Cengage learning COMPLETE solutions manual for multivariable calculus 7th edition 0328693561

INSTRUCTOR
SOLUTIONS
MANUAL


Complete Solutions Manual
for

MULTIVARIABLE CALCULUS
SEVENTH EDITION

DAN CLEGG
Palomar College
BARBARA FRANK
Cape Fear Community College

Australia . Brazil . Japan . Korea . Mexico . Singapore . Spain . United Kingdom . United States

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■

PREFACE
This Complete Solutions Manual contains detailed solutions to all exercises in the text Multivariable
Calculus, Seventh Edition (Chapters 10–17 of Calculus, Seventh Edition, and Calculus: Early

Transcendentals, Seventh Edition) by James Stewart. A Student Solutions Manual is also available,
which contains solutions to the odd-numbered exercises in each chapter section, review section,
True-False Quiz, and Problems Plus section as well as all solutions to the Concept Check questions.
(It does not, however, include solutions to any of the projects.)
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, users of the Early Transcendentals
text should use the references denoted by “ET.”
While we have extended every effort to ensure the accuracy of the solutions presented, we would
appreciate correspondence regarding any errors that may exist. Other suggestions or comments are
also welcome, and can be sent to dan clegg at or in care of the publisher:
Brooks/Cole, Cengage Learning, 20 Davis Drive, Belmont CA 94002-3098.
We would like to thank James Stewart for entrusting us with the writing of this manual and offering suggestions and Kathi Townes of TECH-arts for typesetting and producing this manual as well as
creating the illustrations. We also thank Richard Stratton, Liz Covello, and Elizabeth Neustaetter of
Brooks/Cole, Cengage Learning, for their trust, assistance, and patience.
DAN CLEGG

Palomar College
BARBARA FRANK

Cape Fear Community College

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■

ABBREVIATIONS AND SYMBOLS

CD

concave downward

CU

concave upward

D

the domain of i

FDT

First Derivative Test

HA

horizontal asymptote(s)

I

interval of convergence

I/D

Increasing/Decreasing Test

IP

inÀection point(s)

R

radius of convergence

VA

vertical asymptote(s)

CAS

=

indicates the use of a computer algebra system.

H

indicates the use of l’Hospital’s Rule.

m

indicates the use of Formula m in the Table of Integrals in the back endpapers.

s

indicates the use of the substitution {x = sin {> gx = cos { g{}.

=
=
=
c

=

indicates the use of the substitution {x = cos {> gx = 3 sin { g{}.

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■

CONTENTS
■

10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.1

Curves Defined by Parametric Equations
Laboratory Project

10.2

Polar Coordinates

■

15

18

Bézier Curves

32

33

Laboratory Project

■

Families of Polar Curves

10.4

Areas and Lengths in Polar Coordinates

10.5

Conic Sections

10.6

Conic Sections in Polar Coordinates

Review

1

Running Circles Around Circles

Calculus with Parametric Curves
Laboratory Project

10.3

■

48

51

63
74

80

Problems Plus 93

■

11 INFINITE SEQUENCES AND SERIES
11.1

Sequences

97

97

Laboratory Project

■

Logistic Sequences

110

11.2

Series

11.3

The Integral Test and Estimates of Sums

11.4

The Comparison Tests

11.5

Alternating Series

11.6

Absolute Convergence and the Ratio and Root Tests

11.7

Strategy for Testing Series

11.8

Power Series

11.9

Representations of Functions as Power Series

114
138

143
156

160

11.10 Taylor and Maclaurin Series
Laboratory Project

■

Applied Project

Problems Plus

■

169

179

An Elusive Limit

11.11 Applications of Taylor Polynomials
Review

129

194

195

Radiation from the Stars

210

223

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209

149

1


viii

■

CONTENTS

■

12 VECTORS AND THE GEOMETRY OF SPACE
12.1

Three-Dimensional Coordinate Systems

12.2
12.3
12.4

Vectors 242
The Dot Product 251
The Cross Product 260
Discovery Project

12.5

Equations of Lines and Planes

■

13 VECTOR FUNCTIONS

273

Putting 3D in Perspective

285

287

313

13.1

Vector Functions and Space Curves

13.2
13.3
13.4

Derivatives and Integrals of Vector Functions 324
Arc Length and Curvature 333
Motion in Space: Velocity and Acceleration 348
Review

Problems Plus

271

307

Applied Project

■

■

Cylinders and Quadric Surfaces
Review
297

Problems Plus

235

The Geometry of a Tetrahedron

■

Laboratory Project

12.6

235

■

Kepler’s Laws

313

359

360

367

14 PARTIAL DERIVATIVES

373

14.1

Functions of Several Variables

14.2
14.3
14.4
14.5

14.6
14.7

Limits and Continuity 391
Partial Derivatives 398
Tangent Planes and Linear Approximations 416
The Chain Rule 425
Directional Derivatives and the Gradient Vector 437
Maximum and Minimum Values 449
Applied Project

■

Discovery Project

373

Designing a Dumpster
■

469

Quadratic Approximations and Critical Points

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471

CONTENTS

14.8

Lagrange Multipliers
■

Rocket Science

Applied Project

■

Hydro-Turbine Optimization

Review

Problems Plus

■

474

Applied Project

485

490

505

15 MULTIPLE INTEGRALS

511

15.1

Double Integrals over Rectangles

15.2
15.3
15.4
15.5
15.6
15.7

Iterated Integrals 516
Double Integrals over General Regions 521
Double Integrals in Polar Coordinates 534
Applications of Double Integrals 542
Surface Area
553
Triple Integrals 557
Discovery Project

15.8

■

15.9

511

Volumes of Hyperspheres

Triple Integrals in Cylindrical Coordinates
Discovery Project

■

■

Roller Derby

575
584

594

15.10 Change of Variables in Multiple Integrals
Review
601

Problems Plus

595

615

16 VECTOR CALCULUS

623

16.1

Vector Fields

16.2
16.3
16.4
16.5
16.6
16.7
16.8
16.9

Line Integrals 628
The Fundamental Theorem for Line Integrals
Green’s Theorem 643
Curl and Divergence 650
Parametric Surfaces and Their Areas 659
Surface Integrals 673
Stokes’ Theorem 684
The Divergence Theorem 689
Review
694

Problems Plus

574

The Intersection of Three Cylinders

Triple Integrals in Spherical Coordinates
Applied Project

■

488

623

705

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637

582

■

ix


x

■

CONTENTS

■

17 SECOND-ORDER DIFFERENTIAL EQUATIONS
17.1

Second-Order Linear Equations

17.2

Nonhomogeneous Linear Equations

17.3

Applications of Second-Order Differential Equations

17.4

Series Solutions
Review

■

711

APPENDIX
H

711
715

725

729

735

Complex Numbers

735

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720


NOT FOR SALE
10

PARAMETRIC EQUATIONS AND POLAR COORDINATES

10.1 Curves Defined by Parametric Equations
1. { = w2 + w,

| = w2 3 w, 32 $ w $ 2

w

32

31

0

1

2

{

2

0

0

2

6

|

6

2

0

0

2

2. { = w2 ,

w

| = w3 3 4w, 33 $ w $ 3
±3

±2

±1

0

{

9

4

1

0

|

±15

0

~3

0

3. { = cos2 w,

| = 1 3 sin w, 0 $ w $ @2

w

0

@6

{

1

3@4

|

1

1@2

4. { = h3w + w,

13

@3

@2

1@4

0

I

3
2

E 0=13

0

| = hw 3 w, 32 $ w $ 2

w

32

31

0

1

2

{

h2 3 2

h31

1

h31 + 1

h32 + 2

1=37

2=14

|

32

h31

h2 3 2

1=72

5=39

5=39
h

+2

2=14

1=72
31

h

+1

1=37

1

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1

2

¤

NOT FOR SALE
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

5. { = 3 3 4w, | = 2 3 3w

(a)
w

31

0

1

2

{

7

3

31

35

|

5

2

31

34

(b) { = 3 3 4w i 4w = 3{ + 3 i w = 3 14 { + 34 , so


| = 2 3 3w = 2 3 3 3 14 { + 34 = 2 + 34 { 3 94 i | = 34 { 3
6. { = 1 3 2w, | =

1
w
2

1
4

3 1, 32 $ w $ 4

(a)
w

32

0

2

4

{

5

1

33

37

|

32

31

0

1

(b) { = 1 3 2w i 2w = 3{ + 1 i w = 3 12 { + 12 , so



| = 12 w 3 1 = 12 3 12 { + 12 3 1 = 3 14 { + 14 3 1 i | = 3 14 { 3 34 ,
with 37 $ { $ 5

7. { = 1 3 w2 , | = w 3 2, 32 $ w $ 2

(a)
w

32

31

0

1

2

{

33

0

1

0

33

|

34

33

32

31

0

(b) | = w 3 2 i w = | + 2, so { = 1 3 w2 = 1 3 (| + 2)2
2

i

2

{ = 3(| + 2) + 1, or { = 3| 3 4| 3 3, with 34 $ | $ 0
8. { = w 3 1, | = w3 + 1, 32 $ w $ 2

(a)
w

32

31

0

1

2

{

33

32

31

0

1

|

37

0

1

2

9

(b) { = w 3 1 i w = { + 1, so | = w3 + 1 i | = ({ + 1)3 + 1,

or | = {3 + 3{2 + 3{ + 2, with 33 $ { $ 1

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NOT FOR SALE
SECTION 10.1

9. { =

(a)

CURVES DEFINED BY PARAMETRIC EQUATIONS

I
w, | = 1 3 w

(b) { =

w

0

1

2

3

4

{

0

1

1=414

1=732

2

|

1

0

31

32

I
w i w = {2

33

i | = 1 3 w = 1 3 {2 . Since w D 0, { D 0.

So the curve is the right half of the parabola | = 1 3 {2 .
10. { = w2 , | = w3

(a)
w

32

31

0

1

2

{

4

1

0

1

4

|

38

31

0

1

8

(b) | = w3

i w=

s
3
|

i { = w2 =

 s 2
3
| = | 2@3 . w M R, | M R, { D 0.

11. (a) { = sin 12 , | = cos 12 , 3 $  $ .

(b)

{2 + | 2 = sin2 12  + cos2 12  = 1. For 3 $  $ 0, we have
31 $ { $ 0 and 0 $ | $ 1. For 0 ?  $ , we have 0 ? { $ 1
and 1 A | D 0. The graph is a semicircle.
12. (a) { =

1
2

cos , | = 2 sin , 0 $  $ .
 2
(2{)2 + 12 | = cos2  + sin2  = 1 i 4{2 + 14 | 2 = 1 i

(b)

{2
|2
+ 2 = 1, which is an equation of an ellipse with
2
(1@2)
2

{-intercepts ± 12 and |-intercepts ±2. For 0 $  $ @2, we have
1
2

D { D 0 and 0 $ | $ 2. For @2 ?  $ , we have 0 A { D 3 12

and 2 A | D 0. So the graph is the top half of the ellipse.

13. (a) { = sin w> | = csc w, 0 ? w ?

For 0 ? w ?


2,


.
2

| = csc w =

1
1
= .
sin w
{

(b)

we have 0 ? { ? 1 and | A 1. Thus, the curve is the

portion of the hyperbola | = 1@{ with | A 1.

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3


4

¤

NOT FOR SALE
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

14. (a) { = hw 3 1, | = h2w .
w 2

(b)
2

| = (h ) = ({ + 1) and since { A 31, we have the right side of the
parabola | = ({ + 1)2 .

15. (a) { = h2w

1
2

i 2w = ln { i w =

| =w+1 =

1
2

ln {.

(b)

ln { + 1.

I
w + 1 i {2 = w + 1 i w = {2 3 1.
s
I
I
| = w 3 1 = ({2 3 1) 3 1 = {2 3 2. The curve is the part of the
I
hyperbola {2 3 | 2 = 2 with { D 2 and | D 0.

16. (a) { =

17. (a) { = sinh w, | = cosh w

(b)

i |2 3 {2 = cosh2 w 3 sinh2 w = 1. Since

(b)

| = cosh w D 1, we have the upper branch of the hyperbola | 2 3 {2 = 1.

18. (a) { = tan2 , | = sec , 3@2 ?  ? @2.

1 + tan2  = sec2 

i

1 + { = |2

(b)
i

{ = | 2 3 1. For

3@2 ?  $ 0, we have { D 0 and | D 1. For 0 ?  ? @2, we have
0 ? { and 1 ? |. Thus, the curve is the portion of the parabola { = |2 3 1
in the first quadrant. As  increases from 3@2 to 0, the point ({> |)
approaches (0> 1) along the parabola. As  increases from 0 to @2, the
point ({> |) retreats from (0> 1) along the parabola.
19. { = 3 + 2 cos w, | = 1 + 2 sin w, @2 $ w $ 3@2.

By Example 4 with u = 2, k = 3, and n = 1, the motion of the particle

takes place on a circle centered at (3> 1) with a radius of 2. As w goes from
2


2

to

3
,
2

the particle starts at the point (3> 3) and

2

moves counterclockwise along the circle ({ 3 3) + (| 3 1) = 4 to (3> 31) [one-half of a circle].

 { 2
{
, cos w = | 3 4. sin2 w + cos2 w = 1 i
+ (| 3 4)2 = 1. The motion
2
2
, the particle starts at the point (0> 5) and
of the particle takes place on an ellipse centered at (0> 4). As w goes from 0 to 3
2

20. { = 2 sin w, | = 4 + cos w

i sin w =

moves clockwise to (32> 4) [three-quarters of an ellipse].
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SECTION 10.1

CURVES DEFINED BY PARAMETRIC EQUATIONS

 { 2

¤

5

 | 2

|
{
, cos w = . sin2 w + cos2 w = 1 i
+
= 1. The motion of the
5
2
5
2
particle takes place on an ellipse centered at (0> 0). As w goes from 3 to 5, the particle starts at the point (0> 32) and moves

21. { = 5 sin w, | = 2 cos w

i sin w =

clockwise around the ellipse 3 times.

22. | = cos2 w = 1 3 sin2 w = 1 3 {2 . The motion of the particle takes place on the parabola | = 1 3 {2 . As w goes from 32 to

3, the particle starts at the point (0> 1), moves to (1> 0), and goes back to (0> 1). As w goes from 3 to 0, the particle moves

to (31> 0) and goes back to (0> 1). The particle repeats this motion as w goes from 0 to 2.

23. We must have 1 $ { $ 4 and 2 $ | $ 3. So the graph of the curve must be contained in the rectangle [1> 4] by [2> 3].
24. (a) From the first graph, we have 1 $ { $ 2. From the second graph, we have 31 $ | $ 1= The only choice that satisfies

either of those conditions is III.
(b) From the first graph, the values of { cycle through the values from 32 to 2 four times. From the second graph, the values
of | cycle through the values from 32 to 2 six times. Choice I satisfies these conditions.

(c) From the first graph, the values of { cycle through the values from 32 to 2 three times. From the second graph, we have
0 $ | $ 2. Choice IV satisfies these conditions.

(d) From the first graph, the values of { cycle through the values from 32 to 2 two times. From the second graph, the values of
| do the same thing. Choice II satisfies these conditions.

25. When w = 31, ({> |) = (0> 31). As w increases to 0, { decreases to 31 and |

increases to 0. As w increases from 0 to 1, { increases to 0 and | increases to 1.
As w increases beyond 1, both { and | increase. For w ? 31, { is positive and
decreasing and | is negative and increasing. We could achieve greater accuracy
by estimating {- and |-values for selected values of w from the given graphs and
plotting the corresponding points.
26. For w ? 31, { is positive and decreasing, while | is negative and increasing (these

points are in Quadrant IV). When w = 31, ({> |) = (0> 0) and, as w increases from
31 to 0, { becomes negative and | increases from 0 to 1. At w = 0, ({> |) = (0> 1)
and, as w increases from 0 to 1, | decreases from 1 to 0 and { is positive. At
w = 1> ({> |) = (0> 0) again, so the loop is completed. For w A 1, { and | both
become large negative. This enables us to draw a rough sketch. We could achieve greater accuracy by estimating {- and
|-values for selected values of w from the given graphs and plotting the corresponding points.
27. When w = 0 we see that { = 0 and | = 0, so the curve starts at the origin. As w

increases from 0 to 12 , the graphs show that | increases from 0 to 1 while {
increases from 0 to 1, decreases to 0 and to 31, then increases back to 0, so we
arrive at the point (0> 1). Similarly, as w increases from

1
2

to 1, | decreases from 1

to 0 while { repeats its pattern, and we arrive back at the origin. We could achieve greater accuracy by estimating {- and
|-values for selected values of w from the given graphs and plotting the corresponding points.
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NOT FOR SALE
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

28. (a) { = w4 3 w + 1 = (w4 + 1) 3 w A 0 [think of the graphs of | = w4 + 1 and | = w] and | = w2 D 0, so these equations

are matched with graph V.
(b) | =

I
w D 0.

{ = w2 3 2w = w(w 3 2) is negative for 0 ? w ? 2, so these equations are matched with graph I.

(c) { = sin 2w has period 2@2 = . Note that
|(w + 2) = sin[w + 2 + sin 2(w + 2)] = sin(w + 2 + sin 2w) = sin(w + sin 2w) = |(w), so | has period 2.
These equations match graph II since { cycles through the values 31 to 1 twice as | cycles through those values once.
(d) { = cos 5w has period 2@5 and | = sin 2w has period , so { will take on the values 31 to 1, and then 1 to 31, before |
takes on the values 31 to 1. Note that when w = 0, ({> |) = (1> 0). These equations are matched with graph VI=
(e) { = w + sin 4w, | = w2 + cos 3w. As w becomes large, w and w2 become the dominant terms in the expressions for { and
|, so the graph will look like the graph of | = {2 , but with oscillations. These equations are matched with graph IV.
(f) { =

cos 2w
sin 2w
, |=
. As w < ", { and | both approach 0. These equations are matched with graph III.
4 + w2
4 + w2

29. Use | = w and { = w 3 2 sin w with a w-interval of [3> ].

30. Use {1 = w, |1 = w3 3 4w and {2 = w3 3 4w, |2 = w with a w-interval of

[33> 3]. There are 9 points of intersection; (0> 0) is fairly obvious. The point
in quadrant I is approximately (2=2> 2=2), and by symmetry, the point in
quadrant III is approximately (32=2> 32=2). The other six points are
approximately (~1=9> ±0=5), (~1=7> ±1=7), and (~0=5> ±1=9).

31. (a) { = {1 + ({2 3 {1 )w, | = |1 + (|2 3 |1 )w, 0 $ w $ 1. Clearly the curve passes through S1 ({1 > |1 ) when w = 0 and

through S2 ({2 > |2 ) when w = 1. For 0 ? w ? 1, { is strictly between {1 and {2 and | is strictly between |1 and |2 . For
every value of w, { and | satisfy the relation | 3 |1 =

|2 3 |1
({ 3 {1 ), which is the equation of the line through
{2 3 {1

S1 ({1 > |1 ) and S2 ({2 > |2 ).
Finally, any point ({> |) on that line satisfies

{ 3 {1
| 3 |1
=
; if we call that common value w, then the given
|2 3 |1
{2 3 {1

parametric equations yield the point ({> |); and any ({> |) on the line between S1 ({1 > |1 ) and S2 ({2 > |2 ) yields a value of

w in [0> 1]. So the given parametric equations exactly specify the line segment from S1 ({1 > |1 ) to S2 ({2 > |2 ).
(b) { = 32 + [3 3 (32)]w = 32 + 5w and | = 7 + (31 3 7)w = 7 3 8w for 0 $ w $ 1.
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SECTION 10.1

CURVES DEFINED BY PARAMETRIC EQUATIONS

¤

7

32. For the side of the triangle from D to E, use ({1 > |1 ) = (1> 1) and ({2 > |2 ) = (4> 2).

Hence, the equations are
{ = {1 + ({2 3 {1 ) w = 1 + (4 3 1) w = 1 + 3w,
| = |1 + (|2 3 |1 ) w = 1 + (2 3 1) w = 1 + w.
Graphing { = 1 + 3w and | = 1 + w with 0 $ w $ 1 gives us the side of the
triangle from D to E. Similarly, for the side EF we use { = 4 3 3w and | = 2 + 3w, and for the side DF we use { = 1
and | = 1 + 4w.
33. The circle {2 + (| 3 1)2 = 4 has center (0> 1) and radius 2, so by Example 4 it can be represented by { = 2 cos w,

| = 1 + 2 sin w, 0 $ w $ 2. This representation gives us the circle with a counterclockwise orientation starting at (2> 1).
(a) To get a clockwise orientation, we could change the equations to { = 2 cos w, | = 1 3 2 sin w, 0 $ w $ 2.
(b) To get three times around in the counterclockwise direction, we use the original equations { = 2 cos w, | = 1 + 2 sin w with

the domain expanded to 0 $ w $ 6.
(c) To start at (0> 3) using the original equations, we must have {1 = 0; that is, 2 cos w = 0. Hence, w =
{ = 2 cos w, | = 1 + 2 sin w,


2

$w$


.
2

So we use

3
.
2

Alternatively, if we want w to start at 0, we could change the equations of the curve. For example, we could use
{ = 32 sin w, | = 1 + 2 cos w, 0 $ w $ .
34. (a) Let {2 @d2 = sin2 w and | 2 @e2 = cos2 w to obtain { = d sin w and

| = e cos w with 0 $ w $ 2 as possible parametric equations for the ellipse
{2 @d2 + | 2 @e2 = 1.
(b) The equations are { = 3 sin w and | = e cos w for e M {1> 2> 4> 8}.
(c) As e increases, the ellipse stretches vertically.

35. Big circle: It’s centered at (2> 2) with a radius of 2, so by Example 4, parametric equations are

{ = 2 + 2 cos w>

| = 2 + 2 sin w>

0 $ w $ 2

Small circles: They are centered at (1> 3) and (3> 3) with a radius of 0=1. By Example 4, parametric equations are

and

(left)

{ = 1 + 0=1 cos w>

| = 3 + 0=1 sin w>

0 $ w $ 2

(right)

{ = 3 + 0=1 cos w>

| = 3 + 0=1 sin w>

0 $ w $ 2

Semicircle: It’s the lower half of a circle centered at (2> 2) with radius 1. By Example 4, parametric equations are
{ = 2 + 1 cos w>

| = 2 + 1 sin w>

 $ w $ 2

To get all four graphs on the same screen with a typical graphing calculator, we need to change the last w-interval to[0> 2] in
order to match the others. We can do this by changing w to 0=5w. This change gives us the upper half. There are several ways to
get the lower half—one is to change the “+” to a “3” in the |-assignment, giving us
{ = 2 + 1 cos(0=5w)>

| = 2 3 1 sin(0=5w)>

0 $ w $ 2

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NOT FOR SALE
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

36. If you are using a calculator or computer that can overlay graphs (using multiple w-intervals), the following is appropriate.

Left side: { = 1 and | goes from 1=5 to 4, so use

{ = 1>

| = w>

1=5 $ w $ 4

| = w>

1=5 $ w $ 4

| = 1=5>

1 $ w $ 10

Right side: { = 10 and | goes from 1=5 to 4, so use
{ = 10>
Bottom: { goes from 1 to 10 and | = 1=5, so use
{ = w>
Handle: It starts at (10> 4) and ends at (13> 7), so use
{ = 10 + w>

| = 4 + w>

0$w$3

Left wheel: It’s centered at (3> 1), has a radius of 1, and appears to go about 30 above the horizontal, so use
{ = 3 + 1 cos w>

| = 1 + 1 sin w>

5
6

$w$

13
6

5
6

$w$

13
6

Right wheel: Similar to the left wheel with center (8> 1), so use
{ = 8 + 1 cos w>

| = 1 + 1 sin w>

If you are using a calculator or computer that cannot overlay graphs (using one w-interval), the following is appropriate.
We’ll start by picking the w-interval [0> 2=5] since it easily matches the w-values for the two sides. We now need to find
parametric equations for all graphs with 0 $ w $ 2=5.
Left side: { = 1 and | goes from 1=5 to 4, so use
{ = 1>

| = 1=5 + w>

0 $ w $ 2=5

| = 1=5 + w>

0 $ w $ 2=5

Right side: { = 10 and | goes from 1=5 to 4, so use
{ = 10>
Bottom: { goes from 1 to 10 and | = 1=5, so use
{ = 1 + 3=6w>

| = 1=5>

0 $ w $ 2=5

To get the x-assignment, think of creating a linear function such that when w = 0, { = 1 and when w = 2=5,
{ = 10. We can use the point-slope form of a line with (w1 > {1 ) = (0> 1) and (w2 > {2 ) = (2=5> 10).
{31 =

10 3 1
(w 3 0) i { = 1 + 3=6w.
2=5 3 0

Handle: It starts at (10> 4) and ends at (13> 7), so use
{ = 10 + 1=2w>

| = 4 + 1=2w>

(w1 > {1 ) = (0> 10) and (w2 > {2 ) = (2=5> 13) gives us { 3 10 =
(w1 > |1 ) = (0> 4) and (w2 > |2 ) = (2=5> 7) gives us | 3 4 =

0 $ w $ 2=5

13 3 10
(w 3 0) i { = 10 + 1=2w.
2=5 3 0

734
(w 3 0) i | = 4 + 1=2w.
2=5 3 0

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SECTION 10.1

CURVES DEFINED BY PARAMETRIC EQUATIONS

¤

Left wheel: It’s centered at (3> 1), has a radius of 1, and appears to go about 30 above the horizontal, so use
{ = 3 + 1 cos

 8
15

w+

5
6


>


| = 1 + 1 sin 8
w+
15





and (w2 > 2 ) = 52 > 13
gives us  3
(w1 > 1 ) = 0> 5
6
6

5
6

=

13
6
5

2

5
6


>

0 $ w $ 2=5

3 5
6
(w 3 0) i  =
30

5
6

+

8
w.
15

Right wheel: Similar to the left wheel with center (8> 1), so use
{ = 8 + 1 cos
37. (a) { = w3

 8

15 w

+

i w = {1@3 , so | = w2 = {2@3 .

5
6


>


| = 1 + 1 sin 8
15 w +

5
6

(b) { = w6


>

0 $ w $ 2=5

i w = {1@6 , so | = w4 = {4@6 = {2@3 .

We get the entire curve | = {2@3 traversed in a left to

Since { = w6 D 0, we only get the right half of the

right direction.

curve | = {2@3 .

(c) { = h33w = (h3w )3

[so h3w = {1@3 ],

| = h32w = (h3w )2 = ({1@3 )2 = {2@3 .
If w ? 0, then { and | are both larger than 1. If w A 0, then { and |
are between 0 and 1. Since { A 0 and | A 0, the curve never quite
reaches the origin.
38. (a) { = w, so | = w32 = {32 . We get the entire curve | = 1@{2 traversed in a

left-to-right direction.

(b) { = cos w, | = sec2 w =

1
1
= 2 . Since sec w D 1, we only get the
cos2 w
{

parts of the curve | = 1@{2 with | D 1. We get the first quadrant portion of
the curve when { A 0, that is, cos w A 0, and we get the second quadrant
portion of the curve when { ? 0, that is, cos w ? 0.
(c) { = hw , | = h32w = (hw )32 = {32 . Since hw and h32w are both positive, we

only get the first quadrant portion of the curve | = 1@{2 .

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10

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NOT FOR SALE
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

39. The case


2

?  ?  is illustrated. F has coordinates (u> u) as in Example 7,

and T has coordinates (u> u + u cos( 3 )) = (u> u(1 3 cos ))

[since cos( 3 ) = cos  cos  + sin  sin  = 3 cos ], so S has

coordinates (u 3 u sin( 3 )> u(1 3 cos )) = (u( 3 sin )> u(1 3 cos ))

[since sin( 3 ) = sin  cos  3 cos  sin  = sin ]. Again we have the

parametric equations { = u( 3 sin ), | = u(1 3 cos ).
40. The first two diagrams depict the case  ?  ?

3
,
2

g ? u. As in Example 7, F has coordinates (u> u). Now T (in the second

diagram) has coordinates (u> u + g cos( 3 )) = (u> u 3 g cos ), so a typical point S of the trochoid has coordinates
(u + g sin( 3 )> u 3 g cos ). That is, S has coordinates ({> |), where { = u 3 g sin  and | = u 3 g cos . When
g = u, these equations agree with those of the cycloid.

41. It is apparent that { = |RT| and | = |TS | = |VW |. From the diagram,

{ = |RT| = d cos  and | = |VW | = e sin . Thus, the parametric equations are
{ = d cos  and | = e sin . To eliminate  we rearrange: sin  = |@e i
sin2  = (|@e)2 and cos  = {@d i cos2  = ({@d)2 . Adding the two
equations: sin2  + cos2  = 1 = {2 @d2 + | 2 @e2 . Thus, we have an ellipse.

42. D has coordinates (d cos > d sin ). Since RD is perpendicular to DE, {RDE is a right triangle and E has coordinates

(d sec > 0). It follows that S has coordinates (d sec > e sin ). Thus, the parametric equations are { = d sec , | = e sin .
43. F = (2d cot > 2d), so the {-coordinate of S is { = 2d cot . Let E = (0> 2d).

Then _RDE is a right angle and _RED = , so |RD| = 2d sin  and

D = ((2d sin ) cos > (2d sin ) sin ). Thus, the |-coordinate of S
is | = 2d sin2 .
44. (a) Let  be the angle of inclination of segment RS . Then |RE| =

2d
.
cos 

(b)

Let F = (2d> 0). Then by use of right triangle RDF we see that |RD| = 2d cos .
Now
|RS | = |DE| = |RE| 3 |RD|


1
1 3 cos2 
sin2 
= 2d
3 cos  = 2d
= 2d
= 2d sin  tan 
cos 
cos 
cos 
So S has coordinates { = 2d sin  tan  · cos  = 2d sin2  and | = 2d sin  tan  · sin  = 2d sin2  tan .
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SECTION 10.1

45. (a)

CURVES DEFINED BY PARAMETRIC EQUATIONS

¤

11

There are 2 points of intersection:
(33> 0) and approximately (32=1> 1=4).

(b) A collision point occurs when {1 = {2 and |1 = |2 for the same w. So solve the equations:
3 sin w = 33 + cos w (1)
2 cos w = 1 + sin w

(2)

From (2), sin w = 2 cos w 3 1. Substituting into (1), we get 3(2 cos w 3 1) = 33 + cos w i 5 cos w = 0 (B) i
cos w = 0 i w =
occurs when w =

3
2 ,



2

or

3
.
2

We check that w =

3
2

satisfies (1) and (2) but w =


2

does not. So the only collision point

and this gives the point (33> 0). [We could check our work by graphing {1 and {2 together as

functions of w and, on another plot, |1 and |2 as functions of w. If we do so, we see that the only value of w for which both
pairs of graphs intersect is w =

3
.]
2

(c) The circle is centered at (3> 1) instead of (33> 1). There are still 2 intersection points: (3> 0) and (2=1> 1=4), but there are

no collision points, since (B) in part (b) becomes 5 cos w = 6 i cos w =

6
5

A 1.

46. (a) If  = 30 and y0 = 500 m@s, then the equations become { = (500 cos 30 )w = 250

I
3w and

| = (500 sin 30 )w 3 12 (9=8)w2 = 250w 3 4=9w2 . | = 0 when w = 0 (when the gun is fired) and again when
w=

250
4=9

I  

E 51 s. Then { = 250 3 250
E 22,092 m, so the bullet hits the ground about 22 km from the gun.
4=9

The formula for | is quadratic in w. To find the maximum |-value, we will complete the square:
k
2

 2 l 1252



+ 4=9 = 34=9 w 3 125
w = 34=9 w2 3 250
w + 125
+
| = 34=9 w2 3 250
4=9
4=9
4=9
4=9
with equality when w =

125
4=9

s, so the maximum height attained is

1252
4=9

1252
4=9

$

1252
4=9

E 3189 m.

As  (0 ?  ? 90 ) increases up to 45 , the projectile attains a

(b)

greater height and a greater range. As  increases past 45 , the
projectile attains a greater height, but its range decreases.

(c) { = (y0 cos )w i w =
| = (y0 sin )w 3 12 jw2

{
.
y0 cos 

i | = (y0 sin )

j
{
3
y0 cos 
2



{
y0 cos 

2

= (tan ){ 3




j
{2 ,
2y02 cos2 

which is the equation of a parabola (quadratic in {).

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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

47. { = w2 > | = w3 3 fw. We use a graphing device to produce the graphs for various values of f with 3 $ w $ . Note that all

the members of the family are symmetric about the {-axis. For f ? 0, the graph does not cross itself, but for f = 0 it has a
cusp at (0> 0) and for f A 0 the graph crosses itself at { = f, so the loop grows larger as f increases.

48. { = 2fw 3 4w3 > | = 3fw2 + 3w4 . We use a graphing device to produce the graphs for various values of f with 3 $ w $ .

Note that all the members of the family are symmetric about the |-axis. When f ? 0, the graph resembles that of a polynomial
of even degree, but when f = 0 there is a corner at the origin, and when f A 0, the graph crosses itself at the origin, and has
two cusps below the {-axis. The size of the “swallowtail” increases as f increases.

49. { = w + d cos w> | = w + d sin w> d A 0. From the first figure, we see that

curves roughly follow the line | = {, and they start having loops when d
is between 1=4 and 1=6. The loops increase in size as d increases.

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 d for which there exist parameter values w and x such that w ? x and
(w + d cos w> w + d sin w) = (x + d cos x> x + d sin x).
In the diagram at the left, W denotes the point (w> w), X the point (x> x),
and S the point (w + d cos w> w + d sin w) = (x + d cos x> x + d sin x).
Since S W = S X = d, the triangle S W X is isosceles. Therefore its base
angles,  = _S W X and  = _S X W are equal. Since  = w 3
 = 2 3
x+w=

3
4

3
2

3x=

5
4


4

and

3 x, the relation  =  implies that

(1).

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SECTION 10.1

CURVES DEFINED BY PARAMETRIC EQUATIONS

¤

13

s
I
Since W X = distance((w> w)> (x> x)) = 2(x 3 w)2 = 2 (x 3 w), we see that
I
1

I
WX
(x 3 w)@ 2
=
cos  = 2
, so x 3 w = 2 d cos , that is,
d
SW
I










x 3 w = 2 d cos w 3 4 (2). Now cos w 3 4 = sin 2 3 w 3 4 = sin 3
4 3w ,
I

 0
0
so we can rewrite (2) as x 3 w = 2 d sin 3
4 3 w (2 ). Subtracting (2 ) from (1) and
dividing by 2, we obtain w =

3

4

3

I



2
d sin 3
2
4


3 w , or

3
4

3w=

d
I
2



sin 3
3 w (3).
4



Since d A 0 and w ? x, it follows from (20 ) that sin 3
4 3 w A 0. Thus from (3) we see that w ?

3
4 .

[We have

implicitly assumed that 0 ? w ?  by the way we drew our diagram, but we lost no generality by doing so since replacing w

by w + 2 merely increases { and | by 2. The curve’s basic shape repeats every time we change w by 2.] Solving for d in
I  3

I
I
2 4 3w
2}
 3
 . Write } = 3
3
w.
Then
d
=
, where } A 0. Now sin } ? } for } A 0, so d A 2.
(3), we get d =
4

sin }
sin 4 3 w
l
k
I
 3
,d< 2 .
As } < 0+ , that is, as w < 3
4
50. Consider the curves { = sin w + sin qw, | = cos w + cos qw, where q is a positive integer. For q = 1, we get a circle of

radius 2 centered at the origin. For q A 1, we get a curve lying on or inside that circle that traces out q 3 1 loops as w
ranges from 0 to 2.
{2 + | 2 = (sin w + sin qw)2 + (cos w + cos qw)2

Note:

= sin2 w + 2 sin w sin qw + sin2 qw + cos2 w + 2 cos w cos qw + cos2 qw
= (sin2 w + cos2 w) + (sin2 qw + cos2 qw) + 2(cos w cos qw + sin w sin qw)
= 1 + 1 + 2 cos(w 3 qw) = 2 + 2 cos((1 3 q)w) $ 4 = 22 ,
with equality for q = 1. This shows that each curve lies on or inside the curve for q = 1, which is a circle of radius 2 centered
at the origin.

q=1

q=2

q=3

q=5

51. Note that all the Lissajous figures are symmetric about the {-axis. The parameters d and e simply stretch the graph in the

{- and |-directions respectively. For d = e = q = 1 the graph is simply a circle with radius 1. For q = 2 the graph crosses

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NOT FOR SALE
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

itself at the origin and there are loops above and below the {-axis. In general, the figures have q 3 1 points of intersection,
all of which are on the |-axis, and a total of q closed loops.

d=e=1
52. { = cos w, | = sin w 3 sin fw.

q=2

q=3

If f = 1, then | = 0, and the curve is simply the line segment from (31> 0) to (1> 0). The

graphs are shown for f = 2> 3> 4 and 5.

It is easy to see that all the curves lie in the rectangle [31> 1] by [32> 2]. When f is an integer, {(w + 2) = {(w) and
|(w + 2) = |(w), so the curve is closed. When f is a positive integer greater than 1, the curve intersects the x-axis f + 1 times
and has f loops (one of which degenerates to a tangency at the origin when f is an odd integer of the form 4n + 1).
I


As f increases, the curve’s loops become thinner, but stay in the region bounded by the semicircles | = ± 1 + 1 3 {2

and the line segments from (31> 31) to (31> 1) and from (1> 31) to (1> 1). This is true because
I
||| = |sin w 3 sin fw| $ |sin w| + |sin fw| $ 1 3 {2 + 1. This curve appears to fill the entire region when f is very large, as
shown in the figure for f = 1000.

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