C 3 34 5 5. F x 4 x x C1 if x 0 ,
3 F x 4 x 34 x5 C2 if x 0 1 7. F u 3 u 3 6u12 C 9. G sin 5 cos C 1 11. F x x 2 5 sin1x C 13. Fx x 5 3 x 6 4 1 2 25 145 3 4 Cx D 15. x x Cx D 17. 2 x 126 x 19. 4x 32 2x 52 4 21. 2 sin t tan t 4 2s3
1
9
7.
x
12
y
y=2
x
y=_2
9. (a) None
y
(b) Dec. on , (c) None (d) CU on , 0; CD on 0, ; IP 0, 2 (e) See graph at right.
2 x
11. (a) None
(b) Inc. on 2n, 2n 1, n an integer; dec. on 2n 1, 2n 2 (c) Loc. max. f 2n 1 2; loc. min. f 2n 2 (d) CU on 2n 3, 2n 3; CD on 2n 3, 2n 53; IPs (2n 3, 14 ) y (e) 2 _2π
2π _π
π _2
x
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
(b) Inc. on ( ln 3, ), dec. on ( , 14 ln 3) (c) Loc. min. 1 f ( 4 ln 3) 314 334 (d) CU on , (e) See graph at right.
loc. max. f 0 2; loc. min. f 0.23 1.96, f 1.62 19.2; CU on , 0.12, 1.24, ; CD on 0.12, 1.24; IPs 0.12, 1.98, 1.24, 12.1
1 4
2
f 0
y
_0.5
27.
1
0
1
2
_5
5 0
0
19. A. 2,
y
0.82, 0.22; (s23, e32 ) 29. Max. at x 0, min. at x 0.87, IPs at x 0.52
x
31. For C 1, f is periodic with period 2 and has local maxima at 2n 2 , n an integer. For C 1, f has no graph. For 1 C 1, f has vertical asymptotes. For C 1, f is continuous on ⺢. As C increases, f moves upward and its oscillations become less pronounced. 37. 500, 125 39. 3 s3 r 2 41. 4s3 cm from D; at C 43. L C 45. $11.50 47. 1.16718557 49. Fx e x 4 sx C 51. 2 arctan x 1 1 53. 2 x 2 x 3 4x 4 2x 1 1 55. st 6 t 3 t 2 3t 1 57. No 59. (b) About 8.5 in. by 2 in. (c) 20s3 in. by 20 s23 in.
x"3
x 4
”_ 3 , _
4œ„ 6 ’ 9
y
B. None C. About (0, 0) D. HA y 0 E. Dec. on , 1, 1, F. None G. CU on 1, ; CD on , 1 H. See graph at right.
1
x
y
{ x x 1}
0.4
1.5
_20
B. None C. None D. HA y 0; VAs x 0, x 3 E. Inc. on 1, 3; dec. on , 0, 0, 1, 3, F. Loc. min. f 1 14 G. CU on 0, 3, 3, ; CD on , 0 H. See graph at right. B. y-int. 0; x-int. 2, 0 C. None D. None E. Inc. on ( 43 , ), dec. on (2, 43 ) F. Loc. min. f ( 43 ) 49 s6 G. CU on 2, H. See graph at right.
f 2.1
_1
2
17. A. x x 0, 3
23. Inc. on (s3, 0), (0, s3 ); dec. on ( , s3 ), (s3, ); loc. max. f (s3 ) 29 s3, loc. min. f (s3 ) 29 s3; CU on (s6, 0), (s6, ); CD on ( , s6 ), (0, s6 ); IPs (s6, 365 s6 ), (s6, 365 s6 )
x
1
15. A. ⺢
21. A.
2.5
15
B. y-int. 0; x-int. 0, 1 C. None D. None E. Inc. on ( 14 , ); dec. on ( , 14 ) 27 F. Loc. min. f ( 14 ) 256 1 G. CU on ( , 2 ), 1, ; CD on ( 12 , 1); IPs ( 12 , 161 ), 1, 0 H. See graph at right.
A55
25. Inc. on 0.23, 0, 1.62, ; dec. on , 0.23, 0, 1.62;
y
13. (a) None
■
CHAPTER 5
π 2
Exercises 5.1
_1 0
x
1 _
π 2
■
page 260
1. (a) 40, 52 y
y
y=ƒ
y=ƒ
5
5
1.5 0
ƒ _5
5
5
(b) 43.2, 49.2 _1.5
10
x
0
5
10
x
A56
■
3. (a)
77 60
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
, underestimate
(b)
y
y
1.5
1.5
1.0
1.0
0.5
0.5
0
1
2
3
5. (a) 8, 6.875
3. 0.856759
, overestimate
y
The Riemann sum represents the sum of the areas of the two rectangles above the x-axis minus the sum of the areas of the three rectangles below the x-axis.
0
5 x
4
25 12
1
2
3
4
5 x
y
y
5. 0.028 11. 124.1644 17.
x18 s2x x 2 dx
n
n
27. lim
n l i1
0
(b) 5, 5.375
0
x
1
y
1
x
29. (a) 4
2
2
5 i n (b) 10 sin
(c) 5.75, 5.9375
x
1
0
39.
5 x1 f x dx
47.
1 2
51.
x01 x 4 dx
1
1.
x
page 282
■
5 9
5.
3. 18 55 63 2
13.
7. 2 1e
15. 1
9. 52 19.
17. ln 3
21. 23. e 1 25. 1 4 27. 2 e 2 e 2 29. The function f x 1x 2 is not continuous on the interval 1
1
1, 3 , so the Evaluation Theorem cannot be applied. 31. 2 33. 0, 1.32; 0.84 y 35. 3.75
2
2
38
43. 3 3 49.
y tan x dx s3 4 12 12
1
11.
37. 3
41. 122
y dxx 1
364 3
(d) 2
35. 2.5
2
63 4 256 5
y
y
2 n 5 (c) 3
9 4
Exercises 5.3 0
4 3
21.
2 4in 4 1 2 4in5 n
33. 3
31.
y
n l i1
3 4
1
19. 42
25. lim
2
x
7. (a) 4 (b) 6 (c) 10 9. 475, 85 13. 0.3084 15. x0 x sin x dx
23. 3.75 2
1 0
y=˛ 0
0
x
1
1
x
(d) M6
1
7. 34.7 ft, 44.8 ft
9. 63.2 L, 70 L
11. 155 ft
0
2
x
n
13. lim
s1 15in 15n 4
n l i1
39.
15. The region under the graph of y tan x from 0 to 4 17. (a) lim
nl
(c)
64 n6
n
i
5
(b)
2
_2 12
_1
19. sin b, 1 ■
53 0
_4
2
32 3
Exercises 5.2
x 52 C
n n 1 2n 2n 1 12 2
i1
2 5
page 272
5
_6
1. 0.25
y
The Riemann sum represents the sum of the areas of the two rectangles above the x-axis minus the sum of the areas of the two rectangles below the x-axis.
2
41. 2t t 2 t 4 t 4 C 1 3 3
2 0
_2
x
1
43. sec x C
45.
4 3
47. Increase in the child’s weight (in pounds) between the ages of 5 and 10 49. Number of gallons of oil leaked in the first 2 hours 51. Increase in revenue when production is increased from 1000 to 5000 units
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
55. (a) 2 m (b) 6 m 53. Newton-meters 1 57. (a) vt 2 t 2 4t 5 ms (b) 416 23 m 59. 1.4 mi 61. 1800 L 63. 3 3
41
Exercises 5.5
Exercises 5.4
■
1. sin 3x C 3. 9 x 3 132 C 1 5. 11 2x2 C 7. 5 x 2 35 C
page 291
(b) (0, 3) (c) x 3
g
0
y
37.
182 9
5 2 t 1 cos 4 5
59.
Chapter 5 Review
© 0
x
1. True 11. False
7. t y y 2 sin y
arctan1x cos sx 11. y x2 2x 24x 2 1 39x 2 1 13. tx 15. 4x 2 1 9x 2 1 17. 2 9. hx
(b) 2, 4
16 15
51. 1 e 49. 2 57. All three areas are equal.
61. 5
L
page 300
(c)
3. True 13. True
5. False
7. True
9. True
Exercises 1. (a) 8
1 3
(b) 5.7 y
y
y=ƒ
2
19. (a) 1
■
43. 25
1 10
True-False Quiz
t
5. tx s1 2x
41. 2e 2 e
39. 4
45. lne 1 47. 0 53. 25 55. 6
tx 1 x 2
y=1+t@
35. 0
x
1
25. 29. 33.
1
3.
1 3x 221 C 11. 3 ln x3 C 1 2 3 ln 5 3x C 15. 3 s3ax bx 3 C 2 (1 cos t C 19. 3 1 e x 32 C 1 2 7 23. 3 cot x32 C 7 sin C 1 27. 3 sec 3x C ln sin1 x C x 31. ln1 cos 2 x C xe C 1 1 tan x 2 ln1 x 2 C
17. 21.
y
2
1 63
13.
(d)
page 298
1 3
9. 1. (a) 0, 2, 5, 7, 3
■
A57
y=ƒ
2
y (5, 4)
6
6
0
x
2
0
2
x
y=(x-3)@
(2, 1)
1
0 1
21. 38 3
1
3
4
x
5
23. 12sin t2 C 27. 2 lncos x 2 C 1
y 1
(b) 9
( 12 , 2), 4, 6, 8, 9
(d) See graph at right.
3. 2 4 5. f c, f b, x0x f t dt a 9 21 1 7. 37 9. 10 11. 76 13. 4 15. 3 sin 1 7 17. 1e 1 19. ln 2 4 21. sx 2 4x C 1
23. s257
25. (a) Loc. max. at 1 and 5; loc. min. at 3 and 7
(c)
2
(4, 1)
2 0
4
6
8 x
_1 _2
27. 29 29. (a) 2 sn, s4n 2, n an integer 0
(b) 0, 1, (s4n 1, s4n 3 ), and (s4n 1, s4n 1 ), n an integer 0 (c) 0.74 31. f x x 32, a 9 33. (b) Average expenditure over 0, t ; minimize average expenditure
25. 2e sx C 29.
1 4
ln1 x 4 C
33. ln 1 sec C 37. y (2e x e sx )2x Fx s1 x 4 4 x13 sx 2 3 dx 4 s3 41. 1.11 Number of barrels of oil consumed from Jan. 1, 2000, through Jan. 1, 2003 e 2x1 2x 45. 72,400 47. f x 49. c 1.62 51. 1 ex 31. 35. 39. 43.
64 5
CHAPTER 6 Exercises 6.1
■
page 309
1 1 x 2 ln x 14 x 2 C 3. 5 x sin 5x 25 cos 5x C 5. 2r 2e r2 C 1 2 2 7. x 2 cos x 2 x sin x 3 cos x C
1.
1 2
A58
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
1
11. t arctan 4t 8 ln1 16t 2 C 1
e 22 sin 3 3 cos 3 C 1 1 1 3 17. 2 2 ln 2 19. 4 4 e2 3 1 23. 2ln 22 4 ln 2 2 12 ( 12 6s3 ) 1 27. 2 4 2(sin sx sx cos sx ) C 1 3 3 3 (b) 4 cos x sin x 8 x 16 sin 2x C 2 8 (b) 3, 15 x ln x3 3ln x2 6 ln x 6 C 9 13 41. 2 ett 2 2t 2 m 43. 2 2 ln 3 9 1 13
13. 15. 21. 25. 29. 31. 37. 39.
Exercises 6.2
■
page 319
11 cos 5x 13 cos 3x C 3. 384 3 1 5. 4 7. 38 9. 2 2 sin 4 sin 2 C 1 1 11. 192 3 4 13. 2 cos 2x ln cos x C 1 15. ln1 sin x C 17. 2 tan 2 x C 1 2 19. tan x x C 21. 5 tan 5 t 3 tan 3 t tan t C 117 1 3 23. 8 25. 3 sec x sec x C 1 4 27. 4 sec x tan 2x ln sec x C
1.
1 5
1 6
35. (b)
33. ln csc x cot x C
sin 3x 141 sin 7x C
sx 2 9 C 9x 1 41. 24 s38 4
29. s3 3 1 1 31. 3 csc 3 5 csc 5 C
39. 3 x 2 18 sx 2 9 C 1
37.
ln 2 95 ln 3 (or 95 ln 83) 1 1 19. 36 ln x 5 361 ln x 1 C 6x 5 21. 2 ln x 3 ln x 2 1x C 17.
27 5
23. ln x 1 2 lnx 2 9 3 tan1x3 C 1
1
25.
1 2
lnx 2 1 (1s2 ) tan1(xs2 ) C
27.
1 2
lnx 2 2x 5 32 tan1
29.
1 3
ln x 1 16 lnx 2 x 1
31.
1 12 ln x
33.
1 2 s3 x1 tan1 2x 2 2x 4 9 s3
35. 2 ln
39. ln
x1 x1
25 9
e 2 ex 1
2
C
3 10
37.
x
x1 2
C
1 2x 1 tan1 C s3 s3
2x 1 C 3x 2 2x 4
x 2 153 34 x 2 123 C
C
1 2
) lnx 2 x 2 2x s7 tan1
Exercises 6.4
page 333
■
1. 4
1 1 sec x tan x ln sec x tan x C 2 2 s4x 2 9 C 5. 7. 3 6 9x 1 9. 2 tan21z ln cos1z C
3.
x x sin1 C 53. sx 2 7 C a sa x 2 s1 x 2 1 s1 x 2 C 55. ln x
2y 1 2y 1 7 11. s6 4y 4y 2 8 sin1 8 s7 121 6 4y 4y 2 32 C
57. 59.
1 4 1 3
2
sin1 x 2 14 x 2 s1 x 4 C ln 3x 1 s9x 2 6x 8 C
1 x1 tan1x 1 2 61. 2 x 2x 2 63. s 1 cos 3 t3 Exercises 6.3
■
65.
1 6
C
(s48 sec1 7)
page 327
A A B B C (b) x3 3x 1 x x1 x 12 A B Bx C A 2 3. (a) (b) x4 x1 x1 x x1 A B Cx D 2 5. (a) 1 x1 x1 x 1 At B Ct D Et F (b) 2 2 2 t 1 t 4 t 4 2
1. (a)
2x 1 C s7 1 43. t ln P 9 ln0.9P 900 C, where C 10.23 41. ( x
s25 x 2 43. 45. ln (sx 2 16 x) C C 25x 1 1 47. 4 sin12x 2 x s1 4x 2 C 1 49. 6 sec1x3 sx 2 92x 2 C 51.
7. x 6 ln x 6 C 9. 2 ln x 5 ln x 2 C 1 3 1 11. 2 ln 2 13. a ln x b C 15. 2 ln 2 2
9. 2 2x 1 ln2x 1 x C
sin 3x 3 lnsin x 1 C
13.
1 9
15.
e x s3 1 ln x C 2s3 e s3
17.
1 5
ln x 5 sx 10 2 C
[
]
ln xs4 ln x 2 ln ln x s4 ln x 2 C se 2x 1 cos1ex C 14 x 5 x 2 32 58 x s5 x 2 258 sin1( xs5 ) C 15 sin 2x cos 3x 152 cos3x C 1 1 29. 10 1 2x52 6 1 2x32 C 1 1 31. ln cos x 2 tan 2x 4 tan 4x C 19. 21. 25. 27.
1 2
2
33.
2
ln(s2 2x 1 2 x s2 1 C ln 2 2 ln 2
x1
2x
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 6.5
■
1
page 343
1. (a) L 2 6, R2 12, M2 9.6
(b) L 2 is an underestimate, R2 and M2 are overestimates. (c) T2 9 I (d) L n Tn I Mn Rn 3. (a) T4 0.895759 (underestimate) (b) M4 0.908907 (overestimate) T4 I M4 5. (a) 5.932957, E M 0.063353 (b) 5.869247, E S 0.000357 7. (a) 2.413790 (b) 2.411453 (c) 2.412232 9. (a) 0.146879 (b) 0.147391 (c) 0.147219 11. (a) 4.513618 (b) 4.748256 (c) 4.675111 13. (a) 0.495333 (b) 0.543321 (c) 0.526123 15. (a) 1.064275 (b) 1.067416 (c) 1.074915 17. (a) T10 0.881839, M10 0.882202 (b) ET 0.013, EM 0.006 (c) n 366 for Tn , n 259 for Mn 19. (a) T10 1.719713, ET 0.001432; S10 1.718283, ES 0.000001 (b) ET 0.002266, ES 0.0000016 (c) n 151 for Tn , n 107 for Mn , n 8 for Sn 21. (a) 2.8 (b) 7.954926518 (c) 0.2894 (d) 7.954926521 (e) The actual error is much smaller. (f ) 10.9 (g) 7.953789422 (h) 0.0593 (i) The actual error is smaller. ( j) n 50
3. 2 12t 2 ; 0.495, 0.49995, 0.4999995; 0.5 1 5. 12 7. D 9. 2e2 11. D 13. 0 19. 1 21. 9 23. D 25. 17. D 75 8 8 31. 3 ln 2 9 27. 4 29. D y
2 9
Rn
Tn
4 8 16
0.140625 0.191406 0.219727
n
EL
ER
ET
EM
4 8 16
0.109375 0.058594 0.030273
0.140625 0.066406 0.032227
0.015625 0.003906 0.000977
0.007813 0.001953 0.000488
0.390625 0.316406 0.282227
0.265625 0.253906 0.250977
0
The observations are the same as after Example 1. 25. (a) 11.5 (b) 12 (c) 11.6 27. 37.73 fts 29. 10,177 megawatt-hours (b) 0.3413 33. 59.4 31. (a) 23.44
2 ≈+9
0
20
π 2
0
39. (a)
y
t
1
2 5 10 100 1,000 10,000
(c)
t
sin 2xx 2 dx
It appears that the integral is convergent.
0.447453 0.577101 0.621306 0.668479 0.672957 0.673407
1
1 ≈
©=
sin@ x ≈
1
10
0.1
41. C 53. (a)
43. C
45. D
47.
49. p 1, 11 p
y 1
y=F(t)
y 1 0
0
0.5
Exercises 6.6
1
■
1.5
2
x
page 352
Abbreviations: C, convergent; D, divergent 1. (a) Infinite interval (b) Infinite discontinuity (c) Infinite discontinuity (d) Infinite interval
7
y=sec@ x
ƒ=
35.
_7
x
1
37. Infinite area
Mn 0.242188 0.248047 0.249512
y=
x"1
y " ex
23. Ln
1 25
0.5
n
15. 32 3
35. 23
33. e
A59
700
t (in hours)
(b) The rate at which the fraction Ft increases as t increases (c) 1; all bulbs burn out eventually 55. 8264.5 years 57. 1000 61. C 1; ln 2
A60
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Chapter 6 Review
■
Exercises 7.2
page 354
■
page 370
1. 2
True-False Quiz
y
y
1 y= x
1. False 3. False 5. False 7. False 9. (a) True (b) False 11. False 13. False
x=1
Exercises
0
x=2 x
0
x
y=0
1. 5 10 ln 3 3. ln 2 1 3 3 1 5. 9 sec 9x 7 sec 7x 5 sec 5x 3 sec 3x C 2
7. cosln t C 9. 5 ln 4 1 13. ln x 2 lnx 2 1 C 64
11. s3 3
124 25
3. 162
y
15. 3 sin 3 5 sin 5 7 sin 7 C 17. x sec x ln sec x tan x C 1
19. 21. 23. 25. 27. 33. 35. 39. 45. 51.
2
1
(6, 9)
ln9x 2 6x 5 19 tan1 ( 12 3x 1) C ln x 2 sx 2 4x C 1 12 cot 3 4x 3 cot 4x C 3 2 1 1 2 lnx 1 3 tan x s2 tan (xs2 ) C 2 29. 0 31. 6 32 5 x x sin1 C 2 2 s4 x 1 1 4s1 sx C 37. 2 sin 2x 8 cos 4x C 1 1 1 41. 36 43. D 8e 4 4 ln 4 8 47. D 49. /4 1 x 2x sin1e x ] C [ e e s1 2 1 18
y
y=9
x=0 x=2 œ„ y
53. 4 2x 1 sx 2 x 1 8 ln ( x 2 sx 2 x 1 ) C 55. No 57. (a) 1.090608 (overestimate) 1
3
0
5. 4 21
x
0
y
x
y (1, 1)
y=x y=˛ 0
x
0
x
1
(b) 1.088840 (underestimate) (c) 1.089429 (unknown) 59. (a) 0.006 , n 259 (b) 0.003 , n 183 61. 8.6 mi 63. (a) 3.8 (b) 1.7867, 0.000646 (c) n 30
7. 6415
y
y
¥=x
(4, 2)
x=2y x
0
9. 6
x
0 y
y
CHAPTER 7 Exercises 7.1 32 1. 3 9 11. 8
■
page 361
3. e 1e 3 32 13. 3 15. ln 2
1
10
5. 19.5 7. 6 17. 0, 0.90; 0.04
y=1
y=œ„ x
9. 72
(1, 1)
y=x
19. 0, 0.70; 0.08 1 21. 2 y
0 x
y=cos x
x
0
11. 2930
y=sin 2x
y
y
x=¥
(1, 1)
0 π 6
π 2
x
y=≈ _1
23. 27. 29. 33.
118 ft 25. 84 m2 8868; increase in population over a 10-year period r sR 2 r 2 r 22 R 2 arcsinrR 31. 6 35. f t 3t 2 37. 0 m 1; m ln m 1 4 23
0
x
0 x=_1
13. 1330
15. x04 1 tan3 x2 dx
17. 0, 0.747; 0.132
19.
11 8
2
x
21. (a) Solid obtained by rotating the region 0 y cos x, 0 x 2 about the x-axis (b) Solid obtained by rotating the region y 4 x y 2, 0 y 1 about the y-axis 1 1 23. 1110 cm 3 25. 3 r 2h 27. h2 (r 3 h) 2 2 3 29. 3 b h 31. 10 cm 33. 24 35. 2 37. 3 r 2 2 2 2 41. (a) 8R x0 sr y dy (b) 2 r R 43. (b) r 2h 5 r 3 2 2 2 2 45. 12 r 47. 8 x0 sR y sr y dy Exercises 7.3
■
page 376
1. Circumference 2 x, height x x 12; 15 y
1
x
3. 2 5. 1 1e 7. 16 9. 212 11. 7687 13. 2503 15. 176 17. 676 19. 24 21. x12 2 x ln x dx 23.
x01 2 x 1 sin x2 x 4 dx 27. 1.142 x0 2 4 y ssin y dy
0 x 3 about the y-axis 31. Solid obtained by rotating the region bounded by (i) x 1 y 2, x 0, and y 0, or (ii) x y 2, x 1, and y 0 about the line y 3 33. 8110 35. 8 3 ln 4 37. 43 4 1 39. 3 r 3 41. 3 r 2h ■
page 383
1. 3 s10 3. 243 (82 s82 1) 5. 240 7. 3 9. ln(s2 1) 11. sinh 1 13. s1 e 2 s2 ln(s1 e 2 1) 1 ln(s2 1) 2
15.
1261
x02 s1 sin 2 x dx
19. 5.115840
17.
y
4
1
1 1 a b
21. (a) Gm1 m2
23. 6.5 10 6 N 25. 3.5 10 4 lb 27. 5.27 10 5 N 29. (a) 5.63 10 3 lb (b) 5.06 10 4 lb
(c) 4.88 10 4 lb (d) 3.03 10 5 lb 1 10 31. 2.5 10 5 N 33. 10; 1; (21 , 21 ) 35. 0, 1.6 1 2 1 37. (1e 1, 4 e 1) 39. (5 , 2)
(
s2 4 1 , 4 s2 1) 4 (s2 1) 4 2 1 1 43. 3 , 0, (0, 3 ) 47. (0, 12 ) 49. 3 r 2h 41.
■
page 404
1. y Kx 3. y Ksx 2 1 1 5. y ln sec y 3 x 3 x C
9. u st tan t 25 1 11. cos x x sin x y 2 3 e 3y 2
2
2
1
2 3
32
1
y=sin ≈
_œ„„„ _œπ/2
œπ/2 „„„
0
(c) No 19. cos y cos x 1
5
s9y 4 6y 2 2 dy
21. 1.569619
23. ln 3
1 2
2.5
21. III 0
7. u Ae 2tt
4a 4 15. y 7e x sin x a s3 17. (a) sin1y x 2 C (b) y sinx 2 , s2 x s2
y
25. 6
(b) 8.50 109 J
13. y
25. 29. Solid obtained by rotating the region 0 y x 4,
Exercises 7.4
11. 650,000 ft-lb 13. 3857 J 15. 2450 J 17. (a) 1.06 10 6 J (b) 2.0 m
Exercises 7.6
x
A61
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
25.
1 x
23. IV 27.
y
y 3
(b) (a)
27. sx 27 [1 9x32 10 s10 ] 31. 29.36 in. 2
Exercises 7.5 ■ page 394 15 1. 9 ft-lb 3. 180 J 5. 4 ft-lb 25 7. (a) 24 1.04 J (b) 10.8 cm 9. (a) 625 ft-lb (b) 1875 4 ft-lb
29. 209.1 m
2.5
0
3 x
_3
_3
3x
_3
(c) _3
A62
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
29.
31.
y 3
CHAPTER 8
y 3
Exercises 8.1
page 418
■
Abbreviations: C, convergent; D, divergent 3x
_3
3x
_3
_3
_3
(b) At M
and A0 A0
Ce 1 sM kt Ce 1
Chapter 7 Review
■
, where C
sM sA0 sM sA0
page 408
Exercises 1.
125 6
3. e
7. 16565
5. 64 15
11 6
5
6
2 n1
5. a n ( 3 )
1
3. 3 , 5 , 7 , 9 , 11 , 13 ; yes; 2
(b) rk; the concentration approaches rk regardless of the value of C0 37. P40 732, P80 985; t 55 39. (a) dydt ky1 y y0 (b) y (c) 3:36 PM y0 1 y0 ekt t100 0.2 43. (a) 15e kg (b) 15e 12.3 kg 45. About 4.9% 47. tk 49. (a) dAdt k sA M A 2
defined as a function whose domain is the set of positive integers. (b) The terms an approach 8 as n becomes large. (c) The terms an become large as n becomes large. 1 2 3 4
33. (b) Pt M Mekt; M 35. (a) Ct C0 rkekt rk
sM kt
1. (a) A sequence is an ordered list of numbers. It can also be
9. 3 2ah h 2 32 4
11. x01 1 x 32 1 x 2 2 dx 13. (a) 215 (b) 6 (c) 815 15. (a) 0.38 (b) 0.87 17. Solid obtained by rotating the region 0 y cos x ,
0 x 2 about the y-axis 19. Solid obtained by rotating the region in the first quadrant bounded by x 4 y 2 and the axes about the x-axis 125 15 21. 36 23. 3 s3 m 3 25. 2 27. 3.292287 29. 3.2 J 31. (a) 80003 8378 ft-lb (b) 2.1 ft 1 12 33. 458 lb 35. ( 2 , 5 ) 37. 2 2 3 2 39. y y cos x x sin x C 2 41. rt 5e tt 43. (a)
7. a n 5n 3 9. 5 11. 0 13. D 15. 0 17. 0 19. 0 21. 0 23. e 2 25. D 27. ln 2 29. (a) 1060, 1123.60, 1191.02, 1262.48, 1338.23 (b) D 31. Convergent by the Monotonic Sequence Theorem;
5 L8 33. Decreasing; yes 35. Not monotonic; yes 37. 2 39. (3 s5 )2 41. (b) (1 s5 )2 43. 62 Exercises 8.2
page 427
■
1. (a) A sequence is an ordered list of numbers whereas a
series is the sum of a list of numbers. (b) A series is convergent if the sequence of partial sums is a convergent sequence. A series is divergent if it is not convergent. 5 3. 3 5. 15 7. D 9. D 11. D 13. 2 3 11 2 1138 15. D 17. D 19. 2 21. 6 23. 9 25. 333 27. 3 x 3; x3 x
2 2 cos x
29. All x ;
31. a1 0, an 2 nn 1 for n 1, sum 1 33. (a) Sn D1 c n 1 c
(b) 5
35. (s3 1)2 37. 1 nn 1 39. The series is divergent. 45. sn is bounded and increasing. 1 2 1 2 7 8
47. (a) 0, 9 , 9 , 3 , 3 , 9 , 9 , 1 49. (a) 2 , 6 , 24 , 120 ; n 1! 1 n 1! 1 5 23 119
Exercises 8.3 1. C
(c) 1
page 436
■
y
y=
a™ 0
1
1 x 1.3
a£ 2
a¢ 3
a∞ 4
...
x
3. (a) Nothing (b) C 5. p-series; geometric series; b 1; 1 b 1 7. C 9. C 11. C 13. C 15. D 17. C 19. C 21. D 23. C 25. D 27. p 1 31. (a) 1.54977, error 0.1 (b) 1.64522, error 0.005
(b) The pair of lines y x, for C 0; the hyperbola x 2 y 2 C for C 0.
(c) n 1000 37. Yes
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 8.4
page 446
■
Abbreviations: AC, absolutely convergent; CC, conditionally convergent 1. (a) A series whose terms are alternately positive and negative (b) 0 bn1 bn and lim n l bn 0, where bn an (c) Rn bn1 3. C 5. C 7. D 9. 7 11. 5 13. 0.9721 15. 0.0676 17. An underestimate 19. D 21. AC 23. CC 25. AC 27. AC 29. AC 31. D 33. AC 35. AC 37. D 39. (a) and (d)
19. s∞ s£ s¡
3
f
s™ s¸ 4
6
s¢ 1.2
21.
n0
Exercises 8.5
A63
2n1
2x ,R1 2n 1
s∞=sß 3
f
page 451
■
1. A series of the form n0 cnx an, where x is a variable
and a and the cn’s are constants 3. 1, 1, 1 5. 1, 1, 1 7. , , 1 1 1 9. 2 , ( 2 , 2 11. 4, 4, 4 13. 2, 4, 0 1 15. b, a b, a b 17. 0, { 2 } 19. (a) Yes (b) No 21. k k 23. (a) , s¸ s™ s¢ 2 (b), (c)
s£=s¢ s¡=s™
s¸
2
2
]
3
23. C
t 8n2 ,R1 8n 2
n0
25. C
1
n1
27. 0.199989 33. (b) 0.920
J¡ 8
_8
Exercises 8.7 1. b 8 f
s¡ s£ s∞
_2
25. 1, 1, f x 1 2x1 x 2
29. 2
■
page 456
1. 10
1 x , 1, 1
n n
3.
5.
n0
1
7.
5
n0
11.
n0
n1
x n, 5, 5
9.
1
n0
1n1 1 x n, 1, 1 2 n1 n
n0
1 2 1 (c) 2 (b)
1 n 2n 1x , R 1 n
1 nn 1x , R 1 n
n2
15. ln 5
n1
n
n0
n
n
1 2n1 x , 3, 3 9 n1
xn ,R5 n5 n
n2 n 17. n1 x , R 2 n3 2 1n1 n ln 3 x ,R3 n3 n n1
n 1x , R 1 n
x 2n ,R 2n!
1
n
7.
n0
5n n x ,R n!
x 2n1 ,R 9. n0 2n 1! 11. 7 5x 2 x 22, R e3 x 3n, R 13. n0 n! 1 x 2n, R 15. 1n1 2n! n0
1 3 5 2n 1 x 9 n, R 9 2 n 3 2n1 n! 1 3 5 2n 3 n x x ,R1 23. 1 1n1 2 2 nn! n2 n 1n 2 n x ,R2 25. 1n 2 n4 n0 17.
1 n 1x , R 1 n
, 1, 1
3n
n0
13. (a)
x
3.
n0
n0
Exercises 8.6
page 469
58!
5.
■
8
x 2n1 ,R1 4n 2 1 29. 0.000983 31. 0.09531 37. 1, 1 , 1, 1, 1, 1
n1
1
n
n0
2n 2n x ,R 2n! n0 1 x 2n2, R 1 29. 1n 2n 1 n0 1 n2 x ,R 31. 1n n! n0
27.
1
n
A64
33.
1 2
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
1
x
n
n1
1 3 5 2n 1 2n1 x ,R2 n! 2 3n1
(b)
1n12 2n1x 2n ,R 35. 2n! n1 1 x 4n, R 37. 1n 2n! n0
4 2
1.5
T¸=T¡=T™=T£
_1.5
T0 T1
T2 T3
T4 T5
0.7071
1
0.6916
0.7074
0.7071
0
1
0.2337
0.0200
0.0009
1
1
3.9348
0.1239
1.2114
T6
(c) As n increases, Tnx is a good approximation to f x on a larger and larger interval.
1.5
Tˆ=T˜=T¡¸=T¡¡
f
x
3.
1 s3 2 2
f
x
6
1 4
x
6
2
s3 12
x
6
1.1
f T£
T¢=T∞=Tß=T¶
_1.5
39. 0.81873
41. (a) 1
n1
(b) x
1 3 5 2n 1 2n x 2 nn!
1
1 3 5 2n 1 x 2n1 2 nn! 2n 1
n1
x 6n2 43. C 1 n 6n 22n! n0 x4 1 3 5 2n 3 3n1 45. C x 1n1 x 8 2 nn!3n 1 n2 1 1 49. 0.09998750 51. 3 53. 120 47. 0.440 55. 1 2 x 2 24 x 4 57. 1 6 x 2 x 4 59. e 61. 1s2 63. e 3 1 3
25
1
7 360
1
5. x x 1 6
3
1.6
_1
x4
nx
n
1
T£
65. (a)
2
f
(b) 2
_1.6
n1
2
67. (a) 1
x 2
1
n1
n2
1 3 5 2n 3 2n x 2 nn!
7. x 2x 2 2x 3 3
(b) 99,225 Exercises 8.8
■
page 476
_1
1.5
1. (a) T0 x 1 T1x, T2x 1 2 x 2 T3x, 1
T4x 1 12 x 2 241 x 4 T5x, 1 T6x 1 12 x 2 241 x 4 720 x6
T¢=T∞
T£ f
2
f _2π
_2
T™=T£
(b) 0.00125
11. (a) x 3 x3 1
(b) 0.06
13. (a) 1 x (b) 0.00006 3 3 15. (a) 8 8 x 16 1024 x 16 2 2
2π
Tß
9. (a) 1 2 x 1
T¸=T¡
_4
5 65,536
(b) 0.0000034 17. 0.57358 19. Four 21. 1.037 x 1.037 23. 21 m, no 27. (c) They differ by about 8 109 km.
x 16 3
3
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Chapter 8 Review
■
page 479
7. (a)
A65
(b) y 1 x 2, x 0
y (0, 1) t=0
True-False Quiz 1. False 11. True
3. True 13. True
(1, 0) t=1
5. False 7. False 9. False 15. False 17. True
x
0
Exercises 1
5. 0 7. e 12 9. C 11. C 1. 2 3. D 13. D 15. C 17. C 19. C 21. CC 23. AC 25. 8 27. 4 29. e e 33. 0.9721 37. 4, 6, 2 39. 0.5, [2.5, 3.5) 1 41. 2
1
n
n0
43.
1 x 2n! 6
1 x
2n
45.
,R1
n n2
n0
s3 x 2n 1! 6 n1
2 n1
(2, _3) t=4
9. (a) x 2 y 2 1, x 0
11. (a) y 1x, y 1
(b)
(b)
y (0, 1)
xn ,R1 n
(1, 1) x
0
49. 51. 53.
(b)
x
0
47.
y
x 8n4 1 ,R 2n 1! n0 1 5 9 4n 3 n 1 x , R 16 2 n! 2 6n1 n1 xn C ln x n1 n n! (a) 1 12 x 1 18 x 12 161 x 13 (c) 0.000006 1.5 n
(0, _1)
13. (a) y 2 ln x 1 1
(b)
y 1 0
x
1
15. Moves counterclockwise along the circle
x 32 y 12 4 from 3, 3 to 3, 1
T£
17. Moves 3 times clockwise around the ellipse
x 225 y 24 1, starting and ending at 0, 2
f
55.
19.
2
0
(0, 1) t=1
CHAPTER 9 Exercises 9.1 1.
y
1 6
x
(_1, 0) t=0 ■
(0, _1) t=_1
page 486 3.
y
y t=π {0, π@}
t=5 5, 5} {1+œ„
21.
y 1
t=0 (1, 0)
5
t=4 (3, 0)
t=0 (0, 0)
5. (a)
(b) y 23 x 133
y
(1, 5) t=2
(_2, 3) t=1 (_5, 1) t=0 0
x
3
3 t=0
x
(_8, _1) t=_1
23.
1
t= 2
5 x
1
3
x
3
25. (b) x 2 5t, y 7 8t, 0 t 1 27. (a) x 2 cos t, y 1 2 sin t, 0 t 2
(b) x 2 cos t, y 1 2 sin t, 0 t 6 (c) x 2 cos t, y 1 2 sin t, 2 t 32 31. The curve y x 23 is generated in (a). In (b), only the portion with x 0 is generated, and in (c) we get only the portion with x 0. 35. x a cos , y b sin ; x 2a 2 y 2b 2 1, ellipse
A66
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
41. s2 e 1
37. (a) Two points of intersection
8
4
6
6 25 4
(b) One collision point at 3, 0 when t 32 (c) There are still two intersection points, but no collision point. 39. For c 0, there is a cusp; for c 0, there is a loop whose size increases as c increases. 3
43. e 3 11 e8
0
2.5
21
1
_1
1 2
1
1
0 0
1.5
0
1.5
47. 6 s2, s2
45. 612.3053 51. (a) _3
21 1
t 0, 4
15
_1
41. As n increases, the number of oscillations increases; a and
b determine the width and height. Exercises 9.2
■
15
15
page 494
1. 53t 2 1 3. y x 5. y 2ex 3 7. y 2 x 3 3 9. 1 2 t, 34t, t 0 11. et, et1 e t , t 0
15
(b) 294
13. Horizontal at 6, 16, vertical at 10, 0 15. Horizontal at (s2 , 1) (four points), vertical at 2, 0 1 1 17. 0.25, 0.36, ( 4 , (1s2 ) 2 ln 2) 19. 21. y x, y x
Exercises 9.3
■
page 502
1. (a)
(b) π
”1, 2 ’
π 4
y
7.5
O π
”_2, 4 ’ x
0 8.5
3
1, 52, 1, 32 (c)
2, 54, 2, 94
(3, 2)
1 2
23. (a) d sin r d cos 27. ab 29. e 2 12 33.
x12 s1 4t 2 dt
37. 4s2 2
35.
25. 5, 6, ( 27 , 31. 2r 2 d 2
208 32 3
)
O
x02 s3 2 sin t 2 cos t dt
7
3, 2 2, 3, 2 3. (a)
π
”3, 2 ’
0
(b)
3π
”2œ„ 2, 4 ’ 3π 4
π 2
5
O
O
39. s103 ln(3 s10 ) s2 ln(1 s2 )
0, 3
2, 2
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
(c)
35. π 3
■
A67
37.
π
¨= 4
(12, s32)
π
”_1, 3 ’
5. (a) (i) (s2, 4) (ii) (s2, 54) (b) (i) 4, 116 (ii) 4, 56 7.
39.
¨=
2π 3
41.
π
¨= 3
2
3π
9.
”_1, 2 ’
π ¨= 6
r=2 r=1
(3, π)
1
(3, 0)
O
O O
π
¨=_ 2
r=4
43. 11.
¨=
1
π ”_1, 2 ’
45.
7π 3
(2, 0)
r=3
(6, 0) O
r=2 O
¨=
47. s3 49. 51. Horizontal at (3s2, 4), (3s2, 34);
5π 3
13. Circle, center (0, 2 ), radius 2 3
3
15. Horizontal line, 1 unit
above the x-axis 17. r cot csc 19. r 2c cos 21. (a) 6 (b) x 3 23. 25.
vertical at 3, 0, 0, 2 3 3 53. Horizontal at ( 2 , 3), ( 2 , 53), and the pole; vertical at (2, 0), ( 12 , 23), ( 12 , 43) 55. Center b2, a2, radius sa 2 b 22 57.
59.
3.5
7
π
”1, 2 ’
_3
O π
¨=_ 6
3
7
O _2.5
27.
29.
about the origin 63. (a) A rose with n loops if n is odd and 2n loops if n is even (b) Number of loops is always 2n 65. For 0 a 1, the curve is an oval, which develops a dimple as a l 1. When a 1, the curve splits into two parts, one of which has a loop.
O
7
2
¨=π 4
π
1
5
4 6
33. ¨= 8
8
7
61. By counterclockwise rotation through angle 6, 3, or
O
31.
7
Exercises 9.4
■
1. 64 9. 4
3. 12 s38
2
3
O
page 508 5. 36
7. 414
A68
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
(d)
13. 3
11.
3
r=2 cos 3¨
3
π
¨= 6
1
”- 4 , 0 ’
” 4 , π’ O
3
3
3
3
15. 8 17. 2 s3 4 19. 3 2 s3 21. 3
23. 8 2
25. 2 1
1
27. 31.
1
( 3 s3 ) 29. ( , 3), ( , 53), and the pole (s32, 3), (s32, 23), and the pole 1 4
1 2
1 2
35. 3 2 132 1 8
33.
Exercises 9.5
37. 29.0653
page 514
■
1. r 424 7 sin 3. r 154 3 cos 5. r 81 sin 7. r 42 cos (b) Parabola (c) y 1 9. (a) 1
(d)
x=_ 8
17. The ellipse is nearly circular
23. 25. 27. 29.
e=1.0 e=0.6 e=0.8
(b) r 1.49 10 8 1 0.017 cos 35.64 AU 7.0 10 7 km 3.6 10 8 km
Chapter 9 Review
■
page 515
True-False Quiz 1. False
1 π
” 2 , 2 ’
e=0.4
when e is close to 0 and becomes more elongated as e l 1. At e 1, the curve becomes a parabola.
3. False
5. True
7. False
y=1
Exercises 1. x y 2 8y 12
O
3. y 1x y
y (0, 6), t=_4
11. (a)
1 4
(b) Ellipse
(d)
0
(5, 1), t=1
(c) y 12
x
π
”4, 2 ’
(3, π)
x
5. x t, y st, t 0; x t 4, y t 2;
O
x tan 2 t, y tan t, 0 t 2
(3, 0) 12 3π ” 5 , 2 ’
7.
9. π ”1, 2 ’
1 (2, π)
y=_12
(2, 0)
(2, π)
O O
13. (a)
1 3
(b) Ellipse
(d)
(c) x
9 2
_1 3π ”1, 2 ’
9
x= 2 3 π
” 2 , 2 ’ 9 ” 4 , π’
11.
9
13. 1
” 8 , 0’
” 2 , 0’
O
O 3 3π
” 2 , 2 ’
15. (a) 2
(b) Hyperbola
x"1
(c) x 38
15. r 2cos sin
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
17.
A69
21. A plane parallel to the xz-plane and 4 units to the left of it 23. A half-space consisting of all points in front of the plane
x3
_0.3
25. All points on or between the horizontal planes z 0 and
1.2
z6
27. All points on or inside a sphere with radius s3 and center O 29. All points on or inside a circular cylinder of radius 3 with 21. 1 19. 2 23. sin t t cos tcos t t sin t, t 2 2cos t t sin t3 11 3 25. ( 8 , 4 ) y 27. Vertical tangent at
(3a2, s3 a2), 3a, 0; horizontal tangent at a, 0, (a2, 3s3 a2)
(3a, 0)
axis the y-axis 31. y 0 33. r 2 x 2 y 2 z 2 R 2 35. 14x 6y 10z 9, a plane perpendicular to AB Exercises 10.2
page 529
■
l l l l l l l l 1. AB DC, DA CB, DE EB, EA CE 3. (a) (b) u-v u
(a, 0) x
0
u+v
_v
u
v
(c) 29. 18 31. 2, 3 35. 2(5 s5 1)
33. 12
2
37.
2
Exercises 10.1
7. a 2, 0, 2
y
z A(2, 3)
A (0, 3, 1) 0 B(_2, 1)
y 0
a
a
x
B (2, 3, _1)
x
11. 0, 1, 1 z
y
2
13. 2, 18 , 1, 42 , 13, 10 15. i j 2 k, 4 i j 9 k, s14 , s82 17. 9 i 9 j 9 k 8
y
y=2-x, z=0
0
k0, 1, _1l
x
k3, _1l
x
0
7. (a) PQ 6 , QR 2 s10 , RP 6 ; isosceles triangle (b) PQ 3 , QR 3 s5 , RP 6 ; right triangle 9. (a) No (b) Yes 11. x 32 y 82 z 12 30 1 1 1 13. 3, 2, 1, 5 15. ( 2 , 2 , 2 ), s32 5 1 1 17. (b) 2 , 2 s94, 2 s85 19. (a) x 22 y 32 z 62 36 (b) x 22 y 32 z 62 4 (c) x 22 y 32 z 62 9
k0, 0, _3l
k_2, 4l
y=2-x
x
k0, 1, 2l
k1, 3l z
2
5. a 4, 2
page 521
■
intersects the xy-plane in the line y 2 x, z 0 (see graph at right)
u
y
1. 4, 0, 3 3. Q; R 5. A vertical plane that
w+v+u
9. 1, 3
CHAPTER 10
v w
w
v
2 s4 1 2 s 1 s4 1 ln 2 s 2 1 39. All curves have the vertical asymptote x 1. For c 1, the curve bulges to the right. At c 1, the curve is the line x 1. For 1 c 0, it bulges to the left. At c 0 there is a cusp at (0, 0). For c 0, there is a loop. 41. r 43 cos 43. x acot sin cos , y a1 sin 2 2
(d)
v+w
1
4
19.
2, 2s3
21. F (6 s3 5 s2 ) i (6 5 s2 ) j 3.32 i 13.07 j,
F 13.5 lb, 76
23. s493 22.2 mih, N8 W 25. T1 196 i 3.92 j, T2 196 i 3.92 j y 27. (a), (b) (d) s 97 , t sa
a 0
b
c x
tb
29. A sphere with radius 1, centered at x0, y0, z0
11 7
A70
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 10.3
■
page 535
1. (b), (c), (d) are meaningful 3. 15 5. 19 7. 32
9. u v 2 , u w 2 1
1
9 4 s7 1 95 101 15. cos1 20 2 s7 17. (a) Neither (b) Orthogonal (c) Orthogonal (d) Parallel 19. Yes 21. i j ks3 or i j ks3 9 12 9 27 54 18 23. 3, 5 , 5 25. 7 , 49 , 49 , 49 29. 0, 0, 2 s10 or any vector of the form s, t, 3s 2 s10 , s, t ⺢ 13 31. 38 J 33. 250 cos 20 235 ft-lb 35. 5 1 37. cos (1s3 ) 55 13. cos1
[
]
41. x 3t, y 1 t, z 2 2t 43. P1 and P3 are parallel, P2 and P4 are identical 25 45. s225 47. 3 49. 7 s618 53. 1s6 Exercises 10.6
■
page 558
1. (a) Parabola
(b) Parabolic cylinder with rulings parallel to the z-axis (c) Parabolic cylinder with rulings parallel to the x-axis 3. Elliptic cylinder 5. Parabolic cylinder z
z
y
x
Exercises 10.4
■
1. 2 i j 3 k 3. 3 i 4 j 2 k 5. 2 i 13 j 8 k 7. t 4 i 2t 3 j t 2 k (c) Vector 9. (a) Scalar (b) Meaningless
(d) Meaningless (e) Meaningless (f) Scalar 11. 24; into the page 13. 5, 3, 1 , 5, 3, 1 15. 12s209, 1s209, 8s209 , 12s209, 1s209, 8s209 23. 16 25. (a) 6, 3, 2 (b) 72 1 27. (a) 13, 14, 5 (b) 2 s390 29. 82 31. 3 35. 10.8 sin 80 10.6 N m 37. 417 N 39. (b) s973 45. (a) No (b) No (c) Yes Exercises 10.5 1. (a) True
■
x
y
page 543
page 551
(b) False (c) True (d) False (e) False (f ) True (g) False (h) True (i) True ( j) False (k) True 3. r 2 i 4 j 10 k t3 i j 8 k; x 2 3t, y 4 t, z 10 8t 5. r i 6 k ti 3 j k; x 1 t, y 3t, z 6 t 1 7. x 2 2t, y 1 2 t, z 3 4t; x 22 2y 2 z 34 9. x 1 t, y 1 2t, z 1 t ; x 1 y 12 z 1 11. Yes 13. (a) x2 y 23 z 17 (b) ( 27 , 117 , 0), ( 43 , 0, 113 ), 0, 2, 1 15. rt 2 i j 4 k t2 i 7 j 3 k, 0 t 1 17. Parallel 19. Skew 21. 2x y 5z 1 23. 2x y 3z 0 25. x y z 2 27. 33x 10y 4z 190 29. x 2y 4z 1 31. 2, 3, 5 33. Neither, 70.5 35. Parallel 37. (a) x 2 y8 z7 (b) cos1(s65) 119 or 61 39. xa yb zc 1
7. Cylindrical surface z
y
x
9. (a) x k, y 2 z 2 1 k 2, hyperbola k 1;
y k, x 2 z 2 1 k 2, hyperbola k 1; z k, x 2 z 2 1 k 2, circle (b) The hyperboloid is rotated so that it has axis the y-axis (c) The hyperboloid is shifted one unit in the negative y-direction
11. x k, y 2 4z 2 4 4k 29, ellipse ( k 3)
y k, x 2 9z 2 9 9k 24, ellipse ( k 2) z k, 4x 2 9y 2 361 k 2 , ellipse ( k 1) z
Ellipsoid
(0, 0, 1)
(_3, 0, 0) (0, 2, 0)
(0, _2, 0)
y (3, 0, 0)
(0, 0, _1)
x
13. x k, y 2 z 2 k 2, hyperbola k 0
y k, x 2 z 2 k 2, circle k 0 z k, y 2 x 2 k 2, hyperbola k 0 x 0, y z, lines z 0, y x, lines
z
0
Cone x
y
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
15. x k, 4y 2 z 2 4 k 2,hyperbola
z
29.
A71
z=2
y k, x 2 z 2 4k 2 4, circle ( k 1) z k, 4y 2 x 2 4 k 2, hyperbola z
Hyperboloid of two sheets
z=œ„„„„„ ≈+¥ 0 y
x
31. 4x y z , paraboloid 33. 2
y x
17. x k, y 4z k , 2
2
2 z1 0
z
parabola y k, x 2 4z 2 k, ellipse k 0 z k, y x 2 4k 2, parabola
1
0
1 y
Elliptic paraboloid with axis the y-axis
0 1
1
x
x
19. x k, y z 2 k 2,
1. [1, 5] 5.
z
parabola y k, z 2 x 2 k, hyperbola k 0 z k, y k 2 x 2, parabola
■
page 568
3. 1, 0, 0 7.
y
z
π (0, 1, 0)
y
Hyperbolic paraboloid
y
x
1
x
(π, 1, 0)
x
x2 y2 z2 1 9 4 36
Hyperboloid of two sheets with axis the z-axis
0
y
Exercises 10.7
21.
2
z
9.
(0, 0, 6)
x
(0, 0, _6)
11.
z
z
y y
y=≈ x
1 x
23.
x y2 z2 6 3 2
y
z
Elliptic paraboloid with vertex 0, 0, 0 and axis the x-axis
13. rt t, 2t, 3t , 0 t 1;
x t, y 2t, z 3t, 0 t 1 15. rt 3t 1, 2t 1, 5t 2 , 0 t 1; x 3t 1, y 2t 1, z 5t 2, 0 t 1 17. VI 19. IV 21. V z 23. 25. 0, 0, 0, 1, 0, 1
y x
25. x 2
y 22 z 32 1 4
z (0, 0, 3)
Ellipsoid with center 0, 2, 3
(0, 4, 3)
0
0
y
y
x
x
27. y 12 x 22 z 12 (2, _1, 1)
z
29. rt t i 2 t 2 1 j 2 t 2 1 k 1
Circular cone with vertex 2, 1, 1 and axis parallel to the y-axis
1
31. x 2 cos t, y 2 sin t, z 4 cos 2t x
y
A72
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
33. (a), (c)
37. y 6x , x 6y 6 5 2 81 39. ( x 2 ) y 2 4 , x 2 ( y
(b) rt 1, 2t
y
)
5 2 3
5
(_3, 2)
r(_1)
rª(_1)
0
35. (a), (c)
169
x
7.5
37. (a), (c)
y
2.5
y
2 ” œ„ , œ„2 ’ 2
5
π
rª” 4 ’ 0
rª(0)
x
π
r” 4 ’
Exercises 10.9
0
(b) rt cos t i 2 sin t j
x
1
(b) rt e t i 3e 3t j
39. rt 2t, 1, 1(2 st ) 2
41. rt 2te t i [31 3t k
43. rt b 2t c
47. 1, 2t, 3t 2 , 1s14, 2s14, 3s14 , j 45 k 0, 2, 6t , 6t 2, 6t, 2 49. x 1 5t, y 1 4t, z 1 3t 51. x 1 t, y t, z 1 t 53. (a) Not smooth (b) Smooth (c) Not smooth 55. 66° 57. 4 i 3 j 5 k 59. i j k 61. e t i t 2 j t ln t t k C 2 2 63. t 2 i t 3 j ( 3 t 32 3) k 65. Yes 2 73. 1 4t cos t 11t sin t 3t 3 cos t 45.
49. 2.07 1010 Å 2 m
41. 1, 3, 1
(1, 1)
r(0) 1
page 586
■
1. vt t, 1
y
v(2)
at 1, 0 vt st 2 1
(_2, 2)
a(2) 0 x
3 5
Exercises 10.8
■
1. 20 s29
3.
3. vt 3 sin t i 2 cos t j
y
at 3 cos t i 2 sin t j vt s5 sin 2 t 4
3
” 2 , œ„ 3’ (3, 0)
a” π3 ’
x
0
page 576 1 27
13 32 8
5. 9.5706
2 3 4 si 1 s j 5 s k s29 s29 s29 9. 3 sin 1, 4, 3 cos 1 7. rts
(2s29 ) cos t, 5s29, (2s29 ) sin t, (b) 292 sin t, 0, cos t 1 13. (a) 2t 1 e 2t, s2 e t, s2 e t e 1 (b) s2 e 2te 2t 12 1 19 4 15. 24t 2 132 17. 25 19. 7 s14 x 2 2x 32 21. e x 2 1 x 1 e 23. 15sx 1 100x 332
5. vt i 2t j
z
at 2 j vt s1 4t 2
a(1) (1, 1, 2)
11. (a)
v” π3 ’
(0, 2)
v(1) y x
7. 2t, 3t 2, 2t , 2, 6t, 2 , t s9t 2 8 9. s2 i e t j et k, e t j et k, e t et
11. vt t i 2t j k, rt
25. ( ln 2, 1s2 ); approaches 0 27. (a) P (b) 1.3, 0.7 29. 4 1 2
( 12 t 2 1 i t 2 j t k
13. (a)
rt (b)
( 13 t 3 t ) i t sin t 1 j ( 14 14 cos 2t) k
y=x –@ y=k(x) z 4
_4
x
31. a is y f x, b is y $x 33. 1(s2e )
35.
_10 200
_1
t
0.6 0.4 0.2 0
2 2 1 3 3 3
, ,
,
0
0 y _200
10
, , , , , 2 3
2 1 2 3 3 3
15. t 4 17. rt t i t j 2 t 2 k, vt s25t 2 2 19. (a) 22 km (b) 3.2 km (c) 500 ms 5
1 2 3 3
21. 30 ms 23. 10.2 , 79.8 25. 13.0 36.0 , 55.4 85.5 27. (a) 16 m (b) 23.6 upstream 20
12
40
0 40
0 _4
_12 t
29. 0, 1
Chapter 10 Review
■
3. True 5. True 7. True 9. True 13. False 15. False 17. True 21. False 23. False
Exercises 1. (a) x 1 y 2 z 1 69 (b) y 22 z 12 68, x 0 (c) Center 4, 1, 3, radius 5 3. u v 3 s2; u v 3 s2; out of the page 5. 2, 4 7. (a) 2 (b) 2 (c) 2 (d) 0 1 9. cos1( 3 ) 71 11. (a) 4, 3, 4 (b) s412 13. 166 N, 114 N 15. x 4 3t, y 1 2t, z 2 3t 17. x 2 2t, y 2 t, z 4 5t 19. 4x 3y z 14 21. (1, 4, 4) 23. Skew 25. 22s26 27. Plane 29. Cone 2
2
page 588
True-False Quiz 1. True 11. False 19. False
(b) rt i sin t j cos t k, r t 2 cos t j 2 sin t k 39. rt 4 cos t i 4 sin t j 5 4 cos t k, 0 t 2 1 41. 3 i 2 2 j 2 k 43. 86.631 45. 2 47. (a) t 2, t, 1st 4 t 2 1 (b) 2t, 1 t 4, 2t 3 t st 8 4t 6 2t 4 5t 2 (c) st 8 4t 6 2t 4 5t 2t 4 t 2 12 49. 1217 32 51. vt 1 ln t i j et k, vt s2 2 ln t ln t2 e2t, at 1t i et k 53. (a) About 3.8 ft above the ground, 60.8 ft from the athlete (b) 21.4 ft (c) 64.2 ft from the athlete 55. t
33. t 1
31. e e , s2 t
A73
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
2
CHAPTER 11 Exercises 11.1
■
page 599
(b) ⺢ 2 (c) 0, 1. (a) 4 3. (a) e (b) x, y, z z x 2 y 2 5. x, y y x
(c) 1, y
y=_x 0
7. x, y
1 9
x
x 2 y 2 1
y 1 9 ≈+¥=1
0
x
z
z
9. x, y y x 2, x 1
y
0
y=≈
y x
y
x
31. Hyperboloid of two sheets
_1
33. Ellipsoid
z
z
11. x, y, z x 2 y 2 z 2 1
(0, 1, 2)
x
1
z
0
y
(1, 1, 0)
(0, 2, 0)
x
0
(0, 2, 0)
y
y
x
x (0, 1, _2)
35. 4x y z 16 37. (a) 2
2
13. 3x 2y z 6 , plane
z (0, 0, 6)
2
z
(0, 1, 0) y x
(2, 1, 0)
0 (2, 0, 0) x
(0, 3, 0) y
A74
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
15. z y 2 1 , parabolic cylinder
y
35.
z
0
y
x
17. z 4x y 1
19. z sx y ,
elliptic paraboloid
top half of cone
2
2
2
37.
2
5
z
z
x
z 0 _5 _2
_5 y0
5
0 y
x
(0, 0, 1) y
21. 56, 35 23. Steep; nearly flat 25. y 2x 2 k 27. y ln x k y
2 1 0 _1
0
x
_2
1 234
29. y kex
31. y 2 x 2 k 2
y 1 2
z 0
y
x
0
y 3
3
2 1 0
0 0
_2 _1
2
39.
0 x
43 2 1
0x
x
x
0 1 2
_3
3
_2
0
y
z
y
z=4 1 0
2
3
4 x
z=3
z=1
■
_2
page 608
1. Nothing; if f is continuous, f 3, 1 6 3. 2025 5. Does not exist 7. Does not exist 9. 0 11. Does not exist 13. 2 15. Does not exist 17. The graph shows that the function approaches different
numbers along different lines. 19. hx, y 2 x 3y 6 2 s2x 3y 6 ;
x, y 2x 3y 6 21. x, y y e x2 23. x, y x 2 y 2 4
25. x, y, z y 0, y sx 2 z 2
27. x, y x, y 0, 0 y x
x
(a) C (b) II 43. (a) F (b) I (a) B (b) VI Family of parallel planes Family of hyperboloids of one or two sheets with axis the y-axis 51. (a) Shift the graph of f upward 2 units (b) Stretch the graph of f vertically by a factor of 2 (c) Reflect the graph of f about the xy-plane (d) Reflect the graph of f about the xy-plane and then shift it upward 2 units 53. If c 0, the graph is a cylindrical surface. For c 0, the level curves are ellipses. The graph curves upward as we leave the origin, and the steepness increases as c increases. For c 0, the level curves are hyperbolas. The graph curves upward in the y-direction and downward, approaching the xy-plane, in the xdirection giving a saddle-shaped appearance near (0, 0, 1).
z=2
0
41. 45. 47. 49.
Exercises 11.2
33. x 2 9y 2 k
2 2
29. 0
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 11.3
page 614
■
1. (a) The rate of change of temperature as longitude varies,
with latitude and time fixed; the rate of change as only latitude varies; the rate of change as only time varies. (b) Positive, negative, positive 3. (a) Positive (b) Negative 5. fx 1, 2 8 slope of C1 , fy1, 2 4 slope of C2 z
z 16
16
(1, 2, 8)
(1, 2, 8)
C¡ 0
4
2
y
x
C™
0
4
2
y
x
(1, 2)
(1, 2)
A75
&z 3yz 2x &z 3xz 2y , &x 2z 3xy &y 2z 3xy &z &z 1 y 2z 2 z 39. , &x 1 y y 2z 2 &y 1 y y 2z 2 41. (a) f x, t y (b) f x y, f x y 2 3 43. fxx 12x 6y , fxy 18xy 2 fyx , fyy 18x 2y 45. zxx 2yx y3, zxy x yx y3 zyx , zyy 2xx y3 47. uss es sin t, ust es cos t uts , utt es sin t 51. 12xy, 72xy 53. 24 sin4x 3y 2z, 12 sin4x 3y 2z 55. e r2 sin cos r sin 65. R 2R 21 73. x 1 t, y 2, z 2 2t 77. 2 71. No 37.
79. (a) 0.2
7. fx x, y 3, fyx, y 8y 3
z 0
9. &z&x e 3y, &z&y 3xe 3y
_0.2
11. fx x, y 2yx y2, fyx, y 2xx y2
_1 y 0
13. &w& cos cos , &w& sin sin
2r 2 2rs lnr 2 s 2 , fsr, s 2 r s2 r s2 17. &u&t e wt (1 wt), &u&w e wt 19. fx y 2z 3, fy 2xyz 3 3z, fz 3xy 2z 2 3y 21. &w&x 1x 2y 3z, &w&y 2x 2y 3z, &w&z 3x 2y 3z 23. &u&x et sin , &u&t xet sin , &u& xet cos 25. fx yz 2 tan yt, fy xyz 2 t sec 2 yt xz 2 tan yt, fz 2xyz tan yt, ft xy 2z 2 sec 2 yt 3 1 27. &u&xi xisx 12 x 22 x n2 29. 5 31. 3 33. fx x, y y 2 3x 2 y , fy x, y 2xy x 3 35. fx 2x 2xy, fy 2y x 2 15. frr, s
2
1
f
z
x
0
x 4y 4x 2y 3 y 5 x 5 4x 3y 2 xy 4 , fyx, y 2 2 2 x y x 2 y 2 2 (e) No, since fxy and fyx are not continuous.
(c) 0, 0
Exercises 11.4
page 624
■
1. z 8x 2y
400
1 z 0
z 200
_1 0 x 0 _10
0
2 y
10
19. 21. 23. 29.
0
5
0 y
_5
0 2
2
x
y
x y 14 12 3 2 6 23 x 73 y 203 ; 2.846 17. 7 x 7 y 7 z; 6.9914 2 2 3 dz 3x ln y dx 2x y dy dR 2 cos ' d 2 cos ' d 2 sin ' d ' 25. 5.4 cm 2 27. 16 cm 3 z 0.9225, dz 0.9 1 2.3% 31. 17 0.059 33. 1 x, 2 y
11. 2x y 1
_2
5. z y
9.
1 4
2
3. x 2y z 4
7.
15. 0 _2
_1 x
(b) fx x, y
10 10
0
1
13.
1 2
fx
z 0
Exercises 11.5
_10 _2 x
0
2
0
_2
2 y
fy 0
2
_2
&z s e r t cos sin , &s ss 2 t 2 t &z e r s cos sin 2 &t ss t 2 9. 62 11. 7, 2
z 0
x
page 631
1. cos x cos y sin x sin y(2 st ) 3. e yz 2t xz 2xyz 2 5. &z&s 2x y xt 2yt, &z&t 2x y xs 2ys 7.
_2
■
0
2 y
A76
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
&u &u &x &u &y &u &u &x &u &y , , &r &x &r &y &r &s &x &s &y &s &u &x &u &y &u &t &x &t &y &t &v &p &v &q &v &r &v 15. , &x &p &x &q &x &r &x &v &p &v &q &v &r &v , &y &p &y &q &y &r &y &v &p &v &q &v &r &v &z &p &z &q &z &r &z 9 9 17. 85, 178, 54 19. 7 , 7 21. 36, 24, 30 4xy 32 y 3yz 2x 3xz 2y 23. 25. , x 2x 2sxy 2z 3xy 2z 3xy 2 2 z 1y z , 27. 1 y y 2z 2 1 y y 2z 2 29. 2 Cs 31. 0.33 ms per minute (b) 10 m 2s (c) 0 ms 33. (a) 6 m3s 35. 0.27 Ls 37. (a) &z&r &z&x cos &z&y sin , &z& &z&xr sin &z&yr cos 43. 4rs &2z&x 2 4r 2 4s 2 &2z&x &y 4rs &2z&y 2 2 &z&y 13.
Exercises 11.7
■
page 650
1. (a) f has a local minimum at (1, 1).
(b) f has a saddle point at (1, 1). 1 3. Maximum f (1, 2 ) 11 5. Minima f 1, 1 0, f 1, 1 0, saddle point at 0, 0 7. Saddle points at 1, 1, 1, 1 9. None 11. Saddle points 0, n, n an integer 13. Minimum f 0, 0 0, saddle points at 1, 0 15. Maximum f 0, 0 2, minimum f 0, 2 2, saddle points 1, 1 17. Maximum f 3, 3 3 s32, minimum f 53, 53 3 s32 19. Minima f 1.714, 0 9.200, f 1.402, 0 0.242, saddle point (0.312, 0), lowest point 1.714, 0, 9.200 21. Maxima f 1.267, 0 1.310, f 1.629, 1.063 8.105, saddle points 0.259, 0, 1.526, 0, highest points 1.629, 1.063, 8.105 23. Maximum f 2, 0 9, minimum f 0, 3 14 25. Maximum f 1, 1 7, minimum f 0, 0 4 27. Maximum f 3, 0 83, minimum f 1, 1 0 29.
(_1, 0, 0)
(1, 2, 0)
0
Exercises 11.6
page 642
■
_1
1. 16 s3 4 3. (a) ( f x, y 5y 2 12x 2y, 10xy 4x 3 5
1
z _2
(b) 4, 16 (c) 17213 5. (a) e 2yz, 2xze 2yz, 2xye 2yz (b) 1, 12, 0 (c) 223 7. 2310 9. 4 s2 11. 9 (2s5 ) 13. 25 17. s3, 1, 1, 1 15. 4s2, 1, 1 21. All points on the line y x 1 19. (b) 12, 92 23. (a) 40(3 s3 ) 327 25. (a) 32s3 (b) 38, 6, 12 (c) 2 s406 27. 13 y1 z1 x2 31. (a) 4x 5y z 4 (b) 4 5 1 33 (a) x y z 1 (b) x 1 y z 35. 37. 4, 8 , x 2y 4 y 2
0 x
1
31. s3 33. (2, 1, s5 ), (2, 1, s5 ) 35. 3 , 3 , 3 4 37. 16s3 39. 3 41. Cube, edge length c12 43. Square base of side 40 cm, height 20 cm 45. L 3 (3s3 ) 100 100 100
Exercises 11.8
■
page 658
1. No maximum, minima f 1, 1 f 1, 1 2 3. Maxima f 2, 1 4, minima f 2, 1 4 5. Maximum f 1, 3, 5 70, minimum
f 1, 3, 5 70
9. Maximum s3, minimum 1
x+2y=4 (2, 1) z 1
_1
4 2y _2
7. Maximum 2s3, minimum 2s3
±f(2, 1)
0
_3
11. Maximum f ( 2 , 2 , 2 , 2 ) 2, 1 1 1 1
x
minimum f ( 12 , 12 , 12 , 12 ) 2 13. Maximum f (1, s2, s2 ) 1 2 s2,
0
_1 1 x
2
1
y
2
41. (s63, )2s63, s62) 45. x 1 10t, y 1 16t, z 2 12t 49. If u a, b and v c, d , then afx bfy and c fx d fy
are known, so we solve linear equations for fx and fy .
minimum f (1, s2, s2 ) 1 2 s2 3 1 15. Maximum 2 , minimum 2 17. Maxima f (1s2, )1(2 s2 )) e 14, minima f (1s2, 1(2 s2 )) e14 19. 59, 30 25–35. See Exercises 31–41 in Section 11.7. 37. L 3 (3s3 ) 1 1 1 39. Nearest ( 2 , 2 , 2 ), farthest 1, 1, 2 41. Maximum 9.7938, minimum 5.3506 43. (a) cn (b) When x1 x2 xn
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Chapter 11 Review
■
CHAPTER 12
page 660
True-False Quiz 1. True 11. True
3. False
5. False
7. True
Exercises 12.1
9. False
3.
y
_1
x
■
page 672
z
1. (a) 288 (b) 144 3. (a) 22 4.935 (b) 0 5. 248 7. 60 9. 3 21 15. 261,63245 17. 2 ln 2
1
21. 9 ln 2
Exercises 1. x, y y x 1
A77
■
1
x
13. 2
(s3 1) 12 166 4 31. 15 (2 s2 1) 27
23.
27. 47.5 29. 35. 21e 57
(c) 0 11. 10 19. 6
25. 2 e 2 3
1 2
1
33. 36
y
_1
2 z
y=_x-1 0 0
5.
7.
y
y
y 2
1 0
2
34
0
x
1 1
5
37. 6 39. Fubini’s Theorem does not apply. The integrand has an infi-
nite discontinuity at the origin.
5 1
Exercises 12.2
x
0
1
9. 3 11. f x 1s2x y 2, f y ys2x y 2 13. tu tan1v, tv u1 v 2 15. Tp lnq er , Tq pq er , Tr perq er 17. f xx 24x, f xy 2y f yx, f yy 2x 19. f xx kk 1x k2 y lz m, f xy klx k1 y l1z m f yx,
2
x
2
f xz kmx k1 y lz m1 f zx, f yy ll 1x k y l2z m, f yz lmx k y l1z m1 f zy, f zz mm 1x k y lz m2 y2 x1 23. (a) z 8x 4y 1 (b) 1z 8 4 y1 z1 x2 25. (a) 2x 2y 3z 3 (b) 4 4 6 y1 z 27. (a) x 2y 5z 0 (b) x 2 2 5 1 1 29. (2, 2 , 1), (2, 2 , 1) 24 32 31. 60x 5 y 5 z 120; 38.656 3 33. 2xy 1 6p 3x 2 y 2 pe p e p 4z 3 p cos p sin p 43 35. 47, 108 41. ze x sy z sy, xz(2 sy ), 2 43. 5 9 45. s1452, 4, 2 47. Minimum f 4, 1 11 49. Maximum f 1, 1 1; saddle points (0, 0), (0, 3), (3, 0) 51. Maximum f 1, 2 4, minimum f 2, 4 64 53. Maximum f 1, 0 2, minima f 1, 1 3, saddle points 1, 1, 1, 0 55. Maximum f (s23, 1s3 ) 2(3 s3 ), minimum f (s23, 1s3 ) 2(3 s3 ) 57. Maximum 1, minimum 1 59. (314, 314s2, 314 ), (314, 314s2, 314 ) 61. P(2 s3 ), P(3 s3 )6, P(2 s3 3)3
■
page 680
1. 3. e 32 45 5. e 1 1 9. ln 2 11. 2 1 cos 1 13. 7 31 1 15. 0 17. 18 19. 8 21. 6 9 20 1 2
25. 31.
4 9
1 3
x x
2 4 0 y2
27.
32
29. 2 33.
64 3
f x, y dx dy
7. 147 20
256 21 128 15
23.
x33 x0s9x
f x, y dy dx
2
y
y
3
y=œ„ x
2
≈+¥=9
x=4 0
35.
y=0
x
4
0
–3
x0ln 2 xe2 f x, y dx dy
y
y
y=0
3
y=ln x or x=e †
ln 2
x=2 y=0 0
37. e 9 1 1 6
45. 0
1. 5.
x x
5 2
39.
sin 81
xxD sx 3 y 3 dA s2
Exercises 12.3 2 0
1 4
■
41.
1 1 3
2
(2 s2 1)
49. 8
x
43. 1
51. 23
page 686
f r cos , r sin r dr d
3.
x22 xx2 f x, y dy dx
33 2
y
4 0
7 x
R
x
A78
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
9. 2 sin 9
16 2 13. 3 a 3 17. 23[1 (1s2 )] 83(64 24 s3 ) 21. 12 1 1 cos 9 2s23 25. 2 15 1800 ft 3 (b) s 2 29. 16 31. (a) s 4 1
7. 0
11.
3 64
4 3
15. 19. 23. 27.
x01 xsx1 x01y f x, y, z dz dy dx x01 x0y x01y f x, y, z dz dx dy x01 x01z x0y f x, y, z dx dy dz x01 x01y x0y f x, y, z dx dz dy 1z f x, y, z dy dz dx x01 x01sx xsx 1 1z 1z xsx f x, y, z dy dx dz x0 x0 31.
2
2
2
2
Exercises 12.4 1.
64 3
page 693
■
3. 3 , ( 3 , 0) 4
C
5. 6, ( 4 , 2 )
4
3 3
e 2 1 4e 3 1 , 2e 2 1 9e 2 1 27 8 1 3 9. 2 , ( 5 , 2 ) 11. ( 8 , 316) 13. 2a5, 2a5 if vertex is (0, 0) and sides are along positive axes 1 1 1 15. 16 e 4 1, 8 e 2 1, 16 e 4 2e 2 3 189 1269 1917 17. 20 , 28 , 35 7. 4 e 2 1, 1
2 1 16 19. m 28, x, y , , Ix 3 264, 3 9 Iy 161 4 3 2 , I0 416 9 264 21. a 43, a 43; as3, as3 Exercises 12.5
5. 3 e 3 1 7. 4 9. 3. 1 16 13. 83e 15. 163 17. 3 1
21. (a)
x01 x0x x0s1y
2
35.
79 30
33 571 , ( 358 553 , 79 , 553 )
39. (a) m
dz dy dx
65 28
(b) 14 13
23. 60.533 z 25.
11.
1 10
19. 36
37. a 5, 7a12, 7a12, 7a12
s9x x33 xs9x x15y sx 2 y 2 dz dy dx 2
2
(b) x, y, z , where 3 s9x x 1m x3 xs9x 2
x x x
5y 2 1 3 s9x 2 5y 3 s9x 2 1 3 s9x 2 5y 3 s9x 2 1
y 1m x
xsx 2 y 2 dz dy dx
x ysx 2 y 2 dz dy dx z 1m x x zsx 2 y 2 dz dy dx 3 s9x 5y 2 (c) x3 xs9x x1 x y 2 32 dz dy dx 2
2
41. (a)
page 700
■
x01 xy1 x0y f x, y, z dz dx dy x01 x0x x0y f x, y, z dz dy dx x01 xz1 xy1 f x, y, z dx dy dz x01 x0y xy1 f x, y, z dx dz dy x01 x0x xzx f x, y, z dy dz dx x01 xz1 xzx f x, y, z dy dx dz
33.
11 24
3 32
28 30 128 45 208 , , 9 44 45 220 135 660 1 (c) 240 68 15 2 43. Ix Iy Iz 3 kL5 45. L38 47. The region bounded by the ellipsoid x 2 2y 2 3z 2 1 (b) x, y, z
1
Exercises 12.6 1. (a)
2
0
1
page 706
■
(b)
z
y
x s4x x22 x06 xs4x f x, y, z dz dy dx 6 2 s4x x0 x2 xs4x f x, y, z dz dx dy 2 s4z x2 x06 xs4z f x, y, z dx dy dz 6 2 s4z x0 x2 xs4z f x, y, z dx dz dy 2 s4x x2 xs4x x06 f x, y, z dy dz dx 2 s4z xs4z x06 f x, y, z dy dx dz x2
π
” 2, 4 , 1 ’
2
27.
2
5
0
2
2
π 4
2
2
2
z π ”4, _ 3 , 5’
4
1
0
π
y
_3
x
y
2
x
2
2
2
2
2
1 x1 x01x x0y f x, y, z dz dy dx s1y x01 xs1y x0y f x, y, z dz dx dy s1y x01 xz1 xs1y f x, y, z dx dy dz 1 y s1y x0 x0 xs1y f x, y, z dx dz dy 1 x1 x01x xz1x f x, y, z dy dz dx s1z x01 xs1z xz1x f x, y, z dy dx dz 2
29.
2
(s2, s2, 1) 3. (a) (s2, 74 , 4)
(b)
(2, 2 s3, 5) (2, 43 , 2)
5. Circular cylinder, radius 3, axis the z-axis 7. Circular paraboloid 9. (a) z r 2 (b) r 2 sin z 11. 13. Cylindrical coordinates: z=2-r@
2
6 r 7, 0 2, 0 z 20
1
2
z=r@
2
0 x
y
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
15.
A79
29. (a) (0, 0, 8 a)
(b) 4Ka 515 31. 23[1 (1s2 )], (0, 0, 3[8(2 s2 )]) 33. 56 35. (4 s2 5)15 37. 39. 13699 3
z
64 3
4
4
4
x
■
y
384 19. e e 5 21. 25 (a) 162 (b) 0, 0, 15 Ka 28, 0, 0, 2a3 27. 0 (a) xxxC hPtP dV , where C is the cone (b) 3.1 1019 ft-lb 6
17. 23. 25. 29.
Exercises 12.7
■
Exercises 12.8
page 711
y sx
11. 3 13. 6 15. 2 ln 3 4 17. (a) 3 abc (b) 1.083 10 12 km 3 3 8 19. 5 ln 8 21. 2 sin 1 23. e e1
(0, 0, 1)
(1, 0, 0)
1 0
Chapter 12 Review
■
page 722
y
x
True-False Quiz z
(b)
page 721
3. 0 5. 2u vw 1. 14 7. The parallelogram with vertices (0, 0), (6, 3), (12, 1), (6, 2) 9. The region bounded by the line y 1, the y-axis, and
z
1. (a)
■
1. True
3. True
5. True
7. False
( 12 s2, 12 s6, s2 ) Exercises
π π
π 4
0
”2, 3 , 4 ’
1. 64.0 3. 4e 2 4e 3 5. 2 sin 1 7. 3 4 9. x0 x2 f r cos , r sin r dr d 11. The region inside the loop of the four-leaved rose r sin 2 1
2
π 3
y
x
3. (a) 4, 3, 6 (b) (s2, 32, 34) 5. Half-cone 7. Circular cylinder, radius 2, axis the z-axis 9. (a) sin2 cos (b) sin 2 sin z z 11. 13. 2
∏=2
2 2
0 y
x
x
15. 0 4, 0 cos z 94 (2 s3 ) 17. 3
y
in the first quadrant 1 1 7 1 15. 2 e 6 2 17. 4 ln 2 2 sin 1 19. 8 21. 815 23. 40.5 25. 96 64 2 27. 15 29. 176 31. 3 33. 2ma 39 1 35. (a) 4 (b) ( 13 , 158 ) 1 1 (c) Ix 12 , Iy 24 ; y 1s3, x 1s6 37. (a) 0, 0, h4 (b) a 4h10 39. (s3, 3, 2), 4, 3, 3 41. (2s2, 2s2, 4s3 ), (4, 4, 4s3 ) 43. r 2 z 2 4, 2 45. 97.2 sy 47. x01 x01z xsy f x, y, z dx dy dz 49. ln 2 13.
CHAPTER 13 Exercises 13.1 1.
page 730
1
_2
19.
■
y
π 6
x
_1
0
y
x02 x03 x02 f r cos , r sin , z r dz dr d
21. 45 23. 1516 25. 156215 27. (a) 10 (b) (0, 0, 2.1)
2
_1
_2
1
x
51. 0
A80
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
3.
5.
y
31. (a)
y
(b) y 1x, x 0
y
2
0
2
x
x
0
x
0
_2
7.
9.
z
y Cx
z
Exercises 13.2 1. x y
x
11. II 19.
13. I
■
page 740
(17 s17 1)
3. 1638.4
5.
17 3
11. 13. s14 e 1 59 15. (a) Positive (b) Negative 17. 105 6 19. 5 cos 1 sin 1 2 21. 3 3 2.5
7. 320
y
15. IV
1 12
17. III
9.
1 12
1 5
6
97 3
The line y 2x
4.5
2.5 4.5
2
.5
4.5
2.5 4.5
21. 23.
25.
23. (a)
1 2 ( f x, y i j x 2y x 2y x ( f x, y, z i sx 2 y 2 z 2 z y j k sx 2 y 2 z 2 sx 2 y 2 z 2 ( f x, y y 2 i x j
0
y 8
_2
_4
6
6
29. 2.04, 1.03
1.6
4
6 x
945 16,777,216
Exercises 13.3
6
1
27. 2k, 4, 0 29. (a) x 1m xC x x, y, z ds, y 1m xC y x, y, z ds, z 1m xC z x, y, z ds, where m xC x, y, z ds (b) 0, 0, 3 4 2 31. Ix k (2 3 ), Iy k (2 3 ) 2 4 33. 2 35. 26 37. 1.67 10 ft-lb 39. (b) Yes 25.
2 _2
F(r(0))
0.2
0
6
1.6
2 œ„
2
27.
(b)
1 F ” r ” ’’
4
_4
1e
F(r(1))
6
_6
11 8
■
page 749
1. 40 3. f x, y 3x 2 5xy 2y 2 K 5. Not conservative 7. f x, y x 2 cos y y sin x K x 9. f x, y ye x sin y K 1 11. (a) f x, y 4 x 4y 4 (b) 4 13. 15. 17. 27. 29.
(a) (a) 2 (a) (a)
(b) 77 f x, y, z xyz z 2 f x, y, z xy 2 cos z (b) 0 19. 30 21. No 23. Conservative Yes (b) Yes (c) Yes Yes (b) Yes (c) No
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 13.4 1. 6
■
3.
7. e 1
15. 2 19. 3 21. (c) 13.
27. (a) Direction reverses (b) Number of coils doubles 29. 3x y 3z 3 31. x y z 2 33. 3 s14 4 35. 15 3 52 2 72 1 37. 23(2 s2 1) 39. 6(17 s17 5 s5 )
page 756
2 3
4 3
625 2 9 2
11. 24
1 3
9.
17. 12 1
23.
A81
■
( 13 , 13 )
41. 2 s21 1
17 4
[ln(2 s21 ) ln s17 ]
43. (2 s6 3 ) 45. 13.9783 47. (a) 1.83 (b) 1.8616 8
Exercises 13.5
■
page 763
1. (a) x 2 i 3xy j xz k (b) yz 3. (a) x y i y j k (b) z 1 (2sz ) 5. (a) 0 (b) 1 7. (a) 1y, 1x, 1x (b) 1x 1y 1z 9. (a) Zero (b) curl F points in the negative z-direction 11. f x, y, z xyz K 13. f x, y, z x 2y y 2z K 15. Not conservative 17. No Exercises 13.6
■
49. 8 s14 51. (b) 45
[
]
ln (11s5 3s70 )(3s5 s70 )
15 16
2
z 0
2
page 773
1. Plane through 0, 3, 1 containing vectors 1, 0, 4 ,
2
1, 1, 5 3. Hyperbolic paraboloid
y
0
2
1 1 0x
(c) x02 x0 s36 sin 4u cos 2v 9 sin 4u sin 2v 4 cos 2u sin 2u du dv 53. 4 57. 6(37s37 17s17 )
5. √ constant 2
Exercises 13.7
1
_2 0
u constant
1.5
y1 2
x
713
9. 1
z 0 z
u constant
0
√ constant
_1 _1
_1 _1 _1
x 0 1 _1
0
1
y
y
1
4
where D projection of S on xz-plane
u constant
1
page 784
15. 16 17. 16 19. 180 21. 6 23. 3 8 25. 0 27. 2 3 29. 3.4895 31. xxS F dS xxD P&h&x Q R&h&z dA,
2
7.
■
1. 8(1 s2 s3 ) 33.17 3. 900 5. 5s548 1240 7. 171 s14 9. s324 11. 364 s2 3 13. 60(391s17 1)
z 0
0
0 x 1
1
√ constant
I 13. II x 1 u v, y 2 u v, z 3 u v x x, z z, y s1 x 2 z 2 x 2 sin cos , y 2 sin sin , z 2 cos , 0 4, 0 2 or x x, y y, z s4 x 2 y 2, x 2 y 2 2 21. x x, y 4 cos , z 4 sin , 0 x 5, 0 2 25. x x, y ex cos , 1 z ex sin , 0 x 3, 0 2
33. 0, 0, a2 (b) 4329 s2 5 35. (a) Iz xxS x 2 y 2 x, y, z dS 8 37. 0 kgs 39. 3 a 30 41. 1248 Exercises 13.8 1. 0 3. 0 9. (a) 812
page 790 7. 80
5. 1
(b) 5
11. 15. 17. 19.
[
■
z 0 5 2
]
0 y
(c) x 3 cos t, y 3 sin t, z 1 3cos t sin t, 0 t 2
2
0
2 x
4 z
z 0
2
2 0
_2
1 1 y
0
1 0
x
_2
2
y
13. 3
0
2
2
0
_2 x
A82
■
APPENDIX E ANSWERS TO ODD-NUMBERED EXERCISES
Exercises 13.9
■
31. sin s53, cos 3 , tan s52, 2
page 796
5. 2 7. 92 9. 0 11. 323 13. 0 81 15. 341 s260 20 arcsin(s33) 17. 1320 19. Negative at P1 , positive at P2 21. div F 0 in quadrants I, II; div F 0 in quadrants III, IV Chapter 13 Review
■
page 798
True-False Quiz 1. False
3. True
5. False
csc 3s5, cot 2s5
33. sin 1s10, cos 3s10, tan 3 , 1
csc s10, sec s103 35. 5.73576 cm 37. 24.62147 cm 59. (4 6 s2 )15 61. (3 8 s2 )15 24 63. 25 65. 3, 53 67. 4, 34, 54, 74 69. 6, 2, 56, 32 71. 0, , 2 73. 0 x 6 and 56 x 2 75. 0 x 4, 34 x 54, 74 x 2 77.
7. True
y
(b) Positive 3. 6 s10 5. 15 11 110 7. 3 9. 12 4e 11. f x, y e y xe xy 13. 0 1 17. 8 25. 6 (27 5 s5 ) 27. 60(391 s17 1) 1 29. 643 33. 2 37. 21 4
1. (a) Negative
1. 76
π 3
0
x
5π 6
y
79.
APPENDIXES Exercises A
1
1 2
Exercises
0
π 2
3π 2π 2
π
x
5π 3π 2
page A8
■
5. 5
3. 20
13. 3 cm
11. 67.5 17.
y
9. 75° rad 120
7. 720
15. 19.
2 3
y
81.
y
1
0
_π
π
x
2π
0 0
x
x
315°
89. 14.34457 cm2
3π _ 4
Exercises C 21.
■
1. s1 s2 s3 s4 s5
y
5. 1 3 5 7 1
2 rad 0
3
5
3. 3 4 3 5 3 6
7. 110 210 310 n10
7 9
10
9. 1 1 1 1 1n1
11.
i
i1
x
19
13.
i1
csc34 s2, sec34 s2, cot34 1 25. sin92 1, cos92 0, csc92 1, cot92 0, tan92 and sec92 undefined
2i
1 3
27. 61
19.
x
29. nn 1
33. nn 6n 113 2
45. 14
Exercises D
1. (b) 0.405
■
97 300
(d) an a0
49. 2 n1 n 2 n 2
page A38
i
i1
2
29. cos 5 , tan 4 , csc 3 , sec 4 , cot
4 3
n
i
i0
25. 0
31. nn 6n 173
43.
2
17.
i1
23. 3276
csc56 2, sec56 2s3, cot56 s3 5
5
35. nn 2n n 104 (b) 5 100 1 (c) 41. (a) n 4
1
5
15.
2 3
27. sin56 2 , cos56 s32, tan56 1s3, 3
n
i i1
21. 80 23. sin34 1s2, cos34 1s2, tan34 1,
4
page A30
INDEX
RP
denotes Reference Page numbers.
absolutely convergent series, 441 absolute maximum and minimum, 198, 645 absolute maximum and minimum values, 198, 648, 652, 656 absolute value, RP1 absolute value function, 5 acceleration, 90, 280, 578 vector components of, 582 addition formulas for sine and cosine, A5, RP2 addition of vectors, 523 Airy, Sir George, 452 Airy function, 452 Algebra Review, RP1 alternating harmonic series, 439 alternating series, 437 Alternating Series Estimation Theorem, 440 Alternating Series Test, 438 angle(s), A1 between vectors, 531, 532 negative and positive, A2 standard position, A2 angular momentum, 587 angular speed, 580 antiderivative, 241 general vs. particular, 277, 330 antidifferentiation formulas, 243, RP5 aphelion, 512 approximate integration, 268, 333 approximating cylinder, 365 approximation by differentials, 135
to e , 145 to an integral, 268, 333 by the Midpoint Rule, 268, 334 by Newton’s method, 236 by Riemann sums, 263 by Simpson’s Rule, 338 tangent line, 133 by Taylor polynomials, 471 by the Trapezoidal Rule, 335 approximation, linear, 133 to a tangent plane, 619, 623 arc curvature, 572, 573, 574, 576 Archimedes, 37, 390 arc length, 378, 381 of a parametric curve, 491 of a polar curve, 507 of a space curve, 570, 571 arc length formula, 379 arc length function, 571 area between curves, 357, 358 of a circle, 316 cross-sectional, 368 enclosed by a parametric curve, 491 by exhaustion, 37 by Green’s Theorem, 754 in polar coordinates, 505 of a region, 256 of a sector of a circle, 504 of a surface, 771, 772 under a curve, 256 under a graph of f from a to b , 284 Area Problem, 251 arrow diagram for a function, 2
astroid, 126, 141, 494 asymptote(s) horizontal, 60 of a hyperbola, 510 slant, 225 vertical, 57 average cost function, 234 average rate of change, 78 units for, 80 average speed of molecules, 353 average value of a function, 289, 673, 702 average velocity, 24, 76, 578 axes, coordinate, 517 axis of a parabola, 509 Barrow, Isaac, 37, 284 base of an exponential function, 16, 144 of a logarithm, 16, 154 change of, 157 base of a cylinder, 362 basis vectors, 527 Bernoulli, James, 398 Bernoulli, John, 398, 459, 486 Bessel, Friedrich, 448 Bessel function, 127, 448, 452 binomial coefficient, 465 binomial series, 464, 465, 466 Binomial Theorem, RP1 binormal vector, 574, 575, 576 blood flow, 137 boundary curve, 786 bounded sequence, 413 A83
A84
■
INDEX
bounded set, 649 Boyle’s Law, 5 Brache, Tycho, 584 brachistochrone problem, 486 branches of a hyperbola, 510 bullet-nose curve, 120 C 1 transformation, 714 cable, hanging, 181 calculator, graphing, 27, 416, 484, 502 calculus, differential vs. integral, 284 cancellation equations, 150, RP4 Cantor, Georg, 428 Cantor set, 428 cardioid, 126, 499 carrying capacity, 401 CAS. See computer algebra system Cassini, Giovanni, 504 catenary, 181 Cauchy, Augustin-Louis, 37, 670, A18 Cauchy-Schwarz Inequality, 537 Cauchy’s Mean Value Theorem, A18 Cavalieri, Bonaventura, 37, 340 Cavalieri’s Principle, 372 center of gravity, 390 center of mass, 390, 391, 393, 689, 699, 733, 777 of a lamina, 689 of a solid, 699 of a wire, 733 centripetal force, 580 centroid of a plane region, 391, 393 of a solid, 699 Chain Rule, 113, 114, 116 for several variables, 625, 627, 628 change of base in a logarithm, 157 change of variables: in a double integral, 717 in an integral, 293 in a triple integral, 719, 720 charge, electric, 160, 688, 699 charge density, 688 circle area of, 316 equation of, RP1 circle of curvature, 576 circular cylinder, 362 circulation of a velocity field, 789 Cissoid of Diocles, 503 Clairaut, Alexis, 613 Clairaut’s Theorem, 613, A23 closed curve, 744 Closed Interval Method, 202
closed set, 649 closed surface, 780 Cobb-Douglas production function, 658 cochleoid, 516 coefficient binomial, 465 of friction, 113, 204 of a polynomial, 13 of a power series, 447 combinations of functions, 18 common ratio, 421 comparison properties of the integral, 271 Comparison Test, 433 comparison tests for series, 432 Comparison Theorem for integrals, 351 Completeness Axiom, 418 component function, 559, 726 component of b along a, 534 components of a vector, 524 composite function, 19 continuity of, 51, 606 derivative of, 113 limit of, 51 composition of functions, 19 compound interest, 172, 173, 193 computer algebra system, 28 for finding a root, 236 for graphing a function of two variables, 593, 594, 597 for graphing an implicit equation, 400 for graphing a parametric surface, 768 for graphing a space curve, 562 for graphing a sequence, 416 for integration, 326, 330, 342, 456 for plotting a vector field, 727 concavity, 214 Concavity Test, 215 conchoid, 503 conductivity (of a substance), 783 cone, 557 parametrization of, 768 conic section, 509, 511 directrix, 509, 511 eccentricity, 511 focus (foci), 509, 510, 511 polar equations for, 510, 511, 512 vertex (vertices), 509, 510 connected region, 744 conservation of energy, 748, 749 conservative vector field, 729, 746, 749, 759
constant function, 94 derivative of, 95 Constant Multiple Law of limits, 36 Constant Multiple Rule, 98 constraint, 652, 656 continuity of a composite function, 51, 606 of a function, 45, 46, 50 on an interval, 48 of inverse functions, 152 from the left or right, 47 of a vector function, 560, 605, 606, 607 continuous compounding of interest, 172, 173, 193 continuous function, 46 contour curve, 594 contour map, 594 convergence absolute, 441 conditional, 442 of an improper integral, 346, 349 interval of, 449, 450 radius of, 449, 450 of a sequence, 412 of a series, 421 convergent improper integral, 346, 349 convergent sequence, 412 convergent series, 421 properties of, 425 coordinate axes, 517 coordinate planes, 517 coordinates cylindrical, 703, 705 polar, 496, 497 spherical, 707, 708 three-dimensional Cartesian, 518 coplanar vectors, 542 Cornu’s spiral, 495 cosine function, A2 derivative of, 102 graph, A7 inverse, 177 power series for, 464, 466 cost function, 103, 231 Coulomb’s Law, 225 critical number, 202 critical point, 645 cross product, 537 geometric characterization of, 540 magnitude of, 540 properties of, 541 cross-section, 363
INDEX
cross-sectional area, 368 cubic function, 13 curl of a vector field, 758 current, electric, 23, 29, 320 curvature, 572, 573, 574, 576 curve(s) boundary, 786 bullet-nose, 120 cissoid of Diocles, 503 closed, 744 cochleoid, 516 Cornu’s spiral, 495 Devil’s, 126 dog saddle, 601 epitrochoid, 495 equipotential, 601 guidelines for sketching, 220 grid, 766 helix, 560 involute, 495 kampyle of Eudoxus, 126 length of, 378, 507, 570, 571 level, 594 longbow, 516 monkey saddle, 601 orientation of, 736, 751 orthogonal, 126 ovals of Cassini, 504 parametric, 482 piecewise-smooth, 565, 733 plane, 560 polar, 498 serpentine, 112 simple, 746 smooth, 378, 565 space, 560, 562 strophoid, 516 swallowtail catastrophe, 488 toroidal spiral, 562 trefoil knot, 562 Tschirnhausen cubic, 126 twisted cubic, 562 witch of Maria Agnesi, 112, 488 curve sketching, 220 with technology, 223 curve of steepest ascent, 642 cusp, 565 cycloid, 485, 493 cylinder, 362 parabolic, 554 parametrization of, 768 cylindrical coordinates, 703 integration in, 704, 705 cylindrical shell, 373
decay, law of natural, 167 decay, radioactive, 169 decomposition of a function, 20 decreasing function, 7 decreasing sequence, 413 definite integral, 263 properties of, 269, 270 Substitution Rule for, 296 of a vector function, 567 definite integration by parts, 304, 305 by substitution, 296 degree of a polynomial, 13 del ((), 636 delta notation, 78 demand function, 231 density of a lamina, 688 liquid, 388 mass vs. weight, 388 dependent variable, 2, 591, 628 derivative(s), 73, 78, 83 of a composite function, 113 of a constant function, 95 directional, 633, 634, 637 domain of, 84 of exponential functions, 164 as a function, 83 of a function f at a number a, 77 higher, 90, 612 of hyperbolic functions, 183 of an integral, 286 of an inverse function, 152 of inverse hyperbolic functions, 184 of inverse trigonometric functions, 176, 177, 179 of logarithmic functions, 160 normal, 764 notation, 86 partial, 609, 612 of a polynomial, 99 of a power function, 95 of a power series, 454 of a product, 106 of a quotient, 108 as a rate of change, 79 second, 90, 565 as the slope of a tangent, 78 third, 91 of trigonometric functions, 100, 101, 110, 111 of a vector function, 563, 564 Descartes, René, 37, 200
■
A85
determinant, 537 Devil’s curve, 126 Difference Law of limits, 36 Difference Rule, 99 differentiable function, 87, 620 differential, 135, 621, 623 differential calculus, 284 differential equation, 106, 168, 243, 397 family of solutions to, 399 first-order, 398 general solution of, 399 logistic, 401 order of, 398 partial, 613 separable, 398 solution of, 168, 398 differentiation, 86 of a continuous function, 88 formulas for, 94, RP5 formulas for vector functions, 566 and integration as inverse processes, 288 implicit, 121, 122, 630 logarithmic, 163 partial, 609, 612 of power series, 454 term by term, 454 of a vector function, 566 differentiation operator, 86 directional derivative, 633, 634, 637 maximum value of of, 638 direction field, 403 direction numbers, 546 Direct Substitution Property, 37 directrix of a conic, 511 of a parabola, 509 discontinuity of a function, 46, 47 discontinuous integrand, 349 disk method, 364, 365, 368 displacement, 279 displacement vector, 522, 534 distance between planes, 551, 553 between point and line, 544 between point and plane, 545, 551 between points in space, 519 distance formula, RP1 in three dimensions, 519 Distance Problem, 258 divergence of an improper integral, 346, 349 of an infinite series, 421
A86
■
INDEX
divergence (continued) of a sequence, 412 of a vector field, 760, 761 Divergence, Test for, 425 Divergence Theorem, 791 divergent improper integral, 346, 349 divergent sequence, 412 divergent series, 421 division of power series, 467 DNA, helical shape of, 561, 578 domain of a function, 2, 591 Doppler effect, 633 dot product, 530, 531 properties of, 531 double-angle formulas, A6, RP2 double integral, 665, 674 change of variable in, 717 Midpoint Rule for, 667 over general regions, 674, 675 in polar coordinates, 682, 684, 685 properties of, 671, 672, 679, 680 over rectangles, 665 double Riemann sum, 666 e (the number): as a limit, 145, A35 as a sum of an infinite series, 462 , definition of a limit, 31, 64 eccentricity, 511 electric charge, 160, 688, 699 electric current, 23, 29, 320 electric field, 729, 765, 795 electric flux, 783 elementary function, 332 elimination constant of a drug, 197 ellipse, 510 ellipsoid, 555, 557 elliptic paraboloid, 555, 557 end behavior of a function, 68 endpoint extreme values, 202 energy conservation of, 748, 749 kinetic, 749 potential, 749 epitrochoid, 495 equation(s) cancellation, 150 differential (see differential equation) of an ellipse, 510 heat conduction, 616 of a hyperbola, 510 Laplace’s, 614, 762 of a line in space, 545, 546 logistic (differential), 401
Maxwell’s 765 of a parabola, 509 parametric, 482, 546, 560, 765 of a plane, 549 point-slope, 74 polar, 498 slope-intercept, 11 of a space curve, 560 of a sphere, 520 wave, 614 equipotential curves, 601 equivalent vectors, 522 error in approximate integration, 335, 341 percentage, 137 relative, 136 in Taylor approximation, 472 error bounds, 337, 341 error estimate for alternating series, 440 for the Midpoint Rule, 335, 337 for Simpson’s Rule, 341 for the Trapezoidal Rule, 335, 337 error function, 292 escape velocity, 396 Euclid, 37 Eudoxus, 37 Euler, Leonhard, 145, 208, 430, 462 Evaluation Theorem, 275 even function, 6, 220 exponential decay, 167 exponential function(s), 16, 142, A33, RP4 derivative of, 164 integration of, 276, 277, 466 graph, 144 limits involving, 146 natural, 146, A31 power series for, 462, 466 properties of, A32, A33 exponential graph, 144 exponential growth, 167 Exponents, Laws of, 144, A32, A33 extreme value, 198 Extreme Value Theorem, 199, 649 family of exponential functions, 144 family of solutions, 399 fat circle, 125 Fermat, Pierre, 37, 200 Fermat’s Principle, 235 Fermat’s Theorem, 200 Fibonacci, 411, 419 Fibonacci sequence, 411, 419, 428
field conservative, 729, 746, 749, 759 electric, 729, 765, 794 force, 725, 728 gradient, 729 gravitational, 728 incompressible, 751 irrotational, 760 magnetic, 765 scalar, 726 vector, 725, 762 velocity, 728 First Derivative Test, 213 for Absolute Extreme Values, 229 first octant, 517 first-order differential equation, 398 fixed point of a function, 211 flow lines, 731 floor function, 40 fluid flow, 728, 782, 794 flux, 137, 781, 782, 783 focus (foci) of a conic, 511 of an ellipse, 510 of a hyperbola, 510 of a parabola, 509 folium of Descartes, 122 foot-pound (unit of work), 385 force, 384 exerted by liquid, 387 force field, 728 four-leaved rose, 498 fractions, partial, 320 Frenet-Serret formulas, 577 Fresnel, Augustin, 287 Fresnel function, 287 Fubini, Guido, 670 Fubini’s Theorem, 670, 694 function(s), 1, 2 absolute value, 5 arc length, 571 Airy, 452 arrow diagram of, 2 average cost, 234 average value of, 289, 673, 702 Bessel, 127, 448, 452 Cobb-Douglas production, 658 combinations of, 18 component, 559, 726 composite, 19 concavity of, 214 constant, 94 continuity of, 46, 560, 605, 606, 607 contour map of, 594
INDEX
cost, 103, 231 cubic, 13 decomposition of, 20 decreasing, 7 demand, 231 derivative of, 77 differentiability of, 620 differentiable, 87, 153 discontinuity of, 46, 47 domain of, 2, 591 elementary, 332 end behavior of, 68 error, 292 even, 6, 220 exponential, 16, 142, A33, RP4 extreme value of, 198 fixed point of, 211 floor, 40 Fresnel, 287 gradient of, 636 graph of, 2, 592 greatest integer, 40 harmonic, 614, 764 Heaviside, 23, 29 hyperbolic, 181, RP4 implicit, 122, 398, 630 increasing, 7 integrable, 263, 665 inverse, 148, 149 inverse hyperbolic, 183, RP4 inverse trigonometric, 175, RP3 limit of, 24, 602, 605, 604, 605 linear, 11, 593 logarithmic, 16, 154, RP4 machine diagram of, 2 marginal cost, 80, 103, 231 marginal profit, 232 marginal revenue, 231 maximum and minimum values of, 198, 644, 645, 648, 652 natural exponential, 146 natural logarithm, 155, A29 nondifferentiable, 89 nonintegrable, 332 normal density, 219 of n variables, 598 odd, 7, 220 one-to-one, 148 piecewise defined, 5 polynomial, 13, 605 position, 76 potential, 729 power, 14, RP3 profit, 232
quadratic, 13 ramp, 23 range of, 2, 591 rational, 15, 605 reciprocal, 15 reflected, 17 representation as a power series, 452 representations of, 2, 594, 598 revenue, 231 root, 14 of several variables, 598 shifted, 17 sine integral, 292 step, 6 stretched, 17 symmetry of, 6 tabular, 4 of three variables, 598 transformation of, 16 translation of, 17 trigonometric, 15 of two variables, 591 value of, 2 vector, 559 vector-valued, 559 Fundamental Theorem of Calculus, 284, 286, 288 for line integrals, 742 for vector functions, 567 Galileo, 24, 37, 487, 491 Gauss, Karl Friedrich, 263, 791, A24 Gauss’s Law, 783 Gauss’s Theorem, 791 geometric series, 421 Geometry Review, RP1 gradient, 636, 637 gradient vector, 636, 637 interpretations of, 641 gradient vector field, 729 graph(s), of a curve, 220 of exponential functions, 16, 144 of a function, 2 of a function of two variables, 592 of logarithmic functions, 16, 154 of a parametric curve, 482 of a parametric surface, 777 polar, 498 of a sequence, 412 of trigonometric functions, A7, RP2 graphing device, used to graph: curve, 223 parametric curve, 484
■
A87
polar curve, 502 sequence, 416 Gravitational acceleration, 384 gravitational field, 728 gravitation law, 396 great circle, 713 greatest integer function, 40 Green, George, 752 Green’s identities, 764 Green’s Theorem, 751, 795 area by, 754 vector forms, 762, 763 Gregory, James, 340, 456, 459 Gregory’s series, 456 grid curve, 766 ground speed, 530 growth, law of natural, 167 growth rate, 280 half-angle formulas, A6, RP2 half-life, 170 half-space, 598 harmonic function, 614, 764 harmonic series, 424, 432 heat conductivity, 783 heat conduction equation, 616 heat flow, 783 Heaviside, Oliver, 29 Heaviside function, 23, 29 Hecht, Eugene, 135, 137 helix, 560 higher derivative(s), 90, 612 Hooke’s Law, 386 horizontal asymptote, 60 Horizontal Line Test, 148 Huygens, Christiaan, 398, 486 hydrostatic pressure and force, 387 hyperbola, 510 asymptotes, 510 branches, 510 hyperbolic function(s), 181, RP4 derivative of, 183 identities, 182 inverse, 183, RP4 hyperbolic paraboloid, 556, 557 hyperboloid, 557 i (standard basis vector), 527 I/D Test, 212 ideal gas law, 113, 616 image, 714 implicit differentiation, 121, 122, 630 implicit function, 122, 398, 630 Implicit Function Theorem, 630, 631
A88
■
INDEX
improper integral, 345, 346, 349 convergence/divergence of, 346, 349 incompressible velocity field, 761 Increasing/Decreasing Test, 212 increasing function, 7 increasing sequence, 413 increment, 78, 620, 623 indefinite integral(s), 276, 277 table of, 277, RP6–10 independence of path, 744 independent variable, 2, 591, 628 indeterminate forms of limits, 187 difference, 191 product, 190 power, 192 quotient, 187, 188 index of summation, A23 inertia (moment of ), 691, 699, 741 infinite discontinuity, 47 infinite interval, 345 infinite limit, 56, 63 infinite sequence. See sequence infinite series. See series inflection point, 215 initial condition, 399 initial point of a parametric curve, 483 of a vector, 522 initial-value problem, 399 instantaneous rate of change, 79 instantaneous velocity, 24, 76 integrable function, 665 integral approximation to, 268, 333 change of variables in, 293, 713, 717, 720 conversion to polar coordinates, 684 definite, 263, 567 derivative of, 286 double, 665, 674, 675, 679 evaluating, 274 improper, 346, 349 indefinite, 276, 277 iterated, 668, 675 line, 731, 732, 734, 736, 738 properties of, 269, 270, 271 surface, 775 of symmetric functions, 297 table of, 304, RP6–10 trigonometric, 310 triple, 693, 694, 695 units for, 281 integral calculus, 284 Integral Test, 429, 431
integrand, 263 discontinuous, 349 integration, 263 approximate, 333 by computer algebra system, 326, 330, 342, 456 of exponential functions, 272, 276, 466 formulas, RP6–10 indefinite, 276, 277 limits of, 263 numerical, 333 partial, 668 by partial fractions, 320 by parts, 304, 305 of power series, 454, 466 of rational functions, 320 reversing order of, 679 by substitution, 293, 294 tables, use of, 328 term by term, 454 by trigonometric substitution, 314, 315 of a vector function, 567 interest compounded continuously, 172, 173, 193 Intermediate Value Theorem, 52 intermediate variable, 628 intersecting planes, 550 intersection of polar graphs, 506 interval of convergence, 449, 450 inverse function(s), 148, 149 continuity of, 152 differentiability of, 153 graph of, 151 steps of finding, 151 inverse hyperbolic functions, 183, RP4 derivatives of, 184 expressed in terms of logarithms, 184 inverse substitution in integration, 315 inverse transformation, 714 inverse trigonometric functions, 175, RP3 derivatives of, 176, 177, 179 involute of a circle, 495 irrotational vector field, 760 isothermal, 595 iterated integral, 668, 675 j (standard basis vector), 527 Jacobi, Carl, 716 Jacobian, 716, 719 Joule (unit of work), 385 jump discontinuity, 47
k (standard basis vector), 527 kampyle of Eudoxus, 126 Kepler, Johannes, 584 Kepler’s Laws, 584 kinetic energy, 749 Kondo, Shigeru, 462 Lagrange, Joseph-Louis, 206, 208, 653 Lagrange multiplier, 652, 653, 657 Lagrange’s form of the remainder term, 461 lamina, 391, 688, 689 Laplace, Pierre, 614 Laplace operator, 762 Laplace’s equation, 614, 762 lattice, 545 Law of Conservation of Energy, 749 law of cosines, A9, RP2 law of gravitation, 106, 396 law of natural growth or decay, 167, 168 law of the lever, 390 laws of exponents, 144, A32, RP1 laws of logarithms, 155, A30, RP4 law of sines, RP2 learning curve, 405 least squares method, 652 left-hand limit, 29 Leibniz, Gottfried Wilhelm, 87, 284, 398 Leibniz notation, 86, 263, 287 lemniscate, 126 length of a curve, 378 of a polar curve, 507 of a space curve, 570, 571 of a vector, 525 level curve, 594 level surface, 598 tangent plane to, 640 l’Hospital, Marquis de, 188, 194 l’Hospital’s Rule, 188 libration point, 241 limaçon, 508 Limit Comparison Test, 434 Limit Laws, 35, A10 for functions of two variables, 604 for sequences, 414 limit(s) calculating, 35 of composite functions, 51 e (the number) as, 145 , definition, 31, 64 of exponential functions, 146 of a function, 24
INDEX
of a function of two variables, 602, 604 of a function of three variables, 607 infinite, 56, 63 at infinity, 58, 63 involving infinity, 56 intuitive definition of, 25 laws, 35 left-hand, 29 of logarithmic function, 155 of natural logarithm, 157 Newton’s discovery of, 37 one-sided, 29 precise definition of, 31, 64 properties of, 35, 41 right-hand, 30 of a sequence, 412 involving sine function, 42 of a vector function, 559 limits of integration, 263 line, RP1 normal, 97, 640 secant, 74 slope of, 74 tangent, 75 line (in space): of intersection (of planes), 550 parametric equations of, 546 skew, 548 symmetric equations of, 547 tangent, 563 vector equation of, 545 linear approximation, 133, 619, 623 applications of, to physics, 135 linear equation of a plane, 549 linear function, 11 of two variables, 593 linearization, 133, 619 linear model, 11 line integral, 731, 732, 734, 736, 738 Fundamental Theorem for, 742 for a plane curve, 731, 734 with respect to arc length, 734 for a space curve, 736 of vector fields, 738, 739 work defined as, 738 liquid force, 388 Lissajous figure, 488 local maximum and minimum, 199 local maximum and minimum values, 198, 645 logarithm(s), 16, 154 defined as an integral, A28 laws of, 155, RP4
natural, 155 notation for, 155 logarithmic differentiation,163 logarithmic function(s), 16, 154, A34 with base a, 154, A34 derivative of, 160, 161, 162 graph of, 16, 154 limit of, 155, 157, A30 natural, 155, A29 properties of, 155 logistic differential equation, 401 logistic growth, 401 longbow curve, 516 machine diagram of a function, 2 Maclaurin, Colin, 459 Maclaurin series, 458, 459 table of, 466 magnetic field, 765 magnitude of a vector, 525 marginal cost function, 80, 103, 231 marginal profit function, 232 marginal propensity to consume or save, 427 marginal revenue function, 231 mass of a sheet, 777 of a solid, 699 of a wire, 733 mass, center of, 390, 391, 393, 689, 699, 733, 777 mass density, 388 mathematical induction, principle of, A25 mathematical model, 3, 11 Cobb-Douglas, 658 exponential, 16 linear, 11 logarithmic, 16 polynomial, 13 power function, 14 for population growth, 401 rational function, 15 trigonometric, 15 maximum and minimum values, 198, 644, 645, 648, 652 Maxwell’s equations, 765 mean life of an atom, 353 Mean Value Theorem, 207 Mean Value Theorem for integrals, 290 method of cylindrical shells, 373 method of exhaustion, 37 method of Lagrange multipliers, 653, 657
■
A89
method of least squares, 652 midpoint formula, RP1 Midpoint Rule, 268, 334 for double integrals, 667 error in using, 335, 337 for triple integrals, 701 mixing problems, 402 Möbius strip, 774, 779 model. See mathematical model moment about an axis, 390, 689 of inertia, 691, 699, 741 of a lamina, 392, 689 of a mass, 390, 391 polar, 691 second, 691 of a solid, 699 of a system of particles, 390, 391 monkey saddle, 601 monotonic sequence, 413 Monotonic Sequence Theorem, 418 motion in space, 578 multiple integral. See double integral; triple integral multiplication, scalar, of vectors, 523 multiplication of power series, 467 multiplier (Lagrange), 652, 653, 657 multiplier effect, 427 natural exponential function, 146, A31 derivative of, 165 graph of, 146 limit of, 146 power series for, 462, 466 natural growth law, 167 natural logarithm function, 155, A29 derivative of, 162 limit of, 157, A30 negative angle, A2 net area, 265 Net Change Theorem, 279 Newton, Sir Isaac, 37, 284, 584 discovery of limit, 37 newton (unit of force), 384 newton-meter (unit of work), 384 Newton’s Law of Cooling, 171 Newton’s Law of Gravitation, 106, 396, 584, 728 Newton’s method, 236 Newton’s Second Law of Motion, 345, 580, 584 nondifferentiable function, 89 nonintegrable function, 332 nonparallel planes, 550
A90
■
INDEX
normal component of acceleration, 582, 583 normal density function, 219 normal derivative, 764 normal line, 97, 640 normal plane, 575 normal vector, 549, 574, 575 nth-degree Taylor polynomial, 460 numerical integration, 333 octant, 517 odd function, 7, 220 one-sided limits, 29, 30 one-to-one function, 148 one-to-one transformation, 714 open region, 744 optimization problems, 226 strategy for solving, 226 orbit of a planet, 584 order of a differential equation, 398 order of integration reversed, 679 Oresme, Nicole, 424 orientation of a curve, 736, 751 of a surface, 779, 780 oriented surface, 779, 780 orthogonal curves, 126 orthogonal projection, 536 orthogonal surfaces, 644 orthogonal trajectories, 126 orthogonal vectors, 533 osculating circle, 576 osculating plane, 575 Ostrogradsky, Mikhail, 791 ovals of Cassini, 504 Pappus of Alexandria, 394 parabola, 509 parabolic cylinder, 554 paraboloid, 555, 556, 557 parallelepiped, 362 volume of, 542 Parallelogram Law, 523 parallel planes, 550 parallel vectors, 524 parameter, 482, 545, 560 parametric curve, 482 arc length of, 491 area under, 491 tangent to, 488, 489 parametric equations, 482 of a line, 546 of a space curve, 560 of a surface, 765 of a trajectory, 580
parametric surface, 765 area of, 770, 771 graph of, 777 parametrization of a curve, 562 with respect to arc length, 571 paraxial rays, 135 partial derivatives, 609, 612 notation for, 610 as rates of change, 610 rule for finding, 610 second, 612 as slopes of tangent lines, 610 partial differential equation, 613 partial fractions, 320 partial fractions, sum of, 321 partial integration, 304, 305, 668 partial sum of a series, 421 partition, 262 regular, 264, 666 parts, integration by, 304, 305 pascal (unit of pressure), 388 path, 744 percentage error, 137 perihelion, 514 periodic function, 220 perpendicular vectors, 533 piecewise defined function, 5 piecewise-smooth curve, 565, 733 plane, 548, 549 coordinate, 517 equation of, 549 horizontal, 518 normal, 575 osculating, 575 parallel, 550 tangent, to a surface, 617, 640, 769 vertical, 519 plane region of type I, 675 plane region of type II, 676 planetary motion, 584 point of inflection, 215 point-slope equation of a line, 74 Poiseuille’s Law, 137 polar axis, 496 polar coordinates, 496, 497 area in, 504 converted to Cartesian coordinates, 497 conic sections in, 510 double integral converted to, 684, 685 polar curve, 498 arc length of, 507 graph of, 498, 502 tangent line to, 500 polar equation, graph of, 498
polar equation of a conic, 510 polar moment of inertia, 691 polar rectangle, 682 polar region, 504 area of, 505 pole, 496 polynomial function, 13 continuity of, 50, 606 of two variables, 605 population growth, 3, 168, 401 position function, 76, 90 position vector, 524, 525 positive angle, A2 positive orientation of a curve, 751 of a boundary curve, 786 of a surface, 780 potential energy, 749 potential function, 729 pound (unit of force), 384 power, 80, 281 power function(s), 14, RP3 derivative of, 95 Power Law of limits, 36 Power Rule, 95, 97, 116 power series, 447 coefficient of, 447 differentiation of, 454 division of, 467 integration of, 454 interval of convergence, 449 multiplication of, 467 radius of convergence, 449 representation of a function, 452 pressure exerted by a liquid, 387 prime notation, 77, 98 principle of mathematical induction, A25 principle unit normal vector, 575 product cross, 537 dot, 530, 531 scalar, 631 scalar triple, 542 vector triple, 542 product formulas for sine and cosine, A6 Product Law of limits, 36 Product Rule, 106 profit function, 232 projection, 518, 533, 534, 536 p-series, 432 quadratic formula, RP1 quadratic function, 13
INDEX
quadric surface, 554 cone, 557 cylinder, 554 ellipsoid, 555 hyperboloid, 557 paraboloid, 555, 556 quadric surfaces graph of, 557 table of, 557 Quotient Law of limits, 36 Quotient Rule, 108 radian measure, 15, A1 radioactive decay, 169 radiocarbon dating, 174 radius of convergence, 449, 450 radius of gyration, 692 ramp function, 23 range of a function, 2, 591 rate of change, 79, 279 average, 79 derivative as, 79 instantaneous, 79 rate of cooling, 171 rate of decay, 169 rate of growth, 168, 280 rate of reaction, 279 rates, related, 127 rational function, 15, 605 continuity of, 50 integration by partial fractions, 320 Ratio Test, 443 reciprocal function, 15 Reciprocal Rule, 113 rectangular coordinates three-dimensional, 518 conversion to cylindrical coordinates, 703 conversion to spherical coordinates, 708 rectilinear motion, 244 reduction formula, 308 reflection transformation a function, 17 region between two curves, 357 connected, 744 open, 744 plane, of type I or II, 675, 676 simple plane, 752 simple solid, 791 simply-connected, 746 solid, of type 1, 2, or 3, 695, 696, 697 under a curve, 251 regular partition, 264
related rates, 127 strategy for solving, 129 relative error, 136 relative growth rate, 168 relative maximum and minimum, 199 remainder estimate for an alternating series, 440 remainder of the Taylor series, 461 Roberval, Gilles de, 276, 491 Rolle, Michel, 205 Rolle’s Theorem, 205 removable discontinuity, 47 representation of a function: in four ways, 3 as a power series, 452, 594 resultant force, 528 revenue function, 231 reversing order of integration, 679 revolution, solid of, 367 Riemann, Georg Bernhard, 262, 263 Riemann sum(s), 262 for multiple integrals, 666, 694 right cylinder, 362 right-hand limit, 30 right-hand rule, 517, 539 root function, 14 continuity of, 50 Root Law of Limits, 36 Root Test, 445 rubber membrane, vibration of, 448 Ruling of a surface, 554 saddle point, 646 sample point, 256, 262, 664 scalar, 523 scalar equation of a plane, 549 scalar field, 726 scalar multiple of a vector, 523 scalar product, 531 scalar projection, 534 scalar triple product, 542 scatter plot, 3 secant line, 74, 75 secant vector, 563 second derivative, 90 of a vector function, 565 Second Derivative Test, 216 Second Derivatives Test, 646 second moment, 691 second partial derivative, 612 sector of a circle, area of, 504 separable differential equation, 398 sequence, 410 bounded, 413 convergent, 412
decreasing, 413 divergent, 412 Fibonacci, 411, 419, 428 graph of, 416 increasing, 413 limit of, 412 limit laws for, 414 monotonic, 413 of partial sums, 421 term of, 410 series, 420 absolutely convergent, 441 alternating, 437 alternating harmonic, 439 binomial, 464, 465 coefficient of, 447 convergent, 421 divergent, 421 geometric, 421 Gregory’s, 456 harmonic, 424, 432 infinite, 420 Maclaurin, 458, 459 p-, 432 partial sum of, 421 power, 447 sum of, 421 Taylor, 458, 459 term of, 420 trigonometric, 447 serpentine, 112 set bounded, 649 closed, 649 shell method, 373 shifts of functions, 17 Sierpinski carpet, 429 sigma notation, 256, A23 simple curve, 746 simple harmonic motion, 120 simple plane region, 752 simple solid region, 791 simply-connected region, 746 Simpson, Thomas, 339, 340 Simpson’s Rule, 338, 340 error bounds for, 341 sine function, A2 derivative of, 101 graph, A7 inverse, 175 limit involving, 42 power series for, 463, 466 sine integral function, 292 sink, 795 skew lines, 548
■
A91
A92
■
INDEX
slant asymptote, 225 slope of a curve, 74 slope field, 403 slope-intercept equation of a line, 11 smooth curve, 378, 565 smooth surface, 769 solid, volume of, 367, 368, 698 solid of revolution, 367 solid region, simple, 791 solution curve, 403 solution of a differential equation, 398 source, 795 space, three-dimensional, 517 space curve, 560 arc length of, 570 speed, 79, 530, 578 sphere equation of, 520 flux across, 781 parametrization of, 767 surface area of, 771, 775 spherical coordinates, 707, 708 integration in, 709 spherical wedge, 709 spring constant, 386 Squeeze Theorem, 41 for sequences, 414 standard basis vectors, 527 standard position of an angle, A2 stellar stereography, 353 step function, 6 Stokes, Sir George, 786 Stokes’ Theorem, 786, 795 strophoid, 516 strategy for optimization problems, 226 strategy for related rates problems, 129 streamlines, 731 stretching transformations of a function, 17 substitution, trigonometric, 314, 315 Substitution Rule, 293, 294 for definite integrals, 296 subtraction formulas for sine and cosine, A5 sum of a geometric series, 422 of an infinite series, 421 of partial fractions, 321 Riemann, 262 telescoping, 424 of vectors, 523 Sum Law of limits, 36 Summation notation, A23 Sum Rule, 98
surface closed, 780 graph of, 777 level, 598 oriented, 779, 780 orthogonal, 644 parametric, 765 quadric (see quadric surface) smooth, 769 surface area of a parametric surface, 770, 771 of a sphere, 771, 772, 776 of a surface of revolution, 772 surface integral, 775 of a parametric surface, 776 of a vector field, 781 surface of revolution, parametrization of, 768 swallowtail catastrophe curve, 488 symmetric equations of a line, 547 symmetric functions, integrals of, 297 symmetry, 6, 297 symmetry principle, 391 system of particles, moment of, 390, 391 T 1 inverse transformation, 714 table of integrals, 328, RP6–10 tabular function, 4 tangent function, A2 derivative of, 110 graph, A7 inverse, 177 tangential component of acceleration, 582, 583 tangent line to a curve, 74, 75 to a parametric curve, 488, 489 to a polar curve, 500 to a space curve, 563 vertical, 89 tangent line approximation, 133 to a space curve, 564 tangent plane to a level surface, 640 to a parametric surface, 769 to a surface z f x, y, 617 tangent plane approximation, 619 Tangent Problem, 73 tangent vector, 563, 769 tautochrone problem, 486 Taylor, Brook, 459 Taylor polynomial, 460, 471 Taylor series, 458, 459 Taylor’s Formula, 461 telescoping sum, 424
term-by-term differentiation and integration, 454 terminal point of a parametric curve, 483 of a vector, 522 terminal velocity, 406 term of a sequence, 410 term of a series, 420 test for convergence/divergence of series Alternating Series Test, 437 Comparison Test, 432, 433 Integral Test, 429, 431 Limit Comparison Test, 434 Ratio Test, 443 Test for Divergence, 425 Thomson, William (Lord Kelvin), 752, 786 third derivative, 91 three-dimensional coordinate system, 517, 518 TNB frame, 575 toroidal spiral, 562 torque, 543, 587 Torricelli, Evangelista, 491 Torricelli’s Law, 82, 105 torsion, 577, 775 torus, 372, 394 total differential, 621 total fertility rate, 139 trace, 553 trajectory, parametric equations for, 580 transformation, 714 of a function, 16 inverse, 714 Jacobian of, 716 one-to-one, 714 translation of a function, 17 Trapezoidal Rule, 335 error in, 335, 337 tree diagram, 628 trefoil knot, 562 Triangle Inequality, A10 for vectors, 537 Triangle Law, 523 trigonometric functions, 15, 100, A22, RP2 continuity of, 50 derivatives of, 100, 101, 110, 111 graphs of, A7, RP2 integrals of, 277 inverse, 175, RP3 trigonometric identities, A4, RP2 trigonometric integrals, 310 trigonometric substitutions in integration, 314, 315
INDEX
Trigonometry Review, A1, RP2 triple integral, 693, 694, 695 applications of, 698 in cylindrical coordinates, 704, 705 over a general bounded region, 695, 696, 697 Midpoint Rule for, 701 in spherical coordinates, 709 triple product, 542 triple Riemann sum, 694 trochoid, 487 Tschirnhausen cubic, 126 twisted cubic, 562 type I plane region, 675 type II plane region, 676 type 1 solid region, 695 type 2 solid region, 696 type 3 solid region, 697 unit normal vector, 575, 576 unit tangent vector, 563, 576 unit vector, 528 value of a function, 2 variable, intermediate, 628 variable(s), change of, 293, 713, 717, 720 variables, dependent and independent, 2, 591, 628 vector(s), 522 acceleration, 578, 582 addition of, 523 angle between, 531, 532 basis, 527 binormal, 575, 576 components of, 524, 534 coplanar, 542 cross product of, 537 difference of, 524 displacement, 522, 534 dot product, 530, 531 equivalent, 522 force, 728 geometric representation of, 524 gradient, 636, 637 i, j , and k, 527 initial point, 522 length of, 525 magnitude of, 525 multiplication of, 523 normal, 549, 575
orthogonal, 533 parallel, 524 perpendicular, 533 position, 524, 525 principle unit normal, 575 projection, 518, 533, 534, 536 properties of, 526 scalar multiple of, 523 secant, 563 standard basis, 527 subtraction of, 524 sum of, 523 tangent, 563 terminal point, 522 three-dimensional, 525 triple product, 542 two-dimensional, 525 unit, 528 unit normal, 575, 576 unit tangent, 563, 576 velocity, 578 zero, 522 vector equation of a line, 545 of a line segment, 548 of a plane, 549 of a plane curve, 560 vector field, 725, 726 conservative, 729, 746, 749 curl of, 758, 759 divergence of, 760, 761 electric, 729, 765, 794 force, 725, 728 flux of, 781, 782, 783 gradient, 729 gravitational, 728 incompressible, 761 irrotational, 760 line integral of, 738, 739 surface integral of, 781 velocity, 728 vector form of Green’s Theorem, 762, 763 vector function, 559 continuity of, 560 derivative of, 563, 564 differentiation formulas for, 566 integration of, 567 limit of, 559 vector product, 537
■
A93
vector projection, 533, 534 vector triple product, 542 vector-valued function. See vector function velocity, 24, 76, 79, 90 average, 24, 76 instantaneous, 24, 76 velocity field, 728 Velocity Problem, 76 velocity vector, 578 Verhulst, Pierre-François, 401 vertical asymptote, 57 Vertical Line Test, 4 vertical plane, 519 vertical tangent line, 89 vertical translation of a graph, 17 visual representations of a function, 2, 592, 594, 598 volume by cross-sections, 364 by cylindrical shells, 373 by disks, 364, 365, 368 by double integrals, 663, 665, 666 of a solid of revolution, 367, 368 by triple integrals, 698 washer method, 366, 368 wave equation, 614 weight (force), 385 weight density, 388 wind-chill index, 616 witch of Maria Agnesi, 112, 488 work, 384, 385 Wren, Sir Christopher, 493 world population growth, 168, 401 work, 384, 534 expressed as an integral, 385 as a line integral, 738 units for, 384, 385 x -axis, 517 x -coordinate, 518 y-axis, 517 y-coordinate, 518 z-axis, 517 z-coordinate, 518 zero vector, 522
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Assignment Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12 Accessing an Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12 Using the Assignment Page . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14
ANSWERING QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17 Numerical Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17 Numerical Questions with Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18 Numerical Questions with Significant Figures . . . . . . . . . . . . . . . . . . . . . . .19 Math Notation: Using the MathPad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20 Math Notation: Using the CalcPad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21 Math Notation: Using the Keyboard . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21
WebAssign works with any recent browser and computer. Some assignments may require an updated browser and/or plugins like Java, Flash, Shockwave, or Adobe Reader. For technical support go to http://webassign.net/student.html or email [email protected].
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Contents USING
THE
GRAPHPAD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22 GraphPad Interface Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23 Drawing Graph Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23 Selecting Graph Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24 Moving and Editing Graph Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25 Using Fractions or Decimals as Coordinates . . . . . . . . . . . . . . . . . . . . . . . .25 Endpoints—Closed or Open? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26 Graph Objects—Solid or Dashed? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26 Specifying a Region with a Fill for Inequalities . . . . . . . . . . . . . . . . . . . . . .26 Erasing One Graph Object . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27 Erasing Everything on Your Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27 Example: Graphing Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28
ADDITIONAL FEATURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29 Calendar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29 Communication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29
GRADES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .30 TECHNICAL TIPS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31
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The Start Smart Guide for Students
Getting Started
GETTING STARTED Welcome to Enhanced WebAssign, the integrated, online learning system that gives you 24/7 access to your math, physics, astronomy, chemistry, biology, and statistics assignments. Now, you can do homework, take quizzes and exams, and receive your scores and graded assignments from any computer with an Internet connection and web browser, any time of the day or night. Note: As a live, web-based program, Enhanced WebAssign is updated regularly with new features and improvements. Please refer to WebAssign’s online Help for the most current information.
Technical Startup Tips Before you start, please note the following important points: ❍ Most standard web connections should work with WebAssign. We recommend using Firefox 1.0 or later, or Internet Explorer 5.5 or later. We do not recommend the AOL browser. ❍ You can use a 56 KBPS modem, broadband, or school network connection. ❍ Your browser needs to have both JavaScript and Java enabled. ❍ You cannot skip the login page. WebAssign must know it is you before delivering your assignments. Note: If you’d like to bookmark WebAssign on your computer, we recommend that you bookmark https://www.webassign.net/login.html or the appropriate address for your school.
Login to WebAssign In order to access WebAssign your instructor will provide you with login information or a Class Key. Login information will consist of a username, institution code, and an initial password. The Class Key will allow you to self-register and create your own login. You will
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Getting Started need to remember the username and initial password you set after self-registering. Please note that Class Keys are not the same as access codes. See pages 8–9 for instructions on registering your access code number. You will need to login first before you are able to register an access code.
➢ To get started 1. If you are using a shared computer, completely exit any browsers that are already open. 2. Open a new web browser and go to https://www.webassign.net/login.html, or the web address provided by your instructor. If your instructor has provided you with a Username, Institution (school code), and Password, continue with step 3. If you have been provided with a Class Key (usually your institution name and a series of numbers), then skip to step 5. 3. Enter your Username, Institution (school code), and Password provided by your instructor. Institution If you do not know your Institution, you can search for it by clicking (what’s this?) above the Institution entry box. In the What’s My Institution Code pop-up window, enter your school name and click go!. The Institution Search Results table will give you choices of the School Names that most closely match your entry, and the Institution Code that you should enter in the Institution entry box on the WebAssign Login screen. Password If you have forgotten or do not know your Password, click (Reset Password) above the Password entry box, and follow the directions on the WebAssign New Password Request screen. You will need to submit your username, institution code, and the email address on file in your WebAssign account. If you are unsure of your username or listed email address, please check with your instructor. WebAssign cannot reset your username or password. 6
The Start Smart Guide for Students
Getting Started 4. Click Log In. 5. If your instructor gave you a Class Key, you will use it to create your account. Click the I have a Class Key button. You will need to use this key only once when you register. 6. Enter the Class Key code in the field provided and click Submit. If your Class Key is recognized, you will be given fields for creating your username and password and for entering basic student information. 7. Enter a username in the field provided and then click Check Availability to determine whether or not your username is already in use. If it is, an available alternate username will be suggested. Remember your username because you will use it every time you login to WebAssign. 8. Enter and then re-enter a password. Remember your password because you will use it every time you login to WebAssign. 9. Under Student Information enter your first and last name, email address, and student ID. 10. Click Create My Account. 11. If you see confirmation of your account creation, you will now be able to login to WebAssign. Click Log in now.
Note: Before starting WebAssign on a shared computer, always exit any browsers and restart your browser application. If you simply close the browser window or open a new window, login information contained in an encrypted key may not be yours.
Logout When you are finished with your work, click the Logout link in the upper right corner of your Home page, and exit the browser completely to avoid the possibility of someone else accessing your work.
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Your Enhanced WebAssign Home Page
YOUR ENHANCED WEBASSIGN HOME PAGE Your personalized Home page is your hub for referencing and managing all of your Enhanced WebAssign assignments.
Using Access Codes Some classes require an access code for admission. Please remember: ❍ An access code is not the same as a Class Key or a login password. ❍ An access code is good for one class only unless the textbook includes a two-term access code. ❍ An access code is an alphanumeric code that is usually packaged with your textbook. It can begin with 2 or 3 letters, followed by an alphanumeric code, or it can have a longer prefix such as BCEnhanced-S followed by four sets of four characters. ❍ If your textbook did not include an access code, you can buy one at your bookstore, or from your personalized Home page by clicking the Purchase an access code online button.
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Your Enhanced WebAssign Home Page
➢ To enter an Access Code 1. Under WebAssign Notices, select the proper prefix from the Choose your access code prefix pull-down menu.
WebAssign notices 2. Click Go. 3. In the entry boxes, type in your access code exactly as it appears on your card. (When you purchase online, the access code is entered automatically.)
Access code entry 4. Click Enter your access code. If you have chosen the wrong prefix from the previous screen, you can click the Choose a different access code button to try again. If your access code is a valid unused code, you will receive a message that you have successfully entered the code for the class. Click the Home or My Assignments button to proceed. Enhanced WebAssign
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Your Enhanced WebAssign Home Page
Customizing Your Home Page Your instructor has initial control over what you see on your Home page to make sure that you have all of the information you need. Your instructor might also set controls so that you can further personalize this page by moving or hiding certain modules.
Student Home page If your instructor has allowed you to personalize your Home page, each module will have markings like this:
Calendar module To move a module On the module’s heading line, click an up, down, or sideways arrow (indicated by white triangles) until the module is where you’d like it placed on the page. To minimize a module On the module’s heading line, click the underscore. To hide a module On the module’s heading line, click the x.
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Your Enhanced WebAssign Home Page
Changing Your Password For your personal security, it’s a good idea to change the initial password provided by your instructor.
➢ To change your password 1. Click the My Options link in the upper right of your Home page. 2. In the My Options pop-up window, under the Personal Info tab: Enter your new password in the Change Password entry box next to (enter new password), then Reenter your new password exactly the same in the entry box next to (reenter for confirmation). 3. Enter your current password in the entry box under If you made any changes above, enter your current password here and then click save:, located at the bottom of the popup window. 4. Click the Save button in the bottom right corner of the pop-up window. If the change was successful, you will see the message Your password has been changed. Note: Passwords are case-sensitive. This means that if you capitalize any of the letters, you must remember to capitalize them the same way each time you sign in to Enhanced WebAssign.
Changing Your Email Address If your instructor provided you with an email address, you can easily change it to your own personal email address any time.
➢ To change your email address 1. Click the My Options link in the upper right of your Home page.
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Working with Assignments 2. In the My Options pop-up window, under the Personal Info tab, enter your valid email address in the Email Address box. 3. Enter your current password in the entry box under If you made any changes above enter your current password here and then click save:, located at the bottom of the popup screen. 4. Click the Save button in the bottom right corner of the pop-up window. A confirmation email will be sent to your new email address. Once you receive the confirmation email, you must click the link in the email to successfully complete and activate this change.
WORKING
WITH
ASSIGNMENTS
The courses that have been set up for you by your instructor(s) appear on your Enhanced WebAssign personalized Home page. If you have more than one course, simply select the course you want to work with from the pull-down menu.
Assignment Summary There are two ways to get a quick summary of your assignments. On the Home page: ❍ Click the My Assignments link in the upper left menu bar, or ❍ Click the Current Assignments link in the My Assignments module on the Home page.
Accessing an Assignment Once your assignments are displayed on your Home page, simply click the name of the assignment you’d like to begin. ❍ If you have previously submitted an assignment, you will see your most recent responses, if your instructor allows this feature. ❍ If you have already submitted the assignment, there will usually be a link to Review All Submissions on the page, if your instructor has allowed it. 12
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Working with Assignments
Assignment summary
Math assignment
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Working with Assignments
Physics assignment
Using the Assignment Page When you click on an assignment name, your assignment will load. Within the About this Assignment page are links to valuable information about your assignment’s score, submission options, and saving your work in progress. Within each question, there might also be “enhanced” action links to useful tutorial material such as book content, videos, animations, active figures, simulations, and practice problems. The links available may vary from one assignment to another.
Actions Click a button or link to take one of the following actions: Current Score This gives you a quick look at your current score versus the maximum possible score. Question Score This gives you a pop-up window showing your score for each question. 14
The Start Smart Guide for Students
Working with Assignments Submission Options This gives you a pop-up window explaining how you can submit the assignment and whether it can be submitted by question part, by whole question, or by the whole assignment. Submissions Made This shows you the number of submissions you’ve made. This information is only displayed on assignments that require submission of the entire assignment. Notes This feature gives you a pop-up window with a text box in which you can enter and save notes or show your work with a particular question. Submit New Answers To Question Use this button when you’re ready to submit your answer for the question. This feature allows you to answer just the parts you want scored. If you leave any part of a question unanswered, the submission will not be recorded for that part. Submit Whole Question Use this button to submit your answer(s) for the entire question. If you leave any part of a question unanswered, the submission will be recorded as if the entire question has been answered, and graded as such. Save Work This button allows you to save the work you’ve done so far on a particular question, but does not submit that question for grading. View Saved Work Located in the question’s header line, this allows you to view work that you previously saved for that question. Show Details Located in the question’s header line, this link shows your score on each part of the question, how many points each part of the question is worth, and how many submissions are allowed for each part if you can submit each part separately. Submit All New Answers This submits all of your new answers for all of the questions in the assignment. Enhanced WebAssign
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Working with Assignments Save All Work This allows you to save all the work you’ve done on all of the questions in the assignment, but does not submit your work for grading. Ask Your Teacher This feature allows you to send a question about the assignment to your instructor. Extension Request This allows you to submit a request to your instructor for an extension of time on an assignment. Home This link takes you to your personalized Home page. My Assignments This link takes you to your assignments page. Open Math Palette This opens a tool to use in writing answers that require math notation. Read it This links to question-specific textbook material in PDF form. Practice Another Version This provides you with an alternate version of the assigned problem. Within the pop-up window you will be able to answer the practice problem and have that answer checked. You will also be able to practice additional versions of your assigned problem. Practice it This links to a practice problem or set of practice problems in a pop-up window. No grade is recorded on the work you do on practice problems. See it This links to a tutorial video. Hint This links to a pop-up window with helpful hints in case you get stuck on a question.
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Answering Questions Hint: Active Figure This links to an animated simulation to help you better understand the concepts being covered. Note: Your instructor has the ability to turn on/off many of the options listed above.
ANSWERING QUESTIONS Enhanced WebAssign uses a variety of question types that you’re probably already familiar with using, such as multiple choice, true/false, free response, etc. Always be sure to pay close attention to any instructions within the question regarding how you are supposed to submit your answers.
Numerical Questions There are a few key points to keep in mind when working on numerical questions: ❍ Numbers can be entered in both scientific notation and numerical expressions, such as fractions. ❍ WebAssign uses the standard scientific notation “E” or “e” for “times 10 raised to the power.” (Note: both uppercase E and lowercase e are acceptable in WebAssign.) For example, 1e3 is the scientific notation for 1000. ❍ Numerical answers may not contain commas ( , ) or equal signs ( = ). ❍ Numerical answers may only contain: • Numbers • E or e for scientific notation • Mathematical operators +, -, *, / ❍ Numerical answers within 1% of the actual answer are counted as correct, unless your instructor chooses a different tolerance. This is to account for rounding errors in calculations. In general, enter three significant figures for numerical answers.
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Answering Questions
➢ Example: Numerical Question Let’s suppose you’re presented a question to which your answer is the fraction “one over sixty-four.” Following are examples of Correct and Incorrect answer formats: Correct Answers Any of these formats would be correct: 1/64 0.015625 0.0156 .0156 1.5625E-2 Incorrect Answers These formats would be graded as incorrect: O.015625
The first character is the letter “O”
0. 015625
There is an improper space in the answer
1.5625 E-2
There is an improper space using E notation
l/64
The first character is lowercase letter “L”
5,400
There is a comma in the answer
1234.5=1230
There is an equal sign in the answer
Numerical Questions with Units Some Enhanced WebAssign questions require a number and a unit, and this is generally, although not always, indicated in the instructions in the question. You will know that a unit is expected when there is no unit after the answer box. When you are expected to enter units and do not, you will get an error message telling you that units are required. Note: Whether omission of the unit counts as a submission depends on the submission options chosen by the instructor.
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Answering Questions
Numerical with units The easiest units to use in this question are m, but the answer converted to yd would also be scored correct.
Numerical Questions with Significant Figures Some numerical questions require a specific number of significant figures (sig figs) in your answer. If a question checks sig figs, you will see a sig fig icon next to the answer box. If you enter the correct value with the wrong number of sig figs, you will not receive credit, but you will receive a hint that your number does not have the correct number of sig figs. The sig fig icon is also a link to the rules used for sig figs in WebAssign.
Check for significant figures
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Answering Questions
Math Notation: Using the MathPad In many math questions, Enhanced WebAssign gives you a MathPad. The MathPad provides easy input of math notation and symbols, even the more complicated ones. If your answer involves math notation or symbols, the MathPad will become available when you click the answer box.
1
2
3
Top Symbols The buttons on the top are single input buttons for frequently used operations. Word Buttons When you click one of the word buttons Functions, Symbols, Relations, or Sets, you are given a drop-down menu with symbols or notation from which to choose. For example, if you click the Sets button, you get set notation (figure 2 above). If you then click a right arrow button, additional symbols become available (figure 3 above). 20
The Start Smart Guide for Students
Answering Questions To insert any available notation or symbol into your answer, simply click it.
Math Notation: Using the CalcPad CalcPad, as its name implies, is designed for use with the more complicated symbol and notation entry in calculus. It functions in a similar manner to the MathPad described above. If your course uses CalcPad, check online for additional instructions.
Math Notation: Using the Keyboard If you use your keyboard to enter math notation (calculator notation), you must use the exact variables specified in the questions. The order is not important, as long as it is mathematically correct.
➢ Example: Math Notation Using Keyboard In the example below, the keyboard is used to enter the answer in the answer field.
Symbolic question
Expression Preview Clicking the eye button allows you to preview the expression you’ve entered in calculator notation. Use this preview feature to help determine if you have properly placed your parentheses. Enhanced WebAssign
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Using the GraphPad
Symbolic Formatting Help If you’re unsure about how to symbolically enter your answer properly, use the symbolic formatting help button to display allowed notation.
Allowed notation for symbolic formatting
USING
+ for addition
x+1
- for subtraction
x-1, or –x
* or nothing for multiplication
4*x, or 4x
/ for division
x/4
** or ^ for exponential
x**3, or x^3
( ) where necessary to group terms
4/(x+1), or 3(x+1)
abs( ) to take the absolute value of a variable or expression
abs(-5) = 5
sin, cos, tan, sec, csc, cot, asin, acos, atan functions (angle x expressed in radians)
sin(2x)
sqrt( ) for square root of an expression
sqrt(x/5)
x^ (1/n) for the nth root of a number
x^ (1/3), or (x-3)^ (1/5)
pi for 3.14159…
2 pi x
e for scientific notation
1e3 = 1000
ln( ) for natural log
ln(x)
exp( ) for “e to the power of”
exp(x) = ex
THE
GRAPHPAD
Introduction The Enhanced WebAssign GraphPad lets you graph one or more mathematical elements directly on a set of coordinate axes. Your graph is then scored automatically when you submit the assignment for grading. 22
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Using the GraphPad The GraphPad currently supports points, rays, segments, lines, circles, and parabolas. Inequalities can also be indicated by filling one or more areas.
GraphPad Interface Overview The middle of GraphPad is the drawing area. It contains labeled coordinate axes, which may have different axis scales and extents depending on the nature of the question you are working on. On the left side of GraphPad is the list of Tools that lets you create graph objects and select objects to edit. The bottom of the GraphPad is the Object Properties toolbar, which becomes active when you have a graph element selected. This toolbar shows you all the details about the selected graph object and also lets you edit properties of that. On the right side of GraphPad is the list of Actions that lets you create fills and delete objects from your graph.
Drawing Graph Objects
To draw a line, first click on the line button in the Tools area. The line button will highlight to blue, and then you can place Enhanced WebAssign
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Using the GraphPad two points (which are on that line) by clicking twice within the drawing area. Don’t worry if you don’t place the line exactly where you want it initially; you can move these points around before submitting for grading. The arrows on the end of the line indicate that the line goes off to infinity on both ends.
To draw a ray or a segment, first click the small arrow on the right side of the line button to open the selection of line-type tools. Choose ray or segment, and then place it by clicking twice within the drawing area. For rays, the first point is the endpoint of the ray. The arrow on the other end of the ray indicates that it goes off to infinity on that end.
Circles, points, and parabolas can be drawn in the same manner. Circles are drawn by placing a point at the center first, then a point on the radius of the circle. Parabolas are drawn by placing the vertex first, then a point on the parabola. Parabolas can be horizontal or vertical. Points are even easier—just click the point button and then click where you want the point to appear.
Selecting Graph Objects
To edit a graph object you have drawn, that object must be “selected” as the active object. (When you first draw an object, it is created in the selected state.) When a graph element is “selected”, the color of the line changes and two “handles” are visible. The handles are the two square points you clicked to create the object. To select an object, click on the object’s line. To deselect the object, click on the object’s line, a blank area on the drawing area, or a drawing tool.
Not Selected 24
Selected The Start Smart Guide for Students
Using the GraphPad
Moving and Editing Graph Objects Once an object is selected, you can modify it by using your mouse or the keyboard. As you move it, you’ll notice that you cannot move the handles off the drawing area. To move an object with the mouse, click and drag the object’s line. Or click and drag one of the handles to move just that handle. On the keyboard, the arrow keys also move the selected object around by one unit. As you move the object or handle you’ll see that the Object Properties toolbar changes to remain up to date. You can also use the coordinate boxes in the Object Properties toolbar to edit the coordinates of the handles directly. Use this method to enter decimal or fractional coordinates.
Using Fractions or Decimals as Coordinates To draw an object with handle coordinates that are fractions or decimals, you must use the Object Properties toolbar. Draw the desired object anywhere on the drawing area, then use the coordinate boxes in the Object Properties toolbar to change the endpoint(s) to the desired value. To enter a fraction just type “3/4”, for example. Note: The points and lines you draw must be exactly correct when you submit for grading. This means you should not round any of your values—if you want a point at 11/3, you should enter 11/3 in the coordinate box rather than 3.667. Also, mixed fractions are not acceptable entries. This means that 3 2/3 is an incorrect entry.
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Using the GraphPad
Endpoints—Closed or Open? If the selected object is a segment or ray, the Endpoint controls in the Object Properties toolbar can be clicked to toggle the endpoint from closed to open. As a shortcut, you can also toggle an endpoint by clicking on the endpoint when the ray or segment is in the unselected state.
Graph Objects—Solid or Dashed? For any selected object other than a point, the Solid/Dash buttons in the Object Properties toolbar can be used to make the object solid or dashed. To change graph objects to solid or dashed (for inequalities, for example), select the object and click the Solid or Dash button.
Specifying a Region with a Fill for Inequalities
To graph an inequality, you must specify a region on the graph. To do this, first draw the line(s), circle(s), or other object(s) that will define the region you want to represent your
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The Start Smart Guide for Students
Using the GraphPad answer. Be sure to specify the objects as either solid or dashed, according to the inequality you are graphing! Then choose the fill button in the Actions area, and click inside the region that you want filled.
If you decide you wanted the fill in a different area, you can use the fill tool again to undo and then redo the fill in a different location. Choose the fill tool, click the filled region that you want to unfill, and then click the region that you do want to fill.
Erasing One Graph Object
To erase a single graph object, first select that element in the drawing area, then click the Delete icon in the Actions area or press the Delete key on your keyboard.
Erasing Everything on Your Graph
The Clear All button in the Actions area will erase all of your graph objects. (If the drawing area is already empty, the Clear All button is disabled.)
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Using the GraphPad
Example: Graphing Question Let’s suppose you’re asked to graph the inequality y > 5 x + 1 , 5
( ) and ( ) . You would
and you want to use the points 0,
1 5
1, 5 1 5
first place any line on the drawing area.
Then, adjust the points using the Object Properties Boxes.
Next, you would define the line as dashed since the inequality does not include the values on the line.
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Additional Features Finally, you would select the Fill Tool and click on the desired region to complete the graph.
ADDITIONAL FEATURES Calendar The Calendar link presents you with a calendar showing all of your assignments on their due dates. You can also click on any date and enter your own personal events.
Communication The Communication link gives you access to Private Messages and course Forums, if your instructor has enabled these features. Forums The Forums are for discussions with all the members of your class. Your instructor can create forums, and you can create topics within a forum or contribute to a current topic. Private Messages Private Messages are for communication between you and your instructor. If your instructor has enabled private messages, click the New Message link to send your instructor a message. Enhanced WebAssign
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Grades
GRADES The Grades link at the top of all your WebAssign pages gives you access to the raw scores and grades that your instructor posts. This page may also include statistics on the whole class, and a histogram of scores for each category of assignment and each individual assignment. It may have your individual average for each category of assignment, as well as the score on each of your assignments. Your instructor will let you know what Scores and Grades will be posted in your course. If your instructor has enabled all of the options, your display will be similar to the one below.
Grades Overall Grade This score is calculated from the various categories of assignments, for example, Homework, Test, In Class, Quiz, Lab, and Exam. Your instructor may have different categories. Category Grades The Category Grades give the contribution to your overall grade from each of the categories. If you click a grade that is a link, you will get a pop-up window explaining how the number was calculated.
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Technical Tips Class Statistics Class Statistics shows the averages, minimum scores, maximum scores, and standard deviation of the class at large. My Scores Summary This link presents a pop-up window with a summary of your raw scores and the class statistics on each assignment, if your teacher has posted these.
My Scores summary
TECHNICAL TIPS Enhanced WebAssign relies on web browsers and other related technology that can lead to occasional technical issues. The following technical tips can help you avoid some common problems. Cookies Allow your browser to accept cookies. WebAssign will work if you set your browser to not accept cookies; however, if an encrypted cookie is not saved to your
For technical support go to http://webassign.net/student.html or email [email protected].
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Technical Tips computer during your session, you may be asked to login again more frequently. Once you logout and exit your browser, the cookie is deleted. Login and Credit If you see an assignment that does not have your name at the top, you have not logged in properly. You will not receive credit for any work you do on an assignment if your name is not associated with it. If you find yourself in the midst of this situation, make notes of your solution(s) and start over. Be aware that any randomized values in the questions will probably change. Logout When You Finish Your Session If someone wants to use your computer for WebAssign, logout and exit the browser before relinquishing control. Otherwise, the work you have just completed may be written over by the next user. Server Although it is very rare, the WebAssign server may occasionally be unavailable. If the WebAssign server is unavailable, instructors will provide instructions for submitting your assignments—possibly including new due dates. The policy for handling server problems will vary from instructor to instructor. Use the Latest Browser Software Use the latest version of Firefox, Mozilla, Netscape, or Internet Explorer browsers. Older versions of browsers may not be supported by WebAssign.
For technical support go to http://webassign.net/student.html or email [email protected].
32
The Start Smart Guide for Students
REFERENCE PAGES
Cut here and keep for reference
TA B L E O F I N T E R VA L S Notation
Set description
a, b
x
a x b
a, b
x
a x b
a, b
x
a x b
a, b
x
Picture a
b
Notation
Set description
a,
x a
x x a
x x b
x x b
x
a, a
a x b
b
a
b
a
b
Picture
, b , b ,
a a b b
⺢ (set of all real numbers)
SPECIAL FUNCTIONS POWER FUNCTIONS f x x a
y
(i) f x x n , n a positive integer
y
y=x$ (1, 1)
y=x^
y=x# y=≈
(_1, 1)
y=x%
(1, 1) x
0
(_1, _1)
x
0
n even n odd (iii) f x x 1
n (ii) f x x 1n s x , n a positive integer
y
y
y
y=Δ
(1, 1) 0
1 x
(1, 1) x
0
1
x
0
x ƒ=œ„
#x ƒ=œ„
INVERSE TRIGONOMETRIC FUNCTIONS arcsin x sin1x y &? sin y x and
y π 2
y 2 2
y=tan–!x=arctan x
lim tan1 x
0
arccos x cos x y &? cos y x and 0 y 1
arctan x tan1x y &? tan y x and
x
1
x
y 2 2
_ π2
3
x l
lim tan1 x
xl
2
2
REFERENCE PAGES SPECIAL FUNCTIONS EXPONENTIAL AND LOGARITHMIC FUNCTIONS log a x y
&?
ln x log e x,
y
y=´
a x y
y=x
ln e 1
where
ln x y &? e y x
1
y=ln x
0
Cancellation Equations
Laws of Logarithms
logaa x
a
1. log axy log a x log a y
lne x x
e ln x x
x
log a x
x
2. loga
x y
loga x loga y
lim e x 0
lim e x
x l
3. logax r r loga x
xl
lim ln x
x l 0
® ” ’ 2 1
® ” ’ 4 1
y
10® 4®
e®
x
1
lim ln x
xl
y
2®
y=log™ x
1.5®
y=ln x y=log∞ x y=log¡¸ x
1
1®
0
x
1
x
0
Exponential functions
Logarithmic functions
y
HYPERBOLIC FUNCTIONS y=cosh x
sinh x
e x ex 2
csch x
1 sinh x
y=tanh x
cosh x
e x ex 2
sech x
1 cosh x
x
tanh x
sinh x cosh x
coth x
cosh x sinh x
y=sinh x
INVERSE HYPERBOLIC FUNCTIONS y sinh1x
y cosh1x &? cosh y x y tanh1x
sinh1x ln( x sx 2 1 )
&? sinh y x
&?
tanh y x
and
y0
cosh1x ln( x sx 2 1 )
1 tanh1x 2 ln
4
1x 1x
REFERENCE PAGES
Cut here and keep for reference
D I F F E R E N T I AT I O N R U L E S GENERAL FORMULAS 1.
d c 0 dx
2.
d cf x c f x dx
3.
d f x tx f x tx dx
4.
d f x tx f x tx dx
5.
d f xtx f xtx txf x (Product Rule) dx
6.
d dx
7.
d f tx f txtx (Chain Rule) dx
8.
d x n nx n1 (Power Rule) dx
f x tx
txf x f xtx tx 2
(Quotient Rule)
EXPONENTIAL AND LOGARITHMIC FUNCTIONS 9. 11.
d e x e x dx
10.
d a x a x ln a dx
d 1 ln x dx x
12.
1 d log a x dx x ln a
TRIGONOMETRIC FUNCTIONS 13.
d sin x cos x dx
14.
d cos x sin x dx
15.
d tan x sec 2x dx
16.
d csc x csc x cot x dx
17.
d sec x sec x tan x dx
18.
d cot x csc 2x dx
INVERSE TRIGONOMETRIC FUNCTIONS 19.
d 1 sin1x dx s1 x 2
20.
d 1 cos1x dx s1 x 2
21.
d 1 tan1x dx 1 x2
22.
d 1 csc1x dx x sx 2 1
23.
d 1 sec1x dx x sx 2 1
24.
1 d cot1x dx 1 x2
HYPERBOLIC FUNCTIONS 25.
d sinh x cosh x dx
26.
d cosh x sinh x dx
27.
d tanh x sech 2x dx
28.
d csch x csch x coth x dx
29.
d sech x sech x tanh x dx
30.
d coth x csch 2x dx
INVERSE HYPERBOLIC FUNCTIONS 31.
d 1 sinh1x dx s1 x 2
32.
d 1 cosh1x dx sx 2 1
33.
d 1 tanh1x dx 1 x2
34.
d 1 csch1x dx x sx 2 1
35.
d 1 sech1x dx x s1 x 2
36.
d 1 coth1x dx 1 x2
5
REFERENCE PAGES TA B L E O F I N T E G R A L S BASIC FORMS 1.
y u dv uv y v du
2.
yu
3.
y
4.
ye
5.
y
6.
n
du
u n1 C, n 1 n1
du ln u C u
u
du e u C
a u du
au C ln a
y sin u du cos u C
11.
y csc u cot u du csc u C
12.
y tan u du ln sec u C
13.
y cot u du ln sin u C
14.
y sec u du ln sec u tan u C
15.
y csc u du ln csc u cot u C
16.
y sa
17.
ya
y u su
7.
y cos u du sin u C
8.
y sec u du tan u C
18.
9.
y csc u du cot u C
19.
ya
y sec u tan u du sec u C
20.
yu
10.
2
2
2
2
2
FORMS INVOLVING sa 2 u 2 , a 0 u a2 ln(u sa 2 u 2 ) C sa 2 u 2 2 2
21.
y sa
22.
yu
23.
y
a sa 2 u 2 sa 2 u 2 du sa 2 u 2 a ln u u
24.
y
sa 2 u 2 sa 2 u 2 du ln(u sa 2 u 2 ) C 2 u u
25.
y sa
26.
y sa
27.
y u sa
28.
y u sa
29.
y a
2
u 2 du
2
sa 2 u 2 du
u 2 a4 a 2u 2 sa 2 u 2 ln(u sa 2 u 2 ) C 8 8
du u2
2
u 2 du u2
2
u
2
du
2
2
2
C
ln(u sa 2 u 2 ) C
du 2
u2
u a2 ln(u sa 2 u 2 ) C sa 2 u 2 2 2
1 sa 2 u 2 a ln a u
C
sa 2 u 2 C a 2u
du u 2 C u 2 32 a sa 2 u 2 6
du 2
u
2
sin1
u C, a 0 a
du 1 u tan1 C u2 a a du 2
a2
1 u sec1 C a a
du 1 ua ln u2 2a ua du 1 ua ln a2 2a ua
C C
REFERENCE PAGES
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TA B L E O F I N T E G R A L S FORMS INVOLVING sa 2 u 2, a 0 u a2 u sin1 C sa 2 u 2 2 2 a
30.
y sa
31.
y u sa
32.
y
a sa 2 u 2 sa 2 u 2 du sa 2 u 2 a ln u u
33.
y
1 u sa 2 u 2 du sa 2 u 2 sin1 C 2 u u a
34.
y sa
35.
y u sa
36.
y u sa
37.
y a
38.
y a
u 2 du
2
2
u 2 du u2
2
u a4 u 2u 2 a 2 sa 2 u 2 sin1 C 8 8 a
u 2 du
2
du u
2
2
2
u2
1 a sa 2 u 2 ln a u
C
1 sa 2 u 2 C a 2u
u 2 32 du
2
C
u a2 u sin1 C sa 2 u 2 2 2 a
du
2
u 3a 4 u 2u 2 5a 2 sa 2 u 2 sin1 C 8 8 a
du u 2 C u 2 32 a sa 2 u 2
2
FORMS INVOLVING su 2 a 2, a 0 u a2 ln u su 2 a 2 C su 2 a 2 2 2
39.
y su
40.
y u su
41.
y
a su 2 a 2 du su 2 a 2 a cos1 C u u
42.
y
su 2 a 2 su 2 a 2 du ln u su 2 a 2 C u2 u
43.
y su
44.
y su
45.
y u su
46.
y u
a 2 du
2
2
2
a 2 du
u a4 2u 2 a 2 su 2 a 2 ln u su 2 a 2 C 8 8
du 2
a2
u 2 du 2
a
2
2
2
ln u su 2 a 2 C
du
2
a2
u a2 ln u su 2 a 2 C su 2 a 2 2 2
su 2 a 2 C a 2u
du u C a 2 32 a 2 su 2 a 2
7
REFERENCE PAGES TA B L E O F I N T E G R A L S FORMS INVOLVING a bu 47.
y a bu b
48.
u du 1 y a bu 2b [a bu
49.
y ua bu a ln
50.
y u a bu au a
51.
y a bu
52.
y ua bu
53.
y a bu
54.
y u sa bu du 15b
55.
y sa bu 3b
1
u du
(a bu a ln a bu ) C
2
2
2
3
du
1
du
u a bu
1
b
2
u du
du
2
u 2 du
2
2
C
ln
2
y sa bu 15b
57.
y u sa bu sa ln
2
3
C
3bu 2aa bu32 C
2
2 sa
sa bu sa C, if a 0 sa bu sa
tan1
a bu C, a
58.
y
sa bu du 2 sa bu a u
59.
y
b sa bu sa bu du u2 u 2
60.
y u sa bu du b2n 3
61.
y sa bu
62.
y u sa bu an 1u
if a 0
y u sa bu du
2
du
8a 2 3b 2u 2 4abu sa bu C
1
C
bu 2a sa bu C
56.
du
a2 2a ln a bu a bu
a bu
2
u 2 du
1 b3
u du
n
C
1 1 a bu 2 ln aa bu a u
2
u n du
a bu u
n
] C
a 1 2 ln a bu C b 2a bu b
2
4aa bu 2a 2 ln a bu
y u sa bu du
u na bu32 na
2u nsa bu 2na b2n 1 b2n 1 sa bu n1
yu
n1
sa bu du
u n1 du
y sa bu
b2n 3 2an 1
yu
du sa bu
n1
8
REFERENCE PAGES
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TA B L E O F I N T E G R A L S TRIGONOMETRIC FORMS 63.
y sin u du
1 2
64.
y cos u du
1 2
65.
y tan u du tan u u C
2
2
u 14 sin 2u C u 14 sin 2u C
1
76.
y cot u du n 1 cot
77.
y sec u du n 1 tan u sec
78.
y csc u du n 1 cot u csc y sin au sin bu du
n
n
y cot
y sec
n2
u
n2 n1
y csc
n2
1
n2
y cot u du cot u u C
67.
y sin u du 2 sin u cos u C
79.
68.
y
80.
y cos au cos bu du
69.
y tan u du
81.
y sin au cos bu du
70.
y cot u du
82.
71.
y sec u du
y u sin u du sin u u cos u C
83.
72.
y csc u du
y u cos u du cos u u sin u C
84.
yu
n
sin u du u n cos u n
73.
y sin u du n sin
u du
85.
yu
n
cos u du u n sin u n
74.
y cos u du n cos
86.
y sin u cos u du
75.
y
1 3
3
2
cos 3u du 13 2 cos 2u sin u C 3
3
3
1 2
cot 2u ln sin u C
1 2
sec u tan u 12 ln sec u tan u C
3
1 2
csc u cot u 12 ln csc u cot u C
1
n
1
n
tan 2u ln cos u C
1 2
u cos u
n1
u sin u
n1
1 tan u du tan n1u n1 n
n1 n
n1 n
y tan
y sin
y cos
n2
n2
u du
n
m
n2
u du
u du
n2 n1
n2
66.
2
n2
u
1
n
2
u
n1
u du
u du
sina bu sina bu C 2a b 2a b sina bu sina bu C 2a b 2a b cosa bu cosa bu C 2a b 2a b
yu
yu
n1
n1
cos u du
sin u du
sin n1u cos m1u n1 nm nm
sin n1u cos m1u m1 nm nm
y sin
n2
u cosmu du
y sin u cos n
m2
u du
INVERSE TRIGONOMETRIC FORMS 87.
y sin
88.
y cos
u du u cos1u s1 u 2 C
y tan
u du u tan1u ln1 u 2 C
89.
90.
91.
1
u du u sin1u s1 u 2 C
92.
y u tan
93.
y
94.
yu
95.
y
1
1
1
u du
u2 1 u tan1u C 2 2
u n sin1u du
1 n1
1 2
2u 1 u s1 u sin1u C 4 4 2
y
u sin1u du
y
2u 2 1 u s1 u 2 u cos1u du cos1u C 4 4
2
9
cos1u du
1 n1
u n tan1u du
1 n1
n
u n1 sin1u
y s1 u
u n1 cos1u
u n1 tan1u
u n1 du
, n 1
2
u n1 du
y s1 u y
2
, n 1
u n1 du , n 1 1 u2
REFERENCE PAGES TA B L E O F I N T E G R A L S EXPONENTIAL AND LOGARITHMIC FORMS 96.
y ue
97.
y
98.
ye
au
99.
ye
au
au
du
1 au 1e au C a2 1 n au n u e a a
u ne au du
y
u n1e au du
sin bu du
e au a sin bu b cos bu C a b2
cos bu du
e au a cos bu b sin bu C a b2
100.
y ln u du u ln u u C
101.
yu
102.
y u ln u du ln ln u C
2
n
ln u du
u n1 n 1 ln u 1 C n 12
1
2
HYPERBOLIC FORMS
y sinh u du cosh u C 104. y cosh u du sinh u C 105. y tanh u du ln cosh u C 106. y coth u du ln sinh u C 107. y sech u du tan sinh u C
y csch u du ln tanh u C 109. y sech u du tanh u C 110. y csch u du coth u C 111. y sech u tanh u du sech u C 112. y csch u coth u du csch u C
103.
2
2
1
FORMS INVOLVING s2au u 2 , a 0
ua a2 au cos1 s2au u 2 2 2 a
113.
y s2au u
114.
y u s2au u
115.
y
au s2au u 2 du s2au u 2 a cos1 u a
116.
y
2 s2au u 2 au s2au u 2 du cos1 2 u u a
117.
y s2au u
118.
y s2au u
119.
y s2au u
120.
y u s2au u
2
du 2
du
2
u du
2
u 2 du
2
du
au a
du
2
C C
C
s2au u 2 a cos1
C
2u 2 au 3a 2 a3 au cos1 s2au u 2 6 2 a
cos1
au a
C
u 3a 3a 2 au cos1 s2au u 2 2 2 a
1 2
108.
s2au u 2 C au
10
C
C
WEB PROJECTS These projects can be completed anytime after you have studied the corresponding section in the textbook. To select a project, go to www.stewartcalculus.com and click on PROJECTS. 2
DERIVATIVES
Early Methods for Finding Tangents 2.3 APPLIED PROJECT Building a Better Roller Coaster 2.5 APPLIED PROJECT Where Should a Pilot Start Descent? 2.8 LABORATORY PROJECT Taylor Polynomials
2.1 WRITING PROJECT
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3
INVERSE FUNCTIONS
3.7 WRITING PROJECT
■
The Origins of L’Hospital’s Rule
4
APPLICATIONS OF DIFFERENTIATION
4.1 APPLIED PROJECT
■
4.5 APPLIED PROJECT
■
The Calculus of Rainbows The Shape of a Can
5
INTEGRALS
5.3 DISCOVERY PROJECT 5.4 WRITING PROJECT APPLIED PROJECT
■
6
TECHNIQUES OF INTEGRATION
6.4 DISCOVERY PROJECT
7
APPLICATIONS OF INTEGRATION
7.2 DISCOVERY PROJECT
Area Functions Newton, Leibniz, and the Invention of Calculus Where To Sit at the Movies
■ ■
Patterns in Integrals
■
Rotating on a Slant 7.4 DISCOVERY PROJECT Arc Length Contest 7.5 APPLIED PROJECT Calculus and Baseball 7.6 APPLIED PROJECT How Fast Does a Tank Drain? APPLIED PROJECT Which Is Faster, Going Up or Coming Down? ■ ■
■ ■ ■
8
SERIES
Logistic Sequences 8.7 LABORATORY PROJECT An Elusive Limit WRITING PROJECT How Newton Discovered the Binomial Series 8.8 APPLIED PROJECT Radiation from the Stars
8.1 LABORATORY PROJECT
■ ■
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■
9
PARAMETRIC EQUATIONS
9.1 LABORATORY PROJECT
■
AND POLAR COORDINATES
9.2 LABORATORY PROJECT
■
10 VECTORS AND
THE GEOMETRY OF SPACE
Running Circles Around Circles Bézier Curves
The Geometry of a Tetrahedron 10.5 LABORATORY PROJECT Putting 3D in Perspective 10.9 APPLIED PROJECT Kepler’s Laws 10.4 DISCOVERY PROJECT
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11 PARTIAL DERIVATIVES
Designing a Dumpster DISCOVERY PROJECT Quadratic Approximations and Critical Points 11.8 APPLIED PROJECT Rocket Science APPLIED PROJECT Hydro-Turbine Optimization
11.7 APPLIED PROJECT
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12 MULTIPLE INTEGRALS
Volumes of Hyperspheres The Intersection of Three Cylinders 12.7 LABORATORY PROJECT Families of Surfaces APPLIED PROJECT Roller Derby
12.5 DISCOVERY PROJECT
■
12.6 DISCOVERY PROJECT
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13 VECTOR CALCULUS
13.8 WRITING PROJECT
■
Three Men and Two Theorems
