INSTRUCTOR’S SOLUTIONS MANUAL SINGLE VARIABLE Collin County Community College
WILLIAM ARDIS
THOMAS’ CALCULUS TWELFTH E...
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INSTRUCTOR’S SOLUTIONS MANUAL SINGLE VARIABLE Collin County Community College
WILLIAM ARDIS
THOMAS’ CALCULUS TWELFTH EDITION BASED ON THE ORIGINAL WORK BY
George B. Thomas, Jr. Massachusetts Institute of Technology
AS
REVISED BY
Maurice D. Weir Naval Postgraduate School
Joel Hass
University of California, Davis
This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson Addison-Wesley from electronic files supplied by the author. Copyright © 2010, 2005, 2001 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-60807-9 ISBN-10: 0-321-60807-0 1 2 3 4 5 6 BB 12 11 10 09
PREFACE TO THE INSTRUCTOR This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because the CAS command templates would give them all away). In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or rewritten every solution which appeared in previous solutions manuals to ensure that each solution ì conforms exactly to the methods, procedures and steps presented in the text ì is mathematically correct ì includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra ì includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation ì is formatted in an appropriate style to aid in its understanding Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within the text grouping require a change only in the input function or other numerical input parameters associated with the problem (such as the interval endpoints or the number of iterations). For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com.
TABLE OF CONTENTS 1 Functions 1 1.1 1.2 1.3 1.4
Functions and Their Graphs 1 Combining Functions; Shifting and Scaling Graphs 8 Trigonometric Functions 19 Graphing with Calculators and Computers 26 Practice Exercises 30 Additional and Advanced Exercises 38
2 Limits and Continuity 43 2.1 2.2 2.3 2.4 2.5 2.6
Rates of Change and Tangents to Curves 43 Limit of a Function and Limit Laws 46 The Precise Definition of a Limit 55 One-Sided Limits 63 Continuity 67 Limits Involving Infinity; Asymptotes of Graphs 73 Practice Exercises 82 Additional and Advanced Exercises 86
3 Differentiation 93 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9
Tangents and the Derivative at a Point 93 The Derivative as a Function 99 Differentiation Rules 109 The Derivative as a Rate of Change 114 Derivatives of Trigonometric Functions 120 The Chain Rule 127 Implicit Differentiation 135 Related Rates 142 Linearizations and Differentials 146 Practice Exercises 151 Additional and Advanced Exercises 162
4 Applications of Derivatives 167 4.1 4.2 4.3 4.4 4.5 4.6 4.7
Extreme Values of Functions 167 The Mean Value Theorem 179 Monotonic Functions and the First Derivative Test 188 Concavity and Curve Sketching 196 Applied Optimization 216 Newton's Method 229 Antiderivatives 233 Practice Exercises 239 Additional and Advanced Exercises 251
5 Integration 257 5.1 5.2 5.3 5.4 5.5 5.6
Area and Estimating with Finite Sums 257 Sigma Notation and Limits of Finite Sums 262 The Definite Integral 268 The Fundamental Theorem of Calculus 282 Indefinite Integrals and the Substitution Rule 290 Substitution and Area Between Curves 296 Practice Exercises 310 Additional and Advanced Exercises 320
6 Applications of Definite Integrals 327 6.1 6.2 6.3 6.4 6.5 6.6
Volumes Using Cross-Sections 327 Volumes Using Cylindrical Shells 337 Arc Lengths 347 Areas of Surfaces of Revolution 353 Work and Fluid Forces 358 Moments and Centers of Mass 365 Practice Exercises 376 Additional and Advanced Exercises 384
7 Transcendental Functions 389 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8
Inverse Functions and Their Derivatives 389 Natural Logarithms 396 Exponential Functions 403 Exponential Change and Separable Differential Equations 414 ^ Indeterminate Forms and L'Hopital's Rule 418 Inverse Trigonometric Functions 425 Hyperbolic Functions 436 Relative Rates of Growth 443 Practice Exercises 447 Additional and Advanced Exercises 458
8 Techniques of Integration 461 8.1 8.2 8.3 8.4 8.5 8.6 8.7
Integration by Parts 461 Trigonometric Integrals 471 Trigonometric Substitutions 478 Integration of Rational Functions by Partial Fractions 484 Integral Tables and Computer Algebra Systems 491 Numerical Integration 502 Improper Integrals 510 Practice Exercises 518 Additional and Advanced Exercises 528
9 First-Order Differential Equations 537 9.1 9.2 9.3 9.4 9.5
Solutions, Slope Fields and Euler's Method 537 First-Order Linear Equations 543 Applications 546 Graphical Solutions of Autonomous Equations 549 Systems of Equations and Phase Planes 557 Practice Exercises 562 Additional and Advanced Exercises 567
10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642
11 Parametric Equations and Polar Coordinates 647 11.1 11.2 11.3 11.4 11.5 11.6 11.7
Parametrizations of Plane Curves 647 Calculus with Parametric Curves 654 Polar Coordinates 662 Graphing in Polar Coordinates 667 Areas and Lengths in Polar Coordinates 674 Conic Sections 679 Conics in Polar Coordinates 689 Practice Exercises 699 Additional and Advanced Exercises 709
CHAPTER 1 FUNCTIONS 1.1 FUNCTIONS AND THEIR GRAPHS 1. domain œ (_ß _); range œ [1ß _)
2. domain œ [0ß _); range œ (_ß 1]
3. domain œ Ò2ß _); y in range and y œ È5x 10 ! Ê y can be any positive real number Ê range œ Ò!ß _). 4. domain œ (_ß 0Ó Ò3, _); y in range and y œ Èx2 3x ! Ê y can be any positive real number Ê range œ Ò!ß _). 5. domain œ (_ß 3Ñ Ð3, _); y in range and y œ Ê3 t!Ê
4 3t
4 3t,
now if t 3 Ê 3 t ! Ê
4 3t
!, or if t 3
! Ê y can be any nonzero real number Ê range œ Ð_ß 0Ñ Ð!ß _).
6. domain œ (_ß %Ñ Ð4, 4Ñ Ð4, _); y in range and y œ 2
% t 4 Ê 16 Ÿ t 16 ! Ê nonzero real number Ê range œ Ð_ß
# "' 18 Ó
Ÿ
2 t2 16
2 t2 16 ,
2 t2 16
now if t % Ê t2 16 ! Ê 2
!, or if t % Ê t 16 ! Ê
2 t2 16
!, or if
! Ê y can be any
Ð!ß _).
7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. #
9. base œ x; (height)# ˆ #x ‰ œ x# Ê height œ
È3 #
x; area is a(x) œ
" #
(base)(height) œ
" #
(x) Š
È3 # x‹
œ
È3 4
x# ;
perimeter is p(x) œ x x x œ 3x. 10. s œ side length Ê s# s# œ d# Ê s œ
d È2
; and area is a œ s# Ê a œ
" #
d#
11. Let D œ diagonal length of a face of the cube and j œ the length of an edge. Then j# D# œ d# and D# œ 2j# Ê 3j# œ d# Ê j œ
d È3
. The surface area is 6j# œ
6d# 3
12. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ ˆx, Èx‰ œ ˆ m"# ,
#
œ 2d# and the volume is j$ œ Š d3 ‹ Èx x
œ
" Èx
$Î#
œ
(x 0). Thus,
"‰ m .
13. 2x 4y œ 5 Ê y œ "# x 54 ; L œ ÈÐx 0Ñ2 Ðy 0Ñ2 œ Éx2 Ð "# x 54 Ñ2 œ Éx2 4" x2 54 x œ É 54 x2 54 x
25 16
œ É 20x
2
20x 25 16
œ
È20x2 20x 25 4
14. y œ Èx 3 Ê y2 3 œ x; L œ ÈÐx 4Ñ2 Ðy 0Ñ2 œ ÈÐy2 3 4Ñ2 y2 œ ÈÐy2 1Ñ2 y2 œ Èy4 2y2 1 y2 œ Èy4 y2 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
d$ 3È 3
25 16
.
2
Chapter 1 Functions
15. The domain is a_ß _b.
16. The domain is a_ß _b.
17. The domain is a_ß _b.
18. The domain is Ð_ß !Ó.
19. The domain is a_ß !b a!ß _b.
20. The domain is a_ß !b a!ß _b.
21. The domain is a_ß 5b Ð5ß 3Ó Ò3, 5Ñ a5, _b 22. The range is Ò2, 3Ñ. 23. Neither graph passes the vertical line test (a)
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.1 Functions and Their Graphs 24. Neither graph passes the vertical line test (a)
(b)
Ú xyœ" Þ Ú yœ1x Þ or or kx yk œ 1 Í Û Í Û ß ß Ü x y œ " à Ü y œ " x à 25.
x y
0 0
1 1
27. Faxb œ œ
2 0
26.
x y
0 1
1 0
2 0
" , x0 28. Gaxb œ œ x x, 0 Ÿ x
4 x2 , x Ÿ 1 x2 2x, x 1
29. (a) Line through a!ß !b and a"ß "b: y œ x; Line through a"ß "b and a#ß !b: y œ x 2 x, 0 Ÿ x Ÿ 1 f(x) œ œ x 2, 1 x Ÿ 2 Ú Ý 2, ! Ÿ x " Ý !ß " Ÿ x # (b) f(x) œ Û Ý Ý 2ß # Ÿ x $ Ü !ß $ Ÿ x Ÿ % 30. (a) Line through a!ß 2b and a#ß !b: y œ x 2 " Line through a2ß "b and a&ß !b: m œ !& # œ x #, 0 x Ÿ # f(x) œ œ " $ x &$ , # x Ÿ &
f(x) œ œ
œ "$ , so y œ "$ ax 2b " œ "$ x
$ ! ! Ð"Ñ œ " $ % #! œ #
(b) Line through a"ß !b and a!ß $b: m œ Line through a!ß $b and a#ß "b: m œ
" $
& $
$, so y œ $x $ œ #, so y œ #x $
$x $, " x Ÿ ! #x $, ! x Ÿ #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
Chapter 1 Functions
31. (a) Line through a"ß "b and a!ß !b: y œ x Line through a!ß "b and a"ß "b: y œ " Line through a"ß "b and a$ß !b: m œ !" $" œ Ú x " Ÿ x ! " !xŸ" f(x) œ Û Ü "# x $# "x$
" #
(b) Line through a2ß 1b and a0ß 0b: y œ 12 x Line through a0ß 2b and a1ß 0b: y œ 2x 2 Line through a1ß 1b and a3ß 1b: y œ 1 32. (a) Line through ˆ T# ß !‰ and aTß "b: m œ faxb œ
(b)
"! TaTÎ#b
œ "# , so y œ "# ax "b " œ "# x
Ú
1 2x
faxb œ Û 2x 2 Ü 1
$ #
2 Ÿ x Ÿ 0 0xŸ1 1xŸ3
œ T# , so y œ T# ˆx T# ‰ 0 œ T# x "
!, 0 Ÿ x Ÿ T# # T T x ", # x Ÿ T
Ú A, Ý Ý Ý Aß faxb œ Û Aß Ý Ý Ý Ü Aß
! Ÿ x T# T # Ÿx T T Ÿ x $#T $T # Ÿ x Ÿ #T
33. (a) ÚxÛ œ 0 for x − [0ß 1)
(b) ÜxÝ œ 0 for x − (1ß 0]
34. ÚxÛ œ ÜxÝ only when x is an integer. 35. For any real number x, n Ÿ x Ÿ n ", where n is an integer. Now: n Ÿ x Ÿ n " Ê Ðn "Ñ Ÿ x Ÿ n. By definition: ÜxÝ œ n and ÚxÛ œ n Ê ÚxÛ œ n. So ÜxÝ œ ÚxÛ for all x − d . 36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.1 Functions and Their Graphs 37. Symmetric about the origin Dec: _ x _ Inc: nowhere
38. Symmetric about the y-axis Dec: _ x ! Inc: ! x _
39. Symmetric about the origin Dec: nowhere Inc: _ x ! !x_
40. Symmetric about the y-axis Dec: ! x _ Inc: _ x !
41. Symmetric about the y-axis Dec: _ x Ÿ ! Inc: ! x _
42. No symmetry Dec: _ x Ÿ ! Inc: nowhere
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5
6
Chapter 1 Functions
43. Symmetric about the origin Dec: nowhere Inc: _ x _
44. No symmetry Dec: ! Ÿ x _ Inc: nowhere
45. No symmetry Dec: ! Ÿ x _ Inc: nowhere
46. Symmetric about the y-axis Dec: _ x Ÿ ! Inc: ! x _
47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. 48. faxb œ x& œ
" x&
and faxb œ axb& œ
" a x b&
œ ˆ x"& ‰ œ faxb. Thus the function is odd.
49. Since faxb œ x# " œ axb# " œ faxb. The function is even. 50. Since Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ and Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ the function is neither even nor odd. 51. Since gaxb œ x$ x, gaxb œ x$ x œ ax$ xb œ gaxb. So the function is odd. 52. gaxb œ x% $x# " œ axb% $axb# " œ gaxbß thus the function is even. 53. gaxb œ
" x# "
54. gaxb œ
x x# " ;
55. hatb œ
" t ";
œ
" axb# "
œ gaxb. Thus the function is even.
gaxb œ x#x" œ gaxb. So the function is odd.
hatb œ
" t " ;
h at b œ
" " t.
Since hatb Á hatb and hatb Á hatb, the function is neither even nor odd.
56. Since l t$ | œ l atb$ |, hatb œ hatb and the function is even.
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Section 1.1 Functions and Their Graphs 57. hatb œ 2t ", hatb œ 2t ". So hatb Á hatb. hatb œ 2t ", so hatb Á hatb. The function is neither even nor odd. 58. hatb œ 2l t l " and hatb œ 2l t l " œ 2l t l ". So hatb œ hatb and the function is even. 59. s œ kt Ê 25 œ kÐ75Ñ Ê k œ
" 3
Ê s œ 3" t; 60 œ 3" t Ê t œ 180
60. K œ c v# Ê 12960 œ ca18b2 Ê c œ 40 Ê K œ 40v# ; K œ 40a10b# œ 4000 joules 61. r œ 62. P œ
k s
Ê6œ
k v
k 4
Ê k œ 24 Ê r œ
Ê 14.7 œ
k 1000
24 s ;
10 œ
24 s
Ê k œ 14700 Ê P œ
Êsœ
14700 v ;
12 5
23.4 œ
14700 v
Êvœ
24500 39
¸ 628.2 in3
63. v œ f(x) œ xÐ"% 2xÑÐ22 2xÑ œ %x$ 72x# $!)x; ! x 7Þ 64. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB # AB # œ 2# Ê AB œ È2Þ So, #
h# "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ " Ê The equation of AB is y œ f(x) œ x "; x − Ò!ß "Ó. (b) AÐxÑ œ 2x y œ 2xÐx "Ñ œ 2x# #x; x − Ò!ß "Ó. 65. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 66. (a) Graph f because it is linear. (b) Graph g because it contains a!ß "b. (c) Graph h because it is a nonlinear odd function. x #
67. (a) From the graph, (b)
x #
1
x 0:
x #
x 0:
x 2
4 x
1
Ê 4 x
x #
1
4 x
Ê x − (2ß 0) (%ß _)
1 4x 0 # 2x8 0 Ê x 2x
0 Ê
(x4)(x2) #x
0
(x4)(x2) #x
0
Ê x 4 since x is positive; 1
4 x
0 Ê
x# 2x8 2x
0 Ê
Ê x 2 since x is negative; sign of (x 4)(x 2) ïïïïïðïïïïïðïïïïî 2 % Solution interval: (#ß 0) (%ß _)
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7
8
Chapter 1 Functions 3 2 x 1 x 1 3 2 x 1 x 1
68. (a) From the graph, (b) Case x 1:
Ê x − (_ß 5) (1ß 1) Ê
3(x1) x 1
2
Ê 3x 3 2x 2 Ê x 5. Thus, x − (_ß 5) solves the inequality. Case 1 x 1:
3 x 1
2 x 1
Ê
3(x1) x 1
2
Ê 3x 3 2x 2 Ê x 5 which is true if x 1. Thus, x − (1ß 1) solves the inequality. 3 Case 1 x: x1 x2 1 Ê 3x 3 2x 2 Ê x 5 which is never true if 1 x, so no solution here. In conclusion, x − (_ß 5) (1ß 1). 69. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax, yb lie on the same vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !, for any x. 70. price œ 40 5x, quantity œ 300 25x Ê Raxb œ a40 5xba300 25xb 71. x2 x2 œ h2 Ê x œ
h È2
œ
È2 h 2 ;
cost œ 5a2xb 10h Ê Cahb œ 10Š
È2 h 2 ‹
10h œ 5hŠÈ2 2‹
72. (a) Note that 2 mi = 10,560 ft, so there are È800# x# feet of river cable at $180 per foot and a10,560 xb feet of land cable at $100 per foot. The cost is Caxb œ 180È800# x# 100a10,560 xb. (b) Ca!b œ $"ß #!!ß !!! Ca&!!b ¸ $"ß "(&ß )"# Ca"!!!b ¸ $"ß ")'ß &"# Ca"&!!b ¸ $"ß #"#ß !!! Ca#!!!b ¸ $"ß #%$ß ($# Ca#&!!b ¸ $"ß #()ß %(* Ca$!!!b ¸ $"ß $"%ß )(! Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. Df : _ x _, Dg : x 1 Ê Df
g
œ Dfg : x 1. Rf : _ y _, Rg : y 0, Rf g : y 1, Rfg : y 0
2. Df : x 1 0 Ê x 1, Dg : x 1 0 Ê x 1. Therefore Df Rf œ Rg : y 0, Rf g : y È2, Rfg : y 0
g
œ Dfg : x 1.
3. Df : _ x _, Dg : _ x _, DfÎg : _ x _, DgÎf : _ x _, Rf : y œ 2, Rg : y 1, RfÎg : 0 y Ÿ 2, RgÎf : "# Ÿ y _ 4. Df : _ x _, Dg : x 0 , DfÎg : x 0, DgÎf : x 0; Rf : y œ 1, Rg : y 1, RfÎg : 0 y Ÿ 1, RgÎf : 1 Ÿ y _ 5. (a) 2 (d) (x 5)# 3 œ x# 10x 22 (g) x 10
(b) 22 (e) 5 (h) (x# 3)# 3 œ x% 6x# 6
(c) x# 2 (f) 2
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Section 1.2 Combining Functions; Shifting and Scaling Graphs 6. (a) "3 (d)
(b) 2
" x
(c)
(e) 0
(g) x 2
(h)
(f)
" " x 1 1
œ
x x
"
# 1
x" x#
œ
" x 1 3 4
1œ
x x1
7. af‰g‰hbaxb œ fagahaxbbb œ faga4 xbb œ fa3a4 xbb œ fa12 3xb œ a12 3xb 1 œ 13 3x 8. af‰g‰hbaxb œ fagahaxbbb œ fagax2 bb œ fa2ax2 b 1b œ fa2x2 1b œ 3a2x2 1b 4 œ 6x2 1 9. af‰g‰hbaxb œ fagahaxbbb œ fˆgˆ 1x ‰‰ œ fŠ 1 1 % ‹ œ fˆ 1 x 4x ‰ œ É 1 x 4x " œ É 15x4x" x
2
10. af‰g‰hbaxb œ fagahaxbbb œ fŠgŠÈ2 x‹‹ œ f
ŠÈ2 x‹ 2
ŠÈ2 x‹
œ fˆ $ x ‰ œ 1 2x
2 x $ x 2 3 $2 xx
8 3x 7 2x
œ
11. (a) af‰gbaxb (d) a j‰jbaxb
(b) a j‰gbaxb (e) ag‰h‰f baxb
(c) ag‰gbaxb (f) ah‰j‰f baxb
12. (a) af‰jbaxb (d) af‰f baxb
(b) ag‰hbaxb (e) a j‰g‰f baxb
(c) ah‰hbaxb (f) ag‰f‰hbaxb
g(x)
f(x)
(f ‰ g)(x)
(a)
x7
Èx
Èx 7
(b)
x2
3x
3(x 2) œ 3x 6
(c)
x#
Èx 5
Èx# 5
(d)
x x1
x x1
" x1 " x
1
13.
(e) (f)
" x
gaxb" g ax b
œ
x x (x1)
œx
x
" x
x
" lx "l .
14. (a) af‰gbaxb œ lgaxbl œ (b) af‰gbaxb œ
x x 1 x x 1 1
x x"
œ
Ê"
" g ax b
œ
x x"
Ê"
x x"
œ
" g ax b
Ê
" x"
œ
" gaxb ß so
gaxb œ x ".
(c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x . (d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.) #
The completed table is shown. Note that the absolute value sign in part (d) is optional. gaxb faxb af‰gbaxb " " lxl x" lx "l x" x# Èx
x" x
Èx #
x
15. (a) faga1bb œ fa1b œ 1 (d) gaga2bb œ ga0b œ 0
x x"
lxl lxl (b) gafa0bb œ ga2b œ 2 (e) gafa2bb œ ga1b œ 1
(c) fafa1bb œ fa0b œ 2 (f) faga1bb œ fa1b œ 0
16. (a) faga0bb œ fa1b œ 2 a1b œ 3, where ga0b œ 0 1 œ 1 (b) gafa3bb œ ga1b œ a1b œ 1, where fa3b œ 2 3 œ 1 (c) gaga1bb œ ga1b œ 1 1 œ 0, where ga1b œ a1b œ 1
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9
10
Chapter 1 Functions (d) fafa2bb œ fa0b œ 2 0 œ 2, where fa2b œ 2 2 œ 0 (e) gafa0bb œ ga2b œ 2 1 œ 1, where fa0b œ 2 0 œ 2 (f) fˆgˆ "# ‰‰ œ fˆ #" ‰ œ 2 ˆ #" ‰ œ 5# , where gˆ "# ‰ œ "# 1 œ "#
17. (a) af‰gbaxb œ fagaxbb œ É 1x 1 œ É 1 x x ag‰f baxb œ gafaxbb œ
1 Èx 1
(b) Domain af‰gb: Ð_, 1Ó Ð0, _Ñ, domain ag‰f b: Ð1, _Ñ (c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ 18. (a) af‰gbaxb œ fagaxbb œ 1 2Èx x ag‰f baxb œ gafaxbb œ 1 kxk (b) Domain af‰gb: Ò0, _Ñ, domain ag‰f b: Ð_, _Ñ (c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð_, 1Ó 19. af‰gbaxb œ x Ê fagaxbb œ x Ê
g ax b g ax b 2
œ x Ê gaxb œ agaxb 2bx œ x † gaxb 2x
Ê gaxb x † gaxb œ 2x Ê gaxb œ 1 2x x œ
2x x1
20. af‰gbaxb œ x 2 Ê fagaxbb œ x 2 Ê 2agaxbb3 4 œ x 2 Ê agaxbb3 œ 21. (a) y œ (x 7)#
(b) y œ (x 4)#
22. (a) y œ x# 3
(b) y œ x# 5
x6 2
3 x6 Ê gaxb œ É 2
23. (a) Position 4
(b) Position 1
(c) Position 2
(d) Position 3
24. (a) y œ (x 1)# 4
(b) y œ (x 2)# 3
(c) y œ (x 4)# 1
(d) y œ (x 2)#
25.
26.
27.
28.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.2 Combining Functions; Shifting and Scaling Graphs 29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
11
12
Chapter 1 Functions
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.2 Combining Functions; Shifting and Scaling Graphs 53.
54.
55. (a) domain: [0ß 2]; range: [#ß $]
(b) domain: [0ß 2]; range: [1ß 0]
(c) domain: [0ß 2]; range: [0ß 2]
(d) domain: [0ß 2]; range: [1ß 0]
(e) domain: [2ß 0]; range: [!ß 1]
(f) domain: [1ß 3]; range: [!ß "]
(g) domain: [2ß 0]; range: [!ß "]
(h) domain: [1ß 1]; range: [!ß "]
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13
14
Chapter 1 Functions
56. (a) domain: [0ß 4]; range: [3ß 0]
(b) domain: [4ß 0]; range: [!ß $]
(c) domain: [4ß 0]; range: [!ß $]
(d) domain: [4ß 0]; range: ["ß %]
(e) domain: [#ß 4]; range: [3ß 0]
(f) domain: [2ß 2]; range: [3ß 0]
(g) domain: ["ß 5]; range: [3ß 0]
(h) domain: [0ß 4]; range: [0ß 3]
58. y œ a2xb# 1 œ %x# 1
57. y œ 3x# 3 59. y œ "# ˆ"
"‰ x#
œ
" #
" #x#
60. y œ 1
" axÎ$b#
œ1
61. y œ È%x 1
62. y œ 3Èx 1
# 63. y œ É% ˆ x# ‰ œ "# È16 x#
64. y œ "$ È% x#
65. y œ " a3xb$ œ " 27x$
66. y œ " ˆ x# ‰ œ "
$
* x#
x$ )
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Section 1.2 Combining Functions; Shifting and Scaling Graphs 67. Let y œ È#x " œ faxb and let gaxb œ x"Î# , "Î# "Î# haxb œ ˆx " ‰ , iaxb œ È#ˆx " ‰ , and #
#
"Î# jaxb œ ’È#ˆx "# ‰ “ œ faBb. The graph of
haxb is the graph of gaxb shifted left
" #
unit; the
graph of iaxb is the graph of haxb stretched vertically by a factor of È#; and the graph of jaxb œ faxb is the graph of iaxb reflected across the x-axis. 68. Let y œ È"
x #
œ faxbÞ Let gaxb œ axb"Î# ,
haxb œ ax #b"Î# , and iaxb œ œ È"
x #
" È # a x
#b"Î#
œ faxbÞ The graph of gaxb is the
graph of y œ Èx reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of È#. 69. y œ faxb œ x$ . Shift faxb one unit right followed by a shift two units up to get gaxb œ ax "b3 #.
70. y œ a" Bb$ # œ Òax "b$ a#bÓ œ faxb. Let gaxb œ x$ , haxb œ ax "b$ , iaxb œ ax "b$ a#b, and jaxb œ Òax "b$ a#bÓ. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis.
71. Compress the graph of faxb œ of 2 to get gaxb œ unit to get haxb œ
" #x . Then " #x ".
" x
horizontally by a factor
shift gaxb vertically down 1
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15
16
Chapter 1 Functions
72. Let faxb œ œ
"
#
ŠxÎÈ#‹
" x#
and gaxb œ
"œ
# x#
"œ
" # ’Š"ÎÈ#‹B“
" # Š B# ‹
"
"Þ Since
È# ¸ "Þ%, we see that the graph of faxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb.
$ 73. Reflect the graph of y œ faxb œ È x across the x-axis $ to get gaxb œ Èx.
74. y œ faxb œ a#xb#Î$ œ Òa"ba#bxÓ#Î$ œ a"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph of faxb is the graph of gaxb œ x#Î$ compressed horizontally by a factor of 2.
75.
76.
77. *x# #&y# œ ##& Ê
x# &#
y# $#
œ"
78. "'x# (y# œ ""# Ê
x# # È Š (‹
y# %#
œ"
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Section 1.2 Combining Functions; Shifting and Scaling Graphs 79. $x# ay #b# œ $ Ê
x# "#
a y #b # #
ŠÈ$‹
80. ax "b# #y# œ % Ê
œ"
Ê
83.
x# "'
#
ŠÈ#‹
y# *
y a#b‘# #
ŠÈ$‹
# # 82. 'ˆx $# ‰ *ˆy "# ‰ œ &%
81. $ax "b# #ay #b# œ ' ax " b #
x a"b‘# ##
#
œ"
Ê
’xˆ $# ‰“ $#
ˆy "# ‰# #
ŠÈ'‹
œ"
œ " has its center at a!ß !b. Shiftinig 4 units
left and 3 units up gives the center at ah, kb œ a%ß $b. # x a4b‘# ay 3#3b œ " 4# a y $b # œ ". Center, C, is a%ß 3#
So the equation is Ê
ax % b # 4#
$b, and
major axis, AB, is the segment from a)ß $b to a!ß $b.
84. The ellipse
x# %
y# #&
œ " has center ah, kb œ a!ß !b.
Shifting the ellipse 3 units right and 2 units down produces an ellipse with center at ah, kb œ a$ß #b and an equation
ax 3 b# %
y a#b‘# #&
œ ". Center,
C, is a3ß #b, and AB, the segment from a$ß $b to a$ß (b is the major axis.
85. (a) (fg)(x) œ f(x)g(x) œ f(x)(g(x)) œ (fg)(x), odd (b) Š gf ‹ (x) œ
f(x) g(x)
œ
f(x) g(x)
œ Š gf ‹ (x), odd
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y# # È Š #‹
œ"
17
18
Chapter 1 Functions (c) ˆ gf ‰ (x) œ (d) (e) (f) (g) (h) (i)
g(x) f(x)
œ
g(x) f(x)
œ ˆ gf ‰ (x), odd
f # (x) œ f(x)f(x) œ f(x)f(x) œ f # (x), even g# (x) œ (g(x))# œ (g(x))# œ g# (x), even (f ‰ g)(x) œ f(g(x)) œ f(g(x)) œ f(g(x)) œ (f ‰ g)(x), even (g ‰ f)(x) œ g(f(x)) œ g(f(x)) œ (g ‰ f)(x), even (f ‰ f)(x) œ f(f(x)) œ f(f(x)) œ (f ‰ f)(x), even (g ‰ g)(x) œ g(g(x)) œ g(g(x)) œ g(g(x)) œ (g ‰ g)(x), odd
86. Yes, f(x) œ 0 is both even and odd since f(x) œ 0 œ f(x) and f(x) œ 0 œ f(x). 87. (a)
(b)
(c)
(d)
88.
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Section 1.3 Trigonometric Functions 1.3 TRIGONOMETRIC FUNCTIONS 1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m radians and
51 4
1 ‰ 3. ) œ 80° Ê ) œ 80° ˆ 180° œ
41 9
2. ) œ
s r
œ
101 8
œ
51 4
1 ‰ (b) s œ r) œ (10)(110°) ˆ 180° œ
1
)
231
0
1 #
s r
œ
30 50
31 4 " È2 È" 2
sin )
0
cos )
1
tan )
0
È3
0
und.
"
und.
" È3
und.
0
1
und.
È 2
1
#
und.
È23
sec ) csc )
0
"
"
0
" und.
7. cos x œ 45 , tan x œ 34 9. sin x œ
È8 3
, tan x œ È8
"
6.
È2
3#1
)
1'
sin )
"
cos )
!
" #
tan )
und.
È 3
cot )
!
È"3
sec )
und.
#
csc )
"
È23
8. sin x œ
2 È5
10. sin x œ
12 13
13.
14.
period œ 1
13
È #3
12. cos x œ
, cos x œ
" È2
&1 ' " # È #3
È"3
"
È"3
È 3
"
È 3
2 È3
È2
È23
#
È2
#
"# È3 #
" È5
, tan x œ 12 5 È3 #
, tan x œ
" È3
period œ 41 16.
period œ 2
m
‰ ¸ 34° œ 0.6 rad or 0.6 ˆ 180° 1
11. sin x œ È"5 , cos x œ È25
15.
551 9
Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.)
È #3 "#
cot )
œ
ˆ 180° ‰ œ 225° 1
4. d œ 1 meter Ê r œ 50 cm Ê ) œ 5.
1101 18
period œ 4
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1 % " È2
19
20
Chapter 1 Functions
17.
18.
period œ 6
period œ 1
19.
20.
period œ 21
period œ 21
21.
22.
period œ 21
period œ 21
23. period œ 1# , symmetric about the origin
24. period œ 1, symmetric about the origin
25. period œ 4, symmetric about the s-axis
26. period œ 41, symmetric about the origin
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.3 Trigonometric Functions 27. (a) Cos x and sec x are positive for x in the interval ˆ 12 , 12 ‰; and cos x and sec x are negative for x in the intervals ˆ 321 , 12 ‰ and ˆ 12 , 321 ‰. Sec x is undefined when cos x is 0. The range of sec x is (_ß 1] ["ß _); the range of cos x is ["ß 1]. (b) Sin x and csc x are positive for x in the intervals ˆ 321 , 1‰ and a!, 1b; and sin x and csc x are negative for x in the intervals a1, !b and ˆ1, 321 ‰. Csc x is undefined when sin x is 0. The range of csc x is (_ß 1] [1ß _); the range of sin x is ["ß "].
28. Since cot x œ
" tan x
, cot x is undefined when tan x œ 0
and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values.
29. D: _ x _; R: y œ 1, 0, 1
30. D: _ x _; R: y œ 1, 0, 1
31. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x 32. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x 33. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x 34. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x 35. cos (A B) œ cos (A (B)) œ cos A cos (B) sin A sin (B) œ cos A cos B sin A (sin B) œ cos A cos B sin A sin B 36. sin (A B) œ sin (A (B)) œ sin A cos (B) cos A sin (B) œ sin A cos B cos A (sin B) œ sin A cos B cos A sin B 37. If B œ A, A B œ 0 Ê cos (A B) œ cos 0 œ 1. Also cos (A B) œ cos (A A) œ cos A cos A sin A sin A œ cos# A sin# A. Therefore, cos# A sin# A œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
21
22
Chapter 1 Functions
38. If B œ 21, then cos (A 21) œ cos A cos 21 sin A sin 21 œ (cos A)(1) (sin A)(0) œ cos A and sin (A 21) œ sin A cos 21 cos A sin 21 œ (sin A)(1) (cos A)(0) œ sin A. The result agrees with the fact that the cosine and sine functions have period 21. 39. cos (1 x) œ cos 1 cos B sin 1 sin x œ (1)(cos x) (0)(sin x) œ cos x 40. sin (21 x) œ sin 21 cos (x) cos (21) sin (x) œ (0)(cos (x)) (1)(sin (x)) œ sin x 41. sin ˆ 3#1 x‰ œ sin ˆ 3#1 ‰ cos (x) cos ˆ 3#1 ‰ sin (x) œ (1)(cos x) (0)(sin (x)) œ cos x 42. cos ˆ 3#1 x‰ œ cos ˆ 3#1 ‰ cos x sin ˆ 3#1 ‰ sin x œ (0)(cos x) (1)(sin x) œ sin x œ sin ˆ 14 13 ‰ œ sin
44. cos
111 1#
45. cos
1 12
œ cos ˆ 13 14 ‰ œ cos
46. sin
51 1#
œ sin ˆ 231 14 ‰ œ sin ˆ 231 ‰ cos ˆ 14 ‰ cos ˆ 231 ‰ sin ˆ 14 ‰ œ Š
21 ‰ 3
È2
1 8
œ
1 cos ˆ 281 ‰ #
œ
1 #
49. sin#
1 1#
œ
1 cos ˆ 211# ‰ #
œ
1 #
3 4
Ê sin ) œ „
52. sin2 ) œ cos2 ) Ê
sin2 ) cos2 )
1 4
œ cos
47. cos#
51. sin2 ) œ
cos
#
È3 #
È3 2
œ
1 3
cos
cos
21 3
1 4
sin
sin
1 4
cos ˆ 14 ‰ sin
1 3
1 3
È2 È3 # ‹Š # ‹
71 1#
œ cos ˆ 14
1 4
È2 ˆ"‰ # ‹ #
43. sin
œŠ
sin 1 3
21 3
œŠ
Š
È2 ˆ "‰ # ‹ #
sin ˆ 14 ‰ œ ˆ "# ‰ Š
Š
È2 # ‹
œ
È2 È3 # ‹Š # ‹
Š
È3 È2 # ‹Š # ‹
2 È2 4
48. cos#
51 1#
œ
1‰ 1 cos ˆ 10 1# #
œ
2 È3 4
50. sin#
31 8
œ
1 cos ˆ 681 ‰ #
Ê tan2 ) œ 1 Ê tan ) œ „ 1 Ê ) œ 14 ,
31 51 71 4 , 4 , 4
cos2 ) cos2 )
œ
È3 È2 # ‹ Š # ‹
œ
Ê ) œ 13 ,
È 6 È 2 4 È 2 È 6 4 1 È 3 2È 2
œ
ˆ "# ‰ Š œ
œ
1 Š
È3 ‹ #
#
1 Š #
È2 ‹ #
È2 # ‹
œ
œ
œ
1 È 3 2È 2
2 È3 4
2 È2 4
21 41 51 3 , 3 , 3
53. sin 2) cos ) œ 0 Ê 2sin ) cos ) cos ) œ 0 Ê cos )a2sin ) 1b œ 0 Ê cos ) œ 0 or 2sin ) 1 œ 0 Ê cos ) œ 0 or sin ) œ "# Ê ) œ 12 , 321 , or ) œ 16 , 561 Ê ) œ 16 , 12 , 561 , 321 54. cos 2) cos ) œ 0 Ê 2cos2 ) 1 cos ) œ 0 Ê 2cos2 ) cos ) 1 œ 0 Ê acos ) 1ba2cos ) 1b œ 0 Ê cos ) 1 œ 0 or 2cos ) 1 œ 0 Ê cos ) œ 1 or cos ) œ "# Ê ) œ 1 or ) œ 13 , 531 Ê ) œ 13 , 1, 531 55. tan (A B) œ
sin (AB) cos (AB)
œ
sin A cos Bcos A cos B cos A cos Bsin A sin B
œ
sin A cos B cos A sin B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos B cos A cos B
œ
tan Atan B 1tan A tan B
56. tan (A B) œ
sin (AB) cos (AB)
œ
sin A cos Bcos A cos B cos A cos Bsin A sin B
œ
sin A cos B cos A sin B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos B cos A cos B
œ
tan Atan B 1tan A tan B
57. According to the figure in the text, we have the following: By the law of cosines, c# œ a# b# 2ab cos ) œ 1# 1# 2 cos (A B) œ 2 2 cos (A B). By distance formula, c# œ (cos A cos B)# (sin A sin B)# œ cos# A 2 cos A cos B cos# B sin# A 2 sin A sin B sin# B œ 2 2(cos A cos B sin A sin B). Thus c# œ 2 2 cos (A B) œ 2 2(cos A cos B sin A sin B) Ê cos (A B) œ cos A cos B sin A sin B.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.3 Trigonometric Functions 58. (a) cosaA Bb œ cos A cos B sin A sin B sin ) œ cosˆ 1# )‰ and cos ) œ sinˆ 1# )‰ Let ) œ A B
sinaA Bb œ cos’ 1# aA Bb“ œ cos’ˆ 1# A‰ B“ œ cos ˆ 1# A‰ cos B sin ˆ 1# A‰ sin B œ sin A cos B cos A sin B (b) cosaA Bb œ cos A cos B sin A sin B cosaA aBbb œ cos A cos aBb sin A sin aBb Ê cosaA Bb œ cos A cos aBb sin A sin aBb œ cos A cos B sin A asin Bb œ cos A cos B sin A sin B Because the cosine function is even and the sine functions is odd. 59. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (60°) œ 4 9 12 cos (60°) œ 13 12 ˆ "# ‰ œ 7. Thus, c œ È7 ¸ 2.65. 60. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (40°) œ 13 12 cos (40°). Thus, c œ È13 12 cos 40° ¸ 1.951. 61. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand, if C is obtuse (as in the figure on the right), then sin C œ sin (1 C) œ hb . Thus, in either case, h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B. a # b # c # 2ab
By the law of cosines, cos C œ
and cos B œ
a # c # b # . 2ac
Moreover, since the sum of the
interior angles of a triangle is 1, we have sin A œ sin (1 (B C)) œ sin (B C) œ sin B cos C cos B sin C #
#
#
#
#
#
b c c b ˆ h ‰ h ‰ œ ˆ hc ‰ ’ a 2ab a2a# b# c# c# b# b œ “ ’ a 2ac “ b œ ˆ 2abc
ah bc
Ê ah œ bc sin A.
Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives h sin A sin C sin B bc œ ðóóóóóóóñóóóóóóóò a œ c œ b . law of sines 62. By the law of sines,
sin A #
œ
sin B 3
œ
È3/2 c .
By Exercise 61 we know that c œ È7. Thus sin B œ
3È 3 2È 7
¶ 0.982.
63. From the figure at the right and the law of cosines, b# œ a# 2# 2(2a) cos B œ a# 4 4a ˆ "# ‰ œ a# 2a 4. Applying the law of sines to the figure, Ê
È2/2 a
œ
È3/2 b
sin A a
œ
sin B b
Ê b œ É 3# a. Thus, combining results,
a# 2a 4 œ b# œ
3 #
a# Ê 0 œ
" #
a# 2a 4
Ê 0 œ a# 4a 8. From the quadratic formula and the fact that a 0, we have aœ
4È4# 4(1)(8) #
œ
4 È 3 4 #
¶ 1.464.
64. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator is in radians mode). (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
23
24
Chapter 1 Functions
65. A œ 2, B œ 21, C œ 1, D œ 1
66. A œ "# , B œ 2, C œ 1, D œ
" #
67. A œ 12 , B œ 4, C œ 0, D œ
68. A œ
L 21 ,
" 1
B œ L, C œ 0, D œ 0
69-72. Example CAS commands: Maple f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#69 (Section 1.3)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[21/b (x c)] + d Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x, 41, 41 }]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.3 Trigonometric Functions 69. (a) The graph stretches horizontally.
(b) The period remains the same: period œ l B l. The graph has a horizontal shift of
" #
period.
70. (a) The graph is shifted right C units.
(b) The graph is shifted left C units. (c) A shift of „ one period will produce no apparent shift. l C l œ ' 71. (a) The graph shifts upwards l D lunits for D ! (b) The graph shifts down l D lunits for D !Þ
72. (a) The graph stretches l A l units.
(b) For A !, the graph is inverted.
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25
26
Chapter 1 Functions
1.4 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4.
The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space.
1. d.
2. c.
3. d.
4. b.
5-30.
For any display there are many appropriate display widows. The graphs given as answers in Exercises 530 are not unique in appearance.
5. Ò2ß 5Ó by Ò15ß 40Ó
6. Ò4ß 4Ó by Ò4ß 4Ó
7. Ò2ß 6Ó by Ò250ß 50Ó
8. Ò1ß 5Ó by Ò5ß 30Ó
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Section 1.4 Graphing with Calculators and Computers 9. Ò4ß 4Ó by Ò5ß 5Ó
10. Ò2ß 2Ó by Ò2ß 8Ó
11. Ò2ß 6Ó by Ò5ß 4Ó
12. Ò4ß 4Ó by Ò8ß 8Ó
13. Ò1ß 6Ó by Ò1ß 4Ó
14. Ò1ß 6Ó by Ò1ß 5Ó
15. Ò3ß 3Ó by Ò0ß 10Ó
16. Ò1ß 2Ó by Ò0ß 1Ó
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Chapter 1 Functions
17. Ò5ß 1Ó by Ò5ß 5Ó
18. Ò5ß 1Ó by Ò2ß 4Ó
19. Ò4ß 4Ó by Ò0ß 3Ó
20. Ò5ß 5Ó by Ò2ß 2Ó
21. Ò"!ß "!Ó by Ò'ß 'Ó
22. Ò&ß &Ó by Ò#ß #Ó
23. Ò'ß "!Ó by Ò'ß 'Ó
24. Ò$ß &Ó by Ò#ß "!Ó
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Section 1.4 Graphing with Calculators and Computers 25. Ò0Þ03ß 0Þ03Ó by Ò1Þ25ß 1Þ25Ó
26. Ò0Þ1ß 0Þ1Ó by Ò3ß 3Ó
27. Ò300ß 300Ó by Ò1Þ25ß 1Þ25Ó
28. Ò50ß 50Ó by Ò0Þ1ß 0Þ1Ó
29. Ò0Þ25ß 0Þ25Ó by Ò0Þ3ß 0Þ3Ó
30. Ò0Þ15ß 0Þ15Ó by Ò0Þ02ß 0Þ05Ó
31. x# #x œ % %y y# Ê y œ # „ Èx# #x ). The lower half is produced by graphing y œ # Èx# #x ).
32. y# "'x# œ " Ê y œ „ È" "'x# . The upper branch is produced by graphing y œ È" "'x# .
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30
Chapter 1 Functions
33.
34.
35.
36.
37.
38Þ
39.
40.
CHAPTER 1 PRACTICE EXERCISES 1. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ 2. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰ surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰
"Î#
#Î$
C #1
#
Ê A œ 1ˆ #C1 ‰ œ
C# %1 .
$ $V . The volume is V œ %$ 1 r$ Ê r œ É %1 . Substitution into the formula for
.
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Chapter 1 Practice Exercises 3. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies the equation tan ) œ 4. tan ) œ
rise run
œ
h &!!
x# x
œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b.
Ê h œ &!! tan ) ft.
5.
6.
Symmetric about the origin.
Symmetric about the y-axis.
7.
8.
Neither
Symmetric about the y-axis.
9. yaxb œ axb# " œ x# " œ yaxb. Even. 10. yaxb œ axb& axb$ axb œ x& x$ x œ yaxb. Odd. 11. yaxb œ " cosaxb œ " cos x œ yaxb. Even. 12. yaxb œ secaxb tanaxb œ 13. yaxb œ
axb% " axb$ #axb
œ
x% " x$ #x
sinaxb cos# axb
œ
sin x cos# x
œ sec x tan x œ yaxb. Odd.
%
" œ xx$ # x œ yaxb. Odd.
14. yaxb œ axb sinaxb œ axb sin x œ ax sin xb œ yaxb. Odd. 15. yaxb œ x cosaxb œ x cos x. Neither even nor odd. 16. yaxb œ axbcosaxb œ x cos x œ yaxb. Odd. 17. Since f and g are odd Ê faxb œ faxb and gaxb œ gaxb. (a) af † gbaxb œ faxbgaxb œ ÒfaxbÓÒgaxbÓ œ faxbgaxb œ af † gbaxb Ê f † g is even (b) f 3 axb œ faxbfaxbfaxb œ ÒfaxbÓÒfaxbÓÒfaxbÓ œ faxb † faxb † faxb œ f 3 axb Ê f 3 is odd. (c) fasinaxbb œ fasinaxbb œ fasinaxbb Ê fasinaxbb is odd. (d) gasecaxbb œ gasecaxbb Ê gasecaxbb is even. (e) lgaxbl œ lgaxbl œ lgaxbl Ê lgl is evenÞ
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Chapter 1 Functions
18. Let faa xb œ faa xb and define gaxb œ fax ab. Then gaxb œ faaxb ab œ faa xb œ faa xb œ fax ab œ gaxb Ê gaxb œ fax ab is even. 19. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since l x l attains all nonnegative values, the range is Ò#ß _Ñ. 20. (a) Since the square root requires " x !, the domain is Ð_ß "Ó. (b) Since È" x attains all nonnegative values, the range is Ò#ß _Ñ. 21. (a) Since the square root requires "' x# !, the domain is Ò%ß %Ó. (b) For values of x in the domain, ! Ÿ "' x# Ÿ "', so ! Ÿ È"' x# Ÿ %. The range is Ò!ß %Ó. 22. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since $#x attains all positive values, the range is a"ß _b. 23. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since #ex attains all positive values, the range is a$ß _b. 24. (a) The function is equivalent to y œ tan #x, so we require #x Á
k1 #
for odd integers k. The domain is given by x Á
k1 %
for
odd integers k. (b) Since the tangent function attains all values, the range is a_ß _b. 25. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) The sine function attains values from " to ", so # Ÿ #sina$x 1b Ÿ # and hence $ Ÿ #sina$x 1b " Ÿ ". The range is Ò3ß 1Ó. 26. (a) The function is defined for all values of x, so the domain is a_ß _b. & (b) The function is equivalent to y œ È x# , which attains all nonnegative values. The range is Ò!ß _Ñ. 27. (a) The logarithm requires x $ !, so the domain is a$ß _b. (b) The logarithm attains all real values, so the range is a_ß _b. 28. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) The cube root attains all real values, so the range is a_ß _b. 29. (a) (b) (c) (d)
Increasing because volume increases as radius increases Neither, since the greatest integer function is composed of horizontal (constant) line segments Decreasing because as the height increases, the atmospheric pressure decreases. Increasing because the kinetic (motion) energy increases as the particles velocity increases.
30. (a) Increasing on Ò2, _Ñ (c) Increasing on a_, _b
(b) Increasing on Ò1, _Ñ (d) Increasing on Ò "# , _Ñ
31. (a) The function is defined for % Ÿ x Ÿ %, so the domain is Ò%ß %Ó. (b) The function is equivalent to y œ Èl x l, % Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The range is Ò!ß #Ó.
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Chapter 1 Practice Exercises
33
32. (a) The function is defined for # Ÿ x Ÿ #, so the domain is Ò#ß #Ó. (b) The range is Ò"ß "Ó. 33. First piece: Line through a!ß "b and a"ß !b. m œ Second piece: Line through a"ß "b and a#ß !b. m faxb œ œ
" x, ! Ÿ x " # x, " Ÿ x Ÿ #
34. First piece: Line through a!ß !b and a2ß 5b. m œ Second piece: Line through a2ß 5b and a4ß !b. m faxb œ
10
5 2 x, 5x 2 ,
!" " "! œ " œ " " œ !# " œ "
" Ê y œ x " œ " x œ " Ê y œ ax "b " œ x # œ # x
5! 5 5 2! œ 2 Ê y œ 2x !5 5 œ 4 2 œ 2 œ 52 Ê
y œ 52 ax 2b 5 œ 52 x 10 œ 10
!Ÿx2 (Note: x œ 2 can be included on either piece.) 2ŸxŸ4
35. (a) af‰gba"b œ faga"bb œ fŠ È"" # ‹ œ fa"b œ (b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ (c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ
" É "# #
" "Îx
œ
" È#Þ&
" "
œ"
or É &#
œ x, x Á !
(d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ
"
" É Èx # #
œ
% x# È É " #È x #
$ 36. (a) af‰gba"b œ faga"bb œ fˆÈ " "‰ œ fa!b œ # ! œ # $ (b) ag‰f ba#b œ faga#bb œ ga# #b œ ga!b œ È !"œ"
(c) af‰f baxb œ fafaxbb œ fa# xb œ # a# xb œ x $ $ $ È (d) ag‰gbaxb œ gagaxbb œ gˆÈ x "‰ œ É x""
#
37. (a) af‰gbaxb œ fagaxbb œ fˆÈx #‰ œ # ˆÈx #‰ œ x, x #. ag‰f baxb œ fagaxbb œ ga# x# b œ Èa# x# b # œ È% x# (b) Domain of f‰g: Ò#ß _ÑÞ Domain of g‰f: Ò#ß #ÓÞ
(c) Range of f‰g: Ð_ß #ÓÞ Range of g‰f: Ò!ß #ÓÞ
% 38. (a) af‰gbaxb œ fagaxbb œ fŠÈ" x‹ œ ÉÈ" x œ È " x.
ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É" Èx (b) Domain of f‰g: Ð_ß "ÓÞ Domain of g‰f: Ò!ß "ÓÞ 39.
y œ faxb
(c) Range of f‰g: Ò!ß _ÑÞ Range of g‰f: Ò!ß "ÓÞ y œ af‰f baxb
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5x 2
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Chapter 1 Functions
40.
41.
42.
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. 43.
It does not change the graph.
44.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.
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Chapter 1 Practice Exercises 45.
46.
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis. 47.
48.
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. 49. (a) y œ gax 3b (c) y œ gaxb (e) y œ 5 † gaxb
" #
The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x 0 across the y-axis. (b) y œ gˆx 3# ‰ 2 (d) y œ gaxb (f) y œ ga5xb
50. (a) Shift the graph of f right 5 units (b) Horizontally compress the graph of f by a factor of 4 (c) Horizontally compress the graph of f by a factor of 3 and a then reflect the graph about the y-axis (d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left "# unit. (e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units. (f) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the graph up "4 unit. 51. Reflection of the grpah of y œ Èx about the x-axis followed by a horizontal compression by a factor of 1 2 then a shift left 2 units.
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Chapter 1 Functions
52. Reflect the graph of y œ x about the x-axis, followed by a vertical compression of the graph by a factor of 3, then shift the graph up 1 unit.
53. Vertical compression of the graph of y œ
1 x2
by a
factor of 2, then shift the graph up 1 unit.
54. Reflect the graph of y œ x1Î3 about the y-axis, then compress the graph horizontally by a factor of 5.
55.
56.
period œ 1 57.
period œ 41 58.
period œ 2
period œ 4
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Chapter 1 Practice Exercises 59.
60.
period œ 21
period œ 21 1 3
61. (a) sin B œ sin
œ
b c
œ
b #
Ê b œ 2 sin
1 3
œ 2Š
È3 # ‹
œ È3. By the theorem of Pythagoras,
a# b# œ c# Ê a œ Èc# b# œ È4 3 œ 1. 1 3
(b) sin B œ sin
œ
b c
œ
2 c
Ê cœ
2 sin 13
œ È23 œ Š ‹ #
4 È3
#
. Thus, a œ Èc# b# œ ÊŠ È43 ‹ (2)# œ É 34 œ
62. (a) sin A œ
a c
Ê a œ c sin A
(b) tan A œ
a b
Ê a œ b tan A
63. (a) tan B œ
b a
Ê aœ
(b) sin A œ
a c
Ê cœ
64. (a) sin A œ
a c
(c) sin A œ
a c
b tan B
œ
a sin A
È c # b # c
65. Let h œ height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50° œ hc , tan 35° œ hb , and b c œ 10. Thus, h œ c tan 50° and h œ b tan 35° œ (c 10) tan 35° Ê c tan 50° œ (c 10) tan 35° Ê c (tan 50° tan 35°) œ 10 tan 35° tan 35° Ê c œ tan10 50°tan 35° Ê h œ c tan 50° œ
10 tan 35° tan 50° tan 50°tan 35°
¸ 16.98 m.
66. Let h œ height of balloon above ground. From the figure at the right, tan 40° œ ha , tan 70° œ hb , and a b œ 2. Thus, h œ b tan 70° Ê h œ (2 a) tan 70° and h œ a tan 40° Ê (2 a) tan 70° œ a tan 40° Ê a(tan 40° tan 70°) 70° œ 2 tan 70° Ê a œ tan 240°tantan 70° Ê h œ a tan 40° œ
2 tan 70° tan 40° tan 40°tan 70°
¸ 1.3 km.
67. (a)
(b) The period appears to be 41.
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2 È3
.
37
38
Chapter 1 Functions (c) f(x 41) œ sin (x 41) cos ˆ x#41 ‰ œ sin (x 21) cos ˆ x# 21‰ œ sin x cos since the period of sine and cosine is 21. Thus, f(x) has period 41.
x #
68. (a)
(b) D œ (_ß 0) (!ß _); R œ [1ß 1] (c) f is not periodic. For suppose f has period p. Then f ˆ #"1 kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all integers k. Choose k so large that
" #1
kp
" 1
Ê 0
" (1/21)kp
1. But then
f ˆ #"1 kp‰ œ sin Š (1/#1")kp ‹ 0 which is a contradiction. Thus f has no period, as claimed. CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)). 2. Yes, there are many such function pairs. For example, if g(x) œ (2x 3)$ and f(x) œ x"Î$ , then (f ‰ g)(x) œ f(g(x)) œ f a(2x 3)$ b œ a(2x 3)$ b
"Î$
œ 2x 3.
3. If f is odd and defined at x, then f(x) œ f(x). Thus g(x) œ f(x) 2 œ f(x) 2 whereas g(x) œ (f(x) 2) œ f(x) 2. Then g cannot be odd because g(x) œ g(x) Ê f(x) 2 œ f(x) 2 Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is even, then g(x) œ f(x) 2 is also even: g(x) œ f(x) 2 œ f(x) 2 œ g(x). 4. If g is odd and g(0) is defined, then g(0) œ g(0) œ g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0. 5. For (xß y) in the 1st quadrant, kxk kyk œ 1 x Í x y œ 1 x Í y œ 1. For (xß y) in the 2nd quadrant, kxk kyk œ x 1 Í x y œ x 1 Í y œ 2x 1. In the 3rd quadrant, kxk kyk œ x 1 Í x y œ x 1 Í y œ 2x 1. In the 4th quadrant, kxk kyk œ x 1 Í x (y) œ x 1 Í y œ 1. The graph is given at the right.
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Chapter 1 Additional and Advanced Exercises 6. We use reasoning similar to Exercise 5. (1) 1st quadrant: y kyk œ x kxk Í 2y œ 2x Í y œ x. (2) 2nd quadrant: y kyk œ x kxk Í 2y œ x (x) œ 0 Í y œ 0. (3) 3rd quadrant: y kyk œ x kxk Í y (y) œ x (x) Í 0 œ 0 Ê all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y kyk œ x kxk Í y (y) œ 2x Í 0 œ x. Combining these results we have the graph given at the right: 7. (a) sin# x cos# x œ 1 Ê sin# x œ 1 cos# x œ (1 cos x)(1 cos x) Ê (1 cos x) œ Ê
1cos x sin x
œ
sin# x 1cos x
sin x 1cos x
(b) Using the definition of the tangent function and the double angle formulas, we have #
tan ˆ x# ‰ œ
sin# ˆ x# ‰ cos# ˆ #x ‰
œ
"
x ‹‹ cos Š2 Š #
#
"cos Š2 Š #x ‹‹ #
œ
1cos x 1cos x
.
8. The angles labeled # in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled ! are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which ) b implies ab c œ 2a cos a c Ê (a c)(a c) œ b(2a cos ) b) Ê a# c# œ 2ab cos ) b# Ê c# œ a# b# 2ab cos ).
9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah œ bc sin A œ ab sin C œ ac sin B Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B. 10. As in Section 1.3, Exercise 61, (Area of ABC)# œ œ
" 4
(base)# (height)# œ
" 4
a# h # œ
" 4
a# b# sin# C
a# b# a" cos# Cb . By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ
Thus, (area of ABC)# œ œ
" 4
" 16
" 4
a# b# a" cos# Cb œ #
Š4a# b# aa# b# c# b ‹ œ
" 16
" 4
a# b# Œ" Š a
#
b# c# ‹ #ab
#
œ
a# b# 4
a# b# c# 2ab
Š"
.
aa # b # c # b 4a# b#
#
‹
ca2ab aa# b# c# bb a2ab aa# b# c# bbd
" ca(a b)# c# b ac# (a b)# bd œ 16 c((a b) c)((a b) c)(c (a b))(c (a b))d a b c a b c a b c a b c œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s a)(s b)(s c), where s œ a#bc .
œ
" 16
Therefore, the area of ABC equals Ès(s a)(s b)(s c) . 11. If f is even and odd, then f(x) œ f(x) and f(x) œ f(x) Ê f(x) œ f(x) for all x in the domain of f. Thus 2f(x) œ 0 Ê f(x) œ 0.
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40
Chapter 1 Functions f(x) f((x)) œ f(x) #f(x) œ E(x) Ê E # even function. Define O(x) œ f(x) E(x) œ f(x) f(x) #f(x) œ f(x) #f(x) . Then O(x) œ f(x) #f((x)) œ f(x)# f(x) œ Š f(x) #f(x) ‹ œ O(x) Ê O is an odd function
12. (a) As suggested, let E(x) œ
f(x) f(x) #
Ê E(x) œ
is an
Ê f(x) œ E(x) O(x) is the sum of an even and an odd function. (b) Part (a) shows that f(x) œ E(x) O(x) is the sum of an even and an odd function. If also f(x) œ E" (x) O" (x), where E" is even and O" is odd, then f(x) f(x) œ 0 œ aE" (x) O" (x)b (E(x) O(x)). Thus, E(x) E" (x) œ O" (x) O(x) for all x in the domain of f (which is the same as the domain of E E" and O O" ). Now (E E" )(x) œ E(x) E" (x) œ E(x) E" (x) (since E and E" are even) œ (E E" )(x) Ê E E" is even. Likewise, (O" O)(x) œ O" (x) O(x) œ O" (x) (O(x)) (since O and O" are odd) œ (O" (x) O(x)) œ (O" O)(x) Ê O" O is odd. Therefore, E E" and O" O are both even and odd so they must be zero at each x in the domain of f by Exercise 11. That is, E" œ E and O" œ O, so the decomposition of f found in part (a) is unique. 13. y œ ax# bx c œ a Šx# ba x
b# 4a# ‹
b# 4a
c œ a ˆx
b ‰# 2a
b# 4a
c
(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the right. If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward to the right. If b 0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c 0, and downward ?c units if ?c 0. 14. (a) If a 0, the graph rises to the right of the vertical line x œ b and falls to the left. If a 0, the graph falls to the right of the line x œ b and rises to the left. If a œ 0, the graph reduces to the horizontal line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 15. Each of the triangles pictured has the same base b œ v?t œ v(1 sec). Moreover, the height of each triangle is the same value h. Thus "# (base)(height) œ
" #
bh
œ A" œ A# œ A$ œ á . In conclusion, the object sweeps out equal areas in each one second interval.
16. (a) Using the midpoint formula, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope of OP œ
?y ?x
œ
b/2 a/2
œ
b a
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 1 Additional and Advanced Exercises (b) The slope of AB œ
b 0 0 a
41
œ ba . The line segments AB and OP are perpendicular when the product #
of their slopes is " œ ˆ ba ‰ ˆ ba ‰ œ ba# . Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB is perpendicular to OP when a œ b. 17. From the figure we see that 0 Ÿ ) Ÿ cos ) œ
and AB œ AD œ 1. From trigonometry we have the following: sin ) œ sin ) cos ) .
œ CD, and tan ) œ œ We can see that: w " area ˜AEB area sector DB area ˜ADC Ê # aAEbaEBb "# aADb2 ) "# aADbaCDb AE AB
œ AE, tan ) œ
1 2
CD AD
EB AE
Ê "# sin ) cos ) "# a"b2 ) "# a"batan )b Ê "# sin ) cos ) "# )
" sin ) # cos )
18. af‰gbaxb œ fagaxbb œ aacx db b œ acx ad b and ag‰f baxb œ gafaxbb œ caax bb d œ acx cb d Thus af‰gbaxb œ ag‰f baxb Ê acx ad b œ acx bc d Ê ad b œ bc d. Note that fadb œ ad b and gabb œ cb d, thus af‰gbaxb œ ag‰f baxb if fadb œ gabb.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
EB AB
œ EB,
42
Chapter 1 Functions
NOTES:
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND TANGENTS TO CURVES 1. (a)
?f ?x
œ
f(3) f(2) 3#
2. (a)
?g ?x
œ
g(1) g(1) 1 (1)
3. (a)
?h ?t
œ
h ˆ 341 ‰ h ˆ 14 ‰ 1 31 4 4
œ
?g ?t
œ
g(1) g(0) 10
(2 1) (2 1) 10
4. (a) 5.
?R ?)
œ
R(2) R(0) 20
6.
?P ?)
œ
P(2) P(1) 21
7. (a)
?y ?x
œ
?y ?x
œ
?y ?x
œ
?y ?x
œ
?y ?x
œ
?y ?x
œ
?y ?x
œ
œ œ
28 9 1
œ
œ
œ
1 1 2
œ 19
(b)
?f ?x
œ
f(1) f(") 1 (1)
œ
20 #
œ1
œ0
(b)
?g ?x
œ
g(0)g(2) 0(2)
œ
04 #
œ 2
(b)
?h ?t
œ
h ˆ 1# ‰ h ˆ 16 ‰ 11 # 6
œ
?g ?t
œ
g(1) g(1) 1 (1)
œ
1 1 1 #
È 8 1 È 1 #
œ 14 œ 12
3" #
œ
(8 16 10)(" % &) 1
ˆa2 h b2 3 ‰ ˆ 2 2 3 ‰ h
œ
(b)
0 È3 1 3
œ
3 È 3 1
(2 1) (2 ") #1
œ0
œ1 œ22œ0
4 4h h2 3 1 h
œ
4h h2 h
œ 4 h. As h Ä 0, 4 h Ä 4 Ê at Pa2, 1b the slope is 4.
(b) y 1 œ 4ax 2b Ê y 1 œ 4x 8 Ê y œ 4x 7 8. (a)
ˆ 5 a1 h b 2 ‰ ˆ 5 1 2 ‰ h
œ
5 1 2h h2 4 h
œ
2h h2 h
œ 2 h. As h Ä 0, 2 h Ä 2 Ê at Pa1, 4b the
slope is 2. (b) y 4 œ a2bax 1b Ê y 4 œ 2x 2 Ê y œ 2x 6 9. (a)
ˆa2 h b2 2 a 2 h b 3 ‰ ˆ 2 2 2 a 2 b 3 ‰ h
œ
4 4h h2 4 2h 3 a3b h
œ
2h h2 h
œ 2 h. As h Ä 0, 2 h Ä 2 Ê at
Pa2, 3b the slope is 2. (b) y a3b œ 2ax 2b Ê y 3 œ 2x 4 Ê y œ 2x 7. 10. (a)
ˆa1 h b2 4 a 1 h b ‰ ˆ 1 2 4 a 1 b ‰ h
œ
1 2h h2 4 4h a3b h
œ
h2 2h h
œ h 2. As h Ä 0, h 2 Ä 2 Ê at
Pa1, 3b the slope is 2. (b) y a3b œ a2bax 1b Ê y 3 œ 2x 2 Ê y œ 2x 1. 11. (a)
a2 h b 3 2 3 h
œ
8 12h 4h2 h3 8 h
œ
12h 4h2 h3 h
œ 12 4h h2 . As h Ä 0, 12 4h h2 Ä 12, Ê at
Pa2, 8b the slope is 12. (b) y 8 œ 12ax 2b Ê y 8 œ 12x 24 Ê y œ 12x 16. 12. (a)
2 a1 h b3 ˆ 2 1 3 ‰ h
œ
2 1 3h 3h2 h3 1 h
œ
3h 3h2 h3 h
œ 3 3h h2 . As h Ä 0, 3 3h h2 Ä 3, Ê at
Pa1, 1b the slope is 3. (b) y 1 œ a3bax 1b Ê y 1 œ 3x 3 Ê y œ 3x 4. 13. (a)
a1 hb3 12a1 hb ˆ13 12a"b‰ h 2
œ
1 3h 3h2 h3 12 12h a11b h
œ
9h 3h2 h3 h
œ 9 3h h2 . As h Ä 0,
9 3h h Ä 9 Ê at Pa1, 11b the slope is 9. (b) y a11b œ a9bax 1b Ê y 11 œ 9x 9 Ê y œ 9x 2.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
44
Chapter 2 Limits and Continuity
14. (a)
?y ?x
œ
a2 h b 3 3 a 2 h b 2 4 ˆ 2 3 3 a 2 b 2 4 ‰ h 2
8 12h 6h2 h3 12 12h 3h2 4 0 h
œ
œ
3h2 h3 h
œ 3h h2 . As h Ä 0,
3h h Ä 0 Ê at Pa2, 0b the slope is 0. (b) y 0 œ 0ax 2b Ê y œ 0. 15. (a)
?p ?t
Slope of PQ œ
Q
650 225 20 10 650 375 20 14 650 475 20 16.5 650 550 20 18
Q" (10ß 225) Q# (14ß 375) Q$ (16.5ß 475) Q% (18ß 550)
œ 42.5 m/sec œ 45.83 m/sec œ 50.00 m/sec œ 50.00 m/sec
(b) At t œ 20, the sportscar was traveling approximately 50 m/sec or 180 km/h. 16. (a)
Slope of PQ œ
Q Q" (5ß 20) Q# (7ß 39) Q$ (8.5ß 58) Q% (9.5ß 72)
80 20 10 5 80 39 10 7 80 58 10 8.5 80 72 10 9.5
?p ?t
œ 12 m/sec œ 13.7 m/sec œ 14.7 m/sec œ 16 m/sec
(b) Approximately 16 m/sec 17. (a)
(b)
?p ?t
œ
174 62 2004 2002
œ
112 #
œ 56 thousand dollars per year
(c) The average rate of change from 2001 to 2002 is ??pt œ
62 27 20022 2001 œ 35 thousand dollars per year. 111 62 The average rate of change from 2002 to 2003 is ??pt œ 2003 2002 œ 49 thousand dollars per year. So, the rate at which profits were changing in 2002 is approximatley "# a35 49b œ 42 thousand dollars
18. (a) F(x) œ (x 2)/(x 2) x 1.2 F(x) 4.0 ?F ?x ?F ?x ?F ?x
œ
?g ?x ?g ?x
œ
œ œ
1.1 3.4
1.01 3.04
1.001 3.004
1.0001 3.0004
1 3
4.0 (3) œ 5.0; 1.2 1 3.04 (3) œ 4.04; 1.01 1 3.!!!% (3) œ 4.!!!%; 1.0001 1
?F ?x ?F ?x
œ œ
3.4 (3) œ 4.4; 1.1 1 3.004 (3) œ 4.!!%; 1.001 1
È g(2) g(1) œ #21" ¸ 0.414213 21 È1 h" g(1 h) g(1) (1 h) 1 œ h
?g ?x
œ
g(1.5) g(1) 1.5 1
(b) The rate of change of F(x) at x œ 1 is 4. 19. (a)
œ
œ
È1.5 " 0.5
¸ 0.449489
(b) g(x) œ Èx 1h È1 h ŠÈ1 h 1‹ /h
1.1 1.04880
1.01 1.004987
1.001 1.0004998
1.0001 1.0000499
1.00001 1.000005
1.000001 1.0000005
0.4880
0.4987
0.4998
0.499
0.5
0.5
(c) The rate of change of g(x) at x œ 1 is 0.5.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
per year.
Section 2.1 Rates of Change and Tangents to Curves (d) The calculator gives lim hÄ! 20. (a) i) ii) (b)
f(3) f(2) 32
œ
f(T) f(2) T#
œ
"" 3 #
1 " " T # T#
T f(T) af(T) f(2)b/aT 2b
"
œ
6
1
È1 h" h
œ "# .
œ "6
#TT T#
2 #T
œ
45
œ
2T #T(T 2)
2.1 0.476190 0.2381
œ
2T #T(2 T)
2.01 0.497512 0.2488
œ #"T , T Á 2
2.001 0.499750 0.2500
2.0001 0.4999750 0.2500
2.00001 0.499997 0.2500
2.000001 0.499999 0.2500
(c) The table indicates the rate of change is 0.25 at t œ 2. " ‰ (d) lim ˆ #T œ 4" TÄ#
NOTE: Answers will vary in Exercises 21 and 22. s 15 0 ˜s 20 15 10 ˜s 30 20 21. (a) Ò0, 1Ó: ˜ ˜t œ 1 0 œ 15 mphà Ò1, 2.5Ó: ˜t œ 2.5 1 œ 3 mphà Ò2.5, 3.5Ó: ˜t œ 3.5 2.5 œ 10 mph (b) At Pˆ "# , 7.5‰: Since the portion of the graph from t œ 0 to t œ 1 is nearly linear, the instantaneous rate of change
will be almost the same as the average rate of change, thus the instantaneous speed at t œ
" #
is
15 7.5 1 0.5
œ 15 mi/hr.
At Pa2, 20b: Since the portion of the graph from t œ 2 to t œ 2.5 is nearly linear, the instantaneous rate of change will 20 be nearly the same as the average rate of change, thus v œ 20 2.5 2 œ 0 mi/hr. For values of t less than 2, we have Slope of PQ œ
Q
?s ?t
15 20 1 2 œ 5 mi/hr 19 20 1.5 2 œ 2 mi/hr 19.9 20 1.9 2 œ 1 mi/hr
Q" (1ß 15) Q# (1.5ß 19) Q$ (1.9ß 19.9)
Thus, it appears that the instantaneous speed at t œ 2 is 0 mi/hr. At Pa3, 22b: s Q Slope of PQ œ ? ?t 35 22 43 30 22 3.5 3 23 22 3.1 3
Q" (4ß 35) Q# (3.5ß 30) Q$ (3.1ß 23)
Slope of PQ œ
Q
œ 13 mi/hr
Q" (2ß 20)
œ 16 mi/hr
Q# (2.5ß 20)
œ 10 mi/hr
Q$ (2.9ß 21.6)
20 22 2 3 œ 2 mi/hr 20 22 2.5 3 œ 4 mi/hr 21.6 22 2.9 3 œ 4 mi/hr
Thus, it appears that the instantaneous speed at t œ 3 is about 7 mi/hr. (c) It appears that the curve is increasing the fastest at t œ 3.5. Thus for Pa3.5, 30b s Q Slope of PQ œ ? Q Slope of PQ œ ?t 35 30 4 3.5 œ 10 mi/hr 34 30 3.75 3.5 œ 16 mi/hr 32 30 3.6 3.5 œ 20 mi/hr
Q" (4ß 35) Q# (3.75ß 34) Q$ (3.6ß 32)
˜A ˜t
œ
10 15 30
(b) At Pa1, 14b: Q Q" (2ß 12.2) Q# (1.5ß 13.2) Q$ (1.1ß 13.85)
¸ 1.67
gal day à
Ò0, 5Ó:
Q# (3.25ß 25) Q$ (3.4ß 28)
Slope of PQ œ
˜A ˜t
?A ?t
12.2 14 2 1 œ 1.8 gal/day 13.2 14 1.5 1 œ 1.6 gal/day 13.85 14 1.1 1 œ 1.5 gal/day
œ
3.9 15 50
¸ 2.2
gal day à Ò7,
10Ó:
?s ?t
22 30 3 3.5 œ 16 mi/hr 25 30 3.25 3.5 œ 20 mi/hr 28 30 3.4 3.5 œ 20 mi/hr
Q" (3ß 22)
Thus, it appears that the instantaneous speed at t œ 3.5 is about 20 mi/hr. 22. (a) Ò0, 3Ó:
?s ?t
˜A ˜t
Q Q" (0ß 15) Q# (0.5ß 14.6) Q$ (0.9ß 14.86)
œ
0 1.4 10 7
¸ 0.5
Slope of PQ œ
gal day ?A ?t
15 14 0 1 œ 1 gal/day 14.6 14 0.5 1 œ 1.2 gal/day 14.86 14 0.9 1 œ 1.4 gal/day
Thus, it appears that the instantaneous rate of consumption at t œ 1 is about 1.45 gal/day. At Pa4, 6b:
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
46
Chapter 2 Limits and Continuity Q Q" (5ß 3.9) Q# (4.5ß 4.8) Q$ (4.1ß 5.7)
Slope of PQ œ 3.9 6 54 4.8 6 4.5 4 5.7 6 4.1 4
?A ?t
Q
œ 2.1 gal/day
Q" (3ß 10)
œ 2.4 gal/day
Q# (3.5ß 7.8)
œ 3 gal/day
Q$ (3.9ß 6.3)
Slope of PQ œ
10 6 3 4 œ 4 gal/day 7.8 6 3.5 4 œ 3.6 gal/day 6.3 6 3.9 4 œ 3 gal/day
Thus, it appears that the instantaneous rate of consumption at t œ 1 is 3 gal/day. At Pa8, 1b: Q Slope of PQ œ ??At Q Slope of PQ œ Q" (9ß 0.5) Q# (8.5ß 0.7) Q$ (8.1ß 0.95)
0.5 1 9 8 œ 0.5 gal/day 0.7 1 8.5 8 œ 0.6 gal/day 0.95 1 8.1 8 œ 0.5 gal/day
Q" (7ß 1.4)
4.8 7.8 4.5 3.5 œ 3 gal/day 6 7.8 4 3.5 œ 3.6 gal/day 7.4 7.8 3.6 3.5 œ 4 gal/day
Q" (2.5ß 11.2)
Q# (7.5ß 1.3) Q$ (7.9ß 1.04)
?A ?t
?A ?t
1.4 1 7 8 œ 0.6 gal/day 1.3 1 7.5 8 œ 0.6 gal/day 1.04 1 7.9 8 œ 0.6 gal/day
Thus, it appears that the instantaneous rate of consumption at t œ 1 is 0.55 gal/day. (c) It appears that the curve (the consumption) is decreasing the fastest at t œ 3.5. Thus for Pa3.5, 7.8b s Q Slope of PQ œ ??At Q Slope of PQ œ ? ?t Q" (4.5ß 4.8) Q# (4ß 6) Q$ (3.6ß 7.4)
Q# (3ß 10) Q$ (3.4ß 8.2)
Thus, it appears that the rate of consumption at t œ 3.5 is about 4 gal/day.
11.2 7.8 2.5 3.5 œ 3.4 gal/day 10 7.8 3 3.5 œ 4.4 gal/day 8.2 7.8 3.4 3.5 œ 4 gal/day
2.2 LIMIT OF A FUNCTION AND LIMIT LAWS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1. (b) 1 (c) 0 (d) 0.5 2. (a) 0 (b) 1 (c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t) approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0. (d) 1 3. (a) True (d) False (g) True
(b) True (e) False
(c) False (f) True
4. (a) False (d) True
(b) False (e) True
(c) True
5.
x
lim x Ä 0 kx k x kx k
does not exist because
x kx k
œ
x x
œ 1 if x 0 and
approaches 1. As x approaches 0 from the right,
x kx k
x kxk
œ
x x
œ 1 if x 0. As x approaches 0 from the left,
approaches 1. There is no single number L that all the
function values get arbitrarily close to as x Ä 0. 6. As x approaches 1 from the left, the values of
" x 1
become increasingly large and negative. As x approaches 1
from the right, the values become increasingly large and positive. There is no one number L that all the function values get arbitrarily close to as x Ä 1, so lim x" 1 does not exist. xÄ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.2 Limit of a Function and Limit Laws
47
7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the definition of f(x) at x! itself. 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of the xÄ0
value f(0) itself. 9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5. xÄ1
10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x), xÄ1
xÄ1
whether it exists or what its value is if it does exist, from knowing the value of f(1) alone. 11.
lim (2x 5) œ 2(7) 5 œ 14 5 œ 9
x Ä (
12. lim ax# 5x 2b œ (2)# 5(2) 2 œ 4 10 2 œ 4 xÄ#
13. lim 8(t 5)(t 7) œ 8(6 5)(6 7) œ 8 tÄ'
14.
lim ax$ 2x# 4x 8b œ (2)$ 2(2)# 4(2) 8 œ 8 8 8 8 œ 16
x Ä #
15. lim
x3
œ
x Ä # x6
17.
23 26
16. lim# 3s(2s 1) œ 3 ˆ 23 ‰ 2 ˆ 23 ‰ 1‘ œ 2 ˆ 43 1‰ œ
5 8
sÄ
$
lim 3(2x 1)# œ 3(2(1) 1)# œ 3(3)# œ 27
x Ä "
y2
18. lim
# y Ä # y 5y 6
19.
œ
œ
22 (2)# 5(#) 6
œ
4 4 10 6
œ
4 #0
œ
" 5
%
lim (5 y)%Î$ œ [5 (3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16
y Ä $
20. lim (2z 8)"Î$ œ (2(0) 8)"Î$ œ (8)"Î$ œ 2 zÄ!
21. lim
3
22. lim
È5h 4 2 h
h Ä ! È3h 1 1
hÄ0
œ
5 È4 2
23. lim
œ
x5
# x Ä & x 25
24.
œ
3 È3(0) 1 1
œ lim
hÄ0
œ
3 È1 1
È5h 4 2 h
†
œ
3 2
È5h 4 2 È5h 4 2
œ lim
a5h 4b 4
h Ä 0 hŠÈ5h 4 2‹
œ lim
5h
h Ä 0 hŠÈ5h 4 2‹
œ lim
5 4 x5
œ lim
x3
x Ä & (x 5)(x 5)
lim # x Ä $ x 4x 3
œ lim
œ lim
x3
1
x Ä & x5
x Ä $ (x 3)(x 1)
œ lim
œ
" 55
œ
" 10
1
œ
" 3 1
x Ä $ x 1
5
h Ä 0 È5h 4 2
œ "2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2 3
48 25.
Chapter 2 Limits and Continuity x# 3x "0 x5
lim
x Ä &
(x 5)(x 2) x5
œ lim
x Ä &
(x 5)(x 2) x2
26. lim
x# 7x "0 x#
œ lim
27. lim
t# t 2 t# 1
t Ä " (t 1)(t 1)
xÄ#
tÄ"
t# 3t 2 # t Ä " t t 2
29.
lim $ # x Ä # x 2x
lim
2x 4
5y$ 8y#
1 x 1 x x Ä 1 1
32. lim
xÄ0
33. lim
1 xc1
u% "
y# (5y 8)
œ lim
uÄ1
œ lim
4x x# x Ä % 2 Èx
œ lim
36. lim
lim
x Ä "
œ
(v 2) av# 2v 4b
Èx 3
x Ä * ˆÈ x 3 ‰ ˆ È x 3 ‰
x1 x Ä 1 Èx 3 2
37. lim
œ
Èx# 12 4 x2 xÄ2
x Ä "
œ lim
xÄ2
(x 2)(x 2)
x2 x Ä 2 È x # 5 3
lim
x Ä 2
8 16
1 ‰ x1
œ lim
xÄ1
œ lim
uÄ1
au# "b (u 1) u# u 1
"
œ lim
(x 2)(x 2)
444 (4)(8)
œ
x1
x2
œ 2
4 3
12 32
œ
3 8
xÄ%
œ lim ŠÈx 3 #‹ œ È4 2 œ 4 xÄ1
ax # 8 b *
œ lim
x Ä 1 (x 1) ŠÈx# 8 $‹
2 33
œ
(x 2) ŠÈx# 12 4‹
x Ä 2 Èx# 12 4
œ
2 1
œ lim x ˆ2 Èx‰ œ 4(2 2) œ 16
(x 1) ˆÈx 3 #‰ (x 3) 4 xÄ1
x Ä 1 È x # ) $
œ
œ
" 6
œ lim
(x 1) ŠÈx# 8 $‹
œ "3
ax# 12b 16
œ lim
x Ä 2 (x 2) ŠÈx# 12 4‹
œ
4 È16 4
ax 2b ŠÈx# 5 3‹
È x# 5 3 x2 x Ä 2
œ lim
" È9 3
œ
x Ä * Èx 3
(1 1)(1 1) 111
œ
# v Ä # (v 2) av 4b
x ˆ2 È x ‰ ˆ 2 È x ‰ 2 Èx xÄ%
œ lim
œ
v# 2v 4
œ lim
œ lim
œ lim
2
x Ä 1 ax 1bax 1b
x Ä 2 ŠÈx# 5 3‹ ŠÈx# 5 3‹
ax 2b ŠÈx# 5 3‹
œ #"
œ lim x1 œ 1
ŠÈx# 12 4‹ ŠÈx# 12 4‹
x Ä 2 (x 2) ŠÈx# 12 4‹
œ
œ
ŠÈx# 8 $‹ ŠÈx# 8 $‹
lim
(x 1)(x 1) lim x Ä 1 (x 1) ŠÈx# ) $‹
lim
5y 8
(x 1) ˆÈx 3 2‰ È ˆ x 3 #‰ ˆ È x 3 #‰ xÄ1
39. lim
40.
œ 21
œ lim
È x# 8 3 x1
œ lim
2 4
xÄ1
# v Ä # (v 2)(v 2) av 4b
x(4 x) x Ä % 2 Èx
œ
2
œ 13
1 œ lim Š ax 12x bax 1b † x ‹ œ lim
au# "b (u 1)(u 1) au# u 1b (u 1)
œ lim
Èx 3 x9
xÄ*
x
xÄ1
œ lim
35. lim
b 1b b ax c 1b c 1bax b 1b
ax
3 #
1 2 1 2
# y Ä ! 3y 16
xÄ1
œ
œ
# x Ä # x
œ lim ˆ 1 x x † ax
œ lim
t2
t Ä " t 2
œ lim
1cx x
12 11
œ
œ lim
œ lim
x Ä 1 x1
% v Ä # v 16
t2
2(x 2)
v$ 8
34. lim
xÄ#
œ lim
œ lim
$ u Ä 1 u 1
38.
t Ä " (t 2)(t 1)
# # y Ä ! y a3y 16b
x b1 1 x
œ lim (x 5) œ 2 5 œ 3
t Ä " t1
(t 2)(t 1)
œ lim
x Ä &
œ lim
# x Ä # x (x 2)
% # y Ä 0 3y 16y
31. lim
(t 2)(t 1)
œ lim
28.
30. lim
xÄ#
œ lim (x 2) œ & # œ 7
œ
œ
œ lim
" 2
ax 2b ŠÈx# 5 3‹
x Ä 2
È9 3 4
ax # 5 b 9
œ 23
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.2 Limit of a Function and Limit Laws 41.
2 È x# 5 x3 x Ä 3
lim
œ
Š2 Èx# 5‹ Š2 Èx# 5‹
œ lim
(x 3) Š2 Èx# 5‹
x Ä 3
9 x#
lim
œ
x Ä 3 (x 3) Š2 Èx# 5‹
4x x Ä 4 5 È x# 9
œ lim
x Ä 4 Š5 Èx# 9‹ Š5 Èx# 9‹
a4 xb Š5 Èx# 9‹
xÄ4
x Ä 3 (x 3) Š2 Èx# 5‹
a4 xb Š5 Èx# 9‹
œ lim
42. lim
16 x#
x Ä 3 (x 3) Š2 Èx# 5‹
(3 x)(3 x)
lim
œ lim
(4 x)(4 x)
xÄ4
œ lim
3x
x Ä 3 2 È x # 5
œ
6 2 È4
œ
3 2
a4 xb Š5 Èx# 9‹ 25 ax# 9b
xÄ4
a4 xb Š5 Èx# 9‹
œ lim
4 ax # 5 b
œ lim
œ lim
xÄ4
5 È x# 9 4x
œ
5 È25 8
œ
5 4
2
43. lim a2sin x 1b œ 2sin 0 1 œ 0 1 œ 1
44. lim sin2 x œ Š lim sin x‹ œ asin 0b2 œ 02 œ 0
45. lim sec x œ
46. lim tan x œ
xÄ0
xÄ0
47. lim
xÄ0
1 x sin x 3cos x
1
lim x Ä 0 cos x œ
œ
1 0 sin 0 3cos 0
1 cos 0
œ
œ
1 1
xÄ0
œ1
100 3
œ
xÄ0
xÄ0
sin x
lim x Ä 0 cos x
œ
sin 0 cos 0
œ
0 1
œ0
1 3
48. lim ax2 1ba2 cos xb œ a02 1ba2 cos 0b œ a1ba2 1b œ a1ba1b œ 1 xÄ0
49. x Ä lim1Èx 4 cosax 1b œ x Ä lim1Èx 4 † x Ä lim1cosax 1b œ È1 4 † cos 0 œ È4 1 † 1 œ È4 1 50. lim È7 sec2 x œ É lim a7 sec2 xb œ É7 lim sec2 x œ È7 sec2 0 œ É7 a1b2 œ 2È2 xÄ0
xÄ0
xÄ0
51. (a) quotient rule (c) sum and constant multiple rules
(b) difference and power rules
52. (a) quotient rule (c) difference and constant multiple rules
(b) power and product rules
53. (a) xlim f(x) g(x) œ ’xlim f(x)“ ’ x lim g(x)“ œ (5)(2) œ 10 Äc Äc Äc (b) xlim 2f(x) g(x) œ 2 ’xlim f(x)“ ’ xlim g(x)“ œ 2(5)(2) œ 20 Äc Äc Äc (c) xlim [f(x) 3g(x)] œ xlim f(x) 3 xlim g(x) œ 5 3(2) œ 1 Äc Äc Äc lim f(x) f(x) 5 5 xÄc (d) xlim œ lim f(x) lim g(x) œ 5(2) œ 7 Ä c f(x) g(x) x
54. (a) (b) (c) (d) 55. (a) (b)
Äc
x
Äc
lim [g(x) 3] œ lim g(x) lim 3 œ $ $ œ !
xÄ%
xÄ%
xÄ%
lim xf(x) œ lim x † lim f(x) œ (4)(0) œ 0
xÄ%
xÄ%
xÄ%
#
lim [g(x)] œ ’ lim g(x)“ œ [3]# œ 9
xÄ%
#
g(x) x Ä % f(x) 1
lim
xÄ%
œ
Ä%
lim g(x)
x
lim f(x) lim 1
xÄ%
xÄ%
œ
3 01
œ3
lim [f(x) g(x)] œ lim f(x) lim g(x) œ 7 (3) œ 4
xÄb
xÄb
xÄb
lim f(x) † g(x) œ ’ lim f(x)“ ’ lim g(x)“ œ (7)(3) œ 21
xÄb
xÄb
xÄb
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
49
50
Chapter 2 Limits and Continuity lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(3) œ 12
(c)
xÄb
xÄb
xÄb
xÄb
xÄb
7 3
œ 37
lim [p(x) r(x) s(x)] œ lim p(x) lim r(x) lim s(x) œ 4 0 (3) œ 1
56. (a)
x Ä #
x Ä #
x Ä #
x Ä #
lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(3) œ 0
(b)
x Ä #
x Ä #
x Ä #
x Ä #
lim [4p(x) 5r(x)]/s(x) œ ’4 lim p(x) 5 lim r(x)“ ‚ lim s(x) œ [4(4) 5(0)]/3 œ
(c)
x Ä #
57. lim
hÄ!
x Ä #
(1 h)# 1# h
œ lim
hÄ!
(2 h)# (2)# h hÄ!
58. lim 59. lim
hÄ!
ˆ #" h ‰ ˆ "# ‰ h hÄ! È7 h È7 h hÄ!
61. lim
1 2h h# 1 h
œ lim
hÄ!
44hh# 4 h hÄ!
œ lim
[3(2 h) 4] [3(2) 4] h
60. lim
œ
xÄb
lim f(x)/g(x) œ lim f(x)/ lim g(x) œ
(d)
œ lim
2 2 h "
œ lim
h(2 h) h
œ lim
hÄ!
hÄ!
œ lim (h 4) œ 4 hÄ!
2 (2 h) 2h(# h)
hÄ!
h ŠÈ7 h È7‹
h
œ lim
œ "4
h Ä ! h(4 2h)
œ lim
(7 h) 7
h Ä ! h ŠÈ7 h È7‹
œ lim
h
h Ä ! h ŠÈ7hÈ7‹
œ lim
È3(0 h) 1 È3(0) 1 h hÄ!
œ lim
3
h Ä ! È3h 1 1
œ
œ lim
ŠÈ3h 1 "‹ ŠÈ3h 1 "‹ h ŠÈ3h 1 "‹
hÄ!
(3h 1) "
œ lim
h Ä ! h ŠÈ3h 1 1 ‹
œ lim
3h
h Ä ! h ŠÈ3h 1 "‹
3 #
63. lim È5 2x# œ È5 2(0)# œ È5 and lim È5 x# œ È5 (0)# œ È5; by the sandwich theorem, xÄ!
xÄ!
lim f(x) œ È5
xÄ!
64. lim a2 x# b œ 2 0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2 xÄ!
65. (a)
xÄ!
lim Š1
xÄ!
x# 6‹
œ1
0 6
xÄ!
œ 1 and lim 1 œ 1; by the sandwich theorem, lim
(b) For x Á 0, y œ (x sin x)/(2 2 cos x) lies between the other two graphs in the figure, and the graphs converge as x Ä 0.
66. (a)
lim Š "#
xÄ!
lim
xÄ!
1cos x x#
x# 24 ‹
œ lim
1
xÄ! #
lim
x#
x Ä ! #4
œ
" #
x sin x
x Ä ! 22 cos x
xÄ!
0œ
" #
and lim
"
xÄ! #
œ1
œ "# ; by the sandwich theorem,
œ "# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
h Ä ! È 7 h È 7
" #È 7
62. lim
"6 3
œ lim (2 h) œ 2
h(h 4) h
ŠÈ7 h È7‹ ŠÈ7 h È7‹
hÄ!
x Ä #
œ3
œ lim
2h
hÄ!
œ lim
3h
hÄ! h
x Ä #
Section 2.2 Limit of a Function and Limit Laws (b) For all x Á 0, the graph of f(x) œ (1 cos x)/x# lies between the line y œ "# and the parabola yœ
" #
x# /24, and the graphs converge as x Ä 0.
67. (a) f(x) œ ax# *b/(x 3) x 3.1 f(x) 6.1 2.9 5.9
x f(x)
3.01 6.01
3.001 6.001
3.0001 6.0001
3.00001 6.00001
3.000001 6.000001
2.99 5.99
2.999 5.999
2.9999 5.9999
2.99999 5.99999
2.999999 5.999999
The estimate is lim f(x) œ 6. x Ä $
(b)
(c) f(x) œ
x# 9 x3
œ
(x 3)(x 3) x3
œ x 3 if x Á 3, and lim (x 3) œ 3 3 œ 6. x Ä $
68. (a) g(x) œ ax# #b/ Šx È2‹ x g(x)
1.4 2.81421
1.41 2.82421
1.414 2.82821
1.4142 2.828413
1.41421 2.828423
1.414213 2.828426
(b)
(c) g(x) œ
x# 2 x È2
œ
Šx È2‹ Šx È2‹ Šx È2‹
œ x È2 if x Á È2, and
69. (a) G(x) œ (x 6)/ ax# 4x 12b x 5.9 5.99 G(x) .126582 .1251564 x G(x)
6.1 .123456
6.01 .124843
5.999 .1250156 6.001 .124984
lim
x Ä È#
5.9999 .1250015 6.0001 .124998
Šx È2‹ œ È2 È2 œ 2È2.
5.99999 .1250001 6.00001 .124999
5.999999 .1250000 6.000001 .124999
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
51
52
Chapter 2 Limits and Continuity (b)
(c) G(x) œ
x6 ax# 4x 12b
œ
x6 (x 6)(x 2)
œ
" x#
70. (a) h(x) œ ax# 2x 3b / ax# 4x 3b x 2.9 2.99 h(x) 2.052631 2.005025 x h(x)
3.1 1.952380
3.01 1.995024
"
if x Á 6, and lim
x Ä ' x 2
œ
" ' 2
œ "8 œ 0.125.
2.999 2.000500
2.9999 2.000050
2.99999 2.000005
2.999999 2.0000005
3.001 1.999500
3.0001 1.999950
3.00001 1.999995
3.000001 1.999999
(b)
(c) h(x) œ
x# 2x 3 x# 4x 3
œ
(x 3)(x 1) (x 3)(x 1)
œ
x1 x1
71. (a) f(x) œ ax# 1b / akxk 1b x 1.1 1.01 f(x) 2.1 2.01 .9 1.9
x f(x)
.99 1.99
if x Á 3, and lim
x1
x Ä $ x1
œ
31 31
œ
4 #
œ 2.
1.001 2.001
1.0001 2.0001
1.00001 2.00001
1.000001 2.000001
.999 1.999
.9999 1.9999
.99999 1.99999
.999999 1.999999
(b)
(c) f(x) œ
x# " kx k 1
(x 1)(x 1)
1 œ (x x1)(x 1) (x 1)
œ x 1, x 0 and x Á 1 , and lim (1 x) œ 1 (1) œ 2. x Ä 1 œ 1 x, x 0 and x Á 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.2 Limit of a Function and Limit Laws 72. (a) F(x) œ ax# 3x 2b / a2 kxkb x 2.1 2.01 F(x) 1.1 1.01 1.9 .9
x F(x)
1.99 .99
2.001 1.001
2.0001 1.0001
2.00001 1.00001
2.000001 1.000001
1.999 .999
1.9999 .9999
1.99999 .99999
1.999999 .999999
(b)
(c) F(x) œ
x# 3x 2 2 kx k
(x 2)(x 1)
œ (x 2)(x# x") 2x
73. (a) g()) œ (sin ))/) ) .1 g()) .998334
, x 0 , and lim (x 1) œ 2 1 œ 1. x Ä # œ x 1, x 0 and x Á 2
.01 .999983
.001 .999999
.0001 .999999
.00001 .999999
.000001 .999999
.1 .998334
.01 .999983
.001 .999999
.0001 .999999
.00001 .999999
.000001 .999999
74. (a) G(t) œ (1 cos t)/t# t .1 G(t) .499583
.01 .499995
.001 .499999
.0001 .5
.00001 .5
.000001 .5
.1 .499583
.01 .499995
.001 .499999
.0001 .5
.00001 .5
.000001 .5
) g()) lim g()) œ 1
)Ä!
(b)
t G(t)
lim G(t) œ 0.5
tÄ!
(b)
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53
54
Chapter 2 Limits and Continuity
75. xlim f(x) exists at those points c where xlim x% œ xlim x# . Thus, c% œ c# Ê c# a1 c# b œ 0 Ê c œ 0, 1, or 1. Äc Äc Äc Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1. xÄ!
xÄ!
x Ä 1
xÄ1
76. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f(x) œ 5 Á 0. xÄ#
lim f(x) lim 5 % xÄ% œ xÄlim x lim 2 œ
f(x)5 x Ä % x 2
77. 1 œ lim
xÄ%
lim f(x) 5
xÄ%
%#
xÄ%
xÄ%
78. (a) 1 œ lim
f(x) x#
lim f(x) lim f(x) œ xÄlim# x# œ xÄ %# Ê
(b) 1 œ lim
f(x) x#
œ ’ lim
x Ä # x Ä #
xÄ
#
x Ä #
79. (a) 0 œ 3 † 0 œ ’ lim
xÄ#
Ê lim f(x) 5 œ 2(1) Ê lim f(x) œ 2 5 œ 7. xÄ%
lim f(x) œ 4.
x Ä #
f(x) lim x" “ x “ ’x Ä #
œ ’ lim
x Ä #
f(x) ˆ " ‰ x “ #
Ê
lim
x Ä #
f(x) x
œ 2.
f(x) 5 x # “ ’xlim Ä#
5 (x 2)“ œ lim ’Š f(x) x # ‹ (x 2)“ œ lim [f(x) 5] œ lim f(x) 5
f(x) 5 x # “ ’xlim Ä#
(x 2)“ Ê lim f(x) œ 5 as in part (a).
xÄ#
Ê lim f(x) œ 5.
xÄ#
xÄ#
xÄ#
(b) 0 œ 4 † 0 œ ’ lim
xÄ#
80. (a) 0 œ 1 † 0 œ ’ lim
f(x) # “ ’ lim xÄ! x xÄ!
(b) 0 œ 1 † 0 œ 81. (a)
lim x sin
xÄ!
(b) 1 Ÿ sin
82. (a)
" x
’ lim f(x) # “ ’ lim xÄ! x xÄ!
" x
xÄ#
#
x“ œ ’ lim
f(x)
# xÄ! x
x“ œ
lim ’ f(x) x# xÄ!
# “ ’ lim x# “ œ lim ’ f(x) x# † x “ œ lim f(x). That is, lim f(x) œ 0.
xÄ!
† x“ œ
xÄ!
lim f(x) . xÄ! x
That is,
xÄ!
lim f(x) xÄ! x
œ 0.
œ0
Ÿ 1 for x Á 0:
x 0 Ê x Ÿ x sin
" x
Ÿ x Ê lim x sin
" x
œ 0 by the sandwich theorem;
x 0 Ê x x sin
" x
x Ê lim x sin
" x
œ 0 by the sandwich theorem.
xÄ! xÄ!
lim x# cos ˆ x"$ ‰ œ 0
xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
xÄ!
Section 2.3 The Precise Definition of a Limit (b) 1 Ÿ cos ˆ x"$ ‰ Ÿ 1 for x Á 0 Ê x# Ÿ x# cos ˆ x"$ ‰ Ÿ x# Ê lim x# cos ˆ x"$ ‰ œ 0 by the sandwich xÄ!
theorem since lim x# œ 0. xÄ!
83-88. Example CAS commands: Maple: f := x -> (x^4 16)/(x 2); x0 := 2; plot( f(x), x œ x0-1..x0+1, color œ black, title œ "Section 2.2, #83(a)" ); limit( f(x), x œ x0 ); In Exercise 85, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be overcome in Maple by entering the function as f := x -> (surd(x+1, 3) 1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f, x] f[x_]:=(x3 x2 5x 3)/(x 1)2 x0= 1; h = 0.1; Plot[f[x],{x, x0 h, x0 h}] Limit[f[x], x Ä x0] 2.3 THE PRECISE DEFINITION OF A LIMIT 1. Step 1: Step 2:
kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5 $ 5 œ 7 Ê $ œ 2, or $ 5 œ 1 Ê $ œ 4. The value of $ which assures kx 5k $ Ê 1 x 7 is the smaller value, $ œ 2.
Step 1: Step 2:
kx 2k $ Ê $ x 2 $ Ê $ # x $ 2 $ 2 œ 1 Ê $ œ 1, or $ 2 œ 7 Ê $ œ 5. The value of $ which assures kx 2k $ Ê 1 x 7 is the smaller value, $ œ 1.
Step 1: Step 2:
kx (3)k $ Ê $ x $ $ Ê $ 3 x $ 3 $ 3 œ 7# Ê $ œ "# , or $ $ œ "# Ê $ œ 5# .
2.
3.
The value of $ which assures kx (3)k $ Ê 7# x "# is the smaller value, $ œ "# . 4.
Step 1:
¸x ˆ 3# ‰¸ $ Ê $ x
3 #
$ Ê $
3 #
x$
3 #
Step 2:
œ 7# Ê $ œ #, or $ 3# œ "# Ê $ œ 1. The value of $ which assures ¸x ˆ 3# ‰¸ $ Ê 7# x "# is the smaller value, $ œ ".
Step 1:
¸x "# ¸ $ Ê $ x
$
3 #
5. " #
$ Ê $
" #
x$
" #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
55
56
Chapter 2 Limits and Continuity Step 2:
" or $ #" œ 47 Ê $ œ 14 . "¸ 4 ¸ The value of $ which assures x # $ Ê 9 x
$
" #
œ
4 9
Ê $œ
" 18 ,
4 7
is the smaller value, $ œ
" 18 .
6.
Step 1: Step 2:
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3 $ $ œ 2.7591 Ê $ œ 0.2409, or $ $ œ 3.2391 Ê $ œ 0.2391. The value of $ which assures kx 3k $ Ê 2.7591 x 3.2391 is the smaller value, $ œ 0.2391.
7. Step 1: Step 2:
kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5 From the graph, $ 5 œ 4.9 Ê $ œ 0.1, or $ 5 œ 5.1 Ê $ œ 0.1; thus $ œ 0.1 in either case.
8. Step 1: Step 2:
kx (3)k $ Ê $ x 3 $ Ê $ 3 x $ 3 From the graph, $ 3 œ 3.1 Ê $ œ 0.1, or $ 3 œ 2.9 Ê $ œ 0.1; thus $ œ 0.1.
9. Step 1: Step 2:
kx 1k $ Ê $ x 1 $ Ê $ 1 x $ 1 9 7 From the graph, $ 1 œ 16 Ê $ œ 16 , or $ 1 œ 25 16 Ê $ œ
10. Step 1: Step 2:
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3 From the graph, $ 3 œ 2.61 Ê $ œ 0.39, or $ 3 œ 3.41 Ê $ œ 0.41; thus $ œ 0.39.
11. Step 1:
kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2 From the graph, $ 2 œ È3 Ê $ œ 2 È3 ¸ 0.2679, or $ 2 œ È5 Ê $ œ È5 2 ¸ 0.2361; thus $ œ È5 2.
Step 2:
12. Step 1: Step 2:
9 16 ;
thus $ œ
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1 From the graph, $ 1 œ thus $ œ
È5 2 # .
È5 #
Ê $œ
È5 2 #
¸ 0.1180, or $ 1 œ
13. Step 1: Step 2:
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1 7 16 From the graph, $ 1 œ 16 9 Ê $ œ 9 ¸ 0.77, or $ 1 œ 25 Ê
14. Step 1:
¸x "# ¸ $ Ê $ x
Step 2:
7 16 .
From the graph, $ thus $ œ 0.00248.
" #
œ
" # 1 2.01
$ Ê $ Ê $œ
1 2
" #
x$
" #.01
" #
¸ 0.00248, or $
" #
œ
È3 #
9 25
Ê $œ
2 È3 #
œ 0.36; thus $ œ
1 1.99
Ê $œ
1 1.99
¸ 0.1340;
9 25
" #
œ 0.36.
¸ 0.00251;
15. Step 1: Step 2:
k(x 1) 5k 0.01 Ê kx 4k 0.01 Ê 0.01 x 4 0.01 Ê 3.99 x 4.01 kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4 Ê $ œ 0.01.
16. Step 1:
k(2x 2) (6)k 0.02 Ê k2x 4k 0.02 Ê 0.02 2x 4 0.02 Ê 4.02 2x 3.98 Ê 2.01 x 1.99 kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2 Ê $ œ 0.01.
Step 2: 17. Step 1: Step 2:
¹Èx 1 "¹ 0.1 Ê 0.1 Èx 1 " 0.1 Ê 0.9 Èx 1 1.1 Ê 0.81 x 1 1.21 Ê 0.19 x 0.21 kx 0k $ Ê $ x $ . Then, $ œ !Þ"* Ê $ œ !Þ"* or $ œ !Þ#"; thus, $ œ 0.19.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.3 The Precise Definition of a Limit 18. Step 1: Step 2:
¸Èx "# ¸ 0.1 Ê 0.1 Èx "# 0.1 Ê 0.4 Èx 0.6 Ê 0.16 x 0.36 ¸x "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" . Then, $
19. Step 1: Step 2:
20. Step 1: Step 2:
21. Step 1: Step 2:
22. Step 1: Step 2:
57
" 4
œ 0.16 Ê $ œ 0.09 or $
" 4
œ 0.36 Ê $ œ 0.11; thus $ œ 0.09.
¹È19 x $¹ " Ê " È19 x $ 1 Ê 2 È19 x % Ê 4 19 x 16 Ê % x 19 16 Ê 15 x 3 or 3 x 15 kx 10k $ Ê $ x 10 $ Ê $ 10 x $ 10. Then $ 10 œ 3 Ê $ œ 7, or $ 10 œ 15 Ê $ œ 5; thus $ œ 5. ¹Èx 7 4¹ 1 Ê " Èx 7 % 1 Ê 3 Èx 7 5 Ê 9 x 7 25 Ê 16 x 32 kx 23k $ Ê $ x 23 $ Ê $ 23 x $ 23. Then $ 23 œ 16 Ê $ œ 7, or $ 23 œ 32 Ê $ œ 9; thus $ œ 7. ¸ "x 4" ¸ 0.05 Ê 0.05
" x
" 4
0.05 Ê 0.2
" x
0.3 Ê
kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4. 2 2 Then $ % œ 10 3 or $ œ 3 , or $ 4 œ 5 or $ œ 1; thus $ œ 3 .
10 #
x
10 3
or
10 3
x 5.
kx# 3k !.1 Ê 0.1 x# 3 0.1 Ê 2.9 x# 3.1 Ê È2.9 x È3.1 ¹x È3¹ $ Ê $ x È3 $ Ê $ È3 x $ È3. Then $ È3 œ È2.9 Ê $ œ È3 È2.9 ¸ 0.0291, or $ È3 œ È3.1 Ê $ œ È3.1 È3 ¸ 0.0286; thus $ œ 0.0286.
23. Step 1: Step 2:
kx# 4k 0.5 Ê 0.5 x# 4 0.5 Ê 3.5 x# 4.5 Ê È3.5 kxk È4.5 Ê È4.5 x È3.5, for x near 2. kx (2)k $ Ê $ x 2 $ Ê $ # x $ 2. Then $ # œ È4.5 Ê $ œ È4.5 # ¸ 0.1213, or $ # œ È3.5 Ê $ œ # È3.5 ¸ 0.1292; thus $ œ È4.5 2 ¸ 0.12.
24. Step 1: Step 2:
25. Step 1: Step 2:
¸ "x (1)¸ 0.1 Ê 0.1
" x
1 0.1 Ê 11 10
" x
9 10 10 10 10 Ê 10 11 x 9 or 9 x 11 .
kx (1)k $ Ê $ x 1 $ Ê $ " x $ ". " 10 " Then $ " œ 10 9 Ê $ œ 9 , or $ " œ 11 Ê $ œ 11 ; thus $ œ
" 11 .
kax# 5b 11k " Ê kx# 16k 1 Ê " x# 16 1 Ê 15 x# 17 Ê È15 x È17. kx 4k $ Ê $ x 4 $ Ê $ % x $ %. Then $ % œ È15 Ê $ œ % È15 ¸ 0.1270, or $ % œ È17 Ê $ œ È17 % ¸ 0.1231; thus $ œ È17 4 ¸ 0.12.
26. Step 1: Step 2:
27. Step 1: Step 2:
¸ 120 ¸ x 5 " Ê "
120 x
&1 Ê 4
120 x
6 Ê
" 4
x 120
" 6
Ê 30 x 20 or 20 x 30.
kx 24k $ Ê $ x 24 $ Ê $ 24 x $ 24. Then $ 24 œ 20 Ê $ œ 4, or $ 24 œ 30 Ê $ œ 6; thus Ê $ œ 4. kmx 2mk 0.03 Ê 0.03 mx 2m 0.03 Ê 0.03 2m mx 0.03 2m Ê 0.03 2 0.03 m x2 m . kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2. 0.03 0.03 Then $ 2 œ 2 0.03 m Ê $ œ m , or $ 2 œ # m Ê $ œ
0.03 m .
In either case, $ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
0.03 m .
58
Chapter 2 Limits and Continuity kmx 3mk c Ê c mx 3m c Ê c 3m mx c 3m Ê 3
28. Step 1:
kx 3k $ Ê $ x 3 $ Ê $ $ B $ $. Then $ $ œ $ mc Ê $ œ mc , or $ $ œ $ mc Ê $ œ
Step 2:
¸(mx b) ˆ m# b‰¸ - Ê c mx m# c Ê c ¸x "# ¸ $ Ê $ x "# $ Ê $ "# x $ "# .
29. Step 1: Step 2:
Then $
" #
œ
" #
c m
Ê $œ
c m,
or $
" #
œ
" #
c m
c m. m #
Ê $œ
c m
x 3
In either case, $ œ
c m.
c m. " #
In either case, $ œ
c m.
mx c
m #
Ê
c m
c m
x
" #
c m.
k(mx b) (m b)k 0.05 Ê 0.05 mx m 0.05 Ê 0.05 m mx 0.05 m 0.05 Ê 1 0.05 m x" m .
30. Step 1:
kx 1k $ Ê $ x 1 $ Ê $ " x $ ". 0.05 0.05 Then $ " œ " 0.05 m Ê $ œ m , or $ " œ " m Ê $ œ
Step 2:
0.05 m .
In either case, $ œ
0.05 m .
31. lim (3 2x) œ 3 2(3) œ 3 xÄ3
ka3 2xb (3)k 0.02 Ê 0.02 6 2x 0.02 Ê 6.02 2x 5.98 Ê 3.01 x 2.99 or 2.99 x 3.01. 0 k x 3k $ Ê $ x 3 $ Ê $ $ x $ $ . Then $ $ œ 2.99 Ê $ œ 0.01, or $ $ œ 3.01 Ê $ œ 0.01; thus $ œ 0.01.
Step 1: Step 2:
32.
lim (3x #) œ (3)(1) 2 œ 1
x Ä 1
k(3x 2) 1k 0.03 Ê 0.03 3x 3 0.03 Ê 0.01 x 1 0.01 Ê 1.01 x 0.99. kx (1)k $ Ê $ x 1 $ Ê $ " x $ 1. Then $ " œ 1.01 Ê $ œ 0.01, or $ " œ 0.99 Ê $ œ 0.01; thus $ œ 0.01.
Step 1: Step 2:
33. lim
x# 4
x Ä # x#
34.
35.
œ lim
xÄ#
#
(x 2)(x 2) (x 2)
œ lim (x 2) œ # # œ 4, x Á 2 xÄ#
(x 2)(x 2) (x 2)
Step 1:
¹Š xx 24 ‹
Step 2:
Ê 1.95 x 2.05, x Á 2. kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2. Then $ 2 œ 1.95 Ê $ œ 0.05, or $ 2 œ 2.05 Ê $ œ 0.05; thus $ œ 0.05.
lim
x Ä &
x# 6x 5 x5
4¹ 0.05 Ê 0.05
œ lim
x Ä &
(x 5)(x 1) (x 5)
% 0.05 Ê 3.95 x 2 4.05, x Á 2
œ lim (x 1) œ 4, x Á 5. x Ä &
(x 5)(x ") (x 5)
Step 1:
# ¹Š x x 6x5 5 ‹
Step 2:
Ê 5.05 x 4.95, x Á 5. kx (5)k $ Ê $ x 5 $ Ê $ & x $ &. Then $ & œ 5.05 Ê $ œ 0.05, or $ & œ 4.95 Ê $ œ 0.05; thus $ œ 0.05.
(4)¹ 0.05 Ê 0.05
4 0.05 Ê 4.05 x 1 3.95, x Á 5
lim È1 5x œ È1 5(3) œ È16 œ 4
x Ä $
Step 1:
¹È1 5x 4¹ 0.5 Ê 0.5 È1 5x 4 0.5 Ê 3.5 È1 5x 4.5 Ê 12.25 1 5x 20.25
Step 2:
Ê 11.25 5x 19.25 Ê 3.85 x 2.25. kx (3)k $ Ê $ x 3 $ Ê $ $ x $ $. Then $ $ œ 3.85 Ê $ œ 0.85, or $ $ œ 2.25 Ê 0.75; thus $ œ 0.75.
36. lim
4
xÄ# x
Step 1:
œ
4 #
œ2
¸ 4x 2¸ 0.4 Ê 0.4
4 x
2 0.4 Ê 1.6
4 x
2.4 Ê
10 16
x 4
10 24
Ê
10 4
x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
10 6
or
5 3
x 25 .
Section 2.3 The Precise Definition of a Limit Step 2:
59
kx 2k $ Ê $ x 2 $ Ê $ # x $ #. Then $ # œ 53 Ê $ œ "3 , or $ # œ #5 Ê $ œ "# ; thus $ œ 3" .
37. Step 1: Step 2:
k(9 x) 5k % Ê % 4 x % Ê % 4 x % 4 Ê % % x 4 % Ê % % x 4 %. kx 4k $ Ê $ x 4 $ Ê $ % x $ %. Then $ 4 œ % 4 Ê $ œ %, or $ % œ % % Ê $ œ %. Thus choose $ œ %.
38. Step 1:
k(3x 7) 2k % Ê % 3x 9 % Ê 9 % 3x * % Ê 3
Step 2:
39. Step 1: Step 2:
40. Step 1: Step 2:
41. Step 1: Step 2:
42. Step 1: Step 2:
43. Step 1: Step 2:
x 3 3% .
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3. Then $ 3 œ $ 3% Ê $ œ 3% , or $ 3 œ 3 3% Ê $ œ 3% . Thus choose $ œ 3% . ¹Èx 5 2¹ % Ê % Èx 5 2 % Ê 2 % Èx 5 2 % Ê (2 %)# x 5 (2 %)# Ê (2 %)# & x (2 %)# 5. kx 9k $ Ê $ x 9 $ Ê $ 9 x $ 9. Then $ * œ %# %% * Ê $ œ %% %# , or $ * œ %# %% * Ê $ œ %% %# . Thus choose the smaller distance, $ œ %% %# . ¹È4 x 2¹ % Ê % È4 x 2 % Ê 2 % È4 x 2 % Ê (2 %)# % x (2 %)# Ê (2 %)# x 4 (2 %)# Ê (2 %)# % x (2 %)# %. k x 0k $ Ê $ x $ . Then $ œ (2 %)# 4 œ %# %% Ê $ œ %% %# , or $ œ (2 %)# 4 œ 4% %# . Thus choose the smaller distance, $ œ 4% %# . For x Á 1, kx# 1k % Ê % x# " % Ê " % x# " % Ê È1 % kxk È1 % Ê È" % x È1 % near B œ ". kx 1k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ È1 % Ê $ œ " È1 %, or $ 1 œ È" % Ê $ œ È" % 1. Choose $ œ min š" È1 %ß È1 % "›, that is, the smaller of the two distances. For x Á 2, kx# 4k % Ê % x# 4 % Ê 4 % x# 4 % Ê È4 % kxk È4 % Ê È4 % x È4 % near B œ 2. kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2. Then $ 2 œ È% % Ê $ œ È% % #, or $ # œ È% % Ê $ œ # È% %. Choose $ œ min šÈ% % #ß # È% %› . ¸ "x 1¸ % Ê %
" x
"% Ê "%
" x
"% Ê
" 1%
% "%,
" 1%.
x
kx 1k $ Ê $ x 1 $ Ê " $ x " $ . Then " $ œ " " % Ê $ œ " " " % œ " % % , or " $ œ " " % Ê $ œ Choose $ œ
44. Step 1:
% 3
" "%
"œ
% "%.
the smaller of the two distances.
¸ x"# "3 ¸ % Ê %
" x#
" 3
% Ê
" 3
%
" x#
" 3
% Ê
1 3% 3
" x#
1 $% 3
3 È $. Ê É 1 3 $% kxk É " 3 $% , or É " 3 $% x É "$ % for x near
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ê
3 "
%$ x#
3 "
%$ 60
Chapter 2 Limits and Continuity Step 2:
¹x È3¹ $ Ê $ x È3 $ Ê È3 $ x È3 $ . Then È3 $ œ É " 3 $% Ê $ œ È3 É " 3 $% , or È3 $ œ É " 3 $% Ê $ œ É " 3 $% È3. Choose $ œ min šÈ3 É " 3 $% ß É " 3 $% È3›.
45. Step 1: Step 2:
46. Step 1: Step 2:
47. Step 1:
#
¹Š xx 3* ‹ (6)¹ % Ê % (x 3) 6 %, x Á 3 Ê % x 3 % Ê % $ x % $. kx (3)k $ Ê $ x 3 $ Ê $ $ x $ 3. Then $ $ œ % $ Ê $ œ %, or $ $ œ % $ Ê $ œ %. Choose $ œ %. #
¹Š xx 11 ‹ 2¹ % Ê % (x 1) 2 %, x Á 1 Ê " % x " %. kx 1k $ Ê $ x 1 $ Ê " $ x " $ . Then " $ œ " % Ê $ œ %, or " $ œ " % Ê $ œ %. Choose $ œ %. x 1: l(4 2x) 2l % Ê ! 2 2x % since x 1Þ Thus, 1
% #
x !;
x 1: l(6x 4) 2l % Ê ! Ÿ 6x 6 % since x 1. Thus, " Ÿ x 1 6% . Step 2:
48. Step 1: Step 2:
kx 1k $ Ê $ x 1 $ Ê " $ x 1 $ . Then 1 $ œ " #% Ê $ œ #% , or " $ œ 1 6% Ê $ œ 6% . Choose $ œ 6% . x !: k2x 0k % Ê % 2x ! Ê #% x 0; x 0: ¸ x# !¸ % Ê ! Ÿ x #%.
k x 0k $ Ê $ x $ . Then $ œ #% Ê $ œ #% , or $ œ #% Ê $ œ #%. Choose $ œ #% .
49. By the figure, x Ÿ x sin
" x
Ÿ x for all x 0 and x x sin
then by the sandwich theorem, in either case, lim x sin xÄ!
50. By the figure, x# Ÿ x# sin
" x
" x
" x
x for x 0. Since lim (x) œ lim x œ 0, xÄ!
œ 0.
xÄ!
Ÿ x# for all x except possibly at x œ 0. Since lim ax# b œ lim x# œ 0, then
by the sandwich theorem, lim x# sin xÄ!
" x
xÄ!
œ 0.
xÄ!
51. As x approaches the value 0, the values of g(x) approach k. Thus for every number % 0, there exists a $ ! such that ! kx 0k $ Ê kg(x) kk %. 52. Write x œ h c. Then ! lx cl $ Í $ x c $ , x Á c Í $ ah cb c $ , h c Á c Í $ h $ , h Á ! Í ! lh !l $ . Thus, limfaxb œ L Í for any % !, there exists $ ! such that lfaxb Ll % whenever ! lx cl $ x Äc
Í lfah cb Ll % whenever ! lh !l $ Í lim fah cb œ L. hÄ!
53. Let f(x) œ x# . The function values do get closer to 1 as x approaches 0, but lim f(x) œ 0, not 1. The function f(x) œ x# never gets arbitrarily close to 1 for x near 0.
xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.3 The Precise Definition of a Limit
61
54. Let f(x) œ sin x, L œ "# , and x! œ 0. There exists a value of x (namely, x œ 16 ) for which ¸sin x "# ¸ % for any given
% 0. However, lim sin x œ 0, not "# . The wrong statement does not require x to be arbitrarily close to x! . As another\ xÄ!
example, let g(x) œ sin "x , L œ #" , and x! œ 0. We can choose infinitely many values of x near 0 such that sin you can see from the accompanying figure. However, lim sin xÄ!
" x
" x
œ
" #
as
fails to exist. The wrong statement does not require all
values of x arbitrarily close to x! œ 0 to lie within % 0 of L œ "# . Again you can see from the figure that there are also infinitely many values of x near 0 such that sin "x œ 0. If we choose % ¸sin "x #" ¸ % for all values of x sufficiently near x! œ 0.
#
55. kA *k Ÿ 0.01 Ê 0.01 Ÿ 1 ˆ x# ‰ 9 Ÿ 0.01 Ê 8.99 Ÿ Ê
2É 8.99 1
ŸxŸ
2É 9.01 1
1 x# 4
" 4
we cannot satisfy the inequality
Ÿ 9.01 Ê
4 1
(8.99) Ÿ x# Ÿ
4 1
(9.01)
or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right
endpoint was rounded down. 56. V œ RI Ê (120)(10) 51
V R
ŸRŸ
œ I Ê ¸ VR 5¸ Ÿ 0.1 Ê 0.1 Ÿ (120)(10) 49
120 R
5 Ÿ 0.1 Ê 4.9 Ÿ
120 R
Ÿ 5.1 Ê
10 49
R 1#0
10 51
Ê
Ê 23.53 Ÿ R Ÿ 24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 57. (a) $ x 1 0 Ê " $ x 1 Ê f(x) œ x. Then kf(x) 2k œ kx 2k œ 2 x 2 1 œ 1. That is, kf(x) 2k 1 "# no matter how small $ is taken when " $ x 1 Ê lim f(x) Á 2. xÄ1
(b) 0 x 1 $ Ê " x " $ Ê f(x) œ x 1. Then kf(x) 1k œ k(x 1) 1k œ kxk œ x 1. That is, kf(x) 1k 1 no matter how small $ is taken when " x " $ Ê lim f(x) Á 1. xÄ1
(c) $ x 1 ! Ê " $ x 1 Ê f(x) œ x. Then kf(x) 1.5k œ kx 1.5k œ 1.5 x 1.5 1 œ 0.5. Also, ! x 1 $ Ê 1 x " $ Ê f(x) œ x 1. Then kf(x) 1.5k œ k(x 1) 1.5k œ kx 0.5k œ x 0.5 " 0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that $ x 1 $ but kf(x) 1.5k "# Ê lim f(x) Á 1.5. xÄ1
58. (a) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 4k œ 2. Thus for % 2, kh(x) 4k % whenever 2 x 2 $ no matter how small we choose $ 0 Ê lim h(x) Á 4. xÄ#
(b) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 3k œ 1. Thus for % 1, kh(x) 3k % whenever 2 x 2 $ no matter how small we choose $ 0 Ê lim h(x) Á 3. xÄ#
(c) For 2 $ x 2 Ê h(x) œ x# so kh(x) 2k œ kx# 2k . No matter how small $ 0 is chosen, x# is close to 4 when x is near 2 and to the left on the real line Ê kx# 2k will be close to 2. Thus if % 1, kh(x) 2k % whenever 2 $ x 2 no mater how small we choose $ 0 Ê lim h(x) Á 2. xÄ#
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62
Chapter 2 Limits and Continuity
59. (a) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 4k 0.8. Thus for % 0.8, kf(x) 4k % whenever 3 $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 4. xÄ$
(b) For 3 x 3 $ Ê f(x) 3 Ê kf(x) 4.8k 1.8. Thus for % 1.8, kf(x) 4.8k % whenever 3 x 3 $ no matter how small we choose $ 0 Ê lim f(x) Á 4.8. xÄ$
(c) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 3k 1.8. Again, for % 1.8, kf(x) 3k % whenever $ $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 3. xÄ$
60. (a) No matter how small we choose $ 0, for x near 1 satisfying " $ x " $ , the values of g(x) are near 1 Ê kg(x) 2k is near 1. Then, for % œ "# we have kg(x) 2k "# for some x satisfying " $ x " $ , or ! kx 1k $ Ê
lim g(x) Á 2.
x Ä 1
(b) Yes, lim g(x) œ 1 because from the graph we can find a $ ! such that kg(x) 1k % if ! kx (1)k $ . x Ä 1
61-66. Example CAS commands (values of del may vary for a specified eps): Maple: f := x -> (x^4-81)/(x-3);x0 := 3; plot( f(x), x=x0-1..x0+1, color=black, # (a) title="Section 2.3, #61(a)" ); L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q); plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e) q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 ); delta := abs(x0-q); head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta, color=black, linestyle=[1,3,3], title=head )); end do: Mathematica (assigned function and values for x0, eps and del may vary): Clear[f, x] y1: œ L eps; y2: œ L eps; x0 œ 1; f[x_]: œ (3x2 (7x 1)Sqrt[x] 5)/(x 1) Plot[f[x], {x, x0 0.2, x0 0.2}] L: œ Limit[f[x], x Ä x0] eps œ 0.1; del œ 0.2; Plot[{f[x], y1, y2},{x, x0 del, x0 del}, PlotRange Ä {L 2eps, L 2eps}]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.4 One-Sided Limits 2.4 ONE-SIDED LIMITS 1. (a) True (e) True (i) False
(b) True (f) True (j) False
(c) False (g) False (k) True
(d) True (h) False (l) False
2. (a) True (e) True (i) True
(b) False (f) True (j) False
(c) False (g) True (k) True
(d) True (h) True
3. (a)
lim f(x) œ
x Ä #b
2 #
" œ #, lim c f(x) œ $ # œ " xÄ#
(b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 4# 1 œ 3, lim b f(x) œ 4# " œ $ xÄ%
xÄ%
(d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x) xÄ% xÄ% xÄ% 4. (a)
lim f(x) œ
x Ä #b
2 #
œ 1, lim c f(x) œ $ # œ ", f(2) œ 2 xÄ#
(b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 3 (1) œ 4, lim b f(x) œ 3 (1) œ 4 x Ä "
x Ä "
(d) Yes, lim f(x) œ 4 because 4 œ x Ä "
lim
x Ä "c
f(x) œ
lim
x Ä "b
f(x)
5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0 xÄ! (b) lim c f(x) œ lim c 0 œ 0 xÄ!
(c)
xÄ!
lim f(x) does not exist because lim b f(x) does not exist xÄ! xÄ!
6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since Èx Ÿ g(x) Ÿ Èx when x 0 xÄ! (b) No, lim c g(x) does not exist since Èx is not defined for x 0 xÄ!
(c) No, lim g(x) does not exist since lim c g(x) does not exist xÄ! xÄ! 7. (aÑ
lim f(x) œ " œ lim b f(x) xÄ1 (c) Yes, lim f(x) œ 1 since the right-hand and left-hand (b)
x Ä 1c
xÄ1
limits exist and equal 1
8. (a)
(b)
lim f(x) œ 0 œ lim c f(x) xÄ1
x Ä 1b
(c) Yes, lim f(x) œ 0 since the right-hand and left-hand xÄ1
limits exist and equal 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
63
64
Chapter 2 Limits and Continuity
9. (a) domain: 0 Ÿ x Ÿ 2 range: 0 y Ÿ 1 and y œ 2 (b) xlim f(x) exists for c belonging to Äc (0ß 1) ("ß #) (c) x œ 2 (d) x œ 0
10. (a) domain: _ x _ range: " Ÿ y Ÿ 1 (b) xlim f(x) exists for c belonging to Äc (_ß 1) ("ß ") ("ß _) (c) none (d) none
11.
x Ä !Þ&c
lim
13.
x Ä #b
14.
x Ä 1c
15.
h Ä !b
lim
2 0.5 2 È3 É 3/2 É xx É 1 œ 0.5 1 œ 1/2 œ
lim
x Ä 1b
" 1 È0 œ ! É "1 É xx # œ # œ
5‰ ˆ x x 1 ‰ ˆ 2x ˆ 2 ‰ 2(2) 5 ˆ1‰ x# x œ 2 1 Š (2)# (2) ‹ œ (2) 2 œ 1
lim
ˆ x " 1 ‰ ˆ x x 6 ‰ ˆ 3 7 x ‰ œ ˆ 1 " 1 ‰ ˆ 1 1 6 ‰ ˆ 3 7 1 ‰ œ ˆ "# ‰ ˆ 71 ‰ ˆ 27 ‰ œ 1
lim
Èh# 4h 5 È5 h
œ lim b hÄ! 16.
12.
lim
h Ä !c
(b)
18. (a)
(b)
ah# 4h 5b 5
h ŠÈh# 4h 5 È5‹
È6 È5h# 11h 6 h
œ lim c hÄ! 17. (a)
œ lim b Š hÄ!
x Ä #c
lim
x Ä 1b
lim
x Ä 1c
œ lim c Š hÄ!
6 a5h# 11h 6b
lim
lim
œ lim b hÄ!
h ŠÈ6 È5h# 11h 6‹
x Ä #b
kx 2 k x 2
(x 3)
œ
kx 2 k x2
œ
lim
lim
x Ä #c
È2x (x 1) kx 1 k
È2x (x 1) kx 1 k
œ lim b xÄ1
h(5h 11)
h ŠÈ6 È5h# 11h 6‹
(x 3)
x Ä #b
œ
œ
04 È5 È5
œ
2 È5
È5h# 11h 6 È6 È5h# 11h 6 È ‹ Š È66 ‹ h È5h# 11h 6
x Ä #b
lim
h(h 4)
h ŠÈh# 4h 5 È5‹
œ lim c hÄ!
œ
(x 3)
Èh# 4h 5 È5 È # 4h 5 È5 ‹ Š Èhh# ‹ h 4h 5 È5
(x2) (x#)
(0 11) È6 È6
11 œ 2È 6
akx 2k œ ax 2b for x 2b
(x 3) œ a(2) 3b œ 1
(x 3) ’ (x(x#2) ) “
lim
œ
x Ä #c
akx 2k œ (x 2) for x 2b
(x 3)(1) œ (2 3) œ 1
È2x (x 1) (x 1)
akx 1k œ x 1 for x 1b
œ lim b È2x œ È2 xÄ1
œ lim c xÄ1
È2x (x 1) (x 1)
akx 1k œ (x 1) for x 1b
œ lim c È2x œ È2 xÄ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.4 One-Sided Limits 19. (a)
) Ä $b
20. (a)
t Ä %b
Ú) Û )
lim
œ1
3 3
lim at ÚtÛb œ 4 4 œ 0
21. lim
sin È2) È 2)
22. lim
sin kt t
23. lim
sin 3y 4y
)Ä!
tÄ!
yÄ!
24.
œ
tan 2x x xÄ!
25. lim
2t
27. lim
xÄ!
)Ä!
œ lim c ˆ "3 † hÄ! sin 2x ‰ ˆ cos 2x x xÄ!
œ lim
œ 2 lim
t
sin t t Ä ! ˆ cos t ‰
x csc 2x cos 5x
œ
3h ‰ sin 3h
x x cos x
2 3
lim at ÚtÛb œ 4 3 œ 1
" ‰ cos 5x
xÄ!
œ
" 3
Œ
œ
" lim
)Ä!c
(where ) œ kt) (where ) œ 3y)
3 4
" 3
œ
sin ) )
"
‹ Š lim x Ä ! cos 2x xÄ!
œ Š #" lim
†1œ
2 sin 2x #x ‹
t
Ä!
" 3
(where ) œ 3h)
œ1†2œ2
t
"
‹ Š lim cos 5x ‹ x Ä ! sin 2x xÄ!
6x# cos x
œ lim ˆ sin xxcos x
œk†1œk
tÄ!
x Ä ! sin x sin 2x
x Ä ! sin x cos x
t Ä %c
œ
œ 2 Š lim cos t‹ Œ lim" sin t œ 2 † " † " œ 2
t cos t sin t
tÄ!
xÄ!
(b)
Ú) Û )
lim
3 sin ) 4 )lim Ä! )
œ
œ Š lim
sin 2x
x Ä ! x cos 2x
œ 2 lim
xÄ!
)Ä!
" " sin 3h 3 h lim Ä !c ˆ 3h ‰
œ
œ lim
œ lim ˆ sinx2x †
sin ) )
œ k lim
sin 3y 3 4 ylim Ä ! 3y
28. lim 6x# (cot x)(csc 2x) œ lim 29. lim
k sin ) )
œ lim
3 sin 3y " 4 ylim 3y Ä!
œ
h
t Ä ! tan t
k sin kt kt
tÄ!
) Ä $c
(where x œ È2))
œ1
sin x x
xÄ!
œ lim
lim h Ä !c sin 3h
26. lim
œ lim
(b)
2x
œ lim ˆ3 cos x †
x sin x
xÄ!
x cos x ‰ sin x cos x
œ lim ˆ sinx x † xÄ!
†
œ ˆ #" † 1‰ (1) œ
2x ‰ sin 2x
" ‰ cos x
" #
œ3†"†1œ3
lim
x
x Ä ! sin x
œ lim Š sin" x ‹ † lim ˆ cos" x ‰ lim Š sin" x ‹ œ (1)(1) 1 œ 2 xÄ!
30. lim
xÄ!
xÄ!
x
x# x sin x #x
1 cos ) ) Ä ! sin 2)
31. lim
œ lim
)Ä!
)Ä!
34. lim
sin (sin h) sin h sin )
) Ä ! sin 2)
36. lim
sin 5x
x Ä ! sin 4x
œ
œ lim
)Ä!
œ lim
)Ä!
sin ) )
sin ) )
)Ä!
"# (1) œ 0
1 cos2 ) a2sin ) cos )ba1 cos )b
œ lim
)Ä!
sin2 ) a2sin ) cos )ba1 cos )b
œ lim
xÄ!
xa1 c cos xb 9x2 sin2 3x 9x2
œ lim
1 c cos x 9x
2 x Ä ! ˆ sin3x3x ‰
œ
" lim ˆ 1 9 x
Ä!
cos x ‰ x
2 lim ˆ sin3x3x ‰
xÄ!
œ
" 9 a0 b 12
œ0
œ 1 since ) œ sin h Ä 0 as h Ä 0 2) ‰ #)
5x œ lim ˆ sin sin 4x †
4x 5x
)Ä!
œ lim
" #
œ 1 since ) œ 1 cos t Ä 0 as t Ä 0
sin ) œ lim ˆ sin 2) †
xÄ!
"# ˆ sinx x ‰‰ œ 0
œ0
0 a2ba2b
xa1 cos xb sin2 3x xÄ!
sin(1 cos t) 1cos t
x
a1 cos )ba1 cos )b a2sin ) cos )ba1 cos )b
œ lim
33. lim
35. lim
" #
xÄ!
œ lim
x x cos x sin2 3x xÄ!
hÄ!
œ lim ˆ #x
sin ) a2cos )ba1 cos )b
32. lim
tÄ!
xÄ!
œ
" # )lim Ä!
† 54 ‰ œ
ˆ sin) ) †
5 4 xlim Ä!
2) ‰ sin 2)
ˆ sin5x5x †
œ
" #
4x ‰ sin 4x
†1†1œ œ
5 4
" #
†1†1œ
5 4
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65
66
Chapter 2 Limits and Continuity
37. lim ) cos ) œ 0 † 1 œ 0 )Ä!
38. lim sin ) cot 2) œ lim sin ) )Ä!
)Ä!
tan 3x
39. lim
x Ä ! sin 8x
œ
3 8 xlim Ä!
40. lim
yÄ!
œ
sin 3x œ lim ˆ cos 3x †
xÄ!
" ‰ sin 8x
œ lim sin ) )Ä!
œ lim
xÄ!
3 8
†1†1†1œ
sin 3y sin 4y cos 5y y cos 4y sin 5y
yÄ!
œ lim
) cot 4) 2 2 ) Ä ! sin ) cot 2)
42. lim
)Ä!
sin ) cos ) 3) 2 ) cos sin 3)
œ lim
)Ä!
2 lim 4) cos2 4) cos ) ) Ä ! cos 2) sin 4)
œ lim
cos 4) sin 4) 2 2) 2 sin ) cos sin2 2)
" sin 8x
cos 2)
) Ä ! 2 cos )
†
8x 3x
œ
1 2
† 83 ‰
3 8
yÄ!
sin ) sin 3)
2 ) Ä ! ) cos ) cos 3)
)
œ lim
sin 4y cos 5y 3†4†5y œ lim Š siny3y ‹ Š cos 4y ‹ Š sin 5y ‹ Š 3†4†5y ‹
cos 5y ˆ 3†4 ‰ lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos 4y ‹ 5 yÄ!
tan ) 2 ) Ä ! ) cot 3)
cos 2) 2sin ) cos )
sin 3x œ lim ˆ cos 3x †
ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ
sin 3y cot 5y y cot 4y
41. lim
œ
cos 2) sin 2)
œ1†1†1†1†
œ
12 5
œ lim ˆ sin) ) ‰ˆ sin3)3) ‰ˆ cos ) 3cos 3) ‰ œ a1ba1bˆ 13†1 ‰ œ 3 )Ä!
) cos 4) sin2 2) 2 2 ) Ä ! sin ) cos 2) sin 4)
œ lim
12 5
) cos 4) a2sin ) cos )b2 2 2 ) Ä ! sin ) cos 2) sin 4)
œ lim
) cos 4) ˆ4sin2 ) cos2 )‰ 2 2 ) Ä ! sin ) cos 2) sin 4)
œ lim
4) cos ) 4) cos ) œ lim ˆ sin4)4) ‰Š coscos ‹ œ lim Š sin14) ‹Š coscos ‹ œ ˆ 11 ‰Š 11†12 ‹ œ 1 2 2) 2 2) 2
)Ä!
2
)Ä!
2
4)
43. Yes. If lim b f(x) œ L œ lim c f(x), then xlim f(x) œ L. If lim b f(x) Á lim c f(x), then xlim f(x) does not exist. Äa Äa xÄa xÄa xÄa xÄa 44. Since xlim f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim f(x) can be found by calculating Äc Äc xÄc xÄc lim b f(x). xÄc
45. If f is an odd function of x, then f(x) œ f(x). Given lim b f(x) œ 3, then lim c f(x) œ $. xÄ! xÄ! 46. If f is an even function of x, then f(x) œ f(x). Given lim c f(x) œ 7 then xÄ#
can be said about
lim
x Ä #c
lim
x Ä #b
f(x) œ 7. However, nothing
f(x) because we don't know lim b f(x). xÄ#
47. I œ (5ß 5 $ ) Ê 5 x & $ . Also, Èx 5 % Ê x 5 %# Ê x & %# . Choose $ œ %# Ê lim Èx 5 œ 0. x Ä &b
48. I œ (% $ ß %) Ê % $ x 4. Also, È% x % Ê % x %# Ê x % %# . Choose $ œ %# Ê lim È% x œ 0. x Ä %c
49. As x Ä 0 the number x is always negative. Thus, ¹ kxxk (1)¹ % Ê ¸ xx 1¸ % Ê 0 % which is always true independent of the value of x. Hence we can choose any $ 0 with $ x ! Ê
2 ¸ x 2 ¸ 50. Since x Ä # we have x 2 and kx 2k œ x 2. Then, ¹ kxx 2 k " ¹ œ x 2 " % Ê 0 %
which is always true so long as x #. Hence we can choose any $ !, and thus # x # $ 2 Ê ¹ kxx 2k "¹ % . Thus,
x 2
lim x Ä #b kx2k
x
lim x Ä ! c kx k
œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ 1.
Section 2.5 Continuity 51. (a) (b)
lim
x Ä %!!b
67
ÚxÛ œ 400. Just observe that if 400 x 401, then ÚxÛ œ 400. Thus if we choose $ œ ", we have for any
number % ! that 400 x 400 $ Ê lÚxÛ 400l œ l400 400l œ ! %. lim c ÚxÛ œ 399. Just observe that if 399 x 400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any
x Ä %!!
number % ! that 400 $ x 400 Ê lÚxÛ 399l œ l399 399l œ ! %. (c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist. x Ä %!!
x Ä %!!
52. (a)
x Ä %!!
lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx 0¸ % Ê % Èx % Ê ! x %# for x positive. Choose $ œ %# xÄ! Ê lim b f(x) œ 0.
x Ä !b
xÄ!
(b)
lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0. x Ä !c xÄ! Since kx# 0k œ kx# 0k œ x# % whenever kxk È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰ 0¸ %
if $ x 0. (c) The function f has limit 0 at x! œ 0 since both the right-hand and left-hand limits exist and equal 0. 2.5 CONTINUITY 1. No, discontinuous at x œ 2, not defined at x œ 2 2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5 xÄ$ 3. Continuous on [1ß 3] 4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ ! xÄ" xÄ" 5. (a) Yes
(b) Yes,
(c) Yes
(d) Yes
6. (a) Yes, f(1) œ 1
lim
x Ä "b
f(x) œ 0
(b) Yes, lim f(x) œ 2 xÄ1
(c) No
(d) No
7. (a) No
(b) No
8. ["ß !) (!ß ") ("ß #) (#ß $) 9. f(2) œ 0, since lim c f(x) œ 2(2) 4 œ 0 œ lim b f(x) xÄ# xÄ# 10. f(1) should be changed to 2 œ lim f(x) xÄ1
11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0). xÄ" xÄ1 xÄ"
Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than f(0) œ 1. xÄ!
12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1). xÄ" xÄ1 xÄ" Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than f(2) œ 2. xÄ#
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68
Chapter 2 Limits and Continuity
13. Discontinuous only when x 2 œ 0 Ê x œ 2
14. Discontinuous only when (x 2)# œ 0 Ê x œ 2
15. Discontinuous only when x# %x $ œ ! Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1 16. Discontinuous only when x# 3x 10 œ 0 Ê (x 5)(x 2) œ 0 Ê x œ 5 or x œ 2 17. Continuous everywhere. ( kx 1k sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. ( kxk " Á 0 for all x; limits exist and are equal to function values.) 19. Discontinuous only at x œ 0 20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n ") 1# , n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ
n1 # ,
n an integer, but
continuous at all other x. 22. Discontinuous when
1x #
is an odd integer multiple of 1# , i.e.,
1x #
œ (2n 1) 1# , n an integer Ê x œ 2n 1, n an
integer (i.e., x is an odd integer). Continuous everywhere else. 23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n 1) 1# , n an integer, but continuous at all other x. 24. Continuous everywhere since x% 1 1 and " Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1 sin# x 1; limits exist and are equal to the function values. 25. Discontinuous when 2x 3 0 or x 3# Ê continuous on the interval 3# ß _‰ . 26. Discontinuous when 3x 1 0 or x
" 3
Ê continuous on the interval 3" ß _‰ .
27. Continuous everywhere: (2x 1)"Î$ is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2 x)"Î& is defined for all x; limits exist and are equal to function values. 29. Continuous everywhere since lim xÄ3 30. Discontinuous at x œ 2 since
x2 x 6 x3
œ lim xÄ3
ax 3bax 2b x3
œ lim ax 2b œ 5 œ ga3b xÄ3
lim faxb does not exist while fa2b œ 4. x Ä 2
31. xlim sin (x sin x) œ sin (1 sin 1) œ sin (1 0) œ sin 1 œ 0, and function continuous at x œ 1. Ä1 32. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !. tÄ!
33. lim sec ay sec# y tan# y 1b œ lim sec ay sec# y sec# yb œ lim sec a(y 1) sec# yb œ sec a(" ") sec# 1b yÄ1
yÄ1
yÄ1
œ sec 0 œ 1, and function continuous at y œ ". 34. lim tan 14 cos ˆsin x"Î$ ‰‘ œ tan 14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !. xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.5 Continuity 35. lim cos ’ È19 13 sec 2t “ œ cos ’ È19 13 sec 0 “ œ cos tÄ!
1 È16
œ cos
1 4
œ
È2 # ,
and function continuous at t œ !.
36. lim1 Écsc# x 5È3 tan x œ Écsc# ˆ 16 ‰ 5È3 tan ˆ 16 ‰ œ Ê4 5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at xÄ
'
x œ 1' . 37. g(x) œ
x# 9 x3
(x 3)(x 3) (x 3)
œ
38. h(t) œ
t# 3t 10 t#
39. f(s) œ
s$ " s# 1
40. g(x) œ
œ
œ
œ x 3, x Á 3 Ê g(3) œ lim (x 3) œ 6 xÄ$
(t 5)(t 2) t#
as# s 1b (s 1) (s 1)(s 1)
x# 16 x# 3x 4
œ
œ t 5, t Á # Ê h(2) œ lim (t 5) œ 7 tÄ#
s# s " s1 ,
œ
(x 4)(x 4) (x 4)(x 1)
œ
x4 x1
s Á 1 Ê f(1) œ lim Š s sÄ1
#
s1 s1 ‹
4‰ , x Á 4 Ê g(4) œ lim ˆ xx 1 œ
xÄ%
œ
3 #
8 5
41. As defined, lim c f(x) œ (3)# 1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have xÄ$ xÄ$ 6a œ 8 Ê a œ 43 .
42. As defined,
lim
x Ä #c
g(x) œ 2 and
4b œ 2 Ê b œ "# .
lim
x Ä #b
g(x) œ b(2)# œ 4b. For g(x) to be continuous we must have
43. As defined, lim c f(x) œ 12 and lim b f(x) œ a# a2b 2a œ 2a# 2a. For f(x) to be continuous we must have xÄ# xÄ# 12 œ 2a# 2a Ê a œ 3 or a œ 2. 44. As defined, lim c g(x) œ b b1
xÄ0
0b b1
œ
b b1
œ b Ê b œ 0 or b œ 2.
45. As defined,
lim
x Ä 1 c
f(x) œ 2 and
and lim b g(x) œ a0b2 b œ b. For g(x) to be continuous we must have xÄ0
lim
x Ä 1 b
f(x) œ aa1b b œ a b, and
lim f(x) œ aa1b b œ a b and
x Ä 1c
lim f(x) œ 3. For f(x) to be continuous we must have 2 œ a b and a b œ 3 Ê a œ
x Ä 1b
5 #
and b œ "# .
46. As defined, lim c g(x) œ aa0b 2b œ 2b and lim b g(x) œ a0b2 3a b œ 3a b, and xÄ0 xÄ0 lim g(x) œ a2b2 3a b œ 4 3a b and lim b g(x) œ 3a2b 5 œ 1. For g(x) to be continuous we must xÄ0
x Ä 2c
have 2b œ 3a b and 4 3a b œ 1 Ê a œ 3# and b œ 3# .
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69
70
Chapter 2 Limits and Continuity
47. The function can be extended: f(0) ¸ 2.3.
48. The function cannot be extended to be continuous at x œ 0. If f(0) ¸ 2.3, it will be continuous from the right. Or if f(0) ¸ 2.3, it will be continuous from the left.
49. The function cannot be extended to be continuous at x œ 0. If f(0) œ 1, it will be continuous from the right. Or if f(0) œ 1, it will be continuous from the left.
50. The function can be extended: f(0) ¸ 7.39.
51. f(x) is continuous on [!ß "] and f(0) 0, f(1) 0 Ê by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(1) Ê the equation f(x) œ 0 has at least one solution between x œ 0 and x œ 1.
52. cos x œ x Ê (cos x) x œ 0. If x œ 1# , cos ˆ 1# ‰ ˆ 1# ‰ 0. If x œ 1# , cos ˆ 1# ‰ for some x between
1 #
and
1 #
1 #
0. Thus cos x x œ 0
according to the Intermediate Value Theorem, since the function cos x x is continuous.
53. Let f(x) œ x$ 15x 1, which is continuous on [4ß 4]. Then f(4) œ 3, f(1) œ 15, f(1) œ 13, and f(4) œ 5. By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals % x 1, " x 1, and " x 4. That is, x$ 15x 1 œ 0 has three solutions in [%ß 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 54. Without loss of generality, assume that a b. Then F(x) œ (x a)# (x b)# x is continuous for all values of x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value Theorem, since a a # b b, there is a number c between a and b such that F(x) œ a # b .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.5 Continuity 55. Answers may vary. Note that f is continuous for every value of x. (a) f(0) œ 10, f(1) œ 1$ 8(1) 10 œ 3. Since $ 1 10, by the Intermediate Value Theorem, there exists a c so that ! c 1 and f(c) œ 1. (b) f(0) œ 10, f(4) œ (4)$ 8(4) 10 œ 22. Since 22 È3 10, by the Intermediate Value Theorem, there exists a c so that 4 c 0 and f(c) œ È3. (c) f(0) œ 10, f(1000) œ (1000)$ 8(1000) 10 œ 999,992,010. Since 10 5,000,000 999,992,010, by the Intermediate Value Theorem, there exists a c so that ! c 1000 and f(c) œ 5,000,000.
56. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f(x) œ x$ 3x 1 is a point c where f(c) œ 0. (b) The points where y œ x$ crosses y œ 3x 1 have the same y-coordinate, or y œ x$ œ 3x 1 Ê f(x) œ x$ 3x 1 œ 0. (c) x$ 3x œ 1 Ê x$ 3x 1 œ 0. The solutions to the equation are the roots of f(x) œ x$ 3x 1. (d) The points where y œ x$ 3x crosses y œ 1 have common y-coordinates, or y œ x$ 3x œ 1 Ê f(x) œ x$ 3x 1 œ !. (e) The solutions of x$ 3x 1 œ 0 are those points where f(x) œ x$ 3x 1 has value 0. 57. Answers may vary. For example, f(x) œ
sin (x 2) x2
is discontinuous at x œ 2 because it is not defined there.
However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2. 58. Answers may vary. For example, g(x) œ
" x1
has a discontinuity at x œ 1 because lim g(x) does not exist. x Ä "
Š lim c g(x) œ _ and lim b g(x) œ _.‹ x Ä " x Ä " 59. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $ 0 there is an irrational number x (actually infinitely many) in the interval (x! $ ß x! $ ) Ê f(x) œ 0. Then 0 kx x! k $ but kf(x) f(x! )k œ 1 "# œ %, so x lim f(x) fails to exist Ê f is discontinuous at x! rational. Äx !
On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x! $ ß x! $ ) Ê f(x) œ 1. Again x lim f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at Äx !
every point. (b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x! $ ß x! ) or (x! ß x! $ ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and x Ä x!
lim f(x) exist by the same arguments used in part (a).
x Ä x !
60. Yes. Both f(x) œ x and g(x) œ x g ˆ "# ‰
œ0 Ê
f(x) g(x)
" #
are continuous on [!ß "]. However
is discontinuous at x œ
f(x) g(x)
is undefined at x œ
" #
since
" #.
61. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not. 62. Let f(x) œ œ
" x1
" (x 1) 1
œ
and g(x) œ x 1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x)) " x
is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be
continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1. 63. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [aß b].
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72
Chapter 2 Limits and Continuity
64. Let f(x) be the new position of point x and let d(x) œ f(x) x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is then in its original position. 65. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a 0 and f(1) œ b 1 because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x) x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0) 0 œ a 0 and g(1) œ f(1) 1 œ b 1 0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that g(c) œ 0 Ê f(c) c œ 0 or f(c) œ c. 66. Let % œ
kf(c)k #
0. Since f is continuous at x œ c there is a $ 0 such that kx ck $ Ê kf(x) f(c)k %
Ê f(c) % f(x) f(c) %. If f(c) 0, then % œ "# f(c) Ê " #
" #
If f(c) 0, then % œ f(c) Ê
f(c) f(x) 3 #
3 #
f(c) f(x)
f(c) Ê f(x) 0 on the interval (c $ ß c $ ). " #
f(c) Ê f(x) 0 on the interval (c $ ß c $ ).
67. By Exercises 52 in Section 2.3, we have xlim faxb œ L Í lim fac hb œ L. Äc hÄ0
Thus, faxb is continuous at x œ c Í xlim faxb œ facb Í lim fac hb œ facb. Äc hÄ0
68. By Exercise 67, it suffices to show that lim sinac hb œ sin c and lim cosac hb œ cos c. hÄ0
hÄ0
Now lim sinac hb œ lim asin cbacos hb acos cbasin hb‘ œ asin cbŠ lim cos h‹ acos cbŠ lim sin h‹ hÄ0
hÄ0
hÄ0
hÄ0
By Example 11 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac hb œ sin c and thus faxb œ sin x is hÄ0
continuous at x œ c. Similarly,
hÄ0
hÄ0
lim cosac hb œ lim acos cbacos hb asin cbasin hb‘ œ acos cbŠ lim cos h‹ asin cbŠ lim sin h‹ œ cos c.
hÄ0
hÄ0
Thus, gaxb œ cos x is continuous at x œ c.
hÄ0
69. x ¸ 1.8794, 1.5321, 0.3473
70. x ¸ 1.4516, 0.8547, 0.4030
71. x ¸ 1.7549
72. x ¸ 1.5596
73. x ¸ 3.5156
74. x ¸ 3.9058, 3.8392, 0.0667
75. x ¸ 0.7391
76. x ¸ 1.8955, 0, 1.8955
hÄ0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2.6 LIMITS INVOLVING INFINITY; ASMYPTOTES OF GRAPHS 1. (a) (c) (e) (g)
lim f(x) œ 0
(b)
xÄ2
lim
x Ä 3 c
f(x) œ 2
lim
x Ä 3 b
f(x) œ 2
(d) lim f(x) œ does not exist xÄ3
lim b f(x) œ 1
(f)
lim f(x) œ does not exist
(h) x lim f(x) œ 1 Ä_
xÄ0 xÄ0
lim f(x) œ _
x Ä 0c
(i) x Ä lim f(x) œ 0 _ 2. (a) (c)
lim f(x) œ 2
(b)
lim f(x) œ 1
(d) lim f(x) œ does not exist
xÄ4
x Ä 2c
xÄ2
lim f(x) œ _ x Ä 3 b (g) lim f(x) œ _ (e)
x Ä 3
(i)
lim f(x) œ 3
x Ä 2b
lim f(x) œ _
(f)
x Ä 3 c
lim
(h)
x Ä 0b
(k) x lim f(x) œ 0 Ä_
lim f(x) œ _
lim f(x) œ does not exist
(j)
x Ä 0c
xÄ0
(l) x Ä lim f(x) œ 1 _
Note: In these exercises we use the result
"
lim mÎn xÄ „_ x
Theorem 8 and the power rule in Theorem 1:
lim
xÄ „_
œ 0 whenever
ˆ xm"În ‰ œ
lim
(b) 3
4. (a) 1
(b) 1
5. (a)
" #
(b)
" #
6. (a)
" 8
(b)
" 8
7. (a) 53
(b)
10. 3") Ÿ 11.
lim
tÄ_
12. r Ä lim_
0. This result follows immediately from ˆ x" ‰mÎn œ Š
"
lim ‹ xÄ „_ x
mÎn
(b) 53
3 4
9. "x Ÿ
m n
xÄ „_
3. (a) 3
8. (a)
f(x) œ _
sin 2x x
Ÿ
" x
cos ) 3)
Ÿ
" 3)
2 t sin t t cos t
Ê x lim Ä_ Ê
13. (a) x lim Ä_
2x 3 5x 7
lim
) Ä _
œ lim
2 t
tÄ_
r sin r 2r 7 5 sin r
sin 2x x
œrÄ lim_
œ x lim Ä_
œ 0 by the Sandwich Theorem
cos ) 3)
œ 0 by the Sandwich Theorem
1 ˆ sint t ‰ 1 ˆ cost t ‰
œ
1 ˆ sinr r ‰ 2 7r 5 ˆ sinr r ‰ 2 3x 5 7x
3 4
œ
2 5
010 10
œ 1
œrÄ lim_
10 200
œ
(b)
" #
2 5
(same process as part (a))
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ 0mÎn œ 0.
73
74
Chapter 2 Limits and Continuity 2 Š x7$ ‹
$
2x 7 14. (a) x lim œ x lim Ä _ x$ x# x 7 Ä_ (b) 2 (same process as part (a)) " x
x"#
15. (a) x lim Ä_
x1 x# 3
œ x lim Ä_
1 x3#
16. (a) x lim Ä_
3x 7 x# 2
œ x lim Ä_
1 x2#
17. (a) x lim Ä_
7x$ x$ 3x# 6x
18. (a) x lim Ä_
x$
20. (a) x lim Ä_ 9 #
2x%
x7#
œ x lim Ä_
œ x lim Ä_
10x& x% 31 x'
19. (a) x lim Ä_
(b)
" 4x 1
3 x
œ2
œ0
(b) 0 (same process as part (a))
œ0
(b) 0 (same process as part (a)) œ(
7 1 3x x6# " x$ 4 x"$ x#
1
œ x lim Ä_
2
(b) 7 (same process as part (a))
œ!
x"# x31' 1
10 x
œ x lim Ä_
9x% x 5x# x 6
1 "x x"# x7$
(b) 0 (same process as part (a))
œ0
(b) 0 (same process as part (a))
9 x"$
5 x#
x"$ x6%
œ
9 #
(same process as part (a)) 2x$ 2x 3 3x$ 3x# 5x
21. (a) x lim Ä_
2 x2# x3$
œ x lim Ä_
œ 23
3 3x x5#
(b) 23 (same process as part (a)) x % x% 7x$ 7x# 9
22. (a) x lim Ä_
" 1 7x x7# x9%
œ x lim Ä_
œ 1
(b) 1 (same process as part (a)) 8
8
3
3
3 x2 x2 É 8x 23. x lim œ Êx lim œ É 82 00 œ È4 œ 2 2x2 x œ x lim Ä_ Ä _ Ê 2 1x Ä _ 2 1x 2
24. x Ä lim Šx x1‹ _ 8x2 3 2
1 Î3
œxÄ lim _Œ
5
1
" 1x x12 8
x
3 x2
1 Î3
œ Œx Ä lim _
5
1
" 1x x12 8
3 x2
5
x
1 Î3
œ ˆ " 8 0 0 0 ‰
1 Î3
œ ˆ "8 ‰
1 Î3
œ
x2 x2 ‰ œ_ 25. x Ä lim lim œ Œx Ä lim œ ˆ 01_ Š 1x ‹ œ x Ä 0 _ x2 7x _Œ 1 7x _ 1 7x 3
1
5
1
5
5
x x x2 x2 É x 5x œ x lim 26. x lim œ Êx lim œ É 1 0 0 0 0 œ È0 œ 0 Ä _ x3 x 2 Ä _Ê 1 x12 x23 Ä _ 1 x12 x23 2
27. x lim Ä_
2Èx xc" 3x 7
29. x Ä lim _
30. x lim Ä_
œ x lim Ä_
3 xÈ 5 x È 3 xÈ 5 x È
x "x % x #x $
Š
œxÄ lim _
œ x lim Ä_
2 ‹ Š x"# ‹ x"Î# 3 7x
œ0
1 xÐ"Î&Ñ Ð"Î$Ñ 1 xÐ"Î&Ñ Ð"Î$Ñ
x x"# 1 x"
œxÄ lim _
28. x lim Ä_ " ‹ 1 Š #Î"& x
" ‹ 1 Š #Î"&
2 Èx 2 Èx
œ x lim Ä_
2 ‹" x"Î# 2 Š "Î# ‹1 x
Š
œ 1
œ1
x
œ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" #
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2x&Î$ x"Î$ 7 x)Î& 3x Èx
31. x lim Ä_
3 x 5x 3 È 2x x#Î$ 4
32. x Ä lim _ 33. x lim Ä_
È x2 1 x1
34. x Ä lim _
œ x lim Ä_ œxÄ lim _
œ x lim Ä_
È x2 1 x1
1
" x#Î$
2
È x 2 1 ÎÈ x 2 a x 1 b ÎÈ x 2
œxÄ lim _
" 7 x"*Î"& x)Î& 3 " x$Î& x""Î"!
2x"Î"&
5 3x " x"Î$
È
3 œ 5# È x
4x
œ x lim Ä_
È x 2 1 ÎÈ x 2 a x 1 bÎ È x 2
œ_
È a x 2 1 bÎx 2 ax 1bÎx
œ x lim Ä_
È 1 1 Îx 2 a1 1 Îx b
È a x 2 1 bÎ x 2
È1 0 a1 0 b
œ
È 1 1 Îx 2
œxÄ lim œ x lim œ _ ax 1bÎaxb Ä _ a 1 1 Î x b
œ1
È1 0 a1 0b
a x 3 bÎ x a x 3 bÎ x a1 3 Îx b x3 35. x lim œ x lim œ x lim œ x lim œ Ä _ È4x2 25 Ä _ È4x2 25ÎÈx2 Ä _ Èa4x2 25bÎx2 Ä _ È4 25Îx2 ˆ4 3x3 ‰ÎÈx6
4 3x 36. x Ä lim œxÄ lim œxÄ lim _ Èx6 9 _ Èx6 9ÎÈx6 _ 3
"
œ_
37.
lim x Ä !b 3x
39.
lim x Ä #c x 2
41.
lim x Ä )b x8
3
2x
4
43. lim
# x Ä ( (x7)
œ _ œ _ œ_
lim "Î$ x Ä !b 3x
46. (a)
lim "Î& x Ä !b x 4
47. lim
#Î& xÄ! x
49.
51. 52.
lim
x Ä ˆ 1# ‰
œ lim
4
# x Ä ! ax"Î& b
œ x lim Ä_
ˆ 4 Îx 3 3‰ È 1 9 Îx 6
Š positive positive ‹
40.
lim x Ä $b x 3
Š negative positive ‹
42.
lim x Ä &c 2x10
positive Š positive ‹
44. lim
œ_
"
3x
"
# x Ä ! x (x1)
2
(b)
lim "Î$ x Ä !c 3x
(b)
lim "Î& x Ä !c x
48. lim
"
#Î$ xÄ! x
tan x œ _
50.
œ3
œ_
positive Š negative ‹
2
a0 3 b È1 0
" #
positive Š negative ‹
lim x Ä !c 2x
5
œ
œ
œ _
38.
œ_
2
ˆ4 3x3 ‰Îˆx3 ‰ Èax6 9bÎx6
positive Š positive ‹
œ_
2
45. (a)
a1 0 b È4 0
2
œ 1
œ_
Š negative negative ‹
œ _
negative Š positive †positive ‹
œ _ œ _
œ lim
"
# x Ä ! ax"Î$ b
œ_
lim sec x œ _
x Ä ˆ #1 ‰
lim (1 csc )) œ _
)Ä!
lim (2 cot )) œ _ and lim c (2 cot )) œ _, so the limit does not exist )Ä!
) Ä !b
"
œ lim b xÄ#
" (x2)(x2)
œ_
Š positive"†positive ‹
"
œ lim c xÄ#
" (x2)(x2)
œ _
Š positive†"negative ‹
53. (a)
lim # x Ä # b x 4
(b)
lim # x Ä # c x 4
(c)
lim # x Ä #b x 4
(d)
lim # x Ä #c x 4
"
œ
lim x Ä #b (x2)(x2)
"
œ _
Š positive†"negative ‹
"
œ
lim x Ä #c (x2)(x2)
"
œ_
Š negative"†negative ‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
75
76
Chapter 2 Limits and Continuity
54. (a)
lim # x Ä "b x 1
(b)
lim # x Ä "c x 1
(c)
lim # x Ä "b x 1
(d)
lim # x Ä "c x 1
x
œ lim b xÄ"
x (x1)(x1)
œ_
positive Š positive †positive ‹
x
œ lim c xÄ"
x (x1)(x1)
œ _
positive Š positive †negative ‹
x
œ
lim x Ä "b (x1)(x1)
x
œ_
negative Š positive †negative ‹
x
œ
lim x Ä "c (x1)(x1)
x
œ _
negative Š negative †negative ‹
55. (a)
lim x Ä !b #
x#
" x
œ 0 lim b xÄ!
" x
œ _
" Š negative ‹
(b)
lim x Ä !c #
x#
" x
œ 0 lim c xÄ!
" x
œ_
" Š positive ‹
(c)
lim # x Ä $È2
(d)
lim x Ä 1 #
56. (a)
x#
x#
lim x Ä #b
" x
" x
œ
x# 1 2x 4 x# 1
2#Î$ #
œ
" #
(d)
lim x Ä !c 2x 4
œ
(b) (c) (d) (e) 58. (a)
x# 3x 2 x$ 2x#
lim b
x# 3x 2 x$ 2x#
lim
xÄ#
#
x 3x 2 x$ 2x# x 3x 2 x$ 2x#
lim
œ lim c xÄ#
lim x Ä #b
(c)
x Ä 0c
(d)
x Ä "b
(e)
lim x Ä !b x(x #)
x# 1 2x 4
lim x Ä #c
œ _
positive Š negative ‹
œ0
(x 2)(x 1) x# (x 2)
œ _
(x 2)(x 1) x# (x 2)
œ lim b xÄ#
(x 2)(x 1) x# (x 2)
œ lim c xÄ#
œ lim
œ lim
(x 2)(x 1) x# (x 2)
œ _
xÄ!
lim
x# 3x 2 x$ 4x
lim
x# 3x 2 x$ 4x x"
2†0 #4
(x 2)(x 1) x# (x 2)
xÄ#
œ lim b xÄ#
x# 3x 2 x$ 4x
(b)
œ
œ lim
x# 3x 2 x$ 4x
lim
x Ä #b
and
œ lim b xÄ#
x 3x 2 x$ 2x#
#
xÄ!
œ lim b xÄ!
#
lim
x Ä #c
(x 1)(x 1) 2x 4
(b)
" 4
lim b
xÄ#
3 #
Š positive positive ‹
œ lim b xÄ"
xÄ!
œ 2"Î$ 2"Î$ œ 0
œ_
lim x Ä "b 2x 4
57. (a)
" #"Î$
ˆ "1 ‰ œ
(c)
x# 1
œ
xÄ#
(x 2)(x ") x(x #)(x 2)
(x 2)(x ")
œ lim c xÄ! œ lim b xÄ"
(x 2)(x ") x(x #)(x 2) (x 2)(x ") x(x #)(x 2)
œ
x1 x#
œ
" 4
,xÁ2
x1 x#
œ
" 4
,xÁ2
x1 x#
œ
" 4
,xÁ2 †negative Š negative positive†negative ‹
œ lim b xÄ#
lim x Ä #b x(x #)(x 2)
†negative Š negative positive†negative ‹
(x 1) x(x #)
œ
(x 1)
lim x Ä #b x(x #)
œ lim c xÄ! œ lim b xÄ"
œ_
(x 1) x(x #)
œ
negative Š positive †positive ‹
x"
negative Š negative †positive ‹
œ_
œ
" 8
œ_
(x 1) x(x #)
œ _
lim x Ä !c x(x #)
" #(4)
0 (1)(3)
negative Š negative †positive ‹ negative Š negative †positive ‹
œ0
so the function has no limit as x Ä 0. lim 2
59. (a)
t Ä !b
60. (a)
t Ä !b
61. (a)
x Ä !b
(c)
x Ä "b
3 ‘ t"Î$
œ _
" lim t$Î& 7‘ œ _
lim
2
lim
lim
" ’ x#Î$
2 “ (x 1)#Î$
œ_
lim
" ’ x#Î$
2 “ (x 1)#Î$
œ_
(b)
t Ä !c
(b)
t Ä !c
lim
" ’ x#Î$
2 “ (x 1)#Î$
œ_
(b)
x Ä !c
lim
" ’ x#Î$
2 “ (x 1)#Î$
œ_
(d)
x Ä "c
" t$Î&
3 ‘ t"Î$
œ_
7‘ œ _
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs lim
" ’ x"Î$
1 “ (x 1)%Î$
œ_
(b)
x Ä !c
lim
" ’ x"Î$
1 “ (x 1)%Î$
œ _
(d)
x Ä "c
62. (a)
x Ä !b
(c)
x Ä "b
lim
" ’ x"Î$
1 “ (x 1)%Î$
œ _
lim
" ’ x"Î$
1 “ (x 1)%Î$
œ _
63. y œ
" x1
64. y œ
" x1
65. y œ
" #x 4
66. y œ
3 x3
67. y œ
x3 x2
68. y œ
2x x1
œ1
" x#
69. Here is one possibility.
œ#
2 x1
70. Here is one possibility.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
77
78
Chapter 2 Limits and Continuity
71. Here is one possibility.
72. Here is one possibility.
73. Here is one possibility.
74. Here is one possibility.
75. Here is one possibility.
76. Here is one possibility.
77. Yes. If x lim Ä_
f(x) g(x)
œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä lim _
f(x) g(x)
œ 2 as well.
78. Yes, it can have a horizontal or oblique asymptote. 79. At most 1 horizontal asymptote: If x lim Ä_ f(x) lim x Ä _ g(x)
f(x) g(x)
œ L, then the ratio of the polynomials' leading coefficients is L, so
œ L as well.
Èx 9 Èx 4 80. x lim Š Èx 9 Èx 4‹ œ x lim ’Èx 9 Èx 4“ † ’ Èx 9 Èx 4 “ œ x lim Ä_ Ä_ Ä_ 5 Èx 5 0 œ x lim œ x lim œ 11 œ 0 9 4 Ä _ Èx 9 Èx 4 Ä_
ax 9 b a x 4 b Èx 9 Èx 4
É1 x É1 x
È 2 È 2 81. x lim Š Èx2 25 Èx2 "‹ œ x lim ’Èx2 25 Èx2 "“ † ’ Èx2 25 Èx2 " “ œ x lim Ä_ Ä_ Ä_ x 25 x "
œ x lim Ä_
26 Èx2 25 Èx2 "
œ x lim Ä_
26 x
É1 x252 É1 x12
œ
0 11
œ0
È 2 ˆx 3 ‰ ˆ x ‰ x 82. x Ä lim lim lim Š Èx2 3 x‹ œ x Ä ’Èx2 3 x“ † ’ Èxx2 33 “œxÄ _ _ _ Èx2 3 x x 3 È x2 3x 3 œxÄ lim œ lim œ lim œ 1 0 1 œ 0 2 È 3 x _ x Ä _ É1 2 È x Ä _ É1 32 1 x 3x x x x2 2
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ˆx2 25‰ ˆx2 "‰ Èx2 25 Èx2 "
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs È
ˆ4x2 ‰ ˆ4x2 3x 2‰
4x 3x 2 83. x Ä lim lim lim Š 2x È4x2 3x 2‹ œ x Ä ’2x È4x2 3x 2“ † ’ 2x “œxÄ _ _ _ 2x È4x2 3x 2 2x È4x2 3x 2 c3x b 2 c3x b 2 3 x2 Èx2 3x 2 cx œxÄ lim œxÄ lim œxÄ lim œxÄ lim _ 2x È4x2 3x 2 _ È2x É4 3 22 _ 2x É4 3 22 _ 2 É4 3 22 x x x cx x x x x2 œ 3202 œ 43 2
È 2 84. x lim Š È9x2 x 3x‹ œ œ x lim ’È9x2 x 3x“ † ’ È9x2 x 3x “ œ x lim Ä_ Ä_ Ä_ 9x x 3x
œ x lim Ä_
x È9x2 x 3x
œ x lim Ä_
xx 2 É 9x2 x
xx2 3x x
1 É9 "x 3
œ x lim Ä_
œ
1 33
ˆ9x2 x‰ ˆ9x2 ‰ È9x2 x 3x
œ "6
È 2 È 2 85. x lim Š Èx2 3x Èx2 2x‹ œ x lim ’ Èx2 3x Èx2 2x“ † ’ Èx2 3x Èx2 2x “ œ x lim Ä_ Ä_ Ä_ x 3x x 2x 5x 5 5 5 œ x lim œ x lim œ 11 œ # È 2 3 2 Ä_ È 2 Ä_ x 3x
x 2x
É1 x É1 x
Èx# x Èx# x œ lim ’Èx# x Èx# x“ † ’ Èx# x Èx# x “ œ lim 86. x lim È x# x È x# x Ä_ xÄ_ xÄ_ 2x 2 2 œ x lim œ x lim œ 11 œ 1 È # " " Ä_ È # Ä_ x x
ˆx2 3x‰ ˆx2 2x‰ Èx2 3x Èx2 2x
x x
ax # x b a x # x b È x# x È x# x
É1 x É1 x
87. For any % 0, take N œ 1. Then for all x N we have that kf(x) kk œ kk kk œ 0 %. 88. For any % 0, take N œ 1. Then for all y N we have that kf(x) kk œ kk kk œ 0 %. " x#
89. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 0k $ Ê
" x#
Ê
B ! Í " x#
" x#
#
B0 Í x
" B
" ÈB
Í kxk
. Choose $ œ
" ÈB
, then 0 kxk $ Ê kxk
xÄ!
B ! Í lxl B" . Choose $ œ B" . Then ! kx 0k $ Ê lxl
" B
Ê
" lx l
" lx l
2 (x 3)#
B ! Í
2 (x 3)#
$ œ É B2 , then 0 kx 3k $ Ê
B0 Í 2 (x 3)#
(x 3) 2
#
" B
Í (x 3)#
B 0 so that lim
2
# x Ä $ (x 3)
2 B
B. Now,
x Ä ! lx l 2 (x 3)#
Now,
#
B ! Í (x 5)
Ê kx 5k
" ÈB
Ê
" (x 5)#
" B
Í kx 5k
B so that lim
"
" ÈB
# x Ä & (x 5)
. Choose $ œ
œ _.
B.
Í ! kB $k É B2 . Choose
œ _.
92. For every real number B 0, we must find a $ 0 such that for all x, 0 kx (5)k $ Ê 1 (x 5)#
"
B so that lim
91. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 3k $ Ê Now,
" ÈB
B so that lim x"# œ _.
90. For every real number B 0, we must find a $ 0 such that for all x, ! kx 0k $ Ê " lx l
B. Now,
" ÈB
1 (x 5)#
B.
. Then 0 kx (5)k $
œ _.
93. (a) We say that f(x) approaches infinity as x approaches x! from the left, and write lim c f(x) œ _, if x Ä x! for every positive number B, there exists a corresponding number $ 0 such that for all x, x! $ x x! Ê f(x) B. (b) We say that f(x) approaches minus infinity as x approaches x! from the right, and write lim b f(x) œ _, x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such that for all x, x! x x! $ Ê f(x) B.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
79
80
Chapter 2 Limits and Continuity (c) We say that f(x) approaches minus infinity as x approaches x! from the left, and write lim f(x) œ _, x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such that for all x, x! $ x x! Ê f(x) B. 94. For B 0,
" x
B 0 Í x B" . Choose $ œ B" . Then ! x $ Ê 0 x
95. For B 0,
" x
B 0 Í x" B 0 Í x
Ê B" x Ê 96. For B !,
" x#
" x
B so that lim c xÄ!
" x
" B
Ê
" x#
" x#
Ê 99. y œ
101. y œ
" x#
B ! so that lim b xÄ#
œx1
x# % x"
B so that lim b xÄ!
" x
œ _.
" B
Í x 2 B" Í x 2 B" . Choose $ œ B" . Then " x#
B 0 so that lim c xÄ#
" x#
œ _.
B Í ! x 2 B" . Choose $ œ B" . Then # x # $ Ê ! x # $ Ê ! x 2
" 1 x#
" x"
œx"
$ x"
œ _.
B Í 1 x#
" #B . Then " $ x " Ê " 1 x# B for ! x 1 and x# x"
" x
œ _.
B Í x " # B Í (x 2)
98. For B 0 and ! x 1, $
Ê
Í B" x. Choose $ œ B" . Then $ x !
2 $ x 2 Ê $ x 2 ! Ê B" x 2 0 Ê 97. For B 0,
" B
" B
Í (" x)(" x) B" . Now
$ x 1 0 Ê " x $ x near 1 Ê
"
lim # x Ä "c " x
" #B
1x #
1 since x 1. Choose Ê (" x)(" x) B" ˆ 1 # x ‰ B"
œ _.
100. y œ
x# " x1
œx"
102. y œ
x2 " #x %
œ #" x "
# x1
$ #x %
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" B
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 103. y œ
x# 1 x
105. y œ
x È 4 x#
œx
107. y œ x#Î$
" x
104. y œ
x$ 1 x#
106. y œ
" È 4 x#
œx
" x#
108. y œ sin ˆ x# 1 1 ‰
" x"Î$
109. (a) y Ä _ (see accompanying graph) (b) y Ä _ (see accompanying graph) (c) cusps at x œ „ 1 (see accompanying graph)
110. (a) y Ä 0 and a cusp at x œ 0 (see the accompanying graph) (b) y Ä 32 (see accompanying graph) (c) a vertical asymptote at x œ 1 and contains the point Š1,
3 ‹ 3 2È 4
(see accompanying graph)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
81
82
Chapter 2 Limits and Continuity
CHAPTER 2 PRACTICE EXERCISES 1. At x œ 1: Ê
f(x) œ
lim
x Ä "c
lim
x Ä "b
f(x) œ 1
lim f(x) œ 1 œ f(1)
x Ä 1
Ê f is continuous at x œ 1. At x œ 0: lim c f(x) œ lim b f(x) œ 0 Ê lim f(x) œ 0. xÄ!
xÄ!
xÄ!
But f(0) œ 1 Á lim f(x) xÄ!
Ê f is discontinuous at x œ 0. If we define fa!b œ !, then the discontinuity at x œ ! is removable. At x œ 1: lim c f(x) œ 1 and lim f(x) œ 1 xÄ"
Ê lim f(x) does not exist
xÄ"
xÄ1
Ê f is discontinuous at x œ 1. 2. At x œ 1: Ê
f(x) œ 0 and
lim
x Ä "
lim
x Ä "
f(x) œ 1
lim f(x) does not exist
x Ä "
Ê f is discontinuous at x œ 1. At x œ 0: lim f(x) œ _ and lim f(x) œ _ xÄ!
Ê lim f(x) does not exist
xÄ!
xÄ!
Ê f is discontinuous at x œ 0. At x œ 1: lim f(x) œ lim f(x) œ 1 Ê lim f(x) œ 1. xÄ"
xÄ1
xÄ"
But f(1) œ 0 Á lim f(x) xÄ1
Ê f is discontinuous at x œ 1. If we define fa"b œ ", then the discontinuity at x œ " is removable. 3. (a) (b) (c) (d) (e) (f)
lim a3fatbb œ 3 lim fatb œ 3(7) œ 21
t Ä t!
t Ä t!
#
lim afatbb# œ Š lim fatb‹ œ a(b# œ 49
t Ä t!
t Ä t!
lim afatb † gatbb œ lim fatb † lim gatb œ (7)(0) œ 0
t Ä t!
t Ä t!
lim fatb t Ä t! g(t)7
Ät
t Ä t!
lim fatb
œ
Ät
t
œ
!
lim agatb 7b
t
t
!
Ät
lim fatb
Ät
t
!
Ät
lim gatb lim 7 t
!
!
œ
7 07
œ1
lim cos agatbb œ cos Š lim gatb‹ œ cos ! œ 1
t Ä t!
t Ä t!
lim kfatbk œ ¹ lim fatb¹ œ k7k œ 7
t Ä t!
t Ä t!
(g) lim afatb gatbb œ lim fatb lim gatb œ 7 0 œ 7 t Ä t!
(h)
4. (a) (b) (c) (d)
t Ä t!
lim Š " ‹ t Ä t! fatb
œ
" lim fatb
t
Ät
t Ä t!
" 7
œ
!
œ 71
lim g(x) œ lim g(x) œ È2
xÄ!
xÄ!
lim ag(x) † f(x)b œ lim g(x) † lim f(x) œ ŠÈ2‹ ˆ "# ‰ œ
xÄ!
xÄ!
xÄ!
lim af(x) g(x)b œ lim f(x) lim g(x) œ
xÄ!
"
lim x Ä ! f(x)
œ
" lim f(x)
xÄ!
xÄ!
œ
" " #
œ2
xÄ!
" #
È2 #
È2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 2 Practice Exercises (e) (f)
" #
lim ax f(x)b œ lim x lim f(x) œ 0
xÄ!
xÄ!
f(x)†cos x x 1 xÄ!
lim
xÄ!
lim f(x)† lim cos x
œ
xÄ!
xÄ!
lim x lim 1
xÄ!
xÄ!
œ
ˆ "# ‰ (1) 01
œ
83
" #
œ #"
5. Since lim x œ 0 we must have that lim (4 g(x)) œ 0. Otherwise, if lim (% g(x)) is a finite positive xÄ!
xÄ!
xÄ!
’ 4xg(x) “
’ 4xg(x) “
œ _ and lim b œ _ so the limit could not equal 1 as xÄ! x Ä 0. Similar reasoning holds if lim (4 g(x)) is a finite negative number. We conclude that lim g(x) œ 4. number, we would have lim c xÄ!
xÄ!
6. 2 œ lim
x Ä %
xÄ!
’x lim g(x)“ œ lim x † lim xÄ!
x Ä %
’ lim g(x)“ œ 4 lim xÄ!
x Ä %
(since lim g(x) is a constant) Ê lim g(x) œ xÄ!
xÄ!
2 %
x Ä %
œ #" .
’ lim g(x)“ œ 4 lim g(x) xÄ!
xÄ!
7. (a) xlim faxb œ xlim x"Î$ œ c"Î$ œ facb for every real number c Ê f is continuous on a_ß _b. Äc Äc (b) xlim gaxb œ xlim x$Î% œ c$Î% œ gacb for every nonnegative real number c Ê g is continuous on Ò!ß _Ñ. Äc Äc " c#Î$ " c"Î'
(c) xlim haxb œ xlim x#Î$ œ Äc Äc (d) xlim kaxb œ xlim x"Î' œ Äc Äc
œ hacb for every nonzero real number c Ê h is continuous on a_ß !b and a_ß _b. œ kacb for every positive real number c Ê k is continuous on a!ß _b
8. (a) - ˆˆn "# ‰1ß ˆn "# ‰1‰, where I œ the set of all integers. n−I (b) - an1ß an 1b1b, where I œ the set of all integers. n−I (c) a_ß 1b a1ß _b (d) a_ß !b a!ß _b 9.
(a)
x# 4x 4 x2
lim x Ä !c x(x 7) (b) 10. (a)
#
x# x
œ lim
1 x# (x 1)
x# x
x # a# x % a%
œ xlim Äa
13. lim
(x h)# x# h
œ lim
(x h)# x# h xÄ!
œ lim
xÄ!
x
œ _ and lim b xÄ!
" Èx
ax # a # b ax # a # b a x # a # b
hÄ!
œ lim
xÄ!
1 x# (x 1)
2 (2 x) 2x(# x)
œ lim
#
x x
"
"
œ
œ
œ
0 2(9)
œ0
, x Á 0 and x Á 1.
œ _.
, x Á 0 and x Á 1. The limit does not
1
lim # x Ä "b x (x 1)
"
# x Ä 0 x (x 1)
& % $ x Ä ! x 2x x
# x Ä " x (x 1)
" x # a#
œ xlim Äa
œ lim
œ _ Ê lim
œ lim
x2
x Ä # x(x 7)
x1
x Ä 1 1 Èx
ax# 2hx h# b x# h xÄ!
, x Á 2, and lim
# x Ä ! x (x 1)(x 1)
œ lim
ax# 2hx h# b x# h
x2
œ lim
œ _ and
x Ä 1 ˆ1 È x ‰ ˆ 1 È x ‰
, x Á 2; the limit does not exist because
x Ä # x(x 7)
x(x 1)
"
12. xlim Äa
" " x #
x(x 1)
lim # x Ä "c x (x 1)
œ lim
#
œ lim
$ # x Ä " x ax 2x 1b
1 Èx 1x
15. lim
(x 2)(x 2)
œ lim
11. lim
14. lim
œ _
$ # x Ä ! x ax 2x 1b
lim & % $ x Ä " x 2x x
hÄ!
x2 x(x 7)
x2
x Ä ! x(x 7)
x Ä # x(x 7)(x #)
œ lim
lim & % $ x Ä ! x 2x x
xÄ1
œ lim
œ _ and lim b xÄ!
x 4x 4
exist because
(x 2)(x 2)
x Ä ! x(x 7)(x 2)
lim $ # x Ä # x 5x 14x
Now lim c xÄ! (b)
œ lim
lim $ # x Ä ! x 5x 14x
œ _.
" #
" #a #
œ lim (2x h) œ 2x hÄ!
œ lim (2x h) œ h xÄ!
"
x Ä ! 4 #x
œ "4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
84
Chapter 2 Limits and Continuity (# x)$ 8 x
16. lim
xÄ!
ax$ 6x# 12x 8b 8 x
œ lim
xÄ!
ˆx1Î3 1‰ x1Î3 1 œ lim ˆÈx 1‰ È x 1 xÄ1 xÄ1 œ 1 1 1 1 1 œ 23
17. lim
18.
tan 2x
œ lim
sin 2x
x Ä ! cos 2x
lim csc x œ limc x1
20.
x Ä 1c
21.
xÄ1
22.
xÄ1
ax 1bˆÈx 1‰
œ lim
2Î3 1Î3 x Ä 1 ax 1bax x 1b
Èx 1
œ lim
2Î3 1Î3 x Ä 1 x x 1
1 sin x
†
1x ‰ˆ 1x ‰ˆ 2x ‰ œ lim ˆ sin2x2x ‰ˆ cos cos 2x sin 1x 1x œ 1 † 1 † 1 †
cos 1x sin 1x
xÄ!
2 1
œ
2 1
œ_
lim sin ˆ x2 sin x‰ œ sin ˆ 12 sin 1‰ œ sin ˆ 12 ‰ œ 1
lim cos2 ax tan xb œ cos2 a1 tan 1b œ cos2 a1b œ a1b2 œ 1
23. lim xÄ0 24. lim xÄ0
8x 3sin x x
œ lim xÄ0
cos 2x 1 sin x
2x 1 œ lim ˆ cossin † x xÄ0
8 3 sinx x 1
œ
8 3 a1 b 1
œ4
cos 2x 1 ‰ cos 2x 1
œ lim xÄ0
"Î$
lim [4 g(x)]"Î$ œ 2 Ê ’ lim b 4 g(x)“ x Ä !b xÄ! lim
x Ä È&
27. lim
xÄ1
28.
ˆx2Î3 x1Î3 1‰ˆÈx 1‰ ˆÈx 1‰ax2Î3 x1Î3 1b
ˆx1Î3 4‰ˆx1Î3 4‰ ˆx1Î3 4‰ˆx1Î3 4‰ ˆx2Î3 4x1Î3 16‰ˆÈx )‰ x2Î3 16 œ lim œ lim † ˆÈx )‰ax2Î3 4x1Î3 16b È È Èx 8 x 8 x 8 x Ä 64 x Ä 64 x Ä 64 ˆx1Î3 4‰ˆÈx )‰ ax 64bˆx1Î3 4‰ˆÈx )‰ 4 4ba8 8b 8 œ lim ax 64bax2Î3 4x1Î3 16b œ lim x2Î3 4x1Î3 16 œ a16 16 16 œ 3 x Ä 64 x Ä 64
x Ä ! tan 1x
26.
xÄ!
lim
19. lim
25.
†
œ lim ax# 6x 12b œ 12
" x g(x)
3x# 1 g(x)
xÄ1
5 x#
œ
(x g(x)) œ
lim
x Ä È&
œ2 Ê " #
œ lim xÄ0
sin2 2x sin xacos 2x 1b
œ lim xÄ0
4sin x cos2 x cos 2x 1
œ0 Ê
Ê È5 lim
x Ä È5
g(x) œ
" #
Ê
lim
x Ä È5
g(x) œ
" #
È5
xÄ1
lim g(x) œ _ since lim a5 x# b œ 1
x Ä #
x Ä #
lim f(x) œ lim c x Ä "c x Ä "
lim x Ä "c
x ax # 1 b x# 1
œ
lim
x Ä "c
x ax # 1 b kx # 1 k
x œ 1, and
#
lim f(x) œ lim b xkaxx# 11k b œ lim b x Ä "b x Ä " x Ä " œ lim (x) œ (1) œ 1. Since
x ax # 1 b a x # "b
x Ä 1
lim f(x) Á lim b f(x) x Ä "c x Ä " Ê
lim f(x) does not exist, the function f cannot be
x Ä 1
extended to a continuous function at x œ 1. At x œ 1:
lim f(x) œ lim c
x Ä "c
xÄ"
x ax # 1 b kx # 1 k
x ax # 1 b kx # 1 k
œ lim c xÄ"
x ax # 1 b x# "
x ax # 1 b ax # 1 b
œ lim c (x) œ 1, and xÄ"
lim f(x) œ lim b œ lim b œ lim b x œ 1. Again lim f(x) does not exist so f xÄ1 xÄ" xÄ" xÄ1 cannot be extended to a continuous function at x œ 1 either.
x Ä "b
œ
4a0ba1b2 11
lim 4 g(x) œ 8, since 2$ œ 8. Then lim b g(x) œ 2. xÄ!
x Ä !b
œ _ Ê lim g(x) œ 0 since lim a3x# 1b œ 4
lim x Ä # Èg(x)
29. At x œ 1:
œ2 Ê
cos2 2x 1 sin xacos 2x 1b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ0
Chapter 2 Practice Exercises 30. The discontinuity at x œ 0 of f(x) œ sin ˆ "x ‰ is nonremovable because lim sin xÄ!
" x
does not exist.
31. Yes, f does have a continuous extension to a œ 1: " define f(1) œ lim xxÈ œ 43 . % x xÄ1
32. Yes, g does have a continuous extension to a œ 1# : ) 5 g ˆ 1# ‰ œ lim1 45)cos #1 œ 4 . )Ä #
33. From the graph we see that lim h(t) Á lim h(t) tÄ! tÄ! so h cannot be extended to a continuous function at a œ 0.
34. From the graph we see that lim c k(x) Á lim b k(x) xÄ! xÄ! so k cannot be extended to a continuous function at a œ 0.
35. (a) f(1) œ 1 and f(2) œ 5 Ê f has a root between 1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1.32471795724 36. (a) f(2) œ 2 and f(0) œ 2 Ê f has a root between 2 and 0 by the Intermediate Value Theorem. (b), (c) root is 1.76929235424 # $
#x $ x 37. x lim œ x lim œ Ä _ &x ( Ä _ & (x
x 39. x Ä lim _
#
%x ) $x $
#! &!
ˆ" œxÄ lim _ $x
œ
#
# &
% $x#
#
#x $ 38. x Ä lim œxÄ lim _ &x# ( _ &
) ‰ $x$
$ x# ( x#
œ
#! &!
œ!!!œ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
# &
85
86
Chapter 2 Limits and Continuity "
" x# 40. x lim œ x lim œ Ä _ x # (x " Ä _ " (x x"#
! "!!
œ!
#
%
x (x x( 41. x Ä lim œxÄ lim œ _ _ x 1 _ " "x
$
x x x" 42. x lim œxÄ lim œ_ Ä _ "#x$ "#) _ "# "#) x$
sin x " sin x 43. x lim Ÿ x lim œ ! since int x Ä _ as x Ä _ Êx lim œ !. Ä _ gx h Ä _ gx h Ä _ gx h
44.
lim
)Ä_
45. x lim Ä_
cos ) " )
Ÿ lim
#
) Ä _)
x sin x #Èx x sin x
#Î$
œ ! Ê lim
)Ä_
œ x lim Ä_
cos ) " )
" sinx x È#x " sinx x
"
œ
&Î$
x x " x 46. x lim œ x lim #x œ Ä _ x#Î$ cos# x Ä _Œ " cos#Î$
œ !.
"!! "!
"! "!
œ"
œ"
x
47. (a) y œ (b) y œ
x2 4 x3
is undefined at x œ 3: lim c xx 34 œ _ and lim b xx 34 œ _, thus x œ 3 is a vertical asymptote. xÄ3 xÄ3 2
x2 x 2 x2 2x 1
2
is undefined at x œ 1: lim c xÄ1
x2 x 2 x2 2x 1
œ _ and lim b xÄ1
x2 x 2 x2 2x 1
œ _, thus x œ 1 is a vertical
asymptote. (c) y œ
x2 x ' x2 2x 8
is undefined at x œ 2 and 4: lim
x x' 2
œ lim b x Ä %
lim 2 x Ä %b x 2x 8 48. (a) y œ
1 x2 1 x2 x2 " : x lim Ä _ x2 "
x3 x4
x2 x '
2 x Ä 2 x 2x 8
x3
œ lim
x Ä 2 x4
œ 56 ; lim c x Ä %
x2 x ' x2 2x 8
œ lim c x Ä %
x3 x4
œ_
œ _. Thus x œ 4 is a vertical asymptote. 1
1
x2 œ x lim Ä _ 1
1 x2
œ
1
1
1 1
1x œ 1 and x Ä lim œ lim x2 œ _ x2 " x Ä _ 1 x12
œ
10 È1 0
2
1 1
œ 1, thus y œ 1 is a
horizontal asymptote. (b) y œ
Èx 4 Èx 4 Èx 4 : x lim Ä _ Èx 4
(c) y œ
È x2 4 È x2 4 : x lim x x Ä_
œx Ä lim _
É1 x42 1
œ
œ x lim Ä_
È1 0 1
1 È4x É1 B4
œ x lim Ä_ œ
1 1
É1 x42 1
œ
œ 1 , thus y œ 1 is a horizontal asymptote.
È1 0 1
œ 1 and x lim Ä _
thus y œ
1
2
" 3
œx Ä lim_
É1 x42
È
9
" 3
x xx
2
is a horizontal asymptote.
0.1 0.7943
œx Ä lim _
É1 x42 x
cx
0.01 0.9550
0.001 0.9931
0.0001 0.9991
1
9
x 9 0 " x2 É 9x and x Ä lim lim œ É 19 21 œ 0 œ 3, _ x Ä _ Ê 9 x12
CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES 1. (a)
x x2
œ 1, thus y œ 1 and y œ 1 are horizontal asymptotes.
x 9 x 9 0 x2 (d) y œ É 9x lim É 9x lim œ É 91 21: 21 œ 0 œ xÄ_ x Ä _ Ê 9 x12 2
È x2 4 x
0.00001 0.9999
Apparently, lim b xx œ 1 xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 2 Additional and Advanced Exercises 87 (b)
2. (a)
x ˆ "x ‰"ÎÐln xÑ Apparently,
10
100
1000
0.3679
0.3679
0.3679
"ÎÐln xÑ lim ˆ " ‰ xÄ_ x
œ 0.3678 œ
" e
(b)
3.
lim L œ lim c L! É" v Ä cc vÄc
v# c#
lim v œ L! É1 vÄcc # œ L! É1 #
c# c#
œ0
The left-hand limit was needed because the function L is undefined if v c (the rocket cannot move faster than the speed of light). 4. (a) ¹
Èx #
1¹ 0.2 Ê 0.2
Èx #
1 0.2 Ê 0.8
Èx #
1.2 Ê 1.6 Èx 2.4 Ê 2.56 x 5.76.
(b) ¹
Èx #
1¹ 0.1 Ê 0.1
Èx #
1 0.1 Ê 0.9
Èx #
1.1 Ê 1.8 Èx 2.2 Ê 3.24 x 4.84.
5. k10 (t 70) ‚ 10% 10k 0.0005 Ê k(t 70) ‚ 10% k 0.0005 Ê 0.0005 (t 70) ‚ 10% 0.0005 Ê 5 t 70 5 Ê 65° t 75° Ê Within 5° F. 6. We want to know in what interval to hold values of h to make V satisfy the inequality lV "!!!l œ l$'1h "!!!l Ÿ "!. To find out, we solve the inequality: **! l$'1h "!!!l Ÿ "! Ê "! Ÿ $'1h "!!! Ÿ "! Ê **! Ÿ $'1h Ÿ "!"! Ê $' 1 Ÿ hŸ
"!"! $'1
Ê )Þ) Ÿ h Ÿ )Þ*. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. The interval in which we should hold h is about )Þ* )Þ) œ !Þ" cm wide (1 mm). With stripes 1 mm wide, we can expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.
7. Show lim f(x) œ lim ax# 7b œ ' œ f(1). xÄ1
xÄ1
Step 1: kax 7b 6k % Ê % x# 1 % Ê 1 % x# 1 % Ê È1 % x È1 %. #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
88
Chapter 2 Limits and Continuity Step 2: kx 1k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ È1 % or $ " œ È1 %. Choose $ œ min š1 È1 %ß È1 % 1› , then 0 kx 1k $ Ê kax# (b 6k % and lim f(x) œ 6. By the continuity test, f(x) is continuous at x œ 1. xÄ1
8. Show lim" g(x) œ lim" xÄ
xÄ
%
" 2x
œ 2 œ g ˆ 4" ‰ .
%
Step 1: ¸ #"x 2¸ % Ê % #"x # % Ê # % #"x # % Ê Step 2: ¸B "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" . Then $ Choose $ œ
" 4
œ
" 4 #%
% 4(#%)
Ê $œ
" 4
" 4 #%
œ
% 4(2 %)
, or $
" 4
œ
, the smaller of the two values. Then 0 ¸x
By the continuity test, g(x) is continuous at x œ
" 4
" 4 #% Ê 4" ¸ $
" 4#%
x
" 4 #% ¸ #"x
" 4#%
.
" 4
% 4(2 %)
$œ
œ
Ê
2¸ % and lim"
.
xÄ
%
" #x
œ 2.
.
9. Show lim h(x) œ lim È2x 3 œ " œ h(2). xÄ#
xÄ#
Step 1: ¹È2x 3 1¹ % Ê % È2x 3 " % Ê " % È2x 3 " % Ê
(1 %)# $ #
x
(" %)# 3 . #
Step 2: kx 2k $ Ê $ x 2 $ or $ # x $ #. (" % )# $ Ê $œ # (" % Ñ # $ (" %Ñ# " #œ # #
Then $ # œ
#
Ê $œ
œ%
# (" %)# $ œ " (1# %) # # %# . Choose $ œ %
œ%
%# #,
%# #
, or $ # œ
(" %)# $ #
the smaller of the two values . Then,
! kx 2k $ Ê ¹È2x 3 "¹ %, so lim È2x 3 œ 1. By the continuity test, h(x) is continuous at x œ 2. xÄ#
10. Show lim F(x) œ lim È9 x œ # œ F(5). xÄ&
xÄ&
Step 1: ¹È9 x 2¹ % Ê % È9 x # % Ê 9 (2 %)# x * (# %)# . Step 2: 0 kx 5k $ Ê $ x & $ Ê $ & x $ &. Then $ & œ * (# %)# Ê $ œ (# %)# % œ %# #%, or $ & œ * (# %)# Ê $ œ % (# %)# œ %# #%. Choose $ œ %# #%, the smaller of the two values. Then, ! kx 5k $ Ê ¹È9 x #¹ %, so lim È9 x œ #. By the continuity test, F(x) is continuous at x œ 5.
xÄ&
11. Suppose L" and L# are two different limits. Without loss of generality assume L# L" . Let % œ
" 3
(L# L" ).
Since x lim f(x) œ L" there is a $" 0 such that 0 kx x! k $" Ê kf(x) L" k % Ê % f(x) L" % Äx !
Ê "3 (L# L" ) L" f(x)
" 3
(L# L" ) L" Ê 4L" L# 3f(x) 2L" L# . Likewise, x lim f(x) œ L# Ä x! so there is a $# such that 0 kx x! k $# Ê kf(x) L# k % Ê % f(x) L# % Ê "3 (L# L" ) L# f(x) 3" (L# L" ) L# Ê 2L# L" 3f(x) 4L# L" Ê L" 4L# 3f(x) 2L# L" . If $ œ min e$" ß $# f both inequalities must hold for 0 kx x! k $ : 4L" L# 3f(x) 2L" L# Ê 5(L" L# ) 0 L" L# . That is, L" L# 0 and L" L# 0, L" %L# 3f(x) 2L# L" a contradiction. 12. Suppose xlim f(x) œ L. If k œ !, then xlim kf(x) œ xlim 0 œ ! œ ! † xlim f(x) and we are done. Äc Äc Äc Äc % If k Á 0, then given any % !, there is a $ ! so that ! lx cl $ Ê lfaxb Ll l5l Ê lkllfaxb Ll % Ê lkafaxb Lb| % Ê lakfaxbb akLbl %. Thus, xlim kf(x) œ kL œ kŠxlim f(x)‹. Äc Äc
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 2 Additional and Advanced Exercises 89 13. (a) Since x Ä 0 , 0 x$ x 1 Ê ax$ xb Ä 0 Ê
lim f ax$ xb œ lim c f(y) œ B where y œ x$ x. yÄ!
x Ä !b
(b) Since x Ä 0 , 1 x x$ 0 Ê ax$ xb Ä 0 Ê
(c) Since x Ä 0 , 0 x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax$ xb œ lim b f(y) œ A where y œ x$ x. yÄ!
x Ä !c
lim f ax# x% b œ lim b f(y) œ A where y œ x# x% . yÄ!
x Ä !b
(d) Since x Ä 0 , 1 x 0 Ê ! x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax# x% b œ A as in part (c).
x Ä !b
14. (a) True, because if xlim (f(x) g(x)) exists then xlim (f(x) g(x)) xlim f(x) œ xlim [(f(x) g(x)) f(x)] Äa Äa Äa Äa œ xlim g(x) exists, contrary to assumption. Äa " x
(b) False; for example take f(x) œ
and g(x) œ x" . Then neither lim f(x) nor lim g(x) exists, but xÄ!
lim (f(x) g(x)) œ lim ˆ "x x" ‰ œ lim 0 œ 0 exists.
xÄ!
xÄ!
xÄ!
xÄ!
(c) True, because g(x) œ kxk is continuous Ê g(f(x)) œ kf(x)k is continuous (it is the composite of continuous functions). 1, x Ÿ 0 Ê f(x) is discontinuous at x œ 0. However kf(x)k œ 1 is (d) False; for example let f(x) œ œ 1, x 0 continuous at x œ 0. x# "
15. Show lim f(x) œ lim
x Ä 1 x 1
x Ä 1
(x 1)(x ") (x 1)
œ lim
x Ä 1
Define the continuous extension of f(x) as F(x) œ œ
œ #, x Á 1.
x# 1 x1 ,
2
x Á " . We now prove the limit of f(x) as x Ä 1 , x œ 1
exists and has the correct value. #
Step 1: ¹ xx 1" (#)¹ % Ê %
(x 1)(x ") (x 1)
# % Ê % (x 1) # %, x Á " Ê % " x % ".
Step 2: kx (1)k $ Ê $ x 1 $ Ê $ " x $ ". Then $ " œ % " Ê $ œ %, or $ " œ % " Ê $ œ %. Choose $ œ %. Then ! kx (1)k $ #
Ê ¹ xx 1" a#b¹ % Ê
lim F(x) œ 2. Since the conditions of the continuity test are met by F(x), then f(x) has a
x Ä 1
continuous extension to F(x) at x œ 1. 16. Show lim g(x) œ lim xÄ$
xÄ$
x# 2x 3 2x 6
œ lim
xÄ$
(x 3)(x ") 2(x 3)
œ #, x Á 3. #
Define the continuous extension of g(x) as G(x) œ œ
x 2x 3 2x 6 ,
2
xÁ3 . We now prove the limit of g(x) as , xœ3
x Ä 3 exists and has the correct value. Step 1: ¹ x
#
2x 3 #x 6
2¹ % Ê %
(x 3)(x ") 2(x 3)
# % Ê %
x" #
# % , x Á $ Ê $ #% x $ #% .
Step 2: kx 3k $ Ê $ x 3 $ Ê $ $ x $ $. Then, $ $ œ $ #% Ê $ œ #%, or $ $ œ $ #% Ê $ œ #%. Choose $ œ #%. Then ! kx 3k $ Ê ¹x
#
2x 3 2x 6
2¹ % Ê lim
xÄ$
(x 3)(x ") #(x 3)
œ 2. Since the conditions of the continuity test hold for G(x),
g(x) can be continuously extended to G(x) at B œ 3. 17. (a) Let % ! be given. If x is rational, then f(x) œ x Ê kf(x) 0k œ kx 0k % Í kx 0k %; i.e., choose $ œ %. Then kx 0k $ Ê kf(x) 0k % for x rational. If x is irrational, then f(x) œ 0 Ê kf(x) 0k % Í ! % which is true no matter how close irrational x is to 0, so again we can choose $ œ %. In either case, given % ! there is a $ œ % ! such that ! kx 0k $ Ê kf(x) 0k %. Therefore, f is continuous at x œ 0. (b) Choose x œ c !. Then within any interval (c $ ß c $ ) there are both rational and irrational numbers. If c is rational, pick % œ #c . No matter how small we choose $ ! there is an irrational number x in (c $ ß c $ ) Ê kf(x) f(c)k œ k0 ck œ c
c #
œ %. That is, f is not continuous at any rational c 0. On
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90
Chapter 2 Limits and Continuity the other hand, suppose c is irrational Ê f(c) œ 0. Again pick % œ #c . No matter how small we choose $ ! there is a rational number x in (c $ ß c $ ) with kx ck œ kxk
c #
œ% Í
œ % Ê f is not continuous at any irrational c 0.
If x œ c 0, repeat the argument picking % œ nonzero value x œ c. 18. (a) Let c œ
c #
kc k #
œ
c # .
x
c #
Then kf(x) f(c)k œ kx 0k
3c #.
Therefore f fails to be continuous at any
m n
be a rational number in [0ß 1] reduced to lowest terms Ê f(c) œ "n . Pick % œ
" #n
œ %. Therefore f is discontinuous at x œ c, a rational number.
" #n .
No matter how small $ ! is taken, there is an irrational number x in the interval (c $ ß c $ ) Ê kf(x) f(c)k œ ¸0 "n ¸ œ
" n
(b) Now suppose c is an irrational number Ê f(c) œ 0. Let % 0 be given. Notice that number reduced to lowest terms with denominator 2 and belonging to [0ß 1]; denominator 3 belonging to [0ß 1];
" 4
and
[0ß 1]; etc. In general, choose N so that
" N
3 4
with denominator 4 in [0ß 1];
" 3
and
" 2 3 5, 5, 5
2 3
and
" #
is the only rational
the only rationals with 4 5
with denominator 5 in
% Ê there exist only finitely many rationals in [!ß "] having
denominator Ÿ N, say r" , r# , á , rp . Let $ œ min ekc ri k : i œ 1ß á ß pf . Then the interval (c $ ß c $ ) contains no rational numbers with denominator Ÿ N. Thus, 0 kx ck $ Ê kf(x) f(c)k œ kf(x) 0k œ kf(x)k Ÿ N" % Ê f is continuous at x œ c irrational. (c) The graph looks like the markings on a typical ruler when the points (xß f(x)) on the graph of f(x) are connected to the x-axis with vertical lines.
19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator Ê 0 1R represents the midnight point (at the same exact time). Suppose x" is a point on the equator “just after" noon Ê x" 1R is simultaneously “just after" midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T(x" ) T(x" 1R) 0. At exactly the same moment in time pick x# to be a point just before midnight Ê x# 1R is just before noon. Then T(x# ) T(x# 1R) 0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between 0 (noon) and 1R (simultaneously midnight) such that T(c) T(c 1R) œ 0; i.e., there is always a pair of antipodal points on the earth's equator where the temperatures are the same. #
#
# # " 20. xlim f(x)g(x) œ xlim af(x) g(x)b‹ Šxlim af(x) g(x)b‹ “ ’af(x) g(x)b af(x) g(x)b “ œ "% ’Šxlim Äc Äc % Äc Äc œ "% ˆ$# a"b# ‰ œ #.
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Chapter 2 Additional and Advanced Exercises 91 21. (a) At x œ 0: lim r (a) œ lim aÄ!
œ lim
1 (" a)
aÄ!
a Ä ! a ˆ" È1 a‰
At x œ 1: (b) At x œ 0:
lim
a Ä "b
œ
r (a) œ
" È1 a a 1 " È1 0
œ lim c aÄ!
1 (" a) a ˆ" È1 a‰
" #
œ
aÄ!
1 (1 a)
lim
a Ä "b a ˆ1 È1 a‰
lim r (a) œ lim c aÄ!
a Ä !c
È1 a
œ lim Š " a
" È1 a a
œ lim c aÄ!
a
œ lim
a Ä 1 a ˆ" È1 a‰ È1 a
œ lim c Š " a aÄ!
a a ˆ 1 È 1 a ‰
" È1 a
‹ Š " È1 a ‹
œ lim c aÄ!
œ
" " È0
œ1
" È1 a
‹ Š " È1 a ‹
" œ _ (because the " È1 a " œ _ (because the " È1 a
denominator is always negative); lim b r (a) œ lim b aÄ! aÄ! is always positive). Therefore, lim r (a) does not exist.
denominator
aÄ!
At x œ 1:
lim
a Ä "b
r (a) œ
lim
a Ä "b
1 È 1 a a
œ
lim
"
a Ä 1b " È1 a
œ1
(c)
(d)
22. f(x) œ x 2 cos x Ê f(0) œ 0 2 cos 0 œ 2 0 and f(1) œ 1 2 cos (1) œ 1 # 0. Since f(x) is continuous on [1ß !], by the Intermediate Value Theorem, f(x) must take on every value between [1 #ß #]. Thus there is some number c in [1ß !] such that f(c) œ 0; i.e., c is a solution to x 2 cos x œ 0. 23. (a) The function f is bounded on D if f(x) M and f(x) Ÿ N for all x in D. This means M Ÿ f(x) Ÿ N for all x in D. Choose B to be max ekMk ß kNkf . Then kf(x)k Ÿ B. On the other hand, if kf(x)k Ÿ B, then B Ÿ f(x) Ÿ B Ê f(x) B and f(x) Ÿ B Ê f(x) is bounded on D with N œ B an upper bound and M œ B a lower bound. (b) Assume f(x) Ÿ N for all x and that L N. Let % œ L # N . Since x lim f(x) œ L there is a $ ! such that Äx !
0 kx x! k $ Ê kf(x) Lk % Í L % f(x) L % Í L Í
LN #
f(x)
3L N # .
But L N Ê
LN #
LN #
f(x) L
LN #
N Ê N f(x) contrary to the boundedness assumption
f(x) Ÿ N. This contradiction proves L Ÿ N.
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92
Chapter 2 Limits and Continuity (c) Assume M Ÿ f(x) for all x and that L M. Let % œ Ê L
ML #
f(x) L
ML #
3L M #
Í
f(x)
ML # . As in part (b), 0 kx L M M, a contradiction. #
24. (a) If a b, then a b 0 Ê ka bk œ a b Ê max Öaß b× œ
ab #
ka b k #
If a Ÿ b, then a b Ÿ 0 Ê ka bk œ (a b) œ b a Ê max Öaß b× œ œ
2b #
xÄ0
ab #
sina" cos xb x
œ lim
lim b sinsinÈxx œ
xÄ0
†
sin x
xÄ0
sinasin xb x
xÄ0
sinax# xb x xÄ0
œ lim
29. lim
sinax# %b x2
œ lim
30. lim
sinˆÈx $‰ x9
28. lim
xÄ2
xÄ9
sin x " cos x
xÄ0
œ lim
ka b k #
.
sina" cos xb " cos x
lim b sinB x †
xÄ0
27. lim
œ lim
xÄ0 x
26.
ab ab 2a # # œ # œ a. ka b k ab œ a # b b # a # #
œ
œ b.
(b) Let min Öaß b× œ 25. lim œ
x! k $
sinasin xb sin x
†
" cos x x
Èx sin Èx
†
†
œ lim
sin x x
x Èx
xÄ0
sinasin xb sin x
sinax# %b x# %
† ax 2b œ lim
xÄ9
xÄ0
sinˆÈx $‰ Èx $
† lim
" cos# x
x Ä 0 xa" cos xb
" Èx $
† lim
sin x
xÄ0 x
œ " † lim
sin# x
x Ä 0 xa" cos xb
œ " † " œ ".
sinax# xb # x Ä 0 x x
† lim ax "b œ " † " œ "
sinax# %b x# %
† lim ax 2b œ " † % œ %
xÄ2
†
sina" cos xb " cos x
œ " † lim b sin"Èx † lim b Èx œ " † ! † ! œ !. x Ä 0 Š Èx ‹ x Ä 0
† ax "b œ lim
œ lim
œ lim
œ " † ˆ #! ‰ œ !.
sinax# xb # x Ä 0 x x
xÄ2
" cos x " cos x
†
œ lim
xÄ9
xÄ0
xÄ2
sinˆÈx $‰ Èx $
† lim
"
x Ä 9 Èx $
œ"†
" '
œ
" '
31. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is an oblique asymptote. y œ 32. As x Ä „ _,
2x3Î2 2x 3 Èx 1 1 x
œ 2x
3 Èx 1 ,
thus the oblique asymptote is y œ 2x.
Ä 0 Ê sinˆ 1x ‰ Ä 0 Ê 1 sinˆ 1x ‰ Ä 1, thus as x Ä „ _, y œ x x sinˆ 1x ‰ œ xˆ1 sinˆ 1x ‰‰ Ä x;
thus the oblique asymptote is y œ x. 33. As x Ä „ _, x2 1 Ä x2 Ê Èx2 1 Ä Èx2 ; as x Ä _, Èx2 œ x, and as x Ä _, Èx2 œ x; thus the oblique asymptotes are y œ x and y œ x. 34. As x Ä „ _, x 2 Ä x Ê Èx2 2x œ Èxax 2b Ä Èx2 ; as x Ä _, Èx2 œ x, and as x Ä _, Èx2 œ x; asymptotes are y œ x and y œ x.
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CHAPTER 3 DIFFERENTIATION 3.1 TANGENTS AND THE DERIVATIVE AT A POINT 1. P" : m" œ 1, P# : m# œ 5
2. P" : m" œ 2, P# : m# œ 0
3. P" : m" œ 5# , P# : m# œ "#
4. P" : m" œ 3, P# : m# œ 3
5. m œ lim
hÄ!
c4 (" h)# d a4 (1)# b h
a1 2h h# b1 h hÄ!
œ lim
œ lim
hÄ!
h(# h) h
œ 2;
at ("ß $): y œ $ #(x (1)) Ê y œ 2x 5, tangent line
6. m œ lim
hÄ!
c(1 h 1)# 1d c(" ")# 1d h
h#
œ lim
hÄ! h
œ lim h œ 0; at ("ß "): y œ 1 0(x 1) Ê y œ 1, hÄ!
tangent line
È 2È 1 h 2È 1 œ lim 2 1 h h 2 h hÄ! hÄ! 4(1 h) 4 œ lim œ lim È1 2h 1 h Ä ! 2h ŠÈ1 h 1‹ hÄ!
7. m œ lim
†
2È 1 h 2 2È 1 h #
œ 1;
at ("ß #): y œ 2 1(x 1) Ê y œ x 1, tangent line
8. m œ lim
hÄ!
"
( 1 h)#
( "")#
h
a2h h# b # h Ä ! h(1 h)
œ lim
1 (1 h)# # h Ä ! h(1h) 2h lim # œ 2; h Ä ! (1 h)
œ lim œ
at ("ß "): y œ 1 2(x (1)) Ê y œ 2x 3, tangent line
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94
Chapter 3 Differentiation (2 h)$ (2)$ h
9. m œ lim
hÄ!
8 12h 6h# h$ 8 h
œ lim
hÄ!
œ lim a12 6h h# b œ 12; hÄ!
at (2ß 8): y œ 8 12(x (2)) Ê y œ 12x 16, tangent line
10. m œ lim
(
h
hÄ!
œ
"2 8(8)
hÄ!
hÄ!
at ˆ#ß "8 ‰ : y œ 8" Ê yœ 11. m œ lim
hÄ!
x
" #,
12 6h h# 8(2 h)$
œ lim
3 œ 16 ;
3 16
8 (# h)$ 8h(# h)$
œ lim
a12h 6h# h$ b 8h(# h)$
œ lim
hÄ!
" " # h)$ ( #)$
3 16 (x
(2))
tangent line
c(2 h)# 1d 5 h
œ lim
hÄ!
a5 4h h# b 5 h
hÄ!
at (2ß 5): y 5 œ 4(x 2), tangent line 12. m œ lim
hÄ!
c(" h) 2(1 h)# d (1) h
œ lim
hÄ!
h(4 h) h
œ lim
a1 h 2 4h 2h# b 1 h
3 (3
h h) 2
3
h
hÄ!
œ lim
hÄ!
(3 h) 3(h 1) h(h 1)
h Ä ! h(h 1)
at ($ß $): y 3 œ 2(x 3), tangent line 14. m œ lim
hÄ!
8 (2
h)#
2
h
hÄ!
(2 h)$ 8 h hÄ!
œ lim
a8 12h 6h# h$ b 8 h hÄ!
œ lim
16. m œ lim
hÄ!
c(1 h)$ 3(1 h)d 4 h
hÄ!
at ("ß %): y 4 œ 6(t 1), tangent line 17. m œ lim
hÄ!
È4 h 2 h
œ lim
hÄ!
œ "4 ; at (%ß #): y 2 œ 18. m œ lim
hÄ!
œ
" È9 3
È(8 h) 1 3 h
" 4
È4 h 2 h
†
hÄ!
h a12 6h h# b h hÄ!
a1 3h 3h# h$ 3 3hb 4 h
œ lim
œ lim
œ lim
at (2ß )): y 8 œ 12(t 2), tangent line
È4 h 2 È4 h 2
œ 3;
œ 2;
8 2 a4 4h h# b h(2 h)# hÄ!
8 2(2 h)# # h Ä ! h(2 h)
œ lim
at (2ß 2): y 2 œ 2(x 2) 15. m œ lim
2h
œ lim
h(3 2h) h
œ lim
at ("ß "): y 1 œ 3(x 1), tangent line 13. m œ lim
œ %;
œ lim
2h(4 h) h(2 h)#
œ
8 4
œ 2;
œ 12;
œ lim
hÄ!
(4 h) 4
h Ä ! h ŠÈ4 h #‹
h a6 3h h# b h
œ lim
œ 6;
h
h Ä ! h ŠÈ4 h #‹
œ
" È4 #
(x 4), tangent line
œ lim
hÄ!
È9 h 3 h
œ 6" ; at (8ß 3): y 3 œ
19. At x œ 1, y œ 5 Ê m œ lim
hÄ!
" 6
†
È9 h 3 È9 h 3
œ lim
(9 h) 9
h Ä ! h ŠÈ9 h 3‹
œ lim
h
h Ä ! h ŠÈ9 h 3‹
(x 8), tangent line
5(" h)# 5 h
œ lim
hÄ!
5 a1 2h h# b 5 h
œ lim
hÄ!
5h(2 h) h
œ 10, slope
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Section 3.1 Tangents and the Derivative at a Point c1 (2 h)# d (3) h
20. At x œ 2, y œ 3 Ê m œ lim
hÄ!
" #
21. At x œ 3, y œ
Ê m œ lim
"
h) 1
(3
#"
h1 h 1
22. At x œ 0, y œ 1 Ê m œ lim
hÄ!
hÄ!
2 (2 h) 2h(2 h)
œ lim
h
hÄ!
œ lim
hÄ!
(1) h
a1 4 4h h# b 3 h
hÄ!
hÄ!
h
œ lim
œ lim
h(4 h) h
œ 4, slope
œ "4 , slope
h Ä ! 2h(2 h)
(h 1) (h ") h(h 1)
œ lim
œ lim
2h
h Ä ! h(h 1)
œ 2, slope
c(x h)# 4(x h) 1d ax# 4x 1b h hÄ! a2xh h# 4hb lim œ lim (2x h 4) œ 2x h hÄ! hÄ!
23. At a horizontal tangent the slope m œ 0 Ê 0 œ m œ lim ax# 2xh h# 4x 4h 1b ax# 4x 1b h hÄ!
œ lim
œ
4;
2x 4 œ 0 Ê x œ 2. Then f(2) œ 4 8 1 œ 5 Ê (2ß 5) is the point on the graph where there is a horizontal tangent. c(x h)$ 3(x h)d ax$ 3xb h
24. 0 œ m œ lim
hÄ!
œ lim
hÄ!
3x# h 3xh# h$ 3h h
ax$ 3x# h 3xh# h$ 3x 3hb ax$ 3xb h
œ lim
hÄ!
œ lim a3x# 3xh h# 3b œ 3x# 3; 3x# 3 œ 0 Ê x œ 1 or x œ 1. Then hÄ!
f(1) œ 2 and f(1) œ 2 Ê ("ß 2) and ("ß 2) are the points on the graph where a horizontal tangent exists. 25. 1 œ m œ lim
"
h) 1
(x
x " 1
h
hÄ!
œ lim
hÄ!
(x 1) (x h 1) h(x 1)(x h 1)
h
œ lim
h Ä ! h(x 1)(x h 1)
œ (x " 1)#
Ê (x 1)# œ 1 Ê x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ 1 and m œ 1 Ê y œ 1 (x 0) œ (x 1). If x œ 2, then y œ 1 and m œ 1 Ê y œ 1 (x 2) œ (x 3). 26.
" 4
Èx h Èx
œ m œ lim œ lim
h
y œ 2 "4 (x 4) œ
hÄ!
f(2 h) f(2) h
x 4
Èx h Èx h
hÄ!
h Ä ! h ŠÈx h Èx‹
27. lim
œ lim
h
hÄ!
œ
" #È x
. Thus,
" 4
œ
†
Èx h Èx Èx h Èx
" #Èx
(x h) x
œ lim
h Ä ! h ŠÈx h Èx‹
Ê Èx œ 2 Ê x œ 4 Ê y œ 2. The tangent line is
1.
œ lim
hÄ!
a100 4.9(# h)# b a100 4.9(2)# b h
4.9 a4 4h h# b 4.9(4) h
œ lim
hÄ!
œ lim (19.6 4.9h) œ 19.6. The minus sign indicates the object is falling downward at a speed of 19.6 m/sec. hÄ!
28. lim
hÄ!
f(10 h) f(10) h
hÄ!
1(3 h)# 1(3)# h hÄ!
f(3 h) f(3) h hÄ!
œ lim
f(2 h) f(2) h hÄ!
œ lim
29. lim
30. lim
3(10 h)# 3(10)# h
œ lim
hÄ!
41 3
3 a20h h# b h
œ lim
hÄ!
œ 60 ft/sec.
1 c9 6h h# 9d h hÄ!
œ lim
(2 h)$ 431 (2)$ h
œ lim
41 3
hÄ!
31. At ax0 , mx0 bb the slope of the tangent line is lim
hÄ!
œ lim 1(6 h) œ 61 hÄ!
c12h 6h# h$ d h
œ lim
hÄ!
amax0 hb bb am x0 bb ax 0 h b x 0
41 3
c12 6h h# d œ 161
œ lim
hÄ!
mh h
The equation of the tangent line is y am x0 bb œ max x0 b Ê y œ mx b. 32. At x œ 4, y œ
1 È4 È
œ
" #
œ lim – 22hÈ44hh † hÄ!
and m œ lim
hÄ!
2 È4 h 2 È4 h —
È4
1 h
h
"#
œ lim – hÄ!
È4
1 h
h
"#
†
2È 4 h 2È 4 h —
œ lim m œ m. hÄ!
È
œ lim Š 22hÈ44hh ‹ hÄ!
hb h œ lim È 4 a4 È œ lim È 2h 4 hŠ2 4 h‹ 2h 4 hŠ2 È4 h‹ hÄ! hÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
95
96
Chapter 3 Differentiation 1 1 œ lim È œ È 1 È œ 16 2 4 hŠ2 È4 h‹ 2 4Š2 4‹ hÄ!
f(0 h) f(0) h hÄ!
33. Slope at origin œ lim
h# sin ˆ "h ‰ h hÄ!
œ lim
œ lim h sin ˆ "h ‰ œ 0 Ê yes, f(x) does have a tangent at hÄ!
the origin with slope 0. g(0 h) g(0) h
34. lim
hÄ!
œ lim
hÄ!
h sin ˆ "h ‰ h
œ lim sin h" . Since lim sin hÄ!
hÄ!
" h
does not exist, f(x) has no tangent at
the origin. 35.
lim
h Ä !c
f(0 h) f(0) h
lim f(0 h)h f(0) hÄ! 36.
œ lim c hÄ!
1 0 h
œ _, and lim b hÄ!
f(0 h) f(0) h
10 h
œ lim b hÄ!
œ _ Ê yes, the graph of f has a vertical tangent at the origin.
œ _, and lim b U(0 h)h U(0) œ lim b hÄ! hÄ! does not have a vertical tangent at (!ß ") because the limit does not exist. lim
h Ä !c
œ _. Therefore,
U(0 h) U(0) h
œ lim c hÄ!
01 h
11 h
œ 0 Ê no, the graph of f
37. (a) The graph appears to have a cusp at x œ 0.
(b)
lim c
hÄ!
f(0 h) f(0) h
œ lim c hÄ!
h#Î& 0 h
œ lim c hÄ!
" h$Î&
œ _ and lim b hÄ!
" h$Î&
œ _ Ê limit does not exist
Ê the graph of y œ x#Î& does not have a vertical tangent at x œ 0.
38. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0) h
œ lim c hÄ!
h%Î& 0 h
œ lim c hÄ!
" h"Î&
œ _ and lim b hÄ!
" h"Î&
œ _ Ê limit does not exist
Ê y œ x%Î& does not have a vertical tangent at x œ 0.
39. (a) The graph appears to have a vertical tangent at x œ !.
(b)
lim
hÄ!
f(0 h) f(0) h
œ lim
hÄ!
h"Î& 0 h
œ lim
"
%Î& hÄ! h
œ _ Ê y œ x"Î& has a vertical tangent at x œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.1 Tangents and the Derivative at a Point 40. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0) h
œ lim
hÄ!
h$Î& 0 h
"
œ lim
#Î& hÄ! h
œ _ Ê the graph of y œ x$Î& has a vertical tangent at x œ 0.
41. (a) The graph appears to have a cusp at x œ 0.
(b)
lim c
hÄ!
f(0 h) f(0) h
œ lim c hÄ!
4h#Î& 2h h
œ lim c hÄ!
4 h$Î&
2 œ _ and lim b hÄ!
4 h$Î&
#œ_
Ê limit does not exist Ê the graph of y œ 4x#Î& 2x does not have a vertical tangent at x œ 0.
42. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0) h
œ lim
hÄ!
h&Î$ 5h#Î$ h
œ lim h#Î$ hÄ!
5 h"Î$
œ 0 lim
y œ x&Î$ 5x#Î$ does not have a vertical tangent at x œ !.
5
"Î$ hÄ! h
does not exist Ê the graph of
43. (a) The graph appears to have a vertical tangent at x œ 1 and a cusp at x œ 0.
(b) x œ 1:
lim
hÄ!
(1 h)#Î$ (1 h 1)"Î$ " h
œ lim
hÄ!
(1 h)#Î$ h"Î$ " h
œ _
Ê y œ x#Î$ (x 1)"Î$ has a vertical tangent at x œ 1;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
97
98
Chapter 3 Differentiation x œ 0:
lim
hÄ!
f(0 h) f(0) h
œ lim
hÄ!
h#Î$ (h 1)"Î$ (1)"Î$ h
" œ lim ’ h"Î$
hÄ!
(h ")"Î$ h
h" “
does not exist Ê y œ x#Î$ (x 1)"Î$ does not have a vertical tangent at x œ 0. 44. (a) The graph appears to have vertical tangents at x œ 0 and x œ 1.
(b) x œ 0:
lim
hÄ!
f(0 h) f(0) h
œ lim
hÄ!
h"Î$ (h 1)"Î$ (")"Î$ h
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ 0;
x œ 1:
lim
hÄ!
f(1 h) f(1) h
œ lim
hÄ!
(1 h)"Î$ (" h 1)"Î$ 1 h
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ ".
45. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim b
hÄ!
f(0 h) f(0) h
œ lim b xÄ!
Èh 0 h
œ lim
"
h Ä ! Èh
È kh k 0
f(0 h) f(0) h
œ lim c œ lim c h hÄ! hÄ! Ê y has a vertical tangent at x œ 0. lim
h Ä !c
œ _; È kh k kh k
œ lim c hÄ!
" È kh k
œ_
46. (a) The graph appears to have a cusp at x œ 4.
(b)
lim b
f(4 h) f(4) h
œ lim b hÄ!
Èk4 (4 h)k 0 h
lim
f(4 h) f(4) h
œ lim c
Èk4 (4 h)k h
hÄ!
h Ä !c
hÄ!
œ lim b hÄ!
œ lim c hÄ!
È kh k h
È kh k lhl
œ lim b hÄ!
œ lim c hÄ!
" Èh
" È kh k
œ _;
œ _
Ê y œ È% x does not have a vertical tangent at x œ 4. 47-50. Example CAS commands: Maple: f := x -> x^3 + 2*x;x0 := 0; plot( f(x), x=x0-1/2..x0+3, color=black, title="Section 3.1, #47(a)" ); q := unapply( (f(x0+h)-f(x0))/h, h );
# part (a) # part (b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.2 The Derivative as a Function L := limit( q(h), h=0 ); # part (c) sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 ); # part (d) tan_line := f(x0) + L*(x-x0); plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black, linestyle=[1,2,5,6,7], title="Section 3.1, #47(d)", legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)", "Secant line (h=2)","Secant line (h=3)"] ); Mathematica: (function and value for x0 may change) Clear[f, m, x, h] x0 œ p; f[x_]: œ Cos[x] 4Sin[2x] Plot[f[x], {x, x0 1, x0 3}] dq[h_]: œ (f[x0+h] f[x0])/h m œ Limit[dq[h], h Ä 0] ytan: œ f[x0] m(x x0) y1: œ f[x0] dq[1](x x0) y2: œ f[x0] dq[2](x x0) y3: œ f[x0] dq[3](x x0) Plot[{f[x], ytan, y1, y2, y3}, {x, x0 1, x0 3}] 3.2 THE DERIVATIVE AS A FUNCTION 1. Step 1: f(x) œ 4 x# and f(x h) œ 4 (x h)# f(x h) f(x) h
Step 2:
œ
c4 (x h)# d a4 x# b h
œ
a4 x# 2xh h# b 4 x# h
œ
2xh h# h
œ
h(2x h) h
œ 2x h Step 3: f w (x) œ lim (2x h) œ 2x; f w ($) œ 6, f w (0) œ 0, f w (1) œ 2 hÄ!
2. F(x) œ (x 1)# 1 and F(x h) œ (x h 1)# " Ê Fw (x) œ lim œ lim
hÄ!
hÄ!
ax# 2xh h# 2x 2h 1 1b ax# 2x 1 1b h w
w
œ lim
w
œ 2(x 1); F (1) œ 4, F (0) œ 2, F (2) œ 2 3. Step 1: g(t) œ
" t#
and g(t h) œ "
Step 2:
"
# # g(t h) g(t) œ (t h)h t h 2t h) 2t h œ h( (t h)# t# h œ (t h)# t#
Step 3: gw (t) œ lim
2t h
# # h Ä ! (t h) t
4. k(z) œ
1 z #z
and k(z h) œ
œ œ
" 2z#
œ
Œ
2t t# †t#
1 (z h) 2(z h)
(" z)(z h) lim (1 z h)z #(z h)zh hÄ!
2xh h# 2h h
œ lim (2x h 2) hÄ!
" (t h)#
œ
œ
hÄ!
c(x h 1)# 1d c(x 1)# 1d h
t# (t h)# (t h)# †t#
h
œ
2 t$
t# at# 2th h# b (t h)# †t# †h
œ
œ
2th h# (t h)# t# h
2 ; gw (1) œ 2, gw (2) œ "4 , gw ŠÈ3‹ œ 3È 3
Ê kw (z) œ lim
hÄ!
Š
"
(z h) " z #(z h) #z ‹
# z h z# zh lim z z zh 2(z h)zh hÄ!
h
œ lim
h
h Ä ! 2(z h)zh
œ lim
"
h Ä ! #(z h)z
; kw (") œ "# , kw (1) œ "# , kw ŠÈ2‹ œ 4"
5. Step 1: p()) œ È3) and p() h) œ È3() h)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
99
100
Chapter 3 Differentiation
Step 2:
p() h) p()) h
œ
œ
È3() h) È3) h
3h h ŠÈ3) 3h È3)‹
Step 3: pw ()) œ lim
œ
ŠÈ3) 3h È3)‹ h
œ
3 È3) 3h È3)
3
œ
h Ä ! È3) 3h È3)
†
œ
3 È 3) È 3)
3 2È 3 )
ŠÈ3) 3h È3)‹ ŠÈ3) 3h È3)‹
; pw (1) œ
œ
(3) 3h) 3) h ŠÈ3) 3h È3)‹
, pw (3) œ "# , pw ˆ 32 ‰ œ
3 2È 3
3 #È2
È2s 2h 1 È2s 1 h hÄ!
6. r(s) œ È2s 1 and r(s h) œ È2(s h) 1 Ê rw (s) œ lim œ lim
ŠÈ2s h 1 È2s 1‹ h
hÄ!
œ lim
ŠÈ2s 2h 1 È2s 1‹
†
œ lim
2h
œ
" È2s 1
; rw (0) œ 1, rw (1) œ
2
" È3
, rw ˆ #" ‰ œ
hÄ!
6x# h 6xh# 2h$ h
hÄ!
9. s œ r(t) œ œ lim
t 2t1
Š
dr ds
œ
2 2È2s 1
2 ax$ 3x# h 3xh# h$ b 2x$ h hÄ!
2(x h)$ 2x$ h hÄ!
œ lim
œ lim
œ lim a6x# 6xh 2h# b œ 6x# hÄ!
œ lim
th 2(th)1
and r(t h) œ
(t b h)(2t b 1) c t(2t b 2h b 1) ‹ (2t b 2h b 1)(2t b 1)
h
hÄ!
œ lim
œ
2t# t 2ht h 2t# 2ht t (2t 2h 1)(2t 1)h hÄ! " " (2t 1)(2t 1) œ (2t 1)#
dv dt
œ lim
hÄ!
œ lim
10.
2 È2s 1 È2s 1
3 2 2 3 2 2 3 2 ˆas hb3 2as hb2 3‰ ˆs3 2s2 3‰ œ lim s 3s h 3sh h 2s h 4sh h 3 s 2s 3 h hÄ! hÄ! 2 2 3 2 hˆ3s2 3sh h2 4s h‰ lim 3s h 3sh hh 4sh h œ lim œ lim a3s2 3sh h2 4s hb œ 3s2 2s h hÄ! hÄ! hÄ!
8. r œ s3 2s2 3 Ê œ
dy dx
h a6x# 6xh 2h# b h
œ lim
œ
" È2
7. y œ f(x) œ 2x$ and f(x h) œ 2(x h)$ Ê œ lim
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
h Ä ! È2s 2h 1 È2s 1
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
(2s 2h 1) (2s 1)
œ lim
ŠÈ2s 2h 1 È2s 1‹
’(t h)
hÄ!
ht# h# t h h(t h)t hÄ!
œ lim
Œ
" “ ˆt " ‰ t h h
œ lim
11. p œ f(q) œ
t
" Èq 1
h
hÄ!
(t h)(2t 1) t(2t 2h 1) (2t 2h 1)(2t 1)h h
h Ä ! (2t 2h 1)(2t 1)h
œ lim
hÄ!
t# ht 1 h Ä ! (t h)t
and f(q h) œ œ lim
œ
h
t
" " t h h
t# 1 t#
" È(q h) 1
Ê
h Ä ! (2t 2h 1)(2t 1)
Š
œ lim
h(t
h)t t (t (t h)t
dp dq
h)
‹
h
hÄ!
œ1
"
œ lim
" t#
œ lim
Š È(q
hÄ!
"
h)
1
‹ Š Èq"
1
‹
h
Èq 1 Èq h 1
h Ä ! hÈ q h 1 È q 1
h
hÄ!
t ‰ Š 2(t bt bh)hb 1 ‹ ˆ 2t b 1
œ lim
ds dt
œ lim
œ lim
Èq b 1 c Èq b h b 1 Èq b h b 1 Èq b 1
Ê
œ
ˆÈ q 1 È q h 1 ‰ ˆ È q 1 È q h 1 ‰ 1) (q h 1) † ˆÈq 1 Èq h 1‰ œ lim hÈq h 1(qÈq 1 ˆÈ q 1 È q h 1 ‰ h Ä ! h Èq h 1 Èq 1 hÄ! h " lim œ lim Èq h 1 Èq 1 ˆÈq 1 Èq h 1‰ h Ä ! h È q h 1 È q 1 ˆÈ q 1 È q h 1 ‰ hÄ! " " œ È q 1 È q 1 ˆÈ q 1 È q 1 ‰ 2(q 1) Èq 1
dz dw
œ lim
œ lim œ
12.
Š È3(w " h) 2 h
hÄ!
œ lim
hÄ!
"
È3w 2 ‹
ŠÈ3w 2 È3w 3h 2‹ hÈ3w 3h 2 È3w 2
œ lim
œ lim
È3w 2 È3w 3h 2
h Ä ! hÈ3w 3h 2 È3w 2
†
ŠÈ3w2È3w3h2‹ ŠÈ3w 2 È3w 3h 2‹
3
h Ä ! È3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
œ
œ
œ lim
(3w 2) (3w 3h 2)
h Ä ! hÈ3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
3 È3w 2 È3w 2 ŠÈ3w 2 È3w 2‹
3 2(3w 2) È3w 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.2 The Derivative as a Function 13. f(x) œ x œ
x(x h)# 9x x# (x h) 9(x h) x(x h)h
œ
h(x# xh 9) x(x h)h
" #x
14. k(x) œ w
hÄ!
œ œ 16.
dy dx
œ
œ
x# 9 x#
œ
9 9 (x b h) “ ’x x “
’(x h)
h
x$ 2x# h xh# 9x x$ x# h 9x 9h x(x h)h
œ
x# xh 9 h Ä ! x(x h)
and k(x h) œ
x# h xh# 9h x(x h)h
œ
œ1
9 x#
; m œ f w (3) œ 0
Š # "x h k(x h) k(x) œ lim h h hÄ! hÄ! h " " lim œ lim (2 x)(# x h) œ (2 x)# ; h Ä ! h(2 x)(2 x h) hÄ! " 2 (x h)
Ê kw (x) œ lim
$ # # $ # # $ # c(t h)$ (t h)# d at$ t# b œ lim at 3t h 3th h b h at 2th h b t t h hÄ! hÄ! # # # # $ # lim 3t h 3th hh 2th h œ lim h a3t 3th h h 2t hb œ lim a3t# 3th hÄ! hÄ! hÄ! ¸ 3t# 2t; m œ ds œ 5 dt tœ"
#
" ‹ x
œ lim
ax b hb b 3 1 c ax b hb
œ lim
b3 1x c x
h
hÄ!
œ lim
17. f(x) œ
4h
b h b 3ba1 c xb c ax b 3ba1 c x c hb a1 c x c hba1 c xb h
œ lim
8 ŠÈx 2 Èx h 2‹ hÈ x h 2 È x 2
4
†
8 È(x h) 2
ŠÈx 2 Èx h 2‹
œ
8h hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹
œ
8 Èx 2 Èx 2 ŠÈx 2 Èx 2‹
œ
œ
œ
h# 2t hb
x h 3 x2 xh 3x x 3 x2 3x xh 3h h a1 x h b a 1 x b
4 ; dy ¹ a1 xb2 dx xœ2
f(x h) f(x) h
Ê
ŠÈx 2 Èx h 2‹
œ lim
hÄ!
h Ä ! a1 x hba1 xb
and f(x h) œ
8 Èx 2
ax
œ lim
hÄ!
h Ä ! ha1 x hba1 xb
œ
f(x h) f(x) h
Ê
9 (x h)
; f w (x) œ lim
" 16
k (2) œ ds dt
x# xh 9 x(x h)
œ
(# x) (2 x h) h(2 x)(2 x h)
œ lim
15.
and f(x h) œ (x h)
9 x
œ
œ
4 a3 b 2
4 9
È(x b h) c 2 Èx c 2 8
œ
8
h
8[(x 2) (x h 2)] hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹ 8
Ê f w (x) œ lim
h Ä ! Èx h 2 Èx 2 ŠÈx 2 Èx h 2‹
4 (x 2)Èx 2
; m œ f w (6) œ
4 4È 4
œ "# Ê the equation of the tangent
line at (6ß 4) is y 4 œ "# (x 6) Ê y œ "# x $ % Ê y œ "# x (. ˆ1 È4 (z h)‰ Š1 È4 z‹
18. gw (z) œ lim
h
hÄ!
œ
h
hÄ!
(4 z h) (4 z) lim h Ä ! h ŠÈ4 z h È4 z‹ " œ "# 2È 4 3 "# z $# # Ê w
ŠÈ4 z h È4 z‹
œ lim
œ
h lim h Ä ! h ŠÈ4 z h È4 z‹
†
ŠÈ4 z h È4 z‹ ŠÈ4 z h È4 z‹ "
œ lim
h Ä ! ŠÈ4 z h È4 z‹
œ
" 2È 4 z
m œ gw (3) œ
Ê the equation of the tangent line at ($ß #) is w 2 œ "# (z 3)
Êwœ
œ "# z (# .
19. s œ f(t) œ 1 3t# and f(t h) œ 1 3(t h)# œ 1 3t# 6th 3h# Ê a1 3t# 6th 3h# b a1 3t# b h hÄ!
œ lim
20. y œ f(x) œ " œ lim
" " x
21. r œ f()) œ hÄ!
h
h
hÄ!
œ lim
x
" x
2 È4 )
œ lim (6t 3h) œ 6t Ê hÄ!
and f(x h) œ 1 œ lim
h
h Ä ! x(x h)h
œ
œ lim
Ê
dy dx "
h Ä ! x(x h)
œ
œ lim
" x#
" 3
Ê
È œ
dy dx ¹x= 3
2 È4 () h)
Ê
dr d)
œ lim
hÄ!
f(t h) f(t) h
œ6
f(x h) f(x) h hÄ!
œ lim
œ lim
Š1
x
" ‹ Š1 " ‹ h x h
hÄ!
f() h) f()) œ lim h hÄ! hÄ! È È 2È4 ) #È% ) h Š2 % ) 2 4 ) h‹ lim † È Š2 4 ) #È4 ) h‹ h Ä ! hÈ 4 ) È 4 ) h
and f() h) œ
2È 4 ) 2È 4 ) h hÈ 4 ) È 4 ) h
" xh
ds ¸ dt t=c"
ds dt
È4 c ) c h È4 c ) 2
2
h
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
101
102
Chapter 3 Differentiation 4(% )) 4(% ) h)
œ lim
h Ä ! 2hÈ4 ) È4 ) h ŠÈ4 ) È4 ) h‹
œ
2 (4 )) Š2È4 )‹
œ
" (4 ))È4 )
Ê
dr ¸ d) )œ!
œ lim
2
h Ä ! È4 ) È4 ) h ŠÈ4 ) È% ) h‹
œ
" 8
22. w œ f(z) œ z Èz and f(z h) œ (z h) Èz h Ê œ lim
Šz h Èz h‹ ˆz Èz‰
hÄ!
h
œ 1 lim
(z h) z
h Ä ! h ŠÈz h Èz‹
h Èz h Èz h hÄ!
œ lim
œ 1 lim
"
h Ä ! Èz h Èz
"
dw dz
œ lim
hÄ!
œ lim –1 hÄ!
œ"
" 2È z
Ê
f(z h) f(z) h Èz h Èz h
dw ¸ dz zœ4
œ
†
ŠÈz h Èz‹ ŠÈz h Èz‹ —
5 4
"
fazb faxb a x #b a z # b xz " z #x # 23. f w axb œ zlim œ zlim œ zlim œ zlim œ zlim œ Äx zx Ä x zx Ä x az xbaz #bax #b Ä x az xbaz #bax #b Ä x az #bax #b ˆz2 3z 4‰ ˆx2 3x 4‰
" ax #b #
fazb faxb z 3z x 3x z x 3z 3x 24. f w axb œ zlim œ zlim œ zlim œ zlim zx zx zx Ä x zx Äx Äx Äx az xbaz xb 3‘ az xbaz xb 3az xb az xb 3‘ œ 2x 3 œ zlim œ zlim œ zlim zx zx Äx Äx Äx z
2
2
2
2
x
gazb gaxb z a x "b x a z " b z x " zc" x " 25. gw axb œ zlim œ zlim œ zlim œ zlim œ zlim œ Äx zx Äx zx Ä x az xbaz "bax "b Ä x az xbaz "bax "b Ä x az "bax "b g az b g a x b 26. gw axb œ zlim œ zlim Äx zx Äx
ˆ" Èz‰ˆ" Èx‰ zx
œ zlim Äx
Èz Èx zx
†
Èz Èx Èz Èx
" a x "b #
zx " œ zlim œ zlim œ Ä x az x bˆÈ z È x ‰ Ä x Èz Èx
" #È x
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x œ 0), then positive Ê the slope is always increasing which matches (b). 28. Note that the slope of the tangent line is never negative. For x negative, f#w (x) is positive but decreasing as x increases. When x œ 0, the slope of the tangent line to x is 0. For x 0, f#w (x) is positive and increasing. This graph matches (a). 29. f$ (x) is an oscillating function like the cosine. Everywhere that the graph of f$ has a horizontal tangent we expect f$w to be zero, and (d) matches this condition. 30. The graph matches with (c). 31. (a) f w is not defined at x œ 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For example, lim c xÄ!
f(x) f(0) x0
œ slope of line joining (%ß 0) and (!ß #) œ
" #
but lim b xÄ!
line joining (0ß 2) and ("ß 2) œ 4. Since these values are not equal, f w (0) œ
f(x) f(0) x0
f(0) lim f(x)x 0 xÄ!
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ slope of
does not exist.
Section 3.2 The Derivative as a Function 32. (a)
103
(b) Shift the graph in (a) down 3 units
33.
(b) The fastest is between the 20th and 30th days; slowest is between the 40th and 50th days.
34. (a)
35. Answers may vary. In each case, draw a tangent line and estimate its slope. ‰F (a) i) slope ¸ 1.54 Ê dT ii) slope ¸ 2.86 Ê dt ¸ 1.54 hr iii) slope ¸ 0 Ê
dT dt
¸ 0‰ hrF
iv) slope ¸ 3.75
dT ‰F dt ¸ 2.86 hr ‰F Ê dT dt ¸ 3.75 hr
(b) The tangent with the steepest positive slope appears to occur at t œ 6 Ê 12 p.m. and slope ¸ 7.27 Ê The tangent with the steepest negative slope appears to occur at t œ 12 Ê 6 p.m. and ‰F slope ¸ 8.00 Ê dT dt ¸ 8.00 hr (c)
36. Answers may vary. In each case, draw a tangent line and estimate the slope. lb (a) i) slope ¸ 20.83 Ê dW ii) slope ¸ 35.00 Ê dt ¸ 20.83 month iii) slope ¸ 6.25 Ê
dW dt
dW dt
lb ¸ 35.00 month
lb ¸ 6.25 month
(b) The tangentwith the steepest positive slope appears to occur at t œ 2.7 months. and slope ¸ 7.27 lb Ê dW dt ¸ 53.13 month
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dT dt
¸ 7.27‰ hrF .
104
Chapter 3 Differentiation
(c)
37. Left-hand derivative: For h 0, f(0 h) œ f(h) œ h# (using y œ x# curve) Ê œ lim c hÄ!
h# 0 h
œ lim c h œ 0; hÄ!
Right-hand derivative: For h 0, f(0 h) œ f(h) œ h (using y œ x curve) Ê œ lim b hÄ! Then lim c hÄ!
h0 h
lim
h Ä !c
œ lim b 1 œ 1; hÄ!
f(0 h) f(0) h
Á lim b hÄ!
f(0 h) f(0) h
lim
h Ä !b
œ lim c 0 œ 0; hÄ!
f(1 h) f(1) h
lim
h Ä !c
Right-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2(1 h) œ 2 2h Ê
Then lim c hÄ!
(2 2h)2 h
œ lim b hÄ!
f(1 h) f(1) h
2h h
hÄ!
È1 h " h
œ lim c
lim
h Ä !b
ŠÈ1 h "‹ h
hÄ!
†
ŠÈ1 h "‹ ŠÈ1 h 1‹
œ lim c hÄ!
lim
h Ä !c
Then lim c hÄ!
(2h 1) " h f(1 h) f(1) h
40. Left-hand derivative:
lim
h Ä !c
Right-hand derivative: œ lim b hÄ! Then lim c hÄ!
h h(1 h)
œ lim b 2 œ 2; hÄ!
f(1 h) f(") h
lim b
hÄ!
œ lim b hÄ!
f(1 h) f(1) h
f(1 h) f(1) h
Á lim b hÄ!
(1 h) " h ŠÈ1 h "‹
œ lim c hÄ!
Á lim b hÄ!
" È1 h 1
lim
h Ä !b
Ê the derivative f w (1) does not exist. (1 h) " h
œ lim c hÄ!
f(1 h) f(") h " 1h
f(1 h) f(1) h
f(" h) f(1) h
Right-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ 2(1 h) 1 œ 2h 1 Ê œ lim b hÄ!
œ lim c 1 œ 1; hÄ!
Š 1 " h "‹
œ lim b hÄ!
h
œ lim b hÄ!
Š
1 (1 h) 1 h ‹
h
œ 1; f(1 h) f(1) h
Ê the derivative f w (1) does not exist.
41. f is not continuous at x œ 0 since lim faxb œ does not exist and fa0b œ 1 xÄ!
42. Left-hand derivative: Right-hand derivative: Then lim c hÄ!
g(h) g(0) h
lim
h Ä !c
g(h) g(0) h
lim
h Ä !b
œ lim b hÄ!
22 h
Ê the derivative f w (1) does not exist.
39. Left-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ È1 h Ê œ lim c
œ lim c hÄ!
œ lim b 2 œ 2; hÄ!
f(1 h) f(1) h
Á lim b hÄ!
f(0 h) f(0) h
Ê the derivative f w (0) does not exist.
38. Left-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2 Ê
œ lim b hÄ!
f(0 h) f(0) h
œ lim c hÄ!
g(h) g(0) h
œ lim b hÄ!
g(h) g(0) h
h1Î3 0 h h
2Î3
œ lim c hÄ!
0 h
1 h2Î3
œ lim b hÄ!
œ +_;
1 h1Î3
œ +_;
œ _ Ê the derivative gw (0) does not exist.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ #" ;
f("h)f(1) h
Section 3.2 The Derivative as a Function 43. (a) The function is differentiable on its domain $ Ÿ x Ÿ 2 (it is smooth) (b) none (c) none 44. (a) The function is differentiable on its domain # Ÿ x Ÿ 3 (it is smooth) (b) none (c) none 45. (a) The function is differentiable on $ Ÿ x 0 and ! x Ÿ 3 (b) none (c) The function is neither continuous nor differentiable at x œ 0 since lim c f(x) Á lim b f(x) xÄ! xÄ! 46. (a) f is differentiable on # Ÿ x 1, " x 0, 0 x 2, and 2 x Ÿ 3 (b) f is continuous but not differentiable at x œ 1: lim f(x) œ 0 exists but there is a corner at x œ 1 since x Ä 1
œ 3 and lim b f(" h)h f(1) œ 3 Ê f w (1) does not exist hÄ! hÄ! (c) f is neither continuous nor differentiable at x œ 0 and x œ 2: at x œ 0, lim c f(x) œ 3 but lim b f(x) œ 0 Ê lim f(x) does not exist; lim c
f(1 h) f(") h
xÄ!
xÄ0
xÄ!
at x œ 2, lim f(x) exists but lim f(x) Á f(2) xÄ#
xÄ#
47. (a) f is differentiable on " Ÿ x 0 and 0 x Ÿ 2 (b) f is continuous but not differentiable at x œ 0: lim f(x) œ 0 exists but there is a cusp at x œ 0, so f(0 h) f(0) h hÄ!
f w (0) œ lim
xÄ!
does not exist
(c) none 48. (a) f is differentiable on $ Ÿ x 2, 2 x 2, and 2 x Ÿ 3 (b) f is continuous but not differentiable at x œ 2 and x œ 2: there are corners at those points (c) none 49. (a) f w (x) œ lim
hÄ!
f(x h) f(x) h
œ lim
hÄ!
(x h)# ax# b h
œ lim
hÄ!
x# 2xh h# x# h
œ lim (2x h) œ 2x hÄ!
(b)
(c) yw œ 2x is positive for x 0, yw is zero when x œ 0, yw is negative when x 0 (d) y œ x# is increasing for _ x 0 and decreasing for ! x _; the function is increasing on intervals where yw 0 and decreasing on intervals where yw 0 f(x h) f(x) h hÄ!
50. (a) f w (x) œ lim
œ lim
hÄ!
Š xc" h h
1 x ‹
œ lim
hÄ!
x (x h) x(x h)h
œ lim
"
h Ä ! x(x h)
œ
" x#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
105
106
Chapter 3 Differentiation
(b)
(c) yw is positive for all x Á 0, yw is never 0, yw is never negative (d) y œ "x is increasing for _ x 0 and ! x _ 51. (a) Using the alternate formula for calculating derivatives: f w (x) œ zlim Äx œ
$ $ lim z x z Ä x 3(z x)
œ
az# zx x# b lim (z x)3(z x) zÄx
œ
# # lim z zx3 x zÄx
f(z) f(x) zx
#
w
$
Š z3
œ zlim Äx
œ x Ê f (x) œ x
x$ 3 ‹
zx
#
(b)
(c) yw is positive for all x Á 0, and yw œ 0 when x œ 0; yw is never negative (d) y œ
x$ 3
is increasing for all x Á 0 (the graph is horizontal at x œ 0) because y is increasing where yw 0; y is
never decreasing
52. (a) Using the alternate form for calculating derivatives: f w (x) œ zlim Äx œ
% % lim z x z Ä x 4(z x)
œ
$ xz# x# z x$ b lim (z x) az 4(z x) zÄx
œ
f(z) f(x) zx
$ # # $ lim z xz 4 x z x zÄx
œ zlim Äx $
Œ
z% 4
x% 4
zx
w
œ x Ê f (x) œ x$
(b)
(c) yw is positive for x 0, yw is zero for x œ 0, yw is negative for x 0 (d) y œ
x% 4
is increasing on 0 x _ and decreasing on _ x 0
# # # a2(x h)# 13(x h) 5b a2x# 13x 5b œ lim 2x 4xh 2h 13x h13h 5 2x 13x 5 h hÄ! hÄ! # lim 4xh 2hh 13h œ lim (4x 2h 13) œ 4x 13, slope at x. The slope is 1 when hÄ! hÄ!
53. yw œ lim œ
4x 13 œ "
Ê 4x œ 12 Ê x œ 3 Ê y œ 2 † 3# 13 † 3 5 œ 16. Thus the tangent line is y 16 œ (1)(x 3) Ê y œ x "$ and the point of tangency is (3ß 16). 54. For the curve y œ Èx, we have yw œ lim
hÄ!
œ lim
"
h Ä ! Èx h Èx
œ
" #Èx
ŠÈx h Èx‹ h
†
ŠÈx h Èx‹ ŠÈx h Èx‹
œ lim
(x h) x
h Ä ! ŠÈx h Èx‹ h
. Suppose ˆ+ß Èa‰ is the point of tangency of such a line and ("ß !) is the point
on the line where it crosses the x-axis. Then the slope of the line is
Èa 0 a (1)
œ
Èa a1
which must also equal
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.2 The Derivative as a Function " ; 2È a
using the derivative formula at x œ a Ê
exist: its point of tangency is ("ß "), its slope is
Èa a1
œ
" #È a
œ
" Ê 2a œ a 1 Ê a œ 1. #Èa " # ; and an equation of the line is
Thus such a line does y1œ
" #
(x 1)
Ê y œ "# x "# . 55. Yes; the derivative of f is f w so that f w (x! ) exists Ê f w (x! ) exists as well. 56. Yes; the derivative of 3g is 3gw so that gw (7) exists Ê 3gw (7) exists as well. 57. Yes, lim
g(t)
t Ä ! h(t)
can exist but it need not equal zero. For example, let g(t) œ mt and h(t) œ t. Then g(0) œ h(0)
œ 0, but lim
g(t)
t Ä ! h(t)
œ lim
tÄ!
mt t
œ lim m œ m, which need not be zero. tÄ!
58. (a) Suppose kf(x)k Ÿ x# for " Ÿ x Ÿ 1. Then kf(0)k Ÿ 0# Ê f(0) œ 0. Then f w (0) œ lim œ lim
hÄ!
f(h) 0 h
œ lim
hÄ!
f(h) h .
For khk Ÿ 1, h# Ÿ f(h) Ÿ h# Ê h Ÿ
hÄ!
f(h) h
f(0 h) f(0) h
Ÿ h Ê f w (0) œ lim
hÄ!
f(h) h
œ0
by the Sandwich Theorem for limits. (b) Note that for x Á 0, kf(x)k œ ¸x# sin "x ¸ œ kx# k ksin xk Ÿ kx# k † 1 œ x# (since " Ÿ sin x Ÿ 1). By part (a), f is differentiable at x œ 0 and f w (0) œ 0.
59. The graphs are shown below for h œ 1, 0.5, 0.1. The function y œ y œ Èx so that
" #È x
œ lim
hÄ!
Èx h Èx h
" 2È x
. The graphs reveal that y œ
is the derivative of the function
Èx h Èx h
gets closer to y œ
" #È x
as h gets smaller and smaller.
60. The graphs are shown below for h œ 2, 1, 0.5. The function y œ 3x# is the derivative of the function y œ x$ so that 3x# œ lim
hÄ!
(xh)$ x$ h
. The graphs reveal that y œ
(xh)$ x$ h
gets closer to y œ 3x# as h
gets smaller and smaller.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
107
108
Chapter 3 Differentiation
61. The graphs are the same. So we know that for f(x) œ kxk , we have f w (x) œ
kx k x
.
62. Weierstrass's nowhere differentiable continuous function.
63-68. Example CAS commands: Maple: f := x -> x^3 + x^2 - x; x0 := 1; plot( f(x), x=x0-5..x0+2, color=black, title="Section 3.2, #63(a)" ); q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b) L := limit( q(x,h), h=0 ); # (c) m := eval( L, x=x0 ); tan_line := f(x0) + m*(x-x0); plot( [f(x),tan_line], x=x0-2..x0+3, color=black, linestyle=[1,7], title="Section 3.2 #63(d)", legend=["y=f(x)","Tangent line at x=1"] ); Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ): # (e) Yvals := map( f, Xvals ): evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >); plot( L, x=x0-5..x0+3, color=black, title="Section 3.2 #63(f)" ); Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ): <<Miscellaneous`RealOnly` Clear[f, m, x, y, h] x0= 1 /4; f[x_]:=x2 Cos[x] Plot[f[x], {x, x0 3, x0 3}]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.3 Differentiation Rules q[x_, h_]:=(f[x h] f[x])/h m[x_]:=Limit[q[x, h], h Ä 0] ytan:=f[x0] m[x0] (x x0) Plot[{f[x], ytan},{x, x0 3, x0 3}] m[x0 1]//N m[x0 1]//N Plot[{f[x], m[x]},{x, x0 3, x0 3}] 3.3 DIFFERENTIATION RULES 1. y œ x# 3 Ê
œ
dy dx
2. y œ x# x 8 Ê
dy dx
3. s œ 5t$ 3t& Ê
œ
ds dt
ax# b
d dx
x$ x Ê
4 3
d dt
a5t$ b
6. y œ
x$ 3
x# #
d dt
d# y dx# " 4
œ x# x
7. w œ 3z# z" Ê
dw dz
œ 6z$ z# œ
x 4
8. s œ 2t" 4t# Ê
ds dt
10. y œ 4 2x x$ Ê 11. r œ
" # 3 s
5# s" Ê
dy dx dr ds
24 )$
48 )&
d# s dt#
d# w dz#
œ
d# y dx#
a15t# b
d dt
a15t% b œ 30t 60t$
6 z$ 2 t#
œ 2x 1 0 œ 2x 1
" z#
d# w dz#
Ê
d# s dt#
Ê
8 t$
œ # 3x% œ # œ 23 s$ 5# s# œ dr d)
d dt
œ 126z& 42z 42
œ 18z% 2z$ œ
œ 4t$ 24t% œ
œ 12x 10 10x$ œ 12x 10
dy dx
12. r œ 12)" 4)$ )% Ê œ
Ê
œ 2t# 8t$ œ
9. y œ 6x# 10x 5x# Ê
œ 2
œ 8x
dy dx
d# y dx#
œ#
a3t& b œ 15t# 15t% Ê
Ê
d# y dx#
œ 21z' 21z# 42z Ê
dw dz
œ 4x# 1 Ê
dy dx
(3) œ 2x 0 œ #B Ê
œ 2x 1 0 œ 2x 1 Ê
4. w œ 3z( 7z$ 21z# Ê 5. y œ
d dx
3 x%
2 3s$
Ê
d# y dx#
5 2s#
Ê
10 x$
Ê
d# y dx#
œ 0 12x& œ d# r ds#
œ 12)# 12)% 4)& œ
18 z%
4 t$
12 )%
2 z$
24 t%
œ 12 0 30x% œ 12
30 x%
12 x&
œ 2s% 5s$ œ 12 )#
4 )&
Ê
2 s%
d# r d) #
5 s$
œ 24)$ 48)& 20)'
20 )'
13. (a) y œ a3 x# b ax$ x 1b Ê yw œ a3 x# b † #
#
d dx
$
ax$ x 1b ax$ x 1b † %
d dx
a3 x# b
#
œ a3 x b a3x 1b ax x 1b (2x) œ 5x 12x 2x 3 (b) y œ x& 4x$ x# 3x 3 Ê yw œ 5x% 12x# 2x 3 14. (a) y œ (2x 3) a5x# 4xb Ê yw œ (2x 3)(10x 4) a5x# 4xb (2) œ 30x# 14x 12 (b) y œ (2x 3) a5x# 4xb œ 10x$ 7x2 12x Ê yw œ 30x# 14x 12 15. (a) y œ ax# 1b ˆx 5 "x ‰ Ê yw œ ax# 1b †
d dx
ˆx 5 "x ‰ ˆx 5 x" ‰ †
d dx
ax# 1b
œ ax# 1b a1 x# b ax 5 x" b (2x) œ ax# 1 1 x# b a2x# 10x 2b œ 3x# 10x 2 (b) y œ x$ 5x# 2x 5
" x
Ê yw œ 3x# 10x 2
" x#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" x#
109
110
Chapter 3 Differentiation
16. y œ a1 x2 b ˆx3Î4 x3 ‰ (a) yw œ a1 x2 b † ˆ 34 x1Î4 3x4 ‰ ˆx3Î4 x3 ‰ a2xb œ (b) y œ x3Î4 x3 x11Î4 x1 Ê yw œ
3 4x1Î4
2x 5 3x 2 ; use the quotient rule: u œ 2x 5 and (3x 2)(2) (2x 5)(3) 4 6x 15 œ 6x (3x œ (3x192)# (3x 2)# #)#
17. y œ œ
4 3x 3x2 x ;use the quotient ˆ3x2 x‰a3b a4 3xba6x1b a3x2 xb2
18. y œ œ
t# " t# t 2
œ
at "bat "b at #bat "b
21. v œ (1 t) a1 t# b 22. w œ
x5 2x 7
Ê ww œ
Ès " Ès 1
23. f(s) œ
NOTE:
d ds
"
œ
œ
9x2 3x 18x2 21x 4 a3x2 xb2
ˆÈ s ‰ œ
24. u œ
5x " #È x
25. v œ
1 x 4È x x
Ê
du dx
œ
œ
1 t 1t#
t" t2,
Ê
" #È s
ˆ È s "‰ Š
" ax # 1 b a x # x 1 b #
œ
1 x2
vuw uvw v#
v œ x 0.5 Ê uw œ 2x and vw œ 1 Ê gw (x) œ
at #ba"b at "ba"b at 2 b 2
2x 7 2x 10 (2x 7)#
œ
"
"
Ès ‹ ˆÈs 1‰ Š #Ès ‹
#
#
vuw uvw v#
9x2 24x 4 a3x2 xb2
a1 t# b (") (1 t)(2t) a1 t# b#
ˆÈ s 1 ‰
œ
œ
œ
t#t" at 2 b 2
œ
" t# 2t 2t# a1 t# b#
œ
œ
vuw uvw v#
" at 2 b 2
t# 2t " a1 t# b#
17 (2x 7)#
ˆ È s "‰ ˆ È s 1 ‰ 2 È s ˆÈ s 1 ‰
#
œ
" È s ˆÈ s 1‰#
from Example 2 in Section 3.2
4x
x Š1
œ
5x 1 4x$Î#
È x ‹ ˆ1 x 4 È x ‰ 2
x#
26. r œ 2 Š È" È)‹ Ê rw œ 2 ) 27. y œ
dv dt
ˆ2Èx‰ (5) (5x 1) Š È" ‹ x
Ê vw œ
œ
t Á " Ê f w (t) œ
(2x 7)(1) (x 5)(2) (2x 7)#
Ê f w (s) œ
11 7Î4 4 x
rule: u œ 4 3x and v œ 3x2 x Ê uw œ 3 and vw œ 6x 1 Ê yw œ
x# 4 # x0.5 ; use the quotient rule: u œ x 4 and # # # # (x 0.5)(2x) ax 4b (") x 4 x4 œ 2x (xx 0.5) œ x(x # (x 0.5)# 0.5)#
20. f(t) œ
v œ 3x 2 Ê uw œ 2 and vw œ 3 Ê yw œ
19. g(x) œ œ
3 x4
3 x34 4x1Î4 11 7Î4 x12 4 x
È)(0) 1 Š )
"
È) ‹
#
œ
2È x " x#
" #È )
" œ )$Î#
" )"Î#
; use the quotient rule: u œ 1 and v œ ax# 1b ax# x 1b Ê uw œ 0 and
vw œ ax 1b (2x 1) ax# x 1b (2x) œ 2x$ x# 2x 1 2x$ 2x# 2x œ 4x$ 3x# 1 Ê
dy dx
œ
vuw uvw v#
28. y œ
(x 1)(x 2) (x 1)(x #)
29. y œ
" #
30. y œ
" 1 #0
x%
3 #
œ
œ
0 1 a4x$ 3x# 1b ax # 1 b # a x # x 1 b #
x# 3x 2 x# 3x 2
Ê yw œ
œ
4x$ 3x# 1 ax # 1 b# a x # x 1 b #
ax# 3x 2b (2x 3) ax# 3x 2b (2x 3) (x 1)# (x 2)#
œ
6x# 12 (x 1)# (x 2)#
œ
6 ax # 2b (x 1)# (x 2)#
x# x Ê yw œ 2x$ 3x 1 Ê yww œ 6x# 3 Ê ywww œ 12x Ê yÐ%Ñ œ 12 Ê yÐnÑ œ 0 for all n 5
x& Ê yw œ
" #4
x% Ê yww œ
" 6
x$ Ê ywww œ
" #
x# Ê yÐ%Ñ œ x Ê yÐ&Ñ œ 1 Ê yÐnÑ œ 0 for all n 6
31. y œ ax 1bax2 3x 5b œ x3 2x2 8x 5 Ê yw œ 3x2 4x 8 Ê yww œ 6x 4 Ê ywww œ 6 Ê yÐnÑ œ 0 for all n 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.3 Differentiation Rules
111
32. y œ a4x3 3xba2 xb œ 4x4 8x3 3x2 6x Ê yw œ 16x3 24x2 6x 6 Ê yww œ 48x2 48x 6 Ê ywww œ 96x 48 Ê yÐ%Ñ œ 96 Ê yÐnÑ œ 0 for all n 5 33. y œ
x $ 7 x
34. s œ
t# 5t 1 œ 1 5t t"# œ 1 5t" t# # ' d s $ 6t% œ "! dt# œ 10t t$ t%
Ê
œ x# 7x" Ê
() " ) a ) # ) 1 b )$
35. r œ 36. u œ Ê
œ
dy dx
)$ " )$
" )$
œ"
t# Ê
Ê
œ 1 )$ Ê
dr d)
d# w dz#
"Ê
œ 2z$ 0 œ 2z$ œ
" 3
1
#
%
Ê 40. p œ Ê
d# p dq#
œ
" 6
#" q% 5q' œ
q# 3 (q 1)$ (q 1)$ dp dq
q' q# 3q% 3 12q%
œ
" 6
œ
" #q %
q#
" 1#
q#
d# p dq#
& t#
# t$
$ )%
Ê
d# r d) #
z Ê
dw dz
œ z# 0 1 œ z# 1
œ 12)& œ
"# )&
œ 1 x$
8 3
" 4
dw dz
œ 4z$ 0 œ 4z$ Ê
4" q% Ê
dp dq
œ
" 6
d# w dz#
œ 12z#
q 6" q$ q& œ
" 6
q
" 6q$
" q&
& q6
q# 3 aq$ 3q# 3q 1b aq$ 3q# 3q 1b
œ "# q# œ #"q# Ê
"% x$
# z$
" 1#
x x%
3 z œ z"
38. w œ (z 1)(z 1) az# 1b œ az# 1b az# 1b œ z% 1 Ê 3 39. p œ Š q12q ‹ Š q q$ 1 ‹ œ
œ 2 14x$ œ #
œ 0 $)% œ $)% œ
# $ % ax # x b a x # x 1 b œ x(x 1) axx% x "b œ x axx% 1b œ x x% x œ x% # du % œ 3x% œ $ Ê ddxu# œ 12x& œ "# dx œ 0 3x x& x%
" z#
d# y dx#
œ 0 5t# 2t$ œ 5t# 2t$ œ
ds dt
37. w œ ˆ 13z3z ‰ (3 z) œ ˆ "3 z" 1‰ (3 z) œ z" œ
( x#
œ 2x 7x# œ #x
œ q$ œ
œ
q# 3 2q$ 6q
œ
q# 3 2q aq# 3b
œ
" #q
œ
" #
q"
" q$
41. u(0) œ 5, uw (0) œ 3, v(0) œ 1, vw (0) œ 2 d d (a) dx (uv) œ uvw vuw Ê dx (uv)¸ x = 0 œ u(0)vw (0) v(0)uw (0) œ 5 † 2 (1)(3) œ 13 (b)
d dx
ˆ vu ‰ œ
vuw uvw v#
Ê
d dx
(c)
d dx d dx
ˆ uv ‰
œ
uvw vuw u#
Ê
d dx
(d)
(7v 2u) œ 7vw
ˆ vu ‰¸
x=0
ˆ uv ‰¸ x=0 d 2uw Ê dx
œ œ
v(0)uw (0) u(0)vw (0) (v(0))# u(0)vw (0) v(0)uw (0) (u(0))#
(")(3) (5)(2) (1)# (5)(2) (1)(3) œ (5)# w w
œ
œ 7 œ
7 25
(7v 2u)¸ x = 0 œ 7v (0) 2u (0) œ 7 † 2 2(3) œ 20
42. u(1) œ 2, uw (1) œ 0, v(1) œ 5, vw (1) œ 1 d (a) dx (uv)¸ x = 1 œ u(1)vw (1) v(1)uw (1) œ 2 † (1) 5 † 0 œ 2 (b)
d dx
ˆ vu ‰¸
x=1
œ
(c)
d dx d dx
ˆ uv ‰¸
x=1
œ
(d)
v(1)uw (")u(1)vw (1) (v(1))# u(1)vw (")v(1)uw (1) (u(1))# w
œ œ
5†02†(1) (5)# 2†(1)5†0 (2)#
œ
2 25
œ 12
(7v 2u)¸ x = 1 œ 7v (1) 2uw (1) œ 7 † (1) 2 † 0 œ 7
43. y œ x$ 4x 1. Note that (#ß ") is on the curve: 1 œ 2$ 4(2) 1 (a) Slope of the tangent at (xß y) is yw œ 3x# 4 Ê slope of the tangent at (#ß ") is yw (2) œ 3(2)# 4 œ 8. Thus the slope of the line perpendicular to the tangent at (#ß ") is "8 Ê the equation of the line perpendicular to the tangent line at (#ß ") is y 1 œ "8 (x 2) or y œ 8x 45 .
(b) The slope of the curve at x is m œ 3x# 4 and the smallest value for m is 4 when x œ 0 and y œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
112
Chapter 3 Differentiation
(c) We want the slope of the curve to be 8 Ê yw œ 8 Ê 3x# 4 œ 8 Ê 3x# œ 12 Ê x# œ 4 Ê x œ „ 2. When x œ 2, y œ 1 and the tangent line has equation y 1 œ 8(x 2) or y œ 8x 15; when x œ 2, y œ (2)$ 4(2) 1 œ 1, and the tangent line has equation y 1 œ 8(x 2) or y œ 8x 17. 44. (a) y œ x$ 3x 2 Ê yw œ 3x# 3. For the tangent to be horizontal, we need m œ yw œ 0 Ê 0 œ 3x# 3 Ê 3x# œ 3 Ê x œ „ 1. When x œ 1, y œ 0 Ê the tangent line has equation y œ 0. The line perpendicular to this line at ("ß !) is x œ 1. When x œ 1, y œ 4 Ê the tangent line has equation y œ 4. The line perpendicular to this line at ("ß %) is x œ 1. (b) The smallest value of yw is 3, and this occurs when x œ 0 and y œ 2. The tangent to the curve at (!ß 2) has slope 3 Ê the line perpendicular to the tangent at (!ß 2) has slope "3 Ê y 2 œ "3 (x 0) or yœ 45. y œ
" 3
4x x# 1
x 2 is an equation of the perpendicular line. Ê
dy dx
œ
ax# 1b(4) (4x)(2x) ax # 1 b #
œ
4x# 4 8x# ax # 1 b#
œ
4 ax# "b ax # 1 b#
. When x œ 0, y œ 0 and yw œ
4(0 1) 1
œ %, so the
tangent to the curve at (!ß !) is the line y œ 4x. When x œ 1, y œ 2 Ê yw œ 0, so the tangent to the curve at ("ß 2) is the line y œ 2. 46. y œ
8 x# 4
Ê yw œ
ax# 4b(0) 8(2x) ax # 4 b #
œ
16x ax # 4 b #
. When x œ 2, y œ 1 and yw œ
16(2) a2 # 4 b#
œ "# , so the tangent
line to the curve at (2ß ") has the equation y 1 œ "# (x 2), or y œ x# 2. 47. y œ ax# bx c passes through (!ß !) Ê 0 œ a(0) b(0) c Ê c œ 0; y œ ax# bx passes through ("ß #) Ê 2 œ a b; yw œ 2ax b and since the curve is tangent to y œ x at the origin, its slope is 1 at x œ 0 Ê yw œ 1 when x œ 0 Ê 1 œ 2a(0) b Ê b œ 1. Then a b œ 2 Ê a œ 1. In summary a œ b œ 1 and c œ 0 so the curve is y œ x# x. 48. y œ cx x# passes through ("ß !) Ê 0 œ c(1) 1 Ê c œ 1 Ê the curve is y œ x x# . For this curve, yw œ 1 2x and x œ 1 Ê yw œ 1. Since y œ x x# and y œ x# ax b have common tangents at x œ 0, y œ x# ax b must also have slope 1 at x œ 1. Thus yw œ 2x a Ê 1 œ 2 † 1 a Ê a œ 3 Ê y œ x# 3x b. Since this last curve passes through ("ß !), we have 0 œ 1 3 b Ê b œ 2. In summary, a œ 3, b œ 2 and c œ 1 so the curves are y œ x# 3x 2 and y œ x x# . 49. y œ 8x 5 Ê m œ 8; faxb œ 3x2 4x Ê f w axb œ 6x 4; 6x 4 œ 8 Ê x œ 2 Ê fa2b œ 3a2b2 4a2b œ 4 Ê a2, 4b 50. 8x 2y œ 1 Ê y œ 4x
" #
Ê m œ 4; gaxb œ 13 x3 32 x2 1 Ê g w axb œ x2 3x; x2 3x œ 4 Ê x œ 4 or x œ 1
Ê ga4b œ 13 a4b3 32 a4b2 1 œ 53 , ga1b œ 13 a1b3 32 a1b2 1 œ 56 Ê ˆ4, 53 ‰ or ˆ1, 56 ‰ 51. y œ 2x 3 Ê m œ 2 Ê m¼ œ 12 ; y œ
x x2
Ê „ 2 œ x 2 Ê x œ 4 or x œ 0 Ê if x œ 4, y œ 52. m œ
y8 x3;
ax 2ba1b xa1b 2 œ ax ; 2 ax 2 b 2 2 b2 ax 2 b 2 4 0 4 2 œ 2, and if x œ 0, y œ 0 2
Ê yw œ
faxb œ x2 Ê f w axb œ 2x; m œ f w axb Ê
y8 x3
œ 2x Ê
x2 8 x3
œ 12 Ê 4 œ ax 2b2 œ 0 Ê a4, 2b or a0, 0b.
œ 2x Ê x2 8 œ 2x2 6x Ê x2 6x 8 œ 0
Ê x œ 4 or x œ 2 Ê fa4b œ 4 œ 16, fa2b œ 2 œ 4 Ê a4, 16b or a2, 4b. 2
2
53. (a) y œ x$ x Ê yw œ 3x# 1. When x œ 1, y œ 0 and yw œ 2 Ê the tangent line to the curve at ("ß !) is y œ 2(x 1) or y œ 2x 2.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.3 Differentiation Rules
113
(b)
(c)
y œ x$ x Ê x$ x œ 2x 2 Ê x$ 3x 2 œ (x 2)(x 1)# œ 0 Ê x œ 2 or x œ 1. Since y œ 2x 2 y œ 2a2b 2 œ 6; the other intersection point is (2ß 6)
54. (a) y œ x$ 6x# 5x Ê yw œ 3x# 12x 5. When x œ 0, y œ 0 and yw œ 5 Ê the tangent line to the curve at (0ß 0) is y œ 5x. (b)
(c)
y œ x$ 6x# 5x $ # $ # # Ê x 6x 5x œ 5x Ê x 6x œ 0 Ê x (x 6) œ 0 Ê x œ 0 or x œ 6. y œ 5x Since y œ 5a6b œ $!, the other intersection point is (6ß 30).
55. lim xÄ1 56.
x50 1 x1
lim x Ä 1
œ 50 x49 ¹
x2Î9 1 x1
xœ1
œ 92 x7Î9 ¹
œ 50 a1b49 œ 50
x œ 1
œ
2 9a1b7Î9
œ 92
57. gw axb œ œ
2x 3 a
x0 , since g is differentiable at x œ 0 Ê lim b a2x 3b œ 3 and lim c a œ a Ê a œ 3 x0 xÄ0 xÄ0
58. f w axb œ œ
a x 1 , since f is differentiable at x œ 1 Ê lim b a œ a and lim c a2bxb œ 2b Ê a œ 2b, and 2bx x 1 x Ä 1 x Ä 1
since f is continuous at x œ 1 Ê lim b aax bb œ a b and lim c abx2 3b œ b 3 Ê a b œ b 3 x Ä 1 x Ä 1 Ê a œ 3 Ê 3 œ 2b Ê b œ 32 . 59. Paxb œ an xn an" xn" â a# x# a" x a! Ê P w axb œ nan xn" an "ban" xn# â #a# x a" 60. R œ M# ˆ C#
M‰ 3
œ
C #
61. Let c be a constant Ê
M# "3 M$ , where C is a constant Ê dc dx
œ0 Ê
d dx
(u † c) œ u †
dc dx
c†
du dx
dR dM
œ CM M#
œu†0c
du dx
œc
du dx
. Thus when one of the
functions is a constant, the Product Rule is just the Constant Multiple Rule Ê the Constant Multiple Rule is a special case of the Product Rule. 62. (a) We use the Quotient rule to derive the Reciprocal Rule (with u œ 1):
d dx
ˆ "v ‰ œ
v†0 1† dv dx v#
œ
"† dv dx v#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ v"# †
dv dx
.
114
Chapter 3 Differentiation
(b) Now, using the Reciprocal Rule and the Product Rule, we'll derive the Quotient Rule: d ˆ"‰ " du dx v v † dx (Product Rule) dv v du dx u dx , the Quotient Rule. v#
œu† œ 63. (a)
d dx
(uvw) œ
d dx w
dv œ u † ˆ v#1 ‰ dx
((uv) † w) œ (uv) dw dx w †
d dx
" du v dx
(uv) œ uv
œ uvww uv w uw vw d d % (b) dx au" u# u$ u% b œ dx aau" u# u$ b u% b œ au" u# u$ b du dx u% du$ du# du" ‰ % ˆ œ u" u# u$ du dx u% u" u# dx u$ u" dx u$ u# dx
dw dx d dx
(Reciprocal Rule) Ê
w ˆu
dv dx
v
a u" u# u$ b Ê
d dx
du ‰ dx
d dx
œ uv
d dx
d ˆ "‰ dx u † v du u dv dx v dx v#
ˆ vu ‰ œ
ˆ vu ‰ œ
dw dx
wu
dv dx
wv
du dx
a u" u# u$ u% b
(using (a) above)
du$ du# du" % Ê au" u# u$ u% b œ u" u# u$ du dx u" u# u% dx u" u$ u% dx u# u$ u% dx œ u" u# u$ u%w u" u# u$w u% u" u#w u$ u% u"w u# u$ u% d Generalizing (a) and (b) above, dx au" âun b œ u" u# âun" unw u" u# âun# unw " un d dx
(c) 64.
d m b dx ax
65. P œ Ê
œ
œ
d ˆ 1 ‰ dx xm
xm †01ˆm†xm 1 ‰ ax m b 2
œ
m†xm x2m
1
œ m † xm12m œ m † xm1
nRT an# Vnb V# . We are holding T constant, and a, b, n, R are # # (V nb)†0 (nRT)(1) dP 2an# V (0) aVa#anb# b (2V) œ (VnRT dV œ (Vnb)# nb)# V$
66. Aaqb œ
km q
cm
hq #
á u"w u# âun
œ akmbq" cm ˆ h# ‰q Ê
dA dq
also constant so their derivatives are zero
œ akmbq# ˆ h# ‰ œ km q#
h #
Ê
d# A dt#
œ #akmbq$ œ
#km q$
3.4 THE DERIVATIVE AS A RATE OF CHANGE 1. s œ t# $t #, 0 Ÿ t Ÿ # (a) displacement œ ?s œ s(#) s(0) œ !m #m œ # m, vav œ (b) v œ aœ
ds dt œ #t d# s dt# œ #
?s ?t
œ
Ê a(0) œ # m/sec# and a(#) œ # m/sec#
changes direction at t œ
$ #.
2. s œ 't t# , ! Ÿ t Ÿ ' (a) displacement œ ?s œ s(') s(0) œ ! m, vav œ aœ
œ " m/sec
$ Ê kv(0)k œ l$l œ $ m/sec and kv(#)k œ 1 m/sec;
(c) v œ 0 Ê #t $ œ 0 Ê t œ $# . v is negative in the interval ! t
(b) v œ
# #
ds dt œ ' d# s dt# œ #
?s ?t
œ
! '
$ #
and v is positive when
$ #
t # Ê the body
œ ! m/sec
#> Ê kv(0)k œ l 'l œ ' m/sec and kv(')k œ l'l œ ' m/sec; Ê a(0) œ # m/sec# and a(') œ # m/sec#
(c) v œ 0 Ê ' #t œ 0 Ê t œ $. v is positive in the interval ! t $ and v is negative when $ t ' Ê the body changes direction at t œ $. 3. s œ t$ 3t# 3t, 0 Ÿ t Ÿ 3 (a) displacement œ ?s œ s(3) s(0) œ 9 m, vav œ (b) v œ
ds dt
#
?s ?t
œ
9 3
œ 3 m/sec
œ 3t 6t 3 Ê kv(0)k œ k3k œ 3 m/sec and kv(3)k œ k12k œ 12 m/sec; a œ #
#
d# s dt#
œ 6t 6
Ê a(0) œ 6 m/sec and a(3) œ 12 m/sec (c) v œ 0 Ê 3t# 6t 3 œ 0 Ê t# 2t 1 œ 0 Ê (t 1)# œ 0 Ê t œ 1. For all other values of t in the interval the velocity v is negative (the graph of v œ 3t# 6t 3 is a parabola with vertex at t œ 1 which opens downward Ê the body never changes direction).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.4 The Derivative as a Rate of Change 4. s œ
t% 4
115
t$ t# , 0 Ÿ t Ÿ $
(a) ?s œ s($) s(0) œ $
m, vav œ
?s ?t
œ
* %
$
œ
$ %
m/sec
(b) v œ t 3t 2t Ê kv(0)k œ 0 m/sec and kv($)k œ ' m/sec; a œ 3t# 6t 2 Ê a(0) œ 2 m/sec# and a($) œ "" m/sec# (c) v œ 0 Ê t$ 3t# 2t œ 0 Ê t(t 2)(t 1) œ 0 Ê t œ 0, 1, 2 Ê v œ t(t 2)(t 1) is positive in the interval for 0 t 1 and v is negative for 1 t 2 and v is positive for # t $ Ê the body changes direction at t œ 1 and at t œ #. 5. s œ
25 t#
#
* %
5t , 1 Ÿ t Ÿ 5
(a) ?s œ s(5) s(1) œ 20 m, vav œ (b) v œ
50 t$
a(5) œ
4 25
(c) v œ 0 Ê 6. s œ
25 t5
5 t#
20 4
œ 5 m/sec
Ê kv(1)k œ 45 m/sec and kv(5)k œ
" 5
m/sec; a œ
150 t%
10 t$
Ê a(1) œ 140 m/sec# and
m/sec# 50 5t t$
œ 0 Ê 50 5t œ 0 Ê t œ 10 Ê the body does not change direction in the interval
, % Ÿ t Ÿ 0
(a) ?s œ s(0) s(4) œ 20 m, vav œ 20 4 œ 5 m/sec (b) v œ a(0) (c) v œ
25 (t 5)# Ê kv(4)k œ 25 œ 25 m/sec# 0 Ê (t255)# œ 0 Ê v is
m/sec and kv(0)k œ " m/sec; a œ
50 (t5)$
Ê a(4) œ 50 m/sec# and
never 0 Ê the body never changes direction
7. s œ t$ 6t# 9t and let the positive direction be to the right on the s-axis. (a) v œ 3t# 12t 9 so that v œ 0 Ê t# 4t 3 œ (t 3)(t 1) œ 0 Ê t œ 1 or 3; a œ 6t 12 Ê a(1) œ 6 m/sec# and a(3) œ 6 m/sec# . Thus the body is motionless but being accelerated left when t œ 1, and motionless but being accelerated right when t œ 3. (b) a œ 0 Ê 6t 12 œ 0 Ê t œ 2 with speed kv(2)k œ k12 24 9k œ 3 m/sec (c) The body moves to the right or forward on 0 Ÿ t 1, and to the left or backward on 1 t 2. The positions are s(0) œ 0, s(1) œ 4 and s(2) œ 2 Ê total distance œ ks(1) s(0)k ks(2) s(1)k œ k4k k2k œ 6 m. 8. v œ t# 4t 3 Ê a œ 2t 4 (a) v œ 0 Ê t# 4t 3 œ 0 Ê t œ 1 or 3 Ê a(1) œ 2 m/sec# and a(3) œ 2 m/sec# (b) v 0 Ê (t 3)(t 1) 0 Ê 0 Ÿ t 1 or t 3 and the body is moving forward; v 0 Ê at 3bat 1b 0 Ê " t 3 and the body is moving backward (c) velocity increasing Ê a 0 Ê 2t 4 0 Ê t 2; velocity decreasing Ê a 0 Ê 2t 4 0 Ê ! Ÿ t 2 9. sm œ 1.86t# Ê vm œ 3.72t and solving 3.72t œ 27.8 Ê t ¸ 7.5 sec on Mars; sj œ 11.44t# Ê vj œ 22.88t and solving 22.88t œ 27.8 Ê t ¸ 1.2 sec on Jupiter. 10. (a) v(t) œ sw (t) œ 24 1.6t m/sec, and a(t) œ vw (t) œ sw w (t) œ 1.6 m/sec# (b) Solve v(t) œ 0 Ê 24 1.6t œ 0 Ê t œ 15 sec (c) s(15) œ 24(15) .8(15)# œ 180 m (d) Solve s(t) œ 90 Ê 24t .8t# œ 90 Ê t œ
30 „ 15È2 #
¸ 4.39 sec going up and 25.6 sec going down
(e) Twice the time it took to reach its highest point or 30 sec 11. s œ 15t "# gs t# Ê v œ 15 gs t so that v œ 0 Ê 15 gs t œ 0 Ê gs œ
15 t
. Therefore gs œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
15 20
œ
3 4
œ 0.75 m/sec#
116
Chapter 3 Differentiation
12. Solving sm œ 832t 2.6t# œ 0 Ê t(832 2.6t) œ 0 Ê t œ 0 or 320 Ê 320 sec on the moon; solving se œ 832t 16t# œ 0 Ê t(832 16t) œ 0 Ê t œ 0 or 52 Ê 52 sec on the earth. Also, vm œ 832 5.2t œ 0 Ê t œ 160 and sm (160) œ 66,560 ft, the height it reaches above the moon's surface; ve œ 832 32t œ 0 Ê t œ 26 and se (26) œ 10,816 ft, the height it reaches above the earth's surface. 13. (a) s œ 179 16t# Ê v œ 32t Ê speed œ kvk œ 32t ft/sec and a œ 32 ft/sec# (b) s œ 0 Ê 179 16t# œ 0 Ê t œ É 179 16 ¸ 3.3 sec È É 179 (c) When t œ É 179 16 , v œ 32 16 œ 8 179 ¸ 107.0 ft/sec 14. (a)
lim1 v œ lim1 9.8(sin ))t œ 9.8t so we expect v œ 9.8t m/sec in free fall
)Ä
(b) a œ
#
dv dt
)Ä
#
œ 9.8 m/sec# (b) between 3 and 6 seconds: $ Ÿ t Ÿ 6 (d)
15. (a) at 2 and 7 seconds (c)
16. (a) P is moving to the left when 2 t 3 or 5 t 6; P is moving to the right when 0 t 1; P is standing still when 1 t 2 or 3 t 5 (b)
17. (a) (c) (e) (f)
190 ft/sec at 8 sec, 0 ft/sec From t œ 8 until t œ 10.8 sec, a total of 2.8 sec Greatest acceleration happens 2 sec after launch
(g) From t œ 2 to t œ 10.8 sec; during this period, a œ
(b) 2 sec (d) 10.8 sec, 90 ft/sec
v(10.8) v(2) 10.8 2
¸ 32 ft/sec#
18. (a) Forward: 0 Ÿ t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6; Slows down: 0 Ÿ t 1, 3 t 5, and 6 t 7 (b) Positive: 3 t 6; negative: 0 Ÿ t 2 and 6 t 7; zero: 2 t 3 and 7 t 9 (c) t œ 0 and 2 Ÿ t Ÿ 3 (d) 7 Ÿ t Ÿ 9 19. s œ 490t# Ê v œ 980t Ê a œ 980 (a) Solving 160 œ 490t# Ê t œ
4 7
sec. The average velocity was
s(4/7) s(0) 4/7
œ 280 cm/sec.
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Section 3.4 The Derivative as a Rate of Change (b) At the 160 cm mark the balls are falling at v(4/7) œ 560 cm/sec. The acceleration at the 160 cm mark was 980 cm/sec# . 17 (c) The light was flashing at a rate of 4/7 œ 29.75 flashes per second. 20. (a)
(b)
21. C œ position, A œ velocity, and B œ acceleration. Neither A nor C can be the derivative of B because B's derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position. Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves B for acceleration. 22. C œ position, B œ velocity, and A œ acceleration. Curve C cannot be the derivative of either A or B because C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is negative where B has negative slopes and positive where B has positive slopes. 23. (a) c(100) œ 11,000 Ê cav œ #
11,000 100
œ $110
(b) c(x) œ 2000 100x .1x Ê cw (x) œ 100 .2x. Marginal cost œ cw (x) Ê the marginal cost of producing 100 machines is cw (100) œ $80 (c) The cost of producing the 101st machine is c(101) c(100) œ 100 201 10 œ $79.90 24. (a) r(x) œ 20000 ˆ1 "x ‰ Ê rw (x) œ w
(b) r a"!"b œ $"Þ*'Þ (c) x lim rw (x) œ x lim Ä_ Ä_
20000 x#
20000 x#
, which is marginal revenue. rw a"!!b œ
20000 100#
œ $#Þ
œ 0. The increase in revenue as the number of items increases without bound
will approach zero. 25. b(t) œ 10' 10% t 10$ t# Ê bw (t) œ 10% (2) a10$ tb œ 10$ (10 2t) (a) bw (0) œ 10% bacteria/hr (b) bw (5) œ 0 bacteria/hr w % (c) b (10) œ 10 bacteria/hr 26. Q(t) œ 200(30 t)# œ 200 a900 60t t# b Ê Qw (t) œ 200(60 2t) Ê Qw (10) œ 8,000 gallons/min is the rate the water is running at the end of 10 min. Then
Q(10) Q(0) 10
œ 10,000 gallons/min is the average rate the water flows
during the first 10 min. The negative signs indicate water is leaving the tank.
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117
118
Chapter 3 Differentiation
27. (a) y œ 6 ˆ1
t ‰# 1#
œ 6 Š1
(b) The largest value of value of
dy dt
dy dt
t 6
t# 144 ‹
Ê
dy dt
œ
t 12
1
is 0 m/h when t œ 12 and the fluid level is falling the slowest at that time. The smallest
is 1 m/h, when t œ 0, and the fluid level is falling the fastest at that time.
(c) In this situation,
Ÿ 0 Ê the graph of y is
dy dt
always decreasing. As
dy dt
increases in value,
the slope of the graph of y increases from 1 to 0 over the interval 0 Ÿ t Ÿ 12.
28. (a) V œ
4 3
1 r$ Ê
(b) When r œ 2,
dV dr
dV dr
œ 41 r # Ê
dV ¸ dr r=2
œ 41(2)# œ 161 ft$ /ft
œ 161 so that when r changes by 1 unit, we expect V to change by approximately 161. Therefore
when r changes by 0.2 units V changes by approximately (161)(0.2) œ 3.21 ¸ 10.05 ft$ . Note that V(2.2) V(2) ¸ 11.09 ft$ . 29. 200 km/hr œ 55 59 m/sec œ t œ 25, D œ
10 9
#
(25) œ
500 9 m/sec, 6250 9 m
and D œ
10 # 9 t
30. s œ v! t 16t# Ê v œ v! 32t; v œ 0 Ê t œ
v! 32
Ê Vœ
20 9
t. Thus V œ
500 9
Ê
; 1900 œ v! t 16t# so that t œ
Ê v! œ È(64)(1900) œ 80È19 ft/sec and, finally,
80È19 ft sec
†
60 sec 1 min
†
60 min 1 hr
†
1 mi 5280 ft
v! 32
20 9
tœ
500 9
Ê t œ 25 sec. When
Ê 1900 œ
v!# 3#
v!# 64
¸ 238 mph.
31.
v œ 0 when t œ 6.25 sec v 0 when 0 Ÿ t 6.25 Ê body moves right (up); v 0 when 6.25 t Ÿ 12.5 Ê body moves left (down) body changes direction at t œ 6.25 sec body speeds up on (6.25ß 12.5] and slows down on [0ß 6.25) The body is moving fastest at the endpoints t œ 0 and t œ 12.5 when it is traveling 200 ft/sec. It's moving slowest at t œ 6.25 when the speed is 0. (f) When t œ 6.25 the body is s œ 625 m from the origin and farthest away. (a) (b) (c) (d) (e)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.4 The Derivative as a Rate of Change 32.
(a) v œ 0 when t œ
3 #
sec
(b) v 0 when 0 Ÿ t 1.5 Ê body moves left (down); v 0 when 1.5 t Ÿ 5 Ê body moves right (up) (c) body changes direction at t œ 3# sec (d) body speeds up on ˆ 3# ß &‘ and slows down on !ß 3# ‰ (e) body is moving fastest at t œ 5 when the speed œ kv(5)k œ 7 units/sec; it is moving slowest at t œ
3 #
when the
speed is 0 (f) When t œ 5 the body is s œ 12 units from the origin and farthest away. 33.
6 „ È15 3 6 È15 t 3
(a) v œ 0 when t œ
sec
(b) v 0 when
6 È15 3
Ê body moves left (down); v 0 when 0 Ÿ t
6 È15 3
or
6 È15 3
tŸ4
Ê body moves right (up) 6 „ È15 sec 3 6 È15 6 È15 Š 3 ß #‹ Š 3 ß %“
(c) body changes direction at t œ (d) body speeds up on
È15
and slows down on ’0ß 6 3
È15
‹ Š#ß 6 3
‹.
(e) The body is moving fastest at t œ 0 and t œ 4 when it is moving 7 units/sec and slowest at t œ (f) When t œ
6È15 3
the body is at position s ¸ 6.303 units and farthest from the origin.
34.
(a) v œ 0 when t œ
6 „ È15 3
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6„È15 3
sec
119
120
Chapter 3 Differentiation È 6 È15 or 6 3 15 t Ÿ 4 Ê body is moving left (down); v 0 when 3 È 6 È15 t 6 3 15 Ê body is moving right (up) 3 È body changes direction at t œ 6 „ 3 15 sec È È È È body speeds up on Š 6 3 15 ß #‹ Š 6 3 15 ß %“ and slows down on ’!ß 6 3 15 ‹ Š#ß 6 3 15 ‹
(b) v 0 when 0 Ÿ t
(c) (d)
(e) The body is moving fastest at 7 units/sec when t œ 0 and t œ 4; it is moving slowest and stationary at t œ (f) When t œ
6 È15 3
6 „ È15 3
the position is s ¸ 10.303 units and the body is farthest from the origin.
3.5 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 1. y œ 10x 3 cos x Ê 2. y œ
3 x
5 sin x Ê
3. y œ x2 cos x Ê
dy dx
œ
dy dx
5
(cos x) œ 10 3 sin x
(sin x) œ
d dx
œ Èx sec x tan x
dy dx
5. y œ csc x 4Èx 7 Ê " x#
3 x#
d dx
3 x#
5 cos x
œ x2 asin xb 2x cos x œ x2 sin x 2x cos x
4. y œ Èx sec x 3 Ê
6. y œ x# cot x
œ 10 3
dy dx
Ê
dy dx
dy dx
œ csc x cot x
œ x#
œ x# csc# x 2x cot x
sec x 2È x
d dx
0 œ Èx sec x tan x 4 #È x
(cot x) cot x †
d dx
sec x 2È x
0 œ csc x cot x ax# b
2 x$
2 Èx
œ x# csc# x (cot x)(2x)
2 x$
2 x$
sin x 2 7. faxb œ sin x tan x Ê f w axb œ sin x sec2 x cos x tan x œ sin x sec2 x cos x cos x œ sin xasec x 1b
8. gaxb œ csc xcot x Ê gw axb œ csc xacsc2 xb acsc xcot xbcot x œ csc3 x csc x cot2 x œ csc xacsc2 x cot2 xb 9. y œ (sec x tan x)(sec x tan x) Ê #
dy dx
œ (sec x tan x)
d dx
(sec x tan x) (sec x tan x) #
d dx
(sec x tan x)
œ (sec x tan x) asec x tan x sec xb (sec x tan x) asec x tan x sec xb œ asec# x tan x sec x tan# x sec$ x sec# x tan xb asec# x tan x sec x tan# x sec$ x tan x sec# xb œ 0. ŠNote also that y œ sec# x tan# x œ atan# x 1b tan# x œ 1 Ê 10. y œ (sin x cos x) sec x Ê
œ (sin x cos x)
dy dx
d dx
sin# x cos x sin x cos# x cos x sin x cos# x
œ
" cos# x
d dx (sin x cos x) (sin x cos x) sin x x sin x cos cos cos# x x
œ sec# x
ŠNote also that y œ sin x sec x cos x sec x œ tan x 1 Ê 11. y œ œ
Ê
dy dx
œ
(1 cot x)
csc# x csc# x cot x csc# x cot x (1 cot x)#
12. y œ œ
cot x 1 cot x
cos x 1 sin x
Ê
dy dx
œ
sin x sin# x cos# x (1 sin x)#
(1 sin x)
œ
d dx
œ
œ 0.‹
(sec x) sec x
œ (sin x cos x)(sec x tan x) (sec x)(cos x sin x) œ œ
dy dx
(cot x) (cot x) (1 cot x)#
d dx
(1 cot x)
œ
dy dx
œ sec# x.‹
(1 cot x) acsc# xb (cot x) acsc# xb (1 cot x)#
csc# x (1 cot x)#
d (cos x) (cos x) dx (1 sin x) b (cos x) acos xb œ (1 sin x) a(1sin xsin (1 sin x)# x)# (1 sin x) sin x 1 " (1 sin x)# œ (1 sin x)# œ 1 sin x d dx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.5 Derivatives of Trigonometric Functions 13. y œ
4 cos x
" tan x
14. y œ
cos x x
x cos x
œ 4 sec x cot x Ê Ê
œ
dy dx
œ 4 sec x tan x csc# x
dy dx
x(sin x) (cos x)(1) x#
15. y œ x# sin x 2x cos x 2 sin x Ê
121
(cos x)(1) x(sin x) cos# x
œ
x sin x cos x x#
cos x x sin x cos# x
dy dx
œ ax# cos x (sin x)(2x)b a(2x)(sin x) (cos x)(2)b 2 cos x
dy dx
œ ax# (sin x) (cos x)(2x)b a2x cos x (sin x)(2)b 2(sin x)
œ x# cos x 2x sin x 2x sin x 2 cos x 2 cos x œ x# cos x 16. y œ x# cos x 2x sin x 2 cos x Ê
œ x# sin x 2x cos x 2x cos x 2 sin x 2 sin x œ x# sin x 17. faxb œ x3 sin x cos x Ê f w axb œ x3 sin xasin xb x3 cos xacos xb 3x2 sin x cos x œ x3 sin2 x x3 cos2 x 3x2 sin x cos x 18. gaxb œ a2 xbtan2 x Ê gw axb œ a2 xba2 tan x sec2 xb a1btan2 x œ 2a2 xbtan x sec2 x tan2 x œ 2a2 xbtan xa sec2 x tan xb 19. s œ tan t t Ê 1 csc t 1 csc t
21. s œ œ
Ê
œ sec# t 1
ds dt ds dt
20. s œ t# sec t 1 Ê
sin t 1 cos t
Ê
ds dt
œ 2t sec t tan t
(1 csc t)(csc t cot t) (" csc t)(csc t cot t) (1 csc t)#
œ
csc t cot t csc# t cot t csc t cot t csc# t cot t (1 csc t)#
22. s œ
ds dt
(1 cos t)(cos t) (sin t)(sin t) (1 cos t)#
œ
23. r œ 4 )# sin ) Ê
2 csc t cot t (1 csc t)#
œ
dr d)
œ ˆ) #
24. r œ ) sin ) cos ) Ê
dr d)
d d)
œ
cos t cos# t sin# t (1 cos t)#
œ
cos t " (1 cos t)#
œ 1 "cos t œ
" cos t 1
(sin )) (sin ))(2))‰ œ a)# cos ) 2) sin )b œ )() cos ) # sin ))
œ () cos ) (sin ))(1)) sin ) œ ) cos )
dr 25. r œ sec ) csc ) Ê d) œ (sec ))(csc ) cot )) (csc ))(sec ) tan )) " " cos ) sin ) ‰ " " # # œ ˆ cos ) ‰ ˆ sin ) ‰ ˆ sin ) ‰ ˆ sin" ) ‰ ˆ cos" ) ‰ ˆ cos ) œ sin# ) cos# ) œ sec ) csc )
26. r œ (1 sec )) sin ) Ê 27. p œ &
" cot q
œ
sin q cos q cos q
Ê
dp dq
œ
dp dq
dp dq
tan q 1 tan q
31. p œ
q sin q q2 1
Ê
Ê dp dq
œ sec# q
œ (1 csc q)(sin q) (cos q)(csc q cot q) œ (sin q 1) cot# q œ sin q csc# q
(cos q)(cos q sin q) (sin q cos q)(sin q) cos# q
cos# q cos q sin q sin# q cos q sin q cos# q
30. p œ
œ
œ (" sec )) cos ) (sin ))(sec ) tan )) œ (cos ) ") tan# ) œ cos ) sec# )
œ 5 tan q Ê
28. p œ (1 csc q) cos q Ê 29. p œ
dr d)
œ
" cos# q
œ sec# q
(1 tan q) asec# qb (tan q) asec# qb (1 tan q)#
dp dq
œ
œ
ˆq2 1‰aq cos q sin qa1bb aq sin qba2qb aq 2 1 b 2
œ
sec# q tan q sec# q tan q sec# q (1 tan q)#
œ
œ
sec# q (1 tan q)#
q3 cos q q2 sin q q cos q sin q 2q2 sin q aq 2 1 b 2
q3 cos q q2 sin q q cos q sin q aq 2 1 b 2
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122
Chapter 3 Differentiation aq sec qbˆ3 sec2 q‰ a3q tan qbaq sec q tan q sec qa1bb aq sec qb2 3 2 ˆ 3q sec q q sec q 3q sec q tan q 3q sec q q sec q tan2 q sec q tan q‰ q sec3 q 3q2 sec q tan q q sec q tan2 q sec q tan q
32. p œ œ
3q tan q q sec q
Ê
dp dq
œ
œ
aq sec qb2
aq sec qb2
33. (a) y œ csc x Ê yw œ csc x cot x Ê yww œ a(csc x) acsc# xb (cot x)(csc x cot x)b œ csc$ x csc x cot# x œ (csc x) acsc# x cot# xb œ (csc x) acsc# x csc# x 1b œ 2 csc$ x csc x (b) y œ sec x Ê yw œ sec x tan x Ê yww œ (sec x) asec# xb (tan x)(sec x tan x) œ sec$ x sec x tan# x œ (sec x) asec# x tan# xb œ (sec x) asec# x sec# x 1b œ 2 sec$ x sec x 34. (a) y œ 2 sin x Ê yw œ 2 cos x Ê yww œ 2(sin x) œ 2 sin x Ê ywww œ 2 cos x Ê yÐ%Ñ œ 2 sin x (b) y œ 9 cos x Ê yw œ 9 sin x Ê yww œ 9 cos x Ê ywww œ 9(sin x) œ 9 sin x Ê yÐ%Ñ œ 9 cos x 35. y œ sin x Ê yw œ cos x Ê slope of tangent at x œ 1 is yw (1) œ cos (1) œ "; slope of tangent at x œ 0 is yw (0) œ cos (0) œ 1; and slope of tangent at x œ 3#1 is yw ˆ 3#1 ‰ œ cos 3#1
œ 0. The tangent at (1ß !) is y 0 œ 1(x 1), or y œ x 1; the tangent at (0ß 0) is y 0 œ 1(x 0), or y œ x; and the tangent at ˆ 31 ‰ # ß 1 is y œ 1.
36. y œ tan x Ê yw œ sec# x Ê slope of tangent at x œ 13 is sec# ˆ 13 ‰ œ 4; slope of tangent at x œ 0 is sec# (0) œ 1; and slope of tangent at x œ
1 3
is sec# ˆ 13 ‰ œ 4. The tangent
at ˆ 13 ß tanˆ 13 ‰‰ œ Š 13 ß È3‹ is y È3 œ 4ˆx 13 ‰ ; the tangent at (0ß 0) is y œ x; and the tangent at ˆ 13 ß tan ˆ 13 ‰‰ œ Š 13 ß È3‹ is y È3 œ 4 ˆx 13 ‰ . 37. y œ sec x Ê yw œ sec x tan x Ê slope of tangent at x œ 13 is sec ˆ 13 ‰ tan ˆ 13 ‰ œ 2È3 ; slope of tangent is sec ˆ 14 ‰ tan ˆ 14 ‰ œ È2 . The tangent at the point ˆ 1 ß sec ˆ 1 ‰‰ œ ˆ 1 ß #‰ is y 2 œ #È3 ˆx 1 ‰ ;
at x œ
1 4
3
3
3
the tangent at the point
3
ˆ 14 ß sec ˆ 14 ‰‰
œ
Š 14 ß È2‹
is y È2
œ È2 ˆx 14 ‰ .
38. y œ 1 cos x Ê yw œ sin x Ê slope of tangent at È
x œ 13 is sin ˆ 13 ‰ œ #3 ; slope of tangent at x œ ‰ œ 1. The tangent at the point is sin ˆ 31 #
31 #
ˆ 13 ß " cos ˆ 13 ‰‰ œ ˆ 13 ß 3# ‰ È
is y 3# œ #3 ˆx 13 ‰ ; the tangent at the point ˆ 3#1 ß " cos ˆ 3#1 ‰‰ œ ˆ 3#1 ß 1‰ is y 1 œ x 3#1 39. Yes, y œ x sin x Ê yw œ " cos x; horizontal tangent occurs where 1 cos x œ 0 Ê cos x œ 1 Ê x œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.5 Derivatives of Trigonometric Functions 40. No, y œ 2x sin x Ê yw œ 2 cos x; horizontal tangent occurs where 2 cos x œ 0 Ê cos x œ #. But there are no x-values for which cos x œ #. 41. No, y œ x cot x Ê yw œ 1 csc# x; horizontal tangent occurs where 1 csc# x œ 0 Ê csc# x œ 1. But there are no x-values for which csc# x œ 1. 42. Yes, y œ x 2 cos x Ê yw œ 1 2 sin x; horizontal tangent occurs where 1 2 sin x œ 0 Ê 1 œ 2 sin x Ê "# œ sin x Ê x œ 16 or x œ 561 43. We want all points on the curve where the tangent line has slope 2. Thus, y œ tan x Ê yw œ sec# x so that yw œ 2 Ê sec# x œ 2 Ê sec x œ „ È2 Ê x œ „ 14 . Then the tangent line at ˆ 14 ß "‰ has equation y 1 œ 2 ˆx 14 ‰ ; the tangent line at ˆ 14 ß "‰ has equation y 1 œ 2 ˆx 14 ‰ .
44. We want all points on the curve y œ cot x where the tangent line has slope 1. Thus y œ cot x Ê yw œ csc# x so that yw œ 1 Ê csc# x œ 1 Ê csc# x œ 1 Ê csc x œ „ 1 Ê x œ 1# . The tangent line at ˆ 1# ß !‰ is y œ x 12 .
2 cos x ‰ 45. y œ 4 cot x 2 csc x Ê yw œ csc# x 2 csc x cot x œ ˆ sin" x ‰ ˆ 1 sin x
(a) When x œ 1# , then yw œ 1; the tangent line is y œ x w
1 #
2.
(b) To find the location of the horizontal tangent set y œ 0 Ê 1 2 cos x œ 0 Ê x œ then y œ % È3 is the horizontal tangent. 46. y œ 1 È2 csc x cot x Ê yw œ È2 csc x cot x csc# x œ ˆ sin" x ‰ Š
1 3
È2 cos x 1 ‹ sin x
(a) If x œ 14 , then yw œ 4; the tangent line is y œ 4x 1 4. (b) To find the location of the horizontal tangent set yw œ 0 Ê È2 cos x 1 œ 0 Ê x œ xœ
31 4 ,
radians. When x œ 13 ,
31 4
radians. When
then y œ 2 is the horizontal tangent.
47. lim sin ˆ "x #" ‰ œ sin ˆ #" #" ‰ œ sin 0 œ 0 xÄ2
48.
lim
x Ä 16
49. lim1 )Ä
6
È1 cos (1 csc x) œ É1 cos ˆ1 csc ˆ 1 ‰‰ œ È1 cos a1 † a2bb œ È2
sin ) "# ) 16
6
œ
d d) asin )b¹)œ 1 6
œ cos )¹
)œ 16
œ cosˆ 16 ‰ œ
È3 2
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123
124
Chapter 3 Differentiation tan ) 1 ) 14
50. lim1 )Ä
4
œ
d d) atan )b¹)œ 1
œ sec2 )¹
4
œ sec2 ˆ 14 ‰ œ 2
)œ 14
1 ‰ ‘ ˆ 1 ‰ ‘ ˆ 1 ‰‘ œ sec 1 œ 1 51. lim sec cos x 1 tan ˆ 4 sec x 1 œ sec 1 1 tan 4 sec 0 1 œ œ sec 1 tan 4
xÄ!
x ‰ ˆ 1 tan 0 ‰ ˆ 1‰ 52. lim sin ˆ tan1xtan 2 sec x œ sin tan 0 2 sec 0 œ sin # œ 1
xÄ!
53. lim tan ˆ1 tÄ!
sin t ‰ t
œ tan Š1 lim
tÄ!
1) ‰ 54. lim cos ˆ sin ) œ cos Š1 lim
)
sin t t ‹
‹ ) Ä ! sin )
)Ä!
œ tan (1 1) œ 0 "
œ cos Œ1 †
lim
sin ) )
)Ä!
" œ cos ˆ1 † 1 ‰ œ 1
dv da ˆ1‰ 55. s œ 2 2 sin t Ê v œ ds dt œ 2 cos t Ê a œ dt œ 2 sin t Ê j œ dt œ 2 cos t. Therefore, velocity œ v 4 œ È2 m/sec; speed œ ¸v ˆ 1 ‰¸ œ È2 m/sec; acceleration œ a ˆ 1 ‰ œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ È2 m/sec$ . 4
56. s œ sin t cos t Ê v œ velocity œ
v ˆ 14 ‰
58.
œ cos t sin t Ê a œ
ds dt
œ 0 m/sec; speed œ
sin# 3x x# xÄ!
57. lim f(x) œ lim xÄ!
4
¸v ˆ 14 ‰¸
4
œ sin t cos t Ê j œ
dv dt
œ 0 m/sec; acceleration œ
œ cos t sin t. Therefore œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ 0 m/sec$ .
da dt
a ˆ 14 ‰
4
œ lim 9 ˆ sin3x3x ‰ ˆ sin3x3x ‰ œ 9 so that f is continuous at x œ 0 Ê lim f(x) œ f(0) Ê 9 œ c. xÄ!
xÄ!
lim g(x) œ lim c (x b) œ b and lim b g(x) œ lim b cos x œ 1 so that g is continuous at x œ 0 Ê lim c g(x) xÄ! xÄ! xÄ! xÄ! œ lim b g(x) Ê b œ 1. Now g is not differentiable at x œ 0: At x œ 0, the left-hand derivative is
x Ä !c
xÄ!
d dx
(x b)¸ x = 0 œ 1, but the right-hand derivative is
(cos x)¸ x=0 œ sin 0 œ 0. The left- and right-hand
d dx
derivatives can never agree at x œ 0, so g is not differentiable at x œ 0 for any value of b (including b œ 1). 59.
d*** dx***
d% dx%
(cos x) œ sin x because
(cos x) œ cos x Ê the derivative of cos x any number of times that is a
multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 œ 249 † 4 3 Ê œ
$
d dx$
#%*†%
d ’ dx #%* % (cos x)“ œ †
60. (a) y œ sec x œ Ê
d dx
" cos x
$
d dx$
Ê
d*** dx***
(cos x)
(cos x) œ sin x.
dy dx
œ
(cos x)(0) (1)(sin x) (cos x)#
œ
(sin x)(0) (1)(cos x) (sin x)#
œ
sin x cos# x
sin x ‰ œ ˆ cos" x ‰ ˆ cos x œ sec x tan x
(sec x) œ sec x tan x
(b) y œ csc x œ Ê
d dx
d dx
Ê
dy dx
cos x sin# x
œ
" ‰ ˆ cos x ‰ œ ˆ sin x sin x œ csc x cot x
(csc x) œ csc x cot x
(c) y œ cot x œ Ê
" sin x
cos x sin x
Ê
dy dx #
œ
(sin x)(sin x) (cos x)(cos x) (sin x)#
œ
sin# xcos# x sin# x
œ
" sin# x
œ csc# x
(cot x) œ csc x
61. (a) t œ 0 Ä x œ 10 cosa0b œ 10 cm; t œ cm (b) t œ 0 Ä v œ 10 sina0b œ 0 sec ;tœ
1 3 1 3
Ä x œ 10 cosˆ 13 ‰ œ 5 cm; t œ 341 Ä x œ 10 cosˆ 341 ‰ œ 5È2 cm cm cm Ä v œ 10 sinˆ 13 ‰ œ 5È3 sec ; t œ 341 Ä v œ 10 sinˆ 341 ‰ œ 5È2 sec
62. (a) t œ 0 Ä x œ 3 cosa0b 4 sina0b œ 3 ft; t œ
1 2
Ä x œ 3 cosˆ 12 ‰ 4 sinˆ 12 ‰ œ 4 ft;
t œ 1 Ä x œ 3 cosa1b 4 sina1b œ 3 ft ft (b) t œ 0 Ä v œ 3 sina0b 4 cosa0b œ 4 sec ;tœ t œ 1 Ä v œ 3 sina1b 4 cosa1b œ 4
1 2
Ä v œ 3 sinˆ 12 ‰ 4 cosˆ 12 ‰ œ 3
ft sec ;
ft sec
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.5 Derivatives of Trigonometric Functions
125
63.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y œ closer and closer to the black curve y œ cos x because
d dx
(sin x) œ
is true as h takes on the values of 1, 0.5, 0.3 and 0.1.
sin x lim sin (x h) h hÄ!
sin (x h) sin x h
get
œ cos x. The same
64.
cos (x h) cos x h cos x lim cos (x h) œ sin x. h hÄ!
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y œ
get
closer and closer to the black curve y œ sin x because
The
d dx
(cos x) œ
same is true as h takes on the values of 1, 0.5, 0.3, and 0.1. 65. (a)
The dashed curves of y œ
sinax hb sinax hb #h
are closer to the black curve y œ cos x than the corresponding dashed
curves in Exercise 63 illustrating that the centered difference quotient is a better approximation of the derivative of this function. (b)
The dashed curves of y œ
cosax hb cosax hb #h
are closer to the black curve y œ sin x than the corresponding dashed
curves in Exercise 64 illustrating that the centered difference quotient is a better approximation of the derivative of this function. 66. lim
hÄ!
k0 h k k 0 h k 2h
œ lim
xÄ!
k h k k hk 2h
œ lim 0 œ 0 Ê the limits of the centered difference quotient exists even hÄ!
though the derivative of f(x) œ kxk does not exist at x œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
126
Chapter 3 Differentiation
67. y œ tan x Ê yw œ sec# x, so the smallest value yw œ sec# x takes on is yw œ 1 when x œ 0; yw has no maximum value since sec# x has no largest value on ˆ 1# ß 1# ‰ ; yw is never negative since sec# x 1.
68. y œ cot x Ê yw œ csc# x so yw has no smallest value since csc# x has no minimum value on (!ß 1); the largest value of yw is 1, when x œ 1# ; the slope is never positive since the largest value yw œ csc2 x takes on is 1. 69. y œ
sin x x appears to cross the y-axis at y œ 1, since lim sin x œ 1; y œ sinx2x appears to cross the y-axis xÄ! x
at y œ 2, since lim sinx2x œ 2; y œ sinx4x appears to xÄ! cross the y-axis at y œ 4, since lim sinx4x œ 4. xÄ!
However, none of these graphs actually cross the y-axis since x œ 0 is not in the domain of the functions. Also, lim
xÄ!
sin 5x x
sin (3x) x
œ 5, lim
xÄ!
œ k Ê the graphs of y œ yœ
sin kx x
œ 3, and lim
sin kx x
xÄ!
yœ
sin 5x x ,
sin (3x) , x
and
approach 5, 3, and k, respectively, as
x Ä 0. However, the graphs do not actually cross the y-axis. 70. (a)
sin h h
h 1 0.01 0.001 0.0001 lim
hÄ!
sin h° h
ˆ sinh h ‰ ˆ 180 ‰ 1 .99994923 1 1 1
.017452406 .017453292 .017453292 .017453292 œ lim
xÄ!
1 ‰ sin ˆh† 180 h
œ lim
1
180
hÄ!
1 ‰ sin ˆh† 180 1 180 †h
œ lim
)Ä!
1 sin )
180
)
œ
1 180
() œ h †
1 180 )
(converting to radians) (b)
cos h1 h
h 1 0.01 0.001 0.0001 lim
hÄ!
0.0001523 0.0000015 0.0000001 0
cos h1 h
(c) In degrees,
œ 0, whether h is measured in degrees or radians.
d dx
(sin x) œ lim
hÄ!
œ lim ˆsin x † hÄ!
cos h 1 ‰ h
sin (x h) sin x h
lim ˆcos x † hÄ!
1 ‰ œ (sin x)(0) (cos x) ˆ 180 œ
1 180
œ lim
hÄ!
sin h ‰ h
(sin x cos h cos x sin h) sin x h
œ (sin x) † lim ˆ cos hh 1 ‰ (cos x) † lim ˆ sinh h ‰ hÄ!
hÄ!
cos x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.6 The Chain Rule (d) In degrees, œ lim
hÄ!
d dx
hÄ!
(cos x)(cos h 1) sin x sin h h
œ (cos x) lim ˆ hÄ!
(e)
d# dx# d# dx#
cos (x h) cos x h
(cos x) œ lim cos h 1 ‰ h
œ lim
hÄ!
œ lim ˆcos x † hÄ!
127
(cos x cos h sin x sin h) cos x h
cos h 1 ‰ h
lim ˆsin x † hÄ!
sin h ‰ h
1 ‰ 1 (sin x) lim ˆ sinh h ‰ œ (cos x)(0) (sin x) ˆ 180 œ 180 sin x
hÄ!
1 ‰# ˆ 180
d$ dx$
(sin x) œ
d dx
1 ˆ 180
(cos x) œ
d dx
1 1 ‰# ˆ 180 sin x‰ œ ˆ 180 cos x;
cos x‰ œ
sin x;
(sin x) œ d$ dx$
d dx
#
$
1 ‰ 1 ‰ sin x‹ œ ˆ 180 cos x; Š ˆ 180
(cos x) œ
d dx
#
$
1 ‰ 1 ‰ cos x‹ œ ˆ 180 sin x Š ˆ 180
3.6 THE CHAIN RULE 1. f(u) œ 6u 9 Ê f w (u) œ 6 Ê f w (g(x)) œ 6; g(x) œ
" #
x% Ê gw (x) œ 2x$ ; therefore
dy dx
œ f w (g(x))gw (x) œ 6 † 2x$ œ 12x$
2. f(u) œ 2u$ Ê f w (u) œ 6u# Ê f w (g(x)) œ 6(8x 1)# ; g(x) œ 8x 1 Ê gw (x) œ 8; therefore œ 6(8x 1)# † 8 œ 48(8x 1)#
dy dx
œ f w (g(x))gw (x)
3. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (3x 1); g(x) œ 3x 1 Ê gw (x) œ 3; therefore
œ f w (g(x))gw (x)
dy dx
œ (cos (3x 1))(3) œ 3 cos (3x 1) 4. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin ˆ 3x ‰ ; g(x) œ ‰ œ "3 sin ˆ 3x ‰ œ sin ˆ 3x ‰ † ˆ " 3
x 3
Ê gw (x) œ "3 ; therefore
dy dx
œ f w (g(x))gw (x)
5. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin (sin x); g(x) œ sin x Ê gw (x) œ cos x; therefore dy dx
œ f w (g(x))gw (x) œ (sin (sin x)) cos x
6. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (x cos x); g(x) œ x cos x Ê gw (x) œ 1 sin x; therefore dy dx
œ f w (g(x))gw (x) œ (cos (x cos x))(1 sin x)
7. f(u) œ tan u Ê f w (u) œ sec# u Ê f w (g(x)) œ sec# (10x 5); g(x) œ 10x 5 Ê gw (x) œ 10; therefore dy dx
œ f w (g(x))gw (x) œ asec# (10x 5)b (10) œ 10 sec# (10x 5)
8. f(u) œ sec u Ê f w (u) œ sec u tan u Ê f w (g(x)) œ sec ax# 7xb tan ax# 7xb ; g(x) œ x# 7x Ê gw (x) œ 2x 7; therefore
dy dx
œ f w (g(x))gw (x) œ (2x 7) sec ax# 7xb tan ax# 7xb
9. With u œ (2x 1), y œ u& :
dy dx
œ
dy du du dx
œ 5u% † 2 œ 10(2x 1)%
10. With u œ (4 3x), y œ u* :
dy dx
œ
dy du du dx
œ 9u) † (3) œ 27(4 3x))
11. With u œ ˆ1 x7 ‰ , y œ u( : 12. With u œ ˆ x# 1‰ , y œ u"! :
dy dx
œ
dy dx
œ
#
13. With u œ Š x8 x "x ‹ , y œ u% : 14. With u œ 3x2 4x 6, y œ u1Î2 :
) œ 7u) † ˆ "7 ‰ œ ˆ" x7 ‰
dy du du dx dy du du dx
dy dx
dy dx
œ œ
œ 10u"" † ˆ "# ‰ œ 5 ˆ x# 1‰
dy du du dx
dy du du dx
œ 4u$ † ˆ x4 1
"‰ x#
""
#
$
œ 4 Š x8 x "x ‹ ˆ x4 1
œ "# u1Î2 † a6x 4b œ
3x 2 È3x2 4x6
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"‰ x#
128
Chapter 3 Differentiation
15. With u œ tan x, y œ sec u: 16. With u œ 1 "x , y œ cot u: 17. With u œ sin x, y œ u$ :
œ
dy dx dy dx
œ
dy du du dx
dy dx
œ
dy du du dx
18. With u œ cos x, y œ 5u% :
dy dx
œ
19. p œ È3 t œ (3 t)"Î# Ê 3 20. q œ È 2r r# œ a2r r# b
21. s œ œ
4 31 4 1
sin 3t
4 51
dp dt
"Î3
œ
œ acsc# ub ˆ x"# ‰ œ x"# csc# ˆ1 x" ‰
œ 3u# cos x œ 3 asin# xb (cos x)
dy du du dx
Ê
cos 5t Ê
œ (sec u tan u) asec# xb œ (sec (tan x) tan (tan x)) sec# x
dy du du dx
" #
dq dr
œ a20u& b (sin x) œ 20 acos& xb (sin x)
(3 t)"Î# † œ
" 3
d dt
a2r r# b
(3 t) œ "# (3 t)"Î# œ
2 Î3
†
d dr
a2r r# b œ
ds dt
œ
cos 3t †
d dt
(3t)
ds dt
œ cos ˆ 3#1t ‰ †
d dt
ˆ 3#1t ‰ sin ˆ 3#1t ‰ †
4 31
4 51
(sin 5t) †
d dt
" 3
" 2È 3 t
a2r r# b
(5t) œ
4 1
2 Î3
2 2r 3a2rr# b2Î3
(2 2r) œ
cos 3t
4 1
sin 5t
(cos 3t sin 5t)
22. s œ sin ˆ 3#1t ‰ cos ˆ 3#1t ‰ Ê œ 321 ˆcos 3#1t sin 3#1t ‰ 23. r œ (csc ) cot ))" Ê
dr d)
24. r œ 6(sec ) tan ))3Î2 Ê
œ (csc ) cot ))# dr d)
d d)
d dt
ˆ 3#1t ‰ œ
(csc ) cot )) œ
œ 6 † 3# asec ) tan )b1Î#
d d) asec
31 2
cos ˆ 3#1t ‰
csc ) cot ) csc# ) (csc ) cot ))#
œ
31 2
sin ˆ 3#1t ‰
csc ) (cot ) csc )) (csc ) cot ))#
œ
csc ) csc ) cot )
) tan )b œ 9Èsec ) tan )asec ) tan ) sec2 )b
d d # d % % # # 25. y œ x# sin% x x cos# x Ê dy xb cos# x † dx œ x dx asin xb sin x † dx ax b x dx acos d d œ x# ˆ4 sin$ x dx (sin x)‰ 2x sin% x x ˆ2 cos$ x † dx (cos x)‰ cos# x
d dx
(x)
œ x# a4 sin$ x cos xb 2x sin% x xa a2 cos$ xb (sin x)b cos# x œ 4x# sin$ x cos x 2x sin% x 2x sin x cos$ x cos# x d ˆ"‰ x d $ $ asin& xb sin& x † dx x 3 dx acos xb cos x † œ "x a5 sin' x cos xb asin& xb ˆ x"# ‰ 3x a a3 cos# xb (sin x)b acos$ xb ˆ 3" ‰
26. y œ
" x
sin& x
x 3
cos$ x Ê yw œ
œ 5x sin' x cos x
" x#
" d x dx
sin& x x cos# x sin x
" 3
ˆ x3 ‰
cos$ x
" " ‰" 7 d ( ' ˆ Ê dy 21 (3x 2) 4 #x# dx œ 21 (3x 2) † dx (3x 2) 7 " ‰# ˆ " ‰ " ' ' ˆ # 21 (3x 2) † 3 (1) 4 #x# x$ œ (3x 2) $ x Š4 "# ‹
27. y œ œ
d dx
(1) ˆ4
" ‰# #x #
†
d dx
ˆ4
" ‰ #x#
#x
% 28. y œ (5 2x)$ "8 ˆ x2 1‰ Ê
dy dx
$ $ œ 3(5 2x)% (2) 84 ˆ x2 1‰ ˆ x2# ‰ œ 6(5 2x)% ˆ x"# ‰ ˆ x2 1‰
$
œ
6 (5 2x)%
Š 2x 1‹ x#
29. y œ (4x 3)% (x 1)$ Ê %
dy dx
œ (4x 3)% (3)(x 1)% †
%
$
$
d dx
(x 1) (x 1)$ (4)(4x 3)$ † %
œ (4x 3) (3)(x 1) (1) (x 1) (4)(4x 3) (4) œ 3(4x 3) (x 1) œ
$
(4x 3) (x 1)%
c3(4x 3) 16(x 1)d œ
%
d dx $
(4x 3)
16(4x 3) (x 1)$
$
(4x 3) (4x 7) (x 1)%
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.6 The Chain Rule '
30. y œ (2x 5)" ax# 5xb Ê &
2 ax# 5xb (2x 5)#
œ 6 ax# 5xb
dy dx
&
'
œ (2x 5)" (6) ax# 5xb (2x 5) ax# 5xb (1)(2x 5)# (2)
'
d ˆ d 31. h(x) œ x tan ˆ2Èx‰ 7 Ê hw (x) œ x dx tan ˆ2x"Î# ‰‰ tan ˆ2x"Î# ‰ † dx (x) 0 d ˆ "Î# ‰ " # ˆ "Î# ‰ "Î# ‰ #ˆ È ‰ ˆ ˆ † dx 2x œ x sec 2x tan 2x œ x sec 2 x † È tan 2Èx‰ œ Èx sec# ˆ2Èx‰ tan ˆ2Èx‰ x
d ˆ d 32. k(x) œ x# sec ˆ "x ‰ Ê kw (x) œ x# dx sec x" ‰ sec ˆ x" ‰ † dx ax# b œ x# sec ˆ x" ‰ tan ˆ x" ‰ † œ x# sec ˆ "x ‰ tan ˆ x" ‰ † ˆ x"# ‰ 2x sec ˆ x" ‰ œ 2x sec ˆ x" ‰ sec ˆ x" ‰ tan ˆ x" ‰
33. faxb œ È7 x sec x Ê f w axb œ "# a7 x sec xb1Î2 ax † asec x tan xb asec xb † "b œ 34. gaxb œ œ
tan 3x ax 7 b 4
ax 7b4 ˆsec2 3x†$‰ atan 3xb4ax 7b3 †1 ax 7b4 ‘2
Ê gw axb œ
x sec x tan x sec x #È7x sec x
ax 7b3 ˆ$ax 7bsec2 3x 4tan 3x‰ ax 7 b 8
ˆ$ax 7bsec2 3x 4tan 3x‰ ax 7 b5 #
35. f()) œ ˆ 1 sincos) ) ‰ Ê f w ()) œ 2 ˆ 1 sincos) ) ‰ † œ
œ
ˆ x" ‰ 2x sec ˆ x" ‰
d dx
(2 sin )) acos ) cos# ) sin# )b (1 cos ))$
sin 3t ‰ 36. g(t) œ ˆ 1 3 2t
"
(2 sin )) (cos ) 1) (1 cos ))$
œ
3 2t 1 sin 3t
œ
37. r œ sin a)# b cos (2)) Ê
d d)
Ê gw (t) œ
ˆ 1 sincos) ) ‰ œ œ
†
(1 cos ))(cos )) (sin ))(sin )) (1 cos ))#
2 sin ) (1 cos ))#
a1 sin 3tba2b a3 2tba3 cos 3tb a1 sin 3tb2
œ sin a)# b (sin 2))
dr d)
2 sin ) 1 cos )
d d)
œ
2 2sin 3t 9 cos 3t 6t cos 3t a1 sin 3tb2
(2)) cos (2)) acos a)# bb †
d d)
a) # b
œ sin a)# b (sin 2))(2) (cos 2)) acos a)# bb (2)) œ 2 sin a)# b sin (#)) 2) cos (2)) cos a)# b 38. r œ Šsec È)‹ tan ˆ ") ‰ Ê
dr d)
œ )"# sec È) sec# ˆ ") ‰ 39. q œ sin Š Ètt 1 ‹ Ê œ cos Š Ètt 1 ‹ †
2
t1
dq dt
41. y œ sin# (1t 2) Ê
" #È )
tan ˆ ") ‰ sec È) tan È) œ Šsec È)‹ ”
œ cos Š Ètt 1 ‹ †
dq dt
Èt 1
40. q œ cot ˆ sint t ‰ Ê
œ Šsec È)‹ ˆ sec# ") ‰ ˆ )"# ‰ tan ˆ ") ‰ Šsec È) tan È)‹ Š
È
t t
1
d dt
Èt 1 (1)t †
d dt
sec# ˆ )" ‰ )# •
ˆÈ t 1 ‰
ˆÈ t 1 ‰
#
1) t œ cos Š Ètt 1 ‹ Š 2(t ‹ œ Š 2(tt1)2$Î# ‹ cos Š Ètt 1 ‹ 2(t 1)$Î#
œ csc# ˆ sint t ‰ †
d dt
ˆ sint t ‰ œ ˆcsc# ˆ sint t ‰‰ ˆ t cos tt# sin t ‰
œ 2 sin (1t 2) †
dy dt
Š Ètt 1 ‹ œ cos Š Ètt 1 ‹ †
tan È) tan ˆ ") ‰ #È )
" ‹ #È )
d dt
sin (1t 2) œ 2 sin (1t 2) † cos (1t 2) †
d dt
(1t 2)
œ 21 sin (1t 2) cos (1t 2) 42. y œ sec# 1t Ê
dy dt
œ (2 sec 1t) †
43. y œ (1 cos 2t)% Ê 44. y œ ˆ1 cot ˆ #t ‰‰ œ
csc# ˆ #t ‰ $ ˆ1 cot ˆ t ‰‰ #
#
dy dt
Ê
d dt
(sec 1t) œ (2 sec 1t)(sec 1t tan 1t) †
œ 4(1 cos 2t)& † dy dt
œ 2 ˆ1 cot ˆ #t ‰‰
d dt $
d dt
(1t) œ 21 sec# 1t tan 1t
(1 cos 2t) œ 4(1 cos 2t)& (sin 2t) † †
dˆ dt 1
cot ˆ #t ‰‰ œ 2 ˆ1 cot ˆ #t ‰‰
$
d dt
(2t) œ
8 sin 2t (1 cos 2t)&
† ˆcsc# ˆ #t ‰‰ †
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dˆt‰ dt #
129
130
Chapter 3 Differentiation
45. y œ at tan tb10 Ê 46. y œ ˆt3Î4 sin t‰ œ
4 Î3
œ t1 asin tb4Î3 Ê
dy dt
œ t1 ˆ 34 ‰asin tb1Î3 cost t2 asin tb4Î3 œ
4asin tb1Î3 cos t 3t
asin tb4Î3 t2
asin tb1Î3 a4t cos t 3cos tb 3t2 3
2
47. y œ Š t3 t 4t ‹ Ê 4‰ 48. y œ ˆ 3t 5t 2
œ
œ 10at tan tb9 at † sec2 t 1 † tan tb œ 10 t9 tan9 tat sec2 t tan tb œ 10 t10 tan9 t sec2 t 10 t9 tan10 t
dy dt
5
2
2
dy dt
œ 3Š t3 t 4t ‹ †
dy dt
4‰ œ 5ˆ 3t 5t 2
Ê
ˆt3 4t‰a2tb t2 ˆ3t2 4‰
6
at3 4tb2
†
a5t 2b†3 a3t 4b†5 a5t 2b2
œ
3t4 at3 4tb2
†
2t4 8t2 3t4 4t2 at3 4tb2 6
2‰ œ 5ˆ 5t † 3t 4
œ
15t 6 15t 20 a5t 2b2
3t4 ˆt4 4t2 ‰ t4 at2 4b4
3t2 ˆt2 4‰ at2 4b4
œ 6
5t 2b œ 5 aa3t † 4 b6
26 a5t 2b2
130a5t 2b4 a3t 4b6
49. y œ sin acos (2t 5)b Ê
dy dt
œ cos (cos (2t 5)) †
d dt
cos (2t 5) œ cos (cos (2t 5)) † (sin (2t 5)) †
d dt
(2t 5)
œ 2 cos (cos (2t 5))(sin (2t 5)) ˆ ˆ t ‰‰ † 50. y œ cos ˆ5 sin ˆ 3t ‰‰ Ê dy dt œ sin 5 sin 3 œ 53 sin ˆ5 sin ˆ 3t ‰‰ ˆcos ˆ 3t ‰‰
d dt
ˆ5 sin ˆ 3t ‰‰ œ sin ˆ5 sin ˆ 3t ‰‰ ˆ5 cos ˆ 3t ‰‰ †
$
dy % ˆ t ‰‘# d † dt 1 dt œ 3 1 tan 1# " ‘ % ˆ t ‰‘# $ˆ t ‰ #ˆ t ‰ tan 1# tan 1# sec 1# † 1# œ 1
51. y œ 1 tan% ˆ 1t# ‰‘ Ê œ 12 1 52. y œ
" 6
$
c1 cos# (7t)d Ê "Î#
Ê
œ "# a1 cos at# bb
"Î#
53. y œ a1 cos at# bb
dy dt
2 cos ŒÉ1 Èt É1 Èt†2Èt
# tan% ˆ 1t# ‰‘ tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰‘
#
d dt
tan ˆ 1t# ‰‘
#
c1 cos# (7t)d † 2 cos (7t)(sin (7t))(7) œ 7 c1 cos# (7t)d (cos (7t) sin (7t))
3 6 " #
# tan% ˆ 1t# ‰‘ œ 3 1 tan% ˆ 1t# ‰‘ 4 tan$ ˆ 1t# ‰ †
a1 cos at# bb
"Î#
†
d dt
a1 cos at# bb œ
#
" #
a1 cos at# bb
"Î#
ˆsin at# b †
d dt
a t# b ‰
dy dt
œ 4 cos ŒÉ1 Èt †
d dt
"
ŒÉ1 Èt œ 4 cos ŒÉ1 Èt †
# É 1 È t
†
d dt
ˆ1 Èt‰
cos ŒÉ1 Èt
œ
55. y œ tan2 asin3 tb Ê
œ
ˆ 3t ‰
at b asin at# bb † 2t œ È1t sin cos at# b
54. y œ 4 sin ŒÉ1 Èt Ê œ
œ
dy dt
d dt
Ét tÈt
dy dt
56. y œ cos4 asec2 3tb Ê
œ 2 tanasin3 tb † sec2 asin3 tb † a3sin2 t † acos tbb œ 6 tanasin3 tbsec2 asin3 tbsin2 t cos t dy dt
œ 4 cos3 asec2 a3tbbasinasec2 a3tbb † 2 aseca3tbbaseca3tb tana3tb † 3bb
œ 24 cos3 asec2 a3tbbsinasec2 a3tbbsec2 a3tb tana3tb 4
57. y œ 3ta2t2 5b Ê
dy dt
3
58. y œ Ê3t É2 È1 t Ê œ
" #Ê3tÉ2È1t
3
4
3
3
œ 3t † 4a2t2 5b a4tb 3 † a2t2 5b œ 3a2t2 5b ’16t2 2t2 5“ œ 3a2t2 5b a18t2 5b
dy dt
" #É2È1t
1 Î2
œ "# Œ3t É2 È1 t †
1 #È 1 t
œ
" #Ê3tÉ2È1t
" Œ3 # Š2 È1 t‹
12È1tÉ2È1t " 4È1tÉ2È1t
œ
1 Î2
" # a1
tb1Î2 a1b
12È1tÉ2È1t " 8È1tÉ2È1tÊ3tÉ2È1t
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.6 The Chain Rule $ # # 59. y œ ˆ1 "x ‰ Ê yw œ 3 ˆ1 x" ‰ ˆ x"# ‰ œ x3# ˆ1 x" ‰ Ê yww œ ˆ x3# ‰ † # œ ˆ x3# ‰ ˆ2 ˆ1 x" ‰ ˆ x"# ‰‰ ˆ x6$ ‰ ˆ1 x" ‰ œ œ x6$ ˆ1 x" ‰ ˆ1 2x ‰
60. y œ ˆ1 Èx‰ Ê yww œ œ
" #
œ
" #x
61. y œ
" #
"
Ê yw œ ˆ1 Èx‰
œ
ˆ1 x" ‰
ˆ "# x"Î# ‰ œ
" #
6 x$
ˆ1 x" ‰# œ
ˆ1 x" ‰# ˆ1 x" ‰# †
ˆ1 Èx‰$ Š
ˆ x3# ‰
ˆ1 x" ‰ ˆ x" 1 x" ‰
ˆ1 Èx‰# x"Î#
#
" #È x
$
x" ˆ1 Èx‰ “ œ
" #
1‹ œ
" #x
" #
x" ˆ1 Èx‰
ˆ1 Èx‰$ Š 3#
$
"# x"Î# ˆ1 Èx‰ 1‘
" ‹ #Èx
cot (3x 1) Ê yw œ 9" csc# (3x 1)(3) œ 3" csc# (3x 1) Ê yww œ ˆ 32 ‰ (csc (3x 1) † 2 3
d dx
# $ ’ˆ1 Èx‰ ˆ "# x$Î# ‰ x"Î# (2) ˆ1 Èx‰ ˆ "# x"Î# ‰“
$Î# ˆ 1 Èx‰ ’ " # x
" 9
#
6 x%
d dx 6 x$
131
csc (3x 1)(csc (3x 1) cot (3x 1) †
d dx
csc (3x 1))
d dx
#
(3x 1)) œ 2 csc (3x 1) cot (3x 1)
62. y œ 9 tan ˆ x3 ‰ Ê yw œ 9 ˆsec# ˆ x3 ‰‰ ˆ 3" ‰ œ 3 sec# ˆ x3 ‰ Ê yww œ 3 † 2 sec ˆ x3 ‰ ˆsec ˆ x3 ‰ tan ˆ x3 ‰‰ ˆ "3 ‰ œ 2 sec# ˆ 3x ‰ tan ˆ 3x ‰ 63. y œ xa2x 1b4 Ê yw œ x † 4a2x 1b3 a2b 1 † a2x 1b4 œ a2x 1b3 a8x a2x 1bb œ a2x 1b3 a10x 1b Ê yww œ a2x 1b3 a10b 3a2x 1b2 a2ba10x 1b œ 2a2x 1b2 a5a2x 1b 3a10x 1bb œ 2a2x 1b2 a40x 8b œ 16a2x 1b2 a5x 1b 5
4
5
4
4
64. y œ x2 ax3 1b Ê yw œ x2 † 5ax3 1b a3x2 b 2xax3 1b œ xax3 1b ’15x3 2ax3 1b“ œ ax3 1b a17x4 2xb 4
3
3
Ê yww œ ax3 1b a68x3 2b 4ax3 1b a3x2 ba17x4 2xb œ 2ax3 1b ’ax3 1ba34x3 1b 6x2 a17x4 2xb“ 3
œ 2ax3 1b a136x6 47x3 1b 65. g(x) œ Èx Ê gw (x) œ
" #È x
Ê g(1) œ 1 and gw (1) œ
therefore, (f ‰ g)w (1) œ f w (g(1)) † gw (1) œ 5 †
" #
œ
" u#
; f(u) œ u& 1 Ê f w (u) œ 5u% Ê f w (g(1)) œ f w (1) œ 5;
5 #
66. g(x) œ (1 x)" Ê gw (x) œ (1 x)# (1) œ Ê f w (u) œ
" #
" (1x)#
Ê g(1) œ
" # w
and gw (1) œ
" 4
; f(u) œ 1
Ê f w (g(1)) œ f w ˆ #" ‰ œ 4; therefore, (f ‰ g)w (1) œ f (g(1))gw (1) œ 4 †
67. g(x) œ 5Èx Ê gw (x) œ Ê f w (g(1)) œ f w (5) œ
5 Ê g(1) œ 5 and gw (1) œ #5 #È x 1 1 # ˆ1‰ 10 csc # œ 10 ; therefore,
" 4
" u
œ1
1‰ ; f(u) œ cot ˆ 110u ‰ Ê f w (u) œ csc# ˆ 110u ‰ ˆ 10 œ
1 10
csc# ˆ 110u ‰
1 (f ‰ g)w (1) œ f w (g(1))gw (1) œ 10 † 5# = 14
68. g(x) œ 1x Ê gw (x) œ 1 Ê g ˆ "4 ‰ œ 14 and gw ˆ 4" ‰ œ 1; f(u) œ u sec# u Ê f w (u) œ 1 2 sec u † sec u tan u œ 1 2 sec# u tan u Ê f w ˆg ˆ "4 ‰‰ œ f w ˆ 14 ‰ œ 1 2 sec# 14 tan 14 œ 5; therefore, (f ‰ g)w ˆ 4" ‰ œ f w ˆg ˆ 4" ‰‰ gw ˆ 4" ‰ œ 51 69. g(x) œ 10x# x 1 Ê gw (x) œ 20x 1 Ê g(0) œ 1 and gw (0) œ 1; f(u) œ œ
2u# 2 au # 1 b #
Ê f w (u) œ
au# 1b(2) (2u)(2u) au # 1 b #
Ê f w (g(0)) œ f w (1) œ 0; therefore, (f ‰ g)w (0) œ f w (g(0))gw (0) œ 0 † 1 œ 0
" 2 w x# 1 Ê g (x) œ x$ Ê g(1) œ 0 and 4(u 1) 1 ‰ (u 1)(1) (u 1)(1) 2 ˆ uu œ 2(u(u1)(2) 1 † (u 1)# 1)$ œ (u 1)$ w w w
70. g(x) œ œ
2u u # 1
#
1‰ 1‰ gw (1) œ 2; f(u) œ ˆ uu Ê f w (u) œ 2 ˆ uu 1 1
Ê f w (g(1)) œ f w (0) œ 4; therefore,
(f ‰ g) (1) œ f (g(1))g (1) œ (4)(2) œ 8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
d du
1‰ ˆ uu 1
132
Chapter 3 Differentiation
71. y œ fagaxbb, f w a3b œ 1, gw a2b œ 5, ga2b œ 3 Ê y w œ f w agaxbbg w axb Ê y w ¹
x œ2
œ a1b † 5 œ 5 72. r œ sinafatbb, fa0b œ 13 , f w a0b œ 4 Ê 73. (a) y œ 2f(x) Ê
dy dx
œ 2f w (x) Ê
(b) y œ f(x) g(x) Ê (c) y œ f(x) † g(x) Ê
dr dt
œ cosafatbb † f w atb Ê
œ f w (x) gw (x) Ê
dy dx
dy dx
œ
g(x)f w (x) f(x)gw (x) [g(x)]#
(e) y œ f(g(x)) Ê
dy dx
œ f w (g(x))gw (x) Ê
(d) y œ
f(x) g(x)
Ê
" #
Ê
dy dx ¹ x=2
dy dx
œ
(g) y œ (g(x))# Ê
dy dx
œ 2(g(x))$ † gw (x) Ê
Ê
dy dx ¹ x=2
œ
" #
(f(x))"Î# † f w (x) œ
"Î#
Ê
" # # "Î#
dy dx
a(f(2))# (g(2)) b
74. (a) y œ 5f(x) g(x) Ê (b) y œ f(x)(g(x))$ Ê
f (x) #Èf(x)
œ
dy dx ¹ x=3
dy dx ¹ x=2
œ
" 3
f (2) #Èf(2)
dy dx ¹ x=1
ˆ "3 ‰ #È 8
œ
œ
37 6
(3) œ 1 œ
" 6È 8
" 1 #È 2
œ
œ 2(g(3))$ gw (3) œ 2(4)$ † 5 œ "Î#
œ
È2 24
5 3#
a2f(x) † f w (x) 2g(x) † gw (x)b " #
a8# 2# b
"Î#
ˆ2 † 8 †
" 3
2 † 2 † (3)‰ œ 3È517
œ 5f w (1) gw (1) œ 5 ˆ 3" ‰ ˆ 38 ‰ œ 1
œ f(x) a3(g(x))# gw (x)b (g(x))$ f w (x) Ê
dy dx
(2) ˆ "3 ‰ (8)(3) ##
w
a2f(2)f w (2) 2g(2)gw (2)b œ
œ 5f w (x) gw (x) Ê
dy dx
Ê
a(f(x))# (g(x))# b
œ
œ f(3)gw (3) g(3)f w (3) œ 3 † 5 (4)(21) œ 15 81
œ f w (g(2))gw (2) œ f w (2)(3) œ w
(f) y œ (f(x))"Î# Ê
(h) y œ a(f(x))# (g(x))# b
dy dx ¹ x=3
g(2)f w (2) f(2)gw (2) [g(2)]#
œ
dy dx ¹ x=2
2 3
œ f w (3) gw (3) œ 21 5
dy dx ¹ x=3
œ f(x)gw (x) g(x)f w (x) Ê
dy dx
œ cosafa0bb † f w a0b œ cosˆ 13 ‰ † 4 œ ˆ "# ‰ † 4 œ 2
dr dt ¹tœ0
œ 2f w (2) œ 2 ˆ "3 ‰ œ
dy dx ¹ x=2
œ f w aga2bbgw a2b œ f w a3b † 5
dy dx ¹ x = 0
œ $f(0)(g(0))# gw (0) (g(0))$ f w (0)
œ 3(1)(1)# ˆ 3" ‰ (1)$ (5) œ 6 (c) y œ œ
f(x) g(x) 1
Ê
(g(x) 1)f w (x) f(x) gw (x) (g(x) 1)#
œ
dy dx
(4") ˆ "3 ‰(3) ˆ 83 ‰ (41)#
Ê
dy dx ¹ x = 1
(g(1) 1)f w (1) f(1)gw (1) (g(1) 1)#
œ
œ1
(d) y œ f(g(x)) Ê
dy dx
œ f w (g(x))gw (x) Ê
dy dx ¹ x = 0
œ f w (g(0))gw (0) œ f w (1) ˆ "3 ‰ œ ˆ "3 ‰ ˆ 3" ‰ œ 9"
(e) y œ g(f(x)) Ê
dy dx
œ gw (f(x))f w (x) Ê
dy dx ¹ x = 0
œ gw (f(0))f w (0) œ gw (1)(5) œ ˆ 83 ‰ (5) œ 40 3
(f) y œ ax"" f(x)b
#
Ê
œ 2 ax"" f(x)b
dy dx
$
a11x"! f w (x)b Ê
dy dx ¹ x=1
œ 2(1 f(1))$ a11 f w (1)b
" ‰ œ 2(1 3)$ ˆ11 "3 ‰ œ ˆ 42$ ‰ ˆ 32 3 œ 3
(g) y œ f(x g(x)) Ê
dy dx
œ f w (x g(x)) a1 gw (x)b Ê
dy dx ¹ x = 0
œ f w (0 g(0)) a1 gw (0)b œ f w (1) ˆ1 "3 ‰
œ ˆ "3 ‰ ˆ 43 ‰ œ 49 75.
ds dt
œ
ds d)
†
d) dt :
s œ cos ) Ê
76.
dy dt
œ
dy dx
†
dx dt :
y œ x# 7x 5 Ê
77. With y œ x, we should get (a) y œ (b)
u 5
7 Ê
y œ 1 "u Ê " œ " u# † (x 1)#
dy du dy du
œ
œ
" 5
œ sin ) Ê
ds d)
dy dx
dy dx
œ 2x 7 Ê
œ
dy dx ¹ x = 1
œ 9 so that
dy dt
œ
ds dt
dy dx
œ †
dx dt
ds d)
†
d) dt
œ9†
œ 1†5œ 5 " 3
œ3
œ 1 for both (a) and (b):
; u œ 5x 35 Ê " u#
œ sin ˆ 3#1 ‰ œ 1 so that
ds ¸ d) ) = 321
; u œ (x 1)
" a(x 1) " b#
†
" (x 1)#
"
du dx
œ 5; therefore,
Ê
du dx
œ (x 1)# †
dy dy dx œ du #
†
œ (x 1) (1) œ
" (x 1)#
du " dx œ 5 " (x 1)# ;
† 5 œ 1, as expected therefore
dy dx
œ
œ 1, again as expected
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dy du
†
du dx
Section 3.6 The Chain Rule 78. With y œ x$Î# , we should get (a) y œ u$ Ê
dy du
as expected. (b) y œ Èu Ê
œ
dy dx
3 #
x"Î# for both (a) and (b):
œ 3u# ; u œ Èx Ê
dy du
œ
" #È u
du dx
; u œ x$ Ê
du dx
" #È x
; therefore,
dy dx
œ
dy du
†
du dx
œ 3u# †
œ 3x# ; therefore,
dy dx
œ
dy du
†
du dx
œ
œ
" #Èu
" #È x
# œ 3 ˆÈx‰ †
† 3x# œ
" #È x $
" #Èx
† 3x# œ
3 #
œ
3 #
Èx,
x"Î# ,
again as expected. 2
2
1‰ 1‰ 1‰ 79. y œ ˆ xx and x œ 0 Ê y œ ˆ 00 œ a1b2 œ 1. yw œ 2ˆ xx 1 1 1 †
yw ¹
xœ0
œ
4 a0 1 b a0 1 b 3
œ
4 13
ax 1b†1 ax 1b†1 ax 1 b 2
1b 2 œ 2 aaxx 1 b ax 1 b 2 œ
œ 4 Ê y 1 œ 4ax 0b Ê y œ 4x 1
80. y œ Èx2 x 7 and x œ 2 Ê y œ Éa2b2 a2b 7 œ È9 œ 3. y w œ "# ax2 x 7b y w¹
xœ2
œ
2 a2 b 1 2 É a2 b 2 a 2 b 7
81. y œ 2 tan ˆ 14x ‰ Ê (a)
dy dx ¹ x = 1
œ
1 #
dy dx
4 ax 1 b ax 1 b 3
œ
3 6
œ
" #
œ ˆ2 sec#
1 Î2
a2x 1b œ
2x 1 2 È x 2 x 7
Ê y 3 œ "# ax 2b Ê y œ "# x 2 1x ‰ ˆ 1 ‰ 4 4
œ
1 #
sec#
1x 4
sec# ˆ 14 ‰ œ 1 Ê slope of tangent is 2; thus, y(1) œ 2 tan ˆ 14 ‰ œ 2 and yw (1) œ 1 Ê tangent line is
given by y 2 œ 1(x 1) Ê y œ 1x 2 1 (b) yw œ 1# sec# ˆ 14x ‰ and the smallest value the secant function can have in # x 2 is 1 Ê the minimum value of yw is 1# and that occurs when 1# œ 1# sec# ˆ 14x ‰ Ê 1 œ sec# ˆ 14x ‰ Ê „ 1 œ sec ˆ 14x ‰ Ê x œ 0. 82. (a) y œ sin 2x Ê yw œ 2 cos 2x Ê yw (0) œ 2 cos (0) œ 2 Ê tangent to y œ sin 2x at the origin is y œ 2x; y œ sin ˆ x# ‰ Ê yw œ "# cos ˆ x# ‰ Ê yw (0) œ "# cos 0 œ "# Ê tangent to y œ sin ˆ x# ‰ at the origin is y œ "# x. The tangents are perpendicular to each other at the origin since the product of their slopes is 1. (b) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m cos 0 œ m; y œ sin ˆ mx ‰ Ê yw œ m" cos ˆ mx ‰ Ê yw (0) œ m" cos (0) œ m" . Since m † ˆ m" ‰ œ 1, the tangent lines are perpendicular at the origin.
(c) y œ sin (mx) Ê yw œ m cos (mx). The largest value cos (mx) can attain is 1 at x œ 0 Ê the largest value yw can attain is kmk because kyw k œ km cos (mx)k œ kmk kcos mxk Ÿ kmk † 1 œ kmk . Also, y œ sin ˆ mx ‰ ˆ x ‰¸ Ÿ ¸ m" ¸ ¸cos ˆ mx ‰¸ Ÿ km" k Ê the largest value yw can attain is ¸ m" ¸ . Ê yw œ m" cos ˆ mx ‰ Ê kyw k œ ¸ " m cos m (d) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m Ê slope of curve at the origin is m. Also, sin (mx) completes m periods on [0ß 21]. Therefore the slope of the curve y œ sin (mx) at the origin is the same as the number of periods it completes on [0ß 21]. In particular, for large m, we can think of “compressing" the graph of y œ sin x horizontally which gives more periods completed on [0ß 21], but also increases the slope of the graph at the origin. 83. s œ A cos (21bt) Ê v œ
ds dt
œ A sin (21bt)(21b) œ 21bA sin (21bt). If we replace b with 2b to double the
frequency, the velocity formula gives v œ 41bA sin (41bt) Ê doubling the frequency causes the velocity to # # double. Also v œ #1bA sin (21bt) Ê a œ dv dt œ 41 b A cos (21bt). If we replace b with 2b in the
acceleration formula, we get a œ 161# b# A cos (41bt) Ê doubling the frequency causes the acceleration to $ $ quadruple. Finally, a œ 41# b# A cos (21bt) Ê j œ da dt œ 81 b A sin (21bt). If we replace b with 2b in the jerk formula, we get j œ 641$ b$ A sin (41bt) Ê doubling the frequency multiplies the jerk by a factor of 8. 21 21 21 ‰ 84. (a) y œ 37 sin 365 (x 101)‘ 25 Ê yw œ 37 cos 365 (x 101)‘ ˆ 365 œ
741 365
21 cos 365 (x 101)‘ .
The temperature is increasing the fastest when yw is as large as possible. The largest value of 21 21 cos 365 (x 101)‘ is 1 and occurs when 365 (x 101) œ 0 Ê x œ 101 Ê on day 101 of the year
( µ April 11), the temperature is increasing the fastest.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
133
134
Chapter 3 Differentiation
(b) yw (101) œ
741 365
21 cos 365 (101 101)‘ œ
85. s œ (" 4t)"Î# Ê v œ v œ 2(" 4t)
"Î#
ds dt
œ
Ê aœ
dv dt
86. We need to show a œ œ
k 2È s
† kÈs œ
87. v proportional to œ 2sk$Î# † dx dt
88. Let
k Ès
kT 2
k #
cos (0) œ
741 365
¸ 0.64 °F/day
(1 4t)"Î# (4) œ 2(1 4t)"Î# Ê v(6) œ 2(" % † 6)"Î# œ " #
œ † 2(1 4t)
is constant: a œ
dv dt
œ
$Î#
dv ds
(4) œ 4(1 4t)
†
ds dt
and
dv ds
œ
" Ès
Ê vœ
k Ès
for some constant k Ê
dv ds
m/sec;
Ê a(6) œ 4(1 4 † 6)$Î# œ 14#5 m/sec#
ˆkÈs‰ œ
k 2È s
œ 2sk$Î# . Thus, a œ
#
œ f(x). Then, a œ dT dL
d ds
$Î#
2 5
Ê aœ
dv ds
†
†
ds dt
ds dt
œ
dv ds
†v
which is a constant.
œ k# ˆ s"# ‰ Ê acceleration is a constant times
89. T œ 21É Lg Ê œ
#
dv dt
" #
741 365
dv dt
œ 21 †
œ
" #É Lg
dv dx
†
dx dt
œ
†
" g
œ
1 gÉ Lg
dv dx
† f(x) œ œ
1 ÈgL
d dx
" s#
dT du
œ
dv ds
œ
dv ds
†v
so a is inversely proportional to s# .
ˆ dx ‰ dt † f(x) œ
. Therefore,
dv dt
œ
d dx
(f(x)) † f(x) œ f w (x)f(x), as required.
dT dL
†
dL du
œ
1 ÈgL
† kL œ
1 kÈ L Èg
œ
" #
† 21kÉ Lg
, as required.
90. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g(0), then f ‰ g is differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0 so there is no contradiction. 91. As h Ä 0, the graph of y œ
sin 2(xh)sin 2x h
approaches the graph of y œ 2 cos 2x because lim
hÄ!
sin 2(xh)sin 2x h
œ
d dx
(sin 2x) œ 2 cos 2x.
92. As h Ä 0, the graph of y œ
cos c(x h)# dcos ax# b h #
approaches the graph of y œ 2x sin ax b because cos c(x h)# dcos ax# b h hÄ!
lim
œ
d dx
ccos ax# bd œ 2x sin ax# b.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.7 Implicit Differentiation 93. (a)
(b)
df dt
œ 1.27324 sin 2t 0.42444 sin 6t 0.2546 sin 10t 0.18186 sin 14t
(c) The curve of y œ
approximates y œ
df dt
dg dt
the best when t is not 1, 1# , 0, 1# , nor 1.
94. (a)
(b)
dh dt
œ 2.5464 cos (2t) 2.5464 cos (6t) 2.5465 cos (10t) 2.54646 cos (14t) 2.54646 cos (18t)
(c)
3.7 IMPLICIT DIFFERENTIATION 1. x# y xy# œ 6: Step 1:
Šx#
Step 2:
x#
dy dx
Step 3:
dy dx dy dx
ax# 2xyb œ 2xy y#
Step 4:
dy dx
œ
y † 2x‹ Šx † 2y
2xy
dy dx
dy dx
y# † 1‹ œ 0
œ 2xy y#
2xy y# x# 2xy
2. x$ y$ œ 18xy Ê 3x# 3y#
dy dx
œ 18y 18x
dy dx
Ê a3y# 18xb
dy dx
œ 18y 3x# Ê
dy dx
œ
6y x# y# 6x
3. 2xy y# œ x y: Step 1:
Š2x
Step 2:
2x
dy dx
dy dx
2y‹ 2y
2y
dy dx
dy dx
dy dx
œ1
dy dx
œ 1 2y
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
135
136
Chapter 3 Differentiation dy dx dy dx
Step 3: Step 4:
(2x 2y 1) œ " 2y œ
1 2y 2x 2y 1
4. x$ xy y$ œ 1 Ê 3x# y x
dy dx
3y#
œ 0 Ê a3y# xb
dy dx
dy dx
œ y 3x# Ê
dy dx
y 3x# 3y# x
œ
5. x# (x y)# œ x# y# : Step 1:
x# ’2(x y) Š1
Step 2:
2x# (x y)
Step 3:
dy dx dy dx
Step 4:
dy dx
dy dx ‹“
2y
(x y)# (2x) œ 2x 2y
dy dx
œ 2x 2x# (x y) 2x(x y)#
c2x# (x y) 2yd œ 2x c1 x(x y) (x y)# d œ
œ
2x c1 x(x y) (x y)# d 2x# (x y) 2y
œ
dy dx
x c1 x(x y) (x y)# d y x# (x y)
œ
x a1 x# xy x# 2xy y# b x# y x$ y
x 2x$ 3x# y xy# x# y x$ y
6. (3xy 7)# œ 6y Ê 2(3xy 7) † Š3x Ê
dy dx
dy dx
3y‹ œ 6
[6x(3xy 7) 6] œ 6y(3xy 7) Ê (x 1) (x 1) (x 1)#
dy dx
dy dx
Ê 2(3xy 7)(3x)
y(3xy 7) œ x(3xy 7) 1 œ
dy dx
6
dy dx
œ 6y(3xy 7)
#
3xy 7y 1 3x# y 7x
7. y# œ
x" x1
Ê 2y
8. x3 œ
2x y x 3y
Ê x4 3x3 y œ 2x y Ê 4x3 9x2 y 3x3 y w œ 2 y w Ê a3x3 1by w œ 2 4x3 9x2 y
Ê yw œ
œ
dy dx
œ
Ê
2 (x 1)#
dy dx
œ
" y(x 1)#
2 4x3 9x2 y 3x3 1
9. x œ tan y Ê 1 œ asec# yb
dy dx
Ê
dy dx
œ
" sec# y
œ cos# y
dy dy dy # # # 10. xy œ cot axyb Ê x dy dx y œ csc (xy)Šx dx y‹ Ê x dx x csc (xy) dx œ y csc (xy) y
Ê
dy dx x
x csc# (xy)‘ œ y csc# (xy) "‘ Ê
11. x tan (xy) œ ! Ê 1 csec# (xy)d Šy x œ
1 x sec# (xy)
y x
œ
cos# (xy) x
y x
œ
dy dx ‹
œ
y csc# (xy) "‘ x" csc# (xy)‘
œ yx
œ 0 Ê x sec# (xy)
dy dx
œ 1 y sec# (xy) Ê
dy dx
œ
" y sec# (xy) x sec# (xy)
cos# (xy) y x
12. x4 sin y œ x3 y2 Ê 4x3 (cos y)
dy dx
œ 3x2 y2 x3 † 2y
13. y sin Š "y ‹ œ 1 xy Ê y ’cos Š y" ‹ † (1) dy dx
dy dx
’ "y cos Š "y ‹ sin Š y" ‹ x“ œ y Ê
" y# dy dx
†
dy dx “
œ
dy dx
Ê acos y 2x3 yb
sin Š y" ‹ †
dy dx
œ x
y "y cos Š "y ‹ sin Š "y ‹ x
œ
dy dx
dy dx
œ 3x2 y2 4x3 Ê
dy dx
œ
3x2 y2 4x3 cos y 2x3 y
y Ê y #
y sin Š "y ‹ cos Š "y ‹ xy
14. x cosa2x 3yb œ y sin x Ê x sina2x 3yba2 3y w b cosa2x 3yb œ y cos x y w sin x Ê 2x sina2x 3yb 3x y w sina2x 3yb cosa2x 3yb œ y cos x y w sin x Ê cosa2x 3yb 2x sina2x 3yb y cos x œ asin x 3x sina2x 3ybby w Ê y w œ 15. )"Î# r"Î# œ 1 Ê 16. r 2È) œ
3 #
" #
)"Î# "# r"Î# †
)#Î$ 34 )$Î% Ê
dr d)
dr d)
œ0 Ê
dr d)
" ’ #È “œ r
)"Î# œ )"Î$ )"Î% Ê
" #È )
dr d)
Ê
dr d)
œ
2È r 2È )
cosa2x 3yb 2x sina2x 3yb y cos x sin x 3x sina2x 3yb Èr
œÈ
)
œ )"Î# )"Î$ )"Î%
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.7 Implicit Differentiation 17. sin (r )) œ
" #
Ê [cos (r ))] ˆr )
18. cos r cot ) œ r ) Ê (sin r)
dr d)
œ0Ê
dr ‰ d)
dr d)
[) cos (r ))] œ r cos (r )) Ê
csc# ) œ r )
Ê
dr d)
19. x# y# œ 1 Ê 2x 2yyw œ 0 Ê 2yyw œ 2x Ê y(1) xyw y#
Ê yww œ
œ
20. x#Î$ y#Î$ œ 1 Ê
2 3
y x Š xy ‹ y#
since yw œ xy Ê
x"Î$ 23 y"Î$
œ0 Ê
dy dx
dy dx
d# y dx#
œ
" 3
2x 2 2y
21. y# œ x# 2x Ê 2yyw œ 2x 2 Ê yw œ
d# y dx#
œ yww œ
d# y dx#
Ê
y# (x 1)# y$
ww
œy œ
y # x # y$
d# y dx#
d dx
,
y# a" y# b y$
œ
dr d)
#
csc ) œ rsin r)
ayw b œ
œ
dy dx
œ )r , cos (r )) Á 0
d dx
Š xy ‹
" y$
œ yx
"Î$ "Î$
"Î$
œ ˆ yx ‰
y (x 1)yw y#
; then yww œ
" y1
œ
y (x 1) Š x y 1 ‹ y#
œ (y 1)" ; then yww œ (y 1)# † yw
" (y 1)$
œ yww œ
23. 2Èy œ x y Ê y"Î# yw œ 1 yw Ê yw ˆy"Î# 1‰ œ 1 Ê
dy dx
œ yw œ
" y "Î# 1
Èy Èy 1
œ
; we can
" #
differentiate the equation yw ˆy"Î# 1‰ œ 1 again to find yww : yw ˆ y$Î# yw ‰ ˆy"Î# 1‰ yww œ 0 Ê ˆy"Î# 1‰ yww œ
" #
w # $Î#
cy d y
Ê
d# y dx#
" #
œ yww œ
#
Œy
" $Î# "Î# 1 y
ay "Î# 1b
" $ 2y$Î# ay "Î# 1b
œ
œ
" $ # ˆ1 È y ‰
24. xy y# œ 1 Ê xyw y 2yyw œ 0 Ê xyw 2yyw œ y Ê yw (x 2y) œ y Ê yw œ œ
(x 2y)yw y(1 2yw ) (x 2y)#
œ
2y(x 2y) 2y# (x 2y)$
œ
y
œ
y
(x 2y) ’ (x 2y) “ y ’1 2 Š (x 2y) ‹“ (x 2y)#
2y# 2xy (x 2y)$
œ
œ
"
(x 2y)
y (x2y)
;
d# y dx#
œ yww
cy(x 2y) y(x 2y) 2y# d (x 2y)#
2y(x y) (x 2y)$ #
25. x$ y$ œ 16 Ê 3x# 3y# yw œ 0 Ê 3y# yw œ 3x# Ê yw œ xy# ; we differentiate y# yw œ x# to find yww : # ww
w
w
w #
# ww
y y y c2y † y d œ 2x Ê y y œ 2x 2y cy d œ
2xy$ 2x% y&
Ê
d# y dx# ¹ (2ß2)
œ
32 32 32
ww
Ê y œ
2x 2y Š y#
26. xy y# œ 1 Ê xyw y 2yyw œ 0 Ê yw (x 2y) œ y Ê yw œ since yw k (0
ß
1)
œ "# we obtain yww k (0ß
1)
œ
27. y# x# œ y% 2x at (#ß ") and (#ß 1) Ê 2y Ê
dy dx
a2y 4y$ b œ 2 2x Ê
#
x# ‹ y#
œ
2x y#
2x% y$
œ 2
(2) ˆ "# ‰ (1)(0) 4
dy dx
œ
x" #y $ y
Ê
;
"Î$ x"Î$ †ˆ "3 y #Î$ ‰ Œ y"Î$ y"Î$ ˆ "3 x #Î$ ‰ x x#Î$
22. y# 2x œ 1 2y Ê 2y † yw 2 œ 2yw Ê yw (2y 2) œ 2 Ê yw œ œ (y 1)# (y 1)" Ê
d# y dx#
23 y"Î$ ‘ œ 23 x"Î$ Ê yw œ
x1 y
œ
[sin r )] œ r csc# ) Ê
œ yw œ xy ; now to find
x"Î$ †ˆ 3" y #Î$ ‰ yw y"Î$ ˆ "3 x #Î$ ‰ œ x#Î$ "Î$ y " "Î$ %Î$ " x œ 3x %Î$ 3y"Î$ x#Î$ 3 y
x#Î$ y"Î$
r cos (r )) ) cos (r ))
œ
dy dx
Differentiating again, yww œ Ê
dr d)
dr d)
y (x2y)
Ê yww œ
(x 2y) ayw b (y) a1 2yw b (x 2y)#
œ 4"
dy dx 2x dy dx ¹ ( 2ß1)
œ 4y$
dy dx
2 Ê 2y
œ 1 and
dy dx ¹ ( 2ß 1)
dy dx
4y$
dy dx
œ 2 2x
œ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
137
138
Chapter 3 Differentiation #
28. ax# y# b œ (x y)# at ("ß !) and ("ß 1) Ê 2 ax# y# b Š2x 2y Ê and
dy dx
c2y ax# y# b (x y)d œ 2x ax# y# b (x y) Ê
dy dx ¹ (1ßc1)
dy dx
dy dx ‹
œ
œ 2(x y) Š1
2x ax# y# b (x y) 2y ax# y# b (x y)
dy dx ‹
Ê
dy dx ¹ (1ß0)
œ 1
œ1
29. x# xy y# œ 1 Ê 2x y xyw 2yyw œ 0 Ê (x 2y)yw œ 2x y Ê yw œ (a) the slope of the tangent line m œ yw k (2 3) œ ß
Ê the tangent line is y 3 œ
7 4
(b) the normal line is y 3 œ 47 (x 2) Ê y œ 47 x
7 4
2x y 2y x
;
(x 2) Ê y œ
7 4
x
" #
29 7
30. x# y# œ 25 Ê 2x 2yyw œ 0 Ê yw œ xy ; (a) the slope of the tangent line m œ yw k (3
4)
ß
œ xy ¹
(3ß 4)
œ
Ê the tangent line is y 4 œ
3 4
3 4
(x 3) Ê y œ
3 4
x
25 4
(b) the normal line is y 4 œ 43 (x 3) Ê y œ 43 x 31. x# y# œ 9 Ê 2xy# 2x# yyw œ 0 Ê x# yyw œ xy# Ê yw œ yx ; (a) the slope of the tangent line m œ yw k ( 1ß3) œ yx ¸ ( 1ß3) œ 3 Ê the tangent line is y 3 œ 3(x 1) Ê y œ 3x 6 (b) the normal line is y 3 œ "3 (x 1) Ê y œ 3" x
8 3
32. y# 2x 4y " œ ! Ê 2yyw 2 4yw œ 0 Ê 2(y 2)yw œ 2 Ê yw œ
" y#
;
(a) the slope of the tangent line m œ yw k ( 2ß1) œ 1 Ê the tangent line is y 1 œ 1(x 2) Ê y œ x 1 (b) the normal line is y 1 œ 1(x 2) Ê y œ x 3 33. 6x# 3xy 2y# 17y 6 œ 0 Ê 12x 3y 3xyw 4yyw 17yw œ 0 Ê yw (3x 4y 17) œ 12x 3y 12x 3y Ê yw œ 3x 4y 17 ; (a) the slope of the tangent line m œ yw k ( 1ß0) œ Ê yœ
6 7
x
"2x 3y 3x 4y 17 ¹ ( 1ß0)
œ
6 7
Ê the tangent line is y 0 œ
6 7
(x 1)
6 7
(b) the normal line is y 0 œ 76 (x 1) Ê y œ 76 x
7 6
34. x# È3xy 2y# œ 5 Ê 2x È3xyw È3y 4yyw œ 0 Ê yw Š4y È3x‹ œ È3y 2x Ê yw œ (a) the slope of the tangent line m œ yw k ŠÈ3 2‹ œ ß
È3y 2x ¹ 4y È3x ŠÈ3ß2‹
œ 0 Ê the tangent line is y œ 2
(b) the normal line is x œ È3 35. 2xy 1 sin y œ 21 Ê 2xyw 2y 1(cos y)yw œ 0 Ê yw (2x 1 cos y) œ 2y Ê yw œ (a) the slope of the tangent line m œ yw k ˆ1 12 ‰ œ ß
2y 2x 1 cos y ¹ ˆ1ß 1 ‰ 2
y
1 #
œ 1# (x 1) Ê y œ 1# x 1
(b) the normal line is y
1 #
œ
2 1
(x 1) Ê y œ
2 1
x
2 1
2y 2x 1 cos y
œ 1# Ê the tangent line is 1 #
36. x sin 2y œ y cos 2x Ê x(cos 2y)2yw sin 2y œ 2y sin 2x yw cos 2x Ê yw (2x cos 2y cos 2x) œ sin 2y 2y sin 2x Ê yw œ
sin 2y 2y sin 2x cos 2x 2x cos 2y
(a) the slope of the tangent line m œ yw k ˆ 14
ß
1‰ 2
œ
; sin 2y 2y sin 2x cos 2x 2x cos 2y ¹ ˆ 1 ß 1 ‰ 4 2
y
1 #
œ 2 ˆx 14 ‰ Ê y œ 2x
(b) the normal line is y
1 #
œ "# ˆx 14 ‰ Ê y œ "# x
œ
1 1 #
œ 2 Ê the tangent line is
51 8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
È3y 2x 4y È3x
;
Section 3.7 Implicit Differentiation 37. y œ 2 sin (1x y) Ê yw œ 2 [cos (1x y)] † a1 yw b Ê yw [1 2 cos (1x y)] œ 21 cos (1x y) Ê yw œ (a) the slope of the tangent line m œ yw k (1 0) œ ß
21 cos (1x y) 1 2 cos (1x y) ¹(1ß0)
y 0 œ 21(x 1) Ê y œ 21x 21 (b) the normal line is y 0 œ #"1 (x 1) Ê y œ 2x1
139
21 cos (1x y) 1 # cos (1x y)
;
œ 21 Ê the tangent line is
" #1
38. x# cos# y sin y œ 0 Ê x# (2 cos y)(sin y)yw 2x cos# y yw cos y œ 0 Ê yw c2x# cos y sin y cos yd 2x cos# y 2x# cos y sin y cos y
œ 2x cos# y Ê yw œ
;
(a) the slope of the tangent line m œ yw k (0 1) œ ß
2x cos# y 2x# cos y sin y cos y ¹ (0ß1)
œ 0 Ê the tangent line is y œ 1
(b) the normal line is x œ 0 39. Solving x# xy y# œ 7 and y œ 0 Ê x# œ 7 Ê x œ „ È7 Ê ŠÈ7ß !‹ and ŠÈ7ß !‹ are the points where the curve crosses the x-axis. Now x# xy y# œ 7 Ê 2x y xyw 2yyw œ 0 Ê (x 2y)yw œ 2x y 2 È 7 y 2x y È È Ê yw œ 2x x 2y Ê m œ x 2y Ê the slope at Š 7ß !‹ is m œ È7 œ 2 and the slope at Š 7ß !‹ is È
m œ 2È77 œ 2. Since the slope is 2 in each case, the corresponding tangents must be parallel. 40. xy 2x y œ 0 Ê x
dy dx
y2
dy dx
œ0 Ê
dy dx
œ
y2 1x
; the slope of the line 2x y œ 0 is 2. In order to be
parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope of 2 " the tangent is "# . Therefore, y1 x œ # Ê 2y 4 œ 1 x Ê x œ 3 2y. Substituting in the original equation,
y(3 2y) 2(3 2y) y œ 0 Ê y# 4y 3 œ 0 Ê y œ 3 or y œ 1. If y œ 3, then x œ 3 and y 3 œ 2(x 3) Ê y œ 2x 3. If y œ 1, then x œ 1 and y 1 œ 2(x 1) Ê y œ 2x 3.
41. y% œ y# x# Ê 4y$ yw œ 2yyw 2x Ê 2 a2y$ yb yw œ 2x Ê yw œ y x2y$ ; the slope of the tangent line at È3 " È È È Š 43 ß #3 ‹ is y x2y$ ¹ È3 È3 œ È3 4 6È3 œ " 4 3 œ # " 3 œ 1; the slope of the tangent line at Š 43 ß #" ‹ # 4 Œ
is
x y2y$ ¹
Œ
È3 4
ß
œ
1 2
È3 " #
4
28
4
ß
#
2
œ
2È 3 42
8
œ È3
42. y# (2 x) œ x$ Ê 2yyw (2 x) y# (1) œ 3x# Ê yw œ œ
4 #
y# 3x# 2y(2 x)
; the slope of the tangent line is m œ
œ 2 Ê the tangent line is y 1 œ 2(x 1) Ê y œ 2x 1; the normal line is y 1 œ "# (x 1) Ê y œ "# x
43. y% 4y# œ x% 9x# Ê 4y$ yw 8yyw œ 4x$ 18x Ê yw a4y$ 8yb œ 4x$ 18x Ê yw œ œ
y# 3x# 2y(2 x) ¹ (1ß1)
x a2x# 9b y a2y# 4b
œ m; (3ß 2): m œ
(3)(18 9) 2(8 4)
œ 27 8 ; ($ß #): m œ
27 8
; (3ß #): m œ
27 8
ß
5 4
(b) yw œ 0 Ê
and yw k (2 4) œ ß
#
3y x y# 3x
4 5
œ
2x$ 9x 2y$ 4y
; (3ß #): m œ 27 8
44. x$ y$ 9xy œ 0 Ê 3x# 3y# yw 9xyw 9y œ 0 Ê yw a3y# 9xb œ 9y 3x# Ê yw œ (a) yw k (4 2) œ
4x$ 18x 4y$ 8y
3 #
9y 3x# 3y# 9x
œ
3y x# y# 3x
;
œ 0 Ê 3y x# œ 0 Ê y œ
x# 3
#
$
#
Ê x$ Š x3 ‹ 9x Š x3 ‹ œ 0 Ê x' 54x$ œ 0
Ê x$ ax$ 54b œ 0 Ê x œ 0 or x œ $È54 œ 3 $È2 Ê there is a horizontal tangent at x œ 3 $È2 . To find the corresponding y-value, we will use part (c). (c)
dx dy
œ0 Ê
y# 3x 3y x#
$
œ 0 Ê y# 3x œ 0 Ê y œ „ È3x ; y œ È3x Ê x$ ŠÈ3x‹ 9xÈ3x œ 0
3 3 Ê x$ 6È3 x$Î# œ 0 Ê x$Î# Šx$Î# 6È3‹ œ 0 Ê x$Î# œ 0 or x$Î# œ 6È3 Ê x œ 0 or x œ È 108 œ 3 È 4.
Since the equation x$ y$ 9xy œ 0 is symmetric in x and y, the graph is symmetric about the line y œ x. That is, if
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140
Chapter 3 Differentiation (aß b) is a point on the folium, then so is (bß a). Moreover, if yw k (a b) œ m, then yw k (b a) œ ß
ß
" m
. Thus, if the folium has a
horizontal tangent at (aß b), it has a vertical tangent at (bß a) so one might expect that with a horizontal tangent at 3 3 3 3 3 xœÈ 54 and a vertical tangent at x œ 3 $È4, the points of tangency are ŠÈ 54ß 3 È 4‹ and Š3 È 4ß È 54‹, respectively. One can check that these points do satisfy the equation x$ y$ 9xy œ 0. 45. x# 2xy 3y# œ 0 Ê 2x 2xyw 2y 6yyw œ 0 Ê yw (2x 6y) œ 2x 2y Ê yw œ xy 3y x ¹ (1ß1)
line m œ yw k (1 1) œ ß
xy 3y x
Ê the slope of the tangent
œ 1 Ê the equation of the normal line at (1ß 1) is y 1 œ 1(x 1) Ê y œ x 2. To find
where the normal line intersects the curve we substitute into its equation: x# 2x(2 x) 3(2 x)# œ 0 Ê x# 4x 2x# 3 a4 4x x# b œ 0 Ê 4x# 16x 12 œ 0 Ê x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 and y œ x 2 œ 1. Therefore, the normal to the curve at (1ß 1) intersects the curve at the point (3ß 1). Note that it also intersects the curve at (1ß 1). 46. Let p and q be integers with q 0 and suppose that y œ Èxp œ xpÎq . Then yq œ xp . Since p and q are integers and q
d q dx ay b xp 1ap pÎqb œ qp
assuming y is a differentiable function of x, œ
p q
†
xp
axpÎq b
1 q
47. y# œ x Ê
1
œ
p q
†
dy dx
œ
" #y
xp 1 xp pÎq
œ
p q
†
œ
p1 Ê qyq 1 dy Ê dx œ px
d p dx ax b
dy dx
pxp qyq
œ
1
œ
1
p q
xp yq
†
1 1
a p Îq b 1
†x
. If a normal is drawn from (aß 0) to (x" ß y" ) on the curve its slope satisfies
Ê y" œ 2y" (x" a) or a œ x" "# . Since x" 0 on the curve, we must have that a
" #
y" 0 x" a
œ 2y"
. By symmetry, the two Èx
Èx
points on the parabola are ˆx" ß Èx" ‰ and ˆx" ß Èx" ‰ . For the normal to be perpendicular, Š x" "a ‹ Š a x"" ‹ œ 1 Ê
x" (a x" )#
œ 1 Ê x" œ (a x" )# Ê x" œ ˆx"
" #
#
2x w 48. 2x# 3y# œ 5 Ê 4x 6yyw œ 0 Ê yw œ 2x 3y Ê y k (1 1) œ 3y ¹ 3x# 2y
Ê yw k (1 1) œ ß
3x# 2y ¹ (1ß1)
œ
and yw k (1
3 #
and y" œ „ #" . Therefore, ˆ 4" ß „ #" ‰ and a œ
œ 23 and yw k (1
(1ß1)
ß
y# œ x$ Ê 2yyw œ 3x# Ê yw œ
" 4
x" ‰ Ê x " œ
ß
1)
œ
3x# 2y ¹ (1ß 1)
ß
1)
œ 2x 3y ¹
(1ß 1)
œ
2 3
; also,
œ #3 . Therefore the
tangents to the curves are perpendicular at (1ß 1) and (1ß 1) (i.e., the curves are orthogonal at these two points of intersection). 49. (a) x2 y2 œ 4, x2 œ 3y2 Ê a3y2 b y2 œ 4 Ê y2 œ 1 Ê y œ „ 1. If y œ 1 Ê x2 a1b2 œ 4 Ê x2 œ 3 Ê x œ „ È3. If y œ 1 Ê x2 a1b2 œ 4 Ê x2 œ 3 Ê x œ „ È3. x2 y2 œ 4 Ê 2x 2y dy dx œ 0 Ê m1 œ At ŠÈ3ß 1‹: m1 œ
œ
dy dx
È3 1
dy dx
œ
At ŠÈ3ß 1‹: m1 œ
dy dx
œ
At ŠÈ3ß 1‹: m1 œ
dy dx
1
È3 2
Ê x œ 1 Š
dy Ê 1 œ 23 y dx Ê m2 œ
At Š 14 ß
È3 2 ‹:
At Š 14 ß
m1 œ
È3 2 ‹:
dy dx
m1 œ
dy dx
È3 2 2 ‹
œ
œ dy dx
œ
œ
œ È3 and m2 œ
dy dx
œ
ŠÈ3‹ a 1 b
œ È3 and m2 œ 3 4
È3 3
œ
È 3 3 a1 b
œ
dy dx
Êyœ „
È 33
œ
Ê m1 † m2 œ
È3 3
ŠÈ3‹ 3 a 1 b
È3 2 .
œ
dy dx
Ê m1 † m2 œ ŠÈ3‹Š
œ
If y œ
œ 14 . x œ 1 y2 Ê 1 œ 2y dy dx Ê m1 œ
x 3y
È3 3 ‹
œ 1
È ŠÈ3‹Š 33 ‹
Ê m1 † m2 œ ŠÈ3‹Š È3 3 È3 2 dy dx
œ 1
È3 3 ‹
Ê m1 † m2 œ ŠÈ3‹Š Êxœ1Š
2
È3 2 ‹
œ 1
È3 3 ‹
œ 1
œ 41 . If
1 œ 2y and x œ 13 y2
3 2y
1 2ŠÈ3Î2‹
œ
È3 3 a1 b
œ
dy dx
(b) x œ 1 y2 , x œ 13 y2 Ê ˆ 13 y2 ‰ œ 1 y2 Ê y2 œ yœ
dy dx
È3 3 a 1 b
œ È3 and m2 œ
ŠÈ3‹
œ
œ xy and x2 œ 3y2 Ê 2x œ 6y dy dx Ê m2 œ
œ È3 and m2 œ
È a13b
At ŠÈ3ß 1‹: m1 œ
dy dx
œ È13 and m2 œ
1 2ŠÈ3Î2‹
œ
1 È3
dy dx
and m2 œ
œ
dy dx
3 2ŠÈ3Î2‹
œ
œ
3 2ŠÈ3Î2‹
3 È3
Ê m1 † m2 œ Š È13 ‹Š È33 ‹ œ 1
œ È33 Ê m1 † m2 œ Š È13 ‹Š È33 ‹ œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3 4
.
Section 3.7 Implicit Differentiation 50. y œ 13 x b, y2 œ x3 Ê Ê
x4 4
dy dx
dy œ 13 and 2y dx œ 3x2 Ê
51. xy$ x# y œ 6 Ê x Š3y#
dy dx ‹
a4 b 2 2
3x2 2y
2
Ê ˆ 13 ‰Š 3x 2y ‹ œ 1 Ê
y$ x#
dy dx
2xy œ 0 Ê
$
#
#
x œ 3xy y$ 2xy ; thus
dx dy
a0b2 2
x2 2
appears to equal
"
dy dx
dx dy
dy dx
2
œ ! and ˆ 13 ‰Š 3x 2y ‹ œ 1 is
a3xy# x# b œ y$ 2xy Ê
x# y Š2x
dx dy ‹
œ0 Ê
dx dy
2
2
œ y Ê Š x2 ‹ œ x3
œ 8. At a4ß 8b, y œ 31 x b Ê 8 œ 31 a4b b Ê b œ
y 2xy $ # # $ œ 3xy # x# ; also, xy x y œ 6 Ê x a3y b y dx dy
œ
œ x3 Ê x4 4x3 œ 0 Ê x3 ax 4b œ 0 Ê x œ 0 or x œ 4. If x œ 0 Ê y œ
indeterminant at a0, 0b. If x œ 4 Ê y œ
Ê
dy dx
dy dx
œ
28 3 . y$ 2xy 3xy# x#
ay$ 2xyb œ 3xy# x#
. The two different treatments view the graphs as functions
symmetric across the line y œ x, so their slopes are reciprocals of one another at the corresponding points (aß b) and (bß a). 52. x$ y# œ sin# y Ê 3x# 2y œ
3x# 2 sin y cos y 2y
appears to equal
dy dx
œ (2 sin y)(cos y)
; also, x$ y# œ sin# y Ê 3x# "
dy dx
dx dy
dy dx
Ê
dy dx
(2y 2 sin y cos y) œ 3x# Ê
2y œ 2 sin y cos y Ê
dx dy
œ
2 sin y cos y 2y 3x#
dy dx
œ
3x# 2y 2 sin y cos y
; thus
dx dy
. The two different treatments view the graphs as functions symmetric across the line
y œ x so their slopes are reciprocals of one another at the corresponding points (aß b) and (bß a). 53-60. Example CAS commands: Maple: q1 := x^3-x*y+y^3 = 7; pt := [x=2,y=1]; p1 := implicitplot( q1, x=-3..3, y=-3..3 ): p1; eval( q1, pt ); q2 := implicitdiff( q1, y, x ); m := eval( q2, pt ); tan_line := y = 1 + m*(x-2); p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ): p3 := pointplot( eval([x,y],pt), color=blue ): display( [p1,p2,p3], ="Section 3.7 #57(c)" ); Mathematica: (functions and x0 may vary): Note use of double equal sign (logic statement) in definition of eqn and tanline. <
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
141
142
Chapter 3 Differentiation
3.8 RELATED RATES 1. A œ 1r# Ê 2. S œ 41r# Ê 3. y œ 5x,
dx dt
dx dt
dS dt
dy dt
7. x2 y2 œ 25,
dy dt
œ 5 dx dt Ê
dy dt
œ 5a2b œ 10
œ 2x dx dt ; when x œ 1 Ê
dy dt
œ5Ê dx dt
dr dt
dy dx œ 2 Ê 2 dx dt 3 dt œ 0 Ê 2 dt 3a2b œ 0 Ê
dy dt
œ3Ê
6. x œ y3 y,
dr dt
œ 81r
œ2Ê
4. 2x 3y œ 12, 5. y œ x2 ,
œ 21r
dA dt
œ 3y2 dy dt
dx dt
dy dt ;
dL dt
dx dt
œ 1,
a5ba1b a12ba3b
œ
dx dt
œ 3a2b2 a5b a5b œ 55
dy dy œ 2 Ê 2x dx dt 2y dt œ 0; when x œ 3 and y œ 4 Ê 2a3ba2b 2a4b dt œ 0 Ê
4 dy " 2 2 dy 3 dx 27 , dt œ # Ê 3x y dt 2x y dt œ 0; 2 3 dx 9 3a2b2 ˆ 13 ‰ ˆ "# ‰ 2a2b ˆ 13 ‰ dx dt œ 0 Ê dt œ 2
Ê
œ3
œ 2a1ba3b œ 6
when y œ 2 Ê
8. x2 y3 œ
9. L œ Èx2 y2 ,
dy dt
dx dt
Éa5b2 a12b2
10. r s2 v3 œ 12,
dr dt
Ê 4 2a1ba3b
œ
dy dt
œ3Ê
dL dt
œ
when x œ 2 Ê a2b2 y3 œ
1 Š2x dx dt 2È x2 y2
2y dy dt ‹ œ
dy dt
Ê y œ 13 . Thus
4 27
dy x dx dt y dt È x2 y2 ;
when x œ 5 and y œ 12
31 13
œ 4,
ds dt 3a2b2 dv dt
œ 3 Ê œ0Ê
dr ds dt 2s dt dv 1 dt œ 6
2 3 3v2 dv dt œ 0; when r œ 3 and s œ 1 Ê a3b a1b v œ 12 Ê v œ 2
dx m dS dx dS m2 dt œ 5 min Ê dt œ 12x dt ; when x œ 3 Ê dt œ 12a3ba5b œ 180 min 2 m dV dV m3 2 dx x3 , dx dt œ 5 min Ê dt œ 3x dt ; when x œ 3 Ê dt œ 3a3b a5b œ 135 min
11. (a) S œ 6x2 , (b) V œ 12. S œ 6x2 , Ê
dV dt
dS dt
2
in œ 72 sec Ê
œ 3a3b2 a2b œ 54
13. (a) V œ 1r# h Ê
dV dt dV dt
(c) V œ 1r# h Ê
dS dt œ in3 sec
dx 12x dx dt Ê 72 œ 12a3b dt Ê
œ 1 r# œ
14. (a) V œ "3 1r# h Ê
dh dt 1r# dh dt
dV dt œ " # dh 2 3 1r dt 3 1rh
dx dt
œ2
in sec ;
V œ x3 Ê
(b) V œ 1r# h Ê 21rh
dV dt
" # dh 3 1r dt dr dt
(b) V œ "3 1r# h Ê
œ
15. (a)
dV dt dV dt dR dt
" œ 1 volt/sec (b) dI dt œ 3 amp/sec " ˆ dV dI ‰ " ˆ dV V dI ‰ ‰ ˆ dR ‰ Ê dR œ R ˆ dI Ê dR dt I dt dt œ I dt R dt dt œ I dt I dt ˆ " ‰‘ œ ˆ #" ‰ (3) œ 3# ohms/sec, R is increasing œ "# 1 12 # 3
16. (a) P œ RI# Ê
dP dt
œ I#
(b) P œ RI# Ê 0 œ
dP dt
dR dt
2RI
œ I#
dR dt
œ 3x2 dx dt ; when x œ 3
œ 21rh
dV dt
dV dt
(d)
dV dt
dr dt
dr dt
(c)
(c)
œ 32
œ 23 1rh
dr dt
dI dt
2RI
dI dt
Ê
dR dt
œ 2RI I#
dI dt
œ
2 ˆ PI ‰ dI I# dt
œ 2P I$
dI dt
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.8 Related Rates 17. (a) s œ Èx# y# œ ax# y# b
"Î#
Ê
ds dt
œ
x dx Èx# y# dt
(b) s œ Èx# y# œ ax# y# b
"Î#
Ê
ds dt
œ
x dx Èx# y# dt
(c) s œ Èx# y# Ê s# œ x# y# Ê 2s
ds dt
œ 2x
ds dt
œ
x dx Èx# y# z# dt
y dy Èx# y# z# dt
(b) From part (a) with
dx dt
œ0 Ê
(c) From part (a) with
ds dt
œ 0 Ê 0 œ 2x
19. (a) A œ (c) A œ
" # " #
ab sin ) Ê ab sin ) Ê
20. Given A œ 1r# ,
(b) (c)
œ œ
" # " #
œ
ab cos ) ab cos )
d) dt d) dt
2y
dy dt
2z
œ
Ê
dy dt
dx dt
œ yx
dy dt
dz dt
z dz Èx# y# z# dt
2z
dy dt
Ê
dz dt
(b) A œ "# b sin )
z dz x dt
ab sin ) Ê
dA dt
dx dt " #
"# a sin )
da dt
œ 0.01 cm/sec, and r œ 50 cm. Since
dr dt
2y
dx dt
2y
dx dt
z dz Èx# y# z# dt
y dy Èx# y# z# dt dx dt
œ 2x
ds dt
Ê 2s † 0 œ 2x
dy dt
œ 21r
dA dt
y dy x dt
œ0 œ
" #
ab cos )
d) dt
"# b sin )
da dt
db dt
dr dt
, then
dA ¸ dt r=50
" ‰ œ 21(50) ˆ 100 œ 1 cm# /min.
dw dt œ 2 cm/sec, j œ 12 cm and w œ 5 cm. dA dj dA # A œ jw Ê dt œ j dw dt w dt Ê dt œ 12(2) 5(2) œ 14 cm /sec, increasing dj dw P œ 2j 2w Ê dP dt œ 2 dt 2 dt œ 2(2) 2(2) œ 0 cm/sec, constant "Î# " dj ‰ # # "Î# ˆ D œ Èw# j# œ aw# j# b Ê dD 2w dw Ê dD dt œ # aw j b dt 2j dt dt
21. Given (a)
dj dt
dA dt dA dt
ds dt
y dy Èx# y# dt
2y
dx dt
18. (a) s œ Èx# y# z# Ê s# œ x# y# z# Ê 2s Ê
143
œ 2 cm/sec,
(5)(2) (12)(2) È25 144
22. (a) V œ xyz Ê
œ
dj w dw dt j dt Èw# j#
œ 14 13 cm/sec, decreasing œ yz
dV dt
dx dt
xz
xy
dy dt
dz dt
Ê
dV ¸ dt (4ß3ß2)
œ (3)(2)(1) (4)(2)(2) (4)(3)(1) œ 2 m$ /sec
dx (b) S œ 2xy 2xz 2yz Ê dS dt œ (2y 2z) dt (2x 2z) ¸ Ê dS œ (10)(1) (12)(2) (14)(1) œ 0 m# /sec dt
dy dt
(2x 2y)
dz dt
(4ß3ß2)
"Î# (c) j œ Èx# y# z# œ ax# y# z# b Ê
Ê 23. Given:
dj ¸ dt (4ß3ß2)
dx dt
dj dt
œ
x dx Èx# y# z# dt
y dy Èx# y# z# dt
z dz Èx# y# z# dt
œ Š È429 ‹ (1) Š È329 ‹ (2) Š È229 ‹ (1) œ 0 m/sec
œ 5 ft/sec, the ladder is 13 ft long, and x œ 12, y œ 5 at the instant of time
(a) Since x# y# œ 169 Ê
dy dt
œ xy
dx dt
‰ œ ˆ 12 5 (5) œ 12 ft/sec, the ladder is sliding down the wall
(b) The area of the triangle formed by the ladder and walls is A œ is changing at (c) cos ) œ
x 13
" #
xy Ê
dA dt
œ ˆ "# ‰ Šx
dy dt
y
dx dt ‹ .
The area
# [12(12) 5(5)] œ 119 # œ 59.5 ft /sec.
Ê sin )
24. s# œ y# x# Ê 2s
" #
ds dt
œ 2x
d) dt
œ
" 13
dx dt
2y
†
dx dt
dy dt
Ê
Ê
ds dt
d) dt
" œ 13 sin ) †
dx dt
œ
" s
dy dt ‹
Šx
dx dt
y
œ ˆ 5" ‰ (5) œ 1 rad/sec Ê
ds dt
œ
" È169
[5(442) 12(481)] œ 614 knots
25. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the girl and kite Ê s# œ (300)# x# Ê
ds dt
œ
x dx s dt
œ
400(25) 500
œ 20 ft/sec.
" # 26. When the diameter is 3.8 in., the radius is 1.9 in. and dr dt œ 3000 in/min. Also V œ 61r Ê $ ˆ " ‰ Ê dV dt œ 121(1.9) 3000 œ 0.00761. The volume is changing at about 0.0239 in /min.
dV dt
œ 121r
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dr dt
144
Chapter 3 Differentiation
27. V œ
" 3 3r 4h # 3 1r h, h œ 8 (2r) œ 4 Ê r œ 3 Ê dh ¸ 90 ˆ 9 ‰ dt h = 4 œ 1614# (10) œ 2561 ¸ 0.1119 dr 4 dh 4 ˆ 90 ‰ 15 r œ 4h 3 Ê dt œ 3 dt œ 3 2561 œ 321
(a) (b)
28. (a) V œ
" 3
1r# h and r œ
Ê Vœ
15h #
" 3
" 3
Vœ
#
‰ hœ 1 ˆ 4h 3
1 3
y# (3R y) Ê
y œ 8 we have
" 1441 (6)
œ
dy dt
1 3
œ
dV dt
#
751h$ 4
œ
161h# dh 9 dt
2251h# dh 4 dt
œ
dV dt
Ê
dh ¸ dt h = 5
œ
4(50) 2251(5)#
8 2251
œ
¸ 0.0849 m/sec œ 8.49 cm/sec Ê
dy dt
dy dt
" dV dt
œ 13 a6Ry 3y# b‘
Ê at R œ 13 and
m/min
(b) The hemisphere is on the circle r (13 y)# œ 169 Ê r œ È26y y# m (c) r œ a26y y# b 5 2881
œ 30. If V œ
4 3
"Î#
Ê
" #
œ
dr dt
#
Ê
4 151
c2y(3R y) y# (1)d œ
dV dt
¸ 0.1492 m/sec œ 14.92 cm/sec
‰ hœ 1 ˆ 15h #
" 241
Ê
m/sec œ 11.19 cm/sec
¸ 0.0113 m/min œ 1.13 cm/min dr 15 dh dr ¸ 8 ‰ ˆ 15 ‰ ˆ 225 (b) r œ 15h # Ê dt œ # dt Ê dt h = 5 œ # 1 œ 29. (a) V œ
161h$ 27
a26y y# b
"Î#
(26 2y)
dy dt
Ê
dr dt
œ
13 y dy È26y y# dt
Ê
dr ¸ dt y = 8
œ
13 8 È26†8 64
ˆ #" ‰ 41
m/min
1r$ , S œ 41r# , and
dV dt
œ kS œ 4k1r# , then
dV dt
œ 41r#
Ê 4k1r# œ 41r#
dr dt
dr dt
Ê
dr dt
œ k, a constant.
Therefore, the radius is increasing at a constant rate. 4 dV dV dr $ $ # dr 3 1r , r œ 5, and dt œ 1001 ft /min, then dt œ 41r dt Ê dt dr # dt œ 81(5)(1) œ 401 ft /min, the rate at which the surface area
31. If V œ œ 81r
œ 1 ft/min. Then S œ 41r# Ê
dS dt
is increasing.
32. Let s represent the length of the rope and x the horizontal distance of the boat from the dock. (a) We have s# œ x# 36 Ê dx ¸ dt s = 10
œ
(b) cos ) œ d) dt
Ê
6 r
œ
10 È10# 36
œ
s ds x dt
œ
s ds Ès# 36 dt
. Therefore, the boat is approaching the dock at
(2) œ 2.5 ft/sec.
Ê sin ) 6 8 ‰ 10# ˆ 10
dx dt
d) dt
œ r6#
dr dt
Ê
d) dt
œ
6 dr r# sin ) dt
. Thus, r œ 10, x œ 8, and sin ) œ
8 10
3 † (2) œ 20 rad/sec
33. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal distance between the balloon and the bicycle. The relationship between the variables is s# œ h# x# " ˆ dh dx ‰ " Ê ds Ê ds dt œ s h dt x dt dt œ 85 [68(1) 51(17)] œ 11 ft/sec. 34. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is dh dh " dV 10 V œ 91h Ê dV dt œ 91 dt Ê the rate the coffee is rising is dt œ 91 dt œ 91 in/min. (b) Let h be the height of the coffee in the pot. From the figure, the radius of the filter r œ œ
$
1h 1#
, the volume of the filter. The rate the coffee is falling is
35. y œ QD" Ê
dy dt
œ D"
dQ dt
QD#
dD dt
œ
" 41
(0)
233 (41)#
(2) œ
dh dt
466 1681
œ
4 dV 1h# dt
œ
4 #5 1
h #
Ê Vœ
" 3
1r# h
(10) œ 581 in/min.
L/min Ê increasing about 0.2772 L/min
36. Let P(xß y) represent a point on the curve y œ x# and ) the angle of inclination of a line containing P and the origin. Consequently, tan ) œ #
and cos )kx=3 œ
#
x y # x #
œ
#
3 9 # 3 #
y x
œ
Ê tan ) œ " 10
, we have
x# d) # x œ x Ê sec ) dt d) ¸ dt x=3 œ 1 rad/sec.
œ
dx dt
Ê
d) dt
œ cos# )
dx dt
. Since
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dx dt
œ 10 m/sec
Section 3.8 Related Rates 37. The distance from the origin is s œ Èx# y# and we wish to find œ
(5)(1) (12)(5) È25 144
ds ¸ dt (5ß12)
" #
œ
ax# y# b
d) dt
œ
dx dt
Š2x
2y
dy dt ‹¹ (5ß12)
œ 5 m/sec
38. Let s œ distance of car from foot of perpendicular in the textbook diagram Ê tan ) œ Ê
"Î#
145
cos# ) ds 132 dt
;
ds dt
d) dt
œ 264 and ) œ 0 Ê
Ê sec# )
s 13#
d) dt
œ
" ds 13# dt
œ 2 rad/sec. A half second later the car has traveled 132 ft
right of the perpendicular Ê k)k œ 14 , cos# ) œ "# , and
ds dt
œ 264 (since s increases) Ê
d) dt
œ
ˆ "# ‰ 132
(264) œ 1 rad/sec.
39. Let s œ 16t# represent the distance the ball has fallen, h the distance between the ball and the ground, and I the distance between the shadow and the point directly beneath the ball. Accordingly, s h œ 50 and since the triangle LOQ and # triangle PRQ are similar we have I œ 5030h h Ê h œ 50 16t and I œ Ê
30 a50 16t# b 50 a50 16t# b
dI ¸ dt t= 12
œ
30 Ê
1500 16t#
dI dt
œ 1500 8t$
œ 1500 ft/sec.
40. When x represents the length of the shadow, then tan ) œ given that
d) dt
31 #000
œ 0.27° œ
Ê sec# )
80 x
rad/min. At x œ 60, cos ) œ
d) dt
œ 80 x# #
x sec ¸ Ê ¸ dx dt œ ¹ 80
3 5
#
dx dt
Ê
) d) dt ¹¹ Š d) = dt
dx dt 31 2000
x# sec# ) d) 80 dt
œ
and sec ) = 35 ‹
œ
31 16
. We are ft/min
¸ 0.589 ft/min ¸ 7.1 in./min. 41. The volume of the ice is V œ
4 3
1r$ 43 14$ Ê
thickness of the ice is decreasing at
5 721
œ 41r#
dV dt
Ê
dr dt
dr ¸ dt r=6
œ
5 721
in./min when
in/min. The surface area is S œ 41r# Ê
# œ 10 3 in /min, the outer surface area of the ice is decreasing at
10 3
dS dt
œ 81r
dr dt
in# /min.
œ 10 in$ /min, the 5 ‰ ¸ œ 481 ˆ 72 Ê dS dt 1
dV dt
r=6
42. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between the car and r dr ds ¸ 5 plane Ê 9 s# œ r# Ê ds dt œ È # dt Ê dt r=5 œ È16 (160) œ 200 mph Ê speed of plane speed of car r 9
œ 200 mph Ê the speed of the car is 80 mph. 43. Let x represent distance of the player from second base and s the distance to third base. Then #
#
(a) s œ x 8100 Ê 2s œ 2x Ê ds 60 È Ê s œ 30 13 and dt œ 30È13 (16) œ ds dt
(b) sin )" œ Ê
(c)
d)" dt
†
90 s
Ê cos )"
œ
90 Š30È13‹ (60)
œ xs dx dt 32 È13 ¸
ds dt
d)" dt
90 s†x
d)" dt
90 œ s# cos )" †
ds dt
8.875 ft/sec
œ 90 s# †
ds dt
Ê
32 † ŠÈ ‹œ 13
8 65
rad/sec; cos )# œ
90 œ s# cos )" †
xÄ!
œ ˆ x# 908100 ‰
dx dt
d)# dt
90 s
œ
ds dt
œ s90†x †
Ê sin )#
90 Š30È13‹ (60)
s
Ê
lim d)# x Ä ! dt
" 6
rad/sec;
œ
"6
d)# dt
œ
90 s# sin )#
†
ds dt
ds dt
. Therefore, x œ 60 and s œ 30È13
d)# dt
œ 90 s# †
ds dt
Ê
d)# dt
œ
90 s# sin )#
†
ds dt
8 32 † ŠÈ rad/sec. ‹ œ 65 13
90 ˆ xs ‰ † ˆ dx ‰ ˆ 90 ‰ ˆ dx ‰ ˆ ‰ œ ˆs90 #† x ‰ † dt œ s# dt œ x# 8100
œ lim ˆ x# 908100 ‰ (15) œ
œ 16 ft/sec
. When the player is 30 ft from first base, x œ 60
. Therefore, x œ 60 and s œ 30È13 Ê
œ
ds dt
d)" dt
dx dt
dx dt
dx dt
Ê lim
xÄ!
d)" dt
ˆ xs ‰ ˆ dx ‰ ˆ 90 ‰ ˆ dx ‰ œ Š s90 #† x ‹ dt œ s# dt s
rad/sec
44. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the distance " da db db da ‘ between the ships. By the Law of Cosines, D# œ a# b# 2ab cos 120° Ê dD dt œ #D 2a dt 2b dt a dt b dt . When a œ 5,
da dt
œ 14, b œ 3, and
db dt
œ 21, then
dD dt
œ
413 2D
where D œ 7. The ships are moving
dD dt
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ 29.5 knots apart.
146
Chapter 3 Differentiation
3.9 LINEARIZATION AND DIFFERENTIALS 1. f(x) œ x$ 2x 3 Ê f w (x) œ 3x# 2 Ê L(x) œ f w (2)(x 2) f(2) œ 10(x 2) 7 Ê L(x) œ 10x 13 at x œ 2 2. f(x) œ Èx# 9 œ ax# 9b
"Î#
Ê f w (x) œ ˆ "# ‰ ax# 9b
œ 45 (x 4) 5 Ê L(x) œ 45 x 3. f(x) œ x
" x
9 5
"Î#
(2x) œ
x È x# 9
Ê L(x) œ f w (4)(x 4) f(4)
at x œ 4
Ê f w (x) œ 1 x# Ê L(x) œ f(1) f w (1)(x 1) œ # !(x 1) œ #
4. f(x) œ x"Î$ Ê f w (x) œ
" $x#Î$
" 1#
Ê L(x) œ f w (8)ax a8bb fa8b œ
(x 8) 2 Ê L(x) œ
" 1#
x
4 3
5. f(x) œ tan x Ê f w axb œ sec2 x Ê Laxb œ fa1b f w a1bax 1b œ 0 1ax 1b œ x 1 6. (a) f(x) œ sin x Ê f w axb œ cos x Ê Laxb œ fa0b f w a0bax 0b œ x Ê Laxb œ x (b) f(x) œ cos x Ê f w axb œ sin x Ê Laxb œ fa0b f w a0bax 0b œ 1 Ê Laxb œ 1 (c) f(x) œ tan x Ê f w axb œ sec2 x Ê Laxb œ fa0b f w a0bax 0b œ x Ê Laxb œ x 7. f(x) œ x# 2x Ê f w (x) œ 2x 2 Ê L(x) œ f w (0)(x 0) f(0) œ 2(x 0) 0 Ê L(x) œ 2x at x œ 0 8. f(x) œ x" Ê f w (x) œ x# Ê L(x) œ f w (1)(x 1) f(1) œ (1)(x 1) 1 Ê L(x) œ x 2 at x œ 1 9. f(x) œ 2x# 4x 3 Ê f w (x) œ 4x 4 Ê L(x) œ f w (1)(x 1) f(1) œ 0(x 1) (5) Ê L(x) œ 5 at x œ 1 10. f(x) œ 1 x Ê f w (x) œ 1 Ê L(x) œ f w (8)(x 8) f(8) œ 1(x 8) 9 Ê L(x) œ x 1 at x œ 8 3 11. f(x) œ È x œ x"Î$ Ê f w (x) œ ˆ "3 ‰ x#Î$ Ê L(x) œ f w (8)(x 8) f(8) œ
12. f(x) œ
x x1
Ê L(x) œ
Ê f w (x) œ " 4
x
" 4
(1)(x 1) (")(x) (x 1)#
œ
" (x 1)#
" 1#
(x 8) 2 Ê L(x) œ
Ê L(x) œ f w (1)(x 1) f(1) œ
" 4
(x 1)
" 1#
x
4 3
at x œ 8
" #
at x œ 1
13. f w axb œ ka" xbk" . We have fa!b œ " and f w a!b œ k. Laxb œ fa!b f w a!bax !b œ " kax !b œ " kx '
14. (a) faxb œ a" xb' œ " axb‘ ¸ " 'axb œ " 'x (b) faxb œ
# " x
"
œ #" axb‘
(c) faxb œ a" xb
"Î#
¸ " ˆ "# ‰x œ "
(d) faxb œ È2 x# œ È#Š"
x# #‹
(e) faxb œ a% $xb"Î$ œ %"Î$ ˆ" (f) faxb œ ˆ"
¸ #" a"baxb‘ œ # #x
" ‰2Î$ #x
"Î#
x #
¸ È#Š"
$x ‰"Î$ %
" x# # #‹
¸ %"Î$ ˆ" 2Î$
œ ’" ˆ # " x ‰“
œ È#Š"
" $x ‰ $ %
x# %‹
œ %"Î$ ˆ" x% ‰
¸ " $# ˆ # " x ‰ œ "
# ' $x
15. (a) (1.0002)&! œ (1 0.0002)&! ¸ 1 50(0.0002) œ 1 .01 œ 1.01 3 (b) È 1.009 œ (1 0.009)"Î$ ¸ 1 ˆ " ‰ (0.009) œ 1 0.003 œ 1.003 3
16. f(x) œ Èx 1 sin x œ (x 1)"Î# sin x Ê f w (x) œ ˆ "# ‰ (x 1)"Î# cos x Ê Lf (x) œ f w (0)(x 0) f(0) œ 3 (x 0) 1 Ê Lf (x) œ 3 x 1, the linearization of f(x); g(x) œ Èx 1 œ (x 1)"Î# Ê gw (x) #
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.9 Linearization and Differentials œ ˆ "# ‰ (x 1)"Î# Ê Lg (x) œ gw (0)(x 0) g(0) œ w
w
" #
(x 0) 1 Ê Lg (x) œ
" #
x 1, the linearization of g(x);
h(x) œ sin x Ê h (x) œ cos x Ê Lh (x) œ h (0)(x 0) h(0) œ (1)(x 0) 0 Ê Lh (x) œ x, the linearization of h(x). Lf (x) œ Lg (x) Lh (x) implies that the linearization of a sum is equal to the sum of the linearizations. 17. y œ x$ 3Èx œ x$ 3x"Î# Ê dy œ ˆ3x# #3 x"Î# ‰ dx Ê dy œ Š3x#
3 ‹ 2È x
dx
"Î# "Î# "Î# 18. y œ xÈ1 x# œ x a1 x# b Ê dy œ ’(1) a1 x# b (x) ˆ "# ‰ a1 x# b (2x)“ dx
œ a1 x# b
"Î#
#
a1 2x# b È 1 x#
Ê dy œ Š (2) a1 a1xb x# b(2x)(2x) ‹ dx œ #
19. y œ
2x 1 x #
20. y œ
2È x 3 ˆ1 È x ‰
Ê dy œ
ca1 x# b x# d dx œ
œ
2x"Î# 3 a1 x"Î# b
" # 3 È x ˆ1 È x ‰
Ê dy œ Š
dx 2 2x# a1 x # b #
dx
x "Î# ˆ3 ˆ1 x"Î# ‰‰ 2x"Î# ˆ #3 x "Î# ‰ 9 a1 x"Î# b
#
‹ dx œ
3x "Î# 3 3 # 9 a1 x"Î# b
dx
dx
21. 2y$Î# xy x œ 0 Ê 3y"Î# dy y dx x dy dx œ 0 Ê ˆ3y"Î# x‰ dy œ (1 y) dx Ê dy œ
1y 3 È y x
22. xy# 4x$Î# y œ 0 Ê y# dx 2xy dy 6x"Î# dx dy œ 0 Ê (2xy 1) dy œ ˆ6x"Î# y# ‰ dx Ê dy œ
6È x y# 2xy 1
dx
23. y œ sin ˆ5Èx‰ œ sin ˆ5x"Î# ‰ Ê dy œ ˆcos ˆ5x"Î# ‰‰ ˆ 5# x"Î# ‰ dx Ê dy œ
5 cos ˆ5Èx‰ 2È x
dx
24. y œ cos ax# b Ê dy œ csin ax# bd (2x) dx œ 2x sin ax# b dx $
$
$
25. y œ 4 tan Š x3 ‹ Ê dy œ 4 Šsec# Š x3 ‹‹ ax# b dx Ê dy œ 4x# sec# Š x3 ‹ dx 26. y œ sec ax# 1b Ê dy œ csec ax# 1b tan ax# 1bd (2x) dx œ 2x csec ax# 1b tan ax# 1bd dx 27. y œ 3 csc ˆ1 2Èx‰ œ 3 csc ˆ1 2x"Î# ‰ Ê dy œ 3 ˆcsc ˆ1 2x"Î# ‰‰ cot ˆ1 2x"Î# ‰ ˆx"Î# ‰ dx Ê dy œ È3 csc ˆ1 2Èx‰ cot ˆ1 2Èx‰ dx x
28. y œ 2 cot Š È"x ‹ œ 2 cot ˆx"Î# ‰ Ê dy œ 2 csc# ˆx"Î# ‰ ˆ #" ‰ ˆx$Î# ‰ dx Ê dy œ
" È x$
csc# Š È"x ‹ dx
29. f(x) œ x# 2x, x! œ 1, dx œ 0.1 Ê f w (x) œ 2x 2 (a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ 3.41 3 œ 0.41 (b) df œ f w (x! ) dx œ [2(1) 2](0.1) œ 0.4 (c) k?f df k œ k0.41 0.4k œ 0.01 30. f(x) œ 2x# 4x 3, x! œ 1, dx œ 0.1 Ê f w (x) œ 4x 4 (a) ?f œ f(x! dx) f(x! ) œ f(.9) f(1) œ .02 (b) df œ f w (x! ) dx œ [4(1) 4](.1) œ 0 (c) k?f df k œ k.02 0k œ .02
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dx
147
148
Chapter 3 Differentiation
31. f(x) œ x$ x, x! œ 1, dx œ 0.1 Ê f w (x) œ 3x# 1 (a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ .231 (b) df œ f w (x! ) dx œ [3(1)# 1](.1) œ .2 (c) k?f df k œ k.231 .2k œ .031 32. f(x) œ x% , x! œ 1, dx œ 0.1 Ê f w (x) œ 4x$ (a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ .4641 (b) df œ f w (x! ) dx œ 4(1)$ (.1) œ .4 (c) k?f df k œ k.4641 .4k œ .0641 33. f(x) œ x" , x! œ 0.5, dx œ 0.1 Ê f w (x) œ x# (a) ?f œ f(x! dx) f(x! ) œ f(.6) f(.5) œ "3 " ‰ (b) df œ f w (x! ) dx œ (4) ˆ 10 œ 52 (c) k?f df k œ ¸ "3 25 ¸ œ
" 15
34. f(x) œ x$ 2x 3, x! œ 2, dx œ 0.1 Ê f w (x) œ 3x# 2 (a) ?f œ f(x! dx) f(x! ) œ f(2.1) f(2) œ 1.061 (b) df œ f w (x! ) dx œ (10)(0.10) œ 1 (c) k?f df k œ k1.061 1k œ .061 35. V œ
4 3
1r$ Ê dV œ 41r!# dr
36. V œ x$ Ê dV œ 3x!# dx
37. S œ 6x# Ê dS œ 12x! dx 38. S œ 1rÈr# h# œ 1r ar# h# b Ê
dS dr
œ
1 a r # h # b 1 r# È r# h #
"Î#
Ê dS œ
, h constant Ê
1 a2r#! h# b Ér#! h#
dS dr
œ 1 ar# h# b
"Î#
1r † r ar# h# b
"Î#
dr, h constant
39. V œ 1r# h, height constant Ê dV œ 21r! h dr
40. S œ 21rh Ê dS œ 21r dh
41. Given r œ 2 m, dr œ .02 m (a) A œ 1r# Ê dA œ 21r dr œ 21(2)(.02) œ .081 m# 1‰ (b) ˆ .08 41 (100%) œ 2% 42. C œ 21r and dC œ 2 in. Ê dC œ 21 dr Ê dr œ œ 21(5) ˆ 1" ‰ œ 10 in.#
" 1
Ê the diameter grew about
43. The volume of a cylinder is V œ 1r# h. When h is held fixed, we have
dV dr
2 1
in.; A œ 1r# Ê dA œ 21r dr
œ #1rh, and so dV œ #1rh dr. For h œ $! in.,
r œ ' in., and dr œ !Þ& in., the volume of the material in the shell is approximately dV œ #1rh dr œ #1a'ba$!ba!Þ&b œ ")!1 ¸ &'&Þ& in$ . 44. Let ) œ angle of elevation and h œ height of building. Then h œ $!tan ), so dh œ $!sec# ) d). We want ldhl !Þ!%h, &1 &1 sin ) which gives: l$!sec# ) d)l !Þ!%l$!tan )l Ê cos"# ) ld)l !Þ!% cos ) Ê ld)l !Þ!%sin ) cos ) Ê ld)l !Þ!%sin "# cos "# œ !Þ!" radian. The angle should be measured with an error of less than !Þ!" radian (or approximatley !Þ&( degrees), which is a percentage error of approximately !Þ('%.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.9 Linearization and Differentials 45. The percentage error in the radius is (a) Since C œ 21 r Ê ˆ21 dr ‰ dt 21 r
œ
ˆ21 r dr ‰ dt 1 r2
œ
‚ 100 Ÿ 2%.
dC dr dt œ 21 dt . The percentage ˆ dr ‰ dt r ‚ 100 Ÿ 2%.
‚ 100 œ
(b) Since A œ 1 r2 Ê
ˆ dr ‰ dt r
‚ 100 œ 2
ˆ dr ‰ dt r
(a) Since S œ 6x Ê
dS dt
œ
ˆ dx ‰ 2 dtx
œ
œ 12x
dx dt .
ˆ dx ‰ dt x
‚ 100 Ÿ 0.5%.
The percentage error in the cube's surface area is
œ 3x2 dx dt . The percentage error in the cube's volume is
dV dt
Ê k31h# dhk Ÿ of h is
ˆ dS ‰ dt S
ˆ12x dx ‰ dt 6x2
‚ 100 œ
‚ 100
ˆ dV ‰ dt V
‚ 100 œ
ˆ3x2 dx ‰ dt x3
‚ 100
‚ 100 Ÿ 3a0.5%b œ 1.5%
47. V œ 1h$ Ê dV œ 31h# dh; recall that ?V ¸ dV. Then k?Vk Ÿ (1%)(V) œ " 3
‚ 100
‚ 100 Ÿ 2a0.5%b œ 1%
(b) Since V œ x3 Ê ˆ dx ‰ 3 dtx
ˆ dA ‰ dt A
‚ 100
‚ 100 Ÿ 2a2%b œ 4%.
46. The percentage error in the edge of the cube is 2
ˆ dC ‰ dt C
error in calculating the circle's circumference is
œ 21 r dr dt . The percentage error in calculating the circle's area is given by
dA dt
149
(1) a1h$ b 100
Ê kdhk Ÿ
" 300
(1) a1h$ b 100
Ê kdVk Ÿ
(1) a1h$ b 100
h œ ˆ 3" %‰ h. Therefore the greatest tolerated error in the measurement
%. #
48. (a) Let Di represent the interior diameter. Then V œ 1r# h œ 1 ˆ D#i ‰ h œ
1D#i h 4
#
" ‰ 5 1 Di dV œ 51Di dDi . Recall that ?V ¸ dV. We want k?Vk Ÿ (1%)(V) Ê kdVk Ÿ ˆ 100 Š # ‹
Ê 51Di dDi Ÿ
1D#i 40
Ê
dDi Di
51D#i # 1Di# œ 40
and h œ 10 Ê V œ
Ê
Ÿ 200. The inside diameter must be measured to within 0.5%.
(b) Let De represent the exterior diameter, h the height and S the area of the painted surface. S œ 1De h Ê dS œ 1hdDe dDe Ê dS S œ De . Thus for small changes in exterior diameter, the approximate percentage change in the exterior diameter is equal to the approximate percentage change in the area painted, and to estimate the amount of paint required to within 5%, the tanks's exterior diameter must be measured to within 5%. 49. Given D œ 100 cm, dD œ 1 cm, V œ œ” 50. V œ œ
10% 1 # 10' 1 6
4 3
1 D$ 200
# • a10 %b œ ”
1 r$ œ
4 3
$
10' 1 # 10' 1 6
1 ˆ D# ‰ œ
Ê kdVk Ÿ
1 D$ 200
4 3
$
1 ˆ D# ‰ œ
1 D$ 6
Ê dV œ
1 #
D# dD œ
1 #
(100)# (1) œ
10% 1 #
. Then
dV V
(100%)
• % œ 3% 1 D$ 6
Ê dV œ #
Ê ¹ 1D# dD¹ Ÿ
1 D# #
3 ‰ 1D dD; recall that ?V ¸ dV. Then k?Vk Ÿ (3%)V œ ˆ 100 Š 6 ‹
$
1 D$ #00
Ê kdDk Ÿ
D 100
œ (1%) D Ê the allowable percentage error in
measuring the diameter is 1%. 51. W œ a
b g
œ a bg" Ê dW œ bg# dg œ bgdg Ê #
dWmoon dWearth
œ
b dg ‹ (5.2)# b dg Š # ‹ (32)
Š
#
32 ‰ œ ˆ 5.2 œ 37.87, so a change of
gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth. 52. (a) T œ 21 Š Lg ‹
"Î#
Ê dT œ 21ÈL ˆ "# g$Î# ‰ dg œ 1ÈL g$Î# dg
(b) If g increases, then dg 0 Ê dT 0. The period T decreases and the clock ticks more frequently. Both the pendulum speed and clock speed increase. (c) 0.001 œ 1È100 ˆ980$Î# ‰ dg Ê dg ¸ 0.977 cm/sec# Ê the new g ¸ 979 cm/sec#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
150
Chapter 3 Differentiation
53. E(x) œ f(x) g(x) Ê E(x) œ f(x) m(x a) c. Then E(a) œ 0 Ê f(a) m(a a) c œ 0 Ê c œ f(a). Next f(x) m(x a) c f(a) œ 0 Ê xlim œ 0 Ê xlim ’ f(x)x xa a m“ œ 0 (since c œ f(a)) Äa Äa Ê f w (a) m œ 0 Ê m œ f w (a). Therefore, g(x) œ m(x a) c œ f w (a)(x a) f(a) is the linear approximation, as claimed.
we calculate m: xlim Äa
E(x) xa
54. (a) i. Qaab œ faab implies that b! œ faab. ii. Since Qw axb œ b" #b# ax ab, Qw aab œ f w aab implies that b" œ f w aab. iii. Since Qww axb œ #b# , Qww aab œ f ww aab implies that b2 œ In summary, b! œ faab, b" œ f w aab, and b2 œ
ww
f aa b # .
ww
f aa b # .
(b) faxb œ a" xb" ; f w axb œ "a" xb# a"b œ a" xb# ; f ww axb œ #a" xb$ a"b œ #a" xb$ Since fa!b œ ", f w a!b œ ", and f ww a!b œ #, the coefficients are b! œ ", b" œ ", b# œ ## œ ". The quadratic approximation is Qaxb œ " x x# .
(c)
As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical.
(d) gaxb œ x" ; gw axb œ "x# ; gww axb œ #x$ Since ga"b œ ", gw a"b œ ", and gww a"b œ # , the coefficients are b! œ ", b" œ ", b# œ #
# #
œ ". The quadratic
approximation is Qaxb œ " ax "b ax "b . As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical.
(e) haxb œ a" xb"Î# ; hw axb œ "# a" xb"Î# ; hww axb œ "% a" xb$Î# Since ha!b œ ", hw a!b œ "# , and hww a!b œ "% , the coefficients are b! œ ", b" œ "# , b# œ approximation is Qaxb œ "
x #
#
x 8
"% 2
œ "8 . The quadratic
. As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical.
(f) The linearization of any differentiable function uaxb at x œ a is Laxb œ uaab uw aabax ab œ b! b" ax ab, where b! and b" are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization for faxb at x œ ! is " x; the linearization for gaxb at x œ " is " ax "b or # x; and the linearization for haxb at x œ ! is " x# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Practice Exercises
151
55-58. Example CAS commands: Maple: with(plots): a:= 1: f:=x -> x • 3 x • 2 2*x; plot(f(x), x=1..2); diff(f(x),x); fp := unapply (ww ,x); L:=x -> f(a) fp(a)*(x a); plot({f(x), L(x)}, x=1..2); err:=x -> abs(f(x) L(x)); plot(err(x), x=1..2, title = #absolute error function#); err(1); Mathematica: (function, x1, x2, and a may vary): Clear[f, x] {x1, x2} = {1, 2}; a = 1; f[x_]:=x3 x2 2x Plot[f[x], {x, x1, x2}] lin[x_]=f[a] f'[a](x a) Plot[{f[x], lin[x]}, {x, x1, x2}] err[x_]=Abs[f[x] lin[x]] Plot[err[x], {x, x1,x 2}] err//N After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del) eps = 0.5; del = 0.4 Plot[{err[x], eps},{x, a del, a del}] CHAPTER 3 PRACTICE EXERCISES 1. y œ x& 0.125x# 0.25x Ê 2. y œ 3 0.7x$ 0.3x( Ê 3. y œ x$ 3 ax# 1# b Ê 4. y œ x( È7x
" 1 1
Ê
dy dx
œ 3x# 3(2x 0) œ 3x# 6x œ 3x(x 2)
dy dx
œ 7x' È7
œ 2(x 1) a2x# 4x 1b 6. y œ (2x 5)(4 x)" Ê œ 3(4 x)# $
8. y œ Š1
csc ) #
)# 4‹
#
œ 5x% 0.25x 0.25
œ 2.1x# 2.1x'
dy dx
5. y œ (x 1)# ax# 2xb Ê
7. y œ a)# sec ) 1b Ê
dy dx
dy dx
œ (x 1)# (2x 2) ax# 2xb (2(x 1)) œ 2(x 1) c(x 1)# x(x 2)d
dy dx
œ (2x 5)(1)(4 x)# (1) (4 x)" (2) œ (4 x)# c(2x 5) 2(4 x)d
dy d)
Ê
#
œ 3 a)# sec ) 1b (2) sec ) tan )) dy d)
œ 2 Š1
csc ) #
)# ˆ csc ) cot ) 4‹ #
#) ‰ œ Š1
csc ) #
)# 4 ‹ (csc
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
) cot ) ))
152
Chapter 3 Differentiation
9. s œ
Èt 1 Èt
Ê
ds dt
œ
10. s œ
" Èt 1
Ê
ds dt
œ
ˆ1 Èt‰†
" sin# x
"
ˆ1 Èt‰
ˆÈ t 1 ‰
#
"
Èt ‹
#
œ
13. s œ cos% (1 2t) Ê 14. s œ cot$ ˆ 2t ‰ Ê
ds dt
ˆ1 Èt‰ Èt 2Èt ˆ1 Èt‰
#
" # #Èt ˆ1 Èt‰
" # 2 È t ˆÈ t 1 ‰
dy dx
œ (2 csc x)(csc x cot x) 2( csc x cot x) œ (2 csc x cot x)(1 csc x)
œ 4 cos$ (1 2t)(sin (1 2t))(2) œ 8 cos$ (1 2t) sin (1 2t)
ds dt
œ 3 cot# ˆ 2t ‰ ˆcsc# ˆ 2t ‰‰ ˆ t#2 ‰ œ
15. s œ (sec t tan t)& Ê
œ
œ (4 tan x) asec# xb (2 sec x)(sec x tan x) œ 2 sec# x tan x
dy dx
œ csc# x 2 csc x Ê
2 sin x
œ
#
ˆÈt 1‰ (0) 1 Š
11. y œ 2 tan# x sec# x Ê 12. y œ
"
Èt Èt Š #Èt ‹
#
ds dt
16. s œ csc& a1 t 3t# b Ê
6 t#
cot# ˆ 2t ‰ csc# ˆ 2t ‰
œ 5(sec t tan t)% asec t tan t sec# tb œ 5(sec t)(sec t tan t)& ds dt
œ 5 csc% a1 t 3t# b acsc a1 t 3t# b cot a1 t 3t# bb (1 6t)
œ 5(6t 1) csc& a1 t 3t# b cot a1 t 3t# b " #
) cos ) sin ) È2) sin )
17. r œ È2) sin ) œ (2) sin ))"Î# Ê
dr d)
œ
(2) sin ))"Î# (#) cos ) 2 sin )) œ
18. r œ 2)Ècos ) œ 2) (cos ))"Î# Ê
dr d)
œ 2) ˆ "# ‰ (cos ))"Î# (sin )) 2(cos ))"Î# œ
) sin ) Ècos )
2Ècos )
2 cos ) ) sin ) Ècos )
œ
19. r œ sin È2) œ sin (2))"Î# Ê 20. r œ sin Š) È) 1‹ Ê
œ cos (2))"Î# ˆ "# (2))"Î# (2)‰ œ
œ cos Š) È) 1‹ Š1
cos È2) È 2)
" ‹ 2È ) 1
œ
2È)"1 #È ) "
œ
" #
22. y œ 2Èx sin Èx Ê
dy dx
" 2 œ 2Èx ˆcos Èx‰ Š 2È ‹ ˆsin Èx‰ Š 2È ‹ œ cos Èx x x
x# csc
2 x
Ê
x# ˆcsc
2 x
cot x2 ‰ ˆ x#2 ‰ ˆcsc x2 ‰ ˆ "# † 2x‰ œ csc
cos Š) È) 1‹
dy dx
21. y œ
" #
dr d)
dr d)
2 x
cot
2 x
x csc
2 x
sin Èx Èx
dy "Î# sec (2x)# tan (2x)# (2(2x) † 2) sec (2x)# ˆ "# x$Î# ‰ dx œ x 8x"Î# sec (2x)# tan (2x)# "# x$Î# sec (2x)# œ "# x"Î# sec (2x)# c16 tan (2x)# x# d or #x"$Î# seca#xb2 16x# tana2xb#
23. y œ x"Î# sec (2x)# Ê œ
24. y œ Èx csc (x 1)$ œ x"Î# csc (x 1)$ Ê
dy dx
œ x"Î# acsc (x 1)$ cot (x 1)$ b a3(x 1)# b csc (x 1)$ ˆ "# x"Î# ‰
œ 3Èx (x 1)# csc (x 1)$ cot (x 1)$ or
" csc(x #È x
csc (x 1)$ 2È x
œ
" #
Èx csc (x 1)$ x" 6(x 1)# cot (x 1)$ ‘
1)$ c1 6x(x 1)# cot (x 1)$ d
25. y œ 5 cot x# Ê 26. y œ x# cot 5x Ê
dy dx
œ 5 acsc# x# b (2x) œ 10x csc# ax# b
dy dx
œ x# acsc# 5xb (5) (cot 5x)(2x) œ 5x# csc# 5x 2x cot 5x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"‘
Chapter 3 Practice Exercises
153
27. y œ x# sin# a2x# b Ê
dy dx
œ x# a2 sin a2x# bb acos a2x# bb (4x) sin# a2x# b (2x) œ 8x$ sin a2x# b cos a2x# b 2x sin# a2x# b
28. y œ x# sin# ax$ b Ê
dy dx
œ x# a2 sin ax$ bb acos ax$ bb a3x# b sin# ax$ b a2x$ b œ 6 sin ax$ b cos ax$ b 2x$ sin# ax$ b
29. s œ ˆ t 4t 1 ‰ 30. s œ
#
Ê
" 15(15t 1)$
œ 2 ˆ t 4t 1 ‰
$
(4t)(1) Š (t 1)(4) ‹ œ 2 ˆ t 4t 1 ‰ (t 1)#
" œ 15 (15t 1)$ Ê
#
Èx
ds dt
31. y œ Š x 1 ‹ Ê
dy dx
#
2È x
32. y œ Š 2Èx 1 ‹ Ê
dy dx
(x 1)#
2È x
œ 2 Š 2È x 1 ‹
# "Î# 33. y œ É x x# x œ ˆ1 "x ‰ Ê
dy dx
œ
34. y œ 4xÉx Èx œ 4x ˆx x"Î# ‰ œ ˆx Èx‰
"Î#
#
’2x Š1
" ‹ #È x
"Î#
(x 1) 2x (x 1)$
œ
ˆ2Èx 1‰ Š È" ‹ ˆ2Èx‰ Š È" ‹ x x ˆ2 È x 1 ‰
#
ˆ1 "x ‰"Î# ˆ x"# ‰ œ
" #
4 (t 1)#
" œ 15 (3)(15t 1)% (15) œ
ds dt
" (x 1) Š #È ‹ ˆÈx‰ (1) x
Èx
œ 2 Šx1‹ †
$
Ê
dy dx
œ "
"Î#
"Î#
3 (15t 1)%
1x (x 1)$
#x # É 1
œ 4x ˆ "# ‰ ˆx x"Î# ‰
4 ˆx Èx‰“ œ ˆx Èx‰
œ
œ (t 8t$1)
4Èx Š È"x ‹
ˆ2 È x 1 ‰ $
œ
" x
ˆ1 "# x"Î# ‰ ˆx x"Î# ‰"Î# (4)
ˆ2x Èx 4x 4Èx‰ œ
35. r œ ˆ cossin) ) 1 ‰ Ê
dr d)
)) (sin ))(sin )) œ 2 ˆ cossin) ) 1 ‰ ’ (cos ) 1)(cos “ œ 2 ˆ cossin) ) " ‰ Š cos (cos ) 1)#
(2 sin )) (1 cos )) (cos ) 1)$
œ
2 sin ) (cos ) ")#
dr d)
1 ‰ (1 cos ))(cos )) (sin ) ")(sin )) œ 2 ˆ 1sin )cos “œ ) ’ (1 cos ))#
œ
#
1 ‰ 36. r œ ˆ 1sin )cos Ê )
œ
4 ˆ2 È x 1‰$
2(sin ) ") (1 cos ))$
#
6x 5Èx É x Èx
) cos ) sin# ) ‹ (cos ) ")#
acos ) cos# ) sin# ) sin )b
2(sin ) 1)(cos ) sin ) 1) (1 c os ))$
37. y œ (2x 1) È2x 1 œ (2x 1)$Î# Ê
dy dx
œ
3 #
(2x 1)"Î# (2) œ 3È2x 1
38. y œ 20(3x 4)"Î% (3x 4)"Î& œ 20(3x 4)"Î#! Ê 39. y œ 3 a5x# sin 2xb 40. y œ a3 cos$ 3xb
$Î#
"Î$
Ê
Ê dy dx
dy dx
dy dx
" ‰ œ 20 ˆ 20 (3x 4)"*Î#! (3) œ
œ 3 ˆ 3# ‰ a5x# sin 2xb
œ "3 a3 cos$ 3xb
%Î$
&Î#
[10x (cos 2x)(2)] œ
a3 cos# 3xb (sin 3x)(3) œ
3 (3x 4)"*Î#!
9(5x cos 2x) a5x# sin 2xb&Î#
3 cos# 3x sin 3x a3 cos$ 3xb%Î$
2 41. xy 2x 3y œ 1 Ê axyw yb 2 3yw œ 0 Ê xyw 3yw œ 2 y Ê yw (x 3) œ 2 y Ê yw œ yx 3
42. x# xy y# 5x œ 2 Ê 2x Šx Ê
dy dx
œ
dy dx
dy dx
dy dx
5œ!Êx
dy dx
2y
dy dx
œ 5 2x y Ê
dy dx
(x 2y) œ 5 2x y
5 2x y x 2y
43. x$ 4xy 3y%Î$ œ 2x Ê 3x# Š4x Ê
y‹ 2y
dy dx
ˆ4x 4y"Î$ ‰ œ 2 3x# 4y Ê
4y‹ 4y"Î$ dy dx
œ
dy dx
œ 2 Ê 4x
dy dx
4y"Î$
dy dx
œ 2 3x# 4y
2 3x# 4y 4x 4y"Î$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
154
Chapter 3 Differentiation
44. 5x%Î& 10y'Î& œ 15 Ê 4x"Î& 12y"Î& " #
45. (xy)"Î# œ 1 Ê
(xy)"Î# Šx
46. x# y# œ 1 Ê x# Š2y 47. y# œ
Ê 2y
x x 1
x‰ 48. y# œ ˆ 11 x
"Î#
dy dx
dy dx ‹
y# (2x) œ 0 Ê 2x# y
Ê y% œ
"x 1x
49. p$ 4pq 3q# œ 2 Ê 3p# Ê
dp dq
œ
dp dq
Ê
Ê 4y$ dp dq
dy dx
œ x"Î# y"Î# Ê
dy dx
œ 2xy# Ê
dy dx
œ yx
Ê
dy dx
œ
6q œ 0 Ê 3p#
dp dq
dy dx
œ
œ
(1 x)(1) (1 x)Ð") (" x)#
Ê
$Î#
dp dq ‹
Ê 1 œ 3# a5p# 2pb
dy dx
œ yx
4q
dp dq
œ 6q 4p Ê
dp dq
a3p# 4qb œ 6q 4p
&Î#
Š10p
dp dq
2
dp dq ‹
Ê 23 a5p# 2pb
&Î#
œ
dp dq
(10p 2)
#
œ a5p3(5p 2p1)b
2r sin 2s sin 2s cos 2s
œ
œ
(2r 1)(sin 2s) cos 2s
53. (a) x$ y$ œ 1 Ê 3x# 3y# Ê
d# y dx#
œ
(b) y# œ 1 Ê
d# y dx#
2 x
œ
2xy# a2yx# b Š y%
Ê 2y
dy dx
2xy x# Š y# x%
œ
" ‹ yx#
œ
x y
Ê
d# y dx#
œ
dy dx
x# ‹ y#
œ
y(1) x y#
(cos 2s) œ 2r sin 2s 2 sin s cos s
dr ds
2xy# y% dy dx
1 2s œ 0 Ê
dr ds
d# y dx#
œ
#
œ yx# Ê
dy dx 2x% y
œ
" yx#
œ
(2s 1) œ 1 2s 2r Ê y# (2x) ax# b Š2y
dr ds
œ
" 2s 2r 2s 1
dy dx ‹
y%
2xy$ 2x% y&
Ê
dy dx
œ ayx# b
"
Ê
dy dx
œ
d# y dx#
œ ayx# b
#
’y(2x) x#
dy dx “
2xy# 1 y$ x%
dy dx dy dx
œ0 Ê
œ
dr ds
œ (2r 1)(tan 2s)
dr ‰ ds
Ê
2 x#
54. (a) x# y# œ 1 Ê 2x 2y dy dx
œ x" y Ê
&Î#
52. 2rs r s s# œ 3 Ê 2 ˆr s
(b)
dy dx
" 2y$ (1 x)#
dr ‰ 51. r cos 2s sin# s œ 1 Ê r(sin 2s)(2) (cos 2s) ˆ ds 2 sin s cos s œ 0 Ê dr ds
" œ "3 x"Î& y"Î& œ 3(xy) "Î&
dy dx
" #y(x 1)#
dy dx
4 Šp q
œ 4x"Î& Ê
dy dx
6q 4p 3p# 4q
50. q œ a5p# 2pb Ê
y‹ œ 0 Ê x"Î# y"Î#
dy dx
(x 1)(1) (x)(1) (x 1)#
œ
œ 0 Ê 12y"Î&
dy dx
œ 0 Ê 2y œ
y x Š xy ‹ y#
œ
œ 2x Ê
dy dx
y# x# y$
œ
" y$
x y
(since y# x# œ 1)
55. (a) Let h(x) œ 6f(x) g(x) Ê hw (x) œ 6f w (x) gw (x) Ê hw (1) œ 6f w (1) gw (1) œ 6 ˆ "# ‰ a%b œ (
(b) Let h(x) œ f(x)g# (x) Ê hw (x) œ f(x) a#g(x)b gw (x) g# (x)f w (x) Ê hw (0) œ #f(0)g(0)gw (0) g# (0)f w (0) œ #(1)(1) ˆ "# ‰ (1)# ($) œ # (c) Let h(x) œ
f(x) g(x) 1
Ê hw (x) œ
(g(x) 1)f (x) f(x)g (x) (g(x) 1)# w w w
w
(& 1) ˆ "# ‰ 3 a%b (g(1) ")f (1) f(1)g (1) œ (g(1) 1)# (& 1)# " " w w w f (g(0))g (0) œ f (1) ˆ # ‰ œ ˆ # ‰ ˆ "# ‰ œ "% w w w w
Ê hw (1) œ
(d) Let h(x) œ f(g(x)) Ê hw (x) œ f (g(x))g (x) Ê hw (0) œ
w
w
œ
& "#
(e) Let h(x) œ g(f(x)) Ê hw (x) œ gw (f(x))f w (x) Ê hw (0) œ g (f(0))f (0) œ g (1)f (0) œ a%b ($) œ "# (f) Let h(x) œ (x f(x))$Î# Ê hw (x) œ 3# (x f(x))"Î# a1 f w (x)b Ê hw (1) œ 3# (1 f(1))"Î# a1 f w (1)b œ 3# (1 3)"Î# ˆ1 "# ‰ œ *# (g) Let h(x) œ f(x g(x)) Ê hw (x) œ f w (x g(x)) a1 gw (x)b Ê hw (0) œ f w (g(0)) a1 gw (0)b œ f w (1) ˆ1 "# ‰ œ ˆ "# ‰ ˆ $# ‰ œ
%$Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Practice Exercises " #È x
56. (a) Let h(x) œ Èx f(x) Ê hw (x) œ Èx f w (x) f(x) † " #
(b) Let h(x) œ (f(x))"Î# Ê hw (x) œ
(f(x))"Î# af w (x)b Ê hw (0) œ
(c) Let h(x) œ f ˆÈx‰ Ê hw (x) œ f w ˆÈx‰ †
" #È x
" œ 5" (3) ˆ #" ‰ #È 1 " "Î# (2) œ 3" # (9)
Ê hw (1) œ È1 f w (1) f(1) † " #
(f(0))"Î# f w (0) œ " #È 1 w
Ê hw (1) œ f w ŠÈ1‹ †
" 5
œ
†
" #
œ
œ 13 10
" 10
(d) Let h(x) œ f(1 5 tan x) Ê hw (x) œ f w (1 5 tan x) a5 sec# xb Ê h (0) œ f w (1 5 tan 0) a5 sec# 0b œ f w (1)(5) œ "5 (5) œ 1 (2 cos x)f (x) f(x)(sin x) f(0)(0) Ê hw (0) œ (2 1)f(2(0) œ 3(9 2) œ (2 cos x)# 1)# h(x) œ 10 sin ˆ 1#x ‰ f # (x) Ê hw (x) œ 10 sin ˆ 1#x ‰ a2f(x)f w (x)b f # (x) ˆ10 cos ˆ 1#x ‰‰ ˆ 1# ‰ hw (1) œ 10 sin ˆ 1# ‰ a2f(1)f w (1)b f # (1) ˆ10 cos ˆ 1# ‰‰ ˆ 1# ‰ œ 20(3) ˆ "5 ‰ ! œ 12
(e) Let h(x) œ (f) Let Ê
57. x œ t# 1 Ê dy dt
œ
dy dx
†
"Î$
œ 2 au# 2ub ds ¸ du u=2
Ê
"Î$
" 3
5; thus
œ ’2 a2# 2(2)b
cos É8 sin ˆs 16 ‰ 2 2É8 sin ˆs
Ê
œ
dt du
59. r œ 8 sin ˆs 16 ‰ Ê œ
œ
dw ¸ ds s = 0
1‰
and
2 sin (0) 3 1
Ê
œ
"Î$
5“ ˆ 23 ‰ a2# 2(2)b
; thus,
d# y dx#
œ
œ
"3 8#Î$
œ
" 3
7b
#Î$
2 3
d ) ‰‰ dt
(2)) œ
(1 7)#Î$ œ
d# y dx#
œ
†
dw dr
dy dx
" 6
2 3
d) dt
" 3
2 # 3 au "Î$
#Î$
2ub
dr ¸ dt t = 0
d) dt
#Î$
œ
1b
x#Î$ 3" y#Î$ dy ˆ #Î$ ‰ ˆ 23 dx ‹ y
# ax#Î$ b
2ub
#Î$
#
dw dr
dy dx
† 8 cos ˆs 16 ‰‘
d) dt
œ
) # 2)t1
; r œ a)# 7b
œ
ˆ 6" ‰ (1)
œ
a3y# 1b œ 2 sin x Ê
dy dx
œ
2 sin x 3y# 1
Ê
dy dx ‹
x "Î$ ‰
Ê
dy dx
#Î$
œ yx#Î$ Ê
d# y dx# ¹ (8ß8)
œ
dy dx ¹ (8ß8)
œ 1;
dy dx
œ
y#Î$ x#Î$
ˆ8#Î$ ‰ 23 †8 "Î$ †(1)‘ ˆ8#Î$ ‰ ˆ 23 †8 "Î$ ‰ 8%Î$
" 6
" " f(t h) f(t) 2t 1 (2t 2h 1) " " œ #(t h) 1h #t 1 œ (2t 2t 1 and f(t h) œ #(t h) 1 Ê h 2h 1)(2t 1)h f(t h) f(t) 2h 2 w œ lim (2t 2h 21)(#t 1) (2t 2h 1)(2t 1)h œ (2t 2h 1)(2t 1) Ê f (t) œ hlim h Ä! hÄ! # # (2t 1)
63. f(t) œ œ œ
œ
" 6
œ #" œ0 Ê
"Î$
d) ¸ dt t= 0, ) = 1
#
dy dx
9 #
œ È3
d) ¸ dt t = 0
†
œ 2t 5
" œ cos ˆÈr 2‰ Š #È ‹ r
(2)t 1) œ )# Ê
dr ¸ d) t = 0
ds dt
(u 1)
; now t œ 0 and )# t ) œ 1 Ê ) œ 1 so that
œ 2 sin x Ê a3y#
È3 ‹
2È4
) a)# 7b
Ê dy dx
œ
(u 1); s œ t# 5t Ê
(2 1) œ 2 ˆ2 † 8"Î$ 5‰ ˆ8#Î$ ‰ œ 2(2 † 2 5) ˆ 4" ‰ œ
# É8 sinˆ s 16 ‰
(cos 0)(8) Š
#Î$
5“ ˆ 32 ‰ au#
cos ŠÉ8 sin ˆs 16 ‰ 2‹
œ0 Ê
(3 1)(2 cos 0) (2 sin 0)(6†0) (3 1)#
œ
œ
dr ds
a3y# 1b (2 cos x) (2 sin x) Š6y
ˆx#Î$ ‰ Š 23 y "Î$
4
œ
dw ds
6
62. x"Î$ y"Î$ œ 4 Ê Ê
(2u 2) œ
œ ’2 au# 2ub
dt du
2É8 sin ˆ 16 ‰
œ 0;
d# y dx# ¹ (0ß1)
#Î$
œ
ds dt
†
œ 3(cos 2x)(2) œ 6 cos 2x œ 6 cos a2t# 21b œ 6 cos a2t# b ; thus,
œ 8 cos ˆs 16 ‰ ; w œ sin ˆÈr 2‰ Ê
dr ds
61. y$ y œ 2 cos x Ê 3y# œ
au# 2ub
32
œ 6 cos (0) † 0 œ 0
dy dt ¹ t = 0
cos ŠÉ8 sin ˆ 16 ‰ 2‹†8 cos ˆ 16 ‰
dr " # d ) œ 3 a) dr ¸ 2 d) ) = 1 œ 3
dy dx
w
ds du
60. )# t ) œ 1 Ê ˆ)# t ˆ2) Ê
w
œ 2t; y œ 3 sin 2x Ê
dx dt
œ 6 cos a2t# b † 2t Ê
dx dt
58. t œ au# 2ub
Ê
Ê hw (x) œ
f(x) 2 cos x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dy dx ¹ (0ß1)
1 1
œ 1
155
156
Chapter 3 Differentiation
64. g(x) œ 2x# 1 and g(x h) œ 2(x h)# 1 œ 2x# 4xh 2h# 1 Ê œ
4xh 2h# h
g(x h) g(x) h
œ 4x 2h Ê gw (x) œ lim
hÄ!
g(x h) g(x) h
œ
a2x# 4xh 2h# 1b a2x# 1b h
œ lim (4x 2h) œ 4x hÄ!
65. (a)
lim f(x) œ lim c x# œ 0 and lim b f(x) œ lim b x# œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0) it xÄ! xÄ! xÄ! xÄ! xÄ! follows that f is continuous at x œ 0. (c) lim c f w (x) œ lim c (2x) œ 0 and lim b f w (x) œ lim b (2x) œ 0 Ê lim f w (x) œ 0. Since this limit exists, it (b)
x Ä !c
xÄ!
xÄ!
xÄ!
follows that f is differentiable at x œ 0.
xÄ!
xÄ!
66. (a)
lim f(x) œ lim c x œ 0 and lim b f(x) œ lim b tan x œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0), it xÄ! xÄ! xÄ! xÄ! xÄ! follows that f is continuous at x œ 0. (c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b sec# x œ 1 Ê lim f w (x) œ 1. Since this limit exists it (b)
x Ä !c
xÄ!
xÄ!
xÄ!
follows that f is differentiable at x œ 0.
xÄ!
xÄ!
67. (a)
lim f(x) œ lim c x œ 1 and lim b f(x) œ lim b (2 x) œ 1 Ê lim f(x) œ 1. Since lim f(x) œ 1 œ f(1), it xÄ" xÄ" xÄ" xÄ" xÄ" follows that f is continuous at x œ 1. (c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b 1 œ 1 Ê lim c f w (x) Á lim b f w (x), so lim f w (x) does (b)
x Ä "c
xÄ"
xÄ"
xÄ"
not exist Ê f is not differentiable at x œ 1. 68. (a)
xÄ"
xÄ"
xÄ"
lim f(x) œ lim c sin 2x œ 0 and lim b f(x) œ lim b mx œ 0 Ê lim f(x) œ 0, independent of m; since xÄ! xÄ! xÄ! xÄ! f(0) œ 0 œ lim f(x) it follows that f is continuous at x œ 0 for all values of m. x Ä !c
xÄ!
(b)
xÄ1
lim f w (x) œ lim c (sin 2x)w œ lim c 2 cos 2x œ 2 and lim b f w (x) œ lim b (mx)w œ lim b m œ m Ê f is x Ä !c xÄ! xÄ! xÄ! xÄ! xÄ! differentiable at x œ 0 provided that lim c f w (x) œ lim b f w (x) Ê m œ 2. xÄ!
xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Practice Exercises 69. y œ
x #
" #x 4
" #
œ
x (2x 4)" Ê
dy dx
œ
Ê (2x 5)(2x 3) œ 0 Ê x " 2x
Ê
Ê xœ „
" #
œ1
dy dx
2 (2x)#
2(2x 4)# ; the slope of the tangent is 3# Ê 3# œ
" #
2(2x 4)#
" # # # (2x 4)# Ê (2x 4) œ 1 Ê 4x 16x 16 œ 1 Ê 4x 16x 15 œ 0 œ 5# or x œ 3# Ê ˆ 5# ß 49 ‰ and ˆ 3# ß 4" ‰ are points on the curve where the slope
Ê 2 œ 2(2x 4)# Ê 1 œ
70. y œ x
" #
" " #x# ; the slope of the tangent is 3 Ê 3 œ 1 #x# " "‰ # ß # are points on the curve where the slope is 3.
œ1
Ê ˆ "# ß "# ‰ and ˆ
71. y œ 2x$ 3x# 12x 20 Ê
dy dx
Ê 2œ
œ 6x# 6x 12; the tangent is parallel to the x-axis when
dy dx
" #x #
157
is 3# .
Ê x# œ
" 4
œ0
Ê 6x# 6x 12 œ 0 Ê x# x 2 œ 0 Ê (x 2)(x 1) œ 0 Ê x œ 2 or x œ 1 Ê (#ß !) and ("ß #7) are points on the curve where the tangent is parallel to the x-axis. 72. y œ x$ Ê
œ 3x# Ê
dy dx
œ 12; an equation of the tangent line at (#ß )) is y 8 œ 12(x 2)
dy dx ¹ ( 2ß 8)
Ê y œ 12x 16; x-intercept: 0 œ 12x 16 Ê x œ 43 Ê ˆ 43 ß !‰ ; y-intercept: y œ 12(0) 16 œ 16 Ê (0ß 16) 73. y œ 2x$ 3x# 12x 20 Ê
dy dx
œ 6x# 6x 12
(a) The tangent is perpendicular to the line y œ 1
x 24
when
dy dx
œ Š ˆ" " ‰ ‹ œ 24; 6x# 6x 12 œ 24 #4
Ê x# x 2 œ 4 Ê x# x 6 œ 0 Ê (x 3)(x 2) œ 0 Ê x œ 2 or x œ 3 Ê (#ß 16) and ($ß 11) are x points where the tangent is perpendicular to y œ 1 24 . dy È (b) The tangent is parallel to the line y œ 2 12x when dx œ 12 Ê 6x# 6x 12 œ 12 Ê x# x œ 0 Ê x(x 1) œ 0 Ê x œ 0 or x œ 1 Ê (!ß 20) and ("ß () are points where the tangent is parallel to y œ È2 12x. 74. y œ
1 sin x x
Ê
dy dx
œ
x(1 cos x) (1 sin x)(1) x#
Ê m" œ
dy dx ¹ x=1
œ
1 # 1#
œ 1 and m# œ
dy 1# dx ¹ x=c1 1#
œ 1. Since m" œ m"# the
tangents intersect at right angles. 75. y œ tan x, 1# x of y yœ Ê
1 #
Ê
dy dx
œ sec# x; now the slope
œ is "# Ê the normal line is parallel to " # x# when dy dx œ 2. Thus, sec x œ 2 Ê cos# x œ 2 „" cos# x œ "# Ê cos x œ È Ê x œ 14 and x œ 14 2 x #
for 1# x
1 #
Ê ˆ 14 ß 1‰ and ˆ 14 ß "‰ are points
where the normal is parallel to y œ x# .
76. y œ 1 cos x Ê
dy dx
œ sin x Ê
dy dx ¹ ˆ 1 ß1‰
œ 1
2
Ê the tangent at ˆ 1# ß 1‰ is the line y 1 œ ˆx 1# ‰ Ê y œ x 1# 1; the normal at ˆ 1# ß 1‰ is
y 1 œ (1) ˆx 1# ‰ Ê y œ x
77. y œ x# C Ê " #
œ
ˆ "# ‰#
dy dx
1 #
1
œ 2x and y œ x Ê
CÊCœ
dy dx
œ 1; the parabola is tangent to y œ x when 2x œ 1 Ê x œ
" 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" #
Êyœ
" #
; thus,
158
Chapter 3 Differentiation
78. y œ x$ Ê
dy dx
œ 3x# Ê
dy dx ¹ x œ a
œ 3a# Ê the tangent line at aaß a$ b is y a$ œ 3a# (x a). The tangent line
intersects y œ x$ when x$ a$ œ 3a# (x a) Ê (x a) ax# xa a# b œ 3a# (x a) Ê (x a) ax# xa 2a# b œ 0 Ê (x a)# (x 2a) œ 0 Ê x œ a or x œ 2a. Now
dy dx ¹ x œ c2a
œ 3(2a)# œ 12a# œ 4 a3a# b, so the slope at
x œ 2a is 4 times as large as the slope at aaß a$ b where x œ a. 79. The line through a0ß 3b and a5ß 2b has slope m œ y œ x 3; y œ
c x1
Ê
dy dx
c (x 1)# ,
œ
3 (2) 05
œ 1 Ê the line through a0ß 3b and a5ß 2b is
so the curve is tangent to y œ x 3 Ê
Ê (x 1)# œ c, x Á 1. Moreover, y œ
c x1
intersects y œ x 3 Ê #
dy dx
œ 1 œ
c (x 1)#
œ x 3, x Á 1
c x 1
Ê c œ (x 1)(x 3), x Á 1. Thus c œ c Ê (x 1) œ (x 1)(x 3) Ê (x 1)[x 1 (x 3)] œ !, x Á 1 Ê (x 1)(2x 2) œ 0 Ê x œ 1 (since x Á 1) Ê c œ 4. 80. Let Šbß „ Èa# b# ‹ be a point on the circle x# y# œ a# . Then x# y# œ a# Ê 2x 2y Ê
dy dx ¹ x œ b
œ
b „È a # b #
y Š „ Èa# b# ‹ œ
Ê normal line through Šbß „ Èa# b# ‹ has slope „È a # b # b
(x b) Ê y … Èa# b# œ
„È a # b # b
„È a # b # b
œ0 Ê
dy dx
dy dx
œ xy
Ê normal line is
x … Èa# b# Ê y œ „
È a# b# b
x
which passes through the origin. 81. x# 2y# œ 9 Ê 2x 4y œ "4 x
9 4
5 #
œ0 Ê
dy dx
x œ 2y Ê
œ "4 Ê the tangent line is y œ 2 "4 (x 1)
dy dx ¹ (1ß2)
and the normal line is y œ 2 4(x 1) œ 4x 2.
82. x$ y# œ 2 Ê 3x# 2y œ 3# x
dy dx
dy dx
œ0 Ê
œ
dy dx
3x# 2y
Ê
dy dx ¹ (1ß1)
and the normal line is y œ 1 32 (x 1) œ
83. xy 2x 5y œ 2 Ê Šx
y‹ 2 5
dy dx
dy dx
œ0 Ê
(x 5) œ y 2 Ê
Ê the tangent line is y œ 2 2(x 3) œ 2x 4 and the normal line is y œ 2 84. (y x)# œ 2x 4 Ê 2(y x) Š dy dx 1‹ œ 2 Ê (y x) Ê the tangent line is y œ 2 34 (x 6) œ 85. x Èxy œ 6 Ê 1
" #Èxy
Šx
dy dx
3 4
x
dy dx
œ 1 (y x) Ê
y 2 x 5
dy dx
œ
Ê
1 #
(x 3) œ "# x 7# .
dy dx
œ
1yx yx
3 2
x"Î# 3y"Î#
y œ 4 "4 (x 1) œ 4" x
17 4
87. x$ y$ y# œ x y Ê ’x$ Š3y# Ê
dy dx
dy dx
œ0 Ê
dy dx ¹ (3ß2)
Ê
dy dx
y œ 2Èxy Ê
dy dx
œ
2Èxy y x
Ê
œ2
dy dx ¹ (6ß2)
dy dx ¹ (4ß1)
Ê the tangent line is y œ 1 54 (x 4) = 54 x 6 and the normal line is y œ " 45 (x 4) œ 86. x$Î# 2y$Î# œ 17 Ê
(x 1)
œ
and the normal line is y œ 2 43 (x 6) œ 43 x 10.
5 #
y‹ œ 0 Ê x
3 #
x 3" .
2 3
dy dx
œ #3 Ê the tangent line is y œ 1
dy dx
œ
x"Î# 2y"Î#
Ê
dy dx ¹ (1ß4)
4 5
œ
x
5 4
11 5
.
œ "4 Ê the tangent line is
and the normal line is y œ 4 4(x 1) œ 4x. dy dx ‹
y$ a3x# b“ 2y
a3x$ y# 2y 1b œ 1 3x# y$ Ê
dy dx
œ
dy dx
œ1
1 3x# y$ 3x$ y# 2y 1
Ê
dy dx
Ê 3x$ y#
dy dx ¹ (1ß1)
dy dx
2y
œ 24 , but
Therefore, the curve has slope "# at ("ß ") but the slope is undefined at ("ß 1).
dy dx
dy dx ¹ (1ßc1) is
dy dx
œ " 3x# y$
undefined.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3 4
Chapter 3 Practice Exercises 88. y œ sin (x sin x) Ê
dy dx
œ [cos (x sin x)](1 cos x); y œ 0 Ê sin (x sin x) œ 0 Ê x sin x œ k1,
k œ 2, 1, 0, 1, 2 (for our interval) Ê cos (x sin x) œ cos (k1) œ „ 1. Therefore,
dy dx
œ 0 and y œ 0 when
1 cos x œ 0 and x œ k1. For #1 Ÿ x Ÿ 21, these equations hold when k œ 2, 0, and 2 (since cos (1) œ cos 1 œ 1). Thus the curve has horizontal tangents at the x-axis for the x-values 21, 0, and 21 (which are even integer multiples of 1) Ê the curve has an infinite number of horizontal tangents. 89. B œ graph of f, A œ graph of f w . Curve B cannot be the derivative of A because A has only negative slopes while some of B's values are positive. 90. A œ graph of f, B œ graph of f w . Curve A cannot be the derivative of B because B has only negative slopes while A has positive values for x 0. 91.
92.
93. (a) 0, 0
(b) largest 1700, smallest about 1400
94. rabbits/day and foxes/day sin x
95. lim
# x Ä ! 2x x
96. lim
3x tan 7x #x
97. lim
sin r
98. lim
sin (sin )) )
xÄ!
r Ä ! tan 2r
)Ä!
Ê lim
)Ä!
99.
lim c ) Ä ˆ1‰
xÄ!
" (#x 1) “
œ lim ˆ 3x 2x
sin 7x ‰ 2x cos 7x
œ lim ’ˆ sinx x ‰ †
xÄ!
œ lim ˆ sinr r † rÄ!
101.
lim b
)Ä!
œ lim
xÄ!
lim
œ
1 2 cot# ) 5 cot# ) 7 cot ) 8
1cos ) )#
œ lim b )Ä!
x sin x x Ä ! 2(1 cos x)
œ lim
)Ä!
2 sin# ˆ #) ‰ )#
sin (sin )) sin )
" ˆ 27 ‰
†
‹œ
œ ˆ "# ‰ (1) ˆ 1" ‰ œ
3 #
ˆ1 † 1 † 27 ‰ œ 2
" #
. Let x œ sin ). Then x Ä 0 as ) Ä 0
œ1
lim c ) Ä ˆ1‰
œ lim
cos 2r
sin 2r r Ä ! ˆ 2r ‰
sin x x
sin 7x 7x
xÄ!
† "# ‰ œ ˆ "# ‰ (1) lim
2
x sin x x Ä ! 2 2 cos x
)Ä!
lim Š cos"7x †
)Ä!
4 tan# ) tan ) 1 tan# ) &
œ (1)(1)(1) œ 1
102.
3 #
)Ä!
sin (sin )) sin )
lim
œ
(sin )) ˆ sin ) ‰ œ lim Š sinsin œ lim ) ‹ )
2
100.
2r tan 2r
œ (1) ˆ "1 ‰ œ 1
Š4 tan" ) tan"# ) ‹ Š" tan5# ) ‹ Š cot"# ) 2‹ Š5 cot7 ) cot8# ) ‹
œ
œ
(4 0 0) (1 0)
(0 2) (5 0 0)
œ lim
x sin x # x x Ä ! 2 ˆ2 sin ˆ # ‰‰
œ lim ’ )Ä!
sin ˆ #) ‰ ˆ #) ‰
†
sin ˆ #) ‰ ˆ #) ‰
œ4
œ 52 †
x x
œ lim ’ sin## ˆ# x ‰ † xÄ!
#
sin x x “
† "# “ œ (1)(1) ˆ "# ‰ œ
ˆx‰
œ lim ’ sin #ˆ x ‰ † xÄ!
#
" #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ˆ #x ‰
sin ˆ #x ‰
†
sin x x “
159
160 103.
Chapter 3 Differentiation tan x x
lim
xÄ!
œ lim
)Ä!
104.
œ lim ˆ cos" x †
tan ) )
œ 1; let ) œ tan x Ê ) Ä 0 as x Ä 0 Ê lim g(x) œ lim xÄ!
(tan x) œ lim ’ tantan † x
tan (tan x)
x Ä ! sin (sin x)
105. (a) S œ 21r# 21rh and h constant Ê (b) S œ 21r# 21rh and r constant Ê (d) S constant Ê
dh dt
(b) r constant Ê
dr dt
108. V œ s$ Ê 109.
dR" dt
dr dt
œ 3s# †
ds dt
œ 1 ohm/sec,
R# œ 50 ohms Ê " dR (30)# dt
110.
dR dt
œ
" (75)#
" R
œ
dZ dt
œ
dr dt
" dV 3s# dt
Ê (2r
; so s œ 20 and " R
œ
" R"
" R#
dV dt
Ê
dA dt
œ 2x
dx dt
2y
Ê D
dD dt
Ê
dr dt
œ
r dh 2rh dt
1 r# dr Èr# h# “ dt
œ
" È5
œ (21)(10) ˆ 12 ‰ œ 40 m# /sec
œ 1200 cm$ /min Ê " dR R# dt
œ
" dR" R"# dt
ds dt
" dR# R## dt
œ
" 3(20)#
(1200) œ 1 cm/min
. Also, R" œ 75 ohms and
Therefore, from the derivative equation, Ê
dR dt
5625 ‰ œ (900) ˆ 5000 5625†5000 œ
œ
dy dt
dh dt dh dt
dh 1rh Èr# h# dt
(10)(3)(20)(2) È10# 20#
œ 10 m/sec and
dr dt 21r h) dr dt œ r
œ ’1Èr# h#
dr dt
œ 2 ohms/sec; Z œ ÈR# X# Ê
dx dt
dr dt
;
œ 12 m/sec Ê
" " 75 50 Ê R œ 30 ohms. " " ‰ ˆ " (50)# (0.5) œ 5625 5000
111. Given
dD dt
ds dt
(using the result of #105);
œ 1. Therefore, to make f continuous at the origin,
œ (41r 21h)
dr dt
dh dt
1Èr# h#
dr 1 r# Èr# h# “ dt
; so r œ 10 and Ê
dr dt
sin x
x Ä ! sin (sin x)
œ (41r 21h)
dr ‰ dt
21r
dr dt
œ 1 † lim
dX dt
œ 3 ohms/sec and
X œ 20 ohms Ê
5
tan (tan x) tan x
1rh dh Èr# h# dt
œ
dS dt
1r# dr dt È r# h #
œ
" cos x “
lim ) ) Ä ! sin )
1Èr# h#
œ 0.5 ohm/sec; and
dR# dt
(1)
dy dt
dS dt
œ0 Ê
œ 21 r
dA dt dV dt
œ0 Ê
dh ‰ ˆr dr dt h dt È r# h #
œ ’1Èr# h#
dS dt
(c) In general,
œ 1r †
dS dt
(a) h constant Ê
œ 21r dh dt #1 ˆr dh h dt
dr dt
†
œ 41r dr dt 21 h
dS dt dS dt
œ 0 Ê 0 œ (41r 21h)
dS dt
106. S œ 1rÈr# h# Ê
œ 41r
dS dt
œ
sin x lim x Ä ! sin (sin x)
define f(0) œ 1.
(c) S œ 21r# 21rh Ê
sin x sin (sin x)
xÄ!
let ) œ sin x Ê ) Ä 0 as x Ä 0 Ê
107. A œ 1r# Ê
xÄ!
œ 1. Therefore, to make g continuous at the origin, define g(0) œ 1.
lim f(x) œ lim
xÄ!
sin x ‰ x
xÄ!
dZ dt
œ
dX R dR dt X dt È R # X#
9(625) 50(5625)
œ
" 50
œ 0.02 ohm/sec.
so that R œ 10 ohms and
¸ 0.45 ohm/sec.
œ 5 m/sec, let D be the distance from the origin Ê D# œ x# y# Ê 2D œx
œ (3)(10) (%)(5) Ê
dy dx dt y dt dD 10 dt œ 5 œ
dD dt
. When (xß y) œ ($ß %), D œ É$# a%b# œ & and 2. Therefore, the particle is moving away from the origin at 2 m/sec
(because the distance D is increasing). # œ 11 units/sec. Then D# œ x# y# œ x# ˆx$Î# ‰ È3# 3$ œ 6 and substitution in the œ x(2 3x) dx dt ; x œ 3 Ê D œ
112. Let D be the distance from the origin. We are given that œ x# x$ Ê 2D
dD dt
œ 2x
dx dt
3x#
dx dt
derivative equation gives (2)(6)(11) œ (3)(2 9) 113. (a) From the diagram we have (b) V œ
" 3
#
1r h œ
" 3
1 ˆ 52
#
10 h
h‰ h œ
œ
4 r 41 h$ 75
Ê rœ Ê
dV dt
2 5
œ
dx dt
Ê
dD dt
dx dt
œ 4 units/sec.
h. 41h# dh 25 dt
, so
dV dt
œ 5 and h œ 6 Ê
dh dt
125 œ 144 1 ft/min.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Practice Exercises 114. From the sketch in the text, s œ r) Ê Therefore,
ds dt
œ 6 ft/sec and r œ 1.2 ft Ê
ds dt
115. (a) From the sketch in the text, Ê)œ0Ê
d) dt
d) dt
œr d) dt
)
dr dt
. Also r œ 1.2 is constant Ê
dr dt
œ0 Ê
dx dt
œ sec# )
œr
d) dt
œ (1.2)
d) dt
œ 0.6 rad/sec and x œ tan ). Also x œ tan ) Ê
œ asec 0b (0.6) œ 0.6. Therefore the speed of the light is 0.6 œ
3 5
d) dt ;
at point A, x œ 0
km/sec when it reaches
point A. (b)
(3/5) rad sec
†
1 rev 21 rad
116. From the figure,
a r
†
60 sec min
œ
b BC
œ
18 1
Ê
a r
revs/min œ
b È b # r#
. We are given
that r is constant. Differentiation gives, " r
†
da dt
œ
‰ ŠÈb# r# ‹ ˆ db dt (b) Š È b # r#
b œ 2r and Ê œ
da dt
db dt
b ‰ ‹ ˆ db dt b# r#
. Then,
œ 0.3r
Ô È(2r)# r# (0.3r) (2r) É2r( #0.3r)# × (2r) r Ù œ rÖ # # (2r) r
Õ
È3r# (0.3r)
4r# (0.3r) 3r#
È
3r
œ
Ø
a3r# b (0.3r) a4r# b (0.3r) 3 È 3 r#
œ
0.3r 3È 3
œ
r 10È3
m/sec. Since
da dt
is positive, the distance OA is increasing
when OB œ 2r, and B is moving toward O at the rate of 0.3r m/sec. 117. (a) If f(x) œ tan x and x œ 14 , then f w (x) œ sec# x, f ˆ 14 ‰ œ 1 and f w ˆ 14 ‰ œ 2. The linearization of f(x) is L(x) œ 2 ˆx 14 ‰ (1) œ 2x
1 2 #
.
(b) If f(x) œ sec x and x œ 14 , then f w (x) œ sec x tan x, f ˆ 1 ‰ œ È2 and f w ˆ 1 ‰ œ È2. The 4
4
linearization of f(x) is L(x) œ È2 ˆx 14 ‰ È2 œ È2x
118. f(x) œ
" 1 tan x
È2(% 1) . 4
Ê f w (x) œ
sec# x (1 tan x)#
. The linearization at x œ 0 is L(x) œ f w (0)(x 0) f(0) œ 1 x.
119. f(x) œ Èx 1 sin x 0.5 œ (x 1)"Î# sin x 0.5 Ê f w (x) œ ˆ "# ‰ (x 1)"Î# cos x
Ê L(x) œ f w (0)(x 0) f(0) œ 1.5(x 0) 0.5 Ê L(x) œ 1.5x 0.5, the linearization of f(x).
120. f(x) œ œ
2 1 x
2 (1 x)#
È1 x 3.1 œ 2(1 x)" (1 x)"Î# 3.1 Ê f w (x) œ 2(1 x)# (1) "# (1 x)"Î#
" 2È 1 x
.
œ 5 rad/sec
#
dx dt
ds dt
161
Ê L(x) œ f w (0)(x 0) f(0) œ 2.5x 0.1, the linearization of f(x).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
162
Chapter 3 Differentiation
121. S œ 1 rÈr# h# , r constant Ê dS œ 1 r † "# ar# h# b Ê dS œ
"Î#
#h dh œ
1rh Èr# h# dh.
Height changes from h! to h! dh
1 r h! adhb Ér# h#!
122. (a) S œ 6r# Ê dS œ 12r dr. We want kdSk Ÿ (2%) S Ê k12r drk Ÿ
12r# 100
Ê kdrk Ÿ
r 100
. The measurement of the
edge r must have an error less than 1%. #
3r dr ‰ (b) When V œ r$ , then dV œ 3r# dr. The accuracy of the volume is ˆ dV V (100%) œ Š r$ ‹ (100%) r ‰ œ ˆ 3r ‰ (dr)(100%) œ ˆ 3r ‰ ˆ 100 (100%) œ 3%
123. C œ 21r Ê r œ dV œ
C# 21 #
C 21
, S œ 41 r # œ
C# 1
, and V œ
4 3
1 r$ œ
C$ 61 #
. It also follows that dr œ
" #1
dC, dS œ
2C 1
dC and
dC. Recall that C œ 10 cm and dC œ 0.4 cm. 0.2 ˆ drr ‰ (100%) œ ˆ 0.2 ‰ ˆ 2101 ‰ (100%) œ (.04)(100%) œ 4% (a) dr œ 0.4 21 œ 1 cm Ê 1 8 1 ‰ ˆ dS ‰ ˆ 8 ‰ ˆ 100 (b) dS œ 20 (100%) œ 8% 1 (0.4) œ 1 cm Ê S (100%) œ 1 10# 21 #
#
(0.4) œ
20 1#
‰ ˆ 20 ‰ 61 cm Ê ˆ dV V (100%) œ 1# Š 1000 ‹ (100%) œ 12%
124. Similar triangles yield
35 h
œ
(c) dV œ
Ê dh œ 120a# da œ
15 6 120 a#
Ê h œ 14 ft. The same triangles imply that 20h a œ 6a Ê h œ 120a" 6 " ‰ 2 ‰ ˆ „ 1"# ‰ œ ˆ "#! ‰ˆ „ "# da œ ˆ 120 œ „ 45 ¸ „ .0444 ft œ „ 0.53 inches. a# "&#
CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES 1. (a) sin 2) œ 2 sin ) cos ) Ê #
#
d d)
Ê cos 2) œ cos ) sin ) (b) cos 2) œ cos# ) sin# ) Ê
(sin 2)) œ d d)
d d)
(cos 2)) œ
(2 sin ) cos )) Ê 2 cos 2) œ 2[(sin ))(sin )) (cos ))(cos ))] d d)
acos# ) sin# )b Ê 2 sin 2) œ (2 cos ))(sin )) (2 sin ))(cos ))
Ê sin 2) œ cos ) sin ) sin ) cos ) Ê sin 2) œ 2 sin ) cos ) 2. The derivative of sin (x a) œ sin x cos a cos x sin a with respect to x is cos (x a) œ cos x cos a sin x sin a, which is also an identity. This principle does not apply to the equation x# 2x 8 œ 0, since x# 2x 8 œ 0 is not an identity: it holds for 2 values of x (2 and 4), but not for all x. 3. (a) f(x) œ cos x Ê f w (x) œ sin x Ê f ww (x) œ cos x, and g(x) œ a bx cx# Ê gw (x) œ b 2cx Ê gww (x) œ 2c; also, f(0) œ g(0) Ê cos (0) œ a Ê a œ 1; f w (0) œ gw (0) Ê sin (0) œ b Ê b œ 0; f ww (0) œ gww (0) Ê cos (0) œ 2c Ê c œ "# . Therefore, g(x) œ 1 "# x# . (b) f(x) œ sin (x a) Ê f w (x) œ cos (x a), and g(x) œ b sin x c cos x Ê gw (x) œ b cos x c sin x; also, f(0) œ g(0) Ê sin (a) œ b sin (0) c cos (0) Ê c œ sin a; f w (0) œ gw (0) Ê cos (a) œ b cos (0) c sin (0) Ê b œ cos a. Therefore, g(x) œ sin x cos a cos x sin a. (c) When f(x) œ cos x, f www (x) œ sin x and f Ð%Ñ (x) œ cos x; when g(x) œ 1 "# x# , gwww (x) œ 0 and gÐ%Ñ (x) œ 0. Thus f www (0) œ 0 œ gwww (0) so the third derivatives agree at x œ 0. However, the fourth derivatives do not agree since f Ð%Ñ (0) œ 1 but gÐ%Ñ (0) œ 0. In case (b), when f(x) œ sin (x a) and g(x) œ sin x cos a cos x sin a, notice that f(x) œ g(x) for all x, not just x œ 0. Since this is an identity, we have f ÐnÑ (x) œ gÐnÑ (x) for any x and any positive integer n.
4. (a) y œ sin x Ê yw œ cos x Ê yww œ sin x Ê yww y œ sin x sin x œ 0; y œ cos x Ê yw œ sin x Ê yww œ cos x Ê yww y œ cos x cos x œ 0; y œ a cos x b sin x Ê yw œ a sin x b cos x Ê yww œ a cos x b sin x Ê yww y œ (a cos x b sin x) (a cos x b sin x) œ 0
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Chapter 3 Additional and Advanced Exercises
163
(b) y œ sin (2x) Ê yw œ 2 cos (2x) Ê yww œ 4 sin (2x) Ê yww 4y œ 4 sin (2x) 4 sin (2x) œ 0. Similarly, y œ cos (2x) and y œ a cos (2x) b sin (2x) satisfy the differential equation yw w 4y œ 0. In general, y œ cos (mx), y œ sin (mx) and y œ a cos (mx) b sin (mx) satisfy the differential equation yww m# y œ 0. 5. If the circle (x h)# (y k)# œ a# and y œ x# 1 are tangent at ("ß #), then the slope of this tangent is m œ 2xk (1 2) œ 2 and the tangent line is y œ 2x. The line containing (hß k) and ("ß #) is perpendicular to ß
y œ 2x Ê
k2 h1
œ "# Ê h œ 5 2k Ê the location of the center is (5 2kß k). Also, (x h)# (y k)# œ a#
Ê x h (y k)yw œ 0 Ê 1 ayw b# (y k)yw w œ 0 Ê yww œ w
1 ay b# ky w
. At the point ("ß #) we know
ww
y œ 2 from the tangent line and that y œ 2 from the parabola. Since the second derivatives are equal at ("ß #) we obtain 2 œ
1 (2)# k#
Ê kœ
9 #
# . Then h œ 5 2k œ 4 Ê the circle is (x 4)# ˆy 9# ‰ œ a# . Since ("ß #)
lies on the circle we have that a œ
5È 5 2
.
6. The total revenue is the number of people times the price of the fare: r(x) œ xp œ x ˆ3
x ‰# , where 40 x ‰ ˆ x ‰ 40 3 40
" ‰ dr x ‰# x ‰ˆ dr ‘ 0 Ÿ x Ÿ 60. The marginal revenue is dx œ ˆ3 40 2x ˆ3 40 40 Ê dx œ ˆ3 2x 40 x x dr œ 3 ˆ3 40 ‰ ˆ1 40 ‰ . Then dx œ 0 Ê x œ 40 (since x œ 120 does not belong to the domain). When 40 people
are on the bus the marginal revenue is zero and the fare is p(40) œ ˆ3 7. (a) y œ uv Ê
dy dt
œ
du dt
x ‰# 40 ¹ x œ 40
œ $4.00.
v u dv dt œ (0.04u)v u(0.05v) œ 0.09uv œ 0.09y Ê the rate of growth of the total production is
9% per year. (b) If
du dt
œ 0.02u and
dv dt
œ 0.03v, then
dy dt
œ (0.02u)v (0.03v)u œ 0.01uv œ 0.01y, increasing at 1% per year.
8. When x# y# œ 225, then yw œ xy . The tangent line to the balloon at (12ß 9) is y 9 œ Ê yœ
4 3
4 3
(x 12)
x 25. The top of the gondola is 15 8
œ 23 ft below the center of the balloon. The intersection of y œ 23 and y œ 43 x 25 is at the far right edge of the gondola Ê 23 œ Ê xœ
3 #
4 3
x 25
. Thus the gondola is 2x œ 3 ft wide.
9. Answers will vary. Here is one possibility.
10. s(t) œ 10 cos ˆt 14 ‰ Ê v(t) œ 10 (a) s(0) œ 10 cos ˆ 14 ‰ œ È
ds dt
œ 10 sin ˆt 14 ‰ Ê a(t) œ
dv dt
œ
d# s dt#
œ 10 cos ˆt 14 ‰
2
(b) Left: 10, Right: 10
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164
Chapter 3 Differentiation
(c) Solving 10 cos ˆt 14 ‰ œ 10 Ê cos ˆt 14 ‰ œ 1 Ê t œ 341 when the particle is farthest to the left. Solving 10 cos ˆt 14 ‰ œ 10 Ê cos ˆt 14 ‰ œ 1 Ê t œ 14 , but t 0 Ê t œ 21 41 œ 741 when the particle is farthest to the right. Thus, v ˆ 341 ‰ œ 0, v ˆ 741 ‰ œ 0, a ˆ 341 ‰ œ 10, and a ˆ 741 ‰ œ 10. (d) Solving 10 cos ˆt 14 ‰ œ 0 Ê t œ 11. (a) s(t) œ 64t 16t# Ê v(t) œ
ds dt
1 4
Ê v ˆ 14 ‰ œ 10, ¸v ˆ 14 ‰¸ œ 10 and a ˆ 14 ‰ œ !.
œ 64 32t œ 32(2 t). The maximum height is reached when v(t) œ 0
Ê t œ 2 sec. The velocity when it leaves the hand is v(0) œ 64 ft/sec. (b) s(t) œ 64t 2.6t# Ê v(t) œ ds dt œ 64 5.2t. The maximum height is reached when v(t) œ 0 Ê t ¸ 12.31 sec. The maximum height is about s(12.31) œ 393.85 ft. 12. s" œ 3t$ 12t# 18t 5 and s# œ t$ 9t# 12t Ê v" œ 9t# 24t 18 and v# œ 3t# 18t 12; v" œ v# Ê 9t# 24t 18 œ 3t# 18t 12 Ê 2t# 7t 5 œ 0 Ê (t 1)(2t 5) œ 0 Ê t œ 1 sec and t œ 2.5 sec. 13. m av# v#! b œ k ax#! x# b Ê m ˆ2v substituting
dx dt
œv Ê m
dv dt
dv ‰ dt
œ k ˆ2x
dx ‰ dt
Ê m
dv dt
2x ‰ œ k ˆ 2v
dx dt
Ê m
dv dt
œ kx ˆ "v ‰
dx dt
œ 2At B Ê v ˆ t" # t# ‰ œ 2A ˆ t" # t# ‰ B œ A at" t# b B is the
instantaneous velocity at the midpoint. The average velocity over the time interval is vav œ œ
Bt# Cb aAt#" t# t"
. Then
œ kx, as claimed.
14. (a) x œ At# Bt C on ct" ß t# d Ê v œ aAt##
dx dt
Bt" Cb
œ
at# t" b cA at# t" b Bd t# t" #
?x ?t
œ A at# t" b B.
(b) On the graph of the parabola x œ At Bt C, the slope of the curve at the midpoint of the interval ct" ß t# d is the same as the average slope of the curve over the interval. 15. (a) To be continuous at x œ 1 requires that lim c sin x œ lim b (mx b) Ê 0 œ m1 b Ê m œ 1b ; xÄ1 xÄ1 (b) If yw œ œ
cos x, x 1 is differentiable at x œ 1, then lim c cos x œ m Ê m œ 1 and b œ 1. xÄ1 m, x 1
16. faxb is continuous at ! because lim
xÄ!
œ
x ‰ ˆ 1 cos x ‰ lim ˆ 1 xcos # 1 cos x xÄ!
œ
" cos x x
# lim ˆ sinx x ‰ xÄ!
œ ! œ fa!b. f w (0) œ lim
f(x) f(0) x0
ˆ 1 "cos x ‰
w
xÄ!
œ
" #
œ lim
xÄ!
1 c cos x 0 x
x
. Therefore f (0) exists with value
" #
.
17. (a) For all a, b and for all x Á 2, f is differentiable at x. Next, f differentiable at x œ 2 Ê f continuous at x œ 2 Ê lim c f(x) œ f(2) Ê 2a œ 4a 2b 3 Ê 2a 2b 3 œ 0. Also, f differentiable at x Á 2 xÄ2
Ê f w (x) œ œ
a, x 2 . In order that f w (2) exist we must have a œ 2a(2) b Ê a œ 4a b Ê 3a œ b. 2ax b, x 2
Then 2a 2b 3 œ 0 and 3a œ b Ê a œ
3 4
and b œ
9 4
.
(b) For x #, the graph of f is a straight line having a slope of
$ %
and passing through the origin; for x #, the graph of f
is a parabola. At x œ #, the value of the y-coordinate on the parabola is
$ #
which matches the y-coordinate of the point
on the straight line at x œ #. In addition, the slope of the parabola at the match up point is
$ %
which is equal to the
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 18. (a) For any a, b and for any x Á 1, g is differentiable at x. Next, g differentiable at x œ 1 Ê g continuous at x œ 1 Ê lim b g(x) œ g(1) Ê a 1 2b œ a b Ê b œ 1. Also, g differentiable at x Á 1 x Ä "
Ê gw (x) œ œ
a, x 1 . In order that gw (1) exist we must have a œ 3a(1)# 1 Ê a œ 3a 1 3ax# 1, x 1
Ê a œ "# .
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Chapter 3 Additional and Advanced Exercises
165
(b) For x Ÿ ", the graph of g is a straight line having a slope of "# and a y-intercept of ". For x ", the graph of g is a cubic. At x œ ", the value of the y-coordinate on the cubic is
$ #
which matches the y-coordinate of the point
on the straight line at x œ ". In addition, the slope of the cubic at the match up point is "# which is equal to the slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 19. f odd Ê f(x) œ f(x) Ê 20. f even Ê f(x) œ f(x) Ê
d dx
(f(x)) œ
(f(x)) œ
d dx
d dx d dx
(f(x)) Ê f w (x)(1) œ f w (x) Ê f w (x) œ f w (x) Ê f w is even.
(f(x)) Ê f w (x)(1) œ f w (x) Ê f w (x) œ f w (x) Ê f w is odd.
21. Let h(x) œ (fg)(x) œ f(x) g(x) Ê hw (x) œ x lim Äx œ x lim Äx œ
!
f(x) g(x) f(x) g(x! ) f(x) g(x! ) f(x! ) g(x! ) x x!
g(x! ) f(x! ) x lim ’ g(x)x x! “ Ä x!
!
h(x) h(x! ) x x!
œ x lim Äx
!
f(x) g(x) f(x! ) g(x! ) x x!
f(x! ) !) œ x lim ’f(x) ’ g(x)x xg(x ““ x lim ’g(x! ) ’ f(x)x x! ““ Äx Äx ! !
w
g(x! ) f (x! ) œ 0 †
g(x! ) lim ’ g(x)x x! “ x Ä x!
!
w
g(x! ) f (x! ) œ g(x! ) f w (x! ), if g is
continuous at x! . Therefore (fg)(x) is differentiable at x! if f(x! ) œ 0, and (fg)w (x! ) œ g(x! ) f w (x! ). 22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f(0) œ 0 and g is continuous at 0. (a) If f(x) œ sin x and g(x) œ kxk , then kxk sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0 and g(x) œ kxk is continuous at x œ 0. (b) If f(x) œ sin x and g(x) œ x#Î$ , then x#Î$ sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0 and g(x) œ x#Î$ is continuous at x œ 0. (c) If f(x) œ 1 cos x and g(x) œ $Èx, then $Èx (1 cos x) is differentiable because f w (0) œ sin (0) œ 0, f(0) œ 1 cos (0) œ 0 and g(x) œ x"Î$ is continuous at x œ 0. (d) If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable because f w (0) œ 1, f(0) œ 0 and sin ˆ "x ‰
lim x sin ˆ "x ‰ œ lim
xÄ!
" x
xÄ!
œ lim
tÄ_
sin t t
œ 0 (so g is continuous at x œ 0).
23. If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable at x œ 0 because f w (0) œ 1, f(0) œ 0 and lim x sin ˆ "x ‰ œ lim
xÄ!
sin ˆ "x ‰ " x
xÄ!
œ lim
tÄ_
sin t t
œ 0 (so g is continuous at x œ 0). In fact, from Exercise 21,
hw (0) œ g(0) f w (0) œ 0. However, for x Á 0, hw (x) œ x# cos ˆ "x ‰‘ ˆ x"# ‰ 2x sin ˆ x" ‰ . But lim hw (x) œ lim cos ˆ "x ‰ 2x sin ˆ x" ‰‘ does not exist because cos ˆ x" ‰ has no limit as x Ä 0. Therefore, xÄ!
xÄ!
the derivative is not continuous at x œ 0 because it has no limit there. 24. From the given conditions we have f(x h) œ f(x) f(h), f(h) 1 œ hg(h) and lim g(h) œ 1. Therefore, hÄ!
f(xh) f(x) h hÄ! w
f w (x) œ lim
f(x) f(h) f(x) h hÄ!
œ lim
œ lim f(x) ’ f(h)h 1 “ œ f(x) ’ lim g(h)“ œ f(x) † 1 œ f(x) hÄ!
Ê f w (x) œ f(x) and f axbexists at every value of x.
hÄ!
25. Step 1: The formula holds for n œ 2 (a single product) since y œ u" u# Ê
dy dx
œ
du" dx
u# u"
du# dx
Step 2: Assume the formula holds for n œ k: y œ u" u# âuk Ê
du# duk dx u$ âuk á u" u# âuk-1 dx d(u" u# âuk ) If y œ u" u# âuk ukb1 œ au" u# âuk b ukb1 , then dy ukb1 u" u# âuk dudxkb1 dx œ dx dukb1 du# duk ‰ " œ ˆ du dx u# u$ âuk u" dx u$ âuk â u" u# âukc1 dx ukb1 u" u# âuk dx dukb1 du# duk " œ du dx u# u$ âukb1 u" dx u$ â ukb1 â u" u# âukc1 dx ukb1 u" u# âuk dx . dy dx
œ
du" dx
u# u$ âuk u"
.
Thus the original formula holds for n œ (k1) whenever it holds for n œ k.
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.
166
Chapter 3 Differentiation
26. Recall ˆ mk ‰ œ œ
m! m! m! m! ˆm‰ ˆm‰ ˆ m ‰ k! (m k)! . Then 1 œ 1! (m 1)! œ m and k k 1 œ k! (m k)! (k 1)! (m k 1)! m! (k 1) m! (m k) (m 1)! ˆm1‰ œ (k m!1)!(m(m 1)k)! œ (k 1)! ((m (k 1)! (m k)! 1) (k 1))! œ k 1 . Now, we prove
Leibniz's rule by mathematical induction. Step 1: If n œ 1, then Step 2:
d(uv) dv du dx œ u dx v dx . Assume that the statement is true for n œ k, that is: # k # k " dk (uv) dk u dk " u dv dk v ˆk‰ d u d v ˆ k ‰ du d v dxk œ dxk v k dxk " dx 2 dxk # dx# á k 1 dv dxk " u dxk . kb" k k " k (uv) d dk u dv dk" u d# v ddxk u" v ddxuk dv ‘ If n œ k 1, then d dx(uv) œ dx Š d dx k " k ‹ œ dx ’k dxk dx k dxk" dx# “
’ˆ k2 ‰
dk " u d# v dxk " dx#
ˆ k2 ‰
dk # u d$ v dxk # dx$ “
á ’ˆ k k 1 ‰
d# u dk " v dx# dxk "
ˆ kk 1 ‰
du dk u dx dxk
v“
k" # dk v dkb" u ‘ dk " u dk u dv ˆ k1 ‰ ˆ k2 ‰‘ ddxk"u ddxv# á dxk u dxk " œ dxk " v (k 1) dxk dx du dk v dkb" v dk " u dk u dv dk " u d# v ˆ k k 1 ‰ ˆ kk ‰‘ dx ˆ k 2 1 ‰ dx k " dxk u dxk" œ dxk" v (k 1) dxk dx dx# du dk v dkb" v ˆ k k 1 ‰ dx dxk u dxk " .
du dx
á
Therefore the formula (c) holds for n œ (k 1) whenever it holds for n œ k. 27. (a) T# œ
41 # L g
(b) T# œ
41 # L g
ÊLœ
T# g 41 #
ÊTœ
#1 È L; Èg
ÊLœ
a1 sec# ba32.2 ft/sec# b 41 #
dT œ
#1 Èg
†
" dL #È L
Ê L ¸ 0.8156 ft œ
1 ÈLg dL;
dT œ
1 Èa!Þ)"&' ftba32.2 ft/sec# b a!Þ!"
ftb ¸ 0.00613 sec.
(c) Since there are 86,400 sec in a day, we have a0.00613 secba86,400 sec/dayb ¸ 529.6 sec/day, or 8.83 min/day; the clock will lose about 8.83 min/day. 28. v œ s$ Ê
dv dt
# œ $s# ds dt œ ka's b Ê
ds dt
œ #k. If s! œ the initial length of the cube's side, then s" œ s! #k
Ê #k œ s! s" . Let t œ the time it will take the ice cube to melt. Now, t œ œ
" "Î$ " ˆ $% ‰
s! #k
œ
s! s ! s "
œ
av! b"Î$ "Î$ av! b ˆ $% v! ‰ "Î$
¸ "" hr.
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CHAPTER 4 APPLICATIONS OF DERIVATIVES 4.1 EXTREME VALUES OF FUNCTIONS 1. An absolute minimum at x œ c# , an absolute maximum at x œ b. Theorem 1 guarantees the existence of such extreme values because h is continuous on [aß b]. 2. An absolute minimum at x œ b, an absolute maximum at x œ c. Theorem 1 guarantees the existence of such extreme values because f is continuous on [aß b]. 3. No absolute minimum. An absolute maximum at x œ c. Since the function's domain is an open interval, the function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values. 4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill the conclusions of Theorem 1. 5. An absolute minimum at x œ a and an absolute maximum at x œ c. Note that y œ g(x) is not continuous but still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 6. Absolute minimum at x œ c and an absolute maximum at x œ a. Note that y œ g(x) is not continuous but still has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 7. Local minimum at a"ß !b, local maximum at a"ß !b 8. Minima at a#ß !b and a#ß !b, maximum at a!ß #b 9. Maximum at a!ß &b. Note that there is no minimum since the endpoint a#ß !b is excluded from the graph. 10. Local maximum at a$ß !b, local minimum at a#ß !b, maximum at a"ß #b, minimum at a!ß "b 11. Graph (c), since this the only graph that has positive slope at c. 12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c. 13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a. 14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b. 15. f has an absolute min at x œ 0 but does not have an absolute max. Since the interval on which f is defined, 1 x 2, is an open interval, we do not meet the conditions of Theorem 1.
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Chapter 4 Applications of Derivatives
16. f has an absolute max at x œ 0 but does not have an absolute min. Since the interval on which f is defined, 1 x 1, is an open interval, we do not meet the conditions of Theorem 1.
17. f has an absolute max at x œ 2 but does not have an absolute min. Since the function is not continuous at x œ 1, we do not meet the conditions of Theorem 1.
18. f has an absolute max at x œ 4 but does not have an absolute min. Since the function is not continuous at x œ 0, we do not meet the conditions of Theorem 1.
19. f has an absolute max at x œ xœ
31 2 .
1 2
and an absolute min at
Since the interval on which f is defined,
0 x 21, is an open interval, we do not meet the conditions of Theorem 1.
20. f has an absolute max at x œ 0 and an absolute min at x œ 12 and x œ 1. Since f is continuous on the closed interval on which it is defined, 1 Ÿ x Ÿ 21, we do meet the conditions of Theorem 1.
21. f(x) œ
2 3
x 5 Ê f w (x) œ
f(2) œ
19 3 ,
2 3
Ê no critical points;
f(3) œ 3 Ê the absolute maximum
is 3 at x œ 3 and the absolute minimum is 19 3 at x œ 2
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Section 4.1 Extreme Values of Functions 22. f(x) œ x 4 Ê f w (x) œ 1 Ê no critical points; f(4) œ 0, f(1) œ 5 Ê the absolute maximum is 0 at x œ 4 and the absolute minimum is 5 at x œ "
23. f(x) œ x# 1 Ê f w (x) œ 2x Ê a critical point at x œ 0; f(1) œ 0, f(0) œ 1, f(2) œ 3 Ê the absolute maximum is 3 at x œ 2 and the absolute minimum is 1 at x œ 0
24. f(x) œ % x# Ê f w (x) œ 2x Ê a critical point at x œ 0; f(3) œ 5, f(0) œ 4, f(1) œ 3 Ê the absolute maximum is 4 at x œ 0 and the absolute minimum is 5 at x œ 3
25. F(x) œ x"# œ x# Ê Fw (x) œ 2x$ œ
2 x$
, however
x œ 0 is not a critical point since 0 is not in the domain; F(0.5) œ 4, F(2) œ 0.25 Ê the absolute maximum is 0.25 at x œ 2 and the absolute minimum is 4 at x œ 0.5
26. F(x) œ "x œ x" Ê Fw (x) œ x# œ
" x#
, however
x œ 0 is not a critical point since 0 is not in the domain; F(2) œ "# , F(1) œ 1 Ê the absolute maximum is 1 at x œ 1 and the absolute minimum is
27. h(x) œ $Èx œ x"Î$ Ê hw (x) œ
" 3
" #
at x œ 2
x#Î$ Ê a critical point
at x œ 0; h(1) œ 1, h(0) œ 0, h(8) œ 2 Ê the absolute maximum is 2 at x œ 8 and the absolute minimum is 1 at x œ 1
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Chapter 4 Applications of Derivatives
28. h(x) œ 3x#Î$ Ê hw (x) œ #x"Î$ Ê a critical point at x œ 0; h(1) œ 3, h(0) œ 0, h(1) œ 3 Ê the absolute maximum is 0 at x œ 0 and the absolute minimum is 3 at x œ 1 and at x œ 1
29. g(x) œ È4 x# œ a4 x# b Ê gw (x) œ
" #
a4 x# b
"Î#
"Î#
(2x) œ
x È 4 x#
Ê critical points at x œ 2 and x œ 0, but not at x œ 2 because 2 is not in the domain; g(2) œ 0, g(0) œ 2, g(1) œ È3 Ê the absolute maximum is 2 at x œ 0 and the absolute minimum is 0 at x œ 2 30. g(x) œ È5 x# œ a& x# b a5 x# b (2x) x " w Ê g (x) œ ˆ # ‰ œ È # Ê critical points at x œ È5 "Î#
"Î#
&x
and x œ 0, but not at x œ È5 because È5 is not in the domain; f ŠÈ5‹ œ 0, f(0) œ È5 Ê the absolute maximum is 0 at x œ È5 and the absolute minimum is È5 at x œ 0
31. f()) œ sin ) Ê f w ()) œ cos ) Ê ) œ
1 #
is a critical point,
is not a critical point because #1 is not interior to the domain; f ˆ #1 ‰ œ 1, f ˆ 1# ‰ œ 1, f ˆ 561 ‰ œ "# but ) œ
1 #
Ê the absolute maximum is 1 at ) œ
minimum is 1 at ) œ
1 #
1 #
and the absolute
32. f()) œ tan ) Ê f w ()) œ sec# ) Ê f has no critical points in 1‰ ˆ 1 3 ß 4 . The extreme values therefore occur at the ‰ œ È3 and f ˆ 14 ‰ œ 1 Ê the absolute endpoints: f ˆ 1 3 maximum is 1 at ) œ 14 and the absolute minimum is È3 at ) œ 1 3
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Section 4.1 Extreme Values of Functions
171
33. g(x) œ csc x Ê gw (x) œ (csc x)(cot x) Ê a critical point at x œ 1# ; g ˆ 13 ‰ œ È23 , g ˆ 1# ‰ œ 1, g ˆ 231 ‰ œ È23 Ê the absolute maximum is
at x œ
2 È3
absolute minimum is 1 at x œ
1 3
and x œ
21 3 ,
and the
1 #
34. g(x) œ sec x Ê gw (x) œ (sec x)(tan x) Ê a critical point at x œ 0; g ˆ 13 ‰ œ 2, g(0) œ 1, g ˆ 16 ‰ œ È23 Ê the absolute maximum is 2 at x œ 13 and the absolute minimum is 1 at x œ 0
35. f(t) œ 2 ktk œ # Èt# œ # at# b Ê f w (t) œ "# at# b
"Î#
"Î#
(2t) œ Èt # œ kttk t
Ê a critical point at t œ 0; f(1) œ 1, f(0) œ 2, f(3) œ 1 Ê the absolute maximum is 2 at t œ 0 and the absolute minimum is 1 at t œ 3
36. f(t) œ kt 5k œ È(t 5)# œ a(t 5)# b œ
" #
a(t 5)# b
"Î#
(2(t 5)) œ
t5 È(t 5)#
"Î#
œ
Ê f w (t) t5 kt 5 k
Ê a critical point at t œ 5; f(4) œ 1, f(5) œ 0, f(7) œ 2 Ê the absolute maximum is 2 at t œ 7 and the absolute minimum is 0 at t œ 5 37. f(x) œ x%Î$ Ê f w (x) œ
4 3
x"Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 16 Ê the absolute
maximum is 16 at x œ 8 and the absolute minimum is 0 at x œ 0 38. f(x) œ x&Î$ Ê f w (x) œ
5 3
x#Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 32 Ê the absolute
maximum is 32 at x œ 8 and the absolute minimum is 1 at x œ 1 39. g()) œ )$Î& Ê gw ()) œ
3 5
)#Î& Ê a critical point at ) œ 0; g(32) œ 8, g(0) œ 0, g(1) œ 1 Ê the absolute
maximum is 1 at ) œ 1 and the absolute minimum is 8 at ) œ 32 40. h()) œ 3)#Î$ Ê hw ()) œ 2)"Î$ Ê a critical point at ) œ 0; h(27) œ 27, h(0) œ 0, h(8) œ 12 Ê the absolute maximum is 27 at ) œ 27 and the absolute minimum is 0 at ) œ 0 41. y œ x2 6x 7 Ê y w œ 2x 6 Ê 2x 6 œ 0 Ê x œ 3. The critical point is x œ 3. 42. faxb œ 6x2 x3 Ê f w axb œ 12x 3x2 Ê 12x 3x2 œ 0 Ê 3xa4 xb œ 0 Ê x œ 0 or x œ 4. The critical pointss are x œ 0 and x œ 4.
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172
Chapter 4 Applications of Derivatives
43. faxb œ xa4 xb3 Ê f w axb œ x3a4 xb2 a1b‘ a4 xb3 œ a4 xb2 3x a4 xb‘ œ a4 xb2 a4 4xb œ 4a4 xb2 a1 xb Ê 4a4 xb2 a1 xb œ 0 Ê x œ 1 or x œ 4. The critical points are x œ 1 and x œ 4. 44. gaxb œ ax 1b2 ax 3b2 Ê g w axb œ ax 1b2 † 2ax 3ba1b 2ax 1ba1b † ax 3b2 œ 2ax 3bax 1bax 1b ax 3b‘ œ 4ax 3bax 1bax 2b Ê 4ax 3bax 1bax 2b œ 0 Ê x œ 3 or x œ 1 or x œ 2. The critical points are x œ 1, x œ 2, and x œ 3. 45. y œ x2
2 x
Ê y w œ 2x
2 x2
œ
2x3 2 x2
2x3 2 x2
Ê
œ ! Ê 2x3 2 œ ! Ê x œ 1;
2x3 2 x2
œ undefined Ê x2 œ 0 Ê x œ 0.
The domain of the function is a_, 0b a0, _b, thus x œ 0 is not in the domain, so the only critical point is x œ 1. 46. faxb œ
Ê f w axb œ
x2 x2
ax 2†b2x x2 a1b ax 2 b 2
œ
x2 4x ax 2 b 2
Ê
x2 4x ax 2 b 2
œ ! Ê x2 4x œ ! Ê x œ 0 or x œ 4;
x2 4x ax 2 b 2
œ undefined
2
Ê ax 2b œ 0 Ê x œ 2. The domain of the function is a_, 2b a2, _b, thus x œ 2 is not in the domain, so the only critical points are x œ 0 and x œ 4 47. y œ x2 32Èx Ê y w œ 2x
16 Èx
œ
2x3Î2 16 Èx
Ê
2x3Î2 16 Èx
œ ! Ê 2x3Î2 16 œ 0 Ê x œ 4;
2x3Î2 16 Èx
œ undefined
Ê Èx œ 0 Ê x œ 0. The critical points are x œ 4 and x œ 0. 48. gaxb œ È2x x2 Ê g w axb œ
1x È2x x2
Ê
1x È2x x2
œ 0 Ê 1 x œ 0 Ê x œ 1;
1x È2x x2
œ undefined Ê È2x x2 œ 0
2x x2 œ 0 Ê x œ 0 or x œ 2. The critical points are x œ 0, x œ 1, and x œ 2. 49. Minimum value is 1 at x œ 2.
50. To find the exact values, note that yw œ 3x2 2, which is zero when x œ „ É 23 . Local maximum at ŠÉ 23 ß 4
4È 6 9 ‹
¸ a0Þ816ß 5Þ089b; local
minimum at ŠÉ #$ ß %
%È ' * ‹
¸ a0.816ß 2.911b
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Section 4.1 Extreme Values of Functions 51. To find the exact values, note that that yw œ 3x2 2x 8 œ a3x 4bax 2b, which is zero when x œ 2 or x œ %$ . ‰ Local maximum at a2ß 17b; local minimum at ˆ %$ ß %" #(
52. Note that yw œ 5x# ax 5bax 3b, which is zero at x œ 0, x œ 3, and x œ 5. Local maximum at a3, 108b; local minimum at a5, 0b; a0, 0b is neither a maximum nor a minimum.
53. Minimum value is 0 when x œ " or x œ ".
54. Note that yw œ
Èx 2 Èx ,
which is zero at x œ 4 and is
undefined when x œ 0. Local maximum at a0, 0b; absolute minimum at a4, 4b
55. The actual graph of the function has asymptotes at x œ „ ", so there are no extrema near these values. (This is an example of grapher failure.) There is a local minimum at a!ß "b.
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173
174
Chapter 4 Applications of Derivatives
56. Maximum value is 2 at x œ "; minimum value is 0 at x œ " and x œ $.
" # at x œ "à "# as x œ ".
57. Maximum value is minimum value is
" # at x œ 0à "# as x œ 2.
58. Maximum value is minimum value is
59. yw œ x#Î$ a"b #$ x"Î$ ax #b œ crit. pt. x œ %& xœ!
derivative ! undefined
&x % $ x $È
extremum local max local min
value "# "Î$ œ "Þ!$% #& "! 0
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Section 4.1 Extreme Values of Functions 60. yw œ x#Î$ a#xb #$ x"Î$ ax# %b œ crit. pt. x œ " xœ! xœ"
derivative ! undefined !
extremum minimum local max minimum
" a #xb a"bÈ% #È % x # x# a% x# b % #x # œÈ È % x# % x#
61. yw œ x œ
crit. pt. x œ # x œ È # x œ È# xœ#
)x# ) $ x $È
derivative undefined ! ! undefined
value $ 0 $
x#
extremum local max minimum maximum local min
value ! # # !
62. yw œ x# #È$" x a 1b #xÈ$ x œ
x# a%xba$ xb #È $ x
crit. pt. xœ0 x œ "# & xœ$
œ
_5x# "#x #È $ x
derivative ! ! undefined
extremum minimum local max minimum
value ! "%% "Î# ¸ %Þ%'# "#& "& !
extremum minimum
value #
#, x " 63. yw œ œ ", x " crit. pt. xœ"
64. yw œ œ crit. pt. xœ! xœ"
derivative undefined
", # #x,
x! x!
derivative undefined !
extremum local min local max
value $ %
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175
176
Chapter 4 Applications of Derivatives
2x 2, 65. yw œ œ 2x 6, crit. pt. x œ 1 xœ1 xœ3
x1 x1
derivative ! undefined !
extremum maximum local min maximum
value 5 1 5
"% x# "# x "& % , xŸ" x$ 'x# )x, x" w w # if x ", and limc f a" hb œ ". Also, f axb œ $x "#x ) if x ", and
66. We begin by determining whether f w axb is defined at x œ ", where faxb œ œ Clearly, f w axb œ "# x
" #
hÄ!
limb f w a" hb œ ". Since f is continuous at x œ ", we have that f w a"b œ ". Thus,
hÄ!
f w axb œ œ
"# x "# , $x "#x ) , #
Note that "# x But #
#È $ $
crit. pt. x œ " x ¸ $Þ"&&
" #
xŸ" x"
œ ! when x œ ", and $x# "#x ) œ ! when x œ
¸ !Þ)%& ", so the critical points occur at x œ " and x œ derivative ! !
extremum local max local min
È "# „ È"## %a$ba)b œ "# „' %) #a$b È # # $ $ ¸ $Þ"&&.
œ#„
#È$ $ .
value 4 ¸ $Þ!(*
67. (a) No, since f w axb œ #$ ax #b"Î$ , which is undefined at x œ #.
(b) The derivative is defined and nonzero for all x Á #. Also, fa#b œ ! and faxb ! for all x Á #. (c) No, faxb need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form Òa, bÓ would have both a maximum value and minimum value on the interval. (d) The answers are the same as (a) and (b) with 2 replaced by a. x$ *x, x Ÿ $ or ! Ÿ x $ $x$ *, x $ or ! x $ w . Therefore, f a x b œ . œ x$ *x, $ x ! or x $ $x$ *, $ x ! or x $ No, since the left- and right-hand derivatives at x œ !, are * and *, respectively. No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively. No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively. The critical points occur when f w axb œ ! (at x œ „ È$) and when f w axb is undefined (at x œ ! and x œ „ $). The
68. Note that faxb œ œ (a) (b) (c) (d)
minimum value is ! at x œ $, at x œ !, and at x œ $; local maxima occur at ŠÈ$ß 'È$‹ and ŠÈ$ß 'È$‹.
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Section 4.1 Extreme Values of Functions 69. Yes, since f(x) œ kxk œ Èx# œ ax# b
"Î#
Ê f w (x) œ
" #
ax# b
"Î#
(2x) œ
x ax# b"Î#
œ
x kx k
177
is not defined at x œ 0. Thus it is
not required that f w be zero at a local extreme point since f w may be undefined there. 70. If f(c) is a local maximum value of f, then f(x) Ÿ f(c) for all x in some open interval (aß b) containing c. Since f is even, f(x) œ f(x) Ÿ f(c) œ f(c) for all x in the open interval (bß a) containing c. That is, f assumesa local maximum at the point c. This is also clear from the graph of f because the graph of an even function is symmetric about the y-axis. 71. If g(c) is a local minimum value of g, then g(x) g(c) for all x in some open interval (aß b) containing c. Since g is odd, g(x) œ g(x) Ÿ g(c) œ g(c) for all x in the open interval (bß a) containing c. That is, g assumes a local maximum at the point c. This is also clear from the graph of g because the graph of an odd function is symmetric about the origin. 72. If there are no boundary points or critical points the function will have no extreme values in its domain. Such functions do indeed exist, for example f(x) œ x for _ x _. (Any other linear function f(x) œ mx b with m Á 0 will do as well.) 73. (a) Vaxb œ "'!x &#x# %x$ Vw axb œ "'! "!%x "#x# œ %ax #ba$x #!b The only critical point in the interval a!ß &b is at x œ #. The maximum value of Vaxb is 144 at x œ #. (b) The largest possible volume of the box is 144 cubic units, and it occurs when x œ # units. 74. (a) f w axb œ $ax# #bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The function faxb œ x$ $x has two critical points at x œ " and x œ ". The function faxb œ x$ " has one critical point at x œ !Þ The function faxb œ x$ x has no critical points.
(b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the cubic function has no extreme values.) 75. s œ "# gt# v! t s! Ê
ds dt
œ gt v! œ ! Ê t œ
2
Thus sŠ vg! ‹ œ "# gŠ vg! ‹ v0 Š vg! ‹ s0 œ 76.
Now satb œ s0 Í tˆ gt2 v0 ‰ œ 0 Í t œ 0 or t œ
s0 s0 is the maximum height over the interval 0 Ÿ t Ÿ
œ ! Ê tan t œ " Ê t œ never negative) Ê the peak current is #È# amps. dI dt
œ #sin t #cos t, solving
v!2 2g
v! g.
dI dt
1 %
2v0 g . 2v0 g .
n1 where n is a nonnegative integer (in this exercise t is
77. Maximum value is 11 at x œ 5; minimum value is 5 on the interval Ò3ß 2Ó; local maximum at a5ß 9b
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178
Chapter 4 Applications of Derivatives
78. Maximum value is 4 on the interval Ò5ß 7Ó; minimum value is 4 on the interval Ò2ß 1Ó.
79. Maximum value is 5 on the interval Ò3ß _Ñ; minimum value is 5 on the interval Ð_ß 2Ó.
80. Minimum value is 4 on the interval Ò"ß $Ó
81-86. Example CAS commands: Maple: with(student): f := x -> x^4 - 8*x^2 + 4*x + 2; domain := x=-20/25..64/25; plot( f(x), domain, color=black, title="Section 4.1 #81(a)" ); Df := D(f); plot( Df(x), domain, color=black, title="Section 4.1 # 81(b)" ) StatPt := fsolve( Df(x)=0, domain ) SingPt := NULL; EndPt := op(rhs(domain)); Pts :=evalf([EndPt,StatPt,SingPt]); Values := [seq( f(x), x=Pts )]; Maximum value is 2.7608 and occurs at x=2.56 (right endpoint). %
Minimum value $ is -6.2680 and occurs at x=1.86081 (singular point). Mathematica: (functions may vary) (see section 2.5 re. RealsOnly ): <<Miscellaneous `RealOnly` Clear[f,x] a = 1; b = 10/3;
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Section 4.2 The Mean Value Theorem
179
f[x_] =2 2x 3 x2/3 f'[x] Plot[{f[x], f'[x]}, {x, a, b}] NSolve[f'[x]==0, x] {f[a], f[0], f[x]/.%, f[b]//N In more complicated expressions, NSolve may not yield results. In this case, an approximate solution (say 1.1 here) is observed from the graph and the following command is used: FindRoot[f'[x]==0,{x, 1.1}] 4.2 THE MEAN VALUE THEOREM 1. When f(x) œ x# 2x 1 for 0 Ÿ x Ÿ 1, then 2. When f(x) œ x#Î$ for 0 Ÿ x Ÿ 1, then 3. When f(x) œ x
" x
for
" #
fa1b fa0b 10
Ÿ x Ÿ 2, then
fa3b fa1b 31
5. When f(x) œ x3 x2 for 1 Ÿ x Ÿ 2, then ¸ 1.22 and
1 È7 3
œ f w acb Ê 3 œ 2c 2 Ê c œ #" .
œ f w acb Ê 1 œ ˆ 32 ‰ c"Î$ Ê c œ
fa2b fa1Î2b 2 1 Î2
4. When f(x) œ Èx 1 for 1 Ÿ x Ÿ 3, then
1 È7 3
fa1b fa0b 10
œ f w acb Ê 0 œ " œ f w acb Ê
fa2b fa1b 2 a1b
È2 #
œ
" c#
8 #7 .
Ê c œ 1.
" #È c 1
Ê c œ #3 .
œ f w acb Ê 2 œ 3c2 2c Ê c œ
1 „ È7 . 3
¸ 0.549 are both in the interval 1 Ÿ x Ÿ 2.
x3 2 Ÿ x Ÿ 0 g a 2 b w w w 2 w , then ga22ba 2b œ g acb Ê 3 œ g acb. If 2 Ÿ x 0, then g (x) œ 3x Ê 3 œ g acb x2 ! x Ÿ 2 Ê 3c2 œ 3 Ê c œ „ 1. Only c œ 1 is in the interval. If ! x Ÿ 2, then gw (x) œ 2x Ê 3 œ g w acb Ê 2c œ 3 Ê c œ 32 .
6. When g(x) œ œ
7. Does not; f(x) is not differentiable at x œ 0 in ("ß 8). 8. Does; f(x) is continuous for every point of [0ß 1] and differentiable for every point in (0ß 1). 9. Does; f(x) is continuous for every point of [0ß 1] and differentiable for every point in (0ß 1). 10. Does not; f(x) is not continuous at x œ 0 because lim c f(x) œ 1 Á 0 œ f(0). xÄ! 11. Does not; f is not differentiable at x œ 1 in (2ß 0). 12. Does; f(x) is continuous for every point of [0ß 3] and differentiable for every point in (0ß 3). 13. Since f(x) is not continuous on 0 Ÿ x Ÿ 1, Rolle's Theorem does not apply:
lim f(x) œ lim c x œ 1 Á 0 œ f(1).
x Ä 1c
xÄ1
14. Since f(x) must be continuous at x œ 0 and x œ 1 we have lim b f(x) œ a œ f(0) Ê a œ 3 and xÄ! lim c f(x) œ lim b f(x) Ê 1 3 a œ m b Ê 5 œ m b. Since f(x) must also be differentiable at xÄ1
xÄ1
x œ 1 we have lim c f w (x) œ lim b f w (x) Ê 2x 3k x=1 œ mk x=1 Ê 1 œ m. Therefore, a œ 3, m œ 1 and b œ 4. xÄ1 xÄ1
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180
Chapter 4 Applications of Derivatives
15. (a) i ii iii iv (b) Let r" and r# be zeros of the polynomial P(x) œ xn an-1 xn-1 á a" x a! , then P(r" ) œ P(r# ) œ 0. Since polynomials are everywhere continuous and differentiable, by Rolle's Theorem Pw (r) œ 0 for some r between r" and r# , where Pw (x) œ nxn-1 (n 1) an-1 xn-2 á a" . 16. With f both differentiable and continuous on [aß b] and f(r" ) œ f(r# ) œ f(r$ ) œ 0 where r" , r# and r$ are in [aß b], then by Rolle's Theorem there exists a c" between r" and r# such that f w (c" ) œ 0 and a c# between r# and r$ such that f w (c# ) œ 0. Since f w is both differentiable and continuous on [aß b], Rolle's Theorem again applies and we have a c$ between c" and c# such that f ww (c$ ) œ 0. To generalize, if f has n1 zeros in [aß b] and f ÐnÑ is continuous on [aß b], then f ÐnÑ has at least one zero between a and b. 17. Since f ww exists throughout [aß b] the derivative function f w is continuous there. If f w has more than one zero in [aß b], say f w (r" ) œ f w (r# ) œ 0 for r" Á r# , then by Rolle's Theorem there is a c between r" and r# such that f ww (c) œ 0, contrary to f ww 0 throughout [aß b]. Therefore f w has at most one zero in [aß b]. The same argument holds if f ww 0 throughout [aß b]. 18. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f w (x) has three or more zeros, f ww (x) has 2 or more zeros and f www (x) has at least one zero. This is a contradiction since f www (x) is a non-zero constant when f(x) is a cubic polynomial. 19. With f(2) œ 11 0 and f(1) œ 1 0 we conclude from the Intermediate Value Theorem that f(x) œ x% 3x 1 has at least one zero between 2 and 1. Then 2 x 1 Ê ) x$ 1 Ê 32 4x$ 4 Ê 29 4x$ 3 1 Ê f w (x) 0 for 2 x 1 Ê f(x) is decreasing on [#ß 1] Ê f(x) œ 0 has exactly one solution in the interval (#ß 1). 20. f(x) œ x$
4 x#
7 Ê f w (x) œ 3x#
8 x$
0 on (_ß 0) Ê f(x) is increasing on (_ß 0). Also, f(x) 0 if x 2 and
f(x) 0 if 2 x 0 Ê f(x) has exactly one zero in (_ß !). 21. g(t) œ Èt Èt 1 4 Ê gw (t) œ
" #È t
" 2Èt1
0 Ê g(t) is increasing for t in (!ß _); g(3) œ È3 2 0 and
g(15) œ È15 0 Ê g(t) has exactly one zero in (!ß _)Þ 22. g(t) œ
" "t
È1 t 3.1 Ê gw (t) œ
" ("t)#
" 2 È 1 t
0 Ê g(t) is increasing for t in (1ß 1); g(0.99) œ 2.5 and
g(0.99) œ 98.3 Ê g(t) has exactly one zero in (1ß 1). 23. r()) œ ) sin# ˆ 3) ‰ 8 Ê rw ()) œ 1 32 sin ˆ 3) ‰ cos ˆ 3) ‰ œ 1 "3 sin ˆ 23) ‰ 0 on (_ß _) Ê r()) is increasing on (_ß _); r(0) œ 8 and r(8) œ sin# ˆ 83 ‰ 0 Ê r()) has exactly one zero in (_ß _). 24. r()) œ 2) cos# ) È2 Ê rw ()) œ 2 2 sin ) cos ) œ 2 sin 2) 0 on (_ß _) Ê r()) is increasing on (_ß _); r(#1) œ 41 cos (#1) È2 œ 41 1 È2 0 and r(21) œ 41 1 È2 0 Ê r()) has exactly one zero in (_ß _). 5 Ê rw ()) œ (sec ))(tan )) )3% 0 on ˆ!ß 1# ‰ Ê r()) is increasing on ˆ!ß 1# ‰ ; r(0.1) ¸ 994 and r(1.57) ¸ 1260.5 Ê r()) has exactly one zero in ˆ!ß 1# ‰ .
25. r()) œ sec )
" )$
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Section 4.2 The Mean Value Theorem
181
26. r()) œ tan ) cot ) ) Ê rw ()) œ sec# ) csc# ) 1 œ sec# ) cot# ) 0 on ˆ!ß 1# ‰ Ê r()) is increasing on ˆ0ß 1# ‰ ; r ˆ 14 ‰ œ 14 0 and r(1.57) ¸ 1254.2 Ê r()) has exactly one zero in ˆ!ß 1# ‰ . 27. By Corollary 1, f w (x) œ 0 for all x Ê f(x) œ C, where C is a constant. Since f(1) œ 3 we have C œ 3 Ê f(x) œ 3 for all x. 28. g(x) œ 2x 5 Ê gw (x) œ 2 œ f w (x) for all x. By Corollary 2, f(x) œ g(x) C for some constant C. Then f(0) œ g(0) C Ê 5 œ 5 C Ê C œ 0 Ê f(x) œ g(x) œ 2x 5 for all x. 29. g(x) œ x# Ê gw (x) œ 2x œ f w (x) for all x. By Corollary 2, f(x) œ g(x) C. (a) f(0) œ 0 Ê 0 œ g(0) C œ 0 C Ê C œ 0 Ê f(x) œ x# Ê f(2) œ 4 (b) f(1) œ 0 Ê 0 œ g(1) C œ 1 C Ê C œ 1 Ê f(x) œ x# 1 Ê f(2) œ 3 (c) f(2) œ 3 Ê 3 œ g(2) C Ê 3 œ 4 C Ê C œ 1 Ê f(x) œ x# 1 Ê f(2) œ 3 30. g(x) œ mx Ê gw (x) œ m, a constant. If f w (x) œ m, then by Corollary 2, f(x) œ g(x) b œ mx b where b is a constant. Therefore all functions whose derivatives are constant can be graphed as straight lines y œ mx b. 31. (a) y œ
x# #
C
(b) y œ
32. (a) y œ x# C
x$ 3
C
(b) y œ x# x C
33. (a) yw œ x# Ê y œ
" x
C
(b) y œ x
" x
35. (a) y œ "# cos 2t C (c) y œ
cos 2t 2 sin
(c) y œ 5x
(b) y œ 2 sin t #
C
" x
C
(b) y œ 2Èx C
" #
" #
x% 4
(c) y œ x$ x# x C
C
x"Î# Ê y œ x"Î# C Ê y œ Èx C (c) y œ 2x# 2Èx C
34. (a) yw œ
(c) y œ
C
t #
C (b) yw œ )"Î# Ê y œ
36. (a) y œ tan ) C
2 3
)$Î# C
(c) y œ
2 3
)$Î# tan ) C
37. f(x) œ x# x C; 0 œ f(0) œ 0# 0 C Ê C œ 0 Ê f(x) œ x# x 38. g(x) œ "x x# C; 1 œ g(1) œ "1 (1)# C Ê C œ 1 Ê g(x) œ x" x# 1 39. r()) œ 8) cot ) C; 0 œ r ˆ 14 ‰ œ 8 ˆ 14 ‰ cot ˆ 14 ‰ C Ê 0 œ 21 1 C Ê C œ 21 1 Ê r()) œ 8) cot ) 21 1 40. r(t) œ sec t t C; 0 œ r(0) œ sec (0) 0 C Ê C œ 1 Ê r(t) œ sec t t 1 41. v œ
ds dt
œ *Þ)t & Ê s œ %Þ*t# &t C; at s œ "! and t œ ! we have C œ "! Ê s œ %Þ*t# &t "!
42. v œ
ds dt
œ $#t # Ê s œ "'t# #t C; at s œ % and t œ
43. v œ
ds dt
œ sina1tb Ê s œ 1" cosa1tb C; at s œ ! and t œ ! we have C œ
44. v œ
ds dt
œ 12 cosˆ #1t ‰ Ê s œ sinˆ #1t ‰ C; at s œ " and t œ 1# we have C œ " Ê s œ sinˆ #1t ‰ "
" #
we have C œ " Ê s œ "'t# #t " " 1
Êsœ
" cosa1tb 1
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182
Chapter 4 Applications of Derivatives
45. a œ $# Ê v œ $#t C" ; at v œ #! and t œ ! we have C" œ #! Ê v œ $#t #! Ê s œ "'t# #!t C# ; at s œ & and t œ ! we have C# œ & Ê s œ "'t# #!t & 46. a œ 9.8 Ê v œ 9.8t C" ; at v œ $ and t œ ! we have C" œ $ Ê v œ *Þ)t $ Ê s œ %Þ*t# $t C# ; at s œ ! and t œ ! we have C# œ ! Ê s œ %Þ*t# $t 47. a œ %sina#tb Ê v œ #cosa#tb C" ; at v œ # and t œ ! we have C" œ ! Ê v œ #cosa#tb Ê s œ sina#tb C# ; at s œ $ and t œ ! we have C# œ $ Ê s œ sina#tb $ Ê v œ 1$ sinˆ $1t ‰ C" ; at v œ ! and t œ ! we have C" œ ! Ê v œ 1$ sinˆ $1t ‰ Ê s œ cosˆ $1t ‰ C# ; at s œ " and t œ ! we have C# œ ! Ê s œ cosˆ $1t ‰
48. a œ
* ˆ $t ‰ 1# cos 1
49. If T(t) is the temperature of the thermometer at time t, then T(0) œ 19° C and T(14) œ 100° C. From the Mean Value Theorem there exists a 0 t! 14 such that
T(14) T(0) 14 0
œ 8.5° C/sec œ Tw (t! ), the rate at which the temperature was
changing at t œ t! as measured by the rising mercury on the thermometer. 50. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been going that speed at least once during the trip. 51. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip. 52. The runner's average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value Theorem, the runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 0 mph and the runner's speed is continuous, by the Intermediate Value Theorem, the runner's speed must have been 11 mph at least twice. 53. Let d(t) represent the distance the automobile traveled in time t. The average speed over 0 Ÿ t Ÿ 2 is w
Value Theorem says that for some 0 t! 2, d (t! ) œ
d(2) d(0) #0 .
d(2) d(0) #0 .
The Mean
w
The value d (t! ) is the speed of the automobile at time t!
(which is read on the speedometer). 54. aatb œ vw atb œ "Þ' Ê vatb œ "Þ't C; at a!ß !b we have C œ ! Ê vatb œ "Þ't. When t œ $!, then va$!b œ %) m/sec. 55. The conclusion of the Mean Value Theorem yields 56. The conclusion of the Mean Value Theorem yields
" b
"a ba
b # a# ba
b‰ œ c"# Ê c# ˆ a ab œ a b Ê c œ Èab.
œ 2c Ê c œ
a b # .
57. f w (x) œ [cos x sin (x 2) sin x cos (x 2)] 2 sin (x 1) cos (x 1) œ sin (x x 2) sin 2(x 1) œ sin (2x 2) sin (2x 2) œ 0. Therefore, the function has the constant value f(0) œ sin# 1 ¸ 0.7081 which explains why the graph is a horizontal line. 58. (a) faxb œ ax #bax "bxax "bax #b œ x& &x$ %x is one possibility.
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Section 4.2 The Mean Value Theorem
183
(b) Graphing faxb œ x& &x$ %x and f w axb œ &x% "&x# % on Ò$ß $Ó by Ò(ß (Ó we see that each x-intercept of f w axb lies between a pair of x-intercepts of faxb, as expected by Rolle's Theorem.
(c) Yes, since sin is continuous and differentiable on a _ß _b. 59. faxb must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that faxb is zero twice between a and b. Then by the Mean Value Theorem, f w axb would have to be zero at least once between the two zeros of faxb, but this can't be true since we are given that f w axb Á ! on this interval. Therefore, faxb is zero once and only once between a and b. 60. Consider the function kaxb œ faxb gaxb. kaxb is continuous and differentiable on Òa, bÓ, and since kaab œ faab gaab and kabb œ fabb gabb, by the Mean Value Theorem, there must be a point c in aa, bb where kw acb œ !. But since kw acb œ f w acb gw acb, this means that f w acb œ gw acb, and c is a point where the graphs of f and g have tangent lines with the same slope, so these lines are either parallel or are the same line. 61. f w axb Ÿ 1 for 1 Ÿ x Ÿ 4 Ê faxb is differentiable on 1 Ÿ x Ÿ 4 Ê f is continuous on 1 Ÿ x Ÿ 4 Ê f satisfies the conditions of the Mean Value Theorem Ê Ê fa4b fa1b Ÿ 3 62. 0 f w axb
" #
fa4b fa1b 41
œ f w acb for some c in 1 x 4 Ê f w acb Ÿ 1 Ê
fa4b fa1b 3
Ÿ1
for all x Ê f w axb exists for all x, thus f is differentiable on a1, 1b Ê f is continuous on Ò1, 1Ó
Ê f satisfies the conditions of the Mean Value Theorem Ê
fa1b fa1b 1 a1b
œ f w acb for some c in Ò1, 1Ó Ê 0
fa1b fa1b 2
Ê 0 fa1b fa1b 1. Since fa1b fa1b 1 Ê fa1b 1 fa1b 2 fa1b, and since 0 fa1b fa1b we have fa1b fa1b. Together we have fa1b fa1b 2 fa1b. 63. Let fatb œ cos t and consider the interval Ò0, xÓ where x is a real number. f is continuous on Ò0, xÓ and f is differentiable on a0, xb since f w atb œ sin t Ê f satisfies the conditions of the Mean Value Theorem Ê Ò0, xÓ Ê
cos x 1 x
œ sin c. Since 1 Ÿ sin c Ÿ 1 Ê 1 Ÿ sin c Ÿ 1 Ê 1 Ÿ
Ê x Ÿ cos x 1 Ÿ x Ê lcos x 1l Ÿ x œ l x l. If x 0, 1 Ÿ
cos x 1 x
cos x 1 x
faxb fa0b x a0 b
œ f w acb for some c in
Ÿ 1. If x 0, 1 Ÿ
cos x 1 x
Ÿ 1 Ê x cos x 1 x
Ê x Ÿ cos x 1 Ÿ x Ê axb Ÿ cos x 1 Ÿ x Ê lcos x 1l Ÿ x œ l x l. Thus, in both cases, we have lcos x 1l Ÿ l x l. If x œ 0, then lcos 0 1l œ l1 1l œ l0l Ÿ l0l, thus lcos x 1l Ÿ l x l is true for all x. 64. Let f(x) œ sin x for a Ÿ x Ÿ b. From the Mean Value Theorem there exists a c between a and b such that sin b sin a sin a sin a ¸ œ cos c Ê 1 Ÿ sin bb Ÿ 1 Ê ¸ sin bb Ÿ 1 Ê ksin b sin ak Ÿ kb ak. ba a a 65. Yes. By Corollary 2 we have f(x) œ g(x) c since f w (x) œ gw (x). If the graphs start at the same point x œ a, then f(a) œ g(a) Ê c œ 0 Ê f(x) œ g(x).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ÿ1
" #
184
Chapter 4 Applications of Derivatives
66. Assume f is differentiable and lfawb faxbl Ÿ lw xl for all values of w and x. Since f is differentiable, f w axb exists and f w axb œ lim wÄx
fawb faxb wx
using the alternative formula for the derivative. Let gaxb œ lxl, which is continuous for all x.
By Theorem 10 from Chapter 2, ¸f w axb¸ œ º lim wÄx lfawb faxbl Ÿ lw xl for all w and x Ê ¸f w axb¸ œ lim wÄx
lfawb faxbl lw x l
fawb faxb wx º
lfawbfaxbl lw x l
faxb œ lim ¹ fawwb x ¹ œ wlim wÄx Äx
lfawb faxbl lw x l .
Since
Ÿ 1 as long as w Á x. By Theorem 5 from Chapter 2,
Ÿ lim 1 œ 1 Ê ¸f w axb¸ Ÿ 1 Ê 1 Ÿ f w axb Ÿ 1. wÄx
67. By the Mean Value Theorem we have we have f(b) f(a) 0 Ê f w (c) 0.
f(b) f(a) ba
œ f w (c) for some point c between a and b. Since b a 0 and f(b) f(a),
68. The condition is that f w should be continuous over [aß b]. The Mean Value Theorem then guarantees the existence of a point c in (aß b) such that w
f(b) f(a) ba w
œ f w (c). If f w is continuous, then it has a minimum and
maximum value on [aß b], and min f Ÿ f (c) Ÿ max f w , as required. 69. f w (x) œ a1 x% cos xb
"
Ê f ww (x) œ a1 x% cos xb
#
a4x$ cos x x% sin xb
#
œ x$ a1 x% cos xb (4 cos x x sin x) 0 for 0 Ÿ x Ÿ 0.1 Ê f w (x) is decreasing when 0 Ÿ x Ÿ 0.1 f(0.1) " 0.1
Ê min f w ¸ 0.9999 and max f w œ 1. Now we have 0.9999 Ÿ
Ÿ 1 Ê 0.09999 Ÿ f(0.1) 1 Ÿ 0.1
Ê 1.09999 Ÿ f(0.1) Ÿ 1.1. 70. f w (x) œ a1 x% b
"
Ê f ww (x) œ a1 x% b
#
a4x$ b œ
4x$ $ a1 x % b
0 for 0 x Ÿ 0.1 Ê f w (x) is increasing when
0 Ÿ x Ÿ 0.1 Ê min f w œ 1 and max f w œ 1.0001. Now we have 1 Ÿ
f(0.1) 2 0.1
Ÿ 1.0001
Ê 0.1 Ÿ f(0.1) 2 Ÿ 0.10001 Ê 2.1 Ÿ f(0.1) Ÿ 2.10001. 71. (a) Suppose x 1, then by the Mean Value Theorem (b)
f(1) Mean Value Theorem f(x)x 0 1 f(x) f(1) Yes. From part (a), lim c x 1 xÄ1
f(x) f(1) x1
0 Ê f(x) f(1). Suppose x 1, then by the
Ê f(x) f(1). Therefore f(x) 1 for all x since f(1) œ 1.
f(1) Ÿ 0 and lim b f(x)x 0. Since f w (1) exists, these two one-sided 1 xÄ1 limits are equal and have the value f w (1) Ê f w (1) Ÿ 0 and f w (1) 0 Ê f w (1) œ 0.
72. From the Mean Value Theorem we have
f(b) f(a) ba
œ f w (c) where c is between a and b. But f w (c) œ 2pc q œ 0
has only one solution c œ #qp . (Note: p Á 0 since f is a quadratic function.) 4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST 1. (a) f w (x) œ x(x 1) Ê critical points at 0 and 1 (b) f w œ ± ± Ê increasing on (_ß !) and ("ß _), decreasing on (!ß ") ! " (c) Local maximum at x œ 0 and a local minimum at x œ 1 2. (a) f w (x) œ (x 1)(x 2) Ê critical points at 2 and 1 (b) f w œ ± ± Ê increasing on (_ß #) and ("ß _), decreasing on (2ß ") # " (c) Local maximum at x œ 2 and a local minimum at x œ 1 3. (a) f w (x) œ (x 1)# (x 2) Ê critical points at 2 and 1 (b) f w œ ± ± Ê increasing on (2ß 1) and ("ß _), decreasing on (_ß 2) # "
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.3 Monotonic Functions and the First Derivative Test (c) No local maximum and a local minimum at x œ 2 4. (a) f w (x) œ (x 1)# (x 2)# Ê critical points at 2 and 1 (b) f w œ ± ± Ê increasing on (_ß 2) (#ß ") ("ß _), never decreasing # " (c) No local extrema 5. (a) f w (x) œ (x 1)(x 2)(x 3) Ê critical points at 2, 1 and 3 (b) f w œ ± ± ± Ê increasing on (2ß 1) and ($ß _), decreasing on (_ß 2) and ("ß $) # " $ (c) Local maximum at x œ 1, local minima at x œ 2 and x œ 3 6. (a) f w (x) œ (x 7)(x 1)(x 5) Ê critical points at 5, 1 and 7 (b) f w œ ± ± ± Ê increasing on (5ß 1) and (7ß _), decreasing on (_ß 5) and ("ß 7) & " ( (c) Local maximum at x œ 1, local minima at x œ 5 and x œ 7 7. (a) f w (x) œ
x 2 ax 1 b ax 2 b
Ê critical points at x œ 0, x œ 1 and x œ 2
w
(b) f œ )( ± ± Ê increasing on a_ß 2b and a1ß _b, decreasing on a2ß 0b and a0ß 1b 2 0 1 (c) Local minimum at x œ 1 8. (a) f w (x) œ
ax 2bax 4b ax 1bax 3b
Ê critical points at x œ 2, x œ 4, x œ 1, and x œ 3
w
(b) f œ ± )( ± )( Ê increasing on a_ß 4b, a1ß 2b and a3ß _b, decreasing on 1 3 4 2 a4ß 1b and a2ß 3b (c) Local maximum at x œ 4 and x œ 2 9. (a) f w (x) œ 1 w
4 x2
œ
x2 4 x2
Ê critical points at x œ 2, x œ 2 and x œ 0.
(b) f œ ± )( ± Ê increasing on a_ß 2b and a2ß _b, decreasing on a2ß 0b and a0ß 2b 0 2 2 (c) Local maximum at x œ 2, local minimum at x œ 2 10. (a) f w (x) œ 3
6 Èx
œ
3È x 6 Èx
Ê critical points at x œ 4 and x œ 0
(b) f w œ ( ± Ê increasing on a4ß _b, decreasing on a0ß 4b 0 4 (c) Local minimum at x œ 4 11. (a) f w (x) œ x"Î$ (x 2) Ê critical points at x œ 2 and x œ 0 (b) f w œ ± )( Ê increasing on (_ß 2) and (0ß _), decreasing on (2ß 0) 0 2 (c) Local maximum at x œ 2, local minimum at x œ 0 12. (a) f w (x) œ x"Î# (x 3) Ê critical points at x œ 0 and x œ 3 (b) f w œ ( ± Ê increasing on (3ß _), decreasing on (0ß 3) 0 3 (c) No local maximum and a local minimum at x œ 3 13. (a) f w (x) œ asin x 1ba2cos x 1b, 0 Ÿ x Ÿ 21 Ê critical points at x œ 12 , x œ 231 , and x œ 431 (b) f w œ Ò ± ± ± Ó Ê increasing on ˆ 231 ß 431 ‰, decreasing on ˆ0ß 12 ‰, ˆ 12 ß 231 ‰ and ˆ 431 ß 21‰ 1 21 41 0 21 2 3
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
185
186
Chapter 4 Applications of Derivatives
(c) Local maximum at x œ
41 3
and x œ 0, local minimum at x œ
21 3
and x œ 21
14. (a) f w (x) œ asin x cos xbasin x cos xb, 0 Ÿ x Ÿ 21 Ê critical points at x œ 14 , x œ 341 , x œ 541 , and x œ 741 (b) f w œ Ò ± ± ± ± Ó Ê increasing on ˆ 14 ß 341 ‰ and ˆ 541 ß 741 ‰, decreasing on ˆ0ß 14 ‰, 1 51 71 31 0 21 4 4
ˆ 341 ß 541 ‰ and ˆ 741 ß 21‰
4
(c) Local maximum at x œ 0, x œ
4
31 4
and x œ
71 4 ,
local minimum at x œ 14 , x œ
51 4
and x œ 21
15. (a) Increasing on a2ß 0b and a2ß 4b, decreasing on a4ß 2b and a0ß 2b (b) Absolute maximum at a4ß 2b, local maximum at a0ß 1b and a4ß 1b; Absolute minimum at a2ß 3b, local minimum at a2ß 0b 16. (a) Increasing on a4ß 3.25b, a1.5ß 1b, and a2ß 4b, decreasing on a3.25ß 1.5b and a1ß 2b (b) Absolute maximum at a4ß 2b, local maximum at a3.25ß 1b and a1ß 1b; Absolute minimum at a1.5ß 1b, local minimum at a4ß 0b and a2ß 0b 17. (a) Increasing on a4ß 1b, a0.5ß 2b, and a2ß 4b, decreasing on a1ß 0.5b (b) Absolute maximum at a4ß 3b, local maximum at a1ß 2b and a2ß 1b; No absolute minimum, local minimum at a4ß 1band a0.5ß 1b 18. (a) Increasing on a4ß 2.5b, a1ß 1b, and a3ß 4b, decreasing on a2.5ß 1b and a1ß 3b (b) No absolute maximum, local maximum at a2.5ß 1b, a1ß 2b and a4ß 2b; No absolute minimum, local minimum at a1ß 0b and a3ß 1b 19. (a) g(t) œ t# 3t 3 Ê gw (t) œ 2t 3 Ê a critical point at t œ 3# ; gw œ ± , increasing on $Î# ˆ_ß 3# ‰ , decreasing on ˆ 3# ß _‰ 3 21 3 (b) local maximum value of g ˆ 3# ‰ œ 21 4 at t œ # , absolute maximum is 4 at t œ # 20. (a) g(t) œ 3t# 9t 5 Ê gw (t) œ 6t 9 Ê a critical point at t œ
3 #
; gw œ ± , increasing on ˆ_ß 32 ‰ , $Î#
decreasing on ˆ 3# ß _‰
(b) local maximum value of g ˆ 3# ‰ œ
47 4
at t œ 3# , absolute maximum is
47 4
at t œ
3 #
21. (a) h(x) œ x$ 2x# Ê hw (x) œ 3x# 4x œ x(4 3x) Ê critical points at x œ 0, 43 Ê hw œ ± ± , increasing on ˆ0ß 43 ‰ , decreasing on (_ß !) and ˆ 43 ß _‰ ! %Î$ 4 (b) local maximum value of h ˆ 43 ‰ œ 32 27 at x œ 3 ; local minimum value of h(0) œ 0 at x œ 0, no absolute extrema 22. (a) h(x) œ 2x$ 18x Ê hw (x) œ 6x# 18 œ 6 Šx È3‹ Šx È3‹ Ê critical points at x œ „ È3 Ê hw œ | | , increasing on Š_ß È3‹ and ŠÈ$ß _‹ , decreasing on ŠÈ$ß È3‹ È$ È $ (b) a local maximum is h ŠÈ3‹ œ 12È3 at x œ È3; local minimum is h ŠÈ3‹ œ 12È3 at x œ È3, no absolute extrema
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.3 Monotonic Functions and the First Derivative Test 23. (a) f()) œ 3)# 4)$ Ê f w ()) œ 6) 12)# œ 6)(1 2)) Ê critical points at ) œ 0,
" #
187
Ê f w œ ± ± , ! "Î#
increasing on ˆ0ß "# ‰ , decreasing on (_ß !) and ˆ "# ß _‰ (b) a local maximum is f ˆ "# ‰ œ 4" at ) œ #" , a local minimum is f(0) œ 0 at ) œ 0, no absolute extrema 24. (a) f()) œ 6) )$ Ê f w ()) œ 6 3)# œ 3 ŠÈ2 )‹ ŠÈ2 )‹ Ê critical points at ) œ „ È2 Ê f w œ ± ± , increasing on ŠÈ2ß È2‹, decreasing on Š_ß È2‹ and ŠÈ2ß _‹ È# È # (b) a local maximum is f ŠÈ2‹ œ 4È2 at ) œ È2, a local minimum is f ŠÈ2‹ œ %È2 at ) œ È2, no absolute extrema 25. (a) f(r) œ 3r$ 16r Ê f w (r) œ 9r# 16 Ê no critical points Ê f w œ , increasing on (_ß _), never decreasing (b) no local extrema, no absolute extrema 26. (a) h(r) œ (r 7)$ Ê hw (r) œ 3(r 7)# Ê a critical point at r œ 7 Ê hw œ ± , increasing on ( (_ß 7) ((ß _), never decreasing (b) no local extrema, no absolute extrema 27. (a) f(x) œ x% 8x# 16 Ê f w (x) œ 4x$ 16x œ 4x(x 2)(x 2) Ê critical points at x œ 0 and x œ „ 2 Ê f w œ ± ± ± , increasing on (#ß !) and (#ß _), decreasing on (_ß 2) and (!ß #) # ! # (b) a local maximum is f(0) œ 16 at x œ 0, local minima are f a „ 2b œ 0 at x œ „ 2, no absolute maximum; absolute minimum is 0 at x œ „ 2 28. (a) g(x) œ x% 4x$ 4x# Ê gw (x) œ 4x$ 12x# )x œ 4x(x 2)(x 1) Ê critical points at x œ 0, 1, 2 Ê gw œ ± ± ± , increasing on (0ß 1) and (#ß _), decreasing on (_ß 0) and (1ß #) ! " # (b) a local maximum is g(1) œ 1 at x œ 1, local minima are g(0) œ 0 at x œ 0 and g(2) œ 0 at x œ 2, no absolute maximum; absolute minimum is 0 at x œ 0, 2 29. (a) H(t) œ
3 % # t
t' Ê Hw (t) œ 6t$ 6t& œ 6t$ (1 t)(" t) Ê critical points at t œ 0, „ 1
Ê Hw œ ± ± ± , increasing on (_ß 1) and (0ß 1), decreasing on ("ß 0) and ("ß _) " ! " (b) the local maxima are H(1) œ "# at t œ 1 and H(1) œ "# at t œ 1, the local minimum is H(0) œ 0 at t œ 0, absolute maximum is
" #
at t œ „ 1; no absolute minimum
30. (a) K(t) œ 15t$ t& Ê Kw (t) œ 45t# 5t% œ 5t# (3 t)(3 t) Ê critical points at t œ 0, „ 3 Ê Kw œ ± ± ± , increasing on (3ß 0) (0ß 3), decreasing on (_ß 3) and (3ß _) $ ! $ (b) a local maximum is K(3) œ 162 at t œ 3, a local minimum is K(3) œ 162 at t œ 3, no absolute extrema 31. (a) f(x) œ x 6Èx 1 Ê f w (x) œ 1
3 Èx 1
œ
Èx 1 3 Èx 1
Ê critical points at x œ 1 and x œ 10
Ê f w œ ( ± , increasing on a10, _b, decreasing on a1, 10b 1 10 (b) a local minimum is fa10b œ 8, a local and absolute maximum is fa1b œ 1, absolute minimum of 8 at x œ 10
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Chapter 4 Applications of Derivatives
32. (a) g(x) œ 4Èx x2 3 Ê gw (x) œ
2 Èx
2 2x3Î2 Èx
2x œ
Ê critical points at x œ 1 and x œ 0
Ê g w œ Ð ± , increasing on a0, 1b, decreasing on a1, _b 0 1 (b) a local minimum is fa0b œ 3, a local maximum is fa1b œ 6, absolute maximum of 6 at x œ 1 33. (a) g(x) œ xÈ8 x# œ x a8 x# b
"Î#
Ê gw (x) œ a8 x# b
"Î#
x ˆ "# ‰ a) x# b
"Î#
(2x) œ
2(2 x)(2 x) ÊŠ2È2 x‹ Š2È2 x‹
Ê critical points at x œ „ 2, „ 2È2 Ê gw œ ( ± ± Ñ , increasing on (#ß #), decreasing on # # #È# #È # Š#È2ß #‹ and Š#ß #È2‹ (b) local maxima are g(2) œ 4 at x œ 2 and g Š2È2‹ œ 0 at x œ 2È2, local minima are g(2) œ 4 at x œ 2 and g Š2È2‹ œ 0 at x œ 2È2, absolute maximum is 4 at x œ 2; absolute minimum is 4 at x œ 2 34. (a) g(x) œ x# È5 x œ x# (5 x)"Î# Ê gw (x) œ 2x(5 x)"Î# x# ˆ "# ‰ (5 x)"Î# (1) œ
5x(4x) 2 È 5 x
Ê critical points at
x œ 0, 4 and 5 Ê gw œ ± ± Ñ , increasing on (0ß 4), decreasing on (_ß !) and (%ß &) & ! % (b) a local maximum is g(4) œ 16 at x œ 4, a local minimum is 0 at x œ 0 and x œ 5, no absolute maximum; absolute minimum is 0 at x œ 0, 5 35. (a) f(x) œ
x# 3 x#
Ê f w (x) œ
2x(x 2) ax# 3b (1) (x 2)#
œ
(x 3)(x ") (x #)#
Ê critical points at x œ 1, 3
Ê f w œ ± )( ± , increasing on (_ß 1) and ($ß _), decreasing on ("ß #) and (#ß $), # " $ discontinuous at x œ 2 (b) a local maximum is f(1) œ 2 at x œ 1, a local minimum is f(3) œ 6 at x œ 3, no absolute extrema 36. (a) f(x) œ
x$ 3x# 1
Ê f w (x) œ
3x# a3x# 1b x$ (6x) a3x# 1b#
œ
3x# ax# 1b a3x# 1b#
Ê a critical point at x œ 0
Ê f w œ ± , increasing on (_ß !) (!ß _), and never decreasing ! (b) no local extrema, no absolute extrema 37. (a) f(x) œ x"Î$ (x 8) œ x%Î$ 8x"Î$ Ê f w (x) œ
4 3
x"Î$ 83 x#Î$ œ
w
4(x 2) 3x#Î$
Ê critical points at x œ 0, 2
Ê f œ ± )( , increasing on (#ß !) (!ß _), decreasing on (_ß 2) ! # (b) no local maximum, a local minimum is f(2) œ 6 $È2 ¸ 7.56 at x œ 2, no absolute maximum; absolute minimum is 6 $È2 at x œ 2 38. (a) g(x) œ x#Î$ (x 5) œ x&Î$ 5x#Î$ Ê gw (x) œ
5 3
x#Î$
10 3
x"Î$ œ
5(x 2) $ 3È x
Ê critical points at x œ 2 and
w
x œ 0 Ê g œ ± )( , increasing on (_ß 2) and (!ß _), decreasing on (2ß !) ! # (b) local maximum is g(2) œ 3 $È4 ¸ 4.762 at x œ 2, a local minimum is g(0) œ 0 at x œ 0, no absolute extrema 39. (a) h(x) œ x"Î$ ax# 4b œ x(Î$ 4x"Î$ Ê hw (x) œ x œ 0,
„2 È7
7 3
x%Î$ 43 x#Î$ œ
ŠÈ7x 2‹ ŠÈ7x #‹ $ # 3È x
Ê critical points at
2 Ê hw œ ± )( ± , increasing on Š_ß È ‹ and Š È27 ß _‹ , decreasing on 7 ! #ÎÈ( #ÎÈ(
2 ß !‹ and Š!ß È27 ‹ ŠÈ 7
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Section 4.3 Monotonic Functions and the First Derivative Test 2 (b) local maximum is h Š È ‹œ 7
$ 24 È 2 7(Î'
¸ 3.12 at x œ
2 È7 ,
the local minimum is h Š È27 ‹ œ
$ 24 È 2 7(Î'
189
¸ 3.12, no absolute
extrema 40. (a) k(x) œ x#Î$ ax# 4b œ x)Î$ 4x#Î$ Ê kw (x) œ
8 3
x&Î$ 83 x"Î$ œ
8(x 1)(x 1) $ 3È x
Ê critical points at
w
x œ 0, „ 1 Ê k œ ± )( ± , increasing on ("ß !) and ("ß _), decreasing on (_ß 1) ! " " and (!ß 1) (b) local maximum is k(0) œ 0 at x œ 0, local minima are k a „ 1b œ 3 at x œ „ 1, no absolute maximum; absolute minimum is 3 at x œ „ 1 41. (a) f(x) œ 2x x# Ê f w (x) œ 2 2x Ê a critical point at x œ 1 Ê f w œ ± ] and fa1b œ 1 and fa2b œ 0 2 1 a local maximum is 1 at x œ 1, a local minimum is 0 at x œ 2. (b) There is an absolute maximum of 1 at x œ 1; no absolute minimum. (c)
42. (a) f(x) œ (x 1)# Ê f w (x) œ 2(x 1) Ê a critical point at x œ 1 Ê f w œ ± ] and f(1) œ 0, f(0) œ 1 ! " Ê a local maximum is 1 at x œ 0, a local minimum is 0 at x œ 1 (b) no absolute maximum; absolute minimum is 0 at x œ 1 (c)
43. (a) g(x) œ x# 4x 4 Ê gw (x) œ 2x 4 œ 2(x 2) Ê a critical point at x œ 2 Ê gw œ [ ± and " # g(1) œ 1, g(2) œ 0 Ê a local maximum is 1 at x œ 1, a local minimum is g(2) œ 0 at x œ 2 (b) no absolute maximum; absolute minimum is 0 at x œ 2 (c)
44. (a) g(x) œ x# 6x 9 Ê gw (x) œ 2x 6 œ 2(x 3) Ê a critical point at x œ 3 Ê gw œ [ ± and % $ g(4) œ 1, g(3) œ 0 Ê a local maximum is 0 at x œ 3, a local minimum is 1 at x œ 4
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190
Chapter 4 Applications of Derivatives
(b) absolute maximum is 0 at x œ 3; no absolute minimum (c)
45. (a) f(t) œ 12t t$ Ê f w (t) œ 12 3t# œ 3(2 t)(2 t) Ê critical points at t œ „ 2 Ê f w œ [ ± ± $ # # and f(3) œ 9, f(2) œ 16, f(2) œ 16 Ê local maxima are 9 at t œ 3 and 16 at t œ 2, a local minimum is 16 at t œ 2 (b) absolute maximum is 16 at t œ 2; no absolute minimum (c)
46. (a) f(t) œ t$ 3t# Ê f w (t) œ 3t# 6t œ 3t(t 2) Ê critical points at t œ 0 and t œ 2 Ê f w œ ± ± ] and f(0) œ 0, f(2) œ 4, f(3) œ 0 Ê a local maximum is 0 at t œ 0 and t œ 3, a $ ! # local minimum is 4 at t œ 2 (b) absolute maximum is 0 at t œ 0, 3; no absolute minimum (c)
x$ 3
2x# 4x Ê hw (x) œ x# 4x 4 œ (x 2)# Ê a critical point at x œ 2 Ê hw œ [ ± and ! # h(0) œ 0 Ê no local maximum, a local minimum is 0 at x œ 0 (b) no absolute maximum; absolute minimum is 0 at x œ 0 (c)
47. (a) h(x) œ
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Section 4.3 Monotonic Functions and the First Derivative Test
191
48. (a) k(x) œ x$ 3x# 3x 1 Ê kw (x) œ 3x# 6x 3 œ 3(x 1)# Ê a critical point at x œ 1 Ê kw œ ± ] and k(1) œ 0, k(0) œ 1 Ê a local maximum is 1 at x œ 0, no local minimum ! " (b) absolute maximum is 1 at x œ 0; no absolute minimum (c)
49. (a) faxb œ È25 x2 Ê f w axb œ
x È25 x2
Ê critical points at x œ 0, x œ 5, and x œ 5
w
Ê f œ Ð ± Ñ , fa5b œ 0, fa0b œ 5, fa5b œ 0 Ê local maximum is 5 at x œ 0; local minimum of 0 at 5 5 0 x œ 5 and x œ 5 (b) absolute maximum is 5 at x œ 0; absolute minimum of 0 at x œ 5 and x œ 5 (c)
50. (a) faxb œ Èx2 2x 3, 3 Ÿ x _ Ê f w axb œ
2x 2 Èx2 2x 3
Ê only critical point in 3 Ÿ x _ is at x œ 3
w
Ê f œ [ , fa3b œ 0 Ê local minimum of 0 at x œ 3, no local maximum 3 (b) absolute minimum of 0 at x œ 3, no absolute maximum (c)
51. (a) gaxb œ
x2 x2 1 ,
0 Ÿ x 1 Ê g w axb œ
x2 4x 1 ax 2 1 b 2
Ê only critical point in 0 Ÿ x 1 is x œ 2 È3 ¸ 0.268 È
Ê g w œ [ l ), gŠ2 È3‹ œ 4È3 3 6 ¸ 1.866 Ê local minimum of 0 1 0.268 maximum at x œ 0. È (b) absolute minimum of 4È3 3 6 at x œ 2 È3, no absolute maximum
È3 4È 3 6
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at x œ 2 È3, local
192
Chapter 4 Applications of Derivatives
(c)
52. (a) gaxb œ
x2 4 x2 ,
2 x Ÿ 1 Ê g w axb œ
Ê only critical point in 2 x Ÿ 1 is x œ 0
8x a4 x 2 b 2
w
Ê g œ ( l ], ga0b œ 0 Ê local minimum of 0 at x œ 0, local maximum of 1 2 0 (b) absolute minimum of 0 at x œ 0, no absolute maximum (c)
1 3
at x œ 1.
53. (a) faxb œ sin 2x, 0 Ÿ x Ÿ 1 Ê f w axb œ 2cos 2x, f w (x) œ 0 Ê cos 2x œ 0 Ê critical points are x œ 14 and x œ 341 Ê f w œ [ ± ± ] , fa0b œ 0, fˆ 14 ‰ œ 1, fˆ 341 ‰ œ 1, fa1b œ 0 Ê local maxima are 1 at x œ 14 and 0 1 1 31 0 4
4
at x œ 1, and local minima are 1 at x œ w
31 4
and 0 at x œ 0.
(b) The graph of f rises when f 0, falls when f w 0, and has local extreme values where f w œ 0. The function f has a local minimum value at x œ 0 and x œ 341 , where the values of f w change from negative to positive. The function f has a local maximum value at x œ 1 and x œ 14 , where the values of f w change from positive to negative. 54. (a) faxb œ sin x cos x, 0 Ÿ x Ÿ 21 Ê f w axb œ cos x sin x, f w (x) œ 0 Ê tan x œ 1 Ê critical points are x œ 341 and x œ 741 Ê f w œ [ ± ± ] , fa0b œ 1, fˆ 341 ‰ œ È2, fˆ 741 ‰ œ È2, fa21b œ 1 Ê local maxima are 71 31 0 21 È2 at x œ
4
4
31 4
and 1 at x œ 21, and local minima are È2 at x œ w
w
71 4
and 1 at x œ 0.
(b) The graph of f rises when f 0, falls when f 0, and has local extreme values where f w œ 0. The function f has a local minimum value at x œ 0 and x œ 741 , where the values of f w change from negative to positive. The function f has a local maximum value at x œ 21 and x œ 341 , where the values of f w change from positive to negative.
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Section 4.3 Monotonic Functions and the First Derivative Test 55. (a) faxb œ È3cos x sin x, 0 Ÿ x Ÿ 21 Ê f w axb œ È3sin x cos x, f w (x) œ 0 Ê tan x œ xœ
1 6
and x œ
71 6
1 È3
193
Ê critical points are
Ê f w œ [ ± ± ] , fa0b œ È3, fˆ 16 ‰ œ 2, fˆ 761 ‰ œ 2, fa21b œ È3 Ê local 1 71 0 21 6 6
maxima are 2 at x œ
1 6
and È3 at x œ 21, and local minima are 2 at x œ w
w
71 6
and È3 at x œ 0.
(b) The graph of f rises when f 0, falls when f 0, and has local extreme values where f w œ 0. The function f has a local minimum value at x œ 0 and x œ 761 , where the values of f w change from negative to positive. The function f has a local maximum value at x œ 21 and x œ 16 , where the values of f w change from positive to negative. 56. (a) faxb œ 2x tan x, 12 x x œ 14 and x œ maximum is
1 2
1 4
1 2
Ê f w axb œ 2 sec2 x, f w (x) œ 0 Ê sec2 x œ 2 Ê critical points are Ê f w œ ( ± ± ) , fˆ 14 ‰ œ 12 1, fˆ 14 ‰ œ 1 12 Ê local 1 1 12 14 2 4
1 at x œ 14 , and local minimum is 1
(b) The graph of f rises when f w 0, falls when f w 0, and has local extreme values where f w œ 0. The function f has a local minimum value at x œ 14 , where the values
1 2
at x œ 14 .
of f w change from negative to positive. The function f has a local maximum value at x œ 14 , where the values of f w change from positive to negative.
57. (a) f(x) œ
x #
2 sin ˆ x# ‰ Ê f w (x) œ
Ê f w œ [ ± ] ! #1 #1Î$
cos ˆ x# ‰ , f w (x) œ 0 Ê cos ˆ x# ‰ œ "# Ê a critical point at x œ 231 and f(0) œ 0, f ˆ 231 ‰ œ 13 È3, f(21) œ 1 Ê local maxima are 0 at x œ 0 and 1 " #
at x œ 21, a local minimum is 13 È3 at x œ 231 (b) The graph of f rises when f w 0, falls when f w 0, and has a local minimum value at the point where f w changes from negative to positive.
58. (a) f(x) œ 2 cos x cos# x Ê f w (x) œ 2 sin x 2 cos x sin x œ 2(sin x)(1 cos x) Ê critical points at x œ 1, 0, 1 Ê f w œ [ ± ] and f(1) œ 1, f(0) œ 3, f(1) œ 1 Ê a local maximum is 1 at x œ „ 1, a local 1 1 ! minimum is 3 at x œ 0
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194
Chapter 4 Applications of Derivatives
(b) The graph of f rises when f w 0, falls when f w 0, and has local extreme values where f w œ 0. The function f has a local minimum value at x œ 0, where the values of f w change from negative to positive.
59. (a) f(x) œ csc# x 2 cot x Ê f w (x) œ 2(csc x)(csc x)(cot x) 2 acsc# xb œ 2 acsc# xb (cot x 1) Ê a critical point at x œ 14 Ê f w œ ( ± ) and f ˆ 14 ‰ œ 0 Ê no local maximum, a local minimum is 0 at x œ 14 1 ! 1Î% w (b) The graph of f rises when f 0, falls when f w 0, and has a local minimum value at the point where f w œ 0 and the values of f w change from negative to positive. The graph of f steepens as f w (x) Ä „ _.
60. (a) f(x) œ sec# x 2 tan x Ê f w (x) œ 2(sec x)(sec x)(tan x) 2 sec# x œ a2 sec# xb (tan x 1) Ê a critical point at x œ 14 Ê f w œ ( ± ) and f ˆ 14 ‰ œ 0 Ê no local maximum, a local minimum is 0 at x œ 14 1Î# 1Î# 1Î% w (b) The graph of f rises when f 0, falls when f w 0, and has a local minimum value where f w œ 0 and the values of f w change from negative to positive.
61. h()) œ 3 cos ˆ #) ‰ Ê hw ()) œ 3# sin ˆ #) ‰ Ê hw œ [ ] , (!ß $) and (#1ß 3) Ê a local maximum is 3 at ) œ 0, ! #1 a local minimum is 3 at ) œ 21 62. h()) œ 5 sin ˆ #) ‰ Ê hw ()) œ minimum is 0 at ) œ 0 63. (a)
5 #
cos ˆ #) ‰ Ê hw œ [ ] , (!ß 0) and (1ß 5) Ê a local maximum is 5 at ) œ 1, a local 1 !
(b)
(c)
(d)
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Section 4.3 Monotonic Functions and the First Derivative Test 64. (a)
(b)
(c)
(d)
65. (a)
(b)
66. (a)
(b)
195
67. The function faxb œ x sinˆ 1x ‰ has an infinite number of local maxima and minima. The function sin x has the following properties: a) it is continuous on a_, _b; b) it is periodic; and c) its range is Ò1, 1Ó. Also, for a 0, the function a range of Ð_, aÓ Òa, _Ñ on
’ 1a , 1a “.
In particular, if a œ 1, then
1 x
Ÿ 1 or
1 x
1 x
has
1 when x is in Ò1, 1Ó. This means
sinˆ 1x ‰ takes on the values of 1 and 1 infinitely many times in times on the interval Ò1, 1Ó, which occur when 1 1 31 51 2 2 2 ˆ1‰ x œ „ 2 , „ 2 , „ 2 ,. . . . Ê x œ „ 1 , „ 31 , „ 51 , . . . . Thus sin x has infinitely many local maxima and minima
in the interval Ò1, 1Ó. On the interval Ò0, 1Ó, 1 Ÿ sinˆ 1x ‰ Ÿ 1 and since x 0 we have x Ÿ x sinˆ 1x ‰ Ÿ x. On the interval Ò1, 0Ó, 1 Ÿ sinˆ 1x ‰ Ÿ 1 and since x 0 we have x x sinˆ 1x ‰ x. Thus faxb is bounded by the lines y œ x
and y œ x. Since sinˆ 1x ‰ oscillates between 1 and 1 infinitely many times on Ò1, 1Ó then f will oscillate between y œ x and y œ x infinitely many times. Thus f has infinitely many local maxima and minima. We can see from the graph (and verify later in Chapter 7) that lim x sinˆ 1x ‰ œ 1 and lim x sinˆ 1x ‰ œ 1. The graph of f does not have any absolute xÄ_
xÄ_
maxima., but it does have two absolute minima.
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196
Chapter 4 Applications of Derivatives
68. f(x) œ a x# b x c œ a ˆx# ba x‰ c œ a Šx# ba x
b# 4a# ‹
b# 4a
c œ a ˆx
b ‰# 2a
b# 4ac 4a
ˆ 2ab ß _‰
, a parabola whose
vertex is at x œ . Thus when a 0, f is increasing on and decreasing on ˆ_ß when a 0, b b w f is increasing on ˆ_ß #a ‰ and decreasing on ˆ #a ß _‰ . Also note that f (x) œ 2ax b œ 2a ˆx #ba ‰ Ê for b 2a
b ‰ #a ;
a 0, f w œ | ; for a 0, f w œ ± . bÎ2a bÎ2a
69. f(x) œ a x# b x Ê f w axb œ 2a x b, fa1b œ 2 Ê a b œ 2, f w a1b œ 0 Ê 2a b œ 0 Ê a œ 2, b œ 4 Ê f(x) œ 2x# 4x 70. f(x) œ a x3 b x# c x d Ê f w axb œ 3a x2 2b x c, fa0b œ 0 Ê d œ 0, fa1b œ 1 Ê a b c d œ 1, f w a0b œ 0 Ê c œ 0, f w a1b œ 0 Ê 3a 2b c œ 0 Ê a œ 2, b œ 3, c œ 0, d œ 0 Ê f(x) œ 2x3 3x# 4.4 CONCAVITY AND CURVE SKETCHING 1. y œ
x$ 3
x# #
2. y œ
x% 4
2x# 4 Ê yw œ x$ 4x œ x ax# 4b œ x(x 2)(x 2) Ê yww œ 3x# 4 œ ŠÈ3x 2‹ ŠÈ3x 2‹ . The
Ê yw œ x# x 2 œ (x 2)(x 1) Ê yww œ 2x 1 œ 2 ˆx "# ‰ . The graph is rising on (_ß 1) and (#ß _), falling on ("ß #), concave up on ˆ "# ß _‰ and concave down on ˆ_ß "# ‰ . Consequently, a local maximum is 3# at x œ 1, a local minimum is 3 at x œ 2, and ˆ "# ß 34 ‰ is a point of inflection. 2x
" 3
graph is rising on (2ß 0) and (#ß _), falling on (_ß #) and (!ß #), concave up on Š_ß È23 ‹ and Š È23 ß _‹ and concave down on Š È23 ß È23 ‹ . Consequently, a local maximum is 4 at x œ 0, local minima are 0 at x œ „ 2, and 2 16 Š È23 ß 16 9 ‹ and Š È3 ß 9 ‹ are points of inflection.
3 4
ax# 1b
#Î$
Ê yw œ ˆ 34 ‰ ˆ 23 ‰ ax# 1b
"Î$
minima are 0 at x œ „ 1; yww œ ax# 1b
"Î$
(2x) œ x ax# 1b
"Î$
, yw œ ) ( ± )( " " ! Ê the graph is rising on ("ß !) and ("ß _), falling on (_ß ") and (!ß ") Ê a local maximum is 34 at x œ 0, local
3. y œ
(x) ˆ 3" ‰ ax# 1b
%Î$
(2x) œ
x # 3 3
$ É
ax # 1 b %
,
yww œ ± ) ( )( ± Ê the graph is concave up on Š_ß È3‹ and ŠÈ3ß _‹, concave " " È$ È $ $ È down on ŠÈ3ß È3‹ Ê points of inflection at Š „ È3ß $ % ‹ %
x#Î$ ax# 1b, yw œ ± )( ± ! " " Ê the graph is rising on (_ß 1) and ("ß _), falling on (1ß ") Ê a local maximum is 27 7 at x œ 1, a local
4. y œ
9 14
x"Î$ ax# 7b Ê yw œ
3 14
x#Î$ ax# 7b
9 14
x"Î$ (2x) œ
3 #
ww &Î$ minimum is 27 ax# 1b 3x"Î$ œ 2x"Î$ x&Î$ œ x&Î$ a2x# 1b , 7 at x œ 1; y œ x
yww œ )( Ê the graph is concave up on (!ß _), concave down on (_ß !) Ê a point of inflection at (!ß !). ! 5. y œ x sin 2x Ê yw œ 1 2 cos 2x, yw œ [ ± ± ] Ê the graph is rising on ˆ 13 ß 13 ‰ , falling #1Î$ 1Î$ #1Î$ 1Î$ on ˆ #31 ß 13 ‰ and ˆ 13 ß 231 ‰ Ê local maxima are 231 13
È3 #
at x œ 13 and
21 3
È3 #
È3 #
at x œ 231 and
1 3
È3 #
at x œ
1 3
21 3
, local minima are
; yww œ 4 sin 2x, yww œ [ ± ± ± ] Ê the ! #1Î$ 1Î# #1Î$ 1Î# graph is concave up on ˆ 1# ß !‰ and ˆ 1# ß 231 ‰ , concave down on ˆ 231 ß 1# ‰ and ˆ!ß 1# ‰ Ê points of inflection at
at x œ
ˆ 1# ß 1# ‰ , (!ß !), and ˆ 1# ß 1# ‰
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Section 4.4 Concavity and Curve Sketching
197
6. y œ tan x 4x Ê yw œ sec# x 4, yw œ ( ± ± ) Ê the graph is rising on ˆ 12 ß 13 ‰ and 1 1Î# 1Î$ Î# 1Î$ ˆ 13 ß 1# ‰ , falling on ˆ 13 ß 13 ‰ Ê a local maximum is È3 431 at x œ 13 , a local minimum is È3 431 at x œ 13 ; yww œ 2(sec x)(sec x)(tan x) œ 2 asec# xb (tan x), yww œ ( ± ) Ê the graph is concave up on ˆ0ß 1# ‰ , ! 1Î# 1Î# 1 ˆ ‰ concave down on 2 ß 0 Ê a point of inflection at (0ß 0)
7. If x 0, sin kxk œ sin x and if x 0, sin kxk œ sin (x) œ sin x. From the sketch the graph is rising on ˆ 3#1 ß 1# ‰ , ˆ!ß 1# ‰ and ˆ 3#1 ß #1‰ , falling on ˆ21ß 3#1 ‰ , ˆ 1# ß !‰ and ˆ 1# ß 3#1 ‰ ; local minima are 1 at x œ „ 3#1 and 0 at x œ !; local maxima are 1 at x œ „
1 #
and 0 at
x œ „ #1; concave up on (#1ß 1) and (1ß #1), and concave down on (1ß 0) and (!ß 1) Ê points of inflection are (1ß !) and (1ß !) 8. y œ 2 cos x È2 x Ê yw œ 2 sin x È2, yw œ [ ± ± ± ] Ê rising on 1 $1Î# $1Î% 1Î% &1Î% ˆ 341 ß 14 ‰and ˆ 541 ß 3#1 ‰ , falling on ˆ1ß 341 ‰ and ˆ 14 ß 541 ‰ Ê local maxima are 2 1È2 at x œ 1, È2 È2
at x œ 14 and 31#
at x œ
31 # ,
È at x œ 341 and È2 514 2 at x œ 541 ; Ê concave up on ˆ1ß 1# ‰ and ˆ 1# ß 3#1 ‰ , concave down on
and local minima are È2
yww œ 2 cos x, yww œ [ ± ± ] 1 $1Î# 1Î# 1Î# ˆ 1# ß 1# ‰ Ê points of inflection at Š 1# ß
È 21 # ‹
and Š 1# ß
31 È 2 4
1È2 4
È 21 # ‹
9. When y œ x# 4x 3, then yw œ 2x 4 œ 2(x 2) and yww œ 2. The curve rises on (#ß _) and falls on (_ß #). At x œ 2 there is a minimum. Since yww 0, the curve is concave up for all x.
10. When y œ ' 2x x# , then yw œ # 2x œ 2(" x) and yww œ 2. The curve rises on (_ß 1) and falls on (1ß _). At x œ 1 there is a maximum. Since yw w 0, the curve is concave down for all x.
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Chapter 4 Applications of Derivatives
11. When y œ x$ 3x 3, then yw œ 3x# 3 œ 3(x 1)(x 1) and yww œ 6x. The curve rises on (_ß 1) ("ß _) and falls on (1ß 1). At x œ 1 there is a local maximum and at x œ 1 a local minimum. The curve is concave down on (_ß 0) and concave up on (!ß _). There is a point of inflection at x œ 0.
12. When y œ x(6 2x)# , then yw œ 4x(6 2x) (' 2x)# œ 12(3 x)(" x) and yww œ 12(3 x) 12(" x) œ 24(x 2). The curve rises on (_ß ") ($ß _) and falls on ("ß $). The curve is concave down on (_ß #) and concave up on (#ß _). At x œ 2 there is a point of inflection. 13. When y œ 2x$ 6x# 3, then yw œ 6x# 12x œ 6x(x 2) and yww œ 12x 12 œ 12(x 1). The curve rises on (!ß #) and falls on (_ß 0) and (#ß _). At x œ 0 there is a local minimum and at x œ 2 a local maximum. The curve is concave up on (_ß ") and concave down on ("ß _). At x œ 1 there is a point of inflection.
14. When y œ 1 9x 6x# x$ , then yw œ 9 12x 3x# œ $(x 3)(B 1) and yww œ 12 6x œ 6(x 2). The curve rises on ($ß ") and falls on (_ß 3) and ("ß _). At x œ 1 there is a local maximum and at x œ 3 a local minimum. The curve is concave up on (_ß 2) and concave down on (#ß _). At x œ 2 there is a point of inflection.
15. When y œ (x 2)$ 1, then yw œ 3(x 2)# and yww œ 6(x 2). The curve never falls and there are no local extrema. The curve is concave down on (_ß #) and concave up on (#ß _). At x œ 2 there is a point of inflection.
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Section 4.4 Concavity and Curve Sketching 16. When y œ 1 (x 1)$ , then yw œ 3(x 1)# and yww œ 6(x 1). The curve never rises and there are no local extrema. The curve is concave up on (_ß 1) and concave down on ("ß _). At x œ 1 there is a point of inflection.
17. When y œ x% 2x# , then yw œ 4x$ 4x œ 4x(x 1)(x 1) and yww œ 12x# 4 œ 12 Šx
" È 3 ‹ Šx
" È3 ‹ .
The curve
rises on ("ß !) and ("ß _) and falls on (_ß 1) and (!ß "). At x œ „ 1 there are local minima and at x œ 0 a local maximum. The curve is concave up on Š_ß È"3 ‹ and Š È"3 ß _‹ and concave down on Š È"3 ß È"3 ‹ . At x œ
„" È3
there are points of inflection. 18. When y œ x% 6x# 4, then yw œ 4x$ 12x œ 4x Šx È3‹ Šx È3‹ and yww œ 12x# 12 œ 12(x 1)(x 1). The curve rises on Š_ß È3‹ and Š!ß È3‹ , and falls on ŠÈ3ß !‹ and ŠÈ3ß _‹ . At x œ „ È3there are local maxima and at x œ 0 a local minimum. The curve is concave up on ("ß ") and concave down on (_ß 1) and ("ß _). At x œ „ 1 there are points of inflection. 19. When y œ 4x$ x% , then yw œ 12x# 4x$ œ 4x# ($ x) and yww œ 24x 12x# œ 12x(2 x). The curve rises on a_ß $b and falls on a$ß _b. At x œ 3 there is a local maximum, but there is no local minimum. The graph is concave up on a!ß #b and concave down on a_ß !b and a#ß _b. There are inflection points at x œ 0 and x œ 2. 20. When y œ x% 2x$ , then yw œ 4x$ 6x# œ 2x# (2x 3) and yww œ 12x# 12x œ 12x(x 1). The curve rises on ˆ 3# ß _‰ and falls on ˆ_ß 32 ‰ . There is a local minimum at x œ 3# , but no local maximum. The curve is
concave up on (_ß 1) and (!ß _), and concave down on (1ß 0). At x œ 1 and x œ 0 there are points of inflection.
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Chapter 4 Applications of Derivatives
21. When y œ x& 5x% , then yw œ 5x% 20x$ œ 5x$ (x 4) and yww œ 20x$ 60x# œ 20x# (x 3). The curve rises on (_ß !) and (%ß _), and falls on (!ß %). There is a local maximum at x œ 0, and a local minimum at x œ 4. The curve is concave down on (_ß 3) and concave up on (3ß _). At x œ 3 there is a point of inflection. % % $ 22. When y œ x ˆ x# 5‰ , then yw œ ˆ x# 5‰ x(4)ˆ x# 5‰ ˆ "# ‰ $ ww ‰ ˆx ‰# ˆ "# ‰ ˆ 5x ‰ œ ˆ x# 5‰ ˆ 5x # 5 , and y œ 3 # 5 # 5
$ # ˆ x# 5‰ ˆ 5# ‰ œ 5 ˆ x# 5‰ (x 4). The curve is rising
on (_ß #) and (10ß _), and falling on (#ß 10). There is a local maximum at x œ 2 and a local minimum at x œ 10. The curve is concave down on (_ß %) and concave up on (%ß _). At x œ 4 there is a point of inflection. 23. When y œ x sin x, then yw œ " cos x and yww œ sin x. The curve rises on (!ß 21). At x œ 0 there is a local and absolute minimum and at x œ 21 there is a local and absolute maximum. The curve is concave down on (!ß 1) and concave up on (1ß #1). At x œ 1 there is a point of inflection.
24. When y œ x sin x, then yw œ " cos x and yww œ sin x. The curve rises on (!ß 21). At x œ 0 there is a local and absolute minimum and at x œ 21 there is a local and absolute maximum. The curve is concave up on (!ß 1) and concave down on (1ß #1). At x œ 1 there is a point of inflection.
25. When y œ È3x 2 cos x, then yw œ È3 2 sin x and yww œ 2 cos x. The curve is increasing on ˆ!ß 431 ‰ and ˆ 531 ß 21‰, and decreasing on ˆ 431 ß 531 ‰. At x œ 0 there is a local and absolute minimum, at x œ maximum, at x œ
51 3
41 3
there is a local
there is a local minimum, and and at
x œ 21 there is a local and absolute maximum. The curve is concave up on ˆ!ß 12 ‰ and ˆ 321 ß 21‰, and is concave down onˆ 12 ß 321 ‰. At x œ 12 and x œ 321 there are points of inflection.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching 26. When y œ 43 x tan x, then yw œ
sec2 x and
4 3
yww œ 2 sec2 x tan x. The curve is increasing on ˆ 16 ß 16 ‰, and decreasing on ˆ 12 ß 16 ‰ and ˆ 16 ß 12 ‰. At x œ 16 there is a local minimum, at x œ
1 6
there is a local
maximum,there are no absolute maxima or absolute minima. The curve is concave up on ˆ 12 ß 0‰, and is concave down onˆ0ß 12 ‰. At x œ 0 there is a point of inflection. 27. When y œ sin x cos x, then yw œ sin2 x cos2 x œ cos 2x and yww œ 2 sin 2x. The curve is increasing on ˆ0ß 14 ‰ and ˆ 341 ß 1‰, and decreasing on ˆ 14 ß 341 ‰. At x œ 0 there is a local minimum, at x œ maximum, at x œ
31 4
1 4
there is a local and absolute
there is a local and absolute minimum,
and at x œ 1 there is a local maximum. The curve is concave down on ˆ0ß 12 ‰, and is concave up onˆ 12 ß 1‰. At x œ 12 there is a point of inflection. 28. When y œ cos x È3sin x, then yw œ sin x È3cos x and yww œ cos x È3sin x. The curve is increasing on ˆ0ß 13 ‰ and ˆ 431 ß 21‰, and decreasing on ˆ 13 ß 431 ‰. At
x œ 0 there is a local minimum, at x œ and absolute maximum, at x œ
41 3
1 3
there is a local
there is a local and
absolute minimum, and at x œ 21 there is a local maximum. The curve is concave down on ˆ0ß 561 ‰ and ˆ 11'1 ß 21‰, and is concave up on ˆ 561 ß 11'1 ‰. At x œ 561 and x œ 1161 there are points of inflection. 29. When y œ x"Î& , then yw œ
" 5
4 *Î& x%Î& and yww œ 25 x .
The curve rises on (_ß _) and there are no extrema. The curve is concave up on (_ß !) and concave down on (!ß _). At x œ 0 there is a point of inflection.
30. When y œ x#Î& , then yw œ
2 5
6 )Î& x$Î& and yww œ 25 x .
The curve is rising on (0ß _) and falling on (_ß !). At x œ 0 there is a local and absolute minimum. There is no local or absolute maximum. The curve is concave down on (_ß !) and (!ß _). There are no points of inflection, but a cusp exists at x œ 0.
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Chapter 4 Applications of Derivatives
31. When y œ yww œ
x È x2 1 ,
3x . The ax2 1b5Î2
then yw œ
1 ax2 1b3Î2
and
curve is increasing on a_ß _b.
There are no local or absolute extrema. The curve is concave up on a_ß !b and concave down on a!ß _b. At x œ 0 there is a point of inflection. 32. When y œ
È 1 x2 2x 1 ,
then yw œ
ax 2 b a2x 1b2 È1 x2
and
4x3 12x2 7 . The curve is decreasing on a2x 1b3 a1 x2 b3Î2 ˆ1ß "# ‰ and ˆ "# ß 1‰. There are no absolute extrrema,
yww œ
there is a local maximum at x œ 1 and a local minimum at x œ 1. The curve is concave up on a1ß 0.92b and ˆ "# ß 0.69‰, and concave down on ˆ0.92ß "# ‰ and a0.69ß 1b. At x ¸ 0.92 and x ¸ 0.69 there are points of inflection.
33. When y œ 2x 3x#Î$ , then yw œ 2 2x"Î$ and yww œ 23 x%Î$ . The curve is rising on (_ß !) and ("ß _), and falling on (!ß "). There is a local maximum at x œ 0 and a local minimum at x œ 1. The curve is concave up on (_ß !) and (!ß _). There are no points of inflection, but a cusp exists at x œ 0.
34. When y œ 5x#Î& 2x, then yw œ 2x$Î& 2 œ 2 ˆx$Î& 1‰ and yww œ 65 x)Î& . The curve is rising on (0ß 1) and falling on (_ß 0) and ("ß _). There is a local minimum at x œ 0 and a local maximum at x œ 1. The curve is concave down on (_ß !) and (!ß _). There are no points of inflection, but a cusp exists at x œ 0. 35. When y œ x#Î$ ˆ #5 x‰ œ
5 #Î$ x&Î$ , then # x yw œ 53 x"Î$ 53 x#Î$ œ 53 x"Î$ (1 x) and "Î$ yww œ 59 x%Î$ 10 œ 95 x%Î$ (1 2x). 9 x
The curve is rising on (!ß ") and falling on (_ß !) and ("ß _). There is a local minimum at x œ 0 and a local maximum at x œ 1. The curve is concave up on ˆ_ß "# ‰ and concave down on ˆ "# ß !‰ and (0ß _). There is a point of inflection at x œ "# and a cusp at x œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching 36. When y œ x#Î$ (x 5) œ x&Î$ 5x#Î$ , then "Î$ yw œ 53 x#Î$ 10 œ 35 x"Î$ (x 2) and 3 x yww œ
10 9
x"Î$
10 9
x%Î$ œ
10 9
x%Î$ (x 1). The curve
is rising on (_ß !) and (#ß _), and falling on (!ß #). There is a local minimum at x œ 2 and a local maximum at x œ 0. The curve is concave up on ("ß 0) and (!ß _), and concave down on (_ß 1). There is a point of inflection at x œ 1 and a cusp at x œ 0. 37. When y œ xÈ8 x# œ x a8 x# b # "Î#
yw œ a 8 x b
# "Î#
œ a8 x b
œ
2x ax 12b É a8 x # b $
, then
# "Î#
(x) ˆ "# ‰ a8 x b a8 2x b œ $#
(#x)
2(2 x)(2 x)
#
yww œ ˆ "# ‰a8 x# b #
"Î#
ÊŠ2È2 x‹ Š2È2 x‹
and
(2x)a8 2x# b a8 x# b
#"
(4x)
. The curve is rising on (#ß #), and falling
on Š2È2ß 2‹ and Š#ß 2È2‹ . There are local minima x œ 2 and x œ 2È2, and local maxima at x œ 2È2 and x œ 2. The curve is concave up on Š2È2ß !‹ and concave down on Š!ß #È2‹ . There is a point of inflection at x œ 0. 38. When y œ a2 x# b
$Î#
, then yw œ ˆ 3# ‰ a2 x# b
"Î#
(2x)
œ 3xÈ2 x# œ 3xÊŠÈ2 x‹ ŠÈ2 x‹ and yww œ (3) a2 x# b œ
"Î#
6(" x)(1 x) ÊŠÈ2 x‹ ŠÈ2 x‹
(3x) ˆ "# ‰ a2 x# b
"Î#
(2x)
. The curve is rising on
ŠÈ2ß !‹ and falling on Š!ß È2‹ . There is a local maximum at x œ 0, and local minima at x œ „ È2. The curve is concave down on ("ß ") and concave up on ŠÈ2ß "‹ and Š"ß È2‹ . There are points of inflection at x œ „ 1. 39. When y œ È16 x# , then yw œ yww œ
16 a16 x# b3Î2
x È16 x#
and
. The curve is rising on a4ß 0b and falling
on a0ß 4b. There is a local and absolute maximum at x œ 0 and local and absolute minima at x œ 4 and x œ 4. The curve is concave down on a4ß 4b. There are no points of inflection.
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Chapter 4 Applications of Derivatives
40. When y œ x# #x , then yw œ 2x ww
y œ2
4 x3
œ
2x3 4 x3 . The
# x#
œ
2x3 # x#
and
curve is falling on a_ß 0b
and a0ß 1b, and rising on a1ß _b. There is a local minimum at x œ 1. There are no absolute maxima or absolute minima. 3 The curve is concave up on Š_ß È 2‹ and a0, _b, and 3 concave down on ŠÈ 2, 0‹ . There is a point of 3 inflection at x œ È 2.
x# 3 x # , then (x 3)(x 1) and (x 2)#
41. When y œ œ
yww œ
yw œ
2x(x 2) ax# 3b (") (x2)#
(2x 4)(x 2)# ax# 4x 3b# (x 2) (x 2)%
œ
2 (x 2)$
.
The curve is rising on (_ß ") and ($ß _), and falling on ("ß #) and (#ß $). There is a local maximum at x œ 1 and a local minimum at x œ 3. The curve is concave down on (_ß #) and concave up on (#ß _). There are no points of inflection because x œ 2 is not in the domain. 3 42. When y œ È x3 1, then yw œ
yww œ
2x . The ax3 1b5Î3
x# ax3 1b2Î3
and
curve is risng on a_ß 1b,
a1, 0b, and a0ß _b. There is are no local or absolute extrema. The curve is concave up on a_ß 1b and a0, _b, and concave down on a1, 0b . There are points of inflection at x œ 1 and x œ 0. 43. When y œ yww œ
8x x2 4 ,
then yw œ
16xˆx 12‰ . The ax 2 4 b 3 2
8 ˆx 2 4 ‰ ax 2 4 b 2
and
curve is fallng on a_ß 2b
and a2ß _b, and is rising on a2ß 2b. There is a local and absolute minimum at x œ 2, and a local and absolute maximum at x œ 2. The curve is concave down on Š_ß 2È3‹ and Š0, 2È3‹, and concave up on Š2È3, 0‹and Š2È3, _‹. There are points of inflection at x œ 2È3, x œ 0, and x œ 2È3. y œ 0 is a horizontal asymptote. 44. When y œ yww œ
5 x4 5 ,
then yw œ
100x2 ˆx4 3‰ . The ax 4 5 b 3
20x3 ax 4 5 b 2
and
curve is risng on a_ß 0b,
and is falling on a0ß _b. There is a local and absolute maximum at x œ 0, and there is no local or aboslute minimum. The curve is concave up on 4 4 4 4 4 3‹ and ŠÈ 3, _‹, and concave down on ŠÈ 3, 0‹ and Š0, È 3‹. There are points of inflection at x œ È 3 Š_ß È 4 and x œ È 3. There is a horizontal asymptote of y œ 0.
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Section 4.4 Concavity and Curve Sketching
205
x# 1, kxk 1 , then 1 x# , kxk 1 2x, kxk 1 2, kxk " yw œ œ and yww œ œ . The 2x, kxk " #, kxk "
45. When y œ kx# 1k œ œ
curve rises on ("ß !) and ("ß _) and falls on (_ß 1) and (0ß 1). There is a local maximum at x œ 0 and local minima at x œ „ 1. The curve is concave up on (_ß 1) and ("ß _), and concave down on ("ß "). There are no points of inflection because y is not differentiable at x œ „ 1 (so there is no tangent line at those points). Ú x# 2x, x 0 46. When y œ kx 2xk œ Û 2x x# , 0 Ÿ x Ÿ 2 , then Ü x# 2x, x 2 Ú 2x 2, x 0 Ú 2, x 0 w ww 2 2x, 0 x 2 y œÛ , and y œ Û 2, 0 x 2 . Ü 2x 2, x 2 Ü 2, x 2 #
The curve is rising on (!ß 1) and (#ß _), and falling on (_ß !) and ("ß #). There is a local maximum at x œ 1 and local minima at x œ 0 and x œ 2. The curve is concave up on (_ß !) and (#ß _), and concave down on (!ß #). There are no points of inflection because y is not differentiable at x œ 0 and x œ 2 (so there is no tangent at those points). 47. When y œ Èkxk œ
Èx , x 0 , then È x , x 0
Ú
" x $Î# , x0 , x0 #È x y œ Û " and yww œ (x)4 $Î# . , x 0 , x0 Ü 2 È x 4 w
Since lim c yw œ _ and lim b yw œ _ there is a xÄ! xÄ!
cusp at x œ 0. There is a local minimum at x œ 0, but no local maximum. The curve is concave down on (_ß !) and (!ß _). There are no points of inflection. 48. When y œ Èkx 4k œ Ú w
y œÛ Ü
" 2È x 4 " #È 4 x
, x4
Èx 4 , x 4 , then È4 x , x 4 ww
and y œ , x4
(x 4) $Î# 4 (4 x) $Î# 4
, x4 , x4
.
Since lim c yw œ _ and lim b yw œ _ there is a cusp xÄ4 xÄ4
at x œ 4. There is a local minimum at x œ 4, but no local maximum. The curve is concave down on (_ß %) and (%ß _). There are no points of inflection. 49. yw œ 2 x x# œ (1 x)(# x), yw œ ± ± " # Ê rising on ("ß #), falling on (_ß 1) and (#ß _) Ê there is a local maximum at x œ 2 and a local minimum at x œ 1; yww œ 1 2x, yww œ ± "Î# "‰ ˆ Ê concave up on _ß # , concave down on ˆ "# ß _‰ Ê a point of inflection at x œ
" #
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Chapter 4 Applications of Derivatives
50. yw œ x# x 6 œ (x 3)(x 2), yw œ ± ± # $ Ê rising on (_ß #) and (3ß _), falling on (2ß 3) Ê there is a local maximum at x œ 2 and a local minimum at x œ 3; yww œ 2x 1, yww œ ± "Î# " ˆ ‰ Ê concave up on # ß _ , concave down on ˆ_ß "# ‰ Ê a point of inflection at x œ
" #
51. yw œ x(x 3)# , yw œ ± ± Ê rising on ! $ (!ß _), falling on (_ß !) Ê no local maximum, but there is a local minimum at x œ 0; yww œ (x 3)# x(2)(x 3) œ 3(x 3)(x 1), yww œ ± ± Ê concave " $ up on (_ß ") and ($ß _), concave down on ("ß $) Ê points of inflection at x œ 1 and x œ 3 52. yw œ x# (2 x), yw œ ± ± Ê rising on ! # (_ß #), falling on (2ß _) Ê there is a local maximum at x œ 2, but no local minimum; yww œ 2x(2 x) x# (1) œ x(4 3x), yww œ ± ± Ê concave up ! %Î$ on ˆ!ß 43 ‰ , concave down on (_ß !) and ˆ 43 ß _‰ Ê points of inflection at x œ 0 and x œ
4 3
53. yw œ x ax# 12b œ x Šx 2È3‹ Šx 2È3‹ , yw œ ± ± ± Ê rising on ! #È$ #È $ Š2È3ß !‹ and Š#È3ß _‹ , falling on Š_ß #È3‹ and Š!ß #È3‹ Ê a local maximum at x œ 0, local minima at x œ „ 2È3 ; yww œ 1 ax# 12b xa2xb œ 3ax 2bax 2b, yww œ ± ± Ê concave up on (_ß #) and (#ß _), concave down on (#ß #) Ê points of inflection # # at x œ „ 2 54. yw œ (x 1)# (2x 3), yw œ ± ± " $Î# 3 3‰ ˆ ‰ ˆ Ê rising on # ß _ , falling on _ß # Ê no local maximum, a local minimum at x œ 3# ;
yww œ 2(x 1)(2x 3) (x 1)# (2) œ 2(x 1)(3x 2), yww œ ± ± Ê concave up on " #Î$ 2 ˆ_ß 3 ‰ and ("ß _), concave down on ˆ 23 ß "‰ Ê points of inflection at x œ 23 and x œ 1
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Section 4.4 Concavity and Curve Sketching
207
55. yw œ a8x 5x# b (4 x)# œ x(8 5x)(% x)# , yw œ ± ± ± Ê rising on ˆ!ß 85 ‰ , ! % )Î& 8 ˆ falling on (_ß !) and 5 ß _‰ Ê a local maximum at xœ
8 5
, a local minimum at x œ 0;
yww œ (8 10x)(4 x)# a8x 5x# b (2)(% x)(1) œ 4(4 x) a5x# 16x 8b , yww œ ±
)#È' & 8 2È 6 8 2È 6 Š 5 ß 5 ‹
±
)#È' &
± Ê concave up on Š_ß 8 52 %
and (4ß _) Ê points of inflection at x œ
8 „ 2È 6 5
È6
‹ and Š 8 52
È6
ß %‹ , concave down on
and x œ 4
56. yw œ ax# 2xb (x 5)# œ x(x 2)(x 5)# , yw œ ± ± ± Ê rising on (_ß !) and ! # & a2ß _b, falling on a!ß 2b Ê a local maximum at x œ 0, a local minimum at x œ 2; yww œ a2x 2bax 5b# 2ax# 2xbax 5b œ 2ax 5ba2x# 8x 5b , yww œ ± ± ± Ê concave up on & %È' %È' # 4 È 6 4 È 6 Š # ß 2 ‹
#
È
È
and a5ß _b, concave down on Š_ß %2 6 ‹ and Š 4# 6 ß 5‹ Ê points of inflection at x œ
57. yw œ sec# x, yw œ ( ) Ê rising on ˆ 1# ß 1# ‰ , 1Î# 1Î# never falling Ê no local extrema; yww œ 2(sec x)(sec x)(tan x) œ 2 asec# xb (tan x), yww œ ( ± ) Ê concave ! 1Î# 1Î# 1‰ 1 ˆ ˆ ‰ up on !, # , concave down on # ß ! , ! is a point of inflection. 58. yw œ tan x, yw œ ( ± ) Ê rising on ˆ0ß 1# ‰ , ! 1Î# 1Î# 1 falling on ˆ # ß !‰ Ê no local maximum, a local minimum at x œ 0; yww œ sec# x, yww œ ( ) Ê concave up 1Î# 1Î# 1 1 on ˆ # ß # ‰ Ê no points of inflection
59. yw œ cot
) #
, yw œ ( ± ) Ê rising on (!ß 1), 1 ! #1 falling on (1ß #1) Ê a local maximum at ) œ 1, no local minimum; yww œ "# csc# #) , yww œ ( ) Ê never ! #1 concave up, concave down on (!ß #1) Ê no points of inflection
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4 „È 6 2
and x œ 5
208
Chapter 4 Applications of Derivatives
60. yw œ csc#
) #
, yw œ ( ) Ê rising on (!ß 21), never ! #1 falling Ê no local extrema; yww œ 2 ˆcsc #) ‰ ˆcsc #) ‰ ˆcot #) ‰ ˆ "# ‰ œ ˆcsc# #) ‰ ˆcot #) ‰, yww œ ( ± ) 1 ! #1 Ê concave up on (1ß #1), concave down on (!ß 1) Ê a point of inflection at ) œ 1
61. yw œ tan# ) 1 œ (tan ) 1)(tan ) 1), yw œ ( | ± ) Ê rising on 1Î# 1Î% 1Î# 1Î% 1 1 1 1 ˆ # ß 4 ‰ and ˆ 4 ß # ‰ , falling on ˆ 14 ß 14 ‰
Ê a local maximum at ) œ 14 , a local minimum at ) œ 14 ;
yww œ 2 tan ) sec# ), yww œ ( ± ) ! 1Î# 1Î# 1‰ ˆ Ê concave up on !ß # , concave down on ˆ 1# ß !‰ Ê a point of inflection at ) œ 0
62. yw œ 1 cot# ) œ (" cot ))(1 cot )), yw œ ( | ± ) Ê rising on ˆ 14 ß 341 ‰ , 1 ! 1Î% $1Î% 1 3 1 falling on ˆ0ß 4 ‰ and ˆ 4 ß 1‰ Ê a local maximum at )œ ww
31 4 ,
a local minimum at ) œ #
ww
1 4
;
y œ 2(cot )) acsc )b, y œ ( ± ) 1 ! 1Î# 1 1 Ê concave up on ˆ!ß # ‰ , concave down on ˆ # ß 1‰ Ê a point of inflection at ) œ
1 #
63. yw œ cos t, yw œ [ ± ± ] Ê rising on ! #1 1Î# $1Î# ˆ!ß 1# ‰ and ˆ 3#1 ß 21‰ , falling on ˆ 1# ß 3#1 ‰ Ê local maxima at tœ
1 #
and t œ 21, local minima at t œ 0 and t œ
yww œ sin t, yww œ [ ± ] 1 ! #1 Ê concave up on (1ß #1), concave down on (!ß 1) Ê a point of inflection at t œ 1
31 #
;
64. yw œ sin t, yw œ [ ± ] Ê rising on (!ß 1), 1 ! #1 falling on (1ß 21) Ê a local maximum at t œ 1, local minima at t œ 0 and t œ 21; yww œ cos t, yww œ [ ± ± ] Ê concave up on ˆ!ß 1# ‰ ! #1 1Î# $1Î# 3 1 and ˆ # ß #1‰ , concave down on ˆ 1# ß 3#1 ‰ Ê points of inflection at t œ
1 #
and t œ
31 #
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Section 4.4 Concavity and Curve Sketching 65. yw œ (x 1)#Î$ , yw œ ) ( Ê rising on " (_ß _), never falling Ê no local extrema; yww œ 23 (x 1)&Î$ , yww œ ) ( " Ê concave up on (_ß 1), concave down on ("ß _) Ê a point of inflection and vertical tangent at x œ 1 66. yw œ (x 2)"Î$ , yw œ )( Ê rising on (2ß _), # falling on (_ß #) Ê no local maximum, but a local minimum at x œ 2; yww œ 13 (x 2)%Î$ , yww œ )( Ê concave down on (_ß 2) and # (#ß _) Ê no points of inflection, but there is a cusp at xœ2
67. yw œ x#Î$ (x 1), yw œ )( ± Ê rising on ! " ("ß _), falling on (_ß ") Ê no local maximum, but a local minimum at x œ 1; yww œ "3 x#Î$ 32 x&Î$ " 3
x&Î$ (x 2), yww œ ± )( ! # Ê concave up on (_ß 2) and (!ß _), concave down on (#ß !) Ê points of inflection at x œ 2 and x œ 0, and a vertical tangent at x œ 0 œ
68. yw œ x%Î& (x 1), yw œ ± )( Ê rising on ! " ("ß 0) and (!ß _), falling on (_ß ") Ê no local maximum, but a local minimum at x œ 1; yww œ "5 x%Î& 54 x*Î& œ "5 x*Î& (x 4), yww œ )( ± Ê concave up on (_ß 0) and ! % (4ß _), concave down on (0ß 4) Ê points of inflection at x œ 0 and x œ 4, and a vertical tangent at x œ 0
69. yw œ œ
#x, x Ÿ 0 w , y œ ± Ê rising on 2x, x 0 !
(_ß _) Ê no local extrema; yww œ œ
2, x 0 , 2, x 0
yww œ )( Ê concave up on (!ß _), concave ! down on (_ß !) Ê a point of inflection at x œ 0 70. yw œ œ
x# , x Ÿ 0 w , y œ ± Ê rising on x# , x 0 !
(!ß _), falling on (_ß !) Ê no local maximum, but a 2x, x Ÿ 0 local minimum at x œ 0; yww œ œ , 2x, x 0 yww œ ± Ê concave up on (_ß _) ! Ê no point of inflection
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209
210
Chapter 4 Applications of Derivatives
71. The graph of y œ f ww (x) Ê the graph of y œ f(x) is concave up on (!ß _), concave down on (_ß !) Ê a point of inflection at x œ 0; the graph of y œ f w (x) Ê yw œ ± ± Ê the graph y œ f(x) has both a local maximum and a local minimum
72. The graph of y œ f ww (x) Ê yww œ ± Ê the graph of y œ f(x) has a point of inflection, the graph of y œ f w (x) Ê yw œ ± ± Ê the graph of y œ f(x) has both a local maximum and a local minimum
73. The graph of y œ f ww (x) Ê yww œ ± ± Ê the graph of y œ f(x) has two points of inflection, the graph of y œ f w (x) Ê yw œ ± Ê the graph of y œ f(x) has a local minimum
74. The graph of y œ f ww (x) Ê yww œ ± Ê the graph of y œ f(x) has a point of inflection; the graph of y œ f w (x) Ê yw œ ± ± Ê the graph of y œ f(x) has both a local maximum and a local minimum
75. y œ
2x# x 1 x# 1
76. y œ
x# 49 x# 5x 14
œ1
5 x2
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Section 4.4 Concavity and Curve Sketching 77. y œ
x% " x#
79. y œ
" x# 1
œ x#
#
2 81. y œ xx# 1 œ "
83. y œ
x# x1
78. y œ
x# 4 2x
œ
80. y œ
x# x# 1
œ"
" x# 1
" x# 1
82. y œ
x# 4 x# 2
œ"
2 x# 2
" x1
84. y œ xx 14 œ 1 x
" x#
œx1
#
x #
2 x
3 x1
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211
212
Chapter 4 Applications of Derivatives
85. y œ
x# x 1 x1
87. y œ
x$ 3x# 3x 1 x# x 2
89. y œ
91. y œ
93.
œx
" x1
86. y œ x
#
x1 x1
88. y œ
x$ x 2 x x#
x x# 1
90. y œ
x1 x # ax 2 b
8 x# 4
92. y œ
4x x# 4
Point P Q R S T
yw !
œx4
5x 7 x# x #
œ x
" x1
œ x 1
2x 2 x x#
yww !
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Section 4.4 Concavity and Curve Sketching 94.
213
95.
96.
97. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here. (a) The body is moving away from the origin when kdisplacementk is increasing as t increases, 0 t 2 and 6 t 9.5; the body is moving toward the origin when kdisplacementk is decreasing as t increases, 2 t 6 and 9.5 t 15 (b) The velocity will be zero when the slope of the tangent line for y œ s(t) is horizontal. The velocity is zero when t is approximately 2, 6, or 9.5 sec. (c) The acceleration will be zero at those values of t where the curve y œ s(t) has points of inflection. The acceleration is zero when t is approximately 4, 7.5, or 12.5 sec. (d) The acceleration is positive when the concavity is up, 4 t 7.5 and 12.5 t 15; the acceleration is negative when the concavity is down, 0 t 4 and 7.5 t 12.5 98. (a) The body is moving away from the origin when kdisplacementk is increasing as t increases, 1.5 t 4, 10 t 12 and 13.5 t 16; the body is moving toward the origin when kdisplacementk is decreasing as t increases, 0 t 1.5, 4 t 10 and 12 t 13.5 (b) The velocity will be zero when the slope of the tangent line for y œ s(t) is horizontal. The velocity is zero when t is approximately 0, 4, 12 or 16 sec. (c) The acceleration will be zero at those values of t where the curve y œ s(t) has points of inflection. The acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 sec. (d) The acceleration is positive when the concavity is up, 0 t 1.5, 6 t 8 and 10 t 13.5, the acceleration is negative when the concavity is down, 1.5 t 6, 8 t 10 and 13.5 t 16. 99. The marginal cost is
dc dx
which changes from decreasing to increasing when its derivative
d# c dx#
is zero. This is a
point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units. 100. The marginal revenue is
dy dx
d# y dx# is positive Ê the curve is concave up # when ddxy# 0 Ê the curve is concave down
and it is increasing when its derivative
Ê ! t 2 and 5 t 9; marginal revenue is decreasing Ê 2 t 5 and 9 t 12.
101. When yw œ (x 1)# (x 2), then yww œ 2(x 1)(x 2) (x 1)# . The curve falls on (_ß 2) and rises on (#ß _). At x œ 2 there is a local minimum. There is no local maximum. The curve is concave upward on (_ß ") and ˆ 53 ß _‰ , and concave downward on ˆ"ß 53 ‰ . At x œ 1 or x œ 53 there are inflection points.
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214
Chapter 4 Applications of Derivatives
102. When yw œ (x 1)# (x 2)(x 4), then yww œ 2(x 1)(x 2)(x 4) (x 1)# (x 4) (x 1)# (x 2) œ (x 1) c2 ax# 6x 8b ax# 5x 4b ax# 3x 2bd œ 2(x 1) a2x# 10x 11b. The curve rises on (_ß 2) and (%ß _) and falls on (#ß %). At x œ 2 there is a local maximum and at x œ 4 a local minimum. The È3 5 È3 ß # ‹
curve is concave downward on (_ß ") and Š 5 2 È3
Š 5 #
ß _‹ . At x œ 1,
5 È3 #
and
5 È3 #
È3
and concave upward on Š1ß 5 #
‹ and
there are inflection points.
103. The graph must be concave down for x 0 because f ww (x) œ x"# 0.
104. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always be concave up or concave down so it will have no inflection points and no cusps or corners. 105. The curve will have a point of inflection at x œ 1 if 1 is a solution of yww œ 0; y œ x$ bx# cx d Ê yw œ 3x# 2bx c Ê yw w œ 6x 2b and 6(1) 2b œ 0 Ê b œ 3. 106. (a) f(x) œ ax# bx c œ a ˆx# ba x‰ c œ a Šx# ba B
b# 4a# ‹
b# 4a
c œ a ˆx
b b whose vertex is at x œ 2a Ê the coordinates of the vertex are Š 2a ß b
#
b ‰# #a
b# 4ac 4a
a parabola
4ac 4a ‹
(b) The second derivative, f ww (x) œ 2a, describes concavity Ê when a 0 the parabola is concave up and when a 0 the parabola is concave down. 107. A quadratic curve never has an inflection point. If y œ ax# bx c where a Á 0, then yw œ 2ax b and yww œ 2a. Since 2a is a constant, it is not possible for yww to change signs. 108. A cubic curve always has exactly one inflection point. If y œ ax$ bx# cx d where a Á 0, then yw œ 3ax# 2bx c and yww œ 6ax 2b. Since 3ab is a solution of yww œ 0, we have that yww changes its sign b b at x œ 3a and yw exists everywhere (so there is a tangent at x œ 3a ). Thus the curve has an inflection b point at x œ 3a . There are no other inflection points because yww changes sign only at this zero.
109. y ww œ ax 1bax 2b, when y ww œ 0 Ê x œ 1 or x œ 2; y ww œ ± l Ê points of inflection at x œ 1 1 2 and x œ 2 110. y ww œ x2 ax 2b3 ax 3b, when y ww œ 0 Ê x œ 3, x œ 0, or x œ 2; y ww œ ± l l Ê points of 0 2 3 inflection at x œ 3 and x œ 2 111. y œ a x3 b x2 c x Ê y w œ 3a x2 2b x c and y ww œ 6a x 2b; local maximum at x œ 3 Ê 3a a3b2 2ba3b c œ 0 Ê 27a 6b c œ 0; local mimimum at x œ 1 Ê 3a a1b2 2ba1b c œ 0 Ê 3a 2b c œ 0; point of inflection at a1, 11b Ê aa1b3 ba1b2 ca1b œ 11 Ê a b c œ 11 and 6aa1b 2b œ 0 Ê 6a 2b œ 0. Solving 27a 6b c œ 0, 3a 2b c œ 0, a b c œ 11, and 6a 2b œ 0 Ê a œ 1,b œ 3, and c œ 9 Ê y œ x3 3x2 9x
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Section 4.4 Concavity and Curve Sketching 112. y œ
x2 a bxc
2 a3 b a b b x2 2c x a b à local maximum at x œ 3 Ê ba3baba32c œ ab x c b 2 b c b2 ba1b2 2ca1b a b a1b2 a œ 0 Ê b 2c a b œ 0 and ba1b c œ 2 Ê ab a 1 b c b 2
Ê yw œ
a1 ,2b Ê
0 Ê 9b 6c a b œ 0; local minimum at a 2b 2c œ 1. Solving
9b 6c a b œ 0, b 2c a b œ 0, and a 2b 2c œ 1 Ê a œ 3,b œ 1, and c œ 1 Ê y œ 113. If y œ x& 5x% 240, then yw œ 5x$ (x 4) and yww œ 20x# (x 3). The zeros of yw are extrema, and there is a point of inflection at x œ $Þ
114. If y œ x$ 12x# , then yw œ 3x(x 8) and yww œ 6(x 4). The zeros of yw and yww are extrema and points of inflection, respectively.
115. If y œ
4 5
x& 16x# 25, then yw œ 4x ax$ 8b and
yww œ 16 ax$ 2b . The zeros of yw and yww are extrema and points of inflection, respectively.
116. If y œ
x% 4 $
x$ 3 #
215
4x# 12x 20, then
yw œ x x )x "# œ (x 3)(x 2)# Þ So y has a local minimum at x œ $ as its only extreme value. Also yww œ $x# #x ) œ (3x 4)(x 2) and there are inflection points at both zeros, %$ and 2, of yww .
117. The graph of f falls where f w 0, rises where f w 0, and has horizontal tangents where f w œ 0. It has local minima at points where f w changes from negative to positive and local maxima where f w changes from positive to negative. The graph of f is concave down where f ww 0 and concave up where f ww 0. It has an inflection point each time f ww changes sign, provided a tangent line exists there.
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x2 3 x1 Þ
216
Chapter 4 Applications of Derivatives
118. The graph f is concave down where f ww 0, and concave up where f ww 0. It has an inflection point each time f ww changes sign, provided a tangent line exists there.
4.5 APPLIED OPTIMIZATION 1. Let j and w represent the length and width of the rectangle, respectively. With an area of 16 in.# , we have that (j)(w) œ 16 Ê w œ 16j" Ê the perimeter is P œ 2j 2w œ 2j 32j" and Pw (j) œ 2 w
Solving P (j) œ 0 Ê
2(j 4)(j 4) j#
32 j#
œ
2 aj# 16b j#
.
œ 0 Ê j œ 4, 4. Since j 0 for the length of a rectangle, j must be 4 and
w œ 4 Ê the perimeter is 16 in., a minimum since Pww (j) œ
16 j$
0.
2. Let x represent the length of the rectangle in meters (! x %) Then the width is % x and the area is Aaxb œ xa% xb œ %x x# . Since Aw axb œ % #x, the critical point occurs at x œ #. Since, Aw axb ! for ! x # and Aw axb ! for # x %, this critical point corresponds to the maximum area. The rectangle with the largest area measures # m by % # œ # m, so it is a square. Graphical Support:
3. (a) The line containing point P also contains the points (!ß ") and ("ß !) Ê the line containing P is y œ 1 x Ê a general point on that line is (xß 1 x). (b) The area A(x) œ 2x(1 x), where 0 Ÿ x Ÿ 1. (c) When A(x) œ 2x 2x# , then Aw (x) œ 0 Ê 2 4x œ 0 Ê x œ "# . Since A(0) œ 0 and A(1) œ 0, we conclude that A ˆ "# ‰ œ "# sq units is the largest area. The dimensions are " unit by "# unit. 4. The area of the rectangle is A œ 2xy œ 2x a12 x# b , where 0 Ÿ x Ÿ È12 . Solving Aw (x) œ 0 Ê 24 6x# œ 0 Ê x œ 2 or 2. Now 2 is not in the domain, and since A(0) œ 0 and A ŠÈ12‹ œ 0, we conclude that A(2) œ 32 square units is the maximum area. The dimensions are 4 units by 8 units. 5. The volume of the box is V(x) œ x(15 2x)(8 2x) œ 120x 46x# 4x$ , where 0 Ÿ x Ÿ 4. Solving Vw (x) œ 0 Ê 120 92x 12x# œ 4(6 x)(5 3x) œ 0 Ê x œ 53 or 6, but 6 is not in the domain. Since V(0) œ V(4) œ 0, $ V ˆ 53 ‰ œ #%&! #( ¸ *" in must be the maximum volume of the box with dimensions
14 3
‚
35 3
‚
5 3
inches.
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Section 4.5 Applied Optimization
217
6. The area of the triangle is A œ "# ba œ b# È400 b# , where b# 0 Ÿ b Ÿ 20. Then dA œ " È400 b# db
œ
#
200 b È400 b#
#
2È400 b#
œ 0 Ê the interior critical point is b œ 10È2.
When b œ 0 or 20, the area is zero Ê A Š10È2‹ is the maximum area. When a# b# œ 400 and b œ 10È2, the value of a is also 10È2 Ê the maximum area occurs when a œ b. 7. The area is A(x) œ x(800 2x), where 0 Ÿ x Ÿ 400. Solving Aw (x) œ 800 4x œ 0 Ê x œ 200. With A(0) œ A(400) œ 0, the maximum area is A(200) œ 80,000 m# . The dimensions are 200 m by 400 m. 8. The area is 2xy œ 216 Ê y œ
108 x
. The amount of fence
needed is P œ 4x 3y œ 4x 324x" , where ! x; dP 324 # dx œ 4 x# œ 0 Ê x 81 œ 0 Ê the critical points are 0 and „ 9, but 0 and 9 are not in the domain. Then Pww (9) 0 Ê at x œ 9 there is a minimum Ê the dimensions of the outer rectangle are 18 m by 12 m Ê 72 meters of fence will be needed. 9. (a) We minimize the weight œ tS where S is the surface area, and t is the thickness of the steel walls of the tank. The surface area is S œ x# %xy where x is the length of a side of the square base of the tank, and y is its depth. The ˆ # #!!! ‰. Treating the volume of the tank must be &!!ft$ Ê y œ &!! x# . Therefore, the weight of the tank is waxb œ t x x ‰ . The critical value is at x œ "!. Since www a"!b œ tˆ# %!!! ‰ !, thickness as a constant gives ww axb œ tˆ#x #!!! x# "!$ there is a minimum at x œ "!. Therefore, the optimum dimensions of the tank are "! ft on the base edges and & ft deep. (b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel walls would likely be determined by other considerations such as structural requirements.
10. (a) The volume of the tank being ""#& ft$ , we have that yx# œ ""#& Ê y œ ""#& x# . The cost of building the tank is ""#& $$(&! # w caxb œ &x $!xˆ x# ‰, where ! x. Then c axb œ "!x x# œ ! Ê the critical points are ! and "&, but ! is not
in the domain. Thus, cww a"&b ! Ê at x œ "& we have a minimum. The values of x œ "& ft and y œ & ft will minimize the cost. (b) The cost function c œ &ax# %xyb "!xy, can be separated into two items: (1) the cost of the materials and labor to fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of the tanks is ax# %xyb, it can be deduced that the unit cost to fabricate the tanks is $&/ft# . Normally, excavation costs are per unit volume of ‰ # excavated material. Consequently, the total excavation cost can be taken as "!xy œ ˆ "! x ax yb. This suggests that the unit cost of excavation is
$"!Îft# x
where x is the length of a side of the square base of the tank in feet. For the least
expensive tank, the unit cost for the excavation is
$"!Îft# "& ft
œ
$!Þ'( ft$
œ
$") yd$ .
The total cost of the least expensive tank is
$$$(&, which is the sum of $#'#& for fabrication and $(&! for the excavation.
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218
Chapter 4 Applications of Derivatives
11. The area of the printing is (y 4)(x 8) œ 50. ‰ Consequently, y œ ˆ x 50 8 4. The area of the paper is 50 A(x) œ x ˆ x 8 4‰ , where 8 x. Then 50 ‰ Aw (x) œ ˆ x 50 8 4 x Š (x 8)# ‹ œ
4(x 8)# 400 (x 8)#
œ0
Ê the critical points are 2 and 18, but 2 is not in the domain. Thus Aww (18) 0 Ê at x œ 18 we have a minimum Therefore the dimensions 18 by 9 inches minimize the amount minimize the amount of paper. 12. The volume of the cone is V œ V(y) œ
1 3
1 3
a9 y# b (y 3) œ
" 3
1r# h, where r œ x œ È9 y# and h œ y 3 (from the figure in the text). Thus,
a27 9y 3y# y$ b Ê Vw (y) œ
points are 3 and 1, but 3 is not in the domain. Thus Vww (1) œ volume of V(1) œ
1 3
321 3
(8)(4) œ
13. The area of the triangle is A()) œ
cubic units. ab sin ) #
, where 0 ) 1.
Solving A ()) œ 0 Ê œ 0 Ê ) œ 1# . Since Aww ()) ) œ ab sin Ê Aww ˆ 1# ‰ 0, there is a maximum at ) œ 1# . # w
ab cos ) #
1 # 3 a9 6y 3y b œ 1(1 y)(3 y). The critical 1 3 (' 6(1)) 0 Ê at y œ 1 we have a maximum
14. A volume V œ 1r# h œ 1000 Ê h œ
1000 1 r#
. The amount of
material is the surface area given by the sides and bottom of # the can Ê S œ 21rh 1r# œ 2000 r 1r , 0 r. Then dS dr
œ 2000 r# 21r œ ! Ê
are 0 and d# S dr#
rœ
10 $ È 1
1r$ 1000 r#
œ 0. The critical points
, but 0 is not in the domain. Since
œ
4000 r$ #1 0, we 10 cm and h œ 1000 $ È 1 r# 1
have a minimum surface area when œ
10 $ È 1
cm. Comparing this result to
the result found in Example 2, if we include both ends of the can, then we have a minimum surface area when the can is shorter-specifically, when the height of the can is the same as its diameter. 15. With a volume of 1000 cm and V œ 1r# h, then h œ A œ 8r# 21rh œ 8r#
2000 r
. Then Aw (r) œ
but r œ 0 results in no can. Since Aww (r) œ 16 16. (a) The base measures "! #x in. by
"&#x #
1000 1r# . The amount of aluminum used per can is 8r$ 1000 16r 2000 œ 0 Ê the critical points are 0 and 5, r# œ 0 Ê r# 1000 0 we have a minimum at r œ 5 Ê h œ 40 r$ 1 and h:r œ 8:1.
in., so the volume formula is Vaxb œ
xa"!#xba" #
œ #x$ #&x# (&x.
(b) We require x !, #x "!, and #x "&. Combining these requirements, the domain is the interval a!ß &b.
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Section 4.5 Applied Optimization
219
(c) The maximum volume is approximately 66.02 in.$ when x ¸ "Þ*' in. (d) Vw axb œ 'x# &!x (&. The critical point occurs when Vw axb œ !, at x œ œ ww
#& „ &È( , that '
&! „ Éa&!b# %a'ba(&b # a 'b
œ
&! „ È(!! "#
is, x ¸ "Þ*' or x ¸ 'Þ$(. We discard the larger value because it is not in the domain. Since
V axb œ "#x &!, which is negative when x ¸ "Þ*' , the critical point corresponds to the maximum volume. The maximum volume occurs when x œ
#& &È( '
¸ "Þ*', which comfimrs the result in (c).
17. (a) The "sides" of the suitcase will measure 24 2x in. by 18 2x in. and will be 2x in. apart, so the volume formula is Vaxb œ 2xa24 2xba18 2xb œ 8x$ 168x# 862x. (b) We require x !, 2x 18, and 2x 12. Combining these requirements, the domain is the interval a!ß *b.
(c) The maximum volume is approximately 1309.95 in.$ when x ¸ 3Þ3* in. (d) Vw axb œ #%x# $$'x )'% œ #%ax# "%x $'b. The critical point is at x œ
"% „ Éa"%b# %a"ba$'b # a" b
œ
"% „ È&# #
œ ( „ È"$, that is, x ¸ $Þ$* or x ¸ "!Þ'". We discard the larger value because it is not in the domain. Since Vww axb œ #%a#x "%b which is negative when x ¸ $Þ$*, the critical point corresponds to the maximum volume. The maximum value occurs at x œ ( È"$ ¸ $Þ$*, which confirms the results in (c). (e) )x$ "')x# )'#x œ ""#! Ê 8ax$ #"x# "!)x "%!b œ ! Ê )ax #bax &bax "%b œ !. Since "% is not in the fomain, the possible values of x are x œ # in. or x œ & in. (f) The dimensions of the resulting box are #x in., a#% #xb in., and a") #xb. Each of these measurements must be positive, so that gives the domain of a!ß *b. 18. If the upper right vertex of the rectangle is located at axß % cos !Þ& xb for ! x 1, then the rectangle has width #x and height % cos !Þ&x, so the area is Aaxb œ )x cos !Þ&x. Solving Aw axb œ ! graphically for ! x 1, we find that x ¸ #Þ#"%. Evaluating #x and % cos !Þ&x for x ¸ #Þ#"%, the dimensions of the rectangle are approximately %Þ%$ (width) by "Þ(* (height), and the maximum area is approximately (Þ*#$. 19. Let the radius of the cylinder be r cm, ! r "!. Then the height is #È"!! r# and the volume is Varb œ #1r# È"!! r# cm$ . Then, Vw arb œ #1r# Š " ‹a#rb Š#1È"!! r# ‹a#rb È"!! r#
œ
#1r$ %1ra"!! r# b È"!! r#
œ
#1ra#!! $r# b È"!! r# .
É $# . Since Vw arb ! for The critical point for ! r "! occurs at r œ É #!! $ œ "!
! r "!É #$ and Vw arb ! for "!É #$ r "!, the critical point corresponds to the maximum volume. The dimensions are r œ "!É #$ ¸ )Þ"' cm and h œ
#! È$
¸ ""Þ&& cm, and the volume is
%!!!1 $È $
¸ #%")Þ%! cm$ .
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220
Chapter 4 Applications of Derivatives
20. (a) From the diagram we have 4x j œ 108 and V œ x# j. The volume of the box is V(x) œ x# (108 4x), where 0 Ÿ x 27. Then Vw (x) œ 216x 12x# œ 12x(18 x) œ 0 Ê the critical points are 0 and 18, but x œ 0 results in no box. Since Vww (x) œ 216 24x 0 at x œ 18 we have a maximum. The dimensions of the box are 18 ‚ 18 ‚ 36 in. #
(b) In terms of length, V(j) œ x# j œ ˆ 1084 j ‰ j. The graph indicates that the maximum volume occurs near j œ 36, which is consistent with the result of part (a).
21. (a) From the diagram we have 3h 2w œ 108 and V œ h# w Ê V(h) œ h# ˆ54 #3 h‰ œ 54h# 3# h$ . Then Vw (h) œ 108h 9# h# œ
9 #
h(24 h) œ 0
Ê h œ 0 or h œ 24, but h œ 0 results in no box. Since Vww (h) œ 108 9h 0 at h œ 24, we have a maximum volume at h œ 24 and w œ 54 3# h œ 18.
(b)
22. From the diagram the perimeter is P œ 2r 2h 1r, where r is the radius of the semicircle and h is the height of the rectangle. The amount of light transmitted proportional to A œ 2rh "4 1r# œ r(P 2r 1r) 4" 1r# œ rP 2r# 34 1r# . Then
dA dr
œ P 4r 3# 1r œ 0
(4 1)P 2P 4P 21 P 8 31 Ê 2h œ P 8 31 8 31 œ 8 31 . Therefore, 2rh œ 4 8 1 gives the proportions that admit the # most light since ddrA# œ 4 3# 1 0.
Ê rœ
23. The fixed volume is V œ 1r# h 23 1r$ Ê h œ
V 1 r#
2r 3
, where h is the height of the cylinder and r is the radius
of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the 8 # surface area of the hemisphere. Thus, we minimize C œ 21rh 41r# œ 21r ˆ 1Vr# 2r3 ‰ 41r# œ 2V r 3 1r . Then œ
dC dr
œ 2V r#
4V"Î$ 1"Î$ †3#Î$
16 3
2†3"Î$ †V"Î$ 3†#†1"Î$
1r œ 0 Ê V œ œ
8 3
‰ 1r$ Ê r œ ˆ 3V 81
3"Î$ †2†4†V"Î$ 2†3"Î$ †V"Î$ 3†#†1"Î$
‰ œ ˆ 3V 1
"Î$
"Î$
. From the volume equation, h œ
. Since
d# C dr#
œ
4V r$
16 3
V 1 r#
1 0, these
dimensions do minimize the cost.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2r 3
Section 4.5 Applied Optimization 24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram the area of the cross section is A()) œ cos ) sin ) cos ), 0 ) 1# . Then Aw ()) œ sin ) cos# ) sin# ) œ a2 sin# ) sin ) 1b œ (2 sin ) 1)(sin ) 1) so Aw ()) œ 0 Ê sin ) œ
sin ) Á 1 when 0 )
1 #
w
. Also, A ()) 0 for 0 )
1 6
w
and A ()) 0 for
1 6
" #
or sin ) œ 1 Ê ) œ
)
1 #
1 6
because
. Therefore, at ) œ
1 6
there is a maximum. 25. (a) From the diagram we have: AP œ x, RA œ ÈL x# , PB œ 8.5 x, CH œ DR œ 11 RA œ 11 ÈL x# , QB œ Èx# (8.5 x)# , HQ œ 11 CH QB œ 11 ’11 ÈL x# Èx# (8.5 x)# “ #
#
#
œ ÈL x# Èx# (8.5 x)# , RQ œ RH HQ #
œ (8.5)# ŠÈL x# Èx# (8.5 x)# ‹ . It #
#
#
#
follows that RP œ PQ RQ Ê L# œ x# ŠÈL# x# Èx# (x 8.5)# ‹ (8.5)# Ê L# œ x# L# x# 2ÈL# x# È17x (8.5)# 17x (8.5)# (8.5)# Ê 17# x# œ 4 aL# x# b a17x (8.5)# b Ê L# œ x# œ
4x$ 4x 17
(b) If f(x) œ
2x$ 2x 8.5
œ
$
4x 4x17
51 8 ,
is minimized, then L# is minimized. Now f w (x) œ 51 8
. Thus L# is minimized when x œ
cylinder is formed, x œ 21r Ê r œ Then Vww (x)
17x$ 17x (8.5)#
œ
17x$ ‰# 17x ˆ 17 #
4x# (8x 51) (4x 17)#
Ê f w (x) 0 when x
51 8
51 8 .
then L ¸ 11.0 in.
26. (a) From the figure in the text we have P œ 2x 2y Ê y œ Ê V(x) œ
œ
.
and f w (x) 0 when x (c) When x œ
17# x# 4 c17x (8.5)# d
x #1
P #
x. If P œ 36, then y œ 18 x. When the
and h œ y Ê h œ 18 x. The volume of the cylinder is V œ 1r# h
x) 18x# x$ . Solving Vw (x) œ 3x(12 œ0 Ê 41 41 œ 13 ˆ3 x# ‰ Ê Vww (12) 0 Ê there is a
x œ 0 or 12; but when x œ 0, there is no cylinder. maximum at x œ 12. The values of x œ 12 cm and
y œ 6 cm give the largest volume. (b) In this case V(x) œ 1x# (18 x). Solving Vw (x) œ 31x(12 x) œ 0 Ê x œ 0 or 12; but x œ 0 would result in no cylinder. Then Vww (x) œ 61(6 x) Ê Vww (12) 0 Ê there is a maximum at x œ 12. The values of x œ 12 cm and y œ 6 cm give the largest volume. 27. Note that h# r# œ $ and so r œ È$ h# . Then the volume is given by V œ 1$ r# h œ 1$ a$ h# bh œ 1h 1$ h$ for dV # # ! h È$, and so dV dh œ 1 1r œ 1a" r b. The critical point (for h !) occurs at h œ ". Since dh ! for ! h ", and dV ! for " h È$, the critical point corresponds to the maximum volume. The cone of greatest dh
volume has radius È# m, height "m, and volume 28. Let d œ Éax 0b2 ay 0b2 œ Èx2 y2 and 2
#1 $
x a
m$ .
y b
œ 1 Ê y œ ba x b. We can minimize d by minimizing
2
2b2 a2 x
a b2 a2 b2
is the critical point Ê y œ ba Š a2ab b2 ‹ b œ
D œ ˆÈx2 y2 ‰ œ x2 ˆ ba x b‰ Ê D w œ 2x 2ˆ ba x b‰ˆ ba ‰ œ 2x Ê 2Šx
b2 a2 x
b2 a ‹
œ0Êxœ
2
2b2 a .
Dw œ 0
a2 b a2 b2 .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
221
222
Chapter 4 Applications of Derivatives
D ww œ 2 line
x a
y b
2b2 a2
Ê D ww Š a2ab b2 ‹ œ 2 2
2b2 a2
a2 b a2 b2 ‹
is the point on the
œ 1 that is closest to the origin.
29. Let Saxb œ x 1x , x 0 Ê S w axb œ 1 we only consider x œ 1. S ww axb œ 30. Let Saxb œ S ww axb œ
2
0 Ê the critical point is local minimum Ê Š a2ab b2 ,
2 x3
2 x3
1 x2 ww
œ
Ê S a1b
x2 1 x2 1 w x2 . S axb œ 0 Ê x2 œ œ 123 0 Ê local minimum
0 Ê x2 1 œ 0 Ê x œ „ 1. Since x 0, when x œ 1
4x2 , x 0 Ê S w axb œ x12 8x œ 8x x2 1 . S w axb œ 0 Ê 8x x2 1 œ 0 Ê 8x3 1 œ 0 Ê x œ "# . 8 Ê S ww ˆ "# ‰ œ a1Î22b3 8 0 Ê local minimum when x œ "# . 3
1 x
3
31. The length of the wire b œ perimeter of the triangle circumference of the circle. Let x œ length of a side of the equilateral triangle Ê P œ 3x, and let r œ radius of the circle Ê C œ 21r. Thus b œ 3x 21 r Ê r œ b 213x . The area of the circle is 1 r2 and the area of an equilateral triangle whose sides are x is "# axbŠ given by A œ Aw œ 0 Ê
È3 2 4 x
È3 2 x
3b 21
P œ 3Š È33b ‹œ 19 œb
9b È3 1 9
1 r2 œ
9 21 x
œ
È3 2 4 x .
Thus, the total area is
È3 2 È È ab 3xb2 Ê Aw œ 23 x 231 ab 3xb œ 23 x 23b1 4 x 41 È3 3b 9 ww È3 1 9 . A œ 2 21 0 Ê local minimum at the critical point. 2
1 ˆ b 213x ‰ œ
œ0Êxœ
9 21 x
m is the length of the trianglular segment and C œ 2 1ˆ b 213x ‰ œ b 3x
9b È3 1 9
È3 1 b È3 1 9
œ
È3 2 4 x
È3 2 x‹
m is the length of the circular segment.
32. The length of the wire b œ perimeter of the square circunference of the circle. Let x œ length of a side of the square Ê P œ 4x, and let r œ radius of the circle Ê C œ 21r. Thus b œ 4x 21 r Ê r œ b 214x . The area of the circle is 1 r2 and the area of a square whose sides are x is x2 . Thus, the total area is given by A œ x2 1 r2 2
œ x2 1 ˆ b 214x ‰ œ x2
ab 4xb2 41
Ê Aw œ 2x
8 18 x, Aw œ 0 Ê 2x 2b 1 1x œ 0 0 Ê local minimum at the critical point. P œ 4ˆ 4 b 1 ‰ œ 4 4b 1 m is the length of the square
Aww œ 2 18 segment and C œ 2 1ˆ b 214x ‰ œ b 4x œ b Êxœ
b 41.
4b 41
4 2 1 ab
œ
4xb œ 2x
b1 41
2b 1
m is the length of the circular segment.
33. Let ax, yb œ ˆx, 43 x‰ be the coordinates of the corner that intersects the line. Then base œ 3 x and height œ y œ 43 x, thus the area of therectangle is given by A œ a3 xbˆ 43 x‰ œ 4x 43 x2 , 0 Ÿ x Ÿ 3. Aw œ 4 83 x, Aw œ ! Ê x œ 32 . Aww œ 43 Ê Aww ˆ 32 ‰ 0 Ê local maximum at the critical point. The base œ 3
3 2
œ
3 2
and the height œ 43 ˆ 32 ‰ œ 2.
34. Let ax, yb œ Šx, È9 x2 ‹ be the coordinates of the corner that intersects the semicircle. Then base œ 2x and height œ y œ È9 x2 , thus the area of the inscribed rectangle is given by A œ a2xbÈ9 x2 , 0 Ÿ x Ÿ 3. Then 2 È 2ˆ9 x2 ‰ 2x2 Aw œ 2È9 x2 a2xb x œ œ 18 4x , Aw œ 0 Ê 18 4x2 œ 0 Ê x œ „ 3 2 , only x œ È 9 x2
È 9 x2
È 4 x2
2
3È 2 2
lies in
0 Ÿ x Ÿ 3. A is continuous on the closed interval 0 Ÿ x Ÿ 3 Ê A has an absolute maxima and absolute minima. È È Aa0b œ 0, Aa3b œ 0, and AŠ 3 2 ‹ œ Š3È2‹Š 3 2 ‹ œ 9 Ê absolute maxima. Base of rectangle is 3È2 and height 2
is
2
3È 2 2 .
35. (a) f(x) œ x# (b) f(x) œ x#
a x a x
Ê f w (x) œ x# a2x$ ab , so that f w (x) œ 0 when x œ 2 implies a œ 16 Ê f ww (x) œ 2x$ ax$ ab , so that f ww (x) œ 0 when x œ 1 implies a œ 1
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Section 4.5 Applied Optimization
223
36. If f(x) œ x$ ax# bx, then f w (x) œ 3x# 2ax b and f ww (x) œ 6x 2a. (a) A local maximum at x œ 1 and local minimum at x œ 3 Ê f w (1) œ 0 and f w (3) œ 0 Ê 3 2a b œ 0 and 27 6a b œ 0 Ê a œ 3 and b œ 9. (b) A local minimum at x œ 4 and a point of inflection at x œ 1 Ê f w (4) œ 0 and f ww (1) œ 0 Ê 48 8a b œ 0 and 6 2a œ 0 Ê a œ 3 and b œ 24. 37. (a) satb œ "'t# *'t ""# Ê vatb œ sw atb œ $#t *'. At t œ !, the velocity is va!b œ *' ft/sec. (b) The maximum height ocurs when vatb œ !, when t œ $. The maximum height is sa$b œ #&' ft and it occurs at t œ $ sec. (c) Note that satb œ "'t# *'t ""# œ "'at "bat (b, so s œ ! at t œ " or t œ (. Choosing the positive value of t, the velocity when s œ ! is va(b œ "#) ft/sec. 38.
Let x be the distance from the point on the shoreline nearest Jane's boat to the point where she lands her boat. Then she needs to row È% x# mi at 2 mph and walk ' x mi at 5 mph. The total amount of time to reach the village is faxb œ
È % x# #
'x &
have: #È%x x# œ
" &
hours (! Ÿ x Ÿ '). Then f w axb œ
" " # #È% x# a#xb
" &
œ
"& . Solving f w axb œ !, we
x #È% x#
Ê &x œ #È% x# Ê #&x# œ %a% x# b Ê #"x# œ "' Ê x œ „
% È#" .
We discard the negative
value of x because it is not in the domain. Checking the endpoints and critical point, we have fa!b œ #Þ#, fŠ È%#" ‹ ¸ #Þ"#, and fa'b ¸ $Þ"'. Jane should land her boat
% È#"
¸ !Þ)( miles down the shoreline from the point
nearest her boat. 39.
) x
œ
h x #(
Êhœ)
œ Ɉ)
#"' ‰# x
#"' x
and Laxb œ Éh# ax #(b#
ax #(b# when x !. Note that Laxb is
minimized when faxb œ ˆ)
#"' ‰# x
ax #(b# is
minimized. If f w axb œ !, then ‰ˆ #"' ‰ #ˆ) #"' x x# #ax #(b œ ! "(#) ‰ x$
Ê ax #(bˆ"
œ ! Ê x œ #( (not acceptable
since distance is never negative or x œ "#. Then La"#b œ È#"*( ¸ %'Þ)( ftÞ 40. (a) s" œ s# Ê sin t œ sin ˆt 13 ‰ Ê sin t œ sin t cos Êtœ
1 3
1 3
sin
1 3
" #
cos t Ê sin t œ
41 3
or
sin t
(b) The distance between the particles is s(t) œ ks" s# k œ ¸sin t sin ˆt 13 ‰¸ œ Ê sw (t) œ then s(0) œ
Šsin t È3 cos t‹ Šcos t È3 sin t‹ 2 ¹sin t È3 cos t¹ È3 #
since
d dx
kxk œ
x kx k
" #
È3 #
cos t Ê tan t œ È3
¹sin t È3 cos t¹
Ê critical times and endpoints are 0, 13 ,
, s ˆ 13 ‰ œ 0, s ˆ 561 ‰ œ 1, s ˆ 431 ‰ œ 0, s ˆ 1161 ‰ œ 1, s(21) œ
È3 #
51 41 111 6 , 3 , 6 ,
21;
Ê the greatest distance between the
particles is 1. (c) Since sw (t) œ
Šsin t È3 cos t‹ Šcos t È3 sin t‹ 2 ¹sin t È3 cos t¹
we can conclude that at t œ
1 3
and
41 w 3 , s (t)
has cusps and the distance
between the particles is changing the fastest near these points.
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224
Chapter 4 Applications of Derivatives
41. I œ
k d2 ,
let x œ distance the point is from the stronger light source Ê 6 x œ distance the point is from the other light
source. The intensity of illumination at the point from the stronger light is I1 œ point from the weaker light is I2 œ light Ê k1 œ 8k2 . Ê I1 œ
8k2 x2 .
k2 . a6 x b 2
The total intensity is given by I œ I1 I2 œ
3 3 16a6 xb3 k2 2x3 k2 and I w œ 0 Ê 16a6x3 ax6bk2xb3 2x k2 œ 0 Ê x 3 a6 x b 3 6k2 2 Ê I ww a4b œ 48k 44 a6 4b4 0 Ê local minimum. The point
v20 g sin 2!
œ
4v20 g
Ê
dR d!
œ
2v20 g cos 2!
and
dR d!
and intensity of illumination at the
Since the intensity of the first light is eight times the intensity of the second
œ
42. R œ
k1 x2 ,
œ0Ê
2v20 g cos 2!
8k2 x2
k2 a6 xb2
2 Ê I w œ 16k x3
2k2 a6 x b 3
16a6 xb3 k2 2x3 k2 œ 0 Ê x œ 4 m. I ww œ
48k2 x4
6k2 a6 x b 4
should be 4 m from the stronger light source.
œ 0 Ê ! œ 14 .
0 Ê local maximum. Thus, the firing angle of ! œ
1 4
d2 R d !2
œ
4v20 g sin 2!
Ê
d2 R d ! 2 ¹! œ 1
œ
4v20 ˆ1‰ g sin 2 4
4
œ 45‰ will maximize the range R.
43. (a) From the diagram we have d# œ 4r# w# . The strength of the beam is S œ kwd# œ kw a4r# w# b . When r œ 6, then S œ 144kw kw$ . Also, Sw (w) œ 144k 3kw# œ 3k a48 w# b so Sw (w) œ 0 Ê w œ „ 4È3 ; Sww Š4È3‹ 0 and 4È3 is not acceptable. Therefore S Š4È3‹ is the maximum strength. The dimensions of the strongest beam are 4È3 by 4È6 inches. (b)
(c)
Both graphs indicate the same maximum value and are consistent with each other. Changing k does not change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce the strongest beam). 44. (a) From the situation we have w# œ 144 d# . The stiffness of the beam is S œ kwd$ œ kd$ a144 d# b # # where 0 Ÿ d Ÿ 12. Also, Sw (d) œ 4kd a108 d b Ê critical points at 0, 12, and 6È3. Both d œ 0 and
"Î#
,
È144 d#
d œ 12 cause S œ 0. The maximum occurs at d œ 6È3. The dimensions are 6 by 6È3 inches. (b) (c)
Both graphs indicate the same maximum value and are consistent with each other. The changing of k has no effect. 45. (a) s œ 10 cos (1t) Ê v œ 101 sin (1t) Ê speed œ k101 sin (1t)k œ 101 ksin (1t)k Ê the maximum speed is 101 ¸ 31.42 cm/sec since the maximum value of ksin (1t)k is 1; the cart is moving the fastest at t œ 0.5 sec, 1.5 sec, 2.5 sec and 3.5 sec when ksin (1t)k is 1. At these times the distance is s œ 10 cos ˆ 1# ‰ œ 0 cm and a œ 101# cos (1t) Ê kak œ 101# kcos (1t)k Ê kak œ 0 cm/sec# (b) kak œ 101# kcos (1t)k is greatest at t œ 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec and 4.0 sec, and at these times the magnitude of the cart's position is ksk œ 10 cm from the rest position and the speed is 0 cm/sec.
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Section 4.5 Applied Optimization
225
46. (a) 2 sin t œ sin 2t Ê 2 sin t 2 sin t cos t œ 0 Ê (2 sin t)(1 cos t) œ 0 Ê t œ k1 where k is a positive integer (b) The vertical distance between the masses is s(t) œ ks" s# k œ ˆas" s# b# ‰ Ê sw (t) œ ˆ "# ‰ a(sin 2t 2 sin t)# b œ
4(2 cos t 1)(cos t ")(sin t)(cos t 1) ksin 2t 2 sin tk
"Î#
Ê the greatest distance is
3È 3 #
3È 3 #
at t œ
21 3
ds dt
œ
ds ¸ dt t=1
" #
a(12 12t)# 64t# b
"Î#
and
1,
41 3 ,
21; then s(0) œ 0, 3È 3 2 ,
s(21) œ 0
41 3 "Î#
[2(12 12t)(12) 128t] œ
208t 144 È(12 12t)# 64t#
Ê
ds ¸ dt t=0
œ 12 knots and
œ 8 knots
(c) The graph indicates that the ships did not see each other because s(t) 5 for all values of t.
(e)
21 3 ,
"Î#
2(cos 2t cos t)(sin 2t 2 sin t) ksin 2t 2 sin tk
, s(1) œ 0, s ˆ 431 ‰ œ ¸sin ˆ 831 ‰ 2 sin ˆ 431 ‰¸ œ
47. (a) s œ È(12 12t)# (8t)# œ a(12 12t)# 64t# b (b)
œ a(sin 2t 2 sin t)# b
(2)(sin 2t 2 sin t)(2 cos 2t 2 cos t) œ
Ê critical times at 0,
s ˆ 231 ‰ œ ¸sin ˆ 431 ‰ 2 sin ˆ 231 ‰¸ œ
"Î#
lim ds t Ä _ dt
œ
(208t 144)# É lim 144( " t)# 64t# tÄ_
(d) The graph supports the conclusions in parts (b) and (c).
Š208
œ Ë lim
#
144 t ‹ #
t Ä _ 144 Š " 1‹ 64 t
#
œ É 144208 64 œ È208 œ 4È13
which equals the square root of the sums of the squares of the individual speeds. 48. The distance OT TB is minimized when OB is a straight line. Hence n! œ n" Ê )" œ )# .
49. If v œ kax kx# , then vw œ ka 2kx and vww œ 2k, so vw œ 0 Ê x œ v
ww
ˆ #a ‰
œ 2k 0. The maximum value of v is
ka 4
#
a #
. At x œ
a #
there is a maximum since
.
50. (a) According to the graph, yw a!b œ !. (b) According to the graph, yw aLb œ !. (c) ya!b œ !, so d œ !. Now yw axb œ $ax# #bx c, so yw a!b œ ! implies that c œ !. There fore, yaxb œ ax$ bx# and yw axb œ $ax# #bx. Then yaLb œ aL$ bL# œ H and yw aLb œ $aL# #bL œ !, so we have two linear equations in two unknowns a and b. The second equation gives b œ $aL # . Substituting into the first equation, we have aL$
$aL$ #
œ H, or
aL$ #
œ H, so a œ # LH$ . Therefore, b œ $ LH# and the equation for y is $
#
yaxb œ # LH$ x$ $ LH# x# , or yaxb œ H’#ˆ Lx ‰ $ˆ Lx ‰ “.
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226
Chapter 4 Applications of Derivatives
51. The profit is p œ nx nc œ n(x c) œ ca(x c)" b(100 x)d (x c) œ a b(100 x)(x c) œ a (bc 100b)x 100bc bx# . Then pw (x) œ bc 100b 2bx and pww (x) œ 2b. Solving pw (x) œ 0 Ê x œ At x œ
c #
ww
50 there is a maximum profit since p (x) œ 2b 0 for all x.
c #
50.
52. Let x represent the number of people over 50. The profit is p(x) œ (50 x)(200 2x) 32(50 x) 6000 œ 2x# 68x 2400. Then pw (x) œ 4x 68 and pww œ 4. Solving pw (x) œ 0 Ê x œ 17. At x œ 17 there is a maximum since pww (17) 0. It would take 67 people to maximize the profit. 53. (a) A(q) œ kmq" cm h# q, where q 0 Ê Aw (q) œ kmq#
h #
œ
hq# 2km 2q#
and Aww (q) œ 2kmq$ . The
ww É 2km É 2km É 2km critical points are É 2km h , 0, and h , but only h is in the domain. Then A Š h ‹ 0 Ê at
q œ É 2km h there is a minimum average weekly cost. (b) A(q) œ
(kbq)m q
cm h# q œ kmq" bm cm h# q, where q 0 Ê Aw (q) œ 0 at q œ É 2km h as in (a).
Also Aww (q) œ 2kmq$ 0 so the most economical quantity to order is still q œ É 2km h which minimizes the average weekly cost. 54. We start with caxb œ the cost of producing x items, x !, and
c ax b x
œ the average cost of producing x items, assumed to
be differentiable. If the average cost can be minimized, it will be at a production level at which Ê
x c w ax b c a x b x#
d c ax b dx Š x ‹
œ ! (by the quotient rule) Ê x cw axb caxb œ ! (multiply both sides by x# ) Ê cw axb œ
œ!
c ax b x
where cw axb is
the marginal cost. This concludes the proof. (Note: The theorem does not assure a production level that will give a minimum cost, but rather, it indicates where to look to see if there is one. Find the production levels where the average cost equals the marginal cost, then check to see if any of them give a mimimum.) 55. The profit p(x) œ r(x) c(x) œ 6x ax$ 6x# 15xb œ x$ 6x# 9x, where x 0. Then pw (x) œ 3x# 12x 9 œ 3(x 3)(x 1) and pw w (x) œ 6x 12. The critical points are 1 and 3. Thus pww (1) œ 6 0 Ê at x œ 1 there is a local minimum, and pww (3) œ ' 0 Ê at x œ 3 there is a local maximum. But p(3) œ 0 Ê the best you can do is break even. 56. The average cost of producing x items is caxb œ
c ax b x
œ x# #!x #!ß !!! Ê c w axb œ #x #! œ ! Ê x œ "!, the
only critical value. The average cost is ca"!b œ $"*ß *!! per item is a minimum cost because c ww a"!b œ # !. 57. Let x œ the length of a side of the square base of the box and h œ the height of the box. V œ x2 h œ 48 Ê h œ total cost is given by C œ 6 † x 4a4 † x hb œ 6x 2
3 C w œ 0 Ê 12x x2 768 x œ 4 Ê h œ 48 42 œ
2
‰ 16xˆ 48 x2 ww
œ 6x
œ 0 Ê 12x3 768 œ 0 Ê x œ 4; C œ 12 2
3 and Ca4b œ 6a4b
768 4
w
x 0 Ê C œ 12x
2
768 x ,
1536 x2
Ê C ww a4b œ 12
1536 42
768 x2
œ
48 x2 . 12x3 768 x2
The
0 Ê local minimum.
œ 288 Ê the box is 4 ft ‚ 4 ft ‚ 3 ft, with a minimum cost of $288
58. Let x œ the number of $10 increases in the charge per room, then price per room œ 50 10x, and the number of rooms filled each night œ 800 40x Ê the total revenue is Raxb œ a50 10xba800 40xb œ 400x2 6000x 40000, ww 0 Ÿ x Ÿ 20 Ê Rw axb œ 800x 6000; Rw axb œ 0 Ê 800x 6000 œ 0 Ê x œ 15 2 ; R axb œ 800 ‰ ˆ 15 ‰ Ê Rww ˆ 15 2 œ 800 0 Ê local maximum. The price per room is 50 10 2 œ $125. 59. We have
dR dM
œ CM M# . Solving
d# R dM#
œ C #M œ ! Ê M œ C# . Also,
d$ R dM$
œ # ! Ê at M œ
maximum.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
C #
there is a
Section 4.5 Applied Optimization
227
60. (a) If v œ cr! r# cr$ , then vw œ 2cr! r 3cr# œ cr a2r! 3rb and vww œ 2cr! 6cr œ 2c ar! 3rb . The solution of vw œ 0 is r œ 0 or 2r3! , but 0 is not in the domain. Also, vw 0 for r 2r3! and vw 0 for r 2r3! Ê at r œ 2r3! there is a maximum. (b) The graph confirms the findings in (a).
61. If x 0, then (x 1)# 0 Ê x# 1 2x Ê then
# # # # Š a a 1 ‹ Š b b 1 ‹ Š c c 1 ‹ Š d d " ‹
62. (a) f(x) œ
x È a# x#
Ê f w (x) œ
aa # x # b
"Î#
x # aa # x # b aa # x # b
Ê gw (x) œ
ab# (d x)# b (d x)# ab# (d x)# b$Î#
ab# (d x)# b
"Î#
œ
a# x# x# aa# x# b$Î#
œ
a# aa# x# b$Î#
0
"Î# "Î# (d x)# ab# (d x)# b b# (d x)#
b # 0 Ê g(x) is a decreasing function of x ab# (d x)# b$Î# dt Since c" , c# 0, the derivative dx is an increasing function of x (from part (a)) minus a decreasing dt d# t " w " w w function of x (from part (b)): dx œ c"" f(x) c"# g(x) Ê dx # œ c" f (x) c# g (x) 0 since f (x) dt gw (x) 0 Ê dx is an increasing function of x.
œ
(c)
dx Èb# (d x)#
2. In particular if a, b, c and d are positive integers,
16.
Ê f(x) is an increasing function of x (b) g(x) œ
x# 1 x
œ
0 and
63. At x œ c, the tangents to the curves are parallel. Justification: The vertical distance between the curves is D(x) œ f(x) g(x), so Dw (x) œ f w (x) gw (x). The maximum value of D will occur at a point c where Dw œ 0. At such a point, f w (c) gw (c) œ 0, or f w (c) œ gw (c). 64. (a) f(x) œ 3 4 cos x cos 2x is a periodic function with period 21 (b) No, f(x) œ 3 4 cos x cos 2x œ 3 4 cos x a2 cos# x 1b œ 2 a1 2 cos x cos# xb œ 2(1 cos x)# 0 Ê f(x) is never negative. 65. (a) If y œ cot x È2 csc x where 0 x 1, then yw œ (csc x) ŠÈ2 cot x csc x‹. Solving yw œ 0 Ê cos x œ Ê x œ 14 . For 0 x
1 4
we have yw 0, and yw 0 when
1 4
x 1. Therefore, at x œ
value of y œ 1. (b)
The graph confirms the findings in (a).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 4
" È2
there is a maximum
228
Chapter 4 Applications of Derivatives
66. (a) If y œ tan x 3 cot x where 0 x
1 #
, then yw œ sec# x 3 csc# x. Solving yw œ 0 Ê tan x œ „ È3
Ê x œ „ 13 , but 13 is not in the domain. Also, yww œ 2 sec# x tan x 6 csc# x cot x 0 for all 0 x Therefore at x œ 1 there is a minimum value of y œ 2È3.
1 2
.
3
(b)
The graph confirms the findings in (a). # # 67. (a) The square of the distance is Daxb œ ˆx $# ‰ ˆÈx !‰ œ x# #x *% , so Dw axb œ #x # and the critical
point occurs at x œ ". Since Dw axb ! for x " and Dw axb ! for x ", the critical point corresponds to the minimum distance. The minimum distance is ÈDa"b œ
È& # .
(b)
The minimum distance is from the point ˆ $# ß !‰ to the point a"ß "b on the graph of y œ Èx, and this occurs at the
value x œ " where Daxb, the distance squared, has its minimum value. 68. (a) Calculus Method:
The square of the distance from the point Š"ß È$‹ to Šxß È"' x# ‹ is given by #
Daxb œ ax "b# ŠÈ"' x# È$‹ œ x# #x " "' x# #È%) $x# $ œ #x #! #È%) $x# . Then Dw axb œ #
" #
†
# È%) $x# a'xb
œ #
'x È%) $x# .
Solving Dw axb œ ! we have: 'x œ #È%) $x#
Ê $'x# œ %a%) $x# b Ê *x# œ %) $x# Ê "#x# œ %) Ê x œ „ #. We discard x œ # as an extraneous solution, leaving x œ #. Since Dw axb ! for % x # and Dw axb ! for # x %, the critical point corresponds to the minimum distance. The minimum distance is ÈDa#b œ #. Geometry Method: The semicircle is centered at the origin and has radius %. The distance from the origin to Š"ß È$‹ is #
Ê"# ŠÈ$‹ œ #. The shortest distance from the point to the semicircle is the distance along the radius containing the point Š"ß È$‹. That distance is % # œ #.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.6 Newton's Method
229
(b)
The minimum distance is from the point Š"ß È$‹ to the point Š#ß #È$‹ on the graph of y œ È"' x# , and this occurs at the value x œ # where Daxb, the distance squared, has its minimum value. 4Þ6 NEWTON'S METHOD 1. y œ x# x 1 Ê yw œ 2x 1 Ê xnb1 œ xn Ê x# œ
2 3
4 9
23 1 4 3 1
Ê x# œ 2
Ê x# œ
42" 4 1
œ
2 3
4 6 9 129
œ
x#n xn 1 # xn 1
2 3
" #1
#"7 1 1 " 3 3
œ 3"
" 90
6 5
1296 6 625 5 3 864 1 125
œ 2 Ê x# œ 2
œ
6 5
16 2 3 32 1
1296 750 1875 4320 625
œ 2
11 31
1 "4 " œ #" #1 20 25 4 29 œ #5 1"# œ 12 1 #
Ê x# œ "#
œ
¸ 2.41667
" 1#
5 4
113 2000
œ
2500113 2000
œ
2387 #000
6. From Exercise 5, xnb1 œ xn œ 54
625512 2000
œ 54
111 # 1
œ 2
" 3
œ "3
œ
6 5
171 4945
; x! œ 1 Ê x " œ 1
œ
5763 4945
1 1 3 4 1
œ
6 5
¸ 1.16542; x! œ 1 Ê x" œ 1
"13 4 1
œ 51 31 ¸ 1.64516 2xn x#n 1 2 2xn
; x! œ 0 Ê x" œ 0
x%n 2 4xn$ ; x!
œ 1 Ê x" œ 1
"2 4
œ
5 4
00" #0
œ "#
44" #4
œ
5 #
Ê x# œ
5 4
Ê x# œ
5 #
625 256 2 125 16
œ
5 4
5 25 4 1 #5
625512 2000
¸ 1.1935
x%n 2 4xn$
113 2000
; x! œ 0 Ê x" œ 0
œ 15# ¸ .41667; x! œ 2 Ê x" œ 2
5. y œ x% 2 Ê yw œ 4x$ Ê xnb1 œ xn œ
2 3
¸ .61905; x! œ 1 Ê x" œ 1
x$n 3xn 1 3xn# 3
x%n xn 3 4xn$ 1
4. y œ 2x x# 1 Ê yw œ 2 2x Ê xnb1 œ xn 5 #
œ
29 œ 90 ¸ 0.32222
3. y œ x% x 3 Ê yw œ 4x$ 1 Ê xnb1 œ xn Ê x# œ
13 21
111 #1
¸ 1.66667
5 3
2. y œ x$ 3x 1 Ê yw œ 3x# 3 Ê xnb1 œ xn Ê x# œ "3
œ
; x! œ 1 Ê x " œ 1
; x! œ 1 Ê x" œ 1
"2 4
œ 1
" 4
œ 54 Ê x# œ 54
625 256 2 125 16
¸ 1.1935
7. f(x! ) œ 0 and f w (x! ) Á ! Ê xnb1 œ xn
f axn b f axn b w
gives x" œ x! Ê x# œ x! Ê xn œ x! for all n 0. That is, all of
the approximations in Newton's method will be the root of f(x) œ 0. 8. It does matter. If you start too far away from x œ x! œ 0.5, for instance, leads to x œ
1 #
1 #
, the calculated values may approach some other root. Starting with
as the root, not x œ
1 #
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
230
Chapter 4 Applications of Derivatives
9. If x! œ h 0 Ê x" œ x! œh
Èh Š
f(x! ) f w (x! )
Èh
Š È" ‹ 2
f(h) f w (h)
œ h ŠÈh‹ Š2Èh‹ œ h;
"
‹ #Èh
if x! œ h 0 Ê x" œ x! œ h
œh
f(x! ) f w (x! )
œ h
f(h) f w (h)
œ h ŠÈh‹ Š2Èh‹ œ h.
h
"Î$
10. f(x) œ x"Î$ Ê f w (x) œ ˆ "3 ‰ x#Î$ Ê xnb1 œ xn " xn #Î$ ˆ ‰ xn 3
œ 2xn ; x! œ 1 Ê x" œ 2, x# œ 4, x$ œ 8, and x% œ 16 and so forth. Since kxn k œ 2lxnc1 l we may conclude that n Ä _ Ê kxn k Ä _.
11. i) is equivalent to solving x$ $x " œ !. ii) is equivalent to solving x$ $x " œ !. iii) is equivalent to solving x$ $x " œ !. iv) is equivalent to solving x$ $x " œ !. All four equations are equivalent. 12. f(x) œ x 1 0.5 sin x Ê f w (x) œ 1 0.5 cos x Ê xnb1 œ xn 13. f(x) œ tan x 2x Ê f w (x) œ sec# x 2 Ê xnb1 œ xn
xn 1 0.5 sin xn 1 0.5 cos xn
tan axn b 2xn sec# axn b
; if x! œ 1.5, then x" œ 1.49870
; x! œ 1 Ê x" œ 1.2920445
Ê x# œ 1.155327774 Ê x16 œ x17 œ 1.165561185 14. f(x) œ x% 2x$ x# 2x 2 Ê f w (x) œ 4x$ 6x# 2x 2 Ê xnb1 œ xn
x%n 2xn$ xn# 2xn 2 4xn$ 6xn# 2xn 2
if x! œ 0.5, then x% œ 0.630115396; if x! œ 2.5, then x% œ 2.57327196 15. (a) The graph of f(x) œ sin 3x 0.99 x# in the window 2 Ÿ x Ÿ 2, 2 Ÿ y Ÿ 3 suggests three roots. However, when you zoom in on the x-axis near x œ 1.2, you can see that the graph lies above the axis there. There are only two roots, one near x œ 1, the other near x œ 0.4. (b) f(x) œ sin 3x 0.99 x# Ê f w (x) œ 3 cos 3x 2x Ê xnb1 œ xn
sin (3xn ) 0.99xn# 3 cos (3xn ) 2xn
and the solutions
are approximately 0.35003501505249 and 1.0261731615301
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
Section 4.6 Newton's Method 16. (a) Yes, three times as indicted by the graphs (b) f(x) œ cos 3x x Ê f w (x) œ 3 sin 3x 1 Ê xnb1 œ xn
cos a3xn b xn 3 sin a3xn b 1
; at
approximately 0.979367, 0.887726, and 0.39004 we have cos 3x œ x 17. f(x) œ 2x% 4x# 1 Ê f w (x) œ 8x$ 8x Ê xnb1 œ xn
2x%n 4xn# 1 8xn$ 8xn
; if x! œ 2, then x' œ 1.30656296; if
x! œ 0.5, then x$ œ 0.5411961; the roots are approximately „ 0.5411961 and „ 1.30656296 because f(x) is an even function. 18. f(x) œ tan x Ê f w (x) œ sec# x Ê xnb1 œ xn
tan axn b sec# axn b
; x! œ 3 Ê x" œ 3.13971 Ê x# œ 3.14159 and we
approximate 1 to be 3.14159. 19. From the graph we let x! œ 0.5 and f(x) œ cos x 2x Ê xnb1 œ xn
cos axn b 2xn sin axn b 2
Ê x" œ .45063
Ê x# œ .45018 Ê at x ¸ 0.45 we have cos x œ 2x.
20. From the graph we let x! œ 0.7 and f(x) œ cos x x Ê xnb1 œ xn
xn cos axn b 1 sin axn b
Ê x" œ .73944
Ê x# œ .73908 Ê at x ¸ 0.74 we have cos x œ x.
21. The x-coordinate of the point of intersection of y œ x2 ax 1b and y œ Ê x3 x2
1 x
Ê xnb1 œ xn
1 x
œ 0 Ê The x-coordinate is the root of faxb œ x3 x2 xn xn x1n 3x2n 2xn 12 x 3
2
is the solution of x2 ax 1b œ 1 x
Ê f w axb œ 3x2 2x
1 x2 .
1 x
Let x0 œ 1
Ê x" œ 0.83333 Ê x2 œ 0.81924 Ê x3 œ 0.81917 Ê x7 œ 0.81917 Ê r ¸ 0.8192
n
22. The x-coordinate of the point of intersection of y œ Èx and y œ 3 x2 is the solution of Èx œ 3 x2 1 Ê Èx 3 x2 œ 0 Ê The x-coordinate is the root of faxb œ Èx 3 x2 Ê f w axb œ 2È 2x. Let x0 œ 1 x Ê xnb1 œ xn
Èxn 3 x2n
Èxn 2xn 1
2
Ê x" œ 1.4 Ê x2 œ 1.35556 Ê x3 œ 1.35498 Ê x7 œ 1.35498 Ê r ¸ 1.3550
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
231
232
Chapter 4 Applications of Derivatives
23. If f(x) œ x$ 2x 4, then f(1) œ 1 0 and f(2) œ 8 0 Ê by the Intermediate Value Theorem the equation x$ 2x 4 œ 0 has a solution between 1 and 2. Consequently, f w (x) œ 3x# 2 and xnb1 œ xn
x$n 2xn 4 3x#n 2
.
Then x! œ 1 Ê x" œ 1.2 Ê x# œ 1.17975 Ê x$ œ 1.179509 Ê x% œ 1.1795090 Ê the root is approximately 1.17951. 24. We wish to solve 8x% 14x$ 9x# 11x 1 œ 0. Let f(x) œ 8x% 14x$ 9x# 11x 1, then f w (x) œ 32x$ 42x# 18x 11 Ê xnb1 œ xn x! 1.0 0.1 0.6 2.0
8x%n 14xn$ 9xn# 11xn 1 3#xn$ 42xn# 18xn 11
.
approximation of corresponding root 0.976823589 0.100363332 0.642746671 1.983713587
25. f(x) œ 4x% 4x# Ê f w (x) œ 16x$ 8x Ê xib1 œ xi
faxi b f axi b w
œ xi
xi$ xi . %x#i #
Iterations are performed using the
procedure in problem 13 in this section. (a) For x! œ # or x! œ !Þ), xi Ä " as i gets large. (b) For x! œ !Þ& or x! œ !Þ#&, xi Ä ! as i gets large. (c) For x! œ !Þ) or x! œ #, xi Ä " as i gets large. (d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.) For x! œ x! œ
È21 7
È 721
or x! œ
or x! œ
È 721
È21 7 ,
Newton's method does not converge. The values of xi alternate between
as i increases.
26. (a) The distance can be represented by # D(x) œ É(x 2)# ˆx# "# ‰ , where x 0. The
distance D(x) is minimized when # f(x) œ (x 2)# ˆx# "# ‰ is minimized. If
# f(x) œ (x 2)# ˆx# "# ‰ , then
f w (x) œ 4 ax$ x 1b and f ww (x) œ 4 a3x# 1b 0. Now f w (x) œ 0 Ê x$ x 1 œ 0 Ê x ax# 1b œ 1 Ê x œ x#"1 . (b) Let g(x) œ
" x # 1
Ê xnb1 œ xn
x œ ax# 1b " xn
Πx#
n 1
Î
2xn Ñ # x# 1 1
ÏŠ
‹
n
"
#
x Ê gw (x) œ ax# 1b (2x) 1 œ
2x ax # 1 b #
1
; x! œ 1 Ê x% œ 0.68233 to five decimal places.
Ò
27. f(x) œ (x 1)%! Ê f w (x) œ 40(x 1)$* Ê xnb1 œ xn
axn 1b%! 40 axn 1b$*
39xn " 40
œ
. With x! œ 2, our computer
gave x)( œ x)) œ x)* œ â œ x#!! œ 1.11051, coming within 0.11051 of the root x œ 1. 28. Since s œ r ) Ê 3 œ r ) Ê ) œ 3r . Bisect the angle ) to obtain a right tringle with hypotenuse r and opposite side of length 1. Then sin farb œ sinˆ 2r3 ‰
1 r
) 2
œ
1 r
Ê sin
ˆ 3r ‰ 2
œ
1 r
Ê sinˆ 2r3 ‰ œ
1 r
Ê sin
3 2r
Ê f w arb œ 2r32 cosˆ 2r3 ‰ r12 ; r0 œ 1 Ê rn 1 œ rn
Ê r2 œ 1.00282 Ê r3 œ 1.00282 Ê r ¸ 1.0028 Ê ) œ
3 1.00282
1 r
œ 0. Thus the solution r is a root of
sinˆ 2r3n ‰ r1n 32 cosˆ 2r3n ‰ 2rn
1 r2 n
Ê r1 œ 1.00280
¸ 2.9916
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.7 Antiderivatives 4.7 ANTIDERIVATIVES 1. (a) x#
(b)
x$ 3
(c)
x$ 3
x# x
2. (a) 3x#
(b)
x) 8
(c)
x) 8
3x# 8x
3. (a) x$
(b) x3
4. (a) x#
(b) x4
$
$
(c) x3 x# 3x
#
x$ 3
(c)
x # #
x# 2
x
5. (a)
" x
(b)
5 x
(c) 2x
6. (a)
" x#
(b)
" 4x#
(c)
x% 4
(c)
2 3
Èx$ 2Èx
(c)
3 4
x%Î$ 3# x#Î$
5 x " #x #
7. (a) Èx$
(b) Èx
8. (a) x%Î$
(b)
9. (a) x#Î$
(b) x"Î$
(c) x"Î$
10. (a) x"Î#
(b) x"Î#
(c) x$Î#
11. (a) cos (1x)
(b) 3 cos x
(c)
12. (a) sin (1x)
(b) sin ˆ 1#x ‰
(c) ˆ 12 ‰ sin ˆ 1#x ‰ 1 sin x
13. (a) tan x
(b) 2 tan ˆ x3 ‰
‰ (c) 23 tan ˆ 3x #
14. (a) cot x
‰ (b) cot ˆ 3x #
(c) x 4 cot (2x)
15. (a) csc x
(b)
" 5
csc (5x)
(c) 2 csc ˆ 1#x ‰
16. (a) sec x
(b)
4 3
sec (3x)
(c)
17.
' (x 1) dx œ
19.
'
ˆ3t# #t ‰ dt œ t$
21.
'
a2x$ 5x 7b dx œ
23.
'
ˆ x"# x# 3" ‰ dx œ ' ˆx# x# 3" ‰ dx œ
24.
'
ˆ "5
2 x$
x# #
" #
x#Î$
xC t# 4
C " #
x% 5# x# 7x C x " 1
2x‰ dx œ ' ˆ 5" 2x$ 2x‰ dx œ
" 5
x$ 3
cos (1x) 1
2 1
cos (3x)
sec ˆ 1#x ‰
18.
' (5 6x) dx œ 5x 3x# C
20.
'
Š t# 4t$ ‹ dt œ
22.
'
a1 x# 3x& b dx œ x "3 x$ #" x' C
#
3" x C œ x" #
x Š 2x# ‹
2x# #
Cœ
x 5
x$ 3
t$ 6
" x#
t% C
x 3
C
x# C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
233
234
Chapter 4 Applications of Derivatives x#Î$
25.
' x"Î$ dx œ
27.
'
ˆÈx $Èx‰ dx œ ' ˆx"Î# x"Î$ ‰ dx œ
28.
'
Š
29.
'
Š8y
30.
'
Èx #
Š "7
2 3
2 Èx ‹
2 ‹ y"Î%
1 ‹ y&Î%
Cœ
3 #
x#Î$ C
' x&Î% dx œ
26. x$Î#
dx œ ' ˆ "# x"Î# 2x"Î# ‰ dx œ dy œ ' ˆ8y 2y"Î% ‰ dy œ
dy œ ' ˆ "7 y&Î% ‰ dy œ
" 7
8y# #
x%Î$
3 #
" #
Cœ
4 3
2 3
x "Î% "4
Cœ
4 % x È
C
x$Î# 43 x%Î$ C
$Î#
"Î#
#
#
Šx3 ‹ 2 Šx" ‹ C œ
" 3
x$Î# 4x"Î# C
$Î%
2 Š y 3 ‹ C œ 4y# 83 y$Î% C 4
"Î%
y Š y 1 ‹ C œ 4
y 7
4 y"Î%
C
31.
'
2x a1 x$ b dx œ ' a2x 2x# b dx œ
32.
'
x$ (x 1) dx œ ' ax# x$ b dx œ
33.
'
tÈtÈt t#
34.
'
4 È t t$
35.
' 2 cos t dt œ 2 sin t C
36.
' 5 sin t dt œ 5 cos t C
37.
' 7 sin 3) d) œ 21 cos 3) C
38.
' 3 cos 5) d) œ 35 sin 5) C
39.
' 3 csc# x dx œ 3 cot x C
40.
' sec3# x dx œ tan3 x C
41.
'
42.
'
43.
' a4 sec x tan x 2 sec# xb dx œ 4 sec x 2 tan x C
44.
'
45.
' asin 2x csc# xb dx œ "# cos 2x cot x C
46.
' (2 cos 2x 3 sin 3x) dx œ sin 2x cos 3x C
47.
'
1 cos 4t #
dt œ ' ˆ "#
" #
cos 4t‰ dt œ
" #
t "# ˆ sin4 4t ‰ C œ
t 2
sin 4t 8
C
48.
'
1 cos 6t #
dt œ ' ˆ "#
" #
cos 6t‰ dt œ
" #
t "# ˆ sin6 6t ‰ C œ
t 2
sin 6t 12
C
49.
' a1 tan# )b d) œ ' sec# ) d) œ tan ) C
50.
' a2 tan# )b d) œ ' a1 1 tan# )b d) œ ' a1 sec# )b d) œ ) tan ) C
51.
' cot# x dx œ ' acsc# x 1b dx œ cot x x C
$Î#
dt œ ' Š t4$
csc ) cot ) #
" 2
dt œ ' Š t t#
t"Î# t# ‹
t"Î# t$ ‹
2x# #
x " 1
"
2 Š x1 ‹ C œ x# #
Š x# ‹ C œ "x
dt œ ' ˆt"Î# t$Î# ‰ dt œ
t"Î#
" #
2 x
C
" #x #
C
"Î# Š t " ‹ C œ 2Èt #
t dt œ ' ˆ4t$ t&Î# ‰ dt œ 4 Š # ‹ Š t 3 ‹ C œ t2#
#
$Î# #
d) œ "# csc ) C
acsc# x csc x cot xb dx œ #" cot x
" #
2 5
sec ) tan ) d) œ
2 5
2 Èt
C
2 3t$Î#
C
sec ) C
csc x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.7 Antiderivatives 52.
' a1 cot# xb dx œ ' a1 acsc# x 1bb dx œ ' a2 csc# xb dx œ 2x cot x C
53.
' cos ) (tan ) sec )) d) œ ' (sin ) 1) d) œ cos ) ) C
54.
'
csc ) csc ) sin )
55.
d dx
2) C‹ œ Š (7x28
56.
d dx
Š (3 x3 5)
57.
d dx
ˆ "5 tan (5x 1) C‰ œ
58.
d dx
ˆ3 cot ˆ x 3 " ‰ C‰ œ 3 ˆcsc# ˆ x 3 " ‰‰ ˆ "3 ‰ œ csc# ˆ x 3 " ‰
59.
d dx
# ˆ x" ‰ œ 1 C œ (1)(1)(x 1)
) )‰ ' ‰ ˆ sin d) œ ' ˆ csc csc ) sin ) sin ) d) œ
%
"
4(7x 2)$ (7) 28
d) œ '
# (3)
‹ œ (3x 5)#
asec# (5x 1)b (5) œ sec# (5x 1)
#
" (x 1)# x# #
d dx
Š x# sin x C‹ œ
(b) Wrong:
d dx
(x cos x C) œ cos x x sin x Á x sin x
(c) Right:
d d) d d)
(b) Right: (c) Right:
2x #
d dx
ˆ xx 1 C‰ œ x# #
cos x œ x sin x
$
Š sec3 ) C‹ œ
3 sec# ) 3
ˆ "# tan# ) C‰ œ "# (2 tan )) sec# ) œ tan ) sec# ) ˆ "# sec# ) C‰ œ "# (2 sec )) sec ) tan ) œ tan ) sec# ) $
3(2x 1)# (2) 3
Š (2x 3 1) C‹ œ
(b) Wrong:
d dx
a(2x 1)$ Cb œ 3(2x 1)# (2) œ 6(2x 1)# Á 3(2x 1)#
d dx
ax# x Cb
(b) Wrong:
d dx
Šax# xb
65. Right:
66. Wrong:
d dx
Š
"Î#
"Î#
œ
" #
ax# x Cb
C‹ œ $
" #
ax# xb
" Œ 3 ŠÈ2x 1‹ C œ
d ˆ x 3 ‰3 dx Š x 2
d dx
œ 2(2x 1)# Á (2x 1)#
a(2x 1)$ Cb œ 6(2x 1)#
64. (a) Wrong:
(c) Right:
" (x 1)#
(sec ) tan )) œ sec$ ) tan ) Á tan ) sec# )
d dx
d dx
œ
cos x Á x sin x
63. (a) Wrong:
(c) Right:
(x 1)(") x(1) (x 1)#
(x cos x sin x C) œ cos x x sin x cos x œ x sin x
d d)
62. (a) Wrong:
sin x
60.
61. (a) Wrong:
d dx
d) œ ' sec# ) d) œ tan ) C
" cos# )
œ (7x 2)$
C‹ œ Š (3x 35) " 5
" 1sin# )
sinˆx2 ‰ x
3‰ C‹ œ 3ˆ xx 2
C‹ œ
d dx
"Î#
"Î#
(2x 1) œ
(2x 1) œ
2x 1 2 È x# x C
2x 1 2 È x# x
ˆ 3" (2x 1)$Î# C‰ œ
2 ax 2b†1 ax 3b†1 ax 2 b
2
x†cosˆx2 ‰a2xb sinˆx2 ‰†1 x2
œ
3 6
2
Á
Á È2x 1
(2x 1)"Î# (2) œ È2x 1
3b 5 œ 3 aaxx œ 2 b2 ax 2 b 2
2x2 cosˆx2 ‰ sinˆx2 ‰ x2
Á È2x 1
15ax 3b2 ax 2 b 4
x cosˆx2 ‰ sinˆx2 ‰ x2
67. Graph (b), because
dy dx
œ 2B Ê y œ x# C. Then y(1) œ 4 Ê C œ 3.
68. Graph (b), because
dy dx
œ B Ê y œ "# x# C. Then y(1) œ 1 Ê C œ
3 #
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
235
236
Chapter 4 Applications of Derivatives
69.
dy dx
œ 2x 7 Ê y œ x# 7x C; at x œ 2 and y œ 0 we have 0 œ 2# 7(2) C Ê C œ 10 Ê y œ x# 7x 10
70.
dy dx
œ 10 x Ê y œ 10x
71.
dy dx
œ
" x#
C; at x œ 0 and y œ 1 we have 1 œ 10(0)
x œ x# x Ê y œ x"
Ê y œ x" 72.
x# #
x# #
" #
or y œ x"
x# #
0# #
C Ê C œ 1 Ê y œ 10x
x# #
C; at x œ 2 and y œ 1 we have 1 œ 2"
" #
2# #
x# #
1
C Ê C œ "#
dy dx
œ 9x# 4x 5 Ê y œ 3x$ 2x# 5x C; at x œ 1 and y œ 0 we have 0 œ 3(1)$ 2(1)# 5(1) C
dy dx
œ $x#Î$ Ê y œ
Ê C œ 10 Ê y œ 3x$ 2x# 5x 10
73.
$x"Î$ " $
C œ *; at x œ 9x"Î$ C; at x œ " and y œ & we have & œ *(")"Î$ C Ê C œ %
Ê y œ 9x"Î$ % " #È x
œ
" #
x"Î# Ê y œ x"Î# C; at x œ 4 and y œ 0 we have 0 œ 4"Î# C Ê C œ 2 Ê y œ x"Î# 2
74.
dy dx
œ
75.
ds dt
œ 1 cos t Ê s œ t sin t C; at t œ 0 and s œ 4 we have 4 œ 0 sin 0 C Ê C œ 4 Ê s œ t sin t 4
76.
ds dt
œ cos t sin t Ê s œ sin t cos t C; at t œ 1 and s œ 1 we have 1 œ sin 1 cos 1 C Ê C œ 0
Ê s œ sin t cos t 77.
dr d)
œ 1 sin 1) Ê r œ cos (1)) C; at r œ 0 and ) œ 0 we have 0 œ cos (10) C Ê C œ " Ê r œ cos (1)) 1
78.
dr d)
œ cos 1) Ê r œ
79.
dv dt
œ
80.
dv dt
œ 8t csc# t Ê v œ 4t# cot t C; at v œ 7 and t œ
" #
" 1
sin(1)) C; at r œ 1 and ) œ 0 we have 1 œ
sec t tan t Ê v œ
" #
sec t C; at v œ 1 and t œ 0 we have 1 œ 1 #
Ê v œ 4t# cot t 7 1#
81.
d# y dx#
Ê
œ 2 6x Ê dy dx
œ 2x 3x# C" ; at
dy dx #
" 1
dy dx
sin (10) C Ê C œ " Ê r œ " #
sec (0) C Ê C œ
" #
" 1
Ê vœ
sin (1)) 1 " #
sec t
" #
#
we have 7 œ 4 ˆ 1# ‰ cot ˆ 1# ‰ C Ê C œ 7 1#
œ 4 and x œ 0 we have 4 œ 2(0) 3(0)# C" Ê C" œ 4
œ 2x 3x 4 Ê y œ x# x$ 4x C# ; at y œ 1 and x œ 0 we have 1 œ 0# 0$ 4(0) C# Ê C# œ 1
Ê y œ x# x$ 4x 1 82.
d# y dx#
œ0 Ê
dy dx
œ C" ; at
dy dx
œ 2 and x œ 0 we have C" œ 2 Ê
dy dx
œ 2 Ê y œ 2x C# ; at y œ 0 and x œ 0 we
have 0 œ 2(0) C# Ê C# œ 0 Ê y œ 2x 83.
d# r dt#
œ
d# s dt#
œ
2 t$
œ 2t$ Ê
dr dt
œ t# C" ; at
dr dt
œ 1 and t œ 1 we have 1 œ (1)# C" Ê C" œ 2 Ê
dr dt
œ t# 2
Ê r œ t" 2t C# ; at r œ 1 and t œ 1 we have 1 œ 1" 2(1) C# Ê C# œ 2 Ê r œ t" 2t 2 or r œ "t 2t 2 84.
3t 8
Ê
ds dt
œ
3t# 16
C" ; at
s œ 4 and t œ 4 we have 4 œ
ds dt $
4 16
œ 3 and t œ 4 we have 3 œ
C# Ê C# œ 0 Ê s œ
3(4)# 16
C" Ê C" œ 0 Ê
ds dt
œ
3t# 16
$
t 16
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ê sœ
t$ 16
C# ; at
Section 4.7 Antiderivatives 85.
d$ y dx$
œ6 Ê
Ê
dy dx
d# y dx# #
œ 6x C" ; at
œ 3x 8x C# ; at
d$ ) dt$
œ0 Ê
#
œ 8 and x œ 0 we have 8 œ 6(0) C" Ê C" œ 8 Ê
œ 0 and x œ 0 we have 0 œ 3(0)# 8(0) C# Ê C# œ 0 Ê
d# y dx# œ 6x 8 dy # dx œ 3x 8x $ #
Ê y œ x 4x C$ ; at y œ 5 and x œ 0 we have 5 œ 0$ 4(0)# C$ Ê C$ œ 5 Ê y œ x 4x 5 86.
$
dy dx
d# y dx#
237
d# ) dt#
œ C" ; at
d# ) dt#
d# ) d) dt# œ 2 Ê dt " d) # dt œ 2t # Ê ) œ t ) œ t# "# t È2
œ 2 and t œ 0 we have
have "# œ 2(0) C# Ê C# œ "# Ê È2 œ 0# " (0) C$ Ê C$ œ È2 Ê #
œ "# and t œ 0 we "# t C$ ; at ) œ È2 and t œ 0 we have œ 2t C# ; at
d) dt
87. yÐ%Ñ œ sin t cos t Ê ywww œ cos t sin t C" ; at ywww œ 7 and t œ 0 we have 7 œ cos (0) sin (0) C" Ê C" œ 6 Ê ywww œ cos t sin t 6 Ê yww œ sin t cos t 6t C# ; at yww œ 1 and t œ 0 we have 1 œ sin (0) cos (0) 6(0) C# Ê C# œ 0 Ê yww œ sin t cos t 6t Ê yw œ cos t sin t 3t# C$ ; at yw œ 1 and t œ 0 we have 1 œ cos (0) sin (0) 3(0)# C$ Ê C$ œ 0 Ê yw œ cos t sin t 3t# Ê y œ sin t cos t t$ C% ; at y œ 0 and t œ 0 we have 0 œ sin (0) cos (0) 0$ C% Ê C% œ 1 Ê y œ sin t cos t t$ 1 88. yÐ%Ñ œ cos x 8 sin (2x) Ê ywww œ sin x 4 cos (2x) C" ; at ywww œ 0 and x œ 0 we have 0 œ sin (0) % cos (2(0)) C" Ê C" œ 4 Ê ywww œ sin x 4 cos (2x) 4 Ê yww œ cos x 2 sin (2x) 4x C# ; at yww œ 1 and x œ 0 we have 1 œ cos (0) 2 sin (2(0)) 4(0) C# Ê C# œ 0 Ê yww œ cos x 2 sin (2x) 4x Ê yw œ sin x cos (2x) 2x# C$ ; at yw œ 1 and x œ 0 we have 1 œ sin (0) cos (2(0)) 2(0)# C$ Ê C$ œ 0 Ê yw œ sin x cos (2x) 2x# Ê y œ cos x "# sin (2x) 23 x$ C% ; at y œ 3 and x œ 0 we have 3 œ cos (0)
" #
sin (2(0)) 23 (0)$ C% Ê C% œ 4 Ê y œ cos x
" #
sin (2x) 23 x$ 4
89. m œ yw œ 3Èx œ 3x"Î# Ê y œ 2x$Î# C; at (*ß 4) we have 4 œ 2(9)$Î# C Ê C œ 50 Ê y œ 2x$Î# 50 90. Yes. If F(x) and G(x) both solve the initial value problem on an interval I then they both have the same first derivative. Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that F(x) œ G(x) C for all x. In particular, F(x! ) œ G(x! ) C, so C œ F(x! ) G(x! ) œ 0. Hence F(x) œ G(x) for all x. 91.
dy dx
œ 1 34 x"Î$ Ê y œ ' ˆ1 34 x"Î$ ‰ dx œ x x%Î$ C; at (1ß 0.5) on the curve we have 0.5 œ 1 1%Î$ C
Ê C œ 0.5 Ê y œ x x%Î$ 92.
dy dx
œ x 1 Ê y œ ' (x 1) dx œ
Ê yœ 93.
dy dx
" #
x# #
x
x# #
x C; at (1ß 1) on the curve we have 1 œ
(")# #
(1) C Ê C œ "#
" #
œ sin x cos x Ê y œ ' (sin x cos x) dx œ cos x sin x C; at (1ß 1) on the curve we have
" œ cos (1) sin (1) C Ê C œ 2 Ê y œ cos x sin x 2 94.
dy dx
œ
" #È x
1 sin 1x œ
" #
x"Î# 1 sin 1x Ê y œ ' ˆ #" x"Î# sin 1x‰ dx œ x"Î# cos 1x C; at (1ß #) on the
curve we have 2 œ 1"Î# cos 1(1) C Ê C œ 0 Ê y œ Èx cos 1x 95. (a)
ds dt
œ 9.8t 3 Ê s œ 4.9t# 3t C; (i) at s œ 5 and t œ 0 we have C œ 5 Ê s œ 4.9t# 3t 5;
displacement œ s(3) s(1) œ ((4.9)(9) 9 5) (4.9 3 5) œ 33.2 units; (ii) at s œ 2 and t œ 0 we have C œ 2 Ê s œ 4.9t# 3t 2; displacement œ s(3) s(1) œ ((4.9)(9) 9 2) (4.9 3 2) œ 33.2 units; (iii) at s œ s! and t œ 0 we have C œ s! Ê s œ 4.9t# 3t s! ; displacement œ s(3) s(1) œ ((4.9)(9) 9 s! ) (4.9 3 s! ) œ 33.2 units
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238
Chapter 4 Applications of Derivatives
(b) True. Given an antiderivative f(t) of the velocity function, we know that the body's position function is s œ f(t) C for some constant C. Therefore, the displacement from t œ a to t œ b is (f(b) C) (f(a) C) œ f(b) f(a). Thus we can find the displacement from any antiderivative f as the numerical difference f(b) f(a) without knowing the exact values of C and s. 96. a(t) œ vw (t) œ 20 Ê v(t) œ 20t C; at (0ß 0) we have C œ 0 Ê v(t) œ 20t. When t œ 60, then v(60) œ 20(60) œ 1200 97. Step 1:
d# s dt#
œ k Ê # k Š t# ‹
sœ Step 2:
ds dt
d# s dt#
œ k Ê
Ê
ds dt
œ kt C" ; at
ds dt
‰ k ˆ 88 k #
#
#
œ kt 44 Ê s œ #
Ê s œ kt# 44t. Then Ê 968 k
1936 k
œ 45 Ê
#
kt #
kt# #
ds dt
œ kt 88 Ê
88t
88 k
‰ Ê 242 œ (88) 88 ˆ 88 k 2k
œ ' k dt œ kt C; at
ds dt
œ 88 and t œ 0 we have C" œ 88 Ê
88t C# ; at s œ 0 and t œ 0 we have C# œ 0 Ê s œ
œ 0 Ê 0 œ kt 88 Ê t œ
Step 3: 242 œ 98.
ds dt
m sec .
ds dt
(88)# k
Ê 242 œ
(88)# 2k
Ê k œ 16
œ 44 when t œ 0 we have 44 œ k(0) C Ê C œ 44 #
44t C" ; at s œ 0 when t œ 0 we have 0 œ k(0) # 44(0) C" Ê C" œ 0
ds 44 dt œ 0 Ê kt 44 œ 0 Ê t œ k 968 968 ft k œ 45 Ê k œ 45 ¸ 21.5 sec2 .
99. (a) v œ ' a dt œ ' ˆ15t"Î# 3t"Î# ‰ dt œ 10t$Î# 6t"Î# C;
‰ and s ˆ 44 k œ
ds dt
‰ k ˆ 44 k #
#
‰ 44 ˆ 44 k œ 45
(1) œ 4 Ê 4 œ 10(1)$Î# 6(1)"Î# C Ê C œ 0
Ê v œ 10t$Î# 6t"Î#
(b) s œ ' v dt œ ' ˆ10t$Î# 6t"Î# ‰ dt œ 4t&Î# 4t$Î# C; s(1) œ 0 Ê 0 œ 4(1)&Î# 4(1)$Î# C Ê C œ 0 Ê s œ 4t&Î# 4t$Î# 100.
d# s dt#
œ 5.2 Ê
ds dt
œ 5.2t C" ; at
ds dt
œ 0 and t œ 0 we have C" œ 0 Ê
ds dt
œ 5.2t Ê s œ 2.6t# C# ; at s œ 4
4 and t œ 0 we have C# œ 4 Ê s œ 2.6t# 4. Then s œ 0 Ê 0 œ 2.6t# 4 Ê t œ É 2.6 ¸ 1.24 sec, since t 0
101.
d# s dt#
œa Ê
ds dt
œ ' a dt œ at C;
when t œ 0 Ê s! œ
a(0)# #
ds dt
œ v! when t œ 0 Ê C œ v! Ê
v! (0) C" Ê C" œ s! Ê s œ
at# #
102. The appropriate initial value problem is: Differential Equation: s œ s! when t œ 0. Thus, Ê
ds dt
ds dt
œ'
œ gt v! . Thus s œ ' " #
Thus s œ gt# v! t s!.
ds dt
œ at v! Ê s œ
at# #
v! t C" ; s œ s!
v! t s! d# s dt#
œ g with Initial Conditions:
g dt œ gt C" ; ds dt (0) œ v! Ê v! œ (g)(0) agt v! b dt œ "# gt# v! t C# ; s(0) œ s! œ "#
ds dt
œ v! and
C" Ê C" œ v! (g)(0)# v! (0) C# Ê C# œ s!
103 106 Example CAS commands: Maple: with(student): f := x -> cos(x)^2 + sin(x); ic := [x=Pi,y=1]; F := unapply( int( f(x), x ) + C, x ); eq := eval( y=F(x), ic ); solnC := solve( eq, {C} ); Y := unapply( eval( F(x), solnC ), x ); DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=1]], color=black, linecolor=black, stepsize=0.05, title="Section 4.7 #103" );
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Chapter 4 Practice Exercises
239
Mathematica: (functions and values may vary) The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution of the initial value problems for exercises 103 - 105. Clear[x, y, yprime] yprime[x_] = Cos[x]2 Sin[x]; initxvalue = 1; inityvalue = 1; y[x_] = Integrate[yprime[t], {t, initxvalue, x}] inityvalue If the solution satisfies the differential equation and initial condition, the following yield True yprime[x]==D[y[x], x] //Simplify y[initxvalue]==inityvalue Since exercise 106 is a second order differential equation, two integrations will be required. Clear[x, y, yprime] y2prime[x_] = 3 Exp[x/2] 1; initxval = 0; inityval = 4; inityprimeval = 1; yprime[x_] = Integrate[y2prime[t],{t, initxval, x}] inityprimeval y[x_] = Integrate[yprime[t], {t, initxval, x}] inityval Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue). y2prime[x]==D[y[x], {x, 2}]//Simplify y[initxval]==inityval yprime[initxval]==inityprimeval Plot[{y[x], yprime[x]}, {x, initxval 3, initxval 3}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,0,1]}] CHAPTER 4 PRACTICE EXERCISES 1. No, since f(x) œ x$ 2x tan x Ê f w (x) œ 3x# 2 sec# x 0 Ê f(x) is always increasing on its domain cos x 2. No, since g(x) œ csc x 2 cot x Ê gw (x) œ csc x cot x 2 csc# x œ sin #x
2 sin# x
œ sin"# x (cos x 2) 0
Ê g(x) is always decreasing on its domain 3. No absolute minimum because x lim (7 x)(11 3x)"Î$ œ _. Next f w (x) œ Ä_ (11 3x)"Î$ (7 x)(11 3x)#Î$ œ
(11 3x) (7 x) (11 3x)#Î$
œ
4(1 x) (11 3x)#Î$
Ê x œ 1 and x œ
11 3
are critical points.
Since f w 0 if x 1 and f w 0 if x 1, f(1) œ 16 is the absolute maximum. 4. f(x) œ
ax b x# 1
Ê f w (x) œ
We require also that f(3) w
#a$x "bax $b ax # 1 b #
# a ax# 1b 2x(ax b) ab œ aaxax#2bx 1 b# ax # 1 b# œ 1. Thus " œ 3a8b Ê 3a b œ w
" ; f w (3) œ 0 Ê '% (*a 'b a) œ ! Ê &a $b œ !.
). Solving both equations yields a œ 6 and b œ 10. Now,
so that f œ ± ± ± ± . Thus f w changes sign at x œ $ from 1 1 3 1/3 positive to negative so there is a local maximum at x œ $ which has a value f(3) œ 1. f (x) œ
5. Yes, because at each point of [!ß "Ñ except x œ 0, the function's value is a local minimum value as well as a local maximum value. At x œ 0 the function's value, 0, is not a local minimum value because each open interval around x œ 0 on the x-axis contains points to the left of 0 where f equals 1. 6. (a) The first derivative of the function f(x) œ x$ is zero at x œ 0 even though f has no local extreme value at x œ 0. (b) Theorem 2 says only that if f is differentiable and f has a local extreme at x œ c then f w (c) œ 0. It does not assert the (false) reverse implication f w (c) œ 0 Ê f has a local extreme at x œ c.
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Chapter 4 Applications of Derivatives
7. No, because the interval 0 x 1 fails to be closed. The Extreme Value Theorem says that if the function is continuous throughout a finite closed interval a Ÿ x Ÿ b then the existence of absolute extrema is guaranteed on that interval. 8. The absolute maximum is k1k œ 1 and the absolute minimum is k0k œ 0. This is not inconsistent with the Extreme Value Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half closed, such as Ò"ß "Ñ, so there is nothing to contradict. 9. (a) There appear to be local minima at x œ 1.75 and 1.8. Points of inflection are indicated at approximately x œ 0 and x œ „ 1.
(b) f w (x) œ x( 3x& 5x% 15x# œ x# ax# 3b ax$ 5b. The pattern yw œ ± ± ± ± 3 ! È È$ È $ 5 3 indicates a local maximum at x œ È5 and local minima at x œ „ È3 . (c)
10. (a) The graph does not indicate any local extremum. Points of inflection are indicated at approximately x œ $% and x œ 1.
(b) f w (x) œ x( 2x% 5
10 x$
œ x$ ax$ 2b ax( 5b . The pattern f w œ )( ± ± indicates 7 3 ! È È 5 2
3 7 a local maximum at x œ È 5 and a local minimum at x œ È 2.
(c)
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Chapter 4 Practice Exercises
241
11. (a) g(t) œ sin# t 3t Ê gw (t) œ 2 sin t cos t 3 œ sin (2t) 3 Ê gw 0 Ê g(t) is always falling and hence must decrease on every interval in its domain. (b) One, since sin# t 3t 5 œ 0 and sin# t 3t œ 5 have the same solutions: f(t) œ sin# t 3t 5 has the same derivative as g(t) in part (a) and is always decreasing with f(3) 0 and f(0) 0. The Intermediate Value Theorem guarantees the continuous function f has a root in [$ß 0]. 12. (a) y œ tan ) Ê
dy d)
œ sec# ) 0 Ê y œ tan ) is always rising on its domain Ê y œ tan ) increases on every interval
in its domain (b) The interval 14 ß 1‘ is not in the tangent's domain because tan ) is undefined at ) œ
1 #
. Thus the tangent need not
increase on this interval. 13. (a) f(x) œ x% 2x# 2 Ê f w (x) œ 4x$ 4x. Since f(0) œ 2 0, f(1) œ 1 0 and f w (x) 0 for 0 Ÿ x Ÿ 1, we may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when 0 Ÿ x Ÿ 1. È (b) x# œ 2 „ 4 8 0 Ê x# œ È3 1 and x 0 Ê x ¸ È.7320508076 ¸ .8555996772 #
14. (a) y œ
x x1 $
Ê yw œ
" (x 1)# w #
0, for all x in the domain of
x x1
Êyœ
x x1
is increasing in every interval in its domain.
$
(b) y œ x 2x Ê y œ 3x 2 0 for all x Ê the graph of y œ x 2x is always increasing and can never have a local maximum or minimum 15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) œ a! be the initial amount and V(1440) œ a! (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir after the rain, where 24 hr œ 1440 min. Assume that V(t) is continuous on [!ß 1440] and differentiable on (!ß 1440). The Mean Value Theorem says that for some t! in (!ß 1440) we have Vw (t! ) œ
V(1440) V(0) 1440 0
œ
a! (1400)(43,560)(7.48) a! 1440
œ
456,160,320 gal 1440 min
œ 316,778 gal/min. Therefore at t! the reservoir's volume was increasing at a rate in excess of 225,000 gal/min. 16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the d difference 3x g(x) is a constant K because gw (x) œ 3 œ dx (3x). Thus g(x) œ 3x K, the same form as F(x). x 1 x 1 x 1 œ 1 x 1 Ê x 1 differs from x 1 (x 1) x(1) d ˆ x ‰ d ˆ " ‰ œ (x " 1)# œ dx dx x 1 œ (x 1)# x1 .
17. No,
18. f w (x) œ gw (x) œ
2x ax # 1 b #
by the constant 1. Both functions have the same derivative
Ê f(x) g(x) œ C for some constant C Ê the graphs differ by a vertical shift.
19. The global minimum value of
" #
occurs at x œ #.
20. (a) The function is increasing on the intervals Ò$ß #Ó and Ò"ß #Ó. (b) The function is decreasing on the intervals Ò#ß !Ñ and Ð!ß "Ó. (c) The local maximum values occur only at x œ #, and at x œ #; local minimum values occur at x œ $ and at x œ " provided f is continuous at x œ !. 21. (a) t œ 0, 6, 12
(b) t œ 3, 9
(c) 6 t 12
(d) 0 t 6, 12 t 14
22. (a) t œ 4
(b) at no time
(c) 0 t 4
(d) 4 t 8
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Chapter 4 Applications of Derivatives
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
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Chapter 4 Practice Exercises
243
33. (a) yw œ 16 x# Ê yw œ ± ± Ê the curve is rising on (%ß %), falling on (_ß 4) and (%ß _) % % Ê a local maximum at x œ 4 and a local minimum at x œ 4; yww œ 2x Ê yww œ ± Ê the curve ! is concave up on (_ß !), concave down on (!ß _) Ê a point of inflection at x œ 0 (b)
34. (a) yw œ x# x 6 œ (x $)(x 2) Ê yw œ ± ± Ê the curve is rising on (_ß 2) and ($ß _), # $ falling on (#ß $) Ê local maximum at x œ 2 and a local minimum at x œ 3; yww œ 2x 1 Ê yww œ ± Ê concave up on ˆ "# ß _‰ , concave down on ˆ_ß "# ‰ Ê a point of inflection at x œ "# "Î# (b)
35. (a) yw œ 6x(x 1)(x 2) œ 6x$ 6x# 12x Ê yw œ ± ± ± Ê the graph is rising on ("ß !) " ! # and (#ß _), falling on (_ß 1) and (!ß #) Ê a local maximum at x œ 0, local minima at x œ 1 and x œ 2; yww œ 18x# 12x 12 œ 6 a3x# 2x 2b œ 6 Šx yww œ ± on
±
"È( $ 1 È7 1 È7 Š 3 ß 3 ‹
"È( $
1 È7 3 ‹ Šx
1 È7 3 ‹ È7
Ê the curve is concave up on Š_ß 1 3
Ê points of inflection at x œ
Ê È7
‹ and Š 1 3
ß _‹ , concave down
1 „ È7 3
(b)
36. (a) yw œ x# (6 4x) œ 6x# 4x$ Ê yw œ ± ± Ê the curve is rising on ˆ_ß #3 ‰, falling on ˆ #3 ß _‰ ! $Î# 3 ww # Ê a local maximum at x œ # ; y œ 12x 12x œ 12x(" x) Ê yww œ ± ± Ê concave up on ! " (!ß "), concave down on (_ß !) and ("ß _) Ê points of inflection at x œ 0 and x œ 1 (b)
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244
Chapter 4 Applications of Derivatives
37. (a) yw œ x% 2x# œ x# ax# 2b Ê yw œ ± ± ± Ê the curve is rising on Š_ß È2‹ and ! È# È # ŠÈ2ß _‹ , falling on ŠÈ2ß È2‹ Ê a local maximum at x œ È2 and a local minimum at x œ È2 ; yww œ 4x$ 4x œ 4x(x 1)(x 1) Ê yww œ ± ± ± Ê concave up on ("ß 0) and ("ß _), " ! " concave down on (_ß 1) and (0ß 1) Ê points of inflection at x œ 0 and x œ „ 1 (b)
38. (a) yw œ 4x# x% œ x# a4 x# b Ê yw œ ± ± ± Ê the curve is rising on (2ß 0) and (0ß 2), # ! # falling on (_ß 2) and (#ß _) Ê a local maximum at x œ 2, a local minimum at x œ 2; yww œ 8x 4x$ œ 4x a2 x# b Ê yww œ ± ± ± Ê concave up on Š_ß È2‹ and Š0ß È2‹ , concave ! È# È # down on ŠÈ2ß 0‹ and ŠÈ2ß _‹ Ê points of inflection at x œ 0 and x œ „ È2 (b)
39. The values of the first derivative indicate that the curve is rising on (!ß _) and falling on (_ß 0). The slope of the curve approaches _ as x Ä ! , and approaches _ as x Ä 0 and x Ä 1. The curve should therefore have a cusp and local minimum at x œ 0, and a vertical tangent at x œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Practice Exercises 40. The values of the first derivative indicate that the curve is rising on ˆ!ß "# ‰ and ("ß _), and falling on (_ß !) and ˆ "# ß "‰ . The derivative changes from positive to negative at x œ "# , indicating a local maximum there. The slope of the curve approaches _ as x Ä 0 and x Ä 1 , and approaches _ as x Ä 0 and as x Ä 1 , indicating cusps and local minima at both x œ 0 and x œ 1.
41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches _ as x Ä 0 and as x Ä 1, indicating vertical tangents at both x œ 0 and x œ 1.
È33
42. The graph of the first derivative indicates that the curve is rising on Š!ß 17 16 on (_ß !) and xœ
17 È33 16
È È Š 17 16 33 ß 17 16 33 ‹
Ê a local maximum at x œ
17 È33 16
È33
‹ and Š 17 16
ß _‹ , falling
, a local minimum at
. The derivative approaches _ as x Ä 0 and x Ä 1, and approaches _ as x Ä 0 ,
indicating a cusp and local minimum at x œ 0 and a vertical tangent at x œ 1.
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245
246
Chapter 4 Applications of Derivatives
43. y œ
x1 x3
45. y œ
x# 1 x
œx
47. y œ
x$ 2 #x
œ
49. y œ
x# 4 x# 3
œ1
œ1
x# #
4 x3
" x
" x
" x# 3
44. y œ
2x x5
œ2
46. y œ
x# x 1 x
48. y œ
x% 1 x#
œ x#
50. y œ
x# x# 4
œ1
10 x5
œx1
" x
" x#
4 x# 4
51. (a) Maximize f(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36 Ê f w (x) œ
" #
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx #Èx È36 x
Ê derivative fails to exist at 0 and 36; f(0) œ 6,
and f(36) œ 6 Ê the numbers are 0 and 36
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Chapter 4 Practice Exercises (b) Maximize g(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36 Ê gw (x) œ
" #
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx #Èx È36 x
Ê critical points at 0, 18 and 36; g(0) œ 6,
g(18) œ 2È18 œ 6È2 and g(36) œ 6 Ê the numbers are 18 and 18 52. (a) Maximize f(x) œ Èx (20 x) œ 20x"Î# x$Î# where 0 Ÿ x Ÿ 20 Ê f w (x) œ 10x"Î# 3# x"Î# œ
20 3x #È x
œ 0 Ê x œ 0 and x œ
œ
40È20 3È 3
Ê the numbers are
20 3
20 3
‰ É 20 ˆ are critical points; f(0) œ f(20) œ 0 and f ˆ 20 3 œ 3 20
and
40 3
.
(b) Maximize g(x) œ x È20 x œ x (20 x)"Î# where 0 Ÿ x Ÿ 20 Ê gw (x) œ Ê È20 x œ
" #
Ê xœ
the numbers must be " #
53. A(x) œ
79 4
and
79 4 . " 4 .
The critical points are x œ
79 4
2È20 x 1 #È20 x
‰ and x œ 20. Since g ˆ 79 4 œ
(2x) a27 x# b for 0 Ÿ x Ÿ È27
Ê Aw (x) œ 3(3 x)(3 x) and Aw w (x) œ 6x. The critical points are 3 and 3, but 3 is not in the domain. Since Aw w (3) œ 18 0 and A ŠÈ27‹ œ 0, the maximum occurs at x œ 3 Ê the largest area is A(3) œ 54 sq units. 54. The volume is V œ x# h œ 32 Ê h œ 32 x# . The 32 ‰ # ˆ surface area is S(x) œ x 4x x# œ x# 128 x , where x 0 Ê Sw (x) œ
2(x 4) ax# 4x 16b x#
Ê the critical points are 0 and 4, but 0 is not in the domain. Now Sw w (4) œ 2 256 4$ 0 Ê at x œ 4 there is a minimum. The dimensions 4 ft by 4 ft by 2 ft minimize the surface area. #
55. From the diagram we have ˆ h# ‰ r# œ ŠÈ3‹ Ê r# œ
12h# 4
#
. The volume of the cylinder is #
V œ 1r h œ 1 Š 12 4 h ‹ h œ #
1 4
0 Ÿ h Ÿ 2È3 . Then Vw (h) œ
a12h h$ b , where 31 4
20 ‰ 3
(2 h)(2 h)
Ê the critical points are 2 and 2, but 2 is not in the domain. At h œ 2 there is a maximum since Vw w (2) œ 31 0. The dimensions of the largest cylinder are radius œ È2 and height œ 2. 56. From the diagram we have x œ radius and y œ height œ 12 2x and V(x) œ "3 1x# (12 2x), where
0 Ÿ x Ÿ 6 Ê Vw (x) œ 21x(4 x) and Vw w (4) œ 81. The critical points are 0 and 4; V(0) œ V(6) œ 0 Ê x œ 4 gives the maximum. Thus the values of r œ 4 and h œ 4 yield the largest volume for the smaller cone.
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81 4
œ0 and g(20) œ 20,
247
248
Chapter 4 Applications of Derivatives
‰ , where p is the profit on grade B tires and 0 Ÿ x Ÿ 4. Thus 57. The profit P œ 2px py œ 2px p ˆ 40510x x Pw (x) œ
2p (5 x)#
ax# 10x 20b Ê the critical points are Š5 È5‹, 5, and Š5 È5‹ , but only Š5 È5‹ is in
the domain. Now Pw (x) 0 for 0 x Š5 È5‹ and Pw (x) 0 for Š5 È5‹ x 4 Ê at x œ Š5 È5‹ there is a local maximum. Also P(0) œ 8p, P Š5 È5‹ œ 4p Š5 È5‹ ¸ 11p, and P(4) œ 8p Ê at x œ Š5 È5‹ there is an absolute maximum. The maximum occurs when x œ Š5 È5‹ and y œ 2 Š5 È5‹ , the units are hundreds of tires, i.e., x ¸ 276 tires and y ¸ 553 tires. 58. (a) The distance between the particles is lfatbl where fatb œ cos t cosˆt 1% ‰. Then, f w atb œ sin t sinˆt 1% ‰. Solving f w atb œ ! graphically, we obtain t ¸ "Þ"(), t ¸ %Þ$#!, and so on.
Alternatively, f w atb œ ! may be solved analytically as follows. f w atb œ sin’ˆt 1) ‰ 1) “ sin’ˆt 1) ‰ 1) “ œ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ œ #sin 1) cosˆt 1) ‰ so the critical points occur when cosˆt 1) ‰ œ !, or t œ
$1 )
k1. At each of these values, fatb œ „ cos $)1
¸ „ !Þ('& units, so the maximum distance between the particles is !Þ('& units. (b) Solving cos t œ cos ˆt 1% ‰ graphically, we obtain t ¸ #Þ(%*, t ¸ &Þ)*!, and so on.
Alternatively, this problem can be solved analytically as follows. cos t œ cos ˆt 1% ‰ cos’ˆt 1) ‰ 1) “ œ cos’ˆt 1) ‰ 1) “ cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1) œ cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1) #sin ˆt 1) ‰sin 1) œ ! sin ˆt 1) ‰ œ ! tœ The particles collide when t œ
(1 )
(1 )
k1
¸ #Þ(%*. (plus multiples of 1 if they keep going.)
59. The dimensions will be x in. by "! #x in. by "' #x in., so Vaxb œ xa"! #xba"' #xb œ %x$ &#x# "'!x for ! x &. Then Vw axb œ "#x# "!%x "'! œ %ax #ba$x #!b , so the critical point in the correct domain is x œ #. This critical point corresponds to the maximum possible volume because Vw axb ! for ! x # and Vw axb ! for 2 x &. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in.$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Practice Exercises Graphical support:
60. The length of the ladder is d" d# œ 8 sec ) 6 csc ). We wish to maximize I()) œ 8 sec ) 6 csc ) Ê Iw ()) œ 8 sec ) tan ) 6 csc ) cot ). Then Iw ()) œ 0 Ê 8 sin$ ) 6 cos$ ) œ 0 Ê tan ) œ
3 È 6 #
Ê
3 3 3 d" œ 4 É 4 È 36 and d# œ È 36 É4 È 36
Ê the length of the ladder is about 3 3 3 36‹ É4 È 36 œ Š4 È 36‹ Š4 È
$Î#
¸ "*Þ( ft.
61. g(x) œ 3x x$ 4 Ê g(2) œ 2 0 and g(3) œ 14 0 Ê g(x) œ 0 in the interval [#ß 3] by the Intermediate Value Theorem. Then gw (x) œ 3 3x# Ê xnb1 œ xn
3xn x$n 4 33xn#
; x! œ 2 Ê x" œ 2.22 Ê x# œ 2.196215, and
so forth to x& œ 2.195823345. 62. g(x) œ x% x$ 75 Ê g(3) œ 21 0 and g(4) œ 117 0 Ê g(x) œ 0 in the interval [$ß %] by the Intermediate Value Theorem. Then gw (x) œ 4x$ 3x# Ê xnb1 œ xn
x%n x$n 75 4xn$ 3xn#
; x! œ 3 Ê x" œ 3.259259
Ê x# œ 3.229050, and so forth to x& œ 3.22857729. 63.
' ax$ 5x 7b dx œ
64.
' Š8t$ t# t‹ dt œ 8t4% t6$ t## C œ 2t% t6$ t## C
65.
' ˆ3Èt t4# ‰ dt œ ' ˆ3t"Î# 4t# ‰ dt œ 3t$Î# 4t1" C œ 2t$Î# 4t C
66.
' Š #È" t t3% ‹ dt œ ' ˆ #" t"Î# 3t% ‰ dt œ #" Œ t"Î# (3t3)$ C œ Èt t"$ C
x% 4
5x# #
7x C
#
Š 3# ‹
" #
67. Let u œ r 5 Ê du œ dr
' ar dr5b
#
œ'
du u#
œ ' u# du œ
u " 1
C œ u" C œ ar " 5b C
68. Let u œ r È2 Ê du œ dr
'
6 dr
$
Šr È2‹
œ 6'
dr
$
Šr È2‹
œ 6'
du u$
#
œ 6' u$ du œ 6 Š u# ‹ C œ 3u# C œ
3
#
ŠrÈ2‹
C
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250
Chapter 4 Applications of Derivatives
69. Let u œ )# 1 Ê du œ 2) d) Ê
'
3)È)# 1 d) œ ' Èu ˆ #3 du‰ œ
70. Let u œ 7 )2 Ê du œ 2) d) Ê
'È)
d) œ '
7 ) 2
" Èu
ˆ #" du‰ œ
" #
x$ a 1 x % b
"Î%
du œ ) d) 3 #
" #
$Î# $Î# ' u"Î# du œ 3# Œ u$Î# C œ a ) # 1b C 3 C œ u #
du œ ) d)
"Î# ' u"Î# du œ #" Œ u"Î# C œ È7 )2 C " C œ u #
71. Let u œ 1 x% Ê du œ 4x$ dx Ê
'
" #
" 4
du œ x$ dx
dx œ ' u"Î% ˆ "4 du‰ œ
" 4
$Î% " $Î% ' u"Î% du œ 4" Œ u$Î% C œ 3" a1 x% b C 3 C œ 3 u 4
72. Let u œ 2 x Ê du œ dx Ê du œ dx
' (2 x)$Î& dx œ ' u$Î& ( du) œ ' u$Î& du œ u
)Î&
Š 85 ‹
73. Let u œ
'
sec# 10s
" 10
Ê du œ
s 10
C œ 85 u)Î& C œ 85 (2 x))Î& C
ds Ê 10 du œ ds
ds œ ' asec# ub (10 du) œ 10 ' sec# u du œ 10 tan u C œ 10 tan
74. Let u œ 1s Ê du œ 1 ds Ê
" 1
s 10
C
du œ ds
' csc# 1s ds œ ' acsc# ub ˆ 1" du‰ œ 1" ' csc# u du œ 1" cot u C œ 1" cot 1s C
75. Let u œ È2 ) Ê du œ È2 d) Ê
' csc È2) cot È2) d) œ ' ) 3
76. Let u œ
'
sec
) 3
tan
77. Let u œ
'
Ê du œ
x 4
) 3
" 3
" È2
du œ d)
(csc u cot u) Š È"2 du‹ œ
" È2
(csc u) C œ È"2 csc È2) C
d) Ê 3 du œ d)
d) œ ' (sec u tan u)(3 du) œ 3 sec u C œ 3 sec
Ê du œ
" 4
) 3
C
dx Ê 4 du œ dx
2u ‰ dx œ ' asin# ub (4 du) œ ' 4 ˆ 1 cos du œ 2' (1 cos 2u) du œ 2 ˆu # œ 2u sin 2u C œ 2 ˆ x4 ‰ sin 2 ˆ x4 ‰ C œ #x sin x# C
sin#
x 4
78. Let u œ
'
cos#
79. y œ '
x # x #
Ê du œ
" #
dx œ ' a1 x# b dx œ x x" C œ x
Ê C œ 1 Ê y œ x
" x
81.
dr dt
œ ' Š15Èt
3 Èt ‹
" 3
" x
" #
Cœ
x #
C; y œ 1 when x œ 1 Ê 1
1 1
C œ 1
sin 2u #
sin x C
1
# 80. y œ ' ˆx "x ‰ dx œ ' ˆx# 2
y œ 1 when x œ 1 Ê
C
dx Ê 2 du œ dx
2u ‰ dx œ ' acos# ub (2 du) œ ' 2 ˆ 1 cos du œ ' (1 cos 2u) du œ u #
x# " x#
sin 2u ‰ #
2
1 1
"‰ x#
dx œ ' ax# 2 x# b dx œ
C œ 1 Ê C œ 3" Ê y œ
$
x 3
x$ 3
2x x" C œ
2x
dt œ ' ˆ15t"Î# 3t"Î# ‰ dt œ 10t$Î# 6t"Î# C;
Ê 10(1)$Î# 6(1)"Î# C œ 8 Ê C œ 8. Thus
dr dt
dr dt
" x
x$ 3
2x
" x
C;
" 3
œ 8 when t œ 1
œ 10t$Î# 6t"Î# 8 Ê r œ ' ˆ10t$Î# 6t"Î# 8‰ dt
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Additional and Advanced Exercises œ 4t&Î# 4t$Î# 8t C; r œ 0 when t œ 1 Ê 4(1)&Î# 4(1)$Î# 8(1) C" œ 0 Ê C" œ 0. Therefore, r œ 4t&Î# 4t$Î# 8t 82.
d# r dt#
Ê
œ ' cos t dt œ sin t C; rw w œ 0 when t œ 0 Ê sin 0 C œ 0 Ê C œ 0. Thus, dr dt
œ ' sin t dt œ cos t C" ; rw œ 0 when t œ 0 Ê 1 C" œ 0 Ê C" œ 1. Then
d# r dt# œ sin t dr dt œ cos t
1
Ê r œ ' (cos t 1) dt œ sin t t C# ; r œ 1 when t œ 0 Ê 0 0 C# œ 1 Ê C# œ 1. Therefore, r œ sin t t 1 CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES 1. If M and m are the maximum and minimum values, respectively, then m Ÿ f(x) Ÿ M for all x − I. If m œ M then f is constant on I. 3x 6, 2 Ÿ x 0 has an absolute minimum value of 0 at x œ 2 and an absolute 9 x# , 0 Ÿ x Ÿ 2 maximum value of 9 at x œ 0, but it is discontinuous at x œ 0.
2. No, the function f(x) œ œ
3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the closed endpoint. Extreme values occur only where f w œ 0, f w does not exist, or at the endpoints of the interval. Thus the extreme points will not be at the ends of an open interval. 4. The pattern f w œ ± ± ± ± indicates a local maximum at x œ 1 and a local " # $ % minimum at x œ 3. 5. (a) If yw œ 6(x 1)(x 2)# , then yw 0 for x 1 and yw 0 for x 1. The sign pattern is f w œ ± ± Ê f has a local minimum at x œ 1. Also yww œ 6(x 2)# 12(x 1)(x 2) " # œ 6(x 2)(3x) Ê yw w 0 for x 0 or x 2, while yww 0 for 0 x 2. Therefore f has points of inflection at x œ 0 and x œ 2. There is no local maximum. (b) If yw œ 6x(x 1)(x 2), then yw 0 for x 1 and 0 x 2; yw 0 for " x 0 and x 2. The sign sign pattern is yw œ ± ± ± . Therefore f has a local maximum at x œ 0 and " ! # È7
local minima at x œ 1 and x œ 2. Also, yww œ ") ’x Š 1 $ 1 È 7 $
x
1 È 7 $
È7
‹“ ’x Š 1 $
‹“ , so yww 0 for
and yww 0 for all other x Ê f has points of inflection at x œ
6. The Mean Value Theorem indicates that
f(6) f(0) 60
1 „È 7 $
.
œ f w (c) Ÿ 2 for some c in (0ß 6). Then f(6) f(0) Ÿ 12 indicates the
most that f can increase is 12. 7. If f is continuous on [aß c) and f w (x) Ÿ 0 on [aß c), then by the Mean Value Theorem for all x − [aß c) we have f(c) f(x) cx
Ÿ 0 Ê f(c) f(x) Ÿ 0 Ê f(x) f(c). Also if f is continuous on (cß b] and f w (x) 0 on (cß b], then for
all x − (cß b] we have
f(x) f(c) xc
0 Ê f(x) f(c) 0 Ê f(x) f(c). Therefore f(x) f(c) for all x − [aß b].
8. (a) For all x, (x 1)# Ÿ 0 Ÿ (x 1)# Ê a1 x# b Ÿ 2x Ÿ a1 x# b Ê "# Ÿ (b) There exists c − (aß b) such that Ê kf(b) f(a)k Ÿ
" #
c 1 c#
œ
f(b) f(a) ba
Ê
f(a) ¹ f(b)b a ¹
œ ¸ 1 c c# ¸ Ÿ
" #
x 1 x#
Ÿ
" #
, from part (a)
kb ak .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
.
251
252
Chapter 4 Applications of Derivatives
9. No. Corollary 1 requires that f w (x) œ 0 for all x in some interval I, not f w (x) œ 0 at a single point in I. 10. (a) h(x) œ f(x)g(x) Ê hw (x) œ f w (x)g(x) f(x)gw (x) which changes signs at x œ a since f w (x), gw (x) 0 when x a, f w (x), gw (x) 0 when x a and f(x), g(x) 0 for all x. Therefore h(x) does have a local maximum at x œ a. (b) No, let f(x) œ g(x) œ x$ which have points of inflection at x œ 0, but h(x) œ x' has no point of inflection (it has a local minimum at x œ 0). " a bc#
11. From (ii), f(1) œ lim
xÄ „_
x" # x Ä „ _ bx cx #
f(x) œ
1 "x x c 2x xÄ „_ dy dx
œ ! and if c œ 0,
œ 3x# 2kx 3 œ 0 Ê x œ
13. The area of the ?ABC is A(x) œ w
where 0 Ÿ x Ÿ 1. Thus A (x) œ
1 "x 2 x Ä „ _ bx c x " 1 x then lim 2 x Ä „ _ bx x
œ
lim
lim
12.
œ 0 Ê a œ 1; from (iii), either 1 œ x lim f(x) or 1 œ x Ä lim f(x). In either case, Ä_ _ lim
2k „ È4k# 36 6 " #
œ " Ê b œ 0 and c œ ". For if b œ ", then œ
2 x
xÄ „_
œ „ _. Thus a œ 1, b œ 0, and c œ 1.
Ê x has only one value when 4k# 36 œ 0 Ê k# œ 9 or k œ „ 3.
(2) È1 x# œ a1 x# b
x È 1 x#
1 x"
lim
"Î#
,
Ê 0 and „ 1 are
critical points. Also A a „ 1b œ 0 so A(0) œ 1 is the maximum. When x œ 0 the ?ABC is isosceles since AC œ BC œ È2 .
f (c h) f (c) œ f ww (c) Ê for % œ "# kf ww (c)k 0 h hÄ0 Ê ¹ f (ch)h f (c) f ww (c)¹ "# kf ww (c)k . Then f w (c) œ w
14. lim
w
w
w
3 #
f ww (c) ww
0 Ê "# kf ww (c)k
f (c h) f ww (c) "# kf ww (c)k . If f ww (c) 0, then h f (c h) "# f ww (c) 0; likewise if f ww (c) 0, then 0 "# h w
Ê f ww (c) "# kf ww (c)k Ê
there exists a $ 0 such that 0 khk $
w
w
f (c h) h
" #
f ww (c)
w
kf ww (c)k
kf ww (c)k œ f ww (c) f ww (c) w
f (c h) h w
3 #
f ww (c).
(a) If f (c) 0, then $ h 0 Ê f (c h) 0 and 0 h $ Ê f (c h) 0. Therefore, f(c) is a local maximum. (b) If f ww (c) 0, then $ h 0 Ê f w (c h) 0 and 0 h $ Ê f w (c h) 0. Therefore, f(c) is a local minimum. 15. The time it would take the water to hit the ground from height y is É 2y g , where g is the acceleration of gravity. The product of time and exit velocity (rate) yields the distance the water travels: È64(h y) œ 8 É 2 ahy y# b D(y) œ É 2y g g
"Î#
, 0 Ÿ y Ÿ h Ê Dw (y) œ 4 É 2g ahy y# b # "Î#
are critical points. Now D(0) œ 0, D ˆ h# ‰ œ 8 É g2 Šhˆ h# ‰ ˆ h# ‰ ‹ the hole is at y œ
h #
ba h
œ
(h 2y) Ê 0,
tan " 1 ha tan " a h
œ
h tan " a h a tan "
ba h ;
tan (" )) œ
tan " tan ) 1 tan " tan )
. Solving for tan " gives tan " œ
; and tan ) œ
bh h# a(b a)
a h
. These equations
or
#
ah a(b a)b tan " œ bh. Differentiating both sides with respect to h gives 2h tan " ah# a(b a)b sec# "
d" dh
œ b. Then
d" dh
h #
and h
œ 4hÉ g2 and D(h) œ 0 Ê the best place to drill
.
16. From the figure in the text, tan (" )) œ give
"Î#
bh œ 0 Ê 2h tan " œ b Ê 2h Š h# a(b a) ‹ œ b
Ê 2bh# œ bh# ab(b a) Ê h# œ a(b a) Ê h œ Èa(a b) .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Additional and Advanced Exercises
253
17. The surface area of the cylinder is S œ 21r# 21rh. From the diagram we have Rr œ H H h Ê h œ RH R rH and S(r) œ 21r(r h) œ 21r ˆr H r HR ‰ œ 21 ˆ1 HR ‰ r# 21Hr, where 0 Ÿ r Ÿ R.
Case 1: H R Ê S(r) is a quadratic equation containing the origin and concave upward Ê S(r) is maximum at r œ R. Case 2: H œ R Ê S(r) is a linear equation containing the origin with a positive slope Ê S(r) is maximum at r œ R. Case 3: H R Ê S(r) is a quadratic equation containing the origin and concave downward. Then dS H‰ dS H‰ RH ˆ ˆ dr œ 41 1 R r 21H and dr œ 0 Ê 41 1 R r 21H œ 0 Ê r œ 2(H R) . For simplification we let r‡ œ
RH 2(H R)
.
(a) If R H 2R, then 0 H 2R Ê H 2(H R) Ê r*= 2(HRH R) R. Therefore, the maximum occurs at the right endpoint R of the interval 0 Ÿ r Ÿ R because S(r) is an increasing function of r. (b) If H œ 2R, then r‡ œ
2R# 2R
œ R Ê S(r) is maximum at r œ R.
(c) If H 2R, then 2R H 2H Ê H 2(H R) Ê S(r) is a maximum at r œ r‡ œ
RH 2(H R)
H 2(H R)
1 Ê
RH 2(H R)
R Ê r‡ R. Therefore,
.
Conclusion: If H − (0ß 2R], then the maximum surface area is at r œ R. If H − (2Rß _), then the maximum is at r œ r‡ œ 2(HRH R) . 18. f(x) œ mx 1
" x
Ê f w (x) œ m
" x#
and f w w (x) œ
2 x$
0 when x 0. Then f w (x) œ 0 Ê x œ
If f Š È"m ‹ 0, then Èm 1 Èm œ 2Èm 1 0 Ê m
" 4
" Èm
yields a minimum.
. Thus the smallest acceptable value for m is
" 4
.
19. (a) The profit function is Paxb œ ac exbx aa bxb œ ex# ac bbx a. Pw axb œ #ex c b œ ! Ê x œ c#eb . Pww axb œ #e ! if e ! so that the profit function is maximized at x œ c #e b . (b) The price therefore that corresponds to a production level yeilding a maximum profit is p¹
xœ c#eb
œ c eˆ c #e b ‰ œ
c b #
dollars. #
(c) The weekly profit at this production level is Paxb œ eˆ c #e b ‰ ac bbˆ c #e b ‰ a œ
ac b b # %e #
a.
(d) The tax increases cost to the new profit function is Faxb œ ac exbx aa bx txb œ ex ac b tbx a. bc cbt ww Now Fw axb œ #ex c b t œ ! when x œ t # #e . Since F axb œ #e ! if e !, F is maximized e œ when x œ c #be t units per week. Thus the price per unit is p œ c eˆ c #be t ‰ œ c #b t dollars. Thus, such a tax increases the cost per unit by
cbt #
The x-intercept occurs when
" x
cb #
œ
t #
dollars if units are priced to maximize profit.
20. (a)
$œ!Ê
" x
œ $ Ê x œ $" .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
254
Chapter 4 Applications of Derivatives
(b) By Newton's method, xn" œ xn
faxn b f ax n b . w
Here f w axn b œ x# n œ
" x#n .
"
xn $
So xn" œ xn
"
x# n
œ xn Š x"n $‹x#n
œ xn xn $x#n œ #xn $xn# œ xn a# $xn b. 21. x" œ x! and
a q " x!
fax! b f w ax ! b
q
qx!q xq! a qxq! "
x a œ x! ! q " œ qx
!
with weights m! œ
In the case where x! œ
a xq! "
q" q
q
x aq "b a œ ! q " œ x! Š q q " ‹
a Š"‹ xq! " q
qx!
and m" œ "q .
we have xq! œ a and x" œ
a Š q q " ‹ qa " Š q" ‹ xq! " x!
œ
so that x" is a weighted average of x!
a Š q q " xq! "
q" ‹ œ
a . xq! " #
dy d y 22. We have that ax hb# ay hb# œ r# and so #ax hb #ay hb dy dx œ ! and # # dx #ay hb dx# œ ! hold. dy x y dx dy . " dx
dy Thus #x #y dy dx œ #h #h dx , by the former. Solving for h, we obtain h œ #
d y equation yields # # dy dx #y dx# #Œ
dy x y dx dy " dx
œ !. Dividing by 2 results in "
Substituting this into the second
dy dx
#
y ddxy# Œ
dy x y dx dy " dx
œ !.
23. (a) aatb œ sww atb œ k ak !b Ê sw atb œ kt C" , where sw a!b œ )) Ê C" œ )) Ê sw atb œ kt )). So satb œ #
kt #
kt# #
))t C# where sa!b œ ! Ê C# œ ! so satb œ
))t œ "!!. Solving for t we obtain t œ
kŠ ))
È))# #!!k ‹ k ))# #!!
so that k œ
)) œ ! or kŠ ))
)) „ È))# #!!k . k
È))# #!!k ‹ k
kt# #
))t. Now satb œ "!! when
At such t we want sw atb œ !, thus
)) œ !. In either case we obtain ))# #!!k œ !
¸ $)Þ(# ft/sec# .
(b) The initial condition that sw a!b œ %% ft/sec implies that sw atb œ kt %% and satb œ w
The car is stopped at a time t such that s atb œ kt %% œ ! Ê t œ ‰ sˆ %% k
œ
k ˆ %% ‰# # k
‰ %%ˆ %% k
œ
%%# #k
œ
*') k
œ
‰ *')ˆ #!! ))#
%% k .
kt# #
%%t where k is as above.
At this time the car has traveled a distance
œ #& feet. Thus halving the initial velocity quarters
stopping distance. 24. haxb œ f # axb g# axb Ê hw axb œ #faxbf w axb #gaxbgw axb œ #faxbf w axb gaxbgw axb‘ œ #faxbgaxb gaxbafaxbb‘ œ # † ! œ !. Thus haxb œ c, a constant. Since ha!b œ &, haxb œ & for all x in the domain of h. Thus ha"!b œ &. 25. Yes. The curve y œ x satisfies all three conditions since
dy dx
œ " everywhere, when x œ !, y œ !, and
d# y dx#
œ ! everywhere.
26. yw œ $x# # for all x Ê y œ x$ #x C where " œ "$ # † " C Ê C œ % Ê y œ x$ #x %. 27. sww atb œ a œ t# Ê v œ sw atb œ maximum for this t‡ . Since satb t‡ œ a$Cb"Î$ . So ÊCœ
a%bb$Î% $ .
a$Cb"Î$ ‘% 12
t$ w ‡ ‡ $ C. We seek v! œ s a!b œ C. We know that sat b œ b for some t and s is at a % % œ 12t Ct k and sa!b œ ! we have that satb œ 12t Ct and also sw at‡ b œ ! so that
Ca$Cb"Î$ œ b Ê a$Cb"Î$ ˆC
Thus v! œ sw a!b œ
a%bb$Î% $
œ
#È# $Î% . $ b
$C ‰ "#
œ b Ê a$Cb"Î$ ˆ $%C ‰ œ b Ê $"Î$ C%Î$ œ
28. (a) sww atb œ t"Î# t"Î# Ê vatb œ sw atb œ #$ t$Î# #t"Î# k where va!b œ k œ (b) satb œ
% &Î# "& t
% %$ t$Î# %$ t k# where sa!b œ k# œ "& . Thus satb œ
% # $Î# #t"Î# $ Ê vatb œ $ t % &Î# % %$ t$Î# %$ t "& . "& t
%b $
%$ Þ
29. The graph of faxb œ ax# bx c with a ! is a parabola opening upwards. Thus faxb ! for all x if faxb œ ! for at most one real value of x. The solutions to faxb œ ! are, by the quadratic equation #
#b „ Éa#bb# %ac . #a
Thus we require
#
a#bb %ac Ÿ ! Ê b ac Ÿ !.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Additional and Advanced Exercises 30. (a) Clearly faxb œ aa" x b" b# Þ Þ Þ aan x bn b# ! for all x. Expanding we see faxb œ aa#" x# #a" b" x b"# b Þ Þ Þ aan# x# #an bn x bn# b œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b !.
Thus aa" b" a# b# Þ Þ Þ an bn b# aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b Ÿ ! by Exercise 29.
Thus aa" b" a# b# Þ Þ Þ an bn b# Ÿ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b. (b) Referring to Exercise 29: It is clear that faxb œ ! for some real x Í b# %ac œ !, by quadratic formula. Now notice that this implies that faxb œ aa" x b" b# Þ Þ Þ aan x bn b# œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b œ ! Í aa" b" a# b# Þ Þ Þ an bn b# aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b œ !
Í aa" b" a# b# Þ Þ Þ an bn b# œ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b But now faxb œ ! Í ai x bi œ ! for all i œ "ß #ß Þ Þ Þ ß n Í ai x œ bi œ ! for all i œ "ß #ß Þ Þ Þ ß n.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
255
256
Chapter 4 Applications of Derivatives
NOTES
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 5 INTEGRATION 5.1 AREA AND ESTIMATING WITH FINITE SUMS 1. faxb œ x#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.
"
#
iœ! $
" #
œ "# Š!# ˆ "# ‰ ‹ œ
#
" 4
œ 4" Š!# ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
(a) ˜x œ
"! #
œ
" #
and xi œ i˜x œ
i #
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"! %
œ
" %
and xi œ i˜x œ
i %
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"! #
œ
" #
and xi œ i˜x œ
i #
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"! %
œ
" %
and xi œ i˜x œ
i %
Ê an upper sum is !ˆ 4i ‰ †
2. faxb œ x$
iœ! 2
iœ1 %
#
" )
#
#
#
#
" #
# œ "# Šˆ "# ‰ +1# ‹ œ
#
" 4
# # # œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1# ‹ œ
iœ"
" %
†
( )
" %
$! ‰ † ˆ "' œ
œ
( $#
& ) "& $#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.
"
$
iœ! $
" #
œ "# Š!$ ˆ "# ‰ ‹ œ
$
" 4
œ 4" Š!$ ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
œ
" #
and xi œ i˜x œ
i #
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"! %
œ
" %
and xi œ i˜x œ
i %
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"! #
œ
" #
and xi œ i˜x œ
i #
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"! %
œ
" %
and xi œ i˜x œ
i %
Ê an upper sum is !ˆ 4i ‰ †
(a) ˜x œ
"! #
iœ! 2
iœ1 %
iœ"
$
" "'
$
$
$
$' #&'
$
" #
$ œ "# Šˆ "# ‰ +1$ ‹ œ
$
" 4
$ $ $ œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1$ ‹ œ œ
" #
†
* )
œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
* '%
"!! #&'
œ
* "' #& '%
258
Chapter 5 Integration
3. faxb œ
" x
Since f is decreasing on Ò1ß 5Ó, we use left endpoints to obtain upper sums and right endpoints to obtain lower sums.
#
(a) ˜x œ
&" #
œ # and xi œ " i˜x œ " #i Ê a lower sum is ! x"i † # œ #ˆ "$ "& ‰ œ
(b) ˜x œ
&" %
œ 1 and xi œ " i˜x œ " i Ê a lower sum is
(c) ˜x œ
&" #
œ # and xi œ " i˜x œ " #i Ê an upper sum is ! x"i † # œ #ˆ" $" ‰ œ
(d) ˜x œ
&" %
œ 1 and xi œ " i˜x œ " i Ê an upper sum is
4. faxb œ % x#
iœ" % !" xi iœ" "
† " œ "ˆ #"
iœ! $ !" xi iœ!
" $
† " œ "ˆ"
" #
&" ‰ œ
" %
"' "&
" $
(( '!
) $
"% ‰ œ
#& "#
Since f is increasing on Ò#ß !Ó and decreasing on Ò!ß #Ó, we use left endpoints on Ò#ß !Ó and right endpoints on Ò!ß #Ó to obtain lower sums and use right endpoints on Ò#ß !Ó and left endpoints on Ò!ß #Ó to obtain upper sums.
(a) ˜x œ
# a#b #
œ # and xi œ # i˜x œ # #i Ê a lower sum is # † ˆ% a#b# ‰ # † a% ## b œ !
(b) ˜x œ
# a#b %
œ " and xi œ # i˜x œ # i Ê a lower sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
"
%
iœ!
iœ$
œ "ˆˆ% a#b# ‰ ˆ% a"b# ‰ a% "# b a% ## b‰ œ ' (c) ˜x œ
# a#b #
œ # and xi œ # i˜x œ # #i Ê a upper sum is # † ˆ% a!b# ‰ # † a% !# b œ "'
(d) ˜x œ
# a#b %
œ " and xi œ # i˜x œ # i Ê a upper sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
#
$
iœ"
iœ#
œ "ˆˆ% a"b# ‰ a% !# b a% !# b a% "# b‰ œ "% 5. faxb œ x#
œ
" #
Using 4 rectangles Ê ˜x œ " % ! œ Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
" %
Using 2 rectangles Ê ˜x œ #
#
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
#
#
"! $#
#
œ
"! #
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
& "'
#
œ "% Šˆ ") ‰ ˆ $) ‰ ˆ &) ‰ ˆ () ‰ ‹ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
#" '%
Section 5.1 Area and Estimating with Finite Sums 6. faxb œ x$
œ
" #
Using 4 rectangles Ê ˜x œ " % ! œ Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
" %
$
$
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
$
$
$
$
( œ "% Š " $ )& ‹œ $
7. faxb œ
" x
"! #
Using 2 rectangles Ê ˜x œ #) # † '%
%*' % † )$
œ
œ
Using 2 rectangles Ê ˜x œ œ #ˆ "# "% ‰ œ $#
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
( $#
"#% )$
&" #
œ
$" "#)
œ # Ê #afa#b fa%bb
Using 4 rectangles Ê ˜x œ & % " œ " Ê "ˆfˆ $# ‰ fˆ &# ‰ fˆ (# ‰ fˆ *# ‰‰ œ "ˆ #$
8. faxb œ % x#
# &
# (
#* ‰ œ
"%)) $†&†(†*
Using 2 rectangles Ê ˜x œ œ #a$ $b œ "#
259
œ
# a#b #
%*' &†(†*
œ
%*' $"&
œ # Ê #afa"b fa"bb
Using 4 rectangles Ê ˜x œ # %a#b œ " Ê "ˆfˆ $# ‰ fˆ "# ‰ fˆ "# ‰ fˆ $# ‰‰ # # # # œ "ŠŠ% ˆ $# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ $# ‰ ‹‹ œ "' ˆ *% † # "% † #‰ œ "' "! # œ ""
9. (a) D ¸ (0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) œ 87 inches (b) D ¸ (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) œ 87 inches 10. (a) D ¸ (1)(300) (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) œ 5220 meters (NOTE: 5 minutes œ 300 seconds) (b) D ¸ (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) (0)(300) œ 4920 meters (NOTE: 5 minutes œ 300 seconds) 11. (a) D ¸ (0)(10) (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10) (35)(10) (44)(10) (30)(10) œ 3490 feet ¸ 0.66 miles (b) D ¸ (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10) (35)(10) (44)(10) (30)(10) (35)(10) œ 3840 feet ¸ 0.73 miles 12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the midpoints of each time interval to approximate this area using rectangles. Thus, D ¸ (20)(0.001) (50)(0.001) (72)(0.001) (90)(0.001) (102)(0.001) (112)(0.001) (120)(0.001) (128)(0.001) (134)(0.001) (139)(0.001) ¸ 0.967 miles (b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours œ 22.7 sec. At 22.7 sec, the velocity was approximately 120 mi/hr.
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260
Chapter 5 Integration
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing acceleration † ?t. Thus, ?t œ 1 and speed ¸ [32.00 19.41 11.77 7.14 4.33](1) œ 74.65 ft/sec (b) Using right end-points we obtain a lower estimate: speed ¸ [19.41 11.77 7.14 4.33 2.63](1) œ 45.28 ft/sec (c) Upper estimates for the speed at each second are: t 0 1 2 3 4 5 v 0 32.00 51.41 63.18 70.32 74.65 Thus, the distance fallen when t œ 3 seconds is s ¸ [32.00 51.41 63.18](1) œ 146.59 ft. 14. (a) The speed is a decreasing function of time Ê right end-points give an lower estimate for the height (distance) attained. Also t 0 1 2 3 4 5 v 400 368 336 304 272 240 gives the time-velocity table by subtracting the constant g œ 32 from the speed at each time increment ?t œ 1 sec. Thus, the speed ¸ 240 ft/sec after 5 seconds. (b) A lower estimate for height attained is h ¸ [368 336 304 272 240](1) œ 1520 ft. 15. Partition [!ß #] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of these subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating 1 125 343 $ $ rectangles are f(m" ) œ (0.25)$ œ 64 , f(m# ) œ (0.75)$ œ 27 64 , f(m$ ) œ (1.25) œ 64 , and f(m% ) œ (1.75) œ 64 Notice that the average value is approximated by œ
" length of [!ß#]
†”
" #
$ $ $ $ ’ˆ 4" ‰ ˆ #" ‰ ˆ 43 ‰ ˆ #" ‰ ˆ 45 ‰ ˆ #" ‰ ˆ 47 ‰ ˆ #" ‰“ œ
$" "'
approximate area under • . We use this observation in solving the next several exercises. curve f(x) œ x$
16. Partition [1ß 9] into the four subintervals ["ß $], [3ß &], [&ß (], and [(ß *]. The midpoints of these subintervals are m" œ 2, m# œ 4, m$ œ 6, and m% œ 8. The heights of the four approximating rectangles are f(m" ) œ "# , f(m# ) œ "4 , f(m$ ) œ 6" , and f(m% ) œ 8" . The width of each rectangle is ?x œ 2. Thus, Area ¸ 2 ˆ "# ‰ 2 ˆ 4" ‰ 2 ˆ 6" ‰ 2 ˆ 8" ‰ œ
Ê average value ¸
25 1#
area length of ["ß*]
œ
ˆ 25 ‰ 12 8
œ
25 96 .
17. Partition [0ß 2] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of the subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating rectangles are " #
f(m" ) œ œ
" #
" #
sin#
1 4
œ
" #
" #
œ 1, and f(m% ) œ
œ 1, f(m# ) œ
" 2
sin#
71 4
œ
sin#
" #
Š È"2 ‹ œ 1. The width of each rectangle is ?x œ #" . Thus,
31 4
œ
" #
" #
œ 1, f(m$ ) œ
" 2
sin#
51 4
œ
" #
Š È"2 ‹
#
" 2
#
Area ¸ (1 1 1 1) ˆ "# ‰ œ 2 Ê average value ¸
area length of [0ß2]
œ
2 #
œ 1.
18. Partition [0ß 4] into the four subintervals [0ß 1], [1ß 2ß ], [2ß 3], and [3ß 4]. The midpoints of the subintervals are m" œ "# , m# œ #3 , m$ œ 5# , and m% œ 7# . The heights of the four approximating rectangles are f(m" ) œ 1 Šcos Š
% 1 ˆ "# ‰ 4 ‹‹
œ 1 ˆcos ˆ 18 ‰‰ œ 0.27145 (to 5 decimal places),
f(m# ) œ 1 Šcos Š
% 1 ˆ 3# ‰ 4 ‹‹
œ 1 ˆcos ˆ 381 ‰‰ œ 0.97855, f(m3 ) œ 1 Šcos Š
%
œ 0.97855, and f(m% ) œ 1 Šcos Š
%
% 1 ˆ 7# ‰ 4 ‹‹
% 1 ˆ #5 ‰ 4 ‹‹
œ 1 ˆcos ˆ 581 ‰‰
%
%
œ 1 ˆcos ˆ 781 ‰‰ œ 0.27145. The width of each rectangle is
?x œ ". Thus, Area ¸ (0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1) œ 2.5 Ê average 2.5 5 value ¸ lengtharea of [0ß4] œ 4 œ 8 .
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Section 5.1 Area and Estimating with Finite Sums
261
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate œ (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) œ 758 gal, lower estimate œ (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) œ 543 gal. (b) upper estimate œ (70 97 136 190 265 369 516 720) œ 2363 gal, lower estimate œ (50 70 97 136 190 265 369 516) œ 1693 gal. (c) worst case: 2363 720t œ 25,000 Ê t ¸ 31.4 hrs; best case: 1693 720t œ 25,000 Ê t ¸ 32.4 hrs 20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate œ (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) œ 60.9 tons lower estimate œ (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) œ 46.8 tons (b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) œ 126.6 tons, so near the end of September 125 tons of pollutants will have been released. #
21. (a) The diagonal of the square has length 2, so the side length is È#. Area œ ŠÈ#‹ œ # (b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring #1 1 "' œ ) . Area œ "'ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ % sin 1 œ #È# ¸ #Þ)#) #
)
)
%
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring #1 1 $# œ "' . Area œ $#ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ ) sin 1 œ #È# ¸ $Þ!'" #
"'
"'
)
(d) Each area is less than the area of the circle, 1. As n increases, the area approaches 1. 22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring #1 1 1 ‰ˆ ˆ " ‰ˆ cos 1n ‰ œ "# sin #n1 . #n œ n . The area of each isosceles triangle is AT œ # # sin n (b) The area of the polygon is AP œ nAT œ
n #
sin
#1 n ,
n nÄ_ #
so lim
sin
#1 n
œ lim 1 † nÄ_
sin #n1 ˆ #n1 ‰
œ1
(c) Multiply each area by r# . AT œ "# r# sin #n1 AP œ n# r# sin
lim AP œ 1r
#
#1 n
nÄ_
23-26. Example CAS commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi; plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); N := [ 100, 200, 1000 ]; # (b) for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
262
Chapter 5 Integration
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); Mathematica: (assigned function and values for a and b may vary): Symbols for 1, Ä , powers, roots, fractions, etc. are available in Palettes (under File). Never insert a space between the name of a function and its argument. Clear[x] f[x_]:=x Sin[1/x] {a,b}={1/4, 1} Plot[f[x],{x, a, b}] The following code computes the value of the function for each interval midpoint and then finds the average. Each sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell. n =100; dx = (b a) /n; values = Table[N[f[x]], {x, a dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n n =200; dx = (b a) /n; values = Table[N[f[x]],{x, a + dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n n =1000; dx = (b a) /n; values = Table[N[f[x]],{x, a dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n FindRoot[f[x] == average,{x, a}] 5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS 2
1. ! kœ1
3
2. ! kœ1
6k k1
œ
6(1) 11
6(2) 21
œ
6 2
k1 k
œ
11 1
21 2
31 3
12 3
œ7
œ0
1 2
2 3
œ
7 6
4
3. ! cos k1 œ cos (11) cos (21) cos (31) cos (41) œ 1 1 1 1 œ 0 kœ1
5
4. ! sin k1 œ sin (11) sin (21) sin (31) sin (41) sin (51) œ 0 0 0 0 0 œ 0 kœ1
3
5. ! (1)kb1 sin kœ1
1 k
œ (1)"" sin
1 1
(1)#" sin
1 #
(")$" sin
1 3
œ 01
È3 #
œ
È3 2 #
4
6. ! (1)k cos k1 œ (1)" cos (11) (1)# cos (21) (1)$ cos (31) (1)% cos (41) kœ1
œ (1) 1 (1) 1 œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.2 Sigma Notation and Limits of Finite Sums 6
7. (a) ! 2kc1 œ 2"" 2#" 2$" 2%" 2&" 2'" œ 1 2 4 8 16 32 kœ1 5
(b) ! 2k œ 2! 2" 2# 2$ 2% 2& œ 1 2 4 8 16 32 kœ0 4
(c) ! 2k1 œ 2"" 2!" 2"" 2#" 2$" 2%" œ 1 2 4 8 16 32 kœ
"
All of them represent 1 2 4 8 16 32 6
8. (a) ! (2)k
œ (2)"" (2)#" (2)$" (2)%" (2)&" (2)'" œ 1 2 4 8 16 32
1
kœ1 5
(b) ! (1)k 2k œ (1)! 2! Ð")" 2" (1)# 2# (1)$ 2$ (1)% 2% (1)& 2& œ 1 2 4 8 16 32 kœ0 3
(c) ! (1)k1 2k2 œ Ð")#" 2## (")"" 2"# (")!" 2!# (1)"" 2"# (")#" 2## kœ 2
(1)$" 2$# œ 1 2 4 8 16 32; (a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two 4
(")k " k1
9. (a) ! kœ2 2
(b) ! kœ0 1
(c) ! kœ
"
(1)# " 21
œ
(")k k1
œ
(1)! 01
(")k k2
œ
(1) " 1 2
(")$ " 31
(")" 11
(")! 02
(")# 21
(")% " 41
œ 1
œ1
(")" 12
" #
œ 1
" #
" #
" 3
" 3
" 3
(a) and (c) are equivalent; (b) is not equivalent to the other two. 4
10. (a) ! (k 1)# œ (1 1)# (2 1)# (3 1)# (4 1)# œ 0 1 4 9 kœ1 3
(b) ! (k 1)# œ (1 1)# (0 1)# (1 1)# (2 1)# (3 1)# œ 0 1 4 9 16 kœ 1
"
(c) ! k# œ (3)# (2)# (1)# œ 9 4 1 kœ 3
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two. 6
4
11. ! k
4
12. ! k#
kœ1
13. !
kœ1
5
5
15. ! (1)k1
14. ! 2k kœ1
kœ1
n
kœ1
" k
5
16. ! (1)k kœ1
n
17. (a) ! 3ak œ 3 ! ak œ 3(5) œ 15 kœ1 n
(b) ! kœ1 n
bk 6
œ
" 6
kœ1 n
! bk œ kœ1
" 6
(6) œ 1
n
n
kœ1 n
kœ1 n
kœ1 n
kœ1 n
kœ1 n
kœ1
(c) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 1 (d) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 11 n
(e) ! (bk 2ak ) œ ! bk 2 ! ak œ 6 2(5) œ 16 kœ1
kœ1
" #k
kœ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
k 5
263
264
Chapter 5 Integration n
n
kœ1 n
kœ1
n
18. (a) ! 8ak œ 8 ! ak œ 8(0) œ 0 n
n
kœ1
kœ1
(c) ! (ak 1) œ ! ak ! 1 œ 0 n œ n kœ1
10
19. (a) ! k œ kœ1
10(10 1) #
n
(b) ! 250bk œ 250 ! bk œ 250(1) œ 250 kœ1 n
n
kœ1
n
kœ1
kœ1
kœ1
(d) ! (bk 1) œ ! bk ! 1 œ " n 10
(b) ! k# œ
œ 55
kœ1
10(10 1)(2(10) 1) 6
œ 385
13(13 1)(2(13) 1) 6
œ 819
#
10
(c) ! k$ œ ’ 10(10# 1) “ œ 55# œ 3025 kœ1
13
20. (a) ! k œ kœ1
13(13 1) #
13
(b) ! k# œ
œ 91
kœ1
#
13
(c) ! k$ œ ’ 13(13# 1) “ œ 91# œ 8281 kœ1
7
7
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
5
21. ! 2k œ 2 ! k œ 2 Š 7(7 # ") ‹ œ 56 23. ! a3 k# b œ ! 3 ! k# œ 3(6)
24. ! ak# 5b œ ! k# ! 5 œ
22. ! kœ1
6(6 ")(2(6) 1) 6
6(6 ")(2(6) 1) 6
1k 15
1 15
œ
5
!kœ
1 15
kœ1
Š 5(5 # 1) ‹ œ 1
œ 73
5(6) œ 61
5
5
5
5
kœ1
kœ1
kœ1
kœ1
7
7
7
7
kœ1
kœ1
kœ1
kœ1
1) 25. ! k(3k 5) œ ! a3k# 5kb œ 3 ! k# 5 ! k œ 3 Š 5(5 1)(2(5) ‹ 5 Š 5(5 # 1) ‹ œ 240 6
1) 26. ! k(2k 1) œ ! a2k# kb œ 2 ! k# ! k œ 2 Š 7(7 1)(2(7) ‹ 6
5
27. !
k$ 225
kœ1
kœ1
#
7
$
5
Œ! k œ 7
28. Œ! k ! kœ1
kœ1
k$ 4
" 2 #5
7
5
5
kœ1
kœ1
$
! k $ Œ! k œ #
œ Œ! k kœ1
" 4
" #25
#
7(7 1) #
œ 308
$
Š 5(5 # 1) ‹ Š 5(5 # 1) ‹ œ 3376 #
7
! k$ œ Š 7(7 1) ‹ # kœ1
7
" 4
#
Š 7(7 # 1) ‹ œ 588 500
29. (a) ! 3 œ 3a7b œ 21
(b) ! 7 œ 7a500b œ 3500
kœ1
kœ1
264
262
kœ3
jœ1
(c) Let j œ k 2 Ê k œ j 2; if k œ 3 Ê j œ 1 and if k œ 264 Ê j œ 262 Ê ! 10 œ ! 10 œ 10a262b œ 2620
36
28
28
28
kœ9
jœ1
jœ1
jœ1
30. (a) Let j œ k 8 Ê k œ j 8; if k œ 9 Ê j œ 1 and if k œ 36 Ê j œ 28 Ê ! k œ ! a j 8b œ ! j !8 œ
28a28 1b 2
)a28b œ 630 17
15
kœ3
jœ1
(b) Let j œ k 2 Ê k œ j 2; if k œ 3 Ê j œ 1 and if k œ 17 Ê j œ 15 Ê ! k2 œ ! a j 2b2 15
15
15
15
jœ1
jœ1
jœ1
jœ1
œ ! a j2 4j 4b œ ! j2 ! 4j ! 4 œ
15a15 1ba2a15b 1b 6
4†
15a15 1b 2
4a15b
œ 1240 480 60 œ 1780
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Section 5.2 Sigma Notation and Limits of Finite Sums 71
(c) Let j œ k 17 Ê k œ j 17; if k œ 18 Ê j œ 1 and if k œ 71 Ê j œ 54 Ê !kak 1b kœ3
54
54
54
54
54
jœ1
jœ1
jœ1
jœ1
œ ! a j 17baa j 17b 1b œ ! a j2 33j 272b œ ! j2 ! 33j ! 272 jœ1
œ
54a54 1ba2a54b 1b 6
33 †
54a54 1b 2
272a54b œ 53955 49005 14688 œ 117648
n
n
31. (a) ! 4 œ 4n
(b) ! c œ cn
kœ1 n
n
n
kœ1
kœ1
kœ1
kœ1
(c) ! ak 1b œ ! k ! 1 œ
n an 1 b 2
nœ
n
n
32. (a) ! ˆ 1n 2n‰ œ ˆ 1n 2n‰n œ 1 2n2 kœ1 n
(c) ! kœ1
k n2
œ
1 n an 1 b n2 2
œ
n2 n 2
(b) ! kœ1
c n
œ
c n
†nœc
n1 2n
33. (a)
(b)
(c)
34. (a)
(b)
(c)
35. (a)
(b)
(c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
265
266
Chapter 5 Integration
36. (a)
(b)
(c)
37. kx" x! k œ k1.2 0k œ 1.2, kx# x" k œ k1.5 1.2k œ 0.3, kx$ x# k œ k2.3 1.5k œ 0.8, kx% x$ k œ k2.6 2.3k œ 0.3, and kx& x% k œ k3 2.6k œ 0.4; the largest is lPl œ 1.2. 38. kx" x! k œ k1.6 (2)k œ 0.4, kx# x" k œ k0.5 (1.6)k œ 1.1, kx$ x# k œ k0 (0.5)k œ 0.5, kx% x$ k œ k0.8 0k œ 0.8, and kx& x% k œ k1 0.8k œ 0.2; the largest is lPl œ 1.1. 39. faxb œ " x#
Let ˜x œ
"! n
œ
n
!a" c#i b " œ n $
œ
n n$
# $n n"# '
. Thus,
nÄ_
Let ˜x œ
# '
$! n
œ
" n#
$ n
n
!#ci ˆ $ ‰ œ ! 'i † n n
iœ"
iœ"
n
Thus,
41. faxb œ x# "
lim ! 'i nÄ_ iœ" n
Let ˜x œ
$! n
n
œ"
œ
n #( ! # i n iœ"
nÄ_ iœ1
œ"
" $
œ
# $
and ci œ i˜x œ $ n
œ
") n#
n
!i œ
iœ"
†
# œ lim *n n# *n nÄ_
œ
$ n n
and ci œ i˜x œ
†nœ
$i n.
") n#
$ n
iœ"
$ n
#n$ $n# n 'n$
lim !a" ci# b n"
The right-hand sum is n an " b #
†
œ
*n# *n n#
œ lim ˆ* n* ‰ œ *.
!ac#i "b $ œ !Šˆ $i ‰# "‹ $ œ n n n
iœ"
iœ1
n
$ n
n
nan "ba#n "b 'n $
n
!an# i# b
" n$
n
iœ1
œ"
œ lim Œ" 40. faxb œ #x
!Š" ˆ i ‰# ‹ œ
"! # i n$ iœ1
œ"
and ci œ i˜x œ ni . The right-hand sum is n
" n
iœ1
n
" n
nÄ_
$i n. $ n
#( nan "ba#n "b ‹ n$ Š '
The right-hand sum is n
!Š *i## "‹ n
iœ"
$
* ") #( *a#n$ $n# nb n n# $œ $Þ Thus, #n $ # n #( ") n n*# lim !ac#i "b $n œ lim Œ $ œ # nÄ_ iœ" nÄ_
œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
* $ œ "#.
Section 5.2 Sigma Notation and Limits of Finite Sums 42. faxb œ $x#
"! n
Let ˜x œ n
œ
" n
and ci œ i˜x œ ni . The right-hand sum is
n
!$c#i ˆ " ‰ œ !$ˆ i ‰# ˆ " ‰ œ n
iœ"
#n$ $n# n #n$
œ
œ lim Œ nÄ_
43. faxb œ x x# œ xa" xb
n
iœ"
n
# $n n"# #
œ
# $n n"# #
œ
"! n
" n
Let ˜x œ
œ
n
# #
$ n$
n
iœ"
n
. Thus, lim !$c#i ˆ "n ‰ nÄ_ iœ"
œ ".
and ci œ i˜x œ ni . The right-hand sum is
n
n "! i n# iœ"
44. faxb œ $x #x#
iœ"
œ
" n a n "b ‹ n# Š #
œ
" "n #
$ nan "ba#n "b ‹ n$ Š '
! i# œ
!aci ci# b " œ !Š i ˆ i ‰# ‹ " œ n n n n
iœ"
œ lim ”Š nÄ_
" "n # ‹
Œ
Let ˜x œ
"! n
" n
œ
# $n n"# '
• œ
" #
# '
n $! i # n iœ"
n #! # i $ n iœ"
$n# $n #n#
#n# $n " $n#
!a$ci #c#i b " œ !Š $i #ˆ i ‰# ‹ " œ n n n n
iœ"
iœ"
œ
$ n a n "b ‹ n# Š #
œ
$ $n #
n#$ Š nan "ba' #n "b ‹ n # $ " $
nÄ_ iœ"
œ lim ”Š
$ $n # ‹
Œ
Let ˜x œ
"! n n
" n
œ
# $n n"# $
iœ"
iœ"
œ
#n# an# 2n "b 4n4 n
œ
• œ
$ #
# $
œ
"$ ' .
and ci œ i˜x œ ni . The right-hand sum is
!a2c3i b " œ !Š2ˆ i ‰3 ‹ " œ n n n
Thus,
œ
. Thus, lim !a$ci #c#i b "n
n#
n
œ &' .
and ci œ i˜x œ ni . The right-hand sum is n
n
n "! # i n$ iœ"
n
45. faxb œ 2x3
# $ # " nan "ba#n "b Š ‹ œ n #n# n #n '$nn$ n ' n$ n # $n n"# . Thus, lim !aci c#i b "n ' nÄ_ iœ"
nÄ_
267
n# 2n " #n #
n #! 3 i n4 iœ"
œ
œ
" #n n"# #
" # " lim !a2c3i b "n œ lim ” n# n# nÄ_ iœ" nÄ_
# n a n "b Š # ‹ n4
.
" • œ #.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
268
Chapter 5 Integration
46. faxb œ x2 x3
Let ˜x œ
! a"b n n
œ
" n
and ci œ " i˜x œ " ni . The rightn
2 $ hand sum is !ac#i ci3 b "n œ !Šˆ" ni ‰ ˆ" ni ‰ ‹ n" iœ1
n
œ ! Š2 œ
iœ1 n
!2 n iœ"
œ 2n anb
n# 2n " 4n#
4 6n n#2
1 4
5n 5 #n
4n2 6n# 3n2
œ lim ”2
5 #
5 n
œ2 nÄ_
3
2 n
œ2 " n#
5 5n #
•œ2
5 2
4 6n n#2 3
1 2n 4
œ
7 12 .
4 3
1 4
5i n
n 5! i n2 iœ"
%i n2
2
n %! 2 i n3 iœ"
5 n an "b ‹ n2 Š # " n#
i " n3 ‹ n 3
iœ1 n
œ !Š n2
iœ1 n 1! 3 i n4 iœ"
5i n2
% nan "ba#n "b ‹ n3 Š '
%i2 n3
i3 n4 ‹
1 nan "b ‹ n4 Š #
2
n
. Thus, lim !ac#i ci3 b n" nÄ_ iœ"
5.3 THE DEFINITE INTEGRAL
1.
'02 x# dx
2.
'"! 2x$ dx
3.
'(& ax# 3xb dx
4.
'"% "x dx
5.
'#$ 1 " x dx
6.
'0" È4 x# dx
7.
' ! Î% (sec x) dx
8.
'0 Î% (tan x) dx 1
1
9. (a) (c) (e) (f)
10. (a) (b) (c) (d) (e) (f)
11. (a) (c)
" & '#2 g(x) dx œ 0 (b) ' g(x) dx œ ' g(x) dx œ 8 & " & & 2 '"2 3f(x) dx œ 3'"2 f(x) dx œ 3(4) œ 12 (d) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 (4) œ 10 # " " & & & '" [f(x) g(x)] dx œ '" f(x) dx '" g(x) dx œ 6 8 œ 2 '"& [4f(x) g(x)] dx œ 4 '"& f(x) dx '"& g(x) dx œ 4(6) 8 œ 16
'"* 2f(x) dx œ 2 '"* f(x) dx œ 2(1) œ 2 '(* [f(x) h(x)] dx œ '(*f(x) dx '(* h(x) dx œ 5 4 œ 9 '(* [2f(x) 3h(x)] dx œ 2 '(* f(x) dx 3 '(* h(x) dx œ 2(5) 3(4) œ 2 '*"f(x) dx œ '"* f(x) dx œ (1) œ 1 '"( f(x) dx œ '"* f(x) dx '(* f(x) dx œ 1 5 œ 6 '*( [h(x) f(x)] dx œ '(* [f(x) h(x)] dx œ '(* f(x) dx '(* h(x) dx œ 5 4 œ 1 '"2 f(u) du œ '"2 f(x) dx œ 5 '#" f(t) dt œ '"2 f(t) dt œ 5
(b) (d)
'"2 È3 f(z) dz œ È3 '"2 f(z) dz œ 5È3 '"2 [f(x)] dx œ '"2 f(x) dx œ 5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral
'!$ g(t) dt œ '$! g(t) dt œ È2 '$! [g(x)] dx œ '$! g(x) dx œ È2
12. (a) (c)
(b)
'"$ h(r) dr œ '"$ h(r) dr '"" h(r) dr œ 6 0 œ 6 " $ $ ' h(u) du œ Œ ' h(u) du œ ' h(u) du œ 6 $ " "
14. (a) (b)
15. The area of the trapezoid is A œ " #
(5 2)(6) œ 21 Ê
œ 21 square units
" #
(3 1)(1) œ 2 Ê
" #
(B b)h
" #
(B b)h
'# ˆ #x 3‰ dx %
16. The area of the trapezoid is A œ œ
(d)
'$! g(u) du œ '$! g(t) dt œ È2 '$! Èg(r)2 dr œ È"2 '$! g(t) dt œ Š È"2 ‹ ŠÈ2‹ œ 1
'$% f(z) dz œ '!% f(z) dz '!$ f(z) dz œ 7 3 œ 4 '%$ f(t) dt œ '$% f(t) dt œ 4
13. (a)
œ
(b)
'"Î#
$Î#
(2x 4) dx
œ 2 square units
17. The area of the semicircle is A œ œ
9 #
1 Ê
" #
1r# œ
" #
1(3)#
'$$ È9 x# dx œ 9# 1 square units
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
269
270
Chapter 5 Integration
18. The graph of the quarter circle is A œ œ 41 Ê
" 4
1 r# œ
" 4
1(4)#
'%! È16 x# dx œ 41 square units
19. The area of the triangle on the left is A œ
" #
bh œ
œ 2. The area of the triangle on the right is A œ œ
" #
Ê
(1)(1) œ
" #.
(2)(2) bh
Then, the total area is 2.5
'# kxk dx œ 2.5 square units "
20. The area of the triangle is A œ Ê
" # " #
" #
bh œ
'" a1 kxkb dx œ 1 square unit "
21. The area of the triangular peak is A œ
" #
(2)(1) œ 1
" #
bh œ
" #
(2)(1) œ 1.
The area of the rectangular base is S œ jw œ (2)(1) œ 2. Then the total area is 3 Ê
'"" a2 kxkb dx œ 3 square units
22. y œ 1 È1 x# Ê y 1 œ È1 x# Ê (y 1)# œ 1 x# Ê x# (y 1)# œ 1, a circle with center (!ß ") and radius of 1 Ê y œ 1 È1 x# is the upper semicircle. The area of this semicircle is A œ "# 1r# œ "# 1(1)# œ 1# . The area of the rectangular base is A œ jw œ (2)(1) œ 2. Then the total area is 2 Ê
'"" Š1 È1 x# ‹ dx œ 2 1# square units
1 #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral 23.
'!b x2 dx œ "# (b)( b2 ) œ b4
24.
'!b 4x dx œ "# b(4b) œ 2b#
25.
'ab 2s ds œ "# b(2b) "# a(2a) œ b# a#
26.
'ab 3t dt œ "# b(3b) "# a(3a) œ 3# ab# a# b
(b)
'02 È4 x2 dx œ "4 1a2b2 ‘ œ 1
#
27. (a)
'22 È4 x2 dx œ "# 1a2b2 ‘ œ 21
28. (a)
'01 Š3x È1 x2 ‹ dx œ '01 3x dx '01 È1 x2 dx œ "# a1ba3b‘ 4" 1a1b2 ‘ œ 14 3#
(b)
271
'01 Š3x È1 x2 ‹ dx œ '01 3x dx '01 3x dx '11 È1 x2 dx œ "# a1ba3b‘ "# a1ba3b‘ 2" 1a1b2 ‘ œ 12
È#
#
29.
'"
31.
'1#1 ) d) œ (2#1)
33.
'0
35.
'!"Î# t# dt œ ˆ 3‰
œ
37.
'a#a x dx œ (2a)#
39.
'!
3 È 7
x dx œ
ŠÈ2‹ #
#
(1)# #
œ
1# #
œ
31 # #
" #
30.
'!Þ&#Þ& x dx œ (2.5)#
32.
'È& # # r dr œ Š5È#2‹
x dx œ
3 7‹ ŠÈ
œ
3
#
7 3
34.
'!!Þ$ s# ds œ (0.3)3
" 24
36.
'!1Î# )# d) œ ˆ 3‰
a# #
œ
3a# #
38.
'a
$ b‹ ŠÈ
3
œ
b 3
40.
'!$b x# dx œ (3b)3
(0.5)# #
#
$
1 $ #
È$a
$
x# dx œ
#
È
$
#
" $ #
$ È b
œ3 #
ŠÈ2‹ #
œ 24
œ 0.009
œ
1$ #4
#
x dx œ
ŠÈ3a‹ #
$
a# #
œ a#
œ 9b$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
272
Chapter 5 Integration
41.
'$" 7 dx œ 7(1 3) œ 14
43.
'!2 (2t 3) dt œ 2 '"" t dt '!2 3 dt œ 2 ’ 2#
44.
'!
45.
'#" ˆ1 #z ‰ dz œ '#" 1 dz '#" #z dz œ '#" 1 dz "# '"# z dz œ 1[1 2] "# ’ 2# 1# “ œ " "# ˆ 3# ‰ œ 74
46.
'$! (2z 3) dz œ '$! 2z dz '$! 3 dz œ 2 '!$ z dz '$! 3 dz œ 2 ’ 3#
47.
'"# 3u# du œ 3 '"# u# du œ 3 ”'!# u# du '!" u# du• œ 3 Š’ 23
48.
'"Î#" 24u# du œ 24 '"Î#"
49.
'!# a3x# x 5b dx œ 3 '!# x# dx '!# x dx '!# 5 dx œ 3 ’ 23
50.
'"! a3x# x 5b dx œ '!" a3x# x 5b dx œ ”3 '!" x# dx '!" x dx '!" 5 dx•
#
È2
Št È2‹ dt œ
È2
'!
'!2 5x dx œ 5 '!2 x dx œ 5 ’ 2#
#
42.
t dt
È2
'!
È2 dt œ
0# #“
ŠÈ2‹
#
0# #—
#
$
'!"
u# du
'!"Î#
0$ 3“
$
0$ 3‹
51. Let ?x œ
b0 n
œ
b n
#
Š 1#
0# #‹
$
’ "3
$
u# du— œ 24 ” 13
$
œ ’3 Š 13
5(1 0)“ œ ˆ 3# 5‰ œ
0$ 3“
0# #“
#
3[0 3] œ 9 9 œ 0
0$ 3 “‹
ˆ "# ‰$ 3 •
#
’ 2#
$
œ 3 ’ 23
œ 24 ’
0# #“
ˆ 78 ‰ 3
1$ 3“
5[2 0] œ (8 2) 10 œ 0
7 #
x# œ 2?xß á ß xn " œ (n 1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 3(?x)# ?x œ 3(?x)$ f(c# ) ?x œ f(2?x) ?x œ 3(2?x)# ?x œ 3(2)# (?x)$ f(c$ ) ?x œ f(3?x) ?x œ 3(3?x)# ?x œ 3(3)# (?x)$ ã f(cn ) ?x œ f(n?x) ?x œ 3(n?x)# ?x œ 3(n)# (?x)$ n
Then Sn œ ! f(ck ) ?x œ ! 3k# (?x)$ kœ1 n
kœ1
1) œ 3(?x)$ ! k# œ 3 Š bn$ ‹ Š n(n 1)(2n ‹ 6 $
kœ1
œ
$
b #
ˆ2
3 n
"‰ n#
Ê
'!b 3x# dx œ n lim Ä_
b$ #
ˆ2
3 n
"‰ n#
œ 3 ˆ 37 ‰ œ 7
“œ7
and let x! œ 0, x" œ ?x,
n
œ 10
È2 ’È2 0“ œ 1 2 œ 1
#
u# du œ 24 –
0# #“
3(2 0) œ 4 6 œ 2 #
–
œ b$ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral 52. Let ?x œ
b0 n
œ
and let x! œ 0, x" œ ?x,
b n
x# œ 2?xß á ß xn " œ (n 1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 1(?x)# ?x œ 1(?x)$ f(c# ) ?x œ f(2?x) ?x œ 1(2?x)# ?x œ 1(2)# (?x)$ f(c$ ) ?x œ f(3?x) ?x œ 1(3?x)# ?x œ 1(3)# (?x)$ ã f(cn ) ?x œ f(n?x) ?x œ 1(n?x)# ?x œ 1(n)# (?x)$ n
n
Then Sn œ ! f(ck ) ?x œ ! 1k# (?x)$ kœ1 n
kœ1
1) œ 1(?x)$ ! k# œ 1 Š bn$ ‹ Š n(n 1)(2n ‹ 6 $
kœ1
œ
1b 6
$
ˆ2
53. Let ?x œ
3 n
b0 n
"‰ n#
œ
b n
Ê
'!b 1x# dx œ n lim Ä_
1 b$ 6
ˆ2
3 n
"‰ n#
œ
1 b$ 3 .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn " œ (n 1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 2(?x)(?x) œ 2(?x)# f(c# ) ?x œ f(2?x) ?x œ 2(2?x)(?x) œ 2(2)(?x)# f(c$ ) ?x œ f(3?x) ?x œ 2(3?x)(?x) œ 2(3)(?x)# ã f(cn ) ?x œ f(n?x) ?x œ 2(n?x)(?x) œ 2(n)(?x)# n
n
Then Sn œ ! f(ck ) ?x œ ! 2k(?x)# kœ1 n
kœ1
œ 2(?x)# ! k œ kœ1
œ b# ˆ1 "n ‰ Ê 54. Let ?x œ
b0 n
œ
# 2 Š bn# ‹ Š n(n 2 1) ‹
'!b 2x dx œ n lim Ä_ b n
b# ˆ1 n" ‰ œ b# .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn " œ (n 1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: " # ‰ f(c" ) ?x œ f(?x) ?x œ ˆ ?x # 1 (?x) œ # (?x) ?x f(c# ) ?x œ f(2?x) ?x œ ˆ 2?# x 1‰ (?x) œ "# (2)(?x)# ?x f(c$ ) ?x œ f(3?x) ?x œ ˆ 3?# x 1‰ (?x) œ
" #
(3)(?x)# ?x
f(cn ) ?x œ f(n?x) ?x œ ˆ n?# x 1‰ (?x) œ
" #
(n)(?x)# ?x
ã
n
n
kœ1
kœ1
Then Sn œ ! f(ck ) ?x œ ! ˆ "# k(?x)# ?x‰ œ œ
" 4
b# ˆ1 1n ‰ b Ê
'! ˆ x# 1‰ dx œ n lim Ä_ b
" #
n
n
kœ1
kœ1
(?x)# ! k ?x ! 1 œ
ˆ 4" b# ˆ1 n" ‰ b‰ œ
" 4
" #
Š bn# ‹ Š n(n 2 1) ‹ ˆ bn ‰ (n) #
b# b.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
273
274
Chapter 5 Integration
55. av(f) œ Š È3" 0 ‹ œ
È$
'!
" È3
È$
'!
x# dx
" È3
ŠÈ3‹ 3
56. av(f) œ ˆ 3 " 0 ‰
È$
'!
" È3
$
œ
ax# 1b dx
" È3
1 dx
ŠÈ3 0‹ œ 1 1 œ 0.
'!$ Š x# ‹ dx œ 3" ˆ #" ‰ '!$ x# dx #
$
#
œ "6 Š 33 ‹ œ 3# ; x# œ 3# .
'!" a3x# 1b dx œ " " œ 3 ' x# dx ' 1 dx œ 3 Š 13 ‹ (1 0) ! !
57. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!" a3x# 3b dx œ " " œ 3 ' x# dx ' 3 dx œ 3 Š 13 ‹ 3(1 0) ! !
58. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!$ (t 1)# dt $ $ $ œ 3" ' t# dt 32 ' t dt 3" ' 1 dt ! ! !
59. av(f) œ ˆ 3 " 0 ‰
œ
" 3
$
#
Š 33 ‹ 32 Š 3#
0# #‹
3" (3 0) œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral 60. av(f) œ Š 1 1(2) ‹
'#" at# tb dt
'#" t# dt 3" '#" t dt " # œ "3 ' t# dt 3" ' t# dt 3" Š 1# ! ! œ
" 3
#
œ
" 3
$
Š 13 ‹ 3" Š (32) ‹ $
61. (a) av(g) œ Š 1 "(1) ‹
" #
œ
3 #
(2)# # ‹
.
'"" akxk 1b dx
'"! (x 1) dx "# '!" (x 1) dx ! ! " " œ "# ' x dx "# ' 1 dx "# ' x dx "# ' 1 dx " " ! ! œ
" #
#
œ "# Š 0#
(1)# # ‹
#
"# (0 (1)) "# Š 1#
0# #‹
"# (1 0)
œ "# .
'"$ akxk 1b dx œ #" '"$ (x 1) dx $ $ œ "# ' x dx "# ' 1 dx œ "# Š 3# 1# ‹ "# (3 1) " "
(b) av(g) œ ˆ 3 " 1 ‰
#
#
œ 1.
(c) av(g) œ Š 3 "(1) ‹ œ
" 4 " 4
'"$ akxk 1b dx
'"" akxk 1b dx 4" '"$ akxk 1b dx " 4
(see parts (a) and (b) above).
62. (a) av(h) œ Š 0 "(1) ‹
'"0 kxk dx œ '"0 (x) dx
œ
œ
(1 2) œ
'"0 x dx œ 0#
#
(1)# #
œ "# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
275
276
Chapter 5 Integration
(b) av(h) œ ˆ 1 " 0 ‰ #
œ Š "#
'0" kxk dx œ '0" x dx
0# #‹
œ "# .
(c) av(h) œ Š 1 "(1) ‹
'"" kxk dx
'"0 kxk dx '0" kxk dx
œ
" #
Œ
œ
" #
ˆ "# ˆ "# ‰‰ œ "# (see parts (a) and (b)
above).
ba n and let ck be the right n ab a b × and ck œ a kabn ab . n
63. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ endpoint of each subinterval. So the partition is P œ Öa, a n
n
kœ"
kœ" b
We get the Riemann sum ! fack b˜x œ ! c † this expression remains cab ab. Thus,
ba n
ba n , n
c ab a b ! " n kœ"
œ
#a b a b , n
a
œ
c ab a b n
...,a
† n œ cab ab. As n Ä _ and mPm Ä !
'a c dx œ cab ab.
64. Consider the partition P that subdivides the interval Ò0, 2Ó into n subintervals of width ˜x œ right endpoint of each subinterval. So the partition is P œ Ö0, n
n
kœ"
kœ"
‰ Riemann sum ! fack b˜x œ ! ’2ˆ 2k n 1“ † As n Ä _ and mPm Ä ! the expression
œ
2 n
4 an "b n
2 n
2 n,
2†
2 n,
n
! ˆ 4k 1‰ œ n
kœ"
...,n†
8 n2
n
!k
kœ"
2 n
n
!1 œ
kœ"
8 n2
†
n
n
œ
n
ba ! # a n Œ kœ"
kœ" n
#a a b a b ! k n kœ"
œ ab aba# aab ab# †
n" n
kœ" n ab a b ! # k n# kœ" #
ab a b $ '
†
œ
b a n
n an " b #
2 n
and let ck be the
2 n
œ
2 n
2k n .
We get the
†nœ
4 an " b n
2.
'02 a2x 1b dx œ 6.
65. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ We get the Riemann sum ! fack b˜x œ ! c#k ˆ b n a ‰
œ
œ 2× and ck œ k †
2 n
2 has the value 4 2 œ 6. Thus,
endpoint of each subinterval. So the partition is P œ
20 n
ba n
and let ck be the right
Öa, a b n a , a #abn ab , . . ., a nabn ab × and ck œ a kabn ab . n n # # # œ b n a ! Ša kabn ab ‹ œ bn a ! Ša# #akabn ab k abn# ab ‹ kœ" kœ" † na#
an "ba#n "b n#
#a a b a b # n#
†
n a n "b #
ab a b $ n$
†
nan "ba#n "b '
$ " " n" ab ab$ # n n# † " ' " ab a b $ †# ' b $ $ x# dx œ b$ a$ . a
œ ab aba# aab ab# †
As n Ä _ and mPm Ä ! this expression has value ab aba# aab ab# † " œ ba# a$ ab# #a# b a$ "$ ab$ $b# a $ba# a$ b œ
b$ $
a$ $.
Thus,
'
66. Consider the partition P that subdivides the interval Ò1, 0Ó into n subintervals of width ˜x œ the right endpoint of each subinterval. So the partition is P œ Ö1, 1
1 n,
1 2 †
1 n,
0 a 1 b n
. . ., 1 n †
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ 1 n
1 n
and let ck be
œ 0× and
Section 5.3 The Definite Integral ck œ 1 k † œ
1 n
n
kœ" n
kœ"
2 œ 1 kn . We get the Riemann sum ! fack b˜x œ ! Šˆ1 kn ‰ ˆ1 kn ‰ ‹ †
k n
1
n
! Š1
kœ"
3 an " b 2n
œ 2
n
1 n
n
2
ˆ kn ‰ ‹ œ 2n ! 1
2k n
kœ"
an "ba#n "b . 'n 2
3 n2
!k
1 n3
kœ"
n
! k2 œ 2 † n n
kœ"
3 n2
n an " b #
†
As n Ä _ and mPm Ä ! this expression has value 2
'01 ax x# bdx œ 56 .
3 2
1 n3
the right endpoint of each subinterval. So the partition is P œ Ö1, 1 ck œ 1 k † œ
3 n
n
! Š3
kœ"
œ 18
3 n
18k n
36an "b n
œ 1 27k2 n2
3k n .
n
n
kœ" n
kœ"
3 n,
1 2 †
We get the Riemann sum ! fack b˜x œ ! Š3ˆ1
2
6k n
27an "ba#n "b . 2n2
1‹ œ
18 n
n
!1
kœ"
72 n2
!k
kœ"
81 n3
n
! k2 œ
kœ"
18 n
†n
3 n,
3k ‰2 n 72 n2
†
1 n
nan "ba#n "b '
œ 56 . Thus,
1 3
67. Consider the partition P that subdivides the interval Ò1, 2Ó into n subintervals of width ˜x œ
†
2 a 1 b n
. . ., 1 n † 2ˆ1 n an " b #
277
œ 3 n
3k ‰ n
81 n3
†
3 n
and let ck be
œ 2× and 1‹ †
3 n
nan "ba#n "b '
As n Ä _ and mPm Ä ! this expression has value 18 36 27 œ 9. Thus,
'21 a3x# 2x 1bdx œ 9.
68. Consider the partition P that subdivides the interval Ò1, 1Ó into n subintervals of width ˜x œ the right endpoint of each subinterval. So the partition is P œ Ö1, 1 ck œ 1 k † œ
2 n
œ 1
2 n
n
! Š1
kœ"
6k n
œ 2n † n
12 n#
†
œ 2 6 †
" "n "
12k2 n2
n an " b #
4†
'1 x3 dx œ 0.
2k n .
24 n$
n
2 n,
1 2 †
n
We get the Riemann sum ! fack b˜x œ ! c3k ˆ 2n ‰ œ kœ"
8k3 n3 ‹
†
# $n n"# "
n
n
kœ"
kœ"
œ n2 Œ! 1 n6 ! k
nan "ba#n "b '
4†
" 2n n12 1
16 n4
†
12 n2
nn Š a # "b ‹
n
! k2
kœ" 2
kœ" n 8 ! 3 k 3 n kœ"
œ 2 6 †
n" n
2 n
4†
# n,
1 a 1 b n
. . ., 1 n †
n
! ˆ1
kœ"
œ 2 n
2 n
and let ck be
œ 1× and
2k ‰3 n
an "ba#n "b n#
4†
an "b2 n2
. As n Ä _ and mPm Ä ! this expression has value 2 6 8 4 œ 0.
1
Thus,
ba n and let ck be the # a b a b n a b a b endpoint of each subinterval. So the partition is P œ Öa, a b n a , a n , . . . , a n œ b× and n n n 3 ck œ a kabn ab . We get the Riemann sum ! fack b˜x œ ! c3k ˆ b n a ‰ œ b n a ! Ša kabn ab ‹ kœ" kœ" kœ" n n n 2 3 2 n 3 n 2 2 3 2 œ b n a ! Ša3 3a kanb ab 3ak anb2 ab k abn3 ab ‹ œ b n a Œ ! a3 3a abn ab ! k 3aabn2 ab ! k2 ab n3ab ! k3 kœ" kœ" kœ" kœ" kœ" 2 # $ 4 2 3a b a n n " 3a b a n n " # n " b a n n " a b a b a b a ba b a b a b œ b n a † na3 † # n$ † n4 † Š # ‹ n# '
69. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ
3a2 ab ab# 2
a ab a b $ 2
4 2 an "ba#n "b ab 4 ab † an n2"b n# 2 1 # $ 2 # $n n"# " " ab ab4 " n n2 œ ab aba3 3a ab2 ab † " n aab 2 ab † † . As n " 4 1 b # 4 2 4 4 4 ab aba3 3a ab2 ab aab ab$ ab 4 ab œ b4 a4 . Thus, x3 dx œ b4 a
œ ab aba3
†
n" n
right
†
'
Ä _ and mPm Ä ! this expression has value
a4 4. 10 n
70. Consider the partition P that subdivides the interval Ò0, 1Ó into n subintervals of width ˜x œ right endpoint of each subinterval. So the partition is P œ Ö0, 0 n
n
kœ"
kœ"
We get the Riemann sum ! fack b˜x œ ! a3ck c3k bˆ 1n ‰ œ
1 n
n
1 n,
02†
! Š3 †
kœ"
k n
1 n,
. . ., 0 n †
1 n n
œ
1 n
and let ck be the
œ 1× and ck œ 0 k †
3 ˆ kn ‰ ‹ œ 1n Œ 3n ! k kœ"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 n3
n
! k3
kœ"
1 n
œ kn .
278
Chapter 5 Integration œ
3 n2
†
n an " b #
has value
3 2
1 4
1 n4
2
† Š nan # "b ‹ œ
œ 54 . Thus,
3 2
†
n" n
1 4
†
a n "b 2 n2
œ
3 2
" "n "
†
" 2n n12 1
†
1 4
. As n Ä _ and mPm Ä ! this expression
'01 a3x x3 bdx œ 54 .
71. To find where x x# 0, let x x# œ 0 Ê x(1 x) œ 0 Ê x œ 0 or x œ 1. If 0 x 1, then 0 x x# Ê a œ 0 and b œ 1 maximize the integral. 72. To find where x% 2x# Ÿ 0, let x% 2x# œ 0 Ê x# ax# 2b œ 0 Ê x œ 0 or x œ „ È2. By the sign graph, 0 0 0 , we can see that x% 2x# Ÿ 0 on ’È2ß È2“ Ê a œ È2 and b œ È2 ! È# È # minimize the integral. 73. f(x) œ
" 1 x#
is decreasing on [0ß 1] Ê maximum value of f occurs at 0 Ê max f œ f(0) œ 1; minimum value of f occurs
at 1 Ê min f œ f(1) œ
" 1 1#
œ
" #
. Therefore, (1 0) min f Ÿ " #
That is, an upper bound œ 1 and a lower bound œ 74. See Exercise 73 above. On [0ß 0.5], max f œ (0.5 0) min f Ÿ min f œ Then
" 4
" 1 1#
2 5
'0
0.5
" 1 0#
Ÿ
'00.5 1 " x
#
dx
'0.5" 1 " x
#
dx Ÿ
75. 1 Ÿ sin ax# b Ÿ 1 for all x Ê (1 0)(1) Ÿ
" #
œ 1, min f œ Ÿ
2 5
'0
'0.5" 1 " x
#
dx Ÿ (1 0) max f Ê
" #
'0" 1 " x
Ÿ
#
dx Ÿ 1.
.
f(x) dx Ÿ (0.5 0) max f Ê
œ 0.5. Therefore (1 0.5) min f Ÿ
'0" 1 " x
2 5
Ê
0.5
#
" 1 (0.5)#
" 1 x#
œ 0.8. Therefore
dx Ÿ
" #
. On [0.5ß 1], max f œ
dx Ÿ (1 0.5) max f Ê
13 20
Ÿ
'0" 1 " x
#
dx Ÿ
9 10
" 4
Ÿ
" 1 (0.5)#
'0.5" 1 1 x
#
dx Ÿ
œ 0.8 and 2 5
.
.
'0" sin ax# b dx Ÿ (1 0)(1) or '0"sin x# dx Ÿ 1
Ê
'0"sin x# dx cannot
equal 2. 76. f(x) œ Èx 8 is increasing on [!ß "] Ê max f œ f(1) œ È1 8 œ 3 and min f œ f(0) œ È0 8 œ 2È2 . Therefore, (1 0) min f Ÿ
'0" Èx 8 dx Ÿ (1 0) max f
Ê 2È 2 Ÿ
'0" Èx 8 dx Ÿ 3.
77. If f(x) 0 on [aß b], then min f 0 and max f 0 on [aß b]. Now, (b a) min f Ÿ Then b a Ê b a 0 Ê (b a) min f 0 Ê
'ab f(x) dx 0.
78. If f(x) Ÿ 0 on [aß b], then min f Ÿ 0 and max f Ÿ 0. Now, (b a) min f Ÿ b a Ê b a 0 Ê (b a) max f Ÿ 0 Ê
'a f(x) dx Ÿ 0.
'ab f(x) dx Ÿ (b a) max f.
'ab f(x) dx Ÿ (b a) max f.
'0" (sin x x) dx Ÿ 0 (see Exercise 78) Ê '0" sin x dx '0" x dx Ÿ 0 '0" sin x dx Ÿ Š 1# 0# ‹ Ê '0" sin x dx Ÿ "# . Thus an upper bound is "# .
79. sin x Ÿ x for x 0 Ê sin x x Ÿ 0 for x 0 Ê Ê
'0" sin x dx Ÿ '0" x dx
Ê
Then
b
#
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral 80. sec x 1
x# #
on ˆ 1# ß 1# ‰ Ê sec x Š1
since [0ß 1] is contained in ˆ 1# ß 1# ‰ Ê Ê
x# #‹
0 on ˆ 1# ß 1# ‰ Ê
279
'0" ’sec x Š1 x# ‹“ dx 0 (see Exercise 77) #
'0" sec x dx '0" Š1 x# ‹ dx 0 Ê '0" sec x dx '0" Š1 x# ‹ #
#
'0" sec x dx '0" 1 dx "# '0" x# dx Ê '0" sec x dx (1 0) "# Š 13 ‹ Ê '0" sec x dx 76 . Thus a lower bound $
is 76 .
'ab f(x) dx is a constant K. b b b Ê ' av(f) dx œ (b a)K œ (b a) † b " a ' f(x) dx œ ' f(x) dx. a a a
81. Yes, for the following reasons: av(f) œ
" ba
'ab av(f) dx œ 'ab K dx
Thus
œ K(b a)
82. All three rules hold. The reasons: On any interval [aß b] on which f and g are integrable, we have: (a) av(f g) œ
" ba
'ab [f(x) g(x)] dx œ b " a ”'ab f(x) dx 'ab g(x) dx• œ b " a 'ab f(x) dx b " a 'ab g(x) dx
œ av(f) av(g) (b) av(kf) œ (c) av(f) œ
" ba
" ba
'ab kf(x) dx œ b " a ”k 'ab f(x) dx• œ k ” b " a 'ab f(x) dx• œ k av(f)
'ab f(x) dx Ÿ b " a 'ab g(x) dx since f(x) Ÿ g(x) on [aß b], and b " a 'ab g(x) dx œ av(g).
Therefore, av(f) Ÿ av(g). 83. (a) U œ max" ?x max# ?x á maxn ?x where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is increasing on [aß b]; L œ min" ?x min# ?x á minn ?x where min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore U L œ (max" min" ) ?x (max# min# ) ?x á (maxn minn ) ?x œ (f(x" ) f(x! )) ?x (f(x# ) f(x" ))?x á (f(xn ) f(xnc1 )) ?x œ (f(xn ) f(x! )) ?x œ (f(b) f(a)) ?x. (b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is increasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn œ (f(x" ) f(x! )) ?x" (f(x# ) f(x" ))?x# á (f(xn ) f(xnc1 )) ?xn Ÿ (f(x" ) f(x! )) ?xmax (f(x# ) f(x" )) ?xmax á (f(xn ) f(xnc1 )) ?xmax . Then U L Ÿ (f(xn ) f(x! )) ?xmax œ (f(b) f(a)) ?xmax œ kf(b) f(a)k ?xmax since f(b) f(a). Thus lim (U L) œ lim (f(b) f(a)) ?xmax œ 0, since ?xmax œ lPl . lPl Ä 0
lPl Ä 0
84. (a) U œ max" ?x max# ?x á maxn ?x where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xnc" ) since f is decreasing on [aß b]; L œ min" ?x min# ?x á minn ?x where min" œ f(x" ), min# œ f(x# )ß á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore U L œ (max" min" ) ?x (max# min# ) ?x á (maxn minn ) ?x œ (f(x! ) f(x" )) ?x (f(x" ) f(x# ))?x á (f(xn " ) f(xn )) ?x œ (f(x! ) f(xn )) ?x œ (f(a) f(b)) ?x.
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280
Chapter 5 Integration
(b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xn " ) since f is decreasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where min" œ f(x" ), min# œ f(x# ), á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn œ (f(x! ) f(x" )) ?x" (f(x" ) f(x# ))?x# á (f(xn " ) f(xn )) ?xn Ÿ (f(x! ) f(xn )) ?xmax œ (f(a) f(b) ?xmax œ kf(b) f(a)k ?xmax since f(b) Ÿ f(a). Thus lim (U L) œ lim kf(b) f(a)k ?xmax œ 0, since ?xmax œ lPl . lPl Ä 0
lPl Ä 0
85. (a) Partition 0ß 1# ‘ into n subintervals, each of length ?x œ x# œ 2?x, á , xn œ n?x œ
1 #n
with points x! œ 0, x" œ ?x, Since sin x is increasing on 0ß 1# ‘ , the upper sum U is the sum of the areas
1 #.
of the circumscribed rectangles of areas f(x" ) ?x œ (sin ?x)?x, f(x# ) ?x œ (sin 2?x) ?x, á , f(xn ) ?x œ (sin n?x) ?x. Then U œ (sin ?x sin 2?x á sin n?x) ?x œ ” œ”
1 cos ˆˆn " ‰ 1 ‰ cos 4n 1 # 2n 1 • ˆ #n ‰ # sin 4n
1 cos ˆ 1 1 ‰‰ 1 ˆcos 4n # 4n 1 4n sin 4n
œ
1 cos ˆ 1 1 ‰ cos 4n # 4n sin 1
Š 14n ‹ 4n
'!
1Î#
(b) The area is
œ
cos ?#x cosˆ ˆn #" ‰ ?x‰ • ?x # sin ?#x
sin x dx œ n lim Ä_
1 cos ˆ 1 1 ‰ cos 4n # 4n sin 1
Š 14n ‹
œ
1 cos 1# 1
œ 1.
4n
n
86. (a) The area of the shaded region is !˜xi † mi which is equal to L. iœ" n
(b) The area of the shaded region is !˜xi † Mi which is equal to U. iœ"
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure and the first part of the figure. Thus this area is U L. n
n
iœ"
iœ"
87. By Exercise 86, U L œ !˜xi † Mi !˜xi † mi where Mi œ maxÖfaxb on the ith subinterval× and n
n
iœ"
iœ"
mi œ minÖfaxb on the ith subinterval×. Thus U L œ !aMi mi b˜xi !% † ˜xi provided ˜xi $ for each n
n
iœ"
iœ"
i œ "ß Þ Þ Þ , n. Since !% † ˜xi œ % !˜xi œ %ab ab the result, U L %ab ab follows.
88. The car drove the first 150 miles in 5 hours and the second 150 miles in 3 hours, which means it drove 300 miles in 8 hours, for an average of 300 8 mi/hr œ 37.5 mi/hr. In terms of average values of functions, the function whose average value we seek is 30, 0 Ÿ t Ÿ 5 v(t) œ œ , and the average value is 50, 5 1 Ÿ 8 (30)(5) (50)(3) 8
œ 37.5.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral 89-94. Example CAS commands: Maple: with( plots ); with( Student[Calculus1] ); f := x -> 1-x; a := 0; b := 1; N :=[ 4, 10, 20, 50 ]; P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]: display( P, insequence=true ); 95-98. Example CAS commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi; plot( f(x), x=a..b, title="#95(a) (Section 5.3)" ); N := [ 100, 200, 1000 ]; # (b) for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c) Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); 89-98. Example CAS commands: Mathematica: (assigned function and values for a, b, and n may vary) Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a, b dx, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals dx, yvals}]; Plot[f, {x, a, b}, Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b a)/n; f = Sin[x]2 ;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
281
282
Chapter 5 Integration
xvals =Table[N[x], {x, a dx, b, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx,xvals, yvals}]; Plot[f, {x, a, b}, Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1,Length[yvals]}]//N Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a dx/2, b dx/2, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx/2, xvals dx/2, yvals}]; Plot[f, {x, a, b},Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N 5.4 THE FUNDAMENTAL THEOREM OF CALCULUS 1.
'c (2x 5) dx œ cx# 5xd#! œ a0# 5(0)b a(2)# 5(2)b œ 6
2.
'c ˆ5 x# ‰ dx œ ’5x x4 “ %
0
2
4
3.
#
$
3
'
2
0
xax 3b dx œ
'
2
0
5.
3
'c ax2 2x 3b dx œ ’ x3
3
1
'
4
0
Š3x
x$ 4‹
#
dx œ ’ 3x#
'c ax$ 2x 3b dx œ ’ x4 2
6.
%
2
7.
'
8.
' '
1Î3
9.
'
1
10.
'
31Î4
11.
'
1Î3
12.
1
% x% 16 “ !
1
3
œ Š a23b
œ
3 a2 b 2 2 ‹
133 4
3
Š a03b
3
1 #
# #
3 a0 b 2 2 ‹
œ 10 3
œ Š a13b a1b# 3a1b‹ Š (31) (1)# 3(1)‹ œ
œ Š 3(4) #
x# 3x“
2 3x2 2 “!
(3)# 4 ‹
4% 16 ‹
3
#
Š 3(0) #
(0)% 16 ‹
œ8 %
œ Š 24 2# 3(2)‹ Š (42) (2)# 3(2)‹ œ 12 %
"
$
ˆx# Èx‰ dx œ ’ x3 32 x$Î# “ œ ˆ "3 32 ‰ 0 œ 1
32
1
$#
x'Î& dx œ 5x"Î& ‘ " œ ˆ #5 ‰ (5) œ
0
1Î$
2 sec# x dx œ [2 tan x]!
5 #
œ ˆ2 tan ˆ 13 ‰‰ (2 tan 0) œ 2È3 0 œ 2È3
a1 cos xb dx œ [x sin x]1! œ a1 sin 1b a0 sin 0b œ 1
1Î4
0
x# 3x“
Š5(3)
!
0
0
4# 4‹
ax2 3xb dx œ ’ x3
1
4.
œ Š5(4)
$1Î%
csc ) cot ) d) œ [csc )]1Î% œ ˆcsc ˆ 341 ‰‰ ˆcsc ˆ 14 ‰‰ œ È2 ŠÈ2‹ œ 0 1Î$
4 sec u tan u du œ [4 sec u]!
œ 4 sec ˆ 13 ‰ 4 sec 0 œ 4(2) 4(1) œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
20 3
Section 5.4 The Fundamental Theorem of Calculus 13.
14.
'
0
" cos 2t # 1Î2
'
dt œ
0
ˆ"
1Î2 #
2t 'c ÎÎ " cos dt œ ' # 1 3
1Î3
1 3
1Î3 #
œ ˆ "# ˆ 13 ‰
'
1Î4
15.
'
1Î6
16.
0
0
" 4
" #
cos 2t‰ dt œ "# t
ˆ"
" #
'
1Î4
0
sin 2t‘ 1Î# œ ˆ "# (0)
cos 2t‰ dt œ "# t
sin 2 ˆ 13 ‰‰ ˆ #" ˆ 13 ‰
tan# x dx œ
!
" 4
" 4
1Î6
0
sin 2 ˆ 13 ‰‰ œ
1 6
" 4
sin 231
1 6
" 4
sin ˆ 321 ‰ œ
1 3
œ ˆtan ˆ 14 ‰ 14 ‰ atan a0b 0b œ 1
asec2 x 2sec x tan x tan2 xbdx œ
'
1Î6
0
'
1Î8
1 Î8
sin 2x dx œ ’ "# cos 2x“
!
0
œ ˆ "# cos 2ˆ 18 ‰‰ ˆ "# cos 2a0b‰ œ
a2sec2 x 2sec x tan x 1bdx 1 6
2
2 È 2 4
1
#
1 3
1 3
œ Š4 tan ˆ 14 ‰
1 ˆ 14 ‰ ‹
Š4 tan ˆ 13 ‰
1 ˆ 13 ‰ ‹
œ (4(1) 4) Š4 ŠÈ3‹ 3‹ œ 4È3 3
19.
'"" (r 1)# dr œ '"" ar# 2r 1b dr œ ’ r3 r# r“ " œ Š (31)
20.
'È (t 1) at# 4b dt œ 'È at$ t# 4t 4b dt œ ’ t4 t3 2t# 4t“ÈÈ$
$
"
È3
È3
3
œ
%
ŠÈ3‹ 4
$
ŠÈ3‹
2 ŠÈ3‹ 4È3
3
'È" Š u#
22.
'cc y y 2y dy œ 'cc ay2 2y2 b dy œ ’ y3
(
2
3
'
È2
1
$
(1)# (1)‹ Š 13 1# 1‹ œ 38
$
%
#
21.
1
$
3
%
23.
È3 4
'cÎ Î% ˆ4 sec# t t1 ‰ dt œ ' Î Î% a4 sec# t 1t# b dt œ 4 tan t 1t ‘ 11Î%Î$ 1
18.
sin 2 ˆ 1# ‰‰ œ 14
1 4
œ [2 tan x 2sec x x]1!Î6 œ ˆ2 tanˆ 16 ‰ 2secˆ 16 ‰ ˆ 16 ‰‰ a2 tan 0 2sec 0 0b œ 2È3 17.
" 4
sin 2t‘ 1Î$
1 Î4
'
sin 2(0)‰ ˆ "# ˆ 1# ‰
1Î$
" 4
asec# x 1bdx œ [tan x x]!
asec x tan xb2 dx œ
" 4
" u& ‹
du œ
'È" Š u#
(
2
)
u u& ‹ du œ ’ 16
1
5
3
3
3
s# È s s#
ds œ
'
1
È2
ŠÈ3‹ 4
$
ŠÈ3‹ 3
$ #
2 ŠÈ3‹ 4 ŠÈ3‹ œ 10È3 )
" " 4u% “È#
)
1 œ Š 16
c1
" 4(1)% ‹
ŠÈ2‹ 16
"
œ Š ac31b ac21b ‹ Š ac33b ac23b ‹ œ c3 3
2y1 “
ˆ1 s$Î# ‰ ds œ ’s
È# 2 “ Ès "
œ È 2
3
2 É È2
Š1
2 È1 ‹
%
4 ŠÈ2‹
œ4 3
22 3
œ È2 2$Î% 1
4 œ È2 È 81
24.
'
8
ˆx1Î3 1‰ˆ2 x2Î3 ‰ x1Î3 1
dx œ
'
8
2x1Î3 x 2 x2Î3 x1Î3 1
dx œ
'
1
8
ˆ2 x2Î3 2x1Î3 x1Î3 ‰ dx œ
2x 35 x5Î3 3x2Î3 34 x4Î3 ‘3 œ Š2a8b 35 a8b5Î3 3a8b2Î3 34 a8b4Î3 ‹ Š2a1b 35 a1b5Î3 3a1b2Î3 34 a1b4Î3 ‹ 1 œ 137 20 25.
'
1
sin 2x 1Î2 2 sin x
dx œ
'
1
2 sin x cos x 1Î2 2 sin x
dx œ
'
1
1Î2
1
cos x dx œ ’sin x“
1Î2
œ asin a1bb ˆsin ˆ 12 ‰‰ œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
283
284 26.
Chapter 5 Integration
'
1Î3
0
'
œ
'
acos x sec xb2 dx œ 1Î3
0
ˆ 12 cos 2x
5 2
1Î3
0
'
acos2 x 2 sec2 xbdx œ
1Î3
0
ˆ cos 2x2 1 2 sec2 x‰dx
1Î3
sec2 x‰dx œ ’ 14 sin 2x 52 x tan x“
!
œ ˆ 14 sin 2ˆ 13 ‰ 52 ˆ 13 ‰ tanˆ 13 ‰‰ ˆ 14 sin 2a0b 52 a0b tana0b‰ œ
'c% kxk dx œ ' !% kxk dx '! 4
27.
28.
'
1
" ! #
acos x kcos xk b dx œ 1 #
œ sin
29. (a)
!
d dx
30. (a)
'
(b)
d dx
31. (a)
'
(b)
d dt
32. (a)
'
'
1Î#
!
" # (cos
' !% x dx '!
x cos x) dx
4
#
x dx œ ’ x# “
'
1
" 1Î# #
!
#
%
%
#
’ x# “ œ Š 0# !
(cos x cos x) dx œ
'
1Î#
!
(4)# # ‹
Èx
cos t dt œ [sin t]! œ sin Èx sin 0 œ sin Èx Ê
Èx
'
Œ
!
sin x
d dx
Œ
'
Èx
!
cos t dt œ
d ˆÈ ‰‰ cos t dt œ ˆcos Èx‰ ˆ dx x œ ˆcos Èx ‰ ˆ "# x"Î# ‰ œ
3t# dt œ ct$ d "
sin x
1
'
Πt%
sin x
1
Œ'
t%
!
tan )
!
d d)
œ sin$ x 1 Ê
d dx
Œ
'
sin x
1
3t# dt œ
d dx
d dx
ˆsin Èx‰ œ cos Èx ˆ "# x"Î# ‰
cos Èx 2È x
asin$ x 1b œ 3 sin# x cos x
'
t%
!
t%
u"Î# du œ 23 u$Î# ‘ ! œ
2 3
at% b
$Î#
0œ
2 ' 3 t
Ê
d dt
Œ'
Œ
'
t%
!
Èu du œ
ˆ 23 t' ‰ œ 4t&
d dt
Èu du œ Èt% ˆ dtd at% b‰ œ t# a4t$ b œ 4t&
) sec# y dy œ [tan y]tan œ tan (tan )) 0 œ tan (tan )) Ê !
'
Œ
!
tan )
d d)
tan )
!
sec# y dy œ
d d)
(tan (tan )))
sec# y dy œ asec# (tan ))b ˆ dd) (tan ))‰ œ asec# (tan ))b sec# )
33. y œ
'
35. y œ
'È sin t# dt œ '
x
!
È1 t# dt Ê
'
x2
2
sin t3 dt Ê
'
2
x
34. y œ
sin t# dt Ê
œx†
dy dx
'
x2
dy dx
œ
d dx Œ
2
'
1
x
" t
dt Ê
dy dx
œ
" x
,x0
# d ˆÈ ‰‰ œ Šsin ˆÈx‰ ‹ ˆ dx x œ (sin x) ˆ "# x"Î# ‰ œ 2sinÈxx
sin t3 dt 1 †
'
x2
2
sin t3 dt œ x † sin ax# b
3 d # dx ax b
'
2
x2
sin t3 dt
sin t3 dt x
#
#
dy dx
œ È1 x#
x2
'c t t 4 dt ' 1
Èx
!
x
œ 2x# sin x6 37. y œ
dy dx
!
36. y œ x
œ 16
1Î#
œ asec# (tan ))b sec# ) (b)
0# #‹
cos x dx œ [sin x]!
d 3t# dt œ a3 sin# xb ˆ dx (sin x)‰ œ 3 sin# x cos x
Èu du œ
!
#
Š 4#
cos Èx 2È x
œ (b)
kxk dx œ
9È 3 8
sin 0 œ 1
Èx
'
4
51 6
3
t# t# 4
dt Ê
x# x# 4
x# x# 4
œ0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.4 The Fundamental Theorem of Calculus
'
38. y œ Œ
x
0
39. y œ
'
40. y œ
'
sin x
dt È1 t#
!
tan x
dt 1 t#
!
, kxk Ê
'
3
10
at3 "b dt Ê
dy dx
1 #
dy dx
Ê
œ 3Œ
dy dx
0
x
" È1 sin# x
œ
'
10
d at3 "b dt dx Œ
d ˆ dx (sin x)‰ œ
0
x
10
10
'
at3 "b dt œ 3ax3 "b Œ
" Ècos# x
(cos x) œ
cos x kcos xk
œ
cos x cos x
" d ‰ ˆ dx œ ˆ 1 tan (tan x)‰ œ ˆ sec"# x ‰ asec# xb œ 1 #x
'$# ax# 2xbdx '#! ax# 2xbdx '!# ax# 2xbdx $
œ ’ x3 x# “
# $
$
’ x3 x# “
!
$
#
$
’ x3 x# “
#
!
$
œ ŠŠ (32) (2)# ‹ Š (33) (3)# ‹‹ $
ŠŠ 03 0# ‹ Š (32) (2)# ‹‹ $
$
$
ŠŠ 23 2# ‹ Š 03 0# ‹‹ œ
28 3
42. 3x# 3 œ 0 Ê x# œ 1 Ê x œ „ 1; because of symmetry about the y-axis, Area œ 2 Œ
'!" a3x# 3bdx '"# a3x# 3bdx
"
#
2 Š cx$ 3xd ! cx$ 3xd " ‹ œ 2 c aa1$ 3(1)b a0$ 3(0)bb aa2$ 3(2)b a1$ 3(1)bd œ 2(6) œ 12
43. x$ 3x# 2x œ 0 Ê x ax# 3x 2b œ 0 Ê x(x 2)(x 1) œ 0 Ê x œ 0, 1, or 2; Area œ
'!" ax$ 3x# 2xbdx '"# ax$ 3x# 2xbdx "
%
%
œ ’ x4 x$ x# “ ’ x4 x$ x# “ !
œ
% Š 14
$
#
1 1 ‹ %
% Š 04
$
# "
#
0 0 ‹ %
’Š 24 2$ 2# ‹ Š 14 1$ 1# ‹“ œ
10
at3 "b dt
œ 1 since kxk
41. x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2; Area œ
0
x
" #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 #
285
286
Chapter 5 Integration
44. x"Î$ x œ 0 Ê x"Î$ ˆ1 x#Î$ ‰ œ 0 Ê x"Î$ œ 0 or 1 x#Î$ œ 0 Ê x œ 0 or 1 œ x#Î$ Ê x œ 0 or 1 œ x# Ê x œ 0 or „ 1; Area œ œ
'c! ˆx"Î$ x‰dx '!" ˆx"Î$ x‰dx '") ˆx"Î$ x‰dx "
%Î$
’ 34
x
! x# # “ "
œ ’Š 34 (0)%Î$
’ 34 x%Î$
0# #‹
" x# # “!
’ 43 x%Î$ (1)# # ‹“
Š 34 (1)%Î$
’Š 34 (1)%Î$
1# #‹
Š 34 (0)%Î$
0# # ‹“
’Š 34 (8)%Î$
8# #‹
Š 34 (1)%Î$
1# # ‹“
œ
" 4
" 4
ˆ2!
$ 4
#" ‰ œ
) x# # “"
83 4
45. The area of the rectangle bounded by the lines y œ 2, y œ 0, x œ 1, and x œ 0 is 21. The area under the curve y œ 1 cos x on [0ß 1] is
'!
1
(1 cos x) dx œ [x sin x]!1 œ (1 sin 1) (0 sin 0) œ 1. Therefore the area of
the shaded region is 21 1 œ 1. 46. The area of the rectangle bounded by the lines x œ 16 , x œ " #
51 6 ,
y œ sin
ˆ 561 16 ‰ œ 13 . The area under the curve y œ sin x on 16 ß 561 ‘ is
œ ˆcos
51 ‰ 6
È3 # ‹
ˆcos 16 ‰ œ Š
È3 #
'
1 6
œ
51Î6
1Î6
" #
œ sin
51 6
, and y œ 0 is &1Î'
sin x dx œ [cos x]1Î'
œ È3. Therefore the area of the shaded region is È3 13 .
47. On 14 ß 0‘ : The area of the rectangle bounded by the lines y œ È2, y œ 0, ) œ 0, and ) œ 14 is È2 ˆ 14 ‰ œ
1È2 4
. The area between the curve y œ sec ) tan ) and y œ 0 is
'c
!
1Î4
sec ) tan ) d) œ [sec )]!1Î%
œ (sec 0) ˆsec ˆ 14 ‰‰ œ È2 1. Therefore the area of the shaded region on 14 ß !‘ is
1È2 4
On 0ß 14 ‘ : The area of the rectangle bounded by ) œ 14 , ) œ 0, y œ È2, and y œ 0 is È2 ˆ 14 ‰ œ under the curve y œ sec ) tan ) is of the shaded region on !ß 14 ‘ is È
'
1Î4
!
1È2 4
1Î%
sec ) tan ) d) œ [sec )]!
œ sec
1 4
Š È 2 1‹ .
1È2 4
. The area
sec 0 œ È2 1. Therefore the area
ŠÈ2 1‹ . Thus, the area of the total shaded region is
È
1È2 #
Š 1 4 2 È2 1‹ Š 1 4 2 È2 1‹ œ
.
48. The area of the rectangle bounded by the lines y œ 2, y œ 0, t œ 14 , and t œ 1 is 2 ˆ1 ˆ 14 ‰‰ œ 2 area under the curve y œ sec# t on 14 ß !‘ is under the curve y œ 1 t# on [!ß "] is
'c
!
1Î4
"
$
œ
dt 3 œ 0 3 œ 3 Ê (d) is a solution to this problem.
dy dx
œ
50. y œ
'c sec t dt 4
Ê
dy dx
œ sec x and y(1) œ
51. y œ
'
sec t dt 4 Ê
dy dx
œ sec x and y(0) œ
x
1
!
x
and y(1) œ
'
" t
dt 3 Ê
" x
1
1
Thus, the total
. Therefore the area of the shaded region is ˆ2 1# ‰
'
" 1 t
$
5 3
49. y œ
x
$
!
2 3
'
!
. The
! ˆ 1‰ sec# t dt œ [tan t] 1Î% œ tan 0 tan 4 œ 1. The area
'! a1 t# b dt œ ’t t3 “ " œ Š1 13 ‹ Š0 03 ‹ œ 32 .
area under the curves on 14 ß "‘ is 1
1 #
'cc sec t dt 4 œ 0 4 œ 4 1
1
!
5 3
œ
" 3
1 #
.
Ê (c) is a solution to this problem.
sec t dt 4 œ 0 4 œ 4 Ê (b) is a solution to this problem.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.4 The Fundamental Theorem of Calculus 52. y œ
'
53. y œ
'
x
" " t x
#
55. Area œ
dt 3 Ê
dy dx
œ
" x
and y(1) œ
'
"
" t
"
dt 3 œ 0 3 œ 3 Ê (a) is a solution to this problem.
sec t dt 3
'c ÎÎ
b 2 b 2
54. y œ
$
4h ˆ b# ‰ 3b#
œ ˆ bh #
ˆ bh #
bh ‰ 6
Œh ˆ # ‰ b
bh ‰ 6
œ bh
x
"
È1 t# dt 2
bÎ2
4h ˆ #b ‰ 3b#
œ
bh 3
'
4hx$ 3b# “ bÎ2
ˆh ˆ 4h ‰ # ‰ dx œ ’hx b# x
œ Œhˆ #b ‰
$
2 3
bh
56. k 0 Ê one arch of y œ sin kx will occur over the interval 0ß 1k ‘ Ê the area œ œ "k cos ˆk ˆ 1k ‰‰ ˆ k" cos (0)‰ œ 57.
dc dx
œ
58. r œ
" #È x
'
$
!
œ
Š2
" #
x"Î# Ê c œ
'
dx œ 2
'
2 (x 1)# ‹
x
!
$
!
'
1Îk
!
sin kx dx œ
" k
œ t"Î# ‘ 0 œ Èx; c(100) c(1) œ È100 È1 œ $9.00 x
" (x 1)# ‹
$
dx œ 2 x ˆ x11 ‰‘ ! œ 2 ’Š3
" (3 1) ‹
Š0
" (0 1) ‹“
œ 2 3 "4 1‘ œ 2 ˆ2 4" ‰ œ 4.5 or $4500 59. (a) t œ 0 Ê T œ 85 3È25 0 œ 70‰ F; t œ 16 Ê T œ 85 3È25 16 œ 76‰ F; t œ 25 Ê T œ 85 3È25 25 œ 85‰ F (b) average temperatuve œ œ
1 25 Š85a25b
1 25 0
2a25 25b
' 25 Š85 3È25 t‹ dt œ 251 ’85t 2a25 tb3Î2 “ 25 !
0
3 Î2
‹
1 25 Š85a0b
2a25 0b
3Î2
‰
‹ œ 75 F
3 60. (a) t œ 0 Ê H œ È0 1 5a0b1Î3 œ 1 ft; t œ 4 Ê H œ È4 1 5a4b1Î3 œ È5 5È 4 ¸ 10.17 ft; 1 Î 3 t œ 8 Ê H œ È8 1 5a8b œ 13 ft
(b) average height œ œ
61.
'
62.
'
x
1
!
x
1 2 8 Š 3 a8
1b
3 Î2
1 80
' 8 ŠÈt 1 5 t1Î3 ‹ dt œ 18 ’ 23 at 1b3Î2 154 t4Î3 “ 8
15 4
!
0
a8b
4Î3
‹
f(t) dt œ x# 2x 1 Ê f(x) œ f(t) dt œ x cos 1x Ê f(x) œ
63. f(x) œ 2
'#
x
"
9 1t
d dx
1 2 8 Š 3 a0
'
d dx
'
!
1
x
x
1b
3Î2
f(t) dt œ
d dx
4Î3 15 ‹ 4 a0b
œ
29 3
¸ 9.67 ft
ax# 2x 1b œ 2x 2
f(t) dt œ cos 1x 1x sin 1x Ê f(4) œ cos 1(4) 1(4) sin 1(4) œ 1
dt Ê f w (x) œ 1 (x9 1) œ
9 x 2
L(x) œ 3(x 1) f(1) œ 3(x 1) 2 œ 3x 5
Ê f w (1) œ 3; f(1) œ 2
'#" " 1 9 t dt œ 2 0 œ 2;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 Îk
cos kx‘ !
2 k
" "Î# dt # t
Š1
287
288
Chapter 5 Integration
64. g(x) œ 3
'
1
x#
sec (t 1) dt Ê gw (x) œ asec ax# 1bb (2x) œ 2x sec ax# 1b Ê gw (1) œ 2(1) sec a(1)# 1b #
a"b " œ 2; g(1) œ 3 ' sec (t 1) dt œ 3 ' sec (t 1) dt œ 3 0 œ 3; L(x) œ 2(x (1)) g(1) 1
1
œ 2(x 1) 3 œ 2x 1 65. (a) (b) (c) (d) (e) (f) (g)
True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus. True: g is continuous because it is differentiable. True, since gw (1) œ f(1) œ 0. False, since gww (1) œ f w (1) 0. True, since gw (1) œ 0 and gww (1) œ f w (1) 0. False: gww (x) œ f w (x) 0, so gww never changes sign. True, since gw (1) œ f(1) œ 0 and gw (x) œ f(x) is an increasing function of x (because f w (x) 0).
66. Let a œ x0 x1 x2 â xn œ b be any partition of Òa, bÓ and let F be any antiderivative of f. n
(a) !Faxi b Faxi1 b‘ iœ1
œ Fax1 b Fax0 b‘ Fax2 b Fax1 b‘ Fax3 b Fax2 b‘ â Faxn1 b Faxn2 b‘ Faxn b Faxn1 b‘ œ Fax0 b Fax1 b Fax1 b Fax2 b Fax2 b â Faxn1 b Faxn1 b Faxn b œ Faxn b Fax0 b œ Fabb Faab (b) Since F is any antiderivative of f on Òa, bÓ Ê F is differentiable on Òa, bÓ Ê F is continuous on Òa, bÓ. Consider any subinterval Òxi1 , xi Ó in Òa, bÓ, then by the Mean Value Theorem there is at least one number ci in Ðxi1 , xi Ñ such that n
Faxi b Faxi1 b‘ œ Fw aci baxi xi1 b œ faci baxi xi1 b œ faci b?xi . Thus Fabb Faab œ !Faxi b Faxi1 b‘ iœ1
n
œ !faci b?xi . iœ1
n
(c) Taking the limit of Fabb Faab œ !faci b?xi we obtain Ê Fabb Faab œ 'a faxb dx
iœ1
lim aFabb Faabb œ
mPmÄ0
n
lim Œ!faci b?xi
mPmÄ0
iœ1
b
67-70. Example CAS commands: Maple: with( plots ); f := x -> x^3-4*x^2+3*x; a := 0; b := 4; F := unapply( int(f(t),t=a..x), x ); # (a) p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#67(a) (Section 5.4)" ): p1; dF := D(F); # (b) q1 := solve( dF(x)=0, x ); pts1 := [ seq( [x,f(x)], x=remove(has,evalf([q1]),I) ) ]; p2 := plot( pts1, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '(x)=0" ): display( [p1,p2], title="81(b) (Section 5.4)" ); incr := solve( dF(x)>0, x ); # (c) decr := solve( dF(x)<0, x ); df := D(f); # (d) p3 := plot( [df(x),F(x)], x=a..b, legend=["y = f '(x)","y = F(x)"], title="#67(d) (Section 5.4)" ): p3; q2 := solve( df(x)=0, x );
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.4 The Fundamental Theorem of Calculus pts2 := [ seq( [x,F(x)], x=remove(has,evalf([q2]),I) ) ]; p4 := plot( pts2, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where f '(x)=0" ): display( [p3,p4], title="81(d) (Section 5.4)" ); 71-74. Example CAS commands: Maple: a := 1; u := x -> x^2; f := x -> sqrt(1-x^2); F := unapply( int( f(t), t=a..u(x) ), x ); dF := D(F); # (b) cp := solve( dF(x)=0, x ); solve( dF(x)>0, x ); solve( dF(x)<0, x ); d2F := D(dF); # (c) solve( d2F(x)=0, x ); plot( F(x), x=-1..1, title="#71(d) (Section 5.4)" ); 75.
Example CAS commands: Maple: f := `f`; q1 := Diff( Int( f(t), t=a..u(x) ), x ); d1 := value( q1 );
76.
Example CAS commands: Maple: f := `f`; q2 := Diff( Int( f(t), t=a..u(x) ), x,x ); value( q2 );
67-76. Example CAS commands: Mathematica: (assigned function and values for a, and b may vary) For transcendental functions the FindRoot is needed instead of the Solve command. The Map command executes FindRoot over a set of initial guesses Initial guesses will vary as the functions vary. Clear[x, f, F] {a, b}= {0, 21}; f[x_] = Sin[2x] Cos[x/3] F[x_] = Integrate[f[t], {t, a, x}] Plot[{f[x], F[x]},{x, a, b}] x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}] x/.Map[FindRoot[f'[x]==0, {x, #}] &,{1, 2, 4, 5, 6}] Slightly alter above commands for 75 - 80. Clear[x, f, F, u] a=0; f[x_] = x2 2x 3 u[x_] = 1 x2 F[x_] = Integrate[f[t], {t, a, u(x)}] x/.Map[FindRoot[F'[x]==0,{x, #}] &,{1, 2, 3, 4}] x/.Map[FindRoot[F''[x]==0,{x,#}] &,{1, 2, 3, 4}]
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289
290
Chapter 5 Integration
After determining an appropriate value for b, the following can be entered b = 4; Plot[{F[x], {x, a, b}] 5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE " 2
1. Let u œ 2x 4 Ê du œ 2 dx Ê
' 2a 2x 4b dx œ ' 5
2u5
" 2
du œ ' u5 du œ " 7
2. Let u œ 7x 1 Ê du œ 7 dx Ê
' 7È7x 1 dx œ ' 7a7x 1b ' 2xa x2 5b 4x3 dx ax 4 1 b 2
u6 C œ
1 2È x
dx œ ' ˆ1 Èx‰
" 4
" 3
sin 3x dx œ '
" 3
x sin a2x b dx œ '
sec 2t tan 2t dt œ '
" 4
C
" 2
'
" 10
u4 du œ 1 Èx
" 2
1 x4 1
C
du œ a3x 2b dx
u5 C œ
" 10
a 3x2 4xb
5
C
dx
dx œ ' u1Î3 2 du œ 2' u1Î3 du œ 2 †
3 4
u4Î3 C œ
3 2
ˆ1 Èx‰4Î3 C
du œ dx
" 4
" #
du œ x dx
sin u du œ 4" cos u C œ 4" cos 2x# C " #
du œ dt
sec u tan u du œ
10. Let u œ 1 cos 2t Ê du œ ˆ1 cos
a7x 1b3Î2 C
sin u du œ 3" cos u C œ 3" cos 3x C
#
t ‰# #
2 3
du œ x3 dx
dx Ê 2 du œ
1 Î3 1 Èx
9. Let u œ 2t Ê du œ 2 dt Ê
'
u3Î2 C œ
3
u4 "2 du œ
8. Let u œ 2x# Ê du œ 4x dx Ê
'
2 3
2
7. Let u œ 3x Ê du œ 3 dx Ê
'
C
œ ' 4x3 ax4 1b dx œ ' 4 u2 4" du œ ' u2 du œ u1 C œ
6. Let u œ 1 Èx Ê du œ
'
6
dx œ ' 2 u4 2" du œ ' u4 du œ 3" u3 C œ 3" a x2 5b
' a3x 2ba3 x2 4xb4 dx œ ' ' ˆ1ÈÈxx‰
a 2x 4b
du œ x dx
5. Let u œ 3x2 4x Ê du œ a6x 4bdx œ 2a3x 2b dx Ê
1Î3
" 6
du œ dx
" 2
4. Let u œ x4 1 Ê du œ 4x3 dx Ê
'
" 6
dx œ ' 7u1Î2 "7 du œ ' u1Î2 du œ
1 Î2
3. Let u œ x2 5 Ê du œ 2x dx Ê 4
du œ dx
" #
sin
t #
" #
sec u C œ
dt Ê 2 du œ sin
ˆsin #t ‰ dt œ ' 2u# du œ
2 3
u$ C œ
t 2 2 3
" #
sec 2t C
dt
ˆ1 cos #t ‰$ C
11. Let u œ 1 r$ Ê du œ 3r# dr Ê 3 du œ 9r# dr
' È9r
# dr 1 r$
œ ' 3u"Î# du œ 3(2)u"Î# C œ 6 a1 r$ b
"Î#
C
12. Let u œ y% 4y# 1 Ê du œ a4y$ 8yb dy Ê 3 du œ 12 ay$ 2yb dy
'
12 ay% 4y# 1b ay$ 2yb dy œ ' 3u# du œ u$ C œ ay% 4y# 1b C #
$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.5 Indefinite Integrals and the Substitution Rule 13. Let u œ x$Î# 1 Ê du œ
'
x"Î# dx Ê
Èx sin# ˆx$Î# 1‰ dx œ '
14. Let u œ "x Ê du œ
'
3 #
" x#
2 3
2 3
du œ Èx dx
sin# u du œ
2 3
ˆ #u
" 4
sin 2u‰ C œ
" 3
ˆx$Î# 1‰
" 6
sin ˆ2x$Î# 2‰ C
dx
cos# ˆ x" ‰ dx œ ' cos# aub du œ " œ 2x 4" sin ˆ x2 ‰ C " x#
'
cos# aub du œ ˆ u#
" 4
" sin 2u‰ C œ 2x
" 4
sin ˆ x2 ‰ C
15. (a) Let u œ cot 2) Ê du œ 2 csc# 2) d) Ê "# du œ csc# 2) d)
'
csc# 2) cot 2) d) œ '
" #
#
#
u du œ "# Š u# ‹ C œ u4 C œ 4" cot# 2) C
(b) Let u œ csc 2) Ê du œ 2 csc 2) cot 2) d) Ê "# du œ csc 2) cot 2) d)
'
csc# 2) cot 2) d) œ ' "# u du œ "# Š u# ‹ C œ u4 C œ 4" csc# 2) C #
16. (a) Let u œ 5x 8 Ê du œ 5 dx Ê
'
dx È5x 8
œ'
" 5
Š È"u ‹ du œ " #
(b) Let u œ È5x 8 Ê du œ
'
dx È5x 8
œ'
2 5
du œ
2 5
" 5
'
" 5
#
du œ dx
u"Î# du œ
" 5
ˆ2u"Î# ‰ C œ
(5x 8)"Î# (5) dx Ê
uCœ
2 5
2 5
du œ
2 5
u"Î# C œ
2 5
È5x 8 C
dx È5x8
È5x 8 C
17. Let u œ 3 2s Ê du œ 2 ds Ê "# du œ ds
'
È3 2s ds œ ' Èu ˆ " du‰ œ " ' u"Î# du œ ˆ " ‰ ˆ 2 u$Î# ‰ C œ " (3 2s)$Î# C # # # 3 3
18. Let u œ 5s 4 Ê du œ 5 ds Ê
'
" È5s 4
ds œ '
" Èu
ˆ 5" du‰ œ
" 5
" 5
'
du œ ds u"Î# du œ ˆ 5" ‰ ˆ2u"Î# ‰ C œ
2 5
È5s 4 C
19. Let u œ 1 )# Ê du œ 2) d) Ê "# du œ ) d)
'
&Î% 4 4 )È 1 ) # d) œ ' È u ˆ "# du‰ œ "# ' u"Î% du œ ˆ "# ‰ ˆ 45 u&Î% ‰ C œ 25 a1 )# b C
20. Let u œ 7 3y# Ê du œ 6y dy Ê "# du œ 3y dy
'
$Î# 3yÈ7 3y# dy œ ' Èu ˆ "# du‰ œ "# ' u"Î# du œ ˆ "# ‰ ˆ 23 u$Î# ‰ C œ 3" a7 3y# b C
21. Let u œ 1 Èx Ê du œ
'
" È x ˆ" È x ‰ #
dx œ '
2 du u#
" 2È x
dx Ê 2 du œ
œ 2u C œ
22. Let u œ 3z 4 Ê du œ 3 dz Ê
'
" 3
2 1 È x
dx
C
du œ dz
cos (3z 4) dz œ ' (cos u) ˆ "3 du‰ œ
23. Let u œ 3x 2 Ê du œ 3 dx Ê
'
" 3
" Èx
" 3
' cos u du œ 3" sin u C œ 3" sin (3z 4) C
du œ dx
sec# (3x 2) dx œ ' asec# ub ˆ "3 du‰ œ
" 3
'
sec# u du œ
" 3
tan u C œ
" 3
tan (3x 2) C
24. Let u œ tan x Ê du œ sec# x dx
'
tan# x sec# x dx œ ' u# du œ
" 3
u$ C œ
" 3
tan$ x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
291
292
Chapter 5 Integration
25. Let u œ sin ˆ x3 ‰ Ê du œ
'
r$ 18
1 Ê du œ
r# 6
$
sec# ˆ x# ‰ dx Ê 2 du œ sec# ˆ x# ‰ dx " 4
tan) ˆ x# ‰ C
dr Ê 6 du œ r# dr
r % Š7
r& 10 $
r& 10 ‹
'
'
$
Ê du œ "# r% dr Ê 2 du œ r% dr
dr œ ' u$ (2 du) œ 2 ' u$ du œ 2 Š u4 ‹ C œ "# Š7 %
29. Let u œ x$Î# 1 Ê du œ
'
sin' ˆ x3 ‰ C
r r r# Š 18 1‹ dr œ ' u& (6 du) œ 6 ' u& du œ 6 Š u6 ‹ C œ Š 18 1‹ C &
28. Let u œ 7
'
" #
" #
tan( ˆ x# ‰ sec# ˆ x# ‰ dx œ ' u( (2 du) œ 2 ˆ 8" u) ‰ C œ
27. Let u œ
'
cos ˆ x3 ‰ dx Ê 3 du œ cos ˆ x3 ‰ dx
sin& ˆ x3 ‰ cos ˆ x3 ‰ dx œ ' u& (3 du) œ 3 ˆ 6" u' ‰ C œ
26. Let u œ tan ˆ x# ‰ Ê du œ
'
" 3
3 #
x"Î# dx Ê
2 3
r& 10 ‹
%
C
du œ x"Î# dx
x"Î# sin ˆx$Î# 1‰ dx œ ' (sin u) ˆ 23 du‰ œ
2 3
'
sin u du œ
2 3
(cos u) C œ 23 cos ˆx$Î# 1‰ C
30. Let u œ csc ˆ v # 1 ‰ Ê du œ "# csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv Ê 2 du œ csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv
'
csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv œ ' 2 du œ 2u C œ 2 csc ˆ v # 1 ‰ C
31. Let u œ cos (2t 1) Ê du œ 2 sin (2t 1) dt Ê "# du œ sin (2t 1) dt
'
sin (2t 1) cos# (2t 1)
dt œ ' #"
du u#
œ
" #u
Cœ
" # cos (2t 1)
C
32. Let u œ sec z Ê du œ sec z tan z dz
'
sec z tan z Èsec z
33. Let u œ
'
" t#
" t
dz œ '
" Èu
du œ ' u"Î# du œ 2u"Î# C œ 2Èsec z C
1 œ t" 1 Ê du œ t# dt Ê du œ
" Èt
" )#
" "Î# # t
dt Ê 2 du œ
sin
" )
" )
cos
Ê du œ ˆcos ") ‰ ˆ )"# ‰ d) Ê du œ " )
cos È) È) sin# È)
d) œ '
" È)
dt
t$ a1 t% b dt œ ' u$ ˆ "4 du‰ œ $
cos
" ‹ #È )
" )
" )
d)
C
d) Ê 2 du œ
" È)
cot È) csc È) d)
cot È) csc È) d) œ ' 2 du œ 2u C œ 2 csc È) C œ
37. Let u œ 1 t% Ê du œ 4t$ dt Ê
'
" )#
d) œ ' u du œ #" u# C œ #" sin#
36. Let u œ csc È) Ê du œ Šcsc È) cot È)‹ Š
'
" Èt
cos ˆÈt 3‰ dt œ ' (cos u)(2 du) œ 2 ' cos u du œ 2 sin u C œ 2 sin ˆÈt 3‰ C
35. Let u œ sin
'
dt
cos ˆ "t 1‰ dt œ ' (cos u)(du) œ ' cos u du œ sin u C œ sin ˆ "t 1‰ C
34. Let u œ Èt 3 œ t"Î# 3 Ê du œ
'
" t#
" 4 " 4
2 sin È)
du œ t$ dt
ˆ 4" u% ‰ C œ
" 16
%
a 1 t% b C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
C
Section 5.5 Indefinite Integrals and the Substitution Rule 38. Let u œ 1
'
" É2 x#
" x
" x
40. Let u œ 1
" x#
Ê du œ
É x x& 1 dx œ '
39. Let u œ 2
'
" x
" x#
dx
É x x 1 dx œ ' " x#
Ê du œ
Ê du œ
2 x3
#
'É
3 x11
x3
#
3
3 x3
Ê du œ
dx œ '
1É x4
9 x4
3 x3
x3
" #
dx Ê
#
41. Let u œ 1
É1
4
3
x2 È x3 1
du œ
" 9
dx Ê
du œ
dx œ '
dx œ '
1 É1 x4
" 3
u$Î# C œ
2 3
1 x3
2 3
u$Î# C œ
2 3
ˆ1 "x ‰$Î# C
2 3
ˆ2 "x ‰$Î# C
dx
dx œ ' Èu
42. Let u œ x3 1 Ê du œ 3x2 dx Ê
' É x x 1 dx œ '
dx œ ' Èu du œ ' u"Î# du œ
" x
dx
' x1 É x x 1 dx œ ' x1 É1 x" 3
" x#
dx œ ' Èu du œ ' u"Î# du œ
" x#
293
" #
1 x4
du œ
" #
' u"Î# du œ 13 u$Î# C œ 13 ˆ1 x" ‰$Î# C #
dx
3 x3
dx œ ' Èu
" 9
du œ
" 9
' u"Î# du œ 272 u$Î# C œ 272 ˆ1 x3 ‰$Î# C 3
du œ x2 dx
1 " Èu 3
du œ
" 3
' u"Î# du œ 23 u1Î# C œ 23 ax3 1b$Î# C
43. Let u œ x ". Then du œ dx and x œ u ". Thus ' xax "b10 dx œ ' au "bu10 du œ ' au11 u10 b du œ
1 12 12 u
1 11 11 u
Cœ
1 12 ax
"b12
1 11 ax
"b11 C
44. Let u œ 4 x. Then du œ 1 dx and a1b du œ dx and x œ 4 u. Thus ' x È4 xdx œ ' a4 ubÈu a1bdu œ ' a4 ubˆu1Î2 ‰du œ ' au3Î2 4u1Î2 b du œ 25 u5Î2 83 u3Î2 C œ 25 a4 xb5Î2 83 a4 xb3Î2 C 45. Let u œ " x. Then du œ 1 dx and a1b du œ dx and x œ 1 u. Thus ' ax 1b2 a" xb5 dx œ ' a2 ub2 u5 a1b du œ ' au7 4u6 4u5 b du œ 18 u8 47 u7 23 u6 C œ 18 a" xb8 47 a" xb7 23 a" xb6 C 46. Let u œ x 5. Then du œ dx and x œ u 5. Thus ' ax 5bax 5b1Î3 dx œ ' au 10bu1Î3 du œ ' ˆu4Î3 10u1Î3 ‰ du œ 37 u7Î3
15 4Î3 2 u
C œ 37 ax 5b7Î3
15 2 ax
5b4Î3 C
47. Let u œ x# ". Then du œ #x dx and "# du œ x dx and x# œ u ". Thus ' x$ Èx# " dx œ ' au "b "# Èu du œ
" #
' au$Î# u"Î# bdu œ "# ’ #& u&Î# #$ u$Î# “ C œ "& u&Î# "$ u$Î# C œ "& ax# "b&Î# "$ ax# "b$Î# C
48. Let u œ x3 " Ê du œ 3x# dx and x3 œ u ". So ' 3B& Èx3 " dx œ ' au "bÈu du œ ' au$Î# u"Î# bdu œ #& u&Î# #$ u$Î# C œ #& ax$ "b 49. Let u œ x2 4 Ê du œ 2x dx and œ "4 u2 C œ 4" ax2 4b
2
&Î#
" #
#$ ax$ "b
$Î#
C
du œ x dx . Thus '
x ax 2 4 b 3
dx œ ' ax2 4b x dx œ ' u3 "# du œ 3
" #
' u3 du
C
50. Let u œ x 4 Ê du œ dx and x œ u 4 . Thus '
x ax 4 b 3
dx œ ' ax 4b3 x dx œ ' u3 au 4bdu œ ' au2 4u3 b du
œ u1 2u2 C œ ax 4b1 2ax 4b2 C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
294
Chapter 5 Integration
51. (a) Let u œ tan x Ê du œ sec# x dx; v œ u$ Ê dv œ 3u# du Ê 6 dv œ 18u# du; w œ 2 v Ê dw œ dv
'
18 tan# x sec# x dx œ a2 tan$ xb# œ 2 6 u$ C $
'
'
18u# du œ a2 u $ b# 6 2 tan $x C # #
œ
6 dv (2 v)#
œ'
6 dw w#
œ 6 ' w# dw œ 6w" C œ # 6 v C
(b) Let u œ tan x Ê du œ 3 tan x sec x dx Ê 6 du œ 18 tan# x sec# x dx; v œ 2 u Ê dv œ du
'
dx œ '
18 tan# x sec# x a2 tan$ xb#
œ'
6 du (2 u)#
6 dv v#
6 œ v6 C œ 2 6 u C œ # tan $x C
(c) Let u œ 2 tan$ x Ê du œ 3 tan# x sec# x dx Ê 6 du œ 18 tan# x sec# x dx
'
dx œ '
18 tan# x sec# x a2 tan$ xb#
6 du u#
6 œ u6 C œ 2 tan $x C
52. (a) Let u œ x 1 Ê du œ dx; v œ sin u Ê dv œ cos u du; w œ 1 v# Ê dw œ 2v dv Ê
'
" #
Èw dw œ
" 3
w$Î# C œ
" 3
a 1 v# b
$Î#
Cœ
" 3
a1 sin# ub
$Î#
Cœ " #
#
(b) Let u œ sin (x 1) Ê du œ cos (x 1) dx; v œ 1 u Ê dv œ 2u du Ê
'
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ ' u È1 u# du œ ' œ ˆ "# ˆ 23 ‰ v$Î# ‰ C œ
" 3
v$Î# C œ
" 3
a1 u # b
$Î#
Cœ
" 3
(c) Let u œ 1 sin (x 1) Ê du œ 2 sin (x 1) cos (x 1) dx Ê
'
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ '
œ
" 3
a1 sin# (x 1)b
$Î#
œ
" 6
dr œ ' Š
cos Èu ˆ" È u ‹ 1#
sin È) É) cos$ È)
d) œ
4 Écos È)
'
sin È) È) Écos$ È)
Èv dv œ ' $Î#
" #
" 1#
du œ sin (x 1) cos (x 1) dx
" #
u"Î# du œ
du‰ œ ' (cos v) ˆ 6" dv‰ œ
" ‹ #È )
d) œ '
d) Ê 2 du œ
2 du u$Î#
$
" #
ˆ 23 u$Î# ‰ C
" 6
sin v C œ
" #È u
" 6
du Ê
sin Èu C
sin È) È)
d)
œ 2 ' u$Î# du œ 2 ˆ2u"Î# ‰ C œ
4 Èu
C
" #
u% C œ
" #
%
a3t# 1b C;
(3 1)% C Ê 3 œ 8 C Ê C œ 5 Ê s œ
" #
%
a3t# 1b 5
56. Let u œ x# 8 Ê du œ 2x dx Ê 2 du œ 4x dx "Î$
dx œ ' u"Î$ (2 du) œ 2 ˆ 3# u#Î$ ‰ C œ 3u#Î$ C œ 3 ax# 8b
y œ 0 when x œ 0 Ê 0 œ 3(8) 57. Let u œ t
1 1#
#Î$
#
C Ê C œ 12 Ê y œ 3 ax 8b
#Î$
#Î$
C;
12
Ê du œ dt
s œ ' 8 sin# ˆt
" 1 #È u
C
" #
du œ (2r 1) dr; v œ Èu Ê dv œ
s œ ' 12t a3t# 1b dt œ ' u$ (2 du) œ 2 ˆ "4 u% ‰ C œ
y œ ' 4x ax# 8b
dv œ
v"Î# dv
55. Let u œ 3t# 1 Ê du œ 6t dt Ê 2 du œ 12t dt " #
C
dv œ u du
C
s œ 3 when t œ 1 Ê 3 œ
$Î#
a1 sin# (x 1)b
sin È3(2r 1)# 6 C
54. Let u œ cos È) Ê du œ Šsin È)‹ Š
œ
Èu du œ '
" 3
C
53. Let u œ 3(2r 1)# 6 Ê du œ 6(2r 1)(2) dr Ê (2r 1) cos È3(2r 1)# 6 È3(2r 1)# 6
" #
" #
a1 sin# (x 1)b
#
'
dw œ v dv
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ ' È1 sin# u sin u cos u du œ ' vÈ1 v# dv œ'
'
" #
dt œ ' 8 sin# u du œ 8 ˆ u# "4 sin 2u‰ C œ 4 ˆt 11# ‰ 2 sin ˆ2t 16 ‰ C; s œ 8 when t œ 0 Ê 8 œ 4 ˆ 11# ‰ 2 sin ˆ 16 ‰ C Ê C œ 8 13 1 œ 9 13 Ê s œ 4ˆt 11# ‰ 2 sin ˆ2t 16 ‰ 9 13 œ 4t 2 sin ˆ2t 16 ‰ 9 1‰ 1#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" 6
du
Section 5.5 Indefinite Integrals and the Substitution Rule 58. Let u œ
1 4
) Ê du œ d)
r œ ' 3 cos# ˆ 14 )‰ d) œ ' 3 cos# u du œ 3 ˆ u#
sin 2u‰ C œ 3# ˆ 14 )‰ 34 sin ˆ 1# 2)‰ C; C Ê C œ 1# 43 Ê r œ 3# ˆ 14 )‰ 43 sin ˆ 1# 2)‰ 1#
when ) œ 0 Ê 18 œ 381 43 sin 1# Ê r œ 3# ) 34 sin ˆ 1# 2)‰ 18 43 Ê r œ
rœ
1 8
1 #
59. Let u œ 2t ds dt
295
3 2
)
3 4
" 4
cos 2)
1 8
3 4
3 4
Ê du œ 2 dt Ê 2 du œ 4 dt
œ ' 4 sin ˆ2t 1# ‰ dt œ ' (sin u)(2 du) œ 2 cos u C" œ 2 cos ˆ2t 1# ‰ C" ;
at t œ 0 and
œ 100 we have 100 œ 2 cos ˆ 1# ‰ C" Ê C" œ 100 Ê
ds dt
œ 2 cos ˆ2t 1# ‰ 100
ds dt
Ê s œ ' ˆ2 cos ˆ2t 1# ‰ 100‰ dt œ ' (cos u 50) du œ sin u 50u C# œ sin ˆ2t 1# ‰ 50 ˆ2t 1# ‰ C# ; at t œ 0 and s œ 0 we have 0 œ sin ˆ 1# ‰ 50 ˆ 1# ‰ C# Ê C# œ 1 251 Ê s œ sin ˆ2t 1# ‰ 100t 251 (1 251) Ê s œ sin ˆ2t 1# ‰ 100t 1
60. Let u œ tan 2x Ê du œ 2 sec# 2x dx Ê 2 du œ 4 sec# 2x dx; v œ 2x Ê dv œ 2 dx Ê dy dx
œ ' 4 sec# 2x tan 2x dx œ ' u(2 du) œ u# C" œ tan# 2x C" ;
at x œ 0 and
dy dx
œ 4 we have 4 œ 0 C" Ê C" œ 4 Ê
Ê y œ ' asec# 2x 3b dx œ ' asec# v 3b ˆ "# dv‰ œ
at x œ 0 and y œ 1 we have 1 œ
" #
dy dx " #
" #
dv œ dx
œ tan# 2x 4 œ asec# 2x 1b 4 œ sec# 2x 3
tan v 3# v C# œ " #
(0) 0 C# Ê C# œ 1 Ê y œ
" #
tan 2x 3x C# ;
tan 2x 3x 1
61. Let u œ 2t Ê du œ 2 dt Ê 3 du œ 6 dt
s œ ' 6 sin 2t dt œ ' (sin u)(3 du) œ 3 cos u C œ 3 cos 2t C; at t œ 0 and s œ 0 we have 0 œ 3 cos 0 C Ê C œ 3 Ê s œ 3 3 cos 2t Ê s ˆ 1# ‰ œ 3 3 cos (1) œ 6 m
62. Let u œ 1t Ê du œ 1 dt Ê 1 du œ 1# dt
v œ ' 1# cos 1t dt œ ' (cos u)(1 du) œ 1 sin u C" œ 1 sin (1t) C" ; at t œ 0 and v œ 8 we have 8 œ 1(0) C" Ê C" œ 8 Ê v œ
ds dt
œ 1 sin (1t) 8 Ê s œ ' (1 sin (1t) 8) dt
œ ' sin u du 8t C# œ cos (1t) 8t C# ; at t œ 0 and s œ 0 we have 0 œ 1 C# Ê C# œ 1 Ê s œ 8t cos (1t) 1 Ê s(1) œ 8 cos 1 1 œ 10 m
63. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, sin# x C" œ 1 cos# x C" Ê C# œ 1 C" ; also cos# x C# œ cos#2x "# C# Ê C$ œ C# "# œ C" "# . 64. (a) Š
" 60
" ‹ 0
'
0
1Î60
"Î'!
Vmax sin 1201t dt œ 60 Vmax ˆ 120" 1 ‰ cos (1201t)‘ !
œ V#1max [1 1] œ 0 (b) Vmax œ È2 Vrms œ È2 (240) ¸ 339 volts (c)
'
0
œ
1Î60
aVmax b# sin# 1201t dt œ aVmax b#
aVmax b# #
t ˆ 240" 1 ‰ sin
"Î'! 2401t‘ !
'
œ
1Î60
0
ˆ 1 cos# 2401t ‰ dt œ
aVmax b# #
aVmax b# #
œ V#max 1 [cos 21 cos 0]
'
0
1Î60
(1 cos 2401t) dt
" ˆ 60 ˆ 240" 1 ‰ sin (41)‰ ˆ0 ˆ #40" 1 ‰ sin (0)‰‘ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
aVmax b# 1#0
296
Chapter 5 Integration
5.6 SUBSTITUTION AND AREA BETWEEN CURVES 1. (a) Let u œ y 1 Ê du œ dy; y œ 0 Ê u œ 1, y œ 3 Ê u œ 4
'
3
0
Èy 1 dy œ
'
4
1
%
u"Î# du œ 23 u$Î# ‘ " œ ˆ 23 ‰ (4)$Î# ˆ 23 ‰ (1)$Î# œ ˆ 23 ‰ (8) ˆ 23 ‰ (1) œ
14 3
(b) Use the same substitution for u as in part (a); y œ 1 Ê u œ 0, y œ 0 Ê u œ 1
'c Èy 1 dy œ ' 0
1
1
0
"
u"Î# du œ 23 u$Î# ‘ ! œ ˆ 23 ‰ (1)$Î# 0 œ
2 3
2. (a) Let u œ 1 r# Ê du œ 2r dr Ê "# du œ r dr; r œ 0 Ê u œ 1, r œ 1 Ê u œ 0
'
1
0
r È1 r# dr œ
'
0
!
"# Èu du œ 3" u$Î# ‘ " œ 0 ˆ 3" ‰ (1)$Î# œ
1
" 3
(b) Use the same substitution for u as in part (a); r œ 1 Ê u œ 0, r œ 1 Ê u œ 0
'c r È1 r# dr œ ' 1
1
0
0
"# Èu du œ 0
3. (a) Let u œ tan x Ê du œ sec# x dx; x œ 0 Ê u œ 0, x œ
'
1Î4
0
'
tan x sec# x dx œ
1
"
#
1# #
u du œ ’ u# “ œ !
0
0œ
1 4
Ê uœ1
" #
(b) Use the same substitution as in part (a); x œ 14 Ê u œ 1, x œ 0 Ê u œ 0
'c Î 0
1 4
'
tan x sec# x dx œ
0
#
1
u du œ ’ u# “
!
œ0
"
" #
œ "#
4. (a) Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 1
'
1
0
$ $ ' 3u# du œ cu$ d " " œ (1) a(1) b œ 2 1
3 cos# x sin x dx œ
1
(b) Use the same substitution as in part (a); x œ 21 Ê u œ 1, x œ 31 Ê u œ 1
'
31
3 cos# x sin x dx œ
21
'
1
1
3u# du œ 2
5. (a) u œ 1 t% Ê du œ 4t$ dt Ê
'
1
0
'
$
2
" 1 4
t$ a1 t% b dt œ
" 4
du œ t$ dt; t œ 0 Ê u œ 1, t œ 1 Ê u œ 2 %
#
u u$ du œ ’ 16 “ œ "
2% 16
1% 16
œ
15 16
(b) Use the same substitution as in part (a); t œ 1 Ê u œ 2, t œ 1 Ê u œ 2
'c
1
'
$
t$ a1 t% b dt œ
1
2
" 4
2
u$ du œ 0
6. (a) Let u œ t# 1 Ê du œ 2t dt Ê
'
È7
0
t at# 1b
"Î$
dt œ
'
8
1
" #
" #
du œ t dt; t œ 0 Ê u œ 1, t œ È7 Ê u œ 8 )
u"Î$ du œ ˆ "# ‰ ˆ 34 ‰ u%Î$ ‘ " œ ˆ 38 ‰ (8)%Î$ ˆ 38 ‰ (1)%Î$ œ
45 8
(b) Use the same substitution as in part (a); t œ È7 Ê u œ 8, t œ 0 Ê u œ 1
'cÈ 0
7
t at# 1b
"Î$
dt œ
'
1
" 8 #
u"Î$ du œ
7. (a) Let u œ 4 r# Ê du œ 2r dr Ê
' a4 5rr b 1
1
# #
dr œ 5
'
5
5
" #
" #
'
1
8
" #
u"Î$ du œ 45 8
du œ r dr; r œ 1 Ê u œ 5, r œ 1 Ê u œ 5
u# du œ 0
(b) Use the same substitution as in part (a); r œ 0 Ê u œ 4, r œ 1 Ê u œ 5
'
0
1
5r a4 r# b #
dr œ 5
'
4
5
" #
&
u# du œ 5 "# u" ‘ % œ 5 ˆ "# (5)" ‰ 5 ˆ "# (4)" ‰ œ
" 8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.6 Substitution and Area Between Curves 8. (a) Let u œ 1 v$Î# Ê du œ
'
1
10Èv a1 v$Î# b
0
#
dv œ
'
2
" u#
1
3 #
v"Î# dv Ê
ˆ 20 ‰ 3 du œ
'
20 3
du œ 10Èv dv; v œ 0 Ê u œ 1, v œ 1 Ê u œ 2
20 3 2
20 " 1‘ "‘# u# du œ 20 3 u " œ 3 # 1 œ
1
297
10 3
(b) Use the same substitution as in part (a); v œ 1 Ê u œ 2, v œ 4 Ê u œ 1 4$Î# œ 9
'
4
10Èv
# 1 a1 v$Î# b
dv œ
'
9
20 " ‘ * 20 ˆ " 1‰ 20 ˆ 7 ‰ ˆ 20 ‰ 3 du œ 3 u # œ 3 9 2 œ 3 18 œ
" u#
2
70 #7
9. (a) Let u œ x# 1 Ê du œ 2x dx Ê 2 du œ 4x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4
'
È3
4x È x# 1
0
dx œ
'
4
'
du œ
2 1 Èu
4
1
%
2u"Î# du œ 4u"Î# ‘ " œ 4(4)"Î# 4(1)"Î# œ 4
(b) Use the same substitution as in part (a); x œ È3 Ê u œ 4, x œ È3 Ê u œ 4
È3
'cÈ
4x 3 È x# 1
'
dx œ
4
du œ 0
2 Èu
4
10. (a) Let u œ x% 9 Ê du œ 4x$ dx Ê
'
1
x$ 0 È x% 9
dx œ
'
10
9
" 4
du œ x$ dx; x œ 0 Ê u œ 9, x œ 1 Ê u œ 10 "!
" 4
u"Î# du œ 4" (2)u"Î# ‘ * œ
" #
(10)"Î# #" (9)"Î# œ
È10 3 #
(b) Use the same substitution as in part (a); x œ 1 Ê u œ 10, x œ 0 Ê u œ 9
'c
0
x$ 1 È x% 9
dx œ
'
9
" 10 4
u"Î# du œ
'
10
9
" 4
" 3
11. (a) Let u œ 1 cos 3t Ê du œ 3 sin 3t dt Ê
'
1Î6
0
(1 cos 3t) sin 3t dt œ
'
1
" 3
0
'
1Î6
(1 cos 3t) sin 3t dt œ
12. (a) Let u œ 2 tan
'c
0
1Î2
t #
t #
dt œ
" 3
1
" #
Ê du œ
ˆ2 tan #t ‰ sec#
'
2
2
1
'c
1Î2
ˆ2 tan #t ‰ sec#
t #
'
dt œ 2
1
'
0
cos z È4 3 sin z
dz œ
'
4
4
" Èu
" 6
#
#
"
t #
(1)# 6" (0)# œ
Ê u œ 1, t œ
u du œ ’ 3" Š u# ‹ “ œ dt Ê 2 du œ sec#
1 3
1 6
Ê u œ 1 cos
1 #
œ1
" 6
Ê u œ 1 cos 1 œ 2
" 6
(2)# 6" (1)# œ
t #
dt; t œ
1 #
" 2
Ê u œ 2 tan ˆ 41 ‰ œ 1, t œ 0 Ê u œ 2
#
u (2 du) œ cu# d " œ 2# 1# œ 3 1 #
3
Ê u œ 1, t œ
1 #
Ê uœ3
$
u du œ cu# d " œ 3# 1# œ 8 " 3
13. (a) Let u œ 4 3 sin z Ê du œ 3 cos z dz Ê 21
!
1 6
(b) Use the same substitution as in part (a); t œ 1Î2
"
#
sec#
'
du œ sin 3t dt; t œ 0 Ê u œ 0, t œ
u du œ ’ 3" Š u# ‹ “ œ
(b) Use the same substitution as in part (a); t œ 1Î3
3 È10 #
u"Î# du œ
du œ cos z dz; z œ 0 Ê u œ 4, z œ 21 Ê u œ 4
ˆ 3" du‰ œ 0
(b) Use the same substitution as in part (a); z œ 1 Ê u œ 4 3 sin (1) œ 4, z œ 1 Ê u œ 4
'c
1
cos z 1 È4 3 sin z
dz œ
'
4
4
" Èu
ˆ 3" du‰ œ 0
14. (a) Let u œ 3 2 cos w Ê du œ 2 sin w dw Ê "# du œ sin w dw; w œ 1# Ê u œ 3, w œ 0 Ê u œ 5
'c
0
sin w # 1Î2 (3 2 cos w)
dw œ
'
5
3
u# ˆ #" du‰ œ
" #
&
cu" d $ œ
" #
" ˆ "5 "3 ‰ œ 15
(b) Use the same substitution as in part (a); w œ 0 Ê u œ 5, w œ
'
!
1Î2
sin w (3 2 cos w)#
dw œ
'
5
3
u# ˆ #" du‰ œ
" #
5
' u# du œ
1 #
Ê uœ3
" 15
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
298
Chapter 5 Integration
15. Let u œ t& 2t Ê du œ a5t% 2b dt; t œ 0 Ê u œ 0, t œ 1 Ê u œ 3
'
1
0
'
Èt& 2t a5t% 2b dt œ
0
16. Let u œ 1 Èy Ê du œ
'
4
dy # 1 2 È y ˆ1 È y ‰
œ
'
3
" # 2 u
3
$
u"Î# du œ 23 u$Î# ‘ ! œ
; y œ 1 Ê u œ 2, y œ 4 Ê u œ 3
dy 2È y
'
du œ
(3)$Î# 23 (0)$Î# œ 2È3
2 3
3
2
$
u# du œ cu" d # œ ˆ 13 ‰ ˆ 12 ‰ œ
" 6
17. Let u œ cos 2) Ê du œ 2 sin 2) d) Ê "# du œ sin 2) d); ) œ 0 Ê u œ 1, ) œ
'
1Î6
!
cos$ 2) sin 2) d) œ
18. Let u œ tan ˆ 6) ‰ Ê du œ
'
31Î2
1
'
1Î2
1
" 6
'
'
cot& ˆ 6) ‰ sec# ˆ 6) ‰ d) œ
1
#
u$ du œ ’ 2" Š u# ‹“
"Î# "
œ
1
"
%
" # 4 ˆ 1# ‰
'
!
" È3
" 4(1)#
,)œ
31 #
9
1
" 4
Ê u œ tan
"
'
1
1 4
œ1
œ 12
%
du œ sin t dt; t œ 0 Ê u œ 5 4 cos 0 œ 1, t œ 1 Ê u œ 5 4 cos 1 œ 9
5u"Î% ˆ "4 du‰ œ
0
" #
3 4
È3 ‹
'
(1 sin 2t)$Î# cos 2t dt œ
œ
u 3 3 3 & È3 u (6 du) œ ’6 Š 4 ‹“ "ÎÈ$ œ 2u% ‘ "ÎÈ$ œ 2(1)% # Š "
'
5 4
9
1
*
u"Î% du œ 54 ˆ 45 u&Î% ‰‘ " œ 9&Î% 1 œ $&Î# "
20. Let u œ 1 sin 2t Ê du œ 2 cos 2t dt Ê "# du œ cos 2t dt; t œ 0 Ê u œ 1, t œ 1Î4
Ê u œ cos 2 ˆ 16 ‰ œ
1Î
5 (5 4 cos t)"Î% sin t dt œ
!
1Î2
sec# ˆ 6) ‰ d) Ê 6 du œ sec# ˆ 6) ‰ d); ) œ 1 Ê u œ tan ˆ 16 ‰ œ
19. Let u œ 5 4 cos t Ê du œ 4 sin t dt Ê 1
'
u$ ˆ "# du‰ œ "#
1 6
" ˆ 2 &Î# ‰‘ ! 2 5 u "
"# u$Î# du œ
1 4
Ê uœ0
œ ˆ 15 (0)&Î# ‰ ˆ 15 (1)&Î# ‰ œ
" 5
21. Let u œ 4y y# 4y$ 1 Ê du œ a4 2y 12y# b dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 4(1) (1)# 4(1)$ 1 œ 8
'
1
!
a4y y# 4y$ 1b
#Î$
a12y# 2y 4b dy œ
'
8
1
)
u#Î$ du œ 3u"Î$ ‘ " œ 3(8)"Î$ 3(1)"Î$ œ 3
22. Let u œ y$ 6y# 12y 9 Ê du œ a3y# 12y 12b dy Ê
'
1
!
ay$ 6y# 12y 9b
23. Let u œ )$Î# Ê du œ
'
È1 3
!
1Î2
1
œ
" #
'
È) cos# ˆ)$Î# ‰ d) œ " t
" 4
1
!
2 3
'
9
du œ ay# 4y 4b dy; y œ 0 Ê u œ 9, y œ 1 Ê u œ 4 %
" 3
u"Î# du œ 3" ˆ2u"Î# ‰‘ * œ
2 3
(4)"Î# 32 (9)"Î# œ
2 3
(2 3) œ 32
3 du œ È) d); ) œ 0 Ê u œ 0, ) œ È 1# Ê u œ 1
cos# u ˆ 23 du‰ œ 23 ˆ #u
" 4
1
sin 2u‰‘ ! œ
2 3
ˆ 1#
" 4
sin 21‰ 32 (0) œ
1 3
Ê du œ t# dt; t œ 1 Ê u œ 0, t œ #" Ê u œ 1
t# sin# ˆ1 "t ‰ dt œ
ay# 4y 4b dy œ
)"Î# d) Ê
#
24. Let u œ 1
'c
3 #
"Î#
4
" 3
'!
1
sin# u du œ ˆ u2
" 4
" sin 2u‰‘ ! œ ’ˆ #"
" 4
sin (2)‰ ˆ #0
" 4
sin 0‰“
sin 2
25. Let u œ 4 x# Ê du œ 2x dx Ê "# du œ x dx; x œ 2 Ê u œ 0, x œ 0 Ê u œ 4, x œ 2 Ê u œ 0 Aœ
'c
0 2
xÈ4 x# dx %
œ 23 u$Î# ‘ ! œ
2 3
'
2
!
xÈ4 x# dx œ
(4)$Î# 23 (0)$Î# œ
'
4
!
"# u"Î# du
'
4
0
"# u"Î# du œ 2
'
4
" ! #
u"Î# du œ
16 3
26. Let u œ 1 cos x Ê du œ sin x dx; x œ 0 Ê u œ 0, x œ 1 Ê u œ 2
'
!
1
(1 cos x) sin x dx œ
'
!
2
#
#
u du œ ’ u2 “ œ !
2# #
0# #
œ2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
'
!
4
u"Î# du
Section 5.6 Substitution and Area Between Curves
299
27. Let u œ 1 cos x Ê du œ sin x dx Ê du œ sin x dx; x œ 1 Ê u œ 1 cos (1) œ 0, x œ 0 Ê u œ 1 cos 0 œ 2 Aœ
'c
0
1
3 (sin x) È1 cos x dx œ
'
2
!
28. Let u œ 1 1 sin x Ê du œ 1 cos x dx Ê Because of symmetry about x œ 1# , A œ 2 œ
'
!
1
3u"Î# (du) œ 3 " 1
'c
'
2
!
#
u"Î# du œ 2u$Î# ‘ ! œ 2(2)$Î# 2(0)$Î# œ 2&Î#
du œ cos x dx; x œ 1# Ê u œ 1 1 sin ˆ 1# ‰ œ 0, x œ 0 Ê u œ 1
0
1 1Î2 #
(cos x) (sin (1 1 sin x)) dx œ 2
'
1
(" cos 2x) # !
dx œ
'
" #
1
!
(1 cos 2x) dx œ
" #
x
30. For the sketch given, a œ 13 , b œ 13 ; f(t) g(t) œ Aœ œ
" #
1
!
1 #
(sin u) ˆ 1" du‰
sin u du œ [cos u]1! œ (cos 1) (cos 0) œ 2 1 cos 2x ; #
29. For the sketch given, a œ 0, b œ 1; f(x) g(x) œ 1 cos# x œ sin# x œ Aœ
'
'c ÎÎ ˆ "# sec# t 4 sin# t‰ dt œ "# ' 1 3
1Î3
1 3 1Î3
1Î3
1Î3
1 3
1Î3
'c Î sec# t dt 2'
" #
" #
" #
œ
'
1Î3
1Î3
sin# t dt œ
" #
" #
'
1Î3
1Î3
1Î$ sin 2t # ]1Î$
1Î$
[tan t]1Î$ 2[t
1 #
[(1 0) (0 0)] œ
sec# t a4 sin# tb œ
sec# t dt 4
(1 cos 2t) dt œ
sin 2x ‘ 1 # !
sec# t 4 sin# t;
'
sec# t dt 4
œ È3 4 †
1 3
1Î3
(" cos 2t) #
1Î3
È3 œ
dt
41 3
31. For the sketch given, a œ 2, b œ 2; f(x) g(x) œ 2x# ax% 2x# b œ 4x# x% ; Aœ
'c
2
$
2
a4x# x% b dx œ ’ 4x3
# x& 5 “ #
œ ˆ 32 3
32 ‰ 5
ˆ 32 ‰‘ œ 32 3 5
64 5
œ
(" 0) 4
œ
" 3
64 3
320192 15
œ
128 15
32. For the sketch given, c œ 0, d œ 1; f(y) g(y) œ y# y$ ; Aœ
'
1
!
ay# y$ b dy œ
'
1
!
y# dy
'
!
1
$
"
%
"
(" 0) 3
y$ dy œ ’ y3 “ ’ y4 “ œ !
!
" 4
œ
" 1#
33. For the sketch given, c œ 0, d œ 1; f(y) g(y) œ a12y# 12y$ b a2y# 2yb œ 10y# 12y$ 2y; Aœ
'
1
!
a10y# 12y$ 2yb dy œ
'
‰ œ ˆ 10 3 0 (3 0) (1 0) œ
!
1
10y# dy
'
!
1
12y$ dy
'
1
!
"
"
"
$‘ 12 % ‘ 2 #‘ 2y dy œ 10 3 y ! 4 y ! # y !
4 3
34. For the sketch given, a œ 1, b œ 1; f(x) g(x) œ x# a2x% b œ x# 2x% ; Aœ
'c ax# 2x% b dx œ ’ x3 1
$
1
" 2x& 5 “ "
œ ˆ 3" 52 ‰ 3" ˆ 52 ‰‘ œ
35. We want the area between the line y œ 1, 0 Ÿ x Ÿ 2, and the curve y œ (formed by y œ x and y œ 1) with base 1 and height 1. Thus, A œ œ ˆ2
8 ‰ 1#
" #
œ2
2 3
" #
œ
'
!
2
2 3
x# 4,
4 5
œ
10 12 15
œ
22 15
738?= the area of a triangle
Š1
x# 4‹
dx "# (1)(1) œ ’x
# x$ 1# “ !
" #
5 6
36. We want the area between the x-axis and the curve y œ x# , 0 Ÿ x Ÿ 1 :6?= the area of a triangle (formed by x œ 1, x y œ 2, and the x-axis) with base 1 and height 1. Thus, A œ
'
!
1
$
"
x# dx "# (1)(1) œ ’ x3 “ !
" #
œ
" 3
" #
œ
5 6
37. AREA œ A1 A2 A1: For the sketch given, a œ 3 and we find b by solving the equations y œ x# 4 and y œ x# 2x simultaneously for x: x# 4 œ x# 2x Ê 2x# 2x 4 œ 0 Ê 2(x 2)(x 1) Ê x œ 2 or x œ 1 so b œ 2: f(x) g(x) œ ax# 4b ax# 2xb œ 2x# 2x 4 Ê A1 œ
'cc a2x# 2x 4b dx 2
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
300
Chapter 5 Integration $
œ ’ 2x3
2x# #
4x“
# $
‰ œ ˆ 16 3 4 8 (18 9 12) œ 9
œ
16 3
11 3 ;
A2: For the sketch given, a œ 2 and b œ 1: f(x) g(x) œ ax# 2xb ax# 4b œ 2x# 2x 4
'c a2x# 2x 4b dx œ ’ 2x3 1
Ê A2 œ
$
2
œ 23 1 4
16 3
x# 4x“
"
‰ œ ˆ 23 1 4‰ ˆ 16 3 48
#
4 8 œ 9;
Therefore, AREA œ A1 A2 œ
11 3
9œ
38 3
38. AREA œ A1 A2 A1: For the sketch given, a œ 2 and b œ 0: f(x) g(x) œ a2x$ x# 5xb ax# 3xb œ 2x$ 8x Ê A1 œ
'c a2x$ 8xb dx œ ’ 2x4 0
%
2
! 8x# # “ #
œ 0 (8 16) œ 8;
A2: For the sketch given, a œ 0 and b œ 2: f(x) g(x) œ ax# 3xb a2x$ x# 5xb œ 8x 2x$ Ê A2 œ
'
2
0
#
a8x 2x$ b dx œ ’ 8x2
Therefore, AREA œ A1 A2 œ 16
# 2x% “ 4 !
œ (16 8) œ 8;
39. AREA œ A1 A2 A3 A1: For the sketch given, a œ 2 and b œ 1: f(x) g(x) œ (x 2) a4 x# b œ x# x 2 Ê A1 œ
'cc ax# x 2b dx œ ’ x3 1
$
2
x# #
2x“
" #
œ ˆ "3
" #
2‰ ˆ 83
4‰ œ
4 2
7 3
" #
œ
143 6
œ
1" 6 ;
" #
œ 9# ;
A2: For the sketch given, a œ 1 and b œ 2: f(x) g(x) œ a4 x# b (x 2) œ ax# x 2b Ê A2 œ
'c
2
$
1
ax# x 2b dx œ ’ x3
x# #
2x“
# "
œ ˆ 83
4 #
4‰ ˆ 13
1 2
2‰ œ 3 8
A3: For the sketch given, a œ 2 and b œ 3: f(x) g(x) œ (x 2) a4 x# b œ x# x 2 Ê A3 œ
'
3
$
x# #
ax# x 2b dx œ ’ x3
2
Therefore, AREA œ A1 A2 A3 œ
11 6
9 #
$
2x“ œ ˆ 27 3 # 9 #
ˆ9
9 #
6‰ ˆ 38
83 ‰ œ 9
5 6
œ
4 2
4‰ œ 9
9 #
38 ;
49 6
40. AREA œ A1 A2 A3 $
A1: For the sketch given, a œ 2 and b œ 0: f(x) g(x) œ Š x3 x‹ Ê A1 œ
" 3
'c
0 2
ax$ 4xb dx œ
" 3
%
’ x4 2x# “
!
x 3
for x:
xœ
f(x) g(x) œ œ
" 3
x 3
x 3
Ê
x$ 3
xœ0 Ê 4 3
x 3
43 x œ
" 3
ax$ 4xb
œ 0 3" (4 8) œ 34 ;
#
A2: For the sketch given, a œ 0 and we find b by solving the equations y œ x$ 3
x$ 3
œ
x$ 3
x and y œ
x 3
simultaneously
(x 2)(x 2) œ 0 Ê x œ 2, x œ 0, or x œ 2 so b œ 2:
$
Š x3 x‹ œ 3" ax$ 4xb Ê A2 œ 3"
'
0
2
ax$ 4xb dx œ
" 3
'
0
2
a4x x$ b œ
(8 4) œ 43 ; $
A3: For the sketch given, a œ 2 and b œ 3: f(x) g(x) œ Š x3 x‹ Ê A3 œ
" 3
'
3
2
ax$ 4xb dx œ
" 3
Therefore, AREA œ A1 A2 A3 œ
$
%
’ x4 2x# “ œ #
4 3
4 3
25 12
œ
" 3
x 3
œ
" 3
ax$ 4xb
ˆ 81 ‰ ˆ 16 ‰‘ œ 4 2†9 4 8
3225 1#
œ
" 3
ˆ 81 ‰ 4 14 œ
19 4
41. a œ 2, b œ 2; f(x) g(x) œ 2 ax# 2b œ 4 x# Ê Aœ
" 3
'c a4 x# bdx œ ’4x x3 “ # 2
$
#
2
8‰ œ 2 † ˆ 24 3 3 œ
œ ˆ8 83 ‰ ˆ8 83 ‰
32 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
25 12 ;
’2x#
# x% 4 “!
Section 5.6 Substitution and Area Between Curves 42. a œ 1, b œ 3; f(x) g(x) œ a2x x# b (3) œ 2x x# 3
'c a2x x# 3b dx œ ’x# x3 3
Ê Aœ œ ˆ9
$
3x“
1
9‰ ˆ1
27 3
1 3
43. a œ 0, b œ 2; f(x) g(x) œ 8x x% Ê A œ #
œ ’ 8x2
# x& 5 “!
œ 16
32 5
" 3
3‰ œ 11
œ
'
2
0
œ
$ "
32 3
a8x x% b dx
80 32 5
œ
48 5
44. Limits of integration: x# 2x œ x Ê x# œ 3x Ê x(x 3) œ 0 Ê a œ 0 and b œ 3; f(x) g(x) œ x ax# 2xb œ 3x x# Ê Aœ œ
27 #
'
3
0
#
a3x x# b dx œ ’ 3x2 27 18 #
9œ
œ
$ x$ 3 “!
9 #
45. Limits of integration: x# œ x# 4x Ê 2x# 4x œ 0 Ê 2x(x 2) œ 0 Ê a œ 0 and b œ 2; f(x) g(x) œ ax# 4xb x# œ 2x# 4x Ê Aœ
'
2
0
œ 16 3
$
a2x# 4xb dx œ ’ 2x 3 œ
16 #
32 48 6
œ
# 4x# 2 “!
8 3
46. Limits of integration: 7 2x# œ x# 4 Ê 3x# 3 œ 0 Ê 3(x 1)(x 1) œ 0 Ê a œ 1 and b œ 1; f(x) g(x) œ a7 2x# b ax# 4b œ 3 3x# Ê Aœ
'c a3 3x# b dx œ 3 ’x x3 “ " 1
$
"
1
œ 3 ˆ1 "3 ‰ ˆ1 3" ‰‘ œ 6 ˆ 32 ‰ œ 4
47. Limits of integration: x% 4x# 4 œ x# Ê x% 5x# 4 œ 0 Ê ax# 4b ax# 1b œ 0 Ê (x 2)(x 2)(x 1)(x 1) œ ! Ê x œ 2, 1, 1, 2; f(x) g(x) œ ax% 4x# 4b x# œ x% 5x# 4 and g(x) f(x) œ x# ax% 4x# 4b œ x% 5x# 4 Ê Aœ
'
1
2
'cc ax% 5x# 4bdx 'c ax% 5x# 4bdx 1
1
ax% 5x# 4bdx &
œ ’ x5 œ ˆ "5 œ
1
2
60 5
5 3
5x$ 3
4x“
"
&
’ x5
#
4‰ ˆ 32 5 60 3
œ
300180 15
40 3
5x$ 3
4x“
8‰ ˆ 5"
5 3
" "
&
’ 5x
5x$ 3
4‰ ˆ 5"
4x“ 5 3
# "
4‰ ˆ 32 5
40 3
8‰ ˆ 5"
œ8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
5 3
4‰
301
302
Chapter 5 Integration
48. Limits of integration: xÈa# x# œ 0 Ê x œ 0 or Èa# x# œ 0 Ê x œ 0 or a# x# œ 0 Ê x œ a, 0, a; Aœ œ
" #
œ
" 3
'c xÈa# x# dx ' 0
a
a
0
’ 23 aa# x# b # $Î#
aa b
$Î# !
“
" 3
a
xÈa# x# dx
"# ’ 23 aa# x# b
# $Î#
’ aa b
$Î# a
“
!
2a$ 3
“œ
49. Limits of integration: y œ Èkxk œ
Èx, x Ÿ 0 and Èx, x 0
5y œ x 6 or y œ x5 65 ; for x Ÿ 0: Èx œ x5 65 Ê 5Èx œ x 6 Ê 25(x) œ x# 12x 36 Ê x# 37x 36 œ 0 Ê (x 1)(x 36) œ 0 Ê x œ 1, 36 (but x œ 36 is not a solution); for x 0: 5Èx œ x 6 Ê 25x œ x# 12x 36 Ê x# 13x 36 œ 0 Ê (x 4)(x 9) œ 0 Ê x œ 4, 9; there are three intersection points and Aœ
'c ˆ x 5 6 Èx‰dx ' 0
1
4
0
#
œ ’ (x 106) 23 (x)$Î# “ œ ˆ 36 10
25 10
!
ˆ x 5 6 Èx‰dx
4
9
ˆÈ x
%
#
"
'
’ (x 106) 23 x$Î# “ ’ 23 x$Î#
23 ‰ ˆ 100 10
2 3
† 4$Î#
!
36 10
0‰ ˆ 32 † 9$Î#
x6‰ 5
dx
* (x 6)# 10 “ % 225 10
2 3
† 4$Î#
100 ‰ 10
œ 50 10
20 3
50. Limits of integration: x# 4, x Ÿ 2 or x 2 y œ kx# 4k œ œ 4 x# , 2 Ÿ x Ÿ 2 for x Ÿ 2 and x 2: x# 4 œ
x# 2
4
Ê 2x# 8 œ x# 8 Ê x# œ 16 Ê x œ „ 4; for 2 Ÿ x Ÿ 2: 4 x# œ
x# #
4 Ê 8 2x# œ x# 8
Ê x# œ 0 Ê x œ 0; by symmetry of the graph,
Aœ2
'
0
2
#
’Š x2 4‹ a4 x# b“dx 2
œ 2 ˆ 8# 0‰ 2 ˆ32
64 6
'
4
2
#
$
#
’Š x2 4‹ ax# 4b“dx œ 2 ’ x2 “ 2 ’8x
16 68 ‰ œ 40
!
56 3
œ
% x$ 6 “#
64 3
51. Limits of integration: c œ 0 and d œ 3; f(y) g(y) œ 2y# 0 œ 2y# Ê Aœ
'
0
3
$
$
2y# dy œ ’ 2y3 “ œ 2 † 9 œ 18 !
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
5 3
Section 5.6 Substitution and Area Between Curves 52. Limits of integration: y# œ y 2 Ê (y 1)(y 2) œ 0 Ê c œ 1 and d œ 2; f(y) g(y) œ (y 2) y#
'c ay 2 y# b dy œ ’ y# 2
Ê Aœ
#
1
2y
œ ˆ 4# 4 83 ‰ ˆ "# 2 3" ‰ œ 6
8 3
" #
# y$ 3 “ "
2
" 3
œ
9 #
53. Limits of integration: 4x œ y# 4 and 4x œ 16 y Ê y# 4 œ 16 y Ê y# y 20 œ 0 Ê (y 5)(y 4) œ 0 Ê c œ 4 and d œ 5; #
f(y) g(y) œ ˆ 164y ‰ Š y 44 ‹ œ " 4
Ê Aœ
y$ 3
'c ay# y 20b dy 5
4
y# #
œ
" 4
’
œ
" 4 " 4
ˆ 125 3 189 ˆ 3
œ
y# y20 4
20y“
&
% 25 ‰ 100 4" ˆ 64 2 3 9 243 ‰ 180 œ 2 8
80‰
16 #
54. Limits of integration: x œ y# and x œ 3 2y# Ê y# œ 3 2y# Ê 3y# œ 3 Ê 3(y 1)(y 1) œ 0 Ê c œ 1 and d œ 1; f(y) g(y) œ a3 2y# b y# œ 3 3y# œ 3 a1 y# b Ê A œ 3 " y$ 3 “ "
œ 3 ’y
œ 3 ˆ1
"‰ 3
'c a1 y# b dy 1
1
3 ˆ1 3" ‰
œ 3 † 2 ˆ1 "3 ‰ œ 4 55. Limits of integration: x œ y# y and x œ 2y# 2y 6 Ê y# y œ 2y# 2y 6 Ê y# y 6 œ 0 Ê ay 3bay 2b œ 0 Ê c œ 2 and d œ 3; f(y) g(y) œ ay# yb a2y# 2y 6b œ y# y 6 Ê Aœ
'c ay# y 6b dy œ ’ y3 3
$
2
œ ˆ9
18‰ ˆ 83 2 12‰ œ
9 2
"# y2 6y“
3 2
125 6
56. Limits of integration: x œ y#Î$ and x œ 2 y% Ê y#Î$ œ 2 y% Ê c œ 1 and d œ 1; f(y) g(y) œ a2 y% b y#Î$ Ê Aœ œ ’2y
'c ˆ2 y% y#Î$ ‰ dy 1
1
y& 5
53 y&Î$ “
"
"
œ ˆ2 "5 35 ‰ ˆ2 œ 2 ˆ2 "5 35 ‰ œ 12 5
" 5
35 ‰
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303
304
Chapter 5 Integration
57. Limits of integration: x œ y# 1 and x œ kyk È1 y# Ê y# 1 œ kyk È1 y# Ê y% 2y# 1 œ y# a1 y# b Ê y% 2y# 1 œ y# y% Ê 2y% 3y# 1 œ 0 Ê a2y# 1b ay# 1b œ 0 Ê 2y# 1 œ 0 or y# 1 œ 0 " #
Ê y# œ
or y# œ 1 Ê y œ „ „È 2 #
È2 #
or y œ „ 1.
are not solutions Ê y œ „ 1; for 1 Ÿ y Ÿ 0, f(x) g(x) œ yÈ1 y# ay# 1b Substitution shows that œ 1 y# y a1 y# b Aœ2
"Î#
, and by symmetry of the graph,
'c ’1 y# y a1 y# b"Î# “ dy 0
1
'c a1 y# b dy 2 'c y a1 y# b"Î# dy œ 2 ’y y3 “ !
œ2
0
0
1
$
"
1
œ 2 (! 0) ˆ1
" ‰‘ 3
ˆ 23
# $Î#
2 ˆ "# ‰ ” 2 a1 3y b •
! "
0‰ œ 2
58. AREA œ A1 A2 Limits of integration: x œ 2y and x œ y$ y# Ê y$ y# œ 2y Ê y ay# y 2b œ y(y 1)(y 2) œ 0 Ê y œ 1, 0, 2: for 1 Ÿ y Ÿ 0, f(y) g(y) œ y$ y# 2y Ê A1 œ
'c
0 1
œ 0 ˆ "4
" 3
%
y$ 3
ay$ y# 2yb dy œ ’ y4 1‰ œ
y# “
! "
5 12 ;
for 0 Ÿ y Ÿ 2, f(y) g(y) œ 2y y$ y#
'! a2y y$ y# b dy œ ’y# y4 2
Ê A2 œ Ê ˆ4
16 4
%
# y$ 3 “!
38 ‰ 0 œ 38 ;
Therefore, A1 A2 œ
5 12
8 3
œ
37 12
59. Limits of integration: y œ 4x# 4 and y œ x% 1 Ê x% 1 œ 4x# 4 Ê x% 4x# 5 œ 0 Ê ax# 5b (x 1)(x 1) œ 0 Ê a œ 1 and b œ 1; f(x) g(x) œ 4x# 4 x% 1 œ 4x# x% 5 Ê Aœ
'c a4x# x% 5b dx œ ’ 4x3 1
1
œ ˆ 43
" 5
5‰ ˆ 43
" 5
$
5‰ œ 2 ˆ 43
"
x& 5
5x“
" 5
5‰ œ
" 104 15
60. Limits of integration: y œ x$ and y œ 3x# 4 Ê x$ 3x# 4 œ 0 Ê ax# x 2b (x 2) œ 0 Ê (x 1)(x 2)# œ 0 Ê a œ 1 and b œ 2; f(x) g(x) œ x$ a3x# 4b œ x$ 3x# 4 Ê Aœ
'c ax$ 3x# 4b dx œ ’ x4
œ ˆ 16 4
24 3
2
%
1
8‰ ˆ 41 " 4‰ œ
3x$ 3
4x“
# "
27 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.6 Substitution and Area Between Curves 61. Limits of integration: x œ 4 4y# and x œ 1 y% Ê 4 4y# œ 1 y% Ê y% 4y# 3 œ 0 Ê Šy È3‹ Šy È3‹ (y 1)(y 1) œ 0 Ê c œ 1 and d œ 1 since x 0; f(y) g(y) œ a4 4y# b a1 y% b
'c a3 4y# y% b dy 1
œ 3 4y# y% Ê A œ 4y$ 3
œ ’3y
" y& 5 “ "
1
œ 2ˆ3
4 3
5" ‰ œ
56 15
#
62. Limits of integration: x œ 3 y# and x œ y4 #
Ê 3 y# œ y4 Ê
3y# 4
3œ0 Ê
3 4
(y 2)(y 2) œ 0 #
Ê c œ 2 and d œ 2; f(y) g(y) œ a3 y# b Š y4 ‹ y# 4‹
œ 3 Š1 œ 3 ˆ2
8 ‰ 12
'c Š1 y4 ‹ dy œ 3 ’y 1y# “ # 2
Ê Aœ3 ˆ 2
#
$
#
2
8 ‰‘ 12
œ 3 ˆ4
16 ‰ 12
œ 12 4 œ 8
63. a œ 0, b œ 1; f(x) g(x) œ 2 sin x sin 2x Ê Aœ
'
1
0
(2 sin x sin 2x) dx œ 2 cos x
cos 2x ‘ 1 2 !
œ 2(1) "# ‘ ˆ2 † 1 "# ‰ œ 4
64. a œ 13 , b œ 13 ; f(x) g(x) œ 8 cos x sec# x Ê Aœ œ Š8 †
'c
1Î3 1Î3
È3 #
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x] 1Î$ È3 #
È3‹ Š8 †
È3‹ œ 6È3
‰ 65. a œ 1, b œ 1; f(x) g(x) œ a1 x# b cos ˆ 1x # Ê Aœ œ ˆ1
" 3
'c 1 x# cos ˆ 1#x ‰‘ dx œ ’x x3 1
$
1
12 ‰ ˆ1
" 3
12 ‰ œ 2 ˆ 23 12 ‰ œ
2 1 4 3
sin ˆ 1#x ‰“
" "
4 1
66. A œ A1 A2 a" œ 1, b" œ 0 and a# œ 0, b# œ 1; f" (x) g" (x) œ x sin ˆ 1#x ‰ and f# (x) g# (x) œ sin ˆ 1#x ‰ x Ê by symmetry about the origin,
A" A# œ 2A" Ê A œ 2 œ 2 ’ 12 cos ˆ 1#x ‰
" x# # “!
œ 2 ˆ 12 "# ‰ œ 2 ˆ 4211 ‰ œ
'
0
1
sin ˆ 1x ‰ ‘ # x dx
œ 2 ˆ 12 † 0 "# ‰ ˆ 12 † 1 0‰‘ 4 1 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
305
306
Chapter 5 Integration
67. a œ 14 , b œ 14 ; f(x) g(x) œ sec# x tan# x Ê Aœ œ œ
'c
1Î4 1Î4 1Î4
'c
1Î4
'c
1Î4 1Î4
asec# x tan# xb dx
csec# x asec# x 1bd dx 1Î%
1 † dx œ [x]1Î% œ
1 4
ˆ 14 ‰ œ
1 #
68. c œ 14 , d œ 14 ; f(y) g(y) œ tan# y a tan# yb œ 2 tan# y œ 2 asec# y 1b Ê A œ
'c
1Î4 1Î4
2 asec# y 1b dy
1Î% œ 2[tan y y]1Î% œ 2 ˆ1 14 ‰ ˆ1 14 ‰‘
œ 4 ˆ1 14 ‰ œ 4 1
69. c œ 0, d œ 1# ; f(y) g(y) œ 3 sin yÈcos y 0 œ 3 sin yÈcos y Ê Aœ3
'
1Î2
0
1Î#
sin yÈcos y dy œ 3 23 (cos y)$Î# ‘ !
œ 2(0 1) œ 2
"Î$ ‰ 70. a œ 1, b œ 1; f(x) g(x) œ sec# ˆ 1x 3 x
Ê Aœ
'c sec# ˆ 13x ‰ x"Î$ ‘ dx œ 13 tan ˆ 13x ‰ 43 x%Î$ ‘ "" 1
1
œ Š 13 È3 34 ‹ ’ 13 ŠÈ3‹ 34 “ œ
6È 3 1
71. A œ A" A# Limits of integration: x œ y$ and x œ y Ê y œ y$ Ê y$ y œ 0 Ê y(y 1)(y 1) œ 0 Ê c" œ 1, d" œ 0 and c# œ 0, d# œ 1; f" (y) g" (y) œ y$ y and f# (y) g# (y) œ y y$ Ê by symmetry about the origin, A" A# œ 2A# Ê A œ 2 œ 2 ˆ "# 4" ‰ œ
'
1
0
#
ay y$ b dy œ 2 ’ y#
" #
" y% 4 “!
72. A œ A" A# Limits of integration: y œ x$ and y œ x& Ê x$ œ x& Ê x& x$ œ 0 Ê x$ (x 1)(x 1) œ 0 Ê a" œ 1, b" œ 0 and a# œ 0, b# œ 1; f" (x) g" (x) œ x$ x& and f# (x) g# (x) œ x& x$ Ê by symmetry about the origin, A" A# œ 2A# Ê A œ 2 œ 2 ˆ "4 6" ‰ œ
" 6
'
0
1
%
ax$ x& b dx œ 2 ’ x4
" x' 6 “!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.6 Substitution and Area Between Curves 73. A œ A" A# Limits of integration: y œ x and y œ
" x#
$
Ê xœ
" x# ,
307
xÁ0
Ê x œ 1 Ê x œ 1 , f" (x) g" (x) œ x 0 œ x Ê A" œ
'
1
0
#
"
x dx œ ’ x2 “ œ "# ; f# (x) g# (x) œ
'
œ x# Ê A# œ A œ A" A# œ
!
2
1
" #
and b œ
Ê Aœ œŠ
È2 #
#
" #
œ1 1 4
Ê aœ0
f(x) g(x) œ cos x sin x
'
1Î4
0
0
" " ‘ x# dx œ " x " œ # 1 œ #;
74. Limits of integration: sin x œ cos x Ê x œ 1 4;
" x#
1Î%
(cos x sin x) dx œ [sin x cos x]!
È2 # ‹
(0 1) œ È2 1
75. (a) The coordinates of the points of intersection of the line and parabola are c œ x# Ê x œ „ Èc and y œ c (b) f(y) g(y) œ Èy ˆÈy‰ œ 2Èy Ê the area of the lower section is, AL œ œ2
'
0
c
'
0
c
[f(y) g(y)] dy
Èy dy œ 2 23 y$Î# ‘ ! œ c
4 3
c$Î# . The area of the
entire shaded region can be found by setting c œ 4: A œ ˆ 43 ‰ 4$Î# œ 43†8 œ 32 3 . Since we want c to divide the region 32 4 $Î# into subsections of equal area we have A œ 2AL Ê 3 œ 2 ˆ 3 c ‰ Ê c œ 4#Î$ (c) f(x) g(x) œ c x# Ê AL œ œ
4 3
Èc
Èc
c
c
'cÈ [f(x) g(x)] dx œ 'cÈ ac x# b dx œ ’cx x3 “ È $
c
cÈc
œ 2 ’c$Î#
c$Î# . Again, the area of the whole shaded region can be found by setting c œ 4 Ê A œ
condition A œ 2AL , we get
4 3
$Î#
c
œ
#Î$
Ê cœ4
32 3
32 3 .
c$Î# 3 “
From the
as in part (b).
76. (a) Limits of integration: y œ 3 x# and y œ 1 Ê 3 x# œ 1 Ê x# œ 4 Ê a œ 2 and b œ 2; f(x) g(x) œ a3 x# b (1) œ 4 x# Ê Aœ
'c a4 x# b dx œ ’4x x3 “ # 2
$
# 32 3
2
œ ˆ8 83 ‰ ˆ8 83 ‰ œ 16
16 3
œ
(b) Limits of integration: let x œ 0 in y œ 3 x# Ê y œ 3; f(y) g(y) œ È3 y ˆÈ3 y‰ œ 2(3 y)"Î# Ê Aœ2
'c (3 y)"Î# dy œ 2 'c (3 y)"Î# (1) dy œ (2) ’ 2(3 3y)
œ ˆ 43 ‰ (8) œ
3
1
3
1
$Î#
“
$ "
œ ˆ 43 ‰ 0 (3 1)$Î# ‘
32 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
308
Chapter 5 Integration
77. Limits of integration: y œ 1 Èx and y œ Ê 1 Èx œ
2 Èx
2 Èx
, x Á 0 Ê Èx x œ 2 Ê x œ (2 x)#
Ê x œ 4 4x x# Ê x# 5x 4 œ 0 Ê (x 4)(x 1) œ 0 Ê x œ 1, 4 (but x œ 4 does not satisfy the equation); y œ È2x and y œ x4 Ê È2x œ x4 Ê 8 œ xÈx Ê 64 œ x$ Ê x œ 4. Therefore, AREA œ A" A# : f" (x) g" (x) œ ˆ1 x"Î# ‰ Ê A" œ
'
œ ˆ1
"8 ‰ 0 œ
2 3
1
0
œ ˆ4 † 2
" x# 8 “!
ˆ1 x"Î# x4 ‰ dx œ ’x 23 x$Î#
16 ‰ 8
37 24 ; f# (x)
g# (x) œ 2x"Î#
ˆ4 "8 ‰ œ 4
œ
15 8
17 8 ;
x 4
x 4
Ê A# œ
'
1
4
ˆ2x"Î# 4x ‰ dx œ ’4x"Î#
Therefore, AREA œ A" A# œ
37 24
17 8
œ
3751 24
œ
88 24
% x# 8 “"
œ
11 3
78. Limits of integration: (y 1)# œ 3 y Ê y# 2y 1 œ 3 y Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0 Ê y œ 2 since y 0; also, 2Èy œ 3 y Ê 4y œ 9 6y y# Ê y# 10y 9 œ 0 Ê (y 9)(y 1) œ 0 Ê y œ 1 since y œ 9 does not satisfy the equation; AREA œ A" A# f" (y) g" (y) œ 2Èy 0 œ 2y"Î# Ê A" œ 2 Ê A# œ
'
1
0
'
1
2
"
$Î#
y"Î# dy œ 2 ’ 2y3 “ œ 34 ; f# (y) g# (y) œ (3 y) (y 1)# !
# c3 y (y 1)# d dy œ 3y "# y# "3 (y 1)$ ‘ " œ ˆ6 2 3" ‰ ˆ3
Therefore, A" A# œ
4 3
7 6
œ
œ
15 6
" #
80. A œ
'
a
b
2f(x) dx
'
a
b
'
0
a
aa# x# b dx œ 2 a# x "3 x$ ‘ ! œ 2 Ša$ a
a$ $ Š 4a3 ‹
(2a) aa# b œ a$ ; limit of ratio œ lim b aÄ!
'
f(x) dx œ 2
a
b
0‰ œ 1
" 3
" #
œ 67 ;
5 2
79. Area between parabola and y œ a# : A œ 2 Area of triangle AOC:
" #
f(x) dx
'
a
b
f(x) dx œ
'
a
b
œ
3 4
a$ 3‹
0œ
4a$ 3 ;
which is independent of a.
f(x) dx œ 4
81. The lower boundary of the region is the line through the points azß 1 z2 b and Šz 1ß 1 az 1b2 ‹. The equation of this line is y a1 z2 b œ
ˆ 1 az 1 b 2 ‰ ˆ 1 z 2 ‰ ax z1z z1
The area of theregion is given by œ
'
z
1b œ a2z 1bax 1b Ê y œ a2z 1bx az2 z 1b.
aa1 x2 b aa2z 1bx az2 z 1bbbdy
' z ax2 a2z 1bx z2 zbdy œ 13 x3 "# a2z 1bx2 az2 zbx‘ zz 1 1
z
œ Š 13 az 1b3 "# a2z 1baz 1b2 az2 zbaz 1b‹ ˆ 13 z3 "# a2z 1bz2 az2 zbz‰ œ 16 . No matter where we choose z, the area of the region bounded by y œ 1 x2 and the line through the points azß 1 z2 b and Šz 1ß 1 az 1b2 ‹ is always 16 . 82. It is sometimes true. It is true if f(x) g(x) for all x between a and b. Otherwise it is false. If the graph of f lies below the graph of g for a portion of the interval of integration, the integral over that portion will be negative and the integral over [aß b] will be less than the area between the curves (see Exercise 71).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.6 Substitution and Area Between Curves " #
83. Let u œ 2x Ê du œ 2 dx Ê
'
3
sin 2x x
1
'
dx œ
6
ˆ "# du‰ œ
sin u ˆ u# ‰
2
du œ dx; x œ 1 Ê u œ 2, x œ 3 Ê u œ 6
'
6
du œ cF(u)d '# œ F(6) F(2)
sin u u
2
84. Let u œ 1 x Ê du œ dx Ê du œ dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 0
'
1
0
'
f(1 x) dx œ
0
1
f(u) ( du) œ
'
0
1
f(u) du œ
'
1
f(u) du œ
0
'
1
0
f(x) dx
85. (a) Let u œ x Ê du œ dx; x œ 1 Ê u œ 1, x œ 0 Ê u œ 0
'c f(x) dx œ ' 0
f odd Ê f(x) œ f(x). Then
1
0
1
f(u) ( du) œ
'
1
œ 3 (b) Let u œ x Ê du œ dx; x œ 1 Ê u œ 1, x œ 0 Ê u œ 0 f even Ê f(x) œ f(x). Then
'c f(x) dx œ ' 0
1
1
0
0
f(u) ( du) œ
'
f(u) ( du) œ
1
0
f(u) du œ
'
0
1
'
1
0
f(u) du œ
'
0
1
f(u) du
f(u) du œ 3
'c f(x) dx when f is odd. Let u œ x Ê du œ dx Ê du œ dx and x œ a Ê u œ a and x œ ! Ê u œ !. Thus ' f(x) dx œ ' f(u) du œ ' f(u) du œ ' f(u) du œ ' f(x) dx. c Thus ' f(x) dx œ ' f(x) dx ' f(x) dx œ ' f(x) dx ' f(x) dx œ !. c c 'c sin x dx œ [cos x]1Î#1Î# œ cos ˆ 1# ‰ cos ˆ 1# ‰ œ ! ! œ !. 0
86. (a) Consider
a
0
0
a
0
a
a
0
a
a
a
a
a
a
0
a
0
0
a
0
0
1/2
(b)
1/2
87. Let u œ a x Ê du œ dx; x œ 0 Ê u œ a, x œ a Ê u œ 0 Iœ
'
' f(u)f(af(au) duu) œ ' f(x)f(af(ax) dxx) dx f(x)f(ax) ' f(x)f(af(ax) dxx) œ ' f(x) ' dx œ [x]! œ a 0 œ a. I I œ ' f(x)f(x) f(ax) f(ax) dx œ a
0
f(x) dx f(x)f(ax)
œ
'
0
a
f(au) f(au)f(u)
a
Ê
a
0
0
a
a
a
a
0
0
Therefore, 2I œ a Ê I œ 88. Let u œ
'
a
( du) œ
xy
x
" t
xy t
dt œ
a #
0
.
t Ê du œ xy t# dt Ê xy du œ
'
y
1
u" du œ
0
'
y
1
" u
du œ
'
y
1
" t
" u
dt Ê u" du œ
du œ
'
1
y
" t
" t
dt; t œ x Ê u œ y, t œ xy Ê u œ 1. Therefore,
dt
89. Let u œ x c Ê du œ dx; x œ a c Ê u œ a, x œ b c Ê u œ b
' cc
b c
a c
90. (a)
f(x c) dx œ
'
a
b
f(u) du œ
'
a
b
f(x) dx (b)
(c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
309
310
Chapter 5 Integration
91-94. Example CAS commands: Maple: f := x -> x^3/3-x^2/2-2*x+1/3; g := x -> x-1; plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" ); q1 := [ -5, -2, 1, 4 ]; # (b) q2 := [seq( fsolve( f(x)=g(x), x=q1[i]..q1[i+1] ), i=1..nops(q1)-1 )]; for i from 1 to nops(q2)-1 do # (c) area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+1] ); end do; add( area[i], i=1..nops(q2)-1 ); # (d) Mathematica: (assigned functions may vary) Clear[x, f, g] f[x_] = x2 Cos[x] g[x_] = x3 x Plot[{f[x], g[x]}, {x, 2, 2}] After examining the plots, the initial guesses for FindRoot can be determined. pts = x/.Map[FindRoot[f[x]==g[x],{x, #}]&, {1, 0, 1}] i1=NIntegrate[f[x] g[x], {x, pts[[1]], pts[[2]]}] i2=NIntegrate[f[x] g[x], {x, pts[[2]], pts[[3]]}] i1 i2 CHAPTER 5 PRACTICE EXERCISES 1. (a) Each time subinterval is of length ?t œ 0.4 sec. The distance traveled over each subinterval, using the midpoint rule, is ?h œ "# avi vib1 b ?t, where vi is the velocity at the left endpoint and vib1 the velocity at
the right endpoint of the subinterval. We then add ?h to the height attained so far at the left endpoint vi to arrive at the height associated with velocity vib1 at the right endpoint. Using this methodology we build the following table based on the figure in the text: t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90 76 65 h (ft) 0 2 9 25 56 114 188 257 320 378 432 481 525 564 592 620.2 t (sec) v (fps) h (ft)
6.4 50 643.2
6.8 37 660.6
7.2 25 672
7.6 12 679.4
8.0 0 681.8
NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. Remember that some shifting of the graph occurs in the printing process. The total height attained is about 680 ft. (b) The graph is based on the table in part (a).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Practice Exercises
311
2. (a) Each time subinterval is of length ?t œ 1 sec. The distance traveled over each subinterval, using the midpoint rule, is ?s œ "# avi vib1 b ?t, where vi is the velocity at the left, and vib1 the velocity at the
right, endpoint of the subinterval. We then add ?s to the distance attained so far at the left endpoint vi to arrive at the distance associated with velocity vib1 at the right endpoint. Using this methodology we build the table given below based on the figure in the text, obtaining approximately 26 m for the total distance traveled: t (sec) 0 1 2 3 4 5 6 7 8 9 10 v (m/sec) 0 0.5 1.2 2 3.4 4.5 4.8 4.5 3.5 2 0 s (m) 0 0.25 1.1 2.7 5.4 9.35 14 18.65 22.65 25.4 26.4
(b) The graph shows the distance traveled by the moving body as a function of time for 0 Ÿ t Ÿ 10.
10
!
3. (a)
kœ1 10
ak 4
œ
" 4
10
" 4
! ak œ
kœ1
(2) œ #"
10
10
kœ1 10
!
(d)
kœ1
kœ1
ˆ 5#
10
bk ‰ œ !
5 #
kœ1
20
20
kœ1 20
kœ1
! ˆ" #
(c)
kœ1 20
kœ1
10
! bk œ kœ1
2bk ‰ 7
20
œ !
kœ1 20
" #
2 7
kœ1
kœ1
kœ1
kœ1
5 #
(10) 25 œ 0
(b)
20
! bk œ
kœ1 20
kœ1
kœ1
" #
20
20
20
kœ1
kœ1
kœ1
! (ak bk ) œ ! ak ! bk œ 0 7 œ 7
(20) 27 (7) œ 8
kœ1
5. Let u œ 2x 1 Ê du œ 2 dx Ê 5
1
(2x 1)"Î# dx œ
'
9
1
3
1
x ax# 1b
7. Let u œ
x 2
"Î$
dx œ
'
8
0
" #
du œ dx; x œ 1 Ê u œ 1, x œ 5 Ê u œ 9 *
u"Î# ˆ "# du‰ œ u"Î# ‘ " œ 3 1 œ 2
6. Let u œ x# 1 Ê du œ 2x dx Ê
" #
du œ x dx; x œ 1 Ê u œ 0, x œ 3 Ê u œ 8 )
u"Î$ ˆ "# du‰ œ 38 u%Î$ ‘ ! œ
3 8
(16 0) œ 6
Ê 2 du œ dx; x œ 1 Ê u œ 1# , x œ 0 Ê u œ 0
'c cos ˆ x# ‰ dx œ 'c 0
0
1
1Î2
(cos u)(2 du) œ [2 sin u]!1Î# œ 2 sin 0 2 sin ˆ 1# ‰ œ 2(0 (1)) œ 2
8. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ 0
10
! aak 2b œ ! ak ! 2 œ 0 2(20) œ 40
(d)
'
10
10
! 3ak œ 3 ! ak œ 3(0) œ 0
4. (a)
'
10
! (bk 3ak ) œ ! bk 3 ! ak œ 25 3(2) œ 31
! (ak bk 1) œ ! ak ! bk ! " œ 2 25 (1)(10) œ 13
(c)
'
(b)
1Î2
(sin x)(cos x) dx œ
'
0
1
#
"
u du œ ’ u2 “ œ !
1 #
Ê uœ1
" #
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312
Chapter 5 Integration
'c f(x) dx œ "3 'c 3 f(x) dx œ 3" (12) œ 4 (b) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 4 œ 2 c c c ' g(x) dx œ 'c g(x) dx œ 2 ' ' (d) (1 g(x)) dx œ 1 g(x) dx œ 1(2) œ 21 c c 'c Š f(x) 5 g(x) ‹ dx œ 5" 'c f(x) dx 5" 'c g(x) dx œ 5" (6) 5" (2) œ 85 2
9. (a)
2
2
5
2
2
2
(c)
5
5
(e)
' ' '
10. (a)
2
5
5
5
2
2
2
2
0
(c)
0
2
(e)
' 7 g(x) dx œ "7 (7) œ 1 f(x) dx œ ' f(x) dx œ 1 [g(x) 3 f(x)] dx œ ' g(x) dx 3' g(x) dx œ
2
" 7
(b)
0 2
(d)
0
2
0
5
2
0
2
0
' '
2
2
2
5
5
2
2
2
1 2
0
g(x) dx œ
'
0
2
g(x) dx
È2 f(x) dx œ È2
'
0
2
'
0
1
g(x) dx œ 1 2 œ 1
f(x) dx œ È2 (1) œ 1È2
f(x) dx œ 1 31
11. x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1; Area œ
'
1
0
ax# 4x 3b dx "
$
'
1
3
ax# 4x 3b dx
$
œ ’ x3 2x# 3x“ ’ x3 2x# 3x“ !
œ
$ ’Š "3
$ "
#
2(1) 3(1)‹ 0“ $
$
’Š 33 2(3)# 3(3)‹ Š 13 2(1)# 3(1)‹“ œ ˆ "3 1‰ 0 ˆ 3" 1‰‘ œ 12. 1
x# 4
8 3
œ 0 Ê 4 x# 0 Ê x œ „ 2;
Area œ
'c Š1 x4 ‹ dx ' 2
œ ’x
2
# x$ 12 “ # 2$ 12 ‹
œ ’Š2
3
#
2
’x
Š1
x# 4‹
dx
$ x$ 12 “ #
Š2
(2)$ 12 ‹“
œ 43 ˆ 43 ‰‘ ˆ 34 43 ‰ œ
’Š3
3$ 12 ‹
Š2
2$ 12 ‹“
13 4
13. 5 5x#Î$ œ 0 Ê 1 x#Î$ œ 0 Ê x œ „ 1; Area œ
'c ˆ5 5x#Î$ ‰ dx ' 1
1
1
8
ˆ5 5x#Î$ ‰ dx
" ) œ 5x 3x&Î$ ‘ " 5x 3x&Î$ ‘ " œ ˆ5(1) 3(1)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘
ˆ5(8) 3(8)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘ œ [2 (2)] [(40 96) 2] œ 62
14. 1 Èx œ 0 Ê x œ 1; Area œ
'
0
1
ˆ1 Èx‰ dx
'
1
4
ˆ1 Èx‰ dx
" % œ x 23 x$Î# ‘ ! x 23 x$Î# ‘ " œ ˆ1 23 (1)$Î# ‰ 0‘ ˆ4 23 (4)$Î# ‰ ˆ1 23 (1)$Î# ‰‘ ‰ "‘ œ "3 ˆ4 16 3 3 œ 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Practice Exercises " x# ,
15. f(x) œ x, g(x) œ œ
'
2
1
"‰ x#
ˆx
'
œ
Š 42
2
1
Šx
'
b
[f(x) g(x)] dx
a
#
#
dx œ ’ x# "x “ œ ˆ #4 #" ‰ ˆ #" 1‰ œ 1 "
" Èx
16. f(x) œ x, g(x) œ œ
a œ 1, b œ 2 Ê A œ
" Èx ‹dx
, a œ 1, b œ 2 Ê A œ #
œ ’ x# 2Èx“
'
b
a
[f(x) g(x)] dx
#
" 7 4 È 2 #
2È2‹ ˆ "# 2‰ œ
'
#
17. f(x) œ ˆ1 Èx‰ , g(x) œ 0, a œ 0, b œ 1 Ê A œ œ
'
1
ˆ1 2x"Î# x‰ dx œ ’x 43 x$Î#
0
" x# # “!
#
x% #
" x( 7 “!
œ1
" #
" 7
œ
b
a
œ1
18. f(x) œ a1 x$ b , g(x) œ 0, a œ 0, b œ 1 Ê A œ œ ’x
'
a
b
'
[f(x) g(x)] dx œ 4 3
" #
œ
" 6
1
0
ˆ1 Èx‰# dx œ
(6 8 3) œ
[f(x) g(x)] dx œ
'
0
1
œ2
'
0
3
'
d
c
[f(y) g(y)] dy œ
y# dy œ
2 3
'
#
a1 x$ b dx œ
'
1
0
9 14
3
0
a2y# 0b dy
$
cy$ d ! œ 18
20. f(y) œ 4 y# , g(y) œ 0, c œ 2, d œ 2 Ê Aœ œ ’4y
'
c
d
[f(y) g(y)] dy œ
# y$ 3 “ #
œ 2 ˆ8
8‰ 3
œ
'
0
1
ˆ1 2Èx x‰ dx
" 6
19. f(y) œ 2y# , g(y) œ 0, c œ 0, d œ 3 Ê Aœ
313
'c a4 y# b dy 2
2
32 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
a1 2x$ x' b dx
314
Chapter 5 Integration y# 4
21. Let us find the intersection points:
y2 4
œ
Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0 Ê y œ 1 or y œ 2 Ê c œ 1, d œ 2; f(y) œ Ê Aœ
'
d
c
y2 4
2
#
1
œ
" 4
'c ay 2 y# b dy œ 4" ’ y#
œ
" 4
ˆ 4# 4 38 ‰ ˆ "# 2 3" ‰‘ œ
#
1
y# 4
'c Š y 4 2 y4 ‹ dy
[f(y) g(y)] dy œ
2
, g(y) œ
2y 9 8
y# 4 4
22. Let us find the intersection points:
# y$ 3 “ "
œ
y 16 4
Ê y# y 20 œ 0 Ê (y 5)(y 4) œ 0 Ê y œ 4 or y œ 5 Ê c œ 4, d œ 5; f(y) œ Ê Aœ
'
d
c
[f(y) g(y)] dy œ
y 16 4
, g(y) œ
y# 4 4
'c Š y 416 y 4 4 ‹ dy 5
#
4
œ
" 4
'c ay 20 y# b dy œ "4 ’ y#
œ
" 4 " 4
125 ‰ ˆ 25 ‰‘ ˆ "#6 80 64 # 100 3 3 ˆ 9# 180 63‰ œ 4" ˆ 9# 117‰ œ 8" (9 234) œ
œ
5
#
20y
4
23. f(x) œ x, g(x) œ sin x, a œ 0, b œ Ê Aœ
'
b
a
#
[f(x) g(x)] dx œ
œ ’ x# cos x“
1Î% !
#
œ Š 31#
'
1Î4
(x sin x) dx
1
24. f(x) œ 1, g(x) œ ksin xk , a œ 1# , b œ Ê Aœ œ
'c
œ2
0
'
b
a
[f(x) g(x)] dx œ
(1 sin x) dx
1Î2 1Î2
'
0
'
0
1Î2
243 8
1 4
0
È2 # ‹
& y$ 3 “ %
'c
1Î2
1Î2
1 2
a1 ksin xkb dx
(1 sin x) dx 1Î#
(1 sin x) dx œ 2[x cos x]!
œ 2 ˆ 1# 1‰ œ 1 2 25. a œ 0, b œ 1, f(x) g(x) œ 2 sin x sin 2x Ê Aœ
'
0
1
(2 sin x sin 2x) dx œ 2 cos x
cos 2x ‘ 1 # !
œ 2 † (1) "# ‘ ˆ2 † 1 "# ‰ œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Practice Exercises
315
26. a œ 13 , b œ 13 , f(x) g(x) œ 8 cos x sec# x
'c
1Î3
Ê Aœ œ Š8 †
1Î3
È3 #
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x]1Î$
È3‹ Š8 †
È3 #
È3‹ œ 6È3
27. f(y) œ Èy, g(y) œ 2 y, c œ 1, d œ 2
'
Ê Aœ œ
'
1
2
d
c
[f(y) g(y)] dy œ
'
2
1
Èy (2 y)‘ dy
ˆÈy 2 y‰ dy œ ’ 23 y$Î# 2y
œ Š 43 È2 4 2‹ ˆ 23 2 "# ‰ œ
4 3
# y# # “"
È2
7 6
œ
8 È 2 7 6
28. f(y) œ 6 y, g(y) œ y# , c œ 1, d œ 2 Ê Aœ
'
œ ’6y
y# #
œ4
c
7 3
d
[f(y) g(y)] dy œ
" #
# y$ 3 “"
œ
'
2
1
a6 y y# b dy
œ ˆ12 2 38 ‰ ˆ6
24 14 3 6
œ
" #
3" ‰
13 6
29. f(x) œ x$ 3x# œ x# (x 3) Ê f w (x) œ 3x# 6x œ 3x(x 2) Ê f w œ ± ± ! # Ê f(0) œ 0 is a maximum and f(2) œ 4 is a minimum. A œ 30. A œ
'
a
0
ˆa"Î# x"Î# ‰# dx œ
œ a# ˆ1
4 3
"# ‰ œ
a# 6
'
a
0
&Î$
A# œ
" y# “ # !
œ
" 10
(6 8 3) œ
'
0
1
%
&Î$
1
Ê the total area is A" A# œ
!
a
x# # “0
œ a# 34 Èa † aÈa
a# 6
ˆy#Î$ y‰ dy
! y# # “ "
œ
$
‰ ax$ 3x# b dx œ ’ x4 x$ “ œ ˆ 81 4 27 œ
; the area below the x-axis is
'c ˆy#Î$ y‰ dy œ ’ 3y5 0
0
3
ˆa 2Èa x"Î# x‰ dx œ ’ax 43 Èa x$Î#
31. The area above the x-axis is A" œ œ ’ 3y5
'
11 10
6 5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
a# #
27 4
316
Chapter 5 Integration
32. A œ
'
1Î4
0
'
31Î2
51Î4
(cos x sin x) dx
'
51Î4
1Î4
(sin x cos x) dx 1Î%
(cos x sin x) dx œ [sin x cos x]! &1Î%
$1Î#
[ cos x sin x]1Î% [sin x cos x]&1Î% œ ’Š
È2 #
È2 # ‹
(0 1)“ ’Š
’(1 0) Š
33. y œ x# 34. y œ
'
x
0
'
x
1
" t
È2 #
dt Ê
dy dx
È2 # ‹“
xœ0 Ê yœ
'
0
0
È2 # ‹
È2 #
Š
œ
8È 2 #
2 œ 4È2 2
" x
Ê
d# y dx#
œ 2x
ˆ1 2Èsec t‰ dt Ê
È2 #
œ2
" x#
; y(1) œ 1
œ 1 2Èsec x Ê
dy dx
d# y dx#
ˆ1 2Èsec t‰ dt œ 0 and x œ 0 Ê
dy dx
36. y œ
'c È2 sin# t dt 2 so that dydx œ È2 sin# x; x œ 1
dt 3 Ê
dy dx
œ
;xœ5 Ê yœ
sin x x
5
5
sin t t
1
1
" t
dt œ 1 and yw (1) œ 2 1 œ 3
œ 1 2Èsec 0 œ 3
'
sin t t 5
'
œ 2 ˆ "# ‰ (sec x)"Î# (sec x tan x) œ Èsec x (tan x);
35. y œ
x
'
È2 # ‹“
dt 3 œ 3
x
Ê yœ
1
'cc È2 sin# t dt 2 œ 2 1
1
37. Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx
' 2(cos x)"Î# sin x dx œ ' 2u"Î# ( du) œ 2 ' u"Î# du œ 2 Š u"Î#" ‹ C œ 4u"Î# C œ 4(cos x)"Î# C #
38. Let u œ tan x Ê du œ sec# x dx
' (tan x)$Î# sec# x dx œ ' u$Î# du œ ˆu"Î#"‰ C œ 2u"Î# C œ (tanx)2 "Î# C #
39. Let u œ 2) 1 Ê du œ 2 d) Ê
" #
du œ d)
' [2) 1 2 cos (2) 1)] d) œ ' (u 2 cos u) ˆ "# du‰ œ u4
#
#
œ ) ) sin (2) 1) C, where C œ C" 40. Let u œ 2) 1 Ê du œ 2 d) Ê
'Š œ
41.
42.
" #
" È 2 ) 1
" #
2 sec# (#) 1)‹ d) œ
" 4
sin u C" œ
(2)1)# 4
sin (2) 1) C"
is still an arbitrary constant
du œ d)
' Š È"u 2 sec# u‹ ˆ #" du‰ œ #" ' ˆu"Î# 2 sec# u‰ du
"Î#
Š u " ‹ "# (2 tan u) C œ u"Î# tan u C œ (2) 1)"Î# tan (2) 1) C #
' ˆt 2t ‰ ˆt 2t ‰ dt œ ' ˆt# t4 ‰ dt œ ' at# 4t# b dt œ t3$ 4 Š t 1" ‹ C œ t3$ 4t C #
t # ' (t1)t%#1 dt œ ' t#t%2t dt œ ' ˆ t"# t2$ ‰ dt œ ' at# 2t$ b dt œ (t 1)" 2 Š # ‹ C œ "t t"# C
43. Let u œ #t$Î# Ê du œ $Èt dt Ê "$ du œ Èt dt
' Èt sin ˆ#t$Î# ‰dt œ "$ ' sin u du œ "$ cos u C œ "$ cosˆ#t$Î# ‰ C Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Practice Exercises 44. Let u œ " sec ) Ê du œ sec ) tan) d) Ê
317
' sec ) tan) È" sec ) d) œ ' u"Î# du œ #$ u$Î# C œ #$ a" sec )b$Î# C
'c a3x# 4x 7b dx œ cx$ 2x# 7xd "" œ c1$ 2(1)# 7(1)d c(1)$ 2(1)# 7(1)d œ 6 (10) œ 16 1
45.
1
46.
'
47.
'
48.
'
49.
'
1
0
"
a8s$ 12s# 5b ds œ c2s% 4s$ 5sd ! œ c2(1)% 4(1)$ 5(1)d 0 œ 3
2
4 # 1 v 27
1
'
dv œ
2
#
4v# dv œ c4v" d " œ ˆ #4 ‰ ˆ 14 ‰ œ 2
1
#(
x%Î$ dx œ 3x"Î$ ‘ " œ 3(27)"Î$ ˆ3(1)"Î$ ‰ œ 3 ˆ "3 ‰ 3(1) œ 2
4
dt 1 tÈt
œ
'
4
œ
dt
t$Î#
1
'
4
1
%
50. Let x œ 1 Èu Ê dx œ
'
4
ˆ1 Èu‰"Î# Èu
1
'
du œ
3
2
2 È4
t$Î# dt œ 2t"Î# ‘ " œ " #
u"Î# du Ê 2 dx œ
du Èu
(2) È1
œ1
; u œ 1 Ê x œ 2, u œ 4 Ê x œ 3
$
x"Î# (2 dx) œ 2 ˆ 23 ‰ x$Î# ‘ # œ
4 3
ˆ3$Î# ‰ 43 ˆ2$Î# ‰ œ 4È3 83 È2 œ
4 3
Š3È3 2È2‹
51. Let u œ 2x 1 Ê du œ 2 dx Ê 18 du œ 36 dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 3
'
'
1
36 dx $ 0 (2x1)
œ
3
# $
$
9 ‘ ˆ 9 ‰ ˆ 9 ‰ 18u$ du œ ’ "8u 2 “ œ u # " œ 3 # 1 # œ 8 "
1
52. Let u œ 7 5r Ê du œ 5 dr Ê "5 du œ dr; r œ 0 Ê u œ 7, r œ 1 Ê u œ 2
'
1
dr 3 # 0 È(7 5r)
œ
'
1
0
(7 5r)#Î$ dr œ
'
2
#
u#Î$ ˆ 5" du‰ œ 5" 3u"Î$ ‘ ( œ
7
&Î#
" 8 !
#
$Î%
53. Let u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê #3 du œ x"Î$ dx; x œ
'
1
1Î8
œ
x"Î$ ˆ1 x#Î$ ‰
$Î#
'
dx œ
0
3Î4
u$Î# ˆ #3 du‰ œ ’ˆ 23 ‰ Š u 5 ‹“
1Î2
0
x$ a1 9x% b
" ˆ 25 ‰ œ 18 16
"Î#
$Î#
dx œ
'
0
1
sin# 5r dr œ
56. Let u œ 4t
'
1Î%
0
œ
1 8
1 4
" 36
#Î$
œ
3 4
, x œ 1 Ê u œ 1 1#Î$ œ 0
! &Î# œ 53 u&Î# ‘ $Î% œ 53 (0)&Î# ˆ 53 ‰ ˆ 43 ‰
'
51
0
" 16
" 16
" 5
"Î# #
œ
1 8
1Î4
"
%
Ê u œ 1 9 ˆ #" ‰ œ
25 16
#&Î"'
" "Î# ‘ œ 18 u "
du œ dr; r œ 0 Ê u œ 0, r œ 1 Ê u œ 51
Ê du œ 4 dt Ê
'
#&Î"'
" #
" 90
asin# ub ˆ "5 du‰ œ
31Î4
du œ x$ dx; x œ 0 Ê u œ 1, x œ
" " u$Î# ˆ 36 du‰ œ ’ 36 Š u " ‹“
" ˆ 18 (1)"Î# ‰ œ
cos# ˆ4t 14 ‰ dt œ
25Î16
1
55. Let u œ 5r Ê du œ 5 dr Ê
'
Ê u œ 1 ˆ 8" ‰
27È3 160
54. Let u œ 1 9x% Ê du œ 36x$ dx Ê
'
3 3 7È 2‹ ŠÈ
3 5
" 4
" 5
2u
sin 2u ‘ &1 4 !
œ ˆ 1#
sin 101 ‰ #0
du œ dt; t œ 0 Ê u œ 14 , t œ
acos# ub ˆ "4 du‰ œ
" 4
u2
sin 2u ‘ $1Î% 4 1Î%
œ
ˆ0 1 4 " 4
sin 0 ‰ 20
Ê uœ Š 381
œ
1 #
31 4
sin ˆ 3#1 ‰ ‹ 4
4" Š 18
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
sin ˆ 1# ‰ ‹ 4
318
Chapter 5 Integration
'
1Î$
57.
'
31Î4
58.
0
1Î$
sec# ) d) œ [tan )]!
1Î4
31
1
1 3
tan 0 œ È3
$1Î%
csc# x dx œ [cot x]1Î% œ ˆ cot
59. Let u œ
'
œ tan
cot#
x 6 x 6
Ê du œ dx œ
'
" 6
ˆ cot 14 ‰ œ 2
dx Ê 6 du œ dx; x œ 1 Ê u œ 16 , x œ 31 Ê u œ
1Î2
1Î6
31 ‰ 4
'
6 cot# u du œ 6
1Î2
1 # 1Î#
acsc# u 1b du œ [6(cot u u)]1Î' œ 6 ˆ cot
1Î6
1 #
1# ‰ 6 ˆcot
1 6
16 ‰
œ 6È3 21 60. Let u œ
'
1
0
tan#
) 3 ) 3
Ê du œ d) œ
'
1
0
" 3
d) Ê 3 du œ d); ) œ 0 Ê u œ 0, ) œ 1 Ê u œ
ˆsec#
) 3
1‰ d) œ
'
1Î3
0
1 3 1Î$
3 asec# u 1b du œ [3 tan u 3u]!
œ 3 tan
1 3
3 ˆ 13 ‰‘ (3 tan 0 0)
œ 3È3 1
'c
sec x tan x dx œ [sec x]!1Î$ œ sec 0 sec ˆ 13 ‰ œ 1 2 œ 1
'
csc z cot z dz œ [csc z]1Î% œ ˆ csc
0
61.
62.
1Î3
31Î4
1Î4
$1Î%
31 ‰ 4
ˆ csc 14 ‰ œ È2 È2 œ 0
63. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
'
1Î2
0
'
5(sin x)$Î# cos x dx œ
1
0
1 #
Ê uœ1
"
"
5u$Î# du œ 5 ˆ 25 ‰ u&Î# ‘ ! œ 2u&Î# ‘ ! œ 2(1)&Î# 2(0)&Î# œ 2
64. Let u œ 1 x# Ê du œ 2x dx Ê du œ 2x dx; x œ 1 Ê u œ 0, x œ 1 Ê u œ 0
'c
1 1
'
2x sin a1 x# b dx œ
0
sin u du œ 0
0
65. Let u œ sin 3x Ê du œ 3 cos 3x dx Ê
'c ÎÎ
1 2
15 sin% 3x cos 3x dx œ
1 2
'
1
1
" 3
du œ cos 3x dx; x œ 1# Ê u œ sin ˆ 3#1 ‰ œ 1, x œ
15u% ˆ "3 du‰ œ
'
1
1
1 #
"
5u% du œ cu& d " œ (1)& (1)& œ 2
66. Let u œ cos ˆ x# ‰ Ê du œ "# sin ˆ x# ‰ dx Ê 2 du œ sin ˆ x# ‰ dx; x œ 0 Ê u œ cos ˆ 0# ‰ œ 1, x œ
'
21Î3
0
cos% ˆ x# ‰ sin ˆ x# ‰ dx œ
'
1
1Î2
$
u% (2 du) œ ’2 Š u3 ‹“
67. Let u œ 1 3 sin# x Ê du œ 6 sin x cos x dx Ê
'
1Î2
0
3 sin x cos x È1 3 sin# x
dx œ
'
4
1
" Èu
ˆ #" du‰ œ
'
4
1
68. Let u œ 1 7 tan x Ê du œ 7 sec# x dx Ê Ê u œ 1 7 tan
'
0
1Î4
#
sec x (1 7 tan x)#Î$
1 4
" #
Ê u œ sin ˆ 3#1 ‰ œ 1
"Î# "
œ
2 3
ˆ "# ‰$ 32 (1)$ œ
2 3
"Î# #
"
1 #
21
Ê u œ cos Š #3 ‹ œ
(8 1) œ
du œ 3 sin x cos x dx; x œ 0 Ê u œ 1, x œ %
21 3
14 3
Ê u œ 1 3 sin#
%
" #
u"Î# du œ ’ 2" Š u " ‹“ œ u"Î# ‘ " œ 4"Î# 1"Î# œ 1
" 7
du œ sec# x dx; x œ 0 Ê u œ 1 7 tan 0 œ 1, x œ
1 4
œ8
dx œ
'
1
8
"
u#Î$
ˆ 7" du‰ œ
'
1
8
" 7
"Î$
)
3
"
)
u#Î$ du œ ’ 7" Š u " ‹“ œ 37 u"Î$ ‘ " œ
3 7
(8)"Î$ 37 (1)"Î$ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" #
3 7
1 #
œ4
Chapter 5 Practice Exercises 69. Let u œ sec ) Ê du œ sec ) tan ) d); ) œ 0 Ê u œ sec 0 œ 1, ) œ
'
1Î3
0
tan ) È2 sec )
" È2
œ
'
d) œ
"Î#
#
#
"
'
1Î3
0
’ ˆu " ‰ “ œ ’
1Î3
sec ) tan ) sec ) tan ) d) œ È2 (sec ))$Î# sec ) È2 sec ) 0 # 2 2 2 È2u “ œ È2(2) Š È2(1) ‹ œ " cos Èt 2È t
70. Let u œ sin Èt Ê du œ ˆcos Èt‰ ˆ "# t"Î# ‰ dt œ #
1 4
tœ
'
1 Î4 #
Ê u œ sin
71. (a) av(f) œ
'
" 1 (1)
'c
" k (k)
(b) av(f) œ
1
" 1Î2 Èu
1 1
'c
k k
(2 du) œ 2
" 30
73. favw œ
" #k
0
3
0
0
" ba
'
a
b
1
" È2 u$Î#
du œ
" È2
'
1
1 3 2
œ2 u$Î# du
È2 1 dt Ê 2 du œ
cos Èt Èt
dt; t œ
1# 36
Ê u œ sin
1 6
œ
" #
,
a
" b a
" b a
"
#
’ mx2 bx“
Èa x"Î# dx œ
[f(x)]ab œ
"
’ mx2 bx“
È3 x"Î# dx œ
0
f w (x) dx œ
"
u"Î# du œ 4Èu‘ "Î# œ 4È1 4É #" œ 2 Š2 È2‹ #
(mx b) dx œ
3
1
1Î2
" #
a
(b) yav
'
(mx b) dx œ
' È3x dx œ "3 ' œ a " 0 ' Èax dx œ "a '
72. (a) yav œ
'
Ê u œ sec
œ1
dt œ
cos Èt Ét sin Èt
1# Î36
1 #
d) œ
2
1 3
319
k
ck
#
#
m(1) b(1)‹“ œ ’Š m(1) 2 b(1)‹ Š #
œ
" #
œ
" #k
#
" #
#
m(k) b(k)‹“ œ ’Š m(k) 2 b(k)‹ Š #
(2b) œ b " #k
(2bk) œ b
È3 3
23 x$Î# ‘ $ œ !
È3 3
23 (3)$Î# 23 (0)$Î# ‘ œ
È3 3
Š2È3‹ œ 2
Èa a
32 x$Î# ‘ a œ !
Èa a
ˆ 32 (a)$Î# 32 (0)$Î# ‰ œ
Èa a
ˆ 32 aÈa‰ œ
[f(b) f(a)] œ
f(b) f(a) ba
2 3
a
so the average value of f w over [aß b] is the
slope of the secant line joining the points (aß f(a)) and (bß f(b)), which is the average rate of change of f over [aß b]. 74. Yes, because the average value of f on [aß b] is and the average value of the function is
" #
'
a
" ba
'
a
b
f(x) dx. If the length of the interval is 2, then b a œ 2
b
f(x) dx.
75. We want to evaluate " $'& !
'
$'&
!
f(x) dx œ
" $'&
'
$'&
!
#1 Œ$(sin” $'& ax "!"b• #&dx œ
#1 Notice that the period of y œ sin” $'& ax "!"b• is
length 365. Thus the value of
76.
" '(&#!
œ
'#
'(& !
$( $'&
'
$'&
!
" '&& Œ”)Þ#(a'(&b
#'a'(&b #†"!&
#1 $'&
"Þ)(a'(&b $†"!&
$
'
!
$'&
#1 sin” $'& ax "!"b•dx
#& $'&
'
$'&
!
dx
œ $'& and that we are integrating this function over an iterval of
#1 ax "!"b•dx sin” $'&
a)Þ#( "!& a#'T "Þ)(T# bbdT œ #
#1
$( $'&
" '&& ”)Þ#(T
• ”)Þ#(a#!b
#& $'&
#'T# #†"!&
#'a#!b #†"!&
#
'
!
$'&
dx is
"Þ)(T$ $†"!& • "Þ)(a#!b $†"!&
$
$( $'&
†!
#& $'&
† $'& œ #&.
'(& #!
• ¸
" '&& a$(#%Þ%%
"'&Þ%!b
œ &Þ%$ œ the average value of Cv on [20, 675]. To find the temperature T at which Cv œ &Þ%$, solve &Þ%$ œ )Þ#( "!& a#'T "Þ)(T# b for T. We obtain "Þ)(T# #'T #)%!!! œ ! ÊTœ
#' „ Éa#'b# %a"Þ)(ba#)%!!!b È#"#%**' œ #' „ $Þ(% . #a"Þ)(b ‰
So T œ $)#Þ)# or T œ $*'Þ(#. Only T œ $*'Þ(# lies in the
interval [20, 675], so T œ $*'Þ(# C. 77.
dy dx
œ È2 cos$ x
78.
dy dx
œ È2 cos3 a7x2 b †
d 2 dx a7x b
œ 14xÈ2 cos3 a7x2 b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
320
Chapter 5 Integration
79.
dy dx
œ
d dx Œ
80.
dy dx
œ
d dx Œ
' x $ ' t dt œ $'x %
1
%
'sec# x t " " dt œ dxd Œ'#sec x t " " dt œ sec "x " dxd asec xb œ sec" xsectan xx #
#
#
#
81. Yes. The function f, being differentiable on [aß b], is then continuous on [aß b]. The Fundamental Theorem of Calculus says that every continuous function on [aß b] is the derivative of a function on [aß b]. 82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on [aß b], then
'
a
b
f(x) dx œ F(b) F(a). In particular, if F(x) is an antiderivaitve of È1 x% on [0ß 1], then
'
0
1
È1 x% dx
œ F(1) F(0). 83. y œ
'x1 È1 t# dt œ '1x È1 t# dt
84. y œ
'
0
" # cos x 1 t
dt œ
'
cos x
0
" 1 t#
dt Ê
Ê
dy dx
œ
d dx
”
dy dx
œ
d dx
”
'
'1x È1 t# dt• œ dxd ”'1x È1 t# dt• œ È1 x#
cos x
0
" d ‰ ˆ dx œ ˆ 1 cos (cos x)‰ œ ˆ sin"# x ‰ ( sin x) œ #x
" 1 t#
" sin x
d dt• œ dx ”
'
cos x
0
" 1 t#
dt•
œ csc x
85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each interval by averaging the widths at top and bottom. This gives the estimate A ¸ "& † ˆ ! # $' $' # &% &% # &" &" #%*Þ& %*Þ&# &% &% #'%Þ% '%Þ% # '(Þ& '(Þ&# %# ‰
A ¸ &*'" ft# . The cost is Area † ($2.10/ft# ) ¸ a5961 ft# b a$2.10/ft# b œ $12,518.10 Ê the job cannot be done for $11,000.
86. (a) Before the chute opens for A, a œ 32 ft/sec# . Since the helicopter is hovering, v! œ 0 ft/sec
Ê v œ ' 32 dt œ 32t v! œ 32t. Then s! œ 6400 ft Ê s œ ' 32t dt œ 16t# s! œ 16t# 6400.
At t œ 4 sec, s œ 16(4)# 6400 œ 6144 ft when A's chute opens;
(b) For B, s! œ 7000 ft, v! œ 0, a œ 32 ft/sec# Ê v œ ' 32 dt œ 32t v! œ 32t Ê s œ ' 32t dt œ 16t# s! œ 16t# 7000. At t œ 13 sec, s œ 16(13)# 7000 œ 4296 ft when B's chute opens;
(c) After the chutes open, v œ 16 ft/sec Ê s œ ' 16 dt œ 16t s! . For A, s! œ 6144 ft and for B,
s! œ 4296 ft. Therefore, for A, s œ 16t 6144 and for B, s œ 16t 4296. When they hit the ground, 4296 s œ 0 Ê for A, 0 œ 16t 6144 Ê t œ 6144 16 œ 384 seconds, and for B, 0 œ 16t 4296 Ê t œ 16 œ 268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opens Ê B hits the ground first. CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES 1. (a) Yes, because
'
0
1
f(x) dx œ
(b) No. For example,
'
0
1
(b) True:
'
0
1
7f(x) dx œ
" 7
(7) œ 1
"
8x dx œ c4x# d ! œ 4, but
'
0
1
"
È8x dx œ ’2È2 Š x$Î# œ 3 ‹“ !
#
' f(x) dx œ ' f(x) dx œ 3 'c [f(x) g(x)] dx œ 'c f(x) dx 'c g(x) dx œ 'c f(x) dx ' 2
2. (a) True:
" 7
5
4È 2 3
ˆ1$Î# 0$Î# ‰ œ
4È 2 3
Á È4
5
2
5
5
5
2
2
2
2
2
2
5
f(x) dx
'c g(x) dx œ 4 3 2 œ 9 5
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Additional and Advanced Exercises
'c f(x) dx œ 4 3 œ 7 2 œ 'c g(x) dx
(c) False:
5
5
2
2
'c [f(x) g(x)] dx 0 Ê 'c [g(x) f(x)] dx 0. Ê ' [g(x) f(x)] dx 0 which is a contradiction. c
œ
" a
'
0
sin ax a
x
sin ax a
'
0
'
d
Πdx
œ cos ax
0
'
x
2
5
2
x
x
0
0
x
0
0
'
f(t) cos at dt sin ax
f(t) cos at dt
0
2
x
f(t) cos at dt x
5
' f(t) sin ax cos at dt "a ' f(t) cos ax sin at dt cos ax ' ' f(t) cos at dt f(t) sin at dt Ê dy a dx œ cos ax Œ " a
f(t) sin a(x t) dt œ x
5
Ê
On the other hand, f(x) Ÿ g(x) Ê [g(x) f(x)] 0 3. y œ
x
0
f(t) sin at dt
'
(f(x) cos ax) sin ax
sin ax a
cos ax a x
'
d
Πdx
x
f(t) sin at dt
0
f(t) sin at dt
0
cos ax a
(f(x) sin ax)
' f(t) cos at dt sin ax ' f(t) sin at dt. Next, d y ' f(t) cos at dt (cos ax) Œ dxd ' f(t) cos at dt a cos ax ' dx œ a sin ax Ê
x
œ cos ax
dy dx
x
0 x
#
#
0
x
0
'
d (sin ax) Πdx
a cos ax
'
0
x
0
x
0
4. x œ
'
" #
œ
'
x
0
" 0 È1 4t#
'
d dx
(x) œ
Š dy dx ‹ Ê
"Î#
x
0
f(t) cos at dt
dt Ê
" È14y#
a1 4y# b
x
0
'
'
d dx
x
0
y
0
'
" È1 4t#
dy 4y Š dx ‹
(8y) Š dy dx ‹ œ
È1 4y#
0
'
x
0
x
f(t) cos at dt a cos ax
'
0
x
f(t) sin at dt f(x).
f(t) cos at dt f(x)
f(t) sin at dt œ f(x). Note also that yw (0) œ y(0) œ 0.
dt œ
œ È1 4y# . Then
dy dx
f(t) sin at dt
f(t) cos at dt (cos ax)f(x) cos ax
f(t) sin at dt a sin ax
cos ax a
x
0
f(t) sin at dt (sin ax)f(x) sin ax œ a sin ax
y
Ê 1œ
'
f(t) sin at dt œ a sin ax
Therefore, yww a# y œ a cos ax a# Œ sinaax
321
œ
'
d dy
”
d# y dx#
œ
4y ˆÈ1 4y# ‰ È1 4y#
y
0
" È1 4t#
d dx
ˆÈ1 4y# ‰ œ
dt• Š dy dx ‹ from the chain rule
œ 4y. Thus
d# y dx#
ˆÈ1 4y# ‰ Š dy dx ‹
d dy
œ 4y, and the constant of
proportionality is 4.
5. (a)
'
0
x#
f(t) dt œ x cos 1x Ê
Ê f ax# b œ
(b)
'
0
f(x)
d dx
cos 1x 1x sin 1x . 2x $
t# dt œ ’ t3 “
f(x)
!
œ
" 3
'
x#
0
f(t) dt œ cos 1x 1x sin 1x Ê f ax# b (2x) œ cos 1x 1x sin 1x cos 21 21 sin 21 4
Thus, x œ 2 Ê f(4) œ
(f(x))$ Ê
" 3
" 4
œ
3 (f(x))$ œ x cos 1x Ê (f(x))$ œ 3x cos 1x Ê f(x) œ È 3x cos 1 x
3 3 Ê f(4) œ È 3(4) cos 41 œ È 12
6.
'
a
0
f(x) dx œ
a# #
a #
sin a
Ê f(a) œ Fw (a) œ a 7.
'
1
b
" #
1 #
cos a. Let F(a) œ
sin a
a #
cos a
f(x) dx œ Èb# 1 È2 Ê f(b) œ
d db
1 #
'
a
0
f(t) dt Ê f(a) œ Fw (a). Now F(a) œ
sin a Ê f ˆ 1# ‰ œ
'
b
1
f(x) dx œ
" #
side of the equation is:
d dx
”
'
0
x
f(u)(x u) du• œ
d dx
”
' ”'
d dx
0
'
0
x
ab# 1b
x
8. The derivative of the left side of the equation is:
1 #
0
u
" #
1 #
sin
"Î#
d dx
#
#
(2b) œ
f(t) dt• du• œ
f(u) x du
ˆ1‰
'
0
'
0
cos
b È b# 1
1 #
a# #
1 #
a #
sin a
sin
1 #
Ê f(x) œ
œ
1 #
1 #
cos a
" #
1 #
x È x# 1
x
f(t) dt; the derivative of the right
x
u f(u) du
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
" #
322
Chapter 5 Integration œ
d dx
œ
'
'
”x
0
x
f(u) du•
'
d dx
x
0
u f(u) du œ
'
x
0
d f(u) du x ” dx
dy dx
0
x
f(u) du• xf(x) œ
'
0
x
f(u) du xf(x) xf(x)
x
0
f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0
when x œ 0, the constant must be 0. Therefore,
9.
'
' ”' x
0
0
u
f(t) dt• du œ
'
0
x
f(u)(x u) du.
œ 3x# 2 Ê y œ ' a3x# 2b dx œ x$ 2x C. Then (1ß 1) on the curve Ê 1$ 2(1) C œ 1 Ê C œ 4
Ê y œ x$ 2x 4 10. The acceleration due to gravity downward is 32 ft/sec# Ê v œ ' 32 dt œ 32t v! , where v! is the initial
velocity Ê v œ 32t 32 Ê s œ ' (32t 32) dt œ 16t# 32t C. If the release point, at t œ !, is s œ 0, then C œ 0 Ê s œ 16t# 32t. Then s œ 17 Ê 17 œ 16t# 32t Ê 16t# 32t 17 œ 0. The discriminant of this quadratic equation is 64 which says there is no real time when s œ 17 ft. You had better duck.
11.
'c f(x) dx œ 'c x#Î$ dx ' œ œ œ
12.
3
0
8
8
3
4 dx
0
35 x&Î$ ‘ ! [4x]!$ ) ˆ0 35 (8)&Î$ ‰ (4(3) 36 5
0) œ
'c f(x) dx œ 'c Èx dx ' 3
0
4
4
3
$
$
œ 23 (x)$Î# ‘ % ’ x3 4x“ œ 0 ˆ (4) 2 3
œ
13.
'
3œ
7 3
g(t) dt œ
'
16 3
2
0
#
1
0
$Î# ‰‘
t dt
"
$ ’ Š 33
'
1
2
12
ax# 4b dx
0
!
96 5
!
4(3)‹ 0 “
sin 1t dt
#
œ ’ t2 “ 1" cos 1t‘ " !
œ ˆ "# 0‰ 1" cos 21 ˆ 1" cos 1‰‘ " #
œ
14.
'
0
2
2 1
h(z) dz œ
'
0
1
È1 z dz
'
1
2
(7z 6)"Î$ dz
" # 3 œ 23 (1 z)$Î# ‘ ! 14 (7z 6)#Î$ ‘ " œ 23 (1 1)$Î# ˆ 23 (1 0)$Î# ‰‘ 3 14 (7(2) 6)#Î$ 3 ‰ 55 œ 23 ˆ 67 14 œ 42
3 14
(7(1) 6)#Î$ ‘
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Additional and Advanced Exercises
'c f(x) dx œ 'cc dx 'c a1 x# b dx ' 2
15.
1
2
1
2
1
" x$ 3 “ "
œ [x]" # ’x
1$ 3‹
16.
ˆ 23 ‰ 4 2 œ
2 3
'c h(r) dr œ 'c r dr ' 2
0
1
1
#
œ ’ r2 “
!
"
" #
’2(2) 2(1)“
13 3
a1 r# b dr
'
2
dr
1
"
!
2 3
Š Š1
1œ
17. Ave. value œ œ
(1)$ 3 ‹•
Š1
’r r3 “ [r]#"
(1)# # ‹
œ "#
1
0
$
œ Š0
2 dx
1
[2x]#"
œ a1 (2)b ”Š1 œ1
2
" ba
1$ 3‹
0‹ a2 1b
7 6
'
b
a
#
f(x) dx œ
" #0
#
'
2
0
#
’Š 1# 0‹ Š 2# 2‹ Š 1# 1‹“ œ
18. Ave. value œ
" ba
'
b
a
f(x) dx œ
" 30
'
3
0
" #
f(x) dx œ
”
'
1
0
x dx
'
2
1
(x 1) dx• œ
" #
"
#
#
’ x2 “ #" ’ x2 x“ !
# "
" #
f(x) dx œ
" 3
”
'
0
1
dx
'
1
2
'
0 dx
19. Let f(x) œ x& on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
2
10 n
3
dx• œ
" 3
[1 0 0 3 2] œ
œ "n . Then "n , n2 , á ,
_
&
n n
2 3
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ n" ‰ is the upper sum for
œ
j 1
_
&
! Š j ‹ ˆ " ‰ œ lim f(x) œ x& on [0ß 1] Ê n lim n n Ä_ nÄ_ jœ1 œ
'
1
'
"
x& dx œ ’ x6 “ œ !
0
" n
& & & ’ˆ n" ‰ ˆ n2 ‰ á ˆ nn ‰ “ œ n lim ’1 Ä_
&
2& á n& “ n'
" 6
20. Let f(x) œ x$ on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10 n
œ "n . Then "n , n2 , á ,
_
$
n n
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ n" ‰ is the upper sum for
_
$
! f(x) œ x on [0ß 1] Ê n lim Ä_ jœ1 œ
'
0
1
%
"
x$ dx œ ’ x4 “ œ !
œ
j 1
Š nj ‹
$
ˆ "n ‰
œ
lim " nÄ_ n
$ ’ˆ n" ‰
ˆ n2 ‰$
$
á ˆ nn ‰ “ œ n lim ’1 Ä_
$
2$ á n$ “ n%
" 4
21. Let y œ f(x) on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10 n
œ "n . Then "n , n2 , á ,
_
right-hand endpoints of the subintervals. Since f is continuous on [!ß 1], !
œ
j 1
_
! f Š j ‹ ˆ " ‰ œ lim y œ f(x) on [0ß 1] Ê n lim n n Ä_ nÄ_ jœ1 " 22. (a) n lim [2 4 6 á 2n] œ n lim Ä _ n# Ä_ on [0ß 1] (see Exercise 21)
" n
" n
f Š nj ‹ ˆ n" ‰
4 n
6 n
á
2n ‘ n
œ
'
0
1
are the
is a Riemann sum of
f ˆ n" ‰ f ˆ 2n ‰ á f ˆ nn ‰‘ œ
2n
n n
'
0
1
f(x) dx
"
2x dx œ cx# d! œ 1, where f(x) œ 2x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
323
324
Chapter 5 Integration
" (b) n lim c1"& 2"& á n"& d œ n lim Ä _ n"' Ä_ "& f(x) œ x on [0ß 1] (see Exercise 21)
"& "& "& ’ˆ n1 ‰ ˆ n2 ‰ á ˆ nn ‰ “ œ
" n
'
'
1
0
"'
"
" 16 ,
x"& dx œ ’ x16 “ œ !
where
1
" " (c) n lim sin 1n sin 2n1 á sin nn1 ‘ œ sin n1 dx œ 1" cos 1x‘ ! œ 1" cos 1 ˆ 1" cos 0‰ Ä_ n 0 œ 12 , where f(x) œ sin 1x on [0ß 1] (see Exercise 21) " (d) n lim c1"& 2"& á n"& d œ Šn lim Ä _ n"( Ä_ " ‰ œ 0 ˆ 16 œ 0 (see part (b) above) " n"&
(e) n lim Ä_
c1"& 2"& á n"& d œ n lim Ä_
œ Šn lim n‹ Šn lim Ä_ Ä_
" n"'
" " n ‹ Šn lim Ä _ n"'
c1"& 2"& á n"& d‹ œ Šn lim Ä_
" n‹
'
1
0
x"& dx
c1"& 2"& á n"& d
n n"'
c1"& 2"& á n"& d‹ œ Šn lim n‹ Ä_
'
1
0
x"& dx œ _ (see part (b) above)
23. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal to r, the radius of the circle) and a vertex angle of )n where )n œ 2n1 . The area of each triangle is An œ
" # # r
sin )n Ê the area of the polygon is A œ nAn œ nr# #
(b) n lim A œ n lim Ä_ Ä_
sin
21 n
œ n lim Ä_
n 1 r# 21
sin
21 n
nr# #
nr# 21 # sin n . 2 1 sin ˆ n ‰ # ˆ 2n1 ‰ œ a1r b
sin )n œ
œ n lim a1 r # b Ä_
lim
2 1 În Ä 0
sin ˆ 2n1 ‰ ˆ 2n1 ‰
œ 1 r#
24. Partition [!ß 1] into n subintervals, each of length ?x œ 1n with the points x! œ 0, x" œ n" , x# œ n2 , á , xn œ nn œ 1. The inscribed rectangles so determined have areas # # # f(x! ) ?x œ (0)# ?x, f(x" ) ?x œ ˆ "n ‰ ?x, f(x# ) ?x œ ˆ n2 ‰ ?x, á , f(xnc1 ) œ ˆ n n 1 ‰ ?x. The sum of these areas #
#
#
#
is Sn œ Š0# ˆ "n ‰ ˆ n2 ‰ á ˆ n n 1 ‰ ‹ ?x œ Š "n# lim
nÄ_
' ga$b œ '
25. (a) ga"b œ (b)
#
Sn œ n lim Š "n$ Ä_ 1
1
$
1
(c) ga"b œ
2# n$
á
(n ")# n$ ‹
œ
'
1
0
2# n#
x# dx œ
á 1$ 3
(n 1)# " n# ‹ n
œ
1# n$
2# n$
á
(n ")# n$ .
Then
œ 3" .
fatb dt œ ! fatb dt œ "# a#ba"b œ "
' fatb dt œ ' fatb dt œ "% a1 ## b œ 1 1
1
1
1
(d) gw axb œ faxb œ ! Ê x œ $, ", $ and the sign chart for gw axb œ faxb is relative maximum at x œ ". (e) gw a"b œ fa"b œ # is the slope and ga"b œ
± ± ± . So g has a 3 1 3
' fatb dt œ 1, by (c). Thus the equation is y 1 œ #ax "b 1
1
y œ #x # 1 . (f) gww axb œ f w axb œ ! at x œ " and gww axb œ f w axb is negative on a$ß "b and positive on a"ß "b so there is an inflection point for g at x œ ". We notice that gww axb œ f w axb ! for x on a"ß #b and gww axb œ f w axb ! for x on a#ß %b, even though gww a#b does not exist, g has a tangent line at x œ #, so there is an inflection point at x œ #. (g) g is continuous on Ò$ß %Ó and so it attains its absolute maximum and minimum values on this interval. We saw in (d) that gw axb œ ! Ê x œ $, ", $. We have that ga$b œ
' $ fatb dt œ '$" fatb dt œ 1## 1
#
œ #1
' fatb dt œ ! $ ga$b œ ' fatb dt œ " % ga%b œ ' fatb dt œ " "# † " † " œ "#
ga"b œ
1
1
1
1
Thus, the absolute minimum is #1 and the absolute maximum is !. Thus, the range is Ò#1ß !Ó.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Additional and Advanced Exercises
325
'x cos 2t dt " œ sin x ' x cos 2t dt " Ê yw œ cos x cosa2xb; when x œ 1 we have 1 yw œ cos 1 cosa21b œ " " œ #. And yww œ sin x 2sina2xb; when x œ 1, y œ sin 1 ' cos 2t dt " x 1
26. y œ sin x
1
œ ! ! " œ ". 27. f(x) œ
'
28. f(x) œ
'
œ
x
" t
1/x
dt Ê f w (x) œ
sin x
" 1 t# " sin x
cos x
" cos x
'ÈÈ sin t# dt
30. f(x) œ
'
y
x
xb3
y
dx ‰ d ˆ " ‰‰ ˆ dx Š "" ‹ ˆ dx œ x x
" x
x ˆ x"# ‰ œ
" x
" x
œ
2 x
" d " d ‰ ˆ dx ‰ ˆ dx dt Ê f w (x) œ ˆ 1 sin (sin x)‰ ˆ 1 cos (cos x)‰ œ #x #x
29. g(y) œ
2
" x
cos x cos# x
sin x sin# x
# # d ˆ d ˆ Èy‰‹ œ Ê gw (y) œ Šsin ˆ2Èy‰ ‹ Š dy 2Èy‰‹ Šsin ˆÈy‰ ‹ Š dy
sin 4y Èy
sin y 2È y
d ‰ t(5 t) dt Ê f w (x) œ (x 3)(& (x 3)) ˆ dx (x 3)‰ x(5 x) ˆ dx dx œ (x 3)(2 x) x(5 x)
œ 6 x x# 5x x# œ 6 6x. Thus f w (x) œ 0 Ê 6 6x œ 0 Ê x œ 1. Also, f ww (x) œ 6 ! Ê x œ 1 gives a maximum.
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326
Chapter 5 Integration
NOTES
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CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6.1 VOLUMES USING CROSS-SECTIONS (diagonal)# #
1. A(x) œ
œ
ˆ È x ˆ È x ‰ ‰ # #
œ 2x; a œ 0, b œ 4;
V œ 'a A(x) dx œ '0 2x dx œ cx# d ! œ 16 b
4
1(diameter)# 4
2. A(x) œ
œ
%
1 c a2 x # b x # d 4
#
œ
1c2 a1 x# bd 4
#
œ 1 a1 2x# x% b ; a œ 1, b œ 1;
V œ 'a A(x) dx œ 'c1 1 a1 2x# x% b dx œ 1 ’x 23 x$ b
1
" x& 5 “ "
#
œ 21 ˆ1
2 3
"5 ‰ œ
161 15
#
3. A(x) œ (edge)# œ ’È1 x# ŠÈ1 x# ‹“ œ Š2È1 x# ‹ œ 4 a1 x# b ; a œ 1, b œ 1; V œ 'a A(x) dx œ 'c1 4a1 x# b dx œ 4 ’x b
1
#
(diagonal)# #
4. A(x) œ
œ
œ
#
Š2È1 x# ‹
V œ 'a A(x) dx œ 2'c1 a1 x# b dx œ 2 ’x 1
" #
5. (a) STEP 1) A(x) œ
#
" x$ 3 “ "
(side) † (side) † ˆsin 13 ‰ œ
STEP 2) a œ 0, b œ 1
œ 8 ˆ1 "3 ‰ œ
16 3
#
’È1 x# ŠÈ1 x# ‹“
b
" x$ 3 “ "
" #
œ 2 a1 x# b; a œ 1, b œ 1; œ 4 ˆ1 "3 ‰ œ
8 3
† Š2Èsin x‹ † Š2Èsin x‹ ˆsin 13 ‰ œ È3 sin x
STEP 3) V œ 'a A(x) dx œ È3 '0 sin x dx œ ’È3 cos x“ œ È3(1 1) œ 2È3 1
1
b
!
(b) STEP 1) A(x) œ (side)# œ Š2Èsin x‹ Š2Èsin x‹ œ 4 sin x STEP 2) a œ 0, b œ 1
STEP 3) V œ 'a A(x) dx œ '0 4 sin x dx œ c4 cos xd 1! œ 8 1
b
#
6. (a) STEP 1) A(x) œ 1(diameter) œ 14 (sec x tan x)# œ 4 1 sin x ‘ # œ 4 sec x asec# x 1b 2 cos #x STEP 2) a œ 13 , b œ
1Î3
b
1 4
’2È3
asec# x tan# x 2 sec x tan xb
1 3
STEP 3) V œ 'a A(x) dx œ 'c1Î3 œ
1 4
1 3
1 4
ˆ2 sec# x 1
2 sin x ‰ cos# x
2 Š ˆ "" ‰ ‹ Š2È3 #
1 3
1 3
STEP 3) V œ 'a A(x) dx œ 'c1Î3 ˆ2 sec# x 1 b
1Î3
2 tan x x 2 ˆ cos" x ‰‘1Î$ 1Î$
2 Š ˆ "" ‰ ‹‹“ œ #
(b) STEP 1) A(x) œ (edge)# œ (sec x tan x)# œ ˆ2 sec# x 1 2 STEP 2) a œ 13 , b œ
1 4
dx œ
2 sin x ‰ cos# x
1 4
Š4È3
21 3 ‹
sin x ‰ cos# x
dx œ 2 Š2È3 13 ‹ œ 4È3
21 3
7. (a) STEP 1) A(x) œ alengthb † aheightb œ a6 3xb † a10b œ 60 30x STEP 2) a œ 0, b œ 2 STEP 3) V œ 'a A(x) dx œ b
'02 a60 30xb dx œ c60x 15x2 d 2! œ a120 60b 0 œ 60
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328
Chapter 6 Applications of Definite Integrals
(b) STEP 1) A(x) œ alengthb † aheightb œ a6 3xb † Š 20 2a26 3xb ‹ œ a6 3xba4 3xb œ 24 6x 9x2 STEP 2) a œ 0, b œ 2
STEP 3) V œ 'a A(x) dx œ '0 a24 6x 9x2 bdx œ c24x 3x# 3x3 d ! œ a48 12 24b 0 œ 36 b
2
2
8. (a) STEP 1) A(x) œ "# abaseb † aheightb œ ˆÈx x2 ‰ † a6b œ 6Èx 3x STEP 2) a œ 0, b œ 4
STEP 3) V œ 'a A(x) dx œ b
" #
(b) STEP 1) A(x) œ
'04 ˆ6x1Î2 3x‰ dx œ 4x3Î2 32 x2 ‘ 4! œ a32 24b 0 œ 8 2
‰ œ † 1ˆ diameter 2
STEP 2) a œ 0, b œ 4
STEP 3) V œ 'a A(x) dx œ b
1 4
9. A(y) œ
(diameter)# œ
1 4
#
ŠÈ5y# 0‹ œ
d
&
# !
" #
10. A(y) œ
1 4
(leg)(leg) œ
† 1Š
2
51 4
51 4
œ
1 2
†
x x3Î2 14 x2 4
œ 18 ˆx x3Î2 41 x2 ‰
41 x2 ‰ dx œ "# x2 52 x5Î2
1 3‘ 4 12 x !
œ 18 ˆ8
64 5
16 ‰ 3
18 a0b œ
1 "&
y% ;
y% dy
a2& 0b œ 81
" #
# È1 y# ˆÈ1 y# ‰‘ œ
V œ 'c A(y) dy œ 'c1 2a1 y# b dy œ 2 ’y d
Èx x2 2 ‹ 2
'04 ˆx x3Î2
1 8
c œ 0, d œ 2; V œ 'c A(y) dy œ '0 œ ’ˆ 541 ‰ Š y5 ‹“ œ
" #
1
" y$ 3 “ "
" #
#
ˆ2È1 y# ‰ œ 2 a1 y# b ; c œ 1, d œ 1;
œ 4 ˆ1 "3 ‰ œ
8 3
11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have
b h
œ
4 3
Ê h œ 34 b. The
equation of the line through a5, 0b and a0, 4b is y œ 45 x 4, thus the length of the base œ 45 x 4 and the 6 2 height œ 34 ˆ 45 x 4‰ œ 35 x 3.Thus Aaxb œ "# abaseb † aheightb œ "# ˆ 45 x 4‰ † ˆ 35 x 3‰ œ 25 x 12 5 x6 6 2 and V œ 'a Aaxb dx œ '0 ˆ 25 x b
5
12 5 x
5 2 3 6‰ dx œ 25 x 65 x2 6x‘ 0 œ a10 30 30b 0 œ 10
12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar triangles we have b h
œ
3 5
2 Ê b œ 35 h. Thus Aayb œ abaseb2 œ ˆ 35 y‰ œ
9 2 25 y
Ê V œ 'c Aayb dy œ '0 d
5
9 2 25 y
3 3‘ 5 dy œ 25 y 0 œ 15 0 œ 15
13. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) œ (side length)# œ s# ; STEP 2) a œ 0, b œ h; STEP 3) V œ 'a A(x) dx œ '0 s# dx œ s# h b
h
(b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the prism described above, regardless of the number of turns Ê V œ s# h
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.1 Volumes Using Cross-Sections 14. 1) The solid and the cone have the same altitude of 12. 2) The cross sections of the solid are disks of diameter x ˆ x# ‰ œ x# . If we place the vertex of the cone at the origin of the coordinate system and make its axis of symmetry coincide with the x-axis then the cone's cross sections will be circular disks of diameter x ˆ x‰ x 4 4 œ # (see accompanying figure). 3) The solid and the cone have equal altitudes and identical parallel cross sections. From Cavalieri's Principle we conclude that the solid and the cone have the same volume. 15. R(x) œ y œ 1 œ 1 ˆ2
4 2
16. R(y) œ x œ
3y #
x #
8 ‰ 12
# Ê V œ '0 1[R(x)]# dx œ 1'0 ˆ1 x# ‰ dx œ 1'0 Š1 x 2
2
2
x# 4‹
dx œ 1 ’x
x# #
21 3
œ
‰ dy œ 1' Ê V œ '0 1[R(y)]# dy œ 1'0 ˆ 3y # 0 2
17. R(y) œ tan ˆ 14 y‰ ; u œ
1 4
#
2
y Ê du œ
1 4
2
9 4
#
y# dy œ 1 34 y$ ‘ ! œ 1 †
3 4
dy Ê 4 du œ 1 dy; y œ 0 Ê u œ 0, y œ 1 Ê u œ
# x$ 12 “ !
† 8 œ 61 1 4
;
# 1Î% V œ '0 1[R(y)]# dy œ 1'0 tan ˆ 14 y‰‘ dy œ 4 '0 tan# u du œ 4 '0 a1 sec# ub du œ 4[u tan u]! 1
1Î4
1
1Î4
œ 4 ˆ 14 1 0‰ œ 4 1 1 #
18. R(x) œ sin x cos x; R(x) œ 0 Ê a œ 0 and b œ œ 1'0
1Î2
xœ
1 #
(sin x cos x)# dx œ 1 '0
1Î2
Ê u œ 1‘ Ä V œ 1'0
1
" 8
(sin 2x)# 4
are the limits of integration; V œ '0
1Î2
dx; u œ 2x Ê du œ 2 dx Ê
sin# u du œ
1 8
#u
" 4
sin
1 2u‘ !
œ
1 8
du 8
œ
dx 4
1[R(x)]# dx
; x œ 0 Ê u œ 0,
ˆ 1# 0‰ 0‘ œ
1# 16
19. R(x) œ x# Ê V œ '0 1[R(x)]# dx œ 1 '0 ax# b dx 2
2
œ 1 '0 x% dx œ 1 ’ x5 “ œ 2
#
&
!
#
321 5
20. R(x) œ x$ Ê V œ '0 1[R(x)]# dx œ 1'0 ax$ b dx 2
2
œ 1 '0 x' dx œ 1 ’ x7 “ œ 2
(
# !
#
1281 7
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329
330
Chapter 6 Applications of Definite Integrals
21. R(x) œ È9 x# Ê V œ 'c3 1[R(x)]# dx œ 1 'c3 a9 x# b dx 3
$ x$ 3 “ $
œ 1 ’9x
3
œ 21 9(3)
27 ‘ 3
œ 2 † 1 † 18 œ 361
22. R(x) œ x x# Ê V œ '0 1[R(x)]# dx œ 1'0 ax x# b dx 1
1
œ 1'0 ax# 2x$ x% b dx œ 1 ’ x3 1
$
œ 1 ˆ 13
" #
5" ‰ œ
1 30
(10 15 6) œ
23. R(x) œ Ècos x Ê V œ '0
1Î2
1Î#
œ 1 csin xd !
2x% 4
1 30
#
" x& 5 “!
1[R(x)]# dx œ 1'0 cos x dx 1Î2
œ 1(1 0) œ 1
1Î4
1Î4
24. R(x) œ sec x Ê V œ 'c1Î4 1[R(x)]# dx œ 1 ' 1Î4 sec# x dx 1Î%
œ 1 ctan xd 1Î% œ 1[1 (1)] œ 21
25. R(x) œ È2 sec x tan x Ê V œ œ1
'01Î4 1[R(x)]# dx
'01Î4 ŠÈ2 sec x tan x‹# dx
œ 1 '0 Š2 2È2 sec x tan x sec# x tan# x‹ dx 1Î4
œ 1 Œ'0 2 dx 2È2 '0 sec x tan x dx 1Î4
1Î%
œ 1 Œ[2x]!
1Î4
1Î%
2È2 [sec x]!
$
’ tan3 x “
'01Î4 (tan x)# sec# x dx
1Î% !
œ 1 ’ˆ 1# 0‰ 2È2 ŠÈ2 1‹ "3 a1$ 0b“ œ 1 Š 1# 2È2
11 3 ‹
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Section 6.1 Volumes Using Cross-Sections 26. R(x) œ 2 2 sin x œ 2(1 sin x) Ê V œ '0 1[R(x)]# dx 1Î2
œ 1 '0 4(1 sin x)# dx œ 41 '0 a1 sin# x 2 sin xb dx 1Î2
1Î2
œ 41'0 1 "# (1 cos 2x) 2 sin x‘ dx 1Î2
œ 41'0 ˆ 3# 1Î2
cos 2x 2
2 sin x‰
1Î# œ 41 3# x sin42x 2 cos x‘ ! œ 41 ˆ 341 0 0‰ (0 0 2)‘ œ 1(31 8)
27. R(y) œ È5 y# Ê V œ 'c1 1[R(y)]# dy œ 1 'c1 5y% dy 1
1
"
œ 1 cy& d " œ 1[1 (1)] œ 21
28. R(y) œ y$Î# Ê V œ '0 1[R(y)]# dy œ 1'0 y$ dy 2
2
#
%
œ 1 ’ y4 “ œ 41 !
29. R(y) œ È2 sin 2y Ê V œ '0 1[R(y)]# dy 1Î2
œ 1'0 2 sin 2y dy œ 1 c cos 2yd ! 1Î2
1Î#
œ 1[1 (1)] œ 21
30. R(y) œ Écos
1y 4
Ê V œ 'c2 1[R(y)]# dy 0
œ 1 'c2 cos ˆ 14y ‰ dy œ 4 sin 0
31. R(y) œ
2 y1
1y ‘ ! 4 #
œ 4[0 (1)] œ 4
Ê V œ '0 1[R(y)]# dy œ 41 '0 3
3
" ay 1 b 2
dy
$
œ 41 ’ y 1 1 “ œ 41 4" a1b‘ œ 31 !
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331
332
Chapter 6 Applications of Definite Integrals È2y y # 1
32. R(y) œ
Ê V œ '0 1[R(y)]# dy œ 1'0 2y ay# 1b 1
1
#
dy;
#
cu œ y 1 Ê du œ 2y dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 2d Ä V œ 1'1 u# du œ 1 "u ‘ " œ 1 #" (1)‘ œ 2
#
1 #
33. For the sketch given, a œ 1# , b œ 1# ; R(x) œ 1, r(x) œ Ècos x; V œ 'a 1 a[R(x)]# [r(x)]# b dx b
œ 'c1Î2 1(1 cos x) dx œ 21'0 (1 cos x) dx œ 21[x sin x]! 1Î2
1Î2
1Î#
œ 21 ˆ 1# 1‰ œ 1# 21
34. For the sketch given, c œ 0, d œ 14 ; R(y) œ 1, r(y) œ tan y; V œ 'c 1 a[R(y)]# [r(y)]# b dy d
œ 1'0 a1 tan# yb dy œ 1 '0 a2 sec# yb dy œ 1[2y tan y]! 1Î4
1Î4
35. r(x) œ x and R(x) œ 1 Ê V œ œ '0 1 a1 x# b dx œ 1 ’x 1
1Î%
œ 1 ˆ 1# 1‰ œ
1# #
1
'01 1 a[R(x)]# [r(x)]# b dx
" x$ 3 “!
œ 1 ˆ1 "3 ‰ 0‘ œ
21 3
36. r(x) œ 2Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]# [r(x)]# b dx 1
œ 1'0 (4 4x) dx œ 41’x 1
" x# # “!
œ 41 ˆ1 "# ‰ œ 21
37. r(x) œ x# 1 and R(x) œ x 3
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx 2
œ 1'c1 ’(x 3)# ax# 1b “ dx 2
#
œ 1 'c1 cax# 6x 9b ax% 2x# 1bd dx 2
œ 1 'c1 ax% x# 6x 8b dx 2
&
œ 1 ’ x5
x$ 3
œ 1 ˆ 32 5
8 3
#
6x# #
8x“
24 #
16‰ ˆ 5"
"
" 3
6 #
‰ ˆ 5†30533 ‰ œ 8‰‘ œ 1 ˆ 33 5 3 28 3 8 œ 1
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1171 5
Section 6.1 Volumes Using Cross-Sections 38. r(x) œ 2 x and R(x) œ 4 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx 2
œ 1'c1 ’a4 x# b (2 x)# “ dx 2
#
œ 1 'c1 ca16 8x# x% b a4 4x x# bd dx 2
œ 1'c1 a12 4x 9x# x% b dx 2
œ 1 ’12x 2x# 3x$ œ 1 ˆ24 8 24
# x& 5 “ "
32 ‰ 5
ˆ12 2 3 5" ‰‘ œ 1 ˆ15
33 ‰ 5
œ
1081 5
39. r(x) œ sec x and R(x) œ È2
Ê V œ 'c1Î4 1 a[R(x)]# [r(x)]# b dx 1Î4
œ 1 'c1Î4 a2 sec# xb dx œ 1[2x tan x]1Î% 1Î4
1Î%
œ 1 ˆ 1# 1‰ ˆ 1# 1‰‘ œ 1(1 2)
40. R(x) œ sec x and r(x) œ tan x
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx 1
œ 1 '0 asec# x tan# xb dx œ 1 '0 1 dx œ 1[x]!" œ 1 1
1
41. r(y) œ 1 and R(y) œ 1 y
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 1
œ 1'0 c(1 y)# 1d dy œ 1 '0 a1 2y y# 1b dy 1
1
œ 1 '0 a2y y# b dy œ 1 ’y#
" y$ 3 “!
1
œ 1 ˆ1 "3 ‰ œ
41 3
42. R(y) œ 1 and r(y) œ 1 y Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 1
œ 1'0 c1 (1 y)# d dy œ 1'0 c1 a1 2y y# bd dy 1
1
œ 1'0 a2y y# b dy œ 1 ’y# 1
" y$ 3 “!
œ 1 ˆ1 "3 ‰ œ
21 3
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333
334
Chapter 6 Applications of Definite Integrals
43. R(y) œ 2 and r(y) œ Èy Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 4
œ 1'0 (4 y) dy œ 1 ’4y 4
% y# 2 “!
œ 1(16 8) œ 81
44. R(y) œ È3 and r(y) œ È3 y#
È3
Ê V œ '0
È3
œ 1 '0
$
œ 1 ’ y3 “
1 a[R(y)]# [r(y)]# b dy
È3
c3 a3 y# bd dy œ 1'0 È$ !
y# dy
œ 1È3
45. R(y) œ 2 and r(y) œ 1 Èy Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 1
œ 1'0 ’4 ˆ1 Èy‰ “ dy 1
#
œ 1 '0 ˆ4 1 2Èy y‰ dy 1
œ 1 '0 ˆ3 2Èy y‰ dy 1
œ 1 ’3y 43 y$Î# œ 1 ˆ3
" y# # “!
"# ‰ œ 1 ˆ 18683 ‰ œ
4 3
71 6
46. R(y) œ 2 y"Î$ and r(y) œ 1
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 1
# œ 1'0 ’ˆ2 y"Î$ ‰ 1“ dy 1
œ 1'0 ˆ4 4y"Î$ y#Î$ 1‰ dy 1
œ 1 '0 ˆ3 4y"Î$ y#Î$ ‰ dy 1
œ 1 ’3y 3y%Î$
" 3y&Î$ 5 “!
œ 1 ˆ3 3 53 ‰ œ
31 5
47. (a) r(x) œ Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]# [r(x)]# b dx 4
œ 1'0 (4 x) dx œ 1 ’4x 4
(b) r(y) œ 0 and R(y) œ y#
% x# # “!
œ 1(16 8) œ 81
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 2
œ 1'0 y% dy œ 1 ’ y5 “ œ 2
&
# !
321 5
# (c) r(x) œ 0 and R(x) œ 2 Èx Ê V œ '0 1 a[R(x)]# [r(x)]# b dx œ 1'0 ˆ2 Èx‰ dx 4
œ 1'0 ˆ4 4Èx x‰ dx œ 1 ’4x 4
8x$Î# 3
4
% x# “ # !
œ 1 ˆ16
64 3
16 ‰ #
œ
81 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.1 Volumes Using Cross-Sections (d) r(y) œ 4 y# and R(y) œ 4 Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’16 a4 y# b “ dy 2
2
œ 1 '0 a16 16 8y# y% b dy œ 1 '0 a8y# y% b dy œ 1 ’ 83 y$ 2
2
48. (a) r(y) œ 0 and R(y) œ 1
# y& “ 5 !
#
œ 1 ˆ 64 3
32 ‰ 5
œ
2241 15
y #
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy 2
# œ 1'0 ˆ1 y# ‰ dy œ 1'0 Š1 y 2
2
y# #
œ 1 ’y
#
y$ 12 “ !
œ 1 ˆ#
(b) r(y) œ 1 and R(y) œ 2
4 2
8 ‰ 12
y# 4‹
œ
dy
21 3
y #
# Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’ˆ2 y# ‰ 1“ dy œ 1 '0 Š4 2y 2
2
œ 1'0 Š3 2y 2
y# 4‹
dy œ 1 ’3y y#
2
# y$ 12 “ !
œ 1 ˆ6 4
8 ‰ 12
œ 1 ˆ2 23 ‰ œ
y# 4
1‹ dy
81 3
49. (a) r(x) œ 0 and R(x) œ 1 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx 1
œ 1 'c1 a1 x# b dx œ 1 'c1 a1 2x# x% b dx 1
1
#
2x$ 3
œ 1 ’x
" x& 5 “ "
103 ‰ œ 21 ˆ 1515 œ
œ 21 ˆ1
2 3
15 ‰
161 15
(b) r(x) œ 1 and R(x) œ 2 x# Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’a2 x# b 1“ dx 1
1
#
œ 1 'c1 a4 4x# x% 1b dx œ 1'c1 a3 4x# x% b dx œ 1 ’3x 43 x$ 1
œ
21 15
1
(45 20 3) œ
561 15
" x& 5 “ "
œ 21 ˆ3
4 3
15 ‰
2 3
15 ‰
(c) r(x) œ 1 x# and R(x) œ 2 Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’4 a1 x# b “ dx 1
1
#
œ 1 'c1 a4 1 2x# x% b dx œ 1'c1 a3 2x# x% b dx œ 1 ’3x 23 x$ 1
œ
21 15
1
(45 10 3) œ
641 15
" x& 5 “ "
œ 21 ˆ3
50. (a) r(x) œ 0 and R(x) œ hb x h
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx b
# œ 1 '0 ˆ hb x h‰ dx b
œ 1'0 Š hb# x# b
#
$
x œ 1h# ’ 3b #
x# b
2h# b
x h# ‹ dx b
x“ œ 1h# ˆ b3 b b‰ œ !
1 h# b 3
# (b) r(y) œ 0 and R(y) œ b ˆ1 yh ‰ Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1b# '0 ˆ1 yh ‰ dy h
œ 1b# '0 Š1 h
2y h
y# h# ‹
dy œ 1b# ’y
y# h
h
h
y$ 3h# “ !
œ 1b# ˆh h 3h ‰ œ
1 b# h 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
335
336
Chapter 6 Applications of Definite Integrals
51. R(y) œ b Èa# y# and r(y) œ b Èa# y# Ê V œ 'ca 1 a[R(y)]# [r(y)]# b dy a
œ 1 'ca ’ˆb Èa# y# ‰ ˆb Èa# y# ‰ “ dy #
a
#
œ 1 'ca 4bÈa# y# dy œ 4b1'ca Èa# y# dy a
a
1a# #
œ 4b1 † area of semicircle of radius a œ 4b1 †
œ 2a# b1#
52. (a) A cross section has radius r œ È#y and area 1r# œ #1y. The volume is '0 #1ydy œ 1 cy# d ! œ #&1. &
(b) Vahb œ ' Aahbdh, so
dV dh
œ Aahb. Therefore
For h œ %, the area is #1a%b œ )1, so
dh dt
œ
dV dt
" )1
œ
dV dh
†
œ Aahb †
dh dt
$
$ )1
† $ units sec œ
hca
53. (a) R(y) œ Èa# y# Ê V œ 1'ca aa# y# b dy œ 1 ’a# y a$ 3“
œ 1 ’a# h "3 ah$ 3h# a 3ha# a$ b dV $ dt œ 0.2 m /sec dV # dh œ 101h 1h
(b) Given Ê
and a œ 5 m, find Ê
dV dt
œ
dV dh
†
dh dt
œ 1 Ša# h
†
so
dh dt
œ
" A ah b
†
dV dt .
units$ sec . hca
y$ 3 “ ca
h$ 3
dh dt ,
&
œ 1 ’a# h a$
h# a ha# ‹ œ
(h a)$ 3
Ša$
a$ 3 ‹“
1h# (3a h) 3
#
From part (a), V(h) œ 1h (153 h) œ 51h# 13h dh ¸ 0.2 " " œ 1h(10 h) dh dt Ê dt hœ4 œ 41(10 4) œ (201)(6) œ 1#01 m/sec. dh ¸ dt hœ4 .
$
54. Suppose the solid is produced by revolving y œ 2 x about the y-axis. Cast a shadow of the solid on a plane parallel to the xy-plane. Use an approximation such as the Trapezoid Rule, to #
estimate 'a 1cRaybd# dy ¸ ! 1Œ #k ˜y. n
b
d^
kœ"
55. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius h has been removed. Thus its area is A" œ 1R# 1h# œ 1 aR# h# b . The cross section of the hemisphere is a disk of #
radius ÈR# h# . Therefore its area is A# œ 1 ŠÈR# h# ‹ œ 1 aR# h# b . We can see that A" œ A# . The altitudes of both solids are R. Applying Cavalieri's Principle we find Volume of Hemisphere œ (Volume of Cylinder) (Volume of Cone) œ a1R# b R "3 1 aR# b R œ 56. R(x) œ œ
1 144
x 1#
È36 x# Ê V œ ' 1[R(x)]# dx œ 1' 0 0
’12x$
6
' x& 5 “!
œ
1 144
Š12 † 6$
6
6& 5‹
œ
1 †6 $ 144
x# 144
ˆ12
a36 x# b dx œ
36 ‰ 5
1 144
2 3
1 R$ .
'06 a36x# x% b dx
1 ‰ ˆ 6036 ‰ œ ˆ 196 œ 144 5
361 5
cm$ . The plumb bob will
weigh about W œ (8.5) ˆ 3651 ‰ ¸ 192 gm, to the nearest gram. c7
c7
57. R(y) œ È256 y# Ê V œ 'c16 1[R(y)]# dy œ 1'c16 a256 y# b dy œ 1 ’256y œ 1 ’(256)(7)
7$ 3
Š(256)(16)
16$ 3 ‹“
$
œ 1 Š 73 256(16 7)
16$ 3 ‹
( y$ 3 “ "'
œ 10531 cm$ ¸ 3308 cm$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.2 Volume Using Cylindrical Shells
337
58. (a) R(x) œ kc sin xk , so V œ 1'0 [R(x)]# dx œ 1'0 (c sin x)# dx œ 1'0 ac# 2c sin x sin# xb dx 1
1
œ 1'0 ˆc# 2c sin x
'1
1
(b)
1
1cos 2x ‰ dx œ 1 0 ˆc# "# 2c sin x cos#2x ‰ dx # 1 œ 1 ˆc# "# ‰ x 2c cos x sin42x ‘ ! œ 1 ˆc# 1 1# 2c 0‰ (0 2c 0)‘ œ 1 ˆc# 1 1# 4c‰ . Let 2 V(c) œ 1 ˆc# 1 1# 4c‰ . We find the extreme values of V(c): dV dc œ 1(2c1 4) œ 0 Ê c œ 1 is a critical # # point, and V ˆ 12 ‰ œ 1 ˆ 14 1# 18 ‰ œ 1 ˆ 1# 14 ‰ œ 1# 4; Evaluate V at the endpoints: V(0) œ 1# and # # V(1) œ 1 ˆ 3# 1 4‰ œ 1# (4 1)1. Now we see that the function's absolute minimum value is 1# 4, taken on at the critical point c œ 12 . (See also the accompanying graph.) # From the discussion in part (a) we conclude that the function's absolute maximum value is 1# , taken on at
the endpoint c œ 0. (c) The graph of the solid's volume as a function of c for 0 Ÿ c Ÿ 1 is given at the right. As c moves away from [!ß "] the volume of the solid increases without bound. If we approximate the solid as a set of solid disks, we can see that the radius of a typical disk increases without bounds as c moves away from [0ß 1].
59. Volume of the solid generated by rotating the region bounded by the x-axis and y œ faxb from x œ a to x œ b about the x-axis is V œ 'a 1[f(x)]# dx œ 41, and the volume of the solid generated by rotating the same region about the line b
'ab 1[f(x) 1]# dx œ 81. Thus 'ab 1[f(x) 1]# dx 'ab 1[f(x)]# dx œ 81 41 b b b b Ê 1'a a[f(x)]# 2f(x) " [f(x)]# b dx œ 41 Ê 'a a2f(x) "b dx œ 4 Ê 2'a f(x) dx 'a dx œ 4 b b Ê 'a f(x) dx "# ab ab œ 2 Ê 'a f(x) dx œ 4 #b a
y œ 1 is V œ
60. Volume of the solid generated by rotating the region bounded by the x-axis and y œ faxb from x œ a to x œ b about the x-axis is V œ 'a 1[f(x)]# dx œ 61, and the volume of the solid generated by rotating the same region about the line b
'ab 1[f(x) 2]# dx œ 101. Thus 'ab 1[f(x) 2]# dx 'ab 1[f(x)]# dx œ 101 61 b b b b Ê 1'a a[f(x)]# 4f(x) 4 [f(x)]# b dx œ 41 Ê 'a a4f(x) 4b dx œ 4 Ê 4'a f(x) dx 4'a dx œ 4 b b Ê 'a f(x) dx ab ab œ 1 Ê 'a f(x) dx œ 1 b a
y œ 2 is V œ
6.2 VOLUME USING CYLINDRICAL SHELLS 1. For the sketch given, a œ 0, b œ 2;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š1 b
2
x# 4‹
dx œ 21'0 Šx
x# 4‹
dx œ 21'0 Š2x
2
x$ 4‹
#
dx œ 21 ’ x#
œ 21 † 3 œ 61 2. For the sketch given, a œ 0, b œ 2;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š2 b
2
3. For the sketch given, c œ 0, d œ È2;
È2
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 d
2
È2
21y † ay# b dy œ 21'0
x$ 4‹
# x% 16 “ !
dx œ 21 ’x#
%
y$ dy œ 21 ’ y4 “
È# !
œ 21 ˆ 4#
# x% 16 “ !
16 ‰ 16
œ 21(4 1) œ 61
œ 21
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
338
Chapter 6 Applications of Definite Integrals
4. For the sketch given, c œ 0, d œ È3;
È3
È3
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y † c3 a3 y# bd dy œ 21 '0 d
%
y$ dy œ 21 ’ y4 “
È3 !
œ
91 #
5. For the sketch given, a œ 0, b œ È3;
È3
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † ŠÈx# 1‹ dx; b
’u œ x# 1 Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4“ Ä V œ 1'1 u"Î# du œ 1 23 u$Î# ‘ " œ %
4
21 3
ˆ4$Î# 1‰ œ ˆ 231 ‰ (8 1) œ
141 3
6. For the sketch given, a œ 0, b œ 3;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š Èx9x ‹ dx; $9 b
3
cu œ x$ 9 Ê du œ 3x# dx Ê 3 du œ 9x# dx; x œ 0 Ê u œ 9, x œ 3 Ê u œ 36d Ä V œ 21 '9 3u"Î# du œ 61 2u"Î# ‘ * œ 121 ŠÈ36 È9‹ œ 361 $'
36
7. a œ 0, b œ 2;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x x ˆ x2 ‰‘ dx b
2
œ '0 21x# † 2
3 #
dx œ 1 '0 3x# dx œ 1 cx$ d ! œ 81 2
#
8. a œ 0, b œ 1;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ˆ2x x2 ‰ dx b
1
œ 1 '0 2 Š 3x# ‹ dx œ 1 ' 3x# dx œ 1 cx$ d ! œ 1 1
1
#
"
0
9. a œ 0, b œ 1;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x c(2 x) x# d dx b
1
œ 21'0 a2x x# x$ b dx œ 21 ’x# 1
œ 21 ˆ1
" 3
4" ‰ œ 21 ˆ 12 124 3 ‰ œ
x$ 3
101 12
œ
" x% 4 “!
51 6
10. a œ 0, b œ 1;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ca2 x# b x# d dx b
1
œ 21'0 x a2 2x# b dx œ 41'0 ax x$ b dx 1
#
œ 41 ’ x#
1
" x% 4 “!
œ 41 ˆ "2 4" ‰ œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.2 Volume Using Cylindrical Shells 11. a œ 0, b œ 1;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Èx (2x 1)‘ dx b
1
" œ 21'0 ˆx$Î# 2x# x‰ dx œ 21 25 x&Î# 23 x$ "# x# ‘ ! 1
œ 21 ˆ 25
2 3
15 ‰ "# ‰ œ 21 ˆ 12 20 œ 30
71 15
12. a œ ", b œ 4;
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21x ˆ 32 x"Î# ‰ dx b
4
œ 31'1 x"Î# dx œ 31 23 x$Î# ‘ " œ 21 ˆ4$Î# "‰ %
4
œ 21(8 1) œ 141
sin x, 0 x Ÿ 1 0xŸ1 Ê xf(x) œ œ ; since sin 0 œ 0 we have 0, x œ 0 x, x œ 0 sin x, 0 x Ÿ 1 Ê xf(x) œ sin x, 0 Ÿ x Ÿ 1 xf(x) œ œ sin x, x œ 0
13. (a) xf(x) œ œ
x†
sin x x ,
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † f(x) dx and x † f(x) œ sin x, 0 Ÿ x Ÿ 1 by part (a) 1
b
Ê V œ 21'0 sin x dx œ 21[ cos x]1! œ 21( cos 1 cos 0) œ 41 1
tan# x x ,
tan# x, 0 x Ÿ 1/4 0 x Ÿ 14 Ê xg(x) œ œ ; since tan 0 œ 0 we have 0, x œ 0 x † 0, x œ 0 tan# x, 0 x Ÿ 1/4 Ê xg(x) œ tan# x, 0 Ÿ x Ÿ 1/4 xg(x) œ œ tan# x, x œ 0
14. (a) xg(x) œ œ
x†
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † g(x) dx and x † g(x) œ tan# x, 0 Ÿ x Ÿ 1/4 by part (a) 1Î4
b
Ê V œ 21'0 tan# x dx œ 21'0 asec# x 1b dx œ 21[tan x x]! 1Î4
1Î4
1Î%
œ 21 ˆ1 14 ‰ œ
41 1 # #
15. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y Èy (y)‘ dy d
2
œ 21'0 ˆy$Î# y# ‰ dy œ 21 ’ 2y5 2
&Î#
&
œ 21 ” 25 ŠÈ2‹ œ
161 15
2$ 3•
È
# y$ 3 “!
œ 21 Š 8 5 2 83 ‹ œ 161 Š
È2 5
3" ‹
Š3È2 5‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
339
340
Chapter 6 Applications of Definite Integrals
16. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y cy# (y)ddy d
2
œ 21'0 ay$ y# b dy œ 21 ’ y4 2
%
œ 161 ˆ 56 ‰ œ
401 3
# y$ 3 “!
œ 161 ˆ 42 3" ‰
17. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y a2y y# bdy d
2
œ 21'0 a2y# y$ b dy œ 21 ’ 2y3 2
$
œ 321 ˆ "3 4" ‰ œ
321 12
œ
81 3
# y% 4 “!
œ 21 ˆ 16 3
"6 ‰ 4
18. c œ 0, d œ 1;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y a2y y# ybdy d
1
œ 21'0 y ay y# b dy œ 21'0 ay# y$ b dy 1
1
$
œ 21 ’ y3
" y% 4 “!
œ 21 ˆ 13 "4 ‰ œ
1 6
19. c œ 0, d œ 1;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ 21'0 y[y (y)]dy d
1
œ 21'0 2y# dy œ 1
41 3
"
cy$ d ! œ
41 3
20. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 yˆy y2 ‰dy d
2
œ 21 '0
2
y2 2 dy
2
œ 13 c y3 d ! œ
81 3
21. c œ 0, d œ 2;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y c(2 y) y# d dy d
2
œ 21 '0 a2y y# y$ b dy œ 21 ’y# 2
œ 21 ˆ4
8 3
16 ‰ 4
œ
1 6
y$ 3
(48 32 48) œ
# y% 4 “!
161 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.2 Volume Using Cylindrical Shells
341
22. c œ 0, d œ 1;
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y c(2 y) y# d dy d
1
œ 21'0 a2y y# y$ b dy œ 21 ’y# 1
œ 21 ˆ1
1 3
14 ‰ œ
1 6
(12 4 3) œ
y$ 3
51 6
" y% 4 “!
shell ‰ shell 23. (a) V œ 'a 21 ˆ radius Š height ‹dx œ '0 21 x a3xbdx œ 61'0 x2 dx œ 21 cx3 d ! œ 161 b
2
2
2
2 shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹dx œ '0 21 a4 xb a3xbdx œ 61'0 a4x x2 bdx œ 61 2x2 13 x3 ‘ ! œ 61ˆ8 83 ‰ œ 321 b
2
2
2 shell ‰ shell (c) V œ 'a 21 ˆ radius Š height ‹dx œ '0 21 ax 1b a3xbdx œ 61'0 ax2 xbdx œ 61 13 x3 12 x2 ‘ ! œ 61ˆ 83 2‰ œ 281 b
2
2
6 shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 y ˆ2 13 y‰dy œ 21'0 ˆ2y 13 y2 ‰dy œ 21 y2 19 y3 ‘ ! œ 21a36 24b œ 241 d
6
6
shell ‰ shell (e) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 a7 yb ˆ2 13 y‰dy œ 21'0 ˆ14 d
6
6
13 3 y
13 y2 ‰dy œ 21 14y
13 2 6 y
6
19 y3 ‘ !
œ 21a84 78 24b œ 601
6 shell ‰ shell (f) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 ay 2b ˆ2 13 y‰dy œ 21'0 ˆ4 43 y 13 y2 ‰dy œ 21 4y 23 y2 19 y3 ‘ ! d
6
6
œ 21a24 24 24b œ 481 shell ‰ shell 24. (a) V œ 'a 21 ˆ radius Š height ‹dx œ '0 21 x a8 x3 bdx œ 21'0 a8x x4 bdx œ 21 4x2 15 x5 ‘ ! œ 21ˆ16 b
2
2
2
32 ‰ 5
œ
961 5
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹dx œ '0 21 a3 xb a8 x3 bdx œ 21'0 a24 8x 3x3 x4 bdx b
2
2
2
œ 21 24x 4x2 34 x4 15 x5 ‘ ! œ 21ˆ48 16 12
32 ‰ 5
œ
2641 5
shell ‰ shell (c) V œ 'a 21 ˆ radius Š height ‹dx œ '0 21 ax 2b a8 x3 bdx œ 21'0 a16 8x 2x3 x4 bdx b
2
2
2
œ 21 16x 4x2 12 x4 15 x5 ‘ ! œ 21ˆ32 16 8
32 ‰ 5
œ
3361 5
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 y † y1Î3 dy œ 21'0 y4Î3 dy œ d
8
8
61 7
y7Î3 ‘ 8 œ !
61 7 a128b
œ
7681 7
8 shell ‰ shell (e) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 a8 yb y1Î3 dy œ 21'0 ˆ8y1Î3 y4Î3 ‰dy œ 21 6y4Î3 37 y7Î3 ‘ ! d
8
œ 21ˆ96
384 ‰ 7
œ
8
5761 7
8 shell ‰ shell (f) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 ay 1b y1Î3 dx œ 21'0 ˆy4Î3 y1Î3 ‰dy œ 21 37 y7Î3 34 y4Î3 ‘ ! d
8
‰ œ 21ˆ 384 7 12 œ
8
9361 7
shell ‰ shell 25. (a) V œ 'a 21 ˆ radius Š height ‹dx œ 'c1 21 a2 xb ax 2 x2 bdx œ 21'c1 a4 3x2 x3 bdx œ 21 4x x3 14 x4 ‘ 1 b
2
œ 21a8 8 4b 21ˆ4 1 14 ‰ œ
2
271 2
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹dx œ 'c1 21 ax 1b ax 2 x2 bdx œ 21'c1 a2 3x x3 bdx œ 21 2x 32 x2 14 x4 ‘ 1 b
.
2
2
œ 21a4 6 4b 21ˆ2
3 2
14 ‰ œ
2
271 2
shell ‰ shell (c) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 yˆÈy ˆÈy‰‰dy '1 21 yˆÈy ay 2b‰dy d
4
1
œ 41'0 y3Î2 dy 21'1 ˆy3Î2 y2 2y‰dy œ 4
1
œ
81 5 a1b
21ˆ 64 5
64 3
16‰ 21ˆ 52
1 3
81 5
1‰ œ
y5Î2 ‘ 1 21 52 y5Î2 31 y3 y2 ‘ 4 ! 1 721 5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
342
Chapter 6 Applications of Definite Integrals
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 a4 ybˆÈy ˆÈy‰‰dy '1 21 a4 ybˆÈy ay 2b‰dy d
4
1
œ 41'0 ˆ4Èy y3Î2 ‰dy 21'1 ˆy2 y3Î2 6y 4Èy 8‰dy 4
1
1 4 œ 41 83 y3Î2 25 y5Î2 ‘ ! 21 13 y3 25 y5Î2 3y2 83 y3Î2 8y‘ 1 64 64 ‰ ˆ1 2 œ 41ˆ 83 25 ‰ 21ˆ 64 3 5 48 3 32 21 3 5 3
8‰ œ
8 3
1081 5
shell ‰ shell 26. (a) V œ 'a 21 ˆ radius Š height ‹dx œ 'c1 21 a1 xb a4 3x2 x4 bdx œ 21'c1 ax5 x4 3x3 3x2 4x 4bdx b
1
1
1
œ 21 16 x6 15 x5 34 x4 x3 2x2 4x‘ 1 œ 21ˆ 16
1 5
3 4
1 2 4‰ 21ˆ 16
1 5
1 2 4‰ œ
3 4
shell ‰ shell 4 y ˆÈ 4 y‰‰dy ' 21 y É 4 y ŠÉ 4 y ‹ dy (b) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 yˆÈ ” • 3 3 1 d
œ 41'0 y5Î4 dy 1
œ
161 9
œ
161 9
4
1
y9Î4 ‘ 1 !
41 È3
41 È3
41 È È 3 Š8 3
'14 yÈ4 ydy cu œ 4 y Ê y œ 4 u Ê du œ dy; y œ 1 Ê u œ 3, y œ 4 Ê u œ 0d
'30 a4 ubÈu du œ 1691 a1b È41 '03 ˆ4Èu u3Î2 ‰du œ 1691 È41 38 u3Î2 52 u5Î2 ‘ 30 3
18 È 3‹ 5
161 9
œ
881 5
œ
3
8721 45
shell ‰ shell 27. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y † 12 ay# y$ b dy œ 241 '0 ay$ y% b dy œ 241 ’ y4 d
1
œ 241 ˆ 14 15 ‰ œ
241 20
œ
1
" y& 5 “!
%
61 5
shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(1 y) c12 ay# y$ bd dy œ 241'0 (1 y) ay# y$ b dy d
1
1
œ 241'0 ay# 2y$ y% b dy œ 241 ’ y3 1
$
y% 2
" y& 5 “!
œ 241 ˆ "3
1 2
" ‰ 15 ‰ œ 241 ˆ 30 œ
41 5
shell ‰ shell (c) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆ 85 y‰ c12 ay# y$ bd dy œ 241 '0 ˆ 85 y‰ ay# y$ b dy d
1
œ 241'0 ˆ 85 y# 1
œ
241 12
13 5
1
8 $ y$ y% ‰ dy œ 241 ’ 15 y
13 20
y%
" y& 5 “!
8 œ 241 ˆ 15
13 20
241 60
15 ‰ œ
(32 39 12)
œ 21
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆy 25 ‰ c12 ay# y$ bd dy œ 241'0 ˆy 25 ‰ ay# y$ b dy d
1
1
2 $ œ 241'0 ˆy$ y% 25 y# 25 y$ ‰ dy œ 241'0 ˆ 25 y# 35 y$ y% ‰ dy œ 241 ’ 15 y 1
1
2 œ 241 ˆ 15
3 20
15 ‰ œ
241 60
(8 9 12) œ
241 12
2
%
œ 21 ’ y4
# y' 24 “ !
%
2' 24 ‹
œ 21 Š 24
#
%
œ 321 ˆ 4"
4 ‰ 24
dy œ '0 21y Šy# 2
y# # ‹“
2
œ 21 '0 Š2y# 2
y% 2
y$
y& 4‹
#
$
dy œ 21 ’ 2y3
%
y& 10
y% 4
#
y' #4 “ !
2
œ 21'0 Š5y# 54 y% y$ 2
y& 4‹
#
$
dy œ 21 ’ 5y3
%
5y& 20
y% 4
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆy 58 ‰ ’ y# Š y4 d
2
œ 21'0 Šy$ 2
y& 4
85 y#
5 32
#
%
y% ‹ dy œ 21 ’ y4
%
y' #4
5y$ #4
81 3 y% 4‹
œ 21 ˆ 16 3
œ
32 10
16 4
64 ‰ 24
dy œ '0 21(5 y) Šy#
# y' #4 “ ! y# # ‹“
2
2
y# # ‹“
dy œ 21'0 Šy$
dy œ '0 21(2 y) Šy#
y# # ‹“
shell ‰ shell (c) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(5 y) ’ y# Š y4 d
y% 4‹
2 ‰ œ 321 ˆ 4" 6" ‰ œ 321 ˆ 24 œ
shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(2 y) ’ y# Š y4 d
" y& 5 “!
y%
œ 21
shell ‰ shell 28. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y ’ y# Š y4 d
3 20
2
œ 21 ˆ 40 3
160 20
16 4
dy œ '0 21 ˆy 58 ‰ Šy# 2
#
5y& 160 “ !
œ 21 ˆ 16 4
64 24
40 24
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
81 5
y% 4‹
64 ‰ 24
dy
dy
œ 81 y% 4‹
160 ‰ 160
dy
œ 41
y& 4‹
dy
561 5
Section 6.2 Volume Using Cylindrical Shells
343
shell ‰ shell 29. (a) About x-axis: V œ 'c 21 ˆ radius Š height ‹dy d
œ '0 21yˆÈy y‰dy œ 21'0 ˆy$Î# y# ‰dy 1
1
" œ 21 #& y&Î# "$ y$ ‘ ! œ 21ˆ #& "$ ‰ œ
#1 "&
shell ‰ shell About y-axis: V œ 'a 21 ˆ radius Š height ‹dx b
œ '0 21xax x# bdx œ 21'0 ax2 x3 bdx 1
1
$
œ 21’ x$
" x% % “!
œ 21ˆ "$ "% ‰ œ
1 '
(b) About x-axis: Raxb œ x and raxb œ x# Ê V œ 'a 1Raxb# raxb# ‘dx œ '0 1cx# x% ddx b
$
œ 1’ x$
" x& & “!
œ 1ˆ "$ "& ‰ œ
1
#1 "&
About y-axis: Rayb œ Èy and rayb œ y Ê V œ 'c 1Rayb# rayb# ‘dy œ '0 1cy y2 ddy d
#
œ 1’ y#
" y$ $ “!
œ 1ˆ "# "$ ‰ œ
1
1 '
# 30. (a) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’ˆ #x #‰ x# “dx %
b
œ 1'0 ˆ $% x# #x %‰dx œ 1’ x% x# %x“ %
$
œ 1a"' "' "'b œ "'1
% !
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹dx œ '0 #1xˆ x# # x‰dx %
b
œ '0 #1xˆ# x# ‰dx œ #1'0 Š#x %
%
œ #1’x#
% x$ ' “!
œ #1ˆ"'
'% ‰ '
x# # ‹dx $#1 $
œ
shell ‰ shell (c) V œ 'a 21 ˆ radius Š height ‹dx œ '0 #1a% xbˆ x# # x‰dx œ '0 #1a% xbˆ# x# ‰dx œ #1'0 Š) %x %
b
œ #1’)x #x#
% x$ “ ' !
%
œ #1ˆ$# $#
'% ‰ '
%
x# # ‹dx
'%1 $
œ
# (d) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’a) xb# ˆ' #x ‰ “dx œ 1'0 ’a'% "'x x# b Š$' 'x x% ‹“dx %
b
%
#
1'0 ˆ $% x# "!x #)‰dx œ 1’ x% &x# #)x“ œ 1"' a&ba"'b a(ba"'b‘ œ 1a$ba"'b œ %)1 %
%
$
!
shell ‰ shell 31. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '1 21y(y 1) dy d
2
œ 21'1 ay# yb dy œ 21 ’ y3 2
$
# y# # “"
œ 21 ˆ 83 42 ‰ ˆ "3 #" ‰‘ œ 21 ˆ 73 2 "# ‰ œ 13 (14 12 3) œ
51 3
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21x(2 x) dx œ 21'1 a2x x# b dx œ 21 ’x# b
2
2
œ 21 ˆ4 83 ‰ ˆ1 3" ‰‘ œ 21 ˆ 12 3 8 ‰ ˆ 3 3 " ‰‘ œ 21 ˆ 34 32 ‰ œ
41 3
shell ‰ shell ' ˆ 203 ‰ (c) V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21 ˆ 10 3 x (2 x) dx œ 21 1 b
2
#
" $‘ 8 # ˆ 40 œ 21 20 3 x 3 x 3 x " œ 21 3
2
32 3
38 ‰ ˆ 20 3
8 3
16 3
# x$ 3 “"
x x# ‰ dx
3" ‰‘ œ 21 ˆ 33 ‰ œ 21
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '1 21(y 1)(y 1) dy œ 21'1 (y 1)# œ 21 ’ (y31) “ œ d
2
2
$
# "
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
21 3
344
Chapter 6 Applications of Definite Integrals
shell ‰ shell 32. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21yay# 0b dy d
2
œ 21'0 y$ dy œ 21 ’ y4 “ œ 21 Š 24 ‹ œ 81 2
%
#
%
!
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx b
œ '0 21x ˆ2 Èx‰ dx œ 21'0 ˆ2x x$Î# ‰ dx 4
4
%
2 †2 & 5 ‹
œ 21 x# 25 x&Î# ‘ ! œ 21 Š16 œ 21 ˆ16
64 ‰ 5
21 5
œ
321 5
(80 64) œ
shell ‰ shell (c) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(4 x) ˆ2 Èx‰ dx œ 21'0 ˆ8 4x"Î# 2x x$Î# ‰ dx b
4
4
%
œ 21 8x 83 x$Î# x# 25 x&Î# ‘ ! œ 21 ˆ32
64 3
16
64 ‰ 5
œ
21 15
(240 320 192) œ
21 15
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(2 y) ay# b dy œ 21 '0 a2y# y$ b dy œ 21 ’ 23 y$ d
2
œ 21 ˆ 16 3
16 ‰ 4
œ
321 12
2
81 3
(4 3) œ
(112) œ
2241 15
# y% 4 “!
shell ‰ shell 33. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21yay y$ b dy d
1
œ '0 21 ay# y% b dy œ 21 ’ y3 1
œ
$
41 15
" y& “ 5 !
œ 21 ˆ "3 5" ‰
shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy d
œ '0 21(1 y) ay y$ b dy 1
œ 21 '0 ay y# y$ y% b dy œ 21 ’ y# 1
#
y$ 3
y% 4
" y& 5 “!
œ 21 ˆ "#
" 3
" 4
5" ‰ œ
21 60
(30 20 15 12) œ
71 30
shell ‰ shell 34. (a) V œ 'c 21 ˆ radius Š height ‹dy d
œ '0 21y c1 ay y$ bddy 1
œ 21 '0 ay y# y% b dy œ 21 ’ y# 1
#
œ 21 ˆ "# œ
" 3
5" ‰ œ
21 30
y$ 3
" y& 5 “!
(15 10 6)
111 15
(b) Use the washer method: V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’1# ay y$ b “ dy œ 1 '0 a1 y# y' 2y% b dy œ 1 ’y d
1
œ 1 ˆ1
" 3
" 7
52 ‰ œ
1 105
1
#
(105 35 15 42) œ
y$ 3
y( 7
971 105
" 2y& 5 “!
(c) Use the washer method: V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’c1 ay y$ bd 0“ dy œ 1'0 ’1 2 ay y$ b ay y$ b “ dy d
1
œ 1'0 a1 y# y' 2y 2y$ 2y% b dy œ 1 ’y 1
œ
1 210
(70 30 105 2 † 42) œ
1
#
y$ 3
y( 7
y#
#
y% #
1211 210
" 2y& 5 “!
œ 1 ˆ1
" 3
" 7
1
" #
25 ‰
shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(1 y) c1 ay y$ bd dy œ 21 '0 (1 y) a1 y y$ b dy d
1
1
œ 21'0 a1 y y$ y y# y% b dy œ 21'0 a1 2y y# y$ y% b dy œ 21 ’y y# 1
œ 21 ˆ 1 1
1
" 3
" 4
5" ‰ œ
21 60
(20 15 12) œ
231 30
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
y$ 3
y% 4
" y& 5 “!
Section 6.2 Volume Using Cylindrical Shells shell ‰ shell 35. (a) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21y ˆÈ8y y# ‰ dy d
2
œ 21'0 Š2È2 y$Î# y$ ‹ dy œ 21 ’ 4 5 2 y&Î# 2
È
# y% 4 “!
&
œ 21
4È2†ŠÈ2‹
2% 4
5
81 5
œ 21 † 4 ˆ 85 1‰ œ
$
œ 21 Š 4†52
(8 5) œ
4 †4 4 ‹
241 5
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ŠÈx b
4
&
œ 21 Š 2†52
4% 3# ‹
'
œ 21 Š 25
2) 32 ‹
œ
1†2( 160
x# 8‹
dx œ 21'0 Šx$Î#
(32 20) œ
4
1†2* †3 160
œ
1†2% †3 5
œ
x$ 8‹
dx œ 21 ’ 25 x&Î#
481 5
shell ‰ shell 36. (a) V œ 'a 21 ˆ radius Š height ‹ dx b
œ '0 21x ca2x x# b xd dx 1
œ 21 '0 x ax x# b dx œ 21'0 ax# x$ b dx 1
$
1
œ 21 ’ x3
" x% 4 “!
œ 21 ˆ "3 4" ‰ œ
1 6
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(1 x) ca2x x# b xd dx œ 21'0 (1 x) ax x# b dx b
1
1
œ 21 '0 ax 2x# x$ b dx œ 21 ’ x2 32 x$ 1
" x% “ 4 !
#
œ 21 ˆ 12
2 3
"4 ‰ œ
21 1#
(6 8 3) œ
37. (a) V œ 'a 1 cR# (x) r# (x)d dx œ 1 '1Î16 ˆx"Î# 1‰ dx b
1
"
œ 1 2x"Î# x‘"Î"' œ 1 (2 1) ˆ2 † œ 1 ˆ1
7 ‰ 16
œ
" 4
" ‰‘ 16
91 16
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dy œ '1 21y Š y"% b
2
œ 21'1 ˆy$ 2
y ‰ 16
dy œ 21 ’ 12 y#
œ 21 ˆ "8 8" ‰ ˆ #" œ
21 32
91 16
(8 1) œ
" ‰‘ 3#
d
2
œ 1 "3 y$ œ
1 48
y ‘# 16 "
" ‰ 32
" 16 ‹
dy
" œ 1 ˆ 24 8" ‰ ˆ 3"
(2 6 16 3) œ
dy
# y# 32 “ "
œ 21 ˆ 4"
38. (a) V œ 'c 1 cR# (y) r# (y)d dy œ '1 1 Š y"%
" 16 ‹
" ‰‘ 16
111 48
shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '1Î4 21x Š È"x "‹ dx b
1
œ 21 '1Î4 ˆx"Î# x‰ dx œ 21 ’ 23 x$Î# 1
œ 21 ˆ 23 "# ‰ ˆ 23 †
" 8
" ‰‘ 3#
" x# “ 2 "Î%
œ 1 ˆ 43 1
" 6
" ‰ 16
œ
1 48
(4 † 16 48 8 3) œ
111 48
39. (a) H3=5 : V œ V" V#
V" œ 'a 1[R" (x)]# dx and V# œ 'a 1[R# (x)]# with R" (x) œ É x 3 2 and R# (x) œ Èx, b"
b#
"
#
a" œ 2, b" œ 1; a# œ 0, b# œ 1 Ê two integrals are required
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 6
% x% “ 32 !
345
346
Chapter 6 Applications of Definite Integrals
(b) [ +=2/<: V œ V" V#
V" œ 'a 1 a[R" (x)]# [r" (x)]# b dx with R" (x) œ É x 3 2 and r" (x) œ 0; a" œ 2 and b" œ 0; b"
"
b V# œ 'a
#
1 a[R# (x)]# [r# (x)]# b dx with R# (x) œ É x 3 2 and r# (x) œ Èx; a# œ 0 and b# œ 1
#
Ê two integrals are required
shell ‰ shell shell (c) W2/66: V œ 'c 21 ˆ radius Š height ‹ dy œ 'c 21y Š height ‹ dy where shell height œ y# a3y# 2b œ 2 2y# ; d
d
c œ 0 and d œ 1. Only 98/ integral is required. It is, therefore preferable to use the =2/66 method. However, whichever method you use, you will get V œ 1. 40. (a) H3=k: V œ V" V# V$
Vi œ 'c 1[Ri (y)]# dy, i œ 1, 2, 3 with R" (y) œ 1 and c" œ 1, d" œ 1; R# (y) œ Èy and c# œ 0 and d# œ 1; di
i
R$ (y) œ (y)"Î% and c$ œ 1, d$ œ 0 Ê three integrals are required (b) [ +=2/<: V œ V" V#
Vi œ 'c 1a[Ri (y)]# [ri (y)]# b dy, i œ 1, 2 with R" (y) œ 1, r" (y) œ Èy, c" œ 0 and d" œ 1; di
i
R# (y) œ 1, r# (y) œ (y)"Î% , c# œ 1 and d# œ 0 Ê two integrals are required
shell ‰ shell shell (c) W2/66: V œ 'a 21 ˆ radius Š height ‹dx œ 'a 21xŠ height ‹dx, where shell height œ x# ax% b œ x# x% , b
b
a œ 0 and b œ 1 Ê only one integral is required. It is, therefore preferable to use the =2/66 method. However, whichever method you use, you will get V œ 561 . 41. (a) V œ 'a 1 cR# axb r# axbd dx œ 'c4 1 ”ŠÈ25 x2 ‹ a3b# •dx œ 1'c4 c25 x2 9d dx œ 1'c4 a16 x2 bdx b
4
%
4
64 ‰ 64 ‰ 2561 ˆ 3 1 64 3 œ 3 5001 3 Ê Volume of portion removed
œ 116x 13 x3 ‘ 4 œ 1ˆ64 (b) Volume of sphere œ 43 1a5b3 œ
È" 1
shell ‰ shell 42. V œ 'a 21 ˆ radius Š height ‹ dx œ '1 b
#
œ
4
5001 3
2561 3
œ
2441 3
21 x sinax2 1b dx; Òu œ x2 1 Ê du œ 2x dx; x œ 1 Ê u œ 0,
x œ È" 1 Ê u œ 1Ó Ä 1'0 sin u du œ 1 ccos ud 1! œ 1a1 1b œ 21 1
r shell ‰ shell 43. V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21 x ˆ hr x h‰dx œ 21'0 ˆ hr x2 h x‰dx œ 21 3rh x3 h2 x2 ‘ ! b
r
2
œ 21Š r3h
r2 h # ‹
r
œ 13 1 r2 h
shell ‰ shell 44. V œ 'c 21 ˆ radius Š height ‹dy œ '0 21 y”Èr2 y2 ˆÈr2 y2 ‰•dy œ 41'0 yÈr2 y2 dy r
d
r
Òu œ r2 y2 Ê du œ 2y dy; y œ 0 Ê u œ r2 , y œ r Ê u œ 0] Ä 21'r2 Èu du œ 21'0 u1Î2 du 0
œ
2 4 1 3 Î2 ‘ r 3 u !
œ
r2
41 3 3 r
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.3 Arc Lengths 6.3 ARC LENGTHS 1.
dy dx
œ
" 3
† 3# ax# 2b
"Î#
† 2x œ Èax# 2b † x
Ê L œ '0 È1 ax# 2b x# dx œ '0 È1 2x# x% dx $
3
œ '0 Éa1 x# b# dx œ '0 a1 x# b dx œ ’x $
œ3
2.
dy dx
œ
3 #
œ 12
27 3
Èx Ê L œ ' É1 49 x dx; u œ 1 49 x 0 4
Ê du œ
dx Ê
9 4
du œ dx; x œ 0 Ê u œ 1; x œ 4
4 9
Ê u œ 10d Ä L œ '1 u"Î# ˆ 49 du‰ œ 10
œ
3.
dx dy
8 27
œ y#
#
" 4y#
% Ê Š dx dy ‹ œ y
3
œ '1 Éy% 3
" #
œ '1 ÊŠy# 3
$
œ ’ y3 dx dy
œ
4 9
23 u$Î# ‘ "! "
Š10È10 1‹
Ê L œ '1 É1 y%
4.
$ x$ “ 3 !
$
" #
" 16y%
" 4y# ‹
$ y " “ 4 "
#
" #
" 16y%
" #
" 16y%
dy
dy
dy œ '1 Šy# 3
" ‰ 1#
œ ˆ 27 3
" 4y# ‹
dy
ˆ 3" 4" ‰ œ 9 #
y"Î# "# y"Î# Ê Š dx dy ‹ œ
" 4
" 1#
" 3
" 4
œ9
(1 4 3) 1#
œ9
(2) 1#
œ
53 6
Šy 2 y" ‹
Ê L œ '1 Ê1 "4 Šy 2 y" ‹ dy 9
œ '1 Ê "4 Šy 2 y" ‹ dy œ '1 9
œ
" #
9
" #
" Èy ‹
Ê ŠÈ y
*
$Î#
$
"
dx dy
dy
'19 ˆy"Î# y"Î# ‰ dy œ "# 23 y$Î# 2y"Î# ‘ *"
œ ’ y 3 y"Î# “ œ Š 33 3‹ ˆ "3 1‰ œ 11
5.
#
œ y$
" 4y$
#
' Ê Š dx dy ‹ œ y
Ê L œ '1 É1 y' 2
œ '1 Éy' 2
œ '1 Šy$ 2
œ Š 16 4
" 2
y $ 4 ‹
" (16)(2) ‹
" 16y'
" 2
" 16y'
" #
32 3
dy
2
%
œ
" 16y'
dy œ '1 ÊŠy$
dy œ ’ y4
" 3
y $ 4 ‹
#
dy
# y # “ 8 "
ˆ "4 "8 ‰ œ 4
" 32
" 4
" 8
œ
128184 32
œ
123 32
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
347
348 6.
Chapter 6 Applications of Definite Integrals dx dy
œ
y# #
" #y #
#
Ê Š dx dy ‹ œ
" 4
ay% 2 y% b
Ê L œ '2 É1 "4 ay% 2 y% b dy 3
œ '2 É "4 ay% 2 y% b dy 3
7.
œ
" #
'23 Éay# y# b# dy œ "# '23 ay# y# b dy
œ
" #
’ y3 y" “ œ
dy dx
$
$
" #
#
"‰ ˆ 27 ˆ 8 " ‰‘ œ 3 3 3 # #
#Î$ œ x"Î$ "4 x"Î$ Ê Š dy dx ‹ œ x
Ê L œ '1 É1 x#Î$ 8
œ '1 Éx#Î$ 8
" #
x #Î$ 16
" #
x #Î$ 16
" #
" #
ˆ 26 3
8 3
#" ‰ œ
" #
ˆ6 #" ‰ œ
13 4
x #Î$ 16
dx
dx
# œ '1 Ɉx"Î$ "4 x"Î$ ‰ dx œ '1 ˆx"Î$ "4 x"Î$ ‰ dx 8
8
)
œ 34 x%Î$ 38 x#Î$ ‘ " œ œ 8.
dy dx
3 8
2x%Î$ x#Î$ ‘ ) "
3 8
ca2 † 2% 2# b (2 1)d œ
œ x# 2x 1
œ (1 x)#
#
2
œ '0 É(1 x)%
" #
œ '0 Ê’(1 x)# 2
œ '0 ’(1 x)# 2
" #
(1x) % 16
# (1x) # “ 4
(1x) # “ 4
(1x) % 16
99 8
" " 4 (1x)#
% Ê Š dy dx ‹ œ (1 x)
Ê L œ '0 É1 (1 x)% 2
(32 4 3) œ
œ x# 2x 1
4 (4x4)#
" " 4 (1x)#
3 8
" #
" 16(1x)%
dx
dx
dx
dx; cu œ 1 x Ê du œ dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 3d
Ä L œ '1 ˆu# "4 u# ‰ du œ ’ u3 "4 u" “ œ ˆ9 3
$
$
"
9.
dx dy
" ‰ 1#
ˆ 3" 4" ‰ œ
108143 12
œ
106 12
œ
53 6
#
% œ Èsec% y 1 Ê Š dx dy ‹ œ sec y 1 1Î4
1Î4
Ê L œ 'c1Î4 È1 asec% y 1b dy œ ' 1Î4 sec# y dy 1Î%
œ ctan yd 1Î% œ 1 (1) œ 2
10.
dy dx
#
% œ È3x% 1 Ê Š dy dx ‹ œ 3x 1
c1
c1
Ê L œ 'c2 È1 a3x% 1b dx œ 'c2 È3 x# dx $
œ È3 ’ x3 “
" #
œ
È3 3
c1 (2)$ d œ
È3 3
(" 8) œ
7È 3 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.3 Arc Lengths 11. (a)
dy dx
#
(b)
# œ 2x Ê Š dy dx ‹ œ 4x
Ê L œ 'c1 Ê1 Š dy dx ‹ dx #
2
œ 'c1 È1 4x# dx 2
(c) L ¸ 6.13
12. (a)
dy dx
#
(b)
% œ sec# x Ê Š dy dx ‹ œ sec x
Ê L œ 'c1Î3 È1 sec% x dx 0
(c) L ¸ 2.06
13. (a)
dx dy
#
(b)
# œ cos y Ê Š dx dy ‹ œ cos y
Ê L œ '0 È1 cos# y dy 1
(c) L ¸ 3.82
14. (a)
dx dy
#
œ È1y y# Ê Š dx dy ‹ œ 1Î2
Ê L œ 'c1Î2 É1
œ 'c1Î2 a1 y# b 1Î2
"Î#
y# a1 y # b
y# 1 y#
(b) 1Î2
dy œ ' 1Î2 É 1 " y# dy
dy
(c) L ¸ 1.05
15. (a) 2y 2 œ 2
dx dy
#
# Ê Š dx dy ‹ œ (y 1)
(b)
Ê L œ 'c1 È1 (y 1)# dy 3
(c) L ¸ 9.29
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349
350
Chapter 6 Applications of Definite Integrals
16. (a)
dy dx
#
(b)
# # œ cos x - cos x + x sin x Ê Š dy dx ‹ œ x sin x
Ê L œ '0 È1 x# sin# x dx 1
(c) L ¸ 4.70
17. (a)
dy dx
#
(b)
# œ tan x Ê Š dy dx ‹ œ tan x
# x cos# x Ê L œ '0 È1 tan# x dx œ '0 É sin cos dx #x
1Î6
œ '0
1Î6
1Î6
œ '0 sec x dx 1Î6
dx cos x
(c) L ¸ 0.55
18. (a)
dx dy
#
(b)
# œ Èsec# y 1 Ê Š dx dy ‹ œ sec y 1
Ê L œ 'c1Î3 È1 asec# y 1b dy 1Î4
1Î4
1Î4
œ 'c1Î3 ksec yk dy œ ' 1Î3 sec y dy (c) L ¸ 2.20
#
" 19. (a) Š dy dx ‹ corresponds to 4x here, so take So y œ Èx from ("ß ") to (4ß 2).
dy dx
" . #È x
as
Then y œ Èx C and since ("ß ") lies on the curve, C œ 0.
(b) Only one. We know the derivative of the function and the value of the function at one value of x. #
20. (a) Š dx dy ‹ corresponds to So y œ
" y%
here, so take
dy dx
" y# .
as
Then x œ y" C and, since (!ß ") lies on the curve, C œ 1
" "x.
(b) Only one. We know the derivative of the function and the value of the function at one value of x. 21. y œ '0 Ècos2t dt Ê x
œ '0
1Î4
œ Ècos2x Ê L œ '0
1Î4
Ê1 ’Ècos2x“ dx œ '0 #
1Î4
È1 cos2x dx œ '
1Î4
0
È2cos2 x dx
È2cos x dx œ È2csin xd 1Î4 œ È2sinˆ 1 ‰ È2sina!b œ 1 0 4
22. y œ ˆ1 x2Î3 ‰
3 Î2 È 2 , 4
œ 'È2Î4 É1 1
dy dx
ŸxŸ1Ê
1 x2Î3 dx x2Î3
œ 32 a1b2Î3 32 Š
È2 2Î3 4 ‹
dy dx
1Î2 œ 32 ˆ1 x2Î3 ‰ ˆ 23 x1Î3 ‰ œ
œ 'È2Î4 É1
1 x2Î3
œ
3 4
1
3 2
32 ˆ 12 ‰ œ
ˆ1 x2Î3 ‰1Î2 x1Î3
1
1 dx œ 'È2Î4 É x12Î3 dx œ 'È2Î4 x11Î3 dx œ 'È2Î4 x1Î3 dx œ 1
1
#
Ê L œ 'È2Î4 Ë1 ” a1 xx1Î3 b • dx 1
Ê total length œ 8ˆ 34 ‰ œ 6
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2Î3 1Î2
3 2
" x2Î3 ‘ È 2Î4
Section 6.3 Arc Lengths 23. y œ 3 2x, 0 Ÿ x Ÿ 2 Ê
dy dx
351
œ 2 Ê L œ '0 É1 a2b# dx œ '0 È5 dx œ ’È5 x“ œ 2È5. 2
2
2
0
d œ Éa2 0b a3 a1bb œ 2È5 2
2
24. Consider the circle x2 y2 œ r2 , we will find the length of the portion in the first quadrant, and multiply our result by 4. y œ Èr2 x2 , 0 Ÿ x Ÿ r Ê œ 4'0
r
r È r2 x 2
dx œ 4r'0
r
25. 9x2 œ yay 3b2 Ê Ê dx œ
2
œ ’ ay 4y1b 1“dy2 œ 26. 4x2 y2 œ 64 Ê
x È r2 x 2
#
Ê L œ 4'0 Ë1 ” È 2 x 2 • dx œ 4'0 É1 r x r
œ
d dy ’yay
2
y2 2y 1 4y dy2 4y
2
y2 “ œ
2 œ dx2 ’ 4x y dx“ œ dx
16x2 2 y2 dx
œ
dx œ 4'0 É r2 r x2 dx r
2
ay 3 b2 ay 1 b 2 dy2 36x2
dy2 œ
dx dy
œ
ay 3bay 1b 6x
ay 3 b2 ay 1b 2 dy2 4yay3b2
dy2
ay 1 b 2 2 4y dy
d dx ’64“
Ê 8x 2y dy dx œ 0 Ê
œ Š1
16x2 2 y2 ‹dx
œ
dy dx
œ
4x y
y2 16x2 dx2 y2
œ
4x2 64 16x2 dx2 y2
È2 œ Ê1 Š dy ‹ Ê 27. È2 x œ '0 Ê1 Š dy dt ‹ dt, x 0 Ê dx #
x
x2 r2 x 2
dx 3b2 “ Ê 18x dy œ 2yay 3b ay 3b2 œ 3ay 3bay 1b Ê
bay 1b ds2 œ dx2 dy2 œ ’ ay 36x dy“ dy2 œ
d 2 dx ’4x
r
dx È r2 x 2
d 2 dy ’9x “
ay 3bay 1b dy; 6x
œ
dy dx
#
dy dx
Ê dy œ
4x y dx;
ds2 œ dx2 dy2
œ
20x2 64 dx2 y2
œ
4 2 y2 a5x
16bdx2
œ „ 1 Ê y œ f(x) œ „ x C where C is any real
number. 28. (a) From the accompanying figure and definition of the differential (change along the tangent line) we see that dy œ f w (xkc1 ) ˜ xk Ê length of kth tangent fin is È( ˜ xk )# (dy)# œ È( ˜ xk )# [f w (xkc1 ) ˜ xk ]# .
n
n
! (length of kth tangent fin) œ lim ! È( ˜ xk )# [f w (xk 1 ) ˜ xk ]# (b) Length of curve œ n lim Ä_ nÄ_ kœ1
! È1 [f w (xk 1 )]# ˜ xk œ ' È1 [f w (x)]# dx œ n lim Ä_ a n
kœ1
b
kœ1
4
2
29. x2 y2 œ 1 Ê y œ È1 x2 ; P œ Ö0, 14 , 12 , 34 , 1× Ê L ¸ ! Éaxi xi1 b2 ayi yi1 b2 œ ʈ 14 0‰ Š kœ1
ʈ 1#
1 ‰2 4
È Š 23
È15 2 4 ‹
ʈ 43
1 ‰2 2
È Š 47
È3 2 2 ‹
2 ʈ1 34 ‰ Š0
30. Let ax1 , y1 b and ax2 , y2 b, with x2 x1 , lie on y œ m x b, where m œ #
y2 y1 x2 x1 ,
then
dy dx
È7 2 4 ‹
œ m Ê L œ 'x È1 m# dx x2 1
#
#
2
ax 2 x 1 b
1
ax2 x1 b œ Éax2 x1 b# ay2 y1 b# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
1‹
¸ 1.55225
ax 2 x 1 b a y 2 y 1 b y1 œ È1 m# c xd xx21 œ È1 m# ax2 x1 b œ Ê1 Š yx22 ax2 x1 b x1 ‹ ax2 x1 b œ Ê ax x b # É ax 2 x 1 b# a y 2 y 1 b #
È15 4
352
Chapter 6 Applications of Definite Integrals
31. y œ 2x3Î2 Ê
# œ 3x1Î2 ; Laxb œ '0 É1 3t1Î2 ‘ dt œ '0 È1 9t dt; Òu œ 1 9t Ê du œ 9dt, t œ 0 Ê u œ 1, x
dy dx
t œ x Ê u œ 1 9x] Ä 32. y œ
x3 3
x2 x
1 4x 4
1 9
È
'119x Èu du œ 272 u3Î2 ‘ 119x œ 272 a1 9xb3Î2 272 ; La1b œ 272 a10b3Î2 272 œ 2Š10 2710 1‹
Ê
dy dx
Laxb œ '0 Ë1 ”at 1b2 x
œ '0 Ê 16at 1b x
œ '0
x
4at1b4 1 4 at 1 b 2
Ä '1
x 1
La1b œ
8 3
4
œ x2 2x 1 # 1 4 at 1 b 2 •
16at 1b8 8at 1b4 " 16at 1b4
dt œ '0 ”at 1b2 x
x
x
1 12
œ
1 ; 4 ax 1 b 2
#
4
b 8 at 1b dt œ '0 Ê 16at 116 at 1 b 4 x
1 4 at 1 b 2
8
4
"
x
dt œ '0 Ê x
4at1b4 1‘# 16at 1b4
4at1b4 1‘# 16at 1b4
dt
dt
•dt; Òu œ t 1 Ê du œ dt, t œ 0 Ê u œ 1, t œ x Ê u œ x 1Ó
x 1
1 8
œ ax 1b2
1 dt œ '0 Ë1 ” 4a4tat 1b 1 dt œ '0 Ê1 b2 •
1 2 1 3 1 1 2 ”u 4 u •du œ 3 u 4 u ‘ 1
1 4 ax 1 b 2
œ Š 31 ax 1b3
1 4 ax 1 b ‹
ˆ 13 14 ‰ œ 13 ax 1b3
59 24
33-38. Example CAS commands: Maple: with( plots ); with( Student[Calculus1] ); with( student ); f := x -> sqrt(1-x^2);a := -1; b := 1; N := [2, 4, 8 ]; for n in N do xx := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x,f(x)],x=xx)]; L := simplify(add( distance(pts[i+1],pts[i]), i=1..n )); T := sprintf("#33(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [f(x),pts], x=a..b, title=T ): end do: display( [seq(P[n],n=N)], insequence=true, scaling=constrained ); L := ArcLength( f(x), x=a..b, output=integral ): L = evalf( L );
# (b) # (a)
# (c)
33-38. Example CAS commands: Mathematica: (assigned function and values for a, b, and n may vary) Clear[x, f] {a, b} = {1, 1}; f[x_] = Sqrt[1 x2 ] p1 = Plot[f[x], {x, a, b}] n = 8; pts = Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / N Show[{p1,Graphics[{Line[pts]}]}] Sum[ Sqrt[ (pts[[i 1, 1]] pts[[i, 1]])2 (pts[[i 1, 2]] pts[[i, 2]])2 ], {i, 1, n}] NIntegrate[ Sqrt[ 1 f'[x]2 ],{x, a, b}]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 4 ax 1 b
1 12 ;
Section 6.4 Areas of Surfaces of Revolution 6.4 AREAS OF SURFACES OF REVOLUTION 1.
(a)
dy dx
#
(b)
% œ sec# x Ê Š dy dx ‹ œ sec x
Ê S œ 21'0
1Î4
(tan x) È1 sec% x dx
(c) S ¸ 3.84
2. (a)
dy dx
#
(b)
2 œ 2x Ê Š dy dx ‹ œ 4x
Ê S œ 21'0 x# È1 4x# dx 2
(c) S ¸ 53.23
3. (a) xy œ 1 Ê x œ Ê S œ 21'1
2
" y
" y
Ê
dx dy
#
dx œ y"# Ê Š dy ‹ œ
" y%
(b)
È1 y% dy
(c) S ¸ 5.02
4. (a)
dx dy
#
# œ cos y Ê Š dx dy ‹ œ cos y
(b)
Ê S œ 21'0 (sin y) È1 cos# y dy 1
(c) S ¸ 14.42
# 5. (a) x"Î# y"Î# œ 3 Ê y œ ˆ3 x"Î# ‰ "Î# ‰ ˆ ˆ Ê dy "# x"Î# ‰ dx œ 2 3 x #
"Î# ‰ ˆ Ê Š dy dx ‹ œ 1 3x
(b)
#
# # Ê S œ 21'1 ˆ3 x"Î# ‰ É1 a1 3x"Î# b dx 4
(c) S ¸ 63.37
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353
354
Chapter 6 Applications of Definite Integrals #
"Î# ‰ ˆ œ 1 y"Î# Ê Š dx dy ‹ œ 1 y
dx dy
6. (a)
#
(b)
# Ê S œ 21 '1 ˆy 2Èy‰ É1 a1 y"Î# b dx 2
(c) S ¸ 51.33
#
(b)
# œ tan y Ê Š dx dy ‹ œ tan y
dx dy
7. (a)
Ê S œ 21'0 Š'0 tan t dt‹ È1 tan# y dy 1Î3
y
œ 21'0 Š'0 tan t dt‹ sec y dy 1Î3
y
(c) S ¸ 2.08
#
(b)
# œ Èx# 1 Ê Š dy dx ‹ œ x 1
dy dx
8. (a)
È5
Ê S œ 21'1 Š'1 Èt# 1 dt‹ È1 ax# 1b dx
È5
x
œ 21'1 Š'1 Èt# 1 dt‹ x dx x
(c) S ¸ 8.55
9. y œ œ
Ê
x #
1È5 #
dy dx
' ˆ x ‰ É1 œ "# ; S œ 'a 21y Ê1 Š dy dx ‹ dx Ê S œ 0 21 #
" 4
dx œ
1È5 #
'04 x dx
% !
Ê x œ 2y Ê
x #
4
# ’ x# “ œ 41È5; Geometry formula: base circumference œ 21(2), slant height œ È4# 2# œ 2È5
Ê Lateral surface area œ
10. y œ
#
b
dx dy
" #
(41) Š2È5‹ œ 41È5 in agreement with the integral value
# È È ' È # ' œ 2; S œ 'c 21x Ê1 Š dx dy ‹ dy œ 0 21 † 2y 1 2 dy œ 41 5 0 y dy œ 21 5 cy d ! #
d
2
2
#
œ 21È5 † 4 œ 81È5; Geometry formula: base circumference œ 21(4), slant height œ È4# 2# œ 2È5 Ê Lateral surface area œ " (81) Š2È5‹ œ 81È5 in agreement with the integral value #
11.
dy dx
' œ "# ; S œ 'a 21yÊ1 Š dy dx ‹ dx œ 1 21 #
b
3
(x 1) #
É1 ˆ "# ‰# dx œ
1È5 #
'13 (x 1) dx
œ
1È5 #
#
’ x# x“
$ "
1È5 #
È ˆ 9# 3‰ ˆ "# 1‰‘ œ 1 # 5 (4 2) œ 31È5; Geometry formula: r" œ "# "# œ 1, r# œ 3# "# œ 2, œ slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(r" r# ) ‚ slant height œ 1(1 2)È5
œ 31È5 in agreement with the integral value 12. y œ
x #
" #
Ê x œ 2y 1 Ê
dx dy
dx œ 2; S œ 'c 21x Ê1 Š dy ‹ dy œ '1 21(2y 1)È1 4 dy œ 21È5 '1 (2y 1) dy d
#
2
#
œ 21È5 cy# yd " œ 21È5 [(4 2) (1 1)] œ 41È5; Geometry formula: r" œ 1, r# œ 3,
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
Section 6.4 Areas of Surfaces of Revolution slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(1 3)È5 œ 41È5 in agreement with the integral value 13.
dy dx
#
x# 3
œ
’u œ 1
x% 9
Ê S œ '0
2
x% 9
Ê Š dy dx ‹ œ Ê du œ
4 9
x œ 0 Ê u œ 1, x œ 2 Ê u œ Ä S œ 21 '1
25Î9
14.
œ
1 3
dy dx
œ
1 2 " 4 du œ # 3 1 ˆ 12527 ‰ œ 98811 3 27 #
Ê S œ '3Î4 21Èx É1 15Î4
15Î4
15.
œ
’ˆ 15 4
œ
41 3
(8 1) œ
dy dx
" (2 2x) # È2x x#
œ
ˆ 34
dx;
dx;
#&Î*
u$Î# ‘ "
" 4x
dx
$Î# dx œ 21 ’ 23 ˆx 4" ‰ “
" 4
" ‰$Î# 4
41 3
x$ 9
du œ
x% 9
" 4x
x"Î# Ê Š dy dx ‹ œ
œ 21'3Î4 Éx
É1
25 ‘ 9
u"Î# †
ˆ 125 ‰ 27 1 œ " #
" 4
x$ dx Ê
21 x$ 9
" ‰$Î# “ 4
"&Î% $Î%
41 3
$ ’ˆ 42 ‰
Ê Š dy dx ‹ œ
(1 x)# 2x x#
œ
1“
281 3
œ
#
1x È2x x#
Ê S œ '0 5 21È2x x# É1 1Þ5 Þ
œ 21'0 5 È2x x#
È2x x#
1Þ5
(1 x)# 2x x#
1 2x x#
È2x x#
Þ
dx
dx
œ 21'0 5 dx œ 21[x]"Þ& !Þ& œ 21 1Þ5 Þ
16.
dy dx
" 2È x 1
œ
#
dy Ê Š dx ‹ œ
" 4(x 1)
Ê S œ '1 21Èx 1 É1 5
œ 21'1 É(x 1) 5
œ 21 ’ 23 ˆx
17.
œ
41 3
œ
1 6
dx dy
" 4
& 5 ‰$Î# “ 4 "
981 6
dx
dx œ 21'1 Éx 5
œ
41 3
‰$Î# ˆ 94 ‰$Î# “ œ ’ˆ 25 4
(125 27) œ
" 4(x 1)
œ
5 4
dx
$Î# $Î# ’ˆ5 45 ‰ ˆ1 45 ‰ “ 41 3
$
Š 52$
3$ 2$ ‹
491 3
% ' œ y# Ê Š dx dy ‹ œ y Ê S œ 0 #
1
u œ 1 y% Ê du œ 4y$ dy Ê
" 4
21 y$ 3
È1 y% dy;
du œ y$ dy; y œ 0
Ê u œ 1, y œ 1 Ê u œ 2d Ä S œ '1 21 ˆ "3 ‰ u"Î# ˆ 4" du‰ 2
œ
1 6
'12 u"Î# du œ 16 32 u$Î# ‘ #" œ 19 ŠÈ8 1‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
355
356
Chapter 6 Applications of Definite Integrals
18. x œ ˆ "3 y$Î# y"Î# ‰ Ÿ 0, when 1 Ÿ y Ÿ 3. To get positive area, we take x œ ˆ "3 y$Î# y"Î# ‰ Ê
dx dy
#
œ "# ˆy"Î# y"Î# ‰ Ê Š dx dy ‹ œ
" 4
ay 2 y" b
Ê S œ '1 21 ˆ "3 y$Î# y"Î# ‰ É1 4" ay 2 y" b dy 3
œ 21'1 ˆ 3" y$Î# y"Î# ‰ É 4" ay 2 y" b dy 3
Éay"Î# y "Î# b#
œ 21'1 ˆ "3 y$Î# y"Î# ‰ 3
œ 1'1 ˆ "3 y# 3
2 3
dx dy
œ
" È 4 y
œ 41 '0
15Î4
œ
20.
dx dy
81 3
œ
3
$
y# 3
$
y“ œ 1 ˆ 27 9 "
161 9
#
Ê S œ '0
15Î4
" 4 y
Ê Š dx dy ‹ œ
21 † 2È4 y É1
È5 y dy œ 41 23 (5 y)$Î# ‘ "&Î% œ 831 ’ˆ5 !
Š 5È 5 " È2y 1
dy œ 1'1 y"Î# ˆ 3" y 1‰ Šy"Î#
y 1‰ dy œ 1 ’ y9
œ 19 (18 1 3) œ 19.
#
5È 5 8 ‹
œ
81 3
#
Š 40
" 2y1
Ê Š dx dy ‹ œ
41 È 2 3
"
œ 21È2 23 y$Î# ‘ &Î) œ
È 5 5 È 5 ‹ 8
œ 1
$Î#
3
3‰ ˆ "9
9 3
" 4y
15 ‰$Î# 4
" 3
dy œ 41'0
15Î4
1‰‘ œ 1 ˆ3
" 9
" 3
1‰
È(4 y) 1 dy
$Î# 5$Î# “ œ 831 ’ˆ 45 ‰ 5$Î# “
“œ
41 È 2 3
Š1
#
dy œ 21'5Î8 È(2y 1) 1 dy œ 21'5Î8 È2 y"Î# dy 1
" 2y 1
5È 5 ‹ 8È 8
41 È 2 3
œ
21. S œ 21'1Î2 È2y 1 Ê1 Š È2y" 1 ‹ dy œ 21 '1Î2 È2y 1 É1 1
dy œ 1 '1 ˆ 3" y 1‰ (y 1) dy
351È5 3
Ê S œ '5Î8 21È2y 1 É1
’1$Î# ˆ 85 ‰
" ‹ y"Î#
1
" 2y1
1
È
5 Š 8†2 8†22È 2
È5
‹œ
1 12
Š16È2 5È5‹
dy œ 21'1Î2 È2y 1 É 2y2y1 dy 1
1 3 œ 21 '1Î2 È2y dy œ 2È2 1 '1Î2 Èy dy œ 2È2 1 23 y3Î2 ‘1Î2 œ 2È2 1 ”Š 23 È13 ‹ Œ 23 Ɉ "# ‰ • œ 2È2 1 Š 23 1
1
È œ 2È2 1 Š 2 È21 ‹ œ 3 2
22. y œ
" 3
ax# 2b
21 3
Š2È2 1‹
È2
Ê dy œ xÈx# 2 dx Ê ds œ È1 a2x# x% b dx Ê S œ 21'0 x È1 2x# x% dx
$Î#
È2
È2
œ 21'0 xÉax# 1b# dx œ 21'0 23. ds œ Èdx# dy# œ ÊŠy$ œ ÊŠy$
" 4y$ ‹
#
dy œ Šy$
" 4y$ ‹
#
" 4y$ ‹
È2
x ax# 1b dx œ 21'0 ax$ xb dx œ 21 ’ x4
1 dy œ ÊŠy'
" #
%
" 16y' ‹
dy; S œ '1 21y ds œ 21'1 y Šy$ 2
2
#
&
1 dy œ ÊŠy'
"‰ "‰ ˆ " " ‰‘ œ 21 ˆ 31 œ 21 ’ y5 4" y" “ œ 21 ˆ 32 5 8 5 4 5 8 œ "
24. y œ cos x Ê
dy dx
" #
œ 21 ˆ 44 22 ‰ œ 41
" 16y' ‹
dy
dy œ 21'1 ˆy% "4 y# ‰ dy 2
(8 † 31 5) œ
2531 20
# È1 sin# x dx ' œ sin x Ê Š dy dx ‹ œ sin x Ê S œ 21 c1Î2 (cos x)
25. y œ Èa# x# Ê
#
dy dx
œ
" #
aa# x# b
Ê S œ 21'ca Èa# x# É1 a
21 40
" 4y$ ‹
È# x# # “!
"Î#
x# aa # x # b
1Î2
(2x) œ
x È a# x#
#
Ê Š dy dx ‹ œ
x# aa # x # b
dx œ 21'ca Èaa# x# b x# dx œ 21'ca a dx œ 21a[x]ca a a
a
œ 21a[a (a)] œ (21a)(2a) œ 41a#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" ‹ 3È 2
Section 6.4 Areas of Surfaces of Revolution 26. y œ œ
r h
21 r h
x Ê
dy dx
#
É h h# r
#
œ
r h
#
Ê Š dy dx ‹ œ
Ê S œ 21 '0
h
r# h#
r h
x É1
r# h#
dx œ 21'0
h
r h
#
#
x É h h# r dx
'0h x dx œ 2h1r Èh# r# ’ x# “ h œ 2h1r Èh# r# Š h# ‹ œ 1rÈh# r# #
#
#
#
0
# # # È16# y# 27. The area of the surface of one wok is S œ 'c 21x Ê1 Š dx dy ‹ dy. Now, x y œ 16 Ê x œ #
d
Ê
dx dy
œ c7
y È16# y#
#
Ê Š dx dy ‹ œ
c7
; S œ 'c16 21È16# y# É1
y# 16# y#
y# 16# y#
c7
dy œ 21'c16 Èa16# y# b y# dy
œ 21'c16 16 dy œ 321 † 9 œ 2881 ¸ 904.78 cm# . The enamel needed to cover one surface of one wok is
V œ S † 0.5 mm œ S † 0.05 cm œ (904.78)(0.05) cm$ œ 45.24 cm$ . For 5000 woks, we need 5000 † V œ 5000 † 45.24 cm$ œ (5)(45.24)L œ 226.2L Ê 226.2 liters of each color are needed. 28. y œ Èr# x# Ê œ 21'a
abh
œ 21'a
œ "#
2x È r# x #
œ
Èar# x# b x# dx œ 21r' a
29. y œ ÈR# x# Ê abh
dy dx
dy dx
œ "#
2x È R # x#
abh
œ
x# r# x # ;
S œ 21 'a
abh
Èr# x# É1
x# r# x #
dx
dx œ 21rh, which is independent of a.
abh
30. (a) x# y# œ 45# Ê x œ È45# y# Ê 45
Ê Š dx dy ‹ œ
#
x È R # x#
ÈaR# x# b x# dx œ 21R ' a
S œ 'c22Þ5 21 È45# y# É1
#
x È r# x #
y# 45# y#
dx Ê Š dy ‹ œ
x# R # x# ;
S œ 21'a
abh
ÈR# x# É1
x# R # x#
dx œ 21Rh dx dy
œ
y È45# y#
dy œ 21 '
45 22Þ5
#
Ê Š dx dy ‹ œ
y# 45# y#
;
Èa45# y# b y# dy œ 21 † 45'
45 22Þ5
dy
œ (21)(45)(67.5) œ 60751 square feet (b) 19,085 square feet 31. (a) An equation of the tangent line segment is (see figure) y œ f(mk ) f w (mk )(x mk ). When x œ xkc1 we have r" œ f(mk ) f w (mk )(x5 1 mk ) œ f(mk ) f w (mk ) ˆ ?#xk ‰ œ f(mk ) f w (mk ) when x œ xk we have r# œ f(mk ) f w (mk )(x5 mk ) k œ f(mk ) f w (mk ) ?x # ;
(b) L#k œ (?xk )# (r# r" )#
?x k #
;
#
ˆf w (mk ) ?#xk ‰‘ œ (?xk )# [f w (mk )?xk ]# Ê Lk œ È(?xk )# [f w (mk )?xk ]# , as claimed (c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent œ (?xk )# f w (mk )
?x k #
line segment about the x-axis is given by ?Sk œ 1(r" r# )Lk œ 1[2f(mk )] Éa?xk b# [f w (mk )?xk ]# using parts (a) and (b) above. Thus, ?Sk œ 21f(mk ) È1 [f w (mk )]# ?xk . ! ?Sk œ lim ! 21f(mk ) È1 [f w (mk )]# ?xk œ ' 21f(x) È1 [f w (x)]# dx (d) S œ n lim Ä_ nÄ_ a kœ1 kœ1 n
32. y œ ˆ1 x#Î$ ‰
$Î#
n
Ê
dy dx
œ
Ê S œ 2'0 21 ˆ1 x#Î$ ‰ 1
3 #
b
ˆ1 x#Î$ ‰"Î# ˆ 32 x"Î$ ‰ œ
$Î#
ˆ1x#Î$ ‰"Î# x"Î$
#
Ê Š dy dx ‹ œ
1x#Î$ x#Î$
œ
" x#Î$
$Î# " É1 ˆ x#Î$ 1‰ dx œ 41'0 ˆ1 x#Î$ ‰ Èx#Î$ dx 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
dx
357
358
Chapter 6 Applications of Definite Integrals $Î# œ 41'0 ˆ1 x#Î$ ‰ x"Î$ dx; u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê 32 du œ x"Î$ dx; 1
! x œ 0 Ê u œ 1, x œ 1 Ê u œ 0d Ä S œ 41'1 u$Î# ˆ 3# du‰ œ 61 52 u&Î# ‘ " œ 61 ˆ0 52 ‰ œ 0
121 5
6.5 WORK AND FLUID FORCES 1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) œ kx. The work done by F is W œ '0 F(x) dx œ k '0 x dx œ 3
3
k #
$
cx# d ! œ
9k # .
This work is equal to 1800 J Ê
2. (a) We find the force constant from Hooke's Law: F œ kx Ê k œ
Ê kœ
F x
800 4
9 #
k œ 1800 Ê k œ 400 N/m
œ 200 lb/in.
(b) The work done to stretch the spring 2 inches beyond its natural length is W œ '0 kx dx œ 200 '0 x dx œ 200 ’ x# “ 2
2
#
œ 200(2 0) œ 400 in † lb œ 33.3 ft † lb (c) We substitute F œ 1600 into the equation F œ 200x to find 1600 œ 200x Ê x œ 8 in.
# !
3. We find the force constant from Hooke's law: F œ kx. A force of 2 N stretches the spring to 0.02 m Ê 2 œ k † (0.02) N 4N Ê k œ 100 m . The force of 4 N will stretch the rubber band y m, where F œ ky Ê y œ Fk Ê y œ 100 N Ê y œ 0.04 m m
œ 4 cm. The work done to stretch the rubber band 0.04 m is W œ '0
0Þ04
œ
(100)(0.04)# #
kx dx œ 100 '0
0Þ04
#
x dx œ 100 ’ x# “
!Þ!% !
œ 0.08 J
4. We find the force constant from Hooke's law: F œ kx Ê k œ
F x
Ê kœ
90 1
Ê k œ 90
N m.
The work done to stretch the
‰ spring 5 m beyond its natural length is W œ '0 kx dx œ 90 '0 x dx œ 90 ’ x# “ œ (90) ˆ 25 # œ 1125 J 5
5
#
& !
5. (a) We find the spring's constant from Hooke's law: F œ kx Ê k œ
œ
F x
21,714 8 5
œ
21,714 3
Ê k œ 7238
lb in
(b) The work done to compress the assembly the first half inch is W œ '0 kx dx œ 7238 '0 x dx œ 7238 ’ x# “ 0Þ5
#
œ (7238) (0.5) # œ
(7238)(0.25) #
1Þ0
1Þ0
Þ
Þ
#
"Þ! !Þ&
œ
7238 #
c1 (0.5)# d œ
6. First, we find the force constant from Hooke's law: F œ kx Ê k œ compresses the scale x œ this far is W œ '0
1Î8
#
!Þ& !
¸ 905 in † lb. The work done to compress the assembly the second half inch is:
W œ '0 5 kx dx œ 7238 '0 5 x dx œ 7238 ’ x# “
" 8
0Þ5
in, he/she must weigh F œ kx œ #
kx dx œ 2400 ’ x# “
"Î) !
œ
2400 2†64
œ
F x
2,400 ˆ 8" ‰
œ 18.75 lb † in. œ
(7238)(0.75) #
150 " ‰ ˆ 16
¸ 2714 in † lb
œ 16 † 150 œ 2,400
lb in .
If someone
œ 300 lb. The work done to compress the scale
25 16
ft † lb
7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to x, the length of the rope still hanging: F(x) œ 0.624x. The work done is: W œ '0 F(x) dx œ '0 0.624x dx œ 0.624 ’ x# “ 50
50
#
œ 780 J
&! !
8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the ground is Faxb œ "%% %x. The work done is: W œ 'a F(x) dx œ '0 a"%% %xbdx œ c144x 2x# d ! œ 1944 ft † lb b
18
")
9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) œ (4.5)(180 x) where x is the position of the car off the first floor. The work done is: W œ '0
180
F(x) dx œ 4.5'0
180
(180 x) dx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.5 Work and Fluid Forces œ 4.5 ’180x
")! x# # “!
180# # ‹
œ 4.5 Š180#
œ
4.5†180# #
359
œ 72,900 ft † lb
10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F(x) œ xk# . The work done b is W œ 'a xk# dx œ k 'a x"# dx œ k x" ‘ a œ k ˆ b" "a ‰ œ b
b
k(a b) ab
11. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is F œ 0.8a#! xb. So: W œ '0 0.8a#! xb dx œ 0.8 ’20x 20
#! x# # “!
œ 160 ft † lb.
12. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is F œ 2a#! xb. So: W œ '0 2a#! xb dx œ 2 ’20x 20
#! x# # “!
œ 400 ft † lb.
Note that since the force in Exercise 12 is 2.5 times the force in Exercise 11 at each elevation, the total work is also 2.5 times as great. 13. We will use the coordinate system given. (a) The typical slab between the planes at y and y ?y has a volume of ?V œ (10)(12) ?y œ 120 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 62.4 ?V œ 62.4 † 120 ?y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ?W œ force ‚ distance œ 62.4 † 120 † y † ?y ft † lb. The work it takes to lift all 20
the water is approximately W ¸ ! ?W 0 20
œ ! 62.4 † 120y † ?y ft † lb. This is a Riemann sum for 0
the function 62.4 † 120y over the interval 0 Ÿ y Ÿ 20. The work of pumping the tank empty is the limit of these sums: W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “ 20
#
#! !
‰ œ (62.4)(120)(200) œ 1,497,600 ft † lb œ (62.4)(120) ˆ 400 #
5 ‰ (b) The time t it takes to empty the full tank with ˆ 11 –hp motor is t œ
W †lb 250 ftsec
œ
1,497,600 ft†lb †lb 250 ftsec
œ 5990.4 sec œ 1.664 hr
Ê t ¸ 1 hr and 40 min (c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “ 10
#
œ 1497.6 sec œ 0.416 hr ¸ 25 min (d) In a location where water weighs 62.26
"! !
‰ œ 374,400 ft † lb and the time is t œ œ (62.4)(120) ˆ 100 #
lb ft$ :
a) W œ (62.26)(24,000) œ 1,494,240 ft † lb. b) t œ 1,494,240 œ 5976.96 sec ¸ 1.660 hr Ê t ¸ 1 hr and 40 min 250 In a location where water weighs 62.59
lb ft$
a) W œ (62.59)(24,000) œ 1,502,160 ft † lb b) t œ 1,502,160 œ 6008.64 sec ¸ 1.669 hr Ê t ¸ 1 hr and 40.1 min 250
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
W †lb 250 ftsec
360
Chapter 6 Applications of Definite Integrals
14. We will use the coordinate system given. (a) The typical slab between the planes at y and y ?y has a volume of ?V œ (20)(12) ?y œ 240 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 62.4 ?V œ 62.4 † 240 ?y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ?W œ force ‚ distance 20
œ 62.4 † 240 † y † ?y ft † lb. The work it takes to lift all the water is approximately W ¸ ! ?W 10 20
œ ! 62.4 † 240y † ?y ft † lb. This is a Riemann sum for the function 62.4 † 240y over the interval 10
10 Ÿ y Ÿ 20. The work it takes to empty the cistern is the limit of these sums: W œ '10 62.4 † 240y dy 20
#
œ (62.4)(240) ’ y# “ (b) t œ
W †lb 275 ftsec
œ
#!
œ (62.4)(240)(200 50) œ (62.4)(240)(150) œ 2,246,400 ft † lb
"! 2,246,400 ft†lb 275
¸ 8168.73 sec ¸ 2.27 hours ¸ 2 hr and 16.1 min
(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is W œ '10 62.4 † 240y dy œ (62.4)(240) ’ y# “ 15
#
Then the time is t œ
W †lb 275 ftsec
œ
936,000 #75
"& "!
œ (62.4)(240) ˆ 225 #
100 ‰ #
‰ œ 936,000 ft. œ (62.4)(240) ˆ 125 #
¸ 3403.64 sec ¸ 56.7 min lb ft$ :
(d) In a location where water weighs 62.26
a) W œ (62.26)(240)(150) œ 2,241,360 ft † lb. b) t œ 2,241,360 œ 8150.40 sec œ 2.264 hours ¸ 2 hr and 15.8 min 275 ‰ œ 933,900 ft † lb; t œ 933,900 c) W œ (62.26)(240) ˆ 125 # #75 œ 3396 sec ¸ 0.94 hours ¸ 56.6 min lb ft$
In a location where water weighs 62.59
a) W œ (62.59)(240)(150) œ 2,253,240 ft † lb. b) t œ 2,253,240 œ 8193.60 sec œ 2.276 hours ¸ 2 hr and 16.56 min 275 ‰ œ 938,850 ft † lb; t œ 938,850 c) W œ (62.59)(240) ˆ 125 # 275 ¸ 3414 sec ¸ 0.95 hours ¸ 56.9 min #
15. The slab is a disk of area 1x# œ 1ˆ y# ‰ , thickness ˜y, and height below the top of the tank a"! yb. So the work to pump #
the oil in this slab, ˜W, is 57a"! yb1ˆ y# ‰ . The work to pump all the oil to the top of the tank is W œ '0
10
571 # 4 a"!y
y$ bdy œ
571 4
$
’ "!$y
"! y% % “!
œ 11,8751 ft † lb ¸ 37,306 ft † lb. #
16. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is a"% yba1bˆ y# ‰ and since the tank is half full and the volume of the original cone is V œ "$ 1r# h œ "$ 1a&# ba"!b œ with half the volume the cone is filled to a height y, œ
571 "%y$ 4 ’ $
$ È &!! y% “ % !
#&!1 '
#&!1 $
ft3 , half the volume œ $ È
$ œ $" 1 y% y Ê y œ È &!! ft. So W œ '0 #
&!!
#&!1 '
571 # 4 a"%y
ft3 , and y$ b dy
¸ 60,042 ft † lb. #
‰ ?y 17. The typical slab between the planes at y and and y ?y has a volume of ?V œ 1(radius)# (thickness) œ 1 ˆ 20 # œ 1 † 100 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 51.2 ?V œ 51.2 † 1001 ?y lb Ê F œ 51201 ?y lb. The distance through which F must act is about (30 y) ft. The work it takes to lift all the 30
30
kerosene is approximately W ¸ ! ?W œ ! 51201(30 y) ?y ft † lb which is a Riemann sum. The work to pump the 0
0
tank dry is the limit of these sums: W œ '0 51201(30 y) dy œ 51201 ’30y 30
¸ 7,238,229.48 ft † lb
$! y# # “!
‰ œ (5120)(4501) œ 51201 ˆ 900 #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.5 Work and Fluid Forces 18. (a) Follow all the steps of Example 5 but make the substitution of 64.5 W œ '0
8
œ
64.51 4
64.51†8$ 3
(10 y)y# dy œ
64.51 4
$
’ 10y 3
%
y 4
)
“ œ !
64.51 4
$
Š 103†8
lb ft$ %
8 4
lb ft$ .
for 57
361
Then,
1‰ ‰ a8$ b ˆ 10 ‹ œ ˆ 64.5 4 3 2
œ 21.51 † 8$ ¸ 34,582.65 ft † lb
(b) Exactly as done in Example 5 but change the distance through which F acts to distance ¸ (13 y) ft. Then W œ '0
8
571 4
(13 y)y# dy œ
571 4
$
’ 13y 3
œ (191) a8# b (7)(2) ¸ 53,482.5 ft † lb
) y% 4 “!
œ
571 4
$
Š 133†8
8% 4‹
‰ œ ˆ 5741 ‰ a8$ b ˆ 13 3 2 œ
571†8$ †7 3 †4
#
19. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness) œ 1 ˆÈy‰ ?y ft$ . #
The force F(y) required to lift this slab is equal to its weight: F(y) œ 73 † ?V œ 731 ˆÈy‰ ?y œ 731 y ?y lb. The distance through which F(y) must act to lift the slab to the top of the reservoir is about a4 yb ft, so the work done is approximately ?W ¸ 731 y a4 yb?y ft † lb. The work done lifting all the slabs from y œ 0 ft to y œ 4 ft is n
approximately W ¸ ! 731 yk a4 yk b?y ft † lb. Taking the limit of these Riemann sums as n Ä _, we get kœ0
4 4 4 1 ‰ 2336 W œ '0 731 y a4 ybdy œ 731'0 a4y y2 bdy œ 731 2y# 13 y$ ‘ 0 œ 731ˆ32 64 ft † lb. 3 œ 3
20. The typical slab between the planes at y and y?y has a volume of about ?V œ (length)(width)(thickness) œ ˆ2È25 y2 ‰a10b ?y ft$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 53 † ?V œ 53ˆ2È25 y2 ‰a10b ?y œ 1060È25 y2 ?y lb. The distance through which F(y) must act to lift the slab to the level of 15 m above the top of the reservoir is about a20 yb ft, so the work done is approximately ?W ¸ 1060È25 y2 a20 yb?y ft † lb. The work done lifting all the slabs from y œ 5 ft to y œ 5 ft is n
approximately W ¸ ! 1060É25 y2k a20 yk b?y ft † lb. Taking the limit of these Riemann sums as n Ä _, we get kœ0
W œ 'c5 1060È25 y2 a20 ybdy œ 1060'c5 a20 ybÈ25 y2 dy œ 1060”'c5 20 È25 y2 dy 'c5 y È25 y2 dy• 5
5
5
5
To evaluate the first integral, we use we can interpret 'c5 È25 y2 dy as the area of the semicircle whose radius is 5, thus 5
'c55 20È25 y2 dy œ 20'c55 È25 y2 dy œ 20 "# 1a5b2 ‘ œ 2501. To evaluate the second integral let u œ 25 y2 5 0 Ê du œ 2y dy; y œ 5 Ê u œ 0, y œ 5 Ê u œ 0, thus 'c5 y È25 y2 dy œ "# '0 Èu du œ 0. Thus,
1060”'c5 20 È25 y2 dy 'c5 y È25 y2 dy• œ 1060a2501 0b œ 2650001 ¸ 832522 ft † lb. 5
5
21. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈ25 y# ‰ ?y m$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 9800 † ?V #
œ 98001 ˆÈ25 y# ‰ ?y œ 98001 a25 y# b ?y N. The distance through which F(y) must act to lift the slab to the level of 4 m above the top of the reservoir is about (4 y) m, so the work done is approximately ?W ¸ 98001 a25 y# b (4 y) ?y N † m. The work done lifting all the slabs from y œ 5 m to y œ 0 m is 0
approximately W ¸ ! 98001 a25 y# b (4 y) ?y N † m. Taking the limit of these Riemann sums, we get c5
W œ 'c5 98001 a25 y# b (4 y) dy œ 98001 'c5 a100 25y 4y# y$ b dy œ 98001 ’100y 0
0
œ 98001 ˆ500
25†25 #
4 3
† 125
625 ‰ 4
25 #
y# 34 y$
¸ 15,073,099.75 J
22. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈ100 y# ‰ ?y œ 1 a100 y# b ?y ft$ . The force is F(y) œ 56ft$lb † ?V œ 561 a100 y# b ?y lb. The distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
! y% 4 “ &
362
Chapter 6 Applications of Definite Integrals
(12 y) ft, so the work done is ?W ¸ 561 a100 y# b (12 y) ?y lb † ft. The work done lifting all the slabs 10
from y œ 0 ft to y œ 10 ft is approximately W ¸ ! 561 a100 y# b (12 y) ?y lb † ft. Taking the limit of these 0
Riemann sums, we get W œ '0 561 a100 y b (12 y) dy œ 561'0 a100 y# b (12 y) dy 10
10
#
œ 561'0 a1200 100y 12y# y$ b dy œ 561 ’1200C 10
œ 561 ˆ12,000
10,000 #
4 † 1000
10,000 ‰ 4
100y# #
12y$ 3
"! y% 4 “!
œ (561) ˆ12 5 4 5# ‰ (1000) ¸ 967,611 ft † lb.
It would cost (0.5)(967,611) œ 483,805¢ œ $4838.05. Yes, you can afford to hire the firm. 23. F œ m œ
" #
dv dt
œ mv
dv dx #
by the chain rule Ê W œ 'x mv x#
m cv# (x# ) v (x" )d œ
24. weight œ 2 oz œ
dv dx
"
" #
weight 32
œ
" 8
œ
3#
" #56
26. weight œ 1.6 oz œ 0.1 lb Ê m œ
0.1 lb 32 ft/sec#
œ
ft 27. v1 œ 0 mph œ 0 sec , v2 œ 153 mph œ 224.4
W œ 'x Faxb dx œ "
28. weight œ 6.5 oz œ
" #
6.5 16
mv##
"
dv ‰ dx
dx œ m "# v# (x)‘ x" x#
" slugs; W œ ˆ #" ‰ ˆ #56 slugs‰ (160 ft/sec)# ¸ 50 ft † lb
hr 1 min 5280 ft 25. 90 mph œ 901 hrmi † 601 min † 60 sec † 1 mi œ 132 ft/sec; m œ 0.3125 lb ‰ # W œ ˆ "# ‰ ˆ 32 ft/sec# (132 ft/sec) ¸ 85.1 ft † lb
x#
x#
mv## "# mv"# , as claimed.
lb; mass œ
2 16
dx œ m'x ˆv
" #
mv1#
lb Ê m œ
œ
" 3 #0
ft sec ;
œ
0.3125 32
slugs;
slugs; W œ ˆ "# ‰ ˆ 3"#0 slugs‰ (280 ft/sec)# œ 122.5 ft † lb
2 oz œ 0.125 lb Ê m œ
2 "ˆ " ‰ # 256 a224.4b
6.5 (16)(32)
0.3125 lb 32 ft/sec#
2 "ˆ " ‰ # 256 a!b
0.125 lb 32 ft/sec#
œ
" 256
slugs;
œ 98.35 ft-lb.
6.5 slugs; W œ ˆ "# ‰ Š (16)(32) slugs‹ (132 ft/sec)# ¸ 110.6 ft † lb
29. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [!ß (]. The typical slab between the planes at y and y ?y has a volume of about?V œ 1(radius)# (thickness) #
œ 1 ˆ y 1417.5 ‰ ?y in$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ
4 9
?V œ
41 9
#
ˆ y 1417.5 ‰ ?y oz.
The distance through which F(y) must act to lift this slab to the level of 1 inch above the top is about (8 y) in. The work #
b done lifting the slab is about ?W œ ˆ 491 ‰ ay 1417.5 a8 yb?y in † oz. The work done lifting all the slabs from y œ 0 to # 7
y œ 7 is approximately W œ ! 0
41 9†14# (y
17.5)# (8 y) ?y in † oz which is a Riemann sum. The work is the limit of these
sums as the norm of the partition goes to zero: W œ '0
7
œ
41 9†14#
œ
41 9†14#
'07 a2450 26.25y 27y# y$ bdy œ 9†4141 7% 4
’
$
9†7
30. Work œ '6 370 000
35ß780ß000 ß
ß
1000 MG r#
26.25 #
41 9†14# ay %
#
17.5b# a8 ybdy
’ y4 9y$
26.25 #
y# 2450y“
( !
#
† 7 2450 † 7“ ¸ 91.32 in † oz
dr œ 1000 MG '6 370 000
35ß780ß000 ß
ß
" œ (1000) a5.975 † 10#% b a6.672 † 10"" b Š 6,370,000
dr r#
$&ß()!ß!!!
œ 1000 MG "r ‘ 'ß$(!ß!!!
" 35,780,000 ‹
¸ 5.144 ‚ 10"! J
31. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's right-hand edge: y œ x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then x œ 5 y and the total width is L(y) œ 2x œ 2(5 y). The depth of the strip is (y). The force exerted by the c2
c2
water against one side of the plate is therefore F œ 'c5 w(y) † L(y) dy œ 'c5 62.4 † (y) † 2(5 y) dy
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.5 Work and Fluid Forces c2
œ 124.8 'c5 a5y y# b dy œ 124.8 5# y# "3 y$ ‘ & œ 124.8 ˆ 5# † 4 œ
(124.8) ˆ 105 #
#
117 ‰ 3
œ
(124.8) ˆ 315 6 234 ‰
" 3
† 8‰ ˆ 5# † 25
" 3
363
† 125‰‘
œ 1684.8 lb
32. An equation for the line of the plate's right-hand edge is y œ x 3 Ê x œ y 3. Thus the total width is L(y) œ 2x œ 2(y 3). The depth of the strip is (2 y). The force exerted by the water is F œ 'c3 w(2 y)L(y) dy œ 'c3 62.4 † (2 y) † 2(3 y) dy œ 124.8'c3 a6 y y# b dy œ 124.8 ’6y 0
0
œ (124.8) ˆ18
9 #
0
y# #
! y$ 3 “ $
‰ 9‰ œ (124.8) ˆ 27 # œ 1684.8 lb
strip 33. (a) The width of the strip is Layb œ 4, the depth of the strip is a10 yb Ê F œ 'a w † Š depth ‹Faybdy b
œ '0 62.4a10 yba4bdy œ 249.6'0 a10 ybdy œ 249.6’10y 3
3
3 y# “ # 0
œ 249.6ˆ30 92 ‰ œ 6364.8 lb
strip (b) The width of the strip is Layb œ 3, the depth of the strip is a10 yb Ê F œ 'a w † Š depth ‹Faybdy b
œ '0 62.4a10 yba3bdy œ 187.2'0 a10 ybdy œ 187.2’10y 4
4
4 y# # “0
œ 187.2a40 8b œ 5990.4 lb
strip 34. The width of the strip is Layb œ 2È25 y2 , the depth of the strip is a6 yb Ê F œ 'a w † Š depth ‹Faybdy b
œ '0 62.4a6 ybˆ2È25 y2 ‰dy œ 124.8'0 a6 ybÈ25 y2 dy œ 124.8”'0 6 È25 y2 dy '0 y È25 y2 dy• 5
5
5
5
To evaluate the first integral, we use we can interpret '0 È25 y2 dy as the area of a quarter circle whose radius is 5, thus 5
'05 6È25 y2 dy œ 6'05 È25 y2 dy œ 6 4" 1a5b2 ‘ œ 7521 . To evaluate the second integral let u œ 25 y2 Ê du œ 2y dy; y œ 0 Ê u œ 25, y œ 5 Ê u œ 0, thus '0 y È25 y2 dy œ "# '25 Èu du œ 5
25 œ 13 u3Î2 ‘ 0 œ
125 3 .
0
Thus, 124.8”'0 6 È25 y2 dy '0 y È25 y2 dy• œ 124.8ˆ 7521 5
5
125 ‰ 3
" #
'025 u1Î2 du
¸ 9502.7 lb.
35. Using the coordinate system of Exercise 32, we find the equation for the line of the plate's right-hand edge to be y œ 2x 4 Ê x œ y # 4 and L(y) œ 2x œ y 4. The depth of the strip is (1 y). (a) F œ 'c4 w(1 y)L(y) dy œ 'c4 62.4 † (1 y)(y 4) dy œ 62.4 'c4 a4 3y y# b dy œ 62.4 ’4y 0
0
œ (62.4) ’(4)(4)
(3)(16) #
(b) F œ (64.0) ’(4)(4)
(3)(16) #
0
64 3 “
œ (62.4) ˆ16 24
64 3 “
œ
(64.0)(120 64) 3
64 ‰ 3
œ
(62.4)(120 64) 3
œ 1164.8 lb
¸ 1194.7 lb
36. Using the coordinate system given, we find an equation for the line of the plate's right-hand edge to be y œ 2x 4 Ê x œ 4#y and L(y) œ 2x œ 4 y. The depth of the strip is (1 y) Ê F œ '0 w(1 y)(4 y) dy 1
œ 62.4'0 ay# 5y 4b dy œ 62.4 ’ y3 1
œ (62.4) ˆ "3
$
5 #
5y# #
4‰ œ (62.4) ˆ 2 156 24 ‰ œ
4y“
"
! (62.4)(11) 6
œ 114.4 lb
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3y# #
! y$ 3 “ %
364
Chapter 6 Applications of Definite Integrals
37. Using the coordinate system given in the accompanying figure, we see that the total width is L(y) œ 63 and the depth of the strip is (33.5 y) Ê F œ '0 w(33.5 y)L(y) dy 33
œ '0
33
64 ‰ † (33.5 y) † 63 dy œ ˆ 12 (63)'0 (33.5 y) dy $ 33
64 1 #$
64 ‰ œ ˆ 12 (63) ’33.5y $
œ
(64)(63)(33)(67 33) (#) a12$ b
$$ y# # “!
‰ ’(33.5)(33) œ ˆ 641#†63 $
33# # “
œ 1309 lb
38. Using the coordinate system given in the accompanying figure, we see that the right-hand edge is x œ È1 y# so the total width is L(y) œ 2x œ 2È1 y# and the depth of the strip is (y). The force exerted by the water is therefore F œ 'c1 w † (y) † 2È1 y# dy 0
œ 62.4'c1 È1 y# d a1 y# b œ 62.4 ’ 23 a1 y# b 0
39. (a) F œ ˆ62.4
lb ‰ ft3 a8
$Î# !
“
"
œ (62.4) ˆ 23 ‰ (1 0) œ 41.6 lb
ftba25 ft2 b œ 12480 lb
strip (b) The width of the strip is Layb œ 5, the depth of the strip is a8 yb Ê F œ 'a w † Š depth ‹Faybdy b
œ '0 62.4a8 yba5bdy œ 312'0 a8 ybdy œ 312’8y 5
5
5 y# “ # 0
œ 312ˆ40
25 ‰ 2
œ 8580 lb
(c) The width of the strip is Layb œ 5, the depth of the strip is a8 yb, the height of the strip is È2 dy 5ÎÈ2
strip Ê F œ 'a w † Š depth ‹Faybdy œ '0 b
40 œ 312È2 Š È 2
25 4 ‹
5ÎÈ2
62.4a8 yba5bÈ2 dy œ 312È2 '0
a8 ybdy œ 312È2 ’8y
œ 9722.3
40. The width of the strip is Layb œ 34 Š2È3 y‹, the depth of the strip is a6 yb, the height of the strip is 2È 3
strip Ê F œ 'a w † Š depth ‹Faybdy œ '0 b
œ
93.6 È È3 ’12y 3
5 ÎÈ 2 y# “ # 0
3y2 y2 È3
62.4a6 yb † 34 Š2È3 y‹ È23 dy œ
2È 3 y3 “ 3 0
œ
93.6 È3 Š72
93.6 È3
'02
È3
2 È3
dy
Š12È3 6y 2yÈ3 y2 ‹dy
36 12È3 8È3 ‹ ¸ 1571.04 lb
41. The coordinate system is given in the text. The right-hand edge is x œ Èy and the total width is L(y) œ 2x œ 2Èy. (a) The depth of the strip is (2 y) so the force exerted by the liquid on the gate is F œ '0 w(2 y)L(y) dy 1
" œ '0 50(2 y) † 2Èy dy œ 100 '0 (2 y)Èy dy œ 100'0 ˆ2y"Î# y$Î# ‰ dy œ 100 43 y$Î# 25 y&Î# ‘ ! 1
1
1
‰ œ 100 ˆ 43 25 ‰ œ ˆ 100 15 (20 6) œ 93.33 lb
2‰ (b) We need to solve 160 œ '0 w(H y) † 2Èy dy for h. 160 œ 100 ˆ 2H 3 5 Ê H œ 3 ftÞ 1
42. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for the line of the end plate's right-hand edge is y œ 5# x Ê x œ 52 y. The total width is L(y) œ 2x œ 54 y and the depth of the typical horizontal strip at level y is (h y). Then the force is F œ '0 w(h y)L(y) dy œ Fmax , h
where Fmax œ 6667 lb. Hence, Fmax œ w'0 (h y) † 45 y dy œ (62.4) ˆ 45 ‰'0 ahy y# b dy h
#
œ (62.4) ˆ 45 ‰ ’ hy#
h
y$ 3 “0
$
œ (62.4) ˆ 45 ‰ Š h#
h
h$ 3‹
3 max ‰ ˆ 54 ‰ ˆ F10.4 œ (62.4) ˆ 45 ‰ ˆ "6 ‰ h$ œ (10.4) ˆ 45 ‰ h$ Ê h œ É
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.6 Moments and Centers of Mass 3 ˆ 54 ‰ ˆ 6667 ‰ œÉ 10.4 ¸ 9.288 ft. The volume of water which the tank can hold is V œ
Height œ h and
" #
(Base) œ
2 5
h Ê Vœ
ˆ 25
#‰
h
#
" #
365
(Base)(Height) † 30, where
#
(30) œ 12h ¸ 12(9.288) ¸ 1035 ft$ .
43. The pressure at level y is p(y) œ w † y Ê the average pressure is p œ #
œ ˆ wb ‰ Š b# ‹ œ
'0b p(y) dy œ b" '0b w † y dy œ b" w ’ y# “ b #
" b
0
wb #
. This is the pressure at level
b #
, which
is the pressure at the middle of the plate. 44. The force exerted by the fluid is F œ '0 w(depth)(length) dy œ '0 w † y † a dy œ (w † a)'0 y dy œ (w † a) ’ y# “ b
œ
# w Š ab# ‹
b
b
#
b 0
‰ œ ˆ wb # (ab) œ p † Area, where p is the average value of the pressure.
45. When the water reaches the top of the tank the force on the movable side is 'c2 (62.4) ˆ2È4 y# ‰ (y) dy 0
œ (62.4)'c2 a4 y# b 0
"Î#
(2y) dy œ (62.4) ’ 23 a4 y# b
$Î# !
“
#
œ (62.4) ˆ 23 ‰ ˆ4$Î# ‰ œ 332.8 ft † lb. The force
compressing the spring is F œ 100x, so when the tank is full we have 332.8 œ 100x Ê x ¸ 3.33 ft. Therefore the movable end does not reach the required 5 ft to allow drainage Ê the tank will overflow. 46. (a) Using the given coordinate system we see that the total width is L(y) œ 3 and the depth of the strip is (3 y).
Thus, F œ '0 w(3 y)L(y) dy œ '0 (62.4)(3 y) † 3 dy 3
3
œ (62.4)(3)'0 (3 y) dy œ (62.4)(3) ’3y 3
$ y# # “!
œ (62.4)(3) ˆ9 9# ‰ œ (62.4)(3) ˆ 9# ‰ œ 842.4 lb
(b) Find a new water level Y such that FY œ (0.75)(842.4 lb) œ 631.8 lb. The new depth of the strip is (Y y) and Y is the new upper limit of integration. Thus, FY œ '0 w(Y y)L(y) dy œ 62.4'0 (Y y) † 3 dy Y
œ (62.4)(3)'0 (Y y) dy œ (62.4)(3) ’Yy Y
Y
y# # “0
Y
œ (62.4)(3) ŠY#
Y# # ‹
#
œ (62.4)(3) Š Y# ‹ . Therefore,
2FY È6.75 ¸ 2.598 ft. So, ?Y œ 3 Y ¸ 3 2.598 ¸ 0.402 ft ¸ 4.8 in Y œ É (62.4)(3) œ É 1263.6 187.2 œ
6.6 MOMENTS AND CENTERS OF MASS 1. Since the plate is symmetric about the y-axis and its density is constant, the distribution of mass is symmetric about the y-axis and the center of mass lies on the y-axis. This means that x œ 0. It remains to find y œ MMx . We model the distribution of mass with @/<>3-+6 strips. The typical strip has center of mass: # (µ x ßµ y ) œ Šxß x 4 ‹ , length: 4 x# , width: dx, area: #
dA œ a4 x# b dx, mass: dm œ $ dA œ $ a4 x# b dx. The moment of the strip about the x-axis is # µ C dm œ Š x #4 ‹ $ a4 x# b dx œ #$ a16 x% b dx. The moment of the plate about the x-axis is Mx œ ' µ C dm œ 'c2 #$ a16 x% b dx œ 2
$ #
’16x
# x& “ 5 #
œ
$ #
’Š16 † 2
2& 5‹
Š16 † 2
2& 5 ‹“
œ
$ †2 #
ˆ32
32 ‰ 5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
128$ 5 .
The mass of the
366
Chapter 6 Applications of Definite Integrals
plate is M œ ' $ a4 x# b dx œ $ ’4x
# x$ “ 3 #
32$ 3 .
œ 2$ ˆ8 83 ‰ œ
Therefore y œ
Mx M
œ
$ Š 128 5 ‹
Š 323$ ‹
œ
12 5 .
The plate's center of
‰ mass is the point (xß y) œ ˆ!ß 12 5 . 2. Applying the symmetry argument analogous to the one in Exercise 1, we find x œ 0. To find y œ MMx , we use the @/<>3-+6 strips technique. The typical strip has center of # mass: (µ x ßµ y ) œ Šxß 25 x ‹ , length: 25 x# , width: dx, #
#
area: dA œ a25 x bdx, mass: dm œ $ dA œ $ a25 x# b dx. The moment of the strip about the x-axis is # µ y dm œ Š 25 # x ‹ $ a25 x# b dx œ
œ 'c5 #$ a25 x# b dx œ 5
#
œ $ † 625 ˆ5 œ 2$ Š5$
10 3
5$ 3‹
$ #
'c55
$ #
a25 x# b dx. The moment of the plate about the x-axis is Mx œ ' µ y dm #
$ #
a625 50x# x% b dx œ
’625x
50 3
x$
& x& 5 “ &
œ 2 † #$ Š625 † 5
50 3
† 5$
1‰ œ $ † 625 † ˆ 38 ‰ . The mass of the plate is M œ ' dm œ 'c5 $ a25 x# b dx œ $ ’25x
œ
5
4 3
$ † 5$ . Therefore y œ
Mx M
œ
$ †5% † ˆ 83 ‰ $ †5$ †ˆ 43 ‰
5& 5‹ & x$ “ 3 &
œ 10. The plate's center of mass is the point (xß y) œ (!ß 10).
3. Intersection points: x x# œ x Ê 2x x# œ 0 Ê x(2 x) œ 0 Ê x œ 0 or x œ 2. The typical @/<>3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß ax x b (x) ‹ #
#
œ Šxß x# ‹ , length: ax x# b (x) œ 2x x# , width: dx, area: dA œ a2x x# b dx, mass: dm œ $ dA œ $ a2x x# b dx. The moment of the strip about the x-axis is # µ y dm œ Š x# ‹ $ a2x x# b dx; about the y-axis it is µ x dm œ x † $ a2x x# b dx. Thus, Mx œ ' µ y dm œ '0 ˆ #$ x# ‰ a2x x# b dx œ #$ '0 a2x$ x% b dx œ #$ ’ x# 2
2
%
# x& 5 “!
œ #$ Š2$
œ 45$ ; My œ ' µ x dm œ '0 x † $ a2x x# b dx œ $ '0 a2x# x$ b œ $ ’ 23 x$ 2
2
M œ ' dm œ '0 $ a2x x# b dx œ $ '0 a2x x# b dx œ $ ’x# 2
2
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1 and y œ
Mx M
# x$ 3 “!
# x% 4 “!
2& 5‹
œ #$ † 2$ ˆ1 45 ‰
œ $ Š2 †
œ $ ˆ4 38 ‰ œ
4$ 3
2$ 3
2% 4‹
œ
. Therefore, x œ
œ ˆ 45$ ‰ ˆ 43$ ‰ œ 53 Ê (xß y) œ ˆ1ß 53 ‰ is the center of mass.
4. Intersection points: x# 3 œ 2x# Ê 3x# 3 œ 0 Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Applying the symmetry argument analogous to the one in Exercise 1, we find x œ 0. The typical @/<>3-+6 strip has center of mass: # # # (µ x ßµ y ) œ Šxß 2x ax 3b ‹ œ Šxß x 3 ‹ , #
#
length: 2x# ax# 3b œ 3 a1 x# b, width: dx, area: dA œ 3 a1 x# b dx, mass: dm œ $ dA œ 3$ a1 x# b dx. The moment of the strip about the x-axis is µ y dm œ 3 $ ax# 3b a1 x# b dx œ 3 $ ax% 3x# x# 3b dx œ #
œ
3 #
$ 'c1 ax% 2x# 3b dx œ 1
#
3 #
&
$ ’ x5
2x$ 3
3x“
" "
œ
3 #
3 #
$ ax% 2x# 3b dx; Mx œ ' µ y dm
† $ † 2 ˆ 5"
2 3
45 ‰ 3‰ œ 3$ ˆ 310 œ 325$ ; 15
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
$ †2% 1# My M
œ
4$ 3
;
Section 6.6 Moments and Centers of Mass M œ ' dm œ 3$ 'c1 a1 x# b dx œ 3$ ’x
" x$ 3 “ "
1
œ 3$ † 2 ˆ1 3" ‰ œ 4$ . Therefore, y œ
Mx M
367
œ 5$††$32†4 œ 58
Ê (xß y) œ ˆ0ß 85 ‰ is the center of mass. 5. The typical 29<3D98>+6 strip has center of mass: $ (µ x ßµ y ) œ Š y y ß y‹ , length: y y$ , width: dy, #
area: dA œ ay y$ b dy, mass: dm œ $ dA œ $ ay y$ b dy. The moment of the strip about the y-axis is $ # µ x dm œ $ Š y y ‹ ay y$ b dy œ $ ay y$ b dy #
œ
$ #
#
ay# 2y% y' b dy; the moment about the x-axis is
1 $ µ y dm œ $ y ay y$ b dy œ $ ay# y% b dy. Thus, Mx œ ' µ y dm œ $ '0 ay# y% b dy œ $ ’ y3
My œ ' µ x dm œ
$ #
'01 ay# 2y% y' b dy œ #$ ’ y3
$
œ $ '0 ay y$ b dy œ $ ’ y# 1
œ
#
" y% 4 “!
2y& 5
œ $ ˆ "# 4" ‰ œ
$ 4
" y( 7 “!
œ
$ #
ˆ "3
. Therefore, x œ
2 5
7" ‰ œ
$ #
œ $ ˆ "3 "5 ‰ œ
15 ‰ ˆ 35 3†42 œ 5†7
4$ ‰ ˆ 4 ‰ œ ˆ 105 $ œ
My M
" y& 5 “!
16 105
2$ 15
;
4$ 105
; M œ ' dm
Mx M
2$ ‰ ˆ 4 ‰ œ ˆ 15 $
and y œ
16 8 ‰ Ê (xß y) œ ˆ 105 ß 15 is the center of mass.
8 15
6. Intersection points: y œ y# y Ê y# 2y œ 0 Ê y(y 2) œ 0 Ê y œ 0 or y œ 2. The typical 29<3D98>+6 strip has center of mass: # # (µ x ßµ y ) œ Š ay yb y ß y‹ œ Š y ß y‹ , #
2
#
#
length: y ay yb œ 2y y , width: dy, area: dA œ a2y y# b dy, mass: dm œ $ dA œ $ a2y y# b dy. The moment about the y-axis is µ x dm œ #$ † y# a2y y# b dy œ #$ a2y$ y% b dy; the moment about the x-axis is µ y dm œ $ y a2y y# b dy œ $ a2y# y$ b dy. Thus, Mx œ ' µ y dm œ '0 $ a2y# y$ b dy œ $ ’ 2y3 2
œ '0
2
$ #
$
a2y$ y% b dy œ
œ $ ’y#
# y$ 3 “!
$ #
%
’ y2
œ $ ˆ4 83 ‰ œ
# y& “ 5 !
4$ 3
œ
$ #
ˆ8
# y% 4 “!
16$ 1#
ˆ 405 32 ‰ œ
4$ 5
; M œ ' dm œ '0 $ a2y y# b dy
œ
$ #
My M
œ ˆ 45$ ‰ ˆ 43$ ‰ œ
32 ‰ 5
. Therefore, x œ
œ
(4 3) œ
4$ 3
; My œ ' µ x dm
16 ‰ 4
œ $ ˆ 16 3
2
3 5
and y œ
Mx M
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1
Ê (xß y) œ ˆ 35 ß "‰ is the center of mass. 7. Applying the symmetry argument analogous to the one used in Exercise 1, we find x œ 0. The typical @/<>3-+6 strip has center of mass: (µ x ßµ y ) œ ˆxß cos# x ‰ , length: cos x, width: dx,
area: dA œ cos x dx, mass: dm œ $ dA œ $ cos x dx. The moment of the strip about the x-axis is µ y dm œ $ † cos# x † cos x dx 2x ‰ œ #$ cos# x dx œ #$ ˆ 1 cos dx œ 4$ (1 cos 2x) dx; thus, # 1Î2
Mx œ ' µ y dm œ 'c1Î2 4$ (1 cos 2x) dx œ 1Î#
œ $ [sin x]1Î# œ 2$ . Therefore, y œ
Mx M
œ
$ 4
x $1 4 †# $
œ
sin 2x ‘ 1Î# # 1Î# 1 8
œ
$ 4
ˆ 1# 0‰ ˆ 1# ‰‘ œ
$1 4
1Î2
; M œ ' dm œ $ ' 1Î2 cos x dx
Ê (xß y) œ ˆ!ß 18 ‰ is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
368
Chapter 6 Applications of Definite Integrals
8. Applying the symmetry argument analogous to the one used in Exercise 1, we find x œ 0. The typical vertical strip has # center of mass: (µ x ßµ y ) œ Šxß sec# x ‹ , length: sec# x, width: dx, area: dA œ sec# x dx, mass: dm œ $ dA œ $ sec# x dx. The # moment about the x-axis is µ y dm œ Š sec x ‹ a$ sec# xb dx #
1Î4
sec% x dx. Mx œ 'c1Î4 µ y dm œ
$ #
' 11ÎÎ44 sec% x dx
œ
$ #
œ
$ #
'c11ÎÎ44 atan# x 1b asec# xb dx œ #$ ' 11ÎÎ44 (tan x)# asec# xb dx #$ ' 11ÎÎ44 sec# x dx œ 2$ ’ (tan3x) “ 1Î4
œ
$ 2
3" ˆ 3" ‰‘ #$ [1 (1)] œ
$
1 Î4
Therefore, y œ
Mx M
œ ˆ 43$ ‰ ˆ 2"$ ‰ œ
2 3
$ 3
$ œ
4$ 3
1Î%
#$ [tan x]1Î%
; M œ ' dm œ $ 'c1Î4 sec# x dx œ $ [tan x]1Î4 œ $ [1 (1)] œ 2$ . 1Î4
1Î4
Ê (xß y) œ ˆ!ß 32 ‰ is the center of mass.
9. Since the plate is symmetric about the line x œ 1 and its density is constant, the distribution of mass is symmetric about this line and the center of mass lies on it. This means that x œ 1. The typical @/<>3-+6 strip has center of mass: # # # (µ x ßµ y ) œ Šxß a2x x ba2x 4xb ‹ œ Šxß x 2x ‹ , #
#
#
#
#
length: a2x x b a2x 4xb œ 3x 6x œ 3 a2x x# b , width: dx, area: dA œ 3 a2x x# b dx, mass: dm œ $ dA œ 3$ a2x x# b dx. The moment about the x-axis is # µ y dm œ 3 $ ax# 2xb a2x x# b dx œ 3 $ ax# 2xb dx #
#
œ $ ax 4x 4x b dx. Thus, Mx œ ' µ y dm œ '0 %
3 #
œ $ 3 2
& Š 25
$
%
2
2
#
4 3
&
$ ax% 4x$ 4x# b dx œ 23 $ ’ x5 x% 34 x$ “
10 ‰ † 2 ‹ œ $ † 2 ˆ 25 1 23 ‰ œ 3# $ † 2% ˆ 6 15 œ 85$ ; M œ ' dm 15 $
œ '0 3$ a2x x# b dx œ 3$ ’x# 2
3 2
3 #
# x$ 3 “!
# !
%
œ 3$ ˆ4 83 ‰ œ 4$ . Therefore, y œ
Mx M
œ ˆ 85$ ‰ ˆ 4"$ ‰ œ 25
Ê (xß y) œ ˆ1ß 25 ‰ is the center of mass. 10. (a) Since the plate is symmetric about the line x œ y and its density is constant, the distribution of mass is symmetric about this line. This means that x œ y. The typical @/<>3-+6 strip has center of mass: È # (µ x ßµ y ) œ Šxß 9 x ‹ , length: È9 x# , width: dx, #
area: dA œ È9 x# dx, mass: dm œ $ dA œ $ È9 x# dx. The moment about the x-axis is È # µ y dm œ $ Š 9# x ‹ È9 x# dx œ
$ #
a9 x# b dx. Thus, Mx œ ' µ y dm œ '0
3
$ #
a9 x# b dx œ
$ #
’9x
$ x$ “ 3 !
(27 9) œ 9$ ; M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a quarter of a circle of radius 3) œ $ ˆ 941 ‰ œ 4 ‰ Therefore, y œ MMx œ (9$ ) ˆ 91$ œ 14 Ê (xß y) œ ˆ 14 ß 14 ‰ is the center of mass. œ
$ #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
91$ 4
.
Section 6.6 Moments and Centers of Mass
369
(b) Applying the symmetry argument analogous to the one used in Exercise 1, we find that x œ 0. The typical vertical strip has the same parameters as in part (a). 3 Thus, M œ ' µ y dm œ ' $ a9 x# b dx c3 #
x
œ #'
3
$ 0 #
a9 x# b dx œ 2(9$ ) œ 18$ ;
M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a semi-circle of radius 3) œ $ ˆ 921 ‰ œ 91$ 2 . Therefore, y œ 4‰ ˆ as in part (a) Ê (xß y) œ 0ß 1 is the center of mass.
Mx M
2 ‰ œ (18$ ) ˆ 91$ œ
4 1
, the same y
11. Since the plate is symmetric about the line x œ y and its density is constant, the distribution of mass is symmetric about this line. This means that x œ y. The typical @/<>3-+6 strip has È # center of mass: (µ x ßµ y ) œ Šxß 3 9 x ‹ , #
length: 3 È9 x# , width: dx, area: dA œ Š3 È9 x# ‹ dx, mass: dm œ $ dA œ $ Š3 È9 x# ‹ dx. The moment about the x-axis is µ y dm œ $
Š3 È9 x# ‹ Š3 È9 x# ‹ #
dx œ
$ #
c9 a9 x# bd dx œ
$ x# #
dx. Thus, Mx œ '0
3
$ x# #
dx œ
$ 6
$
cx$ d ! œ #
9$ #
equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 Ê A œ 3 œ
9 4
9$ 4
(4 1) Ê M œ $ A œ
(4 1). Therefore, y œ
œ ˆ 9#$ ‰ ’ 9$(44 1) “ œ
Mx M
2 41
. The area 19 4
Ê (xß y) œ ˆ 4 2 1 ß 4 2 1 ‰ is the
center of mass. 12. Applying the symmetry argument analogous to the one used in Exercise 1, we find that y œ 0. The typical @/<>3-+6 strip has center of mass: (µ x ßµ y ) œ Œxß length:
" x$
ˆ x"$ ‰ œ
2 x$
" x$
x"$ #
œ (xß 0),
, width: dx, area: dA œ
2 x$
dx,
2$ x$
mass: dm œ $ dA œ dx. The moment about the y-axis is a µ x dm œ x † 2x$$ dx œ 2x$# dx. Thus, My œ ' µ x dm œ '1 2x$# dx œ 2$ x" ‘ " œ 2$ ˆ "a 1‰ œ a
xœ
My M
2$ (a1) a
œ ’ 2$(aa 1) “ ’ $ aa#a 1b “ œ #
13. Mx œ ' µ y dm œ '1
2
Š x2# ‹ #
2a a1
; M œ ' dm œ '1
a
2$ x$
dx œ $ x"# ‘ " œ $ ˆ a"# 1‰ œ a
$ aa# 1b a#
. Therefore,
‰ Ê (xß y) œ ˆ a 2a x œ 2. 1 ß 0 . Also, a lim Ä_
† $ † ˆ x2# ‰ dx
œ '1 ˆ x"# ‰ ax# b ˆ x2# ‰ dx œ '1 2
2
2 x#
dx œ 2'1 x# dx 2
#
œ 2 cx" d " œ 2 ˆ "# ‰ (1)‘ œ 2 ˆ "# ‰ œ 1;
My œ ' µ x dm œ '1 x † $ † ˆ x2# ‰ dx 2
œ '1 x ax# b ˆ x2# ‰ dx œ 2'1 x dx œ 2 ’ x# “ 2
2
#
# "
œ 2 ˆ2 "# ‰ œ 4 1 œ 3; M œ ' dm œ '1 $ ˆ x2# ‰ dx œ '1 x# ˆ x2# ‰ dx œ 2'1 dx œ 2[x]"# œ 2(2 1) œ 2. So xœ
My M
œ
3 #
and y œ
Mx M
œ
" #
2
2
2
Ê (xß y) œ ˆ 3# ß "# ‰ is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
370
Chapter 6 Applications of Definite Integrals
14. We use the @/<>3-+6 strip approach: 1 # M œ'µ y dm œ ' ax x b ax x# b † $ dx x
#
0
'0 ax# x% b † 12x dx 1
" #
œ
œ 6'0 ax$ x& b dx œ 6 ’ x4 1
%
œ 6 ˆ "4 6" ‰ œ
6 4
1œ
" #
" x' 6 “!
;
My œ ' µ x dm œ '0 x ax x# b † $ dx œ '0 ax# x$ b † 12x dx œ 12'0 ax$ x% b dx œ 12 ’ x4 1
œ
12 #0
xœ
œ
My M
3 5
œ
1
1
%
; M œ ' dm œ '0 ax x# b † $ dx œ 12'0 ax# x$ b dx œ 12 ’ x3 1
3 5
and y œ
Mx M
œ
" #
1
$
" x% 4 “!
" x& 5 “!
œ 12 ˆ "4 5" ‰
œ 12 ˆ 3" 4" ‰ œ
12 12
œ 1. So
Ê ˆ 53 ß "# ‰ is the center of mass.
shell ‰ shell 15. (a) We use the shell method: V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21x ’ È4x Š È4x ‹“ dx œ 161'1 b
4
% œ 161'1 x"Î# dx œ 161 23 x$Î# ‘ " œ 161 ˆ 23 † 8 32 ‰ œ 4
321 3
4
(8 1) œ
(b) Since the plate is symmetric about the x-axis and its density $ (x) œ
" x
x Èx
dx
2241 3
is a function of x alone, the distribution of its
mass is symmetric about the x-axis. This means that y œ 0. We use the vertical strip approach to find x: 4 4 4 % My œ ' µ x dm œ '1 x † ’ È4x Š È4x ‹“ † $ dx œ '1 x † È8x † x" dx œ 8'1 x"Î# dx œ 8 2x"Î# ‘ " œ 8(2 † 2 2) œ 16; 4 M œ ' dm œ '1 ’ È4x Š È ‹“ † $ dx œ 8'1 Š È"x ‹ ˆ "x ‰ dx œ 8'1 x$Î# dx œ 8 2x"Î# ‘ " œ 8[1 (2)] œ 8. x 4
So x œ
My M
œ
16 8
4
%
4
œ 2 Ê (xß y) œ (2ß 0) is the center of mass.
(c)
‘ 16. (a) We use the disk method: V œ 'a 1R# (x) dx œ '1 1 ˆ x4# ‰ dx œ 41'1 x# dx œ 41 x" ‘ " œ 41 " 4 (1) b
4
4
%
œ 1[1 4] œ 31
(b) We model the distribution of mass with vertical strips: Mx œ ' µ y dm œ '1
4
2 œ 2'1 x$Î# dx œ 2 ’ È x dm œ '1 x † “ œ 2[1 (2)] œ 2; My œ ' µ x %
4
4
"
2 16 3
2‘ 3
xœ
My M
œ
2 x
ˆ 2x ‰ 2
† ˆ x2 ‰ † $ dx œ '1
4
2 x#
† Èx dx
† $ dx œ 2'1 x"Î# dx œ 2 ’ 2x3 “ œ 4
4 4 4 ' dm œ ' x2 † $ dx œ 2' Èxx dx œ 2' x"Î# dx œ 2 2x"Î# ‘ %" œ 2(4 2) œ 4. 1 1 1
œ 28 3 ;Mœ ˆ 28 ‰ 7 3 4 œ 3 and
yœ
Mx M
œ
2 4
œ
" #
Ê (xß y) œ ˆ 73 ß "# ‰ is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
$Î#
% "
So
Section 6.6 Moments and Centers of Mass (c)
17. The mass of a horizontal strip is dm œ $ dA œ $ L dy, where L is the width of the triangle at a distance of y above its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have Ê Lœ
b h
(h y). Thus, Mx œ ' µ y dm œ '0 $ y ˆ bh ‰ (h y) dy œ h
œ
$b h
Š h#
$
h$ 3‹
œ
$b h
#
h# 2‹
Šh
œ $ bh# ˆ "# 3" ‰ œ œ
$ bh 2
. So y œ
Mx M
$ bh# 6
œ
$b h
# ˆ 2 ‰ Š $bh 6 ‹ $ bh
œ
h 3
œ
hy h
'0h ahy y# b dy œ $hb ’ hy#
; M œ ' dm œ '0 $ ˆ hb ‰ (h y) dy œ h
L b
$b h
#
h
y$ 3 “!
'0h ah yb dy œ $hb ’hy y2 “ h
Ê the center of mass lies above the base of the
triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the medians, as claimed. 18. From the symmetry about the y-axis it follows that x œ 0. It also follows that the line through the points (!ß !) and (!ß $) is a median Ê y œ "3 (3 0) œ 1 Ê (xß y) œ (!ß ").
19. From the symmetry about the line x œ y it follows that x œ y. It also follows that the line through the points (!ß !) and ˆ "# ß "# ‰ is a median Ê y œ x œ 23 † ˆ "# 0‰ œ 3" Ê (xß y) œ ˆ "3 ß 3" ‰ . 20. From the symmetry about the line x œ y it follows that x œ y. It also follows that the line through the point (!ß !) and ˆ #a ß #a ‰ is a median Ê y œ x œ 32 ˆ #a 0‰ œ "3 a Ê (xß y) œ ˆ 3a ß 3a ‰ . 21. The point of intersection of the median from the vertex (0ß b) to the opposite side has coordinates ˆ!ß #a ‰ Ê y œ (b 0) † "3 œ 3b and x œ ˆ #a !‰ † 32 œ 3a Ê (xß y) œ ˆ 3a ß 3b ‰ .
22. From the symmetry about the line x œ
a #
it follows that
xœ It also follows that the line through the points ˆ #a ß !‰ and ˆ #a ß b‰ is a median Ê y œ "3 (b 0) œ 3b a #.
Ê (xß y) œ ˆ #a ß b3 ‰ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
#
!
371
372
Chapter 6 Applications of Definite Integrals " #
23. y œ x"Î# Ê dy œ
x"Î# dx " 4x
Ê ds œ È(dx)# (dy)# œ É1 Mx œ $ '0 Èx É1 2
œ $ '0 Éx 2
" 4
dx œ
" 4x
dx
2$ 3
$Î# ’ˆx 4" ‰ “
dx ;
# !
$Î# $Î# ’ˆ2 "4 ‰ ˆ 4" ‰ “
œ
2$ 3
œ
2$ ˆ 9 ‰$Î# 3 ’ 4
ˆ 4" ‰
$Î#
2$ 3
“œ
"‰ ˆ 27 8 8 œ
13$ 6
24. y œ x$ Ê dy œ 3x# dx Ê dx œ É(dx)# a3x# dxb# œ È1 9x% dx; Mx œ $ '0 x$ È1 9x% dx; 1
" 36
[u œ 1 9x% Ê du œ 36x$ dx Ê
du œ x$ dx;
x œ 0 Ê u œ 1, x œ 1 Ê u œ 10] Ä Mx œ $ '1
10
" 36
u"Î# du œ
$ 36
32 u$Î# ‘ "! œ "
$ 54
ˆ10$Î# 1‰
25. From Example 4 we have Mx œ '0 a(a sin ))(k sin )) d) œ a# k'0 sin# ) d) œ
'01 (1 cos 2)) d) œ a#k ) sin#2) ‘ !1 1 1 1 1 œ a #k1 ; My œ '0 a(a cos ))(k sin )) d) œ a# k '0 sin ) cos ) d) œ a#k csin# )d ! œ 0; M œ '0 ak sin ) d) œ ak[ cos )]1! 1
1
#
a# k #
#
#
œ 2ak. Therefore, x œ
My M
œ 0 and y œ
Mx M
#
" ‰ œ Š a 2k1 ‹ ˆ 2ak œ
a1 4
Ê ˆ!ß a41 ‰ is the center of mass.
26. Mx œ ' µ y dm œ '0 (a sin )) † $ † a d) 1
œ '0 aa# sin )b a1 k kcos )kb d) 1
œ a# '0 (sin ))(1 k cos )) d) 1Î2
a# '1Î2 (sin ))(1 k cos )) d) 1
œ a# '0 sin ) d) a# k'0 sin ) cos ) d) a# '1Î2 sin ) d) a# k '1Î2 sin ) cos ) d) 1Î2
1Î2
1Î#
#
1
a# k ’ sin# ) “
œ a# [ cos )]! #
#
k ˆ "#
1Î#
1
#
a# [ cos )]11Î# a# k ’ sin# ) “
!
1Î# #
"‰ #
œ a [0 (1)] a 0‰ a [(1) 0] a k ˆ0 œ a# a#k a# 1 1 M œ'µ x dm œ ' (a cos )) † $ † a d) œ ' aa# cos )b a1 k kcos )kb d) y
#
1
0
#
a# k #
œ 2a# a# k œ a# (2 k);
0
1Î2 1 œ a '0 (cos ))(1 k cos )) d) a# '1Î2 (cos ))(1 k cos )) d)
#
œ a#
1Î2
'01Î2 cos ) d) a# k ' 0
#
œ a [sin
1Î# ) ]!
œ a# (1 0)
a# k #
a# k #
)
2) ‰ 2) ‰ ˆ 1 cos d) a# '1Î2 cos ) d) a# k'1Î2 ˆ 1 cos d) # #
sin 2) ‘ 1Î# # !
1
a# [sin )]11Î#
1
a# k #
ˆ 1# 0‰ (! 0)‘ a# (0 1)
)
a# k #
sin 2) ‘ 1 # 1Î#
(1 0) ˆ 1# 0‰‘ œ a#
a# k1 4
a#
M œ '0 $ † a d) œ a'0 (1 k kcos )k) d) œ a '0 (1 k cos )) d) a'1Î2 (1 k cos )) d) 1
1Î#
œ a[) k sin )]! œ
a1 #
1
1Î2
1
a[) k sin )]11Î# œ a ˆ 1# k‰ 0‘ a (1 0) ˆ 1# k‰‘
ak a ˆ 1# k‰ œ a1 2ak œ a(1 2k). So x œ
My M
œ 0 and y œ
Mx M
œ
a# (2 k) a(1 #k)
œ
a(2 k) 1 #k
ka ‰ Ê ˆ0ß 2a 1 #k is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
a# k1 4
œ 0;
Section 6.6 Moments and Centers of Mass
373
27. faxb œ x 6, gaxb œ x2 , faxb œ gaxb Ê x 6 œ x2 Ê x2 x 6 œ 0 Ê x œ 3, x œ 2; $ œ 1 M œ 'c2 cax 6b x2 ddx œ "# x2 6x 13 x3 ‘2 3
3
œ ˆ 92 18 9‰ ˆ2 12 83 ‰ œ
xœ
1 125Î6
125 6
3 6 ' 'c32 xcax 6b x2 ddx œ 125 cx2 6x x3 ddx c2
œ
6 1 3 125 3 x
œ
6 ˆ 125 9
3
3x2 14 x4 ‘2
27
81 ‰ 4
6 ˆ 8 125 3
12 4‰ œ 12 ; y œ
3 1 3 1 5 ‘3 2 125 3 x 6x 36x 5 x 2 Ê ˆ 12 , 4‰ is the center of mass.
œ
œ
3 ˆ 125 9
1 125Î6
54 108
3 3 ' 'c32 12 ’ax 6b2 ax2 b2 “dx œ 125 cx2 12x 36 x4 ddx c2 243 ‰ 5
3 ˆ 8 125 3
24 72
32 ‰ 5
œ4
28. faxb œ 2, gaxb œ x2 ax 1b, faxb œ gaxb Ê 2 œ x2 ax 1b Ê x3 x2 2 œ 0 Ê x œ 1; $ œ 1 M œ '0 c2 x2 ax 1bd dx œ '0 c2 x3 x2 d dx 1
1
1
œ 2x "4 x4 13 x3 ‘0 œ ˆ2 xœ
1 17Î12
œ
12 2 17 x
œ
12 ˆ 17 1
œ
6 17 4x
" 4
13 ‰ 0 œ
17 12
'01 xc2 x2 ax 1bddx œ 1217 '01 c2x x4 x3 ddx 1
15 x5 14 x4 ‘0
1 5
14 ‰ 0 œ
33 85 ;
yœ
1 17Î12
1
17 x7 13 x6 15 x5 ‘0 œ
'01 12 ’22 ax2 ax 1bb2 “dx œ 176 '01 c4 x6 2x5 x4 ddx
6 ˆ 17 4
1 7
1 3
15 ‰ 0 œ
698 595
Ê ˆ 33 85 ,
698 ‰ 595
is the center of mass.
29. faxb œ x2 , gaxb œ x2 ax 1b, faxb œ gaxb Ê x2 œ x2 ax 1b Ê x3 2x2 œ 0 Ê x œ 0, x œ 2; $ œ 1 M œ '0 cx2 x2 ax 1bddx œ '0 c2x2 x3 ddx 2
2
2
‰ œ 23 x3 "4 x4 ‘0 œ ˆ 16 3 4 0œ xœ
1 4 Î3
'02 xcx2 x2 ax 1bddx œ 43 '02 c2x3 x4 ddx
2 œ 34 12 x4 5" x5 ‘0 œ 34 ˆ8
yœ
1 4 Î3
4 3
32 ‰ 5
0 œ 56 ;
'02 12 ’ax2 b2 ax2 ax 1bb2 “dx œ 38 '02 c2x5 x6 ddx œ 38 13 x6 7" x7 ‘20 œ 38 ˆ 643 "728 ‰ 0 œ 78
Ê ˆ 65 , 87 ‰ is the center of mass. 30. faxb œ 2 sin x, gaxb œ 0, x œ 0, x œ 21; $ œ 1; M œ '0 c2 sin xddx œ c2x cos xd201 21
œ a41 1b a0 1b œ 41 xœ
1 41
'021 xc2 sin x 0ddx œ 411 '021 c2x x sin xddx
'021 2x dx 411 '021 x sin xdx
œ
1 41
œ
1 2 21 4 1 cx d 0
œ
1 2 4 1 a 41 b
0
œ
1 81
œ
1 81 c4x
1 41 csin x
x cos xd201
1 4 1 a0
21b 0 œ
21 1 2 ;
yœ
1 41
'021 21 ’a2 sin xb2 a0b2 “dx œ 811 '021 c4 4 sin x sin2 xddx
2x ‘ '021 c4 4 sin xddx 811 '021 csin2 xddx œ 811 '021 c4 4 sin xddx 811 '021 1 cos dx 2
4cos x d201
1 161
'021 dx 1611 '021 cos 2x dx [u œ 2x Ê du œ 2dx, x œ 0 Ê u œ 0, x œ 21 Ê u œ 41]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
374
Chapter 6 Applications of Definite Integrals Ä œ
21 21 1 1 1 81 c4x 4cos xd0 161 cxd0 321 1 1 1 81 a81 4b 81 a0 4b 161 a21b
'041 cos u du œ 811 c4x 4cos xd201 1611 cxd201 3211 csin ud401 00œ
9 8
Ê ˆ 212 1 , 89 ‰ is the center of mass.
31. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we have that the length of a particular segment is ds œ Éadxb# adyb# . This implies that Mx œ ' $ y ds, My œ ' $ x ds and M œ ' $ ds. If $ is constant, then x œ
My M
' x ds
œ ' ds œ
' x ds length
and y œ
Mx M
' y ds
œ ' ds œ
' y ds length
.
32. Applying the symmetry argument analogous to the one used in Exercise 1, we find that x œ 0. The typical vertical strip x#
a has center of mass: (µ x ßµ y ) œ Œxß 2 4p , length: a
œ $ Ša œ
$ #
x# 4p ‹
’a# x
2
#Èpa x& 80p# “ c2 pa
È
‰ œ 2a# $ Èpa ˆ 64 80 œ œ 2$ Š2aÈpa œ
3 5
Èpa
dx. Thus, Mx œ ' µ y dm œ 'c2Èpa "# Ša œ 2 † #$ ’a# x
8a# $Èpa 5
2$ paÈpa 12p ‹
2
x& 80p# “ 0
Èpa
x# 4p ,
x# 4p ‹ Ša
œ $ Š2a# Èpa
Èpa
; M œ ' dm œ $ 'c2Èpa Ša 2
œ 4a$ Èpa ˆ1
width: dx, area: dA œ Ša
4 ‰ 12
x# 4p ‹
x# 4p ‹ $
dx œ
2& p# a# Èpa ‹ 80p#
$ #
dx, mass: dm œ $ dA
x 'c22ÈÈpapa Ša# 16p ‹ dx %
#
œ 2a# $ Èpa ˆ1 2
16 ‰ 80
"6 ‰ œ 2a# $ Èpa ˆ 8080
Èpa œ 2 † $ ’ax È
dx œ $ ’ax
x$ 12p “ c2 pa
8a$ Èpa 3
. So y œ
œ 4a$ Èpa ˆ 121#4 ‰ œ
x# 4p ‹
Mx M
œŠ
2
Èpa
x$ 12p “ !
8a# $ Èpa 3 ‹ Š 8a$È 5 pa ‹
a, as claimed.
33. The centroid of the square is located at (#ß #). The volume is V œ (21) ayb (A) œ (21)(2)(8) œ 321 and the surface area is S œ (21) ayb (L) œ (21)(2) Š4È8‹ œ 32È21 (where È8 is the length of a side). 34. The midpoint of the hypotenuse of the triangle is ˆ 3# ß 3‰ Ê y œ 2x is an equation of the median Ê the line y œ 2x contains the centroid. The point ˆ 3# ß $‰ is 3È 5 #
units from the origin Ê the x-coordinate of the
# centroid solves the equation Ɉx 3# ‰ (2x 3)#
œ
È5 #
Ê ˆx# 3x 94 ‰ a4x# 12x 9b œ
5 4
Ê 5x# 15x 9 œ 1 Ê x# 3x 2 œ (x 2)(x 1) œ 0 Ê x œ 1 since the centroid must lie inside the triangle Ê y œ 2. By the Theorem of Pappus, the volume is V œ (distance traveled by the centroid)(area of the region) œ 21 a5 xb "# (3)(6)‘ œ (21)(4)(9) œ 721
35. The centroid is located at (#ß !) Ê V œ (21) axb (A) œ (21)(2)(1) œ 41# 36. We create the cone by revolving the triangle with vertices (0ß 0), (hß r) and (hß 0) about the x-axis (see the accompanying figure). Thus, the cone has height h and base radius r. By Theorem of Pappus, the lateral surface area swept out by the hypotenuse L is given by S œ 21yL œ 21 ˆ #r ‰ Èh# r# œ 1rÈr# h# . To calculate the volume we need the position of the centroid of the triangle. From the diagram we see that the centroid lies on the line y œ
r 2h
# x. The x-coordinate of the centroid solves the equation É(x h)# ˆ 2hr x #r ‰
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.6 Moments and Centers of Mass œ
" 3
Éh#
r# 4
#
#
#
#
Ê Š 4h4h# r ‹ x# Š 4h 2h r ‹ x
inside the triangle Ê y œ
r 2h
xœ
r 3.
39. V œ 21 yA Ê
4 3
2a ‰‘ (1a) 1
2a 1,
2 ar# 4h# b 9
œ0 Ê xœ
2h 3
or
21 ˆ 3r ‰‘ ˆ "#
4h 3
Ê xœ
hr‰ œ
" 3
2h 3 ,
since the centroid must lie
1r# h.
and by symmetry x œ 0
œ 21a# (1 2)
1ab# œ a21yb ˆ 1#ab ‰ Ê y œ
40. V œ 213A Ê V œ 21 ˆa
By the Theorem of Pappus, V œ
37. S œ 21 y L Ê 41a# œ a21yb (1a) Ê y œ 38. S œ 213 L Ê 21 ˆa
r# 4
375
4a ‰‘ 1a# Š # ‹ 31
œ
4b 31
and by symmetry x œ 0
1a$ (31 4) 3
41. V œ 213 A œ (21)(area of the region) † (distance from the centroid to the line y œ x a). We must find the distance from 4a ‰ ˆ0ß 31 to y œ x a. The line containing the centroid and perpendicular to y œ x a has slope 1 and contains the point 4a ˆ!ß 31 ‰ . This line is y œ x 34a1 . The intersection of y œ x a and y œ x 34a1 is the point ˆ 4a 613a1 ß 4a 613a1 ‰ . Thus, #
the distance from the centroid to the line y œ x a is Ɉ 4a 613a1 ‰ ˆ 34a1 Ê V œ (21) Š
È2 (4a 3a1) # ‹ Š 1#a ‹ 61
œ
4a 61
3a1 ‰# 61
œ
È2 (4a 3a1) 61
È2 1a$ (4 31) 6
‰ 42. The line perpendicular to y œ x a and passing through the centroid ˆ!ß 2a 1 has equation y œ x of the two perpendicular lines occurs when x a œ x #
2a 1
centroid to the line y œ x a is Ɉ 2a 2 1a 0‰ ˆ 2a 2 1a
Ê xœ 2a ‰# #
œ
2a a1 21
a(21) È 21
Ê yœ
2a a1 21 .
2a 1.
The intersection
Thus the distance from the
. Therefore, by the Theorem of Pappus the
1 ) surface area is S œ 21 ’ a(2 “ (1a) œ È21a# (2 1). È 21
43. If we revolve the region about the y-axis: r œ a, h œ b Ê A œ 12 ab, V œ 13 1 a2 bß and 3 œ x. By the Theorem of Pappus: 1 a 1 1 2 2 ˆ1 ‰ 3 1 a b œ 21 x 2 ab Ê x œ 3 ; If we revolve the region about the x-axis: r œ b, h œ a Ê A œ 2 ab, V œ 3 1 b aß and 3 œ y. By the Theorem of Pappus: 13 1 b2 a œ 21 y ˆ 12 ab‰ Ê y œ
b 3
Ê ˆ 3a , 3b ‰ is the center of mass.
44. Let Oa0, 0b, Paa, cb, and Qaa, bb be the vertices of the given triangle. If we revolve the region about the x-axis: Let R be the point Raa, 0b. The volume is given by the volume of the outer cone, radius œ RP œ c, minus the volume of the inner cone, radius œ RQ œ b, thus V œ 13 1 c2 a 13 1 b2 a œ 13 1 aac2 b2 b, the area is given by the area of triangle OPR minus area of triangle OQR, A œ "# ac "# ab œ "# aac bb, and 3 œ y. By the Theorem of Pappus: 13 1 aac2 b2 b œ 21 y ’ "# aac bb“ Ê y œ
cb 3 ;
If we revolve the region about the y-axis: Let S and T be the points Sa0, cb and Ta0, bb,
respectively. Then the volume is the volume of the cylinder with radius OR œ a and height RP œ c, minus the sum of the volumes of the cone with radius œ SP œ a and height œ OS œ c and the portion of the cylinder with height œ OT œ b and radius œ TQ œ a with a cone of height œ OT œ b and radius œ TQ œ a removed. Thus V œ 1 a2 c ’ 13 1 a2 c ˆ1 a2 , 13 1 a2 b‰“ œ 23 1 a2 c 23 1 a2 b œ 23 1 a2 aa bb. The area of the triangle is the same as before, A œ "# ac "# ab œ "# aac bb, and 3 œ x. By the Theorem of Pappus: 23 1 a2 aa bb œ 21 x ’ "# aac bb“ Êxœ
2aaa bb 3 ac b b
aa b b Ê Š 2a 3 ac b b ,
cb 2 ‹
is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
376
Chapter 6 Applications of Definite Integrals
CHAPTER 6 PRACTICE EXERCISES # 1. A(x) œ 14 (diameter)# œ 14 ˆÈx x# ‰ œ 14 ˆx 2Èx † x# x% ‰ ; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ b
#
œ
1 4
œ
1 4†70
x& 5 “!
’ x# 74 x(Î#
È3 4
" #
ˆ "#
4 7
5" ‰
91 280 È3 4
ˆ2Èx x‰#
ˆ4x 4xÈx x# ‰ ; a œ 0, b œ 4 È3 4
b
œ
È3 4
œ
32È3 4
ˆ1 1 4
8 5
'04 ˆ4x 4x$Î# x# ‰ dx
% x$ “ 3 !
’2x# 58 x&Î#
3. A(x) œ 1 4
1 4
œ
(side)# ˆsin 13 ‰ œ
Ê V œ 'a A(x) dx œ
œ
'01 ˆx 2x&Î# x% ‰ dx
(35 40 14) œ
2. A(x) œ œ
1 4 "
œ
8È 3 15
32 ‰ œ
(diameter)# œ
1 4
È3 4
8†32 5
ˆ32
(15 24 10) œ
64 ‰ 3
8È 3 15
(2 sin x 2 cos x)#
† 4 asin# x 2 sin x cos x cos# xb
œ 1(1 sin 2x); a œ
1 4
,bœ
51 4
Ê V œ 'a A(x) dx œ 1 '1Î4 (1 sin 2x) dx 51Î4
b
œ 1 x
cos 2x ‘ &1Î% # 1Î%
œ 1 ’Š 541
cos 5#1 # ‹
Š 14
cos 1# # ‹“
œ 1# #
#
%
4. A(x) œ (edge)# œ ŒŠÈ6 Èx‹ 0 œ ŠÈ6 Èx‹ œ 36 24È6 Èx 36x 4È6 x$Î# x# ; a œ 0, b œ 6 Ê V œ 'a A(x) dx œ '0 Š36 24È6 Èx 36x 4È6 x$Î# x# ‹ dx b
6
œ ’36x 24È6 † 23 x$Î# 18x# 4È6 † 25 x&Î# œ 216 576 648 5. A(x) œ
(diameter)# œ
1 4
72 œ 360
Š2Èx
x# 4‹
#
1728 5
œ
1 4
œ
œ 216 16 † È6 È6 † 6 18 † 6# 58 È6 È6 † 6#
1800 1728 5
Š4x x&Î#
œ
72 5
x% 16 ‹ ;
a œ 0, b œ 4 Ê V œ 'a A(x) dx b
'04 Š4x x&Î# 16x ‹ dx œ 14 ’2x# 27 x(Î# 5x†16 “ % œ 14 ˆ32 32 † 87 25 † 32‰
œ
1 4
œ
321 4
%
ˆ1
6. A(x) œ œ
1 4
1728 5
' x$ 3 “!
È3 4
" #
8 7
52 ‰ œ
81 35
&
(35 40 14) œ
(edge)# sin ˆ 13 ‰ œ
È3 4
!
721 35
2Èx ˆ2Èx‰‘#
ˆ4Èx‰# œ 4È3 x; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ '0 4È3 x dx œ ’2È3 x# “ b
1
" !
œ 2È3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
6$ 3
Chapter 6 Practice Exercises
377
7. (a) .3=5 7/>29. :
V œ 'a 1R# (x) dx œ 'c1 1 a3x% b dx œ 1 'c1 9x) dx b
1
1
#
"
œ 1 cx* d " œ 21
(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x a3x% b dx œ 21 † 3'0 x& dx œ 21 † 3 ’ x6 “ œ 1 b
1
1
'
!
Note: The lower limit of integration is 0 rather than 1. (c) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'c1 (1 x) a3x% b dx œ 21 ’ 3x5 b
"
1
&
(d) A+=2/< 7/>29. :
" x' 2 “ "
œ 21 ˆ 35 "# ‰ ˆ 35 "# ‰‘ œ
121 5
R(x) œ 3, r(x) œ 3 3x% œ 3 a1 x% b Ê V œ 'a 1 cR# (x) r# (x)d dx œ 'c1 1 ’9 9 a1 x% b “ dx b
1
œ 91 'c1 c1 a1 2x% x) bd dx œ 91 'c1 a2x% x) b dx œ 91 ’ 2x5 1
1
8. (a) A+=2/< 7/>29. : R(x) œ
, r(x) œ
4 x$
" #
&
" x* 9 “ "
b
21†13 5
œ
261 5
2
(b) =2/66 7/>29. :
V œ 21'1 x ˆ x4$ "# ‰ dx œ 21 ’4x" 2
(c) =2/66 7/>29. :
" #
# x# 4 “"
16 5
4" ‰ œ
1 20
(2 10 64 5) œ
b
2
x
(d) A+=2/< 7/>29. :
# x# 4 “"
571 #0
œ 21 ˆ 4# 1‰ ˆ4 4" ‰‘ œ 21 ˆ 54 ‰ œ
shell ‰ shell V œ 21'a ˆ radius Š height ‹ dx œ 21'1 (2 x) ˆ x4$ "# ‰ dx œ 21'1 ˆ x8$ 4 x
œ 181 25 "9 ‘ œ
# # # & Ê V œ 'a 1cR# (x) r# (x)d dx œ '1 1 ’ˆ x4$ ‰ ˆ "# ‰ “ dx œ 1 16 x4 ‘ " 5 x
"‰ " ˆ 16 " ‰‘ œ 1 ˆ 10 œ 1 ˆ 5†16 32 # 5 4
œ 21 ’ x4#
#
2
4 x#
51 #
1 x# ‰ dx
œ 21 (1 2 2 1) ˆ4 4 1 4" ‰‘ œ
31 #
V œ 'a 1cR# (x) r# (x)d dx b
# œ 1 '1 ’ˆ 7# ‰ ˆ4 2
dx
œ
491 4
161'1 a1 2x$ x' b dx
œ
491 4
161 ’x x#
œ
491 4 491 4 491 4
161 ˆ2 4" 5†"3# ‰ ˆ1 1 5" ‰‘ " 161 ˆ 4" 160 5" ‰
œ œ 9.
4 ‰# x$ “
2
161 160
# x & 5 “"
(40 1 32) œ
491 4
711 10
œ
1031 20
(a) .3=5 7/>29. :
# V œ 1 '1 ŠÈx 1‹ dx œ 1'1 (x 1) dx œ 1 ’ x# x“
#
5
5
‰ ˆ" ‰‘ œ 1 ˆ 24 ‰ œ 1 ˆ 25 # 5 # 1 # 4 œ 81
& "
(b) A+=2/< 7/>29. :
R(y) œ 5, r(y) œ y# 1 Ê V œ 'c 1 cR# (y) r# (y)d dy œ 1 'c2 ’25 ay# 1b “ dy d
2
œ 1'c2 a25 y% 2y# 1b dy œ 1 'c2 a24 y% 2y# b dy œ 1 ’24y 2
2
#
y& 5
32 y$ “
# #
œ 21 ˆ24 † 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
32 5
2 3
† 8‰
378
Chapter 6 Applications of Definite Integrals œ 321 ˆ3
2 5
"3 ‰ œ
321 15
(45 6 5) œ
10881 15
(c) .3=5 7/>29. : R(y) œ 5 ay# 1b œ 4 y#
Ê V œ 'c 1R# (y) dy œ 'c2 1 a4 y# b dy d
2
#
œ 1 'c2 a16 8y# y% b dy 2
œ 1 ’16y
8y$ 3
œ 641 ˆ1
2 3
# y& 5 “ #
"5 ‰ œ
œ 21 ˆ32
641 15
64 3
(15 10 3) œ
32 ‰ 5 5121 15
10. (a) =2/66 7/>29. :
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y Šy d
4
œ 21'0 Šy# 4
œ
21 1#
† 64 œ
y$ 4‹
$
dy œ 21 ’ y3
321 3
%
y% 16 “ !
y# 4‹
dy
œ 21 ˆ 64 3
64 ‰ 4
(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ˆ2Èx x‰ dx œ 21'0 ˆ2x$Î# x# ‰ dx œ 21 ’ 45 x&Î# b
4
œ 21 ˆ 45 † 32
64 ‰ 3
œ
4
1281 15
% x$ 3 “!
(c) =2/66 7/>29. :
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(4 x) ˆ2Èx x‰ dx œ 21'0 ˆ8x"Î# 4x 2x$Î# x# ‰ dx b
4
$Î# œ 21 ’ 16 2x# 54 x&Î# 3 x
œ 641 ˆ1 45 ‰ œ
641 5
4
% x$ 3 “!
œ 21 ˆ 16 3 † 8 32
(d) =2/66 7/>29. :
shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(4 y) Šy d
4
œ 21'0 Š4y 2y# 4
y$ 4‹
dy œ 21 ’2y# 32 y$
y# 4‹
% y% 16 “ !
4 5
† 32
64 ‰ 3
œ 641 ˆ 34 1
dy œ 21'0 Š4y y# y# 4
œ 21 ˆ32
2 3
4 5
y$ 4‹
32 ‰
dy
† 64 16‰ œ 321 ˆ2
8 3
1‰ œ
321 3
11. .3=5 7/>29. : R(x) œ tan x, a œ 0, b œ
1 3
Ê V œ 1 '0 tan# x dx œ 1'0 asec# x 1b dx œ 1[tan x x]! 1Î3
12. .3=5 7/>29. :
1Î3
1Î$
V œ 1'0 (2 sin x)# dx œ 1 '0 a4 4 sin x sin# xb dx œ 1'0 ˆ4 4 sin x 1
1
œ 1 4x 4 cos x
x #
sin 2x ‘ 1 4 !
1
œ 1 ˆ41 4
1 #
0‰ (0 4 0 0)‘ œ
œ
1cos 2x ‰ dx # 9 1 1 ˆ # 8‰ œ 1#
1 Š3È31‹ 3
(91 16)
13. (a) .3=5 7/>29. :
V œ 1'0 ax# 2xb dx œ 1'0 ax% 4x$ 4x# b dx œ 1 ’ x5 x% 34 x$ “ œ 1 ˆ 32 5 16 2
œ
161 15
2
#
(6 15 10) œ
#
&
!
161 15
32 ‰ 3
(b) A+=2/< 7/>29. :
V œ '0 1’1# ax# 2x "b “ dx œ '0 1 dx '0 1 ax "b% dx œ #1 ’1 2
2
#
2
(c) =2/66 7/>29. :
# ax"b& & “!
œ #1 1 †
shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'0 (2 x) c ax# 2xbd dx œ 21'0 (2 x) a2x x# b dx b
2
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
# &
œ
)1 &
Chapter 6 Practice Exercises œ 21'0 a4x 2x# 2x# x$ b dx œ 21'0 ax$ 4x# 4xb dx œ 21 ’ x4 34 x$ 2x# “ œ 21 ˆ4 2
œ
21 3
2
#
%
81 3
(36 32) œ
32 3
!
8‰
(d) A+=2/< 7/>29. :
V œ 1 '0 c2 ax# 2xbd dx 1'0 2# dx œ 1'0 ’4 4 ax# 2xb ax# 2xb “ dx 81 2
2
#
2
#
œ 1'0 a4 4x# 8x x% 4x$ 4x# b dx 81 œ 1'0 ax% 4x$ 8x 4b dx 81 2
2
#
&
‰ œ 1 ’ x5 x% 4x# 4x“ 81 œ 1 ˆ 32 5 16 16 8 81 œ !
1 5
(32 40) 81 œ
721 5
401 5
œ
321 5
14. .3=5 7/>29. :
V œ 21'0 4 tan# x dx œ 81'0 asec# x 1b dx œ 81[tan x x]! 1Î4
1Î4
1Î%
œ 21(4 1)
15. The material removed from the sphere consists of a cylinder and two "caps." From the diagram, the height of the cylinder #
is 2h, where h# ŠÈ$‹ œ ## , i.e. h œ ". Thus #
Vcyl œ a#hb1ŠÈ$‹ œ '1 ft$ . To get the volume of a cap, use the disk method and x# y# œ ## : Vcap œ '" 1x# dy 2
œ '" 1a% y# bdy œ 1’%y 2
œ 1ˆ8 83 ‰ ˆ% 3" ‰‘ œ
# y3 3 “"
&1 3
Vremoved œ Vcyl #Vcap œ '1
ft$ . Therefore, "!1 3
œ
#)1 3
ft$ .
16. We rotate the region enclosed by the curve y œ É12 ˆ1
4x# ‰ 121
and the x-axis around the x-axis. To find the
11Î2
volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1 b
œ 121'c11Î2 Š1 11Î2
4x# 121 ‹
œ 1321 ˆ1 "3 ‰ œ 17. y œ x"Î#
x$Î# 3
Ê
dx œ 121 ’x
2641 3
dy dx
œ
""Î# 4x$ 363 “ ""Î#
ˆ4
2 3
18. x œ y#Î$ Ê
" #
#
x"Î# "# x"Î# Ê Š dy dx ‹ œ
œ '1
8
† 8‰ ˆ2 23 ‰‘ œ
È9y#Î$ 4 3y"Î$
œ
" 4
" 3
5 12
12 Š1
4x# 121 ‹
dx
#
4
ˆ2
14 ‰ 3
œ
4y #Î$ 9
" #
ˆx"Î# x"Î# ‰ dx œ
" #
2x"Î# 23 x$Î# ‘ % "
10 3 dx Ê L œ '1 Ê1 Š dy ‹ dy œ '1 É1 #
8
8
4 9y#Î$
dy
'18 È9y#Î$ 4 ˆy"Î$ ‰ dy; u œ 9y#Î$ 4 Ê du œ 6y"Î$ dy; y œ 1 Ê u œ 13,
y œ 8 Ê u œ 40d Ä L œ 19. y œ
" #
#
dy œ
11Î2
ˆ x" 2 x‰ Ê L œ ' É1 4" ˆ x" 2 x‰ dx 1 4
y"Î$ Ê Š dx dy ‹ œ
2 3
11Î2
$
4
dx dy
dx œ 1 '
œ 881 ¸ 276 in$
4
" #
#
4 ‰ 11 ˆ 4 ‰ ˆ 11 ‰ “ œ 1321 ’1 ˆ 363 œ 241 ’ 11 Š 4 ‹“ 2 363 #
# Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1
œ
4x# ‰ 121 ‹
x'Î& 58 x%Î& Ê
dy dx
" 18
œ
'1340 u"Î# du œ 18" 32 u$Î# ‘ %! œ #"7 40$Î# 13$Î# ‘ ¸ 7.634 "$ " #
#
dy x"Î& "# x"Î& Ê Š dx ‹ œ
" 4
ˆx#Î& 2 x#Î& ‰
# Ê L œ '1 É1 "4 ax#Î& 2 x#Î& b dx Ê L œ '1 É 4" ax#Î& 2 x#Î& b dx œ ' É 4" ax"Î& x"Î& b dx 32
32
32
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
379
380
Chapter 6 Applications of Definite Integrals œ '1
32
œ
" 48
20. x œ
" #
ˆx"Î& x"Î& ‰ dx œ
(1260 450) œ
" 1#
y$
" y
Ê
" % œ '1 É 16 y 2
" #
œ
1710 48
" 5 # 6 285 8
$#
x'Î& 45 x%Î& ‘ " œ
" y#
#
dx dy
œ
" 4
" y%
dy œ '1 ÊŠ 4" y#
y#
2
8 œ ˆ 12 "# ‰ ˆ 1"# 1‰ œ
7 1#
" #
21. S œ 'a 21y Ê1 Š dy dx ‹ dx;
dy dx
œ
#
b
" 16
Ê Š dx dy ‹ œ
œ
" y# ‹
#
" #
ˆ 65 † 2'
y%
" #
" y%
5 4
† 2% ‰ ˆ 56 54 ‰‘ œ
" #
ˆ 315 6
" % Ê L œ '1 Ê1 Š 16 y 2
dy œ '1 Š 4" y# 2
" y# ‹
dy œ ’ 1"# y$ y" “
" #
75 ‰ 4
" y% ‹
dy
# "
13 12 #
" È2x 1
Ê Š dy dx ‹ œ
" #x 1
Ê S œ '0 21È2x 1 É1 3
" #x 1
dx
2 È ' Èx 1 dx œ 2È21 2 (x 1)$Î# ‘ $ œ 2È21 † 2 (8 1) œ œ 21'0 È2x 1 É 2x 2x1 dx œ 2 21 0 3 3 ! 3
3
22. S œ 'a 21y Ê1 Š dy dx ‹ dx; #
b
œ
1 6
dy dx
% ' œ x# Ê Š dy dx ‹ œ x Ê S œ 0 21 † #
1
x$ 3
È1 x% dx œ
1 6
281È2 3
'01 È1 x% a4x$ b dx
'01 È1 x% d a1 x% b œ 16 ’ 32 a1 x% b$Î# “ " œ 19 ’2È2 1“ !
23. S œ 'c 21x Ê1 Š dx dy ‹ dy; #
d
dx dy
ˆ "# ‰ (4 2y) È4y y#
œ
œ
2y È4y y#
#
Ê 1 Š dx dy ‹ œ
4y y# 4 4y y# 4y y#
œ
4 4y y#
Ê S œ '1 21 È4y y# É 4y 4 y# dy œ 41'1 dx œ 41 2
2
24. S œ 'c 21x Ê1 Š dx dy ‹ dy; #
d
œ 1'2 È4y 1 dy œ 6
1 4
dx dy
œ
1 2È y
#
Ê 1 Š dx dy ‹ œ 1
32 (4y 1)$Î# ‘ ' œ #
1 6
(125 27) œ
1 6
" 4y
œ
4y 1 4y
(98) œ
Ê S œ '2 21Èy † 6
È4y 1 È4y
dy
491 3
25. The equipment alone: the force required to lift the equipment is equal to its weight Ê F" (x) œ 100 N.
The work done is W" œ 'a F" (x) dx œ '0 100 dx œ [100x]%! ! œ 4000 J; the rope alone: the force required b
40
to lift the rope is equal to the weight of the rope paid out at elevation x Ê F# (x) œ 0.8(40 x). The work done is W# œ 'a F# (x) dx œ '0 0.8(40 x) dx œ 0.8 ’40x b
40
the total work is W œ W" W# œ 4000 640 œ 4640 J
%! x# # “!
œ 0.8 Š40#
40# # ‹
œ
(0.8)(1600) #
œ 640 J;
26. The force required to lift the water is equal to the water's weight, which varies steadily from 8 † 800 lb to 8 † 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is x‰ F(x) œ 8 † 800 † ˆ 2†24750 œ (6400) ˆ1 †4750
œ '0
4750
6400 ˆ1
x ‰ 9500
dx œ 6400 ’x
œ 22,800,000 ft † lb
x ‰ 9500
lb. The work done is W œ 'a F(x) dx
%(&! x# 2†9500 “ !
b
œ 6400 Š4750
4750# 4†4750 ‹
œ ˆ 34 ‰ (6400)(4750)
27. Force constant: F œ kx Ê 20 œ k † 1 Ê k œ 20 lb/ft; the work to stretch the spring 1 ft is W œ '0 kx dx œ k'0 x dx œ ’20 x# “ œ 10 ft † lb; the work to stretch the spring an additional foot is 1
1
#
"
! #
W œ '1 kx dx œ k '1 x dx œ 20 ’ x# “ œ 20 ˆ 4# "# ‰ œ 20 ˆ 3# ‰ œ 30 ft † lb 2
2
#
"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 6 Practice Exercises 28. Force constant: F œ kx Ê 200 œ k(0.8) Ê k œ 250 N/m; the 300 N force stretches the spring x œ œ
300 250
381
F k
œ 1.2 m; the work required to stretch the spring that far is then W œ '0 F(x) dx œ '0 250x dx 1Þ2
1Þ2
œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J 29. We imagine the water divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0ß 8]. The typical slab between the planes at y and y ?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆ 54 y‰ ?y œ 25161 y# ?y ft$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 62.4 ?V œ
(62.4)(25) 16
1y# ?y lb. The distance through which F(y)
must act to lift this slab to the level 6 ft above the top is about (6 8 y) ft, so the work done lifting the slab is about ?W œ
(62.4)(25) 16
1y# (14 y) ?y ft † lb. The work done
lifting all the slabs from y œ 0 to y œ 8 to the level 6 ft above the top is approximately 8
W¸! !
(62.4)(25) 16
1y# (14 y) ?y ft † lb so the work to pump the water is the limit of these Riemann sums as the norm of
the partition goes to zero: W œ '0
8
$ œ (62.4) ˆ 25161 ‰ Š 14 3 †8
8% 4‹
(62.4)(25) (16)
(62.4)(25)1 16
1y# (14 y) dy œ
'08 a14y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 143 y$ y4 “ ) %
!
¸ 418,208.81 ft † lb
30. The same as in Exercise 29, but change the distance through which F(y) must act to (8 y) rather than (6 8 y). Also change the upper limit of integration from 8 to 5. The integral is:W œ '0
5
œ (62.4) ˆ 25161 ‰'0 a8y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 83 y$ 5
& y% 4 “!
(62.4)(25)1 16
y# (8 y) dy
œ (62.4) ˆ 25161 ‰ Š 83 † 5$
5% 4‹
31. The tank's cross section looks like the figure in Exercise 29 with right edge given by x œ #
slab has volume ?V œ 1(radius)# (thickness) œ 1 ˆ #y ‰ ?y œ F(y) œ 60 †
1 4
y œ y# . A typical horizontal
y# ?y. The force required to lift thisslab is its weight:
y# ?y. The distance through which F(y) must act is (2 10 y) ft, so the work to pump the liquid is
W œ 60'0 1a12 ybŠ y4 ‹dy œ 151 ’ 12y 3 10
22,5001 ft†lb 275 ft†lb/sec
1 4
5 10
¸ 54,241.56 ft † lb
#
$
"! y% 4 “!
œ 22,5001 ft † lb; the time needed to empty the tank is
¸ 257 sec
32. A typical horizontal slab has volume about ?V œ (20)(2x)?y œ (20) ˆ2È16 y# ‰ ?y and the force required to lift this slab is its weight F(y) œ (57)(20) ˆ2È16 y# ‰ ?y. The distance through which F(y) must act is (6 4 y) ft, so the work to pump the olive oil from the half-full tank is W œ 57'c4 (10 y)(20) ˆ2È16 y# ‰ dy 0
"Î# œ 2880 'c4 10È16 y# dy 1140'c4 a16 y# b (2y) dy 0
0
œ 22,800 † (area of a quarter circle having radius 4) 23 (1140) ’a16 y# b œ 335,153.25 ft † lb
$Î# !
“
%
œ (22,800)(41) 48,640
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
382
Chapter 6 Applications of Definite Integrals
33. Intersection points: 3 x# œ 2x# Ê 3x# 3 œ 0 Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Symmetry suggests that x œ 0. The typical @/<>3-+6 strip has # # # center of mass: (µ x ßµ y ) œ Šxß 2x a3 x b ‹ œ Šxß x 3 ‹ , #
#
#
#
#
length: a3 x b 2x œ 3 a1 x b, width: dx, area: dA œ 3 a1 x# b dx, and mass: dm œ $ † dA œ 3$ a1 x# b dx Ê the moment about the x-axis is µ y dm œ œ
3 #
$ ax# 3b a1 x# b dx œ
3 #
&
2x$ 3
$ ’ x5
" x$ 3 “ "
œ 3$ ’x
3x“
" "
3 #
$ ax% 2x# 3b dx Ê Mx œ ' µ y dm œ
œ 3$ ˆ 5"
2 3
3$ 15
3‰ œ
œ 6$ ˆ1 "3 ‰ œ 4$ Ê y œ
Mx M
œ
(3 10 45) œ
32$ 5 †4 $
œ
8 5
32$ 5
3 #
$ 'c1 ax% 2x# 3b dx "
; M œ ' dm œ 3$ 'c1 a1 x# b dx "
. Therefore, the centroid is (xß y) œ ˆ!ß 85 ‰ .
34. Symmetry suggests that x œ 0. The typical @/<>3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß x# ‹ , length: x# , width: dx, area: dA œ x# dx, mass: dm œ $ † dA œ $ x# dx Ê the moment about the x-axis is µ y dm œ #$ x# † x# dx œ
$ #
x% dx Ê Mx œ ' µ y dm œ
$ #
'c22 x% dx œ 10$ cx& d ##
35. The typical @/<>3-+6 strip has: center of mass: (µ x ßµ y ) œ Œxß
#
4 x4 #
, length: 4
area: dA œ Š4 œ $ Š4
x# 4‹
x# 4 ‹dx,
width: dx,
mass: dm œ $ † dA
dx Ê the moment about the x-axis is #
Š4 x4 ‹
µ y dm œ $ †
x# 4,
#
Š4
x# 4‹
dx œ
$ #
Š16
moment about the y-axis is µ x dm œ $ Š4 œ
$ 2
’16x
% x& 5†16 “ !
$ #
œ
64
64 ‘ 5
œ
x% 16 ‹ x# 4‹
œ
16†$ †3 32†$
œ
3 2
and y œ
Mx M
œ
dx. Thus, Mx œ ' µ y dm œ
4
4
My M
x$ 4‹
† x dx œ $ Š4x
; My œ ' µ x dm œ $ '0 Š4x
128$ 5
œ $ (32 16) œ 16$ ; M œ ' dm œ $ '0 Š4 Ê xœ
dx; the
128†$ †3 5†32†$
x# 4‹
œ
dx œ $ ’4x
12 5
% x$ 12 “ !
x$ 4‹
dx œ $ ’2x#
œ $ ˆ16
64 ‰ 1#
œ
$ #
'04 Š16 16x ‹ dx %
% x% 16 “ !
32$ 3
‰ . Therefore, the centroid is (xß y) œ ˆ 3# ß 12 5 .
36. A typical 29<3D98>+6 strip has: # center of mass: (µ x ßµ y ) œ Š y # 2y ß y‹ , length: 2y y# , width: dy, area: dA œ a2y y# b dy, mass: dm œ $ † dA œ $ a2y y# b dy; the moment about the x-axis is µ y dm œ $ † y † a2y y# b dy œ $ a2y# y$ b ; the moment # about the y-axis is µ x dm œ $ † ay 2yb † a2y y# b dy #
œ a4y y b dy Ê Mx œ ' µ y dm œ $ '0 a2y# y$ b dy $ #
#
œ $ ’ 23 y$ œ
$ #
yœ
ˆ 43†8 Mx M
2
%
œ
# y% 4 “!
32 ‰ 5
4†$ †3 3 † 4 †$
œ
œ $ ˆ 23 † 8 32$ 15
16 ‰ 4
œ $ ˆ 16 3
16 ‰ 4
œ
$ †16 12
œ
4$ 3
; My œ ' µ x dm œ
$ #
'02 a4y# y% b dy œ #$ ’ 34 y$ y5 “ #
$ ; M œ ' dm œ $ '0 a2y y# b dy œ $ ’y# y3 “ œ $ ˆ4 83 ‰ œ
2
# !
&
!
4$ 3
Ê xœ
œ 1. Therefore, the centroid is (xß y) œ ˆ 58 ß 1‰ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
My M
œ
$ †32†3 15†$ †4
œ
8 5
and
Chapter 6 Practice Exercises
383
37. A typical horizontal strip has: center of mass: (µ x ßµ y ) #
œ Š y #2y ß y‹ , length: 2y y# , width: dy, area: dA œ a2y y# b dy, mass: dm œ $ † dA œ (1 y) a2y y# b dy Ê the moment about the x-axis is µ y dm œ y(1 y) a2y y# b dy # œ a2y 2y$ y$ y% b dy œ a2y# y$ y% b dy; the moment about the y-axis is # µ x dm œ Š y 2y ‹ (1 y) a2y y# b dy œ " a4y# y% b (1 y) dy œ #
Ê Mx œ ' µ y dm œ '0 a2y# y$ y% b dy œ ’ 23 y$ 2
œ
16 60
œ
" #
" #
#
(20 15 24) œ $
Š 4†32 2%
2& 5
(11) œ
4 15
2' 6‹
2
44 40
œ
11 10
y$ 3
; My œ ' µ x dm œ '0
2
œ 4 ˆ 43 2
œ '0 a2y y# y$ b dy œ ’y# ‰ ˆ 38 ‰ œ œ ˆ 44 15
44 15
y% 4
4 5
œ ˆ4
8 3
16 ‰ 4
œ ˆ 16 3
16 4
32 ‰ 5
œ 16 ˆ "3
a4y# 4y$ y% y& b dy œ
86 ‰ œ 4 ˆ2 45 ‰ œ
# y% 4 “!
" #
# y& 5 “!
a4y# 4y$ y% y& b dy
œ
8 3
24 5
" #
" 4
25 ‰
’ 43 y$ y%
y& 5
; M œ ' dm œ '0 (1 y) a2y y# b dy
# y' 6 “!
2
Ê xœ
My M
‰ ˆ 83 ‰ œ œ ˆ 24 5
9 5
and y œ
Mx M
‰ . Therefore, the center of mass is (xß y) œ ˆ 95 ß 11 10 .
3 3 38. A typical vertical strip has: center of mass: (µ x ßµ y ) œ ˆxß 2x3$Î# ‰ , length: x$Î# , width: dx, area: dA œ x$Î# dx, µ 3 3 3 9$ mass: dm œ $ † dA œ $ † dx Ê the moment about the x-axis is y dm œ †$ dx œ dx; the moment about x$Î#
3 the y-axis is µ x dm œ x † $ x$Î# dx œ
(a) Mx œ $ '1
9
M œ $ '1
9
(b) Mx œ '1
9
" #
3 x$Î# x #
9$ #
ˆ x9$ ‰ dx œ
3$ x"Î#
#x$Î#
# *
20$ 9
’ x# “ œ "
3 ‰ ; My œ $ '1 x ˆ x$Î# dx œ 3$ 2x"Î# ‘ " œ 12$ ;
*
*
9 #
*
9
dx œ 6$ x"Î# ‘ " œ 4$ Ê x œ
ˆ x9$ ‰ dx œ
2x$
x$Î#
dx.
My M
œ
12$ 4$
œ 3 and y œ
Mx M
œ
ˆ 209$ ‰ 4$
œ
5 9
* 3 ‰ 3 ‰ x" ‘ * œ 4; My œ ' x# ˆ $Î# dx œ 2x$Î# ‘ " œ 52; M œ '1 x ˆ x$Î# dx " x 1
œ 6 x"Î# ‘ " œ 12 Ê x œ
9
My M
œ
13 3
and y œ
Mx M
9
œ
" 3
strip 39. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 2 '0 (62.4)(2 y)(2y) dy œ 249.6'0 a2y y# b dy œ 249.6 ’y# b
2
2
# y$ 3 “!
œ (249.6) ˆ4 83 ‰ œ (249.6) ˆ 43 ‰ œ 332.8 lb strip 40. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ '0 75 ˆ 56 y‰ (2y 4) dy œ 75'0 ˆ 53 y 5Î6
b
5Î6
10 3
2y# 4y‰ dy
7 7 # 2 $ ‘ &Î' 50 ‰ 25 ‰ 125 ‰‘ #‰ ˆ 18 œ 75 '0 ˆ 10 dy œ 75 10 ˆ 67 ‰ ˆ 36 ˆ 32 ‰ ˆ 216 3 3 y 2y 3 y 6 y 3 y ! œ (75) 5Î6
œ (75) ˆ 25 9
175 216
250 ‰ 3†#16
‰ œ ˆ 9†75 #16 (25 † 216 175 † 9 250 † 3) œ
strip 41. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 62.4'0 (9 y) Š2 † b
4
%
œ 62.4 6y$Î# 25 y&Î# ‘ ! œ (62.4) ˆ6 † 8
2 5
Èy 2 ‹
(75)(3075) 9†#16
¸ 118.63 lb.
dy œ 62.4'0 ˆ9y"Î# 3y$Î# ‰ dy 4
‰ (48 † 5 64) œ † 32‰ œ ˆ 62.4 5
(62.4)(176) 5
œ 2196.48 lb
strip 42. Place the origin at the bottom of the tank. Then F œ '0 W † Š depth ‹ † L(y) dy, h œ the height of the mercury column, h
strip depth œ h y, L(y) œ 1 Ê F œ '0 849(h y) " dy œ (849)'0 (h y) dy œ 849’hy h
œ
849 # 2 h .
Now solve
849 # 2 h
h
h
y# # “!
œ 849 Šh#
h# #‹
œ 40000 to get h ¸ 9.707 ft. The volume of the mercury is s2 h œ 12 † 9.707 œ 9.707 ft$ Þ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
384
Chapter 6 Applications of Definite Integrals
CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES 1. V œ 1 'a cf(x)d# dx œ b# ab Ê 1'a cf(t)d# dt œ x# ax for all x a Ê 1 [f(x)]# œ 2x a Ê f(x) œ É 2x1 a b
x
2. V œ 1 '0 [f(x)]# dx œ a# a Ê 1 '0 [f(t)]# dt œ x# x for all x a Ê 1[f(x)]# œ 2x 1 Ê f(x) œ É 2x1 1 a
x
3. s(x) œ Cx Ê '0 È1 [f w (t)]# dt œ Cx Ê È1 [f w (x)]# œ C Ê f w (x) œ ÈC# 1 for C 1 x
Ê f(x) œ '0 ÈC# 1 dt k. Then f(0) œ a Ê a œ 0 k Ê f(x) œ '0 ÈC# 1 dt a Ê f(x) œ xÈC# 1 a, x
x
where C 1. 4. (a) The graph of f(x) œ sin x traces out a path from (!ß !) to (!ß sin !) whose length is L œ '0 È1 cos# ) d). !
The line segment from (0ß 0) to (!ß sin !) has length È(! 0)# (sin ! 0)# œ È!# sin# !. Since the shortest distance between two points is the length of the straight line segment joining them, we have ! immediately that ' È1 cos# ) d) È!# sin# ! if 0 ! Ÿ 1 . #
0
(b) In general, if y œ f(x) is continuously differentiable and f(0) œ 0, then '0 È1 [f w (t)]# dt È!# f # (!) !
for ! 0. 5. We can find the centroid and then use Pappus' Theorem to calculate the volume. faxb œ x, gaxb œ x2 , faxb œ gaxb 1 Ê x œ x2 Ê x2 x œ 0 Ê x œ 0, x œ 1; $ œ 1; M œ '0 cx x2 ddx œ "# x2 13 x3 ‘0 œ ˆ "# 13 ‰ 0 œ 1
1 6
xœ
1 1 Î6
'01 xcx x2 ddx œ 6'01 cx2 x3 ddx œ 6 31 x3 41 x4 ‘10 œ 6ˆ 31 41 ‰ 0 œ 21
yœ
1 1 Î6
'01 12 ’x2 ax2 b2 “dx œ 3'01 cx2 x4 ddx œ 3 13 x3 15 x5 ‘10 œ 3ˆ 13 15 ‰ 0 œ 25 Ê The centroid is ˆ 12 , 25 ‰.
3 is the distance from ˆ 12 , 25 ‰ to the axis of rotation, y œ x. To calculate this distance we must find the point on y œ x that also lies on the line perpendicular to y œ x that passes through ˆ 12 , 25 ‰. The equation of this line is y 25 œ 1ˆx 12 ‰ Êxyœ
9 10 .
9 3 œ Ɉ 20
The point of intersection of the lines x y œ
1 ‰2 2
9 ˆ 20
2 ‰2 5
œ
1 . 10È2
9 and y œ x is ˆ 20 ,
9 10
Thus V œ 21Š 101È2 ‹ˆ 16 ‰ œ
9 ‰ 20 .
Thus,
1 . 30È2
6. Since the slice is made at an angle of 45‰ , the volume of the wedge is half the volume of the cylinder of radius height 1. Thus, V œ
" ˆ " ‰2 # ’1 # a1b
“œ
" #
and
1 8.
7. y œ 2Èx Ê ds œ É "x 1 dx Ê A œ '0 2Èx É "x 1 dx œ 3
4 3
(1 x)$Î# ‘ $ œ !
28 3
8. This surface is a triangle having a base of 21a and a height of 21ak. Therefore the surface area is " # # # (21a)(21ak) œ 21 a k. d# x dt#
9. F œ ma œ t# Ê
œaœ
t# m
Ê vœ
x œ 0 when t œ 0 Ê C" œ 0 Ê x œ W œ ' F dx œ '0
Ð12mhÑ"Î%
œ
(12mh)$Î# 18m
œ
F(t) †
12mh†È12mh 18m
œ
2h 3
dx dt
dx t$ dt œ 3m C; v œ 0 when t œ 0 Ê t% "Î% . 12m . Then x œ h Ê t œ (12mh)
dt œ '0
Ð12mhÑ"Î%
† 2È3mh œ
4h 3
t# †
$
t 3m
dt œ
" 3m
'
’ t6 “
Ð12mh)"Î%
0
Cœ0 Ê
dx dt
œ
t$ 3m
The work done is
" ‰ œ ˆ 18m (12mh)'Î%
È3mh
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ê xœ
t% 12m
C" ;
Chapter 6 Additional and Advanced Exercises 10. Converting to pounds and feet, 2 lb/in œ
†
2 lb 1 in
œ 24 lb/ft. Thus, F œ 24x Ê W œ '0
1Î2
12 in 1 ft
24x dx
"Î# " " ‰ œ c12x# d ! œ 3 ft † lb. Since W œ "# mv!# "# mv"# , where W œ 3 ft † lb, m œ ˆ 10 lb‰ ˆ 3# ft/sec # " œ 320 slugs, and v" œ 0 ft/sec, we have 3 œ ˆ #" ‰ ˆ 3#"0 v#! ‰ Ê v!# œ 3 † 640. For the projectile height, s œ 16t# v! t (since s œ 0 at t œ 0) Ê ds dt œ v œ 32t v! . At the top of the ball's path, v œ 0 Ê #
and the height is s œ 16 ˆ 3v#! ‰ v! ˆ 3v#! ‰ œ
v!# 64
3†640 64
œ
385
tœ
v! 3#
œ 30 ft.
11. From the symmetry of y œ 1 xn , n even, about the y-axis for 1 Ÿ x Ÿ 1, we have x œ 0. To find y œ MMx , we n use the vertical strips technique. The typical strip has center of mass: (µ x ßµ y ) œ ˆxß 1 2 x ‰ , length: 1 xn , width: dx, area: dA œ a1 xn b dx, mass: dm œ 1 † dA œ a1 xn b dx. The moment of the strip about the 1 1 n # n # " nb1 2n b 1 x-axis is µ y dm œ a1 x b dx Ê M œ ' a1 x b dx œ 2' " a1 2xn x2n b dx œ x 2x x ‘ #
œ1
2 n1
" #n 1
œ
x c1 # (n 1)(2n 1) 2(2n ") (n 1) (n 1)(#n 1)
œ
0 # 2n# 3n 1 4n 2 n 1 (n 1)(#n 1)
Also, M œ 'c1 dA œ 'c1 a1 xn b dx œ 2 '0 a1 xn b dx œ 2 x 1
yœ
Mx M
œ
1
#
†
2n (n 1)(2n 1)
1
(n 1) 2n
œ
xn b 1 ‘ " n1 !
n1
œ
2n# (n 1)(#n 1)
œ 2 ˆ1
" ‰ n1
#n 1 !
. œ
2n n1.
Therefore,
Ê ˆ!ß #n n 1 ‰ is the location of the centroid. As n Ä _, y Ä
n 2n 1
the limiting position of the centroid is ˆ!ß
" #
so
"‰ # .
12. Align the telephone pole along the x-axis as shown in the accompanying figure. The slope of the top length of pole is 9 ‰ ˆ 14.5 " 81 81 œ 8"1 † 40 † (14.5 9) œ 815.5 †40 40 11 ‰ y œ 891 8111†80 x œ 8"1 ˆ9 80 x is an
œ
11 81†80 .
Thus,
equation of the
line representing the top of the pole. Then, My œ 'a x † 1y# dx œ 1 '0 x 8"1 ˆ9 b
40
11 80
# x‰‘ dx
b 11 ‰# '040 x ˆ9 80 x dx; M œ 'a 1y# dx 40 40 11 ‰‘# ‰# dx. œ 1 '0 8"1 ˆ9 80 x dx œ 64"1 '0 ˆ9 11 80 x
œ
" 641
Thus, x œ
My M
¸
129,700 5623.3
¸ 23.06 (using a calculator to compute
the integrals). By symmetry about the x-axis, y œ 0 so the center of mass is about 23 ft from the top of the pole. 13. (a) Consider a single vertical strip with center of mass (µ x ßµ y ). If the plate lies to the right of the line, then µ µ b) $ dA Ê the plate's first moment the moment of this strip about the line x œ b is (x b) dm œ (x about x œ b is the integral ' (x b)$ dA œ ' $ x dA ' $ b dA œ My b$ A. (b) If the plate lies to the left of the line, the moment of a vertical strip about the line x œ b is ab µ x b dm œ ab µ x b $ dA Ê the plate's first moment about x œ b is ' (b x)$ dA œ ' b$ dA ' $ x dA œ b$ A My . 14. (a) By symmetry of the plate about the x-axis, y œ 0. A typical vertical strip has center of mass: (µ x ßµ y ) œ (xß 0), length: 4Èax, width: dx, area: 4Èax dx, mass: dm œ $ dA œ kx † 4Èax dx, for some a proportionality constant k. The moment of the strip about the y-axis is M œ ' µ x dm œ ' 4kx# Èax dx y
œ 4kÈa'0 x a
&Î#
dx œ
4kÈa 27
x
(Î# ‘ a
0
œ 4ka
"Î#
† a 2 7
(Î#
œ
œ 4kÈa'0 x$Î# dx œ 4kÈa 25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ a
Ê (xß y) œ
a
ˆ 5a ‰ 7 ß0
0
8ka% 7
. Also, M œ ' dm œ '0 4kxÈax dx
8ka$ 5
. Thus, x œ
a
My M
œ
8ka% 7
†
5 8ka$
œ
5 7
is the center of mass. y#
# # a (b) A typical horizontal strip has center of mass: (µ x ßµ y ) œ Œ 4a # ß y œ Š y 8a4a ß y‹ , length: a
width: dy, area: Ša œ 'c2a y kyk Ša 2a
a
y# 4a ‹
y# 4a ‹
dy, mass: dm œ $ dA œ kyk Ša
dy œ 'c2a y# Ša 0
y# 4a ‹
y# 4a ‹
dy '0 y# Ša 2a
dy. Thus, Mx œ ' µ y dm y# 4a ‹
dy
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
y# 4a
,
386
Chapter 6 Applications of Definite Integrals œ 'c2a Šay# 0
%
32a& 20a
œ 8a3
dy '0 Šay# 2a
y% 4a ‹ 8a% 3
y% 4a ‹
œ 0; My œ ' µ x dm œ 'c2a Š y 2a
32a& #0a
'c2a2a kyk ay# 4a# b Š 4a 4a y ‹ dy œ 32a" #
! y& #0a “ #a
dy œ ’ 3a y$ #
4a# 8a ‹
#
" 8a
œ
" 3 #a #
'c02a a16a% y y& b dy 32a" '02a a16a% y y& b dy œ 3#"a
œ
" 32a#
’8a% † 4a#
" 32a#
2a
œ
#
#
64a' 6 “
œ
" 16a#
#
’8a% y# 32a' 3 ‹
Š32a'
œ
#a
œ2†
#
16a% 4 ‹
#
Š2a † 4a
" #a
œ
! y' 6 “ #a
1 3#a#
’8a% y#
† 32 a32a' b œ
4 3
#a y' 6 “!
a% ;
'c2a2a kyk a4a# y# b dy
" 4a
%
" 4a
dy
" 16a#
'c02a a4a# y y$ b dy 4a" '02a a4a# y y$ b dy œ 4a" ’2a# y# y4 “ !
" 4a
yœ
c2a
’8a% † 4a#
M œ ' dm œ 'c2a kyk Š 4a 4ay ‹ dy œ
y# 4a ‹
kyk Ša
' kyk a16a% y% b dy
#
#
#a y& #0a “ !
2a
œ
64a' 6 “
’ 3a y$
%
%
$
a8a 4a b œ 2a . Therefore, x œ
" 4a
’2a# y#
œ ˆ 34 a% ‰ ˆ 2a"$ ‰ œ
My M
#a y% 4 “!
2a 3
and
œ 0 is the center of mass.
Mx M
15. (a) On [0ß a] a typical @/<>3-+6 strip has center of mass: (µ x ßµ y ) œ Šx,
È b# x # È a# x# ‹, #
length: Èb# x# Èa# x# , width: dx, area: dA œ ŠÈb# x# Èa# x# ‹ dx, mass: dm œ $ dA œ $ ŠÈb# x# Èa# x# ‹ dx. On [aß b] a typical @/<>3-+6 strip has center of mass: È # # (µ x ßµ y ) œ Šxß b # x ‹ , length: Èb# x# , width: dx, area: dA œ Èb# x# dx,
mass: dm œ $ dA œ $ Èb# x# dx. Thus, Mx œ ' µ y dm œ '0
a
" #
ŠÈb# x# Èa# x# ‹ $ ŠÈb# x# Èa# x# ‹ dx 'a
b
" #
Èb# x# $ Èb# x# dx
œ
$ #
'0a cab# x# b aa# x# bd dx #$ 'ab ab# x# b dx œ #$ '0a ab# a# b dx #$ 'ab ab# x# b dx
œ
$ #
cab# a# b xd ! #$ ’b# x
œ
$ #
aab# a$ b #$ Š 23 b$ ab#
a
b
x$ 3 “a
œ a$ 3‹
$ #
b$ 3‹
cab# a# b ad #$ ’Šb$
œ
$ b$ 3
$ a$ 3
œ $ Šb
$
a$ 3 ‹;
a$ 3 ‹“
Š b# a
My œ ' µ x dm
œ '0 x$ ŠÈb# x# Èa# x# ‹ dx 'a x$ Èb# x# dx a
b
œ $ '0 x ab# x# b a
œ
$ #
”
2 ab # x # b 3
$Î#
"Î#
dx $ '0 x aa# x# b a
a
$ 2 aa • #”
#
x# b 3
$Î#
# $Î#
#
œ ’ab a b
# $Î#
ab b
a
dx $ 'a x ab# x# b
$ 2 ab • #”
b
#
!
0
$ 3
"Î#
# $Î#
$ 3
“ ’0 aa b
x# b 3
"Î#
$Î#
• a
$ 3
“ ’0 ab# a# b #
#
$Î#
We calculate the mass geometrically: M œ $ A œ $ Š 14b ‹ $ Š 14a ‹ œ œ
$ ab $ a $ b 3
yœ (b) lim
œ
Mx M 4
b Ä a 31
†
4 $1 ab# a# b #
#
4 aa abb b 31(ab)
Ša
#
ab b ab
#
œ
4 31
$
$
a Š bb# a# ‹ œ
dx
b
4 (b a) aa# ab b# b 31 (b a)(b a)
“œ
$1 4
$ b$ 3
$ a$ 3
œ
$ ab $ a $ b 3
ab# a# b . Thus, x œ
œ Mx ; My M
œ
4 aa# ab b# b 31(a b)
2a 1
2a ‰ Ê (xß y) œ ˆ 2a 1 ß 1 is the limiting
; likewise
.
‹ œ ˆ 341 ‰ Š a
#
a# a# ‹ aa
#
œ ˆ 341 ‰ Š 3a 2a ‹ œ
position of the centroid as b Ä a. This is the centroid of a circle of radius a (and we note the two circles coincide when b œ a).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 6 Additional and Advanced Exercises 16. Since the area of the traingle is 36, the diagram may be labeled as shown at the right. The centroid of the triangle is ˆ 3a , 24 ‰ a . The shaded portion is 144 36 œ 108. Write ax, yb for the centroid of the remaining region. The centroid of the whole square is obviously a6, 6b. Think of the square as a sheet of uniform density, so that the centroid of the square is the average of the centroids of the two regions, weighted by area: 'œ
$'ˆ 3a ‰ "!)axb "%%
and ' œ
‰ $'ˆ 24 a "!)ayb "%%
which we solve to get x œ )
a *
and y œ
)a a " b . a
Set
x œ 7 in. (Given). It follows that a œ *, whence y œ œ
7 "*
'% *
in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.
above, we will find x œ 7 "* .) 17. The submerged triangular plate is depicted in the figure at the right. The hypotenuse of the triangle has slope 1 Ê y (2) œ (x 0) Ê x œ (y 2) is an equation of the hypotenuse. Using a typical horizontal strip, the fluid strip strip pressure is F œ ' (62.4) † Š depth ‹ † Š length ‹ dy
c2
c2
œ 'c6 (62.4)(y)[(y 2)] dy œ 62.4 'c6 ay# 2yb dy $
œ 62.4 ’ y3 y# “
#
‰‘ œ (62.4) ˆ 83 4‰ ˆ 216 3 36
'
‰ œ (62.4) ˆ 208 3 32 œ
(62.4)(112) 3
¸ 2329.6 lb
18. Consider a rectangular plate of length j and width w. The length is parallel with the surface of the fluid of weight density =. The force on one side of the plate is F œ ='cw (y)(j) dy œ =j ’ y# “ 0
#
! w
œ
=jw# #
. The
average force on one side of the plate is Fav œ œ
= w
#
’ y# “
! w
œ
=w #
. Therefore the force
= w
'c0w (y)dy
=jw# #
‰ œ ˆ =w # (jw) œ (the average pressure up and down) † (the area of the plate).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
387
388
Chapter 6 Applications of Definite Integrals
NOTES:
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 7 TRANSCENDENTAL FUNCTIONS 7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1. Yes one-to-one, the graph passes the horizontal line test. 2. Not one-to-one, the graph fails the horizontal line test. 3. Not one-to-one since (for example) the horizontal line y œ 2 intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal line test. 5. Yes one-to-one, the graph passes the horizontal line test 6. Yes one-to-one, the graph passes the horizontal line test 7. Not one-to-one since the horizontal line y œ 3 intersects the graph an infinite number of times. 8. Yes one-to-one, the graph passes the horizontal line test 9. Yes one-to-one, the graph passes the horizontal line test 10. Not one-to-one since (for example) the horizontal line y œ 1 intersects the graph twice. 11. Domain: 0 x Ÿ 1, Range: 0 Ÿ y
13. Domain: 1 Ÿ x Ÿ 1, Range: 1# Ÿ y Ÿ
12. Domain: x 1, Range: y 0
1 #
14. Domain: _ x _, Range: 1# y Ÿ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 #
390
Chapter 7 Transcendental Functions
15. Domain: 0 Ÿ x Ÿ 6, Range: 0 Ÿ y Ÿ 3
16. Domain: 2 Ÿ x Ÿ 1, Range: 1 Ÿ y 3
17. The graph is symmetric about y œ x.
(b) y œ È1 x# Ê y# œ 1 x# Ê x# œ 1 y# Ê x œ È1 y# Ê y œ È1 x# œ f " (x) 18. The graph is symmetric about y œ x.
yœ
" x
Ê xœ
" y
Ê yœ
" x
œ f " (x)
19. Step 1: y œ x# 1 Ê x# œ y 1 Ê x œ Èy 1 Step 2: y œ Èx 1 œ f " (x) 20. Step 1: y œ x# Ê x œ Èy, since x Ÿ !. Step 2: y œ Èx œ f " (x) 21. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$ Step 2: y œ $Èx 1 œ f " (x) 22. Step 1: y œ x# 2x 1 Ê y œ (x 1)# Ê Èy œ x 1, since x 1 Ê x œ 1 Èy Step 2: y œ 1 Èx œ f " (x) 23. Step 1: y œ (x 1)# Ê Èy œ x 1, since x 1 Ê x œ Èy 1 Step 2: y œ Èx 1 œ f " (x) 24. Step 1: y œ x#Î$ Ê x œ y$Î# Step 2: y œ x$Î# œ f " (x)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.1 Inverse Functions and Their Derivatives 25. Step 1: y œ x& Ê x œ y"Î& 5 Step 2: y œ È x œ f " (x); Domain and Range of f " : all reals; &
f af " (x)b œ ˆx"Î& ‰ œ x and f " (f(x)) œ ax& b
"Î&
œx
"Î%
œx
26. Step 1: y œ x% Ê x œ y"Î% Step 2: y œ %Èx œ f " (x); Domain of f " : x 0, Range of f " : y 0; %
f af " (x)b œ ˆx"Î% ‰ œ x and f " (f(x)) œ ax% b
27. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$ Step 2: y œ $Èx 1 œ f " (x); Domain and Range of f " : all reals; $
f af " (x)b œ ˆ(x 1)"Î$ ‰ 1 œ (x 1) 1 œ x and f " (f(x)) œ aax$ 1b 1b 28. Step 1: y œ
" #
x
7 #
Ê "
" #
xœy
7 #
Step 2: y œ
" x#
Ê x# œ
" Èx
œ f " (x)
" y
Ê xœ
7 #
30. Step 1: y œ
" x$
" x"Î$ "
Step 2: y œ Domain of f
f af " axbb œ
" y
Ê xœ
Step 2: y
32. Step 1: y œ
œ
" Š "x ‹
œ x since x 0
" y"Î$
: x Á 0, Range of f " : y Á 0;
" $ ax "Î$ b
œ
" x "
œ x and f " afaxbb œ ˆ x"$ ‰
x3 x 2 Ê yax 2b 3 1 œ 2x axb; x1 œ f "
f af " axbb œ
" É x"#
3 " " œÉ (x); x œ f
31. Step 1: y œ Domain of f
œx
" Èy
Šx‹
Ê x$ œ
"Î$
œ x and f " (f(x)) œ 2 ˆ "# x 7# ‰ 7 œ (x 7) 7 œ x
Domain of f " : x 0, Range of f " : y 0; f af " (x)b œ "" # œ "" œ x and f " (f(x)) œ Š Èx ‹
œ ax$ b
Ê x œ 2y 7
Step 2: y œ 2x 7 œ f (x); Domain and Range of f " : all reals; f af " (x)b œ "# (2x 7) 7# œ ˆx 7# ‰ 29. Step 1: y œ
"Î$
"Î$
œ ˆ x" ‰
"
œx
œ x 3 Ê x y 2y œ x 3 Ê x y x œ 2y 3 Ê x œ
: x Á 1, Range of f " : y Á 2;
b3‰ ˆ 2x xc1 3 b3‰ ˆ 2x xc1 2 Èx Èx 3
œ
a2x 3b 3ax 1b a2x 3b 2ax 1b
œ
5x 5
œ x and f " afaxbb œ
3 2ˆ xx b c2‰ 3 3 ˆ xx b c2‰ 1
œ
2y 3 y1
2 ax 3 b 3 a x 2 b ax 3 b ax 2 b
œ
5x 5
œx
Ê yˆÈx 3‰ œ Èx Ê yÈx 3y œ Èx Ê yÈx Èx œ 3y Ê x œ Š y 3y 1‹
2
2
‰ œ f 1 a xb ; Step 2: y œ ˆ x 3x 1
Domain of f " : Ð_, 0Ó a1, _b, Range of f " : Ò0, 9Ñ a9, _b; f af " axbb œ f
"
2 Ɉ x 3x c1‰ 2
Ɉ x 3x c1‰ 3
afaxbb œ
; If x 1 or x Ÿ 0 Ê
Èx
3Š È x c 3 ‹
Èx
Š Èx c 3 ‹ 1
3x x1
0Ê
2 Ɉ x 3x c1‰ 2
Ɉ x 3x c1‰ 3
œ
3x xc1 3x xc1 3
œ
3x 3x 3ax 1b
œ
2
œ
9x ˆÈ x ˆÈ x 3 ‰‰ 2
œ
9x 9
œx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3x 3
œ x and
391
392
Chapter 7 Transcendental Functions
33. Step 1: y œ x2 2x, x Ÿ 1 Ê y 1 œ ax 1b2 , x Ÿ 1 Ê Èy 1 œ x 1, x Ÿ 1 Ê x œ 1 Èy 1 Step 2: y œ 1 Èx 1 œ f 1 axb; Domain of f " : Ò1, _Ñ, Range of f " : Ð_, 1Ó; 2
f af " axbb œ Š1 Èx 1‹ 2Š1 Èx 1‹ œ 1 2Èx 1 x 1 2 2Èx 1 œ x and f " afaxbb œ 1 Èax2 2xb 1, x Ÿ 1 œ 1 Éax 1b2 , x Ÿ 1 œ 1 lx 1l œ 1 a1 xb œ x 34. Step 1: y œ a2x3 1b 3 x Step 2: y œ É
5
1 2
1 Î5
Ê y5 œ 2x3 1 Ê y5 1 œ 2x3 Ê
y5 1 2
3 y œ x3 Ê x œ É
5
1 2
œ f 1 ax b ;
Domain of f " : a_, _b, Range of f " : a_, _b; 3 x f af " axbb œ Œ2ŠÉ
5
1 2 ‹
3
1 Î5
1
œ Š 2Š x
5
1 2 ‹
1‹
1 Î5
œ aax5 1b 1b
1 Î5
œ ax5 b
1 Î5
œ x and
5
1Î5 ’a2x3 1b “ 1
3
f " afaxbb œ Ê
2
3 a2x σ
35. (a) y œ 2x 3 Ê 2x œ y 3 Ê x œ y# 3# Ê f " (x) œ (c)
df ¸ dx xœ 1
36. (a) y œ
" 5
œ 2,
df c" dx ¹ xœ1
x7 Ê
" 5
œ
(c)
œ
" df c" 5 , dx ¹ xœ$%Î&
(c)
œ 4,
df c" dx ¹ xœ3
38. (a) y œ 2x# Ê x# œ Ê xœ (c)
df ¸ dx xœ&
" È2
" #
3
3 #
" #
"
(b)
(x) œ 5x 35
œ5
œ
(b) 5 4
x 4
" 4
(b)
y
Èy Ê f
3 2x œÉ 2 œ x
(b) x #
37. (a) y œ 5 4x Ê 4x œ 5 y Ê x œ 54 y4 Ê f " (x) œ df ¸ dx xœ1Î#
1 b 1 2
xœy7
Ê x œ 5y 35 Ê f df ¸ dx xœ 1
3
"
(x) œ
È x#
œ 4xk xœ5 œ 20,
df c" dx ¹ xœ&0
œ
" #È 2
x"Î# ¹
xœ50
œ
" #0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.1 Inverse Functions and Their Derivatives $
3 3 39. (a) f(g(x)) œ ˆÈ x‰ œ x, g(f(x)) œ È x3 œ x
w
#
w
(b)
w
(c) f (x) œ 3x Ê f (1) œ 3, f (1) œ 3; gw (x) œ 3" x#Î$ Ê gw (1) œ 3" , gw (1) œ
" 3
(d) The line y œ 0 is tangent to f(x) œ x$ at (!ß !); the line x œ 0 is tangent to g(x) œ $Èx at (0ß 0)
40. (a) h(k(x)) œ
" 4
ˆ(4x)"Î$ ‰$ œ x,
k(h(x)) œ Š4 † (c) hw (x) œ w
k (x) œ
x$ 4‹
"Î$
(b)
œx
3x# w w 4 Ê h (2) œ 3, h (2) 4 #Î$ Ê kw (2) œ "3 , 3 (4x)
œ 3; kw (2) œ
(d) The line y œ 0 is tangent to h(x) œ
x$ 4
" 3
at (!ß !);
the line x œ 0 is tangent to k(x) œ (4x)"Î$ at (!ß !) œ 3x# 6x Ê
41.
df dx
43.
df " dx ¹ x œ 4
df c" dx ¹ x œ f(3)
df " dx ¹ x œ f(2)
œ
45. (a) y œ mx Ê x œ
" m
œ
(b) The graph of y œ f 46. y œ mx b Ê x œ
y m
"
df dx
º
œ
xœ2
"
df dx
œ
º
xœ3
" ˆ 3" ‰
œ3
y Ê f " (x) œ "
" 9
œ
" m
œ 2x 4 Ê
42.
df dx
44.
dg " dx ¹x œ 0
b m
dg " dx ¹ x œ f(0)
œ
"
dg dx
º
œ
xœ0
"
df dx
º
œ
xœ5
œ
" 6
" 2
x
(x) is a line through the origin with slope
œ
df " dx ¹ x œ f(5)
Ê f " (x) œ
" m
x
b m;
" m.
the graph of f " (x) is a line with slope
" m
and y-intercept mb .
47. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ x 1 (b) y œ x b Ê x œ y b Ê f " (x) œ x b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y œ x equidistant from that line.
48. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ 1 x; the lines intersect at a right angle (b) y œ x b Ê x œ y b Ê f " (x) œ b x; the lines intersect at a right angle (c) Such a function is its own inverse.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
393
394
Chapter 7 Transcendental Functions
49. Let x" Á x# be two numbers in the domain of an increasing function f. Then, either x" x# or x" x# which implies f(x" ) f(x# ) or f(x" ) f(x# ), since f(x) is increasing. In either case, f(x" ) Á f(x# ) and f is one-to-one. Similar arguments hold if f is decreasing. 50. f(x) is increasing since x# x" Ê
" 3
x#
5 6
" 3
x" 56 ;
df dx
œ
" 3
51. f(x) is increasing since x# x" Ê 27x$# 27x"$ ; y œ 27x$ Ê x œ df dx
œ 81x# Ê
df " dx
œ
" ¸ 81x# 13 x"Î$
œ
" 9x#Î$
œ
" 9
df c" dx
Ê " 3
œ
df dx
œ 24x# Ê
dx
œ
" ¸ 24x# 12 Ð1 xÑ"Î$
œ
œ3
y"Î$ Ê f " (x) œ
" 3
x"Î$ ;
x#Î$
52. f(x) is decreasing since x# x" Ê 1 8x$# 1 8x"$ ; y œ 1 8x$ Ê x œ df c"
" ˆ "3 ‰
" 6(" x)#Î$
" #
(1 y)"Î$ Ê f " (x) œ
" #
(1 x)"Î$ ;
œ "6 (1 x)#Î$
53. f(x) is decreasing since x# x" Ê (1 x# )$ (1 x" )$ ; y œ (1 x)$ Ê x œ 1 y"Î$ Ê f " (x) œ 1 x"Î$ ; df dx
œ 3(1 x)# Ê
df c" dx
œ
" 3(1 x)# ¹ 1cx"Î$ &Î$
54. f(x) is increasing since x# x" Ê x# df dx
œ
5 3
x#Î$ Ê
df c" dx
œ
5 3
" ¹ x#Î$ x$Î&
œ
" 3x#Î$
œ "3 x#Î$
&Î$
x" ; y œ x&Î$ Ê x œ y$Î& Ê f " (x) œ x$Î& ; œ
3 5x#Î&
œ
3 5
x#Î&
55. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so f(x" ) Á f(x# ) and therefore g(x" ) Á g(x# ). Therefore g(x) is one-to-one as well. 56. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so " " f(x" ) Á f(x# ) , and therefore h(x" ) Á h(x# ). 57. The composite is one-to-one also. The reasoning: If x" Á x# then g(x" ) Á g(x# ) because g is one-to-one. Since g(x" ) Á g(x# ), we also have f(g(x" )) Á f(g(x# )) because f is one-to-one. Thus, f ‰ g is one-to-one because x" Á x# Ê f(g(x" )) Á f(g(x# )). 58. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x" Á x# in the domain of g with g(x" ) œ g(x# ). For these numbers we would also have f(g(x" )) œ f(g(x# )), contradicting the assumption that f ‰ g is one-to-one. 59. (g ‰ f)(x) œ x Ê g(f(x)) œ x Ê gw (f(x))f w (x) œ 1 60. W(a) œ 'f(a) 1 ’af " (y)b a# “ dy œ 0 œ 'a 21x[f(a) f(x)] dx œ S(a); Ww (t) œ 1’af " (f(t))b a# “ f w (t) f(a)
a
#
#
œ 1 at# a# b f w (t); also S(t) œ 21f(t)'a x dx 21'a xf(x) dx œ c1f(t)t# 1f(t)a# d 21'a xf(x) dx Ê Sw (t) t
t
t
œ 1t# f w (t) 21tf(t) 1a# f w (t) 21tf(t) œ 1 at# a# b f w (t) Ê Ww (t) œ Sw (t). Therefore, W(t) œ S(t) for all t − [aß b]. 61-68. Example CAS commands: Maple: with( plots );#63 f := x -> sqrt(3*x-2); domain := 2/3 .. 4; x0 := 3; Df := D(f);
# (a)
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Section 7.1 Inverse Functions and Their Derivatives
395
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#61(a) (Section 7.1)" ); q1 := solve( y=f(x), x ); # (b) g := unapply( q1, y ); m1 := Df(x0); # (c) t1 := f(x0)+m1*(x-x0); y=t1; m2 := 1/Df(x0); # (d) t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f,domain); # (e) p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [p1,p2,p3,p4,p5], scaling=constrained, title="#63(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and x0 may vary) If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica to do this. See section 2.5 for details. <<Miscellaneous `RealOnly` Clear[x, y] {a,b} = {2, 1}; x0 = 1/2 ; f[x_] = (3x 2) / (2x 11) Plot[{f[x], f'[x]}, {x, a, b}] solx = Solve[y == f[x], x] g[y_] = x /. solx[[1]] y0 = f[x0] ftan[x_] = y0 f'[x0] (x-x0) gtan[y_] = x0 1/ f'[x0] (y y0) Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b}, Epilog Ä Line[{{x0, y0},{y0, x0}}], PlotRange Ä {{a,b},{a,b}}, AspectRatio Ä Automatic] 69-70. Example CAS commands: Maple: with( plots ); eq := cos(y) = x^(1/5); domain := 0 .. 1; x0 := 1/2; f := unapply( solve( eq, y ), x ); # (a) Df := D(f); plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#70(a) (Section 7.1)" ); q1 := solve( eq, x ); # (b) g := unapply( q1, y ); m1 := Df(x0); # (c) t1 := f(x0)+m1*(x-x0); y=t1; m2 := 1/Df(x0); # (d)
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396
Chapter 7 Transcendental Functions
t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f,domain); # (e) p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [p1,p2,p3,p4,p5], scaling=constrained, title="#70(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and x0 may vary) For problems 69 and 70, the code is just slightly altered. At times, different "parts" of solutions need to be used, as in the definitions of f[x] and g[y] Clear[x, y] {a,b} = {0, 1}; x0 = 1/2 ; eqn = Cos[y] == x1/5 soly = Solve[eqn, y] f[x_] = y /. soly[[2]] Plot[{f[x], f'[x]}, {x, a, b}] solx = Solve[eqn, x] g[y_] = x /. solx[[1]] y0 = f[x0] ftan[x_] = y0 f'[x0] (x x0) gtan[y_] = x0 1/ f'[x0] (y y0) Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b}, Epilog Ä Line[{{x0, y0},{y0, x0}}], PlotRange Ä {{a, b}, {a, b}}, AspectRatio Ä Automatic] 7.2 NATURAL LOGARITHMS 1. (a) ln 0.75 œ ln (b) ln
4 9 " #
3 4
œ ln 3 ln 4 œ ln 3 ln 2# œ ln 3 2 ln 2
œ ln 4 ln 9 œ ln 2# ln 3# œ 2 ln 2 2 ln 3
3 œ ln 1 ln 2 œ ln 2 (d) ln È 9 œ "3 ln 9 œ (e) ln 3È2 œ ln 3 ln 2"Î# œ ln 3 "# ln 2 (f) ln È13.5 œ " ln 13.5 œ " ln 27 œ " aln 3$ ln 2b œ " (3 ln 3 ln 2)
(c) ln
#
2. (a) ln
" 125
#
(e) ln 0.056 œ ln
7 125
3 #
ln 3# œ
2 3
ln 3
#
(b) ln 9.8 œ ln
49 5
œ ln 7# ln 5 œ 2 ln 7 ln 5
(d) ln 1225 œ ln 35# œ 2 ln 35 œ 2 ln 5 2 ln 7
ln 7
œ ln 7 ln 5$ œ ln 7 3 ln 5
3. (a) ln sin ) ln ˆ sin5 ) ‰ œ ln " #
#
œ ln 1 3 ln 5 œ 3 ln 5
(c) ln 7È7 œ ln 7$Î# œ
(c)
#
" 3
sin ) Š sin5 ) ‹
œ ln 5
(f)
ln 35 ln ln 25
" 7
œ
ln 5 ln 7 ln 7 # ln 5
œ
" #
#
" ‰ (b) ln a3x# 9xb ln ˆ 3x œ ln Š 3x 3x 9x ‹ œ ln (x 3) #
ln a4t% b ln 2 œ ln È4t% ln 2 œ ln 2t# ln 2 œ ln Š 2t# ‹ œ ln at# b
4. (a) ln sec ) ln cos ) œ ln [(sec ))(cos ))] œ ln 1 œ 0 (b) ln (8x 4) ln 2# œ ln (8x 4) ln 4 œ ln ˆ 8x 4 4 ‰ œ ln (2x 1) $ "Î$ ") (c) 3 ln Èt# 1 ln (t 1) œ 3 ln at# 1b ln (t 1) œ 3 ˆ "3 ‰ ln at# 1b ln (t 1) œ ln Š (t (t1)(t ‹ 1)
œ ln (t 1)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.2 Natural Logarithms 1 ‰ 5. y œ ln 3x Ê yw œ ˆ 3x (3) œ
7. y œ ln at# b Ê
œ ˆ t"# ‰ (2t) œ
dy dt
œ ln 3x" Ê
9. y œ ln
3 x
10. y œ ln
10 x
dy dx
œ ln 10x" Ê
11. y œ ln () 1) Ê 13. y œ ln x$ Ê
dy dt
17. y œ
x% 4
ln x
dy dx
œ ˆ ) " 1 ‰ (1) œ
dy d)
4
18. y œ ax2 ln xb Ê Ê
" )1
dy dt
14. y œ (ln x)$ Ê d dt
(ln t) œ (ln t)#
œ (ln t)"Î# "# t(ln t)"Î# †
x% 4
4x$ 16
œ
t ˆ "t ‰ (ln t)(1) t#
dy dt
œ
ln x 1 ln x
Ê yw œ
(1 ln x) ˆ "x ‰ (ln x) ˆ x" ‰ (1 ln x)#
22. y œ
x ln x 1 ln x
Ê yw œ
(1ln x) ˆln x x† "x ‰ (x ln x) ˆ x" ‰ (1ln x)#
œ
t ˆ "t ‰ (" ln t)(1) t#
23. y œ ln (ln x) Ê yw œ ˆ ln"x ‰ ˆ "x ‰ œ 24. y œ ln (ln (ln x)) Ê yw œ
" ln (ln x)
25. y œ )[sin (ln )) cos (ln ))] Ê
œ x$ ln x
†
œ
œ
" x
œ lnt# t
lnxx lnxx (1 ln x)#
œ
œ
" x(1 ln x)#
(" ln x)# ln x (1 ln x)#
œ1
ln x (1 ln x)#
" x ln x
d dx
dy d)
" 1 ln t t#
(ln (ln x)) œ
" ln (ln x)
†
" ln x
†
d dx
(ln x) œ
" x (ln x) ln (ln x)
œ [sin (ln )) cos (ln ))] ) cos (ln )) †
œ sin (ln )) cos (ln )) cos (ln )) sin (ln )) œ 2 cos (ln )) 26. y œ ln (sec ) tan )) Ê
dy d)
" xÈ x 1
" #
œ ln x œ
" #
3(ln x)# x
t(ln t)c"Î# #t
(ln t) œ (ln t)"Î#
1 ln t t#
21. y œ
1x 1x
(ln x) œ
2x ln x‰ œ 4x6 aln xb3 ax 2x ln xb œ 4x7 aln xb3 8x7 aln xb4
Ê
ln
d dx
1 x
" ln t t
" #
œ 3(ln x)# †
" )1
3 œ 4ax2 ln xb ˆx2 †
20. y œ
28. y œ
œ ˆ #) " 2 ‰ (2) œ
œ (ln t)# 2 ln t
2t ln t t
d dt
dy dx
dy d)
ln t t
27. y œ ln
3 2t
" x
œ x$ ln x
19. y œ
dy dt
" ‰ ˆ 3 "Î# ‰ œ ˆ t$Î# œ # t
†
dy dx
dy dx
12. y œ ln (2) 2) Ê
3 x
œ (ln t)# 2t(ln t) †
Ê
dy dt
œ ˆ 10x" " ‰ a10x# b œ x"
" #(ln t)"Î# x% 16
8. y œ ln ˆt$Î# ‰ Ê
2 t
" x
œ ˆ 3x" " ‰ a3x# b œ x"
16. y œ tÈln t œ t(ln t)"Î# Ê œ (ln t)"Î#
" ‰ 6. y œ ln kx Ê yw œ ˆ kx (k) œ
œ ˆ x"$ ‰ a3x# b œ
dy dx
15. y œ t(ln t)# Ê
" x
œ
sec ) tan ) sec# ) sec ) tan )
œ
sec )(tan ) sec )) tan ) sec )
" )
sin (ln )) † ") ‘
œ sec )
1) x 3x 2 ln (x 1) Ê yw œ x" #" ˆ x " 1 ‰ œ 2(x 2x(x 1) œ 2x(x 1)
cln (1 x) ln (1 x)d Ê yw œ
" #
1 " x ˆ 1 " x ‰ (1)‘ œ
" #
1x1x ’ (1 x)(" x) “ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" 1 x#
397
398
Chapter 7 Transcendental Functions
29. y œ
1 ln t 1 ln t
Ê
dy dt
(1ln t) ˆ "t ‰ (1 ln t) ˆ t" ‰
œ
(1 ln t)#
30. y œ Éln Èt œ ˆln t"Î# ‰ œ
" #
ˆln t"Î# ‰"Î# †
" t"Î#
Èsin ) cos ) 1 2 ln )
œ
" #
dy d)
" #
œ
œ
œ
dy dt
" sec (ln ))
" #
t
t
ˆln t"Î# ‰"Î# †
œ
t
(1 ln t)#
2 t(1 ln t)#
ˆln t"Î# ‰ œ
" #
ˆln t"Î# ‰"Î# †
sec (ln )) tan (ln )) sec (ln ))
†
d d)
d dt
" t"Î#
†
d dt
ˆt"Î# ‰
†
d d)
(sec (ln ))) œ
œ
dy d)
" #
(ln )) œ
) ˆ cos sin )
tan (ln )) )
sin ) ‰ cos )
2 )
1 # ln )
4 )(1 2 ln )) “
&
&
t
" 4tÉln Èt
x 1b 33. y œ ln Š aÈ ‹ œ 5 ln ax# 1b 1x
34. y œ ln É (x(x2)1)#! œ
" #
" ln t " ln t
(ln sin ) ln cos )) ln (1 2 ln )) Ê
’cot ) tan ) #
Ê
† #" t"Î# œ
31. y œ ln (sec (ln ))) Ê 32. y œ ln
"Î#
œ
" #
ln (1 x) Ê yw œ
5†2x x# 1
[5 ln (x 1) 20 ln (x 2)] Ê yw œ
" #
#" ˆ 1 " x ‰ (1) œ
ˆ x 5 1
20 ‰ x#
œ
5 #
10x x# 1
" #(1 x)
4(x 1) ’ (x(x2)1)(x 2) “
2 œ 5# ’ (x 3x1)(x #) “
35. y œ 'x#Î2 ln Èt dt Ê x#
36. y œ 'Èx ln t dt Ê 3 x È
œ
3 x ln È 3 3È x2
dy dx
dy dx
œ Šln Èx# ‹ †
3 œ ˆln È x‰ †
d dx
d dx
#
ax# b Šln É x# ‹ †
3 ˆÈ x‰ ˆln Èx‰ †
d dx
d dx
#
Š x# ‹ œ 2x ln kxk x ln
kx k È2
3 ˆÈx‰ œ ˆln È x‰ ˆ 3" x#Î$ ‰ ˆln Èx‰ ˆ #" x"Î# ‰
ln Èx 2È x
37.
2 'cc32 x" dx œ cln kxkd # $ œ ln 2 ln 3 œ ln 3
38.
'c01 3x3# dx œ cln k3x 2kd !" œ ln 2 ln 5 œ ln 52
39.
' y 2y25 dy œ ln ky# 25k C
40.
' 4r8r5 dr œ ln k4r# 5k C
41.
42.
#
#
'01 2sincost t dt œ cln k2 cos tkd !1 œ ln 3 ln 1 œ ln 3; or let u œ 2 cos t Ê du œ sin t dt with t œ 0 1 3 Ê u œ 1 and t œ 1 Ê u œ 3 Ê '0 2sincost t dt œ '1 u" du œ cln kukd $" œ ln 3 ln 1 œ ln 3 '01Î3 14 4sincos) ) d) œ cln k1 4 cos )kd !1Î$ œ ln k1 2k œ ln 3 œ ln "3 ; or let u œ 1 4 cos ) Ê du œ 4 sin ) d) 1Î3 c1 " with ) œ 0 Ê u œ 3 and ) œ 13 Ê u œ 1 Ê '0 14 4sincos) ) d) œ 'c3 u" du œ cln kukd " $ œ ln 3 œ ln 3
43. Let u œ ln x Ê du œ
'1
2
2 ln x x
dx œ '0
ln 2
'2
dx x ln x
œ 'ln 2
ln 4
" u
dx; x œ 1 Ê u œ 0 and x œ 2 Ê u œ ln 2;
2u du œ cu# d 0 œ (ln 2)# ln 2
44. Let u œ ln x Ê du œ 4
" x
" x
dx; x œ 2 Ê u œ ln 2 and x œ 4 Ê u œ ln 4; #
4‰ ln 2 ˆ 2 ln 2 ‰ œ ln 2 du œ cln ud lnln 42 œ ln (ln 4) ln (ln 2) œ ln ˆ ln ln 2 œ ln Š ln 2 ‹ œ ln ln 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.2 Natural Logarithms 45. Let u œ ln x Ê du œ
'2
4
dx x(ln x)#
'2
dx; x œ 2 Ê u œ ln 2 and x œ 4 Ê u œ ln 4;
œ 'ln 2 u# du œ "u ‘ ln 2 œ ln"4 ln 4
ln 4
46. Let u œ ln x Ê du œ 16
" x
dx 2xÈln x
œ
'ln 2
ln 16
" #
" x
" ln #
œ ln"##
" ln 2
œ 2 ln" #
" ln #
" # ln 2
œ
œ
" ln 4
dx; x œ 2 Ê u œ ln 2 and x œ 16 Ê u œ ln 16;
u"Î# du œ u"Î# ‘ ln 2 œ Èln 16 Èln 2 œ È4 ln 2 Èln 2 œ 2Èln 2 Èln 2 œ Èln 2 ln 16
47. Let u œ 6 3 tan t Ê du œ 3 sec# t dt;
t ' 6 3sec ' duu œ ln kuk C œ ln k6 3 tan tk C 3 tan t dt œ #
48. Let u œ 2 sec y Ê du œ sec y tan y dy;
' sec# ysectanyy dy œ ' duu œ ln kuk C œ ln k2 sec yk C
49. Let u œ cos
x #
Ê du œ "# sin
sin '01Î2 tan x# dx œ '01Î2 cos
x #
dx Ê 2 du œ sin 1Î
50. Let u œ sin t Ê du œ cos t dt; t œ
'11ÎÎ42 cot t dt œ '11ÎÎ42 51. Let u œ sin
) 3
cos t sin t
Ê du œ
'1Î2 2 cot 3) d) œ '1Î2 1
1 4
du u
dx; x œ 0 Ê u œ 1 and x œ
È2
œ c2 ln kukd 11Î
Ê uœ
" È2
and t œ
1 #
" dt œ '1ÎÈ2 du u œ cln kukd "ÎÈ# œ ln 1
" 3
1 2 cos
sin
È2
dx œ 2 '1
x # x #
x #
)
3
cos )
3
) 3
d) Ê 6 du œ 2 cos
È3Î2
d) œ 6 '1Î2
du u
) 3
" È2
œ 2 ln
1 #
Ê uœ
œ 2 ln È2 œ ln 2
" È2
d) ; ) œ
È3Î2
œ ln È2 1 #
" #
Ê uœ È3 #
and ) œ 1 Ê u œ
53.
'
œ'
dx 2Èx 2x
dx ; 2 È x ˆ1 È x ‰
du u
È2
œ 2 cln kukd 11Î
let u œ 1 Èx Ê du œ
" #È x
dx;
È3 #
;
ln "# ‹ œ 6 ln È3 œ ln 27
52. Let u œ cos 3x Ê du œ 3 sin 3x dx Ê 2 du œ 6 sin 3x dx; x œ 0 Ê u œ 1 and x œ
È2
;
Ê u œ 1;
œ 6 cln kukd 1Î2 œ 6 Šln
'01Î12 6 tan 3x dx œ '01Î12 6cossin3x3x dx œ 2 '11Î
" È2
œ 2 ln
" È2
1 1#
Ê uœ
" È2
;
ln 1 œ 2 ln È2 œ ln 2
' 2Èx ˆdx1 Èx‰ œ '
du u
œ ln kuk C
œ ln ¸1 Èx¸ C œ ln ˆ1 Èx‰ C 54. Let u œ sec x tan x Ê du œ asec x tan x sec# xb dx œ (sec x)(tan x sec x) dx Ê sec x dx œ
'
sec x dx Èln (sec x tan x)
œ'
du uÈln u
œ ' (ln u)"Î# †
55. y œ Èx(x 1) œ (x(x 1))"Î# Ê ln y œ Ê yw œ ˆ "# ‰ Èx(x 1) ˆ x"
" ‰ x 1
œ " #
56. y œ Èax# 1b (x 1)# Ê ln y œ Ê yw œ Èax# +1b (x 1)# ˆ x# x 1 57. y œ É t t 1 œ ˆ t t 1 ‰ Ê
dy dt
œ
" #
"Î#
É t t 1 ˆ "t
Ê ln y œ " ‰ t1
œ
" #
" #
" #
" u
du œ 2(ln u)"Î# C œ 2Èln (sec x tan x) C
ln (x(x 1)) Ê 2 ln y œ ln (x) ln (x 1) Ê
Èx(x 1) (2x 1) 2x(x 1)
œ
w
y y
" #
œ
#
ˆ x#2x 1
#
œ Èax# 1b (x 1)# ’ axx#x1b (xx 1)1 “ œ
[ln t ln (t 1)] Ê
É t t 1 ’ t(t " 1) “ œ
2y y
w
2x " 2Èx(x 1)
cln ax# 1b 2 ln (x 1)d Ê
" ‰ x1
du u ;
" dy y dt
œ
" #
ˆ "t
2 ‰ x1
a2x# x 1b kx 1k Èx# 1 (x 1)
" ‰ t1
" 2Èt (t 1)$Î#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
" x
" x 1
399
400
Chapter 7 Transcendental Functions " #
58. y œ É t(t 1 1) œ [t(t 1)]"Î# Ê ln y œ Ê
dy dt
" œ "# É t(t 1 1) ’ t(t2t 1) “œ
dy d)
" #
ln () 3) ln (sin )) Ê
dy d)
#
) œ (tan )) È2) 1 Š sec tan )
" #) 1 ‹
" #
œ asec# )b È2) 1
62. y œ Ê
dy dt
œ t(t 1)(t+2) ˆ "t
" t(t 1)(t 2) dy dt
" dy y d)
œ
" #() 3)
cos ) sin )
œ
" t1
" ‰ t#
"t
" t1
œ
sec# ) tan )
ˆ #" ‰ ˆ #) 2 1 ‰
œ
" t
" t1
" t#
t(t 2) t(t 1) œ t(t 1)(t 2) ’ (t 1)(t t(t2)1)(t “ œ 3t# 6t 2 2) " dy y dt
Ê ln y œ ln 1 ln t ln (t 1) ln (t 2) Ê
" t(t 1)(t 2)
" dy y d)
ln (2) 1) Ê
tan ) È 2) 1
" dy y dt
61. y œ t(t 1)(t 2) Ê ln y œ ln t ln (t 1) ln (t 2) Ê Ê
" ‰ t1
œ È) 3 (sin )) ’ 2() " 3) cot )“
60. y œ (tan )) È2) 1 œ (tan ))(2) 1)"Î# Ê ln y œ ln (tan )) Ê
œ #" ˆ "t
2t 1 2 at# tb$Î#
59. y œ È) 3 (sin )) œ () 3)"Î# sin ) Ê ln y œ Ê
" dy y dt
[ln t ln (t 1)] Ê
" ‘ t#
œ
" t(t 1)(t #)
œ "t
" t1
" t#
t(t 2) t(t 1) ’ (t 1)(t t(t2)1)(t “ 2)
#
œ at$3t3t#6t2t2b# 63. y œ
)5 ) cos )
64. y œ
) sin ) Èsec ) Ê ln y dy ) sin ) ˆ " d) œ Èsec ) )
Ê 65. y œ Ê
Ê ln y œ ln () 5) ln ) ln (cos )) Ê œ ln ) ln (sin )) cot )
" #
"!
" #
"!
3 x(x 2) " ˆ Ê yw œ 3" É x# 1 x
" 3
" )5
" dy y d)
" )
œ ’ ")
1b
2 3
ln (x 1) Ê
[10 ln (x 1) 5 ln (2x 1)] Ê
(x 1) ˆ 5 Ê yw œ É (2x x1 1)&
3 x(x 2) 67. y œ É x# 1 Ê ln y œ
ln (sec )) Ê
œ
sin ) cos )
cos ) sin )
Ê
dy d)
5 ‰ ˆ " œ ˆ ) )cos ) )5
" )
tan )‰
(sec ))(tan )) 2 sec ) “
tan )‰
xÈ x# 1 Ê ln y œ ln x "# ln ax# (x 1)#Î$ È # yw œ x(x x1)#Î$1 ’ "x x# x 1 3(x 2 1) “
(x 1) 66. y œ É (2x 1)& Ê ln y œ
" #
" dy y d)
yw y
yw y
œ
œ
" x
5 x1
x x# 1
2 3(x 1)
5 2x 1
5 ‰ 2x 1
cln x ln (x 2) ln ax# 1bd Ê
" x#
x(x 1)(x 2) 3 68. y œ É ax# 1b (2x 3) Ê ln y œ
" 3
x(x 1)(x 2) ˆ " 3 Ê yw œ "3 É ax# 1b (2x 3) x
yw y
œ
" 3
ˆ x"
" x#
2x ‰ x# 1
2x ‰ x# 1
cln x ln (x 1) ln (x 2) ln ax# 1b ln (2x 3)d " x1
" x#
2x x# 1
2 ‰ 2x 3
sin x 1 w w 69. (a) f(x) œ ln (cos x) Ê f w (x) œ cos x œ tan x œ 0 Ê x œ 0; f (x) 0 for 4 Ÿ x 0 and f (x) 0 for 0 x Ÿ 13 Ê there is a relative maximum at x œ 0 with f(0) œ ln (cos 0) œ ln 1 œ 0; f ˆ 14 ‰ œ ln ˆcos ˆ 14 ‰‰
œ ln Š È"2 ‹ œ #" ln 2 and f ˆ 13 ‰ œ ln ˆcos ˆ 13 ‰‰ œ ln xœ
1 3
with
f ˆ 13 ‰
" #
œ ln 2. Therefore, the absolute minimum occurs at
œ ln 2 and the absolute maximum occurs at x œ 0 with f(0) œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.2 Natural Logarithms sin (ln x) x
(b) f(x) œ cos (ln x) Ê f w (x) œ
401
" #
œ 0 Ê x œ 1; f w (x) 0 for
Ÿ x 1 and f w (x) 0 for 1 x Ÿ 2 Ê there is a relative maximum at x œ 1 with f(1) œ cos (ln 1) œ cos 0 œ 1; f ˆ "# ‰ œ cos ˆln ˆ "# ‰‰
œ cos ( ln 2) œ cos (ln 2) and f(2) œ cos (ln 2). Therefore, the absolute minimum occurs at x œ x œ 2 with f ˆ "# ‰ œ f(2) œ cos (ln 2), and the absolute maximum occurs at x œ 1 with f(1) œ 1.
" #
and
70. (a) f(x) œ x ln x Ê f w (x) œ 1 "x ; if x 1, then f w (x) 0 which means that f(x) is increasing (b) f(1) œ 1 ln 1 œ 1 Ê f(x) œ x ln x 0, if x 1 by part (a) Ê x ln x if x 1 71.
'15 (ln 2x ln x) dx œ '15 ( ln x ln 2 ln x) dx œ (ln 2)'15 dx œ (ln 2)(5 1) œ ln 2% œ ln 16
72. A œ
'c01Î4 tan x dx '01Î3 tan x dx œ ' 01Î4 cossinxx dx '01Î3 cossinxx dx œ cln kcos xkd !1Î% cln kcos xkd !1Î$ " È2 ‹
œ Šln 1 ln
ˆln
" #
ln 1‰ œ ln È2 ln 2 œ
73. V œ 1'0 Š Èy2 1 ‹ dy œ 41 '0 #
3
3
74. V œ 1 '1Î6 cot x dx œ 1 '1Î6 1Î2
1Î2
" y 1
cos x sin x
3 #
ln 2
dy œ 41 cln ky 1kd $! œ 41(ln 4 ln 1) œ 41 ln 4
1Î# dx œ 1 cln (sin x)d 1Î' œ 1 ˆln 1 ln "# ‰ œ 1 ln 2
75. V œ 21'1Î2 x ˆ x"# ‰ dx œ 21 '1Î2 x" dx œ 21 cln kxkd #"Î# œ 21 ˆln 2 ln #" ‰ œ 21(2 ln 2) œ 1 ln 2% œ 1 ln 16 2
2
76. V œ 1 '0 Š Èx9x ‹ dx œ 271'0 dx œ 271 cln ax$ 9bd ! œ 271(ln 36 ln 9) œ 271(ln 4 ln 9 ln 9) $9 #
3
3
$
œ 271 ln 4 œ 541 ln 2 77. (a) y œ
x# 8
œ '4
# ln x Ê 1 ayw b# œ 1 ˆ 4x x" ‰ œ 1 Š x 4x 4 ‹ œ Š x 4x 4 ‹ Ê L œ '4 É1 ayw b# dx #
8 # x 4
4x
#
#
#
8
dx œ '4 ˆ x4 "x ‰ dx œ ’ x8 ln kxk“ œ (8 ln 8) (2 ln 4) œ 6 ln 2 8
)
#
%
# (b) x œ ˆ y4 ‰ 2 ln ˆ y4 ‰ Ê
dx dy
œ
y 8
2 y
' Ê L œ '4 Ê1 Š dx dy ‹ dy œ 4 #
12
Ê 1
12 # y 16
8y
Š dx dy ‹
#
#
œ 1 Š y8 2y ‹ œ 1 Š y
#
y dy œ '4 Š y8 2y ‹ dy œ ’ 16 2 ln y“ 12
#
16 8y ‹ "# %
#
œ Šy
#
16 8y ‹
#
œ (9 2 ln 12) (1 2 ln 4)
œ 8 2 ln 3 œ 8 ln 9 78. L œ '1 É1 2
" x#
dx Ê
dy dx
œ
" x
Ê y œ ln kxk C œ ln x C since x 0 Ê 0 œ ln 1 C Ê C œ 0 Ê y œ ln x
" ‰ ˆ"‰ 79. (a) My œ '1 x ˆ "x ‰ dx œ 1, Mx œ '1 ˆ 2x x dx œ 2
Ê xœ
2
My M
œ
" ln 2
¸ 1.44 and y œ
Mx M
œ
ˆ "4 ‰ ln 2
" #
'12 x"
#
" ‘ dx œ 2x œ 4" , M œ '1 " #
2
" x
dx œ cln kxkd #" œ ln 2
¸ 0.36
(b)
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402
Chapter 7 Transcendental Functions
80. (a) My œ '1 x Š È"x ‹ dx œ '1 x"Î# dx œ 16
œ
16
' cln kxkd "' " œ ln 4, M œ 1
16
" #
" Èx
2 3
" x$Î# ‘ "' œ 42; Mx œ ' Š È ‹ Š È"x ‹ dx œ " 2 x 1 16
"'
dx œ 2x"Î# ‘ " œ 6 Ê x œ
My M
œ 7 and y œ
Mx M
" #
œ
'116 x" dx
ln 4 6
" (b) My œ '1 x Š È"x ‹ Š È4x ‹ dx œ 4'1 dx œ 60, Mx œ '1 Š 2È ‹ Š È"x ‹ Š È4x ‹ dx œ #'1 x$Î# dx x 16
16
16
œ 4 x"Î# ‘" œ 3, M œ '1 Š È"x ‹ Š È4x ‹ dx œ 4'1 "'
yœ
Mx M
œ
16
16
" x
16
dx œ c4 ln kxkd "' " œ 4 ln 16 Ê x œ
My M
œ
15 ln 16
and
3 4 ln 16
81. faxb œ lnax3 1b, domain of f: a1, _b Ê f w axb œ
3x2 x3 1 ;
f w axb œ 0 Ê 3x2 œ 0 Ê x œ 0, not in the domain;
f w axb œ undefined Ê x3 1 œ 0 Ê x œ 1, not in domain. On a1, _b, f w axb 0 Ê f is increasing on a1, _b Ê f is one-to-one 82. gaxb œ Èx2 ln x, domain of g: x 0.652919 Ê g w axb œ
2x 1x
2Èx2 ln x
œ
2x2 1 ; 2xÈx2 ln x
g w axb œ 0 Ê 2x2 1 œ 0 Ê no real
solutions; g w axb œ undefined Ê 2xÈx2 ln x œ 0 Ê x œ 0 or x ¸ 0.652919, neither in domain. On x 0.652919, g w axb 0 Ê g is increasing for x 0.652919 Ê g is one-to-one 83.
dy dx
84.
d# y dx#
œ1
" x
at ("ß 3) Ê y œ x ln kxk C; y œ 3 at x œ 1 Ê C œ 2 Ê y œ x ln kxk 2
œ sec# x Ê
dy dx
œ tan x C and 1 œ tan 0 C Ê
dy dx
œ tan x 1 Ê y œ ' (tan x 1) dx
œ ln ksec xk x C" and 0 œ ln ksec 0k 0 C" Ê C" œ 0 Ê y œ ln ksec xk x 85. (a) L(x) œ f(0) f w (0) † x, and f(x) œ ln (1 x) Ê f w (x)k xœ0 œ ww
(b) Let faxb œ lnax "b. Since f axb œ
" ax" b#
" ¸ 1x xœ0
œ 1 Ê L(x) œ ln 1 1 † x Ê L(x) œ x
! on Ò!ß !Þ"Ó, the graph of f is concave down on this interval and the
largest error in the linear approximation will occur when x œ !Þ". This error is !Þ" lna"Þ"b ¸ !Þ!!%'* to five decimal places. (c) The approximation y œ x for ln (1 x) is best for smaller positive values of x; in particular for 0 Ÿ x Ÿ 0.1 in the graph. As x increases, so does the error x ln (1 x). From the graph an upper bound for the error is 0.5 ln (1 0.5) ¸ 0.095; i.e., kE(x)k Ÿ 0.095 for 0 Ÿ x Ÿ 0.5. Note from the graph that 0.1 ln (1 0.1) ¸ 0.00469 estimates the error in replacing ln (1 x) by x over 0 Ÿ x Ÿ 0.1. This is consistent with the estimate given in part (b) above. 86. For all positive values of x,
d ln a x dx c
dœ
1 a x
† xa2 œ 1x and
d ln a dx c
ln x d œ 0
1 x
œ 1x . Since ln xa and ln a ln x have
the same derivative, then ln xa œ ln a ln x C for some constant C. Since this equation holds for all positve values of x, it must be true for x œ 1 Ê ln 1x œ ln 1 ln x C œ 0 ln x C Ê ln 1x œ ln x C. By part 3 we know that ln 1x œ ln x Ê C œ 0 Ê ln xa œ ln a ln x.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.3 Exponential Functions 403 (b) yw œ
87. (a)
cos x asin x .
Since lsin xl and lcos xl are less than
or equal to 1, we have for a " " " w a" Ÿ y Ÿ a" for all x.
Thus, lim yw œ ! for all x Ê the graph of y looks aÄ_
more and more horizontal as a Ä _.
88. (a) The graph of y œ Èx ln x appears to be concave upward for all x 0.
(b) y œ Èx ln x Ê yw œ
" #È x
" x
Ê yww œ 4x"$Î#
" x#
œ
" x#
Š
Èx 4
1‹ œ 0 Ê Èx œ 4 Ê x œ 16.
Thus, yww 0 if 0 x 16 and yww 0 if x 16 so a point of inflection exists at x œ 16. The graph of y œ Èx ln x closely resembles a straight line for x 10 and it is impossible to discuss the point of inflection visually from the graph. 7.3 EXPONENTIAL FUNCTIONS 1. (a) e0Þ3t œ 27 Ê ln e0Þ3t œ ln 3$ Ê (0.3t) ln e œ 3 ln 3 Ê 0.3t œ 3 ln 3 Ê t œ 10 ln 3 (b) ekt œ "# Ê ln ekt œ ln 2" œ kt ln e œ ln 2 Ê t œ lnk2 t (c) eÐln 0Þ2Ñt œ 0.4 Ê ˆeln 0Þ2 ‰ œ 0.4 Ê 0.2t œ 0.4 Ê ln 0.2t œ ln 0.4 Ê t ln 0.2 œ ln 0.4 Ê t œ
ln 0.4 ln 0.2
2. (a) e0Þ01t œ 1000 Ê ln e0Þ01t œ ln 1000 Ê (0.01t) ln e œ ln 1000 Ê 0.01t œ ln 1000 Ê t œ 100 ln 1000 " (b) ekt œ 10 Ê ln ekt œ ln 10" œ kt ln e œ ln 10 Ê kt œ ln 10 Ê t œ lnk10 (c) eÐln 2Ñt œ Èt
3. e
" #
t Ê ˆeln 2 ‰ œ 2" Ê 2t œ 2" Ê t œ 1 Èt
œ x# Ê ln e
#
œ ln x# Ê Èt œ 2 ln x Ê t œ 4(ln x)# #
#
4. ex e2x1 œ et Ê ex 2x1 œ et Ê ln ex 2x1 œ ln et Ê t œ x# 2x 1 5. y œ e5x Ê yw œ e5x
d dx
6. y œ e2xÎ3 Ê yw œ e2xÎ3 7. y œ e57x Ê yw œ e57x #
(5x) Ê yw œ 5e5x
d dx d dx
ˆ 2x ‰ Ê yw œ 3
2 3
e2xÎ3
(5 7x) Ê yw œ 7e57x #
8. y œ eˆ4Èxx ‰ Ê yw œ eˆ4Èxx ‰
d dx
ˆ4Èx x# ‰ Ê yw œ Š È2 2x‹ eˆ4Èxx# ‰ x
9. y œ xex ex Ê yw œ aex xex b ex œ xex
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404
Chapter 7 Transcendental Functions
10. y œ (1 2x) e2x Ê yw œ 2e2x (1 2x)e2x
d dx
(2x) Ê yw œ 2e2x 2(1 2x) e2x œ 4xe2x
11. y œ ax# 2x 2b ex Ê yw œ (2x 2)ex ax# 2x 2b ex œ x# ex 12. y œ a9x# 6x 2b e3x Ê yw œ (18x 6)e3x a9x# 6x 2b e3x
(3x) Ê yw œ (18x 6)e3x 3 a9x# 6x 2b e3x
d dx
œ 27x# e3x 13. y œ e) (sin ) cos )) Ê yw œ e) (sin ) cos )) e) (cos ) sin )) œ 2e) cos ) 14. y œ ln ˆ3)e) ‰ œ ln 3 ln ) ln e) œ ln 3 ln ) ) Ê #
15. y œ cos Še) ‹ Ê
dy d)
16. y œ )$ e#) cos 5) Ê
#
œ sin Še) ‹ dy d)
d d)
dy d)
#
" )
œ #
#
Še) ‹ œ Š sin Še) ‹‹ Še) ‹
œ a3)# b ˆe#) cos 5)‰ a)$ cos 5)b e#)
œ )# e#) (3 cos 5) 2) cos 5) 5) sin 5)) 17. y œ ln a3tet b œ ln 3 ln t ln et œ ln 3 ln t t Ê
dy dt
œ
" t
d d)
#
#
a)# b œ 2)e) sin Še) ‹
1t t
dy dt
œ 1 ˆ sin" t ‰
d dt
(sin t) œ 1
cos t sin t
cos t sin t sin t
19. y œ ln
e) 1 e)
20. y œ ln
È) 1 È)
œ ln e) ln ˆ1 e) ‰ œ ) ln ˆ1+e) ‰ Ê œ ln È) ln Š1 È)‹ Ê
" " œ Š È" ‹ Š È ‹ Š 1 "È) ‹ Š #È ‹œ ) # ) )
21. y œ eÐcos tln tÑ œ ecos t eln t œ tecos t Ê 22. y œ esin t aln t# 1b Ê 23.
d d)
(2)) 5(sin 5)) ˆ)$ e#) ‰
1œ
18. y œ ln a2et sin tb œ ln 2 ln et ln sin t œ ln 2 t ln sin t Ê œ
1
'0ln x sin et dt
dy dt
œ Š È" ‹ )
Š1 È)‹ È)
dy dt
2) Š1 È)‹
œ
œ 1 ˆ 1 " e) ‰
d d)
d dt
ˆ 1 e) ‰ œ 1
ŠÈ)‹ Š 1 "È) ‹
" #) Š1 È)‹
œ ecos t tecos t
d d)
œ
d d)
e) 1 e)
œ
d dx
24. y œ 'e4Èx ln t dt Ê yw œ aln e2x b †
Š1 È ) ‹
" #) a1)"Î# b
(cos t) œ (1 t sin t) ecos t
(ln x) œ
d dx
sin x x
ae2x b Šln e4Èx ‹ †
d dx
Še4Èx ‹ œ (2x) a2e2x b ˆ4Èx‰ Še4Èx ‹ †
œ 4xe2x 4Èx e4Èx Š È2x ‹ œ 4xe2x 8e4Èx 25. ln y œ ey sin x Ê Š y" ‹ yw œ ayw ey b (sin x) ey cos x Ê yw Š y" ey sin x‹ œ ey cos x Ê yw Š 1 yey sin x ‹ œ ey cos x Ê yw œ y
26. ln xy œ exy Ê ln x ln y œ exy Ê Ê yw Š 1 ye y
xby
‹œ
xex b y " x
" 1 e)
œ esin t (cos t) aln t# 1b 2t esin t œ esin t aln t# 1b (cos t) 2t ‘
Ê yw œ ˆsin eln x ‰ †
e2x
dy d)
dy d)
Ê yw œ
yey cos x 1 yey sin x
" x
Š y" ‹ yw œ a1 yw b exy Ê yw Š y" exy ‹ œ exy
y axex b y "b x a1 yex b y b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" x
d dx
ˆ4Èx‰
Section 7.3 Exponential Functions 405 27. e2x œ sin (x 3y) Ê 2e2x œ a1 3yw b cos (x 3y) Ê 1 3yw œ 28. tan y œ ex ln x Ê asec# yb yw œ ex 29.
' ae3x 5ex b dx œ e3
31.
" x
Ê yw œ
2e2x cos (x 3y)
Ê 3yw œ
2e2x cos (x 3y)
1 Ê yw œ
2e2x cos (x 3y) 3 cos (x 3y)
axex "b cos# y x
30.
' a2ex 3e2x b dx œ 2ex #3 e2x C
'lnln23 ex dx œ cex d lnln 32 œ eln 3 eln 2 œ 3 2 œ 1
32.
'0ln 2 ex dx œ cex d 0 ln 2 œ e! eln 2 œ 1 2 œ 1
33.
' 8eÐx1Ñ dx œ 8eÐx1Ñ C
34.
' 2eÐ2x1Ñ dx œ eÐ2x1Ñ C
35.
'lnln49 exÎ2 dx œ 2exÎ2 ‘ lnln 94 œ 2 eÐln 9ÑÎ2 eÐln 4)Î2 ‘ œ 2 ˆeln 3 eln 2 ‰ œ 2(3 2) œ 2
36.
'0ln 16 exÎ4 dx œ 4exÎ4 ‘ ln0 16 œ 4 ˆeÐln 16ÑÎ4 e0 ‰ œ 4 ˆeln 2 1‰ œ 4(2 1) œ 4
3x
5ex C
37. Let u œ r"Î# Ê du œ "# r"Î# dr Ê 2 du œ r"Î# dr; ' eÈÈrr dr œ ' er"Î# † r"Î# dr œ 2 ' eu du œ 2eu C œ 2er"Î# C œ 2eÈr C 38. Let u œ r"Î# Ê du œ "# r"Î# dr Ê 2 du œ r"Î# dr; ' eÈcÈrr dr œ ' er"Î# † r"Î# dr œ 2 ' eu du œ 2er"Î# C œ 2eÈr C 39. Let u œ t# Ê du œ 2t dt Ê du œ 2t dt;
' 2tet
#
dt œ ' eu du œ eu C œ et C #
" 4
40. Let u œ t% Ê du œ 4t$ dt Ê
'
%
t$ et dt œ
41. Let u œ
' ex#
1Îx
" x
" 4
du œ t$ dt;
' eu du œ 4" et% C
Ê du œ x"# dx Ê du œ
dx;
dx œ ' eu du œ eu C œ e1Îx C
42. Let u œ x# Ê du œ 2x$ dx Ê
' e x$Î # 1 x
" x#
dx œ ' ex † x$ dx œ #
" #
" #
du œ x$ dx;
' eu du œ "# eu C œ "# ex # C œ "# e1Îx# C
43. Let u œ tan ) Ê du œ sec# ) d); ) œ 0 Ê u œ 0, ) œ
'0
1 Î4
ˆ1 etan ) ‰ sec# ) d) œ '
1Î4
0
1 4
Ê u œ 1;
sec# ) d) '0 eu du œ ctan )d 0 1
1 Î4
ceu d "! œ tan ˆ 14 ‰ tan (0)‘ ae" e! b
œ (1 0) (e 1) œ e 44. Let u œ cot ) Ê du œ csc# ) d); ) œ
'1Î4 ˆ1 ecot ) ‰ csc# ) d) œ '1Î4 1 Î2
1Î2
1 4
Ê u œ 1, ) œ
1 2
Ê u œ 0;
csc# ) d) '1 eu du œ c cot )d 1Î4 ceu d !" œ cot ˆ 12 ‰ cot ˆ 14 ‰‘ ae! e" b 0
1 Î2
œ (0 1) (1 e) œ e
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
406
Chapter 7 Transcendental Functions
45. Let u œ sec 1t Ê du œ 1 sec 1t tan 1t dt Ê
' esec Ð1tÑ sec (1t) tan (1t) dt œ 1" ' eu du œ e1
u
œ sec 1t tan 1t dt;
du 1
Cœ
esec a1tb 1
C
46. Let u œ csc (1 t) Ê du œ csc (1 t) cot (1 t) dt;
' ecsc Ð1tÑ csc (1 t) cot (1 t) dt œ ' eu du œ eu C œ ecsc Ð1tÑ C
47. Let u œ ev Ê du œ ev dv Ê 2 du œ 2ev dv; v œ ln
1 6
Ê u œ 16 , v œ ln
1 #
Ê u œ 1# ;
'lnlnÐÐ11ÎÎ62ÑÑ 2ev cos ev dv œ 2 '11ÎÎ62 cos u du œ c2 sin ud 11ÎÎ26 œ 2 sin ˆ 1# ‰ sin ˆ 16 ‰‘ œ 2 ˆ1 "# ‰ œ 1 #
#
48. Let u œ ex Ê du œ 2xex dx; x œ 0 Ê u œ 1, x œ Èln 1 Ê u œ eln 1 œ 1; Èln 1
'0
2xex cos Šex ‹ dx œ '1 cos u du œ csin ud 1" œ sin (1) sin (1) œ sin (1) ¸ 0.84147 #
1
#
49. Let u œ 1 er Ê du œ er dr;
' 1 e e
r
dr œ '
x
dx œ '
r
50.
' 1 " e
x
let u œ e
" u
du œ ln kuk C œ ln a1 er b C
ecx ecx 1
dx;
1 Ê du œ ex dx Ê du œ ex dx;
' ecec 1 dx œ ' "u du œ ln kuk C œ ln aex 1b C x
x
51.
dy dt
œ et sin aet 2b Ê y œ ' et sin aet 2b dt;
dy dt
œ et sec# a1et b Ê y œ ' et sec# a1et b dt;
let u œ et 2 Ê du œ et dt Ê y œ ' sin u du œ cos u C œ cos aet 2b C; y(ln 2) œ 0 Ê cos ˆeln 2 2‰ C œ 0 Ê cos (2 2) C œ 0 Ê C œ cos 0 œ 1; thus, y œ 1 cos aet 2b 52.
let u œ 1et Ê du œ 1et dt Ê 1" du œ et dt Ê y œ 1" ' sec# u du œ 1" tan u C œ 1" tan a1et b C; y(ln 4) œ 12 Ê 1" tan ˆ1eln 4 ‰ C œ 12 Ê 1" tan ˆ1 † 4" ‰ C œ 12 Ê 1" (1) C œ
53.
d# y dx#
œ 2ex Ê
dy dx
2 1
Ê C œ 13 ; thus, y œ
œ 2ex C; x œ 0 and
3 1
dy dx
" 1
tan a1et b
œ 0 Ê 0 œ 2e! C Ê C œ 2; thus
dy dx x
Ê y œ 2ex 2x C" ; x œ 0 and y œ 1 Ê 1 œ 2e! C" Ê C" œ 1 Ê y œ 2e
54.
d# y dy " 2t " # 2t Ê dy dt# œ 1 e dt œ t # e C; t œ 1 and dt œ 0 Ê 0 œ 1 # e C Ê dy " 2t " # " # " 2t ˆ" # ‰ dt œ t # e # e 1 Ê y œ # t 4 e # e 1 t C" ; t œ 1 and y œ " " # " # " 2t " # Ê C" œ # 4 e Ê y œ # t 4 e ˆ # e 1‰ t ˆ #" 4" e# ‰
#
58. y œ 2s Ê
dy ds
dy ds
2x 1 œ 2 aex xb 1
" #
e# 1; thus
1 Ê " œ
œ 5Ès (ln 5) ˆ "# s"Î# ‰ œ Š 2lnÈ5s ‹ 5Ès #
" #
4" e# #" e# 1 C"
56. y œ 3cx Ê y w œ 3cx (ln 3)(1) œ 3cx ln 3
55. y œ 2x Ê y w œ 2x ln 2 57. y œ 5Ès Ê
Cœ
œ 2ex 2
#
œ 2s (ln 2)2s œ aln 2# b Šs2s ‹ œ (ln 4)s2s
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.3 Exponential Functions 407 59. y œ x1 Ê y w œ 1xÐ1 1Ñ 61. y œ (cos ))
È2 Ê
62. y œ (ln ))1 Ê
60. y œ t1
œ (1 e) t
dy dt
œ 1(ln ))Ð1 1Ñ ˆ ") ‰ œ
1(ln ))Ð1 1Ñ )
63. y œ 7sec ) ln 7 Ê
dy d)
œ a7sec ) ln 7b(ln 7)(sec ) tan )) œ 7sec ) (ln 7)# (sec ) tan ))
64. y œ 3tan ) ln 3 Ê
dy d)
œ a3tan ) ln 3b(ln 3) sec# ) œ 3tan ) (ln 3)# sec# )
65. y œ 2sin 3t Ê
dy dt
œ a2sin 3t ln 2b(cos 3t)(3) œ (3 cos 3t) a2sin 3t b (ln 2)
66. y œ 5c cos 2t Ê
dy dt
67. y œ log2 5) œ
ln 5) ln #
œ a5c cos 2t ln 5b(sin 2t)(2) œ (2 sin 2t) a5c cos 2t b (ln 5) Ê
ln x ln 4
70. y œ
x ln e ln #5
ln x# ln 4
œ
ln x 2 ln 5
œ
ln (1 ) ln 3) ln 3
2
ln x ln 4
œ ˆ ln"# ‰ ˆ 5") ‰ (5) œ
dy d)
68. y œ log3 (1 ) ln 3) œ 69. y œ
x # ln 5
ln x ln 4
Ê
œ3
ln x 2 ln 5
x ‰ 71. y œ x3 log10 x œ x3 ˆ lnln10 œ
1‰ 73. y œ log3 Šˆ xx ‹œ 1 ln 3
dy dx
œ
" x1
" x1
74. y œ log5 Ɉ 3x7x 2 ‰ œ
" #
ln 7x
" #
ln 5
œ
ln 3
ln 3
cos ) (sin ))(ln 7)
77. y œ log10 ex œ 78. y œ
) †5 ) 2 log5 )
œ
ln ex ln 10
ln# r (ln 3)(ln 9)
dy dr
Ê
1 (ln 3) ln Š xx b c1‹
ln 3
†
1 x
3x2 ln x‰ œ
1 2 ln 10 x
" ˆ"‰ œ ’ (ln 3)(ln 9) “ (2 ln r) r œ
Ðln 5ÑÎ2 œ log5 ˆ 3x7x 2 ‰ œ
dy dx
œ
7 2†7x dy d)
ln ˆ 3x7x 2 ‰
Ðln 5ÑÎ2
ln 5
3 2†(3x 2)
œ
œ ˆ ln#5 ‰ ”
(3x 2) 3x 2x(3x 2)
œ
ln ˆ 3x7x 2 ‰ ln 5
1 2 ln 10 x
3x2 log10 x
2 ln r r(ln 3)(ln 9)
x ln 10
Ê yw œ
" #
ln ˆ 3x7x # ‰
ln ) ‰ ˆ ln ) ‰‘ ˆ ) ln" 7 ‰ œ sin (log7 )) œ sin ˆ ln 7 ) cos ln 7
œ
Ê yw œ
•œ
" x(3x 2)
) †5 ) ) 2 ln ln 5
x 3x2 lnln10 œ
1‰ œ ln ˆ xx 1 œ ln (x 1) ln (x 1)
ln (sin )) ln (cos )) ln e) ln 2) )) ) ) ln 2 œ ln (sin )) ln (cos ln 7 ln 7 sin ) " ln 2 ˆ " ‰ (cos ))(ln 7) ln 7 ln 7 œ ln 7 (cot ) tan ) 1 ln
)
œ
1 ˆ 3 ln 10 x
x1 2x ln 5
2 (x 1)(x 1)
ln (3x 2) Ê
)‰ 76. y œ log7 ˆ sin e) #cos œ ) dy d)
" 1 ) ln 3
3 x ln 4
ln x Ê y w œ
œ
ln ) ‰ 75. y œ ) sin (log7 )) œ ) sin ˆ ln Ê 7
Ê
œ ˆ ln"3 ‰ ˆ 1 )" ln 3 ‰ (ln 3) œ
œ ˆ # ln" 5 ‰ (x ln x) Ê y w œ ˆ # ln" 5 ‰ ˆ1 "x ‰ œ
1 3 ln 10 x
1 ln ˆ xx b c1‰
dy d)
" ) ln #
Ê yw œ
ln x ln 4
72. y œ log3 r † log9 r œ ˆ lnln 3r ‰ ˆ lnln 9r ‰ œ
Ê
e
È œ È2 (cos ))Š 2c1‹ (sin ))
dy d)
dy d)
Ê
e
" ln 7
cos (log7 ))
2)
" ln 10
) ‰ˆ ) ) ˆ2 ln ‰ ˆ ) ‰ˆ ) ln1 5 ‰ ln 5 )†5 ln 5 5 a1b )†5 ) ‰2 ˆ2 ln ln 5
œ
5) ln 5a2 log5 )ba) ln 5 1b 5) ln 5a2 log5 )b2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
408
Chapter 7 Transcendental Functions
79. y œ 3log2 t œ 3Ðln tÑÎÐln 2Ñ Ê 80. y œ 3 log8 (log2 t) œ
83.
t ln Šˆeln 3 ‰
sin t
‹
ln 3
3 ln (log2 t) ln 8
ln 8 ln ˆtln 2 ‰ ln #
81. y œ log2 a8tln 2 b œ 82. y œ
œ c3Ðln tÑÎÐln 2Ñ (ln 3)d ˆ t ln" 2 ‰ œ
dy dt
t ln ˆ3sin t ‰ ln 3
œ
Ê
dy dt
3 ln 2 (ln 2)(ln t) ln #
œ
œ
3 ln ˆ lnln 2t ‰ ln 8
œ
t(sin t)(ln 3) ln 3
" t
alog2 3b 3log2 t
" ˆ " ‰ œ ˆ ln38 ‰ ’ (ln t)/(ln 2) “ t ln # œ
œ 3 ln t Ê
œ t sin t Ê
dy dt
œ
dy dt
3 t(ln t)(ln 8)
œ
" t(ln t)(ln #)
" t
œ sin t t cos t
' 5x dx œ ln5 5 C x
84. Let u œ 3 3x Ê du œ 3x ln 3 dx Ê ln13 du œ 3x dx;
' 3 3 3 dx œ ln13 ' u1 du œ ln13 lnlul C œ lnl3ln33 l C x
x
x
85.
'01 2c
)
'c2 0
86.
5c) d) œ
'
0
ˆ " ‰) 2 5
'1
d) œ –
88. Let u œ x"Î# Ê du œ "Î#
" #
" #
ln Š "# ‹
" ln Š "# ‹
!
ln Š 5" ‹ —
# x2ax b dx œ '1 ˆ "# ‰ 2u du œ
x
!
)
2
'14 È2Èx dx œ '14 2x
— œ
Š "5 ‹
87. Let u œ x# Ê du œ 2x dx Ê
È2
"
)
" #‹ ln Š "# ‹
1 Š ) d) œ '0 ˆ "# ‰ d) œ –
œ
" #
œ
ln Š "# ‹
" 2(ln 1 ln 2)
c#
" ln Š 5" ‹
œ #
Š 5" ‹
ln Š 5" ‹
œ
" ln Š 5" ‹
(1 25) œ
'0
24 ln 1 ln 5
" #
du œ x dx; x œ 1 Ê u œ 1, x œ È2 Ê u œ 2;
" #
ln2 # ‘ # œ ˆ 2 ln" 2 ‰ a2# 2" b œ "
x"Î# dx Ê 2 du œ
dx Èx
Ðu 1Ñ # "
7cos t sin t dt œ '1 7u du œ 0
7 ‘! ln 7 " u
œ ˆ ln"7 ‰ a7! 7b œ
90. Let u œ tan t Ê du œ sec# t dt; t œ 0 Ê u œ 0, t œ
'01Î4 ˆ 3" ‰tan t sec# t dt œ '01 ˆ 3" ‰u du œ – 91. Let u œ x2x Ê ln u œ 2x ln x Ê
" du u dx
"
u
Š "3 ‹ ln Š 3" ‹
24 ln 5
; x œ 1 Ê u œ 1, x œ 4 Ê u œ 2;
† x"Î# dx œ 2'1 2u du œ ’ 2ln # “ œ ˆ ln"# ‰ a2$ 2# b œ 2
œ
" ln #
u
89. Let u œ cos t Ê du œ sin t dt Ê du œ sin t dt; t œ 0 Ê u œ 1, t œ 1Î2
" # ln 2
œ
1 4
1 #
4 ln #
Ê u œ 0;
6 ln 7
Ê u œ 1; !
"
" " " — œ ˆ ln 3 ‰ ’ˆ 3 ‰ ˆ 3 ‰ “ œ
2 3 ln 3
!
œ 2 ln x (2x) ˆ x" ‰ Ê
du dx
œ 2u(ln x 1) Ê
x œ 2 Ê u œ 2% œ 16, x œ 4 Ê u œ 4) œ 65,536;
" #
du œ x2x (1 ln x) dx;
'24 x2x (1 ln x) dx œ "# '1665 536 du œ "# cud 6516 536 œ "# (65,536 16) œ 65,520 œ 32,760 # ß
2
2
92. Let u œ 1 2x Ê du œ 2x a2xbln 2 dx Ê
' 93.
'
2
x 2x dx 1 2x2
3x
œ
È3 dx œ
ß
1 2 ln 2 du
2
œ 2x x dx
' 1u du œ 2 ln1 2 lnlul C œ lnŠ12 ln 22
x2
1 2 ln 2
3xŠ 3b1‹ È 3 1 È
C
‹
C 94.
'
È xŠ 2c1‹ dx œ
È
x 2 È2
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.3 Exponential Functions 409 95.
'03 ŠÈ2 1‹ xÈ2 dx œ ’xŠÈ2
97.
'
98.
$
"‹
È2
“ œ 3Š
96.
!
x ‰ ˆ"‰ dx œ ' ˆ lnln10 x dx; u œ ln x Ê du œ
log10 x x
Ä
"‹
" x
'1e xÐln 2Ñ1 dx œ lnx # ‘ e1 œ e ln 2
1ln 2 ln 2
ln 2
œ
" ln #
' ˆ lnln10x ‰ ˆ x" ‰ dx œ ln"10 ' u du œ ˆ ln"10 ‰ ˆ #" u# ‰ C œ 2(lnlnx)10 C #
'14 logx x dx œ '14 ˆ lnln #x ‰ ˆ x" ‰ dx; u œ ln x Ê du œ x" dx; x œ 1 Ê u œ 0, x œ 4 Ê u œ ln 4‘ 4 x‰ ˆ"‰ ' ln 4 ˆ ln"# ‰ u du œ ˆ ln"# ‰ #" u# ‘ ln0 4 œ ˆ ln"# ‰ #" (ln 4)# ‘ œ (ln2 ln4)# œ (lnln4)4 œ ln 4 Ä '1 ˆ ln ln # x dx œ 0 2
#
4 x " " " ln x ‰ #‘ % # # # # # '14 ln 2 log ' 4 ln x " dx œ '1 ˆ lnx2 ‰ ˆ ln # dx œ 1 x dx œ # (ln x) " œ # c(ln 4) (ln 1) d œ # (ln 4) œ # (2 ln 2) œ 2(ln 2) x 2
100.
'1e 2 ln 10 x(log
101.
'02 logx (x # 2) dx œ ln"# '02 cln (x 2)d ˆ x " # ‰ dx œ ˆ ln"# ‰ ’ (ln (x # 2)) “ # œ ˆ ln"# ‰ ’ (ln#4)
10
x)
dx œ '1
e
(ln 10)(2 ln x) (ln 10)
ˆ x" ‰ dx œ c(ln x)# d e1 œ (ln e)# (ln 1)# œ 1 #
2
#
!
#
œ ˆ ln"# ‰ ’ 4(ln# 2) 102.
2 1 ln 2
dx‘
#
99.
œ
'110Î10 log
10
(10x) x
dx œ
(ln 2)# # “
"0 ln 10
œ
3 #
(ln 2)# # “
ln 2
"! '110Î10 cln (10x)d ˆ 10x" ‰ dx œ ˆ ln"010 ‰ ’ (ln (10x)) “ #0 #
"Î"!
#
œ ˆ ln"010 ‰ ’ (ln #100) 0
(ln 1)# # “
#
œ ˆ ln"010 ‰ ’ 4(ln#010) “ œ # ln 10 103.
'09 2 logx (x1 1) dx œ ln210 '09 ln (x 1) ˆ x " 1 ‰ dx œ ˆ ln210 ‰ ’ (ln (x# 1)) “ * œ ˆ ln210 ‰ ’ (ln 210)
104.
'23 2 logx (x1 1) dx œ ln22 '23 ln (x 1) ˆ x" 1 ‰ dx œ ˆ ln22 ‰ ’ (ln (x#1)) “ $ œ ˆ ln22 ‰ ’ (ln22)
105.
'
#
10
!
#
2
#
#
dx x log10 x
‰ ˆ x" ‰ dx œ (ln 10) ' ˆ ln"x ‰ ˆ x" ‰ dx; u œ ln x Ê du œ œ ' ˆ lnln10 x
Ä (ln 10) ' ˆ ln"x ‰ ˆ "x ‰ dx œ (ln 10) '
'
107.
'1ln x "t dt œ cln ktkd ln1 x œ ln kln xk ln 1 œ ln (ln x), x 1
108.
'1e "t dt œ cln ktkd e1
109.
'11/x "t dt œ cln ktkd 1"Îx œ ln ¸ x" ¸ ln 1 œ aln 1 ln kxkb ln 1 œ ln x, x 0
110.
" ln a
dx x ‰# x ˆ ln ln 8
œ (ln 8)#
'
(ln x) # x
(ln ")# # “
(ln ")# # “
œ ln 10
œ ln 2
dx‘
du œ (ln 10) ln kuk C œ (ln 10) ln kln xk C
106.
dx x (log8 x)#
œ'
" u
" x
#
(ln x) " 1
dx œ (ln 8)#
#
C œ (lnln 8)x C
x
x
œ ln ex ln 1 œ x ln e œ x
'1x "t dt œ ln"a ln ktk‘ x1 œ lnln xa lnln 1a œ loga x, x 0
111. y œ (x 1)x Ê ln y œ ln (x 1)x œ x ln (x 1) Ê
w
y y
œ ln (x 1) x †
" (x 1)
Ê yw œ (x 1)x x x 1 ln (x 1)‘
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
410
Chapter 7 Transcendental Functions
112. y œ x2 x2x Ê y x2 œ x2x Ê lnay x2 b œ ln x2x œ 2x ln x Ê w
w
1 w y x 2 ay
2xb œ 2x †
1 x 2x
2 † ln x œ 2 2ln x
Ê y 2x œ ay x ba2 2ln xb Ê y œ aax x b x ba2 2ln xb 2x œ 2ax x x2x ln xb 2
t
2
2x
2
113. y œ ˆÈt‰ œ ˆt"Î# ‰ œ ttÎ# Ê ln y œ ln ttÎ# œ ˆ #t ‰ ln t Ê Ê
dy dt
t
" dy y dt
œ ˆ #" ‰ (ln t) ˆ #t ‰ ˆ "t ‰ œ
" #
ln t #
t
œ ˆÈt‰ ˆ ln# t "# ‰
114. y œ tÈt œ tˆt
"Î# ‰
Ê ln y œ ln tˆt
œ ˆt"Î# ‰ (ln t) Ê
"Î# ‰
" dy y dt
115. y œ (sin x)x Ê ln y œ ln (sin x)x œ x ln (sin x) Ê 116. y œ xsin x Ê ln y œ ln xsin x œ (sin x)(ln x) Ê
w
y y
w
y y
œ ˆ #" t"Î# ‰ (ln t) t"Î# ˆ "t ‰ œ
ln t2 2È t
Ê
t2 œ Š ln2È ‹ tÈ t t
dy dt
x‰ œ ln (sin x) x ˆ cos Ê yw œ (sin x)x cln (sin x) x cot xd sin x
œ (cos x)(ln x) (sin x) ˆ x" ‰ œ
sin x x (ln x)(cos x) x
Ê yw œ xsin x ’ sin x x(lnx x)(cos x) “ 117. y œ sin xx Ê y w œ cos xx w
d x dx ax b; w
if u œ xx Ê ln u œ ln xx œ x ln x Ê
w
u u
œ x†
1 x
1 † ln x œ 1 ln x
Ê u œ x a1 ln xb Ê y œ cos x † x a1 ln xb œ x cos x a1 ln xb x
x
x
118. y œ (ln x)ln x Ê ln y œ (ln x) ln (ln x) Ê
x
w
y y
x
œ ˆ "x ‰ ln (ln x) (ln x) ˆ ln"x ‰
d dx
(ln x) œ
ln (ln x) x
" x
Ê yw œ Š ln (ln xx) " ‹ (ln x)ln x 119. f(x) œ ex 2x Ê f w (x) œ ex 2; f w (x) œ 0 Ê ex œ 2 Ê x œ ln 2; f(0) œ 1, the absolute maximum; f(ln 2) œ 2 2 ln 2 ¸ 0.613706, the absolute minimum; f(1) œ e 2 ¸ 0.71828, a relative or local maximum since f ww (x) œ ex is always positive. 120. The function f(x) œ 2esin ÐxÎ2Ñ has a maximum whenever sin
x #
œ 1 and a minimum whenever sin
œ 1. Therefore the
x #
maximums occur at x œ 1 2k(21) and the minimums occur at x œ 31 2k(21), where k is any integer. The maximum is 2e ¸ 5.43656 and the minimum is 2e ¸ 0.73576. 121. faxb œ x ex Ê f w axb œ x ex a1b ex œ ex x ex Ê f ww axb œ ex ax ex a1b ex b œ x ex 2ex (a) f w axb œ 0 Ê ex x ex œ ex a1 xb œ 0 Ê ex œ 0 or 1 x œ 0 Ê x œ 1, fa1b œ a1be1 œ 1e ; using second derivative test, f ww a1b œ a1be1 2e1 œ 1e 0 Ê absolute maximum at ˆ1, 1e ‰ (b) f ww axb œ 0 Ê x ex 2ex œ ex ax 2b œ 0 Ê ex œ 0 or x 2 œ 0 Ê x œ 2, fa2b œ a2be2 œ f ww a1b 0 and f ww a3b œ e3 a3 2b œ e13 0 Ê point of inflection at ˆ2, e22 ‰ 122. faxb œ œ
ex 1 e2x
Ê f w axb œ
ˆ1 e2x ‰ex ex ˆ2e2x ‰ a1 e2x b2
œ
ex e3x a1 e2x b2
Ê f ww axb œ
2 e2 ;
since
ˆ1 e2x ‰2 ˆex 3e3x ‰ ˆex e3x ‰2ˆ1 e2x ‰ˆ2e2x ‰ 2
2 ’a1 e2x b “
ex ˆ1 6e2x e4x ‰ a1 e2x b3
(a) f w axb œ 0 Ê ex e3x œ 0 Ê ex a1 e2x b œ 0 Ê e2x œ 1 Ê x œ 0; fa0b œ
e0 1 e2a0b
œ 12 ;
2
f w axb œ undefined Ê a1 e2x b œ 0 Ê e2x œ 1 Ê no real solutions. Using the second derivative test, f ww a0b œ
e0 ˆ1 6e2a0b e4a0b ‰ a1 e2a0b b
3
œ
4 8
0 Ê absolute maximum at ˆ0, 12 ‰
(b) f ww axb œ 0 Ê ex a1 6e2x e4x b Ê ex œ 0 or 1 6e2x e4x œ 0 Ê e2x œ Êxœ
lnŠ32È2‹ 2
or x œ
lnŠ32È2‹ 2
. f
lnŠ32È2‹ 2
œ
É 3 2È 2 4 2È 2
and f
a6b „ È36 4 2
lnŠ32È2‹ 2
œ
œ 3 „ 2È2,
É 3 2È 2 4 2È 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
Section 7.3 Exponential Functions 411 since f ww a1b 0, f ww a0b 0, and f ww a1b 0 Ê points of inflection at lnŠ32È2‹
2
123. f(x) œ x# ln
" x
É 3 2È 2
,
4 2È 2
" x
x# Š "" ‹ ax# b œ 2x ln x
Since x œ 0 is not in the domain of f, x œ e"Î# œ " e
2
É 3 2È 2
,
4 2È 2
and
.
Ê f w (x) œ 2x ln
Therefore, f Š È"e ‹ œ
lnŠ32È2‹
ln Èe œ
" e
" #e
ln e"Î# œ
" Èe
" x
x œ x(2 ln x 1); f w (x) œ 0 Ê x œ 0 or ln x œ "# .
. Also, f w (x) 0 for 0 x
ln e œ
" #e
" Èe
" Èe
and f w (x) 0 for x
is the absolute maximum value of f assumed at x œ
" Èe
.
.
124. f(x) œ (x 3)# ex Ê f w (x) œ 2(x 3) ex (x 3)# ex œ (x 3) ex (2 x 3) œ (x 1)(x 3) ex ; thus f w (x) 0 for x 1 or x 3, and f w (x) 0 for 1 x 3 Ê f(1) œ 4e ¸ 10.87 is a local maximum and f(3) œ 0 is a local minimum. Since f(x) 0 for all x, f(3) œ 0 is also an absolute minimum.
125.
'0ln 3 ae2x ex b dx œ ’ e#
126.
'02 ln 2 ˆexÎ2 exÎ2 ‰ dx œ 2exÎ2 2exÎ2 ‘ 20 ln 2 œ ˆ2eln 2 2e ln 2 ‰ a2e! 2e! b œ (4 1) (2 2) œ 5 4 œ 1
2x
127. L œ '0 É1
ex 4
128. S œ 21'0 ˆ e
ecy ‰ #
1
ln 2
œ 21'0
ln 2
œ œ
1 # 1 #
ˆe
y
y
dx Ê
ecy ‰ #
ex “
œ
dy dx
ln 3 0
exÎ2 #
!
œ Š e # eln 3 ‹ Š e# e! ‹ œ ˆ 9# 3‰ ˆ "# 1‰ œ 2 ln 3
8 #
2œ2
Ê y œ exÎ2 C; y(0) œ 0 Ê 0 œ e0 C Ê C œ 1 Ê y œ exÎ2 1
É1 ˆ ey #ecy ‰# dy œ 21 ' 0
ln 2
Ɉ ey #ecy ‰# dy œ 21 '0ln 2 ˆ e
y
ˆe
y
ecy ‰ #
ecy ‰# #
É1 "4 ae2y 2 e2y b dy
dy œ
1 #
'0ln 2 ae2y 2 e2y bdy
"# e2y 2y "# e2y ‘ ln 2 œ 1# ˆ "# e2 ln 2 2 ln 2 "# e2 ln 2 ‰ ˆ "# 0 "# ‰‘ 0 ˆ "# † 4 2 ln 2 "# † 4" ‰ œ 1# ˆ2 8" 2 ln 2‰ œ 1 ˆ 15 ‰ 16 ln 2
129. y œ "# aex ex b Ê œ '0 É e4 1
2x
" #
dy dx
ec2x 4
2 œ "# aex ex b; L œ '0 É1 ˆ "# aex ex b‰ dx œ '0 É1 1
ln 3
œ 'ln 2
4x
2e2x 1 4e2x ae2x 1b2
ln 3 x e ecx
ex ecx
" #
ec2x 4
dy dx
œ
ex ex 1
ex ex 1
œ
2ex e2x 1 ;
dy dx
œ
dx
x 2 L œ 'ln 2 É1 ˆ e2x2e 1 ‰ dx œ 'ln 2 É1
ln 3
ln 3
ln 3
4x
2x
ln 3
2x
ln 3 2x e 1
2
e2x 1
ln 3 e2x b 1
dx œ 'ln 2
dx; ’let u œ ex ex Ê du œ aex ex bdx, x œ ln 2 Ê u œ eln 2 eln 2 œ 2
131. y œ ln cos x Ê 1Î4
1
1b dx œ 'ln 2 É e ae2x2e 1b2 1 dx œ 'ln 2 Ê aaee2x dx œ 'ln 2 1 b2
Ê u œ eln 3 eln 3 œ 3
œ '0
e2x 4
2 dx œ '0 Ɉ "# aex ex b‰ dx œ '0 "# aex ex b dx œ "# cex ex d 10 œ "# ˆe 1e ‰ 0 œ 1
130. y œ lnaex 1b lnaex 1b Ê œ 'ln 2 É e
1
sin x cos x
1 3
1 2
ex e2x c 1 ex
4e2x ae2x 1b2
œ 32 , x œ ln 3
8 Î3
œ tan x; L œ '0
1Î4
1Î4
sec x dx œ cln lsec x tan xld 0
É1 atan xb2 dx œ ' 0
1Î4
È1 tan2 x dx œ '
œ ˆln lsecˆ 14 ‰ tan ˆ 14 ‰l‰ a0b œ lnŠÈ2 1‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1Î4
0
dx
dx
‰ œ 83 “ Ä '3Î2 1u du œ cln luld 3Î2 œ lnˆ 83 ‰ lnˆ 32 ‰ œ lnˆ 16 9 8Î3
e2 1 2e
Èsec2 x dx
412
Chapter 7 Transcendental Functions
132. y œ ln csc x Ê
dy dx
csc x cot x csc x
œ
œ cot x; L œ '1Î6 É1 acot xb2 dx œ '1Î6 È1 cot2 x dx œ '1Î6 Ècsc2 x dx 1Î4
1Î4
1Î4
œ '1Î6 csc x dx œ cln lcsc x cot xld 1Î6 œ ˆln lcscˆ 14 ‰ cot ˆ 14 ‰l‰ ˆln lcscˆ 16 ‰ cot ˆ 16 ‰l‰ 1 Î4
1 Î4
È
œ lnŠÈ2 1‹ lnŠ2 È3‹ œ lnŠ 2È2 13 ‹ 133. (a)
d dx
(x ln x x C) œ x †
(b) average value œ 134. average value œ
" 2 1
'1
e
" e1
" x
ln x 1 0 œ ln x
ln x dx œ
" e1
cx ln x xd e1 œ
" e1
[(e ln e e) (1 ln 1 1)] œ
" e 1
(e e 1) œ
'12 "x dx œ cln kxkd #" œ ln 2 ln 1 œ ln 2
135. (a) f(x) œ ex Ê f w (x) œ ex ; L(x) œ f(0) f w (0)(x 0) Ê L(x) œ 1 x (b) f(0) œ 1 and L(0) œ 1 Ê error œ 0; f(0.2) œ e0 2 ¸ 1.22140 and L(0.2) œ 1.2 Ê error ¸ 0.02140 (c) Since yww œ ex 0, the tangent line approximation always lies below the curve y œ ex . Thus L(x) œ x 1 never overestimates ex . Þ
136. (a) y œ ex Ê yww œ ex 0 for all x Ê the graph of y œ ex is always concave upward (b) area of the trapezoid ABCD 'ln a ex dx area of the trapezoid AEFD Ê ln b
'ln a ex dx Š e ln b
M œ eÐln a
ln bÑÎ2
ln a
eln b ‹ (ln #
b ln a). Now
" #
'ln a ex dx Š e
ln a
e #
ln b
(AB CD)(ln b ln a)
(AB CD) is the height of the midpoint
since the curve containing the points B and C is linear Ê eÐln a
ln b
" #
ln bÑÎ2
(ln b ln a)
‹ (ln b ln a)
'ln a ex dx œ cex d lnln ba œ eln b eln a œ b a, so part (b) implies that ln b
(c)
eÐln aln bÑÎ2 (ln b ln a) b a Š e Ê eln aÎ2 † eln bÎ2 137. A œ 'c2 1 2xx# dx œ 2'0 2
2
Ä A œ 2'1
5
138.
1 A œ '1 2Ð1
xÑ
" u
ba ln b ln a
2x 1 x#
ab #
ln a
eln b ‹ (ln #
b ln a) Ê eÐln aln bÑÎ2
Ê Èeln a Èeln b
ba ln b ln a
ab #
ba ln b ln a
Ê Èab
ab #
ba ln b ln a
dx; cu œ 1 x# Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 5d
du œ 2 cln kukd &" œ 2(ln 5 ln 1) œ 2 ln 5 "
x
" #‹ ln Š "# ‹
1 Š x dx œ 2 '1 ˆ "# ‰ dx œ 2 –
—
œ ln2# ˆ #" 2‰ œ ˆ ln2# ‰ ˆ 3# ‰ œ
3 ln #
"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ab #
" e1
Section 7.3 Exponential Functions 413 139. From zooming in on the graph at the right, we estimate the third root to be x ¸ 0.76666
140. The functions f(x) œ xln 2 and g(x) œ 2ln x appear to have identical graphs for x 0. This is no accident, because xln 2 œ eln 2 ln x œ aeln 2 bln x œ 2ln x . †
141. (a) f(x) œ 2x Ê f w (x) œ 2x ln 2; L(x) œ a2! ln 2b x 2! œ x ln 2 1 ¸ 0.69x 1 (b)
142. (a) f(x) œ log3 x Ê f w (x) œ
" x ln 3
, and f(3) œ
ln 3 ln 3
Ê L(x) œ
" 3 ln 3
(x 3)
ln 3 ln 3
œ
x 3 ln 3
" ln 3
1 ¸ 0.30x 0.09
(b)
143. (a) The point of tangency is apß ln pb and mtangent œ
" p
since
dy dx
œ x" . The tangent line passes through a!ß !b Ê the
equation of the tangent line is y œ "p x. The tangent line also passes throughapß ln pb Ê ln p œ "p p œ " Ê p œ e, and the tangent line equation is y œ "e x. (b)
d# y dx#
œ x"# for x Á ! Ê y œ ln x is concave downward over its domain. Therefore, y œ ln x lies below the graph of
y œ "e x for all x !, x Á e, and ln x
x e
for x !, x Á e.
(c) Multiplying by e, e ln x x or ln x x. e (d) Exponentiating both sides of ln xe x, we have eln x ex , or xe ex for all positive x Á e. (e) Let x œ 1 to see that 1e e1 . Therefore, e1 is bigger. e
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
414
Chapter 7 Transcendental Functions
144. Using Newton's Method: faxb œ lnaxb " Ê f w axb œ
" x
Ê xn" œ xn
lnaxn b" " x8
Ê xn" œ xn ’# lnaxn b“.
Then, x1 œ 2, x2 œ 2.61370564, x3 œ 2.71624393, and x& œ 2.71828183. Many other methods may be used. For example, graph y œ ln x " and determine the zero of y. 7.4 EXPONENTIAL CHANGE AND SEPARABLE DIFFERENTIAL EQUATIONS 1. (a) y œ ecx Ê y w œ ecx Ê 2y w 3y œ 2 aecx b 3ecx œ ecx (b) y œ ecx ec3xÎ2 Ê y w œ e x 3# e 3xÎ2 Ê 2y w 3y œ 2 ˆe x 3# e 3xÎ2 ‰ 3 ae x e 3xÎ2 b œ e x (c) y œ e x Ce 3xÎ2 Ê y w œ e x 3# Ce 3xÎ2 Ê 2y w 3y œ 2 ˆe x 3# Ce 3xÎ2 ‰ 3 ae x Ce 3xÎ2 b œ e 2. (a) y œ "x Ê y w œ
" x#
(b) y œ x " 3 Ê y w œ (c) y œ 3. y œ
" xC
'1x
" x
et t
Ê yw œ
x
#
œ ˆ x" ‰ œ y# " (x 3)#
#
œ ’ (x " 3) “ œ y# #
" (x C)#
œ x " C ‘ œ y#
dt Ê yw œ x"# '1
x t e
t
dt ˆ x" ‰ ˆ ex ‰ Ê x# y w œ '1
x t e
x
dt ex œ x Š x"
t
'1x et
t
dt‹ ex œ xy ex
Ê x# y w xy œ ex 4. y œ
" È 1 x%
'1x È1 t% dt $
Ê y w œ Š 12xx% ‹ Š È
" 1 x%
Ê y w œ #" –
4x$
$
ŠÈ1 x% ‹
'1x È1 t% dt‹ 1
—
'1x È1 t% dt È "
1 x%
$
ŠÈ 1 x% ‹
Ê y w œ Š 12xx% ‹ y 1 Ê y w
2x$ 1 x%
†yœ1
5. y œ ecx tan" a2ex b Ê y w œ ecx tan" a2ex b ecx ’ 1 a"2ex b# “ a2ex b œ ecx tan" a2ex b Ê y w œ y
2 1 4e2x
Ê yw y œ
2 1 4e2x
; y( ln 2) œ ecÐ
ln 2Ñ
tan" a2e
ln 2
2 1 4e2x
b œ 2 tan" 1 œ 2 ˆ 14 ‰ œ
1 #
# # # # # 6. y œ (x 2) ecx Ê y w œ ecx ˆ2xecx ‰ (x 2) Ê y w œ ecx 2xy; y(2) œ (2 2) ec2 œ 0
7. y œ cosx x Ê y w œ x sin xx# cos x Ê y w œ sinx x "x ˆ cosx x ‰ Ê y w œ sinx x 1/2) y ˆ 1# ‰ œ cos(1(/2) œ0 8. y œ
x ln x
9. 2Èxy Ê
2 3
Ê yw œ
dy dx œ $Î#
y
ln x x Š "x ‹ (ln x)#
Ê yw œ
" ln x
" (ln x)#
Ê x# y w œ
x# ln x
x# (ln x)#
y x
Ê xy w œ sin x y Ê xy w y œ sin x;
Ê x# y w œ xy y# ; y(e) œ
e ln e
œ e.
1 Ê 2x"Î# y"Î# dy œ dx Ê 2y"Î# dy œ x"Î# dx Ê ' 2y"Î# dy œ ' x"Î# dx Ê 2 ˆ 23 y$Î# ‰ œ 2x"Î# C"
x"Î# œ C, where C œ
" #
C"
10.
dy dx
œ x# Èy Ê dy œ x# y"Î# dx Ê y"Î# dy œ x# dx Ê ' y"Î# dy œ ' x# dx Ê 2y"Î# œ
11.
dy dx
œ excy Ê dy œ ex ecy dx Ê ey dy œ ex dx Ê
12.
dy dx
œ 3x# ey Ê dy œ 3x# ey dx Ê ey dy œ 3x# dx Ê ' ey dy œ ' 3x# dx Ê ey œ x3 C Ê ey x3 œ C
' ey dy œ ' ex dx
x$ 3
C Ê 2y"Î# "3 x$ œ C
Ê ey œ ex C Ê ey ex œ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.4 Exponential Change and Separable Differential Equations 13.
dy dx
œ Èy cos# Èy Ê dy œ ˆÈy cos# Èy‰ dx Ê
side, substitute u œ Èy Ê du œ
1 2Èy dy
sec# Èy Èy dy
Ê 2 du œ
1 Èy dy,
œ dx Ê '
415
œ ' dx. In the integral on the left-hand
sec# Èy Èy dy
and we have ' sec# u du œ ' dx Ê 2 tan u œ x C
Ê x 2 tan Èy œ C 14. È2xy dy dx œ 1 Ê dy œ Ê È2 15. Èx
y3/2 3 2
dy œ
" #
dy dx
œ
ey eÈ x Èx
Ê dy œ
hand side, substitute u œ Èx Ê du œ
ey eÈ x Èx dx
" #È x
Ê ecy œ 2eÈx C, where C œ C1
16. asec xb
dy dx
1 Èx
dx Ê È2 y1/2 dy œ x1/2 dx Ê È2 ' y1/2 dy œ ' x1/2 dx
3 C1 Ê È2 y3/2 œ 3Èx 32 C1 Ê È2 ˆÈy‰ 3Èx œ C, where C œ 32 C1
x1/2
œ eybÈx Ê
dy dx
dx Ê È2Èydy œ
1 È2xy
œ eybsin x Ê
dy dx
eÈ x Èx
Ê ecy dy œ
dx Ê 2 du œ
" Èx
dx Ê ' ecy dy œ '
dx, and we have
eÈ x Èx
dxÞ In the integral on the right-
' ecy dy œ 2 ' eu du Ê ecy œ 2eu C1
œ eybsin x cos x Ê dy œ aey esin x cos xbdx Ê ecy dy œ esin x cos x dx
Ê ' ecy dy œ ' esin x cos x dx Ê ecy œ esin x C1 Ê ecy esin x œ C, where C œ C1
17.
dy dx
œ 2xÈ1 y2 Ê dy œ 2xÈ1 y2 dx Ê
dy È 1 y2
œ 2x dx Ê '
dy È 1 y2
œ ' 2x dx Ê sin" y œ x# C since kyk "
Ê y œ sinax2 Cb 18.
dy dx
œ
e2x c y ex b y 2y
Ê dy œ
e2x c y ex b y dx
Ê dy œ
e2x ecy ex ey dx
œ
Ê e2y dy œ ex dx Ê ' e2y dy œ ' ex dx Ê
ex e2y dx
e2y #
œ ex C1
Ê e 2ex œ C where C œ 2C1 2 3 2 2 2 3 19. y2 dy dx œ 3x y 6x Ê y dy œ 3x ay 2bdx Ê
20.
dy dx
œ x y 3x 2y 6 œ ay 3bax 2b Ê
y2 y3 2
1 y 3 dy
dy œ 3x2 dx Ê '
œ ax 2bdx Ê '
y2 y3 2
dy œ ' 3x2 dx Ê 31 lnly3 2l œ x3 C
1 y 3 dy
œ ' ax 2bdx
Ê lnly 3l œ "# x2 2x C 21.
1 dy x dx
2 2 2 œ yex 2Èy ex œ ex ˆy 2Èy‰ Ê
Ê'
22.
dy dx
1 dy È y ˆÈ y 2 ‰
1 y 2Èy dy
ey 1 ey dy
2
2
2
1 ecy 1 dy
ln 0.99 1000
2
1 ecy 1 dy
œ ' aex 1bdx
¸ 0.00001 ln (0.9) 0.00001
6Þ05
¸ 2.389 millibars 900 ‰ (c) 900 œ 1013eÐ 0Þ121Ñh Ê 0.121h œ ln ˆ 1013 Ê hœ dy dt
œ aex 1bdx Ê '
¸ 10,536 years
œ kp Ê p œ p! ekh where p! œ 1013; 90 œ 1013e20k Ê k œ
(b) p œ 1013e
25.
2
2
(b) 0.9 œ eÐ 0Þ00001)t Ê (0.00001)t œ ln (0.9) Ê t œ (c) y œ y! eÐ20ß000Ñk ¸ y! e 0Þ2 œ y! (0.82) Ê 82% dp dh
œ ' x ex dx
œ ' aex 1bdx Ê lnl1 ey l œ ex x C Ê lna1 ey b œ ex x C
23. (a) y œ y! ekt Ê 0.99y! œ y! e1000k Ê k œ
24. (a)
1 y 2Èy dy
œ ' x ex dx Ê 2 lnlÈy 2l œ "# ex C Ê 4 lnlÈy 2l œ ex C Ê 4 lnˆÈy 2‰ œ ex C
œ ex y ex ey 1 œ aey 1baex 1b Ê
Ê'
œ x ex dx Ê '
ln (90) ln (1013) 20
ln (1013) ln (900) 0.121
¸ 0.121
¸ 0.977 km
œ 0.6y Ê y œ y! e 0Þ6t ; y! œ 100 Ê y œ 100e 0Þ6t Ê y œ 100e 0Þ6 ¸ 54.88 grams when t œ 1 hr
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
416
Chapter 7 Transcendental Functions
26. A œ A! ekt Ê 800 œ 1000e10k Ê k œ
ln (0.8) 10
Ê A œ 1000eÐln (0Þ8ÑÎ10Ñt , where A represents the amount of sugar that
remains after time t. Thus after another 14 hrs, A œ 1000eÐln Ð0Þ8ÑÎ10Ñ24 ¸ 585.35 kg
27. L(x) œ L! e
kx
Ê
L! #
œ L! e
Ê ln
18k
one-tenth of the surface value,
L! 10
" #
œ 18k Ê k œ
ln 2 18
¸ 0.0385 Ê L(x) œ L! e 0Þ0385x ; when the intensity is
œ L! ec0Þ0385x Ê ln 10 œ 0.0385x Ê x ¸ 59.8 ft
28. V(t) œ V! e tÎ40 Ê 0.1V! œ V! e tÎ40 when the voltage is 10% of its original value Ê t œ 40 ln (0.1) ¸ 92.1 sec 29. y œ y! ekt and y! œ 1 Ê y œ ekt Ê at y œ 2 and t œ 0.5 we have 2 œ e0Þ5k Ê ln 2 œ 0.5k Ê k œ Therefore, y œ eÐln 4Ñt Ê y œ e24 ln 4 œ 424 œ 2.81474978 ‚ 1014 at the end of 24 hrs
ln 2 0.5
œ ln 4.
30. y œ y! ekt and y(3) œ 10,000 Ê 10,000 œ y! e3k ; also y(5) œ 40,000 œ y! e5k . Therefore y! e5k œ 4y! e3k Ê e5k œ 4e3k Ê e2k œ 4 Ê k œ ln 2. Thus, y œ y! eÐln 2Ñt Ê 10,000 œ y! e3 ln 2 œ y! eln 8 Ê 10,000 œ 8y! Ê y! œ 10,000 œ 1250 8 31. (a) 10,000ekÐ1Ñ œ 7500 Ê ek œ 0.75 Ê k œ ln 0.75 and y œ 10,000eÐln 0Þ75Ñt . Now 1000 œ 10,000eÐln 0Þ75Ñt 0.1 Ê ln 0.1 œ (ln 0.75)t Ê t œ lnln0.75 ¸ 8.00 years (to the nearest hundredth of a year) (b) 1 œ 10,000eÐln 0Þ75Ñt Ê ln 0.0001 œ (ln 0.75)t Ê t œ
ln 0.0001 ln 0.75
¸ 32.02 years (to the nearest hundredth of a year)
32. (a) There are (60)(60)(24)(365) œ 31,536,000 seconds in a year. Thus, assuming exponential growth, P œ 257,313,431ekt and 257,313,432 œ 257,313,431eÐ14kÎ31ß536ß000Ñ Ê ln Š 257,313,432 257,313,431 ‹ œ
14k 31,536,000
Ê k ¸ 0.0087542
(b) P œ 257,313,431eÐ0.0087542Ña"&b ¸ 293,420,847 (to the nearest integer). Answers will vary considerably with the number of decimal places retained. 33. 0.9P! œ P! ek Ê k œ ln 0.9; when the well's output falls to one-fifth of its present value P œ 0.2P! 0.2 Ê 0.2P! œ P! eÐln 0Þ9Ñt Ê 0.2 œ eÐln 0Þ9Ñt Ê ln (0.2) œ (ln 0.9)t Ê t œ ln ln 0.9 ¸ 15.28 yr 34. (a)
dp dx
" œ 100 p Ê
dp p
" " œ 100 dx Ê ln p œ 100 x C Ê p œ eÐ 0Þ01xCÑ œ eC e 0Þ01x œ C" e 0Þ01x ;
p(100) œ 20.09 Ê 20.09 œ C" eÐ 0Þ01ÑÐ100Ñ Ê C" œ 20.09e ¸ 54.61 Ê p(x) œ 54.61e 0Þ01x (in dollars) (b) p(10) œ 54.61eÐ 0Þ01ÑÐ10Ñ œ $49.41, and p(90) œ 54.61eÐ 0Þ01ÑÐ90Ñ œ $22.20 (c) r(x) œ xp(x) Ê rw (x) œ p(x) xpw (x); pw (x) œ .5461e 0Þ01x Ê rw (x) œ (54.61 .5461x)e 0Þ01x . Thus, rw (x) œ 0 Ê 54.61 œ .5461x Ê x œ 100. Since rw 0 for any x 100 and rw 0 for x 100, then r(x) must be a maximum at x œ 100. 35. A œ A! ekt and A! œ 10 Ê A œ 10 ekt , 5 œ 10 ekÐ24360Ñ Ê k œ then 0.2Ð10Ñ œ 10 e0.000028254t Ê t œ 36. A œ A! ekt and Êtœ 37. y œ y! e
ln 0.05 0.00499 kt
" #
A! œ A! e139k Ê
" #
ln 0.2 0.000028454
œ e139k Ê k œ
ln (0.5) 24360
¸ 0.000028454 Ê A œ 10 e0.000028454t ,
¸ 56563 years ln (0.5) 139
¸ 0.00499; then 0.05A! œ A! ec0Þ00499t
¸ 600 days
œ y! e ÐkÑÐ3ÎkÑ œ y! e
3
œ
y! e$
y! 20
œ (0.05)(y! ) Ê after three mean lifetimes less than 5% remains
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.4 Exponential Change and Separable Differential Equations 38. (a) A œ A! eckt Ê (b)
" k
œ ec2Þ645k Ê k œ
" #
417
¸ 0.262
ln 2 #.645
¸ 3.816 years
ln 2 ‰ ln 2 ‰ (c) (0.05)A œ A exp ˆ 2.645 t Ê ln 20 œ ˆ 2.645 t Ê tœ
2.645 ln 20 ln #
¸ 11.431 years
39. T Ts œ (T! Ts ) e kt , T! œ 90°C, Ts œ 20°C, T œ 60°C Ê 60 20 œ 70e
10k
Ê
4 7
œe
10k
Êkœ
ln ˆ 74 ‰ 10
¸ 0.05596
(a) 35 20 œ 70ec0Þ05596t Ê t ¸ 27.5 min is the total time Ê it will take 27.5 10 œ 17.5 minutes longer to reach 35°C (b) T T œ (T T ) e kt , T œ 90°C, T œ 15°C Ê 35 15 œ 105e 0Þ05596t Ê t ¸ 13.26 min s
!
s
!
s
40. T 65° œ (T! 65°) e kt Ê 35° 65° œ (T! 65°) e 10k and 50° 65° œ (T! 65°) e 20k . Solving 30° œ (T! 65°) e 10k and 15° œ (T! 65°) e 20k simultaneously Ê (T! 65°) e 10k œ 2(T! 65°) e 20k ln 2 Ê e10k œ 2 Ê k œ ln102 and 30° œ T!e10k65° Ê 30° e10 ˆ 10 ‰ ‘ œ T! 65° Ê T! œ 65° 30° ˆeln 2 ‰ œ 65° 60° œ 5° 41. T Ts œ (T! Ts ) eckt Ê 39 Ts œ (46 Ts ) ec10k and 33 Ts œ (46 Ts ) ec20k Ê 33Ts 46Ts
œ ec20k œ aec10k b# Ê
33Ts 46Ts
39Ts 46Ts
œ ec10k and
#
Ts # # œ Š 39 46Ts ‹ Ê (33 Ts )(46 Ts ) œ (39 Ts ) Ê 1518 79Ts Ts
œ 1521 78Ts T#s Ê Ts œ 3 Ê Ts œ 3°C 42. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room temperature the silver will be 120 min from now, and t! the time the silver will be 10°C above room temperature. We then have the following time-temperature table: time in min. 0 20 (Now) 35 140 t! temperature Ts 70° Ts 60° Ts x Ts y Ts 10° " ‰ T Ts œ (T! Ts ) eckt Ê (60 Ts ) Ts œ c(70 Ts ) Ts d ec20k Ê 60 œ 70ec20k Ê k œ ˆ 20 ln ˆ 67 ‰ ¸ 0.00771 (a) T Ts œ (T! Ts ) ec0Þ00771t Ê (Ts x) Ts œ c(70 Ts ) Ts d e Ð0Þ00771ÑÐ35Ñ Ê x œ 70e 0Þ26985 ¸ 53.44°C (b) T Ts œ (T! Ts ) e 0Þ00771t Ê (Ts y) Ts œ c(70 Ts ) Ts d e Ð0Þ00771ÑÐ140Ñ Ê y œ 70e 1Þ0794 ¸ 23.79°C (c) T Ts œ (T! Ts ) e 0Þ00771t Ê (Ts 10) Ts œ c(70 Ts ) Ts d e Ð0Þ00771Ñ t! Ê 10 œ 70e 0Þ00771t! " ‰ ln ˆ "7 ‰ œ 252.39 Ê 252.39 20 ¸ 232 minutes from now the Ê ln ˆ "7 ‰ œ 0.00771t! Ê t! œ ˆ 0.00771 silver will be 10°C above room temperature 43. From Example 4, the half-life of carbon-14 is 5700 yr Ê Ê (0.445)c! œ c! e 0Þ0001216t Ê t œ
ln (0.445) 0.0001216
" #
c! œ c! eckÐ5700Ñ Ê k œ
ln 2 5700
¸ 0.0001216 Ê c œ c! e 0Þ0001216t
¸ 6659 years
44. From Exercise 43, k ¸ 0.0001216 for carbon-14. (a) c œ c! e 0Þ0001216t Ê (0.17)c! œ c! e 0Þ0001216t Ê t ¸ 14,571.44 years Ê 12,571 BC (b) (0.18)c! œ c! e 0Þ0001216t Ê t ¸ 14,101.41 years Ê 12,101 BC (c) (0.16)c! œ c! e 0Þ0001216t Ê t ¸ 15,069.98 years Ê 13,070 BC 45. From Exercise 43, k ¸ 0.0001216 for carbon-14 Ê y œ y0 e0.0001216t . When t œ 5000 Ê y œ y0 e0.0001216a5000b ¸ 0.5444y0 Ê yy0 ¸ 0.5444 Ê approximately 54.44% remains 46. From Exercise 43, k ¸ 0.0001216 for carbon-14. Thus, c œ c! e 0Þ0001216t Ê (0.995)c! œ c! e 0Þ0001216t Êtœ
ln (0.995) 0.0001216
¸ 41 years old
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418
Chapter 7 Transcendental Functions
^ 7.5 INDETERMINATE FORMS AND L'HOPITAL'S RULE ^ 1. l'Hopital: lim
x2
œ
" #x ¹xœ#
^ 2. l'Hopital: lim
sin 5x x
œ
5 cos 5x ¹ 1 xœ!
# x Ä 2 x 4
xÄ0
^ 3. l'Hopital: lim xÄ_
œ lim
3x#
œ
x2
lim # x Ä 2 x 4
9.
t$ 4t 15 # t Ä 3 t t 12
œ lim
xÄ!
#x # $ x x$ x 1
"
œ lim
7.
x Ä 2 #x
sin x 2x
3 11
œ lim
œ
3t# 4 t Ä 3 2t 1
œ 3 11
3(3)# 4 2(3) 1
5x3 2x 7x3 3
œ x lim Ä_
"5x2 2 21x2
œ x lim Ä_
30x 42x
12. x lim Ä_
x 8x# 12x# 5x
1 "6x 24x 5
œ x lim Ä_
16 24
15. lim
sin t# t
xÄ0
œ x lim Ä_
œ lim
16x
sin x x x$
œ lim
cos x " 3x#
2) 1
lim ) Ä 1Î2 cos (21 ))
18.
1 ) Ä 1Î3 sin ˆ) 3 ‰
19.
lim ) Ä 1Î2 1 cos 2)
œ
3) 1
1 sin )
œ
xÄ!
œ5†1œ5
5x# 3x 7x# 1
1cos x x#
œ x lim Ä_
5 7
3 x "
œ
x#
5 7
ax "bax# x "b
16
x Ä 0 cos x
œ lim
xÄ0 2
œ x lim Ä_
3
lim
1 ) Ä 1Î3 cos ˆ) 3 ‰
cos )
lim ) Ä 1Î2 2 sin 2)
œ
#x# 3x x$ x 1
œ ! or x lim Ä_ lim
x Ä 5
x# 25 x5
œ x lim Ä_
œ lim
2x
x Ä 5 1
# x
"
" x#
3 x#
30 42
œ
5 7
sin 5t 2t
16 1
tÄ0
5 cos 5t 2
œ
5 2
œ 16
œ lim
2 sin ˆ 3#1 ‰
œ lim
xÄ0
cos x 6
œ "6
œ 2
œ3 sin )
lim ) Ä 1Î2 4 cos 2)
œ
" (4)(1)
œ
" x$
œ 10
œ 23
œ
sin x 6x
œ
xb ˆ " cos x ‰ œ lim ” a" xcos 2 " cos x • xÄ!
" #
tÄ0
lim ) Ä 1Î2 sin (21 ))
œ
" 4
2 x Ä 1 ax "ba4x + 4x + 3b
14. lim
œ lim
x Ä 0 sin x
xÄ0
% 'x
œ0
8x#
17.
lim
acos t# b (2t) 1
tÄ0
x Ä 0 cos x 1
16. lim
œ
# t Ä 1 12t 1
œ lim
œ
œ 23 7
11. x lim Ä_
tÄ0
"
x Ä 2 x#
œ lim
or lim
8.
3t#
13. lim
x $ 1
" #
œ
" 4
œ lim
$ t Ä 1 4t t 3
sin 5x 5x
or x lim Ä_
$ x Ä 1 4x x3
œ x lim Ä_
t$ 1
10. lim
5 7
or lim
cos x 2
xÄ!
%x 3 $x # "
œ x lim Ä_
œ lim
lim
œ
œ
10 14
œ lim ”ˆ sinx x ‰ˆ sinx x ‰ˆ " "cos x ‰• œ xÄ!
2 x Ä ! x a" cos xb
^ 6. l'Hopital: lim xÄ_
œ x lim Ä_
# x Ä 1 12x 1
œ 5 lim 5x Ä 0
œ lim
3 11
1 cos x x#
sin# x
œ lim
x Ä 2 ax #bax #b
sin 5x x
xÄ0
x$ 1
ax # x " b
xÄ!
œ 5 or lim
x2
œ lim
# x Ä 2 x 4
"0x 3 14x
2 x Ä 1 a4x + 4x + 3b
x2
or lim
œ x lim Ä_
$ x Ä 1 4x x 3
^ 5. l'Hopital: lim
" 4
5x# 3x 7x# 1
^ 4. l'Hopital: lim œ lim
œ
" 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
! "
œ!
^ Section 7.5 Indeterminate Forms and L'Hopital's Rule x"
20. lim
x Ä 1 ln x sin (1x) x#
21. lim
ln (csc x) 1 # x Ä 1 Î 2 ˆx ˆ # ‰‰
tÄ0
t(1 cos t) t sin t
24. lim
t sin t
t Ä 0 1 cos t
" x
" 1 cos (1x)
œ
œ
lim
(1 cos t) t(sin t) 1 cos t
œ lim
tÄ0
sin t t cos t sin t
œ lim
tÄ0
œ lim
2x
lim
tÄ0
tÄ0
œ
26.
ˆ 1 x‰ tan x œ lim lim x Ä Ð1 Î 2 Ñ c # x Ä Ð1 Î 2 Ñ c
ˆ 1# x‰ cot x
œ
œ lim
ˆ "# ‰) 1 ) )Ä0
œ lim
28. lim 29. lim
x 2x
x x Ä 0 2 1
30. lim
3x "
x x Ä 0 2 1
)Ä0
ˆln ˆ "# ‰‰ ˆ "# ‰) 1 )Ä0
œ lim
3x ln 3
x x Ä 0 2 ln 2
32. x lim Ä_
log2 x log3 (x 3)
œ x lim Ä_
34.
ln (x1) x‰ ˆ ln ln #
œ x lim Ä_ " 1
3‰ œ ˆ ln ln # x lim Ä_
œ
ln ax# 2xb ln x
œ lim b xÄ!
lim b
ln aex "b ln x
œ lim b xÄ! œ lim
Èay a# a y yÄ0
œ lim
yÄ0
36. lim
œ lim
tÄ0
1 (1 0) 1
œ
" ‰ ˆ sin x œ
lim
" ‰ ˆ csc œ #x
x Ä Ð1 Î 2 Ñ c
œ
1# #
œ
" #
cos t cos t cos t t sin t cos t
œ
11"0 1
œ2
lim
x Ä Ð1 Î 2 Ñ c
csc# x #
" 1
œ 1
lim
x Ä Ð1 Î 2Ñ c
sin# x œ 1
œ ln 3
œ
1 †2 ! 0 (ln 2)†2!
œ
ˆx"1‰ ˆ "x ‰
3‰ œ ˆ ln ln # x lim Ä_
Š ln (xln 3 3) ‹
" ln #
ln 3 ln #
œ (ln 2) x lim Ä_
x‰ ˆ ln ln #
œ
œ (ln 2) x lim Ä_
ln x ln (x 3)
x x 1
œ (ln 2) x lim Ä_
ln 3 ‰ œ ˆ ln # x lim Ä_
ˆ "x ‰ ˆx " 3‰
1 1
œ ln 2
œ ˆ llnn 3# ‰ x lim Ä_
Š x2x #
2 ‹ 2x
ˆ x" ‰
œ lim b xÄ!
2x# 2x x# 2x
4x 2 2x 2
œ lim b xÄ!
œ lim b xÄ!
2 #
œ1
x
È5y 25 5 y
35. lim
lim
x Ä 1 Î2
ln 3 ln #
lim b
xÄ!
3! †ln 3 2! †ln 2
œ
a3! b (ln 3)(1) 1
œ2
œ ln ˆ "# ‰ œ ln 1 ln 2 œ ln 2
(1) a2x b (x)(ln 2) a2x b (ln 2) a2x b
xÄ0
ln (x1) log2 x
xÄ!
œ
œ lim
31. x lim Ä_
33.
3sin ) (ln 3)(cos )) 1
œ
cos t (cos t t sin t) cos t
œ lim
ˆx 1# ‰ cos x
3sin ) " )
cot x
2 1#
sin t (sin t t cos t) sin t
œ lim
ˆx 1# ‰ sec x œ lim lim x Ä Ð1 Î 2 Ñ c x Ä Ð1 Î 2 Ñ c
)Ä0
œ
1 x Ä 1 Î 2 2 ˆ x ˆ # ‰‰
25.
27. lim
2
# x Ä 0 sec x
x Ä 0 tan x
x cot x ‰ ˆ csccsc x ˆx ˆ 1# ‰‰ 2 x Ä 1 Î2
œ
" 11
œ lim
2x
sec x tan x x Ä 0 ˆ sec x ‰
lim
23. lim
xÄ1
œ lim
x Ä 0 ln (sec x)
22.
œ lim
yÄ0
Š exec 1 ‹ ˆ "x ‰
œ lim b xÄ!
(5y 25)"Î# 5 y
aay a# b y yÄ0
"Î#
a
xex ex 1
œ lim
yÄ0
œ lim b xÄ!
ex xex ex
ˆ "# ‰ (5y 25) "Î# (5) 1
ˆ "# ‰ aay a# b "Î# (a) 1 yÄ0
œ lim
‰ 37. x lim [ln 2x ln (x 1)] œ x lim ln ˆ x 2x 1 œ ln Šx lim Ä_ Ä_ Ä_
œ
10 1
œ1
œ lim
5
y Ä 0 2È5y 25
œ lim
a
y Ä 0 2Èay a#
2x x1‹
œ ln Šx lim Ä_
œ
" #
œ "# , a 0 2 1‹
œ ln 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
x3 x
œ3
419
420 38.
Chapter 7 Transcendental Functions lim
x Ä !b
(ln x ln sin x) œ lim b ln ˆ sinx x ‰ œ ln Š lim b xÄ! xÄ! 2
xb 39. lim b lnaln asin xb œ lim b x Ä0
40.
2aln xbˆ 1x ‰ cos x sin x
x Ä0
lim
x Ä !b
ˆ 3x x "
" ‰ sin x
basin xb aln xb œ lim b 2alnxxcos œ lim b ’ 2cos x x † x Ä0
x Ä0
x) x œ lim b Š (3x x1)(sin ‹ œ lim b sin x xÄ! xÄ!
3 cos x (3x 1)( sin x) œ lim b Š 3 cos xcos ‹œ x cos x x sin x xÄ!
41.
lim b ˆ x " 1
xÄ1
" ‰ ln x
lim
x Ä !b
" (0 1) 1
#
cos ) "
#
eh (" h) h# hÄ0
45.
et t#
lim t t Ä _ e 1
sin )
œ lim
44. lim
eh " h Ä 0 #h et 2t et
tÄ_
x sin x xÄ0 x tan x
œ lim
ae x 1 b 2 xÄ0 x sin x
œ lim
48. lim
) sin ) cos ) )Ä0 tan ) )
49. lim
50. lim
x Ä0
eh hÄ0 #
x# ex
œ lim
"
tÄ_
œ x lim Ä_
cos x sin x
" #
œ lim
2 ae x 1 be x x xÄ0 cos x sin x
œ lim
et
œ lim
t tÄ_ e
œ x lim Ä_
2x ex
(sin x)(cos x) cos x‰ œ lim b Š (1 cos x) sin ‹ x xÄ!
œ 1
œ
et 2 et
œ lim
1 cos x 2 xÄ0 x sec x tan x
2 ex
œ1
œ0
sin x 2 2 xÄ0 2x sec x tan x 2sec x
2e2x 2ex xÄ0 x cos x sin x
1 sin2 ) cos2 ) sec2 ) " )Ä0
œ lim
sin 3x 3x x2 sin x sin 2x
œ3
œ1
cos ) e)
)Ä0
œ lim
œ lim
010 1
œ lim
) ) Ä 0 e 1
46. x lim x# ex œ x lim Ä_ Ä_ 47. lim
6 #
œ _ † 1 œ _
œ #"
(csc x cot x cos x) œ lim b ˆ sin" x xÄ!
) ) Ä 0 e )1
œ
œ ln 1 œ 0
3 sin x (3x 1)(cos x) 1 sin x x cos x
"
x sin x œ lim b Š sin x cos ‹œ cos x xÄ!
43. lim
3 3 (1)(0) 110
sin x x “
" cos x ‹
(x 1) 1x x œ lim b Š ln(xx1)(ln x) ‹ œ lim b Š (ln x) (x 1) ˆ "x ‰ ‹ œ lim b Š (x ln x) x 1 ‹ xÄ1 xÄ1 xÄ1
" œ lim b Š (ln x 1) 1 ‹ œ xÄ1
42.
œ ln Š lim b xÄ!
x sin x ‹
0 2
œ0
4e2x 2ex xÄ0 x sin x 2cos x
œ lim
2sin2 ) 2 )Ä0 tan )
œ lim
3cos 3x 3 2x xÄ0 2sin x cos 2x cos x sin 2x
œ
œ lim
2 2
œ1
œ lim 2 cos2 ) œ 2 )Ä0
3cos 3x 3 2x xÄ0 sin x cos 2x sin 3x
œ lim
œ
9sin 3x 2 xÄ0 2sin x sin 2x cos x cos 2x 3cos 3x
œ lim
51. The limit leads to the indeterminate form 1_ . Let faxb œ x1ÎÐ1 xÑ Ê ln faxb œ ln ax1ÎÐ1 xÑ b œ lim
x Ä 1b
ln faxb œ lim b xÄ1
ln x 1x
œ lim b xÄ1
ˆ "x ‰ 1
lim
ln faxb œ lim b xÄ1
ln x x1
œ lim b xÄ1
ˆ "x ‰
lim ln faxb œ
xÄ_
œ x lim Ä_
œ
1 2
. Now
1
ln x x1.
" e
Now
œ 1. Therefore lim b x1ÎÐx 1Ñ œ lim b faxb œ lim b eln faxb œ e" œ e xÄ1 xÄ1 xÄ1
53. The limit leads to the indeterminate form _! . Let faxb œ (ln x)1Îx Ê ln faxb œ ln (ln x)1Îx œ x) lim ln (ln x xÄ_
2 4
œ 1. Therefore lim b x1ÎÐ1 xÑ œ lim b faxb œ lim b eln faxb œ e" œ xÄ1 xÄ1 xÄ1
52. The limit leads to the indeterminate form 1_ . Let faxb œ x1ÎÐx 1Ñ Ê ln faxb œ ln ax1ÎÐx 1Ñ b œ x Ä 1b
ln x 1x
œ
ˆ x ln" x ‰ 1
ln (ln x) x .
Now
œ 0. Therefore x lim (ln x)1Îx œ x lim faxb œ x lim eln faxb œ e! œ 1 Ä_ Ä_ Ä_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
^ Section 7.5 Indeterminate Forms and L'Hopital's Rule 54. The limit leads to the indeterminate form 1_ . Let faxb œ (ln x)1ÎÐx eÑ Ê ln faxb œ œ lim b xÄe
ˆ x ln" x ‰ 1
œ lim b xÄe
ln (ln x) x e
ln (ln x) x e
421
œ lim ln faxb xÄe
œ "e . Therefore (ln x)1ÎÐx eÑ œ lim b faxb œ lim b eln faxb œ e"Îe xÄe xÄe
x 55. The limit leads to the indeterminate form 0! . Let faxb œ xc1Îln x Ê ln faxb œ ln ln x œ 1. Therefore
lim
xÄ!
x1Îln x œ lim
eln faxb œ e" œ
faxb œ lim
xÄ!
xÄ!
" e
56. The limit leads to the indeterminate form _! . Let faxb œ x1Îln x Ê ln faxb œ
ln x ln x
œ x lim faxb œ x lim e1n faxb œ e" œ e Ä_ Ä_
œ 1. Therefore x lim x1Îln x Ä_
ln (1 2x) 2 ln x
57. The limit leads to the indeterminate form _! . Let faxb œ (1 2x)1ÎÐ2 ln xÑ Ê ln faxb œ ln (1 2x) Ê x lim ln faxb œ x lim œ x lim 2 ln x Ä_ Ä_ Ä_ ln faxb "Î# œ x lim faxb œ x lim e œe Ä_ Ä_
x 1 2x
œ x lim Ä_
" #
œ
" #
. Therefore x lim (1 2x)1ÎÐ2 ln xÑ Ä_
58. The limit leads to the indeterminate form 1_ . Let faxb œ aex xb1Îx Ê ln faxb œ ln aex xb x
Ê lim ln faxb œ lim xÄ0
xÄ0
ex 1
œ lim
x x Ä 0 e x
ln aex xb x
œ 2. Therefore lim aex xb1Îx œ lim faxb œ lim eln faxb œ e# xÄ0
xÄ0
xÄ0
59. The limit leads to the indeterminate form 0! . Let faxb œ xx Ê ln faxb œ x ln x Ê ln faxb œ œ lim b ln faxb œ lim b xÄ! xÄ!
ln x ˆ "x ‰
ˆ "x ‰
œ lim b xÄ!
Š x"# ‹
ln x ˆ "x ‰
œ lim b (x) œ 0. Therefore lim b xx œ lim b faxb xÄ! xÄ! xÄ!
œ lim b eln faxb œ e! œ 1 xÄ! x 60. The limit leads to the indeterminate form _! . Let faxb œ ˆ1 "x ‰ Ê ln faxb œ
œ
c# Š cx " ‹ lim b 1xx# xÄ!
" 1 x"
œ lim b xÄ!
œ lim b eln faxb œ e! œ 1 xÄ!
œ lim b xÄ!
x x1
ln a1xc" b x "
Ê
lim
x Ä !b
ln faxb
x œ 0. Therefore lim b ˆ1 x" ‰ œ lim b faxb xÄ! xÄ!
x
x
2‰ 2‰ 2‰ 61. The limit leads to the indeterminate form 1_ . Let faxb œ ˆ xx Ê ln faxb œ ln ˆ xx œ x ln ˆ xx 1 1 1 Ê lim ln faxb 2‰ œ lim x ln ˆ xx 1 œ lim Š xÄ_
xÄ_
2 ln ˆ xx b c1‰ 1 x
‹ œ lim Š
ln ax 2b ln ax 1b 1 x
xÄ_
‹œ
2
œ lim Š ax 23xbax 1b ‹ œ lim ˆ 2x6x 1 ‰ œ lim ˆ 62 ‰ œ 3. Therefore, xÄ_
xÄ_
xÄ_
62. The limit leads to the indeterminate form _! . Let faxb œ Š xx 21 ‹ 2
ln Š xx bb21 ‹ 2
Ê lim ln faxb œ lim 1x ln Š xx 21 ‹ œ lim 2
xÄ_
xÄ_
x2 4x 1 3 2 xÄ_ x 2x x 2
œ lim
xÄ_
2x 4 2 xÄ_ 3x 4 x 1
œ lim
x
œ lim
2 xÄ_ 6 x 4
1
63.
x lim ˆ x 2 ‰ xÄ_ x 1
lim faxb œ
xÄ_
2
ln ˆx2 1‰ ln ax 2b x xÄ_
3
œ
Ê ln faxb œ ln Š xx 21 ‹
œ lim
œ 0. Therefore,
œ xlim Œ Ä_
œ lim
2x x2 b 1
xÄ_ 1 Îx
lim Š xx 21 ‹ 2
xÄ_
1Îx
x b1 2 1
ax
xÄ_
c3
b 2bax c 1b x12
lim eln faxb œ e3
xÄ_
œ x1 ln Š xx 21 ‹ 2
x2 4x 1 2 xÄ_ ax 1bax 2b
œ lim
œ lim faxb œ lim eln faxb œ e0 œ 1 xÄ_
xÄ_
2
x lim x2 ln x œ limb Œ ln1 x œ limb Œ x2 œ limb Š 2x ‹ œ limb Š 3x2 ‹ œ 0 2 3 x Ä0 x Ä0 x Ä0 x Ä0
x Ä0 b
x
x
2
64.
1 Îx
1 lim Œ x b 2 1x c 1 xÄ_ x2 1
lim x aln xb2 œ limb Š aln1xb ‹ œ limb Œ
x Ä0 b
x Ä0
x
x Ä0
2aln xb 1x x12
2
2
œ limb Š 2ln1x ‹ œ limb Œ x1 œ limb Š 2xx ‹ œ limb a2xb œ 0 2 x Ä0 x Ä0 x Ä0 x Ä0 x x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
422 65.
Chapter 7 Transcendental Functions lim x tanˆ 12 x‰ œ limb Š cotˆ 1x x‰ ‹ œ limb Š csc2 ˆ11 x‰ ‹ œ
x Ä0 b
x Ä0
x Ä0
2
2
1 1
œ1
1
66.
ln x ‰ x ˆ sin xxtan x ‰ œ lim Š sin x sec lim sin x † ln x œ limb ˆ csc x œ limb Š csc x cot x ‹ œ limb b
x Ä0 b
x Ä0
67. x lim Ä_
È9x 1 Èx 1
68.
Èx Èsin x
lim b
xÄ!
70.
lim x Ä !b csc x
cot x
71. x lim Ä_
lim b
xÄ!
ex x ex
lim
x
ec1Îx
œ lim b xÄ!
e1Îx 1 x
9 1
x Ä0
2
x cos x tan x ‹ 1
œ
0 1
œ0
œ È9 œ 3
"
œ1
lim x Ä 1Î2c sin x
œ lim b cos x œ 1 xÄ!
ˆ 23 ‰x 1 x 1 ˆ 43 ‰
2 ex c x x
œ x lim Ä_
œ Éx lim Ä_
x‰ ˆ cos" x ‰ ˆ cos sin x œ
x‰ ˆ cos sin x ˆ sin" x ‰
œxÄ lim _
x Ä0
œ É 1" œ 1
sin x x
x Ä 1 Î2 c
œ x lim Ä_
2x 4x 5x 2x
2
73. x lim Ä_
" lim
xÄ!b
œ lim b xÄ!
2x 3x 3x 4x
72. x Ä lim _
74.
œ
sec x
lim x Ä 1Î2c tan x
9x 1 x1
œ Éx lim Ä_
œÊ
69.
x Ä0
œ0 x
1 ˆ 42 ‰ ˆ 52 ‰x 1
1 2x ˆ 52 ‰x 1
œxÄ lim _ exax c 1b x
œ x lim Ä_
x12
2x 2
x Ä 0 2x cos x
œ 1 œ_
œ lim b e1Îx œ _ xÄ!
75. Part (b) is correct because part (a) is neither in the 76. Part (b) is correct; the step lim
10 01
exax c 1b a2x 1b 1
œ x lim Ä_
e1Îx Š x12 ‹
œ lim b xÄ!
œ
0 0
nor
œ lim
2
_ _
x Ä 0 # sin x
^ form and so l'Hopital's rule may not be used.
in part (a) is false because lim
2x 2
x Ä 0 2x cos x
is not an
indeterminate quotient form. 77. Part (d) is correct, the other parts are indeterminate forms and cannot be calculated by the incorrect arithmetic 78. (a) We seek c in a#ß !b so that
f ac b g ac b w
fa!b fa#b ga!b ga#b
œ
w
œ
!# !%
œ #" . Since f w acb œ " and gw acb œ #c we have that
" #c
œ #"
Ê c œ ". (b) We seek c in aaß bb so that Êcœ
œ
fabb faab g ab b g a a b
f ac b g ac b
œ
fa$b fa!b $ ! ga$b ga!b œ * ! È œ " $ $( .
w
w
ba # .
(c) We seek c in a!ß $b so that c# % #c
f ac b g ac b
œ "$ Ê c œ
w
w
" „ È$( $
Êc
œ
ba b # a#
œ
" ba.
Since f w acb œ " and gw acb œ #c we have that
œ lim
xÄ0
27 sin 3x 30x
œ lim
xÄ0
81 cos 3x 30
œ
27 10
œ
œ "$ . Since f w acb œ c# % and gw acb œ #c we have that
79. If f(x) is to be continuous at x œ 0, then lim f(x) œ f(0) Ê c œ f(0) œ lim xÄ0
" #c
xÄ0
9x 3 sin 3x 5x$
œ lim
xÄ0
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
9 9 cos 3x 15x#
" ba
^ Section 7.5 Indeterminate Forms and L'Hopital's Rule 80.
limˆ tanx32x
a x2
x Ä0
sin bx ‰ x
œ limŠ tan 2x axx3 x
2
sin bx
x Ä0
‹ œ limŠ 2sec
2
x Ä0
2x a bx2 cos bx 2x sin bx ‹will 3x2
lima2sec2 2x a bx2 cos bx 2x sin bxb œ a 2 œ 0 Ê a œ 2; limŠ 2sec
x Ä0
x Ä0
œ limŠ 8sec œ
x Ä0 16 6b 6
2
2x tan 2x b2 x2 sin bx 4bx cos bx 2sin bx ‹ 6x
œ limŠ 32sec x Ä0
2
2
be in
0 0
form if
2x 2 bx2 cos bx 2x sin bx ‹ 3x2
2x tan2 2x 16sec4 2x b3 x2 cos bx 6b2 x sin bx 6b cos bx 6
‹
œ 0 Ê 16 6b œ 0 Ê b œ 38
81. (a)
(b) The limit leads to the indeterminate form _ _: È # lim Šx Èx# x‹ œ x lim Šx Èx# x‹Š x Èx# x ‹ œ x lim Š xÄ_ Ä_ Ä_ x
œ x lim Ä_
82.
" " É"
" x
œ
" " È" !
lim ŠÈx2 1 Èx‹ œ lim xŠ
_
_
xÄ
xÄ
x x
x # ax # x b ‹ x È x# x
œ x lim Ä_
x x È x# x
œ "#
È x2 1 x
Èx x ‹
œ lim xŠÉ x x2 1 È xx2 ‹ œ lim xŠÉ1 2
_
_
xÄ
xÄ
1 x2
É 1x ‹ œ _
83. The graph indicates a limit near 1. The limit leads to the 2x# (3x 1) Èx 2 x 1 xÄ1 # $Î# "Î# 4x 9# x"Î# lim 2x 3xx 1 x 2 œ lim 1 xÄ1 xÄ1 4 9# #" 45 œ 1 œ 1 1
indeterminate form 00 : lim œ œ
" #
x "Î#
x 84. (a) The limit leads to the indeterminate form 1_ . Let f(x) œ ˆ1 "x ‰ Ê ln f(x) œ x ln ˆ1 x" ‰ Ê x lim ln f(x) Ä_
œ x lim Ä_
ln ˆ1 "x ‰ ˆ "x ‰
œ x lim Ä_
ln a1 x " b x "
#
œ x lim Ä_
Š 1 xx " ‹ x #
œ x lim Ä_
" 1 ˆ "x ‰
œ
" 10
œ1
ˆ1 "x ‰x œ lim f(x) œ lim eln fÐxÑ œ e" œ e Ê x lim Ä_ xÄ_ xÄ_ ˆ1 "x ‰x (b) x 10 100 1000 10,000 100,000
2.5937424601 2.70481382942 2.71692393224 2.71814592683 2.71826823717
Both functions have limits as x approaches infinity. The function f has a maximum but no minimum while g has no extrema. The limit of f(x) leads to the indeterminate form 1_ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
423
424
Chapter 7 Transcendental Functions " ‰x x#
(c) Let f(x) œ ˆ1
Ê ln f(x) œ x ln a1 x# b c#
Š
2x
$ #‹
#
ln a1 x b 1x 2x Ê x lim ln f(x) œ x lim œ x lim œ x lim œ x lim x " Ä_ Ä_ Ä _ x # Ä _ ax $ x b Ä_ x " ln f Ð x Ñ ! ˆ1 x# ‰ œ lim f(x) œ lim e Therefore x lim e 1 œ œ Ä_ xÄ_ xÄ_ ln a1 rkc" b k "
k 85. Let f(k) œ ˆ1 kr ‰ Ê ln f(k) œ
œ lim
œ lim
rk
k Ä _ kr
r
kÄ_ 1
86. (a) y œ x1Îx Ê ln y œ
œ r. Therefore Ê
ln x x
w
y y
œ x lim Ä_
4 6x
œ 0.
#
Š 1 rkrk " ‹ ln a1 rk " b œ lim œ lim 1 rrk " " k k # kÄ_ kÄ_ kÄ_ r ‰k ln fÐkÑ ˆ lim 1 k œ lim f(k) œ lim e œ er . kÄ_ kÄ_ kÄ_
Ê
lim
ˆ x" ‰ (x) ln x x#
œ
4x a3x# 1b
Ê yw œ ˆ 1 x#ln x ‰ ax1Îx b . The sign pattern is
yw œ ± ± which indicates a maximum value of y œ e1Îe when x œ e e ! w ˆ " ‰ ax# b 2x ln x # # (b) y œ x1Îx Ê ln y œ lnx#x Ê yy œ x Ê yw œ ˆ 1 x2$ln x ‰ ax1Îx b . The sign pattern is x% yw œ ± ± which indicates a maximum of y œ e1Î2e when x œ Èe ! Èe (c) y œ x1Îx Ê ln y œ
ˆ "x ‰ axn b (ln x) ˆnxnc1 ‰ x2n
ln x xn
œ
Ê yw œ
xnc1 (1 n ln x) x2n
† x1Îx . The sign pattern is n yw œ ± ± which indicates a maximum of y œ e1Îne when x œ È e n ! È e n
n ˆeln x ‰1Îx œ lim eÐln xÑÎxn œ exp Š lim (d) x lim x1Îx œ x lim Ä_ Ä_ xÄ_ xÄ_ n
87. (a) y œ x tanˆ 1x ‰, lim ˆx tanˆ 1x ‰‰ œ lim Š xÄ
œ lim Š xÄ
_
tanˆ 1x ‰ 1 x
_
_
xÄ
‹ œ lim xÄ_
sec2 ˆ 1x ‰Š x12 ‹ Š x12 ‹
tanˆ 1x ‰ 1 x
‹ œ lim xÄ_
n
ln x xn ‹
sec2 ˆ 1x ‰Š x12 ‹ Š x12 ‹
ˆ " ‰‹ œ e! œ 1 œ exp Šx lim Ä _ nxn
sec2 ˆ 1x ‰ œ 1; lim ˆx tanˆ 1x ‰‰ œ xlim Ä_ xÄ_
lim sec2 ˆ 1x ‰ œ 1 Ê the horizontal asymptote is y œ 1 as x Ä _ and as œ xÄ _
x Ä _. (b) y œ
3x e2x 2x e3x ,
e lim Š 3x 2x e3x ‹ œ 2x
_
xÄ
2e2x lim Š 32 3e3x ‹ xÄ
œ
_
œ
3 2
h Ä0
2x
_
xÄ
2x
lim Š 4e 9e3x ‹ œ
_
xÄ
3x e lim ˆ 9e4x ‰ œ 0; lim Š 2x e3x ‹ 2x
_
xÄ
xÄ
Ê the horizontal asymptotes are y œ 0 as x Ä _ and y œ
lim fa0 hhb fa0b œ lim e
88. f w a0b œ
2e lim Š 32 3e3x ‹ œ
h Ä0
c1Îh2 0 h
œ lim e h Ä0
1Îh2
h
1
3 2
_
as x Ä _.
1
œ limŠ e1hÎh2 ‹ œ lim 1Îh2 h2 2 œ limŠ 2e1hÎh2 ‹ Š 3 ‹ h Ä0 h Ä0 e h Ä0 h
œ limŠ h2 e1Îh ‹ œ 0 2
h Ä0
89. (a) We should assign the value 1 to f(x) œ (sin x)x to make it continuous at x œ 0.
(b) ln f(x) œ x ln (sin x) œ œ lim
x
#
x Ä 0 tan x
œ lim
ln (sin x) ˆ "x ‰ 2x
# x Ä 0 sec x
Ê
lim ln f(x) œ lim b x Ä !b xÄ!
ln (sin x) ˆ "x ‰
œ lim b xÄ!
ˆ sin" x ‰ (cos x) Š x"# ‹
œ 0 Ê lim f(x) œ e! œ 1 xÄ0
(c) The maximum value of f(x) is close to 1 near the point x ¸ 1.55 (see the graph in part (a)).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.6 Inverse Trigonometric Functions (d) The root in question is near 1.57.
90. (a) When sin x 0 there are gaps in the sketch. The width of each gap is 1.
(b) Let f(x) œ (sin x)tan x Ê ln f(x) œ (tan x) ln (sin x) Ê œ
ˆ sin" x ‰ (cos x) csc# x
lim x Ä 1 Î2 c
Ê
ln f(x) œ
lim
x Ä 1 Î2 c
lim
œ
ln (sin x) cot x cos x
lim x Ä 1Î2c ( csc x)
œ0
f(x) œ e! œ 1. Similarly,
lim
x Ä 1 Î2 c
x Ä 1 Î2 b
lim
x Ä 1 Î2 c
f(x) œ e! œ 1. Therefore,
lim
x Ä 1 Î2
f(x) œ 1.
(c) From the graph in part (b) we have a minimum of about 0.665 at x ¸ 0.47 and the maximum is about 1.491 at x ¸ 2.66. 7.6 INVERSE TRIGONOMETRIC FUNCTIONS 1. (a)
1 4
(b) 13
3. (a) 16
(c)
1 6
(b)
1 4
(c) 13
2. (a) 14
(b)
1 3
(c) 16
4. (a)
1 6
(b) 14
(c)
1 3
(c)
1 6
(c)
21 3
5. (a)
1 3
(b)
31 4
(c)
1 6
6. (a)
1 4
(b) 13
7. (a)
31 4
(b)
1 6
(c)
21 3
8. (a)
31 4
(b)
9. sin Šcos"
È2 # ‹
œ sin ˆ 14 ‰ œ
" È2
11. tan ˆsin" ˆ "# ‰‰ œ tan ˆ 16 ‰ œ È"3
1 6
10. sec ˆcos" #" ‰ œ sec ˆ 13 ‰ œ 2 12. cot Šsin" Š
È3 # ‹‹
œ cot ˆ 13 ‰ œ È"3
sin" x œ
1 #
14.
15. x lim tan" x œ Ä_
1 #
16. x Ä lim tan" x œ 1# _
17. x lim sec" x œ Ä_
1 #
18. x Ä lim sec" x œ x Ä lim cos" ˆ "x ‰ œ _ _
13.
lim
x Ä 1c
19. x lim csc" x œ x lim sin" ˆ "x ‰ œ 0 Ä_ Ä_
lim
x Ä 1 b
cos" x œ 1
1 #
20. x Ä lim csc" x œ x Ä lim sin" ˆ "x ‰ œ 0 _ _
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
425
426
Chapter 7 Transcendental Functions
21. y œ cos" ax# b Ê
dy dx
œ
23. y œ sin" È2t Ê
dy dt
œ
È2 #
Ê1 ŠÈ2t‹
25. y œ sec" (2s 1) Ê
dy ds
26. y œ sec" 5s Ê
5 k5sk È(5s)# 1
dy ds
œ
27. y œ csc" ax# 1b Ê
28. y œ csc" ˆ x# ‰ Ê
dy dx
œ
œ
2x È 1 x%
22. y œ cos" ˆ "x ‰ œ sec" x Ê
œ
È2 È1 2t#
24. y œ sin" (1 t) Ê
œ
dy dt
#
30. y œ sin" ˆ t3# ‰ œ csc" Š t3 ‹ Ê
œ
" kx k É x
#
œ
34. y œ tan" (ln x) Ê
dy dx
œ
35. y œ csc" aet b Ê 36. y œ cos" aet b Ê
dy dt
œ
dy dt
Š
" 1
x#
‹
tanc" x
4
ˆ 2t ‰
œ
œ
ˆ "x ‰
1 (ln x)#
3
Š "# ‹ tc"Î# 1 at"Î# b
dy dt
#
œ
s# È 1 s#
œ
et
e t É 1 ae t b #
" È 1 s#
œ
œ
" k2s 1k Ès# s
2t % t# É t 9
9
œ
6 t Èt% 9
Š "# ‹ (t 1)c"Î# 1 c(t 1)"Î# d
#
œ
" 2Èt 1 (1 t 1)
œ
" 2tÈt 1
" x c1 (ln x)# d
œ
" Èe2t 1
œ
e t È1 e
œ È1 s#
38. y œ Ès# 1 sec" s œ as# 1b
" È2t t#
" #Èt(1 t)
2t
"Î# 37. y œ sÈ1 s# cos" s œ s a1 s# b cos" s Ê
œ È1 s#
œ
" atan " xb a1x# b
ke t k É a e t b # 1
œ
" È1 (1 t)#
œ
2 kx k È x # 4
3 œ # œ # # ¹ t ¹ ÊŠ t ‹ 1
32. y œ cot" Èt 1 œ cot" (t 1)"Î# Ê
dy dx
" kx k È x # 1
" È1 t#
dy dt
dy dt
œ
œ
2x ax# 1b Èx% 2x#
œ
4
3
33. y œ ln atan" xb Ê
œ
2x kx # 1 k É a x # 1 b # 1
¸ x# ¸ Ɉ x# ‰# 1
31. y œ cot" Èt œ cot" t"Î# Ê
2 k2s 1k È4s# 4s
dy dt
dy dx
" ksk È25s# 1
œ
Š "# ‹
29. y œ sec" ˆ "t ‰ œ cos" t Ê
œ
2 k2s 1k È(2s 1)# 1
œ
dy dx
œ
2x É 1 ax # b#
"Î#
s# 1 È 1 s#
sec" s Ê
dy ds
œ dy dx
œ a1 s# b
1 s# s# 1 È 1 s#
"Î#
œ
œ ˆ "# ‰ as# 1b
s ˆ "# ‰ a1 s# b
"Î#
(2s)
" È 1 s#
2s# È 1 s# "Î#
(2s)
" ks k È s # 1
œ
s È s# 1
s ks k 1 ks k È s # 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" k sk È s # 1
Section 7.6 Inverse Trigonometric Functions "Î# 39. y œ tan" Èx# 1 csc" x œ tan" ax# 1b csc" x Ê
œ
" x È x# 1
" kx k È x # 1
1 #
tan" ax" b tan" x Ê
41. y œ x sin" x È1 x# œ x sin" x a1 x# b x È 1 x#
x È 1 x#
'È"
44.
'È
" 1 4x#
46.
' 9 "3x
47.
'
48.
'
dx œ '
#
dx œ
" È2
" #
œ
49.
'01 È4 ds
50.
'03
4 s#
È2Î4
" # ŠÈ17‹ x#
'
" 3
œ'
dx xÈ25x# 2
dx xÈ5x# 4
du uÈ u# 2
œ'
du uÈ u# 4
dx œ
" kx k È x # 1
"Î#
Ê
dy dx
œ0
xc# 1 ax " b #
œ sin" x x Š È
dy dx
" 1 x#
" ‹ 1 x#
œ
" x# 1
" 1 x#
ˆ #" ‰ a1 x# b
œ0
"Î#
(2x)
2x 4
x#
tan" ˆ #x ‰ x –
Š "# ‹ 1 ˆ #x ‰
#
—œ
2x x# 4
tan" ˆ #x ‰
2x 4 x#
œ tan" ˆ #x ‰
" #
'03
'02
œ ’ È"2 †
" È8
œ
È2
1 u #
dx œ
" È17
tan"
" 3È 3
" È2
, where u œ 2x and du œ 2 dx
x È17
C
tan" Š Èx3 ‹ C œ
È3 9
tan" Š Èx3 ‹ C
5x sec" ¹ È ¹C 2
, where u œ È5x and du œ È5 dx " #
sec" ¹
" œ 4 sin" #s ‘ ! œ 4 ˆsin"
œ
' È du
" #
, where u œ 5x and du œ 5 dx
sec" ¸ u# ¸ C œ
" È2
#
œ
dy dx
dx œ
#
sec" ¹ Èu2 ¹ C œ
ds È9 4s#
'02 8 dt2t
(2x)
# 1 ’ax# 1b"Î# “
sin" (2x) C
" # ŠÈ3‹ x#
3 œ "# sin" u3 ‘ 0
51.
" #
sin" u C œ
#
œ
' È1 2(2x)
" #
dx œ " #
œ
' 17 " x
"Î#
dx œ sin" ˆ x3 ‰ C
9 x#
45.
Š "# ‹ ax# 1b
œ sin" x
42. y œ ln ax# 4b x tan" ˆ x# ‰ Ê
43.
œ
œ 0, for x 1
40. y œ cot" ˆ x" ‰ tan" x œ
œ sin" x
dy dx
È2Î4
È2Î2
du 8 u#
tan"
du È 9 u#
œ
" #
È5x # ¹
" #
C
sin" 0‰ œ 4 ˆ 16 0‰ œ
21 3
, where u œ 2s and du œ 2 ds; s œ 0 Ê u œ 0, s œ
Šsin"
È2 #
sin" 0‹ œ
" #
ˆ 14 0‰ œ
3È 2 4
Ê uœ
3È 2 #
1 8
, where u œ È2t and du œ È2 dt; t œ 0 Ê u œ 0, t œ 2 Ê u œ 2È2 u È8 “
#È # !
œ
" 4
Štan"
2È 2 È8
tan" 0‹ œ
" 4
atan" 1 tan" 0b œ
" 4
ˆ 14 0‰ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 16
427
428 52.
Chapter 7 Transcendental Functions
'c22
œ ’ È"3 †
53.
È2Î2
'cc1
È
'c22È33
" È3
œ
dt 4 3t#
" #
tan" u# “
È2
œ 'c2
dy yÈ4y# 1
È #
œ csec" kukd #
54.
È
#
œ 56.
œ
#
' È4 6 (rdr 1) ' 2 (xdx 1) œ
58.
œ 59.
œ'
" È2
" 3
" #
" #
†
u È2
'c11ÎÎ22 12cos(sin) d)))
#
'11ÎÎ64
sec" ¸ u# ¸
'0ln
È3
¸ 5u ¸
œ2'
1
" 3
œ 'È3 1
È3
œ '1
œ 11#
1 3
È2 3
Ê u œ È 2
œ 11#
sin" 2(r 1) C , where u œ r 1 and du œ dr
" È2
tan" Š xÈ1 ‹ C 2
tan" (3x 1) C
du , where u œ x uÈu# 25 " C œ 5 sec" ¸ x5 3 ¸ C
1 and du œ 2 dx
3 and du œ dx
, where u œ sin ) and du œ cos ) d); ) œ 1# Ê u œ ", ) œ
du 1 u# ,
where u œ cot x and du œ csc# x dx; x œ
œ tan" 1 tan" È3 œ 14
du 1 u#
È$
œ ctan" ud "
1 4
Ê u œ È 2
1 #
Êuœ"
œ 2 atan" 1 tan" (1)b œ 2 14 ˆ 14 ‰‘ œ 1
" c tan" ud È$
ex dx 1 e2x
1 3
, where u œ 3y and du œ 3 dy; y œ 32 Ê u œ 2, y œ
du , where u œ 2x uÈ u# 4 " C œ 4 sec" ¸ 2x # 1 ¸ C
du 1 u#
1
" tan" ud "
csc# x dx 1 (cot x)#
È2 #
, where u œ 3x 1 and du œ 3 dx
#
uCœ
sec
1 4
1 3È 3
, where u œ 2(r 1) and du œ 2 dr
Cœ
' 1 duu
"
13 ˆ 13 ‰‘ œ
, where u œ x 1 and du œ dx
#
" 5
" #È 3
, where u œ 2y and du œ 2 dy; y œ 1 Ê u œ 2, y œ
' (x 3)È(xdx 3) 25 œ '
œ
63.
" 3 "
tan
œ c2 62.
tan"
’tan" È3 tan" ŠÈ3‹“ œ
C œ 6 sin" ˆ r# 1 ‰ C
du 2 u#
œ
#
du È 4 u#
#
œ 61.
u #
3 #
' (2x 1)Èdx(2x 1) 4 œ "# ' œ
60.
du È 1 u#
œ6'
#
#
' 1 (3xdx 1)
'
" #È 3
œ sec" ¹È2¹ sec" k2k œ
sin" u C œ
3 #
œ 6 sin" 57.
3 #
œ
du uÈ u# 1
2
È #
' È1 34(rdr 1)
#È$
œ sec" ¹È2¹ sec" k2k œ
œ csec" kukd # 55.
#È $
du uÈ u# 1
È2
'c2Î32Î3 yÈ9ydy 1 œ '
, where u œ È3t and du œ È3 dt; t œ 2 Ê u œ 2È3, t œ 2 Ê u œ 2È3
du 4 u#
1 3
œ
1 6
Ê u œ È3 , x œ
1 4
Ê uœ1
1 1#
, where u œ ex and du œ ex dx; x œ 0 Ê u œ 1, x œ ln È3 Ê u œ È3
œ tan" È3 tan" 1 œ
1 3
1 4
œ
1 1#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.6 Inverse Trigonometric Functions 64.
'1e
1Î%
œ c4 65.
' Èy1 dy y
œ
%
" #
œ 66.
œ 4'0
1Î4
du 1 u# , where u œ ln t and 1Î% tan" ud ! œ 4 ˆtan" 14 tan" 0‰
4 dt t a1 ln# tb
" #
' È du
1 u#
" t
dt; t œ 1 Ê u œ 0, t œ e1Î4 Ê u œ
œ 4 tan"
1 4
1 4
, where u œ y# and du œ 2y dy
sin" u C œ
' Èsec1 ytandy y œ ' È du #
#
du œ
429
" #
sin" y# C
, where u œ tan y and du œ sec# y dy
1 u#
œ sin" u C œ sin" (tan y) C 67.
'È
68.
' È dx
69.
'01
œ'
dx x# 4x 3
2x x#
œ'
dx È1 ax# 4x 4b
œ'
œ'
dx È1 ax# 2x 1b
dx È1 (x 2)#
dx È1 (x 1)#
œ sin" (x 2) C
œ sin" (x 1) C
œ 6 '1 È4 at#dt 2t 1b œ 6 '1 È2# dt(t 1)# œ 6 sin" ˆ t # 1 ‰‘ " 0
6 dt È3 2t t#
0
!
œ 6 sin" ˆ "# ‰ sin" 0‘ œ 6 ˆ 16 0‰ œ 1 70.
'11Î2
6 dt È3 4t 4t#
dt 1 ‰‘ œ 3'1Î2 È4 a4t2#dt 4t 1b œ 3 '1Î2 È2# 2 (2t œ 3 sin" ˆ 2t # "Î# 1)# 1
1
œ 3 sin" ˆ "# ‰ sin" 0‘ œ 3 ˆ 16 0‰ œ
"
1 #
71.
' y dy2y 5 œ ' 4 y dy 2y 1 œ ' # (ydy 1)
72.
' y 6ydy 10 œ ' 1 ay dy 6y 9b œ ' 1 (ydy 3)
73.
'12 x 8 2xdx 2 œ 8'12 1 ax dx 2x 1b œ 8'12 1 (xdx 1)
74.
'24
75.
#
#
#
#
#
#
œ
#
#
œ tan" (y 3) C
#
2 dx x# 6x 10
œ 2'2
4
dx 1 ax# 6x 9b
œ 2'2
4
tan" ˆ y # 1 ‰ C
" #
#
œ 8 ctan" (x 1)d " œ 8 atan" 1 tan" 0b œ 8 ˆ 14 0‰ œ 21
#
dx 1 (x 3)#
%
œ 2 ctan" (x 3)d # œ 2 ctan" 1 tan" (1)d œ 2 14 ˆ 14 ‰‘ œ 1
' xx 44 dx œ ' x x 4 dx ' x 4 4 dx; ' x x 4 dx œ #" ' u1 du where u œ x# 4 Ê du œ 2x dx Ê #" du œ x dx Ê ' xx 44 dx œ 12 lnax# 4b 2 tan1 ˆ x2 ‰ C #
#
#
#
#
76.
' t t 6t # 10 dt œ ' at t3b # 1 dt ’Let w œ t 3 Ê w 3 œ t Ê dw œ dt“ Ä ' ww 1" dw œ ' w w " dw ' w 1 " dw; #
#
#
#
#
' w w " dw œ #" ' 1u du where u œ w# 1 Ê du œ 2w dw Ê #" du œ w dw Ê ' w w " dw ' w 1 " dw #
œ 77.
#
1 # 2 lnaw
1
1b tan awb C œ
1 ˆ 2 ln at
1
3b 1‰ tan at 3b C œ
#
1 # 2 lnat
6t 10b tan1 at 3b C
' x x 2x 9 1 dx œ ' a1 2xx 109 bdx œ ' dx ' x 2x 9 dx 10' x 1 9 dx; ' x 2x 9 dx œ ' u1 du where u œ x# 9 1 ˆ x ‰ Ê du œ 2x dx Ê ' dx ' x 2x 9 dx 10' x 1 9 dx œ x lnax# 9b 10 3 tan 3 C #
#
#
#
#
78.
#
#
#
#
' t 2tt 13t 4 dt œ ' at 2 2tt 12 bdt œ ' at 2bdt ' t 2t 1 dt 2' t 1 1 dt; ' t 2t 1 dt œ ' u1 du where u œ t# 1 1 1 2 # 1 Ê du œ 2t dt Ê ' at 2bdt ' t 2t 1 dt 2' t 1 dt œ 2 t 2t lnat 1b 2 tan atb C 3
2
#
#
#
#
#
#
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
430 79.
Chapter 7 Transcendental Functions
'
œ sec 80.
'
sinc" x
' Èe
1 x#
cosc" x
' Èe
1 x#
"x
1x
84.
du , uÈ u# 1
where u œ x 1 and du œ dx
œ'
" uÈ u# 1
du, where u œ x 2 and du œ dx
kx 2k C
dx È 1 x#
C
c" x
dx È 1 x#
C
u# du, where u œ sin" x and du œ
$ asin " xb 3
Cœ
œ'
dx (x 2)È(x 2)# 1
dx È 1 x#
C
' È1tan cx x dx œ ' u"Î# du, where u œ tan" x and du œ 1 dxx " #
œ
85.
u$ 3
œ'
dx œ ' eu du, where u œ cos" x and du œ
' aÈsin " xb## dx œ ' œ
dx (x 2)Èx# 4x 4 1 "
dx œ ' eu du, where u œ sin" x and du œ
œ eu C œ ecos 83.
dx (x 1)È(x 1)# 1
kx 1k C
kuk C œ sec
œ eu C œ esin 82.
œ'
dx (x 2)Èx# 4x 3 "
œ'
dx (x 1)Èx# 2x 1 1 "
kuk C œ sec
œ sec 81.
œ'
dx (x 1)Èx# 2x "
2 3
u$Î# C œ
2 3
atan" xb
' atanc" yb"a1 y#b dy œ '
$Î#
Cœ
"
‹ y# tan " y
Š
1
2 3
dy œ '
#
$
Éatan" xb C " u
du, where u œ tan" y and du œ
dy 1 y#
œ ln kuk C œ ln ktan" yk C
86.
'
"
" asinc" yb È1 y#
dy œ
' Ésin " y# dy œ ' u" du, where u œ sin" y and du œ È1dy y# 1
y
œ ln kuk C œ ln ksin" yk C 87.
'È22
sec# asec " xb xÈ x# 1
dx œ '1Î4 sec# u du, where u œ sec" x and du œ 1Î3
1Î$
œ ctan ud 1Î4 œ tan 88.
'22ÎÈ3
cos asecc" xb xÈ x# 1
"
' eÈsinc e x
" x
1 e#x
tanc1 Èx ‹ 3
1 6
sin
Èxax 1b ”ˆtanc1 Èx‰2 9•
œ 23 tan1 Š 90.
1 3
; x œ È2 Ê u œ
1 4
,xœ2 Ê uœ
1 3
dx xÈ x# 1
;xœ
1 6
,xœ2 Ê uœ
1 3
œ È3 1
1Î3
1Î$
'
1 4
tan
dx œ '1Î6 cos u du, where u œ sec" x and du œ
œ csin ud 1Î' œ sin 89.
1 3
dx xÈ x# 1
œ
2 È3
Ê uœ
È3 " #
dx œ 2'
" u2 9 du
where u œ tan1 Èx Ê du œ
1 1 dx 2 1 + ˆÈ x ‰ 2 È x
Ê 2du œ
C
dx œ ' u du where u œ sin" ex Ê du œ
1 x È1 e#x e dx
œ "# asin" ex b C 2
sinc" 5x x xÄ0
91. lim
œ lim
ŠÈ
xÄ0
5 1
25x#
1
‹
œ5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 dx a 1 x bÈ x
Section 7.6 Inverse Trigonometric Functions 92.
È x# 1 sec " x
lim b
xÄ1
"Î#
ax # 1 b sec " x
œ lim b xÄ1
tanc" a2x " b x "
93. x lim x tan" ˆ 2x ‰ œ x lim Ä_ Ä_ 2 tanc" 3x# 7x# xÄ0
94. lim
95.
96.
lim
x Ä0
lim
tanc1 x2 x sinc1 x
e tanc1 ex e2x x
98.
œ x
’tanc1 ˆÈx‰“
ˆe2x 3‰ 4ae2x 1b2 —
x Ä0
xÈ x 1
sinc1 ˆx2 ‰ asinc1 xb2
œ limb Š x Ä0
x Ä0
x 1 x#
œ lim
Èx 1
‹œ 2x
1 c x2
" #
ln a1 x# b
" x a1 x # b
tanc" x x
tan " x x# ‹
œ x lim Ä_
x #
œ
œ lim b x kxk œ 1 xÄ1
6 7
2Š3x4 1‹ Š1
Š1
Ñ 2 0 1 12 Ó œ c0 2 œ 2 3Î2 1 0 3Î2 Ò a
2 x4 ‹
x2
x2‹
a
b
b
œ limb x Ä0
È
œ1
ex tanc1 ex
2
œ lim
e2x Še2x b 3‹ 2 Še2x b 1‹
4e2x
xÄ_
ˆ1 3ec2x ‰ 4aex ecx b2 —
tanc1 ˆÈx‰ xa1 b xb 3x b 2 2Èx b 1
2 2
2e2x
Še2x b 1‹
4e2x x
œ2
2 14x #
œ00œ0
È bx 2tanc1 ˆÈx‰ œ limb Š a3x 2b ÈxÈx 1 ‹ œ limb 12x2 xb113x b2 x Ä0 x Ä0 1
a
b
2ÈxÈx
b1
œ1
È œ limb Œ 2asinc11xcbx4 1 È x Ä0 œ
# #‹
2x 1 4x
xÄ_
c1
2 2
1
2x
xÄ_
x 2Èx b 1
x#
(2x)
ex tanc1 ex e2xe b 1
œ lim – tan4ex e
È 1 x2 È 1 x2 ‹ 1 x2 xÈ1 x2 sinc1 x
99. If y œ ln x œ Š "x
6
È
Š
% x Ä 0 7 a1 9x b
tanc1 ˆÈx‰ Èxa11 b xb
œ limb
2 lim Š 2 xÄ0b a12x 13x 2bÈx 1
limb
œ lim
ex tanc1 ex e2xe b 1 lim 2e2x 1 xÄ_
2
œ
x
k k
œ x lim Ä_
2x
c1
x Ä0
Œ
"Î#
"
Î œ lim Œ x 1 sin1 x œ lim Ð x Ä0 x Ä0 È1 x2 Ï
xÄ_
limb
12x ‹ 9x%
14x
xÄ0
œ lim – tan4ex e
97.
1
Š "# ‹ ax# 1b
2x 1 x4
x
xÄ_
Š
œ lim
œ lim b xÄ1
431
1 1
Š c1 x ‹ œ limb Œ sinc1 x† œ xlim 2 Ä0b sin x È1 x x Ä0
1
È1 b x2 È1 c x2 È1 x2 x
1
œ
œ1
C, then dy œ – x" dx œ
x 1 x#
Š
x 1
x a1 x# b x$ x atan " xba1 x# b x # a1 x # b
x#
‹ tan " x x#
dx œ
— dx
tan " x x#
dx,
which verifies the formula 100. If y œ
x% 4
cos" 5x
5 4
'È
x% 1 25x#
%
dx, then dy œ ’x$ cos" 5x Š x4 ‹ Š È
5 ‹ 1 25x#
45 Š È
x% ‹“ 1 25x#
dx
œ ax$ cos" 5xb dx, which verifies the formula #
101. If y œ x asin" xb 2x 2È1 x# sin" x C, then c" # dy œ ’asin" xb 2x asin xb 2 2x sin" x 2È1 x# Š È 1 x#
È 1 x#
" È 1 x # ‹“
#
dx œ asin" xb dx, which verifies
the formula 102. If y œ x ln aa# x# b 2x 2a tan" ˆ xa ‰ C, then dy œ –ln aa# x# b #
2x# a# x#
2
2 # 1 Š x# ‹ —
dx
a
#
x # # œ ’ln aa# x# b 2 Š aa# x# ‹ 2“ dx œ ln aa x b dx, which verifies the formula
103.
dy dx
œ
" È 1 x#
Ê dy œ
dx È 1 x#
Ê y œ sin" x C; x œ 0 and y œ 0 Ê 0 œ sin" 0 C Ê C œ 0 Ê y œ sin" x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
432 104.
Chapter 7 Transcendental Functions dy dx
œ
" x# 1
1 Ê dy œ ˆ 1 " x# 1‰ dx Ê y œ tan" (x) x C; x œ 0 and y œ 1 Ê 1 œ tan" 0 0 C
Ê C œ 1 Ê y œ tan" (x) x 1 105.
dy dx
œ
œ1 106.
dy dx
œ
" Ê dy œ Èdx# Ê y œ sec" xÈ x# 1 x x 1 13 œ 231 Ê y œ sec" (x) 231 , x 1 " 1 x#
2 È 1 x#
Ê dy œ Š 1 " x#
kxk C; x œ 2 and y œ 1 Ê 1 œ sec" 2 C Ê C œ 1 sec" 2
2 È 1 x# ‹
dx Ê y œ tan" x 2 sin" x C; x œ 0 and y œ 2
Ê 2 œ tan" 0 2 sin" 0 C Ê C œ 2 Ê y œ tan" x 2 sin" x 2 107. (a) The angle ! is the large angle between the wall and the right end of the blackboard minus the small angle between x ‰ the left end of the blackboard and the wall Ê ! œ cot" ˆ 15 cot" ˆ 3x ‰ . (b)
d! dt
œ
1 15 x ‰ 1 ˆ 15
2
1 3
1 ˆ 3x ‰
2
œ 22515 x2
œ
3 9 x2
540 12x2 d! a225 x2 ba9 x2 b ; dt
œ 0 Ê 540 12x2 œ 0 Ê x œ „ 3È5
È
È
Since x 0, consider only x œ 3È5 Ê !Š3È5‹ œ cot" Š 3155 ‹ cot" Š 3 3 5 ‹ ¸ 0.729728 ¸ 41.8103‰ . Using the first derivative test,
d! dt ¹xœ1
œ
132 565
0 and
d! dt ¹xœ10
132 ‰ œ (!) 5 0 Ê local maximum of 41.8103 when
x œ 3È5 ¸ 6.7082 ft. 108. V œ 1'0 c2# (sec y)# d dy œ 1 c4y tan yd ! 1Î3
1Î$
œ 1 Š 431 È3‹
109. V œ ˆ "3 ‰ 1r# h œ ˆ 3" ‰ 1(3 sin ))# (3 cos )) œ 91 acos ) cos$ )b, where 0 Ÿ ) Ÿ Ê
dV d)
œ 91(sin )) a1 3 cos# )b œ ! Ê sin ) œ 0 or cos ) œ „
" È3
1 #
Ê the critical points are: 0, cos" Š È"3 ‹ , and
cos" Š È"3 ‹ ; but cos" Š È"3 ‹ is not in the domain. When ) œ 0, we have a minimum and when ) œ cos" Š È"3 ‹ ¸ 54.7°, we have a maximum volume. ‰ 110. 65° (90° " ) (90° !) œ 180° Ê ! œ 65° " œ 65° tan" ˆ 21 50 ¸ 65° 22.78° ¸ 42.22° 111. Take each square as a unit square. From the diagram we have the following: the smallest angle ! has a tangent of 1 Ê ! œ tan" 1; the middle angle " has a tangent of 2 Ê " œ tan" 2; and the largest angle # has a tangent of 3 Ê # œ tan" 3. The sum of these three angles is 1 Ê ! " # œ 1 Ê tan" 1 tan" 2 tan" 3 œ 1. 112. (a) From the symmetry of the diagram, we see that 1 sec" x is the vertical distance from the graph of y œ sec" x to the line y œ 1 and this distance is the same as the height of y œ sec" x above the x-axis at x; i.e., 1 sec" x œ sec" (x). (b) cos" (x) œ 1 cos" x, where 1 Ÿ x Ÿ 1 Ê cos" ˆ "x ‰ œ 1 cos" ˆ "x ‰, where x 1 or x Ÿ 1 Ê sec" (x) œ 1 sec" x
113. sin" (1) cos" (1) œ
1 #
0œ
1 #
; sin" (0) cos" (0) œ 0
1 #
œ
1 #
; and sin" (1) cos" (1) œ 1# 1 œ 1# .
If x − ("ß 0) and x œ a, then sin" (x) cos" (x) œ sin" (a) cos" (a) œ sin" a a1 cos" ab œ 1 asin" a cos" ab œ 1 1# œ 1# from Equations (3) and (4) in the text.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.6 Inverse Trigonometric Functions
Ê tan ! œ x and tan " œ
114.
1 #
115. csc" u œ
sec" u Ê
acsc" ub œ
d dx
d dx
" x
Ê
1 #
" x
œ ! " œ tan" x tan"
ˆ 1# sec" u‰ œ 0
du dx
ku k È u # 1
œ
433
.
du dx
ku k È u # 1
, ku k 1
116. y œ tan" x Ê tan y œ x Ê Ê asec# yb œ
" 1 x#
dy dx
œ1 Ê
d d dx (tan y) œ dx (x) " " # sec# y œ È Š 1 x# ‹
œ
dy dx
, as indicated by the triangle
117. f(x) œ sec x Ê f w (x) œ sec x tan x Ê
df " dx ¹xœb
œ
" df dx ¹x œ f
"abb
œ
" secasec " bbtanasec " bb
Since the slope of sec" x is always positive, we the right sign by writing 1 #
118. cot" u œ
tan" u Ê
acot" ub œ
d dx
d dx
du dx
ˆ 1# tan" u‰ œ 0
xœ
" . lx l È x # "
du
1 u#
119. The functions f and g have the same derivative (for x 0), namely
d " dx sec
" . b Š„ È b # " ‹
œ
œ 1 dxu#
" Èx (x 1)
. The functions therefore differ
by a constant. To identify the constant we can set x equal to 0 in the equation f(x) œ g(x) C, obtaining 1‰ " È sin" (1) œ 2 tan" (0) C Ê 1# œ 0 C Ê C œ 1# . For x 0, we have sin" ˆ xx x 1 œ 2 tan 120. The functions f and g have the same derivative for x 0, namely
" 1 x#
1 #
.
. The functions therefore differ by a
constant for x 0. To identify the constant we can set x equal to 1 in the equation f(x) œ g(x) C, obtaining sin" Š È" ‹ œ tan" 1 C Ê 2
È3
1 4
œ
1 4
C Ê C œ 0. For x 0, we have sin"
È3
121. V œ 1 'cÈ3Î3 Š È1" x# ‹ dx œ 1 ' È3Î3 #
œ 1 13 ˆ 16 ‰‘ œ
r È#
r ÎÈ 2
œ '0
œ
dy dx
x È r2 x 2 ;
(in other words, the length of
É r2 r x2 dx œ ' 2
rÎÈ2
0
r Èr2 x2 dx
œ r sin1 Š È12 ‹ 0 œ rˆ 14 ‰ œ 123. (a) A(x) œ
1 4
œ tan"
È3
dx œ 1 ctan" xd È3Î3 œ 1 ’tan" È3 tan" Š
" x
.
È3 3 ‹“
1# #
122. Consider y œ Èr2 x2 Ê to x œ
" 1 x#
" È x# 1
(diameter)# œ
1 4
"
1r 4 .
" 1 x#
dy dx
is undefined at x œ r and x œ r, we will find the length from x œ 0 rÎÈ2
of a circle) Ê L œ '0 r ÎÈ 2
œ r sin1 ˆ xr ‰‘ 0
rÎÈ2
Ê1 Š È 2 x 2 ‹ dx œ '0 2
r x
É1
È
œ r sin1 Š rÎ r 2 ‹ r sin1 a0b
The total circumference of the circle is C œ 8L œ 8ˆ 14r ‰ œ 21 r. #
’ È1" x# Š È1" x# ‹“ œ
œ 1 ctan" xd " œ (1)(2) ˆ 14 ‰ œ (b) A(x) œ (edge)# œ ’ È
1 8
Since
1# #
Š È
# " ‹ “ # 1x
œ
4 1 x#
1 1 x#
Ê V œ 'a A(x) dx œ 'c1 b
1
1 dx 1 x#
Ê V œ 'a A(x) dx œ 'c1 14dxx# b
1
"
œ 4 ctan" xd " œ 4 ctan" (1) tan" (1)d œ 4 14 ˆ 14 ‰‘ œ 21
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x2 r2 x2 dx
434
Chapter 7 Transcendental Functions
124. (a) A(x) œ
1 4
È2Î2
œ 'cÈ2Î2 (b) A(x) œ
(diameter)# œ 1 È 1 x#
(diagonal)# 2
1 4
Š
2 % È 1 x#
#
0‹ œ
1 4
ŠÈ
È2Î2
4 ‹ 1 x#
dx œ 1 csin" xd È2Î2 œ 1 ’sin" Š œ
È
1 2
Š
2 % È 1 x#
#
0‹ œ
2 È 1 x#
È2 # ‹
œ
1 È 1 x#
Ê V œ 'a A(x) dx
sin" Š
b
È2 # ‹“
œ 1 14 ˆ 14 ‰‘ œ
È2Î2
Ê V œ 'a A(x) dx œ 'cÈ2Î2 b
2 È 1 x#
1# #
dx
2Î2 œ 2 csin" xd È2Î2 œ 2 ˆ 14 † 2‰ œ 1
125. (a) sec" 1.5 œ cos" (c) cot" 2 œ
1 #
tan
" 1.5 "
" ‰ (b) csc" (1.5) œ sin" ˆ 1.5 ¸ 0.72973
¸ 0.84107
2 ¸ 0.46365
126. (a) sec" (3) œ cos" ˆ "3 ‰ ¸ 1.91063 (c) cot" (2) œ
1 #
" ‰ (b) csc" 1.7 œ sin" ˆ 1.7 ¸ 0.62887
tan" (2) ¸ 2.67795
127. (a) Domain: all real numbers except those having the form 1# k1 where k is an integer. Range: 1# y
1 #
(b) Domain: _ x _; Range: _ y _ The graph of y œ tan" (tan x) is periodic, the graph of y œ tan atan" xb œ x for _ Ÿ x _.
128. (a) Domain: _ x _; Range: 1# Ÿ y Ÿ
1 #
(b) Domain: " Ÿ x Ÿ 1; Range: " Ÿ y Ÿ 1 The graph of y œ sin" (sin x) is periodic; the graph of y œ sin asin" xb œ x for " Ÿ x Ÿ 1.
129. (a) Domain: _ x _; Range: 0 Ÿ y Ÿ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.6 Inverse Trigonometric Functions (b) Domain: 1 Ÿ x Ÿ 1; Range: " Ÿ y Ÿ 1 The graph of y œ cos" (cos x) is periodic; the graph of y œ cos acos" xb œ x for " Ÿ x Ÿ 1.
130. Since the domain of sec" x is (_ß 1] ["ß _), we have sec asec" xb œ x for kxk 1. The graph of y œ sec asec" xb is the line y œ x with the open line segment from ("ß ") to ("ß ") removed.
131. The graphs are identical for y œ 2 sin a2 tan" xb œ 4 csin atan" xbd ccos atan" xbd œ 4 Š È œ
4x x# 1
x ‹ Š Èx#" 1 ‹ x# 1
from the triangle
132. The graphs are identical for y œ cos a2 sec" xb œ cos# asec" xb sin# asec" xb œ œ
2 x # x#
" x#
x# 1 x#
from the triangle
133. The values of f increase over the interval ["ß 1] because f w 0, and the graph of f steepens as the values of f w increase towards the ends of the interval. The graph of f is concave down to the left of the origin where f ww 0, and concave up to the right of the origin where f ww 0. There is an inflection point at x œ 0 where f ww œ 0 and f w has a local minimum value.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
435
436
Chapter 7 Transcendental Functions
134. The values of f increase throughout the interval (_ß _) because f w 0, and they increase most rapidly near the origin where the values of f w are relatively large. The graph of f is concave up to the left of the origin where f ww 0, and concave down to the right of the origin where f ww 0. There is an inflection point at x œ 0 where f ww œ 0 and f w has a local maximum value. 7.7 HYPERBOLIC FUNCTIONS #
1. sinh x œ 34 Ê cosh x œ È1 sinh# x œ É1 ˆ 34 ‰ œ É1 coth x œ 2. sinh x œ sech x œ 3. cosh x œ œ
8 17
" tanh x
coth x œ
œ
, and csch x œ
4 5
Ê cosh x œ È1 sinh# x œ É1
4 3
" cosh x
17 15
œ
13 5
3 5
, and csch x œ
" sinh x
œ
16 9
" sin x
œ É 25 9 œ
25 œ É 16 œ
5 4
œ
17 8
5 3
, tanh x œ
œ
15 17
, and csch x œ
" sinh x
œ
É 144 , x 0 Ê sinh x œ Ècosh# x 1 œ É 169 25 1 œ 25 œ
" tanh x
œ
13 12
5. 2 cosh (ln x) œ 2 Š e
6. sinh (2 ln x) œ
" cosh x
, sech x œ ln x
ecln x ‹ #
e2 ln x ec2 ln x #
7. cosh 5x sinh 5x œ
œ
e5x e #
9. (sinh x cosh x)% œ ˆ e
x
5x
ecx #
œ
5 13
œ eln x
" eln x
#
eln x eln x #
#
, and csch x œ
œ
e5x e #
ex ecx ‰% #
5x
œ 35 ,
sinh x cosh x
œ
ˆ 43 ‰ ˆ 53 ‰
4 5
, coth x œ
" tanh x
œ
5 4
8 15
, tanh x œ
sinh x cosh x
œ
8 ‰ ˆ 15 ˆ 17 ‰ 15
œ
,
3 4
" cosh x
, sech x œ
ˆ 34 ‰ ˆ 54 ‰
sinh x cosh x
œ 34
#
" tanh x
œ
, tanh x œ
17 ‰ 289 64 , x 0 Ê sinh x œ Ècosh# x 1 œ Ɉ 15 1 œ É 225 1 œ É 225 œ
, coth x œ
4. cosh x œ
" cosh x
œ 35 , sech x œ
9 16
œx Šx# x"# ‹ #
" sinh x
œ
15 8
12 5
, tanh x œ
sinh x cosh x
œ
ˆ 12 ‰ 5 ˆ 13 ‰ 5
œ
12 13
,
5 12
" x
œ
x% " #x#
œ e5x
8. cosh 3x sinh 3x œ
e3x e #
3x
e3x e #
3x
œ e3x
œ aex b% œ e4x
10. ln (cosh x sinh x) ln (cosh x sinh x) œ ln acosh# x sinh# xb œ ln 1 œ 0 11. (a) sinh 2x œ sinh (x x) œ sinh x cosh x cosh x sinh x œ 2 sinh x cosh x (b) cosh 2x œ cosh (x x) œ cosh x cosh x sinh x sin x œ cosh# x sinh# x 12. cosh# x sinh# x œ ˆ e œ
" 4
!
a4e b œ
13. y œ 6 sinh 14. y œ
" #
x 3
" 4
x
ecx ‰# #
ˆe
x
ecx ‰# #
œ
" 4
caex ex b aex ex bd caex ex b aex ex bd œ
" 4
a2ex b a2ex b
(4) œ 1
Ê
dy dx
œ 6 ˆcosh x3 ‰ ˆ "3 ‰ œ 2 cosh
sinh (2x 1) Ê
dy dx
œ
" #
x 3
[cosh (2x 1)](2) œ cosh (2x 1)
15. y œ 2Èt tanh Èt œ 2t"Î# tanh t"Î# Ê
dy dt
œ sech# ˆt"Î# ‰‘ ˆ "# t"Î# ‰ ˆ2t"Î# ‰ ˆtanh t"Î# ‰ ˆt"Î# ‰ œ sech# Èt
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tanh Èt Èt
Section 7.7 Hyperbolic Functions 437 16. y œ t# tanh
" t
œ t# tanh t" Ê
17. y œ ln (sinh z) Ê
œ
dy dz
œ csech# at" bd at# b at# b (2t) atanh t" b œ sech#
dy dt
œ coth z
cosh z sinh z
19. y œ (sech ))(1 ln sech )) Ê
18. y œ ln (cosh z) Ê
dy dz
œ
" t
2t tanh
sinh z cosh z
" t
œ tanh z
dy d)
) tanh ) ‰ œ ˆ sech (sech )) ( sech ) tanh ))(1 ln sech )) sech )
dy d)
) coth ) ‰ œ (csch )) ˆ csch (1 ln csch ))( csch ) coth )) csch )
œ sech ) tanh ) (sech ) tanh ))(1 ln sech )) œ (sech ) tanh ))[1 (1 ln sech ))] œ (sech ) tanh ))(ln sech )) 20. y œ (csch ))(1 ln csch )) Ê
œ csch ) coth ) (1 ln csch ))(csch ) coth )) œ (csch ) coth ))(1 1 ln csch )) œ (csch ) coth ))(ln csch )) 21. y œ ln cosh v
" #
tanh# v Ê
" #
coth# v Ê
dy dv
œ
dy dv
œ
sinh v cosh v
ˆ "# ‰ (2 tanh v) asech# vb œ tanh v (tanh v) asech# vb
cosh v sinh v
ˆ "# ‰ (2 coth v) a csch# vb œ coth v (coth v) acsch# vb
œ (tanh v) a1 sech# vb œ (tanh v) atanh# vb œ tanh$ v 22. y œ ln sinh v
œ (coth v) a1 csch# vb œ (coth v) acoth# vb œ coth$ v
23. y œ ax# 1b sech (ln x) œ ax# 1b ˆ eln x 2ec ln x ‰ œ ax# 1b ˆ x 2xc" ‰ œ ax# 1b ˆ x#2x1 ‰ œ 2x Ê 24. y œ a4x# 1b csch (ln 2x) œ a4x# 1b ˆ eln 2x 2e 25. y œ sinh" Èx œ sinh" ˆx"Î# ‰ Ê
dy dx
É1 ax"Î# b#
26. y œ cosh" 2Èx 1 œ cosh" ˆ2(x 1)"Î# ‰ Ê 27. y œ (1 )) tanh" ) Ê
dy d)
dy dx
œ
œ
" #È x È 1 x
dy d)
œ
dy dx
œ4
" #Èx(1 x)
(2) Š "# ‹ (x 1)c"Î#
œ
Éc2(x 1)"Î# d# 1
œ (1 )) ˆ 1 " )# ‰ (1) tanh" ) œ
28. y œ a)# 2)b tanh" () 1) Ê
œ2
2 4x # ‰ œ a4x# 1b Š 2x (2x) " ‹ œ a4x 1b ˆ 4x# 1 ‰ œ 4x Ê
ln 2x
Š "# ‹ xc"Î#
œ
dy dx
" 1)
" Èx 1 È4x 3
œ
" È4x# 7x 3
tanh" )
œ a)# 2)b ’ 1 ()"1)# “ (2) 2) tanh" () 1) œ
) # 2) ) # 2 )
(2) 2) tanh" () 1)
œ (2) 2) tanh" () 1) 1 29. y œ (1 t) coth" Èt œ (1 t) coth" ˆt"Î# ‰ Ê
30. y œ a1 t# b coth" t Ê
dy dt
œ
dy dx
" È 1 x#
’x Š
32. y œ ln x È1 x# sech" x œ ln x a1 x# b dy dx
" x
œ
"
33. y œ csch
a1 x# b
ˆ "# ‰)
"Î#
Š
œ (1 t) –
Š "# ‹ tc"Î# 1 at"Î# b
#
"
ˆt"Î# ‰ œ
" È 1 x#
— (1) coth
" ‹ xÈ 1 x#
" ‹ xÈ 1 x#
"Î#
(1) sech" x“ œ
" È 1 x#
Ê
coth" Èt
sech" x œ sech" x
sech" x
ˆ "# ‰ a1 x# b
"Î#
(2x) sech" x œ
" x
" x
x È 1 x#
sech" x œ
)
dy d)
" #È t
œ a1 t# b ˆ 1 " t# ‰ (2t) coth" t œ 1 2t coth" t
31. y œ cos" x x sech" x Ê
Ê
dy dt
œ
’ln Š "# ‹“ Š "# ‹ )
Š "# ‹
) #
" Ë 1 ”Š # ‹
•
œ ln (1) ln (2) œ #) Ê1 Š "# ‹
ln 2 #)
Ê1 Š "# ‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
x È 1 x#
sech" x
438
Chapter 7 Transcendental Functions
34. y œ csch" 2) Ê
dy d)
œ
35. y œ sinh" (tan x) Ê
œ
dy dx
36. y œ cosh" (sec x) Ê
(ln 2) 2) 2 ) É 1 a2 ) b
œ
dy dx
ln 2 È1 2#)
œ
sec# x Èsec# x
(sec x)(tan x) Èsec# x 1
œ
(sec x)(tan x) Ètan# x
(b) If y œ sin" (tanh x) C, then x# #
œ
sec# x È1 (tan x)#
37. (a) If y œ tan" (sinh x) C, then
38. If y œ
#
dy dx dy dx
œ œ
œ
sec# x ksec xk
œ
œ
ksec xk ksec xk ksec xk
(sec x)(tan x) ktan xk
œ ksec xk
œ sec x, 0 x
1 #
cosh x cosh x 1 sinh# x œ cosh# x œ sech x, which verifies the formula # sech x sech# x È1 tanh# x œ sech x œ sech x, which verifies the formula
sech" x "# È1 x# C, then
œ x sech" x
dy dx
x# #
Š
" ‹ xÈ 1 x#
2x 4È 1 x#
œ x sech" x, which verifies the
formula x# " #
39. If y œ
coth" x
40. If y œ x tanh" x 41.
'
sinh 2x dx œ œ
42.
'
sinh
x 5
cosh u #
" #
'
" #
x #
C, then
'
44.
' '
dy dx
Cœ
cosh 2x #
4 cosh (3x ln 2) dx œ
tanh
x 7
sinh u C œ
dx œ 7 '
" ˆ " ‰ # ‹ 1 x#
" #
œ x coth" x, which verifies the formula
sinh u du, where u œ 2x and du œ 2 dx C
dx œ 5 ' sinh u du, where u œ
4 3
#
œ tanh" x x ˆ 1 " x# ‰ #" ˆ 12xx# ‰ œ tanh" x, which verifies the formula
x 5
x 5
" 5
and du œ
sinh u cosh u
4 3
4 3
'
dx
C
6 cosh ˆ x# ln 3‰ dx œ 12 ' cosh u du, where u œ œ 12 sinh u C œ 12 sinh ˆ x# ln 3‰ C
œ 45.
œ x coth" x Š x
ln a1 x# b C, then
œ 5 cosh u C œ 5 cosh 43.
dy dx
x #
ln 3 and du œ
" #
dx
cosh u du, where u œ 3x ln 2 and du œ 3 dx
sinh (3x ln 2) C
du, where u œ
x 7
and du œ
" 7
dx
œ 7 ln kcosh uk C" œ 7 ln ¸cosh x7 ¸ C" œ 7 ln ¹ e
xÎ7
e xÎ7 ¹ #
C" œ 7 ln ¸exÎ7 exÎ7 ¸ 7 ln 2 C"
œ 7 ln kexÎ7 e xÎ7 k C 46.
'
coth
) È3
d) œ È 3 '
cosh u sinh u
du, where u œ
œ È3 ln ksinh uk C" œ È3 ln ¹sinh œ È3 ln ¹e)Î 47.
'
È$
e)Î
È$
) È3
and du œ
) È3 ¹
d) È3
C" œ È3 ln ¹ e
¹ È3 ln 2 C" œ È3 ln ¹e)Î
È3
È$ e )ÎÈ$
)Î
#
e)Î
È3
¹ C"
¹C
sech# ˆx "# ‰ dx œ ' sech# u du, where u œ ˆx "# ‰ and du œ dx œ tanh u C œ tanh ˆx "# ‰ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.7 Hyperbolic Functions 439 48.
'
csch# (5 x) dx œ ' csch# u du, where u œ (5 x) and du œ dx œ ( coth u) C œ coth u C œ coth (5 x) C
49.
'
dt œ 2 ' sech u tanh u du, where u œ Èt œ t"Î# and du œ
sech Èt tanh Èt Èt
dt 2È t
œ 2( sech u) C œ 2 sech Èt C 50.
'
csch (ln t) coth (ln t) t
dt œ ' csch u coth u du, where u œ ln t and du œ
dt t
œ csch u C œ csch (ln t) C 51.
x 'lnln24 coth x dx œ 'lnln24 cosh ' 15Î8 sinh x dx œ 3Î4
" u
"&Î) ¸ ¸3¸ ¸ 15 4 ¸ du œ cln kukd $Î% œ ln ¸ 15 8 ln 4 œ ln 8 † 3 œ ln
where u œ sinh x, du œ cosh x dx, the lower limit is sinh (ln 2) œ limit is sinh (ln 4) œ 52.
eln 4 ec ln 4 #
œ
4 Š "4 ‹ #
œ
17Î8 sinh 2x " ' '0ln 2 tanh 2x dx œ '0ln 2 cosh 2x dx œ # 1
eln 2 ec ln 2 #
œ
2 Š "# ‹
5 #
,
œ
3 4
and the upper
ln ˆ 17 ‰ ‘ 8 ln 1 œ
" #
ln
#
15 8
" u
du œ
" #
"(Î)
cln kukd "
" #
œ
17 8
, where
u œ cosh 2x, du œ 2 sinh (2x) dx, the lower limit is cosh 0 œ 1 and the upper limit is cosh (2 ln 2) œ cosh (ln 4) œ 53.
eln 4 ec ln 4 #
c2 ln 2 #
'0ln 2 4ec
)
ln 2‹ Š e
#
e
2 ln 2
#
ec) ‹ #
c ln 2
d) œ 'c ln 4 ae2) 1b d) œ e# )‘ c ln 4
" ln 4‹ œ ˆ 8" ln 2‰ ˆ 32 ln 4‰ œ
‹ Š0
)
e0 # ‹“
ec) ‹ #
ln 2
2)
3 32
ln 2 2 ln 2 œ
d) œ 2 '0 a1 ec2) b d) œ 2 ’) ln 2
œ 2 ˆln 2
" 8
"# ‰ œ 2 ln 2
" 4
3 3#
ln 2
ec#) # “0
ln 2
1 œ ln 4
3 4
'c11ÎÎ44 cosh (tan )) sec# ) d) œ ' 11 cosh u du œ csinh ud "" œ sinh (1) sinh (1) œ Š e #e " ‹ Š e " # e" ‹ "
e" , where u œ tan ), du œ sec# ) d), the lower limit is tan ˆ 14 ‰ œ 1 and the upper
'01Î2 2 sinh (sin )) cos ) d) œ 2'01 sinh u du "
œee
œ 2 ccosh ud "! œ 2(cosh 1 cosh 0) œ 2 Š e #e
c"
1‹
2, where u œ sin ), du œ cos ) d), the lower limit is sin 0 œ 0 and the upper limit is sin ˆ 1# ‰ œ 1
'12 cosht(ln t) dt œ '0ln 2 cosh u du œ csinh ud ln0 2 œ sinh (ln 2) sinh (0) œ e u œ ln t, du œ
58.
c2 ln 4
ln 2
ee "e "e œe # 1 limit is tan ˆ 4 ‰ œ 1
57.
17 8
sinh ) d) œ '0 4ec) Š e
œ
56.
œ
)
œ 2 ’Šln 2
55.
#
'cclnln42 2e) cosh ) d) œ 'cclnln42 2e) Š e œ Še
54.
4 Š "4 ‹
œ
" t
ln 2
ec ln 2 #
0œ
2 #
" #
œ
3 4
, where
dt, the lower limit is ln 1 œ 0 and the upper limit is ln 2
2 Èx # " '14 8 cosh dx œ 16'1 cosh u du œ 16 csinh ud #" œ 16(sinh 2 sinh 1) œ 16 ’Š e #e ‹ Š e #e ‹“ Èx #
œ 8 ae# e# e e" b , where u œ Èx œ x"Î# , du œ
" #
x"Î# dx œ
dx 2È x
, the lower limit is È1 œ 1 and the upper
limit is È4 œ 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
440 59.
60.
Chapter 7 Transcendental Functions
'c0ln 2 cosh# ˆ x# ‰ dx œ 'c0ln 2 cosh#x " dx œ "# 'c0ln 2 (cosh x 1) dx œ "# csinh x xd c0 ln 2 œ
" #
[(sinh 0 0) (sinh ( ln 2) ln 2)] œ
œ
" #
ˆ1
" 4
ln 2‰ œ
3 8
" #
ln 2 œ
" #
’(0 0) Š e
c ln 2 eln 2 #
ln 2‹“ œ
" #
–
Š "# ‹ 2 #
ln 2—
ln È2
3 8
'0ln 10 4 sinh# ˆ x# ‰ dx œ '0ln 10 4 ˆ cosh#x 1 ‰ dx œ 2'0ln 10 (cosh x 1) dx œ 2 csinh x xd ln0 10 œ 2[(sinh (ln 10) ln 10) (sinh 0 0)] œ eln 10 ec ln 10 2 ln 10 œ 10
5 25 61. sinh" ˆ 1#5 ‰ œ ln Š 12 É 144 1‹ œ ln ˆ 32 ‰
63. tanh" ˆ "# ‰ œ
" #
65. sech" ˆ 35 ‰ œ ln Š
67. (a)
'02
È3
dx È 4 x#
" 10
2 ln 10 œ 9.9 2 ln 10
62. cosh" ˆ 35 ‰ œ ln Š 35 É 25 9 1‹ œ ln 3
(1/2) ln 3 ln Š 11 (1/2) ‹ œ #
64. coth" ˆ 45 ‰ œ
1È1 (9/25) ‹ (3/5)
66. csch" Š È"3 ‹ œ ln È3
2 œ sinh" x# ‘ 0
œ ln 3
È3
" #
ln Š (9/4) (1/4) ‹ œ
" #
ln 9 œ ln 3
È4/3 Š1/È3‹
œ ln ŠÈ3 2‹
œ sinh" È3 sinh 0 œ sinh" È3
(b) sinh" È3 œ ln ŠÈ3 È3 1‹ œ ln ŠÈ3 2‹
68. (a)
'01Î3 È 6 dx
œ 2'0
1
1 9x#
dx È a# u# ,
where u œ 3x, du œ 3 dx, a œ 1
"
œ c2 sinh" ud ! œ 2 asinh" 1 sinh" 0b œ 2 sinh" 1 (b) 2 sinh" 1 œ 2 ln Š1 È1# 1‹ œ 2 ln Š1 È2‹
69. (a)
'52Î4
#
" 1 x#
dx œ ccoth" xd &Î% œ coth" 2 coth"
(b) coth" 2 coth" 70. (a)
'01Î2
71. (a)
'13ÎÎ513
" #
" #
œ
ln 3 ln ˆ 9/4 ‰‘ œ 1/4 "Î#
" 1 x #
(b) tanh"
5 4
dx œ ctanh" xd ! " #
œ
(1/2) ln Š 11 (1/2) ‹ œ
dx xÈ1 16x#
œ '4Î5
12Î13
œ tanh" " #
œ c sech" ud 4Î5 œ sech" (b) sech"
12 13
sech"
œ ln Š 13
72. (a) (b)
73. (a)
'12 " #
dx xÈ 4 x#
ˆcsch"
'01
È169 144 ‹ 1#
" 3
tanh" 0 œ tanh"
" #
where u œ 4x, du œ 4 dx, a œ 1
12 13
œ ln Š
4 5
ln
ln 3
du , u È a# u#
12Î13
" #
" #
5 4
sech"
4 5
1È1 (12/13)# ‹ (12/13)
ln Š 5
È25 16 ‹ 4
ln Š
1È1 (4/5)# ‹ (4/5)
œ ln ˆ 5 4 3 ‰ ln ˆ 1312 5 ‰ œ ln 2 ln
# œ "# csch" ¸ x# ¸‘ " œ "# ˆcsch" 1 csch" "# ‰ œ " #
csch" 1‰ œ
cos x È1 sin# x
dx œ '0
0
" #
’ln Š2
" È 1 u#
È5/4 (1/2) ‹
ln Š1 È2‹“ œ
" #
ˆcsch"
" #
" #
5 ln Š 21 ‹ È2
3 #
œ ln ˆ2 † 23 ‰ œ ln
4 3
csch" 1‰
È
!
du œ csinh" ud ! œ sinh" 0 sinh" 0 œ 0, where u œ sin x, du œ cos x dx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.7 Hyperbolic Functions 441 (b) sinh" 0 sinh" 0 œ ln Š0 È0 1‹ ln Š0 È0 1‹ œ 0
74. (a)
'1e xÈ1 dx(ln x) œ
œ '0
1
#
" csinh" ud !
du È a# u#
, where u œ ln x, du œ
" x
dx, a œ 1
œ sinh" 1 sinh" 0 œ sinh" 1
(b) sinh" 1 sinh" 0 œ ln Š1 È1# 1‹ ln Š0 È0# 1‹ œ ln Š1 È2‹ f(x) f(x) and O(x) œ f(x) #f(x) . # f(x) f((x)) œ f(x) #f(x) œ E(x) #
75. Let E(x) œ E(x) œ
Then E(x) O(x) œ
f(x) f(x) #
Ê E(x) is even, and O(x) œ
f(x) f(x) œ 2f(x) # # œ f(x). f(x) f((x)) œ f(x) #f(x) #
Also, œ O(x)
Ê O(x) is odd. Consequently, f(x) can be written as a sum of an even and an odd function. f(x) œ
f(x) f(x) #
because
Thus, if f is even f(x) œ
f(x) f(x) œ # 2f(x) # 0 and
if f is odd, f(x) œ 0
76. y œ sinh" x Ê x œ sinh y Ê x œ Ê ey œ
2x „ È4x# 4 #
f(x) f(x) # 2f(x) #
0 if f is even and f(x) œ
ey ecy #
Ê 2x œ ey
" ey
because
f(x) f(x) #
œ 0 if f is odd.
Ê 2xey œ e2y 1 Ê e2y 2xey 1 œ 0
Ê ey œ x Èx# 1 Ê sinh" x œ y œ ln Šx Èx# 1‹ . Since ey 0, we cannot
choose ey œ x Èx# 1 because x Èx# 1 0. É gk 77. (a) v œ É mg k tanhŒ m t Ê
dv dt
# É gk # É gk É gk œ É mg k ”sech Œ m t•Œ m œ g sech Œ m t.
# É gk # É gk # Thus m dv dt œ mg sech Œ m t œ mgŒ" tanh Œ m t œ mg kv . Also, since tanh x œ ! when x œ !, v œ !
when t œ !. (b)
mg mg mg É kg lim v œ lim É mg lim tanh ŒÉ kg k tanh Œ m t œ É k m t œ É k (1) œ É k
tÄ_
tÄ_
tÄ_
160 (c) É 0.005 œ É 160,000 œ 5
400 È5
œ 80È5 ¸ 178.89 ft/sec
78. (a) s(t) œ a cos kt b sin kt Ê #
ds dt
œ ak sin kt bk cos kt Ê
œ ak# cos kt bk# sin kt
œ k (a cos kt b sin kt) œ k s(t) Ê acceleration is proportional to s. The negative constant k# implies that the acceleration is directed toward the origin. (b) s(t) œ a cosh kt b sinh kt Ê
#
d# s dt#
ds dt
œ ak sinh kt bk cosh kt Ê
d# s dt#
œ ak# cosh kt bk# sinh kt
œ k# (a cosh kt b sinh kt) œ k# s(t) Ê acceleration is proportional to s. The positive constant k# implies that the acceleration is directed away from the origin. 79. V œ 1'0 acosh# x sinh# xb dx œ 1'0 1 dx œ 21 2
2
80. V œ 21 '0
ln
81. y œ
" #
È3
sech# x dx œ 21 ctanh xd ln0
È3
È3 Š1/È3‹
œ 21 – È
cosh 2x Ê yw œ sinh 2x Ê L œ '0
œ ’ "# Š e
ln
2x
ec2x ‹“ # 0
ln
È5
œ
" 4
ˆ5 "5 ‰ œ
È5
3 Š1/È3‹ —
œ1
È1 (sinh 2x)# dx œ ' 0
ln
È5
ln cosh 2x dx œ "# sinh 2x‘ 0
6 5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
È5
442
Chapter 7 Transcendental Functions
82. (a) x lim tanh x œ x lim Ä_ Ä_
ex ecx ex ecx
(b) x Ä lim tanh x œ x Ä lim _ _
œ x lim Ä_
ex ecx ex ecx
ex e1x ex e1x
ex e1x ex e1x
œxÄ lim _
e ecx
ˆex e1x ‰ ˆex e1x ‰
œ x lim Ä_
œxÄ lim _
†
1 ex 1 ex
œ x lim Ä_
ˆex e1x ‰ ˆex e1x ‰
†
ex ex
1 1 e2x
1 1 e2x
œ
10 10
e2x 1 e2x 1
œxÄ lim _
œ1 01 01
œ
œ 1
e ˆ e 2e1x ‰ œ _ 0 œ _ (c) x lim sinh x œ x lim œ x lim œ x lim 2 2 Ä_ Ä_ Ä_ Ä_ 2 x cx ex ecx ˆ e e2 ‰ œ 0 _ œ _ (d) x Ä lim sinh x œ x Ä lim œxÄ lim 2 _ _ _ 2 1 ex
x
x
x
1 ex 1 ex
(e) x lim sech x œ x lim Ä_ Ä_
2 ex ecx
œ x lim Ä_
2 ex e1x
†
(f) x lim coth x œ x lim Ä_ Ä_
ex ecx ex ecx
œ x lim Ä_
ex e1x ex e1x
œ x lim Ä_
lim coth x œ lim b xÄ0
(g)
x Ä 0b
lim coth x œ lim c xÄ0
(h)
x Ä 0c
ex ecx ex ecx ex ecx ex ecx
(i) x Ä lim csch x œ x Ä lim _ _ 83. (a) y œ
H w
œ x lim Ä_
2 ex
œ
1 1 e2x
ˆex e1x ‰ ˆex e1x ‰
†
1 ex 1 ex
œ!
0 10
œ x lim Ä_
œ lim b xÄ0
ex e1x ex e1x
†
ex ex
œ lim b xÄ0
e2x 1 e2x 1
œ _
œ lim c xÄ0
e ex
†
ex ex
œ lim c xÄ0
e2x 1 e2x 1
œ _
2 ex ecx
‰ cosh ˆ w H x Ê tan 9 œ
x
1 ex 1 ex
œxÄ lim _
dy dx
2 ex e1x
†
x
e ex
œxÄ lim _
2ex e2x 1
œ
1 1 e2x
1 1 e2x
0 01
œ
10 10
œ1
œ!
‰ w ˆ w ‰‘ œ sinh ˆ w ‰ œ ˆH w H sinh H x H x
(b) The tension at P is given by T cos 9 œ H Ê T œ H sec 9 œ HÈ1 tan# 9 œ HÉ1 ˆsinh
w H
# x‰
‰ ˆH‰ ˆw ‰ œ H cosh ˆ w H x œ w w cosh H x œ wy " a
84. s œ œ
" a
sinh ax Ê sinh ax œ as Ê ax œ sinh" as Ê x œ
Èsinh# ax 1 œ
" a
Èa# s# 1 œ És#
85. To find the length of the curve: y œ
" a
œ a"# sinh ax‘ 0 œ
" a#
" a
cosh ax œ
Ècosh# ax
" a#
b
b
b
" a
sinh" as; y œ
cosh ax Ê yw œ sinh ax Ê L œ '0 È1 (sinh ax)# dx
Ê L œ '0 cosh ax dx œ "a sinh ax‘ 0 œ b
" a
" a
sinh ab. The area under the curve is A œ '0
b
" a
sinh ab œ ˆ "a ‰ ˆ "a sinh ab‰ which is the area of the rectangle of height
cosh ax dx " a
and length L
as claimed, and which is illustrated below.
86. (a) Let the point located at (cosh uß 0) be called T. Then A(u) œ area of the triangle ?OTP minus the area under the curve y œ Èx# 1 from A to T Ê A(u) œ (b) A(u) œ œ
" #
" #
cosh u sinh u '1
cosh u
cosh# u
(c) Aw (u) œ
" #
" #
Èx# 1 dx Ê Aw (u) œ
sinh# u sinh# u œ
Ê A(u) œ
u #
" #
" #
cosh u sinh u " #
'1cosh u Èx# 1 dx.
acosh# u sinh# ub ŠÈcosh# u 1‹ (sinh u)
acosh# u sinh# ub œ ˆ "# ‰ (1) œ
" #
C, and from part (a) we have A(0) œ 0 Ê C œ 0 Ê A(u) œ
u #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ê u œ 2A
Section 7.8 Relative Rates of Growth
443
7.8 RELATIVE RATES OF GROWTH lim x 3 œ lim e"x œ 0 x Ä _ ex xÄ_ # x$ sin# x x cos x 2x (b) slower, lim œ lim 3x 2 sin œ lim 6x 2excos 2x œ lim 6 4esin œ 0 by the x ex ex xÄ_ xÄ_ xÄ_ xÄ_ 2 6 4 sin 2x 10 2 "0 Sandwich Theorem because ex Ÿ Ÿ ex for all reals and lim ex œ 0 œ lim ex xÄ_ x Ä _ ex
1. (a) slower,
Š "# ‹ x "Î#
"Î#
Èx
lim œ lim xex œ lim x Ä _ ex xÄ_ xÄ_ 4x 4 ‰x ˆ (d) faster, lim ex œ lim œ _ since xÄ_ xÄ_ e (c) slower,
ex
œ
lim xÄ_
" #Èx ex
œ0
1
4 e
x
(e) slower,
lim xÄ_ (f) slower, lim xÄ_ (g) same,
lim xÄ_
2. (a) slower,
lim xÄ_
(b) slower,
lim xÄ_
Š "x ‹ ex
(c) slower,
xÎ2
e ex x
lim xÄ_ œ È0 œ 0
x lim ˆ 3 ‰ œ 0 since x Ä _ 2e œ lim ex"Î2 œ 0 xÄ_
œ
œ
ex
lim xÄ_
lim xÄ_
ex
Š e# ‹
(h) slower,
œ
Š 3# ‹
lim xÄ_ œ
log10 x ex
lim xÄ_
10x% 30x 1 ex x ln x x ex
œ
œ
lim xÄ_
È 1 x% ex
" #
œ
œ
" xex
" #
ln x (ln 10) ex
lim xÄ_
lim xÄ_
1
3 2e
œ
40x$ 30 ex
x (ln x 1) ex
" x
lim xÄ_
œ
œ
œ
(ln 10) ex
lim xÄ_
lim xÄ_
lim xÄ_
"20x# ex
œ
240x ex
lim xÄ_
ln x 1 x Š "x ‹ ex
" (ln 10)xex
œ
lim xÄ_
œ0 œ
240 ex
lim xÄ_
ln x 1 1 ex
œ
œ0
lim xÄ_
œ0
œ É lim xÄ_
1 x% e2x
œ É lim xÄ_
4x$ 2e2x
œ É lim xÄ_
12x# 4e2x
œ É lim xÄ_
24x 8e2x
œ É lim xÄ_
x
Š5‹
x # 5 lim œ lim ˆ #5e ‰ œ 0 since 2e 1 x Ä _ ex xÄ_ ecx " (e) slower, lim œ lim e2x œ 0 x Ä _ ex xÄ_ x (f) faster, lim xeex œ lim x œ _ xÄ_ xÄ_ (g) slower, since for all reals we have 1 Ÿ cos x Ÿ 1 Ê e" Ÿ ecos x Ÿ e" Ê
(d) slower,
e "
e"
ec" ex
Ÿ
ecos x ex
lim œ 0 œ lim ex , so by the Sandwich Theorem we conclude that lim x Ä _ ex xÄ_ xÄ_ x 1 (h) same, lim e ex œ lim eÐx "x 1Ñ œ lim "e œ "e xÄ_ xÄ_ xÄ_ 3. (a) same,
lim xÄ_ (b) faster, lim xÄ_ (c) same,
lim xÄ_
(d) same,
lim xÄ_
x# 4x x#
x& x# x#
lim 2x 4 œ lim x Ä _ 2x xÄ_ œ lim ax$ 1b œ _ xÄ_
œ
È x% x$ x# (x 3)# x#
œ É lim xÄ_
œ
lim xÄ_
x% x$ x%
2(x 3) #x
œ
Ÿ e
cos x
ex
œ1
2 #
œ É lim ˆ1 "x ‰ œ È1 œ 1 xÄ_ lim xÄ_
2 #
œ1
Š"‹
x lim x ln x œ lim lnxx œ lim œ0 x Ä _ x# xÄ_ xÄ_ 1 x x # x 2) 2 (f) faster, lim 2x# œ lim (ln2x œ lim (ln 2)# 2 œ _ xÄ_ xÄ_ xÄ_ $ x (g) slower, lim x xe# œ lim exx œ lim e"x œ 0 xÄ_ xÄ_ xÄ_ # (h) same, lim 8x œ lim 8 œ 8 x Ä _ x# xÄ_
(e) slower,
# È
x x " ‰ lim œ lim ˆ1 x$Î# œ1 x Ä _ x# xÄ_ # (b) same, lim "0x œ lim 10 œ 10 x Ä _ x# xÄ_ x# e x " (c) slower, lim œ lim œ0 x Ä _ x# x Ä _ ex
4. (a) same,
ln x ex
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
e" ex
and also
œ0
24 16e2x
444
Chapter 7 Transcendental Functions #
Š ln x ‹
#
ln 10 lim log10 x œ lim œ ln"10 lim x Ä _ x# x Ä _ x# xÄ_ $ # (e) faster, lim x x#x œ lim (x 1) œ _ xÄ_ xÄ_
(d) slower,
œ
2 ln x x#
2 ln 10
lim xÄ_
Š x" ‹ 2x
" ln 10
œ
lim xÄ_
" x#
œ0
x
" Š 10 ‹
lim œ lim 10"x x# œ 0 x Ä _ x# xÄ_ x x (g) faster, lim (1.1) œ lim (ln 1.1)(1.1) œ lim #x x Ä _ x# xÄ_ xÄ_ # ‰ (h) same, lim x x100x œ lim ˆ1 100 œ 1 # x xÄ_ xÄ_ (f) slower,
5. (a) same,
lim xÄ_
log3 x ln x
(b) same,
lim xÄ_
ln 2x ln x
(c) same,
lim xÄ_
(d) faster,
lim xÄ_
(e) faster, (f) same,
œ œ
ln Èx ln x
x Š ln ln 3 ‹
lim xÄ_
ln x
ˆ #2x ‰ ˆ x" ‰
lim xÄ_
œ
œ
lim xÄ_
lim xÄ_
x ln x
œ
lim xÄ_
lim xÄ_
5 ln x ln x
œ
Š"‹
lim xÄ_
" ln 3
œ
" ln 3
" #
œ
" #
œ_
œ1
Š "# ‹ ln x
lim xÄ_
Èx ln x
œ
(ln 1.1)# (1.1)x #
œ
ln x
lim xÄ_
Š "# ‹ x "Î#
x"Î# ln x
œ
lim xÄ_
" Š "x ‹
œ
lim x œ _ xÄ_
Š x" ‹
œ
lim xÄ_
x #Èx
œ
Èx #
lim xÄ_
œ_
lim 5 œ 5 xÄ_
x lim œ lim x ln" x œ 0 x Ä _ ln x xÄ_ x x (h) faster, lim lne x œ lim ˆe" ‰ œ lim xex œ _ xÄ_ xÄ_ x xÄ_
(g) slower,
6. (a) same,
lim xÄ_
(b) same,
lim xÄ_
lim xÄ_
(d) slower,
lim xÄ_
lim xÄ_
(g) slower,
lim xÄ_
lim xÄ_
œ
lim xÄ_
œ
lim xÄ_
ln x " ‹ x#
ln x
ecx ln x
œ
œ
" ln #
œ
ln x
lim xÄ_
Š È"x ‹ Š
Š lnlnx2 ‹
lim xÄ_
œ
x2 ln x ln x
lim xÄ_
(f) slower,
(h) same,
œ
log10 10x ln x
(c) slower,
(e) faster,
#
log2 x# ln x
Š lnln10x 10 ‹ ln x
œ
" ˆÈx‰ (ln x) " x# ln x
lim xÄ_
" ln 10
ln x# ln x
lim xÄ_
" ex ln x
ln (ln x) ln x
œ
lim xÄ_
ln (2x5) ln x
œ
lim xÄ_
" ln #
ln 10x ln x
œ
lim xÄ_ " ln 10
2 ln x ln x
lim xÄ_
œ
" ln #
"0 Š 10x ‹
Š x" ‹
lim 2 œ xÄ_ œ
" ln 10
2 ln #
lim 1 œ xÄ_
" ln 10
œ0
œ0
lim ˆ x 2‰ œ Š lim x Ä _ ln x xÄ_
lim xÄ_
œ
x ln x ‹
2 œ lim xÄ_
" Š "x ‹
2 œ Š lim x‹ 2 œ _ xÄ_
œ0
"/x Š ln x‹
Š x" ‹ Š 2x2 5 ‹ Š x" ‹
œ
lim xÄ_
œ
" ln x
œ0
2x 2x5
lim xÄ_
œ
lim xÄ_
2 #
œ
lim 1 œ 1 xÄ_ x
7.
e lim œ lim exÎ2 œ _ Ê ex grows faster than exÎ2 ; since for x ee we have ln x e and lim (lnexx) x Ä _ exÎ2 x Ä _ xÄ_ x x x œ lim ˆ lne x ‰ œ _ Ê (ln x)x grows faster than ex ; since x ln x for all x 0 and lim (lnxx)x œ lim ˆ lnxx ‰ xÄ_ xÄ_ xÄ_ œ _ Ê xx grows faster than (ln x)x . Therefore, slowest to fastest are: exÎ2 , ex , (ln x)x , xx so the order is d, a, c, b
8.
2) lim (ln 2) œ lim (ln (ln 2))(ln œ lim (ln (ln 2))# (ln 2) œ (ln (ln# 2)) lim (ln 2)x œ 0 #x x Ä _ x# xÄ_ xÄ_ xÄ_ # Ê (ln 2)x grows slower than x# ; lim x2x œ lim (ln2x#)2x œ lim (ln 2)2 # #x œ 0 Ê x# grows slower than 2x ; xÄ_ xÄ_ xÄ_ x x lim 2ex œ lim ˆ 2e ‰ œ 0 Ê 2x grows slower than ex . Therefore, the slowest to the fastest is: (ln 2)x , x# , 2x xÄ_ xÄ_ and ex so the order is c, b, a, d
x
x
x
#
x
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.8 Relative Rates of Growth lim x œ 1 xÄ_ x (b) false; lim x x 5 œ 1" œ 1 xÄ_ (c) true; x x 5 Ê xx 5 1 if x 1 (or sufficiently large)
9. (a) false;
(d) true; x 2x Ê (e) true;
lim xÄ_
(f) true;
x ln x x
(g) false;
È x# 5 x
(h) true;
Šx " 3‹
10. (a) true;
Š x" ‹
Š "x
(b) true;
œ lim xÄ0
œ1
œ
"‹
x#
œ
" x
(e) true;
Š "x
" ‹ x# " Šx‹
œ1
(f) true;
lim xÄ_ ln (ln x) (g) true; ln x (h) false;
lim xÄ_
x ex
Š "x ‹
œ
Š #2x ‹
œ1
2 if x 1 (or sufficiently large)
lim 1 œ 1 xÄ_ 6 if x 1 (or sufficiently large)
5 x
lim ˆ1 "x ‰ œ 1 xÄ_
œ
and
2 cos x # x x e Ä
Ÿ
3 #
if x is sufficiently large
0 as x Ä _ Ê 1
lim ln x œ lim xÄ_ x xÄ_ œ 1 if x is sufficiently large
x ln x x# ln x ln x
x5 x
" Èx
œ1
2 if x 1 (or sufficiently large)
(d) true; 2 cos x Ÿ 3 Ê ex x ex
Èx x
1 if x 1 (or sufficiently large)
œ1
lim xÄ_
œ0
lim xÄ_
È(x 5)# x
x x3
" ex
1
ln x x
ln x ln 2x
Š "x ‹
(c) false;
ex e2x
lim xÄ_
1 if x 1 (or sufficiently large)
x 2x
œ
ln x ln ax# 1b
œ
lim xÄ_
Š "x ‹ Š
2x
x#
1
‹
œ
Š "x ‹ 1
x ex
2 if x is sufficiently large
œ0
lim xÄ_
x# " #x #
œ
lim ˆ " xÄ_ #
lim xÄ_
f(x) g(x)
œLÁ0 Ê
lim xÄ_
f(x) L¹ 1 if x is sufficiently large Ê L 1 ¹ g(x)
f(x) g(x)
L1 Ê
f(x) g(x)
11. If f(x) and g(x) grow at the same rate, then
Ê f œ O(g). Similarly,
g(x) f(x)
" ‰ #x#
g(x) f(x)
œ
œ " L
" #
Á 0. Then
Ÿ kLk 1 if x is sufficiently large
Ÿ ¸ L" ¸ 1 Ê g œ O(f).
12. When the degree of f is less than the degree of g since in that case
lim xÄ_
f(x) g(x)
œ 0.
f(x) lim œ 0 when the degree of f is smaller x Ä _ g(x) (the ratio of the leading coefficients) when the degrees are the same.
13. When the degree of f is less than or equal to the degree of g since than the degree of g, and
lim xÄ_
f(x) g(x)
œ
a b
14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the same degree grow at the same rate.
15.
lim xÄ_ œ
16.
ln (x ") ln x
lim xÄ_
lim xÄ_
œ
x x 999
ln (x a) ln x
œ
lim xÄ_
Šx " 1‹
Š x "999 ‹ œ lim xx 1 œ lim "1 œ 1 and lim ln (xlnx999) œ lim " xÄ_ xÄ_ xÄ_ x Ä _ Š x" ‹ Šx‹
œ1
lim xÄ_
Šx " a‹ Š x" ‹
œ
lim xÄ_
x x a
œ
lim xÄ_
" 1
œ 1. Therefore, the relative rates are the same.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
445
446 17.
18.
19.
Chapter 7 Transcendental Functions
È10x 1 È x 1 lim œ É lim "0xx 1 œ È10 and lim œ É lim x x 1 œ È1 œ 1. Since the growth rate Èx xÄ_ xÄ_ x Ä _ Èx xÄ_ È È is transitive, we conclude that 10x 1 and x 1 have the same growth rate ˆthat of Èx‰ . È x% x x#
È
%
$
x x œ É lim x x% x œ 1 and lim œ É lim x x% x œ 1. Since the growth rate is x# xÄ_ xÄ_ xÄ_ È % % $ È transitive, we conclude that x x and x x have the same growth rate athat of x# b .
lim xÄ_
lim xÄ_
xn ex
œ
%
nxnc1 ex
lim xÄ_
œá œ
n! ex
lim xÄ_
%
$
œ 0 Ê xn œ o aex b for any non-negative integer n
20. If p(x) œ an xn an1 xn1 á a" x a! , then
nc1
n
lim p(x) œ an lim xex an1 lim xex á x Ä _ ex xÄ_ xÄ_ x " a" lim ex a! lim ex where each limit is zero (from Exercise 19). Therefore, lim p(x) œ0 xÄ_ xÄ_ x Ä _ ex x Ê e grows faster than any polynomial.
21. (a)
lim xÄ_
x1În ln x
œ
lim xÄ_
xÐ1 nÑÎn n ˆ "x ‰
œ ˆ "n ‰
lim x1În œ _ Ê ln x œ o ˆx1În ‰ for any positive integer n xÄ_ '
(b) ln ae17ß000ß000 b œ 17,000,000 Še"(‚"! ‹
1Î10'
œ e"( ¸ 24,154,952.75
(c) x ¸ 3.430631121 ‚ 10"& (d) In the interval c3.41 ‚ 10"& ß 3.45 ‚ 10"& d we have ln x œ 10 ln (ln x). The graphs cross at about 3.4306311 ‚ 10"& .
22.
lim xÄ_
œ
ln x an xn anc1 xn 1 á a" x a!
lim xÄ_
lim Š ln x ‹ xÄ_ xn anc1 Ša n x á
a" xn 1
a! xn
‹
œ
1Îx
lim ’ xÄ_ nxn an
1“
œ
lim xÄ_
" aa n b a n x n b
œ0
Ê ln x grows slower than any non-constant polynomial (n 1) 23. (a)
lim nÄ_
n log2 n n alog2 nb#
œ
slower than n (log2 n)# ; Š "n ‹
(b) œ 0 Ê n log2 n grows
n log2 n n$Î#
lim nÄ_
œ
lim nÄ_
n Š ln ln 2 ‹
n"Î#
" lim œ ln2# lim n"Î# œ0 n Ä _ ˆ "# ‰ n "Î# nÄ_ $Î# Ê n log2 n grows slower than n . Therefore, n log2 n grows at the slowest rate Ê the algorithm that takes O(n log2 n) steps is the most efficient in the long run.
œ
" ln #
" log2 n
lim nÄ_
#
24. (a)
lim nÄ_
(log2 n)# n
œ
lim nÄ_
œ
2 (ln 2)#
œ
2(ln n) Š "n ‹ (ln 2)#
lim nÄ_ than n; lim nÄ_ œ
lim nÄ_
n Š ln ln 2 ‹
n"Î#
œ
Š "n ‹ 1
œ
lim nÄ_
lim nÄ_
(ln n)# n(ln 2)#
(b)
ln n n
œ 0 Ê (log2 n)# grows slower
" ln #
Š"‹
n
2 (ln 2)#
(log2 n)# Èn log2 n
œ
n Š ln ln 2 ‹
œ
lim nÄ_
lim nÄ_
log2 n Èn
ln n n"Î#
n " lim œ ln2# lim n"Î# œ 0 Ê (log2 n)# grows slower than Èn log2 n. Therefore (log2 n)# grows x Ä _ ˆ "# ‰ n "Î# nÄ_ at the slowest rate Ê the algorithm that takes O a(log2 n)# b steps is the most efficient in the long run.
œ
" ln #
lim nÄ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 7 Practice Exercises 25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because 2"* œ 524,288 1,000,000 1,048,576 œ 2#! . 26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because 2") œ 262,144 450,000 524,288 œ 2"* . CHAPTER 7 PRACTICE EXERCISES 1. y œ 10ecxÎ5 Ê 3. y œ
" 4
" 16
xe4x
dy dx
œ (10) ˆ 5" ‰ e xÎ5 œ 2e xÎ5 dy dx
œ
" 4
Ê
dy dx
œ x# ca2x# b e
e4x Ê
cx a4e4x b e4x (1)d
4. y œ x# ec2Îx œ x# e
2x "
5. y œ ln asin# )b Ê
dy d)
œ
2(sin ))(cos )) sin# )
6. y œ ln asec# )b Ê
dy d)
œ
2(sec ))(sec ) tan )) sec# )
#
#
7. y œ log2 Š x# ‹ œ
ln Š x# ‹ ln #
8. y œ log5 (3x 7) œ 9. y œ 8ct Ê
dy dt
11. y œ 5x$Þ' Ê 12. y œ È2 x
Ê
dy dx
ln (3x7) ln 5
2. y œ È2 e
œ
Ê
2 cos ) sin )
œ
" ln #
2x "
" 16
de
2x "
(2x) œ (2 2x)e
dy dx
Š x# ‹ œ x #
œ 8ct (ln 8)(1) œ 8ct (ln 8)
œ 2e 2Îx (1 x)
3 (ln 5)(3x7)
10. y œ 92t Ê
dy dt
œ 92t (ln 9)(2) œ 92t (2 ln 9)
œ 5(3.6)x#Þ' œ 18x#Þ'
dy dx
È2 Ê
dy dx
œ ŠÈ2‹ ŠÈ2‹ xŠ
È 2 1‹
œ 2xŠ
È 2 1‹
Ê y w œ # ln" x ˆ "# ‰ ln (ln x)‘ 2 (ln x)xÎ2 œ (ln x)xÎ2 ln (ln x) 15. y œ sin" È1 u# œ sin" a1 u# b œ
2x "
2 (ln 2)x
w
y y
œ (x 2)xb2 cln (x 2) 1d
u uÈ 1 u#
È2x œ 2eÈ2x
œ 2 tan )
" È 1 u#
"Î#
Ê
dy du
œ
" #
a1 u # b
1 Î2
" ‰ œ (x 2) ˆ x# (1) ln (x 2)
w
14. y œ 2(ln x)xÎ2 Ê ln y œ ln c2(ln x)xÎ2 d œ ln (2) ˆ x# ‰ ln (ln x) Ê
œ
œ ŠÈ 2 ‹ Š È 2 ‹ e
œ 2 cot )
13. y œ (x 2)xb2 Ê ln y œ ln (x 2)xb2 œ (x 2) ln (x 2) Ê Ê
dy dx
a4e4x b œ xe4x 4" e4x 4" e4x œ xe4x
œ ˆ ln"5 ‰ ˆ 3x37 ‰ œ
dy dx
È2x Ê
y y
ˆ"‰
œ 0 ˆ x# ‰ ’ lnx x “ (ln (ln x)) ˆ "# ‰
" ‘ ln x
(2u)
# "Î# Ê1 ’a1 u# b “
œ
u È 1 u # È 1 a1 u # b
œ
u ku k È 1 u #
,0u1
16. y œ sin" Š È"v ‹ œ sin" v"Î# Ê
17. y œ ln acos" xb Ê y w œ
Š È c" 1
x#
cosc" x
‹
dy dv
œ
œ
"
#v
c$Î#
É1 av "Î# b#
œ
" 2v$Î# È1 v "
œ
"
2v$Î# É v
1 v
œ
È v 2v$Î# Èv 1
œ
" È1 x# cos " x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" 2vÈv 1
447
448
Chapter 7 Transcendental Functions
18. y œ z cos" z È1 z# œ z cos" z a1 z# b œ cos" z
z È 1 z#
z È 1 z#
"Î#
Ê
dy dz
œ cos" z
dy dt
20. y œ a1 t# b cot" 2t Ê
œ 2t cot" 2t a1 t# b ˆ 1 24t# ‰
œ tan" t t ˆ 1 " t# ‰ ˆ #" ‰ ˆ "t ‰ œ tan" t
21. y œ z sec" z Èz# 1 œ z sec" z az# 1b œ
z kz k È z # 1
z È z# 1
ˆ "# ‰ a1 z# b
"Î#
(2z)
œ cos" z
19. y œ t tan" t ˆ "# ‰ ln t Ê dy dt
z È 1 z#
sec" z œ
1z È z# 1
"Î#
Ê
dy dz
œ zŠ
" ‹ kz k È z # 1
t 1 t#
" 2t
asec" zb (1) #" az# 1b
"Î#
(2z)
sec" z, z 1
22. y œ 2Èx 1 sec" Èx œ 2(x 1)"Î# sec" ˆx"Î# ‰ Ê
Š " ‹ xc"Î#
23. y œ csc" (sec )) Ê "x
24. y œ a1 x# b etan 25. y œ
sec " Èx
œ 2 –ˆ "# ‰ (x 1)"Î# sec" ˆx"Î# ‰ (x 1)"Î# È#xÈx 1 — œ 2 Š 2Èx 1
dy dx
2 ax# 1b Ècos 2x
dy d)
œ
sec ) tan ) ksec )k Èsec# ) 1
Ê y w œ 2xetan
"x
tan
#
2 ax # 1 b Ècos 2x
"! 3x 4 "! 3x 4 É 26. y œ É 2x 4 Ê ln y œ ln 2x 4 œ
" 10
"x
a1 x# b Š e1 x# ‹ œ 2xetan
Ê ln y œ ln Š 2Èaxcos2x1b ‹ œ ln (2) ln ax# 1b
Ê y w œ ˆ x#2x 1 tan 2x‰ y œ
Ê yw œ
) œ ktan tan )k œ 1, 0 )
ˆ 3x 3 4
" ‰ x2 y
" #
" #x ‹
œ
sec " Èx Èx 1
"x
etan
"x
ln (cos 2x) Ê
cln (3x 4) ln (2x 4)d Ê
"! 3x 4 ˆ " ‰ ˆ 3x 3 4 œÉ 2x 4 10
yw y
œ0
2x x # 1
2 sin 2x) ˆ #" ‰ (cos 2x
yw y
œ
" 10
ˆ 3x 3 4
2 ‰ 2x 4
" ‰ x2
&
1)(t 1) dy " 27. y œ ’ (t(t 2)(t 3) “ Ê ln y œ 5 cln (t 1) ln (t 1) ln (t 2) ln (t 3)d Ê Š y ‹ Š dt ‹
œ 5 ˆ t " 1 28. y œ Ê
" t1
2u2u Èu# 1 Ê ln y dy 2u2u ˆ " du œ Èu# 1 u È)
29. y œ (sin )) Ê
" t#
dy d)
" ‰ t3
Ê
dy dt
œ ln 2 ln u u ln 2 ln 2
&
1)(t 1) ˆ t " 1 œ 5 ’ (t(t 2)(t 3) “
" #
" t1
" t#
dy ln au# 1b Ê Š "y ‹ Š du ‹œ
" u
" ‰ t3
ln 2 #" ˆ u#2u 1 ‰
u ‰ u# 1
È) ˆ cos ) ‰ " )"Î# ln (sin )) Ê ln y œ È) ln (sin )) Ê Š "y ‹ Š dy d) ‹ œ sin ) #
œ (sin ))
È)
ŠÈ) cot )
ln (sin )) ‹ 2È )
30. y œ (ln x)1Îln x Ê ln y œ ˆ ln"x ‰ ln (ln x) Ê
w
y y
œ ˆ ln"x ‰ ˆ ln"x ‰ ˆ x" ‰ ln (ln x) ’ (ln"x)# “ ˆ x" ‰
ln (ln x) Ê y w œ (ln x)1Îln x ’ 1 x(ln x)# “
31.
" x
1 #
ˆ x#2x 1 tan 2x‰
" 10
' ex sin aex b dx œ ' sin u du, where u œ ex and du œ ex dx œ cos u C œ cos aex b C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 7 Practice Exercises 32.
' et cos a3et 2b dt œ 3" ' cos u du, where u œ 3et 2 and du œ 3et dt " 3
œ 33.
449
" 3
sin u C œ
sin a3et 2b C
' ex sec# aex 7b dx œ ' sec# u du, where u œ ex 7 and du œ ex dx œ tan u C œ tan aex 7b C
34.
' ey csc aey 1b cot aey 1b dy œ ' csc u cot u du, where u œ ey 1 and du œ ey dy œ csc u C œ csc aey 1b C
35.
' asec# xb etan x dx œ ' eu du, where u œ tan x and du œ sec# x dx œ eu C œ etan x C
36.
' acsc# xb ecot x dx œ ' eu du, where u œ cot x and du œ csc# x dx œ eu C œ ecot x C
' c1
37.
'c11
38.
'1e Èlnx x dx œ '01 u"Î# du, where u œ ln x, du œ "x dx; x œ 1
" " " 3x 4 dx œ 3 c7 u du, where u œ 3x 4, du œ 3 dx; x œ 1 Ê " " ln 7 œ "3 cln kukd " ( œ 3 cln k1k ln k7kd œ 3 [0 ln 7] œ 3
œ 39.
23 u$Î# ‘ " !
œ 23 1$Î# 23 0$Î# ‘ œ
Ê u œ 0, x œ e Ê u œ 1
2 3
sin ˆ ‰ " '01 tan ˆ x3 ‰ dx œ '01 cos ' 1Î2 " ˆx‰ ˆx‰ ˆ ‰ dx œ 3 1 u du, where u œ cos 3 , du œ 3 sin 3 dx; x œ 0 x 3 x 3
Ê uœ "Î#
œ 3 cln kukd " 40.
'11ÎÎ64
œ 3 ln ¸ "# ¸ ln k1k‘ œ 3 ln
2 cot 1x dx œ 2'1Î6
1Î4
cos 1x sin 1x
dx œ
2 1
È2
'11ÎÎ2
" u
" #
œ
'04
2t t# 25
2 1
È
cln kukd 11ÎÎ2 2 œ c9
dt œ 'c25
" u
2 1
œ ln 2 œ ln 8
du, where u œ sin 1x, du œ 1 cos 1x dx; x œ
’ln ¹ È"2 ¹ ln ¸ #" ¸“ œ
2 1
ln 1
" #
"Î#
43.
'
tan (ln v) v
2 1
#" ln 2‘ œ
ln 2 1
" u
9 25
du, where u œ 1 sin t, du œ cos t dt; t œ 1# Ê u œ 2, t œ
1 6
Ê uœ
œ ln ¸ "# ¸ ln k2k‘ œ ln 1 ln 2 ln 2 œ 2 ln 2 œ ln 4
dv œ ' tan u du œ '
sin u cos u
du, where u œ ln v and du œ
" v
dv
œ ln kcos uk C œ ln kcos (ln v)k C 44.
'
" v ln v
dv œ '
" u
Ê uœ
du, where u œ t# 25, du œ 2t dt; t œ 0 Ê u œ 25, t œ 4 Ê u œ 9
'c11ÎÎ62 1 cossint t dt œ '21Î2 œ cln kukd #
" 6
" È2
ln 2 ln 1 ln 2‘ œ
œ cln kukd * #& œ ln k9k ln k25k œ ln 9 ln 25 œ ln 42.
Ê u œ 1, x œ 1
" # $
Ê uœ
41.
u œ 7, x œ 1 Ê u œ 1
du, where u œ ln v and du œ
" v
dv
œ ln kuk C œ ln kln vk C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" #
" #
,xœ
" 4
450 45.
Chapter 7 Transcendental Functions
'
(ln x)c$ x
œ 46.
'
u # #
ln (x 5) x5
œ 47.
dx œ ' u$ du, where u œ ln x and du œ
'
" r
" #
C œ (ln x)
#
" x
dx
C
dx œ ' u du, where u œ ln (x 5) and du œ #
Cœ
u #
cln (x5)d 2
#
" x 5
dx
C
csc# (1 ln r) dr œ ' csc# u du, where u œ 1 ln r and du œ
" r
dr
œ cot u C œ cot (1 ln r) C 48.
'
cos (1 ln v) v
dv œ ' cos u du, where u œ 1 ln v and du œ "v dv
œ sin u C œ sin (1 ln v) C 49.
'
#
œ
50.
'
" #
x3x dx œ " # ln 3
'
3u du, where u œ x# and du œ 2x dx
a3u b C œ
" # ln 3
#
Š3x ‹ C
2tan x sec# x dx œ ' 2u du, where u œ tan x and du œ sec# x dx œ
" ln #
2tan x ln #
a2u b C œ
51.
'17
52.
" " È '132 5x" dx œ 5" '132 x" dx œ 5" cln kxkd $# " œ 5 aln 32 ln 1b œ 5 ln 32 œ ln Š 32‹ œ ln 2
53.
15 '14 ˆ x8 #"x ‰ dx œ #" '14 ˆ 4" x x" ‰ dx œ #" 8" x# ln kxk‘ %" œ #" ˆ 168 ln 4‰ ˆ 8" ln 1‰‘ œ 16 #" ln 4
ln È4 œ
15 16
ˆln 8
3 #
15 16
ln 2
12‰ œ
2 3
ˆln 8
21 ‰ #
œ
2 3
(ln 8) 7 œ ln ˆ8#Î$ ‰ 7 œ ln 4 7
'cc21 eÐx1Ñ dx œ '10 eu du, where u œ (x 1), du œ dx; x œ 2 œ ce
'c0ln 2
d !"
e2w dw œ œ
57.
dx œ 3 cln kxkd (" œ 3 aln 7 ln 1b œ 3 ln 7
#
2 3
u
56.
" x
'18 ˆ 3x2 x8 ‰ dx œ 23 '18 ˆ "x 12x# ‰ dx œ 23 cln kxk 12x" d )" œ 23 ˆln 8 128 ‰ (ln 1 12)‘ œ
55.
7
3 x
&
œ 54.
dx œ 3'1
C
" #
!
Ê u œ 1, x œ 1 Ê u œ 0
"
œ ae e b œ e 1 " #
'ln0Ð1Î4Ñ eu du, where u œ 2w, du œ 2 dw; w œ ln 2
ceu d 0ln Ð1Î4Ñ œ
" #
ce! eln Ð1Î4Ñ d œ
" #
ˆ1 4" ‰ œ
Ê u œ ln "4 , w œ 0 Ê u œ 0
3 8
'1ln 5 er a3er 1b$Î# dr œ "3 '416 u$Î# du, where u œ 3er 1, du œ 3er dr; r œ 0
Ê u œ 4, r œ ln 5 Ê u œ 16
"' œ 23 u"Î# ‘ % œ 23 ˆ16"Î# 4"Î# ‰ œ ˆ 23 ‰ ˆ 4" #" ‰ œ ˆ 23 ‰ ˆ 4" ‰ œ
58.
'0ln 9 e
)
" 6
ae) 1b"Î# d) œ '0 u"Î# du, where u œ e) 1, du œ e) d); ) œ 0 Ê u œ 0, ) œ ln 9 Ê u œ 8 8
œ
2 3
u$Î# ‘ ) œ !
2 3
ˆ8$Î# 0$Î# ‰ œ
2 3
ˆ2*Î# 0‰ œ
2""Î# 3
œ
32È2 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 7 Practice Exercises 59.
'1e "x (1 7 ln x)"Î$ dx œ 7" '18 u"Î$ du, where u œ 1 7 ln x, du œ x7 dx, x œ 1 œ
60.
'ee
#
" xÈln x
3 #Î$ ‘ ) 14 u "
œ
3 14
3 ‰ ˆ8#Î$ 1#Î$ ‰ œ ˆ 14 (4 1) œ
dx œ 'e (ln x)"Î# e#
Ê u œ 1, x œ e Ê u œ 8
9 14
dx œ '1 u"Î# du, where u œ ln x, du œ 2
" x
451
" x
dx; x œ e Ê u œ 1, x œ e# Ê u œ 2
# œ 2 u"Î# ‘ " œ 2 ŠÈ2 1‹ œ 2È2 2
61.
'13 [ln (vv 11)]
dv œ '1 [ln (v 1)]# 3
#
" v1
dv œ 'ln 2 u# du, where u œ ln (v 1), du œ ln 4
" v 1
dv;
v œ 1 Ê u œ ln 2, v œ 3 Ê u œ ln 4; œ 62.
" 3
" 3
$ ln 4
cu d ln 2 œ
$
$
c(ln 4) (ln 2) d œ
" 3
$
$
c(2 ln 2) (ln 2) d œ
(ln 2)$ 3
(8 1) œ
7 3
(ln 2)$
'24 (1 ln t)(t ln t) dt œ '24 (t ln t)(1 ln t) dt œ '24lnln24 u du, where u œ t ln t, du œ ˆ(t) ˆ "t ‰ (ln t)(1)‰ dt œ (1 ln t) dt; t œ 2 Ê u œ 2 ln 2, t œ 4 Ê u œ 4 ln 4 œ
63.
" #
cu# d 2 ln 2 œ 4 ln 4
" #
c(4 ln 4)# (2 ln 2)# d œ
c(8 ln 2)# (2 ln 2)# d œ
(2 ln 2)# #
(16 1) œ 30 (ln 2)#
'18 log) ) d) œ ln"4 '18 (ln )) ˆ ") ‰ d) œ ln"4 '0ln 8 u du, where u œ ln ), du œ ") d), ) œ 1 4
œ 64.
" #
'1e
" # ln 4
cu # d
ln 8 !
" ln 16
œ
d) œ '1
8(ln 3)(log3 )) )
c(ln 8)# 0# d œ
e
8(ln 3)(ln )) )(ln 3)
#
#
(3 ln 2)# 4 ln 2
œ
Ê u œ 0, ) œ 8 Ê u œ ln 8
9 ln 2 4
d) œ 8 '1 (ln )) ˆ ") ‰ d) œ 8'0 u du, where u œ ln ), du œ e
1
" )
d) ;
) œ 1 Ê u œ 0, ) œ e Ê u œ 1 œ 65.
'c33ÎÎ44 È
" 4 cu # d !
6 9 4x#
œ 4 a1 0 b œ 4 3Î4
3Î2
dx œ 3 ' 3Î4 È3# 2 (2x)# dx œ 3' 3Î2 È #" # du, where u œ 2x, du œ 2 dx; 3 u x œ 34 Ê u œ 3# , x œ
Ê uœ
3 4
$Î#
3 #
œ 3 sin" ˆ u3 ‰‘ $Î# œ 3 sin" ˆ "# ‰ sin" ˆ "# ‰‘ œ 3 16 ˆ 16 ‰‘ œ 3 ˆ 13 ‰ œ 1 66.
'c11ÎÎ55
6 È4 25x#
œ 67.
'c22
3 4 3t#
6 5
dx œ "
sin
6 5
' 11ÎÎ55
ˆ u2 ‰‘ " "
dt œ È3 'c2
5 È2# (5x)#
œ
6 5
"
sin
È3
È3 œ È3 "# tan" ˆ u2 ‰‘ c2È3 œ
'È33
69.
'
" 3 t#
dt œ 'È3
" yÈ4y# 1
3
" # ŠÈ3‹ t#
dy œ '
' 11 È
ˆ "# ‰
"
sin
" 2# u#
È3
ˆ
dt œ È3 'c2È3 2
#
ŠÈ3t‹
2
68.
6 5
du, where u œ 5x, du œ 5 dx; x œ 15 Ê u œ 1, x œ
2
2#
dx œ
" 2# u#
œ sec" kuk C œ sec
œ
6 5
16
ˆ
1 ‰‘ 6
œ
6 5
ˆ 13 ‰
œ
Ê uœ1
21 5
du, where u œ È3t, du œ È3 dt; t œ 2 Ê u œ 2È3, t œ 2 Ê u œ 2È3
È3 #
’tan" ŠÈ3‹ tan" ŠÈ3‹“ œ
dt œ ’ È"3 tan" Š Èt 3 ‹“
2 (2y)È(2y)# 1 "
" ‰‘ #
1 5
dy œ '
" uÈ u# 1
3
È3 œ
" È3
È3 #
13 ˆ 13 ‰‘ œ
Štan" È3 tan" 1‹ œ
" È3
1 È3
ˆ 13 14 ‰ œ
du, where u œ 2y and du œ 2 dy
k2yk C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
È 31 36
452
Chapter 7 Transcendental Functions
70.
'
71.
'È2Î23Î3
dy œ 24 '
24 yÈy# 16
" kyk È9y# 1
" yÈ y# 4#
dy œ 24 ˆ 4" sec" ¸ 4y ¸‰ C œ 6 sec" ¸ 4y ¸ C
3 'È2 dy œ 'È2Î3 k3yk È(3y) # 1 dy œ 2Î3
2
" ku k È u # 1
du, where u œ 3y, du œ 3 dy; yœ
œ csec" ud È2 œ ’sec" 2 sec" È2“ œ 2
72.
È È5
'cc2/È65Î
" kyk È5y# 3
È6ÎÈ5
dy œ 'c2/È5
1 3
1 4
È6
#
dy œ 'c2
#
È5y ÊŠÈ5y‹ ŠÈ3‹
Ê u œ È2, y œ
2 3
Ê uœ2
1 12
œ
È5
È2 3
" u Ê u #
#
du,
ŠÈ3‹
È
œ ’ È"3 sec" ¹ Èu3 ¹“ 73.
'
dx œ '
" È2x x#
cÈ6 c2
where u œ È5y, du œ È5 dy; y œ È25 Ê u œ 2, y œ È56 Ê u œ È6 œ
" È1 ax# 2x 1b
1 È3
’sec" È2 sec"
dx œ '
" È1 (x 1)#
2 È3 “
dx œ '
œ
" È3
" È 1 u#
ˆ 14 16 ‰ œ
" È3
3121
21 ‘ 1#
œ
1 12È3
È 3 1 36
œ
du, where u œ x 1 and du œ dx
"
"
œ sin 74.
'È
u C œ sin dx œ '
" x# 4x 1
(x 1) C
" È3 ax# 4x 4b
dx œ '
"
#
ÊŠÈ3‹ (x 2)#
dx œ '
"
du
#
ÊŠÈ3‹ u#
where u œ x 2 and du œ dx œ sin" Š Èu3 ‹ C œ sin" Š xÈ32 ‹ C 75.
'cc21 v 4v2 5 dv œ 2 'cc21 1 av " 4v 4b dv œ 2 'cc21 1 (v" 2) #
#
"
œ 2 ctan 76.
77.
'c11
3 4v# 4v 4
dv œ
"
œ 2 atan
3 4
'c11
2u ’ È23 tan" Š È ‹“ 3
œ
3 4
œ
È 31 4
'
" ud !
" (t 1)Èt# 2t 8
3 4
$Î# "Î#
dt œ '
Šv#
dv œ 2'0
1
#
" 1 u#
du,
where u œ v 2, du œ dv; v œ 2 Ê u œ 0, v œ 1 Ê u œ 1 1 ˆ 0b œ 2 4 0‰ œ 1#
1 tan
"
" v 4" ‹
dv œ
3 4
'c11
Œ
È3 #
#
"
Šv
#
" #‹
dv œ
3 4
'c31ÎÎ22
Œ
È3 #
"
du
#
u#
where u œ v "# , du œ dv; v œ 1 Ê u œ "# , v œ 1 Ê u œ œ
È3 #
’tan" È3 tan" Š È"3 ‹“ œ
" (t 1)Èat# 2t 1b 9
dt œ '
È3 #
" (t 1)È(t 1)# 3#
13 ˆ 16 ‰‘ œ
dt œ '
" uÈ u# 3#
È3 #
ˆ 261 16 ‰ œ
È3 #
†
1 #
du
where u œ t 1 and du œ dt œ 78.
'
" 3
"
sec
" (3t 1)È9t# 6t
¸ 3u ¸ C œ
dt œ '
" 3
"
sec
¸ t31 ¸ C
" (3t 1)Èa9t# 6t 1b 1
dt œ '
" (3t 1)È(3t 1)# 1#
dt œ
" 3
'
" uÈ u# 1
du
where u œ 3t 1 and du œ 3 dt œ
" 3
"
sec
ku k C œ
" 3
"
sec
k3t 1k C
79. 3y œ 2y1 Ê ln 3y œ ln 2y1 Ê y(ln 3) œ (y 1) ln 2 Ê (ln 3 ln 2)y œ ln 2 Ê ˆln 3# ‰ y œ ln 2 Ê y œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ln 2 ln Š 3# ‹
3 #
Chapter 7 Practice Exercises
453
80. 4y œ 3y2 Ê ln 4y œ ln 3y2 Ê y ln 4 œ ay 2b ln 3 Ê 2 ln 3 œ aln 3 ln 4by Ê aln 12by œ 2 ln 3 9 Ê y œ lnln12 x# 9
81. 9e2y œ x# Ê e2y œ
#
#
82. 3y œ 3 ln x Ê ln 3y œ ln (3 ln x) Ê y ln 3 œ ln (3 ln x) Ê y œ
#
ln Š x9 ‹ œ ln É x9 œ ln ¸ x3 ¸ œ ln kxk ln 3
ln 3 ln (ln x) ln 3
œ
ln (3 ln x) ln 3
#
" #
Ê ln e2y œ ln Š x9 ‹ Ê 2y(ln e) œ ln Š x9 ‹ Ê y œ
83. ln (y 1) œ x ln y Ê eln Ðy1Ñ œ eÐxln yÑ œ ex eln y Ê y 1 œ yex Ê y yex œ 1 Ê y a1 ex b œ 1 Ê y œ 84. ln (10 ln y) œ ln 5x Ê eln Ð10 ln yÑ œ eln 5x Ê 10 ln y œ 5x Ê ln y œ 85. lim
x # $x % x"
87. xlim Ä1
tan x x
xÄ"
89. lim
œ
sin# x
# x Ä ! tanax b
90. lim
sinamxb
x Ä ! sinanxb
91. lim
x Ä 1Î#c
92.
lim
x Ä !b
œ lim
xÄ"
tan 1 1
#x $ "
œ5
b x Ä " x "
œ!
x Ä ! x sin x
# # x Ä ! #x sec ax b
œ lim
xÄ!
m cosamxb n cosanxb
sina#xb
œ lim
# # x Ä ! #x sec ax b
œ
cosa$xb
x Ä 1Î#c cosa(xb
Èx sec x œ lim x Ä !b
Èx cos x
œ
! "
" cos x sin x
94. lim ˆ x"%
#
xÄ!
"‰ x#
# # # # # x Ä ! #x a#sec ax btanax b†#xb #sec ax b
$sina$xb
œ lim
x Ä 1Î#c (sina(xb
œ lim
sin x
x Ä ! cos x
œ
œ lim Š " x%x ‹ œ lim a" x# b † xÄ!
œ
" ""
œ
# ! #†"
œ"
xÄ!
! "
" x%
œ! "
œ lim a" x# b œ lim
% xÄ! x
xÄ!
œ"†_œ_
Èx# x "Èx# x È x# x " È x# x
x x
Notice that x œ Èx# for x ! so this is equivalent to œ x lim Ä_
#x
# Éx $
x
"
" É x # x x# x# x
$
œ x lim Ä_
96. x lim Š x#x " x#x " ‹ œ lim_ Ä_ xÄ "# " œ x lim œ x lim œ! Ä _ #%x Ä _ #x
œ
$ (
95. x lim ŠÈx# x " Èx# x‹ œ x lim Š È x # x " È x # x‹ † Ä_ Ä_ #x " œ x lim È # Ä_ È # x x"
œ
x Ä ! " cos x
œ!
93. lim acsc x cot xb œ lim xÄ!
a b
sec# x
œ lim
#cosa#xb
œ lim
œ
m n
seca(xbcosa$xb œ lim
xÄ!
axa " b " x Ä " bx
œ lim
tan x
88. lim
#sin x†cos x
œ lim
xa "
86. lim
Ê eln y œ exÎ2 Ê y œ exÎ2
x #
# x" œ È"# È" œ " " É" x x"# É" x"
x $ ax # " b x $ a x # " b ax# "bax# "b
œ x lim Ä_
97. The limit leads to the indeterminate form 00 : lim
10x 1 x
98. The limit leads to the indeterminate form 00 : lim
3) 1 )
xÄ0
)Ä0
œ lim
)Ä0
œ x lim Ä_
(ln 10)10x 1
œ lim xÄ0
2sin x 1 x x Ä 0 e 1
99. The limit leads to the indeterminate form 00 : lim
#x $ x% "
(ln 3)3) 1
œ lim xÄ0
'x # %x $
œ x lim Ä_
"#x "#x#
œ ln 10
œ ln 3
2sin x (ln 2)(cos x) ex
œ ln 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" #
" 1 ex
454
Chapter 7 Transcendental Functions 2c sin x " ex 1
100. The limit leads to the indeterminate form 00 : lim
xÄ0
œ lim
xÄ0
5 5 cos x
101. The limit leads to the indeterminate form 00 : lim
œ lim
x x Ä 0 e x1
x sin x2 tan3 x
102. The limit leads to the indeterminate form 00 : lim
xÄ0
œ lim
6x cos x2 4x3 sin x2
3 2 2 x Ä 0 12tan x sec x 6tan x sec x
6x cos x2 4x3 sin x2
œ lim
103. The limit leads to the indeterminate form 00 :
tÄ!
œ lim xÄ4
1 sin (21x) ex 4 1
21# cos (21x) ex 4
œ lim xÄ4
106. The limit leads to the indeterminate form lim
y Ä !
œ5
5 cos x ex
2x2 cos x2 sin x2 3tan4 x 3tan2 x
œ lim
xÄ0
4 2 2 2 2 x Ä 0 60tan x sec x 54tan x sec x 6sec x
œ
lim t Ä !
œ lim xÄ4
Š1 1 b2 2t ‹
2t
œ
6 6
œ1
œ _
21(sin 1x)(cos 1x) ex 4 1
œ 21 #
105. The limit leads to the indeterminate form 00 :
œ
œ ln 2
ˆ6 8x4 ‰cos x2 24x2 sin x2
œ lim
sin# (1x) x 43x e xÄ4
104. The limit leads to the indeterminate form 00 : lim
2x2 cos x2 sin x2 3tan2 x sec2 x
xÄ0
t ln (1 2t) t#
lim
œ lim xÄ0
5 sin x ex 1
xÄ0
œ lim
5 3 x Ä 0 12tan x 18tan x 6tan x
2c sin x (ln 2)( cos x) ex
Š et "t ‹ œ lim Š e " t ‹ œ lim t
lim
t Ä !
_ _:
t
tÄ!
lim e1Îy ln y œ
yÄ!
tÄ!
ln y eyc"
lim
yÄ!
œ
lim
yÄ!
et 1
œ1 y "
e y
" ˆy #‰
Š eyyc" ‹ œ 0
1‰ 107. Let f(x) œ ˆ eex 1
ln x
1‰ 1‰ Ê ln f(x) œ ln x ln ˆ eex ln f(x) œ x lim ln x ln ˆ eex 1 Ê x lim 1 ; this is limit is currently of Ä_ Ä_ x 1 exÎ2 e xÎ2 ˆx‰ the form 0 † _. Before we put in one of the indeterminate forms, we rewrite eex 1 œ exÎ2 e xÎ2 œ coth 2 ; the limit is x
x
ln cothˆ x2 ‰
lim ln x ln cothˆ x2 ‰ œ x lim xÄ_ Ä_ Î Ð œ x lim Ä_ Ï
csch2 Š x2 ‹ cothŠ x2 ‹
1 ln x
x
; the limit leads to the indeterminate form 00 : x lim Ä_
ˆ "# ‰ Ñ
1 2 ˆ x1 ‰ aln xb
ln cothˆ x2 ‰ 1 ln x
2
2xaln xbˆ x ‰ aln xb a b aln xb Ó œ x lim Š 2 sinhxˆ xln‰ xcoshˆ x ‰ ‹ œ x lim Š xsinh ‹ œ x lim Š ‹ x cosh x Ä_ Ä _ Ä _ 2 2 Ò 2
2
aln xb œ x lim Š 2ln xcosh ‹ œ x lim Š x Ä_ Ä_
1
2
2ˆ 1x ‰ 2aln xbˆ 1x ‰ ‹ sinh x
2
2ln x ‰ ˆ 2xsinh œ x lim Š x cosh xx sinh x ‹ x œ x lim Ä_ Ä_
2 1 ‰ln x ˆ ‰ œ 0 Ê lim ˆ eex œ x lim œ x lim eln fÐxÑ œ e0 œ 1 1 Ä _ x2 cosh x x sinh x xÄ_ Ä_ x
x 108. Let f(x) œ ˆ1 3x ‰ Ê ln f(x) œ x ln ˆ1 3x ‰ Ê
indeterminate form
109. (a) x lim Ä_
log2 x log3 x
_ _:
lim
x Ä !b
œ x lim Ä_
c# Š c3x " ‹ 1 3x
x Š ln ln 2 ‹ x Š ln ln 3 ‹
œ x lim Ä_
xc#
œ lim b xÄ!
œ x lim Ä_
x# x # 1
ln 3 ln 2
œ x lim Ä_
(b) x lim Ä_
x x Š "x ‹
(c) x lim Ä_ (d) x lim Ä_
ˆ x ‰ xex ex 100 œ x lim xe x œ x lim Ä _ 100x Ä _ 100 x tan " x œ _ Ê faster
2x #x
œ
lim
xÄ! 3x x3
csc " x Š "x ‹
œ x lim Ä_
sin " ax " b x "
xÄ!
œ0 Ê
lim
x Ä !b
the limit leads to the
ˆ1 3x ‰x œ lim eln fÐxÑ œ e! œ 1 x Ä !b
Ê same rate
ln 3 ln #
œ x lim " œ 1 Ê same rate Ä_
œ _ Ê faster Š x
(e) x lim Ä_
ln a1 3xc" b ; x "
ln f(x) œ lim
œ x lim Ä_
Ê1 Šx
#‹
x #
" ‹#
œ x lim Ä_
"
"
Ê 1 Š x# ‹
œ 1 Ê same rate
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 7 Practice Exercises
$
10x 2x ex
(c) x lim Ä_
Š "x ‹
(b)
" ‹ x% " Š #‹ x
Š
" x#
œ x lim Ä_
tan " ax " b x "
" " lim sin x a#x b xÄ_
" x#
œ1
" ‹ x% " Š %‹ x
"ec2x #
Š ex b2ecx ‹ ecx
œ x lim Ä_ œ x lim Ä_ œ x lim Ä_
œ x lim Ä_
œ
" #
Ê same rate
"# ‰ œ
60x4 ex Š x 1x
Ê same rate
œ x lim Ä_
# #‹
x # x
É1
" #
ˆx
#
œ x lim Ä_
"‰#
2x $
2 ecx aex ecx b
60 ex
œ 0 Ê slower " 1 "#
œ 1 Ê same rate
x
œ x lim Ä_
x 2É1 x"#
œ _ Ê faster
ˆ 2 ‰ œ 2 Ê same rate œ x lim Ä _ 1 ec2x
Ÿ 2 for x sufficiently large Ê true
œ x# 1 M for any positive integer M whenever x ÈM Ê false œ x lim Ä_
x x ln x
(c) x lim Ä_
30x 4x ex
œ x lim Ä_
sech x ecx
#
œ x lim Ä_
œ
Š "# ‹ x
(f) x lim Ä_
x
sinc" Š "x ‹
(e) x lim Ä_
" x#
#
tanc" Š " ‹
(d) x lim Ä_
Š
œxÄ lim_
3cx ˆ 23 ‰x œ 0 Ê slower 2cx œ x lim Ä_ ln 2x ln 2 ln x ˆ ln 2 œ x lim ln x# œ x lim Ä _ 2 (ln x) Ä _ # ln x
110. (a) x lim Ä_ (b) x lim Ä_
111. (a)
aex ecx b #e x
œxÄ lim_
sinh x ex
(f) x lim Ä_
" 1
" x
œ 1 Ê the same growth rate Ê false
Š x" ‹
(e) (f)
112. (a)
tan " x 1 cosh x ex
œ
" ‹ x% Š "# "% ‹ x x
Š
Š
x
(c) x lim Ä_ (d) lnln2x x œ
(f) 113.
df dx
" x # 1
œ
" %‹
Ÿ 1 if x 0 Ê true
ˆ " ‰ œ 0 Ê true œ x lim Ä _ x# 1
" ‹ x%
cos " Š "x ‹ 1
Ÿ
ˆ1‰
#
114. y œ f(x) Ê y œ 1
f w (x) œ x"# Ê
" #
" Šx
"
x œ fÐln 2Ñ
" x
1‹
Ê
true 2 Ê true
if x 1 Ê true
if x 0 Ê true
c"
œ ex 1 Ê Š dfdx ‹
1 #
œ
1
a1 ec2x b Ÿ
f af " (x)b œ 1
œ 0 Ê grows slower Ê true
1) œ 1 if x 0 Ê true
Š "x ‹ ln x œ lim œ0 Ê x1 xÄ_ 1 ln 2 ln x 1 Ÿ 1 1 œ 2 if x
secc" x œ 1 sinh x " œ ex #
" ln x
œ x lim Ä_
ˆ "x ‰
1 # for all x Ê true " c2x b Ÿ " (1 # a1 e #
Ÿ
x (b) x lim Ä _ Š "#
(e)
œ x lim Ä_
ln (ln x) ln x
(d) x lim Ä_
– ln x —
" x
œ
" df Š dx ‹
x œ ln 2
œy1 Ê xœ
œ 1 (x 1) œ x;
df " dx ¹ fÐxÑ
œ
"
Ê Š dfdx ‹ " y 1
df c" dx ¹ fÐxÑ
œ
x œ fÐln 2Ñ
œ
" aex 1bx œ ln 2
Ê f " (x) œ " (x 1)# ¹ fÐxÑ
œ
" x 1
œ
" #1
œ
; f " (f(x)) œ "
’Š1 x" ‹1“
#
" 3
" Š1 "x ‹1
œ x# ;
" f w (x)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
" Š x" ‹
œ x and
455
456
Chapter 7 Transcendental Functions
2 ‰ 115. y œ x ln 2x x Ê y w œ x ˆ 2x ln (2x) 1 œ ln 2x; " #
" w # and y 0 for x "# Ê relative minimum of "# at x œ "# ; f ˆ #"e ‰ œ "e and f ˆ #e ‰ œ 0 Ê absolute minimum is "# at x œ "# and the absolute maximum is 0 at x œ #e
solving y w œ 0 Ê x œ
; y w 0 for x
116. y œ 10x(2 ln x) Ê y w œ 10(2 ln x) 10x ˆ "x ‰
œ 20 10 ln x 10 œ 10(1 ln x); solving y w œ 0 Ê x œ e; y w 0 for x e and y w 0 for x e Ê relative maximum at x œ e of 10e; y ! on Ð!ß e# Ó and y ae# b œ 10e# (2 2 ln e) œ 0 Ê absolute minimum is 0 at x œ e# and the absolute maximum is 10e at x œ e
117. A œ '1
e
dx œ '0 2u du œ cu# d ! œ 1, where u œ ln x and du œ 1
2 ln x x
118. (a) A" œ '10
20
(b) A" œ 'ka
kb
119. y œ ln x Ê
dy dx
120. y œ 9ecxÎ3 Ê Ê
dx ¸ dt xœ9
dx œ cln kxkd #! "! œ ln 20 ln 10 œ ln
" x " x
kb dx œ cln kxkd ka œ ln kb ln ka œ ln
" x
œ
dy dx
Š
xœ
;
dA dx
dA dx
œ
dx dt " 4
(dy/dt) (dy/dx)
œ ˆ "x ‰ Èx œ
Ê
dx dt
œ
œ ln 2, and A# œ '1
2
œ ln
kb ka
dx; x œ 1 Ê u œ 0, x œ e Ê u œ 1
" Èx
b a
Ê
Š "4 ‹ È9 y 3e xÎ3
#
Ê
dy dt ¹ e#
dx œ cln kxkd #" œ ln 2 ln 1 œ ln 2
œ
" e
b
" x
dx œ cln kxkd ab œ ln b ln a
m/sec
; x œ 9 Ê y œ 9e$
#
" È2
dA dx
and " Èe
œ œ
" x#
dA dx
#
0 for 0 x
" È2
ln x 2x$Î#
œ
dA dx
œ 0 Ê 1 2x# œ 0
Ê absolute maximum of
" È2
e"Î# œ
ln x x#
œ
1ln x x#
. Solving
dA dx
at
œ 0 Ê 1 ln x œ 0 Ê x œ e; ln e e
œ
" e
at x œ e units long and y œ
2 ln x 2xÈx
Ê y ww œ 34 x&Î# (2 ln x) "# x&Î# œ x&Î# ˆ 34 ln x 2‰ ;
solving y w œ 0 Ê ln x œ 2 Ê x œ e# ; y w 0 for x e# and and y w 0 for x e# Ê a maximum of 2e ; y ww œ 0 Ê ln x œ 83 Ê x œ e)Î$ ; the curve is concave down on ˆ0ß e)Î$ ‰ and concave up on ˆe)Î$ ß _‰; so there is an inflection point at ˆe)Î$ ß
" È2e
units high.
0 for x e Ê absolute maximum of
" xÈ x
" x
œ ln b ln a, and A# œ 'a
œ ecx (x)(2x) ecx œ ecx a1 2x# b . Solving
ln x x and dA dx
Ê yw œ
œ
dy dt
20 10
" x
Èe$ Èe$ 1 ¸ 5 ft/sec
units long by y œ e
ln x Èx
Ê
dy dx dx dt
"Î#
0 for x e
123. (a) y œ
œ
0 for x
122. A œ xy œ x ˆ lnx#x ‰ œ dA dx
9 e$
3 ‹ e$
#
" È2
dy dt
Š "4 ‹ É9
œ
" È2
;
œ 3e xÎ3 ;
121. A œ xy œ xecx Ê Ê xœ
"
) ‰ . $e%Î$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" e#
units high.
Chapter 7 Practice Exercises #
#
#
(b) y œ ex Ê y w œ 2xex Ê y ww œ 2ex 4x# ex
457
#
#
œ a4x# 2bex ; solving y w œ 0 Ê x œ 0; y w 0 for x 0 and y w 0 for x 0 Ê a maximum at x œ 0 of e! œ 1; there are points of inflection at x œ „ È" ; the 2
curve is concave down for
" È2
x
" È2
and concave
up otherwise. (c) y œ (1 x) ecx Ê y w œ ecx (1 x) ecx œ xecx Ê y ww œ ecx xecx œ (x 1) ecx ; solving y w œ 0 Ê xecx œ 0 Ê x œ 0; y w 0 for x 0 and y w 0 for x 0 Ê a maximum at x œ 0 of (1 0) e! œ 1; there is a point of inflection at x œ 1 and the curve is concave up for x 1 and concave down for x 1.
124. y œ x ln x Ê y w œ ln x x ˆ "x ‰ œ ln x 1; solving y w œ 0 Ê ln x 1 œ 0 Ê ln x œ 1 Ê x œ e" ; y w 0 for x e" and y w 0 for x e" Ê a minimum of e" ln e" œ "e at x œ e" . This minimum is an absolute minimum since yww œ
125.
dy dx
" x
is positive for all x !.
œ Èy cos2 Èy Ê
126. y w œ
3yax1b2 y 1
Ê
dy Èy cos2 Èy
ay 1 b y dy
127. yy w œ secay2 bsec2 x Ê
œ dx Ê 2tanÈy œ x C Ê y œ ˆtan1 ˆ x 2 C ‰‰
2
œ 3ax 1b2 dx Ê y ln y œ ax 1b3 C œ sec2 x dx Ê
y dy secay2 b
sinˆy2 ‰ 2
œ tan x C Ê sinay2 b œ 2tan x C1
sin x 128. y cos2 axb dy sin x dx œ 0 Ê y dy œ cos 2 axb dx Ê
y2 2
œ cos1axb C Ê y œ „ É cosa2xb C1
129.
œ exy2 Ê ey dy œ eax2b dx Ê ey œ eax2b C. We have ya0b œ 2, so e2 œ e2 C Ê C œ 2e2 and e œ eax2b 2e2 Ê y œ lnˆeax2b 2e2 ‰
130.
dy dx
dy dx y
œ
y ln y 1 x2
Ê etan
Ê
c1 a0bC
dy y ln y
œ
Ê lnaln yb œ tan1 axb C Ê y œ ee
dx 1 x2
tanc1 axbbC
. We have ya0b œ e2 Ê e2 œ ee
œ 2 Ê tan1 a0b C œ ln 2 Ê 0 C œ ln 2 Ê C œ ln 2 Ê y œ ee
131. x dy ˆy Èy‰dx œ 0 Ê
dy ˆy È y ‰
œ
dx x
tanc1 a0bbC
tanc1 axbbln 2
Ê 2lnˆÈy 1‰ œ ln x C. We have ya1b œ 1 Ê 2 lnŠÈ1 1‹ œ ln 1 C
Ê 2 ln 2 œ C œ ln 22 œ ln 4. So 2 lnˆÈy 1‰ œ ln x ln 4 œ lna4xb Ê lnˆÈy 1‰ œ "# lna4xb œ lna4xb1/2 Ê eln
ˆ È y 1 ‰
132. y2 dx dy œ So
y3 3
1/2
œ elna4xb Ê Èy 1 œ 2Èx Ê y œ ˆ2Èx 1‰
ex e2x 1
Ê
e2x 1 ex dx
œ ex ex
1 3
œ
dy yc2
Ê
y3 3
2
œ ex ex C. We have ya0b œ 1 Ê
a1 b 3 3
œ e0 e0 C Ê C œ 13 .
Ê y3 œ 3aex ex b 1 Ê y œ c3aex ex b 1 d1/3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
458
Chapter 7 Transcendental Functions
133. Since the half life is 5700 years and A(t) œ A! ekt we have Ê kœ
ln (0.5) 5700
Ê ln (0.1) œ
A! #
" #
œ A! e5700k Ê
œ e5700k Ê ln (0.5) œ 5700k ln Ð0Þ5Ñ
. With 10% of the original carbon-14 remaining we have 0.1A! œ A! e 5700 ln (0.5) 5700
t Ê tœ
(5700) ln (0.1) ln (0.5)
t
ln Ð0Þ5Ñ
Ê 0.1 œ e 5700
t
¸ 18,935 years (rounded to the nearest year).
134. T Ts œ (To Ts ) eckt Ê 180 40 œ a220 40b eckÎ4 , time in hours, Ê k œ 4 ln ˆ 79 ‰ œ 4 ln ˆ 97 ‰ Ê 70 40 œ a220 40b ec4 ln Ð9Î7Ñ t Ê t œ
¸ 1.78 hr ¸ 107 min, the total time Ê the time it took to cool from 180° F to
ln 6 4 ln ˆ 97 ‰
70° F was 107 15 œ 92 min x ‰ 135. ) œ 1 cot" ˆ 60 cot" ˆ 35
œ 30 ’ 60# 2 x#
" 30# (50 x)# “ ;
x ‰ 30 ,
solving d) dx
100 20È17 is not in the domain;
0 x 50 Ê d) dx
d) dx
œ
1 Š 60 ‹
1ˆ
x ‰# 60
Š
" 30 ‹
cx‹ 1 Š 5030
#
œ 0 Ê x# 200x 3200 œ 0 Ê x œ 100 „ 20È17, but d) dx
0 for x 20 Š5 È17‹ and
0 for 20 Š5 È17‹ x 50
Ê x œ 20 Š5 È17‹ ¸ 17.54 m maximizes ) 136. v œ x# ln ˆ "x ‰ œ x# (ln 1 ln x) œ x# ln x Ê Ê 2 ln x 1 œ 0 Ê ln x œ maximum at x œ e
1Î2
;
r h
Ê x œ ec1Î2 ;
" #
œ 2x ln x x# ˆ "x ‰ œ x(2 ln x 1); solving
dv dx
dv dx
1Î2
œ x and r œ 1 Ê h œ e
1Î2
0 for x e and È œ e ¸ 1.65 cm
dv dx
0 for x e
1Î2
dv dx
œ0
Ê a relative
CHAPTER 7 ADDITIONAL AND ADVANCED EXERCISES lim c '0
b
1.
2.
bÄ1
lim " xÄ_ x
" È 1 x#
dx œ lim c csin" xd 0 œ lim c asin" b sin" 0b œ lim c asin" b 0b œ lim c sin" b œ bÄ1 bÄ1 bÄ1 bÄ1 b
'0x tan" t dt œ x lim Ä_ tanc" x
œ x lim Ä_ 3. y œ ˆcos Èx‰
1
1Îx
œ "# lim b xÄ!
"
#x
œ
"Î# sec# Èx " x "Î# #
4. y œ ax ex b2Îx Ê ln y œ Ê x lim ax e b Ä_ ˆ " 5. x lim Ä _ n1
" n#
x
_ ˆ_ form‰
1 #
Ê ln y œ
x 2Îx
'0x tanc" t dt
" x
ln ˆcos Èx‰ x
ln ˆcos Èx‰ and lim b xÄ!
1Îx lim b ˆcos Èx‰ œ e 1Î2 œ
œ "# Ê
xÄ!
2 ln ax ex b x
Ê x lim ln y œ x lim Ä_ Ä_ y # œ x lim e œ e Ä_
á
" ‰ #n
œ lim b xÄ!
2 a1 e x b x ex
1
6. x lim Ä_
" 1x " n
œ
" #
lim
x Ä !b
tan Èx Èx
" Èe
œ x lim Ä_
2ex 1 ex
œ x lim Ä_
2ex ex
œ2
1 ˆ"‰ œ x lim ˆ n" ‰ – 1 " — á ˆ n" ‰ – 1 " — Ä _ n – 1 Š"‹ — 12Š ‹ 1nŠ ‹ n
which can be interpreted as a Riemann sum with partitioning ?x œ œ '0
sin Èx 2Èx cos Èx
n
" n
n
ˆ " Ê x lim Ä _ n1
" n#
á
" ‰ #n
dx œ cln (1 x)d "! œ ln 2
ˆ n" ‰ eÐ1ÎnÑ ˆ n" ‰ e2Ð1ÎnÑ á ˆ n" ‰ enÐ1ÎnÑ ‘ which can be interpreted as a ce1În e2În á ed œ x lim Ä_
Riemann sum with partitioning ?x œ
" n
Ê x lim Ä_
" n
ce1În e2În á ed œ '0 ex dx œ cex d "! œ e 1 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 #
Chapter 7 Additional and Advanced Exercises t 7. A(t) œ '0 ecx dx œ cecx d t0 œ 1 ect , V(t) œ 1'0 ec2x dx œ 1# ec2x ‘ 0 œ t
(a) (b) (c) 8. (a)
t
lim A(t) œ lim a1 ect b œ 1
tÄ_
tÄ_
lim V(t) t Ä _ A(t) limb V(t) A(t)
œ lim
tÄ_
œ limb tÄ!
tÄ!
1 #
ˆ1 ec2t ‰ 1 ect
1 #
ˆ1 ec2t ‰ 1 ect
œ 0;
lim loga 2 œ lim c aÄ1
ln 2 ln a
œ _;
lim loga 2 œ lim b aÄ1
ln 2 ln 1
œ _;
lim loga 2 œ a lim Ä_
ln 2 ln a
œ0
aÄ_
9. A" œ '1
e
dx œ
2 log2 x x e
#
x) œ ’ (ln 2 ln # “ œ 1
" # ln 2
2 ln 2
" 1 x#
#
lim
x Ä !c
a1 ec2t b
a1 ect b œ 1
(b)
4
x
1
dx œ
2 ln 4
'1e lnxx dx
Ê A" : A# œ 2 : 1
" 1 x#
Š
" ‹ x#
Š1 x"# ‹
œ 0 Ê tan" x tan" ˆ x" ‰ is a 1 #
for x 0; it is 1# for
0 since tan" x tan" ˆ "x ‰ is odd. Next lim b tan" x tan" ˆ "x ‰‘ œ ! 1# œ 1#
xÄ!
1 #
œ limb tÄ!
#
constant and the constant is
and
a1 ect b a1 ect b a1 ect b
'1e lnxx dx œ ’ (lnlnx)2 “ e œ ln"# ; A# œ '1e 2 log4
10. y œ tan" x tan" ˆ "x ‰ Ê yw œ " 1 x#
1
œ limb tÄ!
ln 2 ln a
a Ä 1b
x
1 #
lim loga 2 œ lim b aÄ!
a Ä !b a Ä 1c
œ
œ
1 #
459
the
ˆtan" x tan" ˆ "x ‰‰ œ 0 ˆ 1# ‰ œ 1#
11. ln xax b œ xx ln x and ln axx bx œ x ln xx œ x# ln x; then, xx ln x œ x# ln x Ê axx x# bln x œ ! Ê xx œ x# or ln x œ !Þ x ln x œ ! Ê x œ "; xx œ x# Ê x ln x œ 2 ln x Ê x œ 2. Therefore, xax b œ axx bx when x œ 2 or x œ ". x
12. In the interval 1 x 21 the function sin x 0 Ê (sin x)sin x is not defined for all values in that interval or its translation by 21.
13. f(x) œ egÐxÑ Ê f w (x) œ egÐxÑ gw (x), where gw (x) œ 14. (a) (c)
df dx df dx
œ
2 ln ex ex
x 1 x%
Ê f w (2) œ e! ˆ 1 2 16 ‰ œ (b) f(0) œ '1
1
† ex œ 2x #
2 ln t t
2 17
dt œ 0
#
œ 2x Ê f(x) œ x C; f(0) œ 0 Ê C œ 0 Ê f(x) œ x Ê the graph of f(x) is a parabola
15. (a) g(x) h(x) œ 0 Ê g(x) œ h(x); also g(x) h(x) œ 0 Ê g(x) h(x) œ 0 Ê g(x) h(x) œ 0 Ê g(x) œ h(x); therefore h(x) œ h(x) Ê h(x) œ 0 Ê g(x) œ 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
460
Chapter 7 Transcendental Functions f(x) f(x) # f(x) f(x) #
(b)
cfE (x) fO (x)d b fE (x) fO (x)‘ œ fE (x) fO (x) # fE (x) fO (x) œ fE (x); # cfE (x) fO (x)d cfE (x) fO (x)d œ fE (x) fO (x) # fE (x) fO (x) œ fO (x) #
œ œ
(c) Part b Ê such a decomposition is unique. 16. (a) g(0 0) œ
g(0) g(0) 1 g(0) g(0) #
Ê c1 g# (0)d g(0) œ 2g(0) Ê g(0) g$ (0) œ 2g(0) Ê g$ (0) g(0) œ 0
Ê g(0) cg (0) 1d œ 0 Ê g(0) œ 0
g(x) b g(h)
’ “ g(x) g(x h) g(x) g(x) g# (x) g(h) œ lim 1 c g(x) g(h) œ lim g(x) g(h) h h h c 1 g(x) g(h)d hÄ0 hÄ0 hÄ0 g(h) 1 g# (x) lim ’ h “ ’ 1 g(x) g(h) “ œ 1 † c1 g# (x)d œ 1 g# (x) œ 1 [g(x)]# hÄ0
(b) gw (x) œ lim œ dy dx
(c)
œ 1 y# Ê
dy 1 y# "
Ê C œ 0 Ê tan
œ dx Ê tan" y œ x C Ê tan" (g(x)) œ x C; g(0) œ 0 Ê tan" 0 œ 0 C
(g(x)) œ x Ê g(x) œ tan x
17. M œ '0
1
" 2 " xd ! œ 1# 1 x# dx œ 2 ctan ln 2 ln 4 ˆ 1 ‰ œ 1 ; y œ 0 by symmetry #
œ
and My œ '0
1
2x 1 x#
"
dx œ cln a1 x# bd ! œ ln 2 Ê x œ
My M
" 18. (a) V œ 1 '1Î4 Š #È ‹ dx œ x
1 4
'14Î4 x" dx œ 14 cln kxkd %"Î% œ 14 ˆln 4 ln 4" ‰ œ 14 ln 16 œ 14 ln a2% b œ 1 ln 2
" (b) My œ '1Î4 x Š #È ‹ dx œ x
1 2
63 '14Î4 x"Î# dx œ 3" x$Î# ‘ %"Î% œ ˆ 83 24" ‰ œ 64#4 1 œ 24 ;
#
4
4
" " Mx œ '1Î4 "# Š #È ‹ Š 2È ‹ dx œ x x 4
M œ '1Î4 4
yœ
Mx M
œ
" #È x
ˆ "#
'14Î4
" x
%
dx œ 8" ln kxk‘ "Î% œ
dx œ '1Î4 "2 x"Î# dx œ x"Î# ‘ "Î% œ 2 %
4
ln
œ
ln 2 3
b csc ) ‰ r%
Ê
2‰ ˆ 32 ‰
19. (a) L œ k ˆ a bR%cot ) %
1 8
#
dL d)
#
œ k Š b csc R%
)
" #
œ
3 #
b csc ) cot ) ‹; r%
%
" 8
ln 16 œ
" #
ln 2;
; therefore, x œ
solving
%
dL d)
My M
‰ ˆ 23 ‰ œ œ ˆ 63 24
21 1#
œ
7 4
and
œ0
%
Ê r b csc ) bR csc ) cot ) œ 0 Ê (b csc )) ar csc ) R cot )b œ 0; but b csc ) Á 0 since )Á
1 #
Ê r% csc ) R% cot ) œ 0 Ê cos ) œ
r% R%
%
Ê ) œ cos" Š Rr % ‹ , the critical value of )
%
(b) ) œ cos" ˆ 56 ‰ ¸ cos" (0.48225) ¸ 61° 20. In order to maximize the amount of sunlight, we need to maximize the angle ) formed by extending the two red line segments to their vertex. The angle between the two lines is given by ) œ 1 a)1 a1 )# bb. From trig we have 1 ˆ 200 ‰ tan )1 œ 450350 x Ê )1 œ tan1 ˆ 450350 x ‰ and tan a1 )# b œ 200 x Ê a1 )# b œ tan x ‰ Ê ) œ 1 a)1 a1 )# bb œ 1 tan1 ˆ 450350 x ‰ tan1 ˆ 200 x Ê d) dx
d) dx
œ
œ0Ê
1 2 1 ˆ 450350c x ‰
†
350 a450 xb2
350 a450 xb2 122500
1 ‰2 1 ˆ 200 x
200 x2 40000
‰œ † ˆ 200 x2
350 a450 xb2 122500
200 x2 40000
œ 0 Ê 200Ša450 xb2 122500‹ œ 350ax2 40000b
Ê 3x2 3600x 1020000 œ 0 Ê x œ 600 „ 100È70. Since x 0, consider only x œ 600 100È70. Using the first derivative test,
d) dx ¹x œ 100
œ
9 3500
0 and
d) dx ¹x œ 400
œ
9 5000
0 Ê local max when
x œ 600 100È70 ¸ 236.67 ft.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 8 TECHNIQUES OF INTEGRATION 8.1 INTEGRATION BY PARTS 1. u œ x, du œ dx; dv œ sin
'
x sin
x #
dx œ 2x cos
x #
x #
dx, v œ 2 cos
' ˆ2 cos " 1
2. u œ ), du œ d); dv œ cos 1) d), v œ
'
) cos 1) d) œ
3.
sin 1) '
) 1
" 1
x‰ #
x #
;
dx œ 2x cos ˆ x# ‰ 4 sin ˆ x# ‰ C
sin 1);
sin 1) d) œ
) 1
sin 1)
" 1#
cos 1) C
cos t ÐÑ t# ïïïïî ÐÑ 2t ïïïïî ÐÑ 2 ïïïïî
sin t cos t sin t
0 4.
'
t# cos t dt œ t# sin t 2t cos t 2 sin t C
'
x# sin x dx œ x# cos x 2x sin x 2 cos x C
sin x ÐÑ x# ïïïïî ÐÑ 2x ïïïïî ÐÑ 2 ïïïïî
cos x sin x cos x
0 5. u œ ln x, du œ
dx x ;
dv œ x dx, v œ
'1 x ln x dx œ ’ x# 2
6. u œ ln x, du œ
'1 x e
ln x“ '1 #
#
$
ln x dx œ
dx x ;
2 # x
"
#
x# # dx x
dv œ x$ dx, v œ
% ’ x4
ln x“ ' e
1
e % x
;
x% 4
dx 1 4 x
#
#
œ 2 ln 2 ’ x4 “ œ 2 ln 2 "
3 4
œ ln 4
3 4
; œ
e% 4
%
e
x ’ 16 “ œ 1
3e% 1 16
7. u œ x, du œ dx ; dv œ ex dx, v œ ex ;
' x ex dx œ x ex ' ex dx œ x ex ex C
8. u œ x, du œ dx ; dv œ e3x dx, v œ 13 e3x ;
' x e3x dx œ x3 e3x 13 ' e3x dx œ x3 e3x 19 e3x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
462
Chapter 8 Techniques of Integration ex
9. ÐÑ x# ïïïïî ÐÑ 2x ïïïïî ÐÑ 2 ïïïïî
ex ex ex
'
0
x# ex dx œ x# ex 2x ex 2 ex C
e2x
10.
ÐÑ x# 2x 1 ïïïïî 12 e2x ÐÑ 2x 2 ïïïïî 14 e2x ÐÑ 2 ïïïïî 18 e2x
' ax# 2x 1be2x dx œ 12 ax# 2x 1be2x 14 a2x 2be2x 14 e2x C
0
œ ˆ 12 x# 32 x 54 ‰e2x C 11. u œ tan" y, du œ
dy 1 y #
; dv œ dy, v œ y;
' tan" y dy œ y tan" y ' a1ydyy b œ y tan" y #" ln a1 y# b C œ y tan" y ln È1 y# C #
12. u œ sin" y, du œ
dy È 1 y#
; dv œ dy, v œ y;
' sin" y dy œ y sin" y ' Èy1 dy y
#
œ y sin" y È1 y# C
13. u œ x, du œ dx; dv œ sec# x dx, v œ tan x;
' x sec# x dx œ x tan x ' tan x dx œ x tan x ln kcos xk C
14.
' 4x sec# 2x dx; [y œ 2x]
' y sec# y dy œ y tan y ' tan y dy œ y tan y ln ksec yk C
Ä
œ 2x tan 2x ln ksec 2xk C ex
15. ÐÑ x$ ïïïïî ÐÑ 3x# ïïïïî ÐÑ 6x ïïïïî ÐÑ 6 ïïïïî 0
ex ex ex ex
'
x$ ex dx œ x$ ex 3x# ex 6xex 6ex C œ ax$ 3x# 6x 6b ex C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.1 Integration by Parts ecp
16. ÐÑ ïïïïî ÐÑ 4p$ ïïïïî ÐÑ 12p# ïïïïî ÐÑ 24p ïïïïî ÐÑ 24 ïïïïî p%
ecp ecp ecp ecp ecp
'
0
p% ecp dp œ p% ecp 4p$ ecp 12p# ecp 24pecp 24ecp C œ ap% 4p$ 12p# 24p 24b ecp C
ex
17.
ÐÑ x# 5x ïïïïî ex ÐÑ 2x 5 ïïïïî ex ÐÑ 2 ïïïïî ex
' ax# 5xb ex dx œ ax# 5xb ex (2x 5)ex 2ex C œ x# ex 7xex 7ex C
0
œ ax# 7x 7b ex C er
18.
ÐÑ r# r 1 ïïïïî er ÐÑ 2r 1 ïïïïî er ÐÑ 2 ïïïïî er 0
' ar# r 1b er dr œ ar# r 1b er (2r 1) er 2er C œ car# r 1b (2r 1) 2d er C œ ar# r 2b er C
ex
19. x& 5x% 20x$ 60x# 120x 120 0
ÐÑ ïïïïî ex ÐÑ ïïïïî ex ÐÑ ïïïïî ex ÐÑ ïïïïî ex ÐÑ ïïïïî ex ÐÑ ïïïïî ex
' x& ex dx œ x& ex 5x% ex 20x$ ex 60x# ex 120xex 120ex C œ ax& 5x% 20x$ 60x# 120x 120b ex C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
463
464
Chapter 8 Techniques of Integration e4t
20. ÐÑ t# ïïïïî ÐÑ 2t ïïïïî ÐÑ 2 ïïïïî
" 4
e4t
" 16
e4t
" 64
e4t
' t# e4t dt œ t4 e4t 162t e4t 642 e4t C œ t4 e4t 8t e4t 3"# e4t C #
0
#
œ Š t4
t 8
#
" 4t 3# ‹ e
C
21. I œ ' e) sin ) d); cu œ sin ), du œ cos ) d); dv œ e) d), v œ e) d Ê I œ e) sin ) ' e) cos ) d); cu œ cos ), du œ sin ) d); dv œ e) d), v œ e) d Ê I œ e) sin ) Še) cos ) ' e) sin ) d)‹ œ e) sin ) e) cos ) I Cw Ê 2I œ ae) sin ) e) cos )b Cw Ê I œ
" #
ae) sin ) e) cos )b C, where C œ
w
C #
is
Ê 2I œ "# ecy (sin y cos y) Cw Ê I œ "4 ecy (sin y cos y) C œ e 4 (sin 2x cos 2x) C, where C œ
C #
another arbitrary constant 22. I œ ' ecy cos y dy; cu œ cos y, du œ sin y dy; dv œ ecy dy, v œ ecy d Ê I œ ecy cos y ' aecy b (sin y) dy œ ecy cos y ' ecy sin y dy; cu œ sin y, du œ cos y dy; dv œ ecy dy, v œ ecy d Ê I œ ecy cos y Šecy sin y ' aey b cos y dy‹ œ ecy cos y ecy sin y I Cw Ê 2I œ ecy (sin y cos y) Cw Ê I œ
" #
aecy sin y ecy cos yb C, where C œ
23. I œ ' e2x cos 3x dx; u œ cos 3x; du œ 3 sin 3x dx, dv œ e2x dx; v œ
is another arbitrary constant
e2x ‘
Ê
' e2x sin 3x dx; u œ sin 3x, du œ 3 cos 3x, dv œ e2x dx; v œ "# e2x ‘ I œ "# e2x cos 3x 3# Š "# e2x sin 3x 3# ' e2x cos 3x dx‹ œ "# e2x cos 3x 34 e2x sin 3x 94 I Cw
Ê
13 4
Ê Iœ
24.
" #
w
C #
" #
e2x cos 3x
Iœ
" #
3 #
e2x cos 3x 34 e2x sin 3x Cw Ê
e2x 13
(3 sin 3x 2 cos 3x) C, where C œ
4 13
Cw
' ec2x sin 2x dx; [y œ 2x] Ä "# ' ecy sin y dy œ I; cu œ sin y, du œ cos y dy; dv œ ecy dy, v œ ecyd Ê I œ "# Šecy sin y ' ecy cos y dy‹ cu œ cos y, du œ sin y; dv œ ecy dy, v œ ecy d Ê I œ "# ecy sin y "# Šecy cos y ' aecy b ( sin y) dy‹ œ "# ecy (sin y cos y) I Cw c2x
#
25.
' eÈ3sb9 ds; ” 3s 92 œ x ds œ
2 3
' xex dx œ
2 3
3
x dx •
Ä
' ex † 23 x dx œ 23 ' xex dx; cu œ x, du œ dx; dv œ ex dx, v œ ex d ;
Šxex ' ex dx‹ œ
2 3
axex ex b C œ
2 3
ŠÈ3s 9 eÈ3sb9 eÈ3sb9 ‹ C
26. u œ x, du œ dx; dv œ È1 x dx, v œ 23 È(1 x)$ ;
'01 xÈ1 x dx œ 23 È(1 x)$ x‘ "! 23 '01 È(1 x)$ dx œ 23 25 (1 x)&Î# ‘ "! œ 154
27. u œ x, du œ dx; dv œ tan# x dx, v œ ' tan# x dx œ ' œ tan x x;'0
1Î3
œ
1 3
1Î$
x tan# x dx œ cx(tan x x)d !
ŠÈ3 13 ‹ ln
" #
1# 18
œ
1È3 3
ln 2
sin# x cos# x
dx œ '
" cos# x cos# x
'0 (tan x x) dx œ 1Î3
1 3
dx œ '
dx cos# x
' dx
ŠÈ3 13 ‹ ’ln kcos xk
1# 18
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1Î$ x# # “!
w
Section 8.1 Integration by Parts 28. u œ ln ax x# b, du œ œ x ln ax x# b '
; dv œ dx, v œ x; ' ln ax x# b dx œ x ln ax x# b '
(2x 1) dx x x #
œ x ln ax x# b '
(2x 1) dx x 1
u œ ln x
29.
' sin (ln x) dx; Ô du œ "x dx × œ
" #
Ä
Õ dx œ eu du Ø cx cos (ln x) x sin (ln x)d C u œ ln z
30.
' z(ln z)# dz; Ô du œ "z dz × " 2
e2u
" 4
e2u
" 8
e2u
'
0
u# e2u du œ œ
† x dx
dx œ x ln ax x# b 2x ln kx 1k C
From Exercise 21, ' (sin u) eu du œ eu ˆ sin u # cos u ‰ C
' eu † u# † eu du œ ' e2u † u# du;
Ä
Õ dz œ eu du Ø e2u
ÐÑ u# ïïïïî ÐÑ 2u ïïïïî ÐÑ 2 ïïïïî
31.
' (sin u) eu du.
2(x 1) " x 1
2x " x(x 1)
465
#
z 4
u# # #
e2u #u e2u "4 e2u C œ
e2u 4
c2u# 2u 1d C
c2(ln z) 2 ln z 1d C
' x sec x# dx ’Let u œ x# , du œ 2x dx Ê 12 du œ x dx“ Ä ' x sec x# dx
œ
1 2
' sec u du œ 12 lnlsec u tan ul C
œ 12 lnlsec x# tan x# l C 32.
' cosÈÈx x dx ’Let u œ Èx, du œ 2È1 x dx Ê 2du œ È1x dx“ Ä ' cosÈÈx x dx
33.
' xaln xb# dx; Ô du œ x" dx ×
u œ ln x
Õ dx œ eu du Ø e2u
ÐÑ u# ïïïïî ÐÑ 2u ïïïïî ÐÑ 2 ïïïïî
" 2
e2u
" 4
e2u
" 8
e2u
'
0
Ä
34.
' xaln" xb
#
' lnx x dx 2
36.
' alnxxb
3
x# 4
dx ’Let u œ ln x, du œ
35. u œ ln x, du œ
1 x
dx; dv œ
œ lnx x '
" x#
1 x2
' eu † u# † eu du œ ' e2u † u# du;
u# e2u du œ œ
1 x
œ 2' cos u du œ 2 sin u C œ 2 sinÈx C
u# # #
e2u #u e2u "4 e2u C œ
c2(ln x) 2 ln x 1d C œ dx“ Ä '
" xaln xb#
dx œ '
1 u2
x# 2 aln
e2u 4
c2u# 2u 1d C
xb #
x# 2 ln
x
x# 4
C
du œ 1u C œ ln1x C
dx, v œ x1 ;
dx œ lnx x
dx ’Let u œ ln x, du œ
1 x
1 x
C
dx“ Ä '
aln xb3 x
dx œ ' u3 du œ 41 u4 C œ 41 aln xb4 C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
466 37.
Chapter 8 Techniques of Integration
' x3 ex
4
dx ’Let u œ x4 , du œ 4x3 dx Ê 14 du œ x3 dx“ Ä ' x3 ex dx œ 4
3
1 4
' eu du
4
œ 14 eu C œ 14 ex C
3
38. u œ x3 , du œ 3x2 dx; dv œ x2 ex dx, v œ 13 ex ;
' x5 ex
3
dx œ ' x3 ex x2 dx œ 13 x3 ex 3
3
1 3
' ex
3
3
39. u œ x2 , du œ 2x dx; dv œ Èx2 1 x dx, v œ 13 ax2 1b
' x3 Èx2 1 dx
40.
œ
1 2 2 3 x ax
1b
3Î2
1 3
3
3x2 dx œ 13 x3 ex 13 ex C
' ax2 1b
3Î2
3Î2
; 3Î2
œ
1 3
2x dx œ 13 x2 ax2 1b
' x2 sin x3 dx ’Let u œ x3 , du œ 3x2 dx Ê 13 du œ x2 dx“ Ä ' x2 sin x3 dx
2 2 15 ax
1b
' sin u du
5Î2
C
œ 13 cos u C
œ 13 cos x3 C 41. u œ sin 3x, du œ 3cos 3x dx; dv œ cos 2x dx, v œ 12 sin 2x ;
' sin 3x cos 2x dx œ 12 sin 3x sin 2x 32 ' cos 3x sin 2x dx
u œ cos 3x, du œ 3sin 3x dx; dv œ sin 2x dx, v œ 12 cos 2x ;
' sin3x cos 2x dx œ 12 sin 3x sin 2x 32 ” 12 cos 3x cos 2x 32 ' sin3x cos 2x dx• œ 12 sin 3x sin 2x 34 cos 3x cos 2x
9 4
' sin 3x cos 2x dx Ê 54 ' sin 3x cos 2x dx œ 12 sin 3x sin 2x 34 cos 3x cos 2x
Ê ' sin 3x cos 2x dx œ 25 sin 3x sin 2x 35 cos 3x cos 2x C 42. u œ sin 2x, du œ 2cos 2x dx; dv œ cos 4x dx, v œ 14 sin 4x ;
' sin 2x cos 4x dx œ 14 sin 2x sin 4x 12 ' cos 2x sin 4x dx
u œ cos 2x, du œ 2sin 2x dx; dv œ sin 4x dx, v œ 14 cos 4x ;
' sin 2x cos 4x dx œ 14 sin 2x sin 4x 12 ” 14 cos 2x cos 4x 12 ' sin 2x cos 4x dx• œ 14 sin 2x sin 4x 18 cos 2x cos 4x
1 4
' sin 2x cos 4x dx Ê 34 ' sin 2x cos 4x dx œ 14 sin 2x sin 4x 18 cos 2x cos 4x
Ê ' sin 2x cos 4x dx œ 13 sin 2x sin 4x 16 cos 2x cos 4x C 43.
' ex sin ex dx ’Let u œ ex , du œ ex dx“ Ä ' ex sin ex dx
44.
'
45.
' cosÈx dx; Ö dy œ 2È" x dx Ù
eÈ x Èx
dx ’Let u œ Èx, du œ Ô
y œ Èx
1 2È x
×
dx Ê 2du œ
Ä
1 Èx
œ ' sin u du œ cos u C œ cos ex C
dx“ Ä '
eÈ x Èx
dx œ 2' eu du œ 2eu C œ 2eÈx C
' cos y 2y dy œ ' 2y cos y dy;
Õ dx œ 2y dy Ø u œ 2y, du œ 2 dy; dv œ cos y dy, v œ sin y ;
' 2y cos y dy œ 2y sin y ' 2 sin y dy œ 2y sin y 2 cos y C œ 2Èx sin Èx 2 cos Èx C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.1 Integration by Parts Ô
46.
y œ Èx
×
' Èx eÈx dx; Ö dy œ 2È" x dx Ù
'
Ä
467
y ey 2y dy œ ' 2y2 ey dy;
Õ dx œ 2y dy Ø ey
ÐÑ 2y# ïïïïî ey ÐÑ 4y ïïïïî ey ÐÑ 4 ïïïïî ey
' 2y2 ey dy œ 2y# ey 4y ey 4 ey C œ 2x eÈx 4Èx eÈx 4eÈx C
0 47.
sin 2) ÐÑ )# ïïïïî "2 cos 2) ÐÑ 2) ïïïïî "4 sin 2) ÐÑ 2 ïïïïî "8 cos 2)
'01Î2 )# sin 2) d) œ ’ )#
#
0
œ ’ 48.
1# 8
† (1)
1 4
cos 2)
†0
" 4
) #
sin 2)
" 4
cos 2)“
† (1)“ 0 0
" 4
1Î# !
† 1‘ œ
1# 8
" #
1# 4 8
œ
cos 2x ÐÑ x$ ïïïïî "2 sin 2x ÐÑ 3x# ïïïïî "4 cos 2x ÐÑ 6x ïïïïî "8 sin 2x ÐÑ " 6 ïïïïî 16 cos 2x
'01Î2 x$ cos 2x dx œ ’ x#
$
0
œ 49. u œ sec" t, du œ
dt tÈt# 1
'22ÎÈ3 t sec" t dt œ ’ t#
#
œ
51 9
’ "# Èt# 1“
50. u œ sin" ax# b , du œ
È2
'01Î
1$ ’ 16
†0
; dv œ t dt, v œ
sec" t“
# #ÎÈ$
œ
2x dx È 1 x %
#
#
’È1 x% “
!
œ
†0
1 1#
3 8
œ ˆ2 †
"# ŠÈ3 É 43 1‹ œ
"ÎÈ#
1 1#
dt tÈt# 1
; dv œ 2x dx, v œ x# ;
"ÎÈ#
31 8
3x# 4
cos 2x
3x 4
sin 2x
3 8
cos 2x“
† (1)“ 0 0 0
3 8
1Î# ! #
1 † 1‘ œ 316
;
'2ÎÈ3 Š t# ‹
2x sin" ax# b dx œ cx# sin" ax# bd !
œ
t# #
† (1)
2
#ÎÈ$
51 9
31 # 16
sin 2x
'0
É 34 1 œ
È2
1Î
x# †
51 9
1 3
2 3
† 16 ‰ '2ÎÈ3
"# ŠÈ3
2x dx È 1 x%
2
È3 3 ‹
œ
œ ˆ "# ‰ ˆ 16 ‰ '0
t dt 2Èt# 1
51 9
È3 3
œ
51 3È 3 9
È 2 d ˆ1 x % ‰
1Î
2È 1 x%
16È312 1#
51. (a) u œ x, du œ dx; dv œ sin x dx, v œ cos x;
S" œ '0 x sin x dx œ [x cos x]!1 '0 cos x dx œ 1 [sin x]1! œ 1 1
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3 4
œ
3 a4 1 # b 16
468
Chapter 8 Techniques of Integration
(b) S# œ '1 x sin x dx œ ”[x cos x]#11 '1 cos x dx• œ c31 [sin x]1#1 d œ 31 21
21
(c) S$ œ '21 x sin x dx œ [x cos x]$#11 '21 cos x dx œ 51 [sin x]$#11 œ 51 31
31
Ðn 1Ñ1
(d) S8" œ (1)nb1 'n1
x sin x dx œ (1)nb1 c[x cos x]Ðnn1 1Ñ1 [sin x]Ðnn1 1Ñ1 d
œ (1)nb1 c(n 1)1(1)n n1(1)nb1 d 0 œ (2n 1)1
52. (a) u œ x, du œ dx; dv œ cos x dx, v œ sin x; 31Î2 1‰ S" œ '1Î2 x cos x dx œ ”[x sin x]311Î2Î2 '1Î2 sin x dx• œ ˆ 31 # # [cos x]1Î2 œ 21 31Î2
31Î2
(b) S# œ '31Î2 x cos x dx œ [x sin x]&$11ÎÎ22 '31Î2 sin x dx œ 5#1 ˆ 3#1 ‰‘ [cos x]&$11ÎÎ22 œ 41 51Î2
51Î2
(c) S$ œ '51Î2 x cos x dx œ ”[x sin x](&11ÎÎ22 '51Î2 sin x dx• œ ˆ 7#1 71Î2
71Î2
Ð2n1Ñ1Î2
51 ‰ #
[cos x](&11ÎÎ22 œ 61
Ð2n1Ñ1Î2
n1Ñ1Î2 n' (d) Sn œ (1)n 'Ð2n 1Ñ1Î2 x cos x dx œ (1)n ”[x sin x]Ð# sin x dx• Ð2n 1Ñ1Î2 Ð2n 1Ñ1Î2
œ (1)n ’ (2n# 1)1 (1)n 53. V œ '0
ln 2
(2n1)1 #
n 1Ñ1Î2 (1)nc1 “ [cos x]Ð# Ð2n1Ñ1Î2 œ
" #
(2n1 1 2n1 1) œ 2n1
21(ln 2 x) ex dx œ 21 ln 2 '0 ex dx 21'0 xex dx ln 2
ln 2
œ (21 ln 2) cex d ln0 2 21 Œcxex d ln0 2 '0 ex dx ln 2
œ 21 ln 2 21 ˆ2 ln 2 cex d ln0 2 ‰ œ 21 ln 2 21 œ 21(1 ln 2)
54. (a) V œ '0 21xecx dx œ 21 Œcxecx d "! '0 ecx dx 1
1
œ 21 Š "e cecx d "! ‹ œ 21 ˆ "e œ 21
" e
1‰
41 e
(b) V œ '0 21(1 x)ecx dx; u œ 1 x, du œ dx; dv œ ecx dx, 1
v œ ecx ; V œ 21 ”c(1 x) aecx bd "! '0 ecx dx• 1
œ 21 ’[0 1(1)] cecx d "! “ œ 21 ˆ1
55. (a) V œ '0 21x cos x dx œ 21 Œ[x sin x] ! 1Î2
" e
1Î#
1‰ œ
21 e
'0 sin x dx 1Î2
1Î#
œ 21 Š 1# [cos x] ! ‹ œ 21 ˆ 1# 0 1‰ œ 1(1 2)
(b) V œ '0 21 ˆ 1# x‰ cos x dx; u œ 1Î2
1Î#
V œ 21 ˆ 1# x‰ sin x‘ !
1 #
x, du œ dx; dv œ cos x dx, v œ sin x;
21'0 sin x dx œ 0 21[ cos x] ! 1Î2
1Î#
œ 21(0 1) œ 21
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.1 Integration by Parts
469
56. (a) V œ '0 21x(x sin x) dx; 1
sin x ÐÑ x# ïïïïî cos x ÐÑ 2x ïïïïî sin x ÐÑ 2 ïïïïî cos x
Ê V œ 21'0 x# sin x dx œ 21 cx# cos x 2x sin x 2 cos xd ! œ 21 a1# 4b 1
0
1
(b) V œ '0 21(1 x)x sin x dx œ 21# '0 x sin x dx 21 '0 x# sin x dx œ 21# [x cos x sin x]1! a21$ 81b 1
1
1
œ 81 57. (a) A œ '1 ln x dx œ ’x ln x“ '1 dx e
e
e
1 e
œ ae ln e 1 ln 1b ’x“ œ e ae 1b œ 1 1
(b) V œ '1 1aln xb dx œ 1Š’x aln xb2 “ '1 2 ln x dx‹ e
e
2
e
1
œ 1”Šealn eb 1aln 1b ‹ Š’2x ln x“ '1 2 dx‹• 2
e
2
e
1
e
œ 1”e Ša2e ln e 2a1b ln 1b ’2x“ ‹• 1
œ 1”e a2e a2e 2bb• œ 1ae 2b e
(c) V œ '1 21ax 2b ln x dx œ 21'1 ax 2b ln x dx œ 21Œ”ˆ 12 x2 2x‰ln x• '1 ˆ 12 x 2‰ dx e
e
e
1
e
œ 21Œˆ 12 e2 2e‰ln e ˆ 12 2‰ln 1 ”ˆ 14 x2 2x‰• œ 21ˆˆ 12 e2 2e‰ ˆˆ 14 e2 2e‰ 94 ‰‰ œ 12 ae2 9b 1
(d) M œ '1 ln x dx œ 1 (from part (a)); x œ e
1 1
'1 x ln x dx œ ” 12 x2 ln x•
œ 12 e2 Š 14 e2 14 a1b2 ‹ œ 14 ae2 1b; y œ
e
1 1
œ œ
1 2 ae
e
1
2e 2e 2b œ
2b Ê axß yb œ 1
58. (a) A œ '0 tan1 x dx œ ”x tan1 x• '0 1
1
0
e
e
1 † aln 1b ‹ Œ”2x ln x• '1 2 dx œ 2
1 2 ae
1
'1 12 x dx œ Š 12 e2 ln e 12 a1b2 ln 1‹ ” 14 x2 •
'1 12 aln xb2 dx œ 12 Œ”x aln xb2 •
e
2 1 2 ŒŠe aln eb
e
2 Š e 4 1 ß e 2 2 ‹
1 2 Še
e 1
'1 2 ln x dx e
e
Ša2e ln e 2a1b ln "b ’2x“ ‹‹ 1
is the centroid.
x 1 x2 dx
1
œ atan1 1 0b 12 ”lna1 x2 b• œ
1 4
21 aln 2 ln 1b œ
(b) V œ '0 21 x tan1 x dx
1 4
0
21 ln 2
1
1
œ 21” x2 tan1 x•
" #
'01 1 x x dx
œ 21Œ 12 tan1 1 0
" #
'01 ˆ1 1 1 x ‰dx œ 21 18
2
0
œ
21ˆ 18
"ˆ # 1
1 ‰‰ 4
2
2
œ
2
1
#" ”x tan1 x• œ 21ˆ 18 #" a1 tan1 1 a0 0bb‰ 0
1a1 2b 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
e 1
470
Chapter 8 Techniques of Integration
59. av(y) œ œ
" 1
" #1
'021 2ect cos t dt
ect ˆ sin t # cos t ‰‘ #1 !
(see Exercise 22) Ê av(y) œ
" #1
a1 ec21 b
'021 4ect (sin t cos t) dt 21 21 œ 12 '0 ect sin t dt 12 '0 ect cos t dt
60. av(y) œ
œ œ
2 1 2 1
" #1
ect ˆ sin t# cos t ‰ ect ˆ sin t # cos t ‰‘ #1 ! cect sin td #!1 œ 0
61. I œ ' xn cos x dx; cu œ xn , du œ nxn" dx; dv œ cos x dx, v œ sin xd Ê I œ xn sin x ' nxn" sin x dx 62. I œ ' xn sin x dx; cu œ xn , du œ nxn" dx; dv œ sin x dx, v œ cos xd Ê I œ xn cos x ' nxn" cos x dx 63. I œ ' xn eax dx; u œ xn , du œ nxn" dx; dv œ eax dx, v œ "a eax ‘ ÊIœ
xn eax ax a e
n a
' xn" eax dx, a Á !
64. I œ ' aln xbn dx; ’u œ aln xbn , du œ
naln xbn " x
dx; dv œ " dx, v œ x“
Ê I œ xaln xbn ' naln xbn" dx 65.
'ab ax ab faxb dx; ’u œ x a, du œ dx; dv œ faxb dx, v œ 'bx fatb dt œ 'xb fatb dt“ b
œ ”ax ab'b fatb dt• 'a Š'b fatb dt‹ dx œ Œab ab'b fatb dt aa ab'b fatb dt 'a Œ'x fatb dt dx x
b
x
b
a
b
b
a
œ 0 'a Œ'x fatb dt dx œ 'a Œ'x fatb dt dx b
66.
b
b
b
' È1 x2 dx; ’u œ È1 x2 , du œ È x œ x È1 x2 '
1 x2
x 2 È 1 x2
dx œ x È1 x2 '
œ x È1 x2 ' È1 x2 dx ' Ê ' È1 x2 dx œ x È1 x2 ' Ê ' È1 x2 dx œ
x 2
dx; dv œ dx, v œ x“
È1 x2
1 2
" È 1 x2 " È 1 x2
'È"
" x2 " È 1 x2
dx œ x È1 x2 Š'
" x2 È 1 x2
dx '
" È 1 x2
dx‹
dx dx ' È1 x2 dx Ê 2' È1 x2 dx œ x È1 x2 '
1 x2
dx C
67.
' sin" x dx œ x sin" x ' sin y dy œ x sin" x cos y C œ x sin" x cos asin" xb C
68.
' tan" x dx œ x tan" x ' tan y dy œ x tan" x ln kcos yk C œ x tan" x ln kcos atan" xbk C Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" È 1 x2
dx
Section 8.2 Trigonometric Integrals 69.
471
' sec" x dx œ x sec" x ' sec y dy œ x sec" x ln ksec y tan yk C œ x sec" x ln ksec asec" xb tan asec" xbk C œ x sec" x ln ¹x Èx# 1¹ C
70.
' log2 x dx œ x log2 x ' 2y dy œ x log2 x ln2 # C œ x log2 x lnx# C y
71. Yes, cos" x is the angle whose cosine is x which implies sin acos" xb œ È1 x# . 72. Yes, tan" x is the angle whose tangent is x which implies sec atan" xb œ È1 x# . 73. (a)
' sinh" x dx œ x sinh" x ' sinh y dy œ x sinh" x cosh y C œ x sinh" x cosh asinh" xb C; check: d cx sinh" x cosh asinh" xb Cd œ ’sinh" x
x È 1 x#
œ sinh" x dx
(b)
' sinh" x dx œ x sinh" x ' œ x sinh" x a1 x# b
"Î#
" ‹ 1 x#
dx œ x sinh" x
"Î#
C“ œ ’sinh" x
x È 1 x#
dx
' a1 x# b"Î# 2x dx
x È 1 x# “
dx œ sinh" x dx
' tanh" x dx œ x tanh" x ' tanh y dy œ x tanh" x ln kcosh yk C œ x tanh" x ln kcosh atanh" xbk C; check: d cx tanh" x ln kcosh atanh" xbk Cd œ ’tanh" x œ tanh" x
(b)
" #
" È 1 x# “
C
check: d ’x sinh" x a1 x# b 74. (a)
x ŠÈ
sinh asinh" xb
x 1 x#
x ‘ 1 x#
" #
sinh atanh " xb " cosh atanh " xb 1 x# “
dx
dx œ tanh" x dx
' tanh" x dx œ x tanh" x ' 1 x x check: d x tanh" x
x 1 x#
dx œ x tanh" x #" ' 12xx# dx œ x tanh" x #" ln k1 x# k C ln k1 x# k C‘ œ tanh" x 1 x x# 1 x x# ‘ dx œ tanh" x dx #
8.2 TRIGONOMETRIC INTEGRALS 1.
' cos 2x dx œ "# ' cos 2x † 2dx œ "# sin 2x C
2.
'01 3 sin x3 dx œ 9'01 sin x3 †
3.
' cos3 x sin x dx œ ' cos3 x asin xbdx œ 14 cos4 x C
4.
' sin4 2x cos 2x dx œ "# ' sin4 2x cos 2x † 2dx œ 101 sin5 2x C
5.
' sin3 x dx œ ' sin2 x sin x dx œ ' a1 cos2 xb sin x dx œ ' sin x dx ' cos2 x sin x dx œ cos x 13 cos3 x C
6.
' cos3 4x dx œ ' cos2 4x cos 4x dx œ 14 ' a1 sin2 4xb cos 4x † 4dx œ 14 ' cos 4x † 4dx 14 ' sin2 4x cos 4x † 4dx œ 14 sin 4x
7.
1 3 12 sin 4x
1 3 dx
1
œ 9’cos x3 “ œ 9ˆcos 13 cos 0‰ œ 9ˆ "# 1‰ œ 0
9 #
C
' sin5 x dx œ ' asin# xb# sin x dx œ ' a" cos# xb# sin x dx œ ' a" 2cos# x cos4 xbsin x dx œ ' sin x dx ' 2cos# x sin x dx ' cos4 x sin x dx œ cos x #3 cos3 x 51 cos5 x C Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
472 8.
Chapter 8 Techniques of Integration
'01 sin5 ˆ x2 ‰dx (using Exercise 7) œ '01 sinˆ x2 ‰dx '01 2cos# ˆ x2 ‰sinˆ x2 ‰dx '01 cos4 ˆ x2 ‰sinˆ x2 ‰dx 1
œ #cos ˆ x2 ‰ 3% cos3 ˆ 2x ‰ 5# cos5 ˆ 2x ‰‘ 0 œ a!b ˆ#
% $
&# ‰ œ
"' "&
9.
' cos3 x dx œ ' acos# xbcos x dx œ ' a" sin# xbcos x dx œ ' cos x dx ' sin# x cos x dx œ sin x 13 sin3 x C
10.
'01/6 3cos5 3x dx œ '01/6 acos# 3xb# cos 3x † 3dx œ '01/6 a" sin# 3xb# cos 3x † 3dx œ '01/6 a" #sin# 3x sin% 3xbcos 3x † 3dx œ '0 cos 3x † 3dx #'0 sin# 3x cos 3x † 3dx '0 sin% 3x cos 3x † 3dx œ ’sin 3x # sin33x 1/6
œ ˆ" 11.
1/6
2 $
"& ‰ a!b œ
1/6
3
1Î6 sin5 3x 5 “0
) "&
' sin3 x cos3 x dx œ ' sin3 x cos2 x cos xdx œ ' sin3 x a1 sin2 xbcos x dx œ ' sin3 x cos x dx ' sin5 x cos x dx œ 14 sin4 x 16 sin6 x C
12.
13.
' cos3 2x sin5 2x dx œ "# ' cos3 2x sin5 2x † 2dx œ "# ' cos 2x cos2 2x sin5 2x † 2dx œ "# ' a1 sin2 2xb sin5 2x cos 2x † 2dx 1 1 œ "# ' sin5 2x cos 2x † 2dx "# ' sin7 2x cos 2x † 2dx œ 12 sin6 2x 16 sin8 2x C 2x ' cos2 x dx œ ' 1 cos dx œ "# ' a1 cos 2xbdx œ "# ' dx "# ' cos 2x dx œ "# ' dx 4" ' cos 2x † 2dx 2
œ "# x 4" sin 2x C
14.
1/2 1/2 1/2 1/2 1/2 2x '01/2 sin2 x dx œ '01/2 1 cos dx œ "# '0 a1 cos 2xbdx œ "# '0 dx "# '0 cos 2x dx œ "# '0 dx 4" '0 cos 2x † 2dx 2 1 Î2
œ "# x "4 sin 2x‘ 0 15.
œ ˆ #" ˆ 12 ‰ 4" sin 2ˆ 12 ‰‰ ˆ #" a0b 4" sin 2a0b‰ œ ˆ 14 0‰ a0 0b œ
'01/2 sin7 y dy œ '01/2 sin6 y sin y dy œ '01/2 a" cos2 yb$ sin y dy œ '01/2 sin y dy $'01/2 cos2 y sin y dy $'0 cos4 y sin y dy '0 cos6 y sin y dy œ ’cos y $ cos3 y $ cos5 y 1/2
16.
1/2
3
18.
19.
3
&
1Î# cos( y ( “0
œ a!b ˆ" "
$ &
"( ‰ œ
' 7cos7 t dt (using Exercise 15) œ 7’' cos t dt $' sin2 t cos t dt $' sin4 t cos t dt ' sin6 t cos t dt“ &
œ 7Šsin t $ sin3 t $ sin5 t 17.
1 4
sin( t ( ‹
C œ 7sin t 7sin3 t
21 & 5 sin t
sin( t C
1 1 1 1 4x #x ‰ # '01 )sin4 x dx œ )'01 ˆ " cos dx œ #'0 a" #cos #x cos# #xbdx œ #'0 dx #'0 cos #x † #dx #'0 " cos dx # # 1 1 1 1 " œ c#x #sin #xd 0 '0 dx '0 cos 4x dx œ #1 x # sin 4x‘ 0 œ #1 1 œ $1
' )cos4 21x dx œ )' ˆ " cos# 41x ‰# dx œ #' a" #cos 41x cos# 41xbdx œ #' dx %' cos 41x dx #' " cos# )1x dx œ 3' dx %' cos 41x dx ' cos )1x dx œ 3x 1" sin 41 x )"1 sin )1 x C #x ‰ˆ " cos #x ‰ 4x ‰ ' 16 sin# x cos# x dx œ 16' ˆ " cos dx œ %' a" cos# 2xbdx œ %' dx %' ˆ " cos dx # # 2 œ 4x 2' dx 2' cos 4x dx œ 4x 2x "2 sin 4x C œ 2x 2" sin 4x C œ 2x sin 2x cos 2x C
œ 2x 2sin x cos x a2cos2 x 1b C œ 2x 4sin x cos3 x 2sin x cos x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"' $&
Section 8.2 Trigonometric Integrals 20.
473
1 1 1 1 #y ‰# ˆ " cos #y ‰ '01 8 sin4 y cos# y dy œ 8'01 ˆ " cos dy œ '0 dy '0 cos #y dy '0 cos# #y dy '0 cos$ #y dy # # 1 1 1 1 1 1 4y ‰ "' "' # ' ' œ y 2" sin 2y‘ 0 '0 ˆ " cos dy a " sin # y b cos # y dy œ 1 dy cos 4y dy cos #y dy # # 0 # 0 0 0 1 1 sin 2y '0 sin# #y cos #y dy œ 1 ’ "# y ") sin 4y "# sin 2y "# † 3 “ œ 1 1# œ 1# 3
0
21.
'
22.
'01/2 sin2 2) cos3 2) d) œ '01/2 sin2 2)a" sin2 2)bcos 2) d) œ '01/2 sin2 2) cos 2) d) '01/2 sin4 2) cos 2) d)
8cos3 2) sin 2) d) œ 8ˆ "# ‰ cos4 2) C œ cos4 2) C 4
œ ’ "# †
sin3 2) 3
" #
†
1/2 sin5 2) 5 “0
œ!
23.
'021 É " #cos x dx œ '021 ¹ sin x# ¹dx œ '021 sin x# dx œ #cos x# ‘ 20 1 œ # # œ %
24.
'01 È" cos 2x dx œ '01 È# lsin 2x ldx œ '01 È# sin 2x dx œ ’È#cos 2x“ 1 œ È# È# œ #È# 0
25.
'01 È" sin# t dt œ '01 l cos t ldt œ '01/2 cos t dt '11/2 cos t dt œ csin td 10 /2 csin td 11/2 œ " ! ! " œ #
26.
'01 È" cos# ) d) œ '01 l sin ) ld) œ '01 sin ) d) œ ccos )d 10 œ " " œ #
27.
'11ÎÎ32 È sin x 2
" cos x 1 Î2
dx œ '1Î3
1Î2
È" cos x sin2 x È" cos x È" cos x
dx œ '1Î3
1Î2
œ '1Î3 sin x È" cos x dx œ ’ 23 a" cos xb3Î2 “
1 Î2 1 Î3
œ
28.
É 32
1Î2
sin2 x È" cos x Èsin2 x
2
" sin x
'511Î6 Ècos x
" sin x
dx œ '0
1Î6
cos x È" sin x
dx
dx œ '51Î6
'
" sin x
œ'
1
'
1 1 4 È cos4 x È" sin x x " sin x cos4 x È" sin x dx œ 51Î6 cos È dx È" sin x È" sin x dx œ 51Î6 È" sin2 x cos2 x 1 1 1 cos4 x È" sin x dx œ 51Î6 cos3 x È" sin x dx œ 51Î6 cos xa" sin2 xb È" sin x dx cos x 5 1 Î6 1 1 51Î6 cos x È" sin x dx 51Î6 cos x sin2 x È" sin x dx; u2 Èu du 4
'
'
'
'
’Let u œ 1 sin x Ê u 1 œ sin x Ê du œ cos x dx, x œ 1
œ ’ 23 a" sin xb3Î2 “ Š 23 a"
sin 1b
3Î2
51 6
Ê u œ 1 sinˆ 561 ‰ œ 32 , x œ 1 Ê u œ 1 sin 1 œ 1 “
'3Î2 au 1b2 Èu du œ ’ 23 a" sin xb3Î2 “
1
1
5 1 Î6
œ
2
" sin x
œ 2É" sin ˆ 16 ‰ 2È" sin 0 œ 2É 1# 2È1 œ 2 È2
0
œ
dx
3 Î2 3 Î2 3Î2 œ 23 ˆ" cosˆ 12 ‰‰ 23 ˆ" cos ˆ 13 ‰‰ œ 32 32 ˆ 23 ‰
'01Î6 È" sin x dx œ '01Î6 È" 1 sin x ÈÈ" sin x dx œ '01Î6 ÈÈ" sin x dx œ '01Î6 ÈÈcos x 1Î6
2ˆ 3 "
3 Î2 œ Š 23 23 ˆ 32 ‰ ‹ ˆ 27
30.
dx œ '1Î3
2 3
œ ’2a" sin xb1Î2 “
29.
sin2 x È" cos x È" cos2 x
5 1 Î6
'3Î2 ˆu5Î2 2u3Î2 Èu‰ du 1
1 3Î2 sin ˆ 561 ‰‰ ‹ 72 u7Î2 54 u5Î2 32 u3Î2 ‘3Î2 4 5
7Î2 5Î2 3Î2 5Î2 7Î2 23 ‰ Š 27 ˆ 32 ‰ 45 ˆ 32 ‰ 23 ˆ 32 ‰ ‹ œ 45 ˆ 32 ‰ 27 ˆ 32 ‰
'17Î12Î12 È1 sin 2x dx œ '17Î12Î12 È1 1sin 2x ÈÈ1 sin 2x dx œ '17Î12Î12 ÈÈ1 sin 2x dx œ '17Î12Î12 ÈÈcos 2x 2
œ '1Î2
71Î12
cos 2x È1 sin 2x
1 sin 2x 71Î12
dx œ ’È1 sin 2x“
1 Î2
1 sin 2x
2
1 sin 2x
18 35
dx
œ É1 sin 2ˆ 7121 ‰ É1 sin 2ˆ 12 ‰ œ É "# 1 œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 È2
œ
È2 1 È2
474
Chapter 8 Techniques of Integration
31.
'01/2 )È" cos 2) d) œ '01/2 )È# l sin ) l d) œ È#'01/2 ) sin ) d) œ È# c)cos ) sin )d 10 /2 œ È#a"b œ È#
32.
'11 a" cos# tb$Î# dt œ '11 asin# tb$Î# dt œ '11 ¸ sin$ t¸ dt œ '!1 sin$ t dt '!1 sin$ t dt œ '!1 a" cos# tbsin t dt '! a" cos# tbsin t dt œ '1 sin t dt '1 cos# t sin t dt '! sin t dt '! cos# t sin t dt œ ’cos t 1
’cos t
33. 34.
!
1 cos3 t 3 “!
œ ˆ"
" $
!
" "$ ‰ ˆ"
1
" $
" "$ ‰ œ
1
! cos3 t 3 “ 1
) $
' sec2 x tan x dx œ ' tan x sec2 x dx œ 12 tan 2 x C ' sec x tan2 x dx œ ' sec x tan x tan xdx; u œ tan x, du œ sec2 x dx, dv œ sec x tan x dx, v œ sec x; œ sec x tan x ' sec3 x dx œ sec x tan x ' sec2 x sec xdx œ sec x tan x ' atan2 x 1bsec xdx œ sec x tan x Š' tan2 x sec xdx ' sec xdx‹ œ sec x tan x lnlsec x tan xl ' tan2 x sec xdx Ê ' sec x tan2 x dx œ sec x tan x lnlsec x tan xl ' tan2 x sec xdx Ê 2' tan2 x sec xdx œ sec x tan x lnlsec x tan xl Ê ' tan2 x sec xdx œ "# sec x tan x "# lnlsec x tan xl C
35. 36.
' sec3 x tan x dx œ ' sec2 x sec x tan x dx œ 13 sec 3 x C ' sec3 x tan3 x dx œ ' sec2 x tan2 x sec x tan x dx œ ' sec2 xasec2 x 1bsec x tan x dx œ ' sec4 x sec x tan x dx ' sec2 x sec x tan x dx œ 15 sec 5 x 13 sec 3 x C
37.
' sec2 x tan2 x dx œ ' tan2 x sec2 x dx œ 13 tan 3 x C
38.
' sec4 x tan2 x dx œ ' sec2 x tan2 x sec2 x dx œ ' atan2 x 1btan2 x sec2 x dx œ ' tan4 x sec2 x dx ' tan2 x sec2 x dx œ 15 tan 5 x 13 tan 3 x C
39.
'!1/3 2 sec$ x dx; u œ sec x, du œ sec x tan x dx, dv œ sec# x dx, v œ tan x;
'!1/3 2 sec$ x dx œ c2 sec x tan xd !1Î$ #'!1/3 sec x tan2 x dx œ # † " † ! # † # † È$ #'!1/3 sec x asec2 x "bdx œ %È$ #'1/3 sec$ x dx #'1/3 sec x dx; 2'1/3 2 sec$ x dx œ %È$ c#ln l sec x + tan xld !1Î$ !
!
!
2'1/3 2 sec$ x dx œ %È$ #ln l " + !l #ln l # È$ l œ %È$ # ln Š# È$‹ !
'!1/3 2 sec$ x dx œ #È$ ln Š# 40.
È $‹
' ex sec$ aex bdx; u œ secaex b, du œ secaex btanaex bex dx, dv œ sec# aex bex dx, v œ tanaex b. ' ex sec$ aex b dx œ secaex btanaex b ' secaex btan# aex bex dx œ secaex btanaex b ' secaex basec# aex b "bex dx œ secaex btanaex b ' sec$ aex bex dx ' secaex bex dx 2' ex sec$ aex b dx œ secaex btanaex b ln¸secaex b tanaex b¸ C ' ex sec$ aex b dx œ "# ˆsecaex btanaex b ln¸secaex b tanaex b¸‰ C Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.2 Trigonometric Integrals 41.
475
' sec4 ) d) œ ' a" tan2 )bsec2 ) d) œ ' sec2 ) d) ' tan2 ) sec2 ) d) œ tan ) 13 tan3 ) C œ tan ) 13 tan )asec2 ) 1b C œ 13 tan ) sec2 ) 23 tan ) C
42.
'
3sec4 a3xb dx œ ' a" tan2 a3xbbsec2 a3xb3dx œ ' sec2 a3xb3dx ' tan2 a3xb sec2 a3xb3dx œ tan a3xb 13 tan3 a3xb C
43.
'11/4/2 csc4 ) d) œ '11/4/2 a" cot# )bcsc# ) d) œ '11/4/2 csc# ) d) '11/4/2 cot# ) csc# ) d) œ ’cot ) cot3 ) “ 1Î2 $
œ a!b ˆ" "$ ‰ œ 44.
45.
1/4
% $
' sec6 x dx œ ' sec4 x sec2 x dx œ ' atan2 x 1b2 sec2 x dx œ ' atan4 x 2tan2 x 1bsec2 x dx œ ' tan4 x sec2 x dx 2' tan2 x sec2 x dx ' sec2 x dx œ 15 tan 5 x 23 tan 3 x tan x C ' 4 tan3 x dx œ 4'
asec# x "btan x dx œ 4' sec# x tan x dx 4' tan x dx œ % tan# x % ln lsec xl C #
œ 2 tan# x 4 ln lsec xl C œ 2 tan# x 2 ln lsec2 xl C œ 2 tan# x 2 ln a1 tan2 xb C 46.
'11/4/4 6 tan4 x dx œ 6'11/4/4
asec# x "btan2 x dx œ 6'1/4 sec# x tan2 x dx 6'1/4 tan2 x dx 1/4
1/4
œ 6'1/4 sec2 x tan2 x dx 6'1/4 asec2 x 1bdx œ ’' tan$ x “ 1/4
œ #a" a"bb c'tan 47.
1 Î4 xd 1Î4
1/4
$
1Î4 c'xd 1Î4
49.
œ % 'a" a"bb
1/4
$1 #
$1 #
1/4
œ $1 )
1 4
atan2 x 1b
2
atan2 x 1b lnlsec xl C œ 14 tan4 x "# tan2 x lnlsec xl C
' cot6 2x dx œ ' cot4 2x cot2 2x dx œ ' cot4 2x acsc2 2x 1b dx œ ' cot4 2x csc2 2x dx ' cot4 2x dx œ ' cot4 2x csc2 2x dx ' cot2 2x cot2 2x dx œ ' cot4 2x csc2 2x dx ' cot2 2xacsc2 2x 1bdx œ ' cot4 2x csc2 2x dx ' cot2 2x csc2 2x dx ' cot2 2x dx œ ' cot4 2x csc2 2x dx ' cot2 2x csc2 2xdx ' acsc2 2x 1bdx 1 œ ' cot4 2x csc2 2x dx ' cot2 2x csc2 2xdx ' csc2 2x dx ' dx œ 10 cot5 2x 16 cot3 2x "# cot 2x x C '11/6/3 cot3 x dx œ '11/6/3 acsc2 x " bcot x dx œ '11/6/3 csc2 x cot x dx '11/6/3 cot x dx œ ’ cot# x ln l csc xl“ 1Î3 #
1Î6
œ "# ˆ "$ $‰ Šln È#$ ln #‹ œ 50.
1 Î4
6'1/4 sec# x dx 6'1/4 dx
' tan5 x dx œ ' tan4 x tan x dx œ ' asec2 x 1b2 tan x dx œ ' asec4 x 2sec2 x 1btan x dx œ ' sec4 x tan x dx 2' sec2 x tan x dx ' tan x dx œ ' sec3 x sec x tan x dx 2' sec x sec x tan x dx ' tan x dx œ 14 sec4 x sec2 x lnlsec xl C œ
48.
1Î4
% $
lnÈ$
' 8 cot4 t dt œ 8' acsc2 t " bcot2 t dt œ 8' csc2 t cot2 t dt 8' cot2 t dt œ 83 cot3 t 8' acsc2 t " bdt œ 83 cot3 t 8 cot t 8t C
51.
' sin 3x cos 2x dx œ "# ' asin x sin 5xb dx œ "# cos x 10" cos 5x C
52.
' sin 2x cos 3x dx œ "# ' asinaxb sin 5xb dx œ "# ' asin x sin 5xb dx œ "# cos x 10" cos 5x C Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
476
Chapter 8 Techniques of Integration
53.
'11 sin 3x sin 3x dx œ "# '11 acos ! cos 6xb dx œ "# '11 dx "# '11 cos 6x dx œ "# x "#" sin 6x‘ 11 œ 1# 1# ! œ 1
54.
'!1Î2 sin x cos x dx œ "# '!1Î2 asin ! sin 2xb dx œ "# '!1Î2 sin 2x dx œ "% ccos 2xd 1! Î# œ "% a" "b œ "#
55.
' cos 3x cos 4x dx œ "# ' acosaxb cos 7xb dx œ "# ' acos x cos 7xb dx œ 2" sin x 14" sin 7x C
56.
'11ÎÎ22 cos 7x cos x dx œ "# '11ÎÎ22 acos 6x cos 8xb dx œ 2" 6" sin 6x 8" sin 8x‘ 1Î12Î2 œ 0
57.
58.
59.
60.
2) ' sin2 ) cos 3) d) œ ' 1 cos cos 3) d) œ "# ' cos 3) d) "# ' cos 2) cos 3) d) 2 œ "# ' cos 3) d) "# ' "# acosa 2 3b) cosa 2 3b)b d) œ "# ' cos 3) d) 4" ' acosa)b 1 œ "# ' cos 3) d) 4" ' cos ) d) 4" ' cos 5) d) œ 16 sin 3) 4" sin ) 20 sin 5) C
cos 5)b d)
' cos2 2) sin ) d) œ ' a2cos2 ) 1b2 sin ) d) œ ' a4cos4 ) 4cos2 ) 1b sin ) d) œ ' 4cos4 ) sin ) d) ' 4cos2 ) sin ) d) ' sin ) d) œ 45 cos5 ) 43 cos3 ) cos ) C ' cos3 ) sin 2) d) œ ' cos3 ) a2sin ) cos )b d) œ 2' cos4 ) sin ) d) œ 25 cos5 ) C ' sin3 ) cos 2) d) œ ' sin2 ) cos 2) sin ) d) œ ' a1 cos2 )ba2cos2 ) 1bsin ) d) œ ' a2cos4 ) 3cos2 ) 1bsin ) d) œ 2' cos4 ) sin ) d) 3' cos2 ) sin ) d) ' sin ) d) œ 25 cos5 ) cos3 ) cos ) C
61.
62.
' sin ) cos ) cos 3) d) œ "# ' 2sin ) cos ) cos 3) d) œ "# ' sin 2) cos 3) d) œ "# ' "# asina2 3b) sina2 3b)b d) 1 œ "4 ' asina)b sin 5)b d) œ 4" ' asin ) sin 5)b d) œ 4" cos ) 20 cos 5) C ' sin ) sin 2) sin 3) d) œ ' "# acosa1 2b) cosa1 2b)b sin 3) d) œ "# ' acosa)b cos 3)b sin 3) d) œ "# ' sin 3) cos ) d) "# ' sin 3) cos 3) d) œ "# ' "# asina3 1b) sina3 1b)bd) 4" ' 2sin 3) cos 3) d) œ "4 ' asin 2) sin 4)bd) 4" ' sin 6) d) œ 4" ' asin 2) sin 4)bd) 4" ' sin 6) d) œ 18 cos 2)
63.
'
sec3 x tan x
dx œ '
1 16 cos 4)
sec2 x sec x tan x
dx œ '
6) C
1 24 cos
atan2 x 1bsec x tan x
dx œ '
tan2 x sec x tan x
dx '
sec x tan x
dx œ ' tan x sec x dx ' csc x dx
œ sec x lnlcsc x cot xl C
64.
'
sin3 x cos4 x
dx œ '
tan2 x csc x
dx œ '
sin# x sin x cos4 x
dx œ '
a1cos# xb sin x cos4
x
dx œ '
sin x cos4 x
dx '
œ ' sec2 x sec x tan x dx ' sec x tan x dx œ 13 sec3 x sec x C 65.
'
sin2 x sin x dx cos2 x
œ'
a1cos# xb cos2 x
sin x dx œ '
cos# x sin x cos4 x
1 sin x dx cos2 x
'
dx œ ' sec3 x tan x dx ' sec x tan x dx
cos# x sin x dx cos2 x
œ ' sec x tan x dx ' sin x dx
œ sec x cos x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.2 Trigonometric Integrals 66.
67.
'
dx œ '
cot x cos2 x
cos x sin x
dx œ '
1 cos2 x
†
2 2sin x cos x
dx œ '
2 sin 2x
477
dx œ ' csc 2x 2dx œ lnl csc 2x cot 2x l C
2x ' x sin2 x dx œ ' x 1 cos dx œ "# ' x dx "# ' x cos 2x dx ’u œ x, du œ dx, dv œ cos 2x dx, v œ "# sin 2x“ 2
œ "4 x2 #" ’ #" x sin 2x ' #" sin 2x dx“ œ 4" x2 4" x sin 2x 18 cos 2x C 68.
' x cos3 x dx œ ' x cos2 x cos x dx œ ' xa1 sin2 xb cos x dx œ ' x cos x dx ' x sin2 x cos x dx; ' x cos x dx œ x sin x ' sin x dx œ x sin x cos x; ’u œ x, du œ dx, dv œ cos x dx, v œ sin x“
' x sin2 x cos x dx
œ 13 x sin3 x ' 13 sin3 x dx;
’u œ x, du œ dx, dv œ sin2 x cos x dx, v œ 13 sin3 x“
' a1 cos2 xb sin x dx œ 13 x sin3 x 13 ' sin x dx 13 ' cos2 x sin x dx œ 13 x sin3 x 13 cos x 19 cos3 x; Ê ' x cos x dx ' x sin2 x cos x dx œ ax sin x cos xb ˆ 13 x sin3 x 13 cos x 19 cos3 x‰ C œ 13 x sin3 x
1 3
œ x sin x 13 x sin3 x 23 cos x 19 cos3 x C 69. y œ lnasec xb; y w œ
sec x tan x sec x
œ tan x;ay w b# œ tan# x; '!
1Î4
È" tan# x dx œ '
1Î4
!
lsec xl dx œ clnlsec x tan xld !1/4
œ lnŠÈ# "‹ lna! "b œ lnŠÈ# "‹ " 70. M œ '1Î4 sec x dx œ clnlsec x tan xld 1/41Î4 œ lnŠÈ# "‹ ln lÈ# "l œ ln È## " 1Î4
yœ
È
È ' ln È# " 1Î4 #" 1Î4
"
sec# x #
Ê ax, yb œ Œ!ß Šln
dx œ
1Î4 " ctan xd 1Î4 œ È # " #ln È #"
È#" " È#" ‹
71. V œ 1'! sin# x dx œ 1'! 1
1
"
È a" a"bb #ln È# " #"
1 #
dx œ
È# #
#
1
1Î%
csin 2xd !
È# #
1Î%
$1Î%
csin 2xd 1Î%
È# #
73. M œ '0 ax cos xbdx œ ’ "# x2 sin x“
0
1 21 2
csin 2xd 1$1Î% œ
21
21
xœ
#"
'!1 dx 1# '!1 cos 2x dx œ 1# cxd 1! 14 csin 2xd 1! œ 1# a1 !b 14 a! !b œ 1#
72. A œ '! È" cos 4x dx œ '! È# lcos 2xldx œ È# '! œ
"
È ln È# "
" cos 2x 2
1
œ
cos 2x dx È# '1Î% cos 2x dx È# '$1Î% cos 2x dx
È# #
$1Î%
a" !b
È# #
1
a" "b
È# #
a ! "b œ È # È # œ # È #
œ Š "# a21b2 sin a21b‹ Š "# a0b2 sin a0b‹ œ 212 ;
'021 xax cos xbdx œ 211 '021 ax2 x cos xbdx œ 211 '021 x2 dx 211 '021 x cos xdx 2
2
2
’u œ x, du œ dx, dv œ cos x dx, v œ sin x“ 0 '0 sin xdx
21 1 3 612 ’x “0
œ
41 3
œ
1 41 2
'021 ax2 2x cos x cos2 xbdx œ 411 '021 x2 dx 211 '021 x cos xdx 411 '021 cos2 x dx
21 1 212 Œ’x sin x“0
'0 sin xdx œ
œ
21 1 cos x ’ “ 2 21 0
œ
41 3
21
1 212 acos 21
1 3 612 a81
cos 0b œ 2
41 3
0b
0œ 2
41 3 ;
1 212 Œ21 sin 21
yœ
1 21 2
21
'021 "# ax cos xb2 dx 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
478
Chapter 8 Techniques of Integration œ
21 1 3 1212 ’x “0
œ
21 3
1Î3
œ 1'!
1 Î3
œ
'!
1Î3
1 2
dx
1 Î3
œ
21 3
1Î3
0
12 6
0
21 1 812 ’x“0
sin# x dx 1'!
1Î3
21
cos x“
1asin x sec xb2 dx œ 1'!
œ 12 ’ x“ œ
1 212 ’x sin x
21 1 1612 ’sin 2x“0
74. V œ '!
1 2
'!
1Î3
'021 cos 2x2 1 dx œ 231 0 811 '021 cos 2x dx 811 '021 dx 2
0
1Î3
cos 2x dx 21ˆlnlsec 1Î3 0
1 41
œ
81 2 3 121
2
Ê The centroid is Š 431 ,
81 2 3 121 ‹.
asin# x 2sin x sec x sec2 xb dx
2tan x dx 1'!
14 ’ sin 2x“
1È3 8
1 41 2
sec2 x dx œ 1'!
1Î3
1 3l
1 cos 2x 2
lnlsec 0l‰ 1ˆtan
1 3
1Î3
dx 21’ lnlsec xl“
0
1Î3
1’ tan x“
0
tan 0‰
21 ln 2 1È3 œ 12 ˆ 13 0‰ 14 ˆsin 2ˆ 13 ‰ sin 2a0b‰ 21 ln 2 1È3 1Š41 21È3 48 ln 2‹
21 ln 2 1È3 œ
24
8.3 TRIGONOMETRIC SUBSTITUTIONS 1. x œ 3 tan ), 1# ) 1# , dx œ
3 d) cos# )
, 9 x# œ 9 a1 tan# )b œ 9 sec# ) Ê
ˆbecause cos ) 0 when 1# ) 1# ‰ ;
' È dx
œ 3'
' È 3 dx
; c3x œ ud Ä
' È du
œ'
3.
'22
# œ #" tan" #x ‘ # œ
4.
'02 8 dx2x
5.
'03Î2 È dx
6.
'01Î2
9 x#
2.
1 9x# 1 u#
dx 4 x#
#
œ
9 x#
È2
" #
cos ) d) 3 cos# )
dt cos# t asec tb
'02 4dxx
#
œ'
'
d) cos )
œ ln ksec ) tan )k Cw œ ln ¹
du È 1 u#
; u œ tan t, 1# t
, du œ
œ
" #
" #
tan" 1
#" tan" #x ‘ # œ !
; ct œ 2xd Ä
'01Î2
È2
" #
" #
25 #
8. t œ
" 3
1 3ksec )k
œ
kcos )k 3
œ
cos ) 3 ;
x3 ¹ Cw œ ln ¹È9 x# x¹ C , È1 u# œ lsec tl œ sec t ;
" #
tan" (1) œ ˆ #" ‰ ˆ 14 ‰ ˆ #" ‰ ˆ 14 ‰ œ ˆ #" tan" 1
sin" 0 œ
dt È1 t#
1 6
" #
0œ
È2
1Î
œ csin" td 0
a) sin ) cos )b C œ
25 #
sin ), 1# ) 1# , dt œ
" 3
’sin" ˆ 5t ‰
1 4
tan" 0‰ œ ˆ #" ‰ ˆ #" ‰ ˆ 14 ‰ 0 œ
1 16
1 6
œ sin"
7. t œ 5 sin ), 1# ) 1# , dt œ 5 cos ) d), È25 t# œ 5 cos ); ' È25 t# dt œ ' (5 cos ))(5 cos )) d) œ 25 ' cos# ) d) œ 25 ' œ
dt cos# t
œ
œ ' sec t dt œ ln ksec t tan tk C œ ln ¹Èu# 1 u¹ C œ ln ¹È1 9x# 3x¹ C
$Î# œ sin" 3x ‘ ! œ sin"
2 dx È1 4x#
1 #
È 9 x# 3
" È 9 x#
È ˆ 5t ‰ Š 255 t# ‹“
" È2
sin" 0 œ
1 cos 2) #
Cœ
25 #
1 4
0œ
d) œ 25 ˆ #)
sin" ˆ 5t ‰
1 4
sin 2) ‰ 4
tÈ25 t# #
C
C
cos ) d), È1 9t# œ cos );
' È1 9t# dt œ "3 ' (cos ))(cos )) d) œ 3" ' cos# ) d) œ 6" a) sin ) cos )b C œ 6" ’sin" (3t) 3tÈ1 9t# “ C 9. x œ
'È
7 #
sec ), 0 ) 1# , dx œ
dx 4x# 49
œ'
ˆ 7#
sec ) tan )‰ d) 7 tan )
7 #
œ
sec ) tan ) d), È4x# 49 œ È49 sec# ) 49 œ 7 tan ); " #
' sec ) d) œ "# ln ksec ) tan )k C œ "# ln ¹ 2x7 È4x7# 49 ¹ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.3 Trigonometric Substitutions 10. x œ
sec ), 0 ) 1# , dx œ
3 5
' È 5 dx
25x# 9
œ'
sec ) tan ) d), È25x# 9 œ È9 sec# ) 9 œ 3 tan );
3 5
œ ' sec ) d) œ ln ksec ) tan )k C œ ln ¹ 5x 3
5 ˆ 35 sec ) tan )‰ d) 3 tan )
È25x# 9 ¹ 3
C
11. y œ 7 sec ), 0 ) 1# , dy œ 7 sec ) tan ) d), Èy# 49 œ 7 tan );
) tan )) d) ' Èy y 49 dy œ ' (7 tan ))(77 sec œ 7 ' tan# ) d) œ 7 ' asec# ) 1b d) œ 7(tan ) )) C sec ) #
œ 7’
Èy# 49 7
sec" ˆ y7 ‰“ C
12. y œ 5 sec ), 0 ) 1# , dy œ 5 sec ) tan ) d), Èy# 25 œ 5 tan ); sec ) tan )) d) ' Èyy 25 dy œ ' (5 tan ))(5 œ "5 ' 125 sec ) #
$
œ
" 10
$
a) sin ) cos )b C œ
" 10
tan# ) cos# ) d) œ
’sec" ˆ y5 ‰ Š
" 5
Èy# 25 ‹ Š 5y ‹“ y
'
sin# ) d) œ
Cœ’
sec " 10
" 10
ˆ 5y ‰
' (1 cos 2)) d)
Èy# 25 “ #y #
C
13. x œ sec ), 0 ) 1# , dx œ sec ) tan ) d), Èx# 1 œ tan );
'
dx x# È x# 1
œ'
sec ) tan ) d) sec# ) tan )
œ'
d) sec )
œ sin ) C œ
È x# 1 x
C
14. x œ sec ), 0 ) 1# , dx œ sec ) tan ) d), Èx# 1 œ tan );
'
2 dx x$ È x# 1
œ'
2 tan ) sec ) d) sec$ ) tan )
2) ‰ œ 2 ' cos# ) d) œ 2 ' ˆ 1 cos d) œ ) sin ) cos ) C # #
œ ) tan ) cos# ) C œ sec" x Èx# 1 ˆ "x ‰ C œ sec" x
È x# 1 x#
C
15. u œ 9 x# Ê du œ 2x dx Ê "# du œ x dx;
' Èx dx
9 x#
œ "# '
1 Èu du
œ Èu C œ È9 x# C
16. x œ 2 tan ), 1# ) 1# , dx œ 2sec# ) d) , 4 x# œ 4sec# )
' 4xdxx 2
#
œx
œ'
ˆ4tan2 )‰a2sec# )bd) 4sec# ) 1 ˆ x ‰ 2 tan 2 C
œ ' 2 tan2 ) d) œ 2' asec# ) 1bd) œ 2' sec# ) d) 2' d) œ 2 tan ) 2) C
17. x œ 2 tan ), 1# ) 1# , dx œ
' Èx
$ dx x# 4
œ'
œ 8 Œ
œ 8'
a8 tan$ )b (cos )) d) cos# )
ct œ cos )d Ä 8 ' È x# 4 #
t# 1 t%
2 d) cos# )
, Èx# 4 œ
sin$ ) d) cos% )
œ 8'
2 cos ) ; acos# ) 1b ( sin )) d) cos% )
dt œ 8' ˆ t"# t"% ‰ dt œ 8 ˆ "t
ax # 4 b 8 †3
$Î#
Cœ
" 3
ax# 4b
$Î#
" ‰ 3t$
;
C œ 8 Š sec )
sec$ ) 3 ‹
C
4Èx# 4 C œ 13 ax# 8bÈx# 4 C
18. x œ tan ), 1# ) 1# , dx œ sec# ) d), Èx# 1 œ sec );
'
dx x# È x# 1
œ'
sec# ) d) tan# ) sec )
œ'
cos ) d) sin# )
œ sin" ) C œ
È x # 1 x
C
19. w œ 2 sin ), 1# ) 1# , dw œ 2 cos ) d), È4 w# œ 2 cos );
'
8 dw w# È 4 w#
œ'
8†2 cos ) d) 4 sin# )†2 cos )
œ 2'
d) sin# )
œ 2 cot ) C œ
2 È 4 w # w
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
479
480
Chapter 8 Techniques of Integration
20. w œ 3 sin ), 1# ) 1# , dw œ 3 cos ) d), È9 w# œ 3 cos );
' È9w w
#
#
dw œ '
) ' acsc# ) 1b d) œ ' cot# ) d) œ ' Š 1 sinsin # ) ‹ d) œ #
3 cos )†3 cos ) d) 9 sin# )
œ cot ) ) C œ
È 9 w# w
sin" ˆ w3 ‰ C
21. u œ 5x Ê du œ 5dx, a œ 6
1 10 " ˆ u ‰ " ˆ 5x ‰ ' 36 100 ' 1 ' 1 25x dx œ 20 a6b a5xb 5dx œ 20 a u du œ 20 † 6 tan 6 C œ 3 tan 6 C #
#
2
#
2
22. u œ x2 4 Ê du œ 2x dx Ê "# du œ x dx
' x Èx# 4 dx œ "# ' Èu du œ 13 u3Î2 C œ 13 ax# 4b3Î2 C
23. x œ sin ), 0 Ÿ ) Ÿ 13 , dx œ cos ) d), a1 x# b
È3Î2
'0
4x# dx a1 x# b$Î#
œ '0
1Î3
1Î$
œ cos$ );
) '0 asec# ) 1b d) œ 4 '0 Š 1 coscos # ) ‹ d) œ 4 1Î3
4 sin# ) cos ) d) cos$ )
24. x œ 2 sin ), 0 Ÿ ) Ÿ 16 , dx œ 2 cos ) d), a4 x# b
'01
dx a4 x# b$Î#
œ '0
1Î6
2 cos ) d) 8 cos$ )
œ
$Î#
dx ax# 1b$Î#
œ'
sec ) tan ) d) tan$ )
#
œ'
cos ) d) sin# )
x# dx ax# 1b&Î#
œ'
sec# )†sec ) tan ) d) tan& )
27. x œ sin ), 1# )
' a1 xxb
# $Î# '
dx
œ'
' a1 xxb %
29. x œ
'
" 3
œ'
œ'
8 ˆ "# sec# )‰ d) sec% )
6 dt a9t# 1b#
œ'
sec )‰ d) sec% ) #
x x# 1
$Î#
C
œ tan& ); x$ 3 ax# 1b$Î#
"Î#
È 1 x# & ‹ x
C
œ cos );
œ ' cot# ) csc# ) d) œ cot3 ) C œ "3 Š $
" #
C
œ cos$ ); &
È 1 x# $ ‹ x
C
#
sec# ) d), a4x# 1b œ sec% );
œ 4 ' cos# ) d) œ 2() sin ) cos )) C œ 2 tan" 2x
tan ), 1# ) 1# , dt œ 6 ˆ "3
œ tan$ );
d) œ 3 sin" $ ) C œ
, dx œ cos ) d), a1 x# b
cos )†cos ) d) sin% )
&Î#
" 4È 3
œ ' cot% ) csc# ) d) œ cot5 ) C œ "5 Š
tan ), 1# ) 1# , dx œ
8 dx a4x# 1b#
30. t œ
'
" #
dx
1 #
cos ) sin% )
, dx œ cos ) d), a1 x# b
cos$ )†cos ) d) sin' )
28. x œ sin ), 1# ) # "Î#
1 #
œ'
$Î#
œ sin" ) C œ È
26. x œ sec ), 0 ) 12 , dx œ sec ) tan ) d), ax# 1b
'
œ 8 cos$ );
'01Î6 cosd) ) œ "4 ctan )d 1! Î' œ È123 œ
" 4
25. x œ sec ), 0 ) 12 , dx œ sec ) tan ) d), ax# 1b
'
1Î3
#
41 3
œ 4È3
œ 4 ctan ) )d !
$Î#
" 3
4x a4x# 1b
C
sec# ) d), 9t# 1 œ sec# );
œ 2 ' cos# ) d) œ ) sin ) cos ) C œ tan" 3t
3t a9t# 1b
C
31. u œ x2 1 Ê du œ 2x dx Ê "# du œ x dx
' x x 1 dx œ ' ax x x 1 bdx œ ' x dx ' x x 1 dx œ #" x2 #" ' u1 du œ #" x2 #" ln lul C œ #" x2 #" ln lx2 1l C $
2
2
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.3 Trigonometric Substitutions 32. u œ 25 %x2 Ê du œ 8x dx Ê "8 du œ x dx
' 25 x %x dx œ 8" ' u1 du œ 8" ln lul C œ 8" ln a25 %x2 b C 2
33. v œ sin ), 1# ) 1# , dv œ cos ) d), a1 v# b
'
œ'
v# dv a1 v# b&Î#
sin# ) cos ) d) cos& )
œ ' tan# ) sec# ) d) œ
34. r œ sin ), 1# ) 1# ;
' a1 r rb
# &Î# )
œ'
dr
&Î#
œ cos& );
tan$ ) 3
Cœ
" 3
'0
et dt Èe2t 9
tan " Ð4Î3Ñ
œ 'tanc" Ð1Î3Ñ
œ ln ˆ 53 43 ‰ ln Š
(
tan " Ð4Î3Ñ
œ 'tan
3 tan )†sec# ) d) tan )†3 sec )
È10 3
" Ð1Î3Ñ
37.
'11ÎÎ124
2 dt Èt 4tÈt
'11ÎÈ3
2 du 1 u#
tan " Ð4Î3Ñ
#
) (tan )) Š sec tan ) ‹ d)
; ’u œ 2Èt, du œ
œ '1Î6
1Î4
C
È 1 r# ( “ r
C
d), Èe2t 9 œ È9 tan# ) 9 œ 3 sec ); "
3" ‹ œ ln 9 ln Š1 È10‹
œ 'tanc" Ð3Î4Ñ
et dt a1 e2t b$Î#
sec# ) tan )
$
Ð4Î3Ñ sec ) d) œ cln ksec ) tan )kd tan tanc" Ð1Î3Ñ
36. Let et œ tan ), t œ ln (tan )), tan" ˆ 34 ‰ Ÿ ) Ÿ tan" ˆ 43 ‰ , dt œ
'lnlnÐ3Ð4ÎÎ43Ñ Ñ
v ‹ 1 v#
œ ' cot' ) csc# ) d) œ cot7 ) C œ "7 ’
cos& )†cos ) d) sin) )
35. Let et œ 3 tan ), t œ ln (3 tan )), tan" ˆ "$ ‰ Ÿ ) Ÿ tan" ˆ %$ ‰, dt œ ln 4
ŠÈ
2 sec# ) d) sec# )
tan " Ð4Î3Ñ
sec$
)
œ 'tan
" Ð3Î4Ñ
" Èt
dt“ Ä '1ÎÈ3
2 du 1 u#
1
sec# ) tan )
d), 1 e2t œ 1 tan# ) œ sec# ); "
Ð4Î3Ñ cos ) d) œ csin )d tan œ tanc" Ð$Î%Ñ
; u œ tan ),
1Î% œ c2)d 1Î' œ 2 ˆ 14 16 ‰ œ
1 6
4 5
3 5
œ
" 5
Ÿ ) Ÿ 14 , du œ sec# ) d), 1 u# œ sec# );
1 6
38. y œ etan ) , 0 Ÿ ) Ÿ 14 , dy œ etan ) sec# ) d), È1 (ln y)# œ È1 tan# ) œ sec );
'1e yÈ1 dy(ln y)
œ '0
1Î4
#
etan ) sec# ) etan ) sec )
d) œ '0 sec ) d) œ cln ksec ) tan )kd ! 1Î4
1Î%
œ ln Š1 È2‹
39. x œ sec ), 0 ) 1# , dx œ sec ) tan ) d), Èx# 1 œ Èsec# ) 1 œ tan );
'
dx xÈ x# 1
œ'
sec ) tan ) d) sec ) tan )
œ ) C œ sec" x C
40. x œ tan ), dx œ sec# ) d), 1 x# œ sec# );
' x dx1 œ ' secsec) )d) œ ) C œ tan" x C #
#
#
41. x œ sec ), dx œ sec ) tan ) d), Èx# 1 œ Èsec# ) 1 œ tan ); ' x dx œ ' sec )†sec ) tan ) d) œ ' sec# ) d) œ tan ) C œ Èx# 1 C È x# 1
tan )
42. x œ sin ), dx œ cos ) d), 1# )
' È dx
1 x#
œ'
cos ) d) cos )
1 #
;
œ ) C œ sin" x C
43. Let x2 œ tan ), 0 Ÿ ) 1# , 2x dx œ sec2 ) d) Ê x dx œ "# sec2 ) d); È1 x4 œ È1 tan2 ) œ sec )
'Èx
1 x4
dx œ
" #
) "' " " 2 4 È ' sec sec ) d) œ # sec )d) œ # lnlsec ) tan )l C œ # lnl 1 x x l C 2
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481
482
Chapter 8 Techniques of Integration
44. Let ln x œ sin ), 1# Ÿ ) 0 or 0 ) Ÿ 1# , 1x dx œ cos ) d), É1 aln xb2 œ cos ) ) ' É1xlnalnx xb dx œ ' cos ' 1 sinsin) ) d) œ ' csc ) d) ' sin ) d) œ sin ) d) œ 2
2
œ ln» ln1x
É1aln xb2 ln x
2
2 » É1 aln xb C œ ln»
1 É1aln xb2 » ln x
lnlcsc ) cot )l cos ) C
É1 aln xb2 C
45. Let u œ Èx Ê x œ u2 Ê dx œ 2u du Ê ' É 4 x x dx œ ' É 4 u2u 2u du œ 2' È4 u2 du; 2
1 #
u œ 2 sin ), du œ 2 cos ) d), 0 ) Ÿ
, È4 u2 œ 2 cos )
2' È4 u2 du œ 2' a2 cos )b a2 cos )b d) œ 8' cos2 ) d) œ 8'
'
œ 4) 2 sin 2) C œ 4) 4 sin ) cos ) C œ 4 sin1 ˆ u2 ‰
œ 4 sin1 Š
œ 4 sin1 Š
Èx 2 ‹
1 cos 2) d) œ 4 2 È 2 4ˆ u2 ‰Š 42u ‹ C
d) 4' cos 2) d) Èx 2 ‹
ÈxÈ4 x C
È4x x2 C
46. Let u œ x3Î2 Ê x œ u2Î3 Ê dx œ 23 u1Î3 du
' È 1 x x dx œ ' É 3
u2Î3 ˆ 2 u1Î3 ‰du 3 1 au2Î3 b 3
œ'
u1Î3 ˆ 2 ‰ È1 u# 3u1Î3 du
œ
2 3
'È1
1 u#
du œ
2 3
sin1 u C œ
2 3
sin1 ˆx3Î2 ‰ C
47. Let u œ Èx Ê x œ u2 Ê dx œ 2u du Ê ' ÈxÈ1 x dx œ ' u È1 u2 2u du œ 2' u2 È1 u2 du; u œ sin ), du œ cos ) d), 1 ) Ÿ 1 , È1 u2 œ cos ) #
#
4) 2' u2 È1 u2 du œ 2' sin2 ) cos ) cos ) d) œ 2' sin2 ) cos2 ) d) œ 12 ' sin2 2) d) œ 12 ' 1 cos d) 2 1' 1' 1 1 1 1 1 1 œ 4 d) 4 cos 4) d) œ 4 ) 16 sin 4) C œ 4 ) 8 sin 2) cos 2) C œ 4 ) 4 sin ) cos ) a2cos2 ) 1b C
œ 14 ) 12 sin ) cos3 ) 14 sin ) cos ) C œ 14 sin1 u 12 u a1 u2 b œ 1 sin1 Èx 1 Èx a1 xb3Î2 1 Èx È1 x C 4
2
3Î2
14 u È1 u2 C
4
48. Let w œ Èx 1 Ê w2 œ x 1 Ê 2w dw œ dx Ê ' w œ sec ), dx œ sec ) tan ) d), 0 )
1 #
Èx 2 Èx 1 dx
œ'
È w2 1 2w dw w
œ 2' Èw2 1 dw
, Èw2 1 œ tan )
2' Èw2 1 dw œ 2' tan ) sec ) tan ) d); u œ tan ), du œ sec2 ) d), dv œ sec ) tan ) d), v œ sec ) 2' tan ) sec ) tan ) d) œ 2 sec ) tan ) 2' sec3 ) d) œ 2 sec ) tan ) 2' sec2 ) sec )d) œ 2 sec ) tan ) 2' atan2 ) 1bsec ) d) œ 2 sec ) tan ) 2Š' tan2 ) sec ) d) ' sec ) d)‹ œ 2sec ) tan ) 2lnlsec ) tan )l 2' tan2 ) sec ) d) Ê 2' tan2 ) sec ) d) œ sec ) tan ) lnlsec ) tan )l C œ w Èw2 1 lnlw Èw2 1l C œ Èx 1 Èx 2 lnlÈx 1 Èx 2l C
49. x
dy dx
œ Èx# 4; dy œ Èx# 4
Ä yœ' œ 2’
(2 tan ))(2 sec ) tan )) d) 2 sec )
È x# 4 #
dx x
;yœ'
1
È x# 4 x
Ô x œ 2 sec ), 0 ) # × dx; Ö dx œ 2 sec ) tan ) d) Ù Õ Èx# 4 œ 2 tan ) Ø
œ 2' tan# ) d) œ 2 ' asec# ) 1b d) œ 2(tan) )) C
sec" ˆ x# ‰“ C; x œ 2 and y œ 0 Ê 0 œ 0 C Ê C œ 0 Ê y œ 2 ’
È x# 4 #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
sec" x# “
Section 8.3 Trigonometric Substitutions 50. Èx# 9
œ 1, dy œ
dy dx
;yœ'
dx È x# 9
1
Ô x œ 3 sec ), 0 ) # × dx Ö dx œ 3 sec ) tan ) d) Ù Ä y œ ' È x# 9 ; Õ Èx# 9 œ 3 tan ) Ø
œ ' sec ) d) œ ln ksec ) tan )k C œ ln ¹ x3 Ê y œ ln ¹ x3 51. ax# 4b
yœ'
' x dx 4 œ #3 tan" #x C; x œ 2 and y œ 0
3 dx x # 4 ; y œ 3 œ 3# tan" ˆ x# ‰
Ê C œ 381 Ê y 52. ax# 1b
# dy dx
œ Èx# 1, dy œ
sec# ) d) sec$ )
3
È 9 x# 3
A œ '0
1Î2
54.
x2 a2
y2 b2
#
dx ax# 1b$Î#
; x œ tan ), dx œ sec# ) d), ax# 1b
È x# 1
œ 3'0
1Î2
œ 1 Ê y œ „ bÉ1
1 #
a
œ 2ab'0
x2 a2
dx œ 4b'0
1Î2
d) 2ab'0
1 Î2
1Î2
55. (a) A œ '0
1 Î2
cos# ) d) œ
x2 a2 ;
1Î#
c) sin ) cos )d !
3 #
A œ 4'0 bÉ1 a
œ ’x sin1 x“
0
(b) M œ '0
1 Î2
1 Î2 0
'0
sin1 x dx œ
œ sec$ );
x È x# 1
x2 a2
x È 1 x2
œ
a
x2 a2
dx
œ cos ), x œ 0 œ a sin ) Ê ) œ 0, x œ a œ a sin ) Ê ) œ 12 “ cos2 ) d) œ 4ab'0
1Î2
1Î2
ab’sin 2)“
0
1 È1 x2 dx,
dv œ dx, v œ x“ 1 Î2
dx œ œ ˆ "# sin1 "# !‰ ’È1 x2 “
0
1 6È3 12 ; 1#
xœ
1
È
1 6 3 12 1#
12 ’ " x2 1 6È3 12 Œ #
1 Î2
sin1 x“
0
" #
'01Î2
x2 È 1 x2
1 cos2) d) #
œ 2abˆ 12 0‰ abasin 1 sin 0b œ 1ab
'01Î2 x sin1 x dx œ
œ
1 6È3 12 1#
12 1 6È3 12
’u œ sin1 x, du œ œ
C; x œ 0 and y œ 1
31 4
dx œ 4b'0 É1 1Î2
cos2) d) œ 2ab’ )“
1 Î2
x2 a2
cos ) aa cos )b d) œ 4ab'0
sin1 x dx ’u œ sin1 x, du œ 1 Î2
Cœ
tan" 1 C
, dx œ 3 cos ) d), È9 x# œ È9 9 sin# ) œ 3 cos );
’x œ a sin ), 1# Ÿ ) Ÿ 1# , dx œ a cos ) d), É1 4b'0 É1
tan ) sec )
$Î#
3 #
1
x
dx; x œ 3 sin ), 0 Ÿ ) Ÿ
3 cos )†3 cos ) d) 3
Ê 0œ
31 8
œ ' cos ) d) œ sin ) C œ tan ) cos ) C œ
Ê 1œ0C Ê yœ 53. A œ '0
C; x œ 5 and y œ ln 3 Ê ln 3 œ ln 3 C Ê C œ 0
È x# 9 ¹ 3
œ 3, dy œ
dy dx
È x# 9 ¹ 3
3 sec ) tan ) d) 3 tan )
'01Î2 x sin1 x dx
1 È1 x2 dx,
dv œ x dx, v œ "# x2 “
dx
’x œ sin ), 1# ) 1# , dx œ cos ) d), È1 x2 œ cos ), x œ 0 œ sin ) Ê ) œ 0, x œ œ
2 12 Š"ˆ"‰ 1 6È3 12 Œ # #
œ
12 1 1 6È3 12 Œ 48
œ
1 12 1 6È3 12 Œ 48
’ 4) "8 sin 2)“
" #
sin1 ˆ "# ‰ 0‹
'01Î6
1 cos 2) 2
" #
'01Î6
d) œ 1 Î6 0
œ
sin2 ) cos )
cos ) d) œ
12 1 1 6È3 12 Œ 48 3È 3 1 ; 4Š1 6È3 12‹
" 4
12 1 1 6È3 12 Œ 48
" #
" #
œ sin ) Ê ) œ 16 “
'01Î6 sin2 ) d)
'01Î6 d) "4 '01Î6 cos 2) d)
yœ
1
È
1 6 3 12 1#
1
2
'01Î2
’u œ asin xb , du œ
" 1 2 # asin xb dx 2 sinc1 x È1 x2 dx,
dv œ dx, v œ x“
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483
484
Chapter 8 Techniques of Integration œ
2
xasin1 x dxb •
6 1 6È3 12 ”
1 Î2
'0
1 Î2
2x sinc1 x È1 x2 dx
0
’u œ sin1 x, du œ œ œ
2 6 Š " ˆsin1 ˆ "# ‰‰ 1 6È3 12 #
0‹ ”2È1 x2 sin1 x•
Œ2É1 ˆ "# ‰ sin1 ˆ "# ‰ 0 ’2x“
0
56. V œ '0 1ŠÈx tan1 x‹ dx œ 1'0 x tan1 x dx 2
1
1
œ 1Œ’ "# x2 tan1 x“ 0
œ 1Œ 18
" #
" #
1
'01
x2 1 x2
'0
1 Î2
0
1 Î2
2
6 12 1 6È3 12 Œ 72
1 Î2
1 È1 x2 dx,
œ
" #
2È 1 x2 È1 x2 dx
1 1 x2 dx,
1‹ œ
12 121È3 72
12Š1 6È3 12‹
'01 ˆ 1 1 1 x ‰dx œ 1Œ 18 #" '01 ˆ 1 1 1 x ‰dx 2
2
2
0
x3 È1 x2 dx œ 13 x2 a1 x2 b
3Î2
1 3
'
a1 x2 b
3Î2
2
x3 È1 x2 dx œ ' x2 È1 x2 x dx œ
œ 13 a1 x2 b
3 Î2
5" a1 x2 b
(c) Trig substitution: x œ sin ),
1 #
5Î2
"#
3Î2
2x dx œ 13 x2 a1 x2 b
(b) Substitution: u œ 1 x Ê x œ 1 u Ê du œ 2x dx Ê
'
1È3 6
'01 dx "# '01 1 1 x dx œ 1Œ 18 ’ #" x #" tan1 x“1 œ 1ˆ 18 ˆ #" #" tan1 1 0 0‰‰ œ 1a14 2b
2
'
dv œ x dx, v œ #" x2 “
57. (a) Integration by parts: u œ x2 , du œ 2x dx, dv œ x È1 x2 dx, v œ 13 a1 x2 b
'
v œ 2È1 x2 “
2x È1 x2 dx,
2 6 Š1 1 6È3 12 72
’u œ tan1 x, du œ
dx œ 1Œˆ #" tan1 1 0‰
dv œ
"# du
3Î2
2 15 a1
x2 b
5Î2
C
œ x dx
' a1 ub Èu du œ "# ' ˆÈu u3Î2 ‰ du œ 13 u3Î2 5" u5Î2 C
C
Ÿ ) Ÿ 1# , dx œ cos ) d), È1 x2 œ cos )
x3 È1 x2 dx œ ' sin3 ) cos ) cos ) d) œ ' sin2 ) cos2 ) sin ) d) œ ' a1 cos2 )bcos2 ) sin ) d)
œ ' cos2 ) sin ) d) ' cos4 ) sin ) d) œ 13 cos3 ) 5" cos5 ) C œ 13 a1 x2 b
3Î2
5" a1 x2 b
5Î2
C
58. (a) The slope of the line tangent to y œ faxb is given by f w axb. Consider the triangle whose hypotenuse is the 30 ft rope, È 2 the length of the base is x and the height h œ È900 x2 Þ The slope of the tangent line is also 900 x , thus x
È900 x2 . x È900 x2 dx x
f w axb œ
(b) faxb œ ' œ '
30 cos ) 30 sin ) 30 cos
’x œ 30 sin ), 0 ) Ÿ 1# , dx œ 30 cos ) d), È900 x2 œ 30 cos )“
) d) œ 30'
cos2 ) sin )
d) œ 30'
œ 30 lnlcsc ) cot )l 30 cos ) C œ 30 lnº 30 x Ê 0 œ 30 lnº 30 30
È900 302 º 30
ˆ1 sin2 )‰ sin ) È900 x2 º x
d) œ 30' csc ) d) 30' sin ) d) È900 x2 C; fa30b œ 0
È900 302 C œ C Ê faxb œ 30 lnº 30 x
È900 x2 º x
È900 x2
8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 1.
5x 13 (x 3)(x 2)
Ê 2.
A x3
B x2
Ê 5x 13 œ A(x 2) B(x 3) œ (A B)x (2A 3B)
ABœ5 Ê B œ (10 13) Ê B œ 3 Ê A œ 2; thus, 2A 3B œ 13
5x 7 x# 3x 2
Ê
œ
œ
5x 7 (x 2)(x 1)
œ
A x2
B x1
5x 13 (x 3)(x 2)
œ
2 x3
3 x#
Ê 5x 7 œ A(x 1) B(x 2) œ (A B)x (A 2B)
ABœ5 Ê B œ 2 Ê A œ 3; thus, A 2B œ 7
5x 7 x# 3x 2
œ
3 x2
2 x1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.4 Integration of Rational Functions by Partial Fractions 3.
x4 (x 1)#
thus,
4.
œ
A x1
x4 (x 1)#
2x 2 x# 2x 1
œ
B (x 1)#
1 x1
2x 2 (x 1)#
œ
œ
3 (x 1)#
A x1
z1 z# (z 1)
Ê 2x 2 œ A(x 1) B œ Ax (A B) Ê
B (x 1)#
2x 2 x# 2x 1
Ê A œ 2 and B œ 4; thus, 5.
œ
2 x1
z C 1 Ê z 1 œ Az(z 1) B(z 1) Cz# Ê z 1 œ (A C)z# (A B)z B ACœ0 Þ œ
A z
B z#
" z# z 6
œ
z z$ z# 6z
œ
" (z 3)(z 2)
œ
A z3
B z#
t# 8 t# 5t 6
œ1
5t 2 t# 5t 6
(after long division);
Ê B œ 12 Ê A œ 17; thus, t% 9 t% 9t#
9t# 9 t% 9t#
œ1
œ1
t# 8 t# 5t 6
5t 2 t# 5t 6
9t# 9 t# at# 9b #
œ1
œ
2 z
" z#
2 z1
Ê 1 œ A(z 2) B(z 3) œ (A B)z (2A 3B)
œ
17 t3
œ
z z$ z# 6z
5t 2 (t 3)(t 2)
Ê 5t 2 œ A(t 2) B(t 3) œ (A B)t (2A 3B) Ê
8.
z" z# (z 1)
ABœ0 Ê 5B œ 1 Ê B œ "5 Ê A œ 5" ; thus, 2A 3B œ 1
Ê
7.
Aœ2 A B œ 2
4 (x 1)#
Ê A B œ 1 ß Ê B œ 1 Ê A œ 2 Ê C œ 2; thus, B œ 1 à 6.
Aœ1 Ê A œ 1 and B œ 3; A B œ 4
Ê x 4 œ A(x 1) B œ Ax (A B) Ê
œ
A t3
" 5
z3
"5 z2
B t2
ABœ5 Ê B œ (10 2) œ 12 2A 3B œ 2
12 t2
(after long division);
9t# 9 t# at# 9b
œ
A t
B t#
CtD t# 9 #
Ê 9t# 9 œ At at# 9b B at 9b (Ct D)t# œ (A C)t$ (B D)t 9At 9B ACœ0 Þ á á % B D œ 9 Ê Ê A œ 0 Ê C œ 0; B œ 1 Ê D œ 10; thus, t%t 9t9# œ 1 t"# t#109 ß 9A œ 0 á á 9B œ 9 à 9.
" 1 x#
œ
' 1 dxx 10.
" x# 2x
#
A 1x " #
œ
œ
A x
Ê 1 œ A(1 x) B(1 x); x œ 1 Ê A œ
B 1x
' 1 dx x #" ' 1 dx x œ #" cln k1 xk ln k1 xkd C
Ê 1 œ A(x 2) Bx; x œ 0 Ê A œ
B x2
' x dx 2x œ #" ' dxx #" ' x dx 2 œ #" cln kxk ln kx 2kd C
" #
" #
; x œ 1 Ê B œ
" #
;
; x œ 2 Ê B œ #" ;
#
11.
x4 x# 5x 6
'x 12.
x4 5x 6
2x 1 x# 7x 12
'x 13.
#
#
'4
œ
A x6
dx œ
2 7
œ
2x 1 7x 12
y y# 2y 3 8
œ
œ
A x4
B x1
'
; x œ 6 Ê A œ
5 7
2 7
œ
2 7
;
œ 7 ; x œ 4 Ê A œ
9 1
' x dx 6 75 ' x dx 1 œ 72 ln kx 6k 75 ln kx 1k C œ 7" ln k(x 6)# (x 1)& k C B x3
dx œ 9 ' A y3
Ê x 4 œ A(x 1) B(x 6); x œ 1 Ê B œ
8
Ê 2x 1 œ A(x 3) B(x 4); x œ 3 Ê B œ
dx x4
B y1
7'
dx x3
y dy dy " 3 y# 2y 3 œ 4 4 y 3 4 " " ln 15 # ln 5 # ln 3 œ #
'4
dy y1
œ 9;
*
4) œ 9 ln kx 4k 7 ln kx 3k C œ ln ¹ (x C (x 3)( ¹
Ê y œ A(y 1) B(y 3); y œ 1 Ê B œ 8
7 1
œ 43 ln ky 3k
" 4
)
1 4
œ
" 4
;yœ3 Ê Aœ
ln ky 1k‘ % œ ˆ 43 ln 5
" 4
3 4
;
ln 9‰ ˆ 43 ln 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" 4
ln 5‰
485
486 14.
Chapter 8 Techniques of Integration y4 y# y
'1Î2 1
œ " 8
" t$ t# 2t
ln
" 16
ln
œ
B t2
A t 1 6
dy y
œ ln ˆ 27 8 †
27 8
3 '1Î2 1
œ "# ln ktk
" 6
x3 2x$ 8x
B x2
" 3
ln kt 2k
œ
A x
Ê Bœ
" 16
;xœ2 Ê Cœ " 16
x$ x# 2x 1
;'
" 3
" 8
œ c4 ln kyk 3 ln ky 1kd ""Î# œ (4 ln 1 3 ln 2) ˆ4 ln
† 16‰ œ ln
3x 2 (x 1)#
dt t
" 6
' tdt2 3" ' tdt1
(x 3) œ A(x 2)(x 2) Bx(x 2) Cx(x 2); x œ 0 Ê A œ
;'
5 16
dx œ 83 '
x3 2x$ 8x
ln kx 2k C œ
(after long division);
" 16
dx x
" 16
' x dx 2 165 ' x dx 2
dx x1
'0
1
dx (x 1)#
3x 2 (x 1)#
œ
A x1
x$ x# 2x 1
3x 2 (x 1)#
œ (x 2)
x$ dx x# 2x 1
#
œ ’ x# 2x 3 ln kx 1k
(after long division);
3x 2 (x 1)#
œ
A x1
œ Ax (A B) Ê A œ 3, A B œ 2 Ê A œ 3, B œ 1; 'c1 x ' dx œ 'c1 (x 2) dx 3 'c1 x dx 1 c1 (x 1)# œ ’ # 2x 3 ln kx 1k 0
œ Š0 0 3 ln 1 19.
" ax # 1 b #
œ
A x1
x œ 1 Ê C œ
C (x 1)#
D (x 1)# " 4
;xœ1 Ê Dœ
" (#) ‹
œ 2 3 ln 2
Ê 1 œ A(x 1)(x 1)# B(x 1)(x 1)# C(x 1)# D(x 1)# ;
; coefficient of x$ œ A B Ê A B œ 0; constant œ A B C D " #
; thus, A œ
" 4
Ê B œ 4" ; '
' x dx 1 4" ' x dx 1 4" ' (x dx1)
" 4
' (x dx1)
" 4
1¸ ln ¸ xx 1
" 4
x# (x 1) ax# 2x 1b
œ
A x1
B x1
Ê C œ "2 ; x œ 1 Ê A œ œ 21.
#
Š #" 2 3 ln 2
" 4
" 4
œ
A x1
Bx C x# 1
#
œ
dx ax # 1 b #
x 2 ax# 1b
C
Ê x# œ A(x 1)# B(x 1)(x 1) C(x 1); x œ 1
C (x 1)#
; coefficient of x# œ A B Ê A B œ 1 Ê B œ
' x dx 1 43 ' x dx 1 2" ' (x dx1)
1 (x 1) ax# 1b
#
Ê 3x 2 œ A(x 1) B
x$ dx x# 2x 1 ! x " 1 “ "
Ê ABCDœ1 Ê ABœ œ 20.
" (1) ‹
B x1 " 4
0
" " x 1“!
B (x 1)# 0
0
Ê 3x 2 œ A(x 1) B
B (x 1)#
œ ˆ "# 2 3 ln 2 "# ‰ (1) œ 3 ln 2 2 18.
#
œ
" 4
ln kx 1k
3 4
ln kx 1k
" #(x 1)
Cœ
22.
" 2
3 4
;'
x# dx (x 1) ax# 2x 1b
ln k(x 1)(x 1)$ k 4 " #
Ê 1 œ A ax# 1b (Bx C)(x 1); x œ 1 Ê A œ
œ A B Ê A B œ 0 Ê B œ 12 ; constant œ A C Ê A C œ 1 Ê C œ œ
'01 x dx 1 #" '01 (x x 11) dx œ #" ln kx 1k 4" ln ax# 1b #" tan" x‘ "!
" #
" 4
3t# t 4 t$ t
œ
A t
ln 2 Bt C t# 1
" #
tan" 1‰ ˆ #" ln 1
" 4
ln 1
" #
tan" 0‰ œ
" 4
ln 2 #" ˆ 14 ‰ œ
; '0
1
È3
3t# t 4 t$ 1
" 2(x 1)
dx (x 1) ax# 1b
(1 2 ln 2) 8
Ê 3t# t 4 œ A at# 1b (Bt C)t; t œ 0 Ê A œ 4; coefficient of t#
œ A B Ê A B œ 3 Ê B œ 1; coefficient of t œ C Ê C œ 1; '1
; coefficient of x#
#
œ ˆ "# ln 2
; x œ 2
&
œ Ax (A B) Ê A œ 3, A B œ 2 Ê A œ 3, B œ 1; '0 1
3 8
ln ¹ (x 2)x'(x 2) ¹ C
1
1
3 ln 3# ‰
27 4
œ #" '
dt t$ t# 2t
" #
5 16
ln kx 2k
œ '0 (x 2) dx 3 '0
" #
ln kt 1k C
Ê
C x#
œ (x 2)
dy y1
œ 3 ;
3 1
Ê 1 œ A(t 2)(t 1) Bt(t 1) Ct(t 2); t œ 0 Ê A œ "# ; t œ 2
C t1
;tœ1 Ê Cœ
œ 38 ln kxk 17.
Ê y 4 œ A(y 1) By; y œ 0 Ê A œ 4; y œ 1 Ê B œ
B y1
dy œ 4 '1Î2
Ê Bœ
16.
1
y4 y# y
œ ln 15.
A y
dt
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
C
Section 8.4 Integration of Rational Functions by Partial Fractions È3
œ 4 '1
È3
'1
dt t
œ Š4 ln È3 " #
œ 2 ln 3 23.
y# 2y 1 ay # 1 b # $
" #
dt œ 4 ln ktk
" 2
È$
ln at# 1b tan" t‘ "
ln 4 tan" È3‹ ˆ4 ln 1 1 1#
ln 2
Ay B y# 1
œ
(t 1) t# 1
œ ln Š È92 ‹
Cy D ay # 1 b#
487
" #
ln 2 tan" 1‰ œ 2 ln 3 ln 2
1 3
" #
ln 2
1 4
1 1#
Ê y# 2y 1 œ (Ay B) ay# 1b Cy D
œ Ay By# (A C)y (B D) Ê A œ 0, B œ 1; A C œ 2 Ê C œ 2; B D œ 1 Ê D œ 0;
'y 24.
dy œ '
#
2y 1 ay # 1 b #
8x# 8x 2 a4x# 1b# $
œ
Ax B 4x# 1
dy 2 '
" y# 1
Cx D a4x# 1b#
y ay # 1 b#
" y# 1
dy œ tan" y
C
Ê 8x# 8x 2 œ (Ax B) a4x# 1b Cx D
œ 4Ax 4Bx# (A C)x (B D); A œ 0, B œ 2; A C œ 8 Ê C œ 8; B D œ 2 Ê D œ 0;
' 8x
dx œ 2 '
2s 2 as# 1b (s 1)$
œ
# 8x 2 a4x# 1b#
25.
As B s# 1 $
dx 4x# 1
C s1 #
8'
x dx a4x# 1b#
D (s 1)#
œ tan" 2x
E (s 1)$ #
" 4x# 1
C
Ê 2s 2
œ (As B)(s 1) C as 1b (s 1) D as# 1b (s 1) E as# 1b œ cAs% (3A B)s$ (3A 3B)s# (A 3B)s Bd C as% 2s$ 2s# 2s 1b D as$ s# s 1b E as# 1b œ (A C)s% (3A B 2C D)s$ (3A 3B 2C D E)s# (A 3B 2C D)s (B C D E) C œ0 Þ A á á á 3A B 2C D œ0á Ê 3A 3B 2C D E œ 0 ß summing all equations Ê 2E œ 4 Ê E œ 2; á A 3B 2C D œ2á á á B CDEœ2 à summing eqs (2) and (3) Ê 2B 2 œ 0 Ê B œ 1; summing eqs (3) and (4) Ê 2A 2 œ 2 Ê A œ 0; C œ 0 from eq (1); then 1 0 D 2 œ 2 from eq (5) Ê D œ 1;
' as
26.
#
2s 2 1b (s 1)$
s% 81 œ As s as # 9 b # %
ds œ '
Bs C s# 9
ds s# 1
'
ds (s 1)#
2'
ds (s 1)$
œ (s 1)# (s 1)" tan" s C #
Ds E as # 9 b# %
Ê s% 81 œ A as# 9b (Bs C)s as# 9b (Ds E)s
œ A as 18s# 81b aBs Cs$ 9Bs# 9Csb Ds# Es œ (A B)s% Cs$ (18A 9B D)s# (9C E)s 81A Ê 81A œ 81 or A œ 1; A B œ 1 Ê B œ 0;
C œ 0; 9C E œ 0 Ê E œ 0; 18A 9B D œ 0 Ê D œ 18; '
27.
œ ln ksk
9 as # 9 b
x2 x 2 x3 1
A x1
œ
s% 81 s as # 9 b #
ds œ '
ds s
18 '
s ds as # 9 b #
C Bx C x2 x 1
Ê x2 x 2 œ Aax2 x 1b aBx Cbax 1b œ aA Bbx2 aA B Cbx aA Cb
Ê A B œ 1ß A B C œ 1, A C œ 2 Ê adding eq(2) and eq(3) Ê 2A B œ 1, add this equation to eq(1) Ê 3A œ 2 Ê A œ œ œ
2 3
2 3
' x 1 1 dx 13 ' ˆ
2 3
' x 1 1 dx 13 '
ÊBœ1Aœ x4 dx 2 x "# ‰ 34
u 92 u2 34
du œ
2 3
1 3
Ê C œ 1 A B œ 43 ; '
’u œ x
" #
Êu
' x 1 1 dx 31 ' u u
2 œ 23 lnlx 1l 16 ln¹ˆx "# ‰ 34 ¹
2
3 4 " #
" #
œ ' Š x2Î31
a1Î3bx 4Î3 x2 x 1 ‹dx
œ x Ê du œ dx“
du
3 1 x È3 tan Š È3Î2 ‹
x2 x 2 x3 1 dx
3 2
' u 1 2
3 4
du
C œ 23 lnlx 1l 16 lnlx2 x 1l È3tan1 Š 2xÈ3 1 ‹ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
488 28.
Chapter 8 Techniques of Integration 1 x4 x
œ
A x
B x1
Cx D x2 x 1
Ê 1 œ Aax 1bax2 x 1b B xax2 x 1b aC x Dbxax 1b
œ aA B Cbx3 aB C Dbx2 aB Dbx A Ê A œ 1, B D œ 0 Ê D œ B, B C D œ 0 Ê 2B C œ 0 Ê C œ 2B, A B C œ 0 Ê 1 B 2B œ 0 Ê B œ 13 Ê C œ 23 Ê D œ 13 ;
' x 1 x dx œ ' a 1x x1Î31 ax2Î3bxx 11Î3 bdx œ ' 1x dx 13 ' x 1 1 dx 13 ' x 2x x 1 1 dx 4
2
œ lnlxl 29.
x2 x4 1
œ
1 3 lnlx
A x1
1l
B x1 3
1 2 3 lnlx
Cx D x2 1
2
x 1l C
Ê x2 œ Aax 1bax2 1b Bax 1bax2 1b+aCx Dbax 1bax 1b
œ aA B Cbx aA B Dbx2 aA B Cbx A B D Ê A B C œ 0, A B D œ 1, A B C œ 0, A B D œ 0 Ê adding eq(1) to eq (3) gives 2A 2B œ 0, adding eq(2) to eq(4) gives 2A 2B œ 1, adding these two equations gives 4B œ 1 Ê B œ 14 , using 2A 2B œ 0 Ê A œ 14 , using A B D œ 0 Ê D œ 12 , and using A B C œ 0 Ê C œ 0; ' œ 30.
14
'
x2 x x4 3x2 4
1 x 1 dx
œ
A x2
1 4
'
1 x 1 dx
B x2
1 2
Cx D x2 1
'
œ
1 x2 1 dx
41 lnlx
1 4 lnlx
1l
x2 x4 1 dx
1l
1 Î4 œ ' a x 1
1 1 2 tan x
Cœ
1Î4 x1
1Î2 x2 1
1 x1 4 ln¹ x 1 ¹
bdx
21 tan1 x C
Ê x2 x œ Aax 2bax2 1b Bax 2bax2 1b+aCx Dbax 2bax 2b
œ aA B Cbx3 a2A 2B Dbx2 aA B 4Cbx 2A 2B 4D Ê A B C œ 0, 2A 2B D œ 1, A B 4C œ 1, 2A 2B 4D œ 0 Ê subtractin eq(1) from eq (3) gives 5C œ 1 Ê C œ 15 , subtacting eq(2) from eq(4) gives 5D œ 1 Ê D œ 15 , substituting for C in eq(1) gives A B œ 15 , and substituting for D in eq(4) gives 2A 2B œ
'
x2 x x4 3x2 4 dx
3 10 lnlx
31.
Ê A B œ 25 , adding this equation to the previous equatin gives 2A œ
4 5
2l
œ ' Š x3Î102
1 10 lnlx
2) $ 5) # 8) 4 a) # 2 ) 2 b# $
œ
2l
A) B ) # #) 2
1Î10 x2
1 2 10 lnlx
a 1 Î5 b x 1 Î 5 ‹dx x2 1
1l
C) D a ) # 2 ) 2 b#
1 1 5 tan x
œ
3 10
3 5
ÊAœ
3 10
1 Ê B œ 10 ;
' x 1 2 dx 101 ' x 1 2 dx 15 ' x x 1 dx 51 ' x 1 1 dx 2
2
C
Ê 2)$ 5)# 8) 4 œ (A) B) a)# 2) 2b C) D
œ A) (2A B))# (2A 2B C)) (2B D) Ê A œ 2; 2A B œ 5 Ê B œ 1; 2A 2B C œ 8 Ê C œ 2;
2B D œ 4 Ê D œ 2; ' œ' œ 32.
2) 2 ) # 2) 2
" ) # 2) #
d) '
2) $ 5) # 8) 4 a) # 2 ) 2 b#
d) ) # 2) #
'
d a) 2 ) 2 b a) # 2 ) 2 b # "
œ
A) B )# 1
C) D a) # 1 b#
d) '
2) 1 a) # 2 ) 2 b
œ'
#
ln a)# 2) 2b tan
) % 4) $ 2) # 3) 1 a) # 1 b $
d) œ '
#
d a) 2 ) 2 b ) # 2) #
2) 2 a ) # 2 ) 2 b#
'
d) () 1)# 1
d)
" ) # 2) #
() 1) C
E) F a ) # 1 b$
Ê ) % 4) $ 2 ) # 3 ) 1
#
œ (A) B) a)# 1b (C) D) a)# 1b E) F œ (A) B) a)% 2)# 1b aC)$ D)# C) Db E) F œ aA)& B)% 2A)$ 2B)# A) Bb aC)$ D)# C) Db E) F œ A)& B)% (2A C))$ (2B D))# (A C E)) (B D F) Ê A œ 0; B œ 1; 2A C œ 4 Ê C œ 4; 2B D œ 2 Ê D œ 0; A C E œ 3 Ê E œ 1; B D F œ 1 Ê F œ 0;
') 33.
%
4) $ 2) # 3) 1 a) # 1 b $
2x$ 2x# 1 x# x
œ 2x
" x# x
x œ 1 Ê B œ 1; ' 34.
x% x# 1
œ ax# 1b
d) œ '
$
d) )# 1
œ 2x #
2x 2x 1 x# x " x# 1
4'
" x(x 1)
" 3
x$ x
" #
œ ax# 1b
ln kx 1k
" #
;
" x(x 1)
œ ' 2x dx '
x œ 1 Ê A œ "# ; x œ 1 Ê B œ œ
) d) a) # 1 b#
'
œ
dx x
" (x 1)(x 1) " #
;'
ln kx 1k
%
A x
Ê 1 œ A(x 1) Bx; x œ 0 Ê A œ 1;
B x1
dx x1
C
œ x# ln kxk ln kx 1k C œ x# ln ¸ x x 1 ¸ C
" (x 1)(x 1)
'
"4 a)# 1b
#
œ tan" ) 2 a)# 1b
' ;
"
) d) a ) # 1 b$
x ax# x# 1 dx œ $ C œ x3 x #" ln
œ
A x1
1b dx
B x1 " #
Ê 1 œ A(x 1) B(x 1);
' x dx 1 #" ' x dx 1
1¸ ¸ xx 1 C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.4 Integration of Rational Functions by Partial Fractions 35.
9x$ 3x 1 x$ x# #
œ9
9x# 3x " x# (x 1)
(after long division);
9x# 3x 1 x# (x 1) #
œ
A x
B x#
489
C x1
Ê 9x 3x 1 œ Ax(x 1) B(x 1) Cx ; x œ 1 Ê C œ 7; x œ 0 Ê B œ 1; A C œ 9 Ê A œ 2;
' 9x x3xx 1 dx œ ' 9 dx 2 ' dxx ' dxx 7 ' xdx1 œ 9x 2 ln kxk x" 7 ln kx 1k C $
$
36.
#
16x$ 4x# 4x 1
#
œ (4x 4)
12x 4 4x# 4x 1
12x 4 (2x 1)#
;
œ
Ê A œ 6; A B œ 4 Ê B œ 2; ' œ 2(x 1)# 3 ln k2x 1k
37.
y% y# 1 y$ y
œy
" y ay # 1 b
" 2x 1
" y ay # 1 b
;
œ
A 2x 1 $
16x 4x# 4x 1 C" œ 2x#
A y
By C y# 1
y# #
38.
ln kyk
2y% y$ y# y 1
" #
B (2x 1)#
Ê 12x 4 œ A(2x 1) B
dx œ 4 ' (x 1) dx 6 '
dx 2x 1 "
4x 3 ln k2x 1k (2x 1)
2'
dx (2x 1)#
C, where C œ 2 C"
Ê 1 œ A ay# 1b (By C)y œ (A B)y# Cy A
7 Ê A œ 1; A B œ 0 Ê B œ 1; C œ 0; ' œ
y% y# 1 y$ y
dy œ ' y dy '
'
dy y
y dy y# 1
ln a1 y# b C
œ 2y 2
2 y$ y# y 1
2 y$ y# y 1 #
;
œ
2 ay# 1b (y 1) #
œ
A y1
By C y# 1
Ê 2 œ A ay# 1b (By C)(y 1) œ aAy Ab aBy Cy By Cb œ (A B)y# (B C)y (A C) Ê A B œ 0, B C œ 0 or C œ B, A C œ A B œ 2 Ê A œ 1, B œ 1, C œ 1;
'y
$
2y% y# y 1 #
dy œ 2 ' (y 1) dy '
œ (y 1) ln ky 1k where C œ C" 1 39.
'e
40.
'e œ
41.
y2 2
" #
dt œ '
dy y# 3y 2
e3t 2et 1 t e2t 1 e dt;
œ'
dy '
y y# 1
dy y1
'
dy y2
y œ et ” dy œ et dt • Ä
ln ay# 1b tan" y C œ
#
dy y# 1
" 2
e2t Ä
" #
" #
ln ay# 1b tan" y C,
1 e 1 œ ln ¹ yy 2 ¹ C œ ln Š et # ‹ C t
'y
3
2y 1 y# 1
dy œ ' Šy
y1 y# 1 ‹
dy œ
y2 2
'
y y# 1
ln ae2t 1b tan" aet b C
't
#
œ
" 3
dy t6
œ
" 5
' ˆ t " 2 t " 3 ‰ dt œ "5 ln ¸ tt 23 ¸ C
#
" 3
Ä '
dy y# y 2
' y dy 2 3" '
dy y1
œ
" 3
2 ln ¹ yy 1¹ C œ
" 3
cos ) 2 ¸ ln ¸ cos )1 C
cos ) ¸ " ¸ cos ) 1 ¸ ln ¸ 12 cos ) C œ 3 ln cos ) 2 C
" (2x) 12x$ 3x ' (x 2)a4xtan# " 1(2x) ' (x x 2)# dx dx œ ' tan b (x 2)# 4x# 1 dx 3 ' (x dx2) œ atan 4" 2xb# 3 ln kx 2k x 6 2 C œ "# ' tan" (2x) d atan" (2x)b 3' x dx 2 6 #
#
44.
" (3x) 9x$ x " (3x) ' (x 1)a9xtan ' (x x 1)# dx dx œ ' tan # 1b (x 1)# 9x# 1 dx ' (x dx1) œ atan 6" 3xb# ln kx 1k x 1 1 C œ "3 ' tan" (3x) d atan" (3x)b ' x dx 1 #
#
45.
' x Î 1 Èx dx œ ' Èxax1 1b dx ’Let u œ Èx Ê du 3 2
2 u2 1
dy '
sin y 2 ln ¹ sin y 3¹ C
" 5
' cos )sin )cosd)) 2 ; ccos ) œ yd œ
43.
2e2t et e2t 1
4t
œ cet œ yd'
'
ln ay# 1b tan" y C" œ y# 2y ln ky 1k
' sin ycos ysindyy 6 ; [sin y œ t, cos y dy œ dt] œ
42.
et dt 3et 2
2t
" #
dy y1
œ
A u1
B u1
œ
1 dx 2È x
Ê 2 du œ
1 Èx dx“
Ä'
2 u2 1 du;
Ê 2 œ Aau 1b Bau 1b œ aA Bbu A B Ê A B œ 0, A B œ 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dy y# 1
490
Chapter 8 Techniques of Integration Ê B œ 1 Ê A œ 1; '
2 u2 1 du
œ ' a u11
1 u1
bdu œ ' u 1 1 du ' u 1 1 du œ lnlu 1l lnlu 1l C
Èx 1
œ lnº Èx 1 º C 46.
' ax
1Î3
’Let x œ u6 Ê dx œ 6u5 du“ Ä '
1 dx 1 bÈ x
œ 6' du '
œ
6 6 u2 1 du; u2 1
A u1
B u1
1 5 au2 1bu3 6u du
œ'
6u2 u2 1 du
œ ' a6
6 u2 1
b du
Ê 6 œ Aau 1b Bau 1b œ aA Bbu A B Ê A B œ 0,
A B œ 6 Ê B œ 3 Ê A œ 3; 6' du '
œ 6u ' a u31
6 u2 1 du
3 u1
bdu œ 6u 3' u 1 1 du 3' u 1 1 du
1Î6
1 œ 6u 3 lnlu 1l 3 lnlu 1l C œ 6x1Î6 3 lnº xx1Î6 C 1º
47.
' Èxx 1 dx ’Let x 1 œ u2 Ê dx œ 2' du '
œ
2 2 u2 1 du; u2 1
œ 2u du“ Ä '
A u1
B u1
u u2 1
2u du œ '
2u2 u2 1
du œ ' a2
2 u2 1
b du
Ê 2 œ Aau 1b Bau 1b œ aA Bbu A B Ê A B œ 0,
A B œ 2 Ê B œ 1 Ê A œ 1; 2' du '
2 u2 1 du
œ 2u ' a u" 1
" u1
bdu œ 2u ' u 1 1 du ' u 1 1 du
È 11 œ 2u lnlu 1l lnlu 1l C œ 2Èx 1 lnº Èxx C 11º
48.
' xÈx1 9 dx ’Let x 9 œ u2 Ê dx
œ 2u du“ Ä '
2u du œ '
1 au 2 9 b u
2 u2 9
du;
2 u2 9
œ
A u3
Ê 2 œ Aau 3b Bau 3b œ aA Bbu 3A 3B Ê A B œ 0, 3A 3B œ 2 Ê A œ
' u 2 9 2
49.
du œ ' a u1Î33
1Î3 u3
b du œ 31 ' u 1 3
du
' xax 1 1b dx œ ' x axx 1b dx ’Let u œ x4 Ê du 3
4
4
4
1 3
' u 1 3
50.
1 4
Ê B œ 13 ;
œ 4x3 dx“ Ä
1 4
1 3
È
93 lnº Èxx C 93º
' uau 1 1b du; uau 1 1b œ Au u B 1 1 4
' uau 1 1b du œ 41 ' a u1 u " 1 bdu
' 1u du 14 ' u 1 1 du œ 41 lnlul 14 lnlu 1l C œ 14 lnŠ x x1 ‹ C 4
4
' x ax 1 4b dx œ ' x 6
1 3
B u3
du œ 31 lnlu 3l 31 lnlu 3l C œ
Ê 1 œ Aau 1b Bu œ aA Bbu A Ê A œ 1 Ê B œ 1; œ
5
x4
10 ax5
4b dx
œ ’Let u œ x5 Ê du œ 5x4 dx“ Ä
1 5
' u au1 4b du; u au1 4b œ Au uB 2
2
2
C u4
Ê 1 œ Auau 4b Bau 4b Cu2 œ aA Cbu2 a4A Bbu 4B Ê A C œ 0, 4A B œ 0, 4B œ 1 Ê B œ 1 Ê A œ 16 ÊCœ 1 œ 80 lnlul
51. at# 3t 2b Ê
" #
1 20u
dx dt
œC Ê
52. a3t% 4t# 1b
1 1 16 ; 5
' u au1 4b du œ 51 ' Š 1Îu16 1uÎ4 u1Î164 ‹du œ 801 ' u1 du 201 ' u1 du 801 ' u 1 4 du
1 80 lnlu
œ 1; x œ ' t2 t1
dx dt
2
2
1 4l C œ 80 lnlx5 l
1 20x5
2
1 5 80 lnlx
4l C œ
1 x5 4 80 ln¹ x5 ¹
1 20x5
C
t2 2¸ x œ ' t dt 2 ' t dt 1 œ ln ¸ tt 1 C; t 1 œ Ce ; t œ 3 and x œ 0 2 ‰¸ œ "# ex Ê x œ ln ¸2 ˆ tt œ ln kt 2k ln kt 1k ln 2 1 dt t# 3t 2
œ 2È3; x œ 2È3'
dt 3t% 4t# 1
œ È 3'
œ 3 tan" ŠÈ3t‹ È3 tan" t C; t œ 1 and x œ
1 È 3 4
dt t#
" 3
È 3'
Ê
È 31 4
dt t# 1
œ1
È3 4
1 C Ê C œ 1
Ê x œ 3 tan" ŠÈ3t‹ È3 tan" t 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 4
Section 8.5 Integral Tables and Computer Algebra Systems 53. at# 2tb
dx dt
œ 2x 2;
' x dx 1 œ ' t
" #
" 3
t œ 1 and x œ 1 Ê ln 2 œ ln Ê xœ 54. (t 1)
6t t#
dt 2t
ln kx 1k œ
" #
'
#" '
dt t
dt t2
1, t 0
' x dx 1 œ ' t dt 1
œ x# 1 Ê
dx dt
" #
Ê ln kx 1k œ ln ¸ t t # ¸ C; C Ê C œ ln 2 ln 3 œ ln 6 Ê ln kx 1k œ ln 6 ¸ t t 2 ¸ Ê x 1 œ t 6t # #
Ê
491
Ê tan" x œ ln kt 1k C; t œ 0 and x œ 0 Ê tan" 0 œ ln k1k C
#
Ê C œ tan" 0 œ 0 Ê tan" x œ ln kt 1k Ê x œ tan (ln (t 1)), t 1 55. V œ 1 '0Þ5 y# dx œ 1 '0Þ5 #Þ&
#Þ&
#Þ& dx œ 31 Š'0Þ5 ˆ x " 3 x" ‰‹ dx œ 31 ln ¸ x x 3 ¸‘ !Þ& œ 31 ln 25 #Þ&
9 3x x#
56. V œ 21 '0 xy dx œ 21'0 1
'1
1
œ 431 aln kx 1k 2
2x ˆ " (x 1)(2 x) dx œ 41 0 3 " ln k2 xkb‘ ! œ 431 (ln 2)
È3
È3
È3
1È3 3
ln 2;
57. A œ '0 tan" x dx œ cx tan" xd 0 '0 œ
È3
1È3 3
xœ
" A
"# ln ax# 1b‘ 0 œ
È3
'0
x tan" x dx
È
È3
" A
3 " # " Œ # x tan x‘ 0
œ
" A
’ 1# "# ax tan" xb‘ 0 “
œ
" A
Š 1# 5
59. (a)
x# 1 x#
dx
È3 # ‹
µ 1.10 œ
È3 #
16 ‹ œ
4x# 13x 9 x$ 2x# 3x
" A
5
'3
dx dt
œ kx(N x) Ê '
#
x a4x 13x 9b x$ 2x# 3x
" 250 ,
Š 231
dx œ 3'3
" A
5
kœ
dx œ
dx x " A
'3
5
dx x3
2'3
5
Šc4xd $ 3 '3
5
&
dx x(N x)
N œ 1000, t œ 0 and x œ 2 Ê
(b) x œ dx dt
" #
N œ 500 Ê 500
œ k(a x)(b x) Ê
(a) a œ b: Ê
' (a dxx)
" ax
(b) a Á b:
œ
#
œ ' k dt Ê
akt " a
dx ' (a x)(b ' k dt x) œ
t œ 0 and x œ 0 Ê Ê xœ
ab 1 eÐb aÑkt ‘ a beÐb aÑkt
" ba
ln
b a
œ c3 ln kxk ln kx 3k 2 ln kx 1kd &$ œ ln
2 '3
5
dx x 1‹
œ
" A
" 1000
Ê
" N
x ¸ ln ¸ N x œ kt C;
ln ¸ 1000x x ¸ œ
t 250
" 1000
† 499 500e4t œ 1000e4t Ê e4t œ 499 Ê t œ
Ê
" b a
Ê xœa
a akt 1
œ
" a
œC Ê
'
dx xÈ x 3
œ
2 È3
" 4
œ C Ê ln
œ kt
œ (b a)kt
ln ˆ ba ‰
Ê
" a
Ê
" ¸bx¸ ba ln a x œ kt bx b ÐbaÑkt ax œ a e
tan" É x 3 3 C
(We used FORMULA 13(a) with a œ 1, b œ 3)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1000e4t 499 e4t
ln 499 ¸ 1.55 days
a# kt akt 1
' a dx x b " a ' b dx x œ ' k dt x¸ ¸ ba x
" ax
8.5 INTEGRAL TABLES AND COMPUTER ALGEBRA SYSTEMS 1.
" ‰ ln ˆ 499
499x œ e4t (1000 x) Ê a499 e4t b x œ 1000e4t Ê x œ
œ kt C; t œ 0 and x œ 0 Ê
a akt 1
;
µ 3.90 (8 11 ln 2 3 ln 6) œ
' dxx N" ' Ndx x œ ' k dt 2 ¸ ln ¸ 998 œC Ê
125 9
œ k dt " a x
Ê axœ
" N
" 1000
499x 4t Ê 1000 x œ e 4t 1000e œ 499 e4t Ê 500
dx (a x)(bx)
dx x1
dx x3
œ ' k dt Ê
499x ¸ Ê ln ¸ 1000 x œ 4t Ê
60.
dx
È3
58. A œ '3 xœ
" #
'0
œ
x 1 x#
ˆ x " 1 ‰ 23 ˆ 2 " x ‰‰ dx
C;
492 2.
Chapter 8 Techniques of Integration
' xÈdxx4 œ È"
È
È
4 4 ln ¹ Èxx ¹Cœ 4 È4
4
" #
È
42 ln ¹ Èxx ¹C 42
(We used FORMULA 13(b) with a œ 1, b œ 4) 3.
' Èx dx
œ'
x2
2'
(x 2) dx Èx 2 $
œ ˆ 21 ‰
ŠÈx 2‹
œ ' ŠÈx 2‹ dx 2 ' ŠÈx 2‹ "
"
dx
"
2 ˆ 12 ‰
3
dx Èx 2
ŠÈx 2‹ 1
œ Èx 2 ’ 2(x 3 2) 4“ C
(We used FORMULA 11 with a œ 1, b œ 2, n œ 1 and a œ 1, b œ 2, n œ 1) 4.
' (2xxdx3)
" #
œ
$Î#
3) dx ' (2x 3# ' (2x dx3) (2x 3) $Î#
œ
$Î#
" #
' È2xdx 3 3# '
dx $ ˆÈ2x 3‰
' ŠÈ2x 3‹" dx 3# ' ŠÈ2x 3‹$ dx œ ˆ "# ‰ ˆ 2# ‰ ˆÈ2x1 3‰
œ
" #
œ
" #È2x 3
(x3) È2x 3
(2x 3 3) C œ
"
ˆ 3# ‰ ˆ 2# ‰
ˆÈ2x 3‰ (1)
"
C
C
(We used FORMULA 11 with a œ 2, b œ 3, n œ 1 and a œ 2, b œ 3, n œ 3) 5.
' xÈ2x 3 dx œ "# ' (2x 3)È2x 3 dx 3# ' È2x 3 dx œ "# ' ŠÈ2x 3‹$ dx 3# ' ŠÈ2x 3‹" dx &
œ ˆ "# ‰ ˆ 2# ‰
ŠÈ2x 3‹
$
ˆ 3# ‰ ˆ 2# ‰
5
ŠÈ2x 3‹
Cœ
3
(2x 3)$Î# #
2x 5 3 1‘ C œ
(2x 3)$Î# (x 1) 5
C
(We used FORMULA 11 with a œ 2, b œ 3, n œ 3 and a œ 2, b œ 3, n œ 1)
6.
' x(7x 5)$Î# dx œ "7 ' (7x 5)(7x 5)$Î# dx 57 ' (7x 5)$Î# dx œ 7" ' ˆÈ7x 5‰& dx 57 ' ˆÈ7x 5‰$ dx œ ˆ "7 ‰ ˆ 27 ‰
ˆÈ7x 5‰ 7
(
ˆ 57 ‰ ˆ 27 ‰
ˆÈ7x 5‰ 5
&
&Î#
C œ ’ (7x 495) “ ’ 2(7x7 5) 2“ C
&Î#
œ ’ (7x 495) “ ˆ 14x7 4 ‰ C (We used FORMULA 11 with a œ 7, b œ 5, n œ 5 and a œ 7, b œ 5, n œ 3) 7.
' È9x 4x dx œ È9x 4x (#4) ' #
dx xÈ9 4x
C
(We used FORMULA 14 with a œ 4, b œ 9) œ
È9 4x x
È
È
4x 9 2 Š È"9 ‹ ln ¹ È99 ¹C 4x È9
(We used FORMULA 13(b) with a œ 4, b œ 9) œ
8.
'
È9 4x x
dx x# È4x 9
2 3
œ
È
4x 3 ln ¹ È99 ¹C 4x 3 È4x 9 (9)x
4 18
'
dx xÈ4x 9
C
(We used FORMULA 15 with a œ 4, b œ 9) œ
È4x 9 9x
ˆ 29 ‰ Š È29 ‹ tan" É 4x 9 9 C
(We used FORMULA 13(a) with a œ 4, b œ 9) œ 9.
' œ
È4x 9 9x
4 27
tan" É 4x 9 9 C
xÈ4x x# dx œ ' xÈ2 † 2x x# dx œ (x 2)(2x 6)È4x x# 6
4 sin" ˆ x # 2 ‰ C
(We used FORMULA 51 with a œ 2)
$ (x 2)(2x 3†2)È2†2†x x# 2# sin" ˆ x # 2 ‰ 6 È # œ (x 2)(x 33) 4x x 4 sin" ˆ x # 2 ‰ C
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.5 Integral Tables and Computer Algebra Systems 10.
' Èxx x
#
dx œ '
É2† "# x x# x
dx œ É2 † "# x x#
" #
sin" Š
x "# " #
ˆWe used FORMULA 52 with a œ "# ‰
11.
'
dx xÈ 7 x#
œ'
dx #
xÊŠÈ7‹ x#
œ È"7
‹ C œ Èx x#
" #
sin" (2x 1) C
â â â È7 ÊŠÈ7‹# x# â È 7 È 7 x# â â ln â ¹C â C œ È"7 ln ¹ x x â â â â
ŠWe used FORMULA 26 with a œ È7‹
12.
'
dx xÈ 7 x#
œ'
dx #
xÊŠÈ7‹ x#
œ È"7
â â â È7 ÊŠÈ7‹# x# â È 7 È 7 x# â â ln â ¹C â C œ È"7 ln ¹ x x â â â â
ŠWe used FORMULA 34 with a œ È7‹ 13.
' È4x x
#
dx œ '
È 2# x# x
dx œ È2# x# 2 ln ¹ 2
È 2# x# ¹ x
C œ È4 x# 2 ln ¹ 2
È 4 x# ¹ x
C
(We used FORMULA 31 with a œ #) 14.
' Èxx 4 dx œ ' Èx x 2 #
#
#
dx œ Èx# 2# 2 sec" ¸ #x ¸ C œ Èx# 4 2 sec" ¸ x# ¸ C
(We used FORMULA 42 with a œ #) 15.
' e2t cos 3t dt œ 2 e 3 2t
#
#
(2 cos 3t 3 sin 3t) C œ
e2t 13
(2 cos 3t 3 sin 3t) C
(We used FORMULA 108 with a œ 2, b œ 3) 16.
' ec3t sin 4t dt œ (3)ec 4 3t
#
#
(3 sin 4t 4 cos 4t) C œ
ec3t 25
($ sin 4t 4 cos 4t) C
(We used FORMULA 107 with a œ 3, b œ 4) 17.
' x cos" x dx œ '
x" cos" x dx œ
x1b1 11
cos" x
" 1 1
' Èx b
(We used FORMULA 100 with a œ 1, n œ 1) # œ x cos" x " ˆ " sin" x‰ " Š " xÈ1 x# ‹ C œ #
#
#
#
#
1 1
dx 1 x#
x# #
œ
x# #
cos" x
cos" x " 4
" #
' Èx
# dx 1 x#
sin" x 4" xÈ1 x# C
(We used FORMULA 33 with a œ 1) 18.
' x tan" x dx œ '
x" tan" a"xb dx œ
x1b1 11
tan" a"xb
" 11
' 1 xba"bdxx 1 1
# #
œ
x# #
tan" x
" #
' 1xdxx #
(We used FORMULA 101 with a œ 1, n œ 1)
19.
œ
x# #
œ
x# #
' ˆ" 1 " x ‰dx (after long division) tan" x "# ' dx "# ' 1 " x dx œ x# tan" x #" x #" tan" x C œ #" aax# "btan" x xb C tan" x
" #
#
#
#
' x# tan" x dx œ 2x 1 tan" x 2 " 1 ' 1x x# dx œ x3$ tan" x 3" ' 1 x$x# dx 2 1
2 1
(We used FORMULA 101 with a œ 1, n œ 2);
' 1 x$x# dx œ ' œ
x$ 3
#
tan" x
x dx '
x# 6
" 6
x dx 1 x#
œ
x# #
" #
ln a1 x# b C Ê
' x# tan" x dx
ln a1 x# b C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
493
494 20.
Chapter 8 Techniques of Integration
' '
tanc" x x#
xÐ 21Ñ (2 1)
tan" x
" (2 1)
' 1xÐx#Ñ dx œ (x1)" tan" x ' a1 x "x#b dx 2 1
(We used FORMULA 101 with a œ 1, n œ 2); xc" dx 1 x #
Ê' 21.
dx œ ' x# tan" x dx œ œ'
tanc" x x#
œ'
dx x a1 x# b
'
dx x
x dx 1 x #
œ ln kxk
dx œ x" tan" x ln kxk
" #
" #
ln a1 x# b C
ln a1 x# b C
' sin 3x cos 2x dx œ cos105x cos# x C (We used FORMULA 62(a) with a œ 3, b œ 2)
22.
' sin 2x cos 3x dx œ cos105x cos# x C (We used FORMULA 62(a) with a œ 2, b œ 3)
23.
'
8 sin 4t sin
t #
dx œ
8 7
sin ˆ 7t# ‰
8 9
sin ˆ 9t# ‰ C œ 8 –
sin Š 7t #‹ 7
sin Š 9t #‹ 9
—C
(We used FORMULA 62(b) with a œ 4, b œ "# ) 24.
' sin 3t sin 6t dt œ 3 sin ˆ 6t ‰ sin ˆ #t ‰ C
(We used FORMULA 62(b) with a œ "3 , b œ 6" )
25.
' cos 3) cos 4) d) œ 6 sin ˆ 12) ‰ 76 sin ˆ 17#) ‰ C
(We used FORMULA 62(c) with a œ "3 , b œ 4" )
26.
13) # ‹
' cos 2) cos 7) d) œ 13" sin ˆ 132) ‰ 151 sin ˆ 15#) ‰ C œ sin Š13 " 2,
(We used FORMULA 62(c) with a œ 27.
'x
$ x1 ax # 1 b # œ "# lnax#
dx œ ' 1b
x dx x# 1
'
x 2 a1 x # b
dx ax # 1 b# " " # tan
œ
" #
sin Š 15#) ‹ 15
C
b œ 7)
' d axx 11b ' #
#
dx ax # 1 b#
xC
(For the second integral we used FORMULA 17 with a œ 1) 28.
'
x# 6x ax # 3 b #
œ
" È3
dx œ '
dx x# 3
tan" Š Èx3 ‹
'
3 ax # 3 b
6x dx ax # 3 b#
3
'
3 dx ax # 3 b#
Î
œ'
x # Ï 2 ŠÈ3‹ ŒŠÈ3‹ x# #
x#
dx # ŠÈ3‹
" $ 2 ŠÈ3‹
3'
d ax # 3 b ax # 3 b#
tan" Š Èx3 ‹
Ñ Ò
3'
dx
# #
”x# ŠÈ3‹ •
C
ŠFor the first integral we used FORMULA 16 with a œ È3; for the third integral we used FORMULA 17 with a œ È3‹ œ
29.
'
" 2È 3
tan" Š Èx3 ‹
sin" Èx dx;
3 x # 3
x 2 ax # 3 b
C
Ô u œ Èx × 1b1 Ä 2 ' u" sin" u du œ 2 Š 1u1 sin" u x œ u# Õ dx œ 2u du Ø
" 1 1
' Èu b
1 1
1 u#
du‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.5 Integral Tables and Computer Algebra Systems œ u# sin" u '
u# du È 1 u#
(We used FORMULA 99 with a œ 1, n œ 1) œ u# sin" u Š "# sin" u "# uÈ1 u# ‹ C œ ˆu# "# ‰ sin" u "# uÈ1 u# C (We used FORMULA 33 with a œ 1) œ ˆx "# ‰ sin" Èx "# Èx x# C
30.
u œ Èx × " Ä ' cosu u † 2u du œ 2' cos" u du œ 2 Šu cos" u 1" È1 u# ‹ C x œ u# Õ dx œ 2u du Ø (We used FORMULA 97 with a œ 1) œ 2 ŠÈx cos" Èx È1 x‹ C
' cosÈc"xÈx dx; Ô
Ô u œ Èx × 31. dx; Ä x œ u# 1x Õ dx œ 2u du Ø œ sin" u uÈ1 u# C
' ÈÈx
' Èu†2u
1 u#
du œ 2 '
u# È 1 u#
du œ 2 Š "# sin" u "# uÈ1 u# ‹ C
(We used FORMULA 33 with a œ 1) œ sin" Èx Èx È1 x C œ sin" Èx Èx x# C
32.
u œ Èx × Ä x œ u# Õ dx œ 2u du Ø
' ÈÈ2 x x dx; Ô
' È2 u u
#
† 2u du œ 2 ' ÊŠÈ2‹ u# du #
#
#
œ 2 – u# ÊŠÈ2‹ u#
ŠÈ2‹ #
sin" Š Èu2 ‹— C œ uÈ2 u# 2 sin" Š Èu2 ‹ C
ŠWe used FORMULA 29 with a œ È2‹ œ È2x x# 2 sin" È x# C 33.
' (cot t) È1 sin# t dt œ ' È1 sinsintt(cos t) dt ; ” #
œ È1 u# ln ¹ 1
È 1 u# ¹ u
u œ sin t Ä du œ cos t dt •
' È1 u u
#
du
C
(We used FORMULA 31 with a œ 1) È # œ È1 sin# t ln ¹ 1 1 sin t ¹ C sin t
34.
'
dt (tan t) È4 sin# t
œ'
cos t dt (sin t) È4 sin# t
;”
u œ sin t Ä' du œ cos t dt •
du uÈ4 u#
œ "# ln ¹ 2
È 4 u# ¹ u
C
(We used FORMULA 34 with a œ 2) œ "# ln ¹ 2
35.
È4 sin# t ¹ sin t
C
Ô u œ ln y × y œ eu Ä Õ dy œ eu du Ø œ ln ¸ln y È3 (ln y)# ¸ C
' yÈ3 dy(ln y)
#
;
'
eu du eu È 3 u#
œ'
du È 3 u#
œ ln ¹u È3 u# ¹ C
ŠWe used FORMULA 20 with a œ È3‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
495
496 36.
Chapter 8 Techniques of Integration t œ Èy × # Ä 2 ' t tan" t dt œ 2 ’ t# tan" t y œ t# Õ dy œ 2t dt Ø (We used FORMULA 101 with n œ 1, a œ 1)
' tan" Èy dy; Ô œ t# tan" t '
37.
'È
1 dx x2 2x5
t# 1 t# 1
œ'
dt '
" #
' 1 t t #
#
dt“ œ t# tan" t '
t# 1 t#
dt
œ t# tan" t t tan" t C œ y tan" Èy tan" Èy Èy C
dt 1 t#
tœx1 1 dx;” dt œ dx • É ax 1 b 2 4
Ä'
1 Èt2 4 dt
(We used FORMULA 20 with a œ 2) œ ln¹t Èt2 4¹ C œ ln¹ax 1b Éax 1b2 4 C œ ln¹ax 1b Èx2 2x 5¹ C
38.
'È
œ'
x2 dx x2 4x 5
x2 É ax 2 b + 1 2
tœx2 Ä' dt œ dx •
dx; ”
(We used FORMULA 25 with a œ 1) œ – 12 ln¹t Èt2 1¹
tÈt2 1 — 2
œ'
t2 4t 2 Èt2 + 1 dt
œ'
t2 Èt2 + 1 dt
'
4t Èt2 + 1 dt
'
4 Èt2 + 1 dt
(We used FORMULA 20 with a œ 1)
4Èt2 1 ”4 ln¹t Èt2 1¹• C
œ 12 ln¹ax 2b Éax 2b2 1¹ œ 72 ln¹ax 2b Èx2 4x 5¹
39.
at 2 b 2 Èt2 + 1 dt
ax 2 b É a x 2 b 2 1 2
ax 6bÈx2 4x 5 2
4Éax 2b2 1 4 ln¹ax 2b Éax 2b2 1¹ C
C
' È5 4x x2 dx œ ' É9 ax 2b2 dx; ” t œ x 2 • dt œ dx
Ä ' È9 t2 dt;
(We used FORMULA 29 with a œ 3) œ 2t È9 t2 40.
32 1 ˆ t ‰ 2 sin 3
x2É 9 2
Cœ
ax 2b2 92 sin1 ˆ x 3 2 ‰ C œ
' x2 È2x x2 dx œ ' x2 É1 ax 1b2 dx;
x2È 5 2
4x x2 92 sin1 ˆ x 3 2 ‰ C
2 tœx1 2 ” dt œ dx • Ä ' at 1b È1 t2 dt œ ' at 2t 1bÈ1 t2 dt
œ ' t2 È1 t2 dt ' 2tÈ1 t2 dt ' È1 t2 dt (We used FORMULA 30 with a œ 1)
(We used FORMULA 29 with a œ 1) 2 3 Î2
œ ” 18 sin1 ˆ 1t ‰ 81 t È1 t2 a12 2t2 b• 32 a1 t b 4
” 2t È1 t2
12 1 ˆ t ‰ 2 sin 1 •
œ "8 sin1 ax 1b 81 ax 1b É1 ax 1b2 Š12 2ax 1b2 ‹ 32 Š1 ax 1b2 ‹ 12 sin1 ax 1b C œ 58 sin1 ax 1b 23 a2x x2 b 41.
3 Î2
x1 8
x1É 1 2
ax 1b2
È2x x2 a2x2 4x 5b C
' sin& 2x dx œ sin 2x5†#cos 2x 5 5 1 ' sin$ 2x dx œ sin 2x10cos 2x 45 ’ sin 2x3†#cos 2x 3 3 1 ' sin 2x dx“ %
%
#
(We used FORMULA 60 with a œ 2, n œ 5 and a œ 2, n œ 3) % % 2 8 ˆ œ sin 2x10cos 2x 15 sin# 2x cos 2x 15 "# ‰ cos 2x C œ sin 2x10cos 2x 42.
3 Î2
C
'
8 cos% 21t dt œ 8 Š cos
$
21t sin 21t 4 †2 1
41 4
2 sin# 2x cos 2x 15
4 cos 2x 15
' cos# 21t dt‹
(We used FORMULA 61 with a œ 21, n œ 4) œ
cos$ 21t sin 21t 1
6 ’ #t
sin (2†21†t) 4†21 “
C
(We used FORMULA 59 with a œ 21) œ
cos$ 21t sin 21t 1
3t
3 sin 41t 41
Cœ
cos$ 21t sin 21t 1
3 cos 21t sin 21t 21
3t C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
C
Section 8.5 Integral Tables and Computer Algebra Systems 43.
' sin# 2) cos$ 2) d) œ sin #2(2) cos3) 2) 33 #1 ' sin# 2) cos 2) d) $
#
(We used FORMULA 69 with a œ 2, m œ 3, n œ 2) œ
44.
'
sin$ 2) cos# 2) 10
2 5
' sin# 2) cos 2) d) œ sin 2)10cos 2) 25 ’ "# ' sin# 2) d(sin 2))“ œ sin 2)10cos 2) sin152) C $
#
$
2 sin# t sec% t dt œ ' 2 sin# t cos% t dt œ 2 Š sin2t cos 4
c$ t
21 24
#
$
' cos% t dt‹
(We used FORMULA 68 with a œ 1, n œ 2, m œ 4)
t œ sin t cos$ t ' cos% t dt œ sin t cos$ t ' sec% t dt œ sin t cos$ t Š sec4 t tan 1 #
42 41
'
sec# t dt‹
(We used FORMULA 92 with a œ 1, n œ 4) œ sin t cos$ t Š sec œ
2 3
#
t tan t ‹ 3
2 3
tan t C œ
2 3
sec# t tan t
2 3
tan t C œ
2 3
tan t asec# t 1b C
tan$ t C
An easy way to find the integral using substitution:
' 2 sin# t cos% t dt œ '
45.
2 tan# t sec# t dt œ ' 2 tan# t d(tan t) œ
2 3
tan$ t C
' 4 tan$ 2x dx œ 4 Š tan2†#2x ' tan 2x dx‹ œ tan# 2x 4' tan 2x dx #
(We used FORMULA 86 with n œ 3, a œ 2) œ tan# 2x 4# ln ksec 2xk C œ tan# 2x 2 ln ksec 2xk C 46.
' 8 cot% t dt œ 8 Š cot3 t ' cot# t dt‹ $
(We used FORMULA 87 with a œ 1, n œ 4) œ 8 ˆ "3 cot$ t cot t t‰ C (We used FORMULA 85 with a œ 1) 47.
' 2 sec$ 1x dx œ 2 ’ sec11(3xtan1) 1x 33 12 ' sec 1x dx “ (We used FORMULA 92 with n œ 3, a œ 1) œ 1" sec 1x tan 1x 1" ln ksec 1x tan 1xk C (We used FORMULA 88 with a œ 1)
48.
' 3 sec% 3x dx œ 3 ’ sec3(43xtan1) 3x 44 12 ' sec# 3x dx“ #
(We used FORMULA 92 with n œ 4, a œ 3) œ
sec# 3x tan 3x 3
2 3
tan 3x C
(We used FORMULA 90 with a œ 3) 49.
' csc& x dx œ csc5 xcot1 x 55 12 ' csc$ x dx œ csc x4 cot x 34 Š csc3xcot1 x 33 21 ' csc x dx‹ $
$
(We used FORMULA 93 with n œ 5, a œ 1 and n œ 3, a œ 1) œ "4 csc$ x cot x 38 csc x cot x 38 ln kcsc x cot xk C (We used FORMULA 89 with a œ 1) 50.
' 16x$ (ln x)# dx œ 16 ’ x (ln4 x) %
#
2 4
'
%
#
%
x) x$ ln x dx“ œ 16 ’ x (ln4 x) "# ’ x (ln 4
" 4
'
x$ dx““
(We used FORMULA 110 with a œ 1, n œ 3, m œ 2 and a œ 1, n œ 3, m œ 1) %
#
œ 16 Š x (ln4 x)
x% (ln x) 8
x% 3# ‹
C œ 4x% (ln x)# 2x% ln x
x% #
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
497
498
Chapter 8 Techniques of Integration
' et sec$ aet "b dt; ” x œ e
" x 2 ' Ä ' sec$ x dx œ sec3xtan 33 sec x dx 1 1 dx œ et dt • (We used FORMULA 92 with a œ 1, n œ 3) œ sec x#tan x "# ln ksec x tan xk C œ "# csec aet 1b tan aet 1b ln ksec aet 1b tan aet 1bkd C t
51.
t œ È) × t cot t 32 ' 52. csc t dt“ Ä 2 ' csc$ t dt œ 2 ’ csc3 ) œ t# 1 31 ) Õ d) œ 2t dt Ø (We used FORMULA 93 with a œ 1, n œ 3) œ 2 csc t cot t " ln kcsc t cot tk‘ C œ csc È) cot È) ln ¹csc È) cot È)¹ C
' cscÈÈ) d); Ô $
#
53.
#
'01 2Èx# 1 dx; cx œ tan td
Ä 2 '0
1Î4
sec t † sec# t dt œ 2'0
1Î4
1Î%
t‘ sec$ t dt œ 2 ” sec3 t†tan 1 !
32 31
'01Î4 sec t dt•
(We used FORMULA 92 with n œ 3, a œ 1) 1Î% œ csec t † tan t ln ksec t tan tkd œ È2 ln ŠÈ2 1‹ !
54.
È3Î2
'0
dy a1 y# b&Î#
'01Î3
; cy œ sin xd Ä
cos x dx cos& x
x œ '0 sec% x dx œ ’ sec4 xtan 1 “ 1Î3
#
(We used FORMULA 92 with a œ 1, n œ 4) œ ’ sec
55.
#
x tan x 3
'12 ar r1b #
$Î#
2 3
tan x“
1Î$
œ ’ tan3 ) tan ) )“
1Î$ !
!
42 41
'01Î3 sec# x dx
œ ˆ 34 ‰ È3 ˆ 32 ‰ È3 œ 2È3
!
dr; cr œ sec )d Ä
$
1Î$
œ
1Î$ 1Î3 ) ) '01Î3 tan ' 1Î3 tan% ) d) œ ’ tan '0 tan# ) d) sec ) (sec ) tan )) d) œ 0 41 “ $
$
!
3È 3 3
È3
1 3
œ
1 3
(We used FORMULA 86 with a œ 1, n œ 4 and FORMULA 84 with a œ 1) 56.
È3
'01Î
dt at# 1b(Î#
œ ’ cos
%
; ct œ tan )d Ä '0
1Î6
1Î' ) sin ) “ 5 !
45 ”’ cos
#
sec# ) d) sec( )
1Î' ) sin ) “ 3 !
œ '0
1Î6
ˆ 3 3 1 ‰'0
cos& ) d) œ ’ cos
1Î6
%
1Î' ) sin ) “ 5 !
cos ) d)• œ ’ cos
%
) sin ) 5
(We used FORMULA 61 with a œ 1, n œ 5 and a œ 1, n œ 3) È % # Š #3 ‹ ˆ #" ‰ 4 ‰ È3 8 ‰ ˆ"‰ 9 " 4 3†9 48 32†4 œ ˆ 15 œ Š # ‹ ˆ "# ‰ ˆ 15 5 # œ 160 10 15 œ 480 57. S œ
È2
'0
ˆ 5 5 1 ‰ '0
1Î6
4 15
cos$ ) d)
cos# ) sin )
8 15
sin )“
203 480
21yÈ1 (yw )# dx
È2
œ 21 '0
Èx# 2 É1
È2
œ 2È21 '0 œ 2È21 ’ x
x# x# 2
dx
Èx# 1 dx
È x# 1 #
" #
ln ¹x Èx# 1¹“
È2 0
(We used FORMULA 21 with a œ 1) œ È21 ’È6 ln ŠÈ2 È3‹“ œ 21È3 1È2 ln ŠÈ2 È3‹
È3Î2
58. L œ '0
È3Î2
È1 (2x)# dx œ 2 ' 0
É "4 x# dx œ 2 ’ x# É 4" x# ˆ 4" ‰ ˆ #" ‰ ln Šx É 4" x# ‹“
ˆWe used FORMULA 2 with a œ "# ‰
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
È3Î2 0
1Î' !
Section 8.5 Integral Tables and Computer Algebra Systems œ ’ x# È1 4x# œ
È3 4
(2)
59. A œ '0
3
" 4
dx È x 1
ln Š
" 4
ln Šx #" È1 4x# ‹“
È3 #
" 4
1‹
ln 2 œ
È3 #
$
œ ’2Èx 1“ œ 2; x œ !
œ
" A
'03 Èx 1 dx A" '03 Èxdx 1
œ
" #
$ † 23 (x 1)$Î# ‘ ! 1 œ
4 3
" A
È3Î2 0
" 4
œ
È3 4
É1 4 ˆ 34 ‰
" 4
ln Š
È3 #
#" É1 4 ˆ 34 ‰‹
" 4
ln
" #
ln ŠÈ3 2‹
'03 Èxxdx 1
;
(We used FORMULA 11 with a œ 1, b œ 1, n œ 1 and a œ 1, b œ 1, n œ 1) " #A
yœ
'03 x dx 1 œ 4" cln (x 1)d $! œ 4" ln 4 œ #" ln 2 œ ln È2
60. My œ '0 x ˆ 2x36 3 ‰ dx œ 18 '0 3
3
2x 3 2x 3
dx 54'0
3
dx 2x 3
œ c18x 27 ln k2x 3kd $!
œ 18 † 3 27 ln 9 (27 ln 3) œ 54 27 † 2 ln 3 27 ln 3 œ 54 27 ln 3 61. S œ 21 'c1 x# È1 4x# dx; 1
u œ 2x ” du œ 2 dx • Ä œ
1 4
1 4
'c22 u# È1 u# du
’ 8u a1 2u# bÈ1 u#
" 8
ln Šu È1 u# ‹“
(We used FORMULA 22 with a œ 1) œ 1 ’ 2 (1 2 † 4)È1 4 " ln Š2 È1 4‹ 4
8
1 4
#
8
28 (1 2 † 4)È1 4 œ
#
’ #9 È5
" 8
" 8
ln Š2 È1 4‹“
È
ln Š 22È55 ‹“ ¸ 7.62
62. (a) The volume of the filled part equals the length of the tank times the area of the shaded region shown in the accompanying figure. Consider a layer of gasoline of thickness dy located at height y where r y r d. The width of this layer is crbd
2Èr# y# . Therefore, A œ 2 'cr crbd
and V œ L † A œ 2L 'cr
crbd
Èr# y# dy
Èr# y# dy
crbd
# yÈ r y (b) 2L 'cr Èr# y# dy œ 2L ’ # r# sin" yr “ cr (We used FORMULA 29 with a œ r) # # # r) È ‰ È2rd d# Š r# ‹ ˆsin" ˆ d r r ‰ 1# ‰“ œ 2L ’ (d 2rd d# r# sin" ˆ d r r ‰ r# ˆ 1# ‰“ œ 2L ’ˆ d-r # # #
#
63. The integrand f(x) œ Èx x# is nonnegative, so the integral is maximized by integrating over the function's entire domain, which runs from x œ 0 to x œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
499
500
Chapter 8 Techniques of Integration Ê
'01 Èx x# dx œ '01 É2 † "# x x# dx œ ” ˆx #
"‰ #
É2 † "# x x#
ˆWe used FORMULA 48 with a œ "# ‰ œ’
ˆx "# ‰ #
Èx x#
" 8
"
" 8
sin" (2x 1)“ œ !
†
1 #
8" ˆ 1# ‰ œ
ˆ "# ‰# #
sin" Š
x "# " #
"
‹•
!
1 8
64. The integrand is maximized by integrating g(x) œ xÈ2x x# over the largest domain on which g is nonnegative, namely [!ß 2] Ê
'02 xÈ2x x# dx œ ’ (x 1)(2x 63)È2x x
#
" #
sin" (x 1)“
(We used FORMULA 51 with a œ ") œ "# † 1# "# ˆ 1# ‰ œ 1#
# !
CAS EXPLORATIONS 65.
66.
Example CAS commands: Maple: q1 := Int( x*ln(x), x ); q1 = value( q1 ); q2 := Int( x^2*ln(x), x ); q2 = value( q2 ); q3 := Int( x^3*ln(x), x ); q3 = value( q3 ); q4 := Int( x^4*ln(x), x ); q4 = value( q4 ); q5 := Int( x^n*ln(x), x ); q6 := value( q5 ); q7 := simplify(q6) assuming n::integer; q5 = collect( factor(q7), ln(x) ); Example CAS commands: Maple: q1 := Int( ln(x)/x, x ); q1 = value( q1 ); q2 := Int( ln(x)/x^2, x ); q2 = value( q2 ); q3 := Int( ln(x)/x^3, x ); q3 = value( q3 ); q4 := Int( ln(x)/x^4, x ); q4 = value( q4 ); q5 := Int( ln(x)/x^n, x ); q6 := value( q5 ); q7 := simplify(q6) assuming n::integer; q5 = collect( factor(q7), ln(x) );
# (a) # (b) # (c) # (d) # (e)
# (a) # (b) # (c) # (d) # (e)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.5 Integral Tables and Computer Algebra Systems 67.
Example CAS commands: Maple: q := Int( sin(x)^n/(sin(x)^n+cos(x)^n), x=0..Pi/2 ); q = value( q ); q1 := eval( q, n=1 ): q1 = value( q1 ); for N in [1,2,3,5,7] do q1 := eval( q, n=N ); print( q1 = evalf(q1) ); end do: qq1 := PDEtools[dchange]( x=Pi/2-u, q, [u] ); qq2 := subs( u=x, qq1 ); qq3 := q + q = q + qq2; qq4 := combine( qq3 ); qq5 := value( qq4 ); simplify( qq5/2 );
# (a) # (b)
# (c)
65-67. Example CAS commands: Mathematica: (functions may vary) In Mathematica, the natural log is denoted by Log rather than Ln, Log base 10 is Log[x,10] Mathematica does not include an arbitrary constant when computing an indefinite integral, Clear[x, f, n] f[x_]:=Log[x] / xn Integrate[f[x], x] For exercise 67, Mathematica cannot evaluate the integral with arbitrary n. It does evaluate the integral (value is 1/4 in each case) for small values of n, but for large values of n, it identifies this integral as Indeterminate 65. (e)
' xn ln x dx œ x n ln1 x n " 1 ' xn dx, n Á 1 n 1
(We used FORMULA 110 with a œ 1, m œ 1) n 1 n 1 n 1 œ x n ln1 x (nx 1)# C œ nx 1 ˆln x n " 1 ‰ C 66. (e)
' xn ln x dx œ xcnb ln1 x (n)" 1 ' xn dx, n Á 1 n 1
(We used FORMULA 110 with a œ 1, m œ 1, n œ n) œ
x1cn ln x 1n
" 1 n
1cn
Š 1x n ‹ C œ
x1cn 1n
ˆln x
" ‰ 1n
C
67. (a) Neither MAPLE nor MATHEMATICA can find this integral for arbitrary n. (b) MAPLE and MATHEMATICA get stuck at about n œ 5. (c) Let x œ 1# u Ê dx œ du; x œ 0 Ê u œ 1# , x œ 1# Ê u œ 0; I œ '0
1Î2
Ê
sinn x dx sinn xcosn x
œ '1Î2 sinn ˆ 1 u‰ # cosn ˆ 1 u‰ œ '0 sinn ˆ 1 u‰ du
0
#
#
1Î2 x cos x ‰ ' 1Î2 dx œ 1# I I œ '0 ˆ sin sin x cos x dx œ 0 n
n
n
n
1Î2
cosn u du cosn u sinn u
Ê Iœ
œ '0
1Î2
cosn x dx cosn x sinn x
1 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
501
502
Chapter 8 Techniques of Integration
8.6 NUMERICAL INTEGRATION 1.
'12 x dx I.
(a) For n œ 4, ?x œ b n a œ 2 4 1 œ "4 Ê ! mf(xi ) œ 12 Ê T œ " (12) œ 3 ; 8 #
?x #
" 8
œ
;
f(x) œ x Ê f w (x) œ 1 Ê f ww œ 0 Ê M œ 0 Ê kE T k œ 0
(b)
'12 x dx œ ’ x2 “ # œ 2 "# œ 3#
(c)
kET k True Value
#
"
(b)
'12 x dx œ 3#
(c)
kES k True Value
21 " ?x 4 œ 4 Ê 3 " 3 12 (18) œ # ;
Ê kES k œ '1 x dx S œ 2
œ
" 12
3 #
3 #
œ0
" 2
Ê
?x #
" 4
œ
;
f(x) œ 2x 1 Ê f w (x) œ 2 Ê f ww œ 0 Ê M œ 0 Ê kE T k œ 0
(b)
'13 (2x 1) dx œ cx# xd $" œ (9 3) (1 1) œ 6
(c)
kET k True Value
" 2
Ê
?x 3
" 6
œ
m 1 4 2 4 1
mf(xi ) 1 5 3 7 2
x! x" x# x$ x%
xi 1 3/2 2 5/2 3
f(xi ) 1 2 3 4 5
m 1 2 2 2 1
mf(xi ) 1 4 6 8 5
'13 (2x 1) dx œ 6
x! x" x# x$ x%
xi 1 3/2 2 5/2 3
f(xi ) 1 2 3 4 5
m 1 4 2 4 1
mf(xi ) 1 8 6 16 5
x! x" x# x$ x%
xi 1 1/2 0 1/2 1
f(xi ) 2 5/4 1 5/4 2
m 1 2 2 2 1
mf(xi ) 2 5/2 2 5/2 2
;
Ê kES k œ '1 (2x 1) dx S 3
œ66œ0 ‚ 100 œ 0%
'11 ax# 1b dx
(a) For n œ 4, ?x œ b n a œ 1 4(1) œ 42 œ ! mf(xi ) œ 11 Ê T œ " (11) œ 2.75à 4
" #
Ê
?x #
œ
" 4
à
f(x) œ x# 1 Ê f w (x) œ 2x Ê f ww (x) œ 2 Ê M œ 2 1) ˆ " ‰# " Ê kET k Ÿ 11( # # (2) œ 1# or 0.08333
(b)
f(xi ) 1 5/4 3/2 7/4 2
3
6
I.
xi 1 5/4 3/2 7/4 2
Ê kET k œ '1 (2x 1) dx T œ 6 6 œ 0
f Ð%Ñ (x) œ 0 Ê M œ 0 Ê kES k œ 0
3.
x! x" x# x$ x%
‚ 100 œ 0%
II. (a) For n œ 4, ?x œ b n a œ 3 4 1 œ 42 œ ! mf(xi ) œ 36 Ê S œ " (36) œ 6 ;
kES k True Value
mf(xi ) 1 5/2 3 7/2 2
;
‚ 100 œ 0%
(a) For n œ 4, ?x œ b n a œ 3 4 1 œ 42 œ ! mf(xi ) œ 24 Ê T œ " (24) œ 6 ; 4
(c)
m 1 2 2 2 1
2
'13 (2x 1) dx
(b)
f(xi ) 1 5/4 3/2 7/4 2
Ê kET k œ '1 x dx T œ 0
f Ð%Ñ (x) œ 0 Ê M œ 0 Ê kES k œ 0
I.
xi 1 5/4 3/2 7/4 2
‚ 100 œ 0%
II. (a) For n œ 4, ?x œ b n a œ ! mf(xi ) œ 18 Ê S œ
2.
x! x" x# x$ x%
'11 ax# 1b dx œ ’ x3
$
x“
"
"
œ ˆ 13 1‰ ˆ 13 1‰ œ
8 3
Ê ET œ '1 ax# 1b dx T œ 1
Ê kET k œ ¸ 1"# ¸ ¸ 0.08333 (c)
kET k True Value
"
‚ 100 œ Š 18# ‹ ‚ 100 ¸ 3% 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
8 3
11 4
œ 1"#
Section 8.6 Numerical Integration II. (a) For n œ 4, ?x œ b n a œ 1 4(1) œ 42 œ "# Ê ?3x œ "6 à ! mf(xi ) œ 16 Ê S œ " (16) œ 8 œ 2.66667 ; 6
3
f Ð$Ñ (x) œ 0 Ê f Ð%Ñ (x) œ 0 Ê M œ 0 Ê kES k œ 0 (b)
'11 ax# 1b dx œ ’ x3
$
x“
" "
œ
8 3
Ê kES k œ '1 ax# 1b dx S œ 1
(c) 4.
kES k True Value
8 3
(a) For n œ 4, ?x œ b n a œ 0 4(2) œ ! mf(xi ) œ 3 Ê T œ " (3) œ 3 ; 4
2 4
" #
œ
?x #
Ê
œ
4
" 4
f(x) œ x# 1 Ê f w (x) œ 2x Ê f ww (x) œ 2 2) ˆ " ‰# " Ê M œ 2 Ê kET k Ÿ 01( # # (2) œ 1# œ 0.08333
(b)
'02 ax# 1b dx œ ’ x3
$
kET k True Value
x“
" 12
Ê kE T k œ (c)
!
#
œ 0 ˆ 83 2‰ œ
3
f
2 3
5.
6
6
(x) œ 0 Ê f
Ð%Ñ
;
(x) œ 0 Ê M œ 0 Ê kES k œ 0
'2 ax# 1b dx œ 23 2 2 3 3 œ0 kES k True Value ‚ 100
2 3
Ê kES k œ '2 ax# 1b dx S 0
(a) For n œ 4, ?x œ b n a œ 2 4 0 œ 42 œ "2 " ! Ê ?x mf(ti ) œ 25 Ê T œ 4" (25) œ # œ 4 ; $
w
#
ww
25 4
f(t) œ t t Ê f (t) œ 3t 1 Ê f (t) œ 6t 0 ˆ " ‰# Ê M œ 12 œ f ww (2) Ê kET k Ÿ 212 # (12) œ
(b) (c) II. (a)
'02 at$ tb dt œ ’ t4 t# “ # œ Š 24 %
'0 at
$
2# #‹
m 1 2 2 2 1
mf(xi ) 3 5/2 0 3/2 1
Ê ET œ '2 ax# 1b dx T œ 0
x! x" x# x$ x%
xi 2 3/2 1 1/2 0
t! t" t# t$ t%
ti 0 1/2 1 3/2 2
f(xi ) 3 5/4 0 3/4 1
2 3
m 1 4 2 4 1
; " #
f(ti ) 0 5/8 2 39/8 10
m 1 2 2 2 1
0 œ 6 Ê kET k œ '0 at$ tb dt T œ 6 2
Ê
?x 3
œ
" 6
;
(t) œ 0 Ê M œ 0 Ê kES k œ 0
tb dt œ 6 Ê kES k œ '0 at$ tb dt S 2
œ66œ0 kES k True Value
f(xi ) 3 5/4 0 3/4 1
3 4
" œ 12
mf(xi ) 3 5 0 3 1
mf(ti ) 0 5/4 4 39/4 10 25 4
œ 4" Ê kET k œ
‚ 100 œ
(t) œ 6 Ê f
2
(c)
%
¸ "4 ¸ 6 ‚ 100 ¸ 4% ba For n œ 4, ?x œ n œ 2 4 0 œ 42 œ "2 ! mf(ti ) œ 36 Ê S œ " (36) œ 6 ; 6 Ð$Ñ Ð%Ñ
f
(b)
#
!
kET k True Value
x! x" x# x$ x%
xi 2 3/2 1 1/2 0
œ 0%
'02 at$ tb dt I.
mf(xi ) 2 5 2 5 2
3
œ
(c)
m 1 4 2 4 1
"
0
(b)
f(xi ) 2 5/4 1 5/4 2
‚ 100 œ Š 122 ‹ ‚ 100 ¸ 13%
II. (a) For n œ 4, ?x œ b n a œ 0 4(2) œ 42 œ "# Ê ?x œ " ; ! mf(xi ) œ 4 Ê S œ " (4) œ Ð$Ñ
xi 1 1/2 0 1/2 1
‚ 100 œ 0%
'02 ax# 1b dx I.
œ0
8 3
x! x" x# x$ x%
503
t! t" t# t$ t%
ti 0 1/2 1 3/2 2
f(ti ) 0 5/8 2 39/8 10
m 1 4 2 4 1
‚ 100 œ 0%
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
mf(ti ) 0 5/2 4 39/2 10
" 4
504 6.
Chapter 8 Techniques of Integration
'11 at$ 1b dt I.
t! t" t# t$ t%
(a) For n œ 4, ?x œ b n a œ 1 4(1) œ 42 œ "# " ! mf(ti ) œ 8 Ê T œ " (8) œ 2 ; Ê ?x # œ 4 ; 4
f(t) œ t$ 1 Ê f w (t) œ 3t# Ê f ww (t) œ 6t 1) ˆ " ‰# Ê M œ 6 œ f ww (1) Ê kET k Ÿ 11( # # (6) œ
(b)
'11 at$ 1b dt œ ’ t4 t“ "
(c)
kET k True Value
%
3
6
6
'11 at$ 1b dt œ 2 œ22œ0
kES k True Value
7.
'12 s"
#
I.
Ê kES k œ '1 at$ 1b dt S 1
t! t" t# t$ t%
ti 1 1/2 0 1/2 1
f(ti ) 0 7/8 1 9/8 2
s! s" s# s$ s%
si 1 5/4 3/2 7/4 2
f(si ) 1 16/25 4/9 16/49 1/4
m 1 4 2 4 1
mf(ti ) 0 7/2 2 9/2 2
‚ 100 œ 0%
ds ba n
(a) For n œ 4, ?x œ ! mf(si ) œ
179,573 44,100
œ
21 4
Ê Tœ
œ
" 8
" 4
Ê
?x #
œ
Š 179,573 44,100 ‹ œ
" 8
;
179,573 352,800
" 2 w s# Ê f (s) œ s$ 6 Ê f ww (s) œ s% Ê M œ 6 œ f ww (1) # Ê kET k Ÿ 2 1# " ˆ "4 ‰ (6) œ 3"# œ 0.03125 2 2 # " ds œ s# ds œ 1s ‘ " œ #" ˆ 1" ‰ # 1 s 1
¸ 0.50899; f(s) œ
'
(b)
mf(ti ) 0 7/4 2 9/4 2
‚ 100 œ 0%
f Ð$Ñ (t) œ 6 Ê f Ð%Ñ (t) œ 0 Ê M œ 0 Ê kES k œ 0
(c)
m 1 2 2 2 1 1
%
II. (a) For n œ 4, ?x œ b n a œ 1 4(1) œ 42 œ "# Ê ?x œ " ; ! mf(ti ) œ 12 Ê S œ " (12) œ 2 ; (b)
f(ti ) 0 7/8 1 9/8 2
œ Š 14 1‹ Š (41) (1)‹ œ 2 Ê kET k œ '1 at$ 1b dt T œ 2 2 œ 0 %
"
" 4
ti 1 1/2 0 1/2 1
'
œ
1 #
Ê ET œ '1
2
" s#
ds T œ
m 1 2 2 2 1
" #
mf(si ) 1 32/25 8/9 32/49 1/4
0.50899 œ 0.00899
Ê kET k œ 0.00899 kET k True Value
(c)
‚ 100 œ
II. (a) For n œ 4, ?x œ ! mf(si ) œ
0.00899 0.5 ‚ 100 ¸ ba 21 " n œ 4 œ 4
264,821 44,100
Ê Sœ
" 12
2% Ê
?x 3
Š 264,821 44,100 ‹
Ð%Ñ ¸ 0.50042; f Ð$Ñ (s) œ 24 (s) œ s& Ê f
" 12
œ œ
;
264,821 529,200
120 s'
%
"¸ ˆ"‰ Ê M œ 120 Ê kES k Ÿ ¸ 2180 4 (120)
œ
'1
" 384
¸ 0.00260
'2
2
" 1 " " s# ds œ # Ê ES œ 1 s# ds S œ # kES k 0.0004 True Value ‚ 100 œ 0.5 ‚ 100 ¸ 0.08%
(b) (c) 8.
'24 (s "1) I.
#
ba n
œ
42 4
œ
" 2
Ê
450
Ê Tœ
" 4
ˆ 1269 ‰ 450 œ
f(s) œ (s 1) Ê f ww (s) œ Ê kE T k Ÿ
#
œ 0.70500;
Ê f (s) œ (s 2 1)$
6 (s 1)% 42 1#
1269 1800 w
si 1 5/4 3/2 7/4 2
f(si ) 1 16/25 4/9 16/49 1/4
m 1 4 2 4 1
?x #
œ
" 4
;
s! s" s# s$ s%
si 2 5/2 3 7/2 4
f(si ) 1 4/9 1/4 4/25 1/9
m 1 2 2 2 1
Ê Mœ6
ˆ "# ‰# (6) œ
" 4
mf(si ) 1 64/25 8/9 64/49 1/4
0.50042 œ 0.00042 Ê kES k œ 0.00042
ds
(a) For n œ 4, ?x œ ! mf(si ) œ 1269
s! s" s# s$ s%
œ 0.25
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
mf(si ) 1 8/9 1/2 8/25 1/9
Section 8.6 Numerical Integration (b)
'24 (s "1)
%
#
ˆ " ‰ ˆ " ‰ ds œ ’ (s" 1) “ œ 4 1 #1 œ
Ê kET k ¸ 0.03833 (c)
kET k True Value
‚ 100 œ
II. (a) For n œ 4, ?x œ ! mf(si ) œ 1813
#
0.03833 ˆ 23 ‰ ‚ 100 ¸ ba 42 " n œ 4 œ 2
Ê ET œ '2
4
2 3
" 6
Ð$Ñ
24 (s1)&
f
(s) œ
ˆ 1813 ‰ 450 œ 42 180
Ê kE S k Ÿ
'2
Ê f
1813 2700 Ð%Ñ
Ê
?x 3
œ
" 6
; s! s" s# s$ s%
ˆ "# ‰% (120)
4
(b) (c)
9.
¸ 0.67148;
(s) œ œ
" 2 (s 1)# ds œ 3 Ê ES œ kES k 0.00481 True Value ‚ 100 œ ˆ 23 ‰ ‚
Ê M œ 120
120 (s1)'
'2
" 12
4
¸ 0.08333
" (s 1)#
ds S ¸
2 3
1 8
1 4
Ê
?x #
œ
1 8
t! t" t# t$ t%
;
Š2 2È2‹ ¸ 1.89612;
f(t) œ sin t Ê f w (t) œ cos t Ê f ww (t) œ sin t # 1$ Ê M œ 1 Ê kET k Ÿ 11# 0 ˆ 14 ‰ (1) œ 19 # ¸ 0.16149
(b)
'01 sin t dt œ [cos t] 1! œ (cos 1) (cos 0) œ 2
(c)
kET k True Value
‚ 100 œ
0.10388 2
1 12
1 4
Ê
?x 3
œ
1 12
Š2 4È2‹ ¸ 2.00456;
f Ð$Ñ (t) œ cos t Ê f Ð%Ñ (t) œ sin t 0 ˆ 1 ‰% Ê M œ 1 Ê kES k Ÿ 1180 4 (1) ¸ 0.00664
10.
(b)
'01 sin t dt œ 2
(c)
kES k True Value
‚ 100 œ
1 8
0.00456 2
ti 0 1/4 1/2 31/4 1
f(ti ) 0 È2/2 1 È2/2 0
m 1 2 2 2 1
mf(ti ) 0 È2 2 È2 0
Ê kET k œ '0 sin t dt T ¸ 2 1.89612 œ 0.10388 1
; t! t" t# t$ t%
ti 0 1/4 1/2 31/4 1
f(ti ) 0 È2/2 1 È2/2 0
m 1 4 2 4 1
1 4
Ê
?x #
œ
1 8
;
Š2 2È2‹ ¸ 0.60355; f(t) œ sin 1t
t! t" t# t$ t%
ti 0 1/4 1/2 3/4 1
f(ti ) 0 È2/2 1 È2/2 0
'01 sin 1t dt œ [ 1" cos 1t] "! œ ˆ 1" cos 1‰ ˆ 1" cos 0‰ œ 12 ¸ 0.63662 2 1 0.60355 œ 0.03307 kET k 0.03307 True Value ‚ 100 œ ˆ 12 ‰ ‚ 100
m 1 2 2 2 1
mf(ti ) 0 È 2 2 2 È 2 2 0
mf(ti ) 0 È2 2 È2 0
Ê kET k œ '0 sin 1t dt T
¸
(c)
mf(si ) 1 16/9 1/2 16/25 1/9
‚ 100 ¸ 0%
Ê f w (t) œ 1 cos 1t Ê f ww (t) œ 1# sin 1t Ê M œ 1# # Ê kET k Ÿ 1 1# 0 ˆ 14 ‰ a1# b ¸ 0.05140
(b)
m 1 4 2 4 1
1
(a) For n œ 4, ?x œ b n a œ 1 4 0 œ ! mf(ti ) œ 2 2È2 ¸ 4.828 Ê Tœ
f(si ) 1 4/9 1/4 4/25 1/9
Ê ES œ '0 sin t dt S ¸ 2 2.00456 œ 0.00456 Ê kES k ¸ 0.00456
'01 sin 1t dt I.
si 2 5/2 3 7/2 4
‚ 100 ¸ 5%
II. (a) For n œ 4, ?x œ b n a œ 1 4 0 œ ! mf(ti ) œ 2 4È2 ¸ 7.6569 Ê Sœ
0.705 ¸ 0.03833
100 ¸ 1%
(a) For n œ 4, ?x œ b n a œ 1 4 0 œ ! mf(ti ) œ 2 2È2 ¸ 4.8284 Ê Tœ
2 3
0.67148 œ 0.00481 Ê kES k ¸ 0.00481
'01 sin t dt I.
ds T œ
6%
450
Ê Sœ
" (s 1)#
¸ 5%
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
505
506
Chapter 8 Techniques of Integration
II. (a) For n œ 4, ?x œ b n a œ 1 4 0 œ 41 Ê ! mf(ti ) œ 2 4È2 ¸ 7.65685 Ê Sœ
1 12
?x 3
œ
1 12
t! t" t# t$ t%
Š2 4È2‹ ¸ 0.63807;
f Ð$Ñ (t) œ 1$ cos 1t Ê f Ð%Ñ (t) œ 1% sin 1t 0 ˆ 1 ‰% Ê M œ 1% Ê kES k Ÿ 1180 a1% b ¸ 0.00211 4 (b)
'01 sin 1t dt œ 12 ¸ 0.63662
(c)
kES k True Value
‚ 100 œ
0.00145 ˆ 12 ‰
ti 0 1/4 1/2 3/4 1
;
Ê ES œ '0 sin 1t dt S ¸ 1
2 1
f(ti ) 0 È2/2 1 È2/2 0
m 1 4 2 4 1
mf(ti ) 0 È 2 2 2 È 2 2 0
0.63807 œ 0.00145 Ê kES k ¸ 0.00145
‚ 100 ¸ 0%
11. (a) M œ 0 (see Exercise 1): Then n œ 1 Ê ?x œ 1 Ê kET k œ
" 1#
(1)# (0) œ 0 10%
(b) M œ 0 (see Exercise 1): Then n œ 2 (n must be even) Ê ?x œ 12. (a) M œ 0 (see Exercise 2): Then n œ 1 Ê ?x œ 2 Ê kET k œ
2 1#
" #
Ê kE S k œ
Ê kET k Ÿ
2 n
2 12
ˆ 2n ‰# (2) œ
4 3n#
Ê kET k Ÿ
2 n
2 12
ˆ 2n ‰# (2) œ
4 3n#
Ê kET k Ÿ
2 n
2 12
ˆ 2n ‰# (12) œ
8 n#
2 n
Ê kET k Ÿ
2 12
ˆ 2n ‰# (6) œ
4 n#
1 n
Ê kET k Ÿ
1 12
ˆ 1n ‰# (6) œ
1 2n#
2 180
a10% b Ê n É 43 a10% b
(1)% (0) œ 0 10% 4 3
a10% b Ê n É 43 a10% b
(1)% (0) œ 0 10%
2 180
(1)% (0) œ 0 10%
10% Ê n# 4 a10% b Ê n È4 a10% b
œ 200, so let n œ 201 (b) M œ 0 (see Exercise 6): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 17. (a) M œ 6 (see Exercise 7): Then ?x œ
2 180
4 3
10% Ê n# 8 a10% b Ê n È8 a10% b
Ê n 282.8, so let n œ 283 (b) M œ 0 (see Exercise 5): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 16. (a) M œ 6 (see Exercise 6): Then ?x œ
(1)% (0) œ 0 10%
10% Ê n#
Ê n 115.4, so let n œ 116 (b) M œ 0 (see Exercise 4): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 15. (a) M œ 12 (see Exercise 5): Then ?x œ
2 180
10% Ê n#
Ê n 115.4, so let n œ 116 (b) M œ 0 (see Exercise 3): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 14. (a) M œ 2 (see Exercise 4): Then ?x œ
ˆ #" ‰% (0) œ 0 10%
(2)# (0) œ 0 10%
(b) M œ 0 (see Exercise 2): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 13. (a) M œ 2 (see Exercise 3): Then ?x œ
" 180
2 180
(1)% (0) œ 0 10%
10% Ê n#
" #
a10% b Ê n É "# a10% b
Ê n 70.7, so let n œ 71 (b) M œ 120 (see Exercise 7): Then ?x œ
" n
Ê kES k œ
1 180
ˆ "n ‰% (120) œ
2 3n%
10% Ê n%
2 3
a10% b
4 2 % Ê n É 3 a10 b Ê n 9.04, so let n œ 10 (n must be even)
18. (a) M œ 6 (see Exercise 8): Then ?x œ
2 n
Ê kET k Ÿ
2 12
ˆ 2n ‰# (6) œ
4 n#
10% Ê n# 4 a10% b Ê n È4 a10% b
Ê n 200, so let n œ 201 (b) M œ 120 (see Exercise 8): Then ?x œ
2 n
Ê kES k Ÿ
2 180
ˆ 2n ‰% (120) œ
64 3n%
10% Ê n%
4 64 % Ê n É 3 a10 b Ê n 21.5, so let n œ 22 (n must be even)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
64 3
a10% b
Section 8.6 Numerical Integration 19. (a) f(x) œ Èx 1 Ê f w (x) œ Then ?x œ
Ê kET k Ÿ
3 n
" 2
3 12
(x 1)"Î2 Ê f ww (x) œ 4" (x 1)3Î2 œ
ˆ 3n ‰2 ˆ 14 ‰ œ
9 16n2
104 Ê n2
so let n œ 76 (Î# (b) f Ð$Ñ (x) œ 38 (x 1)&Î# Ê f Ð%Ñ (x) œ 15 œ 16 (x 1) Ê kE S k Ÿ
3 180
15 ‰ ˆ n3 ‰% ˆ 16 œ
35 (15) 16(180)n%
Ê Mœ
1 3
4 ŠÈ1‹
œ
1 4
.
9 a104 b Ê n É 16 a104 b Ê n 75,
Ê Mœ
15 ( 16 ˆÈx 1‰
35 (15) ˆ10% ‰ 16(180)
10% Ê n%
9 16
1 3 4 ˆÈ x 1 ‰
507
%
15 ( 16 ŠÈ1‹
œ
. Then ?x œ
15 16
3 n
%
(15) a10 b Ê n É 3 16(180) Ê n 10.6, so let 5
n œ 12 (n must be even) 20. (a) f(x) œ
Ê f w (x) œ #" (x 1)$Î# Ê f w w (x) œ
" È x 1
3 3 ˆ 3 ‰# ˆ 3 ‰ 3% n Ê kET k Ÿ 12 n 4 œ 48n# 15 (Î# Ê f Ð%Ñ (x) œ 105 8 (x 1) 16
Then ?x œ (b) f Ð$Ñ (x) œ Ê kE S k Ÿ
3 180
ˆ n3 ‰% ˆ 105 ‰ 16 œ
3& (105) 16(180)n%
3 4
(x 1)&Î# œ
3 & 4 ˆÈ x 1 ‰
3% ˆ10% ‰ Ên 48 105 * Ê 16 ˆÈx1‰
É3
(x 1)*Î# œ
Mœ
3& (105) ˆ10% ‰ 16(180)
%
3
&
4 ŠÈ1‹
œ
3 4
.
a10% b Ê n 129.9, so let n œ 48 105 105 3 * œ 16 . Then ?x œ n 16 ŠÈ1‹
10% Ê n#
10% Ê n%
Ê Mœ
%
&
130
%
a10 b Ê n É 3 (105) Ê n 17.25, so 16(180)
let n œ 18 (n must be even) 21. (a) f(x) œ sin (x 1) Ê f w (x) œ cos (x 1) Ê f ww (x) œ sin (x 1) Ê M œ 1. Then ?x œ œ
8 12n#
10% Ê n#
8 ˆ10% ‰ 12
32 ˆ10% ‰ 180
%
2 12
ˆ n2 ‰# (1) œ
23.
5 2 a6.0
32 ˆ10% ‰ 180
ˆ 2n ‰# (1)
2 n
Ê kES k Ÿ
2 180
ˆ 2n ‰% (1) œ
32 180n%
10%
%
a10 b Ê n É 32180 Ê n 6.49, so let n œ 8 (n must be even)
8 12n#
10% Ê n#
8 ˆ10% ‰ 12
%
2 n
%
Ê n É 8 a110# b Ê n 81.6, so let n œ 82
(b) f Ð$Ñ (x) œ sin (x 1) Ê f Ð%Ñ (x) œ cos (x 1) Ê M œ 1. Then ?x œ Ê n%
2 12
%
22. (a) f(x) œ cos (x 1) Ê f w (x) œ sin (x 1) Ê f ww (x) œ cos (x 1) Ê M œ 1. Then ?x œ Ê kE T k Ÿ
Ê k ET k Ÿ
Ê n É 8 a110# b Ê n 81.6, so let n œ 82
(b) f Ð$Ñ (x) œ cos (x 1) Ê f Ð%Ñ (x) œ sin (x 1) Ê M œ 1. Then ?x œ Ê n%
2 n
2 n
Ê kE S k Ÿ
2 180
ˆ 2n ‰% (1) œ
32 180n%
10%
%
a10 b Ê n É 32180 Ê n 6.49, so let n œ 8 (n must be even)
2a8.2b 2a9.1bÞ Þ Þ 2a12.7b 13.0ba30b œ 15,990 ft3 .
24. Use the conversion 30 mph œ 44 fps (ft per sec) since time is measured in seconds. The distance traveled as the car accelerates from, say, 40 mph œ 58.67 fps to 50 mph œ 73.33 fps in (4.5 3.2) œ 1.3 sec is the area of the trapezoid (see figure) associated with that time interval: "# (58.67 73.33)(1.3) œ 85.8 ft. The total distance traveled by the Ford Mustang Cobra is the sum of all these eleven trapezoids (using ?t # and the table below): v (mph) v (fps) t (sec) ?t/2
0 0 0 0
30 44 2.2 1.1
40 58.67 3.2 0.5
50 73.33 4.5 0.65
60 88 5.9 0.7
70 102.67 7.8 0.95
80 117.33 10.2 1.2
90 132 12.7 1.25
100 146.67 16 1.65
110 161.33 20.6 2.3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
120 176 26.2 2.8
130 190.67 37.1 5.45
508
Chapter 8 Techniques of Integration
s œ (44)(1.1) (102.67)(0.5) (132)(0.65) (161.33)(0.7) (190.67)(0.95) (220)(1.2) (249.33)(1.25) (278.67)(1.65) (308)(2.3) (337.33)(2.8) (366.67)(5.45) œ 5166.346 ft ¸ 0.9785 mi " 25. Using Simpson's Rule, ?x œ 1 Ê ?x 3 œ 3 ; ! myi œ 33.6 Ê Cross Section Area ¸ " (33.6) 3
œ 11.2 ft# . Let x be the length of the tank. Then the Volume V œ (Cross Sectional Area) x œ 11.2x. Now 5000 lb of gasoline at 42 lb/ft$ $ Ê V œ 5000 42 œ 119.05 ft Ê 119.05 œ 11.2x Ê x ¸ 10.63 ft
26.
24 0.019 2 c
xi 0 1 2 3 4 5 6
x! x" x# x$ x% x5 x'
yi 1.5 1.6 1.8 1.9 2.0 2.1 2.1
m 1 4 2 4 2 4 1
myi 1.5 6.4 3.6 7.6 4.0 8.4 2.1
2a0.020b 2a0.021b Þ Þ Þ 2a0.031b 0.035 d œ 4.2 L
27. (a) kES k Ÿ
ba % 180 a?x b M; n 1 ?x 1 8 Ê 3 œ #4 ;
œ 4 Ê ?x œ
1 #
0 4
œ
1 8
; ¸f Ð%Ñ ¸ Ÿ 1 Ê M œ 1 Ê kES k Ÿ
(b) ?x œ ! mf(xi ) œ 10.47208705 Ê Sœ
1 #4
x! x" x# x$ x%
(10.47208705) ¸ 1.37079
xi 0 1/8 1/4 31/8 1/2
ˆ 1# 0‰ 180
ˆ 18 ‰% (1) ¸ 0.00021
f(xi ) 1 0.974495358 0.900316316 0.784213303 0.636619772
m 1 4 2 4 1
mf(x1i ) 1 3.897981432 1.800632632 3.136853212 0.636619772
‰ (c) ¸ ˆ 0.00021 1.37079 ‚ 100 ¸ 0.015% 28. (a) ?x œ
ba n
2 ae0 30È1
(b) kES k Ÿ 29. T œ Tœ
10 10
œ
0.01
4e
10 180
œ 0.1 Ê erfa1b œ 0.04
2e
0.09
4e
2 ˆ 0.1 ‰ È3 3 ay0
4y1 2y2 4y3 Þ Þ Þ 4y9 y10 b
0.81
Þ Þ Þ 4e
e1 b ¸ 0.843
a0.1b4 a12b ¸ 6.7 ‚ 106
?x ba 2 ay0 2y1 2y2 2y3 Þ Þ Þ 2yn1 yn b where ?x œ n and f b a ay0 y1 y1 y2 y2 Þ Þ Þ ync1 ync1 yn b œ b n a Š fax0 b 2 fax1 b fax1 b 2 fax2 b n 2
Since f is continuous on each interval [xk1 , xk ], and [xk1 , xk ] with fack b œ
faxkc1 b faxk b ; #
faxkc1 b faxk b #
is continuous on [a, b]. So ÞÞÞ
faxnc1 b faxn b ‹. 2
is always between faxk1 b and faxk b, there is a point ck in
this is a consequence of the Intermediate Value Theorem. Thus our sum is
n
n
k œ1
k œ1
! ˆ b a ‰fack b which has the form ! ?xk fack b with ?xk œ n
ba n
for all k. This is a Riemann Sum for f on [a, b].
?x b a 3 ay0 4y1 2y2 4y3 Þ Þ Þ 2yn2 4yn1 yn b where n is even, ?x œ n and S œ b n a ˆ y0 4y3 1 y2 y2 4y33 y4 y4 4y3 5 y6 Þ Þ Þ ync2 4y3 nc1 yn ‰ œ b n a Š fax0 b 4fa6x1 b fax2 b fax2 b 4fa6x3 b fax4 b fax4 b 4fa6x5 b fax6 b Þ Þ Þ faxnc2 b 4fa6xnc1 b faxn b ‹ 2
30. S œ
fax2k b 4fax2kb1 b fax2kb2 b 6
f is continuous on [a, b]. So
is the average of the six values of the continuous function on the interval [x2k , x2k2 ], so it is between
the minimum and maximum of f on this interval. By the Extreme Value Theorem for continuous functions, f takes on its maximum and minimum in this interval, so there are xa and xb with x2k Ÿ xa , xb Ÿ x2k2 and faxa b Ÿ
fax2k b 4fax2kb1 b fax2kb2 b 6
Ÿ faxb b. By the Intermediate Value Theorem, there is ck in [x2k , x2k2 ] with
fack b œ
fax2k b 4fax2kb1 b fax2kb2 b . 6
So our sum has the form ! ?xk fack b with ?xk œ
n/2
k œ1
ba an/2b ß
a Riemann sum for f on [a, b].
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.6 Numerical Integration 31. (a) a œ 1, e œ
" #
Ê Length œ 4 '0 É1 1Î2
œ 2 '0 È4 cos# t dt œ '0 1Î2
1Î2
1 #0 .
cos# t dt
f(t) dt; use the
Trapezoid Rule with n œ 10 Ê ?t œ œ
" 4
ba n
œ
ˆ 1# ‰ 0 10
'01Î2 È4 cos# t dt ¸ ! mf(xn ) œ 37.3686183 10
Ê Tœ
?t #
nœ0
(37.3686183) œ
1 40
(37.3686183)
œ 2.934924419 Ê Length œ 2(2.934924419) ¸ 5.870 (b) kf ww (t)k 1 Ê M œ 1 Ê kE T k Ÿ 32. ?x œ ÊS
ba 1#
a?t# Mb Ÿ
ˆ 1# ‰ 0 1#
ˆ #10 ‰# 1 Ÿ 0.0032
10 1 ?x 1 ! mf(xi ) 8 œ 8 Ê 3 œ 24 ; 1 œ 24 a29.18480779b ¸ 3.82028
x! x" x# x$ x% x5 x' x( x) x* x"!
xi 0 1/20 1/10 31/20 1/5 1/4 31/10 71/20 21/5 91/20 1/2
f(xi ) 1.732050808 1.739100843 1.759400893 1.790560631 1.82906848 1.870828693 1.911676881 1.947791731 1.975982919 1.993872679 2
m 1 2 2 2 1 1 2 2 2 2 1
mf(xi ) 1.732050808 3.478201686 3.518801786 3.581121262 3.658136959 3.741657387 3.823353762 3.895583461 3.951965839 3.987745357 2
x! x" x# x$ x% x5 x' x( x)
xi 0 1/8 1/4 31/8 1/2 51/8 31/4 71/8 1
f(xi ) 1.414213562 1.361452677 1.224744871 1.070722471 1 1.070722471 1.224744871 1.361452677 1.414213562
m 1 4 2 4 2 4 2 4 1
mf(xi ) 1.414213562 5.445810706 2.449489743 4.282889883 2 4.282889883 2.449489743 5.445810706 1.414213562
œ 29.184807792
dy 33. The length of the curve y œ sin ˆ 3#10 x‰ from 0 to 20 is: L œ '0 Ê1 Š dx ‹ dx; #
20
œ
91 # 400
cos# ˆ 3#10 x‰ Ê L œ '0 É1 20
91 # 400
#
25
#
1# 4
'c2525 É1 14
#
sin# ˆ 150x ‰ Ê L œ
œ
31 #0
dy cos ˆ 3#10 x‰ Ê Š dx ‹
#
cos# ˆ 3#10 x‰ dx. Using numerical integration we find L ¸ 21.07 in
34. First, we'll find the length of the cosine curve: L œ 'c25 Ê1 Š dy dx ‹ dx; Ê Š dy dx ‹ œ
dy dx
dy dx
œ 25501 sin ˆ 150x ‰
sin# ˆ 150x ‰ dx. Using a numerical integrator we find
L ¸ 73.1848 ft. Surface area is: A œ length † width ¸ (73.1848)(300) œ 21,955.44 ft. Cost œ 1.75A œ (1.75)(21,955.44) œ $38,422.02. Answers may vary slightly, depending on the numerical integration used. 35. y œ sin x Ê
dy dx
# È1 cos# x dx; a numerical integration gives ' œ cos x Ê Š dy dx ‹ œ cos x Ê S œ 0 21(sin x)
œ
x #
#
1
S ¸ 14.4 36. y œ
x# 4
Ê
dy dx
#
dy Ê Š dx ‹ œ
x# 4
Ê S œ '0 21 Š x4 ‹ É1 2
#
x# 4
dx; a numerical integration gives S ¸ 5.28
37. A calculator or computer numerical integrator yields sin" 0.6 ¸ 0.643501109. 38. A calculator or computer numerical integrator yields 1 ¸ 3.1415929.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
509
510
Chapter 8 Techniques of Integration
8.7 IMPROPER INTEGRALS 1.
'0_ x dx 1 œ
2.
'1_ xdx
3.
'01 Èdxx œ
bÄ!
4.
'04 È dx
œ lim c bÄ4
5.
'c11
6.
'18 xdx
7.
'01 È dx
8.
'01 r dr
9.
'c_2 x2 dx1 œ '
'1b xdx
1Þ001
bÄ_
4x
#
bÄ_
œ lim
1Þ001
'0b x dx 1 œ
lim
#
b
bÄ_
0œ
1 #
Þ
lim
b Ä 0b
"
c2x1Î2 d b œ lim b Š2 2Èb‹ œ 2 0 œ 2 bÄ0
'0b a4 xb"Î# dx œ '0
1 #
‰ œ 1000 œ lim c1000x 0Þ001 d "b œ lim ˆ b1000 0 001 1000 bÄ_ bÄ_
'b1 x"Î# dx œ
lim b
œ 'c1
lim ctan" xd 0 œ lim atan" b tan" 0b œ
bÄ_
lim
b Ä 4c
’2È4 b Š2È4‹“ œ 0 4 œ 4
" b œ lim c 3x"Î$ ‘ c1 lim b 3x"Î$ ‘ c bÄ! cÄ! œ lim c 3b"Î$ 3(1)"Î$ ‘ lim 3(1)"Î$ 3c"Î$ ‘ œ (0 3) (3 0) œ 6 bÄ! cÄ! dx x#Î$
0
1
dx x#Î$
'0 œ '8 xdx "Î$
dx x#Î$
" b œ lim #3 x#Î$ ‘ ) lim 3# x#Î$ ‘ c bÄ! cÄ! œ lim c 3# b#Î$ 3# (8)#Î$ ‘ lim b 3# (1)#Î$ 3# c#Î$ ‘ œ 0 3# (4)‘ ˆ 3# 0‰ œ 9# bÄ! cÄ! "Î$
0
œ lim c csin" xd 0 œ lim c asin" b sin" 0b œ bÄ1
bÄ1
1 #
0œ
1 #
œ lim b c1000r0 001 d 1b œ lim b a1000 1000b0 001 b œ 1000 0 œ 1000 bÄ! bÄ! Þ
#
œ
dx x"Î$
b
1 x#
0Þ999
1
lim
b Ä _
1 ¸‘ # ' _ x dx lim cln kx 1kd b 2 lim cln kx 1kd b 2 œ lim ln ¸ xx 1 œ bÄ 1 b _ b Ä _ b Ä _ 3¸ " ˆln ¸ ¸ b 1 ¸‰ œ ln 3 ln Š lim bb 1 ln b 1 1 ‹ œ ln 3 ln 1 œ ln 3 2
_
dx x1
'c2_ x2 dx4 œ
b Ä _
11.
'2_ v2 dvv œ
bÄ_
12.
'2_ t 2dt1 œ
13.
_ 'c_
#
#
2x dx ax # 1 b #
2
b Ä _
10.
#
Þ
tan" #x ‘ 2 œ b
lim
lim
b Ä _
ˆtan" 1 tan" b2 ‰ œ
1 4
ˆ 1# ‰ œ
31 4
b lim 2 ln ¸ v v " ¸‘ # œ lim ˆ2 ln ¸ b b " ¸ 2 ln ¸ 2 # " ¸‰ œ 2 ln (1) 2 ln ˆ "# ‰ œ 0 2 ln 2 œ ln 4
bÄ_
" ¸‘ b "¸ ˆln ¸ bb ¸ 2 " ¸‰ œ ln (1) ln ˆ "3 ‰ œ 0 ln 3 œ ln 3 lim ln ¸ tt 1 # œ lim 1 ln # 1
bÄ_
bÄ_
_
dx œ ' _ ax2x '0 # 1 b# 0
2x dx ax # 1 b#
;”
u œ x# 1 Ä du œ 2x dx •
'_1 duu '1_ duu œ #
1 u" ‘ c lim u" ‘ b c lim 1 Ä_
#
bÄ_
"c (1)‘ œ (1 0) (0 1) œ 0 œ lim ˆ1 "b ‰ c lim Ä_ bÄ_
14.
_ 'c_
ax #
x dx 4b$Î#
œ' _ 0
%
ax #
x dx 4b$Î#
_
'0
ax # c
x dx 4b$Î#
;”
u œ x# 4 Ä du œ 2x dx •
œ lim ’ È"u “ c lim ’ È"u “ œ lim Š #" Ä_ bÄ_ bÄ_ % b
" Èb ‹
'_4 2udu
$Î#
_
'4
du 2u$Î#
c lim Š È"c #" ‹ œ ˆ #" 0‰ ˆ0 #" ‰ œ 0 Ä_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.7 Improper Integrals 15.
'01 È) " )#
d); ”
2)
u œ ) # 2) Ä du œ 2() ") d) •
'03 2Èduu œ
lim b
bÄ!
'b3 2Èduu œ
Èu‘ 3 œ lim ŠÈ3 Èb‹ œ È3 0 b b Ä !b
lim
b Ä !b
œ È3 16.
'02 Ès " 4
s#
" #
'02 È2s ds
'0
du 2È u
lim c cÄ2
'0c È ds
ds œ
œ lim b 'b bÄ!
4
4
2
s#
ds È 4 s#
'0_ (1 dxx)Èx ; –
c #
sin" 0‰ œ (2 0) ˆ 1# 0‰ œ
'0_ u2 du1 œ
u œ Èx — Ä du œ 2dx Èx
0 c u œ 4 s# du ds ' Ä "# '4 È lim • c u 0 È 4 s# cÄ2 du œ 2s ds
4 c œ lim b Èu‘ b lim c sin" #s ‘ ! c Ä 2 bÄ!
4 s#
œ lim b Š2 Èb‹ lim c ˆsin" cÄ2 bÄ!
17.
;”
#
lim
bÄ_
'0b u2 du1 œ #
41 #
lim c2 tan" ud 0 b
bÄ_
œ lim a2 tan" b 2 tan" 0b œ 2 ˆ 1# ‰ 2(0) œ 1 bÄ_
18.
'1_
dx xÈ x# 1
œ '1
2
dx xÈ x# 1
_
'2
dx xÈ x# 1
œ lim b 'b bÄ1
2
dx xÈ x# 1
c lim Ä_
'2c
dx xÈ x# 1
œ lim b csec" kxkd b c lim csec" kxkd 2 œ lim b asec" 2 sec" bb c lim asec" c sec" 2b Ä_ Ä_ bÄ1 bÄ1 œ ˆ 13 0‰ ˆ 1# 13 ‰ œ 1# 2
19.
'0_ a1 v b adv1 tan " vb œ #
c
lim cln k1 tan" vkd 0 œ lim cln k1 tan" bkd ln k1 tan" 0k b
bÄ_
bÄ_
œ ln ˆ1 1# ‰ ln (1 0) œ ln ˆ1 1# ‰ 20.
21.
c x '0_ 161tan dx œ x
# b
"
'c0_ )e) d) œ œ 1
#
#
#
lim ’8 atan" xb “ œ lim ’8 atan" bb “ 8 atan" 0b œ 8 ˆ 1# ‰ 8(0) œ 21#
#
bÄ_
lim
b Ä _
)e) e) ‘ 0 œ a0 † e! e! b b
ˆ e"cb ‰
lim
b Ä _
bÄ_
0
^ (l'Hopital's rule for
_ _
lim
b Ä _
cbeb eb d œ 1
lim
b Ä _
ˆ becb" ‰
form)
œ 1 0 œ 1 22.
'0_ 2e) sin ) d) œ œ lim 2 bÄ_
œ lim
bÄ_
23.
0 'c_ e
24.
_ 'c_ 2xe
œ
25.
x
k k
c) ’ 1e 1
'0b 2e) sin ) d)
lim
bÄ_
( sin ) cos ))“
2(sin b cos b) 2eb
b 0
2(sin 0 cos 0) 2e!
(FORMULA 107 with a œ 1, b œ 1) œ0
2(0 1) 2
œ1
dx œ ' _ ex dx œ lim cex d !b œ lim a1 eb b œ (1 0) œ 1 b Ä _ b Ä _ 0
_
dx œ ' _ 2xe x dx '0 2xe x dx œ lim ce x d b c lim ce Ä_ b Ä _ # # cecc (1)d œ (1 0) (0 1) œ 0 lim c1 aecb bd c lim Ä_ 0
x#
#
#
#
0
x# c 0
d
b Ä _
'01 x ln x dx œ
#
lim b ’ x# ln x
bÄ!
œ "4 lim b bÄ!
Š "b ‹ Š
4 ‹ b$
1
x# 4 “b
#
œ ˆ "# ln 1 4" ‰ lim b Š b# ln b bÄ!
511
b# 4‹
œ 4" lim b bÄ!
#
œ 4" lim b Š b4 ‹ œ 4" 0 œ 4" bÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ln b Š b2# ‹
0
512 26.
Chapter 8 Techniques of Integration
'01 ( ln x) dx œ
lim cx x ln xd b" œ c1 1 ln 1d lim b cb b ln bd œ 1 0 lim b b Ä !b bÄ! bÄ!
ln b Š "b ‹
œ 1 lim b bÄ!
Š b" ‹ Š
" ‹ b#
œ 1 lim b b œ 1 0 œ 1 bÄ! 27.
'02 È ds
28.
'01 È4r dr
29.
'12
30.
'24 È dt
31.
'41 Èdxkxk œ
b œ lim c sin" #s ‘ 0 œ lim c ˆsin" b# ‰ sin" 0 œ bÄ2 bÄ2
4 s#
0œ
1 #
œ lim c c2 sin" ar# bd 0 œ lim c c2 sin" ab# bd 2 sin" 0 œ 2 † b
1 r%
bÄ1
t# 4
bÄ1
#
œ lim b csec" sd b œ sec" 2 lim b sec" b œ bÄ1 bÄ1
1 3
% œ lim b "# sec" #t ‘ b œ lim b ˆ "# sec" 4# ‰ bÄ2 bÄ2
sec" ˆ b# ‰‘ œ
ds sÈ s# 1
t
1 #
lim c
bÄ!
'b1 Èdxx
lim b 'c
4
cÄ!
dx Èx
" #
0œ
1 #
0œ1
" #
ˆ 13 ‰
1 3
" #
†0œ
1 6
b 4 œ lim c 2Èx‘ c1 lim b 2Èx‘ c bÄ! cÄ!
œ lim c Š2Èb‹ ˆ2È(1)‰ 2È4 lim b 2Èc œ 0 2 2 † 2 0 œ 6 bÄ! cÄ! 32.
'02 Èkxdx 1k œ '01 È dx
1x
'1
2
dx Èx 1
#
b
œ lim c ’2È1x“ lim ’2Èx 1“ bÄ1 ! cÄ1 c
œ lim Š2È1b‹ Š2È1 0‹ 2È2 1 lim Š2Èc 1‹ œ 0 2 2 0 œ 4 bÄ1 cÄ1 33.
'c_1 )
34.
'0_ (x 1)dxax
d) 5) 6
#
2 ¸‘ b 2 ¸‘ " 2 ¸ ˆ"‰ œ lim ln ¸ )) œ lim ln ¸ bb ln ¸ 3 3 1 3 œ 0 ln # œ ln 2 1
bÄ_
1b
lim ’ " bÄ_ #
œ
35.
#
œ lim #" ln kx 1k bÄ_
" ln Š Èbb ‹ # 1
'01Î2 tan ) d) œ
lim b Ä 12 c
36.
'01Î2 cot ) d) œ
37.
'01 Èsin ) d) ; c1 ) œ xd '0
1
38.
1)
sin x Èx
'c11ÎÎ22 '021
39.
bÄ_
lim
bÄ0
" #
tan
"
b“
" 4
" #
ln ax# 1b
’ "#
ln
" È1
" #
1 tan" x‘ 0 œ lim ’ #" ln Š Èxx ‹ # 1 b
bÄ_
"
tan
0“ œ
" #
ln 1
" #
†
1 #
" #
ln 1
" #
" #
tan" x“
†0œ
b 0
1 4
c ln kcos )kd b0 œ lim1 c c ln kcos bkd ln 1 œ lim1 c c ln kcos bkd œ _, the integral diverges bÄ bÄ 2
2
cln ksin )kd 1b Î# œ ln 1 lim
cln ksin bkd œ lim
bÄ0
Ä '1
0
œ '0
1
sin x dx Èx
bÄ0
Since 0 Ÿ
sin x dx Èx .
sin x Èx
Ÿ
" Èx
cln ksin bkd œ _, the integral diverges for all 0 Ÿ x Ÿ 1 and '0
1
dx Èx
converges, then
dx converges by the Direct Comparison Test.
cos ) d) (1 2))"Î$ dx 2x"Î$
Ô x œ 11 2x) × Ä ; )œ ## Õ d) œ dx Ø #
converges, then '0
21
'0ln 2 x# ec1Îx dx; "x œ y‘
Ä
sin x# dx #x"Î$
'201 cos 2xˆ 1 c #
"Î$
x‰ #
dx
œ '0
21
sin ˆ x# ‰ dx #x"Î$
. Since 0 Ÿ
sin x# 2x"Î$
Ÿ
" 2x"Î$
for all 0 Ÿ x Ÿ 21 and
converges by the Direct Comparison Test.
'_1Îln 2 y ey#dy œ '1_Îln 2 e #
y
œ 0 ec1Îln 2 œ e 1Îln 2 , so the integral converges.
y
dy œ lim ce y d b1Îln 2 œ lim ce b d ce 1Îln 2 d bÄ_ bÄ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.7 Improper Integrals 40.
'01 eÈcÈx
41.
'01 Èt dtsin t .
513
dx; y œ Èx‘ Ä 2'0 ecy dy œ 2 2e , so the integral converges. 1
x
" Èt sin t
Since for 0 Ÿ t Ÿ 1, 0 Ÿ
" Èt
Ÿ
and '0
1
dt Èt
converges, then the original integral converges as well by
the Direct Comparison Test. 42.
'01 t dtsin t ; let f(t) œ t "sin t and g(t) œ t" , then $
œ lim
œ 6. Now,'0
1
6
t Ä 0 cos t
dt t$
t$
œ lim
f(t)
lim t Ä 0 g(t)
t Ä 0 t sin t
3t#
œ lim
t Ä 0 1 cos t
œ lim
6t
t Ä 0 sin t
1 œ lim b #"t# ‘ b œ #" lim b #"b# ‘ œ _, which diverges Ê '0 bÄ! bÄ!
1
dt t sin t
diverges
by the Limit Comparison Test. 43.
'02 1 dxx
œ '0
1
#
'
diverges Ê 44.
45.
'1
2
dx 1 x#
dx 1x
œ '0
Ê
'02
dx 1x
1
and'0
1
x ¸‘ b " ¸ " b ¸‘ 0 œ _, which œ lim c #" ln ¸ 11 x 0 œ lim c # ln 1 b
dx 1 x#
bÄ1
diverges as well.
'1
2
dx 1x
and '0
1
dx 1x
dx 1x
œ lim c c ln (1 x)d b0 œ lim c c ln (1 b)d 0 œ _, which diverges bÄ1 bÄ1
diverges as well.
'c11 ln kxk dx œ 'c01 ln (x) dx '01 ln x dx; '01 ln x dx œ œ 1 0 œ 1; 'c1 ln (x) dx œ 1 Ê
lim
b Ä !b
'c11 ax ln kxk b dx œ 'c01 cx ln (x)d dx '01 (x ln x) dx œ #
cx ln x xd 1b œ [1 † 0 1] lim b [b ln b b] bÄ!
'c11 ln kxk dx œ 2 converges.
0
46.
bÄ1
2
dx # 0 1x
'02
dx 1 x#
#
lim b ’ x# ln x
bÄ!
#
#
œ "# ln 1 4" ‘ lim b ’ b# ln b b4 “ "# ln 1 4" -‘ lim b ’ c# ln c bÄ! cÄ! converges (see Exercise 25 for the limit calculations). 47.
'1_ 1 dxx
48.
'4_ Èxdx 1 ; x lim Ä_
$
;0Ÿ
" x$ 1
Š Èx1
1
_
'4
'2_ È dv
ŠÈ 1
_
Ê '0 51.
d) 1 e )
'0_ È dx
x' 1
Èx Èx 1
x$
œ x lim Ä_
converges Ê " 1 È"x
œ
" 1 0
'1_ 1 dxx
)
Ÿ
" e)
œ 4" 0
" 4
œ '0
1
" #b#
dx È x' 1
"‰ #
_
œ 1 and '4
dx Èx
b œ lim 2Èx‘ 4 œ _,
bÄ_
'2_ Èdvv œ
b lim 2Èv‘ # œ _,
bÄ_
diverges by the Limit Comparison Test.
_ d)
for 0 Ÿ ) _ and '0
_
œ
" #
0 œ 0 Ê the integral
converges by the Direct Comparison Test.
$
Èv
'1
1
x# 4 “c
diverges by the Limit Comparison Test.
‹
dv Èv 1
#
lim b ’ x# ln x cÄ!
e)
_ d)
œ lim cec) d b0 œ lim aecb 1b œ 1 Ê '0 bÄ_
bÄ_
converges by the Direct Comparison Test.
œ lim ˆ bÄ_
œ x lim Ä_
dx Èx 1
1
_
'0_ 1 d)e ; 0 Ÿ 1"e )
_ dx
for 1 Ÿ x _ and '1
" œ v lim œ v lim œ È1" 0 œ 1 and Ä _ Èv 1 Ä _ É1 " Š È"v ‹ v v
; lim v1 v Ä _
which diverges Ê '2 50.
‹
Š È"x ‹
which diverges Ê
49.
" x$
Ÿ
c# 4“
1
x# 4 “b
dx È x' 1
_
Ê '0
'0
dx È x' 1
1
dx È x' 1
_ dx
'1
x$
_ dx
and '1
x$
b œ lim 2x" # ‘ 1
bÄ_
converges by the Direct Comparison Test.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
e)
converges
514 52.
Chapter 8 Techniques of Integration
'2_ È dx
x# 1
ŠÈ
; x lim Ä_
" x#
Š x" ‹
_
which diverges Ê '2
53.
'1_ Èxx 1 dx; x lim Ä_
Œ
‹
œ x lim Ä_
dx È x# 1
Œ
#
1
Èx x#
x#
1
Èx
" É1
œ x lim Ä_
x È x# 1
54.
œ x lim Ä_
Èx Èx 1
"
œ x lim Ä_
œ 1; É1 "
x%
ŠÈ
; x lim Ä_
1
_
'2
_ 2 cos x
57.
x dx È x% 1
#
#
'1_ 2xdx converges #
'4_ t 2 dt 1 ; $Î#
_
t$Î#
lim t Ä _ t$Î# 1
2 dt t$Î# 1
_
#
for x 1 and '1
" œ 1; x%
'2_ Èx dx
_ dx
œ '2
x%
x
œ lim cln xd b2 œ _, bÄ_
lim cln xd 1b œ _, which diverges
bÄ_
2 x#
b dx œ lim x2 ‘ 1 œ lim ˆ b2 12 ‰ œ
bÄ_
bÄ_
2 1
#
_ 2 dt
œ 1 and '4
t$Î#
b 4 œ lim 4t"Î# ‘ 4 œ lim Š È 2‹ œ 2 Ê b
bÄ_
bÄ_
'4_ 2t dt converges $Î#
converges by the Limit Comparison Test.
'2_ lndxx ; 0 "x ln"x for x 2 and '2_ dxx diverges
59.
'1_ ex
60.
'e_ ln (ln x) dx; cx œ eyd
dx; 0
$Î#
'1_ 1 xsin x dx converges by the Direct Comparison Test.
Ê
58.
x
#
dx diverges by the Direct Comparison Test.
x
Ê '4
" x
ex x
e
_ dx
for x 1 and '1
_
Ä
x
'1_ Èedx x ; x lim Ä_
ŠÈ
" ex
_
Ê '2
dx ln x
diverges by the Direct Comparison Test.
'1_ e xdx diverges by the Direct Comparison Test.
diverges Ê
x
'e_ (ln y) ey dy; 0 ln y (ln y) ey for y e and 'e_ ln y dy œ
which diverges Ê 'e ln ey dy diverges Ê
61.
" É1
œ x lim Ä_
'1_ Èx x dx œ '1_ xdx
diverges by the Limit Comparison Test.
'1_ 1 xsin x dx; 0 Ÿ " xsin x Ÿ x2 Ê
lim cln bd b2 œ _,
bÄ_
dx converges by the Limit Comparison Test.
x#
'1_ 2 xcos x dx; 0 "x Ÿ 2 xcos x for x 1 and '1_ dxx œ Ê '1
56.
È x% È x% 1
œ x lim Ä_
ŠÈ
which diverges Ê
55.
x ‹ x% 1 x ‹ x%
x
_ Èx 1
bÄ_
'2_ Èx dx
'2_ x" dx œ
diverges by the Limit Comparison Test.
b 2 œ lim 2x"Î# ‘ 1 œ lim Š È 2‹ œ 2 Ê '1 b
bÄ_
" œ 1; x#
lim cy ln y yd be œ _,
bÄ_
'e_ ln (ln x) dx diverges by the Direct Comparison Test. e
'_
‹
'_
È ex x " œ x lim œ x lim œ È1" 0 œ 1; 1 Èdxex œ 1 e xÎ2 dx Ä _ È ex x Ä _ É1 xx Š È" x ‹ e e _ _ 2 c xÎ2 b bÎ2 1Î2 œ lim c2e d " œ lim a2e 2e b œ Èe Ê 1 e xÎ2 dx converges Ê 1 Èedx converges xx b Ä _ bÄ_ x
'
'
by the Limit Comparison Test. 62.
'1_ e dx 2 x
x
ˆ ex " 2x ‰ ˆ e"x ‰
; x lim Ä_
œ x lim Ä_
œ lim aecb e" b œ
" e
bÄ_
63.
' __ È dx _ dx
'1
x#
_
x% 1
œ 2 '0
dx È x% 1
Ê _
; '0
ex e x 2 x
_
'1
dx ex
dx È x% 1
œ x lim Ä_
" x 1Š 2e ‹
converges Ê œ '0
1
dx È x% 1
b œ lim x" ‘ 1 œ lim ˆ b" 1‰ œ 1 Ê
bÄ_
bÄ_
œ _
'1
_
'1
'
_
" 1 0
dx e x 2 x
_ dx
œ 1 and '1
dx
œ lim cecx d b1 bÄ_
converges by the Limit Comparison Test.
dx È x% 1
_ È x% 1
ex
'0
1
dx È x% 1
_
'1
dx x#
and
converges by the Direct Comparison Test.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.7 Improper Integrals 64.
'c__ e dxec
x
x
_
œ 2'0
dx ex ecx
" ex ecx
;0
" ex
_
for x 0; '0
dx ex
_
converges Ê 2 '0
dx ex ecx
converges by the
Direct Comparison Test.
'12 x(lndxx)
; ct œ ln xd Ä '0
ln 2
ln 2
œ lim b ’ p" 1 t1cp “ œ lim b b bÄ! bÄ! Ê the integral converges for p 1 and diverges for p 1
65. (a)
p
'2_ x(lndxx)
(b)
dt tp
_
p
; ct œ ln xd Ä 'ln 2
dt tp
b1cp p1
" 1p
(ln 2)1cp
and this integral is essentially the same as in Exercise 65(a): it converges
for p 1 and diverges for p Ÿ 1 66.
'0_ x2xdx1 œ
lim cln ax# 1bd 0 œ lim cln ab# 1bd 0 œ lim ln ab# 1b œ _ Ê the integral b
#
bÄ_
'c_ x2xdx1 œ b
diverges. But lim
#
bÄ_
œ lim (ln 1) œ 0
bÄ_
bÄ_
lim cln ax# 1bd
#
#
b
bÄ_
' __ x 2x 1 dx
b
" œ lim cln ab# 1b ln ab# 1bd œ lim ln Š bb# 1‹
bÄ_
bÄ_
bÄ_
_
67. A œ '0 ecx dx œ lim cecx d b0 œ lim aecb b aec0 b bÄ_
œ01œ1
68. x œ yœ
" A " 2A
'0_ xecx dx œ
bÄ_
lim cxecx ecx d b0 œ lim abecb ecb b a0 † ec0 ec0 b œ 0 1 œ 1;
bÄ_
'0_ aecx b# dx œ #" '0_ ec2x dx œ
_
bÄ_ "
lim bÄ_ #
#" ec2x ‘ b œ lim 0
"
bÄ_ #
ˆ #" ec2b ‰ #" ˆ #" ec2 0 ‰ œ 0 †
" 4
œ
" 4
_
69. V œ '0 21xecx dx œ 21 '0 xecx dx œ 21 lim cxecx ecx d b0 œ 21 ’ lim abecb ecb b 1“ œ 21 bÄ_
_
bÄ_
_
b 70. V œ '0 1 aecx b# dx œ 1 '0 ec2x dx œ 1 lim "# ec2x ‘ 0 œ 1 lim ˆ "# ec2b "# ‰ œ
bÄ_
bÄ_
71. A œ '0 (sec x tan x) dx œ lim1 c cln ksec x tan xk ln ksec xkd b0 œ lim1 c ˆln ¸1 bÄ 2 bÄ 2 œ lim1 c ln k1 sin bk œ ln 2 1Î2
bÄ
1 #
tan b ¸ sec b
ln k1 0k‰
2
72. (a) V œ '0
1Î2
1 sec# x dx '0 1 tan# x dx œ 1 '0 asec# x tan# xb dx œ '0 1Î2
œ 1 '0 dx œ 1Î2
1Î2
1Î2
1 csec# x asec# x 1bd dx
1# #
(b) Souter œ '0 21 sec xÈ1 sec# x tan# x dx '0 21 sec x(sec x tan x) dx œ 1 lim1 c ctan# xd 0 bÄ 1Î2
1Î2
b
2
œ 1 ” lim1 c ctan bd 0• œ 1 lim1 c atan bb œ _ Ê Souter diverges; Sinner œ '0 21 tan xÈ1 sec% x dx bÄ bÄ #
1Î2
#
2
2
'0 21 tan x sec x dx œ 1 lim1 c ctan# xd 0 œ 1 ” lim1 c ctan# bd 0• œ 1 lim1 c atan# bb œ _ bÄ bÄ bÄ 1Î2
#
b
2
2
2
Ê Sinner diverges
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
515
516
Chapter 8 Techniques of Integration
73. (a)
'3_ ec3x dx œ
b lim "3 ec3x ‘ 3 œ lim ˆ 3" ec3b ‰ ˆ 3" ec3 3 ‰ œ 0 †
bÄ_
bÄ_
_
" 3
† ec9 œ
" 3
ec9
¸ 0.0000411 0.000042. Since ecx Ÿ ec3x for x 3, then '3 ecx dx 0.000042 and therefore #
(b)
'0_ ecx '03 ecx
#
#
#
dx can be replaced by '0 ecx dx without introducing an error greater than 0.000042. 3
#
dx µ = 0.88621
_
# b 74. (a) V œ '1 1 ˆ "x ‰ dx œ 1 lim x" ‘ 1 œ 1 ’ lim ˆ b" ‰ ˆ 1" ‰“ œ 1(0 1) œ 1
bÄ_
bÄ_
(b) When you take the limit to _, you are no longer modeling the real world which is finite. The comparison step in the modeling process discussed in Section 4.2 relating the mathematical world to the real world fails to hold. 75. (a)
(b) > int((sin(t))/t, t=0..infinity); ˆanswer is 1# ‰
(b) > f:= 2*exp(t^2)/sqrt(Pi); > int(f, t=0..infinity); (answer is 1)
76. (a)
77. (a) faxb œ
1 x2 /2 È 21 e
f is increasing on (_, 0]. f is decreasing on [0, _). f has a local maximum at a0, fa0bb œ Š!ß
1 È 21 ‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.7 Improper Integrals (b) Maple commands: >f: œ exp(x^2/2)(sqrt(2*pi); >int(f, x œ 1..1); >int(f, x œ 2..2); >int(f, x œ 3..3);
517
¸ 0.683 ¸ 0.954 ¸ 0.997
(c) Part (b) suggests that as n increases, the integral approaches 1. We can take 'cn f(x) dx as close to 1 as we want by n
_
choosing n 1 large enough. Also, we can make 'n f(x) dx and ' _ f(x) dx as small as we want by choosing n large enough. This is because 0 faxb ex/2 for x 1. (Likewise, 0 faxb ex/2 for x 1.)
_
n
_
Thus, 'n f(x) dx 'n ex/2 dx.
'n_ ex/2 dx œ lim 'nc ex/2 dx œ lim c 2ex/2 dcn œ lim c 2ec/2 2en/2 d œ 2en/2 cÄ_ n/2
As n Ä _, 2e
'
cÄ_
cÄ_
_ Ä 0, for large enough n, 'n f(x) dx is as small as we want. Likewise for large enough n,
n
_
f(x) dx is as small as we want. _
_
78. (a) The statement is true since ' _ f(x) dx œ ' _ f(x) dx 'a f(x) dx, 'b f(x) dx œ 'a f(x) dx 'a f(x) dx b
a
b
b
and 'a f(x) dx exists since f(x) is integrable on every interval [aß b]. b
(b)
' a_ f(x) dx 'a_ f(x) dx œ ' a_ f(x) dx 'ab f(x) dx 'ab f(x) dx 'a_ f(x) dx b a b _ _ œ ' _ f(x) dx 'b f(x) dx 'a f(x) dx œ ' _ f(x) dx 'b f(x) dx
79.
Example CAS commands: Maple: f := (x,p) -> x^p*ln(x); domain := 0..exp(1); fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=domain, y=-50..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p= -2","p = -1","p = 0","p = 1","p = 2"], title="#79 (Section 8.7)" ); q1 := Int( f(x,p), x=domain ); q2 := value( q1 ); q3 := simplify( q2 ) assuming p>-1; q4 := simplify( q2 ) assuming p<-1; q5 := value( eval( q1, p=-1 ) ); i1 := q1 = piecewise( p<-1, q4, p=-1, q5, p>-1, q3 );
80.
Example CAS commands: Maple: f := (x,p) -> x^p*ln(x); domain := exp(1)..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=exp(1)..10, y=0..100, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#80 (Section 8.7)" ); q6 := Int( f(x,p), x=domain ); q7 := value( q6 ); q8 := simplify( q7 ) assuming p>-1; q9 := simplify( q7 ) assuming p<-1;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
518
Chapter 8 Techniques of Integration q10 := value( eval( q6, p=-1 ) ); i2 := q6 = piecewise( p<-1, q9, p=-1, q10, p>-1, q8 );
81.
Example CAS commands: Maple: f := (x,p) -> x^p*ln(x); domain := 0..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=0..10, y=-50..50, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#81 (Section 8.7)" ); q11 := Int( f(x,p), x=domain ): q11 = lhs(i1+i2); `` = rhs(i1+i2); `` = piecewise( p<-1, q4+q9, p=-1, q5+q10, p>-1, q3+q8 ); `` = piecewise( p<-1, -infinity, p=-1, undefined, p>-1, infinity );
82.
Example CAS commands: Maple: f := (x,p) -> x^p*ln(abs(x)); domain := -infinity..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=-4..4, y=-20..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#82 (Section 8.7)" ); q12 := Int( f(x,p), x=domain ); q12p := Int( f(x,p), x=0..infinity ); q12n := Int( f(x,p), x=-infinity..0 ); q12 = q12p + q12n; `` = simplify( q12p+q12n );
79-82. Example CAS commands: Mathematica: (functions and domains may vary) Clear[x, f, p] f[x_]:= xp Log[Abs[x]] int = Integrate[f[x], {x, e, 100)] int /. p Ä 2.5 In order to plot the function, a value for p must be selected. p = 3; Plot[f[x], {x, 2.72, 10}] CHAPTER 8 PRACTICE EXERCISES 1. u œ ln (x 1), du œ
dx x1
; dv œ dx, v œ x;
' ln (x 1) dx œ x ln (x 1) ' x x 1 dx œ x ln (x 1) ' dx ' x dx 1 œ x ln (x 1) x ln (x 1) C" œ (x 1) ln (x 1) x C" œ (x 1) ln (x 1) (x 1) C, where C œ C" 1 2.
u œ ln x, du œ
dx x
; dv œ x# dx, v œ
" 3
x$ ;
' x# ln x dx œ 3" x$ ln x ' 3" x$ ˆ x" ‰ dx œ x3
$
ln x
x$ 9
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 8 Practice Exercises 3. u œ tan" 3x, du œ
; dv œ dx, v œ x;
3 dx 1 9x#
' tan" 3x dx œ x tan" 3x ' 13x 9xdx
#
œ x tan" (3x)
" 6
;”
y œ 1 9x# Ä x tan" 3x dy œ 18x dx •
" 6
' dyy
ln a1 9x# b C
4. u œ cos" ˆ x# ‰ , du œ
dx È 4 x#
; dv œ dx, v œ x;
' cos" ˆ x# ‰ dx œ x cos" ˆ x# ‰ ' Èx dx
4 x #
;”
y œ 4 x# Ä x cos" ˆ x# ‰ dy œ 2x dx •
" #
' Èdyy
# œ x cos" ˆ x# ‰ È4 x# C œ x cos" ˆ x# ‰ 2É1 ˆ x# ‰ C
ex
5.
ÐÑ (x 1)# ïïïïî ex ÐÑ 2(x 1) ïïïïî ex ÐÑ 2 ïïïïî ex Ê
0
' (x 1)# ex dx œ c(x 1)# 2(x 1) 2d ex C
sin (1 x)
6. ÐÑ x# ïïïïî ÐÑ 2x ïïïïî ÐÑ 2 ïïïïî
cos (1 x) sin (1 x) cos (1 x) Ê
0
' x# sin (1 x) dx œ x# cos (1 x) 2x sin (1 x) 2 cos (1 x) C
7. u œ cos 2x, du œ 2 sin 2x dx; dv œ ex dx, v œ ex ; I œ ' ex cos 2x dx œ ex cos 2x 2 ' ex sin 2x dx; u œ sin 2x, du œ 2 cos 2x dx; dv œ ex dx, v œ ex ;
I œ ex cos 2x 2 ’ex sin 2x 2 ' ex cos 2x dx“ œ ex cos 2x 2ex sin 2x 4I Ê I œ
ex cos 2x 5
2ex sin 2x 5
8. u œ sin 3x, du œ 3 cos 3x dx; dv œ ec2x dx, v œ "# ec2x ; I œ ' ec2x sin 3x dx œ "# ec2x sin 3x
3 2
' ec2x cos 3x dx;
u œ cos 3x, du œ 3 sin 3x dx; dv œ ec2x dx, v œ "# ec2x ; I œ "# ec2x sin 3x 3# ’ "# ec2x cos 3x Ê Iœ
4 13
3 #
' ec2x sin 3x dx“ œ "# ec2x sin 3x 34 ec2x cos 3x 94 I
2 c2x ˆ "# ec2x sin 3x 43 ec2x cos 3x‰ C œ 13 e sin 3x
3 13
ec2x cos 3x C
9.
' x x 3xdx 2 œ ' x2dx2 ' x dx 1 œ 2 ln kx 2k ln kx 1k C
10.
' x x 4xdx 3 œ #3 ' xdx3 #" ' x dx 1 œ #3 ln kx 3k #" ln kx 1k C
11.
' x(xdx 1)
12.
' x x(x11) dx œ ' ˆ x2 1 x2 x" ‰ dx œ 2 ln ¸ x" ¸ "x C œ 2 ln kxk "x 2 ln kx 1k C x
#
#
#
#
œ ' Š "x
1 x1
1 (x 1)# ‹
dx œ ln kxk ln kx 1k
" x1
C
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
C
519
520 13.
Chapter 8 Techniques of Integration
' cos )sin )cosd)) 2 ; ccos ) œ yd #
œ
" 3
Ä '
dy y1
" 3
' y dy # œ 3" ln ¹ yy 12 ¹ C
cos ) 2 ¸ " ¸ cos ) 1 ¸ ln ¸ cos ) 1 C œ 3 ln cos ) 2 C
14.
) d) ' sin )cos sin ) 6 ; csin ) œ xd
15.
' 3x x 4xx 4 dx œ '
16.
' x4xdx4x œ ' x4 dx4 œ 2 tan" ˆ #x ‰ C
17.
' (v2v3)8vdv œ #" '
Ä
#
#
$
dx '
4 x
$
'
dx x# x 6
x4 x# 1
œ
" 5
)2¸ ' xdx 2 5" ' xdx 3 œ 5" ln ¸ sin sin ) 3 C
dx œ 4 ln kxk
" #
ln ax# 1b 4 tan" x C
#
$
œ
œ "3 '
dy y# y 2
" 16
3 Š 4v
5 8(v 2)
" 8(v #) ‹
dv œ 38 ln kvk
5 16
ln kv 2k
" 16
ln kv 2k C
&
ln ¹ (v 2)v'(v 2) ¹ C
18.
7) dv ' (v (3v ' (v 2) 1dv ' 1)(v 2)(v 3) œ
19.
' t 4tdt 3 œ "# ' t dt 1 #" ' t dt 3 œ #" tan" t
20.
' t t tdt 2 œ "3 ' t t dt2 3" ' t t dt1 œ 6" ln kt# 2k 6" ln at# 1b C
21.
' x xx x 2 dx œ ' ˆx x 2xx 2 ‰ dx œ ' x dx 32 ' x dx 1 34 ' x dx 2 œ x#
22.
' xx 1x dx œ ' ˆ1 xx "x ‰ dx œ ' ’1 x(x " 1) “ dx œ ' dx ' x dx 1 ' dxx œ x
23.
' x x4x4x 3 dx œ ' ˆx x 3x4x 3 ‰ dx œ '
24.
'
%
#
#
%
$
dv v3
3) œ ln ¹ (v (v2)(v 1)# ¹ C
" #È 3
" #
tan" Š Èt 3 ‹ C œ
tan" t
#
#
#
4 3
$
#
#
'
2x$ x# 21x 24 dx œ x# 2x 8 2 x# 3x 3 ln kx 4k
(2x 3)
" 3
Ô u œ Èx 1 ×
' x ˆ3Èdxx 1‰ ; Ö du œ 2Èdxx 1 Ù
Ä
x dx
x ‘ x# 2x 8
3 #
' x dx 1 #9 ' xdx3 œ x#
#
dx œ ' (2x 3) dx
" 3
9 #
2 3
ln kx 1k C
ln kx 3k
3 #
ln kx 1k C
' x dx # 32 ' x dx 4
2 3
' au udu1b u œ 3" ' u du 1 3" ' u du 1 œ 3" ln ku 1k 3" ln ku 1k C #
È
1" ln ¹ Èxx ¹C 11
3 Ô u œ Èx × dx Ö du œ dx Ù Ä ' 3 x‰ ; x ˆ1 È 3x#Î$ Õ dx œ 3u# du Ø
26.
'
27.
' e ds 1 ; Ô du œ es ds ×
u œ es 1
s
C
ln kx 2k C
Õ dx œ 2u du Ø
" 3
t È3
ln kx 1k ln kxk C
$
#
œ
tan"
ln kx 2k
$
$
È3 6
#
#
œ
25.
'
#
#
#
dv v2
Õ ds œ
du u1
Ø
Ä
3u# du u$ (1 u)
œ 3'
du u(1 u)
3 x È
œ 3 ln ¸ u u 1 ¸ C œ 3 ln ¹ 1 È 3 x¹ C
' u(udu 1) œ ' udu1 ' duu œ ln ¸ u u 1 ¸ C œ ln ¸ e e " ¸ C œ ln k1 ecs k C s
s
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 8 Practice Exercises
28.
'
È s Ôu œ es 1× ds Ö du œ Èe ds Ù Ä È es 1 ; 2 e s 1 Õ ds œ 2u# du Ø
du ' u a2uu du 1b œ 2 ' (u 1)(u ' u du 1 ' u du 1 œ ln ¸ uu 1" ¸ C 1) œ #
u 1
È
1" œ ln ¹ Èees ¹C 11 s
' È16y dy y
29. (a)
' È16y dy y
(b)
#
œ "# '
#
; cy œ 4 sin xd Ä 4 '
d a16 y# b È16 y#
30. (a)
' Èx dx
œ
(b)
' Èx dx
; cx œ 2 tan yd Ä
4 x# 4 x#
' 4xdxx ' 4xdxx
31. (a) (b)
#
#
" #
œ È16 y# C sin x cos x dx cos x
œ 4 cos x C œ
4È16 y# 4
C œ È16 y# C
' dÈa4 x b œ È4 x# C #
4 x#
œ #" '
d a4 x # b 4 x#
; cx œ 2 sin )d Ä
' 2 tan y2†2secsecy y dy œ 2 ' sec y tan y dy œ 2 sec y C œ È4 x# C #
œ #" ln k4 x# k C
È 2 cos ) d) ' 2 sin4)†cos œ ' tan ) d) œ ln kcos )k C œ ln Š 4 2 x ‹ C ) #
#
œ "# ln k4 x# k C 32. (a)
' È t dt
œ
(b)
' È t dt
; t œ
4t# 1 4t# 1
" 8
' dÈa4t
# 1b 4t# 1
" #
œ
" 4
sec )‘ Ä
È4t# 1 C
'
u œ 9 x# Ä #" ' du œ 2x dx •
" #
sec ) tan )† "# sec ) d) tan )
œ
" 4
' sec# ) d) œ tan4 ) C œ È4t4 1 C #
33.
' 9xdxx
34.
' x a9dx x b œ 9" ' dxx 18" ' 3 dx x 18" ' 3 dx x œ 9" ln kxk 18" ln k3 xk 18" ln k3 xk C
#
;”
du u
œ #" ln kuk C œ ln
" Èu
C œ ln
" È 9 x#
C
#
œ
" 9
ln kxk
" 18
ln k9 x# k C
35.
' 9 dxx
36.
' È dx
37.
' sin3 x cos4 x dx œ ' cos4 xa1 cos2 xbsin x dx œ ' cos4 x sin x dx ' cos6 x sin x dx œ cos5 x cos7 x C
38.
#
œ
9 x#
" 6
;”
' 3 dx x 6" ' 3dxx œ 6" ln k3 xk 6" ln k3 xk C œ 6" ln ¸ xx 33 ¸ C x œ 3 sin ) Ä dx œ 3 cos ) d) •
) ' 33 cos ' d) œ ) C œ sin" x C cos ) d) œ 3
5
7
' cos5 x sin5 x dx œ ' sin5 x cos4 x cos x dx œ ' sin5 x a1 sin2 xb2 cos x dx œ ' sin5 x cos x dx 2' sin7 x cos x dx ' sin9 x cos x dx œ sin6 x 2sin8 x sin10 x C 6
8
10
39.
' tan4 x sec2 x dx œ tan5 x C
40.
' tan3 x sec3 x dx œ ' asec2 x 1b sec2 x † sec x † tan x dx œ ' sec4 x † sec x † tan x dx ' sec2 x † sec x † tan x dx
5
œ
sec5 x 5
sec3 x 3
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
521
522 41.
Chapter 8 Techniques of Integration
' sin 5) cos 6) d) œ "# ' asina)b sina11)bb d) œ "# ' sina)b d) "# ' sina11)b d) œ "# cosa)b ##" cos 11) C œ "# cos )
" ## cos
11) C
42.
' cos 3) cos 3) d) œ "# ' acos 0 cos 6)b d) œ "# '
d)
43.
' É1 cosˆ 2t ‰ dt œ ' È2¸ cos
t¸ 4
44.
' et Ètan2 et 1 dt œ ' k sec et k et dt œ lnk sec et tan et k C 3" 180 Ð%Ñ
45. kEs k Ÿ
t¸ 4
(˜x)% M where ˜x œ
dt œ 4È2 ¸ sin
3" n
œ
2 n
; f(x) œ
" x
" #
' cos 6) d) œ "# ) 121 sin 6) C
C
œ x" Ê f w (x) œ x# Ê f ww (x) œ 2x$ Ê f'''(x) œ 6x%
(x) œ 24x& which is decreasing on [1ß 3] Ê maximum of f Ð%Ñ (x) on [1ß 3] is f Ð%Ñ (1) œ 24 Ê M œ 24. Then " ‰ ˆ 2 ‰% 768 ‰ ˆ " ‰ 180 ‰ 768 ‰ ˆ 180 kEs k Ÿ 0.0001 Ê ˆ 3180 Ÿ 0.0001 Ê n"% Ÿ (0.0001) ˆ 768 Ê n% 10,000 ˆ 180 n (24) Ÿ 0.0001 Ê n% Ê f
Ê n 14.37 Ê n 16 (n must be even) 10 12
46. kET k Ÿ Ê
2 3n#
47. ˜x œ
(˜x)# M where ˜x œ
Ÿ 10$ Ê
ba n
œ
10 6
3n# #
œ
1 6
10 n #
œ
1000 Ê n Ê
˜x #
œ
1 1#
" n ;0 2000 3
Ÿ f ww (x) Ÿ 8 Ê M œ 8. Then kET k Ÿ 10$ Ê
! mf(xi ) œ 12 Ê T œ ˆ 1 ‰ (12) œ 1 ; 12 iœ0
6
! mf(xi ) œ 18 and iœ0
˜x 3
œ
1 18
Ê
1‰ S œ ˆ 18 (18) œ 1 .
48. ¸f Ð%Ñ (x)¸ Ÿ 3 Ê M œ 3; ˜x œ
2" n
œ
" n
ˆ "n ‰# (8) Ÿ 10$
Ê n 25.82 Ê n 26
x! x" x# x$ x% x& x'
xi 0 1/6 1/3 1/2 21/3 51/6 1
f(xi ) 0 1/2 3/2 2 3/2 1/2 0
m 1 2 2 2 2 2 1
mf(xi ) 0 1 3 4 3 1 0
x! x" x# x$ x% x& x'
xi 0 1/6 1/3 1/2 21/3 51/6 1
f(xi ) 0 1/2 3/2 2 3/2 1/2 0
m 1 4 2 4 2 4 1
mf(xi ) 0 2 3 8 3 2 0
;
6
" 12
%
"‰ ˆ"‰ & . Hence kEs k Ÿ 10& Ê ˆ 2180 Ê n (3) Ÿ 10
" 60n%
Ÿ 10& Ê n%
10& 60
Ê n 6.38 Ê n 8 (n must be even) 49. yav œ œ œ ¸
" 365 0
" ˆ ‰ 21 ‘ ‰ ˆ ˆ 365 ‰ 21 ‘ ‰‘ 37 ˆ 365 365 21 cos 365 (365 101) 25(365) 37 21 cos 365 (0 101) 25(0) 37 21 37 21 37 ˆ 21 21 ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ 21 cos 365 (264) 25 21 cos 365 (101) œ 21 cos 365 (264) cos 365 (101) ‰ 37 #1 (0.16705 0.16705) 25 œ 25° F
50. av(Cv ) œ ¸
21 " '0365 37 sin ˆ 365 ˆ 21 ‰ ‰‘ $'& (x 101)‰ 25‘ dx œ 365 37 ˆ 365 21 cos 365 (x 101) 25x !
" 655
" 67520
" "3 '20675 c8.27 10& a26T 1.87T# bd dT œ 655 8.27T 10
&
T#
0.62333 10&
25
'(&
T$ ‘ #!
[(5582.25 59.23125 1917.03194) (165.4 0.052 0.04987)] ¸ 5.434;
8.27 10& a26T 1.87T# b œ 5.434 Ê 1.87T# 26T 283,600 œ 0 Ê T ¸
26È676 4(1.87)(283,600) #(1.87)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
¸ 396.45° C
Chapter 8 Practice Exercises 51. (a) Each interval is 5 min œ
1 12
hour.
2a2.4b 2a2.3b Þ Þ Þ 2a2.4b 2.3 d œ ‰ (b) a60 mphbˆ 12 29 hours/gal ¸ 24.83 mi/gal 1 2.5 24 c
29 12
¸ 2.42 gal
52. Using the Simpson's rule, ˜x œ 15 Ê ˜x 3 œ 5; ! mf(xi ) œ 1211.8 Ê Area ¸ a1211.8ba5b œ 6059 ft2 ;
x! x" x# x$ x% x5 x' x( x)
The cost is Area † a$2.10/ft2 b ¸ a6059 ft2 ba$2.10/ft2 b œ $12,723.90 Ê the job cannot be done for $11,000.
53.
'03 È dx
œ lim c '0 bÄ3
54.
'01 ln x dx œ
b
9 x#
xi 0 15 30 45 60 75 90 105 120
f(xi ) 0 36 54 51 49.5 54 64.4 67.5 42
b œ lim c sin" ˆ x3 ‰‘ 0 œ lim c sin" ˆ b3 ‰ sin" ˆ 03 ‰ œ bÄ3 bÄ3
dx È 9 x#
lim cx ln x xd 1b œ (1 † ln 1 1) lim b cb ln b bd œ 1 lim b b Ä !b bÄ! bÄ!
m 1 4 2 4 2 4 2 4 1 1 #
mf(xi ) 0 144 108 204 99 216 128.8 270 42 1 #
0œ
ln b Š "b ‹
œ 1 lim b bÄ!
œ 1 0 œ 1 55.
'c11 ydy
56.
'c_2 () d1))
#Î$
œ 'c1 0
'0
1
dy y#Î$
œ 'c2
1
$Î&
)$Î&
lim $Î& ) Ä _ ()1)
dy y#Î$
d) () 1)$Î&
1
_
d) )$Î&
diverges Ê
'3_ u 2du2u œ '3_ u du 2 '3_ duu œ
58.
'1_ 4v3v v1
#
_
#
dv œ '1 ˆ "v
1 œ 2 † 3 lim b y"Î$ ‘ b œ 6 Š1 lim b b"Î$ ‹ œ 6 bÄ! bÄ!
2
57.
$
dy y#Î$
) '2 'c1 () d1) $Î&
_
œ 1 and '2
œ 2 '0
" v#
'
d) () 1)$Î&
_ 2
d) () 1)$Î&
b lim ln ¸ u u 2 ¸‘ 3 œ lim ln ¸ b b 2 ¸‘ ln ¸ 3 3 2 ¸ œ 0 ln ˆ "3 ‰ œ ln 3
4 ‰ 4v 1
bÄ_
dv œ lim ln v bÄ_
bÄ_
59.
'0_ x# ecx dx œ
60.
'c0_ xe3x dx œ
61.
'c__ 4xdx 9 œ 2 '0_ 4xdx 9 œ #" '0_ x dx
1 ln 3 œ 1 ln
3 4
lim
b Ä _
bÄ_
x3 e3x "9 e3x ‘ 0 œ 9" b
#
#
1 6
62.
'c__ x 4dx16 œ 2'0_ x 4dx16 œ 2
63.
) ) Ä _ È)# 1
lim
b
ln (4v 1)‘ 1
b
ˆ 23 † 1# ‰ 0 œ #
" 4
" v
lim cx# ecx 2xecx 2ecx d 0 œ lim ab# ecb 2becb 2ecb b (2) œ 0 2 œ 2
bÄ_
#
" #
diverges
bÄ_
œ lim ln ˆ 4b b 1 ‰ b" ‘ (ln 1 1 ln 3) œ ln
œ
converges if each integral converges, but
#
_ d)
œ 1 and '6
)
9 4
œ
lim
b Ä _
" # b lim Ä_
ˆ b3 e3b 9" e3b ‰ œ 9" 0 œ 9"
32 tan" ˆ 2x ‰‘ b œ 3 0
" # b lim Ä_
32 tan" ˆ 2b ‰‘ 3
" 3
tan" (0)
b lim tan" ˆ 4x ‰‘ 0 œ 2 Š lim tan" ˆ 4b ‰‘ tan" (0)‹ œ 2 ˆ 1# ‰ 0 œ 1
bÄ_
bÄ_
_
diverges Ê '6
d) È)# 1
diverges
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Š b" ‹ Š
" ‹ b#
523
524
Chapter 8 Techniques of Integration _
_
_
64. I œ '0 ecu cos u du œ lim cecu cos ud b0 '0 ecu sin u du œ 1 lim cecu sin ud b0 '0 aecu b cos u du bÄ_
Ê I œ 1 0 I Ê 2I œ 1 Ê I œ 65.
bÄ_
" #
converges
'1_ lnz z dz œ '1e lnz z dz 'e_ lnz z dz œ ’ (ln2z) “ e #
1
Ê diverges 66. 0
_
67.
' __ e 2dxec
68.
_ dx 'c_ ' x a1 e b œ
69.
_
œ 2'0
x
x
#
x
Š x"# ‹
lim
xÄ0
’ # " x “ x a1 e b
Ê
#
2 dx ex ecx 1
_
_
'0
dx x # a1 e x b
4 dx ex
'
x # a1 e x b x# xÄ0
œ lim
2x$Î# 3
converges Ê
0
dx # x 1 x a1 e b
'0
1
#
#
bÄ_
e
'1_ Èect dt converges t
' __ e 2dxec x
dx x # a1 e x b
x
converges _
'1
œ lim a1 ex b œ 2 and '0
'c__ x a1dx e b diverges
' 1 x dxÈx ; – œ
bÄ_
Ÿ ect for t 1 and '1 ect dt converges Ê
ect Èt
b
#
lim ’ (ln#z) “ œ Š 12 0‹ lim ’ (ln2b) "# “ œ _
1
xÄ0
dx x # a1 e x b
dx x#
;
diverges Ê
'01 x a1dx e b diverges #
x
x
u œ Èx — Ä du œ #dx Èx
' u 1†2u udu œ ' ˆ2u# 2u 2 1 2 u ‰ du œ 32 u$ u# 2u 2 ln k1 uk C #
x 2Èx 2 ln ˆ1 Èx‰ C
70.
' x4 x2 dx œ ' ˆx 4xx 42 ‰ dx œ '
71.
'
$
#
#
dx x ax # 1 b #
;”
x œ tan ) Ä dx œ sec# ) d) •
œ ln ksin )k
" #
œ'
3 #
' x dx 2 #5 ' x dx 2 œ x#
#
3 #
ln kx 2k
5 #
) d) ' tansec) sec œ ' cossin))d) œ ' Š 1 sinsin) ) ‹ d(sin )) ) #
$
#
%
#
'È
73.
' 2 cossinx x sin x dx œ '
d(x 1) È1 (x 1)#
#
œ sin" (x 1) C
2 csc# x dx '
cos x dx sin# x
' csc x dx œ 2 cot x
" sin x
ln kcsc x cot xk C
œ 2 cot x csc x ln kcsc x cot xk C 74.
sin ) ' cos ' 1 coscos) ) d) œ ' sec# ) d) ' d) œ tan ) ) C ) d) œ
75.
' 819dvv
76.
'2_ (x dx1)
#
#
#
#
%
œ
#
" #
' v dv 9 12" ' 3 dv v 1"# ' 3 dv v œ 12" ln ¸ 33 vv ¸ "6 tan" v3 C #
b œ lim 1 " x ‘ 2 œ lim 1 " b (1)‘ œ 0 1 œ 1
bÄ_
bÄ_
cos (2) 1)
77. ÐÑ ) ïïïïî ÐÑ 1 ïïïïî 0
" #
ln kx 2k C
sin# ) C œ ln ¹ Èxx# 1 ¹ "# Š Èxx# 1 ‹ C
72.
dx 2x x#
x dx
sin (2) 1) "4 cos (2) 1) Ê
' ) cos (2) 1) d) œ #) sin (2) 1) "4 cos (2) 1) C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 8 Practice Exercises 78.
' x x 2xdx 1 œ ' ˆx 2 x 3x 2x 2 1 ‰ dx œ '
(x 2) dx 3 '
$
#
#
x# #
œ
" x1
2x 3 ln kx 1k œ #" '
dx x1
'
C
79.
2) d) ' (1sin cos 2) )
80.
'11ÎÎ42 È1 cos 4x dx œ È2 '11ÎÎ42 cos 2x dx œ ’ È#2 sin 2x“ 1Î2 œ È#2
81.
' Èx dx
#
d(1 cos 2)) (1 cos 2))#
dx (x 1)#
œ
" #(1 cos 2))
Cœ
" 4
sec# ) C
1Î4
;”
2x
yœ2x Ä ' dy œ dx •
(2 y) dy Èy
œ
y$Î# 4y"Î# C œ
2 3
2 3
(2 x)$Î# 4(2 x)"Î# C
$
ŠÈ2 x‹
œ 2–
82.
3
' È1v v
#
#
2È2 x— C
' cos )sin†cos)) d) œ ' a1 sinsin ))b d) œ ' csc# ) d) ' d) œ cot ) ) C #
dv; cv œ sin )d Ä
œ sin" v
È 1 v# v
#
#
C
83.
1) " ' y dy2y 2 œ ' (yd(y (y 1) C 1) 1 œ tan
84.
'È
85.
' z azz1 4b dz œ "4 ' ˆ "z z"
86.
' x$ ex
87.
' È t dt
#
#
x dx 8 2x# x%
#
œ
" #
'
d ax # 1 b
#
#
#
dx œ
9 4t#
" #
' x# ex
œ "8 '
88. u œ tan" x, du œ
œ
É 9 ax # 1 b#
d a9 4t# b È9 4t# dx 1 x#
sin" Š x
z1 ‰ z# 4
d a x# b œ
#
" #
" #
dz œ
" 4
#
1 3 ‹
C
ln kzk
" 4z
ˆx# ex# ex# ‰ C œ
" 8
ln az# 4b #
ax # 1 b e x #
" 8
tan"
z #
C
C
œ 4" È9 4t# C
; dv œ
dx x#
, v œ x" ;
' tancx x dx œ x" tan" x ' x a1dx x b œ x" tan" x ' dxx ' 1xdxx "
#
œ 89.
'e œ
" x
#
"
tan
et dt 3et 2 ; t "‰ ln ˆ eet # 2t
x ln kxk cet œ xd Ä
' tan$ t dt œ '
91.
'1_ ln yy dy ; Ö
92.
ln a1 x b C œ
tanc" x x
dx ' (x 1)(x ' x dx 1 ' 2) œ
(tan t) asec# t 1b dt œ
Ô x œ ln y × Ù Ä dx œ dy y x Õ dy œ e dx Ø
b œ lim ˆ 2e 2b
bÄ_
#
#
ln kxk ln È1 x# C dx x#
"¸ œ ln kx 1k ln kx 2k C œ ln ¸ xx # C
C
90.
$
" #
" ‰ 4e2b
'0_ xe†e
x
3x
ˆ0 "4 ‰ œ
tan# t #
' tan t dt œ
ln ksec tk C
_
b dx œ '0 xec2x dx œ lim x# ec2x 4" ec2x ‘ 0
bÄ_
" 4
u œ ln (sin v) v dv ' lncot(sinv dvv) œ ' (sincos cos v dv • v) ln (sin v) ; ” du œ
tan# t #
sin v
Ä
' duu œ ln kuk C œ ln kln (sin v)k C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
525
526
Chapter 8 Techniques of Integration
93.
' eln Èx dx œ ' Èx dx œ 23 x$Î# C
94.
' e) È3 4e) d); ”
95.
' 1 sin(cos5t dt5t)
96.
' È dv
97.
' 1 drÈr ; –
98.
d ˆx 10x 9‰ 20x ' x 4x 10x dx œ ' x 10x 9 œ ln kx% 10x# 9k C 9
99.
' 1 x x
e2v
#
;”
; 1 ”
u œ 4e) Ä du œ 4e) d) •
u œ Èr dr — Ä du œ 2È r
100.
' 1 x x
101.
' 11 xx
dx œ 3'
2
3
2
dx;
3
1 x2 1 x3 2
#
x 1 x2
3x2 1 x3
œ
œ sec" x C œ sec" aev b C
#
%
dx œ ' ax
dx xÈ x# 1
œ 5" tan" u C œ 5" tan" (cos 5t) C
du 1 u#
' 2u1 duu œ ' ˆ2 1 2 u ‰ du œ 2u 2 ln k1 uk C œ 2Èr 2 ln ˆ1 Èr‰ C
%
2
'
x œ ev Ä dx œ ev dv •
#
3
' È3 u du œ 4" † 23 (3 u)$Î# C œ 6" a3 4e) b$Î# C
u œ cos 5t Ä 5" ' du œ 5 sin 5t dt •
$
%
" 4
b dx œ ' x dx 12 ' 1 2xx
2
dx œ 21 x2 21 lna1 x2 b C
dx œ 3lnl1 x3 l C
A 1x
Bx C 1 x x2
Ê 1 x2 œ Aa1 x x2 b aBx Cba1 xb
œ aA Bbx aA B Cbx aA Cb Ê A B œ 1ß A B C œ 0, A C œ 1 Ê A œ 23 , B œ 13 , C œ 13 ;
' 11 xx ”
dx œ ' a 12Î3x
2 3
u œ x "# • Ä du œ dx
1 3
'
a 1 Î3 b x 1 Î 3 1 x x2
u 32 u2
2 œ 16 ln¹ 34 ˆx "# ‰ ¹
Ê
2 3
2
3
dx; ”
2 1x
' ÈxÉ1 Èx dx;
103.
1 a1 x b 2
”
'
3 4
u u2
du
1 2
'
3 4
1 u2
2
dx œ
1 au 1 b 2 u3
œ 16 lnl1 x x2 l
du œ '
2 3
' 1 1 x dx 31 '
du œ 16 ln¹ 34 u2 ¹
?2 2u 2 u3
3 4
x1 2 ˆx "# ‰
dx;
1 u 1 È3 tan Š È3Î2 ‹
1 1 2x 1 È3 tan Š È3 ‹
dx œ 32 lnl1 xl 61 lnl1 x x2 l
2
uœ1x Ä' du œ dx •
œ lnl1 xl
1 3
" 1 1 x # È3 tan Š È3Î2 ‹
' 1 1 x dx 31 ' 1 x x1 x
' a11 xxb
102.
du œ
3 4
b dx œ 32 ' 1 1 x dx 31 ' 1 x x1 x
du œ '
1 u
1 1 2x 1 È3 tan Š È3 ‹
du '
2 u2
C
du '
2 u3
du œ lnlul
2 u
C
w œ Èx Ê w2 œ x 2 • Ä ' 2w È1 w dw 2w dw œ dx
È1 w ÐÑ 2w2 ïïïïî ÐÑ 4w ïïïïî ÐÑ 4 ïïïïî
2 3 a1
wb3Î2
4 15 a1 8 105 a1
0 3 Î2 œ 43 xˆ1 Èx‰
wb5Î2 wb7Î2
Ê ' 2w2 È1 w dw œ 43 w2 a1 wb3Î2
16 È ˆ x 1 15
5Î2 Èx‰
32 ˆ 105 1
16 15 wa1
wb5Î2
32 105 a1
7Î2 Èx‰ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
wb7Î2 C
1 u2
C
Chapter 8 Practice Exercises 104.
' É1 È1 x dx; ” w œ È1 x Ê w2 œ 1 x • Ä ' 2w È1 w dw; 2w dw œ dx
3 Î2 ’u œ 2w, du œ 2 dw, dv œ È1 w dw, v œ 23 a1 wb “
' 2w È1 w dw œ 43 wa1 wb3Î2 ' 43 a1 wb3Î2 dw œ 43 wa1 wb3Î2 158 a1 wb5Î2 C œ 43 È1 xŠ1 È1 x‹
105.
3 Î2
' ÈxÈ11 x dx; ” u œ Èx Ê u 'È2
du œ '
1 u2
2
2u du œ dx
2sec2 ) sec )
8 15 Š1
œx
È1 x‹
•Ä'
2 È 1 u2
5Î2
C
du; ’u œ tan ), 12 ) 12 , du œ sec2 ) d), È1 u2 œ sec )“
d) œ ' 2 sec ) d) œ 2lnlsec ) tan )l C œ 2 ln¹È1 u2 u¹ C
œ 2 ln¹È1 x Èx¹ C
106.
' 1Î2 É1 È1 x2 dx; 0
’x œ sin ), 12 ) 12 , dx œ cos ) d), È1 x2 œ cos ), x œ 0 œ sin ) Ê ) œ 0, x œ Ä '0
1 Î6
È1 cos ) cos ) d) œ '
1Î6 È1cos2 ) È1 cos )
0
cos ) d) œ '0
1Î6
sin ) cos ) È1 cos )
’u œ cos ), du œ sin ) d), dv œ œ œ œ œ
1 Î6
limb ”’2 cos ) a1 cos )b1Î2 “
c Ä0
c
limb ”Š2 cosˆ 16 ‰ ˆ1 cosˆ 16 ‰‰
1 Î2
cÄ0
1 Î6
sin ) È1 cos )
œ sin ) Ê ) œ 16 “
' 1Î6 Èsin ) cos )
lim
cÄ0b c
1 cos )
2a1 cos )b1Î2 sin ) d)• 1 Î6
2 cos c a1 cos cb1Î2 ‹ ’ 34 a1 cos )b3Î2 “
c
•
È 3 1 Î2 2 ‹
3 Î2 2 cos c a1 cos cb1Î2 Š 43 ˆ1 cosˆ 16 ‰‰ 43 a1 cos cb3Î2 ‹•
limb ”È3 Š1
È 3 1 Î2 2 ‹
2 cos c a1 cos cb1Î2 34 Š1
c Ä0
È 3 1 Î2 2 ‹
34 Š1
È3 3Î2 2 ‹
d) ;
d), v œ 2a1 cos )b1Î2 “
lim ”È3 Š1
c Ä0 b
œ È 3 Š1
107.
'c
d) œ
" #
œ Š1
È 3 3 Î2 2 ‹
È3 1Î2 4 È3 Š 3 ‹ 2 ‹
œ
34 a1 cos cb3Î2 •
Š4 È3‹É2È3 3È 2
' x lnxxln x dx œ ' xa1 lnxln xb dx; ” u œ 1 1 ln x • Ä ' u u 1 du œ ' du ' 1u du œ u lnlul C du œ x dx
œ a1 ln xb lnl1 ln xl C œ ln x lnl1 ln xl C 108.
' x ln x†ln1 aln xb dx; ” u œ lna1ln xb • Ä ' u1 du œ lnlul C œ lnllnaln xbl C
109.
'x
du œ
ln x
ln x dx; x
x ln x dx
2 ’u œ xln x Ê ln u œ ln xln x œ aln xb Ê u1 du œ
2 ln x x
dx Ê du œ
2u ln x x
dx œ
2xln x ln x x
dx“ Ä
" #
' du
œ "# u C œ "# xln x C 110.
' aln xbln x ’ 1x lnalnx xb “dx; ’u œ aln xbln x Ê ln u œ ln aln xbln x œ aln xb ln aln xb Ê 1u du œ Š axlnlnxxb lnalnx xb ‹dx Ê du œ u ’ 1x
lnaln xb x “dx
œ aln xbln x ’ x1
lnaln xb x “dx“
Ä ' du œ u C œ aln xbln x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
527
528
Chapter 8 Techniques of Integration
'
111.
1 xÈ 1 x4
œ
1 2
dx œ '
x x2 È 1 x4
dx; ’x2 œ sin ), 0 Ÿ ) 12 , 2x dx œ cos ) d), È1 x4 œ cos )“ Ä
' csc ) d) œ 12 lnlcsc ) cot )l C œ 12 ln» x1
2
È 1 x4 » x2
C œ 12 ln» 1
' È1x x dx; ’u œ È1 x Ê u2 œ 1 x Ê 2u du œ dx“ Ä ' 12uu
du œ '
2
112.
2 u2 1
œ
A u1
B u1
2
È 1 x4 » x2
2u2 u2 1
1 2
' sincos) cos) ) d)
C
du œ ' ˆ2
2 ‰ u2 1 du;
Ê 2 œ Aau 1b Bau 1b œ aA Bbu A B Ê A B œ 0, A B œ 2
Ê A œ 1 Ê B œ 1; ' ˆ2
2 ‰ u2 1 du
œ ' 2 du ' a u 1 1
1 u1
b du
È
x1 œ 2u lnlu 1l lnlu 1l C œ 2È1 x 12 lnº È11 C x1º
' a faa xb dx; ’u œ a x Ê du œ dx, x œ 0 Ê u œ a, x œ a Ê u œ 0“ Ä ' 0 faub du œ ' a faub du, which is
113. (a)
0
a
the same integral as '0 faxb dx.
0
a
sinˆ 1 ‰ cos x cosˆ 1 ‰ sin x ' 1Î2 sin xsin xcos x dx œ ' 1Î2 sin ˆ 1 sinx‰ˆ 1 cosx‰ˆ 1 x‰ dx œ ' 1Î2 sinˆ 1 ‰ cos x cos ˆ 1 ‰ sin x cos ˆ 1 ‰ cos x sinˆ 1 ‰ sin x dx
(b)
2
0
0
œ' œ
'
1 Î2
2
2
0
2
1Î2 cos x sin x cos x sin x dx Ê 2 0 sin x cos x dx œ 0 1Î2 1Î2 x 1 ’x“ œ 12 Ê 2 0 sin xsin cos x dx œ 2 Ê 0
'
' '
1Î2
0 1Î2
0
2
sin x sin x cos x
dx '0
sin x sin x cos x
dx œ
2
2
1Î2
2
cos x cos x sin x
dx œ '0
sin x cos x sin x cos x
dx œ '0
1Î2
dx
1 4
x sin x sin x sin x cos x x ' sin xsin xcos x dx œ ' sin x cos xsinxcos cos ' sincosxxcossinxx dx ' sinxsin dx œ ' sin x x cos x dx cos x dx x sin x ' sin xsinxcos x dx œ x lnlsin x cos xl ' sin xsinxcos x dx œ ' dx ' cos sin x cos x dx Ê 2' sin xsinxcos x dx œ x lnlsin x cos xl Ê ' sin xsinxcos x dx œ x2 12 lnlsin x cos xl C
114.
' 1 sinsinx x dx œ ' 2
115.
2
œ ' dx '
sin2 x cos2 x 1 sin2 x cos 2x cos2 x
sec2 x 1 2tan2 x
dx œ '
dx œ x
tan2 x sec2 x tan2 x
dx œ '
1 1 È È2 tan Š 2 tan x‹
tan2 x sec2 x sec2 x sec2 x tan2 x
dx œ '
tan2 x sec2 x sec2 x tan2 x
dx '
sec2 x sec2 x tan2 x
dx
C
cos xb cos x x x ' 11 cos ' a11 cos ' 1 2cossinx x cos x dx œ ' sin1 x dx ' 2cos ' cos x dx œ x dx œ sin x dx sin x dx œ ' csc2 x dx 2' csc x cot x dx ' cot2 x dx œ cot x 2csc x ' acsc2 x 1bdx œ œ 2cot x 2csc x x C 2
116.
2
2
2
2
2
2
2
CHAPTER 8 ADDITIONAL AND ADVANCED EXERCISES #
1. u œ asin" xb , du œ
'
2 sinc" x dx È 1 x#
; dv œ dx, v œ x;
asin" xb dx œ x asin" xb ' #
#
u œ sin" x, du œ
dx È 1 x#
2x sinc" x dx È 1 x# 1x
"
' " x
;
; dv œ È2x dx # , v œ 2È1 x# ;
c ' 2xÈsin x dx œ 2 asin" xb È1 x# '
2.
2
1Î2
1 x#
#
2 dx œ 2 asin" xb È1 x# 2x C; therefore
#
asin" xb dx œ x asin" xb 2 asin" xb È1 x# 2x C œ
" x
,
" " x(x 1) œ x " x(x 1)(x 2)
œ
" x1 , " " 2x x 1
" #(x 2)
,
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 8 Additional and Advanced Exercises " " " " " x(x 1)(x 2)(x 3) œ 6x #(x 1) #(x 2) 6(x 3) , " " " " " x(x 1)(x 2)(x 3)(x 4) œ #4x 6(x 1) 4(x #) 6(x 3) m
" x(x 1)(x 2) â (x m) m
œ! kœ0
(")k (k!)(m k)!(x k) ;
therefore '
" 24(x 4)
Ê the following pattern:
dx x(x 1)(x 2) â (x m)
(") œ ! ’ (k!)(m k)! ln kx kk“ C k
kœ0
3. u œ sin" x, du œ
dx È 1 x#
' x sin" x dx œ x#
sin" x '
#
4.
œ
x# #
sin" x
œ
x# #
sin" x
6.
'
z œ Èy
'
dt t È1 t#
;”
x# #
;”
;
x "# ˆ #)
Õ d) œ
œ
" #
ln Št È1 t# ‹
#
" ax# 2b# 4x#
u# 1
¹
" #
sin ) cos ) ) 4
Ø
du u# 1
tan" u C œ
sin# ) cos ) d) 2 cos )
C
" Èy
C
" #
Ä
' (u 1)duau# 1b
ln ¸ tansec) ) " ¸ #" ) C
sin" t C dx œ '
" ax# 2x 2b ax# 2x 2b
dx
œ
" 16
' ’ x 2x 2x 2 2 (x 1)2 1 x 2x 2x 2# (x 1)2 1 “ dx
œ
" 16
2x 2 " " ln ¹ xx# (x 1) tan" (x 1)d C 2x # ¹ 8 ctan
#
sin" x
u œ tan )
' u du 1 #" ' u du 1 #" ' uu du1 œ #" ln ¹ Èu 1
%
x# #
sin" x '
C
" #
' x " 4 dx œ '
Cœ
x# #
' sin" Èy dy œ y sin" Èy Èy È1 y# sin
œ
" #
sin 2) ‰ 4
x sin" x dx œ
d) ' sincos) ) cos ' tan d)) 1 ; Ô du œ sec# ) d) × ) œ
t œ sin ) Ä dt œ cos ) d) • #
'
x œ sin ) Ä dx œ cos ) d) •
' 2z sin" z dz; from Exercise 3, ' z sin" z dz
— Ä
dy 2È y
zÈ1 z# sin " z C Ê 4 " Èy # È sin y y sin" Èy # #
z# sin " z #
œy
x# dx 2È 1 x#
# " sin# ) d) œ x# sin" # xÈ1 x# sin " x C 4
' sin" Èy dy; – dz œ œ
5.
; dv œ x dx, v œ
#
#
#
#
7. x lim Ä_
'cxx sin t dt œ x lim c cos td xcx œ x lim c cos x cos (x)d œ x lim a cos x cos xb œ x lim 0œ0 Ä_ Ä_ Ä_ Ä_
8.
'x1 cost t dt;
lim b
xÄ!
#
lim b x 'x
1
xÄ!
lim b
xÄ!
cos t t#
limb
tÄ!
#
! ln nÉ1 9. n lim Ä_ kœ1 Ä
t Š cos# t ‹ t
œ limb tÄ!
" cos t
œ1 Ê
lim b 'x
1
xÄ!
cos t t#
dt diverges since '0
1
dt t#
diverges; thus
^ dt is an indeterminate 0 † _ form and we apply l'Hopital's rule:
1 t x 'x cos dt œ t
n
Š "# ‹
k n
'1
x
lim b
xÄ!
cos t t#
" x
dt
œ lim b xÄ!
Š cos# x ‹ Š
x " ‹ x#
œ lim b cos x œ 1 xÄ!
! ln ˆ1 k ˆ " ‰‰ ˆ " ‰ œ ' ln (1 x) dx; ” œ n lim n n Ä_ 0 n
kœ1
1
u œ 1 x, du œ dx x œ 0 Ê u œ 1, x œ 1 Ê u œ 2 •
'12 ln u du œ cu ln u ud #" œ (2 ln 2 2) (ln 1 1) œ 2 ln 2 1 œ ln 4 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
529
530
Chapter 8 Techniques of Integration nc1
! 10. n lim Ä_ kœ0 œ '0
1
11.
dy dx
" È n# k#
dx œ csin" xd ! œ
n ‹ ˆ n" ‰ n# k#
nc1
! œ n lim Ä_ kœ0
Î Ï
"
Ê 1 Š dy dx ‹ œ
1Î4
1Î2
#
#
a1 x# b 4x# a1 x # b #
x ' ˆ1 œ '0 Š "1 x# ‹ dx œ 0 1Î2
ˆ n" ‰
Ê1 ŠÈcos 2t‹ dt œ È2 '0 #
1Î4
œ1 #
2x 1 x#
Ò
1 #
#
œ
Ñ #
Ê1 ’k Š "n ‹“
# ' œ Ècos 2x Ê 1 Š dy dx ‹ œ 1 cos 2x œ 2 cos x; L œ 0 1Î%
dy dx
kœ0
"
" È 1 x #
œ È2 csin td ! 12.
nc1
! ŠÈ œ n lim Ä_
2 ‰ 1 x#
œ ˆ "# ln 3‰ (0 ln 1) œ ln 3
œ
1 2x# x% a1 x # b#
x ' œ Š 11 x# ‹ ; L œ 0
dx œ '0 ˆ1 1Î2
#
" 1x
#
1Î2
" ‰ 1x
#
dy ‹ dx Ê1 Š dx "Î#
x ¸‘ dx œ x ln ¸ 11 x !
" #
shell ‰ shell 13. V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21xy dx b
1
œ 61 '0 x# È1 x dx; 1
Ô uœ1x × du œ dx Õ x# œ (1 u)# Ø
Ä 61 '1 (1 u)# Èu du 0
œ 61 '1 ˆu"Î# 2u$Î# u&Î# ‰ du 0
!
œ 61 23 u$Î# 45 u&Î# 27 u(Î# ‘ " œ 61 ˆ 23 84 30 ‰ 16 ‰ œ 61 ˆ 70 105 œ 61 ˆ 105 œ 32351 14. V œ 'a 1y# dx œ 1 '1 b
œ 1 '1 ˆ dx x 4
4
5 dx x#
151 4
27 ‰
25 dx x# (5 x)
dx ‰ 5x
% œ 1 ln ¸ 5 x x ¸ 5x ‘ " œ 1 ˆln 4 54 ‰ 1 ˆln
œ
4 5
" 4
5‰
21 ln 4
shell ‰ shell 15. V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21xex dx b
1
œ 21 cxex ex d "! œ 21
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ècos# t dt
Chapter 8 Additional and Advanced Exercises 16. V œ '0
ln 2
21(ln 2 x) aex 1b dx
œ 21 '0
ln 2
c(ln 2) ex ln 2 xex xd dx ln 2
œ 21 ’(ln 2) ex (ln 2)x xex ex
x# # “0
œ 21 ’2 ln 2 (ln 2)# 2 ln 2 2
(ln 2)# # “
21(ln 2 1)
#
œ 21 ’ (ln#2) ln 2 1“ 17. (a) V œ '1 1 c1 (ln x)# d dx e
œ 1 cx x(ln x)# d 1 21'1 ln x dx e
e
(FORMULA 110) e œ 1 cx x(ln x)# 2(x ln x x)d 1 e œ 1 cx x(ln x)# 2x ln xd 1 œ 1 ce e 2e (1)d œ 1
(b) V œ '1 1(1 ln x)# dx œ 1'1 c1 2 ln x (ln x)# d dx e
e
œ 1 cx 2(x ln x x) x(ln x)# d 1 21'1 ln x dx e
e
œ 1 cx 2(x ln x x) x(ln x)# 2(x ln x x)d 1 e œ 1 c5x 4x ln x x(ln x)# d 1 œ 1 c(5e 4e e) (5)d œ 1(2e 5) e
18. (a) V œ 1 '0 aey b# 1‘ dy œ 1'0 ae2y 1b dy œ 1 e# y‘ ! œ 1 ’ e# 1 ˆ "# ‰“ œ 1
1
"
2y
#
1 ae # 3 b #
(b) V œ 1'0 aey 1b# dy œ 1'0 ae2y 2ey 1b dy œ 1 e# 2ey y‘ ! œ 1 ’Š e# 2e 1‹ ˆ "# 2‰“ 1
1
#
œ 1 Š e# 2e 5# ‹ œ 19. (a)
lim
x Ä !b
x ln x œ 0 Ê
"
2y
#
1 ae# 4e 5b #
lim
x Ä !b
f(x) œ 0 œ f(0) Ê f is continuous
# Ô u œ (ln x) × Ö du œ (2 ln x) dx Ù # 2 2 $ x Ù x$ # ' (b) V œ '0 1x# (ln x)# dx; Ö (ln x) ’ “ Š x3 ‹ (2 ln x) Ö dv œ x# dx Ù Ä 1 Œb lim 3 0 b Ä !b $ Õ Ø v œ x3 $
œ 1 ”ˆ 83 ‰ (ln 2)# ˆ 23 ‰ lim b ’ x3 ln x bÄ!
2
x$ 9 “ b•
#
œ 1 ’ 8(ln3 2)
16(ln 2) 9
16 27 “
20. V œ '0 1( ln x)# dx 1
œ 1 Œ lim b cx(ln x)# d b 2'0 ln x dx bÄ! 1
1
œ 21 lim b cx ln x xd 1b œ 21 bÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dx x
531
532
Chapter 8 Techniques of Integration
21. M œ '1 ln x dx œ cx ln x xd e1 œ (e e) (0 1) œ 1; e
Mx œ '1 (ln x) ˆ ln#x ‰ dx œ e
œ
" #
e
e
e
" #
'1e (ln x)# dx
Šcx(ln x)# d 1 2 '1 ln x dx‹ œ
My œ '1 x ln x dx œ ’ x œ
" #
’x# ln x
My M
therefore, x œ 22. M œ '0
1
e
x# # “1
œ
#
e
" #
œ
ln x # “1
’Še#
e# 1 4
" #
" #
(e 2);
'1e x dx
e# #‹
and y œ
"# “ œ
Mx M
œ
" 4
ae# 1b ;
e2 #
"
œ 2 csin" xd ! œ 1;
2 dx È 1 x#
My œ '0
" 2x dx È # È1 x# œ 2 ’ 1 x “ œ 2; ! My 2 therefore, x œ M œ 1 and y œ 0 by symmetry 1
23. L œ '1 É1 e
tanc" e
œ '1Î4
" x#
dx œ '1
e
dx; ”
tan " e
d) œ '1Î4
(sec )) atan# ) 1b tan )
œ ŠÈ1 e# ln ¹
È x# 1 x
È 1 e# e
tanc" e x œ tan ) Ä L œ '1Î4 • # dx œ sec ) d)
sec )†sec# ) d) tan )
(tan ) sec ) csc )) d) œ csec ) ln kcsc ) cot )kd 1tanÎ4
"e ¹‹ ’È2 ln Š1 È2‹“ œ È1 e# ln Š
È 1 e# e
"e
"e ‹ È2 ln Š1 È2‹
# # ' È ' yÈ1 e2y dy; ” 24. y œ ln x Ê 1 Š dx dy ‹ œ 1 x Ê S œ 21 c x 1 x dy Ê S œ 21 0 e #
d
Ä S œ 21'1 È1 u# du; ” e
c" e
È 1 e# e È2 1 ‹
œ 1 ’ŠÈ1 e# ‹ e ln ¹È1 e# e¹“ 1 ’È2 † 1 ln ŠÈ2 1‹“
È 2“
25. S œ 21 'c1 f(x) É1 cf w (x)d# dx; f(x) œ ˆ1 x#Î$ ‰ 1
$Î# " ‰ dx; – œ 41'0 ˆ1 x#Î$ ‰ ˆ x"Î$ 1
" œ 61 † 25 (1 u)&Î# ‘ ! œ
26. y œ '1 ÉÈt 1 dt Ê x
dy dx 16 1
27.
'1_ ˆ x ax 1 #"x ‰ dx œ
lim
#
œ
lim " bÄ_ #
’ln
ab # 1 b b
integral diverges if a
Ê S œ 21 'c1 ˆ1 x#Î$ ‰ 1
" x#Î$
$Î#
†
dx Èx#Î$
1 u œ x#Î$ 3 $Î# '1 du œ 61'0 (1 u)$Î# d(1 u) 2 dx — Ä 4 † # 1 0 (1 u) du œ 3 x"Î$
2
16
œ 45 a16b5Î4 45 a1b5Î4 œ
'1b ˆ x ax 1 #"x ‰ dx œ #
ln 2 “ ; a
" #
Ê cf w (x)d# 1 œ
121 5
bÄ_ a
$Î#
œ ÉÈx 1 Ê L œ '1 Ê1 ŠÉÈx 1‹ dx œ '1 É1 Èx 1 dx
4 œ '1 È x dx œ ’ 45 x5Î4 “
16
u œ ey du œ ey dy •
tanc" e u œ tan ) Ä 21 '1Î4 sec ) † sec# ) d) • # du œ sec ) d)
œ 21 ˆ "# ‰ csec ) tan ) ln ksec ) tan )kd tan 1Î% œ 1 ’eÈ1 e# ln Š
1
# lim ab b 1b bÄ_
; for a œ "# :
a
124 5
lim #a ln ax# 1b
bÄ_
2a lim b bÄ_ b
È b# 1 b bÄ_
lim
16
œ lim b bÄ_
œ lim É1 bÄ_
" b#
" #
ln x‘ 1 œ lim ’ #" ln
2 ˆa c 12 ‰
b
bÄ_
œ _ if a
œ1 Ê
" bÄ_ #
lim
" #
”ln
ax # 1 b x
b
a
“
1
Ê the improper ab # 1 b b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"Î#
ln 2"Î# •
Chapter 8 Additional and Advanced Exercises " #
œ Ê
_
'1
ˆln 1
" #
lim ln
bÄ_
ˆ x#ax 1
28. G(x) œ lim
_ dx xp
ab # 1 b b
" ‰ #x
bÄ_
29. A œ '1
ln 2‰ œ ln42 ; if a a
ab # 1 b b bÄ_
" #
: 0 Ÿ lim
" #
" #
(b1)2a b 1
lim
bÄ_
" #
œ _ Ê the improper integral diverges if a " #
dx converges only when a œ
'0b ecxt dt œ
œ lim (b 1)2ac1 œ 0 bÄ_
; in summary, the improper integral
and has the value ln42 cxb
b lim "x ecxt ‘ 0 œ lim Š 1 xe ‹ œ
bÄ_
bÄ_
10 x
œ
" x
if x 0 Ê xG(x) œ x ˆ "x ‰ œ 1 if x 0
converges if p 1 and diverges if p Ÿ 1. Thus, p Ÿ 1 for infinite area. The volume of the solid of revolution
_
_ dx
about the x-axis is V œ '1 1 ˆ x"p ‰ dx œ 1 '1 p
a
#
x2p
which converges if 2p 1 and diverges if 2p Ÿ 1. Thus we want
for finite volume. In conclusion, the curve y œ xcp gives infinite area and finite volume for values of p satisfying
p Ÿ 1.
30. The area is given by the integral A œ '0
1
dx xp
;
p œ 1: A œ lim b cln xd œ lim b ln b œ _, diverges; bÄ! bÄ! 1 b
p 1: A œ lim b cx1cp d 1b œ 1 lim b b1cp œ _, diverges; bÄ! bÄ!
p 1: A œ lim b cx1cp d 1b œ " lim b b1cp œ 1 0, converges; thus, p 1 for infinite area. bÄ! bÄ! The volume of the solid of revolution about the x-axis is Vx œ 1'0
1
p
" #
, and diverges if p
" #
dx x2p
which converges if 2p 1 or
. Thus, Vx is infinite whenever the area is infinite (p 1).
_
_
The volume of the solid of revolution about the y-axis is Vy œ 1 '1 cR(y)d# dy œ 1'1 converges if
2 p
dy y2Îp
which
1 Í p 2 (see Exercise 29). In conclusion, the curve y œ xcp gives infinite area and finite
volume for values of p satisfying 1 Ÿ p 2, as described above. _
31. (a) >(1) œ '0 e t dt œ lim
bÄ_
'0b e
t
dt œ lim ce t d b0 œ lim e"b (1)‘ œ 0 1 œ 1 bÄ_
bÄ_
(b) u œ tx , du œ xtxc1 dt; dv œ ect dt, v œ ect ; x œ fixed positive real _
_
Ê >(x 1) œ '0 tx e t dt œ lim ctx e t d b0 x '0 tx 1 e t dt œ lim ˆ beb 0x e! ‰ x>(x) œ x>(x) bÄ_
(c) >(n 1) œ n>(n) œ n!: n œ 0: >(0 1) œ >(1) œ 0!; n œ k: Assume >(k 1) œ k! n œ k 1: >(k 1 1) œ (k 1) >(k 1) œ (k 1)k! œ (k 1)! Thus, >(n 1) œ n>(n) œ n! for every positive integer n.
x
bÄ_
for some k 0; from part (b) induction hypothesis definition of factorial
x n n 32. (a) >(x) ¸ ˆ xe ‰ É 2x1 and n>(n) œ n! Ê n! ¸ n ˆ ne ‰ É 2n1 œ ˆ ne ‰ È2n1
(b)
533
n 10 20 30 40 50 60
ˆ ne ‰n È2n1 3598695.619 2.4227868 ‚ 10") 2.6451710 ‚ 10$# 8.1421726 ‚ 10%( 3.0363446 ‚ 10'% 8.3094383 ‚ 10)"
calculator 3628800 2.432902 ‚ 10") 2.652528 ‚ 10$# 8.1591528 ‚ 10%( 3.0414093 ‚ 10'% 8.3209871 ‚ 10)"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
534
Chapter 8 Techniques of Integration
(c)
ˆ ne ‰n È2n1 3598695.619
n 10 ÐÑ
cos 3x
2e2x
ÐÑ
1 3 sin
4e2x
ÐÑ
19 cos 3x
33. e2x
Iœ
e2x 3
sin 3x
2e2x 9
13 9
Iœ
e2x 9
(3 sin 3x 2 cos 3x) Ê I œ
e2x 13
(3 sin 3x 2 cos 3x) C
sin 4x
3e3x
ÐÑ
4" cos 4x
9e3x
ÐÑ
" 16 sin 4x
3x
I œ e4 cos 4x 35.
calculator 3628800
3x
cos 3x 49 I Ê
ÐÑ
34. e3x
ˆ ne ‰n È2n1 e1Î12n 3628810.051
3e3x 16
sin 4x
I Ê
9 16
sin 3x
ÐÑ
sin x
3 cos 3x
ÐÑ
cos x
9 sin 3x
ÐÑ
sin x
Iœ
25 16
e3x 16
(3 sin 4x 4 cos 4x) Ê I œ
e3x 25
(3 sin 4x 4 cos 4x) C
I œ sin 3x cos x 3 cos 3x sin x 9I Ê 8I œ sin 3x cos x 3 cos 3x sin x Ê I œ sin 3x cos x83 cos 3x sin x C 36.
cos 5x
ÐÑ
sin 4x
sin 5x
ÐÑ
"4 cos 4x
25cos 5x
ÐÑ
" 16 sin 4
I œ "4 cos 5x cos 4x Ê Iœ
" 9
sin 5x sin 4x
9 I Ê 16 I œ "4 cos 5x cos 4x
5 16
sin 5x sin 4x
sin bx
aeax
ÐÑ
"b cos bx
a# eax
ÐÑ
b"# sin bx
ax
I œ eb cos bx Ê Iœ
25 16
(4 cos 5x cos 4x 5 sin 5x sin 4x) C ÐÑ
37. eax
5 16
eax a# b#
aeax b#
sin bx
a# b#
I Ê Ša
#
b# b# ‹ I
œ
eax b#
(a sin bx b cos bx)
(a sin bx b cos bx) C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 8 Additional and Advanced Exercises ÐÑ
38. eax
cos bx
aeax
ÐÑ
" b
a# eax
ÐÑ
b"# cos bx
Iœ
eax b
sin bx
Ê Iœ
" x
cos bx
a# b#
I Ê Ša
#
b# b# ‹ I
eax b#
œ
(a cos bx b sin bx)
(a cos bx b sin bx) C
eax a# b#
39. ln (ax)
aeax b#
sin bx
ÐÑ
1
ÐÑ
x
I œ x ln (ax) ' ˆ "x ‰ x dx œ x ln (ax) x C 40. ln (ax) " x
Iœ
" 3
ÐÑ
x#
ÐÑ
" 3
x$
x$ ln (ax) ' ˆ "x ‰ Š x3 ‹ dx œ $
41.
' 1 dxsin x œ '
42.
' 1 sin dxx cos x œ '
œ'
2 dz ‹ 1 z# 1 Š 2z # ‹ 1 z
Š
2 dz (1 z)#
" 3
x$ ln (ax) 9" x$ C
œ
2 1z
œ'
2 dz ‹ 1 z# 2z 1 z# 1 Š ‹ 1 z# 1 z#
Š
Cœ
2 1 tan ˆ #x ‰
2 dz 1 z# 2z 1 z#
C
œ'
œ ln k1 zk C
dz 1z
œ ln ¸tan ˆ x# ‰ 1¸ C 43.
'01Î2 1 dxsin x œ '01
44.
'11ÎÎ32
45.
) '01Î2 2 dcos '1 ) œ 0
œ
46.
dx 1 cos x
1 3È 3
'12Î12Î3
œ
" È2
1
Š
2 dz (1 z)#
œ '1ÎÈ3
2 dz ‹ 1 z# 1 z# 1Š ‹ 1 z#
Š
2 dz ‹ 1 z# 1 z# 2Š ‹ 1 z#
cos ) d) sin ) cos ) sin )
È$ z# “ 4 "
' sin t dt cos t œ ' œ
'11ÎÈ3
œ '0
1
œ '0
1
"
œ 1 2 z ‘ ! œ (1 2) œ 1
dz z#
" œ 1z ‘ "ÎÈ$ œ È3 1
2 dz 2 2z# 1 z#
œ '0
1
œ
2 dz z# 3
2 È3
’tan"
z È3 “
" !
œ
2 È3
tan"
" È3
È 31 9
œ ’ "# ln z
47.
œ
2 dz ‹ 1 z# 2z 1Š ‹ 1 z#
Š
È3
œ '1
È3
z# ‹ Š 2 dz# ‹ 1 z# 1z
Š1 Ô 2z Š1
z# ‹
× 2z # Š 1 z# ‹Ø Õ Š1 z# ‹
œ '1
œ Š "# ln È3 43 ‹ ˆ0 4" ‰ œ
2 dz ‹ 1 z# 2z 1 z# Š ‹ 1 z# 1 z#
Š
œ'
2 dz 2z 1 z#
œ'
2 a1 z# b dz 2z 2z$ 2z 2z$
ln 3 4
1 z# 2z
œ
" 4
œ
" È2
1 2 ln ¹ zz ¹C 1 È2
(ln 3 2) œ
" #
dz
" #
2 dz (z 1)# 2
È3
œ '1
Šln È3 1‹
È
tan ˆ t ‰ 1 È2
ln º tan ˆ #t ‰ 1 È2 º C #
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535
536 48.
Chapter 8 Techniques of Integration
t dt ' 1cos cos 'Š t œ
œ'
49.
a1 z# b dz a1 z # b z #
1 z# ‹ Š 2 dz# ‹ 1 z# 1 z # 1 Š 1 z# ‹ 1z
œ'
dz z # a1 z # b
œ' '
' sec ) d) œ ' cosd) ) œ ' Š
2 dz ‹ 1 z# # Š 1 z# ‹ 1 z
2 a1 z# b dz a1 z # b # a 1 z # b a 1 z # b dz 1 z#
œ'
dz z#
œ'
2 dz 1 z#
œ'
œ ln k1 zk ln k1 zk C œ ln »
50.
' csc ) d) œ ' sind)) œ ' Š
2 dz ‹ 1 z# 2z Š ‹ 1 z#
1 tan Š )# ‹
2'
1 tan Š )# ‹ »
œ'
dz z
œ'
dz z# 1
2 dz (1 z)(1 z)
2 a1 z# b dz a1 z # b a 1 z # 1 z # b
œ "z 2 tan" z C œ cot ˆ #t ‰ t C œ'
dz 1z
'
dz 1z
C
œ ln kzk C œ ln ¸tan #) ¸ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 9 FIRST-ORDER DIFFERENTIAL EQUATIONS 9.1 SOLUTIONS, SLOPE FIELDS AND EULER'S METHOD 1. y w œ x y Ê slope of 0 for the line y œ x. For x, y 0, y w œ x y Ê slope 0 in Quadrant I. For x, y 0, y w œ x y Ê slope 0 in Quadrant III. For kyk kxk, y 0, x 0, y w œ x y Ê slope 0 in Quadrant II above y œ x. For kyk kxk, y 0, x 0, y w œ x y Ê slope 0 in Quadrant II below y œ x. For kyk kxk, x 0, y 0, y w œ x y Ê slope 0 in Quadrant IV above y œ x. For kyk kxk, x 0, y 0, y w œ x y Ê slope 0 in Quadrant IV below y œ x. All of the conditions are seen in slope field (d). 2. y w œ y 1 Ê slope is constant for a given value of y, slope is 0 for y œ 1, slope is positive for y 1 and negative for y 1. These characteristics are evident in slope field (c).
3. y w œ xy Ê slope œ 1 on y œ x and 1 on y œ x. y w œ xy Ê slope œ 0 on the y-axis, excluding a0, 0b, and is undefined on the x-axis. Slopes are positive for x 0, y 0 and x 0, y 0 (Quadrants II and IV), otherwise negative. Field (a) is consistent with these conditions.
4. y w œ y2 x2 Ê slope is 0 for y œ x and for y œ x. For kyk kxk slope is positive and for kyk kxk slope is negative. Field (b) has these characteristics.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
538
Chapter 9 First-Order Differential Equations
5.
6.
7. y œ 1 '1 at yatbbdt Ê x
8. y œ '1 1t dt Ê x
dy dx
dy dx
œ x yaxb; ya1b œ 1 '1 at yatbbdt œ 1; 1
œ x1 ; ya1b œ '1 1t dt œ 0; 1
9. y œ 2 '0 a1 yatbbsin t dt Ê x
dy dx
dy dx
dy dx
œ x y, ya1b œ 1
œ x1 , ya1b œ 0
œ a1 yaxbbsin x; ya0b œ 2 '0 a1 yatbbsin t dt œ 2; 0
dy dx
œ a1 ybsin x,
ya0b œ 2 10. y œ 1 '0 yatb dt Ê x
dy dx
œ yaxb; ya0b œ 1 '0 yatb dt œ 1; 0
1 ‰ # (.5)
11. y" œ y! Š1
y! x! ‹
dx œ 1 ˆ1
y# œ y " Š 1
y" x" ‹
dx œ 0.25 ˆ1
y$ œ y # Š 1
y# x# ‹
dx œ 0.3 ˆ1
dy dx
ˆ x" ‰ y œ 1 Ê P(x) œ
Ê yœ
" x
" x
' x † 1 dx œ x" Š x#
Ê y(3.5) œ
#
3.5 #
4 3.5
œ
4.25 7
, Q(x) œ 1 Ê
œ y, ya0b œ 1
œ 0.25,
0.25 ‰ (.5) 2.5
0.3 ‰ 3 (.5)
dy dx
œ 0.3,
œ 0.75;
' P(x) dx œ ' x" dx œ ln kxk œ ln x, x 0 Ê v(x) œ eln x œ x
C‹ ; x œ 2, y œ 1 Ê 1 œ 1
C 2
Ê C œ 4 Ê y œ
x #
4 x
¸ 0.6071
12. y" œ y! x! (1 y! ) dx œ 0 1(1 0)(.2) œ .2, y# œ y" x" (1 y" ) dx œ .2 1.2(1 .2)(.2) œ .392, y$ œ y# x# (1 y# ) dx œ .392 1.4(1 .392)(.2) œ .5622; dy 1 y
œ x dx Ê ln k1 yk œ
x# #
C; x œ 1, y œ 0 Ê ln 1 œ
" #
#
C Ê C œ #" Ê ln k1 yk œ x#
" #
#
Ê y œ 1 ea1x bÎ2 Ê y(1.6) ¸ .5416 13. y" œ y! (2x! y! 2y! ) dx œ 3 [2(0)(3) 2(3)](.2) œ 4.2, y# œ y" (2x" y" 2y" ) dx œ 4.2 [2(.2)(4.2) 2(4.2)](.2) œ 6.216, y$ œ y# (2x# y# 2y# ) dx œ 6.216 [2(.4)(6.216) 2(6.216)](.2) œ 9.6969; dy dx
œ 2y(x 1) Ê
dy y
œ 2(x 1) dx Ê ln kyk œ (x 1)# C; x œ 0, y œ 3 Ê ln 3 œ 1 C Ê C œ ln 3 1
Ê ln y œ (x 1)# ln 3 1 Ê y œ eÐx1Ñ ln 31 œ eln 3 ex 2x œ 3exÐx2Ñ Ê y(.6) ¸ 14.2765 #
#
14. y" œ y! y#! (1 2x! ) dx œ 1 1# [1 2(1)](.5) œ .5, y# œ y" y#" (1 2x" ) dx œ .5 (.5)# [1 2(.5)](.5) œ .5, y$ œ y# y## (1 2x# ) dx œ .5 (.5)# [1 2(0)](.5) œ .625; dy y#
œ (1 2x) dx Ê y" œ x x# C; x œ 1, y œ 1 Ê 1 œ 1 (1)# C Ê C œ 1 Ê
Ê yœ
" 1 x x#
Ê y(.5) œ
" 1 .5 (.5)#
œ4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" y
œ 1 x x#
Section 9.1 Solutins, Slope Fields and Euler's Method
539
#
15. y" œ y! 2x! ex! dx œ 2 2(0)(.1) œ 2, # # y# œ y" 2x" ex" dx œ 2 2(.1) eÞ1 (.1) œ 2.0202, # # y$ œ y# 2x# ex# dx œ 2.0202 2(.2) eÞ2 (.1) œ 2.0618, # # # # dy œ 2xex dx Ê y œ ex C; y(0) œ 2 Ê 2 œ 1 C Ê C œ 1 Ê y œ ex 1 Ê y(.3) œ eÞ3 1 ¸ 2.0942 16. y" œ y! ay! ex! b dx œ 2 a2 † e0 b (.5) œ 3, y2 œ y1 ay1 ex1 b dx œ 3 a3 † e0.5 b (.5) œ 5.47308, y3 œ y2 ay2 ex2 b dx œ 5.47308 a5.47308 † e1.0 b (.5) œ 12.9118, dy dx
œ y ex Ê
Ê y œ 2ee 17. y" y# y$ y% y& dy y
x
œ ex dx Ê lnlyl œ ex C; x œ 0, y œ 2 Ê ln 2 œ 1 C Ê C œ ln 2 1 Ê lnlyl œ ex ln 2 1
dy y
"
Ê ya1.5b œ 2ee
1.5
"
¸ 65.0292
œ 1 1(.2) œ 1.2, œ 1.2 (1.2)(.2) œ 1.44, œ 1.44 (1.44)(.2) œ 1.728, œ 1.728 (1.728)(.2) œ 2.0736, œ 2.0736 (2.0736)(.2) œ 2.48832; œ dx Ê ln y œ x C" Ê y œ Cex ; y(0) œ 1 Ê 1 œ Ce! Ê C œ 1 Ê y œ ex Ê y(1) œ e ¸ 2.7183
18. y" œ 2 ˆ 21 ‰ (.2) œ 2.4, ‰ y# œ 2.4 ˆ 2.4 1.2 (.2) œ 2.8, ‰ y$ œ 2.8 ˆ 2.8 1.4 (.2) œ 3.2, 3.2 y% œ 3.2 ˆ 1.6 ‰ (.2) œ 3.6, ‰ y& œ 3.6 ˆ 3.6 1.8 (.2) œ 4; dy y
œ
dx x
Ê ln y œ ln x C Ê y œ kx; y(1) œ 2 Ê 2 œ k Ê y œ 2x Ê y(2) œ 4 #
19. y" œ 1 ’ (È1) “ (.5) œ .5, 1 #
.5) y# œ .5 ’ (È “ (.5) œ .39794, 1.5 #
y$ œ .39794 ’ (.39794) “ (.5) œ .34195, È2 #
y% œ .34195 ’ (.34195) È2.5 “ (.5) œ .30497, y& œ .27812, y' œ .25745, y( œ .24088, y) œ .2272; dy " dx È y# œ Èx Ê y œ 2 x C; y(1) œ 1 Ê 1 œ 2 C Ê C œ 1 Ê y œ
" 1 #È x
Ê y(5) œ
" 1 #È 5
20. y" œ 1 a0 † sin 1b ˆ "3 ‰ œ 1, y# œ 1 ˆ "3 † sin 1‰ ˆ 3" ‰ œ 1.09350, y$ œ 1.09350 ˆ 23 † sin 1.09350‰ ˆ 3" ‰ œ 1.29089, y% œ 1.29089 ˆ 33 † sin 1.29089‰ ˆ 3" ‰ œ 1.61125, y& œ 1.61125 ˆ 43 † sin 1.61125‰ ˆ 3" ‰ œ 2.05533,
y' œ 2.05533 ˆ 53 † sin 2.05533‰ ˆ 3" ‰ œ 2.54694; y w œ x sin y Ê csc y dy œ x dx Ê lnlcsc y cot yl œ 12 x2 C Ê csc y cot y œ e 2 x C œ Ce 2 x 1 2
Ê
1 cos y sin y
1 2
2 2 2 œ Ce 2 x Ê cotˆ 2y ‰ œ Ce 2 x ; ya0b œ 1 Ê cotˆ 21 ‰ œ Ce0 œ C Ê cotˆ 2y ‰ œ cotˆ 21 ‰e 2 x 1
1
2 Ê y œ 2 cot1 Šcotˆ 12 ‰e 2 x ‹, ya2b œ 2 cot1 ˆcotˆ 12 ‰e2 ‰ œ 2.65591 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
¸ .2880
540
Chapter 9 First-Order Differential Equations
21. y œ 1 x a1 x0 y0 bex x0 Ê yax0 b œ 1 x0 a1 x0 y0 bex0 x0 œ 1 x0 a1 x0 y0 ba1b œ y0 dy dx
œ 1 a1 x0 y0 bex x0 Ê y œ 1 x a1 x0 y0 bex x0 œ
dy dx
xÊ
dy dx
œ xy
22. y w œ faxb, yax0 b œ y0 Ê y œ 'x fatbdt C, yax0 b œ 'x fatbdt C œ C Ê C œ y0 Ê y œ 'x fatbdt y0 x
x0
x
0
0
0
23-34. Example CAS commands: Maple: ode := diff( y(x), x ) = y(x); icA := [0, 1]; icB := [0, 2]; icC := [0,-1]; DEplot( ode, y(x), x=0..2, [icA,icB,icC], arrows=slim, linecolor=blue, title="#23 (Section 9.1)" ); Mathematica: To plot vector fields, you must begin by loading a graphics package. <
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 9.1 Solutins, Slope Fields and Euler's Method 23.
24.
25.
26.
27.
28.
35.
dy dx
œ 2xex , ya0b œ 2 Ê yn1 œ yn 2xn exn dx œ yn 2xn exn a0.1b œ yn 0.2xn exn 2
2
2
2
On a TI-84 calculator home screen, type the following commands: 2 STO > y: 0 STO > x: y (enter) y 0.2*x*e^(x^2) STO > y: x 0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a1b ¸ 3.45835 2 2 2 The exact solution: dy œ 2xex dx Ê y œ ex C; ya0b œ 2 œ e0 C Ê C œ 1 Ê y œ 1 ex Ê yexact a1b œ 1 e ¸ 3.71828 36.
dy dx
œ 2y2 ax 1b, ya2b œ 12 Ê yn1 œ yn 2yn2 axn 1bdx œ yn 0.2 yn2 axn 1b
On a TI-84 calculator home screen, type the following commands: 0.5 STO > y: 2 STO > x: y (enter) y 0.2*y2 ax 1b STO > y: x 0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a2b ¸ 0.19285 œ 2y2 ax 1b Ê
dy dx
ya2b œ 12 Ê
œ a2b2 2a2b C œ C Ê C œ 2 Ê
ya3b œ
1 1 Î2
1 a3 b 2 2 a 3 b 2
dy y2
œ a2x 2bdx Ê 1y œ x2 2x C Ê
The exact solution:
1 y
1 y
œ x2 2x C
œ x2 2x 2 Ê y œ
1 x2 2x 2
œ 0.2
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541
542 37.
Chapter 9 First-Order Differential Equations dy dx
œ
Èx y ,y
0ß ya0b œ " Ê yn1 œ yn
È xn yn dx
œ yn
È xn yn a0.1b
œ yn 0."
È xn yn
On a TI-84 calculator home screen, type the following commands: 1 STO > y: 0 STO > x: y (enter) y 0.1*(Èx /y) STO > y: x 0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a1b ¸ 1.5000 The exact solution: dy œ Ê 38.
dy dx
y2 2
œ 23 x3/2
1 2
Èx y dx
Ê y dy œ Èx dx Ê
y2 2
2
œ 32 x3/2 C; aya20bb œ
12 2
œ
1 2
œ 23 a0b3/2 C Ê C œ
1 2
Ê y œ É 43 x3/2 1 Ê yexact a"b œ É 43 a1b3/2 1 ¸ 1.5275
œ 1 y2 , ya0b œ 0 Ê yn1 œ yn a1 yn2 bdx œ yn a1 yn2 ba0.1b œ yn 0.1a1 yn2 b
On a TI-84 calculator home screen, type the following commands: 0 STO > y: 0 STO > x: y (enter) y 0.1*(1 y2 ) STO > y: x 0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a1b ¸ 1.3964 The exact solution: dy œ a1 y2 bdx Ê
dy 1 y2
œ dx Ê tan1 y œ x C; tan1 ya0b œ tan1 0 œ 0 œ 0 C Ê C œ 0
Ê tan1 y œ x Ê y œ tan x Ê yexact a"b œ tan 1 ¸ 1.5574 39.
Example CAS commands: Maple: ode := diff( y(x), x ) = x + y(x);ic := y(0)=-7/10; x0 := -4;x1 := 4;y0 := -4; y1 := 4; b := 1; P1 := DEplot( ode, y(x), x=x0..x1, y=y0..y1, arrows=thin, title="#39(a) (Section 9.1)" ): P1; Ygen := unapply( rhs(dsolve( ode, y(x) )), x,_C1 ); # (b) P2 := seq( plot( Ygen(x,c), x=x0..x1, y=y0..y1, color=blue ), c=-2..2 ): # (c) display( [P1,P2], title="#39(c) (Section 9.1)" ); CC := solve( Ygen(0,C)=rhs(ic), C ); # (d) Ypart := Ygen(x,CC); P3 := plot( Ypart, x=0..b, title="#39(d) (Section 9.1)" ): P3; euler4 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/4 ): # (e) P4 := odeplot( euler4, [x,y(x)], x=0..b, numpoints=4, color=blue ): display( [P3,P4], title="#39(e) (Section 9.1)" ); euler8 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/8 ): # (f) P5 := odeplot( euler8, [x,y(x)], x=0..b, numpoints=8, color=green ): euler16 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/16 ): P6 := odeplot( euler16, [x,y(x)], x=0..b, numpoints=16, color=pink ): euler32 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/32 ): P7 := odeplot( euler32, [x,y(x)], x=0..b, numpoints=32, color=cyan ): display( [P3,P4,P5,P6,P7], title="#39(f) (Section 9.1)" ); << N|h | `percent error` >, # (g) < 4 | (x1-x0)/ 4 | evalf[5]( abs(1-eval(y(x),euler4(b))/eval(Ypart,x=b))*100 ) >, < 8 | (x1-x0)/ 8 | evalf[5]( abs(1-eval(y(x),euler8(b))/eval(Ypart,x=b))*100 ) >, < 16 | (x1-x0)/16 | evalf[5]( abs(1-eval(y(x),euler16(b))/eval(Ypart,x=b))*100 ) >, < 32 | (x1-x0)/32 | evalf[5]( abs(1-eval(y(x),euler32(b))/eval(Ypart,x=b))*100 ) > >;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 9.2 First-Order Linear Differential Equations 39-42. Example CAS commands: Mathematica: (assigned functions, step sizes, and values for initial conditions may vary) Problems 39 - 42 involve use of code from Problems 23 - 34 together with the above code for Euler's method. 9.2 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 1. x
dy dx
y œ ex Ê
dy dx
ˆ x" ‰ y œ
ex x
, P(x) œ
" x
, Q(x) œ
ex x
' P(x) dx œ ' "x dx œ ln kxk œ ln x, x 0 Ê v(x) œ e' PÐxÑ dx œ eln x œ x " ' C y œ v(x) v(x) Q(x) dx œ "x ' x ˆ ex ‰ dx œ "x aex Cb œ e x ,x0 x
x
2. ex
dy dx
2ex y œ 1 Ê
dy dx
2y œ ex , P(x) œ 2, Q(x) œ ex
' P(x) dx œ ' 2 dx œ 2x Ê v(x) œ e' PÐxÑ dx œ e2x y œ e" ' e2x † ecx dx œ e" ' ex dx œ e" aex Cb œ ecx Cec2x 2x
2x
3. xyw 3y œ
sin x x#
,x0Ê
2x
dy dx
ˆ x3 ‰ y œ
, P(x) œ
sin x x$
3 x
, Q(x) œ
sin x x$
' 3x dx œ 3 ln kxk œ ln x$ , x 0 Ê v(x) œ eln x œ x$ y œ x" ' x$ ˆ sinx x ‰ dx œ x" ' sin x dx œ x" ( cos x C) œ Cxcos x , x 0 $
$
$
$
$
4. yw (tan x) y œ cos# x, 1# x
1 #
Ê
dy dx
$
(tan x) y œ cos# x, P(x) œ tan x, Q(x) œ cos# x
sin x 1 1 " ln Ðcos xÑ " ' tan x dx œ ' cos œ (cos x)" x dx œ ln kcos xk œ ln (cos x) , # x # Ê v(x) œ e y œ (cos"x) " ' (cos x)" † cos# x dx œ (cos x)' cos x dx œ (cos x)(sin x C) œ sin x cos x C cos x
5. x
dy dx
2y œ 1 "x , x 0 Ê
dy dx
" x
ˆ 2x ‰ y œ
" x#
, P(x) œ
2 x
, Q(x) œ
" x
" x#
' 2x dx œ 2 ln kxk œ ln x# , x 0 Ê v(x) œ eln x œ x# y œ x" ' x# ˆ x" x" ‰ dx œ x" ' (x 1) dx œ x" Š x# x C‹ œ #" x" xC , x 0 #
#
#
#
#
6. (1 x) yw y œ Èx Ê
dy dx
#
ˆ 1 " x ‰ y œ
Èx 1 x
, P(x) œ
#
" 1x
, Q(x) œ
Èx 1x
' 1 " x dx œ ln (1 x), since x 0 Ê v(x) œ eln Ð1 xÑ œ 1 Èx 2x C y œ 1 " x ' (1 x) Š 1 x ‹ dx œ 1 " x ' Èx dx œ ˆ 1 " x ‰ ˆ 32 x$Î# C‰ œ 3(1 x) 1 x $Î#
7.
dy dx
"# y œ
Ê yœ 8.
dy dx
dy dx
" e xÎ2
exÎ2 Ê P(x) œ "# , Q(x) œ
" #
exÎ2 Ê ' P(x) dx œ "# x Ê v(x) œ e xÎ2
' ecxÎ2 ˆ #" exÎ2 ‰ dx œ exÎ2 ' #" dx œ exÎ2 ˆ #" x C‰ œ #" xexÎ2 CexÎ2
2y œ 2xec2x Ê P(x) œ 2, Q(x) œ 2xec2x Ê ' P(x) dx œ ' 2 dx œ 2x Ê v(x) œ e2x
Ê yœ 9.
" #
" e2x
' e2x a2xec2x b dx œ e" ' 2x dx œ ec2x ax# Cb œ x# ec2x Cec2x 2x
ˆ x" ‰ y œ 2 ln x Ê P(x) œ x" , Q(x) œ 2 ln x Ê ' P(x) dx œ '
Ê v(x) œ ec ln x œ
" x
" x
dx œ ln x, x 0
Ê y œ x' ˆ x" ‰ (2 ln x) dx œ x c(ln x)# Cd œ x (ln x)# Cx
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543
544 10.
Chapter 9 First-Order Differential Equations dy dx
ˆ x2 ‰ y œ
, x 0 Ê P(x) œ
cos x x#
" x#
Ê v(x) œ eln x œ x# Ê y œ #
11.
ds dt
t1 (t 1)$
ˆ t 4 1 ‰ s œ
2 x
#
Ê P(t) œ
4 t1
12. (t 1)
" (t1)%
dr d)
2s œ 3(t 1)
ds dt
" (t 1)#
sin x C x#
' P(t) dt œ ' t41 dt œ 4 ln kt 1k œ ln (t 1)% ' (t 1)% ’ (tt1)" “ dt œ (t "1) ' at# 1b dt t1 (t 1)$
Ê
$
t (t 1)%
Ê
ˆ t 2 1 ‰ s œ 3
ds dt
(sin x C) œ
%
C (t 1)% " (t 1)$
Ê P(t) œ
2 t1
, Q(t) œ 3 (t 1)$
#
Ê
13.
t$ 3(t 1)%
#
dx œ 2 ln kxk œ ln x# , x 0
2 x
' P(t) dt œ ' t 2 1 dt œ 2 ln kt 1k œ ln (t 1)# Ê v(t) œ eln Ðt1Ñ œ (t 1)# s œ (t "1) ' (t 1)# c3 (t 1)$ d dt œ (t "1) ' c3(t 1)# (t 1)" d dt
Ê
#
" (t 1)#
œ
$
Š t3 t C‹ œ
#
, Q(t) œ
%
" (t 1)%
Ê ' P(x) dx œ '
cos x x#
' x# ˆ cosx x ‰ dx œ x" ' cos x dx œ x"
Ê v(t) œ eln Ðt1Ñ œ (t 1)% Ê s œ œ
, Q(x) œ
#
c(t 1)$ ln kt 1k Cd œ (t 1) (t 1)# ln (t 1)
, t 1
C (t 1)#
' P()) d) œ ' cot ) d) œ ln ksin )k Ê v()) œ eln sin r œ sin" ) ' (sin ))(sec )) d) œ sin" ) ' tan ) d) œ sin" ) aln ksec )k Cb
(cot )) r œ sec ) Ê P()) œ cot ), Q()) œ sec ) Ê 1 #
œ sin ) because 0 )
Ê
k
)k
œ (csc )) aln ksec )k Cb 14. tan ) Ê
r œ sin# ) Ê
r tan )
œ
sin# ) tan )
Ê
dr d)
(cot )) r œ sin ) cos ) Ê P()) œ cot ), Q()) œ sin ) cos )
' P()) d) œ ' cot ) d) œ ln ksin )k œ ln (sin )) since 0 ) 1# Ê v()) œ eln Ðsin )Ñ œ sin ) r œ sin" ) ' (sin )) (sin ) cos )) d) œ sin" ) ' sin# ) cos ) d) œ ˆ sin" ) ‰ Š sin3 ) C‹ œ sin3 ) sinC ) #
2y œ 3 Ê P(t) œ 2, Q(t) œ 3 Ê ' P(t) dt œ ' 2 dt œ 2t Ê v(t) œ e2t Ê y œ œ e"2t ˆ 3# e2t C‰ ; y(0) œ 1 Ê 3# C œ 1 Ê C œ "# Ê y œ 3# "# ec2t
15.
dy dt
16.
dy dt
œ
" t#
dy d)
2y t
œ t# Ê P(t) œ
$
" k) k
#
sin ) )
" )
Ê P()) œ
, Q()) œ
sin ) )
Ê
8 5
; y ˆ 1# ‰ œ 1 Ê C œ
1 #
C 4
œ 1 Ê C œ 12 5 Ê yœ
" ) #
" ) 1 2)
Ê yœ
Ê y œ ") cos )
#
Q(x) œ œ ex
#
dy dx
#
2 ax# xb y œ x#
e (x 1)#
' (x " 1)
Ê
#
ex x1
Ê
dy dx
18 1#
' sin ) d) œ ") ( cos ) C) ' P()) d) œ 2 ln k)k
Ê v()) œ ec2 ln )
dx œ ex ’ #
(x 1) "
Ê C œ 6 Ê y œ 6ex
1
‰ # 2 Ê y œ )# sec ) ˆ 18 1# 2 )
1) 2 ’ x(x x1 “ y œ
' P(x) dx œ ' 2x dx œ x# #
#
12 5t#
' a)# b a)# sec ) tan )b d) œ )# ' sec ) tan ) d) œ )# (sec ) C) œ )# sec ) C)# ;
y ˆ 13 ‰ œ 2 Ê 2 œ Š 19 ‹ (2) C Š 19 ‹ Ê C œ 19. (x 1)
k k
ˆ 2) ‰ y œ )# sec ) tan ) Ê P()) œ 2) , Q()) œ )# sec ) tan ) Ê
œ )# Ê y œ
t$ 5
' at# b at# b dt
Ê ' P()) d) œ ln k)k Ê v()) œ eln ) œ k)k
' k)k ˆ sin) ) ‰ d) œ ") ' ) ˆ sin) ) ‰ d) for ) Á 0 C )
" t#
#
&
ˆ ") ‰ y œ
' 3e2t dt
" e2t
, Q(t) œ t# Ê ' P(t) dt œ 2 ln ktk Ê v(t) œ eln t œ t# Ê y œ
#
œ ") cos ) dy d)
2 t
' t% dt œ t" Š t5 C‹ œ t5 tC ; y(2) œ 1
Ê yœ
18.
dr d)
$
Ê
17.
dr d)
x#
#
ex (x 1)#
Ê
dy dx
2xy œ
#
Ê v(x) œ ex Ê y œ
" e
x#
#
ex (x 1)#
Ê P(x) œ 2x,
' ex# ’ (x e 1) “ dx x#
#
#
C“ œ xe 1 Cex ; y(0) œ 5 Ê 0 1 1 C œ 5 Ê " C œ 5
#
ex x1
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k k
Section 9.2 First-Order Linear Differential Equations 20.
dy dx
xy œ x Ê P(x) œ x, Q(x) œ x Ê ' P(x) dx œ ' x dx œ " ex# Î2
œ
21.
dy dt
#
Šex Î2 C‹ œ 1
22. (a)
; y(0) œ 6 Ê 1 C œ 6 Ê C œ 7 Ê y œ 1
C ex# Î2
du dt
' ˆekt ‰ (0) dt œ ekt (0 C) œ Cekt ; y(0) œ y!
" eckt
k m
u œ 0 Ê P(t) œ " ektÎm
Ê yœ (b)
23. x ' " cos x
" x
" ex# Î2
' ex Î2 † x dx #
7 ex# Î2
; u(0) œ u! Ê
C ekÐ0ÑÎm
Ê u(t) œ ektÎm
œ u! Ê C œ u! Ê u œ u! eÐkÎmÑt
œ mk dt Ê ln u œ mk t C Ê u œ eakÎmbtC Ê u œ eÐkÎmÑt † eC . Let eC œ C1 Þ
† C1 and ua0b œ u0 œ
† C1 œ C1 . So u œ u0 eÐkÎmÑt
1 eÐkÎmÑa0b
dx œ x aln kxk Cb œ x ln kxk Cx Ê (b) is correct
' cos x dx œ cos" x (sin x C) œ tan x cosC x
Ê ln
(b)
du u
Ê C œ y! Ê y œ y! ekt
' P(t) dt œ ' mk dt œ mk t œ mkt
, Q(t) œ 0 Ê ktÎm
1 eÐkÎmÑt
25. Steady State œ
26. (a)
k m
' ektÎm † 0 dt œ e C
œ mk u Ê
du dt
Then u œ
di dt
" #
œ
R L
V R Rt L
and we want i œ " #
Ê RL ln " i
iœ0 Ê
27. (a) t œ (b) t œ di dt
L R
Ê iœIe Ê iœ
3L R 2L R
R L
Ê iœ iœ
Ê iœ
V L
" eRtÎL
(b) i(0) œ 0 Ê (c) i œ
V R
Ê
di dt
" #
ˆ VR ‰ Ê
œt Ê tœ
" #
ÐRtÎLÑÐLÎRÑ
V R V R
" #
L R
Ê (b) is correct
ˆ VR ‰ œ
V R
a1 eRtÎL b Ê
" #
œ 1 eRtÎL Ê "# œ eRtÎL
ln 2 sec
C" cRtÎL di œ RL dt Ê ln i œ Rt œ CeRtÎL ; i(0) œ I Ê I œ C L C" Ê i œ e e
Ê i œ IeRtÎL amp œ I eRtÎL Ê eRtÎL œ
" #I
(c) t œ
28. (a)
#
Ê v(x) œ ex Î2 Ê y œ
ky œ 0 Ê P(t) œ k, Q(t) œ 0 Ê ' P(t) dt œ ' k dt œ kt Ê v(t) œ ekt
Ê yœ
24.
x# #
Ê Rt L œ ln
R L
L R
ln 2 sec
œ I e amp
ÐR/LÑÐ2LÎRÑ
Ê P(t) œ
œ ln 2 Ê t œ
t
a1 eÐR/LÑÐ3LÎRÑ b œ a1 e
" #
bœ
, Q(t) œ
' eRtÎL ˆ VL ‰ dt œ e "
RtÎL
V L
V R V R
a1 ec3 b ¸ 0.9502
V R V R
a1 ec2 b ¸ 0.8647
Ê
'
P(t) dt œ '
RL eRtÎL ˆ VL ‰ C‘ œ
V R
R L
amp, or about 95% of the steady state value amp, or about 86% of the steady state value
dt œ
Rt L
Ê v(t) œ eRtÎL
Ce ÐRÎLÑt
C œ 0 Ê C œ VR Ê i œ VR VR ecRtÎL R ˆ R ‰ ˆ VR ‰ œ VL Ê i œ œ 0 Ê di dt L i œ 0 L
V R
V R
is a solution of Eq. (11); i œ Ce ÐRÎLÑt
29. y w y œ y2 ; we have n œ 2, so let u œ y12 œ y1 . Then y œ u1 and
du dx
dy œ 1y2 dx Ê
dy dx
œ y2 du dx
' dx 1 Ê u2 du œ u2 Ê du œ ex as the integrating factor, we have dx u dx u œ 1. With e d C 1 1 x x x x ‰ ex ˆ du dx u œ dx ae ub œ e . Integrating, we get e u œ e C Ê u œ 1 ex œ y Ê y œ 1 C œ ex
30. y w y œ xy2 ; we have n œ 2, so let u œ y1 . Then y œ u1 and 1 Substituting:u2 du œ xu2 Ê dx u
‰ ex ˆ du dx u œ
d x dx ae ub
du dx
' dx
u œ x. Using e
œ x ex Ê ex u œ ex a1 xb C Ê u œ
du dx
œ y2 dy dx Ê
dy dx
ex ex C
2 du œ y2 du dx œ u dx .
œ ex as an integrating factor: e x a1 x b C ex
Ê y œ u 1 œ
ex ex xex C
31. xy w y œ y2 Ê y w ˆ 1x ‰y œ ˆ 1x ‰y2 . Let u œ y1a2b œ y3 Ê y œ u1/3 and y2 œ u2/3 . dy du 2 dy w ˆ 1 ‰ˆ du ‰ 2 ˆ 1 ‰ˆ du ‰ˆ 2/3 ‰. Thus we have dx œ 3y dx Ê y œ dx œ 3 dx ay b œ 3 dx u
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Chapter 9 First-Order Differential Equations
ˆ 13 ‰ˆ du ‰ˆ 2/3 ‰ ˆ 1x ‰u1/3 œ ˆ 1x ‰u2/3 Ê dx u ' 3x dx
e
3
œ e3ln x œ eln x œ x3 . Thus
Ê y œ ˆ1
du dx
d 3 dx ax ub
ˆ 3x ‰u œ ˆ 3x ‰1. The integrating factor, vaxb, is
œ ˆ 3x ‰x3 œ 3x2 Ê x3 u œ x3 C Ê u œ 1
C x3
œ y3
C ‰1/3 x3
32. x2 y w 2xy œ y3 Ê y w ˆ 2x ‰y œ ˆ x12 ‰y3 . Paxb œ ˆ 2x ‰, Qaxb œ ˆ x12 ‰, n œ 3. Let u œ y13 œ y2 . du ˆ2‰ ˆ1‰ ˆ x4 ‰u œ x22 . Let the integrating factor, vaxb, be Substituting gives du dx a2b x u œ 2 x2 Ê dx c4
e' a x bdx œ eln x
c4
œ x4 . Thus
2 Ê y œ ˆ 5x Cx4 ‰
d 4 dx ax
ub œ 2x6 Ê x4 u œ 25 x5 C Ê u œ
2 4 5x Cx
œ y2
1/2
9.3 APPLICATIONS 1. Note that the total mass is 66 7 œ 73 kg, therefore, v œ v0 eakÎmbt Ê v œ 9e3.9tÎ73 3.9tÎ73 (a) satb œ ' 9e3.9tÎ73 dt œ 2190 C 13 e
Since sa0b œ 0 we have C œ
2190 13
3.9tÎ73 ‰ ˆ and lim satb œ lim 2190 œ 13 1 e tÄ_
tÄ_
2190 13
¸ 168.5
The cyclist will coast about 168.5 meters. 73 ln 9 (b) 1 œ 9e3.9tÎ73 Ê 3.9t 73 œ ln 9 Ê t œ 3.9 ¸ 41.13 sec It will take about 41.13 seconds. 2. v œ v0 eakÎmbt Ê v œ 9ea59,000Î51,000,000bt Ê v œ 9e59tÎ51,000 (a) satb œ ' 9e59tÎ51,000 dt œ 459,0000 e59tÎ51,000 C 59 Since sa0b œ 0 we have C œ
459,0000 59
ˆ1 e59tÎ51,000 ‰ œ and lim satb œ lim 459,0000 59 tÄ_
tÄ_
459,0000 59
¸ 7780 m
The ship will coast about 7780 m, or 7.78 km. ln 9 59t (b) 1 œ 9e59tÎ51,000 Ê 51,000 œ ln 9 Ê t œ 51,000 ¸ 1899.3 sec 59 It will take about 31.65 minutes. 3. The total distance traveled œ v0km Ê a2.75bak39.92b œ 4.91 Ê k œ 22.36. Therefore, the distance traveled is given by the function satb œ 4.91ˆ1 ea22.36/39.92bt ‰. The graph shows satb and the data points.
4.
a0.80ba49.90b œ 1.32 Ê k œ 998 k 33 20 We know that v0km œ 1.32 and mk œ 33a998 œ . 49.9b 33 v0 m ˆ ak/mbt ‰ Using Equation 3, we have: satb œ k 1 e œ 1.32ˆ1 v0 m k
œ coasting distance Ê
e20t/33 ‰ ¸ 1.32a1 e0.606t b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 9.3 Applications 5. y œ mx Ê
œmÊ
y x
orthogonals:
dy dx
xy y x2 w
œ 0 Ê y w œ yx . So for
œ yx Ê y dy œ x dx Ê
y2 2
x2 2
œC
Ê x y œ C1 2
2
6. y œ cx2 Ê Ê yw œ
y x2
2y x .
œcÊ
x2 y 2xy x4
œ 0 Ê x2 y w œ 2xy
w
So for the orthogonals:
dy dx
x œ 2y
2
Ê 2ydy œ xdx Ê y2 œ x2 C Ê y œ „ É x2 C, 2
C0
7. kx2 y2 œ 1 Ê 1 y2 œ kx2 Ê
1 y2 x2
œk
x2 a2yby ˆ1 y2 ‰2x Ê œ 0 Ê 2yx2 y w œ a1 y2 ba2xb x% ˆ1 y2 ‰a2xb ˆ1 y 2 ‰ Ê y w œ 2xy2 œ xy . So for the orthogonals: 2 ˆ1 y 2 ‰ 2 dy xy dy œ x dx Ê ln y y2 œ x2 C dx œ 1y2 Ê y w
2x 8. 2x2 y2 œ c2 Ê 4x 2yy w œ 0 Ê y w œ 4x 2y œ y . For
orthogonals:
dy dx
œ
y 2x
Ê
dy y
œ
dx 2x
Ê ln y œ "# ln x C
Ê ln y œ ln x1/2 ln C1 Ê y œ C1 kxk1/2
9. y œ cex Ê
y ecx
œcÊ
e x y w c yae x bac1b ae x b2
œ!
Ê ex y w œ yex Ê y w œ y. So for the orthogonals: dy dx
œ
1 y 2
Ê y dy œ dx Ê
y2 2
œxC
Ê y œ 2x C1 Ê y œ „ È2x C1
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Chapter 9 First-Order Differential Equations xŠ 1y ‹y c ln y w
10. y œ ekx Ê ln y œ kx Ê
œkÊ
ln y x
Ê Š xy ‹ y w ln y œ 0 Ê y w œ dy dx
x2
y ln y x .
x y ln y Ê y ln y dy œ x dx " 2 1 2 ˆ " 2‰ # y ln y 4 ay b œ # x 2 y2 ln y y2 œ x2 C1
œ0
So for the orthogonals:
œ
Ê Ê
C
2 w 11. 2x2 3y2 œ 5 and y2 œ x3 intersect at a1, 1b. Also, 2x2 3y2 œ 5 Ê 4x 6y y w œ 0 Ê y w œ 4x 6y Ê y a1, 1b œ 3
y21 œ x3 Ê 2y1 y1w œ 3x2 Ê y1w œ x2 2
12. (a) x dx y dy œ 0 Ê
y2 2 w
3x2 2y1
Ê y1w a1, 1b œ 32 . Since y w † y1w œ ˆ 23 ‰ˆ 32 ‰ œ 1, the curves are orthogonal.
œ C is the general equation
of the family with slope y œ xy . For the orthogonals: yw œ
y x
Ê
dy y
œ
Ê ln y œ ln x C or y œ C1 x
dx x
(where C1 œ e Ñ is the general equation of the orthogonals. C
(b) x dy 2y dx œ 0 Ê 2y dx œ x dy Ê Ê "# Š dy y ‹œ
dy 2y
œ
dx x
Ê "# ln y œ ln x C Ê y œ C1 x2 is
dx x
the equation for the solution family. " # ln
"y # y
w
y ln x œ C Ê
Ê slope of orthogonals is
1 x dy dx
œ 0 Ê yw œ
2y x
x œ 2y 2
Ê 2y dy œ x dx Ê y2 œ x2 C is the general equation of the orthogonals.
13. Let y(t) œ the amount of salt in the container and V(t) œ the total volume of liquid in the tank at time t. Then, the departure rate is (a) Rate entering œ
2 lb gal
†
y(t) V(t) 5 gal min
(the outflow rate). œ 10 lb/min
(b) Volume œ V(t) œ 100 gal (5t gal 4t gal) œ (100 t) gal (c) The volume at time t is (100 t) gal. The amount of salt in the tank at time t is y lbs. So the concentration at any time t is 100y t lbs/gal. Then, the rate leaving œ 100y t (lbs/gal) † 4 (gal/min) œ (d)
dy dt
4y 100t
lbs/min
œ 10
4y 100 t
Ê
dy dt
ˆ 1004 t ‰ y œ 10 Ê P(t) œ
4 100t
œ 4 ln (100 t) Ê v(t) œ e4 ln Ð100 tÑ œ (100 t)% Ê y œ œ
10 (100 t)%
&
Š (1005t) C‹ œ 2(100 t)
C (100 t)%
Ê C œ (150)(100)% Ê y œ 2(100 t)
, Q(t) œ 10 Ê " (100t)%
' P(t) dt œ ' 1004 t dt
' (100 t)% (10 dt)
; y(0) œ 50 Ê 2(100 0) %
(150)(100) (100 t)%
Ê y œ 2(100 t)
C (100 0)%
œ 50
150 t ‰% ˆ1 100
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 9.4 Graphical Solutions of Autonomous Equations (e) y(25) œ 2(100 25) 14. (a)
dV dt
(150)(100)% (10025)%
¸ 188.56 lbs Ê concentration œ
y(25) volume
¸
188.6 125
¸ 1.5 lb/gal
œ a5 3b œ 2 Ê V œ 100 2t
The tank is full when V œ 200 œ 100 2t Ê t œ 50 min (b) Let yatb be the amount of concentrate in the tank at time t. dy dt
gal gal lb lb œ Š "# gal ‹Š5 min ‹ Š 100y 2t gal ‹Š3 min ‹Ê
Q(t) œ 52 ; P(t) œ 32 ˆ 501 t ‰ Ê ' P(t) dt
vatb œ e yatb œ
1 at 50b3/2
dy dt
œ
5 2
32 ˆ 50y t ‰ Ê
dy dt
3 2at 50b y
œ
5 2
' P(t) dt œ 32 ' t 150 dt œ 32 ln at 50b since t 50 0
œ e 2 ln at50b œ at 50b3/2 3
' 25 at 50b3/2 dt œ at 50b3/2 at 50b5/2 C ‘ Ê yatb œ t 50
C at 50b3/2
Apply the initial condition (i.e., distilled water in the tank at t œ 0): ya0b œ 0 œ 50 ya50b œ 100
C 503/2
505/2 1003/2
Ê C œ 505/2 Ê yatb œ t 50
505/2 . at 50b3/2
When the tank is full at t œ 50,
¸ 83.22 pounds of concentrate.
15. Let y be the amount of fertilizer in the tank at time t. Then rate entering œ 1
lb gal
†1
gal min
œ1
lb min
and the
volume in the tank at time t is V(t) œ 100 (gal) [1 (gal/min) 3 (gal/min)]t min œ (100 2t) gal. Hence 3y ‰ dy ˆ ˆ 3 ‰ rate out œ ˆ 100y 2t ‰ 3 œ 1003y2t lbs/min Ê dy dt œ 1 100 2t lbs/min Ê dt 100 2t y œ 1 Ê P(t) œ
3 1002t
2t) ' P(t) dt œ ' 10032t dt œ 3 ln (100 Ê v(t) œ eÐ 3 ln Ð100 2tÑÑ2 # 2t) "Î# y œ (100 "2t)c$Î# ' (100 2t)$Î# dt œ (100 2t)$Î# ’ 2(100# C“
, Q(t) œ 1 Ê
œ (100 2t)$Î# Ê
œ (100 2t) C(100 2t)$Î# ; y(0) œ 0 Ê [100 2(0)] C[100 2(0)]$Î# Ê C(100)$Î# œ 100 " Ê C œ (100)"Î# œ 10 Ê y œ (100 2t)
œ 2
3È100 2t 10
(100 2t)$Î# 10
. Let
dy dt
œ0 Ê
dy dt
œ 2
ˆ #3 ‰ (100 2t)"Î# (2) 10
œ 0 Ê 20 œ 3È100 2t Ê 400 œ 9(100 2t) Ê 400 œ 900 18t Ê 500 œ 18t
Ê t ¸ 27.8 min, the time to reach the maximum. The maximum amount is then y(27.8) œ [100 2(27.8)]
[100 2(27.8)]$Î# 10
¸ 14.8 lb
16. Let y œ y(t) be the amount of carbon monoxide (CO) in the room at time t. The amount of CO entering the y ‰ˆ 3 ‰ y 4 3 ‰ 12 $ room is ˆ 100 ‚ 10 œ 1000 ft$ /min, and the amount of CO leaving the room is ˆ 4500 10 œ 15,000 ft /min. Thus,
dy dt
Ê yœ
œ
12 1000
" etÎ15ß000
y 15,000
Ê
dy dt
12 ' 1000 etÎ15ß000 dt
" 15,000
yœ
12 1000
Ê P(t) œ
" 15,000
, Q(t) œ
12 1000
Ê v(t) œ etÎ15ß000
15,000 tÎ15,000 Ê y œ ectÎ15ß000 ˆ 12†1000 e C‰ œ etÎ15ß000 a180etÎ15ß000 Cb ;
y(0) œ 0 Ê 0 œ 1(180 C) Ê C œ 180 Ê y œ 180 180ectÎ15ß000 . When the concentration of CO is 0.01% y .01 in the room, the amount of CO satisfies 4500 œ 100 Ê y œ 0.45 ft$ . When the room contains this amount we tÎ15ß000 ‰ ¸ 37.55 min. have 0.45 œ 180 180e tÎ15ß000 Ê 179.55 Ê t œ 15,000 ln ˆ 179.55 180 œ e 180 9.4 GRAPHICAL SOUTIONS OF AUTONOMOUS EQUATIONS 1. y w œ ay 2bay 3b (a) y œ 2 is a stable equilibrium value and y œ 3 is an unstable equilibrium. (b) yww œ a2y 1by w œ 2ay 2bˆy 12 ‰ay 3b
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Chapter 9 First-Order Differential Equations
(c)
2. y w œ ay 2bay 2b (a) y œ 2 is a stable equilibrium value and y œ 2 is an unstable equilibrium. (b) yww œ 2yy w œ 2ay 2byay 2b
(c)
3. y w œ y3 y œ ay 1byay 1b (a) y œ 1 and y œ 1 is an unstable equilibrium and y œ 0 is a stable equilibrium value. (b) yww œ a3y2 1by w œ 3ay 1bŠy
1 È3 ‹yŠy
1 È3 ‹ay
1b
(c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 9.4 Graphical Solutions of Autonomous Equations 4. y w œ yay 2b (a) y œ 0 is a stable equilibrium value and y œ 2 is an unstable equilibrium. (b) yww œ a2y 2by w œ 2yay 1bay 2b
(c)
5. y w œ Èy, y 0 (a) There are no equilibrium values. 1 1 w Èy œ "# (b) yww œ 2È y y œ 2È y
(c)
6. y w œ y Èy, y 0 (a) y œ 1 is an unstable equilibrium. (b) yww œ Š1
1 2È y ‹
y w œ Š1
1 ˆ 2È y ‹ y
Èy‰ œ ˆÈy "# ‰ˆÈy 1‰
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552
Chapter 9 First-Order Differential Equations
(c)
7. y w œ ay 1bay 2bay 3b (a) y œ 1 and y œ 3 is an unstable equilibrium and y œ 2 is a stable equilibrium value. (b) yww œ a3y2 12y 11bay 1bay 2bay 3b œ 3ay 1bŠy
6 È3 ‹ay 3
2bŠy
6 È3 3 ‹ay
(c)
8. y w œ y3 y2 œ y2 ay 1b (a) y œ 0 and y œ 1 is an unstable equilibrium. (b) yww œ a3y2 2ybay3 y2 b œ y3 a3y 2bay 1b
(c)
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3b
Section 9.4 Graphical Solutions of Autonomous Equations 9.
dP dt
œ 1 2P has a stable equilibrium at P œ "# .
10.
dP dt œ Pa1 2Pb has an unstable equilibrium d2 P dP dt2 œ a1 4Pb dt œ Pa1 4Pba1 2Pb
11.
dP dt œ 2PaP 3b has a d2 P dP dt2 œ 2a2P 3b dt œ
d2 P dt2
œ 2 dP dt œ 2a1 2Pb
at P œ 0 and a stable equilibrium at P œ "# .
stable equilibrium at P œ 0 and an unstable equilibrium at P œ 3. 4Pa2P 3baP 3b
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553
554 12.
Chapter 9 First-Order Differential Equations dP dt d2 P dt2
œ 3Pa1 PbˆP "# ‰ has a stable equilibria at P œ 0 and P œ 1 an unstable equilibrium at P œ "# . 3 œ #3 a6P2 6P+1b dP dt œ # PŠP
3 È3 ˆ ‹ P 6
"# ‰ŠP
3 È3 ‹aP 6
1b
13.
Before the catastrophe, the population exhibits logistic growth and Patb Ä M0 , the stable equilibrium. After the catastrophe, the population declines logistically and Patb Ä M1 , the new stable equilibrium. 14.
dP dt
œ rPaM PbaP mb, r, M, m 0
The model has 3 equilibrium points. The rest point P œ 0, P œ M are asymptotically stable while P œ m is unstable. For initial populations greater than m, the model predicts P approaches M for large t. For initial populations less than m, the model predicts extinction. Points of inflection occur at P œ a and P œ b where a œ "3 M m ÈM2 mM m2 ‘ and b œ "3 M m ÈM2 mM m2 ‘. (a) The model is reasonable in the sense that if P m, then P Ä 0 as t Ä _; if m P M, then P Ä M as t Ä _; if P M, then P Ä M as t Ä _. (b) It is different if the population falls below m, for then P Ä 0 as t Ä _ (extinction). If is probably a more realistic model for that reason because we know some populations have become extinct after the population level became too low.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 9.4 Graphical Solutions of Autonomous Equations (c) For P M we see that
dP dt
555
œ rPaM PbaP mb is negative. Thus the curve is everywhere decreasing. Moreover,
P ´ M is a solution to the differential equation. Since the equation satisfies the existence and uniqueness conditions, solution trajectories cannot cross. Thus, P Ä M as t Ä _. (d) See the initial discussion above. (e) See the initial discussion above. 15.
dv dt
œg
k 2 mv ,
Equilibrium: Concavity:
g, k, m 0 and vatb 0
dv dt
d2 v dt2
œg
k 2 mv
œ 0 Ê v œ É mg k
ˆ k ‰ˆ œ 2ˆ mk v‰ dv dt œ 2 m v g
k 2‰ mv
(a)
(b)
160 (c) vterminal œ É 0.005 œ 178.9
ft s
œ 122 mph
16. F œ Fp Fr ma œ mg kÈv dv kÈ v, va0b œ v0 dt œ g m Thus,
dv dt
2
2
‰ , the terminal velocity. If v0 ˆ mg ‰ , the object will fall faster and faster, approaching the œ 0 implies v œ ˆ mg k k 2
‰ , the object will slow down to the terminal velocity. terminal velocity; if v0 ˆ mg k 17. F œ Fp Fr ma œ 50 5kvk dv 1 dt œ m a50 5kvkb The maximum velocity occurs when
dv dt
œ 0 or v œ 10
ft sec .
18. (a) The model seems reasonable because the rate of spread of a piece of information, an innovation, or a cultural fad is proportional to the product of the number of individuals who have it (X) and those who do not (N X). When X is small, there are only a few individuals to spread the item so the rate of spread is slow. On the other hand, when (N X) is small the rate of spread will be slow because there are only a few indiciduals who can receive it during the interval of time. The rate of spread will be fastest when both X and (N X) are large because then there are a lot of individuals to spread the item and a lot of individuals to receive it. (b) There is a stable equilibrium at X œ N and an unstable equilibrium at X œ 0. d2 X dt2
dX 2 œ k dX dt aN Xb kX dt œ k XaN XbaN 2Xb Ê inflection points at X œ 0, X œ
N 2,
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and X œ N.
556
Chapter 9 First-Order Differential Equations
(c)
(d) The spread rate is most rapid when x œ 19. L di dt Ri œ V Ê Concavity:
Eventually all of the people will receive the item.
œ VL RL i œ RL ˆ VR i‰, V, L, R 0 œ RL ˆ VR i‰ œ 0 Ê i œ VR
di dt d2 i dt2 œ
Equilibrium:
N 2.
di dt
ˆ R ‰2 ˆ VR i‰ ˆ RL ‰ di dt œ L
Phase Line:
If the switch is closed at t œ 0, then ia0b œ 0, and the graph of the solution looks like this:
As t Ä _, it Ä isteady state œ
V R.
(In the steady state condition, the self-inductance acts like a simple wire connector and, as
a result, the current throught the resistor can be calculated using the familiar version of Ohm's Law.) 20. (a) Free body diagram of the pearl:
(b) Use Newton's Second Law, summing forces in the direction of the acceleration: ˆ m m P ‰g mk v. mg Pg kv œ ma Ê dv dt œ (c) Equilibrium: Ê vterminal œ Concavity:
dv dt
œ
k amPbg mŠ k
v‹ œ 0
am Pbg k
d2 v dt2
ˆ k ‰ am k Pbg v‹ œ mk dv dt œ m Š 2
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Section 9.5 Systems of Equations and Phase Planes
557
(d)
(e) The terminal velocity of the pearl is
am Pbg . k
9.5 SYSTEMS OF EQUATIONS AND PHASE PLANES 1. Seasonal variations, nonconformity of the environments, effects of other interactions, unexpected disasters, etc. 2. x œ r cos ) Ê y œ r sin ) Ê Solve for cos ) 2
Ê
dr dt
dr dt
dr dt
dx dt dy dt
œ r sin ) œ r cos )
d) dt
d) dt
cos )
dr dt
œ y x xax2 y2 b œ r sin ) r cos ) r3 cos )
2 2 3 sin ) dr dt œ x y xax y b œ r cos ) r sin ) r sin )
by adding cos ) ‚ eq(1) to sin ) ‚ eq(2):
sin2 )
dr dt
œ cos )ar sin ) r cos ) r3 cos )b sin )ar cos ) r sin ) r3 sin )b
œ r sin ) cos ) r cos2 ) r3 cos2 ) r sin ) cos ) r sin ) r3 sin2 ) œ r r3 œ ra1 r2 b
Solve for
d) dt
by adding asin )b ‚ eq(1) to cos ) ‚ eq(2):
r sin ) ddt) r cos2 ) ddt) œ sin )ar sin ) r cos ) r3 cos )b cos )ar cos ) r sin ) r3 sin )b Ê r ddt) œ r sin2 ) r sin ) cos ) r3 sin ) cos ) r cos2 ) r sin ) cos ) r3 sin ) cos ) œ r Ê ddt) œ 1 2 If r œ 1 (that is, the trajectory starts on the circle x2 y2 œ 1), then dr dt ¹rœ1 œ a1bŠ1 a1b ‹ œ 0, thus the trajectory 2
remains on the circle, and rotates around the circle in a clockwise direction, since
d) dt
œ 1. The solution is periodic since
at any point ax, yb on the trajectory, ax, yb œ ar cos ), r sin )b œ a1 cos ), 1 sin )b œ acos ), sin )b Ê both x and y are periodic. 3. This model assumes that the number of interactions is porportional to the product of x and y: dy y‰ dx m ˆ ˆ ‰ dt œ aa b ybx, a 0, dt œ m 1 M y n x y œ y m M y n x Þ To find the equilibrium points: dx dt œ 0 Ê aa b ybx œ 0 Ê x œ 0 or y œ
a b a b
0);
Mn m x
M;
(remember dy dt
œ 0 Ê yˆ m
m M
y n x‰ Ê y œ 0 or y œ
Thus there are two equlibrium points, both occur when x œ 0, a0, 0b and a0, Mb.
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558
Chapter 9 First-Order Differential Equations
Implies coexistence is not possible because eventually trout die out and bass reach their population limit.
4. The coefficients a, b, m, and n need to be determined by sampling or by analyzing historical data. Then, more specific graphical predictions can be made. These predictions would then have to be compared to actual population growth patterns. If the predictions match actual results, we have partially validated our model. If necessary, more tests could be run. However, it should be remembered that the primary purpose of a graphical analysis is to analyze the behavior qualitatively. With reference to Figure 9.29, attempt to maintain the fish populations in Region B through stocking and regulation (open and closed seasons). For example, should Regions A or D be entered, restocking the appropriate species can cause a return to Region B. 5. (a) Logistic growth occurs in the absence of the competitor, and simple interaction of the species: growth dominates the competition when either population is small so it is difficult to drive either species to extinction. (b) a œ per capita growth rate for trout m œ per capita growth rate for bass b œ intensity of competition to the trout n œ intensity of competition to the bass k1 œ environmental carrying capacity for the trout k2 œ environmental carrying capacity for the bass (c)
dx dt
œ 0 Ê aŠ1
yœ
a b
dy a b k1 x; dt
mŠ1
y k2 ‹
Case I:
a b
x k1 ‹x
b x y œ ’aŠ1
œ 0 Ê mŠ1
y k 2 ‹y
x k1 ‹
n x y œ ’mŠ1
n x œ 0 Ê y œ 0 or y œ k2
k2 and
By picking
a b
m n
b y“x œ 0 Ê x œ 0 or aŠ1
n k2 m x.
y k2 ‹
x k1 ‹
b y œ 0 Ê x œ 0 or
n x“y œ 0 Ê y œ 0 or
There are five cases to consider.
k1 .
k2 and
m n
k1 we ensure an equilibrium point exists inside the first quadrant.
k1 b m k1 k2 Graphical analysis implies four equilibrium points exist: a0, 0b, ak1 , 0b, a0, k2 b, and Š a m a m b n k1 k2 ,
a m k2 a n k1 k2 a m b n k1 k2 ‹
(the point of intersection of the two boundaries in the first quadrant). All of these equilibrium points are unstable except for the point of intersection. The possibility of coexistence is predicted by this model.
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Section 9.5 Systems of Equations and Phase Planes Case II:
k2 and
a b
m n
559
k1 .
a0, k2 b: unstable ak1 , 0b: stable a0, 0b: unstable Trout wins: ak1 , 0b Not sensitive No coexistence
Case III:
a b
k2 and
m n
k1 .
a0, k2 b: stable ak1 , 0b: unstable a0, 0b: unstable Bass wins: a0, k2 b Not sensitive No coexistence
Case IV:
a b
k2 and
m n
k1 .
a0, k2 b: stable ak1 , 0b: stable a0, 0b: unstable k1 b m k1 k2 Šam a m b n k1 k2 ,
a m k2 a n k1 k2 a m b n k 1 k 2 ‹:
unstable
Bass or trout: a0, k2 b or ak1 , 0b Very sensitive Coexistence is possible but not predicted If we assume
a b
k2 and
m n
k1 then graphical analysis implies four equilibrium poins exist: a0, k2 b, ak1 , 0b,
k1 b m k1 k2 a m k2 a n k1 k2 a0, 0b, and Š a m a m b n k1 k2 , a m b n k1 k2 ‹ (the Case V: ba œ k2 and b ak1 œ nmk2 (lines coincide).
point of intersection of the two boundaries in the first quadrant).
a0, k2 b: stable ak1 , 0b: stable a0, 0b: unstable Line segment joining a0, k2 b and ak1 , 0b: stable Bass wins: a0, k2 b Not sensitive Coexistence is likely outcome Note that all points on the line segment joining a0, k2 b and ak1 , 0b are rest points. 6. For a fixed price, as Q increases,
dP dt
gest smaller and, possibly, becomes negative. This observation implies that as the
quantity supplied increases, the price will not rise as fast. If Q gets high enough, then the price will decrease. Next, dQ consider dQ dt : For a fixed quantity, as P increases, dt gets larger. Thus, as the market price increases, the quantity supplied will increase at a faster rate. If P is too small,
dQ dt
will be negative and the quantity supplied will decrease.
This observation is the traditional explanation of the effect of market price levels on the quantity supplied.
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560
(a)
Chapter 9 First-Order Differential Equations
dP dt
œ 0 and
Now
dP dt
œ 0 gives the equilibrium points aP, Qb: a0, 0b and a25.8, 775b.
dQ dt
0 when PQ 20,000 and P 0;
dP dt
0 otherwise.
dQ dt
0 when P
Q 30
and Q 0;
dQ dt
0 otherwise.
(b) These considerations give the following graphical analysis: Region I II III IV
dQ dt
dP dt
0 0 0 0
0 0 0 0
The equilibrium point a0, 0b is unstable. The graphical analysis for the point a25.8, 775b is inconclusive: trajectories near the point may be periodic, or may spiral toward or away from the point. dQ (c) The curve dP dt œ 0 or PQ œ 20000 can be thought of as the demand curve; dt œ 0 or Q œ 30P can be viewed as the supply curve. 7. (a)
dx dt
œ a x b x y œ aa b ybx and
(b)
dy dx
œ
am n x b y aa b y b x
dy dt
œ m y n x y œ am n xby Ê
dy dt
œ
dy dx dx dt
Ê
dy dx
œ
dy dt dx dt
œ
am n x b y aa b y b x
Ê Š ya b‹dy œ ˆ mx n‰dx Ê ' Š ya b‹dy œ ' ˆ mx n‰dx Ê a lnlyl b y œ m lnlxl n x C
Ê lnlya l ln eb y œ lnlxm l ln en x ln eC Ê lnlya eb y l œ lnlxm en x eC l Ê ya eb y œ xm en x eC , let K œ eC Ê ya eb y œ Kxm en x (c) fayb œ ya eb y Ê f w ayb œ a ya1 eb y b ya eb y œ ya1 eb y aa b yb and f w ayb œ 0 Ê y œ 0 or y œ ba ; f ww ˆ ba ‰ œ bˆ ba ‰
a 1 a
e
a
0 Ê fayb has a unique max of My œ ˆ eab ‰ when y œ ba . gaxb œ xm en x
Ê g w axb œ m xm1 en x n xm en x œ xm1 en x am n xb and g w axb œ 0 Ê x œ 0 or x œ g ww ˆ mn ‰ œ nˆ mn ‰
m1 m
e
0 Ê gaxb has a unique max of Mx œ ˆ emn ‰
(d) Consider trajectory ax, yb Ä ˆ mn , ba ‰. ya eb y œ Kxm en x Ê Ê
a
lim Š eyb y †
xÄmÎn y Äa Îb
en x xm ‹
œ
lim K Ê
xÄmÎn yÄaÎb
My Mx
œ K. Thus,
ya eb y
œ
m
when x œ
m n;
m n.
nx ya † exm œ K, taking the limit of both sides eb y My xm Mx en x represents the equation any solution
trajectory must satisfy if the trajectory approaches the rest point asymptotically.
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Section 9.5 Systems of Equations and Phase Planes
561
(e) Pick initial condition y0 ba . Then, from the figure at right, fay0 b My implies xm en x
My xm Mx en x
ya0 eb y0
œ
My and thus
Mx . From the figure for gaxb, there exists a
unique x0 y
a b
m n
satisfying
xm en x
Mx . That is, for each
there is a unique x satisfying
ya eb y
œ
My xm Mx en x .
Thus,
there can exist only one trajectory solution approaching ˆ mn , ba ‰. (You can think of the point ax0 , y0 b as the initial condition for that trajectory.) (f) Likewise there exists a unique trajectory when y0 ba . Again, fay0 b My implies xm en x
Mx . From the figure for gaxb, there exists a unique x0
a unique x satisfying 8. Let z œ y w œ
dy dx
Ê
dz dx
a
y eb y
œ
My x Mx en x . m
m n
satisfying
xm en x
My xm Mx en x
œ
ya0 eb y0
My and thus
Mx . That is, for each y
Thus, there can exist only one trajectory solution
dy dx
œz dz dx
œ Fax, y, zb
In general, for the n order differential equation given by yanb œ Fˆx, y, yw , yww , . . ., yan1b ‰, let z1 œ y w œ th
dz1 dx
there is
approaching ˆ mn , ba ‰.
œ z w œ y ww , then given the differential equation y ww œ Fax, y, y w b, we can write it as the following
system of first order differential equations:
Ê
a b
œ z1w œ y ww , let z2 œ z1w œ y ww , Ê
œ z2w œ y www , . . ., let zn1 œ znw 2 œ y an 1b Ê znw 1 œ yanb . This gives us the
dz2 dx
following system of first order differential equations:
dy dx
œ z1 dz1 dx dz2 dx
œ z2 œ z3
dznc2 dx dznc1 dx
9. In the absence of foxes Ê b œ 0 Ê
dx dt
dy dx
ã œ zn1 œ Fax, y, z1 , z2 , . . ., zn1 b
œ a x and the population of rabbits grows at a rate proportional to the number of
rabbits. 10. In the absence of rabbits Ê d œ 0 Ê
dy dt
œ c y and the population of foxes decays (since the foxes have no food source)
at a rate proportional to the number of foxes. 11.
dx dt œ aa b ybx œ 0 ˆ dc , ba ‰. For the point
Êyœ
a b
or x œ 0;
dy dt
œ ac d xby œ 0 Ê x œ
c d
or y œ 0 Ê equilibrium points at a0, 0b or
a0, 0b, there are no rabbits and no foxes. It is an unstable equilibrium point, if there are no foxes, but
a few rabbits are introduced, then
dx dt
œ a Ê the rabbit population will grow exponentially away from a0, 0b
12. Let xatb and yatb both be positive and suppose that they satisfy the differential equations dy dt
œ ac d xby. Let Catb œ a ln yatb b yatb d xatb c ln xatb Ê C w atb œ a
y atb yatb w
dx dt w
œ aa b ybx and
b y atb d x w atb c xxaattbb w
œ Š yaatb b‹y w atb Š xactb d‹x w atb œ Š yaatb b‹ac d xatbbxatb Š xactb d‹aa b yatbbyatb œ 0 Since C w atb œ 0 Ê Catb œ constant. 13. Consider a particular trajectory and suppose that ax0 , y0 b is such that x0
c d
and y0 ba , then
dx dt
0 and
dy dt
0 Ê the
rabbit population is increasing while the fox population is decreasing, points on the trajectory are moving down and to the right; if x0
c d
and y0 ba , then
dx dt
0 and
dy dt
trajectory are moving up and to the right; if x0
0 Ê both the rabbit and fox populations are increasing, points on the c d
and y0 ba , then
dx dt
0 and
dy dt
0 Ê the rabbit population is
decreasing while the fox population is increasing, points on the trajectory are moving up and to the left; and finally if
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
562
Chapter 9 First-Order Differential Equations
x0
c d
and y0 ba , then
dx dt
0 and
dy dt
0 Ê both the rabbit and fox populations are decreasing, points on the trajectory
are moving down and to the left. Thus, points travel around the trajectory in a counterclockwise direction. Note that we will follow the same trajectory if ax0 , y0 b starts at a different point on the trajectory. 14. There are three possible cases: If the rabbit population begins (before the wolf) and ends (after the wolf) at a value larger than the equilibrium level of x œ dc , then the trajectory moves closer to the equilibrium and the maximum value of the foxes is smaller. If the rabbit population begins (before the wolf) and ends (after the wolf) at a value smaller than the equilibrium level of x œ dc , but greater than 0, then the trajectory moves further from the equilibrium and the maximum value of the foxes is greater. If the rabbit population begins and ends very near the equilibrium value, then the trajectory will stay near the equilibrium value, since it is a stable equilibrium, and the fox population will remain roughly the same. CHAPTER 9 PRACTICE EXERCISES 2ax 2b3/2 a3x 4b 15 3/2 2ax 2b a3x 4b C“ 15
1. y w œ xey Èx 2 Ê ey dy œ xÈx 2 dx Ê ey œ Ê y œ ln’ 2. y w œ xyex Ê 2
3/2
2ax 2b a3x 4b 15
dy y
C“ Ê y œ ln’
C Ê ey œ
2ax 2b3/2 a3x 4b 15
C
œ ex x dx Ê ln y œ "# ex C 2
2
3. sec x dy x cos2 y dx œ 0 Ê
dy cos2 y
x dx œ sec x Ê tan y œ cos x x sin x C
4. 2x2 dx 3Èy csc x dy œ 0 Ê 3Èy dy œ
2x2 csc x dx
Ê 2y3/2 œ 2a2 x2 bcos x 4x sin x C
Ê y3/2 œ a2 x2 bcos x 2x sin x C1 5. y w œ
ey xy
Ê yey dy œ
Ê ay 1bey œ ln kxk C
dx x
6. y w œ xexy csc y Ê y w œ
x ex ey csc
yÊ
ey csc y dy
œ x ex dx Ê
7. xax 1bdy y dx œ 0 Ê xax 1bdy œ y dx Ê
dy y
œ
ey 2 asin
dx x ax 1 b
y cos yb œ ax 1bex C
Ê ln y œ lnax 1b lnaxb C
Ê ln y œ lnax 1b lnaxb ln C1 Ê ln y œ lnŠ C1 axx 1b ‹ Ê y œ
8. y w œ ay2 1bax1 b Ê
dy
y 2 1
œ
dx x
Ê
1 lnŠ yy c b1‹
2
C1 ax 1b x
1 œ ln x C Ê lnŠ yy 1 ‹ œ 2ln x ln C1 Ê
y1 y1
œ C1 x2
9. 2y w y œ xex/2 Ê y w "# y œ x2 ex/2 .
' ˆ "‰ paxb œ "# , vaxb œ e c # dx œ ex/2 .
ex/2 y w "# ex/2 y œ ˆex/2 ‰ˆ x2 ‰ˆex/2 ‰ œ 10.
w
y 2
x 2
Ê
d ˆ x/2 dx e
y‰ œ
x 2
Ê ex/2 y œ
x2 4
2
C Ê y œ ex/2 Š x4 C‹
y œ ex sin x Ê y w 2y œ 2ex sin x.
paxb œ 2, vaxb œ e' 2dx œ e2x . e2x y w 2e2x y œ 2e2x ex sin x œ 2ex sin x Ê
d 2x dx ae
yb œ 2ex sin x Ê e2x y œ ex asin x cos xb C
Ê y œ ex asin x cos xb Ce2x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 9 Practice Exercises 11. xy w 2y œ 1 x1 Ê y w ˆ 2x ‰y œ vaxb œ e2'
dx x
1 x
563
1 x2 .
2
œ e2ln x œ eln x œ x2 .
x2 y w 2xy œ x 1 Ê
d 2 dx ax yb
œ x 1 Ê x2 y œ
x2 2
" #
xCÊyœ
1 x
C x2
12. xy w y œ 2x ln x Ê y w ˆ 1x ‰y œ 2 ln x. vaxb œ e' d ˆ1 dx x
dx x
2 œ eln x œ 1x . ˆ 1x ‰y w ˆ 1x ‰ y œ 2x ln x Ê
† y‰ œ 2x ln x Ê
† y œ c ln x d2 C Ê y œ xc ln x d2 Cx
1 x
13. a1 ex bdy ayex ex bdx œ 0 Ê a1 ex by w ex y œ ex Ê y w œ ' a exdx b
vaxb œ e 1 b ex œ elnae 1b œ ex 1. x cx aex 1by w aex 1bˆ 1 e ex ‰y œ a1e ex b aex 1b Ê Êyœ
ecx C ex 1
d x dx ay e b
œ
ecx a1 e x b .
x
œ
ecx C 1 ex
14. ex dy aex y 4xbdx œ 0 Ê Ê
ex 1 ex y
dy dx
1by d œ ex Ê aex 1by œ ex C
d aex dx c
' y œ 4x ex Ê paxb œ 1, vaxb œ e 1dx œ ex Ê ex
y ex œ 4x e2x
dy dx x
œ 4x e2x Ê y ex œ ' 4x e2x dx Ê y ex œ 2x e2x e2x C Ê y œ 2x ex e C ex
15. ax 3y2 b dy y dx œ 0 Ê x dy y dx œ 3y2 dy Ê
d dx axyb
œ 3y2 dy Ê xy œ y3 C '
16. x dy a3y x2 cos xb dx œ 0 Ê y w ˆ 3x ‰y œ x3 cos x. Let vayb œ e
3dx x
3
œ e3ln x œ eln x œ x3 .
Then x3 y w 3x2 y œ cos x and x3 y œ ' cos x dx œ sin x C. So y œ x3 asin x Cb
w ˆ 2 ‰ 17. ax 1b dy dx 2y œ x Ê y x 1 y œ
So y w ax 1b2 Ê yax 1b2 œ
x x1.
' x b2 1 dx
Let vaxb œ e
2 ax 1 b a x
1b2 y œ
x3 3
C Ê y œ ax 1b2 Š x3
y œ ax 1b2 Š x3 3
x2 2
x2 2
x ax 1 b a x
1b2 Ê
d dx yax
3
x2 2
2 1b2 ‘ œ xax 1b Ê yax 1b œ ' xax 1bdx
C‹. We have ya0b œ 1 Ê 1 œ C. So
1‹ ' ˆ 2 ‰dx
1 2 w ˆ2‰ 18. x dy dx 2y œ x 1 Ê y x y œ x x . Let vaxb œ e
Ê So 19.
dy dx
2
œ e2lnax1b œ elnax1b œ ax 1b2 .
d x4 2 3 2 dx ax yb œ x x Ê x y œ 4 2 4 2x2 1 y œ x4 4x1 2 "# œ x 4x 2
x2 2
CÊyœ
x2 4
x
C x2
œ eln x œ x2 . So x2 y w 2xy œ x3 x 2
"# . We have ya1b œ 1 Ê 1 œ
3x2 y œ x2 . Let vaxb œ e' 3x dx œ ex . So ex y w 3x2 ex y œ x2 ex Ê 2
3
3
3
3
We have ya0b œ 1 Ê e0 a1b œ 13 e0 C Ê 1 œ
3
1 3
d dx axyb
d x3 dx Še y‹
3
3
cos x x .
C
" #
3
Ê C œ 41 .
3
œ x2 ex Ê ex y œ 13 ex C. 3
C Ê C œ 43 and ex y œ 13 ex
20. xdy ay cos xbdx œ 0 Ê xy w y cos x œ 0 Ê y w ˆ 1x ‰y œ So xy w xˆ 1x ‰y œ cos x Ê
3
1 4
4 3
Êyœ
1 3
43 ex
' 1 dx x œ eln x œ x.
Let vaxb œ e
œ cos x Ê xy œ ' cos x dx Ê xy œ sin x C. We have yˆ 12 ‰ œ 0 Ê ˆ 12 ‰0 œ 1 C
Ê C œ 1. So xy œ 1 sin x Ê y œ
1 sin x x xc2
21. xy w ax 2by œ 3x3 ex Ê y w ˆ x x 2 ‰y œ 3x2 ex . Let vaxb œ e' ˆ x ‰dx œ ex2ln x œ xe2 . So x x ex w ex ˆ x 2 ‰ d ˆ y œ 3 Ê dx y † xe2 ‰ œ 3 Ê y † xe2 œ 3x C. We have ya1b œ 0 Ê 0 œ 3a1b C Ê C œ 3 x2 y x2 x Êy†
ex x2
3
x
œ 3x 3 Ê y œ x2 ex a3x 3b
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564
Chapter 9 First-Order Differential Equations
22. y dx a3x xy 2bdy œ 0 Ê Payb œ
3 y
dx dy
3x xy 2 y
œ0Ê
dx dy
3x y
x œ 2y Ê
dx dy
Š 3y 1‹x œ 2y .
1 Ê ' Paybdy œ 3ln y y Ê vayb œ e3ln yy œ y3 ey
y3 ey x w y3 ey Š 3y 1‹x œ 2y2 ey Ê y3 ey x œ ' 2y2 ey dy œ 2ey ay2 2y 2b C Ê y3 œ Êy œ 3
2ˆy2 2y 2‰ Cey . x 2ˆy2 2y 2‰ 4eyb1 x
We have ya2b œ 1 Ê 1 œ
2a1 2 2b Cec1 2
Ê C œ 4e and
23. To find the approximate values let yn œ yn1 ayn1 cos xn1 ba0.1b with x0 œ 0, y0 œ 0, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. x y x y 1.1 1.6241 0 0 1.2 1.8319 0.1 0.1000 1.3 2.0513 0.2 0.2095 1.4 2.2832 0.3 0.3285 1.5 2.5285 0.4 0.4568 1.6 2.7884 0.5 0.5946 1.7 3.0643 0.6 0.7418 1.8 3.3579 0.7 0.8986 1.9 3.6709 0.8 1.0649 2.0 4.0057 0.9 1.2411 1.0 1.4273 24. To find the approximate values let yn œ yn1 a2 yn1 ba2xn1 3ba0.1b with x0 œ 3, y0 œ 1, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. x y x y 1.9 5.3172 3.0 1.0000 1.8 5.9026 2.9 0.7000 1.7 6.3768 2.8 0.3360 1.6 6.7119 2.7 0.0966 1.5 6.8861 2.6 0.5998 1.4 6.8861 2.5 1.1718 1.3 6.7084 2.4 1.8062 1.2 6.3601 2.3 2.4913 1.1 5.8585 2.2 3.2099 1.0 5.2298 2.1 3.9393 2.0 4.6520 2ync1 25. To estimate ya3b, let y œ yn1 Š xncxn1c1 1 ‹a0.05b with initial values x0 œ 0, y0 œ 1, and 60 steps. Use a spreadsheet,
graphing calculator, or CAS to obtain ya3b ¸ 0.8981. 26. To estimate ya4b, let zn œ yn1 Š
x2nc1 2ync1 1 ‹a0.05b xnc1
with initial values x0 œ 1, y0 œ 1, and 60 steps. Use a
spreadsheet, graphing calculator, or CAS to obtain ya4b ¸ 4.4974. 27. Let yn œ yn1 ˆ exnc1 b1ync1 b 2 ‰adxb with starting values x0 œ 0 and y0 œ 2, and steps of 0.1 and 0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs. (a)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 9 Practice Exercises
565
(b) Note that we choose a small interval of x-values because the y-values decrease very rapidly and our calculator cannot handle the calculations for x Ÿ 1. (This occurs because the analytic solution is y œ 2 lna2 ex b, which has an asymptote at x œ ln 2 ¸ 0.69. Obviously, the Euler approximations are misleading for x Ÿ 0.7.)
y
28. Let yn œ yn1 Š eynncc11 xnncc11 ‹adxb with starting values x0 œ 0 and y0 œ 0, and steps of 0.1 and 0.1. Use a spreadsheet, x2
programmable calculator, or CAS to generate the following graphs. (a) (b)
29.
x y dy dx
1 1.2 1.4 1.6 1 0.8 0.56 0.28
œ x Ê dy œ x dx Ê y œ
Ê 1 œ
" #
Ê ya2b œ
30.
x y dy dx
œ
CÊCœ 22 2
3 2
œ
" #
32
x2 2
1.8 0.04
2.0 0.4
C; x œ 1 and y œ 1
Ê yaexactb œ
x2 2
3 2
is the exact value.
1 1.2 1.4 1.6 1.8 2.0 1 0.8 0.6333 0.4904 0.3654 0.2544 1 x
Ê dy œ x1 dx Ê y œ lnkxk C; x œ 1 and y œ 1
Ê 1 œ ln 1 C Ê C œ 1 Ê yaexactb œ lnkxk 1 Ê ya2b œ ln 2 1 ¸ 0.3069 is the exact value.
31.
x y dy dx
1 1.2 1.4 1.6 1.8 2.0 1 1.2 0.488 1.9046 2.5141 3.4192
œ xy Ê
Êyœe
dy y
x2 2 C
œ x dx Ê lnkyk œ x2 2
x2 2
C
x2 2
œ e † eC œ C1 e ; x œ 1 and y œ 1 x2
Ê 1 œ C1 e1/2 Ê C1 œ e1/2 yaexactb œ e1/2 † e 2 œ eˆx 1‰/2 Ê ya2b œ e3/2 ¸ 4.4817 is the exact value. 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
566
Chapter 9 First-Order Differential Equations
32.
x y
1 1.2 1.4 1.6 1.8 2.0 1 1.2 1.3667 1.5130 1.6452 1.7688
dy y2 1 dx œ y Ê y dy œ dx Ê 2 œ x " " 2 # œ 1 C Ê C œ # Ê y œ
C; x œ 1 and y œ 1
2x 1 È Ê yaexactb œ 2x 1 Ê ya2b œ È3 ¸ 1.7321 is the exact value.
33.
dy dx
œ y2 1 Ê y w œ ay 1bay 1b. We have y w œ 0 Ê ay 1b œ 0, ay 1b œ 0 Ê y œ 1, 1.
(a) Equilibrium points are 1 (stable) and 1 (unstable) (b) y w œ y2 1 Ê y ww œ 2yy w Ê y ww œ 2yay2 1b œ 2yay 1bay 1b. So y ww œ 0 Ê y œ 0, y œ 1, y œ 1.
(c)
34.
dy dx
œ y y2 Ê y w œ ya1 yb. We have y w œ 0 Ê ya1 yb œ 0 Ê y œ 0, 1 y œ 0 Ê y œ 0, 1.
(a) The equilibrium points are 0 and 1. So, 0 is unstable and 1 is stable. (b) Let ïî œ increasing, íï œ decreasing. yw ! yw ! yw ! qqíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqpy 0 1 y w œ y y2 Ê y ww œ y w 2yy w Ê y ww œ ay y2 b 2yay y2 b œ y y2 2y2 2y3 Ê y ww œ 2y3 3y2 y œ ya2y2 3y 1b Ê y ww œ ya2y 1bay 1b. So, y ww œ 0 Ê y œ 0, 2y 1 œ 0, y 1 œ 0 Ê y œ 0, y œ "# , y œ 1. Let ïî œ concave up, íï œ concave down. y ww ! y ww ! y ww ! y ww ! qíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqñqqïïïïïîqpy 0 1 1/2
(c)
35. (a) Force œ Mass times Acceleration (Newton's Second Law) or F œ ma. Let a œ
dv dt
œ
dv ds
†
ds dt
œ v dv ds . Then
2 2 ma œ mgR2 s2 Ê a œ gR2 s2 Ê v dv Ê v dv œ gR2 s2 ds Ê ' v dv œ ' gR2 s2 ds ds œ gR s
Ê
v2 2
œ
ÊCœ
gR2 s C1 2 v0 2gR
Ê v2 œ Êv œ 2
2gR2 s 2gR2 s
2C1 œ
v20
2gR2 s
C. When t œ 0, v œ v0 and s œ R Ê v20 œ
2gR
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2gR2 R
C
Chapter 9 Additional and Advanced Exercises (b) If v0 œ È2gR, then v2 œ
2gR2 s
2
È2gR. Then Ê v œ É 2gR s , since v 0 if v0
ds dt
œ
È2gR2 Ès
567
Ê Ès ds œ È2gR2 dt
Ê ' s1/2 ds œ ' È2gR2 dt Ê 23 s3/2 œ È2gR2 t C1 Ê s3/2 œ ˆ 32 È2gR2 ‰t C; t œ 0 and s œ R Ê R3/2 œ ˆ 32 È2gR2 ‰a0b C Ê C œ R3/2 Ê s3/2 œ ˆ 32 È2gR2 ‰t R3/2 œ ˆ 32 RÈ2g‰t R3/2 3 œ R3/2 ˆ 32 R1/2 È2g‰t 1 ‘ œ R3/2 ’ Š
36.
v0 m k
a0.86ba30.84b k 0.8866t
œ coasting distance Ê
Ê satb œ 0.97a1 e
È2gR 2R ‹t
2/3 0‰ 0‰ ‘ ‘ Ê s œ R 1 ˆ 3v 1 “ œ R3/2 ˆ 3v 2R t 1 2R t
œ 0.97 Ê k ¸ 27.343. satb œ
v0 m ˆ k 1
eak/mbt ‰ Ê satb œ 0.97ˆ1 ea27.343/30.84bt ‰
b. A graph of the model is shown superimposed on a graph of the data.
CHAPTER 9 ADDITIONAL AND ADVANCED EXERCISES 1. (a)
dy dt
A œ kA V ac yb Ê dy œ k V ay cbdt Ê
dy yc
' œ k A V dt Ê
dy yc
A œ ' k A V dt Ê lnky ck œ k V t C1
Ê y c œ „ eC1 ek V t . Apply the initial condition, ya0b œ y0 Ê y0 œ c C Ê C œ y0 c A
Ê y œ c ay0 cbek V t . A (b) Steady state solution: y_ œ lim yatb œ lim c ay0 cbek V t ‘ œ c ay0 cba0b œ c A
tÄ_
2.
tÄ_
damvb damvb dm dm dv dm dm dm dv dm dt œ F av ub dt Ê F œ dt av ub dt Ê F œ m dt v dt v dt u dt Ê F œ m dt u dt . dm dt œ b Ê m œ kbkt C. At t œ 0, m œ m0 , so C œ m0 and m œ m0 kbkt. u kb k m0 kbkt dv Thus, F œ am0 kbktb dv dt ukbk œ am0 kbktbkgk Ê dt œ g m0 kbkt Ê v œ gt u lnŠ m0 ‹ C1
v œ 0 at t œ 0 Ê C1 œ 0. So v œ gt u lnŠ m0 m0kbkt ‹ œ t œ 0 Ê y œ "# gt2 c’ t Š
m0 kbkt m0 kbkt kbk ‹ lnŠ m0 ‹
dy dt
Ê y œ ' ’ gt u lnŠ
m0 kbkt m0 ‹
“dt and u œ c, y œ 0 at
“
' 3. (a) Let y be any function such that vaxby œ ' vaxbQaxb dx C, vaxb œ e Paxb dx . Then ' Paxb dx ' d w w Ê v w axb œ œ e Paxb dx Paxb œ vaxbPaxb. dx avaxb † yb œ vaxb † y y † v axb œ vaxbQaxb. We have vaxb œ e
Thus vaxb † y w y † vaxb Paxb œ vaxbQaxb Ê y w y Paxb œ Qaxb Ê the given y is a solution.
(b) If v and Q are continuous on c a, b d and x − aa, bb, then
d dx ’
'xx vatbQatb dt“ œ vaxbQaxb 0
Ê ' vatbQatb dt œ ' vaxbQaxb dx. So C œ y0 vax0 b ' vaxbQaxb dx. From part (a), vaxby œ ' vaxbQaxb dx C. x Substituting for C: vaxby œ ' vaxbQaxb dx y0 vax0 b ' vaxbQaxb dx Ê vaxby œ y0 vax0 b when x œ x0 . x 0
4. (a) y w Paxby œ 0, yax0 b œ 0. Use vaxb œ e' Paxb dx as an integrating factor. Then ' Paxb dx
Ê y œ Ce œ C3 e (b)
' Paxb dx
d y axb dx avaxbc 1
' Paxb dx
and y1 œ C1 e
' Paxb dx
, y2 œ C# e
d dx avaxbyb
œ 0 Ê vaxby œ C
, y1 ax0 b œ y2 ax0 b œ 0, y1 y2 œ aC1 C2 be' Paxb dx
and y1 y2 œ 0 0 œ 0. So y1 y2 is a solution to y w Paxby œ 0 with yax0 b œ 0.
y2 axb db œ
' Paxb dx ' Paxb dx d e aC1 dx Še
C2 b ‘‹ œ
d dx aC1
C2 b œ
d dx aC3 b
œ !.
' dxd avaxbc y1 axb y2 axb dbdx œ avaxbc y1 axb y2 axb db œ ' ! dx œ C Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
568
Chapter 9 First-Order Differential Equations '
'
(c) y1 œ C1 e Paxb dx , y2 œ C# e Paxb dx , y œ y1 y2 . So yax0 b œ 0 Ê C1 e Ê C1 C2 œ 0 Ê C1 œ C2 Ê y1 axb œ y2 axb for a x b. 5. ax2 y2 bdx x y dy œ 0 Ê Ê
dx x
dv v ˆ 1v v‰
œ
dy dx
œ0Ê'
ˆx 2 y 2 ‰ xy
'
dx x
œ yx
y x
œ y1Îx
y x
' Paxb dx
C# e
œ Fˆ yx ‰ Ê Favb œ v1 v Ê
œ C Ê lnlxl 41 lnl2v2 1l œ C Ê 4lnlxl
v dv 2v2 1
' Paxb dx
2 lnl2ˆ yx ‰
œ!
dx x
dv v Favb
œ0
1l œ C
Ê lnlx4 l lnº 2y x2 x º œ C Ê ln¹x2 a2y2 x2 b¹ œ C Ê x2 a2y2 x2 b œ eC Ê x2 a2y2 x2 b œ C 2
2
6. x2 dy ay2 x ybdx œ 0 Ê Ê'
dx x
'
dv v2
œ C Ê lnlxl
7. ˆx eyÎx y‰dx x dy œ 0 Ê Ê'
dx x
'
dv ev
dx x
'
Ê
dy dx
œ C Ê lnlxl
1 v
dy dx
œ
x eyÎx y x
2
œ ˆ yx ‰ 1 yÎx
œ eyÎx
y x
œ Fˆ yx ‰ Ê Favb œ v2 v Ê
œ C Ê lnlxl y x
œ Fˆ yx ‰ Ê Favb œ ev v Ê
a1 vbdv v2 1
œ0Ê'
ax y b xy
dy dx
œ
dx x
'
dv v2 1
œ
y x 1 1 xy
'
œ Fˆ yx ‰ Ê Favb œ
v dv v2 1
Ê'
cosˆ y x x ‰ œ
dx x
y x
dx x
v1 1v
Ê
2
x2 x2 º
cosˆ yx 1‰ œ Fˆ yx ‰ Ê Favb œ v cosav 1b Ê
dx x
dv v ae v v b
œ0
dx x
dv 1 v ˆ v1 c bv‰
œ0
œ C Ê 2 tan1 ˆ yx ‰ ln¹y2 x2 ¹ œ C dx x
dv v av cosav 1bb
œ0
' secav 1b dv œ 0 Ê lnlxl lnlsecav 1b tanav 1bl œ C Ê lnlxl ln¹secˆ yx 1‰ tanˆ yx 1‰¹ œ C
10. ˆx sin yx y cos yx ‰ dx x cos yx dy œ 0 Ê Ê
œ0
œ 0 Ê lnlxl tan1 v 21 lnlv2 1l œ C
2
y x
dv v av2 vb
œC
x y
Ê 2 lnlxl 2 tan1 v ln¹ˆ yx ‰ 1¹ œ C Ê lnlx2 l 2 tan1 ˆ yx ‰ lnº y 9. y w œ
dx x
œ C Ê lnlxl ev œ C Ê lnlxl eyÎx œ C
8. ax ybdy ax ybdx œ 0 Ê Ê'
ˆy 2 x y ‰ x2
œ
dy dx
dv v av tan vb
œ0Ê'
dx x
dy dx
œ
ˆx sin xy y cos xy ‰ x cos xy
œ
y x
tan yx œ Fˆ yx ‰ Ê Favb œ v tan v
' cot v dv œ 0 Ê lnlxl lnlsin vl œ C Ê lnlxl ln¹sin yx ¹ œ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 SEQUENCES 1. a" œ
1 1 1#
2. a" œ
1 1!
3.
a" œ
"2 ##
œ 0, a# œ
œ 1, a# œ (1)# #1
" #!
œ
œ 1, a# œ
œ "4 , a$ œ
1 3 3#
" 2
1 6
, a$ œ
(")$ 41
œ
1 3!
, a% œ
œ 3" , a$ œ
1 4 4#
œ 92 , a% œ œ
1 4!
(1)% 61
œ
" 5
3 œ 16
1 24 (1)& 81
, a% œ
œ 7"
4. a" œ 2 (1)" œ 1, a# œ 2 (1)# œ 3, a$ œ 2 (1)$ œ 1, a% œ 2 (1)% œ 3 5. a" œ
2 ##
œ
6. a" œ
2" #
" #
, a# œ
œ
" #
, a# œ " # 255 128
7. a" œ 1, a# œ 1 a( œ
127 64
, a) œ
8. a" œ 1, a# œ a* œ
" 362,880
" #
2# 2$
œ
œ
" #
2# 1 2# 3 #
œ
511 256
, a$ œ 3 #
" #
œ
" ##
, a"! œ
ˆ #" ‰ " 3 œ 6 " 3,628,800
, a$ œ
, a"! œ
3 4
, a$ œ
, a* œ
2$ #%
, a$ œ
, a% œ
, a% œ
2$ 1 2$
œ
7 4
œ
2% 2& 7 8
œ
" #
, a% œ
, a% œ
7 4
2% " 2%
" #$
œ
a' œ
,
15 8
ˆ "6 ‰ 4
œ
" #4
, a& œ
ˆ #"4 ‰ 5
œ
$ (1)% ˆ "# ‰ (1)# (2) œ 1, a$ œ (1)2 (1) œ "# , a% œ # # " " a( œ 3"# , a) œ 64 , a* œ 1#"8 , a"! œ 256
1†(2) œ 1, a$ œ 2†(31) œ 32 , a% # a) œ "4 , a* œ 29 , a"! œ "5
10. a" œ 2, a# œ a( œ 27 ,
15 16
, a& œ
15 8
" #%
œ
œ
31 16 , a'
63 32
,
1023 512
9. a" œ 2, a# œ " 16
œ
œ
3†ˆ 23 ‰ 4
" 1#0
, a' œ
" 7#0
œ 4" , a& œ
œ "# , a& œ
, a( œ
" 5040
(1)& ˆ 4" ‰ #
4†ˆ "# ‰ 5
, a) œ
œ
" 8
" 40,320
,
,
œ 52 , a' œ 3" ,
11. a" œ 1, a# œ 1, a$ œ 1 1 œ 2, a% œ 2 1 œ 3, a& œ 3 2 œ 5, a' œ 8, a( œ 13, a) œ 21, a* œ 34, a"! œ 55 12. a" œ 2, a# œ 1, a$ œ "# , a% œ
ˆ "# ‰ 1
œ
" #
, a& œ
ˆ "# ‰ ˆ "# ‰
œ 1, a' œ 2, a( œ 2, a) œ 1, a* œ "# , a"! œ
13. an œ (1)n1 , n œ 1, 2, á
14. an œ (1)n , n œ 1, 2, á
15. an œ (1)n1 n# , n œ 1, 2, á
16. an œ
(")n n#
1
, n œ 1, 2, á
18. an œ
2n 5 n an 1 b
, n œ 1, 2, á
17. an œ
2n 1 3 an 2 b ,
n œ 1, 2, á
19. an œ n# 1, n œ 1, 2, á
20. an œ n 4 , n œ 1, 2, á
21. an œ 4n 3, n œ 1, 2, á
22. an œ 4n 2 , n œ 1, 2, á
23. an œ
3n 2 n! ,
n œ 1, 2, á
24. an œ
n3 5n 1
, n œ 1, 2, á
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" #
570
Chapter 10 Infinite Sequences and Series
25. an œ
1 (1)n #
1
, n œ 1, 2, á
26. an œ
27. n lim 2 (0.1)n œ 2 Ê converges Ä_ n (")n n
29. n lim Ä_
" 2n 1 #n
30. n lim Ä_
2n " 1 3È n
œ n lim Ä_
31. n lim Ä_
" 5n% n% 8n$
œ n lim Ä_
32. n lim Ä_
n3 n# 5n 6
œ n lim Ä_
n3 (n 3)(n 2)
œ n lim Ä_
33. n lim Ä_
n# 2n 1 n1
œ n lim Ä_
(n 1)(n 1) n1
œ n lim (n 1) œ _ Ê diverges Ä_
34 n lim Ä_
" n$ 70 4n#
ˆ "n ‰ 2 ˆ "n ‰ 2
œ n lim Ä_
œ 1 Ê converges
2Èn Š È"n ‹
1 ˆ 8n ‰
" ‹n n# 70 Š #‹4 n
Š
œ 1 Ê converges
œ _ Ê diverges
Š È"n 3‹
œ n lim Ä_
2 #
œ n lim Ä_
Š n"% ‹ 5
œ 5 Ê converges " n#
œ 0 Ê converges
œ _ Ê diverges 36. n lim (1)n ˆ1 "n ‰ does not exist Ê diverges Ä_
35. n lim a1 (1)n b does not exist Ê diverges Ä_ ˆ n #n " ‰ ˆ1 "n ‰ œ lim ˆ "# 37. n lim Ä_ nÄ_ ˆ2 38. n lim Ä_
" ‰ˆ 3 #n
"‰ #n
ˆ "# ‰n œ lim 40. n lim Ä_ nÄ_
É n 2n 41. n lim 1 œ É n lim Ä_ Ä_ 42. n lim Ä_
" (0.9)n
" ‰ˆ 1 #n
n" ‰ œ
œ 6 Ê converges
(")n #n
œ Ú n# Û, n œ 1, 2, á
(Theorem 5, #4)
28. n lim Ä_
œ n lim 1 Ä_
(1)n n
n "# (1)n ˆ "# ‰ #
" #
Ê converges 39. n lim Ä_
(")nb1 #n 1
œ 0 Ê converges
œ 0 Ê converges
2n n1
œ Ên lim Š 2 ‹ œ È2 Ê converges Ä _ 1 " n
ˆ "0 ‰n œ _ Ê diverges œ n lim Ä_ 9
ˆ 1 n" ‰‹ œ sin 43. n lim sin ˆ 1# "n ‰ œ sin Šn lim Ä_ Ä_ #
1 #
œ 1 Ê converges
44. n lim n1 cos (n1) œ n lim (n1)(1)n does not exist Ê diverges Ä_ Ä_ 45. n lim Ä_
sin n n
46. n lim Ä_
sin# n #n
47. n lim Ä_
n #n
œ 0 because n" Ÿ œ 0 because 0 Ÿ
œ n lim Ä_
" #n ln 2
sin n n
sin# n #n
Ÿ
Ÿ
" n
" #n
Ê converges by the Sandwich Theorem for sequences Ê converges by the Sandwich Theorem for sequences
^ œ 0 Ê converges (using l'Hopital's rule)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.1 Sequences 48. n lim Ä_
3n n$
49. n lim Ä_
ln (n ") Èn
50. n lim Ä_
ln n ln 2n
œ n lim Ä_
3n ln 3 3n#
œ n lim Ä_
œ n lim Ä_
œ n lim Ä_ ˆn " 1‰
" ‹ Š #È n
ˆ "n ‰ 2 ‰ ˆ 2n
3n (ln 3)# 6n
œ n lim Ä_
œ n lim Ä_
2È n n1
3n (ln 3)$ 6
œ n lim Ä_
^ œ _ Ê diverges (using l'Hopital's rule)
Š È2n ‹
1 Š n" ‹
œ 0 Ê converges
œ 1 Ê converges
51. n lim 81În œ 1 Ê converges Ä_
(Theorem 5, #3)
52. n lim (0.03)1În œ 1 Ê converges Ä_
(Theorem 5, #3)
ˆ1 7n ‰n œ e( Ê converges 53. n lim Ä_ ˆ1 "n ‰n œ lim ’1 54. n lim Ä_ nÄ_
(") n “
(Theorem 5, #5) n
œ e" Ê converges
(Theorem 5, #5)
n È 55. n lim 10n œ n lim 101În † n1În œ 1 † 1 œ 1 Ê converges Ä_ Ä_
# n n È ˆÈ 56. n lim n# œ n lim n‰ œ 1# œ 1 Ê converges Ä_ Ä_
ˆ 3 ‰1În œ nÄ_ 1În œ 57. n lim lim n Ä_ n nÄ_ lim 31În
" 1
œ 1 Ê converges
(Theorem 5, #3 and #2)
(Theorem 5, #2)
(Theorem 5, #3 and #2)
58. n lim (n 4)1ÎÐn4Ñ œ x lim x1Îx œ 1 Ê converges; (let x œ n 4, then use Theorem 5, #2) Ä_ Ä_ 59. n lim Ä_
ln n n1În
lim Ä_ ln1Înn œ œ nlim n n
Ä_
_ 1
œ _ Ê diverges
(Theorem 5, #2)
60. n lim cln n ln (n 1)d œ n lim ln ˆ n n 1 ‰ œ ln Šn lim Ä_ Ä_ Ä_ n n È 61. n lim 4n n œ n lim 4È n œ 4 † 1 œ 4 Ê converges Ä_ Ä_
n n1‹
œ ln 1 œ 0 Ê converges
(Theorem 5, #2)
n È 62. n lim 32n1 œ n lim 32 a1Înb œ n lim 3# † 31În œ 9 † 1 œ 9 Ê converges Ä_ Ä_ Ä_
œ n lim Ä_
"†2†3â(n 1)(n) n†n†nân†n
63. n lim Ä_
n! nn
64. n lim Ä_
(4)n n!
65. n lim Ä_
n! 106n
œ n lim Ä_
" 'n Š (10n! ) ‹
66. n lim Ä_
n! 2n 3n
œ n lim Ä_
" ˆ 6n!n ‰
œ 0 Ê converges
ˆ " ‰ œ 0 and Ÿ n lim Ä_ n
n! nn
0 Ê n lim Ä_
n! nn
(Theorem 5, #3)
œ 0 Ê converges
(Theorem 5, #6)
œ _ Ê diverges
œ _ Ê diverges
(Theorem 5, #6)
(Theorem 5, #6)
ˆ " ‰1ÎÐln nÑ œ lim exp ˆ ln"n ln ˆ n" ‰‰ œ lim exp ˆ ln 1lnnln n ‰ œ e" Ê converges 67. n lim Ä_ n nÄ_ nÄ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
571
572
Chapter 10 Infinite Sequences and Series
n ˆ1 n" ‰n ‹ œ ln e œ 1 Ê converges 68. n lim ln ˆ1 "n ‰ œ ln Šn lim Ä_ Ä_
(Theorem 5, #5)
" ‰‰ ˆ 3n " ‰n œ lim exp ˆn ln ˆ 3n 69. n lim œ n lim exp Š ln (3n 1) " ln (3n 1) ‹ 3n 1 Ä _ 3n 1 nÄ_ Ä_ n 3
3
6n #Î$ ˆ6‰ œ n lim exp 3n 1 "3n 1 œ n lim exp Š (3n 1)(3n Ê converges 1) ‹ œ exp 9 œ e Ä_ Ä_ Š ‹ #
n#
"
"
ˆ n ‰n œ lim exp ˆn ln ˆ n n 1 ‰‰ œ lim exp Š ln n ln" (n 1) ‹ œ lim exp n n 1 70. n lim ˆn‰ Ä _ n1 nÄ_ nÄ_ nÄ_ Š "# ‹ n
œ n lim exp Š Ä_
n# n(n 1) ‹
"
œe
Ê converges
1) ˆ x ‰1În œ lim x ˆ #n " 1 ‰1În œ x lim exp ˆ n" ln ˆ #n " 1 ‰‰ œ x lim exp Š ln (2n 71. n lim ‹ n Ä _ 2n 1 nÄ_ nÄ_ nÄ_ 2 ! œ x n lim exp ˆ 2n1 ‰ œ xe œ x, x 0 Ê converges Ä_ n
ˆ1 72. n lim Ä_
" ‰n n#
œ n lim exp ˆn ln ˆ1 Ä_
" ‰‰ n#
œ n lim exp Ä_
ln Š1 n"# ‹
exp – œ n lim Ä_
ˆ n" ‰
Š n2$ ‹‚Š1 n"# ‹ Š n"# ‹
—
œ n lim exp ˆ n# 2n1 ‰ œ e! œ 1 Ê converges Ä_ 73. n lim Ä_
3 n †6 n 2cn †n!
œ n lim Ä_
36n n!
œ 0 Ê converges
ˆ 10 ‰n
ˆ 12 ‰n ˆ 10 ‰n 11 11 n 9 n 12 ‰n ˆ 11 ‰n ˆ 12 ‰ ˆ ‰ ˆ 11 11 10 12
11 74. n lim œ n lim n ‰n Ä _ ˆ 109 ‰ ˆ 11 Ä_ 12 (Theorem 5, #4)
75. n lim tanh n œ n lim Ä_ Ä_
en e en e
76. n lim sinh (ln n) œ n lim Ä_ Ä_
77. n lim Ä_
n# sin ˆ "n ‰ 2n 1
œ n lim Ä_
(Theorem 5, #6)
n n
œ n lim Ä_
eln n e 2
ln n
sin ˆ "n ‰
Èn sinŠ È1 ‹ œ lim 79. n lim n Ä_ nÄ_
ˆ" cos "n ‰ ˆ n" ‰
sinŠ È1n ‹
Èn 1
œ n lim Ä_ n ˆ "n ‰ #
œ n lim Ä_
œ n lim Ä_
Š 2n n"# ‹
78. n lim n ˆ1 cos "n ‰ œ n lim Ä_ Ä_
e2n " e2n 1
ˆ 120 ‰n 121 n ˆ 108 ‰ 1 110
œ n lim Ä_
2e2n 2e2n
Š n2# n2$ ‹
œ n lim Ä_
œ n lim " œ 1 Ê converges Ä_
œ _ Ê diverges
ˆcos ˆ n" ‰‰ Š n"# ‹
œ n lim Ä_
œ 0 Ê converges
œ n lim Ä_
sin ˆ n" ‰‘ Š "# ‹ n Š n"# ‹
cos Š È1n ‹Š
1 2n3Î2
1 ‹ 2n3Î2
cos ˆ n" ‰ # ˆ 2n ‰
œ
" #
Ê converges
sin ˆ "n ‰ œ 0 Ê converges œ n lim Ä_
œ n lim cos Š È1n ‹ œ cos 0 œ 1 Ê converges Ä_
80. n lim a3n 5n b1În œ n lim exp’lna3n 5n b1În “ œ n lim exp’ lna3 n 5 b “ œ n lim exp– Ä_ Ä_ Ä_ Ä_ n
n
œ n lim exp’ Ä_
Š 35n ‹ln 3 ln 5
81. n lim tan" n œ Ä_
ˆ 35nn ‰ 1 1 #
exp’ “ œ n lim Ä_
Ê converges
ˆ 35 ‰n ln 3 ln 5 ˆ 35 ‰n 1 “
n
3n ln 3 b 5n ln 5 3n b 5n
1
—
œ expaln 5b œ 5 82. n lim Ä_
" Èn
tan" n œ 0 †
1 #
œ 0 Ê converges
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.1 Sequences ˆ " ‰n 83. n lim Ä_ 3
" È 2n
573
n
n œ n lim Šˆ 3" ‰ Š È"2 ‹ ‹ œ 0 Ê converges Ä_
(Theorem 5, #4)
#
n 1‰ ! È 84. n lim n# n œ n lim exp ’ ln ann nb “ œ n lim exp ˆ 2n n# n œ e œ 1 Ê converges Ä_ Ä_ Ä_
85. n lim Ä_
(ln n)#!! n
86. n lim Ä_
(ln n)& Èn
œ n lim Ä_
200 (ln n)"** n
œ n lim Ä_
200†199 (ln n)"*) n
œ á œ n lim Ä_
200! n
œ 0 Ê converges
%
œ n lim Ä_ –
Š 5(lnnn) ‹ "
Š #Èn ‹
— œ n lim Ä_
10(ln n)% Èn
œ n lim Ä_ È
80(ln n)$ Èn
œ á œ n lim Ä_
#
87. n lim Šn Èn# n‹ œ n lim Šn Èn# n‹ Š n Èn# n ‹ œ n lim Ä_ Ä_ Ä_ n n n œ
" #
88. n lim Ä_
œ 0 Ê converges
œ n lim Ä_
" 1 É1
" n
Ê converges " È n# 1 È n# n
œ n lim Š Ä_ È
É1 n"# É1 "n
œ n lim Ä_ 89. n lim Ä_
n n È n# n
3840 Èn
ˆ "n 1‰
' 90. n lim Ä_ 1
n
" xp
œ n lim Ä_
È n# 1 È n# n 1 n
œ 2 Ê converges
'1n x" dx œ n lim Ä_
" n
È # È # " ‹ Š Èn# 1 Èn# n ‹ n# 1 È n# n n 1 n n
ln n n
dx œ n lim ’ " Ä _ 1 p
œ n lim Ä_ n
" xpc1 “ 1
" n
œ 0 Ê converges
œ n lim Ä_
" 1 p
ˆ np"c1 1‰ œ
(Theorem 5, #1) " p 1
if p 1 Ê converges
72 91. Since an converges Ê n lim a œ L Ê n lim a œ n lim ÊLœ Ä_ n Ä _ n1 Ä _ 1 an Ê L œ 9 or L œ 8; since an 0 for n 1 Ê L œ 8
72 1L
Ê La1 Lb œ 72 Ê L2 L 72 œ 0
an 6 92. Since an converges Ê n lim a œ L Ê n lim a œ n lim ÊLœ Ä_ n Ä _ n1 Ä _ an 2 Ê L œ 3 or L œ 2; since an 0 for n 2 Ê L œ 2
L6 L2
Ê LaL 2b œ L 6 Ê L2 L 6 œ 0
È8 2an Ê L œ È8 2L Ê L2 2L 8 œ 0 Ê L œ 2 93. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ or L œ 4; since an 0 for n 3 Ê L œ 4 È8 2an Ê L œ È8 2L Ê L2 2L 8 œ 0 Ê L œ 2 94. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ or L œ 4; since an 0 for n 2 Ê L œ 4 È5an Ê L œ È5L Ê L2 5L œ 0 Ê L œ 0 or L œ 5; since 95. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ an 0 for n 1 Ê L œ 5 ˆ12 Èan ‰ Ê L œ Š12 ÈL‹ Ê L2 25L 144 œ 0 96. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ Ê L œ 9 or L œ 16; since 12 Èan 12 for n 1 Ê L œ 9 97. an 1 œ 2
n 1, a1 œ 2. Since an converges Ê n lim a œ L Ê n lim a œ n lim Š2 Ä_ n Ä _ n1 Ä_ Ê L2 2L 1 œ 0 Ê L œ 1 „ È2; since an 0 for n 1 Ê L œ 1 È2 1 an ,
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 an ‹
ÊLœ2
1 L
574
Chapter 10 Infinite Sequences and Series
È 1 an Ê L œ È 1 L 98. an 1 œ È1 an , n 1, a1 œ È1. Since an converges Ê n lim a œ L Ê n lim a œ n lim Ä_ n Ä _ n1 Ä_ Ê L2 L 1 œ 0 Ê L œ
1 „ È5 ; 2
since an 0 for n 1 Ê L œ
1 È5 2
99. 1, 1, 2, 4, 8, 16, 32, á œ 1, 2! , 2" , 2# , 2$ , 2% , 2& , á Ê x" œ 1 and xn œ 2nc2 for n 2 100. (a) 1# 2(1)# œ 1, 3# 2(2)# œ 1; let f(aß b) œ (a 2b)# 2(a b)# œ a# 4ab 4b# 2a# 4ab 2b# œ 2b# a# ; a# 2b# œ 1 Ê f(aß b) œ 2b# a# œ 1; a# 2b# œ 1 Ê f(aß b) œ 2b# a# œ 1 #
‰ 2œ (b) r#n 2 œ ˆ aa2b b
a# 4ab 4b# 2a# 4ab 2b# (a b)#
In the first and second fractions, yn n. Let
a b
œ
aa# 2b# b (a b)#
œ
„" y#n
Ê rn œ Ê2 „ Š y"n ‹
represent the (n 1)th fraction where
for n a positive integer 3. Now the nth fraction is lim rn œ È2.
a 2b ab
a b
#
1 and b n 1
and a b 2b 2n 2 n Ê yn n. Thus,
nÄ_
101. (a) f(x) œ x# 2; the sequence converges to 1.414213562 ¸ È2 (b) f(x) œ tan (x) 1; the sequence converges to 0.7853981635 ¸
1 4
(c) f(x) œ ex ; the sequence 1, 0, 1, 2, 3, 4, 5, á diverges 102. (a) n lim nf ˆ "n ‰ œ lim b f(??xx) œ lim b f(0??x)x f(0) œ f w (0), where ?x œ Ä_ ?x Ä ! ?x Ä ! " " ˆ " ‰ w " (b) n lim n tan œ f (0) œ x # œ 1, f(x) œ tan n 1 0 Ä_
" n
(c) n lim n ae1În 1b œ f w (0) œ e! œ 1, f(x) œ ex 1 Ä_ (d) n lim n ln ˆ1 2n ‰ œ f w (0) œ 1 22(0) œ 2, f(x) œ ln (1 2x) Ä_ #
103. (a) If a œ 2n 1, then b œ Ú a# Û œ Ú 4n
#
4n 1 Û # #
#
œ Ú2n# 2n "# Û œ 2n# 2n, c œ Ü a# Ý œ Ü2n# 2n "# Ý #
œ 2n# 2n 1 and a# b# œ (2n 1) a2n# 2nb œ 4n# 4n 1 4n% 8n$ 4n# #
œ 4n% 8n$ 8n# 4n 1 œ a2n# 2n 1b œ c# . (b) a lim Ä_
# Ú a# Û # Ü a# Ý
œ a lim Ä_
2n# 2n 2n# 2n 1
œ 1 or a lim Ä_
#
Ú a# Û #
Ü a# Ý
œ a lim sin ) œ Ä_
2n1 ‰ 104. (a) n lim (2n1)1Î a2nb œ n lim exp ˆ ln2n œ n lim exp Ä_ Ä_ Ä_
21 Š 2n 1‹
#
(b)
n 40 50 60
15.76852702 19.48325423 23.19189561
sin ) œ 1
exp ˆ #"n ‰ œ e! œ 1; œ n lim Ä_
n n n! ¸ ˆ ne ‰ È 2n1 , Stirlings approximation Ê È n! ¸ ˆ ne ‰ (2n1)1Î a2nb ¸ n È n!
lim
) Ä 1 Î2
n e
for large values of n
n e
14.71517765 18.39397206 22.07276647
ˆ"‰
ln n " n 105. (a) n lim œ n lim œ n lim œ0 Ä _ nc Ä _ cncc1 Ä _ cnc Ðln %ÑÎc (b) For all % 0, there exists an N œ e such that n eÐln %ÑÎc Ê ln n lnc % Ê ln nc ln ˆ "% ‰ Ê nc "% Ê n"c % Ê ¸ n"c 0¸ % Ê lim n"c œ 0 nÄ_
106. Let {an } and {bn } be sequences both converging to L. Define {cn } by c2n œ bn and c2nc1 œ an , where n œ 1, 2, 3, á . For all % 0 there exists N" such that when n N" then kan Lk % and there exists N# such that when n N# then kbn Lk %. If n 1 2max{N" ß N# }, then kcn Lk %, so {cn } converges to L.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.1 Sequences
575
107. n lim n1În œ n lim exp ˆ "n ln n‰ œ n lim exp ˆ n" ‰ œ e! œ 1 Ä_ Ä_ Ä_ 108. n lim x1În œ n lim exp ˆ "n ln x‰ œ e! œ 1, because x remains fixed while n gets large Ä_ Ä_ 109. Assume the hypotheses of the theorem and let % be a positive number. For all % there exists a N" such that when n N" then kan Lk % Ê % an L % Ê L % an , and there exists a N# such that when n N# then kcn Lk % Ê % cn L % Ê cn L %. If n max{N" ß N# }, then L % an Ÿ bn Ÿ cn L % Ê kbn Lk % Ê n lim b œ L. Ä_ n 110. Let % !. We have f continuous at L Ê there exists $ so that kx Lk $ Ê kf(x) f(L)k %. Also, an Ä L Ê there exists N so that for n N kan Lk $ . Thus for n N, kf(an ) f(L)k % Ê f(an ) Ä f(L). 111. an1 an Ê
3(n 1) 1 (n 1) 1
3n 1 n1
3n 4 n#
Ê
3n 1 n1
Ê 3n# 3n 4n 4 3n# 6n n 2
Ê 4 2; the steps are reversible so the sequence is nondecreasing;
3n " n1
3 Ê 3n 1 3n 3
Ê 1 3; the steps are reversible so the sequence is bounded above by 3 112. an1 an Ê
(2(n 1) 3)! ((n 1) 1)!
(2n 3)! (n 1)!
Ê
(2n 5)! (n 2)!
(2n 3)! (n 1)!
Ê
(2n 5)! (2n 3)!
(n 2)! (n 1)!
Ê (2n 5)(2n 4) n 2; the steps are reversible so the sequence is nondecreasing; the sequence is not bounded since 113. an1 Ÿ an Ê
(2n 3)! (n 1)!
œ (2n 3)(2n 2)â(n 2) can become as large as we please
2nb1 3nb1 (n 1)!
Ÿ
2n 3n n!
2nb1 3nb1 2n 3n
Ê
(n 1)! n!
Ÿ
Ê 2 † 3 Ÿ n 1 which is true for n 5; the steps are
reversible so the sequence is decreasing after a& , but it is not nondecreasing for all its terms; a" œ 6, a# œ 18, a$ œ 36, a% œ 54, a& œ 324 5 œ 64.8 Ê the sequence is bounded from above by 64.8 114. an1 an Ê 2
2 n 1
" #nb1
2
2 n
" #n
Ê
reversible so the sequence is nondecreasing; 2 115. an œ 1
" n
converges because
116. an œ n
" n
diverges because n Ä _ and
117. an œ
2 n 1 2n
œ1
" #n
and 0
" #n
" n
2 n 2 n
2 " n1 #nb1 " #n Ÿ 2 Ê
" #n
Ê
2 n(n 1)
#n"b1 ; the steps are
the sequence is bounded from above
Ä 0 by Example 1; also it is a nondecreasing sequence bounded above by 1
" n
; since
" n " n
Ä 0 by Example 1, so the sequence is unbounded Ä 0 (by Example 1) Ê
" #n
Ä 0, the sequence converges; also it is
a nondecreasing sequence bounded above by 1 118. an œ
2 n 1 3n
n
œ ˆ 23 ‰
" 3n
; the sequence converges to ! by Theorem 5, #4
119. an œ a(1)n 1b ˆ nn 1 ‰ diverges because an œ 0 for n odd, while for n even an œ 2 ˆ1 n" ‰ converges to 2; it diverges by definition of divergence 120. xn œ max {cos 1ß cos 2ß cos 3ß á ß cos n} and xn1 œ max {cos 1ß cos 2ß cos 3ß á ß cos (n 1)} xn with xn Ÿ 1 so the sequence is nondecreasing and bounded above by 1 Ê the sequence converges. 121. an an1 Í and
1 È2n Èn
1 È2n Èn
" È2(n 1) Èn 1
Í Èn 1 È2n# 2n Èn È2n# 2n Í Èn 1 Èn
È2 ; thus the sequence is nonincreasing and bounded below by È2 Ê it converges
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576
Chapter 10 Infinite Sequences and Series
122. an an1 Í
n1 n
(n 1) " n1
Í n# 2n 1 n# 2n Í 1 0 and
n1 n
1; thus the sequence is
nonincreasing and bounded below by 1 Ê it converges 123.
4nb1 3n œ4 4n 3 ‰n ˆ 4 4 4;
ˆ 43 ‰n so an an1 Í 4 ˆ 43 ‰n 4 ˆ 43 ‰n" Í ˆ 43 ‰n ˆ 43 ‰n1 Í 1
3 4
and
thus the sequence is nonincreasing and bounded below by 4 Ê it converges
124. a" œ 1, a# œ 2 3, a$ œ 2(2 3) 3 œ 2# a22 "b † 3, a% œ 2 a2# a22 "b † 3b 3 œ 2$ a2$ 1b 3, a& œ 2 c2$ a2$ 1b 3d 3 œ 2% a2% 1b 3, á , an œ 2n" a2n" 1b 3 œ 2n" 3 † 2n1 3 œ 2n1 (1 3) 3 œ 2n 3; an an1 Í 2n 3 2n1 3 Í 2n 2n1 Í 1 Ÿ 2 so the sequence is nonincreasing but not bounded below and therefore diverges 125. Let 0 M 1 and let N be an integer greater than Ê n M nM Ê n M(n 1) Ê
n n1
M 1M
. Then n N Ê n
M.
M 1M
Ê n nM M
126. Since M" is a least upper bound and M# is an upper bound, M" Ÿ M# . Since M# is a least upper bound and M" is an upper bound, M# Ÿ M" . We conclude that M" œ M# so the least upper bound is unique. 127. The sequence an œ 1
(")n #
is the sequence
" #
,
3 #
,
" #
,
3 #
, á . This sequence is bounded above by
3 #
,
but it clearly does not converge, by definition of convergence. 128. Let L be the limit of the convergent sequence {an }. Then by definition of convergence, for corresponds an N such that for all m and n, m N Ê kam Lk kam an k œ kam L L an k Ÿ kam Lk kL an k
% #
% #
% #
% #
there
and n N Ê kan Lk #% . Now
œ % whenever m N and n N.
129. Given an % 0, by definition of convergence there corresponds an N such that for all n N, kL" an k % and kL# an k %. Now kL# L" k œ kL# an an L" k Ÿ kL# an k kan L" k % % œ 2%. kL# L" k 2% says that the difference between two fixed values is smaller than any positive number 2%. The only nonnegative number smaller than every positive number is 0, so kL" L# k œ 0 or L" œ L# . 130. Let k(n) and i(n) be two order-preserving functions whose domains are the set of positive integers and whose ranges are a subset of the positive integers. Consider the two subsequences akÐnÑ and aiÐnÑ , where akÐnÑ Ä L" , aiÐnÑ Ä L# and L" Á L# . Thus ¸akÐnÑ aiÐnÑ ¸ Ä kL" L# k 0. So there does not exist N such that for all m, n N Ê kam an k %. So by Exercise 128, the sequence Öan × is not convergent and hence diverges. 131. a2k Ä L Í given an % 0 there corresponds an N" such that c2k N" Ê ka2k Lk %d . Similarly, a2k1 Ä L Í c2k 1 N# Ê ka2k1 Lk %d . Let N œ max{N" ß N# }. Then n N Ê kan Lk % whether n is even or odd, and hence an Ä L. 132. Assume an Ä 0. This implies that given an % 0 there corresponds an N such that n N Ê kan 0k % Ê kan k % Ê kkan kk % Ê kkan k 0k % Ê kan k Ä 0. On the other hand, assume kan k Ä 0. This implies that given an % 0 there corresponds an N such that for n N, kkan k 0k % Ê kkan kk % Ê kan k % Ê kan 0k % Ê an Ä 0. 133. (a) f(x) œ x# a Ê f w (x) œ 2x Ê xn1 œ xn
x#n a #xn
Ê xn1 œ
2x#n ax#n ab 2xn
œ
x#n a 2xn
œ
ˆxn xa ‰ #
n
(b) x" œ 2, x# œ 1.75, x$ œ 1.732142857, x% œ 1.73205081, x& œ 1.732050808; we are finding the positive number where x# 3 œ 0; that is, where x# œ 3, x 0, or where x œ È3 .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.2 Infinite Series
577
134. x" œ 1, x# œ 1 cos (1) œ 1.540302306, x$ œ 1.540302306 cos (1 cos (1)) œ 1.570791601, x% œ 1.570791601 cos (1.570791601) œ 1.570796327 œ 1# to 9 decimal places. After a few steps, the arc axnc1 b and line segment cos axnc1 b are nearly the same as the quarter circle. 135-146. Example CAS Commands: Mathematica: (sequence functions may vary): Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n, 1, 25}] Limit[a[n], n Ä 8] Mathematica: (sequence functions may vary): Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n, 1, 25}] Limit[a[n], n Ä 8] The last command (Limit) will not always work in Mathematica. You could also explore the limit by enlarging your table to more than the first 25 values. If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01 of the limit, do the following. Clear[minN, lim] lim= 1 Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}] minN For sequences that are given recursively, the following code is suggested. The portion of the command a[n_]:=a[n] stores the elements of the sequence and helps to streamline computation. Clear[a, n] a[1]= 1; a[n_]; = a[n]= a[n 1] (1/5)(n1) first25= Table[N[a[n]], {n, 1, 25}] The limit command does not work in this case, but the limit can be observed as 1.25. Clear[minN, lim] lim= 1.25 Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}] minN 10.2 INFINITE SERIES 1. sn œ
a a1 r n b (1 r)
œ
n 2 ˆ1 ˆ "3 ‰ ‰ " 1 ˆ3‰
2. sn œ
a a1 r n b (1 r)
œ
9 ‰ˆ " ‰n ‰ ˆ 100 1 ˆ 100 " 1 ˆ 100 ‰
3. sn œ
a a1 r n b (1 r)
œ
1 ˆ "# ‰ 1 ˆ "# ‰
4. sn œ
1 (2)n 1 (2)
, a geometric series where krk 1 Ê divergence
5.
" (n 1)(n #)
œ
" n1
n
Ê n lim s œ Ä_ n
Ê n lim s œ Ä_ n
Ê n lim s œ Ä_ n
" n#
2 1 ˆ "3 ‰
" ˆ #3 ‰
œ3 9 ‰ ˆ 100
" ‰ 1 ˆ 100
œ
œ
" 11
2 3
Ê sn œ ˆ #" 3" ‰ ˆ 3" 4" ‰ á ˆ n " 1
" ‰ n#
œ
" #
" n#
Ê n lim s œ Ä_ n
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" #
578 6.
Chapter 10 Infinite Sequences and Series œ
5 n(n 1)
5 n
5 n1
Ê sn œ ˆ5 52 ‰ ˆ 52 53 ‰ ˆ 53 54 ‰ á ˆ n 5 1 n5 ‰ ˆ n5
5 ‰ n1
œ 5
5 n1
Ê n lim s œ5 Ä_ n 7. 1
8.
" 16
9.
7 4
" 4
10. 5
" 16
" 64
7 16
5 4
" 256
" 64
7 64
5 16
á , the sum of this geometric series is
á , the sum of this geometric series is
á , the sum of this geometric series is
5 64
5 1 ˆ "# ‰
" 1 ˆ "3 ‰
œ 10
œ
3 #
" 1 ˆ "3 ‰
œ 10
œ
3 #
14. 2
4 5
" 1 ˆ "5 ‰
8 25
œ2
16 125
5 6
" ‰ 25 œ 17 6
œ
4 5
" 1#
7 3
5 1 ˆ "4 ‰
œ4
" ‰ #7
á , is the sum of two geometric series; the sum is
" ‰ #7
á , is the difference of two geometric series; the sum is
ˆ 18
á œ 2 ˆ1
15. Series is geometric with r œ
œ
" 1 ˆ "4 ‰
17 #
13. (1 1) ˆ 1# "5 ‰ ˆ 41 1 1 ˆ "# ‰
ˆ 74 ‰
1 ˆ "4 ‰
œ
œ
23 #
12. (5 1) ˆ 5# "3 ‰ ˆ 45 9" ‰ ˆ 85 5 1 ˆ "# ‰
" ‰ ˆ 16 1 ˆ 4" ‰
á , the sum of this geometric series is
11. (5 1) ˆ 5# "3 ‰ ˆ 45 9" ‰ ˆ 85
" 1 ˆ "4 ‰
2 5
" ‰ 1#5
4 25
á , is the sum of two geometric series; the sum is
8 125
á ‰ ; the sum of this geometric series is 2 Š 1 "ˆ 2 ‰ ‹ œ 5
Ê ¹ 25 ¹ 1 Ê Converges to
2 5
1 1 25
œ
5 3
1 8
œ
1 7
16. Series is geometric with r œ 3 Ê ¹3¹ 1 Ê Diverges 17. Series is geometric with r œ
Ê ¹ 18 ¹ 1 Ê Converges to
1 8
1 18
18. Series is geometric with r œ 23 Ê ¹ 23 ¹ 1 Ê Converges to _
19. 0.23 œ !
nœ0
_
21. 0.7 œ !
nœ0
23 100
7 10
ˆ 10" # ‰n œ
" ‰n ˆ 10 œ
23 Š 100 ‹
"
1 ˆ 100 ‰
7 Š 10 ‹
1
" Š 10 ‹
œ
œ
_
nœ0
nœ0
_
nœ0
414 1000
nœ0
22. 0.d œ !
" 1 Š 10 ‹
24. 1.414 œ 1 !
_
7 9
6 Š 100 ‹
ˆ 10" $ ‰n œ 1
œ 25
20. 0.234 œ !
23 99
_
1 ‰ ˆ 6 ‰ ˆ " ‰n 23. 0.06 œ ! ˆ 10 œ 10 10
23 1 ˆ 23 ‰
œ
6 90
414 Š 1000 ‹
" 1 Š 1000 ‹
œ
d 10
234 1000
ˆ 10" $ ‰n œ
" ‰n ˆ 10 œ
234 Š 1000 ‹
" 1 Š 1000 ‹
d Š 10 ‹
" 1 Š 10 ‹
œ
d 9
" 15
œ1
414 999
œ
"413 999
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
234 999
10 3
Section 10.2 Infinite Series 25. 1.24123 œ
124 100
_
!
123 10&
nœ0
_
26. 3.142857 œ 3 !
nœ0
œ lim
124 100
28.
lim nan 1b nÄ_ an 2ban 3b
29.
lim 1 nÄ_ n 4
œ 0 Ê test inconclusive
30.
lim 2 n nÄ_ n 3
œ lim
33. 34.
10
1Š
" ‹ 10$
Š 142,857 ' ‹ 10
1Š
" ‹ 10'
œ
124 100
œ3
123 10& 10#
142,857 10' 1
œ
œ
124 100
3,142,854 999,999
123 99,900
œ
œ
123,999 99,900
œ
41,333 33,300
116,402 37,037
œ 1 Á 0 Ê diverges
lim n nÄ_ n 10
32.
Š 123& ‹
ˆ 10" ' ‰n œ 3
142,857 10'
27.
31.
1 nÄ_ 1
ˆ 10" $ ‰n œ
579
n2 n 2 nÄ_ n 5n 6
2n 1 nÄ_ 2n 5
œ lim
1 nÄ_ 2n
œ lim
œ lim
2 nÄ_ 2
œ 1 Á 0 Ê diverges
œ 0 Ê test inconclusive
lim cos 1n œ cos 0 œ 1 Á 0 Ê diverges
nÄ_
n lim ne nÄ_ e n
œ
n lim n e nÄ_ e 1
en n nÄ_ e
œ lim
œ lim
1 nÄ_ 1
œ 1 Á 0 Ê diverges
lim ln 1n œ _ Á 0 Ê diverges
nÄ_
lim cos n 1 œ does not exist Ê diverges
nÄ_
35. sk œ ˆ1 2" ‰ ˆ 2" 3" ‰ ˆ 3" 4" ‰ á ˆ k " 1 k" ‰ ˆ k" œ lim ˆ1 kÄ_
" ‰ k1
kÄ_
œ 1
" k1
Ê
œ 1, series converges to 1
36. sk œ ˆ 31 34 ‰ ˆ 34 39 ‰ ˆ 39 œ lim Š3
" ‰ k1
3 ‹ ak 1 b 2
3 ‰ 16
á Š ak 3 1b2
3 k2 ‹
Š k32
3 ‹ ak 1b2
œ 3
lim sk
kÄ_
3 ak 1b2
Ê
lim sk
kÄ_
œ 3, series converges to 3
37. sk œ ŠlnÈ2 lnÈ1‹ ŠlnÈ3 lnÈ2‹ ŠlnÈ4 lnÈ3‹ á ŠlnÈk lnÈk 1‹ ŠlnÈk 1 lnÈk‹ œ lnÈk 1 lnÈ1 œ lnÈk 1 Ê
lim sk œ lim lnÈk 1 œ _; series diverges
kÄ_
kÄ_
38. sk œ atan 1 tan 0b atan 2 tan 1b atan 3 tan 2b á atan k tan ak 1bb atan ak 1b tan kb œ tan ak 1b tan 0 œ tan ak 1b Ê lim sk œ lim tan ak 1b œ does not exist; series diverges kÄ_
kÄ_
39. sk œ ˆcos1 ˆ 12 ‰ cos1 ˆ 13 ‰‰ ˆcos1 ˆ 13 ‰ cos1 ˆ 14 ‰‰ ˆcos1 ˆ 14 ‰ cos1 ˆ 15 ‰‰ á ˆcos1 ˆ 1k ‰ cos1 ˆ k 1 1 ‰‰ ˆcos1 ˆ k 1 1 ‰ cos1 ˆ k 1 # ‰‰ œ 13 cos1 ˆ k 1 # ‰ Ê
lim sk œ lim ’ 13 cos1 ˆ k 1 # ‰“ œ
kÄ_
kÄ_
1 3
1 2
œ 16 , series converges to
1 6
40. sk œ ŠÈ5 È4‹ ŠÈ6 È5‹ ŠÈ7 È6‹ á ŠÈk 3 Èk 2‹ ŠÈk 4 Èk 3‹ œ Èk 4 2 Ê
lim sk œ lim ’Èk 4 2“ œ _; series diverges
kÄ_
kÄ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
580 41.
42.
Chapter 10 Infinite Sequences and Series 4 " " "‰ " ‰ ˆ ˆ" "‰ ˆ" (4n 3)(4n 1) œ 4n 3 4n 1 Ê sk œ 1 5 5 9 9 13 ˆ 4k " 3 4k " 1 ‰ œ 1 4k " 1 Ê lim sk œ lim ˆ1 4k " 1 ‰ œ 1 kÄ_ kÄ_
œ
6 (2n 1)(2n 1)
A 2n 1
A(2n 1) B(2n 1) (2n 1)(2n 1)
œ
B 2n 1
á ˆ 4k " 7
" ‰ 4k 3
Ê A(2n 1) B(2n 1) œ 6 Ê (2A 2B)n (A B) œ 6
k k 2A 2B œ 0 ABœ0 6 Ê œ Êœ Ê 2A œ 6 Ê A œ 3 and B œ 3. Hence, ! (2n 1)(2n œ 3 ! ˆ #n " 1 1) A Bœ6 ABœ6 nœ1 nœ1
œ 3 Š "1
" 3
lim 3 ˆ1
kÄ_
43.
40n (2n1)# (2n1)#
" 3
" 5
" ‰ #k 1
œ
A (2n1)
" 5
" 7
á
" #(k 1) 1
" 2k 1
" #k 1 ‹
œ
A(2n1)(2n1)# B(2n1)# C(2n1)(2n1)# D(2n1)# (2n1)# (2n1)# # #
œ 3 ˆ1
" ‰ #k 1
" ‰ #n 1
Ê the sum is
œ3
B (2n1)#
C (2n1) #
D (2n1)#
Ê A(2n 1)(2n 1)# B(2n 1) C(2n 1)(2n 1) D(2n 1) œ 40n Ê A a8n$ 4n# 2n 1b B a4n# 4n 1b C a8n$ 4n# 2n 1b œ D a4n# 4n 1b œ 40n Ê (8A 8C)n$ (4A 4B 4C 4D)n# (2A 4B 2C 4D)n (A B C D) œ 40n Ú Ú 8A 8C œ 0 8A 8C œ 0 Ý Ý Ý Ý 4A 4B 4C 4D œ 0 A BC Dœ 0 B Dœ 0 Ê Û Ê Û Ê œ Ê 4B œ 20 Ê B œ 5 œ 2A 4B 2C 4D 40 A 2 œ 2D œ 20 B C 2D 20 2B Ý Ý Ý Ý Ü A B C D œ 0 Ü A B C D œ 0 k ACœ0 Ê C œ 0 and A œ 0. Hence, ! ’ (#n1)40n and D œ 5 Ê œ # (2n1)# “ A 5 C 5 œ 0 nœ1 k
œ 5 ! ’ (#n" 1)# nœ1
44.
" (#n1)# “
œ 5 Š1
" (2k1)# ‹
2n 1 n# (n 1)#
" n#
Ê
œ
45. sk œ Š1 Ê
Š È" 2
kÄ_
" ‰ #"Î#
" ˆ #"Î#
lim sk œ
kÄ_
47. sk œ ˆ ln"3 œ ln"#
" ‰ ln #
" #
" 1
œ
" 9
" #5
" #5
á
" (2k1)# ‹
Ê
" (#k1)#
" (#k1)# ‹
œ5 " ‰ 16
á ’ (k " 1)#
" k# “
’ k"#
" (k 1)# “
" È4 ‹
á ŠÈ "
k1
" Èk ‹
Š È" k
" Èk 1 ‹
œ1
" Èk 1
œ1
" ˆ #"Î$
" ‰ ln 3
" (2(k1) 1)#
œ1
Š È"3
" Èk 1 ‹
" ‰ #"Î$ #"
ˆ ln"4
" ln (k 2)
" (k 1)# “
" È3 ‹
lim sk œ lim Š1
kÄ_
46. sk œ ˆ "# Ê
kÄ_
" È2 ‹
Ê sk œ ˆ1 4" ‰ ˆ 4" 9" ‰ ˆ 9"
lim sk œ lim ’1
kÄ_
" 9
Ê the sum is n lim 5 Š1 Ä_
" (n 1)#
œ 5 Š 1"
" ‰ #"Î%
ˆ ln"5
á ˆ #1ÎÐ"k
" ‰ ln 4
1Ñ
" ‰ #1Îk
á Š ln (k" 1)
ˆ #1"Îk
" ln k ‹
" ‰ #1ÎÐk1Ñ
Š ln (k" 2)
œ
" #
" #1ÎÐk1Ñ
" ln (k 1) ‹
lim sk œ ln"#
kÄ_
48. sk œ ctan" (1) tan" (2)d ctan" (2) tan" (3)d á ctan" (k 1) tan" (k)d ctan" (k) tan" (k 1)d œ tan" (1) tan" (k 1) Ê lim sk œ tan" (1) kÄ_
49. convergent geometric series with sum
" 1 Š È" ‹ 2
50. divergent geometric series with krk œ È2 1
œ
È2 È 2 1
1 #
œ
1 4
1 #
œ 14
œ 2 È2
51. convergent geometric series with sum
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Š 3# ‹ 1 Š "# ‹
œ1
Section 10.2 Infinite Series 52. n lim (1)n1 n Á 0 Ê diverges Ä_
53. n lim cos (n1) œ n lim (1)n Á 0 Ê diverges Ä_ Ä_
54. cos (n1) œ (1)n Ê convergent geometric series with sum " 1Š
55. convergent geometric series with sum
56. n lim ln Ä_
" 3n
" ‹ e#
2 " 1 Š 10 ‹
58. convergent geometric series with sum
" 1 Š "x ‹
59. difference of two geometric series with sum ˆ1 "n ‰n œ lim ˆ1 60. n lim Ä_ nÄ_
_
63. ! nœ1
n! 1000n
2n 3n 4n
since r œ _
! nœ1
64.
2n 3n 4n
_
nœ1
5 6
" ‰n n
2œ œ
Ê
2n 4n
_
!
¹ 12 ¹
nœ1
3n 4n
_
20 9
œ
18 9
2 9
x x1
" 1 Š 23 ‹
" 1 Š 3" ‹
œ3
œ
3 #
3 #
œ e" Á 0 Ê diverges 62. n lim Ä_ _
n
_
n
nn n!
œ n lim Ä_ _
n
n†nân 1†#ân
n lim n œ _ Ê diverges Ä_
n
œ ! ˆ 21 ‰ ! ˆ 43 ‰ ; both œ ! ˆ 21 ‰ and ! ˆ 43 ‰ are geometric series, and both converge nœ1
1 and r œ
nœ1
3 4
Ê
¹ 34 ¹
nœ1
1, respectivley Ê
nœ1
_
! ˆ 1 ‰n 2
nœ1
œ
1 2
1 12
_
n
œ 1 and ! ˆ 34 ‰ œ nœ1
3 4
1 34
œ3Ê
œ 1 3 œ 4 by Theorem 8, part (1)
2n 4n n n nÄ_ 3 4
lim
œ
e# e # 1
œ _ Á 0 Ê diverges
œ! 1 2
œ
" 1 Š "5 ‹
œ _ Á 0 Ê diverges
57. convergent geometric series with sum
61. n lim Ä_
581
œ
lim
nÄ_
_
_
nœ1
nœ1
2n 4n 3n 4n
" "
ˆ 12 ‰n " 3 n nÄ_ ˆ 4 ‰ "
œ lim
œ
1 1
œ 1 Á 0 Ê diverges by nth term test for divergence
65. ! ln ˆ n n 1 ‰ œ ! cln (n) ln (n 1)d Ê sk œ cln (1) ln (2)d cln (2) ln (3)d cln (3) ln (4)d á cln (k 1) ln (k)d cln (k) ln (k 1)d œ ln (k 1) Ê
lim sk œ _, Ê diverges
kÄ_
66. n lim a œ n lim ln ˆ 2n n 1 ‰ œ ln ˆ #" ‰ Á 0 Ê diverges Ä_ n Ä_ 67. convergent geometric series with sum 68. divergent geometric series with krk œ _
_
nœ0
nœ0
" 1 ˆ 1e ‰ e1 1e
¸
œ
23.141 22.459
1 1e
1
69. ! (1)n xn œ ! (x)n ; a œ 1, r œ x; converges to _
_
nœ0
nœ0
" 1 (x)
n 70. ! (1)n x2n œ ! ax# b ; a œ 1, r œ x# ; converges to
œ
" 1 x#
" 1x
for kxk 1
for kxk 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
582
Chapter 10 Infinite Sequences and Series
71. a œ 3, r œ _
72. ! nœ0
œ
(1)n #
x1 #
; converges to _
ˆ 3 "sin x ‰n œ !
nœ0
3 sin x 2(4 sin x)
œ
3 sin x 8 2 sin x
3 1 Šx
ˆ 3 "sin x ‰n ; a œ
" #
" 1 2x
74. a œ 1, r œ x"# ; converges to
for k2xk 1 or kxk
" #
; converges to
77. a œ 1, r œ sin x; converges to
_
79. (a) ! nœ2 _
80. (a) ! nœ1
" 1 (x 1)
" 1 Š3
x # ‹
" 1 sin x
œ
œ
" #x
for kx 1k 1 or 2 x 0
for kln xk 1 or e" x e _
5 (n 2)(n 3)
(b) !
nœ0 _
nœ3
" 4
(b) one example is 3# (c) one example is 1
" #
for all x‰
for x Á (2k 1) 1# , k an integer
(b) !
" #
" ‹ 1 Š 3 sin x
for ¸ 3 # x ¸ 1 or 1 x 5
2 x1
" (n 4)(n 5)
81. (a) one example is
ˆ "# ‰
#
" 1 ln x
78. a œ 1, r œ ln x; converges to
" 3 sin x
Ÿ
; converges to
x ¸1¸ " ‹ œ x# 1 for x# 1 or kxk 1. # x
75. a œ 1, r œ (x 1)n ; converges to 3x #
" 3 sin x
,rœ
" #
" 1Š
" 4
" #
Ÿ
for all x ˆsince
73. a œ 1, r œ 2x; converges to
76. a œ 1, r œ
6 x" " œ 3 x for 1 # 1 or 1 x 3 # ‹
" 8
" 16
á œ
3 4
3 8
3 16
" 4
" 8
Š "# ‹ 1 Š "# ‹
á œ " 16
_
" (n 2)(n 3)
(c) !
5 (n 2)(n 1)
(c) !
nœ5 _
nœ20
" (n 3)(n #)
5 (n 19)(n 18)
œ1
Š 3# ‹ 1 Š "# ‹
á œ1
œ 3 Š "# ‹ 1 Š "# ‹
œ 0.
_
Š k# ‹
nœ0
1 Š "# ‹
n 1 82. The series ! kˆ 12 ‰ is a geometric series whose sum is
œ k where k can be any positive or negative number.
_
_
_
_
_
nœ1
nœ1
nœ1
nœ1
nœ1
_
_
_
_
_
nœ1
nœ1
nœ1
nœ1
nœ1
n n 83. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! Š bann ‹ œ ! (1) diverges.
n n n 84. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! aan bn b œ ! ˆ 4" ‰ œ
n
n
_
85. Let an œ ˆ "4 ‰ and bn œ ˆ #" ‰ . Then A œ ! an œ nœ1
" 3
_
_
_
nœ1
nœ1
nœ1
" 3
Á AB.
n , B œ ! bn œ 1 and ! Š bann ‹ œ ! ˆ #" ‰ œ 1 Á
86. Yes: ! Š a"n ‹ diverges. The reasoning: ! an converges Ê an Ä 0 Ê
" an
A B
.
Ä _ Ê ! Š a"n ‹ diverges by the
nth-Term Test.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.3 The Integral Test 87. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series that diverges does not change the divergence of the series. 88. Let An œ a" a# á an and n lim A œ A. Assume ! aan bn b converges to S. Let Ä_ n Sn œ (a" b" ) (a# b# ) á (an bn ) Ê Sn œ (a" a# á an ) (b" b# á bn ) Ê b" b# á bn œ Sn An Ê n lim ab" b# á bn b œ S A Ê ! bn converges. This Ä_ contradicts the assumption that ! bn diverges; therefore, ! aan bn b diverges. 89. (a) (b)
2 1r
œ5 Ê
Š 13 2 ‹ 1r
2 5
œ5 Ê
œ1r Ê rœ
#
; 2 2 ˆ 53 ‰ 2 ˆ 53 ‰ á
3 5
3 œ 1 r Ê r œ 10 ;
13 10
90. 1 eb e2b á œ
" 1 e b
" 9
œ9 Ê
13 2
13 #
3 ‰ ˆ 10
œ 1 eb Ê eb œ
13 #
3 ‰# ˆ 10
13 #
3 ‰$ ˆ 10 á
Ê b œ ln ˆ 98 ‰
8 9
91. sn œ 1 2r r# 2r$ r% 2r& á r2n 2r2n1 , n œ 0, 1, á Ê sn œ a1 r# r% á r2n b a2r 2r$ 2r& á 2r2n1 b Ê n lim s œ Ä_ n 1 2r œ 1 r# , if kr# k 1 or krk 1 92. L sn œ
a 1r
a a1 r n b 1r
œ
#
#
#
94. (a) L" œ 3, L# œ 3 ˆ 43 ‰ , L$ œ 3 ˆ 43 ‰ , á , Ln œ 3 ˆ 43 ‰
nc1
" #
á œ
4 1
" #
An œ lim
È3 4
È3 ˆ " ‰2 4 ‹ 3
nÄ_
! 3a4bk2 Š
È3 8 ˆ5‰ 4
œ
kœ2
An œ
È3 4
œ
È3 ˆ " ‰2 4 ‹ 33
n
2r 1 r#
È3 1#
, A$ œ A# 3a4bŠ
, A5 œ A4 3a4b3 Š
È3 ˆ " ‰ k 1 4 ‹ 32
È3 lim nÄ_ Œ 4
œ
n
3È3Œ! kœ2
È3 4
œ 8 m#
Ê n lim L œ n lim 3 ˆ 43 ‰ Ä_ n Ä_
(b) Using the fact that the area of an equilateral triangle of side length s is
A% œ A$ 3a4b2 Š
arn 1 r
93. area œ 2# ŠÈ2‹ (1)# Š È" ‹ á œ 4 2 1 2
A# œ A" 3Š
" 1 r#
È3 ˆ " ‰2 4 ‹ 32
È3 ˆ " ‰2 4 ‹ 34 ,
œ
kœ2
œ
È3 4 ,
A" œ
...,
n
È3 4
œ_
È3 2 4 s , we see that È3 È3 È3 4 12 #7 ,
k 1 ! 3È3a4bk$ ˆ 9" ‰ œ
4kc$ 9k 1
nc1
1 36
3È 3 Œ 1 4 œ 9
n
4kc$ . 9k 1
È3 4
3È3Œ!
È3 4
1 ‰ 3È3ˆ 20 œ
kœ2
È3 ˆ 4 1
53 ‰
œ 85 A"
10.3 THE INTEGRAL TEST 1. faxb œ
1 x2
œ lim
bÄ_
2. faxb œ
1 x0.2
œ lim
bÄ_
_1
is positive, continuous, and decreasing for x 1; '1
_1
ˆ 1b 1‰ œ 1 Ê ' 1
_
x2
dx converges Ê !
n œ1
1 n2
x2
_
ˆ 54 b0.8 54 ‰ œ _ Ê ' 1
1 x0.2
_
bÄ_
'1b x1
2
dx œ lim
bÄ_
b
’ 1x “
1
converges
is positive, continuous, and decreasing for x 1; '1
_
dx œ lim
1 x0.2
dx œ lim
bÄ_
'1b x1
0.2
dx œ lim
bÄ_
dx diverges Ê ! n10.2 diverges n œ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
b
’ 45 x0.8 “
1
583
584
Chapter 10 Infinite Sequences and Series
3. faxb œ œ
1 x2 4
_
is positive, continuous, and decreasing for x 1; '1
lim ˆ 1 tan1 b2 bÄ_ 2
4. faxb œ
1 1 1 ‰ 2 tan 2
œ
1 4
bÄ_
2
_
bÄ_
_
œ lim
ˆ 2e12b
bÄ_
_
'3
x x2 4
x x2 4
1 2e2
_
_ ln x2
'3
œ
1 ‰ ln 2
1 ln 2
_
Ê '2
_
dx converges Ê !
1 xaln xb2
'3
b
bÄ_
x x2 4
_
n œ3
n œ1
bÄ_
dx œ lim
bÄ_
'3
ln x x
2
_
'7
2
x exÎ3
dx œ lim _
n œ3
n œ2
n enÎ3
n œ1
10. faxb œ
2
x exÎ3
18b Š 3a6b ‹ ebÎ3 2
Ê !
'7
b
bÄ_
bÄ_
œ
1 e1Î3
x4 x2 2x 1
œ
x4 ax 1 b 2
_
bÄ_ _
Ê !
n œ8
’lnlx 1l
bÄ_
9 e1
œ lim
bÄ_
_
ln 4 2
n4 n2 2n 1 diverges
16 e4Î3
x4 ax 1 b 2 b
1
2 ln x2 x2
bÄ_
3
36 e2
18x exÎ3
_
Ê !
n œ2
!
n œ7
11. converges; a geometric series with r œ
b
b
54 “ exÎ3 7 327 e7Î3
2
n enÎ3
x1 ax 1 b 2
ˆlnlb 1l
n4 n2 2n 1
" 10
2
converges
ˆ 21 lnab2 4b 21 lna13b‰ œ _ Ê ' 3
x x2 4
dx
0 for x e, thus f is decreasing for x 3;
_ ln x2
a2aln bb 2aln 3bb œ _ Ê '3
x ax 6 b 3exÎ3
œ
_
”'8 bÄ_
bÄ_
bÄ_
’ ln1x “
n œ3
327 e7Î3
e
dx œ lim
œ lim
b
dx œ lim
2
x
dx
n œ3
ˆ b54 ‰ Î3
bÄ_
'2b xaln1xb
2 ! lnnn diverges
’ e3xxÎ3
25 e5Î3
dx œ lim
0 for x 6, thus f is decreasing for x 7;
œ lim
bÄ_
_
Ê '7
x2 exÎ3
Š 3b
2
18b 54 ebÎ3
œ 2
1 4
b
dx converges Ê !
n œ7
13. diverges; by the nth-Term Test for Divergence, n lim Ä_
3 ax 1 b 2
œ
n2 converges enÎ3
1 16
2 25
3 36
7x ax 1 b 3
dx• œ lim ”'8 bÄ_
b
_
ln 7 37 ‰ œ _ Ê '8
0
1
327 ‹ e7Î3
converges
dx '8
3 b1
_
is continuous for x 2, f is positive for x 4, and f w axb œ
3 x 1 “8
b
’ 12 e2x “
lim
! n2 n 4 diverges
_
2
dx œ lim
327 e7Î3
4 e2Î3
decreasing for x 8; '8 œ lim
2 8
’2aln xb“ œ lim
bÄ_
_
dx œ lim
œ lim
1 5
is positive and continuous for x 1, f w axb œ
x2 exÎ3
_
1
_
bÄ_
3
b
2 2 diverges Ê ! lnnn diverges Ê ! lnnn œ
9. faxb œ
bÄ_
b
’lnlx 4l“
lim
0 for x 2, thus f is decreasing for x 3;
’ 21 lnax2 4b“ œ lim
is positive and continuous for x 2, f w axb œ b
4 x2 ax 2 4 b 2 b
dx œ lim
_
1 xaln xb2
1 naln nb2
n œ2
is positive and continuous for x 1, f w axb œ
dx œ lim
ln x2 x
x
1
n œ1
diverges Ê ! n2 n 4 diverges Ê ! n2 n 4 œ 8. faxb œ
'1b e2x dx œ
Ê '1 e2x dx converges Ê ! e2n converges
_
ˆ ln1b
bÄ_
œ
1 ‰ 2e2
_
bÄ_
is positive, continuous, and decreasing for x 2; '2
1 xaln xb2
œ lim 7. faxb œ
’ 21 tan1 2x “
n œ1
5. faxb œ e2x is positive, continuous, and decreasing for x 1; '1 e2x dx œ lim
6. faxb œ
bÄ_
bÄ_
_
bÄ_
'1b x 1 4 dx œ
dx œ lim
1 x4
b
lim
2
n œ1
_ alnlb 4l ln 5b œ _ Ê '1 x 1 4 dx diverges Ê ! n 1 4 diverges
œ lim
2
_
is positive, continuous, and decreasing for x 1; '1
1 x4
'1b x 1 4 dx œ
dx œ lim
_ Ê '1 x 1 4 dx converges Ê ! n 1 4 converges
1 1 1 2 tan 2
1 x2 4
_
!
n œ8
x4 ax 1 b 2
n4 n2 2n 1
0 for x 7, thus f is
1 x1
dx '8
3 ax 1 b 2
dx diverges
diverges
12. converges; a geometric series with r œ n n1
b
œ1Á0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" e
1
dx•
Section 10.3 The Integral Test 14. diverges by the Integral Test; '1
n
_
15. diverges; ! nœ1
3 Èn
_
16. converges; ! nœ1
_
" Èn
œ3!
nœ1
2 nÈ n
_
dx œ 5 ln (n 1) 5 ln 2 Ê '1
5 x1
_
nœ1
8 n
_
œ 2 !
nœ1
_
œ 8 !
nœ1
dx Ä _
, which is a divergent p-series (p œ #" ) " n$Î#
, which is a convergent p-series (p œ 3# )
17. converges; a geometric series with r œ 18. diverges; !
5 x1
" 8
1 _
and since !
1 n
nœ1
19. diverges by the Integral Test:
_
" n
diverges, 8 !
nœ1
'2n lnxx dx œ "# aln# n ln 2b
Ê
t œ ln x × dt œ dx Ä x Õ dx œ et dt Ø œ lim 2ebÎ2 (b 2) 2eÐln 2ÑÎ2 (ln 2 2)‘ œ _
20. diverges by the Integral Test:
'2_ lnÈxx dx; Ô
1 n
diverges
'2_ lnxx dx
'ln_2 tetÎ2 dt œ
Ä _
b
lim 2tetÎ2 4etÎ2 ‘ ln 2
bÄ_
bÄ_
21. converges; a geometric series with r œ 22. diverges; n lim Ä_ _
23. diverges; ! nœ0
2 n 1
5n 4n 3
_
œ 2 !
nœ0
" n1
1
ˆ ln 5 ‰ ˆ 54 ‰n œ _ Á 0 œ n lim Ä _ ln 4
5n ln 5 4n ln 4
œ n lim Ä_
2 3
, which diverges by the Integral Test
24. diverges by the Integral Test:
'1n 2xdx 1 œ #" ln (2n 1)
25. diverges; n lim a œ n lim Ä_ n Ä_
2n n1
26. diverges by the Integral Test:
'1n Èx ˆÈdxx 1‰ ; – u œ
27. diverges; n lim Ä_
Èn ln n
œ n lim Ä_
œ n lim Ä_
2n ln 2 1
œ_Á0 Èx "
du œ
" Š 2È ‹ n
Š "n ‹
œ n lim Ä_
Èn #
Ä _ as n Ä _
dx Èx
Ènb1 du
' —Ä 2
u
œ ln ˆÈn 1‰ ln 2 Ä _ as n Ä _
œ_Á0
ˆ1 n" ‰n œ e Á 0 28. diverges; n lim a œ n lim Ä_ n Ä_ 29. diverges; a geometric series with r œ
" ln #
30. converges; a geometric series with r œ
31. converges by the Integral Test:
¸ 1.44 1
" ln 3
¸ 0.91 1
'3_ (ln x) ÈŠ(ln‹x) 1 dx; ” " x
#
u œ ln x Ä du œ x" dx •
'ln_3
" uÈ u# 1
du
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
585
586
Chapter 10 Infinite Sequences and Series b œ lim csec" kukd ln 3 œ lim csec" b sec" (ln 3)d œ lim cos" ˆ "b ‰ sec" (ln 3)‘
bÄ_
bÄ_
œ cos" (0) sec" (ln 3) œ
1 #
32. converges by the Integral Test:
'1_ x a1 "ln xb dx œ '1_ 1 Š(ln‹x) " x
#
œ lim ctan" ud 0 œ lim atan" b tan" 0b œ b
bÄ_
bÄ_
sec" (ln 3) ¸ 1.1439
bÄ_
1 #
0œ
dx; ”
#
'0_ 1"u
u œ ln x Ä du œ "x dx •
#
du
1 #
33. diverges by the nth-Term Test for divergence; n lim n sin ˆ "n ‰ œ n lim Ä_ Ä_
sin ˆ "n ‰ ˆ "n ‰
œ lim
34. diverges by the nth-Term Test for divergence; n lim n tan ˆ "n ‰ œ n lim Ä_ Ä_
tan ˆ "n ‰ ˆ "n ‰
œ n lim Ä_
xÄ0
œ1Á0
sin x x
Š n"# ‹ sec# ˆ n" ‰ Š n"# ‹
œ n lim sec# ˆ "n ‰ œ sec# 0 œ 1 Á 0 Ä_ 35. converges by the Integral Test:
'1_ 1 e e x
1 #
œ lim atan" b tan" eb œ bÄ_
36. converges by the Integral Test: œ lim 2 ln bÄ_
u ‘b u1 e
dx; ”
2x
'e_
u œ ex Ä du œ ex dx •
" 1 u#
ctan" ud e du œ n lim Ä_
b
tan" e ¸ 0.35
_
'1
u œ ex × _ _ dx; du œ ex dx Ä 'e u(1 2 u) du œ 'e ˆ u2 Õ dx œ " du Ø u Ô
2 1 ex
2 ‰ u1
du
œ lim 2 ln ˆ b b 1 ‰ 2 ln ˆ e e 1 ‰ œ 2 ln 1 2 ln ˆ e e 1 ‰ œ 2 ln ˆ e e 1 ‰ ¸ 0.63 bÄ_
37. converges by the Integral Test:
38. diverges by the Integral Test:
'1_ 81tancx x dx; ” u œ tan dx x • "
"
#
du œ
1 x#
'1_ x x1 dx; ” u œ x
39. converges by the Integral Test:
#
#
1 Ä du œ 2x dx •
'1_ sech x dx œ 2
Ä
x
x #
bÄ_
#
'2_ du4 œ
" #
'1b 1 eae b
lim
'11ÎÎ42 8u du œ c4u# d 11ÎÎ24 œ 4 Š 14
b lim #" ln u‘ 2 œ lim
1# 16 ‹
"
bÄ_ #
bÄ_
œ
31 # 4
(ln b ln 2) œ _
dx œ 2 lim ctan" ex d 1 b
bÄ_
œ 2 lim atan" eb tan" eb œ 1 2 tan" e ¸ 0.71 bÄ_
40. converges by the Integral Test:
'1_ sech# x dx œ
œ 1 tanh 1 ¸ 0.76 41.
'1_ ˆ x a 2 x " 4 ‰ dx œ a lim (bb2)4 bÄ_
lim
bÄ_
'1b sech# x dx œ
lim ca ln kx 2k ln kx 4kd 1 œ lim ln b
bÄ_
œ a lim (b 2) bÄ_
bÄ_
a 1
lim ctanh xd b1 œ lim (tanh b tanh 1)
bÄ_
(b 2)a b4
bÄ_
ln ˆ 35 ‰ ; a
_, a 1 œœ Ê the series converges to ln ˆ 53 ‰ if a œ 1 and diverges to _ if 1, a œ 1
a 1. If a 1, the terms of the series eventually become negative and the Integral Test does not apply. From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges. 42.
'3_ ˆ x " 1 x 2a 1 ‰ dx œ " 2ac1 b Ä _ #a(b 1)
œ lim
b
lim ’ln ¹ (xx1)12a ¹“ œ lim ln
bÄ_
œ
3
bÄ_
b1 (b 1)2a
b"
ln ˆ 422a ‰ ; lim
2a b Ä _ (b 1)
1, a œ "# Ê the series converges to ln ˆ #4 ‰ œ ln 2 if a œ _, a "#
" #
and diverges to _ if
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.3 The Integral Test if a
" #
. If a
" #
587
, the terms of the series eventually become negative and the Integral Test does not apply.
From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges. 43. (a)
(b) There are (13)(365)(24)(60)(60) a10* b seconds in 13 billion years; by part (a) sn Ÿ 1 ln n where n œ (13)(365)(24)(60)(60) a10* b Ê sn Ÿ 1 ln a(13)(365)(24)(60)(60) a10* bb œ 1 ln (13) ln (365) ln (24) 2 ln (60) 9 ln (10) ¸ 41.55 _
44. No, because ! nœ1
" nx
œ
" x
_
! nœ1
" n
_
and ! nœ1
" n
diverges
_
_
_
nœ1
nœ1
nœ1
45. Yes. If ! an is a divergent series of positive numbers, then ˆ "# ‰ ! an œ ! ˆ a#n ‰ also diverges and
an #
an .
_
There is no “smallest" divergent series of positive numbers: for any divergent series ! an of positive numbers nœ1
_
! ˆ an ‰ has smaller terms and still diverges. #
nœ1
_
_
_
nœ1
nœ1
nœ1
46. No, if ! an is a convergent series of positive numbers, then 2 ! an œ ! 2an also converges, and 2an an . There is no “largest" convergent series of positive numbers. 47. (a) Both integrals can represent the area under the curve faxb œ
1 Èx 1 ,
and the sum s50 can be considered an 50
approximation of either integral using rectangles with ?x œ 1. The sum s50 œ !
nœ1
integral
1 Èn 1
is an overestimate of the
'151 Èx1 1 dx. The sum s50 represents a left-hand sum (that is, the we are choosing the left-hand endpoint of
each subinterval for ci ) and because f is a decreasing function, the value of f is a maximum at the left-hand endpoint of each sub interval. The area of each rectangle overestimates the true area, thus '1
51
manner, s50 underestimates the integral '0
50
1 Èx 1 dx.
1 Èx 1 dx
50
!
nœ1
1 Èn 1 .
In a similar
In this case, the sum s50 represents a right-hand sum and because
f is a decreasing function, the value of f is aminimum at the right-hand endpoint of each subinterval. The area of each 50
rectangle underestimates the true area, thus ! nœ1
1 Èn 1
œ ’2Èx 1“ œ 2È52 2È2 ¸ 11.6 and '0 51
50
1
50
11.6 !
nœ1
1 Èn 1
1 Èx 1 dx
50
1 Èx 1 dx.
Evaluating the integrals we find '1
51
1 Èx 1 dx
50
œ ’2Èx 1“ œ 2È51 2È1 ¸ 12.3. Thus, 0
12.3.
nb1
(b) sn 1000 Ê '1
'0
1 Èx 1 dx
nb1
œ ’2Èx 1“
1
2
œ 2Èn 1 2È2 1000 Ê n Š500 2È2‹ ¸ 251414.2
Ê n 251415.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
588
Chapter 10 Infinite Sequences and Series 30
48. (a) Since we are using s30 œ !
nœ1
1 n4
_
to estimate ! nœ1
of the area under the curve faxb œ
_
the error is given by !
1 n4 ,
nœ31
1 n4 .
We can consider this sum as an estimate
when x 30 using rectangles with ?x œ 1 and ci is the right-hand endpoint of
1 x4
each subinterval. Since f is a decreasing function, the value of f is a minimum at the right-hand endpoint of each _
subinterval, thus ! nœ31
'30
_
1 n4
1 x4 dx
œ lim '30
b
bÄ_
1 x4 dx
b
œ lim ’ 3x1 3 “ œ lim Š 3b1 3 bÄ_
bÄ_
30
1 ‹ 3a30b3
¸ 1.23 ‚ 105 .
Thus the error 1.23 ‚ 105 Þ (b) We want S sn 0.000001 Ê 'n
_
œ
lim ˆ 3b1 3 bÄ_
1 ‰ 3n3
œ
1 3n3
49. We want S sn 0.01 Ê 'n
_
œ
1 2n2
0.000001 Ê n
1 x3 dx
0.01 Ê 'n
1 x2 4 dx
bÄ_
1 n2 4
_
10 n0.1
0.1 Ê lim 'n
b
1 4
bÄ_ 1 1 ˆ n ‰ 2 tan 2
b
bÄ_
1 x4 dx
b
œ lim ’ 3x1 3 “ bÄ_
n
œ lim 'n
b
1 x3 dx
bÄ_
bÄ_
bÄ_
n
1 ‰ 2n2
¸ 1.195
1 n3
1 x2 4 dx
b
œ lim ’ 2x1 2 “ œ lim ˆ 2b1 2
b
œ lim ’ 21 tan1 ˆ 2x ‰“ bÄ_
n
0.1 Ê n 2tanˆ 12 0.2‰ ¸ 9.867 Ê n 10 Ê S ¸ s10
1 x1.1 dx
0.00001 Ê 'n
_
1 x1.1 dx
œ lim 'n
b
bÄ_
1 x1.1 dx
b
œ lim ’ x10 lim ˆ b10 0.1 “ œ 0.1 bÄ_
bÄ_
n
10 ‰ n0.1
0.00001 Ê n 100000010 Ê n 1060
52. S sn 0.01 Ê 'n
_
œ
1 x3 dx
œ lim 'n
¸ 0.57
51. S sn 0.00001 Ê 'n œ
_
1 x4 dx
¸ 69.336 Ê n 70.
É 1000000 3 3
nœ1
œ lim ˆ 12 tan1 ˆ b2 ‰ 12 tan1 ˆ n2 ‰‰ œ nœ1
_
8
_
10
0.000001 Ê 'n
0.01 Ê n È50 ¸ 7.071 Ê n 8 Ê S ¸ s8 œ !
50. We want S sn 0.1 Ê 'n
œ!
1 x4 dx
lim Š 2aln1bb2 bÄ_
1 dx xaln xb3
1 ‹ 2aln nb2
n
n
kœ1
kœ1
0.01 Ê 'n
_
œ
1 2aln nb2
œ lim 'n
b
1 dx xaln xb3
bÄ_
È50
0.01 Ê n e
1 dx xaln xb3
b
œ lim ’ 2aln1xb2 “ bÄ_
n
¸ 1177.405 Ê n 1178
53. Let An œ ! ak and Bn œ ! 2k aa2k b , where {ak } is a nonincreasing sequence of positive terms converging to 0. Note that {An } and {Bn } are nondecreasing sequences of positive terms. Now, Bn œ 2a# 4a% 8a) á 2n aa2n b œ 2a# a2a% 2a% b a2a) 2a) 2a) 2a) b á ˆ2aa2n b 2aa2n b á 2aa2n b ‰ Ÿ 2a" 2a# a2a$ 2a% b a2a& 2a' 2a( 2a) b á ðóóóóóóóóóóóóóóñóóóóóóóóóóóóóóò 2n1 terms _
ˆ2aa2nc1 b 2aa2nc1 1b á 2aa2n b ‰ œ 2Aa2n b Ÿ 2 ! ak . Therefore if ! ak converges, kœ1
then {Bn } is bounded above Ê ! 2k aa2k b converges. Conversely, _
An œ a" aa# a$ b aa% a& a' a( b á an a" 2a# 4a% á 2n aa2n b œ a" Bn a" ! 2k aa2k b . kœ1
_
Therefore, if ! 2 aa2k b converges, then {An } is bounded above and hence converges. k
kœ1
54. (a) aa2n b œ _
Ê !
nœ2
" 2n ln a2n b " n ln n
œ
" 2n †n(ln 2)
_
_
nœ2
nœ2
Ê ! 2 n a a2 n b œ ! 2 n
" #n †n(ln 2)
œ
" ln #
_
! nœ2
" n
, which diverges
diverges.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.3 The Integral Test " #np
(b) aa2n b œ
55. (a)
_
_
nœ1
nœ1
Ê ! 2 n a a2 n b œ ! 2 n † "
#pc1
'2_ x(lndxx)
u œ ln x • Ä du œ dx x
œœ
;”
_
n œ ! ˆ #p"c1 ‰ , a geometric series that nœ1
'2_ x dxln x œ
p œ 1.
cpb1
lim ’ up 1 “
bÄ_
b ln 2
œ lim Š 1 " p ‹ cbp1 (ln 2)p1 d bÄ_
Ê the improper integral converges if p 1 and diverges if p 1.
_, p "
For p œ 1:
" a2n bpc1
nœ1
'ln_2 ucp du œ
(ln 2)cpb1 , p 1
" p1
_
œ!
1 or p 1, but diverges if p Ÿ 1.
converges if
p
" #np
589
lim cln (ln x)d b2 œ lim cln (ln b) ln (ln 2)d œ _, so the improper integral diverges if
bÄ_
bÄ_
_
" n(ln n)p
(b) Since the series and the integral converge or diverge together, ! nœ2
converges if and only if p 1.
56. (a) p œ 1 Ê the series diverges (b) p œ 1.01 Ê the series converges _
(c) ! nœ2
" n aln n$ b
" 3
œ
_
" n(ln n)
! nœ2
; p œ 1 Ê the series diverges
(d) p œ 3 Ê the series converges 57. (a) From Fig. 10.11(a) in the text with f(x) œ Ÿ 1 '1 f(x) dx Ê ln (n 1) Ÿ 1 n
Ÿ ˆ1
" #
" 3
á
"‰ n
" #
" x
and ak œ
" 3
á
(b) From the graph in Fig. 10.11(b) with f(x) œ Ê 0
cln (n 1) ln nd œ ˆ1
If we define an œ 1
" #
œ
nb1
, we have '1 " n
" 3
" n
" x
" n1 " " # 3
,
" x
dx Ÿ 1
" #
" 3
á
" n
Ÿ 1 ln n Ê 0 Ÿ ln (n 1) ln n
ln n Ÿ 1. Therefore the sequence ˜ˆ1
1 and below by 0. " n1
" k
nb1
'n
" x
á
" n 1
" #
" 3
á n" ‰ ln n™ is bounded above by
dx œ ln (n 1) ln n ln (n 1)‰ ˆ1
" #
" 3
á
" n
ln n‰ .
ln n, then 0 an1 an Ê an1 an Ê {an } is a decreasing sequence of
nonnegative terms.
_
_
# # b 58. ex Ÿ ex for x 1, and '1 ecx dx œ lim cex d " œ lim ˆeb e1 ‰ œ ec1 Ê '1 ecx dx converges by
bÄ_
bÄ_
_
n #
the Comparison Test for improper integrals Ê ! e nœ0
10
59. (a) s10 œ !
'10_ x1
nœ1
" n3
dx œ lim
3
bÄ_
Ê 1.97531986 _
(b) s œ !
nœ1
" n3
10
60. (a) s10 œ !
'10_ x1
nœ1
4
¸
" n4
(b) s œ !
nœ1
" n4
¸
c2 b
lim ’ x2 “
bÄ_
'11_ x1
'10b x4 dx œ 1 3993
10
4
dx œ lim c3 b
bÄ_
c2 b
lim ’ x2 “
bÄ_
œ lim ˆ 2b1 2 bÄ_
bÄ_ 10
s 1.082036583
1.08229 1.08237 2
'11b x3 dx œ
1 ‰ 200
œ
11
œ lim ˆ 2b1 2 bÄ_
1 ‰ 242
œ
1 242
and
1 200
Ê 1.20166 s 1.20253
1 200
lim ’ x3 “
#
nœ1
œ 1.202095; error Ÿ
œ 1.082036583;
Ê 1.082036583
bÄ_
s 1.97531986
1.20166 1.20253 2
bÄ_
dx œ lim
3
'10b x3 dx œ
1 242
dx œ lim
_
'11_ x1
œ 1.97531986;
_
œ 1 ! en converges by the Integral Test.
1.20253 1.20166 2
'11b x4 dx œ
c3 b
lim ’ x3 “
bÄ_
œ lim ˆ 3b1 3 bÄ_
1 3000
œ 1.08233; error Ÿ
œ 0.000435
1 ‰ 3000
œ
11
œ lim ˆ 3b1 3 bÄ_
1 3000
Ê 1.08229 s 1.08237
1.08237 1.08229 2
œ 0.00004
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 ‰ 3993
œ
1 3993
and
590
Chapter 10 Infinite Sequences and Series
10.4 COMPARISON TESTS _
1. Compare with ! nœ1
" n2 ,
which is a convergent p-series, since p œ 2 1. Both series have nonnegative terms for n 1. For
n 1, we have n2 Ÿ n2 30 Ê _
2. Compare with ! nœ1
" n3 ,
1 n2
1 n2 30 .
_
Then by Comparison Test, !
1 n2 30
nœ1
converges.
which is a convergent p-series, since p œ 3 1. Both series have nonnegative terms for n 1. For
n 1, we have n4 Ÿ n4 2 Ê
1 n4
Ê
1 n 4 2
n n4
Ê
n n 4 2
1 n3
n n 4 2
_
n1 n 4 2 .
Then by Comparison Test, ! nœ1
n1 n 4 2
converges. _
3. Compare with ! nœ2
" Èn ,
which is a divergent p-series, since p œ
n 2, we have Èn 1 Ÿ Èn Ê _
4. Compare with ! nœ2
" n,
_
nœ1
" , n3Î2
n2
1 n
_
nœ1
" 3n ,
_
nœ1 _
œ È5 !
nœ1
1 n3Î2
Ê
1 n2
cos2 n n3Î2
nœ2
1 Èn 1
diverges.
n2
n n
n n2
œ
1 n
Ê
Ÿ
1 . n3Î2
n2 n2 n
3 2
n2
n n
_
n1 . Thus !
nœ2
n 3 an 4 b n4 4
diverges.
1. Both series have nonnegative terms for n 1. _
cos2 n n3Î2
Then by Comparison Test, ! nœ1
converges.
È5 . n3Î2
_
The series ! nœ1
1 n †3 n
1 n3Î2
Ÿ
1 3n .
_
Then by Comparison Test, ! nœ1
is a convergent p-series, since p œ
3 2
1 n †3 n
converges. _
È5 n3Î2
1, and the series !
nœ1
converges by Theorem 8 part 3. Both series have nonnegative terms for n 1. For n 1, we have
n3 Ÿ n4 Ê 4n3 Ÿ 4n4 Ê n4 4n3 Ÿ n4 4n4 œ 5n4 Ê n4 4n3 Ÿ 5n4 20 œ 5an4 4b Ê Ê
n2 n2 n
which is a convergent geometric series, since lrl œ ¹ 13 ¹ 1. Both series have nonnegative terms for
n 1. For n 1, we have n † 3n 3n Ê
7. Compare with !
_
Then by Comparison Test, !
1 Èn .
which is a convergent p-series, since p œ
For n 1, we have 0 Ÿ cos2 n Ÿ 1 Ê
6. Compare with !
Ÿ 1. Both series have nonnegative terms for n 2. For
which is a divergent p-series, since p œ 1 Ÿ 1. Both series have nonnegative terms for n 2. For
n 2, we have n2 n Ÿ n2 Ê
5. Compare with !
1 Èn 1
" #
Ÿ5Ê _
8. Compare with ! nœ1
n4 n4 4
" Èn ,
Ÿ
5 n3
Ê É nn444 Ÿ É n53 œ
È5 n3Î2
_
n4 4n3 n4 4
Ÿ 5.
Then by Comparison Test, ! É nn444 converges. nœ1
which is a divergent p-series, since p œ
" #
Ÿ 1. Both series have nonnegative terms for n 1. For
n 1, we have Èn 1 Ê 2Èn 2 Ê 2Èn 1 3 Ê nˆ2Èn 1‰ 3n 3 Ê 2 nÈn n 3 Ê n2 2 nÈn n n2 3 Ê Ê
Èn 1 È n2 3
1 Èn .
n ˆn 2 È n 1 ‰ n2 3
1Ê
n 2È n 1 n2 3
1 n
Ê
ˆÈ n 1 ‰ 2 n2 3
1 n
ÊÊ
ˆÈ n 1‰ 2 n2 3
_
Èn 1 È n2 3 nœ1
Then by Comparison Test, !
diverges.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
É 1n
Section 10.4 Comparison Tests _
" n2 ,
9. Compare with ! nc2 n3 c n2 b 3 1 În 2
œ lim _
! nœ1
nœ1
nÄ_
n2 n3 n2 3
which is a convergent p-series, since p œ 2 1. Both series have positive terms for n 1. lim
an nÄ_ bn
n3 2n2 3 2 nÄ_ n n 3
œ lim
3n2 4n 2 nÄ_ 3n 2n
œ lim
6n 4 nÄ_ 6n 2
œ lim
œ lim
6 nÄ_ 6
œ 1 0. Then by Limit Comparison Test,
converges. _
" Èn ,
10. Compare with ! nœ1
É nn2bb12
œ lim
591
which is a divergent p-series, since p œ
n œ lim É nn2 2 œ É lim
n2 n 2 nÄ_ n 2
2
nÄ_ 1ÎÈn
nÄ_
œ É lim
nÄ_
" #
2n 1 2n
Ÿ 1. Both series have positive terms for n 1. lim
an nÄ_ bn
œ É lim
2 nÄ_ 2
œ È1 œ 1 0. Then by Limit Comparison
_
Test, ! É nn212 diverges. nœ1
_
" n,
11. Compare with ! nan b 1b Šn2
œ lim
nœ2
b 1‹an c 1b
_
Test, ! nœ2
n3 + n2 3 2 nÄ_ n n n 1
n an 1 b an2 1ban 1b
_
nœ1
lim an nÄ_ bn
nÄ_
_
nœ1 5n
Èn 4n
œ lim
†
nÄ_ 1ÎÈn
6n 2 nÄ_ 6n 2
œ lim
œ lim
6 nÄ_ 6
œ 1 0. Then by Limit Comparison
which is a convergent geometric series, since lrl œ ¹ 21 ¹ 1. Both series have positive terms for
œ lim
13. Compare with !
3n2 2n 2 nÄ_ 3n 2n 1
œ lim
diverges.
" 2n ,
12. Compare with ! n 1.
an nÄ_ bn
œ lim
1 În
nÄ_
which is a divergent p-series, since p œ 1 Ÿ 1. Both series have positive terms for n 2. lim
" Èn ,
2n 3 b 4n 1 Î2 n
4n 3 4n nÄ_
œ lim
4n ln 4 n 4 nÄ_ ln 4
œ lim
_
œ 1 0. Then by Limit Comparison Test, !
which is a divergent p-series, since p œ
nœ1
1 2
an nÄ_ bn
_
nÄ_
converges.
Ÿ 1. Both series have positive terms for n 1. lim
n œ lim ˆ 54 ‰ œ _. Then by Limit Comparison Test, !
5n n nÄ_ 4
œ lim
2n 3 4n
nœ1
5n È n †4 n
diverges.
_
n 14. Compare with ! ˆ 25 ‰ , which is a convergent geometric series, since lrl œ ¹ 25 ¹ 1. Both series have positive terms for nœ1
n 1. œ exp
b 3 ‰n ˆ 2n 5n b 4
n 15 ‰n 15 ‰ lim ˆ 10n 15 ‰ œ exp lim lnˆ 10n œ exp lim n lnˆ 10n n œ 10n 8 10n 8 nÄ_ a2Î5b nÄ_ 10n 8 nÄ_ nÄ_ b 15 ‰ 10 lnˆ 10n 10b 8 70n2 70n2 10n b 8 lim œ exp lim 10n b151În10n œ exp lim a10n 15 2 2 1 În ba10n 8b œ exp nlim nÄ_ nÄ_ nÄ_ Ä_ 100n 230n 120
lim an nÄ_ bn
œ lim
œ exp lim
œ exp lim
140n nÄ_ 200n 230 _
15. Compare with ! nœ2
œ lim
" ln n
nÄ_ 1În
nœ1
lnŠ1 n"2 ‹ 1 În 2
_
3 ‰n œ e7Î10 0. Then by Limit Comparison Test, ! ˆ 2n converges. 5n 4 nœ1
which is a divergent p-series, since p œ 1 Ÿ 1. Both series have positive terms for n 2. lim
an nÄ_ bn
n nÄ_ ln n _
nÄ_
" n,
œ lim
16. Compare with ! œ lim
140 nÄ_ 200
" n2 ,
œ lim
1 nÄ_ 1În
_
œ lim n œ _. Then by Limit Comparison Test, ! nÄ_
nœ2
" ln n
diverges.
which is a convergent p-series, since p œ 2 1. Both series have positive terms for n 1. lim
an nÄ_ bn
1
œ lim
nÄ_
1
2 " Š n3 ‹
n2
Š n23 ‹
œ lim
1 " nÄ_ 1 n2
_
œ 1 0. Then by Limit Comparison Test, ! lnˆ1 nœ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"‰ n2
converges.
592
Chapter 10 Infinite Sequences and Series _
" Èn
17. diverges by the Limit Comparison Test (part 1) when compared with ! nœ1
, a divergent p-series:
"
lim
Œ #Èn È $ n Š È"n ‹
nÄ_
Èn $ n 2È n È
œ n lim Ä_
ˆ " œ n lim Ä _ #n
1Î6
‰œ
" #
18. diverges by the Direct Comparison Test since n n n n Èn 0 Ê
_
" n
term of the divergent series ! nœ1
3 n Èn
" n
, which is the nth
" n
or use Limit Comparison Test with bn œ
19. converges by the Direct Comparison Test;
sin# n 2n
Ÿ
" #n
, which is the nth term of a convergent geometric series
20. converges by the Direct Comparison Test;
1 cos n n#
Ÿ
2 n#
21. diverges since n lim Ä_
2n 3n 1
œ
2 3
Š nn# È"n ‹
" n#
converges
Á0
22. converges by the Limit Comparison Test (part 1) with lim nÄ_
and the p-series !
" n$Î#
, the nth term of a convergent p-series:
ˆ n n " ‰ œ 1 œ n lim Ä_
" ‹ Š $Î# n
23. converges by the Limit Comparison Test (part 1) with lim
Š n(n 10n1)(n" 2) ‹ Š n"# ‹
nÄ_
10n# n n# 3n 2
œ n lim Ä_
œ n lim Ä_
20n 1 2n 3
24. converges by the Limit Comparison Test (part 1) with lim
n# (n
" n#
, the nth term of a convergent p-series:
œ n lim Ä_
" n#
œ 10
20 2
, the nth term of a convergent p-series:
5n$
3n 2) Šn# 5‹
Š n"# ‹
nÄ_
œ n lim Ä_
5n$ 3n n$ 2n# 5n 10
15n# 3 3n# 4n 5
œ n lim Ä_ n
œ n lim Ä_
30n 6n 4
œ5
n
n
n ‰ 25. converges by the Direct Comparison Test; ˆ 3n n 1 ‰ ˆ 3n œ ˆ "3 ‰ , the nth term of a convergent geometric series
26. converges by the Limit Comparison Test (part 1) with "
Š $Î# ‹ n
lim nÄ_ Š " È$ n
2
$
‹
É n n$ 2 œ lim É1 œ n lim Ä_ nÄ_
" n$Î#
, the nth term of a convergent p-series:
œ1
2 n$
27. diverges by the Direct Comparison Test; n ln n Ê ln n ln ln n Ê
" n
_
28. converges by the Limit Comparison Test (part 2) when compared with ! nœ1 #
lim nÄ_
’ (lnn$n) “ Š n"# ‹
œ n lim Ä_
(ln n)# n
œ n lim Ä_
2(ln n) Š n" ‹ 1
œ 2 n lim Ä_
29. diverges by the Limit Comparison Test (part 3) with lim
nÄ_
’È
1 “ n ln n ˆ n" ‰
œ n lim Ä_
Èn ln n
" n
Š 2È n ‹ ˆ n" ‰
" n#
" ln n
" ln (ln n)
_
and ! nœ3
" n
, a convergent p-series:
œ0
, the nth term of the divergent harmonic series:
"
œ n lim Ä_
ln n n
œ n lim Ä_
Èn 2
œ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
diverges
Section 10.4 Comparison Tests " n&Î%
30. converges by the Limit Comparison Test (part 2) with lim
n)# ’ (ln$Î# “ n
nÄ_ Š
" ‹ n&Î%
(ln n)# n"Î%
œ n lim Ä_
œ n lim Ä_
ˆ 2 lnn n ‰
" n
31. diverges by the Limit Comparison Test (part 3) with lim
nÄ_
ˆ 1 "ln n ‰ ˆ "n ‰
œ n lim Ä_
n 1 ln n
32. diverges by the Integral Test:
" Š n" ‹
œ n lim Ä_
, the nth term of a convergent p-series:
œ 8 n lim Ä_
" Š $Î% ‹ 4n
ln n n"Î%
œ 8 n lim Ä_
ˆ n" ‰ Š
" ‹ 4n$Î%
œ 32 n lÄ im_
" n"Î%
œ 32 † 0 œ 0
, the nth term of the divergent harmonic series:
œ n lim nœ_ Ä_
'2_ lnx(x11) dx œ 'ln_3 u du œ
" u# ‘ b œ lim ln 3
lim bÄ_ 2
"
bÄ_ #
ab# ln# 3b œ _
" 33. converges by the Direct Comparison Test with n$Î# , the nth term of a convergent p-series: n# 1 n for " " n 2 Ê n# an# 1b n$ Ê nÈn# 1 n$Î# Ê $Î# or use Limit Comparison Test with nÈ n# 1
n
" n$Î# Èn n# 1
34. converges by the Direct Comparison Test with n# 1 Èn
Ê n# 1 Ènn$Î# Ê _
35. converges because ! nœ1 _
! nœ1
" n2n
"n n2n
n$Î# Ê
_
œ!
nœ1
" n2n
_
" #n
!
nœ1
593
, the nth term of a convergent p-series: n# 1 n# " n$Î#
_
nœ1
or use Limit Comparison Test with
" . n$Î#
which is the sum of two convergent series:
converges by the Direct Comparison Test since
36. converges by the Direct Comparison Test: !
1 n# .
" n #n
" #n
_
n 2n n# 2n
_
, and !
œ ! ˆ n2" n nœ1
nœ1
"‰ n#
" 2n
and
is a convergent geometric series
" n2n
" n#
Ÿ
" #n
" n#
, the sum of
the nth terms of a convergent geometric series and a convergent p-series 37. converges by the Direct Comparison Test: 38. diverges; n lim Š3 Ä_
nc1
" 3n ‹
ˆ" œ n lim Ä_ 3
" 3nc1 1 "‰ 3n
" 3
œ
" 3nc1
, which is the nth term of a convergent geometric series
Á0 _
n 39. converges by Limit Comparison Test: compare with ! ˆ 15 ‰ , which is a convergent geometric series with lrl œ nœ1
lim nÄ_
1 1 Š n2n b b 3n † 5n ‹
a 1 Î5 b n
œ n lim Ä_
n1 n2 3n
œ n lim Ä_
1 2n 3
_
nœ1
3 Š 23n b b 4n ‹ n
n
a 3 Î4 b n
œ n lim Ä_
8n 12n 9n 12n
œ n lim Ä_
8 ‰n ˆ 12 1 9 ‰n ˆ 12 1
œ
1 1
_
nœ1
œ
œ
2 n lim 2 aln 2b n Ä _ 2n aln 2b2
1 5
1,
œ 1 0.
41. diverges by Limit Comparison Test: compare with ! n lim 2 ln 2 1 n Ä _ 2n ln 2
1,
œ 0.
n 40. converges by Limit Comparison Test: compare with ! ˆ 34 ‰ , which is a convergent geometric series with lrl œ
lim nÄ_
1 5
Š 2n 2cnn ‹ n
1 n,
which is a divergent p-series, n lim Ä_
†
1În
2 n œ n lim Ä _ 2n n
œ 1 0. _
_
nœ1
nœ1
42. diverges by the definition of an infinite series: ! lnˆ n n 1 ‰ œ ! ln n ln an 1b‘, sk œ aln 1 ln 2b aln 2 ln 3b Þ Þ Þ alnak 1b ln kb aln k ln ak 1bb œ ln ak 1b Ê lim sk œ _ kÄ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
594
Chapter 10 Infinite Sequences and Series _
43. converges by Comparison Test with ! nœ2
_
which converges since !
1 n an 1 b
sk œ ˆ1 12 ‰ ˆ 12 13 ‰ Þ Þ Þ ˆ k 1 2
nœ2
1 ‰ k1
ˆ k 1 1 1k ‰ œ 1
Ê nan 1ban 2b! nan 1b Ê n! nan 1b Ê
1 n!
_
1 n3 ,
nœ1
œ
œ
n2 lim n Ä _ n2 3n 2
œ n lim Ä_
45. diverges by the Limit Comparison Test (part 1) with "‰ n
ˆsin lim n Ä _ ˆ "n ‰
œ lim
xÄ0
sin x x
nœ2
Ê lim sk œ 1; for n 2, an 2b! 1
1 k
kÄ_
1 n an 1 b
an
which is a convergent p-series, n lim Ä_
2n 2n 3
œ n lim Ä_
c 1bx 2bx
1În3
œ10
2 2
" n
, the nth term of the divergent harmonic series:
" n
, the nth term of the divergent harmonic series:
œ1
46. diverges by the Limit Comparison Test (part 1) with ˆtan "n ‰ lim n Ä _ ˆ "n ‰
Ÿ
_
œ ! ’ n 1 1 n1 “, and
an
44. converges by Limit Comparison Test: compare with ! n 3 a n 1 bx lim n Ä _ an 2ban 1bnan 1bx
1 n an 1 b
œ n lim Š " ‹ Ä _ cos " n
ˆsin n" ‰ ˆ n" ‰
œ lim ˆ cos" x ‰ ˆ sinx x ‰ œ 1 † 1 œ 1 xÄ0
tanc" n n1.1
47. converges by the Direct Comparison Test:
_
1 #
n1.1
1
and ! nœ1
1 #
œ
#
n1.1
_
! nœ1
" n1.1
is the product of a
convergent p-series and a nonzero constant 48. converges by the Direct Comparison Test: sec" n
1 #
Ê
secc" n n1 3 Þ
ˆ 1# ‰ n1 3 Þ
_
and ! nœ1
ˆ 1# ‰ n1 3 Þ
œ
1 #
_
! nœ1
" n1 3 Þ
is the
product of a convergent p-series and a nonzero constant
49. converges by the Limit Comparison Test (part 1) with œ n lim Ä_
" ec2n 1 ec2n
" ec2n 1 ec2n
: n lim Ä_
" n#
: n lim Ä_
52. converges by the Limit Comparison Test (part 1) with " 123án
lim nÄ_ 54.
œ
Š nan 2b 1b ‹ Š n"# ‹
" 1 2# 3# á n#
Š n"# ‹
œ n lim coth n œ n lim Ä_ Ä_
en ecn en ecn
n Š tanh ‹ n#
Š n"# ‹
œ n lim tanh n œ n lim Ä_ Ä_
en e en e
n n
œ1
51. diverges by the Limit Comparison Test (part 1) with 1n : n lim Ä_
53.
n Š coth ‹ n#
œ1
50. converges by the Limit Comparison Test (part 1) with œ n lim Ä_
" n#
" ˆ n(n #
1) ‰
œ
œ n lim Ä_ œ
"
2 n(n 1) .
2n# n# n
n(n b 1)(2n b 1) 6
œ
1 Š nÈ n n‹
ˆ 1n ‰
" n# : n lim Ä_
Š
œ n lim Ä_
Èn n ‹ n#
Š n"# ‹
1 n n È
œ 1.
n È œ n lim nœ1 Ä_
The series converges by the Limit Comparison Test (part 1) with
œ n lim Ä_
4n 2n 1
6 n(n 1)(2n 1)
Ÿ
œ n lim Ä_ 6 n$
4 2
" n# :
œ 2.
Ê the series converges by the Direct Comparison Test
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.4 Comparison Tests
595
an 55. (a) If n lim œ 0, then there exists an integer N such that for all n N, ¹ bann 0¹ 1 Ê 1 bann 1 Ä _ bn Ê an bn . Thus, if ! bn converges, then ! an converges by the Direct Comparison Test. an (b) If n lim œ _, then there exists an integer N such that for all n N, bann 1 Ê an bn . Thus, if Ä _ bn ! bn diverges, then ! an diverges by the Direct Comparison Test. _
56. Yes, ! nœ1
an n
converges by the Direct Comparison Test because
an n
an
an 57. n lim œ _ Ê there exists an integer N such that for all n N, Ä _ bn ! then bn converges by the Direct Comparison Test
an bn
1 Ê an bn . If ! an converges,
58. ! an converges Ê n lim a œ 0 Ê there exists an integer N such that for all n N, 0 Ÿ an 1 Ê an# an Ä_ n Ê ! a#n converges by the Direct Comparison Test 59. Since an 0 and n lim a œ _ Á 0, by nth term test for divergence, ! an diverges. Ä_ n 60. Since an 0 and n lim a n2 † an b œ 0, compare !an with ! n"# , which is a convergent p-series; n lim Ä_ Ä_
an 1În2
œ n lim a n2 † an b œ 0 Ê !an converges by Limit Comparison Test Ä_ _
61. Let _ q _ and p 1. If q œ 0, then !
nœ2
_
! nœ2
1 nr
where 1 r p, then n lim Ä_
œ n lim Ä_
1 aln nbcq npcr
qc1 lim qaln nb n Ä _ ap rbnpcr
aln nbq np 1Înr
œ 0. If q 0, n lim Ä_
œ n lim Ä_ qc2
q ap rbnpcr aln nb1cq
aln nbq np
œ n lim Ä_ aln nbq npcr
_
œ!
nœ2
aln nbq npcr ,
œ n lim Ä_
1 np ,
which is a convergent p-series. If q Á 0, compare with
and p r 0. If q 0 Ê q 0 and n lim Ä_
qaln nbqc1 ˆ 1n ‰ ap rbnpcrc1
œ n lim Ä_
qaln nbqc1 ap rbnpcr .
aln nbq npcr
If q 1 Ÿ 0 Ê 1 q 0 and
œ 0, otherwise, we apply L'Hopital's Rule again. n lim Ä_ qc2
qaq 1baln nbqc2 ˆ 1n ‰ ap rb2 npcrc1
qaq 1baln nb qaq 1baln nb q aq 1 b œ n lim . If q 2 Ÿ 0 Ê 2 q 0 and n lim œ n lim œ 0; otherwise, we Ä _ ap rb2 npcr Ä _ ap rb2 npcr Ä _ ap rb2 npcr aln nb2cq apply L'Hopital's Rule again. Since q is finite, there is a positive integer k such that q k Ÿ 0 Ê k q 0. Thus, after k qaq 1bâaq k 1baln nbqck qaq 1bâaq k 1b œ n lim Ä _ ap rbk npcr aln nbkcq ap rbk npcr _ q series ! alnnnpb converges. n œ1
applications of L'Hopital's Rule we obtain n lim Ä_ 0 in every case, by Limit Comparison Test, the
_
62. Let _ q _ and p Ÿ 1. If q œ 0, then !
nœ2
_
! nœ2
1 np ,
aln nbq
np
which is a divergent p-series. Then n lim Ä_
where 0 p r Ÿ 1. n lim Ä_ lim
aln nbq np
ar p b n
rcpc1
n Ä _ aqbaln nbcqc1 ˆ 1n ‰
œ n lim Ä_
aln nbq
np
1 În r
œ
q lim aln nb n Ä _ npcr rcp
ar p bn . aqbaln nbcqc1
1Înp
œ
_
œ!
nœ2
1 np ,
which is a divergent p-series. If q 0, compare with _
œ n lim aln nbq œ _. If q 0 Ê q 0, compare with ! Ä_
nœ2
nrcp lim cq n Ä _ aln nb
otherwise, we apply L'Hopital's Rule again to obtain n lim Ä_
1 nr ,
since r p 0. Apply L'Hopital's to obtain
If q 1 Ÿ 0 Ê q 1 0 and n lim Ä_
a r pb2 nrcp aqbaq 1baln nbcqc2
œ 0. Since the limit is
2 rcpc1
a r pb n aqbaq 1baln nbcqc2 ˆ 1n ‰ 2 rcp
ar pbnrcp aln nbqb1 a q b
œ n lim Ä_
qb2
œ _,
a r pb2 nrcp . aqbaq 1baln nbcqc2
If
a r pb n aln nb œ n lim œ _, otherwise, we aqbaq 1b Ä_ apply L'Hopital's Rule again. Since q is finite, there is a positive integer k such that q k Ÿ 0 Ê q k 0. Thus, after
q 2 Ÿ 0 Ê q 2 0 and n lim Ä_
k applications of L'Hopital's Rule we obtain n lim Ä_
a r pbk nrcp aqbaq 1bâaq k 1baln nbcqck
œ n lim Ä_
a r pbk nrcp aln nbqbk aqbaq 1bâaq k 1b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ _.
596
Chapter 10 Infinite Sequences and Series _
q
Since the limit is _ if q 0 or if q 0 and p 1, by Limit comparison test, the series ! alnnpncbr diverges. Finally if q 0 _
q
and p œ 1 then !
aln nb np
Ê aln nbq 1 Ê
aln nbq n
nœ2
_
_
œ !
nœ2
aln nb n
q
_
. Compare with ! nœ2 _
1n . Thus !
nœ2
aln nbq n
n œ1
1 n,
which is a divergent p-series. For n 3, ln n 1
diverges by Comparison Test. Thus, if _ q _ and p Ÿ 1,
q
the series ! alnnpncbr diverges. n œ1
63. Converges by Exercise 61 with q œ 3 and p œ 4. 64. Diverges by Exercise 62 with q œ
1 2
and p œ 12 .
65. Converges by Exercise 61 with q œ 1000 and p œ 1.001. 66. Diverges by Exercise 62 with q œ
1 5
and p œ 0.99.
67. Converges by Exercise 61 with q œ 3 and p œ 1.1. 68. Diverges by Exercise 62 with q œ 12 and p œ 12 . 69. Example CAS commands: Maple: a := n -> 1./n^3/sin(n)^2; s := k -> sum( a(n), n=1..k ); # (a)] limit( s(k), k=infinity ); pts := [seq( [k,s(k)], k=1..100 )]: # (b) plot( pts, style=point, title="#69(b) (Section 10.4)" ); pts := [seq( [k,s(k)], k=1..200 )]: # (c) plot( pts, style=point, title="#69(c) (Section 10.4)" ); pts := [seq( [k,s(k)], k=1..400 )]: # (d) plot( pts, style=point, title="#69(d) (Section 10.4)" ); evalf( 355/113 ); Mathematica: Clear[a, n, s, k, p] a[n_]:= 1 / ( n3 Sin[n]2 ) s[k_]= Sum[ a[n], {n, 1, k}] points[p_]:= Table[{k, N[s[k]]}, {k, 1, p}] points[100] ListPlot[points[100]] points[200] ListPlot[points[200] points[400] ListPlot[points[400], PlotRange Ä All] To investigate what is happening around k = 355, you could do the following. N[355/113] N[1 355/113] Sin[355]//N a[355]//N N[s[354]]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.5 The Ratio and Root Tests
597
N[s[355]] N[s[356]] _
_
70. (a) Let S œ ! n12 , which is a convergent p-series. By Example 5 in Section 10.2, ! nan 1 1b converges to 1. By Theorem 8, nœ1
_
Sœ!
nœ1 _
1 n2
nœ1
_
œ!
_
!
1 n an 1 b
nœ1
_
!
1 n2
nœ1
_
œ!
1 n an 1 b
nœ1
nœ1
1 n an 1 b
_
!
nœ1
Š n12
1 n an 1 b ‹
also converges.
_
(b) Since ! nan 1 1b converges to 1 (from Example 5 in Section 10.2), S œ 1 ! Š n12 nœ1
nœ1
_
(c) The new series is comparible to
! 13 , n
nœ1
1 n an 1 b ‹
_
œ 1 ! n2 an1 1b nœ1
_
so it will converge faster because its terms Ä 0 faster than the terms of ! n12 . nœ1
1000
1000
(d) The series 1 ! n2 an1 1b gives a better approximation. Using Mathematica, 1 ! n2 an1 1b œ 1.644933568, while nœ1
1000000
!
nœ1
1 n2
nœ1
œ 1.644933067. Note that
1 6
œ 1.644934067. The error is 4.99 ‚ 107 compared with 1 ‚ 106 .
2
10.5 THE RATIO AND ROOT TESTS
1.
2.
3.
2n n!
0 for all n 1; lim Œ nÄ_
n2 3n
2nb" "b! 2n n!
an
0 for all n 1; lim Œ
n
b1b b 2 3nb1 nb2 n 3
3 lim ˆ n3 n †3 †
œ
nÄ_
b1bc1b! b1bb1b2 c1b! anb1b2
aan
0 for all n 1; lim Œ nÄ_
aan
an
_
n! 2n ‹
nÄ_
an
nÄ_
an 1 b ! an 1 b2
2 †2 lim Š an" b†n! †
œ
œ
_
n
œ lim ˆ n 2 " ‰ œ 0 1 Ê ! 2n! converges nÄ_
3n ‰ n2
n3 ‰ ˆ1‰ œ lim ˆ 3n 6 œ lim 3 œ nÄ_
lim Š na†nan21b2b! †
nÄ_
n œ1
nÄ_
a n "b 2 an 1 b ! ‹
_
1 Ê ! n 3n 2 converges
1 3
n œ1
n 3n 4n 1 œ lim Š nn22n 4n 4 ‹ œ lim Š 2n 4 ‹ 3
2
2
nÄ_
nÄ_
1 b! œ lim ˆ 6n 2 4 ‰ œ _ 1 Ê ! aann diverges 1 b2 nÄ_
4.
2nb1 n †3 n 1
n œ1
0 for all n 1; lim nÄ_
_
2an1b1 1b†3an1b 2n1 n†3n 1
1
an
nb1
lim Š an21b†3†n2 1 †3 †
œ
nÄ_
n †3 n 1 2n1 ‹
œ lim ˆ 3n2n 3 ‰ œ lim ˆ 23 ‰ œ nÄ_
nÄ_
2 3
1
nb1
Ê ! n2†3nc1 converges n œ1
5.
n4 4n
0 for all n 1; lim Œ nÄ_
œ lim ˆ 14 nÄ_
6.
3nb2 ln n
1 n
3 2n2
1 n3
b1b4 4nb1 n4 4n
an
1 ‰ 4n4
0 for all n 2; lim Œ nÄ_ _
œ œ
3anb1bb2 ln anb1b 3nb2 ln n
1 4
4
lim Š an4n †14b †
4n n4 ‹
nÄ_
_
œ lim Š n
4
nÄ_
4n3 6n2 4n 1 ‹ 4n4
4
1 Ê ! n4n converges
œ
n œ1
nb2
lim Š ln3an †31b †
nÄ_
ln n 3nb2 ‹
œ lim Š ln 3anlnn1b ‹ œ lim Š nÄ_
nÄ_
3 n 1
nb1
‹ œ lim ˆ 3n n 3 ‰ nÄ_
nb2
œ lim ˆ 31 ‰ œ 3 1 Ê ! 3ln n diverges nÄ_
7.
n 2 an 2 b ! nx32n
n œ2
an
0 for all n 1; lim nÄ_
b 1b2aan b 1b b 2b! b 1bx32an 1b
an
n2 an 2b! nx32n
œ
7 ˆ 6n 15 ‰ ˆ6‰ œ lim Š 3n27n2 15n 18n ‹ œ lim 54n 18 œ lim 54 œ 2
nÄ_
nÄ_
nÄ_
2
3ban 2b! lim Š an an1ba1nb † †nx32n †32
nÄ_ 1 9
_
nx32n n 2 an 2 b ! ‹
œ lim Š n nÄ_
1 Ê ! n annx32n2b! converges 2
n œ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3
5n2 7n 3 ‹ 9n3 9n2
598 8.
Chapter 10 Infinite Sequences and Series n †5 n a2n 3b lnan 1b
0 for all n 1;
1b†a2n 3b lim Š 5an † na2n 5b
œ
nÄ_
lim Œ
a2an
nÄ_
lnan 1b lnan 2b ‹
an b 1b†5n 1 1b 3b lnaan 1b n†5n a2n 3b lnan 1b
n 1 b† 5 † 5 lim Š a2na 5b lnan 2b † n
œ
1b
nÄ_
a2n 3b lnan 1b ‹ n †5 n
a b 15 25 ‰ lim Š 10n 2n2 25n † lim Š ‹ † lim Š lnlnann 21b ‹ œ lim ˆ 20n 4n 5 5n 2
œ
nÄ_
nÄ_ _
nÄ_
nÄ_
1 nb1 1 nb2
‹
n †5 ! ‰ ˆ n2‰ ˆ 1‰ œ lim ˆ 20 4 † lim n 1 œ 5 † lim 1 œ 5 † 1 œ 5 1 Ê a2n 3b lnan 1b diverges nÄ_
9.
10.
7 a2n 5bn
4n a3n bn
nÄ_
nÄ_
n œ2
7 n 0 for all n 1; lim É œ a2n 5bn nÄ_
nÄ_
nÄ_
3‰ lim ˆ 4n 3n 5 œ
nÄ_
n 1
nÄ_
n 1
lim ˆ 43 ‰ œ
nÄ_
n 0 for all n 1; lim Ê’lnˆe2 1n ‰“
Ê ! ’lnˆe2 1n ‰“
n
n œ1
n 3 ‰n 3 ‰n ˆ 4n 11. ˆ 4n 0 for all n 2; lim É œ 3n 5 3n 5
_
n œ1
4 ‰ lim ˆ 3n œ 0 1 Ê ! a3n4 bn converges
n
n 1
_
n È
lim Š 2n 7 5 ‹ œ 0 1 Ê ! a2n 7 5bn converges
nÄ_
_
4 n 0 for all n 1; lim É œ a3n bn
12. ’lnˆe2 1n ‰“
n
œ
4 3
nÄ_
_
n
3‰ 1 Ê ! ˆ 4n diverges 3n 5 n œ1
1 1 În
lim ’lnˆe2 1n ‰“
nÄ_
œ lnae2 b œ 2 1
diverges
n œ1
13.
8 ˆ3 1n ‰2n
8 n 0 for all n 1; lim É œ ˆ3 1 ‰2n nÄ_
lim Œ ˆ
nÄ_
n
n
n È 8
3 1n ‰
n " 0 for all n 2; lim É n1bn œ
nÄ_
17. converges by the Ratio Test:
converges
nÄ_
n
n œ1
nÄ_
2
n œ1
n È
nÄ_
”
œ n lim Ä_
nÄ_
È
(n b 1) 2 2nb1 •
”
_
n È
1 ! 1"bn converges lim Š n È n n‹ œ 0 1 Ê n
lim Š n1În 1 1 ‹ œ
lim anb1 n Ä _ an
8 1 ‰2n n œ1 3 n
n n lim ˆ1 1n ‰ œ e1 1 Ê ! ˆ1 1n ‰ converges
nÄ_
"
_
1Ê !ˆ
_
2
n n n 15. ˆ1 1n ‰ 0 for all n 1; lim Ɉ1 1n ‰ œ
n1bn
1 9
lim sinŠ È1n ‹ œ sina0b œ 0 1 Ê ! ’sinŠ È1n ‹“ converges
nÄ_
16.
œ
_
n
n 14. ’sinŠ È1n ‹“ 0 for all n 1; lim Ê ’sinŠ È1n ‹“ œ
2
2
È n 2 #n
•
Š (nenbb1)1 ‹
n œ2
È È n (n 1) 2 ˆ1 n" ‰ 2 ˆ #" ‰ œ œ n lim † 2È2 œ n lim Ä _ #nb1 Ä_ n
" #
1
2
18. converges by the Ratio Test:
lim anb1 n Ä _ an
œ n lim Ä_
19. diverges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
20. diverges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
# Š nen ‹
Š (nenbb1)! 1 ‹ ˆ en!n ‰
b 1)! ‹ Š (n 10nb1 ˆ 10n!n ‰
œ n lim Ä_
21. converges by the Ratio Test:
œ n lim Ä_
(n ")! enb1
†
en n!
œ n lim Ä_
(n ")! 10nb1
†
10n n!
Š (n10bn1)1 ‹
"!
Š n10n ‹
†
œ n lim Ä_
"!
lim anb1 n Ä _ an
(n 1)2 enb1
œ n lim Ä_
(n ")"! 10n 1
†
en lim n2 œ n Ä _
œ n lim Ä_
œ n lim Ä_
10n n"!
ˆ1 n" ‰# ˆ "e ‰ œ
n" e
n 10
" e
1
œ_
œ_
ˆ1 "n ‰"! ˆ 1"0 ‰ œ œ n lim Ä_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" 10
1
Section 10.5 The Ratio and Root Tests ˆ nn 2 ‰n œ lim ˆ1 22. diverges; n lim a œ n lim Ä_ n Ä_ nÄ_ 23. converges by the Direct Comparison Test:
2(1)n (1.25)n
2 ‰ n n
599
œ e# Á 0 n
n
œ ˆ 45 ‰ c2 (1)n d Ÿ ˆ 45 ‰ (3) which is the nth term of a convergent
geometric series 24. converges; a geometric series with krk œ ¸ 23 ¸ 1 ˆ1 3n ‰n œ lim ˆ1 25. diverges; n lim a œ n lim Ä_ n Ä_ nÄ_ ˆ1 26. diverges; n lim a œ n lim Ä_ n Ä_
" ‰n 3n
3 ‰ n n
œ n lim 1 Ä_
27. converges by the Direct Comparison Test:
ln n n$
n n$
œ
œ e$ ¸ 0.05 Á 0
Š "3 ‹ n
" n#
n "Î$ ¸ 0.72 Á 0 œe
for n 2, the nth term of a convergent p-series. n
n (ln n) n È É 28. converges by the nth-Root Test: n lim an œ n lim nn œ n lim Ä_ Ä_ Ä_
29. diverges by the Direct Comparison Test: with "n .
" n
" n#
œ
n1 n#
ln n n
" n
" ‰n n#
ˆˆ n" œ n lim Ä_
anb1 an
œ n lim Ä_
(n 1) ln (n 1) #nb1
†
33. converges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
(n 2)(n 3) (n 1)!
†
n! (n 1)(n 2)
34. converges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
(n 1)$ en 1
œ
35. converges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
(n 4)! 3! (n 1)! 3nb1
anb1 an
œ n lim Ä_
†
anb1 an
œ n lim Ä_
(n 1)! (2n 3)!
38. converges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
(n 1)! (n 1)nb1
"
ˆ1 "n ‰n
œ
" e
en n$
†
(n 1)2nb1 (n 2)! 3nb1 (n 1)!
37. converges by the Ratio Test: n lim Ä_
œ n lim Ä_
ln n n
œ n lim Ä_
Š "n ‹ 1
œ01
" ‰n ‰1În n#
ˆ" œ n lim Ä_ n
"‰ n#
for n 3
32. converges by the Ratio Test: n lim Ä_
36. converges by the Ratio Test: n lim Ä_
œ n lim Ä_
"# ˆ "n ‰ for n 2 or by the Limit Comparison Test (part 1)
n n ˆ n" È É 30. converges by the nth-Root Test: n lim an œ n lim Ä_ Ä_
31. diverges by the Direct Comparison Test:
a(ln n)n b1În ann b1În
2n n ln (n)
" e
†
nn n!
1
œ01
œ n lim Ä_
3n n! n2n (n 1)!
(2n 1)! n!
†
" #
1
3! n! 3n (n 3)!
†
œ
n4 3(n 1)
" 3
œ
1
2‰ ˆ n n 1 ‰ ˆ 32 ‰ ˆ nn œ n lim 1 œ Ä_
œ n lim Ä_
n" (2n 3)(2n 2)
œ01
ˆ n ‰n œ lim œ n lim Ä _ n1 nÄ_
"
ˆ n bn " ‰n
1
n n n È 39. converges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É (ln n)n Ä_
n n È ln n
œ n lim Ä_
" ln n
œ01
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2 3
1
œ01
600
Chapter 10 Infinite Sequences and Series n n È Èln n
n n n È 40. converges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É (ln n)nÎ2 Ä_
Ä_ Ä_
n n lim È n Èln n lim n
œ
œ01
n È n œ 1‹ Šn lim Ä_
41. converges by the Direct Comparison Test:
œ
n! ln n n(n 2)!
ln n n(n 1)(n 2)
" (n 1)(n #)
œ
n n(n 1)(n 2)
" n#
which is the nth-term of a convergent p-series an 1 an
42. diverges by the Ratio Test: n lim Ä_
œ n lim Ä_
3n 1 (n 1)$ 2n
1
†
n$ 2n 3n
†
a2nbx nx‘2
43. converges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
an1bx‘2 2(n 1)‘x
44. converges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
a2n 5bˆ2nb1 3‰ 3nb1 2
œ n lim ’ 2n 5 “ † n lim ’ 2 †6 4 † 2 3 † 3 6 “ œ 1 † Ä _ 2n 3 Ä _ 3 †6 n 9 † 3 n 2 † 2 n 6 n
n
n
45. converges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
2 3
œ
ˆ 1 b nsin n ‰ an an
"n
Š 1 b tan n
47. diverges by the Ratio Test: n lim Ä_
anb1 an
œ n lim Ä_
œ n lim Ä_ †
ˆ #3 ‰ œ
an 1 b 2 (2n 2)(2n 1)
3n 2 a2n 3ba2n 3b
3 #
1
œ n lim Ä_
œ n lim ’ 2n 5 † Ä _ 2n 3
n2 2n 1 4n2 6n 2
œ
œ n lim Ä_
3n 1 2n 5
œ n lim Ä_
" tan " n n
œ
3 #
œ 0 since the numerator
1
2 ‰ 48. diverges; an1 œ n n 1 an Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 an1 ‰ Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 ‰ ˆ nn 1 an2 a " n n 1 n 2 3 " Ê an1 œ ˆ n 1 ‰ ˆ n ‰ ˆ n 1 ‰ â ˆ # ‰ a" Ê an1 œ n 1 Ê an1 œ n 1 , which is a constant times the
general term of the diverging harmonic series
49. converges by the Ratio Test: n lim Ä_
50. converges by the Ratio Test:
n ln n n 10
0 and a" œ
Ê an1 œ
n ln n n 10
" #
œ n lim Ä_
lim anb1 n Ä _ an
œ n lim Ä_
anb1 an
œ n lim Ä_
51. converges by the Ratio Test: n lim Ä_ 52.
anb1 an
Š 2n ‹ an an
Œ
Èn n #
œ n lim Ä_
an
an
œ n lim Ä_
Š 1 bnln n ‹ an an
2 n
œ01 n n È
œ n lim Ä_
n
œ
"ln n n
" #
1
œ n lim Ä_
Ê an 0; ln n 10 for n e"! Ê n ln n n 10 Ê
an an ; thus an1 an
53. diverges by the nth-Term Test: a" œ
" 3
" #
" n
œ01
n ln n n 10
1
Ê n lim a Á 0, so the series diverges by the nth-Term Test Ä_ n
3 3 6 " %! " 2 " 2 " 2 " É É É É , a# œ É 3 , a$ œ Ê 3 œ 3 , a% œ ËÊ 3 œ 3 ,á ,
%
n! " n! " n " É an œ É a œ 1 because šÉ 3 Ê n lim 3 › is a subsequence of š 3 › whose limit is 1 by Table 8.1 Ä_ n
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 4
1
2 †6 n 4 † 2 n 3 † 3 n 6 3 †6 n 9 † 3 n 2 † 2 n 6 “
œ01
‹ an
an
c1‰ ˆ 3n 2n b 5 an an
n3 (n 1)3
1
2 3
anb1 46. converges by the Ratio Test: n lim œ n lim Ä _ an Ä_ 1 approaches 1 # while the denominator tends to _
œ n lim Ä_
Section 10.5 The Ratio and Root Tests 54. converges by the Direct Comparison Test: a" œ n!
" #
# $
#
' %
'
#%
, a# œ ˆ "# ‰ , a$ œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , a% œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , á
n
Ê an œ ˆ "# ‰ ˆ "# ‰ which is the nth-term of a convergent geometric series anb1 an
55. converges by the Ratio Test: n lim Ä_ n" " œ n lim œ 1 # Ä _ 2n 1
2nb1 (n 1)! (n 1)! (2n 2)!
œ n lim Ä_
†
(2n)! 2n n! n!
2(n 1)(n 1) (2n #)(2n 1)
œ n lim Ä_
(3n 3)! 1)! (n 2)! anb1 56. diverges by the Ratio Test: n lim œ n lim † n! (n (3n)! Ä _ an Ä _ (n 1)! (n 2)! (n 3)! (3n 3)(3 2)(3n 1) 2 ‰ ˆ 3n 1 ‰ œ n lim œ n lim 3 ˆ 3n n# n 3 œ 3 † 3 † 3 œ 27 1 Ä _ (n 1)(n 2)(n 3) Ä_ n
n (n!) n È 57. diverges by the Root Test: n lim an ´ n lim œ n lim Ä_ Ä _ É an n b # Ä_
n
œ_1
n! n#
n
n (n!) n (n!) É 58. converges by the Root Test: n lim œ n lim œ n lim É an n b n Ä_ Ä_ Ä_ nn# " Ÿ n lim œ01 Ä_ n
n! nn
ˆ " ‰ ˆ 2n ‰ ˆ 3n ‰ â ˆ n n 1 ‰ ˆ nn ‰ œ n lim Ä_ n
" #n ln 2
n n n È 59. converges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É 2n# Ä_
n #n
œ n lim Ä_
n n n È 60. diverges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É a#n b # Ä_
n 4
œ_1
n
n
anb1 an
61. converges by the Ratio Test: n lim Ä_
1†3 â (2n 1) (2†4 â #n) a3n 1b
62. converges by the Ratio Test: an œ Ê n lim Ä_
(2n 2)! c2nb1 (n 1)!d# a3nb1 1b #
œ n lim Š 4n 6n 2 ‹ Ä _ 4n# 8n 4 63. Ratio: n lim Ä_
anb1 an
†
a1 3cn b a3 3cn b
œ n lim Ä_
œ n lim Ä_
a2n n!b# a3n 1b (2n)!
œ1†
" (n 1)p
†
" 3
np 1
œ
" 3
anb1 an
œ n lim Ä_
" (ln (n 1))p
†
†
1†2†3†4 â (2n 1)(2n) (2†4 â 2n)# a3n 1b
œ
œ n lim Ä_
4n 2n n! 1†3† â †(2n 1)
œ
œ n lim Ä_
2n " (4†#)(n 1)
(2n)! a2n n!b# a3n 1b
(2n ")(2n 2) a3n 1b 2# (n 1)# a3n 1 1b
1
(ln n)p 1
" n n ‰p ˆÈ
" (1)p
œ
œ ’n lim Ä_
œ 1 Ê no conclusion
ln n ln (n 1) “
p
œ ”n lim Ä_
ˆ "n ‰
p
ˆ n b 1 ‰ • œ Šn lim Ä_ "
n" n ‹
œ (1)p œ 1 Ê no conclusion " n n È Root: n lim an œ n lim É (ln n)p œ Ä_ Ä_
"
p
lim (ln n)1În ‹ ŠnÄ_ ˆ
"
‰
; let f(n) œ (ln n)1În , then ln f(n) œ
ln (ln n) n ln n Ê n lim ln f(n) œ n lim œ n lim œ n lim n 1 Ä_ Ä_ Ä_ Ä_ " ln fÐnÑ ! n È an œ œ n lim e œ e œ 1; therefore lim Ä_ nÄ_
" n ln n p
lim (ln n)1În ‹ ŠnÄ_
65. an Ÿ
n 2n
_
_
for every n and the series ! nœ1
n #n
œ
ˆ n ‰p œ 1p œ 1 Ê no conclusion œ n lim Ä_ n1
n " n È É Root: n lim an œ n lim np œ n lim Ä_ Ä_ Ä_
64. Ratio: n lim Ä_
1†3† â †(2n 1)(2n 1) 4nb1 2nb1 (n 1)!
œ01
ln (ln n) n
œ 0 Ê n lim (ln n)1În Ä_ œ (1)" p œ 1 Ê no conclusion
converges by the Ratio Test since n lim Ä_
(n ") 2nb1
†
2n n
œ
" #
Ê ! an converges by the Direct Comparison Test nœ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
p
" 4
1
601
602
Chapter 10 Infinite Sequences and Series 2
66.
2n n!
0 for all n 1; lim nÄ_ _
2 2anb1b anb1b! 2 2n n!
œ
n2 b2nb1
lim Š a2n1b†n! †
nÄ_
n! ‹ 2n2
2nb1
†4 ‰ ˆ 2†4 1ln 4 ‰ œ lim Š 2n1 ‹ œ lim ˆ n2 1 œ lim n
nÄ_
n
nÄ_
nÄ_
n2
œ _ 1 Ê ! 2n! diverges n œ1
10.6 ALTERNATING SERIES, ABSOLUTE AND CONDITIONAL CONVERGENCE 1.
converges by the Alternating Convergence Test since: un œ Ê
1 È n 1
Ÿ
1 Èn
Ê un1 Ÿ un ;
lim un œ
nÄ_
lim 1 nÄ_ Èn
1 Èn
0 for all n 1; n 1 Ê n 1 n Ê Èn 1 Èn
œ 0. _
_
nœ1
nœ1
2. converges absolutely Ê converges by the Alternating Convergence Test since ! kan k œ !
" n$Î#
which is a
convergent p-series 3. converges Ê converges by Alternating Series Test since: un œ Ê an 1b3n1 n 3n Ê
1 an1b3nb1
Ÿ
1 n 3n
Ê un1 Ÿ un ;
1 n3n
0 for all n 1; n 1 Ê n 1 n Ê 3n1 3n
lim un œ
4. converges Ê converges by Alternating Series Test since: un œ
œ 0.
lim 1 n nÄ_ n 3
nÄ_
4 aln nb2
0 for all n 2; n 2 Ê n 1 n
Ÿ
1 aln nb2
Ê
5. converges Ê converges by Alternating Series Test since: un œ
n n2 1
0 for all n 1; n 1 Ê 2n2 2n n2 n 1
Ê ln an 1b ln n Ê aln an 1bb2 aln nb2 Ê lim un œ
nÄ_
lim 4 2 nÄ_ aln nb
1 aln an1bb2
4 aln an1bb2
Ÿ
4 aln nb2
Ê un1 Ÿ un ;
œ 0.
Ê n3 2n2 2n n3 n2 n 1 Ê nan2 2n 2b n3 n2 n 1 Ê nŠan 1b2 1‹ an2 1ban 1b Ê
n n 2 1
n 1 an 1 b 2 1
Ê un1 Ÿ un ;
lim un œ
nÄ_
lim 2 n nÄ_ n 1
œ 0.
6. diverges Ê diverges by nth Term Test for Divergence since:
2 lim n2 5 nÄ_ n 4
7. diverges Ê diverges by nth Term Test for Divergence since:
2n 2 nÄ_ n
lim
œ1Ê
œ_Ê
5 lim a1bn1 nn2 4 œ does not exist 2
nÄ_
lim a1bn1 2n2 œ does not exist n
nÄ_
_
_
nœ1
nœ1
8. converges absolutely Ê converges by the Absolute Convergence Test since ! kan k œ ! anb1 nÄ_ an
Ratio Test, since lim
œ
lim 10 nÄ_ n 2
10n a n 1 bx ,
which converges by the
œ01
9. diverges by the nth-Term Test since for n 10 Ê
n 10
_
n ‰n ˆ n ‰n Á 0 Ê ! (1)n1 ˆ 10 1 Ê n lim diverges Ä _ 10 nœ1
10. converges by the Alternating Series Test because f(x) œ ln x is an increasing function of x Ê Ê un un1 for n 1; also un 0 for n 1 and
" lim n Ä _ ln n
11. converges by the Alternating Series Test since f(x) œ
ln x x
is decreasing
œ0
Ê f w (x) œ
Ê un un1 ; also un 0 for n 1 and n lim u œ n lim Ä_ n Ä_
" ln x
ln n n
1 ln x x#
œ n lim Ä_
0 when x e Ê f(x) is decreasing Š "n ‹ 1
œ0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.6 Alternating Series, Absolute and Conditional Convergence 12. converges by the Alternating Series Test since f(x) œ ln a1 x" b Ê f w (x) œ
" x(x 1)
0 for x 0 Ê f(x) is decreasing
ˆ1 n" ‰‹ œ ln 1 œ 0 Ê un un1 ; also un 0 for n 1 and n lim u œ n lim ln ˆ1 "n ‰ œ ln Šn lim Ä_ n Ä_ Ä_ 13. converges by the Alternating Series Test since f(x) œ Ê un unb1 ; also un 0 for n 1 and n lim u œ Ä_ n 3È n 1 Èn 1
14. diverges by the nth-Term Test since n lim Ä_
_
_
nœ1
nœ1
Èx " x1
1 x 2È x 2Èx (x 1)#
Ê f w (x) œ
Èn " lim n Ä _ n1
0 Ê f(x) is decreasing
œ0
3É 1
œ n lim Ä_
" n
"
1 Š Èn ‹
œ3Á0
" ‰n 15. converges absolutely since ! kan k œ ! ˆ 10 a convergent geometric series
16. converges absolutely by the Direct Comparison Test since ¹ (1)
nb1
(0.1)n
n
¹œ
" (10)n n
n
" ‰ ˆ 10 which is the nth term
of a convergent geometric series 17. converges conditionally since
" Èn
" Èn 1
" Èn
0 and n lim Ä_
_
_
nœ1
nœ1
œ 0 Ê convergence; but ! kan k œ !
" n"Î#
is a divergent p-series 18. converges conditionally since _
_
! kan k œ !
nœ1
nœ1
" 1 Èn
" 1 Èn
" 1 Èn 1
is a divergent series since _
_
nœ1
nœ1
19. converges absolutely since ! kan k œ !
n n $ 1
n! #n
20. diverges by the nth-Term Test since n lim Ä_ 21. converges conditionally since _
œ!
nœ1
" n3
diverges because
" n3 " n3
" (n 1) 3
" 4n
" 1 Èn
0 and n lim Ä_
" 1 È n
and
" #È n
n n $ 1
_
and !
" n#
nœ1
" n"Î#
œ 0 Ê convergence; but is a divergent p-series
which is the nth-term of a converging p-series
œ_ 0 and n lim Ä_
_
and ! nœ1
" n
" n 3
_
œ 0 Ê convergence; but ! kan k nœ1
is a divergent series
_
22. converges absolutely because the series ! ¸ sinn# n ¸ converges by the Direct Comparison Test since ¸ sinn# n ¸ Ÿ nœ1
3n 5n
23. diverges by the nth-Term Test since n lim Ä_
œ1Á0 nb1
24. converges absolutely by the Direct Comparison Test since ¹ (n2)5n ¹ œ
2nb1 n 5 n
n
2 ˆ 25 ‰ which is the nth term
of a convergent geometric series 25. converges conditionally since f(x) œ
" x#
" x
Ê f w (x) œ ˆ x2$
"‰ x#
0 Ê f(x) is decreasing and hence _
_
nœ1
nœ1
ˆ " n" ‰ œ 0 Ê convergence; but ! kan k œ ! un unb1 0 for n 1 and n lim Ä _ n# _
œ!
nœ1
" n#
_
!
nœ1
" n
603
1 n n#
is the sum of a convergent and divergent series, and hence diverges
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" n#
604
Chapter 10 Infinite Sequences and Series
26. diverges by the nth-Term Test since n lim a œ n lim 101În œ 1 Á 0 Ä_ n Ä_ 27. converges absolutely by the Ratio Test: n lim Š uunbn 1 ‹ œ n lim Ä_ Ä_ ” 28. converges conditionally since f(x) œ
'2_ x dxln x œ
lim
Š "x ‹
bÄ_
_
_
nœ1
nœ1
Ê ! kan k œ !
'2b ln x dx œ " n ln n
n 1
•œ
2 3
1
1d Ê f w (x) œ cln(x(x) ln x)# 0 Ê f(x) is decreasing
" x ln x
" n ln n
Ê un unb1 0 for n 2 and n lim Ä_
(n")# ˆ 23 ‰ n n# ˆ 23 ‰
œ 0 Ê convergence; but by the Integral Test,
lim cln (ln x)d b2 œ lim cln (ln b) ln (ln 2)d œ _
bÄ_
bÄ_
diverges
_
" x b#
29. converges absolutely by the Integral Test since '1 atan" xb ˆ 1 " x# ‰ dx œ lim ’ atan # bÄ_
œ lim ’atan bÄ_
"
#
"
bb atan
#
1b “ œ
30. converges conditionally since f(x) œ œ
1 Š lnxx ‹ ln
x Š lnxx ‹
(x ln x)#
œ n lim Ä_
Š "n ‹ 1 Š n" ‹
_
_
nœ1
nœ1
! kan k œ !
œ
" #
1 ln x (x ln x)#
# # ’ˆ 1# ‰ ˆ 14 ‰ “ œ
ln x x ln x
Ê f w (x) œ
1
31 # 32
Š "x ‹ (x ln x) (ln x) Š1 x" ‹ (x ln x)#
0 Ê un un1 0 when n e and n lim Ä_
œ 0 Ê convergence; but n ln n n Ê
ln n n ln n
b
“
" nln n
" n
Ê
ln n n ln n
ln n nln n
" n
so that
diverges by the Direct Comparison Test
31. diverges by the nth-Term Test since n lim Ä_ _
_
nœ1
nœ1
n n1
œ1Á0
n 32. converges absolutely since ! kan k œ ! ˆ 5" ‰ is a convergent geometric series
33. converges absolutely by the Ratio Test: n lim Š uunbn 1 ‹ œ n lim Ä_ Ä_
("00)nb1 (n1)!
_
_
nœ1
nœ1
34. converges absolutely by the Direct Comparison Test since ! kan k œ !
†
n! (100)n
œ n lim Ä_
" n# 2n 1
and
"00 n1
œ01
" n# 2n 1
" n#
nth-term of a convergent p-series _
_
nœ1
nœ1
_
35. converges absolutely since ! kan k œ ! ¹ (nÈ1)n ¹ œ ! _
36. converges conditionally since ! nœ1 _
_
nœ1
nœ1
! kan k œ !
" n
cos n1 n
n
nœ1
_
œ!
nœ1
(1)n n
" n$Î#
is a convergent p-series
is the convergent alternating harmonic series, but
diverges
1) n È kan k œ n lim 37. converges absolutely by the Root Test: n lim Š (n(2n) n ‹ Ä_ Ä_ n
1 În
œ n lim Ä_
n" #n
œ
" #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
which is the
Section 10.6 Alternating Series, Absolute and Conditional Convergence 38. converges absolutely by the Ratio Test: n lim ¹ anb1 ¹ œ n lim Ä _ an Ä_
a(n 1)!b# ((2n 2)!)
39. diverges by the nth-Term Test since n lim kan k œ n lim Ä_ Ä_
œ n lim Ä_
ˆ n # 1 ‰n1 œ _ Á 0 n lim Ä_
(n 1)(n 2)â(n (n 1)) #nc1
œ n lim Ä_
(2n)! 2n n! n
(n 1)# 3 (2n 2)(2n 3)
œ
3 4
Èn 1 Èn 1
†
Èn 1 Èn Èn 1 Èn
œ
" Èn 1 Èn _
decreasing sequence of positive terms which converges to 0 Ê !
nœ1
_
_
nœ1
nœ1
" Èn 1 Èn
Èn
lim nÄ_
" 1
Èn
"
Èn
(n 1)# (2n 2)(2n 1)
œ n lim Ä_
œ
" 4
1
(n ")(n 2)â(2n) 2n n
†
(2n 1)! n! n! 3n
1
41. converges conditionally since
! kan k œ !
(2n)! (n!)#
(n 1)! (n 1)! 3nb1 (2n 3)!
40. converges absolutely by the Ratio Test: n lim ¹ anabn 1 ¹ œ n lim Ä_ Ä_ œ n lim Ä_
†
605
and š Èn 1" Èn › is a (")n Èn 1 Èn
diverges by the Limit Comparison Test (part 1) with
œ n lim Ä_
Èn Èn 1 Èn
œ n lim Ä_
1 É1 1n 1
œ
converges; but " Èn ;
a divergent p-series:
" #
È
#
n n 42. diverges by the nth-Term Test since n lim ŠÈn# n n‹ œ n lim ŠÈn# n n‹ † Š Ènn# ‹ Ä_ Ä_ n n
œ n lim Ä_
n È n # n n
œ n lim Ä_
" É1 "n 1
" #
œ
Á0
É n Èn Èn
43. diverges by the nth-Term Test since n lim ŠÉn Èn Èn‹ œ n lim ŠÉn Èn Èn‹ Ä_ Ä_ – É n È n È n — Èn
œ n lim Ä_
É n Èn Èn
œ n lim Ä_
" É1
"
Èn 1
" #
œ
Á0
44. converges conditionally since š Èn "Èn 1 › is a decreasing sequence of positive terms converging to 0 _
(")n Èn Èn 1
Ê !
nœ1
_
so that ! nœ1
converges; but n lim Ä_
" Èn Èn 1
"
Èn Š È"n ‹
Š Èn
1
‹
Èn È n È n 1
œ n lim Ä_
_
diverges by the Limit Comparison Test with ! nœ1
45. converges absolutely by the Direct Comparison Test since sech (n) œ
" Èn
œ n lim Ä_
" 1É1 "n
œ
" #
which is a divergent p-series
2 en ecn
œ
2en e2n 1
2en e2n
œ
2 en
which is the
nth term of a convergent geometric series _
_
nœ1
nœ1
46. converges absolutely by the Limit Comparison Test (part 1): ! kan k œ ! Apply the Limit Comparison Test with lim
nÄ_
47.
1 4
1 6
Œ
2 en c ecn 1 en
1 8
1 10
œ n lim Ä_
1 12
1 14
2en en ecn
1 en ,
the n-th term of a convergent geometric series:
œ n lim Ä_ _
ÞÞÞ œ !
nœ1
2 1 ec2n
(")nb1 2 an 1 b ;
n 2 n 1 Ê 2an 2b 2an 1b Ê
2 en ecn
œ2
converges by Alternating Series Test since: un œ
1 2 aa n 1 b 1 b
Ÿ
1 2 an 1 b
Ê un1 Ÿ un ;
lim un œ
nÄ_
1 2 an 1 b
lim 1 nÄ_ 2an1b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
0 for all n 1;
œ 0.
606
Chapter 10 Infinite Sequences and Series
48. 1
1 4
1 9
1 16
1 25
1 36
1 49
1 64
_
_
_
nœ1
nœ1
nœ1
Þ Þ Þ œ ! an ; converges by the Absolute Convergence Test since ! kan k œ !
" n#
which is a convergent p-series 49. kerrork ¸(1)' ˆ "5 ‰¸ œ 0.2 51. kerrork ¹(1)'
(0.01)& 5 ¹
50. kerrork ¸(1)' ˆ 10" & ‰¸ œ 0.00001
œ 2 ‚ 10""
52. kerrork k(1)% t% k œ t% 1
53. kerrork 0.001 Ê un1 0.001 Ê
1 an 1 b 2 3
0.001 Ê an 1b2 3 1000 Ê n 1 È997 ¸ 30.5753 Ê n 31
54. kerrork 0.001 Ê un1 0.001 Ê
n1 an 1 b 2 1
0.001 Ê an 1b2 1 1000an 1b Ê n
998È9982 4a998b 2
¸ 998.9999 Ê n 999 55. kerrork 0.001 Ê un1 0.001 Ê
1 3 ˆ an 1 b 3 È n 1 ‰
3
0.001 Ê Šan 1b 3Èn 1‹ 1000
2
È Ê ŠÈn 1‹ 3Èn 1 10 0 Ê Èn 1 œ 3 29 40 œ 2 Ê n œ 3 Ê n 4
56. kerrork 0.001 Ê un1 0.001 Ê
1 lnalnan 3bb
0.001 Ê lnalnan 3bb 1000 Ê n 3 ee
1000
¸ 5.297 ‚ 10323228467
which is the maximum arbitrary-precision number represented by Mathematica on the particular computer solving this problem.. 57.
" (2n)!
58.
" n!
Ê (2n)!
5 10'
10' 5
Ê
5 10'
10' 5
n! Ê n 9 Ê 1 1
59. (a) an an1 fails since _
_
nœ1
nœ1
" 3
" #!
œ 200,000 Ê n 5 Ê 1
" #
" #!
" 3!
" 4!
_
_
nœ1
nœ1
" 4!
" 5!
" 6!
" 6!
" 7!
" 8!
¸ 0.54030 " 8!
¸ 0.367881944
n n n n (b) Since ! kan k œ ! ˆ 3" ‰ ˆ #" ‰ ‘ œ ! ˆ 3" ‰ ! ˆ #" ‰ is the sum of two absolutely convergent
series, we can rearrange the terms of the original series to find its sum: ˆ "3
" 9
" 27
60. s#! œ 1
" #
" 3
á ‰ ˆ #"
" 4
á
" 19
" 4
" 20
" 8
በœ
ˆ "3 ‰
1 ˆ "3 ‰
ˆ "# ‰
1 ˆ "# ‰
œ
" #
1 œ #"
" #
†
" #1
¸ 0.6687714032 Ê s#!
¸ 0.692580927
_
61. The unused terms are ! (1)j 1 aj œ (1)n 1 aan 1 an 2 b (1)n 3 aan 3 an 4 b á jœn 1
œ (1)n 1 caan 1 an 2 b aan 3 an 4 b á d . Each grouped term is positive, so the remainder has the same sign as (1)n 1 , which is the sign of the first unused term. 62. sn œ
" 1 †2
" #†3
" 3†4
á
" n(n 1)
n
œ!
kœ1
" k(k 1)
n
œ ! ˆ k" kœ1
œ ˆ1 "# ‰ ˆ "# 3" ‰ ˆ 3" 4" ‰ ˆ 4" 5" ‰ á ˆ n"
" ‰ k1
" ‰ n1
which are the first 2n terms
of the first series, hence the two series are the same. Yes, for n
sn œ ! ˆ k" kœ1
" ‰ k 1
œ ˆ1 #" ‰ ˆ #" 3" ‰ ˆ 3" 4" ‰ ˆ 4" 5" ‰ á ˆ n " 1 n" ‰ ˆ n"
" ‰ n1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ 1
" n1
Section 10.6 Alternating Series, Absolute and Conditional Convergence
607
ˆ1 n " 1 ‰ œ 1 Ê both series converge to 1. The sum of the first 2n 1 terms of the first Ê n lim s œ n lim Ä_ n Ä_ ˆ1 n " 1 ‰ œ 1. series is ˆ1 n " 1 ‰ n " 1 œ 1. Their sum is n lim s œ n lim Ä_ n Ä_ _
_
_
_
nœ1
nœ1
nœ1
nœ1
63. Theorem 16 states that ! kan k converges Ê ! an converges. But this is equivalent to ! an diverges Ê ! kan k diverges _
_
nœ1
nœ1
64. ka" a# á an k Ÿ ka" k ka# k á kan k for all n; then ! kan k converges Ê ! an converges and these imply that _
_
nœ1
nœ1
º ! an º Ÿ ! kan k _
65. (a) ! kan bn k converges by the Direct Comparison Test since kan bn k Ÿ kan k kbn k and hence nœ1 _
! aan bn b converges absolutely
nœ1 _
_
_
(b) ! kbn k converges Ê ! bn converges absolutely; since ! an converges absolutely and nœ1 _
nœ1
nœ1 _
_
! bn converges absolutely, we have ! can (bn )d œ ! aan bn b converges absolutely by part (a)
nœ1 _
_
_
nœ1
nœ1
nœ1
nœ1
nœ1 _
(c) ! kan k converges Ê kkk ! kan k œ ! kkan k converges Ê ! kan converges absolutely
66. If an œ bn œ (1)n
" Èn
_
, then ! (1)n nœ1
67. s" œ "# , s# œ "# 1 œ " #
s$ œ 1 s% œ s$ s& œ s% s' œ s& s( œ s'
" 4
" 6
" 8
" 3 ¸ 0.1766, " " " #4 #6 #8 " 5 ¸ 0.312, " " " 46 48 50
" #
" Èn
nœ1
_
_
nœ1
nœ1
converges, but ! an bn œ !
" n
diverges
,
" 10
" 1#
" 14
" 16
" 18
" #0
" 2#
¸ 0.5099,
" 30
" 3#
" 34
" 36
" 38
" 40
" 42
" 44
¸ 0.512,
" 52
" 54
" 56
" 58
" 60
" 62
" 64
" 66
¸ 0.51106
N" 1
68. (a) Since ! kan k converges, say to M, for % 0 there is an integer N" such that º ! kan k Mº nœ1
N" 1
N" 1
_
nœ1
nœ1
nœN"
Í » ! kan k ! kan k ! kan k »
% #
_
Í » ! kan k» nœN"
% #
_
Í ! kan k nœN"
% #
% #
. Also, ! an
converges to L Í for % 0 there is an integer N# (which we can choose greater than or equal to N" ) such that ksN# Lk
% #
_
. Therefore, ! kan k nœN"
% #
and ksN# Lk
% #
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
608
Chapter 10 Infinite Sequences and Series _
k
nœ1
nœ1
(b) The series ! kan k converges absolutely, say to M. Thus, there exists N" such that º ! kan k Mº % whenever k N" . Now all of the terms in the sequence ekbn kf appear in ekan kf. Sum together all of the N terms in ekbn kf, in order, until you include all of the terms ekan kf nœ" 1 , and let N# be the largest index in the N#
N#
_
nœ1
nœ1
nœ1
sum ! kbn k so obtained. Then º ! kbn k Mº % as well Ê ! kbn k converges to M. 10.7 POWER SERIES _
nb1 1. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ xxn ¹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! (1)n , a divergent Ä_ Ä_ nœ1
_
series; when x œ 1 we have ! 1, a divergent series nœ1
(a) the radius is 1; the interval of convergence is 1 x 1 (b) the interval of absolute convergence is 1 x 1 (c) there are no values for which the series converges conditionally nb1
2. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x(x5)5)n ¹ 1 Ê kx 5k 1 Ê 6 x 4; when x œ 6 we have Ä_ Ä_ _
_
nœ1
nœ1
! (1)n , a divergent series; when x œ 4 we have ! 1, a divergent series
(a) the radius is 1; the interval of convergence is 6 x 4 (b) the interval of absolute convergence is 6 x 4 (c) there are no values for which the series converges conditionally nb1
1) " " 3. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (4x (4x 1)n ¹ 1 Ê k4x 1k 1 Ê 1 4x 1 1 Ê # x 0; when x œ # we Ä_ Ä_ _
_
_
_
_
nœ1
nœ1
nœ1
nœ1
nœ1
have ! (1)n (1)n œ ! (1)2n œ ! 1n , a divergent series; when x œ 0 we have ! (1)n (1)n œ ! (1)n , a divergent series (a) the radius is "4 ; the interval of convergence is #" x 0 (b) the interval of absolute convergence is "# x 0
(c) there are no values for which the series converges conditionally nb1
4. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (3xn2)1 Ä_ Ä_ Ê 1 3x 2 1 Ê
" 3
†
n (3x 2)n ¹
ˆ n ‰ 1 Ê k3x 2k 1 1 Ê k3x 2k n lim Ä _ n1
x 1; when x œ
" 3
_
nœ1
(b) the interval of absolute convergence is
" 3
(c) the series converges conditionally at x œ nb1
2) 5. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 10 nb1 Ä_ Ä_
nœ1
" n
conditionally convergent; when x œ 1 we have ! (a) the radius is "3 ; the interval of convergence is
_
we have !
" 3
(")n n
which is the alternating harmonic series and is
, the divergent harmonic series
Ÿx1
x1 " 3
10n (x 2)n ¹
1 Ê
kx 2 k 10
1 Ê kx 2k 10 Ê 10 x 2 10
_
_
nœ1
nœ1
Ê 8 x 12; when x œ 8 we have ! (")n , a divergent series; when x œ 12 we have ! 1, a divergent series (a) the radius is "0; the interval of convergence is 8 x 12 (b) the interval of absolute convergence is 8 x 12 (c) there are no values for which the series converges conditionally
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.7 Power Series nb1
6. n lim k2xk 1 Ê k2xk 1 Ê "# x ¹ uunbn 1 ¹ 1 Ê n lim ¹ (2x) (2x)n ¹ 1 Ê n lim Ä_ Ä_ Ä_ _
! (")n , a divergent series; when x œ
nœ1
" #
" #
; when x œ "# we have
_
we have ! 1, a divergent series nœ1
(a) the radius is "# ; the interval of convergence is "# x (b) the interval of absolute convergence is "# x
" #
" #
(c) there are no values for which the series converges conditionally nb1
1)x 7. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n (n 3) † Ä_ Ä_
(n 2) nxn ¹
1 Ê kxk n lim Ä_
_
Ê 1 x 1; when x œ 1 we have ! (")n nœ1
_
have ! nœ1
n n#,
n n#
(n 1)(n 2) (n 3)(n)
1 Ê kxk 1
, a divergent series by the nth-term Test; when x œ " we
a divergent series
(a) the radius is "; the interval of convergence is " x " (b) the interval of absolute convergence is " x " (c) there are no values for which the series converges conditionally nb1
8. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x n2)1 Ä_ Ä_
†
n (x 2)n ¹
ˆ 1 Ê kx 2k n lim Ä_ _
Ê 1 x 2 1 Ê 3 x 1; when x œ 3 we have !
nœ1
_
! nœ1
(1)n n ,
" n,
n ‰ n1
1 Ê kx 2k 1
a divergent series; when x œ " we have
a convergent series
(a) the radius is "; the interval of convergence is 3 x Ÿ " (b) the interval of absolute convergence is 3 x " (c) the series converges conditionally at x œ 1 nb1
x 9. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ (n 1)Èn 1 3nb1
Ê
kx k 3
nÈ n 3n xn ¹
1 Ê
kx k 3
n n ‹ n 1 ‹ ŠÉ n lim Ä _ n1
Šn lim Ä_ _
(1)(1) 1 Ê kxk 3 Ê 3 x 3; when x œ 3 we have !
nœ1
_
when x œ 3 we have !
nœ1
1 , n$Î#
(")n , n$Î#
1
an absolutely convergent series;
a convergent p-series
(a) the radius is 3; the interval of convergence is 3 Ÿ x Ÿ 3 (b) the interval of absolute convergence is 3 Ÿ x Ÿ 3 (c) there are no values for which the series converges conditionally nb1
10. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (xÈn1) 1 † Ä_ Ä_
Èn (x 1)n ¹
1 Ê kx 1k Én lim Ä_ _
Ê 1 x 1 1 Ê 0 x 2; when x œ 0 we have !
nœ1
_
we have ! nœ1
1 , n"Î#
(")n , n"Î#
n n1
609
1 Ê kx 1k 1
a conditionally convergent series; when x œ 2
a divergent series
(a) the radius is 1; the interval of convergence is 0 Ÿ x 2 (b) the interval of absolute convergence is 0 x 2 (c) the series converges conditionally at x œ 0 nb1
ˆ " ‰ 1 for all x 11. n lim † n! ¹ 1 Ê kxk n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ x Ä_ Ä _ (n 1)! xn Ä _ n1 (a) the radius is _; the series converges for all x
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610
Chapter 10 Infinite Sequences and Series
(b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1
nb1
ˆ " ‰ 1 for all x 12. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ 3 x † 3nn!xn ¹ 1 Ê 3 kxk n lim Ä_ Ä _ (n 1)! Ä _ n1 (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally 13. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹4 Ä_ Ä_
nb1 2nb2
x n1
†
n 4n x2n ¹
ˆ 4n ‰ œ 4x# 1 Ê x# 1 Ê x# n lim Ä _ n1
_
_
nœ1
nœ1
n 2n Ê 12 x 12 ; when x œ 12 we have ! 4n ˆ 12 ‰ œ !
_
! nœ1
4n ˆ 1 ‰2n n 2
_
œ!
nœ1
1 n,
1 n
1 4
, a divergent p-series; when x œ
1 2
we have
a divergent p-series
(a) the radius is 12 ; the interval of convergence is 12 x (b) the interval of absolute convergence is 12 x
1 2
1 2
(c) there are no values for which the series converges conditionally nb1
14. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 1) Ä_ Ä _ an 1b2 3nb1
n2 3n (x 1)n ¹
_
2
1 Ê lx 1l n lim Š n ‹ œ 31 lx 1l 1 Ä _ 3 an 1 b 2 _
Ê 2 x 4; when x œ 2 we have ! (n2 3)3n œ ! (n1) , an absolutely convergent series; when x œ 4 we have 2 n
nœ1
_
n
nœ1
_
n
! (3) ! 12 , an absolutely convergent series. n2 3n œ n
nœ1
nœ1
(a) the radius is 3; the interval of convergence is 2 Ÿ x Ÿ 4 (b) the interval of absolute convergence is 2 Ÿ x Ÿ 4 (c) there are no values for which the series converges conditionally nb1
x 15. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ È(n 1)# 3
È n# 3 ¹ xn
_
Ê 1 x 1; when x œ 1 we have !
nœ1
_
! nœ1
" È n# 3
1 Ê kxk Én lim Ä_
(")n È n# 3
n# 3 n# 2n 4
" Ê kxk 1
, a conditionally convergent series; when x œ 1 we have
, a divergent series
(a) the radius is 1; the interval of convergence is 1 Ÿ x 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 n 1
x 16. n lim † ¹ uun n 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ È(n 1)# 3
È n# 3 ¹ xn
_
Ê 1 x 1; when x œ 1 we have !
nœ1
1 Ê kxk Én lim Ä_
" È n# 3
n# 3 n# 2n 4
" Ê kxk 1 _
, a divergent series; when x œ 1 we have !
nœ1
a conditionally convergent series (a) the radius is 1; the interval of convergence is 1 x Ÿ 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
(")n È n# 3
,
Section 10.7 Power Series 3) 17. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n 1)(x 5nb1 Ä_ Ä_
nb1
†
5n n(x 3)n ¹
1 Ê
kx 3 k lim 5 nÄ_
ˆ n n " ‰ 1 Ê _
Ê kx 3k 5 Ê 5 x 3 5 Ê 8 x 2; when x œ 8 we have !
nœ1
_
series; when x œ 2 we have !
nœ1
n5n 5n
n(5)n 5n
kx 3 k 5
611
1
_
œ ! (1)n n, a divergent nœ1
_
œ ! n, a divergent series nœ1
(a) the radius is 5; the interval of convergence is 8 x 2 (b) the interval of absolute convergence is 8 x 2 (c) there are no values for which the series converges conditionally nb1
18. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n 1)x Ä_ Ä _ 4nb1 an# 2n 2b _
Ê 4 x 4; when x œ 4 we have !
nœ1
4 n an # 1 b ¹ nxn n(1)n n# 1
1 Ê
kx k 4 n lim Ä_
#
(n 1) n 1 ¹ n an# a2n 2bb ¹ 1 Ê kxk 4 _
, a conditionally convergent series; when x œ 4 we have !
nœ1
n n# 1
,
a divergent series (a) the radius is 4; the interval of convergence is 4 Ÿ x 4 (b) the interval of absolute convergence is 4 x 4 (c) the series converges conditionally at x œ 4 19. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä_
Èn 1 xnb1 3nb1
†
3n È n xn ¹
1 Ê
kx k 3
ˆ n n 1 ‰ 1 Ê Én lim Ä_
kx k 3
1 Ê kxk 3
_
_
nœ1
nœ1
Ê 3 x 3; when x œ 3 we have ! (1)n Èn , a divergent series; when x œ 3 we have ! Èn, a divergent series (a) the radius is 3; the interval of convergence is 3 x 3 (b) the interval of absolute convergence is 3 x 3 (c) there are no values for which the series converges conditionally 20. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä_
nbÈ 1
n 1 (2x5)nb1 ¹ n n (2x5)n È
1 Ê k2x 5k n lim Š Ä_
nbÈ 1
n1 ‹ n n È
1
t È
lim t Ä_ Ê k2x 5k Œ tlim n n 1 Ê k2x 5k 1 Ê 1 2x 5 1 Ê 3 x 2; when x œ 3 we have È n
_
Ä_
_
n n n ! (1) È È n, a divergent series since n lim n œ 1; when x œ 2 we have ! È n, a divergent series Ä_ nœ1 nœ1
(a) the radius is "# ; the interval of convergence is 3 x 2
(b) the interval of absolute convergence is 3 x 2 (c) there are no values for which the series converges conditionally _
_
_
nœ1
nœ1
21. First, rewrite the series as ! a2 (1)n bax 1bn1 œ ! 2ax 1bn1 ! (1)n ax 1bn1 . For the series nœ1
_
n
! 2ax 1bn1 : lim ¹ unb1 ¹ 1 Ê lim ¹ 2ax1nbc1 ¹ 1 Ê lx 1l lim 1 œ lx 1l 1 Ê 2 x 0; For the un nÄ_ n Ä _ 2 ax 1 b nÄ_ nœ1 _
nb1
n
series ! (1)n ax 1bn1 : n lim 1 œ lx 1l 1 ¹ uunbn 1 ¹ 1 Ê n lim ¹ ( 1) ax 1b ¹ 1 Ê lx 1ln lim Ä_ Ä _ (1)n ax1bnc1 Ä_ nœ1 _
Ê 2 x 0; when x œ 2 we have ! a2 (1)n ba1bn1 , a divergent series; when x œ 0 we have nœ1
_
! a2 (1)n b, a divergent series
nœ1
(a) the radius is 1; the interval of convergence is 2 x 0 (b) the interval of absolute convergence is 2 x 0 (c) there are no values for which the series converges conditionally
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612
Chapter 10 Infinite Sequences and Series
( 1) 22. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä_
Ê _
! nœ1
17 9
x
19 9 ;
(1)n 32n ˆ 1 ‰n 3n 9
when x œ _
œ!
nœ1
(1)n 3n ,
17 9
3 ax 2bnb1 3 an 1 b
nb1 2nb2
_
we have ! nœ1
†
(1)n 32n ˆ 1 ‰n 9 3n
(b) the interval of absolute convergence is
17 9
(c) the series converges conditionally at x œ
23.
_
œ!
nœ1
1 3n ,
9n n1
œ 9lx 2l 1
a divergent series; when x œ
19 9
we have
a conditionally convergent series.
(a) the radius is 19 ; the interval of convergence is
lim ¹ uunbn 1 ¹ nÄ_
1 Ê lx 2ln lim Ä_
3n ¹ (1)n 32n ax 2bn
1 Ê n lim Ä_ »
Š1
n
"
nb1
1‹
xnb1
Š1 "n ‹ xn n
17 9
x
xŸ
19 9
19 9
19 9 "
t
lim Š1 t ‹ e Ä_ » 1 Ê kxk lim Š1 " ‹n 1 Ê kxk ˆ e ‰ 1 Ê kxk 1 n nÄ_ t
_
n Ê 1 x 1; when x œ 1 we have ! (1)n ˆ1 "n ‰ , a divergent series by the nth-Term Test since nœ1
lim ˆ1 nÄ_
" ‰n n
_
n œ e Á 0; when x œ 1 we have ! ˆ1 n" ‰ , a divergent series nœ1
(a) the radius is "; the interval of convergence is 1 x 1 (b) the interval of absolute convergence is 1 x 1 (c) there are no values for which the series converges conditionally 24. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ ln (nxnln1)xn Ä_ Ä_
nb1
¹ 1 Ê kxk n lim Ä_ º
ˆn " 1‰ ˆ n" ‰ º
ˆ n ‰ 1 Ê kxk 1 1 Ê kxk n lim Ä _ n1
_
Ê 1 x 1; when x œ 1 we have ! (1)n ln n, a divergent series by the nth-Term Test since n lim ln n Á 0; Ä_ nœ1
_
when x œ 1 we have ! ln n, a divergent series nœ1
(a) the radius is 1; the interval of convergence is 1 x 1 (b) the interval of absolute convergence is 1 x 1 (c) there are no values for which the series converges conditionally nb1 nb1
x ˆ1 n" ‰n ‹ Š lim (n 1)‹ 1 25. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n 1) ¹ 1 Ê kxk Šn lim nn xn Ä_ Ä_ Ä_ nÄ_ Ê e kxk n lim (n 1) 1 Ê only x œ 0 satisfies this inequality Ä_
(a) the radius is 0; the series converges only for x œ 0 (b) the series converges absolutely only for x œ 0 (c) there are no values for which the series converges conditionally nb1
26. n lim (n 1) 1 Ê only x œ 4 satisfies this inequality ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n n!1)!(x(x4)4)n ¹ 1 Ê kx 4k n lim Ä_ Ä_ Ä_ (a) the radius is 0; the series converges only for x œ 4 (b) the series converges absolutely only for x œ 4 (c) there are no values for which the series converges conditionally nb1
27. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 2) Ä_ Ä _ (n 1) 2nb1
n2n (x 2)n ¹
1 Ê
kx 2 k lim # nÄ_
ˆ n n 1 ‰ 1 Ê
kx 2 k #
1 Ê kx 2k 2
_
_
nœ1
nœ1
! (1) Ê 2 x 2 2 Ê 4 x 0; when x œ 4 we have ! " n , a divergent series; when x œ 0 we have n the alternating harmonic series which converges conditionally (a) the radius is 2; the interval of convergence is 4 x Ÿ 0 (b) the interval of absolute convergence is 4 x 0 (c) the series converges conditionally at x œ 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
nb1
,
Section 10.7 Power Series nb1
613
nb1
(n 2)(x 1) ˆ n 2 ‰ 1 Ê 2 kx 1k 1 28. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ ((2)2)n (n 1)(x 1)n ¹ 1 Ê 2 kx 1k n lim Ä_ Ä_ Ä _ n1
Ê kx 1k _
" #
Ê "# x 1
" #
" #
Ê
x 3# ; when x œ
" #
_
we have ! (n 1) , a divergent series; when x œ nœ1
we have ! (1) (n 1), a divergent series n
n œ1
(a) the radius is "# ; the interval of convergence is (b) the interval of absolute convergence is
" #
" #
x
x
3 #
3 #
(c) there are no values for which the series converges conditionally nb1
x 29. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ (n 1) aln (n 1)b#
Ê kxk (1) Œn lim Ä_ _
! nœ1
(1)n n(ln n)#
#
ˆ "n ‰ ˆ nb" 1 ‰
n(ln n)# xn ¹
1 Ê kxk Šn lim Ä_
n1 n ‹
1 Ê kxk Šn lim Ä_
#
n ln n ‹ n 1 ‹ Šn lim Ä _ ln (n 1)
#
1
1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have _
which converges absolutely; when x œ 1 we have !
nœ1
" n(ln n)#
which converges
(a) the radius is "; the interval of convergence is 1 Ÿ x Ÿ 1 (b) the interval of absolute convergence is 1 Ÿ x Ÿ 1 (c) there are no values for which the series converges conditionally nb1
x 30. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ (n 1) ln (n 1)
n ln (n) xn ¹
1 Ê kxk Šn lim Ä_
ln (n) n ‹ n 1 ‹ Šn lim Ä _ ln (n 1) _
(1)n n ln n
Ê kxk (1)(1) 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have !
nœ2
_
when x œ 1 we have !
nœ2
" n ln n
1
, a convergent alternating series;
which diverges by Exercise 38, Section 9.3
(a) the radius is "; the interval of convergence is 1 Ÿ x 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 2nb3
5) 31. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ (4x (n 1)$Î# Ä_ Ä_
n$Î# (4x 5)2n
1
¹ 1 Ê (4x 5)# Šn lim Ä_
Ê k4x 5k 1 Ê 1 4x 5 1 Ê 1 x absolutely convergent; when x œ
3 #
_
we have ! nœ1
(")2nb1 n$Î#
3 #
_
; when x œ 1 we have !
nœ1
$Î#
1 Ê (4x 5)# 1
(1)2nb1 n$Î#
_
œ!
nœ1
" n$Î#
which is
, a convergent p-series
(a) the radius is "4 ; the interval of convergence is 1 Ÿ x Ÿ (b) the interval of absolute convergence is 1 Ÿ x Ÿ
n n1‹
3 #
3 #
(c) there are no values for which the series converges conditionally nb2
32. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (3x2n1)4 Ä_ Ä_
†
2n 2 (3x 1)nb1 ¹
ˆ 2n 2 ‰ 1 Ê k3x 1k 1 1 Ê k3x 1k n lim Ä _ 2n 4 _
Ê 1 3x 1 1 Ê 23 x 0; when x œ 23 we have !
nœ1
_
when x œ 0 we have !
nœ1
(")nb1 2n 1
_
œ!
nœ1
" #n 1
(1)nb1 2n 1
, a conditionally convergent series;
, a divergent series
(a) the radius is "3 ; the interval of convergence is 23 Ÿ x 0 (b) the interval of absolute convergence is 23 x 0 (c) the series converges conditionally at x œ 23
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3 #
614
Chapter 10 Infinite Sequences and Series nb1
x ˆ 1 ‰ 1 for all x 33. n lim † 2†4†6xân a2nb ¹ 1 Ê kxk n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ 2†4†6âa2nba2an 1bb Ä _ 2n 2 (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb2
a2n 3bn 3 5 7 a2n 1ba2an 1b 1bx 34. n lim † 3†5†7âan2n2 1bxnb1 ¹ 1 Ê kxk n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ † † â an 1b2 2nb1 Š 2an 1b2 ‹ 1 Ê only Ä_ Ä_ Ä_ x œ 0 satisfies this inequality (a) the radius is 0; the series converges only for x œ 0 (b) the series converges absolutely only for x œ 0 (c) there are no values for which the series converges conditionally _
35. For the series ! nœ1
12ân n 12 22 â n2 x ,
recall 1 2 â n œ
nan b 1b
_
2
2 n
n an 1 b 2
and 12 22 â n2 œ
nan 1ba2n 1b 6
_
nb1 rewrite the series as ! Œ n n b 1 2 2n b 1 xn œ ! ˆ 2n 3 1 ‰xn ; then n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ a2an3x 1 b 1 b † Ä _ Ä _ 6 nœ1 nœ1 a
ba
b
so that we can
a2n 1b 3xn ¹
1
_
Ê kxk n lim ¹ a2n 1b ¹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! ˆ 2n 3 1 ‰a1bn , a conditionally Ä _ a2n 3b nœ1 _
convergent series; when x œ 1 we have ! ˆ 2n 3 1 ‰, a divergent series. nœ1
(a) the radius is 1; the interval of convergence is 1 Ÿ x 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 _
36. For the series ! ŠÈn 1 Èn‹ax 3bn , note that Èn 1 Èn œ nœ1
_
can rewrite the series as ! nœ1
Ê lx 3ln lim Ä_
ax 3 b n Èn 1 Èn ;
Èn 1 Èn Èn 2 Èn 1
Èn 1 Èn 1
†
nb1
Èn 1 Èn Èn 1 Èn
x3 then n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ a b Ä_ Ä _ Èn 2 Èn 1
ax 3 b n
_
nœ1
1 Èn 1 Èn ,
so that we
¹1
a 1 b n Èn 1 Èn ,
nœ1
_
1 Èn 1 Èn
Èn 1 Èn
1 Ê lx 3l 1 Ê 2 x 4; when x œ 2 we have !
convergent series; when x œ 4 we have !
œ
a conditionally
a divergent series;
(a) the radius is 1; the interval of convergence is 2 Ÿ x 4 (b) the interval of absolute convergence is 2 x 4 (c) the series converges conditionally at x œ 2 nb1
a n 1 bx x 37. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ Ä_ Ä _ 3†6†9âa3nba3an 1bb
3†6†9âa3nb ¹ nx xn 2 nb1
2 4 6 2n 2 n 1 x 38. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ a † † âa ba a bbb Ä_ Ä _ a2†5†8âa3n 1ba3an 1b 1bb2 9 9 Ê lxl 4 Ê R œ 4 2 nb1
n 1 x 39. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ a a bx b Ä_ Ä _ 2nb1 a2an 1bbx
2n a2nbx ¹ an xb 2 x n
an 1 b 1 Ê lxln lim ¹ ¹1Ê Ä _ 3 an 1 b
a2†5†8âa3n 1bb2 ¹ a2†4†6âa2nbb2 xn
lx l 3
1 Ê lxl 3 Ê R œ 3 2
1 Ê lxln lim ¹ a2n 2b ¹ 1 Ê Ä _ a3n 2b2
2
an 1 b 1 Ê lxln lim ¹ ¹1Ê Ä _ 2a2n 2ba2n 1b
lx l 8
4 lx l 9
1
1 Ê lxl 8 Ê R œ 8
2
n n ‰n n n Ɉ ˆ n ‰n 1 Ê lxle1 1 Ê lxl e Ê R œ e È 40. n lim un 1 Ê n lim x 1 Ê lxl n lim n1 Ä_ Ä_ Ä _ n1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.7 Power Series nb1
nb1
41. n lim 3 1 Ê lxl ¹ uunbn 1 ¹ 1 Ê n lim ¹ 3 3n xxn ¹ 1 Ê lxl n lim Ä_ Ä_ Ä_ _
_
! 3n ˆ 1 ‰n œ ! a1bn , which diverges; at x œ 3
nœ0
nœ0
1 3
1 3
_
_
nœ0
nœ0
615
Ê 31 x 31 ; at x œ 31 we have _
n we have ! 3n ˆ 13 ‰ œ ! 1 , which diverges. The series ! 3n xn
_
œ ! a3xbn is a convergent geometric series when 13 x nœ0
1 3
and the sum is
nœ0
1 1 3x .
nb1
e 4 42. n lim 1 1 Ê lex 4l 1 Ê 3 ex 5 Ê ln 3 x ln 5; ¹ uunbn 1 ¹ 1 Ê n lim ¹ a aex 4b bn ¹ 1 Ê lex 4l n lim Ä_ Ä_ Ä_ x
_
_
_
_
nœ0 _
nœ0
nœ0
nœ0
n n at x œ ln 3 we have ! ˆeln 3 4‰ œ ! a1bn , which diverges; at x œ ln 5 we have ! ˆeln 5 4‰ œ ! 1, which
diverges. The series ! aex 4bn is a convergent geometric series when ln 3 x ln 5 and the sum is nœ0
2nb2
43. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 4n1)1 Ä_ Ä_
4n (x 1)2n ¹
†
1 Ê
(x 1)# lim 4 nÄ_ _
Ê 2 x 1 2 Ê 1 x 3; at x œ 1 we have !
nœ0
_
we have ! nœ0 _
! nœ0
(x ")2n 4n "
#
œ!
2 4n
nœ0
_
œ!
nœ0
œ
" 1 Šxc # ‹
_
2n
4
4 4n
(x 4
n
9n (x 1)2n ¹
†
1 Ê
(x 1)# lim 9 nÄ_
nœ0
!
nœ0
nœ0
nœ0
k1k 1 Ê (x 1)# 9 Ê kx 1k 3
(3)2n 9n
_
œ ! 1 which diverges; at x œ 2 we have nœ0
_
œ ! " which also diverges; the interval of convergence is 4 x 2; the series
(x 1) 9n "
_
4 4 ")# “ œ 4 x# 2x 1 œ 3 2x x#
_
!
_
n œ ! 44n œ ! 1, which diverges; at x œ 3
is a convergent geometric series when 1 x 3 and the sum is
Ê 3 x 1 3 Ê 4 x 2; when x œ 4 we have !
nœ0 _
k1k 1 Ê (x 1)# 4 Ê kx 1k 2
nœ0
2nb2
32n 9n
1 5 ex .
œ ! 1, a divergent series; the interval of convergence is 1 x 3; the series
44. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 9n1)1 Ä_ Ä_
_
œ
_
# Šˆ x # 1 ‰ ‹ "
’
n
(2)2n 4n
1 1 ae x 4 b
nœ0
2n
_
n
# œ ! Šˆ x3 1 ‰ ‹ is a convergent geometric series when 4 x 2 and the sum is nœ0
1 1 Šxb 3 ‹
#
œ
"
’
9
(x 1)# “ 9
œ
9 9 x# 2x 1
45. n lim ¹ uunbn 1 ¹ 1 Ê n lim Ä_ Ä_ º
œ
9 8 2x x#
ˆÈx 2‰nb1 2nb1
†
2n ˆÈ x 2 ‰ n º
1 Ê ¸È x 2 ¸ 2 Ê 2 È x 2 2 Ê 0 È x 4
_
_
Ê 0 x 16; when x œ 0 we have ! (1)n , a divergent series; when x œ 16 we have ! (1)n , a divergent nœ0
nœ0
_
series; the interval of convergence is 0 x 16; the series !
nœ0
0 x 16 and its sum is
1Œ
" Èx c 2 œ
#
Œ
2c
"
Èx #
2
œ
Èx 2 n Š # ‹
is a convergent geometric series when
2 4 Èx
nb1
46. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (ln(lnx)x)n ¹ 1 Ê kln xk 1 Ê 1 ln x 1 Ê e" x e; when x œ e" or e we Ä_ Ä_ _
_
_
nœ0
nœ0
nœ0
obtain the series ! 1n and ! (1)n which both diverge; the interval of convergence is e" x e; ! (ln x)n œ when e" x e
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" 1 ln x
616
Chapter 10 Infinite Sequences and Series
47. n lim ¹ uunbn 1 ¹ 1 Ê n lim Šx Ä_ Ä_ º
#
1 3 ‹
n 1
n † ˆ x# 3 1 ‰ º 1 Ê
ax # 1 b lim 3 nÄ_
x# " 3
k1k 1 Ê
1 Ê x# 2
_
Ê kxk È2 Ê È2 x È2 ; at x œ „ È2 we have ! (1)n which diverges; the interval of convergence is nœ0
_
È2 x È2 ; the series !
nœ0
" # 1 Š x 3b 1 ‹
œ
" # Š 3 c x3 c 1 ‹
œ
# Š x 3 1 ‹
n
is a convergent geometric series when È2 x È2 and its sum is
3 # x#
ax 48. n lim ¹ uun n 1 ¹ 1 Ê n lim ¹ Ä_ Ä_
# 1 bn 2n
1
2n ¹ ax # 1 b n
†
1
1 Ê kx# 1k 2 Ê È3 x È3 ; when x œ „ È3 we
_
_
nœ0
nœ0
have ! 1n , a divergent series; the interval of convergence is È3 x È3 ; the series ! Š x convergent geometric series when È3 x È3 and its sum is
nb1
49. n lim ¹ (x #n3) b1 Ä_
†
2n (x 3)n ¹
" # 1 Šx 2 1‹
"
œ
2
œ
Šx# 1 ‹
#
#
1 2 ‹
n
is a
2 3 x#
_
1 Ê kx 3k 2 Ê 1 x 5; when x œ 1 we have ! (1)n which diverges; nœ1
_
when x œ 5 we have ! (1) which also diverges; the interval of convergence is 1 x 5; the sum of this n
nœ1
convergent geometric series is œ
2 x1
" 3 1 Šxc # ‹
œ
2 x 1
n
. If f(x) œ 1 #" (x 3) 4" (x 3)# á ˆ #" ‰ (x 3)n á n
then f w (x) œ #" #" (x 3) á ˆ #" ‰ n(x 3)n1 á is convergent when 1 x 5, and diverges 2 (x 1)#
when x œ 1 or 5. The sum for f w (x) is
, the derivative of
2 x1
.
n
50. If f(x) œ 1 "# (x 3) 4" (x 3)# á ˆ #" ‰ (x 3)n á œ œx
(x 3)# 4
(x 3)$ 12
_
á ˆ "# ‰
n (x 3)n n 1
1
2 x1
then ' f(x) dx _
á . At x œ 1 the series ! n21 diverges; at x œ 5 nœ1
2 the series ! (n1) 1 converges. Therefore the interval of convergence is 1 x Ÿ 5 and the sum is n
nœ1
2 ln kx 1k (3 ln 4), since '
dx œ 2 ln kx 1k C, where C œ 3 ln 4 when x œ 3.
2 x1
51. (a) Differentiate the series for sin x to get cos x œ 1 œ
x# #!
x% 4!
x' 6!
) x"! 1 x8! 10! á . 2nb2 n ! a b # lim ¹ x † x#8 ¹ œ x2 n lim n Ä _ (2n 2)! Ä_
(b) sin 2x œ 2x
2$ x$ 3!
2& x& 5!
2( x( 7!
" 6!
œ 2x 52. (a) (b)
d x
5x% 5!
7x' 7!
9x) 9!
11x"! 11!
á
The series converges for all values of x since Š a2n 1ba" 2n 2b ‹ œ 0 1 for all x.
1†00†
" 4!
$ $
( (
2 x 3!
aex b œ 1
& &
2 x 5!
2x 2!
3x# 3!
2 x 7!
0†
4x$ 4!
' ex dx œ ex C œ x x#
#
#
(c) ex œ 1 x x#! ˆ1 † 3!" 1 † #"! ˆ1 † 5!" 1 † 4!"
x$ 3! " #! " #!
" 3!
* *
2 x 9!
2* x* 9!
2"" x"" 11!
" #
0†
"" ""
á
0†
5x% 5!
2 x 11!
á œ 2x
8x$ 3!
&
(
*
""
128x 512x 2048x 32x 5! 7! 9! 11! á " "‰ $ ‰ # ˆ (c) 2 sin x cos x œ 2 (0 † 1) (0 † 0 1 † 1)x ˆ0 † " # 1 † 0 0 † 1 x 0 † 0 1 † # 0 † 0 1 † 3! x ˆ0 † 4!" 1 † 0 0 † #" 0 † 3!" 0 † 1‰ x% ˆ0 † 0 1 † 4!" 0 † 0 #" † 3!" 0 † 0 1 † 5!" ‰ x&
ˆ0 †
3x# 3!
" 5!
0 † 1‰ x' á ‘ œ 2 ’x
á œ1x
x# #!
x$ 3!
x% 4!
4x$ 3!
16x& 5!
á“
á œ ex ; thus the derivative of ex is ex itself
x$ x% x& x 3! 4! 5! á C, which is the general antiderivative of e % & x4! x5! á ; ecx † ex œ 1 † 1 (1 † 1 1 † 1)x ˆ1 † #"! 1 † 1 #"! † 1 3!" † 1‰ x$ ˆ1 † 4!" 1 † 3!" #"! † #"! 3!" † 1 4!" † 1‰ x% † 3!" 3!" † #"! 4!" † 1 5!" † 1‰ x& á œ 1 0 0 0 0 0 á
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
† 1‰ x#
Section 10.8 Taylor and Maclaurin Series 53. (a) ln ksec xk C œ ' tan x dx œ ' Šx #
œ
x #
%
x 1#
'
)
x 45
17x 2520 1 #
converges when #
(b) sec x œ
d(tan x) dx
œ
when 1# x
d dx
x
x$ 3
œ1x
x$ 6
œx
$
x 6
x& 24
&
x 24
(b) sec x tan x œ when
1 #
17x' 45
61x( 5040
(
" 4 62x) 315
x# #
17x( 315
5x% 24
5 ‰ % 24 x
62x* 2835
61x' 720
62x* 2835
á ‹ dx
á ‹ œ 1 x#
d(sec x) dx
*
œ
277x 72,576
d dx
á ‹ Š1
61 ˆ 720
á ,
277x* 72,576
61x 5040
2x% 3
x# #
17x' 45
x% 12
62x) 315
x' 45
17x) 2520
31x"! 14,175
á ,
á , converges
1 #
x x# 2
5 48 1 #
5 48
5x% 24
x# #
5x% 24
61 ‰ ' 720 x
61x' 720
61x' 720
á‹
á
á ‹ dx
á C; x œ 0 Ê C œ 0 Ê ln ksec x tan xk á , converges when 1# x
Š1
x# #
5x% 24
61x' 720
5x$ 6
á‹ œ x
1 #
61x& 120
277x( 1008
á , converges
1 #
(c) (sec x)(tan x) œ Š1
x# #
2 œ x ˆ "3 #" ‰ x$ ˆ 15
1# x
1 #
x
17x( 315
2x& 15
54. (a) ln ksec x tan xk C œ ' sec x dx œ ' Š1 œx
5 œ 1 ˆ "# "# ‰ x# ˆ 24 2x% 3
2x& 15
á C; x œ 0 Ê C œ 0 Ê ln ksec xk œ
(c) sec# x œ (sec x)(sec x) œ Š1 #
"!
31x 14,175 1#
Šx
x$ 3
617
1 #
5x% 24 " 6
_
61x' 720
á ‹ Šx
5 ‰ & 24 x
17 ˆ 315
" 15
x$ 3
5 72
2x& 15
17x( 315
61 ‰ ( 720 x
á‹ á œ x
5x$ 6
61x& 120
277x( 1008
á ,
_
55. (a) If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n 1)(n 2)â(n (k 1)) an xnk and f ÐkÑ (0) œ k!ak nœ0
Ê ak œ
f ÐkÑ (0) k!
nœk _
; likewise if f(x) œ ! bn xn , then bk œ nœ0
f ÐkÑ (0) k!
Ê ak œ bk for every nonnegative integer k
_
(b) If f(x) œ ! an xn œ 0 for all x, then f ÐkÑ (x) œ 0 for all x Ê from part (a) that ak œ 0 for every nonnegative integer k nœ0
10.8 TAYLOR AND MACLAURIN SERIES 1. f(x) œ e2x , f w (x) œ 2e2x , f ww (x) œ 4e2x , f www (x) œ 8e2x ; f(0) œ e2a0b œ ", f w (0) œ 2, f ww (0) œ 4, f www (0) œ 8 Ê P! (x) œ 1, P" (x) œ 1 2x, P# (x) œ 1 x 2x# , P$ (x) œ 1 x 2x# 43 x3 2. f(x) œ sin x, f w (x) œ cos x , f ww (x) œ sin x , f www (x) œ cos x; f(0) œ sin 0 œ 0, f w (0) œ 1, f ww (0) œ 0, f www (0) œ 1 Ê P! (x) œ 0, P" (x) œ x, P# (x) œ x, P$ (x) œ x 16 x3 3. f(x) œ ln x, f w (x) œ
" x
, f ww (x) œ x"# , f www (x) œ
2 x$ ;
f(1) œ ln 1 œ 0, f w (1) œ 1, f ww (1) œ 1, f www (1) œ 2 Ê P! (x) œ 0,
P" (x) œ (x 1), P# (x) œ (x 1) "# (x 1)# , P$ (x) œ (x 1) "# (x 1)# "3 (x 1)$ 4. f(x) œ ln (1 x), f w (x) œ f w (0) œ 5. f(x) œ
œ 1, f ww (0) œ (1)
1 1 " x
(1 x)" , f ww (x) œ (1 x)# , f www (x) œ 2(1 x)$ ; f(0) œ ln 1 œ 0,
œ 1, f www (0) œ 2(1)$ œ 2 Ê P! (x) œ 0, P" (x) œ x, P# (x) œ x
œ x" , f w (x) œ x# , f ww (x) œ 2x$ , f www (x) œ 6x% ; f(2) œ
Ê P! (x) œ P$ (x) œ
" 1x œ #
" #
" " " " # , P" (x) œ # 4 (x 2), P# (x) œ # " " " # $ 4 (x 2) 8 (x 2) 16 (x 2)
" #
x# #,
P$ (x) œ x
, f w (2) œ 4" , f ww (2) œ 4" , f www (x) œ 83
4" (x 2) 8" (x 2)# ,
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
x# #
x$ 3
618
Chapter 10 Infinite Sequences and Series
6. f(x) œ (x 2)" , f w (x) œ (x 2)# , f ww (x) œ 2(x 2)$ , f www (x) œ 6(x 2)% ; f(0) œ (2)" œ œ 4" , f ww (0) œ 2(2)$ œ P$ (x) œ
" #
x 4
x# 8
" 4
, f www (0) œ 6(2)% œ 83 Ê P! (x) œ
x$ 16
" #
, P" (x) œ
f ww ˆ 14 ‰ œ sin P# (x) œ
È2 #
1 4 œ È2 ˆx #
È , f www ˆ 14 ‰ œ cos 14 œ #2 Ê È2 È 1‰ ˆx 14 ‰# , P$ (x) œ #2 4 4
4x , P# (x) œ
" #
, f w (0) œ (2)#
x 4
È2 È2 1 w ˆ1‰ 4 œ cos 4 œ # # ,f È È È P! œ #2 , P" (x) œ #2 #2 ˆx 14 ‰ , È2 È È ˆx 14 ‰ 42 ˆx 14 ‰# 1#2 ˆx 14 ‰$ #
7. f(x) œ sin x, f w (x) œ cos x, f ww (x) œ sin x, f www (x) œ cos x; f ˆ 14 ‰ œ sin È2 #
" #
" #
1 4
œ
x# 8
,
,
8. f(x) œ tan x, f w (x) œ sec2 x, f ww (x) œ 2sec2 x tan x, f www (x) œ 2sec4 x 4sec2 x tan2 x; f ˆ 14 ‰ œ tan 14 œ 1 , f w ˆ 14 ‰ œ sec2 ˆ 14 ‰ œ 2 , f ww ˆ 14 ‰ œ 2sec2 ˆ 14 ‰ tan ˆ 14 ‰ œ 4 , f www ˆ 14 ‰ œ 2sec4 ˆ 14 ‰ 4sec2 ˆ 14 ‰ tan2 ˆ 14 ‰ œ 16 Ê P! (x) œ 1 , 2 2 3 P" (x) œ 1 2 ˆx 14 ‰ , P# (x) œ 1 2 ˆx 14 ‰ 2 ˆx 14 ‰ , P$ (x) œ 1 2 ˆx 14 ‰ 2 ˆx 14 ‰ 83 ˆx 14 ‰
9. f(x) œ Èx œ x"Î# , f w (x) œ ˆ "# ‰ x"Î# , f ww (x) œ ˆ 4" ‰ x$Î# , f www (x) œ ˆ 38 ‰ x&Î# ; f(4) œ È4 œ 2, " 3 f w (4) œ ˆ "# ‰ 4"Î# œ 4" , f ww (4) œ ˆ 4" ‰ 4$Î# œ 32 ,f www (4) œ ˆ 38 ‰ 4&Î# œ 256 Ê P! (x) œ 2, P" (x) œ 2 "4 (x 4), P# (x) œ 2 4" (x 4)
" 64
(x 4)# , P$ (x) œ 2 4" (x 4)
" 64
(x 4)#
" 51#
(x 4)$
10. f(x) œ (1 x)"Î# , f w (x) œ "# (1 x)"Î# , f ww (x) œ 4" (1 x)$Î# , f www (x) œ 38 (1 x)&Î# ; f(0) œ (1)"Î# œ 1, f w (0) œ "# (1)"Î# œ "# , f ww (0) œ 4" (1)$Î# œ 4" , f www (0) œ 83 (1)&Î# œ 83 Ê P! (x) œ 1, P" (x) œ 1 2" x, P# (x) œ 1 2" x 8" x# , P$ (x) œ 1 2" x 8" x#
1 16
x$
11. f(x) œ ex , f w (x) œ ex , f ww (x) œ ex , f www (x) œ ex Ê á f ÐkÑ (x) œ a1bk ex ; f(0) œ ea0b œ ", f w (0) œ 1, _
f ww (0) œ 1, f www (0) œ 1, á ß f ÐkÑ (0) œ (1)k Ê ex œ 1 x 12 x# 16 x3 á œ !
nœ0
(1)n n n! x
12. f(x) œ x ex , f w (x) œ x ex ex , f ww (x) œ x ex 2ex , f www (x) œ x ex 3ex Ê á f ÐkÑ (x) œ x ex k ex ; f(0) œ a0bea0b œ 0, _
f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3, á ß f ÐkÑ (0) œ k Ê x x# 12 x3 á œ !
nœ0
1 n an 1 b ! x
13. f(x) œ (1 x)" Ê f w (x) œ (1 x)# , f ww (x) œ 2(1 x)$ , f www (x) œ 3!(1 x)% Ê á f ÐkÑ (x) œ (1)k k!(1 x)k1 ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3!, á ß f ÐkÑ (0) œ (1)k k! _
_
nœ0
nœ0
Ê 1 x x# x$ á œ ! (x)n œ ! (1)n xn 14. f(x) œ
2x 1x
Ê f w (x) œ
œ 6(1 x)$ , f www (x) œ 18(1 x)% Ê á f ÐkÑ (x) œ 3ak!b(1 x)
3 ww (1 x)# , f (x)
_
f w (0) œ 3, f ww (0) œ 6, f www (0) œ 18, á ß f ÐkÑ (0) œ 3ak!b Ê 2 3x 3x# 3x$ á œ 2 ! 3xn nœ1
_
15. sin x œ !
nœ0 _
16. sin x œ !
nœ0
_
(")n x2nb1 (#n1)!
Ê sin 3x œ !
(")n x2nb1 (#n1)!
Ê sin
nœ0
_
17. 7 cos (x) œ 7 cos x œ 7 !
nœ0
x #
_
œ!
nœ0
(")n x2n (2n)!
(")n (3x)2nb1 (#n1)!
2n 1
(")n ˆ #x ‰ (#n1)!
œ7
7x# #!
_
(")n 32nb1 x2nb1 (#n1)!
œ 3x
(")n x2nb1 #2n 1 (2n1)!
x #
œ!
nœ0
_
œ!
nœ0
7x% 4!
7x' 6!
œ
3$ x$ 3!
x$ 2$ †3!
3& x& 5!
x& 2& †5!
á
á
á , since the cosine is an even function
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
k 1
; f(0) œ 2,
Section 10.8 Taylor and Maclaurin Series _
18. cos x œ !
nœ0
19. cosh x œ _
œ!
nœ0
œ!
nœ0
ex ecx #
Ê 5 cos 1x œ 5 !
nœ0
œ
" #
’Š1 x#
œ
" #
’Š1 x
x# #!
x$ 3!
(1)n (1x)2n (#n)!
x% 4!
œ5
51 # x# 2!
51 % x% 4!
á ‹ Š1 x
x# #!
51 ' x' 6!
x$ 3!
á
x% 4!
á ‹“ œ 1
x# #!
x% 4!
x' 6!
á
x2n (2n)!
20. sinh x œ _
_
(1)n x2n (2n)!
619
ex ecx #
x# #!
x$ 3!
x% 4!
á ‹ Š1 x
x# #!
x$ 3!
x% 4!
á ‹“ œ x
x$ 3!
x& 5!
x' 6!
á
x2n 1 (2n 1)!
21. f(x) œ x% 2x$ 5x 4 Ê f w (x) œ 4x$ 6x# 5, f ww (x) œ 12x# 12x, f www (x) œ 24x 12, f Ð4Ñ (x) œ 24 Ê f ÐnÑ (x) œ 0 if n 5; f(0) œ 4, f w (0) œ 5, f ww (0) œ 0, f www (0) œ 12, f Ð4Ñ (0) œ 24, f ÐnÑ (0) œ 0 if n 5 24 % $ % $ Ê x% 2x$ 5x 4 œ 4 5x 12 3! x 4! x œ x 2x 5x 4 22. f(x) œ
x# x1
Ê f w (x) œ
2x x# ; f ww (x) ax 1 b 2
œ
2 ; ax 1 b 3
f www (x) œ
6 ax 1 b 4
Ê f ÐnÑ (x) œ
a1 b n n x ; ax 1bnb1
f(0) œ 0, f w (0) œ 0, f ww (0) œ 2,
_
f www (0) œ 6, f ÐnÑ (0) œ a1bn nx if n 2 Ê x# x3 x4 x5 Þ Þ Þ œ ! a1bn xn nœ2
23. f(x) œ x$ 2x 4 Ê f w (x) œ 3x# 2, f ww (x) œ 6x, f www (x) œ 6 Ê f ÐnÑ (x) œ 0 if n 4; f(2) œ 8, f w (2) œ 10, 6 # $ f ww (2) œ 12, f www (2) œ 6, f ÐnÑ (2) œ 0 if n 4 Ê x$ 2x 4 œ 8 10(x 2) 12 2! (x 2) 3! (x 2) œ 8 10(x 2) 6(x 2)# (x 2)$
24. f(x) œ 2x$ x# 3x 8 Ê f w (x) œ 6x# 2x 3, f ww (x) œ 12x 2, f www (x) œ 12 Ê f ÐnÑ (x) œ 0 if n 4; f(1) œ 2, f w (1) œ 11, f ww (1) œ 14, f www (1) œ 12, f ÐnÑ (1) œ 0 if n 4 Ê 2x$ x# 3x 8 12 # $ # $ œ 2 11(x 1) 14 2! (x 1) 3! (x 1) œ 2 11(x 1) 7(x 1) 2(x 1) 25. f(x) œ x% x# 1 Ê f w (x) œ 4x$ 2x, f ww (x) œ 12x# 2, f www (x) œ 24x, f Ð4Ñ (x) œ 24, f ÐnÑ (x) œ 0 if n 5; f(2) œ 21, f w (2) œ 36, f ww (2) œ 50, f www (2) œ 48, f Ð4Ñ (2) œ 24, f ÐnÑ (2) œ 0 if n 5 Ê x% x# 1 48 24 # $ % # $ % œ 21 36(x 2) 50 2! (x 2) 3! (x 2) 4! (x 2) œ 21 36(x 2) 25(x 2) 8(x 2) (x 2) 26. f(x) œ 3x& x% 2x$ x# 2 Ê f w (x) œ 15x% 4x$ 6x# 2x, f ww (x) œ 60x$ 12x# 12x 2, f www (x) œ 180x# 24x 12, f Ð4Ñ (x) œ 360x 24, f Ð5Ñ (x) œ 360, f ÐnÑ (x) œ 0 if n 6; f(1) œ 7, f w (1) œ 23, f ww (1) œ 82, f www (1) œ 216, f Ð4Ñ (1) œ 384, f Ð5Ñ (1) œ 360, f ÐnÑ (1) œ 0 if n 6 216 384 360 # $ % & Ê 3x& x% 2x$ x# 2 œ 7 23(x 1) 82 2! (x 1) 3! (x 1) 4! (x 1) 5! (x 1) œ 7 23(x 1) 41(x 1)# 36(x 1)$ 16(x 1)% 3(x 1)& 27. f(x) œ x# Ê f w (x) œ 2x$ , f ww (x) œ 3! x% , f www (x) œ 4! x& Ê f ÐnÑ (x) œ (1)n (n 1)! xn2 ; f(1) œ 1, f w (1) œ 2, f ww (1) œ 3!, f www (1) œ 4!, f ÐnÑ (1) œ (1)n (n 1)! Ê x"# _
œ 1 2(x 1) 3(x 1)# 4(x 1)$ á œ ! (1)n (n 1)(x 1)n nœ0
28. f(x) œ
1 a1 x b 3
Ê f w (x) œ 3(1 x)4 , f ww (x) œ 12(1 x)5 , f www (x) œ 60 (1 x)6 Ê f ÐnÑ (x) œ
fa0b œ 1, f w a0b œ 3, f ww a0b œ 12, f www a0b œ 60, á , f ÐnÑ a0b œ _
œ!
nœ0
an 2 b ! 2
Ê
1 a1 x b 3
an 2 b ! 2
(1 x)n3 ;
œ 1 3x 6x# 10x3 á
an 2ban 1b n x 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
620
Chapter 10 Infinite Sequences and Series
29. f(x) œ ex Ê f w (x) œ ex , f ww (x) œ ex Ê f ÐnÑ (x) œ ex ; f(2) œ e# , f w (2) œ e# , á f ÐnÑ (2) œ e# Ê ex œ e# e# (x 2)
e# #
(x 2)#
e$ 3!
_
(x 2)$ á œ !
nœ0
e# n!
(x 2)n
30. f(x) œ 2x Ê f w (x) œ 2x ln 2, f ww (x) œ 2x (ln 2)# , f www (x) œ 2x (ln 2)3 Ê f ÐnÑ (x) œ 2x (ln 2)n ; f(1) œ 2, f w (1) œ 2 ln 2, f ww (1) œ 2(ln 2)# , f www (1) œ 2(ln 2)$ , á , f ÐnÑ (1) œ 2(ln 2)n 2(ln 2)# #
Ê 2x œ 2 (2 ln 2)(x 1)
(x 1)#
2(ln 2)3 3!
_
(x 1)3 á œ !
nœ0
2(ln 2)n (x1)n n!
31. f(x) œ cosˆ2x 12 ‰, f w (x) œ 2 sinˆ2x 12 ‰, f ww (x) œ 4 cosˆ2x 12 ‰, f www (x) œ 8 sinˆ2x 12 ‰, f a4b axb œ 24 cosˆ2x 12 ‰ß f a5b axb œ 25 sinˆ2x 12 ‰ß . . ; fˆ 14 ‰ œ 1, f w ˆ 14 ‰ œ 0, f ww ˆ 14 ‰ œ 4, f www ˆ 14 ‰ œ 0, f a4b ˆ 14 ‰ œ 24 , 2 4 f a5b ˆ 14 ‰ œ 0, . . ., f Ð2nÑ ˆ 14 ‰ œ a1bn 22n Ê cosˆ2x 12 ‰ œ 1 2ˆx 14 ‰ 23 ˆx 14 ‰ . . . _
œ!
nœ0
a1bn 22n ˆ x a2nbx
2n 14 ‰
7 Î2 32. f(x) œ Èx 1, f w (x) œ 12 ax 1b1Î2 , f ww (x) œ 14 ax 1b3Î2 , f www (x) œ 38 ax 1b5Î2 , f a4b (x) œ 15 , . . .; 16 ax 1b 1 1 3 15 1 1 1 5 f(0) œ 1, f w (0) œ , f ww (0) œ , f www (0) œ , f a4b (0) œ , . . . Ê Èx 1 œ 1 x x2 x3 x4 Þ Þ Þ 2
4
8
16
_
a1bn 2n a2nbx x
33. The Maclaurin series generated by cos x is ! nœ0
by _
! nœ0
2 1x
2
8
16
which converges on a_, _b and the Maclaurin series generated
_
is 2 ! xn which converges on a1, 1b. Thus the Maclaurin series generated by faxb œ cos x nœ0
a1bn 2n a2nbx x
128
2 1x
is given by
_
2 ! xn œ 1 2x 25 x2 Þ Þ Þ Þ which converges on the intersection of a_, _b and a1, 1b, so the nœ0
interval of convergence is a1, 1b. _
34. The Maclaurin series generated by ex is ! nœ0
xn nx
which converges on a_, _b. The Maclaurin series generated by _
faxb œ a1 x x2 bex is given by a1 x x2 b !
nœ0
_
35. The Maclaurin series generated by sin x is ! nœ0 _
generated by lna1 xb is !
nœ1
a1bnc1 n x n
xn nx
œ 1 12 x2 23 x3 Þ Þ Þ Þ which converges on a_, _bÞ
a 1 b n 2n1 a2n 1bx x
which converges on a_, _b and the Maclaurin series
which converges on a1, 1b. Thus the Maclaurin series genereated by
_
faxb œ sin x † lna1 xb is given by Œ !
nœ0
_
a 1 b n a1bnc1 n 2n1 Œ ! n x a2n 1bx x nœ1
œ x2 21 x3 61 x4 Þ Þ Þ Þ which converges on
the intersection of a_, _b and a1, 1b, so the interval of convergence is a1, 1b. _
36. The Maclaurin series generated by sin x is ! nœ0
a 1 b n 2n1 a2n 1bx x
_
genereated by faxb œ x sin2 x is given by xŒ !
nœ0
œ x3 13 x5 _
37. If ex œ !
nœ0
f ÐnÑ (a) n!
2 7 45 x
which converges on a_, _b. The Maclaurin series
2 a 1 b n 2n1 a2n 1bx x
_
œ xŒ !
nœ0
_
a 1 b n a 1 b n 2n1 2n1 Œ ! a2n 1bx x a2n 1bx x nœ0
. . . which converges on a_, _bÞ
(x a)n and f(x) œ ex , we have f ÐnÑ (a) œ ea f or all n œ 0, 1, 2, 3, á !
Ê ex œ ea ’ (x 0!a)
(x a)" 1!
(x a)# 2!
á “ œ ea ’1 (x a)
(x a)# 2!
á “ at x œ a
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.9 Convergence of Taylor Series
621
38. f(x) œ ex Ê f ÐnÑ (x) œ ex for all n Ê f ÐnÑ (1) œ e for all n œ 0, 1, 2, á Ê ex œ e e(x 1)
e #!
(x 1)#
e 3!
(x 1)$ á œ e ’1 (x 1)
f ww (a) f www (a) # $ w # (x a) 3! (x a) á Ê f (x) www œ f w (a) f ww (a)(x a) f 3!(a) 3(x a)# á Ê f ww (x) œ f ww (a) f www (a)(x Ðn 2Ñ Ê f ÐnÑ (x) œ f ÐnÑ (a) f Ðn1Ñ (a)(x a) f # (a) (x a)# á w w Ðn Ñ Ðn Ñ
(x 1)# 2!
39. f(x) œ f(a) f w (a)(x a)
Ê f(a) œ f(a) 0, f (a) œ f (a) 0, á , f
(a) œ f
a)
(x 1)$ 3!
f Ð4Ñ (a) 4!
á“
4 † 3(x a)# á
(a) 0
40. E(x) œ f(x) b! b" (x a) b# (x a)# b$ (x a)$ á bn (x a)n Ê 0 œ E(a) œ f(a) b! Ê b! œ f(a); from condition (b), lim
xÄa
Ê Ê
f(x) f(a) b" (x a) b# (x a)# b$ (x a)$ á bn (x a)n (x a)n
œ0
w a)# á nbn (x a)n 1 lim f (x) b" 2b# (x a) n(x3b$ (xa) œ0 n 1 xÄa f ww (x) 2b# 3! b$ (x a) á n(n ")bn (x a)n w b" œ f (a) Ê xlim n(n 1)(x a)n 2 Äa " #
f ww (a) Ê xlim Äa " www œ b$ œ 3! f (a) Ê xlim Äa Ê b# œ
g(x) œ f(a) f w (a)(x a)
f www (x) 3! b$ á n(n 1)(n 2)bn (x a)n n(n 1)(n #)(x a)n
f
ÐnÑ
(x) n! bn n!
f ww (a) 2!
œ 0 Ê bn œ
(x a)# á
3
3
" n!
f ÐnÑ (a) n!
2
œ0
œ0
f ÐnÑ (a); therefore, (x a)n œ Pn (x) #
41. f(x) œ ln (cos x) Ê f w (x) œ tan x and f ww (x) œ sec# x; f(0) œ 0, f w (0) œ 0, f ww (0) œ 1 Ê L(x) œ 0 and Q(x) œ x2 42. f(x) œ esin x Ê f w (x) œ (cos x)esin x and f ww (x) œ ( sin x)esin x (cos x)# esin x ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 1 Ê L(x) œ 1 x and Q(x) œ 1 x 43. f(x) œ a1 x# b
"Î#
x# #
Ê f w (x) œ x a1 x# b
f ww (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1
$Î#
and f ww (x) œ a1 x# b
$Î#
3x# a1 x# b
&Î#
; f(0) œ 1, f w (0) œ 0,
x# #
44. f(x) œ cosh x Ê f w (x) œ sinh x and f ww (x) œ cosh x; f(0) œ 1, f w (0) œ 0, f ww (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1 45. f(x) œ sin x Ê f w (x) œ cos x and f ww (x) œ sin x; f(0) œ 0, f w (0) œ 1, f ww (0) œ 0 Ê L(x) œ x and Q(x) œ x 46. f(x) œ tan x Ê f w (x) œ sec# x and f ww (x) œ 2 sec# x tan x; f(0) œ 0, f w (0) œ 1, f ww œ 0 Ê L(x) œ x and Q(x) œ x 10.9 CONVERGENCE OF TAYLOR SERIES _
1. ex œ 1 x
x# #!
á œ !
2. ex œ 1 x
x# #!
á œ !
nœ0 _
nœ0
xn n!
Ê e5x œ 1 (5x)
(5x)# #!
á œ 1 5x
xn n!
Ê exÎ2 œ 1 ˆ #x ‰
ˆ #x ‰# #!
á œ1
_
3. sin x œ x
x$ 3!
x& 5!
á œ!
4. sin x œ x
x$ 3!
x& 5!
á œ!
nœ0 _
nœ0
(1)n x2n 1 (#n1)!
Ê 5 sin (x) œ 5 ’(x)
(1)n x2n 1 (#n1)!
Ê sin
1x #
œ
1x #
ˆ 1#x ‰$ 3!
(x)$ 3!
ˆ 1#x ‰& 5!
x #
x# 2# #!
(x)& 5!
ˆ 1#x ‰( 7!
5# x# #!
_
5$ x$ 3!
x$ 2$ 3!
á œ!
nœ0
_
á œ !
nœ0
_
(1)n xn 2n n!
x á “ œ ! 5((1) #n1)!
n 1 2n 1
nœ0
_
1 x á œ ! (21) 2n 1 (#n1)! nœ0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
(1)n 5n xn n!
n 2n 1 2n 1
x# #
622
Chapter 10 Infinite Sequences and Series _
5. cos x œ !
nœ0
_
6. cos x œ !
nœ0
x$ 2†2!
œ1
Ê cos 5x2 œ !
a1bn x2n (2n)!
$Î# cos Š xÈ ‹ 2
x' 2# †4!
_
7. lna1 xb œ !
nœ1
_
nœ0
10.
Ê
x* 2$ †6!
1 1x
œ ! xn Ê nœ0
_
nœ0
n œ0
_
nœ0 _
(1)n x2n (2n)!
13. cos x œ !
nœ0
œ
x% 4!
x' 6!
_
(1)n x2nb1 (2n1)!
14. sin x œ !
nœ0
œ Šx
x$ 3!
_
nœ0 _
16. cos x œ !
nœ0
œ1
" #
œ!
nœ1
nœ0
nœ0
_
xnb1 n!
œ!
nœ0
nœ0
x"! 10!
x# #
1 cos x œ
x( 7!
_
x# #
a1bn 32nb1 x8nb4 n
nœ0
x$ #!
œ x x#
_
(1)n x2nb1 (#n1)!
nœ0
_
(1)n x2n (#n)!
nœ0
x& 4!
(1)n x2nb3 (2n1)!
œ!
1!
14 x 18 x2
x% 3!
x# #
œ
x8 4
œ 3x4 9x12
n
" #
x6 3
9 6 16 x
...
243 20 5 x
27 9 64 x
1 3 16 x
...
x( 5!
x* 7!
2187 28 7 x
...
á
œ x$
x& 3!
11
x# 2
x% 4!
á
x' 6!
x) 8!
x"! 10!
á
nœ2
x* 9!
x"" 11!
x$ 3!
_
œ Œ!
nœ0
á‹ x _
(1)n x2n (2n)!
Ê x# cos ax# b œ x# !
nœ0
nœ0
_
" #
(2x)% 2†4!
(2x)' 2†6!
" #
" #
_
œ!
nœ1
" #
! (1) (2x) œ (2n)!
œ
n
2n
nœ0
(2x)) 2†8!
(1)n x2nb1 (#n1)! x$ 3!
(1)n (1x)2n (#n)!
_
cos 2x #
(1)nb1 (2x)2n #†(2n)!
_
œ!
x4 2
n 2n
Ê x cos 1x œ x !
á
x á œ ! ((1) #n)!
Ê sin x x
2x ‰ 18. sin# x œ ˆ 1cos œ # _
_
(1)n x2n (2n)!
(2x)# 2†2!
x& 5!
15. cos x œ !
17. cos# x œ
x) 8!
_
Ê x# sin x œ x# Œ !
Ê
nœ1
œ x2
nœ0
_
(1)n x2nb1 (2n1)!
15625x12 6!
(1)n x3n 2n (2n)!
n œ0
a1bnc1 x2n n
œ!
n n 1 œ #" ! ˆ #" x‰ œ ! ˆ #" ‰ xn œ
xn n!
Ê xex œ x Œ !
12. sin x œ !
_
n
_
n
nœ0
_
xn n!
11. ex œ !
_
" 1 # 1 "# x
œ
1 2x
_
œ!
œ ! a1bn ˆ 34 x3 ‰ œ ! a1bn ˆ 34 ‰ x3n œ 1 34 x3
1 1 34 x3
nœ0
(#n)!
nœ0
2nb1 a1bn ˆ3x4 ‰ 2n 1
nœ0
_
_
nœ1
_
œ ! a1bn xn Ê
œ!
625x8 4!
2n
"Î#
$
a1bn ŒŠ x# ‹
_
a1bnc1 ˆx2 ‰ n
Ê lna1 x2 b œ !
Ê tan1 a3x4 b œ !
1 1x
"Î#
25x4 #!
œ1
á _
a1bnc1 xn n
(1)n 52n x4n (2n)!
œ!
nœ0
$ cos ŒŠ x# ‹
œ
_
2n
(1)n 5x2 ‘ (2n)!
nœ0
a1bn x2nb1 2n 1
8. tan1 x œ !
9.
_
(1)n x2n (2n)!
cos 2x œ
_
nœ1
" #
_
nœ0
2n
"# Š1
x( 7!
_
nœ0
(2x)# 2!
(1)n (2x)2n 2†(2n)!
(2x)# #!
x* 9!
x$ 3!
(")n 12n x2nb1 (#n)!
œ!
"# ’1
á œ1!
œ!
(1)n ax# b (#n)!
" #
x& 5!
œ
x
(")n x4n (#n)!
(2x)% 4!
x"" 11!
œx
2
(2x)' 6!
_
nœ1
(2x)' 6!
nœ2
1 # x$ 2!
œ x#
œ1!
(2x)% 4!
_
á œ!
x' 2!
(2x)) 8!
(1)n x2n 1 (2n1)!
1 % x& 4!
1 ' x( 6!
x"! 4!
x"% 6!
á
á
á“
(1)n 22n 1 x2n (2n)!
á‹ œ
(2x)# 2†2!
(2x)% 2†4!
(2x)' 2†6!
(1)n 22n 1 x2n (2n)!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
á
...
Section 10.9 Convergence of Taylor Series 19.
x# 12x
_
nœ1
22.
" 1x
_
nœ0
nœ0
(1)nc1 (2x)n n
20. x ln (1 2x) œ x !
21.
_
œ x# ˆ 1"2x ‰ œ x# ! (2x)n œ ! 2n xn2 œ x# 2x$ 2# x% 2$ x& á _
(1)nc1 2n xn n
œ!
nœ1
_
œ ! xn œ 1 x x# x$ á Ê
2 a1 x b $
d dx
nœ0
œ
d# dx#
ˆ 1" x ‰ œ
d dx
Š (1"x)# ‹ œ
d dx
1
œ 2x#
ˆ 1" x ‰ œ
2# x$ #
" (1x)#
2$ x% 4
2% x& 5
á _
_
nœ1
n œ0
œ 1 2x 3x# á œ ! nxn1 œ ! (n 1)xn _
a1 2x 3x# á b œ 2 6x 12x# á œ ! n(n 1)xn2 nœ2
_
œ ! (n 2)(n 1)xn nœ0
3
5
7
23. tan1 x œ x 13 x3 15 x5 17 x7 Þ Þ Þ Ê x tan1 x2 œ xŠx2 13 ax2 b 15 ax2 b 17 ax2 b Þ Þ Þ ‹ _
œ x3 13 x7 15 x11 17 x15 Þ Þ Þ œ !
nœ1
x3 3!
24. sin x œ x œx
4 x3 3!
16 x5 5!
x2 2!
25. ex œ 1 x œ Š1 x 26. sin x œ x œ Š1 _
2
x 2!
x5 5!
x2 2!
x3 3!
4
1) x œ ! Š ((2n)! nœ0
x3 3!
x 4!
n 2n
x3 3!
x5 5!
x7 7!
64 x7 7!
a1bn x4nc1 2n 1
á Ê sin x † cos x œ "# sin 2x œ "# Š2x á œx
á and
1 1x
2 x3 3
2x5 15
4 x7 315
_
á œ!
nœ0
a2xb3 3!
x7 7!
œ 1 x x2 x3 á Ê ex
6
x 6!
á and cos x œ 1
á ‹ Šx
3
x 3!
x2 2!
x4 4!
x6 6!
a2xb7 7!
á‹
1 1x 25 4 24 x
_
á œ ! ˆ n!1 a1bn ‰xn nœ0
á Ê cos x sin x
5
x 3
lna1 x2 b œ x3 Šx2 12 ax2 b 13 ax2 b 14 ax2 b á ‹
x 5!
7
(1)n 22n x2nb1 (#n1)!
á ‹ a1 x x2 x3 á b œ 2 32 x2 56 x3
a2xb5 5!
x 7!
á‹ œ 1 x
x2 2!
x3 3!
x4 4!
x5 5!
x6 6!
x7 7!
á
(1)n x2nb1 (#n1)! ‹
27. lna1 xb œ x 12 x2 13 x3 14 x4 á Ê œ 13 x3 16 x5 19 x7
1 9 12 x
_
2
3
4
nc1
á œ ! a13nb x2n1 nœ1
28. lna1 xb œ x 12 x2 13 x3 14 x4 á and lna1 xb œ x 12 x2 13 x3 14 x4 á Ê lna1 xb lna1 xb _
œ ˆx 12 x2 13 x3 14 x4 á ‰ ˆx 12 x2 13 x3 14 x4 á ‰ œ 2x 23 x3 25 x5 á œ ! 2n 2 1 x2n1 nœ0
29. ex œ 1 x œ Š1 x
x2 2! x2 2!
x3 3! x3 3!
á and sin x œ x á ‹Šx
x3 3!
x5 5!
x3 3!
x5 5!
x7 7!
á ‹ œ x x2 13 x3
x7 7!
á Ê ex † sin x 1 5 30 x
ÞÞÞÞ
30. lna1 xb œ x 12 x2 13 x3 14 x4 á and 1 " x œ 1 x x# x$ á Ê ln1a1xxb œ lna1 xb † 7 4 œ ˆx 12 x2 13 x3 14 x4 á ‰a1 x x# x$ á b œ x 12 x2 56 x3 12 x ÞÞÞÞ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" 1x
623
624
Chapter 10 Infinite Sequences and Series 2
31. tan1 x œ x 13 x3 15 x5 17 x7 Þ Þ Þ Ê atan1 xb œ atan1 xbatan1 xb 44 8 6 œ ˆx 13 x3 15 x5 17 x7 Þ Þ Þ‰ˆx 13 x3 15 x5 17 x7 Þ Þ Þ‰ œ x2 23 x4 23 45 x 105 x Þ Þ Þ Þ 32. sin x œ x
x3 3!
x5 5!
x7 7!
á and cos x œ 1
œ cos x † "# sin 2x œ "# Š1 33. sin x œ x
x3 3!
x5 5!
x7 7!
Ê esin x œ 1 Šx
3
x 3!
x2 2!
x4 4!
x6 6!
x2 2!
x 5!
7
x 7!
x2 2!
á ‹ 12 Šx
x6 6!
a2xb 3!
á ‹Š2x
á and ex œ 1 x 5
x4 4!
3
á Ê cos2 x † sin x œ cos x † cos x † sin x
x3 3!
á
3
x5 5!
x 3!
a2xb5 5!
x7 7!
a2xb7 7!
á ‹ œ x 76 x3
2
á ‹ 16 Šx
x3 3!
x5 5!
x7 7!
61 5 120 x
1247 7 5040 x
ÞÞÞ
3
á‹ á
œ 1 x 12 x2 18 x4 Þ Þ Þ Þ x3 x5 x7 1 3 1 5 1 7 1 3 1 5 1 1 ˆ 3! 5! 7! á and tan x œ x 3 x 5 x 7 x Þ Þ Þ Ê sinatan xb œ x 3 x 5 x 3 5 1 ˆ 1 ˆ 16 ˆx 13 x3 15 x5 17 x7 Þ Þ Þ‰ 120 x 13 x3 15 x5 17 x7 Þ Þ Þ‰ 5040 x 13 x3 15 x5 17 x7 5 7 x 12 x3 38 x5 16 x ÞÞÞ
34. sin x œ x œ
17 x7 Þ Þ Þ‰ 7
Þ Þ Þ‰ á
35. Since n œ 3, then f a4b axb œ sin x, lf a4b axbl Ÿ M on Ò0, 0.1Ó Ê lsin xl Ÿ 1 on Ò0, 0.1Ó Ê M œ 1. Then lR3 a0.1bl Ÿ 1 l0.14x 0l
4
œ 4.2 ‚ 106 Ê error Ÿ 4.2 ‚ 106
36. Since n œ 4, then f a5b axb œ ex , lf a5b axbl Ÿ M on Ò0, 0.5Ó Ê lex l Ÿ Èe on Ò0, 0.5Ó Ê M œ 2.7. Then lR4 a0.5bl Ÿ 2.7 l0.55x 0l œ 7.03 ‚ 104 Ê error Ÿ 7.03 ‚ 104 5
kx k & 5!
37. By the Alternating Series Estimation Theorem, the error is less than 5 Ê kxk È 6 ‚ 10# ¸ 0.56968 38. If cos x œ 1
Ê kxk& a5!b a5 ‚ 10% b Ê kxk& 600 ‚ 10%
%
x# #
and kxk 0.5, then the error is less than ¹ (.5) 24 ¹ œ 0.0026, by Alternating Series Estimation Theorem;
since the next term in the series is positive, the approximation 1
x# #
is too small, by the Alternating Series Estimation
Theorem 39. If sin x œ x and kxk 10$ , then the error is less than
a10c$ b 3!
$
¸ 1.67 ‚ 1010 , by Alternating Series Estimation Theorem; $
The Alternating Series Estimation Theorem says R# (x) has the same sign as x3! . Moreover, x sin x Ê 0 sin x x œ R# (x) Ê x 0 Ê 10$ x 0.
40. È1 x œ 1
x #
x# 8
x$ 16
#
á . By the Alternating Series Estimation Theorem the kerrork ¹ 8x ¹
œ 1.25 ‚ 10& c $
3Ð0Þ1Ñ (0.1)$ 3!
c $
(0.1)$ 3!
41. kR# (x)k œ ¹ e3!x ¹ 42. kR# (x)k œ ¹ e3!x ¹
2x ‰ 43. sin# x œ ˆ 1 cos œ #
Ê
d dx
asin# xb œ
œ 2x
(2x)$ 3!
d dx
(2x)& 5!
" #
1.87 ‚ 104 , where c is between 0 and x
œ 1.67 ‚ 10% , where c is between 0 and x #
" #
Š 2x 2!
(2x)( 7!
cos 2x œ 2$ x% 4!
2& x' 6!
" #
"# Š1
(2x)# 2!
á ‹ œ 2x
(2x)% 4!
(2x)$ 3!
(2x)' 6!
(2x)& 5!
á‹ œ (2x)( 7!
2x# #!
2$ x% 4!
2& x' 6!
á
á Ê 2 sin x cos x
á œ sin 2x, which checks
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
(0.01)# 8
Section 10.9 Convergence of Taylor Series 44. cos# x œ cos 2x sin# x œ Š1 œ1
#
2x #!
$ %
2 x 4!
& '
2 x 6!
(2x)# #!
(2x)% 4!
(2x)' 6!
á œ 1 x# "3 x%
2 45
(2x)) 8!
x'
#
á ‹ Š 2x #!
" 315
2$ x% 4!
2& x' 6!
2( x) 8!
625
á‹
x) á
45. A special case of Taylor's Theorem is f(b) œ f(a) f w (c)(b a), where c is between a and b Ê f(b) f(a) œ f w (c)(b a), the Mean Value Theorem. 46. If f(x) is twice differentiable and at x œ a there is a point of inflection, then f ww (a) œ 0. Therefore, L(x) œ Q(x) œ f(a) f w (a)(x a). 47. (a) f ww Ÿ 0, f w (a) œ 0 and x œ a interior to the interval I Ê f(x) f(a) œ Ê f(x) Ÿ f(a) throughout I Ê f has a local maximum at x œ a (b) similar reasoning gives f(x) f(a) œ local minimum at x œ a
f ww (c# ) #
f ww (c# ) #
(x a)# Ÿ 0 throughout I
(x a)# 0 throughout I Ê f(x) f(a) throughout I Ê f has a
48. f(x) œ (1 x)" Ê f w (x) œ (1 x)# Ê f ww (x) œ 2(1 x)$ Ê f Ð3Ñ (x) œ 6(1 x)% Ê f Ð4Ñ (x) œ 24(1 x)& ; therefore
" 1 x
¸ 1 x x# x$ . kxk 0.1 Ê
&
%
Ð4Ñ
10 11
" 1 x
10 9
‰ Ê ¹ (1"x)& ¹ ˆ 10 9
&
%
‰ Ê the error e$ Ÿ ¹ max f 4! (x) x ¹ (0.1)% ˆ 10 ‰ œ 0.00016935 0.00017, since ¹ f Ê ¹ (1x x)& ¹ x% ˆ 10 9 9
Ð4Ñ
&
(x) 4! ¹
œ ¹ (1"x)& ¹ .
49. (a) f(x) œ (1 x)k Ê f w (x) œ k(1 x)k1 Ê f ww (x) œ k(k 1)(1 x)k2 ; f(0) œ 1, f w (0) œ k, and f ww (0) œ k(k 1) Ê Q(x) œ 1 kx k(k # ") x# " (b) kR# (x)k œ ¸ 3†3!2†" x$ ¸ 100 Ê kx$ k
" 100
Ê 0x
" 100"Î$
or 0 x .21544
50. (a) Let P œ x 1 Ê kxk œ kP 1k .5 ‚ 10n since P approximates 1 accurate to n decimals. Then, P sin P œ (1 x) sin (1 x) œ (1 x) sin x œ 1 (x sin x) Ê k(P sin P) 1k œ ksin x xk Ÿ
kx k $ 3!
0.125 3!
‚ 103n .5 ‚ 103n Ê P sin P gives an approximation to 1 correct to 3n decimals.
_
_
nœ0
nœk
51. If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n 1)(n 2)â(n k 1)an xnk and f ÐkÑ (0) œ k! ak Ê ak œ
f ÐkÑ (0) k!
for k a nonnegative integer. Therefore, the coefficients of f(x) are identical with the corresponding
coefficients in the Maclaurin series of f(x) and the statement follows. 52. Note: f even Ê f(x) œ f(x) Ê f w (x) œ f w (x) Ê f w (x) œ f w (x) Ê f w odd; f odd Ê f(x) œ f(x) Ê f w (x) œ f w (x) Ê f w (x) œ f w (x) Ê f w even; also, f odd Ê f(0) œ f(0) Ê 2f(0) œ 0 Ê f(0) œ 0 (a) If f(x) is even, then any odd-order derivative is odd and equal to 0 at x œ 0. Therefore, a" œ a$ œ a& œ á œ 0; that is, the Maclaurin series for f contains only even powers. (b) If f(x) is odd, then any even-order derivative is odd and equal to 0 at x œ 0. Therefore, a! œ a# œ a% œ á œ 0; that is, the Maclaurin series for f contains only odd powers. 53-58. Example CAS commands: Maple: f := x -> 1/sqrt(1+x); x0 := -3/4; x1 := 3/4; # Step 1: plot( f(x), x=x0..x1, title="Step 1: #53 (Section 10.9)" );
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
626
Chapter 10 Infinite Sequences and Series
# Step 2: P1 := unapply( TaylorApproximation(f(x), x = 0, order=1), x ); P2 := unapply( TaylorApproximation(f(x), x = 0, order=2), x ); P3 := unapply( TaylorApproximation(f(x), x = 0, order=3), x ); # Step 3: D2f := D(D(f)); D3f := D(D(D(f))); D4f := D(D(D(D(f)))); plot( [D2f(x),D3f(x),D4f(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 3: #57 (Section 9.9)" ); c1 := x0; M1 := abs( D2f(c1) ); c2 := x0; M2 := abs( D3f(c2) ); c3 := x0; M3 := abs( D4f(c3) ); # Step 4: R1 := unapply( abs(M1/2!*(x-0)^2), x ); R2 := unapply( abs(M2/3!*(x-0)^3), x ); R3 := unapply( abs(M3/4!*(x-0)^4), x ); plot( [R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 4: #53 (Section 10.9)" ); # Step 5: E1 := unapply( abs(f(x)-P1(x)), x ); E2 := unapply( abs(f(x)-P2(x)), x ); E3 := unapply( abs(f(x)-P3(x)), x ); plot( [E1(x),E2(x),E3(x),R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], linestyle=[1,1,1,3,3,3], title="Step 5: #53 (Section 10.9)" ); # Step 6: TaylorApproximation( f(x), view=[x0..x1,DEFAULT], x=0, output=animation, order=1..3 ); L1 := fsolve( abs(f(x)-P1(x))=0.01, x=x0/2 ); # (a) R1 := fsolve( abs(f(x)-P1(x))=0.01, x=x1/2 ); L2 := fsolve( abs(f(x)-P2(x))=0.01, x=x0/2 ); R2 := fsolve( abs(f(x)-P2(x))=0.01, x=x1/2 ); L3 := fsolve( abs(f(x)-P3(x))=0.01, x=x0/2 ); R3 := fsolve( abs(f(x)-P3(x))=0.01, x=x1/2 ); plot( [E1(x),E2(x),E3(x),0.01], x=min(L1,L2,L3)..max(R1,R2,R3), thickness=[0,2,4,0], linestyle=[0,0,0,2], color=[red,blue,green,black], view=[DEFAULT,0..0.01], title="#53(a) (Section 10.9)" ); abs(`f(x)`-`P`[1](x) ) <= evalf( E1(x0) ); # (b) abs(`f(x)`-`P`[2](x) ) <= evalf( E2(x0) ); abs(`f(x)`-`P`[3](x) ) <= evalf( E3(x0) ); Mathematica: (assigned function and values for a, b, c, and n may vary) Clear[x, f, c] f[x_]= (1 x)3/2 {a, b}= {1/2, 2}; pf=Plot[ f[x], {x, a, b}]; poly1[x_]=Series[f[x], {x,0,1}]//Normal poly2[x_]=Series[f[x], {x,0,2}]//Normal poly3[x_]=Series[f[x], {x,0,3}]//Normal Plot[{f[x], poly1[x], poly2[x], poly3[x]}, {x, a, b}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,1,0], RGBColor[0,0,1], RGBColor[0,.5,.5]}];
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.10 The Binomial Series
627
The above defines the approximations. The following analyzes the derivatives to determine their maximum values. f''[c] Plot[f''[x], {x, a, b}]; f'''[c] Plot[f'''[x], {x, a, b}]; f''''[c] Plot[f''''[x], {x, a, b}]; Noting the upper bound for each of the above derivatives occurs at x = a, the upper bounds m1, m2, and m3 can be defined and bounds for remainders viewed as functions of x. m1=f''[a] m2=-f'''[a] m3=f''''[a] r1[x_]=m1 x2 /2! Plot[r1[x], {x, a, b}]; r2[x_]=m2 x3 /3! Plot[r2[x], {x, a, b}]; r3[x_]=m3 x4 /4! Plot[r3[x], {x, a, b}]; A three dimensional look at the error functions, allowing both c and x to vary can also be viewed. Recall that c must be a value between 0 and x, so some points on the surfaces where c is not in that interval are meaningless. Plot3D[f''[c] x2 /2!, {x, a, b}, {c, a, b}, PlotRange Ä All] Plot3D[f'''[c] x3 /3!, {x, a, b}, {c, a, b}, PlotRange Ä All] Plot3D[f''''[c] x4 /4!, {x, a, b}, {c, a, b}, PlotRange Ä All] 10.10 THE BINOMIAL SERIES 1. (1 x)"Î# œ 1 "# x
ˆ "# ‰ ˆ "# ‰ x#
2. (1 x)"Î$ œ 1 "3 x
ˆ "3 ‰ ˆ 32 ‰ x#
"Î#
8. a1 x# b
"Î$
œ 1 "# x$ œ 1 3" x#
"Î# 9. ˆ1 1x ‰ œ 1 "# ˆ 1x ‰
" 16
x$ á
ˆ 3" ‰ ˆ 32 ‰ ˆ 35 ‰ x$
á œ 1 3" x 9" x#
5 81
x$ á
3!
#!
ˆ "# ‰ ˆ "# ‰ (2x)# #!
(2)(3) ˆ x# ‰
4 6. ˆ1 x3 ‰ œ 1 4 ˆ x3 ‰
á œ 1 "# x 8" x#
ˆ "# ‰ ˆ 3# ‰ (x)#
4. (1 2x)"Î# œ 1 "# (2x) # 5. ˆ1 x# ‰ œ 1 # ˆ x# ‰
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ x$ 3!
#!
3. (1 x)"Î# œ 1 "# (x)
7. a1 x$ b
#!
#
#!
(4)(3) ˆ x3 ‰ #!
#
Š "# ‹ Š "# ‹ Š 3# ‹ (2x)$ 3! $
3!
(4)(3)(2) ˆ x3 ‰
#!
ˆ "3 ‰ ˆ 34 ‰ ax# b# #!
#!
3!
(2)(3)(4) ˆ x# ‰
ˆ "# ‰ ˆ 3# ‰ ax$ b#
ˆ "# ‰ ˆ "# ‰ ˆ 1x ‰#
ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ (x)$
3!
$
x$ á
5 16
á œ 1 x 12 x# 12 x$ á
á œ 1 x 34 x# "# x$ (4)(3)(2)(1) ˆ x3 ‰
3!
ˆ 3" ‰ ˆ 34 ‰ ˆ 37 ‰ ax# b$ 3!
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ ˆ 1x ‰$
4
4!
ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ ax$ b$
3!
á œ 1 "# x 83 x#
0 á œ 1 43 x 23 x2
á œ 1 "# x$ 38 x' á œ 1 3" x# 29 x%
á œ1
" #x
1 8x#
" 16x$
x* á
5 16
14 81
x' á
á
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
4 27
x3
1 4 81 x
628 10.
Chapter 10 Infinite Sequences and Series x 3 È 1x
(4)(3)x# #!
11. (1 x)% œ 1 4x $
12. a1 x# b œ 1 3x#
(4)(3)(2)x$ 3!
(3)(2) ax# b #!
#
(3)(2)(1) ax# b 3!
(3)(2)(2x)# #!
% 14. ˆ1 x# ‰ œ 1 4 ˆ x# ‰
(4)(3) ˆ x# ‰
#
'00 2 sin x# dx œ '00 2 Šx# x3! x5! Þ
Þ
kE k Ÿ
(
Þ
x
x# 4
'00 1 È "
x$ 18
Š1 x
á“
kE k Ÿ
(0.1)& 10
œ 0.000001
'!!Þ#&
$È
1 x# dx œ '0
0Þ25
&
(0.25) 45
œ 1 3x# 3x% x' œ 1 6x 12x# 8x$
(4)(3)(2) ˆ x# ‰ 3!
$
(4)(3)(2)(1) ˆ x# ‰
$
á ‹ dx œ ’ x3
%
4!
x( 7†3!
!Þ#
á“
œ 1 2x 32 x# "# x$ $
¸ ’ x3 “
!
!Þ# !
" 16
x%
¸ 0.00267 with error
x# #!
x$ 3!
x% 4!
á 1‹ dx œ '0 Š1 0Þ2
¸ 0.19044 with error kEk Ÿ
(0.2)% 96
x% 2
á“
3x) 8
x# 3
Š1
á ‹ dx œ ’x
x% 9
x& 10
á ‹ dx œ ’x
x$ 9
x# 6
x #
x$ 24
á ‹ dx
¸ 0.00002
!Þ"
¸ [x]!Þ" ! ¸ 0.1 with error
!
x& 45
á“
!Þ#& !
¸ ’x
!Þ#& x$ 9 “!
¸ 0.25174 with error
'00 1 sinx x dx œ '00 1 Š1 x3! x5! x7! á ‹ dx œ ’x 3x†3! 5x†5! 7x†7! á “ !Þ" ¸ ’x 3x†3! 5x†5! “ !Þ" Þ
#
%
'
$
&
(
$
&
!
(0.1)7 7†7!
Þ
%
(0.1)9 216
¸ 0.0996676643, kEk Ÿ 21. a1 x% b Ê
"Î#
œ (1)"Î#
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ 4!
'0
0Þ1
Š1
x% #
x) 8
Š "# ‹ 1
'
)
ˆ "# ‰ ˆ "# ‰
(1)"Î# ax% b
#!
%
x"# 16
$
x& 10
x( 42
á“
!Þ" !
¸ ’x
x$ 3
x& 10
¸ 4.6 ‚ 1012
(1)(Î# ax% b á œ 1
!
¸ 2.8 ‚ 1012
'00 1 exp ax# b dx œ '00 1 Š1 x# x2! x3! x4! á ‹ dx œ ’x x3 Þ
5x"' 128
x% #
#
(1)$Î# ax% b
á ‹ dx ¸ ’x
x) 8
x"# 16 !Þ"
x& 10 “ !
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ 3!
5x"' 128
(1)&Î# ax% b
$
á
¸ 0.100001, kEk Ÿ
(0.1)9 72
¸ 1.39 ‚ 1011
" x '01 ˆ 1 xcos x ‰ dx œ '01 Š "# x4! x6! x8! 10! á ‹ dx ¸ ’ x# 3x†4! 5x†6! 7x†8! 9†x10! “ #
%
'
)
$
&
(
#
¸ 0.4863853764, kEk Ÿ 23.
á
¸ 0.0000217
¸ 0.0999444611, kEk Ÿ
22.
!
0Þ1
Þ
20.
!Þ#
dx œ '0 Š1
kE k Ÿ 19.
" x
1 x%
Þ
18.
"!
14 4 81 x
œ 1 4x 6x# 4x$ x%
(3)(2)(1)(2x)$ 3!
á œ x 13 x# 29 x3
3!
¸ 0.0000003
(.2) 7†3!
œ ’x
17.
'
'00 2 ec x " dx œ '00 2 Þ
16.
#!
$
ˆ 3" ‰ ˆ 34 ‰ ˆ 37 ‰ x$
#!
(4)(3)(2)x% 4!
13. (1 2x)$ œ 1 3(2x)
15.
ˆ "3 ‰ ˆ 34 ‰ x#
œ xa1 xb"Î3 œ xŒ1 ˆc "3 ‰x
1 11†12!
!
10
¸ 1.9 ‚ 10
'01 cos t# dt œ '01 Š1 t# 4!t t6! á ‹ dt œ ’t 10t 9t†4! 13t †6! á “ " %
)
*
"#
&
*
"$
!
Ê kerrork
" 13†6!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
¸ .00011
!Þ" x( 42 “ !
Section 10.10 The Binomial Series 24.
'01 cos Èt dt œ '01 Š1 #t 4!t 6!t 8!t á ‹ dt œ ’t t4 3t†4! 4t†6! 5t†8! á “ " #
Ê kerrork x
Ê kerrork
t' 3!
" 15†7!
t"! 5!
x& 5
x( 7†2!
x* 9†3!
x"" 11†4!
" 33†34
t$ 3
aex (1 x)b œ
" t%
2x% 5!
" )&
t# 6!
) 7!
ay tan" yb œ
" t%
œ
9!
t# #
’1
t% 8!
)%
" y$
x# #
$
%
&
t"& 15†7!
á“
x
x$ 3
¸
!
x( 7†3!
t& 5
#
x #
x# ##
t( 7†2!
t* 9†3!
t"" 11†4!
t"$ 13†5!
x"" 11†5!
á“
x
!
¸ 0.00064 t% 1#
'
%
x 3†4
x
t' 30
x 5†6
x# #
á“ ¸
t# 2†2
t$ 3†3
x$ 3#
x% 4#
!
)
x 7†8
x% 1#
Ê kerrork
t& 5†5
(0.5)' 30
¸ .00052
$#
á (1)"&
t% 4†4
x 31†32
x
á“ ¸ x !
x# ##
x$ 3#
x% 4#
x& 5#
x$" 31#
á (1)$"
x$ 3!
á ‹ 1 x‹ œ
" #
x 3!
x# 4!
á Ê lim
x% 4!
á ‹ Š1 x
x# #!
x$ 3!
x% 4!
á ‹“ œ
2x% 5!
2x' 7!
ex (1 x) x#
xÄ0
" # x$ 3!
xÄ0
x# #!
" 13†5!
F(x) ¸ x
á Ê lim
sin )‹ œ
)Ä0
" y$
2x' 7!
#
$
á ‹ dt œ ’t
á‹ œ
#
)$ 6
t$ 4
x# 4!
Š1 cos t t# ‹ œ
Š )
t"" 11†5!
á ‹ dt œ ’ t3
#
ŠŠ 1 x
’Š1 x
t( 7†3!
á ‹ dt œ ’ t2
" x#
" x
aex ex b œ 2x# 3!
t( 7
" x
œ lim Š 5!"
34.
t# 3
tÄ0
33.
t 2
x 3!
œ lim Š 4!"
32.
œ lim Š "#
œ2
31.
t& 5
t"# 5!
Ê kerrork
(0.5)' 6# ¸ .00043 " 32# ¸ .00097 when
xÄ0
30.
t"! 4!
¸ .00089 when F(x) ¸
x
" x#
$
á ‹ dt œ ’ t3
t) 3!
28. (a) F(x) œ '0 Š1 (b) kerrork
t"% 7!
x
Ê kerrork
t' 2!
27. (a) F(x) œ '0 Št (b) kerrork
#
¸ 0.000013
x
x$ 3
%
¸ 0.000004960
26. F(x) œ '0 Št# t% ¸
$
!
" 5†8!
25. F(x) œ '0 Št#
29.
629
ex e x
x
t# #
t% 4!
Š1
œ x lim Š2 Ä_
2x# 3!
t' 6!
á ‹“ œ 4!"
)& 5!
á‹ œ
t# 6!
" x
Š2x
2x$ 3!
2x& 5!
2x( 7!
á‹
y% 7
á‹ œ 2
t% 8!
#
á Ê lim
" cos t Š t# ‹ t%
tÄ0
" á ‹ œ 24
" )&
)$ 6
Š)
á‹ œ
’y Š y
)
)$ 3!
" 5!
)# 7!
)% 9!
$
á Ê lim
sin ) ) Š )6 ‹ )&
)Ä0
" 1 #0
y$ 3
y& 5
á ‹“ œ
" 3
y# 5
y% 7
á Ê lim
yÄ0
y tan " y y$
œ lim Š 3" yÄ0
y# 5
" 3
tanc" y sin y y$ cos y
Ê lim
yÄ0
œ
Œy
y$ 3
y& 5
á Œy y$
tanc" y sin y y$ cos y
œ lim
Œ
" 6
35. x# Š1 e1Îx ‹ œ x# ˆ1 1 ˆ1 œ x lim Ä_
" #x#
" 6x%
y& 5!
á
cos y
yÄ0
#
y$ 3!
23y# 5!
á
cos y
" x#
" #x %
œ
Œ
y$ 6
23y& 5!
y$
cos y
á
œ
Œ
" 6
23y# 5!
á
cos y
œ "6 " 6x'
á ‰ œ 1
" #x #
" 6x%
#
x# Še1Îx 1‹ á Ê x lim Ä_
á ‰ œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
á‹
630
Chapter 10 Infinite Sequences and Series
36. (x 1) sin ˆ x " 1 ‰ œ (x 1) Š x " 1
" 3!(x 1)$
Ê x lim (x 1) sin ˆ x " 1 ‰ œ x lim Š1 Ä_ Ä_
37.
ln a1 x# b 1 cos x
38.
x# 4 ln (x 1)
# Œx
œ
x% #
x# #!
1 Š1
x' 3
á
x% 4!
á‹
(x c 2)#
’(x 2)
#
x 2
œ lim
2)#
x Ä 2 ’1 x c# 2 (x
3
39. sin 3x2 œ 3x2 92 x6
(x c 2)$ 3
á“
81 10 40 x
10 3x2 92 x6 81 40 x . . . 2 2 x4 4 x6 . . . 2x xÄ0 3 45
œ lim
40. lna1 x3 b œ x3
41. 1 1
x6 2
6
xÄ0
Š #"!
x# 4!
(x 2)(x 2)
œ
œ lim
œ
x# #
Œ1
x3
1 2x
9
x9 3
á“
1 3x
" 3!(x 1)#
á
á‹
x2 ’1
x
#
2
(x c 2)# 3
" 5!(x 1)%
á
á‹ œ 1 x# #
Š #"!
x# 4!
Œ1
œ lim
xÄ0
x% 3
á
á‹
x Ä 2 ln (x 1)
8 3 92 x4 81 40 x . . . 2 2 4 4 2 x x . . . xÄ0 3 45
x12 4
œ
. . . and x sin x2 œ x3 16 x7
12
3
6
9
1 x2 x3 x4 . . . 1 8 1 120 x 5040 x12 Þ Þ Þ x Ä 0 1
œ lim
4 6 45 x
sin 3x2
. . . Ê lim
x Ä 0 1 cos 2x
3 2
1 4 6x
1 11 120 x
1 15 5040 x
Þ Þ Þ Ê lim
xÄ0
œ1
Þ Þ Þ œ e1 œ e
1 4x
3 4 5 3 2 42. ˆ 14 ‰ ˆ 14 ‰ ˆ 14 ‰ Þ Þ Þ œ ˆ 14 ‰ ”1 ˆ 14 ‰ ˆ 14 ‰ Þ Þ Þ • œ
43. 1
32 42 2x
34 44 4x
36 46 6x
ÞÞÞ œ 1
1 ˆ 3 ‰2 2x 4
1 ˆ 3 ‰4 4x 4
1 ˆ 3 ‰6 6x 4
1 1 64 1 1Î4
œ
1 4 64 3
œ
1 48
Þ Þ Þ œ cosˆ 34 ‰
44.
1 2
1 2 †2 2
1 3 †2 3
1 4 †2 4
2 3 4 Þ Þ Þ œ ˆ 12 ‰ 12 ˆ #1 ‰ 31 ˆ #1 ‰ 41 ˆ #1 ‰ Þ Þ Þ œ lnˆ1 21 ‰ œ lnˆ 23 ‰
45.
1 3
13 33 3x
15 35 5x
17 37 7x
ÞÞÞ œ
46.
2 3
23 3 3 †3
25 3 5 †5
27 3 7 †7
3 5 7 Þ Þ Þ œ ˆ 23 ‰ 13 ˆ 23 ‰ 15 ˆ 23 ‰ 17 ˆ 32 ‰ Þ Þ Þ œ tan1 ˆ 32 ‰
1 3
1 ˆ 1 ‰3 3x 3
1 ˆ 1 ‰5 5x 3
1 ˆ 1 ‰7 7x 3
Þ Þ Þ œ sinˆ 13 ‰ œ
47. x3 x4 x5 x6 Þ Þ Þ œ x3 a1 x x2 x3 Þ Þ Þ b œ x3 ˆ 1 1 x ‰ œ 48. 1
32 x2 2x
34 x4 4x
36 x6 6x
ÞÞÞ œ 1
2 1 2x a3xb
4 1 4x a3xb
2
6 1 6x a3xb
22 x4 2x
23 x5 3x
24 x6 4x
Þ Þ Þ œ cosa3xb
3
Þ Þ Þ œ x2 Š1 2x
51. 1 2x 3x2 4x3 5x4 Þ Þ Þ œ 52. 1
x 2
x2 3
x3 4
x4 5
d dx a1
Þ Þ œ x1 Šx
a2xb2 2x
a2xb3 3x
a2xb4 4x
x3 3
x4 4
x5 5
x3 1 + x2
Þ Þ Þ ‹ œ x2 e2x
x x2 x3 x4 x5 Þ Þ Þ b œ
x2 2
È3 2
x3 1x
49. x3 x5 x7 x9 Þ Þ Þ œ x3 Š1 x2 ax2 b ax2 b Þ Þ Þ ‹ œ x3 ˆ 1 +1x2 ‰ œ 50. x2 2x3
œ 2! œ 2
x# 4
Ê lim
á“
. . . and 1 cos 2x œ 2x2 23 x4
œ lim
" 5!(x 1)%
# lim ln a1 x b x Ä 0 1 cos x
Ê
" 3!(x 1)#
á‹ œ 1
œ4
x3 x2 x3 x4 . . . 1 11 1 16 x7 120 x 5040 x15 Þ Þ Þ
œ
x% 3
" 5!(x 1)&
d ˆ 1 ‰ dx 1 x
œ
Þ Þ ‹ œ x1 lna1 xb œ
1 a1 x b 2 lna1xb x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
lnˆ1 x3 ‰ x sin x2
Section 10.10 The Binomial Series x‰ 53. ln ˆ 11 x œ ln (1 x) ln (1 x) œ Šx
54. ln (1 x) œ x " n10n
" 10)
x$ 3
x% 4
á
(1)n 1 xn n
x$ 3
x% 4
á ‹ Šx
á Ê kerrork œ ¹ (")n
x# #
n 1 n
x
x$ 3
¹œ
" n10n
x% 4
á ‹ œ 2 Šx
x$ 3
x& 5
á‹
when x œ 0.1;
Ê n10n 10) when n 8 Ê 7 terms
55. tan" x œ x " #n 1
x# #
x# #
631
" 10$
x$ 3
x& 5
Ê n
56. tan" x œ x
x$ 3
x* 9
á
(")n 1 x2n 2n1
1
á Ê kerrork œ ¹ (1)2nx1
n 1 2n 1
x( 7
x* 9
á
(1)n1 x2n1 2n1
_
(1)n 2n1
nœ1
_
(1)nc1 2n1
nœ1
2n 1 x2n1 ¹
2n 1
á and n lim † ¹x Ä _ 2n 1
Ê tan" x converges for kxk 1; when x œ 1 we have ! we have !
¹œ
" #n 1
when x œ 1;
œ 500.5 Ê the first term not used is the 501st Ê we must use 500 terms
1001 # x& 5
x( 7
¸ 2n 1 ¸ œ x# œ x# n lim Ä _ #n 1
which is a convergent series; when x œ 1
which is a convergent series Ê the series representing tan" x diverges for kxk 1
(1)n 1 x2n 1 x$ x& x( x* á and when the series representing 48 3 5 7 9 á 2n 1 " ' error less than 3 † 10 , then the series representing the sum " ‰ " ‰ " ‰ 48 tan" ˆ 18 32 tan" ˆ 57 20 tan" ˆ #39 also has an error of magnitude less than 10' ;
" ‰ tan" ˆ 18 has an
57. tan" x œ x
thus
2nc1
kerrork œ 48
" Š 18 ‹
#n 1
" 3†10'
Ê n 4 using a calculator Ê 4 terms
58. ln (sec x) œ '0 tan t dt œ '0 Št x
"Î#
x
t$ 3
x# 3x% # 8 2nb3 lim ¹ 1†3†5â(2n 1)(2n 1)x † n Ä _ 2†4†6â(2n)(2n 2)(2n 3)
59. (a) a1 x# b
¸1
2t& 15
á ‹ dt ¸
x# #
x% 12
$ 5x' " x ¸ x x6 16 Ê sin 2†4†6â(2n)(2n ") 1†3†5â(2n 1)x2nb1 ¹ 1 Ê
x' 45
3x& 40
á
5x( 112
; Using the Ratio Test: (2n 1)(2n1)
x# n lim ¹ ¹1 Ä _ (2n 2)(2n 3)
Ê kxk 1 Ê the radius of convergence is 1. See Exercise 69. d dx
(b)
acos" xb œ a1 x# b
60. (a) a1 t# b œ1
"Î# t# #
term, 61.
" 1x
5x 112
Ê cos" x œ
¸ (1)"Î# ˆ "# ‰ (1)$Î# at# b 3t% 2# †2!
(b) sinh" ˆ 4" ‰ ¸ (
"Î#
" 4
3†5t' 2$ †3!
" 384
1 #
sin" x ¸
x
, evaluated at t œ
" 4
Šx
ˆ "# ‰ ˆ #3 ‰ (1) &Î# at# b# #!
Ê sinh" x ¸ '0 Š1
3 40,960
1 #
t# #
3t% 8
5t' 16 ‹
x$ 6
3x& 40
5x( 112 ‹
¸
1 #
x
x$ 6
3x& 40
ˆ #" ‰ ˆ #3 ‰ ˆ #5 ‰ (1) (Î# at# b$ 3!
dt œ x
x$ 6
3x& 40
5x( 112
œ 0.24746908; the error is less than the absolute value of the first unused
since the series is alternating Ê kerrork
œ 1 "(x) œ 1 x x# x$ á Ê
d dx
ˆ 11x ‰ œ
" 1 x#
œ
d dx
5 ˆ "4 ‰ 112
(
¸ 2.725 ‚ 10'
a1 x x# x$ á b
œ 1 2x 3x# 4x$ á 62.
" 1 x#
œ 1 x# x% x' á Ê
d dx
ˆ 1 " x# ‰ œ
2x a1 x # b #
œ
d dx
a1 x# x% x' á b œ 2x 4x$ 6x& á
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
5x( 112
632
Chapter 10 Infinite Sequences and Series
8â(2n 2)†(2n) 63. Wallis' formula gives the approximation 1 ¸ 4 ’ 3†23††45††45††67††67†â (2n 1)†(2n 1) “ to produce the table
n µ1 10 3.221088998 20 3.181104886 30 3.167880758 80 3.151425420 90 3.150331383 93 3.150049112 94 3.149959030 95 3.149870848 100 3.149456425 At n œ 1929 we obtain the first approximation accurate to 3 decimals: 3.141999845. At n œ 30,000 we still do not obtain accuracy to 4 decimals: 3.141617732, so the convergence to 1 is very slow. Here is a Maple CAS procedure to produce these approximations: pie := proc(n) local i,j; a(2) := evalf(8/9); for i from 3 to n do a(i) := evalf(2*(2*i2)*i/(2*i1)^2*a(i1)) od; [[j,4*a(j)] $ (j = n5 .. n)] end _
_
_
64. (a) faxb œ 1 !ˆ mk ‰xk Ê f w axb œ !ˆ mk ‰k xk1 Ê a1 xb † f w axb œ a1 xb!ˆ mk ‰k xk1 kœ1
kœ1
_
_
kœ1
_
_
_
_
kœ2
kœ1
œ !ˆ mk ‰k xk 1 x † !ˆ mk ‰k xk1 œ !ˆ mk ‰k xk 1 !ˆ mk ‰k xk œ ˆ m1 ‰a1b x0 !ˆ mk ‰k xk1 !ˆ mk ‰k xk kœ1
kœ1
_
œ m!
kœ2
ˆ mk ‰k xk 1
kœ1
_
!
kœ1
ˆ mk ‰k xk
kœ1
_
ˆ mk ‰k xk 1
Note that: ! kœ2
_
_
œ!
kœ1
_
m ‰ ˆk 1 ak
1b xk .
_
_
m ‰ k ! ˆ m ‰k xk Thus, a1 xb † f w axb œ m ! ˆ mk ‰k xk 1 ! ˆ mk ‰k xk œ m ! ˆ k 1 ak 1b x k kœ2
_
œ m!
kœ1
‰ ’ˆ k m 1 ak
1b xk
kœ1
ˆ mk ‰k xk “
kœ1
_
œ m!
kœ1
m ‰ ’ˆˆ k 1 ak
kœ1
1b
ˆ mk ‰k ‰xk “.
m†am "bâam ak 1b 1b ak 1b m†am "bâk!am k 1b k ak 1 b ! m†am "bâam k 1b k œ m†am "bâk!am k 1b aam kb kb œ m m†am "bâk!am k 1b k!
‰ ˆm‰ Note that: ˆ k m 1 ak 1b k k œ œ
m†am "bâam kb k!
_
_
_
kœ1
kœ1
kœ1
œ mˆ mk ‰.
m ‰ ! ’ˆmˆ m ‰ ‰xk “ œ m m!ˆ m ‰xk ˆm‰ ‰ k Thus, a1 xb † f w axb œ m ! ’ˆˆ k 1 ak 1b k k x “ œ m k k _
œ mŒ1 !ˆ mk ‰xk œ m † faxb Ê f w axb œ kœ1
m†faxb a1 x b
if " x 1.
(b) Let gaxb œ a1 xbm faxb Ê gw axb œ ma1 xbm1 faxb a1 xbm f w axb œ ma1 xbm1 faxb a1 xbm †
m†faxb a1 x b
œ ma1 xbm1 faxb a1 xbm1 † m † faxb œ 0.
(c) gw axb œ 0 Ê gaxb œ c Ê a1 xbm faxb œ c Ê faxb œ Ê fa0b œ 1
65. a1 x# b
"Î#
_
!ˆ m ‰a0bk k
kœ1
œ a1 ax# bb 3!
_
œ ca1 xbm . Since faxb œ 1 !ˆ mk ‰xk kœ1
m
m
œ 1 0 œ 1 Ê ca1 0b œ 1 Ê c œ 1 Ê faxb œ a1 xb .
"Î#
ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ (1) (Î# ax# b$
c a1xbcm
œ (1)"Î# ˆ "# ‰ (1)$Î# ax# b
á œ1
#
x #
%
1†3x 2# †#!
1†3†5x 2$ †3!
'
ˆ "# ‰ ˆ 3# ‰ (1)c&Î# ax# b# _
á œ1!
nœ1
#!
1†3†5â(2n1)x2n #n †n!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 10.10 The Binomial Series Ê sin" x œ '0 a1 t# b x
dt œ '0 Œ1 ! _
x
"Î#
1†3†5â(2n 1)x2n #n †n!
nœ1
_
dt œ x !
nœ1
1†3†5â(2n 1)x2nb1 #†4â(2n)(2n 1)
633
,
where kxk 1
_
œ 'x ˆ t"#
tan" x œ 'x
" t%
1 #
Ê tan" x œ œ
lim
b Ä _
_
1 #
_
66. ctan" td x œ
" t'
" t)
" x
"t
" 3t$
_
t#
bÄ_
_
Š 1# ‹
œ 'x – t " — dt œ 'x 1Š ‹
á ‰ dt œ lim " 3x$
dt 1 t#
"t
" 3t$
" 5t&
" t#
ˆ1
" 7t(
" t#
b
á ‘x œ
" " x td c_ œ tan" x 1# 5x& á , x 1; ctan " " " " " " ‘x 5t& 7t( á b œ x 3x$ 5x& 7x( á
œ
" t%
" t'
" x
" 3x$
'
x
_
á ‰ dt
" 5x&
" 7x(
dt 1 t# "
Ê tan
x œ 1#
á
" x
" 3x$
" 30
x& á ;
" 5x&
á ,
x 1
67. (a) ei1 œ cos (1) i sin (1) œ 1 i(0) œ 1 (b) ei1Î4 œ cos ˆ 14 ‰ i sin ˆ 14 ‰ œ
" È2
i È2
œ Š È" ‹ (1 i) 2
(c) ei1Î2 œ cos ˆ 1# ‰ i sin ˆ 1# ‰ œ 0 i(1) œ i 68. ei) œ cos ) i sin ) Ê ei) œ ei()) œ cos ()) i sin ()) œ cos ) i sin ); ei) eci) ; # ei) eci) œ #i
ei) ei) œ cos ) i sin ) cos ) i sin ) œ 2 cos ) Ê cos ) œ ei) ei) œ cos ) i sin ) (cos ) i sin )) œ 2i sin ) Ê sin ) 69. ex œ 1 x
x# #!
x$ 3! (i))# 2!
ei) œ 1 i) Ê œ
ei) eci) œ # % )# 1 #! )4!
ei) eci) #i
œ)
œ )$ 3!
)& 5!
Š1 i)
)' 6!
Š1 i)
(i))# #!
(i))$ 3!
(i))% 4!
œ 1 i)
(i))# 2!
á œ 1 i)
á‹ Š1 i)
(i))# #!
(i))$ 3!
(i))$ 3!
(i))# #!
(i))$ 3!
(i))% 4!
(i))% 4!
á and (i))% 4!
á
á‹
#
á œ cos );
(i))# #!
)( 7!
x% i) 4! á Ê e (i))$ (i))% 3! 4!
$
%
#
$
%
)) )) )) )) (i3! (i4! á‹ Š1 i) (i#)!) (i3! (i4! á‹
#i
á œ sin )
70. ei) œ cos ) i sin ) Ê ei) œ eiÐ)Ñ œ cos ()) i sin ()) œ cos ) i sin )
ei) eci) œ cosh i) # i) c i) œ e 2e œ sinh i)
(a) ei) ei) œ (cos ) i sin )) (cos ) i sin )) œ 2 cos ) Ê cos ) œ (b) ei) ei) œ (cos ) i sin )) (cos ) i sin )) œ 2i sin ) Ê i sin ) 71. ex sin x œ Š1 x
x# #! " 6
#
œ (1)x (1)x ˆ
x$ 3!
x% 4!
"‰ $ # x
á ‹ Šx " 6
ˆ
x$ 3!
"‰ % 6 x
x& 5!
x( 7!
á‹
ˆ 1#"0 1"# x
" ‰ & #4 x x
á œ x x# 3" x$
ex † eix œ eÐ1iÑx œ ex (cos x i sin x) œ ex cos x i ae sin xb Ê e sin x is the series of the imaginary part _
of eÐ1iÑx which we calculate next; eÐ1iÑx œ !
nœ0
œ 1 x ix Ð1iÑx
of e
is x
(xix)n n!
œ 1 (x ix)
(x ix)# #!
(x ix)$ 3!
(x ix)% 4!
" " " " " # $ $ % & & ' #! a2ix b 3! a2ix 2x b 4! a4x b 5! a4x 4ix b 6! a8ix b á Ê the imaginary 2 # 2 $ 4 & 8 ' " $ " & " ' # #! x 3! x 5! x 6! x á œ x x 3 x 30 x 90 x á in agreement with our x
product calculation. The series for e sin x converges for all values of x. 72.
d dx
ˆeÐaibÑ ‰ œ
á
d dx
ceax (cos bx i sin bx)d œ aeax (cos bx i sin bx) eax (b sin bx bi cos bx)
œ aeax (cos bx i sin bx) bieax (cos bx i sin bx) œ aeÐaibÑx ibeÐaibÑx œ (a ib)eÐaibÑx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
part
634
Chapter 10 Infinite Sequences and Series
73. (a) ei)" ei)# œ (cos )" i sin )" )(cos )# i sin )# ) œ (cos )" cos )# sin )" sin )# ) i(sin )" cos )# sin )# cos )" ) œ cos()" )# ) i sin()" )# ) œ eiÐ)" )# Ñ " " ) i sin ) ‰ (b) ei) œ cos()) i sin()) œ cos ) i sin ) œ (cos ) i sin )) ˆ cos cos ) i sin ) œ cos ) i sin ) œ ei) 74.
a bi ÐabiÑx C" iC# œ ˆ aa# bib# ‰ eax (cos bx i sin bx) C" iC# a# b# e ax œ a# e b# (a cos bx ia sin bx ib cos bx b sin bx) C" iC# ax œ a# e b# [(a cos bx b sin bx) (a sin bx b cos bx)i] C" iC# ax ax œ e (a cosa#bxb#b sin bx) C" ie (a sina#bxb#b cos bx) iC# ; ÐabiÑx ax ibx ax ax ax
e
'e
œe e
ÐabiÑx
dx œ
œ e (cos bx i sin bx) œ e cos bx ie sin bx, so that given
a bi a# b#
eÐabiÑx C" iC# we conclude that ' eax cos bx dx œ
and ' eax sin bx dx œ
e (a sin bx b cos bx) a# b# ax
eax (a cos bx b sin bx) a# b#
C"
C#
CHAPTER 10 PRACTICE EXERCISES 1. converges to 1, since n lim a œ n lim Š1 Ä_ n Ä_ 2. converges to 0, since 0 Ÿ an Ÿ
2 Èn
(1)n n ‹
œ1
, n lim 0 œ 0, n lim Ä_ Ä_
œ 0 using the Sandwich Theorem for Sequences
2 Èn
ˆ 1 2n2 ‰ œ lim ˆ #"n 1‰ œ 1 3. converges to 1, since n lim a œ n lim Ä_ n Ä_ nÄ_ n
4. converges to 1, since n lim a œ n lim c1 (0.9)n d œ 1 0 œ 1 Ä_ n Ä_ 5. diverges, since ˜sin
n1 ™ #
œ e0ß 1ß 0ß 1ß 0ß 1ß á f
6. converges to 0, since {sin n1} œ {0ß 0ß 0ß á } 7. converges to 0, since n lim a œ n lim Ä_ n Ä_
ln n# n
8. converges to 0, since n lim a œ n lim Ä_ n Ä_
ln (2n") n
œ 2 n lim Ä_
Š "n ‹ 1
Š 2n 2b 1 ‹
œ n lim Ä_
1
ˆ n nln n ‰ œ lim 9. converges to 1, since n lim a œ n lim Ä_ n Ä_ nÄ_ 10. converges to 0, since n lim a œ n lim Ä_ n Ä_
ln a2n$ 1b n
œ0
1Š "n ‹
œ n lim Ä_
1
Š
" e
ˆ1 n" ‰cn œ lim , since n lim a œ n lim Ä_ n Ä_ nÄ_
œ1
6n# ‹ 2n$ 1
1
ˆ n n 5 ‰n œ lim Š1 11. converges to ec5 , since n lim a œ n lim Ä_ n Ä_ nÄ_ 12. converges to
œ0
œ n lim Ä_
(5) n ‹
" ˆ1 "n ‰n
œ
n
12n 6n#
œ n lim Ä_
œ0
œ ec5 by Theorem 5 " e
by Theorem 5
ˆ 3 ‰1În œ lim 13. converges to 3, since n lim a œ n lim Ä_ n Ä_ n nÄ_
3 n1În
œ
3 1
œ 3 by Theorem 5
ˆ 3 ‰1În œ lim 14. converges to 1, since n lim a œ n lim Ä_ n Ä_ n nÄ_
31În n1În
œ
1 1
œ 1 by Theorem 5
n
2 n
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 10 Practice Exercises
21În 1 Š "n ‹
15. converges to ln 2, since n lim a œ n lim n a21În 1b œ n lim Ä_ n Ä_ Ä_
œ n lim Ä_
–
Š 21În ln 2‹ n#
—
635
œ n lim 21În ln 2 Ä_
Š n#" ‹
œ 2! † ln 2 œ ln 2 2
n È 16. converges to 1, since n lim a œ n lim 2n 1 œ n lim exp Š ln (2nn 1) ‹ œ n lim exp Œ 2n1b 1 œ e! œ 1 Ä_ n Ä_ Ä_ Ä_
17. diverges, since n lim a œ n lim Ä_ n Ä_
(n 1)! n!
œ n lim (n 1) œ _ Ä_
18. converges to 0, since n lim a œ n lim Ä_ n Ä_
(4)n n!
Š "# ‹
Š "# ‹
19.
" (2n 3)(2n 1)
œ
#n 3
Š "# ‹
Ê sn œ –
2n 1
Š "# ‹
" Ê n lim s œ n lim Ä_ n Ä _ –6
20.
2 n(n 1)
œ
2 n
2 n1
ˆ1 œ n lim Ä_ 21.
22.
9 (3n 1)(3n 2) œ 3# 3n3#
œ
_
_
nœ0
nœ0
_
nœ1
ˆ 34 ‰ 1ˆ c4" ‰
3 4n
" en
3 3n 2
_
nœ1
—–
Š "# ‹ 5
Š "# ‹
Ê sn œ ˆ #3 35 ‰ ˆ 35 38 ‰ ˆ 38 3 ‰ 3n 2
œ
7
Š "‹
# — á – #n 3
Š "# ‹
2n 1 —
œ
Š "# ‹ 3
Š "# ‹
2n 1
2 ‰ n1
œ #2
2 n1
Ê n lim s Ä_ n
Ê sn œ ˆ 92 ˆ œ n lim Ä_
3 ‰ 11
á ˆ 3n 3 1
3 ‰ 3n 2
3 #
2 ‰ 2 ‰ ˆ 2 13 13 17 2 2 ‰ 2 9 4n1 œ 9
, a convergent geometric series with r œ
" e
ˆ 172
2 ‰ 21
á ˆ 4n2 3
and a œ 1 Ê the sum is
_
‰n a convergent geometric series with r œ "4 and a œ œ ! ˆ 34 ‰ ˆ " 4 nœ0
" 1 Š "e ‹
3 4
2 ‰ 4n 1
œ
e e1
Ê the sum is
œ 35
25. diverges, a p-series with p œ 26. !
5
" 6
ˆ3 Ê n lim s œ n lim Ä_ n Ä_ #
23. ! en œ !
Š "# ‹
œ 1
8 2 2 (4n 3)(4n 1) œ 4n 3 4n 1 œ 29 4n21 Ê n lim s Ä_ n
24. ! (1)n
œ
Ê sn œ ˆ #2 23 ‰ ˆ 32 42 ‰ á ˆ n2
2 ‰ n1
3 3n 1
2n 1 —
3
œ 0 by Theorem 5
5 n
_
œ 5 !
nœ1
" x"Î#
27. Since f(x) œ _
" n,
" #
diverges since it is a nonzero multiple of the divergent harmonic series
Ê f w (x) œ #x"$Î# 0 Ê f(x) is decreasing Ê an1 an , and _
lim a œ lim nÄ_ n nÄ_
1 Èn
œ 0, the
) ! " diverges, the given series converges conditionally. series ! (" Èn converges by the Alternating Series Test. Since Èn n
nœ1
nœ1
28. converges absolutely by the Direct Comparison Test since
" #n$
" n$
for n 1, which is the nth term of a convergent
p-series
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
636
Chapter 10 Infinite Sequences and Series
29. The given series does not converge absolutely by the Direct Comparison Test since the nth term of a divergent series. Since f(x) œ Ê an1 an , and n lim a œ Ä_ n
" lim n Ä _ ln (n 1)
" ln (x 1)
" ln (n 1)
" (ln (x 1))# (x 1)
w
Ê f (x) œ
" n1
, which is
0 Ê f(x) is decreasing
œ 0, the given series converges conditionally by the Alternating
Series Test. 30.
'2_ x(ln" x)
#
dx œ lim
bÄ_
'2b
" x(ln x)#
b dx œ lim c(ln x)" d 2 œ lim ˆ ln"b
bÄ_
bÄ_
" ‰ ln 2
œ
" ln #
Ê the series
converges absolutely by the Integral Test 31. converges absolutely by the Direct Comparison Test since
ln n n$
n n$
" n#
œ
, the nth term of a convergent p-series
n n 32. diverges by the Direct Comparison Test for en n Ê ln ˆen ‰ ln n Ê nn ln n Ê ln nn ln (ln n)
Ê n ln n ln (ln n) Ê
33. n lim Ä_
Š
"
n
È n#
Š n"# ‹
34. Since f(x) œ
1
‹
ln n ln (ln n)
œ Én lim Ä_
3x# x$ 1
n# n# 1
Ê f w (x) œ
" n
, the nth term of the divergent harmonic series
œ È1 œ 1 Ê converges absolutely by the Limit Comparison Test
3x a2 x$ b ax $ 1 b#
0 when x 2 Ê an1 an for n 2 and n lim Ä_
3n# n$ 1
œ 0, the
series converges by the Alternating Series Test. The series does not converge absolutely: By the Limit #
Comparison Test, n lim Ä_
Š n$3n 1 ‹ ˆ n" ‰
œ n lim Ä_
3n$ n$ 1
œ 3. Therefore the convergence is conditional.
35. converges absolutely by the Ratio Test since n lim ’ n 2 † Ä _ (n 1)! 36. diverges since n lim a œ n lim Ä_ n Ä_
(")n an# 1b 2n# n 1
n! n1“
œ n lim Ä_
n2 (n 1)#
does not exist nb1
37. converges absolutely by the Ratio Test since n lim † ’ 3 Ä _ (n 1)!
n! 3n “
œ n lim Ä_
3 n1
œ01
n 2 3 n È É 38. converges absolutely by the Root Test since n lim an œ n lim nn œ n lim Ä_ Ä_ Ä_ n n
39. converges absolutely by the Limit Comparison Test since n lim Ä_
40. converges absolutely by the Limit Comparison Test since n lim Ä_ nb1
41. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 4) † Ä_ Ä _ (n 1)3nb1
œ01
n3n (x 4)n ¹
1 Ê
" Š $Î# ‹ n
Š Èn(n "1)(n Š n"# ‹
Š È "# n n
kx 4 k lim 3 nÄ_
1
‹
‹ 2)
ˆ n n 1 ‰ 1 Ê
nœ1
n
3 n3n
n(n 1)(n 2) n$
n # an # 1 b n%
œ Én lim Ä_
_
harmonic series, which converges conditionally; at x œ 1 we have!
œ01
œ Én lim Ä_
Ê kx 4k 3 Ê 3 x 4 3 Ê 7 x 1; at x œ 7 we have ! _
6 n
nœ1 _
œ!
nœ1
kx 4 k 3
(1)n 3n n3n " n
œ1
œ1
1
_
œ ! ("n ) , the alternating n
nœ1
, the divergent harmonic series
(a) the radius is 3; the interval of convergence is 7 Ÿ x 1 (b) the interval of absolute convergence is 7 x 1 (c) the series converges conditionally at x œ 7
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 10 Practice Exercises 42. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x1) † (2n1)! ¹ 1 Ê (x 1)# n lim Ä_ Ä _ (2n1)! (x1)2nc2 Ä_ (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally 2n
nb1
43. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (3x(n1)1)# † Ä_ Ä_
n# (3x 1)n ¹
1 Ê k3x 1k n lim Ä_
Ê 1 3x 1 1 Ê 0 3x 2 Ê 0 x _
" n#
œ !
nœ1 _
have ! nœ1
" (#n)(2n1)
2 3
n# (n 1)# _
œ 0 1, which holds for all x
1 Ê k3x 1k 1 nc1
_
2nc1
; at x œ 0 we have ! (1) n# (1) œ ! ("n)# n
nœ1
nœ1
, a nonzero constant multiple of a convergent p-series, which is absolutely convergent; at x œ _
(1)n 1 (1)n n#
(a) the radius is
œ ! ("n)#
n 1
2 3
we
, which converges absolutely
nœ1
" 3
; the interval of convergence is 0 Ÿ x Ÿ
(b) the interval of absolute convergence is 0 Ÿ x Ÿ
2 3
2 3
(c) there are no values for which the series converges conditionally 44. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ n2 † Ä_ Ä _ 2n 3 Ê _
! nœ1
k2x 1k #
n1 2n 1
†
(2x 1)nb1 2nb1
†
2n 1 n1
†
2n (2x 1)n ¹
1 Ê
k2x 1k lim 2 nÄ_
n2 ¸ 2n 3 †
2n " ¸ n1
(1) 1 Ê k2x 1k 2 Ê 2 2x 1 2 Ê 3 2x 1 Ê 3# x
(2)n #n
_
œ!
lim ˆ n 1 ‰ œ n Ä _ 2n 1
nœ1
" #
(")n (n1) 2n 1
" #
1
; at x œ 3# we have
which diverges by the nth-Term Test for Divergence since
Á 0; at x œ
" #
_
n1 we have ! 2n 1 † nœ1
2n #n
_
n" œ ! 2n 1 , which diverges by the nth-Term Test nœ1
(a) the radius is 1; the interval of convergence is 3# x (b) the interval of absolute convergence is 3# x
" #
" #
(c) there are no values for which the series converges conditionally nb1
45. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ x Ä_ Ä _ (n 1)nb1 Ê
kx k e
nn xn ¹
¸ˆ n ‰n ˆ n " 1 ‰¸ 1 Ê 1 Ê kxk n lim Ä _ n1
kx k e n lim Ä_
ˆ n " 1 ‰ 1
† 0 1, which holds for all x
(a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1
46. n lim † ¹ uunbn 1 ¹ 1 Ê n lim ¹ x Ä_ Ä _ Èn 1 _
Èn xn ¹
n 1 Ê kxk n lim 1 Ê kxk 1; when x œ 1 we have Ä _ Én1
_
! (È1) , which converges by the Alternating Series Test; when x œ 1 we have ! n n
nœ1
nœ1
" Èn
, a divergent p-series
(a) the radius is 1; the interval of convergence is 1 Ÿ x 1 (b) the interval of absolute convergence is 1 x 1 (c) the series converges conditionally at x œ 1 2nb1
47. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (n 32)x nb1 Ä_ Ä_ _
_
nœ1
nœ1
the series ! nÈ31 and !
n1 È3
†
3n (n 1)x2n
1
¹1 Ê
x# 3 n lim Ä_
2‰ È È3; ˆ nn 1 1 Ê 3x
, obtained with x œ „ È3, both diverge
(a) the radius is È3; the interval of convergence is È3 x È3 (b) the interval of absolute convergence is È3 x È3 (c) there are no values for which the series converges conditionally
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
637
638
Chapter 10 Infinite Sequences and Series 2nb3
48. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ (x 2n1)x 3 Ä_ Ä_
2n 1 (x 1)2nb1 ¹
†
ˆ 2n 1 ‰ 1 Ê (x 1)# (1) 1 1 Ê (x 1)# n lim Ä _ 2n 3 _
Ê (x 1)# 1 Ê kx 1k 1 Ê 1 x 1 1 Ê 0 x 2; at x œ 0 we have ! (1)#n(1)1 n
2nb1
nœ1
_
œ!
nœ1 _
(1)3nb1 2n 1
that ! nœ1
" 2n 1
_
œ!
nœ1
(1)nc1 2n 1
which converges conditionally by the Alternating Series Test and the fact _
(1)n (1)2nb1 2n 1
diverges; at x œ 2 we have !
nœ1
_
œ!
nœ1
(1)n 2n 1
, which also converges conditionally
(a) the radius is 1; the interval of convergence is 0 Ÿ x Ÿ 2 (b) the interval of absolute convergence is 0 x 2 (c) the series converges conditionally at x œ 0 and x œ 2 (n 1)x 49. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹ cschcsch (n)xn Ä_ Ä_
c"
Ê kxk n lim ¹ e1 ee 2n Ä_
2n 1 2
¹1 Ê
nb1
¹ 1 Ê kxk n lim Ä_ »
Š enb1 c2ecnc1 ‹ ˆ en c2ecn ‰
»1
_
kx k e
1 Ê e x e; the series ! a „ ebn csch n, obtained with x œ „ e, nœ1
both diverge since n lim a „ e) csch n Á 0 Ä_ n
(a) the radius is e; the interval of convergence is e x e (b) the interval of absolute convergence is e x e (c) there are no values for which the series converges conditionally 50. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹x Ä_ Ä_
nb1
coth (n 1) xn coth (n) ¹
c2nc2
1 Ê kxk n lim † ¹1e Ä _ 1 ec2nc2
1 ec2n 1 ec2n ¹
1 Ê kxk 1
_
Ê 1 x 1; the series ! a „ 1bn coth n, obtained with x œ „ 1, both diverge since n lim a „ 1bn coth n Á 0 Ä_ nœ1
(a) the radius is 1; the interval of convergence is 1 x 1 (b) the interval of absolute convergence is 1 x 1 (c) there are no values for which the series converges conditionally 51. The given series has the form 1 x x# x$ á (x)n á œ 52. The given series has the form x ln ˆ 53 ‰ ¸ 0.510825624
x# #
x$ 3
á (1)n1
53. The given series has the form x
x$ 3!
x& 5!
á (1)n
x2n 1 (2n 1)!
54. The given series has the form 1
x# 2!
x% 4!
á (1)n
x2n (2n)!
55. The given series has the form 1 x 56. The given series has the form x tan" Š È"3 ‹ œ 57. Consider
" 1 2x
x$ 3
x# 2!
x& 5
x# 3!
á
xn n!
á (1)n
xn n
" 1x
, where x œ
" 4
; the sum is
á œ ln (1 x), where x œ
2 3
nœ0
nœ0
1 3
á œ ex , where x œ ln 2; the sum is eln Ð2Ñ œ 2 á œ tan" x, where x œ
as the sum of a convergent geometric series with a œ 1 and r œ 2x Ê _
4 5
á œ sin x, where x œ 1; the sum is sin 1 œ 0
á œ cos x, where x œ 13 ; the sum is cos
x2n 1 (2n 1)
œ
; the sum is
" È3
1 6
_
" 1 ˆ "4 ‰
œ 1 (2x) (2x)# (2x)$ á œ ! (2x)n œ ! 2n xn where k2xk 1 Ê kxk
" 1 2x
" #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
; the sum is
œ
" #
Chapter 10 Practice Exercises " 1 x$
58. Consider
" 1 x$
as the sum of a convergent geometric series with a œ 1 and r œ x$ Ê
œ
639
" 1 ax$ b
_
# $ œ 1 ax$ b ax$ b ax$ b á œ ! (1)n x3n where kx$ k 1 Ê kx$ k 1 Ê kxk 1 nœ0
_
59. sin x œ !
nœ0
_
60. sin x œ !
nœ0 _
61. cos x œ !
nœ0
_
62. cos x œ !
nœ0
_
63. ex œ !
nœ0 _
64. ex œ !
nœ0
_
(1)n x2nb1 (2n 1)!
nœ0
Ê sin
_
(1)n x2n (2n)!
3 Ê cosŠ Èx 5 ‹ œ !
_ ˆ 1 x ‰n
xn n!
# Ê ex œ !
69.
_
a x # b n!
nœ0
œ
3 8
"Î#
_
œ!
nœ0
nœ0
(1)n x2n n!
Ê f w (x) œ x a3 x# b
&Î#
¸
" #
"
#% † 4
" 9 †2 $
$Î#
a3 x# b
9 32
2†1!
2 †2!
" #
œ
"Î#
3 8
,
2 †3!
œ (1 x)" Ê f w (x) œ (1 x)# Ê f ww (x) œ 2(1 x)$ Ê f www (x) œ 6(1 x)% ; f(2) œ 1, f w (2) œ 1, " 1 x
œ 1 (x 2) (x 2)# (x 2)$ á
(x 1)" Ê f w (x) œ (x 1)# Ê f ww (x) œ 2(x 1)$ Ê f www (x) œ 6(x 1)% ; f(3) œ ww
f (3) œ
Ê
x"" 11†3!
tanc" x x
Ê f ww (x) œ x# a3 x# b
$Î#
2 4$
6 4%
www
, f (2) œ
" x
" a
œ
" a#
" x 1
Ê
" a$
(x a)
"
#( †7†2!
"
2"! †10†3!
"
" 4
œ
(x a)#
x"( 17†5!
x#$ 23†7!
dx œ '1 Š1 1Î2
" 5# †#&
" 7 # †# (
x# 3
*
2"$ †13†4!
*
'11Î2
"Î#
3x a3 x# b ; f(1) œ 2, f w (1) œ "# , f ww (1) œ "8 # $ Ê È3 x# œ 2 (x 1) 3(x$ 1) 9(x& 1) á
'01 x sin ax$ b dx œ '01 x Šx$ x3! x5! &
(1)n x6n 5n (#n)!
1 n xn #n n!
n œ0
n
_
œ!
(1)n x10nÎ3 (#n)!
" 4#
(x 3)
" 4$
" 4%
#
(x 3)
œ x" Ê f w (x) œ x# Ê f ww (x) œ 2x$ Ê f www (x) œ 6x% ; f(a) œ 6 a%
œ ’ x5
71.
_
œ!
'
" #
nœ0
(2n)!
'01Î2 exp ax$ b dx œ '01Î2 Š1 x$ x2! x3! x4! ¸
70.
#
n!
nœ0
" x1 œ œ 4"# ,
f www (a) œ
œ!
5
nœ0
(1)n 22nb1 x2nb1 32nb1 (#n 1)!
_
2n
_ (1)n Š x3 ‹ È
Ê eÐ1xÎ2Ñ œ !
67. f(x) œ
" x
(1)n ˆx5Î3 ‰ (2n)!
nœ0
f ww (2) œ 2, f www (2) œ 6 Ê
68. f(x) œ
nœ0
2n
3 f www (1) œ 32
f (3)
nœ0
Ê cos ˆx5Î3 ‰ œ !
Ê f www (x) œ 3x$ a3 x# b
w
_
œ!
(2n 1)!
(1)n x2n (2n)!
xn n!
" 1x
nœ0
_ (1)n Š 2x ‹ 3
œ!
2x 3
(1)n 12nb1 x2nb1 (#n 1)!
œ!
2nb1
(1)n x2nb1 (2n 1)!
65. f(x) œ È3 x# œ a3 x# b
66. f(x) œ
_
(1)n (1x)2nb1 (2n 1)!
Ê sin 1x œ !
"&
x#* 29†9!
x% 5
" 9# †2*
"
x#" 7!
x#( 9!
x% 4
, f w (a) œ a"# , f ww (a) œ
x( 7†2!
x"! 10†3!
x"! 3!
x"' 5!
x## 7!
x& 25
x( 49
x* 81
" 19# †#"*
" 21# †##"
x"$ 13†4!
á“
"Î# !
¸ 0.484917143 á ‹ dx œ '0 Šx% 1
x#) 9!
á ‹ dx
" !
x' 7
" 11# †2""
x) 9
x"! 11
" 13# †2"$
á ‹ dx œ ’x
" 15# †2"&
" 17# †#"(
x$ 9
2 a$
,
(x 3) á
(x a)$ á
á ‹ dx œ ’x
2"' †16†5!
" a
,
$
á “ ¸ 0.185330149
"#
" a%
" 4
x"" 121
á“
"Î# !
¸ 0.4872223583
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
640 72.
Chapter 10 Infinite Sequences and Series
'01Î64
dx œ '0
1Î64
tan " x Èx
œ 23 x$Î#
2 #1
x(Î#
" Èx 2 55
x$ 3
Šx
x""Î#
2 105
$
7 sin x 2x x Ä 0 e 1
73. lim
74.
x Ä 0 Š2x
œ lim
5! #
á ‹ dx œ '0
1Î64
"Î'%
&
2$ x $ 3!
œ ˆ 3†28$ #
á‹
x Ä 0 Š2
2# x #!
á‹
75.
œ lim
tÄ0
76. lim
"‰ t#
œ
# " 2 Š 4! t6! á‹ # Š1 2t4! á‹
Š sinh h ‹ cos h
#
h# Π#!
$
œ lim
h% 5!
# lim t # 2 2 cos t t Ä 0 2t (1 cos t)
" cos# z z Ä 0 ln (1 z) sin z
77. lim
2 105†8"&
á ‰ ¸ 0.0013020379
œ
7 #
$
&
$
&
2 Š )3! )5! á‹
œ lim
$
&
Š )3! )5! á‹
)Ä0
œ lim
h% 4!
%
t t# 2 2 Œ1 t# 4! á
t Ä 0 2t# Š1 1
t# #
t% 4!
á‹
%
t' 6!
t 2 Π4!
œ lim
tÄ0
Št%
2t' 4!
á á‹
" 1#
œ
Œ1
h# 3!
%
#
%
h5! á Œ1 h#! h4! á h#
h' 6!
h' 7!
á
h#
hÄ0
2 55†8""
œ2
hÄ0
h# 3!
á‹
) Š) )3! )5! á‹
œ lim
h#
hÄ0
2 21†8(
%
2$ x # 3!
#
lim ˆ " t Ä 0 # 2 cos t
ˆx"Î# 3" x&Î# 5" x*Î# 7" x"$Î# á ‰ dx
7 Š1 x3! x5! á‹
œ lim
$
)Ä0
Š 3!" )5! á‹
)Ä0
2# x # 2!
x( 7
Š1 ) )#! )3! á‹ Š1 ) )#! )3! á‹ 2)
œ lim )#
x"&Î# á ‘ !
#
) c) lim e )e sin)2) )Ä0
x& 5
7 Šx x3! x5! á‹
œ lim
2 Š 3!"
œ lim
hÄ0
1 Š1 z#
œ lim
#
z% 3
Š #"!
" 3!
h# 5!
$
h# 4!
h% 6!
h% 7!
á‹ œ
" 3
%
á‹
$
&
z Ä 0 Šz z# z3 á‹ Šz z3! z5! á‹
Šz# z3 á‹
œ lim
#
$
%
z Ä 0 Š z# 2z3 z4 á‹
#
Š1 z3 á‹
œ lim
#
z z Ä 0 Š "# 2z 3 4 á‹
y# cos y cosh y yÄ0
78. lim
"
œ lim
œ 2 y#
œ lim
y Ä 0 Œ1
y Ä 0 Œ1 2y% á 6!
y# #
y% 4!
y' 6!
á Œ1
xÄ0
Ê
r x#
3 x#
r x#
y% 4!
y' 6!
á
y#
œ lim
y Ä 0 Œ
2y# #
'
2y6! á
œ 1
$
79. lim ˆ sinx$3x
y# #!
s‰ œ lim – xÄ0
œ 0 and s
9 #
80. The approximation sin x ¸
&
(3x) Š3x (3x) 6 120 á‹
x$
œ 0 Ê r œ 3 and s œ 6x 6 x#
r x#
s— œ lim Š x3# xÄ0
9 #
81x# 40
á
9 #
is better than sin x ¸ x.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
r x#
s‹ œ 0
Chapter 10 Practice Exercises nb1
(3n 1)(3n 2)x â(2n) ¸ 3n 2 ¸ 1 Ê kxk 81. n lim † #†52†8†4â†6(3n ¹ 2†5†8#â †4†6â(2n)(2n 2) 1)xn ¹ 1 Ê kxk n lim Ä_ Ä _ 2n 2 Ê the radius of convergence is 23
2 3
nb1
(2n1)(2n3)(x1) ¸ 2n 3 ¸ 1 Ê kxk 82. n lim † 34††59††714ââ(2n(5n1)x1)n ¹ 1 Ê kxk n lim ¹ 3†5†47†â 9†14â(5n1)(5n4) Ä_ Ä _ 5n 4 Ê the radius of convergence is 52 n
"‰ k#
83. ! ln ˆ1 kœ2
n
n
kœ2
kœ2
5 2
œ ! ln ˆ1 k" ‰ ln ˆ1 k" ‰‘ œ ! cln (k 1) ln k ln (k 1) ln kd
œ cln 3 ln 2 ln 1 ln 2d cln 4 ln 3 ln 2 ln 3d cln 5 ln 4 ln 3 ln 4d cln 6 ln 5 ln 4 ln 5d á cln (n 1) ln n ln (n 1) ln nd œ cln 1 ln 2d cln (n 1) ln nd after cancellation n
"‰ k#
Ê ! ln ˆ1 kœ2
n
84. ! kœ2
" k# 1 -
ˆ n " 1
_
Ê !
kœ2
" #
œ
n
!ˆ kœ2
" ‰‘ n1
" k # 1
_
1 ‰ œ ln ˆ n2n Ê ! ln ˆ1
"‰ k#
kœ2
" k1
œ
" #
" ‰ k1
ˆ 1"
" #
"ˆ3 œ n lim Ä_ # 2
œ
1 n
" n
" #
1‰ œ n lim ln ˆ n 2n œ ln Ä_
" ‰ n1
1 ‰ n1
œ
œ
" #
ˆ #3
" n
" ‰ n1
œ
" #
2(n 1) 2n ’ 3n(n 1)2n(n “œ 1)
1†4†7â(3n 2) (3n)!
nœ1 _
dy dx
_
œ!
nœ1 _
1†4†7â(3n 2) (3n 1)!
x3nc1
1†4†7â(3n5) (3n3)!
x3nc2
(3n ") (3n 1)(3n 2)(3n 3)
1†4†7â(3n 2) (3n 2)!
x3nc2 œ x !
œ x Œ1 !
1†4†7â(3n 2) (3n)!
x œ xy 0 Ê a œ 1 and b œ 0
x# 1x
œ x# x# (x) x# (x)# x# (x)$ á œ x# x$ x% x& á œ ! (1)n xn which
Ê
d# y dx#
œ!
nœ1 _
nœ1
86. (a)
x3n Ê
3n# n 2 4n(n 1)
3 4
3nb3
_
is the sum
ˆ 11 3" ‰ ˆ #" 4" ‰ ˆ 3" 5" ‰ ˆ 4" 6" ‰ á ˆ n " # n" ‰
1)x $ 85. (a) n lim † 1†4†7â(3n)! ¹ 1†4†7â(3n(3n2)(3n 3)! (3n 2)x3n ¹ 1 Ê kx k n lim Ä_ Ä_ œ kx$ k † 0 1 Ê the radius of convergence is _
(b) y œ 1 !
" #
œ
x# 1 (x)
nœ2
3n
_
nœ2
converges absolutely for kxk 1 _
_
nœ2
nœ2
(b) x œ 1 Ê ! (1)n xn œ ! (1)n which diverges _
_
87. Yes, the series ! an bn converges as we now show. Since ! an converges it follows that an Ä 0 Ê an 1 nœ1
nœ1
_
_
nœ1
nœ1
for n some index N Ê an bn bn for n N Ê ! an bn converges by the Direct Comparison Test with ! bn _
88. No, the series ! an bn might diverge (as it would if an and bn both equaled n) or it might converge (as it would if nœ1
an and bn both equaled "n ). _
_
nœ1
kœ1
!(xk1 xk ) œ lim (xn1 x" ) œ lim (xn1 ) x" Ê both the series and 89. ! (xn1 xn ) œ n lim Ä_ nÄ_ nÄ_ sequence must either converge or diverge.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
641
642
Chapter 10 Infinite Sequences and Series Š 1 banan ‹
90. It converges by the Limit Comparison Test since n lim Ä_
an
" 1 a n
œ n lim Ä_
_
œ 1 because ! an converges nœ1
and so an Ä 0. _
91. ! nœ1
œ a"
an n
ˆ "9 92. an œ œ
" 10
" ln n
" ln #
a# #
" 11
á
a$ 3
a% 4
á a" ˆ "# ‰ a# ˆ 3" 4" ‰ a% ˆ 5"
" ‰ 16 a"'
" #
á
" #
" 3
" 7
8" ‰ a)
(a# a% a) a"' á ) which is a divergent series
for n 2 Ê a# a$ a% á , and
ˆ1
" 6
" ln #
" ln 4
_
á ‰ which diverges so that 1 !
nœ2
" ln 8
" n ln n
á œ
" ln #
" # ln 2
" 3 ln 2
á
diverges by the Integral Test.
CHAPTER 10 ADDITIONAL AND ADVANCED EXERCISES 1. converges since
" (3n #)Ð2n 1ÑÎ2
" Š $Î# ‹ n
lim nÄ_
Š
"
(3n
‹ 2)$Î#
" (3n 2)$Î#
$
1$ 192 ‹
œ
nœ1
" (3n 2)$Î#
converges by the Limit Comparison Test:
ˆ 3n n 2 ‰$Î# œ 3$Î# œ n lim Ä_
2. converges by the Integral Test: 1 œ Š 24
_
and !
'1_ atan" xb# x dx1 œ #
" x b$
lim ’ atan 3
bÄ_
" b b$
b
tan “ œ lim ’ a 3
bÄ_
"
1$ 192 “
71 $ 192
c2n
e 3. diverges by the nth-Term Test since n lim a œ n lim (1)n tanh n œ lim (1)n Š 11 (1)n ec2n ‹ œ n lim Ä_ n Ä_ Ä_ bÄ_ does not exist
4. converges by the Direct Comparison Test: n! nn Ê ln (n!) n ln (n) Ê Ê logn (n!) n Ê
logn (n!) n$
" n#
12 (3)(5)(4)#
, a% œ ˆ 53††64 ‰ ˆ 42††53 ‰ ˆ 31††42 ‰ œ
given series and
12 (n 1)(n 3)(n 2)#
6. converges by the Ratio Test: n lim Ä_
"2 (4)(6)(5)#
12 n%
12 (1)(3)(2)#
, a# œ
_
,á Ê 1!
nœ1
1 †2 3 †4
_
nœ1
" 32nb1
cos x œ
" #
È3 #
1 3
12 (2)(4)(3)#
, a$ œ ˆ 42††53 ‰ ˆ 31††42 ‰
represents the
, which is the nth-term of a convergent p-series
anb1 an
œ n lim Ä_
n (n 1)(n 1)
_
and ! nœ1
2n 32n
œ01
È3 1#
ˆx
n n n È 2È
È3 ww # ,f 1 ‰$ á 3
Ê f ˆ 13 ‰ œ 0.5, f w ˆ 13 ‰ œ
ˆx 13 ‰ 4" ˆx 13 ‰#
" 1L
Ê L# L 1 œ 0 Ê L œ
; the first subseries is a convergent geometric series and the
n 2n É second converges by the Root Test: n lim 32n œ n lim Ä_ Ä_
9. f(x) œ cos x with a œ
œ
12 (n 1)(n 3)(n 2)#
7. diverges by the nth-Term Test since if an Ä L as n Ä _, then L œ 8. Split the given series into !
n
, which is the nth-term of a convergent p-series
5. converges by the Direct Comparison Test: a" œ 1 œ œ
ln (n!) ln (n)
9
œ
" †1 9
œ
" 9
1
ˆ 13 ‰ œ 0.5, f www ˆ 13 ‰ œ
È3 #
, f Ð4Ñ ˆ 13 ‰ œ 0.5;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 „ È 5 #
Á0
Chapter 10 Additional and Advanced Exercises
643
10. f(x) œ sin x with a œ 21 Ê f(21) œ 0, f w (21) œ 1, f ww (21) œ 0, f www (21) œ 1, f Ð4Ñ (21) œ 0, f Ð5Ñ (21) œ 1, f Ð6Ñ (21) œ 0, f Ð7Ñ (21) œ 1; sin x œ (x 21) x# #!
11. ex œ 1 x
x$ 3!
(x 21)$ 3!
(x 21)& 5!
(x 21)( 7!
á
á with a œ 0
12. f(x) œ ln x with a œ 1 Ê f(1) œ 0, f w (1) œ 1, f ww (1) œ 1,f www (1) œ 2, f Ð4Ñ (1) œ 6; (x 1)# #
ln x œ (x 1)
(x 1)$ 3
(x 1)% 4
á
13. f(x) œ cos x with a œ 221 Ê f(221) œ 1, f w (221) œ 0, f ww (221) œ 1, f www (221) œ 0, f Ð4Ñ (221) œ ", f Ð5Ñ (221) œ 0, f Ð6Ñ (221) œ 1; cos x œ 1 "# (x 221)# 4!" (x 221)% 6!" (x 221)' á 14. f(x) œ tan" x with a œ 1 Ê f(1) œ tan
"
1 4
xœ
(x 1) 2
(x 1)# 4
1 4
(x 1)$ 12
, f w (1) œ
" #
, f ww (1) œ "# , f www (1) œ
ˆ ba ‰n ln ˆ ba ‰ a n nÄ_ ˆ b ‰ 1
16. 1
2 10
3 10#
_
œ1!
nœ0
œ1
200 999
n1
17. sn œ !
kœ0
7 10$
'kkb1
2 10%
_
2
103n1
!
30 999
dx 1 x#
7 999
3 10&
7 10'
_
3
103n2
nœ0
0†ln ˆ ba ‰ 01
œ ln b
!
999237 999
œ
Ê sn œ '0
1
_
á œ1!
nœ1
7
œ
1În
n
lnˆˆ ba ‰ 1‰ n nÄ_
Ê n lim c œ ln b lim Ä_ n
œ ln b since 0 a b. Thus, n lim c œ eln b œ b. Ä_ n
103n3
nœ0
;
á
n 15. Yes, the sequence converges: cn œ aan bn b1În Ê cn œ b ˆˆ ba ‰ 1‰
œ ln b lim
" #
œ1
2 ‰ ˆ 10
2 103n
$ 1ˆ " ‰
10
_
2
!
nœ1
Š 103# ‹
$ 1ˆ " ‰
3 103n
_
1
!
nœ1
Š 107$ ‹
"‰ 1 ˆ 10
10
7 103n
$
412 333
dx 1 x#
'1
2
dx 1 x#
Ê n lim s œ n lim atan" n tan" 0b œ Ä_ n Ä_ nb1
á 'nc1 1 dxx# Ê sn œ '0 n
n
dx 1 x#
1 #
#
1) 18. n lim † (n 1)(2x ¹ uunbn 1 ¹ œ n lim ¹ (n 1)x ¹ œ n lim ¹ x † (n 1) ¹ œ ¸ 2x x 1 ¸ 1 nxn Ä_ Ä _ (n 2)(2x 1)n 1 Ä _ 2x 1 n(n 2) Ê kxk k2x 1k ; if x 0, kxk k2x 1k Ê x 2x 1 Ê x 1; if "# x 0, kxk k2x 1k n
Ê x 2x 1 Ê 3x 1 Ê x "3 ; if x #" , kxk k2x 1k Ê x 2x 1 Ê x 1. Therefore,
the series converges absolutely for x 1 and x "3 .
19. (a) No, the limit does not appear to depend on the value of the constant a (b) Yes, the limit depends on the value of b (c) s œ Š1 œ n lim Ä_
cos ˆ na ‰ n n ‹ a n
lim Š1
nÄ_
Ê ln s œ
sin ˆ na ‰ cos ˆ na ‰ 1
cos ˆ na ‰ n
cos ˆ na ‰ n bn ‹
œ
ln Œ1
cos ˆ na ‰ n
ˆ "n ‰ 01 10
Ê n lim ln s œ Ä_
" cos ˆ na ‰ Œ n
Š
a a a n sin ˆ n ‰ cos ˆ n ‰ n#
"‹
n#
œ e1Îb 1În
_
" #
1
œ 1 Ê n lim s œ e" ¸ 0.3678794412; similarly, Ä_
a n ‰n 20. ! an converges Ê n lim a œ 0; n lim ’ˆ 1 sin “ # Ä_ n Ä_ nœ1
œ
an ‰ ˆ 1sin œ n lim œ # Ä_
Ä_
1sin Šnlim an ‹ #
Ê the series converges by the nth-Root Test
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
1sin 0 #
644
Chapter 10 Infinite Sequences and Series nb1 nb1
21. n lim ¹ uunbn 1 ¹ 1 Ê n lim ¹b x † Ä_ Ä _ ln (n 1)
ln n bn xn ¹
1 Ê kbxk 1 Ê "b x
" b
œ5 Ê bœ „
" 5
22. A polynomial has only a finite number of nonzero terms in its Taylor series, but the functions sin x, ln x and ex have infinitely many nonzero terms in their Taylor expansions. sin (ax) sin xx x$ xÄ0
Šax
œ lim
23. lim
a$ x$ 3!
á‹ Šx
xÄ0 xÄ0
24.
25. (a) (b)
26.
un unb1
un unb1
œ
(n 1)# n#
un unb1
œ
n1 n
œ
2n(2n 1) (2n 1)#
Ê Cœ
3 #
Œ1
œ 1 Ê lim
œ1
œ1
œ
a# x# #
1 n
œ lim ’ a x# 2
4n# 2n 4n# 4n 1
á b
" n#
nœ1 _
" n
nœ1
n
5n# 4n# 4n 1
œ
Š4
5
4 n
_
_
_
nœ1
nœ1
nœ1
&
Š a5!
" # 5! ‹ x
á“
a# 4
a# x# 48
á ‹ œ 1
converges
diverges
œ1
5 4n# 4n 1
" 3!
œ 76
xÄ0
_
Š 64 ‹
" 3!
a$ 3!
œ 1 Ê lim Š "2x#b
Ê C œ 2 1 and !
œ1
1 and kf(n)k œ
œ 23!
Ê C œ 1 Ÿ 1 and !
0 n#
xÄ0
$
#x#
" n#
2 n
a% x% 4!
xÄ0
Ê b œ 1 and a œ „ 2
á‹ x
sin 2x sin x x x$
is finite if a 2 œ 0 Ê a œ 2; lim
lim cos#axx# b xÄ0
x$ 3!
x$
Š 3# ‹ n
5n#
– Š4n# c 4n b 1‹ — n#
after long division
_
" ‹ n#
Ÿ 5 Ê ! un converges by Raabe's Test nœ1
27. (a) ! an œ L Ê an# Ÿ an ! an œ an L Ê ! an# converges by the Direct Comparison Test (b) converges by the Limit Comparison Test: n lim Ä_
an Š1c an ‹
an
œ n lim Ä_
" 1 an
_
œ 1 since ! an converges and nœ1
therefore x lim a œ0 Ä_ n 28. If 0 an 1 then kln (1 an )k œ ln (1 an ) œ an
a#n #
a$n 3
á an an# an$ á œ
an 1 an
,
a positive term of a convergent series, by the Limit Comparison Test and Exercise 27b _
29. (1 x)" œ 1 ! xn where kxk 1 Ê nœ1
#
" (1x)#
$
4 œ 1 2 ˆ "# ‰ 3 ˆ "# ‰ 4 ˆ "# ‰ á n ˆ "# ‰ _
30. (a) ! xn1 œ nœ1
_
Ê !
nœ1 _
(b) x œ !
nœ1
x# 1 x
n(n 1) xn n(n ") xn
_
Ê ! (n 1)xn œ nœ1
œ
2 x
$
Š1 "x ‹
Ê xœ
œ
2x# (x 1)$
2x# (x 1)$
2xx# (1x)#
œ
n 1
d dx
_
(1 x)" œ ! nxn1 and when x œ nœ1
" #
we have
á _
Ê ! n(n 1)xn1 œ nœ1
2 (1x)$
_
Ê ! n(n 1)xn œ nœ1
, kxk 1
Ê x$ 3x# x 1 œ 0 Ê x œ 1 Š1
È57 "Î$ 9 ‹
Š1
¸ 2.769292, using a CAS or calculator 31. (a)
" (1x)#
œ
d dx
ˆ 1" x ‰ œ
d dx
(b) from part (a) we have
_
a1 x x# x$ á b œ 1 2x 3x# 4x$ á œ ! nxn1
_
! n ˆ 5 ‰ n 1 6
nœ1
2x (1x)$
nœ1
ˆ "6 ‰ œ ˆ 6" ‰ ’
2
" “ 1 ˆ 56 ‰
œ6
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
È57 "Î$ 9 ‹
Chapter 10 Additional and Advanced Exercises _
(c) from part (a) we have ! npn1 q œ nœ1
_
_
32. (a) ! pk œ ! 2k œ kœ1
kœ1
ˆ "# ‰
1ˆ "# ‰
q (1 p)#
œ
q q#
" q
œ
_
_
kœ1
kœ1
œ 1 and E(x) œ ! kpk œ ! k2k œ
" #
_
! k21k œ ˆ " ‰ #
kœ1
" 1ˆ "# ‰‘#
œ2
by Exercise 31(a) _
_
kœ1
kœ1
(b) ! pk œ ! œ ˆ "6 ‰
" 1ˆ 56 ‰‘#
_
_
(c) ! pk œ ! kœ1
kœ1
_
" k1
œ!
kœ1
5kc1 6k
œ
_
5
e
" k(k1)
kœ1
6
_
œ ! ˆ k" kœ1
" ‰ k1
œ
lim ˆ1 kÄ_
C! e
kt!
" ‰ k1
ˆ1 e 1e kt!
kœ1
" 6
œ
_
_
kœ1
kœ1
nkt! ‰
Ê Rœ
lim R œ nÄ_ n ¸ 0.58195028;
(b) t! œ
C! ekt! 1 " 0.05
kœ1
e " a1 e "! b 1e " R R"! 0.0001 R
C! e kt! 1 e kt!
" e1 ¸ 0.58197671; R R"! ¸ 0.00002643 Ê Þ1n e Þ1 ˆ1 e Þ1n ‰ R Rn œ , # œ #" ˆ eÞ1 " 1 ‰ ¸ 4.7541659; Rn R# Ê 1eÞ1 e 1 ˆ #" ‰ ˆ eÞ1 " 1 ‰ 1 e Þ1 n n Ê 1 enÎ10 "# Ê enÎ10 "# Ê 10 ln ˆ "# ‰ Ê 10 ln ˆ "# ‰ Ê n 6.93
34. (a) R œ
_
! k ˆ 5 ‰ k 1 6
œ 1 and E(x) œ ! kpk œ ! k Š k(k " 1) ‹
Ê R" œ e" ¸ 0.36787944 and R"! œ
Rœ
(c)
kœ1
5kc1 6k
, a divergent series so that E(x) does not exist
a1 e n b 1e "
1
_
œ6
33. (a) Rn œ C! ekt! C! e2kt! á C! enkt! œ (b) Rn œ
_
! ˆ 5 ‰k œ ˆ " ‰ ’ ˆ 6 ‰5 “ œ 1 and E(x) œ ! kpk œ ! k 6 5 1ˆ ‰
" 5
Ê Rekt! œ R C! œ CH Ê ekt! œ
CH CL
Ê t! œ
" k
œ
C! ekt! 1
Ê nœ7
ln Š CCHL ‹
ln e œ 20 hrs
(c) Give an initial dose that produces a concentration of 2 mg/ml followed every t! œ
" 0.0#
2 ‰ ln ˆ 0.5 ¸ 69.31 hrs
by a dose that raises the concentration by 1.5 mg/ml " 0.1 ‰ ‰ (d) t! œ 0.2 ln ˆ 0.03 œ 5 ln ˆ 10 3 ¸ 6 hrs
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
645
646
Chapter 10 Infinite Sequences and Series
NOTES:
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES 11.1 PARAMETRIZATIONS OF PLANE CURVES 1. x œ 3t, y œ 9t# , _ t _ Ê y œ x#
2. x œ Èt , y œ t, t 0 Ê x œ Èy or y œ x# , x Ÿ 0
3. x œ 2t 5, y œ 4t 7, _ t _ Ê x 5 œ 2t Ê 2(x 5) œ 4t Ê y œ 2(x 5) 7 Ê y œ 2x 3
5. x œ cos 2t, y œ sin 2t, 0 Ÿ t Ÿ 1 Ê cos# 2t sin# 2t œ 1 Ê x# y# œ 1
4. x œ 3 3t, y œ 2t, 0 Ÿ t Ÿ 1 Ê y# œ t Ê x œ 3 3 ˆ y# ‰ Ê 2x œ 6 3y Ê y œ 2 23 x, ! Ÿ x Ÿ $
6. x œ cos (1 t), y œ sin (1 t), 0 Ÿ t Ÿ 1 Ê cos# (1 t) sin# (1 t) œ 1 Ê x# y# œ 1, y !
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
648
Chapter 11 Parametric Equations and Polar Coordinates
7. x œ 4 cos t, y œ 2 sin t, 0 Ÿ t Ÿ 21 Ê
16 cos# t 16
4 sin# t 4
œ1 Ê
x# 16
9. x œ sin t, y œ cos 2t, 12 Ÿ t Ÿ
y# 4
8. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21 œ1
1 2
Ê y œ cos 2t œ 1 2sin# t Ê y œ 1 2x2
11. x œ t2 , y œ t6 2t4 , _ t _ 2 3
2 2
Ê y œ at b 2at b Ê y œ x3 2x2
13. x œ t, y œ È1 t# , 1 Ÿ t Ÿ 0 Ê y œ È1 x#
Ê
16 sin# t 16
25 cos# t 25
œ1 Ê
x# 16
y# #5
œ1
10. x œ 1 sin t, y œ cos t 2, 0 Ÿ t Ÿ 1 Ê sin# t cos# t œ 1 Ê ax 1b# ay 2b# œ 1
12. x œ
t t1,
Ê tœ
yœ
x x1
t2 t1,
1 t 1
Êyœ
2x 2x 1
14. x œ Èt 1, y œ Èt, t 0 Ê y# œ t Ê x œ Èy# 1, y 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.1 Parametrizations of Plane Curves 15. x œ sec# t 1, y œ tan t, 1# t Ê sec# t 1 œ tan# t Ê x œ y#
1 #
16. x œ sec t, y œ tan t, 1# t
1 # #
Ê sec# t tan# t œ 1 Ê x# y œ 1
17. x œ cosh t, y œ sinh t, _ 1 _ Ê cosh# t sinh# t œ 1 Ê x# y# œ 1
18. x œ 2 sinh t, y œ 2 cosh t, _ t _ Ê 4 cosh# t 4 sinh# t œ 4 Ê y# x# œ 4
19. (a) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21
20. (a) x œ a sin t, y œ b cos t,
(b) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21 (c) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41 (d) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41
649
1 #
ŸtŸ
51 #
(b) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 21 (c) x œ a sin t, y œ b cos t, 1# Ÿ t Ÿ 9#1 (d) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 41
21. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes through a"ß $b at t œ !. We determine a and b so that the line goes through a%ß "b when t œ ". Since % œ " a Ê a œ &. Since " œ $ b Ê b œ %. Therefore, one possible parameterization is x œ " &t, y œ $ %t, 0 Ÿ t Ÿ ". 22. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes through a"ß $b at t œ !. We determine a and b so that the line goes through a$ß #b when t œ ". Since $ œ " a Ê a œ %. Since # œ $ b Ê b œ &. Therefore, one possible parameterization is x œ " %t, y œ $ &t, 0 Ÿ t Ÿ ". 23. The lower half of the parabola is given by x œ y# " for y Ÿ !. Substituting t for y, we obtain one possible parameterization x œ t# ", y œ t, t Ÿ 0Þ 24. The vertex of the parabola is at a"ß "b, so the left half of the parabola is given by y œ x# #x for x Ÿ ". Substituting t for x, we obtain one possible parametrization: x œ t, y œ t# #t, t Ÿ ". 25. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a#ß $b for t œ ! and passes through a"ß "b at t œ ". Then x œ fatb, where fa!b œ # and fa"b œ ". Since slope œ ??xt œ "# "! œ $, x œ fatb œ $t # œ # $t. Also, y œ gatb, where ga!b œ $ and ga"b œ ". Since slope œ
?y ?t
œ
"3 "!
œ 4. y œ gatb œ %t $ œ $ %t.
One possible parameterization is: x œ # $t, y œ $ %t, t !.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
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Chapter 11 Parametric Equations and Polar Coordinates
26. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a"ß #b for t œ ! and passes through a!ß !b at t œ ". Then x œ fatb, where fa!b œ " and fa"b œ !. Since slope œ Since slope œ
?x ?t ?y ?t
œ œ
! a"b "! !# "! œ
œ ", x œ fatb œ "t a"b œ " t. Also, y œ gatb, where ga!b œ # and ga"b œ !. #. y œ gatb œ #t # œ # #t.
One possible parameterization is: x œ " t, y œ # #t, t !. 27. Since we only want the top half of a circle, y 0, so let x œ 2cos t, y œ 2lsin tl, 0 Ÿ t Ÿ 41 28. Since we want x to stay between 3 and 3, let x œ 3 sin t, then y œ a3 sin tb2 œ 9sin# t, thus x œ 3 sin t, y œ 9sin# t, 0Ÿt_ 29. x# y# œ a# Ê 2x 2y y# t# y# œ a# Ê y œ
dy dx œ 0 a È1t# and
Ê x
dy dy x dx œ y ; let t œ dx Ê œ È1att , _ t _
xy œ t Ê x œ yt. Substitution yields
30. In terms of ), parametric equations for the circle are x œ a cos ), y œ a sin ), 0 Ÿ ) 21. Since ) œ as , the arc length parametrizations are: x œ a cos as , y œ a sin as , and 0 Ÿ
s a
21 Ê 0 Ÿ s Ÿ 21a is the interval for s.
31. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, and from trigonometry we know that tan ) œ yx Ê y œ x tan ). The equation of the line through a0, 2b and a4, 0b is given by y œ 12 x 2. Thus x tan ) œ 12 x 2 Ê x œ
4 2 tan ) 1
and y œ
4 tan ) 2 tan ) 1
where 0 Ÿ ) 12 .
32. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, and from trigonometry we know that tan ) œ yx Ê y œ x tan ). Since y œ Èx Ê y2 œ x Ê ax tan )b2 œ x Ê x œ cot2 ) Ê y œ cot ) where 0 ) Ÿ 12 .
33. The equation of the circle is given by ax 2b2 y2 œ 1. Drop a vertical line from the point ax, yb on the circle to the x-axis, then ) is an angle in a right triangle. So that we can start at a1, 0b and rotate in a clockwise direction, let x œ 2 cos ), y œ sin ), 0 Ÿ ) Ÿ 21. 34. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, whose height is y and whose base is x 2. By trigonometry we have tan ) œ x y 2 Ê y œ ax 2b tan ). The equation of the circle is given by x2 y2 œ 1 Ê x2 aax 2btan )b2 œ 1 Ê x2 sec2 ) 4x tan2 ) 4tan2 ) 1 œ 0. Solving for x we obtain xœ
4tan2 ) „ Éa4tan2 )b2 4 sec2 ) a4tan2 ) 1b 2 sec2 )
œ
4tan2 ) „ 2È1 3tan2 ) 2 sec2 )
œ 2sin2 ) „ cos )Ècos2 ) 3sin2 )
œ 2 2cos2 ) „ cos )È4cos2 ) 3 and y œ Š2 2cos2 ) „ cos )È4cos2 ) 3 2‹ tan ) œ 2sin ) cos ) „ sin )È4cos2 ) 3. Since we only need to go from a1, 0b to a0, 1b, let x œ 2 2cos2 ) cos )È4cos2 ) 3, y œ 2sin ) cos ) sin )È4cos2 ) 3, 0 Ÿ ) Ÿ tan1 ˆ 1 ‰. 2
To obtain the upper limit for ), note that x œ 0 and y œ 1, using y œ ax 2b tan ) Ê 1 œ 2 tan ) Ê ) œ tan1 ˆ 12 ‰. 35. Extend the vertical line through A to the x-axis and let C be the point of intersection. Then OC œ AQ œ x 2 2 and tan t œ OC œ x2 Ê x œ tan2 t œ 2 cot t; sin t œ OA Ê OA œ sin2 t ; and (AB)(OA) œ (AQ)# Ê AB ˆ sin2 t ‰ œ x# # 2 2 2 sin t Ê AB ˆ sin t ‰ œ ˆ tan t ‰ Ê AB œ tan# t . Next y œ 2 AB sin t Ê y œ 2 ˆ 2tansin# tt ‰ sin t œ 2
2 sin# t tan# t
œ 2 2 cos# t œ 2 sin# t. Therefore let x œ 2 cot t and y œ 2 sin# t, 0 t 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.1 Parametrizations of Plane Curves 36. Arc PF œ Arc AF since each is the distance rolled and Arc PF œ nFCP Ê Arc PF œ b(nFCP); ArcaAF œ ) b Ê Arc AF œ a) Ê a) œ b(nFCP) Ê nFCP œ nOCG œ
1 #
a b
);
); nOCG œ nOCP nPCE œ nOCP ˆ 1# !‰ . Now nOCP œ 1 nFCP œ 1 ba ). Thus nOCG œ 1 ba ) œ 1 ba )
1 #
1 #
! Ê
! Ê ! œ 1 ba ) ) œ 1
1 # ) ˆ ab b )‰ .
Then x œ OG BG œ OG PE œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1 œ (a b) cos ) b cos ˆ a b b )‰ . Also y œ EG œ CG CE œ (a b) sin ) b sin !
ab b
)‰
œ (a b) sin ) b sin ˆ1 a b b )‰ œ (a b) sin ) b sin ˆ a b b )‰ . Therefore x œ (a b) cos ) b cos ˆ a b b )‰ and y œ (a b) sin ) b sin ˆ a b b )‰ . If b œ 4a , then x œ ˆa 4a ‰ cos ) œ œ œ œ
3a 4 3a 4 3a 4 3a 4
cos )
œ œ œ
3a 4 3a 4 3a 4 3a 4
cos 3) œ
3a 4
cos Š
a ˆ 4a ‰ ˆ 4a ‰
)‹
cos ) 4a (cos ) cos 2) sin ) sin 2))
cos ) a(cos )) acos# ) sin# )b (sin ))(2 sin ) cos ))b a 2a # # 4 cos ) sin ) 4 sin ) cos ) # $ ) cos$ ) 3a 4 (cos )) a1 cos )b œ a cos ); a a ˆ4‰ a‰ a 3a a 3a 4 sin ) 4 sin Š ˆ 4a ‰ )‹ œ 4 sin ) 4 sin 3) œ 4
cos ) cos
y œ ˆa œ
a 4 a 4 a 4 a 4
a 4
cos$ )
sin ) 4a (sin ) cos 2) cos ) sin 2))
sin ) 4a a(sin )) acos# ) sin# )b (cos ))(2 sin ) cos ))b sin ) sin ) sin )
a 4 3a 4 3a 4
sin ) cos# ) sin ) cos# )
a 4 a 4 #
sin$ )
2a 4
cos# ) sin )
sin$ )
(sin )) a1 sin )b
a 4
sin$ ) œ a sin$ ).
37. Draw line AM in the figure and note that nAMO is a right angle since it is an inscribed angle which spans the diameter of a circle. Then AN# œ MN# AM# . Now, OA œ a, AN AM a œ tan t, and a œ sin t. Next MN œ OP Ê OP# œ AN# AM# œ a# tan# t a# sin# t Ê OP œ Èa# tan# t a# sin# t œ (a sin t)Èsec# t 1 œ x œ OP sin t œ
a sin$ t cos t œ #
a sin# t cos t #
. In triangle BPO,
a sin t tan t and
y œ OP cos t œ a sin t Ê x œ a sin# t tan t and y œ a sin# t.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
651
652
Chapter 11 Parametric Equations and Polar Coordinates
38. Let the x-axis be the line the wheel rolls along with the y-axis through a low point of the trochoid (see the accompanying figure).
Let ) denote the angle through which the wheel turns. Then h œ a) and k œ a. Next introduce xw yw -axes parallel to the xy-axes and having their origin at the center C of the wheel. Then xw œ b cos ! and yw œ b sin !, where ! œ 3#1 ). It follows that xw œ b cos ˆ 3#1 )‰ œ b sin ) and yw œ b sin ˆ 3#1 )‰
œ b cos ) Ê x œ h xw œ a) b sin ) and y œ k yw œ a b cos ) are parametric equations of the trochoid.
# # # 39. D œ É(x 2)# ˆy "# ‰ Ê D# œ (x 2)# ˆy "# ‰ œ (t 2)# ˆt# "# ‰ Ê D# œ t% 4t
Ê
d aD # b dt
17 4
œ 4t$ 4 œ 0 Ê t œ 1. The second derivative is always positive for t Á 0 Ê t œ 1 gives a local
minimum for D# (and hence D) which is an absolute minimum since it is the only extremum Ê the closest point on the parabola is (1ß 1). # # 40. D œ Ɉ2 cos t 34 ‰ (sin t 0)# Ê D# œ ˆ2 cos t 34 ‰ sin# t Ê
d aD # b dt
œ 2 ˆ2 cos t 34 ‰ (2 sin t) 2 sin t cos t œ (2 sin t) ˆ3 cos t 3# ‰ œ 0 Ê 2 sin t œ 0 or 3 cos t Ê t œ 0, 1 or t œ
1 3
,
51 3
. Now
#
#
d aD b dt#
#
#
œ 6 cos t 3 cos t 6 sin t so that
#
#
d aD b dt#
3 #
œ0
(0) œ 3 Ê relative
# # # # maximum, d dtaD# b (1) œ 9 Ê relative maximum, d dtaD# b ˆ 13 ‰ œ 92 Ê relative minimum, and d # aD # b ˆ 5 1 ‰ œ 9# Ê relative minimum. Therefore both t œ 13 and t œ 531 give points on the ellipse dt# 3 È È the point ˆ 34 ß !‰ Ê Š1ß #3 ‹ and Š1ß #3 ‹ are the desired points.
41. (a)
(b)
(c)
42. (a)
(b)
(c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
closest to
Section 11.1 Parametrizations of Plane Curves 43.
44. (a)
(b)
(c)
45. (a)
(b)
46. (a)
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
653
654
Chapter 11 Parametric Equations and Polar Coordinates
47. (a)
(b)
(c)
48. (a)
(b)
(c)
(d)
11.2 CALCULUS WITH PARAMETRIC CURVES 1. t œ Ê
1 4 Ê dy dx ¹ tœ 1
x œ 2 cos
d# y dx#
dyw /dt dx/dt
œ cot
1 4
1 4
œ È2, y œ 2 sin
1 4
œ È2;
dx dt
œ 2 sin t,
dy dt
œ 2 cos t Ê
œ 1; tangent line is y È2 œ 1 Šx È2‹ or y œ x
dy dx
œ
œ
dy/dt dx/dt w È 2 2 ; dy dt
2 cos t 2 sin t
œ cot t
œ csc# t
4
Ê
œ
œ
csc# t 2 sin t
" œ 2 sin $t Ê
d# y dx# ¹ tœ 1
œ È 2
4
2. t œ "6 Ê x œ sin ˆ21 ˆ 6" ‰‰ œ sin ˆ 13 ‰ œ dy dt
œ 21 sin 21t Ê
tangent line is y œ cos$"21t Ê
" #
dy dx
œ
21 sin 21t 21 cos 21t
œ È3 ’x Š
d# y dx# ¹ tœc 1
È3 #
œ tan 21t Ê
, y œ cos ˆ21 ˆ 6" ‰‰ œ cos ˆ 13 ‰ œ dy dx ¹ tœc 1
œ tan ˆ21 ˆ
" ‰‰ 6
" #
;
dx dt
œ tan ˆ
œ 21 cos 21t,
1‰ 3
œ È3;
6
È3 # ‹“
or y œ È3x 2;
dyw dt
œ 21 sec# 21t Ê
d# y dx#
œ
21 sec# 21t 21 cos 21t
œ 8
6
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.2 Calculus With Parametric Curves 1 4
3. t œ
œ "# tan t Ê dyw dt
1 4
Ê x œ 4 sin
21 3 Ê dy dx ¹ tœ 21
4. t œ Ê
œ "# tan
dy dx ¹ tœ 1 4
d# y dx#
œ "# sec# t Ê
dyw /dt dx/dt
œ
1 4
"# sec# t 4 cos t
œ
œ 4 cos t,
dx dt
œ 2 sin t Ê
dy dt
dy dx
œ
dy/dt dx/dt
œ
2 sin t 4 cos t
d# y dx# ¹ tœ 1
" œ 8 cos $t Ê
œ
È2 4
4
21 3
x œ cos
È3 dx 21 3 œ # ; dt È3 È 3 x # ‹œ
œ "# , y œ È3 cos
œ È3 ; tangent line is y Š
d# y dx# ¹ tœ 21
œ È2;
œ "# ; tangent line is y È2 œ "# Šx 2È2‹ or y œ "# x 2È2 ;
3
Ê
1 4
œ 2È2, y œ 2 cos
œ sin t,
dy dt
œ È3 sin t Ê dyw dt
ˆ "# ‰‘ or y œ È3 x;
dy dx
œ0 Ê
œ
È3 sin t sin t
œ È3
d# y dx#
œ
œ0
0 sin t
œ0
3
5. t œ
Ê xœ
1 4
y
" #
1 4
,yœ
" #
œ 1,
dx dt
;
dy dt
œ 1 † ˆx 4" ‰ or y œ x 4" ;
" #Èt
œ
dyw dt
Ê
œ
dy dx
œ
dy/dt dx/dt d# y dx#
œ 4" t$Î# Ê
œ 1; tangent line is d# y dx# ¹ tœ 1
œ 4" t$Î# Ê
œ 2
4
dy dx
œ
sec# t 2 sec# t tan t
" 2 tan t
œ
œ
" #
cot t Ê
d# y dx# ¹ tœc 1
œ
œ
dy dx ¹ tœc 1 4
y (1) œ "# (x 1) or y œ "# x "# ; Ê
" #É "4
œ
dy dx ¹ tœ 1 4
dyw /dt dx/dt
œ
6. t œ 14 Ê x œ sec# ˆ 14 ‰ 1 œ 1, y œ tan ˆ 14 ‰ œ 1; Ê
Ê
1 2È t
dyw dt
" #
œ 2 sec# t tan t,
dx dt
œ sec# t
dy dt
cot ˆ 14 ‰ œ #" ; tangent line is d# y dx#
œ "# csc# t Ê
œ
"# csc# t 2 sec# t tan t
œ "4 cot$ t
" 4
4
7. t œ œ
1 6
Ê x œ sec
sec# t sec t tan t
1 6
y œ tan
1 6
œ
dy dx ¹ tœ 1
œ csc
1 6
œ 2; tangent line is y
d# y dx#
dyw /dt dx/dt
œ
œ csc t Ê
2 È3 ,
" È3
;
dx dt
œ sec t tan t,
6
dyw dt
œ csc t cot t Ê
œ
csc t cot t sec t tan t
œ
dy dt
œ sec# t Ê
" È3
œ 2 Šx
d# y dx# ¹ tœ 1
œ cot$ t Ê
œ
dy dx
2 È3 ‹
dy/dt dx/dt
or y œ 2x È3 ;
œ 3È3
6
8. t œ 3 Ê x œ È3 1 œ 2, y œ È3(3) œ 3; È
œ 3 Èt3t 1 œ dyw dt
Ê
œ
dy dx ¹ tœ3
3 È 3 1 È3(3)
œ
dx dt
œ 4t,
y 1 œ 1 † (x 5) or y œ x 4; 10. t œ 1 Ê x œ 1, y œ 2;
œ
dx dt
dy dt
œ
Ê xœ
sin t 1 cos t
d# y dx#
œ
dy/dt dx/dt
œ
Ê
3 2tÈ3t Èt1
dyw dt
œ t"# ,
œ 4t$ Ê
œ 2t Ê dy dt
y (2) œ 1(x 1) or y œ x 1;
1 3
œ
3 #
(3t)"Î# Ê
dy dx
œ
ˆ 3# ‰ (3t) "Î# ˆ "# ‰ (t1) "Î#
Š 2tÈ3t3Èt 1 ‹ Š 2Èt1 1 ‹
œ tÈ33t
œ 3"
9. t œ 1 Ê x œ 5, y œ 1;
11. t œ
dy dt
œ 2; tangent line is y 3 œ 2[x (2)] or y œ 2x 1;
È3t 3 (t 1) "Î# ‘3Èt 1 3 (3t) "Î# ‘ # # 3t
d# y dx# ¹ tœ3
œ "# (t 1)"Î# ,
dx dt
Ê
1 3
sin
dy dx ¹ tœ 1 3
œ
1 3
œ
1 3
sin ˆ 13 ‰ 1cos ˆ 13 ‰
œ
dyw dt
" t
dy dx
d# y dx#
Ê
œ
œ
dyw /dt dx/dt
dy dx
œ
œ 1 Ê
œ
ˆ "t ‰ Š t"# ‹
d# y dx#
œ
4t$ 4t 2t 4t
œ t# Ê
œ
" #
Ê
œ t Ê
1 Š t"# ‹
dy dx ¹ tœc1
d# y dx# ¹ tœc1 dy dx ¹ tœ1
œ t# Ê
œ (1)# œ 1; tangent line is
œ
" #
œ 1; tangent line is
d# y dx# ¹ tœ1
œ1
È3 #
dy , y œ 1 cos 13 œ 1 #" œ #" ; dx dt œ 1 cos t, dt œ sin t Ê È Š #3 ‹ È œ ˆ " ‰ œ È3 ; tangent line is y "# œ È3 Šx 13 #3 ‹ #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dy dx
œ
dy/dt dx/dt
655
656
Chapter 11 Parametric Equations and Polar Coordinates 1È3 3
Ê y œ È3x 1 (1 cos t)#
œ 12. t œ Ê
1 2 Ê dy dx ¹ tœ 1
2;
d# y dx# ¹ tœ 1
Ê
dyw dt
(1 cos t)(cos t) (sin t)(sin t) (1cos t)#
œ
1 1 cos t
œ
d# y dx#
Ê
œ
dyw /dt dx/dt
œ
œ
cos t sin t
1 ‰ ˆ 1 cos t 1 cos t
œ 4
3
1 2
x œ cos œ cot
2
13. t œ 2 Ê x œ
œ 0, y œ 1 sin
1 #
œ 2;
œ sin t,
dx dt
w
œ 0; tangent line is y œ 2;
œ 13 , y œ
1 21
1 2
tangent line is y œ 9x 1;
œ 2;
2 21
dyw dt
œ
dx dt
14. t œ 0 Ê x œ 0 e0 œ 1, y œ 1 e0 œ 0; dyw dt
dx dt
2y$ 3t# œ 4 Ê 6y#
œ csc t Ê
œ
4 at 1 b 3 at 1 b 3
œ 1 et ,
dx dt
d# y dx#
4t œ 0 Ê 3x2
dx dt
œ 4t Ê
œ
6t 6y#
6t œ 0 Ê
dy dt
dy dt
e t a1 e t b 3
œ
œ
t y#
d# y dx# ¹ tœ2
œ
4 a2 1 b 3 a2 1 b3
d# y dx# ¹ tœ0
Ê
dx dt
œ
4t 3x2
;
; thus
dy dx
œ
dy/dt dx/dt
œ
œ
œ 1
œ
a2 1 b 2 a2 1 b 2
œ 9;
œ
e 0 1 e0
2
at 1 b 2 at 1 b 2
Ê
e t 1 et
œ
d# y dx# ¹ tœ 1
œ csc t Ê
œ
dy dx
œ cot t
$
dy dx
Ê
Ê
dy dx
csc# t sin t
œ
d y dx#
œ et Ê
dy dt
Ê
œ
1 at 1 b 2
œ
e t a1 e t b 2
tangent line is y œ 12 x 12 ; 15. x3 2t# œ 9 Ê 3x2
d# y dx#
#
#
dy dt
1 , dy at 1b2 dt
t 1b œ 4ata Ê 1 b3
œ cos t Ê
dy dt
œ 108 Ê
e 0 a1 e 0 b 3
Š yt# ‹
dy dx ¹ tœ0
œ 21 ;
œ 18
œ
Š c4t ‹
dy dx ¹ tœ2
3x2
t(3x2 ) y# (4t)
œ
3x2 4y#
;tœ2
Ê x3 2(2)# œ 9 Ê x3 8 œ 9 Ê x3 œ 1 Ê x œ 1; t œ 2 Ê 2y$ 3(2)# œ 4 Ê 2y$ œ 16 Ê y$ œ 8 Ê y œ 2; therefore 16. x œ É5 Èt Ê Ê at 1b
dy dt
œ
" #È t
therefore,
dy dx ¹ tœ4
œ
dy dt
dy dt
"Î#
"
Èt y œ #at 1b œ
" #y È t #tÈt 2Èt
œ
œ
"4 dx dt
dy t Èy ‹ dt
3x"Î#
œ
y 2Èt1
Œ 2Èy (t b 1) b 2tÈt b 1 Š
"
2t b 1 ‹ 1 b 3x"Î#
dx dt
œ 2t 1 Ê ˆ1 3x"Î# ‰
2È y Ê
œ
dy dt
dy dt
dx dt
1 #
; therefore
Èt É5 Èt
" #yÈt #Ètat" b
sin t x cos t 2
dy dx ¹ tœ1
œ
†
4Èt É5 Èt "
2 3
Š 2Èct yb 1 2Èy‹ ŠÈt 1 Èy ‹ t
dy dt
œ
dx dt
Èt 1
œ
2t1 13x"Î#
y 2È t 1
yÈy 4yÈt 1 2Èy (t 1) 2tÈt 1
; yÈt 1 2tÈy œ 4
2Èy Š Èt y ‹
; thus
; t œ 0 Ê x 2x$Î# œ 0 Ê x ˆ1 2x"Î# ‰ œ 0 Ê x œ 0; t œ 0
t sin t 2t œ y Ê sin t t cos t 2 œ Ê xœ
4
œ 2t 1 Ê
œ0 Ê
dx dt dy dt ;
sin 1 1 cos 1 2
–
1
Š 1# ‹ cos 1
sin 1 2
dy dx ¹ tœ0
œ
œ 1 Ê (sin t 2) thus œ
dy dx
œ
41 8 21
È4
È0 1
Œ 2È4(0 1) 2(0)È0 1 4
dx dt
" #yÈt
È È œ #t t "2 t œ
dy dt dx dt
œ
dy dx
10È3 9
œ
Ê yÈ0 1 2(0)Èy œ 4 Ê y œ 4; therefore
18. x sin t 2x œ t Ê
" "Î# ; y(t 1) œ Èt Ê y (t 1) dy dt œ # t
t œ 4 Ê x œ É5 È4 œ È3; t œ 4 Ê y † 3 œ È4 Ê y œ
cyÈy c 4yÈt b 1
dy/dt dx/dt
3 œ 16
4È t É 5 È t
; thus
Èt 1 y ˆ " ‰ (t 1)"Î# 2Èy 2t ˆ " y"Î# ‰ # #
Ê ŠÈ t 1 dy dx
3 a "b 2 4a#b#
œ
ˆ "# t"Î# ‰ œ
2Š" 2a 23 bÈ4‹É& È4
17. x 2x$Î# œ t# t Ê Ê
ˆ5 Èt‰
y Ê
#ˆ" #yÈt‰É& Èt ; "t
œ
" #
œ
dx dt
dy dx ¹ tœ2
4(4)
2(0) 1
Œ 1 3(0)"Î# dx dt
œ 6
œ 1 x cos t Ê
sin t t cos t 2 c x cos t ‰ ˆ 1sin tb2
dx dt
œ
1 x cos t sin t2
; t œ 1 Ê x sin 1 2x œ 1
œ 4
—
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
dy dt
œ0
Section 11.2 Calculus With Parametric Curves 19. x œ t3 t, y 2t3 œ 2x t2 Ê Ê
œ
dy dx
2t 2 3t2 1
Ê
dy dx ¹ tœ1
œ
dx 2 dt œ 3t 2 a1 b 2 œ1 3 a1 b 2 1
20. t œ lnax tb, y œ t et Ê 1 œ Ê
œ
dy dx
21. A œ '0
21
t et et xt1;
1 ˆ dx x t dt
1,
6t2 œ 2 dx dt 2t Ê
dy dt
1‰ Ê x t œ
t œ 0 Ê 0 œ lnax 0b Ê x œ 1 Ê
dx dx dt 1 Ê dt œ a0 b e 0 e 0 dy dx ¹ tœ0 œ 1 0 1
dy dt
œ 2a3t2 1b 2t 6t2 œ 2t 2
x t 1, œ
657
dy dt
œ t et et ;
1 2
y dx œ '0 aa1 cos tbaa1 cos tbdt œ a2 '0 a1 cos tb2 dt œ a2 '0 a1 2cos t cos2 tbdt 21
œ a2 '0 ˆ1 2cos t 21
21
1 cos 2t ‰ dt 2
21
œ a2 '0 ˆ 23 2cos t 21 cos 2t‰dt œ a2 ” 23 t 2sin t 41 sin 2t• 21
21 0
œ a2 a31 0 0b 0 œ 31 a2 22. A œ '0 x dy œ '0 at t2 baet bdt ”u œ t t2 Ê du œ a1 2tbdt; dv œ aet bdt Ê v œ et • 1
1
1
œ et at t2 bº '0 et a1 2tbdt ”u œ 1 2t Ê du œ 2dt; dv œ et dt Ê v œ et • 1
0
1
1
0
0
œ et at t2 bº ”et a1 2tbº '0 2et dt• œ ”et at t2 b et a1 2tb 2et •º 1
0
œ ae1 a0b e1 a1b 2e1 b ae0 a0b e0 a1b 2e0 b œ 1 3e1 œ 1 23. A œ 2'1 y dx œ 2'1 ab sin tbaa sin tbdt œ 2ab'0 sin2 t dt œ 2ab'0 0
1
1
0
1
1
3 e
1 cos 2t 2
dt œ ab'0 a1 cos 2tb dt 1
œ ab’t 12 sin 2t“ œ abaa1 0b !b œ 1 ab 0
24. (a) x œ t2 , y œ t6 , 0 Ÿ t Ÿ 1 Ê A œ '0 y dx œ '0 at6 b2t dt œ '0 2t7 dt œ ’ 14 t8 “ œ 1
1
1
1
0
1 4
0œ
(b) x œ t3 , y œ t9 , 0 Ÿ t Ÿ 1 Ê A œ '0 y dx œ '0 at9 b3t2 dt œ '0 3t11 dt œ ’ 14 t12 “ œ 1
1
1
1
0
25.
dx dt
œ sin t and
dy dt
1 4
1 4
0œ
1 4
#
#
‰ Š dy Éasin tb# a1 cos tb# œ È2 2 cos t œ 1 cos t Ê Êˆ dx dt dt ‹ œ
cos t ‰ È2 ' É sin# t dt Ê Length œ '0 È2 2 cos t dt œ È2 '0 Ɉ 11 cos t (1 cos t) dt œ 1 cos t 0 1
œ È2 '0
1
sin t È1 cos t
1
1
dt (since sin t 0 on [0ß 1]); [u œ 1 cos t Ê du œ sin t dt; t œ 0 Ê u œ 0,
# t œ 1 Ê u œ 2] Ä È2 '0 u"Î# du œ È2 2u"Î# ‘ ! œ 4 2
26.
dx dt
œ 3t# and
dy dt
È3
Ê Length œ '0 Ä '1
4
27.
dx dt
3 #
#
#
‰ Š dy Éa3t# b# (3t)# œ È9t% 9t# œ 3tÈt# 1 Šsince t 0 on ’0ß È3“‹ œ 3t Ê Êˆ dx dt dt ‹ œ 3tÈt# 1 dt; ’u œ t# 1 Ê
3 #
du œ 3t dt; t œ 0 Ê u œ 1, t œ È3 Ê u œ 4“
%
u"Î# du œ u$Î# ‘ " œ (8 1) œ 7
œ t and
dy dt
#
#
Èt# a2t 1b œ Éat 1b# œ kt 1k œ t 1 since 0 Ÿ t Ÿ 4 ‰ Š dy œ (2t 1)"Î# Ê Êˆ dx dt dt ‹ œ
Ê Length œ '0 at 1b dt œ ’ t2 t“ œ a8 4b œ 12 4
#
% !
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
658 28.
Chapter 11 Parametric Equations and Polar Coordinates dx dt
œ a2t 3b"Î# and
dy dt
#
#
‰ Š dy Éa2t 3b a1 tb# œ Èt# 4t 4 œ kt 2k œ t 2 œ 1 t Ê Êˆ dx dt dt ‹ œ
since 0 Ÿ t Ÿ 3 Ê Length œ '0 (t 2) dt œ ’ t2 2t“ œ 3
3
#
!
29.
dx dt
œ 8t cos t and
dy dt
dx dt
#
#
‰ Š dy Éa8t cos tb# a8t sin tb# œ È64t# cos# t 64t# sin# t œ 8t sin t Ê Êˆ dx dt dt ‹ œ 1 #
œ k8tk œ 8t since 0 Ÿ t Ÿ
30.
21 #
Ê Length œ '0
1 Î2
1Î#
8t dt œ c4t# d !
œ ˆ sec t" tan t ‰ asec t tan t sec# tb cos t œ sec t cos t and
œ 1# #
‰ Š dy œ sin t Ê Êˆ dx dt dt ‹
dy dt
œ Éasec t cos tb# asin tb# œ Èsec# t 1 œ Ètan# t œ ktan tk œ tan t since 0 Ÿ t Ÿ Ê Length œ '0
1 Î3
31.
dx dt
œ sin t and
dy dt
tan t dt œ '0
1 Î3
1Î$
dt œ c ln kcos tkd !
sin t cos t
" #
œ ln
#
1 3
ln 1 œ ln 2
‰ Š dy Éasin tb# acos tb# œ 1 Ê Area œ ' 21y ds œ cos t Ê Êˆ dx dt dt ‹ œ #
#
œ '0 21a2 sin tba1bdt œ 21 c2t cos td #!1 œ 21[a41 1b a0 1b] œ 81# 21
32.
dx dt
œ t"Î# and
È3
œ '0
dy dt
#
21 ˆ 23 t$Î# ‰ É t
" t
È3
'0
#
21 ˆ 23 t$Î# ‰ É t #
fatb œ 21 ˆ 23 t$Î# ‰ É t Ê
33.
dx dt
È3
'0
281 9
Fatb dt œ
œ 1 and
È2
dy dt
#
" t
dt œ
41 3
È3
'0
1 t
Ê Area œ ' 21x ds
tÈt# 1 dt; cu œ t# 1 Ê du œ 2t dt; t œ 0 Ê u œ 1,
'14 231 Èu du œ 491 u$Î# ‘ %" œ 2891
’t œ È3 Ê u œ 4“ Ä Note:
#
#
Èt t" œ É t ‰ Š dy œ t"Î# Ê Êˆ dx dt dt ‹ œ
1 t
dt is an improper integral but limb fatb exists and is equal to 0, where tÄ!
. Thus the discontinuity is removable: define Fatb œ fatb for t 0 and Fa0b œ 0
. #
#
# È2‹ œ Ét# 2È2 t 3 Ê Area œ ' 21x ds ‰# Š dy œ t È2 Ê Êˆ dx dt dt ‹ œ Ê1 Št
œ 'cÈ2 21 Št È2‹ Ét# 2È2 t 3 dt; ’u œ t# 2È2 t 3 Ê du œ Š2t 2È2‹ dt; t œ È2 Ê u œ 1,
’t œ È2 Ê u œ 9“ Ä '1 1Èu du œ 23 1u$Î# ‘ " œ 9
*
21 3
a27 1b œ
521 3
' 21y ds œ '0 ‰ Š dy 34. From Exercise 30, ʈ dx dt dt ‹ œ tan t Ê Area œ #
#
1Î$
œ 21 c cos td ! 35.
dx dt
œ 2 and
dy dt #
1 Î3
21 cos t tan t dt œ 21 '0
1Î3
sin t dt
œ 21 "# (1)‘ œ 1
È2# 1# œ È5 Ê Area œ ' 21y ds œ ' 21at 1bÈ5 dt ‰ Š dy œ 1 Ê Êˆ dx dt dt ‹ œ 0 #
#
1
"
œ 21È5 ’ t2 t“ œ 31È5. Check: slant height is È5 Ê Area is 1a1 2bÈ5 œ 31È5 . !
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.2 Calculus With Parametric Curves 36.
dx dt
œ h and
659
Èh# r# Ê Area œ ' 21y ds œ ' 21rtÈh# r# dt ‰ Š dy œ r Ê Êˆ dx dt dt ‹ œ 0 #
#
dy dt
œ 21rÈh# r#
1
'01 t dt œ 21rÈh# r# ’ t2 “ " œ 1rÈh# r# . #
!
Check: slant height is Èh# r# Ê Area is
1rÈh# r# . 37. Let the density be $ œ 1. Then x œ cos t t sin t Ê
dx dt
œ t cos t, and y œ sin t t cos t Ê
dy dt
#
#
1 #
È(t cos t)# (t sin t)# œ ktk dt œ t dt since 0 Ÿ t Ÿ ‰ Š dy Ê dm œ 1 † ds œ ʈ dx dt dt ‹ dt œ M œ ' dm œ '0
1Î2
1Î#
œ csin t t cos td ! yœ
Mx M
œ
1#
Š3
4
‹
œ
#
Š 18 ‹
xœ
My M
œ
3‰ #
Š 18 ‹
1Î#
ct# sin t 2 sin t 2t cos td ! 24 1#
1Î#
ct# cos t 2 cos t 2t sin td !
œ
12 1
Therefore axß yb œ ˆ 12 1
38. Let the density be $ œ 1. Then x œ et cos t Ê
dx dt
1 Î2
œ 24 1#
, where we integrated by parts. Therefore,
acos t t sin tb t dt œ '0
1 Î2
24 1# .
1# 4
œ 3
2. Next, My œ ' µ x dm œ '0
. The curve's mass is
1Î2 1Î2 1Î2 . Also Mx œ ' µ y dm œ '0 asin t t cos tb t dt œ '0 t sin t dt '0 t# cos t dt
1Î#
œ ccos t t sin td ! ˆ 3#1
1# 8
t dt œ
œ t sin t
31 #
t cos t dt '0
1Î2
t# sin t dt
3, again integrating by parts. Hence
‰ ß 24 1# 2 .
œ et cos t et sin t, and y œ et sin t Ê
dy dt
œ et sin t et cos t
#
#
‰ Š dy Éaet cos t et sin tb# aet sin t et cos tb# dt œ È2e2t dt œ È2 et dt. Ê dm œ 1 † ds œ ʈ dx dt dt ‹ dt œ 1 1 The curve's mass is M œ ' dm œ '0 È2 et dt œ È2 e1 È2 . Also Mx œ ' µ y dm œ '0 aet sin tb ŠÈ2 et ‹ dt 2t 21 œ '0 È2 e2t sin t dt œ È2 ’ e5 (2 sin t cos t)“ œ È2 Š e5 5" ‹ Ê y œ
1
1
!
Mx M
œ
È2 Š e21 " ‹ 5 5 È 2 e1 È 2
œ
e21 " 5 ae1 1b
.
2t 21 Next My œ ' µ x dm œ '0 aet cos tb ŠÈ2 et ‹ dt œ '0 È2 e2t cos t dt œ È2 ’ e5 a2 cos t sin tb“ œ È2 Š 2e5 25 ‹
1
My M
œ
1 !
21
Ê xœ
1
È2 Š 2e5 52 ‹ È 2 e1 È 2
21
21
21
œ 52eae1 12b . Therefore axß yb œ Š 52eae1 12b ß 5 eae1 11b ‹.
39. Let the density be $ œ 1. Then x œ cos t Ê
dx dt
œ sin t, and y œ t sin t Ê
dy dt
œ 1 cos t
#
#
‰ Š dy Éasin tb# a1 cos tb# dt œ È2 2 cos t dt. The curve's mass Ê dm œ 1 † ds œ ʈ dx dt dt ‹ dt œ is M œ ' dm œ '0 È2 2 cos t dt œ È2'0 È1 cos t dt œ È2 '0 É2 cos# ˆ #t ‰ dt œ 2 '0 ¸cos ˆ #t ‰¸ dt 1
1
œ 2 '0 cos ˆ #t ‰ dt ˆsince 0 Ÿ t Ÿ 1 Ê 0 Ÿ 1
t #
1
1
1 Ÿ 1# ‰ œ 2 2 sin ˆ 2t ‰‘ ! œ 4. Also Mx œ ' µ y dm
œ '0 at sin tb ˆ2 cos #t ‰ dt œ '0 2t cos ˆ #t ‰ dt '0 2 sin t cos ˆ #t ‰ dt 1
1
1
1 1 œ 2 4 cos ˆ 2t ‰ 2t sin ˆ #t ‰‘ ! 2 "3 cos ˆ 3# t‰ cos ˆ "# t‰‘ ! œ 41
16 3
Ê yœ
Next My œ ' µ x dm œ '0 acos tbˆ2 cos #t ‰ dt œ '0 cos t cos ˆ #t ‰ dt œ 2 ’sin ˆ 2t ‰ 1
œ
4 3
Ê xœ
My M
ˆ 43 ‰ 4
œ
œ
1
" 3
ˆ41 16 ‰ Mx 3 œ1 M œ 4 sin ˆ 3# t‰ 1 “ œ 2 32 3 !
43 .
. Therefore axß yb œ ˆ 3" ß 1 34 ‰.
40. Let the density be $ œ 1. Then x œ t$ Ê
dx dt
œ 3t# , and y œ
3t# #
Ê
dy dt
œ 3t Ê dm œ 1 † ds
#
#
‰ Š dy Éa3t# b# (3t)# dt œ 3 ktk Èt# 1 dt œ 3tÈt# 1 dt since 0 Ÿ t Ÿ È3. The curve's mass œ ʈ dx dt dt ‹ dt œ
È3
is M œ ' dm œ '0 œ
9 #
È3
'0
$Î# 3tÈt# 1 dt œ ’at# 1b “
t$ Èt# 1 dt œ
87 5
È3 !
È3
œ 7. Also Mx œ ' µ y dm œ '0
œ 17.4 (by computer) Ê y œ
Mx M
œ
17.4 7
3t# #
Š3tÈt# 1‹ dt
¸ 2.49. Next My œ ' µ x dm
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
660
Chapter 11 Parametric Equations and Polar Coordinates È3
È3
œ '0 t$ † 3t Èat# 1b dt œ 3 '0 t% Èt# 1 dt ¸ 16.4849 (by computer) Ê x œ
My M
œ
¸ 2.35.
16.4849 7
Therefore, axß yb ¸ a2.35ß 2.49b. œ 2 sin 2t and
dx dt
41. (a)
Ê Length œ '0
1Î2
œ 1 cos 1t and
dx dt
(b)
#
#
#
#
‰ Š dy Éa2 sin 2tb# a2 cos 2tb# œ 2 œ 2 cos 2t Ê Êˆ dx dt dt ‹ œ
dy dt
1Î#
2 dt œ c2td !
œ1
‰ Š dy Éa1 cos 1tb# a1 sin 1tb# œ 1 œ 1 sin 1t Ê Êˆ dx dt dt ‹ œ
dy dt
Ê Length œ '1Î2 1 dt œ c1td "Î# œ 1 1 Î2
"Î#
42. (a) x œ gayb has the parametrization x œ gayb and y œ y for c Ÿ y Ÿ d Ê
dx dy
œ gw ayb and
dy dy
œ 1; then
dx dx ' ' È1 [gw ayb]# dy Length œ 'c ÊŠ dy dy ‹ Š dy ‹ dy œ c Ê1 Š dy ‹ dy œ c #
d
(b) x œ y3Î2 , 0 Ÿ y Ÿ œ
3 Î2 8 27 a4b
4 3
#
Ê
3Î2 8 27 a1b
(c) x œ 32 y2Î3 , 0 Ÿ y Ÿ 1 Ê œ lim
3 2
a Ä0
œ 32 y1Î2 Ê L œ '0
4 Î3
dx dy
œ
#
d
56 27 dx dy
d
É1 ˆ 32 y1Î2 ‰# dy œ '
0
1
1
Ê
œ
œ
È3 1 1
È23 œ 2
2È 3 1 È3 2
2 sin )b,
dy dx º )œ0
(b) x œ ˆ1 2 sinˆ 1# ‰‰cosˆ 1# ‰ œ 0, y œ ˆ1 2 sinˆ 1# ‰‰sinˆ 1# ‰ œ 3;
œ
œ
y œ ˆ1 2 sinˆ 431 ‰‰sinˆ 431 ‰ œ
45.
or t œ
31 2
(a) the maximum slope is
dy dx º
(a) the minimum slope is
dy dx º
dx dt
œ cos t and
dy dt
dx dt
œ 1,
d2 y dx2
Ê
tœ1Î2
2 3
œ 2cos ) sin ) cos )a1 2 sin )b
œ
œ
01 20
2 sinˆ2ˆ 1# ‰‰ cosˆ 1# ‰ 2 cosˆ2ˆ 1# ‰‰ sinˆ 1# ‰
3 È3 2
;
dy dx º )œ41/3
œ
œ
œ
1 2
00 2 1
œ0
2 sinˆ2ˆ 431 ‰‰ cosˆ 431 ‰ 2 cosˆ2ˆ 431 ‰‰ sinˆ 431 ‰
dy dt
œ sin t Ê
dy dx
œ
sin t 1
œ sin t Ê
dy dx ,
œ cos t Ê
d2 y dx2
œ
œ cos t. The
cos t 1
in other words, points where
d2 y dx2
œ0
œ ± ± 31Î2 1 Î2
œ sinˆ 321 ‰ œ 1, which occurs at x œ
dy dx
Ê 2 cos# t 1 œ 0 Ê cos t œ „ È2 # ß 1‹
d dy dt Š dx ‹
œ sinˆ 12 ‰ œ 1, which occurs at x œ 12 , y œ 1 cosˆ 12 ‰ œ 1
tœ31Î2
œ 2 cos 2t Ê
y œ sin 2 ˆ 14 ‰ œ 1 Ê Š
dy d)
2 sina2a0bb cosa0b 2 cosa2a0bb sina0b
dy dx º )œ1/2
maximum and minimum slope will occur at points that maximize/minimize Ê cos t œ 0 Ê t œ
2 3
a Ä0
œ Š4 3È3‹
44. x œ t, y œ 1 cos t, 0 Ÿ t Ÿ 21 Ê 1 2
'a1 É y yÎ Î 1 dy
dy œ lim
a
(a) x œ a1 2 sina0bbcosa0b œ 1, y œ a1 2 sina0bbsina0b œ 0;
È3 1 2 ,
0
3 Î2 3Î2 lim ” 32 † 23 ˆy2Î3 1‰ • œ lim Ša2b3Î2 ˆa2Î3 1‰ ‹ œ 2È2 1 a Ä0
a Ä0
dx 2 d) œ 2cos ) sin )a1 4cos ) sin ) cos ) 2 sin 2) cos ) 2cos2 ) 2sin2 ) sin ) œ 2 cos 2) sin )
(c) x œ ˆ1 2 sinˆ 431 ‰‰cosˆ 431 ‰ œ
1 y2Î3
4 Î3
1
'a1 ˆy2Î3 1‰1Î2 ˆ 23 y1Î3 ‰ dy œ
2cos ) sin ) cos )a1 2 sin )b 2cos2 ) sin )a1 2 sin )b
3Î2 É1 94 y dy œ ” 49 † 23 ˆ1 94 y‰ •
# œ y1Î3 Ê L œ '0 É1 ay1Î3 b dy œ '0 É1
43. x œ a1 2 sin )bcos ), y œ a1 2 sin )bsin ) Ê dy dx
4Î3
œ
dy/dt dx/dt
" È2
œ
2 cos 2t cos t
Ê tœ
1 4
,
œ 31 4
,
2 a2 cos# t 1b cos t 51 4
,
71 4
; then
31 2 ,
dy dx
y œ 1 cosˆ 321 ‰ œ 1
œ0 Ê
2 a2 cos# t 1b cos t
. In the 1st quadrant: t œ
1 4
œ0
Ê x œ sin
1 4
œ
È2 #
is the point where the tangent line is horizontal. At the origin: x œ 0 and y œ 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
and
Section 11.2 Calculus With Parametric Curves Ê sin t œ 0 Ê t œ 0 or t œ 1 and sin 2t œ 0 Ê t œ 0, the origin. Tangents at origin: 46.
dx dt
œ dy dx
œ 2 cos 2t and
dy dt
œ 3 cos 3t Ê
3 ca2 cos# t 1b (cos t) 2 sin t cos t sin td 2 a2 cos# t 1b (3 cos t) a4 cos# t3b 2 a2 cos# t 1b
œ0 Ê
œ 2 Ê y œ 2x and
dy dx ¹ tœ0
œ
dy dx
œ
œ
dy/dt dx/dt
and y œ sin 3 ˆ 16 ‰ œ 1 Ê Š
3(cos 2t cos t sin 2t sin t) 2 a2 cos# t1b
(3 cos t) a2 cos# t 1 2 sin# tb 2 a2 cos# t 1b
È3 #
È3 # ß 1‹
Ê tœ
1 6
,
51 6
,
œ
(3 cos t) a4 cos# t 3b 2 a2 cos# t 1b
; then
71 6
,
111 6
. In the 1st quadrant: t œ
1 6
1 #
,
31 #
and
Ê x œ sin 2 ˆ 16 ‰ œ
È3 #
is the point where the graph has a horizontal tangent. At the origin: x œ 0
and y œ 0 Ê sin 2t œ 0 and sin 3t œ 0 Ê t œ 0,
1 #
, 1,
the tangent lines at the origin. Tangents at the origin: 3 cos (31) 2 cos (21)
œ
3 cos 3t 2 cos 2t
1 give the tangent lines at
œ 0 Ê 3 cos t œ 0 or 4 cos# t 3 œ 0: 3 cos t œ 0 Ê t œ
4 cos# t 3 œ 0 Ê cos t œ „
œ
1 31 # , 1, # ; thus t œ 0 and t œ dy dx ¹ tœ1 œ 2 Ê y œ 2x
661
31 #
and t œ 0,
dy dx ¹ tœ0
œ
3 cos 0 2 cos 0
1 3
œ
,
21 3
, 1,
41 3
,
51 3
3 #
x, and
Ê t œ 0 and t œ 1 give
3 #
Ê yœ
dy dt
œ a sin t Ê Length
dy dx ¹ tœ1
œ 3# Ê y œ 3# x
47. (a) x œ aat sin tb, y œ aa1 cos tb, 0 Ÿ t Ÿ 21 Ê
dx dt
œ aa1 cos tb,
œ '0 Éaaa1 cos tbb# aa sin tb# dt œ '0 Èa# 2a# cos t a# cos# t a# sin# t dt 21
21
œ aÈ2'0 È1 cos t dt œ aÈ2'0 É2 sin2 ˆ 2t ‰ dt œ 2a'0 sinˆ 2t ‰ dt œ ’4a cosˆ 2t ‰“ 21
21
21
21
0
œ 4a cos 1 4a cosa0b œ 8a (b) a œ 1 Ê x œ t sin t, y œ 1 cos t, 0 Ÿ t Ÿ 21 Ê
dx dt 21
œ 1 cos t,
dy dt
œ sin t Ê Surface area œ
œ '0 21a1 cos tbÉa1 cos tb# asin tb# dt œ '0 21a1 cos tbÈ1 2 cos t cos# t sin# t dt 21
3 Î2 œ 21'0 a1 cos tbÈ2 2 cos t dt œ 2È21'0 a1 cos tb3Î2 dt œ 2È21'0 ˆ1 cos ˆ2 † 2t ‰‰ dt 21
21
21
3 Î2 œ 2È21'0 ˆ2 sin2 ˆ 2t ‰‰ dt œ 81'0 sin3 ˆ 2t ‰ dt 21
21
’u œ
t 2
Ê du œ 12 dt Ê dt œ 2 du; t œ 0 Ê u œ 0, t œ 21 Ê u œ 1“
œ 161'0 sin3 u du œ 161'0 sin2 u sin u du œ 161'0 a1 cos2 u bsin u du œ 161'0 sin u du 161'0 cos2 u sin u du 1
1
œ ’161cos u
1 161 3 3 cos u“0
1
œ ˆ161
161 ‰ 3
1
ˆ161
161 ‰ 3
œ
641 3
48. x œ t sin t, y œ 1 cos t, 0 Ÿ t Ÿ 21; Volume œ '0 1 y2 dx œ '0 1a1 cos tb2 a1 cos tbdt 21
21
2t ‰ œ 1'0 a1 3cos t 3cos2 t cos3 tbdt œ 1'0 ˆ1 3cos t 3ˆ 1 cos cos2 t cos t‰dt 2 21
21
œ 1'0 ˆ 52 3cos t 32 cos 2t a1 sin2 tb cos t‰dt œ 1'0 ˆ 52 4cos t 32 cos 2t sin2 t cos t‰dt 21
21
21
œ 1’ 52 t 4sin t 34 sin 2t 31 sin3 t “
0
œ 1a51 0 0 0b 0 œ 512
47-50. Example CAS commands: Maple: with( plots ); with( student ); x := t -> t^3/3; y := t -> t^2/2; a := 0; b := 1; N := [2, 4, 8 ]; for n in N do
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1
662
Chapter 11 Parametric Equations and Polar Coordinates tt := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x(t),y(t)],t=tt)]; L := simplify(add( student[distance](pts[i+1],pts[i]), i=1..n )); # (b) T := sprintf("#47(a) (Section 11.2)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ): # (a) end do: display( [seq(P[n],n=N)], insequence=true ); ds := t ->sqrt( simplify(D(x)(t)^2 + D(y)(t)^2) ): # (c) L := Int( ds(t), t=a..b ): L = evalf(L);
11.3 POLAR COORDINATES 1. a, e; b, g; c, h; d, f
2. a, f; b, h; c, g; d, e
3. (a) ˆ2ß 1# 2n1‰ and ˆ2ß 1# (2n 1)1‰ , n an integer
(b) (#ß 2n1) and (#ß (2n 1)1), n an integer (c) ˆ2ß 3#1 2n1‰ and ˆ2ß 3#1 (2n 1)1‰ , n an integer
(d) (#ß (2n 1)1) and (#ß 2n1), n an integer
4. (a) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer (b) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer (c) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer (d) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer
5. (a) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (b) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (c) x œ r cos ) œ 2 cos 21 œ 1, y œ r sin ) œ 2 sin 21 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
(d) x œ r cos ) œ 2 cos
71 3
3
œ 1, y œ r sin ) œ 2 sin
71 3
œ È3 Ê Cartesian coordinates are Š1ß È3‹
(e) x œ r cos ) œ 3 cos 1 œ 3, y œ r sin ) œ 3 sin 1 œ 0 Ê Cartesian coordinates are (3ß 0) (f) x œ r cos ) œ 2 cos 1 œ 1, y œ r sin ) œ 2 sin 1 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
3
(g) x œ r cos ) œ 3 cos 21 œ 3, y œ r sin ) œ 3 sin 21 œ 0 Ê Cartesian coordinates are (3ß 0) (h) x œ r cos ) œ 2 cos ˆ 1 ‰ œ 1, y œ r sin ) œ 2 sin ˆ 1 ‰ œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
6. (a) x œ È2 cos
1 4
œ 1, y œ È2 sin
3
1 4
œ 1 Ê Cartesian coordinates are (1ß 1)
(b) x œ 1 cos 0 œ 1, y œ 1 sin 0 œ 0 Ê Cartesian coordinates are (1ß 0) (c) x œ 0 cos 1# œ 0, y œ 0 sin 1# œ 0 Ê Cartesian coordinates are (!ß 0) (d) x œ È2 cos ˆ 1 ‰ œ 1, y œ È2 sin ˆ 1 ‰ œ 1 Ê Cartesian coordinates are (1ß 1) 4
(e) x œ 3 cos
51 6
œ
4
3È 3 2
, y œ 3 sin
51 6
È
œ 3# Ê Cartesian coordinates are Š 3 # 3 ß 3# ‹
(f) x œ 5 cos ˆtan" 43 ‰ œ 3, y œ 5 sin ˆtan" 43 ‰ œ 4 Ê Cartesian coordinates are ($ß 4)
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Section 11.3 Polar Coordinates
663
(g) x œ 1 cos 71 œ 1, y œ 1 sin 71 œ 0 Ê Cartesian coordinates are (1ß 0) (h) x œ 2È3 cos 231 œ È3, y œ 2È3 sin 231 œ 3 Ê Cartesian coordinates are ŠÈ3ß 3‹ 7. (a) a1, 1b Ê r œ È12 12 œ È2, sin ) œ
1 È2
and cos ) œ
1 È2
1 4
Ê)œ
Ê Polar coordinates are ŠÈ2, 14 ‹
(b) a3, 0b Ê r œ Éa3b2 02 œ 3, sin ) œ 0 and cos ) œ 1 Ê ) œ 1 Ê Polar coordinates are a3, 1b 2
(c) ŠÈ3, 1‹ Ê r œ ÊŠÈ3‹ a1b2 œ 2, sin ) œ 12 and cos ) œ (d) a3, 4b Ê r œ Éa3b2 42 œ 5, sin ) œ
4 5
È3 2
111 6
Ê)œ
Ê Polar coordinates are ˆ2,
111 ‰ 6
and cos ) œ 35 Ê ) œ 1 arctanˆ 43 ‰ Ê Polar coordinates are
ˆ5, 1 arctanˆ 43 ‰‰ 8. (a) a2, 2b Ê r œ Éa2b2 a2b2 œ 2È2, sin ) œ È12 and cos ) œ È12 Ê ) œ 341 Ê Polar coordinates are Š2È2, 341 ‹ (b) a0, 3b Ê r œ È02 32 œ 3, sin ) œ 1 and cos ) œ 0 Ê ) œ 2
(c) ŠÈ3, 1‹ Ê r œ ÊŠÈ3‹ 12 œ 2, sin ) œ
1 2
1 2
Ê Polar coordinates are ˆ3, 12 ‰
and cos ) œ
(d) a5, 12b Ê r œ É52 a12b2 œ 13, sin ) œ 12 13 and cos ) œ
5 12
È3 2
Ê)œ
51 6
Ê Polar coordinates are ˆ2,
51 ‰ 6
‰ Ê ) œ arctanˆ 12 5 Ê Polar coordinates are
ˆ13, arctanˆ 12 ‰‰ 5 9. (a) a3, 3b Ê r œ È32 32 œ 3È2, sin ) œ È1 and cos ) œ È1 Ê ) œ 2 2 Š3È2,
51 4
Ê Polar coordinates are
51 4 ‹
(b) a1, 0b Ê r œ Éa1b2 02 œ 1, sin ) œ 0 and cos ) œ 1 Ê ) œ 0 Ê Polar coordinates are a1, 0b 2
(c) Š1, È3‹ Ê r œ Êa1b2 ŠÈ3‹ œ 2, sin ) œ ˆ2,
È3 2
and cos ) œ
1 2
Ê)œ
51 3
Ê Polar coordinates are
51 ‰ 3
(d) a4, 3b Ê r œ É42 a3b2 œ 5, sin ) œ
3 5
and cos ) œ 45 Ê ) œ 1 arctanˆ 34 ‰ Ê Polar coordinates are
ˆ5, 1 arctanˆ 43 ‰‰ 10. (a) a2, 0b Ê r œ Éa2b2 02 œ 2, sin ) œ 0 and cos ) œ 1 Ê ) œ 0 Ê Polar coordinates are a2, 0b (b) a1, 0b Ê r œ È12 02 œ 1, sin ) œ 0 and cos ) œ 1 Ê ) œ 1 or ) œ 1 Ê Polar coordinates are a1, 1b or a1, 1b (c) a0, 3b Ê r œ É02 a3b2 œ 3, sin ) œ 1 and cos ) œ 0 Ê ) œ (d) Š
È3 1 2 , 2‹
are ˆ1,
Ê r œ ÊŠ 71 ‰ 6
È3 2 2 ‹
2
ˆ 12 ‰ œ 1, sin ) œ 12 and cos ) œ
1 2
Ê Polar coordinates are ˆ3, 12 ‰
È3 2
Ê)œ
71 6
or ) œ 561 Ê Polar coordinates
or ˆ1, 561 ‰
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Chapter 11 Parametric Equations and Polar Coordinates
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
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Section 11.3 Polar Coordinates
665
26.
27. r cos ) œ 2 Ê x œ 2, vertical line through (#ß 0)
28. r sin ) œ 1 Ê y œ 1, horizontal line through (0ß 1)
29. r sin ) œ 0 Ê y œ 0, the x-axis
30. r cos ) œ 0 Ê x œ 0, the y-axis
31. r œ 4 csc ) Ê r œ
4 sin )
32. r œ 3 sec ) Ê r œ
Ê r sin ) œ 4 Ê y œ 4, a horizontal line through (0ß 4)
3 cos )
Ê r cos ) œ 3 Ê x œ 3, a vertical line through (3ß 0)
33. r cos ) r sin ) œ 1 Ê x y œ 1, line with slope m œ 1 and intercept b œ 1 34. r sin ) œ r cos ) Ê y œ x, line with slope m œ 1 and intercept b œ 0 35. r# œ 1 Ê x# y# œ 1, circle with center C œ (!ß 0) and radius 1 36. r# œ 4r sin ) Ê x# y# œ 4y Ê x# y# 4y 4 œ 4 Ê x# (y 2)# œ 4, circle with center C œ (0ß 2) and radius 2 37. r œ
5 sin )2 cos )
Ê r sin ) 2r cos ) œ 5 Ê y 2x œ 5, line with slope m œ 2 and intercept b œ 5
38. r# sin 2) œ 2 Ê 2r# sin ) cos ) œ 2 Ê (r sin ))(r cos )) œ 1 Ê xy œ 1, hyperbola with focal axis y œ x )‰ˆ " ‰ 39. r œ cot ) csc ) œ ˆ cos Ê r sin# ) œ cos ) Ê r# sin# ) œ r cos ) Ê y# œ x, parabola with vertex (0ß 0) sin ) sin )
which opens to the right sin ) ‰ 40. r œ 4 tan ) sec ) Ê r œ 4 ˆ cos Ê r cos# ) œ 4 sin ) Ê r# cos# ) œ 4r sin ) Ê x# œ 4y, parabola with #)
vertex œ (!ß 0) which opens upward
41. r œ (csc )) er cos ) Ê r sin ) œ er cos ) Ê y œ ex , graph of the natural exponential function 42. r sin ) œ ln r ln cos ) œ ln (r cos )) Ê y œ ln x, graph of the natural logarithm function 43. r# 2r# cos ) sin ) œ 1 Ê x# y# 2xy œ 1 Ê x# 2xy y# œ 1 Ê (x y)# œ 1 Ê x y œ „ 1, two parallel straight lines of slope 1 and y-intercepts b œ „ 1 44. cos# ) œ sin# ) Ê r# cos# ) œ r# sin# ) Ê x# œ y# Ê kxk œ kyk Ê „ x œ y, two perpendicular lines through the origin with slopes 1 and 1, respectively. 45. r# œ 4r cos ) Ê x# y# œ 4x Ê x# 4x y# œ 0 Ê x# 4x 4 y# œ 4 Ê (x 2)# y# œ 4, a circle with center C(2ß 0) and radius 2
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Chapter 11 Parametric Equations and Polar Coordinates
46. r# œ 6r sin ) Ê x# y# œ 6y Ê x# y# 6y œ 0 Ê x# y# 6y 9 œ 9 Ê x# (y 3)# œ 9, a circle with center C(0ß 3) and radius 3 47. r œ 8 sin ) Ê r# œ 8r sin ) Ê x# y# œ 8y Ê x# y# 8y œ 0 Ê x# y# 8y 16 œ 16 Ê x# (y 4)# œ 16, a circle with center C(0ß 4) and radius 4 48. r œ 3 cos ) Ê r# œ 3r cos ) Ê x# y# œ 3x Ê x# y# 3x œ 0 Ê x# 3x # Ê ˆx 3# ‰ y# œ
9 4
, a circle with center C ˆ 3# ß !‰ and radius
9 4
y# œ
9 4
3 #
49. r œ 2 cos ) 2 sin ) Ê r# œ 2r cos ) 2r sin ) Ê x# y# œ 2x 2y Ê x# 2x y# 2y œ 0 Ê (x 1)# (y 1)# œ 2, a circle with center C(1ß 1) and radius È2 50. r œ 2 cos ) sin ) Ê r# œ 2r cos ) r sin ) Ê x# y# œ 2x y Ê x# 2x y# y œ 0 # Ê (x 1)# ˆy "# ‰ œ 54 , a circle with center C ˆ1ß "# ‰ and radius
È5 #
È
51. r sin ˆ) 16 ‰ œ 2 Ê r ˆsin ) cos 16 cos ) sin 16 ‰ œ 2 Ê #3 r sin ) "# r cos ) œ 2 Ê Ê È3 y x œ 4, line with slope m œ " and intercept b œ 4 È3
È3 #
È3
È
52. r sin ˆ 231 )‰ œ 5 Ê r ˆsin 231 cos ) cos 231 sin )‰ œ 5 Ê #3 r cos ) "# r sin ) œ 5 Ê Ê È3 x y œ 10, line with slope m œ È3 and intercept b œ 10 53. x œ 7 Ê r cos ) œ 7 55. x œ y Ê r cos ) œ r sin ) Ê ) œ
y "# x œ 2
È3 #
x "# y œ 5
54. y œ 1 Ê r sin ) œ 1 1 4
56. x y œ 3 Ê r cos ) r sin ) œ 3
57. x# y# œ 4 Ê r# œ 4 Ê r œ 2 or r œ 2 58. x# y# œ 1 Ê r# cos# ) r# sin# ) œ 1 Ê r# acos# ) sin# )b œ 1 Ê r# cos 2) œ 1 59.
x# 9
y# 4
œ 1 Ê 4x# 9y# œ 36 Ê 4r# cos# ) 9r# sin# ) œ 36
60. xy œ 2 Ê (r cos ))(r sin )) œ 2 Ê r# cos ) sin ) œ 2 Ê 2r# cos ) sin ) œ 4 Ê r# sin 2) œ 4 61. y# œ 4x Ê r# sin# ) œ 4r cos ) Ê r sin# ) œ 4 cos ) 62. x# xy y# œ 1 Ê x# y# xy œ 1 Ê r# r# sin ) cos ) œ 1 Ê r# (1 sin ) cos )) œ 1 63. x# (y 2)# œ 4 Ê x# y# 4y 4 œ 4 Ê x# y# œ 4y Ê r# œ 4r sin ) Ê r œ 4 sin ) 64. (x 5)# y# œ 25 Ê x# 10x 25 y# œ 25 Ê x# y# œ 10x Ê r# œ 10r cos ) Ê r œ 10 cos ) 65. (x 3)# (y 1)# œ 4 Ê x# 6x 9 y# 2y 1 œ 4 Ê x# y# œ 6x 2y 6 Ê r# œ 6r cos ) 2r sin ) 6 66. (x 2)# (y 5)# œ 16 Ê x# 4x 4 y# 10y 25 œ 16 Ê x# y# œ 4x 10y 13 Ê r# œ 4r cos ) 10r sin ) 13
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Section 11.4 Graphing in Polar Coordinates 67. (!ß )) where ) is any angle 68. (a) x œ a Ê r cos ) œ a Ê r œ (b) y œ b Ê r sin ) œ b Ê r œ
a cos ) b sin )
Ê r œ a sec ) Ê r œ b csc )
11.4 GRAPHING IN POLAR COORDINATES 1. 1 cos ()) œ 1 cos ) œ r Ê symmetric about the x-axis; 1 cos ()) Á r and 1 cos (1 )) œ 1 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin
2. 2 2 cos ()) œ 2 2 cos ) œ r Ê symmetric about the x-axis; 2 # cos ()) Á r and 2 2 cos (1 )) œ 2 2 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin
3. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 )) œ 1 sin ) Á r Ê not symmetric about the x-axis; 1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
4. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 )) œ 1 sin ) Á r Ê not symmetric about the x-axis; 1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
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Chapter 11 Parametric Equatins and Polar Coordinates
5. 2 sin ()) œ 2 sin ) Á r and 2 sin (1 )) œ 2 sin ) Á r Ê not symmetric about the x-axis; 2 sin (1 )) œ 2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
6. 1 2 sin ()) œ 1 2 sin ) Á r and 1 2 sin (1 )) œ 1 2 sin ) Á r Ê not symmetric about the x-axis; 1 2 sin (1 )) œ 1 2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
7. sin ˆ #) ‰ œ sin ˆ #) ‰ œ r Ê symmetric about the y-axis; sin ˆ 21#) ‰ œ sin ˆ 2) ‰ , so the graph is symmetric about the x-axis, and hence the origin.
8. cos ˆ #) ‰ œ cos ˆ #) ‰ œ r Ê symmetric about the x-axis; cos ˆ 21#) ‰ œ cos ˆ 2) ‰ , so the graph is symmetric about the y-axis, and hence the origin.
9. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.4 Graphing in Polar Coordinates 10. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin
11. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin
12. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin
13. Since a „ rß )b are on the graph when (rß )) is on the graph ˆa „ rb# œ 4 cos 2( )) Ê r# œ 4 cos 2)‰ , the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin
14. Since (rß )) on the graph Ê (rß )) is on the graph ˆa „ rb# œ 4 sin 2) Ê r# œ 4 sin 2)‰ , the graph is symmetric about the origin. But 4 sin 2()) œ 4 sin 2) Á r# and 4 sin 2(1 )) œ 4 sin (21 2)) œ 4 sin (2)) œ 4 sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 15. Since (rß )) on the graph Ê (rß )) is on the graph ˆa „ rb# œ sin 2) Ê r# œ sin 2)‰ , the graph is symmetric about the origin. But sin 2()) œ ( sin 2)) sin 2) Á r# and sin 2(1 )) œ sin (21 2)) œ sin (2)) œ ( sin 2)) œ sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis
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Chapter 11 Parametric Equatins and Polar Coordinates
16. Sincea „ rß )b are on the graph when (rß )) is on the graph ˆa „ rb# œ cos 2()) Ê r# œ cos 2)‰, the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin.
Ê r œ 1 Ê ˆ1ß 1# ‰ , and ) œ 1# Ê r œ 1 w )r cos ) Ê ˆ1ß 1# ‰ ; rw œ ddr) œ sin ); Slope œ rrw sin cos )r sin )
17. ) œ
1 #
sin# )r cos ) sin ) cos )r sin ) sin# ˆ 1# ‰(1) cos 1# sin 1# cos 1# (1) sin 1#
œ
Ê Slope at ˆ1ß 1# ‰ is œ 1; Slope at ˆ1ß 1# ‰ is
sin# ˆ 1# ‰(1) cos ˆ 1# ‰ sin ˆ 1# ‰ cos ˆ 1# ‰(1) sin ˆ 1# ‰
œ1
18. ) œ 0 Ê r œ 1 Ê ("ß 0), and ) œ 1 Ê r œ 1 dr Ê ("ß 1); rw œ d) œ cos );
rw sin )r cos ) cos ) sin )r cos ) rw cos )r sin ) œ cos ) cos )r sin ) 0 sin 0(1) cos 0 cos ) sin )r cos ) Ê Slope at ("ß 0) is coscos # 0(1) sin 0 cos# )r sin ) cos 1 sin 1(1) cos 1 1; Slope at ("ß 1) is cos# 1(1) sin 1 œ 1
Slope œ œ œ
Ê r œ 1 Ê ˆ"ß 14 ‰ ; ) œ 14 Ê r œ 1 Ê ˆ1ß 14 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ"ß 341 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ1ß 341 ‰ ;
19. ) œ
rw œ
1 4
dr d)
œ 2 cos 2);
Slope œ
r sin )r cos ) r cos )r sin ) w w
Ê Slope at ˆ1ß 14 ‰ is Slope at ˆ1ß 14 ‰ is Slope at ˆ1ß 341 ‰ is Slope at ˆ1ß 341 ‰ is
2 cos 2) sin )r cos ) 2 cos 2) cos )r sin ) 2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰ 2 cos ˆ 1 ‰ cos ˆ 1 ‰(1) sin ˆ 1 ‰
œ
#
4
4
œ 1;
2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰ 2 cos ˆ 1# ‰ cos ˆ 14 ‰(1) sin ˆ 14 ‰
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1;
œ 1;
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1
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Section 11.4 Graphing in Polar Coordinates 20. ) œ 0 Ê r œ 1 Ê (1ß 0); ) œ 12 Ê r œ 1 Ê ˆ1ß 12 ‰ ; ) œ 1# Ê r œ 1 Ê ˆ"ß 12 ‰ ; ) œ 1 Ê r œ 1 Ê (1ß 1); rw œ
dr d) œ 2 sin 2); )r cos ) 2 sin 2) sin )r cos ) Slope œ rr sin cos )r sin ) œ 2 sin 2) cos )r sin ) 2 sin 0 sin 0cos 0 Ê Slope at (1ß 0) is 2 sin 0 cos 0sin 0 , which is undefined; 2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰ Slope at ˆ1ß 12 ‰ is 2 sin 2 ˆ 12 ‰ cos ˆ21 ‰(1) sin ˆ 21 ‰ œ 0; w w
2
Slope at ˆ1ß 12 ‰ is Slope at ("ß 1) is
2
2
2 sin 2 ˆ 1# ‰ sin ˆ 1# ‰(1) cos ˆ 1# ‰ 2 sin 2 ˆ 1 ‰ cos ˆ 1 ‰(1) sin ˆ 1 ‰ #
2 sin 21 sin 1cos 1 2 sin 21 cos 1sin 1
#
#
œ 0;
, which is undefined
21. (a)
(b)
22. (a)
(b)
23. (a)
(b)
24. (a)
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
671
672
Chapter 11 Parametric Equatins and Polar Coordinates
25.
26. r œ 2 sec ) Ê r œ
27.
2 cos )
Ê r cos ) œ 2 Ê x œ 2
28.
29. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ 1 cos ) Í 1 cos () 1) œ 1 (cos ) cos 1 sin ) sin 1) œ 1 cos ) œ (1 cos )) œ r; therefore (rß )) is on the graph of r œ 1 cos ) Í (rß ) 1) is on the graph of r œ 1 cos ) Ê the answer is (a).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.4 Graphing in Polar Coordinates 30. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ cos 2) Í sin ˆ2() 1)) 1# ‰ œ sin ˆ2) 5#1 ‰ œ sin (2)) cos ˆ 5#1 ‰ cos (2)) sin ˆ 5#1 ‰ œ cos 2) œ r; therefore (rß )) is on the graph of r œ sin ˆ2) 1# ‰ Ê the answer is (a).
31.
33. (a)
34. (a)
32.
(b)
(c)
(b)
(d)
(c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
673
674
Chapter 11 Parametric Equatins and Polar Coordinates
(d)
(e)
11.5 AREA AND LENGTHS IN POLAR COORDINATES 1. A œ '0 "# )# d) œ 16 )3 ‘ ! œ 1
1
13 6
2. A œ '1Î4 "# a2 sin )b# d) œ 2'1Î4 sin2 ) d) œ 2'1Î4 1Î2
1Î2
œ ˆ 12 0‰ ˆ 14 12 ‰ œ 3. A œ '0
21
" #
1 4
1Î2
4. A œ '0
21
(4 2 cos ))# d) œ '0
21
œ
" #
a#
œ '0
" #
21
" # # [a(1 cos ))] d) 21 ˆ #3 2 cos ) "# 0
'
5. A œ 2 '0
1 Î4
6. A œ '1Î6 1 Î6
œ 4" ) 7. A œ '0
1Î2
" #
" #
1 Î4
1 ‘ 1Î6 6 sin 6) 1Î6
" #
1Î2
1Î6
" #
" #
#1
" 2
sin 2)‘ ! œ 181
a# a1 2 cos ) cos# )b d) œ " #
" 4
a# 3# ) 2 sin )
d) œ
" #
cos2 3) d) œ
) " #
sin 4) ‘ 1Î% 4 !
'11ÎÎ66
œ 4" ˆ 16 0‰ 4" ˆ 16 0‰ œ
(4 sin 2)) d) œ '0
8. A œ (6)(2)'0
1 Î2
21
1 cos 4) #
' 11ÎÎ66
" #
1Î2
2 ) ‰‘ a16 16 cos ) 4 cos# )b d) œ '0 8 8 cos ) 2 ˆ 1 cos d) #
cos 2)‰ d) œ
cos# 2) d) œ '0
acos 3)b2 d) œ
d) œ '1Î4 a1 cos 2)bd) œ ) 21 sin 2)‘1Î4
1 2
œ '0 (9 8 cos ) cos 2)) d) œ 9) 8 sin ) 21
1 cos 2) 2
œ
1 cos 6) 2
a#
2) ‰ '021 ˆ1 2 cos ) 1 cos d) # #1
sin 2)‘ ! œ
3 #
1a#
1 8
d) œ
" 4
'11ÎÎ66
a1 cos 6)b d)
1 12
1Î#
2 sin 2) d) œ c cos 2)d !
œ2
(2 sin 3)) d) œ 12 '0 sin 3) d) œ 12 cos3 3) ‘ ! 1Î6
" #
1Î'
œ4
9. r œ 2 cos ) and r œ 2 sin ) Ê 2 cos ) œ 2 sin ) Ê cos ) œ sin ) Ê ) œ 14 ; therefore A œ 2 '0
1Î4
œ '0
1Î4
" #
(2 sin ))# d) œ '0
1Î4
2) ‰ 4 ˆ 1 cos d) œ '0 #
œ c2) sin
1Î4
1Î% 2) d !
œ
1 #
4 sin# ) d)
(2 2 cos 2)) d)
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.5 Area and Lengths in Polar Coordinates 10. r œ 1 and r œ 2 sin ) Ê 2 sin ) œ 1 Ê sin ) œ 1 6
Ê )œ
or
51 6
" #
; therefore
A œ 1(1)# '1Î6
51Î6
" #
c(2 sin ))# 1# d d)
œ 1 '1Î6 ˆ2 sin# ) "# ‰ d) 51Î6
œ 1 '1Î6 ˆ1 cos 2) "# ‰ d) 51Î6
œ 1 '1Î6 ˆ "# cos 2)‰ d) œ 1 2" )
sin 2) ‘ &1Î' # 1Î'
œ 1 ˆ 511#
41 3È 3 6
51Î6
" #
sin
51 ‰ 3
1 ˆ 12
" #
sin 13 ‰ œ
11. r œ 2 and r œ 2(1 cos )) Ê 2 œ 2(1 cos )) Ê cos ) œ 0 Ê ) œ „ 1# ; therefore A œ 2 '0
1Î2
œ '0
1Î2
œ '0
1Î2
œ '0
1Î2
" #
[2(1 cos ))]# d) "# area of the circle
4 a1 2 cos ) cos# )b d) ˆ "# 1‰ (2)# 4 ˆ1 2 cos )
1 cos 2) ‰ #
d) 2 1
(4 8 cos ) 2 2 cos 2)) d) 21 1Î#
œ c6) 8 sin ) sin 2)d !
21 œ 51 8
12. r œ 2(1 cos )) and r œ 2(1 cos )) Ê 1 cos ) œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# or 3#1 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0
1Î2
" #
[2(1 cos ))]# d) 2 '1Î2 "# [2(1 cos ))]# d) 1
œ '0 4a1 2cos ) cos# )bd) 1Î2
'1Î2 4 a1 2 cos ) cos# )bd) 1
œ '0
4 ˆ1 2 cos )
œ '0
(6 8 cos ) 2 cos 2)) d) '1Î2 (6 8 cos ) 2 cos 2)) d)
1Î2 1Î2
1 cos 2) ‰ #
d) '1Î2 4 ˆ1 2 cos ) 1
1 cos 2) ‰ #
d)
1
1Î#
œ c6) 8 sin ) sin 2)d !
c6) 8 sin ) sin 2)d 11Î# œ 61 16
13. r œ È3 and r# œ 6 cos 2) Ê 3 œ 6 cos 2) Ê cos 2) œ 1 6
Ê )œ
" #
(in the 1st quadrant); we use symmetry of the
graph to find the area, so A œ 4 '0 ” "# (6 cos 2)) "# ŠÈ3‹ • d) 1Î6
#
œ 2 '0 (6 cos 2) 3) d) œ 2 c3 sin 2) 3)d ! 1Î6
1Î'
œ 3È3 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
675
676
Chapter 11 Parametric Equatins and Polar Coordinates
14. r œ 3a cos ) and r œ a(1 cos )) Ê 3a cos ) œ a(1 cos )) Ê 3 cos ) œ 1 cos ) Ê cos ) œ "# Ê ) œ 13 or 13 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0
1Î3
" #
c(3a cos ))# a# (1 cos ))# d d)
œ '0 a9a# cos# ) a# 2a# cos ) a# cos# )b d) 1Î3
œ '0
1Î3
a8a# cos# ) 2a# cos ) a# b d)
œ '0 c4a# (1 cos 2)) 2a# cos ) a# d d) 1Î3
œ '0 a3a# 4a# cos 2) 2a# cos )b d) 1Î3
1Î$
œ c3a# ) 2a# sin 2) 2a# sin )d !
œ 1a# 2a# ˆ "# ‰ 2a# Š
È3 # ‹
œ a# Š1 1 È3‹
15. r œ 1 and r œ 2 cos ) Ê 1 œ 2 cos ) Ê cos ) œ "# Ê )œ
A œ 2'
1
21 3
in quadrant II; therefore
" 21Î3 #
c(2 cos ))# 1# d d) œ '21Î3 a4 cos# ) 1b d) 1
œ '21Î3 [2(1 cos 2)) 1] d) œ '21Î3 (1 2 cos 2)) d) 1
1
œ c) sin 2)d 1#1Î$ œ
1 3
È3 #
16. r œ 6 and r œ 3 csc ) Ê 6 sin ) œ 3 Ê sin ) œ Ê )œ
1 6
or
51 6
œ '1Î6 ˆ18 51Î6
9 #
; therefore A œ '1Î6
51Î6
csc# )‰ d) œ 18)
" #
" #
a6# 9 csc# )b d)
9 #
cot )‘ 1Î'
&1Î'
œ Š151 9# È3‹ Š31 9# È3‹ œ 121 9È3
17. r œ sec ) and r œ 4 cos ) Ê 4 cos ) œ sec ) Ê cos2 ) œ Ê ) œ 13 , 231 , 431 , or 531 ; therefore 1Î3 A œ 2 0 "# a16 cos# ) sec# )b d) 1Î3 œ 0 a8 8 cos 2) sec# )b d) 1Î3 œ c8) 4 sin 2) tan )d0
1 4
'
'
œ Š 831 2È3 È3‹ a0 0 0b œ
81 3
È3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.5 Area and Lengths in Polar Coordinates 18. r œ 3 csc ) and r œ 4 sin ) Ê 4 sin ) œ 3 csc ) Ê sin2 ) œ Ê ) œ 13 ,
21 41 51 3 , 3 , or 3 ; therefore 1 Î2 " a16 sin# ) 9 csc# 1 Î3 #
A œ 41 2 '
677
3 4
)b d)
œ 41 '1Î3 a8 8 cos 2) 9 csc# )b d) 1 Î2
1 Î2
œ 41 c8) 4 sin 2) 9 cot )d1Î3 œ 41 ’a41 0 0b Š 831 2È3 3È3‹“ œ
81 3
È3
19. (a) r œ tan ) and r œ Š Ê sin# ) œ Š Ê cos# ) Š È2 #
È2 # ‹
È2 # ‹
csc ) Ê tan ) œ Š
È2 # ‹
cos ) Ê 1 cos# ) œ Š
È2 # ‹cos
csc )
È2 # ‹
cos )
) 1 œ 0 Ê cos ) œ È2 or
(use the quadratic formula) Ê ) œ
1 4
(the solution
in the first quadrant); therefore the area of R" is A" œ '0
1Î4
È2 #
œ
" #
and OB œ Š
'01Î4 asec# ) 1b d) œ "# ctan ) )d !1Î% œ "# ˆtan 14 14 ‰ œ "# 18 ; AO œ Š È#2 ‹ csc 1#
" #
tan# ) d) œ È2 # ‹
1 4
csc
œ 1 Ê AB œ Ê1# Š
È2 # # ‹
therefore the area of the region shaded in the text is 2 ˆ "#
1 8
œ
È2 #
Ê the area of R# is A# œ
4" ‰ œ
3 #
1 4
but does not generate the segment AB of the liner œ _ on the line r œ (b)
lim
) Ä 1 Î2
œ
lim
È2 #
sin ) ˆ cos )
r œ sec ) as ) Ä
œ
" 4
;
1 4
generates the arc OB of r œ tan )
csc ).
" ‰ cos )
1c #
È2 È2 # ‹Š # ‹
csc ). Instead the interval generates the half-line from B to
tan ) œ _ and the line x œ 1 is r œ sec ) in polar coordinates; then
) Ä 1 Î2 c
Š
. Note: The area must be found this way
since no common interval generates the region. For example, the interval 0 Ÿ ) Ÿ È2 #
" #
œ
lim
) Ä 1 Î2 c
ˆ sincos) ) 1 ‰ œ
lim
) Ä 1 Î2 c
(tan ) sec ))
) ‰ ˆ cos sin ) œ 0 Ê r œ tan ) approaches
lim
) Ä 1 Î2 c
Ê r œ sec ) (or x œ 1) is a vertical asymptote of r œ tan ). Similarly, r œ sec ) (or x œ 1)
is a vertical asymptote of r œ tan ). 20. It is not because the circle is generated twice from ) œ 0 to 21. The area of the cardioid is A œ 2 '0
1
œ 32)
" #
2) (cos ) 1)# d) œ '0 acos# ) 2 cos ) 1b d) œ '0 ˆ 1 cos 2 cos ) 1‰ d) #
sin 2) 4
1
1
2 sin )‘ ! œ
21. r œ )# , 0 Ÿ ) Ÿ È5 Ê
#
. The area of the circle is A œ 1 ˆ "# ‰ œ
È5
œ 2); therefore Length œ '0
dr d)
È5
31 #
1
È5
œ '0 k)k È)# 4 d) œ (since ) 0) '0
) œ È5 Ê u œ 9“ Ä '4
9
22. r œ
e) È2
,0Ÿ)Ÿ1 Ê
dr d)
" #
œ
Èu du œ
e) È2
1 4
Ê the area requested is actually 3#1
Éa)# b# (2))# d) œ '
) È ) # 4 d ) ; u œ ) # 4 Ê
" 2 $Î# ‘ * # 3 u %
œ
È5
0
" #
È ) % 4) # d)
du œ ) d); ) œ 0 Ê u œ 4,
19 3
; therefore Length œ '0 ÊŠ Èe 2 ‹ Š Èe 2 ‹ d) œ '0 Ê2 Š e# ‹ d) 1
)
#
)
#
1
œ '0 e) d) œ e) ‘ ! œ e1 1 1
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2)
1 4
œ
51 4
678
Chapter 11 Parametric Equatins and Polar Coordinates
23. r œ 1 cos ) Ê
œ sin ); therefore Length œ '0 È(1 cos ))# ( sin ))# d) 21
dr d)
1 œ 2 '0 È2 2 cos ) d) œ 2'0 É 4(1 #cos )) d) œ 4 '0 É 1 #cos ) d) œ 4 '0 cos ˆ #) ‰ d) œ 4 2 sin 2) ‘ ! œ 8 1
1
24. r œ a sin#
) #
1
, 0 Ÿ ) Ÿ 1, a 0 Ê
œ '0 Éa# sin% 1
) #
) #
a# sin#
dr d) ) #
cos#
œ a sin
) #
cos
) #
1
# ; therefore Length œ '0 Ɉa sin# #) ‰ ˆa sin 1
d) œ '0 a ¸sin #) ¸ Ésin# 1
) #
) #
cos#
1
6 1 cos )
œ '0
1Î2
,0Ÿ)Ÿ
1 #
É (1 36 cos ))#
œ ˆsince
" 1 cos )
Ê
dr d)
œ
; therefore Length œ '0
1Î2
6 sin ) (1 cos ))#
d) œ 6 '0
1Î2
36 sin# ) a1 cos )b%
" ¸ 1cos ¸ ) É1
0
1Î2
) #
sin# ) (1 cos ))#
d)
cos# ) sin# ) 0 on 0 Ÿ ) Ÿ 1# ‰ 6 '0 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos d) ) )#
1Î2
1Î2
#
#
6 sin ) ʈ 1 6cos ) ‰ Š (1 cos ))# ‹ d)
1Î2
cos ) È ' œ 6 '0 ˆ 1 "cos ) ‰ É (12 2cos ) )# d) œ 6 2 0
œ 3'0 sec$
# cos #) ‰ d)
d) œ (since 0 Ÿ ) Ÿ 1) a ' sin ˆ #) ‰ d)
1 œ 2a cos 2) ‘ ! œ 2a
25. r œ
) #
d) œ 6'0
1Î4
d) (1 cos ))$Î#
œ 6È2 '0
1Î2
1Î% sec$ u du œ (use tables) 6 Œ sec u2tan u ‘ !
d) ˆ2 cos# #) ‰$Î# " #
'01Î4
œ 3'0
1Î2
¸sec$ #) ¸ d)
sec u du
1Î% œ 6 Š È"2 2" ln ksec u tan uk‘ ! ‹ œ 3 ’È2 ln Š1 È2‹“
26. r œ
2 1 cos )
1 #
,
Ÿ)Ÿ1 Ê
4 œ '1Î2 Ê (1 cos ) ) # Š1 1
œ
dr d)
2 sin ) (1 cos ))#
sin# ) ‹ a1 cos )b#
œ ˆsince 1 cos ) 0 on
1 #
sin ) ; therefore Length œ '1Î2 ʈ 1 2cos ) ‰ Š (12cos ))# ‹ d) 1
) sin d) œ '1Î2 ¸ 1 2cos ) ¸ É (1 (1cos )cos ) )# 1
#
1
œ '1Î2 csc$ ˆ #) ‰ d) œ ˆsince csc 1
1Î#
" #
'11ÎÎ42
)
d)
#
1
2Œ csc u2cot u ‘ 1Î%
#
cos ) sin Ÿ ) Ÿ 1‰ 2 '1Î2 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos ) )#
cos ) d) È ' È ' œ 2 '1Î2 ˆ 1 "cos ) ‰ É (12 2cos ))# d) œ 2 2 1Î2 (1 cos ))$Î# œ 2 2 1Î2 1
) #
1 #
0 on
1
#
#
#
d) ˆ2 sin# )# ‰$Î#
)
d)
œ '1Î2 ¸csc$ #) ¸ d) 1
Ÿ ) Ÿ 1‰ 2 '1Î4 csc$ u du œ (use tables) 1Î2
1Î#
csc u du œ 2 Š È"2 2" ln kcsc u cot uk‘ 1Î% ‹ œ 2 ’ È"2
" #
ln ŠÈ2 1‹“
œ È2 ln Š1 È2‹ 27. r œ cos$ œ '0
1Î4
œ '0
) 3
Ê
dr d)
œ sin
) 3
cos#
) 3
; therefore Length œ '0
Écos' ˆ 3) ‰ sin# ˆ 3) ‰ cos% ˆ 3) ‰ d) œ '0
1Î4
1Î4 1cos ˆ 2) ‰ 3
#
d) œ
" #
)
3 2
sin
28. r œ È1 sin 2) , 0 Ÿ ) Ÿ 1È2 Ê Length œ '0
È
1 2
œ '0
È
1 2
1Î4
É(1 sin 2))
sin 2) ' É 212sin 2) d) œ 0
È
1 2
2) ‘ 1Î% 3 ! dr d)
œ
cos# 2) (1 sin 2))
œ " #
1 8
Ɉcos$ 3) ‰# ˆ sin
) 3
# cos# 3) ‰ d)
ˆcos# 3) ‰ Écos# ˆ 3) ‰ sin# ˆ 3) ‰ d) œ '
1Î4
0
cos# ˆ 3) ‰ d)
3 8
(1 sin 2))"Î# (2 cos 2)) œ (cos 2))(1 sin 2))"Î# ; therefore
d) œ '0
È2 d) œ ’È2 )“
È
1 2
1È# !
#
sin 2) cos É 1 2 sin 2)1 sin 2)
#
2)
d)
œ 21
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.6 Conic Sections 29. Let r œ f()). Then x œ f()) cos ) Ê
dx d)
679
‰# œ cf w ()) cos ) f()) sin )d# œ f w ()) cos ) f()) sin ) Ê ˆ dx d)
œ cf w ())d# cos# ) 2f w ()) f()) sin ) cos ) [f())]# sin# ); y œ f()) sin ) Ê #
dy d)
œ f w ()) sin ) f()) cos )
# # w w # w # # Ê Š dy d) ‹ œ cf ()) sin ) f()) cos )d œ cf ())d sin ) 2f ())f()) sin ) cos ) [f())] cos ). Therefore #
# # w # # # # # w # # ˆ dx ‰# Š dy ˆ dr ‰# d) d) ‹ œ cf ())d acos ) sin )b [f())] acos ) sin )b œ cf ())d [f())] œ r d) .
' Ér# ˆ ddr) ‰# d). ‰# Š dy Thus, L œ '! ʈ dx d) d) ‹ d) œ ! "
30. (a) r œ a Ê
"
#
œ 0; Length œ '0 Èa# 0# d) œ '0 kak d) œ ca)d #!1 œ 21a 21
dr d)
(b) r œ a cos ) Ê
dr d)
œ a sin ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
dr d)
œ a cos ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
1
œ '0 kak d) œ ca)d 1! œ 1a 1
(c) r œ a sin ) Ê
21
1
1
1
œ '0 kak d) œ ca)d 1! œ 1a 1
'021 a(1 cos )) d) œ 2a1 c) sin )d #!1 œ a 21 rav œ 21"0 '0 a d) œ #"1 ca)d #!1 œ a 1Î2 1Î# rav œ ˆ 1 ‰"ˆ 1 ‰ 'c1Î2 a cos ) d) œ 1" ca sin )d 1Î# œ 2a 1
31. (a) rav œ (b) (c)
" 2 1 0
#
#
32. r œ 2f()), ! Ÿ ) Ÿ " Ê
dr d)
œ 2f w ()) Ê r# ˆ ddr) ‰ œ [2f())]# c2f w ())d# Ê Length œ '! É4[f())]# 4 cf w ())d# d) "
#
œ 2 '! É[f())]# cf w ())d# d) which is twice the length of the curve r œ f()) for ! Ÿ ) Ÿ " . "
11.6 CONIC SECTIONS 1. x œ
y# 8
Ê 4p œ 8 Ê p œ 2; focus is (2ß 0), directrix is x œ 2 #
2. x œ y4 Ê 4p œ 4 Ê p œ 1; focus is (1ß 0), directrix is x œ 1 #
3. y œ x6 Ê 4p œ 6 Ê p œ 4. y œ
x# 2
Ê 4p œ 2 Ê p œ
1 #
3 #
; focus is ˆ!ß 3# ‰ , directrix is y œ
3 #
; focus is ˆ!ß 1# ‰ , directrix is y œ 1#
5.
x# 4
y# 9
œ 1 Ê c œ È4 9 œ È13 Ê foci are Š „ È13ß !‹ ; vertices are a „ 2ß 0b ; asymptotes are y œ „ 3# x
6.
x# 4
y# 9
œ 1 Ê c œ È9 4 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a0ß „ 3b
7.
x# 2
y# œ 1 Ê c œ È2 1 œ 1 Ê foci are a „ 1ß 0b ; vertices are Š „ È2ß !‹
8.
y# 4
x# œ 1 Ê c œ È4 1 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a!ß „ 2b ; asymptotes are y œ „ 2x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
680
Chapter 11 Parametric Equatins and Polar Coordinates
9. y# œ 12x Ê x œ
y# 1#
#
10. x# œ 6y Ê y œ x6 Ê 4p œ 6 Ê p œ focus is ˆ!ß 3# ‰ , directrix is y œ 3#
Ê 4p œ 12 Ê p œ 3;
focus is ($ß !), directrix is x œ 3
11. x# œ 8y Ê y œ
x# 8
focus is ˆ!ß
" ‰ 16 ,
x# ˆ "4 ‰
Ê 4p œ
" 4
Ê pœ
directrix is y œ
#
#
focus is ˆ
" ‰ 1# ß ! ,
directrix is x œ
#
14. y œ 8x# Ê y œ ˆx" ‰ Ê 4p œ
;
8
" 16
15. x œ 3y# Ê x œ ˆy" ‰ Ê 4p œ 3
" 16
" 3
" 1#
focus is ˆ!ß
Ê pœ
;
y 12. y# œ 2x Ê x œ # Ê 4p œ 2 Ê p œ " ˆ ‰ focus is # ß ! , directrix is x œ "#
Ê 4p œ 8 Ê p œ 2;
focus is (!ß 2), directrix is y œ 2
13. y œ 4x# Ê y œ
3 #
" 1#
;
" ‰ 32 ,
16. x œ 2y# Ê x œ focus is
ˆ 8" ß !‰ ,
Ê 4p œ
" #
directrix is x œ
Ê pœ " 8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
Ê pœ
" 3#
directrix is y œ
y# ˆ "# ‰
" 8
" #
" 8
;
" 32
;
Section 11.6 Conic Sections #
#
y 17. 16x# 25y# œ 400 Ê #x5 16 œ1 Ê c œ Èa# b# œ È25 16 œ 3
#
19. 2x# y# œ 2 Ê x# y# œ 1 Ê c œ Èa# b# œ È2 1 œ 1
#
#
21. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b# œ È3 2 œ 1
#
#
x 18. 7x# 16y# œ 112 Ê 16 y7 œ 1 Ê c œ Èa# b# œ È16 7 œ 3
#
#
20. 2x# y# œ 4 Ê x# y4 œ 1 Ê c œ Èa# b# œ È4 2 œ È2
#
#
x 22. 9x# 10y# œ 90 Ê 10 y9 œ 1 Ê c œ Èa# b# œ È10 9 œ 1
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681
682
Chapter 11 Parametric Equations and Polar Coordinates #
#
23. 6x# 9y# œ 54 Ê x9 y6 œ 1 Ê c œ Èa# b# œ È9 6 œ È3
#
#
y x 24. 169x# 25y# œ 4225 Ê 25 169 œ1 Ê c œ Èa# b# œ È169 25 œ 12
#
25. Foci: Š „ È2ß !‹ , Vertices: a „ 2ß 0b Ê a œ 2, c œ È2 Ê b# œ a# c# œ 4 ŠÈ2‹ œ 2 Ê 26. Foci: a!ß „ 4b , Vertices: a0ß „ 5b Ê a œ 5, c œ 4 Ê b# œ 25 16 œ 9 Ê 27. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 ; asymptotes are y œ „ x
# # 29. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b# œ È8 8 œ 4; asymptotes are y œ „ x
x# 9
#
y# #5
x# 4
y# #
œ1 #
x 28. 9x# 16y# œ 144 Ê 16 y9 œ 1 Ê c œ Èa# b# œ È16 9 œ 5; asymptotes are y œ „ 34 x
# # 30. y# x# œ 4 Ê y4 x4 œ 1 Ê c œ Èa# b# œ È4 4 œ 2È2; asymptotes are y œ „ x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ1
Section 11.6 Conic Sections
683
31. 8x# 2y# œ 16 Ê x# y8 œ 1 Ê c œ Èa# b# œ È2 8 œ È10 ; asymptotes are y œ „ 2x
32. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b# œ È3 1 œ 2; asymptotes are y œ „ È3x
# # 33. 8y# 2x# œ 16 Ê y# x8 œ 1 Ê c œ Èa# b# œ È2 8 œ È10 ; asymptotes are y œ „ x
y x 34. 64x# 36y# œ 2304 Ê 36 64 œ 1 Ê c œ Èa# b# œ È36 64 œ 10; asymptotes are y œ „ 4
#
#
#
#
#
#
3
35. Foci: Š!ß „ È2‹ , Asymptotes: y œ „ x Ê c œ È2 and
a b
œ 1 Ê a œ b Ê c# œ a# b# œ 2a# Ê 2 œ 2a#
Ê a œ 1 Ê b œ 1 Ê y# x# œ 1 36. Foci: a „ 2ß !b , Asymptotes: y œ „ Ê 4œ
4a# 3
" È3
x Ê c œ 2 and
Ê a# œ 3 Ê a œ È3 Ê b œ 1 Ê
x# 3
b a
œ
" È3
Ê bœ
a È3
4 3
Ê c# œ a# b# œ a#
y# œ 1
37. Vertices: a „ 3ß 0b , Asymptotes: y œ „ 43 x Ê a œ 3 and
b a
œ
4 3
Ê bœ
(3) œ 4 Ê
38. Vertices: a!ß „ 2b , Asymptotes: y œ „ 12 x Ê a œ 2 and
a b
œ
1 2
Ê b œ 2(2) œ 4 Ê
x# 9 y# 4
y# 16
x# 16
39. (a) y# œ 8x Ê 4p œ 8 Ê p œ 2 Ê directrix is x œ 2, focus is (#ß !), and vertex is (!ß 0); therefore the new directrix is x œ 1, the new focus is (3ß 2), and the new vertex is (1ß 2)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ1 œ1
a# 3
œ
4a# 3
684
Chapter 11 Parametric Equations and Polar Coordinates
40. (a) x# œ 4y Ê 4p œ 4 Ê p œ 1 Ê directrix is y œ 1, focus is (!ß 1), and vertex is (!ß 0); therefore the new directrix is y œ 4, the new focus is (1ß 2), and the new vertex is (1ß 3)
41. (a)
x# 16
y# 9
œ 1 Ê center is (!ß 0), vertices are (4ß 0)
(b)
(b)
and (%ß !); c œ Èa# b# œ È7 Ê foci are ŠÈ7ß 0‹ and ŠÈ7ß !‹ ; therefore the new center is (%ß $), the new vertices are (!ß 3) and (8ß 3), and the new foci are Š4 „ È7ß $‹
42. (a)
x# 9
y# 25
œ 1 Ê center is (!ß 0), vertices are (0ß 5) and (0ß 5); c œ Èa# b# œ È16 œ 4 Ê foci are
(b)
(!ß 4) and (!ß 4) ; therefore the new center is (3ß 2), the new vertices are (3ß 3) and (3ß 7), and the new foci are (3ß 2) and (3ß 6)
43. (a)
x# 16
y# 9
œ 1 Ê center is (!ß 0), vertices are (4ß 0)
(b)
and (4ß 0), and the asymptotes are œ „ or Èa# b# œ È25 œ 5 Ê foci are y œ „ 3x 4 ;cœ x 4
y 3
(5ß 0) and (5ß 0) ; therefore the new center is (2ß 0), the new vertices are (2ß 0) and (6ß 0), the new foci are (3ß 0) and (7ß 0), and the new asymptotes are yœ „
3(x 2) 4
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Section 11.6 Conic Sections y# 4
44. (a)
x# 5
œ 1 Ê center is (!ß 0), vertices are (0ß 2)
and (0ß 2), and the asymptotes are yœ „
2x È5
y 2
œ „
x È5
685
(b)
or
; c œ Èa# b# œ È9 œ 3 Ê foci are
(0ß 3) and (0ß 3) ; therefore the new center is (0ß 2), the new vertices are (0ß 4) and (0ß 0), the new foci are (0ß 1) and (0ß 5), and the new asymptotes are 2x y2œ „ È 5
45. y# œ 4x Ê 4p œ 4 Ê p œ 1 Ê focus is ("ß 0), directrix is x œ 1, and vertex is (0ß 0); therefore the new vertex is (2ß 3), the new focus is (1ß 3), and the new directrix is x œ 3; the new equation is (y 3)# œ 4(x 2) 46. y# œ 12x Ê 4p œ 12 Ê p œ 3 Ê focus is (3ß 0), directrix is x œ 3, and vertex is (0ß 0); therefore the new vertex is (4ß 3), the new focus is (1ß 3), and the new directrix is x œ 7; the new equation is (y 3)# œ 12(x 4) 47. x# œ 8y Ê 4p œ 8 Ê p œ 2 Ê focus is (0ß 2), directrix is y œ 2, and vertex is (0ß 0); therefore the new vertex is (1ß 7), the new focus is (1ß 5), and the new directrix is y œ 9; the new equation is (x 1)# œ 8(y 7) Ê focus is ˆ!ß #3 ‰ , directrix is y œ 3# , and vertex is (0ß 0); therefore the new vertex is (3ß 2), the new focus is ˆ3ß "# ‰ , and the new directrix is y œ 7# ; the new equation is
48. x# œ 6y Ê 4p œ 6 Ê p œ
3 #
(x 3)# œ 6(y 2) 49.
x# 6
y# 9
œ 1 Ê center is (!ß 0), vertices are (0ß 3) and (!ß 3); c œ Èa# b# œ È9 6 œ È3 Ê foci are Š!ß È3‹
and Š!ß È3‹ ; therefore the new center is (#ß 1), the new vertices are (2ß 2) and (#ß 4), and the new foci are Š#ß 1 „ È3‹ ; the new equation is 50.
x# 2
(x 2)# 6
(y 1)# 9
œ1
y# œ 1 Ê center is (!ß 0), vertices are ŠÈ2ß !‹ and ŠÈ2ß !‹ ; c œ Èa# b# œ È2 1 œ 1 Ê foci are
(1ß 0) and ("ß !); therefore the new center is (3ß 4), the new vertices are Š3 „ È2ß 4‹ , and the new foci are (2ß 4) and (4ß 4); the new equation is 51.
x# 3
y# #
(x 3)# #
(y 4)# œ 1
œ 1 Ê center is (!ß 0), vertices are ŠÈ3ß !‹ and ŠÈ3ß !‹ ; c œ Èa# b# œ È3 2 œ 1 Ê foci are
(1ß 0) and ("ß !); therefore the new center is (2ß 3), the new vertices are Š2 „ È3ß 3‹ , and the new foci are (1ß 3) and (3ß 3); the new equation is 52.
x# 16
y# #5
(x 2)# 3
(y 3)# #
œ1
œ 1 Ê center is (!ß 0), vertices are (!ß &) and (!ß 5); c œ Èa# b# œ È25 16 œ 3 Ê foci are
(0ß 3) and (0ß 3); therefore the new center is (4ß 5), the new vertices are (4ß 0) and (4ß 10), and the new foci are (4ß 2) and (4ß 8); the new equation is 53.
x# 4
y# 5
(x 4)# 16
(y 5)# #5
œ1
œ 1 Ê center is (!ß 0), vertices are (2ß 0) and (2ß 0); c œ Èa# b# œ È4 5 œ 3 Ê foci are ($ß !) and
(3ß 0); the asymptotes are „
x #
œ
y È5
Ê yœ „
È5x #
; therefore the new center is (2ß 2), the new vertices are
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686
Chapter 11 Parametric Equations and Polar Coordinates
(4ß 2) and (0ß 2), and the new foci are (5ß 2) and (1ß 2); the new asymptotes are y 2 œ „ equation is 54.
x# 16
y# 9
(x 2)# 4
(y 2)# 5
È5 (x 2) #
; the new
œ1
œ 1 Ê center is (!ß 0), vertices are (4ß 0) and (4ß 0); c œ Èa# b# œ È16 9 œ 5 Ê foci are (5ß !)
and (5ß 0); the asymptotes are „
œ
x 4
Ê yœ „
y 3
3x 4
; therefore the new center is (5ß 1), the new vertices are
(1ß 1) and (9ß 1), and the new foci are (10ß 1) and (0ß 1); the new asymptotes are y 1 œ „ the new equation is
(x 5) 16
#
(y 1) 9
#
3(x 5) 4
;
œ1
55. y# x# œ 1 Ê center is (!ß 0), vertices are (0ß 1) and (0ß 1); c œ Èa# b# œ È1 1 œ È2 Ê foci are Š!ß „ È2‹ ; the asymptotes are y œ „ x; therefore the new center is (1ß 1), the new vertices are (1ß 0) and (1ß 2), and the new foci are Š1ß 1 „ È2‹ ; the new asymptotes are y 1 œ „ (x 1); the new equation is (y 1)# (x 1)# œ 1 56.
y# 3
x# œ 1 Ê center is (!ß 0), vertices are Š0ß È3‹ and Š!ß È3‹ ; c œ Èa# b# œ È3 1 œ 2 Ê foci are (!ß #)
and (!ß 2); the asymptotes are „ x œ
y È3
Ê y œ „ È3x; therefore the new center is (1ß 3), the new vertices are
Š"ß $ „ È3‹ , and the new foci are ("ß &) and (1ß 1); the new asymptotes are y 3 œ „ È3 (x 1); the new equation is (y 3)# 3
(x 1)# œ 1
57. x# 4x y# œ 12 Ê x# 4x 4 y# œ 12 4 Ê (x 2)# y# œ 16; this is a circle: center at C(2ß 0), a œ 4 58. 2x# 2y# 28x 12y 114 œ 0 Ê x# 14x 49 y# 6y 9 œ 57 49 9 Ê (x 7)# (y 3)# œ 1; this is a circle: center at C(7ß 3), a œ 1 59. x# 2x 4y 3 œ 0 Ê x# 2x 1 œ 4y 3 1 Ê (x 1)# œ 4(y 1); this is a parabola: V(1ß 1), F(1ß 0) 60. y# 4y 8x 12 œ 0 Ê y# 4y 4 œ 8x 12 4 Ê (y 2)# œ 8(x 2); this is a parabola: V(#ß 2), F(!ß #) 61. x# 5y# 4x œ 1 Ê x# 4x 4 5y# œ 5 Ê (x 2)# 5y# œ 5 Ê
(x 2)# 5
y# œ 1; this is an ellipse: the
center is (2ß 0), the vertices are Š2 „ È5ß 0‹ ; c œ Èa# b# œ È5 1 œ 2 Ê the foci are (4ß 0) and (!ß 0) 62. 9x# 6y# 36y œ 0 Ê 9x# 6 ay# 6y 9b œ 54 Ê 9x# 6(y 3)# œ 54 Ê
x# 6
(y 3)# 9
œ 1; this is an ellipse:
the center is (0ß 3), the vertices are (!ß 0) and (!ß 6); c œ Èa# b# œ È9 6 œ È3 Ê the foci are Š0ß 3 „ È3‹ 63. x# 2y# 2x 4y œ 1 Ê x# 2x 1 2 ay# 2y 1b œ 2 Ê (x 1)# 2(y 1)# œ 2 # Ê (x1) (y 1)# œ 1; this is an ellipse: the center is (1ß 1), the vertices are Š" „ È2ß "‹ ; 2
c œ Èa# b# œ È2 1 œ 1 Ê the foci are (2ß 1) and (0ß 1) 64. 4x# y# 8x 2y œ 1 Ê 4 ax# 2x 1b y# 2y 1 œ 4 Ê 4(x 1)# (y 1)# œ 4 Ê (x 1)#
(y1)# 4
œ 1; this is an ellipse: the center is (1ß 1), the vertices are (1ß 3) and
(1ß 1); c œ Èa# b# œ È4 1 œ È3 Ê the foci are Š1ß " „ È3‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.6 Conic Sections 65. x# y# 2x 4y œ 4 Ê x# 2x 1 ay# 4y 4b œ 1 Ê (x 1)# (y 2)# œ 1; this is a hyperbola: the center is (1ß 2), the vertices are (2ß 2) and (!ß 2); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š1 „ È2ß #‹ ; the asymptotes are y 2 œ „ (x 1) 66. x# y# 4x 6y œ 6 Ê x# 4x 4 ay# 6y 9b œ 1 Ê (x 2)# (y 3)# œ 1; this is a hyperbola: the center is (2ß 3), the vertices are (1ß 3) and (3ß 3); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š2 „ È2ß 3‹ ; the asymptotes are y 3 œ „ (x 2) (y 3)# 6
67. 2x# y# 6y œ 3 Ê 2x# ay# 6y 9b œ 6 Ê
x# 3
œ 1; this is a hyperbola: the center is (!ß $),
the vertices are Š!ß 3 „ È6‹ ; c œ Èa# b# œ È6 3 œ 3 Ê the foci are (0ß 6) and (!ß 0); the asymptotes are y 3 È6
œ „
x È3
Ê y œ „ È2x 3
68. y# 4x# 16x œ 24 Ê y# 4 ax# 4x 4b œ 8 Ê
y# 8
(x 2)# 2
œ 1; this is a hyperbola: the center is (2ß 0),
the vertices are Š2ß „ È8‹ ; c œ Èa# b# œ È8 2 œ È10 Ê the foci are Š2ß „ È10‹ ; the asymptotes are y È8
x 2 È2
œ „
Ê y œ „ 2(x 2) y# k
69. (a) y# œ kx Ê x œ
; the volume of the solid formed by
Èkx
revolving R" about the y-axis is V" œ '0 œ
1 k#
Èkx
'0
y% dy œ
1x# Èkx 5
#
#
1 Š yk ‹ dy
; the volume of the right
circular cylinder formed by revolving PQ about the y-axis is V# œ 1x# Èkx Ê the volume of the solid formed by revolving R# about the y-axis is V$ œ V# V" œ
41x# Èkx 5
. Therefore we can see the
ratio of V$ to V" is 4:1.
(b) The volume of the solid formed by revolving R# about the x-axis is V" œ '0 1 ŠÈkt‹ dt œ 1k'0 t dt x
œ
1kx# #
#
x
. The volume of the right circular cylinder formed by revolving PS about the x-axis is #
V# œ 1 ŠÈkx‹ x œ 1kx# Ê the volume of the solid formed by revolving R" about the x-axis is V$ œ V# V" œ 1kx# 70. y œ '
w H
x dx œ
w H
#
1kx# #
Š x# ‹ C œ
wx# 2H
œ
1kx# #
. Therefore the ratio of V$ to V" is 1:1.
C; y œ 0 when x œ 0 Ê 0 œ
w(0)# 2H
C Ê C œ 0; therefore y œ
wx# 2H
is the
equation of the cable's curve 71. x# œ 4py and y œ p Ê x# œ 4p# Ê x œ „ 2p. Therefore the line y œ p cuts the parabola at points (2pß p) and (2pß p), and these points are È[2p (2p)]# (p p)# œ 4p units apart. 72. x lim Š b x ba Èx# a# ‹ œ Ä_ a œ
b a x lim Ä_
’
x # ax # a # b “ x È x # a#
œ
b a x lim Ä_
b a x lim Ä_
’
Šx Èx# a# ‹ œ
a# “ x È x # a#
b a x lim Ä_
–
Šx Èx# a# ‹ Šx Èx# a# ‹ x È x # a#
œ0
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—
687
688
Chapter 11 Parametric Equations and Polar Coordinates
73. Let y œ É1
x# 4
on the interval 0 Ÿ x Ÿ 2. The area of the inscribed rectangle is given by x# 4‹
A(x) œ 2x Š2É1 Ê Aw (x) œ 4É1
x# 4
œ 4xÉ1
x# É1 x4#
x# 4
(since the length is 2x and the height is 2y) x# 4
. Thus Aw (x) œ 0 Ê 4É1
x# É1 x4#
œ 0 Ê 4 Š1
x# 4‹
x# œ 0 Ê x# œ 2
Ê x œ È2 (only the positive square root lies in the interval). Since A(0) œ A(2) œ 0 we have that A ŠÈ2‹ œ 4 is the maximum area when the length is 2È2 and the height is È2. 74. (a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root #
Ê V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241 2
2
#
(b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root #
Ê V œ 2'0 1 ŠÉ4 49 y# ‹ dy œ 2 '0 1 ˆ4 49 y# ‰ dy œ 21 4y 3
75. 9x# 4y# œ 36 Ê y# œ œ
91 4
9x# 36 4
'24 ax# 4b dx œ 941 ’ x3
$
3
4 27
$
y$ ‘ ! œ 161
Ê y œ „ #3 Èx# 4 on the interval 2 Ÿ x Ÿ 4 Ê V œ '2 1 Š #3 Èx# 4‹ dx #
4
%
4x“ œ #
91 4
ˆ 64 ‰ ˆ8 ‰‘ œ 3 16 3 8
91 4
ˆ 56 ‰ 3 8 œ
31 4
(56 24) œ 241
76. Let P" (pß y" ) be any point on x œ p, and let P(xß y) be a point where a tangent intersects y# œ 4px. Now y# œ 4px Ê 2y
dy dx
œ 4p Ê
dy dx
œ
2p y
Ê y# yy" œ 2px 2p# . Since x œ Ê
" #
y# yy" 2p# œ 0 Ê y œ
tangents from P" are m" œ
; then the slope of a tangent line from P" is y# 4p
œ
dy dx
œ
#
y , we have y# yy" œ 2p Š 4p ‹ 2p# Ê y# yy" œ
2y" „ È4y#" 16p# #
2p y" Èy#" 4p#
y y" x (p)
and m# œ
" #
2p y
y# 2p#
œ y" „ Èy#" 4p# . Therefore the slopes of the two Ê m" m# œ
2p y" Èy#" 4p#
4p# y#" ay#" 4p# b
œ 1
Ê the lines are perpendicular 77. (x 2)# (y 1)# œ 5 Ê 2(x 2) 2(y 1)
dy dx
œ0 Ê
dy dx
2 # # œ yx 1 ; y œ 0 Ê (x 2) (0 1) œ 5
Ê (x 2)# œ 4 Ê x œ 4 or x œ 0 Ê the circle crosses the x-axis at (4ß 0) and (!ß 0); x œ 0 Ê (0 2)# (y 1)# œ 5 Ê (y 1)# œ 1 Ê y œ 2 or y œ 0 Ê the circle crosses the y-axis at (!ß 2) and (!ß !). At (4ß 0): At (!ß !): At (!ß #):
2 œ 04 1 œ 2 Ê the tangent line is y œ 2(x 4) or y œ 2x 8
dy dx dy dx dy dx
2 œ 00 1 œ 2 Ê the tangent line is y œ 2x
2 œ 20 1 œ 2 Ê the tangent line is y 2 œ 2x or y œ 2x 2
78. x# y# œ 1 Ê x œ „ È1 y# on the interval 3 Ÿ y Ÿ 3 Ê V œ 'c3 1 ˆÈ1 y# ‰ dy œ 2'0 1 ˆÈ1 y# ‰ dy 3
œ 21'0 a1 y# b dy œ 21 ’y 3
79. Let y œ É16 vertical strip:
16 9
$ y$ 3 “!
É16 aµ x ßµ y b œ xß #
# É16 16 9 x
#
#
3
œ 241
x# on the interval 3 Ÿ x Ÿ 3. Since the plate is symmetric about the y-axis, x œ 0. For a
Ê mass œ dm œ $ dA œ $É16 µ y dm œ
#
Š$ É16
16 9
16 9
16 9
x#
, length œ É16
16 9
x# , width œ dx Ê area œ dA œ É16
x# dx. Moment of the strip about the x-axis:
x# ‹ dx œ $ ˆ8 98 x# ‰ dx so the moment of the plate about the x-axis is
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
16 9
x# dx
Section 11.7 Conics in Polar Coordinates Mx œ ' µ y dm œ 'c3 $ ˆ8 89 x# ‰ dx œ $ 8x 3
M œ 'c3 $ É16 3
16 9
8 27
$
x$ ‘ $ œ 32$ ; also the mass of the plate is
x# dx œ 'c3 4$ É1 ˆ 3" x‰ dx œ 4$ 'c1 3È1 u# du where u œ 3
689
1
#
x 3
Ê 3 du œ dx; x œ 3
Ê u œ 1 and x œ 3 Ê u œ 1. Hence, 4$ 'c1 3È1 u# du œ 12$ 'c1 È1 u# du 1
œ 12$ ’ "2 ŠuÈ1 u# sin" u‹“ 80. y œ Èx# 1 Ê
dy dx
" #
œ
È2
1 ' œ É 2x x# 1 Ê S œ 0 #
–
u œ È2x — Ä du œ È2 dx
21 È2
ax# 1b
"
"
1
œ 61$ Ê y œ
"Î#
(2x) œ
x È x# 1
Mx M
œ
32$ 61$
#
œ
Ê Š dy dx ‹ œ
È2
16 31
. Therefore the center of mass is ˆ!ß 3161 ‰ .
x# x # 1
#
dy Ê Ê1 Š dx ‹ œ É1
È2
dy 1 È # ' 21yÊ1 Š dx ‹ dx œ '0 21Èx# 1 É 2x x# 1 dx œ 0 21 2x 1 dx ; #
'02 Èu# 1 du œ È21
2
#
#
’ "2 ŠuÈu# 1 ln Šu Èu# 1‹‹“ œ !
1 È2
’2È5 ln Š2 È5‹“
81. (a) tan " œ mL Ê tan " œ f w (x! ) where f(x) œ È4px ; f w (x) œ œ
2p y!
" #
(4px)"Î# (4p) œ
(c) tan ! œ œ
2p È4px
Ê f w (x! ) œ
2p È4px!
Ê tan " œ
(b) tan 9 œ mFP œ
2p y! . y! 0 y! x! p œ x! p
tan 9 tan " 1 tan 9 tan "
y#! 2p(x! p) y! (x! p 2p)
œ
œ
y Š x ! p c y2p ‹
!
!
y 1 b Š x ! p ‹ Š y2p ‹
!
!
4px! 2px! 2p# y! (x! p)
œ
2p(x! p) y! (x! p)
œ
2p y!
11.7 CONICS IN POLAR COORDINATES # y# 1. 16x# 25y# œ 400 Ê #x5 16 œ 1 Ê c œ Èa# b# œ È25 16 œ 3 Ê e œ ca œ 35 ; F a „ 3ß 0b ;
directrices are x œ 0 „
a e
œ „
5 ˆ 35 ‰
œ „
25 3
# x# 2. 7x# 16y# œ 112 Ê 16 y7 œ 1 Ê c œ Èa# b# œ È16 7 œ 3 Ê e œ ca œ 34 ; F a „ 3ß 0b ;
directrices are x œ 0 „
a e
x# x# 1
œ „
4 ˆ 34 ‰
œ „
16 3
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690
Chapter 11 Parametric Equations and Polar Coordinates
3. 2x# y# œ 2 Ê x# y2 œ 1 Ê c œ Èa# b# œ È2 1 œ 1 Ê e œ ca œ È12 ; F a0ß „ 1b ; #
directrices are y œ 0 „
4. 2x# y# œ 4 Ê
x# #
a e
œ „
a e
œ „2
œ 1 Ê c œ Èa# b#
y# 4
œ È4 2 œ È2 Ê e œ directrices are y œ 0 „
È2 Š È12 ‹
c a
œ
È2 2
; F Š0ß „ È2‹ ;
œ „ È22 œ „ 2È2 Š ‹ 2
# # 5. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b# œ È3 2 œ 1 Ê e œ ca œ È13 ; F a0ß „ 1b ;
directrices are y œ 0 „
a e
œ „
È3
œ „3
Š È13 ‹
# x# 6. 9x# 10y# œ 90 Ê 10 y9 œ 1 Ê c œ Èa# b# œ È10 9 œ 1 Ê e œ ca œ È110 ; F a „ 1ß 0b ;
directrices are x œ 0 „
7. 6x# 9y# œ 54 Ê
x# 9
a e
œ „
y# 6
œ È9 6 œ È3 Ê e œ directrices are x œ 0 „
a e
È10 Š È110 ‹
œ „ 10
œ 1 Ê c œ Èa# b# c a
œ
È3 3
; F Š „ È3ß 0‹ ;
œ „ È33 œ „ 3È3 Š ‹ 3
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Section 11.7 Conics in Polar Coordinates
691
y# x# 8. 169x# 25y# œ 4225 Ê 25 169 œ 1 Ê c œ Èa# b# œ È169 25 œ 12 Ê e œ c œ 12 ; F a0ß „ 12b ; a
directrices are y œ 0 „
a e
œ „
13
13 ˆ 12 ‰ 13
œ „
169 12
x# #7
y# 36
9. Foci: a0ß „ 3b , e œ 0.5 Ê c œ 3 and a œ
c e
œ
3 0.5
œ 6 Ê b# œ 36 9 œ 27 Ê
10. Foci: a „ 8ß 0b , e œ 0.2 Ê c œ 8 and a œ
c e
œ
8 0.#
œ 40 Ê b# œ 1600 64 œ 1536 Ê
œ1 x# 1600
y# 1536
11. Vertices: a0ß „ 70b , e œ 0.1 Ê a œ 70 and c œ ae œ 70(0.1) œ 7 Ê b# œ 4900 49 œ 4851 Ê
œ1
x# 4851
12. Vertices: a „ 10ß 0b , e œ 0.24 Ê a œ 10 and c œ ae œ 10(0.24) œ 2.4 Ê b# œ 100 5.76 œ 94.24 Ê 13. Focus: ŠÈ5ß !‹ , Directrix: x œ Ê eœ
È5 3
. Then PF œ
È5 3
Ê x# 2È5 x 5 y# œ 14. Focus: (%ß 0), Directrix: x œ PF œ
È œ #3 PD 3 ˆ # 32 4 x 3
9 È5
Ê c œ ae œ È5 and
256 ‰ 9
Ê
" 4
œ
#
5 9
Šx#
16 3
18 È5
x
Ê
81 5 ‹
Ê c œ ae œ 4 and
x# y# œ
16 3
È3 #
¸x
Ê
#
x ˆ 64 ‰ 3
4 9
a e
È5 3
16 3
Ê
œ
ae e#
16 3
œ
ae e#
¹x
x# y# œ 4 Ê
œ
16 ¸ 3
Ê
9 È5
PD Ê ÊŠx È5‹ (y 0)# œ
Ê È(x 4)# (y 0)# œ x
a e
9 È5 ¹
x# 9
Ê
Ê (x 4)# y# œ #
y ˆ 16 ‰ 3
9 È5
y# 4
4 e# 3 4
Ê
È5 e#
œ
x# 100
Ê e# œ
9 È5 #
Ê Šx È5‹ y# œ
5 9
y# 4900
y# 94.24
#
1 4
#
ax# 32x 256b Ê
3 4
x# y# œ 48 Ê
Šx
œ
" #
1 È2
. Then PF œ #
1 È2
#
œ1 œ
16 3
ˆx
Ê e# œ
16 ‰# 3
Ê eœ
3 4
È3 #
. Then
Ê x# 8x 16 y#
1 #
. Then
4
x# 64
y# 48
œ1
#
PD Ê ÊŠx È2‹ (y 0)# œ
Šx 2È2‹ Ê x# 2È2 x 2 y# œ
9 È5 ‹
œ1
16. Focus: ŠÈ2ß !‹ , Directrix: x œ 2È2 Ê c œ ae œ È2 and Ê eœ
œ1
5 9
4 " # 15. Focus: (%ß 0), Directrix: x œ 16 Ê c œ ae œ 4 and ae œ 16 Ê ae e# œ 16 Ê e# œ 16 Ê e œ 4 Ê e œ PF œ 1 PD Ê È(x 4)# (y 0)# œ 1 kx 16k Ê (x 4)# y# œ 1 (x 16)# Ê x# 8x 16 y#
œ
œ1
" #
a e
œ 2È 2 Ê
1 È2
ae e#
œ 2È 2 Ê
È2 e#
œ 2 È 2 Ê e# œ
#
¹x 2È2¹ Ê Šx È2‹ y#
Šx# 4È2 x 8‹ Ê
" #
x# y# œ 2 Ê
x# 4
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y# #
œ1
" #
692
Chapter 11 Parametric Equations and Polar Coordinates
17. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 Ê e œ œ
È2 1
c a
œ È2 ; asymptotes are y œ „ x; F Š „ È2 ß !‹ ;
directrices are x œ 0 „
a e
œ „
" È2
# x# 18. 9x# 16y# œ 144 Ê 16 y9 œ 1 Ê c œ Èa# b# œ È16 9 œ 5 Ê e œ ca œ 54 ; asymptotes are
y œ „ 34 x; F a „ 5ß !b ; directrices are x œ 0 „ œ „
a e
"6 5
# # 19. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b# œ È8 8 œ 4 Ê e œ ca œ È48 œ È2 ; asymptotes are
y œ „ x; F a0ß „ 4b ; directrices are y œ 0 „ œ „
È8 È2
a e
œ „2
20. y# x# œ 4 Ê
y# 4
x# 4
œ 1 Ê c œ Èa# b#
œ È 4 4 œ 2È 2 Ê e œ
c a
œ
2È 2 2
œ È2 ; asymptotes
are y œ „ x; F Š0ß „ 2È2‹ ; directrices are y œ 0 „ œ „
2 È2
a e
œ „ È2
21. 8x# 2y# œ 16 Ê
x# 2
y# 8
œ È2 8 œ È10 Ê e œ
œ 1 Ê c œ Èa# b# c a
œ
È10 È2
œ È5 ; asymptotes
are y œ „ 2x; F Š „ È10ß !‹ ; directrices are x œ 0 „ œ „
È2 È5
œ „
a e
2 È10
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Section 11.7 Conics in Polar Coordinates
693
# 22. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b# œ È3 1 œ 2 Ê e œ ca œ È23 ; asymptotes are
y œ „ È3 x; F a0ß „ 2b ; directrices are y œ 0 „ œ „
È3 Š È23 ‹
œ „
a e
3 #
23. 8y# 2x# œ 16 Ê
y# 2
x# 8
œ È2 8 œ È10 Ê e œ
œ 1 Ê c œ Èa# b# c a
È10 È2
œ
œ È5 ; asymptotes
are y œ „ x# ; F Š0ß „ È10‹ ; directrices are y œ 0 „ œ „
È2 È5
œ „
a e
2 È10
y# x# 24. 64x# 36y# œ 2304 Ê 36 64 œ 1 Ê c œ Èa# b# 5 œ È36 64 œ 10 Ê e œ ca œ 10 6 œ 3 ; asymptotes are
y œ „ 43 x; F a „ 10ß !b ; directrices are x œ 0 „ œ „
6 ˆ 53 ‰
œ „
a e
18 5
25. Vertices a!ß „ 1b and e œ 3 Ê a œ 1 and e œ
c a
œ 3 Ê c œ 3a œ 3 Ê b# œ c# a# œ 9 1 œ 8 Ê y#
x# 8
œ1
26. Vertices a „ 2ß !b and e œ 2 Ê a œ 2 and e œ
c a
œ 2 Ê c œ 2a œ 4 Ê b# œ c# a# œ 16 4 œ 12 Ê
x# 4
y# 1#
œ1
œ 3 Ê c œ 3a Ê a œ 1 Ê b# œ c# a# œ 9 1 œ 8 Ê x#
y# 8
œ1
27. Foci a „ 3ß !b and e œ 3 Ê c œ 3 and e œ
c a
28. Foci a!ß „ 5b and e œ 1.25 Ê c œ 5 and e œ œ 25 16 œ 9 Ê
#
y 16
#
x 9
c a
œ 1.25 œ
5 4
Ê cœ
5 4
a Ê 5œ
5 4
a Ê a œ 4 Ê b# œ c# a#
œ1
29. e œ 1, x œ 2 Ê k œ 2 Ê r œ
2(1) 1 (1) cos )
œ
2 1cos )
30. e œ 1, y œ 2 Ê k œ 2 Ê r œ
2(1) 1 (1) sin )
œ
2 1sin )
31. e œ 5, y œ 6 Ê k œ 6 Ê r œ
6(5) 1 5 sin )
32. e œ 2, x œ 4 Ê k œ 4 Ê r œ
4(2) 1 2 cos )
33. e œ "# , x œ 1 Ê k œ 1 Ê r œ
ˆ "# ‰ (1) 1 ˆ "# ‰ cos )
œ
œ
30 15 sin )
8 12 cos )
œ
1 2cos )
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
694
Chapter 11 Parametric Equations and Polar Coordinates ˆ "4 ‰ (2) 1 ˆ "4 ‰ cos )
34. e œ 4" , x œ 2 Ê k œ 2 Ê r œ
35. e œ "5 , y œ 10 Ê k œ 10 Ê r œ 36. e œ "3 , y œ 6 Ê k œ 6 Ê r œ 37. r œ
" 1 cos )
38. r œ
6 2 cos )
œ
ˆ "5 ‰ (10) 1 ˆ "5 ‰ sin )
ˆ "3 ‰ (6) 1 ˆ "3 ‰ sin )
œ
2 4cos )
œ
10 5sin )
6 3sin )
Ê e œ 1, k œ 1 Ê x œ 1
œ
3 1 ˆ "# ‰ cos )
Ê eœ
" #
, k œ 6 Ê x œ 6;
#
a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 3 Ê
3 4
aœ3
Ê a œ 4 Ê ea œ 2
39. r œ
25 10 5 cos )
Ê eœ
" #
Ê rœ #
40. r œ
4 22 cos )
41. r œ
400 16 8 sin ) " #
œ
ˆ #5 ‰
1 ˆ "# ‰ cos )
, k œ 5 Ê x œ 5; a a1 e# b œ ke
Ê a ’1 ˆ "# ‰ “ œ
eœ
ˆ 25 ‰ 10
5 ‰ 1 ˆ 10 cos )
Ê rœ
5 #
Ê
2 1cos )
Ê rœ
3 4
aœ
5 #
Ê aœ
10 3
Ê ea œ
5 3
Ê e œ 1, k œ 2 Ê x œ 2
ˆ 400 ‰ 16
8 ‰ 1 ˆ 16 sin )
Ê rœ
25 1 ˆ "# ‰ sin )
, k œ 50 Ê y œ 50; a a1 e# b œ ke #
Ê a ’1 ˆ "# ‰ “ œ 25 Ê Ê ea œ
3 4
a œ 25 Ê a œ
100 3
50 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 11.7 Conics in Polar Coordinates 42. r œ
12 3 3 sin )
Ê rœ
4 1 sin )
Ê e œ 1,
43. r œ
kœ4 Ê yœ4
44. r œ
4 2 sin )
8 2 2 sin )
Ê rœ
4 1 sin )
Ê e œ 1,
k œ 4 Ê y œ 4
Ê rœ
2 1 ˆ "# ‰ sin )
Ê eœ
" #
,kœ4 #
Ê y œ 4; a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 2 Ê
3 4
aœ2 Ê aœ
8 3
Ê ea œ
4 3
45. r cos ˆ) 14 ‰ œ È2 Ê r ˆcos ) cos 14 sin ) sin 14 ‰ œ È2 Ê " r cos ) " r sin ) œ È2 Ê " x È2
È2
È2
" È2
y
œ È2 Ê x y œ 2 Ê y œ 2 x
46. r cos ˆ) Ê
31 ‰ 4
œ 1 Ê r ˆcos ) cos
È2 2
r cos ) Ê y œ x È 2
47. r cos ˆ)
21 ‰ 3
È3 2
sin ) sin
31 ‰ 4
œ1
r sin ) œ 1 Ê x y œ È2
œ 3 Ê r ˆcos ) cos
Ê r cos ) 1 2
È2 2
31 4
21 3
sin ) sin " #
r sin ) œ 3 Ê x
Ê x È 3 y œ 6 Ê y œ
È3 3
È3 #
21 ‰ 3
œ3
yœ3
x 2È 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
695
696
Chapter 11 Parametric Equations and Polar Coordinates
48. r cos ˆ) 13 ‰ œ 2 Ê r ˆcos ) cos Ê
1 2
r cos )
È3 2
r sin ) œ 2 Ê
Ê x È3 y œ 4 Ê y œ
È3 3
1 3
sin ) sin 13 ‰ œ 2
" #
x
x
4È 3 3
È3 #
yœ2
È 49. È2 x È2 y œ 6 Ê È2 r cos ) È2 r sin ) œ 6 Ê r Š #2 cos )
È2 #
sin )‹ œ 3 Ê r ˆcos
1 4
cos ) sin
œ 3 Ê r cos ˆ) 14 ‰ œ 3 È 50. È3 x y œ 1 Ê È3 r cos ) r sin ) œ 1 Ê r Š #3 cos )
œ
" #
Ê r cos ˆ) 16 ‰ œ
1 #
sin )‹ œ
" #
Ê r ˆcos
1 6
cos ) sin
1 6
sin )‰
" #
51. y œ 5 Ê r sin ) œ 5 Ê r sin ) œ 5 Ê r sin ()) œ 5 Ê r cos ˆ 1# ())‰ œ 5 Ê r cos ˆ) 1# ‰ œ 5 52. x œ 4 Ê r cos ) œ 4 Ê r cos ) œ 4 Ê r cos () 1) œ 4 53.
54.
55.
56.
57. (x 6)# y# œ 36 Ê C œ (6ß 0), a œ 6 Ê r œ 12 cos ) is the polar equation
58. (x 2)# y# œ 4 Ê C œ (2ß 0), a œ 2 Ê r œ 4 cos ) is the polar equation
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 4
sin )‰
Section 11.7 Conics in Polar Coordinates 59. x# (y 5)# œ 25 Ê C œ (!ß 5), a œ 5 Ê r œ 10 sin ) is the polar equation
60. x# (y 7)# œ 49 Ê C œ (!ß 7), a œ 7 Ê r œ 14 sin ) is the polar equation
61. x# 2x y# œ 0 Ê (x 1)# y# œ 1 Ê C œ (1ß 0), a œ 1 Ê r œ 2 cos ) is the polar equation
62. x# 16x y# œ 0 Ê (x 8)# y# œ 64 Ê C œ (8ß 0), a œ 8 Ê r œ 16 cos ) is the polar equation
# 63. x# y# y œ 0 Ê x# ˆy "# ‰ œ 4" Ê C œ ˆ!ß "# ‰ , a œ "# Ê r œ sin ) is the
# 64. x# y# 34 y œ 0 Ê x# ˆy 32 ‰ œ 94 Ê C œ ˆ0ß 23 ‰ , a œ 23 Ê r œ 43 sin ) is the
polar equation
65.
polar equation
66.
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697
698
Chapter 11 Parametric Equations and Polar Coordinates
67.
68.
69.
70.
71.
72.
73.
74.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 11 Practice Exercises 75. (a) Perihelion œ a ae œ a(1 e), Aphelion œ ea a œ a(1 e) (b) Planet Perihelion Aphelion Mercury 0.3075 AU 0.4667 AU Venus 0.7184 AU 0.7282 AU Earth 0.9833 AU 1.0167 AU Mars 1.3817 AU 1.6663 AU Jupiter 4.9512 AU 5.4548 AU Saturn 9.0210 AU 10.0570 AU Uranus 18.2977 AU 20.0623 AU Neptune 29.8135 AU 30.3065 AU (0.3871) a1 0.2056# b 0.3707 œ 1 0.2056 1 0.2056 cos ) cos ) (0.7233) a1 0.0068# b 0.7233 Venus: r œ 1 0.0068 cos ) œ 1 0.0068 cos ) 0.0167# b 0.9997 Earth: r œ 11a10.0167 cos ) œ 1 0.0617 cos ) a1 0.0934# b 1.511 Mars: r œ (1.524) œ 1 0.0934 1 0.0934 cos ) cos ) # a1 0.0484 b 5.191 Jupiter: r œ (5.203) œ 1 0.0484 cos ) 1 0.0484 cos ) (9.539) a1 0.0543# b 9.511 Saturn: r œ 1 0.0543 cos ) œ 1 0.0543 cos ) a1 0.0460# b 19.14 Uranus: r œ (19.18) œ 1 0.0460 cos ) 1 0.0460 cos ) (30.06) a1 0.0082# b 30.06 Neptune: r œ 1 0.0082 cos ) œ 1 0.0082 cos )
76. Mercury: r œ
CHAPTER 11 PRACTICE EXERCISES 1. x œ
t #
and y œ t 1 Ê 2x œ t Ê y œ 2x 1
" # tan t and y# œ "4 # #
3. x œ
and y œ
" #
sec t Ê x# œ
" 4
tan# t
sec# t Ê 4x# œ tan# t and
2. x œ Èt and y œ 1 Èt Ê y œ 1 x
4. x œ 2 cos t and y œ 2 sin t Ê x# œ 4 cos# t and y# œ 4 sin# t Ê x# y# œ 4
4y œ sec t Ê 4x# 1 œ 4y# Ê 4y# 4x# œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
699
700
Chapter 11 Parametric Equations and Polar Coordinates
5. x œ cos t and y œ cos# t Ê y œ (x)# œ x#
6. x œ 4 cos t and y œ 9 sin t Ê x# œ 6 cos# t and x# 16
y# œ 81 sin# t Ê
x# 9
7. 16x# 9y# œ 144 Ê
y# 16
y# 81
œ1
œ 1 Ê a œ 3 and b œ 4 Ê x œ 3 cos t and y œ 4 sin t, 0 Ÿ t Ÿ 21
8. x# y# œ 4 Ê x œ 2 cos t and y œ 2 sin t, 0 Ÿ t Ÿ 61 9. x œ
" #
" #
tan t, y œ
Ê xœ
" #
tan
1 3
sec t Ê
œ
œ 2 cos3 ˆ 13 ‰ œ
È3 #
dy dx
œ
dy/dt dx/dt
œ
" #
sec
1 3
and y œ
" #
sec t tan t " # # sec t
œ
tan t sec t
œ sin t Ê
œ1 Ê yœ
È3 #
x 4" ;
d# y dx#
dy dx ¹ tœ1Î3
œ
dyw /dt dx/dt
œ sin
œ
" #
1 3
cos t sec# t
œ
È3 #
;tœ
1 3
œ 2 cos3 t Ê
d# y dx# ¹ tœ1Î3
" 4
10. x œ "
" t#
,yœ"
yœ1
3 #
œ "# Ê y œ 3x
Ê
3 t
dy dx
œ
11. (a) x œ 4t2 , y œ t3 1 Ê t œ „ (b) x œ cos t, y œ tan t Ê sec t œ
"3 4
;
Èx 2 1 x
Š t3# ‹
œ
dy/dt dx/dt
Š t2$ ‹
d# y dx#
œ
œ 32 t Ê
dyw /dt dx/dt
œ
ÊyœŠ„
ˆ #3 ‰ Š t2$ ‹
Èx 3 2 ‹
dy dx ¹ tœ2
œ
œ 3# (2) œ 3; t œ 2 Ê x œ 1
3 $ 4 t
Ê
1œ „
d# y dx# ¹ tœ2
x3Î2 8
1
1 x2
1œ
Ê tan2 t 1 œ sec2 t Ê y2 œ
œ
3 4
1 x2 x2
" ##
œ
5 4
and
(2)$ œ 6
È 1 x2 x
Êyœ „
12. (a) The line through a1, 2b with slope 3 is y œ 3x 5 Ê x œ t, y œ 3t 5, _ t _ (b) ax 1b2 ay 2b2 œ 9 Ê x 1 œ 3 cos t, y 2 œ 3 sin t Ê x œ 1 3 cos t, y œ 2 3 sin t, 0 Ÿ t Ÿ 21 (c) y œ 4x2 x Ê x œ t, y œ 4t2 t, _ t _ (d) 9x2 4y2 œ 36 Ê 13. y œ x"Î#
x$Î# 3
Ê
dy dx
x2 4
œ
" #
y2 9
œ 1 Ê x œ 2 cos t, y œ 3 sin t, 0 Ÿ t Ÿ 21 #
x"Î# "# x"Î# Ê Š dy dx ‹ œ
" 4
ˆ x" 2 x‰ Ê L œ ' É1 4" ˆ x" 2 x‰ dx 1 4
# Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1 4
œ
" #
ˆ4
2 3
14. x œ y#Î$ Ê œ '1
8
4
† 8‰ ˆ2 23 ‰‘ œ dx dy
È9x#Î$ 4 3x"Î$
œ
15. y œ
5 12
ˆ2 #
14 ‰ 3
x"Î$ Ê Š dx dy ‹ œ
2 3
œ
4x #Î$ 9
" #
ˆx"Î# x"Î# ‰ dx œ
" #
2x"Î# 23 x$Î# ‘ % "
10 3 dx Ê L œ '1 Ê1 Š dy ‹ dy œ '1 É1 #
8
8
4 9x#Î$
dy
'18 È9x#Î$ 4 ˆx"Î$ ‰ dx; u œ 9x#Î$ 4 Ê du œ 6y"Î$ dy; x œ 1 40 " ' " 2 $Î# ‘ %! Ä L œ 18 u"Î# du œ 18 œ #"7 40$Î# 13$Î# ‘ ¸ 7.634 3 u "$ 13
dx œ
x œ 8 Ê u œ 40d
" #
4
" 3
x'Î& 58 x%Î& Ê
dy dx
œ
" #
#
dy x"Î& "# x"Î& Ê Š dx ‹ œ
" 4
Ê u œ 13,
ˆx#Î& 2 x#Î& ‰
# Ê L œ '1 É1 "4 ax#Î& 2 x#Î& b dx Ê L œ '1 É 4" ax#Î& 2 x#Î& b dx œ ' É 4" ax"Î& x"Î& b dx 32
32
32
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 11 Practice Exercises œ '1
32
œ
" 48
16. x œ
" #
ˆx"Î& x"Î& ‰ dx œ
(1260 450) œ
" 1#
y$
" y
Ê
" % œ '1 É 16 y 2
" #
œ
1710 48
dx dt
$#
x'Î& 45 x%Î& ‘ " œ
#
dx dy
œ
" 4
" y#
" y%
dy œ '1 ÊŠ 4" y#
y#
8 œ ˆ 12 "# ‰ ˆ 1"# 1‰ œ
17.
" 5 # 6 285 8
2
7 1#
œ 5 sin t 5 sin 5t and
" #
" 16
Ê Š dx dy ‹ œ
œ
" y# ‹
" #
ˆ 65 † 2'
y%
" #
5 4
2
2
" y# ‹
ˆ 315 6
dy œ ’ 1"# y$ y" “
" #
75 ‰ 4
" y% ‹
dy
# "
13 12
#
‰ Š dy œ 5 cos t 5 cos 5t Ê Êˆ dx dt dt ‹
dy dt
" #
" % Ê L œ '1 Ê1 Š 16 y
" y%
dy œ '1 Š 4" y#
#
† 2% ‰ ˆ 56 54 ‰‘ œ
#
œ Éa5 sin t 5 sin 5tb# a5 cos t 5 cos 5tb# œ 5Èsin# 5t #sin t sin 5t sin# t cos# t #cos t cos 5t cos# 5t œ &È# #asin t sin 5t cos t cos 5 tb œ 5È#a" cos %tb œ 5É%ˆ "# ‰a" cos %tb œ "!Èsin# #t œ "!lsin #tl œ "!sin #t (since ! Ÿ t Ÿ 1# ) Ê Length œ '!
1 Î2
18.
dx dt
1Î#
"!sin #t dt œ c5 cos #td !
œ 3t2 12t and
dy dt
œ a&ba"b a&ba"b œ "! #
#
‰ Š dy Éa3t2 12tb# a3t2 12tb# œ È288t# "8t4 œ 3t2 12t Ê Êˆ dx dt dt ‹ œ
œ 3È2 ktkÈ16 t2 Ê Length œ '! 3È2 ktkÈ16 t2 dt œ 3È2'! t È16 t2 dt; ’u œ 16 t2 Ê du œ 2t dt "
"
Ê "# du œ t dt; t œ 0 Ê u œ 16; t œ 1 Ê u œ 17“; œ
19.
dx d)
3È 2 2
œ $ sin ) and
dy d)
$1Î2
#
20. x œ t and y œ
'
'16"7 Èu du œ 3È2 2 23 u3/2 ‘1617 œ 3È2 2 Š 23 a17b3/2 23 a16b3/2 ‹
† 23 Ša17b3/2 64‹ œ È2Ša17b3/2 64‹ ¸ 8.617.
Ê Length œ '!
œ
3È 2 2
t$ 3
$ d) œ $'!
$1Î2
d) œ $ˆ $#1 !‰ œ
t, È3 Ÿ t Ÿ È3 Ê
È3
Èt% #t# " dt œ
È 3
#
#
‰ Š dy Éa$ sin )b# a$ cos )b# œ È$asin# ) cos# )b œ $ œ $ cos ) Ê Êˆ dx d) d) ‹ œ
'
dx dt
*1 #
œ 2t and
È3
Èt% 2t# " dt œ
È 3
œt
dy dt
'
#
" Ê Length œ '
È3
È 3
È3
È 3
Éat# "b# dt œ
Éa2tb# at# "b# dt
È
'È33 at# "b dt œ ’ t3 t“ 3
È3 È 3
œ 4È3 21. x œ
t# #
and y œ 2t, 0 Ÿ t Ÿ È5 Ê *
œ 21 23 u$Î# ‘ % œ 22. x œ t#
" 2t
761 3
dx dt
" È2
ŸtŸ1 Ê
Ê Surface Area œ '1ÎÈ2 21 ˆt# 1
1
œ 21 Š2
dy dt
È5
œ 2 Ê Surface Area œ '0
21(2t)Èt# 4 dt œ '4 21u"Î# du 9
, where u œ t# 4 Ê du œ 2t dt; t œ 0 Ê u œ 4, t œ È5 Ê u œ 9
and y œ 4Èt ,
œ 21 '1ÎÈ2 ˆt#
œ t and
" ‰ˆ 2t 2t
" ‰ 2t#
"‰ #t
dx dt
œ 2t
ʈ2t
" ‰# 2t#
" 2t#
dy dt
œ
2 Èt
Š È2 t ‹ dt œ 21 '1ÎÈ2 ˆt#
dt œ 21 '1ÎÈ2 ˆ2t$ 1
and
#
3 #
1
" ‰ Ɉ 2t #t
"
" ‰# #t#
4" t$ ‰ dt œ 21 2" t% 3# t 8" t# ‘ "ÎÈ#
3È 2 4 ‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dt
701
702
Chapter 11 Parametric Equations and Polar Coordinates
23. r cos ˆ) 13 ‰ œ 2È3 Ê r ˆcos ) cos
1 3
sin ) sin 13 ‰
È r cos ) #3 r sin ) œ 2È3 Ê r cos ) È3 r sin ) œ 4È3 Ê x È3 y œ 4È3 " #
œ 2È3 Ê Ê yœ
È3 3
24. r cos ˆ) œ
È2 #
x4
31 ‰ 4
Ê
œ È2 #
È2 #
Ê r ˆcos ) cos
r cos )
Ê yœx1
25. r œ 2 sec ) Ê r œ
2 cos )
È2 #
31 4
r sin ) œ
sin ) sin È2 #
31 ‰ 4
Ê x y œ 1
Ê r cos ) œ 2 Ê x œ 2
26. r œ È2 sec ) Ê r cos ) œ È2 Ê x œ È2
27. r œ 3# csc ) Ê r sin ) œ 3# Ê y œ 3#
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Chapter 11 Practice Exercises 28. r œ 3È3 csc ) Ê r sin ) œ 3È3 Ê y œ 3È3
29. r œ 4 sin ) Ê r# œ 4r sin ) Ê x# y# 4y œ 0 Ê x# (y 2)# œ 4; circle with center (!ß 2) and radius 2.
30. r œ 3È3 sin ) Ê r# œ 3È3 r sin ) Ê x# y# 3È3 y œ 0 Ê x# Šy circle with center Š!ß
3È 3 # ‹
and radius
3È 3 # ‹
#
œ
27 4
;
3È 3 #
31. r œ 2È2 cos ) Ê r# œ 2È2 r cos ) #
Ê x# y# 2È2 x œ 0 Ê Šx È2‹ y# œ 2; circle with center ŠÈ2ß 0‹ and radius È2
32. r œ 6 cos ) Ê r# œ 6r cos ) Ê x# y# 6x œ 0 Ê (x 3)# y# œ 9; circle with center (3ß 0) and radius 3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
703
704
Chapter 11 Parametric Equations and Polar Coordinates
# 33. x# y# 5y œ 0 Ê x# ˆy #5 ‰ œ
and a œ
5 #
25 4
Ê C œ ˆ!ß 5# ‰
; r# 5r sin ) œ 0 Ê r œ 5 sin )
34. x# y# 2y œ 0 Ê x# (y 1)# œ 1 Ê C œ (!ß 1) and a œ 1; r# 2r sin ) œ 0 Ê r œ 2 sin )
# 35. x# y# 3x œ 0 Ê ˆx #3 ‰ y# œ
and a œ
3 #
9 4
Ê C œ ˆ 3# ß !‰
; r# 3r cos ) œ 0 Ê r œ 3 cos )
36. x# y# 4x œ 0 Ê (x 2)# y# œ 4 Ê C œ (2ß 0) and a œ 2; r# 4r cos ) œ 0 Ê r œ 4 cos )
37.
38.
39. d
40.
e
41.
l
42.
f
43. k
44.
h
45.
i
46.
j
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 11 Practice Exercises 47. A œ 2'0
1
" # # r
d) œ '0 (2 cos ))# d) œ '0 a4 4 cos ) cos# )b d) œ '0 ˆ4 4 cos ) 1
œ '0 ˆ 9# 4 cos ) 1
48. A œ '0
1Î3
" #
1
cos 2) ‰ #
1
sin 2) ‘ 1 4 !
d) œ 92 ) 4 sin )
asin# 3)b d) œ '0
1Î3
6) ‰ ˆ 1 cos d) œ #
" 4
)
" 6
œ
9 #
A œ 4'0
" #
1Î$
sin 6)‘ !
œ
1 12 1 #
1 4
Ê )œ
c(1 cos 2))# 1# d d) œ 2 '0 a1 2 cos 2) cos# 2) 1b d) 1Î4
œ 2'0 ˆ2 cos 2) 1Î4
" #
cos 4) ‰ #
d) œ 2 sin 2)
" 2
)
d)
1
49. r œ 1 cos 2) and r œ 1 Ê 1 œ 1 cos 2) Ê 0 œ cos 2) Ê 2) œ 1Î4
1 cos 2) ‰ #
sin 4) ‘ 1Î% 8 !
œ 2 ˆ1
1 8
; therefore
0‰ œ 2
1 4
50. The circle lies interior to the cardioid. Thus, 1Î2
A œ 2 ' 1Î2 1Î2
" #
[2(1 sin ))]# d) 1 (the integral is the area of the cardioid minus the area of the circle) 1Î2
œ ' 1Î2 4 a1 2 sin ) sin# )b d) 1 œ ' 1Î2 (6 8 sin ) 2 cos 2)) d) 1 œ c6) 8 cos ) sin 2)d 1Î# 1 œ c31 (31)d 1 œ 51
51. r œ 1 cos ) Ê
dr d)
œ sin ); Length œ '0 È(1 cos ))# ( sin ))# d) œ '0 È2 2 cos ) d) 21
œ '0 É 4(1 #cos )) d) œ '0 2 sin 21
1Î#
21
52. r œ 2 sin ) 2 cos ), 0 Ÿ ) Ÿ
1 #
53. r œ 8 sin$ ˆ 3) ‰ , 0 Ÿ ) Ÿ
1 4
Ê
dr d)
#1 d) œ 4 cos 2) ‘ ! œ (4)(1) (4)(1) œ 8
) #
Ê
œ 8 asin# ) cos# )b œ 8 Ê L œ
21
dr d)
#
œ 2 cos ) 2 sin ); r# ˆ ddr) ‰ œ (2 sin ) 2 cos ))# (2 cos ) 2 sin ))#
'01Î2 È8 d) œ ’2È2 )“ 1Î# œ 2È2 ˆ 1# ‰ œ 1È2 !
#
œ 64 sin% ˆ 3) ‰ Ê L œ '0 É64 sin% ˆ 3) ‰ d) œ '0 1Î4
#
œ 8 sin# ˆ 3) ‰ cos ˆ 3) ‰ ; r# ˆ ddr) ‰ œ 8 sin$ ˆ 3) ‰‘ 8 sin# ˆ 3) ‰ cos ˆ 3) ‰‘ 1Î4
8 sin# ˆ 3) ‰ d) œ '0 8 ’ 1Î4
1cos ˆ 23) ‰ “ #
d)
1Î% œ '0 4 4 cos ˆ 23) ‰‘ d) œ 4) 6 sin ˆ 23) ‰‘ ! œ 4 ˆ 14 ‰ 6 sin ˆ 16 ‰ 0 œ 1 3 1Î4
54. r œ È1 cos 2) Ê
dr d)
œ
" #
(1 cos 2))"Î# (2 sin 2)) œ
#
Ê r# ˆ ddr) ‰ œ 1 cos 2) œ
2 2 cos 2) 1 cos 2)
1Î2
sin# 2) 1 cos 2)
œ
(1 cos 2))# sin# 2) 1 cos 2)
sin 2) È1 cos 2)
œ
# Ê ˆ ddr) ‰ œ
sin# 2) 1 cos 2)
1 2 cos 2) cos# 2) sin# 2) 1cos 2)
œ 2 Ê L œ ' 1Î2 È2 d) œ È2 1# ˆ 1# ‰‘ œ È2 1 #
55. x# œ 4y Ê y œ x4 Ê 4p œ 4 Ê p œ 1; therefore Focus is (0ß 1), Directrix is y œ 1
x# #
œ y Ê 4p œ 2 Ê p œ "# ; therefore Focus is ˆ!ß "# ‰; Directrix is y œ "#
56. x# œ 2y Ê
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#
705
706
Chapter 11 Parametric Equations and Polar Coordinates
57. y# œ 3x Ê x œ
y# 3
Ê 4p œ 3 Ê p œ
3 4
#
58. y# œ 83 x Ê x œ ˆy8 ‰ Ê 4p œ
;
3
therefore Focus is ˆ 34 ß 0‰ , Directrix is x œ 34
x# 7
59. 16x# 7y# œ 112 Ê
y# 16
61. 3x# y# œ 3 Ê x# Ê c œ 2; e œ
c a
œ
2 1
y# 3
x# 4
c a
œ
62. 5y# 4x# œ 20 Ê
y# 4 3 # ;
60. x# 2y# œ 4 Ê c a
œ
Ê pœ
therefore Focus is ˆ 23 ß !‰ , Directrix is x œ
œ1
Ê c# œ 16 7 œ 9 Ê c œ 3; e œ
8 3
Ê c œ È2 ; e œ
3 4
œ 1 Ê c# œ 1 3 œ 4
œ 2; the asymptotes are
Ê c œ 3, e œ
c a
œ
y# # œ È2 #
x# 5
2 3
;
2 3
1 Ê c# œ 4 2 œ 2
œ 1 Ê c# œ 4 5 œ 9
the asymptotes are y œ „
2 È5
x
y œ „ È3 x
#
63. x# œ 12y Ê 1x# œ y Ê 4p œ 12 Ê p œ 3 Ê focus is (!ß 3), directrix is y œ 3, vertex is (0ß 0); therefore new vertex is (2ß 3), new focus is (2ß 0), new directrix is y œ 6, and the new equation is (x 2)# œ 12(y 3) #
y 64. y# œ 10x Ê 10 œ x Ê 4p œ 10 Ê p œ #5 Ê focus is ˆ #5 ß 0‰ , directrix is x œ 5# , vertex is (0ß 0); therefore new vertex is ˆ "# ß 1‰ , new focus is (2ß 1), new directrix is x œ 3, and the new equation is (y 1)# œ 10 ˆx "# ‰
65.
x# 9
y# #5
œ 1 Ê a œ 5 and b œ 3 Ê c œ È25 9 œ 4 Ê foci are a!ß „ 4b , vertices are a!ß „ 5b , center is
(0ß 0); therefore the new center is ($ß 5), new foci are (3ß 1) and (3ß 9), new vertices are ($ß 10) and ($ß 0), and the new equation is
(x 3)# 9
(y 5)# #5
œ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 11 Practice Exercises 66.
x# 169
y# 144
707
œ 1 Ê a œ 13 and b œ 12 Ê c œ È169 144 œ 5 Ê foci are a „ 5ß 0b , vertices are a „ 13ß 0b , center
is (0ß 0); therefore the new center is (5ß 12), new foci are (10ß 12) and (0ß 12), new vertices are (18ß 12) and (8ß 12), and the new equation is 67.
y# 8
x# 2
(x 5)# 169
(y 12)# 144
œ1
œ 1 Ê a œ 2È2 and b œ È2 Ê c œ È8 2 œ È10 Ê foci are Š0ß „ È10‹ , vertices are
Š0ß „ 2È2‹ , center is (0ß 0), and the asymptotes are y œ „ 2x; therefore the new center is Š2ß 2È2‹, new foci are Š2ß 2È2 „ È10‹ , new vertices are Š2ß 4È2‹ and (#ß 0), the new asymptotes are y œ 2x 4 2È2 and #
y œ 2x 4 2È2; the new equation is 68.
x# 36
y# 64
Šy 2È2‹ 8
(x 2)# #
œ1
œ 1 Ê a œ 6 and b œ 8 Ê c œ È36 64 œ 10 Ê foci are a „ 10ß 0b , vertices are a „ 6ß 0b , the center
is (0ß 0) and the asymptotes are
y 8
œ „
x 6
or y œ „ 43 x; therefore the new center is (10ß 3), the new foci are
(20ß 3) and (0ß 3), the new vertices are (16ß 3) and (4ß 3), the new asymptotes are y œ y œ 43 x
49 3
; the new equation is
(x 10)# 36
(y 3)# 64
4 3
x
31 3
and
œ1
69. x# 4x 4y# œ 0 Ê x# 4x 4 4y# œ 4 Ê (x 2)# 4y# œ 4 Ê
(x 2)# 4
y# œ 1, a hyperbola; a œ 2 and
b œ 1 Ê c œ È1 4 œ È5 ; the center is (2ß 0), the vertices are (!ß 0) and (4ß 0); the foci are Š2 „ È5 ß 0‹ and the asymptotes are y œ „
x 2 #
70. 4x# y# 4y œ 8 Ê 4x# y# 4y 4 œ 4 Ê 4x# (y 2)# œ 4 Ê x#
(y 2)# 4
œ 1, a hyperbola; a œ 1 and
b œ 2 Ê c œ È1 4 œ È5 ; the center is (!ß 2), the vertices are (1ß 2) and ("ß 2), the foci are Š „ È5ß 2‹ and the asymptotes are y œ „ 2x 2 71. y# 2y 16x œ 49 Ê y# 2y 1 œ 16x 48 Ê (y 1)# œ 16(x 3), a parabola; the vertex is ($ß 1); 4p œ 16 Ê p œ 4 Ê the focus is (7ß 1) and the directrix is x œ 1 72. x# 2x 8y œ 17 Ê x# 2x 1 œ 8y 16 Ê (x 1)# œ 8(y 2), a parabola; the vertex is (1ß 2); 4p œ 8 Ê p œ 2 Ê the focus is (1ß 4) and the directrix is y œ 0 73. 9x# 16y# 54x 64y œ 1 Ê 9 ax# 6xb 16 ay# 4yb œ 1 Ê 9 ax# 6x 9b 16 ay# 4y 4b œ 144 Ê 9(x 3)# 16(y 2)# œ 144 Ê
(x 3)# 16
(y 2)# 9
œ 1, an ellipse; the center is (3ß 2); a œ 4 and b œ 3
Ê c œ È16 9 œ È7 ; the foci are Š$ „ È7ß 2‹ ; the vertices are (1ß 2) and (7ß 2) 74. 25x# 9y# 100x 54y œ 44 Ê 25 ax# 4xb 9 ay# 6yb œ 44 Ê 25 ax# 4x 4b 9 ay# 6y 9b œ 225 # # Ê (x 2) (y 3) œ 1, an ellipse; the center is (2ß 3); a œ 5 and b œ 3 Ê c œ È25 9 œ 4; the foci are 9
25
(2ß 1) and (2ß 7); the vertices are (2ß 2) and (2ß 8) 75. x# y# 2x 2y œ 0 Ê x# 2x 1 y# 2y 1 œ 2 Ê (x 1)# (y 1)# œ 2, a circle with center (1ß 1) and radius œ È2 76. x# y# 4x 2y œ 1 Ê x# 4x 4 y# 2y 1 œ 6 Ê (x 2)# (y 1)# œ 6, a circle with center (2ß 1) and radius œ È6
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708
Chapter 11 Parametric Equations and Polar Coordinates
77. r œ
2 1 cos )
Ê e œ 1 Ê parabola with vertex at (1ß 0)
78. r œ
8 2 cos )
Ê rœ
ke œ 4 Ê
" #
4 1 ˆ "# ‰ cos )
Ê eœ
k œ 4 Ê k œ 8; k œ
a e
" #
Ê ellipse;
ea Ê 8 œ
a ˆ "# ‰
"# a
ˆ " ‰ ˆ 16 ‰ 8 Ê a œ 16 3 Ê ea œ # 3 œ 3 ; therefore the center is ˆ 83 ß 1‰ ; vertices are ()ß 1) and ˆ 83 ß 0‰
79. r œ
6 1 2 cos )
Ê e œ 2 Ê hyperbola; ke œ 6 Ê 2k œ 6
Ê k œ 3 Ê vertices are (2ß 1) and (6ß 1)
80. r œ Ê
12 3 sin ) " 3
Ê rœ
4 1 ˆ "3 ‰ sin )
Ê eœ
" 3
; ke œ 4 #
k œ 4 Ê k œ 12; a a1 e# b œ 4 Ê a ’1 ˆ 3" ‰ “
œ 4 Ê a œ 9# Ê ea œ ˆ "3 ‰ ˆ 9# ‰ œ 3# ; therefore the center is ˆ 3# ß 3#1 ‰ ; vertices are ˆ3ß 1# ‰ and ˆ6ß 3#1 ‰
81. e œ 2 and r cos ) œ 2 Ê x œ 2 is directrix Ê k œ 2; the conic is a hyperbola; r œ Ê rœ
ke 1 e cos )
83. e œ
" #
84. e œ
" 3
85.
ke 1 e sin )
Ê rœ
(4)(1) 1 cos )
Ê rœ
(2) ˆ "# ‰ 1 ˆ "# ‰ sin )
2 2 sin )
and r sin ) œ 6 Ê y œ 6 is directrix Ê k œ 6; the conic is an ellipse; r œ
Ê rœ
ke 1 e cos )
4 1 cos )
and r sin ) œ 2 Ê y œ 2 is directrix Ê k œ 2; the conic is an ellipse; r œ
Ê rœ
(2)(2) 1 # cos )
4 1 # cos )
82. e œ 1 and r cos ) œ 4 Ê x œ 4 is directrix Ê k œ 4; the conic is a parabola; r œ Ê rœ
Ê rœ
ke 1 e sin )
Ê rœ
(6) ˆ "3 ‰ 1 ˆ "3 ‰ sin )
6 3 sin )
(a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root: #
V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241 2
2
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 11 Additional and Advanced Exercises
709
(b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root: #
V œ 2 '0 1 ŠÉ4 49 y# ‹ dy œ 2'0 1 ˆ4 49 y# ‰ dy œ 21 4y 3
86.
9x# 4y# œ 36, x œ 4 Ê y# œ œ
87.
91 4
%
$
’ x3 4x“ œ #
(a) r œ
k 1 e cos ) #
91 4
3
9x# 36 4
Ê yœ
ˆ 64 ‰ ˆ8 ‰‘ œ 3 16 3 8
3 #
91 4
4 27
$
y$ ‘ ! œ 161
Èx# 4 ; V œ ' 1 Š 3 Èx# 4‹ dx œ # 2 4
ˆ 56 3
24 ‰ 3
œ
31 4
#
91 4
'24 ax# 4b dx
(32) œ 241
Ê r er cos ) œ k Ê Èx# y# ex œ k Ê Èx# y# œ k ex Ê x# y#
œ k 2kex e# x# Ê x# e# x# y# 2kex k# œ 0 Ê a1 e# b x# y# 2kex k# œ 0 (b) e œ 0 Ê x# y# k# œ 0 Ê x# y# œ k# Ê circle; 0 e 1 Ê e# 1 Ê e# 1 0 Ê B# 4AC œ 0# 4 a1 e# b (1) œ 4 ae# 1b 0 Ê ellipse; e œ 1 Ê B# 4AC œ 0# 4(0)(1) œ 0 Ê parabola; e 1 Ê e# 1 Ê B# 4AC œ 0# 4 a1 e# b (1) œ 4e# 4 0 Ê hyperbola 88.
Let (r" ß )" ) be a point on the graph where r" œ a)" . Let (r# ß )# ) be on the graph where r# œ a)# and )# œ )" 21. Then r" and r# lie on the same ray on consecutive turns of the spiral and the distance between the two points is r# r" œ a)# a)" œ a()# )" ) œ 21a, which is constant.
CHAPTER 11 ADDITIONAL AND ADVANCED EXERCISES 1. Directrix x œ 3 and focus (4ß 0) Ê vertex is ˆ 7# ß !‰ Ê pœ
" #
Ê the equation is x
7 #
œ
y# #
2. x# 6x 12y 9 œ 0 Ê x# 6x 9 œ 12y Ê
(x3)# 12
œ y Ê vertex is (3ß 0) and p œ 3 Ê focus is (3ß 3) and the
directrix is y œ 3 3. x# œ 4y Ê vertex is (!ß 0) and p œ 1 Ê focus is (!ß 1); thus the distance from P(xß y) to the vertex is Èx# y# and the distance from P to the focus is Èx# (y 1)# Ê Èx# y# œ 2Èx# (y 1)# Ê x# y# œ 4 cx# (y 1)# d Ê x# y# œ 4x# 4y# 8y 4 Ê 3x# 3y# 8y 4 œ 0, which is a circle 4. Let the segment a b intersect the y-axis in point A and intersect the x-axis in point B so that PB œ b and PA œ a (see figure). Draw the horizontal line through P and let it intersect the y-axis in point C. Let nPBO œ ) Ê nAPC œ ). Then sin ) œ yb and cos ) œ xa Ê
x# a#
y# b#
œ cos# ) sin# ) œ 1.
5. Vertices are a!ß „ 2b Ê a œ 2; e œ
c a
Ê 0.5 œ
c #
Ê c œ 1 Ê foci are a0ß „ 1b
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Chapter 11 Parametric Equations and Polar Coordinates
6. Let the center of the ellipse be (xß 0); directrix x œ 2, focus (4ß 0), and e œ 23 Ê ae c œ 2 Ê ae œ 2 c a Ê a œ 23 (2 c). Also c œ ae œ 23 a Ê a œ 23 ˆ2 23 a‰ Ê a œ 43 49 a Ê 59 a œ 43 Ê a œ 12 5 ;x2œ e 28 28 8 # # # ‰ ˆ #3 ‰ œ 18 ˆ 28 ‰ Ê x 2 œ ˆ 12 5 5 Ê x œ 5 Ê the center is 5 ß 0 ; x 4 œ c Ê c œ 5 4 œ 5 so that c œ a b #
#
‰ ˆ 58 ‰ œ œ ˆ 12 5
80 25
; therefore the equation is
ˆx 28 ‰# 5 ˆ 144 ‰ 25
y# ˆ 80 ‰ 25
œ 1 or
7. Let the center of the hyperbola be (0ß y). (a) Directrix y œ 1, focus (0ß 7) and e œ 2 Ê c ae œ 6 Ê Ê a œ 2(2a) 12 Ê a œ 4 Ê b# œ c# a# œ 64 16 œ (b) e œ 5 Ê c Ê cœ œ
625 16
25 4 25 16
a e
œ6 Ê
; y (1) œ œ
75 #
a e a e
a e
‰ 25 ˆx 28 5 144
#
Ê
144 a4 a # b #
5y# 16
œ1
œ c 6 Ê a œ 2c 12. Also c œ ae œ 2a
Ê c œ 8; y (1) œ œ #4 œ 2 # 48; therefore the equation is (y161) a e
Ê y œ 1 Ê the center is (0ß 1); c# œ a# b#
x# 48
œ1
œ c 6 Ê a œ 5c 30. Also, c œ ae œ 5a Ê a œ 5(5a) 30 Ê 24a œ 30 Ê a œ ˆ 54 ‰ 5
œ
œ
" 4
Ê y œ 43 Ê the center is ˆ!ß 43 ‰ ; c# œ a# b# Ê b# œ c# a#
; therefore the equation is
ˆy 34 ‰# ˆ 25 ‰ 16
x#
ˆ 75 ‰ #
œ 1 or
16 ˆy 34 ‰ 25
#
#
#
#
#
2x# 75
8. The center is (0ß 0) and c œ 2 Ê 4 œ a# b# Ê b# œ 4 a# . The equation is 49 a#
y# a#
#
œ1
x# b#
#
œ1 Ê #
%
49 a#
144 b#
œ1
%
œ 1 Ê 49 a4 a b 144a œ a a4 a b Ê 196 49a 144a œ 4a a Ê a 197a# 196
œ 0 Ê aa 196b aa# 1b œ 0 Ê a œ 14 or a œ 1; a œ 14 Ê b# œ 4 (14)# 0 which is impossible; a œ 1 Ê b# œ 4 1 œ 3; therefore the equation is y# 9. b# x# a# y# œ a# b# Ê
dy dx
x# 3
œ1 #
#
œ ba# yx ; at (x" ß y" ) the tangent line is y y" œ Š ba# yx"" ‹ (x x" )
Ê a# yy" b# xx" œ b# x"# a# y"# œ a# b# Ê b# xx" a# yy" a# b# œ 0 10. b# x# a# y# œ a# b# Ê
dy dx
œ
b# x a# y
#
; at (x" ß y" ) the tangent line is y y" œ Š ba# yx"" ‹ (x x" )
Ê b# xx" a# yy" œ b# x"# a# y"# œ a# b# Ê b# xx" a# yy" a# b# œ 0 11.
12.
13.
14.
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5 4
Chapter 11 Additional and Advanced Exercises 15. a9x# 4y# 36b a4x# 9y# 16b Ÿ 0 Ê 9x# 4y# 36 Ÿ 0 and 4x# 9y# 16 0 or 9x# 4y# 36 0 and 4x# 9y# 16 Ÿ 0
16. a9x# 4y# 36b a4x# 9y# 16b 0, which is the complement of the set in Exercise 15
sin t 17. (a) x œ e2t cos t and y œ e2t sin t Ê x# y# œ e4t cos# t e4t sin# t œ e4t . Also yx œ ee2t cos t œ tan t " ˆ y ‰ # # % tan " ayÎxb # # Ê t œ tan Ê x y œe is the Cartesian equation. Since r œ x y# and x 2t
) œ tan" ˆ yx ‰ , the polar equation is r# œ e4) or r œ e2) for r 0 (b) ds# œ r# d)# dr# ; r œ e2) Ê dr œ 2e2) d) # # Ê ds# œ r# d)# ˆ2e2) d)‰ œ ˆe2) ‰ d)# 4e4) d)# œ 5e4) d)# Ê ds œ È5 e2) d) Ê L œ '0 È5 e2) d) 21
œ’
È5 e2) #1 2 “!
È5 #
œ
ae41 1b
# # 18. r œ 2 sin$ ˆ 3) ‰ Ê dr œ 2 sin# ˆ 3) ‰ cos ˆ 3) ‰ d) Ê ds# œ r# d)# dr# œ 2 sin$ ˆ 3) ‰‘ d)# 2 sin# ˆ 3) ‰ cos ˆ 3) ‰ d)‘ œ 4 sin' ˆ 3) ‰ d)# 4 sin% ˆ 3) ‰ cos# ˆ 3) ‰ d)# œ 4 sin% ˆ 3) ‰‘ sin# ˆ 3) ‰ cos# ˆ 3) ‰‘ d)# œ 4 sin% ˆ 3) ‰ d)#
Ê ds œ 2 sin# ˆ 3) ‰ d). Then L œ '0 2 sin# ˆ 3) ‰ d) œ '0 1 cos ˆ 23) ‰‘ d) œ ) 31
31
3 2
$1
sin ˆ 23) ‰‘ ! œ 31
19. e œ 2 and r cos ) œ 2 Ê x œ 2 is the directrix Ê k œ 2; the conic is a hyperbola with r œ Ê rœ
(2)(2) 1 2 cos )
œ
ke 1 e cos )
4 1 2 cos )
20. e œ 1 and r cos ) œ 4 Ê x œ 4 is the directrix Ê k œ 4; the conic is a parabola with r œ Ê rœ 21. e œ
" #
" 3
œ
4 1 cos )
and r sin ) œ 2 Ê y œ 2 is the directrix Ê k œ 2; the conic is an ellipse with r œ
Ê rœ 22. e œ
(4)(1) 1 cos )
2 ˆ "# ‰ 1 ˆ "# ‰ sin
)
œ
ke 1 e sin )
2 2 sin )
and r sin ) œ 6 Ê y œ 6 is the directrix Ê k œ 6; the conic is an ellipse with r œ
Ê rœ
6 ˆ "3 ‰ 1 ˆ "3 ‰ sin
)
ke 1 e cos )
œ
ke 1 e sin )
6 3 sin )
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711
712
Chapter 11 Parametric Equations and Polar Coordinates
23. Arc PF œ Arc AF since each is the distance rolled; nPCF œ Arcb PF Ê Arc PF œ b(nPCF); ) œ ArcaAF Ê Arc AF œ a) Ê a) œ b(nPCF) Ê nPCF œ ˆ ba ‰ ); nOCB œ
1 #
) and nOCB œ nPCF nPCE œ nPCF ˆ 1# !‰ œ ˆ ba ‰ ) ˆ 1# !‰ Ê 1# ) œ ˆ ba ‰ ) ˆ 1# !‰ Ê 1# ) œ ˆ ba ‰ ) 1# ! Ê ! œ 1 ) ˆ ba ‰ ) Ê ! œ 1 ˆ ab b ‰ ).
Now x œ OB BD œ OB EP œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1 ˆ a b b ‰ )‰ œ (a b) cos ) b cos 1 cos ˆˆ a b b ‰ )‰ b sin 1 sin ˆˆ a b b ‰ )‰ œ (a b) cos ) b cos ˆˆ a b b ‰ )‰ and y œ PD œ CB CE œ (a b) sin ) b sin ! œ (a b) sin ) b sin ˆˆ a b b ‰ )‰ œ (a b) sin ) b sin 1 cos ˆˆ a b b ‰ )‰ b cos 1 sin ˆˆ a b b ‰ )‰ œ (a b) sin ) b sin ˆˆ a b b ‰ )‰ ; therefore x œ (a b) cos ) b cos ˆˆ a b b ‰ )‰ and y œ (a b) sin ) b sin ˆˆ a b b ‰ )‰ 24. x œ a(t sin t) Ê
‰ œ a(1 cos t) and let $ œ 1 Ê dm œ dA œ y dx œ y ˆ dx dt dt
dx dt
œ a(1 cos t) a (1 cos t) dt œ a# (1 cos t)# dt; then A œ '0 a# (1 cos t)# dt 21
'021 a1 2 cos t cos# tb dt œ a# '021 ˆ1 2 cos t "# "# cos 2t‰ dt œ a# 32 t 2 sin t sin4 2t ‘ #!1 œ 31a# ; µ x = x œ a(t sin t) and µ y = "# y œ "# a(1 cos t) Ê Mx œ ' µ y dm œ ' µ y $ dA 21 21 21 œ '0 "# a(1 cos t) a# (1 cos t)# dt œ "# a$ '0 (1 cos t)$ dt œ a# '0 a1 3 cos t 3 cos# t cos$ tb dt œ a#
$
œ œ
a$ #
'021 1 3 cos t 3# 3 cos# 2t a1 sin# tb (cos t)‘ dt œ a# ’ 52 t 3 sin t 3 sin4 2t sin t sin3 t “ #1 $
51 a #
$
!
$
. Therefore y œ
Mx M
œ
$
Š 51#a ‹
œ
31 a#
5 6
a. Also, My œ ' µ x dm œ ' µ x $ dA
œ '0 a(t sin t) a# (1 cos t)# dt œ a$ '0 at 2t cos t t cos# t sin t 2 sin t cos t sin t cos# tb dt 21
21
#
œ a$ ’ t2 2 cos t 2t sin t "4 t# xœ 25.
My M
œ
31 # a$ 31 a#
œ 1a Ê ˆ1aß
5 6
" 8
cos 2t
t 4
sin 2t cos t sin# t
#1 cos$ t 3 “!
œ 31# a$ . Thus
a‰ is the center of mass.
" œ <# <" Ê tan " œ tan (<# <" ) œ
tan <# tan <" 1 tan <# tan <"
;
the curves will be orthogonal when tan " is undefined, or when tan <# œ tan"<" Ê g (r)) œ " r w
’ f ()) “ w
#
w
w
Ê r œ f ( ) ) g ( ) )
26.
r œ sin% ˆ 4) ‰ Ê
27.
r œ 2a sin 3) Ê
dr d)
dr d)
œ sin$ ˆ 4) ‰ cos ˆ 4) ‰ Ê tan < œ œ 6a cos 3) Ê tan < œ
r ˆ ddr) ‰
œ
sin% ˆ )4 ‰ sin$ ˆ 4) ‰ cos ˆ )4 ‰ 2a sin 3) 6a cos 3)
œ
" 3
œ tan ˆ 4) ‰ tan 3); when ) œ
1 6
, tan < œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
" 3
tan
1 #
Ê<œ
1 #
Chapter 11 Additional and Advanced Exercises 28.
(b) r) œ 1 Ê r œ )" Ê
(a)
œ
) " ) #
œ ) Ê
Ê < Ä
1 #
dr d)
œ )# Ê tan
lim tan < œ _
)Ä_
from the right as the spiral winds in
around the origin.
29.
tan <" œ
È3 cos ) È3 sin )
œ cot ) is È"3 at ) œ
1 3
; tan <# œ
sin ) cos )
œ tan ) is È3 at ) œ
1 3
; since the product of
these slopes is 1, the tangents are perpendicular 30.
tan < œ
r ˆ ddr) ‰
œ
a(1 cos )) a sin )
is 1 at ) œ
1 #
Ê <œ
713
1 4
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714
Chapter 11 Parametric Equations and Polar Coordinates
NOTES:
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