This Student Solutions Manual contains detailed solutions to selected exercises in the text Multivariable Calculus, Seve
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English Pages [362] Year 2012
Table of contents :
Cover
10
10.1
10.2
10.3
10.4
10.5
10.6
10 Review
11
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
11.10
11.11
11 Review
12
12.1
12.2
12.3
12.4
12.5
12.6
12 Review
13
13.1
13.2
13.3
13.4
13 Review
14
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
14 Review
15
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
15.9
15.10
15 Review
16
16.1
16.2
16.3
16.4
16.5
16.6
16.7
16.8
16 Review
17
17.1
17.2
17.3
17.4
17 Review

STUDENT SOLUTIONS MANUAL for STEWART' S
DAN CLEGG • BARBARA FRANK www.elsolucionario.net
.Student Solutions Manual for .
MULTIVARIABLE .CALCULUS SEVENTH EDITION
DAN CLEGG Palomar College
BARBARA FRANK · Cape Fear Community College
~ . · BROOKS/COLE taa. C EN GAG ELearning·
Australia · Brazil · j apan · Korea • Mexico · Singapore • Spain · United Kingdom · United States
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~
I
BROOKS/COLE ~~r
CENGAGE Learning·
© 2012 Brooks/Cole, Cengage Learning
ISBN 13·: 9870840049452 ISBN 10: 0840049455
ALL RIGHTS RESERVED. No· part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used In any form or by any means graphic, electronic, or mechanical, including but not limited to photocopyIng, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher. For product information and technology assistance, · contact us at Cengage Learning Customer & Sales Support,
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Printed in the United States of America
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I ·ll
D
PREFACE This Student Solutions Manual contains detailed solutions to selected exercises in the text Multivariable Calculus, Seventh Edition (Chapters 1017 of Calculus, Seventh. Edition, and Calculus: Early Transcendentals, Sevent,h Edition) by James Stewart. Specifically, it includes solutions to the oddnumbered exercises in each chapter section, review section, TrueFalse Quiz, and Problems Plus section. Also included are all solutions to the Concept Check questions. Because of differences between the regular version and the Early Transcendentals version of the text, some references are given in a dual format. In these cases, readers of the Early Transcendentals text should use the references denoted by "ET." Each solution is presented in the context of the corresponding section of the text. In general, solutions to the initial exercises involving a new concept illustrate that concept in more detail; this knowledge is then utilized in subsequent solutions. Thus, while the intermediate steps of a solution are given, you may need to refer back to earlier exercises in the section or prior sections for additional explanation of the concepts involved. Note that, in many cases, different routes to an answer may exist which are equally valid; also, answers can be expressed in different but equivalent forms. Thus, the goaJ of this manual is not to give the definitive solution to each exercise, but rather to assist you as a student in understanc#ng the concepts of the text and learning how to apply them to the. challenge of solving a problem. We would like to thank James Stewart for entrusting us with the writing of this manual and offering suggestions and Kathi Townes of TECHarts for typesetting and producing this manual as well as creati ng the illustrations. We also thank Richard Stratton, Liz Covello, and Elizabeth Neustaetter of Brooks/Cole, Cengage Learning, for their trust, assistance, and patience. DAN C LEGG
Palomar College BARBARA FRANK
Cape Fear Community College
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iii
D
ABBREVIATIONS AND SYMBOLS CD
concave downward
cu
concave upward
D
the doma in off
PDT
First Derivative Test
HA
horizontal asymptote(s)
I
interval of convergence
1/D
Increasing/Decreasing Test
IP
inflection point(s)
R
radius of convergence
VA
vertical asymptote(s)
CAS H
· indicates the use of a computer algebra system. indicates the use of ljHospital's Rule. ' indicates the use ofFormu'Ja j in the Table of Integrals in the back endpapers.
indicates the use of the substitution {u
= sin x, if.u = cos x dx}.
indicates the use of the substitution { u = cos x, du
=
sin x dx}.
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v
D · CONTENTS 0
10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 10.1
Curves Defined by Parametric Equations
10.2
Calculus with Parametric Curves
0
7
10.3
Polar Coordinates
10.4 10.5
Areas and Lengths in Polar Coordinates Conic Sections 26
10.6
Conic Sections in Polar Coordinates Review 35
Problems Plus
13 20
32
43
11 INFINITE SEQUENCES AND SERIES
45
11.1
Sequences
11.2 11.3 11 .4
Series 51 The Integral Test and Estimates of Sums The Comparison Tests 62
11.5 11 .6 11 .7
Alternating Series 65 Absolute Convergence and the Ratio and Root Tests Strategy for Testing Series 72
11.8
Power Series
45 59
Problems Plus
0
68
74
11.9 Representations of Functions as Power Series 11.10 Taylor and Maclaurin Series 83 11.1 1 Applications of Taylor Polynomials 90 Review
1
78
97
105
12 VECTQR S AND THE GEOMETRY OF SPACE 12.1
ThreeDimensional Coordinate Systems
12.2 12.3
Vectors 114 The Dot Product
111
111
119
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vii
viii
o
CONTENTS
12.4
The Cross Product
12.5
Equations of Lines and Planes
. 12.6
147
13 VECTOR FUNCTIONS
151
13.1
Vector Functions and Space Curves
151
13.2
Derivatives and Integrals of Vector Functions
13.3
Arc Length and Curvature
13.4
Motion in Space: Velocity and Acceleration Review
157
161 168
173
Problems Plus
179
14 PARTIAL DERIVATIVES
183
14.1
Functions of Several Variables
14.2
Limits and Continuity
183
192
14.3
Partial Derivatives
14.4
Tangent Planes and Linear.Approximations
14.5
The Chain Rule
14.6
Directional Derivatives and the Gradient Vector
195 203
207
14.7
Maximum and Minimum Values
14.8
Lagrange Multipliers Review
213
· 220
229
234
Problems Plus
0
135
140
Problems Plus
0
128
C,ylinders and Quadric Surfaces Review
0
123
245
15 MULTIPLE INTEGRALS
247
15.1
Double Integrals over Rectangles
15.2
Iterated Integrals
247
249
15.3
Double Integrals over General Regions
15.4
Double Integrals in Polar Coordinates
251 258
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CONTENTS
15.5 15.6 15.7 15.8 15.9 15.10
Applications of Double Integrals Surface Area
267
Triple Integrals
269
Triple Integrals in Cylindrical Coordinates Change ofVariables in Multiple Integrals
Problems Plus
16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9
297
303
Vector Fields
303
Line Integrals
305
The Fundamental Theorem for Line Integrals Green's Theorem
316
Parametric Surfaces and Their Areas Surface Integrals
328
Stokes' Theorem
333
The Divergence Theorem
Problems Plus
321
335 .
337
343
17 SECONDORDER DIFFERENTIAL EQUATIONS
.0
17.1
SecondOrder Linear Equations
17.2 17.3 17.4
Nonhomogeneous Linear Equations
345
345
· 347
Applications Of SecondOrder Differential Equations Series Solutions 352 Review 354
A.PPENDIX H
3 10
313
Curl and Divergence
Review
0
285
289
16 . VECTOR CALCULUS 16.1
276 280
Triple Integrals in Spherical Coordinates Review
0
261 '
350
359
Complex Numbers
359
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D
ix
D PARAMETRIC EQUATIONS AND POLAR COORDINATES
10
Curves Defined by Parametric Equations
10.1 1.
X= t 2 + t, y = t2  t,
2
y
StS 2
t= 2 (2, 6)
t
 2
 1
0
l
2
X
2
0
0
2
6
y
6
2
0
0
2
X
3. x = cos2 t,
y
0 S t S 7r /2
y = 1  sin t,
t=O (1, 1)
t
0
7r/6
7r/3
7r/ 2
X
1
3/ 4
1/ 4
0
y
1
1/ 2
1 ~ 0.13
1
r=JI. 6
0
X
5.
X
=3
4t, y
=2 
y
3t
(7, S) t= 1
(a)
(b)
t
 1
0
1
2
X
7
3
1
5
y
5
2
1
4
X= 3  4t y
=2
1. x = 1 
3t
4t
=}
=2
e, v =
(3, 2) t= O X
= X + 3
3( ~x + ~ )
t  2,  2
t = ~ X +~ ' SO
=}
= 2 +~ X 
£
=}
(5,  4) 1=2
y =~X  ~
sts2
y ( 3, 0)
t=2 X
(a)
t
 2
 1
0
1
2
X
3
0
1
0
3
y
4
3
2
 1
0
= t  2 =} t = y + 2, SO X= 1  t 2 = 1  (y + 2? =} x =  (y + 2) 2 + 1, or x =:=  y 2  4y e 3, with  4 S y S 0
(b) y
©
2012 Certgagc Learning. Ail Rights Reserved. May not be scanned, copied, o r duplicated, o r posted to a publicly accessible website, in ''•hole o r in part.
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(1,  2)
t= O
2
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
D
9.
X
=
Vt,
y
y= 1 t
(0, I) ,f = O
(a)
(b) x =
t
0
1
2
3
4
X
0
1
1.414
1.732
2
y
1
0
=}
t = x2
Vt
 1
X
.:....3
 2
y = 1 t = 1  x 2 . Since t ~ 0, x ~ 0.
=}
(2, 3) t=4
So the curve is the right half of the parabola y = 1  x 2 • 11. (a) x = sin ~0, y =cos ~(:1, 2
1r
~ (:1 ~
y
(b)
1r.
x + y = sin ~8 + cos2 ~8 = 1. For 1r ~ 0 ~ 0, we have 2
2
 1 ~ x ~ 0 and 0 ~ y ~ 1. For 0
y ~ 0. The graph is a semicircle. 0
1
13. (a) x
= sint, y =
For 0
csct, 0 .
< t < ~
~ smt
y = csct = .
< t < ~.we haveO < x < 1 andy
=
!. x
I
(b)
'\,,,,
> 1. Thus, the curve is the
portion of the hyperbola y = 1/x withy> 1.
0
15. (a) x
= e2 t
=}.
2t = ln x
t. = ~ ln x.
=}
X
(b)
X
y
y = t+1 = ~lnx+1.
17. (a) x =sinh t, y = cosh t
' =}
y2

x 2 = cosh2 t  sinh2 t = 1. Since
y = cosh t ~ 1, we have the upper branch of the hyperbola y 2
(b)
y
x 2 = 1.

0
. 19. x
=3+
2 cost, y = 1 + 2 sin t,
1r /
2 ~ t ~ 31r/ 2.
X
By Example 4 with r = 2, h = 3, and k = 1, the motion of the particle
takes place on a circle centered at (3, 1) w ith a radius of2. As t goes from ¥ to 2
3; ,
the particle starts at the point (3, 3) and
2
moves counterclockwise along the circle (x  3) + (y  1) = 4 to (3,  1) [onehalf of a circle]. 21. x = 5sint, y=2 cost
=}
sin t=~, cost=~·
2
2
sin t+cos t =1
particle takes place on an ellipse centered at (0, 0). As t goes from
 1r to
=}
(~r + (~r = l.The motionofthe
511', the particle starts at the point (0, 2) and moves
clockwise around the ellipse 3 times. 23. We must have 1 ~ x ~ 4 and 2 ~ y ~ 3. So the graph of the curve must be contained in the rectangle (1, 4] by [2, 3]. ©
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SECTION 10.1
25. When t =  1,(x,y)
= (0, 
CURVES DEFINED BY PARAMETRIC EQUATIONS
D
3
1). As t increases to 0, x decreases to  1 andy
increases to 0. As t increases from 0 to 1, x increases to 0 and y increases. to 1. As { increases beyo~d 1, both x and y increase. Fort
0, the curve never quite
reaches the origin. 39. The case
¥ < (} < 7T is illustrated. 0
has coordinates ( rO, r) as in Example 7,
y
and Q has coordinates (rB, r + r cos(1r B)) = (r B, 7"(1 cos B)) [sincecos(1r  a) = cos7rc;osa + sin7T sin a· = cos a ], so.P has coordinates (r B rsin(1r  B),r(1 cos·B)) = (r(B : sin B),r(l  cos B)) [sincesin(1r  a) = sin 1rcosa  cos1r.sin a = sin a ]. Again we have the parametric equations x = r( (}  sin 8), y = r(l  cos B).
© 2012 CC=nguge Learning. All Rights Res sin2 B = (y/bl and cosO=
x/a =>
cos2 B = (x/a) 2 • Adding the two
X
equations: sin2 B + cos2 B = 1 = x 2 /a 2 + y 2 /b 2 .,Thus, we have an ellipse.
43. C = (2a cot B, 2a), so the xcoordinate of Pis x
= 2a cot B. Let B
~
= (0, 2a).
2a
Then L.OAB is a rightangle and L.OBA = B, so JOAI '= 2asinB and
A = ( (2a sin B) cos B, (2a sin B) sin B). Thus, theycoordinate of P is y = 2asin 2 B. 45. (a)
There are 2 points of intersection:
4
( 3, 0) and approximately (  2.1, 1.4).
4
(b) A collision point occurs when x1 =.xz and Yl = Y2 for the same t. So solve the equations: 3sint=  3+cost
(1)
2 cost = 1 + sin t
(2)
From (2), sin t = 2 cost 1. Substituting into (1), we get 3(2 cost  1) =  3 +cost cost = 0
=>
occurs when t =
t = 3 ", 2
t
or
3 ;.
We check that t =
3 2..,.
satisfies (1) imd (2) butt =
t
=> 5 cost
=0
(*)
=>
does not. So the on ly collision point
and this gives the point (  3, 0). [We could check our work by graphing Xt and x 2 together as
functions oft and, on another plot, y 1 and y2 as functions oft. If we do so, we see that the only value oft for which both pairs of graphs intersect is t ~
3 ; .]
(c) T he circle is centered at (3, 1) instead of (  3, 1). Th~re are sti ll 2 intersection points: (3, 0) and (2. 1, 1.4), but there are no collision points, since(*) in part (b) becomes 5 cost = 6 47. x = t 2 , y = t 3

=> cost =
~
> 1.
ct. We use a graphing device to produce the graphs for various values of c with 7r
::;
t ::;
7r.
Note that all
the members of the family are symmetric about the x axis. For c < 0, the graph does not cross itself, but for c = 0 it has a cusp at (0, 0) and for c
> 0 the graph crosses itself at x
= c, so the loop grows larger as c increases.
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5
6
0
49. x
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
= t + a cost, y = t +
a sin t, a
> 0. From the first figure, we see that
curves roughly follow the li ne y = x, and they start having loops when a is between 1.4 and 1.6. The loops increase in s ize as a increases. X
While not required, the following is· a solution to determine the exact values for which the curve has a loop, that is, we seek the values of a for which there exist parameter values t and u such that t
< u and
(t+acost,t+asint) = (u + acosu,u+asinu). y
In the diagram at the left, T denotes the point (t, t), U the point (u, u), and P the point (t +a cost, t +a sin t) = (u + acosu, u + asinu).
= PU = a, the triangle PTU is isosceles. Therefore its base angles, a = LPTU and {3 = LPUT are equal. Since a .= t  i and
Since PT
(3 = 27r 
u+t
'!I
4
3 .;' 
= s;
u=
5 .;' 
u, the relation a = /3.implies that
(1).
X
Since TU cos a=
p
= distance{{t, t), (u, u)) = J2(u t) 2 = y'2 (u t), we see that
~TU = (u 
t) / v'2, sou  t
PT
u  t = v'2acos(t
a
i)
(2). Nowcos(t ~) = sin[f 
so we can rewrite (2) as u  t dividing by 2, we obtain t Since a> 0 and t
= v'2acosa, that is,
=
= v'2 asine4,. 3 ,.. 4
sinC4,..

t),
t) (2'). Subtracting (2') from (1) and
~a sine: t), or
< u, it follows from (2') that sin(34"
implicitly assumed that 0
(t .;f)] =

3 4,..
t)
t = 72 sine4,..

T
t) (3).
> 0. Thus from (3) we see that t
0. Now sin z < z for z > 0, so a > v'2. sm z
+
e
,..) , 4
a + y'2] .
51. Note that all the Lissajous figures are symmetric about the x axis. The parameters a and b s imply stretch the graph in the
x and ydirections respectively. For a
= b = n = 1 the graph is simply a circle with radius 1. For n = 2 the graph crosses
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SECTION 10.2
CALCULUS WITH PARAMETRICCURVES
0
itself at the origin and there are loops above and b~low the xaxis. In general, the figures have n  1 points of intersection, all of which are on the yaxis, and a total of n closed loops. 1.1
\ + (a,b) =( ~.2)'
t1 =3 n =2 n=l
3.1
 2.1
 1.1
a=b = l
10.2
= (3, 2)
(a, b) = (2, I)  2.1
1.1
(a, b) = (2, 3) (a, b)
3.1
n=2
n =3
Calculus with Parametric Curves 2
1. x = t sin t, y = t + t
dy . dx . dy dy I dt d = 2t + 1, d = t cost+ sm t , and d = d l d =
::}
z
.
t
t
x
dy =  3t2 dx dt ' dt
3
3. x = 1 + 4t  t , y = 2  t ; t = 1.
= 4 2t, an d dy dx
x
t
= dyldt
dx I dt
2t + 1
. . t cos t + sm t
= 43tz Wh  2t . en t
1
'
(x, y) = ( 4, 1) and dy I dx =  ~, so an equa.t ion of the tangent to the curve at the point corresponding to t = 1 is y 1 = ~(x 4), oq/ =  ~x + 7. 5. x = t cos t, y = t sin t; t = 1r.
When t =
1r,
(x, y) = ( 1r, 0) and dy Idx. =
corresponding to t = 7. (a) x
dy . dx . ' dy dy I dt t cos t + sin t d = tcost+smt, d = t(smt)+cost,andd = d ld = . . t t x x t t s m t+cost
= 1 + ln t,
1r
 1r
I( 1) = 1r, so an equation of the tangent to U1e curve at the point
is y  0 = 1r[x  ( 1r) ) , or y = 1rx +
y = t2
+ 2;
x=l+ ln t = 1 ::}
(1, 3).
ln t = O
2
1r •
dy dx 1 dy dyldt 2t 2 dt = 2t, dt = t' and dx = dxldt = 1l t = 2t · At( 1 • 3), =?
t =1 an~ : = 2,so an equationofthetangentis y3=2(x  l),
=?
t=e"'l,soy=t2.+2=(e"' 1 ) 2 + 2 = e 2 " 
ory = 2x + 1. (b)x = 1 + ln t At (1, 3), y'
=?
ln t=x 1
= e 2(l) z · 2 =
9. x = 6sin t, y = t2+t;
2, so an equation of the tangent is y  3
= 2(x 
2
+2,andy'=e 2 xz . 2.
1), or y = 2x + 1.
(0,0).
dy dy 1dt 2t + 1 .  = d ld =  . The potnt (0, 0) corresponds tot = 0, so the dx x t 6cost slope of the tangent at that point is·lJ. An equation of the tangent is therefore y  0=
i(x 0), or y = ix.
© 2012 Ccnt;!lgc Learning. All Rights Rcscr\'Cd. May not be scnnncd, copied, orduplicotcd. or posted to npublicly ncccssiblc wcbsifc, in whole or in part.
www.elsolucionario.net
.
7
8
D
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
2
11. x = t + 1• Y
= t + t '*
The curve is CU when
13. x
= et, y = tet 2
d(~)
d y dx 2
= dt
(h;
dx/dt
~~
dt dx dx/ dt
> 0, that is, when t < 0.
dy dx
::::}
!!:._ (dy)
dy _ dy I dt _ 2t + 1 _ .!_ dx  dx f dt 2t  1 + 2t
2
= e 2 t(
= dyfdt dxjdt 1)
+ (1 
 te t + e t et
=
2
t)(  2e t)
et
= et (1t) = e 2t( _ 1 et
t)
'*
.
= e  2 t( 1 
2
+ 2t) = e3t(2t 3). The curve is CU when
et
2
d y > 0, that .ts, w hen t > 3 . dx 2 2 15. x = 2sint, y = 3cost, 0
< t < 21r.
. i_OO(dy 3 2 dy _ dy/ dt _ 3sint _ ~ d 2 y _ dt dx _ 2sec t _  ~ 3 dx dxj dt  2cost 2 tant, so dx2 dxjdt  2cos t 4 sec t. The curve is CU when sec3 t < 0
17.
X
= t3 
(x, y)
. 3t, y
= t2
= (0,  3).

d
dy
3.
t
dxd = 3t2 t .
t =  1 or 1 (x, y)

sect < 0
::::}
= 2t, SO '
d dy t
=0
cost < 0
::::}
¢>
t
=0
i
3 = 3(t + 1)(t  1), so d: dx = 0
t
::::}
= (2, 2) or (2, 2). The curve has a horizontal
tangent at (0,  3) and vertical tangents at (2,  2) and ( 2, 2).
19. x
= cos 9, y = cos 39. The whole curve is traced out for 0 ~ 9 ~ 1r.
~~
=  3'sin 39, so
'e =0, ~ . 23.,.,or1r
~~
= 0
2
sin 39 = 0 38 = 0, 1r, 21r, or 37r
(t.1)
(1,1)
9=21T/3
9= 0
(x,y) = (1,1), (~ ,  1), (~ , 1) , or(1, 1 ).
dx . dx d 9 =  sLD 8, so dO = 0 sin 8 = 0 8 = 0 or 1r dy d.x (x,y_) = (1, 1) or (1,1). Both dO and dO equal 0 when 8 = 0 and 1r.
( 1, 1) (J=
1T
2
3 38 11 38 .., .s the same s Iope when 0 = 1r. 10 fi nd th e s Iope wIten 9 = 0, we fi nd lim dy dx = lim  sin . n = lim _g cos ,.. = 9, which 1 o~o o ~o  sm" O•O cos" Thus, the curve has horizontal tangents at ( ~ , 1) and (
t, 1), and there are no vertical tangents.
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www.elsolucionario.net
SECTION 10.2
CALCULUS WITH PARAMETRIC CURVES
0
9
21. From the graph, it appears that the rightmost point on the curve :J: = t  t 6 , y = et is about (0.6, 2). To find the exact coordinates, we find the value oft for which the graph has a vertical tangent, that is, 0 = dxl dt = 1  6t5
{::}
t = 11 W.
Hence, the rightmost point is
23. We graph the curve x = t
4

2t3

2e, y = t 3
tin ~he viewing rectangle [2, 1.1] by [ 0.5, 0.5]. This rectangle

corresponds approximately tot E [ 1, 0.8] .
o.s
 0.5
We estimate that the curve has horizontal tangents at about (  1,  0.4) and (  0:11, 0.39) and vertica·Ltangents at
dy dyl dt· about (0, 0) and ( 0.19, 0.37). We calculate d = dxl d =
t
X
dy I dt = 3t2  1 = 0
1=}
t=
3t2 1 . . The honzontal tangents occur when · 4t 3  6 t2  4t
± ~ , so both horizontal tangents are shown in our graph. The vertical tange~ts occur when
dxldt = 2t(2t2  3t 2) = 0 {::} 2t(2t + 1)(t  . 2) = 0 {::} t = 0,  ~or 2. It seems that we have missed one vertical tangent, and indeed if we plot the curve on the tinterval [1. 2, 2.2] we see that there is another vertical tangent at ( 8, 6). 25. x = cos t , y = sin tcos t . dxl dt =  sint, dyl dt =  sin2 t + cos 2 t = cos 2t.
(x,y )= (O , O)
{::} cost = .O. {::}
tisanoddmultipleoflf.Whent=~,
dxl dt =  1 and dyl dt =  1, so dyl dx = 1. When t =
3 ; ,
dxl dt = 1 and
dy I dt =  1. So dy I dx =  1. Thus, y = x andy =  x are both tangent to the curve at (0, 0). 27. x
=rB 
d sinB, y = r  dcosB.
dx dy , . dy (a) dB = r dcosll, dB = dsme, so dx (b) IfO
d sinB
= r _ dcosB '
< d < r, then jdcosllj:::; d 0. This shows that dxldll never vanishes,
·so the trochoid can have no vertical tangent if d 29.
X
= 2t3' y = 1 + 4t 
6t 2
+ 2t 
4= 0
{::}
t2
'
_,..
< r.
dy  dy I dt = 4  2t N 1. dy = 1 ·dx  dxl dt 6t 2 • ow so ve dx
2(3t  2)(t + 1) = 0
{::}
t = ~ or t
~
4  2t = 1. 6t2
. . . ,._ .....,.
:=  1. 1ft = ~ . the point is(~~ '
2 :) ,
.,_,.
and ift = 1,
the point is (2, 4).
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10
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
0
31. By symmetry of the ellipse about the x andyaxes,
A= 4J0a ydx = 4 j~12 bsinB (a sin B) dB= 4ab I 0" 12 sin2 BdB= 4ab Io" 12 ~(1 cos 28) d8 ,
/2
I
= 2ab[8 ~ sin2B]~ = 2ab(~) = 1rab 33. The curve x
= 1 + et, y = t t 2 = t(1  t)
intersects the xaxis when y
y
= 0,
that is, when t = 0 and t = 1. The corresponding values of x are 2 and 1 + e. 0 I
The shaded area is given by
1::1+e
1::
(YT YB) dx =
1
[y(t) OJ x'(t) dt
J; tet dt I
=
1 2 t et 0
= I;(t 
t2)et dt
[fet] ~ + 2
f01 tet dt
dt =
' = 3 Io1 tet dt (e  0) = 3 [ (t1)et] 01
[Formula 97 or parts]
[Formula 96 or parts]
e
e=3~ e
= 3[0  (1)] 35. x

j; tet dt
= r8 dsin8, y = r dcosB.
A= I;~r y dx
=
= J;rr (r dcosB)(r 
[r 2 B 2dr sine+ ~d2 (8
dcos 8) dB= I;rr (r 2
+ (dyjdt?
B + d~ cos2 B) dB
+ ~ .sin 28) ]~7r = 21rr2 + 1rd2
37. x=t+e~t, y=tet, O~t~· 2.
(dx/dt) 2
 · 2drcos
dx jdt= 1 e tanddyjdt=1+et,so
= (1 et) 2 + (1
+ e t? =
1 2et
+ e 2 t + 1 + 2et + e 2 t = 2 + 2e 2t .
Thus, L = I: './(dx/dt) 2 + (dyjdt) 2 dt =I: v2 + 2e 2 t dt ~ 3.1416. 39. .x
= t 2sint,
y = 1  2 cos t, 0 ? t ~ 47r. &cjdt = 1 2 cost and dyjdt
+ (dyjdt) 2 = (1 2 cos t) 2 +
(dx/dt) 2
Thus, L = I: './(dx/dt) 2
+ (dyjdt) 2 dt = Io4 "
41. x = 1 + 3e, y = 4 + 2t~, Thus,
L
=
Ia
43. x
~ t sin t,
.
'./36t2 + 36t4 dt =
y =.t cost,. 0
(~~ y+ ( ~; y e =
+ 4 cos 2 t + 4sin2 t:::;; 5 4cost.
v5 4cost dt ~ 26.7298.
o ~ t ~ 1. ~x/dt ~ 6t and dyjdt = 6t2 , so (dxjdt? + (dy / dt) 2 = 36t2 + 36t4
1
= 3[~u312 J: =
(2 sin t) 2 = 1 4cos t
= 2sint, so
1
Ia 6t .Jl+i2·dt = 61
2(2 3 ; 2  1)
~
= 2(2J2 
JU
2
[u
= 1 + t 2, du = 2tdt]
1)
~ t ~ 1. ~~ = t cost + sin t and ~~
cos t
Gdu)
=  t sin t
+ cos t, so
+ 2t sin t cost + sin2 t + t 2 sin2 t ~ 2t sin t cost + cos2 t
= t 2 (cos2 t + sin2 t) + sin2 t + cos2 t = t2 + l. Thus, L .·=
®
f 01 vt2 +
1 dt
~ [~tvt~ + 1 + ~ ln(t
+vt + 1 ) ] ~ = 4J2 + ~ ln(1 + J2). 2
2012 Ccngagc Learning. All Righ1s Rescrvcc.l. May no1 be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
www.elsolucionario.net
SECTION 10.2
CALCULUS WITH PARAMETRIC CURVES
0
45.
~~ ) + ( !fJd = [e'(cos t 2
(
 sin t)f + [et(sin t +cos t )f
= (e') 2 (cos2 t  2 cos t sin t + sin 2 t) ' + (et) 2 (sin2 t+2sint cost+cos2 t = e 2 t(2 cos2 t + 2 sin2 t ) = 2e 2 t
f 0" 12 et dt =· 12 [ e' ]~ = /2 (e,. 1).
Thus, L = J~" .J2e2t dt = 1.4
47.
The figure shows the curve x = sin t
+ sin 1. 5t, y =
cost for 0 .:::; t 5 47r.
. dx/dt = cost+ 1.5 cos 1.5t and dy / dt = sin t, so 2.1
J04"
Thus, L =
49. x
=t2
(~~) +
et, y
= t + et,
(!!Jd =
+ 3cost cosl.5t + 2.25cos2 1.5t +
(dx/dt) 2 + (dy / dt? = cos2 t
f>o.o,.£f>'
= 3 /2 [cos 2t]: = 3 /2 ( ~ 12
[by symmetry)
The fu ll curve is traversed as t goes from 0 to ¥, because the curve is.the segment of x + y (since x , y 2:: 0), and this segment is completely traversed as t goes from 0 to¥ · Thus,·£ 53. x
= asin8,
(~7) +
(!!Jd·= (a cosO?+ (  bsin8? = a2
SoL = 4 j 0"12
..jaz (1 e
= 11 cos t 

(a2
2 sin 2

b2 ) sin2
~ f 0" 12 sin 2t dt = /2, as above.
0) dO
4 cos(11t/ 2), y
.,;, a 2 cos2 8 + b2 sin2 8 = a 2 (1 sin2 B)+ b2 sin2 8
8= a
[hy '?·mmctty)
2

c2 sin2
= 4a J;' 12
8= a
.
2
2
(1 ~
sin2
8) = a
2
(1  e2 sin2 8)
Vl  e2 sin2 8 dO.
= 11 sin t 4sin(llt/2).
Notice that 0 5 t 5 21r does not give the complete curve because x(O)
f:. x(21r). In fact, we must take t E
[0, 47r] in order to obtain the
complete curve, since the first term in each of the parametric equations has period 27r and the second has period
1
ii2
= ~~, and the least common
integer multiple of these two numbers is 47r. ®
2012 Ccngngc L 1 ( fl±  xfl ) xy  xif ' . . · dt = 1 + (iJ/x) 2 x2 = xz +iF. Usmg the Cham Rule, and tne
~
(d:r;)2+(!!11..)2dt dt dt
r
~
13
d .• 'Jt 
~
y /(d:r;)2+(!!11..)2 dt dt = ('2+ x y•2)1/2 , we bave t hat
0
d¢> _ d¢>/dt _ ds  ds/dt 
(xii xfl ) :i;2 + y2
(b) x = x_and y = !(x )
~
_ I 1 1(±2xyxy 1 l±ii  xiJI + y2)3/ 2  (±2 + y2)3/2 ·
1 _ xii  xiJ . d¢> (±2 + y2)1/2  (~2 + 1;2)3/2 · So"'  ds 
dy .. d2y . 1 .. 0 d . x = 'x = an y = dx' y = dx2'
 II· (d~y/dx2 )  0 . (dyjcb:) i (1 +
So"' 71. X = 8  sin8
~ 2
"' =
icosB  cos [(1  cos B) 2
:i;
ld2 y/ dx2 1
(1 + (dyjdx)2] 312 •
(dyjdx)2 )31 2

= 1' cosB ~
X = sin8, and y = 1  cos8
e sin el 2
+ sin2 B)S/ 2
=
~
iJ =sine
~
y=
cos8. Therefore,
icosB  (cos 8 + sin 8)1 JcosB  II . . = . The top of the arch 1s (1  2 cos + cos2 + sm2 8)312 (2  2 cos 8)3/2 2
e
2
e
characterized by a horizontal tangent, and from Example 2(b) in Section 10.2, the tangent is horizontal when B = (2n  l )1r, d b · B · ti . fi sotaken=1 an su stttute =1rmto 1e ex~resston or ~~;: ~t=
(
2
Jcos 1r 11 11 11 _ cos1r) 3 / 2 = _ ( ))3/ 2 2 12 2 1
1 =4·
73. The coordinates of T are .(r cosB, r sin B). Since TP was unwound from arc T A, T P has length rB. Also LPTQ = LPTR  LQTR
= t 1r 
8,
soP has coordinates x = r cos B + r8 cos(~1r ~B) = r(cos B + Bsin 8),
y = rsin B  rBsin (~1r  8) = r(sin8 8cos9). X
10.3
Polar Coordinates
1. (a) (2, i)
By adding 21r to i, we obtain the point ( 2, opposite
i
is
4 3.,.,
7 ;).
The direction
so ( 2, ~ ) is a point that satisfies the r < 0
· requirement.
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2012 Cengage ~ing. All Rights Reserved. May not be scanned, copied, or duplicated, or posted too publicly accessible wcbsitc, in whole or in part.
www.elsolucionario.net
14
0
(b)
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(1, 
3 ;)
+ 21r) = (1, 5; )
r
> 0: (1, 
r
< 0 : ( 1, 3r + 7r) =
r
> 0: ( ( 1), 1J +1r) = (1, 3; )
r
< 0 : (1 , ~ + 21r) = (
3 ;
(1,~)
0 _ 311' 4
(c) (  1, 2J)
,.
1,
5 2,.)
2
0
(t,f)
3. (a)
X=
1COS7r = 1( 1) = 1 and
y = 1 sin 1r = 1(0) =
0 give us
'TT
the Cartesian coordin~tes ( 1, 0).
(l, 'TT)
0
(b)
x
= 2cos( 23"' ) = 2(~) = 
y
= 2sin( 
2 ;)
give us ( 1,
y =  2sin
= 2(  ~) ~ ~VS
V3).
x = 2cos 3;
(c)
1 and
3 4"
= 2(4) = J2 and =
~( 4) =  J2
gives us ( J2, J2).
5. (a) x
= 2 andy =
 2 => r = J2 2
+ (2)2 = 2 J2 and 8 =
quadrant, the polar coordinates are (i) (2 J2, (b) x =  1 andy =
V3 =>
r
7 ;)
tan 1 (
and (ii) ( 2 J2,
:n := ~ . Since (2, 2) is in the fourth
3 ;).
= J< 1) 2 + (v'3) 2 = 2 and 8 = tan~1 ( :1) =
quadrant, the polar coordinates are (i)
2 :;.
Since_( 1, VS) is in the second
(2, ~)and (ii) (  2, 6; ) .
'·
© 2012 Cengnge Learning. All Rights Reserved. May no1 be scnnned, copied, or dupll sinO tan8+ oo
sin2 8 = l.Aiso,r+oo
lim 0+'Tr/2
.
lim
0+1fj 2+
x= l
=> sinBtanB>oo =>
sin2 {;I,;, 1. Therefore, lim X= 1 . ,·±oo
a vertical asymptote. Also notice that X =
sin2 e ~ 0 for all
=>
e, and X
=>
X=
1 is
= sin2 {;I $ 1 for all e. And
defined at odd multiples of ~. Therefore, the curve lies entirely within the vertical strip 0 $ x
Xi=
1, since the curve is not
< 1.
53. (a) We see that the curve r = 1 + csin 8 crosses itself at the origi n, where r = 0 (in fact the inner loop corresponds to negative rvalues,) so we solve the equation of the lima~on for r = 0
c sin 8 =  1
lei < 1, then this equation has no solution and hence there is no inner loop. But if c < the equation has the two solutions 8 = sin 1 ( 1 /c) and 8 = (} =
1r
1r
+ sin  1 (1/c) and ()= 21r sin 1 (1(c). Tn each case, r
(b) For 0 < c
sin
1
(
1, then on the interval (0, 27T)
1/c), and if c > 1, the solutions are
< 0 for (} between the two so lutions, indicating a loop.
< 1, the dimple (if it exists) is characterized by the fact that y has a local maximum at 8 =
determ.ine for what cvalu~s
So we
2
~
d2 y
dy
=> dB = cos8 + 2csinfJ cos8 = cosfJ + csin 28 =>. dB~., =  sin8 + 2ccos 28.
At 8 = 32,.., this is equal to  (  1) + 2c( 1) = 1  2c, which is negative only for c for  1 < c < 0, y only has a local minimum at e = ~ (indicating a dimple) for c
=>
3 ; .
d ~ is negative at(} = 32~, since by the Second Derivative Test. tllis indicates a.maximum:
y = rsin8 =sinO + c sin 2 (}
55. r = 2 sin (}
sin 8 =  1/ c. Now if
x = r cos(} = 2 sin fJ cos 8 = sin 28, y = 1· sin 8
= 2 sin 2 B
> %. A similar argument shows that
< ~ . .
=>
dy = dyjd(J = 2 · 2 sin fJ cosO= sin 28 = tan 2B dxjdB cos2fJ·2 cos28
dx
When 8 = 7T , dy d.'t =tan ( 2 · 7r) =tan 7T = vrr; 3.
6
57. r= l / 8
=>
6
3
· 3.] [Another method: Use Equation
x=rcos8 = (cosfJ) j B, y=rsin8=(s infJ) / 8
=>
dy dyjd8 sin.O(  1/ 82 ) + (1/ 8) cos 8 8 2  sin (}+(} cosO dx = dx j dO = cos8( 1/ fJ 2 )  (1/ 8) sinB · (} 2 =  cosO  Osin B .dy  0 + 7r( 1) When 8 = 1r, dx = (1) 1r(O) =
 'Tr
l
= 1r.
® 2012 Cengoge loaming. All Rights Resef'.d. Moy not be: ~ned. copl x = rcosO = 3cos0 cosO,_ y = rsinO = 3cos0 sinO => ,
61 . r = 3cos0
2
~ = 3 sin 9 +3cos 2 ,() = 3cos20 = 0 => 2() = ~ or 3; 0 = %or So the tangent is horizontal at (
~, %)
~~ =  6 sin() cos~= 3 sin 20 = 0 63. r = 1 + cos 0 ·
and (
32,
3 ;)
[same as (
3 ;.
32,:a:)] .
=> 29 = 0 or 1r 0 = 0 or~· So the tangent is vertical at (3, 0) and (o, ~).
=> x = r cos 9 = cos() (1 +cos 9), y = r sin 0 = sin 0 (1 + cos 9) =>
·.~ = (1 +cosO) cosO sin2 0 = 2cos 2 0 + cos9  1 = (2cos9  1) (cos(J + 1)
() =
'i· 1r, or 5;
=> horizontal tangent at(~ , 'i), (0, tr), and(~ , 5;
~~ =  ( 1 + cos 8) sin (}  cos ()sin() =  sin (} ( 1 + 2 cos 9) = 0
() = 0, 1r,
2 ;,
or
4 ;
x
2

bx
2
2
ay + ( ~a) = (~b) + (ta?
) .
=> sin 8 = 0 or cos 9 = 
~
=>
e : j::
= 0.
=>
(x  ~b)
2
+ (y  ~ a? = Ha2 + b2 ) , and this is a circle
(!b, ~a) and radius. ~.Ja 2 + b2 •
67. r = 1 + 2 sin(8 / 2): The parameter inteival is [0, 47r].
69. r
= e•inB 
2cos(40).
The parameter interval is [0, 21l'j.
71. r = 1 + cos999 8. The parameter interval is
[0, 2tr). /
1.1
 1.1
©
=>
=> r 2 = ar sin() + br cos 8 '=> x 2 + y 2 = ay + bx =>
+ (~b? + y 2 
with center
cos()=~ or  1
=> vertical tangent ~t (2, 0), (!, 2; ) , and (!, t;).
Note that the ~gent is horizontal, not vertical when 8 = 7r, since 65. r = a sin() + b cos 9
= 0 =>
20 12 1, there is an internal loop. For c 2: 0, the
75. Consider curves with polar equation r
rightmost point on the cu;ve is (1
+ c, 0). For c < 0, the curves are reflections through the vertical axis of the curves
with c > 0.
c = 0.25
77. tan'lj! = tan(¢ e)=
_
c
= 0.75
tan ¢  tane r/> e 1 + tan tan
dy   tane dx dy 1 + dx ta11e
~  ~ tan e ~ (~ ~in e + r cos e) 
 : + :~ tane
 (:;.cos
c= 1
tan e (
c= 2
dy/dB dxjde  tane dyjde
1 + dx j de tane
~ cos B
r sine)
e r sine) + tan B( :; sine + r cos e)
sin2 e rcosB+r ·  case dr dr sin2 e de case+ de. case
_ rcos2 B +rsin2 e r 2 iJ dr . 2 e = dr I dO  dr de cos . + dB sm
10.4 Areas and Lengths in Polar Coordinates
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www.elsolucionario.net
SECTION 10.4 A~~S AND LENGTHS IN POLAR COORDINATES
3. r 2 = 9sin28, r ~ 0, ·a~ 8::; 7rj2.
A 
7.
1'
1
~/2
0
2 lr d8 2
1~/2
.
= 4 + 3sin8,
0
1(9sin2B)d8 l!. 2  2 [ lcos2B)~ 2 0
/2
l!.(4
1  1)=l!.2
%::; () ::; %·
1
= ~
~/2
2
(16 + 9sin 8) dO
[by Theorem 4.5.6(b) [ET 5.5.7(b)]]
rr/2
r /2
= ~·2}
0
[16 + 9 ·~(1 cos28))d~
1 (";  ~cos28) rr / 2
=
9. The area is bounded by r
'
d8 =
[~1 8  tsin2BJ; 12 = ( 4 ~"  O) (0 _
0) =
4~"
= 2 sin 8 for 8 = 0 to 8 = 7r.
~r 2 d8 := ~ 1 rr (2 sinfJ) 2 dB= ~
A= 1"'
[by Theorem 4.5.6(a) [ET 5.5.7(a)]] .
1"'
(2, 'TT/ 2)
2
4sin Bd8
=21" ~(1  cos2fJ)dB ~ [e  ~sin2eJ: =1r Also, note that th is is a circle with radius 1, so its area is 1r(1) 2 =
1 ~r d8 = l ~1 2
2
11. A =
"
2
1r.
fo rr(9+12.cos8+4cos 8)d8 2
2
rr H3 + 2cos8) d8=!
2
2
=
.,..
= ~
[9 + 12cos8 + 4 · t(1 .+ cos 28)] dB
{2rr
lo
(11 + 12 cos 8 + 2 cos 28) d8 =
Hne +
_(_l,_'TT:)H=

':..._:..1+
12 sin O+ sin 28) ~71'
= H227r) = 111r
13. A = =
r 71' ~r2 d8=
Jo
~ fo
= ~
"
1
~ r"(4+4si~4fJ+sin2 4fJ)d8
· Jo
2
2"
0
r rr t{2 + sin4fJ) 2 dfJ =
lo
I4 + 4sin48 +
3
~(1 cos88)] dB
(~ + 4 sin48  ~ cos88) dB=~ [~B cos4fJ
= ~[(97r 1)  ( 1))
·
.
fs sin8B] ~"'
= ~7('
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www.elsolucionario.net
0
21
22
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
D
1 (39 2 2
=
. + 2o1 sm 109] o21r =
1( ) 2 3'1T
3 27T
=
1.4
17. The curve passes through the pole when r
9=
=0
4 cos 30
==?
=0
==?
cos 39 = 0
39 = ~
==?
+ 1rn
=?
%+·in. The part of the shaded loop above the polar axis is traced out for
9 = 0 to 9 = 7T /6,' so we'll use
19. r = 0
sin49 = 0
==?
7T / 6
48
==?
r
and 7T /6 as our limits of integration.
= 1rn
9=
==?
= 4 cos 30
r=sin 40
~n.
r'4 Hsin48) dfJ. = t Jor'4sin 48d8 = t Jor '4t(1cos89)d8 0
=4
2
2
A = Jo
.
1 (9
.
· ]"'I 4 B SlD 89 0 1

0
=4
0
1 (,.)
.1 Tii7T
=
4
This is a
21.
lima~on,
with inner loop traced
r = 1 + 2 sin 8 (rect.) 3
out between 8 =
7 ;
and 1 ~" [found by
solving r = 0].
17T ·
0=6 31r/2
A = 2[ 71r /6
=
ll l'l'
0= 6
77r /6
=> cos 8 =
~

=(
[8+ sin 28)~ 13 = i +
C; + 2 v'3  4) =
9 ;)
12 ) dB=
= J;; a { 4 [~ (1 + cos 28)] 
37r/2
[1 +4sin8 + 4· t{1 cos29)] d9
71r /6
=> 8 = :g. or
A = 2 f 0,. 13 ~[(2 cos8) 2
=
··••,
. [37f/2 [ !(1 + 2sin8? d8 = (1 + 4sin8 + 4sin2 8) d8 =
[9  4cos0 + 20 ~n28J~:~~
23. 2 cos B = 1
.
5 ;
7T 
.
.
¥
.
J0" 13 (4cos2 8  l) d8
1} d8 =
J; /S ( 1 + 2 cos 28) d8
4
r = 2cos0
@) 2012 Cengogc Lcaming. All Rights Reserved. Moy not be oamned, copied, or duplicated, or posted to a publicly accessible website, in whole or in prut.
www.elsolucionario.net
SECTION 10.4
AREAS AND LENGTHS IN POLAR COORDINATES
0
25. To find the area inside the l~miniscate r 2 = 8 cos 29 and outside the circle r = 2, r=2
we first note that the two curves intersect when r 2 = 8 cos 29 and ~ = 2, .
I
that is, when cos2B = ~For 1r < B ::; 1r, cos2B = ~
{::} 2B = ±7r/ 3
{::} B = ±1r/ 6 or ±57r/ 6. The figure shows that the desired area is
or ±511' /3
4 times the area between the curves from 0 to 1r/6. Thus,
A = 4 J0" /. 6 [~(8 cos2B) ~(2) 2 ] dB = 8 J0" 16 (2 cos 2B  1) dB
]
. =8 [ sin2B  B
rr~
0
.
.=8(VS/27r/6) =4\1'3411'/3
27. 3cosB=1+cosB {::} cosB=~
=> B=t ori.
8 =1!. I 3 ,l'
A= 2 f 0" 13 ~ [(3 cos 8? (1 +cos 0) 2 ) dB =
J0"' 13 (8 cos2 8' 2 cos B 1) dB = f 0" 13 [4(1 +cos 2B) 
2 cos B  1) dB
= J0.,.13 (3 + 4cos28 2cosB) dB= [3B + 2sin2B 2sinf!] ~ 1 3
= 1T +v'3.J3=1T
= sinB. => J3 =
29. v'3cosB
A Jo  f "'13 12 (sin B) 2 dB +
1'"' 12
1
hr/ 3 . 2
(J3 cos B)2
3
 Jo f"' / 1. 1(1  cos28) dB+ 2 2
 1 4 [B

=> tanB = J3 => B =
sinBB cos
i·
d()
f rr/3 "/ 2 12 · 3 · 1(1 + cos2B) dB 2
12 12 sm · 28]"o13 +!!4 [B +.!2 sm · 28]1f7f/3
=
t [(i  4)  o] + ~ [ (~ + o) ( f + 4)]
_ 
.l!.. _ 12
_il 16
+. .!!:8 _
31. sin 2B = cos 2B
=>
1fl _ 16

5rr _
24
2 sin B = 1 cos2B
r =
J3cos8
fl 4
=> tan 28 = 1 => 28 = ~ =>
=>
B=. f
A= 8 · 2 f0"/8'~ sin 2 2B dB= 8 f 0"'18 H1 cos 48) dB = 4[9
t sin48]~
18
= 4(f 
33. sin2B = cos2B => tan2B = 1 =>
i
·1) =
2B =
~1
~
=> (} =
f
A= 4 f 0"' 18 ~sin 2B dB [since r 2 = sin 2B) =
f 0"'18 2 sin2B dB= [; cos 28] ; 18
= ~
v'2 ( 
1)
....···B=i
= 1 ~ Y2.
© 2012 Ccngage learning. All Riglus Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessibk website, in whole or in part.
www.elsolucionario.net
·
23
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
24 • 0
35. The darker shaded region (from 0 = 0 to 0 = 27T / 3) represents ~ of the desired area plus ~ of the area of the inner loop. From this area, we' ll subtract~ of the area of the inner loop (the lighter shaded region from 0 = 27r /3 to 0
= 7r), and then
double that difference to obtain the desired ·area.
A = 2 [J;.,. =
13
H~ +~os0) 2
dO
g.,.13 H~ +cos 0) 2 dB]
£02.,.13 (~+cosO+ cos2 0) dB f2:
.
2 13 (~ + cos 0"+ cos 0) dB
.
= J0211'/S [~+cos 0 +
~(1 + cos28)) dO
 I2:/3
= [~ + 4
~ + sin 20 ]
sin 8 +
2
4
13
2 "
_
a + cosO+ ~(1 + cos28)) d(J
[~ + 4
0
sin 8 +
~ + sin 20 1.,. 2
4
211'/ 3
.:.8::2)  (!!:4 + !!:) 2 + (1!: 6 + .il 2 + 1!: 3  .:.8::2)
 (1!: + .il 2 + 1!: 3 
6
= ~ + ~ J3 = H 7T+3v'3) 37. The pole is a point of intersection. 1 + sin 8 = 3 sin 0 (J
=~ or
=> 1 =
sin 0 = ~
=>
2 sin (J
=>
5 ;.
The other two ~oints of intersec~ion ~e ( ~,
i) and
(
f,
5 6").
r=l + sinO
39. 2 sin 20
=1
sin 20
=?
=
~
=?
o 1!:6' &.,.6 '
2 =
13.,. 6 '
or
11" 6 °
By symmetry, the eight points of intersection are given by r = 2sin20 11' 5,. 13rr 1771" (1, 8) , where (} _ 12 , an d , 12 , 12 , and 12
i;, 1t;, 1: 2" ; and 2[ ; .
(  1, 8), where 0 =
[There are many ways to describe these points.]
41. T he pole is a point of intersection. sin (J sinO (1  2 cos 9)
=0
8 = 0, 7T,
t. or  t
and ( ~,
2 ; )
©
= sin 20 = 2 sin(} cos 0
sinO= 0 or coslJ =
t
=>
=> the other intersectio~ points"are ( ~. t )
[by symmen:'].
r =sin20
2012 Ccngage Learning. All R ights Reserved. May not be scotuled, copied, or dupHcuted, or posted to a publicly accessible website, in whole or in part.
www.elsolucionario.net
SECTION 10.4 . 3.4
43.
3
r=
AREAS AND LENGTHS IN POLAR COORDINATES
0
25
y = 2x
2(}
3
r = 1 +sinO 1.4
1.4
3
 0.3
From th~ first graph, we see that the pole .is one point of intersection. By zooming in or usin~ the cursor, we find the 0values of the intersection points to be a:~ 0.88786 estimated by plotting y = 1 +sin x andy
~
0.89 and
:r  a ~ 2.25. (The first of these values may be more easily
= 2x in rectangular coordinates; see the second graph.) By symmetry, the total
area contained is twice the area contained in the first quadrant; that is,
A= 21o. ~(20) 2 dO+ =
45. L =
2[1r/Z ~(1 +sin 0.)
2
d8 = 12
(~0 3 ] ~ + [8  2cosB + (~0  t sin2B)J:
1"' =
402 .d8 +
l1r/~ [1 + 2sinU +
H1 cos 2B)] d8
~a 3 + [(~+~)(a 2cosa + ~a:  t sin2o:)) ~ 3.4645
1b
Jr2+(d7·jd0) 2 dB= fo7r }(2cos0)2 +(2sinB.)2d8
= 17r
V
4(cos2 B + sin2 B) dB =
1"' ..f4d8 =
[2B];
= 21r
As a check, note that the curve is a circle of radius 1, so its circumference is 27r(1 ) = 21r.
=
1 Je 2
"'
2
(02 + 4)d0 =
1 2
1f
BV02 +4d0
Now let u = 02 + 4, so that du = 20d8 [Ode=
49. The curve r
t du]
= cos 4 (0 / 4) is completely traced with 0 :::; () :::; 47r.
r 2 + (drfd8) 2
= [cos4 (B/4)] 2 + = cos8.(B/4) +
[4cos3 (B/ 4) · (sin(B/4)) · t ]
2
cos6 (8 /4) sin2 (B / 4)
= cos6 (8/4)[cos2 (B/4) +sin2 (0/4)] = L
and
cos6 (B/4)
= f041r Jcos6 (0/4) dO = J04"' jcos3 .(0/4) j rj,B . = 2 f~" cos8 (B/4)dB [sincecos3 (B/4) ~ 0 for O :S B $ 27r]
= 8 J;12(1 
sin 2 u) cosudu = 8 f 01 (1 x 2 ) dx
= 8J;12 cos3 udu
[u = tB]
x = sinu, ·] [ dx = cosudu
= 8[x ~x3) ~ = 8(1 ~) = lf @ 2012 Ccngu~c Learning. All Rights Rcscrv~. May not be scanned, copied, or duplicaled. or posted too public ly accessible w·cbsite, in whole or in part
www.elsolucionario.net
26
0
CHAPTER 111 PARAMETRIC EQUATIONS AND POLAR COORDINATES
51. One loop of the curve r = cos 28 is traced with  1rI 4 ~ 8 ~ 1r I 4. r2 +
(~~)
2
14
= cos2 28 + ( 2 sin 28) 2 = cos2 28 + 4sin2 28 = 1 + 3 sin2 28
L /"'
::::}
Vl + 3 sin
2
?r /4
53. The curver = sin(6sin9) is completely traced with 0
r2
~ 8 ~ 1r.
(~;y =s~n2 (6sin8)+36cos2 8cos2 (6sin0)
:
::::}
L
r = sin{6sin9)
L"
::::}
~
28 d8:::::: 2.4221.

= cos{6sin9)_ 6cos8, so
Jsin2 (6sin8)+36cos2 8pos2(6sinO)d8:::::8.0091.
55. (a) From (10.2.6),
s
=I: 21ryJ(dxld0)2 + (dyld9) 2 dO [from the derivation of Equation 10.4.5]
(b) The curve r 2 20 =
~
::::}
= cos 28 goes through the pole when cos 28 = 0 8=
~.
We'll rotate the curve from () = 0 to 8 =
::::} ~
(o. f)
and double
? = cos28 (1.0)
this value to obtain the total surface area generated. r 2 = cos28 ::::}
· dr 2r dO= 2sin28 ::::}
( dr )
dO
2
2
2
sin 28 sin 28  ~  cos20'
10.5 Conic Sections 1. x 2
= 6y and x 2 = 4py
::::}
4p = 6
::::}
p
= ~
The vertex is (0, 0), the focus is (0, ~),and the directrix
isy
= ~.
3. 2x =  y 2
::::}
y2
= 2x.
4p
= 2
::::} p '\
The vertex is (0, O),the focus is (  ~, 0), and the directrix is x = ~. y
x=4 6
X
X
y= ~
©
2012 Cengage Learning. All
Righ~
Reserved. May not be scanned, copied, ocduplieated, or posted to a publicly acccssible website, in whole or in pan.
www.elsolucionario.net
=  ~.
SECTION 10.5
5. (x + 2) 2 = 8 (y 3). 4p = 8, sop= 2. The vertex is
7. y 2
(  2, 3), the focus is ( 2, 5), and the directrix is y = 1.
y2
CONIC SECTIONS
+ 2y + 12x + 25 = 0 =l> + 2y + 1 =  12x 24 =;.. (y + 1)2 =  12(x + 2). 4p = 
y
0
27
12, sop= 3.
The vertex is (  2, 1), the focus is (5,  1), and the directrix is x = 1.
 __ ,y_~_!_ X
0
:
~
X
Ijx=l 9. The equation has the form y 2 = 4px, where p < 0. Since the parabola passes through ( 1, 1), we have 12 = 4p(  1), so
4p =  1 and an equation is y 2 =  x or x =  y 2 • 4p =  1, sop = . ISX
i and the focus is ( ;i,0) while the directrix
1
= 4· y
2
(0, ±v'2).
ellipse is centered at {0, 0), with vertices at (0, ±2). The foci are
..fi 2
13. x
2
+ 9y2 = 9
x2
Y2
9 +l
b = Vi.= 1, c = ~ =
= 1
=l>
a=
v'9=I = J8 =
v'9 = 3,
15. 9x 2
9(x 2
2v'2.
The ellipse is centered at {0, 0), with vertices (± 3, 0).
18x + 4y2 = 27

+ 1) + 4y 2 =
9(x 1) 2 + 4y2
The foci are (±2v'2, 0).
a = 3, b = 2, c = y
0
= 36
27 + 9
J5 =:
(x 1)2 4
y2
+ g = 1
center (1, 0),
vertices (1 , ± 3}, foci (1, ±J5)
I
3
2x
3
y
X
(1, 3)
 I
3 X
(1, 3) 2
17. The center is (0, 0), a= 3, and b = 2, so an equation is
©
~
2
+ Y9
= 1. c = ~ =
J5, so the foci are (0, =;I=J5).
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www.elsolucionario.net
=}
28
CHAPTER 10 PARAMETRIC EQUATIONS AND POLARCOORDINATES
0
yz 19.
xz
9 =1

25
a = 5, b = 3, c = J25 + 9
::::?
= J34
::::?
center (0, 0), vertices (0, ±5), foci (0, ±J34), asymptotes y = ±~x. Note: It is helpful to draw a 2aby2b rectangle whose center is the center of
the hyperbola. The asymptotes are the extended diagonals of the rectang le.
21 . x 2
y2

= 100
x2
y2
100

100
c = JlOO + 100 = 10J2 => foci (±10 J2, 0), asymptotes y
=1
y
a = b = 10,
::::?
center (0, 0), vertices (±10, 0),
= .±f§x = ±x
X
,.._!",, ,~_;.......
23. 4x2 4(x
y2
2


24x  4y + 28 = 0
6x + 9)  (y
2
+ 4y + 4) =
4(x '_ 3)2  (y + 2)2 = 4 a
28 + 36 4
(x  3)2  (y + 2)2 = 1 => 1
4
= v'f = 1, b = v'4 = 2, c = Vf+4 = J5
::::?
center (3,  2), vertices (4,  2) and (2, 2), foci (3 ± asymptotes y
25. x 2
=y+1
+2=
J5, 2),
±2(x 3).
x 2
= 1(y + 1). This is an equation of a parabola with 4p = 1, so p =
i. The vertex is (0, 1) and the
focus is (0,  ~). 27. x 2
= 4y 2y2
x; + (y ~ 1)
2
x 2
+ 2y2 
4y
=0
x 2 + 2(y2

2y + 1)
=2
x 2
+ 2(y 
1)
2
=2
= l.. This is an equation of an ellipse with vertices a~ (±J2, 1) . The foci ~eat (±.,J2=1, 1)
29. y2 + iy = 4x2 + 3
y 2
+ 2y + 1 = 4x 2 + 4
of a hyperbola with vertices (0,  1 ± 2)
(y
+ 1? 
4x 2
=4
= {0, 1) and (0, 3). The foci are at (0, 
1
(y: ) 1±
= (±1,
1) .
2 
J4+T)
x 2 = 1. This is an equation = (0,  1 ±
J5).
31. The parabola wi~ vertex {0, 0) and focus (1 , 0) ?pens to the right and hasp= 1, so its equation is y 2 = 4px, or y 2 = 4x. ' 33. The distance from. U1e focus ( 4, 0) to the directrix x = 2,is 2  (  4) = 6, so the distance from the focus to the vertex is ~(6) = 3 and the vertex is (  1, 0). Since the focus is to the left of the vertex, p =  3. An equation is y 2 .= 4p(x + 1)
y
2
= 12(x + 1). © 2012 CcnGogc: Lcmning. Al l Rjglns Rest.! rvcd. Mny not he scanned. copied, or duplicutccl, or posted to a publicly occcssiblc website. in whole or in p..1rt.
www.elsolucionario.net
=>
SECTION 10.5
CONIC SECTIONS
29
0
35. A parabola with vertical axis and vertex (2, 3) has equation y 3 = a(x 2?. Since it passes through (I; 5), we have 53=a(I2) 2
'*
a =2,soanequationisy 3 =2(x2? .
37. The ellipse with foci (±2, 0) and vertices (± 5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 2,
so b2 = a2
c 2 = 25 4 = 21. An equation is

x2
12
+ ;1 = 25
1.
39. Since the vertices are (0, 0) and (0, 8), the ellipse has center (0, 4) with a vertical axis and a = 4. The foci at (0, 2) and (0, 6) 2
are 2 units from the center, soc= 2 and b = .,Ja2  c2 = .,j4 2

_41. An equation of an ellipse with center (1,4) and vertex ( 1,0) is (x
from the center, soc= 2. Thus, b2
+ 22
= 42
'*
. 2
2
b = 25  9 = 16, so the equation is
~
~2
!
2
1
2
)
+ (y ~24)
2 ....:. ;
2
a
b
'*
= 1. Foci (±5, 0)
2
= 1
= 1. The focus (1,6) is 2 units
+ (y 164 )
c = 5 and 32
2
= 1
.
+ b2 =52
'*
2
 r6 = 1.
45. The center of a hyperbola with vertices ( 3,  4) and ( 3, 6) is ( 3, 1), so a = 5 and a.n equation is . (x + 3) 2 · · b2 = l.Focl(  3,  7)and( 3,9)* c =8,so5 2 +b2 =82
'*
b2 =6425=39andthe
(  1} 2 {x + 3) 2 equation is y = I 25 39 . .
.
2
47. The center of a hyperbola with vertices (±3, 0) is (0, 0), so a = 3 and an equation is
Asymptotes y
= ±2x
b
'* a =
2
'*
~2
•

2
t
2
= 1.
. 2 2 b = 2(3) = 6 and the equation is ~  JL = 1. 9 36
49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance
a  c from it) while the farthest point is the other vertex (at a distance of a+ c). So for this lunar orbit, (a c)+ (a+ c) = 2a = (1728 + 110) + (1728 + 314), or a = 1940; and (a+ c) .(a  c) = 2c = 314  110, or c = 102. Thus, b2 = a
2

y2
x2
2
c = 3,753,196, and the equation is , , 3 763 600
+ 3,753, 196
=I.
51. (a) Set up the coordinate system so that A is ( 200, 0) and B is (200, 0) .
IPAI\PBI =
(1200)(980) = 1,176,000 ft =
b2  c2
3,339,375 121

2 
a 
121x 2 1,500,625
2
1~ mi = 2a
*
a= 1 ~ 5 , and c ~ 200 so
2
121y  1 3,339,375  .
© 20 12 Ccngoge Leoming. All Rights Rescn'Cil. May not be scanned, copied. or duplicated. or posted to a publi x = 200 =>
(b) Due north of B
(121}(200) 2 1,500,625
2
121y 1 3,339,375 =
=>
y = 133,575 ~ 248 mi 539
53. The function whose graph is the upper branch of this hyperbola is concave upward. The function is
55. (a) If k
> 16, then k 
> 0, and
16
2
2
xk + yk = 1 is an ellipse since it is the sum of two squares on the left side.   16 2
(b) IfO
< k < 16, then k 
16
< 0, and xk
2
+ k
~
= 1 is a hypel"bola since it is the difference of two squares on the
16
left side. (c) If k
< 0, then k 
16
< 0, and there is no cw11e since the left side is the sum of two negative tenns, which cannot equal
(d) In case (a), a 2 = k, b2
b2
= 16 
k , and c2
= 4ptJ =>
57. x 2
2
y :;
= k 16, and c2 = a2 
= a2 +
2x = 4py'
= ~; (x 
b2
b2 = 16, so the foci are at (±4, 0}. Jn case (b), k  16
are
= a2 + p
y
=> y' = ; , so the tangent line at (x0 , y 0 ) is .
p
xo). This line passes through the point (a,  p) on the
4p2
a± ..ja2 + 4p2 2
k,
= 16, and so again the foci are at (±4, 0}.
. . so  p  x6p = xop (a  xo ) =>  4p 2  x 2 = 2axo  2x 2 d1rectnx, 0 0 4 2 x5  2axo  4p2 = 0 x5  2axo + a 2 = a2 + 4p2
(xo a) 2
< 0, so a 2 =
I.
(a,p)
y=  p
x 0 =a± Ja2 + 4p2 • The slopes of the tangent lines at x = a± Ja2 + 4p2
, so the product of the two s lopes ts
a+ Ja2 +
4p2
''. 2p
a  Ja2
+ 4p2
2p
=
a2 _ . (a2 + 4p2) 4p2. =   = 1 4p2 4p2 '
showing that the tangent lines are perpendicular. 59. 9x2 + 4y 2
= 36
2
~
2
+ Y9
= 1. We use the parametrization x
= 2 cost, y = 3 s in t, 0 ~ 't ~ 21r. ·The circumference
is given by
L =I:" J(dx/dt) 2
+ (dyjdt)2 dt = ]~2 " J( 
2 s in t) 2 + (3 cost)2 dt
=I:" )4sin2 t +
9cos2 t dt
= I:" v4 + 5 cos2 t dt Now use Simpson's Rule with n = 8, 11t =
L ~ ss = ~ [J(o) + 4f(*)
+ 2f(~) +
2
1T
8.
0
= ~.and f(t) = 4
4Jen + 2J(1r) +
4/(
v'4 + 5cos2 t to get
5 ;)
+ 21en
+
4tCn
+ J(21T)J ~ 15.9.
® 2012 Ccngage leArning. AJI ftis.hts Rcscn·cd. May nol be scannc:c.l, copicd or t.luplicah!d, or posted to a publicly uccessibJc website, in whole or in pout. 9
www.elsolucionario.net
SECTION 10.5 x2 y2 61. ;_2   2 = 1
a
b
A= 2
la a
r
=?
y2 a:2 _ a2  2 = b a2 a
0
31
b y = ± Jx2 a2. a
=?
J x 2 ·_ a 2 dx ~ 2b [~ J x2 
!!_
CONIC SECTIONS
a2 
2
42
2
In Jx +
J x2 
a2J ] c
a·
b = [cvc2  a 2 a2 ln jc + vc2  a 2 j + a 2 ln!al) a Since a 2 + b2 = c2 , c2  a 2 = b2 , and J c2  a 2 = b.
=!!.a [cb a 2 1n(c + b)+ a 2 !na) = !!_a [ch + a2 (!na = b2 c/a + abln[a/( b +c)), where c2 = a 2 + b2 . 63. 9x2
+ 4y2 =
x2
Y2
'4 + g = 1
36
top half of the ellipse is
H1rab) =
y = 1
A
1b a
=  3 · '2 87T
1 [f(x))2 dx = 12
37T
1 2
0
2
( 4  x ) dx
a= 3, b = 2. By symmetry, x = 0. By Example 2 in Section 7.3, the area o_f the
=?
+ 4y2 =
37T. Solve 9x2
9x2 +4y 2 =36 4y 2 =369x 2
1
~
36 for y to get an equation for the top half of the ell ipse:
y 2 = ~(4x 2 )
2
 2
!n(b +c)))
2
1 ( 3 V 4 x 2 ) 2 2
1 x 3]
3 [4x = 47T
3
dx=  3
87T
y = ~v'4x 2 .Now
=?
12
y
(4 x 2 )dx
 2
2 0
3 ( 16) = 4 = 47T 3 7T
so the centroid is (0, 4/ rr). 2
. . .tmp I'tctt . Iy, •x 65. Dt' ffierenttatmg a2 line at Pis 
tan a
2
= 1
+ yb2
[y
=?
=f. 0]. Thus, the slope ofthe tangent
b:x1 . The slope of F1P is ___)f];_ and of F2P is ____lt!__ By the formula from Problems Plus, we have +
a Y1
=
=
X1
2 Yl b Xt Xt_+_c + a 2_y_1 _.:........;.::.''=1_ b2 x1Y1 2 a yt (Xt +c) b2 (cxt+a2 ) CIJt(CXt + a 2 )
·
_

c
X1 
yt
a2 + b2 x 1(x1 +c) _ a 2 y1 (x1 + c)  b2X tY1 
c
a 2 b2 + b2 cx1 C2 XtY1
+ a 2CIJ1
b2
= cy1
and  a
2
y'f. 
b2 x1 ( x1 c)
 a
2 2
b
+ b2 cx1
= a 2y1 (x1 c)  b2X 1Y 1 = c2x1y1  a 2C1Jt =
b2 (ext  a 2) CIJ1 (cx1 
a 2)
b2
= CIJ1
Thus, a = {3.
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32
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
0
10.6 Conic Sections in Polar Coordinates 1. The.directrix x = 4 is to the right of the focus at the origin, so we use the form with "+ e cos(}" in the denominator.
ed l2 . 4 4 (See Theorem 6 and Figure 2.) An equation is r = (} = (} = (}. 1 2 +cos ' 1+ ecos 1+ 2 cos
3. The directrix y = 2 is above the focus at the origin, so Wf! use the form w ith "+ e sin(}" in the d enominator. An equation is ed _ 1.5(2) _ 6 1 + e sin ()  1 + 1.5 sin(}  2 + 3 sin(}·
_ r 
5. The vertex (4, 3Jr/2) is 4 units below the focus at the origin, so the directrix is 8 units below the focus ( d = 8), and we use the form with " e sin(}" in the denominator. e = 1 for a parabola, so an equation is
ed 1(8) 8 r = 1  e sin () = 1  1 sin (} = 1  sin (} · 7. The directrix r = 4 sec (} (equ ivalent to r cos(} = 4 or x· = 4) is to the right of the focus at the origin, so we w ill use the form with"+ e cos()" in the denominator. The distance from the focus to the directrix is d = 4, so an equation is
~{4)
ed
r = 1 + ecosB = 1 + ~cos(}· 4
9. r =
5  4sinB
2
2=
4 2 +cos((
5 4 5 where e = .1 and ed = .1 = / 1/5 . 1  ~sinB' 5 &
. l/
=> d = 1. y
(4 , 7r/2)
(a) Eccentricity = e = ~ (b) Since e = ~
d= ~. y
y = 2/ 3
(a) Eccentricity= e = 1 (b) Since e
= .1, Lhe conic is a parabola.
(c) Since"+ e sin(}" appears in the denom inator, the directrix is above the focus at the origin. d =
IFll = j , so an equation o flhe directrix is y =
(d) T~e vertex is at ( ~,
i·
¥), midway between .the focus and directrix.
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CONIC SECTIONS IN POLAR COORDINATES
SECTION 10.6
13. r =
:
6+
3 2 1 6 · 1 = ( , where e = cos 9 116 1+ 3 cos0
(a) Eccentricity= e =
(b) Since e =
t
~ and e d = ~
:::}
d=
~
t
< 1, the conic is an ellipse.
(c) Since " + e cos
I
9:
e" appears in the denominator, the directrix is to the right of
the focus at the origin. d =
!Fll = ~.so an equation of the directrix is
x=21
:
(f,1T)
!
x = ~(d) The vertices are ( ~, 0) and ( that is, (
t, 1r), so the center is midway between them~
fn, 1r) .
3 1/ 4 3/ 4 3 15. r = 4  8 cos 9 . 1I 4 = 1  2 cos 9' where e = 2 and ed = 4
d  1s ·
(a) Eccentricity = e = 2
(b) Since e
= 2 > 1, the conic is a hyperbola.
(c) Since " e cos 0" appears in the denominator, the directrix is to the left of the focus at the origin. d'=
IFll = ~.so an equation of the directrix is
3 .X   s·
(d) The vertices are (  ~. 0) and
(i, 1r), so the center is midway between them,
that is, (~ , 1r).
17. (a) r =
1
; .
sm 9
, where e
= 2 and ed =
1 :::}
d
= 12 . The eccentricity
e = 2 > 1, so the conic is a hyperbola. Since" e sin B" appears in the denominator, the directrix is below the focus at the origin. d =
IFll = ~.
so an equation of the directrix is y =  ~ . The vertices are (  1, ~) and 1 (3,
3 2 3 2'~~" ) , sothe center .IS m1"dway between th l'!m, t hat 1.s, ( 3 , 2" ) .
3
(b) By the discussion that precedes Example 4, the equation . IS T
1
=
0
1  2sin ( e
;)•
3
2
® 2012 Cengagc Learning. All Rights Resen·
a = 4.6 x 107/ 0.794. So the maximum distance, which is at aphelion, is
r = a(1 + e) = (4.6 x 107/ 0.794) (1.206} ~ 7.0 x 107 km. 31.
From Exercise 29, we have e = 0.206 and a(l e)= 4.6 x 10 7 km. Thus, a= 4.6 x 10 7/ 0.794. From (7), we can write the .
.
equation of Mercury's orbit as r =a
1
e2
1 + ecos 8
•
. So smce
2
dr a(1 e )esin8 = d9 (1 + ecos8) 2
=>
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CHAPTER 10 REVIEW
0
35
the length of the orbit is
1 2
L=
,.
Jr 2
+ (dr/d0)2 d(J =
a(l e2 )
1
2 +2 +ecos
2
,.
.Jl(t e
0
0
~)~sO dO:::::: 3.6 x 108 km
This seems reasonable, since ,Mercury's orbit is nearly circular, and the circumference of a circle of radius a is 21ra :::::: 3.6 x 108 km.·
10
Review CONCEPT CHECK
1. (a) A parametric curve is a set of points of the form
(x, y) = (J(t), g(t)), where f and g are continuous funct ions of a
variable t. (b) Sketching a parametric curve, like sketching the.graph of a function, is difficult to do in general. We can plot points on the curve by finding f(t) and g(t) for various values oft, either by hand or with a calculator or computer. Sometimes, when
f
and g are given by formulas, we can eliminate t from the equations x = j(t) andy
= g(t) to get a Cartesian equation
relating x andy. It may be easier to graph that equation than to work with the original formulas for x andy in terms oft.
2. (a) You can find :
as a function oft by
calculating~~ = ~~J~~
[if dxjdt
f:
0].
(b) Calculate the area as I: y dx = I: g(t) j'(t)dt [or I; g(t) !' (t)dt ifthe leftmost point is (!({3), g({J)) ~ther than (f(a),g(a))].
3. (a) L = I: J(dxjdt)2 (b)
+ (dy j dt)2 dt =I:
+[.g'(t))2 dt
vlf'(t))2
s =:: I: 27ryJ(dxfdt)2 + (dyfdt) 2 dt =I: 21rg(t) v lf'(t)F + [g'(t)J2 dt
.
4. (a) See Figure 5 in Section 10.3. (b) x
= rcosO,
y = r sinO
(c) To find a polar representation (r, 0) with r ;?. 0 and 0
s 0 < 27r, firs t calculate r =·Jx2 + y2. TI1en 0 is specified by
cosO= xjr and sinO= yjr. d
5. (a) Calculate
Jx =
!:.._ (y)
dy
~ = ~ dO
dO (x) 2
(b) calculate A= I: !r dO= I:
(c) L =I: J(dx/d8.)2
!:.._ (r ::;in 8)
=
. : : l J:;: 
d() (rcosB)
(~) sin(} + r
dr) (dO
cos(}
. ,
wherer = f(O).
cos ()  r sm (}
tlfCOW d8
+ (dy/d0) 2 dO= I:
Jr 2
+ (dr/d0) 2 dO= .r: J[f(0))2 + [!'(0)]2 dO
6. (a) A parabola is a set of points in a plane whose distances from a fixed point F (t11e focus) and a fixed line l (the 'd irectrix) are equal. (b) x 2
= 4py; y 2 = 4px ®
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www.elsolucionario.net
36
PARAMETRIC EQUATIONS AND POLAR COORDINATES
CHAPTER 10
D
7. (a) An ellipse is a set of points in a plane the sum of whose distances from two fixed points (the foci ) is a constant. 2
(b)
X
a2
2
+ _Y_ = 1. a2  c2
8. (a) A hyp~rbola is a set of points in a plane the difference of whose distances from two fixed points (the foci) is a constant.
This difference should be interpreted as the larger distance minus the smaller distance. 2
2
(b) ::__  _ Y_
c2 a2
a2
=±
(c) y
=1
Jc2 _ o,2 a
x
9. (a) Jfa conic section has focus F and corresponding directrix l, then the eccentricity e is the fixed ratio /PFI / /Pl / for points
P of the conic section. (b) e
< 1 for an ellipse; e > 1 for a hyperbola; e = 1 for a parabola.
(c) x = d: r =
ed . ed · . x =  d:.r = . y = d: 1 + e cos (} 1  e cos (}
1'
=
ed ed . . y =  d: r = :::: 1 + e sm 8 1  e sin 8 ·
TRUEFALSEQUIZ 1. False.
3. False.
= (t but its graph bas a vertical tangent when t = 1.
Consider the curve defined by x = f(t)
1) 3 and y= g(t)
= (t 1)2 : Then g'(t) =
2(t  1), so g'(1) = 0,
Note: The statement is true iff' (1) =I= 0 when l (1) = 0.
For example, iff (t) = cost and g( t) = sin t for 0 ::; t :S 4rr, then the curve is a circle of radius I, hence its length 4
is 2rr, but f 0 " J [!'(t)]2
+ [g'(t))2 dt = .{04 " J(
sin t) 2 +(cos t) 2 dt =
f04"
1 dt = 4rr, since as t increases
from 0 to 4rr, the circle is traversed twice. . 5. True.
The curve r = 1  sin 28 is unchanged if we rotate it through 180° about 0 because 1  sin 2(8 + rr) = 1  sin(28 + 2rr) = 1  sin 28. So it's unchanged if we replacer by r. (See the discussion
after Example 8 in Section 10.3.) In other words, it's the same curve as r = (1 sin 20) = sin 28  1. 7. False.
The first pair of equations gives the portion of the parabola y = x 2 with x ;::: 0, whereas the second pair of equations traces out the whole parabola y = x2 •
9. True.
By rotating and translating ~he parabola, we can assume it has an equation of the form y = cx2 , where c > 0. Tile tangent at the point (a, ca2 ) is the line y  ca2 = 2ca(x  a); i.e., y = 2cax  ca2 • This tangent meets the parabola at the points [since c
(x, cx
2
)
(x, cx2 ) where cx 2 = 2cax  ca2 • This equation is equivalent to x2 = 2a~  a 2
> 0]. But x 2 = 2ax  a 2 x 2 =
2
(a, ca
).

2ax + a 2 = 0
(x a) 2 = 0
x =a
This shows that eacb tangent meets the parabola at exactly one point.
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CHAPTER 10 REVIEW
0
37
EXERCISES 1.
+ 4t, y = 2  t,  4 ~ t ~ 1. ~ = 2  y, SO 2 2 2 X = {2  y) + 4{2 y) = 4 4y + y + B  4y = y  By+ 12 {::} x + 4 = y 2  By+ 16 = (y  4) 2 . This is part of a parabola with vertex X
y
= t2
( 4, 4), opening to the right. X
3. y
= sec() = ~ () = .! . Since 0 ~ () ~ 1r /2, 0 < x ~ 1 and y ;::: 1. COS X
y
This is part of the hyperbola y = 1/x.
\ ,1,1),00 0
5. Three different sets of parametric equations for the curve y = (i) (ii)
X X
= t, y
= t
4
,
IX are
= Vt 2 y =t
(iii) x = tan2 t, y = tan t, 0 ~ t
d = ~ and the form "+2 s in(}" imply that the directrix is above the focus at the origin and bas equation y
17. x+y=2
= ~The vertices are (1, ~) and (  3, 3; ) .
rcosll+rsinll = 2
¢:>
r(cos0+sin 0)=2
¢:>
2 .
¢:>
r=   

cos(J + s in£1
19. r = (sin £1)/0. As B. ±oo, r __, 0.
As (} __, 0, r .. 1. In the first figure, there are an infinite number of x intercepts at x
= 1rn, n
a nonzero
integer. These correspond to pole points in the second figure.
0.25  0.75
21. x = lnt, y = 1 +
2
dy
dx
When t = 1, (x , y) = (0, 2) and dyjdx 23. r = e 0
=> y
1
dy
t ; t = 1. dt = 2t and dt = t' so dx
=
dy / dt
2.
= rsin£1 = e 0 s in(} and x
= T"cos O = e  0 cosO
dy dyjd£1 *sin(} +rcosO _ e 0 sin(}+ e 0 cos(}  e 0 dx = dx j d(} = ~;cos(}  rsin(}  e 0 cosB  e 0 sinO ·  e 0 When (}
dy = 1r, dx
=
0  ( 1) 1 = 1 + 0 1
=
2t
= dxjdt = 1/t = 2t 2 · => sine cosO
= cos B+sine ·
1.
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www.elsolucionario.net
CHAPTER 10 REVIEW
25. x
= t + sin t, y = t 
d (dy)
dt dx dxjdt
cost
0
39
dy dyjdt 1 + sint dx = d.xjdt = 1 +cos t
~
{1+cost) cost {1 +sint)( sint) 2 2 _ __ _ ___:(,_1,...+.:..__c_os_ t')2 _ _ _ __ = cost+ cos t +sin t + sin t = 1 + cost + sin t l +cost (l+cost)3 (1 + cost)3 3
2 .
27. We graph the curve x = t  3t, y = t + t + 1 for 2.2 :=::; t :=::; 1.2.
By zooming in or using a cursor, we find that the lowest point is about (1.4, 0. 75). To find the exact values, we find the tvalue at which
dyjdt=2t+1 = 0
t=~~ , (x,y),;,(lj,~).
¢=>
29. x ·= 2acost acos2t sint = 0 or cost = ~
y = 2asin t  a sin 2t
~
~ ~
dx = 2asin t + 2asin2t = 2asint(2cost  1) = 0 dt
¢:>
t = O,·t, 1r, orr;. dy = 2a cost  2a cos 2t = 2a(l +cost 2 cos2 t) = 2a(1 cos t)(1 + 2 cost) = 0 dt
271" 4" t = 0'3,or3.
Thus the graph has vertical tangents where t = what the slope is where t
t.
1r
and ~">3" , and horizontal tangents where t
= 0, we use !'Hospital's Ru;e to evaluate t>0 lim ddy/jddt X t t
X
0
a
0
3"
~a
3
 ~a
~a 2
7r
 3a
0
4 7r
~a
~a 2
51r
~a
_ :.::]:a
2
(3a,O)
(a,O)
2
fu :=::; 8 :=::;
{'0 , there are two petals, one with r
> 0 and one
< 0.
33. The curves intersect when 4 cos 8
for
To determine
)C
31. The curve r 2 = 9 cos 50 has 10 " petals." For instance, for with r
4 ;.
y
2rr
3
and
= 0, so there is a. horizontal tangent there.
y :.::]:a
3
= 2;
1r
:=::;
8 :=::;
1r.
=2
~
cos 8
=~
~
8 = ±i
r = 4cos 0
The points of intersection are (2 , i) a nd (2, ~).
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=}
40
D
CHAPTER 10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
=>
35. The curves intersect where 2 sin () = sin() + cos () sin 0
= cos 0 => 0 = "i, and also at the origin (at which 0 = 34rr 8=f
on the second curve).
A= I 0" 14 ~(2 sin0) 2 dO+ I:;f ~(sinO+ cos 0) 2 dO 4
= Io"
14
(1 cos 20) dO+~ I:;t (1 + sin2fJ) dO 4
L =I~ .J(dxjdt)2 + (dyjdt) 2 dt
= 1~ .J(6t)2 +
{6t2)2 dt =I: V36t_2 + 36t4 dt = I~ y'36t2v'1 + t2 dt
=I~ 6\t! v1 + t 2 dt = 6 I~ t v'f+t2 dt = 6 I 1
5
J:
= 6. ~ · ~ [u312 = 2(53 12 39. L =I:"
'
r = sin 0 + cos 8
r = 2 sin 8
= [0 ~ ~ sin20]~ 14 + [~0: t cos2BJ!~~ = ~(1r 1)
u
1 2 ;
(~du)
1) = 2(5VS 1)
Jr2 + (drjdB)2 dB = I:" }(1/0)2 + (1/0 2 )2 dB=
~ [ #+! +In (B + J B2 + 1 )] 2,. = #+I ()
7r:
L7 2
dB
"'
v' 471"2 + 1 271"
+ In ( 271" + ~ ) 71"
11"
+ v'71"2 + 1
s = I 14 27ry.J(dxfdt)2 t. (dyfdt)2 dt = I 14 21r{~t3 + ~c 2 ) J(2/0) 2 + (t2 t3)2 dt 4 3 + .!c 2) . l(t2 + t  3) 2 dt = 21r f 4 (.!t 5 + !! + l t 5) dt  21r [...!..t6 + l!t = 21r J1 (.!t 3 2 v 1 3 6 2 18 6
4) 4  471 ·295 1r 1c 8 1 1024
43. For all c except  1, the curve ·is asymptotic to the line x = 1. For c
< 1, the curve bulges to the right near y =
bulge becomes smaller, until at c
=
0. As c increases, the
1 the curve is the straight line x
= 1.
As c continues to increase, the curve bulges to the left, until at c = 0 there is a cusp at the origin. For c > 0, there is a loop to the left of the origin, whose size and roundness increase as c increases. Note that the xintercept of the curve is always c.
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CHAPTER 10 REVIEW 2
45. ~ +
9
2
.
47. 6y2
'lL = 1 is an ellipse with center (0, 0). 8
a = 3, b = 2 J2, c = 1
=?
36y
+ 55 = 0
~
6y + 9) =  (x + 1)

41
~
2
(y  3) = [;(x + 1), aparabola with vertex (1, 3),
foci (±1, 0), vertices (±3, 0).
opening to the left, p
J
y
+X 
6(y2
0
2 2
directrix x
=  f.t
=?
focus ( ;~, 3) and
= ~. y
3
3
X
( 1,3)
0
X
49. The ellipse with foci (±4, 0) and vertices (±5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c 2
so b2 = a 2
2
c2 ·= 5

2
4

= 9. An equation is ~5
+ Y
9
= 1. 2
.
The asymptote y 10b2 = 16
::::}
= 3x has slope 3, sob = l  ::::} b2

a
3
*
a= 3b and a 2 + b2
= c2 y2
= " and so a2 = 16 
~ = L52 • Thus, an equation is  " 7215
=?
xb: = 1. ?
51 . The center of a hyperbola with foci {0, ±4) is (0, 0), so c= 4 and an equation is y 2 a
= 4,
2
·
{3b? + b2
x2
 8 5 = 1
= 42
=?
5y2 5x2 1, or  72 8
= 1.
53. x 2 = (y  100) has its vertex at (0,100), so one of the vertices ofthe ellipse is (0, 100). Another form of the equation of a
= 4p(y 
parabola is x 2 found at .(0,
b2
__
a2
_
3 9
100) so 4p(y  100)
=
(y 100)
=?
4p
= 1
__
2
401  399
2
3 9
or 25 +
.
x2
(
,== 25. So the equation of th e e ll ipse is /1 + 2
(8y  399? 160,801
55. Directrix x
p = ~.Therefore the shared focus is 3 9
8
x2
::::}
~ ) so 2c = ~ 0 ::::} c = ~ and the center of the ellipse is (0, ~ ). So a= 100 3~ 9 = 4 ~ 1 and
c2
3 9
=4
=?
y  839!))2 a2
=1
=?
=1
·
d
= 4, so e =
~
=?
ed 4  ..,.. .,  1 +ecosfJ 3+cosfJ'
r 
3 57. (a) Jf (a, b) lies on the curve, then there is some parameter value t1 such that~ = a and tr ~ =b. l ft 1 = 0, 1 + tl 1 +tl the point is (0, 0), which lies on the line y 3(1/tl )
3ti
= x.
lft1 =I 0, then the point corresponding tot=
3(1/td
x = 1 + ( 1 /t ) 3 = t~ + 1 = b, Y = 1 + ( 1 /h) 3 = 1
tt3tl+ 1 = a.
So (b, a) also lies on the curve. [Another way to see
this is to do part (e) first; the result is immediate.] The curve intersects the line y = x when
t=
t
2
::::}
..!. is given by tl
~3 1+ t
2
=
3 t 1 +t3
t = 0 or. 1, so the points are (0, 0) and ( ~, ~).
(1+t3)2
)
horizontal tangents at (0, D) and (
(0, D) and ( ?'4,
t=Dort= 3'2,so thereare y.r.
V2', ?'4). Using the symmetry from part (a), we see that there are vertical tangents at
.v2).
(c) Notice that as t > 1 +,we have x> oo andy . y  {x  1) = y + x + 1 =
oo. As t> 1, we have x . oo andy
>
3t+3t 2 +{1+t3)
1+ t 3
3
=
{t+1) = 1 +t3
{t+1)
>
 oo. Also
2
t 2 t+1
>
0 as t
>
1. Soy = x  1 is a
slant asymptote. .d) dx = (I+ t 3 )(3) 3t(3t 2 ) = 3  6t3 dy 6t 3t4 dy dyjdt t(2  t 3) ( dt (1 + t3)2 (1 + t3)2 and from part (b) we have dt = (1 + t3)2. So dx = dxjdt = 1 2t3 .
y
So the curve is concave upward there and has a minimum point at (D, D) and a maximum point at ( .v2,
W). Using this together with the
X
information from parts (a), (b), and (c), we sketch the curve.
3
3
(e)
3t
x? + y 3 =
( 1+ t3
)
27t3 + 27t6 .(1 + t3)3
+ ( 1 3t2 + t3 )
.
2
'y =x  l
27t3 (1 +t3 ) (1 + t3)3
3
3xy = 3C!\3 )C~t3 ) = ( 1 ~~3 ) 2 ,sox 3 +y3 =3xy. (f) We start with the equation from part (e) and substitute x = r cos 0, y = r sin 0. Then x 3 + y 3 = 3xy r 3 cos3
0+
s in3
1'3
e ~ 3r2 cos e sin e. For
) 3(_1 cosO · cosO
r
=1 0, this gives
r=
=>
.
3 ~OS e s~: Dividing numerator and denominator cos O+ sm 0
sinO
.
by cos3 0, we obtam r =
. sm3 e 1 + cos3 0
3sec0 tan B 1 + tan 3
e.
(g) The loop corresponds to 0 E (D, ~).so its area is 
{"/
2
r
A}0
2
r / (l+ sec e tan e 0  ~ {""' u du tan 0)2d 2} {1+u 2
(3sec0 tan0)  ~ 1+tan3 0 d02}0 2
.! ("/ 2
2d02}0
2
2
2
2
3
0
3)
[let u = tan OJ
= lim l!.[~(l+u3)l]b =;! b+oo 2
3
a
2
(h) By symmetry, the area between the foUum and the line y = x  1 is equal to the enclosed area in the third quadrant, plus twice the enclosed area in the fourth quadrant. The area in the third quadrant is~ . and since y = x  1
r sine =  r cos 0 1 1 /_"'/ 2  rr/2
4 [(

=> r =  . e
. () 1 () ) sm + cos
!;JU
2
1 e' the area in the fourth quadrant is +cos
()  (3sec0 tan0) 1 +tan3
2 ]
. 1 + 2 (1) = 3 . d() CAS 1.. Therefore, the total area IS 2 2 2 2
=
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www.elsolucionario.net
=>
0
PROBLEMS PLUS
1. x =
l
t
cos u   du, y =
1
dx = 0 dt
lt
tL
sin u dx cost dy sin t .   du, so by FTCl, we have dt = t and dt =  t  . Vertical tangent lines occur when
1
tL
cost = 0. The parameter value corresponding to (x, y) = {0, 0) is t = 1, so tl1e nearest vertical tangent .
{:}
occurs when t =
~.
Therefore, tl1e arc length between these points is
(dx)
7r/2
L=
1
dt
1
2
2
+ (dy) dt
dt=
11r/Z
2
cos
1
t
+
t2
. 2t
~dt = t2
j '1f/Zdt ~ [ lnt]7f12 = ln :!!. 1
t
1
2
3. ln. terms ofx andy, we have x =?'cosO= (1 + csin B)cosB =cos O+ csinBcosB =cosO + ~csin2B and 2
y = rsinB = (1 + csinO)sinB =sinO+ csin B. Now  1 $sinO $ 1
=> 1 $sinO t csin2 0 $ 1 + c $2, so
1 $ y $ 2. Furthermore, y = 2 when c = 1 and B = ~ , while y = 1 for c = 0 and 0 =
3 ; .
Therefore, we need a viewing.
rectangle with  1 $ y $ 2. To find the xvalues, iook at the equation x =cos B + ~ csin 20 and use the fact that sin 20 ~ 0 for 0 $ B $ ~ and sin 28 $ 0 for  ~ $ B $ 0. [Because r =::, 1 + csin 0 is symmetric about the yaxis, we only need to consider

~
S 8. $
~ .)
So for 
~
$ B .$ 0, x has a maximum value when c = 0 and then x = cos 0 has a maximum value
of 1 at B = 0. Thus, the maximum value of x must occur on [0, ~] with c = 1. Then x = cos 0 + ~ sin 20 ~~ = sinB +cos 2B= sinB + 12 sin2 9
[but sin B #  1 for 0 $ 9 $
H
=?
If sin 9 = ~, then B =
=>
j~ = (2sin91)(sin9+1) = 0when sin9 =  1or~
*
and
2.1
c=t
x = cos {f + ~sin 3 = ~V3. Thus, the maximum value of xis ~V3, and, by symmetry, the minimum value is  ~ V3. Therefore, the smallest viewing rectangle that contains every member of the famil y of polar curves
r = 1 + csinB, where 0 $ c $ 1, is [~V3, ~V3] x [1 , 2].
5. Without Joss of generality, assume the hyperbola has equation :: 
~: = 1. Use implicit differentiation to get
2 2 :  Yb2y' = 0, soy' = b:x. The tangent line at the point (c, d) on the hyperbola has equation y d = b:d c (x  c). a a y . a ' 2
The tangent line intersects the asymptote y = !!. x when .!!.x  d = bzdc (x c) a a a
=> abdx a 2 d 2 = b2 cx  b2 c 2 =>
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43
44
D
CHAPTER 10 PROBLEMS PLUS 2 2
=>
X =
2 2
a d b c b(ad _be)
ad + bc d h . bad +be b an t e yva 1Ue IS  b
=
a
ad 'tbe
=  a
·
. .1ar1y, t he tangent 1.me mtersects . b at (be bad, aad  be ) . Th e m1.d pomt . oftl1ese mtersect1 . .on pomts . 1 .s S1m1 y = ax 
bead) (ad+be + adb e)) = (12bc 12ad) . a2b , 2;;:= ( c, d), the pomt of tangency.
1 (ad+bc 1 a( 2  b + b • 2  Note: Jf y
=·o, then at (±a, 0), the tangent line is x = ± a, and the points of intetsection are clearly equidistant from the point
of tangency.
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INFINITE SEQUENCES AND SERIES
11 . D 11.1
Sequences
1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers, (b) The terms a,. approach 8 as n becomes large. In fact, we can make an as close to 8 as we like by taking n sufficiently large. (c) The terms an become large as n becomes large. In fact, we can make an as large as we like by taking n sufficiently large.
2n
. {
2 4 6 8 10 } 1 + 1' 4 + 1 ' 9 + 1' 16 + 1' 25 + 1' . . .
3. an
= n2 + 1, so the sequence IS
5. an
=
(1)n l . . { 1 1 1 1 1 } 5n , so the sequence IS 51' 52' 53 ' 54' 55 '...
7. ~
=
} 1 . {1 1 1 1 1 (n + 1)!, so the sequence IS 2!'"3!' 4f•5!' 6!'...
9. a 1
= 1, a,,+1 =
a 3 = 5a2  3
5an 3.
= 5(2) 
{1
= 5' 
= 5a3 
3 = 5(7)  3
{
4 3 8 5 } 1, 5' 5' 17' 13 ' . . . .
1 1 1 1 } · 25' 125 '  625' 3125'... .
{11 1
1
1
= 2' 6' 24 , 120, 720 '...
Each term is defined in terms of the preceding term.
3 = 7. a4
=
= 32.
as
= 5a4 
}
a 2 = 5a1
. 
3 = 5(1) 3 = 2.
3 = 5(32)  3 = 157.
The sequence is {1, 2, 7, 32, 157, ... }.
Q.4
a5
= 1:
4
a
4
=:' 1 ~;/7 = ~
,The sequence is {2,
a3 = = 1 +2/52/5 1 + a3
1 2n1
an=    .
15. {  3, 2, 1, ~, ;~, ... }. The first term is 3 and each term is  ~ times the preceding one, so a,. = 3 ( ~
wegetan
2
7•
~. ~. ~' ~ •... }.
The denominator of the nth term is the nth positive odd integer, so
17. { ~, ~,%, 1: , 25~, ... }.
=
The numerator of the nth term is n 2 and its denominator is n
f 1 .
+ 1. Including the alternating signs,
= (1)n+l _n2_ _
®
n+1
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www.elsolucionario.net
45
46
D
19.
3n an= 1+ 6n
n
'
1
0.4286
2
0.4615
3 4
0.4737
5
0.4839 0.4865
21 .
10
5
n
It appears that lim an = 0.5. n+oo
lim n+oo
~ = lim 1 + 6n
n+oo
(3n)jn (1 + 6n)jn
=
lim n+oo
= ~ = .!
3 1/n + 6
6
2
0.4909 0.4918
.....
an= 1 + (U~
n
1 2
0.5000
3
1.2500 0.8750
4
1.0625
5 6 7
0.9688 1.0156
8 9
1.0039 0.9980 1.0010
0
10
5
n
I
It appears that lim a,. = 1.
I
nco
0.9922
10
=1
.
0
0.4884 0.4898
8 9 10
.........
0.5 0.4
0.4800
6 7
23. an
INFINITE SEQUENCES AND SERIES
CHAPTER 11
lim (1+( ~f) =lim 1+ lim ( ~f=1+0 = 1 since
n oo
ntoo
noo
""
(0.2)", so lim a,. = 1  0 = 1 by (9) . Converges n  oo
3 + 5n2 (3 + 5n2)jn2 5 + 3/n2 5+ 0 25. a,, = n + n 2 = (n + n 2)/n2 = 1 + l /n, so a,, + i + 0 = 5 as n + oo.
Converges
27.. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write lim an= lim e 1 fn = e1imnoo( l /nl = e0 = 1. Converges
,.,_ex:>
noo
2mr h lim' b 1' (2mr)/n lim 271' 271' 71' s· . . .,. b 29. Ifb n =  , t en n = 1m ( )/ = / =  =  . mce tan IS contmuous at 4 , y 1 + 8n n+oo n+oo 1 + 8n n n>oo 1 n + 8 8 4 Theorem 7, lim tan ( n >oo
31. an =
2 2 n1r ) n 71' ) = tan ( lim 1 + 8n n+oo 1 + 8n n2 j,;r;J
n2
vn3 + 4n = ../n3 + 4n/v'n3 =
=tan~ 4
= 1.
Converges
.,jii
Jl + 4jn2 , so an + oo as n+ oo since
lim
.,jii = oo and
n>OO
lim j l + 4/ n 2 = 1. Diverges
1\+00
33. lim
n+oo
Jan! =
lim
n+oo
I(%"I = 21 2vn
lim
n+oo
1 1 =  (0) = 0, so lim an = 0 by (6). Converges n 112 noo 2
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SECTION 11.1
35. a,. = cos( n /2).
SEQUENCES
0
47
This sequence diverges s ince the tenns don 't approach any particular real number as n> oo.
The terms take on val4es between  1 and I. 1 _ (2n  1)! = (2n  1}! · = · ___. 0 as n> oo. 37 · an  (2n + 1)! (2n + 1}(2n}(2n 1)! (2n + 1}(2n) 1+e
2n
' >
en e
41. an = n 2 e
43. 0
< 
n
2
n = . Smce en •
2
x2
•
lim 
X+00 e X
cos n < _..!:.._ · [since 0 2n  2"
. ( 1/ n ) 45. an = n sm
n
0 as n> oo because 1 + e 2 n
>
C onverges
1 and e"  en + oo. Converges
2x 11 2 H = lim  = lim  = 0, it follows from Theorem 3 that lim an= 0 . Converges XtCO eX XtOO e:t noo
82 1 :s; cos2 n :s; 1], so since n>oe> lim = 0, { 'c 0 n n } converges to 0 by the Squeeze Theorem. " ·
2
2
1.tm sin t / n n) . s·mce lim sin(1/x) / = sin(1/ = to+ x ..... oo 1 X t 1
[w here t = 1/ x ]
= 1, .tt .,•OIIows ·from Theorem 3
that {an } converges to I .
= lim x > oo
lim
!too
( 2)"' = 1+ 
X
.
lim e 1n Y = e 2 , so by Theorem 3, lim
:roo
n +oo
2 1 + 2/x
( 2)" = 1+ n
2
=2
::::}
e 2 • Converges
2
2 2 + 1/n 49. an = In(2n + 1) ln(n + 1) =In ( 2n n 2 ++ 1) = In .(2 + / n ) + In 2 as n > oo. 1 1 1 2
Converges
51. a.,, = arct an(ln n). Let f(x) = arctan(ln x). Then lim f(x) = ~since In x + oo as x> oo and arctan is continuous. x  oo
Thus, lim a ,= n+oo
lim f(n) =
n +oo
~
Converges
53. {0, 1, 0 , 0, 1, 0, 0, 0, 1, ... } diverges since the sequence takes on only two values, 0 and I , and never stays arbitrarily close to either one (or any other value) for n sufficiently large. n! 55. an= 2 n
57.
1
2
3
(n 1)
= 2 · 2 · 2 · · · · ·  2
n 1 n · 2 2': 2 · 2
2~~
(for n
> 1]
=
n
4> oo as n+ oo, so {an} diverges.
From the graph, it appears.that the sequence converges to 1. { ( 2/e)n} converges to 0 by (7), and hence {1 + ( 2/e)" }
.. . . .. . .
converges to 1 + 0
= 1.
0 '~~~~ 21
®
2012 Ccng;,gc Lc011T1ing. AU RighiS R.escn·l.d. May not be scanned, copied. or duplicated. or posted to a publicly accessible website, in whole or in part.
www.elsolucionario.net
I
48
D
CHAPTER 11
'
INFINITE SEQUENCES AND SERIES
From the graph, it appears that the sequence converges to t.
59.
As n+ oo,
3/ n 2 +2
.. ...
8 + 1/ n
=>
so lim an =~ · n +oo
0 '~~"'./ 21 2
61.
2.~..
..
From the. graph, it appears 'that the sequence {an} ,;_ {
~ ~o~: } is
divergent, since it osci llates between 1 and 1 (approximately). To n2
, then prove this, suppose that {a,.} converges to L. If bn =  2
1+n
..
{bn} converges to 1, and lim abn n+oo
n
=!::.1 = L. But an bn
= cosn, so
lim abn does not exist. This contradiction shows that {an} diverges.
2
oon
•
From the graph, it appears that the sequence approaches 0.
63.
O < a,.= : at = 1060, a2 = 1123.60, a3 = 1191.02, a4 = 1262.48, and as = 1338.23.
(b) lim a,. = 1000 lim (1.06)", so the sequence diverges by (9) with r = 1.06 > 1. n  oo
n+oo
67. (a) We are given that the init!al population is 5000, so Po = 5000. The number of catfish increases by 8% per month and is
decreased by 300 per month, so Pt P,.
= 1.08Pnt 
= Po + 8%Po 
69.
lim xr"'
300, and so on. Thus,
= 5000, we get P 1 = 5100, P 2 = 520.8, P 3 = 5325 (rounding any portion of a
= 5451, P5 = 5587, and P6 = 5734, which is lhe number of catfish in the pond after six monlhs.
If lrl ~ 1, then {r"} diverges by (9), so {nrn} diverges also, since lnr''l = n lr"l :C  t OO
= l.08P1 
300.
(b) Using the recursive formula with Po catfish), P,.
300 = 1.08Po  300, H
= :coo lim =..._ :lb ~ ( r:Z: X+00

In\
T T%
= lim ___!__I ."' :::: 0, so lim nr" %+00 nT noo
~
lr''l· If lrl
p=
> 1.
y
P
=>
p = 0 or p = b  a.
since 1 + Pn
(~)Po. P2 < ( ~ )P1 _< ( ~
so lim Pn $ lim
a: +f3
 => => a : = 2
= 0, we have Pl 
po
bpo
= a +Po
po =
Now we suppose the assertion is true for n
< b  a, then Pn < b  a arid Pn+l > Pn · Po(b  a a+po
Po) > 0 since Po < b a. So Pl > po. ·
= k, that is, Pk < b a and Pk+l > Pk · Then
a(ba  Pk)  b _ a_ ~ __ a(b  a)+ bp~.; ap~.;  bpk __ __.!.. _ ___...!~ b a  Pk+l a+pk a+pk a + pk bpk+l  Pk+l + Pk+l
Pk+l
< b
Pk+2
> ·Pk+l· Thus, the assertion is true for n = k + 1.
a. And Pk+2  Pk+l =
a
A similar proof by induction shows that ifpo
= Pk+l(b a 
Pk+l) a+ Pk+l
> 0 because p~;
0 smce Pk+l < b  a. Therefore,
It is therefore true for all n by mathematical induction.
> b a, then Pn > b a and {Pn}
is decreasing.
In either case the sequence {Pn } is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem.
rt then follows from part (a) that lim Pn = b a. noo
11.2 Series 1. (a) A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers. (b) A series is convergent if the sequence of partial s ums is a convergent sequence. A series is divergent if it is not convergent. (i) 2012 Ccngagc L 2 or x < 2. In that case, the sum of the series is  a 1
~
X
1 r
=
Jrl
1 for all k 2: 3 ['smce ln k• > 1 "k .or .> _ 3] , so ~ LJ lnkd. k 1verges by compan·son w1t'h~ LJ k , wh'h 1c d'1verges because 1t
9. lnk k
k
k=3 •
k =3
is a pseries with p
=:1 ~ 1 (the harmonic series). Thus, f
lnkk diverges since a finite number of terms doesn't affect the
k=l
I
convergence or divergence of a series. 11 .
{fk
+ 4k + 3
Jk3
1.
arctann 7r/2 fior aII n 2: 1, so ~ arctann . 1 hi h . 1. < ""1.2 LJ ~, converges by comparison w1t. h 1r ~ LJ ""1.2• w c converges because 1t is a 2 n=l n n 2 . n n =l n 2 constant times apseries with p
15.
2: 1, so
4'' + n _
3
1
2
= _1 .2 > 1.
4 . 4" = 4 (4)" > 3n"" 3 for all n 2:
00
4n+l
n=l
3" 2
00
1.
00
n~l 4 (4)" 3 = 4 n~l
(4)" 3
is adivergent geometric series (lrl = ~
> 1), so
L::   diverges by the Comparison Test. 17. Use the Lirni~ Comparison Test with an lim an = lim n bn noo Jn 2 + 1
noo
=
1
= JTi2'+T and bn =
1 )1 + (1 fn2)
lim ' n 0. Since the harmonic series
f
n=l
.!. diverges, so does n
1
00
L::l Jn2+1' n= n +1 1 4" . . Companson . TiestwJtlan· .I 19. Use the L1m1t =  + andbn . 1 + 3n
4
= 3"n :
1 +4" . lim= . a.. . 1 +,,3" . 1 +4" ·3" . 1+ . ( 1 +1 ) · 1 =1 > 0 lim = hm  = hm 4" ·  3"  = hm n__.ob bn n__.oo !._ · noo 1 + 3" 4" n__.oo 4" 1 + 3" n.oo 4" .I_ + ~
Since the geometric series 2::: bn =
1+4" 1+4"   >  >  (4" ) 1 + 3"
3n
©
+ 3"
2 3"
~
'
1
r diverges, so does n=l f 11 ++ 34: . Alternatively, use the Comparison Test with
2::: (1
1(4)" . or use the Test for Divergence.
= 
2
3
201 2 Cengoge Le:uning. All R;ghL< Resel'\'ed, May not be SCIUUled, copied, or duplicated, or posted to a publicly acc'cssihlc website, in whole or in pM.
www.elsolucionario.net
64
D
CHAPTER 11
INFINITE SEQUENCES AND SERIES
f+"2
21 . Use the Limit Comparison Test with an = .
3 2
. n 1 Jn+2 = n.oo hm 2n2 + n + 1
an bn
hm 
n.oo
Since
2n +n+ 1
and bn =
vn)
f: ;
[P = ~ > 1], the series
= (15+ + n2n 2 )2 3
3
and bn
4
f: f+"2
=
1
n3
n2 1 ) =   for all n n n+ 1 n+ 1 ·
2
> 0·
also converges.
:
*
l'
+ 2 . 2 0 s· ~ 1 . l )2 = > . mce n'S::l n 3 IS a convergent
= n!...~ (.l;,. +
...
~
1, so
oo
I:
../n4 + 1
n=l n 3 + n 2
diverges by comparison with
n=2n
~) n
2
e n and bn = e": lim abn = lim .(1 noo
n
noo
f: en = n=l f: ·e:. is a convergent geometric series [lrl = ~ < 1], the series n=l f: (1 + ~) n=l n 1 29. Clearly n! = n(n  1)(n  2) · · · (3)(2) > 2 · 2 · 2 · · .. · 2 · 2 = 2'' 1 so..!_ <  . . ' n!  2nl
[lr I = ~
1], so
0. Since
2
en also converges.
E  1
n =1 2 n 1
is a convergent geometric
E \n . converges by the Comparison Test.
n=l
31. Use the Limit Comparison Test with an
.
1 
= f: ~.which diverges because it is a pseries with p = 1 ~ 1.
27. Use the Limit Comparison Test with an = (1 +
series
= /1 = 2
. L.J ~ ( 5 + 2n) a lso converges. = 3 > 1], t he senes n.=l 1 + n 2 2
2(
f: 
n.oo 2 +
4
li n (5 + 2n) l' 5n + 2n 1ln = n.~ ( 1 + n2)2 = n~ (1 + n2)2 · l l (n2 )2
1 n.=ln+1
) 1 +2ln 1ln + 1l n 2
= lim
. n= l 2n + n + 1
23. Use the Limit Compa'rison Test with a,.
. [p psenes
.
(n 3 12 Jn+2)l(n 3 12 n.oo (2n2 + n + 1)ln 2 .
= lim
is a convergent pseries n=l n 12
lim an n .oo bn.
; : n 12
.
sin(1ln)
I1m  = 1rm n.oo bn n.oo
1ln
== sin ( ; )
and bn
= ; . Then I: an and 2: bn are series with positive terms and
. sinO . ~ . . . . _= hm   = 1 > 0. Smce L.J bn IS the divergent harmontc senes, 0 +D 0 n=l
00
2: sin (1 l n) also diverges. n= 1
[Note that we could also use !' Hospital's Rule to ev~luate the limit:
·
. sin(ll x) H • cos(l/x) · ( l/x 2 ) 1 . hm I = %oo lim I = x_____. hmoo cos X =cos O= 1.) 1 X 1 X 2
x  oo
lO
1
33. n~l ~ =
1
00
Rw ~ Tw ~
1
1
.J2 + ..ff7 + v'82 +
1 10
1 1 · · · + .JI0,05I ~ 1.24856. Now ~
0 we know that either both
.! diverges n
[pserics with p = 1]. Therefore, I; an must be
divergent. 45. Yes. Since
I: an is a convergent series with positive terms,
lim an= 0 by Theorem 11.2.6, and I: bn
n ~oo
=I: sin( an) is a
series with positive terms (for large enough n). We have lim bn = lim sin(an) = 1 > 0 by Tbeor~m 2.4.2 noo
an
n +oo
an
[ET Theorem 3.3.2]. Thus, l::bn is also convergent by the Limit Comparison Test.
11.5
Alternating Series
1. (a) An alternating series is a series whose terms are alternately positive and negative.
(b) An alternating series
f; an= n=l f; (l)"
n=l
1
bn, where bn =ian I, converges if O < b,.+l ~ b,. for all nand lim b,,
= 0.
n+oo
(This is the Alternating Series Test.) (c) The error involved in using the partial sum s.,. as an approximation to the total sum s is the remainder R ,. = s  sn and the size of the error is smaller than bn+l; that is, IR~,I ~ bn+l· (This is the Alternating Series Estimation Theor~m.) ©
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www.elsolucionario.net
66
0
INFINITE SEQUENCES AND SERIES
CHAPTER 11
00 2 4 6 8 10 n 2n 2n . 2 2 . 3.   +   +   +···= :E{1)   .Now limb,. = l i m   = lun / =  =fO.Smce 8 7 5 6 9 n=l n+ 4 noo noo n + 4 ntoo 1 + 4 n 1
lim
n+oo
an =I 0 (in fact the limit does not exist), the series diverges by the Test for Divergence.
00
00
n= l
n=l
5. :E a,. = :E (1f 1
2n
1 +
00
1
n=l
2n + 1
= :E (1)" 1 b,.. Now b,. = 
1
> 0, {bn} is decreasing, and
lim b,. = 0, so the noo
series converges by the Alternating Series Test.
7.
E a,.= E( 1)" 32nn + 11 = f (1)"b,.. Now lim b,. =
n=l
n=l
n=l
lim
n+oo
n+oo
~+ 11jnn =
3 2
=f 0. Since
lim n+00
a,. =f 0
(in fact the limit does not exist), the series diverges by the Test for Divergence. 00
9. :E n= l
a,.=
00
00
n=l
n =l
:E ( 1)"e" = :E (1)"bn. Now b,. =
1
n > 0, {bn} is decreasing, and e
lim b,. = 0, so the series converges n+oo
by the Alternating Series Test. n2
11. b,. = n 3
+ 4 > 0 for n 2:: (x 3
1. {bn} is decreasing for n 2:: 2 since
+ 4)(2x) 
x 2 (3x 2 )
(x3 + 4 )2
lim b,. = lim n +00
13. lim
n+oo
nop
_
+ 8 3x3 )
x(2x 3
(x3 + 4 ) 2

x(8 x 3 )
_
 (x 3 + 4 )2 < 0 for x > 2. Also,
1 = 0. Thus, the series 1)"+1 ~converges by the Alternating Series Test. / n/ 1 + 4 n3 n= l n3 + 4
E(
b~ = lim e2 fn = e0 = 1, so ra.oo
lim ( l)n 1 e21" does not exist. Thus, the series
n+oo
f: (1)"
n=l
1 2
e 1" diverges by the
Test for Divergence. 15. a,. = oo
:E
1
+ .,fti
fon
sin(n +.! )1r 1+
n=O
17.
sin(n + ~)1r
(1)"
=
1
+ .,fii,"
Now b,. =
1
1
+ Vn > 0 for n 2:: q, {b,.} is decreasing, and ..~~~ b,. =
0, so the series
converges by the Alternating Series Test.
n~l (  1)" sin(~J
bn
=sin(~) > 0 for n 2:: 2 and sin(~)
2::
sin(
n : 1 ).
and ,.~ sin(~) =
sinO = 0, so the
series converges by the AJternating Series Test.
n"
19. n!
= n·n···· ·n > n => 1· 2 · ·· · · n 
n" n!
l i m  =oo n.+oo
=>
lim n ..... oo
( 1)" " 
n.1
n
·
oo
by the Test for Divergence. 21.
The graph gives us an estimate for the sum of the series
E (O.~)" of 0.55. n.
n=l
bs
(0 8)" =~ : : : 0.000 004, so
1
©
n
does not exist. So the series :E ( 1}" ~, diverges n=l n.
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www.elsolucionario.net
SECTION 11.5
ALTERNATING SERIES
D
67
~ ( 0.8t :.._  ,;.., (  0.8t ' ~57 LJ '
LJ n=1
n.
n.
n=1
~ 0.8 + o.32  o.oss3 + o.oi 706 o.oo2 731 + o.ooo 364 o.ooo 042 ~ o.sso7
Adding b8 to s 7 does t:~ot change the fourth decimal place of 57, so the sum_of the series, correct to four decimal places, is  0.5507. 1 f: ( 1n);+satisfies (i) of the Alternating Series Test because ( ) · n +1 1
23. The series
n=l
=
series is convergent. Now bs
1 56
= 0.000064 > 0.00005 and b6' =
1 66
~
6
0.00002
oo
bn = 0, so
.
E 5n( 1+t 1 converges by the Alternating
n=O
Series Test. To determine absolute convergence, choose an = ~ to get n l'rm' 5n + 1 = 5 > 0 , so ~ 1 .d . b .h L ' . C . ,.. . h the = n+oo ~   1verges y t e mut ompar1son . est w1t n n=l 5n + 1 ' .
an = 1'tm / ( 1/n ) lim b n n~ 1 5n + 1
n~~
hannonic series. Thus, the series
=
7. lim lak+Il .
k~oo
ak
E k ( ~)
E ( 1)" is conditionally convergent.
n=O 5n+ 1 1
(~_)'"+ ] =
1 lim [(k+: ) koo k(~)k
lim k + k too k
1(~) 3
1
= ~3 k~oo lim (1+ .!..) = ~'(1) = k 3
'
'
13 < ·1,sotheseries
is apsolutely convergent by the Ratio Test. since the terms of this series are positive, absolute convergence is the
k
n=1
same as convergence.
4 1 9. lim 1an+ I I = lim [ (l.l)n+l4 . ~J = lim (1.1)n 4 ·= (1.1) lim · n  1,
f: ( 1)n (l.~)n diverges by the Ratio Test. n
n= l
el/n 3 ~
11 . Since 0 ~ 
n
oo
( 1)"el/n.
n=l
n3
2::
e
3
n
(
1)
= e 3 n
and
oo
2:
n=l
· ' 1 is a convergent pseries [p = 3 3 n
00
e l /n n3
> 1], L: n= l
converges, and so
is absolutely convergent.
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www.elsolucionario.net
SECTION 11.6 13.
.
lan+l l
.
[
2
w n+ l
Ji_~ ~ = J2...~ (n + 2) 42n+3 .
{n+1)4 "+ . 10"
1
]
,
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
D
69
. (10 n + 1) 5 . . 00 10" = n~ 42 . n + 2 = 8 < 1, so the senes n~l (n + 1)42~+1
is absolutely convergent by the Ratio Test. Since the.terms of this series are positive, absolute convergence is the same as convergence. 15.
(1)" arctann\ rr/ 2 . oo rr/ 2 rr <  2 , so smce 2:::  2 = 2 2 n n n = l n \
00
2:::
1 2 converges (p = 2
n =l
n
. . oo > 1), the giVen senes 2::: n=l
(1)" arctann
n2
converges absolutely by the Comparison Test. 17.
f: (Inn1)" converges by the Alte~nating Series Test s.ince
..= 2
 1
Inn
>
.!, and since n
1 lim = 0 and 1nn
n oo
{ln1n } is decreasing. Now Inn < n, so
f: .! is the divergent (partial) harmonic series, f: 1 1 diverges by the Comparison Test. Thus, n=2.n nn n= 2
~ (I1n )".IS condi' w twnaIIy convergent. n=2 n 19.
00 . . 00 cos{nrr/ 3) :5 11 and 2::: 11 converges (use the Ratto Test), so the senes 2::: converges absolutely by the n. n= l n. n=l n.1
icos{nrr/ 3)1
n.1
Comparison Test.
. 21 . lim noo
y/i(lJ =
2 2 00 1 < 1, so the senes . 2::: . n + 1 = hm . 1+ + 1/n / = lim 2 1 n 2 n+oo 2n 2 + 1 n oo 2 n =l
2
(
n + 1 ) " IS . absolutely convergent by the 2n 2 + 1
Root Test. 2
23. lim
n..oo
y'jaJ =
lim " (1 +
n... oo
1)"
.!)" n
= lim (1 + n~ oo
.!)" = e>1 n
[by Equation 7.4.9 (or 7.4*.9) [ ET 3.6.6] ],
2
00 so the series L;:
n =l
(
1+ n
diverges by the Root Test.
.
25.
.
100 10 . Ian+ll . I(n + 1) C!1QO"+l · n! . I lim . 100 1) hm   Inn  (n +10 a,. n co (n + 1)! n 0100" .noo n+ 1 n 
noo
. 100 lim  ( 1 + 1 )100 nco n + 1 n
= 0 ·1 = 0 < 1 so the series
oo
nl00100"
n=l
n1
2:::
. is absolutely convergent by the Ratio Test.
27. Use the Ratio Test with the series 1·3 ·5 1 ·3 ·5 ·7 n  1 1 · 3 · 5 · · · · · (2n  1) _ co ,._ 1 1 · 3 · 5 · · · · · (2n  1) 1 ·3 1  3 ! +  5! 7! + .. · +( 1) {2n  1)! +···n~l(  1 ) · (2n  1)! (2n1)! I lim l an+ll = lim ~ (  1)"~1·3·5 .... ·(2n 1)[2(n +1)l]_ a,. noo (2(n+1)  1]! ( 1)" 1 ·1·3·5·····(2n1)
n  oo
=lim ~ (1)(2n+l)(2n 1)! 1 = lim _.!_=0 < 1, n oo (2n + 1)(2n)(2n  1)! noo 2n.. so the given series is ahsolutely convergent and therefore convergent.
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www.elsolucionario.net
INFINITE SEQUENCES AND SERIES
CHAPTER 11
70
D
29 ·
n~l
00
00
2 · 4 · 6 · · · · · (2n) n!
(2 · 1) · (2 · 2) · (2 · 3) · · · ·. (2. n) nl
= n~l
oo
.2nn!
oo
.
.
= .~1 ;;;[ = n~l 2n, wh1ch diverges by the Test for
Divergence since lim 2n = oo. n oo
I a,. I =
.. lim an+l . de fi mt1on, 31 . B y the recurs1ve noo
00
5 > .1, so the series diverges by the Ratio Test.
4
00
n
n=l
a n+l l . lirn  = 11m
I
15n + 11 = 4n + 3
bn I: b"" cosn1r = I: (1 )" 2':., where bn
•
33. The series
n+oo
lim
n400
an
noo
I( l)n+lb~+l + · ( n
> 0 for n ~ 1 and lim bn
n
n=l
1
n ) b  1 n ~
n oo
I= 1.
1m
n ~o oo
b · n n n +1
1 2
= .
. ~ = 12 ( 1) = 12. < 1, so the senes LJ n= l
b~ cos n 1r iS . n
absolutely convergent by the Ratio Test. 3 1 1 3 35. (a) lim 1 / (n + ) 1 = lim n 3 n _,oo 1/n3 n +oo (n + 1) (b) lim
I(n2 ++ 1) · 2n" I=
(c) lim
Iv~ ·( n+1  3
n  oo
n .... oo
(d) lim n .... oo
lim n + 2n
1
n
V:
1
n t oo
n 
1
=
1 (1 + 1/ n) 3
lim n >oo
1 = nlim ( + +oo 2
1 = 3 lim ·;
n
n>oo
n = + 1
= 1.
= 1 .
1  ) 2n
2
3 lim n + oo
Inconclusive
Conclusive (convergent)
v~/ I+Vn
= 3. Conclusive (divergent)
I
.Jn + 1 · 1 +n21 1/n2 + 1 ]  = lim. [ R +1  · 1 + (n + 1) 2 Vn n>oo n 1jn2 + (1 + 1/n)2
I= lim I(nxn+~)' + . · n! I =
37. (a) lim \ an+l
a,.
n >oo
X"
n >oo
lim 1 x1 n +1
n>oo
. = 1. rnconc lus1ve
1 = lx l nlim   = lxl · 0 = 0 < 1, so by the Ratio Test the > oo n + 1
00 x n series I; I converges for all x .
n =O
n.
(b) Since the series of part (a) always convergt N, we have !ani < 2 , so
.
n
2:
!ani converges by
n>N
I;: .;.. In other words, 2:: an converges absoluiely, contradicting the n>N
n
2: an is conditionally convergent. This contradiction shows that 2:: n 2 an diverges .
Remark: The same argument shows that 2: n'~' a .. diverges for any p > 1. (b)
f (n lnl)nn is conditionally convergent. It converges by the Alternating Series Test, but does not converge absolutely .
n =2
·[by the Integral Test, since the function .
00
1 2
dx  x 1n x
00
n~ nan
2
= 00
= n~
lim
t ~oo
2
(
1t
dxx 1n x

2
=
1 f (x) = X ln X
[ t lim ln{ln x )]
t ~ oo
2
00
1
(
= oo]
. Setting an
= (l}n I n for n ~ 2, we find that n n
1)" ln n converges by the Alternating Series Test.
It is easy to find conditionally convergent series "~
is c~ntinuous, positive, and decreasing on [2, oo) and
00
(1t1
n= l
n
2:: an such·that ~na n diverges. Tw~ examples are 2::
1)"1
and
fo , both of which converge by the Alternating Series Test and fai l to converge absolutely because 2:: Ian i is a
pseries with p ©
~
1. In both cases,
2:: nan diverges by the Test for Divergence.
20l2 Ccngage Le(1111 ing. All Rights Reserved. May not be scanned, copied, or duplicated. or posted to a publ ic ly occcssiblc website, in whole or in part.
www.elsolucionario.net
72
D
INFINITE SEQUENCES AND SERIES
CHAPTER 11
11.7 Strategy for Testing Series 1. 1 < 1 = (1)n for all n
n+3n
3n
f
> 1.
3

n=l
(~)n is a convergent geometric series [lrl = t
< 1), so
.
f
n=l
~ n
+3
converges by the Comparison Test.
f
3. lim la..l = lim _2:_ = 1, so lim a n= lim (1}" _2:_ does not exist. Thus, the series ( 1)n __.!:_diverges by n  oo noo n + 2 n 1). Thus, the interval of convergence is [
13.Ifan = (1)" ~ . then 4" ln n
xn
00
x =  4, 2:(1)" n ln 4
1
% , the series n=l 2:: 00
n
312
S· the series
% When x =
is a convergent pseries
S, t].
lim l a n+t l = lim I xn+l . _4" ln n i= E11im 1n n =1.::1 · 1 an n+oo 4n+l ln(n + 1) x" 4 n~oo ln(n + 1) 4
n~oo
00
n
%,so R =
lx l
2:: n= 2
=
< 1
lx l < 4, so R = 4. When
¢:>
00 00 [(1) ( 4)t 1 . 1 1 1 = 2:  . Smce lnn < n for n ~2.  >  and 2::  is tbe " ln 4 1n n n n = 2 n n n =2 1n n
divergent harmonic series (without then = 1 term),
f 1n1n ' is divergent by the Comparison Test. When x = 4, .
n= 2 00 xn 2:: (  1)n n ln 4 n =2 n
15. If an =
1
00
= 2:: (1)" ln n=2
n
I
, which converges by the Alternating Series Test. Thus, I = (  4, 4] .
I . ·1
2
(x  2)"' , then lim . a n+t (x  2)"+1 n + )1 )   = lim ( · ( n2 + 1 n ~oo an n+ oo n +1 2 +1 X 2 "
Ratio Test, the series
oo
2::
n=O
x = 1, the series
(x 2)" n2
+1
converges when lx 21 < 1
I= lx 21 lim .
[R = 1]
n~ oo
¢:>
2
(
n + 1 ) = n +1 2+1
1 < x 2 < 1
n =O
comparison with the p series
f: n~.
[p = 2 > 1]. Thus, the interval of convergence is I = [1, 3].
1
2

 converges by
+1
·
n= l
00
1
By the Ratio Test, the series .~
3"'(x + 4)"
Vn
converges when 3 lx
+ 41
< 1
¢:>
lx
+ 41
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www.elsolucionario.net
21. By the
1 < x < 3. When
¢:>
f: ( 1)" + converges by the Alternating Series Test; when x = 3, the series f: n +1 n
n=O
lx 
l
76
CHAPTER 11
0
INFINITE SEQUENCES AND SERIES
t < x + 4 < t
 ¥ oo
b"
.
n lx
bn+l
n>oo
By the Ratio Test, the series converges when lx
a b < x < a+ b. When lx  al
23. lfa,
= n! (2x 
for all x :f;
n~oo
~ al < 1
I
1). When x =
n=1
t
= (
~
kn + 1 ,...... ex> kn + 2
J
lxl
x 2 x+ 1 ·
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
D
3 = A(x + 1) + B(x  2). Let x = 2 to get A
1 and
=
79
=  1 to get B =  1. Thus
3 _ _1_ _ _ 1_=~( 1 )1 x2 x  2  X  2 X+ 1  2 1 (x/2) 1  ( x) =
f:
n=O
E (=)" _n=O E(x)n
=~
2 n=O 2
[~2 (1:)" 1(1)"']xn = E [( 1t+l n~l]x" 2 n=O 2
We represented f as the sum of two geometric series; the first converges for x E ( 2, 2) and the second converges for ( 1, ~) . Thus, the sum converges for x E (  1, 1) =I.
=
13. (a) f(x )
1
(1+x) 2
d [ L..... ~ (  1)n x " ] =,  dx n=O
= ·d ( 1) dx
1+x
~
=
L (1)"+1nx" 1
n=l
[from Exercise 3]
~
[from Theorem 2(i)] =
L (1)"(n +
n=O
l)x" with R = ~ .
In the last step, note that we decreased the initial value of the summation variable n by 1, and then increased each occurrence ofn in the term by 1 [also note that (1)"+2 = ( 1)" ].
(b)f(x)=
1 1 =1:j_[ ] =1:j_[E(1)"(n+ 1)x"] (1+x) 3 2dx (1+x) 2 2dx n=O ~
[frompart(a)]
~
= ~ L (1)"(n + 1)nxn 1 = ~ L ( 1)n(n + 2)(n + 1)x" with R = 1. n=l n=O x2 1 2 2 1 ~ (c)f(x) = ( +x) 3 =x · (l+x) 3 = x · 2nJ; { 1)''(n+2)(n+1)x" 1 0
= 1: E (1)"{n+2)(n+ 2 n=O
[frompart(b)]
l)x"+~
To write the power series with x " rather ~an ·x"+ 2 , we will decrease each occurrence of n in the term by 2 and increase 1 .CX>
the initial value of the summation variable by 2. This gives us
15. f(x) = ln(5  x) =
~~ = 5x
Putting x = 0, we get C 1 17. We know that 1 + 4X
= ln 5.
=
1  4X
f(x) = {1+x4x)2 = · 4x. for J4xJ
< 1
Jxl
~~ = 1  x/5
2
1:1[f: (::\"] 5
The series converges for Jx/51
1
(
_1: 5
2 ..~ ( 1)"(n)(n 1)x,. with R =
n=O 5}
dx = C
1.
~
< 1 Jx J < 5, so R = 5.
00
) =
4 2 (1 + 4x)
L (4x)". Differentiating, we get
n=O
f;{  4)n+1 (n+1)xn = f;(1)n4n(n+1)xn+l 4 n=O n=O
=X
< ;}, so R = ;} .
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www.elsolucionario.net
f:
E xn+l = ·C..:... x" 5n=o5"(n+ 1) n=tn5n
80
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
1
19. By Example 5, (
1 X
)2 =
00
L: (n + 1}xn. Thus,
n =O
1+ X _ 1 X _ ~ ( } ,n ~( n+l !(x ) _ (1x)2  (1 x)2 + (1 x)2  ~o n + 1 x + n~o n + 1}x 00
00
= 2::(n+1}xn+
E
nxn
[make the starting values equal]
n =l
n=O
= 1 + f; [(n + 1) + n ]xn = 1 + f; (2n + 1}xn = f; (2n + 1}xn n= l
The series converges when lx2/ 16l
82
= s1 
x3 162, 83
n= l
with R
= 1.
n=O
< 1 x 2 < 16 lxl < 4, so R = 4. The partial sums are 9 1 = ~. 16
5
x = 82 + 163 , 84 = 83 
x1 164 , 8s
= S4 +
x9 , .... Note that 8 1 corresponds to the first term of the infinite 165
sum, regardless of the value of the summation variable and the value of the exponent.
As n increases, 8n(x) approximates
x)
23. f(x) = In ( 1 + 1 X
better on the interval of convergence, which is (  4, 4}.
= ln(1 + x)  ln(1  x) =
=I L~0(1)nxn + n~o =1
f
(2 + 2x2 + 2x4
xn] dx
+ ·· ·) dx
=1
I
 dx 
1 +X
+
I
dx = 1 X
4 2 3 [(1 x + x  x +x
=I f;
2x2 " dx
=C +
n=O
,
oo
But /(0) = ln t = 0, soC = 0 and we have f(x) =
E

• .. )
1
dx (
X
)+
I
+ (1 + x + x
2
 dx 1 X
+ x 3 +x4 + ... )] dx
2 2 X n+l n=O 2n + 1
f;
zx2n+l
with R n = O 2n + 1
which both diverge by the Limit Comparison Test with b,. =
I
= 1. If x =
E 2
n =O
.!.n .
© 2012 Cc1J8'1g< LCllllling. All Rights Rcscn'Cd. Moy not be: SCOilllCd. copied, or duplieotcd. orpost
2:{ l )"'x5 " dx = C + 2:(1)" x 5n ,.=o n=O
  5 dx = 1+x
+ x) dx also has R
x ln{l
5 )"=f:{1)"xu" = nf;(x =O n= o
1  ( x·5 )
 .. dx = 1 + x~
I=
n=1
x"+3 ( )'
+ x) as well. By Theorem 2, the series for
29 · 1 +1x 5 =
I
oo
+ I: (1)"'
)(
2n j 3
)
82
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
x3
33 . By Example 7' arctan x = x   3
5
1
The senes . 1s . aIternatmg, . so 1'f we use t11ree terms, t 11e error ·ts at most (0.2) 7
(0 2? Thus, to five decimal places, arctan0.2 ~ 0.2  · 
3
2
,
1
X Jo (x)
+ XJo(X) +X
2
00
Jo(x)
= n~l 00
= n~l _

00
r~l
(0 2? + · 
( 1)"2n(2n 1)x2" 22"(nl)2
2
J
1;
+ 3~0
= n~O nf ::}
f (x)
00
0.000 002.
0.197 40.
(1t2nx2 "' 22"(n!)2
00
+ 1~1
( 1)"2nx2'' 22"(n!J2
oo
+ ,~1
( 1t2nx2" 22"(n!)2
00
+ n~l
]
+ n~O oo
+ ,~1 00
+ ,~1
( 1)"x 2"+2 22" (n!)2 ( 1t 1 x 2" 22n 2 [(n 1)!] 2 (1)"(1 ) 1 22n 2x 2" 2 2 "(n!)2
x 2" ,
2
 2n + 2n 4n 22" (n.1)2
00
2n
]
X
=0
~ 0.000062, it follows from The Alternating Series Estimation Theorem that, correct to three decimal places,
Jo (x) dx ~ 1 
37. (a) f(x)
~
2 2
Ln=l
.i
7
[2n(2n  1) + 2n  2 n 22 "(n!)2
= ~ ( 1)" [4n
16 28
~
( 1)"2n(2n1)x2 " 22"(n!) 2
n~l
1 0
5
( 1)"2n(2n 1)x2 " 22n(n!)2
= ~ ( 1)"
Since
(0.2? (0.2) 5 (0.2) 7 3 + 5  7 + ... .
+ x5  x7 + · ·' · so arctan 0. 2 =. 02
Xn
I
~ 0.920. 00
nxn l
00
oo
x.,."
n=O
n
Xn1
I:  1 =
= •~1 ~ = •~1 (n 1)!
f( x)
(b) By Theorem 9.4.2, the only solution to the differential equation df(x)/d.'C = f(:c) is f(x) = Ke" , but f(O) K
= 1 and f(x) = e".
Or: We could solve the equation df(x) / dx
39. If a,..
= 1, so
= nXn2' then by the Ratio Test, n.oo lim
convergence, so R
x
= 1. When = ± 1,
= f(x.) as a separable differential equation.
Ian+ I= I(X+ an
I
lim
noo
f Ix: I= f n
n= l
" =1
n
n+l )2
1
21 .~ xn
=. lxl n+oo lim ( ....2::_1 ) 2 = lx l < 1. for n+
~which is a convergentpseries (p = 2 > 1), so the interval of n
convergence for f is [1, 1]. By Theorem 2, the radii of convergence off' and f" are both 1, so we need only check the 00
endpoints. f(x)
x"
= I; 2 n=l
n
00
nxn 1
00
x"
=> . f'( x ) = I:  2  = I;   , and this series diverges for x = 1 (harmonic series) n=l n n=O n + 1
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www.elsolucionario.net
SECTION 11.10
TAYLOR AND MACLAURIN SERIES 00
=  1 (Alternating Series Test), so the interval of convergence is [1, 1). !" (x) = I:
and conve rges for x
n=l
at both 1 and  1 (Test for Divergence) since lim
n.oo
•
41. By Example 7, tan 1 x =
6 = tan
have
1)
 1(
7r=~f
(1)"
y'3 n = O (2n + 1)3"
11.10
(1/ v'3) n +l ~( 2n + 1 = n~O 2
1)n
= n~O ( 
y3
=
2n+ 1
00
=2v'3f n=O
1 )"
n+ 1
1, 1).
r.;• we
y3
(1)" 1
3
n.1
1
y3 2n + 1' SO
(1)" . (2n + 1)3"
Taylor and Maclaurin Series J(o)
f (nl (x)
0
2"'
1
1
2"'(ln 2)
ln2
2
2"'(1n 2) 2
(In 2) 2
3
2"'(1n 2) 3
(In2?
4
2"' (In 2) 4
(In 2) 4
11.
n
f (n)(x)
J (n) (0)
0
sinh x
0
1
cosh x
1
2
sinhx
0
3
cosh x
1
4
sinhx
0
n
J Cnl(x) 4 x  3x2 + 1
lim noo
I a,. I= I (n + = (In = < a ..+!
2 lim ) !xi noo n 11
J (o)
0
so sinh x=
if n is odd
I II
r
r
an+
·I
=0
2
< 1
¢>
lxl cosx  1 = .~ (  1)" {2n)! 1
dx = C + 2:::{1) n =l
x cos(x
n
X 2n
2n
·(
.
cos x X
1
=
oo
2n  1
n=l
(2n) !
2: {1)" .::_____ =>
•
) , wtth R = oo. 2n 1.
@ 20 12 Ccngagc Lcunlin g. All Rights Ri:st:rvc:J. May not be scwmct.l. cupicd, or duplicated, or pO!'Itr.:d to a publicly IU:cc~ siblc website, in whole or in pan.
www.elsolucionario.net
X
wG
88
0
INFINITE SEQUENCES AND SERIES
CHAPTER 11 •
oo
x2n+l
2: ( 1)" 2
51 . arctanx =
n+1
n=O
I
for lxl
5
(1 /2) 1.5 
(1/ 2r
~
x2n+ 4
n =O
2n+ 1
for Jxl < 1 and
x2n+5
oo
3
x arctanx dx
00
< 1, so x3 arctanx = 2: (1)"  
= C + ,..;;:o ( 1)"' (2n + 1)( 2n + 5)' Since ~ < 1, we have
9
11
(1/ 2) · · {1/ 2) +~ ~ 0.0059 and subtracting T i l ~ 6.3 x 10 6 does not affect the fourth decimal place,
so J0112 x 3 arctan x dx ~ 0.0059 by the Alternating Series Estimation Theorem.' 53. v'1 + x4 = (1 + .x 4 ) 112 =
00
(1/ 2) (x 4)", so n
2: n =O
I
}1 + x 4 dx = C + ,
L:
oo n =O
( 1/ 2) x4n+1  and hence, since 0.4 < 1, n 4n + .1
we have I=
1
0.4
o
00
} 1+x4 dx = E
n=O
(1/ 2) (0.4) 4n+l n 4n + 1
(o.4) 1 ~ (o.4) 5 H4) (o.4) 9 Ht)(~) (o.4) 13 H4)(~)(~) (o.4) 17 =( 1) o r + l !  5  +2,9+ 3! ~+ 4! rr +· ·· (0.4) 5
(0.4) 1 3
(0.4) 0
= 0.4 + w  ;:;2 + 208 Now
(O~~)o ~ 3.6 x 10
6
n, so its Taylor series at a is p(x) = L; ~(x  a)i. t.
i =O
Put x a.= 1, so that x = a+ 1. Then p(a + 1) =
n.
p(i) (a)
i=O
t.
2:  .1  . n
This is true for any a, so replace a by x : p(x
p(i)(x)
+ 1) = iL: .1 =O t.
73. Assume that lf" (x)l ~ M , so /" (x) ~ M for a ~ x ~a + d. Now fa"' !'11 (t) dt ~ 1
f"(x)  J"(a) ~ M (x  a)
=>
f"(x) ~ J"(a) + M(x  a). T hus,
J'(x)  f '(a) ~ J"(a)(x  a)+ ~M(x a )2
J: M dt
=>
I: J"(t) dt ~ I: [!"(a)+ M(t a)] dt
=> J'(x) :::; /'(a) + J"(a)(x  a)+ ~M(x a) 2 =>
.r: f(t )dt ~ .r: [!' (a) + J"(a) (t  a)+ tM(t  a?] dt
=>
f(x)  f(a) :::; · f' (a)(x a) + tf"(a)(x  a? + iM(x a) 3 . So f(x)  f(a)  J'(a)(x a) ~f"(a)(x  a) 2 ~ ~ M(x  a)3 . But R 2(x) = f (x) T2(x) = f (x)  f(a)  J'(a)(x a) ~f"(a) (x a) 2 , so R2(x) :::; !M(x  a) 3 • A sim ilar argument using f"'(x ) ~ M shows that R2(x) ~  t M(x a) 3 . So IR2 (x2)l ::; tM lx  a j3 . Although we have assumed that x > a, a similar calculation shows that.this inequality is a lso true if x
< a.
© 2012 Ccngngc Lc:uning. All Righls Rcscr'\'cd. May nol be scanned, copied, or duplicated, or posted lo • publicly accessible wcbsilc. in whole or in pan.
www.elsolucionario.net
=>
89
90
0
CHAPTER 11
75. (a) g(x)
= :L; 00
n~
(1 + x) g'(x)
INFINITE SEQUENCES AND SERIES
(k)n x"'
= (1 +
00 => g'(x) = :L;
n~
00 :L;
x)
n=1
(k)n nxn
1
(k) nx"n
1
,
so
(k)n nx"' + :L; (k)n nxn
= :L;
1
00
n=1
00
n= 1
. .
I
=
f: (n+k1) (n + 1)xn + f: (k)nxn n
n=O 00
_

n~o (n + 1) 00
= n~O
[
n=O
Replace n with n + 1 ] in the first series
k(k1)(k2) .. ·(kn+1)(kn) n (n + 1)! x +
00
(n + l )k(k  '1)(k  2) ... (k n + 1)
(n+ 1)!
= k n~o k(k 
1)(k
.
[
n~O (n)
k(k 1)(k  2)···(kn+ 1) ] ,, nl
n
((k n) _+n]X
2l!... (k n + 1) x" = k fo (~) x" = kg(x)
Thus, g'(x) = k g(x).
1 +x
(b) h(x)
= (1 + x)k g(x)
=>
h'(x) =  k(l + x)  k 1g{x) + (1 + x) k g'(x)
[Product Rule]
= k(l + x)i.: 1 g(x) + (1 + x)  k kg(x)
[from part (a)]
1 +x
= k(l + x)k 1 g(x) +
k{1 + x)"' 1 g(x)
(c) From part (b) we see that h(x) must be constant for x E (1 , 1), so h(x) Thus, h{x) = 1 = (1 + x)k g{x)
11.11
=0
= h(O) = 1 for x E ( 1, 1).
g(x) = {1 + x)k for x E ( 1, 1).
Applications of Taylor Polynomials
.
1. (a)
n
J (x)
0
cosx
1
sinx
0
2
cosx
 1
3
sinx
0
4
cosx
1
5
sin x
0
6
cosx
1
t
lx l 5 0.2
=?
lxl 3
5 (0.2) 3.
f(3 ) (x) is an odd function and it is increasing on (0, 0.2) since secx and tanx are i"ncreasing on [0, 0. 2) ,
I
so J C3>(x) l 5 J (o.2)
3
~ 1.085 158892. Thus, IR2(x) l 5 J< >i~· 2) (0.2) 3 ~ 0.001447.
0.0004
(c)
· From the graph ofiR2(x)l
= isec x 
T2(x) l, it seems that the
error is less than 0.000 339 on [0.2, 0.2). 0
19.
J 1.
n=l 77,1'
3, then lim an = 0 and lim n  oo
Sn
n t ex>
= 3.
5. (a) Test for Divergence: If lim a,, does not exist or if lim an n+oo
n + oo
=I 0, then the series 2:"':.. an is divergent. n 1
(b) Imegral Test: Suppose f is a continuous, positive, decreasing fun ction on
:2:;::'=1 an (i) If
f1
00
(ii) If f 1
00
is convergent if and only if the improper integral
f1
00
[1, oo) and let an= f(n). Then the series
f(x) dx is convergent. In other words:
f(x) dx is convergent, then 2:;::'=1 a, is convergent. f( x) dx is divergent, then
I::;"=1 a,.
(c) Comparison Test: Suppose that E an and
is divergent.
2: b,. are series with positive terms.
(i) If E b,. is convergent and an ::=; bn for all n, then (ii) If E br. is divergent and an ~ bn for a ll n, then
E a,. is also convergent.
2: an is a lso divergent.
(d) Limit Comparison Test: Suppose that E a,. and E bn are series with positive tenns. If lim (a ,/bn) n. oo
= c, where c is a
finite number and c > 0, then either both series converge o r both diverge. (e) Alternating Series Test: Ifthe alternating series :2:;::'= 1 (  1)n 1 bn
= b1
 b2
+ b:3 
b4
+ bo 
ba
+ ···
[b,.
satisfies (i) bn+l :::; bn for all n and (ii) lim b,. = 0, then the series is convergent. n 1 ~r n+oo lim ~ = oo, then the series 2::;:'= 1 an is divergent.
(iii) If lim
~
n co
n.oo
11.+ 00
L
= 1, the Root Test is inconclusive.
6. (a) A series L an is calledab~olutely convergent if the series of absolute values
2:: Iani is convergent.
(b) If a series L an is absolutely convergent, then it is convergent. (c) A series L:al' is called conditionally convergent if it is convergent but not absolutely convergent. 7. (a) Use (3) in Section 11.3. (b) See Example 5 in Section 11.4. (c) By adding terms until you reach the desired accuracy given by the Alternating Series Estimation Theorem. 00
8. (a)
2:: en(x 
a)"
n=O 00
(b) Given the power series
L
en(x a)", the radius of convergence is:
n=O
(i) 0 if the series converges only when x
=a
(ii) oo if the series converges (or all x, or (iii) a positive number R such that the series converges if lx 
al < Rand diverges if lx al > R.
(c) The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. Corresponding to the cases in part (b), the interval of convergence is: (i) the single point {a}, (ii) all real numbers, that is, the reaL number line ( oo, oo ), or (iii) an interval with endpoints a  R and a + R which can contain neither, either, or both of the endpoints. In this case, we must test the series for ·convergence at each endpoint to determine the interval of convergence. 9. (a), (b) See Theorem 11 .9.2. 10. (a) Tn(x)
(b)
n
f(i)(a)
i =O
~.
= L  .1 
oo j (n){ ) I: __a_
(x  a)'
(x  a)"
nl
n=O
®
2012 Cengnge Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ton publicly ncccssiblc website, in wbole or in part.
www.elsolucionario.net
CHAPTER 11 REVIEW
(c)
oo f (n) (O)
L  1 n.
n=O
xn
0
99
[a= 0 in part (b)]
(d) See Theorem 11.10.8. (e) See Tay.lor's Inequality ( 11.10.9).
11 . (a)(f) See Table I on page 786 [ ET 762]. 12. See the binomial series ( 11 .10.17) for the expansion. The radius of convergence for the binomial series is 1.
TRUEFALSE QUIZ 1. False.
See Note 2 after Theorem I 1.2.6.
3. True.
If lim an
5. False.
For example, take en = (  1)" j (n6n) .
n  oo
= L, then as n 7+ 00, 2n + 1 + oo, so a2n + l
. . I_+_ an I= lim . I(n +1 1) a,. 1
7. False, smce hm
noo
n  oo
3
.· · n3 ~ 1 = lun 1
n+oo
+
L.
I(n + 1) . · 1j1/n31  = n n3
3
3
. hm n oo
1 (1 + 1/ n) 3
= 1.
See the note after Example 2 in Section 11.4.
9. False. 11. True.
See (9) in Section II . I .
13. True.
By Theorem 11.10.5 the coefficient of x 3 is ~
111 /
(0)
= 31 => !'" (0) = 2.
Or: Use Theorem 11 .9.2 to differentiate f three times. For example, let a..
15. False.
= bn = (1r.
[2:: (1)" a,.
17. True by Theorem 11.6.3.
+ 0.9(0.1? + 0.9(0.1) 3 + ·· · =
f: (0.9)(0.1)nl = __2.:Q_ = 1 by the formula 1  0.1
n=l
for the sum of a geometric series [S = a 1 / (1 ; r)] with ratio r satisfying lrl 21 . True.
= 1, so {a,. b.. } is convergent.
is absolutely convergent and hence convergent.]
0.99999 . . . = 0.9 + 0.9(0.1) 1
19. True.
Then {an} and {bn} are divergent, but a,.b,.
< 1.
A finite number of terms doesn't affect convergence or divergence of a series.
EXERCISES 2 + n3
1. { 1 + 2n 3
}
.
n +00
3
3. lim a,. n+OC
.
converges smce lim
= n+ lim ~ = lim oo 1 + n n+ oo
2 + n3
1+
2n 3
/ ~
1 n
·
.
= lim n+00
+1 = 11n 3 + 2
2/ n 3
1
 . 2
+ 1 = oo, so .the sequence diverges.
© 2012 C
00
~ =I: {1f'x"+2 withR= 1. 1+x
n=O
© 2012 Ccngnge Learning. All RighiS R=n.:d. May not be S f( x)= l: cn(xt = 2:( 1)"c.,x" n =O
n=O
n=O
(a) Iff is an odd function, then f(  x) =  f(x)
=>
00
00
u=O
n=O
2: ( 1)"cnx" = 2:  c,,x". The coefficients of any power series
are uniquely determined (by Theorem 11.10.5), so ( 1)" Cn =  en.
If n is even, then (  1)" Co
= C2 = C4 =
= 1, so en =
c,.
=> 2en
=0
· · · = 0. 00
(b) If f is even, then f (x) = f(x)
=>
2: {lt
If n is odd, then (  1 )" = 1, so en C3
= C5 =
···
'
00
c,. xn
n=O
that is, Cl =
=> c,. = 0. Thus, all even coefficients are 0, that is,
= 2: Cn x"
=> {1)" Cn = Cn.
n=O
= Cn =>
2c,..
=0
=>
Cn
= 0. Thus, all odd coefficients are 0,
= 0.
@) 2011 Ccngoge Lcorning. All Rigbls Reserved. Moy not be scanned. copied, or 
X
·
2") + cot(x/ ,. 2
[telescoping sum]
, 1
1 =  as n > oo since xf2n > 0 X
is any integer, then
s,. = Iim ( cotx+  1 cot X n+oo
2n
2n
)
= cot x+ 1 X
If x = 0, then all terms in the series are 0, so the sum is 0. 5. (a) At each stage, each side is replaced by four shorter sides, eacb of length
k of the side length at the preceding stage. Writing s 0 and Eo for the number of sides and the length of the side of the initial t1:iangle, we generate the table at right. In general, we have sn = 3 · x = mr,
nan integer. So using the formula for disks (and either .a CAS or sin2 x = ~ (1  cos 2x) and Formula 99 to evaluate the integral),
l+l~~+''r7"""''...,.._t 40
the volume of the nth bead is
(  :r:/10 sm . x )2 dx = rr Jrmr :r:/ 5 sm . 2 x dx v.n = rr Jf"'' (,.l ) .. e (n 1 ) .. e = 2;g,rr (e (nl)lf/5 e mr/5)
 I
(b) The total volume is rr .{000 e :r:/S sin 2 x dx =
f n =l
V,,
= 2~it
f; [e (n1)"/
5 
en"/ 5 ] =
;g;
2
[telescoping sum].
n=l
Another method: If the volume in part (a) has been written as Vn =
2
~g; en"/ 5 (e"/5

1), then we recognize
·
as a geometric series with a= 2~g; (1  e"15 ) and r
©
= e"/ 5 .
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www.elsolucionario.net
f; Vn n=l
108
D
CHAPTER 11 PROBLEMS PLUS
15. If Lis the length of a side of the equilateral triangle, then the area is A= ~L · 4L = 4£2 and so £ 2
=fa A.
Let r be the radius of one of the circles. When there are n rows of circles, the figure shows that L ( y'3 )' 2 n+ 31
L = v'3r+r+(n2)(2r)+r+v'3r=r(2n2+2../3),sor =
1 The number of circles is 1 + 2 + · · · + n = n(n + ) , and so the total area of the circles is 2
An =
=
n(n + 1)
1fT
2
2 n(n+ 1) 2
1f
=
n(n + 1)
· 7r
2
£2 2 4 (n+v'3 1)
4A/J3 n(n+1) 1rA 2 2 4(n+v'3  1) = {n+../31) 2../3
n(n + 1) 1r A = (n+../31) 2 2.;3 1 + 1/n =
=>
An
1r
1r
>   asn >oo
2
[1+(J31)/n) 2v'3
r
2../3
2r
17. As in Section 11.9 we have to integrate the function x"' by integrating series. Writing x"' = (e1"
Maclaurin series for e"', we have x"' integrate this series termbyterm:
= (eln "')"' =
1 ·1
0
f \"(lnx)"dx
=
}0
=
e"' 1" "' =
oo 2:
x"' dx =
11
n=O
withu = (lnx)",dv =x" dx,sodu=
0
n(ln x)" 1 · x
E (x Inn.x) 1
n=O
n
11 1 x"(lnxf dx. We integrate by parts n=O n. 0 oo 2:
I
x"+1
dxand v =   : . n+ 1
1
1
to+ t
= e"' 1""' and using the
n.
n=O
x"(ln xt dx = n.1
"')"'
= E x" (~ x) n . As with power series, we can
lim /. x"(ln xfdx= lim [x"+l (lnx)"] to+ n
.
t x"(Inxf
o _ n _
2r
~L~
1
n+ 1 } 0
+1
1
lim ( ___!!:__ x"(1nx f 1 dx

to+
t
it n + 1
dx
(where !'Hospital's Ru le was used to help evaluate the first limit). Further integration by parts gives
~ 1 n +1 0
1
x"(ln x)k dx =
( }0
1 1
1
x"(ln x)k 1 dx and, combining these steps, we get
1x"(ln x)"dx = (1)"n!11 x"dx= (1)"n! (n+1)"
0
"' d
1
, 0
X
X
=~ L...
n=O
0
..!_ 1 1 "(In ·)"' d n.I
0
X
.
X
X
=
19. By Table 1 in Section 11. 10, tan 1 x =
~ ..!_ ( 1t n! = ~ L...
n=O
n.I ( n
+ 1) n +1
have
6
= tan1
(
1 )
y'3
L... (
n=O
(1)" n
+ 1) " +1
= ~ ( 1)"1 L...
n=l
n" 1
x 2n+l
00
E ( l t 2'   f or JxJ < 1. In particular, for x = n+ 1
n=O
7r
=?
(n+ 1)n+l
·
2
r.;•
v3
·
we
1
00 (1/../3) "+ (1)" i 1 = n~O (  .1t 2n + 1 = n~O ( 1)" 3 y'3 2n + 1' SO 00
6 00 (1)" 00 (1)" ( 00 ' (1)" )  2 3  2 3 1  ../3 n~O (2n + 1)3"  y'3 n~O (2n + 1)3"  y'3 + f1 (2n + 1)3n
·rr  
. =>
E 00
n=1
(1)" 7r =1. (2n + 1)3" 2 y'3
licutcd, or posted to a publicly oeccssiblc website, in whole or in p:tr1.
www.elsolucionario.net
CHAPTER 11
21. Let f (x ) denote the lefthand side of the equation l no solutions oftheequation. Note that / ( x
< 0 are given by x =
x
2
x
+ 2T +
x2 ! 4
x2
= 1  2! +
)
+ 6f + 8f + .. · = 0. lfx;:::: ~· then f(x ) ;::::
x4
4
x4
x3
·
PROBLEMS PLUS
xs
! 
6
!
xs
= 0 are x = 
(i  1rk)
k is a positive integer. 23. Call the series S. We group the terms according to the number of digits in their denominators:
Now in the group gn , since we have 9 choices for each of then digits in the denominator, there are 9n terms.
10 ~_ 1
[except for the first term in gl]. So 9n
00
Now
L: 9 ( 190 ) n  l
is a geometric series witl1 a
= 9 and r
=
< gn · 10~_ 1 = 9( {0 )"'
fa < 1. Therefore, by the Comparison Test,
n=l
S=
oo
oo
n= l
n=l
L: 9n < L:
9
9(w)
n 1
=
9 1 _ 9110
= 90.
Use the Ratio Test to show that the series for u, v, and w have positive radii of convergence (oo in each case), so Theorem 11'.9.2 applies, and hence, we may differentiate each of these series: du 3x 2 6x5  = +
dx
. .
dv
x6
x3
3!
6!
9x8
x2
x5
x8
++··· = 2! ++ 8! +· ·· = w 9! 5!
x9
dw dx
x4
x7
x 10
S imilarly, dx
= 1 + 3! + 6f + 9! + · · · = u, and
So u' = w, v'
= u, and w' = v. Now differentiate the le fthand side of the desired equation: d
dx (u 3
+ v3 + w 3 
3uvw)
2
3
13
3
+ wa 
+ 03 + 03 
4!
+ 7T +
lO!
+ · ·· = v.
= 3u2 u' + 3v2 v' + 3w 2 w'  3(u'vw + uv'w + uvw') = 3u w
u +v
=x+
+ 3v2 u + 3w 2 v 
3(vw 2 + u 2 w
+ u~ 2 ) = 0 =>
3uvw = C. To find the value of the constant C, we put x = 0 in the last equation and get 3(1 · 0 · 0)
= C =>
C
= 1, so u 3 + •v 3 + w3 
109
1 and t~ere are
+ 8f  · · · = cosx. The solutions of cosx =
i  1rk, where k is a positive integer. Thus, tl1e solutions off(x)
Furthermore, each term in 9n is less than
0
3uvw = 1.
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1
•
0 for
2 ,
where
12 12.1
D VECTORS AND THE GEOMETRY OF SPACE ThreeDimensional Coordinate Systems
1. We start at the origin, which has coordinates (0, 0,. 0). First we move 4 units along the positive xaxis, affecting only the xcoordinate, bringing us to the point ( 4, 0, 0). We then move 3 units straight downward, in the negative zdirection. Thus only the zcoordinate is affected, and we arrive at {4, 0,  3). 3. The distance from a point to the yzplane is the absolute value of the xcoordinate of the point. G(2, 4, 6) has the x coordinate with the smallest absolute value, soC is the point closest to the ·y zplane. A(  4, 0, 1) must lie in the xzplane since the distance from A to the xzplane, given by the ycoordinate of A, is 0. 5. The equation x
+ y = 2 represents the set of all points in y=2  x
JR3 whose x andycoordinates have a sum of2, or equivalently where y = 2  x. This is the set
y=2 x,z= O
{(x, 2  x , z) I x E JR, z E R } which is a vertical plane that intersects the xyplane in the line y
=2
x, z
= 0.
X
7. We can find the lengths of the sides of the triangle by using the distance formu la between pairs of vertices:
IPQI = )(7 IQRI =
3)2 + [0  ( 2)]2 + [1  ( 3)] ~
j(l 7)2 + {2  0) 2 + {1  1) 2
IRPI = )(3 
= yf16 + 4 +
16 = 6
= vf36 + 4 + o = .J40 = 2 v'15
1)2 + ( 2  2)2 + ( 3  1) 2
The longest side is QR, but the Pythagorean Theorem is not satisfied:
= vf4+ 16+ 16 = 6
IPQI 2 + IRPI 2
=f:.
IQRI 2 . Thus PQR is not a right
triangle. PQR is isosceles, as two sides have the same length. 9. (a) First we find the distances between points:
IABI = ){3 2)2 + {7 4)2 + (2 IBGI = IACI
2)2
= v'26
){1 3) 2 + (3  7) 2 + [3 (  2)]2 =
= )(1 2)2 + (3  4) 2
v'45 = 3 v'5
+ (3 2) 2 = J3
In order for the points to lie on a straight line, the sum of the two shortest distances,must be equal to the longest distance. Sin
12x
+ 36 + y2 
4y + 4 + z
2
+ 4z + 4 '=>
14x  6y  lOz
= 9.
Thus the set of points is a plane perpendicular to the line segment joining A and B (since this plane must contain the perpendicular bisector of the line segment AB).
43. The sphere x 2 gives (x
2

+ y2 + z 2 =
4x + 4) + (y2
4 has center (0, 0 , 0) and radius 2. Completing squares in x 2 
4y + 4) + (z
2

4z + 4) =  11 + 4 + 4
+ 4 =>

4x + y 2
(x  2) 2
+ (y 

4y
2?
+z2 
4z = 11
+ (z 2) 2 = 1,
so this is the sphere with center (2, 2, 2) and radius 1. The (shortest) distance between the spheres is measured along the line segment connecti ng their centers. The distance between (0, 0, 0) and (2, 2, 2) is
.J(2  0)2 + (2 0)2 + (2  0)2 =
ffi = 2 J3, and subtracting the radius of each circle, the distance between the
spheres is 2 v'3  2  1 = 2 .J3  3.
12.2 Vectors 1. (a) The cost of a theater ticket is a scalar, because it has only magnitude. (b) The current in a river is a vector, because it has both magnitude (the speed of the current) arid direction at any given location. (c) If we assume that the initial path is linear, the initial fli ght path from Houston to Dallas is a vector, because it has both magnitude (distance) and direction. (d) T he population of the world is a scalar, because it has only magnitude. 3. Vectors are equal when they share the same length and direction (but not necessarily location). Using the symmetry,of the .
parallelog~am
®
+
+ +
as a guide, we see that AB = D C, DA
+ +
= CB, DE =
+
__.
EB, and EA
+
= CE.
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SECTION 12.2 VECTORS 5. (a)
un
(b)
w~
uNw
u
(d)
(c)
u
u v ~
(e)
0
v+w
(f)
u
Dw v+u+w
u
:Gv uwv
>
7. Because the taif of dis the midpoint of QR we have QR = 2d, and by .the Triangle Law, a + 2d = b c
=b
 d
=?
2d=ba
= b  ( ~ b  ~a)
=
d=!{ba) = th  ~a. AgainbytheTriangleLawwehavec + d = bso
=?
~ ~ + ~b.
9. a= (3  ( 1), 2  1}=(4, 1}
11.a =(2  (1), 2  3} = (3,1}
y
y
B(3, 2)
0
A( 1, 3)
X
13. a = ·(2  0, 3  .3,1  1} = (2, 0,2}
X
15. (1 , 4} + (6,  2} = (1 + 6, 4 + ( 2)} = (5, 2) y
A(0, 3, 1)
X
X
17. (:~, 0, 1)
+ (0, 8, 0) = (3 + 0, 0 + 8, 1 + 0) = (3, 8, 1)
(3, 0, 1)
X
©
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115
116
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
19. a + b
= (5 + (3) ,12 + (6)) = (2,  18)
2a + 3b = (10, 24) + (9, 18) = (1, 42) lal
= y'52 + (  12) 2 = .jl69 = 13
Ia b l = 1(5 (3),  12 (6))1
= 1(8, 6) 1= y'82 + (  6)2 = v'100 = 10
21. a+ b = (i'+ 2j  3 k ) + ( 2 i  j + 5 k )
2a + 3b lal
= 2 (i +
=
i + j + 2k
2j  3 k ) + 3 (2 i  j + 5 k ) = 2 i + 4j  6 k  6 i  3j + 15 k
=
= y'12 + 22 + (3)2 = v'i4
Ia  b l = l(i+ 2j  3 k)  (2 i j +5 k )l = l3 i + 3 j  8 k l = y'32 +3 2 + (8)2 23. The vector  3 i . . .IS d1rect1on
+ 7j
1
r,:n
v58
25. The vector 8 i  j
has length l3 i
+ ~j I=
y'( 3)2 + 72
= v'82
= v'58, so by Equation 4 the unit vector with the same
(  3.1 + 7 J' ) =  r.;o 3 1. + r.;oJ 7 .·
v58
+ 4 k has length 18 i 
the same direction is ~ (8 i  j
v58
j + 4 k l = y'82 + ( 1) 2 + 4 2 =
+ 4k) = ~ i 
~j
J8I =
9, so by Equation 4 the unit vector with
+ ~ k.
.
y
27.
4 i + j + 9k
From the figure, we sec that t a n 0 =
.j3
 1 = J3 =>
B = 60°.
X
29. From the figure, we see that the xcomponent ofv is v1
= Ivi cos(1r/ 3) = 4 ·. ~ =
V2
= Ivi sin(1r / 3) = 4 · .,;; = 2 v'3. Thus
y
2 and theycomponent is v Vz
V
= (th, V2) = (2, 2 .j3 ). v,
X
31. The velocity vector v makes an angle of 40° with the horizontal and
has magnitude equal to the speed at which the footba ll was thrown. From the figure, we see that the horizontal component of v is
Ivi cos 40° =
60 cos 40°
is Ivi sin 40°
= 60 sin 40° ~ 38.57 fils.
~
45.96 ft/s and the vertical component
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SECTION 12.2 VECTORS
0
117
33. The given force vectors can be expressed in terms of their horizontal and vertical components as 300 i and
~00 cos 6~0 i +
200 sin 60°j
these two vectors: F
= 200 ( ~)
= (300 +
100) i
IFI ~ V(200) 2 + (100v'3) 2 =
i + 200 (
+ (0 +
~) j = 100 i + 100 v'3 j . The resultant force F =
100 v'3) j
.j70,000 = 100 V7
200 i
is the sum of
+ 100J3j. Then we have
~ 264.6 N. Let() be the angle F makes with the
· · xax1·s. Then tan () = lOO . I pomt . of F 11es · m . t he second qua drant, so positive _ J3 = . J3 an d the termma 2 200 () = tan
1
( 
~) + 180° ~ 40.9° f. 180° =
139.1° .
35. With respect to the water's surface, the woman's velocity is the vector sum of the velocity of the ship with respect to the water, and the woman's velocity with respect to the ship. If we let north be the positive ydirection, then
v=
(0, 22) + (3, 0) = (3, 22). The woman 's speed is
with the east, where()= tan
1
(
lvl = .jg + 484 ~ 22.2 mi/ h. The vector v makes an angle()
~;) ~ 98°. Therefore, the woman's direction is about N(98 90)
0
W
= N8°W.
37. Let T 1 and T 2 represent the tension vectors in each side of the clothesline as shown in the figure. T 1 and T 2 have equal vertical components and opposite horizontal components, so we can write
T1
=  a i + bj and T 2 =
a i + bj [a, b > 0]. By si milar triangles, !!_ = a
0 08 ·
=> a = 50b. The force due to gravity
4
=
acting on .Ule shirt has magnitude 0.8g ~ (0.8)(9.8) = 7.84 N, hence we have w of the tensile forces counterbalances w , so T 1 ( 50b i
+ bj ) +
(50b i + bj )
= 2bj
= 7.84j
+T2 =
=> (a i + bj ) + (a i + bj )
w
=> b =
7
·:
7.84j. The resultant T 1 + T 2
4
= 3.92 and a= 50b =
= 7.84 j
=>
196. Thus the tensions are
T 1 =  a i + bj =  196i + 3.92j and T 2 = ai + bj = 196i + 3.92j . Alternatively, we can find the value of() and proceed as in Example 7. y
39 .. (a) Set up coordinate axes so that the boatman is ·at the origin, the canal is bordered by the yaxis and the line x = 3, and the current flows in the negative ydirection. The boatman wants to reach the point (3, 2). Let() be the angle, measured from the positive yaxis, in the direction he should steer. (See the figure.)
X
In still water, the boat has velocity v b = (13 sjn B, 13 cos()) and the velocity of the current is V c (0, 3.5), so the true path of the boat is determined by the velocity vector v
= v b + V c = (13 sin B, 13 cos() 
3.5). Let t be the time (in hours)
after the boat departs; then the position of the boat at time t is given by tv and the boat crosses the canal when tv = (13sin0, 13cos0 3.5) t = (3, 2). Thus 13(sinO)t = 3
=>
3 . ()and (13 cosB 3.5) t = 2. 13 sm ·
t = 
'+
Then p is the angle between vectors PQ and P R , q is the angle >
>
between vectors QP and QR, and r is the angle between vectors +
> .
RPaildRQ. >
>
.
2 4 2 ( • 3 ) · (1 • ) =  + 12 =___!Q_and =cos 1 (___!Q_) .,j( 2)2+32 .j12+ 42 y'l3ffi v'ffi p y'ffi
!?O!!nl
PQ·PR =
Thuscos = p +
r
~
y
+
QP . QR .
cosq =
~48° . Similarl,
· (3, 1) 6 3 3 IQ?IIQRI= v'4(2, 3) + 9 v'9 + 1 = v'13 v'f5 = v'13Q so
 1(
q =cos
3 )
v'l3Q ~ 75
d
o
an
180°  {48°+ 75°) = 57°.
.
.
Alternate solution: Apply the Law of Cosines three times as follows: cos p
=
,___.,2 ,___.,2 1,+12 ~~n ,__. , ,__.
2 PQ 2
2
_i?R( I?QI . IQJil. , and ,___.,, __. ,
cos q ·
23. (a) a . b
2 PQ
= (5){6) +
2
j?Q(  lnl IQ'RI
_
cosr 
.
QR (3)(  8) + {7){2)
1
2
..
2 PR QR 1
PR
11
1
= 40 :/= 0, so a and bare not orthogonal. Also, since a
is not a scalar multiple
ofb, a and bare not parallel. (b) a · b = (4) ( 3) + (6)(2) = 0, so a and bare orthogonal (and not parallel). (c) a· b = (1){3) + (2){4) + {5)(1) = 0, soaand bare orthogonal (and not parallel). (d) Because a =  ~ b , a and b ~e parallel. +
+
+
+
+
+
25. QP = (1,  3, 2), QR = (4,  2,  1), and QP · QR = 4 + 62 = 0. Thus QP and QRare orthogonal, so the angle of the triangle at vertex Q is a right angle.
© 2012 Cengogc Learning. All RighiS Resen' 0], so .cos a = cos/3 =
cos:y =
~
v3c
=
~ and
v3
r;an a ·b 5 · 4 + 12 · 6 v w:.r = 13. The scalar projection of b onto a is comp 11 b =  = 13 1a 1
vector projection ofb onto a is, proj 11 b = (
j~~)
: 1
1
= 4·
f3 (5, 12} =
41 . la l = v'9 + 36 + 4 = 7 so the scalar projection of b onto a is compab = . . of b onto a .IS,. prOJa . b = 7j;f 9 a proJeCtiOn
= 79 · 7l (3 , 6 , 
( ~,  ~ ).
j~~
=
~ (3 + 12..:. 6) =
l
The vector
2} = 4o o (3 , 6 ,  2} = ( 49 27 54 18 ) • 49• 49 ·
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= 4 and the
122
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
43. lal =
J4 + 1 + 16 = J2I so the scalar projection of b onto a is compa b =
0
a · bl =
a
1
. . fb . . b prOJeCtion 0 Onto a IS prOJa
1 a 1 j +4k = J2T jaj = J2T · 2iJ2T
(
•
•
1 = 2l 21 J
; ; 21
2
+ 4 k ) = 212 i 
1 =   while the vector
J2I
1 2T j
+ 2T4 k.
' a ·b a·b 2 45. ( ortha b ) · a = (b  p roja b ) · a = b · a  (proja b } · a = b · a   2 a· a = b · a   2 Ia I = b · a  a· b Ja l lal
= 0.
So they are orthogonal by (7). 47. compa b =
a·b
j;f =
~
2
a · b = 2lal = 2 v'IO. lfb = (b1, b:l, b3) , then we need 3bl
+ Ob2 1b3 = 2 v'IO.
One possible solution is obtained by taking b1 = 0, b2 = 0, bs =  2 v'IO. In general, b = ( s, t, 3s  2 v'10 ), s, t E llt
+ (20 8) k = 6 i + 2j + 12 k so, by Equation 12, the work done is 6j + 9 k ) · (6i + 2j + 12k} = 48  12 + 108 = 144joules.
49. The displacement vector is D
W = F · D = (8 i 
51. Here ID I = 80 ft, IF I
w
=
= (6 
0) i + (12  10} j
30 lb, and B = 40°. Thus
=F . D = IF IIDI cos B = (30)(80) cos 40° = 2400 cos 40° ~ 1839 ftlb.
53. First note that n = (a, b) is perpendicular to the line, because if Q 1 = (a 1, bt) and Q2 +
n · Q1 Q2 = aa2  aa1
+~ 
bb1
= 0, since aa2 + bb:l =
 c = aa1
+ bb1
=
(a2 , b2) lie on the line, then
from the equation of the line.
I
Let P2 = (x2, y2) lie on the line. Then the distance from P1 to the line is tJ:te absolute value of the scalar projection +p. (P t P. ) In · (x2 x 1, Y2  Yl ) l lax2 ax1 + by2 lnJ1 I lax1 + by1 + cl ofp1 2 onto n . compn 1 2 = I I = = n vr:?'7l:? a2 + b2 vr::?7'l:? a2 + b2
since ax2
+ by2 =c. The required distance is 1(3)(  2) + ( 4 )(3) + 51 = .
)32
+ ( 4)2
,I
13 . 5
55. For convenience, consider the unit cube positioned so that its back left corner is at the origin, and its edges lie along the coordinate axes. The diagonal of the cube that begins at the origin and ends at (1, 1, 1} has vector representation (1, 1, 1). The angle B between this vector and the vector of the edge which also begins at the ori~in and runs along the xliXis [that is,
. . b B (1, 1, 1) · (1, 0, 0) 1 (1,0,0)] 1sg1ven ycos = I( , , }I I( , 0, 0) I = J3 1 1 1 1 57. Consider the H CH combination consisting of the sole carbon atom and the two hydrogen atoms that are at (1, 0, 0) and (0, 1, 0) (or any H  C H combination, for that matter). Vector representations of the line segments emanating from the
carbon atom and extending to these two hydrogen atoms are ( 1 
(0 
t,0  t, 0 
t) =
(
t , t, t) and
t.1 t.0 t) ~ ( t, t,  t ). The bond angle, B, is therefore given by
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SECTION 12.4 THE CROSS PRODUCT , 0
123
59. Let a = (a1, a2, a3) and= (b1, b2, b3).
Property 2:
a· b = (a1, a2 , a3 ) · (b1, b2, bs) = = b1a1
Property 4:
+ ~a2 + b3a3 =
a1b1
+ a2b2 + asbs
(b1, ~. bs ) · (a1 , a2, as ) = b ·a
(ca) · b = (ca1, ca2, ca3) · (b1, b2, bs) = (ca1)b1 + (ca2)b2 + (ca3)b3
= c(a1b1 + a2b2 + asbs) = c (a· b) = a1(cb1) + a2(~) + a 3(cbs) = (a1 , a2, as)· (cbt, c~, cbs) = a· (c b )
Property 5:
0 ·a = (0, 0, 0) · (a1, a2, a3) = {O)(al) + (O)(a2) + (O)(a3) = 0
61 . Ia · b l = l lal lbl cos8l = lallbl lcos 81. Since lcos81 ~ 1, Ia · bl = lal lbl lcos 81 ~ lal lbl.
Note: We have equality in the case ofcos8 = · ± 1, so 8 = 0 or 8 =
1r,
thus equality when a and bare parallel.
t he Parallelogram Law states that the sum of the squares of the
63. (a)
lengths of the diagonals of a parallelogram equals the sum of the squares of its (four) sides.
2
2
2
2
(b) Ia + b l =(a + b) · (a + b )= lal + 2(a ·b)+ lbl and Ia  b l = (a b )· (a b)= lal
2

2(a · b )+ lb l2 •
2
Adding these two equations gives Ia + b l2 + Ia b l2 = 2 lal + 2 lbl 2 .
12.4
The Cross Product k
j
1. a x b
=
= I~
6 0 2 0 8 0
2 . 1 01

6  2 . J 0 01 1
+
6 oI k 0 8 1
= [0  ( 16)]i  (0  O)j + (48  O) k = 16 i + 48k Now (a x b)· a = (16, 0, 48) · (6, 0, 2) = 96 + 0  96 = 0 and (a x b)· b = (16, 0, 48) · (0, 8, 0) = 0 + 0 + 0 = 0, so a x b is orthogonal to both a and b . j
3. a x b =
k
1 3  2 1 0 5
= I ~ ~~I i  ~~ ~
·= (15  O)i  (5  2) j
+ [0 
lj l~~ ~I +
k
( 3))k = 15 i 3j +3 k
Since (a x b) · a= {15i  3j + 3k) · (i + 3j  2k) = 15  96 = O, a x bisorthogonal to a . Since (ax b)· b
®
= (15 i 
3j + 3 k) · (  i
+ 5 k) = 
15 + 0 + 15 = 0, a x b is orthogonal to b .
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124
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
k
j
5. ax b
=
1  1 1 1
1
1
2
= ,~
= {~ i 
=
t
e = (1 
j+ ~
1 1/t t
+ ~ k)
2
1
=
2
· (i  j  k)
= ~ + 1 ~
~ lk ~ 0 and
= ~  1 + ~ = 0, so a x
1/t
t2
t 2 ) k = (1  t) i
t2). (t, 1, 1/t) =

2
b is orthogonal to both a and b .
I I t21 t I·J + I t t2 I
+ (t3 
Since {a X b ). a = (1 t,O,t3
 11, J. + 111
1
2
1 1/t i
1t21
t) i  (t t) j
Since (a x b ) · b
11
k) · Gi + j + ~ k)
k
j
7. a x b
li
2
Now {ax b )· a= {~ i  j
(a x b ) · b
~
= ( 1  t, 0, t 3  t 2 )
• (
+ (t3 
t
t 2 , t 2 , 1) = t 2
1 k
t2) k
e +0 +t

2
t

t3 + 0 + t 3

= 0, a t2
X b is orthogonal to a.
= 0, a
x b is orthogonal to b .
9. According to the discussion precedin.g Theorem II , i x "j = k, so (i x j ) x k = k x k = 0 [by Example 2).
11. (j  k)
X
(k i) = (j  k)
X
k
+ (j 
= j X k + (  k)
X
k) .X (  i)
by Property 3 of Theorem I I
k + j X ( i) + ( k) X (  i)
by Property 4 of Theorem 11 2
= (j X k ) + (1){k X k) + ( 1)(j X i)+ (1) (k X i) = i + ( 1) 0 + (1)( k) + J = i + j + k
by Property 2 of Theorem I 1 by Example 2 and the discussion preceeding Theorem II
13. (a) Since b x cis a vector, the dot product a· (b x c) is meaningful and is a scalar. (b) b ·cis a scalar, so a x (b ·c) is meaningless, as the cross product is defined only for two vectors. (c) Since b x cis a vector, the cross 'product a x (b x c) is meaningful and results in another vector. (d) b · cis a scalar, so the dot product a · (b ·c) is meaningless, as the dot product is defined only for two vectors. (e) Since (a . b ) and (c . d) are both scalars, the cross product (a . b )
X (c
. d) is meaningless.
(f) a x band c x d arc both vectors, so the dot product (ax b )· (c x d ) is meaningful and is a scalar.
15. If we sketch u
an~
v starting from the same initial point, we see.that the
angle between them is 60°. Using Theorem 9, we have
lu x vi = lullvl sin 0 = {12)(16) sin 60° = 192 ·
v; =
96 v'3.
v
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SECTION 12.4 THE CROSS PRODUCT
j k 17. axb =
2  1 3 = 4 2 1 j k
b xa=
4
2 1
2 1 3 Notice a x b
~~ :Ii  I~
: lj +
=1 ~ ~ li I: ~ lj +
I~
.0
1, k = ( 16) i  (2 12)j + [4 ( 4)] k =  7i+10j+8 k 2 .
21 k = [6 (  1)] . i {12  2)j + ( 4  4) k = 7 i  lOj 8 k 1: 1
=  b x a here, as we know is always true by Property I of Theorem II .
19. By Theorem 8, the cross product of two vectors is orthogonal to both vectors. So we calculate
3i
2j
k1 = 12
 1
1
0
(3,2,1)x( 1, 1, 0) =
1
11 i 
0
I13
11 · j + I 3
0
 1
21 k=  ij +5 k. 1
. I b h ± (1, 1,5) ±(1, 1,5) h . ( 1 1 5 ) So two umt vectors orthogona to ot are v' + + = ../3 , t at 1s,  M, M ~ 37:3 1 1 25 3
21 . Let a= (a1, a2, as). Then
O x a=
0
j
k
0
0
= Ia: :31i
a1 a2 as
j
k
axO = a1 a2 as =
 ~~
Ia~ I
as i _
0 lk= O,
:sl j + 1:1 a2
I a1 as, . + 0 0 J
0
0
I
a1 a21 0 O k = O.
0 0 0
25. ax (b+c) =a x (b1 +c1,b2 +c2,bs +c3)
= (a2(bs + cs) a3(b2 + c2), as(b1
= ((a2bs  asb2)
+ (a2cs 
+ c1) a1(b3 + cs) , a1(b2 + c2) a2(b1 + q))
asc2), (asb1 a1bs) + (asc1  a1c3), (a1b2  a2b1)
+ (a1c2 
= (a2bs asb2 , aab1  a1bs, a1b2 a2b1 ) + (a2cs  asc2, asc1 a1c3, a1c2  a2c1 ) = (a x b) + (a
X
125
c)
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a2c1))
126
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE y
27. By plotting the vertices, we can see that the parallelogram is determined by the +
vectors AB
+
= (2, 3) and AD= (4, 2}. We know that the area of the parallelogram
determ ined by two vectors is equal to the length of the cross product of these vectors. .
+
In order to compute the cross product, we consider the vector AB as the three+
dimensional vector (2, 3 , 0) (and similarly for AD), and then the area of parallelogram ABCD is k
j
IA13 x ml=
2 3 4  2
o =l(o) i (o) j+ (4  12)kl=l16kl = 16 0 +
+
29. (a) Because the plane through P, Q, and R contains the vectors PQ and P R, a .vector orthogonal to both of these vectors +
(such as their cross product) is also orthogonal to the plane. Here PQ +
+
= ( 3, 1, 2} and P R = (3, 2, 4}, so
+
PQ x PR = ((1)(4) (2)(2), (2)(3) (3)(4) , (3)(2) (1)(3)) = (0, 18,  9) Therefore, (0, 18,  9) (or any nonzero scalar multiple thereof, such as (0, 2,  1)) is orthogonal to the plane through P , Q, andR. (b) Note that the area of the tr~angle determined by P, Q, and R ls equal to half of the area of the parallelogram determined by the three points. From part (a), the area of the parallelogram is
I.PQ x t
31 . (a) PQ
PR I = I(0, 18, 9) I = y'O + 324 + 81 = v'4o5 = 9v'5, so the area of the triangle is ~ · 9v'5 = ~ .J5.
= (4, 3, 
+
+
2) and P R
= (5, 5, 1}, so a vector orthogonal to the plane through P, Q, and R is
+
PQ x P R = ((3)(1)  ( 2)(5), ( 2)(5)  (4)(1), (4)(5)  (3) (5)) = (13,  14, 5) [or any scalar mutiple thereof]. +
+
(b) The area of the parallelogram determined by PQ and PR is
IPQ x P'RI = 1(13,14, 5)1 = yl132 + ( 14)2 +52 = v'395,sotheareaoftriangle PQR is
~v'39Q.
33. By Equation 14, the volume of the parallelepiped detennined by a , b , and cis the magnitude of their scalar triple product, '
which is a · (b x c) .
=
1 2 3
1 1 2
= 1 111
21 4
 2
1 1 2
24 1 + 3
112 111 = 1(42)2( 44) + 3(12) = 9.
2 1 4 Thus the volume of the parallelepiped is 9 cubic units. t
+
+
35. a= PQ = (4, 2, 2}, b = PR = (3, 3, 1), and c = P S = (5,5, 1).
so the volume of the parallelepiped is 16 cubic units. ® 201 2 Cengage LC1tning. All Richts Rescrnd. M.:ty not be scnnncd. copied, or duplicntcd, or posccd to u publicly accessible website, in whole or in p:lrt.
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SECTION 12.4 THE CROSS PRODUCT
1
D
127
5 2
1 0 3 0 3 0 =1 1 · 1 5 1 1+( 2) 1 ll = 4 + 60  64 = 0,whichsaysthatthevolume 9 4 5 4 5 9 . . 9  4
37. u · (v x w)= 31
5
of the parallelepiped determined by u , v and w is 0, and jhus these three vectors are coplanar.
39. The magnitude of the torque is Jr J = Jr x FJ = JrJJF JsinO = (0.18 m)(60 N) sin(70 +lOt = 10.8 sin80° r::::: 10.6 N·m.
41 . Using the notation of the text, r = {0, 0.3, 0) and F has direction (0, 3, 4). The angle(} between them can be determined by
cos 9 =
(0, 0.3, 0) . {0, 3, 4) . J(o, o.3, o)JI{o,3, 4)1
100 = 0.3IFisin53.1°
9
=> cos =
0.9 (0.3)(5)
=> cosO = 0.6 =>
(J
r::::: 53.1°. Then Jr l = lri iFI sin(} =>
=> 1F i r:::::417N.
43. From Theorem 9 we have Ia x bl = la l lb l sin 9, where 0 is the angle between a and b , and from Th,eorem 12.3.3 we have
Ja l lbl = a · be . Substituting the second equation into the first gives Ia x b l = a · b sin B, so cos cos (J
=>
a· b = lal lb l cosO
Ia X b l = tan9. Here Ia . a· b
X
b l = 1(1, 2, 2)1 = V1 + 4 + 4 = 3, so tan() = Ia x bbl = ~ = a· v3
J3 => () =
60° .
The distance between a point and a line is the length of the perpendicular
45. (a)
from the point to th~ line, here
d=
In I= IQP' sin
(J
In I=d. But referring to triangle PQS,
= lbl sin 9. But 8 is the angle between
QP =
+ Ia x bl and QR = ~·Thus ~y Theorem 9, sinO = lai Jb l
a
d an so d +
= lb1
.
'I sm 9
= Jbi Ja x bl = Ia x b J Jai Jbl laJ
+
(b) a= QR = ( 1,  2,  1) and b = QP = (1,  5,  7}. Then a x b = ((2)( 7)  ( 1){7 5), ( 1)(1) (  1)(  7) , ( 1)( 5) (  2)(1)} = (9,  8, 7). Thus the distance is d = Ia [:Ibl =
~ JB l + 64 + 49 =
ji¥ = j¥.
47. From Theorem 9 we have Ia x b l = [al[b[ sin 9 so
Ja x b] 2 = [a[ 2 [b[ 2 sin2 () = [a[ 2 lb[2 (1  cos2 B) 2
= fa i Jb[
2

2
2
2
([ai Jb [cos 9) = [a[ lb[

(a · b)
2
by Theorem 12.3.3.
©
201 2 Cengage l c'3ming. All Rights Reserved. May not be scanned, cor lcd, or duplicated, or j>ost
Bt = 8 => t = 1. Therefore, the point of intersection of the line and the plane is given by x = 3  1 = 2, y = 2 + 1 = 3, and z = 5(1) =5 1 that is, the point (2, 3, 5). 47. Parametric equationS for the line are x = t, y = 1 + t, z = ~t and substituting into the equation of the plane gives 4{t){1+t)+3(~t)=8
=> ~t=9 => t=2.Thusx = 2,y=1+2 .= 3,z=H2)=1andthepointof
intersection is (2, 3 1 1). ®
20 12 CenCJlgc Leaminc. All Rights Reserved. May no1 be scanned, wpl 1, and the traces in z ·= k are 4y 2 
4, a family of
x 2 = 4 + k 2 , a family
of hyperbolas. Thus the surface is a hyperboloid of two sheets with axis the y axis.
17. 36x 2
+ y 2 + 36z2 = 36. The traces in x = k are y 2 + 36z2 = 36(1 
a family of ellipses for and are empty for 36x 2
lkl < 1. (The traces are a single point for lkl = 1
lkl > 1.) The traces in y = k'are the circles
+ 36z2 = 36 k 2
traces in z
x2
{"}
+ z2 = 1 
tok 2 ,
= k are the ellipses 36x 2 + y 2 = 36(1 
lkl < 6, and the
k 2 ), 1~1
graph is an ellipsoid centered at the origin with intercepts x
z
k 2 ),
< 1. The
= ± 1, y = ± 6,
= ±1.
19. y = z 2  x 2 • The traces in x
the traces in y
= k are the parabolas y
= k are k = z 2 
are oriented differently fork the parabolas y
= k 2 .
= z2
:
k2 ;
x 2 , which are hyperbolas (note the hyperbolas
> 0 than fork < 0); and the traces in z = k are
x 2 • Thus,
y
l =
21. This is the equation of an ellipsoid: x
2
z2
12
x2
 ]2 is a hyperbolic paraboloid. X
2
2
+ 4y2 + Dz2 = x 2 + ~ + ___!___, = (1/2)
(1/3t
1, with xintercepts ±1, yintercepts ±i 0

and zintercepts ±~. So the major axis is the xaxis and the only possible graph is Vll. 23. This is the equation of a hyperboloid of one sheet, with a
= b = c = 1.
Since the coefficient of y 2 is negative, the axis of the
hyperboloid is the yaxis, hence the correct graph is II.
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138
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
D
25. There are no real values of x and z that satisfy this equation for y
< 0, so this surface does not extend to the left of the .
xzplane. The surface intersects the plane_y
.
= k > 0 in an ellipse. Notice that y occurs to the first power whereas x and z
occur to the second power. So the surface is an elliptic paraboloid with axis theyaxis. Its graph is VI.
27. This surface is a cylinder because the variable y is missing from the equation.' The intersection of the surface and the xzplane is an ellipse. So the graph is VIII .
= x 2 + ~z
29. y 2
2
or y 2 = x 2
. z2
+g
.
•
represents an elliptic
cone with vertex (0, 0, 0) and axis the yaxis.
31. x 2
+ 2y 
2z 2
= 0 or 2y = 2z2 
x 2 or y
2
= z2  ~
represents a hyperbolic paraboloid with center (0, 0, 0).
z
35. Completing squares in all three variables gives
33. Completing squares in y and z gives
4x
2
+ (y 
2)
2
x z + (y ~ ) 2
2
+ 4(z + (z 
(x  2)2  (y + 1? + (z 1)2 = 0 or
2
3) = 4 or
3) 2 = 1, an ellipsoid with
(y + 1? = (x  2) 2
+ (z 1?, a circular cone with
center (2, 1, 1) and axis the horizontal line x
center (0, 2, 3).
z =l. z
y
37. Solving the equation for z we get z. = ±)1 + 4x2
z
+ y 2 , so we plot separately z = ) 1 + 4x2 + y 2 and
=  ) 1 + 4x2 + y2 .
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= 2,
CYLINDERS AND QUADRIC SURFACES
SECTION 12.6
0
139
To restrict the zrange as in the second graph, we can use the option view=  4 . . 4 in Maple's p l"ot3d command, or Plo.t Range
 > { 4 , 4} in Mathematica's Pl ot 3D command.
39. Solving the equation for z we get z
z
41.
= ±y'4x2 + y 2 , so we plot separate ly z = y'4x2 + y2 and z = 
z=2
y'4x2
+ y2.
43. The surface is a paraboloid of revolution (circular paraboloid) with vertex at the origin, axis the yaxis and opens to the right. Thus the trace in the yzplane is a lso a parabola: y = z 2, x
= 0. The equation is y =
x2
+ z 2.
The parabola y=x2
X
= (x, y, z) be an arbitrary point equidist ant fro m ( 1, 0, 0) and the plane x = 1. Then the distance from P to
45. Let P
( 1, 0, 0) is y'(x + 1) 2 + y 2 + z 2 and the distance from P to the plane x = 1 is lx 11/Vf'i (by Equation 12.5.9). So lx 11 = y'(x + 1)2 + y 2 + z 2
x 2  2x + 1 = x 2
+ 2x + 1 +. y 2 + z 2
¢:>
 4x
¢::>
= lx 
(x 1) 2 = (x + 1) 2 + v. 2 + z 2
11
¢:>
= y 2 + z 2 • Thus the collection of all such points P
is a circular
paraboloid with vertex at the origin, axis the x axis, which opens in the negative direction.
47. (a) An equation for an ellipsoid centered at the origin with intercepts x = ±a, y
= ±b, and z =
2
±cis x 2 a
2
2
+ Y2 + .=.._2 = b
c
Here the poles of the model intersect the zaxis at z = ± 6356.523 and the 'e quator intersects the x andyaxes at
x
= ±6378.137, y = ±6378. 137, so an equation is x2 (6378.137) 2 •
•
(b) Traces to z
2
x +y
2
®
=k
~
z2
~ I
are the Circles (6378.137)2
= (6378.137)2
y2
.
+ (6378. 137) 2 + (6356.523)2 = 1
+ (6378.137)2 = 1 
~
(6356.523)2
2
(6378.137) k2 ~6.~ .
2012 C.ngagc Lc:lming. All Rights Re
a 2 . This tells us that all of £2 also lies on

The curve of intersection looks like a bent ellipse. The projection
51.
of this curve onto the xyplane is the set of points (x, y, 0) which satisfy x x2
12
+
2
+ y2 = 1 y2
{1 //2)2
y
2
x
2
+ 2y 2 = 1
= 1. This is an equation of an ellipse.
Review
CONCEPTCH ECK 1. A scalar is a real number, while a vector is a quantity that has both a realvalued magnitude and a direction. 2. To add two vectors geometrically, we can use either.the Triangle Law or the Parallelogram Law, as illustrated in Figures 3 and 4 in ·section 12.2. Algebraically, we add the corresponding components of the vectors. 3. For c
> 0, c a is a vector with. the same direction as a and length c times the length of a. If c < 0, ca points in the opposite
direction as a and has length JcJ times the length of a. (See Figures 7 and IS in Section 12.2.) Algebraically, to find c a we multiply each component of a by c. 4. See (I) in Section 12.2. 5. See Theorem 12.3.3 and Definition 12.3.1. © 2012 Ccn~a.gc Lcnrnin~. All Rights Rcscr'\'ctl. Muy not be scamtcd: copied , or duplicated, or Jl(r.itcd ton publicly i1Ccess ihlc website.!, in whole or In p.1rt.
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CHAPTER 12 REVIEW
0
141
6. The dot product can be used to find the angle between two vectors and the scalar projection of one vector onto another. In particular, the dot product can determine if two·vectors are orthogonal. A lso, the dot product can be used to determine the work done moving an object given the force and di~placement vectors.
7. See the boxed equations as well as Figures 4 and 5 and the accompanying discussion on page 828 [ET 804].
B. See Theorem 12.4.9 and the preceding discussion; use either (4) or (7) in Section 12.4. 9. The cross product can be used to create a vector orthogonal to two given vectors as well as to determine if two vectors are parallel. The cross product can also be used to find the area of a parallelogram determined by two vectors. In addition, the cross product can be used to determine torque if the force and position vectors are known.
•
10. (a) The area of the parallelogram determined by a and b is the length of the cross product:
Ia x
bl.
(b) The volume of the parallelepiped determined by a, b, and cis the magnitude of their sca lar triple product: 11. If an equation of the plane is known, it can be written as ax
+by + cz + d =
Ia . (b
x c )l.
0. A normal vector, which is' perpendicular to the
plane, is (a , b, c) (or any scalar multiple of (a , b, c)). If an equation is not known, we can use points on the plane to find two nonparallel vectors which lie in the plane. The cross,product of these vectors is a vector perpendicular to the plane. 12. The angle between two intersecting planes is defined as the acute angle between their normal vectors. We can find this angle
using Corollary 12.3.6. 13. See (1), (2), and (3) in Section 12.5. 14. See (5), (6), and (7) in Section 12.5. 15. (a) Two (nonzero) vectors are parallel if and only if one is a scalar multiple of the other. In addition, two nonzero vectors are
parallel if and only if their cross product is 0. ( b) Two vectors are perpendicular if and only if their dot product is 0. (c) Two planes are parallel if and only if their normal vectors are parallel. +
+
16. (a) Determine the vectors PQ = (a1 , a2 , a3 ) and P R = (b1, ~ . b3) . If there is a.scalar t such that . (a 1 , a 2, a 3) = t (b1, ~ . b3 ), then the vectors are parallel and the points must all lie on the same line. ~
+
+
+
Alternatively, if PQ x PR = 0, then PQ and PR are paralle l, ~o P, Q, and Rare collinear. Thirdly, an algebraic method is to determine an equation of the line joining two o.f the points, and then check whether or not the third point satisfies this equation. +
+
+
(b) Find the vectors PQ =a, P R = b , PS = c . a x b is normal to the plane formed by P, Q and R, and so S lies on this plane if a x band care orthogonal, that is, if(a x b)· c = 0. (Or use the reasoning in Example 5 in Section 12.4.) A lternatively, find an equation for the plane determined by three of the points and check whether or not the fourth point satisfies this equation.
®
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142
D
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
17. (a) See Exercise 12.4.45.
(b) See Example 8 in Section 12.5. (c) See Example 10 in Section 12.5. 18. The traces of a surface are the curves of intersection of the surface with planes parallel to the coordinate planes. We can find the trace in the plane x = k (parallel to the yzplane) by setting x = k and determining the curve represented by the resulting equation. Traces in the planes y = k (parallel to the xzplane) and z = k {parallel to the xyplane) are found similarly.
19. See Table I in Section 12.6.
TRUEFALSE QUIZ 1. This is false, as the dot product of two vectors is a scalar, not a vector. 3. False. For example, ifu = i and ·v = j then iu · v i= 101 = 0 but iu l lvl = 1 · 1 = 1. In fact, by Theorem 12.3.3, lu · v i = liu llvl cos 91 . 5. True, by Theorem 12.3.2, property 2. 7. True. If 9 is the angl~ between u and v , then by Theorem 12.4.9,lu x vi = iu l lv l sin 9 = lvllul sin 9 = 1':' x ui. (Or, by Theorem 12.4.1l,lu x v i = 1v x ul = 1 lllv x ul = lv x ui.) 9. Theorem 12.4.11 , property 2 tells us that this is true. 11. This is true by Theorem 12.4.11, property 5. 13. This is true because u x vis orthogonal to u (see Theorem 12.4.8), and the dot product of two orthogonal vectors is 0. 15. This is false. A normal vector to the plane is n = (6,  2, 4) . Because (3,  1, 2) = ~n, the vector is parallel ton and hence perpendicular to the plane. 17. This is fa lse. In IR2 , x 2
+ y2
= 1 represents a circle, but { (x, y, z) I x 2 + y 2 = 1} represents a threedimensional surface,
namely, a circular cylinder with axis the z axis.
19. False. For example, i · j = 0 but i
i 0 and j =f. 0 .
21. This is true. Ifu and v are both nonzero, then by (7) in s~ c tion 12.3, u · v = 0 implies that u and v are orthogonal. But u x v = 0 implies that u and v are parallel (see Corollary 12.4.10). Two nonzero vectors can't be both parallel and orthogonal, so at least one of u, v must be 0 .
®
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CHAPTER 12 REVIEW
0
1~
EXERCISES 1. (a) The radius of the sphere is the distance between the points (  1, 2, 1) and {6, 2, 3), namely, J [6  ( 1)]2 + (  2  2) 2 + (3  1) 2 =
0f!}.
By the formula for an equation of a sphere (see page 813 [ET 789]),
an equation of the sphere with center ( 1, 2, 1) and radius
J69 is (x + 1? +
(y  2) 2
+ (z 1) 2
= 69.
(b) The intersection of this sphere with the yz plane is the set of points on the sphere whose xcoord inate is 0. Putting x = 0 into the equation, we have (y  2) 2 + (z  1) 2 = 68, x = 0 which represents a circle in the yzplane with center (0 , 2, 1) and radius
.J68.
(c) Completing squares gives (x  4) 2 + (y + 1) 2 + (z +
3? =  1 + 16 + 1 +9 =
25. Thus the sphere is centered at
(4, 1, 3) and has radius 5. 3. u . v = Ju J.JvJ c~s45° = (2)(3)
'{! =
3 .;2.
Ju x vJ = JuJJv Jsin45° =
(2)(3)~ = 3 .;2.
By the righthand rule, u x vis directed out of the page.
5. Forthetwo vectors to be orthogonal, we need (3, 2,x} · (2x, 4,x} = 0 ~ + 6x + 8 = 0
. t = 1. When t = 1, the parametric equations give x = 2 
+z =
2(2 t)  (1 + 3t) + 4t
1 = 1, y = 1 + 3
= 4 and z =
=2
=?
4. Therefore,
the point of intersection is (1,4,4). 23. Since the direction vectors (2, 3, 4) and (6, ;1, 2) aren't parallel, neither are the lines. For the lines to intersect, the three equations 1
+ 2t =
equations gives t =
1 + 6s, 2 + 3t = 3  s, 3 + 4t = 5 + 2s must be satisfied simultaneously. Solving the first two
k. s =
~ and checking we see these values don't satisfy the third equation. Thus the lines aren't parallel
and they don't intersect, so they must be skew. 25. n 1 = (1, 0, 1) and n2 = (0, 1, 2). Setting z = 0, it is easy to see that (1, 3, 0) is a point on ,the line of intersection of
= 1 andy+ 2z = 3. The direction of this line is v 1 = n 1 x n 2 = (1, 2, 1). A second vector parallel to the desired plane is v 2 = (1,1, 2), since it is perpendicular to x + y 2z = 1. Therefore, the normal of the plane in question is
x z
n = v 1 x v 2 = (4  1, 1 + 2, 1 + 2) = 3 (1, 1, 1) . Taking (x0 , y 0 , z 0 ) = (1, 3, 0), the equation we are looking for is
(x 1) + (y 3) + z = 0 . 27. By Exercise 12.5.75, D 
29. The equation x
¢*
x
+y +z =
4.
l2 (24)1  22 .  ~· + I2 + (4)2 v26
y'32
= z represents a plane perpendicular to
the xzplane and intersecting the xzplane in the line
31. The equation x 2 = y 2 + 4z2 represents a (right elliptical) cone with vertex at the origin and axis the x axis.
x = z,y = 0.
X X
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www.elsolucionario.net
CHAPTER 12 REVIEW .
.
33. An equivalent equation
.
 x2
IS
y2 +4 ; z 2 = 1, a
0
145
35. Completing the squate in y gives
+ 4(y 
1) 2
+ z 2 = 4 or x 2 + (y 
hyperboloid of two sheets with axis the yax.is. For
4x 2
IYI > 2, traces parallel to the xzplane are circles.
an ellipsoid centered at (0, 1, 0).
1) 2
z2
+4 =.1,
z (0, 1,2)
(O, l , 2)
. x2
y2
4 + 16
2
= l. The equation of the ellipsoid is
~
+
i
2
6
2
+ :2
= l, since the horizontal trace in the
plane z = 0 must be the original ellipse. The traces of the ellipsoid in the yz plane must be circles since the surface is obtained 2
by rotation about the xaxis. Therefore, c2 4x2
+ y 2 + z2 =
= 16 and the equation of the ellipsoid is ~ +
i
2
6
2
+ ~ =1
6
16.
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D
PROBLEMS PLUS
1. Since threedimensional situations are often difficult to visualize and work with, let
us first try to find an analogous problem in two dimensions. The analogue of a cube is a square and the analogue of a sphere is a circle. Thus a similar problem in two dimensions is the following: if five circles with the same radius rare contained in a square of side 1 m so that the circles touch each other and four of the circles touch two sides ofthe square, find r. The diagonal of the square is .J2. The diagonal is a lso 4r + 2x . But x is the diagonal of a smaller square of side r . Therefore
~
x = .J2r
.J2=4r+2x =4r+2.J2r= (4+2.J2)r
~ r=
4
/{,/2.
Let's use these ideas to solve the original threedimensional problem. The diagonal of the cube is
\/12
+ 12 + 12 = v'a.
The diagonal of the cube is also 4r + 2x where x is the diagonal of a smaller cube with edger. Therefore
x = Jr 2 + 1·~ + r 2 = v'3 r
~
The radius of each ball is ( v'3

r;;
r;;
v'a = 4r+2x .= 4r+2v3r = (4+2v3)r.Thus r=
v'3
r;;= 2 v'a  3
4 + 2v3
2
~) m.
3. (a) We find the line of intersection L as in Example 12.5. 7(b). Observe that the point (  1, c, c) lies on both planes. Now since
L lies in both planes, it is perpendicular to both of the normal vectors n1 and n 2, and thus parallel to their cross product
n1 x n2 =
c
j
k
1
1 = (2c, c2 + 1, c2  1). So symmetric equations of L can be written as
1  c
c
x +1 y c z c   = ~ = ~ , provided that c I 0,  2c (..  1 c +1
±1.
If c = 0, then the two planes are given by y + z = 0 and x =  1, so symmetric·equations of L are x =  1, y =  z. If
c =  1, then the two planes are given by x + y + z =  1 and x + y + z = 1, and they intersect in the line x = 0, y =  z  1. If c = 1, then the two planes are given by x + y + z = 1 and x  y + z = 1, and they intersect in the line y = 0,
X =
1 Z.
(b) If we set z = t in the
y  c=
(t c)(c2 c2 +1

symmet~ic equations and solve for x andy separately, we get x 1)
 2ct + (c
2
~
x=
c2 +
1

1)
2
, Y=
(c

+ 1 = (t  2c)(  2c) c +1
1)t + 2c El. . . fr th . . unmatmg c om ese equattons, we
&+ 1
have x 2 + y 2 = t 2 + 1. So the curve traced out by L in the plane z = tis a circle with center~~ (0, 0 , t) and radius J fl + 1.
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147
148
0
CHAPTER 12 PROBLEMS PLUS
+ 1), so V = J; A(z )dz = 1r ( ~ z3 + z] ~ =
(c) The area of a horizontal crosssection of the solid is A (z) = 1r(z2
53
53
54
50
4 ,.. 3
5" 2
jv5 1 = 24 . 32 jv 1j = 23 . 32 . Similarly, jv5j = 24 . , jv.71 :;= 25 . 34 , and in general, jv ,. j = n_ 2 . n_ 3 = 3(ir 33 2 3
2
.
Thus
f
Jv ,.J
n= l
f
= lv1 l + lv 2l + = 5+
00
E
n= l
3(i)"
n=3
2
=2 +3+
f
n=l
3(~)"
ll
1
.2.(!~)"= 5 + ___L_ 2 1 ~
G
[sum of a geometric series]
= 5 + 15 = 20
7. (a) When 8 = 8., the block is not moving, so the sum of the forces on the block I
+ F + W = 0 . This re lationship is illustrated
must be 0, thus N
geometrically in the figure. Since the vectors form a right triangle, we have
 J!l_ 11.•nn
tan(O.)  INI 


fl s·
(b) We place the block at the origin and sketch the force vectors acting on the block, including the additional horizontal force
H , with initial points at the origin. We then rotate this system so that Flies along the positive x axis and the inclined plane is parallel to the xaxis. (See the follo~ing figure.)
N F
w
IF! is maximal, so jFj = 11. n for f) > B_,. Then the vectors, in terms of components parallel and perpendicular to the inclined plane, are
N =n j W
= ( mg sin 0) i + (mg cos fJ) j
F
= (11.n) i H = (hmin cos B) i
+ (
hmin sin B) j
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l CHAPTER 12 PROBLEMS PLUS
0
149
Equ!lting components, we have JJ,n mg sin(}+ hm1n cosO= 0
,=?
hmin cos(}+
n mg cos() hrnin s infJ = 0
=?
hmin
(1)
JL•• n = mgsin(}
sin(}+ mgcos(} = n
(2)
(c) Since (2) is solved for n, we substitute into (1): cos() + JJ~ (hmin sin(}+ mg cos 9) = mg sin 9
hmin
hmin
cos fJ +
hmin/1s
=?
'*
sin fJ = mg sin fJ  mgJJ. cos(}
 JJ.• cos'(} ) 1tmin = mg (sin8 cos (}+ 1t. sin(}
.
= mg
( tau8  /1., ) 1 + JJ. tau(}
h ( tan8  tanfJ8 ) d . . . . . . b +tan (}, tau(} an usmg a tngonomctnc tdentlty, From part () a we know JJ~ = tan (} ., so th IS ecomes min = mg 1
this is mg tau(8 8. ) as desired. Note for fJ = 8., hmin = mgtau 0 = 0, which makes sense since the block is at rest for 8. , thus no additional force H is necessary to prevent it from moving. As(} increases, the factor tau(8 8.), and hence the value of hmin, increases slowly for small values ofB 8. but much more rapidly as 8 steeper the inclined
~lane,
the less the
hori zont~l
e. becomes significant. This seems reasonable, as the
components of the various forces affect the movement of the block, so we
would need a much larger magnitude of horizontal force to keep the block motionless. If we allow f)
+
90°, corresponding
to the inclined plane being placed vertically, the value of hmin is quite large; this is to be expected, as it takes a great amount of horizontal force to keep an object from moving vertically. In fact, without friction (so (} 8 = 0), we would have (} +
90°
:::?
hmin +
oo, and it would be impossible to keep the block from slipping.
(d) Since hmo.x is the largest value of h that keeps the block from s lipping, the force of friction is keeping the block from · moving up the inclined plane; thus, F is directed down the plane. Our system of forces is similar to that in part {b), then, except that we have F = (JJ.n) i. (Note that IFI is again maximal.) Following our procedure in parts (b) and (c), we equate components:  JJ,n mgsin(} + hmux cos(} = 0
:::?
hme.x cos(} /)8 n
n  mg cos (}  hm:l.X sin (} = 0
'*
hmax
= mgsin8
e
sin (} + mg cos = n
Then substituting, hmax
cos 8  /1s ( hmax sin 8 + mg cos B) = mg sin 8 hmax
'*
cos(}  hmn.x/1s sin(} = mg sin(} + mgJJ. cos
e '*
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150
0
CHAPTER 12 PROBLEMS PLUS
· ( tan8 +J.£.) _ mg (sin0+J.£.cos8) _ mg h m axcos 0  J.£8 sin B 1  J.£8 tan B tanB +tan B. )
== mg ( 1 tan B. tanB
= mgtan(B +B.•)
We would expect hmax to increase as 8 increases, with similar behavior as we establ ished for hmin, but with hmax values al~ays
hmin
larger than hmin· We can see that this is. the case if we graph hmo.x as a function of 8, as the curve is the graph of
translated 20. to the left, so the equation does seem reasonable. Notice that the equation predicts hmo.x
+
oo as
. 8+ (90°  0., ). In fact, as hmnx increases, the nonnal force increases as well. When (90°  Bs) :50 :S: 90°, the horizontal force is completely counteracted by the sum of the f!Orlnal and frictional forces, so no part of the horizontal force contributes to moving the block up the plane no matter how large its magnitude:
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(
t 2, 7t  12, t 2)
VECTOR FUNCTIONS AND SPACE CURVES
= .(4t 
3, t2, 5t 
t2, and t 2 = 5t 6. From the first equation, t 2  4t + 3 = 0
D
155
6). Equating components (t 3)(t 1) = 0 sot = 1
¢:>
or t = 3. t = 1 does not satisfy the other two equations, butt = 3 does. The piuticles collide when t = 3, at the point (9, 9, 9).
49. Let u (t) = (u1(t), u2(t ), u3(t)) and v (t) = (v1(t), V2(t), va(t)). In each part of this problem the basic procedure is to use Equation l and then analyze the individual component functions using the limit properties we have already developed for realvalued functions. (a) lim u (t ) + lim v(t) = t+a.
t+a
I\ tlim u1(t), lim u2(t), lim ua(t)) +a t+a t+a
+ I Jim V1(t ), lim v2(t), lim v3(t) ) and the limits of these ta. \t+a
t+a
component functions must each exist since the vector functions both possess limits as.t
+
a. Then adding the two vectors
and using the addition property oflimits for realvalued functions, we have that lim u(t)
t ..o.
+ tlim v (t) = I Jim u 1(t) + lim v1(t) , lim u2(t) + + a \t+ 0 there exists
0, 02 > 0, oa > 0 so that ifO < It  a l < 01 then lf(t)  b1 l < c/3, ifO < it a l < 02 then lg(t)  b2l < c/ 3, and ifO < It al
0 there exists
+ J[g(t) b2]2 + J[h(t) b3)2 =
o > 0 such that if 0
(J(t)  b1 ] 2 + [g(t) b2] 2 + [h(t)  baf < c 2. But each term
on the left side of the last inequality is positive, so if 0 < It  a l
l
1
=>
.r '(t) = ((t +l)et,e'·,../2),
= (.J2et, ../2(t + 2)e1 , 2t + 3),
= . l(t + v
,../2t)
1)2e2t + e zt + 2,
( ) and
"'
t
r"(t) = ((t+2)et,et, O).
· Jr ' (t) x r " (t)l
=
lr'(t) x r "(t) l lr'(t)13
= y'2e 21 + =
Then
2(t + 2) 2 e 2t + (2t + 3)2,
2 2 y'2e t +'2(t + 2)2e t + (2t + 3)2 ((t + 1)2 e2t + e2t + 2]3/2
We plot the space curve and its curvature function for  5 :::; t :::; 5 below. K(l)
0.6
5
y
5
I
From the graph of r;,(t) we see that curvature is maximized fort = 0, so the curve bends most sharply at the point (0, 1, 0). The curve bends more gradually as we move away from this point, becoming almost linear. This is reflected in the curvature graph, where n,(t) becomes nearly 0 as /tl increases. 39. Notice that the curve b has two inflection points at which the graph appears alrnos,t straight. We would expect the curvature to
be 0 or nearly 0 at these values, bu't the curve a isn 't near 0 there. Th~s, a must be the graph of y = of curvature, and b is the graph .of y
f (x) rather than the graph
= K( x).
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www.elsolucionario.net
SECTION 13.3 ARC LENGTH AND CURVATURE
41. Using a CAS, we find (after simplifYing) ~~:(t) =
D
165
K(!)
6 v'4cos2 t  12cost + 13 ( _ cos t)3 12 • (To compute cross 17 12
products in Maple, use the VectorCa1cu1us or Li nearA1gebra package and the Cross Produc t (a, b) 0
command; in Mathematica, use Cross (a, b].) Curvature is
21T
41T
61r
I
largest at integer multiples of2:rr. 43.
X
= t2
=> :i; = 2t => X = 2, y = t 3 => iJ 2
\xy iJx\
Then ~~:(t) = [:i;2 + y2)3/2 45. x
= et cos t
=>
:i;
=
j(2t)(6t)  (3t )(2) j [(2t)2 + (3t2)2j3/ 2
= e1(cos t  sint)
y = et sint => iJ = et (cos t + sin't )
\:i;ij  1ixl . . [x2 + y 2 ] 312
x;(t) =
= 3t2
=>
=
=> ij = 6t.
j12t2 6t21 (4t2 + W)3/2
x = e1( 
6t2
= (4t2 + 9t4)3/2 ·
sint cost)+ et(cost sin t)
=> y = e1( sint +cost)+ et(cos t + sint) = 2et cost. Then 1
jet(cos t  sin t)(2et cost)  e (cos t + sin t)( 2et sin t}j = .:....__;..__ _ ___;_;_::_::':';;;;;~ 32 ([et(cos t sin t)J2 + [et(cos t +sin t)j2) /
j2e21(cos 2t  sin t cost + sin t cos t + sin2 t) j = [e21 (cos2 t  2costsint +sin2 t + cos2 t + 2 cos t sint +sin2 t)] 3 ' 2
47.
( 1,
)
2
=  2et sin t,
2
_
3 ,1 correspondstot  1.
_ r '(t) _ (2t,2t ,1) T (t)\r'(t)\  v'4t 2 + 4t 4 +
1
T '(t) =  4t(2t2 + 1) 2 ( 2t, 2t2, 1) + (2t2 + 1)1 (2,4t, 0) = (2t2 + 1) 2 ( 8t2
+ 4t
2
3
+ 2, 8t + 8t
3
j2e2 t(1) j
= [e2t(1 + 1)]312 =
2e2t 1 e31 (2) 3 / 2 = ..fiet
_ (2t,2t\ 1) _ (2 2 1 2t 2 + 1 ,soT(1)  3•3•3)·
[by Formula 3 of Theorem 13.2.3]
+ 4t, 4t) =
2(2t2 + 1)  2 (1 2t2 , 2t, 2t) ( 1  2t2, 2t, 2t)  (1 2t2, 2t,  2t) v'1  4t2 + 4t4 + 8t2 1 + 2t2
49. (0,7r,?) corresponds tot T(t) = r ' (t) \r' (t) \ T(1r)
=
='lr.
r(t) = (2sin 3t,t,2cos3t)
=>
(6 cos 3t,1,6sin3t) =:  1 (6 cos 3t, 1, 6 sin 3t ). } 36 cos 2 3t + 1 + 36sin2 3t V37
= ~ ( 6, 1, 0) is a normal vector for the normal plane, and so (6, 1, 0) is also normal. Thus an equation for the
plane ·is  6(x  0) + 1(y 1r) + O(z + 2)
= 0 ory ·
6x
= ~· 18
ffi=> N (t) =
~~:m,
= (sin3t, O,  cos 3t). So N(1r) = (0, 0, 1) and B (1r) =
~ (6, 1, 0) x
(0,0, 1) =
~ {1, 6, 0).
Since B(1r) is a normal to the osculating plane, so is (1, 6, 0) . An equation for the plane is 1(x 0) + 6(y 1r) ©
+ O(z + 2) =
0 or x
+ 6y =
61r.
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(c) B = T x N
0
167
T _L N , B _L T and B _LN. SoB, T and N form an orthogonal set of vectors in the three
dimensional space !R3 . From parts (a) and (b), dB fds is perpendicular to both BandT, so dB / ds is parallel to N . Therefore, dB / ds = r(s)N , where r(s) is a scalar.

(d) Since B = T x N , T _L N and both T and N are unit vectors, B is a unit vector mutually perpendicular to both T and N. For a plane curve, T and N always lie in the plane of the curve, so that B is a constant unit vector always perpendicular to the plane. Thus dB / ds = 0, but dB f ds =  r(s)N and N =/= 0, so r(s) = 0.
~~ s' =
=> r " = s" T + s' T ' = s" T + s'
61. (a) r ' = s' T
s" T
+ ~(s') 2 N
by the first SerretFrenet formula.
(b) Using part (a), we have
r'
X
r" = (s' T )
X
(s" T
+ ~cs')
2
N]
= ((si T ) x (s" T )] + ((s'T ) x (~(s'? N))
= (s' s")(T x T ) + ~~;(s')
3
(T x N )
[by Property 3 of Theorem 12.4. 11)
= 0 + ~~;(s'? B =
~~;(s') 3 B
(c) Using part (a), we have r 111 = [s" T
+ ~~;(s')
2
= s"' T + s"dT
ds
N ]'
=s
111
T
+ s" T ' ;t ~'(s'? N + 2~~;s' s" N + ~~;(s'? N '
s' + ~'(s'? N
+ 2~s' s" N + ~(s') 2 dN s' ds
= s"' T + s" s'~ N + K1(s') 2 N + 2~s' s" N + ~~;(s') (~ T + rB) = [s"'  K2 {s'?J T + (3~s' s" + ~~;'(s') ] N + ~~;r(s') B 3
[by the second formu la]
3
2
(d) Using parts (b) and (c) and the facts that B · T = 0, B · N = 0, and B · B = 1, we get (r ' x r ") . r 111 = ~~:(s') 3 B · { (s
111

~2 (s') 3 ] T
2
Jr ' x r " J 63. r = (t, T 

+ (3~s' s" + ~~ (s') 2 ] N + KT(s') 3 B} = ~~:(s') 3 ~r(s')3
=
(~~;(s')3] 2
J~~:(s')3 B J
2
·
T •
tt2, !t3 ) => r' = (1, t, t 2 ), r" = (0, 1, 2t), r 111 = (0, 0, 2) => r ' x r " = (t 2 ,  2t, 1) =>
(r 'x r ")· r "' Jr' x r 11 J2

(t2 ,  2t,1) ·(0,0,2) 2  ,....,..t 4 + 4t2 + 1  t 4 + 4t2 + 1
65. For one helix, the vector equation is r (t )
= (10 cost, 10 sin t, 34t/(2rr)) (measuring in angstroms), because the radius of each
helix is 10 angstroms, and z increases by 34 angstroms for each increase of 271' in t. Using the arc length formula, Jetting t go from 0 to 2.9 x 108 x 21r, we find the appmximate length of each helix to be L=
J;·
9
B
xlO x zrr
Jr'(t)J dt
= j~ ·
9
s
x 10
. /
x zrr
y ( 10 sint) 2 + {10cost) 2
+ U!)2 dt =
8 ] 2.9x l0 x2rr
..j100 + (~!)2 t
0
= 2.9 x 108 x 21r
J10o+ (g! ) ~ 2.07 x 10 2
10
A more than two meters!
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www.elsolucionario.net
168
CHAPTER 13 VECTOR FUNCTIONS
D
13.4 Motion in Space: Velocity and Acceleration 1: (a) Ifr(t) = x(t) i + y (t) j + z(t) k is the position vector of the particle at timet, then the average velocity over the time .interval [0, 1] is v ave
? k s· . I _ r(l) r (O) _ (4.5 i + 6.0j + 3.0k) (2.7 i + 9.8j + 3.7k) _ . . h th  1.8 1  3 . 8 J  0 . . 1m11ar y, over t e o er 1 10 •

intervals we have [0.5, 1] :
[1, 2] : [1, 1.5]:
_ r(1) r (0.5) _ (4.5 i + 6.0j + 3.0 k) (3.5 i + i2j + 3.3 k ) _ ._ k ._ 1  0.5 0.5  2.0 1 2 . 4 J 0.6
Vave
_ r (2) r (1) _ (7.3i+7.8 j+2.7k)  (4.5i+6.0j +3.0k) _ . ._ k .  28 · 1 + 1·8 J 03 · 2_ 1 1
V ave 
_ r(1.5) r(1) _ (5.9i+6.4j+2.8k)(4.5 i +6.0 j+3.0 k ) _ . · ._ k  28 · 1 + 08 · J 04 • 1. 5 1 0.5 . .
Yavo
(b) We can estimate the velocity at t = 1 by averaging the average velocities over the time intervals [0.5, 1] and [1, 1.5]:
v(1) ~ ~{(2 i  2.4 j  0.6k) + (2.8 i +0.8j 0.4k)] = 2.4i  0.8j  o·. 5k. Then the speed is Jv (1)\
3. r(t)
~ )(2.4) 2 + (  0.8)2 + ( 0.5)2 ~ 2.58.
= ( . 21 t 2 ,t)
=>
Att = 2:
= (2, 1)
v{t)
= r'(t) = (t, i)
v{2)
a(t)
= r"(t) = (: 1, 0)
a(2) .= (1,0)
jv(t)\ = Jt2 + 1
5.r(t)=3costi+2sintj
=>
Att = 71'/3:
v (t) = 3sinti+2costj
v(~) =  ¥ i+j
a(t) = 3cost i 2sintj
a(i) = ~ i  J3j
/
(3,0) X
Jv (t) \ = )9sin 2 t + 4cos2 t;:::: .) 4 + 5sin2 t Notice that x 2 /9 + y 2j4 = sin 2 t + cos2 t = 1, so the path is an ellipse. z
Att = 1: v(t) = i + 2tj a(t)
v (1) = i + 2 j
= 2j
a(1) = 2j
Jv(t)l=~
Here x = t, y = t2
Y
=> y =
2
~ and z = 2, so the path of the particle is a X
parabola in the plane z = 2.
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www.elsolucionario.net
.
SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
9. r (t) =
0
169
(t2 + t, e  t , t 3 ) => v(t) = r ' (t) = (2t + 1, 2t  1, 3t 2 ), a(t) = v' (t) = (2, 2, 6t),
lv (t)i = .j(2t + 1)2 + (2t 1)2 + (3t 2 )2 = v'9t4 + 8t 2 + 2. 11. r (t) = v'2t i + e' j + et k
= J(e' + e') 2 =
lv(t)l = v'2 + e2 ' + e zt 13. r (t) = e'(cost, sint, t)
v (t)
=> v (t) = r'(t) = v'2i + e' j  e' k, a(t) = v'(t) = e' j + e' k, e' + e' .
=>
= r'(t) = e'(cost,.sin t, t~ + et (sint,cost, 1) = et (cost 
sint,sint +cost, t + 1)
a(t) = v'(t) = 'e'(cost sint  sin t  cost, sin t +cost+ cost  sin t , t + 1 + 1) = et( 2sin t , 2 cost, t + 2)
lv(t)i = etJcos2 t + sin2 t  2cost sint + sin2 t + cos2 t + 2sin tcost + t 2 + 2t + 1 = e' .Jt2
15. a(t)
=i +
+ 2t + 3 =>
2j
v(t)
= Ja(t) dt = J(i +
.soC = k and v(t) = t i + 2tj + k. , r {t)
2j) dt
=ti+
2tj + C and k
= J v(t)"dt = J (t i +
2t j + k) dt
= v {0) = C, = ~t
2
i+
t2 j + t k + D .
But i = r (0) = D, soD= i and r(t) = (tt2 + 1) i + t 2 j + tk.
=> ·
17.. (a) a(t) = 2t i + sintj +cos 2t k
(b)
= J (2t i + sin t j + cos 2t k) dt = t2 i and i = v (O) = j + C, so C = i + j v(t)
and v(t) = (t 2 + 1) i + (1 cost) j + r (t) = J[(t 2 + 1) i + (1 cost) j +
= (tt3 + t) Butj = r(O) 19. r (t) = (t 2 ' 5t,
cost j
+ t sin 2t k + C
t sin2tk.
%
0.6 0.4 0.2 0
t sin2tk)dt
i + (t sin t)j ~cos 2t k + D
=  i k+ D , soD = j + {kand r (t) = (tt3 +t) i+ (t sint + 1)j + (! { cos2t) k.
e 16t)
=}
v(t)
= (2t, 5, 2t 
,16), lv(t)i = v'4t2 + 25 + 4t2  64t + 256 = v'8t2  64t + 281
and dd lv(t)i = t(se  64t + 281) 1 12 (16t  64) . This is zero if and only if the numerator is zero, that is, t . 16t  64 = 0 or t
= 4. Since
:t
lv(t) l < 0 fort < 4 a~d
:t
lv(t)l > 0 fort > 4, the minimum speed of /153 is atta.ined
at t = 4 units of time. 21 . IF (t)i = 20 N in the direction of the positive zaxis, so F (t) = 20 k. Also m = 4 kg, r{O) = 0 and v {O) Since 20k
= F (t) = 4 a(t), a (t) =
5k. Then v (t) = 5t k +
speed is lv (t)l = v 1 + 1 + 25t 2 = v'25t2
c1
=i
 j.
where c1 = i  j so v(t) = i  j + 5t k and the
+ 2. Also r (t) = t i 
t j + ~t2 k + cz and 0 = r (O), so C 2
=0
and r (t) = t i  t j + ~ t2 k.
®
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170
CHAPTER 13 VECTOR FUNCTIONS
0
23. lv(O) I = 200 m/s and, since the angle of elevation is 60°, a unit vector in the _direction of the velocity is (cos60°)i+ (sin60°)j
= ~i + 4!j. Thusv(O) = 2oo(~i +
4j)
= lOOi+ 100J3j and ifwesetuptheaxessothatthe
projectile starts at the origin, then r(O) = 0. Ignoring air resistance, the only force is that due to gravity, so F(t)
= ma(t) = m9 j where 9 ~ 9.8 m/s2. Thus a(t) = 9.8j and, integrating, we have v (t) = 
100 i + 100 V3 j
= v(O) =
C , so v( t)
r (t)
= lOOt i + (100 V3 t  4.9t2 ) j +
r(t)
= lOO t i +
'
= 100 i + (100 V3 
9.8t) j and then (integrating again)
D where·O = r(O) = D. Thus the position function of the projectile is
(100 V3t 4.9t2 ) j .
(a) Parametric equations for the projectile are x(t) y(t)
9.8t j +C. But
= O(and t > 0) =>
100 V3t 4.9t 2
= lOOt, y(t) =
100 V3 t  4.9t2. The projectile reaches the ground when
= t(100 V3 4.9t) = 0 => t = 104~f
~ 35.3 s. So the range is
x( 4°;0) = 100e 4~f3) ~ 3535 rn. 0
10
(b) The maximum height is reached when y(t) has a critical number (or equivalently, when the vertical component of velocity is 0): y' (t) = 0
=> 100 J3  9.8t = 0 => t
= 100~{3 ~ 17.7 s. Thus the maximum height is
v( 10~{!) = 100 V3 ( 10~{!)  4.9 ( 109~{!r ~ 1531 m. (c) From part (a), impact occurs at t =
°{!) = 1~0 i +
v ( 104
J.;f s. Thus, the velocity at impact is
10
[100 J3  9.8 ( 104
°f)] j = 100 i  100 J3j and the speed is
1.f) I= JIO,OOO + 30,000 = 200 mjs.
lv ( 10
25. As in Example 5, r (t)
= (vo cos45°)t i +
> .0)
velocity is vo
= J90g ~ 30 mjs.
t
= v5 s~ 2a. Thus
1502
(voJ2 t gt2 )
j] . The ball lands when
L.
27. Let a be the angle ,of elevation. Then v 0
d
= t[voJ2 t i +
voJ2 . . Iands 90 m away, . 90 = x = 1 vo vrn2' vo J2 = s. Now smce 1t  or v 02 = 2 9 g
y = 0 (and t
::::>:
[(vo sin45°)t t9t2] j
2 ;in a
Two angles of elevation then are a
'
= 150m/sand from Example 5, the horizontal distance traveled by the projectile is
= 800 => ~
909 and the .. . I m1t1a
sin 2a
10.2° and a
:=:::
= ~~~; ~ 0.3484 =>
2a
~ 20.4° or 180 
20.4
= 159.6°.
79.8°.
29. Place the catapult at the origin and assume the catapult is 100 meters from the city, so the city lies between (100, 0)
and (600, 0). The initial speed is v 0 = 80 m/s and let 0 be the angle the catapult is set' at As in Example 5, the trajectory of the catapulted rock is given byr(t) = (80cosO)ti + [(80sin0)t 4.9t2] j. The topofthe near city wall is at (100, 15), which the rock will hit when (80 cos 0) t = 100 =>
®
5 O and (80 sin O)t  4.9t2 4 cos
t=
= 15 =>
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D
SECTION 13.4 MOTION IN SPACE: VELOCin' AND ACCELERATION
171
2
80 sin 9 . 5  4.9 ( 5) 4cos9 4 cosfJ
=> 100 t a n ()  7.65625 sec2 fJ = 15. Replacing sec2 9 with tan 2 8 + 1 gives
= 15
7.65625 tan2 9  100 tanfJ + 22.65625 = 0. Using the quadratic formula, we have tan 9 ~ 0.230635, 12.8306
=>
B ~ 13.0°, 85.5°. So for 13.0° < 9 < 85.5°, the rock will land beyond the near city wall. The base of the far wall is located at (600, 0) which the rock hits if (80 cos fJ)t
= 600
15 = 2 COSu n and (80sin 9)t 
=> t
4.9t2
=0
=>
2
. 15n  4.9 (  15 (} ) 80sm0 · 2cOS!7 2 COS
=> 600tan9 275.625sec2 fJ = 0 =>
= 0
275.625 tan2 9  600 t an0 + 275.625
= 0. Solutions are tan(}~ 0.658678, 1.51819
rock lands beyond the enclosed city ground for 33.4°
=> (} ~ 33.4°, 56.6°.· Thus the
< (} < 56.6°, and the angles that allow the rock to land on city ground
< () < 33.4~, 56.6° < 9 < 85.5°. If you consider t~at the rock can hit the far wall and bounce back into the city, we
are 13.0°
calculate the angles that cause the rock to hit the top of the wall at (600, 15): (80 cos 9)t = 600 (80sin fJ)t 4.9t2 = 15 =>
600tan0 275.625sec2 (}
Solutions are tan 9 :::::: 0. 727506, 1.44936 13.0°
2 cos(}
=> 275.625 tan2 (}  600 tan(}+ 290.625
= o.
=> (} :::::: 36.0°, 55.4°, so the catapult should be set with angle fJ where
< (} < 36.0°' 55.4° < (} < 85.5° .
31. Here a (t)
=  4j 
32 k so v (t)
= 4tj 
32t k + v o =  4t j  32t k + 50 i +80 k
r (t) = 50t i 2t2 j + (BOt 16e) k (note that r o and t
= 15
=> t = ~ and
> 0: SOt  16t 2
= 16t{5  t) = 0
=>
= 50 i 4t j +
(80  32t) k and
= 0). The ball lands when the zcomponent of r (t) is zero
t = 5. The position of the ball then is
r (5) = 50(5) i  2{5) 2 j + [80(5)  16(5) 2 ] k = 250 i  50 j or equivalently the point (250,  50, 0). This is a distance of J250 2 + ( 50)2 + 02
= )65,000 :::::: 255ft from the origin at an angle oftan 1 (~)
::::::
11.3° from the eastern direction
toward the south. The speed of the ball is lv (5) l = I50 i  20j  80 k l = J50 2 + (  20)2 + ( 80)2 = v'9300:::::: 96.4 ftls. 33. (a) After t seconds, the boat will be 5t meters west of point A. The velo~ity
20
of the water at that location is 4~0 (5t)(40 5t) j . The velocity of the boat in still water is.5 i , so the resultant velocity of the boat is v(t) = 5 i + 4~ 0 (5t)( 40 5t) j = 5i + ( ~t fGt 2 ) j . Integrating, we obtain r (t)
= 5t i +
(~tl
 fGt3 ) j +C. lfwe place the origin at A (and consider j
to coincide with the northern direction) then r (O)
=0
reaches the east bank after 8 s, and it is located at r(8)
=> C
= 0 and we have r (t) = 5t i +
= 5(8)i +
( ~(8) 2

fG (8?) j
= 40 i +
(~t 2

1 16
e) j . The boat
16j. Thus the boat is 16 m
downstream. (b) Let Q be the angle north of east that the boat heads. Then the velocity of the boat in still water is given by
5(cos Q) i + 5(sin Q) j . At t seconds, the boat is 5(cos Q)t meters from the west bank, at which point the velocity of the water is .1 ~ 0 [5(cos Q)t][40 5(cos Q)t] j . The resultant velocity of the boat is given by
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172
D
CHAPTER 13 VECTOR FUNCTIONS
g
v(t) = 5(cos a:) i + [5 sin a:+
4 0 (5t cos a:)( 40
 5t cos a:)] j
Integrating, r (t) = (5t cos a:) i + (5tsin a:+ ~t2 cos a:
=
ftt3 cos2 a:) j
the origin at A). The boat will reach the east bank ~hen 5t cos a: = 40 In order to land at point B ( 40, 0) we need 5t sin a: + ~ 8 5(  ) sin cos a:
8 a:+~ (cos  ) a:
40 sin a: + 16 = 0
=?
2
8
3
sin a: =
 %. Thus a: =
sin 
e cos a:  ft t
ft (cos a:) .cos2 a:= 0
cos a:
l
(%)
fue cos 2 a:) j.
(5 cos a:) i + (5 sin a:+ ~t cos a: (where we have again placed =?
3
t
=~
cos 2 a: = 0
=?
8
=  . cos a:
5 cos a: =?
1   (40sina: + 48  32) cos a:
=0
=?
~ 23.6°, so the boat should head 23.6° south of
If
east (upstream). The path does seem realistic. The boat initially heads upstream to counte~act the effect of the current. Near the center of the river,
0 1=:::==::::::::;::>of'...::::::::=:=::::~ 40
the current is stronger and the boat is pushed downstream. When the boat nears the eastern bank, the current is slower and the boat is able to progress upstream to arrive at point B.
 12
35. Ifr' (t) = c x r(t) then r' (t) is perpendicular to both c and r(t). Remember that r ' (t) points in the direction of motion, so if r' (t) is always perpendicular to c, the path of the particle must lie in a plane perpendicular to c. But r ' (t) is also perpendicular to the position vector r(t) which confines the path to a sphere centered at the origin. Considering both restrictions, the path must be contained in a circle that lies in a plane perpendicular to c, and the circle is centered on a line through the origin in the direction of c. 37. r(t)
= (3t 
t 3) i
+ 3t 2 j
r'(t) = (3  3t2 ) i + 6tj,
=?
lr' (t)i = J(3 3t2 )2 + (6t)2 = v'9 + 18t2 + 9t4
=
J(3  3t 2 ) 2 = 3 + 3t
2
,
r" (t) ·= 6t i + 6j, r' (t) x r" (t) = (18 + 18t2 ) k. Then Equation 9 gives  r ' (t) . r" (t)  (3  3e)( 6t) + (6t)(6)  18t + 18t 3 aTlr '(t)/ 3 +3t2 3 + 3t2
aT = v
I
=
d [ 2] ] dt 3 + 3t = 6t
39. r (t) = cost i +sin t j + t k r" (t) =  cost i  sin tj, Then D.T
an
=?
dE


2
18t(1 + t ) 3(1 + t 2 )


6t
[or by Equation 8, 2
. LO .  lr'(t) X r "(t) i  18 + 18e  18(1 + t ) quat10n g1ves aN ir'(t)i  3 + 3 t 2  3 (1 + t 2)
r '(t) =  sint i + costj + k ,
ir'(t)i = Vsin 2 t + cos2 t + 1 =

6 ·
V'2,
r' (t) x r" (t) =sin t i  cos tj + k.
_ r'(t) · r"(t) _ sintcost  sintcost _ d _lr'(t) x r"(t)l _ Vsin2 t+cos 2 t+ 1 = .J2 = l 0 lr'(t)i  . .j2  an aN lr'(t)i /2 /2 ·
41. r (t)=et i +/2t j +et k
=?
r"(t) = et i + et k. Then aT =
®
r'(t)=et i+/2j e 1 k, e2t _ e 2t
et
+e
t
=
(et
+ e t)(et _ et + e t
ir(t) i=v'e21· +2+ e
e t)
2
t=J(e1· +e t)2=e'· +e t,
.
= et  e: t = 2sinht
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CHAPTER 13 REVIEW D
173
43. The tangential component of a is the length of the projection of a onto T, so we sketch the scalar projection of a in the tangential direction to the curve and estimate its length to be 4.5 (using the fact that a has length 10 as a guide). Similarly, the normal component of
a is the length of the projection of a onto N, so we sketch the scalar projection of a in the normal direction to the curve and estimate its lengttJ to be 9.0. Thus aT
:::::::
4.5 cmf s 2 and
X
45. _If the engines are turned off at timet, then the spacecraft wi ll continue to travel in the direction ofv(t) , so we need at such that for some scalars
4 so 7 t2
> 0, r (t) + s v (t)
24t 12t2
8(3 t)t
+ 1 + (t2 + 1)2
= 9
¢}
v(t) = r ' (t) = i
= (6, 4, 9).

4
 ;:;;: :::.;: = 2 (t2 + 1)2
¢}
t
4
+
+ ~ j + (t2 ~ 1) 2 k =>
+ 8t2 
·a is easily seen that t = 1 is a root of this polynomial. Al~o 2 ln 1 + ~
13
3
1
12t + 3
= 0.
= 4, sot= 1 is the desired solution.
Review CONCEPT CHECK
1. A vector function is a function whose domain is a set of real numbers and whose range is a set of vectors. To find the derivative or integral, we can differentiate or integrate each component of the vector funct ion.
2. The tip of the moving vector r (t) of a continuous vector function traces out a space curve. 3. The tangent vector to a smooth curve at a point P with pos ition vector r(t) is the vector r ' (t). The tangent line at p ' is the line through P parnllel to the tangent vector r ' (t). The unit tangent vector is T (t) =
,::m,.
4. (a) (a)  (f) See Theorem 13.2.3.
5. Use Formula 13.3.2, or equivalently, 13.3.3. 6. (a) TI1e curvature of a curve is"' =
(b) K.(t)
'(t) I = I. Tr '(t)
I~~
I
where Tis the unit tangent vector.
_
(c) K.(t) = lr' (t ) X r" (t)l
T' (t) 7. (a) The unit normal vector: N (t) = IT'(t) l'
The binormal vector: B (t)
if"(x) l
·
(d) K(x) [1 + (f'(x))2j3/2
[r'(t)[ 3
= T (t)
x N (t).
(b) See the discussion preceding Example 7 in Section 13.3. 8. (a) Ifr(t) is the position vector of the particle on the space curve, the velocity v (t) = r'(t), the speed is given by lv (t )i, and the acceleration a (t) = v ' (t) = r " (t). © 2012 Ccngugc Learning. All Rights Reserved. Mny not be SCAnned, copied. or duplicated, or posted to a publicly ncccssib\e website. in wltolc or in part.
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174
D
CHAPTER 13 VECTOR FUNCTIONS
(b) a = a7:T +aN N where aT = v' and aN = 1w 2 .
9. See the statement of Kepler's Laws on page 892 [ET 868].
TRUEFALSE QUIZ 1. True. If we reparametrize the curve by replacing u
with direction vector i
= t 3 , we have r(u) = u i + 2u~ + 3uk, which is a line through the origin
+ 2 j + 3 k.
3. False. The vector function represents a line, but the line does not pass through the origin; the xcomponent is 0 only for t ;, 0 which co~esponds to the point {0, 3, 0) not {0, 0, 0). d 5. False. By Formula 5 ofTheorem 13.2.3, d (u(t) x v(t)]
.
t
7. False.
K
= u '(t ) x
v (t)
+ u (t) x v'(t).
is the magnitude of the rate of change of the unit tangent vector T with respect to arc length s, not with respect tot.
9. True. At an inflection point where f is twice continuously differentiable we must have !" (x) = 0, and by Equation 13.3.11, the curvature is 0 there. 11. False. If r(t) is the position of a moving particle at timet and lr(t)l = 1 then the particle lies on the unit circle or the unit sphere, but this does not mean that the speed lr'(t)l must be constant. As a counterexample, let r (t) = (t, Jl=t2), then r' (t)
= ( 1, t/J1 
t 2 ) and lr(t) l =
Jt 2 + 1 
t 2 = 1 but lr'(t)l
= yf1 + t 2 /(1 t 2 ) =
1/ ~ which is not
constant.
13. True. See the discussion preceding Example 7 in Section 13.3.
EXERCISES 1. (a) The corresponding parametric equations for the curve are x
y = cos 1rt, z = sin rrt. Since y 2 I
+ z2
= t,
= 1, the curve is contained in a
circular cylinder with axis the xaxis. Since x
= t, the curve is a helix.
= t i + cos ri j +sin rrt k => r'(t) = i  rrsin 1rt j + rrcos rrt k ~
(b) r(t)
r" ( t) = rr2 cos 1rtj 
1r
2
X
siri 7rt k
3. The projection of the curve C of interSection onto the xyplane is the circle x 2 x
= 4 cost, y = 4 sin t,
0
S t S 21r. From the equation of the plane, w~ have z
equations for Care x = 4 cost, y r(t) = 4 cost i
©
+ y2 =
+ 4 sin t j + (5 
= 4 sin t, z = 5 
4 cost) k , 0
4 cost, 0
=5
16, z = 0. So we can write x
=5
4 cos t, so parametric
S t S 2rr, and the corresponding vector function is
S t S 21r.
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CHAPTER 13 REVIEW· 0
5.
f~(t 2 i +tcos 1rtj
+sin 1rt k ) dt
= (!01 t 2 dt) i + (I~ tcos 1rtdt) j +(I; sin 1rtdt) k I
= [~t3 ]~ i + (~sin 1rt]~  f 01 ~sin 1rtdt) j +[~cos 1rt]~ k 1 •
t]l .
[ 1
= ]j l +
2
;1fCOS1l" 0 J+ ;r
k ;= 1 1•  ;:lf 2 • 2 k J+ ;r 3
where we integrated by parts in the ycomponent. 1. r(t)
= ( t 2 , t 3 , t 4) .
. L=
=?
1
r (t)
= ( 2t, 3e,4t3 )
=?
Jr (t) J = J 4t 2 + 9t4 + 16t6 and 1
J; Jr (t)J dt = J; J4t2 + 9t4 + 16t6 dt. Using Simpson's Rule with f(t) = J4t 2 + 9t4 + 16t 1
have tlt
6
and n = 6 we
= 3(i0 = ~ and L ~ ';' [f(O) + 4f(~) + 2f{1) + 4Ja) + 2/(2) + 4f(~) + /{3)]
=i
[vo
+o +o + 4 . ..j4(~/
+ 9 (~)
4
+
16(~)6 +
2 . .J4(1)2 + 9{1) 4 + 16(1)6
2 4 + 4. J4G) + 9G) + 16 Gt + 2 . .J4(2)2 + 9(2)4 + 16{2)6
+4· ~
4(~)2 +9(~)4 +16(~t+.J4(3) 2 +9{3)4+ 16(3)6 ]
86.631
9. The angle of intersection of the two curves, 8, is the angle between their respective tangents at the point of intersection.
For both curves the point {1, 0, 0) occurs when t r W)
=
sin t i +cos t j + k
r i {0) · r~ (0) = (j + k ) · i
=?
= 0.
rHO)
= 0.
=j +
k and r 2(t)
= i + 2t j + 3t2 k
=}
r~(O)
Therefore, the curves intersect in a right angle, that is, 8 =
= i.
~.
(b) T'(t) = Ht4 +t2 + 1) 3 12 (4t3 +2t) (t2,t, 1) + (t4 + t 2 + 1) 1 12 (2t, 1, 0)  2t3  t 2 1 (t4 + t2 + 1)3/ 2 (t , t , 1) + (t4 + t2 + 1)1 / 2 (2t, 1, 0)  ( 2t5 1
IT (t)l
=

t 3 ,  2t4

t 2 ,  2t3  t) + (2t5 + 2t3 + 2t, t 4 + t 2 + 1, 0) (t4 + t2 + 1)3/ 2 
Jt 6 + 4t4 + 4t2 + t 8
 2t 4 + 1 + 4t6 + 4t 4 + t 2 (t4 + t2 + 1)3/2
=
(t 3 + 2t,  t 4 + 1, 2t 3 (t4
JtB + 5t6 + 6t4 + 5t2 + 1 (t4 + t2 + 1)3/ 2
+ t2 + 1)3/2 and
(t 3 + 2t 1  t 4  2t3  t ) ' '  Jt!!. + 5t6 + 6t4 + 5t2 + 1 .
N (t) 
(c) "'(t)
=
JT 1 (t) \ Jrl(t) l
=
(t4 + t2 + 1)2
13. y I = 4X 3 , y II = 12 X 2 and It( X )
®
+ 4t2 + 1 + t2 + 1)3/ 2
Jt 4
J t!!. + 5t6 + 6t4 + 5t2 + 1 or
(t4
= (1 + lv"l (y')2)3/ 2
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T '(t)
= ~ ( 4sin 2t, 0, 4cos 2t)
B = T xN =
=> T (t) =
r '(t) = (2cos2t, 1, 2 sin2t)
=> N (t)
js (2 cos2t, 1,  2sin2t) =>
= (sin 2t, 0, cos2t). SoN =
N ('rr) = (0, 0,  1) and
js (1, 2, 0}. So a normal to the osculating plane is ( 1, 2, 0) and an equation is
 l (x 0) + 2(y 1r) + O(z  1) = 0 or x  2y + 27r = 0. 17. r (t) = tlnt i +tj + et k , v (t) = r '(t) = (1 + lnt) i + j  e~ k,
lv (t)i =
J(l + ln t)2 + 12 + (e:t)2 =
vf2 +2 ln t + (ln t)2 +e 2 t,
a(t) = v'(t) =
t i + et k
19. We set up the axes so that the shot leaves the ath lete's hand 7 ft above the origin. Then we are given r (O) = 7j, lv (O)I = 43 ft/ s, and v (O) has direction given by a 45° angle of elevation. Then a unit vector in the direction ofv(O) is ~(i + j )
=> v (O) =
~(i + j ). Assuming air resistance is negligible, the only external force is due to gravity, so .as in
Example 13.4.5 we have a =  g j where here g ~ 32 ft/s 2 . Since v ' (t) = a (t), we integrate, giving v (t) =  gt j + C where C = v (O) = r(t) =
~(i + j ) => v (t) = ~ i + ( ~  gt) j . Since r '(t) = v (t) we integrate again, SO·
~t i + ( ~t  ~gt2) j + D. ButD =
(a) At 2 seconds, the shot is at r(2)
r (O) = 7j
=> r (t) = ~t i + ( ~t ~gt2 + 7)j.
= ~(2) i + ( ~ (2)  ~g(2) 2 + 7) j ~ 60.8 i + 3.8j, so the shot is about 3.8 ft. above
the ground, at a horizontal distance of 60.8 ft from the athlete. (b) The shot reaches its maximum height when the vertical component of velocity is 0: ~  gt = 0
=>
t = ;: ~ 0.95 s. Then r (0.95) ~ 28.9 i + 21.4j , so the maximum height is approximately 21.4 ft. v2g
.
(c) The shot hits the ground when the vertical component ofr(t) is 0, so ~t  ~gt 2 + 7 = 0 => 16t2 + ~t + 7 = 0 => t ~ 2.11 s.
r (2. 11) ~ 64.2 i  0.08j , thus the shot lands approximately 64.2 ft from the
athlete. 21. (a) Instead of proceeding directly, we use formula 3 ofTheorem 13.2.3: r (t) v = r'(t)
= t R (t) =>
= R (t) + t R' (t) = coswt i + sinwtj +t v a.
(b) Using the same method as in part (a) and starting with v = R (t) + t R '(t), we have
a= v ' = R '(t) + R '(t) + t R "(t) = 2 R '(t) + t R "(t)
= 2vd +
tad.
(c) Here we have r(t) = e' coswt i + e t sin wtj ,; et R (t). So, as in parts (a) and (b), v = r '(t) = et R '(t) et R (t) = et[R '(t)  R (t)) => a = v' = et [R"(t) R '(t)) e:t[R '(t)  R (t)) = e~ [R"(t) 2 R ' (t) + R (t)J
Thus, the Coriolis acceleration (the sum of the "extra" terms not involving aa) is 2e~ v .i + et R.
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CHAPTER 13 REVIEW 23. (a) r (t) = R coswt i
v
= wR(
+ R sin wtj =>
sin wti + cos wtj). v · r
v = r ' (t) = wR sinwt i
= wR 2 ( 
coswt sin wt
+ wR coswt j , so r
= R(coswt i
+ sinwtcoswt) = 0, so v
0
177
+ sinwtj) and
..L r . Since r points along a
radius of the circle, and v ..L r , vis tangent to the circle. Because it is a velocity vector, v points in the direction of motion. (b) In (a), we wrote v in the form wR u, where u is the unit vector  sin wt i + cos wt j . Clearly Ivi . Ie campI etes one revo I ut10n, . . T speed wR , the part1c a d.1stance 21r R,.m t1me
= wR lui = wR . At
= 2·nR R = 27r . w
. w
(c) a = dv =  w2 R coswt i  w2 Rsinwtj =  w2 R(coswt i + sinwt j ), so a =  w2 r. This shows that a is proportional
dt
to rand points in the opposite direction (toward the origin). Also, ja j = w 2 jr j
= w2R .
(d) By Newton's Second Law (see Section 13.4), F = ma, so jF j = m jaj = mRw2 = m
(~R)
2
= m ~~
2
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0
PROBLEMS PLUS
1. (a) The projectile reaches maximum vosina t= an.d y
. )(vosiiJ.a) = (vosma  g
g
!
he~ght when 0 = ~~ =
[(Vo sina)t
1 (vosina) g 2 g
2
2
2
°
Vo sm
2g
~gt2] = vo sin a 
gt; that is, when
a . Th'1s IS . the max1mum . 'ght attamed . 11e1 when .
the projecti le is fired with an angle of elevation a. This maximum height is largest when a = ~ . In that case, sin a = 1
. . . vfi and the maximum he1ght IS  . 2g (b) Let R = vfi
/g.
We are asked to consider the parabola x 2 + 2Ry  R 2 = 0 which can be rewritten as y = 
The points on or inside this parabola are those for which  R
~ x ~ Rand 0 ~ y ~ ;~ x 2 + ~.
__!._ x 2 + 2R
!!:. . 2
When the projectile is
fired at angle of elevation a, the points (x, y) along its path satisfy the relations x = (v 0 cos a) t and
tyt 2 , whereO ~ t
y = (v0 sina)t 
lxl
~ (2v0 sin a)/g(as inExan1ple 13.4.5). Thus
~ lvo cos aCvo ;ina) I= I~ sin 2al ~ I~; I =
IRI. This shows that R
~ x ~ R.
vo sin a ~ gt ) = t gt Cvo ;in a t) ;: : 0 and
For t in the specified range, we also have y = t (
y
, X 9 = (vosma)  vo cos a 2 0
 1
2
y ( 2Rx +
R)
2
(
X
 vo cos a
)2= (tana)x 
g
x = .
1 1 2 ( 2 = 2Rcos2 ax + 2Rx + tan a) x
=
~ ( 1  _1_ ) cos2
2R
2
a
1
2
2v 02 cos2 a
2
R x +(tan a) x. Thus 2 cos 2 a
R
2
2
2
+ (tana)x _ R = x (1 sec a) + 2R{tana) x  R 2 2R
2
_ {tan a) x + 2R (tan a) x  R 2 ·_ 2R 
:
2
[{tana) x  R]2 . 2R ~0
We have shown that every target that can be hit by the projecti le lies on or inside the parabola y = _ __!._ x 2 + R. 2R 2 Now let (a, b) be any point on or inside the parabola y = 
2~ x
2
+
~· Then·R ~a~ Rand 0 ~ b ~

2~ a
2
+
~
We seek an angle a such that (a, b) lies in the path of the projectile; that is, we wish to find an angle a such that
b= 
2~c~s2 a a
2
+(tan a) a or equivalently b =
~ (tan2 a ~1 1)a2 +(tan a) a . Rearranging this equation we get
;~ tan2 a a tan a+(;~+ b) = 0 or a 2 (tana? 
2
2aR(tana) + (a + 2bR) = 0 (*) . This quadratic equation
for tan a has real solutions exactly when the discriminant is nonnegative. N~w B 2

4AC ;::: 0
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www.elsolucionario.net
179
180
0
CHAPTER 13 PROBLEMS PLUS
(2aR? 4a2 (a 2
b:::;
2~ (R
2
a2 )

y ·= 
2~ x
tan a
= 2aR ±
2
+ 2bR);:::: 0
0
v ) v2 2 ( t  .9 sin a + 29 cos a dt
t  (v / g) sin o 2
+ [(vl g) cosa]2 ln 2
(t~sino + g
[using Formula 2 I in the Table of lntegrals]
[v. (~.rna)' + (~ oo.o)' +(~ '"'")' m( ~ •ina+ (~ •mo)' +(~=aY)
= g gsma
2
(~ •ina)' + (~=o)' (~oooo)' m(~•ina+ (~•ina)' + (~=oY)]
+~sin o g
[~sino · ~+ v cos2 aln(~ sino+~) +~ sino·~  v cos2 a ln(~ sino+~)] g g2 g g g g g2 g g 2
2
= g_ 2 g 2
2
+ cos v 2 . = v stno a g 2f1 We want to maximize L(a) for 0
vg
1n(  (vl(vIg)sin a+v/g) = sm v v a + cos an 1 (1+sino) g) sin a +vI g g 2g 1  sin a 2
2
.
2
s; a s; 1r/ 2.
. ( . a ln ( 1 + sin = v cos a + v cos a · 2  2 cos a sm . a)] g 2g coso , 1 sma 2
L'(a) =  coso+ v
2
[
2g
2
v
2
9
a

a)]
2cosa smo ln 1 + sin . . 1 sm o
2
•
=
a
2 cos cos2 a · 1  sin · . . 1 +sma (1smo) 2
[
2
2 v COSO! [ 1 SlDO! • COSO!+
g
2
] . ln (l+sina) 1  SlDO!
=
v
9
. •ln (1+ sin COSO! [ 2 SIDQ . a) 1  SJDO! ,
L( a ) has critical points for 0 < a < 1r12 when L' (a) = 0 => 2  sin a ln (
~
+ 
]
s~ a) = 0 [since cos a =1= 0].
SIDQ
Solving by graphing (or using a CAS) gives a~ 0.9855. Compare values at the critical point and the endpoints: L(O)
=
0, L(1r12) = v 2 I g, and £(0.9855) ~ 1.20v 2 I g . Thus the distance traveled by the projectile is maximized
for a
~
0.9855 or
~
56°.
7. We can write the vector equation as r (t) = ·at2 + bt + c where a= (a1, a2 , a a}, b = (b1 , b2 , ba }, and c = (c1 , c 2, c3). Then r ' (t) = 2t a+ b which says that each tangent vector is the sum of a scalar multiple of a and the vector b . Thus the tangent vectors are all parallel to the plane determined by a and b so the curve must be parallel to this plane. [Here we assume that a and bare nonparallel. Otherwise the tangent vectors are all parallel and the curve lies along a single line.] A normal
© 2012 Ccngoge Lcoming. All RighLS Rescm :d. May oot be scanocd, copied, 0< duplicnted. 0< poSied to a publicly :10e 0, the level curves are hyperbolas
with axis the yaxis.
0 X
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,
SECTION 14.1
51. The contour map consists of the level curves k = x 2 ellipses with major axis. the x axis. (Or, if k The graph of f(x, y) is the surface z = x
2
FUNCTIONS OF SEVERAL VARIABLES
0
187
+ 9y2 , a family of
= 0, the origin.)
+ 9y2 , an elliptic paraboloid.
X
If we visualize lifting each eUipse k = x 2
z
+ 9y2 of the contour map to the plane
= k, we have horizontal traces that indicate the shape of the graph of f . X
53. The isothermals arc given by k = 100/(1 + x 2
x2
+ 2y2 =
55. f(x , y)
(100 k)/k [0
= xy2 
0, we have a family of circular cyl inders with axis the xaxis and
radius ,fk. When k = 0 the level surface is the xaxis. (fhere are no level surfaces fork
< 0.)
69. (a) The graph of g is 'the graph off shifted upward 2 units. (~)The graph of g is the graph
off stretched vertically by a factor of2.
(c) The graph of g i~ the graph off reflected about the xyplane. (d) T he graph of g(x, y) =  f(x , y) + 2 is the graph off reflected about the xyplane and then shifted upward 2 units.
71. f(x,y)
= 3x 
x4

4y
2

20r.
lOxy
10
0
y
y
Threedimensional view
Front view
It does appear that tlle function bas a maximum value, at the higher of the two "hilltops." From the front view graph, the maximum value appears to be approximately 15. Bolli h illtop~ could be considered local maximum points, as the values off there are larger than at the neighboring points. There doe~ not appear to be any local minimum point; although tlle valley sh(\pe between the two peaks looks like a minimum of some kind, some neig hboring points have lower function values.
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2012 Ccngage Learning. All Righls H.lscn'Cd. Mny not be scunncd, copied, or duplicated. or posted 10 a publicly accessible
www.elsolucionario.net
wcbsil~.
in whole or in p;lr1.
SECTION 14.1
73.
FUNCTIONS OF SEVERAL VARIABLES
D
189
x+y +y
10
f (x, y) =   • As both x and y become large, the function values 2 2 X
5
appear to approach 0, regardless of which direction is considered. As
0
(x, y) approaches the ~rigin, the graph exhibits asymptotic behavior. From some directions, f(x , y)+ oo, while in others f(x , y) + oo.
5
(These are the vertical spikes visible in the graph.) If the graph is examined carefully, however, one can see that f(x, y) approaches 0 along the line y =  x.
y
75. f( x, y)
z=
2 eY
= ecx 2 +v2 • First, if c =
0, the graph is the cylindrical surface
(whose level curves are parallel lines). When c > 0, the vertical trace
above theyaxis remains fixed whi.le the sides of the surface in the x direction
z
"curl" upward, giving the graph a shape resembling an elliptic paraboloid. The level curves of the surface are ellipses centered at the origin.
c=O For 0
< c < 1, the ellipses have major axis the x axis and the eccentricity increases as c + 0.
y
1.2
c = 0.5 (level curves in increments of 1) For c = 1 the level curves are circles centered at the origin.
1.2
c
= 1 (level curves in increments of 1) [continued)
©
2012 Cengage Le:uning. All Rights Rcscn·al. Mlly not be scanned, copied, or duplicotcd 1or posted to o publicly oot 1, the level curves are ellipses with major axis theyaxis, and the eccentricity increases as c increases.
1.2
c = 2 (level curves in increments of 4) For values of c
< 0, the sides of the surface in the xdirection curl downward and approach the xyplane (while the vertical
trace x = 0 remains fixed), giving a saddleshaped appearance to the graph near the point (0, 0, 1). The level curves consist of a family of hyperbolas. As c decreases, the surface becomes flatter in the x direction and the surface's approach to the curve in the trace x = 0 becomes steeper, as the graphs demonstrate.
1.2
c =  0.5 (level curves in increments of0.25)
1.2
c 77. z = x 2
+ y 2 + =Y· When c
0 is an ellipse (see graph below). The intersection with the s urface and the planes x = 0 and
= y 2 and z =
x 2 respectively, so the surface is an elliptic paraboloid.
> 0 the graphs have the same shape, but are reflected in the plane x = 0, because
x2 + y 2 + cxy = ( x) 2 + y 2 +
c
0
0, z ~ 0 for all x andy. Ifx andy have the same sign, then
+ y 2 + cxy ~ x 2 + 1i 
When c
FUNCTIONS OF SEVERAL VARIABLES
( c)(  x)y. That is, the value of z is the same for c at (x, y) as it is for cat ( x, y).
= 1, z = 2'
c= O
So the surface is an elliptic paraboloid for 0
79. (a) P = bLo: K 1  o:
=?
< c < 2, a parabolic cylinder for c =
= bL"' J{0,0)
f
is
!( x, y ) = 0203 + 0302  5 =  5 , ven'fy'mg 2
2 0 0 0
our guess. 5. f(x, y)
= 5x3 
. 7. f( x, y ) =
x 2 y 2 is a poly nomial, and hence continuous, so
4  xy X2
+3y2
f(x, y)
+

f(x, y)
= f(l, 2) = 5{1) 3 
2
{1) (2)
2
= 1.
.tS a ratwna . .., . dh . . d . 1 ;unctiOn an encc contmuous on tts omam.
.
(2, 1) is in the domain off, so f is continuous there and 9. f(x, y) = (x4
' lim
(x,y) {1 ,2)
4y2 )/(x2
+ 2y 2).
lim
{x, y) < (2,1)
f(x, y) = !(2, 1) =
(~)~ +( 2~~~~ 2 4
First approach (0, 0) along the xaxis. Then.f(x, 0) = x /x
0. Now approach (0, 0) along theyaxis. For y =f. 0, f(O, y)
= 4y2 /2y2 =
2
= x2
2 7 for x
=f. 0, so
2, so f(x , y)+ 2. Since f has
two different limits along two different lines, the limit does not exist. ® 2012 Ccngoge Le3ming. All Rights Reserved. May nol he scanned, copied, urduplicated, or po5tcd ton publicly ncccssiblc website, 'in "'··hole or in p:m.
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SECTION 14.2 11.
j(x , y) = (y 2 sin 2 x)j(x4
+ y 4 ).
On the x axis, f(x, 0) = 0 for x 2
.
x axis. Approaching (0, 0) along the hne y
lim sin x = 1, so f(x, y)+
x
:c o
·
13. f(x, y) = ~·
2
= x, f(x, x) = xx 4sin + x 4x
2
=
sin x x2 2
1 ( sinx) =2 ;
(0, 0) along the
2
for x :/: 0 and
has two different limits along two different lines, the limit does not exist.
·
~ I ~~~ lxl since lvl ~ x2 + y2 x2yeY X4
+ 4y 2
.
)x2
Then f(x, 0) = 0 for x
+.y 2 , and lxl + 0 as (x, y)+
'# 0, so f(x, y) +
(0, 0). So
f(x , y) + e0 /5
lim
(x,y)(o,o)
o:
f(x, y) =
0 as (x , y)+ (0, 0) along the x axis. Approaching 2
(0, 0) along theyaxis or the line y
= x also gives a limit ofO. But f (x, x 2 )
2
= x:: 4
::r2
4
tx)2 = x5: 2
::c2
T a:2
=
4
for x
'# 0, so
= t as (x,y)+ (0, 0) along theparabola y = x 2 . .Thus the limit doesn' t exist.
2
19. eY is a composition of continuous functions and hence continuous . xz is a continl.!ous function and tan t is continuous for
t '#
~
+ mr (nan integer), so the composition tan(a f ,., (x , y) =  8x and fv(x , y) = 2y => f ,., (1, 2) =  8 and fv (1, 2) =  4. The graph
off is the paraboloid z = 16  4x2

y 2 and the vertical plane y = 2 intersects it in the parabola z = 12  4x 2 , y = 2
(the curve C 1 in the first figure). The slope of the tangent line to this parabola at (1, 2, 8) is f x(1, 2) =  8. Similarly the plane x = 1 intersects the paraboloid in the parabola
z ~ 12  y 2 ,
X
= 1 (the curve C2 in the second figure) and
the slope of the tangent line at (1 , 2, 8) is f v(1, 2)
=
4. X
JC
20 0
 20
2
2
Note that traces off in planes parallel to·the xzplane. are parabolas which open downwru:d for y and the traces of ! r:: in these p lanes are straight lines, which have negative slopes for y
< 0 and upward for y > O,
< 0 and positive slopes for y > 0. The
traces off in planes parallel to the yzplane are cubi_c curves, and the traces of fu in these planes are parabolas. 15. f( x, y) = y 5
19.
...:..
z = (2x + 3y) 10
by 23· f (x,y) = ax+ ex +dy x
)
aazX = 10{2a: + 3y) 0
=>
21. f(x , y) = x f y = x y
f Y ( ' 1Y
f x(x, y) = 0  3y =  3y, / y(x, y) = 5y4
3xy =>
1
=> fx( x, y ) =
=>
= (ex+ dy)(b} 
f  (x • )  ,y
=
y
0
1
2 = 20{2x + 3y) 9 ,
azy =
3x
10(2x + 3y) 0 3 = 30(2x + 3y) 0 0
.
= 1/ y, fv (x , y ) = xy 2 =  x / y 2
(ex + dy )(a)  (ax+ by)(c) (cx +dy) 2
(ax+ by)(d} (ex+ dy) 2

=
(ad · bc)y ' (ex + dy )2
= (be 
ad)x (ex + dy} 2
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198
D
PARTIAL DERIVATIVES
CHAPTER 14
25. g(u,v)
= (u2 v v3 ) 5
gu(u,v)
=?
gu(u, v) = 5(u2 v v 3 ) 4 (u2
= 5(u2v 
3v2 ) = 5(u2

v 3 ) 4 • 2uv
0 F11 (x,y) = o·y
Fx(x,y)= :xl" cos (e
=?
1"' cos(e) . dt = oy0 [ I.( cos(e) dt 11
.
t
v3 ) \
3v2 )(u2 v v 3 ) 4

. 29. F(x,y) = l " cos(et)dt
= 10u·u(u2 v 
t
]
1
'
dt=cos(e"' ) by theFundamentaiTheoremofCalculus, Partl;
)
0 =  oy
frv .. cos(et) dt = 
cos(e11 ).
11
33. w
= ln(x + 2y + 3z)
35. u = xysin ~(yz)
ou = xy . az
37. h(x, y, z, t)
1
= x + 2y + 3z'
ow oy
ow oz
2
= x + 2y + 3z'
3
= x + 2y + 3z
=>
1
J 1 (yz)2
xy2 (y) = ;:::.:=:::::::::==:::
J 1 y2z2
= x 2 y cos(z/t) =>
hz(x, y, z, t) 39.
ow ox
=>
h.,(x; y, z, t)
= 2xy cos(z/t),
= x2 y sin(z/t)(1/ t) = ( x2 y/t) sin(z/t),
u = Vx 21 + x 22 + · · · +
2 Xn·
For each t·
h11 (x, y, z, t)
ht(x, y, z, t)
= x2 cos(z/t),
= x2 y sin(z/t)( zC 2 ) = (x2 yz/t2 ) sin.(z/t)
= 1, ..., n, u,, = 21 ( x 21 + X22 + · · · + Xn2) 1/2 ( 2x; ) =
= 1n(x+ Jx2 + y2) '*
41 . f(x, y)
f., ( x,y) _
1 [ 1+21 ( x 2 +y2)  1/ 2(2x)] _ ~
x + Y x2
. so f.,(, 34= )
+ y2
1
3+~
(
3
1+
...f32+42
y
43. f(x,y , z)
= x+y + z
=?
fv(x,y,z)
2+(1) soj11 (2,1,  1) = (2 + 1 +( 1 )) 2
= x1l
45. f(x, y)
 x3 y
=
)
x
1 ( 1+ ~
+ Y x2 + y2
X ) ~ , Y x2 y2
= lim
.
l(x + y + z) y(l ) (x + y + z )2
x +z
= (x+y+ z )2'
1
4·
=>
h(y2

3x2 y  3xyh yh2 ) h
,...... o
f (
) _ li
x ,y 
h5
f(x, y + h)  f(x;y) _ h

· = lim {2xy + xh x /a.+0
+
=11 l . s ( + !s)! = s·
_ . f(x + h,y)  f(x,y) _ 1. (x +h)y2  (x + h) 3 y (xy 2 liD f "' (x, y )  ,.1liD .....o h h0 h .
11
Xi
y X~ + X~ + · · · + x?,
.
3
)
U:U h >0
= 2xy  x

x 3 y)
= lim (y 2 h ..... o

3x 2 y 3xyh yh 2 ) = y 2
x(y + h) 2

x 3 (y + h) (xy 2 h


3x2 y
x 3 y) _ 1'
 "~
h(2xy + xh x 3 ) h
3
© 2012 Ccngnge Learning. All Rights Reserved. Mny not be scanned, copied, orduplicoted, or posted to • publicly ncc=iblc website, in whole or in P."t.
www.elsolucionario.net
.
PARTIAL DERIVATIVES
SECTION 14.3
47. x 2 + 2y2 + 3z2
=1
=?

8
ax
8 (x 2 + 2y2 + 3z 2) =  (1) ox
8z 2x + 0 + 6z ox = 0 8z
8z =  2x =  x , and 8 (x 2 + 2y 2 + 3z 2) = 8 (1) 6z 8y 0y 3z 8x
ez '8 z = y ( X 8z
= xyz =>
49. ez
• 8z (e  xy) 8x
0 + 4y + 6z 8y = 0
ax
ax
=?
8z 6z 0y
_ 8z 8z e   xy OX ox
=?
= 4y
=?
=?
= yz =>
yz = yz, so 8z = e_.  x:y 8x • 8z e· oy
=> 8z
+ Z · 1)
8z 6z ax =  2x
=?
D

xy
8z
( e  xy ) 8 z = xz , so 8y
oy = xz =>
xz
ay = e= xy 51. (a) z = f( x ) + g(y)
(b) z = f(x + y).
. az 8y 53. f( x , y)
=
~:
=> Let u
=x+
df au= df (1) du 8y du
= x 3 y 5 + 2x 4 y
= ,q'(y)
. 8z df QU y. Then ox = du ox
=
df du (1)
=f
I
1
(u) = f (x + y) ,
= J'(u) = f'(x + y).
y) = 3x2 y 5 + Bx3 y , fu(x, y) = 5x 3 y 4 + 2x 4 . Then l rr::c (x, y) = fux(x , y) = 15x2 y 4 + Bx 3 , and f uu (x, y) = 20x3 y 3 • f,(x,
=?
f xu(x, y) = 15x 2 y 4 + 8x 3 ,
~
= /'(x),
6xy 5 + 24x2 y,
55. w = Ju2 + v2
uv (u2 + v 2)3/ 2 ' 2
1 · ~  v · ~(u + v 2 )  1 (2v) 2
W uv
=
(Ju2 +
x+y 57. z = arctan  
1  xy
Zx
v2)2
1
'
=
W uu
= v (21) ( u 2 +v2)  3/ 2 ( 2u) =. (u 2 +uvv2)3I 2'
Juz + v 1  v 2 j .,Ju2 + v 2 u2 + v2
u 2 + v2
=
v2 (u2 + v2)3/2
u2

=
(u2 + v2)3/2.
=?
_ 1 (1)(1xy)(x+y)( y)_ l+y2 1+y2 = 1 + (..:.±1L..) 2 . (1  x y)2  (1  x y)2 + (x +y)2  1 + x2 + y2 + x2y2 1  xy
1 + y2
1 = (1 + x2)(1 + y2) = 1 + x2 '
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201 2 Cengogc Learning. All Righls R=rved. Moy nol be scanned, copied. or duplicnlcd. or posted loa publicly accessible wcbsilc. in whole or in part.
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.
199
200
CHAPTER 14
0
PARTIAL DERIVATIVES
(1)(1xy)(x+y)(x) _ 1+x2 1 + ( x+u )2 . (1 xy)2 (1 xy)2 +(x + y)2
1 1 +x 2 (1 + x 2)(1 + y2) = 1 + y2 ·
1
=
zu
1xy
(
Thus U:cy = 61.
u
= 0, Zyx = 0,
Zyy
2y
= (1 + y 2) 2 . 2y =
=
U:c
sin(x2 y) · 2xy
= 2xysin(x2y),
= 2xy · cos(x2 y) · x 2 +sin(x 2y) · (2x) = 2x 3 ycos(x2y) 2xsin(x2 y) and
Uxy
= sin(x2y) ·x2 =  x 2 sin(x 2y),
63. f(x, y)
= x 4 y 2  x3 y =>
= 8x3 y 3x2 ,
f xv
67. u=er8 sin(J 2 ;
0
1t11 x
=
x 2 ·cos(x2y) ·2xy+sin(x2 y) · ( 2x)
= 2x 3 ycos(x2 y) 2xsin(x 2 y).
= Uux.
Thus Uxy
;
Zxy
Uy:c.
= cos(x2 y) =>
uu
+2xx 2 )2, 1
= 4x3 y 2 
fx
3x2 y, fxx
= 12x2 y 2 
6xy, fxxx = 24xy2

6y and '
2
fxvx = 24x y 6x.
=> :~ . =er 8 cosfJ+sinfJ·er 8 (r)=e'. 0 (cosO+rsinO),
= er8 (sinfJ) +(cosO+ rsinO) erO (0) = ero (sinO+ OcosB + rOsinO),
{)~:~O = ero (0 sin 0) +(sinO+ 0 cosO+ rOsinO) · e''8 X
69.w=y+2z=x(y+2z)1
83w {)z{)y{)x
OW 82 w . 2 2 ax=(y+2z)  I, 8yax = (y+2z) (1)=(y+2z) ,
=>
. .
= (2)(y +
2z) 3 (2)
(0) = Oero (2sin0 + Ocos 0 + rO sinO).
= 4(y +
4
2z) 3 = (y + z)3 and 2
ow {)y
2
= x(1)(y + 2z) (1) = x(y + 2z)
71. Assuming that the third partial derivatives off are continuous (easily verified), we can write
f(x, y, z)
= :J;y 2 z3 + arCSin ( X vz)
=?
73. By Definition 4, fx(3, 2) = lim !(3 + h, h 0
= 2xyz3 + 0, fvx = 2yz3 , and fyxz
fu
fxzy
= 6yz 2 =
=
fyxz·
2
,
Then
fxzy·
l
2
2 !(3 , ) which we can approximate by considering h = 0.5 and h = 0.5:
f (3 2 ) ~ !(3.5, 2)  f(3, 2) _ 22.4  17.5 _ 9 8 f (3 2 ) ~ !(2.5, 2)  f(3, 2) _ 10.2 17.5 _ 14 6 A . "' ' ~ 0.5 0.5  · ' "' ' ~ 0.5 0.5 · · veragmg these values, we estimate f,.(3, 2) to be approximately 12.2. Similarly, f x (3, 2.2) _
3 = h"""""+O lim !( +
22 3 22 h , · } !( , · ) which L
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www.elsolucionario.net
SECTION 14.3
·
· · . . '
dI
we can approximate by cons1denng h = 0.5 an
f x(3, 2.~)
1
f (
f(2.5, 2.2)  f(3, 2.2) 9.3  15.9 _ _ = _ _ 05 05
~
To estimat~
)
=  0.5: "' 3, 2. 2 ::::::
PARTIAL DERIVATIVES
0
201
!(3.5, 2.2)  !(3, 2.2) 26.1 15.9 = = 20.4, 0 .5 0.5
. I I h f ( )' = 13. 2. Averagmg t 1ese va ues, we ave "' 3, 2.2 ~ 16.8.
f, 11 (3, 2), we first need an estimate for .fx(3, 1.8):
~ /(3.5, 1.8)  /(3, 1.8)  20.0  18.1  3 8
.fx ( 3 • 1.8 ) ~
0.5

0.5

.
f (3 1 8) ~ /(2:5, 1.8) !(3, 1.8)  12.5  18.1  1 2 . ' x ' . ~ 0.5  0.5  1. .
.
Averaging these values, we get f,(3, 1.8) ~ 7.5. Now fx 11 (x, y)
..
()a[
variables, so Defimtton 4 says that fx 11 x, y .
= ;::;uy
(
fx x, Y
))
=
a
= ;::;[f.,(x, y)) and fx(x, y) is itself a function of two uy li f,(x,y+h)fx (x,y ) m I · ho a
=>
. I· · 1 · · k ·11 f , 11 (3, 2) = lim f x(3, 2 +h) fx(3, 2) . We can estimate t liS va ue usmg our prev1ous wor Wit 1 a = 0.2 and h = 0.2: h
h o
) ~ j,(3, 2.2) f x(3, 2) = 16.8 12.2 = f ( 2 ) ~ f x (3, 1.8) J,(3, 2) 23 0.2 0.2 ' xy 3 ' 0.2
f xy (3 • 2 ~
= 7.5 
12.2 0 .2
= 23 5
· ·
Av~raging these values, we estimate f, 11 (3, 2) to be appr~ximately 23.25.
=
U:r:x
(x2
+ y2 + z2 )  3/2  x( ~) (x2 + y2 + z2)  5f2(2x) = .l
2y2  x2  z·2
By symmetry, u uu
= .(x2 + y2 + z2 )5/2
Thus Uxo: ·+ u,,y + Uzz .
79. Let v
= x +at,
=
2x2  y2  z2
and
U ::z
=
2z 2  x2  y2 (x2 + y 2 + z2 )5/2 .
+ 2y2  x2  z2 + 2z2 (X 2 +y + Z 2)"/2 "
x2  y2
w = x  at.
Then Ut =
8[f(v) + g(w)] f)t
df(v) 8v
u,
= f'(v) + g'(w) and Ux x = 8z
= in(e"' + e!l) => '8x
f"(v)
e"'
+ g"(w) . Thus Utt 8z 8y
=    and 
e"'
+ e!l
eY
2
= a
·
Ux:r.·
8z
8z
= , so  +e"' + eY Dx 8y
D2z e"'(e"' + eY) . e"'(e'") e"'+ 11 82z 0  e11 (e"') 8x 2 = (e +eY) 2 = (e"' + e11 )2' 8x8y=(e"' + ev) 2 = 8y2
dg(w) 8w
= ~ ot +~7ft = af'(v)  ag'(w) and
o[af'(v)a~ ag'(w)J = a[af"(v) + ag"(w) ) = a 2 [f"(v) + y"(w) ). Similarly, by using the Chain Rule we have
=
82 z
= 0.
?
uu
81. z
2x2  y2 z2 . (x2 + y2 + z2) 5/ 2
.
e11 (e"'
+ e 11 )  e 11 (e11 ) + eY )2
(e"'
(
e" +eY) 2
•
e"
=  e"' + e11
e·u
+ = e"' + eY
e"' e"'
+ eY + eY = 1.

ex+y (e"' +eY)2' and
T hus
e"'+Y (e"' + eu) 2
)2 =
(e"'+u? (e"'+Y? (e"' + eY)1  (ex + eu )1 = 0
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www.elsolucionario.net
202
0
CHAPTER 14
PARTIAL DERIVATIVES
.83. By the C hain Rule, taking the partial derivative of both sides with respect to R 1 gives
8R 1 8R 8R 8R1 85. If we fix K
dP p
=
8[(1/Rl) + (1/R2) + (1/R3)]
=
8R1
or R
 2 8R 8R1
_
= Ko , P.(L,Ko) is a function of a single variable L, and ~~ =a
dL =>aL
Jp = Ja L dP
dL
R2 _Rf'
8R
=  R 1 2. Thus 8R1
I
=
is a separable differentia; equation. Then
=> In IPI = a In 141 + C (Ko), where G_(Ko) can depend on K 0 • Then .

IPI = e lniLIf;C(Ko), andsinceP > 0 andL >0, we have P = ec:dnLeC(Ko) = eC(T
everywhere but f:x:v(x, y)
82 K
~
uv
= m.
f xv(x, y)
·p
8K &2 K ·.8 m 8v 2
Thus 
= 4 and
fv(x, y)
__.._ . V _ mRT d 8V _ mR ..p an 8T p .
= mR.
= 21 v 2 m = K.
= 3x y =>
fvx(x, y)
= 3. Since f xy and f vx are continuous
f f 11.,(x, y), Clairaut's Theorem implies that such a function f(x, y) does not exist.
95. By the geometry of partial derivatives, the slope of the tangent line is f x(l , 2). By implicit differentiation of
4x2
+ 2y2 + z 2 = 16, we get8x + 2z (8zj8x) = 0 =>
8zf8x
=  4x/z, so when x = 1 and z =
8zj8x = 2. So the slope is fx(1, 2) = 2. Thus the tangent line is given by z  2 = 2(x  1), y parameter to bet = x  1, we can write parametric equations for this line: x 97. By C lairaut's Theorem, f xvv = (f:x:y) 11
= (/11x}y = fv xv =
Uv).,'ll
= 1 + t,
y
= 2, z = 2 
2 we have ·
= 2. Taking the 2t.
= Uv)yx = fvv x·
= x(x 2 )  3 12 e0 = ·x ix J' 3 • But we are using the point (1, 0), so near (1, 0), g(x) = x  2 • Then g'(x) =  2x 3 and g'(1) = 2, so using (l) we have fx(l, 0) = g'(l) =  2.
99. Let g(x)
101. (a)
=
f(x, 0)
(b) For (x, y)
f
(0, 0),
f x(x,y) =
(3x2y y3)(x2 + y2)  (x3y xy3)(2x) (x2+y2)2 x4y + 4x2y3 _ ys (x2 + y2)2
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SECTION 14.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
D
203
(c) f (0 0) = lim f(h, 0)  f(O, 0) = lim {O/ h2)  0 = 0 and jy(O, 0) = lim j(O, h)  f(O, 0) = '0. "' ' J•O h h o h ho h 5
Y
4
(0 0) = of, = lim f.,(O, h) f , (O , O) = lim (  h  O) / h = 1 while by (2), ' "'Y ' oy h o h h o h
(d) B (3) f
f ,(0, 0) = ofy = lim jy(h, 0)  jy(O, 0) = lim h5/ h4 = ox
!I
(e) For (x,·y)
h
h .....o
,,..... o
h
1.
i= (0, 0), we use a CAS to compute x6
f xy(x, y)=
+ gx4y2 _ gx2y4 _ y6 · (x2+y2)3
Now as (x , y) > (0, 0) along the xaxis, fxv(x, y)> 1 while~
(x, y) > (0, 0) along theyaxis, fxy(X , y)
>
 1. Thus fx 11 isn't
continuous at (0, 0) and Clairaut's Theorem doesn't apply, so there is no contradiction. The graphs of f, 11 and fy x are identical except at the origin, where we observe the discontinuity.
14.4 Tangent Planes and Linear Approximations 1. z = f(x, y) = 3y 2  2x2 + x
=?
f .,(x, y) = 4x + 1, / 11 (x, y) = 6y, so f,(2, 1) = ~7, / y(2, 1) = 6.
By Equation 2, an equation of the tangent plane is z  (3) = f, (2 ,  1) (x 2) + f v(2, 1)[y ( :1)]
=:
z+ 3=  7(x  2)6(y + 1) or z=  7x  6y + 5.
3. z = f(x , y) =
.,fXY =>
f,(x, y) = ~(xy)  112 · y = ~gx, fu(x , y) = ~(xy) 1 1 2 · x = ~.../XlY, so f.,(1 , 1) = ~
and fv(1 , 1) = ~Thus an equation of the tangent plane is z  1 = f ,(1, 1)(x 1) + jy(1, 1)(y 1)
z 1 = ~(x 1) + Hv  1) or x + y  2z 5. z = f( x, y) = xsin(x + y)
=?
=?
= 0.
f , (x , y) = x · cos(x + y)
+ sin (x +
y) · 1 = x cos(x + y) + sin(x + y),
fv( x, y) = xcos(x + y), so f,( 1, 1) = (1) cosO+ sin 0 =  1, / 11 ( 1, 1) = ( 1) cosO =  1 and an equation of the tangent plane is z  0 = (1) ( x + 1) + (1) (y  1) or x + y 7. z = f(x , y)
= x2 +
xy + 3y2, so f,.(x , y)
= 2x
+y
+z
= 0.
=> f,(1, 1)
= 3, fv(x, y) = x + 6y
=?
fv(1, 1) = 7 and an
equation of the tangent plane is z  5 = 3(x 1) + 1(y 1) or z = 3x + 7y  5. After zooming in, the surface and the
® 2012 Ccngogc Le;uning. All Rights Res
oq
aR oR oR 2 2 . dR = oo: do: + (3 d(3 + O"f d"f = (3 cos '"Y do: + 2o:(3 cos 1 d(3  o:(3 sm 1 d1 0
31. dx = ll.x = 0.05, dy = ll.y
= 0.1, z
•
= 5x
2
+ y 2 , z., = lOx, Zy =
2y . .Thus when x. = 1 and y = 2,
dz
= zx(1, 2) dx + z 11 (1, 2) dy = (10)(0.05) + (4)(0.1) = 0.9 while
ll.z
= ! ,(1.05, 2.1) /(1 , 2) =
33. dA
2
5(1.05? + (2.1) 2
8A = 8A ox dx + oy dy = y dx + x dy and Jll.xJ

5 4 = 0.9225.
:::; 0.1, Jll.yJ :::; ~.1. We use dx
the maximum error in the area is about dA = 24(0.1) + 30(0.1)
. = 0.1, dy = 0.1 w1th x=
30, y
= 24; then
= 5.4 cm2 •
35. The volume of a can is V = 1rr 2 h and ll. V ~ dV is an estimate of the amount of tin. Here dV
= 27rr h dr + 1rr~ dh, so put .
dr = 0.04, dh = 0.08 (0.04 on top, 0.04 on bottom) and then ll. V ~ dV = 27r( 48)(0.04) + 7r(16)(0.08) ~ 16.08 cm3 . Thus the amount of tin is about 16 cm3 .
37. T =
";'gR
2r +R2
,
so the differential ofT is
dT = aT dR 8T d oR + or . r mg(2r2 =
=
2
2
2
2
(2r + R )(mg)  mgR(2R) dR + (2r +R )(0)  mgR(4r) d (2r 2 + R 2 ) 2 (2r2 + R2)2 r
R2 ) 4mgRr (2r2 + R2)2 dR (2r2 + R2)2 dr 
Here we have ll.R = 0.1 and ll.r
= 0.1, so we take dR = 0.1, dr = 0.1 with R = 3, r =
0.7. Then the change in the
tension T is approximately
dT
=
mg [2(0.7? (3) 2 ] 4mg(3)(0.7) (2(0.7) 2 + (3) 2 ]2 (O.l)  (2(0.7) 2 + (3)2)2 (0. 1)
0.84mg 1.642 0.802mg (9.98)2  (9.98) 2 =  99.6004 mg ~  0.0 165mg Because the change is negative, tension· decreases.
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www.elsolucionario.net
SECTION 14.5 THE CHAIN RULE
:~
39. First we find
0
207
implicitly by taking partial derivatives of both sides with respect to R 1 :
oR . R 2 oR = Ri. Then by symmetry, 1
2 2 1 17 oR R oR = R2 . When RI = 25, R2 = 40 and Ra = 50, R = 2 00 oR2  ~· oR3 R3
¢::}
R = 21°7°
n. Since the possible error
for each Ri is 0.5%, the maximum error of R is attained by setting t:J.R, = 0.005R.. So
oR oR oR 2( 1 1 1 ) 1 !:l.R ~=:::: dR = oR !:l.R1 + oR !:l.R2 + oR !:l.R3 = (0.005)R R + R + Ra = (0.005)R = l7 ~=:::: 0.059 st. 2 1 2 3 1
~ so l!:l.wl $ 0..02 and l!:l.hl $ 0.02. The relative . . error m . the calculated surface 41 . The errors in measurement are at ~ost 27o,
w
T
area is
0 425  1 )h0·725 dw + 0.1091w0.4 25 (0.725h0·725 0.1091w0.426h0.725 .
dS 0.1091(0.425w · s!:l.S s= 1'::::
To estimate the maximum relative error, we use :
1
}
dh
dw dh = 0.425:;:; + 0.725/i:
I~w I= 0.02 and ~~i = I~hI = 0.02
=
=>
dS = 0.425 (0.02) + 0. 725 (0.02} = 0.023. Thus the max imum percentage error is approximately 2.3%.
s
43. !:l.z = f(a + !:l.x, b + !:l.y)  f(a, b) =(a+ !:l.x? + (b + !:l.y) 2  (a2 + b2 )
= a 2 + 2a!:l.x + (!:l.x) 2 + b2 + 2b!:l.y + (!:l.y? a 2

b2 = 2a !:l.x + (!:l.x? + 2b!:l.y + (!:l.y?
But f,(a, b) = 2a and /y(a, b) = 2b and so !:l.z = f x(a, b) !:l.x + /y(a, b) !:l.y + !:l.x !:l.x + !:l.y !:l.y, which is Definition 7 with e: 1 = !:l.x and e: 2 = !:l.y. Hence f is differentiable. 45. To show that f is continuous at (a, b) we need to show that
equivalently
lim
(6x ,6y) (O,O )
lim
(z,y)~(a,b)
f(x, y) = f(a, b) or
f(a + !:l.x, b + !:l.y) = f(a , b). Since f is differentiable at (a, b),
f(a + !:l.x, b + !:l.y)  f(a, b)= !:l.z = f,(a, b) 6x + /y(a, b) !:l.y + e:1 !:l.x + e:2 !:l.y, where €1 and (!:l.x, !:l.y}
>
€2 +
0 as
(0, 0). Thus f(a + !:l.x, b + !:l.y) = f(a, b)+ f,(a, b) !:l.x + / 11 (a, b) !:l.y + e:1 !:l.x + e:2 !:l.y. Taking the limit of
both sides as (.C:..x, .C:..y) ___.. (0 , 0) gives
lim
( 6 x,61J) (0 ,0)
f (a
+ .C:..x, b + .C:..y) =
f (a , b). Thus f is continuous at (a, b).
.
14.5 The Chain Rule 1. z =
x? + y 2
+ xy,
3. z =·.J1+ x 2 + y 2 ,
x =sin t, y = et x = In t, y
~
&~
&~
=> d =,.d +,. d =(2x+ y}cost+(2y +x)e t
uX
t
uy
t
t
= cost =>
8zdy= 1 (1+x2 + y) 2 1/ 2 . 2 1/2 dz =oz  dx + (2x)·1 + 21 (1+x2 + y) (2y)(smt) = 2 dt ox dt 8y dt t ·
1
.J1 + x2 + y2
(x . )   ysmt
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www.elsolucionario.net
t
208
D
CHAPTER 14
PARTIAL DERIVATIVES
5. w=xevl =, x = t2, y = lt, z=1+2t =>
(.!) .( 1) +
dw = OW dx + ow dy + ow dz = ey/ z. 2t + xevl= dt ox dt /Jy dt oz dt z
(J!...) .2 = evl=(2t z
xeYI=
2
~  2xy) z
z2
7. z = x 2 y 3 , x = scost, y = ssint => az oz ax oz 8y 3 2 2 • = !i"'"' + a a = 2xy cost+ 3x y sm t us uxus y s .
!i"'"
az . 8z Ox az 8y s . . 2 2 3 2 2 ot = ox at+ oy = (2xy )( ssmt) + (3x y )(scost) = 2sxy smt + 3sx y cost
at
9. z = sin() cos, () = st2 , ¢ = s 2 t
=>
~; = ~; ~~ + ~~ ~! =(cos 8 cos q))(t~) + (~; = ~; ~~ + ~~~~ 11. z = er cos 0, r = st, 8 =
dz dt
=(cos() cos ¢>)(2st) + (sinO sin )(s
J s2 + t 2
/Jf dx
= OX dt
/Jf dy
I
I
•
'
& u& 8y u8y au = e ' au = cos v, au = e ' av =
·
Sill v.
g..,(O, 0)
=
a1
= 62.
=> ~
·
M&
By the Cham Rule (3), au = ox au
Yu(O, 0) = f ,., (x(O , 0), y(O, 0)) Xu(O, 0} + fv(x(O, 0), y(O, 0)} y,.(O, 0)
a1 ax
= 2stcos0 cos¢ s 2 sinO sin
ow at
aw
aw ar
aw as
ow at
fu = &ofu+asax+8tax·~=aroy+asay + Bt~
r
/\ /\ 1\
X
w = f(r, s, t), r
THE CHAIN RULE
J
X
= x 4 + x 2 y, x = s + 2t 
u, y
= stu2 =>
oz = az ox+ az.ay = ( 4xa + 2 xy)( 1) + (x2)(tu2), OS OX OS ay as ' 0
oz
=
oz ax+ az ay
m &m
~m
= (4x3 + 2xy)(2) + (x2)(su2)
·
oz = az ax+ az ay = (4x3 + 2xy)(  1} + (x2)(2stu). au ax au ay a·u  When s = 4, t = 2, and u = 1 we have x = 7 and y = 8, so a z = (1484)(1} + {49)(2} = 1582, aaz = {1484) (2) + (49){4) = 3164, aaz = (1484)(1) + {49)(16} = 700. as t u 23o w = xy + yz + zx, x = r cos 0, y = r sin 8, z = r8 => aw = awax + aawaay + aawaaz ar axar yr zr
~(y +z}(cos8} + (x+z)(sin8}+(y+x)(O), .
aw = aw ax+ aaw aay + aaw aaz8 = (y + z)(  rsinO) + ao ox o8 . y 8 z
(~ + z )(rcosO) +
(y + x}(r·)o
aw When ro = 2 and()= ?r/2 we have x = 0, y = 2, and z = 1r, so ar = (2 + rr}(O) + (0 + rr){1) + (2 + 0)(1r / 2) = 27r and
~~ ..
= (2 + 1r) (  2) + (0 + 1r)(O) + (2 + 0)(2) = 2rro
25o N = p + q, p = u + vw, q = v + uw, r = w + uv =9 p+r aN = aN ap au ap au
'
+ aN aq + aN ar . aq au
8r au
= (p + r){1) (p + q){1) (1) + (p + r)(1)  (p + q)(O) (w) + (p + r) (O)  (p + q)(1) (v) (p+r)2 (p + r)2 (p + r)2 =
(r q) + (p+r)w  (p+q)v (p+r)2
oN = aN ap + aN oq + aN or = r  q (w) + p + ro (1) + (p + q) (u) = (r  q)w + (p + r}  (p + q)u ' av op ov 8q 8v or ov (p+r)2 (p +r) 2 (p+ r)2 . (p+1')2 0
aN _ aN ap ow  ap ow
+ 8N
aq aq ow
+ 8N !!!.._ = or OW
ro  q (v) + p+r (u)+ (p +q) (1) (p + r )2 (p + r) 2 (p + r) 2
· When u = 2, v = 3, and w = 4 we have p = 14, q = 11, and r = aN _ ( 1)(4)+24(25)(2) _ 30 __ ~ d 8N av (24)2  576 96' an aw
~0,
= (r q)v +(p+r)u (p + q) (p + r )2
aN 1 + (24)(4)  (25)(3) 20 5 so ou = . ( )2 . = = , 24 576 144
= ·( 1)(3) + (24)(2)25 =~ = ~
(24)2
576
144 °
© 2012 Ceng;~gc LCJming. AJI Rights Reserved. Mn)' nol be scanned. co ph.d, or duplicated, or posted lo o publicly accessible websitt:, in whole or in p;~rt.
www.elsolucionario.net
l
210
CHAPTER 14
0
PARTIAL DERIVATIVES
= x 2 + y 2 , so let F(x, y) = ycosx  x 2 y sin x  2x 2x + y sin x = :'"
27. y cosx
dy
dx
cos x  2y
y
2
= 0. Then by Equation 6
cosx 2y .
+
[2xy  (1 + y 2 ) (1 xiy 2 )]/(1 + x 4 y 2 ) (x2  2xy(1 + x4y 2 ))/(1 + x4y 2 )
and
(1
_ 
+ y 2 )(1 + x 4 y 2 ) 
2xy x 2  2xy(1+ x4y2)
 1 + x4y2 + y2 + x4y4  2xy x 2  2xy  2x 5 y3
31. x 2
+ 2y2 + 3z 2 = 1, so let F(x, y, z) = x 2 + 2y 2 + 3z2 
= .:_ Fx = _ 2x = ~
f)z ox _ 33. e·
Fz
6z
and
3z •
= x yz, so let F(x, y , z) = e· 
8z
 xz
F11
= e=xy =
 Fz
oy
oz oy xyz
= _ Fy Fz
= _ 4y QZ
oz
1
= 0. Then by Equations 7
= _ 2y
=. 0. Then~ =ux
3z .
F. .
F 'z
yz
yz
= .=e·  xy ez  xy
xz
ez___x_y
35. Since x andy are eac~ functions oft, T( x, y) is a function oft, so by the Chain Rule, 3 seconds, x
Then~~
dx
1
~ = ~: + ~~~;.After
= .Jl+t = .,;r+3 = 2, y = 2 + ~ t = 2 + H 3) = 3, d t = 2 .J1+t 1 +t
= Tx(2, 3)
and
~~ + Ty(2 , 3) ~; = 4(t) + 3(~)
37. C = 1449.2 + 4.6T 0.055T 2
.!.
1
2 v'1 + 3  4,
and dy .!. dt  3 .
= 2. Thus the temperature is rising at a rate of2°C/ s.
ac = 4.6 + 0.00029T 3 + 0.016D, so 8T
ac = 0.016..
O.llT + 0.00087T and oD 2
According to the graph, the diver is experiencing a temperature of approximately 12.5°C at t = 20 minutes, so
.c:;, = 4.6  0.11 (12.5) + 0.00087(12.5) dD
dt
~
1
dT
'2 and (jj
~
2
~ 3.36. By sketching tangent lines_at t = 20 to the graphs given, we est~ate
1 . dC . Then, by the Cham Rule, (jj 10
dT 80 dD = 80 &T (jj + &Ddt ~ (3.36) (Iii) + (0.016) ( 2) r::::: 0.33. 1
1
.
Thus the speed of sound experienced by the diver is decreasing at a rate of approximately 0.33 m/s per minute. 39. (a) V = l!wh, so by the Chain Rule, dV dt
= 8V dl! + 8V dw + oV dh = wh dl! + l!h dw + l!w dh = 81! dt
aw dt
&h dt
dt
dt
dt
2 . 2 . 2 + 1 . 2 . 2 + 1 . 2 . ( _ 3)
= 6 m3/s.
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www.elsolucionario.net
SECTION 14.5
THE CHAIN RULE
D
211
(b) S = 2(£w + £h + wh), so by the Chain Rule,
~~ = ~~: + ~! :~ + ~~ ~~ = 2(w +h):+ 2(£ + h):~ + 2(£+ = 2(2 + 2
2
2
(c) L = £ + w + h
dP
2£
::::?
~~ = 2f. ~ +
2w
:~ +
2h
~: =
~~
= 10 m fs 2(1)(2) + 2(2)(2) + 2(2)(3) = 0
::::?
= 0 mjs.
dL/ dt
41.
2
2)2 + 2(1 + 2)2 + 2(1 + 2)(  3)
w)
2
dT
T
dt = 0.15, V = 8.31?
dt
= 0.05,
dV dt
= 8 31 [ 0.15 .
20
and
dV
8.31 dT
T dP 8.31 p 2 Yt· Thus when P
dt = Pdt 
= 20 andT =
320,
_ (0.05)(320)] ~ _ 0 .27 Lf s. 400
43. Let x be the length of the first side of the triangle andy the lengtl1 of the second side. The area A of the triangle is given by A= ~xy sin B where 0 is the angle between the two sides. Thus A is a function of x, y, and (}, and x, y, and 0 are each in tum
functions of timet. We are given that
dA
dt
8A dx = ax dt
and 0 =
+
8A dy ay dt
+
8A dO ao dt
~~ = 3, ~~ ·= 2, and because A is constant, :~ = 0. By the Chain Rule, dA
.
1 0· dt = 2YSI.Il
::::?
dx dt
.
1 0· + 2xsrn
dy
dt
1 + 2xycos(} ·
d8
dt. When x = 20, y
/6 we have
1r
0=
~(30)(sin ~)(3) + ~(20)(sin ~)(2) + ~(20)(30)(cos ~) ~~·
= 45 · l2
+ 300 · vf3 · dO = ll2 + 150 v,) '3 d(} 2 dt dt
 20 · l2
. .or ., d dO g1.ves ddO = ;;;  25/2 = :r.i· 1 · at a rate o f So Ivmg sot he angIe between the s1·des 1·s decreasmg t t 150 v 3 . 12 v 3
1/ (12
~ 0.048 radj s.
V3)
45. (a) By the Chain Rule, :
or = ({Jz) 2
(b)
2
2
(
 r sin B)
+ ~~ r cos 0.
2
2
47. Letu = x 
~= cos(} + ~; sin B, ~; = ~=
2 ({)z . ( {)z ) • 2 ax ) cos 0 +.2 8{)zx {)z 8 y cosO ~mO + By sm 0,
8 z ) + r2 1 (az 80 ) ( 8r
®
=
2
=
2
[(az 8x ) + (az) 8y
dz dz 8u y.Then 88 zX = d  = d U8X U
and
2 ]
2
2 2 . (cos 0 + sm 0) = (az 8x ) + (oz 8y ) .
8z 8z a8zy = ddzU (1). Thus a + 8 y = 0. X
2012 Ccngogc lc:uning. All Rights Res
= tn f,.(x, y).
Thus f ,. (tx , ty) = t"· 1 J,.(x , y) . 59. Given a function defined implicitly by F(x, y) = 0, where F is differentiable and F11 =J: 0, we know that
.
G(x , y) =  F,. so ddy Fy X 2
d y dx2
=  F,. . Let ~
= G(x, y). Differentiating both sides with respect to x and using the Chain Rule gives
= 8G dx + 8G dy 8x dx
~~
~
ay dx
where 8G 8x
= ,E_ ( 8x
Fx) Fy
= _ F11 Fxx 
F,.F
11 ., ,
FJ
8G ay
= ,E_ 8y
(F.,)
=
Fy
T~us 2
d y= ( dx 2
FuF:.,,.  F,.Fyx) (l) F112
+ ( FyFxll 
FxF,m)
F112
(Fa;) F11
Fxx F;  FyxFxFy  F,.uF11Fx + FyyF; FJ But F has continuous second derivatives, so by Clauraut's Theorem, F11 ,
d211 F.,.,F;  2F.,11 F, Fu dx2 = F.3
+ FuuF;
= F.,11 and we have
d . ed as es1r .
II
14.6 Directional Derivatives and the Gradient Vector 1. We can approximate the directional derivative of the pressure function at K in the direction of S by the average rate of change
of pressure between the points where the red line intersects the contour lines closest to K (extend the red line slightly at the left). In the direction ofS, the pressure changes from 1000 millibars to 996 millibars and we estimate the distance between these two points to be approximately 50 km (using the fact that U1e distance from K to S is 300 km). Then the rate of change of pressure in the direction given is approximately
3. Du /{ 20, 30) /T(  20, 30)
= 'V j
{ 20, 30) · U
996
1000 ; 0
= 0.08 millibar/km.
= /T{ 20, 30) ( 7z) + /v{ 
20 ~ hlim !( + h, 30)t  !( +0
20, 30) (
?z).
20 30 • ), so we can approximate /T( 20, 30) by considering h
= ±5 and
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214
0
CHAPTER 14
PARTIAL DERIVATIVES
using the values given in the table·:
fr( 20, 30) ~
!( 25 30)  f( 20 30) '
Similarly, f,( 20, 30)
f (20 1J
,
_
'
= ye "'
_
'
10
=
34  (  33) 10
=  30_10( 33)
(  33) 5
= 1.4,
= 1.2. Averaging these values gives fr( 
20, 30) ~ 1.3.
=
 01 . '
= 0.3. Averaging these values gives
f ,( 20, 30)
~ 0.2.
~ 1.3(}z) + (0.2>(}z) ~ 0.778. fx (x, y)
:::>
= ye_., and fu(x, y) = e"'. If u is a unit vector in the direction of 8 = 21r / 3, then
from Equation 6, Du /(0, 4) = f x (0, 4) co~ 7. f(x, y)
~· f ( 20 • 30) =  26 
20 30 20 = h0 lim /(  • + h~ !(  • JO), so we can approximate f,( 120, 30) with h = ±10:
!(  20 20)  J( 20 30)
Then Duf( 20,30)
15 30 • )
39  (  33) _5
=
'
5
30) ~ !(  20, 40)  J(  20, 30) 10
f,(  20, 30) ~
5. f(x, y)
h ( 20, 30) ~ f (
en + f v(O, en 4) s in
= 4 · ( ~)
+ 1 · {1
= 2+
4.
= sin(2x + 3y). • I
8 (a) 'il f (x, y) = . 1 i 8X
(b) 'il f( 6,4)
+ ~! j = [cos(2x + 3y) · 2] i + [cos(2x + 3y) · 3]j = 2 cos {2x + 3y) i + 3 cos (2x + 3y)j •
~J
= (2cos0) i + (3cosO)j = 2 i + 3j
(c) By Equation 9, Du f(  6, 4) = 'il f( 6, 4) · u = (2 i
+ 3j ) · Hv'3 i 
= x 2 yz xyz 3 (a) 'il f( x, y, z ) = (f,. (x, y , z), fv(x, y, z), f:(x, y, z)) = (2xyz 
j ) = ~ (2v'3 3) = v'3  ~·
9. f(x , y, z)
(b) 'il/(2, 1, 1) = (4 + 1, 4 2,4 + 6) (c) By Equation 14, Duf(2,  1, 1) 11. f(x , y)
= e"' si.ny =>
= (3, 2, 2)
= 'il /(2,  1, 1) · u = (3, 2, 2) · ( 0, ~.  ~) = 0 + ~ 
unit vector in the direction.ofv is u =
J+8 2
2
( 6, 8)
v ' 0, f
(1? =
7. Since D(1, 1)
> 0 and
has a local minimum at (1, 1) by the Second Derivatives Test. 2
(b) D(1, 1) = f.xx (1, 1) fuy(1, 1) [fxu(1, 1W = (4)(2) (3) =  1. Since D(1, 1)
< 0, f
has a sadd le point.at (1, 1) by
' the Second Derivatives Test.
3. In the figure, a point at approximately (1 , 1) is enclosed by level curves wh'ich are oval in shape and ind icate that as we move away from the point in any direc~ion the values off are increasing. Hence we would expect a local minimum at or near (1, 1) . The level curves near (0, 0) resemble hyperbolas, and as we move away from the origin, the values off increase in some directions and decrease in others, so we would expect to find a saddle point there. To verify our predictions, we have f(x, y)
=4+
x3 + y 3

have critical points where these partial derivatives are equal to 0: 3x2 first equation into the second equation gives 3(x2 ) 2

3x = 0
=> f x (x , y) = 3x 2
3xy
3y = 0, 3y

=> 3x(x 3

2

1) = 0

3y, f v (x, y) = 3y 2

3x. We
2
3x = 0. Substituting y = x from the
=> x = 0 or x = 1. T hen we have
two critical points, (0, 0) and (1, 1). The second partial derivatives are f xx (x, y) = 6x, fxu(x, y) = 3, and fvu(x, y) = 6y, '
'
2
2
so D(x, y) = fxx(x , y) fuv(x, y) [fxu(x, y)] = (6x)(6y) (3) = 36xy 9. Then D(O, 0) = 36(0)(0)  9 =  9, and D (1, 1) = 36(1)(1)  9 = 27. Since D(O, 0) D (1, 1)
> 0 and fxx(1, 1) > 0, f
5. f(x, y) = x 2 + x y + y 2 + y
< 0, f has a saddle point at (0, 0) by the Second D erivatives Test. Since
has a local minimum at (1 , 1).
=> f x
= 2x + y, fv = x + 2y + 1, fxx = 2, fx 11 = 1, fvv = 2. Then fx = 0 implies
y=2x,andsubstitutionintofy=x + 2y + 1=0 gives x+2(~2x) +1=0 Then y =~and the only critical point is
D(x, y) = fxxfvv  (fru) 2 = (2)(2) 
=> 3x=1 => x = i·
(t, ~).
(1? =
3, and since_
D(~,  D = 3 > 0 and fxx(i, ~) = 2 > 0, !(~ , ~) = ~ is a local minimum by the Second Derivatives Test.
7. f(x,y) = (x y)(1 xy) = x  y  x 2 y+ xy 2
=> f x = 1  2xy +y2 , fv =  1 x 2'+ 2xy, fx:n = 2y,
f xv =  2x + 2y, fvv = 2x. Then fx = 0 implies 1  2xy + y 2 = 0 and / 11 = 0 implies 1  x 2 + 2xy = 0. Adding the two equations gives 1 + y 2
©

1 x 2 = 0
=>
y2 = x2
=>
y=
±x, but ify = x then f x = 0 implies
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www.elsolucionario.net
SECTION 14.7
1 '+ 2x 2 + x 2
=0
3x2
=}
then substitution into f,
MAXIMUM AND MINIMUM VALUES
0
221
= 1 which has no real solution. l fy = x
= 0 gives 1 
2x
+ x2 = 0
2
x
=}
2
=1
=}
x = ±1, so the critical points are (1, 1) and ( 1, 1). Now D (1, 1) = (  2)(2)  02 =  4 < 0 and
D(  1,  1) = (2)( 2) 0 2 = 4 < 0, so (1, 1) and ( 1,  1) are saddle points.
9. f(x, y)
= y 3 + 3x 2 y 
fw
= 6y 
3y2

12y
6x 2
12. Then f z
=0
6y 2
+2
=}
f ,. = 6xy  12x, ! 11 = 3y 2 + 3x2  12y,
f x:t
= 6y 12,
f xu
= 6x,
= 0 implies 6x(y 2) = 0, sox = 0 or y = 2. If x = 0 then substitution into j y = 0 gives
3y(y 4)
=}
substitution into jy

=0
=}
= 0 gives 12 + 3x2 
y = 0 or y
24 = 0
=}
= 4, so we have critical points (0, 0) and (0, 4). If y
x2
=4
= 2,
=}
x = ±2, so we have critical points (±2, 2).
= ( 12)( 12)  02 = 144 > 0 and fu(O, 0) =  12 < o•.so f(O, 0) = 2 is a local maximum. D{O, 4) = (12){12)  02 = 144 > 0
D(O, 0)
and f xx(O, 4) = 12
D( ±2, 2)
11. f(x, y)
4y 4
= {0)(0) {±12)2 = 144 < 0, so (±2, 2) are saddle po ints.
= x3 
implies x 2
= 4y
12xy + _8y3
=}
fx
= 3x 2 
12y, f u =  12x + 24y2 , fxx
y
= 6x, fx 11 =  12,/1111 = 48 y . Then fx = 0
·= 4y and / 11 = 0 implies x = 2y 2 • Substituting the second equation into the first gives {2y 2 } 2 = 4y =}
x = 0 and if y D(O, 0)
> 0, so f(O , 4) = 30 is a local minimum.
4y(y 3  1)
=0
=}
y
= 0 or y
= 1. lfy
=}
= 0 then
= 1 then x = 2, so the critical points are (0, 0) and {2, 1).
= {0){0) 
(  12) 2 =  144 < 0, so (0, 0) is a saddle point.·
D(2, 1) = (12)(48) ( 12?
= 432 > 0 and fxx(2 , 1) =
12 > 0 so 4
!(2, 1) = 8 is a local .minimum .
13. f(x,y)=e"'cosy
=}
f x=e"'cosy, / 11 =e"'siny.
= 0 implies cosy = 0 or y = f + mr for nan integer. But sin ( f + mr) =1 0, so there are no critical points.
Now f a:
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www.elsolucionario.net
222
D
CHAPTER 14
PARTIAL DERIVATIVES
l" = ( x 2 + y 2) e'~~ 2  :r2 (  2x) + 2xe'~~2 :r2 = 2xe'~~2  :r 2 (1  x. 2  y 2 ),
fzx = 2xe'fl2  o:2 (  2x) + (1 x 2  y 2 ) ( 2x ( 2xe'~~2 :r2) + 2e'fl2 f xv
= 2xe'112 2 o: (
2y) + 2x(2y)eY
22 :r
2
2
= 0 implies y = 0, and substituting into fx 2
2xe"' (1  x 2 )
=0 ~
=
22
(1  x  y ) = ~4xyeY  x (x 2
f 11 v = 2yeV2  o:2 (2y) + (1 + x 2 + y 2 ) ( 2y ( 2yeV2 :r2) + 2eY2 fv
:r2)
:r
2)
2e'~~2 :r2 ((1
x2

y 2 )(1  2x2 )

2x2 ),
+ y 2 ),
= 2e 712 
:r
2 ((1
+ x 2 + y 2 )( 1 + 2y 2 ) + 2y 2 ).
= 0 gives
= o.or x = ± 1. Thus the critical points are (0, 0) and (±1, 0). Now D(O, 0) = (2)(2)  0 > 0 and f :r:x (O , 0) = 2 > 0, x
so f(O, 0) = 0 is a local minimum. D(±l, 0) = ( 4e 1 )(4e 1 )  0
( 0, ~) and D
0 then substitution into f 11 = 0 gives 4y 3 = 2
= 0 gives 4y3 
8y + 2
D(±1.592, 1.267)
:::::}
•
y=
1  1i2, so
= 0. Using a graph, solutions are
~
x
= ±.J2Y, soy =
± 0.720 andy = ,1.267
1.526 gives no realvalued solution for x, but
=>
x
~
±1.592. Thus to three decimal places, the critical points are
~ (0,  0.794), (±0.720, 0.259), and (±1.592, 1.267). Now since f xx = 12x2
= (12x 2 
= ~ > 0 and
1.526, 0.259, and 1.267. (Alternatively, we could have used a calculator or a CAS to find these roots.)
=> x
5 6 3", ; )
is a local minimum.
is a critical point. Substituting x 2 = 2y into fv
approximately y
y
6 , ;)
= ¥
5 ;)
~)
= 0) there and (1r , 1r) is a saddle point.
*> 0 and fxx (~,·?f) < 0 so/(~,~) = .tf! is a local maximum while D(
D(~, ~) =
is
8y)(12y2 )  64x2 , we have D(O, 0.794)

8y, f ., 11 • =  8x, f 1111 = 12y2 ,
> 0, fx x(O,  0.794) > 0, D(±0.720, 0.259) < 0,
> 0, and f xx (± 1.592, 1.267) > 0. Therefore /(0, _ 0.794)
~  1.191 and /(±1.592, 1.267) ~  1.310
are local minima, and (± 0.720, 0.259) are saddle points. There is no highest point on the graph, but the lowest points are approximately (±1.592, 1.267, 1.310).
20
10
0
6 y
27. f(x , y) = x 4 +y3

3x2 +y2 +x  2y + 1
graphs, we see thatto three decimal places,
=> f x (x , y) = 4x 3
f., = 0 when x
~

6x + 1 and f 11 (x , y) = 3y2 + 2y 2. From the
1.301, 0.170, or 1.131, and / 11 = 0 when y
~
1.215 or
0.549. (Alternatively, we could have used a calculator or a CAS to find these roots. We could also use the quadratic formula to find the solutions of / 11 = 0.) So, to three decimal places, f has critical points at (  1.301,  1.215), (  1.301, 0.549), (0.170, 1.215), (0.170, 0.549), (1.131,  1.215), and (1.131, 0.549). Now since / zz = 12x2 and D = (12x2

6)(6y + 2), we have D( 1.301, 1.215)
6, fx 11.= 0, f 11 y = 6y + 2,
< 0, D(  1.301, 0.549) > 0, f xx( 1.301, 0.549) > 0,
D(0.170,  1.215)
> 0,
/ xz (l.131, 0.549)
> 0. Therefore, to three decimal places, f( 1.301, 0.549)
f xx (0.170,  1.215)

< 0, D(0.170, 0.549) < 0, D(l.131, 1.215) < 0, D(l.131, 0.549) > 0, and ~
3.145 and /(1.131, 0.549)
~
@) 2012 Ccngage Leruning. AU Rights Rc:ser\ro. May nol be scanned., COflied, or duplicated, or po~ed to n publicly accessible website, in whole or in p.vL
www.elsolucionario.net
 0.701 are
SECTION 14.7
MAXIMUM AND MINIMUM VALUES
0
225
local minima, !(0.170,  1.215) ::::: 3.197 is a local maximum, and ( 1.301, 1.215), (0.170, 0.549), and (1.131 , 1.215) are saddle points. There is no highest or lowest point on the graph. 3
z
29. Since f is a polynomial it is contint,Jous on
D, so an absolute maximum and minimum exist. Here f x = 2x 2, jy = 2y, and
setting fx = jy = 0 gives (1, 0) as the only critical point (which is inside D), where f (1, 0) =  1. Along £ 1 : x = 0 and
f(O, y) = y 2 for 2 ~ y ::=:; 2, a quadratic function which attains its minimum at y =
b, where f( O, 0) = 0, and its maximum
·aty = ±2, wheref(0,±2) = 4. Along £2: y = x 2 forO ::=:; x ~ 2, and f(x,x 2) = 2x 2  6x + 4 = 2 (x _ ~) 2 _ ~ .
a quadratic which attains its minimum at x = Along £3: y = 2  x for 0
f(x, 2 x) = 2x 2
~
£, where f (~,  ~) =

~, and its maximum at x = 0, where f (0,  2) = 4. y
x ::::; 2, and
6x + 4 = 2(x ~)  ~.a quadratic which attains 2

its minimum at x =~~where f(~, ~) =~,and its maximum at x = 0,
(0,2)
(2,0)
L,
X
where f (O, 2) = 4. Thus the absolute maximum off on D is f(O, ±2) = 4 (0,  2)
and the absolute minimum is !(1, 0) = 1. 31. f x(x,y)
= 2x +2xy,
jy(x,y) = 2y +x2 , and setting f.,= f" = 0
y
On £ 1: y = 1, j(x,  1)
L•
= 5, a constant.
On £2: x = 1, /(1, y) = y 2
L,
(1 , 1)
gives (0, 0) as the only critical point in D , with .f(O, 0) = 4.
(I, I)
L2 0
+ 11 + 5, a quadratic in y which attai ns its
maximum at (1, 1), /(1 , 1) = 7 and its minimum at (1,  ~) , f(1 , 
t)
(1,1)
X
L,
(1, 1)
=.!f .
On £ 3: f(x , 1) = 2x 2 + 5 which attains its maximum at ( 1, 1) and (1, 1) with f(±l , 1)
=
7 and its minimum at (0, 1), f(O, 1)
= 5.
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226
D
CHAPTER 14 PARTIAL DERIVATIVES
On .i4 : f( 1, y)
= y 2 + y + 5 with maximum at ( 1, 1 ), f( 
= 7 and minimum at ( 1, D. f(  1, ~) = Jf.
1, 1)
Thus the absolute maximum is attained at both (±1, 1) with /(±1, 1) = 7 and the absolute minimum on Dis attained at (0, 0) with f(O, 0) = 4. 33. f(x , y) = x 4 +y 4
4xy + 2 is a polynomial and hence continuous on D, so

it has an absolute maximum and minimum on D. fx(x, y)
fv(x, y) = ·4y 3
=>
fv = 0

x
4x; then f x
= y 3 gives x 9 
x = 0
=>
x(x8  1) = 0
(1, 1) with f(1, 1) = 0 is inside D . On £1: y = 0, f(x, 0)
= 3, /(3, y) = y
On £ 2 : x /(3,

12y
+ 83, 0 ~ y
= 2, f( x, 2) =
x4

=> x
= 0 (0,0)
(3,0)
X
= x 4 + 2, x
= 3, !(3, 0)
= 83, and its minimum at x = 0, f (O, 0) = 2.
~ 2, a polynomial in y which attains its m inimum at y
{13) = 83 9 .{/3 ~ 70.0, and its maximum at y =
On L 3 : y
(0, 2)+   L  '3'1 (3, 2) .
1,  1), but only
~ 3, a polynomial in x which attains its maximum at 4
4y and
= 0 implies y = x 3 , and substitution into
±1. Thus the critical points are (0 , 0), (1, 1), and ( 
or x =
0~ x
= 4x 3 
Y
0, ! (3, 0)
= {/3,
= 83.
Bx + 18, 0 ~ x ~ 3, a polynomial in x which attains its minimum at x = q'2,
f( q'2, 2) = is 6 ij2 ~ 10.4, and its maximum at x
= 3, /(3, 2) =
75. On £ 4: x
= 0, f(O, y) =
polynomial in y which attains its maximum at y = 2, f(O , 2) = 18, and its minimum at y
y
= 0, f(O, 0)
4 + 2, 0 ~ y ~ 2, a = 2. Thus the absolute
maximum off on Dis /(3, 0) = 83 and the absolute minimum is /(1, 1) = 0. 35. f,(x, y) = 6x 2 and fv(x, y) = 4y3 • And so f x = 0 and fv = 0 only occur when x = y = 0. Hence, the only critical point
= 0. Now on the circle x 2 + y 2 = 1, y2 = 1  x 2 so let 2 2 f(x, y) =·2x 3 + (1 x 2 ) 2 = .x 4 + 2x 3  2x + 1,1 ~ x ~ 1. Then g'(x) = 4x3 + 6x  4~ =
.inside the disk is at x = y
g(x) = 2, or
~
/(0,±1) =
:::o 0 where /(0, 0)
g(~) = 1, !(~, ±1 ) = g(~) = ~· and ,(2,3)
0
=> x
= 0,
is not in D. Checking the endpoints, we get
.f(1, 0) = g( 1) = 2 and f(1 , 0) = g(1) = 2. Thus the absolute maximum and minimum off on Dare /(1, 0) = 2 and
=  2.
f(  1,0)
A nother method: On the boundary x 2
+ y 2 = 1 we can write x = cos (;I, y =
sin 8, so
f (cos 8, sin 8)
= 2 cos3 8 + sin4 (;I,
0 ~ () ~ 27T. 37. f(x, y)
=  (x2 
jy(X, y)
=
1) 2

(x 2 y x  1) 2
2(x2 y  x  1)x2 . Setting j y(x, y)
There are no critical points for x so f ,
( x,
f,(x, y)
=> f,(x, y)
x~ 1 ) = ·2(x2 
=  2(x 2

1)(2x)  2(x 2 y x  1)(2xy 1) and
= 0 gives either x = 0 or x 2y 
= O,.since f:.,(O, y)
1){2x) 2(x 2 x ~
= .:_2, so we set x 2y  x 1 1
 x  1) (2x x ~
1
x 1 = 0.
=0
{:}
y= x
~1
X
[x =/= 0],
 1) = 4x(x 2  1). Therefore
= fv(x, y) = 0 at the points (1, 2) and ( 1, 0). To classifY these critical points, we calculate
®
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MAXIMUM AND MINIMUM VAL4ES
SECTION 14.7
f :u:(x, y)
=
and f zv(x , y)
12x 2

12x2 y 2
= 8x3 y
+ 6x + 4x. In order to use the Second Derivatives
Of1~  I
D(1,0) = fxx(1 , 0) / 7171 (1,0)  [f:c 71 ( 1 , 0)]~ = 16 > 0,
2
= 10
0, and f:r:c{ 1, 2) = 26 < 0, so
both ( 1, 0) and (1, 2) give local maxima.
X
39. Let d be the distance fr~m {2, 0, 3) to any point (x, y, z ) on the plane x where z
227
+ 12xy + 4y + 2, f vv(x, y) =  2x",
2
Test we calculate
/ :r:c (  1, 0)
0
= 1 x y, and we minimize d2
j;,(x,y) = 2(x 2) + 2{4 x 
= (x 
+y+ z =
1, sod= J(x  2) 2
+ y 2 + {4  x y) 2 • Then y)( 1) = 4x + 2y 12, f 71 (x,y) = 2y + 2(4  x y){1) = ~ f(x, y)
2)
+ y 2 + (z + 3)2
2
2x + 4y  8. Solving
4x + 2y 12 = 0 and 2x + 4y 8 = 0 simultaneously gives x = ~. y = ~.so the only critical point is(~,~). An absolute minimum exists (since there is a minimum di.s tance from the point to the plane) and it must occur at a critical poi nt, so the shortest distance occurs for x =
~. y = ~ for which d =
J(~ 
2)
+ (j) + (4  ~  ~) 2 = ji = ~2
2
41. Let d be the distance from the point {4, 2, 0) to any point {x, y, z) on the cone, sod = ,j(x  4)2
+ y 2 , and we minimize d 2 = (x  4) 2 + (y fz(x, y) = 2 (x  4) + 2x = 4x  8, f 71 (x , y) = 2 (y 
z2 = x 2
f.,
= 0 =>
x = 2, fu
= 0 =>
+ (y 2)2 + z2 where
+ x 2 + y 2 = f(x, y). Then 2) + 2y = 4y 4, and the critical points occur when
2)
2
y = 1. Thus the only critica l point is (2, 1). An absolute minimum exists (since there is a
minimum distance from the cone to the point) which must occur at a critical point, so the points on the cone closest to ( 4, 2, 0) are (2, 1, ±v'5).
43. x
+y+z
= 100, so maximize f(x , y) = xy(l OO  x y) .
f x = 100y 2xy y 2 , f 71 = 100x  x 2

2xy,
=  2y, f vu = 2x, fxv = 100 2x 2y. Then J,r; = 0 implies y = 0 or y = 100 2x. Substituting y = 0 into fv = 0 gives x = 0 or x = 100 and substi~ting y = 100  2x into fv = 0 gives 3x2  100x = 0 sox = 0 or ~ . f, :r.
1
1 0
1 0
Thus the critical points are (0, 0), (100, 0), (0, 100) and ( ~ , ~ ) .
D(O, 0) = D{100, 0)
= D {O, 100) =
10,000 while D(1~ 0 , 1 ~ 0 )
=
10
·~ 00 and f xx (1~0 , 1 ~0 ) = 
2
~ < 0. Thus (0, 0),
(100, 0) and {0, 100) are saddle points whereas f( 1 ~ 0 , 1 ~0 ) is a local maximum . Thus the numbers are x 45. Center the sphere at the origin so that its equation is x 2
+ y2 +z2
=
2 1· ,
V.,
y 2 ) = 8xy Jr 2  x 2
= (Bxy) . ~(r2 x2 y2)1 f2( 2x) + ,jr2 2
Setting Vx = 0 gives y = 0 or 2x
+ y2

~ .
+ y2 + z 2 = r 2 , so take (x, y, z) to be the vertex
in the fi rst octant. Then the box has length 2x, width 2y, and he ight 2z = 2 ,jr2  x 2 
1 0
and orient the inscribed rectangular box so that its
edges are parallel to the coordinate axes. Any vertex of the box satisfies x 2
V(x , y) = (2x)(2y)(2 Jr 2  x 2
= y =z =

y 2 with volume given by
y 2 forO < x < r, 0 < y < r. Then
x2  y2. By=
8y(r2

,jr2 _
2x 2  y 2 ). ·and Vv = 8x(r 2  x 2  2y 2 ) . x2 _ y2 . J r2 _ x2 _ y2
= r 2 , but y > 0 so only the latter solution applies. Similarly, V71 = 0 with x > 0
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228
0
PARTIAL DERIVATIVES
CHAPTER 14
implies x 2
3x = r
2
+ 2y2 =
2
::::>
r
x=
2
Substituting, we have 2x
.
.jr"i]3 =
2
+ y2 = x 2 + 2y2
::::>
x 2 = y2
f .,
y)
= y = r 1V3 and the maximum
= x; (6 x 2y), then the maximum volume is V = xyz .
= i(6y ~ 2xy y 2 )
= ~x (6  x 
= h(6 2x 2y) and fv
4y). Setting fx =:= 0 and j 11 = 0 gives the critical point
(2, 1} which geometrically must give a maximum. Thus the volume of the largest such box is V 49. Let the dimensions be x, y, and z; then 4x
V
'*
+ 2y2 = r 2
r I vf3 = y. Thus the oni,Y critical point is (r I vf3, r I ,/3 ). There must be a maximum
volume and here it must occur at a critical point, so the maximum volume occurs when x
.47. Maximize f(x,
y = x. Then x 2
::::>
= xyz = xv(ic x 
y)
+ 4y + 4z
= ic:cy x 2 y 
so Vx = 0 = Vy when 2x + y = ic and x
= (2)(1) ( ~) = t·
= c and the volume is
xy 2 , x > 0, y > 0. Then V,
= ~cy 2xy y 2 and Vy
+ 2y =~c. So lving, we get x =fie, y =
1 12 c
= ~ex x 2

2xy,
and z = ic x y = fie. From
the geometrical nature of the ·problem, thi s critical point must give an absolute maximum. Thus the box is a cube with edge length f2c. 51. Let the dimensions be x, y and z, then minimize xy + 2(xz
f(x,y)
+ yz) if xyz = 32,000 cm3 • Then = xy + [64,000(x + y)fxy] = xy + 64,000(x 1 + y 1 ), fx = y 64,000x 2 , f v = x 64,000y 2 .
And fx
= 0 implies y =
= 64,000 or x =
64,000/x2 ; substituting into / y = 0 implies x 3
D(x, y) = [(2)(64,000Wx  3y 3 dimensions of the box are x

1
40 and then y = 40. Now
> 0 for ( 40, 40) and 'f.,, (40, 40) > 0 so this is indeed a minimum.
Thus the
= y = 40 em, z = 20 em.
53. Let x, y, z be the dimensions of the rectangular box. Then the volume of the box is xyz and ' L
= J xz + y2 + z2
::::>
L2
= x 2 + y 2 + z2
::::>
Substituting, we have volume V(x,y) = xy JL2 x 2
2
Vy = xJL2 x2 y2 2x 2
+ y2 =
L 2 (since y

xy
JP  x2 
> 0), and
y2
= L2
::::>
x
= J L2 
x2

y 2 (x,y
> 0).

2 y .
. V., =Oimpliesy(L 2  x 2 y2 )=x2 y ::::> y(L 2 2x 2 y2 ) =0 ::::>
v;1 = o ·implies x(L2 
x 2 + 2y 2 = L 2 (since x > 0). Substituting y 2 = L 2 3x2
z

x2

y 2 ) = xy 2
2x2 into x 2
= L/vf3 (since x > 0) f!nd then y =
::::>
x( L 2
x2


+ 2y2 = L 2 gives x 2 + 2L2 
= 0 ::::> 4x 2 = L 2 ::::>
2y2 )
JL2 2(L/vf3) = L/vf3. 2
So the on ly critical point is (Lf,/3, Llvf3 ) which, from th~ geometrical nature of the problem, must give an absolute maximum. Thus th~ maximum volume is V(LI,/3, Ll,/3) = (LI,/3)
2
j L2 (Livf3)
2
2

(LI ,/3) = L 3 I (3 ,/3)
cubic units .
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2012 Cen,SD.Gc Learning. All Rir)lts Rcsco·cd. May not be scanned. cop ied. or duplicaLCd, or poster.! to a publicly accessible website, in whole or in pnrt.
www.elsolucionario.net
SECTION 14.8 . .
n.
:L:
n
·
(x•Y• mx;  bxi) = 0 or .:L: XiYi = m
i= l
n
t. = l
n
.~ Yi = m i~l Xi
1
n
:L:
Now fmm =
n + i~l b=
2x~, fbb
m
n
·
•=1
229
2xi [Vi  (mxi + b)J = 0
n
x~ + b .:L: Xi and fb =
\= 1
:L:
2[yi  (mxi +b)] = 0 implies
1.=1
n
n
2 = 2n and fmb =
i= l
equations do indeed minimize
n
.:L:
0
(ni~t Xi) + nb. Thus we have the two desired equations.
= :L:
i =l
n
n
:L: [y, (mxi +b)]2 . Then /m. = i=l :L: i =l
55. Note that here the variables are m and b, and f{m, b) = implies
LAGRANGE MULTIPLIERS
:L:
.
2Xi. And fmm{m, b)
> 0 always and
i= l
f= d;.
i = 1
14.8
Lagrange Multipliers
1. At the extreme val';les off, the level curves off just touch the curve g(x, y) and the accompanying discussion.)
= 8 with a common tangent line. (See Figure I
We can observe several such occurrenc~ on the contour map, but the level curve
= c with the largest value of c which still intersects the curve g(x, y) = 8 is approximately c =59, and the smallest value of c corresponding to a level curve which intersects g(x, y) = 8 appears to be c = 30. Thus we estimate the maximum f(x , y)
value off subj ect to the constraint g(x j y) = 8 to be about 59 and the minimum to be 30. 3. f(x , y) = x 2
+ y2 , g(x, y) =
xy = 1, and \1 f = >..\lg =>
=>
From the last equation, x =I= 0 andy =I= 0, so 2x = >.y y2 = x2
=>
y
= ±x. But xy = 1, so X= y =
(2x, 2y)
= (>..y, >..x), so 2x = >..y, 2y = >..x, and xy = 1.
>.. = 2xjy. Substituting, we have 2y = (2xjy) x
=>
± 1 and the possible points for the extreme values off are {1, 1) and
( 1,  1). Here there is no maximum value, since the constraint xy = 1 allows x or y to become arbil:(arily large, and hence
f(x , y) = x 2 5. f(x, y) = y 2 and ix2
+ y 2 can be made arbitrarily large. The mi~imum value is !{1 , 1) = 
+ y2 =
x 2 , g(x, y)
= 1, and \1 f = >.\lg =>
{2x, 2y) = (t>..x, 2>..y) , so 2x = t>.x, 2y = 2>..y,
+ >..) = 0 => x = 0 or>. =  4. If x = 0 then the third equation  4 then the second equation gives 2y = 8y => y = 0, and substituting into the third equation,
1. From the first equation we have x{ 4
gives y = ±1. If), = we have x
= tx2 +y 2
f(  1, 1) = 2.
= ±2. Thus the possible extreme values off occur at the points (0, ± 1) and (± 2, 0). Evaluating fat these points,
we see that the maximum value is f{O, ± 1) = 1 and the minimum is !{±2, 0) = 4. 7. f(x, y, z)
= 2x + 2y + z, g(x, y, z)
2>.y = 2, 2)..z = 1, and x 2
+ y2 + z 2 = 2
the fourth equation gives
= x2
+ y2 + z 2 =
9, and \1 f = >.. \1g
9. The first three equations imply x = 2
(l) + ( l) + (
2
\ ) 2
© 2012 Ccngogc Learning. All Rights Rcsc:rvcd. May not be =
= 9 =>
9
 2 4)..
= 9
=>
(2, 2, 1) = (2>..x, 2>..y, 2>.z), so 2>.x = 2,
l• ·l· y =
and z =
\ . 2
~ut substitution into
=> ).. = ±t, so f has possible extreme values at
ned. copi~; or duplicated, or posted to a publicly occ~ssiblc website, in whole or in p3r1.
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230
0
CHAPTER 14
PARTIAL DERIVATIVES
the points (2, 2, 1) ~nd (  2,  2,  1). The maximum value off on x 2
J( 2, 2,  1)
9 is f(2, 2, 1) = 9, and the minimum is
=  9.
9. f(x, y, z) = xyz, g(x, y, z) = x
zero then x = y
x2
+ y2 + z 2 =
2
+ 2y 2 '+ 3z2 = 6. V f = >.Vg
=:.
(yz, xz, x y)
= .>. {2x, 4y, Gz). If any ofx, y, or z is
= z = 0 which contradicts x 2 + 2y2 + 3z2 = 6. Then.>.= (yz) /(,2x)
= 2y 2 and z 2
= ~y 2 . Thus x 2
+ 2y2 + 3z 2 =
6 implies 6y 2
= 6 or y
= '(xz) / (4y) = (xy)/(6z ) or
= ±1. Then the possible points are
/j), (.,/2, ±1, /j), (.,/2, ± 1, /j), ( .,/2,±1, /j). The maximum valu~ off on the ellipsoid is J:i, occurring when all coordinates are positive or exactly two are negative and the minimum is 7a occurring when 1 or 3 of ( .,/2, ±1,
the coordinates are negative. 11. f(x ,y,z) = x 2 +y2 +z2 , g(x,y, z) Case 1: lf x
f. 0, y f.
0 and z
= x~ +y4 + z 4 = 1
f. 0, then V f
Vf
=?
= >.Vg implies.>. = l /(2x2 )
= 1/(2y2 )
(4>.x 3 , 4>.y3 ,4>.z3 ).
= l /(2z 2 ) or x 2 = y2 = z 2 and
_!_ L) (±L L _ _!_) (±L _ _!_ __!_) VS' V3 ' VS' VJ ' V3 ' VJ' W' V3
3x 4 = 1 or x = ±  1 giving the points (±L L L) (±L _
VJ' VS' V3 ' · VJ'
V3
= (2x,2y,2z)~ >.Vg =
all with an !value of v'3. Case 2: If one of the variables equals zero and the other two are not zero, then the squares of the two nonzero coordinates are
equal with common value ~ and corresponding f value of .,/2. Case 3: If exactly two of the variables are zero, then the third variable has value ± 1 with the corresponding f value ofl. Thus
on x 4
+ y'1 + z4 =
13. f(x, y, z, t)
=
X
1, the maximum value off is
y'3 and the minimum value is 1.
1 y 1 Z 1 t, g(x, y, z, t) = x 2 1 y 2 1 z 2 1
). = 1/(2x) = 1/(2y) = 1/(2z) = 1/(2t) and x
e=
and p.Vh = (0, 2p.y, 2p.z). Then 1 Thus x
+y +z
+z =
t)
1, h(x,y, z ) = y 2
+ z2 =
(2>.x, 2).y, 2).z , 2>.t), SO
+ z 2 = 4 =:.
Vf
~ =
= (1,2,0), .>.Vg =
 p.z or y = 1/ (2p.), z
4 implies p. =±~.Then the possible points are
(>.,>., >.)
= 1/ (2p.).
(1, ±.J2, =FV2)
(1, .J2, .J2) = 1 + 2 .J2 and the minimum value is f (1,  .,/2, .J2) = 1  2 .J2.
17. f(x, y, z ) = yz + xy, g(x, y , z ) = xy = 1, h(x, y, z) = y 2
+ z2 =
1
=:. V f = (y, x + z, y), >.Vg = (.>.y, >.x, 0),
p.Vh = (0, 2p.y, 2p.z). Then y = >.y implies>. = 1 [y f. 0 since g(x, y, z) = 1], x p. = z/(2y) = y/(2y) or y2 = z 2 , and so y 2
©
=
= 2 and the minimum value is
= .>., 2 = .>. + 2p.y and 0 = >. + 2~tz so p.y =
= 1 implies x = 1 and y 2
and the 'mfJ.Ximum value is f
(1, 1, 1, 1)
= y ~ z = t. But x 2 1 y 2 1 z 2 1 t2 = 1, so the possible points are
(±~ , ±~, ±~, ±~). Thus the maximum value off is f(~, ~~ ~~
15. f(x,y, z) = x + 2y, g(x,y,z) = x +y
1 =>
+ z2 =
+z =
.>.x + 2p.y andy= 2p.z. Thus
1 implies y = ±72, z = ±72. Then xy
= 1 implies x =
2012 Ccng.agc Learning. All Rights Rescr\'cd. M:1y not be sc:mncd, copic:c.l, or duplicotcd, or posted to u publicly occcssiblc website, in wll(>Je or in po1rt.
www.elsolucionario.net
±.J2 and
SECTION 14.8
the possible points are ( ±/2, ± 7:i ,
y2
0
231
72), (±V2, ±~,  );) . Hence the maximum off subject to the constraints is
f(±J2,±~,±jz) = ~ andtheminimumisf(±J2,±~, 'f72) Note: Since xy
LAGRANGE MULTIPLIERS
=
~·
= 1 is one of the constraints we could have solved the problem by solving f (y, z) = yz + 1 subject to
+ z2 = 1.
19. f( x, y)
= x 2 + y 2 + 4x 
4y. For the interior of the region, we find the critical points: fx
= 2x + 4, f u = 2y 
4, so the
onlycritical point is (  2, 2) (which is inside the region) and f (  2, 2) = 8. For the boundary, we use Lagrange multipliers.
g(x, y)
= x 2 + y2 = 9, so "iJf = >."ilg
=;.
Adding the two equations gives 2x + 2y = 2>.x
x
+y = 0
x2
+ y2
=;.
y = x or>.  1 = 0 2
=>
= 9 implies 2y = 9
y
=;.
(2x + 4, 2y  4)
= (2>.x, 2>.y). Thus 2x + 4 =
+ 2>.y =:.
+y =
x
>.(x
+ y)
=:.
(x
2>.x and 2y  4 = 2>.y.
+ y) (>. 1) =
0, so
>. = 1. But >. = 1 leads to a contradition in 2x + 4 = 2>.x, soy =  x and
= ±12. We have f ( ~· 72) = 9 + 12/2 ~ 25.97 and
r( 12, ~) = 9 12/2 ~  7.97, so the maximum value off on the disk x
2
~ y2
:s: 9 is f
( 72· 12)
= 9 + 12/2
and the minimum is f(  2, 2) = 8. 21. f(x , y) = e xy . For the interior ofthe region, we find the critical points: f x =  yexu, / 11
= xe"'Y , so the only
critical point is (0, 0), and f(O, 0) = 1. For the boundary, we use Lagrange multipliers. g(x, y) = x 2
+ 4y2
= 1
=;.
>. "iJ g = (2>.x, 8>.y), so setting "iJ f = >."iJ g we get ye"'11 = 2>.x and xe"'11 = 8>.y. The first of these gives e "' 11 = 2>.xjy, and then .the second gives  x(  2>.xjy) = 8>.y constraint x 2
+ 4y2
f ( ±~ , ±;~)
= 1 gives
X=
= >.(4x3
x4

x3

=.0
±72 and y=±~ . Now t(±~, 'f 2~) = e 1
1 4
~ 1.284 and
= e 1/ 4 ~ 0.779. The former arc the maxima on the region and the latter are the min.ima.
23. (a) f(x , y) = x , g(x,y) = y 2
1
=> x 2 = 4y2 • Solving this last equation with.the
+ x4

x3
== 0
=;.
"ilf = (1, 0) = >.Vg = >.(4x3

3x2,2y). Then
3x 2 ) (1) and 0 = 2>.y (2). We have >. i= 0 from (1), so (2) gives y = 0. Then, from the constraint equation, x 3 (3;
=;.

1)
=0
=;.
x
= 0 or x = 1. But x = 0 contradicts (1), so the only possible extreme value
subject to the constraint is /(1, 0) = 1. (The question remains whether this is indeed the minimum of f.) (b) The constraint is y 2
+ x4 
x3 = 0
¢}
y2 = x3

x 4 . The left side is nonnegative, so we must have x 3

x 4 :2: 0
which is true only for 0 :S: x :S: 1. Therefore the minimum possible value for f( x, y) =x is 0 which occurs for x = y = 0. However,>."iJ g(O, 0)
= >. (0 0, 0) =
(0 , 0) and "iJ f(O, 0) = (1, 0), so "iJ f(O, 0)
(c) Here "iJ g(O, 0) = 0 but the method of Lagrange multipliers requires that "iJg
©
#
# >."iJ g(O, 0) for all values of>..
0 everywhere on the constraint curve.
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232
D
CHAPTER 14
PARTIAL DERIVATIVES
'*
25. P(L,K)=bLo.Kl o., g(L,K)=mL+nK=p
'VP=(abLo: 1 K 1  o.,(1a)bL'""JC'""), >..'Vg=(>..m, >..n).
Then ab(K/ L) o. = >..m and (1 a)b(L/ K)"' =>..nand mL + nK = p, so ab(K/ L?"'/m = (1  a)b(L/ K) 0 /n or 1
naj[m(1 a)]= (L/ K)"'(L/ K) 1 "'. or L = Kno/[m{1 a)]. Substituting into mL + nK = p gives K = (1 a)pjn and L = apjm for the maximum production. 27. Let the sides of the rectangle be x andy. Then f(x, y) = xy, g(x, y) = 2x + 2y = p
29. The distance from {2,0,  3) to a point (x,y,z) on the plane is d 2
that g(x, y, z) = x + y + z
+(z + 3)
= 1.
2
= .j(x 2)2 + y 2 + (z +
d=
J(~ 
2) ~· + ( ~ )2 + (
3) 2 , so we seek to minimize
subject to the constraint that (x, y, z) lies on the plane x + y + z = 1, that is,
=*
Then 'V f =A 'Vg
(2(x 2), 2y, 2(z + 3)) = (A, .A, .A), sox= (A+ 4)/2,
y = >./2, z = (>..  6)/2. Substituting into the constraint equation gives
>.. = ~ . sox = ~. y = ~.and z =
'V f(x, y) = (y, x)·,
= y and the rectangle with maximum area is a square with side length iP·
>.. 'Vg = (2>.., 2>..). Then A= ~y = ~x i~plies x
d = f(x, y, z) = (x 2) 2 + y 2
'*
.A~ 4 +
i
+ A;
6
= 1
'*
3A 2 = 2
'*
i· This must correspond to a minimum, so the shortest distance is
i + 3) 2 = ..fi = ~
31. Let f(x, y , z) = d 2 = (x 4) 2 + (y 2) 2 + z 2 • Then we want to minimize f subject to the constraint
g (x,y,z) = x 2 +y2
'*
z 2 = 0. 'Vf = )..'\Jg

(2 (x ; 4) ,2(y 2) ,2z) = (2>..x,2>..y, 2Xz), sox 4 = J..x,
'*
y  2 = .Ay, and z = J..z. From the last equation we have z + >..z = 0 But from the constraint equation we have z two equations. So ).. =  1 and x  4 = >..x
'*
4 + 1  z2 = 0
z
'*
=0
'*
x 2 + y2 = 0
'*
x = 2, y  2 = >..y
z (1 +.A) = 0, so either z = 0 or>..= 1. x = y = 0 which is not possible from the first
'*
y = 1, and x·2 + y 2

z2 = 0
'*
= ±v'S. This must correspond to a minimum, so the points on the cone closest to (4, 2, 0)
are (2, 1, ±v'S). ' 33. f(x,y,z) = xyz,g(x,y,z) = x +y + z = 100 implies x
= y =z
=
1
=* 'Vf = (yz,xz, xy) = )..'\Jg =(>.., >..,>..). Then>.. = yz = xz = xy
~0 .
35. If the dimensions are 2x, 2y, and 2z, then maximize f(x, y, z)
= (2x)(2y)(2z)
g(x, y, z) := x 2 + y 2 + z 2 = r 2 (x > 0, y > 0, z > 0). Then 'V f = .A'V g 8yz
= 2Xx, Bxz = 2
and xy = xz
2
'*
= 8xyz subject to
(8yz , Bxz, 8xy) = .A (2x, 2y, 2z) =*
41JZ 4xz 4xy . . 2 2>.y, and 8xy = 2.Xz , so>.=  · = = .This gives x 2 z = y z X Y Z
'*
z 2 = y2 , so x 2 = y 2 = z 2
'*
equation gives 3x2 = r 2
. f ( ~, ~, ~) = 8 ( ~) (
'*
x = r/../3 = y = z.
7J) (~) = 3 ~ r
3
'*
2
2
x = y (since z
'
x = y = z, and substituting into the constraint
Thus the largest volume of such a box is .
'
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www.elsolucionario.net
t= D)
SECTION 14.8 37. f(x ,y,z)
= xyz,
g(x, y ,z)
Then >. = yz = ~xz
= x +2y + 3z = 6 =>
\lf
= {yz,xz ,xy) =
= ~xy implies x = 2y, z = %Y· But 2y + 2y + 2y =
LAGRANGE MULTIPLIERS
D
233
>..Vg = (>.. , 2>., 3>.).
6 soy = 1, x = 2, z = ~ and the volume
is V = ~39. f( x, y, z) = xyz, g(x, y ,.z) = 4(x + y + z) = c 4>.
= yz = xz =
\7 f = (yz, xz, x y), >.Vg = (4>., 4>., 4>.). Thus
=>
xy or x = y = z = fie are the dimensions giving the maximum volume.
41 . If the dimensions of the box are given by x, y, and z, then we need to fi nd the maximum value of f(x, y, z ) = xyz
> 0) subject to the constraint L
[x, y, z
=
..jx2 + y 2 + z 2 or g(x, y, z ) = x 2 + y 2 + z 2 = £ 2 .
\7 f = )..\lg
yz xz (yz , xz, xy ) = >. (2x, 2y, 2z ), so yz = 2>.x => >. =  , xz = 2>.y => >. =  , and xy = 2.Xz 2X
Thus). = yz = xz .=> ~
x2
~
= y2
[since z =f. 0)
2Y ·
(L/ V3)
3
2x
= ).
+ 2J.LX (1),
+ y~  z = 2y
= >. +
From (l) and (2), 2(x  y)
0. \lf
=>
x
2
= £ 2 / 3 · =>
= x2 +
V3 =
y 2 + z 2 subject to the two constraints g(x, y , z)
= 2>. f.t
(3), x + y + 2z = 2 (4), and x 2
= 2~(x  y), so if x =f. y, ~ = 1. Putting this in (3) give,s 2z
+~
= 0 or z = ~. Then (4) and (5) become x
last equation cannot be true, so this case gives no solution. So we must have x 
x = L/
y = z and the
=
x + y + 2z = 2
= (~x, 2y, 2z}, >.\lg = (>. , >., 2>.) and ~\lh = (2J.Lx_, 2~y, ~). Thus we need
2~y (2), 2z
~ = 1 into (l) says>. = 0. Hence z
2x 2
2Z
2X
= £ 3 / (3 V3).
43. We need to fi nd the extreme values of f(x , y, z) and h(x, y, z) = x 2
=>
=> x = y and >.= y z = x y. => x = z [since y =f. 0].
Substituting into the constraint equation gives x 2 + x 2 + x 2 = L 2 maximum vo lume is
=>
z = 0 which imply z = 1  x and z = 2x2 . Thus 2x 2 = 1  x or 2x2
+ y2  z =
0 (5).
= 2>. 1 or >. = z +~. but p~1tting
+ y 3 = 0 and x 2 + y 2 + ~
= y. Then (4) and (5) become 2x + +x
 1
= (2x 
~ 0. The ·
2z
= 2 and
1)(x + 1) = 0 so x= ~ or
x =  1. The two points to check are (t, ~ . ~)and (  1,  1, 2): /(~. ~. ~) = ~and f(1, 1, 2) = 6. Thus(~,~. ~) is the point on the ellipse nearest the origin and (  1,  1, 2) is the one farthest from the origin. 45. f( x, y , z) = yex::, g(x, y, z)
= 9x2 + 4y 2 + 36z2 =
(yex::, ex  z,  yex z) = >.(18x, 8y, 72z)
36, h(x, y , z)
= xy + y z =
1.
\7 f = >.Vg + JL\lh
=>
+ i'(V, x + z, y), so yexz = 18>.x + J.l.V, ex z = 8>.y + ~(x + z ),
 yex :: = 72>.z + ~y. 9x 2 + 4y 2 + 36z 2 .= 36, x y + yz = 1. Using a CAS to solve these 5 equations s imultaneously for x, y, z , >.,
and ~
(in Maple, use the all valu es command), we get 4 realyalued solutions: X~
0.222444,
y ~· 2.157012,
z
~
 0.686049,
>.
~
 0.200401,
~ ~
2.108584
X~
 1.951921,
y
~
 0.545867,
z
~
0.119973,
).
~
0.003141,
~ ~
 0.076238
X ~
0.155142,
y
~
0.904622,
z
~
0.950293,
>.
~
 0.012447,
f.' ~
0.489938
X~
1.138731,
y
~
1.768057, .
z
~
0.573138,
>.
~
0.317141,
f.' ~
1.862675
Subst ituting these values into f gives /(0.222444, 2.157012,  0.686049) ~ 5.3506,
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234
0
CHAPTER 14 PARTIAL DERIVATIVES
f( 1.951921,  0.545867, 0.119973) !(1.138731, 1.768057, 0.573138)
~
~
0.0688, !(0.155142, 0.904622, 0.950293)
~
0.4084,
9.7938. Thus the maximum is approximately 9.7938, and the minimum is
approximately 5.3506.
47. (a) We wish to maximize j(x1, x2, ... , xn) g(x1,x2, .. . , Xn)
\1 f
=
ytx1x2 · · · Xn subject to
= X1 +x2 + ·· · +xn = c and X.;> 0.
= ( ~ (X!X2 · .. Xn).; l (X2 .. · Xn) , ~(X1X2 .. · Xn)'~l (X!X3 · .. Xn),
and >. V g
= (>., >.,
... ,
... , >.),so we need to solve the system of equations 1/n 1/n
x1
1 ( XtX2 · • • Xn ).l.1( ;;: " XtX3 • · • Xn)
This implies n>.x1 But X !
~(XtX2 · .. Xn) !,l (xl · .. Xn1) ~
= n>.x2 =
· · · = n>.x... Note >.
+ X2 + · ·· + Xn = c
have an extreme value is
=>
(~ , ~, n
n
zero (but not equal) as we like,
f
nx1
... ,
= c =>
~). n
x2
1/n 1 /n
= A
X1
X2
1/n
, = nAx 1
1/n
\ = nAX2
· · · Xn · · · Xn
=I 0, otherwise we can't have all x.; > 0. Thus X1 X1
= x2 = · · · = Xn .
= ~n = X2 = X3 = · · · = Xn  Then the only point where f
Since we can choose yalues for (Xt, x2, ... , Xn) that make
f
can
as close to
has no minimum value. Thus the maximum value is
t(;,;, ....;) = \};; ..... ;=;. (b) From part (a), ~ is the maximum value of f. Thus j(x1; x2, . .. , x n )
n
x1
+ x2 + · ·· + Xn
maximum value when Xt
14
,,,,~·
= c, so ytx1X2 · · · Xn
:::;
X1
+ X2 + · · •+ Xn . These two means are equal when f n
~' but this can occur only at the point(;,*'
= X2 = X3 =

= ytx1x2 · · · Xn :::; ~But n .
.
attams 1ts
. .. , * ) we found in part (a). So the means are equal only
c
· · · = Xn = . n
Review
CONCEPT CHECK 1. (a) A function f of two variables is a rule that assigns to each ordered pair (x , y) of real numbers in its domain a unique real number denoted by f(x, y). (b) One way to visualize a function of two variables is by graphing it, resulting in the surface z = f(x, y). Another method for visualizing a function of two variables is a contour map. The contour map consists of level curves ofthe function which are horizontal traces of the graph of the function projected onto the x yplane. Also, we can use an arrow diagram such as Figure I in Section 14.1.
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www.elsolucionario.net
CHAPTER 14 REVIEW
2. A function
f
0
235
of three variables is a rule that assigns to each ordered triple (x, y, z) in its domain a unique real number
f(x, y , z). We can visualize a function of three variables by examining its level surfaces f(x, y, z) = k, where k is a constant. 3.
f(x , y) = L means the values of f(x, y) approach the nwnber Las the point (x, y) approaches the point (a, b)
lim (:z:,11)+{a ,b)
along any path that is within the domain of f. We can show that a limit at a point does not exist by finding two different paths approaching the point along which f( x, y) has different lim its. 4. (a) See Definition 14.2.4.
(b) Iff is continuous on JR2 , its graph will appear as a surface without holes or breaks. 5. (a) See (2) and (3) in Section 14.3.
(b) See "Interpretations of Partial Derivatives" on page 927 [ET 903]. (c) To find f
., regard 1J as a constant and diffe~entiate f(x, y) with respect to x. To find / y, regard x as a constant and
differentiate f( x, y) with respect toy. 6. See the statement of Clairaut's Theorem on page 931 [ET 907]. 7. (a) See (2) in Section 14.4.
(b) See (19) and the preceding discussion in Section 14.6. 8. See (3) and (4) and the accompanying discussion in Section 14.4. We can interpret the linearization off at (a, b) geometrically
as the linear function whose graph is the tangent plane tothe graph off at (a, b) . Thus it is the linear function which best approximates f near (a, b).
9. (a) See Definition 14.4.7. (b) Use Theorem 14.4.8. 10. See (10) and the associated discussion in Section 14.4. 11. See (2) and (3) in Section 14.5. 12. See (7) and the preceding discussion in Section 14.5. 13. (a) See Definition 14.6 .2. We can interpret it as the rate of change off at ( Xo, yo) in the direction of u . Geometrically, if P is
the point (xo, yo , f(xo, Yo)) on the graph off and C is the curve of intersection of the graph off with the vertical plane that passes through P in the direction u , the directional derivative off at (xo, Yo) in the direction ofu is the ·slope of the tangent line to Cat P. (See Figure 5 in Section 14.6.) (b) See Theorem 14.6.3. 14. (a) See (8) and ( 13) in Section 14.6. (b) Du f(x , y)
©
= "\7 f (x, y) · u or Du f(x, y, z) = "\7 f(x, y, z) · u
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236
0
CHAPTER 14 PARTIAL DERIVATIVES
(c) The gradient vector of a fu nction points in the direction of maximum rate of increase of the function. On a graph of the function, the gradient points in the direction of steepest ascent. 15. (a) f has a local maximum at {a, b) if f(x, y)
:5 f(a, b) when {x, y) is near {a, b).
(b) f has an absolute maximum at {a, b) if f(x, y)
(c) f has a local minimum at (a, b) if f(x, y) (d)
~
:5 f(a , b) tor all points (x, y) in the domain of f .
f (a, b) when (x, y) is near (a, b).
f has an absolute minimum at (a, b) if f(x , y)
f(a, b) for aUpoints (x, y) in the domain of f.
~
(e) f has a saddle point at (a, b) if f(a, b) is a local maximum in one direction but a local minimum in another. 16. (a) By Theorem 14.7.2, iff has a local maximum at (a, b) and the firstorder partial derivatives of j exist there, then / :r(a, b) = 0 and / 11 (a, b)= 0. (b) A critical point off is a point (a, b) such that [.(a, b)
= 0 and j 11 (a, b) = 0 or one of these partial derivatives does
not exist. 17. See (3) in Section 14.7. 18. (a) See Figure II and the accompanying discussion in Section 14.7.
(b) See Theorem 14.7.8. (c) See the procedure outlined in (9) in Section 14.7. 19. See the discussion beginning on page 981 [ET 957]; see "Two Constraints" on page 985 [ET 961].
TRUEFALSE QUIZ 1. True. f v(a, b)
= lim
we get f 11 (a, b)
h~o
= lim
u~ b
f(a, b +h)  f(a , b) from Equation 14.3.3. Let h = ·y b. Ash~ 0, y h
+
b. Then by substituting, ,
f(a , y)f(a,b) b . y
5. False. See Example 14.2.3. 7. Tru~. If f has a local minimum and f is differentiable at (a, b) then by Theorem 14.7.2, f x(a, b) V f(a,b)
= 0 and fu (a, b) = 0, so
= (f,(a,b),fv(a,b)) = (0,0} = 0 .
9. False. V f(x , y)
= (0, 1/ y).
11 . True. V f = (cosx, cosy), so IV / I = v'cos 2 x
Du f(x, y)
= Vf
·u
+ cos2 y . But lcosBI :5 1, so IV!I :5 .J?.. Now
= IVf l l.u lcos{:}, but u is a unit vector, so IDu f (x, y) l :5 .J2 · 1 · 1 = ./2.
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www.elsolucionario.net
l CHAPTER 14 REVIEW
0
237
EXERCISES ',
+ y + 1) is defined only when x + y + 1 > 0 ~ y >  x  1, so the domain off is {(x, y) I y >  x  1}, all those points above the line y = x  1.
''
1. ln(x
3. z = f(x , y) = 1  y 2 , a parabolic cylinder
5. The level curves are
k
::2:::
y
''
''
'
J 4x2 + y2 = k or 4x 2 + y 2 =
k 2,
0, a fam ily of e ll ipses. y
X
7.
9.
y
2
I
(©J (©J (©J (©J
0
I
h o
eva1uate
~
2
~
4
T( 8 , 4) ; T ( 6 , )
=
86
;
2
80
± 2 and
= 3,
4 4 4 72 80 T( , ) ~ T(B , ) = ~ = 4. Averaging these values, we estimate Tx(6, 4) to 4
2
  2

~e approximately
+ h~  T ( 6, 4) , which we can approximate with h =
) _ T(6, 6}  T(6 , 4) _ 75  80 _ _
'4 
2x y 2(1)(1) 2 + 2y 2 = 12 + 2 (1) 2 = 3 .
~  T (B, 4), so we can approximateT:c (6, 4) by considering h =
h 0
.Lv
lim (z,v )  (1 ,1) X 2
4
3.'5 ° C / m . Similarly, Tv (6, 4} = lim T(B ,
,.., (B
th I" . e 1m1t:
X
using the values g iven in the table: T x(6, 4)
. T.,(6, 4)
is a rational function, so it is continuous on its d omain.
Since f is defined at (1, 1), we use d irect substitution to
2
11 . (a) T.,(6 , 4) = lim T(B + h,
f
,.,., ( ) _ T(6 , 2 )  T(6, 4) _ 87 80 _ 2 · 5 ' .L v 6 , 4 _  ~
2
±2: 
3 .. 5
A . th veragmg ese
values, we estimate Tv(6, 4) to be approximately  3.0° C / m.
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238
CHAPTER 14 PARTIAL DERIVATIVES
0
(b) Here u
= ( 12• )?.). so by Equation 14.6.9, Du T(6,4) = VT(6,4) · u = T, (6,4) ~ + Tv(6,4) ~·Using our
estimates from part (a), we have Du T(6, 4)
l':;j
(3.5) ~
+ (3.0) ~ =
~
0.35. This means that as we move
l':;j
through the point (6, 4) in the direction of u , the temperature increases at a rate of approximately 0.35°C/m.
. Alternatively, we can use Definition 14.6.2: Du T( 6, 4)
=
lim
r(6+h~,4+h~)T(6,4)
ho
which ·we can estimate with h = ± 2 ,;2. Then Du T(6, 4)
h
~ T( 8 , 6 ) ~T( 6 , 4)
80  ~O 2v2
=
2 2
D u T(6 , 4)
~
T(4, 2)  T(6,4) tn  2v2
7480 =~ = 2v2
=
,
0,
3 . ( ) tn' Averagmg these values, we have D u T 6, 4 v2
~
3
27z
_ !_ [T. ( )]  lim T,('x , y +h)  T:r. (x,y) T. ( ) _lim T., (6,4 +h)  T.,(6, 4) () C T.:V!/ ( X 1 y )  a x X1y h , SO xy 6 1 4 h . y ho , ho estimate with h
~
0
I
1.1 C m.
. h
W 1liC We Can
= ±2. We have Tx (6, 4) ~ 3.5 from part (a), but we will also need values for T.,(6 , 6) and T,(6, 2). If we
use h = ±2 and the values given in the table, we have
T,(6 , 6 ) l':;j T(8, 6)  T(6 , 6) 2
=
80  75 2
= 2.5 , T, (6 , 6) ~ T(4, 6) T (6, 6) =
68  75 = 3 .5 . 2
2
Averaging these values, we estimate T.,(6, 6) ~ 3.0. Similarly,
Tx( 6 , 2 ) l':;j T(8, 2) ~ Tx(6, 2)
=
90 ; 87
= l.S, T.,(6 , 2) ~
2
T(4, 2) ~ T(6, 2)
=
74 ~ 87
= 6 .5 .
Averaging these values, we estimate T, (6, 2) ~ 4.0. Finally, we esti~ate T., 11 (6, 4): T.:vy (6 ! 4)
~
T.,(6, 6) T.,(6, 4) _ 3.0  3.5 _ _ 2
~

2

0 •25 >
T. ( X)J
) ~ T.,(6, 2) T.,(6, 4) _ 4.0  3.5 __ 0 •25 • 61 4 ~ 2 2 
Averaging these values, we have Txy (6, 4) ~ 0.25. 13. f(x , y) = (5y3
fv
+ 2x2y) 8
=>
f .,= 8(5y3
= 8(5y3 + 2x2 yf (15y2 + 2x2) =
17. S(u, v,w)
Sw =
= uarctan(vyfw)
+ 2x 2 y)7
= a rctan(v y'W),
Su = u ·
S,.
2 Vw
2
= 4x3 
x y2
= 32xy(5y 3
2 2 3 (16x + 120y )(5y
u· 1 + (:JW) (v· ~w 1 12 ) =
19. f(x,y)
+ 2x 2·yf(4xy)
=> f x = 12x 2

+ 2x 2 y) 7 ,
1
1
+ (vfo)2 (fo) =
ufo
1 +v2w '
(~: v2 w)
y 2 , f v = 2xy, f.,., = 24x, fv 11
= 2_:~:,
fxv
= fvx = 2y
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CHAPTER 14 REVIEW
21. f( x, y , z) = xky1z'"
=>
fx
= kxklylzm., !11 =
lxky 1 1 z m, fz
= mxky1z""' 1 ,
0
239
f:"x = k(k 1)xk 2 y 1z"",
f vv = l(l 1)xky1 2 zm, f zz = m (m 1)xky1z"' 2 , f xv = f vcx: = klx k 1y 1 1z"', ! ex::= f ::;r; = kmx" 1y 1zm t, f vz = f ::v = lmxk yl1zm1 23.
z = x y + xe11 1"'
=>
25. (a) zx = 6x + 2 => zx (l , 2) = 8 and z 11 =  2y =>
zv(1, 2) = 4, so an equation of the tangent plane is
z  1 = 8(x  1) + 4(y + 2) or z = 8x + 4y + 1. (b) A normal vector to the tangent plane (imd the surface) at (1,  2, 1) is (8, 4,  1). Then parametric equations for the normal ·
Ime there are x = 1 + 8t, y =  2 + 4t, z = 1 
27. (a) Let F(x , y ,z) = x 2 + 2y2

· d
.
.
x1 y+ 2 z 1 =   = =t· 8 4
t, an symmetnc equations are  
3z2 • Then F.,= 2x, Fy = 4y, Fz =  6z, so F,.,(2,  1, 1) = 4, F11 (2,  1, 1) =  4,
F: (2,  1, 1) =  6. From Equation 14.6.1 9, an equation of thetangent planeis 4(x  2) 4(y + 1)  6(z  1) = 0 or, equivalently, 2x  2y  3z = 3. · 14620 · equatwns · fior the normaII'1 ~e are X(b) From Equatwns . . , symmetnc 4 29.
2
Y + 1 = =fl· z 1 = =4
(a) Let F (x, y , z ) = x + 2y + 3z sin(x yz ).. Then F,., = 1 yz cos(x yz ), Fv = 2  xz cos(xyz ), F: = 3  x y cos(xyz) , so F:z: (2, 1, 0) = 1, Fy{2,  1, 0) = 2,·F: (2,  1, 0) = 5. From Equation 14.6. 19, an equation of the tangent plane is 1(x  2) + 2(y + 1) + 5(z  0) = 0 or x + 2y + 5z = 0. 1 1 '(b) From Equatwns . 14 .6 .20, symmetriC . equatiOns . sino=
..!. v4a2 2a
c2.
. '
c2 +be. (Instead of using 0, we could just have
used the Py1hagorean Theorem.) As a result, by Lagrange's method, we must find a, b, c, and A by solving "il f = A"il g which gives the following equations: ca(4a2

c2 )  1 12 = 2A (1), c = 2A (2),
*(
4a2

c 2 ) 112

i c2 (4a 2
(3),and2a+2b +c = P (4). From(2),A= t candso(l)producesca(4a2  c2 ) ....: 112 =c =>
e2 = a2
4a2

(5),
~  34a
=> · c =
+ b=
.;;a
v'3 a =>
2a + a(l + v'3) + v'3a=P b = ·(2 J3
©
3~ (1 + J3) p
(5). Similarly, since {4a 2
~ 
v;a
= b =>
c 2)

b
112
=> 3a + 2v'3a=P => a =
= 3  6 J3 p and
c=
(2
2
=% (1+ vf3)
c2 )  1 12 + b = A
(4a 2  c2 ) 112 = a
= a and A= tc, (3) gives ~  !:__ 4

4a
+ b = ~ . so from
= 2 J3 
3 P and thus
3
V3)P.
2012 Ccngagc learni ng. Atl Ril:l>ts Rcset'·cd. May nol be SC3M 4y 2W = 0 or y = W / 2. Also
+ (L x?J, and fxv
= (4x 2L)(4y 2W) . Then.
+ (L x) 2 )  (4x 2L) 2 (4y 2W) 2 . Thus when x
> 0. Thus a minimum off occurs at (~L, %W)
=~Landy= ~W, D
> 0 and
and this ·minimum value is f(~L, ~W) = ~L 2 W 2 .
There are no other critical points, so the maximum must occur on the boundary. Now along the width ofthe rectangle let g(y) = j(O, y) = f(L , y) = L 2 [y 2
L 2 [2y  2(W y)) = 0
+ (W Y? ]. 0 :::;: y:::;: W . Then g'(y) =
~
y = ~W.
And g(%) = ~L2 W 2 . Checking the endpoints, we get g(O) = g(W) = L 2 W 2 • Along the length of the rectangle let h (x) = f(x, 0) = f(x , W) = W 2 [x2
+ (L x?]. 0 :::;: x
:::;: L. By symmetry h'(x) = 0 ~
x = ~Land
h( ~L) = %L2 W 2 • At the endpoints we have.h(O) = h(L) = L 2 W 2 • Therefore L 2 W 2 is the maximum value of f . This maximum value off occurs when the "cutting" lines correspond to sides of the rectangle.
3. (a) The area of a trapezoid is ~ h(b1 + b2), where his the height (the distance between the two parallel sides) and b1 , ~ are the Ieng!hs of the bases (the parallel sides). From the figure in the text, we see that h
= x sinO, b1 = w 
2x, and
b2 = w  2x + 2x cos 8. Therefore the crosssectional area of the rain gutter is ~(x, 8) = ~xsin 0 [(w  2x)
+ (w 2x + 2xcos 8)) =
= wx sine  2x2 sin B + x 2 sin B cos
e. 0 < X
(x sinO)(w  2x + xcos8) ~w. 0
::;
< B :::;: i
We look for the critical points of A : oAfax = w sin B  4x sin B + 2x sin B cos B and EJAfEJO = wxcos B 2x2 cosB + x 2 (cos 2 8  sin2 8), so {)A j ax= 0 cos B =
4 x w = 2 w 2x 2x 0
= wxcosB 
=
wx ( 2.
= 2wx ®
(0
~
sinO (w  4x + 2x cosO) = 0
~
sin 8 > 0). If, in addition, oAf EJB = 0, then
2x2 cosO+ x 2 (2cos2 8  1)
~) ~w 2 
2x
2
(
2
~) + x 2 [ 2 ( 2  ~)
4x 2 + wx + x
2
[
2 
1]
8 4: + ~:  1] = wx + 3x
2
= x(3x  w)
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245
246
CHAPTER 14 PROBLEMS PLUS
0
Since x
> 0, we must have x = ~w, in which case cosO=~. soB =
and A=
f, sinO=~ . k =
'{!w, b1 = ~w,ln = ~w.
{!:w 2 • As in Example 14.7.6, we can argue from the physical nature of this problem that we have found a local
maximum of A. Now checking the boundary of A, let
i w 2 sin 26, 0 < B ::; ~. Clearly g is maximized when
g( B) = A(w /2, B) = ~ w 2 sin B  ~w 2 sin B + iw2 sin Bcos 6 = sin2B = 1 in which case A= iw2 . Also along the line B =
h'(x) = w 4x = 0 *? x
~. let h(x) = A(x, ~)
= i w, and h(~w) = w(~w)
 2(tw)
2
= wx 2x2 , 0 < x < ~w .
=>
= iw 2 . Since kw2 < "If w 2 , we conclude that
the local maximum found earlier was an absolute maximum. (b) If the metal were bent into a semicircular gutter of radius r, we would have w
2
=
~7r.
../3w, 1t• would be better to bend the metal ·mto a gutter WI'th a sem1c1rcu · · Iar crosssection. . >
211'
gy (x, y)
2
2
2
. w S mce
5. Letg(x ,y)
= 1rr and A= ~11'r 2 = ~1r(~)
12
= xf(~) ·
Theng,.(:,y)
= !(~) +xf'(~) ( 
:
2)
= !(~ )  ~ f'(~;) and
= xf' ( ~) (;) = f' ( ~ ). Thus the tange?t plane at ( x 0 , y 0 , zo) on the surface has equation
[t (;~ )  y x0 !' ( ~~ )] x + [r (~~)] y  z 1
0
= 0. But any plane whose equation is of the form ax+ by+ cz
=0
passes through the origin. Thus the origin is the ~ommon point of intersection. 7. Since we are minimizing the area of the ellipse, and the circle lies above the x axis,
y
the ellipse will intersect the circle for only one value ofy. This yvalue must satisfY both the equation of the circle and the equation of the ellipse. Now 2
2
+ Y2 = 1 =>
x
a2
2
= ab2 (b2 y2 ). Substitutingintotheequationofthe
x2
b
2
circle gives
~2 (b2

y
2
)
+y2 
2y
= 0 => 2
In order for there to be orlly o ne solution to this quadratic equation, the discrirninant must be 0, so 4 4a2 b b2
a 2 b2

g(a, b)
+ a 4 = 0. The area of the ellipse is A(a, b) =
= b2

2 2
a b
+ a4
1rab, and we minimize this function subject to the constraint ·
= 0. rrb 2) (1), 2a(2a2  b
.>. = 21rb
2
2 b(l7r~ a = 47ra4
2)
:=:>
2
(2), b a
2

2 2
a b
+ a4 =
2
~a =0
0 (3). Comparing (1) and (2) gives a( ; b_ b2).
2 2
= ~ b. Substitute this into (3) to get b = ~
=> a
= jf.
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15.1
MULTIPLE INTEGRALS
D
15
Double Integrals over Rectangles
1. (a) The subrectangles are shown in the figure. The surface is the graph of f(x , y) 3
= xy and 6.A = 4, so we estimate
4
2
I: I:
v~
y
i =l j = l
2
f(x;,y1 )6.A
0
= j(2, 2) 6.A + !(2, 4) 6.A + f(4 , 2) 6.A + /(4,4) 6.A + !(6, 2) LU + !(6, 4) 6.A
2
4
6
X
= 4(4) + 8(4) + 8(4) + 16(4) + 12(4) + 24(4) = 288 (b) V ~
3
2
,L: ,L: j (x.:, v1 ) 6.A = /(1, 1) 6.A + f(1, 3) 6.A + !(3, 1) 6.A + !(3, 3) 6.A + !(5, 1) LU + !(5, 3) 6.A
i=lj=l
= 1(4) + 3(4) + 3(4) + 9(4) + 5(4) + 15(4) = 144 3. (a) The subrectangles are shown in the figure. Since 6.A
Jfn.
xe"'11 dA ~
2
2
I: I: i=l j = l
= /(1, ~)
= 1 · ~ = ~ .we estimate y
f (x;1, Yi;) 6.A
I
6.A + /(1, 1) 6.A + /(2, ~) 6.A + /(2, 1) 6.A
I 2
= e 1 1 2 (~ ) +e 1 (~) +2e 1 (~) + 2e 2 ( ~) ~ 0.990
0
.
(b) ffn xe "' 11 dA ~ .
2
2
I: I: f(x;, v1) 6.A
1
1
= !(~, t) 6.A + !(~ , ~) 6.A + f(~, t) 6.A + ! (~. ~) 6.A = ~e l/8a) + ~e318(~) + ~e318(~) + ~e 9/sa) ~ u
I 2
51
5. (a) Each subrectangle and its midpoint are shown in the figure.
f
at each midpoint and estimate 2
2
X
y 4
3
. .
.
.
0
.ffn.f(x , y)dA ~
X
y
i=l j =l
The area of each subrectangle is 6.A = 2, so we evaluate
2
1
.
. .
.
2
2
I: I: f(x.: ,y1 ) 6.A
i = lj = l
= /(1, 2.5) 6.A + /(1, 3.5) AA + /(3, 2.5) AA + /(3, 3.5) AA =  2(2) + (1)(2) + 2(2) + 3(2) = 4
0
2
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SECTION 15.8 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
3. (a) From Equations 2 we have r 2 = (  1) 2 quadrant of the xyplane, so 8 = (b) r 2 = ( 2) 2
5. Since (;I =
f
v'2; tan{;/=
= 4 r 2 =
~
277
_: 1 = 1 and the point ( 1, 1) is in the second
4'~~' , 1).
3
4; tanB = ~ =  v'3 and the point (  2, 2v'3) is in the second quadrant of the
3. Thus, one set of cylindrical coordinates is ( 4,
but r and z may vary, the surface is a vertical halfplane
halfline y = x, x
1. z
+ 2mr; z =
2 ;
2 so r =
s; + 2mr; z = 1. Thus, one set of cylindrical coordinates is ( v'2,
+ (2J3) 2 = i6 so r =
xyplane, so (} =
+ 12 =
0
i~cluding
2 ; ,
3).
the z axis and intersecting the xyplane in the
0.
4  (x 2
+ y 2 ) or 4 
x2
y2,

SQ
the surface is a circular paraboloid with vertex (0, 0, 4), axis the zaxis, and
opening downward. 9. (a) Substituting x 2 + y 2 = r 2 and x
x
+ y2 + z2 =
=x2 
y becomes
= r cos 8, the equation x 2 
1 becomes r 2

r cos 8 + z 2 = 1 or
z 2 = 1 + r cos 8  r 2 . (b) Substituting x = r cosO andy = r sinO, the equation z
z = (rcos8) 2

2
(rsinB? = r 2 (cos2 8 sin 2 B) or z = r 2 cos 28. 1 0
11.
~
r
~
2 and 0
~
:z
~
1 describe a solid circular cylinder with
radius 2, axis the z axis, and height 1, but  11'/2 ~ 8 ~ 11'/2 restricts the solid to the first and fourth quadrants of the xyplane, so we have a halfcylinder.
13. We can position the cylindrical shell vertically so that its axis coincides with the zaxis and its base lies in the xyplane. If we
use centimeters as the unit of measurement, then cylindrical coordinates conveniently describe the shell as 6
o ~ o ~ 211', o .~ z 15.
~
~
r
~
20. The region of integration is given in cylindrical coordinates by
E
= {(r,B,z) 1
11'/2 ~ 8 ~ 11'/2, 0 ~ r ~ 2, 0 ~ z ~ r 2 } . This
represents the solid region above quadrants I and IV of the xyplane enclosed by the circular cylinder r = 2, bounded above by the circular paraboloid z
= r 2 (z = x 2 + y 2 ), and bounded below by the xyplane (z = 2
/ 2 j '2 rr I 71'/ 2 0 Jo T
d¢> f0" 12 dB J: = [cos¢>] ~
X
=
16
p2 dp
( 8]~ 12 [~Pa]~
(1 ~) G)(9) 9; (2 v'3) =
19. The solid E is most conveniently described if we use cylindrical coordinates:
E
= {(r,B,z) I 0 :s e :s i. O:s r :s 3,0 :s z :s 2}. Then
.1; ./~ f(r cos f)_ , rsinB, z) 1' dz d1·d8.
JJJE f(x , y, z) dV = f0" 12
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282
0
CHAPTER 15 MULTIPLE INTEGRALS
21. In spherical coordinates, B is represented by {(p, 0, 4>) I0 ~ p ~ 5, 0 ~ 0 ~ 27r, 0 ~ cf> ~ 1r}. Thus
JJJ8 (x 2 + y 2 +
z2) 2 dV = J; f 02 "' J:(p 2 ) 2 p2 sin cf>dpdO dl/> = Io"' sin 1/>dl/> J~"' dB I~ p6 dp = [ coscf>]~ [0]~" [tp 7 ]~ = (2)(27r)(1s.~211) = 312tJ07r ~ 140,249.7
23. In spherical coordinates, E is represented by {(p, 0, ¢) 12 ~ p ~ 3, 0 ~ 0 ~ 27r, 0 ~ cf> ~ 1f} and
IIIs(:c 2 + y 2 ) dV
= ]~"' g•r I 23 (p2 s in2 4>) p2 sin cf>dpdO dcf> = Io"sin3 cf>dcf> I~"' dO I: p4 dp
I 0"'(1 cos2 1/>) sinl/>dl/> [ 0]~,. [ip5 )~ = [coscf>+ ~ cos3 4>]~ (27r) : i(243 = (1  ~ + 1  U (27r) Cil) = 16~:" =
I
25. In spherical coordinates, E is represented by { (p, 0, 4>) 0 ~ p ~ 1, 0 ~ 8 ~ 2
2
JJJE xe"' + 112 +z dV = .=
J0"/2 J0Tr12 I:(psin cf>cosO)eP fo"12 H1 
cos 2¢>) d¢
2
IoTr12 cos(} d(} ( ~ p2eP
2
[integrate by parts with u = a¢
p2 sin =
i sin 2¢]~/Z [sin(})~/2
32)
"i· 0 ~ 4> ~ ~}.Thus
Io"'12 sin2 1/>d¢ J;12 cos(} d(} j 01 p3eP
J: I:
2
2
peP dp)
= p2 , dv = peP2 dp] 2
[
~p2 eP
2

~eP J: = (i 0) (1 0) (0 + ~) = i
27. The solid region is given byE= { (p, 0, cf>) I 0 ~ p ~ a, 0 ~ 0 ~ 21r, ~ ~ cf> ~ ~}and its volume is
V
= ffis dV = I:/63 J~"' I; p2 sin cf>dpd(}drjJ = I:/: sincf>dl/> I~11' dO I; p2 dp [coscf>J;~: [0)~11' [~p3 )~
.=
= ( ~ +
4) (27r) (ta3) =
1
3
' \ 7ra
29. (a) Since p = 4 cos 4> implies p2 = 4p cos 1/>, the equation is that of a sphere of radius 2 with center at (0, 0, 2). Thus
(b) By the symmetry of the problem Mv=
= M x= = 0. Then
M, 11 = I:" I; 13 I 04 co• 4> p3 cos cf> sin cf> dp d¢ d(} = J~" I; 13 cos cf> sin r/> ( 64 cos4 rjJ) dcf> dO 
2
f "' 64 Jo
[16 cos() •] =71'13 dO '~' ¢=0
2
f 71' .ll Jo 2
dO  21 7r
Hence (x,y,z) = (0,0, 2.1).
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dp
SECTION 15.9 TRIPLE INTEGRALS IN SPHERICAL COORDINATES
0
283
31. (a) By the symmetry of the region, My: = 0 and Mx: = 0. Assuming constant density K, m =
iJ< (from Example 4). Then ·
fffE K dV = K fffE dV =
M .,y = Jjj~ z K dV = K
J;,. }~,. 14 j~o• ¢>(pcos ¢) p2 sin r/J dp drpd8 =
K
J;" J0,.14 sin ¢cos¢ [tP4] ==~os¢ d¢ d8
=if< J:w f0"' 14sin¢cos¢ (cos4 ifJ) .d(j)d(} = iK J:" d(} J;14 cos5 {j}sin (j)d¢ = iK
(8]~" [i cos6 t/>] ~14 =
[ c~r 
iK(27r)( i)
.. ( __) (My: !vfu M.,u) ( 77rK/96 ) m'm'm = 0, 0, 1rf(jB
I: ~
+ y2
= p2 sin2 ¢ and
JJJE (x2 + y2) J( dV = =
K
J;" fo"/4 focos¢(p2 sin2 ifJ) p2 sin¢ dpd¢d8 =
iK J:" J; 14sin
3
= tK
¢cos"' ¢d¢ d8
K
=if< .r:w d(} j~w/
J:.,.. fo"'/4sin3 t/> [kPsJ::~os ¢ d¢d8 4
cos5 ¢
(1 
cos2 ¢) sin¢d¢
[8]~"' [ t cos6 ¢ + ~cos 8 ¢]~ 14
= 1J ~ .;f,
../2. So Ute integral becomes
f0rr14 .fa"''12 fov"i (p sin¢ cos B) (p sin c/> sin t1) p 2 sin¢ dp dB d¢ · =
J~r/ 4 sin3 1/>d¢ J 0rr/ 2 sinB cost1dB .J0v"i p4 dp =
 [13 cos3
A.. 
'~'
cos A..] rr/4.. 1. '2) '~' o 2 l5 ( V"'
41 . The.region of integration is the solid sphere x 2 + y 2 2 p2 sin2 c/> + (p cos lj>  2)
= p2 
4pcos + 4 ~
4 5 [~ 12  ~ 2  ( la  1) ] . M5 ~ fl15


+ (z  2? ~ 4 or equivalently p ~ 4cos¢, so 0 ~ 8 ~ 211', 0 ~ c/> ~
4 =>
0:::; p ~ 4cos ¢.Also (x2 + y 2 + z2) 312 = (p2) 312
J0" 1 2 J:" J04. cos
(4096 cos6 ¢) dOd¢
2 0 cos ¢sin ifJdifJ J 0 " dB =
20 48
3
( ~cos 7 ¢]~
12
[ B ]~"
= 40i~rr
43. In ·cylindrical coordinates, the equation of the cylinder is r = 3, 0
~
z
~
10.
The hemisphere is the upper part of the sphere radius 3, center (0, 0, 10), equation 1·
2 + (z  10) 2 = 32, z ~ 10. In Maple, we can use Ute coords=cylindrical option
in a regular plot3d command. In Mathematica, we can use Parametri c Plo t 3D.
45. If E is the solid enclosed by the surface p
E = { (p, 8, ¢) I 0 ~ p ~ 1 +
V(E)
= fJ{s
dV
= 1 + t sin 60 sin 5¢, it can be described in spherical coordinates as
t sin 68 sin 5¢, 0 ~ 8 ~ 211', 0 ~ ¢ ~ ,1f}. Its volume is given by
= fo" fo2 rr J~l+(sin60sin5r/>)/u P2 sin cf>dpd(Jd¢ = 1~~"
47. (a) From the diagram, z = r cot ¢ 0 to z = ~. r to r= asin ¢ (or use a 2  r 2 = r 2 cot2 ¢ ). Thus 0
V
[using a CAS).
=0
0
= Jor 2" Jora sin v'o.2  r2
rdzdrdB
0
z = rcot r/> 0 y
=
2:r [
(a2  a2 sin2 c/Jo)3/2  a3 sin3 ¢o cot¢o + a3]
;, ~7ra 3 [1
 (cos3 c/>0 + sin2 ¢ 0 cos 1/>0 )]
= lrra3(1  cos ¢ 0 )
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SECTION 15.1 0 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
0
285
(b) The wedge in question is the shaded area rotated from B = 81 to B = 82. Letting
= volume of the region bounded by the sphere of radius P;
V.i
and the cone with angle ¢; (B = 81 to 82) and letting V be the volume of the wedge, we have
y
V = (1122 1l21) (V12  Vn) · = 1(82  B1)[pg(1 cos¢2 )

p~(l  cos¢1 )  p~(1 cos¢2 )
= 1(82 81) [(P~ p~)(1 cos ¢ 2 ) Or: Show that V =
021P2 s in 2 1 rcot ¢
1
p 1 s in 4> 1
01
(P~

pi) (1 cos ¢ 1)]
+ p~(1 cos¢1 ) ]
= i(B2  (Jt)[(p~
pi) (cos ¢ 1 
cos¢2 )]
1
r dz dr dB.
r cot 4> 2
(c) By the Mean Value Theorem with f(p) = p3 there exists some p with p 1 ~ p ~ p2 such that
f(p 2 )

f(p 1 ) =
!' (p)(p2 
p1 ) or p~  p~ = 3Ji D.p. Similarly there exists¢ with ¢ 1 ~ ~ ~ ¢ 2
such that cos ¢ 2  cos ¢ 1 ·= (sin¢) D.¢. Substituting into the result from (b) gives
D. V
15.10 1. x
= (p2 D.p)(B2 B1)(sin ¢)
Change of Variables in Multiple Integrals
= 5u v, y = u + 3v.
1) TheJacobianis a(x ~( ,y) v
3. x
= er sinB,
8(x,y)
 
8(r, B) 
5. x
D.¢ = p2 sin¢> D.pD.¢ D.B.
= ufv,
y
u,v
:;=
= l8xf8u
8x f 8v I = 5 8y f 8u oyfav 11
ll 3
=5(3) (1)(1) =16.
er cos B.
18xf8r 8xj8fJ I ~ e''sinB ercosBI · · r . = e•·er sin2 B ere'· cos2 B = sin2 fJ cos2 ()or cos 2() ayj8r ayjaB er cos() e smfJ . . y = vfw, z = wfu. oxfou 8xfav 8x/8w a(x,y,z) ayfau ayf8v 8yf8w a(u,v,w)azjou 8z/8v 8z/8w 2
1/ v
 u /v 2
0
0
1/w
v jw 2
 wfu2
0
1/u
2
I
= .!_11/w  vjw ! ( u) 0 vjw ! 0 I 0 1/w I v 0 1/u   V 2 .:..wfu 2 1/ u + wfu2 0 =.!.
1 1 (...!_  o) + ~ (o ~) + o= uvw     = o u w uvw v2
v uw
2
1. The transformation maps the boundary of S to the boundary of the imageR, so we first look at side 8 1 in the uvplane. S 1 is
described by v = 0, 0.::; u.::; 3, sox = 2u + 3v
@ 20 12 Ccng:~ge Learning. All Right5 Resencd.
= 2u andy =
u v = u. Eliminating u, we have.x = 2y, 0 _::; x
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~
6. 8 2 is
286
0
CHAPTER 15 MULTIPLE INTEGRALS
= 3 y =? x = 6 + 3(3 y) = 15 3y, 6 :::; x :::; 12. S3 is the line segment v = 2, 0 :::; u :::; 3, sox = 2u + 6 andy = u 2, giving u = y + 2 =? x = 2y + 10, 6 $ X $ 12. Finally, S4 is the segment U = 0, 0 $ V $ 2, SO X = 3v and y = V :=} X = 3y, 0 $ X $ 6. The image of
the line segment u
:::; 2, sox= 6 + 3v andy = 3  v. Then v
= 3, 0:::; v
set S is the region R shown in the xyplane, a parallelogram bounded by these four segments. IJ
y
sl
(0, 2)
s.
s
0
s,
(3,2)

(6, 3)
T
sl (3, 0)
(12,1) 0
II
X
(6, 2)
9.
= v, 0$ u ~ 1, soy= v = 'U and X = u 2 = y2 • Since 0 ~ u ~ 1, the image is the portion of the 2 2 parabola X = y ' 0 :::; y :::; 1. s2 is the segment v = 1, 0 :::; u ~ 1, thus y = v = 1 and X = u ' so 0 :::; X :::; 1. The image is the l.ine segment y = 1, 0 $ X $ 1. 83 is the segment U = 0, 0 $ V $ 1, SO X = u 2 = 0 and y = V :=} 0 $ y $ 1. The
sl
is the line segment u
image is the segment x = 0, 0 :::; y :::; 1. Thus, the image of S is the region R in the first quadrant bounded by the parabola
x = y 2 , theyaxis, and the line y = 1. y
T
u
0
X
= 2x 1, y = 2x + 1 and the parallel lines y = 1 x , y = 3  x. The first pair of equations can be written as y  2x = 1, y  2x = 1. If we let u = y  2x then these lines are mapped to the ' vertical lines u =  1, u = 1 in the uvplane. Similarly, the second pair of equations can be written as x + y = 1, x + y = 3, and setting v = x + y maps these lines to the horizontal lines v = 1, v = 3 in the uvplane. Boundary curves are mapped to boundary curves under a !ransformation, so here the equations u = y 2x, v = x + y define a transformation r  1 that maps R in the xyplane to the square S enclosed by the lines u = 1, u = 1, v = 1, v = 3 in the uvplane. To find the transformation T that maps S to R we solve u = y  2x, v = x + y for x, y: Subtracting the fi_rst equation from the second gives v  u = 3x =? x = ~ (v  u) and adding twice the second equation to the first gives u + 2v = 3y =? y = Hu + 2v). Thus one possible transformation T (there are many) is given by x = Hv u),y = ~('u + 2v).
11. R is a parallelogram enclosed by the parallel lines y
3
v=3
s
u '= 1
u= 1
T
+1
 I
0
v 1 u
1
y=2t1
'',f' :
•
X
]
J0" sin3 rjJ drjJ f02" cos2 BdB J; p4 dp + ab3 ck fo.,. sin3 rjJdrjJ f~" sin2 BdB J; p4 dp
= a3 bck [ ~ cos3 ¢cos I/>] ~ [~B +
i sin2B]~"
= a3 bck (t) (1r) ( t) + ab3 ck (1)
(1r) (i)
23. Letting u = x  2y and v = 3x  y, we have x
[iP5 ] ~ + ab3 ck [t cos3 rjJ cosr/J]~ [~B i sin 28] ~" [tP 5 ]~
= rt7r (a 2 + b2) abck 8 f(v 3u). Then ((x, y)) = 8 u,v
= f.(2v u) and y = 0
and R is the image of the rectangle enclosed by the lines u = 0, u
0
= 4, v =
~ 3/5 1/5
215 1
1/5
= .!. 5
1, and v = 8. Thus
25. Letting u = y x, v = y + x, we nave y = !(u + v), x = Hv  u). Then
8(xy) 1.:...1/2 1/ 21 1 (u: v) = i = 2 and R is the 8 1 2 112
image of the trapezoidal region with vertices (  1, 1), ( 2, 2), {2, 2), and {1, 1). Thus
!h
R
11"
I
2
y x cos  dA=
Y +X
'1
112[
11
cos u   du dv = 2 2 v V
u]u=v
v sinV
1
u=
v
112
dv = 
2
.
27. Letu =x+y andv=x+y.Thenu+v = 2y => v=Hu+v) andu  v=2x 1 8(x ( 'y) ) = 11/2 1/ 2! =.Now[u[= [x+y[< [x[+[y[
2v sin{1) dv = ~sin 1
1
=> X=~(uv).
=>  lsin0, z ·= pcos¢>. Then
au
au~
8u8v
&&
&.
& ..
au .
op = ox op + oy op + oz ap = OX sm¢> cosO+ {}y sm¢> s m9 + {)z cos rf>, and
. . 1 I 8u 8u ~.. 8u ~.. . (} ou S1m1ar y Br/> = &xp.cos'l'cos 9 + oypc~'l'sm  azpsinrj>,and
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CHAPTER 15 PROBLEMS PLUS
au
au
And afJ =  ax
.
A. •
psm.I' Slll
f)
299
. A. () h'l + au ay pSID'I'COS , W I e a2u
a2u afJ
a2u ·2 2 a 2 p2 sin ¢ sin ()
·
= 2 ya a p2 sin2 if> cos() sin 0 + X
2
D
X
2
. 2 A. au psm'l' . A. cos 0 'l'cos2 0 a + aa yu2 p2 sm X
au p sm . 'l' A. • 0 a sm y
Therefore
2
2
2
2 au cot¢> au 1 8 u 1 a u aap2u +  + p2a¢  + + ~;:opap p2 aq? p2 sin2 if> aB2
= ::~ [(sin2 ¢> cos 2 B) + (cos2 ¢> cos2 B) + sin2 B) + ~~ [(sin2 ¢> s in 2 0) + (cos2 ¢> sin2 0) + cos2 OJ + ~~ + au
[cos2 ¢> .+ sin2 ¢)
[ 2 sin if> cos()+ cos ¢> c~ 0  sin ¢> cos 0  cos 0] 2
2
ax
2
psm¢> 2 2 [ 2sin ¢> sinB + cos ¢> s~n() sin ¢> sin() sinO ] ay psm¢>
+ au
2
But 2 sin2 ¢>cos 0 + cos2 if> cos 0  sin2 ¢>cos 0  cos 0 = (sin2 if>+ cos 2 if>  1) cos 0 = 0 and simi larly the coefficient of
aujay is 0. Also sin2 ¢co?>2 0 + cos2 if>cos2 0 + sin2 e = cos 2 fJ (sin2 ¢ + cos2 ¢) + s in2 () coefficient of
a2uj8y
11. Io"' I~ Io" f(t)dtdz dy
2
= 1, and simi larly the
is 1. So Laplace's Equation in spherical coordinates is as stated.
= IIIE f(t)dV , where.
y
· E = {(t, z, y) I 0 ~ t ~ z, 0 ~ z ~ y, 0 ~ y ~ x}.
t =z
If we let. D be the projection of Eon the ytplane then D = {(y, t)
I0 S t S
x, t S y S x}. And we see from the diagram
that E = { (t, z , y) I t ~ z ~ y, t ~ y ~ x, 0 ~ t ~ x }. So }~"' I~ I; f(t) dt dz dy = Io"' I /' .ftv f (t ) dz dy dt = Io"'
= J;' [(h = Io"' =
©
2

[ft"' (y t) f(t) dy] dt
ty)f(t) J~:; dt
[~x2  tx +
tt
2
]
f(t) dt
2
= Io"'
[~x  t x ~t
= Io"'
ax
2

2
+ t 2 ]f(t) dt
2tx + t 2 )f(t) dt
~ I;(x  t) 2 f(t) dt
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a = (r·  h)/ cos¢, and the line passes through the outer rim of the cap when
= cos 1 ((r h)/r). Thus the cap
a = r =:> cos¢ = (r. h)/r =:> ¢
I (r h)/ cos¢:::; p:::; r, 0:::; (}:::; 21r, 0:::; ¢:::; cos 1 ((r h)/ r)} and its volume is
is described by { (p, 8, ¢)
v= =
1
f27r rcos 
Jo
.J o
f27r
f COS  l
Jo
((rh)/r)
((rh)/r)
Jo
112" 1cos
= 3 _ 1

3
0
0
2
[l
sin¢] P=•·
3 3P
((rh)/r) [ 3
r sin¢ 
· A._
r cos 'I'
1( _ 2
r
sinA.d dd.d(} "~'
P
'I'
d¢ d(}
p = (•·h)/cos¢
1
f27r [  3
.I 0
rr
J(rh)/ cos 4> P
h)3
cos
(r _
h? ] ¢ sin rjJ d¢ dO
cos3
 2 ,;.] ¢=cos~ ¢ =O
1
((r h)/•·) d()
.
(rh) h) +r3+ 2 r h) 3] d()  r   2(rh) 3(r r
r7f' [r3 = 31 ./o
2
1
1(
(This volume can also be computed by treating the cap as a solid of revo lution and using the single variable disk method; see Exercise 5.2.49 [ET 6.2.49].) To determine the height h ofthe cap cut from tl1e unit ball by the plane u
+v +w =
1, note that the line u
= ·u = w passes through the origin with
direction vector (1, 1, 1) which is perpendicular to the plane. Therefore this line coincides with a radius of tile sphere that passes through the center of the cap and
h is measured along this line. The line intersects the plane at ( t, sphere at (
t, t) and the
plane u + u + w = I
'7a, '7a, '7a). (See the figure.)
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CHAPTER 15 PROBLEMS PLUS
The distance between these points is h
V=
IlL
=
V jg  t Y= v'3 ( jg  t) = 3(
dV= 2
Ills 1:(~:~:~~ ~
= abc· 1rh (r  th) = abc· 1r ( 1 
= abC7r ( ~ 
©
dV = abc
1
Ills
73. Thus the volume of R is dV = abcV(S)
7J Y[1  i _(1 73)]
"7a) (j + aTa) = abC7r ( ~ 
~) ~ 0.482abc
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0
301
16
VECTOR CALCULUS
0
16.1 Vector Fields 1. F (x, y) = 0.3 i  0.4j All vectors in this field are identical, w ith length 0.5 and parallel to (3, :4) .
y
3. F (x,y) =  ~ i + (y  x) j T he length of the vector  ~ i
vt. +
+ (y 
x) j is
(y  x) 2 . Vectors along the line y
=X
are
horizontal with length ~.
5 F( )  y i +x j . x,y  Jx2 +y2 The length of the vector
~ is 1. x2 + y2
7. F (x, y , z) = k All vectors in this fie ld are parallel to the z axis and have length 1.
9. F(x,y,z)
=xk
At each point (x, y, z), F (x, y, z) is a vector of length For x
,,
lxl.
> 0, all point in the direction of the positive z axis,
while for x
< 0, a ll are in the direction of the negative
z axis. In each plane x = k, all the vectors are identical.
X
]j
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303
304
0
CHAPTER 16
VECTOR CALCULUS
11. F(x, y) ·= {x, y) corresponds to graph IV. In the first quadrant all the vectors have positive xcomponents and negative
ycomponents, in the second quadrant all vectors have negative x  and ycomponents, in the third quadrant all vectors have negative x components and positive ycomponents, and in the fourth quadrant all vectors have ~ositive x  and ycompdnents. In addition, the vectors get shorter as we approach the origin. 13. F (x, y) = (y, y + 2) corresponds to graph I. As in Exercise 12, aJI vectors in quadrants I and II have positive x~components while all vectors in quadrants III and IV have negative xcompo nents.Vectors along the line y
= 2 are horizontal, and the
vectors are independent of x (vectors along horizontal lines are identical).
15. F (x, y, z) = i
+ 2j +
17. F (x, y, z)' =·xi+ y j
3 k corresponds to graph IV, since all vectors have identical length and direction.'
+ 3 k corresponds to graph III; the projection of each vector onto the xyplane is x i + y j , which points
away from the o rigin, and the vectors point generally upward because their zcomponents are all 3.
19.
4.5'
1~I ~
I  4.5 1
I I
I
l
I
I
'
I
\
.
'
'
.. I
'
.
.. '
.'
I
I
'
' \
'I
I
The vector field seems to have very short vectors near the line y = 2x.
I
I
For F (x, y) = (0,
I
o'} we must have y 2 
2xy = 0 and 3xy 6x 2 = 0.
I
I 4.5 I
The first equation holds if y = 0 or y = 2x, and the second holds if
I
I
x = 0 or y = 2x. So both equations hold [and thus F (x, y) = 0] along
t
the line y = 2x.
I ~ \
\
4.5
21.
f(x, y)
= xe"'Y
=>
'i7 f(x , y) = f:,(x, y) i
+ jy (x, y) j
= (xe'"Y · y + e'"Y) i + (xe"'Y. x) j = (xy + l )e"'Y i + x 2 e'"!l j
23. 'ilf(x,y, z) =f:z:(x,y,z)i+fy(x,y,z) j +f::(x,y, z ) k =..) x i +. J y j +..) z ~k x 2 +y2 +z2 x 2 +y2 +z2 x 2 +y2 +z 25. f(x,y)
= x2 
y => 'il f(x,y)
The length ofV' f(x, y) is ..)4x2
= 2xi 
j.
+ 1. When x =I= 0, the vectors point away
from theyaxis in a slightly downward direction with length that increases as the distance from the y axis increases.
27.Wegraph\lf(x, y)=
2x 2 22 l +x+ y
i+
4y jalongwith 2 2y2 1 +x+
a contour map off. The graph shows that the gradient vectors arc perpendicular to the level curves. Also, the gradient vectors point in the direction in which f is increasing and are longer where the level curves are closer together.
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www.elsolucionario.net
SECTION 16.2
29. f ( x , y)
= x 2 + y 2 =>
V f( x, y )
LINE INTEGRALS
0
305
= 2x i + 2y j . Thus, each vector V f (x, y ) lias the same direction and twice the length of
the position vector of the point (x , y), so the vectors all point directly away from the origin and their lengths increase as we move away from the origin. Hence, V f is graph III.
31 . f (x, y) = (x + y) 2
=> V f (x , y) = 2(x + y) i + 2(x + y) j. The x andycomponents of each vector are equal, so all
vectors are parallel to the line y
= x . The vectors are 0 along the line y = :x and their l~ngth increases as the distance from
this line increases. Thus, V f is graph II.
33. At t = 3 the particle is at (2, 1) so its velocity is V (2, 1)
=
(4, 3). After 0.01 units of time, the particle's change in
location should be approximately 0.01 V(2, 1) = 0.01 (4, 3)
= (0.04, 0.0~) . so the particle should be approximately at the
point (2.04, 1.03) .
35. (a) We sketch the vector field F (x; y )
=xi
y j along with
several approximate flow lines. The flow lines appear to be hyperbolas with shape similar to the graph of
y =
±11x, so we might guess that the flow lines have
equations y
= C / x.
= y(t) are parametric equations of a Aow line, then the velocity vector of the flow line at the point (x , y) is x' (t) i + y' (t) j . Since the velocity vectors coincide with the vectors in the vector field, we have x' (t) i + y' (t) j = x i  y j => dx Nt = x, dy I dt =  y . To solve these differential equations, we kno~ dxldt = x => dx/ x = dt => In lxl = t +. C => x = ±~t + c = Aet for some cons~nt A, and dy I dt =  y => dy / y = dt => In Ivi =  t + I< => y = ±e t + K = B e  t for some constant B. Therefore x y = A et B~t = AB = constant. If the flow line passes through (1, 1) then (1) (1) = constant= 1 => x y = 1 => y = 1l x, x > 0.
(b) If x = x(t) andy
16.2 Line Integrals 1. x = t 3 andy
= t, 0 :::; t :::; 2, so by Formula 3
.Ia y ds ~ 1\ 3
3 (
= ..l. . 1 (9t4 :ill 3
3. Parametric equations for C are x
fc x y 4 ds =
~~ Y+ ( ~~ Ydt =
1 2
3
2
3 2 + 1) 3 / 2) O = ..l. G4 (145 1 
= 4 cos t , y = 4 sin t, 
~
2
t y'(3t )2
+ (1)2 dt
=
1) or ..l. (145 v'I4554
:::; t :::;
~.
1\a
yf9t 4 + 1 dt
1)
Then
J:~~ 2 (4 cos t) (4sin t) 4 y'( 4sin t )2 + (4cos t)2 dt = J:~~ 2 4 5 cost sin4 t yf16(sin2 t + cos2 t ) dt
= 45
2 Jrr/ (sin4 t cos t )(4) dt =  rr/ 2
2
(4) 6 [l5 sinGt) Tr/ =  Tr/ 2
2 · 40 5
= 1638.4
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308
VECTOR CALCULUS
CHAPTER 16
0
5. If we choose x as the parameter, parametric equations for Care x = x, y = v'xfor 1:::; x:::; 4 and
fc ( x2y3 v'x) dy =
v'x] 2 ~ dx =
J;' [xz. (Vx)3 
= 12 [14 x ·1
4J14 (x3 1) dx

x] 41 !2 (64 
4  14 + 1)  ill 8
X
= x, y =
=> dy = ~ dx, 0 :::;
y
7.
(2, 1)
On cl:
~X
X :::;
2.
On Cz: x = x, y = 3  :v => dy = dx, 2 :::; x:::; 3. Then
f 0 (x + 2y) dx + x 2 dy = fc 1 (x + 2y} dx + x 2 dy +.J02 (x + 2y} dx + x 2 dy =J: [x+2(~x)+x 2 (~)]dx+J: [x+2(3x}+x2 (1}]dx =
9. x
= 2sint,
y = t, z
fc xyz ds =
= 2cos t,
f 02
0:::; t:::;
(2x + ~x 2 ) dx +
7T.
J23
(6  x x
2
)
dx
Then by Formula 9,
J0 (2 sin t)(t)(  2 cos th/(!!fd + {!fltf + C¥tf dt = J; 4t sin t cost .J(2 cos t)2 + 11)2 + (2 sin t) 2 dt = J0 7T
7T
= 2 y'5 j~'ff t sin2tdt = 2 y'5 [~ t cos 2t
 2tsin 2t
+'i sin2t] ~
.J4(cos 2 t + sin2t) + 1dt
integrate by parts with ] [u t , dv s in 2t dt
=
=
= 2¥'5(~  0) = J57T 11 . Parametric equations for Care x = t, y = 2t, z = 3t, 0 :::; t :::; l. Then
15. Parametric equations for Care x
=1+
3t, y = t, z = 2t, 0 :::; t :5. 1. Then
fc z2 dx + x 2 dy + y 2 dz = J;(2t? · 3 dt + (1 + 3t}~ dt + t 2 · 2 dt
= [¥tl + 3t2 + t]~ 17. (a) Along the line x
=
= J; (23t2 + 6t +
1) dt
¥ + 3+ 1= ¥
=  3, the vectors ofF have positive ycomponents, so since the path goes upward, the integrand F ·Tis
always positive. Therefore f.e l F · dr = f.e 1 F · T ds is positive. .
© 2012 Cengoge Lc scumH:~I. copic(l, oi· duplicntcd. or posted to 11 publicly ncclsslblc wcbsilc. in who le or in r:1rt.
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SECTION 16.2
LINE INTEGRALS
D
309
37. From Example 3, p(x, y) = k(1 y), x = cost, y = sint, and ds = dt, 0 :S t :S 1r ::::}
1:. = Icy2 p(x,y) ds = IoT< sin2 t [k(1 sint)] dt = ki0" (sin2 t  sin3 t) dt = ~k I 0.,. (1 cos2t) dt k Io"(1 cos2 t)sintdt
Let u [
= cost, du =  sin t dt ] in the second integral
= k[~ + I1 1 (1u2 )du] =k (~  ~)
lv = Icx 2 p(x, y)ds = ki0.,. cos2 t(1 = k(~
sint)dt = ~ I 0.,.(1 +cos2t)dt
kio" cos2 tsintdt
i), using t?e same substitution as above. ·
39. W = I c F · dr = I;.,. (t sint,3  cost)· (1  cost,sint) dt
=
I;.,. (t 
t cost  sin t + sin t cos t + 3 sin t  sin t cos t) dt
=I;,. (t tcost + 2sin t) dt = [~t 2  (tsin t + cost)  2 cos t)~,..
integrate by .parts ] [ in the second term
41. r (t) = (2t, t, 1  t), 0 :::; t :::; 1. 1
W = Ic F · dr = Io (2t t 2 , t (1 1
= Io (4t 2t2 + t 1 + 2t  t 2 43. (a) r (t) =
at2 i +
3
bt j
::::}

t? ,1 1+
t (2t) 2 ) · (2,1,  1) dt
t + 4t2 ) dt =I; (t2 + 8t  2) dt = [lt3 + 4t2  2t)~ = ~
v (t) = r' (t) = 2at i + 3bt2 j
::::}
a(t) = v' (t) = 2a i + 6btj, and force is mass times
acceleration: F(t) = ma(t) = 2mai + 6mbtj. (b) W = f c F · dr =
J01 (2mai + 6mbtj ) · (2at i + 3bt2 j) dt = I 01 (4ma2 t +
18mb2 t 3 ) dt
= [2ma2 t 2 + ~mb2 t 4 )~ = 2ma 2 + ~mb2 45. Let F = 185 k. To parametrize the staircase, let x = 20 cost, y = 20 sin t, z
W = f c F · dr =I:.,. (0, 0, 185) · ( 20 sin t, 20 cost, ~) dt = (185) ~
=
~~ t =
J:.,.. dt =
1! t,
0 :::; t :::; 61r ::::}
(185)(90) :::::: 1.67 x 104 ftlb
47. (a) r (t) = (cost, sin t}, 0 :S t:::; 271", and let F =(a, b). Then W = I c F · dr = I ;" (a, b) · ( sin t,cost) dt = I~'lf(asint + bcost) dt = [a cost + bsin t)~.,.
=a +O  a+ O=O (b) Yes. F (x, y)
W
= kx =
(kx, ky) and
= I c F · d r =I;,. (k cost, k s~ t} · (sin t , cost} dt = J;" ( k sin t cost+ k sin t cost) dt = I~.,.. 0 dt = 0.
49. Let r(t) = (x(t), y(t), z(t)) and v = (v1, v2, va). Then
I c v · dr
= I:
(v1, v2, va) · (x' (t), y' (t) , z' (t)) dt
= I:
[v1
x' (t} + v2 y' (t)
+ va z' (t)] dt
= [v1 x(t) + v2 y(t) + va z(t)) : = [v1 x (?) + V2 y(b) + va z(b)J = v 1 [x(b) x(a)J + v2 [y(b) y(a)] + va [z (b)  z(a)] =: (v1, v2, va) · (x(b) x(a), y(b)  y(a) , z(b) z(a)) =
[v1 x(a)
+ V2 y(a) + vs z(a)]
(v1, v2, va) · [(x(b) , y(b) , z(b)} (x(a), y(a), z (a))] = v · (r(b) r (a)]
@ 201 2 C 0, but P doesn't change m
d' . ap ap O or Z  lreCtiOns, SO ax = az = .
ap aQ + 8y ax
(a) d1v F = 
aR
+ 8z
= 0+0+0
=0
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318
0
CHAPTER 16
({)R  {)Q) i + ({)P  {)R) j + ({)Q  {)p) k = (0 0) i t (0  O) j + (o {)P) k = {)P k 8y .{)z {)z 8x ax 8y 8y 8y
(b) curlF =
fJP : S mce ;:;uy
13. curiF
VECTOR CALCULUS
> 0, 
= "il x F
8P k .IS a vector porntmg . . m . the negative . zd"1rect1on. . .q,, vy
• ·
= 8f8x
j
k
8f8y
8f8z
= (6xyz
2

6xyz 2 ) i  (~y 2 z2

3y2 z 2 )j + (2yz3

2yz 3 ) k
=0
and F is defined on all oflR3 with component functions which have continuous partial derivatives, so by Theorem 4, F is conservativ~. Thus, there exists a· function f such that F
= "il f. Then f., (x, y , z) = y 2 z 3 implies
f(x,y,z) = xy2 z 3
+ g(y,z) and f 11 (x,y,z) =
2xyz3 + g11 (y,z). But f 11 (x,y,z) = 2xyz3 ,sog(y,z) = h(z) and
f(x, y,z) = xy 2 z 3
+ h(z) . Thus fz(x, y, z) =
3xy2 z 2 + h'(z) but f:(x, y, z) = 3xy2 z 2 so· h(z) = K, a constant.
Hence a potential function for F is f (x, y, z) = xy 2 z 3
15. curl F
= "il X F =
8f8x
j
k
8f8y
8f8z
+ K.
3xy2 z 2 2x2 yz3 3x2 y2 z 2
= (6x2 yz 2  6x 2 yz 2 ) i  (6xy2 z 2  6xy 2 z) j .+ (4xyz3 = 6xy2 z(I z) j + 2xyz2 (2z 3) k =1 0
6xyz 2 ) k
so F is not conservative.
k
j
17. cur!F
= "il x F =
8f8x 8f8y
8f8z
F is defined on all oflll3 , and the partial derivatives of the component functi~ns are continuous,'so F is conservative. Thus there exists a function f such that "il f =F. Then f,(x, y, z) = e11 = implies f(x, y, z) = xe11=+ g(y, z)
fu(x, y, z) = xzev= + g11 (y, z). But / 11{x, y, z) = xze11=, so g(y, z ) = h(z) and f(x, y, z) = xe11z
f(x , y, z) =
+ h(z).
= xye11z + h'(z) but f:(x, y, z) = xye11= so h(z) =Kanda potential function for F xe11=+ K .
Thus f:(x, y, z)
.19. No. Assume there is such a G. Then div(curl G )
='!
uX
(x sin y) +
~
uy
(cosy) +
~
uz
=>
is
(z  xy) =sin y siny + 1
which contradicts Theorem 11 . j
k
21 . curlF = 8f8x 8f8y 8f8z
f(x)
g(y)
= {0 0) i + {0 O)j + {0 
0) k = 0. Hence F
= f(x) i + g(y) j + h(z) k
h(z)
is irrotational. ® 2012 Ccnl!llgC Lcnming. All Rights Reserved. Mny not be sclllUlcd. copied, or duplicated. or posted ro a publicly accessible websire, in whole or in par1.
www.elsolucionario.net
f 0,
SECTION 16.5
CURL AND DIVERGENCE
0
319
ForExercises2329, 1etF(x,y,z) = P1 i+ Qd + R1 ~and G(x,y,z) = P2 i + Q2j + R2 k.
23. div(F +G)= div(Pt + P2 , Q1 + Qz,R1 + R 2 ) = o(P1 + P2) + o(Q 1+ Qz )" + o(R1/ R2 ) • 0X 0y Z = oP1 + oPz + oQ1 + oQ2 + oR1 + oR2 = (oP1 + oQ1 + oR1) + ( oP2 + oQ2 + oR2 ) ox ox oy oy oz oz ox oy oz ox oy oz
= div(P1 ,,Q1, R1 ) + div(Pz , Qz , Rz ) = div F + div G
27. div(F X G) ='iJ · (F x G)=
=
[Q
1
oR2 ox
+ R2 oQ1 _ Q2 oR1 _ R1 oQ2] _ ox ox ox
[p oRz + Rz oP1 _ Pz oR1 _ .oy oy ay 1
R 1 oPz]
ay
+ [Pt oQ2 + Q2 oP1 _ p2 oQ1. _ Q1 oP2 ] oz oz . oz oz = [p2 (oR1 _
oy
oQ1) +Q 2 (oP1 _ oR1) +Rz(oQ1 _ 8P1 )] oz oz ox OX ay
_[p (oRz _ oQz) f_ Q (oP2 _ oRz) + R1(oQz ~ aP2 )] oy az az ox ox 1
= G · curl F
1
8y
 F · curl G
29. curl{curl F) = 'iJ x ('iJ x F) =
ajax
j
k
a;ay
a;az
Now let's consider grad(div F )  'iJ2 F and compare with the above. (Note that 'iJ 2 F is defined 'on page J J 19 [ET I 095).) [continued]
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320
0
CHAPTER 16
. F) d(d IV gra
VECTOR CALCULUS
2 2 n 2p  [ ( EPH+fiQ1+fiR1)·1 +( 8 P1 +8 Ch

V
OX2

OXOY
OXOZ
oyox
2 8 R1)·
oy 2 + [)y[)z
(8OZOX
J+ ,
2 P1.
2 8 Q1
2 8 R1)k]
+ Ozoy + OZ 2
2 2 2 2 2 2 _ [( 8 P1 + 8 P1 + B P1) i + (8 Q1 + 8 Q1 + 8 Q1) j 8x2 8y2 8z 2 8x 2 8y2 8z 2
Then applying Clairaut's Theorem to reverse the order of differentiation in the second partial derivatives as needed and comparing, we have curl curl F = grad div F  \l 2F as desired. 31. (a)\lr =V.Jx2+y2 + z2= x i+ y j + z k = xi + yj+z k . .Jx2+y2+z2 .Jx2+y2+z2 .Jx2+y2+z2 .Jx2+y2 + z2 j
(b)\lxr= :x
:V
(c) \l
(.!) = r
\l (
.Jx2
r
k
:z =l:V(z)  !(v)]i + [:z(x)  :x(z)]j + [:x(y):V(x)]k = O
y
X
=~
Z
1
+ y2 + z2
) .
·
1
1
1
(2x) · (2y) (2z) . 2 .Jx2 + y2 + z2 . 2 Vx2 + y2 + z2 2 .Jx2 + y2 + z2 ~~~~~1 Jk x2 + y2 + z2 x2 + y2 + z2 x2 + y2 + z~ 
x i+ y j +z k
r
(x2 + y2 + z2)3/2 = r3 (d) \lin r
= \lln(x2 + y2 + x
.
:::::: 1
x2 + y2 + z2
33. By (13), fc f(\lg) · n ds =
z2)1/2 = ~ \lln(x2 + y2 + z2)
y • z k x i + y j +z k r + x2 + y2 + z2 J + x2 + y2 + z2 = x2 + y2 + z2 = 7.2
ffv div(f\lg) dA = JJ'o!f div(\lg) + \lg · \l f] dA by Exercise 25. But div(\lg) = \l2 g.
Hence ffv f\l 2 gdA = fc f(Vg) · nds  ffv \lg · \lf dA. 35. Let f(x, y)
= 1. Then V f
ffv \l2 gdA V 2g = 0
= 0 and Green's first identity (see Exercise 33) says
= f c (Vg) · n ds 
ffv 0 · \lgdA =>
ffv \l 2 gdA = J~ \lg · n ds. But g is harmonic on D, so
=> f c\lg· nds = OandfcDngds = fc(\lg· n)ds =O.
37. (a) We know thatw = vjd, and from the diagram sin (;I= d/r
=> v = d!JJ = (sinO)rw = lw x rl. But vis perpendicular
to both wand r, so that v = w x r.
®
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SECTION 16.6 PARAMETRIC SURFACES AND THE;IR AREAS
0
321
k
j
(b) From(a), v=w x r= 0 0 w =(O ·z wy)i + (wx~ O ·z) j+ (O · y x· O) k= wyi+ wx j X
y
Z
k
j
(c) cur] v
= "il X v =
ajax a jay ajaz wy
wx
0
= [~ (0)  ~ (wx) ] i + [~ ( wy )  ~ (o)] j + [~ (wx ) ~ (wy )] k ay az az ax ax ay = [w 
( w)] k
= 2wk =
2w
39. For any continuous function f on iR3 , define a vector field G (x , y, z) = (g(x , y , z) , 0, 0) where g( x, y, z ) = fox
a
Then div G = a
X
f (t, y, z) dt.
a a a x (g(x ,.y, z)) +a (0) +a (0) = ~ f0 f (t , y , z) dt = f (x, y, z) by the Fundamental Theorem of y
Z
vX
.
3
Calcu lus. Thus every continuous function f on R is the divergence of some vector field.
16.6
Parametric Surfaces and Their Areas
1. P (7, 10, 4) lies on the parametric surface r (u , v) = (2u + 3v , 1 + 5u  v , 2 + u + v) if and only if there are values for u and v where 2u + 3v = 7, 1 + 5u  v = 10, and 2 + u + v = 4. But solving the first two equations simultaneously gives
u = 2, v = 1 and t~ese values do not satisfy the third equation, so P does not He on the surface. Q(5, 22, 5) lies on the surface if2u + 3v = 5, 1 + 5u v
=
22, and 2 + u + v
= 5 for some values ofu and v. Solving the
first two equations simultaneously gives u = 4, v =  1 and these values satisfy the third equation, so Q lies on the surface.
3. r(u , v) = (u + v ) i + (3  v) j + (1 + 4u + 5v) k = (0,3, 1} + u (1,0, 4) + v(1, 1, 5). From Example 3, we recognize this as a vector equation of a pl.ane through the point (0, 3, 1) and containing vectors a = {1, 0, 4) and b = (1,  1, 5). If we j wish to find a more conventional equation for the plane, a normal vector to the plane is ~
X
b =
1
k
0 4 =4 i  j  k
1 1 5
and an equation of the plane is 4(x 0) (y 3) :._ (z 1) = 0 or 4x y z =  4.
5. r(s, t ) = (s, t ,
es
2
),
so the corresponding parametric equations for the surface are X
point (x , y , z) on the surface, we have z = y 2

=
s, y = t, z = t 2

s 2 . For any
x 2 . With no restrictions on the parameters, the surface is z = y 2
we recognize as a hyperbolic paraboloid.
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x
2,
which
322
0
CHAPTER 16
VECTOR CALCULUS
The surf~e has parametric equations x
= u 2 , y = v 2 , z = u + v, ; 1 ~ u
~ 1, 1 ~ v ~ 1.
In Maple, the surface can be graphed by entering
pl ot3d([u"2,v"2,u+v ) ,u=l .. l,v=l .. l);. In Mathematica we use the Paramet r icPlot3D command. If we keep u constant at u 0 , x = uij, a constant, so the corresponding grid curves must be the curves parallel to the yzplane. If vis constant, we have y
= vij, a constant, so these
grid curves are the curves parallel to the xzplane.
9. r (u,v)
= (ucosv,usinv,u5 ).
The surface has pararnetrid equations x = u cos v, y = u sin v, z = u 5 , 1 ~ u ~ 1, 0 ~ v ~ 21r. Note that ifu
= u 0 is constant
then z· = u8 is constant and x = uo cos v, y = Uo sin v describe a
vconstant
luo 1. so the corresponding grid curves are circles parallel to the xyplane. If v = vo, a constant, the parametric
circle in x, y of radius
equations become x
= u cos vo, y = u sin vo, z =
u 5 . Then
= (tan v0 )x, so these are the grid curves we see that lie in vertical planes y = kx through the zaxis.
y
11. x = sin v, y Note that if v
= cosusin4v,
z = sin2usin4v, 0 ~ u ::; 21!',  ~ ::; v::; ~ 
= vo is constant, then x = sin vo is constant, so the
corresponding grid curves must be parallel to the yzplane. These are the vertically oriented grid curves we see, each shaped like a " figureeight." When u = uo is held constant, the parametric equations become x
= sin v, y =
cos u 0 sin 4v,
z = sin 2u0 sin 4v. Since z is'a constant multiple of y, the corresponding grid curves are the curves contained in planes
z = ky that pass through the xaxis. 13. r( '1.1;• v) = u cos v i + u sin v j
+ v k. The parametric equations for the surface are x = u cos v, y = u sin v, z =
v. We look at
the grid curves first; if we fix v, then x andy parametrize a straight line in the plane z = v which intersects the zaxis. If u is held constant, the projection onto the xyplane is circular; with z = v, each grid curve is a helix. The surface is a spiraling ramp, graph IV.
(1:) 2012 Ccngngc Learning. All RighiS Reserved. May not be scanned, copied, or duplicated, or posted to o publicly ace r(1, ~)
= (~.4. i)·
r u = cos v i + sin v j and r , = u sin v i + u. cos v j + k , so a normal vector to the surface at the point ( ~ , r u ( 1, f) x r , ( 1, i') = ( ~ i +
4 j) x ( 4 i + ~ j + k) = 4 i 
(t .4,f)is4(x· t)  ~(y4) + 1 (z ~ )=0 37. r (u,v)=u 2 i +2u.sinv j +ucos v k
or
~j +
4 ,~) is
k. Thus an equation of the ta~gent plane at
~x  h +z =~ .
=> r (1,0)=(1,0, 1).
r u = 2u. i + 2sinv j + cosv k and r , = 2u.cosv j  usinv k ,
2
so a normal vector to the surface at the.point (1 , 0, 1) is r u(1, 0) X r ,(1, 0) = (2 i + k ) X (2j) =  2 i + 4 k. Thus an equation of the tangent plane at (1, 0 , 1) is  2(x 1) + O(y 0) + 4(z  1) = 0 or x + 2z
= 1.
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SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
=
39. The surfaceS is given by z
f(x , y)
2y which intersects the xyplane in the line 3x + 2y
= 6 3x 
I
triangular region given by { (x, y ) 0:::; x:::; 2, 0 :::;
v$
D
= 6, soD is the
3  ~x}. By Formula 9, the surface area of S is I
!fv
A (S)=
= f(x, y) =
A(S) = =
43. z
(~Y + (~Y dA + (2)~ dA = Ji4J.(v dA = v'I4A(D ) = v'I4 (~ · 2 · 3) = 3 y'l4.
J1 + ( 3) 2
= ffv 41 . Here we can write z
1+
1~v
i : i x jy and Dis the disk x
1+
2
= ViA1r
= f (x, y) = ~(x3 1 2 + y 312 ) and D = {(x , y) I 0 $ A(S) = ffv J1 + (..fi) 1
=j
2
A(S ) = ffv Jl
J:::
dx
1
= j f0
~(x + 1)5/ 2]: =
[Hx+ 2)5/2 _ + y2 $
x $ 1, 0 $ y $ 1 }. Then f,
= x 112 , f v = y 112 and
+ (.jfj) 2 dA = f01 J; y'1 + x + ydydx
[tCx + y + 1)312
45. z = J(x, y) = xy with x 2
$ 3, so by Fonnula 9 the area of the surface is
(:~r + (:~r dA = 1~v J1 + (t/ + (~) 2 dA = 41 1~v dA
:lP A(D) = fl ·1r(y'3)
= fo
+ y2
2
1, so f , = y, fu = x
[(x+ 2)
312
+ 1) 312] dx
 (x
Ji(35/2 _ 2s; 2 _
2 s~2 + 1) = A( 31>/2 _
2112 + 1)
=>
+ Y2 + X 2 dA = J;"' J; .fr2TI rdrd(} = J:"' [~ (r2 + I?' 2
J:::
d(}
= J;"' i(2 V2  1) d8 = 2; (2 .j2  1) 47. A parametric representation of the surface is x
A(S) = =
J; J; ) 17 + 4z
2
dxdz
= J01 )
:::; 1: 0 :::; z :::; 1.
A(S) = ffv lr u x r vl dA =
=
J01 J;
.../v 4
( :~y + ( ~~y dA. Then
4I + ,t(ln(2 + v'2I) 
+ 4u2 v 2 + 4u4 dv du =
1
0 $ U u) $ 1
1+
1
ln v'I7]
(v 2 ,  2uv, 2u2 ). Then
= J0 J; (v2 + 2u2 ) dv du = J; (tv 3 + 2u2 vJ ::~ 51. From Equation 9 we have A(S)
JL
17 + 4z 2 dz
%(z ) 17 + 4z2 + ¥ lni2z + v'4z2 + 17 I)] ~ =
(2u, v , 0), r , = (0, u, v), and r u x r ,.
2
= z with 0 :::; x
~ r : = :~ i  j + ¥z k and A (S) =
Note: In general, ify = f(x , z ) then r ,
=
4x + z 2 , z
= (i + 4 j ) X (2z j + k ) = 4 i  j + 2z k.
Hence r :x: x r :
49. r .,
= x, y =
du
J; J; J(v
2
+ 2u2)2 dv du
= J01 (J + 4u2 ) du = [~u + ~u3 ]~ = 4
= ffv J l + (1, ) 2 + Uu) 2 dA. But if Jf, J:::; 1 and If vi :::; 1 then 0 :::;
(f.,? :::; 1,
=> 1 $ 1 + U:~Y + Cfu)2 $ 3 => 1 $ J 1 + (!:,) 2 + Uu) 2 $ v'3. By Property 15.3 .1 1,
ffv1dA $ ffvJl+(f,.)2+(f11 )2 dA $ ffv ../3dA =>
A(D) $ A(S)$../3A( D ) =>
.
1rR2 $ A(S) $ ../31rR2 •
©
20 12 Ceognge L~iog. All Ri(!lus Reserved. Mny 001 be scanned, copied, or duplic:noed, or posted lo a publicly oc= = I:1< I: v'1+ 4r e  r 2
A(S) =
2
2
2
IIo y'1 + 4(x2 + y2)e2(>= +u > dA r dr dB = I:"dB I: r y'1 + 4r2e 2r dr = 27r I: r V1+ 4r2e 2r
2 Y )
2 2
:=:; 4. 2
+ (  2ye"'  v
2
)
2
dA =
2
2
4x
2
2
dr ~ 13.9783
+ 4y 2
1 + (1 +x 2 + y2)4 dydx.
=
Using the Midpoint Rule with f( x, y) 3
A(S) ~
2
2: 2:
1+ (
.
+ 4y2 ) , m 2 1 +x +Y2 4 4x
2
f(xi ,'il; ) ~A = 4 (!(1, 1) + /(1, 3) +
= 3, n = 2 we have
/(3, 1) + / (3, 3) + /(5, 1) + / (5, 3)] ~ 24.2055
i = lj = l
11 6
(b) Using a CAS we have A(S) =
4
4 2 1 + (l x
4
Y:) 4 +x: +y
dy dx
~ .24.2476. This agrees with the estimate in part (a) 
to the first decimal place. 57. z = 1 + 2x
+ 3y + 4y2 , so
Us ing a CAS, we have
o r !!! 8
v'I4 + ll Jn n3 ~+ V'J! 16 0 + 70 .
59. (a) x = a sin u cos v, y
= bsin u sin v, z = ccos u =>
~~ IZfi
(b)
71.
2
z 0
and s ince the ranges of u and v ~ suffic ient to generate the entire graph,
2
'lii! ~
the parametric equations represent an ellipsoid. 2
(c) From the parametric equatio ns (with a = 1, b = 2, and c we calculate r u
= cosucosv i + 2cos usinv j 
r, = s in usin v i + 2sin ucos v j. So r ,. x r,
61. To find the region D : z = x 2 + y 2 implies z
= 6 sin 2 t! cos v i +
+ z 2 = 4z or z 2 
J4 

3sin u k and 3 sin2 'u sin vj + 2 sin u cos u k , and the surface
3z = 0. Thus z = 0 or z = 3 are the planes where the
surfaces intersect. But x 2 + y 2 + z 2 = 4z implies x 2 + y 2 + (z  2) 2 Thus (z  2) 2 = 4  x 2  y 2 or z = 2 +
y
= 3),
0
= 4. so 'z = 3 intersects the upper hemisphere.
x
x 2  y 2 • Therefore D is the region inside the c~cle 2 + y 2 + (3  2) 2 = 4,
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SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
that is, D
0
= {(x, y) I x 2 + y 2 :
1+ 0 + 211" + 0 =
211" + ~·
8zj8x = e11 , 8zj8y = xe11 , so by Formula 4, a CAS gives
JJ5 (x 2 +y2 +z2 )d8 = J01 f 01 (x2 +y2 +x2 e211 )Je2 Y +x2 e2v + 1dxdy ~ 4.5822. 35. We use Formula 4 with z = 3 2x 2
3 2x2

y2
~ 0 are

y2
/i $ s ji X
=> 8zj8x
and
J3 
= · 4x, 8z j8y = 2y. The boundaries of the region
2x2 $ y $
J3 
2.7:2 • so we use a CAS (with precision reduced to
seven or fewer digits; otherwise the calculation may take a long time) to calculate
37. If 8 is given byy n
= h(x,z), then 8
is also the level surface f(x,y,z)
Vf(x,y,z))I =  h., i+jh: k , and = I~/( v x,y,z y'h~+l+h~
n
=y
h(x,z)
= 0.
. h · 1h · h 1ft N · IS t e umt norma t at pomts tot e e . ow we proceed as m the
derivation of (1 0), using Formula 4 to evaluate
where D is the projection of 8 onto the xzplane. Therefore
©
IIs
F · dS
=
Ii ( ~~
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P
•
 Q +R
~~) dA.
publicly oeccssible websilc, in whole or in port.
332
39.
0
CHAPTER 16
VECTOR CALCULUS
m =· ffs }( dS = }( · 47r(~a 2 ) = 21ra 2 K; by symmetry M xz = lvf11z = 0, and M:cv
= Jj~ z }( dS = }( f~" f 0 12 (a cos ¢)(a2 sin ¢) d¢ d() = 2711 x ro =
= (c/a3 )
+ a cos
g(x) = K.
+ xe"'" + [(is a potential function for F . 2xy3 )
= Bx 3 y 6xy 2 =
Furthermore f( x, y) = x 4 y 2

:, (2x 4 y 3x2 y 2 + 4y3 ) and the domain ofF is IR?, F is conse~ative.
x 2 y 3 + y 4 is a potential fu nction for F . t
= 0 corresponds to the point (0, 1) and t = 1
corresponds to (1, 1), so f c F · dr = /(1 , 1)  /(0,.1) = 1  1 = 0.
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339
340
CHAPTER 16 VECTOR CALCULUS
0
15. C1: r(t)=t i +t2 j, l ~t~ l;
C2: r (t) = :ti + j ,  1 ~ t
~
y
1.
cl
( l , l ) ot~~+..., (1 . 1 )
.
Then
Ic xl dx x 2 ydy = ~
f
2t5 )dt + J~ 1 tdt
5 1 (t 
[.!t6)1 + [.!t2]1 6 1 2 1
X
= '0
Using Green's Theorem, we have
j~ xy
2
2
dx  x ydy =
Ji [
2
:x (x y)  :y (xy
= I~ 1 (2xy 2 J~:: 2 dx 2
2
17. Ic x ydx xy dy. = :r2
2 )]
dA =
= I~ 1 (2x5 
Ji
1
(2xy 2xy) dA = /_
2x)dx = [~x 6  x 2
II [ tfx (xy 2 )  %v (x 2 y)] dA = .f[ (  y 2 + y2 ~ 4 :z:2 + y2 ~ 4

t
1
1:
 4xydydx
1
= 0
x 2 ) dA =  I;" I; r 3 dr dB= 81r
19. If we assume there is such a vector field G, then div( curl G) = 2 + 3z  2xz. But div(curl F)
= 0 for all vector fields F.
Thus such a G cannot exist. .21. For any piecewisesmooth simple closed plane curve C bounding a region D, we can apply Green's Theorem to
F (x, y) = f(x) i
+ g~y) j to get Ic f(x) dx + g(y) dy =
IIo [ /!"' g(y) 
. 23. \7 2 f
/!
11
f(x)] dA = Jj0 0 dA = 0. '
= 0 means that~:{ + ~:; = 0. Now ifF= f v i  f x j and C
is any closed path in D , then applying Green's
Theorem, we get
Ic F · dr= Ic f vdx f x dY = IIo [:x (fx)
/!11 Uv)]dA
= .f[0 (fxx + f.uv) dA =  IJ'v 0 dA = 0 Therefore the line integral is independent of path, by Theorem 16.3.3. 25. z
= f (x, y) =
x2
+ 2y with 0 ~ x ~ 1, 0 ~ y ~ 2x. Thus
27. z = f(x, y) = x2 + y 2 with 0 ~ x2 + y 2 ~ 4 so r., x r 11 = 2x i  2y j + k (using upward orientation). Then
II5 z dS=
II
(x 2 +y2 )J4x 2 +4y2 +ldA
;r2 + y2 ~ 4
(Substitute u = 1 + 4r 2 and use tables.)
29. Since the sphere bounds a simple solid region, the Divergence Theorem applies and
IIsF · dS =IIIE divFdV = f.[f E(z 2) dV = I.ff Ez dV 2III E dV
= O [oddfu~ction in =. and
E
os symmctnc
]
_
?=
2 . V(E) = _ 2 . h( 2 3
_Q17r' 3
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CHAPTER 16 REVIEW
D
Alternate solution: F(r(¢,B)) = 4sin¢ cosB cos ¢ i  4 sin¢ sinBj + 6sin ¢ cosB k,
r q, x r o
= 4sin2 ¢
cos .B i + 4sin2
cos B. Then
Jfs F · dS = f02" f 0" (16 sin3 ¢ cos¢ cos2 B 64 = J:" ~(  16sin B) dB =  3 1r
16 sin3 ¢ sin2 B + 24 sin2 ¢ cos¢ cos B) d¢ dB
2
31 .
Since curlF = 0 , JJ8 (curlF) · dS = 0. We parametrize C: r (t) = costi +sin t j , 0::; f c F · dr =
J:" ( cos2 t sin t + sin2 t cost) dt = 1cos3 t + t sin3 t] ~,. = 0.
+z =
33. The surface. is given by x + y
fcF · dr = ffs curlF · dS 35.
JJJ E
t::; 21r and
=
JJJ
divFdV = ,2
1 or z
=1
x y, 0 ::; x::; 1, 0 ::; y::; 1 x and r, x r 11 = i + j + k. Then
ffv( y i  z j  x k ) · (i + j + k) dA
+ 112 + z2.,; 1
= ffv(1)dA = (area of D) = ~.
3dV = 3(volume of sphere)= 47r. Then
F (r(¢, B)) · (r .p x r o) = sin3