Frederick H.Soon
Student's Guide to by J. Marsden and A. Weinstein Volume II
Springer-Verlag New York Berlin Heidelberg Tokyo
Student's Guide to
Calc~llus by J. Marsden and A. Weinstein Volume II
Copyright 1985 Springer-Verlag. All rights reserved.
Frederick H. Soon
Student's Guide to
by J. Marsden and A. Weinstein
With 103 Illustrations
Springer-Verlag New York Berlin Heidelberg Tokyo Copyright 1985 Springer-Verlag. All rights reserved.
AMS Subiect Classification: 26-01
Library of Congress Cataloging-in-Publication Data Soon, Frederick Id. Student's guide to Calculus by J. Marsiden and A. Weinstein. Vol~~rne 2. Guide to Marsclen and Weinstein's Calculus II. 1. Calculus. I. Vlarsden, Jerrold E. Calculus I!. I I . Title. QA303.S773 1985 515 85-25095
cc: 1985 by Springer-Verlag New York Inc. All rights reserved. No part of this book may be translated or reproduced In any torm without written permission from Springer-Verlag, 175 Fifth Avenue, New York, New York 10010. U.S.A. Printed and bound by Halliday Lithograph, West Hanover, Massachusetts Printed in the Un~tedStates of America.
ISBN 0-387-96234-4 Springer-Verlag New York Berlin Heidelberg Tokyo ISBN 3-540-96234-4 Springer-Verlag Berlin Heidelberg New York Tokyo
Copyright 1985 Springer-Verlag. All rights reserved.
Dedicated to: Henry, Ora, Dennis, and Debbie
Copyright 1985 Springer-Verlag. All rights reserved.
FOREWORD
This Student Guide is exceptional, maybe even unique, among such guides in that its author, Fred Soon, was actually a student user of the textbook during one of the years we were writing and debugging the book. of the best students that year, by the way.)
(He was one
Because of his background, Fred
has taken, in the Guide, the point of view of an experienced student tutor helping you to learn calculus. While we do not always think Fred's jokes are as funny as he does, we appreciate his enthusiasm and his desire to enter into communication with his readers; since we nearly always agree with the mathematical judgements he has made in explaining the material, we believe that this Guide can serve you as a valuable supplement to our text. To get maximum benefit from this Guide, you should begin by spending a few moments to acquaint yourself with its structure.
Once you get started in
the course, take advantage of the many opportunities which the text and Student Guide together provide for learning calculus in the only way that any mathematical subject can truly be mastered - through attempting to solve problems on your own.
As you read the text, try doing each example and exercise your-
self before reading the solution; do the same with the quiz problems provided by Fred. Fred Soon knows our textbook better than anyone with the (possible) exception of ourselves, having spent hundreds of hours over the past ten years assisting us with its creation and proofreading.
We have enjoyed our associa-
tion with him over this period, and we hope now that you, too, will benefit from his efforts. Jerry Marsden Alan Tdeinstfin
Copyright 1985 Springer-Verlag. All rights reserved.
HOW TO USE THIS BOOK As the title implies, thisbook is intended to guide the student's study of calculus. Realizing that calculus is not the only class on the college student's curriculum, my objective in writing this book is to maximize understanding with a minimum of time and effort. For each new section of the text, this student guide contains sections entitled Prerequisites, Prerequisite Quiz, Goals, Study Hints, Solutions to Every Other Odd Exercise, Section Quiz, Answers to Prerequisite Quiz, and Answers to Section Quiz.
For each review section, I have included the solu-
tions to every other odd exercise and a chapter test with solutions. A list of prerequisites, if any, is followed by a short quiz to help you decide if you're ready to continue.
If some prerequisite seems vague to
you, review material can be found in the section or chapter of the text listed after each prerequisite.
If you have any difficulty with the simple prerequi-
site quizzes, you may wish to review. As you study, keep the goals in mind.
They may be used as guidelines
and should help you to grasp the most important points. The study hints are provided to help you use your time efficiently. Comments have been offered to topics in the order in which they appear in the text.
I have tried to point out what is worth memorizing and what isn't.
If
time permits, it is advisable to learn the derivations of formulas rather than just memorizing them.
You will find that the course will be more meaningful
Copyright 1985 Springer-Verlag. All rights reserved.
to you and that critical parts of a formula can be recalled even under the stress of an exam.
Other aspects of the study hints include clarification of
text material and "trickso'which will aid you in solving the exercises. Finally, please be aware that your instructor may choose to emphasize topics which I have considered less important. Detailed solutions to every other odd exercise, i.e., 1 , 5 , 9 , etc. are provided as a study aid.
Some students may find it profitable to try the
exercises fi-rst and then compare the method employed in this book.
Since the
authors of the text wrote most of .the exercises in pairs, the answers jnthis book may also be used as a guide to solving the corresponding even exercises. In order to save space, fractions have been written on one line, so be careful about your interpretations. l/(x
+ y)
Thus,
means the reciprocal of
x
l/x
+
y
.
+
y
means
meaning.
ax
and then divide by
In a12 means half of
b In a
,
whereas
,
plus
l/x , whereas
Transcendental functions such as
cos, sin, In, etc. take precedence over division, so cosine of
y
cos ax/b means take the
cos (axlb) has an unambiguous
not the natural logarithm of
a/2
.
Also, everything in the term after the slash is in the denominator, so 1/2/xdx
+
add
to half of the integral.
1
1 means add
1
to the reciprocal of
2Jxdx
.
It does not mean
The latter would be denoted
(1/2)Jxdx
+
1 .
Section quizzes are included for you to evaluate your mastery of the material.
Some of the questions are intended to be tricky, so do not be dis-
couraged if you miss a few of them.
Theanswerstnthese "hard" questions should
add to your knowledge and prepare you for your exams.
Since most students seem
to fear word problems, each quiz contains at least one word problem to help you gain familiarity with this type of question. Finally, answers have been provided to both the prerequisite and section quizzes.
If you don't understand how to arrive at any of the answers, be sure
Copyright 1985 Springer-Verlag. All rights reserved.
t o ask your i n s t r u c t o r . I n t h e r e v i e w s e c t i o n s , I h a v e w r i t t e n more q u e s t i o n s a n d a n s w e r s w h i c h may a p p e a r
011
a t y p i c a l test.
T h e s e may b e u s e d a l o n g w i t h t h e s e c t i o n q i i i z z r s
t o h e l p you s t u d y f o r y o u r t e s t s . S i n c e C a l c u l u s was i n t e n d e d f o r a t h r e e s e m e s t e r c o u r s e , I h a v e a l s o included three-hour
1 5 , and 18. examinations.
c o m p r e h e n s i v e exams a t t h e end of C h a p t e r s 3 , 6 , 9 ,
12,
T h e s e s h o u l d h e l p you p r e p a r e f o r y o u r m i d t e r m s and f i n a l B e s t of l u c k w i t h a l l o f y o u r s t u d i e s .
ACKNON1,EDGEMEKTS S e v e r a l i n d i v i d u a l s n c e d t o b e t h a n k e d f o r h e l p i n g t o p r o d u c e t h i s book I am m o s t g r a t e f u l t o J e r r o l d Y a r s d e n a n d A l a n W e i n s t e i n f o r p r o v i d i n g t h e i ~ s wtlich I , a s a s t u d e n t , l e a r n e d a b o u t d e r i v a f i r s t e d i L i o n o f C a l ~ i ~ l from t i v i > c . and i n t e g r a l s .
A l s o , I am d e e p l y a p p r e c i a t i v e f o r t h e i r a d v i c r and
e x p e r t i s e which t h e y o f f e r e d d u r i n g t h e p r e p a r t i o n of t h i s book.
Invaluable
a i d a n d k n o w l e d g c a a b l e r e v i e w i n g w e r e p r o v i d e d by my p r i m a r y a s s i s t a n t s : S t e p l l e n Hook, F r e d e r i c k D a n i e l s , a n d K a r e n P a o .
T e r e s a L i n g stiou3d b e
r e c o g n i z e d f o r l a y i n g t h e groundwork w i t h tile f i r s t e d i t i o n of t h e s t u d e r i t guide.
F i n a l l y , my g r a t i t u d e g o e s t o my f a t h e r , H e n r y , who d i d tl-ic a r t w o r k ;
t o C h a r l e s 0 l v i . r and B e t t y H s i , my p r o o f r e a d e r s ; a n d t o Rut11 Edmonds, w h o s e t y p i n g made t h i s p ~ ~ b l i c a t i oan r e a l i t y .
F r e d e r i c k H. Soon Berkclev, C a l i f o r n i ~
Copyright 1985 Springer-Verlag. All rights reserved.
CIlAPTtR 7 - - BASIC METHODS OF IYlEGRATlON 7.1 7.2 7.3 7.4 7.R
Calculating Integrals I n t e g r a t i o n by S u b s t i t u t i o n Changing V a r i a b l e s i n t h e D e f i n i t e I n t e g r a l I n t e g r a t i o n by P a r t s Review E x e r c i s e s f o r C h a p t e r 7
CHAPTER 8 8.1 8.2 8.3 8.4 8.5 8.6 8.R
--
DIFFERENTIAL EQUAIIONS
Oscillations Growth and Decay The H y p e r b o l i c F u n c t i o n s 'The I n v e r s e H y p e r b o l i c F u n c t i o n s Separable D i f f e r e n t i a l Equations Linear First-Order Equations Keview E x e r c i s e s f o r C h a p t e r Y
CHAPTER 9 - - APPLICYl IONS OF IWI'ECKATION 9.1 9.2 9.3 9.4 9.5 9.R
Volumes by t h e S l i c e Method Volumes b y t h e S h e l l Metnod Average V a l u e s and t h e Mean V a l u e Theorem f o r I n t e g r a l s C e n t e r o f Mass E n e r g y , Power, and Work Kevlew t x e r c i s e s f o r C h p a t e r 9
COPIPKEHENSIVE 'rES'1' FOR CHAPTERS 7 -. 9
391 397 402 407 413 418
423
CHAPTkR 1 0 - - FURTHER TECHNIQUkS AND APPLICAIIONS OF IN'IEGKATION 10.1 10.2 10.3 10.4 10.5 10.R
Trlgonometrlc I n t e g r a l s P a r t i a l Fract ~ o n s 4 r c Length and S u r f a c e Area P a r a m e t r i c Curves Length and Area i n P o l a r C o o r d i n a t e s Kevlew k x e r c i s e s f o r L h a p t e r 10
CI-IAPThR 11 - - I,IMITS, L'HOPITAL'S RULE, AND hUMERICAL METHODS 11 1 11.2 11.3 11.4 11.5
L l m l t s o f Functions L ' i l 6 p i t a l ' s Rule Improper I n t e g r a l s L l m l t s o f S e q u e n c e s a n d Newton's Method Numerical Integration Revlew E x e r c i s e s f o r C h a p t e r 1 1
Copyright 1985 Springer-Verlag. All rights reserved. ll.K
CHAPTER 1 2
--
INFINITE SERIES
The Sum o f an I n f i n i t e S e r i e s The Comparison T e s t and A l t e r n a t i n g S e r i e s The I n t e g r a l and R a t i o T e s t s Power S e r i e s T a y l o r ' s Formula Complex Numbers Second-Order L i n e a r D i f f e r e n t i a l E q u a t i o n s S e r i e s Solutions of D i f f e r e n t i a l Equations Review E x e r c i s e s f o r C h a p t e r I f COMPREHENSIVE TEST FOR CHAPTERS 7 - 1 2
Copyright 1985 Springer-Verlag. All rights reserved.
CHAPTER 7 B A S I C METHODS OF INTEGRriTION
7.1
Calculating Integrals
PREREQUISITES 1.
Recall how to integrate polynomials (Sections 2.5, 4.4, and 4.5).
2.
Recall integration formulas involving exponentials and logarithms (Section 6.3).
3.
Recall integration formulas involving trigonometric functions and their inverses (Section 5.2 and 5.4).
4.
Recall the relationship between the integral and area (Section 4.6).
PREREQUISITE QUIZ
1.
Perform the following integrations:
(c) 2.
~ ( C O Sx
- 1/x
+
2 2/x )dx
> g(x)
on
[2,3]
an expression for the area between the graphs of
g(x)
and
Suppose
[0,31
g(x) 2 f(x)
on
[0,21
and
f (x)
.
\;rite
f(x)
on
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 7.1
314
GOALS 1.
Be able to evaluate integrals involving sums of polynomials, trigonometric functions, exponentials, and inverse trigonometric functions.
2.
Be able to use integration for solving area and total change problems.
STUDY HINTS 1.
Definite integrals. placed on a a
and
and
In the box preceding Example 2, restrictions are
.
b
Can you explain why?
x = 0
.
-2 , - 3 , -4 ,
...
If n is not an integer, one must impose condition
that avoid roots of negative numbers. undefined for
x
Checking answers. tion.
=
b must have the same sign to avoid the discontinuity in the
integrand at
2.
If n
0
Finally, recall that
In x
is
.
Remember that integration is the inverse of differentia-
Thus, you should always check your answer by differentiating it to
get the integrand. Differentiation can often detect a wrong sign or a wrong factor. 3.
Word problems.
Be sure all quantities are expressed in compatible units.
(See Example 9.)
4.
Review of integration methods.
The material in this section is review.
If any of the examples didn't make sense, go back to the appropriate sections in Chapters 1-6 and review until you understand the examples.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
By combining the sum rule, the constant multiple rule, and the power rule for antidifferentiation, we get
- 2/(-2) +
x
C = x3
+ x2 -
1/2x2
+
J(3x C
2
+
2x
+
3 ~ - ~ ) d= x 3x I 3
+
2 2x /2 +
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 7.1
IJe guess that the antiderivative of
,
tion gives -2a sin 2x j(sin 2x
+
so
3x)dx = -cos 2x12
a
+
sin 2x
should be
2 3x 12. + C
.
.
is a cos 2x
.
-112
315
Differentia-
Thus,
*
Using the sum rule, the constant multiple rule, and the power rule for antidifferentiation, we get
.
x + C
Note that
fore, jf2(x8
+
F(-a)
2x2
-
F(x)
=
1(x8
= -F(a)
,
so
+
=
x 19
F(a) - F(-a)
=
2F(a)
1)dx = 2F(2) = 2(512/9
+
Using the power rule for antidifferentiation, ~ ~ ~ 1 5 1 1 =4 41 jyTcos x dx = sin x
8
In
-71
=
1
3
-
(
6
l 0
=
3n/4
1613
-
::1
2)
=
ds =
3 2x 13 -
+
.
There-
1084/9
''116'
2 = 86415
.
+
1 3jO[1/(x2
.
'lids
=
o .
By the constant multiple rule, ~;[3/(x 3 tan-'xi
9
2x2 - 1)dx
2
l)]dx
=
+
1)ldx =
.
2 from the basic 'trigonometric antidifferentiation formulas, ~;/*sec x dx t a n x ~ , '= ~ 1
.
5 By the logarithm differentiation formula, jl(dt/t)
=
5 In t / = ln 5
0
+
-
lnl=In5.
By the sum rule, the integral is 200
j-200 1
dx
.
200 j-200(90x21 - 8
and the integral is
(a)
+
Since the first integrand is an odd function, the anti-
derivative will be an even function, F(x)
400
97 5580x ~ ~ )dx ~
0
.
;
therefore F(200) = F(-200)
Thus, we are left with
200 j-200 1
dx
=
x1 -200 200
=
. According to the fundamental theorem of calculus, the derivative of the integral is the integrand. Using the chain rule, we have 2 2 (d/dx) e(X '12 + C = ~ x e ()/2 ~ = xe(x ) . Therefore, the formula is correct.
1
*Throughout the student guide, we take cos 2x/2 to mean (1/2)cos 2x, cos (2x/2) .
=st_
Copyright 1985 Springer-Verlag. All rights reserved.
=
37.
A c c o r d i n g t o Example 7 , 2 1 jy [ ~ x e ' ' ) + 3 I n x j d x
(b)
j l n x dx = x I n x 2 = ' e ' ~ ) + 3x i n x
1
-
x
-
3x
+
C
1;
.
Thus, 2 = e ( e )
+
3e -
2 3 e - e - O + 3 = e ( e ) - e + 3 .
41.
We may w r i t e
&-s i n 5 x 2 d x = - /rx + s i n 5x2 dx
Ii2
g(t) =
I +-
i
f (t2)
, where
f (u) =
theorem of c a l c u l u s ( a l t e r n a t i v e v e r s i o n ) , By t h e c h a i n r u l e ,
45.
+
.
s i n i x 2 dx
as
By t h e f u n d a m e n t a l
f ' ( u ) = -JeU
2
+
2 g l ( t ) = f l ( t ) . ( d / d t ) t 2 = -2t
P r o p e r t y 4 of t h e d e f i n i t e i n t e g r a l , t h e e n d p o i n t a d d i t i v i t y r u l e , i s used i n t h i s e x e r c i s e
(c)
If
Y
0
<x
j'n/2
I
+
sin t dt
(-2 cos t )
;
=
[ 4 [(x2
53.
f(t)g(t)dt =
cos(2) - 2 cos x
S i n c e t h e f u n c t i o n i s p o s i t i v e on '1
+
1;(2
= ~
2)/&]dx
The y - i n t e r c e p t
of
y
:
(
+x
2x-1'2)dx ~ ~ ~
2
=
[1,4] ,
1 - x 14
is
=
1 ,
.
=
.
~f
s i n z ) d t = - c o s t12 712
-cos(2)
cos(2) - 2 cos x -cos x
49.
f(t)g(t)dt
s i n t d t = - c o s tIx = -cos x n/2 2 & x < n , then
1.584
J';/2
j:/2
, then
2
-
2 cos x
+
+
2 cos(2) =
Therefore,
if
O < x S 2
if
2 G x
the area is
:j
f(x)dx
.
1;
[ ~ ~ / ~ / ( 5 ! 2+ ) 2 ~ ' ~ ~ / ( 1 / 2 ) 1=
so the white area i n the
c e n t e r i s a c i r c l e of r a d i u s 1
.
We w i l l f i n d t h e a r e a i n t h e f i r s t
q u a d r a n t and t h e n m u l t i p l y by
4
t o find the t o t a l area.
The x - i n t e r c e p t
Copyright 1985 Springer-Verlag. All rights reserved.
53.
(continued) of
2 y = 1 - x 14
is
.
2
- r2 4Ljo(l
so the area of the entire shaded region is
2
57. Integrating the identity sin x
j;l2
the sum rule gives
714 ,
The area of the white quarter circle is
+
sin2xdx
cos2x = 1
2
- x l4)dx -
from
+ ,fd/2cos2dx
=
0
i 8 / 2 and using
to
1"" 0
=
TI(]
1 dx
.
The two
integrals on the left-hand side are equal by assumption and the righthand side is
-;i/2 ,
~ 6 'sin~2x d x
so
problem without ever finding 61.
/:
The area is given by the integral tan-'a of
.
tan-'x
=
[dx/(l
+
2 x )] = tan-lx;:
tan-lb
-
=
For this problem,
A
=
69. Using the identity,
+
,
-ake-kt
(-~/k)(e-kT - 1)
Therefore,
the difference
SO
, regardless of the'length of
-
Guess that the antiderivative bas the form
l n t 1:-ln(t
=
.
[a,b ]
ate to get
(b)
(Note that we did this
(-1/2,~/2) ,
always lies in the interval
the interval (a)
.
By the definition of the inverse tangent function, the value
between two values must be less than
65.
n/4
2 jsin x dx .)
so
a
=
0.0825 ;
R
+
l)]
l)/: = 1 - 0 - ln(e
+
integral was found by guessing that Then differentiation showed that
=
jT ~ 0
We differenti-
e - ~ ~= d(-R/k)e t
=
4(12)(230)
=
- e-(0'0825)(5))
(11040/0.0825)(1 /;[dt/t(t
A
.
-kt T
.
(R/k) (1 -
k
.
-R/k
=
ae-kt
=
1)
j:(dt/t)
+
-
=
.
T
=
5
$45,231.46
.
jE[dt/(t
ln(2) "0.380
j[dt/(t +
11040 ;
+
1)l
.
=
(The second
111 = a In It + 11
+
C
.
a = 1 .)
Copyright 1985 Springer-Verlag. All rights reserved.
=
318 Section 7.1
SECTION QUIZ 1.
Evaluate the following integrals:
(dl
j:l2(2
if,
3
-, -
(g) 2.
sin
cos ~ ) d p
, d
j[(t4
j[dt/(l
a +
+
+
2 2t
u
+
2)/(t
u
-.
O
2
+
1) - et
2 t ) ] = tan-lt + C
and
=
-cot
-1
csc t cot tldt
j[dt/(l
-1
the integrals are equal, we have
-1 cancel, leaving tan t
+
tan t
.
t
2 -1 t ) I =-cot t
+ +
C
= -cot
Obviously,
-1
t
+
C
+
.
Since
The
C's
C
.
.
# -cot-'t
tan-'t
What is the fallacy in this argument? fex sin x dx # (1/2)ex(sin
+
+
.
3.
Show that
4.
Once again, Guilty Gary had borrowed his neighbor's tools without asking.
x
cos x)
C
This time, he was spray painting his house when a friendly policeman stopped to compliment Guilty Gary on the fine job he was doing.
Startled,
Guilty Gary forgot to turn off the spray before turning around.
\%%en he
realized he was spraying the policeman, he accidentally increased the spray rate.
For
litersjmin., where
15
seconds, paint was coming out at t
is in minutes.
(0.25
+
t)
How much paint did Guilty Gary
spray on the policeman?
*Dear Reader: I realize that many of you hate math but are forced to complete this course for graduation. Thus, I have attempted to maintail; interest with "entertaining" word problems. They are not meant to he insulting to your intelligence. Obviously, most of the situations will never happen; however, calculus has several practical uses and such exampizs arc found throughout Marsden and Weinstein's text. I would appreciate your comments on whether my "unusual" word problems should hc kept for tile next edition.
Copyright 1985 Springer-Verlag. All rights reserved.
ANsldERS TO PREREQUISITE QUIZ 1.
2.
(a)
72
(b)
ex - sin-lx
+
c
(c)
sin x - l n x l
-
2 jo[g(x)
- f (x)] dx
2/x
+
C
3
+ j2 [f(x) -
g(x)] dx
APJSWERS TO SECTION QUIZ 1.
4 (a) x / 4 + x 3 - lnlxl + C (b)
186
(c)
2x5/215
(d)
-1
(el
0
(f)
3 sec-lu
(g)
3
t 13
+
+
t
2"';;
+
ex
+
c
+c +
tan-lt - et - csc t
+c
2.
The arbitrary constants have different values.
3.
Differentiate the right-hand side.
4.
3/32 liters
Copyright 1985 Springer-Verlag. All rights reserved.
320
Section 7.2
7.2
Integration by Substitution
PREREQUISITES 1.
Recall how to differentiate rational, trigonometric, logarithmic, and exponential functions (Sections 1.5, 5.2, and 6.3)
2.
Recall how to differentiate by the chain rule (Section 2.2).
3.
Recall how to integrate basic functions (Section 7.1).
PREREQUISITE QUIZ 1.
Differentiate the following expressions: (a) (b)
(c) 2. 3.
cos x
+
e ~ i nx x6
+
Evaluate Evaluate
see 2x
+
+
+
3x2 - 1n(x3 l(x6 ![1/(1
1/2x
3 ~ / 2 ,-3
+
+
cos x
4x)
+
l/x)dx
.
2 + x ) + e x - 31dx
.
GOALS 1.
Be able to recognize the types of integrals which may be evaluated by substitution.
2.
Be able to evaluate integrals using the technique of substitution.
STUDY HIXTS 1.
Integration by substitution. Be sure you understand this technique well.
2.
It is one of the most important techniques of integration.
Choosing a substitution. Practice and studv the examplec. in the first three examples, u
Note t11:jt
is an expression which is raised to
a power.
Copyright 1985 Springer-Verlag. All rights reserved.
3.
More common substitutions.
If
in the integrand, where
again does not have to be an integer.
n
.
so, try substituting u = xn
xn
appears, see if
xn-'
also appears
Two other good choices for
u
If
are used
to derive the shifting and scaling rules. 4.
Shifting rule. of
5.
It is simply a substitution
u = x + a .
Scaling rule. of
6.
Do not memorize the rule.
Do not memorize the rule.
It is simply a substitution
.
u = bx
Differential notation. As an alternative, you may differentiate both sides of
u
=
g(x)
.
du/gl(x) = dx
to get
lu =
,
gt(x)dx
and then solving to get
All of the methods presented are the same.
Cse the
one which you feel most comfortable with.
7.
Giving an answer.
Remember to express your answer in terms of the
original variable.
And don't forget the arbitrary constant.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Let
u = x
2
+
,
4
+c
so
du = 2x dx
+c
2u5/'/5
=
Differentiating, we get (2/5)(5/2)(x2
+
=
. Then 12x(x 2 + 4)3'2dx
2(x2
+
4)5/2/5
(d/dx) [2(x
41~/~(2x) = 2x(x2
+
2
4I3I2
+ .
=
ju3/'du
=
+c .
4j5/'/5
+
C]
=
Differentiation yields the
integrand, so the answer is verifiesd.
5.
Let
u = tan 0
,
so
j(du/u3) = uW2/(-2)
2 du = sec E dB
+C
=
-1/2 tan20
Upon differentiating, we get 2
2(2) tan 0 sec 0/(2 tan2p)'
.
=
+
Then C
/(set2?/tan36)d?
.
2 (did" [-112 tan P
2 sec e/tan39
=
.
+ C]
=
Differentiation yields the
integrand, so the answer is verified.
Copyright 1985 Springer-Verlag. All rights reserved.
9.
+
u = x4
Let
,
2
so
du
.
4x3dx
=
Then
4 (d/dx) [ ( x
D i f f e r e n t i a t i n g t h e answer g i v e s
+
(112) (112) ( x 4
2)-li2(4x3)
l ( x 3 / K 2 ) d x =
,
= x3/&
+
~ ) ' / ~ /+ 2 C]
which i s t h e i n t e g r a n d .
Thus, t h e a n s w e r i s v e r i f i e d . 13.
Let
u 3
2 cos(r )
=
2
cos ( r ) d r
,
so
3 j-u du
=
du 4
+
-u / 4
=
-sin(r
=
C
2
).2r dr
-cos
=
4
2
( r )/4
D i f f e r e n t i a t i n g t h e answer g i v e s (-1/4)(4)(cos integrand. 17.
Let
u = 0
+
-cos u
3
2
.
Then
+
2
2
)x
.
C
( d i d r ) 1-cos
2
12r sin(r
4
2 (r )/4
3
2
+
,
( r ) ) ( - s i n ( r ) ) ( 2 r ) = 2r s i n ( r )cos ( r )
C] =
which i s t h e
Thus, t h e a n s w e r i s v e r i f i e d
+
4
,
so
C = -cos(P
.
du = dO
+
+
4)
Then
.
C
jsin(8
+
4)d0 = j s i n u du =
(The s h i f t i n g r u l e may h a v e b e e n
applied t o t h i s i n t e g r a l . ) D i f f e r e n t i a t i n g t h e answer y i e l d s
+
sin(Y 21.
Let Then
U1I2
u
=
t2
![(t
+
,
4)
C
=
2t
+
3
,
so
du
I)!=]
dt
GT-3
+
(2t
= =
+
integrand.
25.
2t
+
2)dt
=
2(t
ju-li2(du/2)
4)
+
C]
=
=
+
1)dt
.
( 1 / 2 ) ( U ~ / ~ / ( 1 / 2 )+) C
=
+c .
D i f f e r e n t i a t i n g t h e answer g i v e s (1/2)(t2
+
s o t h e answer i s v e r i f i e d
+
+
( d i d 8 ) [-cos(e
3)-lI2(2t
+
2 ) = (t
(d/dt)
(G + C)
+
,
=
which i s t h e
Thus, t h e answer i s v e r i f i e d . 3 j c o s PdB
Use t h e h i n t t o g e t
,
Now l e t
u
=
u - u3/3
+
C = sin B
sin 8
so
=
2 J ( c o s R ) ( c o s 8 ) d ~= J ( c o s 8) (1
du = c o s P dO
.
Then
2
- s i n 0)d0
3 f c o s 5 dR = j ( 1
-
- s i n 3 @ / 3+ C
Copyright 1985 Springer-Verlag. All rights reserved.
2 u )du
:
Ax Let
x
2 sjn u ,
=
"/j
therefore,
(using the half-angle formula)
+
C
.
+
33. Let
u = In t
37.
Let
(a)
(c)
+
, so du
u = sir, x ,
+c
ju du = u / 2 Let
u = sin
xvC-212
2
(b)
-1
Since
-1 2 sin (x/2)
u = cos x
,
=
so
C
cos x dx
+
=
C
.
sin x dx
=
2 -cos x/2
From
2 sin x cos x ,
sin 2x
=
=
jsin x cos x dx
-cos 2x14
+
C
sin 2 ~ 1 2 )+ C
=
+
2u
cos 2u) du
+
2 sin u
therefore, j[sin(ln t)/t]dt
+
C
2j(1
=
x
y
.
dt/t ;
so
jcx
(x/2) (See the figure), the integral is
2 sin x/2
=
+
2(u
=
ju(-du) =-u2/2
Hence
dx =
2 (2 cos u du) = 4j cos u du
A-7
cos u
dx = 2 cos u du :
so
=
+
du
-du
C
.
jsin x cos x dx
Then
.
=
Then
=
jsin x cos x dx
=
.
we get
(112)Jsin 2x dx
sin x cos x =
=
sin 2x/2
(112) 1-cos 2x121
+
C
. =
by the scaling rule.
To show that the three answers we got are really the same, we need to show that they differ from one another by constants. 2 2 sin x/2 - (-cos x/2) = 112 , -cos2x/2 - (-cos 2x14) which is a constant.
=
which is a constant. Also,
2 -cos x12
+
2 (2 cos x - 1)/4
=
-114 ,
Thus, we have shown that all three answers
Copyright 1985 Springer-Verlag. All rights reserved.
SECTION Q U I Z
1.
Evaluate the following i n t e g r a l s : (a)
j(t
+
+ 5)312dt
4)(t
[ H i n t : Let
u = t
+
5 .]
-
(dl
2.
3.
[ H i n t : Compleie t h e s q u a r e . ]
d
(c)
/sec5t tan t d t
What i s wrong w i t h t h e f o l l o w i n g s t a t e m e n t s ?
(a)
i ( 3 - ~ ) ~ / * d= x2 ( 3
(c)
l c o s 5x dx
=
- : ~ ) ~ / ~+ /C 5 .
5 s i n 5x
+
C
The r a t e a t which S c h i z o p h r e n i c Sam s p e n d s l i s t e n i n g t o v o i c e s a f t e r h i s p s y c h i a t r i c v i s i t i s g i v e n by
t 2 / ( t .t 4 )
, 0
months
t
< t < 2.5
F i n d a f o r m u l a d e s c r i b i n g t h e amount o f t i m e s p e n t i n a u d i t o r y h a l l u c i n a t i o n s i n c e h i s l a s t p s y c h i a t r i c appointment.
(Hint: Let
u
=
t
ANSWERS TO PREREQUISITE QUIZ
1.
+
(a)
-sin x
(b)
(cos x ) e
(c)
6x5
2.
x7/7
+
3.
tan-'x
+
6x
sin x
+
ex
2 s e c 2x t a n 2x
sin x
+
( 1 / 2 ) ( l n 3)3x'2
'3
- (3x2+ 4 ) / ( x .
+
I n 1x1
- 3x
- 1/2xL
+
- 3x-4
4x)
+ c
+c
Copyright 1985 Springer-Verlag. All rights reserved.
+
4
ANSWERS TO SECTION QUIZ 1.
2.
3.
+
- 2(t
5)512/5
+c
let
u - e
2(t
(b)
(213) tan-'[e3~;
(c)
5 sin-'
(dl
5 sec t/5
(el
2 exp(4x ) / O
(a)
Minus sign is missing.
(b)
Exponent and denominator should be
(c)
Answer should be
(d)
Arbitrary constant is missing.
(t
j17"/7
+
(a)
+ 4)2/2 -
- 2)/2]
[ ( I
+
+ c ;
C ;
8(t
+
+
let
C ;
4)
+
C ;
u let
=
+
u = 4x2
16 In t
+
3x
.
u
=
t
+
5
.
.
sec x
+
let
v = u - 2
let
sin 5x/5
;
C
4'
. -1
rather than
-3
.
. .
Copyright 1985 Springer-Verlag. All rights reserved.
7.3
Changing Variables in the Definite Integral
PREREQUISITES 1.
Recall how to integrate by the method of substitution (Section 7.2)
2.
Recall the fundamental theorem of calculus (Section 4.4).
PREREQUISITE QUIZ
t
1.
Express the integral
2.
Perform the following integrations: (a)
jcos x sin x dx
(b)
j[3x/(1 +
(c)
4 j(2x - 3) dx
~'(x)dx in terms of
F
.
4 x ) I dx
GOALS 1.
Be able to change the limits of integration while integrating by substitution.
STUDY HINTS
1.
Changing limits.
Once you have learned integration by substitution for
indefinite integrals, definite integrals can easily be computed by chang-
2.
ing the limitseveqtime you change variables.
Simply choose
and then change the limits from
to
Substitution may not work.
x = a
and
b
u = g(a)
u = g(x) and
g(b)
.
If the integral becomes harder, try a different
substitution or maybe integration by substitution is not the correct method to be used for solving the int~?gral. When you are proficient, going through a mental checklist shouldn't take long.
Copyright 1985 Springer-Verlag. All rights reserved.
3.
Use o f t a b l e s .
By t h e t i m e you f i n i s h t h i s
D o n ' t r e l y on t a b l e s .
c o u r s e , you s h o u l d be a b l e t o d e r i v e t h e i n t e g r a t i o n f o r m u l a s found i n the tables.
They a r e p r o v i d e d f o r y o u r c o n v e n i e n c e .
Check w i t h y o u r
i n s t r u c t o r r e g a r d i n g a c c e s s t o a t a b l e d u r i n g an exam.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Let
u
=
+
x
2 ,
so
du
=
1;
[iu312)/(3~2)~ = 2 ( m 5.
u = x 2 + 2 x + 1 ,
Let
+
+
1)(x2
2x
+
9.
Let
u
x
=
,
(i/2)j;eudu 13.
Let
u
f0 2 - j O 5u du 17.
Let
u
= (eu/2)
cos i
=
("14
,T 1 8 ( s i n
;1
21.
( x 12 25.
-
tan
(e
=
+
2 :~(3x
a5
x exp(x )dx
=
.
=
- s i n x dx ;
therefore,
,
so
iu
=
s i n i di ;
therefore,
-1i
cos(nI8)
2
j:j:2
5 c o s x sin x dx
=
&I2
(du/u)
= ln(&
=
'
d
=
-1n/u1
.
cos(.r/8))
tan
We u s e d t h e i d e n t i t y
tan
"
=
t o s o l v e t h i s problem.
-1
6x)dx
+
2
therefore,
- 1)/2
x)
[(x3 3
=
+
x
4 - tan
- 1)/(x2 -1
I f we make t h e s u b s t i t u t i o n (3x2
therefore,
.
du
Division yields
2
d u = ( 2 x + 2 ) d x = 2 ( x + 1)dx ;
so
l n ( c o s ( ~ / 8 ) )- l n ( f i j 2 )
e
j : I ~dU =
=
.
,
1 1 ~ 0 st ) d i
s i n G/cos
dx
.
0
=
2 6 - 213
=
du = 2xdx ;
so
cos x
=
- 013 so
j t l m
therefore,
~)~'*d= x ( 1 / 2 ) ~ ~ ~ u ~ = / [ ~( 1d / u2 ) ~ ~ / ~ / ( 9 / 4 ) 1=
2 [(25)9/4 - (9)9/4] /9 2
dx ;
,
so
6x)I du =
ll/(x2
[ (x
(3)
+
u = x3
+
+
[x
- l/(x
2
+
l)]dx
=
.
n/4
+
3
l ) ] dx =
2
+
1
,
we h a v e du = 5 li)/ 3 m dx j . =2 ] [ ( x 2 + 3x11 u= 1
+
3x
3 ) / 3v/;;(3x
way t o e x p r e s s t h e q u a n t i t y ( X
+
6 ) ] du
+ 3 ) / ( 3 x +6)
1
.
T h e r e i s no s i m p l e
i n t e r m s of
u
.
(1~!i.
Copyright 1985 Springer-Verlag. All rights reserved.
=
25.
(continued) would h a v e t o s o l v e t h e e q u a t i o n u .)
29.
Let
.
L
Therefore,
+
2
,
so
du
+
1;
=
-116
By s u b s t i t u t i n g
u
=
0
[(x
j0
=
1;
,
0
F
F(g(x))
l)/(x
u(0) = 1
and
(-u 1 3 ) Let
+
(-I/&)
3
(b)
+
2x
113
=
2x
1
for
x
i n t e r m s of
+
and
+
+
- In 2
2)dx
=
2(x
+
2 6
I
1)dx
=
.
b
=
2
+
2
,
(116)
Thus, t h e
+
l/fi
,
Thus,
-615
=
+
612
=
( 5 > 5- 2&)/10
du = - s i n x dx
we have 2
~ ~ 1 2 c 0x s s i n x dx =
,
0 2
-IIU du
.
u(n/2) = =
.
F(g(b))
=
I
3 > 0
=
( l / m in 2(3)x
=
bfi
.
- F(g(a))
.
f
b e t h e a n t i d e r i v a t i v e of
;1
I]
a
2 l 3 l 2 ] dx = ~ ; ( d u / 2 u ~ ' ~=) [ ( 1 / 2 ) u 112,
cos x
.
+
(2x
=
+
+
2x
[ l n 18
= (I//?)
2
r l
+
~,!,[dx/dx~
1;
u = xi
(-11211 (a)
+
We c o n c l u d e t h a t t h e s u b s t i t u t i o n was n o t e f f e c t i v e i n t h i s c a s e
area i s
37.
3x 2
+
x3
=
We u s e t h e f i r s t h a l f of f o r m u l a 65 b e c a u s e c =
33.
u
Then,
~ : f ( g ( ~ ) ) ~ ' (=~ ) d ~
T h i s i s t h e same a s
g(b)f(u)du = 'g(a)
I "b) F ( u ) , ~ ( ~ ) Thus, t h e f o r m u l a i s v a l i d i n d e p e n d e n t o f t h e r e l a t i o n
.
s h i p between
€(a)
and
.
g(b)
S E C I I O N QUIZ 1.
Evaluate the following i n t e g r a l s : (2.1
rl j-l(x
+
5
1 ) x dx
[Hint: L e t
u
=
x
+
1
.I
u
=
1
+
2
(b)
j i 3 x exp (-x ) d x
(c)
li[t5/(1
+
t313]dt
[ H i n t : Let
3
i:
.I
Copyright 1985 Springer-Verlag. All rights reserved.
2.
2
n
Consider the integral
1-,3x
. 3 sln(xZ') [exp(cos(x
2 dx
)I
u = x
3
.
.
(a)
Rewrite the integral by substit:uting
(b)
Rewrite the integral in (a) by substituting v = cos u
(c)
Rewrite the integral in (b) by substituting w
(d)
Evaluate the integral.
(e)
What substitution can be used to evaluate the integral in one
=
2v
.
.
step? 3.
Lost in a magic cave, you read, scribbled on the wall, "RIGHT 72 sin 4x dx , LEFT [2dx/(l
+
2 4x ) I "
.
lock marked "EXIT."
0
jw2h(t
+
3
2) dt , RIGHT
f!{;2(7~/r)
x
From that sign, an arrow points to a combination What combination will probably bring you back to
the outside world?
ANSWERS TO PREREQUISITE QUIZ 1.
F(b) - F(a)
2.
(a)
2 sin x/2
(b)
(312) tan-] (x2)
(c)
+
C
5
(2x - 3) I10
2 -cos x/2
or
+
+
+
C
C
C
ANSWERS TO SECTION QUIZ 1.
(a)
12817 - 3213
(b)
0
(c) (d)
1/24 7116
Copyright 1985 Springer-Verlag. All rights reserved.
3 2.
3.
(a)
J-t3
(d)
i e x p [-2 c o s ( n
(e)
Let
sin u i e x p ( t o s u ) I
u
=
3
11 -
2
du
3 e x p [2 C O S ( T ) ] } / 2
3 2 cos(x )
36 R I G H T , 24 LEFT, 35 RIGHT
(Apologies t o the female readers!)
Copyright 1985 Springer-Verlag. All rights reserved.
7.4 Integration by Parts PREREQUISITES
1.
Recall the product rule for differentiation (Section 1.5).
2.
Recall the definition of an inverse function (Section 5.3).
PREREQUISITE QUIZ
1.
Differentiate the following expressions: (a)
x In x
(b)
(5x - 3)(x2
(c) 2.
If
x
2
-
4x
+
1)
tan x
f(x) = 2x
+
3 ,
find a formula for
f-l(x)
GOALS
1.
Be able to use the technique of integration by parts.
2.
Be able to integrate inverse functions.
STUDY HINTS
1.
Integration by parts.
Memorize the formula
don't forget the minus sign.
ju dv = uv - jv du
and
The formula will be used quite often during
your studles of mathematics. Learn the formula well enough so that it becomes second nature to you.
2.
Choice of
u.
,
When integrating by parts, the factor xn
is a positive integer, is often chosen to be
u
.
After
where
n
n
repetitions
of integration by parts, this factor will be eliminated. Another common situation is when one of the factors is let
u
=
In x
.
Note the choice of
u
In x ,
in which case we
in Examples 1
,
2
,
and
3
.
Copyright 1985 Springer-Verlag. All rights reserved.
3.
I
Example 4 d e m o n s t r a t e s a t e c h n i q u e which i s u s e f u l f o r
method.
i n t e g r a l s i n v o l v i n g e x p o n e n t x a l s and t r i g o n o m e t r i c f u n c t i o n s .
The
method works b e c a u s e r e p e a t e d d i f f e r e n t i a t i o n o f t h e f a c t o r s y i e l d s the original factors.
4.
I
"Other" times.
T r y Example 4 a g a i n b y l e t t i n g
methods.
u = sin x
Now, t r y a g a i n w i t h
letting
u = ex
i n t h e second s t e p . u
t h a t you must l e t
Integrating inverse functions.
-
uv
,
/ v du
=
ex
both
i n t h e f i r s t s t e p and t h e n The l a s t method d e m o n s t r a t e s
be t h e e x p o n e n t i a l o r t h e t r i g o n o m e t r i c f u n c -
t i o n s throughout t h e i n t e g r a t i o n .
5.
u
Don't change i n t h e middle.
Once you l e a r n t h e f o r m u l a
jy
s i m p l y rename t h e v a r i a b l e s t o g e t
A l l we d i d was r e p l a c e
u
with
b e r t o e x p r e s s your answer t o
y
=
dx = xy - l x d y
.
with
x
.
F i n a l l y , remem-
i n terms of
x
,
not
and
v
=
=
(x
and
ly dx
/u d v
v
.
y
SOLUTIONS TO EVERY OTHER ODD E X E R C I S E
1.
Let
+
u = x
5.
Let
+
C
and
+
f(x
therefore, cos x
1
, so
dv = c o s x dx
1 ) c o s x dx
=
+
(x
u = xi
and
dx
1 ) s i n x - / s i n x dx
dv = c o s x dx
,
so
du
2x dx
sin x ;
+
1)sin x
+
v = sin x ;
and
.
For t h e i n t e g r a l
, a p p l y i n t e g r a t i o n by p a r t s a g a i n w i t h , du
d v = s i n x dx
=
2 dx
and
v = -cos x ;
-2x c o s x - 1 - c o s x ( 2 d x ) = -2x c o s x c o s x dx
=
/x2 c o s x d x = x 2 s i n x - /2x s i n x dx
j 2 x s i n x dx
2
=
.
therefore,
/x
du
=
x 2 s i n x - [-2x c o s x
+
+
+ +
C
.
, so
/ 2 x s i n x dx
therefore
2 sin x
2 s i n xl
u = 2x
Hence
C = (x
2
- 2)sin x +
Copyright 1985 Springer-Verlag. All rights reserved.
=
9.
13.
Let
u = ln(1Ox) ,
x ;
therefore, jln(l0x)dx u
Let
fore, 2s ,
2 s
=
,
so
so
dv
=
dx ,
=
x ln(1Ox) - jx(dx/x)
dv = e3sds ,
=
17. Let
3s
u
(9s2
6s
-
t2 ,
=
2
+
2)/27
so dv
.
j2t sin t dt t2 sin t2 Let
and
v
=
x ln(1Ox)
-
x + C
3s v = e 13 ;
and
.
3s dv = e ds/3 , du
=
+
+
C
+
and
j2e3'/9
3s = s2e3s/3 - 2se 19
u = ln(cos x)
e
=
/9 ;
.
du = 2t dt, and
Thus,
Again, we substitute w
cos t2
v
=
.
there-
Let
u
=
therefore,
+
+
2e3'/27
.
2 t )
=
3s
2 ds ,
2 2t cos(t )dt ,
=
found by substituting w
21.
2s ds ,
=
dx/x ,
=
~ s ~ e =~ s~ e d /3 s - ~ 2 s e ~ ~ d s / 3Apply parts again: so
e
du
2 3s
2 3s 3s ~ s ~ e =~ s~ e d 13 s - 2se /9
C
du = (10/10x)dx
j2t =
3
v
=
sin t2 ( v is
2 . 2 cos t2 dt = t sln t -
t2
to get the answer
+c . ,
so du = (-sin x/cos x)dx
=
-tan x dx ;
therefore,
25. iiie will use formula (8) with the role of x and y reversed. Let 2 x = JlIy - 1 , y = l/(x + 1) ; therefore, jdl/~.- 1 dy = jx dy = xy -
29.
jy dx
=
m
If we choose g(x)
=
cos x
/
(
x
f(x) = x
.
2
+
1)
-
j[dx/(x
2
+
G(x) = sin x ,
and
I)]
-1
~.v'l/y - 1 - tan
=
x
i
2 F(x) = x /2 and
then
2 jx sin x dx = (x /2)sin x - (1/2)jx2 cos x dx
We obtain
The new integral on the right is more complicated than the one we started with, so this choice of
33.
Note that
Ln x3 = 3 In x
and
=
x
.
Ue get
3(3 In 3
-
2)
.
v
J:
f
and
,
G
so let
i.s not suitable. u = In x ,
In x3 dx = 3f3 ln x dx 1
-
dv
=
dx ,
3[x In
du
=
-
1:
dx/x , 1 dx] =
Copyright 1985 Springer-Verlag. All rights reserved.
334
37.
Section 7.4
Use t h e f o r m u l a f o r i n t e g r a t i n g i n v e r s e f u n c t i o n s , s o and X
+
c o s -'(4x)l::;
d u = -32x dx
or
-
0
(l/8)(1/) 41.
+c .
.
t3 di
.
2x
,
-
= -7124
so
1::
Hence
.
.
Substitute
u = x
aE7
+
*)I
+
1 - 4x2
I. =
=
, du
s i n a x dx
t h e numerator a r e f i n i t e a s
0
t o get
-
.
0.09
becomes
12;~
+
(-e2'
=
and
=
dx
( 1 / 2 f i ) (7r/4)
+
( 1 / 4 ) 2 ~ ' / ~ / : /=~ T / B &
s i n a x / a 2 i " = (-271 c o s 2 a a ) / a
approaches
n/24
y = sin-'2x
,
1/26
and
x s i n a x dx = -x cos a x / a / i T +
J:"
, so
c~~ 114 .
dy =
T ~ U S , ~ ~ ~ ~ " s i n - l dx 2 x= x s i n - l 2 x /
, s o dv
therefore,
2
l e X s i n x dx = e x ( s i n x - c o s x ) /
From Example 4 ,
dx
(4x)
= (114) ( 0 )
-
j n n e x p ( 2 x ) s i n (2x) dx
Use t h e f o r m u l a f o r i n v e r s e f u n c t i o n s w i t h
Let
cos-'(4x)dx
~ / 4 1 : = ~6 ~/ 8
-1
=
u = 1 - 16x
Let
~ h u s ,we g e t (112) [ ( 1 / 2 ) e i ( s i n 6 - c o s
~ : / ~ ( d u / ~ =; , 7 / 8 d ?
49.
=
.
= x dx
u ll2du/32 6
llI8 114 cos -1 ( 4 x ) d x
Therefore,
( x / m ) d x
-du/32
41314
2 7l (1/2)l-2Te'sin
45.
4~:::
Firsr, substitute
2
.
dy = ( - 4 / m ) d x
y = cos
as
a
Geometrically,
a
v
(114)
1/2 = (n- 4)/8v5
=
-
-cos a x / a ;
cos ax dx/a = - x cos ax/
( s i n 2na)/a
tends t o i n f i n i t y ,
2
.
Since t h e terms i n
,fin
x s i n a x dx
.
approaches s i n ax
+
I
-
+
o s c i l l a t e s more a s
t h a t t h e a r e a of e a c h hump g e t s s m a l l e r a s
a
a gets
.
approaches larger.
Note
Finally,
t h e a r e a of one hump above t h e x - a x i s a l m o s t c a n c e l s t h e a r e a of t h e next hump below t h e x - a x i s ;
53.
(a)
j
c o s n x dx = ] ( c o s n - l x ) ( c o s x ) d x
with
n- 1 u = cos x
( - s i n x)dx
and
, and
dv
v = sin x
.
= cos
.
(1 cos n)dx
=
n- 1 x sin x cos
+
.
We a p p l y i n t e g r a t i o n by p a r t s
x dx ,
Therefore,
SO
du = ( n
-
I)COS~-~X
j c o s n x dx = c o s n - l x
[-(n - 1 ) J c o s n - 2 x ( s i n x dx) ( s i n x ) ] = c o s n - l x s i n x 2
0
t h e r e f o r e , t h e t o t a l a r e a approaches
( n - 1 ) l c o s n-2x
-
(n
+ -
s i n x --
(n-l)jcosn-'x l ) ] c o s n x dx
Copyright 1985 Springer-Verlag. All rights reserved.
x
.
53.
(a)
(continued) Rearrangement y i e l d s
.
l c o ~ ~ -dx~ x D i v i s i o n by (b)
Letting
,
n = 2
((2 - 1 ) / 2 l / c o s (cos x s i n x
57.
(a)
+
2
-at
dt
sin(wt) d t u = a e
-at
=
ae
C = (2/4)[cos x s i n x
. =
o)
u = e
= -cos(ut)/u
therefore,
2
.
x
1
+
ji
dt
+
a2/w2 = ( L 2
Therefore,
Q(t)
0
.
aO = ( l / T ) j i T cos
o
dx
x
=
= )
(
=
EC{ 1
therefore,
;
Q
.
2 -at d u = - a e
.
+a
=
Aje
and
/ b ,
-at
v =
cos(ot)/o
-
ae
-at
gives us
n x
+
C
-
e-at [ c o s ( o t )
n
=
Q(0) = -EC
0
2n 0
2;; ) 0
di -
=
+
sin(!Ut)/iL,
Q
+
usin(ot)
b
=
2
1 +
+
, so
C = 0
: ,.
2
=
/u,l
! .
s e p a r a t e l y because an
! I / : I ) ~ ~=" 0
0 ;
-
Rearrangement y i e l d s
2
.
,
-at
Now, l e t
I t i s always t h e c a s e t h a t
dx = ( l / n ) J i
=
dv = s i n ( d t ) d t
2 . 2 t.. 2 )/LC][w /i~,. + a 2 )I ~ - e - " ~ c o s ( . t , t ) /-~ ~cre, - . ,sin(,, t)/ 1
w i l l appear i n t h e denominator.
0 s x
, then
d t = ~[-e-'~cos(wt)/w
I n e a c h c a s e , wemust c o n s i d e r t h e c a s e
(1/;r)jiirf ( x ) . O
(3/4)
(3/2) cos x s i n x
Q = ~ / e - ~ ~ s i n ( wdtt ) = A[-e
- ~ ~ e - ~ ~ [ c o s (+o at s)i n ( h t ) / u ]
C = EC
(a)
-at
dv=cos(wt) d t ,
2
LC
+
~ [ - e - ' ~ c o s ( w t ) / b - jue-atcos(ut)
/LJ) j e - a t s i n ( w t )
+
4 l c o s x dx =
.
and l e t
, and v
D i v i s i o n by
!3c[(a2 C
2
+
+
Q = l(dQ/dt)dt =
. 2 2 - a t . 2 s ~ n ( o t ) / w - /a e s ~ n ( w t )d t / o 1
A(l + a C
+
/ o , so
sin(ut)/o ; -at
x
= (1/2) x
we h a v e
3
o)e-atsin(wt)dt
A = EC(a / o
du = -ae
61.
,
n = 4
By t h e f u n d a m e n t a l t h e o r e m o f c a l c u l u s ,
Let
1)
.
+c
2 / E C ( ~/w (b)
Letting
i( 1 / 2 ) / d x
2 3 ( 3 / 4 ) l c o s x dx = ( c o s x s i n x ) / 4
+
+ x/2) +
(cos x s i n x/2 3x/21
.
C
-
(n
yields the desired r e s u l t .
x dx = c o s x s i n x / 2
+ x) +
3 (cos x s i n x)/4
n
+
x sin x
2 2-1 l c o s x dx = c o s x s i n x/2
we h a v e
2-2
n- 1
nlcosnx dx = cos
n
bo =
a
= (I/?)
( ~ / r ) ; s ~i n n i d i
-
Copyright 1985 Springer-Verlag. All rights reserved.
61.
(a)
(b)
(continued) (11-) (-sos nx)/ n
l o2r
coefficients are
0
a
I
=
x d
letting so
u
=
= (I:")):'
a
dv
=
/
dv
=
(c)
bn
lo1)
-2/n
=
du
a,
(1/)ji7 x2 dx
sin nxln ,
2 2
lo)
so
a
v
=
dv
.
bn
sin nx/n ,
and
v
=
sin nx/n12' 0
-
j20
sin nx dx/
is determined by letting
v
and
-cos nx/n ,
bn
=
2- - ;2 cos nx/n( cos nx dx/n)
=
=
so
- 0)
.
-2111
=
Thus,
a
u
0
=
=
and all other Fourier coeffirients
3
.2i 1 3 ) ; ~= 8
cos nx dx , du
=
Thus,
a0
-4;-In if
so
+ J,?,'
2
b
2x cos
bn = (1/~)(-4a In) 2 8~ 13 ,
n # 0
a
=
and
- m
lo
=
o
is determined
2x dx ,
=
and
v
=
-(2/n)(-2111) =
2 4111
=
sin nx dx ,
n*
+
.
dx/n)
n
( 1 / ~ )*
=
. bn is
du
=
2x dx , (1177)
=
Y
llsing the result from
2n (1/7)(21n)~~ x cos nx dx if
-
n # 0 , b0
=
0
,
- 4 In .
=
and
bn
=
.
liT sin
sin mx cos nx dx 2 N2"
a
2- 2 , (l/~),/~ x " s ~ nx n dx
4/n2
This problem requires using cas nx dx ,
=
, dv
=
2
=
.
13
Using the result from part (b) , a
-cos nxln ,
part (b),
(11'
is determined by
,
(1/~)(2/n)j~' x sin nx dx
(-x2 cos nx/n 1 '0
,/i
. a
du = dx ,
(1/7)(-27/n
=
determined by letting u = x
(d)
277
27, 2 2= (1/~),/ x ~ cos nx dx = (l/a)(x2 sin nx/n,0
r2n J 0 2x sin nx dx/n)
and
dx ,
(l/)(x
=
u = xL ,
by letting
-
and all other Fourier
.
0
(0)
=
.
0
=
n # 0 ,
if
are =
l o27 =
2
cos nx dx ,
2 2cos nx/n
(11-)(-2~/n - sin nx/n 2" ,
2
x sin nx dx = (11-)(-x
(lls)~:-
2
=
.
sln nx dx ,
=
Thus, . a
x cos nx dx = (l/n)(x
+
n) = (11-)(O x ,
,
x
.
0
=
li
=
,/iT sin sin nx dx , .
mu cos nx dx
(n sln n2 # m2
fZP0
-rx
nu;
.
s1n nx It
n
2
2
By Exercise
+ rn = m
CoS
mx
A
50(a),
cos mu ccs nx)! 2
,
(2n
s i n m x cos nx dx =
Copyright 1985 Springer-Verlag. All rights reserved.
61.
(d)
(continued) 2n (-cos 2mx)/imJO = O
.
[sin(rnx - nx)/(m
.
m # n
;in
- n) - sin(mx
m
=
sin2mx dx =
.
n
If
16
Ton2 l(l
= O
- con 2mx) 121 dx
a = 'l/n)jiT 0
sin 4x14)
+4
if b3
. 1
or
3
lo
=
dx
=
(112)
0
provided
x
lo
,
+
0
=
sin 3x
.
a
=
= (l/r) (n) = 1
2x and
If m
b
]Efcos(mx
sin(mx - nx)/ =
217
j0
+ nx) +
n ,
then the half-
[ ( l + cos 2mx)/21 dx =
.
= n
bn = (I/.)
n = 2 =
2n
.
2 cos mx dx =
(sin 2x
cos Ox) (cos nx dx)] n
m # n
:1
1 (x + sin 2nu(/2m) 121 ion2
+
2n
237 (x - sin 2 m x / ~ m ) / ~ ! ~ -
=
+ nx)/(m + n) +
provided
angle formula implies
cos 3x13
nx)]
x
n , then we apply the half-angle formula:
cos(rnx - nx)] dx = [sin(mx (m - n)]
+
+ nx)/(m + n)]
Applying the product formula, we get
7 r1
1in sin mx
Using the product formula,
2n sin nx dx = (1/2)j0 [ cos(mx - nx) - cos(mx
= 0
+
+
2x/2 -
cos 4x)dx = (I/-)(-cos
(lln) [ji71(sin 2x if
sin 3x
,
and
a
sin 3x
n
=
+
cos 4x) (sin nx dx)]
otherwise.
4
+
=
0
+ if =
1
Therefore, a4 = b 2 =
and all other Fourier coefficients are
0
.
SECTION QUIZ 1.
Which statement is the formula for integration by parts?
-
(a)
juv dx = uv
vju dx
(b)
jy dx
=
xy
- jx dy
(c)
/U
dv
=
uv
-
v/du
Copyright 1985 Springer-Verlag. All rights reserved.
2.
3.
Evaluate the following integrals: (a)
jx51n x dx
(b)
5 2 jx cos(x )dx
(c)
jx4exdx
(d)
jehXsin 4x dx
(e)
jsin-'(3x)dx
Comment on the following: jln x dx = x In x - x j(x In x
4.
-
+
C
Differentiation of both sldes shows that
.
Thus, jx In x dx
=
x(x In x - x) -
.
x)dx
(a)
Is integration by parts lsed correctly? Explain.
(b)
Evaluate
jx In x dx
.
Jumping Janet's new inheritance is a porcupine farm with an odd-shaped
A new fence is needed, so Janet decides to make the holes
plot of land. for the posts.
In preparation to hop away from flylng quills, Janet
jumps along the boundary on a pogo stick. The land is bounded by x2 cos x
and
y = -(x - 3~14)'
+
9 ~ ~ 1 1. 6 If one unit equals
how much land area is allotted to each of Jumping Janet's (Hint:
25
y =
10 meters, porcupines?
Draw a graph to determine the limits of integration.)
ANSWERS TO PREREQUISITE QUIZ 1.
(a)
1 +lnx
(b)
15x2 - 46x
(c)
2x tan x
+
+
17
2 2 x sec x
Copyright 1985 Springer-Verlag. All rights reserved.
ANSWERS TO SECTION QUIZ 1.
h
2.
(a)
(d)
3.
6 x6 I n x / 6 - x / 3 6
sin -1
+
C
4x - e 4 X c o s 4 x ) / 8 (3x) - m
/
3
+ +
C
(2)
x sin
(a)
The i n t e g r a t i o n i s b e i n g performed c o r r e c t l y ; useful t o l e t
u = In x
C
and
dv
=
however, i t i s more
x dx
Copyright 1985 Springer-Verlag. All rights reserved.
340
Section 7.R
7.R
Review Exercises for Chapter 7
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
By the sum rule, power rule, and the basic trigonometric rules for
5.
+
j(x
antidifferentiation, we have
sin x)dx
Using integration formulas for sums, exponentials, logarithms, rational /(ex - x2 - l/x
powers, and trigonometric functions, we get x 3 e - x 13 - In 1x1 + s i n x + C 9.
3
-cos x /3
C
2 . jx sln x3 dx
+
5
2) dx =
Substitute u
5 lu du =
6 u 16
=
cos 2x ,
Factor out
114%
~(1,-
+
t2)dt
, then du
2 3x dx
=
Integrate by parts.
+
v = sin x ;
C
+
(x
=
+
,
2 6
+
2) / 6
du = -2 sin 2x dx ;
so =
3
-u 13
+
C
=
+
Let
u
3 t 13 =
therefore,
+
xL
C
,
sin-l(t12)
=
then
.
C
so
du = dt/2
dv
3 t 13
+
v
=
-cos x ;
therefore,
Integrate by parts. v
=
3
Let
3
x 13 ;
x3 in 3x13 - x /9
+c
u
=
+
2x cos x
In 3x ,
therefore,
x ( ,
then
2
Thus,
.
C
cos x dx , du
=
jx2 cos x dx = x
+
2
. sln x
2s ,
-
=
dv
-
dv
=
2 sin x + C =
In 3x dx
2
2x dx ,
.
sin x dx , 2
. sln x
+
.
x dx , du =
=
12x sin x dx
jx2 cos x dx = x
and
.
2 it dt = (112)
J
+
therefore,
therefore,
3 -cos 2x
du = 2dx ,
and
=
.
C
u
.
du13
C =
du = dx ;
so
Now, repeat the integration by parts with
dxlx
or
+
Isin u du/3 = -cos u/3
u = x
2x cos x - /2 cos x dx = x2 sin x 29.
=
3
(112) [ [ d t / m ] +
=
t3/3 = sin-'u
and
x
=
112 and substitute u = t/2 ,
=
m]+ 25.
cos x)dx =
.
2 2 12 cos 2x sin 2x dx = -ju du 21.
u
Integrate by substitution. Let ((x
17.
+
+
.
Integrate by substitution. Let 2 x dx ; therefore,
13.
2 x 12 - cos x + C.
=
=
3dx13x
=
x3 In 3x13 - 1x2dx/3 =
.
Copyright 1985 Springer-Verlag. All rights reserved.
33.
sin 3x13 ;
+
cos 3x19
Substitute w jx2[e(x2)]
dv = cos 3x dx
therefore, jx cos 3x d-K
x sin 3x13
37.
u = x ,
Integrate by parts with
=
x2 ,
x sin 3x/3
du
=
!sin
-
dx ,
v
and
3x dx/3
=
=
.
C
then
Substitute w
fuewdw/2
2x dx ;
2 /x3e(~)dx =
dw
=
therefore,
.
Now, integrate by parts with
J
.
u = Tr r 2 W so dv = e dw/2 , du = dw , and v = eW/2 ; therefore, 1x3e(x )dx 2 2 J weW/2 - lewdw/2 = weW/2 - eW/2 + C = x2e(X '12 - e(X '12 + C . 41.
(x dx)
+
=
,
=
& , then dw
=
.
leJftdx = jew(2w dw) eWdw, W
2we
du=2dw,
+c
- 2eW
(1/2&)dx
.
W
and
45.
v = e
+
C
=
u
2w ,
=
so dv
=
w
2eL(&
-
1)
+
.
C
Using the formula for integrating inverse func-
Thus, 1
cos y ,
Let
u
=
x ,
=
+
so
C =
-1
x tan x
+
+
=
=
du
then
dv
=
. Let
u
-1x) - j(sin y/cos y)dy
=
dx
x dx
1
x2 tanT1x/2
j:l4
[dx/2(1
+
2
x )]
2 x 2
J =
2 (7
1
+
.
tan y
=
=
,
dx
:
+
and
=
+
x tan-lx
-1
=
(du/u)
+
C
=
x tan
v
=
-cos 5x15 ;
x
=
. ,
du
=
1/(1
+ x2)
Integrating by parts and using long division, we have 7714
x
-x cos 5x/51'"~ 0 + jiJ5cos 5x dx15
sin 5 x 1 2 5 1 =~ 7/25 ~~
-1 tan x ,
and
tan y(tan-lx) - jtan y dy
lnlcos(tan-lx)
,
dv = sin 5x dx
~i~~~ sin 5x dx
-(n/5)(-1/5) u
jtan-.'x dx
tan(tan--'x)(tan
lnlul
therefore,
Let
y = tan- 1x ,
du = -sin y dy ; therefore, !tan-lx
so
+
x tan-'x
53.
therefore,
Thus,
tionr, we have
49.
dx/2w ;
=
Now, integrate by parts with
eW(2w - 2)
=
=
=
2
x )
=
, and v ,/;/4x
2 ( 3 2 t n 1 /
-
=
2 x I2
tan-'x
dx
=
1"
/32)ta11-~(~/4)+ ( - ~ 1 2+ tan-'~12) "I 0I 4
-14(dx12) =
Copyright 1985 Springer-Verlag. All rights reserved.
i
=
342
57.
Section 7.R
u = x ,
Integrate by parts with (2x
+
3)3/2/3
dv
+
+
3)312/31i - fi(2x
+
( 2 + ~ 3)512/151A = 5.513 - 25&/15
,
3 , du = dx
3)3/2dx/3
du = 2x dx ;
so
K
65.
Y
]
=
Let
f(x)
on
u = x2
+
therefore, j:13x
~ ; ~ ( 3du/2/;)
is
4 j2(dx/x)
[0,n/2] , ex
In
+
=
cos x > -a3 - 2x - 6 ,
+ 1 + T4164 +
2 n /4
+
3n = e7112+
73. In the first method, let (n/2)cos(nx/2)dx
,
and
dv = sin(nx)dx/Z
therefore,
v
, du
h[sin(rxlZ)sin(ax)/~
+
.
-n sin(nx)dx
cos(7,x)dx
=
sin(~x)/~ ;
=
16n
2
, and v
+
dv
.
, v
Rearrangement yields
let =
u
=
-
on
=
=
ln(4/2) =
J:"
6x)I0
=
+
.nl2
18.225
cos(i-x)dx ,
-
. du
=
=
/T
u
=
=
cos(ax/2) ,
-cos(i-x)/2r ;
+
cos(nx/2)cos(~x)/2r
jsin(~x12)cos(nx)dx
=
+c .
cos(rx) ,
-2 C O S ( ~ / ~ ) / T;
2 cos(nx)cos(rx/2)/~
+
Now, let
sin(nx/Z)sin(rx)
=
=
therefore, ,fsin(nx/2)cos("x)dx
-(n/2)sin(iix/2)dx
=
x
2
192n)/64
then
cos(~x/2)cos(rx)/2~1/3
By the second method, =
sin(.rx/2) ,
jsin(nx/2)cos(nx)dx
(1/4)jsin(nx/~) cos(,x) dx
du
+
- jsin(ix)cos(nx/2)dx/2
sin(nx/2)sin(nx)ln so
u =
(T4
+
,i5
Thus, the area is
so the area is
cos x) - (-x3 - 2x - 6)]dx = (ex + sin x + x 14
dxl
f(x)
4 In x 2 = In 4 - In 2
4
1
.
jtf (x)dx
9 ,
(3/2)&(2)
=
The area under the curve of [a,b]
=
.
.
/kf (x)dx
v
-
i3"/3
=
9fi/li = 3 6 1 1
is
and
therefore,
The area under the curve of
y 9
[a,b]
69.
+
(using integration by substitution) ;
~ A d 2 x+ 3 dx = x(2x
61.
J2x
=
then
dv
=
sin(nx12)dx
therefore.
/sin(-x/?)
;2 cos(~x/2)sin(nx)dx
.
Now, l e t
Copyright 1985 Springer-Verlag. All rights reserved.
,
-
Section 7.R
73.
(continued) u
=
sin(rx)
,
dv = 2 c o s ( ~ x 1 2 ) d x ,
then
4 sin(nx/Z)/n ; therefore, 4 sin(nx)sin(nx/2)
7 7 . (a)
Let
Letting tan
-1
4 j s i n ( ~ x / 2 ) c o (sn x ) d x
du = d x f x ; 2
C = ( I n X ) 12
,
x = 3 tan u
(x/3) ;
du = n c o s ( n x ) d x
+
.
J
therefore,
j ( l n x/x)dx =
2 dx = 3 s e c u du
we h a v e
[
dx/x2&]
=
1:
and
u =
2 [(3 s e c u d u ) I
J ' , -
2 ( 1 1 9 ) j n / 63 ( s e c u d u / t a n u ) = ( 1 / 9 ) j inr//63 ( c o s u d u / s i n 2 u )
(a)
Let
s i n 5x15 ;
dv = c o s u d u ;
, s o dv
u = e2jx
therefore,
.
5 j ( s i n 5x)e2jxdx du = 25e e
25x
. sln
25x
,
dx
5x15
+
e
je
v
=
e25tcos 5 t / 2 6 - 1/26
Q(1.01)
=
u = e25x
.
=
,
= 25e2jX
e
25x
so
Thus,
+
.
25x
.
s l n 5x1130
0 and
cos 5 t
-
e
(100126) [ ( s i n 5 . 0 5 ) / 5
+
)
v =
,
~ e ~ ~ 5 ~x dx c o= s
.
+
Rearrangement
e 2 5 X c o s 5x126
.
e25tsin 5tl130
M u l t i p l y i n g t h i s by -25 t
and
s l n 5x15 -
, gives us
t
,
.
dv = 5 s i n 5 x
c o s 5 x - 2 5 j e 2 5 x c o s 5 x dx
Evaluating a t t h e l i m i t s ,
(100/26)(sin 5 t l 5
, du
c o s 5 x dx
-cos 5x
j e 2 5 x c o s 5 x dx = e
gives us
(b)
25x
., L e t
therefore,
= c o s 5 x dx
Now, l e t
and 25x
=
- f/ ,n, /63[ 3 s e c 2 u d u / ( 9 t a n 2u ) ( 3 s e c u ) ] 7
9
, so
v
Rearrangement y i e l d s
/ 7 1+ -
v = sin u
and
.
C
therefore,
2 9 tAn u 9 t a n u
,
/sin(~x/~)cos(nx)d= x -2 c o s ( n x ) c o s ( r x / 2 ) / ~
, then
+
/ u du = u 1 2 (b)
IT +
u = In x 2
81.
343
+
gives us
.
cos 5.05
-
e
-25.25
1
(3.8462)
r
Copyright 1985 Springer-Verlag. All rights reserved.
-
344 Section 7.R
85.
Let
,
u = x
jeaxcos(bx) ax ae dx ,
then
.
dx
and
dv
=
ax e cos(bx) dx ax u = e
Now, let
v
sin(bx) lb ;
=
dv = sin(bx)
2 cos(bx) dx/b
.
+
+
(a2
+
b
let
2
2
+b
)
, and v
=
2
+
b ) = -eaxcos(bx)/(a2
problem, we have
+ b 2) +
dx/(a2
ax e cos(bx)/(a2
+ b2) +
C
+ b2) =
+
+
+
2 b )
+
b ) ,
2
b
2
- [2a/(a2
=
)I
cos(bx) - 2 ab sin(bx)] /(a
2
+
=
=
2
-cos(bx)/b
Thus,
+
2
b )
2 b ) -
du =
/beaxsln(bx)dx/
+
2 b )
+
.
aeaxcos(bx)l /
+
+
2 b ) -
axeaxcos(bx)] /(a2
2
+
+
+
ax ae cos(bx)l
2
b )
+
eax[(b
2
;
x
in the first part of this
[beaxsin(bx)
C
+ +
dv = sin(bx) dx ,
a cos(bx)] /(a
+
=
,fbeaxsin(bx)dx/
ax 2 jae cos (bx) dx/ (a
2 2
b )
du
ax 2 ae cos(bx)l /(a
x [beaxsin(bx)
+ b2)] +
=
aeaxcos(bx)l /(a2
ax 2 lae cos(bx)dx/(a
+
b )
v
and
For
.
-cos(bx)/b
+
+
so
b2) = [bxeaxsin(bx)
xeax[b sin(bx)
v
aeaxcos (bx) /b2 - ja2eax
.
jeaxcos(bx) dx
2
and
/eaxcos(bx)dx = [eaxsin(bx) / b
2
jxeaxcos(bx) dx
[eaxcos(bx) / (a
/aeaxcos(bx)
(a2
+
[beaxsin(bx)
=
u = beax/(a2
Uslng the result for
2 (a
)I
aeaxcos(bx)l dx/ (a2
b2) ,
abeaxdx/(a
2 ax du = a e dx ,
/xeaxcos(bx) dx = x [beaxsin(bx)
[beaxsin(bx) (a2
+
2
dx ,
Repeat integration by parts with
eaxsin(bx) /b
=
=
so dv = cos(bx) dx ,
Rearrangement yields
ax 2 ae cos(bx)/b2] [b2/(a Thus,
. ,
dx/b
therefore, jeaxcos(bx)dx
du
therefore, jeaxcos(bx) dx
. e s=n(bx)/b - jaeaxsin(bx)dx/b ax
ax , u = ae
,
,
2 b )
+
/
- a
2
.
Copyright 1985 Springer-Verlag. All rights reserved.
)
.
TEST FOR CHAPTER 7
1.
True or false: (a)
Substituting
(b)
If
f
and
u
-x2
=
a
<
b
b (x)dx + ~Eg(x)dx + jcf (x)dx
f : J
For any constant a > 0 ,
+
)I
-1
-1
(1) - tan
(e)
The area of the region bounded by
3.
Substitute
j[dx/lx
3
=
- x
x = sin 8
2
)I
=
5.
6.
b , c
x = 2 - yL
=
45
dx
=
6
.
and the y-axis is
11 +
C
1 fldx 'f -1
terms of
are
. .
( 0 ) = 45 - 0
into the integral
,
la [f (x) + g(x)l
l/x - lnlx! + lnjx -
integral as a definite integral in 4.
Jzg(x)dx
j0a dx = a - 1
./i[dx/(l
Show that
tan
+
a b
then
r l x
(dl
2.
x
2
,
c
<
~ ~ l e u d. u
yields
are integrable functions and
g
real numbers such that
(c)
1 2 ~oexp(-x )dx
into
. and write the
.
Evaluate the following by making a substitution:
+
2
+ 4 x - 3)ldx
(a)
j[(x
(b)
j[et/(l + e
(c)
2 j[sin x/(l 4- cos x)ldx
2)/(x
2t
)I
dt
Evaluate the following integrals usins integration by parts: (a)
jt2sin t dt
(b)
jln(x4)dx
(c)
jx2e-xdx
Let
u
dv
=
dx , so integration by parts gives
(l/x)x - j(-l/x ) x dx
=
1
=
l/x
and 2
sides yields
O =
1
.
+
j(dx/x)
.
Subtracting j(dx/x)
I(dx/x)
=
from both
Explain what went wrong.
Copyright 1985 Springer-Verlag. All rights reserved.
7.
Find t h e a r e a under (a)
2 x exp(-x )
(b)
3 2 x exp(-x )
t h e g r a p h of
Evaluate
9.
A p a r t i c l e ' s acceleration a t time
jtan-'(3x)dx
t = 0
(Hint : 10.
is
J ( 1-x) l ( 1 - x )
on
[ 0,ll
if
f (x)
is:
.
8.
velocity a t
f (x)
0
=
,
t
is
J(1 - t ) / ( t
+
1)
.
If i t s
what i s t h e v e l o c i t y a s a f u n c t i o n of t ?
1. )
An o d d l y d r e s s e d g e n t l e m a n , nicknamed Odd O l l i e , came i n t o t h e Odd Furniture Store.
He b o u g h t a t a b l e t o p whose e d g e s , a c c o r d i n g t o
t h e s a l e s m a n , a r e g i v e n by x
=
1 , and
y = 0
.
F(x)
=
x/(x
Each u n i t r e p r e s e n t s
+
I ) , and t h e l i n e s 1 meter.
x
=
0,
Odd O l l i e a l s o
h a s a c o l l e c t i o n o f ' 3 0 NOT REMOVE UNDER PENALTY OF LAW" t a g s , which h e w a n t s t o make I n t o a t a b l e c l o t h . by
5
cm.
The a v e r a g e t a g m e a s u r e s
3
cm.
Odd O l l i e i s w i l l i n g t o c u t t h e t a g s t o m a t c h t h e s h a p e
of h i s t a b l e .
How many t a g s d o e s Odd O l l i e mini.mally n e e d f o r h i s
tablecloth?
.kNSWERS TO CHAPTER TEST 1.
2.
(a)
False; substitution yields
(b)
True
(c)
False; i t is
(d)
False;
(e)
True
tan
-1
(In a ) ( a - 1) (1)
-
tan-l(0)
D i f f e r e n t i a t e t h e right-hand
1; 1 ( - e U / 2 f i ) d u
.
. =
n/4
.
s i d e and s i m p l i f y .
Copyright 1985 Springer-Verlag. All rights reserved.
4.
(a) (b)
5.
6.
l n G T+ tan
-1
t
+
(e )
-1
C
C
+
(cos x)
C
(c)
-tan
(a)
-t c o s t + 2 t s i n t + 2 c o s t + C
(b)
4x I n x
(c)
-(x2
2
+
-
4x
+
2x
The i n t e g r a l
+
C
2)edX
j(d,c/x)
+
c
may h a v e d i f f e r e n t a d d i t i v e c o n s t a n t s on e a c h
side.
7.
(a)
8.
x tan
1/2-1/2e
-1
(3x)
+
(l/b)ln(l
+ J;_t2 --
9.
sin-'t
10.
205 t a g s
+
2 9x )
+
C
1
Copyright 1985 Springer-Verlag. All rights reserved.
CKaPTER - 8 DIFFERENTIAL EQUATIONS
8.1
Oscillations
PREREQUISITES
1.
Recall how to use the chain rule for differentiation (Section 2.2).
2.
Recall how to differentiate trigonometric functions (Section 5.2).
3.
Recall how to convert between cartesian and polar coordinates (Section 5.1).
4.
Recall how to graph trigonometric furictions (Section 5.5).
PREREQUISITE QUIZ 1.
Differentiate the following expressions: (a)
sin 3x
(b)
cos(x3
2
+
2)
2.
Convert the polar coordinates
3.
Convert the cartesian coordinates
4.
Sketch the graph of
y
=
(2,;-13)
2 cos(x12)
to cartesian coordinates.
(--5/2,0) to polar coordinat~s.
.
GOALS
1.
Be able to solve differential equations of the form
2.
Be able to convert from
x"
= -u,
2
x
A sin ~ , t+ B cos ~ ) t to - > ~ o s ( -~ t )
. and
sketch the graph.
Copyright 1985 Springer-Verlag. All rights reserved.
STUDY HINTS 1.
-N o t a t i o n
and d e f i n i t i o n s .
spring constant. second, i s 2.
2 d xldt2
=
2 -o x
0
,
k is called the
d e s c r i b e s s i m p l e h a r m o n i c m o t i o n , and
i t i s known a s t h e s p r i n g e q u a t i o n . =
,
-ks
.
Spring equation.
t
=
The f r e q u e n c y o f o s c i l l a t i o n s , measured i n r a d i a n s p c r
=
1*;
F
I n t h e f o r c e law
+
x = x cos o t 0
the solution is
x = xo
If
and
dxldt = v
.
( v / & ) ) s i nw t 0
at
0
You s h o u l d
memorize t h i s s o l u t i o n . 3.
Uniqueness.
J u s t be a w a r e t h a t t h e s p r i n g e q u a t i o n h a s a u n i q u e s o l u -
t i o n i f t h e i n i t i a l v a l u e s f o r p o s i t i o n and v e l o c i t y a r e g i v e n .
The
p r o o f of u n i q u e n e s s i s u s u a l l y n o t n e e d e d f o r s o l v i n g p r o b l e m s .
4.
Graphing B sin w t
.
x = a c o s ( o t - 9)
,
x
convert i t t o and
o, =
8 = tan
maximum o c c u r s and e q u a l s
- 3)
= a cos(wt
-1
x = A cos w t
I n order t o graph
.
(B/A)
by u s i n g t h e r e l a t i o n s h i p s
Knowing t h i s , we a l s o know t h a t t h e
wt
a when
-
0 = 0
,
i.e.,
i s t h e phase s h i f t .
The g r a p h r e p e a t s i t s e l f when
fore, the period i s
2n/w
.
+
t = 610
- 8
ot
= 2n
,
which
;
there-
a
F i n a l l y , t h e amplitude i s
SOLUTIONS TO EVERY OTHER ODD EXERC.LSE
1.
We want t o show t h a t cos(3t
+
2n)
.
cos(3t) = f ( t )
5.
2.~13)= f ( t )
.
f(t
+
2 ~ 1 3 )= c o s 3 ( t 2n
,
+
cos(3t
2n/3)
+
d2x/dt2 and
+ o 2x
=
0
is at
x = x cos w t 0 t = 0
.
vo
=
dx/dt
Here,
x = 1 cos 3 t
+
(.-2/3)sin 3 t = cos 3 t
+
(v /w)sin a t 0 d
= 3
,
so the
- 2 sin 3t/3 .
Copyright 1985 Springer-Verlag. All rights reserved.
:
27) =
.
xo = x ( 0 )
sc7lution i s
+
Since t h e cosine function has period
The s o l u t i o n o f where
f(t
9.
xs
A function of the form x a cos(dt - 8)
amplitude -
/
, and phase shift
a
the amplitude is
phase shift is
is
2 2 d x/dt
at
+
x = xOcos
where
.
Here,
solution is x
=
+
the
u2x = 0
sin
(v 0
x0 = ~ ( 0 ) and
t = 0
3 ;
is
1/3
The solution of
1
2-/L ,
has period
. In this case, the period
J
2713 ;
1
=
ot
v0 = dx/dt
,
2
L =
so the
.
-cos 2t
I
17. Here, w = 2 , yo y cos ot
0
21.
+
1 , and
v0
(v /o)sin ult = cos 2t 0
The frequency (a)
=
o:/2n = 2
By definition, o
=
=
+
3 ,
so
o = 47
4- =
y =
.
(3/2)sin 2t
is given, so
VG ,
so the solution is
. ;
therefore, k
=
16
2 I
is the spring constant. Here
x0
=
v0 = 1 ,
so the
solution of the spring equa-
- tan-' 4n (4,)
x
(1/4n)sin 4;t
.
it, rewrite
-
0.08)
.
=
cos 4at
+
To graph
(1,114-) in
polar coordinates: n , or approximately
a cos(4rt
tion becomes
(1 ,U.OS)
,
x
-
This is used to sketcli the graph.
Copyright 1985 Springer-Verlag. All rights reserved.
25.
(a)
2 2 The equation of motion is m(d x/dt ) 2 2 27(d xldt ) = -3x
equation is
where 6x2
f(x) 0
.
0
=
(c)
f(t)
B sin d a t
+
+
/(fcg)
a,,B cos o(at sin u.(at
+
b)
b)
.
Theref3re,
Now if
fog
Now,
d(fog)/dt
2
d (foS)/dt2
+
d2[*
cos u(at
um2[iI cos '~(ar + b)
+B
must be
0
.
sin u(at
+
(a)
=
local minimum of
>0 . V
f't)
has the form = il cos v (at
+
- a 2 d 2 ~cos s(at 2
,L,
(fog) = 0
+ b) +
2 b)] (1 - a )
a b
=
.
di(fog)/
+
+ b) -
.
a2LL!2~
- a 2 , 2 f cos ~ ..,(at + b)
B sin (-(at + b ) ]
=
(fog)
can be nonzero,
.
=
does not affect the differentiation of b
. ~ ' ( x) 0
=
-f(x ) 0
=
By the second derivative test,
0 xo
and is a
.
By the chain rule, we have
dE/dt
=
+
Since this is zero, one of
# 0 and
By the definition of antiderlvatives 0
, which
-alA sin ,#(at + b)
Thus, there is no restriction on
V r ' ( r , ) = -f '(xo)
(b)
=
Since
Therefore
Notice that the choice of
.
-3+
=
2-
f(g(t))
=
+ b) +
the above factors must be zero.
- 'a
fl(x)
satisfies the spring equation,
+ b) , so d (fog)/dt2
B sin o(at + b)]
33.
fog
2
.
.
.
.
.
= 0
0
=
so the linearized equation is
Hence
fog
f(0)
satisfies the spring equation,
A cos , ~ +t B sin ut
1
2 2 m(d xldr ) = f '(xO) (x - xo)
The period of linearized oscillations is
Since
, so this specific
f(x)
.
We are told that
f '(0) = -3 ,
implies
7 2 27(d6x/dt ) = -3x
dt'
+
The linearized equation of motion is
(b)
29.
=
3 2x
(1/2)m(2)(dx/dt)(d
2 2 x/dt )
+
Copyright 1985 Springer-Verlag. All rights reserved.
33.
(c)
By part (b), we know that
E
are sufficiently small at
t
V(x)
is also small.
dxldt
and
x - xo
=
Since E
two terms which comprise E
constant.
js
0 ,
then
If
E
dx/dt
and
x - x0
(1/2)m(dx1dt)~ +
=
must remain a small constant, the
must both be small also.
Therefore,
will both remain small.
SECTION QUIZ
1.
2 Solve the differential equation d y/dx2 dy/dx = 3 when
2.
=
-2
+
and
y
3 cos(tl4)
2 when
=
2
x
(a)
What differential equation of t:he form
8 sin(tl4)
=
1
.
.
x
=
, assuming that
n y = 0
Let
solve?
3.
x
+
xl'(t) = -kx(t)
x
does
Remember to specify the initial position and velocity.
(b)
Convert the given equation into the form x =
(c)
Sketch the graph of
(d)
What is the spring constant if the mass is 2 ?
x
3
cos(o:t - 0 )
.
.
As a money-saving concept, the latest lines of economical cars are not equipped with shock absorbers. After going over a pothole, it has been determined that these 400,000
.
800,000
gram cars have a spring constant of
Initially, after going over a pothole, the car is
5 crn/sec.
from equilibrium and bouncing with a velocity of (a)
Write an equation of the form
x
=
A cos ot
10 cm.
+
R sin wt
describing
the car's vertical motion. (b)
Sketch the graph of the solution.
Copyright 1985 Springer-Verlag. All rights reserved.
ANSI,JERS TO PREREQUISITE QUIZ 1.
(a)
6x c o s 3xL
(b)
-3x2sin(x3
2.
(1,6)
3.
(-5/2,0)
4.
or
+
2)
(5/2,r)
4.11
ANSWERS TO SECTION QUIZ
+
1.
y = -2 c o s -x
2.
(a)
x l ' = -x/lh;
(b)
x
(dl
1/2
(a)
x = 10 c o s ( f i t / 2 )
3.
=
(3/7)s1n - x xo
=
/'Ec o s ( t / 2 -
3
, vo = 2
tan-'(n/3))
+
5 J Z siin(fitI2)
Copyright 1985 Springer-Verlag. All rights reserved.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 8.2
356
8.2
Growth and Decay
PREREQUISITES 1.
Recall how to differentiate exponential functions (Section 6.3)
PREREQUISITE QUIZ 1.
Differentiate the following: (a)
exp(3t)
(b)
2 exp(x )
(c)
exp(-2t
+ 4)
GOALS 1.
Be able to solve differential equations of the form
2.
Be able to understand the concept of half-life and compute it.
f '(t) = y f (t)
.
STCDY HINTS 1.
Decay and growth.
f '(t)
=
yf (t)
equation depending on whether
y
is known as the decay or growth is negative or positive.
of sign, the solution should be memorized; it is 2.
=
f(O)exp(yt)
.
Half-life. - Rather than meniorizing the formula for half-life, it is easiest to apply the definition. By definition, from the solution, we also have fore
3.
f(t)
Regardless
112 = exp(-rt
Doubling time.
.
f(tl,2)
=
f(tl12) = (1/2)f(0)
f(O)exp(-rt
.
;
There-
See how this is used in Example 5, Method 2.
This is similar to the half-life concept except that
the rate constant is positive rather than negative.
Copyright 1985 Springer-Verlag. All rights reserved.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
This exercise is similar to Example 1. We have
dT/dt
differential equation for the iron's temperature is
5.
2 9.
f '(t)
The solution of
v
and
-3 ,
=
so
The solution of so
y(t) 8
y(0)e
=
=
=
?f(t)
-8 2e
=
210 - 20
find out when -0.llt
=
190
=
T
ln(8119)
.
f(t)
Therefore, f(t)
=
2
=
y(t)
Since
y
=
3
-0.11(T
.
Here,
.
Here
20).
-
f (0)
x
=
1
,
=
gives
=
8 ,
2
=
~ e - ~ ( e =~ 2exp(8t ~ )
. Thus 80
[ln(8/19)]/(-0.11)
=
f(O) = 1 ;
is positive,
=
t
-(O.ll)f(t)
=
190exp(-0.llt)
=
f
Rearrangement yields
:1
and
f '(t)
so
f (t)
80
f(O)exp(~t)
=
Therefore,
, i.e., t
17.
21.
.
100 or
=
is
f(t) = T - 20 ,
13. As in Example 3, we let f (0)
f ( t ) = f (O)exp(jt)
. Substituting y
y(O)exp(8t) =
is
=
f(t) = 2exp(-3t)
ff(t)
, so y(0)
yf (t)
, so a
y = -0.11
8)
and
We want to
190exp(-0.llt)
=
z =
.
-
or
7.86 minutes
3f ,
so
Y
therefore, the solution is
must be an increasing function.
3
=
and
f(t) =
See
Fig. 8.2.1. 25.
If the decay law is tIl2
. Here,
(l/~)ln 2
=
f '(t)
=
-if(t)
K =
, then the half-life equation is
0.000021 ,
so
t
112
(1/0.000021)1n 2
=
33,000 years. 29.
Rearrangement yields
r =
In 2/t
decays,
0.10 gram is left, so
ln(0.10)
=
1.5
x
-(I? 2/450000000)t
9 10 years
112
=
In 2/450000000
0.10 = e
,
i.e.,
.
After
-(ln 2/450000000)t , t
=
90; i.e.,
-ln(0.10)450000000/1n
2
"
.
Copyright 1985 Springer-Verlag. All rights reserved.
.
33.
f(t)/f(0)
=
2
=
eY(lO)
100e(ln 2110jt ln 2
implies
0.075t , (a)
In 30
.
7 = In 2/10
=
Then,
(in 2/10)t ,
i.e.,
3000 = t = 101n30/
49 minutes. P(t) = 4P0 ,
37. As in Example 8, if
41.
implies
i.e.,
S(t)
=
~'(t)
=
e0.075t ,
so
4
=
3 0 0 e - O . ~ .~
2000 - 1000.0=2000 ,
eventually.
=
18.5 years.
t = In 4/(0.075)
Differentiation gives
4
then
so
2000
books will be sold
This is the difference between a constant and a natural
decay. (c)
S(t) 2000
1000
45.
(a)
b>on differentiation of the solution, we get 2 2 s I d s + th(t)/t + C ;
tc (b)
+
2 t h(t)/t2
Here, h ( s )
= P
Substituting r-l/t eUdu J -1
=
=
a
+
so the solution is
=
yields
u = -l/s
eu/-l't -1
=
l j l [h(.;)/s2jds + 1(C) a(t)
t/e-lIt
t(da/dt)
=
j:[h(s)/ 2
tj:[h(s)/s
=
Ids
+
+
tC
h .
-l/s
1
is
thererare,
daldt
du
exp(-l/t) = U
- t/e
+C = + t .
-
=
ds/s
I/e
1 , sc
.
tj:(e-11s/s2)ds
2
, so ;:(e-11s/s2)ds
Also,
C = 1
.
a(1)
. =
=
Thus, the solution
SECTIOK QUIZ 1.
Element
Z decays exponentially.
the half-life
o l
element
Z
801 remains after one month.
What is
?
Copyright 1985 Springer-Verlag. All rights reserved.
2.
A population obeys exponential growth. increases from
500
to
750 ,
population to increase from
In 25 years, the population
How long would it take for the same
3 million to 4 million? y/2 , assuming
3.
Solve the differential equation
4.
Solve and sketch the solution of
5.
Suppose an object shrinks exponentially. grams.
6.
=
5y'
=
-y
Exactly one hour later', it weighs
4
weigh
y'
if
y(0)
=
y(3) = 1
3
.
Initially, it weighs
14
grams.
. 15
When will it
grams?
A stranger is trying to decide what to eat at a Mexican restaurant. He t ' ~- Jalepelio peppers?" asks the waiter, " ~ h ~ this "Try it. You'll like it." likeit'sat 60'~
.
After one bite, the stranger's tongue feels
If ice water requires
tongue temperature back down to 37'~),
The waiter tells him,
38'~
what is the decay constant?
90 seconds to bring his
(normal tongue temperature is
Assume the tongue obeys Newton's
law of cooling.
ANSWERS TO PREREQUISITE QUIZ
1.
(a)
3exp(3t)
(b)
2x exp(x
(c)
-2exp(-2t
2
+
4)
ANSWERS TO SECTION QUIZ
1.
3.11 months
2.
17.74 years
3.
y = e
4.
-x/5 y=3e
j.
19.16 hours
6.
(x-3) I2
-0.0348 sec-I = -2.090 min-'
Copyright 1985 Springer-Verlag. All rights reserved.
Section 8.3
360
8.3
The Hyperbolic Functions
PREREQUISITES 1.
Recall how to differentiate exponential functions (Section 6.3).
2.
Recall how to apply the chain rule for differentiating (Section 2.2).
PREREQUISITE QUIZ
1.
Differentiate the following with respect to
+
(a)
et
(b)'
(et
(c)
exp(t2
t :
e-t
+ e-t)/(et +
e-t)
t)
GOALS
1.
Be able to define the hyperbolic trigonometric functions as a function of exponentials.
2.
Be able to differentiate and integrate expressions involving hyperbolic functions.
3.
2 Be able to solve differential equations of the form x" = o x
.
STUDY H I N T S
1.
Definitions. You should memorize (et
+
e-t)/2
cos ,
sinh
sin11 as "cinch" ,
2.
=
(e
t
-
e
-t
)/2
and
ccs.1 t
=
. They are the same except that sinh t has a minus sign.
Remembering that and
sinh t
sinh O
=
0
is odd and cosh
and cosh
cosh 0
=
1
is even.
as it is written, and
may help.
As with
:;in
One usuall\ pronoilnces tan!]
as " t a n i ~ i i . "
Other hyperbolic functions. Notice the simllarlties of formulas
(:)
their trigonometric counterparts.
Copyright 1985 Springer-Verlag. All rights reserved.
xiltli
3.
Derivatives of hyperbolic functions-. Except for the sign of the derivative and the fact that they are hyperbolic functions, the formulas are the same as their trigonometric counterparts. Note that the " c o m o n l ~ " used functions
sinh
,
,
cosh
and
tanh
have a positive sign in front
of the derivative, whereas the others have a negative sign.
4.
Half-angle formulas.
Formulas (8) are useful for integration. They are
analogous to the trigonometric half-angle formulas; note that the negative sign is associated with
5.
sinLx and
sinhLx
.
Antiderivatives of hyperbolic functiorz. As usual, the simplest antiderivatives are determined by reversing the differentiation formulas.
6.
The equation
2 2 d x/dt =
x
+
=
x cosh ~t 0
t = 0
.
d
2
.
x
Memorize the fact that the solution is
(v /u)sinh wt , 0
where
x
Alternatively, one can memorize
and derive the solution by determining
=
xo
x
=
A
and
dx/dt = v
A cosh wt
+
0 at
B sinh kt
B
and
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Rearrangement of formula (3) yields tion, 2 cosh x
5.
2 sech x
=
2 2 sinh x/cosh x
2 2 cosh x/cosh x
=
1
2 tanh x =
2 cosh x
+
+
=
+
e-x) / 2
=
(ex
-
+
2 l/cosh x
2 sinh x =
.
2 (sinh x
By defini-
+
1)/
.
Prove the identity by the method of Example 3. (d/dx) (ex
1
eVX)/2
=
sinh x
(d/dx)cosh x =
.
9.
Using the fact that
(d/dx)sinh x = coshx and the chain rule, we have
13.
Using the fact that
(d/dx)sinh x = cosh x
we have
(d/dx)sinh(cos(8x))
=
and the chain rule,
cosh(cos(8x))-(d/dx)cos(8x)
=
-8 sin 8s cosh(cos 8s)
Copyright 1985 Springer-Verlag. All rights reserved.
362
Section 8.3
17.
Since
(d/dx)coth x
Since
(d/dx)cosh x = sinh x
rule gives
(l+tanhx)
2
=
xo
solution is
x
=
=
xo
-
d
2
x = 0
dx/dt = vo
and y
solution is
+
[sinh x(1
=
+
tanh x12 = [sinh x(l
2 d xldt2
The solution of where
+
tanh x)]
0 cosh 3t
+
tanh x) - sech x]/
and
when
=
dxldt = vo
x = cosh 3t
+
x = xOcosh wt + (vo/w)sinh ot
is
t = 0
.
Here,
0
x
is
.
o = 3
Here,
Let
u = cosh x ,
2 lu du
=
3 u /3
+C
=
=
sinh 2x14 - x/2
+
du = sinh x dx ,
so
3 cosh x/3
+
C
( v /o)sinh w t
x dy/dx)sech (3xy)
+
, so the
C
y = sinh x
to get
y = 3
+C
+
as 3
unit:
sinh x
.
.
2 lsinh x dx =
.
and
2 jcosh x sinh x dx =
.
Using the technique of implicit differentiation, we have 2
0
Shift it up
lcosh 3x dx = sinh 3x13
2 sinh x =: (cosh 2x - 1)/2
2x - 1)/2]dx
so the
.
to obtain the graph of
Use the identity
,
. +
shown in Fig. 8.3.3.
to get
3
x cosh wt 0
t = 0
when
sinh 3t/3
=
Begin with the graph of
Substitute u = 3x
w =
(1/3)sinh 3t = sinh 3t/3
2 d x/dt2 - u2x
4
j[(cosh
tanh x) -
.
The solution of x
+
(d1dx)coth 3x =
(d/dx)tanh x = sech2x , the quotient
and
(dldx) [cosh x/(l
2 cosh x(sech x)] /(I
where
csch2x , the chain rule gives
=
(cosh y)dy/dx = 0
.
Thus,
3(y +
dy/dx = -3y sech23xyI
(cosh y + 3x sechL3xy)
Copyright 1985 Springer-Verlag. All rights reserved.
Section 8.3
53.
By the definition of [(ex
+
emX)
+
cosh x
sinh nx
=
nx (e
(cosh x
+
sinh x ) ~
+
-nx)/2 e
+
sinh x ,
and
-X (ex - e )ln/2"
(2ex)n/2n
=
=
(cosh x
.
enx
(enx - e-nx)/2 = 2enx/2
C O S ~nx
+
+
sinh x ) =~
Also, =
363
enx
cosh nx
.
+
Therefore,
.
sinh nx = enx
SECTION QUIZ 1.
Differentiate the following functions of (a)
sinh 3x
(b)
cosh x sinh 2x
(c)
tan x sinh 2x
(d)
coth x/csch 2x
(e)
tanh(x/2)
x :
- sech x
2.
Write
3.
Solve the following differential equations:
4.
5.
cosh(x12)
in terms of exponentials.
(a)
f1'(x) = 16f(x)
(b)
f "(x)
(c)
2 d y/dx2 = 9y ;
=
-25f(x)
;
f '(0) ;
= 2
;
f '(0) = 3 ;
(dyldx)
lo
.
f (0) = 2
= 6 ;
f(0)
=
1
y(O) = 0
. .
Perform the following integrations: (a)
jx cosh 2x dx
(b)
jeXcosh x dx
(c)
5 jsinh x cosh x dx
One day, two teen-agers decided to equip their grandfather's electric wheelchair with rocket jets. When the elderly man went for his afternoon ride down the street, the faulty rockets did not work immediately. When Grandpa had ridden
100 m.
down the street, the rocket jets began
firing. At that time, he was at an equilibrium position and
vO
was
1
As he accelerated down the street, the ride became bumpier and bumpier,
Copyright 1985 Springer-Verlag. All rights reserved.
364
5.
Section 8.3
(continued) and his height off the seat can be described
by
d2x/dt2
=
x ,
where
is the position of the chair. (a)
Solve the differential equation.
(b)
Sketch the graph of the solution.
(c)
How fast was Grandpa moving, i . e . , find
dx/dt.
ANSWERS TO PREREQUISITE QUIZ 1.
(a)
et - e-t
(b)
4/(et
-
e
-t 2
ANSWERS TO SECTION QUIZ
1.
(a)
3 cosh 3x
(b)
sinh x sinh 2x
(c)
2 sec x sinh 2x
+
+
2 cosh x cosh 2x
2 tan x cosh 2x
2
2
(d)
(-csch x csch2x-e 2 coth x csch 2x coth 2x)/csch 2x
(e)
2 (1/2)sech (~12) - sech x tanh x
+ e-X'2)
2.
(eXJ2
3.
(a)
f (x) = 2 cosh 4x
(b)
f(x)
(c)
y
(a)
x sinh 2x/2
(b) (c)
e2*/4 + x/2 + 6 sinh x/6 + C
(a)
x = sinh t
4.
5.
=
=
/2
cos 5x
+
+
(514)sinh 4x
(315)sin 5x
2 sinh 3x
-
cosh 2x14
f
C
c
Copyright 1985 Springer-Verlag. All rights reserved.
x
Copyright 1985 Springer-Verlag. All rights reserved.
Section 8.4
366
8.4 =Inverse
Hyperbolic Functions
PREREQUISITES 1.
Recall the definition of an inverse function and how to differentiate them (Section 5.3).
2.
Recall how to differentiate the hyperbolic trigonometric functions (Section 8.3).
PREREQUISITE QUIZ 1.
2.
3.
y = -x2
+
4 invertible?
(a)
On what intervals is
(b)
Find a decreasing function which is an inverse of
(a)
cosh(x
(b)
sinh x
Let
f(x) = x
+
1)
5
+x3+x.
Find
(f-')'(2)
2
+
4
.
x :
Differentiate the following with respect to 2
y = -x
.
GOALS
1.
Be able to differentiate and integrate expressions involving the inverse trigonometric hyperbolic functions.
STUDY HINTS
1.
Inverse hyperbolic derivatives. Study the method of deriving the derivative of
sinh-lx
and note its similarity to that for
sin-'x
(Chapter 5).
A11 of the others are derived analogously. The only difference between (d/dx)sinh-lx and
(d/dx)cosh
-1 x
is that the first has a plus sign in the
denominator and the second has a minus sign. A similar statement may be said for the denominator of the derivatives of The derivatives of
tanh-'x
and
is defined for
<
1 ;
coth-'
1x1
coth-'x for
-1 sech x
and
look the same, but 1x1 > 1
.
csch-'x
.
tanh-'x
Think about the graph
to determine if the sign is correct.
Copyright 1985 Springer-Verlag. All rights reserved.
2.
Inverse hyperbolic logarithmic expressions. Again, study how t ~ derive > the formula
similar method.
fi-.
+
sinh-lx = ln(x
The others are derived by a
)
The formulas are normally not worth memorizing.
Learn
to derive them (for exams), or in many cases (for homework), one can simply look them up.
Consult your instructor to see what is expected
on exams
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
id/dx)cosh-l(x2
+
+
2) = [1/j(x2
- 1]2x
2)'
2 (d/dx)tan--lx = 1/(1 - x ) ,
Since
5.
, the chain rule gives
(d/dx)cosh-lx = I / =
Since
Since
(d/dx)sinh-lx = 1
+
(d/dx)exp(l 13.
1 tanh- x
17. Let
y
=
=
(1/2)ln[(l
h F +
+
)/2
.
+
+
Since
+
- x)]
x)/(l
=
0
.
JY . T ~T US,x
=
4x2
+
3
1)
-
+
2/(2
-
2 x )
.
/E .
sinh-lx)~
, so tanh-l(0.5)
=
(1/2)ln(1.5/0.5)
By the quadratic formula, ex
,
ex > 0
+
=
(2y i
we take the positive square root to get
coshT-'y
=
ln(y +
K;).
Change the
variables to get the desired result. 21.
Differentiate
tanh-lx
=
=
. Multiply through by 2eX and rearrange
e-X)/2
I
4
, the chain rule gives
G
[exp(l
=
cosh x = (ex (ex)2 - 2yeX
to get
=
sinh-'x)
/
2xlk
the chain rule and product rule give
tan-l(x2 - 1) - 2x2/(x4 - 2x2) = tan-l(x2
9.
=
(1/2)ln[(l
+
x)/(l
- x)]
.
By the chain rule,
Copyright 1985 Springer-Verlag. All rights reserved.
368 Section 8.4
25.
[I/(~ + E l ) ] .
By the chain rule, we get (2x)I
+
[l/(x
=
)1
=
+.
cl)][l
.
1
-
[l 4 (1/2)(x2
+
=
1)-'12
x
[(E +
E l ) ]
Differentiation yields the integrand, so the
formula is verified. 29.
Substitute u = 2x
33.
Substitute
ln(sin x
37.
+
u
=
z
,
so
du/2
sin x ,
l
)
C
>0
x
or
l n ( L 7 - x)
-1
f '(x) = l/dx
+ 1 . f '(x)
-1//x2 and
=
=
1
differentiable for all
.
- 4x
2
)I
=
therefore, j(cos x/
p-+ 1 + Jx T -- In(
if
x
+
1
)
<0 .
.
Now, if
<0 ,
If x
is not cclntinuous at
lim ff(x) x+o+
j[dx/(l
Therefore,
In( x
/2
entiation yields
.
.
By definition, c o s h - ' ( K i ) if
dx
du = cos x dx ;
so
+
=
x
= 0 -
Thus, c o s h - l ( K +
1)
x
x
>0
we get
because
+ 1 + x) , differf '(x)
1im x+o-f
=
' (XI
is not even once
x
SECTION QUIZ 1.
2.
Differentiate the following functions: (a)
-1 sinh (x/2)
(b)
cosh-'x
sech
(c)
coth-'x
tan-l(x2)
-1
(2x1 , x
,
-1
1x1 > 1
Perform the following integrations:
(b)
1[(1 + x)/(l 2
- xi)ldx
- 1)1
(c)
j[dx/(x
(d)
~sinh-lxdx
Copyright 1985 Springer-Verlag. All rights reserved.
=
3.
A highway patrolwoman had just stopped a driver with alcoholic breath. The driver explained that he was an astrophysicist and had spotted a flying pink elephant travelling along the path
y = cosh-lx
.
(a)
Use logarithms to determine the elephant's position at
(b)
Sketch the elephant's flight path.
(c)
If the y-axis points north and the x-axis points east, estimate x = 2
the elephant's flight direction at
E ,W
,
NW , SW , NE
,
and
.
x
(Choose from
=
2.
N , S ,
SE).
ANSWERS TO PREREQUISITE QUIZ
1.
2.
3.
(a)
(-=,O)
(b)
f-l(x) =
(a)
2xsinh(x
(b)
cosh x
(0,m)
and
-m 2
+1)
1/93
ANSWERS TO SECTION QUIZ 1.
2.
(a)
I/=
(c)
-1 2 2 tan (x )/(I - x )
+
(a)
-1 2 sinh (x )/2
ln(x2
(b)
-ln/l - x / + C
+
C =
2x coth-lx/(l
+
+
K+1)/2
2 x )
+
C
[Factor the denominator and simplify the integrand
first] . (c)
-tanh-'x
+
C
if
1x1
<
1 ;
-coth-'x
+
C
if
1x1 ' 1 ;
or
Copyright 1985 Springer-Verlag. All rights reserved.
3.
(a)
ln(2+6)=1.32
(c)
NE (The a c t u a l a n g l e i s
s / 6 .)
Copyright 1985 Springer-Verlag. All rights reserved.
8.5
Separable Differential Equations
PREREQUISITES
1.
Recall the geometric interpretation of the linear approximation (Section 1.6).
2.
Recall the rules of integration (Chapter 7).
PREREQUISITE QUIZ 1.
2.
Evaluate the following: (a)
j [dx/(l -
(b)
j(ey
(a)
Write a formula which approximates
(b)
How is the linear approximation related to the tangent line of a
+
x)]
y4 - sin y)dy f(xo
+
Ax)
.
graph? (c)
Use the linear approximation to estimate
2
(0.97)
.
GOALS
1.
Be able to recognize separable differential equations and solve them.
2.
Be able to use direction fields or the Euler method for estimating
solutions of differential equations.
STUDY HINTS 1.
Separable equations.
These must be first-order differential equations.
They are separable in the sense that everything involving
y
can be
placed on one side of the equals sign, and everything involving x can be placed on the other side.
Copyright 1985 Springer-Verlag. All rights reserved.
372
2.
Section 8.5
Method of solution. Begin by thinking of
dx
and
dy
as separate
entities. Multiply and divide to put all of the terms involving and all of the terms involving x
y
on separate sides of the equation.
Then integrate both sides. See Examples 1, 2, and 3 .
3.
Electric circuits.
The result of Example 4 should not be memorized;
however, the result is useful for doing the exercises.
4.
Transforming an equation. of solution.
Example 6 demonstrates an interesting method
By letting a new variable represent a derivative, the
original equation became a separable first-order equation. 5.
Direction fields.
These fields are represented by tiny lines which are
the tangent lines to the solution curves.
y(x)
If
is a solution of
dy/dx = f(x,y)
,
then the slope of the graph at the point
(x,y)
dyldx = f(x,y)
.
Line segments through
f(x,y)
used to depict the direction field.
(x,y)
with slope
is are
By drawing a curve which follows
thedlrection of these lines, one can sketch a solution without knowing an explicit formula.
6.
Euler method.
See Figures 8.5.7 and 8.5.8.
This method is based upon the concept of the linear approx-
imation (Section 1.6).
By starting at a point and moving along the tan-
gent line for a short distance, one should remain near the curve. You may find it more efficient to understand the concept and derive the formula, rather than memorizing it.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Rearrange yields y(0)
=
dy/dx
=
cos x
jdy = jcos x dx
1
to get
,
implies C = 1 ,
so so
y
dy = cos x dx =
sin x
+
y = sin x
C
. Integrating both sides
. The initial condition
+ 1 .
Copyright 1985 Springer-Verlag. All rights reserved.
5.
Rearrange the equation to get lni y
i
=
y(1) 9.
lnlx! 2
=
+
eC
-2 ,
=
so
+
dx/(l
Multiply through by
y
y)
=
ln(1
and
x = O
+ x) +
gives
In 2 ,
-2x
to get
+
Integrating both sides yields l n ( 1 y = 1
.
dx/x
=
Integration yields C
. Exponentiation gives
C
gives
dy/y
y = e x
.
Substituting
is the solution.
+ y)
dy/(l
+ x) +
y) = ln(1
l n 2 = l n l + C = C .
1
or upon exponentiating,
C
Thus,
+
y
. Substituting
+
This is Example 4 with
13.
8 ,
E = 1 0 , and
I
the solution is 1
,
,
,
-? At
.
t = 0
1
. '
=
,
, '
l4
2.1 ;
4
lO/B)exp(-8t/3)
*t
=
.
x)
ln(l+y)
2(1
=
+
dx/(l
=
x) ,
L
i.e.,
3 ,
=
.
10=2.1 =
1018
1.25
=
+
+
R
=
Thus,
(2.1 -
0.85exp(-8t/3).
then the graph drops exponentially toward
E/R = 1.25 .
17. dy/dt -sx and
+ y
=
by - rxy = b(s/c)
exy
=
-s(b/r)
+
- r(b/r)(s,'c)
c(b/r)(s/c)
=bs/c
= -sb/r
+
-
bs/c = 0 and
sb/r = 0
dx/dt
=
. Since both x
are constant, this is the eq~ilibriumpoint which also solves
the predator-prey equations.
21. Substituting into y
=
(TO/9.8) [cosh (9.8/TO)x [sinh (9.8/~,,)x] approaches
0
=
and
(T0/mg)[cosh(mgx/T0) -
11
. Now, yf(x)
.
sinh(9.8/T0)x sinh x = 0
straight, but also constant as
.
As
To
+
-
11
h
yields
=
(T0/9.8)(9.8/T0)
gets large,
y = X
9.8/T0
Thus, the solution is not only To
gets large.
Copyright 1985 Springer-Verlag. All rights reserved.
(b)
Differentiation gives the differential equation,
(c)
The slopes must be negative reciprocals for the curves to be orthogonal, so dy = -dx/3cx2
29.
Starting with
dy/dx = -1/3cx
2
The ten-step Euler procedure gives us YA
y
=
1/3cx
(xO,yO) = (0"l) , we want to find
Y,- - (Y,-~
=
3cxL
Separating variables gives
and integration yields
use the recursive formula
33.
.
dy/dx
2
-xn-,)(0.l)
y(1)
+
y10
C
=
~ ( 1 ).
We
+Yn-l
= 2.2469
This exercise is analogous to Example 10. . . - / - - -
it-
-
- -
-
1
Webave
dy/dx=(y-4)(y-3).
shown by the direction field, for
y ( 0 ) = 3.5
is
As 1im
y(x)
3
Copyright 1985 Springer-Verlag. All rights reserved.
37.
Differentiate to get
f "(x)
2y(dy/dx)ex
=
f "'(x)
2 2 2y(d y/dx ) + 2y(dy/dx)l
+
.
.
dy/dx = 1 f "(0) f"'(x)
hy5
Substitute
+
.
4 = 7
+
4 4x.5y (dy/dx)
L
=
y(0)
= 1
into the given equation giving
Substitute these into the equation for
2 + 1
=
+
eXy2
e"[2y(dy/dx) + y + 2(dY/dx)' i. 4 4 20y (dy/dx) + 20y (dy/dx) + 20x[4y3(dy/dx) +
Differentiate again, so
4 2 y (d y/dx2)]
+
f "(x)
to get
Substitute these into the equation for
ff"(0) = 2 + 1 + 2 + 1 4 + 2 + 2 0 + 2 0 = 61
toget
.
SECTION QUIZ
1.
(a)
Solve the differential equation dy/dx = (x given
2.
y(0) = 1
+
4)(y
-
Find the interval in x
(a)
Solve
(b)
Plot the solution to part (a) for x = x.110 , where
(c)
.
.
7)
,
for which the solution is valid.
assuming
y(0)
=
1
. x.
=
0
,
Then, draw a smooth curve through the plotted points.
Use Euler's method to approximate y(1) 1
(d)
dyldx = y/2 ,
, 10
+
.
(b)
...
2)/(x
for
dy/dx = y/2 ,
y(0)
=
Use ten steps.
Plot the eleven polntsobtained from part (c) onto the same graph
as in ( b ) and connect the points with straight line segments. 3.
Solve the differential equation yl'= 1 and
4.
yo
=
2
+
(y1)2 ,
assuming
(dy/dx) l o
=
.
Which of the following equations are separable? (a)
3 dy/dx = x y
-
y ln x
- xy
(b)
dy/dx
(c)
(Y/(x - l))dy/dx
(dl
(x + y)dy/dx
=
x
2
=
+
y
2 = x(y
2
- y)
xy
Copyright 1985 Springer-Verlag. All rights reserved.
0
5.
At che annual witches' convention, new advances in brewology (the science of brewing magic potions) were discussed. potions prevents broom thefts.
One of the newest (x,y) ,
Depending on the location
broom thieves will be swept off their feet and taken for a ride. ride is designed to take the path given by Determine the path y(x)
y(1) = 1
if
dy/dx = (xi
+
x
+
!)/y
The
.
.
ANSWERS TO P R E R E Q U I S I T E QUIZ
1.
2.
(a)
-1nll-xl + C
(b)
eY
(a)
f(xO)+f'(xo)~x
(b)
The linear approximation and the equation of a tangent line have
+
y5/j
+
cos
y
+c
the same formula. (c)
0.94
ANSWERS TO SECTION Q U I Z
1.
(a)
y
(c)
y(1)
=
-343ex/(x
+
7)
3
+
2
$--z---1.0 x
1.6289
Copyright 1985 Springer-Verlag. All rights reserved.
-1njcos xj
3.
y
4.
a
5.
y 12
=
and 2
=
+
2
c
3 x /3
+
2
x 12
+x
- 413
Copyright 1985 Springer-Verlag. All rights reserved.
378 Section 8.6
8.6
Linear First-Order E q u a t i z
PREREQUISITES 1.
Recall basic rules of differentiation, especially the product and chain rules (Chapters 1 and 2).
2.
Recall basic methods of integration, especially exponential functions and substitution (Chapter 7).
PREREQUISITE QUIZ
1.
2.
Differentiate the following: (a)
exp(-sin x)
(b)
2 texp(t)
Evaluate the following integrals: (a)
2 It exp(t )dt
(b)
/(x
(c)
jtetdt
+
2)exp(2x2
+
8x)dx
GOALS
1.
Be able to solve linear first-order differential eauations.
STUDY HINTS
1.
Linearity defined. in the usual usage,
2.
Method of solution. q(x)
.
Differential equations are linear in the sense that, dy/dx and
y
appear only to the first power.
Equation (6) gives the solution of
dy/dx = P(x)y
+
You should not memorize the solution. Instead, learn the method
of solution, which is summarized in the box in the middle of p. 409. As always, practice is the best way to learn
Copyright 1985 Springer-Verlag. All rights reserved.
3.
Choice of method.
Always look for the simplest solution.
For instance,
Example 1 is separable. You get another (simpler) solution by noting
+
that the right-hand side is x(y
.
1)
(However, note that the equa-
tions in Example 2 are not separable.) 4.
Integrating factor defined.
These are multiplicative factors which
transform an expression into one which can be integrated.
exp(-j~(x)dx)
is the integrating factor which is discussed in this section.
5.
Applications.
You should understand how the results in Examples 3-6
were derived.
There is no need to memorize the results; however, the
results may be useful for solving the exercises.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Here,
P(x) = l/(l - x)
. - j ~ ( x )=~ ~
2/(1 - x) + 3
=
Thus we have
[l/(l - x)] [dy/dx - y / ( l - x)]
-
y/(l-x)
-
C)(l
- x)]
x)
=
2(1 - x)-2
and
Q(x)
-sin x ,
y exp(-sin x)
we get
C exp(sin x)
.
When
Thus, the solution is The equation becomes still the same, sn 1,
=
+
3/(1 - x)
2 cos x
=
- y cos x)
+
[l/(l - x)] [2/(1
.
+
x)
31 =
Integration gives
or
y=2+(-3lnll
-xi
+
.
exp(-sin x)(dy/dx
-2
=
= > / ( I - x ) - 3 l n l l -xi + C
P(x) = cos x
9.
Q(x)
-1nll - xi , dropping the integration constant.
(d/dx)[y/(l
5.
and
(d/dt)exp(tK/L).l
y
y =
=
=
-2
.
=
-sin x ,
2exp(-sin x)cos x
.
-2 exp(-sin x)
+
C
0
=
=
0 , x
+
=
0 ,
2exp(sin x)
dI/dt = -RI/L exp(tR/L)(dI/dt
.
-j~(x)dx
so
so
Letting or
u =
y
-2
+
E1)/L
.
C
=
or
C
=
2
.
+
(EOsin ot
+
RI/I.)
=
+
P(t)
is
exp(tii/~)(i:~sin - t + TI)"
Using formula 82 of the integration ta11le on
the inside back cover of the text, w e integrate to get (E~/L)[ exp (CRIL) I ( (R/L)~+ u2)] (K sin ut/L
-
~llcos lit)
+
Iexp(tR/1,)
=
(F, .il i ( i / R j ~ , s p ( r X . ' L )4
Copyright 1985 Springer-Verlag. All rights reserved.
380
Section 8.6
9.
(continued) C
. Multiplying by exp(-tR/L) 2
wcos ot)/((R/L)
+
I
gives
sin wt/L -
(EO/L)(R
=
2 w ) + E ~ / R+ C exp(-tR/L)
.
(See Example 3 for more
details.) 13.
From the solution to Example 4, y We want to find exp(-2.67
t
10-~t) or
x
7 10- t = ln(O.l)
y(t)
so that
and
t
%
8.63
travelled at
t = 0 , we have
21.
- 57.6164)
rt)
ln(1
15.0
.
-
0.9 = 1
Therefore,
-2.67
x
100 days.
.
e
-y t /m
]
.
The distance
Since no distance has been
As
t
, v approaches
-+
57.6164) = -gt/64
v = 0.9(64)
,
i.e.,
t
=
=
seconds. At this time, the person has
dv/dt
=
[F(M~ - rt) - (-r)~tl/(M~ - rt) 2
- grt) (no - rt) - (-r) ign,t 2
C
.
6 10 )
x
951 meters.
The acceleration is [
10-~t)
i.e.,
Now, we want to know when
or
%
+
c = -g
.
so m/) = 64Ig
64(1 - e-gt/64) ,
(-64/g)ln(l fallen
=
x
10-~t)) (2.51
x
6 10 ) ,
x
lo6 seconds
X
rngt/~+ ge-yt/m
jv dt
=
O.g(Z.51
v = (mg/y)[l
travelled is
57.6
=
0.1 = exp(-2.67
17. Cse the result of Example 5:
mg/y = 64 ,
(1 - exp(-2.67
=
- [g(no - rt)'
+
- grt /2)1 /(Mo - rt)'
grt(1Io - rt/2)1/(Mo
- rt)
2
.
=
2
-
FMo/(M,,
Substituting M I
=
2 2 2 2 Mo - rt , we get F M o i 5 - [gM1 + grt(2M0 - rt)/2] /MI = FM0/M1 2 2 2 2 2 2 [2gM1 + g(?l0 - ?$)(PI0 + M o - rt)1/2~~ = FMO/M1 - 12gM1 + g(Mo - HI)]/ 2 2 2 2M1 = FMO/M1 - g(Mo
25.
Any solution exp(,fP(x)dx){j show that
C
expi-,[l'(x)dx)dxj
y
of
2
2
- 5 )/ 2 5 . y'
=
+
P(x)y
[q(x)exp(-,rp(x)dx)dxI is unique:
IxzO
/ I~(x)exp(-!~(x)dx)dx]
+
yo
=
Q(x)
+
must be of the form Cl
.
exp(]~(x)dx)
C I . Therefore, C
Ix=O
Since
=
y(0) = yo , we can
lx=O{ / [Q(x) =
y
X
yorxp(-,[~(x)dx)
lxZo
-
. Now, the right side of this equation
Copyright 1985 Springer-Verlag. All rights reserved.
is
25.
(continued) a c o n s t a n t , d e t e r m i n e d by o p e r a t i o n s on evaluated f o r
.
x = 0
is e x a c t l y o n e
.
y
Hence
29.
(a)
-
dv/dt
yields
=
- rt)
-
,
L,<
-
v(MO
rt)
(2r
- y)
(MO
- iT)l-y/r
gMo/(y
C
- Zr)
v = F/(y - r )
v
=
0
(b)
- 2r) - F/(y
At b u r n o u t , M,,
I gMO/(y -
2r)
when
C/M~-'"
- rt
=
- g(MO t
=
0
- rt)
-
-
/(y
g
.
-
.
Therefore,
-
l : ~ t ? g : , i t r t o c~~
+
0
so
-
rt)-lIr
g
o
2-311-
-
r t ) / ( y - 2r)
-
,
0
Rearrangement
I--, / r
F(MO
. y)
.
F / ( N ~- r t )
- rt) =
-
(d/dt) rv
-
+
- r ) - g(Mo
v = F/(y
-
-
/
C = [
or
or
rt)
rt) I - y / r
1-y/r
or
Thus, t h e s o l u t i o n i s {[gMO/(Y
-
- ( r - y)v/(MO - r t ) ]
-
+
,
e x p ( - / ~ ( x ) d x ) = (>lo
- r t
a r e mo s o l u t i o r u 5 ,
- j ~ ( x ) d x = ln(?ZO
so
r =
.
y
- rt)dv/dt
yv/(Mo
-
'-dr = (6:dt) [ V ( ? I ~
rt) y
+
rt)
and
- rt)l-y'r[dv/dt
, .' ( 2 -
= (KO
rv/(MO
- rt)
(y/r)ln(MO (MO
- r t ) g - yv
0
- y)/(MO
P(x) = ( r
,,
(M
and
.I
Y L
( d / d t ) ( ? l v ) = F - Mg - r v
The e q u a t i o n i s rt)v] = F
-
w = y
and
y
,
Q(x)
and
n o t a f u n c t i o n , an? t h t > r c
C is a constant,
[ ~ l t e r n a t i v e l y ,i f
look a t the equation f o r
, P(s) ,
yo
=
21)
-
F/(.y
+
- r)
/
C/
-
- F / ( Y - r ) ~ ~ ; - " .~ rt)/(y
- 21-1
+
~ ) I M A - " ~ ) ( M-~ r t ) y / r - 1
Mi ,
v
SO
=
- r ) - gM1/(y
F/(y
-
2r) +
- F / ( Y - r ) l (MO/M1)1 - ~ / r
SEC3lON QLIZ
1.
Solve
2.
Solve
3.
Find
x(dy/dx)
+
2y = xi - x
y ' ( x ) - y = e x(y)
between
if
dy/dx
3x
dy/dx and
if
+
if
y(l)
=
1
.
y ( O ) = l .
l/(x
+
dx/dy?]
.
=
I
y)
.
[ ~ i n t : IJliat i s t h e r e l a t i o n i l i i p
Copyright 1985 Springer-Verlag. All rights reserved.
4.
A stuntwoman i s g o i n g down h'iagnra F a l l s i n s i d e a b a r r e l .
Due t o re-
s i s t a n c e from t h e w a t e r , h e r v e l o c i t y c a n be d e s c r i b e d by
m(dv/dt)
mg
- 0 . 7 ~, where
m
i s h e r mass
(50 k g )
and
g
is
9.8n/s
2
,
t h e a c c e l e r a t i o n due t o g r a v i t y .
1.
2.
t = 0
,
(a)
If
(b)
When d o e s her s p e e d become 25 m/s
(a)
-(COS x ) e x p ( - s i n x )
(b)
(1
(a)
2 exp(t ) / 2 + C
(b)
exp(2xi
(c)
te
v = 0
+
t
when
find the velocity function. (freeway d r i v i n g speed)?
2 2 2t )exp(t )
- e
+ t
8x)/4 +
+
C
C
ANSWERS TO SECTIODT QUIZ 2
1.
y = x / 4 -x/3+1/2+7/12x
2.
y = ( e
(b)
3x
2.m
2
+eX)/2
seconds
Copyright 1985 Springer-Verlag. All rights reserved.
=
8.R
E v i e w Exercises f o r Chapter 8
SOLUTIOXS TO EVERY OTHER ODD EXERCISE 1.
Rearrangement g i v e s u s 3t 1
5.
+ =
.
keO = k
With 3t
, so f o r
C
dy/y = 3 d t
k = e
C
,
y
=
Q(t) = 1
and
and m u l t i p l y by
,
e-3t
giving
.
C = 413
9.
f(0)
y
ye
y(0)
e
=
1
C
=
=
/ ~ ( t ) d t= 3 t
.
= e
=
1
t o get
0
=
v5/2
solution i s
x ( t ) = cos t
dy/dx
=
=
ex+'
c ,
1 - exj
-eiY
c
so
exeY
=
-
1
, so
y = 4
.
1 - t
+
x(t)
=
cos t
, s o v0
v0gT/2
sin t
+
= ex
I - l/e
.
.
C
f(x)
Therefore,
=
+
y
=
, so
C
=
f(x) = e
, x0
= 1
v sin t 0
.
= -1
Ce
4x
4x
. 1
=
.
. , and
Substituting
Therefore, the
.
, so rearrangenent gives Substituting
~ h u s , e-y = l / e
-
dy/eY
y(0) = l
I -ex
=
eXdx gives
or
.
Integra-
-l/e = l
+
y = -1nJ1Ie -
.
Separatevariables t o get ln(p
+
Subtract
.
+
gives
.
1 = (-1/3)
- 1)/3
=
giving
- ( - 1 / 3 ) e - ~ ~+ C , and
v0
The s o l u t i o n i s
,
1
( d y / d t - 3y) = e-3t
w
i s unknown.
In y
-3t -
y = (4e3t
t o get
y(0)
3t
T h i s i s a c a s e of s i m p l e harmonic motion w i t h
tion yields
21.
Substitute
-3 t
T h i s i s a c a s e of n a t u r a l growth w i t h
x ( ~ / 4 )= 0
17.
Substitute
and t h e s o l u t i o n i s
Substitute 13.
.
Ce3t
.
, note that
( d ~ d t ) ( y e - ~ ~ )I n t e g r a t e , s o
+
Then, i n t e g r a t i n g g i v e s
So t h e f i n a l s o l u t i o n i s
P(t) = 3
(-1/3)
ke
.
3t
1) or
=
ln(1 - t )
+
dy/(y C
.
+
1)
=
y(0) = 0
dt/(l - t) implies
C
and i n t e g r a t e t o g e t =
0
, so y
+
1
=
y = - - t .
Copyright 1985 Springer-Verlag. All rights reserved.
384
S e c t i o n 8. R
25.
T h i s is simple harmonic motion with v
x
0
6,
=
LII
,
= 112
yo
+
.
Change
so
(r/3/8,
29.
J3/8 c o s ( f i x -
---- ---
-1
and
is
et
+k:
Q(t)
.
.
C
-3
.
Using the h i n t ,
j ~ ( x ) d x= -x
.
(d/dx)(rbex) yielding -1
+
.
37.
+
Since
D
2
.
,
C = 2
2
meaning
6x c o s h ( 3 x )
=
so
x
=
we h a v e
w = x
.
With
e
.-
t
- 2e-X
=
x
+
D
.
- 3e
0
x)
-t
-
x(t)
.
ex(dw/dx
-
x
+
C
.
'(0)
=
1
=
- e
1
+
2e
=
-x
=
Substitute
xet
+
=
or
C
+
3et
=
C =
l i m e-t
=
t*
.
giving X
3et
=
Since 3
=
3
=
and
S~u b s t i t u t e
w
+
P ( x ) = -1
xe
=
implies 3
P(t)
The i n t e g r a t i n g f a c t o r
t e (dx/dt
1i r n
ex , weX
Therefore
y = x 12 - x
(d/dx)sinh x 2
~
0
,
Integrate t o get
so
+
+
=
0
E I u l t i p l y by
+
C ,
0 = -2
.
w = x - 1
again t o get
2e-X
dw/dx
.
3
Integration gives
x(0)
,
-3
=
Thus,
(d/dt)xet
that
To p l o t t h e
This i s a first-order equation with
x6
- --- - - -
33.
.
tan-'(l/fi))
0.612 c o s ( 1 . 4 1 ~- 0 . 1 9 6 ~ ) .
y(x)
graph,
y(x) =
(1/2+/2)
(1/2,1/214)
t o polar coordinates: tan-'(11~))
and
so the s o l u t i o n i s
y(x) = (1/2)cos(fix) sin(&)
,
112
=
=
x
,
w)
=
xeX
Q(x)
+
D i v i d e by
w(0) dy/dx
note
ex ,
t o get
.
y(0) = 0 2
=
1
=
Integrate t o get
D = 2
.
Then t h e s o l u t i o n i s
y = x 12
cosh x
,
the chain r u l e gives
(dldx)sinh(3xi)
- x
-
.
Copyright 1985 Springer-Verlag. All rights reserved.
=
Section 8.R
41.
(d/dx)cosh x
+ 45.
=
sin11 x , we get
u = sinh x ,
j[du/(l
+
v = cosh x
.
53.
(a)
u
+
Here, m = 10
x
=
the force is
therefore, the integral
tan-'(sinh
C =
dv = sinh x dx ,
h&
and
=
8 ,
.
dx/dt = 80 cos(8t)
f ( t ) = f(0)ekt
-6400 sin(n/2)
f(10) = 200,000 ,
We want to determine
61. Let
In 100 x(t)
=
=
eC
,
k
.
In x
=
so
x
=
Dekt
each of 'these, giving e
8k
for
, so ln(31/41)
or
=
or
t
=
so
18'~ in
.
kt
+
.
Now
x(0)
=
82
=
D
and
62
8k ,
OC
.
.
In this case,
t
C
and thus
.
At
=
and
x(8)
82 egk
.
62
=
=
years. dx/dt = kx
,
.
32 ,
so =
k
Inte-
. ~ e t
so substitute
Therefore =
.
(ln 2/10)t
x = ekteC
(118) ln(31/41)
50-c , x
7/16 ,
dx/x = k dt
Exponentiate to get 82
=
k = ln 2/10
Then
. Take logs to get ln(16141)
8 ln(16/41)11n(31/41)
At
" 66.4
t = 10 In 10OIln 2
=
.
2 = ek(lO) or
Separate variables to get
x = 82 exp[(l/8) ln(31141)tl exp[(l/8) ln(31/41)t]
and
=
2 d ,/dt2
f(t) = 10 million = 100,000e
be the temperature above
gration yields D
t
(In 2/10)t
for some constant
dx
=
i.e., the spring
and
-6400 sin(8t)
This is natural growth and it obeys
i.e.,
du
.
so the force is
100,000 and
C so
so k/10 = 64 ,
-6400 newtons.
=
+
x)
=
f(0)
cosh 3x1
=
.
C
-640 sin(8t) ,
+ and
2 2 The force is m(d x/dt )
(b)
3x)I
jx sinh x dx = x cosh x - /cosh x dx
constant is k = 640
57.
du = cosh x dx ;
so
Therefore,
x cosh x - sinh x
and
(d/dx) [(sinh-'x)(cosh
u2) ] = tan- 1u
Integrate by parts with
& / 1
=
.
3 sinh 3x sinh-'x
Substitute becomes
49.
(d/dx)sinh-'x
Using the product rule with
385
.
31/41
=
Therefore,
32/82 =
(118) ln(31/41)t
27 minutes.
Copyright 1985 Springer-Verlag. All rights reserved.
,
386
Section 8.R
65.
By Example 4 of Section 8.5, the solution
1s
I.--
E/R - - - - -
I0
--- ----
-
I 69.
Let
is
x(t)
0
T~ < E/R , EIR
*
E / R ) ~ - ~ .~ /If ~
+ (I -
I = E/R
then the graph begins below
and increases, rather than decreases,
toward
EIR
be the number of gallons of antifreeze in the radiator at
minutes.
Note that
-(1/2)x/4
=
-x/8
.
x(0) = 413
115
-t/8 ,
or
Now,
dx/dt
t
=
=
952 fresh water, x(t) = (0.05)(4)
(413)e-~/~. Multiply by
=
-(flow out)
=
This is eirponential decay, so x(t)
When the mixture is solve
.
8 ln(20/3)
then add fresh water, you wil-l spend
=
(413)e -t/8
=
115
314 and take logs, so
15.2 minutes.
%
t
,
so ln(3120) =
If you drain the radiator and
[4/(1/2)] 2
=
16 minutes.
Therefore,
draining the radiator first is no faster. 73.
P(x) I
l
l
1
I
I
1
I
/
I
l
l
l
l
l
l
is
e
(xO,y0) = (0.1) ,
77. Starting with y10 ,
yo
1 ;
=
y10
l(l.l)1°
=
yn-l
=
=
l(l.1) =
;
2.5937
=
ye
-3x
~e~~
=
-4e-3X/3
~,-~(0.1)
+
y(O) = 1
=
+
implies C = 0 ;
implies
=
y
=
=
=
~(1)
yn
=
jdx ,
ex ,
and
so y(1)
;
e
;
+
For the
ln/y= x =
...
yn-l(0.05)
.
.
.
y3 = l(l.l)(l.l)(l.l)
;
j(dy/y)
therefore y
or
yn-l = (l.l)~,-~
For the twenty-step method,
y
c
Y~~
(1.05)yn-l , which implies y20 = l(1.05)~~= 2.6533
exact solution, dyldx
=
.
y10 = ~ ( 1 ) and
y2 = l(l.l)(l.l)
.
therefore, e-3X(y ' -3y)
( d / d ~ ) y e - ~. ~ Integration
we want
we use the formula yn
y1
+
so the integrating factor
;
=
gives -413
For
-3x
4e-3X
+
3 ,
=
+
C
.
" 2.718282
Copyright 1985 Springer-Verlag. All rights reserved.
77.
(continued) 4.58% , while the twenty-step
The ten-step method gives us an error of method has a 81.
(a)
(b)
2.39%
Differentiate w aw = ay
+b
so w(t)
=
=
Ce
w
=
.
at
=
dw/dx = aw
subtract
b/a
P( t)
tract
.
ay
at e
.
y =Ceat
y
y(t)
at
=
+
=
to get
.
Ceat - b/a
.
k
+
=
Exponentiation gives
Substitute w
y
=
+
b/a
and
- b/a
e-a t(y' - ay)
e-at to get
-b/a
=
a
a case of natural growth,
. Integrate, giving ye-at to get
by
, note that j ~t)( d t = a t .
Q ( t) = b
and multiply by
(d/dt) (ye-a5 ply by
and
In w
C = ek
to get a
=
for
Thus,
w
. Separate variables to get dw/w
and integrate to get
ekeat = C eat
With
+ b/a
Multiply
w' = aw ,
Therefore
y(t)
.
w' = y '
to get
y'
From part (a), a dt ,
(c)
error.
~
.
e
=
e-at
( - b ~ a ) e - ~+ C
=
Sub-
.
-
Multi-
-All ~three ~ answers are the
same. 85.
(a)
In other words, find i.e.,
y2
1
=
+
(yl)*
y
f (x)
=
.
Now, iE
zero and we are left with -1
y
y2 = 1
y If
is constant, or
y
=
y
1
=
'1
y
becomes
. The solution y
=
is one solution is not constant, then rearrangement yields
y2 - 1 or the differential equation
y'
J2--ating variables yields dy/ y 1 = dx
We recognize
rb ,Iaf(x)dx
on
[a,bl
.
=
.
6 =
:A
f (x) -
+
f
(
1
(y ')
dy/dx
.
2
=
Separ-
Integration gives us
as the area under
Chapter 10, we will derive f(x)
,
=
is not valid since the integrand on the right is positive.
Thus,
(b)
f (x)
such that
x
on
[a,b]
. In
for the lengtll of
Thus the formula equates area and arc length.
Copyright 1985 Springer-Verlag. All rights reserved.
388
Section 8.R
TEST FOR CHAPTER 8
1.
True or false: (a)
If
y
is a function of
x ,
+
y 2/2
integrating to get
0
can be solved by
.
C
(e)
y(0) = 2 ,
is constant and
As long as
y(xo)
is all real
.
The most general solution to the differential equation
0
2 cosh x
x ,
For all
is
x
y
=
2e-ax
is specified for some constant
1
=
+
y'
,
= -ay
.
xo , y " - y
=
.
has a unique solution for y(x)
.
2 sinh x
Solve the following differential equations with the given conditions:
+
(a)
d2x/dt2
9x
=
o ,
X(O)
=
1 ,
(b)
d2x/dt2 - 9x
=
O
,
x(0)
=
1 , xf(0)
(c)
2 2 d x/dt
=
0 ,
x(0)
=
2
(a)
Find a solution of the form
+
assuming (b) $.
=
(c)
a
1)
y
The domain of
(dl
3.
+
cosh-'(x2
=
+
(b)
where
2.
y
then y'
9t
y(0) = 1
and y
.
dx/dt
+
Sketch the graph of
Find the solution of
1
XI(O) =
,
xl(l)
=
1
=
2
y
=
A cos(,,t -0)
y'(~)
=
4
.
sin t
+
x
=
+
2e-t
sin2y
if
for
4 y " = -y ,
x(0) = 1
1 , assuming
.
5.
Solve the differential equation dy/dx
6.
Solve the following differential equations with the given initial
=
y(0)
conditions: (a)
dy/dx
=
-2x ;
(b)
f '(x)
+
2 x f(x) = 0 ;
y(0)
=
1 f(0) = A ,
a constant
Copyright 1985 Springer-Verlag. All rights reserved.
=
0
7.
Evaluate the following:
8.
(a)
(d1dt)Jcosh it
(a)
Find an approximate solution for 0
(b)
9.
y(2)
if
=
+
x2
2y
and
y(0) =
by using a 10-step Euler method.
Compare the answer in (a) with the exact solution.
An electric circuit is governed by the equation I cos(,t) 0 satisfying
10.
y
C , R , I0
where V(0)
=
0
.
and
o)
+
C(dV/dt)
are constants.
V/R
Find
=
V(t)
.
Scientific investigators have recently concluded that stupid question asking obeys the law of exponential decay. At the age of five, the average person's stupid questioning peaks, and then decays exponentially. (a)
Suppose a young boy asked an average of when he was five.
1
stupid question daily
He is now thirteen and asks an average of
stupid questions daily.
Write a formula for
q(t)
0.65
, the average
number of stupid questions asked daily, in terms of
t
,
the
person's age. (b)
How many years does it take for this person's stupid question asking to decrease by
50% ?
ANSWERS TO CHAPTER TEST 1.
(a)
False; integration needs to be done with respect to
(b)
True
(c)
True; it is the only solution.
(d)
False; y'(xl)
(e)
True
must also be specified for some
X1
x ,
not
y
.
.
Copyright 1985 Springer-Verlag. All rights reserved.
390
2.
3.
Section 8 . R
+
(a)
x = cos 3 t
(b)
x = cosh 3 t
(c)
x = -3t3/2
(a)
y = &cos(t/2
(b)
(113) s i n 3 t
+
+
(1/3) sinh 3 t
13t/2
Y
+
ewt/2
+
2te-t
(sin t
x
5.
y = t a n
6.
(a)
y = e
(b)
f(x)
(a)
5 s i n h 5t/2Jcosh 5 t
(b)
3 I n [cosh(x/3)]
(c)
2 6 + 7
+
i
i
7.
-1
cos t ) / 2
-
x
[ I ~ /R
-2x =
3
A exp(-x 13)
9.
V
lo.
(a)
q(t)
(b)
12.9 y e a r s
=
2
-tan-'g)
4.
=
+
('iiC-c*. =
+
C
5 sinh-lx
i
+
+
C
1 ) 1 [ c o s ~t
exp[-0.0538(t
+
( RCLLd)sin, t
-
exp(-t/~~)]
- 5)l
Copyright 1985 Springer-Verlag. All rights reserved.
CHAPTER 9 APPLICATIONS OF IPI'TEGRATION
9.1
Volumes by the Slice Method
PREREQUISITES Recall how to derive the integration formula for area by using the in-
1.
finitesimal argument (Section 4.6). Recall the various methods of integration (Cl~apter7).
2.
PREREQUISITE QUIZ The area under the graph of a positive function f ( x )
1.
Sketch a typical graph of meaning of
f (x) dx
f(x)
is
1: f(x)
dx
and use it to explain the geometric
.
2.
Evaluate
{ ( x + ~ d) x~ .
3.
Evaluate
{e4~ dy
.
GOALS
1.
Be able to compute volumes by using the slice method.
2.
Be able to compute volumes by using the disk method.
Copyright 1985 Springer-Verlag. All rights reserved.
STUDY HINTS
1.
Basic formula. where
A(x)
is a solid's cross-sectional area.
dx ,
V =
All of the volume formulas are based upon
Remembering this and de-
riving the formulas in this section will prove to be more beneficial than just memorizing.
2.
Study Examples 1 , 2, and 3 to see how elementary geometry
Slice method.
is used to compute the cross-sectional area.
3.
A(x)
Note that
in the
volume formula corresponds to
t(x)
in the area formula.
Radius dependent upon height.
Example 2 shows how similar triangles are
commonly used as an aid in ccmputing a cross-sectional area.
4.
Disk method.
By this method, each cross-sectional area is simply a circle
whose radius is
f(x)
;
therefore,
A(x)
=
tion rather than memorizing the formula. if
5.
f(x) < 0
7
Note that the formula works even
The area is the difference between that of two circular
regions; therefore, if
g(x) 2 f (x) ,
then
A(x)
Note that this reduces to the disk method if best to learn the derivation.
6.
2
Learn this deriva-
.
Washer method.
g(x)l
.
[f(x)]
WARNING:
f(x)
= T [g(x)] =
0
.
-
IT
[f(x)]
2
.
Again, it is
The integrand is not
[f(x) -
.
Step function argument (p. 425).
This is just for the theoretically in-
~lined. Except in honors courses, most instructors will not expect their students to reproduce the argument.
7.
Cavalieri's -principle.
Basically, it states that two volumes with equal
cross-sectional areas have equal volumes.
Thus, the "tilted" solids in
Exercises 1-4 have the same volume as those which stand "straight up." Note that
2 ir=lrr.Lx 1 i
in the discussion is simply the disk method formul:
Copyright 1985 Springer-Verlag. All rights reserved.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Let the x-axis be vertical as in Exaniple 2.
Then apply the slice method.
Each infinitesimal slice has a circular base with area a thickness 5.
. Thus, V
dx
=
3
JO ndx
=
nr2 = "(1)'
and
- 6)'
2
3-3 . { [(x
Since the cross-section is a square, its area is
- 11 161
2 Therefore, the volume is V = (1136)~; [(x - 6)4 - 2(x - 6) 5
-
(1136) [(x - 6) 15 - 2(x
9.
3
6) 13
+ XI :l
=
2
2 2 (nr /h )(h x - hx2 from
x
to
2
h
3
V/2 , where
=
rr2(h
lx2 .
.
x)3/3h2
x 1 = (1 - 3 a ) h ,
x2
=
to
xl
to height
x2
is
Vhen we bisect the cone, we want
2 V = Fr h/3 ,
(h - x ) =~ h3/2 ,
-
=
The volume of a short cone with height
' 3 nr (h - x) /3h2
Therefore,
we equate
13)
2
is
2 nr (h - x) /3h cone.
+x
3
11 dx
.
2125/54
As shown in Example 2, the volume from height
+
.
V/4
3 x = (1 - l/ 6 ) h
i.e.,
and
the volume of the entire
(3/4)V
.
S~rnilarly,
. Thus, the cuts are at
(I - 3 ~ ) h , and x3
=
(1 - ' ~ ) i l ,
respectively. 13.
The volume of the entire cylinder is
n(5)
2
(20)
=
500n cm3
since the
The volume of the wedge is determined by r = 5 and 3 3 tan0 = 515 = 1 , so the removed volume is 2(5) 13 = 25013 cm . There3 fore, the entire solid's volume is (500n - 25013) 1487.5 cm .
radius is
5 cm.
17.
Here,
f(x) = cos x
+
1
,
so the
disk method gives the volume as 2 njOm(cos x
2 sin x)
/in
=
3n
2
+
2 1) dx
=
2n
TjO (COS2x
V
+
2 cos x
+
1)dx
=
2 njOT((l
2 cos x
+
1)dx
=
~ ( 3 x 1 2i- sin 2x14
+
=
cos 2x)/2
+
i
.
Copyright 1985 Springer-Verlag. All rights reserved.
21.
The e q u a t i o n o f t h e e n t i r e c i r c l e i s
y4
- a)
(x
2
+
y
2
,
= r2
so the equation
y
of t h e s e m i c i r c l e i s
k
1
2
By t h e d i s k
method,
- (x -
I/a+r[r2 a-r
a ) 2] dx = n [ x r 2
- (x - a)3/3]
a-r a + r = -1(2r3 a-r
-
2 r 313)
4 n r 3I 3
=
The s q u a r e i n E x e r c i s e 23 i s c e n t e r e d a t
25.
.
(912,1312)
@
4
*T~
(9/2,13/2+v5/2)
x-axls, 21
2
- fi/2,9/2]
+ (4 -
y = x
and
fi/2
y
=
x
+
(4
+
and
and t h e o t h e r l i e s between
6 ) 12
on
- [-x + ( 2 2 - f i ) / 2 1 2 1 d x I-
('I2
1 (9-&)I2
(912, (9
+
J?;)121.
(26x
+
y
y =
+
V
1 3 6 ) dx
-x
-x
+
+
(22
(22
+
- J?;)/2 %6)/2
For r e v o l u t i o n around t h e =
+
f 2 { [x
n]::,6)
T [ ( ~ + ~[-x) ~+ ~( 2 2 + 1912
- 117
=
+
fi)/212
+
form t h e
+
-/(9+fi)12(-26x 912
(R
d e s i r e d doughnut, revolve t h e c i r c l e centered a t
with radius
+
(R - r ) / 2 .
2 r)/212 = (R - r ) I4
+ r)/2 .
.
The e q u a t i o n of t h e c i r c l e i s
y
Solving f o r
,
we g e t
*J(R
f(~-r)l2 T h e r e f o r e , t h e volume i s nl R r 1 2 ( R
2
,
[
(4
- [x
A doughnut c a n be made by r e v o l v i n g a c i r c l e a r o u n d t h e x - a x i s .
(R
.
(9121fi/2,13/2)
and
&)I2
t h e d i s k method g i v e s t h e volume a s
~ 5 ) / 2 12 i d ~=
2)
from
The s q u a r e c a n be d i v i d e d i n two r e g i o n s
One i s t h e r e g i o n b e t w e e n [(9
each v e r t e x i s
the center; therefore, the v e r t i c e s a r e
t
on
S i n c e e a c h s i d e of t h e s q u a r e 1 ,
has length
4
29.
.
- (X - a 1 2 dx
=
+
117
(4
-
+
To
(O,(R x2
+ ,611
+
+
r)/
(y -
2
- r ) 14 - x 2
- r ) 14 - x 2 +
Copyright 1985 Springer-Verlag. All rights reserved.
+
29.
(continued)
Note that the integral is the area of a semicircle centered at the origin
,
(R - r)/2
with radius
+
n'(~
the doughnut is
2 n(R - r) 18
which is
Thus, the volume of
.
- r)'/4
r)(R
.
SECTION QUIZ 1.
From elementary geometry, we know that the volume of a cone is 2 (113)~r h , where of the cone.
r
is the radius of the base and
Suppose we revolve
then the base radius is the other hand, x/2 , nI3
2.
.
a = 0 ,
V
=
and
1
and
y h
b
=
2
,
.
2
=
rb (x)] 2dx = 7ja[f
x/2
=
on
[0,2]
is the
.
2n13
For the cone, 3
n(x12) I3 /
so the volume is
height
around the x-axis,
Thus, the volume is
[f(x)l 3131;
I
h
2
.
On
f (x) =
3 2 = (nx 124) /
=
What's wrong?
The line
y
=
x
+
1
is revolved about the x-axis to form a solid of
revolution. (a)
A vertical cut is made at resulting solid between
(b)
x = 9
x = 0
.
and
Ijhat is the volume of the x
=
9 ?
Where should another cut be made parallel to the y-axis to get two equal volumes from the solid in (a)?
3.
The curve
y =
x5
on
[0,1]
is revolved around the y-axis.
Use the
disk method to find the volume of the resulting solid.
4.
The cross-sectional area of a solid at height The solid extends from
h
=
given to compute the volume?
0
to
h
=
3-12
.
h
is given by
h / c o s hi
.
Is enough information
If not, what is missing?
If yes, compute it.
Copyright 1985 Springer-Verlag. All rights reserved.
396
Section 9.1
m
Bruce, the boozing butcher has just returned from his afternoon vodka break.
As he begins
to trim the fat off a piece of rib roast, the alcohol begins to take effect and all he can see before his eyes are stars. the beef is cut into the shape shown.
Consequently,
Define the "radius"
r
of the
cross-sectional stars to be the segment from the center to one of the outer vertices.
When our boozing friend was sober, he determined the
area of the star to be the top base has area
2 5r 14
.
If the bottom base has area
2
,
and the height is
20 cm
2
125 cm
,
6 cm , what is
the volume of the meat that Boozer Bruce cut?
ANSWERS TO PREREQUISITE QUIZ
'.
~4
The shaded region is a very thin "rectangle" Its width is
dx
its height,
f(x)
fore,
f(x)dx
and since it is so thin,
, is "constant". There-
is the area of the region.
ANSWERS TO SECTION QUIZ
1.
Jn performing the integration, a factor of substitution
2.
(a) (b)
u
=
2
was forgotten when the
x/2 was made
333n 3 ~ o i ~ =- h .l9 4
3.
57;/7
4.
Yes; i n 1 2 - 1
5.
3 390 cm
Copyright 1985 Springer-Verlag. All rights reserved.
9.2
Volumes by the Shell Method
PREREQUISITES
1.
Recall how to compute volumes by the disk method (Section 9.1)
PREREQUISITE QUIZ
1.
Find the volume of the solid obtained by revolving the graph of x2 - 1
on
[2,4]
=
around the x-axis.
Repeat Question 1 for
2.
y
y
=
sec x
[O,r/4 ]
on
.
GOALS
1.
Be able to compute volumes by using the shell method.
STUDY HINTS
1.
Shell method.
Learn the derivation.
finitesimal cylindrical volumes, its circumference is
2rx
curves d~ 0
=
f(x)
2irx dx(f(x)
g(x)
Useful trick.
,
- g(x))
, this reduces to V
need to use 2.
and
Each shell has radius x , so
. Finally, for a region between the
the height is to get
=
.
. Multiplying by the thickness dx gives
2nx dx
the area of the base as
dV
It is simply a summation of in-
f(x) - g(x)
v = 2rj:x[f(x)
2njbxf(x)dx
.
-
.
Thus, we sum
g(x)]dx
Note that if
.
f(x)
~f
g(x)
=
O , we
/ f (x) 1 . At this point in your studies, you do not know how to
integrate
;
however, many times it is possible to make a sub-
stitution so that even though you cannot compute an integral directly, you can determine it by computing the area under the curve using elementary geometry.
See how this method is used in Example 6.
Copyright 1985 Springer-Verlag. All rights reserved.
398
3.
Section 9.2
Step function argument.
Again, as with the disk method, you will prob-
ably not need to regurgitate the step function argument unless you are enrolled in an honors course.
4.
Choosing a method.
Ask yourinstructor.
In most cases, if
y = f(x)
is revolved around the
y-axis to generate a solid, the volume is best found by using the shell method.
Similarly, revolution around the x-axis implies the use of the
disk method.
If you have a thorough understanding of these two methods,
it is possible to do the same problem using either method.
However, the
simplest method should be used to promote efficiency.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Yf
By the shell method, the volume is
.
2nli x sin x dx with
u = x
2r(-x
cos x
and
Integration by parts v = -cos x
yields
P
+
sin x)
1;
2~(n)
=
=
2.
2
.
For revolution around the y-axis, the shell method
V
gives
u = 3 - x
2
1 2710x(5
=
2
,
so
3
Zn(5x /2 i- x 13)
dul2
1
lo
~ , 1 u ~ ~ ~ /= :r(i7 i~ 3 I '
Y
6
1
+
+
=
.
- 2 70 ~ ' x ndx -x dx ;
therefore,
21~23& du/2 = 17-13
4JZ - 6fi)/3
Let
V
=
+
.
The Zescriptio~ of the square is the re-
+
8 1
+ x)dx
gion between
izl
[4,5]
.
y = 6
and
y = 7
on
For revolution around the y-axis, 5 V = 2.~1~ 7x dx -
the shell method gives 2.1,
5
2 2 6x dx = 2 ~ ( 7 x 12 - 6x 12)
1:
=
Copyright 1985 Springer-Verlag. All rights reserved.
13.
2jix>; dx
The volume is 4-15
.
'
2-(2xii2/5) 0
=
=
If we put the resulting volume of
Example 5 of the previous section on top of the solid generated here, it will produce
1
a cylinder of radius
and height
17.
.
1
Use the method of Example 6. The volume of the top half is 2'14
(r
+
31-
au
6 - j z 2 L - u2
- (x
du = 2
2 ~
o +
(a)
6~(:-) 2 247
the total volume is
I
21.
=
27~:x[4
-
j
=
3j21 '"dr ~ duu + >
-
L Z - ~. Thus,
.
By symmetry, we only need to revcilbe
4
the upper semicircle. 2 [ a -(x-b)']
v
'I2,
f(x) =
and the volume 1s
27jb+a x [a2 - (x b-a
=
Thus,
-
b) 2I 1/2dx
Uslng the method of Example 6, let I -
=
u 2 2~b(7a ) (b)
a
2 2 27 b(a
,
+
2
-a
um + du
27jaa(u
b
m du
=
+ 0
.
b)
+
. Thus, the total volume is 272a2b .
The volume of for
(c)
=
to get
Ta,b 1s
2 2
2- a b
by part (a).
we compute the volume of 2ah
+h
2
)
Ta+h,b
Substituting
to be
2 2- (a
+
+
a
2 h) b
h
=
. Therefore, the difference in the two volumes
As in Exercise 19, we expect the surface area to b t t11c derivarivi (d/dh) [volume (Ta+h,bj - volume(T Thus, the surface area is
2 27 b(2a
11 ,
a,b + 2h)
evaluated at =
2
47 a b
h
=
0
.
.
Copyright 1985 Springer-Verlag. All rights reserved.
=
~
400
Section 9.2
SECTION QUIZ
1.
The curve
y
=
x4
on
[0,1]
is revolved around the x-axis to generate
a solid of revolution. Use the shell method to compute the volume. Compute the same volume using the disk method. 2.
The region between the curves around the y-axis.
3.
(a)
The line
y
=
y = x
and
y
=
x2
on
[0,2]
is revolved
What is the volume of the resulting solid? x
+
5
[2,3]
on
is revolved around the line
x
=
1 .
What is the volume of the solid generated by revolving the region between (b)
y
=
x
5 and the x-axis?
Find a general formula for the volume generated by revolving the region under
Suppose
f (x)
y = f (x)
, and
A
4.
+
=
cos x
on
[a,b]
f(x) > O
for
on
.
[O,n]
x
in
f(x)
f(x)
and the
Assume
for some
x
in
.
(b)
The volume can be computed; it is negative.
(c)
The volume can be computed; it is positive.
(d)
The volume is r2
.
Which of the following is true?
The volume can not be computed because [O,.lI
x = A
.
[a,b]
The region between
x-axis is revolved around the y-axis. (a)
about the line
2r2 - 4 7
because the volume on
[0,r1
is
- 27; and symmetry can be applied. As a pract icai j o k i ' , you g i v r y o l ~ r hungrv little
Y
cousin a staie ilougtinrit.
She h r i ~ k sh ~ , rtrniit
teeth tr>-in#to bite into it. ti]<' d~irlgilnl~tat yon.
i i ~ c , tiir~ocs
Seeiiig i1i.r irol-it tE,rtii s t i i i
stuck in i t , v o u ri,alizc its
I
Aiigriiy,
sat ion ~ ~ ~ i , c Icn . order
tn
V J ~ U C ;~I S
still~dti](
J
c(~I~\L'I--
d~~~gli~ir~t I J ~ ,
you drill a 1-centimeter diameter hole along the y-axis as shown.
Copyright 1985 Springer-Verlag. All rights reserved.
5.
(a)
(b)
The inner radius of the doughnut is
2
centimeters.
Its outer
radius is
4 centimeters. that was the original volume of the
doughnut?
(Hint:
See Example 6.)
Suppose you revolve the line x around the
=
1
between
y
=
-2
and
y
=
-4
y-axis and you subtract the resulting volume from the
doughnut's original volume.
Neglecting the volume of the teeth,
is your answer an approximation or is it exact for the volume of the doughnut with the drilled hole?
Explain.
(Hint: Take a
coffee break and buy yourself a doughnut.) (c)
What answer did you get for the volume by using the method described in part (b)?
ANSWERS TO PREREQUISITE QUIZ
1.
(163.07)~
2.
-
ANSWERS TO SECTION QUIZ 1
1.
2~,1'~~(1 - y
2.
3ii
3.
(a)
68713
5.
(a)
3rL
(b)
1/ 4
18 )dy = ~ / 9 ; rJOx dx = r/9
Approximate; the "curvature" at
y = -2
is greater than that
Copyright 1985 Springer-Verlag. All rights reserved.
402
9.3
Section 9.3
Average V a l u e s and t h e Mean V a l u e Theorem f o r I n t e g r a l s
PREREQUISITES 1.
R e c a l l how t o u s e t h e summation n o t a t i o n ( S e c t i o n 4 . 1 ) .
2.
R e c a l l t h e i n t e r m e d i a t e v a l u e theorem ( S e c t i o n 3 . 1 ) .
3.
R e c a l l t h e meaning of t h e mean v a l u e t h e o r e m f o r d i f f e r e n t i a t i o n (Section 3.6).
PREREQUISITE QUIZ 1.
S t a t e t h e mean v a l u e t h e o r e m f o r d e r i v a t i v e s .
2.
What c o n d i t i o n s a r e n e c e s s a r y t o a p p l y t h e mean v a l u e t h e o r e m ?
3.
E x p l a i n t h e i n t e r m e d i a t e v a l u e t h e o r e m and s t a t e any n e c e s s a r y conditions.
4.
Suppose
al = 1
, a2
=
4 ,
a3
=
-2
and
b. = j J
.
Compute t h e
following:
GOALS 1.
Be a b l e t o compute t h e a v e r a g e of a f u n c t i o n on a g i v e n i n t e r v a l .
2.
Be a b l e t o s t a t e t h e mean v a l u e t h e o r e m f o r i n t e g r a l s and u n d e r s t a n d i t s meaning.
STUDY HINTS 1.
Notation.
Remember t h a t t h e b a r o v e r t h e f u n c t i o n i n d i c a t e s t h a t t h e
average value i s d e s i r e d .
The i n t e r v a l i s i n d i c a t e d a s a s u b s c r i p t i n
Copyright 1985 Springer-Verlag. All rights reserved.
2.
I f you remember t h a t t h e a v e r a g e i s w e i g h t e d , you
Average v a l u e .
s h o u l d h a v e no p r o b l e m d e r i v i n g t h e f o r m u l a .
It i s s i n p l y t h e a r e a
u n d e r t h e c u r v e d i v i d e d by t h e l e n g t h of t h e i n t e r v a l , i . e . , t e g r a l d i v i d e d by t h e l e n g t h .
3.
Clarification.
The numbers
F i g . 9 . 3 . 1 may h e l p you remember t h i s . m
and
M
a s u s e d on p . 435 a r e o f t e n
c h o s e n t o b e t h e minimum and t h e maximum v a l u e s of 4.
I n t e g r a l mean v a l u e t h e o r e m .
the in-
f(x)
on
.
[a,bl
As w i t h s e v e r a l o t h e r p r e v i o u s t h e o r e m s ,
t h i s i s a n e x i s t e n c e theorem.
It s t a t e s t h a t the average is a t t a i n e d
somewhere i n t h e i n t e r v a l , p o s s i b l y many t i m e s ; however, i t d o e s n ' t s p e c i f y e x a c t l y where i t o c c u r s , n o r d o e s t h e t h e o r e m h e l p you f i n d i t . Note t h a t c o n t i n u i t y on a c l o s e d i n t e r v a l i s r e q u i r e d . 5.
Example 5 .
j:f(x)dx
Notice t h a t t h e s o l u t i o n uses t h e f a c t t h a t
= 0
.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
=
=
~ ~ f ( x ) d x ! ( b- a )
4
( x 1 4 ) ; ; = 114
i
In t h i s case, it
.
0
to
2
.
2 3 [ 1 / ( 2 - o ) ] j o x dx
Thus, t h e a v e r a g e i s
=
4
( l / 2 ) ( x 14)
.
The a v e r a g e i s
[1/(1 - o)] jhsin-lx
dx
.
Using t h e formula f o r i n t e -
g r a t i n g i n v e r s e f u n c t i o n s from S e c t i o n 7 . 4 , x sin
l3.
.
T h i s i s j u s t l i k e E x e r c i s e 1 e x c e p t t h a t t h e l i m i t s of i n t e g r a t i o n range frolli
9.
f (x) [a,bl
1 3 [ 1 / ( 1 - o ) ] / ~ xdx
is
5.
-
The a v e r a g e v a l u e i s
-Ix 1
1,
- j,I2sin
The a v e r a g e i s
4
(112) ( x 14
+
y dy = r s i n - ' X I ;
[1/(3 - l)]/?(x3
2v5)
1;
= (112) ( 2 0
+
+
+
we g e t
-12 cos y/;
;;sin-lx =
112 - 1
m ) d x = (1/2)/:(x3
2 ~ -3 2 )
=
9
+
dx
=
.
x-li2)clx
=
+ "15.
Copyright 1985 Springer-Verlag. All rights reserved.
=
404
17.
Section 9.3
Let
t
be the number of hours after midnight, then
12
j3 (14513 +
5t/9)dt
(1124) [(50t)
1; +
2
24
5t 1 6 ) ll51
(l45tl3
+
+
+
lOtl3)dt
24 j15(90
5t2/18) /!j2+ (15t (435
+
37.5)
+
(45
0
if
=
f (b)
5t/3)dtl
5t2/3)
1 fi +
+
135)
+
(810
55'~.
=
-
+
+
=
(90t -
-
292.5)l
- a)
f(x) :/(b
=
f '(x) ,a,,l
.
[f(b) - f(a)l /(b - a)
-
[l/(b
=
[a,b] such that
[l/(b - a)] f:j
=
[a,b]
.
'(t)dt
0
By the [f (b) - f (a)] /
fundamental theorem of calculus, the right-hand side is
some
x
f (a)
=
-
(b - a)
a)]
This can only be
According to the mean value theorem for integrals, there is some in
=
-
We apply the fundamental theorem of calculus: !if '(x)dx
25.
+
j:(15
(1124) [150
=
(1124) (1320) 21.
+
3 o)] [jo 50 dt
[1/(24 -
The formula for average values gives us
. We have shown that the average derivative is attained at
to
[a,b] ,
in
so
f '(to)
[f(b) - f (a)] /(b
=
-
a)
,
which is
the mean value theorem for derivatives. 29.
We need to show that
lim F(x) = F(xO) x+x 0 Section 11.1 , we must show that
IF(x)
1~
- x0l
jtof (s)ds
/MI /MI
on
<6 .
.
By
By the definition used in
- F(xo)
F ( x ) - F(x0)
/<E
=
how, by the extreme value theorem,
[a,b] ,
/x- x 0
def i n i t i n n ,
.
.
F(X) - F(x~)/
so
/f(s)
1
< IM/
/ H / E / / M=/ c
[a,b]
P = / x- xo/ =
Thus, if we let
<
on
whenever
J:~(s)~s
- J;£(S)~S
1 f( s ) 1
has a maximum
and F//M!
11:
f(s)ds/ 0
<
, we have
.
Copyright 1985 Springer-Verlag. All rights reserved.
=
SECTION QUIZ
1.
Find the average of the following functions on the given intervals: (a)
g(t)
(b)
f (x) =
(c)
y = sin(x12)
t
=
m on
zX
on
[1,3] for
-
2.
Suppose (a) (b)
f (x) [0,31
What is
f(X)
-
Find
l2,31
=
0 G x Gi
8 , f (x) ,3,51
-
3.
=
5
.
, if possible.
f (x)
f (x) i3,*1 ,
Find
, and f (x) [5,
?
[0,101
-
(c)
2
=
.
if possible
Consider the function in Question 2. (a)
Can we conclude that
(b)
If
f
f (xo) = 2
for some
is not differentiable somewhere in
f (x ) 0
tinuous, can we conclude that
=
5
xO
in
[3,5] ?
Explain.
[0,10] , but it is confor some
xo
[O,101 ?
in
Explain. (c)
If
f
is continuous, f (7.5)
conclude that
4.
True or false:
f(5) = 3 ?
=
5 ,
and
f(10)
=
7 ,
then can we
Explain.
A continuously differentiable function can not attain its
average at a critical point 5.
On a typical day, the Smith family, consisting of three talkative teenagers and their parents, uses the telephone for
3
+
2 t hours, where
the number of days before or after Wednesday, 12 noon.
Thus,
t t
=
is 1
on
Tuesday or Thursday, etc. (a)
What is the average number of hours the Smith family talks on the phone daily?
(b)
(Hint:
Integrate on the interval -3.5
t
3.5 . )
Show that the mean value theorem for integrals is valid for this
Copyright 1985 Springer-Verlag. All rights reserved.
406
Section 9.3
ANSWERS TO PREREQUISITE QUIZ 1.
If
f
is continuous on
f '(x,)
=
[a,bl
[f(b) - f(a)] /(b
a)
-
and differentiable in for some xo
2.
Continuity and differentiabilitv.
3.
Continuity is a necessity.
If
f
lie on opposite sides of the line line somewhere in 4.
(a)
3
(b)
14
(xl,x2)
in
(a,b)
is continuous, and y
=
C ,
(a,b) ,
then
.
f(xl) ,
f(x2)
then one must cross the
.
ANSWERS TO SECTION QIJIZ 1.
(a) (b)
2.
3.
(44~5-16)/15 3 / in 2
(c)
21-
(a)
5.3
(b)
5.6
(c)
Not posslble
(a)
No,
(b)
Yes, particularly in
(c)
No, there are numerous functions whichsatisfy and
f may not be continuous.
~:Of(x)d~ = 25
4.
False; consider
5.
(a)
2813 hourslday
(b)
xO = (-3?-)/2 (-3.5,3.5)
f (x)
=
x
*
(5,10)
. 3
by the mean value theorem for integrals f(7.5) = 5 ,
f(10)
T\e function doesn't have to be symmerrlc. on
[-1,1]
-4.90 and
.
1.90 ;
1.90 is in the interval
.
Copyright 1985 Springer-Verlag. All rights reserved.
=
9.4 Center of Mass PREREQUISITES 1.
Recall how integration formulas for areas and volumes were derived by using the infinitesimal argument (Sections 4.6, 9.1, and 9.2).
2.
Recall how to use the summation notation (Section 4.1).
PREREQUISITE QUIZ 1.
2.
Use an infinitesimal argument to derive V Use an infinitesimal argument to derive area between two graphs such that
3.
m.x. 1=1 1 1
f(x)
=
A
2njbxf(x)dx
=
b
(x) - g(x)] dx ,
/,[f
> g(x)
on
(a)
In general, does
yn
(b)
In general, does
En m x.1~: m . = . \ z ? 1 = 1 i 1 1=11 1=11
=
.
( ~ 2 , ~ m (E;=lxi) ~)
[a,bl
the
.
?
GOALS 1.
Be able to state and understand the consolidation principle.
2.
Be able to find the center of mass on a line.
3.
Be able to find the center of mass for a plane region.
STTDY HINTS 1.
Consolidation principle.
This allows you to concentrate the entire mass
of an object at its center of mass.
This principle is important in the
derivation of formulas.
"Negative" masses may also be used if a mass
needs to be subtracted.
The usefulness of this principle is illustrated
quite well in Example 7. 2.
E e r of mass on the line.
By remembering that the center of mass is
simply a weighted average of position, you should be able to recall the -
formula
.
x = ~:=~m~x~/~;=~m~
Copyright 1985 Springer-Verlag. All rights reserved.
408
3.
Section 9.4
Warning.
If you think the m 's cancel in the center of mass formula
(3), you need to review Section 4.1. -
4.
Center of mass in the plane. that one formula uses
5.
Symmetry principle.
x.
The only difference between
and the other uses
yi
-
x
and
y
is
.
Properly used, this saves a lot of needless work.
Remember that uniform density is a requirement. Two axes of symmetry is all that is needed to determine the center of mass of a region in the plane.
6.
Center of mass of a region.
It is recommended that you learn to derive
the formula with the aid of Fig. 9.4.10. average where the mass is density
p
Again, this is a weighted
times the area of the region.
The center of mass of the "infinitesimal rectangle" is simply
(x,f(x)/2)
Now, use the consolidation principle and the fact that integration is a continuous summation. The same formula holds even if
f(x)
is negative.
Exercise 28 gives a more general formula.
7.
Density cancels.
In all of the formulas, mass is used.
plane, mass is proportional to density uniform.
8.
x
area.
For region in the
Density cancels when it is
Be careful when density is not uniform.
Step functions. As with the other step function arguments presented in this chapter, you will probably not be expected to recall the material in the supplement.
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
Inthiscase, m3x3)/(m2
+ m3) .
[mlxl + (m2 m2x2
+
M 1 = m l , M 2 = m 2 + m 3 , X1 = x l , The center of mass is
+ m3) (m2x2 + m3x3)/(m2 + m3)l
m x )/(ml 3 3
+ m2 + m3)
and
X 2 = ( m2x2'
(PIIX1 + M2X2)/(>ll + X2)
/ [ml + (m2
+ m 3) I
=
=
(mlxl +
, which is the same formula derived in
Example 1.
Copyright 1985 Springer-Verlag. All rights reserved.
.
Section 9.4
-
5.
Using t h e formula
9.
Using t h e f o r m u l a s
I?1 = 1m . x .
we g e t
1 1
m. 1=1 1
Z2
=
10
+
x
-
,
=
we g e t
-
-
x =
= lO(1)
.
20 = 30
and
+
y =
-
Thus,
x = (1.7
+
xni = l m1. 1y
. / ~ ~ , ~, m ~
2 Ei=lmiyi=
2 0 ( 1 ) = 30 ;
x = 30130 = 1
409
+
IO(0)
y
and
3.3
=
40 ;
413
.
20(2)
40130
=
i
=
A l l o f t h e s i d e s o f t h e t r i a n g l e must h a v e unit length.
T h u s , t h e t h i r d v e r t e x may be
X
geometry. 1/2)/3m 0 (b)
+
=
+
A x - 1)'
( y - 0)'
112
.
I f t h e mass a t
m(l)l/(2m
+m+
m) = 6 1 8
.
-
i s doubled, then
x
-
m) = 318
and
y
[2m(0)
=
Thus, t h e c e n t e r o f mass i s -
17.
Using t h e f o r m u l a s ~:f(x)dx
, we g e t
x
=
+
m(112)
i
.
=
2 ( 1 1 2 ) ~ ~ [ f ( x )d1x l
3
And
= (1/2)j1(4/x
3 2 3 -4 3 J 1 ( 4 / x )dx = (1/2)J116x d x / ( 8 / 3 ) = 3 ( ~ - ~ 1 - 3 ) = 26/27 ((i/2)1n3,26/27)
i
(318,618)
2 ~ ~ x ( 4 / x ~ ) d x / )~d ~x (= ~~/: (x4 l x ) d x / ( - 4 l x ) 3 I n 312
+
+ m ( A 1 2 ) + m(0)l /(2m + rn +
=
=
(1.65,0.96)
.
1:
=
2 2
) dxl
Thus, t h e
.
As shown i n t h e f i g u r e , w e w a n t t o f i n d t h e c e n t e r of mass o f t h e r e g i o n u n d e r -x/2
+
4 jO(-x/2 2x1
4
= 413
.
-.
y
=
4 (1/2)! ( - x / 2 0
4 3 I O / 4 * ( 8 / 3 ) / 4 = 213
.
+
2
+ 2
.
-
Therefore,
x = itx(-xi2
3 2 ) d x = (-x 16
4 2) d x / j o ( - x / 2
=
(1/2,6/6)
f2m(0)
-
21.
1
y = m(r/5/2
j i x f ( ~ ) d ~ l ~ ~ f ( xand ) d xy
3 4 I n x /1/(8/3) = 4 I n 3/(8/3)
c e n t e r of mass i s
+
m(0
=
-
x
=
i s t h e mass of t h e o b j e c t .
m
Thus, t h e c e n t e r o f mass i s
(0,O)
x
.
-
, where
0)/3m = f i / 6
- 0)'
(y
o r by u s i n g p l a n e
= 1
(112,612)
Thus, t h e t h r i d v e r t e x i s
+
4 x - 0)'
d e t e r m i n e d by s o l v i n g
+
+
2)dx
f(x)
+
2 4 2 x ) I 0 / ( - x 14 =
T h e r e f o r e , t h e c e n t e r of mass i s
-(-x/2
+
=
2)dx/
+ 2)
(413,213)
3
1
.
Copyright 1985 Springer-Verlag. All rights reserved.
410
Section 9.4
-
25.
First, recall that
x
=
~ : = ~ m ~ x ~ / ~. i Since =~m~ the masses are inde-
pendent of time, we differentiate to get the velocity at the center of mass, v
=
m.v. = Z? m. (dxi/dt) , and 1=111 1=11
C?
get
.
dx/dt = C? m.(dxi/dt)l~:=lmi 1=1 1
By definition, P
M = ~i,~m, . Rearrange
-
v = P/M = Zn m.(dxi/dt)/~: m = dx/dt = v 1=1 1 1=1 i Note that
cos x
=
P = MV
to
[0,n/4]
.
.
> sin x
on
On an infinitesimal strip, the center of mass is
(x,(cos x
+
sin x)/2)
We take a weighted average by weigh-
ure).
ing the center of mass against its mass, i.e., area. rectangular infinitesimal strip is x
=
Z l m i x i / Z l m i becomes
~b/~(cosx - sin x)dx
The area of the
.
(cos x - sin x)dx
-
uous sum of
(see fig-
~:"x(cos
Integration by parts with
Now the contin-
x - sin x)dx/ u = x
and
v =
+ cos x yields [x(sin x + cos X) 'I4 - J;I4(sin x + cos x)dxl sin x + cos x) 1 'I4 = [(n/&)fi - (-cos x + sin x) ] ; I 4 ]/(fi - 1) = (a714 - I ) / ( & - 1) 0.27 = x . Similarly, the continuous sum of sin x
-
y
=
l m i y i / Z l m i is ~;'~(1/2)(cos
J:'4(cos
x
-
x
+
sin x)(cos
x - sin x)dx/
2 2 sin x)dx = ( 1 / 2 ) ~ ~ ~ ~ ( cxo-s sin x)dx/(fi
trigonometric substitution, we get (112) (sin 2x12) b'4/(b'?
- I)
=
- 1)
(1/2)lg14cos 2x dl/(&
l!4(fi - 1)
.
/
. - I)
~y a =
Therefore, the center of
SECTION QUIZ 1.
(a)
A square region with vertices at (-1,-l) , (-1,l) , (1,-1) , (1,l)
has density
x2 at
(x,y)
.
and
Explain why the symmetry prin-
ciple applies for a vertical axis
Copyright 1985 Springer-Verlag. All rights reserved.
1.
(b)
(3,-1) , (3,l) , (5,l) , and
A square region with vertices at (5,-1)
x
has density
2
at
.
(x,y)
Explain why the symmetry
principle does not apply for a vertical axis.
2.
Find the centers of mass of each individual square region described in Question 1.
3.
Find the center of mass of the combined regions described in Question 1.
4.
The earth has B radius of The moon has a radius of The moon is (a)
3.9
x
6.4 1.6
105 km
x
x
lo3 km lo3 km
and a mass of and a mass of
6.0 7.4
24 10 kg.
x
x
22 10 kg.
from earth.
Assuming the earth and moonareperfect spheres, where is the center of mass of the earth-moon system?
(b) 5.
What information was riot necessary an3 why?
A famous artist likes to focus on the center of mass.
For her latest
painting, the lovely artist has selected three objects.
All of the
individual centers of mass are located in the same plane. is a circular plate with center at square plate has vertices at Each plate has density
(-3,4)
1 kg/m2
a.
and radius
(2,2) , (4,2) , (2,4) ,
One object
and
A
.
(4,4)
and each unit on the xy-plane is
The third object is a frog whose mass is
2 kg
1 rn.
and whose center of
mass is at the origin. (a)
Where is the center of mass for the three objects?
(b)
At the end of the day, the artist kisses the frog and he turns into a handsome prince whose mass increases to center of mass remains at
(0,O)
70 kg
and whose
. Where is the new center
of
mass for the three objects?
Copyright 1985 Springer-Verlag. All rights reserved.
ANSWERS TO PREREQUISITE QUIZ 1.
This is the shell method.
Revolving a thin rectangle at
x
around the
y-axis gives us a cylinder with base circumference or "length" Its height is
f(x)
and its width is
dx ,
so its volume is
27rx
.
2nxf(x)dx
Integrate to get the entire volume. 2.
Consider a thin rectangle at width is
.
dx
x
.
Its height is
Therefore, its area is
f(x) - g(x)
[f(x) - g(x)]dx
and its
. Integrate to
get the entire area.
3.
(a)
No
(b)
No
ANSWERS TO SECTION QUIZ
1.
2. 3. 4.
(a)
The mass at
(-x,y)
(b)
The mass at
(4 - xO,y)
(0,0)
and
equals the mass at
(x,y)
.
does not equal the mass at
(4
+
xo,y)
(204/49,0) % (4.16,0)
(4.14,O) (a)
4.8 moon
(b)
y
3 10 km
from the earth on the line between the earth and the
.
The radii are not needed because the consolidation principle can be used.
5.
.
(a)
(-3111,32111)
(b)
(-3179,32179)
Copyright 1985 Springer-Verlag. All rights reserved.
.
Section 9.5
413
9.5 Energy, Power, and Work PREREQUISITES
1.
Recall the physical interpretation of integrating rates of change (Section 4.6).
PREREOUISITE QUIZ
1.
Given the following quantities for
f (x)
(a)
Velocity (meters per second)
(b)
Water flow rate (gallors per minute)
(c)
Melting rate of a candle
,
what is
jf(x)dx
?
(grams per minute)
GOALS 1.
Be able to state the relationship between power and energy and apply it for problem solving.
2.
Be able to state the relationship between work and force and apply it for problem solving.
STUDY HINTS
1.
Psychology. Many students fear this section because it is "Physics." Remember that you are enrolled in amath course and being a physics major is not a requirement.
You will find the text self contained in what
you need to know.
2.
Units.
The unit of work is the joule which is
be remembered by using a ,
and
dx
.
dW
=
Fdx
=
ma dx
1 kg.m2/s2 .
This can
and using the units of
Another unit to know is the watt which is
m ,
1 joule/
second.
Copyright 1985 Springer-Verlag. All rights reserved.
414
Section 9.5
3.
Energy vs. power.
Energy is the total sum of power, which is rate.
They are related just as distance (integration) is to velocity (differentiation).
4.
Important formulas. Memorlze the fact that work is the integral of force with respect to position, not time. that
dS/dx
.
F
=
However, note that work is a form of kinetic energy;
.
dW/dx = F
this is a special form of the equation 5.
It is not essential to know
Example 4 should be studied thoroughly.
Good example.
important relationships
dW
=
Fdx ,
F
=
mg ,
and
It uses the
.
m = pV
At this 2 -r dx ,
point, we still need to deternine the volume term, which is by using similar triangles.
Remember that gravity acts in a downward
direction. 6.
Sunshine formula.
This is an interesting application of the calculus Consult your instructor to see if you need
you have learned so far. to study it.
None of the equations should be memorized.
The energy
equation (3) is the important one for doing the exercises.
SOLUTIONS TO EVERY OTHER ODD EXERCISE (SUPPLEMENT) 1.
We use the formula, where
sin D
. tan
f, =
or
D
a = 23.5'
.
= a
cot n ,
-1 cos (-tan Y cot 2) = cos at
-
sin a cos(2~T/365)
=
sin D = sin a , cos o
2 E = Aos Y
so 4y.
, we find E
2
sin D
+
sin L sin D cos-'(-tan
. On June 21, we have T
Also. we have
E
=
J--2 sln
Q
- sin23
-1 sin a cos (-1) 1.15 ,
cos
=
i =
+
cos
,
= 0
,' so
and
sin
sin
(~/2)sin(2<~).
which is about
energy received at the equator on June 21.
sin s
1.25
L tan D )
V
=
. Evalu,itin::
times the
This excess is due, of
course, to the long day at the Arctic Circle.
Copyright 1985 Springer-Verlag. All rights reserved.
An
5.
s i n a cos(2~T1365)
R
k
t a n L(k )
or
+
k (tan &
P
1) = 1
.
= cos-'k
,
-1
(-tan k t a n D)
[ k / m \ ]
=
m
.
k sec2t = 1
Thus,
sin D
2
2
E = jcos L - s i n D
=
(
F-k2+ C k
12
=
=
The = I
- ki . E q u i v a l e n t l y ,
1
Using t h e n o t a t i o n o f t h e t e x t ,
s i n 1 s i n D cos
365) =
=
S
t a n fi [ k / J 1 - k ; i )
2
which i m p l i e s
t u t i n g i n t o t h e energy equation,
[-[-I4
i
i
=
On a day on which
L[ k i n ] ] .
equation requires
2
.
t h e sun does not s e t , (24/2~)cos-'[-tan
2
k
To s i m p l i f y t h e c a l c u l a t i o n s , l e t
=
k
k
.
or
1
Now s u b s t i -
+
2 ( k ) cos-I
) cos-l(-l)
cos
=
= 7
x
s i n .P s i n a c o s ( 2 ~ T 1
.
sin k sin D
SOLUTIONS T O EVERY OTHER ODD EXERCISE 1.
We u s e
,b
E = J Pdt
D u r i n g one hou-:,
E
energy output is
5.
.
=
~
0
~
~
~ s i~n 2 1( 1 200 n 5t ) d 0t
.
formula,
E
= (1/2)(1050),/~~~ - ~c (o 1s 2 4 0 - t ) d t
2 4 0 ~ I)i 6 0 0
=
525 (3600)
=
2
(3x 1 2 ) ' ;
=
312
.
Power is
dE/dt
,
seconds, so t h e
By t h e h a l f - a n g l e = 525(t
W
,b
= J
F dx
.
i n t h i s case,
so the u n i t s a r e joules/second.
T h e r e f o r e , t h e r e q u i r e d power i s Recall t h a t -312
,
AK
=
r b ~dx
'a
.
93 j o u l e s / l s e c o n d
In t h i s case,
s o k i n e t i c energy decreases by
1K
=
B =
.1
,I o 3x
dx
As i n Exa:iiple 2 ,
t h e work done i n one s e c o n d = ( 1 k g ) ( 9 . 8 m / s e c i ) ( 1 0 m)
13.
- s i n 24Ort/
1,890,000 joules.
Work i s t h e i n t e g r a l o f f o r c e ;
9.
< t < 3600
0
=
=
98 j o u l e s .
98 w a t t s .
,1 jO(-3x)dx
=
(-3.-12)
,, , I
=
1.5 joules.
Copyright 1985 Springer-Verlag. All rights reserved.
=
17.
I f t h e d e p t h of t h e w a t e r i s b e t w e e n a
thickness
dx
, is
( 1 5 ) ( 3 0 ) dx
.
0
and
2 ,
t h e n t h e volume, w i t h
I f t h e w a t e r i s below t h e
we u s e s i m i l a r t r i a n g l e s t o f i n d t h a t t h e volume i s S i n c e a c u b i c m e t e r o f w a t e r h a s mass volume by
1000
thickness
dx
so
Iv' 3
=
5
25x ) I 2 ] 21.
S i n c e -3
t o f i n d t h e t o t a l mass.
is l i f t e d
x
1 0 0 0 g [ ~450 ~ x dx =
9800(900
+
+
6300
5 j2(600
- 2925)
5(5 - x ) l d x
Each l a y e r o f w a t e r w i t h
- 7 5 x ) x dxl
=
g = 9.8 mlsecL
9800[225x21i
+
(300x2
= 9800(4275) = 4 1 , 8 9 5 , 0 0 0 j o u l e s .
newtons i s needed t o s t r e t c h t h e s p r i n g 5
+
level
k i l o g r a m s , we m u l t i p l y t h e
meters against gravity with
i s neededtocompress the spririg so
lo3
( 1 5 ) [15
2m
cm.
5 cm,
t3
newtons
Work i s t h e i n t e g r a l of f o r c e ,
j o u l e s i n c e t h e f o r c e is c o m t a n t .
li = F(Ax) = ( 3 ) ( 0 . 0 5 ) = 0 . 1 5
SECTTON QUIZ 1.
Uhile doing pull-ups m u s t l i f t your
70
a t y o u r l o c a l a t h l e t i c c l u b , you r e a l i z e t h a t you
k g body o f f t h e ground by
0.5
meter.
How much
work i s done bv y o u r arm m u s c l e s ? 2.
A l g a e n e e d s t o be removed from a d r y d o c k which is
3.
300 m
long.
The c r o s s - s e c t i o n a l a r e a i s
L a t e l a s t n i g h t , a v a m p i r e was s e e n b r e a k i n g i n t o t h e b l o o d b a n k .
h a d s u c k e d t h e b l o o d o u t of t u b e was s i t v of
10 cm
50
cylindrical t e s t tubes.
l o n g and had a d i a m e t e r of
1 . 0 3 g/cm3 ,
1 cm.
lle
Fach t e s t
I f b l o o d h a s a den-
how much work was done by t h e v a m p i r e w h i l e
sucking up h i s d i n n e r ?
Copyright 1985 Springer-Verlag. All rights reserved.
-
,
ANSWERS TO PREREQUISITE QUIZ
1.
(a)
Net distance travelled
(b)
Volume of water flowing through over time
(c)
Amount of candle which melted
ANSWERS TO SECTION QUIZ 1.
343 joules
2.
9.3
3.
0.2 joules
x
lo9 joules
Copyright 1985 Springer-Verlag. All rights reserved.
9.R
Review E x e r c i s e s f o r C h a p t e r 9
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Use t h e d i s k method and t h e h a l f - a n g l e f o r m u l a , s o
(a)
- cos 2x)/2]dx
nj;[(l (b)
+
j:cos
x dx)
Use t h e r e s u l t o f Example 5 , S e c t i o n 9 . 2 w i t h
t e r i a l is s o l i d is
.
(413)~
. .
119)3 1 2 ~
( 4 / 3 ) ~ ( 8 / 9 ) ~ /= ' 6~v%.r/81
-
[1/(1
The a v e r a g e v a l u e of
5 ( X /5
f
1 2
- O)] l o x dx
~ " f ( t ) - :]dt 3
g(x)
(312) ( 4 )
=
The a v e r a g e of [11(1
=
on =
3 (X 1 3 )
[1/(1
26/15
;= [ 1 / ( 2
.
(
2
.
2 12
R
=
1
1;
27(n)
=
r
and
=
2n
=
.
113
f (x)
is
4
1 t /4)10
=
115
= [l/(b
.I
.
113
=
2 0)],fi[x
on
- o ) ] j i f (x)dx
=
.
In t h i s case,
,fi3f (x)dx/2
=
- I / ? ] dx
+
b
- a)] Jaf(t)dt
=
119
=
= jk(x4
.
4/45
=
[ll(b 2
- 2x 1 3
- a)]
+
119)dx
(1/2)((x/4)
- ]
i O1 +
2
dx
(x/4)
1 (112) [ b d x
[l/(b
The v a r i a n c e i s 2 )[ f )
is
2
=
114
2 j 1 2 dxl
+
- a )[
x
)
2
2/3&
=
- a ) ] !Bf ( x ) d x , s o =
- I d
- 312) dx + ,/:(2
(112) [!;(I
=
[l/(b
1. =
.
t h e s t a n d a r d d e v i a t i o n , which i s
1 (112) [ x i O
, -
+
2x1
2
so variance 312)~dxl
= =
=
The s q u a r e r o o t of t h e v a r i a n c e i s 112
=
The s t a n d a r d d e v i a v m=
.
Copyright 1985 Springer-Verlag. All rights reserved.
.
The
.
514
The v a r i a n c e i s 2
- 219
[s,hl =
2
The volume o f t h e removed ma-
.
The a v e r a g e o f
312
7
.
6
k
s i n x)
i ~ g ( x ) d x / ( 2- 0 )
is
[a,b]
+ x / 9 ) 1;
- 2x 19
=
+
l(t
=
=
V = 2 x 1 ; ~ s i n x dx =
+
t i o n i s t h e s q u a r e r o o t of t h e v a r i a n c e , which i s
21.
n(n/2)
=
- a)] ,fBi(t)dt
f ( t ) [a,bl = [ l / ( b
1 3 0 ) 1 1 ~ ( 1+ t ) d t
'0
Thus, t h e volume of t h e r e s u l t i n g
--
2 (3/2)j0f (x)dx 17.
-
- (1
(4/3)n[l
The a v e r a g e v a l u e i s it is
13.
1:
2n(-x c o s x
=
volume o f t h e o r i g i n a l b a l l i.s
9.
- s i n 2x14)
n(x/2
Use t h e s h e l l method and i n t e g r a t e by p a r t s , s o 2n(-x c o s xi:
5.
=
2
V = . r f P s i n x dx =
25.
-
Use the formulas x = j:xf(x)dx/j;f(x)dx Thus, the center of mass is
[x6/6i]/[x5/513 [x9/9/i]/2(32/5) mass is 29.
(32/3)/(32/5)
=
(513,4019)
2
.
1 4 x 14) Io/(x I4 (X
3
+
+
3
1 x 13)10
2
1 3 x )dx/j0(x
+x
2
(716) = (-2/35)/(7/6)
4019
=
.
(a)
1 x = j0x(x3
l2
=
(9120)/(7/12)
=
-
27/35 : y 4
- x )dx/(7/12)
and
2
5 x )dx = (x I5 = =
+
- x2)
(ll2)j;(x3
1
7 5 (x 17 - x 15) :I
. Thus, the center of mass is
=
b
4 j2 30
F dx
=
(-12011~) (-1 - 0)
=
=
sin(~x14)dx
dhadx
.
Integrate from
slab from
a
to
b
[~i(~)c~gh dhl dx If we rotate
dx
30(4/n)
=
(12011~)joules
pgh dh dx
force on the rectangular slab.
is
g(x) = x3
+
)dx/j;(x3
Consider a rectangular slab with width
an area
(c)
2
-1212i5
the force on the slab is
(b)
+x
)dx = (1/2)jA(x6
=
AK
Use the equation I-cos(nxI4)l
-
.
(27135,-121245)
4
=
.2 4 2 r24 = j (x ) dx/2jOx dr = 0 Therefore, ihe center of
.
Therefore,
4
37.
4 2 4 )dx/jOx dx
/ 0X(X
Use the formulas from Exercise 28, Section 9.4 with f(x) = -x
33.
=
113 and y
=
(512/9)/2(32/5)
=
x
2 dx/
y = (1/2)j;[f(x)1
2
-
.
~if(x)dx
-
and
.
x
. Then, at a depth
because the pressure acts on
0
to
f(x)
to find the total
Then integrate the pressure of each
i(x)]
ja[ig(h 212) ,
=
f(x)
therefore,
We use the formula in (b).
F
=
V
VpgI2n
radius of
100 m
2 2 2(nr hl3) + nr h which means
and heights of
2n(100)
obtained
Rotation of the dam face results in a
and a height of =
.
.
volume consisting of two cones and a circular cylinder. 100 m
=
2 dx = ( 1 1 2 ) ~ ; ~ [f~(x)~ dx
around the x-axis, then the volume
2 b nja [f(x)] dx ;
have radii of
F
to find the total force on the dam. b
h ,
2
(125)/3
125 m.
The cylinder has a
.
V
50 m
+
The cones
"(100)
Therefore, 2
=
(50) = 4,000,000:/3 ,
F = (4,000,000~/3)(1000)(9.8)/2~ = (19.6/3)(10
6
9
)
Z=
lo9) newtons = (2/3)(pg 1U ) newtom . (6.53 Copyright 1985 Springer-Verlag. All rights reserved. x
X
420
Section 9.R
41.
Suppose that 0
,
xO
f (a) < 0
f(b) > 0
and
.
I
Let
=
.
b (x)dx faf
I
If
=
the mean value theorem for integrals tells us that there is an in
[a,bl
with
(b
-
a)f(x) 0
required. If
I > 0 , then let
the interval
[al,b] by setting
=
I
=
0 ,
f(x) = 0 , 0
so that
,
al = a - I/f(a) f(x) = f(a)
,:I
with
f (xo)
=
be in
[a,b]
.
function past
.
0
The case b
f (x)
Since
I
instead.
<
=
0
f (a) < 0
on
xO
[al,a] ,
to
[ al,bl
.
f(x)dx
=
in
This is still continuous, and now one may compute that By the mean value theorem for integrals, there is an
f
and extend
for x
as
in
0
[al,b]
this
xo
must
is handled in a similar way; extend the
(See the hint on p.A.53 for another method).
TEST FOR CHAPTER 9
1.
True or false: (a)
The volume of the solid of revolution obtained by revolving the f (x) 2 0
graph of (b)
If
f(x)
<0 ,
around the x-axis is
V
=
.
2-jbxf(x)dx
it can be revolved around the x-axis to form a
solid of revolution. (c)
The center of mass of a plane region with uniform density always lies on the region.
(d)
The average of in
2.
(a,b)
f(x)
on
[a,b]
f(x)
on [a,b]
is attained at least once by
.
(e)
The average of
(a)
Find the volume which results'from revolving
0 <x (b)
<2 ,
never occurs at y
a =
or
b
.
x2 - 1 ,
around the x-axis.
Does it matter that
.
f(x)
<0
on part of the interval?
lfiy?
Copyright 1985 Springer-Verlag. All rights reserved.
f
A force F(x) 1
<x
10
from x Let
1
=
f(x)
=
=
.
l/x2
. applied to a particle on the interval 1s
Find the work done by the force in moving the particle to
rG
x on
=
10
[0,1]
generated by revolving
.
Find the volume of the solid of revolution
the graph of
f(x)
around the following lines:
3
(a)
x
(b)
y = -1
=
.
Find the center of mass of the region between
g(x)
0
Compute the average of
x cos(x
2
+
T)
on
[o,I'<;].
What is the center of mass of the region bounded by
y = x3
and the
[-1,2] ?
x-axis on
Show that the mean value theorem for integrals applies to x2 - I
(b)
~/x+4and the
<x <4 .
x-axis on the interval
(a)
=
on
[-I,O]
f(x) =
.
What is the average of
f (x)
on
[-l,O] ?
Find the volume of the solid of revolution generated by revolving the region between the graphs of
2x
and
x3
on
[0,1] around each axis.
One hot day in India, an elephant was down at the river cooling itself. The elephant filled its trunk with water to squirt on its back.
Unfor-
tunately, the last trunkful of water had a fly in it, which caused a sneeze attack. of
8 cm
Suppose the elephant's cylindrical trunk has a diameter
and a length of
1.5 m.
How much work is minimally required
to force the water out of its trunk if it is pointing straiglit up into the sky?
Copyright 1985 Springer-Verlag. All rights reserved.
422
Section 9.R
ANSWERS TO CHAPTER TEST
1.
rb
2
.
(a)
False;
V = rrj [f ( x ) ] dx
(b)
True
(c)
F a l s e ; t h e c e n t e r o f mass o f a " r i n g " l i e s a t t h e c e n t e r , i n t h e hole.
2.
(d)
F a l s e ; f s h o u l d be c o n t i n u o u s .
(e)
False; consider
(a)
(46115)-
(b)
No, b e c a u s e
f o r any 3.
-9110
4.
(a)
16-15
(b)
17-716
+
f
[-f(x)]
((40
6.
0
7.
(511185 ,20471238)
8.
(a)
24$'?)/35,(9
=
[f(x)I
[-2,2]
.
.
Thus, t h e f o r m u l a i s t h e same
+
18vrF)/28)
i s i n t e g r a b l e and t h e a v e r a g e i s a t t a i n e d a t
which i s i n (b)
2
on
.
5.
f
f(x) = 1
[-1,0]
x
0
=
-vm ,
.
-213
9.
257121
around x - a x i s ;
10.
55 j o u l e s
1 4 ~ 1 1 5 around y - a x i s
Copyright 1985 Springer-Verlag. All rights reserved.
Comprehensive Test
C0MI"PEHENSIVE TEST FOR CHAPTERS 7-9 (Time limit: 1.
(a)
The differentiai equation dy/dx = y/(x
(b)
2 j[dt/(l + t 11
(c)
j[dt/(l
(d)
Revolving
y = x2
revolving
y = x2
terval
3 hours)
If false, explain why.
True or false.
-
423
tan-lt
=
,
[0,a]
+
+
l)(y
xy)
is separabie.
+
C
is valid for all
t
C=;oth-lt
+
C
is valid for all
t # 1
+C
2 t ) I = tanh-'t
+
-1
-cot
=
t
.
around the x-axis yields the same volume as
+ a
2x
+
1
around the line
y
on any in-
1
=
>0 .
(e)
Force is the derivative of work with respect to time.
(f)
One solution of
(g)
The mean value theorem for integrals requires differentiability
dy/dx = -3x
is y
2 -3x 12
=
.
before it can apply to a function. 5 (jOf(t)dt)g(~)
(h)
5 jOf (t)g(t)dt
(i)
The derivative of
=
cosh x
3 4 (j) jtan P db = tan 814 2.
5
+ f (0)jOg(t)dt
.
exists for all real
x
.
+ C
Multiple choice. (a)
Which is the solution of (i)
cosh 2t
(ii)
cos 2t
+
(114)sin 2t
(iii) sin 2t
+
(1/2)cos 2t
(iv) (b)
+
y " + 4y
+
(ii)
(e - 1)/2
(iii) (x2 (iv)
(e
+
+
y'(0)
=
y(O)
=
1 ?
cos 2t
The average value of x exp(x e/2
0 , where
(112)sinh 2t
(1/2)sin 2t
(i)
=
In X)
2
)
on
[0,1] is:
11
1112
Copyright 1985 Springer-Verlag. All rights reserved.
.
42&
2.
Comprehensive Test
(c)
(d)
hJat is the most efficient way to evaluate
(i)
Integrate by parts with
u = sec
(ii)
Integrate by parts with
u
(iii) Substitute
u
=
sec 3 .
(iv)
u
=
tan 6
Substitute
Suppose
f
=
tan
3
+
,/sec3' tan
.
dv = sec32 d5
.
.
is symmetric with respect to the origin and
Furthermore, sl~ppose f 2 0 (i)
Positive
(ii)
Negative
7
d
= tan 2 dS
, dv
e ,
'J
on
[-b,-a] ,
b
>a .
-
the11 f
I-a,hl
is :
(iii) Zero (iv) (e)
There is not enough information
ii rahblt population srows exponenti,,lly. It takes five years for the population to increase from
1030
to
it take for the population to grow from
(i)
About
(ii)
2-4
50
2000 20,000
.
How long does to
30,000 ?
years
years
(iii) 4-6 years (iv) 3.
There is not enough information.
Perform the following integrations:
Copyright 1985 Springer-Verlag. All rights reserved.
Comprehensive Test
.
425
Short answers. (a)
Solve the differential equation
(b)
Differentiate
(c)
Express
(d)
Sketch the graph of
(e)
If
tanh(x2
sinh t
F' = f
+
3)
y' = -y/4 ,
assuming
y(0) = 3
.
as a sum or difference of exponentials. y
=
cosh x
, what is jf(3t)dt
. ? 1
5
Find the center of mass of the regic>n between
+
T
=
80'
.
Initially,
T(0)
for the temperature to change from
y
= s
3
on
=
50'
50'
.
How long does it take
to
75'
7
Solve the following differential equations with the given conditions (a)
dy/dx = xy ,
y(0)
(b)
dy/dx = sin x
+
y ,
=
0 , ~ ( 0 )= 1 , Y ' ( ~ ) =
2 dy/dx2
(C)
8.
and
The temperature of a swimming pool changes according to the formula dT/dt
7.
y = xL
.
[O,ll
6.
.
Let
R
+
7x
=
1 y(0) = 1
be the region between
x
and
x3
on
[-1,2]
.
volume of the solid of revolution obtained by revolving
What is tlic
R
around
each of the following?
9.
(a)
The x-axis.
(b)
The y-axis.
Suppose the price of land in Manhattan increases at an instantaneous rate of
15%
per year.
How long does it take for the value of the
land to triple?
A volcanic mountain top has a shape which can be described as the solid of revolution formed by revolving the shaded region around the y-axis.
Each unit represents
1
kilo-
Copyright 1985 Springer-Verlag. All rights meter and the mountain top hasreserved. a uniform
426 Comprehensive Test
10.
density of
5000 kg/m
3
.
A violent eruption disintegrates the mountain
top sending the particles to form a cloud cover at
y
=
3 . How much
energy was expended by the volcano in forming the cloud cover?
AKSWERS TO COMPREHENSIVE TEST 1.
(a)
True
(b)
True
(c)
False; it is
(d)
True
(e)
False; it is the derivative with respect to position.
(f)
True
tanh-I
+
C
if
/ti >1
,
coth-'t
+C
if
It
1
False; it only requires integrability. False; the right-hand side should be obtained from integration by parts. (i)
True
(j) False; differentiating the right-hand side yields 2.
(a)
iv
(b)
ii
(c)
iii
tan3 6 sec 2$;
(d)
(el 3.
(a) (b) !c)
(a) (el
Copyright 1985 Springer-Verlag. All rights reserved.
.
.
Comprehensive T e s t
4.
(a)
y = 3 exp(-~/4)
(b)
2 2 2x s e c h (x
(c)
(e
t
- e
-t
+
3)
)/2
[ ~ ( 3 t ) 11 3 + C
(e) 5.
(315,12135)
6.
I n 6
7.
(a)
y
(b)
y = (3eX- s i n x
=
2 exp(x 12)
-
cos x)lL
COSYrj~ + (l/J7)s i n f i x
(c)
8.
427
(a)
340-/21
(b)
124~115
9.
I n 311.15 y e a r
10.
3.3
x
1014 j o u l e s
Copyright 1985 Springer-Verlag. All rights reserved.
CHAPTER 10 FURTHER TECHNIQUES AND APPLTCATIONS OF INTEGRATION -
10.1 Trigonometric Integrals PREREOUISITES
1.
Recall how to differentiate and integrate trigonometric functions (Sections 5.2 and 7.1).
2.
Recall how to integrate by substitution (Section 7.2).
3.
Recall how to define trigonometric functions in terms of the sides of a right triangle (Section 5.1).
4.
Recall how to complete a square (Section R.1).
PREREQUISITE QUIZ 1.
2 Differentiate y = sin 4x - cos x
2.
(a)
Evaluate
jcos 3x dx
(b)
Evaluate
jsin(2x
3.
tan x
.
. 3)dx
. Consider the figure at the left
x 4.
-
+
(a)
What is
sin 6 ?
(b)
What is
cos 6 ?
Complete the square in the following:
+
2x
+
4
(a)
x2
(b)
x2 - 3x - 1
Copyright 1985 Springer-Verlag. All rights reserved.
430
Section 10.1
GOALS 1.
Be able to use trigonometric identitiesfor integrating expressions involving products of sines and cosines.
2.
Be able to use trigonometric substitution for integrating expressions , , , or a2 + x2 . involving
LT?
c'
LT?
STUDY HINTS
1.
Half-angle formulas. You should memorize or learn to derive (1 - cos 2x)/2
2 cos x = (1
and
+
.
cos 2x)/2
If you forget which
sign goes with which formula, substitute x = 0 formulas can be derived by using 2 cos x
+
2 sin x
.
sin2x =
as a check.
2 cos 2x = cos x
-
2 sin x
These
and
1
=
Adding and subtracting yields the desired results.
The half-angle formulas are commonly used for integration. 2.
m Integrating sin x cosnx.
2 identity cos x
+
2 sin x
=
Basically, if one exponent is odd, use the 1
whichever is the odd power
and substitute u = sin x
.
or
u = cos x ,
If both are even, use the half-angle
formulas. The trigonometric integal box on p. 458 is a good one to know.
3.
Substituting u = sec x.
When
tan x
times useful to substitute u = sec x
and
.
See Example 3(c).
Addition and product formulas. Knowing that and that
cos 2x =
2 2 cos x - sin x
and (lb) on p. 460. mulas, along with
appear, it is some-
Often, it is necessary to re-
write the integrand into a useful form. 4.
sec x
sin 2x
=
2 sin x cos x
may help you recall formulas (la)
Simply substitute x sin(-x) = -sin x
for deriving the product formulas.
and
for
y
cos(-x)
. =
The addition forcos x ,
are useful
If you choose to memorize the pro-
duct formulas, which are useful for integration, note that the angle x-y
always appears first as it is written on p. 460.
Also, it may be
Copyright 1985 Springer-Verlag. All rights reserved.
4.
(continued) x = y ?"
helpful to ask yourself, "What if only
5.
sin x cos y
In addition, note that
has sine terms on the right-hand side.
Trigonometric substitutions. This technique is often used when 2 2 (+a ix
appears in the integrand, where
n = integer and
a
are
constants. Notice that this technique is based upon the Pythagorean
2
identities cos x
+
2 sin x
substitution to use for
=
x
2 sec x
Know what
in each of the three cases.
Using the
=
1
+
2
.
and
1
tan x
substitution equation, draw an appropriate triangle and label it like those in the box on p. 4 6 1 .
After the integration is completed, use
the triangle to express your answer in the original variable. 6.
Integrating
sec x
and csc x.
Just note theinterestinp trick used in
the integration on p. 462(lines 3 - 5).
7.
Integrating
3 sec-x.
Often, when
appears in the integrand, 3 sec x
trigonometric substitution will call for the integration of The technique is shown in the solution to Example 8(a), that
8.
3 sec x
p. 4 9 6 .
.
Note
is integrated by parts.
Integrals involving axL
+
bx
+
c.
In many instances, the first step
is to complete the square. Then, use a trigonometric substitution. 9.
Example 5 comment.
10. Practice.
If
a = ?b ,
remember that
cos(0) = 1
.
A lot of material has been covered in this section.
placed in memory are easy to forget.
Items
Practice helps to reinforce what
has been memorized.
Copyright 1985 Springer-Verlag. All rights reserved.
432
S e c t i o n 10.1
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
j(l -
,
6
+
u 16 - u 1 4
- x/2
-
+
.
1)/2
By u s i n g t h e h i n t ,
3
!tan
2
=
+c
A
- c o s S) / c o s 61 dH
figure t o get
.
=
=
-
c o s 2x
= j ( c o s 2x - l ) d x / 2 =
-,/
[(I
+
e
0)
+ s
C
- l)d3
=
,
s i n 6x/12
+
=
so r
1 du
=
+
j (u
du
-4
-
u
-6
C = - s e c 3x / 3
dx = 2 s e c
2 , / t a n r 8 d9 =
2(tan r
2
2
+
- s i n x dx ;
=
)du =
+ s e c 5x /
tan
d
2~(sinL3/cosi3)d0=
=
- 9 + C. We u s e d t h e
cns 9
=
1
.
Now, u s e t h e C
.
2 sec t d~
.
2 j t a n r s e c * db = 2 s e c c
+
(
c
2 tan F
2 sec2* du
so
sin Y
and = 2
) /u
6
115 c o s 5 x
=
=
= 2j(sec20
-
- u
2
2 s e c P t a n S da
x
-
s i n 2x14
=
+
y)]
- cos(x
(1/2)[cos(x - y)
u = cos x ,
Kow, l e t
s i n G/cos
2(tan 0
=
.
Let
2
- cbs x)dx
2 see 9
Let
2
i d e n t i t i e s tan 3
- u )du
=
3 6 b e c o m e s ,((sin x/cos x)dx =
= -113 c o s 3s
2 2: [ ( I
du
3
gives us 2
- cos 6x)/2
/ ( c o s 2x
3 x s e c x dx
6
-5/(-5)
2
cos 2x)/2
sin x sin y
- c o s x ) / c o s x] dx
-
j(u
so
.
C
/ ( c o s 2x
t h e r e f o r e , t h e i n t e g r a l becomes -3 / ( - 3 )
=
5
=
.
C
/ s i n 4x s i n 2x dx
/[sin x(l
+ +
(1
=
Thus,
Using t h e p r o d u c t formula we g e t
13.
cos x
3
- / ( I - u ) u du
- cos x/4
,
u = cos x
2
4
2
formula
( c o s 2x
=
s i n 2x/4
9.
6
C = cos x/6
The h a l f - a n g l e cos x
Now, substitute
and t h e i n t e g r a l becomes
4
2
.
c o s x ) s i n x c o s x dx
3 3 , [ s i n x c o s x dx
2
- c o s x , we g e t
1
=
3
2
-sin x
5.
2 sin x
Using t h e i d e n t i t y ,
From t h e f i g u r e , we g e t t h e f i n a l a n s w e r ,
/
,
2 so
+
- cos-l(2/x)) ds
=
+
2
Then
C
C
Copyright 1985 Springer-Verlag. All rights reserved.
.
25.
1) = 4(x
+
118)
2
+
and letting
Note that -cosnx
1;
.
+
+ ja
+
+ 1 = 4(x + + 118 , so
x
u = x
x)
=
, we have
-cos x ,
jtn cosnx
Hence,
cosnx dx
Let
2
cosn(x
+
1;
=
n
is odd,
cosnx dx
For
n = 0 ,
(b)
For
n
2
=
+
(1/4n)(x (c)
For
For
4 (1/2n)jin cos x dx
+
2 cos 2x) dx
(ll8)sin 4x)
n
2
+
,fO'(1/(16n))(1 3 cos 2x)dx
1;
+
=
=
(1/16n) [(5r) i
(1/~)1:~.100
=
sin(754T)
/in+
(112) (1
3/8
0 dx
.
0
=
.
cos 2x)dx =
+
+
2
2 cos 2x
cos 4x)ldx
2 dx =
+
2 cos 2x)dx
+
(1/8n)(x
=
=
sin 2x
+
.
+
3 cos 2x
+
lo
+
+
(3/2)(1
(312)sin 2x
+
=
=
3 cos 2x
+
2 3 cos 2x
cos 4x)
+
(1 -
+
2 2x - cos 2x sin 2x)dx =
5/16
=
.
and a half-angle formula to get (50~/T)l;(l
- cos(754t))dt
(50R/T) (T - (1/754)sin(754T))
. Now, substitute R
=
2.5
and
T
=
+
( 3 1 2 ) ~+
3 2 ((112)sin 2x - (1/6)sin 2x)
10 sin(377t)
T
+
(1/16~)ji'(cos
2 sin (377t)dt
(1/754)sin(754t))
27 jO
(COS X)
3 2 cos 2x) dx = (l/(16n))jon(1
2 (1/16n)jOn(1
(318)iin 4x)
Substitute
dx =
for four cases.
+
2n
(1:211)j0
=
2n (l/8n)jo (1
=
2 sin 2x)cos 2x)dx = (1/16n)(x
33.
=
2 6 2n 2 3 (1/2n)joncos x dx = (1/2r)j0 (cos x) dx =
6 ,
=
+ x)
lo
(118n)~i~jl+ 2 cos 2x
(dl
Then,
1-dx = 2nI2n = 1
2 2n (1/2ji)jin cos x dx = (1/4n)j0 (1 2n (1/2)sin 2x) = 112 .
n = 4 ,
+
=
n
,
(l/8;r)jin(l
x/2
.
+
j(2 d ~ /
+ jZn c o p x
0 (1/2n)jin cos x dx = (1/2n)ji6
(-1116
11-
cosn(n
- cosnx)dx
n)dx = j:(cosnx
+
du = dx
j(dx/=l)
Therefore, we need to consider only even values of (a)
2
Factoring out
so if
dx =
118)
.
j(du/=)
=
13 = 8 u / m
cos(n
.
15/16
j ( d x / / 4 x )
29.
4x
Completing the square, we have
=
=
P
=
(50R/T) (t -
50R - (50R1754T) *
2 ~ 1 3 7 7to get
P
=
125 -
(125-377/(2~.754)) sin 4n = 125 . Copyright 1985 Springer-Verlag. All rights reserved.
4.34
S e c t i o n 10.1
37.
(a)
31:(x
2
+
3(t sin t
sin x
3 ~ 0 ( xs i n x 3jk(x s i n x Let
s i n t cos t )
+
2
+ +
t
J O x s i n x dx
+
,
2
2
.
s i n x cos x)dx s i n x cos x)dx
.
dv = s i n x dx t
jO cos x
In
.
Thus,
~ ( 0 =) 0~ =
,
let
+
s i n x)i
u = x
.
x
j k ( l - c o s 4x)dx = ( 1 / 8 ) ( x
-
I ( t ) = -t c o s t
S1(t) = 0
cos t
whenever
sin t = 0
-1
or
< cos 2
1
t
, -1
of
S(t)
.
When
+
=
XI:
-x c o s
.
sin t
+
Since
dx = (118)
x
(114)sin 4 x ) / k = (1/8)/(t - (114)sin 4 t )
.
+
sin t
+
t / 8 - ( 1 / 3 2 ) s i n 4 t ) l 113 2
+
2
s i n t cos t = 0 ;
2
t cos t
.
=
, and l e t
t / 8 - (1132) s i n 4 t
sin t = 0
n
A(t)
I(t)
+
< sin
t
Then
du = dx
= (1/4)j;sin22x
s i n t cos t = 0
c r i t i c a l v a l u e s of
.
=
sin t
s i n t cos t >1 - 1 = 0 a r e when
Then
2
2
t >1
A(t)
( ~ ( t ) =) ~
+
t sin t
+
Let
so
-t c o s t
=
0
, B(t)
=
.
Therefore,
2 B ( t ) = jOsm x c o s x dx
= -cos
3m [ 3 ( - t
.
c = 0
2 2 s i n x cos x)dx
, so v
dx = (-x cos x
so
t . 2
A(t)
=
i n t o t h e given
s i n x c o s x = (112) s i n 2x
Then
(c)
2
(s(t)13
t = 0
( S ( 0 ) ) ~ ' ( 0 )= 0 = S ( 0 ) 2
=
Substitute
2
and l e t
B(t)
.
s i n x c o s x)dx
+
3(s(t))'s1(t)
=
Now, i n t e g r a t e t o g e t
2
I ( t ) = jk(x s i n x
A(t)
.
2
equation t o get
(b)
( d / d t ) [ T s ( t ) ) 31
Differentiation yields
.
Since
<1 .
Thus,
.
i . e . , whenever
< s i n t < 1 and if t > 1 , t +
-1
Thus,
s'(t)
T h e r e f o r e , t h e o n l y z e r o s of
, i.e.,
t = nn
S(t) =
.
for
These a r e
, c o r r e s p o n d i n g t o r e l a t i v e maxima and minima i s odd,
sln t
,
and c o r r e s p o n d i n g l y ,
~ ' ( t )
c h a n g e s from p o s i t i v e t o n e g a t i v e , i n d i c a t i n g a r e l a t i v e maximum excursion.
T h e r e i s no a b s o l u t e maximum, s i n c e
S(t
+
27)
> S(t) .
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.1
435
SECTION QUIZ 1.
Use the technique of trigonometric substitution to evaluate x
2.
3.
/
2
.
Did you get the expected result?
Evaluate the following integrals:
(c)
~ s e c ~ tan ~ x dx
(d)
j[sin(3x/2)cos
(e)
jsin 3x sin 4x dx
(f)
j[3/(1 +
2x
t2)3'2~
+
l/Jx2]dx
dt
x-axis, and the lines x = 0
,
2 helpful to integrate x cos ax 4.
x = n/2
.
the
[Hint: You may find it
for a constant
u
.I
Evaluate the following integrals: a
j[(pX
+xd]-/)i
(b)
/[(4x
+
+
2)/(x2
j[(x - l)/(x
(c)
5.
y = x cos2x ,
Find the center of mass of the region bounded by
2
6x
+
10)~'~]dx
- 2x - 5)ldx
Underwater divers have recently discovered a sea monster at the ocean floor. (4
-
Its shape has been described as follows:
x2)ll4
and
sin x
If each unit represents of
1500 kg/m3
,
on
The region between
[O,n/41 , revolved around the x-axis.
10 meters
and the sea monster has a density
how much does it weigh?
ANSWERS TO PREREQUISITE QUIZ 1.
4 cos 4 x + 2 s i n x c o s x + s e c2x
2.
(a)
sin 3x13 4- C
(b)
-cos(2x - 3)/2
+
C
Copyright 1985 Springer-Verlag. All rights reserved.
436
Section 10.1
3.
(a)
1
/
6
(a)
(X
+
112
(b)
(x
-
3 1 2 ) ~ - 1314
(b)
4.
7
llm +
3
ANSWERS TO SECTION QUIZ
+
.
, a s expected
1.
sin-lx
2.
(a)
3 5 7 s i n 0 - s i n 9 + 3 s i n 615 - s i n 8 / 7 + C
(b)
5t/16
(e)
s i n x / 2 - s i n 7x114
C
+
sin2t/4
+
3 s i n 4t/64
+
-
3 s i n 2t/48
+
C
C
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.2 437
10.2 Partial Fractions
PREREQUISITES 1.
Recall how to factor a polynomial (Section R.l).
2.
Recall how to integrate by the method of trigonometric substitution (Section 1 0 . 1 ) .
3.
Recall how to integrate by the method of substitution (Section 7 . 2 ) .
PREREQUISITE QUIZ 1.
Evaluate
J [ x 2 / L 7 + x + 2bx
2.
Evaluate
j1'jl/ 1 x +
3.
Factor the following polynomials: (a)
x3 - 27
(b)
x3 - x2
7+ 2 i)dx
2x
.
.
GOALS
1.
Be able to integrate rational expressions by the technique of partial fractions.
STUDY HINTS 1.
Beginning partial fractions. Look before you leap! easier method.
See Example 7.
There may be an
If partial fractions is the method of
choice, be sure the degree of the denominator is larger than the degree of the numerator.
2.
If not, begin with long division.
Denominator factorization. All factors should be of degree one or two. If not, the denominator can be factored further. Check to be sure quadratic factors do not factor further (by using the quadratic formula,
Copyright 1985 Springer-Verlag. All rights reserved.
438
Section 10.2
2.
(continued) if necessary).
x = x - 0 ,
Don't forget that
the linear factors
(x
-
O).(x
-
0)
.
so
x2
See Example 4 .
is composed of Also, remember
that a factor raised to the nth power must be represented
3.
times.
Determination of coefficients. This method is called comparing coefficients:
Multiply so that both sides of the equation are over a common
denominator.
Rather thanexpanding, it is best to leave the expression
as a sum of factored terms.
Then, substitute values of
many terms as possible become zero. linear equations.
4.
n
If no such x's
x
so that as
The result should be a few, simple are left, choose any other constants.
Differentiating to determine coefficients. To solve Example 2, this author prefers Method 1.
I find that there's a greater chance for error
with the differentiation process; you may disagree.
5.
Comparing integrands.
After you have found the coefficients, use a cal-
culator and an arbitrary number to compare the original integrand with your new one.
6.
Rationalizing substitutions. If
[f(x)] P"
appears in the integrand,
you might be able to make a simplification by substituting u i.e.,
7.
uq = f(x)
.
=
[f (x)] l / q
,
See Example 8 .
_Integrals of rational trigonometric expressions. The technique used in Examples 10 and 11 is useful for rational functions in Rather than memorizing equations ( 8 ) ,
sin x
and
cos 2x
=
cos x
( 9 ) , and (lo), it is suggested that
you reproduce Fig. 10.2.2, and use the identities sin 2x 2 2 cos x - sin x
and
=
2 sin x cos x
to complete the substitution. Often, partial
fractions may be necessary to finish the integration.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.2
439
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
+
~/(x - 2)2(x2
Let
+ F)/(x 2 + I ) ~. Then we need 2 2 ~(x' + 1)2 + (CX+D)(X - 2) (x + (Ex
1
B(2
=
2
+
112
,
i.e.,
0 ;
=
gives:
and
+
-2A
. - 212 + (8x + F
and 5 ! ( x
151125
=
+
+
+
ll21 dx
=
1/25
+
15:[dx/(x2
.
=
81125 , D
+
(x2
+
1)
812
+
sin 2814
+
2(1
5.
2
x )
1
:(sec2q
=
+
4 dA/sec 8)
C
=
+ C = 012 + sin 8 . Hence, j[dx/(x
Since the discriminant
(b
2
+
1)
2
1
+
I)]
(x
+
-
C)/(x
2) (x2
+
2x
+
+
2x
+
2)
equations gives A
=
2)
.
=
2A
+
E
=
201125 ,
- 2) +
~ ) ~ ] d =x (1/125) 11 tan-'x
+
10/ [2x/
. The last integral can be evaluated tan 0
2 jcos 8 dB = cos R/2
- 2)2(x2
- 4ac)
=
1 [(I
[(A+ B)X
+
+
i.e.,
/[dx!
cos 28)/2]d1
=
C
+
1) 21 = (1/125){-8 1 n / x - 21 -
for x2
2
x ,
+
+
(1/2)tan-'x
=
2x + 2
we cannot factor it further. Thus, x /(x - 2)(x (Bx
and
(1/125)j[-8/(x
=
+
2
2
x = 2 ,
D = -1125 ;
111125 ,
=
15)/(x2
by using trigonometric substitution with
2
+
Solving this set of equations
(20x
2 1) 1 1
+
0 ; -2A - 4C
=
+ 4j [2x dx/(x2 +
i-8 In / x - 2 / - 5/(x - 2) (x2
Substituting in B
C
+
1)
+
If
(Ex+F)(x
2 2 jdx/ [(x - 2) (x
Il)/(x
~)/(x~ + 1) +
.
4F = 24/25
2
+
(CX
- 2)
C
Thus,
+
+
1)
+
A = -81125, B = 5/125 ,
2
1 = A(x - 2) (x2
.
+
4D
B/(X - 2)
to solve
B = 1/25
comparing coefficients gives A
4F
+
A/(X - 2)
I)'=
(2A - 2B
+
2
is less than zero,
+ 2x
C)x
+ !x
+
+
2) = A/(x - 2)
(2A - 2~)]/
Comparing coefficients and solving simultaneous
215 ,
B
=
315 ,
and
C = 215
.
Thus,
1 [x2/
Copyright 1985 Springer-Verlag. All rights reserved.
+
440
5.
Section 10.2
(continued) The first term gives /[2/(x - 2)l dx = In [(x - 2)
+
term, note that (3x 2
1/(x
+
2x
+
1
want to have
+
d(x
2
2)/(x
.
1)
+
2
+
2x + 2)
.
Factoring
x4
2x
+
2)
+
+
in the numerator and whatever is left con-
2)ldx
2 + 2x + 2)] dx = (312) x 2)/(x 2 2 1) + l)]d~ = (3/2)ln(x + 2x + 2)
j[(3x +
Thus,
- j[l/((x+ +
(3/2)ln(x2
2x2 - 3
2x
+
2 (x
+
1)
+ D/(x 2
(x - l)]/(x
+
3
results: x : 0
-
gives
gives A = -114
,
-
3)]dx
[-ln(x2 13.
+
+
B
=
+
x/(x4
B)(x~ - 1)
+
1) (x
+
2x2
+
C(x
(1
+
.
+
2u - 2 tan-'u
2 x :
+
+
3)
+
l/(x
+
l n / x - 111
1 - 1)/(u2
C = 2&
- 2 tan-'&
j( [2du/(l + 2J(u
+
u )
2 u ) ] /[1
and
2)ldx =
.
1))
+
C
-
1)
.
Using the tech-
+
3)
(Ax
=
-
3)(x
+
B)/(x~
+
+ D ( X ~+
1)
3)
3)
x
x:-A+3C+3D=1
+ C
+
118
1)
+
l/(x - l)]dx = (118) x
=
+ 2u/(l + u + ~ ) - ~ d=u -2(u + 1)-1 + c
=
1
u
=
C
=
2
- 1)/(x2 + 3)/
& , so u2
+
-
21[1/(u2
2 u )
=
x , 2 2jiu /
+
u = tan(x/2)
=
+
C
and
1)ldu =
+
, so
. Thus, j[dx/(l + sin x)]
j[2du/(l + u2 + 2u)l
-2/(1
/[x/(x4
.
Substitute
dx = 2du/(l 2
(1/8)lnI(x
Thus,
+
=
l)]du = 2jdu
+
.
D
2 J[u/(l + u )]2u du
+x)]dx=
We use the technique of Example 11. sin x = 2u/(l
+
+
B - C + D = O ;
118 , and
=
+
2
2x
Comparing coefficients gives the following
C
Thus, j[&/(1
2 u )]du = 2j[(u2
2
+
-
Solving these four equations simultaneously
0 ,
inlx i 11
3) (x
We apply the method of Example 8 by letting 2u du = dx
17.
.
3D = 0
(1/8)/[-2x/(x2
=
3)
.
- 1)
+
A + C + D = O ;
+
2 2x
1) = [(Ax 2
3)(x
B - 3C
x (1):
2 2 j[x /(x - 2) (x
2) - tan-'(x
nique of partial fractions, we have C/(x
For the second 2 2)/(x + 2x + 2)] -
+
(312) [(2x
=
All this information gives
(1/5)4:ln(x - 2)2
9.
+
2x
+
1 .
The reason we break it up chis way is because we
stitutes another term. 1[(2x + 2)/(x -1 tan (x + 1)
2
2
tan(x/2))
=
+c .
Copyright 1985 Springer-Verlag. All rights reserved.
=
Section 10.2
21.
By the shell method, V
6 2nj5[~/(1
fractions yield B/(1 - 2x) x
for all x
,
x = 1
Let
,
-
B/(1
.
2x))
x = 132
,
so -A = 1
+
x)
- x)(l
.
- 2x))ldx
Now, partial
, with ~ / ( l - x) +
2x)Idx
+
Therefore, A(l - 2x)
so
B = 1 . Let 6 Therefore, V = 2nj5 [-l/(l - x) +
B/2
.
i.e., A = -1
B(l - x) =
112 ,
=
l56 =
In 4 (a)
Integrating the right-hand side and using the method of partial
2x)ldx
=
-
i.e.,
+
1/(1
25.
.
-
-
- x)(l
x/((l
=
6 2nj5[x/((l
=
441
2n[ln(x - 1) - (1/2)ln(2x
I)]
.
In 3) = n ln(2251176)
B(80
-
1/20
.
C
Let
-
l/(x when
(b)
x) = 1
for all
x
80 ,
=
-
(1/20)j[-1/(80
=
kt
for constants
+
x)
+
,
,
B
1 ,
=
,
C
we have
(1/20)ln(4/3)
=
-
.
- 60) /
,
(1/20)ln(4/3)
(1/20)ln/ (x
=
20B = 1 ,
-1120
-
1 (x -
80)/(x
60) /
-
80)/(x
Since we assume
exp(20kt) = x
-
-80
+
If
x = 20 when
.
80exp(20kt)
(a) to get ln(4/3)1
=
.
80
=
(1/20)ln(9/8)
part (b) to get
20kt
-
-
, we = =
+
80) =
0
x
60
60)
or
+
ln(4/3) =
(4/3)exp(20kt)
,
=
the formula
(4/3)(x
Thus, x = 80[ 1 - exp(20kt)] / [ 1
(1/20)1n(3/2)
=
Thus, kt
- 60)
Rearrange again to get [ (4/3)exp(20kt)
t = 10
Now, substitute k
(x - 80)/(x
B =
.
and exponentiation yields
.
- 60) 1
or
and the formula becomes
Rearrangement of the equation in part (a) gives ln/(x
.
Since x
- 60) 1
80)/(x
+
~ ( 6 0- x)
1/(60 - x)]dx = (1/20)/[1/(x
60)I dx = (1/20)ln/ (x - 80)/(x t = 0
=
I [A/(80 - x) +
where
so
A
or
=
,
C
and
x = 60 ,
Let
so -20A
becomes (4/3)exp(20kt)
(c)
.
x
A
+C
kt
fractions for the left-hand side, we get B/(60 - x)]dx
277(ln 5 - l n m -
-
X
- llx
=
(4/3)exP(20kt)l
substitute into the formula In part
IOk 10h.
ln(9/8)/200
+
(1/20)1n(4/3)
or
(1/20)[ln(3/2)
.
Therefore, k
=
and
into the formula In
t = 15
x = 80[ 1 - exp(3ln(9/8)/2)1/[
80(1- (9/81~/~)/(1- (413)(9/8)~/~)
= 26.2 kg
ln(9/8)/200
-
1 - (4/3)exp(3111(9/8)/2]
.
Copyright 1985 Springer-Verlag. All rights reserved.
=
442
S e c t i o n 10.2
SECTION Q U I Z
j[dx/(2x3 +
1.
Evaluate
2.
2 Evaluate ~ : [ d ~ / ( ~ 2
Evaluate
4.
Find t h e average of t < 4 .
(Hint:
. .
2 2) x]
-
- 2) ( x
3.
[dx/(x
+
2 4x ) ]
I)]
.
f(t) = t 8 G / ( 1
let
u
=
+
G
.
-
t6)
on t h e i n t e r v a l
2
<
)
.
5.
Evaluate
6.
When t h e m i n i s t e r asked f o r any o b j e c t i o n s , t h e b r i d e ' s g r a n d f a t h e r
i;I4[tan
6/(1
s e c €I)] d ~
s a i d , "Sonny-boy has t o prove t o me h e ' s smart enough t o be h e r husband. y = 10 - x / ( x 2
Look a t t h a t archway.
-
2x - 3)
on
[0,21
describes
He can marry h e r i f he knows t h e a r e a under t h a t archway."
it.
What
answer would b r i n g happiness t o t h e young couple?
LYSWERS TO PREREQUISITE QUIZ 1.
2 sin-l(x12)
2.
X
3.
(a)
+
l
- x n 1 2
+
E
(x - 3 ) ( x L
+
+
3x
+ x2/2 +
2x
+c
C
+
9)
2
(b)
x ( x - 1)
(c)
(x
+ 3) ( x + 2)
ANSWERS TO SECTION Q U I Z
+
2 ) / x \ - 1/4x
+
1.
(118) l n / ( x
2.
( 1 / 8 ) 1 n ( 3 / 2 ) - 1/48
3.
[(2
4.
- ( 1 / 9 ) ( 6 6 m - 1 0 m ) +(l/l2) ln[(J66
+
f i ) / 4 ] l n / x -I-
(1146) l n [ ( m
+
+
fit +
6 ) ( m
C
[(2 - fi)/4] l n / x -
-
-
6)/(J66 -
fi/ -
1 ) ( m
+
l n l x - 11
1 ) / ( m
6)(a + 6)]
+
+
C
1 ) ( m
-
-56.14
2 tan (n/8)1
5.
ln[l
6.
20-1116
Copyright 1985 Springer-Verlag. All rights reserved.
l ) ] -t
Section 10.3
443
10.3 Arc Length and Surface Area PREREQUISITES
1.
Recall how to integrate by using trigonometric substitution (Section 10.1).
2.
Recall how to compute the distance between two points in the plane (Section R.4).
PREREQUISITE QUIZ
1.
What is the distance between the points
2.
Evaluate
j(dx/&
(3,6)
and (4,2) 7
.
)
GOALS
1.
Be able to express the length of a curve as an integral and perform the integration, if possible.
2.
Be able to express the area of a surface of revolution as an integral and perform the integration, if possible.
STUDY HINTS
1.
Definition. The notation
ds
is introduced in this section.
infinitesimal length defined by 2.
+
Ad,)'
(dy)2
Arc length. You should become familiar with Either memorize this or learn to derive it. ddx)
2
+ (dy)'
, substitute dy
=
L
. =
It is an
See Fig. 10.3.1.
.IbA
+
[f '(x)j
Starting with
dx
.
ds =
, factor out dx , and
f '(x)dx
integrate. Since the integrand is positive, you made a mistake if your answer is negative or zero
3.
Constant with no effect. f
The length of
f
+
C
is the same as that of
. Shifting a graph along the y-axis does not change its length.
pute the length of
(x
-
L ) ~ ' on ~
[1,2]
Com-
and compare with Example 2
Copyright 1985 Springer-Verlag. All rights reserved.
444
4.
Section 10.3
Tricks to find arc length.
In general, many textbook and exam questions
are chosen for their simplicity. Thus, the expression under the radical
If
will often simplify; for example, it may be a perfect square.
it is not a perfect square, a trigonometric substitution will often be helpful.
5.
Square roots in arc length problems.
The appearance of square roots in
arc length questions may sometimes present a special problem. Wnen in doubt, take the absolute value of the expression inside the square root. For exampie, consider finding the arc length of x2I3 on [-8,-21
+
j::/(9x2/3
4)19~'/~ dx
u = 9 ~ "+ ~6
yields
=
-13x~'~jdx
jpz&
dull8 = (1/27)u
Step function derivation (pp. 480-482).
We get
and substituting
I
3
1
which is negative. The correct method is to use
6.
.
2
l
m
3/x1I3
=
.
Except in honors classes, most
instructors will not emphasize these theoretical aspects on their exams.
7.
Area of revolution. Learn ts derive fornula (2) in the box on p. 483. It may be easier to think of the infinitesimal frustums as cylinders, so the area is circumference times width. 2nf(x)
8.
and the width is simply the arc length. Thus,
2 ~ r r f ( ~6) f I ( ~ ) ] dx . a Revolution around y-axis. Here, the circumference is than
2ny
x h is that 9.
The circumference is
+
.
A
2nx
A
The width is still the arc length. Thus,
[f '(x)I2
dx
.
=
2nf(x)Ldx
=
rather than
=
The only difference with the preceding formula
f(x) appears in one and
Mathematical illusion?
=
2nr b
x
appears in the other.
The bands in Fig. 10.3.8 are equal in area.
The smaller radius is compensated for by a larger
ds
.
Copyright 1985 Springer-Verlag. All rights reserved.
Why?
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
4 f ( x ) = x 18
Given t h a t
+
2
1/4x
,
Thus t h e l e n g t h of t h e g r a p h i s
I 5.
+
3 3 1 / 4 x dx = ,f1(x 1 2
+
112
f(x)=xn,
If
+
I:/l
Thus, t h e l e n g t h i s
(0
[ ( I - 0)
=
2
- I)~]
[0,1],
-1
+ =
on
I f , m d x
Given t h a t
+ ,6+ fi + +'%. ( 2 - 0)211/2
, and
[1,2]
+
3
+
112x ) d x and
L
=
+
(x
4 (x 18
=
]
is
a
i
p
+
ll
d
+
x
on
[2,5]
.
f '(x)
=
9219
.
.
f '(x)
+
-
[(5 2
=
2)
2
+
on
Thus, t h e l e n g t h i s 1 6xI0
+
2 6xll
(1/2)(x
+
1)
=
1)li2/l
+
(1 - Z ) ~ I
On t h e o t h e r h a n d ,
= 2nli(x
2 3 II4x )
-
=
Thus, t h e a r e a of t h e s u r f a c e o b t a i - n e d by r e v o l v i n g about t h e x-axis
.
Ey d i r e c t c o m p u t a t i o n , we h a v e
, we have
1)ll2
3
3 2
j t X d o i d x l+ d- : ]
f(x)
1/2x
A + [ f ' ( x ) ~ ~ =
=
[ ( 2 - 112
-113
-
( x 1 2 - 1 / 2 x ) dx =
3
T h i s e x e r c i s e i s a n a l o g o u s t o Example 5 . L
13.
=
n- 1 fl(x)=nx
then
h. 9.
L
6
3 f l ( x ) = x I2
we h a v e
f(x)
[ ( 1 / 2 ) (x
+
+
-112
on
.
[0,2]
1 ) - l i 2 1 2dx
=
J V
+
Z r r ~ i [ ~ / ? ] = d x( i / b ) ( 4 x 17.
p f
= s i n x
u = sin x
, so rhe a r e a so
a s i n Example 3 t o g e t
21.
.
Then
A
=
A = 271 [ ( u / 2 ) K
T h i s e x e r c i s e i s a n a l o g o u s t o Example 9 . [0,1]
and
f(x)
=
+
2)dxI
=
2&1[(x
2
12)
(lI2)lnu
f(x)
idi;-Iil
=
x
on
Thus, t h e a r e a i s
A =
)
m
d
x
] = 2fin[!AX
+
2 jl(-x
2 (-x / 2
+
21:)
, 2I
on 2
=
2~%[(1/2)
dx
+
(-312
.
Let
+
.
+
(f, +
.
[1,2]
-x+2
2 ~ [ / k x m d x
i
+
We h a v e
18.7
m dx
A = 2 i j ~ / ~ 2 c o sx
is
du = c o s x dx
5 ) 3 i 2 i = ( 1 / 6 ) ( 1 3 ~-~ 5312) ~
+
+ 2)] =
Copyright 1985 Springer-Verlag. All rights reserved.
=
446
25.
Section 10.3
2
[ f '(x)~
[i/L+
(8127a2) (1
+
+
c
2 9a (x
Let
.
Note that
=
2 sec x
+
4 sec x
jbI2&4-
2n \;l2(tan
x
+
2
+
,
.
+
+
s
+
.
.
+
so
Changing
c
-.":I
+
2 9a (1
so
du =
+
-
b)/4)312
s = (1/27a2) [(4 + has no effect, since
[f '(x)]
And the area is 2 4 sec x dx
,
b)l4
s =
=
jx=' 4u1I2du/9a2 = x= 0
=
,
8
=
so the length is
4 sec x
2 9a (x
2 (8127a ) [(I
f '(x)
.
, so the length
Then
4312
2 312 9a b) 1
2 4sec x dx
2x)L
1
=
=
dropped out when we found
f '(x)
+ b)/4
u
b)14)~'~ 1;
+
- (4 +
b))312
.
2 dx = 4 du/9a
9a2b/4)312]
+
2 9a (x
=
2 9a ( x i b)/4 dx i.e.,
2 9.3 (1
33.
(3a-12)~
2 9a dxl4 ;
(1
29.
=
2 dx
=
2 dx
2 1 ~ j ~ ~ f ( x [f )~ '(x+ )]
.
2rrlbf(x)h + [f '(x)] 2dx = Ja For a small arc length where f(x) is almost constant,
Thk area of n surface of revolution is znj:f
(x)ds
.
this is approximately by the small length.
2n
multiplied by the function value multiplied
To find the area of the surface obtained by re-
volving the given curve, divide the curve into
1 mm
segments.
On
each segment, measure the distance from the x-axis to each endpoint, and
about 37.
Let
fi(1
s = 1 j0.l
- 0.1)
Thus, 41.
(a)
=
LZZZ
sin x
3
0 . 9 ~ 4Z= 1.28
>sl
.
dx
(0.1,1.0001) to (1,2)
from
Using this method, our answer is
16 cmL
s1 be the length of
Then
.
f(x)
then take the average value as
and since
.
Now
,
so
s2
be the length of
cos x
<1 ,
s1 6
1
+
4 x
1;. 1~'2dx
.
=
s2 must be longer than the line s 2
> 40.9) 2 +
(0.9999)~Z= 1.34
.
.
Cutting and unrolling the frustum as in Fig. 10.3.12, we found the area to be the x-axis.
ns(rl
+
r2)
if a linear function is revolved around
Since the formula is independent of
x
and
y ,
the
formula for the area obtained by revolving a linear segment around
Copyright 1985 Springer-Verlag. All rights reserved.
=
41.
(a)
(continued) the y-axis is also n(a
(b)
+b
-
) m ( b
ns(rl
,
a)
+
r2) , which works out to be
since
From part (a), the area is
2
a 12)
=
~ T ~ ( x ' / Z );1
A
we have
=
s = m
. m ( b 2
(
- a2)
=
.
j ; 2 m m d x
=
.
- a)
b
2.rm(b2/2 -
Since m
=
fl(x)
,
/-.--T-T 2njax 1 + [f (x)] dx . Let the entire surface be b
composed of a finite sequence of frustums of cones and use the additivity property of the integral to obtain formula (3)
SECTION QUIZ
1.
Compute the arc length of the following functions:
2.
5 x 12
+
(a)
y
(b)
y =
(c)
y = (2~+5)~"+ 3
=
on
[l,t] ,
1/30x3 on [ 1,2]
t
>1
(Hint: Substitute on
Suppose the arc length of about the arc length of
.)
=
[0,31
f
g =
u
on
f
+
[a,bl k
on
is L
.
LJat can you say
[a,b] , where
k
is a
constant? 3.
Find the area of the surfaces obtained by revolving the curves in Question 1 around the y-axis.
(Hint: Lise formulas 65 and 66 of the
integral table for part (b).)
4.
Find the area of the surfaces obtained b>- revolving tlle curves in Question 1, parts (a) and (b) , around the x-axis.
5.
Upon close inspection, you notice that your pet tarantula's legs can be described by (a)
y = 1
- t3I2
on
[O,l] .
How long are the tarantula's legs? One day, while you were walking your tarantula, a cyclone suddenl?.
appeared and twirled the spider around so fast that it seemed lihe
3
Copyright 1985 Springer-Verlag. All rights reserved.
5.
(continued) s o l i d s u r f a c e was b e i n g formed by r e v o l v i n g o n e o f t h e l e g s a r o u n d t h e y-axis. (b)
What was t h e a p p a r e n t s u r f a c e a r e a ?
ANSWERS TO PREREQUISITE QUIZ 1.
fi
ANSWERS TO SECTION Q U I Z 1.
5
+
1 / 3 0 t 3 - 7/15
(a)
t 12
(b)
(1/4)[ 6 f i - 2 6
(c)
(1000 - 46vZCiI27
+
ln((&
2.
They a r e e q u a l .
3.
(a)
6 T ( 5 t /6 - 1 / 1 0 t 2 - 1 1 / 1 5 )
(b)
( 5 1 f i - 9&)/16
(c)
2n(35000 - 2116&?)/1215
(a)
2 ? ( t 1 ° / 4 - t 130
(b)
( " / 6 ) ( 2 7 - 5&)
(a)
( 1 3 m - 8)/27
(b)
1 6 ( 8 8 4 m - 4)/1215
4.
5.
+
3)/(&
- (1/64)1n(17
+
+
+
2))1
12Ji7)/(9
+
4&))
1 / 9 0 0 t 6 - 491225)
Copyright 1985 Springer-Verlag. All rights reserved.
10.4 Parametric Curves PREREQUISITES
1.
Recall how to sketch curves described by parametric equations (Section 2.4).
2.
Recall how to compute arc lengths (Section 10.3).
PREREQUISITE QUIZ 1.
Sketch the curve described by
x = 2t
2.
Sketch the curve described by
3.
Compute the length of the graph of
4.
+
x = t2
1
and
and
y = 2x
4
y = t
+
3 on
Write a formula for the length of the graph of
.
[1,2]
y = t + 2
.
. [0,1]
y = x
3
- x
. 2
on
(Do not evaluate.)
GOALS 1.
Be able to find the tangent line of a parametric curve.
2.
Be able to relate speed to arc length for parametric curves.
3.
Be able to express the length of a parametric curve as an integral and perform the integration, if possible.
STUDY HINTS 1.
Direction of motion.
Parametric curves move in a specified direction.
This is illustrated in Example 1. 2.
Converting parametric equations. One of the simplest ways to convert a parametric equation into the form for
3.
t
y
=
f(x)
is to solve one equation
and then substitute into the other equation.
Tangent lines.
You should definitely know how to compute a tangent line
for parametric equations.
Remember that for
y = f(x) ,
the tangent
Copyright 1985 Springer-Verlag. All rights reserved.
450
Section 10.4
3.
(continued) l i n e i s g i v e n by metric and
4.
xo
by
f(tO)
ddxldt)
.
For t h e analogous paraby
+
, we m u l t i p l y and d i v i d e by
(dyldt)
Adxldt)'
+
,
g(tO)
.
2
dt
.
dt
Start-
t o get
ds =
I n t e g r a t i o n y i e l d s the a r c length formula.
Speed. I t i s p r o b a b l y b e s t t o memorize t h a t s p e e d mula
, y0
g ' ( t0) / f l ( t0 )
i s r e p l a c e d by
ds = 2
fi(xo)(x - x ) 0
The i n f i n i t e s i m a l argument i s v e r y e a s y t o f o l l o w .
Arc l e n g t h . ing with
5.
f ' (xo)
form,
+
y = yo
, which
(dyldt)'
i s g i v e n by t h e f o r -
is the a r c length integrand.
SOLUTIONS TO EVERY OTHER ODD EXERCISE From
and s o
(x
+
9)
slope
5.
x
1
= =
becomes and
t
.
cos t
. 114
=
2 = (x
+
1)/4
t
+
g(1)
x = t ,
+ x
1
1
=
m
1
.
then
,
1)/4
2 = (114)
x
The g r a p h i s a s t r a i g h t l i n e w i t h and
y-intercept
x
9/4
.
t
.
=
2x
2
+
.
A n o t h e r one i s
x = sin
tlfi
and
y
=
,
the equation
y = t 3 + 1 .
Wecanalsolet
2 x = + t
,
so
Several other solutions a r e possible.
f ( t ) = (112)t
and =
+
+
Many o t h e r s o l u t i o n s a r e p o s s i b l e .
f t
Given
t = (x
we g e t
y = i m , so a parametric representation is
y = i
Ifwelet y
13.
+
y = t
,
1
One o f t h e s i m p l e s t p a r a m e t r i c r e p r e s e n t a t i o n s i s t o l e t
y2
9.
-
x = 4t
2
+
t
g f ( t ) = (2/3)t1I3 f '(I) = 2
y
=
, and
and
.
At
y = g(t) = to = 1
g f ( l ) = 213
[ ( 2 / 3 ) / 2 ] ( x - 312)
+
1 ,
.
t2I3 ,
, we h a v e
we h a v e
fl(t) =
f ( 1 ) = 312
.
So t h e t a n g e n t l i n e h a s or
y = (113) ( x
+
312)
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.4
17.
451
The parametric form of the tangent line equation is: x = f '(to) (t - to) + 2(2
-
1 ,
f (to)
3t)(-3)
and
and
ff(l) = 6 ,
6(t- 1)+ 1
.
g'(t) = -3 g(1) = -1
and
+
y = gr(tO) (t - to)
and
g'(l) = -3 ;
.
Since
At
sin t
t
=
y+
Here,
f '(t) =
to = 1 , we have
Thus, at
1) - 1
y=-3(t-
.
g(tO)
therefore,
=
3 ,
+J(I -
f(1) =
=
the bead is at
2t)/2
COS
, Y
=
The tangent line should be horizontal when X
at these points. izontal.
21.
Since
=
j i 2 t m d t ji/ldu
.
2t
.
Let
u
=
n
u
and
cos €I)/(-2 sin 8)
.
and
=
0
=
nn
,
i.e.,
y
also
2t2
t = nn/2
If
so
t = mn
=
n/2 - 1
dy/d0
.
is
for all integers m
the length is
.
du = 4t dt
1;
+I
When
n
.
yf/x' shows that the tangent is not
s =
Then
s
j k m =
(1/2)
Integrate as in Example 3, Section 10.3, to get
dx/d8 = -2 sin 8
0 ,
2t
so there is a vertical tangent.
dy/dt = 4t3 ,
and
P--+ 1+ 2
(a)
,
and the limit of
[(u/2) u
29.
is an integer. However, x'
So vertical tangents occur when
dx/dt
t = (2n
l)n/2 , where
n
The graph should be vertical at
odd, y' = 0 , vertical.
i.e.,
+
=
From the graph, we see that the tangent is never hor-
is even, y' # 0
If n
0 ,
y'
=
=
(112) [/5 + (112) in(2
1 - cos 8
8 = n/2 ,
.
x
s = (112)
+ &)I
Then dy/dx = (1 -
dy/dx = I/(-2) = -112 ,
The tangent line is
dt =
x
[y - ( ~ / 2- l ) ] /x
= =
Copyright 1985 Springer-Verlag. All rights reserved.
x
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.4
453
SECTION QUIZ
1.
2.
x
,
Sketch t h e curve
(b)
What i s t h e t a n g e n t l i n e when
A c u r v e i s d e s c r i b e d by t
=
x
t
2
(a)
=
t5
y
+
t
=
t 3t2
3
.
=
0 ?
- t
+
7
3
y = 3 t 4 - 2 t 13
and
- 4 . What i s t h e e q u a t i o n of t h e t a n g e n t l i n e when
t
=
+
1 ?
3.
What i s t h e l e n g t h of t h e c u r v e i n Q u e s t i o n 1 f o r
4.
What i s t h e s p e e d o f a p a r t i c l e moving a c c o r d i n g t o t h e p a r a m e t r i c e q u a t i o n s g i v e n i n Q u e s t i o n 2 when
5.
-1
t = 0 ?
The human c a n n o n b a l l ' s p a t h c a n b e p a r a m e t r i c a l l y d e s c r i b e d by and
y
=
5 6 t - it2
.
?
x = 5r
She it; s h o t t h r o u g h a r i n g of f i r e a t t h e h i g h -
e s t p o i n t of h e r f l i g h t and e n d s h e r f e a t by d i v i n g i n t o a n a r r o w t u b e of w a t e r a t
y = 0
.
The cannon i s l o c a t e d a t
(0,O)
.
(a)
How f a s t i s s h e g o i n g t h r o u g h t h e r i n g of f i r e ?
(b)
A t what s p e e d and a t what a n g l e d o e s s h e e n t e r t h e w a t e r ?
ANSWERS TO PREREQUISITE Q U I Z
Copyright 1985 Springer-Verlag. All rights reserved.
454
S e c t i o n 10.4
ANSWERS TO SECTION Q U I Z
+
350
2.
30y
3.
~(13~/~-8)/27
4.
"5
5.
(a)
5
(b)
10 ; -713 r a d i a n s from g r o u n d
33x
-
=
0
Copyright 1985 Springer-Verlag. All rights reserved.
10.5 Length and Area in Polar Coordinates PREREQUISITES 1.
Recall how to compute arc lengths of parametric curves (Section 10.4).
2.
Recall how to convert from cartesian to polar coordinates (Section 5.1).
3.
Recall how to sketch graphs described in polar coordinates (Section 5.6).
4.
Recall the trigonometric half-angle formulas for sine and cosine (Section 5.1 and inside.front cover).
PREREQUISITE QUIZ 1.
Which of the following is equivalent to (a)
2 sin 8 cos 8
(b)
(1
+
cos 28)/2
(c)
(1
-
cos 28)/2
2 sin 6 ?
2.
Sketch the graph of
3.
What is the length of the curve described by for
4.
0
8
?
r = sin 0
in the xy-plane.
x
=
2 cos 4 , y
=
sin 0
(Write a formula, but don't evaluate it.)
Convert the cartesian coordinates
(3,3)
to polar coordinates.
GOALS 1.
Be able to compute the arc lengths of curves described in polar coordinates.
2.
Be able to compute the area of a region described by polar coordinates.
STUDY HINTS 1.
Arc length. Simply use
The derivation of formula (1) on p. 500 is not difficult.
x
=
r cos 8 = f(8)cos
0 and
y
=
r sin 6
=
differentiate by the product rule and substitute into
f(8)sin
L
" ,
=
Copyright 1985 Springer-Verlag. All rights reserved.
456
1.
Section 10.5
(continued) b ja/(dx/dt)2
+
2 (d~ldt) dt
. Don't forget to change
You may find it easier just to memorize
Arfa.
la,Li
.
.
e
A = (l/2)l~r2d8
You should remember that
to
L = / ~ d + 1
2.
[a,b]
.
A sketch is usually
helpful and noting symmetries will save much work.
3.
Choosing limits.
In many instances, you will have to find the limits
of integration. For arc length problems, be careful the limits are chosen so that the curve is traversed only once. of the circle not 0 to -
r = 1
47 .
should be found by integrating from
0
to
2r
,
For most problems about area, as in Example 3, you will
need to find where
4.
For example, the length
.
f(b) = 0
Area between curves.
This is done by computing the larger area and then
subtracting the smaller area.
[f
(e)]
= v%
*
Thus, the integrand is
2
-
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
We have
j~'h
dr/d8 = 3 cos 8 ,
2 cos 0
+
s(1
+
sin 8)
/ 2 T ~cos(F: - ~ - 112) d ?
'0
=
I,
27i
J18
'I2 '0
+
by the identity
the half-angle formula, this becomes 12 sin[(^ - i./2)/2]
=
L
so the length of
2
12JF
is
18;ms
de
sin 8
=
cos(P - r.12)
v%~i''fi cos [(1/2)
(0
.
By
- d2)] d: =
.
Using the formula A (1/2)j:(3 (9/2)($/2
sin 3)*d3 -
=
sin 29/4) ;1
, we get
(1/2)j:r2di
=
(9:?)1;[(1
=
-
cos 2~)/2]d!
9 ~ / 4.
Copyright 1985 Springer-Verlag. All rights reserved.
=
Section 10.5
A
The area is
2 sin 8)d8
. +
+
+
8 sin 6
Using the half-angle
formula yields
8 sin R
+
(1/2)(in(4
=
2 2 sin 8) dB = (1/2)j 0'(16
457
(1
(l/2)jin [I6
-
+
cos 28)/2] de = (112)
[336/2 - 8 cos 8 - sin 28/41
27i
x
-
lo
.
33n/2
The length is L n/2 2 j-T12dan (912)
~'-7
=
dB =
2 sec (9/2)/4 d6
+
.
The area 2
is A = (l/2)1Br2d8 = (1/2)j:/;~tan
(i/Z)di =
n/2 2 (1/2)1-~/~(sec (812) - l)d6 = (tan(812) 912) 17.
The length is The area is
A
L =
=
+
(1/4)i cos 28 21.
2 21 cos 8
(1/2)j0/~(8~+
2 . (28 sln 9
-
I - 7il4 + 1
=
J~/~/(I + con i - i sin 0)'
2 ~ ~ ~ ~ ( R2 +9 F ~cos ~ 0 24/2)d81 s 3 (112) [n 116
1/j2
+ 48
.
+
2 2 3 8 cos d)di = (112) [(i 13)
cos O - 4 sin
(114) sin lo)]
=
e +
3 (112) [n 116 cos E
=
2 cos 8
6 ~ / 6 =. (2 (112) [j;l63
+
+
+
8 = n/6 ,
when
+ cos2R db 716 2 3 sin 8 do = n/2 - 1116+
.
&)n/6
sin2B di
(112) [(3/2)joi6(l
+
A
=
jT16cos2d di]
=
The area is n/2
- cos 2E)d0
=
.
2 8
=
A
+
2 n 12 - 4 - 1/81
6 sin e
the length is L
j:'6h
2 . (1/4)9 sln 20
dB
10'~+
Integrate each term by parts to get
Since
n/2 jnI6(l
+
- ill4 = 2 - -12 . + 2 cos R + cos28)
02 (1
+
(112)
+
x
+ (c + (l/Z) 1 cos 2a)del = (114) [(so - (312)sin 28) n/2 sir. 28) = (114) [n/2 - 3 6 / 4 + n/2 - n/6 - 6/41 = (1/4)(5i;/6 +
6) .
Copyright 1985 Springer-Verlag. All rights reserved.
.
458
Section 10.5
25.
Substituting into the formula L = j B j f ( 0 )
+
2 di if '(13)l
,
we get
SECTION QUIZ
1.
(a)
Find the area bounded by the curve described by and the rays
(b)
8 = 271 and
8 = 377
r = 8 cos 8
.
Write the length of the curve as an integral, but do not evaluate it.
2.
Find the area and the perimeter of the region enclosed by each of the following:
3.
+
(a)
r = sin 0
(b)
r = 12 cos 8 /
cos 8
The region inside both 2 A = (112) [jbacos (8
+
r = sin R
n/3)d8
+ J:
and
r = cos(8
sin2? d91
.
+
n/3)
has area
Find the limits of inte-
gration and compute the area.
4.
An astropirate on the surface of Mars was about to accelerate into hyperspace when he noticed that the space patrol seemed confused and was flying circles around him. relesing a fence described by (a)
In reality, the space patrol was r
=
1
and
r = 1
+
cos 0
.
1 , how much fence material was used?
If the fence has height
(Assume that the space patrol may penetrate their own fences.) (b)
The astropirate was caught in one of theregionsinside outside
r = 1
+
cos A
.
r = 1
and
How much material was needed to make the
top and bottom of the space prison?
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.5
459
ANSWERS T O PREREQUISITE QUIZ
ANSWERS TO SECTION QUIZ 1.
2.
(a)
1 9 ~ ~ 1 1+2n/8
(b)
1::h2
(a)
Area =
(b)
Area = n/2 ; perimeter
+
;
T
3.
a = n/12 ; b
4.
(a)
8+2n
(b)
n/2
+
2 cos I -
=
e
s i n 28 dc
perimeter = 2fi11
7716 ;
=
c = 0 ;
2n d = n/12 ;
area = n/l2 - 114
4
Copyright 1985 Springer-Verlag. All rights reserved.
460
S e c t i o n 10.R
10.R Review E x e r c i s e s f o r C h a p t e r 10SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Substitute
u
,
sin x
=
so
du
,
c o s x dx
=
2 j 3 s i n x c o s x dx
and
=
2 3 3 J'3udu=u + ~ = s i n x + C .
/-'I
.
j(1
Let
2
.
- c o s 6 ) s i n 8 d3
we g e t
-f(l
2
- u )du
x
Now, l e t
=
+
-u
u /3
9.
13.
Let 2 (X
17.
u = x 2 + l ,
+
2
1) Idx
=
j[(u
- l)du/2u 1
+ c
= (1/2) [ l n l x
Let
u
&,
=
-2JG c o s Using
vG +
so
.
2 sin&
2
+
u2
2
+ =
x
+
11
and
+
tan x
+
C
C
.
+
2 s e c 6 do
x
+
211
and
Using t h e
+
2u du = dx
Thus
2
- l / u )du
c
.
and
J' ldu/!u2 +
=
d u = 2 x d x .
111
,
x 2 + x + 2 = ( x +
=
= (1/2)j(l/u
1/(x2
and
du
I n t e g r a t e by p a r t s t o g e t
sin x = 1
lnlsec x
~ h i ~ s ,/, [ d x / ( x
x 2 = u - 1 ,
so
so 2
and
6-7 + C .
tan 0
,
dB
-du = s i n 3 do
- cos
C = c o s 913
d%%
2
l/uI
21u s i n u du
21.
so
so
3
- x2)3'2 -
(1
dx = c o s
,
cos @
=
u = x + 1 / 2 ,
.
sec 8
+
so
Let
J7/4 =
,
sin B
u
3
f i g u r e , t h e a n s w e r becomes
/?+ 714
=
=
714)l
=
l[x3/
(112) [ l n l u l
+
. Then
-2u c o s u
jsin&dx
+
2 sin u
=
+
C
=
+c . 2
- c o s x , we g e t - sin x
+
C
.
2
j ( s i n x/cos x)dx = I ( s e c x
The t e c h n i q u e f o r i n t e g r a t i n g
- c o s x)dx sec x
shown i n Example 6 ( b ) , S e c t i o n 1 0 . 1 .
Copyright 1985 Springer-Verlag. All rights reserved.
is
=
S e c t i o n 10.R
25.
Let
3
I=j[x/(x
-9)ldx
andlet
2
( x 2 + a x + a ) ) l d x = j [ ~ / ( x- a )
a(aB
-
1 ;
and
.
D)
.
+
.
aB - D = 0
+
Thus,
aB = 1
Substitute
, so B
C
.
= 1/3a
+
+
(afi)]] l + C = (1/3a) [ l n i x - a / - l n L 2
29.
+
/(l
ex)-'dx
tuting 33.
j [ l - ex(ex
=
+
u = ex
+
1 1)- ] d x
+
+
C
& ,
Let
u =
41.
Let
u = x2 + 3
45.
Let
u
Let
u 2
aB
=
Therefore, (-x/a
+
1)/(x2
-1/3a
=
+
ax
a2
+6
+
x - ln(ex
1)
+ ax +
C
+
and
D
=
2
a )]dx
+
1 tan- ((2xIa
+
=
a ( B - C)
into
C
D
=
a/2)/
+
I)/
, by s u b s t i -
+
s i n ( x - y)]
s i n x)dx
=
t o get
- ( l / l O ) c o s 5x - ( 1 / 2 ) x
+
so
,
du = ( l / 2 & ) d x
.
Then
(eh/&)dx
so
.
Then
j[x/(x2
du = 2x dx
C = (1/2)ln(x2
( l n 3x)
+5
, so
+
3)
+
+
C = ln&
du = ( 3 1 3 x ) d x = d x / x
.
+
=
2 j e u d u = 2eu
3)ldx
=
(1/2)j(du/u) =
C = (1/2)in(x Then
+
2
+
[ ( l n 3x
3)
+
+
=
, so
cos 9
cos e ) ] de =
1
-jl
[l/(l
-du = s i n 8 dB
+
u
+
2
u )I du
.
.
Then
jg2n [ s i n
0/(1
+
cos
C
5 )a /
.
186.1 49.
=
D
+
a ( B - C)
C = 0 ;
and
, where
.
37.
(1/2)ln/uj
+
=
+
y)
j s i n 3x c o s 2x dx = ( 1 / 2 ) / ( s i n 5 x cos x
+
2 a x + a )]dx
1
s i n x c o s y = (112) [ s i n ( x
Use
+
I=/[x/((x-a)x
2 a ) - (3a)(~/a&)tan-' [2(x
+
ax
B
-B
=
I = (1/3)j[(l/a)/(x - a)
( 1 / 3 a ) i l n / x - a / - (112) [ l n ( x 2
Then
(Cx + D ) / ( x '
E q u a t i n g c o e f f i c i e n t s , we g e t
D = 1 = a(2B)
113
3fi.
a =
461
:+
S i n c e t h e l i m i t s of i n t e g r a t i o n a r e
b o t h 1 , t h e1985 i n t e g r aSpringer-Verlag. l is 0 . Copyright All rights reserved.
.
462
S e c t i o n 10.R
b /-4 +
L =
2 [ f l ( x ) l dx
,
53.
Use
57.
S i n c e t h e graph i s revolved around t h e y-axi*,
+
2rjb x h 2nj;:O
xJi
I2
[ i t (x)
+
.
dx
( i / x i n 10)
where
Here, 2
dx
f x
A
10) ) ~ n + x
(112(ln
+
51100
L2 + i / ( l n
61.
Solving f o r into
y
.
,
x = 1 i.e.,
,
+
we g e t
t o get
y
213
=
2x13
and
y
2 3t /4t3
In t h i s case, it is
=
The l e n g t h
.
y
=
2
3x14
+
514
and
L
+
.
r2
+
.
R&
Let
The a r e a A =
+
A
TI2 4 (1/2)j0 i dt
The l e n g t h
L
2
I. = r
0'
2
2
+4
,
6 (3
i s g i v e n by =
5 (8 110)
3 cos ( ? / 4 ) s i n ( t / 4 ) 2
,
9 c o s (814) [ c o s ( 0 1 4 )
so
+
+
L r2 2
4)
+
+
i / ( i n 10)
2
2 l / ( i n 10) ) /
.
1
.
Substitute
This i s a l i n e 1
.
dyldx = (dy/dt)/
t = 1
,
,
d y l d x = 314
- 2 ) / ( x - 1)
(y
+
di
=
314
,
and
L = (112)
;:&iF-+[ r '(.)I
'd:
=
(1/2),(: 5
i
(r')'
1320
x
In t h i s case,
. 8
=
In this care,
becomes
.
=
.
c'
and
2 [ r ( t ? ) ] dH
A
=
,
2
)
du = 2P d3 ,
so
lbI2
i s g i v e n by
3
6
=
=
2
.
46'
u
+
y-intercept
When
=
~ntegrate
l / ( l n 10)
t = x/3
Thus, the t a n g e n t l i n e i s
i s g i v e n by
4 ( r ~ =) 8~
.
+
+
A0000
A
=
A =
= 2n [ 5 O h 0 0 0 0
J '"
73.
2
For p a r a m e t r i c e q u a t i o n s , t h e s l o p e of t h e g r a p h i s (dxldt)
69.
:0
t
with slope
65.
1
so
l / ( l n 10) 2 dx
2 ( l l 2 ( l n 10) ) l n [ ( 1 0 0
+
l / ( l n 10)'
10)
2
,
10
2n[(x/2)/x
=
L
Thus,
use t h e equation 1
+
271 I o 0 J x 2 10
=
a s i n Example 3 , S e c t i o n 1 0 . 3 t o g e t 2
x
=
.
- l/4x2
f l ( x ) = x2
9 cos ( r / 4 ) 6
s i n ( 8 / 4 ) 1 = 9 c o s (314)
=
+
.
Here,
r
6
9 cos ("/4)sin 2
9 cos (0/4)(1 -
Copyright 1985 Springer-Verlag. All rights reserved.
=
2
(-14) =
S e c t i o n 10.R
73.
(continued) 2 sin (914))~
.
Then
u = sin(@/&),
The a r e a
:j
+
+
128)Jn(1 0
+
128)(3
+
914
3 sin 8
1:
+
= 9n(1/8
This exercise uses the formulas s i n ( x - y)]
and
- ( 1 / 2 ~ )[ ( 1 / ( 3
+
unless
.
m = 3 =
1
.
31x1
lo
=
m)) s i n ( 3
s i n (812))
+
914
20)dB
+
(1
9718
+
11256)
(112) [ c o s ( x
+
+ m)x +
y)
+
+
4 C O S ( ~ / Z+)
+ +
+
cosa) /4]de
1; +
914 =
+
=
(91
+
914 - 914
914
(91
9 ~ 1 1 2 8+
+
y)
+
.
a
=
c o s ( 3 - m)xldx
( 1 / ( 3 - m ) ) s i n ( 3 - m)xl .l o2 n
hm = ( 1 1 n ) ~ i ~ c o3x s s i n mx dx
+
3))cos(m
+
3)x
+
=
.
3 1 5 1 ~ 1 2 5 6+ 914
cos(x - y)]
+ m)x +
1: +
2
8 s i n (0/2)]
27~132 =
A = (112) x
8 sin(0/2)1
s i n x cos y = (1/2)[sin(x
=
-(1/27) [(l/(rn
=
Thus,
3
-
8 sin(5/2)
11128
=
=
=
0
a 3 = ( 1 1 2 n ) ~ ~ ~ ( c6ox s + 1 ) d x = ( l / Z n ) ( ( l / 6 ) s i n 6 x
Then
Also,
sin(m - 3)xldx 2n
cos x cos y
2
2n cos 3x c o s mx dx = ( 1 1 2 r ) J ' ~ [ ( C O S ( 3
(l/i:)j;IT
x)lin
+
+ +
10
c o s (8/2)1dA = ( 9 1 3 2 )
+ c o s 8)dA = 9 ~ 1 3 2+ 0 ) 1; + ( 9 / 2 5 6 ) j ; ( l + c o s
s i n 2812)
.
(9/32)j: [ i
=
4
-
2
- u )du
.
4 C O S ( ~ / ~do116 ))
2
2 cos 8
+
5fi
Let
12'fi12(1 10
=
2
4 cos(8/2)(1
+
( 9 1 3 2 ) [38
2 sin
(9/256)(8
+
cos 5)/2
=
L
A = ( 1 1 2 ) ~ : [ r ( B ) ] d8
+
+
4 c o s (812)
+
(9/32)j:[6(1 97/32
3
+
6 cos (e/2)
- hiiz)
i s g i v e n by
A
, and
du = ( 1 / 4 ) c o s ( 8 / 4 ) d B = 12(fi12
.
2
- c o s ( 0 / 4 ) s i n ( 8 / 4 ) ] d8
3j;[cos(8/4)
=
8 9 c o s ( e / i ) d e = (9/2)j:(l 2
81.
L
so
- u313) :I2
12(,
77.
463
(l/2n)iET[sin(m
+
+
3)x
(l/(rn - 3))cos(m -
.
0
We u s e t h e p r o d u c t f o r m u l a s l i s t e d on p . 460 and t h e h a l f - a n g l e f o r m u l a 2 sin x
2 a = ( l / n ) j i n s i n x c o s mx d x = ( 1 1 2 ~ )x m 2n ~ ~ ' [ c omx s - c o s 2x c o s m l d x = ( 1 / 2 n ) j 0 [ c o s mx - ( 1 / 2 ) ( c o s ( 2 + m)x cos(2
=
-
(1
-
m)x)] dx = ( 1 1 4 ~ [) ( 2 / m ) s i n mx
m ) ) s i n ( 2 - m)x] 2T
jO
.
cos 2x)/2
[COS 4x
+
( l / 2 7 r ) j i n l dx
l o2 ,r
lldx
+
+
,
0
=
unless
-
m = 0
(1/(2 or
0 = - ( 1 / 4 ~ ) ( ( 1 / 4 ) s i n 4x
0 = 1
.
Also,
bm
=
+
+ m ) ) s i n ( 2 + m)x - ( 1 / ( 2 m = 2 . Then a 2 = - ( 1 / 4 ~ ) . + x ) I 02*I
=
-112 ;
2 2 ( ~ / T ) J ' ~s i' n x s i n ( m x ) d x
and =
a.
(l/27)
= x
Copyright 1985 Springer-Verlag. All rights reserved.
+
464 Section 10.R
81.
(continued)
85.
2cos 2x sinim)] dx = (112n)j0 [sin(mx) - (112) (iin(2
-
[sin(m)
2 2 ( 4 1 ~'0 ) [Om(-2 sin 0m sin 8
2
- 2 sin
=
(Llr)!>(l
+
2 2 sin m )-'I2dc
2 ~in~@~)-''~dl
=
J,(Llsin $m)(l
2 - sin fi)-li2di = ( 2v2/r)j0 r m ' [l/(sin 0 cos i)] d: m
corn
tiate c
sin
@ =
sin
@ =
@
m
sin B
to get
0 ,
cos
,
=
.
sin @ cos B d? -1 m
89.
I
Substituting the~givenformulas yields
.
B = 0
m 2 = sin
here k
When
and when
@
=
to get
=
( ~ / T ~ ~ ) j ~ ' 2 dr> [ .l / ~ ~
(a)
n+ 1 a )/(n
+
b - a (b)
+
1) , unless
A
n
=
.
-1
The iength of this graph is
I.
=
/:
dx
=
J?(h
then the length is b
~:m dx = jlfi
b ja(l +
=
.
ln(b/a)
.I
=
If
n
(2k
J:A
2)
and
+ 1 i2k
dx
-
=
-1 ,
A
=
h
-
- a)
.
=
a
. If n If n
L
=
jb/l +
.
k
is a nonnegative integer, we have
( 9 1 4 ) ~, we get
Since
2)12 x"(~+')
dx
=
cos f
$
(bn+'
+
lb(l 'a
. If
L = (S/Zi)(l
+
-
xW1)dx
n
0 ,
=
L
=
=
L
=
. Vow, jet u
=
=
(2k
+
2 Ln-2 n x
ll(l-n)du/~(217 - 2 ) x
1
.
Then
=
g x j ~ ) ~ =' ~ ~
For the more genpiai case where n
+
Differen-
( 9 1 4 ) ~dx
2
(3-2n)/(2n-2)
+
I,
x
.
x
3)/(2k
sin
2 , then
=
+
? = 712 ;
l , then
=
+
=
. .
so
b - a
=
r lbL + n2x2n-2 dx
312 , we have
=
x = n
SO
(U
+
1
I
Substituting u 1 / 2 4 + 9
n
xn)dx
+
Ja
n = 2 case is Example 3 of Section 10.3 For the case
=
Substitute all this and cancel
The area beneath the graph is
- L
, we get dJ
, sin 6
0
=
(n
m)x
(26/i)
cos G dE
sin Q
@ dq =
=
$
+
1+n2b2n-L Id =/l+n2a2n-2[
T
l/(l-n)d,, ,
Copyright 1985 Springer-Verlag. All rights reserved.
1)/
465
Section 10.R
89.
(b) (continued) k
(2n - 2)(u
1 .
- 1)
(u - l)k
Recall that the binomial expansion for
k li=O[r]ui
is
)
1 /(i +
312)
.
Vx
4b(1
+
Around the x-axis,
(c)
n+ 1 a )/(n
+
n
.
-112
=
2 n
+
1)
x-')dx
If n
n [b - a
=
+ xn+l)dx
2n;:(x
Around the x-axis, Ax :12
x
n
=
2 2n-2 i+3/2 [(l+nb 1 k (Note: recall that ( . ) = k!/
2xn
-1
=
-
I
+
4&
=
7,
Vx
-
+
4&
[h2 - a2
.
+
+
=
x
Note that
2n
)dx = nib -
2
l
If
ax
=
2nj:
=
0
,
n
xnk=
then
a
n+ I
=
2x-I
+ xW2)dx
=
n[b - a
+
Vx = r!~:(l
-1
-
or
+ +
2x-'"
Around the y-axis,
w dx i
the arc length which was analyzed in part (b). iook at
2(b
n
n+2 - a )/(n
)
+
unless
.
n
i
,
-112 ,
ln(b/a)l 2(bn+'
x
1)l
+
= r!~;(l
. If n
2ilb(l
=
m dx
,
+
+
(bznf1 - aZn+l)/(2n
a - a
b
(nl/(l-n) /(2n-2))~
=
k-i i+1/2 - 1 u du = (nl/(l-n)/ (2n-2))x
2 2n-2 it312
(1 + n a
.I
and this gives us
[]
k
+
=
V Y unless n
2)l
2
i
i
~ dx ~is
b
= =
m dx
+
217 X times
Thus, we only need to
dx. =
a)
.
If n
b 2 J 2 2n13(x 1 + 4x )dx
.
The integral
2.;/:
dx
=
2r(b
-
=
1
,
then a
=
a
If n
=
2 , then a_
~b[ix~/J;+ 4x2)dx. u
=
x3
, dv
=
=
The latter integral is integrated by parts with
4 x / m - d x , du = 3x2 dx , and
v
=
. Therefore,
Copyright 1985 Springer-Verlag. All rights reserved.
466
Section 10.R
89.
(d) (continued) ~
:
(
~
~
m
b
= )ja(x d
aeariangement yields
2
2 /~J 1 + 4x )dx + [x31;;1;11:
41: ( x 2 m ) d x = b
3
.
in Example h(b),
(112) tan
=
a
so
dx
2 lnjsec 0
+
tan
-1
011
.
/E:z-l;:
Now if
and similarly, if tan 13
tan 6
m+ 'a/)
/
n = (2k
+
. Finally, we get a
+
3)/(2k
2) where
let x = ((u - l)/n ) 1/(2n-2) (u - 1)
for (4k
(3-2n)/ (2n-2)
n
+
du
l)n (n+l) / (n- 1)
-
[n/(n
in
-
2
.
=
u
-
1
k
A
k , i=O i (-l)k-iui+3/2 /(i
get
a
=
k
=
+
-
+
312)
+C .
a
+
m
2)n l/(n-1))
ax = 2 n j ; x n X d x
=
Substitute
+
4/(2k + 1)
-
2)
x=b k & du - ~)]j~=~(u-l)
(u - lIk = Zi=,, JE:=~[:]
.
1))/(2
[nn(n+l)'(l-n)/(n
=
2
is any nonnegative integer,
dx = 1/[ ((2n
I
2bl -
( h z+
+
Then
=
(r/2)[ (1/16) in]
2) = ((3 - 1 - 2)/(2k
.
-
. This
=
-
=
tan 81
2b, then sec 0
=
u du
a
+
lnjsec 0
x=a
Hence
u
+
@
1
. Substitute this into jx-b (3-n)/(2n-2) (U -
(3 - n)/(2n
2)/4 = k
so
From the binomial expansion formula, Ler
+
=
2a, then sec 0
=
substitution gives us (1116) ( 2 b h x - ln
If
m
Section 10.1 and Example 3, Section 10.3, respectively.
Therefore, , / ~ ( r 2 / ~ ) d =x (L/I6)[scc 0 tan
n
3
2 (112) sec Ode -1 2 3 tan e sec e do = (1/8)jE:z-1::(sec @ 3 The integration of sec 0 and sec e are shown in detail
x
metric substitution. Let -1 This yields (118)j::z-1;;
1
x - a
A
. The latter integral can be done with a trigono-
1b(x'/J1~)dx
sec 8:de
.
- 3]i(x2Az)dx]
r]
i (-l)k-iui
(-i)k-iui+1/2
Substitute for
u
.
du = and
k
to
nn (n+l) 1 (1-n)
/ 2 211-2 i+3/2 2 211-2 i+3/2 ((l+nb ) (l+na ) )/(i+3/2).
Copyright 1985 Springer-Verlag. All rights reserved.
.
=
S e c t i o n 10.R
89.
(d) (continued) Around t h e y - a x i s ,
A" = 2nj:
A = 2n~:xdx=r(b
then
1 + 4x2
yields
( n / 6 ) [ (1
+
.
a x dx
A
Y
,
Then
(1
3
b
A Y
2
K
n = (k
If
+
(2 - n ) / ( n
then k .
Let
1) = k
If
,
If
n = 2
--
n = 1
,
A = 2njb x y 2 " = (n/3)~::2
3)/(k
1)
+
2)
= (2
m dx
-
1 - l/(k
(u
-
+
la
Let
+
+
l))/(l/(k
.
i s any i n t e g e r , t h e n f o r
,
and
sin(n - a) = -sin(n
about t h e x-axis. sin u
Since
the point
Also, since
+
a)
=
.
n
1 - I)/] =
and
A
=
-
(u A
/ (n-1)-1
1)
k
=
Y
/ ( i +3/2)1
x
.
but notice that i f
.
2
Y Since (2 - n)/
.
du
P l o t t h e curves point-by-point,
(-1,O)
+
1 ) ) = (k
Then
-
2 2 n - 2 i + 3 / 2 - ( 1 + n a2 2n-2 ) i + 3 / 2 1 [ ( l + n b )
is
a
i s any n o n n e g a t i v e i n t e g e r ,
(2-n)
, P
=
3b2
- 3
-)I
/ (2n - 2 ) ) du
t = 0
du
1 3a2
this into the integral t o get
1
t = n
=
+
l ) k w i t h t h e binomial formula:
[l] ( - ~ ) ~ - ~ u lS .u b s t i t u t e
=
u = 3x2 s o
(l(2)lnlu
n[ n
(a)
u
2 312 b 4x )
x = ( u - 1 ) 1/(2n-2)nl/(l-n)
(3-2n)/ (211-2)
expand
k
. + +
M+1)/(3a2
, s o
=
Y
I n t e g r a t e a s i n Example 3 ,
du
where
-
A
then s u b s t i t u t i o n of
A = (n/3)[ (u/2)& Y
+ I n [ (3b2 + -
, then
,
n = 0
If
I.
4a2)3'2
2 2n-2 l+n b - I)]! 1+~2,2n-2 ( U
[ nn 2'(1-n)/(n
'i=o
+
2 2n-2 u = l + n x
dx = ( n l / (1-n) ( u
-
.
.
.
m dx
then
Section 10.3 t o get (n:6) (
x
2 - a )
2 n f b x h -t 4 x 2 dx = ( n / 6 ) ( l
=
-
4b ) 3 / 2
n = 3
If
(n
2
Y
b 2 2 2 n f a f i x dx = n f i ( b - a )
93.
467
cos(n
for a l l a
P
-
is
m
=
1
and ;
(1,O)
a ) = cos(n
+
n for
a)
, t h e c u r v e s a r e symmetric
c o s ( n - a) = - c o s u
and
sin(^
-
a) =
, t h e c u r v e s a r e a l s o s y m m e t r i c a b o u t t h e y - a x i s and t h e r e -
Copyright 1985 Springer-Verlag. All rights reserved.
468 Section 10.R
93.
(a)
(continued) fore, the origin. the x-axis at
n
times, but as
t
Finally, since
+
1 points.
goes from n
yinln)
=
(Actually, each curve crosses to
2n
2n
+
2
the curve will recross the
x-axis at t:-.e same points it crossed when
(b)
0 , each curve crosses
Consequently, each curve will consist of
t
n
went from
loops, for
0 to
n
r: .)
odd or
even
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.R
93.
469
(d)
TEST FOR CHAPTER 10
1.
True or false. (a)
Every polynomial is a product of linear and/or quadratic factors
(b)
The area between on the interval
(c)
,
is
r
(112);:
=
2
+
(Cx
>0
on
+ II)/x2
, where
g(8)
[f (9) - g(d)]
f(3) 2
l/(x - llLxL
for some constants
> g(5)
. =
A , B , C
A/(x - 1) +
,
and
D
(d)
[a,b) , the surface area obtained by revolution lb 2 around the x-axis is 21Ja&]2+ [fl(x)f (x)] dx .
(e)
In polar coordinates, the arc length of
If
The line lution.
3.
[a,:]
and
Using the method of partial fractions, B/(x - 1)
2.
r = f(0)
f(x)
y = -x
+
1
on
[0,31
r = f(8)
.
over the interval
is revolved to form a surface of rev,)-
Find the surface area if the line is revolved around:
(a)
the y-axis
(b)
the x-axis
Find the arc length of the curves described by the following equations on the given intervals:
2 sin 9
(b)
r
(c)
y = t2+3
=
in polar coordinates for and
x=2t+l
for
0
2;;
O
Copyright 1985 Springer-Verlag. All rights reserved.
470
4.
Section 10.R
Evaluate the following integrals: 1;[(x3
(b)
j[dt/t(t2
5.
Show that
6.
Suppose b
1)l
and
=
2 ~ i ~ / i~ dO ~ o = nn/4 s
.
y2 = f(x) + 3
If
for all integers n 2 0
f (x)
0 on
(b)
The surface areas of revolution obtained by revolving [a,h]
yl
and
y2
on
[a,b]
The surface areas of revolution obtained by revolving [a,b]
f(t) f (t)
-
1) t
f(t)
=
l/(t
=
2 3 sin (t/2)cos (t/2)
and
Y2
Y1
and
Y2
for the given interval:
on the interval
2
on the interval
Find the area of the region inside cos 2a
Y1
around the y-axis.
2
r
=
.
0 G t G 712
sin B cos
e
.
and outside
r
=
.
Find the area between the following curves and the x-axis in the given intervals:
10.
(a)
y
=
4 sin x
(b)
y
=
x sin x cos x
on 2
10, n/81 on
1-14> 7/21
Dragster Debbie, the 80-year-old grandmother, needed some excitement in her life.
She decided to take her Ferrari for a little spin down at the
speedway.
As she floors the gas petal, her position can be described
parametrically by
x = 2 cos t
(a)
What is her speed at time
(h)
At
t
=
8n
,
and
y
=
4 sin t cos t
.
to ?
she realized she had to hurry to her karate lessons,
so she sped off along the tangent line.
Describe the tangent line
with a set of parametric equations
Copyright 1985 Springer-Verlag. All rights reserved.
.
where
around the x-axis.
Compute the average value of
(b)
,
.
The arc lengths of
(a)
[a,b]
, then which of the following are equal, if any?
(a)
on
9.
+
igni2sin28 di
>a >0
(c)
8.
- t
yl = f(x)
on
7.
2 2 7x - 8)/(x - 4)ldx
+
(a)
S e c t i o n 10.R
471
ANSWERS TO CHAPTER TEST 1.
(a)
True
is)
False;
it i s
(1/2)jEi[f(e)]
+ D)/x~
should be
2
.
- [ g ( e ) ] Id:
+
(d)
True
(e)
F a l s e ; t h e given formula i s f o r Cartesian coordinates
(a)
9nfi
(b)
57ifi
(a)
1512
-
(b)
l
l
3 In 3 - 4 In 2 t
/
G
/
+
(l/fi)tan-I [(2t - 1)/3]
) % / 2 s i n 2 ~ = j ( l / 2 - c o s 2812)d. :
~
D/x
.
False;
~
C/x
2
(c)
n
(Cx
2
2
0/ = ~/ ( 1 /c2
+~ c o~s
+
C
2f3/4)l o11712 - ni/4 ;
= (t?/2 - s i n
2 8 / 2 ) d 8 = (012
+
s i n 2614)
l:n12
= n ~ / 4.
( a ) and ( c ) (a)
ln(314)
(b)
7fi130n
7i
+
112
- 415
(b)
+ ( 3 n 6+ 2 0 a -
(a)
2jsin2t0
(b)
x = 2
(a)
3n164 - a 1 8
+
and
1/32 24n)/144
4 4 s i n to
+
4 4 cos t o - 2
COS
2
2
t s i n to]
112
0
y=4(t-an)
Copyright 1985 Springer-Verlag. All rights reserved.
CHAPTER 11
LIMITS, L'HOPITALI s RULE, AND NUMERICAL METHE
11.1 Limits of Functions PREREQUISITES 1.
Recall the basic properties of limits (Section 1.2).
2.
Recall the concept of horizontal and vertical asymptotes (Sections 1.2 and 3.4) .
3.
Recall how to compute limits at infinity or infinite limits (Section 1.2).
PREREQUISITE QUIZ
1.
Complete the following statements (assuming all limits are finite):
(a) (h) (c)
1im 1im x-tx f (x) + x+x g(x) 0 1im f ' i m x x o 3 g = 1 0 l im xca[f (x)/g(x)] =
x
+
2.
bhat is :z(l/x)
3.
1im Does x , [ ( ~
4.
1im = XiXo
+
?
l)/x
Sketch the graph of
1im
/ xi,g(~)
.
provided
1im x-g(x)
$- 0
Explain your answer. 2
1
exist?
If yes, what is it?
2
2 y = x /(1 - x )
.
Discuss the asymptotes.
GOALS 1.
Be able to state the
E
- 3
definition of the limit and apply it to prove
basic general properties of limits.
Copyright 1985 Springer-Verlag. All rights reserved.
474 Section 11.1
2.
Be able to evaluate simple specific limits.
3.
Be able to compute limits at infinity.
4.
Be able to compute one-sided limits.
STUDY HINTS 1.
- 6 definition of the limit is not useful for
Limit definition.
The
computing limits.
It is, however, very important in theoretical work.
E
Example 1 demonstrates how the definition is used.
tween
6
Ix
-
E
so that
Remember that
E
is given and your job is to find a and
E
f(x) - 1
This will permit us to find a relationship be-
and
definition. Both 2.
.
x - xo
in terns of
We write
6
xo/ < 6 ensures that
/ f (x) - k /
< E
.
6 which fits the
are positive numbers.
Basic properties. The box on p. 511 was already presented in Chapter 1 Again, you are reminded that "common sense" should help you recall all of the properties.
3.
Important methods. computations. x - xo
.
Example 3 shows two important tricks used in limit
If the denominator at
4.
a3 - b
3
=
a2 - b2 = (a - b)(a
(a - b)(a
2
+
2ab
+b
2
)
+ b)
to eliminate the radical.
may be used in some situations.
Limits at infinity. To compute limits, dividenumerator and denominator by the largest degree of 1im X"
at
5.
is zero, try to factor out
A second trick is rationalization. If a radical appears in
the denominator, use Also
xo
. A +-.
(l/x) = 0 -m
or
x
in the numerator.
Then use the fact that
horizontal asymptote, if it exists, is simply the limit
Infinite limits. These have already been discussed when vertical asymptotes were introduced in Chapter 3.
Copyright 1985 Springer-Verlag. All rights reserved.
6.
Reciprocal test for infinite limits. If the limit of the reciprocal is zero, then the limit of the original function may be neither. Which occurs depends on how the sign is positive on one side of
f(x)
xO
+ m ,
behaves about
--,
or
.
If
xo
and negative on the other side,
then no limit exists. 7.
One-sided limits. The definition is just
4
------
I
\
like that for ordinary limits except that (xO,xO + ;)
we now consider the intervals (xO - 6,x ) 0
or
5)
I
.
rather than
(x0 - 6,xo +
The concept of one-sided limits is
important when the value of a function
changes from
to
- m
This is also important when absolute values
+ m .
are involved. The figure demonstrates that one-sided limits may exist even though the (two-sided) limit does not exist at 8.
.
xo
Thinking of this as the "sandwich principlew makes the
Comparison test.
test simple to understand. You should understand why the nickname is appropriate.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Let
c
>0
be given.
.
]x - a1 < 6
if
Ix - a1 < 6
,
We note that
we have
2jajlx - a ] < d2
We have to find
+
>0
x2 - a2 = (x - a)2
2 2 / x - a / = /(x - a)2
21a]6
.
such that
Thus, we want
6
+
2a(x
+
2 1x2 - a
.
2a(x - a)
-
to be such that
"
6
This choice (or a smaller one) guarantees that
2
+
21aI6
< (1 +
2ja/)6
such that 6 <1
E
,
and so
(so
If
2 6 +
choose
6
<E
a)/ G Ix - a 1 2 +
2 / a1 5
S
< E . We may G ~ / ( 1+ 2]a/) .
6
L 6)
jx2 - a 2 / < ~if
and
Ix-al<'.
Copyright 1985 Springer-Verlag. All rights reserved.
5.
1im
6f0 exy ( ( 3 tan 0 1 8 1 = exp :$[3
9.
A
13. We need to find \£(XI - k /
have
lim 2 x,2 [ ( x
1im
-
2 ) t x - 3 ) l = x,2[(x
=
Since : z ( l l x )
1im - 2 x-(l)
lim x,(l/x)
1im x-
[(3
+
+
2/x
+
= 0
2
x /x
Divide through by
=
-
4 ) / (x2 - 5x
=
+
2)/(x
-
2
2
=
+
i l x 2 - 2) = 3
x > A
I
=
lim x-[(3x2
2
+
+
.
7/x )]
2x
0)
=
( 1 , ~ ) + 5[:2(l/x)]
+ 4)/(5x2 + x + (3
+
0
+
7)] =
i:(~/~)
0)/(5
+
+
0
lim(J,T -
Upon rationalizing, we have
lim
(Q'Tl?+x)l
.
= 0
= x-~(x
2 + a2
- x
Geometrically, we can look at
x-im
-
x 211
as the differ-
ence between the hypotenuse and one leg of a right triangle. As
a
gets larger and larger with
x
fixed, the difference is getting smaller
and smaller. 29.
We have
1im
x + 2 ( ~- 2 ) 2 = 0
and
l/(x
-
1im 2 the reciprocal test, x + 2 [ l / ( x - 2) 1 33.
37.
2
however, x
-
Note that and
x
1x1 > 0
x+[(x3 - )
212 =
>0
for all
2
>0
and
x
for
x > 0
as
/ = xy0"f(x3
>2 . x -*
- 1)
x
.
Thus. by
+-.
lim yields x _ > 2 + [ ( ~ + 2)1(x - 2))
-
Factoring out 410 ;
=
.
315
25.
)1
.
x >
Using the fact that
and the other rules for limits, this becomes
2
We
jl/x3 1
whenever
< E
1im X"
0
x
. .
.
-2
=
to get l/x
6)]
e3
3 I/&.
0 - 2
+
41x ) / ( 5
+
3)1 = -4
whenever
E
, we note that l / x 3
E
0 , Xlim(3/x -
=
R/ <
so
1im e x p [ ? 8+o(tan 6 / 8 ) ]
1 ( 1 + x 3) / x 3 - 11 = \ ( l + x 3 - x 3) / x 3
so we may choose A
L/~J;-,
-
If (x)
such that
To make this less than
21.
tan 6 / 0 1 =
Using the replacement rule, we get 1im x-2 [ ( x - 2 ) ( x + 2 ) / ( x
17.
0 ,
Note that the exponential function is continuous at
.
This approaches
Therefore, the limit is 0+
= -1
. Thus,
1irn
x-,
(1x1 1x1
+=. =
1
.
Copyright 1985 Springer-Verlag. All rights reserved.
Vertical asvmototes occur where
It
1 1
.
.
.
Y
We find that
1/(x2 - 5x
6)
l/(x - 3)(x
=
as
Tm
(a)
---t
Given
> 0
E
x > A
so
Thus,
x = 3
,
f(x)
y = 0
so
; f (x) / .
Hence,
Let
=
g(x)
Ig(:<)
g
1
x
i 31- and
f(x)
+
are vertical asymptotes.
1im f(x) = 0 X"
.
If(x)
f(x)
0 as =
x
.
l/x
1
<
Given
whenever x > A
< E
and
as
0 such that
>
has limit
sin(l/x)/x
+-
=
=
is the horizontal asymptote.
, there is an A
also gives
--+
x = 2
and
by the assumption that
same 6
(b)
22 ,
lim f (x) = 0
Also,
45.
x
.
- 2)
f(x)
.
""' f (x) ~lirn
asymptotes occur at
X
+ m
whenever
E
E >
0 ,
this
\g(x) 1
since
<
, as well.
Then
Ig(x)l
1im and the sandwich principle applies. x- f (x) = 0 ,
If(x)l
1im so ,x P(x) =
0 also.
49. The numerator is
. Therefore, lim 2 [(x + x + 1)/(x + xi1 (x
53.
+
x # -1 ,
I)]
1im x+l(x2n
=
I)]
I)
=
312
- x2n-1 + x2n-2 - X2n-3 +
.
+
The limit of the terms with even exponents is
so the answer is
57.
2n
+
1
x-t lim[(x3 1
- l)/(x
2
- 1)1
x
=
.
then long division shows us that
the terms with odd exponents is
(x - 1)
and the denominator is
x2 - 1 # 0 , then
if
1)
If
+x+
(x - 1)(x2
-(-I)
=
1
lim 2n+l xl-l[(x x4 - x3 1 ,
+
+
+
l)/(x
x2 - x
+
1)
.
and the limit of
. There are 2n + 1 terms,
.
The numerator approaches e - 1 , which is finite and nonzero, while the denominator approaches 0 limit is
im
on the positive side; therefore, the
.
Copyright 1985 Springer-Verlag. All rights reserved.
478 Section 11.1
1im f(x) = .P X"
61. By definition, when
f(x)
where a one-sided limit of
E ,
=
at
xo
the horizontal asymptote.
i a.x
N f(x) =:i=O
65. Let
N-1 M a N - l ~ /X
+ ... +
N < M
Since
,
g(x)
.
to get
1im xl,[f(x)/g(x)l
+
xy-",(l/x)
have aN/bM g(x)l
.
=
bMW2/x2
[f(x) lg(x)I
x lim[aNxN/xM -too
+ .. . +
-
M
lim
XM
+
aNx +m
=
b M - l / ~+ bM-2/x2
,
+
=
+
a N - I / ~+ aN-2/~2+
-
N < M ,
M bO/x 1
.
N-M if
1im xi,[f(x)/g(x)l
+
. . . + h0/x M1 . 1im X "
If
(l/x) = 0
N
... +
. .. +
aN < 0
,
M
N
aO/x
I/
1im Since x ( l / x ) = 0
.
2/xM M
,
we
1im x+- _[f (x)/g(x)] to get
lirn a0/x 1 M b [ , /
1im then xi,[f
to
x /x
+
=
1in x- [f(XI/
b M - l / ~-'-
The limit of the denominator is bM
.
then
+
then we can divide by
By a similar method,
aN - lN-1 ~ /XM
and
1im [f(x)Ig(x)] = 0 if N < x1im so xi-r[f n(x) /g(x)l = O/bM =
0 ,
M ,
-1
=
so
N
+ .. . +
aN/bM
is even, then
is odd, then
If
x
1im = x- [a3N/xM
then we can divide by
bO/xM]
and the limit is
N
=
If N > M ,
numerator is
If
1im -[aN
=
-
If
+ bM-l/x + bg-2/x2
1im
0 , which is
the limit of the numerator is 0 because
0 by a similar method.
1im x-[bM
bizi.
xlim ,[f(x)/g(x)l
M lim a0/x 1 /+[bM
We also know that
get
M
=
The limit of the denominat3r is bM ?I
=
The vertical asymptotes are
and let
x"/xM
we can divide by
0
- l/x2) , so x?so 1im f (x)
y = (l/x)/(l
y =
- - . Vertical . Division by
is + m or
asymptotes usually occur where the denominator is x2/x2 yields
the line
A vertical asymptote occurs
is called a horizontal asymptote.
.t
1im x+-m f(x)
or
(x)Is(x)l
.
The =
-
1im aN > 0 . [f(x)/s(x)] depends on N - 1' . X+1irn 1irn xi,{f (x)lg(x)] = ,,-,[f (x)/g(x)] . If N - ?f =
1im
- xi-,[f (x)/q(x)l
.
Copyright 1985 Springer-Verlag. All rights reserved.
.
69.
(a)
To satisfy the definition, the graph of
P. -
and
E
L!
+
E
in the interval
ample of a poorly chosen 6 . chosen
6
f(x)
.
[xO,xO + 61
(i) is an ex-
(ii) shows an example of a well-
.
(i)
(b)
must be between
(ii)
The definition of the statement requires the graph of lie above
B
in the interval
of a poorly chosen
S
.
[xO,xO + 61
x
,
(i) shows an example
(ii)
73. Example 5 tells us that For fixed
to
(ii) shows an example of a well-chosen
(i)
1im
.
f(x)
lim -x -~,t= O *_ e = 0 , so lim tM e
,
lim - 2 t 1im T(x,t) = [ti_ R 1~in!.~xl[~_e 1 ]
.
Since
+
B , ', ,
for
'
0
.
[lim thT B sin12x1 2' and
x
are
Copyright 1985 Springer-Verlag. All rights reserved.
k
480
Section 11.1
73.
(continued)
77.
The composite function rule states that if 1im
,
f(x)
1im x+x g(f(x))
then
E
0
>
be given.
0
p >
g(L)
f(x)
=
g(L)l
,
<
< E
E
, i.e.,
Replacing
l/y
, i.e.,
-E
if
there is a
is defined and
lim f (x) x+xo
L , we can find
=
g(f(x))
LI
If (x) -
is defined and that
limf (x) = 2, , we have, whenever x-" <
by >
1/y > A , 1/A= 6
<
-E
f(x) - V
x
=
x
i.e.,
2
<
gives
.
by
O
i.e.,
< 0
,
p
-
+
-E
-F
E
+
9. < f(x)
gives
y < 1/A
x > A ,
2 < f(x)
0
<
y
< 5
-F
+
< E
+
6 ,
<
limf(x)
X "
whenever
k
=
2 < f(l/y)
if
L F
. This means that xlim ,o+f
+
+
L
If (x) V
.
<
E + L
.
0
<
l/x
<
116
=
A .
whenever
( 1 I?) = p
.
Evaluate the following limits: (State
+m
or
-m
-
If(l/y)
SECTION QUIZ
1.
.
with
Ig(f(x))
, i.e., - ~ + k E(l/y)
<
This means that l/y
<
c ,
means that whenever
f(l/y) < k
x > 116
Now, substituting >
, g(y)
p
x - x0 1 < 6 , we also have
1 im yiO+f(l/y)
Similarly,
E
L ,
is continuous at
.
, i.e.,
k) < c
g
such that
/y- L/ <
to conclude that
By the definition of
2
6
. By the definition of
< E
lim f (x) = L and x+xo /g(f(x)) - g(L) 1 < E
x , we apply the previously obtained property of
For such
81.
1
such that, whenever
6
Since
such that, whenever
-
y
We need to find
.
0
=
is continuous at
o
x - xo/ < 6
whenever
g
(B sink x)(O) 3 3
g [ ~ ~ ~ f ( x ) ] . Let
=
o
X3X0
+
(BlsinXLx)(0) + (BZsinhZx)(0)
constants, the limit is
, ~f approprlatc.)
Copyright 1985 Springer-Verlag. All rights reserved.
1.
(e)
lim[(2r3 - 5r2 r-xo lim 5 x4(l/~ )
(f)
lim 6 xio(l/x )
(9)
1 im xi2- [(x2 - 5x
(d)
(h)
; :
(i)
:$[(81
+
+
1000)/(3r3
6)/(x
-
1)I
- 2)l
1
-
t2)/(3
-
&)I
2.
Use the
E
-
6
definition of the limit to prove that
lim 3 2 x+4(x -x)=48.
3.
Use the
E
-A
definition of the limit to prove that
1im 3 x* l(x3 + l ) / x 1
4.
City ordinance limits the height of residential buildings tc
50 feet.
Hodern Architect, Inc. wants to build a home whose height is
30
x)/(x3
+ x2 +
2) feet , x >/ 0 , where
x
+
=
(20x3 +
is the number of centimeters
from a designated boundary. (a)
If a piece of property extends forever, how high will the building be very far from the designated boundary?
(b)
Will Modern Architect violate the building laws? Explain.
(c)
Use the definition of limits to prove your answer in part (a).
ANSWERS TO PREREQUISITE QUIZ 1.
(a)
If (x) + g(x)l
(b)
3 1im X"f (x)
(c)
2.
1im The limit doesn't exist. x+O- (l/x)
=
- m
and
1im x+O+ (l/x) =
+m
.
Copyright 1985 Springer-Verlag. All rights reserved.
1.
482 Section 11.1
.
The horizontal asymptote is
y
The vertical asymptotes are
x = -1
and
-1
=
.
x = 1
ANSWERS TO SECTION QUIZ
1.
(a)
2
(b)
-112
(c)
-1
(dl
213
(e)
Does not exist
(f)
+-
(g)
-1
(h)
1
(i)
108
2.
Choose
6 = 3~;
3.
Choose
A
4.
>
3~
and show that and show that
lf(x) - L /
c
/f(x) - L I
t
< E
(a)
1im x-_[30
(b)
No, the function is strictly increasing.
(c)
Choose
+
A
(20x3
+
> 61/t
x)/(x3
+ x2 +
2)1
and show that
=
whenever whenever
Ix
- 41
x > A
<
.
6
.
10
If (x) -
LI
< E
whenever
x > A
Copyright 1985 Springer-Verlag. All rights reserved.
.
Section 11.2 483
s 11.2 ~ ' H G ~ i t a l 'Rule
PREREQUISITES 1.
Recall how to compute a limit (Sections 1.2 and 11.1).
2.
Recall the various rules for differentiation (Chapters 1 and 2, Sections
5.2, 5.3, and 6.3).
3.
Recall the mean value theorem for differentiation (Section 3.6).
PREREQUISITE QUIZ f (x) = (x - 3)/(x2
- 5x
+
6)
1.
Let
2.
Differentiate g(x)
=
(x2
3.
Differentiate f (x)
=
sin(lOglOx)
4.
Let
f
+
be differentiable on
eX)/(3X
.
- cos x
[2,3] ,
f (2)
+
In x)
=
-1
.
and
f (3) = 3
.
What
does the mean value theorem tell you about the slope of the graph on [2,31 ?
GOALS
1.
Be able to know when and how to use i'~6pital's ruie for computing limits of functions.
STUDY HINTS 1.
L'~6pital's rule. 010
or
-1- .
This can only be used if the function has the form
The process may be repeated as often as necessary, but
be sure the function has the appropriate form. You should be concerned about applying this rule, rather than learning its proof, unless you are
Copyright 1985 Springer-Verlag. All rights reserved. in an honors class.
484
2.
Section 11.2
Trick number 1.
Example 6 shows how to convert a product into a form
for which 11H8pital's rule applies. ciprocal of one of the factors: uv 3.
The trick is to divide by the re=
.
u/(l/v)
Trick number 2. Many times, the limit of an exponential form can be determined by applying function.
l'HBpital's rule to the logarithm of the desired
See Example 7.
The composite function rule for limits and
the continuity of the exponential function permits us to use this trick.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1. 5.
9.
010 ,
This has the form This has the form
010 ,
which has the form
0/0
lim xq(-9
-9110
cos 3x110)
=
-1- ,
This has the form
lim X "
4 x In x
,
4
=
rule, this is
1im
x-to
.
sin 3x110~)
. SO
.
lim X "
(ex/x375) = 1im(ex/375x374) X"
Continuing with
where
1im xq(ln
lim 3 3)] = x+3(4~ 11) = I08
Applying 1'H6pital1s rule again yields
-4 X/X )
373 more times gives us
.
l'HBpital's rule applies, but
which has the form
[(l/x)/(-41c
-5
, which
l1H6pital's rule yields
375! = 375.374. I : .
does not have a form to which
x + O ( ~In x)
-
lim 2 lim x + O [ ( ~ ~3x~ - 1)/5x 1 = x+0(-3
x (e 1 3 7 5 . 3 7 4 ~ ~ .~ ~Using ) 1'HBpitalts rule
lirn x ,-(e /375!) =
lim
so
.
again has the form -1-
13.
so i2[(x4 - 81)/(x
lim
4
) ] = X+O(-~ 1 4 ) = 0
-/-
.
By l'H8pital's
.
17. We need to transform the function to a form for which 11H8pital's rule applies. (llx)
(tan ' ) x
=
has the form
1im x,O[ln(tan
x)/(l/x)]
exp[x ln(tan x)]
-I= .
Applying
and
x ln(tan x )
=
ln(tan x)/
l'H8pital's rule, we have
2 -2 ) I = x+O(-~ lim 2 ICOS x sin x) , = x+O [(sec xltan x)/(-x
lim
.
which has the form 010 lim 2 2 [-2x/(cos x - sin x)] xi0 lim [(tan x)~] = e0 = I xi0
Using 1'H8pita11s rule again, we get =
011
.
By the continuous function rule,
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 11.2
21.
This has the form
2 car
=
010 ,
012 = 0
This has the form -1-
29.
This has the form
33.
This has the form 011 = 0
37.
Write
lim , so xi,[x/(x
010 ,
Of(-2) = 0
=
1
]
) . 2x - l,/sln
+x
=
lim[[ x+o
r 2 ~ x /I +
j
.
25.
2)/2xl
~m[[r' so x-tO
485
2
1im so xi- 1 l(x2
+
+
l)]
2x
=
+
1im ,,[1/2x]
=
0
2 l)/(x
- 1)l
1im x+-l 1(2x +
=
.
010
,
so
1im xin [(l
+ cos x)/(x
- -)]
=
1im ,+_(-sin
xll) =
.
xPln x
1 im (In x/x-') xi0
in the form
m/m
to use
11H6pital's rule.
xiO lin x In x
=
lim(x-l/-px-p-l) = :$(x~/-~) = 0 because p > 0 xi0 lim x 41. By Example 7(a), x+O+~ = 1 . By logarithmic differentiation, we differentiate
=
In y = x In x ,
1) , which implies
l/e
so
y r / y = In x
+
1
is the critical point.
is drawn on a logarithmic scale.
.
Thus
y'
=
xx(ln x
i
The graph on the left
The one on the right is drawn on a
normal scale
SECTION QUIZ 1.
Use 1'Hopital's rule to compute the following limits: (a) (b)
(c) (d) (e)
(f)
1im xiO+x(ln
2 X)
2 1im (I/X)(~ ) x+O+ lim 2x x+-m xe lim 2 x+O(~ cos X/X sin x) lim ( 1 1 2 ~- l/tan x) x+O+ 1im xio+(l/x - l/tan x)
Copyright 1985 Springer-Verlag. All rights reserved.
486
Section 11.2
2.
The function x=n/4
when
(2x - n/4)]
=
f(x)
.
=
(cos x
-
Thus, we apply
lim x+ri/4(-cos x/2)
2
fi/2)/(x
=
- nx/4)
has the form
1'H8pital1s rule to get
.
-fi/4
(a)
Why isn't the calculation correct?
(b)
What is the limit
0/0
ljm sln x)/ x+n/ii[(- '
This is not correct.
lim (x) ? x+n/4f lim(sin x/x) ?
3.
Can we use
1'H8pital's rule to find
4.
Deep in the jungles of Brazil, archaeologists have found a headhunter's
X "
Why or why not?
recipe for shrinking heads.
In addition, the notes of a famous mathe-
matician were found nearby.
His observations determined that after
seconds of shrinking, the head was shrunk to size, where
x
is
x
100 - (25et - t cos t)/(et
t
percent of the original i
1 - t)
.
How effec-
tive is the recipe, i.e., how much can a head be shrunk after a very long time?
ANSlJERS TO PREREQUISITE QUIZ
1.
(a)
+m
2.
[(2x
+
cos x
eX)(3X
+
In
X)
- cos x
+
In x) - (x2
+
ex)(3xln
3
+
sin x
+
1/x)1 /(3X -
2
3.
cos(loglOx)/x In 10
4.
Somewhere in
(2,3) ,
the slope is
4
ANSWERS TO SECTION QUIZ 1.
(a)
0
(b)
1
(c)
0
(d)
1
Copyright 1985 Springer-Verlag. All rights reserved.
ie)
--
(f)
0
(a)
-sin x/(2x - n/4)
(b)
25/71
3.
No;
sin(x)
4.
75% of the original size.
1.
2.
010
does not have the form
or
m/m
.
. does not approach
0
nor
as
x
approaches
"
.
Copyright 1985 Springer-Verlag. All rights reserved.
488 Section 11.3
11.3 Improper Integrals PREREQUISITES 1.
Recall how to compute limits at infinity (Section 1.2).
2.
Recall how to compute infinite limits (Section 1.2).
3.
Recall the techniques of integration (Chapter 7, Sections 10.1 and 10.2).
PREREQUISITE QUIZ Compute the following limits: (State
1.
(b)
1im x+lll/(x - 11, lim 2 x,[~ /(x- 1)1
( ' )
11m ~ 1 5 [-( 5
(a)
2.
+
x)/(25
zm
lf appropriate.)
2
- x 11
Evaluate the following integrals:
GOALS 1.
Be able to demonstrate the convergence or divergence of improper integrals, and evaluate them.
STUDY HINTS
1.
Improper integrals. approaches
i-
Integrals may be improper because the integrand
for some xo
or the interval goes to
im
.
Note that
if an integral is the sum of several improper integrals, all of them must separatelvconverge for the entire integral to converge.
2.
The important part which needs to be compared is the
Comparison test. region near
im
.
As long as
f(x)
is finite in the interval
rue only need. to apply the comparison rest on the intei-val
[ a , ~ ~, ]
[x0,m)
determine convergence.
Copyright 1985 Springer-Verlag. All rights reserved.
to
489
Section 11.3
3.
2 jr3(1/x )dx
Infinite at midinterval.
may appear to be proper since
the integrand is finite at the endpoints. However, you should notice xO
that the integral becomes improper at
=
.
O
You should always look
for points where an integrand becomes infinite before beginning to evaluate an integral.
-
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
2 j1(3/x )dx
5.
Using the method of partial fractions, j;[dx/(x
=
- 1) - dx/(x + 111
(X
lim ln[(b b-
+
(b
9.
1im bi,,(-3/x)
1)1
+
- l)/(b lim
=
In (1)
We want to find
j;f(x)dx
lim b-
l)] =
g(x)
.
3
=
2
lim bi,,(~/~) ln/(x - l)/(x
+ 0
In 3:
.
Thus,
2
j;[dx/(x
/ f (:<) /
so that
+
- I)]
1)1/:
(1/2)j;[dx/
= =
(112)
x
lim By l'Hbpitalls rule, bIn [ ( b - 1)/
.
<
+
also converges. f(x) = 1/(3
f (x)
j;[dx/(3
=
- 3/b)
- l)]
g(x)
=
.
In 312 I-
and that
jag(x)dx
con-
If this is possible, then the comparison test states that
verges.
and
I bl = blim i,,(3
113
<
+
x3)]
(-1/2x2)/:
=
g2(x)
+
,fl[dx/(3
=
113
+
on
+
.
[0,11 x3)]
<
112 -b2/1(::
3 x )
.:
Thus, ji[dx/(3 + x /A(dx/3)
+
2
.
) = 516
on
l/x3 = gl(x)
3 jl(dx/x ) Thus,
3
)I
=
[l,-)
=
I1 (~13), 0 +
+ x3)1
ji[dx/(3
converges. 13. Use the method of Example 6. g(x)
then, show that
Find
g(x)
diverges. Choose
less than the given g, (x) = 112 < 11-
[D,II and choose g, (x) = l/x = 1/Q< f(x)
on
[ 7) +x l;(dx/fii)
I"
jO dx/ 2
=
1 lim (x/2)IO + b-m ln
XI:
.
since
+
!Y[dx~&~]
,- In b 1im
f(x)
on >
[I,-)
=
.
and f(x)
Thus,
jA(dx/~) -Ii;(dx/x)
diverges, j ; [ d x / K 2 ]
diverges.
Copyright 1985 Springer-Verlag. All rights reserved.
=
490
21.
Section 11.3
The integral is improper at
(x2
+ x)] + ji [dx/(x2 +
1 j0[dx/(x2
+
approaches
x
0 ,
=
x)]
1 [dx/(x 2 + x)l /-1
so
Since x2
<x
=
0 [dx/ /-1
(0,l) , we have
on
XI]
1 lim 1 lim Jo(dx/2x) = a+O+ja(d~/2~)= aXl+(ln
--.
Since part of the original integral diverges, the
x/2)
l a1
, which
entire integral diverges. 25.
tan-lx
lies between
.
+
1im bi-[l/2(2
converges, so does Let
f (x)
=
l/(5x2
T ~ ~ s jyf(x)d~ , 33.
+
j(Dlg(x)dx = (-12) [;Ej;(2
(n/2)(-112
29.
n/2
Note that
+
b)
2
1)
1)2/3
x13
and
so we can apply the
x ) - ~ ~ x I= (~/2):2-(2 -n/4
=
+
g(x) = (n/2)/(2
.
+
x ) - ~ / ~ I ;=
Since
jmlg(x)dx
.
jmlf (x)dx
+
+
(n/2)(-112)
=
x ,
for all
f (x) = tanW1x/(2
comparison test with
x)'
-7112 and
and
.
g(x) = 1 / ( 5 ~ ~ ) ~= / 1/ ~ 3fix4/3
converges by the comparison test.
/La( iIx6f5 - 1/x413)dx 6f5 - l/(-u)4f3jdu =
each exponent is less than
=
2 [ l / ( - ~ ) ~ /- ~11(-u)~/~]du -j_
1im{j;[(-u)-615-
(-u)-~/~]du}
bi-
-1
,
.
=
Since
by the result of Example 2, both terms
converge. ~ h u s , j ~ ? ( l / x ~ /-~ ~ / x ~ / ~ ) c Iconverges. x 37.
The integrand is unbounded at
0 ,
so
j-_(l/x 5/3 - 1/x4/3)dx
lirn lirn a+-m b+o- ~ ( l / -x / ~ x ~~ ~ )+ c x+(l/x53 lirn lirn a+-m b+o- (-3/2xZi3 + 3/x1l3) :1 + ~ / 2 3 ( + : diverges toward
41.
- l l ~ * ~ ~ ) d= x 3 1 ~ ' ~ 1:~ ) .
+
This
-_,
so the integral is divergent. -k The parametric equations 8 = t , r = t are equivalent to so
r1(€l) = - k ~ - ~ - .l From Section 10.5 ,
ordinates is
L
&-k-l
)
=
(43k+1)dl =
k
(do18 )
p2)d8<j:/2(XZi2/8k+1)de J,/~(~B/O~).
=
Thus, we have
.
2
+ =
0
=
F
-k
,
the arc length in polar co-
-k 2 (8 ) do
Since
r
>1
la
= J
n/2 ( m / b p ) d @ >
, we have jn/2 fm
f i ~:/~(o/8~+l)di
~ ~ ~ , / ~ 1 2 ( dko) />iL l
=
(0; 71
E G
k jz12(d~/l ) ,
Copyright 1985 Springer-Verlag. All rights reserved.
r
so
41.
(continued)
L
Imn/2 (dO/Ok )
acts like
this converges for
k
k = 0
The case
. By a method similar to that of Example 2,
>1 . is special because the spiral self-intersects
forming a circle of radius k = O
45.
or
.
1
Thus, the arc length is finite for
k > l .
In Example 11, Luke travels
10 mil.lion miles.
must wake up before he gets within
In this problem, he
1 million miles of the sun, so
the distance travelled is
9
9 million miles is
10~)/(~/2)] (1110~) = 6 &
[(9
x
million miles.
Thus, the time to travel hours
10.39
hours. 49.
The limit is
lim [ (~:(dx/x) A+O+
+
.
!i(dx/x)~
Recall that an improper
integral converges only if each part separately converges. 1im
1im
A+o+ j;(dx/x) 53.
(a)
If
x =
-=
as
p (b)
=
=
If
(T
Since
2
A+O+ in 1x1 / A diverges, the entire integral diverges.
-
lj)/o
, then x
=
xl
when
7
a
=
and
x
approaches
approaches - m . dx = dr/o , so substitution yields 2 2 /X1(l/a-o)e-x -m I2(odx) = jxl(i:~)e-x f2dx . 2 -" 1x1 > 1 , then e-x 1 2 < -x/2 and if 1x1 < 1 , then
-xL/2
T
1
.
Since the integrand is symmetric to the y-axis, 2 l2dx = 2ioe-x l2dx < *lidx + 2 j ~ e - ~ / ~=d 2x + 21im/be-x/2dx = b--, 1 lim -x/2 b limewbf2 + 2/vG) = 2 4/vG < . ~hus, 2fb-(-2e )/l=2-2(b2 I2dx < . the comparison test tells us that j:_eix <
2
+
-
57.
If
0
<
a > 1
.
f '(x)
<
1/x2
on
[O,m) ,
the same is true on
By the comparison test, we have lim ,b < b-m(-l/~) i
i.e.,
o
Thus,
0 < lim f(b) -f(a)
l/a,
Since
fl(a) exists,
also exists.
<
limf(x) b* b--
f (a)
~ZO dx
[a,=)
jmf '(x)dx
for ,-
<
2
j (dx/x
)
,
by the fundamental theorem of calculus i.e.,
£(a)
i/a+fi;') .
;?f(b)
Now, since
f '(x)
,,
C , f
Is an increasing function. Therefore, f(x) is a strictly increasi:.: Copyright 1985 Springer-Verlag. All rights reserved.
492
5
.
Section 11.3
(continued) 1im
f u n c t i o n and
buo f ( b )
i s bounded a b o v e by
l/a
+
f (a)
.
Thus, t h e
limit exists,
SECTION Q U I Z 1.
2.
E v a l u a t e t h e f o l l o w i n g improper i n t e g r a l s , i f t h e y converge: j::[x/(x2
(b)
10 4x j-_e dx
(c)
j:x2
815
10 j-lO[(x
.
5
- 3)/(x
+
2
5) Idx
c a n be e v a l u a t e d by s u b s t i t u t i n g
15 2 / - 5 [ ( ~ - 8)/u ]du = ( l n ( u /
to get
+
8/u)
: '1
= In 3
+
8/15
+
Review t h e s e c a l c u l a t i o n s and e x p l a i n t h e e r r o r s , i f a n y .
D i s c u s s t h e convergence o r d i v e r g e n c e of t h e f o l l o w i n g i n t e g r a l s : (a)
jo5 0 [ d x / ( x 4 +
(b)
1-6i d x i (x3"
(c)
4.
+
l ) l dx
I n x dx
The i n t e g r a l u = x
3.
-
(a)
j!2~x3/(x
4
x3
+
x
2
+
l)j
+ 111 - l ) ] dx
E v e r s i n c e o l d man Anderson p a s s e d away 2 5 y e a r s ago. no o n e has s e t f o o t i n h i s mansion.
Each y e a r , on t h e 1 3 t h of March, n e i g h b o r s h a v e
r e p o r t e d s t r a n g e , s c r e e c h i n g n o i s e s from i n s i d e t h e m a n s i o n . year,NastyNicholas
w a n t e d t o p r o v e h i s m a n l i n e s s by s p e n d i n g t h e
night i n themansion.
r a t e of
25x / ( x
4
On t h e m o r n i n g o f t h e 1 4 t h , N a s t y N i c h o l a s emerged One h o u r l a t e r , h i s h a i r b e g a n t u r n i n g w h i t e a t a
shaking with f e a r . 2
This
+
x)
,
i n s t r a n d s of h a i r p e r h o u r .
W i l l a l l of
N a s t y N i c k ' s h a i r e v e n t u a l l y t u r n w h i t e due t o t h e b i g s c a r e ?
Explain.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 11.3
493
ANSWERS TO PREREQUISITE QUIZ 1.
2.
(a)
+-
(b)
+-
(c)
+-
(a)
ln/x/
(b)
sin x
(c)
xe
x
-
+
C
+ x 3/ 3 + x e + C
C
ANSWERS TO SECTION Q U I Z 1.
2.
(a)
Diverges
(b)
e40/4
(c)
9 In 3 - 3
The i n t e g r a l i s i m p r o p e r a t 5 jIl0[(x - 3)/(x
+
u = x 3.
gives
5)
10
1 dx + j-5 [ ( x -
I-5 [ ( U 0
8) /U
2
3)/(x
C o n v e r g e s ; i t i s n o t e v e n i m p r o p e r on
(b)
Converges; i t i s l e s s t h a n Diverges,
By c o m p a r i s o n ,
1
jOdx
1 3 4 I121x / ( x - l ) / d x
1; [25x
2
/ (x
4
10 [ ( x j-10
+
2
5)
1 du + ji5 [ ( u -
(a)
(c)
4.
5
+
2
, so
x = -5
>
1 dx . 8 ) /u
[0,50]
2
3)/(x
+
5 ) 21 dx
=
Then s u b s t i t u t i n g
1 du ,
which d i v e r g e s .
.
+ ~ ' ; ( d x / x ~ ~. ~ ) 4
~ ~ : l ( x- l ) - ' l d x
+ x ) ] dx < j T ( 2 5 / x
Thus, t h e b i g s c a r e w i l l c a u s e l e s s t h a n
25
2
.
) d x = ( - 2 5 1 ~ ); 1
= 25
.
s t r a n d s of h a i r t o t u r n
white.
Copyright 1985 Springer-Verlag. All rights reserved.
494 Section 11.4
11.4 Limits of Sequences and Newton's Method PREREQUISITES
1.
Recall how to compute a limit by using basic properties (Sections 1.2 and 11.1).
2.
Recall how to compute a limit by using 1'HBpital's rule (Section 11.2).
3.
Recall the formula used for t h e linear approximations (Section 1.6).
PREREQUISITE QUIZ
-
1.
Compute :2(2/x
2.
Compute
:E(z/~)
.
3.
Compute
:E[(cos
x
4.
If
f
f (x
+ Ax)
x)
+
. l)/(n
2
-
nx)]
.
is differentiable, write dorm the linear approximation to
.
GOALS 1.
Be able to find the limit of a sequence.
2.
Be able to find roots by using Newton's method.
STUDY HINTS 1.
Limit of sequences. A sequence is merely a list of numbers. -.
For almost
all sequences that we will consider, a definite pattern, which may be expressible by a formula, will exist. E
> 0
, a
is within
(k - ~ , + k
E)
A limit
k
whenever
exists if for any
n
is large enough.
Example 3 shows how to use the precise definition. 2.
Limit of "functional" sequences. muia, i.e.,
a
=
f(n)
sequence as
n
approaches
If a sequence
for a function
f(x)
,
a n
is given by a h r -
then the limit of the
is the same as the limit of the function
Copyright 1985 Springer-Verlag. All rights reserved.
2.
(continued) approaches "
as x
.
Notice that since n
is restricted to he an
integer, a sequence may have a limit even though the function does not For example, the sequence lim sin nn n-
have a limit. lim x-
3.
sin nx
has a limit, but
does not have a limit.
Limit of powers.
The box located at the bottom of p. 542 1s obvious
if you recall the discussion about exponents in Chapter 6. Alternatively, you may remember simple examples:
2n
gets large with
,
n
while
(112)"
gets small. 4.
Comparison test. Notice how this statement isvery similar to the comparAgain, there is a "sandr~ich"effect.
ison test of Section 11.1. 5.
Newton's method.
This procedure uses the linear approximation. You
should either memorize the iteration formula, which is
,
f '(x,)
.
or learn to derive it
seek an approximation so that
.
=
Xn - f (xn)/
Since we are looking for. a root, we
+
Ax)
= f (xn) + f '(xn)/,x
Ax = -f (xn) /f ' (xn) ,
rearrangement yields f(xn)/f '(x,)
f(x
Xn+l
and so xn+l
The formula is used repeatedly until
=
=
xn
.
0
+
Thus,
cx
xnil a xn ,
=
x
-
i.e.,
Ax-0.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
This exercise is analogous to Example 1. so
loan
-
a
=
9a
=
1 - l/lon ,
loan = 1
i.e., a
=
+
1/10
119 - (1/9)(1/10*)
119 - an = (119) (l/lon) , which we want to be less than n
5.
must be at least
6
k2
= =
n2 - 2i/;; . (2)2 - 2 6
Hence, kg =
1/106
l/lon-l ,
,
i.e.,
. Thus,
.
The sequence is determined by substituting n kn
+ ... +
4 - 2fi
;
=
k
=
0 , 1 , 2 ,
(0)* - 2\/r;= 0 ; kl 3
=
(3)2
-
=
(I)? -
247 = 9 - 2 > 3 ; k 4
...
into
?iri= -1 =
;
2 (4) -
Copyright 1985 Springer-Verlag. All rights reserved.
496
Section 11.4
(continued) 2v% = 1 2 ; -1
,
k5
4 - 2fi
=
,
(5)
2
- 2 6
9 - 2 6
The f i r s t t e r m
a0
,
- 2 6 . Thus, t h e s e q u e n c e i s
= 25
i s given a s
a 3 = a2+1
a 5 = a4+l = ( 1 / 5 ) ( 5 / 2 )
all
.
n
We need t o c h o o s e 3/2n
that i f
3/2n
n2/n2
=
/an
3/(2n n
a1
a l + l = (112) ( a 0 a4
a3+1
=
=
Thus, t h e s e q u e n c e i s
-
&/
+ 1) < E > N , we
<E
n
for a l l
and
l/n
have
aO+l
=
+
13/(2n
al) =
(1/4)(2) 112
>N .
< 2 ~ / 3.
+
=
for
Note
I f we
1 ) - 01
<
.
E
1im [(l/n-3)/(1+1/n2)1=(o-3)/(1-0)= n+n
toget
+
(sinn)'
2/n)]
=
0 ;
forali
n
and
1im 1im [l/(n+2)]= [(l/n)/ n-tm n*
therefore, t h e absolute value o f t h e l i m i t is l e s s than
1i m [l/(n n*
o r equal t o
+
2)]
=
0
.
Thus, we c o n c l u d e t h a t t h e l i m i t must
.
0
nm,
, a 1000 sz 1 . 0 0 6 2 3 1im % 1.00001 T h u s , we g u e s s t h a t a = 1 . By n* n and a l o o ~ o O o 1im 1im 1 ' H 8 p i t a 1 1 s r u l e , x* 1im(x/2)1/x = exp [x-(ln(x/2)/x)l = e x p [,,(1/2x)/ll
For
=
.
Notethat
be
.
112
=
N
Divideby
(1
then
> 3 / 2 ~, t h e n f o r < 3 1 2 ~< ( 3 / 2 ) / ( 3 / 2 ~ ) =
choose
-3
<E ,
112 ; a 2
=
such t h a t
N
The n e x t t e r m i s
( 1 / 3 ) ( 3 / 2 ) = 112 ;
=
112 ;
.
112
0 [ l / ( Q + l ) l ~ ~ = ~ ( 1 =/ 2( 1 ) /1)(1/2) ( 1 / 2 ) ( 1 ) = 112 ;
.. . .
, 25 - 2 ° F ,
12
,
0
a
=
we c a l c u l a t e
alOO
1.03999
,
.
=
e 0 = 1 . We u s e t h e f a c t t h a t 1im n,u(l/8)n
= 0
The l i m i t i s D i v i d e by
. {
lim n n+_r - 0
[:z + iE 3b
( n ' ~ n ' ) ~ t o get
if
O < r < l
( l ~ h ) /~[:E(n2 ]
1im n* [ ( 3 b / n 2 ) / ( 1
.
- l)] l3
-
r = 118,
Here
l/n2)]
=
so
i z ( 3 b / ( n 2 - 1))
=
( 0 1 1 ) ~= 0
Copyright 1985 Springer-Verlag. All rights reserved.
.
.
Section 11.4
37.
The i t e r a t i o n f o r m u l a i s Starting with n o t work.
x0
,
0
=
= Xn
Xn+l
we g e t
Starting with
xl
5
+
- (xn
4
- 3 ) / ( 5 x n + 2xn)
xl
1.607143 ;
=
X2 =
1.325370 ;
x3
=
1.167936 ;
x4
=
1.121778;
x5 = 1.118357 ;
1.118340 ;
x,
=
1.118340
.
x6
=
x7
decimal p l a c e s , s o one root
1.118340
is
.
to
6
3ma n d
The c r i t i c a l p o i n t s a r e
t h e g r a p h r e v e a l s o n l y one r o o t , s o o u r answer i s 41.
.
, s o t h i s method d o e s
0 - (-3)/0
=
, we g e t
x0 = 2
2
Xn
497
A n a l y s i s of t h e g r a p h s of 3n/2 ;
i s near
X2
be u s i n g i s
-
tan x
x,+~
ax
=
and
x4
ci
=
4.6042168
= 2
,
x 2 = 7.7898996 For
u = 3
4.6406836
and
x 2 = 7.8113345 For and
=
and
7.8284393
.
and
,
x3 and
.
we s t a r t w i t h
.
, we s t a r t w i t h
.
x 2 = 4.6695848 and
=
x3
=
.
1.362
Sirlce we want
x3
=
, we h a v e x0 = 7 . 7 9 1
, then x3
x0 = 4.642
1.1655820 X2 =
, we g e t
=
, we h a v e x0 = 7 . 8 1 2
1.3241947 x2 =
, we g e t
.
x0 = 1.444
Starting with
.
.
S t a r t i n g with
x 3 = 7.8113345
~ / 2;
.
, then
x0 = 4.607
Starting with
4.6406836
5
Starting with
x0
is near
hl
i s some number j u s t l a r g e r
1.263
=
.
The i t e r a t i o n f o r m u l a we w i l l
, 3 , and
x 3 = 7.7898838
=
x0
Starting with
.
shows t h a t
2
n = 1
x 3 = 4.6042168
x 3 = 1.4320322
4.6695888 x2
a = 5
5n/2
, and s k e t c h i n g
0
1.118340
- a x ) / ( s e e xn - a )
(tan x
where
.
x4 = 1.3241945
and
ax
, we s t a r t w i t h x 0
1.1655612 and
-
and
, a good c h o i c e f o r
= 0
t h a n tan-'(unn/2) For
i s near
X3
xn
tan x
X6 =
, then
x0 = 4 . 6 7 0
S t a r t i n g with
x 2 = 1.432042
, we h a v e x0 = 7 . 8 2 9
X1 =
, we g e t
7.8284393
( c o n t i n u e d on n e x t p a g e )
Copyright 1985 Springer-Verlag. All rights reserved.
498
Section 11.4
41.
(continued)
45.
The first term is is
(2)
2
=
=
4 4 2 8 2 2 ; a& = (2 ) = 28 ; ai = (2 ) = 216
.
exponent of 2 (2n-l ) a = 2 49.
After
1
=
2'
.
a2
a
The second term is
is doubling each time we press the
half-lives,
'
r = 112 53.
is (a)
.
Since 3/l,
After
2
is left. Ingeneral, after n
114
114 , 118
,
1/16
,
...
lim , 1 1 2 ~. nim a
=
0
taking
in the limits of powers
1im [3n/(4n + l)] n1im n* [3/(!, + l/n)]
57.
,
By the sum rule for limits,
nj
"x2" key, so
is left; therefore, the sequence, beginning with
1 , 112
is
the third
Note that the
1 half-life, only half of the substance is left.
half-lives, half of the half, or
. a
2 2 ;
+
1im n-[(-l)nsin
+
:El(-l)"(sin
1im [3n/(4n n-
\(-l)nsin n / < 1
n/(n
,
+
+
1)
l)]
.
+
(-l)nsin n/(n
Dividing by
nln)l(l + lln)] = 314 lim n-[(-l)nsin njn] = 0
+
.
+
n/n
I)] = yields
:2(-ljn(sin
n/
Thus, the limit
.
1im 1im lim a < L - L = 0 . This n-t.. bn < L , then n* bn - n-ra 1im (bn - an) < 0 , but the hypothesis that an < bn says that n1im - an) > 0 , so we have a contradiction. Thus, implies ,-(bn
Suppose
the original assumption is false, and it must be the case that lim n* b n > L .
Copyright 1985 Springer-Verlag. All rights reserved.
Section 11.4
57.
(b)
According to the definition of the limit, /an - L / < wise,
E
or
. /bn - L /
n > N2
+ L
-E
1im c = L n+== n
E
< E
+ L
-E
+L
means that
By hypothesis, <
a
<
+
-E
L
<
lim a nn
whenever c
<
< E
=
L
means that
n > N1
+
L
499
.
Like-
whenever
an < b n < ~ n < ~ +orL
.
whenever n > N3 = max(N1 ,N2)
Thus, we have
SECTION QUIZ 1.
three decimal places.
Start with
programmable, try starting with if 2.
3 f(x) = x 13 - x2 - 1
Use Newton's method to approximate a root of
xO
1 on a TI-58C.
=
xo
=
xo = 1
to
-3
.
.
(Hint: 30 steps are required
If your calculator is
Your calculator may be different.)
Using Newton's method, find the solution of
y5
+
y3
+
y = 2
accurate
to two decimal places.
3.
(a) (b) (c)
4.
2 f(x) = cos xn is a function, what is
If
lim f (x) ? xl im If a a ? nn Comment about your answers to parts (a) and (b). 2 = cos nn is a sequence, what is
Find the limit of the following sequences if they exist: lim
2n n/(n2 - n + 5)l
(a)
,-[(-I)
(b)
E ; [(2n3
- n
+
1)/(3n3
1im n- [((-l)"sin(nn/4))/(n2 5.
2
- 50n
+
)I
2)l
Bouncing Bobby, the aspiring four-year-old trampolinist, was practicing on his parent's bed. bed is
24
The ceiling is eight feet above the floor.
inches off the ground.
height above the floor is (a)
On his
96 - (1/2)~(72)
The
n t h jump, Bouncing Bobb\'s in inches.
How high does Bobby bounce up to after each of his first six jumps?
Copyright 1985 Springer-Verlag. All rights reserved.
500
5.
Section 11.4
(b)
If Bouncing Bobby c o n t i n u e s t o jump f o r a l o n g t i m e b e f o r e h i s
p a r e n t s d i s c o v e r h i s j o y f u l p l a y i n g , how h i g h will he be jumping when h e i s d i s c o v e r e d ?
ANSWERS TO PREREQUISITE QUIZ 1.
-1
2.
0
3.
0
4.
f(x
+
AX)
=
f(x)
+
[f ' ( x ) ] A x
ANSWERS TO SECTION QUIZ 1.
3.279
2.
0.87
3.
(a)
Does n o t e x i s t .
(b)
1
(c)
The v a r i a b l e f o r s e q u e n c e s o n l y t a k e s on i n t e g e r v a l u e s , s o t h e l i m i t o f a s e q u e n c e may e x i s t e v e n i f t h e l i m i t of t h e c o r r e s p o n d ing f u n c t i o n does not e x i s t .
4.
5.
(a)
0
(b)
213
(c)
0
(a)
60 ; 78 ; 8 1 ; 9 1 . 5 ; 9 3 . 7 5 ; and
(b)
96 i n c h e s
94.875 i n c h e s
Copyright 1985 Springer-Verlag. All rights reserved.
Section 11.5
501
11.5 Numerical I n t e g r a t i o n PREREQUISITES
1.
R e c a l l how t o e x p r e s s a n i n t e g r a l a s a Riemann sum ( S e c t i o n 4 . 3 ) .
2.
R e c a l l t h e r e l a t i o n s h i p b e t w e e n i n t e g r a l s and a r e a s ( S e c t i o n 4 . 2 a n d
4.6).
PREREQUISITE QUIZ
1.
Using
10
i n t e r v a l s and
i n t e r v a l , approximate
f (ci)
j h x 2 dx
where
ci
i s t h e midpoint of e a c h
a s a Riemann sum.
Express your answer
u s i n g summation n o t a t i o n .
2.
What i s t h e r e l a t i o n s h i p b e t w e e n
/f(x)dx
and t h e a r e a u n d e r
f(x) ?
GOALS
1.
Be a b l e t o i n t e g r a t e n u m e r i c a l l y by u s i n g Riemann sums, t h e t r a p e z o i d a l r u l e , o r SimpsonVsr u l e .
2.
Be a b l e t o e s t i m a t e t h e e r r o r s i n c u r r e d b y u s i n g Riemann sums, t h e t r a p e z o i d a l r u l e , o r Simpson's r u l e .
STUDY HINTS
1.
Riemann sums.
A f t e r d i v i s i o n i n t o e q u a l s u b i n t e r v a l s , e v a l u a t e a t one
chosen p o i n t i n e a c h i n t e r v a l . t i p l y by
(b
Add up e a c h e v a l u a t i o n and t h e n mul-
- a ) / n , the length of a subinterval.
Often, the i n t e -
grand is evaluated a t e i t h e r t h e r i g h t endpoint o r t h e l e f t endpoint.
2.
Riemann e m .
The e r r o r made by u s i n g Riemann sums d e p e n d s on t h e f i r s t
d e r i v a t i v e of t h e i n t e g r a n d and t h e l e n g t h o f t h e s u b i n t e r v a l s t o t h e f i r s t power.
Copyright 1985 Springer-Verlag. All rights reserved.
502
3.
Section 11.5
Trapezoidal rule. vided inro
n
In this method, the interval of integration is di-
equal parts.
At each division point, the integrand is
evaluated and counted twice except at the endpoints. This is then multiplied by
4.
(b - a)/2n, one-half of a subinterval's length.
Trapezoidal error.
The error in the trapezoidal rule estimate depends
on the second derivative of the integrand and the square of the subinterval's length. 5.
Simpson's rule.
The method requires an even number of subdivisions.
Notice the pattern of the coefficients at the evaluated points; it is
1 , 4 , 2 , 4, 2 ,
...
,
2
,4 ,
1
.
endpoints are considered only once. by
6.
(b
-
a)/3n
,
Simpson's error.
2
and
4 alternate, while the
The weighted sum is then multiplied
one-third of a subinterval's length. The error in the estimate made by Simpson's rule de-
pends on the fourth derivative of the integrand and the fourth power of the length of the subintervals.
7.
Equalsubdivisionsunnecessary. Until this point, the methods presented have used equal subdivisions as a convenience for deriving the special formulas. Be aware that none of the numerical integration methods presented here require equal subdivisions.
However, Simpson's method does
require an even number of intervals and that the length of the terval equal the length of the
8.
Error estimates.
(i
+
l)th
interval for odd
i
ith in-
.
Most beginningcalculuscourses will not require memori
zation of the error estimates; however, you should consult your
Copyright 1985 Springer-Verlag. All rights reserved.
Section 11.5
503
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
J!~(~~
+
l)dx , with n
+ (1/125)~:?~i~
(2/25)
[10(11) (21)/6] x) 5.
= I
L
813
=
2.68 .
=
10
=
(b
1 /-1(x2
The actual value of
n = 10 , Simpson's rule tells us that
With
is approximately
(111'25)
+
1)dx
x
3 (x I3
=
x//x3
=
and
x
i
.
1
=
+
2f(xg)
J
+
4i(xg)
Thus, the integral
(1/30) [ o / v +~ 0 , 4 / m + 0 . 4 / , @ 3 %
+
+
1.2/-
Simpson's rule needs adjacent equidistant intervals, so let Axl [0,1.2] and let
+
j:[x/~2'dx
... +
+ 4f(x1) + 2f(x2) + 4f (x3) +
, where f(x)
f(xlO)]
+
4 - (2125) [10(11)/2]
.
2.6667
1
(1130) [f(xO)
9.
, we have
- a)/n = 115 and 10 2 x. = -1 + il5 . Thus, the Riemann sum is: (1/5)Zi=1(xi + 1) = 10 2 10 2 (l/5)Ei=1 [(-I + i/5) + 11 = [(1/5)~~=~(1- 2i/5 + i /25)1 t 2 = 4 For
&x2
0.1
=
in
.
[1.2,1.8]
sions fulfilling the requirement that
n
=
6
=
0.2 in
for both subdivi-
be even. Write the integral
n
= as a sum of two integrals and apply Simpson's rule to each. ~h'~f(x)d~
+
~:'*f(x)dx
13.
~::if(x)dx
=
(1.2;18) F2.037
4(1.372)
+
2(1.196)
+ 4(0.977) +
0.6851
Z(1.026)
+
4(0.799)
+
4(0.538)
If
=
x
f(x)
f '"(x)
=
+
-cos x , and
"(x)
/
in
want (~124) (Ax)2 we need
n
=
+
sin x , then f l ( x )
trapezoidal rule is f
2(0.662)
fiv(x)
=
=
sin x
1
+
+
+ .
4(1.980)
< 10-'
.
We have or
(r;/2)/Ax >, 180
Ax
.
<
2(1.843)
(0.6118) 10.685
+
0.5551
=
+
+
4(0.819)
+
(0.2/3)(26.116)
cos x , fl'(x)
=
-sin x
+
,
The maximum error from the
[(b - a ) / 1 2 ] ~ ~ ( ~ x ), ~ where
[a,bl
+
M2 = 1 and (10-~.24/r)"~
b
M2
- a
is the maximum =
-i/2 ,
0.00874
.
so we
Therefore,
(Continued next page.)
Copyright 1985 Springer-Verlag. All rights reserved.
504
Section 11.5
13.
(continued) 4 [ ( b - a ) / 1 8 0 1 ~ ~ ( A x ),
F o r S i m p s o n ' s r u l e , t h e maximum e r r o r i s where
,
b - a = n/2
and
.
in
[a,b]
is
i s t h e maximum of
M2
I
x(4(1
/
+
f "(314) error i s
~
x2)(2x)
+
-2.48
.
=
< 611200
2
=
M2/l200 / 2
We h a v e or
f "(x)
.
fr(x)
2 ( 1 - x )(-2x))/2/(1
[ ( b - a ) 1121 M2(Ax)
.
=
{[j(l
+
+
x2)4
+
.
2
So t h e maximum e r r o r x2)4
+
( 1 - x212
( 1 - x2j2](l
From t h i s d a t a , we c a n g u e s s t h a t = 0.005
<
n = (n/2)1~x
[a,b]
on
Ax
M4 = 1
>9
T h e r e f o r e , we n e e d
The maximum e r r o r i n t h e t r a p e z o i d a l r u l e i s where
.
( n / 3 6 0 ) ( ~ x )<~
s o we want
( 1 0 - ~ . 3 6 0 / n ) ~ / ~ 0.1840 17.
/ f iV(x) 1
i s t h e maximum of
M4
Therefore, the f i r s t
2
M2
+
<6 ,
+
x2j3 -
so the
digits are correct.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 11.5
505
SECTION QUIZ 1.
Use e a c h of t h e methods o f Riemann sums ( e v a l u a t i n g a t t h e r i g h t endp o i n t ) , t h e t r a p e z o i d a l method, a n d S i m p s o n ' s method t o e s t i m a t e
1
jO f ( y ) d y
from t h e g i v e n i n f o r m a t i o n .
(a)
f(0) = 2 ;
f(0.3) = 1 ;
(b)
f(0)
f(0.25) = 0 ;
=
3 ;
I f n o t p o s s i b l e , s t a t e why. ;
f(0.6) = - 0 . 5 f ( 0 . 5 ) = -0.5
f ( 0 . 7 ) = -2 ;
;
f(1) = 1
f(0.75) = 0.3 ;
f (I) = 0 (c)
f ( 0 ) = -1
f(113)
;
=
2.
-1 ;
f(2/3)
=
0 ;
f(1) = 0.5
For t h e f i g u r e a t t h e l e f t , = 3 ;
f(-113)
3
.
f(1/3)
=
3 ;
f(-1)
=
0 ;
and. f(1) =
Can t h e t r a p e z o i d a l r u l e be u s e d t o
e s t i m a t e t h e a r e a under t h e curve on
3.
1
If yes, estimate i t .
Santa Claus' e l v e s produce
x3
+
x2
th x week o f t h e new w o r k i n g s e a s o n .
(a)
They b e g i n work
+
x2)dx
12
weeks a f t e r
Thus, t h e t o t a l t o y
.
E s t i m a t e t o y p r o d u c t i o n t o t h e n e a r e s t c e n t by t h e t r a p e z o i d a l r u l e with
(b)
jt0(x3
I f n o , e x p i a i n why.
d o l l a r s w o r t h of t o y s d u r i n g t h e
C h r i s t m a s and work c o n t i n u o u s l y f o r 40 w e e k s . p r o d u c t i o n i n one y e a r i s
[ -1,1]
n = 4
.
E s t i m a t e t o y p r o d u c t i o n t o t h e n e a r e s t c e n t by S i m p s o n ' s r u l e w i t h n = 4 .
(c)
What i s t h e e x a c t amount of t o y s p r o d u c e d i n one y e a r ?
ANSWERS TO PREREQUISITE QUIZ
1.
2 ~ : = ~ ( 2+i 1 ) 1400
2.
They a r e e q u a l .
Copyright 1985 Springer-Verlag. All rights reserved.
?
506
S e c t i o n 11.5
ANSWERS TO SECTION Q U I Z 1.
(a)
Riemann,
0.25 ;
Trapezoidal,
0.25 ;
Simpson's c a n ' t be used
since the t h i r d i n t e r v a l does not equal the fourth i n t e r v a l . (b)
Riemann, -0.050
(c)
Riemann,
;
Trapezoidal,
-1/6 " - 0 . 1 6 7
;
0.325 ;
Trapezoidal,
Simpson's,
4/15 " 0 . 2 6 7
-5112 e - 0 . 4 1 7
;
Simpson's
c a n ' t b e u s e d s i n c e t h e r e i s a n odd number of i n t e r v a l s . 2.
Yes;
3.
(a)
$702,000.00
(b)
$661,333.33
(c)
$661,333.33
.
5
Copyright 1985 Springer-Verlag. All rights reserved.
Section ll.R
507
ll.R Review Exercises for Chapter 11 SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
1 - 2 = (x - 1) /f(x) 6
-
2
+
3(x - 1)
/ x - 11
21
2
1
or
.
€14
,
E
.
.
d2+36
=
x - 1) -. 1 =
Now, assuming that
S
Therefore, if we choose
to be the
then the definition of the limit is satisfied.
+
- 1)l
l)/(x
=
.
tan (-1)
dx2
Multiply - . numerator and denominator by
+
1im[(2x2 x-
1 -:x2) 1
as
x
x+OI~
sin(31x
2
)I
1im 2 x-tO sin(31x ) limit is
0
=
[:$
tends to O
I. [:z
1 + fix
to get
+ I+)] .
I / (-
=
finite, so the entire limit is 1im
+
(m +fix)] :f~[
denominator tends to
13.
+
By the rational function and the composite function rules,
1im xi0 tan[(x 9.
2 (x
=
By the properties of absolute value,
+ 3 1 x - 11
< 1 , 1 f (x) - 2 / < 46 <
minimum of
5.
f (x) - 2
Use the method of Example 1, Section 11.1.
The
and the numerator is
.
sin(3/x2)]
.
:$
is finite, oscillating between
x
-1
=
0
and
and
1
.
Thus, the
.
17.
The horizontal asymptotes occur at
A ,,
1im lim -1 ~-ti-f(x) . x_, tan (3x + 2) = lim -1 n/2 and x-t-- tan (3x + 2) = - ~ / 2,
---------
Y
=
so
y = in12
are the horizontal
asymptotes. The x-intercept occurs where x 21.
=
-2/3
-
2) ,
i.e.,
. The y-intercept is tan-1 2 .
This has the form 2 (12x
+
0 = tan-'(3x
18x)l
=
m/-
, so ltH6pital's rule gives lirn[(3x2 + 8 ) /
lim[6x/(24x - 18)l
X "
X "
=
1im x,(6/24)
=
114
.
Copyright 1985 Springer-Verlag. All rights reserved.
508
Section ll.R
This has the form
,
010
so
1'Hbpital's rule gives
lim
[(112)(x~+9)-l/~(~x)/cosx = ]o .
This has the form
,
010
so 11H8pital's rule gives us
lim {[cos(x - 2) - 1]/3(x x12 1im again to get x+2[-sin(x gives us
2
- 2)
-
.
}
2)/6(x
1im
x+2 [-COS(X - 2)/6]
=
We can apply l'H6pital's rule
.
- 2)]
-116
Applying the same rule
.
The function x cot x
does not have the appropriate form, so it must
be transformed before
l'Hbpit.alls rule can be used.
lim x10 x cot x
=
1im x+O(x/tan
1im 2 x+O(l/sec x) = 1
llHbpital's rule gives us lim X M
2 -x - lim x e - x , ( ~
rule gives us lim ,(2/eX)
=
2 x
/e )
and has the form
X)
and has the form
.
lim(2x/ex) x-
Applying
Thus,
010 ;
therefore,
.
m/m
;
therefore, 1'H8pita11s
l7H8pital's rule again yields
o . 1im [ln(l x+o+
Find the limit of the logarithm and then exponentiate. sin ZX)~'~I = lim [(l/x) ln(1 x+o+ which has the form (1 i
010
.
n 2 x / = 2
.
=
u = x
lim 2 ~ 1 : = 2 a+O+
-
-
+
Thus, x y : + [ ( l
1 ,
0 = 2
.
so
=
1im = bdr(-l/x)
lim [(2 cos 2x1 x+o+
;1
=
~ f ( d ~ / m= )ji(du/';)
jx In x dx
2 2 x ln x/2 - x / 4
+
C*
(In x)(x
=
.
,
.
tE(-lib) +
.
1
U - ~ I =~ ~
=
2
la .
2
.
2 12) - j(x 1 2 ) ( d x / x )
Therefore,
,/:
x ln x dx
lim 2 a In a12 = a+O+ 3 lim 2 ) I = (1/2)a+o+(-a )
by 1'H0pital1s rule, so the integral converges to (0 - 0) = -114
sin ?x)/xI
sin ~ X ) ~ /=~ e2 I
lim 2 2 1 x dx = a+O+ (X ln x/2 - x 14) 1im 2 lim (l/2)a+o+[ln a/(l/a )I = (1/2)a+0+[(lla)/(-1/a
j; x ln
+
Thus, the integral converges to
Integrat~onby parts gives us 2 x In x/2 - jx dx/2
1im x+O+[ln(l
By 1'Hbpital1s rule, we get
Thus, the integral converges to
Substitute
lim a+o+
sin 2x)I
2 lim b Il(l/x )dx = b+mjlx 'dx
The integral is 1 = 1
.
+
+
(0
-
1/41
=
-
.
Copyright 1985 Springer-Verlag. All rights reserved.
=
0
=
Section ll.R
57.
From Section 9.1, the volume is
nji(xe
2 -2x 2 -2x follows: jx e dx = x e /(-2)
+
-x 2 ) dx
[tE(-x2/~e2x - x/2eZX
i
61.
-
+
C
.
+
xe
-2x
/(-2) +
By l'H8pita11s rule,
1im 2x lim(-2/8eZx) = 0 . Also, x-(-x/2e ) = x-2x = 14) = 0 . Thus, n~o(xe-~)~dx
=
;1
1/4eZX)1
=
.
n14
lim Find the limit of the logarithm and then exponentiate. n+_ [n ln(1 + 1im 8/n)] = ,-[ln(l + 8/n)/(l/n)] . Thishas the form 010 and l'~6pital's
1im n- [((-~n-~)/(l
rule gives us
+
81n))l(-n-~)]
Thus, the limit of the original sequence is n/n
65. Divide by 69.
Integrate by parts as
-2x 2 -2x jxe dx = -x e 12
2 -2x -2x le-2xdx/2 = -x e 12 - xe /2 - e-2X/4 2x lim lim 2 x 2 ) = x-(-~x/ie2x) X" lim lim x-(-1/4e2x) = 0 , and _(-e
.
509
to get
For any integer n , lim(sin n n ~ n - ~=) ncos nn
=
51 ,
1im n-tm[l/(l+ 2111)) sin nn
=
0
.
hE(r
-4
.
))
so the sequence becomes and
+m
111
=
=
+
hE18/(1
8/n)l
8
=
.
. 1
.
Thus, l'HBpitalfs rule gives us
cos rn/(-3n
quence oscillates between
=
8 e
NOW, for any integer
.
?(-1/3)'~~n4 n*
n ,
Thus, the se-
Therefore, the limit does not
- m .
exist. 73.
lim ~ [,,(In By l'H6pital1s rule, lim 2 3nn = exp n*
3n)/2n)l
=
lim exp [,-(l/n)/
21 = e O = l . 77.
Note that
f(-2)
=
-12
iteration formula x,+~ -1.44444 , x2 -1.35530 2 3x
+
.
811 (x
=
and
f(-1) = 4
-1.35916 , x3
Thus, one root is
+
tion gives us
,
3 2 = xn - (xn - 3xn =
so let
+
xo
=
-1 . We use the
2 8)/(3xn - 6xn)
to get
-1.35531 , x4 = -1.35530 , x5
-1.35530
.
By long division,
1.35530) = x2 - 4.35530~+ 5.90274 (4.35530 t J-4.64232)/2
.
x1
=
=
(x3 -
The quadratic equa-
. Therefore, the only root is
Copyright 1985 Springer-Verlag. All rights reserved.
510
Section ll.R
81. We have
(
of
85.
(b - a) /2n = 1/20 and
x
f (x)
0
f(xo
)I
/Ax}
+
Ax)
must exist.
1im
f(xo)
=
f(Ax) = 1 because
=
1im Ax+- { [f(xO
z
. The approximation
+
+
f(x)
1
=
If if
AX) =
+
x
(d/dx)(l)
1im Thus x+o+{ [f(xo
.
Ax)
x
lim = Axi3+~ln(xo + AX)
0
=
<0
<0
therefore,
=
+
Ax)
0
-
Therefore, 011
=
.
-
+ Ax)
, then
.
+ AX)/^]
AX) - f (xo)l /Ax} = Ax+o- [f '(xO
11H8pital's rule applies and
applies again.
.
f (0) = 1
1im
f (x + Ax) = cos(xo 0 1im 3x4+[f '(xO + Ax)/ll
.
1
=
1im f (x) = x+O- 1 = 1 ;
+
f(3)
lim For differentiability, Ax+O{[f(xo
the function is continuous. f(x
/
x+o lim f (x) = f (0) = cos(0)
1im x+O+ cos x = 1 and
=
2
~ by the ~ trapezoidal x rule is (1120) [f(2)
For continuity, we must have lim
f (x) = x
If
because
Ax
>0
f (xo)l /Ax}
,
=
because l'H6pital's rule
Since both one-sided limits are equal, the function is
both continuous and differentiable 89.
(a)
we get
lim = hiO{ [f '(xO
f "(x0)
+
h)
-
+
h) - 0
010 ,
limit also has the form lim h,o{ [f "(x0 (b)
010 ,
The right -hand side has the form
By definition, f "'(x0)
=
h+Oi
f '(xo - h)(-l)]
[f "(xO
[f "(x0)
=
+
-
2f (xO
2 h 1 /hi =
+
h)
kz{
+
[f(xo
- (f (xo
f (xo))/h2
+
2h) - 3f (xO
+
h)
+
+ f "(xO)]
h) - f "(xO)] /h!
tuting the formula from part (a) yields 2h)
/2hl
.
This
=
f "(x0)
so l'H6pital's rule yields
f "(x0 - h) (-I)] 12 1im
+
so by 1'86pital1s rule
fl"(x ) 0
=
h) - 2f (xo)
+
/2
.
Substi-
lim{ [(f(xo hi0
+
+
f (xo - h))/
31 (xO) - f ( ~ -g h)] /h31
.
This can be verified by applying 1'Hepital's rule three times. 93.
Sn
is the Riemann sum for a = to = 0 ,
partition is
c. has been chosen to be
1 jo(x
+
2 x )dx
=
fix)
=
+
x2 , where
tl
l/n ,
t2 = 2/n ,
so as
n
ti
2 3 (X 12 + x 13) 1;
=
, =
x
112
+
+
m
Ati = l/n
...
, tn
=
, fit. -+ 0 and
113 = 516
and the b
=
1
Sn
+
.
Copyright 1985 Springer-Verlag. All rights reserved.
.
97.
(a)
(b)
7
y=L+m(x-a)
f(x) L
+
l i e s between t h e two l i n e s means t h a t
.
m(x - a )
m(x - a )
Subtracting
, which i m p l i e s
/
sion gives us (c)
Given
E
>0
-
[f (x)
6 = ~ / 2 m,
1 f (x)
we h a v e shown t h a t
- L/
f (x) = 0
and
N(x) = x
comes
x - f(x)/fP(x)
, i.e.,
f '(x) (b)
Xd'
,
f '(x) # 0
.
.
0/f '(x) = x
Now, i f
.
[f '(x)l
5
N(;)
=
x
-
x] / ( x - x )
in
.
(x,x)
f(G)/f '(x)
=
[(fr(x))
/N'(L)/
x
o/f1(x)
=
,
,
, i.e.,
then
beN(x) =
0 = f(x)/
2
- ( f ' ( ~ ) )+ ~f ( x ) f H ( x ) ] /
.
and r e a r r a n g e m e n t y i e l d s
Now, by p a r t ( b ) ,
/ <6
Thus,
~ ' ( x )= 1 - [ f ' ( x ) f ' ( x ) - f ( x ) f "(x)] /
Since
G -
/x - a
N(x) = x
0 = f(x)
)I(x)/
<
0
N(x) = x - f ( x ) / f ' ( x )
x = x - f(x)/fl(x)
2
divi-
.
f(x) = L
By t h e mean v a l u e t h e o r e m , we h a v e f o r some
whenever
becomes
This simplifies t o
= f (x)f
[f ' ( x ) l
< f(x) - L . Finally,
Then, t h e L i p s c h i t z c o n d l -
then
By t h e sum and q u o t i e n t r u l e s , [fr(x)12
(c)
-
1im
whlch i s t h e d e f i n i t i o n o f If
--
a/
, <
m(x - a ) S f ( x )
If(x) - L/S m l x - a / S m h = m(~/2m)= c/2
t i o n implies
101. ( a )
sm/x a) / 9 m .
/f(x) - L/
L] / ( x
,
-m(x - a )
yields
L
-
L
N ' ( F ) = [ N ( x ) - N ( X ) ] / ( X - X) i s a r o o t of =
x
.
f
, we h a v e
~ h u s , N'(F)
N(x) -
x
=
2 / f ( ~ ) "f ( t ) / ( f l ( ~ ) ) /
=
x ' ( E ) [x -
[N(x)
-
GI .
jf(~)I~/p
2
.
Copyright 1985 Springer-Verlag. All rights reserved.
512
Section ll.R
101. ( c )
(continued) By Consequence 1 of t h e mean v a l u e t h e o r e m , p G
/
[ f ( ~ )- f ( x ) l / ( t - ;)I
since
f(x)
If(6) G q / x/N(x) (d)
.
0
=
x/
And s i n c e
.
2
.
In
/N(x) -
=
3.0
[1.4,1.5]
x
/x
,
x
and
M = 2
x -
, so
For t h e f o u r t h
so 1.8
x
, so
since
/N(x) -
a r e needed t o g u a r a n t e e
x
so
,
.
C = qM/p
0.1
is
and
fl'(x) =
, so
.
G 0.000045 /N(x)
-
For
x/ G
x / < 1.5
Forthefifthiteration,
.
.
For t h e second
/x -
x
2
, we s u b s t i t u t e
, so
iteration
G 2.5
20
x/
, we h a v e
f l ( x ) = 2x
/N(x) -
< 0.000045
-18 /N(x)-xl<1.8~10 10-l8
x(
f o r the f i r s t iteration.
/x -
implies
.
- G / < 0.00765
.
lo-'
15 -
/ N 1 ( ~ )G I q / x - x / ~ / p ~F i n a l l y ,
, t h e maximum of
the third iteration, 1.538
-GI >
'(x)
G if(<)/ G q / <-
x / G (qM/p 2 ) j x - x / 2 , 2 - 2 / N ( x ) - x / < (qM/p ) / x - x /
I G 0.00765
iteration,
lx
XI
-
-
= ~ N ' ( E )Ix
, q
PIS
or
Therefore
Using t h e f o r m u l a , p = 2.8
G q
p
X
lo-'
,
I x - x / <
Thus, f i v e i t e r a t i o n s
decimal place accuracy.
TEST FOR CHAPTER 1 1 1.
True o r f a l s e . (a)
l im sin x x-
(b)
I f S i m p s o n ' s method i s a p p l i c a b l e , e i t h e r t h e t r a p e z o i d a l r u l e o r t h e
=
0
.
method of Riemann sums may a l s o be u s e d . (c)
I f a n i n t e r v a l i s s u b d i v i d e d i n t o a n odd number o f e q u a l s u b i n t e r v a l s , S i m p s o n ' s method may b e u s e d e v e n i f a d i s c o n t i n u i t y e x i s t s .
(d)
(e)
The i n t e g r a l By l ' H 8 p i t a l ' s
1 2 i-l(dx/x ) rule,
1i m
=
-x
-1 1
+
converges.
x + O [ ( ~ l ) / s i n x]
=
I. i m x , O ( l / c ~ ~X )
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section ll.R
2.
Let
2 f ( x ) = x ( Z x + 4 ) ( x - 7) / ( x + 3 ) ( x -
(a)
F i n d t h e v e r t i c a l and h o r i z o n t a l a s y m p t o t e s of
(h)
For t h e v e r t i c a l asymptotes
3.
Use t h e
+f ( x ) 0 E - S
f o r each
xo
4.
and
found i n p a r t ( a ) .
if
x + - 1
if
x
f (x) =
(a)
:z(tan
(b)
1i m x+O+(l l i m -x e
X"
(d)
=
-1
2
x/x3 - 1/x )
+ e
x
I(
G+5-
l im X + , [ ( ~+ 3 ) l j &
- 6 1 1
E v a l u a t e t h e f o l l o w i n g improper i n t e g r a l s : -x
(a)
j;
(b)
ji(dx/ 3
xe
dx
[et/ (1
(c)
6.
1i m f ( x )
Compute t h e f o l l o w i n g l i m i t s :
(C)
5.
compute
f(x)
l im d e f i n i t i o n o f t h e l i m i t t o p r o v e t h a t xi- f ( x ) = 5
x 2 - x + 3 if
,
2
xixO-
1im X+X
x = x0
7)(7x-2)
513
m
+
) eZt)] d t
&+
Sclve t h e equation
x = 3
a c c u r a t e t o t h r e e d e c i m a l p l a c e s by
u s i n g N e w t o n ' s method. 7.
U s i n g t h e f o l l o w i n g methods w i t h t h e p a r t i t i o n 1.5
,
1.75
,
2.0)
,
(0
, 0 . 5 , 1.0 , 1.25 ,
a p p r o x i m a t e t h e a r c l e n g t h of
x4
if
0 < x < l
x3
if
1 < x < 2
f (x) =
8.
'
(a)
Use S i m p s o n ' s m e t h o d .
(b)
Use t h e t r a p e z o i d a l m e t h o d .
DO t h e f o l l o w i n g i n t e g r a l s c o n v e r g e o r d i v e r g e ?
4
(a)
jm_x exp(-x
(b)
/23[dx/(3&+
J u s t i f y your answer
)dx
5
~ 7&)l +
Copyright 1985 Springer-Verlag. All rights reserved.
514
Section ll.R
J;
4 x(ln x) dx
8.
(c)
9.
Functions of the form
--
may have any value (if the limit exists).
Give a simple example of a limit having the form
10.
(a)
+-
(b)
- m
(c)
3
(Hint:
consider
m
-
whose value 1s:
l/ln x - l/(x - 1) . )
Krazy Karen's Kookie Kangaroos is a unique rent-a-kangaroo firm.
People
moving across town can rent a kangaroo, fill its pouch, and ride the kangaroos with their belongings across town.
Krazy Karen, who used to
work for a mad scientist, has developed energizing tablets for her kookie kangaroos. The average kangaroo can hop for after consuming n
energizing tablets.
100n/(n
+
3)
kilometers
The effects of the tablets last
only one day. (a)
Plot the kilometerage, km, vs. the number of tablets, n ,
for
n = 1 , 2 , 3 , 4 , 5 , 6 . n
.
(b)
What is the limit of the sequence as
(c)
Prove your answer in (b) by using the definition of the limit of
-t m
sequences.
ANSWERS TO CHAPTER TEST
1.
(a)
False; limit doesn't exist.
(b)
True
(c)
False; Simpson's method requires an even number of subintervals.
(d)
1 2 False; j-l(dx/x )
(e)
False; 110
=
0 2 1 2 j-l(dx/x ) + jo(dx/x )
=
(-x- ) I O-1 + ( - x
-1
l o1 .
11H6pital's rule doesn't apply to functions with the form
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section ll.R
(a)
y = 2/49
asymptotes at
6 = ~ / 2,
Choose (x
x = -3
Vertical asymptotes at
+
if
1 - 1) - 21 / x + 11
(a)
113
(b)
1
(c)
0
<
<
6 ,
and
x
515
217 ; horizontal
=
.
/f(x) - L / = / x 2 - x + 3 - 51 = /(x+ 1 - 1)
so
/ x + 112 then
+
Ix
+
11
<
'6
+
6 < 26
E
.
\f(x) - L / < 26 =
if
6
<
1
2 -
. Thus,
1.697 (a)
8.682
(b)
8.955
(a)
jm-x exp(-x
f -m m x (b)
4
(C)
3 &
. XI(^& + 5& + '&)I < ' -1 [dx/3 3&4 3&) = xZi3/2 1 - 1 + x2l3/2 1; = i .
+
5&
+
/I_=o .
Therefore,
converges.
J:~ (
(b)
2 2 )dx = ex?(-x ) /2
The integral is improper around
!;(dx/3
(a)
)dx G /I-x ex?(-x
exp(-x )dx
[
(c)
4
x = 0
If =
-1
<x <1
jy1(dx/3 Therefore,
, then
+
lil [dx/
'&)I converges.
The integral converges because
4 1im x4+ ~ ( l nx).
=
0
.
1im x,o+(2/~ - 11x1 1im x+O+ ( I / x - 21x) 1im x+l+ (61111 x - 6/(1 - x))
Copyright 1985 Springer-Verlag. All rights reserved.
516
lo.
Section l l . R
(a)
f
(n)A
75
-e
50
e
e
--
a
a
25 --
r
, * 2 (b)
100
(c)
Choose
4
N > (300
'
-
6
n
~ E ) / E
Copyright 1985 Springer-Verlag. All rights reserved.
CHAPTER 12 INFINITE SERIES
12.1 The Sum of an Infinite Series PREREQUISITES
1.
Recall the concept of convergence and divergence for infinite sequences and improper integrals (Sections 11.3 and 11.4).
2.
Recall how to compute a limit at infinity (Sections 1.2, 11.1 and 11.2).
PREREQUISITE QUIZ
1;
+
2)]
Show that
2.
Does the sequence
3.
Compute the following limits: lim (a)e(,. 1im (b) soc(,.
x
1.
[dx/(x
2
1.
(-1)"
converges. converge as
n
+
-
?
Explain
4
)
2 x/x )
GOALS 1.
Be able to compute the sum of a geometric series.
2.
Be able to use the
ith
term test to show divergence.
Copyright 1985 Springer-Verlag. All rights reserved.
STUDY HINTS 1.
I n f i n i t e s e r i e s , p a r t i a l sums, and s e q u e n c e s . quenceofnumbers.
Suppose we h a v e a s e -
The sum o f t h e f i r s t n numbers i s c a l l e d t h e
The p a r t i a l sums f o r m a s e q u e n c e .
p a r t i a l sum.
t h e sum o f a n i n f i n i t e number o f t e r m s .
nth
An i n f i n i t e s e r i e s i s
I f t h e p a r t i a l sums h a v e a
l i m i t , i t i s t h e sum o f t h e s e r i e s
2.
Convergence v s . d i v e r g e n c e . otherwise, it diverges.
I f a s e r i e s h a s a f i n i t e sum, i t c o n v e r g e s ;
Convergence r e q u i r e s a f i n i t e sum; t h a t i s , t h e
l i m i t o f t h e p a r t i a l sums must e x i s t . does not e x i s t , t h e s e r i e s diverges. 1
+ ...
I f t h e l i m i t of t h e p a r t i a l sums For example,
h a s p a r t i a l sums a l t e r n a t i n g between
1
1 and
- 1 0
+ .
1 - 1
+
-
1
Thus, t h e
l i m i t o f t h e p a r t i a l sums d o e s n o t e x i s t , and t h e s e r i e s d i v e r g e s .
This
example shows t h a t c o n v e r g e n c e i m p l i e s t h a t a s e r i e s h a s a f i n i t e v a l u e , but divergence does
3.
Notation.
a.
imply t h a t a s e r i e s t e n d s t o i n f i n i t y .
i s t h e symbol u s e d f o r t h e
i s t h e symbol d e n o t i n g t h e
4.
1
,
r
=
and
i
Most o f t h e box on p . 563 i s common If
Ir
1 .
1
or
/ rl
you s h o u l d u n d e r s t a n d t h e r e s u l t from y o u r work w i t h e x p o n e n t s .
1 ,
then
limrn = 1 n-
.
IT
r = -1
,
rn
a l t e r n a t e s between
>
If
1
- 1 , s o no l i m i t e x i s t s .
( f i r s t term)/(l
6.
t e r m of a s e r i e s . .
P r o p e r t y 9 i s m o s t l y r e v i e w from p . 5 4 2 .
Geometric s e r i e s .
0 ,
th
p a r t i a l sum.
P r o p e r t i e s of l i m i t s o f s e q u e n c e s . sense.
5.
i
th
i
so
a
You s h o u l d memorize t h e f a c t t h a t
- r a t i o ) , provided
/r/< 1
.
Z. 1=0a r
i
=
a/(l
The s e r i e s b e g i n s a t
i s t h e f i r s t term.
Algebraic rules. these statements.
A g a i n , common s e n s e s h o u l d t e l l you t h e v a l l d i t y of Note t h a t t h e s t a t e m e n t s a p p l y o n l y f o r c o n v e r g e n t
Copyright 1985 Springer-Verlag. All rights reserved.
- r) i =
Any changes which occur at the beginning of a series
Important tails.
7.
do not affect the convergence or divergence. \\%at is important for convergence is the behavior as
i
approaches infinity.
Note how this
example applies to Example 7. 8.
ith
term test.
as
i
Divergence is guaranteed if
approaches
m
.
Note that
a.
does not approach
not guarantee
approaching 0 does
ai
0
convergence. The harmonic series is a counterexample.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
By definition, 3
5.
=
112
+
113
Sn = :7l=lai
+
114
=
.
13/12 ;
S 4
=
=
112 ; S2
=
112
112
+
113
+
114
+
115 = 77/60.
)-i,oar i
=
a/(l
This is a geometric series, so we use mula may he used since 1/(1 - 117)
the sum is 9.
r
=
117
=
1/(6/7)
1
<
During the first year, you draw out
, then
(3/4)($10,000) so forth.
The first term is
r)
516 ;
=
.
The for-
=
1 ,
a
so
.
$10,000
(3/4)2($10,000)
-
r:~=0(3/4)i(10,000)
The series is
(23i+4
1. 1 4
=
.
, then
Next year, you draw out (314) 3 ($10,000), and
10,0OO/(l - 314)
z3/3'
=
819
z4/33
=
16/27
<
1
.
=
$40,000
2i+5 4 5 7 7 /3 ) = 2 / 3 + 2 / 3 +2lO13~+
is a geometric series beginning with
17.
-
113
For an arbitrarily large life span, the total amount to be
drawn out is 13.
. 716
=
+
S1
Thus,
Thus, the sum is
24/35
.
... .
and having a ratio
( ~ ~ / 3 ~ ) / (-1 8/91
=
4 5 9(2 13 )
This
r = =
.
(zn + 3")/gn = ( 2 1 6 ) ~+ ( 3 1 6 ) ~= (1/3In+ ( 1 1 2 ) ~. z;=~ [(zn + 3n)/6n] is the sum of two geometric series and
We have
~:=~(1/3)~+ 1z=1(1/2)n
=
Thus, it equals
Zm n=O (113)(1/3)~ + ~;=~(1/2)(1/2)~
=
(1/3)/
Copyright 1985 Springer-Verlag. All rights reserved.
520
21.
Section 12.1
in
,
li
C:
1=1
i
.
i
l
.
m
1im i-t,ai # 0
Since
1 1 2 ~ )= c T = +~ ~ ~
r = 112 ;
th the i
,
A/&
Divide by
, the denominator tends to
->
+
(1
however,
r = 1
so it diverges. Thus, the entire integral diverges.
Consider the terms As
1=1
:a1=1( 1 1 2 ) ~ converges since
~ ; = ~ ( 1 / 2 ) ~.
25.
Z?
Using the method of Example 6 , we have
1
f i l m.
to get
and the numerator tends to
term test tells us the series
diverges. 29.
lim a . = 112 # 0 , i- 1 33.
Let
a.
:1 (a. 1=0 1 37.
1
This series is equivalent to
(a)
=
+
Let
1
and
+
+
112
+
112
Then
a. i
112
+ ... .
the series diverges. b.
=
-1
for all
.
i
+
h. = 0 1
b.) = 0
converges, but both
1
and
m
m
tk
Since
Z. 1 1=0
and
Zi,O(-l)
be the carriage transit time for trip
diverge.
. Let dk be
k
k
the distance between the crews at the beginning of trip t2n+l = d2n+1/(20
+
7)
and
t2n+2 = d2n+2/(20
+
speed is a sum of the riding and working speeds. dn
- (5
+
=
t~n+l
7)tn
(12/27)(1
C?1=1t.)] 1 =
+
d1(5
=
1=1 1 2n - CiZlti) , and
(12~27)
(13/25)(1
=
- ?nilti
- L;:;lti)
(b)
=
- (12/25)(1
. . .
(12/25)t2n-1 = rnt 1 = rn.(12/27) rn.(15/25) (12127) = rn+'.
(12/13)
= n lim x c "i=Or
fact that
1
+
r
+
r2
i.(12/27)
+ ... +
lim (12127) [(rnil - l)/(r n --
- I)]
+
rn
=
(15/25)t2n+l 211-1
-
Therefore, Also,
The total time for carriage travel is
C:=ot2i+2) .
-
2n+l
(12125) [l - 6Z?lti - (12127)(1
(13/27)t2n
=
dn+l
2n+2
.
ti) =
-
Similarly, (12127)
=
I
2n+l = (13127) =
(15/25)t2n+1
=
lim,n lim n n m l i = l t i = niz('.i=Ot2i+l
r~=0ri+1.(12/13)] (?+I
=
Therefore,
t2n+2 = (12/25)(1
2n - Zi,lti)
(12/25)(15/27)(1
.
Then
since the
Also,
t. = 12(1 - Zrz1ti)
7)Z:
2n - Zizlti - t2,+,)
(12/25)(1
'=2n+l
=
5) ,
.
-
+ n-r.(12/13). lim
.
+
Using the
l)/(r - 1) , we get [(rni2
-
l)/(r
- I)]
Copyright 1985 Springer-Verlag. All rights reserved.
=
Section 12.1
37.
(b)
521
(continued) (12127) (
+
[(12/27)
+
r - 1
-
r l 2 1 3I
-
r-(12/13)1/(1
r)
- 1
(since
lr/
<
1) =
[(I21271 + (12/27).(15/25)1/(1
=
- r)
(12127). (40/25)/1 [25-27 - (13-15)l 125-271 = [(12.40)/(25.27)1
x
.
[25-2714801 = 1
SECTION QUIZ 1.
(a)
Discuss the convergence or divergence of 1)
(b)
+
+
(1
+
0 - 1)
+
(1
+
0 -
.
+ ...
0 - 1)
(1
What happens if the parentheses are removed from the series in part (a)?
2.
The series 312
ratio
1
.
+
312
+
914
+
2718
+ ...
is a geometric series with
1/(1 - 312) = -2
Thus, the sum is
.
The sum of positive
numbers can't be negative. What's wrong? 3.
The sum of a geometric series is is
4.
2
.
5
and the first term of the series
Write the series in the form
Consider the series ith
2Yz13
Ciz0ar
i
.
.
(a)
Use the
term test to analyze the series.
(b)
Does part (a) tell you anything about the tenth partial sum? Explain.
5.
Mindless Marvin, the mixed-up medical student, wanted to start one of his patients on a drug which would be used for the rest of her life Mindless Marvin knew that drugs obeyed exponential decay in the body so that
A exp(-kt)
second dose,
would be left after the first dose.
A[exp(-kt)
could see that after Alexp(-kt)
+
exp(-2kt)
n
+
exp(-2kt)l
isleft.
After the
Even Mindless Yarvin
doses, the amount in the body would be
+ ... +
exp(--nkt)]
. Seeing this, Mindless
Copyright 1985 Springer-Verlag. All rights reserved.
=
522
5.
S e c t i o n 12.1
(continued) Marvin runs t o t h e pharmacist and a s k s , "Won't t h e drug l e v e l become i n f i n i t e and k i l l t h e p a t i e n t a f t e r s e v e r a l doses?" responds, "No, Mindless Marvin." a r e c o n s t a n t s and
Explain t h i s i f
The pharmacist
,k ,
A
and
t
.
/exp(-kt) / < 1
ANSWERS TO PREREQUISITE Q U I Z 1.
!g[dx/(x
2
+
2)1 = ~ i [ d x / ( x+~2)1
i s p r o p e r , so i t converges.
+
+
j;[dx/(x2
+
jy[dx/(x2
2 ~:[dx/(x
.
2)1
2
/k(dx/x ) = -x
2)1
-1
+
/-
1
2)1
= 1
.
Therefore, t h e e n t l r e i n t e g r a l converges. 2.
It d i v e r g e s ; i t o s c i l l a t e s between
3.
(a)
+m
(b)
0
-1
and
.
1
LVSWERS TO SECTION Q U I Z
1.
(a)
The p a r t i a l s sums a r e
, 0 , 0 ,
0
... .
Thus, t h e s e r i e s converges
t o zero. (b)
1
Without t h e p a r e n t h e s e s , t h e p a r t i a l sums a r e 0
,
... .
,
0
,0 , 1 , 0 ,
Thus, t h e s e r i e s d i v e r g e s s i n c e t h e p a r t i a l sums have
no l i m i t . a / ( l - r)
2.
The formula
3.
~;=~2(3/5)~
4.
(a) (b)
I t d i v e r g e s because
a p p l i e s only i f
a.
=
3
Irl
for a l l
P a r t i a l s u m s always e x i s t ; h e r e
SI0
<
i =
30
1
and
.
. izai = 3 The
i
th
>
1
.
term t e s t
only a p p l i e s t o i n f i n i t e s e r i e s , n o t t o p a r t i a l sums. 5.
This i s a geometric s e r i e s where drug l e v e l i s
A exp(-kt)/[l
r
=
exp(-kt)
- exp(-kt)]
.
Thus, t h e maximum
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.2
12.2 =Comparison
523
Test and Alternating Series
PREREQUISITES 1.
Recall how to use the comparison test for studying improper integrals (Sections 11.3).
2.
Recall that the harmonic series diverges (Section 12.1).
PREREQUISITE QUIZ Discuss the convergence or divergence of
1.
3
2.
Show that
j7[l/(x5
+x +
3.
Show that
jl[l/(&
+
2 2s )]dx
3&)]dx
~ y [i/(i2 = ~ - 4i)I
converges.
diverges.
GOALS Be able to demonstrate convergence or divergence by using the comparison
1.
test. 2.
Be able to show convergence or divergence by using the ratio comparison tests.
3.
Be able to distinguish between absolute and conditional convergence.
4.
Be able to apply the alternating series test to show convergence.
STUDY HINTS 1.
Comparison test.
You can only draw a conclusion if the terms of a series
are smaller than a known convergent series or larger than a known divergent series.
If a series is shown to be termwise larger than a conver-
gent series, you have no information; it may converge or diverge.
Sim-
ilarly, showing that a series is termwise smaller than a divergent series tells you nothing. m
than Zi,ll,
Example:
2 ~ y = ~ ( l / n)
and
~ y = ~ ( l / n ) are both less
but this comparison tells you nothing about the convergence
Copyright 1985 Springer-Verlag. All rights reserved.
524
1.
Section 12.2
(continued) or divergence of
Z:
1= 1
values in the test. 2.
(11,')
or
Zm
1= 1
(1111)
.
Note the use of absolute
All of the terms being compared must be positive.
Using the comparison test.
The first step is to guess convergence or
divergence by resemblance to a known series.
If one wishes to show
convergence,onecan make the numerator larger and the denominator smaller. If the new series is less than a known convergent series, then the original series converges. As an example, consider the series :1
1=1
2 6 i3 - i )/(i
+
.
8 7i ) ]
Note that as
i
[(4i
4
+
becomes very large, the numer-
ator may be made larger by increasing the exponent of positive terms or by deleting negative terms.
Similarly, the denominator may be made
smaller by deleting the positive terms with large exponents. Thus, 4 (4i
+
2 6 i3 - i )/(i
+
8 7i ) S (4i4
+
i4)/i6 = 4/i2
so, the series in question converges.
for large
i
and
Similar techniques may be used
if divergence is suspected.
3.
Ratio convergence tests. parison test. m
These are usually easier to use than the com-
Convince yourself of their validity.
, then eventually,
/ a i /<
chi
Again, convergence is implied.
If
lim i-(
! ai 1 /bi) <
, where c is some finite constant.
Similar reasoning should convince you of
the second part of the'test. 4.
Error estimates. Often, we wish to estimate the sum of a series; however, an estimate is useless if it is far from its true answer. Error analysis helps us to decide if an estimate is useful.
Study
Example 5 to seehow upper bounds of errors are estimated.
5.
Alternating series. All alternating series converge with error no greater than
1
if the partial sum,
is the estimate of the sum.
Sn
=
al
+
a2
+ ... +
an
,
The three conditions inthetopboxonp. 573
Copyright 1985 Springer-Verlag. All rights reserved.
5.
Alternating series (continued). need to be shown before a series can be called alternating.
6.
Absolute vs. conditional convergence. Absolute convergence means a series would still converge if all the minus signs were removed. example,
~;=~(-1)~(3/5)~
converges absolutely as a geometric series.
Conditional convergence means that the minus signs are necessary. example,
For
For
C? ~(-l)~/i] converges, but if the minus signs are removed, 1=1
it would diverge.
7.
Increasing sequence property. tional for most classes.
This is theoretical material and is op-
Ask your instructor.
Simply stated, a
sequence increasing toward an upper bound has a limit.
No statement
is made about its associated series.
SOLUTIONS TO EVERY OTHER ODD EXERCISE For
1.
i
>1
, 81(3~ + 2)
< 8 1 3 ~.
Also,
:1
I= 1
which is a convergent geometric series.
(813~)= 8 C ~ = ~ ( l / 3 ), ~
Thus,
CiZl " [81(3~+2)1
con-
verges by the comparison test.
5.
For
i
>1
,
jail = l/(3
i
< 1 / 3 ~ and 1. [(-1)i/(3i + 2 1=1 +
2)
i ~;=~(1/3)
is a convergent
m
geometric series.
Thus,
converges by the compari-
son test.
9.
For
i
>1 ,
3/(2
+
i)
3/(2i
+
i ) = l/i
, and
C'u
[3/(2
< 3i4"
and
divergent harmonic series. Thus,
1= 1
+
i)]
rn
li=l(l/i)
is the
diverges by the
comparison test. 13.
For
n
>1
,
/ a n \= 31(4"
+
2)
is a convergent geometric series. Thus,
c:=~
In=1
(314") =
[3/(hn + 2)1
(114") converges
by the comparison test.
Copyright 1985 Springer-Verlag. All rights reserved.
526
17.
Section 12.2
Note t h a t /
21.
1/(3i
+
+
1
31
zW [1/(3i + 1=1
l/i)l
Let
+
+
2i+1)
0
or
1
+
a i = I1
Let
Z. b . 1=1
a . = 3i/2
3i/2i m
a
j
1
and
+
.
+
212 i + l
<2
(-l)i/
Zm .
-
l=lai i
.
Since
8i
.
= b.
+
There-
( 3 / 2 j i , we h a v e
f1 = 1b .
ai
=
converges, s o does
1
.
I
Since
jsin j m
Thus,since
j'
Lj,l
/
1)
a
i + ai
1
= 1 =
lim
.
Therefore, since
-
1)/(5"
k In=jan
+
for a l l
a . a l s o c o n v e r g e s by c o m p a r i s o n . J 1i m and bn = 1 / 2 ~ Then ni..(an/bn)
= niG
+
~ : = ~ cbo n~v e r g e s , s o d o e s
.
1) < (2/5)n
j,
jp,/
is l e s s than o r equal t o
1
when
=
.
10
d i v e r g e s by t h e
, and f o r
1 # 0
n
.
i
Thus,
Therefore,
.
(215)~+~/(1 215) =
lo a ZnZ1
,
Z=
.
0.37
Thus,
ith term t e s t .
, a.
> 1
In=lai
The e r r o r i n a p p r o x i m a t i n g t h e
1im 7 im [k/(k+l)]=ii,[l/(l+l/k)l=l#O. k-
k
<1
converges a s a
b
Notethat
l im
Since
.
< 0.01
Let
.
1/2i
< 2/zi+'
zifl)
Since
=
3(2k+1)/5k
k 45.
(.-lti1 / ( 8 i
=
m
.
I)] = 1
s e r i e s by
41.
Then, by
111 '
= 1 / ( 2 ~
a n = (2"
Let
b.
, 11
2
b . = 112'
a j < 1 / 2 ' = b
+
a
(zn + 37.
and
Therefore, s i n c e
and
geometric s e r i e s , Let
Since
diverges also.
bi = 3 ( 3 / 4 ) i
.
= sin j/2j
wehave
33.
.
1))
.
1l.= l a i Let
i
+
(1/3)(l/(i
=
converges, s o does
3 ( 3 / 4 ) i = bi
i
+
/ai] = [l m
fore, since
29.
(-lIil / ( 8 i
( - I ) ~ equals e i t h e r
,
3)
i s a harmonic s e r i e s , i t d i v e r g e s .
comparison,
2i+l> *i+l
25.
> 1/(3i +
l/i)
= (-l)i+l(i
a
- lj/i
d i v e r g e s . by t h e
.
Then ith
term
test.
49.
The d e r i v a t i v e of
ln[(n
creasing function f o r n]
=
ln(1) = 0
.
+
ll/n]
n > 1
.
is
l/n(n
+
1)
, so
The s i g n s a l t e r n a t e and
i t i s a de-
1im
n-,
ln f(n
+
T h e r e f o r e , t h e s e r i e s converges c o n d i t i o n a l l y a s an
alternating series.
Copyright 1985 Springer-Verlag. All rights reserved.
1) /
S e c t i o n 12.2 49.
(continued) ~:=~ln[(n + l ) / n l = C-n= 1 [ l n ( n lim t h i s i s a t e l e s c o p i n g sum, Zn,l = n-tm l n ( n + 1 )
-
Absolutely,
The e r r o r i n e s t i m a t i n g (114) (1/5)"+'
.
1/2n In=1 57.
. +
l)/nl
~:=~(1/5)~ is
+
(114)(1/5)~+'
1 1 5 ~ 1%
a l = f i = 2 ;
61.
+ 2)1 .
(n
+
or
1)
2n2
a
<0 +
For p o s i t i v e
;
therefore,
5>
a
c:=~
increasing. a
+
an
20
,
Since
.
,
n
1)
+
for
115~1 a
a
,
if
=
n
so
n
> 26 ,
and
G
=
2.56155
.
.
12
2 4 a n > an a
0
= 1
.
n >2
<0
matical induction.
For
Thus,
,
n = 0
a
,
4
,
so
.
0
.
>0
1)
a
n+l
> an
or
Rearrangement and
<4 ,
then
a
is
i s a l w a y s a sum o f p o s i t i v e t e r m s , i s bounded above by
n al
=
112
Now, assume t h e s t a t e m e n t h o l d s f o r
Then,
a
4 ;
+
(2n) ( n
0 < a
true.
< 412 + fi =
=
n+ 1
s o t h e sequence i s i n -
and
if
a
> a means -217 2 2n2 > -2n3 - 4n2 + 2n +
52 a n / 2
, an+l
We w i l l now show
IJe
an+l
i s bounded a b o v e by
which i m p l i e s
l/5) =
-0.087
3
-2n3 -
or
-
1
Thus,
is l e s s than
[(-l)n/2n~
i s i n c r e a s i n g , we w i l l show t h a t
squaring implies
so
n)/(Zn)(n
> (1 - n) (2n + 2) (n + 2) > 2n + 4 . T h i s i s t r u e
To show t h a t an/2
-
Since
. .
lima = 2 . 5 6 1 5 5 . . n* n
an = (1
creasing.
65.
-
T h i s i s a n i n c r e a s i n g s e q u e n c e w h i c h i s bounded above.
(n
.
= m
(1/5)(1/5)~+'/(1
2.56155 ; a10 = 2.56155 ; a l l = 2.56155 ; a guess t h a t
.
1) - ln(n)l
- ln(1)
11211 < 1 / 5 0
26 Z.n,l [(-1)"/2n
=-=z;
a2
+
only converges c o n d i t i o n a l l y .
The e r r o r i n e s t i m a t i n g
Therefore,
[(-1)"/2n
(a)
+
~ : = ~ ( - l l) n~[ ( n
the series 53.
527
therefore,
a
+ n
1
312
=
-
1
,
4
<4 i.e.,
u s i n g mathe-
, which a
n-1
is
<4 .
i s i n c r e a s i n g and bounded
Copyright 1985 Springer-Verlag. All rights reserved.
528
Section 12.2
65.
(continued)
.
4
above by
to a limjt, an/2
+
a
=
.
R
To find
, we solve
V
5.
The solution is
x
n
69. For any
lim a nn
By the increasing sequence property,
and any
(314)~$(4~x) .
,
$(4"
lim a = 4 n- n
P.
=
-
x)
converges
lim = i = e / ~ + f i = a nnfl
<1 .
.
Let
bn = ( 3 1 4 ) ~ and let
/ a n ]= (314)~~(4~x) < ( 3 1 4 ) ~= bn
Then we have
.
rn
Therefore, since
InZOb
converges as a geometric series, so does
m
Zn,oan
by the comparison test.
SECTION QUIZ m
a
1.
Let
Ci,lai
be the series to be analyzed and let C . b. be a series 1=1 1
whose convergence or divergence is known. test say? (a)
(More than one may be correct.)
1 bi 1
and
I ~ q , ~1 b ~converges, then :1.,lai
also
/ b i /< lail
and
\ ~ y = ~ converges, b ~ / then :C
also
1 a. /
If
What does the comparison
<
converges. (b)
If
converges. (c)
If
/ail
>
Ibi/ and
:1
1=1b.1
1
diverges, then
Cy=lai also
diverges. (dl
If
Ibi! > /ail and
I ~ y = ~ bdiverges, ~ / then :C l,lai
also
diverges. 2.
Does
3.
Can
3
+2+ 1+
'n=10
[(-l)n(n
112
+
113
- 5)/n]
+ ... +
1/100
converge?
Explain.
be analyzed by the alternating series test?
Explain.
4.
Determine if the following series converge conditionally, absolutely, or not at all.
(a)
Justify your answer.
[-11(3~ - 4511
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.2
5.
Little Lisa was a very bad girl.
She went out last night and slept
today rather than doing her homework. brain, she can only remember of what she learned
(2
+ zn)
2
114
Because Little Lisa has a little
n
416
of what she learned yesterday,
days ago and so forth, i.e., she retains
of what she learned
529
days ago.
2 n /
Explain in terms of con-
vergence or divergence of infinite series whether Little Lisa's little brain is capacity-limited, i.e., does she have a bird-sized brain?
ANSWERS TO PREREQUISITE QUIZ 1.
For
i
>5 ,
i/(i2 - 4i)
> i/i2 =
.
l/i
Thus, ~
i [i/(i2 = ~ - 4i)l
diverges by the comparison test and by the divergence of the harmonic series. 2.
1; [dxl( r5 + x3 + 2x2 )] < j';.[dx/2x21
3.
jT[dx/ (& +
'A)]> jT[dx/ ( & + I)&
= =
(-1121:)
GI; ,
11 ,
which converges.
which diverges.
ANSWERS TO SECTION QUIZ 1.
aandc
2.
This is not an infinite series, so it does converge. If the series continued as +1/101
+
11102
+ ...
, then it would diverge.
.
3.
No,
1irn ,-[(n
4.
(a)
Absolute convergence; 1 / ( 3 ~- 45)
-
5)/nj = 1 # 0
< 1 1 2 ~ if
n
>4 ,
and
Z? (112~) is a geometric series. 1-4
Copyright 1985 Springer-Verlag. All rights reserved.
530
4.
S e c t i o n 12.2
(b)
Converges c o n d i t i o n a l l y ; and
(n2
+
5 ) / (n3
m
T n = O ( l / n ) i s t h e harmonic s e r i e s .
+
3n
2
+
1)
> n2/n3
=
l/n
The o r i g i n a l s e r i e s con-
v e r g e s by t h e a l t e r n a t i n g s e r i e s t e s t . (c)
Absolute convergence;
~,1-_(2~/n~ =)
(l/z"n2)
.:
zmn=O ( 1 / 2 ~ ) ,
which converges a s a geometric s e r i e s . (d)
Diverges; t h e
n = 0
term i s i n f i n i t e .
Note t h a t
CnZ1 [ I /
n ( 3 1 2 ) ~ l does converge. 5.
Her b r a i n i s c a p a c i t y - l i m i t e d ;
Xn=l
2 [n / ( 2
+
2n)1
< ZnZl m
2 (n /2n)
,
which converges by r h e r a t i o t e s t .
Copyright 1985 Springer-Verlag. All rights reserved.
12.3 The Integral and Ratio Tests PREREQUISITES 1.
Recall how to evaluate improper integrals (Section 11.3).
2.
Recall that the geometric series converges (Section 12.1).
3.
Recall how to use the
ith
term test to demonstrate divergence
(Section 12.1). 4.
Recall how to use the comparison test to show convergence or divergence (Section 12.2).
5.
Recall how to find limits by using l'Hbpitalls rule (Section 11.2).
PREREQUISITE QUIZ
1.
Evaluate
2.
Compute
3.
Compute
. :1 ( 2 / ~ ). ~ 1= 1 1im 4 [(x5 + x j;(dx/ex)
+
2)/(x
x4 - 2)/(3x
.
- l)I 5
+
.
4.
Compute
5.
Discuss the convergence or divergence of the following series:
lim[(x5 X"
(a)
z:=~
(b)
4 ~ y [(= i ~+ i
2x)1
+ 1)/2nl
[(n
5
+
2)/(i
3
+
i)]
.
GOALS 1.
Be able to use either the integral test, the p-series test, the ratio test, or the root test for determining the convergence or divergence of a series.
2.
Be able to estimate the error when an infinite series is approximated by a partial sum
.
Copyright 1985 Springer-Verlag. All rights reserved.
532
Section 12.3
STUDY HINTS
1.
Integral test.
J u s t understand t h e c o n c l u s i o n .
v e r g e s , s o does t h e s e r i e s . series.
I f t h e i n t e g r a l d i v e r g e s , s o does t h e
Although t h e t e s t a s k s you t o i n t e g r a t e from
i s just the arbitrary s t a r t i n g point. with t h e t a i l . 2
Z":l/n
1=0
2.
I f t h e i n t e g r a l con-
)
p-series.
1
i = 0
,
1
Again, we a r e concerned mostly
Be s u r e t h a t t h e i n i t i a l terms a r e f i n i t e .
diverges since the
-
to
For example,
term i s i n f i n i t e .
T h i s t e s t f o l l o w s d i r e c t l y from t h e i n t e g r a l t e s t .
Know
A good way t o remember t h i s s t a t e m e n t i s t o
t h i s statement well.
r e a l i z e t h a t any exponent l e c s t h a n one makes t h e s e r i e s l a r g e r t h a n t h e harmonic s e r i e s . than
3.
I/(p
-
Ratio t e s t .
l)bTP-l
I n e s t i m a t i n g t h e sum by
s~ '
the error i s l e s s
.
Do n o t confuse t h i s w i t h t h e r a t i o comparison t e s t i n
S e c t i o n 12.2.
The i d e a i s t o compare a term i n t h e t a i l of t h e s e r i e s
w i t h t h e p r e c e d i n g term.
I t makes s e n s e t h a t i f t h e r a t i o i s a b s o l u t e l y
l e s s t h a n one, t h e n t h e terms a r e g e t t i n g s m a l l e r i n g e o m e t r i c p r o g r e s s i o n , s o we e x p e c t a b s o l u t e convergence ( a t l e a s t , i n t u i t i v e l y ) .
If
t h e terms a r e g e t t i n g l a r g e r , t h e n t h e r a t i o i s g r e a t e r t h a n one, and divergence occurs.
Consider t h e harmonic s e r i e s and t h e a l t e r n a t i n g
harmonic s e r i e s t o n o t e t h a t t h e t e s t f a i l s i f t h e r a t i o i s one. r a t i o t e s t i s easy t o use i f f a c t o r i a l s occur. t h e magic number f o r t h e r a t i o t e s t i s one. estimate i s
4.
Root t e s t . a. < 1
/ a N / r / ( l- r )
See Example 7 .
Remember,
An upper bound f o r t h e e r r o r
.
E s s e n t i a l l y , t h e t e s t uses the following f a c t s :
, t h e n any r o o t of
The
ai
i s l e s s t h a n one.
any of i t s r o o t s i s g r e a t e r t h a n one.
If
If
a. > 1
,
0
-
then
T h i s i s n o t a proof t h a t t h e t e s t
works, b u t i t should h e l p you t o remember i t .
Again, we a r e only i n -
t e r e s t e d i n t h e t a i l i n g terms f o r t h e convergence i s s u e .
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.3
533
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
We need to evaluate
/cos n / < 1
Since Now,
9.
p = 2
>1 ,
2
Z ~ = ~ ( C O n/n S )
.
>4
h-'
series
'n= 1
is
(2&/3").
113
,
=
(2613")
,2N+1
/(2N
2Ni3 2 /(4N 0,013 for
n t n316
+
an = 3"/nn Therefore,
+
2N
N
2
-
n ) (2N
4 , so
=
+
,
1 anllin
n= 1
I
< in= 1 1 l/n2 1 . m
ffi
(COS
'n=~+l 1/2N2
3 n/n )
<
< 0.05 ,
i.e.,
= 0.44 .
-
( 3 n - 1 / 2 ~ )= (1/3)bfn/(n
(3"/nn)
25.
The series Ii=l(l/i4 )
29.
When
1)
+
I)]/[ ((2N)(2N
.
,
1
.
.
1)
Therefore, the
+
+
1) - IT2)/(2N)(2N
1)l =
The error estimate is approximately
+
9 n I362880
3in
+
,
I
+
] 11.54
I)!]
=
.
lim a nn converges by the root test. =
.
jan/an-l < r < 1
so the error is less than
Cnz0 [ 2 n 1 / ( 2 n
+
1-
2
~:=~+~(l/n~) , chich is less
+ I)!
7 n 15040
5 n 1120 so
we
converges (absolutely).
laNlr/(l - r) , where
1)!1[n2/(2N)(2N
+
,
1
, so accord-
~ C O Sn/n
which is less than
2 r = IT /(2n)(2n
The ratio is
+
converges.
The error is less than
,
i:=l
Thus, we need
~ z = ~ ( c on/n s )
1 anlan-lI
1im Thus, n-/an/an-lj
u = x2
diverges.
The error is
3
Therefore,
The ratio
. .
<1
jcos n
l)]
we have
l/(p - l)NP-l
I:~~+~(110 ) since
21.
+
The error in estimating a p-series is
3
17.
,
n
By letting
The integral is
[i/(i2
for all so
than or equal to
13.
l2 . b
u/2)
ing to the integral test, 5.
.
i1)ldx
1im = n l ( , b
(1/2)ji(du/u)
get
]l[x/(x2
and the llmlt
1s
0
.
ffi
k
cos kn = 1 > 0
is even,
Therefore, the terms of ln(k
+
&g(l/ln
,
1) > ln k k)
=
0
.
p
converges by the p-series test with
.
When
(cos kn)/ln k
l/ln(k
+
1)
<
l/ln k
k
is odd,
alternate in
.
iz,
cos kn
sign.
= =
4 > 1
-1
;
0
Since
(cos k)/in k
=
Therefore, the absolute values of the terms oi thls
Copyright 1985 Springer-Verlag. All rights reserved.
.
534
29.
Section 12.3
(continued) series decrease to
0
.
Hence
Z;,~[(COS
kn)/ln k]
converges by the
alternating series test. 33.
+ 3 r ) ~>
Zz=o [2r/(2r
The series
,
which converges because
lim r r l/r = lim (2 13 ) (213) = 213 rr-
of the root test:
<1
.
By the compari-
son test, the original series converges. 37'.
If
iEanl 'In
an
N
n
p
>1 ,
>3 .
nPln n)
ith
then
> 1
.
1im Since n-t_/an/# 0
(a)
l/nPln n
converges if
p = 1
Let
Then, we also have
,
the series
'n= 1an
term test.
< l/nP
for n
.
If
p
>1 .
>3
If p
a
=
c0s(9r)/3~
1
=
,
1im
because
In n
>1
for
the integral test gives
jyn 2(du/u) = b-t_ ln uIrn , SO p < 1 , then l/nPln n > l/n In
diverges. Therefore, ~:=~(l/n~lnn) 45.
.
n > N
for all
Then, by the comparison test and the p-series test, Z:=,(l/
ji(dx/x ln x) = if
then by the definition of the limit, there exists
for a11 n > N
diverges by the If
,
1 an]'In
such that
/an\> 1
41.
1
>
{cos[(Zn
,
and
+
2
1) r] }/(2n
a2
=
the series diverges n , whose series
converges for
+
1)4 ;
cos(2i r)/5
4
then . a
p
>1 .
=
cos r ,
a1 =
.
Copyright 1985 Springer-Verlag. All rights reserved.
45.
(c)
Since 1/(2x
/ c o s [ ( 2 n + 1)
+
1)4
applied.
lLO[ 1 1 ( 2 *
.
2
rll
Since
f
G l
.
+
1)3]
-
1; < .
c o n v e r g e s , and s o d o e s
I)*]
~ . e t f(x) =
i s d e c r e a s i n g , t h e i n t e g r a l t e s t may be
1im jyf (x)dx = [l/6(2x b-
+
.
a n G l / ( ? n + 1)4
Therefore
iz=oan
.
SECTION QUIZ 1.
Show t h a t t h e r a t i o t e s t c a n n o t b e u s e d t o a n a l y z e a p - s e r i e s .
2.
D i s c u s s t h e c o n v e r g e n c e ( c o n d i t i o n a l o r a b s o l u t e ) o r d i v e r g e n c e of t h e following s e r i e s : 2 n )I
I:=-_
(b)
~ : = ~ [ ( - 1 ) ~ ( 3 !r/)( 2 n
[tan-'n/(l
m
3.
+
(a)
+
8)
!I
-1-lln
(c)
Zn=1n
(d)
z;,
(g)
~ : = ~ ( 8 ~ /+ ( n3)*)
(h)
2 ~ ; = - ~ ( n / e x (pn ) )
(
"J;; - l ) n
A c e r t a i n p l a n e t i n a d i s t a n t g a l a x y was on t h e v e r g e o f d e s t r u c t i o n
from r a d i o a c t i v i t y .
The s u r v i v o r s h a v e p r o t e c t e d t h e m s e l v e s w i t h a
s p e c i a l s e r u m w h i c h l i n e s t h e i r e n t i r e body s u r f a c e w i t h l e a d . e v e r , l e a d i s n o t a n u n l i m i t e d r e s o u r c e on t h e i r p l a n e t .
How-
Suppose t h a t
the species is shrinking
a s a n a d a p t i v e mechanism s o t h a t t h e t o t a l
s u r f a c e a r e a of t h e
generation i s proportional t o
nth
3"/n!
a f i n i t e amount o f l e a d p r o t e c t t h e p o p u l a t i o n f o r e v e r , i . e . , In=o
(3n/n!)
finite?
.
\,!ill
is
J u s t i f y your answer.
Copyright 1985 Springer-Verlag. All rights reserved.
536
Section 12.3
ANSWERS TO PREREQUISITE Q U I Z 1.
1
2.
213
3.
9
4.
1/3
5.
(a)
l im nh,,[(n
+
l)/?nl = 1/2
,
s o t h e s e r i e s d i v e r g e s by t h e
ith
term
test. (b)
?[(i
5
+
4
i
+
2)/(i
3
+
i)] = +m
, s o t h e s e r i e s d i v e r g e s by t h e
ith term t e s t .
ANSWERS TO SECTION QUIZ ( i - ~ ) ~ / i= ' [ ( i - l ) / i l P
1.
C?
2.
(a)
Converges a b s o l u t e l y ; use i n t e g r a l t e s t .
(b)
Diverges; use
(c)
Converges a b s o l u t e l y ; u s e r o o t t e s t .
(d)
Converges a b s o l u t e l y ; u s e r o o t t e s t .
(e)
C o n v e r g e s a b s o l u t e l y ; compare t o
ip h a s r a t i o
1=1
ith
, whose
l i m i t is one.
term t e s t .
zrn (2n/n!) n=O
and u s e r a t i o t e s t .
m
3.
Inel ( 1 Ill'I3)
(f)
D i v e r g e s ; compare t o
(g)
Converges; u s e r o o t t e s t .
(h)
Converges a b s o l u t e l y ; u s e i n t e g r a l t e s t .
Yes; by r a t i o t e s t ,
1im (3/n) = O < 1 n+=
and u s e p - s e r i e s .
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.4
537
12.4 Power Series PREREQUISITES 1.
Recall how to apply the ratio test to demonstrate convergence or divergence (Section 12.3).
2.
Recall how to apply the root test for analyzing infinite series (Section 12.3).
3.
Recall how to differentiate and integrate polynomials (Chapters 1, 2, and 4).
PREREQUISITE QUIZ 1.
2.
3.
Differentiate the following: (a)
2 x5 - 3x + x
(b)
6x2
+ x4
Perform the following integrations: (a)
J!~(x~
(b)
j(t5
+ x)dx
+
t2
+
3)dt
Do the following series converge or diverge? (a)
c:=~
Justify your answers.
(312n)~
" 2
(b)
Zn=L(lOOn/n!)
GOALS 1.
Be able to find the radius of convergence of a power series.
2.
Be able to differentiate, integrate, and algebraically manipulate convergent power series.
Copyright 1985 Springer-Verlag. All rights reserved.
538 Section 12.4
STUDY HINTS 1.
Z?
Power series. These series have the form
a.(x 1=0 1
.
xo) i
-
They
behave like regular functions in that they can be added, subtracted, multiplied, divided, differentiated, and integrated, but only in the regions where the power series converges.
2.
Ratio test for power series. Learn the test for the general power series by replacing x
.
x - xo
with
As with the ratio test of Sec-
tion 12.3, you need to find the limit of 1im Iii,(ai/ai-l)l
we want
*lx - xO] < 1
Thus, the radius of convergence, R The series converges if -R
+ x0
xO
,
<
x
<
R
+ xO .
Ix - xO I
<
a./ai-l / x - xOI
or
,
.
For convergence, <
1im
.
11 [i,(ai/(ai-l)l
is the reciprocal of the limit.
R ,
i.e.,
-R
<
To determine convergence at
x - x0 -R
+ x0
<
R ,
i.e.,
and
R
+
substitute into the original series and apply the tests presented
earlier in the chapter.
The ratio test will give a ratio of one at
the endpoints, so it is mandatory to apply other tests to analyze the convergence or divergence at
3.
x0 2 R
.
Root test for power series. As with the root test of Section 1 2 . 3 , lim
-1
a
'
needs to be computed. Again, the radius of convergence is
the reciprocal of the limit.
4.
Differentiation and integration. In the differentiation formula, Cn i m ( d / d ~ ) C ~ = ~ a ~-( x x0) = ZiXliai(x - xO) i- 1 , we would like the derivative to converge at nonnegative.
x = xo
.
Thus, the exponent of
x - x0
should be
Hence, in the differentiktion process, the index on the
riglit begins at the next higher integer in most cases.
The index start-
ing point does not change during the integration process.
For both
processes, the radius of convergence does not change.
Copyright 1985 Springer-Verlag. All rights reserved.
S e c t i o n 12.4
5.
Algebraic manipulations.
539
When two s e r i e s a r e a d d e d , s u b t r a c t e d , o r
t h e new r a d i u s o f c o n v e r g e n c e i s a t l e a s t a s b i g a s t h e
multiplied,
s m a l l e r of t h e o r i g i n a l r a d i i .
The r a d i u s o f c o n v e r g e n c e f o r a q u o t i e n t
must be d e t e r m i n e d a f t e r t h e new s e r i e s i s d e t e r m i n e d .
Example 1 0
shows how t e d i o u s d i v i s i o n c a n b e .
6.
A p p l i c a t i o n s o f power s e r i e s .
S t a r t i n g w i t h a known power s e r i e s ,
m a n i p u l a t i o n s may b e p e r f o r m e d t o d e r i v e power s e r i e s f o r new e x p r e s D i f f e r e n t i a t i o n a n d i n t e g r a t i o n a r e commonly u s e d .
sions.
S e e Example
8.
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
L
Here, 1x1
<1
C:
(l/i)
1= 1
lim
.
=
1-
/
(2/(i
+
/ = 1 . > 1 . If
1)1 [ i / 2 ]
1x1
and d i v e r g e s i f
Ix
-
1im
.
9. =
.
I/L = 1
<1
.
1 "
1 [(5i +
l ) / i ] [ ( i - l ) / ( 5 ( i - 1)
+
,
11 < 1
i.e.,
verges f o r
.
0 < x < 2
+
2n/2(2n
1/4((1/2)"
.
xn/(2n
< x < 2
-1 < x
qn) = 1 / 2 ( 1
+
1) ;
+ +
2")
=
4
and
Since Thus,
ZYzl [ ( j i
+
+
qn) = ( - 4 ) n / ( 2 n
.
Therefore,
bn-l]
4n-1/(2n
1 . +
+
s
,
= -1
, so
1
=
x
l)/i]
+
ni..
hn)
=
x
-4
=
[xn/(2n
Ixl
Now, hn)
+
P = lim[1/2(1
therefore,
so the s e r i e s diverges i f
x
1
l)]
1im [(5i i-
Thus, t h e s e r i e s converges i f
+
x
If
l)] = 2
I ) (x
P, =
such t h a t
, the
#
0
-
lIi/il
con-
.
1im & = 1 [ 1 / ( 2 ~ 4")] [2n-1 n-
Here,
1/4
0
+
1=0 [ 2 / ( i
C:
Thus, t h e s e r i e s c o n v e r g e s a b s o l u t e l y f o r a l l
s e r i e s diverges a t t h e endpoints.
9.
,
1
=
is an a l t e r n a t i n g s e r i e s ; t h u s , i t converges.
Therefore, t h e s e r i e s converges f o r We h a v e
x
i s a harmonic s e r i e s ; t h u s , i t d i v e r g e s .
2 ~ y = ~ [ ( - l ) ~ +/ (Ii) ]
5.
Thus, t h e s e r i e s c o n v e r g e s i f
4
(-l)n/((1/2)n
. +
4n/4(2n
=
2n)
. +
2n-1/(2n
+
If 1)
+
+
hn) =
4n)
=
1 / 4 ( ( 1 / 2 ) ~+ l ) ] =
x
.
=
-4
, then
The l i m i t
# 0 ,
Similarly, the s e r i e s diverges a t qn)]
converges f o r
-4
x < 4
.
Copyright 1985 Springer-Verlag. All rights reserved. 1
540 Section 12.4
13.
Here,
ai
2.5(i
- 1))
.
Ci=l (5ixi/zi)
or :C
5i 1=1
,
+
a n)
=
2
is
.
+
n3)/(1
5
1 in5 /[(n -
.
1)
112 , the radius of convergence
If x
=
Zn,O
,
the series is
~;=~(-l)i51 .
( 5 i ~ ~ 1 2 ~converges )
The
2
1im 1im 3 n-p)lan~an-l1 = n-p)~(n2 + n ) /
+
n)5
+
3 (n - 1) 1 1 = 1 by ltH8pital's rule.
, l
=
Thus,
-
a
m
bn converges, Zn=Oa,l is absolutely convergent. x = 1 and
converges for
.
-1
lim/(-l)nnn/l/n is ,-n, 1im n*
converges only for x
=
O ,
which is
1im l = i- [(i
(c)
Using the geometric series
+
l)/i]
xi+'
1=1
(d/dx)jtf (t)dt 1x1 < 1 212
+
= =
=
1/(1 - x) = 1
314
Therefore, the series
+
x
+
x2
+
...
~ \ ~ x= x2/(1 ~ - x) . Also f(x) = 1=0 2 (dldx) (x /(1 - x)) = x(2 - x)/(l - x)2
+
4/8
+
0
.
R =
, note
for
5/16
+ .. .
=
tan-'a
So
. . . , i.e., tan-'x
c;=~ [(i + (c),
1)(1/2)~]
the sum is
=
f(1/2)
(1/2)(2
.
- 1/2)/
.
(d/dx)tan-'x
'n=o (-l)nx2n
anxn
.
(1 - 1 1 2 ) ~= 3
division.
Cn,O
1 , so the radius of convergence is
Thus, using the result of part
Note that
-.
Thus,
and the radius of convergence is
(a)
that :Z
29.
f2
3 5 When x = il , ~ n = ~ / a= ~zrn x ~[(n2 / + n )/(l + n ) 1 = n=O b = n 3In5 . Then lim(lan1/bn) = 1 < , and since n k
a
(d)
1im 1)) = i-(5i/
-
1= 1
Xi=O an. Let
25.
Lim (5112~)(ii-'/5(i i-p)
which both diverge. Therefore, Zm
(n
R = l/l = 1
21.
=
- 2 < x < 2 .
For (1
so L
Since the limit is
for
if 17.
5 i 1 2 ~,
=
= =
1/(1
jE[l/(l
+
2 x )
+
=
1 - x
2 t )Idt
=
n 2n+l = 6n=O [(-I) x /(2n
2
+ x4 3
- x6
x - x 13
+
l)]
+x
and
+ ... 5
by long 7
15 - x / 7
(d/dx)tan-'x
+ =
.
Copyright 1985 Springer-Verlag. All rights reserved.
33.
3 From Exercise 23, sin x = x - x /3! 2 3 sin x = (x - x /3!
have
x2 37.
-
(a)
4 x /3
+
(2/5!
+
cos x = x
(1/3)x3
+
5 (2115)~ +
2 sec x = (d/dx)tan x = 1
x /3
+ ... , .. . .
+ +
+
Therefore, we
5 x /5!
6 2x 145
5 x /5!
we get
- ... ... .
- ...
) =
and
tan x
+
x2
+
4 (213)~
=
sin x/
... .
+
Using the result of part (b) and long division, we get 1/[1
41. Let
+
4
3 x - x /3!
4 6 x /4! - x /6!
... .
3 x /3!
From the result of part (a) and differentiating term by term, we get
(c)
...)(x ... = x2 -
Using long division with sin x =
+
5 x /5! -
5 x /5! -
2 6 1/(3!) )x -
2 cos x = 1 - x /2!
(b)
+
+
+ x2 ?
f(x) =
lxO1
for
i
>N
i=O
(2/3)x4 a.xl 1
+ ...I
=
1 - x
2
+
have radius of convergence
Now
'fi
-+
1 as
i
-+
and use 1'H8pitaltsrule, as in Example ,(a),
that
lim x l / ~ , x-t_ 1).
Thus,
By the root test,
i m,
igl
? ia.xi-I converges as well. 1 0 i=l convergence shows that
iGy<
(for example, write
illi
enough.
. .. . Rf . Thus, if
this series converges. By the theorem,
.
l/secLx =
4 (113)~ -
l/lxol ip 1
=
p. 525 to show
i
l / x o if i islarge i ia.ixo converges. Thus g(xo) =
Therefore, g(x)
has radius of
R 2 Rf . A similar argument, interchanging f and g g Rf > R so Rf = R . Likewise, one shows that Rf = Rh. '6' g
Copyright 1985 Springer-Verlag. All rights reserved.
542
45.
Section 12.4
Let
1x1
From Exercise 44,
f;;'g(t)dt = f (x)
.
Therefore, using
the alternative version of the fundamental theorem of calculus, ff(x) = g(x)
.
SECTION QUIZ 1.
Suppose
f(x)
=
f(x)
~ ; = ~ ( 2 / 3 ~ ) x ~and
g(x)
- g(x)
Explain.
converge?
=
:1 (413~")~~ 1=2
(a)
Does
(b)
If it converges, write an expression for the difference and find its domain.
2.
True or false: (a)
The derivative of
The integral of i+l
a
zl,o[x
3.
Find all
(c)
4.
z:=~
x
[(x
m Ci,O(ix
i-1
/i!)
+
l)i!]
=
Z- [xi-'/ i=O
.
(i - I)!] (b)
~q=~(x~/i!) is
+ 1
i
F:
1=0
(xi/i!)
=
1 , [xi+l /(i 1=0
=
I .
for which the following power series converge:
+
21ncos n11/4~1
The ancient Greeks believed that an average centaur's speed was given by
f (t)
=
2 [2n /(n3 + n)] tn , where t
does
f(t)
t
is a unit of time.
converge?
(a)
For what
(b)
Within the radius of convergence, find a formula for the distance travelled by the centaur.
(c)
Find a formula for the centaur's acceleration and determine the radius of convergence for the acceleration
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.4 543
ANSWERS TO PREREQUISITE QUIZ 1.
2.
3.
(a)
5x4
-
6x
(b)
12x
+
3 4x
(a)
-314
(b)
6 t 16
a)
1im
-1
+ 3
+
1
+
t3/3 1
3t
lIi
1
+
=
C 1im(3/2i) ia
=
0
<1
, so
rn
Z n = (31211)~ ~ converges
by the root test. (b)
izl
(100i/i!)/ [100(i
- l)/(i
- I) !I 1
=
1im i-[l/(i
- I)]
=
0
,
so
converges by the ratio test.
~:=~(100n/n!)
ANSWERS TO SECTION QUIZ 1.
(a)
(b)
3.
4.
converges if
this case,
f(x)
f(r) - g(x)
=
3i+l 2.
- g(x)
f(x)
)x
i
.
2
+
f(x)
and
g(x)
2x13
+
and
g(x)
both converge. In
both converge if
x
-3
<
3
.
1;=~(2/3~ - 4 1 3 ~ " ) ~=~ 2+2x/3 - ~:,~(2/
Its radius of convergence is
-3
<
x
<
3
(a)
False; the index for the derivative should go from
(b)
True
(a)
2 G x G 4
(b)
-l/e
(c)
-6
(a)
-1Gt<1
(b)
Distance = jf(t)dt
(c)
Acceleration
. i = 1
to
m
.
< x < l/e <x < 2
=
=
f '(t)
zrn n= 1 [2n2/(n3 + n)] [tn+'/(n =
1Zx1 [2n2/(n3
+
n)] ntn-'
+ I)] ;
radius of
Copyright 1985 Springer-Verlag. All rights reserved.
544
Section 12.5
12.5 Taylor's Formula PREREQUISITES
1.
Recall how to differentiate and integrate power series (Section 12.4).
2.
Recall how to calculate the radius of convergence of a power series (Section 12.4).
3.
Recall how to use the linear approximation (Section 1.6).
PREREQUISITE QUIZ
1.
for which the power series Cirn,l [(-I)~(x
Find all x
-
z)~/~I
converges. 2.
3.
zrn [(xn=O
Let
f(x) =
(a)
Compute
jf (x)dx
(b)
Compute
(d/dx)f(x)
3)"/n!]
.
. .
Use the linear apprcximation to estimate (2.11) 3
.
GOALS 1.
Be able to find the Taylor series for a given function.
2.
Be able to estimate the error in using the Taylor approximation.
3.
Be able to state the Taylor series expansions of commonly used functions.
STUDY HINTS 1.
Definitions. of
f
f
I '1=0 ..' [f(i) (x0)/i!] (x - xO)
about the point
computed at
xO
.
i
x = x0 - f(i) (xO) If
xO = 0
,
is called the Taylor series is the
lth
derivative of
the series is known as a P I a c l a u r ~ n
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.5
2.
There are two fonns of the remainder for Taylor's
Error estimates.
I:
series. The integral form is ative form is
[f(n+l)(c)/(n
number between
x
an inequality for 3.
545
and
x0
fn+l
.
+
.
[(x 0
1) !I (x
-
t)"/n!]
f(n+l) (t)dt
-
x
,
0
where
.
c
The derivis some
You estimate the remainder by inserting
Proving convergence. Study Examples 4(a)
and 4(b) well.
They demon-
strate how convergence is often proven for Taylor's serles. 4.
Taylor series limitations. The paragraph following Example 4 describes interesting limitations of Taylor's series.
The first problem is that
we may be interested in a point outside the radius of convergence. The second problem is that all of the derivatives may be zero for the Maclaurin series, yet zhe function isn't zero. 5.
Usefulness of Taylor's series. You may be wondering why you should bother with learning about Taylor's series.
It can be used to approxi-
mate functions whose values could only be found in tables before the invention of calculators. The Taylor approximation improves the linear one.
6.
Important series.
The box on p. 600 lists the most important Taylor
series that you will encounter. Althoughyoushould be able to derive each one, it is recommended that you memorize the series in order to save time, except perhaps the series for rapidly derived by integrating [1/(1
+
x)
1/(1
+ x)
ln(1 =
+
1 - x
is a geometric series whose ratio is
x)
which can be
+ x2 -x .I
x3
+ . ..
.
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Since to get
sin x
=
i 2i+1 ~q=~[(-l)x /(2i
+
[(-l)i(3x)2i+1/(2i sin 3x = :1 1=0
7 243x 11120
l)!]
+
, we replace x with l)!]
=
3 3x - 9x /2
+
3x
5 81x I80
-
+ ... .
Copyright 1985 Springer-Verlag. All rights reserved.
546
5.
Section 12.5
+
x2
+
x4)-2(2~
+
x4)-'(2
12x2) ,
+
x2 + x4)-3(2x
,
4x3)
frl(x) = 2(1 f '"(x)
and
3 4~ )(2
+
+ x2 + x~)-~(~&x).
(1
f "(1)
f '(1) = -213,
=
+ x2 + x4) -1
f(x) = (1
Use the method of Example 2.
=
x2
-6(1
+ x2 + x4)-3(zx +
2(36)/27
-
1419 = 1019 ,
and
- (1
+
f(1)
=
1
yields
+
-x
for x
+
(-x
-
(-x2
+ ...
x4)3
f"11'1(0)/6! ; f(x)
113
=
x1l2 ,
~ I ( x = ) (i~i)~-l'~ ,
so
.
2(n - 2) 1 ,
=
Thus,
&
f(1) =
1
=
+
1
(2n - 3)!
1 - x2
=
=
112
1 [(Zn -
< (2,)"
(112)(-1/2)~-~/~,
~ Y X )=
(n) (XI
(-I)~+' [(2n - 3) !I
=
is infinitely differentiable. At
,
f "(1) = -114
-
3) ! 12 (n
-
1)
2
+
,
f '''(I)
and
- 1)
(1/16)(x
2) !I x-(2n-1)12/2n
, choose M = (2n)/2
n
=
and
3
C
Then let (x - 1)/2
a = 112
+
and
+ ')u
(1 u
=
m
=
(X
2 - 1) 18
liX0 [a(" - I)...
x - 1 to get
[ (112) (-1/2)/21 (x - 112
(x - 1 ) / 2 -
=
3/8
.
.. . . .
to be the
I ,
on
use the binomial series
-
1 <
Alternatively, we can assume that x is in the interval
+
+
1 =
f
- 1) - (118)(x
(1/2)(x
x-(2n-1)/2
maximum of
Thus,
, f '(1)
We need to show that Since
=
.
= . 6! = 720
therefore, f"""(0)
f up(x) = (112) (-112) ( - 3 1 2 ) ~ - ~ /,~... ,
1
Thus,
By using the formula for the Maclaurin series, we know that
(b)
xo
+
,
113
1/(1 - x)
in the formula for
(-x2 - x4)2
-
+ ... .
x6
13.
2
- x4 2 - x4)
Substituting
~
f '"(1) = -6(216)/
+ 4(6)(14)/27 + 2(6)(14)/27 - 2419 = -16 + 5613 - 2419 = 0 . 2 4 2 l/(l + x + x ) 113 + (-2/3)(x - 1) + (10/9)(x - 1) /2 + O(x 2 3 113 - 2(x - 1)/3 + 5(x - 1) 19 + O(x - 1) . (a)
+
x2
2 1 2 )~
81
9.
+
+ ~4(1 )+
~
3 4~ )(2
x0 = 1 , we have
Evaluating at
+
f '(x) = -1(1
+ x4)-3(2x + 4x3)2 + x2 + x4)-4(2x + 4
+
1 2 ~ + ~ ) 2(1
,
+
(X -
+
(1
+
(0.2) and
(a - i
x - 1)'12
=
+
1)ui/i!]
x1I2 = 1
[ (112) (-112) (--3/2)/3!1(x - 1)
3 1) 116 -
3
,.. .
Copyright 1985 Springer-Verlag. All rights reserved.
.
+ + ...
=
Section 12.5
17.
The l l a c l a u r i n s e r i e s f o r 2 4 1 - x 12 + x / 4 ! -
is
Thus,
f 0 (x)
f (x) = 1
4
Since
i+l
1 [ ( - 1 )
+
,
f(")(x)
and o n
=
,
[O,l/2]
l e s s than
'
(-l)n+l(n
0.009
R4
.
n+l
,
, where
f(x)
f"(x) = -(1 + x )
- 1)!(1
< 0.002 ,
Therefore,
+
2
x /2
+
ln(2.5) = ln(3/2)
1(-1)~+'(2/3)~/i]
1 ) !] ( x - x o )
f l ( x ) = (1 + x ) - l
We h a v e
we h a v e
( l / ~ ) ~ / i+] L:=~
[f(n+l)(c)/(n
...
,
2.5 = (3/2).(5/3)
-
.
x)-"
so for ln(2.5)
. =
on
n = 4
x 12
,
4 x /24
.
ln(513)
=
-
+
The e r r o r - i s
-2
+ x)
ln(l
,
" (112 -
and
Rn(s)
and
=
x0 = 0
f l " ( x ) = 2 ( 1 + x ) - ~,
[0,2/31
,
,
2
1
=
cos x
... .
fl(x) = 1
=
f 2 ( x ) = f (x) 3
21.
5&7
, R;
< 0.007
the t o t a l error i s 118
+
1/24 - 1/64)
i
U
t h e t e x t , t h e remainder i s
4 R M3/4!
+
1 f "'
/
(x)
4 R M3/4!
=
4
R M 112
t a k e s on f o r
3
x
, where in
M3
i s t h e maximum v a l u e
[xO - R , X ~ + RJ
.
Copyright 1985 Springer-Verlag. All rights reserved.
548
25.
Section 12.5
Let
(b)
-x(l
+
x ) -3/2
f (0)
=
1
7
- ...
x /7!
(x - s i n x ) / x
7
x /5! 3
x /x
3
...)
.
7
2 .
x
sin
2
- x /5!
(b)
3
1/(1
-
2n+1 for
1x1
1
Multiplying out, Since
+
4(1 O
f(x) 2
x )
for a l l
f "'(0) = 0
+
=
-
+
11 = 0 . 8 7 9
+
+
2f (0)
. - sin x
Now,
5
7
+
+
+ ... )
x /3!
0
so
5
...) / ( x 3
x /7! 2
... ,
x /5! -
=
- x /3!
+
by d i v i d i n g by
as
x
approaches
+
x)
n Zn=o~
=
-
m
Tn=O(-l)nxn =
+ ...I =
2~
+
4
.
+ ...
2x3
=
. 2 2
,
, f "'(x)
.
Thus
, f ( 4 ) (0)
f l ( 0 ) ( x - 0)
+
.
xn t e n d s t o
= 24
+
x2)2 = 1
f '(x) = 2(1 = 24x
+
+
2x
+
2
, and
,
2
+
=
24
2
f "(0)(x - 0 ) /2!
, f "(x) , and
fl(0) = O
f (k) (0) = 0 f(x)
x
x )-2x
, f (4)(x)
f(0) = 1
Hence, t h e P l a c l a u r i n s e r i e s f o r f(O)
+
7
+ x 2 - x3
f ( x ) = (1
k > 4
4f (-114)
s i n x = x /3! - x / 5 !
5
1/(1
(1 -
(1 t x )
8x2
+
.
116
+ ...) -
-
x
+ . . .) / ( I -
x)
x )
3
x3 - x /3!
Since t h e term involving
zn=02x (a)
and
(x3/3! - x / 5 !
=
= (1/6
By t h e sum r u l e ,
rn
(1112) [ f (-112)
...
x /7! -
+
f '(x) = 2 -512
Theref o r e ,
The d e n o m i n a t o r i s
(1 + x + x 2 + x
37.
=
Then
x 2 ) -312 - 3x 2 (1
lim xiO[ ( x - s i n x ) / x 2 s i n jd
, we g e t t h e l i m i t a s
0
.
-1
=
, gives
+
x /5!
.
+
.
1
=
~ ( L / z ) I= (1112) [ 4 / +~ 3 2 1 m
5
-
-x + x / 3 !
x
ftl(x) = -(1
f "(0)
1im 2 x,O(l/~ sin x - l/x ) 3
33.
and
n = 4
+
bf(114)
= 1 1 2 , and
R
and
rule for
29.
,
xo = 0
=
+
(1
f
("(x)
, fl'(0)
= 4
for a l l
+ x2)2
=
is
k
=
,
>4 .
f(x) =
f111(0)x3/3! + f(4)(0)x4/
Copyright 1985 Springer-Verlag. All rights reserved.
549
Section 12.5
41.
Since
f(x) = s e c x , 2
f "(x)
+
= sec x tan x
3
3 sec x
45.
We know t h a t 1=1
c o s x = C:
- i + 1 x 2 i / ~ i
-
(Zi)!] = 1/2
49.
Let f
-
=
I n xO
+
so
1=0
.
Then,
3
= sec x tan x
Thus,
a.
a 3 = f "'(0)/3!
.
Therefore,
+
= f (0) = = 0
i
.
fl(0) = 0
-
.
cos x =
- c o s x / x 2 = l - i+l ~ x 2i-2 /
~hus, 1
x / 4 ! + x /6! -
.
= 0
f "'(0)
[(-l)ix2i/(2i)!]
, f "'(x)
so
... . ;
f ' ( X I = x-'
f "(x) = -x-'
;
flPt(x) = 2x13
;
. ~ The~ T a y~l o r e x p a n s i o n o f d e g r e e 4 f o r I n x i s -3 3 -4 4 - x O- 2 ( X - x O )2/ & ' 1 .+ 2 x o ( X - x 0 ) / 3 ! - 6x0 ( x - xo) / + ( x - x O ) / x O - (X - x O ) 2 /2x02 + ( X - x O )3 /3x03 - ( x - x O )4 / 6
.
(a)
For
xo = 1
(b)
(X (c)
-
e ) /3e
3
xo = 2
4
-
(X - e ) / 4 e
-3 s i n 1
, f(0)
f "(x) = e x c o s e
= eXcos ex
.
sin 1)/2lx
4
-
-
2 1) /2
+
-
(x
I ) ~ /
x
In 2
= sin 1 ;
- e
+
(x - e ) / e - (x
2 x . x sln e
+
fl(x)
(x
=
-
[(sin 1)/2]x
2 2 e ) /2e
i
2 2)/2 - (x - 2) 18
eXcos ex
, fl(0)
+
=
, f "(0) = c o s 1 - s i n 1 ;
e z x s i n e x - 2 e Z x s i n e x - e3Xcos e x ,
3
-
.
The f i r s t f o u r t e r m s a r e
2
1
is
, t h e polynomial is
f(x) = s i n ex
cos 1 ;
-
.
, t h e polynomial
= e
( x - 1) - (x
t h e polynomial i s
1 ) /4
3
For
Since
t '"(x)
xO
For
,
4
-
3 - (x
53.
f "(0) = 1
a 2 = f "(0)/2 = 1 / 2 ;
4
.
fl(x) = s e c x t a n x ,
-1 xO (X - s o )
4! = I n xo 4 4x0
2
f(x) = In x
'"'(x)
,
sec x tan x
1 ; al = f l ( 0 ) = O ;
.
, so
3
+3
2 sec x tan x
f(0) = 1
sin 1
+
(cos 1)x
+
f '"(0)
=
[(cos 1
-
.
Copyright 1985 Springer-Verlag. All rights reserved.
550
57.
Section 12.5
(a)
1im fi(0) = Ax+o{ AX) - f(o)l/ax = nlXyo{ [(sin Ax)/Ax - il/sx; = 1im AX+O [(sin AX - AX)/(AX)'I = i i:o[(cos AX - I)/Z(AX)I = 1im Ax+O[-sin Ax121 = O(l'H6pital's rule). For x # O , f '(x) = (x cos x AX] =
-
2 sin x) /x
.
f "(0) = Ax+O lim i [f '(Ax) - f '(o)] / 2 1im [(Ax cos AX sin Ax)/(Ax) 1 /Ax] = Ax+O
Therefore,
-
nlXyo{ [(AX cos Ax 3
lim 2 sin Ax)/(Ax) 1 = Ax+o [(cos Ax - Ax sin AX - cos AX)/~(AX) ] = lim 1im Ax+O[(-sin AX - AX cos Ax)/6(Ax)l = Ax,O[(-~~~ AX - cos AX + Ax sin Ax)/61 2
[X (COS
=
-1/3(1'H6pita11s rule).
x # 0 , f "(x)
For
x - x sin x - cos x) - (x cos x - sln x)2x1 /x4
=
=
2 . (-x sln x - 2x cos x
+ 2 sin x)/x3 . Therefore, f "'(0) = lim 2 . Ax+o{ [f "(AX) - f "(o)] /AX] = [-(AX) sln Ax - 2Ax cos Ax 3 4 lim 2 2 sin Ax + (Ax) /31/(Ax) I = [-2Ax sln AX - (AX) cos AX
~o:i
2 cos Ax
+
2Ax sin Ax
+
2 cos Ax
1im 2 . Ix+O{ 1-2Ax cos Ax + (Ax) sln Ax 2Ax sin Ax 4 sin Ax (b)
+
+
+
2Ax i n Ax
+
4 Ax cos Ax
+
+
(Ax) 2]/4(Ax) 3 1
+
2 I ) ! ] = 1 - x /3!
+
=
2 ~ x 1/12(Ax) 1 = Ax-ot lim ' [-2 cos Ax
(Ax) 2cos Ax
+
21 /24Ax =
ixyoi
[-2 5ln Ax
. Ax] 124) = 0 2Ax cos Ax - (Ax) 2sln
4 x /5!
-
-
2
sin x = ~ ~ = ~ [ ( - l ) ~ x ~ ~ + + ~I )/! (] ~,i so (2i
+
6 x /7!
+
+
.
sin x/x = :I [(-ijixZi/ 1=0
+ ... .
SECTION QUIZ
1.
2. 3.
(a)
What is the value of
C?
~(-l)~/i!]?
(b)
What is the value of
C:
i 2i+l [(-I) n /(2i)!j
1=0
1=0
?
1im (sin x/x) = 1 . xi0 3 Find the Taylor series expansion of f(x) = x + 2x - 2
Use a Taylor series to prove that
about
x
2
=
and simplify your answer. 4.
Find the sum of what is
[(-l)n/4n(2n
+ l)]
.
(Hint:
If
f (x)
=
-1
tan
(x/2) ,
f '(x) 7)
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.5
5.
551
Find the Taylor series expansion of the following functions around the indicated point:
6.
-1
(a)
y = sin
(b)
y
=
x
about
x
0
=
11 3~~-2 about x = 3
A frustrated calculus student was overheard saying, "Taylor's formula is useless."
That evening, the student was visited by Taylor's ghost.
The ghost explained that he travels throughout the night with speed t cos t
at time
t
f(t) =
to haunt nonbelievers of Taylor's formula. He will
return to haunt calculus students who can't find a fourth-order approximation to his speed at
t
=
0.2
.
He will return tomorrow for the an-
swer and an upper bound for the error.
Find the answers which will
prevent return visits by Taylor's ghost.
ANSWERS TO PREREQUISITE QUIZ
ANSWERS TO SECTION QUIZ 1.
2.
(a)
l/e
(b)
-n
The Taylor expansion of so
sin x/x
3.
f(x)
4.
-1 tan (112)
=
10
+
=
2
1 - x /3!
l4(x - 2)
+
sin x
+x
about
4
/5! -
6(x - 2)
2
... -i-
(x
x = 0
, -
is
3 x - x /3!
and the limit is
+ 1
5
x /5!
- ...
.
2) 3 = x 3 + Z x - 2 .
Copyright 1985 Springer-Verlag. All rights reserved.
,
552
5.
6.
Section 12.5
+
3 x I6
+
T~n=2 [(2n - 1) !/2"+'n!
(a)
x
(b)
~:=~[(-l)~(2n + l)! (1) ( 4 ) ( 7 )
Expand around t = 0.2
.
1/120(5)~
t
Error
=
0 ;
< 1-
...
(n - 1) ! (2n
(2n 3
+
t cos t - t - t I2
+
l)]x 2n+l
1)13~n!1(x - 3)n
,
which is
0.196
3 cos(0.2) - (0.2) sin (0.2)/120(5)~
at
<
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.6
553
12.6 Complex Numbers PREREQUISITES 1.
1/(1
R e c a l l t h e power s e r i e s e x p a n s i o n s f o r and
ex
+
x)
,
cos x
, sin x ,
(Section 12.5).
R e c a l l how t o u s e p o l a r c o o r d i n a t e s ( S e c t i o n 5 . 1 ) .
2.
PREREQUISITE QUIZ (2,-5n/8)
i n t h e xy-plane.
1.
Plot the polar coordinates
2.
Convert t h e p o l a r c o o r d i n a t e s
3.
Convert t h e c a r t e s i a n c o o r d i n a t e s
4.
W r i t e down t h e power s e r i e s e x p a n s i o n s f o r t h e f o l l o w i n g : (a)
c o s 2x
(b)
e-'
3n/4)
(-1,
i n t o cartesian coordinates.
(2,-2)
i n t o polar coordinates.
GOALS
1.
Be a b l e t o e x p l a i n t h e r e l a t i o n s h i p between t r i g o n o m e t r y and t h e i m a g i n a r y numbers.
2.
Be a b l e t o do b a s i c a l g e b r a w i t h complex numbers.
3.
Be a b l e t o w r i t e down t h e p o l a r r e p r e s e n t a t i o n of a complex number and u s e t h i s form t o e x t r a c t r o o t s .
STUDY HINTS
1.
I m a g i n a r y numbers d e f i n e d . s h o u l d know t h a t fore, letting
eix
= cosh i x
cosh i x = cos x
x
=
n
The b a s i c d e f i n i t i o n i s
, we g e t
and
+
sinh i x
=
cos x
sinh ix = i s i n x
eiTT = -1
.
+
fl
=
i sin x
.
i
.
You There-
I n a d d i t i o n , b~
.
Copyright 1985 Springer-Verlag. All rights reserved.
554
2.
Section 12.6
A l g e b r a o f complex numbers.
\ h e n you add and m u l i t p l y ,
i
l i k e a n o r d i n a r y number e x c e p t t h a t you must remember t h a t
3.
Terminology.
acts just
.
i2 = -1
As w i t h a n y new s u b j e c t , a n e s s e n t i a l s t a r t i n g p o i n t i s
learning the vocabulary.
Therefore, learning the vocabulary presented
on p . 611 w i l l a i d you i n u n d e r s t a n d i n g t h e d i s c u s s i o n s a b o u t complex numbers.
4.
Kotation.
Many t i m e s ,
( a , b ) w i l l d e n o t e a complex number.
5.
Removing
from d e n o m i n a t o r .
i
t o remove
i
(a
-
bi)/(a
r e m o v e s t h e symbol 6.
bi
.
To s t a n d a r d i z e a n s w e r s , i t i s d e s i r a b l e
from t h e d e n o m i n a t o r j u s t a s we d e s i r e d t o h a v e denomina-
t o r s without r a d i c a l s . p l y i n g by
+
z = a
understood t o stand f o r
It i s
+
a
If
- bi)
bi
yields
occurs i n t h e denominator, multia2
+
b2
i n t h e d e n o m i n a t o r and
from t h e d e n o m i n a t o r .
i
P r o p e r t i e s of complex numbers. helpful i n manipulations.
The p r o p e r t i e s p r e s e n t e d on p . 612 a r e
Note t h e i r s i m i l a r i t i e s t o p r o p e r t i e s of r e a l
numbers.
7.
E x p o n e n t i a l p r o p e r t ies.
P f u l t i p l i c a t i o n and d i v i s i o n of e x p o n e n t s a r e
done e x a c t l y a s t h e y a r e done w i t h r e a l numbers.
8.
Polar representation. z = a
+
ib
.
(r,B)
z = re
i6
i s c a l l e d t h e p o l a r r e p r e s e n t a t i o n of
a r e t h e p o l a r c o o r d i n a t e s of
(a,b)
.
N o t i c e how
e a s y i t i s t o m u i i t p l y two i m a g i n a r y numbers w h i c h a r e w r i t t e n i n p o l a r representation.
9.
IIeMoivre's f o r m u l a .
The n t h
r o o t s o f a n i m a g i n a r y number
l o c a t e d on a c i r c l e c e n t e r e d a t t h e o r i g i n w i t h r a d i u s
i s l o c a t e d a t an a n g l e of
n
-
2rln
1
other
.
roots
peiq
nc.
@ I n from t h e p o s i t i v e r e a l a x i s .
are
One r o o t There a r e
e q u a l l y s p a c e d on t h e c i r c l e , s e p a r a t e d by a n g l e s of
The v a l i d i t y of ~ e ~ o i v r e f' os r m u l a i s e a s i l y d e m o n s t r a t e d u s i n g
t h e p o l a r r e p r e s e n t a t i o n t o r a i s e t h e claimed r o o t s t o t h e
nttl
power.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.6
555
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1,
5.
Using
e i x = c o s x + i s i n x ,
* Y
2
9.
e
weget
e
The complex number point
(x,y)
=
(1
M u l i t p l y by (5
The complex number
+
3i)/34 2
+
(5
2i)
+
+
(x,y)
+
3i)/(5
25.
By t h e q u a d r a t i c f o r m u l a ,
29.
From Example 4 ( d ) hi&
+ 3 = 0 ,
then
i4i(6/2)(1
=
Since
(1
+
+
4i)
.
(-3
6i)
3i)
~
,
+
z2=-3
fi =
x
+
yi
-14
+
2i
is plotted a t the
Thus,
(-2/3)i
.
(0,-213)
(1
+
2i) -
.
+
8i
(5
+
z
or
3i)/(25 - 9ii)
z = i - f i =
( 7 i -4)/2
=
i(v'7/2) ( 1
+
=
.
i)
i ) = ? ( 6 / 2 ) ( 4 i - 4)
+
i = )(-3 ~
M u l t i p l y by
The complex c o n j u g a t e o f M u l t i p l y by 2i/3
=
4
=
,
4i) (-3
-
(10
+
= i
-
4i)
.
(7 i m ) / 2
so
5i)/(l
4i)/(-3
i6i.
=
mi6J-1 =
( 2 ~ 5 i- 2b6)
.
+
+
2 i ) 2 = (10
5i)/
(-30 - 5 5 i -
t o get
, s o t h e i m a g i n a r y p a r t i s -55125
=
.
-1115
41.
Thus,
i n t h e xy-plane.
t o get
2 2 2 0 i ) / ( 9 - 1 6 i ) = (-10 - 5 5 i ) / 2 5
37.
is plotted a t the
.
If
z
+
(-15
21.
33.
yi
We n e e d t o add t h e r e a l and i m a g i n a r y p a r t s s e p a r a t e l y . 3(5 - 2i)
17.
+
.
(4.2)
i s plotted a t
13.
x
i n t h e xy-plane.
is plotted a t
point
C O S ( - T / ~ +. ) i sin(-~12)=
=
i/i
t o get
a
+
bi
is
a
-
bi
(2 - i ) / 3 i = ( 2 i
+
, so I)/(-3)
-
z
=
5 - 2i
, so
-
.
z = -1/3
+
.
Copyright 1985 Springer-Verlag. All rights reserved.
556
Section 12.6
Here,
y
z = 3
+ Oi ,
-
z = 3
so the complex conjugate is z
l(i
I
2
=
is plotted at the left.
-
Oi = 3
lz
/
is the
2
distance from the origin, which is
-
case.
I
z
+
-5
=
0 in this case. 7i
is plotted at the left.
+
225
=
\'
in this
is the angle from the positive x-axis,
8
which is Y
49
=
lzl
-1
case, z
8 = cos
+
1.2
=
0.7i
L2+
jzl =
z
n
2.19
.
is plotted at the left.
+
0.49 =
,
8
In this
Here,
m . The .
0 = ~ o s - ~ ( 1 . 2 / d m );2 0.53
we know that
(-51v'fZ)
b2 = J1.44
argument is
Prom Exmaple 7(b),
=
fi . The argument ,
is the angle from the positive x-axis. B
-5
.
4
-
,
= zn
(8 - 3i)4
so
(8 - )'13
=
=
Two sides of a parallelogram are determined by line segments joinging the origin to
*
Z2
.
The sum is the fourth vertex of the paral-
lelogram.
ex+iY
We know that (e
2x
2 Cos y
+
=
eX.eiy
is
e
cos-'(excos
=
=
In this case, the sum is
ex(cos y
e2xsin2y)112 = [e2X(cos2v
-1 8 = cos (a/r) , where y/eX)
Use the fact that
e
=
ix
=
z l and
z
=
cos-'(cos cos x
+
a
+
+
i sin y)
2 sin y)]
+ bi
y) = y
i sin x
and
. .
r
Thus,
+
. Thus, leZl =
=
-3
ex
/z/ . e
.
10i
.
=
The argument
Therefore,
l-ni/2
=
e.e - 1 7 i / 2
Copyright 1985 Springer-Verlag. All rights reserved.
-
Section 12.6
77.
We have
f(z)
to get 81.
(3
-
+
1/(2
=
- 16i
4i)/(9
+
,
i sin x
we get
+
i sin(3~12)) = (cos 8
85. If
z =
a
+
ib ,
=
1
+
i ,
~hus, 1
+
i
=
J
If
z =
Here,
a z
+
ib ,
7
=
-
-1 sin ( - 3 / m ) 93.
then
z
Here,
Squaring gives =
25
i0
- e3iii/2 = (cos
and
m= fi
=
Z . ~ then
-0.40 (3
+
and
~
.
r
=
=
= 9
4i)'
8
+
-1
9 = cos
24i
+
(-7125)
8
cos
=
16i2
.
(a/r) .= sin-'(blr)
-1 4 = cos ( I / & )
114
=
.
~ -1 cos (alr)
=
8
e
=
-
.
- i cos 5 -1
and J%
=
. Using eix
i sin Q) (cos(3~1/2)+
~
J58
7 - 3i
+
Q
and
and
Thus,
@
~
r =
so
sin
=
- 4i)
.
- 4i)/25
i sin O ) ( - i )
(3 - 4i)/(3
Multiply by
i(8+3-12) = ei3,e37i/2 e
r
.
4i)
r =
so
3i ,
%
) = (3
e
+
1/(3
=
2
By the law of exponents, cos x
89.
i)2
557
-1 sin (b/r)
-1 cos (7/>;8)
=
=
i(-0.40)
-7
" 1.85
=
+
.
24i
So
r
+
4i)
. Thus, (3
=
2
=
25ei(1 .85) Write the number as peiS
I
of
?eim k
where 3i , sin-'(3/~i&) roots are
1;:
0.93
.
Thus,
[,
For
0
,
1
,
6+
3i =
+
v5i
,
.. .
+ 6i
are lZVG
= v%
,n
=
.
1
d7 +
For
;r = cos
I;--
("3114)
=
and the sixth
O , 1 , 2 , 3 , 4 ,
p =
ei(0'64)
-
-1
and
c o ~ - ~ ( 3 / > =) sin-'(r/5/14) 3
i(o;ln+2-ik/n) e
a ei(0'93)
, where k 3
/p
2 ,
J 5 7=
=
ei(0.155+2nk'6)
n-
are given by =
. Then the n th roots
= Z=
v a
and
0.64
q
and
5
=
. Thus,
and the sixth roots
ei(0.107+ 27:k/6) , where
k = 0
, 1 ,
Copyright 1985 Springer-Verlag. All rights reserved.
558
Section 12.6
101. F o r
z
, w
1
=
= e
0
,
so t h e s i x t h
roots are
r o o t s a r e l o c a t e d on t h e u n i t c i r c l e a t a n g l e s of
,
4n/3
and
.
5n/3
The
tenth
r o o t s of
e0
l o c a t e d on t h e u n i t c i r c l e a t a n g l e s o f
n
,
,
6n/5
,
7n/5
.
, and
8715
share comon points a t
0
9 ~ / 5
and
1 0 5 . The q u a d r a t i c f o r m u l a g i v e s u s t h e r o o t s
K T .
-1 i or
a 2 - b2 = 1
( 4 5 v/i;)/8
(-1
and
+
109. ( a )
+
b
z2
+
.
Since
Also,
R=
+
+
22
b
+
-
cos i B
=
Thus,
t a n if3
(e
B
e )/(e
+
1 = (-1 i f i ) / 2
in12
Notethat n-2
+ z
Z n - l = O ,
+ z
+
zn-'
If 1)
n-2
= Zn
-
...
...
n- 1
- z2
either
zn-2
+
+
+
~ 7 ) " -~ (-1
+
&)'I2
+ +
- (-1
a
2
be-
=
.
Since
+
&)'I2
fi)112i]
.
&)1/2il
x
=
.
cos i 8 = [ e Also,
i(i8)
tanh 0
B
e-i(i8)
112
tan i 6 = s i n i B /
.
-tanh B / i
We r e c o g n i z e t h i s a s i tanh
+
. ,
@ = n/2
and
so
tan i 8 =
.
(z-l)(z
n-2 -
e8)i.
r
Therefore,
(I/&) [ ( I
i
b2 = 1
-
i s the function
U s i n g p a r t ( a ) , we h a v e
-
, a2
-
1
b2 = a 2
=
.
il =
(a+bi)'=
-1/2a
=
1
.
fi)'I2il
-0
=
.
i ( i @ ) - e-i(if3) sin i9 = [ e ]/2i
-8
z
0
- (1
[?
, 3n/5 , 4n/5 ,
2n/5
=
(I/&) [-(I
i = (I/&)
,
n/5
then
U s i n g t h e r e s a l t o f Example 2 , we h a v e
z
(b)
,
&a4 - 4a2 - 1
or
~ 2 ) ' " +~ (-1
(tanh 8)e 113. ( a )
.
-1
=
, which a r e
e nik/5
correspond t o
a r e r e a l numbers, we g e t
and
(h)
1
=
~ i ) l / ~ i o] r
(1
2ab
, n/3 , 2n/3 , n ,
z = (-2 i -)I2
a + b i
-=
= (1 ? 4 ) / 2
Therefore, [z
and
a2 - 1/4a2
comes
a
Now, i f
Thus, t h e
The s i x t h and t e n t h r o o t s o f
, which
n
0
are
,
0
.
e Tik13
(Zn-'
+zn-2+...+z+1)=zn-Zn-1+Zn-1-
+
z2
- z2 - z
z
+
z - I = O
+ ... + z + +
-
zn-2 =
1
1
=
z - 1 = zn - 1
or 0
+ ... +
, z2
z
n- 1
z)
Therefore,
if
+ Z ~ - ~ + . . . + Z + ~ = O
+
then
+
.
=
o
or
zn-2
+ . .. +
zn = -zn-l
-
, from t h e o r i g i n a l e q u a t i o n .
Copyright 1985 Springer-Verlag. All rights reserved.
z
Section 12.6
113. (c)
The four roots of three roots of (b)
117. (a)
(b)
z3
z = x and
+
iy ,
2 z
then
+
y = n
when
+
1
+
1 = 0
z
such that
ez
e
Z
,
-1
are
,
2nn
X e (cos y
=
where
In(-1) = in ,
You might define z
+
,
are
i , -i
.
-1 , i
Therefore, the
,
-i , by part
-
If 0
z4 = 1
559
=
-1
n
+i
sin y) = -1
when
x
=
is an integer.
though there are many values of
.
SECTION QUIZ 1.
+
ib) , where
Let
u = i/(a
(a)
What is the imaginary part of
(b)
What is the real part of
(c)
What is
(d)
What is the polar representation of
a
and
b
are real numbers.
u ?
u ?
u.; ?
2.
What is e l x
3.
Find all of the solutions of
4.
Let
z l = 2 + 2 i and
z2 = 2 6 - 2i
(a)
Convert
zI
z2
(b)
Compute
zlaz2 .
u ?
in terms of sines and cosines?
and
z4
+
1 = 0
.
.
to their polar representation.
Express your answer in both the
a 4- bi
form
and the polar representation.
(c)
Explain how multiplication of complex numbers can be related to the lengths and arguments of complex numbers.
5.
Compute the following: (a) (b) (c)
(3
-
+
z
if
(5
+
2i)' z
=
l/(i - 2)
3i)i - 2e
Copyright 1985 Springer-Verlag. All rights reserved.
560
6.
Section 12.6
Ronnie was struggling through his calculus assignment when his fairy godfather appeared to offer his assistance. Asked what the problem was, Ronnie explained that he needed to do some root extractions.
\\'hen
the dumb fairy godfather heard this, he sent for the tooth fairy. Ronnie explained that he had to find the sixth roots of (a)
2
+
3i
.
Fortunately, the tooth fairy understood Ronnie's problem. answer did she give him?
Express the answer in the form
What a
+
bi .
(b)
Plot ail of the points representing
in the xy-plane.
(c)
Of all the answers in (a), find the complex conjugate of the one
6~~
with the smallest positive argument.
ANSWERS TO PREREQUISITE QUIZ
2.
(fi/2>-fi/2)
3.
(2&,-~/4)
4.
(a)
1
(b)
I - x
-
2 4x /2!
+
+
4 16x /4!- ...
2 3 x /2! - x /3!
+
=
:T
...
1=0 =
[(-1)~(2~)~~/(2i)!]
~~=~[(-x)~/i!]
AKSWERS TO SECTION QUIZ
(c)
l/(aL +bL)
(dl
rei8 ,
where
r = 1/d2
+ b2
and
8 = tan
-1
(a/b)
Copyright 1985 Springer-Verlag. All rights reserved.
Section 1 2 . 6
in14
3ni14
, e
5ni/4
, e
3.
e
4.
(a)
z1 =
(b)
(4fi + 4)
(c)
(rlexp(ial))
2J2
, e
exp(niI4)
+
7ni/4
and
( 4 6 - 4)i
561
=
z 2 = L exp(-ni/6)
8&
exp(nill2)
(r2exp(i8 2) ) = r 1r 2exp[('il
+
~ ~ ) i .l The lengths are
multiplied and the arguments are added. 5.
6.
(a)
5+12i
(b)
(-2
(c)
-3
(a)
1.2
+
-
i)/5
2e
+
0.2i ;
1.2i ; (b)
+ 5i 0.4
+
1.2i ;
-0.8 + 1.0i ; -1.2 - 0.2i ;
-0.4 -
0.8 - 1.0i
't
Copyright 1985 Springer-Verlag. All rights reserved.
562
Section 12.7
12.7 Second-Order Linear Differential Equations PREREQUISITES 1.
Recall the quadratic formula (Section R.l).
2.
Recall the polar representation for complex numbers (Section 12.6).
3.
Recall basic differentiation and integration formulas (Section 7.1).
4.
Recall how to solve the spring equation (Section 8.1).
PREREQUISITE QUIZ
1.
What is the solution of
ax2
-+
bx
+
c = 0
if
a , b , c
are real and
a f O ? 2.
Let of
3.
z be the complex number
5
+
5i
.
What is thc polar representatldn
z ?
In terms of trigonometric functions, what is 2 dy/dx2+4y=0,
4.
Solve
5.
Evaluate the following: (a)
~ e ~ ~ d t
(b)
(dldy)(cos
(c)
j(x2
+
yr(0) = 2 ,
eix ?
y(0) = 3 .
y + sin y)
2x)dx
GOALS ayl'+ by'
+
cy = 0
.
1.
Be able to solve differential equations of the form
2.
Be able to find a solution for nonhomogeneous equations using the methods of variation of parameters or undetermined coefficients.
STUDY HINTS 1.
Characteristic equation. The key to solving
ay"+ b y 1 + cy
finding the roots of the characteristic equation
ar2
+
br
+
=
0
c = 0
is
.
This is easily solved using the quadratic formula.
Copyright 1985 Springer-Verlag. All rights reserved.
563
Section 12.7
2.
Distinct characteristic roots.
If
istic roots, then the solution of c exp(rlx) 1
+
conditions.
3.
c exp(r x) ; 2 2 If
and
r2
r2
ayf'+ by1
and
+
are distinct character-
cy = 0 is
y =
are determined by the initial
c2
are complex, then the exponential part ix -
is the only characteristic root,
b
-
4ac = 0
,
If
rl
-
then the solution of
i sin x
.
Repeated characteristic roots. 2
cos x
+
e
(cl + c2x)exp(rlx)
ayr'+ byt + cy = 0 is
.
Method of reduction of order.
When the roots repeat, simply look for a
solution of the form y = v exp(r x)
1
5.
and
of the solution may be rewritten using
i.e.,
4.
rl
cl
rl
Method of root splitting.
,
where
v
.
x
is a function of
This is an alternative derivation of the
formula for repeated roots given in item 3 above (formula (5) on p. 619). You shouldn't worry if you don't understand it, as it won't be used later. 6.
Damping.
The equation d2x/dt2
harmonic motion.
+
B(dx/dt)
+
w
2 x = 0 describes damped
Whether we are dealing with an overdamped, critically
damped, or underdamped case depends upon whether the characteristic equation has two, one, or no real roots.
The solution can be written
in terms of sine and cosine if no real roots exist and
the solution is
oscillatory. See Fig. 12.7.2.
7.
Nonhomogeneous equations. a y U + by' tion.
+
cy = F(x)
is a nonhomogeneous equa-
Notice that the right-hand side is afunctionof x only.
homogeneous equation if
F(x) = 0
.
It is a
The general solution of the homo-
geneous equation (containing two arbitrary constants) is denoted b y
A specific solution of the nonhomogeneous equation (with no constants) is called a particular solution and 1 s denoted b y yp
+
yh
y
11
.
arbitrary
y
P
is the general solution of the nonhomogeneous equation.
.
Then Note
Copyright 1985 Springer-Verlag. All rights reserved.
564
Section 12.7
7.
(continued)
8.
alone solves it and yh adds zero to F(x) . P Method of undetermined coefficients. Depending on the form of the that
y
right-hand side, guess that the particular solution is a linear combination of sines, cosines, exponentials, and polynomials. tiate and substitute into the left-hand side. stants.
See Example 5.
Differen-
Then, solve for the con-
Make up an equation with
F(x)
as a polynomial.
Your guess for a particular solution should be a polynomial of the same degree.
9.
Variation of parameters. --
If
yl and
y2
are solutions of the homo-
geneous equation, we look for a solution of the £ o m yp = vlyl + v2y2 where
vl
and
v2
are also functions of
stitute into the original equation. the equation becomes vlYl ' ' taneously for
v;
and
v;
+
v;y;
=
We set F/a
.
.
x
,
Differentiate and sub-
vlyl '
+
viY2 = 0
and then
These can be solved simul-
and integration of
v;
and
v;
yields
a particular solution. 10. Damped forced oscillations. The box on p. 628 should not be memorized. If you encounter a forced oscillation question on an exam, look for a
,
particular solution of the form A cos(S2t - 6)
where
0
is the for-
cing frequency, and add it to the general solution of the homogeneous equation. 11. Wronksians.
If
yl
and
y2
are solutions of the homogeneous second-
order linear differential equation, and cly1
+
c2y2
yly;
-
y2y; # 0
,
then
y =
is a general solution of the differential equation. This
is the gist of the supplement.
The expression yly2' - y2y;
is called
a Wronksian.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.7
565
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
The general solution of
,
c exp(r2x) 2 and
cl and
2 = 3 5.
where
or
ay"
+
rl and
r2
+
cy = 0
are the roots of
.
3cl
+
.
Thus, the solution is
Thus, y (0) = 0
.
c2 = 1
and
r = (4 i -112
is
y = clexp[(2
+
i)x]
=
+
+
y'
'(0) = 1 gives
- (1/2)exp(x)
+ br +
)
y = clexp(3x)
This yields cl = 112 and
solution is y = (1/2)exp(3x) We have
arL
c2 are constants. Here, r = (4 i m
1
+
is y = c exp(r x) 1 1
Differentiating the solution to Exercise 1 gives c2exp(x)
9.
by'
/
c = O
2 = (4 1 2)/
.
c exp(x) 2
+
3clexp(3x)
=
c1+c2=0
c2 = -112
.
and
The particular
.
(4 i 2i)/2 = 2 i i
.
Thus, the solution
2x c2exp[(2 - i)x] = e [c (cos x 1
-1-
i sin x)
+
c (cos x - i sin x)] = eZX[c (cos x + i sin x) + c (cos x - i sin x)] = 2 1 2 2x e (alms x + a sin x) , where a l = c + c2 and a2 = i(cl - c2) . 2 13.
The general solution is y = (cl + c x)exp(r x) 1 2 repeated root of the characteristic equation.
9
=
(r
-
3 ) '
.
c2x)exp(3x) 3 exp(3x) and 17.
(a)
1
B
=
2
=
.
0
,
so
r = 3
,
where
rl is the r2 - 6r
Here, we have
and the general solution is y = (c
Differentiation gives y' = c2exp(3x)
Substituting y(0) = O
and
y'(~)
=
+
(c,
+ c2x)
1 yields
1
+
+
x
I
0 =
Thus, the particular solution is y = exp(3x) . 2 2 - 4w2 = n I256 - 4(n 14) < 0 , so the spring is underdamped. c2
.
The general solution is x =
(c
1
cos it
+
c2sin ;t)exp(-rt132) where
;=
mn/32
.
A x'
(-cl; sin ot c2;cos ot)exp(-~t/32) + (c1cos it
+
n
/
2
=
+
c2sin wt)(-n/32)exp(-vti32)
Copyright 1985 Springer-Verlag. All rights reserved.
=
566
Section 12.7
17.
(b)
(continued) Substituting x(0) = 0 c
21.
o
2
, so c2
=
and
.
1/;
The general solution is
y
x'(0)
Thus, =
x
+
yp
1
=
yields
(l/;)(sin
=
0 = cl
yh
=
.
;t)exp(-~t/32)
yh , where
1
and
is the general
solution of the homogeneous equation and
is a particular solution yD The characteristic equation is r2 -
of the nonhomogeneous equation. 4r
+
3 = (r - 3)(r
ous equation is
- 1) y
=
fore,
A = 2
c2exp(x) 25.
+
y =Ax
-4A
+
3(Ax
B
=
6
and
2x
+
6
so the general solution of the homogene-
.
+
+
+
C1 = cl
have the form
+
c2
+
,
so
(2A - 4B
+
,
2i
I
+
=
A
(-4A
and
+
.
y"= 0
3B)
6x
=
+
Sub-
10 ;
there-
+
y = clexp(3x)
+
+5 .
The roots are given by
so the general solution of the character-
+
+
2i)xl
c2exp[(2 - 2i)xl
c (cos 2x - i sin 2x)I 2
and
2A - 4(2Ax
5D) = x
2
+
+x ,
.
- c2)
C = i(c 2
so B)
. Hence, the solution is y
+
13x125
+
=
e
=
2x
5(Ax
e
+
x
C2sin 2x) ,
A particular solution should
2
+
and
+ D)
Bx
A = 1/5 , 2x
eZx
=
(C 1cos 2x
y1 = 2Ax+B
i.e.,
42/125
42/125
y'
r2 - 4r
y = Ax2 + B x + D ,
Substitution yields 5B)x
B
. A particular solution
Iience, the solution is
y = clexp[(2
[c (cos 2x+ i sin 2x) 1 where
c exp(x) 2
.
416 - 20)/2 = 2
istic equation is
+
B) = 3fm
The characteristic equation is (4
29.
0 ,
c exp(3x) 1
should have the form stitution gives
=
(Clcos 2x
yl'= 2.4.
2
(-8A
B = 13/25 ,
and
D
2
+
+
C
5Ax
sin 2x) 2
+
x /5
vlyl + v2y2 , where
I
I
vlyl get
and
+ viy; v;e3'
vl =
=
.
The method of variation of parameters gives a particular solution,
tion,
+
+
=
and i/a
+ viex
. =
y1 v2
and
y2
are found by solvin~ v;yl
and
V;
=
are solutions of the homogeneous equa-
FmmExeicise 21, yl = elx 0
yp
3x (3e )
+
";ex
=
+
and
(6x
+
v;y2
=
y2
ex ,
10)/l
=
0
and so we
. Subtrd~t;~"
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.7 567
29.
(continued) gives
2v;e
u = 3x
+
3x
5
6x
=
10
,
i.e.,
+
v; = (3x
+
+
i.e., v; = -(3x (3x
+
.
5)e-3X/3 - e-3X/3
.
5)e-X
-
5)e-X
+
33.
+
[(3x
Integration by parts with
+
j3e-xdx = (3x
v;exp(x)
=
0
and
i.e.,
.
;(tan x)exp(-3x)dx/2 (-tan x)exp(-x)/2
=
+
3v;exp(3x)
2v1exp(3x) = tan x , 1
v;
exp(3x)
=
+
+
5)e-X
+
and
v;exp(x)
[(-x
+
c2exp(x)
.
, and so v,
=
5)
2)exp(-3x)l
+
2x
+
X
6
. +
so v;exp(3x)
Subtraction yields
, and so vl
v; = (tan x)exp(-3x)/2
Similarly, we get
-
,
tan x
=
,
Therefore, the
+
y2 = exp(x)
10)
+
u = -(3x
.
3e-X
+
5)e-3x/3
Similarly, 2v'ex = -(6x 2
8)exp(-x) lexp(x) = clexp(3x)
From Exercise 21, yl
Thus, by lettins
vl = - O x
general solution is y = clexp(3x) -t c2exp(x) exp(3x)
.
5)e-3X
and integrating by parts, we get
~ e - ~ ~ = d -(3x x
yields
+
2v;exp(x)
=
-tan x
.
j(tan x)exp(-x)dx/2
,
=
i.e.,
Thus, the
&
solution is y
=
[exp(x)/2]j(tan
+
clexp(3x)
=
C cos t ,
the homogeneous equation is
x1 A
=
+
1
and
+
B sin 2t
+
2B cos 2t - sin t ;
xl(0) = 0
=
so
-cos 2t
+
+
C
1
=
.
B sin 2t ,
cos t
.
i.e.,
.
Sub-
The solution of so a general
Differentiation yields
therefore, x(0)
,
gives 0 = 2B
Hence, the solution is x
x " = -C cos t = -x
x = A cos 2t
+
x = A cos 2t
-2A sin 2t
and
3C cos t = 3 cos t ,
stitution yields
solution is
Try a particular solution of the form
x' = -C sin t ,
so
-
[exp(3x)/2] j(tan x)exp(-3x)dx
.
x)exp(-x)dx
37. Use the method of Example 7. x
+
c2exp(x)
A
=
-1
=
and
0
gives 0 B = 0
cos t = 2 sin (3t/2)sin(t/2)
=
. by
the product formula.
41.
(a)
Here,
r2
iv%, and
+
and 6
=
4r
+
25
0
=
F0=2,
-1 tan (8121)
implies r = (-4 i J 1 6 - 1 0 0 ) / 2 = -2 t
m = l ,
.
k = 2 5 , u = 5 , 0 = 2 , -,=4,
In terms of sine and cosine, the solution
is x(t)1985 = exp(-4t) [A s i n m t t B c o s m t ] + (2/m)cos(2t - 6) . Copyright Springer-Verlag. All rights reserved.
568 Section 12.7
4 , (a)
(continued) Differentiation gives x1(t) exp(-4t)
[mA
B
+
421505 ,
i.e.,
B
+
(21rn)cos 6 = B
=
-421505 ; xl(0)
A + ( 4 / m )( 8 / m )
168/505
=
-2OOI505J21 = -4oI101ZE
.
For large
t ,
+
y;(x)
=
0
321505
+
i
gives
0
.
- 6) =
+
-4B
=
A , i.e., A
exp(-4t) [(-40/101,6i)
=
the graph approximates
rlexp(rlx)
the LJronksian is
v%t]
(21m)(21/&m)
= x
.
- tan-l(8/21))
(2/&-?)cos(2t
If the Wronksian does not vanish, then set.
& + B cos
-
(2/v'%)cos(2t
.
-1 tan (8121)) 45.
+
+ =
~ h u s , x(t)
s i n m t - (421505)cosfi t] (b)
-4exp(-4t) [A sin
& B sin i/?it] - 2(2/m)sin(2t
cos m t -
gives 0 = B
x(0) = 0
=
and
yl(x)y;(x)
y;(x)
=
yl
.
y2
=
(1
form a fundamental
+ xr 1exp(r
exp(rlx)
- y2(x)y;(x)
rlx exp(2rlx) = exp(2r x) f 0 1
and
+
Thus,
rlx)exp(2rlx) -
Therefore, y1 and
y2
form a fun-
clyl + c2yZ
damental set and the supplement tells us that
.
x) 1
is the
general solution. 49.
(a)
If y(x)
is a solution, then, by the method of reduction of
order, v(x)y(x) yl(x) = y(x)
is also a solution.
and
assumed that 0
.
However,
- v(x)y(x)yJ(x)
y(x) P 0 , if
, We get W(x)
y2(x) = V(X)Y(X)
[vJ(x)y(x) + v(x)yJ(x)j
By e~cpression (la), with
=
so W(x) = 0
vl(x) = 0 ,
then
is simply a mulitple of
v(x)y(x)
is another solution. Therefore, y(x)
and
[y(x)12vf(x)
.
Ice
vl(x) =
is a constant and
. Thus, vl(x) f 0 if
v(x)y(x)
the fundamental set is
yix)
if and only if
v(x)
y(x)
=
v(x)y(x)
2 [~(x)] vf(x) f 0
and
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.7
49.
We substiture yl(x) = xr
(b)
and
y
2 = 1
x
[ ( I - a)/21x(-1-u)~2} - (In x)x(1-')12(rxr-1)
+
.
(1 - r 1nx)lx r+(-1-n)/2
,
B = 0
and
so
r = ((1
-
NOW, by assumption, r2
- 1)
a) i- /(a
2
.
2 rx 3 rx , y ,,,- r e , r e
+ erx
=
i)/fi ,
0
and
rx y = e
4 (r
=
+
l)e ) ,
1 - i
4 rx yl'"=re rx
.
(-1
fore, the general solution is
+
ex/fi(~3cos(-x/&)
exl&(c
+
=
y(x) = clxr +
and
Solving
+
i)/&
+
so
cly(x)
+
Y'
=
rx re
,
"
r4
+
, and
1 = 0 yields (-1 - i)/v5
r
.
=
There-
+
5cos(x/h) + c6sin(x/
~~sin(-x/h)).
c2sin(x/fi))
=
Therefore, y t " ' + y =
+
sin(x/n) = -sin(x/fi)
1cos(x/a)
.
,
) c2sin(x/h)) eX ~ ~ ( C ~ C O S ( X / A +
c4sin(-x/&))
Z) + ex(ccos(-x) + cos(-x/a)
(u - 1)r
- 48)/2 , and so W(x)
The general solution for part (a) is
We expect a solution of the form
+
+
-
C,V(X)Y(X)
(1
x
(In X)X(~-~)/~ form a fundamental set.
c (In x)x(~-")'~ 2
r erx
into
[(In x) (1 - a)/
=
The general solution for Euler's equation is
53.
2
W(X) = xr{ (l/x)x (1-")/2 + (ln x)
expression (18) to get
2
x
569
,
Since
cos(x/fi)
=
the solution reduces to
+ e-~'~(c 3 cos(x/fi) +
c4sin(x/fi))
y =
.
SECTION QUIZ 1.
Find a general solution to each of the following differential equations: (a)
2y"-2y'+y=0
(b)
y"+6yf+9y=0
(c)
-2y"
+ 3yf +
2y
=
0
Copyright 1985 Springer-Verlag. All rights reserved.
570 Section 12.7
2.
For each of the following differential equations, guess the general form of the particular equation:
3.
(a)
y'\y'+y=cos2x+3
(b)
y1V+y1+y=x5-x3+tanx
(c)
yr'+ y'
+y
= e-2X + x 2 - 2
Find the general solution of
Y'"- y = x
homogeneous equation should be
y
=
.
exp(rx)
[~int: one solution of the
.]
4.
Solve y " - 3 y 1 + 2y
5.
Francis, the fruit fly, expends energy when he accelerates; however, he
=
(1 -
and leave your answer as an integral.
gains energy by speeding between fruit trees and coverins more distance since he gets more fruit juices more rapidly. Thus, his distance can -y " + 2y'
be described by
+
If the distance function y tree is
0.5
y
=
0 , where
y(0) = 0
is a function of time
and
y '(0) = fi
t , and the averaye
units apart, how many trees has Francis the fruit fly
visited after one unit of time (assuming he begins at
t = 0 )?
ANSWERS TO PREREQUISITE QUIZ 1.
x = (-b t 6
3.
cos x
4.
y = 3 cos 2 t + sin 2t
5.
(a)
e3t/3
(b)
-sin y
(c)
+
3
)
/
2
a
i sin x
x /3
+c +
cos y
+ x2 + C
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.7
571
ANSWERS TO SECTION Q U I Z 1.
2.
[c1sin(x/2)
(b)
(cl + c2x)exp(3x)
(c)
clexp(2x)
(a)
Acos 2 x + B s i n 2 x + C
(b)
AX
5
4.
5.
cle
X
+
c2cos(x/2)] exp(x/2)
c2exp(-x/2)
+ BX 4 + cx3 + DX 2 + E + F
A exp(-2x)
(c) 3.
+
(a)
+
BX'
+
cx
+
cos x
+
G sin x
D
+ [ c 2 c o s ( 6 x / 2 ) + c3sin(v'%x/2)l e -XI2 - x . + c2ex + e 2x j [ d x / e 2x ( 1 - x 2 ) 3 ~ - e X j [ d x / e X ( l -
2 3 x ) 1
42 t r e e s
Copyright 1985 Springer-Verlag. All rights reserved.
572
Section 12.8
12.8 Series Solutions of Differential Equations PREREQUISITES 1.
Recall how to solve a second-order linear differential equation of the form
2.
ay"
+ by' +
cy = 0
(Section 12.7).
Recall how to differentiate a power series (Section 12.4).
PREREQUISITE QUIZ 1.
Find the general solution y(x)
2.
Solve
3.
Find
(d/dx) tT=O [(2~)~/i!].
4.
Find
(d/dy)~T=~[(-y)~/(i
y"-
23~'+ y = 0 ,
~'(0)
of =
y"
o ,
-t
3 y f + 2y
y(0) = 1
=
0
.
.
+ 2)] .
GOALS 1.
Be able to solve differential equations by using power series
STUDY HIETS 1.
Series solution. Herewe seek the solution of a differential equation i x in the form y = ~ T = ~ a .~ The
a.'s
need to be determined.
Dif-
ferentiate the series and substitute into the original equation.
Be
sure you have changed the summation index when you differentiated. At this point, write out the first few terms and look for a pattern. Finally, the ratio test may be used to show convergence. 2.
Special equations. Legendre's and Hermite's equation are special in that they are specific equations that arise from physical problems. However, the method of solution is the same.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.8 573
3.
Frobenius method.
look for a solution of the form
y
r-m i x Ziz0aix
=
x = 0 ,
y " vanishes at
If the coefficient of
.
The method of solution r-1 x
is the same except that we now solve for the coefficient of x
r+l
, xr ,
, etc., rather than x , x2 , x3 , etc. We solve for r , which is
generally 4.
then
See Example 5.
an integer.
r
Indicia1 equation. The values of
in the Frobenius method is deter-
x
mined by setting the coefficient of the lowest power of
equal to
zero. This is called the indicial (pronounced "in dish al") equation. 5.
Repeated indicial roots.
If the roots repeat, then the solutions have
r m i the form yl (x) = x ZizOaix
y2 (x) = y1 (x) In x
and
+ xr~m ~
i , ~ .b ~ x
SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
Let
y
i
on
=
Zi,oai)c
,
m . i-1 y' = Zi=lla.x
SO
Then equate the coefficients of
x
i
to
and
.
0
yf'=
y"
xYz2i(i - l)a.x
- xy' - y
i-2
.
=
i(i - l)a,xi-2 i m i m i Zi=2 1 Zi=lia.x - CiZOaix = Z. 1=0(i + 2)(i + l)ai+2x i m i ~ $ = ~ i a -~ xZ. a.x = 0 . Thus, 2a2 - . a = 0 (constant term), 6a3 1=0 1 a1 - a1 = 0 (i
+
+
2)(i
a1/3 , a4 (i
+
1
+
l)a./(i
a0/2"(n!)
a2/4
=
+
2)(i
and
a 18 0 1)
a2n+l =
fore, the solution is (2n 5.
+
I)!]
(x
1
,
a2 = 0 i
l)ai+2 - ia. - a. = 0
=
-
12a4 - 2a2
term),
(X
a5 = a 15 3
term).
ai/(i
y = ao[ Cn=,(x
n
i.e., 1)
=
n /2 (n!))]
term), and in general,
Hence, a : ! = a0/2
a1/15 ,
=
+ 2) , a1/3.5 ... . .(2n + =
(x2
,
a3 =
and in general, a. = 1+2 = a0/2.4. a 2n
a12*(n!)/(2n
+
....2n l)!
=
There-
+ al[ ~:=~(2~(n!)x'~
7
+I
/
.
Let
y
=
Z,,oaix
y(O)
=
0
and
2xyf = :Z i(i 1=2 i ~ y ~ ~ 2 i a=. 0x
i
,
y'(0)
.
y 1 = ),liaixi-l
SO =
1
implies . a
=
~ ) a . x ~+-,:?~ 2ia.xi 1=1 1 Thus, 2a2
=
0
2 (x term),
0
=
and
y r i =~Y=~:i(i - l ) a . ~ ~. - ~
and
al = 1
:1=0 : (i + %)(i
(constant tern),
. +
6a3
-
+
Now, y " + l)ai+2x i 2al = 0
+ (x
and 20a5 + 6a3 0 (x3 term). term), 1 2 a p 4a2 = 0 Copyright 1985 Springer-Verlag. All rights reserved.
574
5.
Section 12.8
(continued) Hence, 1/10
9.
,
a2 = 0
.
-a1/3 = -113 ,
=
Therefore, t h e s o l u t i o n i s i
m
y = CiG0aix
Let
a3
,
y'
SO
(i
- xy = 0 = CiZ2i(i + l)ai+*x i - ai-lx i1 -
3*2a3 0
(X
a.
3
= 0
6*5a6 -
and i n g e n e r a l
,
a3 = a0/6
+
(i
,
a 7 = a I42 = a1/504
4
a3 ,
x 112 13.
+
7 x 1504
-
-
= 0
(x
+ ...) .
+
4
term)
=
y = aO(l
0
, 7.6a7
c;=~[ (i
-
, a4
(xi term).
, a6
x I6
+
x
6
.
2)x
,
-
5*4a5 =
+
(x
0
Hence,
a
5
2
=
,
term)
a2 = 0,
,
a3/30 = a 0 1180
=
+ 2) . 1180 + . . .) + a l ( x +
a 1+3 . = a./(i 1
+
+
(x 2 term)
3
i-2
y" = C. i ( i - 1 ) a . x 1=2
( c o n s t a n t term)
1-1
-3a3/10 =
m
and
l)ai+2 - a .
a 5 = a 2 120
=
... .
x5 110 -
Z l=Oaix . ii-l = zap
and i n g e n e r a l ,
Therefore, t h e s o l u t i o n i s 4
i- 1
4*3a4 - a l = 0
+
2) ( i
a 4 = a1/12
-
=
x - x 3 13
Thus, 2 - l a 2 = 0
,
(x term)
,
term)
1)a.x
.
ia.x 1=1 1 i-2 C.
=
m
yn
y
m
, a5
-- -a2/3 = 0
a
+
3)(i
The r e c u r s i o n formula 1 s
a. = al/(i 1+3 r-1 y ' = raOx + (r
+
?)(i
+
2)
+ alxr+l + a 2xr+2 + ... , so + l)alxr + ( r + 2)a2xr+l + . . . and y = r ( r - l)a0xr-2 + r ( r + l)alxr-' + ( r + 2) x 2 ( r + l ) a 2 x r + ... . Thus, 3x y l ' + 2xyf + y = [ 3 r ( r - l ) a O + 2ra0 + a0]xr + [ 3 r ( r + l ) a l + 2 ( r + l ) a l + a l ] x r + l + ... = 0 . S e t t i n g t h e c o e f f i c i e n t of xr equal t o 0 y i e l d s a 0 ( 3 r 2 - r + 1) = 0 , i . e . ,
Let
y
a xr 0
It
.
r = (1 i f i i ) / 6
For t h e
0 , which y i e l d s a 2 = a3
=
.,.
aoX (1-flri)/6 C X
(l+mi)/6
al = 0
= 0 .
Thus,
xr+l
c o e f f i c i e n t , we s e t
because
c2x (1-Jiii)16
1 (In x)61+fii)/6
i sin(JL1 In x/6))
must be
0
.
The g e n e r a l s o l u t i o n i s
Note t h a t
- e (In x ) l b e ( f i In x ) i / 6
.
x116[biC0s(r'ii i n x16)
(1 i m i ) / 6
.
5r
x( l + a i ) / 6
+
y =
y =
-
= xl/6(cos(fi
In x16)
+
T h e r e f o r e , an e q u i v a l e n t s o l u t i o n i s
+b2sin(a
i n x/6)]
3)
=
Similarly,
i s one s o l u t i o n and
y = a x
i s another solution. +
r
+
al (3r2
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.8
17.
Let
y
~ : = ~ a ~ SO x ~ y,
=
anxn-'1. j = 0
6a
+o
3
yl'+ w y =
then
2
+w
2a2
al = 0
2
a.
x.J = O
= 0
or
a
.
a 3 = -w a1/6
or
+
[(j
2
2
n-1
m
=Cn,lnanx
2
Thus,
,
,
,
+
2)(j
-- w
2
y " = ~ : = ~ [ n ( n - 1)
+
1 ) a . xi 3 +2
a /2 0
2If
and
.
If
J
j = 1
, then
j = 2
.
w2a.xjl
12a4
.
4
575
If
,
then
+w
2
a
x
=
2
2
+
a 112 = w a 124 I f j = 3 , t h e n 20a w a3 = 0 2 0 5 4 2n 2n o r a 5 = -w a 1 2 0 = w a 1120 In g e n e r a l , a = (-1) 10 a0/(2n)! 3 1 2n 2n 2n and a 2n+l = (-1) w a1/(2n l)! Thus, t h e g e n e r a l s o l u t i o n i s 0
or
a 4 = -w 2
y = a (1 0
-
2 2 4 4 w x /2! + w x /4!
. + . - ...) +
a (X 1
-
2 2 1 - w x /2!
4 4 o x /4! -
...
a s t h e Maclaurin s e r i e s of
We r e c o g n i z e cos
LX
,
21.
,
where
Thus,
-
s i n wx/w
.
A = a.
I n Example 3 , we l e t we l e t
x
and we r e c o g n i z e
M a c l a u r i n s e r i e s of B s i n ox
+
y2 = x
+
y1
and =
[(6
-
[ ( I 2 - A)(2 - X)/4.3.2]x4
B = a /w 1
4 5 w x /5! -
4 x /6 have
a s the
+
... .
-
y = A c o s wx
+
. [(6 - ) / 4 . 3 . 2 ] x
4
+ ...
and
.
2
+
...
and
y;=
1
+
[(2 - h)/2]x
S i n c e a l l o f t h e s e r i e s do c o n v e r g e ,
t h e y c a n b e m u l t i p l i e d , s o t h e Wronskian i s
+ x 2 + ( 1 8 - 25A + 7b + . . . I = 1 + ( 1 + h ) x 2 + ... . e v e n e x p o n e n t s , s o W(x) > 1 #
Y 2 ( ~ ) y ; ( ~=) 1
...
...) .
5 [ ( 1 2 - ~ ) ( 2- ~ ) / 5 . 4 - 3 ~ ~ 2 1+x ...
A)/3.2]x3+
+
+
Thus, t h e s o l u t i o n i s
1 - (A/2)x2
[(2 - ;)/3.2]x3
).;=-Ax-
2 3 w x /3!
4 5 2 3 w x /3! + w x /5! -
2
4
) x 124
W(x) = y l ( x ) y ; ( x )
+
... -[-bx2
+
-
(A2 - A
-
6)
Note t h a t a l l o f t h e h i g h e r t e r m s
0
.
Therefore,
yl
and
y2
form
a fundamental s e t .
Copyright 1985 Springer-Verlag. All rights reserved.
X
576
S e c t i o n 12.8
SECTION Q U I Z 1.
Find t h e f i r s t few terms of t h e power s e r i e s s o l u t i o n s f o r
y'l'-
x2y
+
y = O . 2.
-
Find a power s e r i e s s o l u t i o n f o r
(b)
Use t h e methods of S e c t i o n 12.7 t o s o l v e
(c)
What s p e c i a l equation do you g e t by equating your answers t o p a r t s
y"
2y' = 0
.
(a)
y " - 2yr = 0
.
(a) and ( b ) ?
3.
An obnoxious t r a v e l l i n g salesman hasmade you extremely i r r i t a t e d .
Con-
s e q u e n t l y , he i r k s you i n t o slamming t h e s p r i n g door i n t o h i s f a c e . Due t o a d e f e c t , t h e f o r c e e x e r t e d by t h e s p r i n g i s given by 3y'
+
xy = 0
,
where
y
Find a power s e r i e s f o r
i s a f u n c t i o n of y(x)
if
x
2y" -
, the door's position.
y(0) = y l ( 0 ) = 1
LVSWERS TO PREREQUISITE QUIZ
ANSWERS TO SECTION QUIZ 1.
2.
3.
3 6 ) a (1 - x I 6 - 5x /720 0 5 8 a (x2 + x /60 + 19x /20160 + 2
... + a 1 (x . . .)
(a)
a02z=0 1 ( 2 ~ ) ~ / n ! l
(b)
y = clexp(2x)
(c)
eZx
1
+x
=
+
4
7
- x 124 - l l x 15040 -
...) +
c2
~o=~
[(2~)~/n!1
2
- 3x /4
+
3 4 7x 124 - 87x 1576
+ ...
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.R
577
12.R Review Exercises for Chapter 12 SOLUTIONS TO EVERY OTHER ODD EXERCISE 1.
- 1/12)
(1112)~= (1/12)/(1 5.
9.
~7,~
This is a geometric series, so it converges.
1
+2+
2
+
113
2 113
+
,
312 = 712
3 113
+
=
.
1/11
+ ...
=
(1112~)= ~7=~(1/12)r
2
+
~ ; = ~ ( 1 / 3 )=~ 2
+
1/(1 - 113)
=
so it converges.
5-" = 1:=~(1/5)~ Cn=l
converges since it is a geometric series with
r < l . 13. Let
a
=
.
(-l)"n13~
1im 1im Then, n_,,ian/an-l/ = [n/(n - l)] /3i
113
=
.
m
Therefore, Cn,lan 17. Let
a
=
converges by the ratio test.
2n2/n!
.
lim 211-1 (2 /n) = n-
.
Then
lim n-lan/an-l
Therefore,
'n= 1
lim nL-(n-1)' /nl = = n-[2 2 (2'" )/n!) diverges by the ratio
1
test. 21.
Let
a
l/(ln n)
=
In n
,
so
In a
=
is an increasing function, In In n such n
,
2 b n = l/n 25.
<
a
2 l/n
,
.
- In n(ln In n)
>2
if
n
> e(e
2
.
)
Thus, for
so the series converges by comparison with
.
The error made in estimating the sum of an alternating series is no greater than
1 - 114
I 'n+ll
+
1/16 - 1/32
+ ... E0.78 .
so the sum is approximately /1 = 1
r = a to find
N
-
0.78
/
=
1
[(l - ~ ) / 3 ~ r/(l ] - r) /
- 2
< 0.05
less than
2 N (1 - N )/3 (5 - 2N) , which is less than
Thus,
[(I
-
n)13~]
-- 0 -
,
.
n)3n1/(2 - n)3"
such that
Although
1,132 < 0.05
there is no formula for the general term, we note that
29.
Since In n
.
- 3
.
'The error is
0.05
119 - 2/27 - 3/81 - 41243 -
We want
>5 . ... = -0.24 . for
N
33. False; consider
Copyright 1985 Springer-Verlag. All rights reserved.
578
S e c t i o n 12.R
False;
e2X = C:
1=0
1 ( 2 x ) ~ / i ! l= 1
True; by t h e sum r u l e ,
True;
+ bi)
/an/
.
z:=~( l a n ]
+
+
-
119)
118
.
/(i
+
=
i 2i
Let
ai = (-1) x
.
l)!
Then
lim i a i l a i l /
Let
1i m R = n_tm/an/an-l /
.
= (-l)n/2n
Then
112 = 2
r a d i u s of convergence i s Since
ln(1
+
+ Ey=laj +
LE,~~~ =
~ : = ~ ( 1 / 9 " ) = ~ : = ~ ( 1 / 9()1 1 9 ) ~=
Therefore, t h e r a d i u s of convergence i s a
+ ... .
/ b n / ) i m p l i e s t h e convergence of
.
0
3 4x 13
Therefore, by t h e comparison t e s t , t h e
This i s a geometric s e r i e s , so (1/9)1(1
+
2x2
~ 7 = +~ LL=obk a ~ = bo
/bn/
/an/
convergence of
+
2x
.
m
bo + CiZl(ai
+
x) =
=
.
2 [x / ( i + 1)1
=
by the r a t i o t e s t .
m
112
=
lim 1-
.
Therefore, t h e
.
cT=~
i+l i [(-1) x Ii]
, we have
+ x4 )
ln(l
=
.
~1= " 1~ ( - l ) ~ + l x ~ ~ / i l We have (e
t
-
e
t
=
l)/t =
(t
,
/I!)
i-1
Ii!)
,
f ( x ) = x3I2
;
=
therefore,
.
:11= 1 ( t i / i ! ) J;[(e
f ' ( x ) = ( 3 1 2 ) ~ " ~,
then
-312 f "'(x) = (312) (112) ( - 1 1 2 ) ~
[ (3) (1) ( - I ) * .
et - 1
SO
t
- l)/t]dt
(5
-
. .. ( 3 -
f ' ( 1 ) = 312 ; 2n)/zn
+
.
, and i n g e n e r a l ,
.
(3/2)(x - 1)
1
2:=0
[ ( 3 ) (1) (-1).
binomial s e r i e s and
f "(1) = 314 ;
. ..
+
. ... (5
- 2n)/2"n!] (x -
(I
=
+
x3l2
x
i.e.,
f ( n ) ( l ) = (3)(1)(-1)
[(3/4)/2!1 (x - 1 ) 2
co
CiZO [ " ( a - 1 )
,
1
=
f(1)
=
...
about
~ f ( ~ ) ( l ) / n(!x] - 1)"
.
-112
f (") (x) =
2(n - 1 ) ) / 2 " ] ~ ( ~ - ~ ~ )E/v~a l u a t ~ n ga t
Therefore, t h e Taylor expansion of
is
a=3/2
=
f "(x) = ( 3 / 2 ) ( 1 / 2 ) x
we g e t t h e c o e f f i c i e n t s s i n c e one t o any power i s one, 1 ;
Then
.
C: I= 1 [ x i / i ( i ! ) l
If
i .
Ei,O(t
x
=
1
=
A l t e r n a t l v e l v , one mav use t h e
... (a -
i .
i + I ) u / I ! ] with
u = x - 1 .
By t h e r a t i o r e s t ,
an/an-l
1
=
11 ( 3 -
2n) ( 5 - 2n)/2"n?][ 2"-l(n
Copyright 1985 Springer-Verlag. All rights reserved.
- I) !/
S e c t i o n 12.R
69.
579
(continued)
/
( 3 - 2n)l
=
/ (5
1im of n-to/an/an-l/ 73.
4 5 3x I 1 2 8 - 3x 1256 4
x 116
+
3x 1128
3x - x
3
5
/a
... ,
a
a 2 - b
a
and
2
.
so
=
(
m
+
(112(--
1)1'2
+
Thus, i f part is
z =
(m+
part i s
1)l"
(-/qz +
-C
I
(1 - x ) 3 / 2 =
0
.
3
is
b ;
t h e com-
b=-1/2a
2
+
2
.
+
and s o
4a4 - 8a2 = 1
and com-
, i.e.,
$n
, so
b = -112a
(-vm + 1)'l2
and
Thus,
a2- 1/4a2=2,
1) = 5 = 4(a2 -
.
.
2abi
If
=
-
. , then the r e a l
( 1 1 2 ( m + 1)'j2)i
t h e imaginary p a r t i s
;
(a+ 1)"'
+
- ( 1 / 2 ( a
(1/2(J5/4
+
+
1)li2)
;
and
;
.
4~
- ( 1 1 2 1 - d m + ~ ) l / ~ ,) i t h e n t h e r e a l
1)
(-a + 1)'"
the absolute value is
+
3 - x /8 -
gives
b i ) 2 = a2 - b
= i-(+m 1)lI2
the imaginary p a r t i s
;
t h e complex c o n j u g a t e i s
81.
i.e.,
(m+ 1)lI2 -
the absolute value i s z =
+
i = (a
4 ( a 4 - 2a
t h e complex c o n j u g a t e i s
If
-
- ( 1 / 2 ( m + 1)'j2)i l)lI2)i
-
+
2
t h e imaginary p a r t i s
+
a
3 x /16
-
2 3x / 8
(1 + x ) 3/2
approaches
Rearrangement g i v e s
p l e t i n g the square gives
?aa 2 - 1 ,
x
+
1 - 3x/2
is
x
2 3x / 8
+
3x/2
and t h e a b s o l u t e v a l u e i s
2ab=-1, 2
+
Therefore,
a ;
bi ;
-
1
is
D i v i d i n g by
is
then
4a4 - 1 = 8a
i.e.,
... .
+ bi
a
+ bi ,
= 2
+ ... .
3x I256
plex conjugate i s
2
x)312
(1 - x ) ~ "
and
5
.
so the limit a s
The r e a l p a r t of
==
+
(1
+ ...
+
The r a d i u s o f c o n v e r g e n c e i s t h e r e c i p r o c a l
R = 1
so
- 3x 1128 -
4 3x I 1 2 8 -
77.
,
1
=
The F l a c l a u r i n s e r i e s f o r
3
.
- 2n)/2n/
( - m
+
1)'12
(1/2(-dm
+
+
1)l")i
; ;
and
.
4~
exp(ni12) = i 6
+
-(1/2(-m-
=
n
.
Thus,
and
iL= -1
,
so
r = 1
and
z = e x p ( ~ i ).
Copyright 1985 Springer-Verlag. All rights reserved.
580
85.
Section 12.R
The characteristic equation is
.
i2i
+4=
c (cos 2x 1
+
i sin 2x)
i sin 2x)
+
c (cos 2x - i sin 2x) = (cl 2
+
+
given by
+
c (cos(-2x) 2
C sin 2x , where 2
C1
The characteristic equation is r = (-3 i -)I2
=
c
r2
+
1
(-3
=
~
~ cos + Bx + C sin x
y ,
Let
A
and
y " = ~e~ - B cos x - C sin x
-6AeX +(3C - 11B)cos x
,
so
which has the roots
+
+
3C - 11B = 1 and -9Blll
-
=
+
+
c )cos 2x 2
c2
-
3r
11B = -130BIll = 1
=
,
,
10 = 0
y'
=
.
0
i.e.,
.
2
+
Thus, the
AeX - B s i n x
+
Hence,
+
=
+
i.e.,
-116 ,
B = -111130
-eX/6 - 11 cos ~ 1 1 3 0+ 33 sin ~ 1 1 3 0+ c exp(-5x)
1
The characteristic equation for y " + 4y r = +2i
.
C = -3Blll ,
C
and
+
c sin 2x 2
.
,
sin 2x
-2v1sin 2x 1
+
2vrcos 2x = 2 dx
the general solution is
and to
G
1 dx c2
r2
+
Thus, we have
and we must solve v'cos 2x 1
I[-x sin 2x/2/xZ]
2
is
+
=
331130
y
=
.
.
4 = 0 ,
so
To obtain a particular solution, we use the
method of variation of parameters. =
0
c exp(2x) 2
-6A =
Thus, the general solution to the homogeneous equation is
y = c cos 2x 1.
y;
=
+
,
C cos x
3y' - 10y =
cos x , A
.
c2exp(2x)
Thus, the general solution of the nonhomogeneous equation is
93.
.
whose solution is
or
Therefore, y"
=
i(cl - c )sin 2x = 2
y = clexp(-5x)
(-3% - 11C)sin x = ex
-3B - 11C
+
C = i(cl - c2) 2
and
7)/2 = -5
?
so
.
c2exp(-2ix) = c (cos 2x + 1
i sin(-2x))
solution to the homogeneous equation is
1
,
0
Thus, the general solution is y = clexp(2ix)
C ros 2x 1 89.
r2
y
=
=
cos 2x
2x = 0
and
and
simultaneously. This yields
XI=
and
+ v'sin 2
y;
v2
=
j[x
c cos 2x 1
cos Z x l 2 L 1dx
+
c sin 2x 2
sin 2x1 [x cos 2 x / 2 z ] dx
+
.
v
1
=
Therefore,
cos 2x11-x sin 2x1
. Since the constants
1
are incorporated into the integrals, the solution simplifies
y = c o s 2x1 [x sin 2x/2&]
dx
+
sin 2xj [x cos 2 x 1 2 6 1 dr
Copyright 1985 Springer-Verlag. All rights reserved.
97. The equation has the form of the damped forced oscillation equation m = 1 ,
which is discussed in the box on p. 628. Here, we have
k
=
9 , y
,
1
=
istic equation
Fo = 1 , and
+
r2
r
+9
=
Q
0
is
.
2
=
The solution of the character-
r = (-1 i %v
.
i)/2
e-t12(c,cos~t + c iinvq
solution of the homogeneous equation is
- Q
(
2
1 = tan- (2/5)
)
.
cos(2t - 0.38)/& 101. If
y
Zj=l rn
.
or
Thus, y"
,
then
+ 2al =
a = -a /I0 5 2
.
-a3/15 = a0/45 6 x 145 -
...)
SO F0cos(iit - 6)/m - r2)2
n- 1 ~:=~na~x
=
+
,
and
2xy = L-n=2n(n - l)a xn-*
2a0
=
0
or
a4 = -a1/6
0
.
If
.
a
=
.
-a0/3 j = 3 ,
If
j = 4 ,
+
y
t
"
then
30a6
+
.
+
zzz2n(n - 1)
=
y2u2 =
x
2 ~ z = ~ a ~= x2a2 ~ ++ ~ a 2 = 0 . If
If
j = 2
,
then
then
20a 5
+
2a2 = 0
+
=
0
or
2a3
Therefore, the general solution is
4 +al(l - x / 6 +
tan-I [ r l -
F =
This is the limiting behavior as
+
or =
Jl;h-j;;; = 3 and
]+2 + 2 a . 2-1] x J = O . Thisgivesus
6a3
0
.
0.38
, then y '
[(j+Z)(j+l)a.
j = 1 12a4
Cn,Oan~
=
n-2 a x
n
5
=
CL
.
.
This transient part approaches zero as t approaches For the part~cularequation,
Thus, the
a = 6 3 aO('. - x 13 +
...) .
n-1 , then y ' = ~ z = ~ n a ~ ,x and y " = ZZz2n(n - I)& y = ~:,~a,x~ n-2 . Thus, 5x2y" + y ' + y = 0 = 51-n=2n(n - l)anx n + Zn,lnanx n-1 a x
105. If
-
a xn 'n=~ n
=
(ao -k a )
I
+ (9 +
2a2)x
+
zcrZ2[5j(j - 1)a. J
+
a.]xj . This gives us al = -ao , a2 = -a /2 = a012 . J 1 then lla2 + 3a3 = 0 or a3 = -11a2/3 = -lla016 . If 31a 3 a (1
n
109. (a)
+ 4a4 = 0 - x + x2/2
or
a4 = -31a3/4 = 343aO/24
3 - 11x /6
+
4 343x 124
.
If
+
l)aj+l +
j
=
j = 3 ,
2 , then
Thus, one solution is
+ ...) .
The equation may be rewritten as dE/dt
.
(j
+
2 2 L(d I/dt )
=
-I/C - R(dI/dt) +
Thus, according to the box on p. 628, this is a damped
spring equation if
m
=
L , k
=
1 / ~ ,
and
y
=
F.
.
Copyright 1985 Springer-Verlag. All rights reserved.
582
S e c t i o n 12.R
109. (b)
The c h a r a c t e r i s t i c e q u a t i o n of 1/0.1=0 (-20
+
is
2
5r
+lOOr+lO=O=r
.
m 2 ) / 2 = -10 2 %?v
geneous e q u a t i o n is
2
+
100(dI/dt)
+ 2 0 r + 2 ,
r =
so
Thus, t h e s o l u t i o n of t h e homo-
+ c2exp(-10 +
I = clexp(-10 - m ) t
,
d ~ / d t= - 1 2 0 ~s i n ( 6 O a t )
Now,
+
5(d21/dt2)
s o by t h e method of undeter-
I
mined c o e f f i c i e n t s , we guess t h a t a p a r t i c u l a r s o l u t i o n i s
+
A s i n Qt
B cos
equivalent t o
I
entiating yi-elds
51 (n2
-
d I/dt
where
2
+
nt .
Solving f o r
2)
+
400fi2]
A s i n 60nt
+ B cos
and u s i n g
I'(o)
.
+
Therefore,
= 0
2 m )
A
(2B
and
B
-
.
Using
cl
2
and
c2
-
yields
.
and
t
,
let
-
2
+
+
2 400Q 1
+
+
2
Differ-
cl
c2(-10
+
-
.
c2exp(-10
yields
m )
0.0002227 cos 60nt
x
2)
m ) t
c2 = I - 6 0 n ~ - (8
and
+
-
I(0) = 0
cl(-10
A = 2n(R
yields
I = clexp(-10
yields
For any f i x e d v a l u e of cos(inct/L)
.
2 20AS1 - Bn ) c o s Qt=
+
I ( t ) = 0.02217 exp(-19.90t)
0.002099 s l n 60nt 113. ( a )
60nt
nt +
(R
Solving f o r
jCJ8)BI /(-2
The o r i g i n a l e q u a t i o n i s
21 = -24nsin(60nt)
2 B = 8fi /[
and
fore, the solution i s
60nA = 0
+
=
P
and s u b s t i t u t i n g i n t o t h e d i f f e r e n t i a l e q u a t i o n
P
2 20BG - AS1 ) s i n
2
.
= 6011
20(dI/dt)
-
(2A
-(2in/5)sin
at , 2
.
d%)t
2)/
There-
+
+
c
+
6 ) t
2
+
B
=
+
&)
+
c l = [ 6 0 n ~ (6 ~ ) B/(-2 I
+
2 G )
.
0.02245 exp(-0.1005t)
+
Sn =
x
.
I f t h e sequence of p a r t i a l sums
i = 1A is i n ( i n x / l ) S1
,
S2
,
...
converges t o a l i m i t , then t h e s e r i e s converges f o r t h i s v a l u e of x (b)
For
and
t
t = 0
, ,
and
x
iE S n = y ( x , t ) = L/2
.
, t h e d e f l e c t i o n of t h e s t r i n g i s
k ~ n = ~ ~[ (~n ns ~i/ 2n ) / ~=1 i ; = , ~ ~ s i n ( n n / 2 ) = 1;=~(-1) A2k+l
.
Copyright 1985 Springer-Verlag. All rights reserved.
0
Section 12.R
117. ( a )
Using t h e b i n o m i a l s e r i e s , w i t h (1/2)(1/4)
(b)
121. (a)
+
2
2 ( 0 . 0 1 ) = 0.02
.
=
Expanding, sin 0 1 6 =
m
e-lI2 -
cos 0 1 6 =
ell2 -
.
2(1.1.2) = 2.24
(1/3!)e5I2
(714)
- . ..
s/4
jo
and
- ...1 / [ 1 / 3
1
+
. ..
... .
and
T~US,
+ (2/9.4!)
( s i n 8 / & ) d ~ = (213) ( 1 ~ 1 4 ) ~- ' ~( 2 / 7 . 3 ! )
+
4 (~14)
, J5/4"
+ (114!)87/2 -
(1/2!)6312
+ (1/5!)e9/2 -
-
( ~ 1 4 ) ~ ' (~2 / 1 1 . 5 ! ) (n/4)1112 and s i m p l i f y i n g g i v e s
114
This i s accurate within
J ; I ~ ( C O S fJ/&)d)d~ = 2 ( n 1 4 ) ~ / -~ ( ? / i . 2 ! ) ( r / 4 ) 5 1 2 912
=
.
(1/2)(-112) ( 1 / ~ 1 ) ~ = / 2 1.12
J4(5/4)=
v%
, x
a = 112
583
xly
=
... .
x
Taking t h e r a t i o
+
( 4 / n ) ( [ l - ( 1 / 5 . 2 ! ) ( ~ / 4 ) ~ (119-4!)
- ( 1 1 7 - 3 ! ) (714)'
Using j u s t t h e l e a d i n g terms,
x/y
%
. ..I
(1/11.5!) ( 1 1 1 4 ) ~-
+
(417;) ( 1 / ( 1 / 3 ) )
.
)
=
=- 1217
x
3.8
.
U s i n g t h e f i r s t two t e r m s i n t h e n u m e r a t o r and d e n o m i n a t o r g i v e s x/y (b)
%
(4/n)(l
The g r a p h s o f y
2 (1110) ( 1 ~ 1 4 )) / ( I 1 3
-
cos 6 1 6
and
-
2 (1142) ( 1 ~ 1 4 ))
sin 6 1 6
" 3.68 . x
a r e shown below.
and
a r e g i v e n p a r a m e t r i c a l l y as t h e a r e a u n d e r t h e s e c u r v e s when
k = l .
The t r a n s i t i o n a l s p i r a l i s drawn below f o r As s e e n from t h e g r a p h s a b o v e , @ = 31~12
.
The y - v a l u e
x
for
and
0
is j u s t s t a r t i n g t o decrease a t
@ = n18
,
< ;< n
.
w i l l continue t o decrease u n t i l
Some of t h e p o i n t s we h a v e p l o t t e d a r e (1.23,0.16)
k = 1
(1.66,0.44)
(0,O) for
for
4
$ = a
= 0
q = 114
.
,
,
Copyright 1985 Springer-Verlag. All rights reserved.
584 Section 12.R
121. (b)
(continued) (1.95,1.09)
/'
for
=
n/2 ,
1.61)
for
0
=
3-14 ,
1.77)
for
4
=
r
.
(1.75,
and
(1.31,
We approximated
the sine and cosine functions with the first five nonzero terms of the FIaclaurin series.
125. True;
1 [ (an
2 a n Ibn
+
b
=
129. Assume that
e
an integer.
-
+
+
1) )+
sum is
-
a
l)!
a/b
b
.
+
=
-
... -
I
and
b ,
1/2! - 1/3!
.
Since k
e - 2 - 1/2! - 1/3! -
+
2)!
+ ...]
.
l)-n
1)] /[1 - l/(k
+
not an integer, a contradiction.
=
so
e
- ...
>
...
is rational.
- l/k!)
b , a -l/k!
e
,
(k!/k!) [l/(k -t 1)
+
Using the Placlaurin expansion of l/(k
.
+ b2)/2
~ : = ~ a ~converges b ~ absolutely.
k!(e - 2
k!/2! - k!/3! -
. . .] < ~z,~(k +
[l/(k
a
2 an -
Since
(a:
lan lbnl
for some integers a
The quantity
+
0 ,
2
converges or
and let
-
k Cn-O(l/n!)
k! [l/(k (k
=
k >b
k! (a/b) - 2k!
e
converges by the sum rule.
zI=~ /anbn/
Therefore,
Now, let
+ b:)/2]
a:
=
is
is simply becomes
l/((k
+
2)
The latter is a geometric series whose 1)]
=
l/k
.
Thus, a
Therefore, e
< l/k ,
so
a
is irrational.
Copyright 1985 Springer-Verlag. All rights reserved.
is
Section 12.R
585
TEST FOR CHAPTER 12 1.
Trueorfalse. w
Cm
(a)
If
ai
w
+
CiZO(ai
d i v e r g e s and
bi)
b.
CiZ0
also diverges,
then
also diverges. ZiZ1(x
i+l
zT=~
(b)
The d e r i v a t i v e of
(c)
Any s e r i e s t h a t converges a b s o l u t e l y a t
/i!)
is
[(i
+
x = xo
.
l)xi/i!]
a l s o converges
conditionally a t that point. (d)
Any e q u a t i o n of t h e form and
c
at2
+ b(t +
2) - c = 0
,
where
a
,b ,
a r e r e a l c o n s t a n t s , has a t l e a s t one s o l u t i o n i n t h e
complex number system. (e) 2.
The s e r i e s
,
5.
3
j
converges a s an a l t e r n a t i n g s e r i e s .
i s a t h i r d degree polynomial and t h a t h(0) = -2
h"(0)
= -18
,
h"'(0) = 24
.
Find t h e Maclaurin s e r i e s f o r
(b)
Find t h e Taylor s e r i e s expansion f o r
(c)
Find an e q u i v a l e n t e x p r e s s i o n w i t h t h e form
+
Find v a l u e s of
, where
c
x
a
f o r which
(a)
converges a b s o l u t e l y
(b)
converges c o n d i t i o n a l l y
(c)
diverges
h(x)
,b , c 4 ZnZ5n (X
h
and
+
around
xo
an
i s t h e l a s t d i g i t of
(b)
an = n
(a)
Find a s o l u t i o n f o r
w(x)
n
(b)
Solve
+
-
x0)3
. +
are constants.
(ann)-'
, i.e.,
a13
=
which satisfies
(Hint: A p a r t i c u l a r s o l u t i o n h a s t h e form Bx s i n 2x
a(x
x = 1
I ) ~ :
Discuss t h e convergence o r divergence of (a)
,
.
(a)
b(x - x ) 0
4.
1
Suppose t h a t h ( t ) h'(0) = 8
3.
-
3
,
wr'+
if: alh5 = 5
, etc.
= ccs
2x + e x
4w
w = Ax cos Lx
.
+
ceX . )
y "'-I- 2y
"
+
y' = 0
for
y(x)
.
Copyright 1985 Springer-Verlag. All rights reserved.
586
6.
Section 12.R
Do the following situations imply convergence, divergence, or give no information? Unless otherwise specified, the series we are testing is ai , where
is assumed to be finite for all
(a)
1im i+, ai = 0
(b)
ai = i-l. 1
(c)
1im ii,,/ai/ai-l1 = 1
(a)
Find the fourth degree Taylor polynomial for
(b)
x = IT
9.
. 2 f(x) = sec x
.
Find the fourth degree Taylor polynomial for x = n
expanded around
8.
f(x) = tan x
Find the fourth degree Taylor polynomial for expanded around x = TI
(c)
+
2
i) / 2
f(x) = ln(cos x)
.
.
(a)
Simplify
(1
(b)
Simplify
J-2
(c)
If
(d)
Find all solutions of the equation y3 = -1
.
+y
=
z
=
2e
+
in12
6i
,
.
and the signs are alternating.
(el
expanded around
1
i
1im i-/ai/ai+l / > 1 and every third sign is negative. 1im i-(llai) I/i = 112
(d)
7.
a.
.
Express your answer in the form
what is the polar representation of
Find two power series solutions for y
"+
xy'
a
+
bi
.
z ?
0 and find the radius
of convergence for each series. 10.
01' Baldy appeared to lose
20
years after he bought his new toupee.
Since 01' Baldy received so many compliments on his youthful looks, he purchased a wool toupee to keep his head warm during the winter. tunately for O1'Baldy, a hungry moth found the toupee. 50%
of what remained of the hairpiece each day.
Unfor-
The moth ate
Unfortunately for the
moth, toupees lack nutrients and it died of malnutrition after ten full days.
Originally, the wool toupee weighed
60 grams.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 12.R
10.
(a)
587
If the moth could eat forever at the same rate, write the total amount which could be eaten as a geometric series.
(b)
Subtract another geometric series to determine how much of the wig was eaten.
ANSWERS TO CHAPTER TEST 1.
2.
3.
False; let
(b)
True
(c)
True
(d)
True
(e)
False; the signs do not alternate.
(a)
4x3-9x2+8x-2
(b)
4(x - 1)
(c)
4(x
(a)
-2<x
(b)
-2
(c)
x
(a)
Diverges; compare to
(b)
Converges absolutely; ~ Z = ~ ( l / n ~ is ) a p-series
-
3
+
3/4f
a. = 1 and
.
(a)
3(x - 1)'
+
(37/8)(x
b.
=
+ 2(x -
314)
-1
for all
1)
+
+
5918
i
1
<x <0
< -2
and
x
>0 ez
4.
5.
6.
(a) c1 sin 2x
+
+ c2(x
c2 cos 2x
- l)ex
En=1(1/9n)
+x
+ c3 =
sin 2x14 C ex 1
+
+
eX/5
(b)
cleX
c xeX 2
(a)
No information
(b)
Converges absolutely
(c)
No information
(d)
Converges absolutely; note subscripts
+
c 3
Copyright 1985 Springer-Verlag. All rights reserved.
588
Section 12.R
7.
(a)
(x-n)+
8.
(a)
i
(b)
a
+ bi ,
a =
9.
3 ( x - n ) /3
where either
- f iand
and
.q =
b =
b =
-fi
n 2n ao~z=O[(-l)x 1 2 ~ n ! ] has radius of convergence = a
b
zrn
1 n=O
z
[(-l)n+12n(n!)x2n+1/(2n
fi or
+
I)!]
m
;
has radius of convergence =
.
rn
6 0 1 2 ) ~- 11=11 6 0 ( l 1 2 ) ~= 60 - 151256 = 153451256 grams
Copyright 1985 Springer-Verlag. All rights reserved.
.
Comprehensive T e s t
COMPREHENSIVE TEST FOR CINPTERS 7-12 (Time l i m i t :
1.
True o r f a l s e .
589
3 hours)
I f f a l s e , e x p l a i n why.
(a)
The i m a g i n a r y p a r t o f a complex number i s a r e a l number.
(b)
By l ' H b p i t a l l s r u l e , we h a v e
(c)
Every i n f i n i t e s e r i e s must e i t h e r c o n v e r g e o r d i v e r g e , b u t t h e y
lim x,O(sin
1im x / e X ) = x , O ( ~ ~ xs / e X )
=
1
.
c a n n o t do b o t h . (d)
An i m p r o p e r i n t e g r a l may h a v e a n u p p e r sum and a l o w e r sum, y e t diverge.
(el
One form of t h e f o r m u l a f o r i n t e g r a t i o n by p a r t s i s
.
f(x)g(x) - j f '(x)g(x)dx (f)
The T a y l o r s e r i e s a p p r o x i m a t i o n m
zi=,(x (g)
-2i
/(2i)!)
/f(x)gl(x)dx =
of
e - ( ~ ~ f)o r
x
near zero i s
,
U s i n g a n e v e n number of s u b i n t e r v a l s , S i m p s o n ' s method g i v e s a b e t t e r a p p r o x i m a t i o n t h a n t h e t r a p e z o i d a l method.
(h)
The c e n t e r of mass of a r e g i o n may l i e on t h e r e g i o n ' s p e r i m e t e r , assuming uniform d e n s i t y .
(i)
If
f(t)
> g(t)
for a l l
g r e a t e r t h a n t h a t of
(j)
The s o l u t i o n o f
t
g(t)
xl(t)
=
,
t h e n t h e a r c l e n g t h of
f(t)
is
o v e r t h e same i n t e r v a l .
-3x(t)
i a a decreasing function throughout
i t s e n t i r e domain 2.
?lultiple choice. (a)
(Choose t h e b e s t a n s w e r . )
Which o f t h e f o l l o w i n g i s a s o l u t i o n of
C Y = ~1 ( - 1 ) i x 2 i / ( 2 i )
(ii)
~ ~ = ~ [ ( - 1 ) ~ ( 2 ~ )!I ~ ~ 1 ( 2 i )
(iv)
zi,o
+
4y = 0 ?
!I
(i)
(iii) ~
y"
y[ = ( - ~~ ) ~ x ~ ~!I / ( ~ i ) [(-l)i2x2i+1/(~i
+
1) !I
Copyright 1985 Springer-Verlag. All rights reserved.
590
2.
Comprehensive T e s t
(b)
L ' H A p i t a l ' s r u l e may be u s e d f o r a l l of t h e f o l l o w i n g e x c e p t : 1i n t+O+
(i) (ii) (iii) (iv) (c)
(sin t ) (ln t )
lim x+3 [ 21 ( x 2 - 9) lim ( 1 / n ) 3n+2 n*
D/( x 3 - 2 7 ) l
-
None of t h e above
I f one wishes t o i n t e g r a t e
+
g(y) = (y2
3y
2
- 2)/y (y
2
+
2 3) ( y
- 1)
by t h e method of p a r t i a l f r a c t i o n s , t h e i n t e g r a n d s h o u l d h a v e t h e form: (i)
( A +~ B) / y 2
(ti)
( A +~ B) / ( y 2
( i i i ) A/y (iv) (d)
(e)
+
+
(Cy
D)/ ( y
- 2)
3y
+
+ B/Y + ( Q
+
2
+
C/y
2
+
D ) / ( ~ ' + 3)
2 (EY + F) / ( Y + 3 ) '
+
3)
(Dy + E) /(Y
+
(Ey
+
2
+ 3)
2
+
+
G/(y
F/(y
-
-
2 F)/(Y + 3)2 + G / ( Y
-
None of t h e above
The complex c o n j u g a t e of (i)
fie-i-i/i
(ii)
fi e x p ( - ~ 1 4 )
(iii)
fi e T i f 4
(iv)
exp(-ir/4)
1
+
i
has polar representation
The mean v a l u e t h e o r e m s t a t e s t h a t f o r some p o i n t (i)
f(x0) = [ l / ( b
(ii)
f(b
( i i i ) f [(b (iv) 3.
+
f (x,)
- a)
+
=
-
[l/(b
xo
in
(a,b)
:
a)] j;fr(x)dx b
- a)] jaf(xo)dx
b a)/2] = jaf (xo)dx b
= jaf (x)dx/(b
- a)
Short answers (a)
F o r T a y l o r ' s t h e o r e m , what i s t h e r e m a i n d e r i n t h e d e r i v a t i v e form?
(b)
S t a t e the
(c)
Define
(d)
E x p r e s s t h e a r c l e n g t h of
(e)
State the alternating s e r i e s test.
7-6
sinh x
d e f i n i t i o n of t h e l i m i t . i n t e r m s of e x p o n e n t i a l s . f ( x ) = xn
on
[0,1]
a s an i n t e g r a l .
Copyright 1985 Springer-Verlag. All rights reserved.
Comprehensive Test
4.
591
Numerical calculations (a)
Estimate y(1)
(y')
if
2
- xy
0
=
and
y(0)
=
2
.
ljse a four-
step Euler method.
(c)
State the formula used in Newton's method and use it to solve
5.
+ x2
-7
=
to estimate
.
Use Simpson's method with
x3
n = 4
4 2 joexp(-x )dx
(b)
.
Fill in the blanks. m
(a)
According to the ratio test, a series
(b)
If
(-b f f i ) / Z a
equals
roots,then the solution of
(c)
The radius of convergence of
(d)
The integration of L cos X
(e)
6.
7.
ay"
cosLx
r2 , two distinct real
rl and
+
by'
+
converges if
cy = 0
is
is ___
1;
1=0
is based upon the trigonometric identity
=
The Maclaurin expansion of
Integration problems. 2
a
+ 4t
(a)
j(t
(b)
j(t3
( 1
j(z,=,
(d)
2 3 jcos y sin y dy
+
1/(1
-
x)
is
-
Evaluate the integrals.
t 7)-l"dt
+
2t2
t)-ldt
~x"/n~"~))dx
Do the following converge absolutely, converge conditionally, or diverge?
Explain why.
m
2
(a)
Zn=,[n
(b)
4 - 1
(c)
10 j-2(3e/x lnix/)dx
(d)
lz=l(3n/nn)
(e)
c:=~~
+
( n - 1)!/(2n+ 114 - 1/16
[ (4n
+
+
I)!] 1/64 -
5) (5n)/(3n
...
2
+ JE) (n /5 - 111 (-1)"
Copyright 1985 Springer-Verlag. All rights reserved.
592
8.
Comprehensive Test
(a)
LJhat is the Taylor series expansion of
(b)
Use your answer in (a) to estimate el
e2x
.
around
x
=
2 ?
Use a fourth-order
approximation. (c) 9.
10.
How accurate is your answer in ( b ) ?
Miscellaneous calculations.
lim
x+o [(e
l/x - e-1/~)/x21
(a)
Compute
(b)
Find the general solution of
(c)
lim ,(x; Compute 9)//3 x+ 3+
y r ' - y 1 - 2y
=
x2
+
e
XI] .
A former mathematician has given up his career to open a bakery.
Cake
flower decorations have green leaves which can be described in polar coordinates by
r
=
cos 38
.
He uses red frosting to outline the
leaves. (a)
Write a formula to describe how much red frosting he uses to outline the leaves on each flower.
(b)
How much green frosting does he use for each flower's leaves?
TO COFIPREHENSII'E TEST 1.
(a)
True
(b)
False;
(sin x/eX)
has the form
011
,
so l'H8pital's rule can't
be used. (c) True (d)
True; consider
(e)
True
(f)
False:
(g)
True
(h)
True:
jLsin x dx
it should be
.
F:=~[(-x
consider the grapb of
2
r
n 2n )n/n!~ = ~:=~[(-1) x /n!l
=
(cos 0 j
.
in polar coordinates.
Copyright 1985 Springer-Verlag. All rights reserved.
Comprehensive Test
1.
(i)
False;
let
f(t)
=
10
and
g(t)
=
t
on
- LI
< c
whenever
[0,1]
593
.
(j) True
2.
(a)
ii
(b)
iv
(c)
iii
(d)
i
(b)
1im f(x) X+XO
=
(c)
sinh x
(ex - e-')/2
=
L
if
(d) (e)
/f(x)
Ix - x
0
/
<
6
dx
An alternating series converges if (i) the signs are alternating, (ii) the terms are decreasing, and (iii) the limit of the terms is zero.
4.
(a) 2.78 (b) 0.836
+
(dl
(1
(b)
lnlt/(t
(c)
(dl
cos 2x)/2
x
Cn,,[
5
+ n+l
cos y/5
-
1) /
+ (t +
+
c
8/3]
+ l)n 3 cos y/3 + C
/(n
Copyright 1985 Springer-Verlag. All rights reserved.
594
Comprehensive Test
(a)
Converges absolutely; use ratio test.
(b)
Converges absolutely; it's a geometric series.
(c)
Diverges;
/(3e/x ln]x/)= 3e [ln(ln/x/) I
,
which diverges at
x = o .
(d)
Converges absolutely; use root test.
(e)
C~~nverges conditionally; use alternating series test and comparison test.
(a)
c:=~
(b)
el
(c)
Error <
[2"e4(x
- ~)"/n!]
4 e4 - 3e
+
4 4 9e /2 - 9e /2
+
4 4 27e 18 = Ile 18
81e4/40
(a) Does not exist
+
2 c ex?(-x) - x 12 + x/2 - el2 - 314 2
(b)
clexp(2x)
(c)
6
(a)
2 6,f;li6X + 8 sin 36 de
Copyright 1985 Springer-Verlag. All rights reserved.
This book is intended to help the student of Volume II of Marsden and Weinstein's Calouius. It runs in parallel with this text, and reinforces the concepts Introduced there with exercises, study hints, and quizzes, Each chapter begins with a section detailing the prerequisites for its study, and a quiz on those prerequisites. Detailed solutions of every other odd-numbered exercise and sample examinations have been included.
ISBN 0-387-962544 ISBN 3-54M16234-4
