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INSTRUCTOR
SOLUTIONS
MANUAL
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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 dclegg@palomar.edu 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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149
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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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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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582
■
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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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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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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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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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¤
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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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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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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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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¤
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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LABORATORY PROJECT
RUNNING CIRCLES AROUND CIRCLES
¤
When f is a fraction, we get a variety of shapes with multiple loops, but always within the same region. For some fractional
values, such as f = 2=359, the curve again appears to fill the region.
LABORATORY PROJECT Running Circles Around Circles
1. The center T of the smaller circle has coordinates ((d 3 e)cos > (d 3 e)sin ).
Arc S V on circle F has length d since it is equal in length to arc DV
(the smaller circle rolls without slipping against the larger.)
d
d
and _S TW = 3 , so S has coordinates
e
e
d3e
{ = (d 3 e)cos + e cos(_S TW ) = (d 3 e)cos + e cos
e
Thus, _S TV =
and
d3e
.
| = (d 3 e)sin 3 e sin(_S TW ) = (d 3 e)sin 3 e sin
e
2. With e = 1 and d a positive integer greater than 2, we obtain a hypocycloid of d
cusps. Shown in the figure is the graph for d = 4. Let d = 4 and e = 1. Using the
sum identities to expand cos 3 and sin 3, we obtain
{ = 3 cos + cos 3 = 3 cos + 4 cos3 3 3 cos = 4 cos3 and
| = 3 sin 3 sin 3 = 3 sin 3 3 sin 3 4 sin3 = 4 sin3 .
3. The graphs at the right are obtained with e = 1 and
d = 12 , 13 , 14 , and
1
10
with 32 $ $ 2. We
conclude that as the denominator g increases, the graph
gets smaller, but maintains the basic shape shown.
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
Letting g = 2 and q = 3, 5, and 7 with 32 $ $ 2 gives us the following:
So if g is held constant and q varies, we get a graph with q cusps (assuming q@g is in lowest form). When q = g + 1, we
obtain a hypocycloid of q cusps. As q increases, we must expand the range of in order to get a closed curve. The following
graphs have d = 32 , 54 , and
11
.
10
4. If e = 1, the equations for the hypocycloid are
{ = (d 3 1) cos + cos ((d 3 1) )
| = (d 3 1) sin 3 sin ((d 3 1) )
which is a hypocycloid of d cusps (from Problem 2). In general, if d A 1, we get a figure with cusps on the “outside ring” and
if d ? 1, the cusps are on the “inside ring”. In any case, as the values of get larger, we get a figure that looks more and more
like a washer. If we were to graph the hypocycloid for all values of , every point on the washer would eventually be arbitrarily
close to a point on the curve.
d=
I
2,
d = h 3 2,
310 $ $ 10
0 $ $ 446
5. The center T of the smaller circle has coordinates ((d + e) cos > (d + e) sin ).
Arc S V has length d (as in Problem 1), so that _S TV =
and _S TW = 3
d
3 =3
e
d+e
e
d
d
, _S TU = 3
,
e
e
since _UTW = .
Thus, the coordinates of S are
and
d+e
d+e
= (d + e) cos 3 e cos
{ = (d + e) cos + e cos 3
e
e
d+e
d+e
| = (d + e) sin 3 e sin 3
= (d + e) sin 3 e sin
.
e
e
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LABORATORY PROJECT
RUNNING CIRCLES AROUND CIRCLES
6. Let e = 1 and the equations become
{ = (d + 1) cos 3 cos((d + 1))
| = (d + 1) sin 3 sin((d + 1))
If d = 1, we have a cardioid. If d is a positive
integer greater than 1, we get the graph of an
“d-leafed clover”, with cusps that are d units
from the origin. (Some of the pairs of figures are
not to scale.)
d = 3, 32 $ $ 2
d = 10, 32 $ $ 2
d = 14 , 34 $ $ 4
d = 17 , 37 $ $ 7
d = 25 , 35 $ $ 5
d = 75 , 35 $ $ 5
d = 43 , 33 $ $ 3
d = 76 , 36 $ $ 6
If d = q@g with q = 1, we obtain a figure that
does not increase in size and requires
3g $ $ g to be a closed curve traced
exactly once.
Next, we keep g constant and let q vary. As q
increases, so does the size of the figure. There is
an q-pointed star in the middle.
Now if q = g + 1 we obtain figures similar to the
previous ones, but the size of the figure does not
increase.
If d is irrational, we get washers that increase in
size as d increases.
d=
I
2, 0 $ $ 200
d = h 3 2, 0 $ $ 446
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17
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18
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.2 Calculus with Parametric Curves
1. { = w sin w, | = w2 + w
i
g|
g{
g|
g|@gw
2w + 1
= 2w + 1,
= w cos w + sin w, and
=
=
.
gw
gw
g{
g{@gw
w cos w + sin w
g|
32w + 1 g{
1
= w1@2 (3h3w ) + h3w 12 w31@2 = 12 w31@2 h3w (32w + 1) = 1@2 w ,
= 3 2 , and
gw
w
2w h gw
2
w
g|
g|@gw
32w + 1
(2w 3 1)w3@2
=
= 1@2 w 3
=
.
g{
g{@gw
1
2hw
2w h
2. { =
I
1
, | = w h3w
w
i
g|
g|@gw
33w2
g|
g{
= 33w2 ,
= 4 3 2w, and
=
=
. When w = 1,
gw
gw
g{
g{@gw
4 3 2w
3. { = 1 + 4w 3 w2 , | = 2 3 w3 ; w = 1.
({> |) = (4> 1) and g|@g{ = 3 32 , so an equation of the tangent to the curve at the point corresponding to w = 1 is
| 3 1 = 3 32 ({ 3 4), or | = 3 32 { + 7.
2 g{
g|@gw
w
2w3
g|
w2 + 1
g|
= 2w,
= 1 + w32 =
=
=
2w
= 2
.
,
and
2
2
gw
gw
w
g{
g{@gw
w +1
w +1
4. { = w 3 w31 , | = 1 + w2 ; w = 1.
When w = 1, ({> |) = (0> 2) and g|@g{ =
2
2
= 1, so an equation of the tangent to the curve at the point corresponding to
w = 1 is | 3 2 = 1({ 3 0), or | = { + 2.
5. { = w cos w, | = w sin w; w = .
g{
g|
g|@gw
w cos w + sin w
g|
= w cos w + sin w,
= w(3 sin w) + cos w, and
=
=
.
gw
gw
g{
g{@gw
3w sin w + cos w
When w = , ({> |) = (3> 0) and g|@g{ = 3@(31) = , so an equation of the tangent to the curve at the point
corresponding to w = is | 3 0 = [{ 3 (3)], or | = { + 2 .
g|
g{
= 3 cos2 (3 sin ),
= 3 sin2 cos , and
g
g
6. { = sin3 , | = cos3 , = @6.
I
I g|@g
g|
33 cos2 sin = 3 cot . When = @6, ({> |) = 18 > 38 3 and g|@g{ = 3 cot(@6) = 3 3,
=
=
g{
g{@g
3 sin2 cos I
I so an equation of the tangent line to the curve at the point corresponding to = @6 is | 3 38 3 = 3 3 { 3 18 ,
I
I
or | = 3 3 { + 12 3.
7. (a) { = 1 + ln w, | = w2 + 2; (1> 3).
g|
g{
1
g|
g|@gw
2w
= 2w>
= > and
=
=
= 2w2 . At (1> 3),
gw
gw
w
g{
g{@gw
1@w
{ = 1 + ln w = 1 i ln w = 0 i w = 1 and
g|
= 2, so an equation of the tangent is | 3 3 = 2({ 3 1),
g{
or | = 2{ + 1.
(b) { = 1 + ln w i ln w = { 3 1 i w = h{31 , so | = w2 + 2 = (h{31 )2 + 2 = h2{32 + 2, and | 0 = h2{32 · 2.
At (1> 3), | 0 = h2(1)32 · 2 = 2, so an equation of the tangent is | 3 3 = 2({ 3 1), or | = 2{ + 1.
8. (a) { = 1 +
{ =1+
I
2
w, | = hw ; (2> h).
2
2
2
1
g|@gw
2whw
g|
g{
g|
= hw · 2w,
= I , and
=
= I = 4w3@2 hw . At (2> h),
gw
gw
g{
g{@gw
2 w
1@ 2 w
I
I
g|
= 4h, so an equation of the tangent is | 3 h = 4h({ 3 2),
w=2 i
w = 1 i w = 1 and
g{
or | = 4h{ 3 7h.
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SECTION 10.2
(b) { = 1 +
CALCULUS WITH PARAMETRIC CURVES
¤
19
I
I
2
4
4
w i
w = { 3 1 i w = ({ 3 1)2 , so | = hw = h({31) , and |0 = h({31) · 4({ 3 1)3 .
At (2> h), | 0 = h · 4 = 4h, so an equation of the tangent is | 3 h = 4h({ 3 2), or | = 4h{ 3 7h.
9. { = 6 sin w, | = w2 + w; (0> 0).
g|@gw
2w + 1
g|
=
=
. The point (0> 0) corresponds to w = 0, so the
g{
g{@gw
6 cos w
slope of the tangent at that point is 16 . An equation of the tangent is therefore
| 3 0 = 16 ({ 3 0), or | = 16 {.
10. { = cos w + cos 2w, | = sin w + sin 2w; (31> 1).
g|@gw
cos w + 2 cos 2w
g|
=
=
. To find the value of w corresponding to
g{
g{@gw
3 sin w 3 2 sin 2w
the point (31> 1), solve { = 31 i cos w + cos 2w = 31 i
cos w + 2 cos2 w 3 1 = 31 i cos w (1 + 2 cos w) = 0 i cos w = 0 or
cos w = 3 12 . The interval [0> 2] gives the complete curve, so we need only find
the values of w in this interval. Thus, w =
we find that w =
2
2
or w =
2
3
or w =
4
.
3
Checking w =
3 2
, 2, 3,
2
corresponds to (31> 1). The slope of the tangent at (31> 1) with w =
of the tangent is therefore | 3 1 = 2({ + 1), or | = 2{ + 3.
11. { = w2 + 1, | = w2 + w
The curve is CU when
g|@gw
2w + 1
1
g|
=
=
=1+
g{
g{@gw
2w
2w
i
i
2
and
4
3
in the equation for |,
032
is
= 2. An equation
31 3 0
g g|
gw g{
g2 |
31@(2w2 )
1
=
= 3 3.
=
2
g{
g{@gw
2w
4w
g2 |
A 0, that is, when w ? 0.
g{2
12. { = w3 + 1, | = w2 3 w
i
g|@gw
2w 3 1
1
g|
2
=
=
3 2
=
g{
g{@gw
3w2
3w
3w
i
g g|
2 3 2w
2
2
3 2 + 3
2
gw g{
g |
3w
3w
3w3 = 2(1 3 w) . The curve is CU when g | A 0, that is, when 0 ? w ? 1.
=
=
=
g{2
g{@gw
3w2
3w2
9w5
g{2
2
13. { = hw , | = wh3w
i
g|
h3w (1 3 w)
g|@gw
3wh3w + h3w
=
= h32w (1 3 w) i
=
=
g{
g{@gw
hw
hw
g g|
gw g{
g2 |
h32w (31) + (1 3 w)(32h32w )
h32w (31 3 2 + 2w)
=
=
=
= h33w (2w 3 3). The curve is CU when
2
w
g{
g{@gw
h
hw
g2 |
A 0, that is, when w A 32 .
g{2
14. { = w2 + 1, | = hw 3 1
The curve is CU when
i
g|@gw
hw
g|
=
=
g{
g{@gw
2w
i
g g|
2whw 3 hw · 2
2
gw g{
(2w)2
g |
2hw (w 3 1)
hw (w 3 1)
=
=
=
=
.
2
3
g{
g{@gw
2w
(2w)
4w3
g2 |
A 0, that is, when w ? 0 or w A 1.
g{2
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
15. { = 2 sin w, | = 3 cos w, 0 ? w ? 2.
g g|
3 3 sec2 w
gw g{
g|
g|@gw
33 sin w
3
g |
3
=
=
= 3 tan w, so 2 =
= 2
= 3 sec3 w.
g{
g{@gw
2 cos w
2
g{
g{@gw
2 cos w
4
2
The curve is CU when sec3 w ? 0 i sec w ? 0 i cos w ? 0 i
2
?w?
3
.
2
16. { = cos 2w, | = cos w, 0 ? w ? .
g g|
1
sec w tan w
gw g{
g|@gw
3 sin w
sin w
1
1
g2 |
1
g|
=
=
=
=
= sec w, so 2 =
= 4
= 3 sec3 w=
g{
g{@gw
32 sin 2w
2 · 2 sin w cos w
4 cos w
4
g{
g{@gw
34 sin w cos w
16
The curve is CU when sec3 w ? 0 i sec w ? 0 i cos w ? 0 i
17. { = w3 3 3w, | = w2 3 3.
({> |) = (0> 33).
2
? w ? .
g|
g|
= 2w, so
=0 C w=0 C
gw
gw
g{
g{
= 3w2 3 3 = 3(w + 1)(w 3 1), so
=0 C
gw
gw
w = 31 or 1 C ({> |) = (2> 32) or (32> 32). The curve has a horizontal
tangent at (0> 33) and vertical tangents at (2> 32) and (32> 32).
18. { = w3 3 3w, | = w3 3 3w2 .
g|
g|
= 3w2 3 6w = 3w(w 3 2), so
=0 C
gw
gw
w = 0 or 2 C ({> |) = (0> 0) or (2> 34).
so
g{
= 3w2 3 3 = 3(w + 1)(w 3 1),
gw
g{
= 0 C w = 31 or 1 C ({> |) = (2> 34) or (32> 32). The curve
gw
has horizontal tangents at (0> 0) and (2> 34), and vertical tangents at (2> 34)
and (32> 32).
19. { = cos , | = cos 3. The whole curve is traced out for 0 $ $ .
g|
g|
= 33 sin 3, so
= 0 C sin 3 = 0 C 3 = 0, , 2, or 3
g
g
C ({> |) = (1> 1), 12 > 31 , 3 12 > 1 , or (31> 31).
= 0, 3 , 2
3 , or g{
g{
= 3 sin , so
= 0 C sin = 0 C = 0 or g
g
({> |) = (1> 1) or (31> 31). Both
C
C
g{
g|
and
equal 0 when = 0 and .
g
g
g|
33 sin 3 H
39 cos 3
= lim
= 9, which is the same slope when = .
= lim
<0 3 cos g{ <0 3 sin Thus, the curve has horizontal tangents at 12 > 31 and 3 12 > 1 , and there are no vertical tangents.
To find the slope when = 0, we find lim
<0
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SECTION 10.2
CALCULUS WITH PARAMETRIC CURVES
¤
21
20. { = hsin , | = hcos . The whole curve is traced out for 0 $ ? 2.
g|
g|
= 3 sin hcos , so
=
g
g
({> |) = (1> h) or (1> 1@h).
=
2
or
3
2
C sin = 0 C = 0 or C
g{
g{
= cos hsin , so
= 0 C cos = 0 C
g
g
C ({> |) = (h> 1) or (1@h> 1). The curve has horizontal tangents
at (1> h) and (1> 1@h), and vertical tangents at (h> 1) and (1@h> 1).
21. From the graph, it appears that the rightmost point on the curve { = w 3 w6 , | = hw
is about (0=6> 2). To find the exact coordinates, we find the value of w for which the
I
graph has a vertical tangent, that is, 0 = g{@gw = 1 3 6w5 C w = 1@ 5 6.
Hence, the rightmost point is
I
I
I 5
1@5
1@ 5 6 3 1@ 6 5 6 > h1@ 6 = 5 · 636@5 > h6
E (0=58> 2=01).
22. From the graph, it appears that the lowest point and the leftmost point on the curve
{ = w4 3 2w, | = w + w4 are (1=5> 30=5) and (31=2> 1=2), respectively. To find the
exact coordinates, we solve g|@gw = 0 (horizontal tangents) and g{@gw = 0
(vertical tangents).
g|
1
= 0 C 1 + 4w3 = 0 C w = 3 I
, so the lowest point is
3
gw
4
2
1
1
9
3
1
I
I
I
I
I
I
+
>
3
+
=
>
3
E (1=42> 30=47).
3
3
3
3
3
3
256
4
4
256
256
256
1
g{
= 0 C 4w3 3 2 = 0 C w = I
, so the leftmost point is
3
gw
2
3
1
2
1
1
3
I
I
I
I
I
I
=
3
E (31=19> 1=19).
3
>
+
>
3
3
3
3
16
2 32
16
16 3 16
23. We graph the curve { = w4 3 2w3 3 2w2 , | = w3 3 w in the viewing rectangle [32> 1=1] by [30=5> 0=5]. This rectangle
corresponds approximately to w M [31> 0=8].
We estimate that the curve has horizontal tangents at about (31> 30=4) and (30=17> 0=39) and vertical tangents at
about (0> 0) and (30=19> 0=37). We calculate
g|@gw
3w2 3 1
g|
=
= 3
. The horizontal tangents occur when
g{
g{@gw
4w 3 6w2 3 4w
g|@gw = 3w2 3 1 = 0 C w = ± I13 , so both horizontal tangents are shown in our graph. The vertical tangents occur when
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
g{@gw = 2w(2w2 3 3w 3 2) = 0 C 2w(2w + 1)(w 3 2) = 0 C w = 0, 3 12 or 2. It seems that we have missed one vertical
tangent, and indeed if we plot the curve on the w-interval [31=2> 2=2] we see that there is another vertical tangent at (38> 6).
24. We graph the curve { = w4 + 4w3 3 8w2 , | = 2w2 3 w in the viewing rectangle [33=7> 0=2] by [30=2> 1=4]. It appears that there
is a horizontal tangent at about (30=4> 30=1), and vertical tangents at about (33> 1) and (0> 0).
We calculate
g|
g|@gw
4w 3 1
=
= 3
, so there is a horizontal tangent where g|@gw = 4w 3 1 = 0 C w = 14 .
g{
g{@gw
4w + 12w2 3 16w
This point (the lowest point) is shown in the first graph. There are vertical tangents where g{@gw = 4w3 + 12w2 3 16w = 0 C
4w(w2 + 3w 3 4) = 0 C 4w(w + 4)(w 3 1) = 0. We have missed one vertical tangent corresponding to w = 34, and if we
plot the graph for w M [35> 3], we see that the curve has another vertical tangent line at approximately (3128> 36).
25. { = cos w, | = sin w cos w.
g{@gw = 3 sin w, g|@gw = 3 sin2 w + cos2 w = cos 2w.
({> |) = (0> 0) C cos w = 0 C w is an odd multiple of
g{@gw = 31 and g|@gw = 31, so g|@g{ = 1. When w =
2.
3
,
2
When w =
2,
g{@gw = 1 and
g|@gw = 31. So g|@g{ = 31. Thus, | = { and | = 3{ are both tangent to the
curve at (0> 0).
26.
From the graph, we discover that the graph of the curve { = cos w + 2 cos 2w,
| = sin w + 2 sin 2w crosses itself at the point (32> 0). To find w at (32> 0),
solve | = 0 C sin w + 2 sin 2w = 0 C sin w + 4 sin w cos w = 0 C
sin w (1 + 4 cos w) = 0 C sin w = 0 or cos w = 3 14 . We find that
w = ± arccos 3 14 corresponds to (32> 0).
g|
g|@gw
cos w + 4 cos 2w
cos w + 8 cos2 w 3 4
Now
=
=
=3
. When w = arccos 3 14 , cos w = 3 14 , sin w =
g{
g{@gw
3 sin w 3 4 sin 2w
sin w + 8 sin w cos w
I
I
3 1 + 12 3 4
3 15
g|
g|
I
i
= 3 I4
= 15.
= 3 I4 = 3 15. By symmetry, w = 3 arccos 3 14
15
15
15
g{
g{
3 4
4 3 2
I
I
I
I
I
The tangent lines are | 3 0 = ± 15 ({ + 2), or | = 15 { + 2 15 and | = 3 15 { 3 2 15.
and
27. { = u 3 g sin , | = u 3 g cos .
(a)
g|
g|
g sin g{
= u 3 g cos ,
= g sin , so
=
.
g
g
g{
u 3 g cos (b) If 0 ? g ? u, then |g cos | $ g ? u, so u 3 g cos D u 3 g A 0. This shows that g{@g never vanishes,
so the trochoid can have no vertical tangent if g ? u.
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,
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SECTION 10.2
CALCULUS WITH PARAMETRIC CURVES
¤
23
28. { = d cos3 , | = d sin3 .
(a)
g|
g|
g{
sin = 33d cos2 sin ,
= 3d sin2 cos , so
=3
= 3 tan .
g
g
g{
cos (b) The tangent is horizontal C g|@g{ = 0 C tan = 0 C = q
The tangent is vertical C cos = 0 C is an odd multiple of
(c) g|@g{ = ±1 C tan = ±1 C is an odd multiple of
4
[All sign choices are valid.]
29. { = 2w3 , | = 1 + 4w 3 w2
C ({> |) = (0> ±d) =
I
I
C ({> |) = ± 42 d> ± 42 d
g|@gw
4 3 2w
g|
g|
=
=
=1 C
. Now solve
g{
g{@gw
6w2
g{
i
6w2 + 2w 3 4 = 0 C 2(3w 3 2)(w + 1) = 0 C w =
2
3
the point is (32> 34).
30. { = 3w2 + 1, | = 2w3 + 1,
2
C ({> |) = (±d> 0).
4 3 2w
=1 C
6w2
29
, and if w = 31,
or w = 31. If w = 23 , the point is 16
27 > 9
g|
6w2
g{
g|
= 6w,
= 6w2 , so
=
= w [even where w = 0].
gw
gw
g{
6w
So at the point corresponding to parameter value w, an equation of the tangent line is | 3 (2w3 + 1) = w[{ 3 (3w2 + 1)].
If this line is to pass through (4> 3), we must have 3 3 (2w3 + 1) = w[4 3 (3w2 + 1)] C 2w3 3 2 = 3w3 3 3w C
w3 3 3w + 2 = 0 C (w 3 1)2 (w + 2) = 0 C w = 1 or 32. Hence, the desired equations are | 3 3 = { 3 4, or
| = { 3 1, tangent to the curve at (4> 3), and | 3 (315) = 32({ 3 13), or | = 32{ + 11, tangent to the curve at (13> 315).
31. By symmetry of the ellipse about the {- and |-axes,
D=4
Ud
0
| g{ = 4
= 2de 3
1
2
U0
@2
sin 2
e sin (3d sin ) g = 4de
@2
0
= 2de 2 = de
U @2
0
sin2 g = 4de
U @2
0
1
(1
2
3 cos 2) g
I
w intersects the |-axis when { = 0,
I
that is, when w = 0 and w = 2. The corresponding values of | are 0 and 2.
32. The curve { = w2 3 2w = w(w 3 2), | =
The shaded area is given by
] |=I2
]
({U 3 {O ) g| =
|=0
1
I gw
(w2 3 2w)
2 w
w=0
0
k
l2
U2
= 3 0 12 w3@2 3 w1@2 gw = 3 15 w5@2 3 23 w3@2
w=2
[0 3 {(w)] | 0 (w) gw = 3
]
2
0
= 3 15 · 25@2 3
2
3
I 8
=
= 3 2 3 15
· 23@2
8
15
I
2
= 321@2 45 3 43
33. The curve { = 1 + hw , | = w 3 w2 = w(1 3 w) intersects the {-axis when | = 0,
that is, when w = 0 and w = 1. The corresponding values of { are 2 and 1 + h.
The shaded area is given by
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°
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24
¤
]
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
{=1+h
(|W 3 |E ) g{ =
{=2
=
]
w=1
w=0
U1
0
=3
[|(w) 3 0] {0 (w) gw =
U1
whw gw 3
U1
0
0
w2 hw gw =
U1
U1
0
0
(w 3 w2 )hw gw
U1
1
whw gw 3 w2 hw 0 + 2 0 whw gw
1
whw gw 3 (h 3 0) = 3 (w 3 1)hw 0 3 h
[Formula 97 or parts]
[Formula 96 or parts]
= 3[0 3 (31)] 3 h = 3 3 h
34. By symmetry, D = 4
U
so
U @2
0
Ud
0
| g{ = 4
d sin3 (33d cos2 sin ) g = 12d2
@2
1
16 U
1
64
3
sin 4 3
1
48
sin3 2
35. { = u 3 g sin , | = u 3 g cos .
U 2u
U @2
0
sin4 cos2 g. Now
U
sin2 14 sin2 2 g = 18 (1 3 cos 2) sin2 2 g
U 1
1
1
(1 3 cos 4) 3 sin2 2 cos 2 g = 16
3 64
sin 4 3
= 18
2
sin4 cos2 g =
sin4 cos2 g =
U0
@2
0
=
32 .
Thus, D = 12d2
32
1
48
sin3 2 + F
= 38 d2 .
U 2
(u 3 g cos )(u 3 g cos ) g = 0 (u2 3 2gu cos + g2 cos2 ) g
2
= u2 3 2gu sin + 12 g2 + 12 sin 2 0 = 2u2 + g2
D=
0
| g{ =
U 2
0
36. (a) By symmetry, the area of R is twice the area inside R above the {-axis. The top half of the loop is described by
I
{ = w2 , | = w3 3 3w, 3 3 $ w $ 0, so, using the Substitution Rule with | = w3 3 3w and g{ = 2w gw, we find that
U 3I3 3
U 3I3 4
U3
3I3
(w 3 3w)2w gw = 2 0
(2w 3 6w2 ) gw = 2 25 w5 3 2w3 0
area = 2 0 | g{ = 2 0
l
k
I
I I = 2 25 (331@2 )5 3 2(331@2 )3 = 2 25 39 3 3 2 33 3 = 24
3
5
(b) Here we use the formula for disks and use the Substitution Rule as in part (a):
U 3I3 6
1 8
I
3
2
4
2
6
9 4 3 3
(w
3
3w)
2w
gw
=
2
(w
3
6w
+
9w
)w
gw
=
2
w
3
w
+
w
8
4
0
0
0
0
l
k
81
27
1@2 8
1@2 6
1@2 4
1
9
81
= 2 8 (33 ) 3 (33 ) + 4 (33 ) = 2 8 3 27 + 4 = 4 volume = U3
| 2 g{ = U 3I3
(c) By symmetry, the |-coordinate of the centroid is 0. To find the {-coordinate, we note that it is the same as the {-coordinate
I
I
I
of the centroid of the top half of R, the area of which is 12 · 24
3 = 12
3. So, using Formula 8.3.8 with D = 12
3,
5
5
5
we get
{=
=
5
I
12 3
5
I
6 3
k
U3
0
{| g{ =
1
(331@2 )7
7
5
I
12 3
0
w2 (w3 3 3w)2w gw =
3 35 (331@2 )5
So the coordinates of the centroid of R are ({> |) =
37. { = w + h3w , | = w 3 h3w , 0 $ w $ 2.
U 3I3
9
7
l
>0 .
5
I
6 3
27 I
5
= 6I
3 7 3+
3
1
7
3 5
7w 3 5w
I 27
3 =
5
9
7
3I3
0
g{@gw = 1 3 h3w and g|@gw = 1 + h3w , so
(g{@gw)2 + (g|@gw)2 = (1 3 h3w )2 + (1 + h3w )2 = 1 3 2h3w + h32w + 1 + 2h3w + h32w = 2 + 2h32w .
Ues
U2I
Thus, O = d (g{@gw)2 + (g|@gw)2 gw = 0 2 + 2h32w gw E 3=1416.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 10.2
38. { = w2 3 w, | = w4 , 1 $ w $ 4=
CALCULUS WITH PARAMETRIC CURVES
g{@gw = 2w 3 1 and g|@gw = 4w3 , so
(g{@gw)2 + (g|@gw)2 = (2w 3 1)2 + (4w3 )2 = 4w2 3 4w + 1 + 16w6 .
Ues
U4I
Thus, O = d (g{@gw)2 + (g|@gw)2 gw = 1 16w6 + 4w2 3 4w + 1 gw E 255=3756.
39. { = w 3 2 sin w, | = 1 3 2 cos w, 0 $ w $ 4.
g{@gw = 1 3 2 cos w and g|@gw = 2 sin w, so
(g{@gw)2 + (g|@gw)2 = (1 3 2 cos w)2 + (2 sin w)2 = 1 3 4 cos w + 4 cos2 w + 4 sin2 w = 5 3 4 cos w.
U 4 I
Ues
5 3 4 cos w gw E 26=7298.
Thus, O = d (g{@gw)2 + (g|@gw)2 gw = 0
I
I
w, | = w 3 w, 0 $ w $ 1.
1
1
g{
g|
= 1 + I and
= 1 3 I , so
gw
gw
2 w
2 w
2 2
2 2 g|
1
1
1
1
1
1
1
g{
+
= 1+ I
+ 13 I
=1+ I +
+13 I +
= 2+ .
gw
gw
4w
4w
2w
2 w
2 w
w
w
] 1u
] 1u
] es
1
1
2
2
gw = lim
gw E 2=0915.
(g{@gw) + (g|@gw) gw =
2+
2+
Thus, O =
+
2w
2w
w<0
d
0
w
40. { = w +
41. { = 1 + 3w2 , | = 4 + 2w3 , 0 $ w $ 1. g{@gw = 6w and g|@gw = 6w2 , so (g{@gw)2 + (g|@gw)2 = 36w2 + 36w4
Thus, O =
]
1
0
]
s
36w2 + 36w4 gw =
1
6w
0
]
s
1 + w2 gw = 6
1
l2
k
I
= 3 23 x3@2 = 2(23@2 3 1) = 2 2 2 3 1
2
I 1 x 2 gx [x = 1 + w2 , gx = 2w gw]
1
42. { = hw + h3w , | = 5 3 2w, 0 $ w $ 3.
g{@gw = hw 3 h3w and g|@gw = 32, so
(g{@gw)2 + (g|@gw)2 = h2w 3 2 + h32w + 4 = h2w + 2 + h32w = (hw + h3w )2 .
3
U3
Thus, O = 0 (hw + h3w ) gw = hw 3 h3w 0 = h3 3 h33 3 (1 3 1) = h3 3 h33 .
43. { = w sin w, | = w cos w, 0 $ w $ 1.
g{
gw
2
+
Thus, O =
g|
gw
2
g|
g{
= w cos w + sin w and
= 3w sin w + cos w, so
gw
gw
= w2 cos2 w + 2w sin w cos w + sin2 w + w2 sin2 w 3 2w sin w cos w + cos2 w
= w2 (cos2 w + sin2 w) + sin2 w + cos2 w = w2 + 1.
I
U1I
21 w2 + 1 gw = 12 w w2 + 1 +
0
1
2
I
I
1
ln w + w2 + 1 0 = 12 2 +
44. { = 3 cos w 3 cos 3w, | = 3 sin w 3 sin 3w, 0 $ w $ .
g{
gw
2
+
g|
gw
2
1
2
I ln 1 + 2 .
g|
g{
= 33 sin w + 3 sin 3w and
= 3 cos w 3 3 cos 3w, so
gw
gw
= 9 sin2 w 3 18 sin w sin 3w + 9 sin2 (3w) + 9 cos2 w 3 18 cos w cos 3w + 9 cos2 (3w)
= 9(cos2 w + sin2 w) 3 18(cos w cos 3w + sin w sin 3w) + 9[cos2 (3w) + sin2 (3w)]
= 9(1) 3 18 cos(w 3 3w) + 9(1) = 18 3 18 cos(32w) = 18(1 3 cos 2w)
= 18[1 3 (1 3 2 sin2 w)] = 36 sin2 w.
Thus, O =
U
U
U I
36 sin2 w gw = 6 0 |sin w| gw = 6 0 sin w gw = 36 cos w 0 = 36 (31 3 1) = 12.
0
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°
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¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
{ = hw cos w, | = hw sin w, 0 $ w $ .
g{ 2 g| 2
+ gw = [hw (cos w 3 sin w)]2 + [hw (sin w + cos w)]2
gw
45.
= (hw )2 (cos2 w 3 2 cos w sin w + sin2 w)
+ (hw )2 (sin2 w + 2 sin w cos w + cos2 w
= h2w (2 cos2 w + 2 sin2 w) = 2h2w
Thus, O =
46. { = cos w + ln(tan 12 w),
I I
UI
UI
2h2w gw = 0 2 hw gw = 2 hw 0 = 2 (h 3 1).
0
| = sin w, @4 $ w $ 3@4.
1
sec2 (w@2)
g{
1
1
g|
= 3 sin w + 2
= 3 sin w +
= 3 sin w +
and
= cos w, so
gw
tan(w@2)
2 sin(w@2) cos(w@2)
sin w
gw
2 2
g{
g|
1
+
= sin2 w 3 2 +
+ cos2 w = 1 3 2 + csc2 w = cot2 w. Thus,
gw
gw
sin2 w
U 3@4
U @2
|cot w| gw = 2 @4 cot w gw
k
l@2
1
= 2 ln |sin w|
= 2 ln 1 3 ln I
@4
2
I 1
= 2 0 + ln 2 = 2 2 ln 2 = ln 2.
O=
@4
47.
The figure shows the curve { = sin w + sin 1=5w, | = cos w for 0 $ w $ 4.
g{@gw = cos w + 1=5 cos 1=5w and g|@gw = 3 sin w, so
(g{@gw)2 + (g|@gw)2 = cos2 w + 3 cos w cos 1=5w + 2=25 cos2 1=5w + sin2 w.
U 4 I
1 + 3 cos w cos 1=5w + 2=25 cos2 1=5w gw E 16=7102.
Thus, O = 0
48. { = 3w 3 w3 , | = 3w2 .
g{ 2
gw
+
g| 2
gw
g{@gw = 3 3 3w2 and g|@gw = 6w, so
= (3 3 3w2 )2 + (6w)2 = (3 + 3w2 )2
and the length of the loop is given by
I
U I3
(3 + 3w2 ) gw = 2 0 (3 + 3w2 ) gw = 2[3w + w3 ]0 3
I I
I
= 2 3 3 + 3 3 = 12 3.
O=
U I3
I
3 3
49. { = w 3 hw , | = w + hw , 36 $ w $ 6.
g{ 2
gw
+
g| 2
gw
Set i (w) =
= (1 3 hw )2 + (1 + hw )2 = (1 3 2hw + h2w ) + (1 + 2hw + h2w ) = 2 + 2h2w , so O =
I
2 + 2h2w . Then by Simpson’s Rule with q = 6 and {w =
63(36)
6
= 2, we get
U6 I
2 + 2h2w gw.
36
O E 23 [i (36) + 4i (34) + 2i (32) + 4i (0) + 2i (2) + 4i (4) + i (6)] E 612=3053.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 10.2
CALCULUS WITH PARAMETRIC CURVES
¤
i g{@gw = 32d csc2 and | = 2d sin2 i g|@gw = 4d sin cos = 2d sin 2.
U @2 s
U @2 s
So O = @4 4d2 csc4 + 4d2 sin2 2 g = 2d @4 csc4 + sin2 2 g. Using Simpson’s Rule with
50. { = 2d cot s
= 16
, and i () = csc4 + sin2 2, we get
i 4 + 4i 5
+ 2i 3
+ 4i 7
+ i 2 E 2=2605d.
O E 2d · V4 = (2d) 16·3
16
8
16
@23@4
4
q = 4, { =
51. { = sin2 w, | = cos2 w, 0 $ w $ 3.
(g{@gw)2 + (g|@gw)2 = (2 sin w cos w)2 + (32 cos w sin w)2 = 8 sin2 w cos2 w = 2 sin2 2w i
Distance =
l@2
I U @2
I k
I
I
U 3 I
2 |sin 2w| gw = 6 2 0 sin 2w gw [by symmetry] = 33 2 cos 2w
= 33 2 (31 3 1) = 6 2.
0
0
The full curve is traversed as w goes from 0 to
,
2
because the curve is the segment of { + | = 1 that lies in the first quadrant
(since {, | D 0), and this segment is completely traversed as w goes from 0 to
52. { = cos2 w, | = cos w, 0 $ w $ 4.
g{ 2
gw
+
g| 2
gw
.
2
Thus, O =
U @2
0
sin 2w gw =
I
2, as above.
= (32 cos w sin w)2 + (3 sin w)2 = sin2 w (4 cos2 w + 1)
I
I
U
|sin w| 4 cos2 w + 1 gw = 4 0 sin w 4 cos2 w + 1 gw
U 31 I
U1 I
= 34 1
4x2 + 1 gx [x = cos w, gx = 3 sin w gw] = 4 31 4x2 + 1 gx
Distance =
U 4
0
=8
=4
U
Thus, O =
0
U tan1 2
U1I
4x2 + 1 gx = 8 0
sec ·
0
U tan1 2
0
gw
So O = 4
g| 2
gw
[2x = tan >
2 gx = sec2 g]
0
I
I
|sin w| 4 cos2 w + 1 gw = 5 +
+
sec2 g
k
ltan1 2
I
I
71
sec3 g = 2 sec tan + 2 ln |sec + tan |
= 4 5 + 2 ln 5 + 2
53. { = d sin , | = e cos , 0 $ $ 2.
g{ 2
1
2
1
2
I
ln 5 + 2 .
= (d cos )2 + (3e sin )2 = d2 cos2 + e2 sin2 = d2 (1 3 sin2 ) + e2 sin2 f2
2
2
2
2
2
2
2
2
2
= d 3 (d 3 e ) sin = d 3 f sin = d 1 3 2 sin = d2 (1 3 h2 sin2 )
d
U @2 t d2 1 3 h2 sin2 g
0
[by symmetry] = 4d
U @2 s
1 3 h2 sin2 g.
0
54. { = d cos3 , | = d sin3 .
g{ 2
gw
+
g| 2
gw
= (33d cos2 sin )2 + (3d sin2 cos )2
= 9d2 cos4 sin2 + 9d2 sin4 cos2 = 9d2 sin2 cos2 (cos2 + sin2 ) = 9d2 sin2 cos2 .
The graph has four-fold symmetry and the curve in the first quadrant corresponds
to 0 $ $ @2. Thus,
U @2
O = 4 0 3d sin cos g
[since d A 0 and sin and cos are positive for 0 $ $ @2]
@2
= 12d 12 sin2 0 = 12d 12 3 0 = 6d
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
°
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¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
55. (a) { = 11 cos w 3 4 cos(11w@2), | = 11 sin w 3 4 sin(11w@2).
Notice that 0 $ w $ 2 does not give the complete curve because
{(0) 6= {(2). In fact, we must take w M [0> 4] in order to obtain the
complete curve, since the first term in each of the parametric equations has
period 2 and the second has period
2
11@2
=
4
11 ,
and the least common
integer multiple of these two numbers is 4.
(b) We use the CAS to find the derivatives g{@gw and g|@gw, and then use Theorem 6 to find the arc length. Recent versions
U 4 s
I of Maple express the integral 0
(g{@gw)2 + (g|@gw)2 gw as 88H 2 2 l , where H({) is the elliptic integral
] 1I
I
1 3 {2 w2
I
gw and l is the imaginary number 31.
2
1
3
w
0
Some earlier versions of Maple (as well as Mathematica) cannot do the integral exactly, so we use the command
evalf(Int(sqrt(diff(x,t)ˆ2+diff(y,t)ˆ2),t=0..4*Pi)); to estimate the length, and find that the arc
length is approximately 294=03. Derive’s Para_arc_length function in the utility file Int_apps simplifies the
U 4 t
3 4 sin w sin 11w
+ 5 gw.
34 cos w cos 11w
integral to 11 0
2
2
56. (a) It appears that as w < ", ({> |) <
1
2
> 12 , and as w < 3", ({> |) < 3 12 > 3 12 .
(b) By the Fundamental Theorem of Calculus, g{@gw = cos 2 w2 and
g|@gw = sin 2 w2 , so by Formula 4, the length of the curve from the origin
to the point with parameter value w is
Uwt
U w t g{ 2 g| 2
+ gx gx = 0 cos2 2 x2 + sin2 2 x2 gx
O= 0
gx
=
Uw
0
1 gx = w
[or 3w if w ? 0]
We have used x as the dummy variable so as not to confuse it with the upper limit of integration.
57. { = w sin w, | = w cos w, 0 $ w $ @2.
2
V=
2
2
g{@gw = w cos w + sin w and g|@gw = 3w sin w + cos w, so
(g{@gw) + (g|@gw) = w cos w + 2w sin w cos w + sin2 w + w2 sin2 w 3 2w sin w cos w + cos2 w
U
2
= w2 (cos2 w + sin2 w) + sin2 w + cos2 w = w2 + 1
I
U @2
2| gv = 0 2w cos w w2 + 1 gw E 4=7394.
g{@gw = cos w and g|@gw = 2 cos 2w, so (g{@gw)2 + (g|@gw)2 = cos2 w + 4 cos2 2w.
I
2 sin 2w cos2 w + 4 cos2 2w gw E 8=0285.
58. { = sin w, | = sin 2w, 0 $ w $ @2.
V=
U
2| gv =
U @2
0
59. { = 1 + whw , | = (w2 + 1)hw , 0 $ w $ 1.
g{ 2
gw
V=
U
2| gv =
U1
0
+
g| 2
gw
= (whw + hw )2 + [(w2 + 1)hw + hw (2w)]2 = [hw (w + 1)]2 + [hw (w2 + 2w + 1)]2
= h2w (w + 1)2 + h2w (w + 1)4 = h2w (w + 1)2 [1 + (w + 1)2 ],
so
s
I
U1
2(w2 + 1)hw h2w (w + 1)2 (w2 + 2w + 2) gw = 0 2(w2 + 1)h2w (w + 1) w2 + 2w + 2 gw E 103=5999=
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 10.2
CALCULUS WITH PARAMETRIC CURVES
¤
29
60. { = w2 3 w3 , | = w + w4 , 0 $ w $ 1.
(g{@gw)2 + (g|@gw)2 = (2w 3 3w2 )2 + (1 + 4w3 )2 = 4w2 3 12w3 + 9w4 + 1 + 8w3 + 16w6 , so
I
U
U1
V = 2| gv = 0 2(w + w4 ) 16w6 + 9w4 3 4w3 + 4w2 + 1 gw E 12=7176.
g{ 2
61. { = w3 , | = w2 , 0 $ w $ 1.
V=
]
1
2|
0
= 2
]
13
4
k
t g{ 2
gw
2 5@2
5x
=
81
=
2
1215
gw
x34
9
+
U1
0
64.
0
g{ 2
+
g
]
gw =
3 83 x3@2
l13
=
4
1
2w2
0
I 1
x 18 gx
81
g{ 2
+
gw
U1
0
2.
k
l13
3x5@2 3 20x3@2
g
s
w2 (9w2 + 4) gw
=
2
9 · 18
]
13
4
(x3@2 3 4x1@2 ) gx
(w2 + w4 ) gw = 18
g{ 2
g| 2
+
2
1215
I
247 13 + 64
= (3 3 3w2 )2 + (6w)2 = 9(1 + 2w2 + w4 ) = [3(1 + w2 )]2 .
gw
2 · d sin3 · 3d sin cos g = 6d2
g| 2
w2
4
g| 2
g
1
0
x = 9w2 + 4, w2 = (x 4)@9,
1
gx = 18w gw, so w gw = 18
gx
2
15
·
]
s
9w4 + 4w2 gw = 2
I
I 3 · 132 13 3 20 · 13 13 3 (3 · 32 3 20 · 8) =
2 · 3w2 · 3(1 + w2 ) gw = 18
U @2
2
= 3w2 + (2w)2 = 9w4 + 4w2 .
gw
gw
63. { = d cos3 , | = d sin3 , 0 $ $
V=
g| 2
g| 2
62. { = 3w 3 w3 , | = 3w2 , 0 $ w $ 1.
V=
+
g
U @2
0
1
3
w3 + 15 w5
1
0
=
48
5
= (33d cos2 sin )2 + (3d sin2 cos )2 = 9d2 sin2 cos2 .
@2
sin4 cos g = 65 d2 sin5 0 = 65 d2
= (32 sin + 2 sin 2)2 + (2 cos 3 2 cos 2)2
= 4[(sin2 3 2 sin sin 2 + sin2 2) + (cos2 3 2 cos cos 2 + cos2 2)]
= 4[1 + 1 3 2(cos 2 cos + sin 2 sin )] = 8[1 3 cos(2 3 )] = 8(1 3 cos )
We plot the graph with parameter interval [0> 2], and see that we should only
integrate between 0 and . (If the interval [0> 2] were taken, the surface of
revolution would be generated twice.) Also note that
| = 2 sin 3 sin 2 = 2 sin (1 3 cos ). So
V=
U
0
=8
=8
2 · 2 sin (1 3 cos ) 2
I I
2 1 3 cos g
I U
I U2I
2 0 (1 3 cos )3@2 sin g = 8 2 0 x3 gx
I k 2 5@2 l2
2 5 x
=
0
16
5
I
2(25@2 ) =
65. { = 3w2 , | = 2w3 , 0 $ w $ 5
g{ 2
x = 1 cos >
gx = sin g
128
5 g| 2
= (6w)2 + (6w2 )2 = 36w2 (1 + w2 ) i
s
I
U5
U5 I
U5
V = 0 2{ (g{@gw)2 + (g|@gw)2 gw = 0 2(3w2 )6w 1 + w2 gw = 18 0 w2 1 + w2 2w gw
l26
k
U 26
U 26
I
x = 1 + w2 >
= 18 1 (x3@2 3 x1@2 ) gx = 18 25 x5@2 3 23 x3@2
= 18 1 (x 3 1) x gx
i
gw
+
gw
gx = 2w gw
= 18
2
5
I
· 676 26 3
2
3
I · 26 26 3 25 3 23 =
1
I
24
949 26 + 1
5
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
66. { = hw 3 w, | = 4hw@2 , 0 $ w $ 1.
g{ 2
gw
+
g| 2
gw
= (hw 3 1)2 + (2hw@2 )2 = h2w + 2hw + 1 = (hw + 1)2 .
s
U1
2(hw 3 w) (hw 3 1)2 + (2hw@2 )2 gw = 0 2(hw 3 w)(hw + 1)g
1
= 2 12 h2w + hw 3 (w 3 1)hw 3 12 w2 0 = (h2 + 2h 3 6)
V=
U1
0
67. If i 0 is continuous and i 0 (w) 6= 0 for d $ w $ e, then either i 0 (w) A 0 for all w in [d> e] or i 0 (w) ? 0 for all w in [d> e]. Thus, i
is monotonic (in fact, strictly increasing or strictly decreasing) on [d> e]. It follows that i has an inverse. Set I = j i 31 ,
that is, define I by I ({) = j(i 31 ({)). Then { = i(w) i i 31 ({) = w, so | = j(w) = j(i 31 ({)) = I ({).
68. By Formula 8.2.5 with | = I ({), V =
d
s
2I ({) 1 + [I 0 ({)]2 g{. But by Formula 10.2.1,
2
g|@gw
(g{@gw)2 + (g|@gw)2
=
. Using the Substitution Rule with { = {(w),
g{@gw
(g{@gw)2
g{
where d = {() and e = {(), we have since g{ =
gw
gw
1 + [I 0 ({)]2 = 1 +
V=
]
g|
g{
v
2 I ({(w))
=1+
(g{@gw)2 + (g|@gw)2 g{
gw =
(g{@gw)2
gw
]
2|
v
g{
gw
2
+
g|
gw
2
gw, which is Formula 10.2.6.
g|
g g|
g!
g
1
g|
g|@gw
|
=
tan31
=
. But
=
=
i
gw
gw
g{
1 + (g|@g{)2 gw g{
g{
g{@gw
{
g g|
g!
|¨{ 3 {
¨|
{¨
| 3 {
¨|
g |
|¨{ 3 {
¨|
1
i
. Using the Chain Rule, and the
= 2
=
=
=
gw g{
gw {
{ 2
gw
1 + (|@
{)
2
{ 2
{ + | 2
] wt
t g{ 2 g| 2
1@2
2 2
gv
g{ 2
fact that v =
+
gw
i
=
+ g|
= { + | 2
, we have that
gw
gw
gw
gw
gw
69. (a) ! = tan31
2
Ue
g|
g{
i
0
g!
g!@gw
=
=
gv
gv@gw
{¨
| 3 {
¨|
{ 2 + | 2
g! {¨
| 3 {
¨| 1
|{¨
| 3 {
¨||
{¨
| 3 {
¨|
=
= 2
. So = = 2
.
gv
({ 2 + | 2 )1@2
({ + | 2 )3@2
({ + | 2 )3@2 ({ 2 + | 2 )3@2
g|
g2 |
, |¨ =
.
g{
g{2
2
g |@g{2 1 · (g2 |@g{2 ) 3 0 · (g|@g{)
=
.
So =
[1 + (g|@g{)2 ]3@2
[1 + (g|@g{)2 ]3@2
(b) { = { and | = i ({) i { = 1, {
¨ = 0 and | =
70. (a) | = {2
=
(b) 0 =
i
g|
= 2{ i
g{
2
2
= I .
53@2
5 5
2
g |@g{2 g2 |
2
=
2.
So
=
=
, and at (1> 1),
g{2
[1 + (g|@g{)2 ]3@2
(1 + 4{2 )3@2
g
= 33(1 + 4{2 )35@2 (8{) = 0 C { = 0 i | = 0. This is a maximum since 0 A 0 for { ? 0 and
g{
0 ? 0 for { A 0. So the parabola | = {2 has maximum curvature at the origin.
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SECTION 10.2
CALCULUS WITH PARAMETRIC CURVES
¤
31
71. { = 3 sin i { = 1 3 cos i {
¨ = sin , and | = 1 3 cos i | = sin i |¨ = cos . Therefore,
cos 3 cos2 3 sin2 cos 3 (cos2 + sin2 )
|cos 3 1|
=
=
=
. The top of the arch is
(2 3 2 cos )3@2
[(1 3 cos )2 + sin2 ]3@2
(1 3 2 cos + cos2 + sin2 )3@2
characterized by a horizontal tangent, and from Example 2(b) in Section 10.2, the tangent is horizontal when = (2q 3 1),
so take q = 1 and substitute = into the expression for : =
|31 3 1|
1
|cos 3 1|
=
= .
4
(2 3 2 cos )3@2
[2 3 2(31)]3@2
72. (a) Every straight line has parametrizations of the form { = d + yw, | = e + zw, where d, e are arbitrary and y, z 6= 0.
For example, a straight line passing through distinct points (d> e) and (f> g) can be described as the parametrized curve
{ = d + (f 3 d)w, | = e + (g 3 e)w. Starting with { = d + yw, | = e + zw, we compute { = y, | = z, {
¨ = |¨ = 0,
and =
|y · 0 3 z · 0|
= 0.
(y 2 + z2 )3@2
(b) Parametric equations for a circle of radius u are { = u cos and | = u sin . We can take the center to be the origin.
So { = 3u sin i {
¨ = 3u cos and | = u cos i |¨ = 3u sin . Therefore,
2 2
2
u sin + u cos2 1
u2
1
= 2 2
= 3 = . And so for any (and thus any point), = .
2
2
3@2
u
u
u
(u sin + u cos )
73. The coordinates of W are (u cos > u sin ). Since W S was unwound from
arc W D, W S has length u. Also _S W T = _S W U 3 _TW U = 12 3 ,
so S has coordinates { = u cos + u cos 12 3 = u(cos + sin ),
| = u sin 3 u sin 12 3 = u(sin 3 cos ).
74. If the cow walks with the rope taut, it traces out the portion of the
involute in Exercise 73 corresponding to the range 0 $ $ , arriving at
the point (3u> u) when = . With the rope now fully extended, the
cow walks in a semicircle of radius u, arriving at (3u> 3u). Finally,
the cow traces out another portion of the involute, namely the reflection
about the {-axis of the initial involute path. (This corresponds to the
range 3 $ $ 0.) Referring to the figure, we see that the total grazing
area is 2(D1 + D3 ). D3 is one-quarter of the area of a circle of radius u, so D3 = 14 (u)2 = 14 3 u2 . We will compute
D1 + D2 and then subtract D2 = 12 u2 to obtain D1 .
u> u . [To see this, note that g{@g = 0 when
corresponds to 2 u> u .] The leftmost point of the involute is
To find D1 + D2 , first note that the rightmost point of the involute is
= 0 or
.
2
= 0 corresponds to the cusp at (u> 0) and =
2
2
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(3u> u). Thus, D1 + D2 =
U @2
=
| g{ 3
U @2
=0
| g{ =
U0
=
| g{.
Now | g{ = u(sin 3 cos ) u cos g = u2 ( sin cos 3 2 cos2 )g. Integrate:
U
(1@u2 ) | g{ = 3 cos2 3 12 2 3 1 sin cos 3 16 3 + 12 + F. This enables us to compute
0
3
3
2
2
2
2 1
1 3
1
D1 + D2 = u 3 cos 3 2 ( 3 1) sin cos 3 6 + 2 = u 0 3 3 3
+
=u
+
6
2
2
6
2
Therefore, D1 = (D1 + D2 ) 3 D2 = 16 3 u2 , so the grazing area is 2(D1 + D3 ) = 2 16 3 u2 + 14 3 u2 = 56 3 u2 .
LABORATORY PROJECT Bézier Curves
1. The parametric equations for a cubic Bézier curve are
{ = {0 (1 3 w)3 + 3{1 w(1 3 w)2 + 3{2 w2 (1 3 w) + {3 w3
| = |0 (1 3 w)3 + 3|1 w(1 3 w)2 + 3|2 w2 (1 3 w) + |3 w3
where 0 $ w $ 1. We are given the points S0 ({0 > |0 ) = (4> 1), S1 ({1 > |1 ) = (28> 48), S2 ({2 > |2 ) = (50> 42), and
S3 ({3 > |3 ) = (40> 5). The curve is then given by
{(w) = 4(1 3 w)3 + 3 · 28w(1 3 w)2 + 3 · 50w2 (1 3 w) + 40w3
|(w) = 1(1 3 w)3 + 3 · 48w(1 3 w)2 + 3 · 42w2 (1 3 w) + 5w3
where 0 $ w $ 1. The line segments are of the form { = {0 + ({1 3 {0 )w,
| = |0 + (|1 3 |0 )w:
S0 S1
{ = 4 + 24w,
| = 1 + 47w
S1 S2
{ = 28 + 22w,
| = 48 3 6w
S2 S3
{ = 50 3 10w,
| = 42 3 37w
2. It suffices to show that the slope of the tangent at S0 is the same as that of line segment S0 S1 , namely
|1 3 |0
.
{1 3 {0
We calculate the slope of the tangent to the Bézier curve:
33|0 (1 3 w)2 + 3|1 32w(1 3 w) + (1 3 w)2 + 3|2 3w2 + (2w)(1 3 w) + 3|3 w2
g|@gw
=
g{@gw
33{20 (1 3 w) + 3{1 [32w(1 3 w) + (1 3 w)2 ] + 3{2 [3w2 + (2w)(1 3 w)] + 3{3 w2
At point S0 , w = 0, so the slope of the tangent is
33|0 + 3|1
|1 3 |0
=
. So the tangent to the curve at S0 passes
33{0 + 3{1
{1 3 {0
through S1 . Similarly, the slope of the tangent at point S3 [where w = 1] is
of line S2 S3 .
33|2 + 3|3
|3 3 |2
=
, which is also the slope
33{2 + 3{3
{3 3 {2
3. It seems that if S1 were to the right of S2 , a loop would appear.
We try setting S1 = (110> 30), and the resulting curve does indeed have a loop.
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SECTION 10.3
POLAR COORDINATES
4. Based on the behavior of the Bézier curve in Problems 1–3, we suspect that the
four control points should be in an exaggerated C shape. We try S0 (10> 12),
S1 (4> 15), S2 (4> 5), and S3 (10> 8), and these produce a decent C. If you are using
a CAS, it may be necessary to instruct it to make the {- and |-scales the same so as
not to distort the figure (this is called a “constrained projection” in Maple.)
5. We use the same S0 and S1 as in Problem 4, and use part of our C as the top of
an S. To prevent the center line from slanting up too much, we move S2 up to
(4> 6) and S3 down and to the left, to (8> 7). In order to have a smooth joint
between the top and bottom halves of the S (and a symmetric S), we determine
points S4 , S5 , and S6 by rotating points S2 , S1 , and S0 about the center of the
letter (point S3 ). The points are therefore S4 (12> 8), S5 (12> 31), and S6 (6> 2).
10.3 Polar Coordinates
1. (a) 2>
3
. The direction
we obtain the point 2> 7
3
4
4
is 3 , so 32> 3 is a point that satisfies the u ? 0
By adding 2 to
opposite
3
,
3
requirement.
(b) 1> 3 3
4
u A 0: 1> 3 3
+ 2 = 1> 5
4
4
+ = 31> 4
u ? 0: 31> 3 3
4
(c) 31> 2
u A 0: 3(31)> 2 + = 1> 3
2
u ? 0: 31> 2 + 2 = 31> 5
2
2. (a) 1>
7
4
u A 0: 1> 7
3 2 = 1> 3 4
4
3 = 31> 3
u ? 0: 31> 7
4
4
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(b) 33> 6
u A 0: 3(33)> 6 + = 3> 7
6
u ? 0: 33> 6 + 2 = 33> 13
6
(c) (1> 31)
= 31 radian E 357=3
u A 0: (1> 31 + 2)
u ? 0: (31> 31 + )
3. (a)
{ = 1 cos = 1(31) = 31 and
| = 1 sin = 1(0) = 0 give us
the Cartesian coordinates (31> 0).
(b)
= 2 3 12 = 31 and
{ = 2 cos 3 2
3
I I
| = 2 sin 3 2
= 2 3 23 = 3 3
3
I give us 31> 3 3 .
(c)
I I
{ = 32 cos 3
= 32 3 22 = 2 and
4
I I
2
=3 2
| = 32 sin 3
4 = 32
2
gives us
4. (a)
I I
2> 3 2 .
I
I I { = 3 2 cos 5
= 3 2 3 22 = 1 and
4
I I I
= 3 2 3 22 = 1
| = 3 2 sin 5
4
gives us (1> 1).
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SECTION 10.3
POLAR COORDINATES
¤
35
{ = 1 cos 5
2 = 1(0) = 0 and
(b)
| = 1 sin 5
2 = 1(1) = 1
gives us (0> 1).
I I
=
2
3 23 = 3 3 and
{ = 2 cos 3 7
6
(c)
= 2 12 = 1
| = 2 sin 3 7
6
I give us 3 3> 1 .
s
I
22 + (32)2 = 2 2 and = tan31 32
= 3 4 . Since (2> 32) is in the fourth
2
I
I
and (ii) 32 2> 3
.
quadrant, the polar coordinates are (i) 2 2> 7
4
4
t
I I
I I 2
(b) { = 31 and | = 3 i u = (31)2 +
3 = 2 and = tan31 313 = 2
. Since 31> 3 is in the second
3
5. (a) { = 2 and | = 32
i u=
and (ii) 32>
quadrant, the polar coordinates are (i) 2> 2
3
6. (a) { = 3
5
3
.
t I I
I
2
31 I1
3
=
tan
=
3 3 + 32 = 27 + 9 = 6 and = tan31 3 I
3 and | = 3 i u =
3
3
I 3 3> 3 is in the first quadrant, the polar coordinates are (i) 6> 6 and (ii) 36>
.
6
Since
7
.
6
s
I
= 3 tan31 2. Since (1> 32) is in the fourth
12 + (32)2 = 5 and = tan31 32
1
I
I
5> 2 3 tan31 2 and (ii) 3 5> 3 tan31 2 .
quadrant, the polar coordinates are (i)
(b) { = 1 and | = 32 i u =
7. u D 1.
The curve u = 1 represents a circle with center
R and radius 1. So u D 1 represents the region on or
8. 0 $ u ? 2, $ $ 3@2.
This is the region inside the
circle u = 2 in the third quadrant.
outside the circle. Note that can take on any value.
9. u D 0, @4 $ $ 3@4.
10. 1 $ u $ 3, @6 ? ? 5@6
= n represents a line through R.
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
11. 2 ? u ? 3,
5
3
$$
12. u D 1, $ $ 2
7
3
13. Converting the polar coordinates (2> @3) and (4> 2@3) to Cartesian coordinates gives us 2 cos 3 > 2 sin
I > 4 sin 2
= 32> 2 3 . Now use the distance formula.
4 cos 2
3
3
g=
3
I = 1> 3 and
t
t
I 2 I
I
I
I
({2 3 {1 )2 + (|2 3 |1 )2 = (32 3 1)2 + 2 3 3 3 = 9 + 3 = 12 = 2 3
14. The points (u1 > 1 ) and (u2 > 2 ) in Cartesian coordinates are (u1 cos 1 > u1 sin 1 ) and (u2 cos 2 > u2 sin 2 ), respectively.
The square of the distance between them is
(u2 cos 2 3 u1 cos 1 )2 + (u2 sin 2 3 u1 sin 1 )2
= u22 cos2 2 3 2u1 u2 cos 1 cos 2 + u12 cos2 1 + u22 sin2 2 3 2u1 u2 sin 1 sin 2 + u12 sin2 1
= u12 sin2 1 + cos2 1 + u22 sin2 2 + cos2 2 3 2u1 u2 (cos 1 cos 2 + sin 1 sin 2 )
= u12 3 2u1 u2 cos(1 3 2 ) + u22 ,
so the distance between them is
15. u2 = 5
s
u12 3 2u1 u2 cos(1 3 2 ) + u22 .
C {2 + | 2 = 5, a circle of radius
I
5 centered at the origin.
u
= 4 C u cos = 4 C { = 4, a vertical line.
sec 16. u = 4 sec C
17. u = 2 cos i u2 = 2u cos C {2 + | 2 = 2{ C {2 3 2{ + 1 + | 2 = 1 C ({ 3 1)2 + | 2 = 1, a circle of
radius 1 centered at (1> 0). The first two equations are actually equivalent since u2 = 2u cos i
u(u 3 2 cos ) = 0 i
u = 0 or u = 2 cos . But u = 2 cos gives the point u = 0 (the pole) when = 0. Thus, the equation u = 2 cos is
equivalent to the compound condition (u = 0 or u = 2 cos ).
18. =
3
i tan = tan
19. u2 cos 2 = 1
3
i
I
I
|
= 3 C | = 3 {, a line through the origin.
{
C u2 (cos2 3 sin2 ) = 1 C (u cos )2 3 (u sin )2 = 1 C {2 3 | 2 = 1, a hyperbola centered at
the origin with foci on the {-axis.
20. u = tan sec =
sin cos2 i u cos2 = sin C (u cos )2 = u sin C {2 = |, a parabola with vertex at the
origin opening upward. The first implication is reversible since cos = 0 would imply sin = u cos2 = 0, contradicting the
fact that cos2 + sin2 = 1.
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SECTION 10.3
21. | = 2
C u sin = 2 C u =
22. | = {
i
C u sin = 1 + 3u cos C u sin 3 3u cos = 1 C u(sin 3 3 cos ) = 1 C
1
sin 3 3 cos 24. 4| 2 = {
u=
C u = 2 csc 5
|
= 1 [{ 6= 0] i tan = 1 i = tan31 1 i = or =
[either includes the pole]
{
4
4
23. | = 1 + 3{
u=
2
sin ¤
POLAR COORDINATES
C 4(u sin )2 = u cos cos 4 sin2 C u = 0 or u =
1
4
cot csc . u = 0 is included in u =
represented by the single equation u =
25. {2 + | 2 = 2f{
C 4u2 sin2 3 u cos = 0 C u(4u sin2 3 cos ) = 0 C u = 0 or
C u2 = 2fu cos 1
4
cot csc when =
2,
so the curve is
cot csc .
C u2 3 2fu cos = 0 C u(u 3 2f cos ) = 0 C u = 0 or u = 2f cos .
u = 0 is included in u = 2f cos when =
26. {| = 4
1
4
2
C (u cos )(u sin ) = 4 C u2
+ q, so the curve is represented by the single equation u = 2f cos =
1
2
· 2 sin cos = 4 C u2 sin 2 = 8 i u2 = 8 csc 2
27. (a) The description leads immediately to the polar equation =
slightly more difficult to derive.
,
6
and the Cartesian equation | = tan 6 { =
I1
3
{ is
(b) The easier description here is the Cartesian equation { = 3.
28. (a) Because its center is not at the origin, it is more easily described by its Cartesian equation, ({ 3 2)2 + (| 3 3)2 = 52 .
(b) This circle is more easily given in polar coordinates: u = 4. The Cartesian equation is also simple: {2 + | 2 = 16.
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
31. u = 2(1 + cos )
32. u = 1 + 2 cos 33. u = , D 0
34. u = ln , D 1
35. u = 4 sin 3
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SECTION 10.3
POLAR COORDINATES
36. u = cos 5
37. u = 2 cos 4
38. u = 3 cos 6
39. u = 1 3 2 sin 40. u = 2 + sin c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
41. u2 = 9 sin 2
42. u2 = cos 4
43. u = 2 + sin 3
44. u2 = 1
I
C u = ±1@ for A 0
45. u = 1 + 2 cos 2
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SECTION 10.3
POLAR COORDINATES
¤
46. u = 3 + 4 cos 47. For = 0, , and 2, u has its minimum value of about 0=5. For =
2
and
3
2 ,
u attains its maximum value of 2.
We see that the graph has a similar shape for 0 $ $ and $ $ 2.
48.
49. { = u cos = (4 + 2 sec ) cos = 4 cos + 2. Now, u < "
(4 + 2 sec ) < " i <
3
2
consider 0 $ ? 2], so lim { =
u<"
or <
3 +
2
lim { =
[since we need only
lim (4 cos + 2) = 2. Also,
<@2
u < 3" i (4 + 2 sec ) < 3" i <
u<3"
i
+
2
or <
3 3
2
, so
lim (4 cos + 2) = 2. Therefore, lim { = 2 i { = 2 is a vertical asymptote.
u<±"
<@2+
50. | = u sin = 2 sin 3 csc sin = 2 sin 3 1.
u < " i (2 3 csc ) < " i
csc < 3" i < + [since we need
only consider 0 $ ? 2] and so
lim | = lim 2 sin 3 1 = 31.
u<"
< +
Also u < 3" i (2 3 csc ) < 3" i csc < " i < 3 and so lim { = lim 2 sin 3 1 = 31.
u<3"
< Therefore lim | = 31 i | = 31 is a horizontal asymptote.
u<±"
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42
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
51. To show that { = 1 is an asymptote we must prove lim { = 1.
u<±"
{ = (u) cos = (sin tan ) cos = sin2 . Now, u < " i sin tan < " i
3
< 2 , so lim { = lim sin2 = 1. Also, u < 3" i sin tan < 3" i
u<"
<
+
2
<@2
, so lim { =
u<3"
lim
<@2+
sin2 = 1. Therefore, lim { = 1 i { = 1 is
u<±"
a vertical asymptote. Also notice that { = sin2 D 0 for all , and { = sin2 $ 1 for all . And { 6= 1, since the curve is not
defined at odd multiples of
.
2
Therefore, the curve lies entirely within the vertical strip 0 $ { ? 1.
52. The equation is ({2 + | 2 )3 = 4{2 | 2 , but using polar coordinates we know that
{2 + | 2 = u2 and { = u cos and | = u sin . Substituting into the given
equation: u6 = 4u2 cos2 u2 sin2 i u2 = 4 cos2 sin2 i
u = ±2 cos sin = ± sin 2. u = ± sin 2 is sketched at right.
53. (a) We see that the curve u = 1 + f sin crosses itself at the origin, where u = 0 (in fact the inner loop corresponds to
negative u-values,) so we solve the equation of the limaçon for u = 0 C f sin = 31 C sin = 31@f. Now if
|f| ? 1, then this equation has no solution and hence there is no inner loop. But if f ? 31, then on the interval (0> 2)
the equation has the two solutions = sin31 (31@f) and = 3 sin31 (31@f), and if f A 1, the solutions are
= + sin31 (1@f) and = 2 3 sin31 (1@f). In each case, u ? 0 for between the two solutions, indicating a loop.
(b) For 0 ? f ? 1, the dimple (if it exists) is characterized by the fact that | has a local maximum at =
determine for what f-values
g2 |
is negative at =
g2
| = u sin = sin + f sin2 At =
3
,
2
i
3
,
2
3
2 .
So we
since by the Second Derivative Test this indicates a maximum:
g|
= cos + 2f sin cos = cos + f sin 2
g
i
g2 |
= 3 sin + 2f cos 2.
g2
this is equal to 3(31) + 2f(31) = 1 3 2f, which is negative only for f A 12 . A similar argument shows that
for 31 ? f ? 0, | only has a local minimum at =
2
(indicating a dimple) for f ? 3 12 .
I
, 0 $ $ 16. u increases as increases and there are eight full revolutions. The graph must be either II or V.
I
I
When = 2, u = 2 E 2=5 and when = 16, u = 16 E 7, so the last revolution intersects the polar axis at
54. (a) u =
approximately 3 times the distance that the first revolution intersects the polar axis, which is depicted in graph V.
(b) u = 2 , 0 $ $ 16. See part (a). This is graph II.
(c) u = cos(@3). 0 $ 3 $ 2 i 0 $ $ 6, so this curve will repeat itself every 6 radians.
3
cos 3 = 0 i 3 = 2 + q i = 3
2 + 3q, so there will be two “pole” values, 2 and
9
2 .
This is graph VI.
(d) u = 1 + 2 cos is a limaçon [see Exercise 53(a)] with f = 2. This is graph III.
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SECTION 10.3
POLAR COORDINATES
¤
(e) Since 31 $ sin 3 $ 1, 1 $ 2 + sin 3 $ 3, so u = 2 + sin 3 is never 0; that is, the curve never intersects the pole.
This is graph I.
(f) u = 1 + 2 sin 3. Solving u = 0 will give us many “pole” values, so this is graph IV.
55. u = 2 sin i
{ = u cos = 2 sin cos = sin 2, | = u sin = 2 sin2 i
g|@g
2 · 2 sin cos sin 2
g|
=
=
=
= tan 2
g{
g{@g
cos 2 · 2
cos 2
I
g|
,
= tan 2 ·
= tan = 3. [Another method: Use Equation 3.]
6 g{
6
3
When =
56. u = 2 3 sin i { = u cos = (2 3 sin ) cos , | = u sin = (2 3 sin ) sin i
g|
g|@g
(2 3 sin ) cos + sin (3 cos )
2 cos 3 2 sin cos 2 cos 3 sin 2
=
=
=
=
2
2
g{
g{@g
(2 3 sin )(3 sin ) + cos (3 cos )
32
sin 3 cos 2
32 sin + sin 3 cos I
I
I
2(1@2) 3
3@2
23 3
g|
1 3 3@2 2
I .
=
When = ,
I
= I
· =
3 g{
32 3@2 3 (31@2)
3 3 + 1@2 2
132 3
57. u = 1@
i { = u cos = (cos )@, | = u sin = (sin )@
i
g|@g
sin (31@2 ) + (1@) cos 2
3 sin + cos g|
=
=
=
·
g{
g{@g
3 cos 3 sin cos (31@2 ) 3 (1@) sin 2
30 + (31)
3
g|
=
=
= 3.
g{
3(31) 3 (0)
1
When = ,
58. u = cos(@3)
i { = u cos = cos(@3) cos , | = u sin = cos(@3) sin i
cos(@3) cos + sin 3 13 sin(@3)
g|
g|@g
=
=
g{
g{@g
cos(@3) (3 sin ) + cos 3 13 sin(@3)
g|
=
When = ,
g{
59. u = cos 2
1
2
1
2
I
I
(31) + (0) 3 3@6
31@2
3
I
= I
= 3 I = 3 3.
(0) + (31) 3 3@6
3@6
3
i { = u cos = cos 2 cos , | = u sin = cos 2 sin i
g|@g
cos 2 cos + sin (32 sin 2)
g|
=
=
g{
g{@g
cos 2 (3 sin ) + cos (32 sin 2)
When =
I
I
I
0 2@2 +
2@2 (32)
g|
3 2
,
= I
I
= I = 1.
4 g{
0 3 2@2 +
2@2 (32)
3 2
60. u = 1 + 2 cos i { = u cos = (1 + 2 cos ) cos , | = u sin = (1 + 2 cos ) sin i
g|@g
(1 + 2 cos ) cos + sin (32 sin )
g|
=
=
g{
g{@g
(1 + 2 cos )(3 sin ) + cos (32 sin )
When =
I I
I
3@2 3 3
2 1 +
2
g|
3
31
233
I
I =
I =
,
= 2 I
.
1 I · =
3 g{
2
9
2 3 3@2 + 2 3 3
32 3 3 3
33 3
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¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
61. u = 3 cos g|
g
i { = u cos = 3 cos cos , | = u sin = 3 cos sin 2
2
2
3
2
i
4
= 33 sin + 3 cos = 3 cos 2 = 0 i 2 = or
C = or
k
l
So the tangent is horizontal at I32 > 4 and 3 I32 > 3
same as I32 > 3 4 .
4
g{
g
= 36 sin cos = 33 sin 2 = 0 i 2 = 0 or 62. u = 1 3 sin C = 0 or
.
2
3
.
4
So the tangent is vertical at (3> 0) and 0> 2 .
i { = u cos = cos (1 3 sin ), | = u sin = sin (1 3 sin ) i
g|
g
= sin (3 cos ) + (1 3 sin ) cos = cos (1 3 2 sin ) = 0 i cos = 0 or sin =
, and 2> 3
.
, or 3
i horizontal tangent at 12 > 6 , 12 > 5
= 6 , 2 , 5
6
2
6
2
g{
g
sin = 3 12 or 1 i =
7 11
, 6 ,
6
or
2
i
i vertical tangent at
Note that the tangent is vertical, not horizontal, when =
lim
i
= cos (3 cos ) + (1 3 sin )(3 sin ) = 3 cos2 3 sin + sin2 = 2 sin2 3 sin 3 1
= (2 sin + 1)(sin 3 1) = 0
<(@2)
1
2
,
2
since
3
2
3 11 > 2 > 6 , and 0> 2 .
> 7
6
g|@g
cos (1 3 2 sin )
g|@g
= lim
= " and lim
= 3".
g{@g
<(@2) (2 sin + 1)(sin 3 1)
<(@2)+ g{@g
63. u = 1 + cos i { = u cos = cos (1 + cos ), | = u sin = sin (1 + cos ) i
g|
g
= (1 + cos ) cos 3 sin2 = 2 cos2 + cos 3 1 = (2 cos 3 1)(cos + 1) = 0 i cos =
i horizontal tangent at 32 > 3 , (0> ), and 32 > 5
= 3 , , or 5
.
3
3
g{
g
= 3(1 + cos ) sin 3 cos sin = 3 sin (1 + 2 cos ) = 0 i sin = 0 or cos = 3 12
, or 4
i vertical tangent at (2> 0), 12 > 2
= 0, , 2
, and 12 > 4
.
3
3
3
3
Note that the tangent is horizontal, not vertical when = , since lim
<
64. u = h
i
{ = u cos = h cos , | = u sin = h sin 1
2
or 31 i
i
g|@g
= 0.
g{@g
i
g|
g
= h sin + h cos = h (sin + cos ) = 0 i sin = 3 cos i tan = 31 i
= 3 14 + q [q any integer] i horizontal tangents at h(q31@4) > q 3 14 .
= h cos 3 h sin = h (cos 3 sin ) = 0 i sin = cos i tan = 1 i
= 14 + q [q any integer] i vertical tangents at h(q+1@4) , q + 14 .
g{
g
i u2 = du sin + eu cos i {2 + | 2 = d| + e{ i
2
2 2 2
2 2
{2 3 e{ + 12 e + | 2 3 d| + 12 d = 12 e + 12 d
i
{ 3 12 e + | 3 12 d = 14 (d2 + e2 ), and this is a circle
I
with center 12 e> 12 d and radius 12 d2 + e2 .
65. u = d sin + e cos 66. These curves are circles which intersect at the origin and at
1
I
2
d> 4 . At the origin, the first circle has a horizontal
tangent and the second a vertical one, so the tangents are perpendicular here. For the first circle [u = d sin ],
g|@g = d cos sin + d sin cos = d sin 2 = d at =
4
and g{@g = d cos2 3 d sin2 = d cos 2 = 0
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SECTION 10.3
at =
4,
POLAR COORDINATES
¤
so the tangent here is vertical. Similarly, for the second circle [u = d cos ], g|@g = d cos 2 = 0 and
g{@g = 3d sin 2 = 3d at =
,
4
so the tangent is horizontal, and again the tangents are perpendicular.
67. u = 1 + 2 sin(@2). The parameter interval is [0> 4].
68. u =
s
1 3 0=8 sin2 . The parameter interval is [0> 2].
69. u = hsin 3 2 cos(4).
The parameter interval is [0> 2].
|cot |
70. u = |tan |
.
The parameter interval [0> ] produces the heart-shaped valentine curve shown in the first window.
The complete curve, including the reflected heart, is produced by the parameter interval [0> 2], but perhaps you’ll agree
that the first curve is more appropriate.
71. u = 1 + cos999 . The parameter interval is [0> 2].
72. u = sin2 (4) + cos(4). The parameter interval
is [0> 2].
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
73. It appears that the graph of u = 1 + sin 3
6
is the same shape as
the graph of u = 1 + sin , but rotated counterclockwise about the
origin by 6 . Similarly, the graph of u = 1 + sin 3 3 is rotated by
.
3
In general, the graph of u = i ( 3 ) is the same shape as that of
u = i (), but rotated counterclockwise through about the origin.
That is, for any point (u0 > 0 ) on the curve u = i (), the point
(u0 > 0 + ) is on the curve u = i ( 3 ), since u0 = i(0 ) = i ((0 + ) 3 ).
74.
From the graph, the highest points seem to have | E 0=77. To find the exact
value, we solve g|@g = 0. | = u sin = sin sin 2
i
g|@g = 2 sin cos 2 + cos sin 2
= 2 sin (2 cos2 3 1) + cos (2 sin cos )
= 2 sin (3 cos2 3 1)
In the first quadrant, this is 0 when cos =
| = 2 sin2 cos = 2 ·
2
3
·
I1
3
=
4
9
I1
3
I
3 E 0=77.
C sin =
t
2
3
C
75. Consider curves with polar equation u = 1 + f cos , where f is a real number. If f = 0, we get a circle of radius 1 centered at
the pole. For 0 ? f $ 0=5, the curve gets slightly larger, moves right, and flattens out a bit on the left side. For 0=5 ? f ? 1,
the left side has a dimple shape. For f = 1, the dimple becomes a cusp. For f A 1, there is an internal loop. For f D 0, the
rightmost point on the curve is (1 + f> 0). For f ? 0, the curves are reflections through the vertical axis of the curves
with f A 0.
f = 0=25
f = 0=75
f=1
f=2
76. Consider the polar curves u = 1 + cosq , where q is a positive integer. First, let
q be an even positive integer. The first figure shows that the curve has a peanut
shape for q = 2, but as q increases, the ends are squeezed. As q becomes large,
the curves look more and more like the unit circle, but with spikes to the points
(2> 0) and (2> ).
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SECTION 10.3
POLAR COORDINATES
The second figure shows u as a function of in Cartesian coordinates for the same
values of q. We can see that for large q, the graph is similar to the graph of | = 1,
but with spikes to | = 2 for { = 0, , and 2. (Note that when 0 ? cos ? 1,
cos1000 is very small.)
Next, let q be an odd positive integer. The third figure shows that the curve is a
cardioid for q = 1, but as q increases, the heart shape becomes more pronounced.
As q becomes large, the curves again look more like the unit circle, but with an
outward spike to (2> 0) and an inward spike to (0> ).
The fourth figure shows u as a function of in Cartesian coordinates for the same
values of q. We can see that for large q, the graph is similar to the graph of | = 1,
but spikes to | = 2 for { = 0 and , and to | = 0 for { = .
g|@g
g|
3 tan 3 tan g{@g
tan ! 3 tan = g{
77. tan # = tan(! 3 ) =
=
g|
g|@g
1 + tan ! tan 1+
tan tan 1+
g{
g{@g
gu
gu
g|
g{
sin2 sin
+
u
cos
3
tan
cos
3
u
sin
3
tan u
cos
+
u
·
g
g
g
cos =
= = g
g{
g|
gu
gu
gu
gu
sin2 +
tan cos 3 u sin + tan sin + u cos cos +
·
g
g
g
g
g
g cos =
u
u cos2 + u sin2 =
gu
gu
gu@g
2
cos2 +
sin g
g
78. (a) u = h
i gu@g = h , so by Exercise 77, tan # = u@h = 1 i
# = arctan 1 =
4.
(b) The Cartesian equation of the tangent line at (1> 0) is | = { 3 1, and that of
the tangent line at (0> h@2 ) is | = h@2 3 {.
(c) Let d be the tangent of the angle between the tangent and radial lines, that
is, d = tan #. Then, by Exercise 77, d =
u
gu@g
i
gu
1
= u
g
d
i
u = Fh@d (by Theorem 9.4.2).
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
LABORATORY PROJECT Families of Polar Curves
1. (a) u = sin q.
q=2
q=3
q=4
q=5
From the graphs, it seems that when q is even, the number of loops in the curve (called a rose) is 2q, and when q is odd,
the number of loops is simply q. This is because in the case of q odd, every point on the graph is traversed twice, due to
the fact that
u( + ) = sin[q( + )] = sin q cos q + cos q sin q =
+
sin q
3 sin q
if q is even
if q is odd
(b) The graph of u = |sin q| has 2q loops whether q is odd or even, since u( + ) = u().
q=2
q=3
q=4
q=5
2. u = 1 + f sin q. We vary q while keeping f constant at 2. As q changes, the curves change in the same way as those in
Exercise 1: the number of loops increases. Note that if q is even, the smaller loops are outside the larger ones; if q is odd, they
are inside.
f=2
q=2
q=3
q=4
q=5
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LABORATORY PROJECT FAMILIES OF POLAR CURVES
¤
49
Now we vary f while keeping q = 3. As f increases toward 0, the entire graph gets smaller (the graphs below are not to scale)
and the smaller loops shrink in relation to the large ones. At f = 31, the small loops disappear entirely, and for 31 ? f ? 1,
the graph is a simple, closed curve (at f = 0 it is a circle). As f continues to increase, the same changes are seen, but in reverse
order, since 1 + (3f) sin q = 1 + f sin q( + ), so the graph for f = f0 is the same as that for f = 3f0 , with a rotation
through . As f < ", the smaller loops get relatively closer in size to the large ones. Note that the distance between the
outermost points of corresponding inner and outer loops is always 2. Maple’s animate command (or Mathematica’s
Animate) is very useful for seeing the changes that occur as f varies.
q=3
3. u =
f = 34
f = 31=4
f = 31
f = 30=8
f = 30=2
f=0
f = 0=5
f=8
1 3 d cos . We start with d = 0, since in this case the curve is simply the circle u = 1.
1 + d cos As d increases, the graph moves to the left, and its right side becomes flattened. As d increases through about 0=4, the right
side seems to grow a dimple, which upon closer investigation (with narrower -ranges) seems to appear at d E 0=42 [the
actual value is
I
2 3 1]. As d < 1, this dimple becomes more pronounced, and the curve begins to stretch out horizontally,
until at d = 1 the denominator vanishes at = , and the dimple becomes an actual cusp. For d A 1 we must choose our
parameter interval carefully, since u < " as 1 + d cos < 0 C < ± cos31 (31@d). As d increases from 1, the curve
splits into two parts. The left part has a loop, which grows larger as d increases, and the right part grows broader vertically,
and its left tip develops a dimple when d E 2=42 [actually,
I
2 + 1]. As d increases, the dimple grows more and more
pronounced. If d ? 0, we get the same graph as we do for the corresponding positive d-value, but with a rotation through about the pole, as happened when f was replaced with 3f in Exercise 2.
[continued]
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
d=0
d = 0=3
d = 0=42,|| $ 0=5
d = 0=41> || $ 0=5
d = 0=9> || $ 0=5
d = 1> || $ 0=1
d = 2=41, | 3 | $ 0=2
d=2
d=4
d = 2=42, | 3 | $ 0=2
4. Most graphing devices cannot plot implicit polar equations, so we must first find an explicit expression (or expressions) for u
in terms of , d, and f. We note that the given equation, u4 3 2f2 u2 cos 2 + f4 3 d4 = 0, is a quadratic in u2 , so we use the
quadratic formula and find that
u2 =
2f2 cos 2 ±
s
s
4f4 cos2 2 3 4(f4 3 d4 )
= f2 cos 2 ± d4 3 f4 sin2 2
2
t
s
so u = ± f2 cos 2 ± d4 3 f4 sin2 2. So for each graph, we must plot four curves to be sure of plotting all the points
which satisfy the given equation. Note that all four functions have period .
We start with the case d = f = 1, and the resulting curve resembles the symbol for infinity. If we let d decrease, the curve
splits into two symmetric parts, and as d decreases further, the parts become smaller, further apart, and rounder. If instead we
let d increase from 1, the two lobes of the curve join together, and as d increases further they continue to merge, until at
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SECTION 10.4
AREAS AND LENGTHS IN POLAR COORDINATES
¤
51
d E 1=4, the graph no longer has dimples, and has an oval shape. As d < ", the oval becomes larger and rounder, since the
f2 and f4 terms lose their significance. Note that the shape of the graph seems to depend only on the ratio f@d, while the size
of the graph varies as f and d jointly increase.
(d> f) = (1> 1)
(d> f) = (0=99> 1)
(d> f) = (0=9> 1)
(d> f) = (0=6> 1)
(d> f) = (1=01> 1)
(d> f) = (4=04> 4)
(d> f) = (1=3> 1)
(d> f) = (1=5> 1)
(d> f) = (2> 1)
(d> f) = (4> 1)
10.4 Areas and Lengths in Polar Coordinates
1. u = h3@4 , @2 $ $ .
D=
]
@2
1 2
u
2
g =
]
@2
1
(h3@4 )2
2
g =
]
@2
2. u = cos , 0 $ $ @6.
D=
]
@6
0
=
1
4
6
1 2
u
2
+
1
2
g =
]
@6
1
2
0
·
1
2
I 3 =
24
2
cos g =
+
1
16
I
3
1
2
]
0
1 3@2
h
2
g =
1
2
k
l
32h3@2
@2
@6
1
(1
2
+ cos 2) g =
1
4
+
= 31(h3@2 3 h3@4 ) = h3@4 3 h3@2
1
2
sin 2
@6
0
3. u2 = 9 sin 2, u D 0, 0 $ $ @2.
D=
]
0
@2
1 2
u
2
g =
]
0
@2
1
(9 sin 2) g
2
=
9
2
1
@2
3 2 cos 2 0 = 3 94 (31 3 1) =
9
2
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
4. u = tan , @6 $ $ @3.
D=
]
@3
@6
1 2
2u
g =
]
@3
@6
]
tan2 g =
1
2
@3
@6
2
1
2 (sec
3 1) g =
I
I
I
I
= 12
3 3 3 3 13 3 3 6 = 12 23 3 3 6 = 13 3 3
5. u =
I
, 0 $ $ 2. D =
]
2
0
1 2
u
2
D=
1
(1
2
0
1
2
=
]
2
2
0
]
D=
1
2
=
@2
]
]
@2
1
2
0
2
1
g
2
0
2
(1 + 2 cos + cos ) g =
0
cos 2 g =
1 3
2
+ 3 sin )2 g =
1
2 ((4
@2
]
·2
0
D=
]
]
I 2
g =
2
1
4
+ 2 sin +
1
2
]
=
4
2
2
0
= 2
1 + 2 cos + 12 (1 + cos 2) g
0
sin 2
1
0
=
1
2
3
2
+ 0 + 0 3 12 (0) =
1
2
(16 + 9 sin2 ) g
]
@2
(16 + 24 sin + 9 sin2 ) g
3@2
[by Theorem 4.5.6(b) [ET 5.5.7(b)]]
3@2
1
2
=
1
2
12
@3
tan 3 @6
.
2
3@2
=
1
2
+ 2 cos +
7. u = 4 + 3 sin , 3 2 $ $
8. u = sin 2, 0 $ $
1
2
+ cos ) g =
3
2
0
6. u = 1 + cos , 0 $ $ .
]
]
g =
1
2
]
0
@2 @2 41
2
16 + 9 · 12 (1 3 cos 2) g
3
9
2
cos 2 g = 41
2 3
9
4
[by Theorem 4.5.6(a) [ET 5.5.7(a)]]
sin 2
@2
0
=
41
4
3 0 3 (0 3 0) =
41
4
.
2
sin2 2 g =
1
2
]
@2
0
1
2 (1
3 cos 4) g =
1
4
3
1
4
sin 4
@2
0
=
1
4
2
=
8
9. The area is bounded by u = 2 sin for = 0 to = .
D=
]
1 2
u
2
0
]
=2
g =
1
2 (1
0
1
2
]
(2 sin )2 g =
0
k
3 cos 2)g = 3
1
2
]
1
2
4 sin2 g
0
sin 2
l
=
0
Also, note that this is a circle with radius 1, so its area is (1)2 = .
10. D =
]
2
0
=
=
1
2
1
2
]
1 2
u
2
2
0
]
2
0
g =
]
2
0
1
(1
2
3 sin )2 g
(1 3 2 sin + sin2 ) g =
3
2
3 2 sin 3
= 12 [(3 + 2) 3 (2)] =
1
2
3
2
1
2
]
cos 2 g =
2
0
1
2
1 3 2 sin + 12 (1 3 cos 2) g
3
2
+ 2 cos 3
1
4
sin 2
2
0
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SECTION 10.4
11. D =
]
2
0
=
=
1
2
]
1 2
u
2
2
0
]
1
2
2
g =
]
2
+ 2 cos )2 g =
1
(3
2
0
(11 + 12 cos + 2 cos 2) g =
0
1
2
]
2
0
=
=
1
2
]
1 2
u
2
2
0
]
1
2
2
g =
2
]
2
0
=
=
1
2
]
41
1 2
u
2
2
0
]
1
2
2
+ 24 sin 3
g =
]
9
2
2
1
(2
2
0
2
(16 + 24 sin + 9 sin2 ) g
0
cos 2 g = 12 41
2 3 24 cos 3
9
+ 4 sin 4 3
1
2
9
4
+ sin 4)2 g =
1
2
]
2
]
2
0
=
=
1
2
]
1 2
2u
2
0
]
1
2
2
g =
]
2
1
2 (3
0
0
= 12 (22) = 11
15. D =
]
2
0
=
=
1
2
1
2
]
1 2
u
2
2
g =
]
2
1
2
0
1
16
sin 8
2
+
1
20
sin 10
2
0
1
2
]
0
2
2
0
(9 3 12 cos 4 + 4 cos2 4) g
11 3 3 sin 4 +
1
4
sin 8
2
0
2
s
1 + cos2 5 g
(1 + cos2 5) g =
0
3
1
2
9 3 12 cos 4 + 4 · 12 (1 + cos 8) g
(11 3 12 cos 4 + 2 cos 8) g =
0
(4 + 4 sin 4 + sin2 4) g
cos 8 g = 12 92 3 cos 4 3
3 2 cos 4)2 g =
2
0
= 12 [(9 3 1) 3 (31)] = 92 14. D =
sin 2
41
2
4 + 4 sin 4 + 12 (1 3 cos 8) g
2
0
]
1
2
16 + 24 sin + 9 · 12 (1 3 cos 2) g
= 12 [(41 3 24) 3 (324)] =
13. D =
(9 + 12 cos + 4 cos2 ) g
2
11 + 12 sin + sin 2 0
+ 3 sin )2 g =
1
(4
2
0
2
0
]
2
AREAS AND LENGTHS IN POLAR COORDINATES
0
9 + 12 cos + 4 · 12 (1 + cos 2) g
= 12 (22) = 11
12. D =
1
2
]
1
2
]
0
2
1 + 12 (1 + cos 10) g
= 12 (3) = 32 c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
16. D =
]
2
0
=
=
=
=
1
2
]
1 2
u
2
2
g =
]
2
1
(1
2
0
+ 5 sin 6)2 g
(1 + 10 sin 6 + 25 sin2 6) g
0
]
1
2
2
0
]
1
2
2
1 + 10 sin 6 + 25 · 12 (1 3 cos 12) g
27
2
0
27 3 53 3 3 53 =
1
2
cos 12 g = 12 27
3
2
25
2
+ 10 sin 6 3
6
cos 6 3
25
24
sin 12
27
2
17. The curve passes through the pole when u = 0
=
5
3
2
0
i 4 cos 3 = 0 i cos 3 = 0
i
3 =
2
+ q i
+ 3 q. The part of the shaded loop above the polar axis is traced out for
= 0 to = @6, so we’ll use 3@6 and @6 as our limits of integration.
] @6
] @6
2
1
1
D=
(4 cos 3) g = 2
(16 cos2 3) g
2
2
3@6
= 16
]
0
@6
1
(1
2
0
+ cos 6) g = 8 +
1
6
sin 6
@6
0
=8
18. For = 0 to = @2, the shaded loop is traced out by u =
6
= 43 I
sin 2 and the
I
unshaded loop is traced out by u = 3 sin 2.
D=
]
@2
0
1 2
2u
g =
]
@2
1
2
0
@2
= 3 14 cos 2 0 =
1
4
sin 2 g
3 3 14 =
1
2
19. u = 0
i sin 4 = 0 i 4 = q i = 4 q.
] @4
] @4
] @4
2
2
1
1
1
1
D=
(sin
4)
g
=
sin
4
g
=
(1 3 cos 8) g
2
2
2
2
0
0
= 3
1
4
1
8
sin 8
@4
0
=
1
4
0
4
=
1
16
20. u = 0
i 2 sin 5 = 0 i sin 5 = 0 i 5 = q i =
] @5
] @5
2
1
1
(2
sin
5)
g
=
4 sin2 5 g
D=
2
2
0
=2
5 q.
0
]
@5
0
1
(1
2
3 cos 10) g = 3
1
10
sin 10
@5
0
=
5
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SECTION 10.4
AREAS AND LENGTHS IN POLAR COORDINATES
21.
This is a limaçon, with inner loop traced
out between =
7
6
and
11
6
[found by
solving u = 0].
D= 2
]
3@2
1
(1
2
7@6
+ 2 sin )2 g =
]
3@2 1 + 4 sin + 4 sin2 g =
7@6
I
3@2 7
3 2 +2 33
= 3 4 cos + 2 3 sin 2 7@6 = 9
2
I
3
2
=
]
3@2 1 + 4 sin + 4 · 12 (1 3 cos 2) g
7@6
I
3 323
22. To determine when the strophoid u = 2 cos 3 sec passes through the pole, we solve
u = 0 i 2 cos 3
1
1
= 0 i 2 cos2 3 1 = 0 i cos2 =
cos 2
i
1
for 0 $ $ with 6= 2 .
i = 4 or = 3
cos = ± I
4
2
U @4
U @4
D = 2 0 21 (2 cos 3 sec )2 g = 0 (4 cos2 3 4 + sec2 ) g
U @4 1
U @4
4 · 2 (1 + cos 2) 3 4 + sec2 g = 0 (32 + 2 cos 2 + sec2 ) g
= 0
@4 = 32 + sin 2 + tan 0 = 3 2 + 1 + 1 3 0 = 2 3 2
23. 2 cos = 1
i cos =
D= 2
=
U @3
0
U @3
0
1
2
i =
1
[(2 cos )2
2
4
1
2
D=
=
3 12 ] g =
3
+
U @3
0
i sin = 0 i = 0 or U 2
(4 cos2 3 1) g
I
3
2
U 2 1 (1 3 sin )2 3 1 g =
2
1
4
5
.
3
or
U @3
(1 + cos 2) 3 1 g = 0 (1 + 2 cos 2) g
@3
= + sin 2 0 =
24. 1 3 sin = 1
3
1
2
U 2
(1 3 cos 2 3 4 sin ) g =
= 14 + 2
1
4
i
(sin2 3 2 sin ) g
3
1
2
sin 2 + 4 cos 2
25. To find the area inside the leminiscate u2 = 8 cos 2 and outside the circle u = 2,
we first note that the two curves intersect when u2 = 8 cos 2 and u = 2,
that is, when cos 2 = 12 . For 3 ? $ , cos 2 =
1
2
C 2 = ±@3
or ±5@3 C = ±@6 or ±5@6. The figure shows that the desired area is
4 times the area between the curves from 0 to @6. Thus,
U @6 1
U @6
D= 4 0
(8 cos 2) 3 12 (2)2 g = 8 0 (2 cos 2 3 1) g
2
k
l@6
I
I
= 8 3@2 3 @6 = 4 3 3 4@3
= 8 sin 2 3 0
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
26. To find the shaded area D, we’ll find the area D1 inside the curve u = 2 + sin 2
and subtract 32 since u = 3 sin is a circle with radius 32 .
U 2
U 2
D1 = 0 12 (2 + sin )2 g = 12 0 (4 + 4 sin + sin2 ) g
U 2 = 12 0 4 + 4 sin + 12 · (1 3 cos 2) g
U 2 9
= 12 0
+ 4 sin 3 12 cos 2 g
2
2
= 12 92 3 4 cos 3 14 sin 2 0 = 12 [(9 3 4) 3 (34)] =
So D = D1 3
9
4
=
9
2
27. 3 cos = 1 + cos 3
9
4
=
9
2
9
4 .
1
2
C cos =
i =
3
or 3 3 .
U @3
D = 2 0 12 [(3 cos )2 3 (1 + cos )2 ] g
U @3
U @3
= 0 (8 cos2 3 2 cos 3 1) g = 0 [4(1 + cos 2) 3 2 cos 3 1] g
@3
(3 + 4 cos 2 3 2 cos ) g = 3 + 2 sin 2 3 2 sin 0
I
I
=+ 33 3=
=
U @3
0
28. 3 sin = 2 3 sin D= 2
=
=
U @2
@6
U @2
@6
=4
1
[(3 sin )2
2
@6
U @2
i 4 sin = 2 i sin =
1
2
i =
6
or
5
.
6
3 (2 3 sin )2 ] g
(9 sin2 3 4 + 4 sin 3 sin2 ] g
(8 sin2 + 4 sin 3 4) g
U @2 1
2 · 2 (1 3 cos 2) + sin 3 1 g
@6
@2
U @2
= 4 @6 (sin 3 cos 2) g = 4 3cos 3 12 sin 2 @6
k
I
I I l
I
= 4 (0 3 0) 3 3 23 3 43
= 4 343 = 3 3
29.
I
3 cos = sin D=
=
=
U @3
0
U @3
0
1
4
I
I
sin i tan = 3 i =
3=
cos U @2 1 I
2
2
1
(sin ) g + @3 2
3 cos g
2
1
2
3
k
=
1
4
=
12
3
3
i
· 12 (1 3 cos 2) g +
U @2
1
@3 2
· 3 · 12 (1 + cos 2) g
@3
@2
sin 2 0 + 34 + 12 sin 2 @3
l
k
I 3 43 3 0 + 34 2 + 0 3 3 +
1
2
I
3
16
+
8
3
I
3 3
16
=
5
24
3
I
3
4
3.
I
3
4
l
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SECTION 10.4
30. D = 4
=2
=2
U @2
0
1
2 (1
3 cos )2 g = 2
U @2 3
3 2 cos +
2
0
31. sin 2 = cos 2
=
0
(1 3 2 cos + cos2 ) g
U @2 1 3 2 cos + 12 (1 + cos 2) g
0
= 3 3 4 sin +
8
U @2
AREAS AND LENGTHS IN POLAR COORDINATES
i
D= 8·2
1
2
0
=4 3
1
4
sin 2
U @2
cos 2 g = 0 (3 3 4 cos + cos 2) g
@2
0
=
3
2
34
sin 2
= 1 i tan 2 = 1 i 2 =
cos 2
i
U @8
1
2
1
2
sin 2 2 g = 8
sin 4
@8
0
U @8
0
= 4 8 3
1
4
1
(1
2
4
i
3 cos 4) g
·1 =
2
31
32. 3 + 2 cos = 3 + 2 sin D= 2
i cos = sin i = 4 or 5
4 .
U 5@4
1
(3 + 2 cos )2 g = @4 (9 + 12 cos + 4 cos2 ) g
2
U 5@4
@4
U 5@4 = @4 9 + 12 cos + 4 · 12 (1 + cos 2) g
=
=
U 5@4
@4
55
4
5@4
(11 + 12 cos + 2 cos 2) g = 11 + 12 sin + sin 2 @4
36
I
I
I
2 + 1 3 11
+ 6 2 + 1 = 11 3 12 2
4
33. sin 2 = cos 2
U @8
i tan 2 = 1 i 2 =
4
i =
8
[since u2 = sin 2]
@8
U @8
= 0 2 sin 2 g = 3 cos 2 0
I
I
= 3 12 2 3 (31) = 1 3 12 2
D= 4
0
1
2
sin 2 g
34. Let = tan31 (e@d). Then
D=
U
0
U @2
g + 12 (e cos )2 g
@2
sin 2 0 + 14 e2 + 12 sin 2 2
1
2 (d sin )
= 14 d2 3
1
2
= 14 (d2 3 e2 ) + 18 e2 3 14 (d2 + e2 )(sin cos )
= 14 (d2 3 e2 ) tan31 (e@d) + 18 e2 3 14 de
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
35. The darker shaded region (from = 0 to = 2@3) represents
From this area, we’ll subtract
1
2
1
2
of the desired area plus
1
2
of the area of the inner loop.
of the area of the inner loop (the lighter shaded region from = 2@3 to = ), and then
double that difference to obtain the desired area.
kU
2
U
2 l
2@3 1 1
D=2 0
+ cos g 3 2@3 12 12 + cos g
2 2
U 2@3 1
=
4
0
U 2@3 1
=
4
0
U + cos + cos2 g 3 2@3 14 + cos + cos2 g
+ cos + 12 (1 + cos 2) g
U 3 2@3 14 + cos + 12 (1 + cos 2) g
2@3 sin 2
sin 2
3
+ sin + +
+ sin + +
4
2
4
4
2
4
0
2@3
I
I I
I
= 6 + 23 + 3 3 83 3 4 + 2 + 6 + 23 + 3 3 83
=
=
4
36. u = 0
+
3
4
I I
3 = 14 + 3 3
i 1 + 2 cos 3 = 0 i cos 3 = 3 12
0 $ 3 $ 2] i =
1
2
= 2@9) represents
2 4
, 9.
9
i 3 =
2 4
, 3
3
[for
The darker shaded region (from = 0 to
of the desired area plus
loop. From this area, we’ll subtract
1
2
1
2
of the area of the inner
of the area of the inner loop (the lighter
shaded region from = 2@9 to = @3), and then double that difference to
obtain the desired area.
kU
l
U @3
2@9 1
D=2 0
(1 + 2 cos 3)2 g 3 2@9 12 (1 + 2 cos 3)2 g
2
u2 = (1 + 2 cos 3)2 = 1 + 4 cos 3 + 4 cos2 3 = 1 + 4 cos 3 + 4 · 12 (1 + cos 6)
Now
= 1 + 4 cos 3 + 2 + 2 cos 6 = 3 + 4 cos 3 + 2 cos 6
and
U
u2 g = 3 +
4
3
sin 3 +
1
3
sin 6 + F, so
2@9 @3
D = 3 + 43 sin 3 + 13 sin 6 0
3 3 + 43 sin 3 + 13 sin 6 2@9
k
l k
I
I I
4
3
1
3 3
2
= 2
+
·
+
·
+ 43 · 23 +
3
0
3
(
+
0
+
0)
3
3
3
2
3
2
3
=
4
3
+
4
3
I
I
3 3 13 3 3 =
3
+
I
3
1
3
·
I l
3 3
2
37. The pole is a point of intersection.
1 + sin = 3 sin =
6
or
i 1 = 2 sin i sin =
1
2
i
5
.
6
The other two points of intersection are
3
2> 6
and
3
5
2> 6
.
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SECTION 10.4
AREAS AND LENGTHS IN POLAR COORDINATES
38. The pole is a point of intersection.
1 3 cos = 1 + sin =
3
4
or
i 3 cos = sin i 31 = tan i
7
4 .
The other two points of intersection are 1 +
39. 2 sin 2 = 1
i sin 2 =
1
2
i 2 =
I
2 3
> 4
2
and 1 3
5 13
, 6, 6 ,
6
or
I
2 7
> 4
2
.
17
.
6
By symmetry, the eight points of intersection are given by
(1> ), where =
5 13
, , 12 ,
12 12
(31> ), where =
and
7 11 19
, 12 , 12 ,
12
17
,
12
and
and
23
.
12
[There are many ways to describe these points.]
40. Clearly the pole lies on both curves. sin 3 = cos 3
3 =
4
+ q [q any integer] i =
12
i tan 3 = 1 i
+ 3 q i
or 3
, so the three remaining intersection points are
4
1
5
I1 > , 3 I
I1 > 3 .
>
,
and
2 12
2 12
2 4
=
5
, ,
12 12
41. The pole is a point of intersection. sin = sin 2 = 2 sin cos sin (1 3 2 cos ) = 0 C sin = 0 or cos =
= 0, ,
and
I
,
3
3 2
> 3
2
or 3 3
1
2
i
i the other intersection points are
I
3 >3
2
C
[by symmetry].
42. Clearly the pole is a point of intersection. sin 2 = cos 2
tan 2 = 1 i 2 =
4
i
+ 2q [since sin 2 and cos 2 must be
positive in the equations] i = 8 + q i =
1
1
9
and I
.
So the curves also intersect at I
4 > 8
4 > 8
2
2
8
or
9
8 .
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
43.
From the first graph, we see that the pole is one point of intersection. By zooming in or using the cursor, we find the -values
of the intersection points to be E 0=88786 E 0=89 and 3 E 2=25. (The first of these values may be more easily
estimated by plotting | = 1 + sin { and | = 2{ in rectangular coordinates; see the second graph.) By symmetry, the total
area contained is twice the area contained in the first quadrant, that is,
] @2
] ]
] 2
2
2
1
1
(2)
g
+
2
(1
+
sin
)
g
=
4
g
+
D= 2
2
2
0
=
44.
3 0
3
4
0
+ 3 2 cos + 12 3
1
4
sin 2
@2
=
4 3
3
@2
+
2
+
4
1 + 2 sin + 12 (1 3 cos 2) g
3 3 2 cos + 12 3
1
4
sin 2 E 3=4645
We need to find the shaded area D in the figure. The horizontal line
representing the front of the stage has equation | = 4 C
u sin = 4 i u = 4@ sin . This line intersects the curve
u = 8 + 8 sin when 8 + 8 sin =
4
sin i
8 sin + 8 sin2 = 4 i 2 sin2 + 2 sin 3 1 = 0 i
32 ±
sin =
I
I
I
4+8
32 ± 2 3
31 + 3
=
=
4
4
2
[the other value is less than 31] i = sin31
I
331
.
2
This angle is about 21=5 and is denoted by in the figure.
U @2
U @2
U @2
U @2
D = 2 12 (8 + 8 sin )2 g 3 2 21 (4 csc )2 g = 64 (1 + 2 sin + sin2 ) g 3 16 csc2 g
= 64
U @2 1 + 2 sin +
1
2
3
1
2
U @2
cos 2 g + 16 (3 csc2 ) g = 64 32 3 2 cos 3
1
4
sin 2
@2
@d
= 16 6 3 8 cos 3 sin 2 + cot = 16[(3 3 0 3 0 + 0) 3 (6 3 8 cos 3 sin 2 + cot )]
@2
+ 16 cot = 48 3 96 + 128 cos + 16 sin 2 3 16 cot I
I
2
3 3 1 = 22 i {2 = 4 3 3 3 2 3 + 1 i
From the figure, {2 +
{2 = 2
s I
I
I
I
3 = 12, so { = 2 3 = 4 12. Using the trigonometric relationships
for a right triangle and the identity sin 2 = 2 sin cos , we continue:
I
I
I
I
I
4
4
12
3 3 1 4 12
12
3+1
+ 16 · 2 ·
·
3 16 · I
D = 48 3 96 + 128 ·
·I
2
2
2
331
3+1
I
I
I
I
I
I
I
331
= 48 3 96 + 64 4 12 + 8 4 12 3 3 1 3 8 4 12 3 + 1 = 48 + 48 4 12 3 96 sin31
2
E 204=16 m2
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SECTION 10.4
45. O =
]
e
d
=
]
0
s
u2 + (gu@g)2 g =
]
0
AREAS AND LENGTHS IN POLAR COORDINATES
¤
s
(2 cos )2 + (32 sin )2 g
]
t
4(cos2 + sin2 ) g =
0
I
4 g = 2 0 = 2
As a check, note that the curve is a circle of radius 1, so its circumference is 2(1) = 2.
46. O =
]
e
d
]
s
u2 + (gu@g)2 g =
]
s
= 1 + (ln 5)2
0
2
0
s
= 1 + (ln 5)2
47. O =
]
e
d
=
]
2
]
I
s
52 g = 1 + (ln 5)2
2
5
1
3
ln 5
ln 5
0
2
0
4 2 +4
1
2
4
48. O =
]
e
d
=
]
2
0
I
I
8 + 8 cos g = 8
2
0
I
= 8
]
2
0
u
g =
I
x gx =
]
s
u2 + (gu@g)2 g =
]
2
0
s
2 + 4 g
2
0
]
s
2 + 4 g =
2
s
52 [1 + (ln 5)2 ] g
2
s
5
5 g = 1 + (ln 5)2
ln 5 0
]
t
(2 )2 + (2)2 g =
Now let x = 2 + 4, so that gx = 2 g
]
0
2
0
0
]
t
2 (2 + 4) g =
2
s
1 + (ln 5)2 2
=
(5 3 1)
ln 5
]
s
u2 + (gu@g)2 g =
2
]
s
(5 )2 + (5 ln 5)2 g =
1
2
·
2
3
1
2
s
4 + 42 g
gx and
k
l4(2 +1)
x3@2
= 13 [43@2 ( 2 + 1)3@2 3 43@2 ] = 83 [(2 + 1)3@2 3 1]
4
]
s
[2(1 + cos )]2 + (32 sin )2 g =
2
I
I
1 + cos g = 8
0
]
2
0
2
0
s
4 + 8 cos + 4 cos2 + 4 sin2 g
t
2 · 12 (1 + cos ) g
] I I ] 2 cos g = 4 · 2
2 cos2 g = 8 2
cos g
2
2
2
0
0
[by symmetry]
= 8(2) = 16
= 8 2 sin
2 0
49. The curve u = cos4 (@4) is completely traced with 0 $ $ 4.
2
u2 + (gu@g)2 = [cos4 (@4)]2 + 4 cos3 (@4) · (3 sin(@4)) · 14
= cos8 (@4) + cos6 (@4) sin2 (@4)
= cos6 (@4)[cos2 (@4) + sin2 (@4)] = cos6 (@4)
U 4 s
U 4 cos6 (@4) g = 0 cos3 (@4) g
0
U @2
U 2
= 2 0 cos3 (@4) g [since cos3 (@4) D 0 for 0 $ $ 2] = 8 0 cos3 x gx
U @2
U1
{ = sin x>
= 8 0 (1 3 sin2 x) cos x gx = 8 0 (1 3 {2 ) g{
O=
x = 14 g{ = cos x gx
1
= 8 { 3 13 {3 0 = 8 1 3 13 =
16
3
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62
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
50. The curve u = cos2 (@2) is completely traced with 0 $ $ 2.
2
u2 + (gu@g)2 = [cos2 (@2)]2 + 2 cos(@2) · (3 sin(@2)) · 12
= cos4 (@2) + cos2 (@2) sin2 (@2)
= cos2 (@2)[cos2 (@2) + sin2 (@2)]
= cos2 (@2)
U 2 s
U 2
U
cos2 (@2) g = 0 |cos(@2)| g = 2 0 cos(@2) g
0
@2
U @2
= 4 0 cos x gx
x = 12 = 4 sin x 0 = 4(1 3 0) = 4
O=
[since cos(@2) D 0 for 0 $ $ ]
51. One loop of the curve u = cos 2 is traced with 3@4 $ $ @4.
u2 +
52. u2 +
gu
g
2
= cos2 2 + (32 sin 2)2 = cos2 2 + 4 sin2 2 = 1 + 3 sin2 2
gu
g
2
= tan2 + (sec2 )2
i O
]
@3
@6
u +
gu
g
2
= sin2 (6 sin ) + 36 cos2 cos2 (6 sin ) i O
54. The curve u = sin(@4) is completely traced with 0 $ $ 8.
2
u +
gu
g
2
= sin2 (@4) +
55. (a) From (10.2.6),
1
16
]
@4
3@4
s
1 + 3 sin2 2 g E 2=4221.
s
tan2 + sec4 g E 1=2789
53. The curve u = sin(6 sin ) is completely traced with 0 $ $ .
2
i O
cos2 (@4) i O
]
0
8
u = sin(6 sin ) i
]
0
gu
= cos(6 sin ) · 6 cos , so
g
t
sin2 (6 sin ) + 36 cos2 cos2 (6 sin ) g E 8=0091.
u = sin(@4) i
t
sin2 (@4) +
1
16
gu
=
g
1
4
cos(@4), so
cos2 (@4) g E 17=1568.
s
(g{@g)2 + (g|@g)2 g
s
Ue
[from the derivation of Equation 10.4.5]
= d 2| u2 + (gu@g)2 g
t
Ue
= d 2u sin u2 + (gu@g)2 g
V=
Ue
d
2|
(b) The curve u2 = cos 2 goes through the pole when cos 2 = 0 i
2 =
2
i =
4.
We’ll rotate the curve from = 0 to =
4
and double
this value to obtain the total surface area generated.
2
gu
gu
sin2 2
sin2 2
2
= 32 sin 2 i
.
=
=
u = cos 2 i 2u
2
g
g
u
cos 2
u
] @4
] @4 I
t
I
2 cos2 2 + sin2 2
g
V=2
2 cos 2 sin cos 2 + sin 2 @cos 2 g = 4
cos 2 sin cos 2
0
0
] @4
] @4 I
I
I @4
1
I
g = 4
= 4
cos 2 sin sin g = 4 3 cos 0 = 34 22 3 1 = 2 2 3 2
cos 2
0
0
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SECTION 10.5
56. (a) Rotation around =
2
is the same as rotation around the |-axis, that is, V =
CONIC SECTIONS
¤
63
Ue
2{ gv where
d
s
gv = (g{@gw)2 + (g|@gw)2 gw for a parametric equation, and for the special case of a polar equation, { = u cos and
s
s
gv = (g{@g)2 + (g|@g)2 g = u2 + (gu@g)2 g [see the derivation of Equation 10.4.5]. Therefore, for a polar
s
Ue
equation rotated around = 2 , V = d 2u cos u2 + (gu@g)2 g.
(b) As in the solution for Exercise 55(b), we can double the surface area generated by rotating the curve from = 0 to =
4
to obtain the total surface area.
V=2
]
@4
2
0
= 4
]
]
t
I
cos 2 cos cos 2 + (sin2 2)@cos 2 g = 4
@4
0
@4
0
I
cos 2 cos ] @4
I
@4
1
g = 4
cos 2 cos I
cos g = 4 sin 0
cos 2
0
u
cos2 2 + sin2 2
g
cos 2
I
I
2
= 4
30 = 2 2
2
10.5 Conic Sections
1. {2 = 6| and {2 = 4s|
i 4s = 6 i s = 32 .
The vertex is (0> 0), the focus is 0> 32 , and the directrix
is | = 3 32 .
3. 2{ = 3| 2
i | 2 = 32{. 4s = 32 i s = 3 12 .
The vertex is (0> 0), the focus is 3 12 > 0 , and the
directrix is { = 12 .
2. 2| 2 = 5{
i |2 = 52 {. 4s = 52 i s = 58 .
The vertex is (0> 0), the focus is 58 > 0 , and the directrix
is { = 3 58 .
4. 3{2 + 8| = 0
i 3{2 = 38|
i {2 = 3 83 |.
4s = 3 83 i s = 3 23 . The vertex is (0> 0), the focus
is 0> 3 23 , and the directrix is | = 23 .
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64
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
2
2
5. ({ + 2) = 8 (| 3 3). 4s = 8, so s = 2. The vertex is
6. { 3 1 = (| + 5) . 4s = 1, so s =
1
4 . The vertex is
(1> 35), the focus is 4 > 35 , and the directrix is { = 34 .
5
(32> 3), the focus is (32> 5), and the directrix is | = 1.
7. | 2 + 2| + 12{ + 25 = 0
8. | + 12{ 3 2{2 = 16
i
| 2 + 2| + 1 = 312{ 3 24 i
i 2{2 3 12{ = | 3 16 i
2({2 3 6{ + 9) = | 3 16 + 18 i
(| + 1)2 = 312({ + 2). 4s = 312, so s = 33.
2({ 3 3)2 = | + 2 i ({ 3 3)2 = 12 (| + 2).
The vertex is (32> 31), the focus is (35> 31), and the
4s = 12 , so s = 18 . The vertex is (3> 32), the focus is
3> 3 15
, and the directrix is | = 3 17
8
8 .
directrix is { = 1.
9. The equation has the form | 2 = 4s{, where s ? 0. Since the parabola passes through (31> 1), we have 12 = 4s(31), so
4s = 31 and an equation is | 2 = 3{ or { = 3| 2 . 4s = 31, so s = 3 14 and the focus is 3 14 > 0 while the directrix
is { = 14 .
2
10. The vertex is (2> 32), so the equation is of the form ({ 3 2) = 4s(| + 2), where s A 0. The point (0> 0) is on the parabola,
so 4 = 4s(2) and 4s = 2. Thus, an equation is ({ 3 2)2 = 2(| + 2). 4s = 2, so s =
directrix is | = 3 52 .
11.
1
2
and the focus is 2> 3 32 while the
I
I
I
I
I
{2
|2
+
= 1 i d = 4 = 2, e = 2, f = d2 3 e2 = 4 3 2 = 2. The
2
4
I ellipse is centered at (0> 0), with vertices at (0> ±2). The foci are 0> ± 2 .
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SECTION 10.5
12.
CONIC SECTIONS
¤
65
I
I
{2
|2
+
= 1 i d = 36 = 6, e = 8,
36
8
I
I
I
I
f = d2 3 e2 = 36 3 8 = 28 = 2 7. The ellipse is centered at (0> 0), with
I
vertices at (±6> 0). The foci are (±2 7> 0).
I
{2 | 2
+
= 1 i d = 9 = 3,
9
1
I
I
I
I
I
e = 1 = 1, f = d2 3 e2 = 9 3 1 = 8 = 2 2.
13. {2 + 9| 2 = 9
C
The ellipse is centered at (0> 0), with vertices (±3> 0).
I
The foci are (±2 2> 0).
14. 100{2 + 36| 2 = 225
{2
9
4
f=
+
|2
25
4
{2
C
=1 i d=
t
I
d2 3 e2 = 25
3
4
225
100
t
9
4
25
4
+
|2
225
36
=1 C
= 52 , e =
t
9
4
= 32 ,
= 2. The ellipse is centered
at (0> 0), with vertices 0> ± 52 . The foci are (0> ±2).
15. 9{2 3 18{ + 4| 2 = 27
16. {2 + 3| 2 + 2{ 3 12| + 10 = 0
C
9({2 3 2{ + 1) + 4| 2 = 27 + 9 C
2
2
9({ 3 1) + 4| = 36 C
|2
({ 3 1)2
+
=1 i
4
9
I
d = 3, e = 2, f = 5 i center (1> 0),
I vertices (1> ±3), foci 1> ± 5
17. The center is (0> 0), d = 3, and e = 2, so an equation is
2
2
C
{ + 2{ + 1 + 3(| 3 4| + 4) = 310 + 1 + 12 C
({ + 1)2 + 3(| 3 2)2 = 3 C
I
({ + 1)2
(| 3 2)2
+
= 1 i d = 3, e = 1,
3
1
I I
f = 2 i center (31> 2), vertices 31 ± 3> 2 ,
I foci 31 ± 2> 2
I
I I
|2
{2
+
= 1. f = d2 3 e2 = 5, so the foci are 0> ± 5 .
4
9
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¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
18. The ellipse is centered at (2> 1), with d = 3 and e = 2. An equation is
I the foci are 2 ± 5> 1 .
19.
I
I
(| 3 1)2
({ 3 2)2
+
= 1. f = d2 3 e2 = 5, so
9
4
I
I
|2
{2
3
= 1 i d = 5, e = 3, f = 25 + 9 = 34 i
25
9
I center (0> 0), vertices (0> ±5), foci 0> ± 34 , asymptotes | = ± 53 {.
Note: It is helpful to draw a 2d-by-2e rectangle whose center is the center of
the hyperbola. The asymptotes are the extended diagonals of the rectangle.
20.
I
{2
|2
3
= 1 i d = 6, e = 8, f = 36 + 64 = 10 i
36
64
center (0> 0), vertices (±6> 0), foci (±10> 0), asymptotes | = ± 86 { = ± 43 {
{2
|2
3
= 1 i d = e = 10,
100
100
I
I
f = 100 + 100 = 10 2 i center (0> 0), vertices (±10> 0),
I { = ±{
foci ±10 2> 0 , asymptotes | = ± 10
10
21. {2 3 | 2 = 100
22. | 2 3 16{2 = 16
C
C
{2
|2
3
= 1 i d = 4, e = 1,
16
1
I
I
16 + 1 = 17 i center (0> 0), vertices (0> ±4),
I foci 0> ± 17 , asymptotes | = ± 41 { = ±4{
f=
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SECTION 10.5
23. 4{2 3 | 2 3 24{ 3 4| + 28 = 0
2
CONIC SECTIONS
¤
67
C
2
4({ 3 6{ + 9) 3 (| + 4| + 4) = 328 + 36 3 4 C
(| + 2)2
({ 3 3)2
3
=1 i
1
4
I
I
I
I
d = 1 = 1, e = 4 = 2, f = 1 + 4 = 5 i
I
center (3> 32), vertices (4> 32) and (2> 32), foci 3 ± 5> 32 ,
4({ 3 3)2 3 (| + 2)2 = 4 C
asymptotes | + 2 = ±2({ 3 3).
24. | 2 3 4{2 3 2| + 16{ = 31
C
(| 2 3 2| + 1) 3 4({2 3 4{ + 4) = 31 + 1 3 16 C
(| 3 1)2 3 4({ 3 2)2 = 16 C
I
I
(| 3 1)2
({ 3 2)2
3
= 1 i d = 16 = 4, e = 4 = 2,
16
4
I
I
f = 16 + 4 = 20 i center (2> 1), vertices (2> 1 ± 4),
I foci 2> 1 ± 20 , asymptotes | 3 1 = ±2({ 3 2).
25. {2 = | + 1
C {2 = 1(| + 1). This is an equation of a parabola with 4s = 1, so s = 14 . The vertex is (0> 31) and the
focus is 0> 3 34 .
26. {2 = | 2 + 1
C {2 3 | 2 = 1. This is an equation of a hyperbola with vertices (±1> 0). The foci are at
I
I ± 1 + 1> 0 = ± 2> 0 .
27. {2 = 4| 3 2| 2
C {2 + 2| 2 3 4| = 0 C {2 + 2(| 2 3 2| + 1) = 2 C {2 + 2(| 3 1)2 = 2 C
I I
{2
(| 3 1)2
+
= 1. This is an equation of an ellipse with vertices at ± 2> 1 . The foci are at ± 2 3 1> 1 = (±1> 1).
2
1
28. | 2 3 8| = 6{ 3 16
C |2 3 8| + 16 = 6{ C (| 3 4)2 = 6{. This is an equation of a parabola with 4s = 6,
so s = 32 . The vertex is (0> 4) and the focus is 32 > 4 .
(| + 1)2
3 {2 = 1. This is an equation
4
I I
of a hyperbola with vertices (0> 31 ± 2) = (0> 1) and (0> 33). The foci are at 0> 31 ± 4 + 1 = 0> 31 ± 5 .
29. | 2 + 2| = 4{2 + 3
C |2 + 2| + 1 = 4{2 + 4 C (| + 1)2 3 4{2 = 4 C
2
{ + 12
+ | 2 = 1. This is an
1@4
t
I
equation of an ellipse with vertices 3 12 > 0 ± 1 = 3 12 > ±1 . The foci are at 3 12 > 0 ± 1 3 14 = 3 12 > ± 3@2 .
30. 4{2 + 4{ + | 2 = 0
2
C 4 {2 + { + 14 + | 2 = 1 C 4 { + 12 + | 2 = 1 C
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¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
31. The parabola with vertex (0> 0) and focus (1> 0) opens to the right and has s = 1, so its equation is | 2 = 4s{, or | 2 = 4{.
32. The parabola with focus (0> 0) and directrix | = 6 has vertex (0> 3) and opens downward, so s = 33 and its equation is
({ 3 0)2 = 4s(| 3 3), or {2 = 312(| 3 3).
33. The distance from the focus (34> 0) to the directrix { = 2 is 2 3 (34) = 6, so the distance from the focus to the vertex is
1
2 (6)
= 3 and the vertex is (31> 0). Since the focus is to the left of the vertex, s = 33. An equation is |2 = 4s({ + 1) i
| 2 = 312({ + 1).
34. The distance from the focus (3> 6) to the vertex (3> 2) is 6 3 2 = 4. Since the focus is above the vertex, s = 4.
An equation is ({ 3 3)2 = 4s(| 3 2) i ({ 3 3)2 = 16(| 3 2).
35. A parabola with vertical axis and vertex (2> 3) has equation | 3 3 = d({ 3 2)2 . Since it passes through (1> 5), we have
5 3 3 = d(1 3 2)2
i d = 2, so an equation is | 3 3 = 2({ 3 2)2 .
36. A parabola with horizontal axis has equation { = d| 2 + e| + f. Since the parabola passes through the point (31> 0),
substitute 31 for { and 0 for |: 31 = 0 + 0 + f. Now with f = 31, substitute 1 for { and 31 for |: 1 = d 3 e 3 1 (1);
and then 3 for { and 1 for |: 3 = d + e 3 1 (2)= Add (1) and (2) to get 4 = 2d 3 2 i d = 3 and then e = 1.
Thus, the equation is { = 3| 2 + | 3 1.
37. The ellipse with foci (±2> 0) and vertices (±5> 0) has center (0> 0) and a horizontal major axis, with d = 5 and f = 2,
so e2 = d2 3 f2 = 25 3 4 = 21. An equation is
|2
{2
+
= 1.
25
21
38. The ellipse with foci (0> ±5) and vertices (0> ±13) has center (0> 0) and a vertical major axis, with f = 5 and d = 13,
so e =
I
|2
{2
+
= 1.
d2 3 f2 = 12. An equation is
144
169
39. Since the vertices are (0> 0) and (0> 8), the ellipse has center (0> 4) with a vertical axis and d = 4. The foci at (0> 2) and (0> 6)
are 2 units from the center, so f = 2 and e =
I
I
I
({ 3 0)2
(| 3 4)2
d2 3 f2 = 42 3 22 = 12. An equation is
+
=1 i
2
e
d2
{2
(| 3 4)2
+
= 1.
12
16
40. Since the foci are (0> 31) and (8> 31), the ellipse has center (4> 31) with a horizontal axis and f = 4.
The vertex (9> 31) is 5 units from the center, so d = 5 and e =
(| + 1)2
({ 3 4)2
+
=1
2
d
e2
i
I
I
I
d2 3 f2 = 52 3 42 = 9. An equation is
({ 3 4)2
(| + 1)2
+
= 1.
25
9
41. An equation of an ellipse with center (31> 4) and vertex (31> 0) is
from the center, so f = 2. Thus, e2 + 22 = 42
({ + 1)2
(| 3 4)2
+
= 1. The focus (31> 6) is 2 units
e2
42
i e2 = 12, and the equation is
(| 3 4)2
({ + 1)2
+
= 1.
12
16
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SECTION 10.5
42. Foci I1 (34> 0) and I2 (4> 0)
i f = 4 and an equation is
2d = |S I1 | + |S I2 | i 2d = 1=8 +
s
82 + (1=8)2
e2 = d2 3 f2 = 25 3 16 = 9 and the equation is
{2
|2
+ 2 = 1. The ellipse passes through S (34> 1=8), so
2
d
e
i 2d = 1=8 + 8=2 i d = 5.
{2
|2
3 2 = 1. Foci (±5> 0) i f = 5 and 32 + e2 = 52
2
3
e
i
|2
{2
3
= 1.
9
16
44. An equation of a hyperbola with vertices (0> ±2) is
e2 = 25 3 4 = 21, so the equation is
¤
{2
|2
+
= 1.
25
9
43. An equation of a hyperbola with vertices (±3> 0) is
e2 = 25 3 9 = 16, so the equation is
CONIC SECTIONS
|2
{2
3 2 = 1. Foci (0> ±5) i f = 5 and 22 + e2 = 52
2
2
e
i
|2
{2
3
= 1.
4
21
45. The center of a hyperbola with vertices (33> 34) and (33> 6) is (33> 1), so d = 5 and an equation is
(| 3 1)2
({ + 3)2
3
= 1. Foci (33> 37) and (33> 9) i f = 8, so 52 + e2 = 82
2
5
e2
equation is
i e2 = 64 3 25 = 39 and the
(| 3 1)2
({ + 3)2
3
= 1.
25
39
46. The center of a hyperbola with vertices (31> 2) and (7> 2) is (3> 2), so d = 4 and an equation is
Foci (32> 2) and (8> 2) i f = 5, so 42 + e2 = 52
({ 3 3)2
(| 3 2)2
3
= 1.
2
4
e2
i e2 = 25 3 16 = 9 and the equation is
({ 3 3)2
(| 3 2)2
3
= 1.
16
9
47. The center of a hyperbola with vertices (±3> 0) is (0> 0), so d = 3 and an equation is
Asymptotes | = ±2{ i
{2
|2
3 2 = 1.
2
3
e
e
{2
|2
= 2 i e = 2(3) = 6 and the equation is
3
= 1.
d
9
36
48. The center of a hyperbola with foci (2> 0) and (2> 8) is (2> 4), so f = 4 and an equation is
The asymptote | = 3 + 12 { has slope 12 , so
5d2 = 16 i d2 =
16
5
d
1
=
e
2
and so e2 = 16 3
16
5
=
i e = 2d and d2 + e2 = f2
64
.
5
Thus, an equation is
(| 3 4)2
({ 3 2)2
3
= 1.
2
d
e2
i d2 + (2d)2 = 42
i
({ 3 2)2
(| 3 4)2
3
= 1.
16@5
64@5
49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance
d 3 f from it) while the farthest point is the other vertex (at a distance of d + f). So for this lunar orbit,
(d 3 f) + (d + f) = 2d = (1728 + 110) + (1728 + 314), or d = 1940; and (d + f) 3 (d 3 f) = 2f = 314 3 110,
or f = 102. Thus, e2 = d2 3 f2 = 3,753,196, and the equation is
|2
{2
+
= 1.
3,763,600
3,753,196
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70
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
50. (a) Choose Y to be the origin, with {-axis through Y and I . Then I is (s> 0), D is (s> 5), so substituting D into the
equation | 2 = 4s{ gives 25 = 4s2 so s =
(b) { = 11 i | =
5
2
and | 2 = 10{.
I
I
110 i |FG| = 2 110
51. (a) Set up the coordinate system so that D is (3200> 0) and E is (200> 0).
|S D| 3 |S E| = (1200)(980) = 1,176,000 ft =
e2 = f2 3 d2 =
(b) Due north of E
3,339,375
121
i
C
mi = 2d i d =
1225
,
11
and f = 200 so
121{2
121| 2
3
= 1.
1,500,625
3,339,375
i { = 200 i
52. |S I1 | 3 |S I2 | = ±2d
2450
11
(121)(200)2
121|2
133,575
3
=1 i |=
E 248 mi
1,500,625
3,339,375
539
s
s
({ + f)2 + | 2 3 ({ 3 f)2 + | 2 = ±2d C
s
s
s
({ + f)2 + | 2 = ({ 3 f)2 + | 2 ± 2d C ({ + f)2 + | 2 = ({ 3 f)2 + | 2 + 4d2 ± 4d ({ 3 f)2 + | 2
s
4f{ 3 4d2 = ±4d ({ 3 f)2 + | 2 C f2 {2 3 2d2 f{ + d4 = d2 ({2 3 2f{ + f2 + | 2 ) C
(f2 3 d2 ){2 3 d2 | 2 = d2 (f2 3 d2 ) C e2 {2 3 d2 | 2 = d2 e2 [where e2 = f2 3 d2 ] C
C
{2
|2
3 2 =1
2
d
e
53. The function whose graph is the upper branch of this hyperbola is concave upward. The function is
u
{2
dI 2
d
=
e + {2 , so | 0 = {(e2 + {2 )31@2 and
2
e
e
e
l
k
d
(e2 + {2 )31@2 3 {2 (e2 + {2 )33@2 = de(e2 + {2 )33@2 A 0 for all {, and so i is concave upward.
| 00 =
e
| = i({) = d
1+
54. We can follow exactly the same sequence of steps as in the derivation of Formula 4, except we use the points (1> 1) and
s
s
({ 3 1)2 + (| 3 1)2 + ({ + 1)2 + (| + 1)2 = 4
s
will lead (after moving the second term to the right, squaring, and simplifying) to 2 ({ + 1)2 + (| + 1)2 = { + | + 4,
(31> 31) in the distance formula (first equation of that derivation) so
which, after squaring and simplifying again, leads to 3{2 3 2{| + 3| 2 = 8.
55. (a) If n A 16, then n 3 16 A 0, and
|2
{2
+
= 1 is an ellipse since it is the sum of two squares on the left side.
n
n 3 16
(b) If 0 ? n ? 16, then n 3 16 ? 0, and
left side.
|2
{2
+
= 1 is a hyperbola since it is the difference of two squares on the
n
n 3 16
(c) If n ? 0, then n 3 16 ? 0, and there is no curve since the left side is the sum of two negative terms, which cannot equal 1.
(d) In case (a), d2 = n, e2 = n 3 16, and f2 = d2 3 e2 = 16, so the foci are at (±4> 0). In case (b), n 3 16 ? 0, so d2 = n,
e2 = 16 3 n, and f2 = d2 + e2 = 16, and so again the foci are at (±4> 0).
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SECTION 10.5
i 2||0 = 4s i |0 =
56. (a) | 2 = 4s{
| 3 |0 =
CONIC SECTIONS
¤
71
2s
, so the tangent line is
|
2s
({ 3 {0 ) i ||0 3 |02 = 2s({ 3 {0 ) C
|0
||0 3 4s{0 = 2s{ 3 2s{0
i ||0 = 2s({ + {0 ).
(b) The {-intercept is 3{0 .
57. {2 = 4s|
|3
i 2{ = 4s|0
i |0 =
{
, so the tangent line at ({0 > |0 ) is
2s
{0
{20
=
({ 3 {0 ). This line passes through the point (d> 3s) on the
4s
2s
directrix, so 3s 3
{20
{0
=
(d 3 {0 ) i 34s2 3 {20 = 2d{0 3 2{20
4s
2s
{20 3 2d{0 3 4s2 = 0 C {20 3 2d{0 + d2 = d2 + 4s2
C
C
s
s
({0 3 d)2 = d2 + 4s2 C {0 = d ± d2 + 4s2 . The slopes of the tangent lines at { = d ± d2 + 4s2
s
d ± d2 + 4s2
, so the product of the two slopes is
are
2s
s
s
d + d2 + 4s2 d 3 d2 + 4s2
d2 3 (d2 + 4s2 )
34s2
·
=
=
= 31,
2s
2s
4s2
4s2
showing that the tangent lines are perpendicular.
58. Without a loss of generality, let the ellipse, hyperbola, and foci be as shown in the figure.
The curves intersect (eliminate |2 ) i
2
2
{
|2
|2
2 {
+
e
= E 2 + e2 i
E2
3
+
D2
E2
d2
e2
2
E 2 {2
e2 {2
e2
2
2
2 E
+
=
E
+
e
i
{
+
= E 2 + e2
D2
d2
D2
d2
{2 =
i
D2 d2 (E 2 + e2 )
E 2 + e2
.
2
2 2 =
d2 E 2 + e2D2
d E +eD
D2 d2
2
Similarly, | 2 =
E 2 e2 (d2 3 D2 )
.
e2D2 + d2 E 2
Next we find the slopes of the tangent lines of the curves:
0
|H
=3
{2
|2
+ 2 =1 i
2
d
e
2{
2|| 0
+ 2 =0 i
2
d
e
|| 0
{
=3 2
e2
d
i
2{
2||0
|| 0
{
E2 {
0
. The product of the slopes
3
=
0
i
=
i
|
=
K
D2
E2
E2
D2
D2 |
2 2 2
2 2 2 D d (E + e )
E
e
d2 E 2 + e2D2
e2 E 2 {2
E 2 + e2
2 2 2
= 3 2 2 02 = 3
. Since d2 3 e2 = f2 and D2 + E 2 = f2 ,
2 = 3 2
d D |0
d 3 D2
E e (d 3 D )
2
2
dD
e2D2 + d2 E 2
{2
e2 {
|2
and
3
=1 i
d2 |
D2
E2
0
0
at ({0 > |0 ) is |H
|K
we have d2 3 e2 = D2 + E 2
i d2 3 D2 = e2 + E 2 , so the product of the slopes is 31, and hence, the tangent lines at
each point of intersection are perpendicular.
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¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
59. 9{2 + 4| 2 = 36
C
|2
{2
+
= 1. We use the parametrization { = 2 cos w, | = 3 sin w, 0 $ w $ 2. The circumference
4
9
is given by
U 2 s
U 2 s
U 2 s
(g{@gw)2 + (g|@gw)2 gw = 0
(32 sin w)2 + (3 cos w)2 gw = 0
4 sin2 w + 9 cos2 w gw
O= 0
U 2 I
4 + 5 cos2 w gw
= 0
I
2 3 0
= , and i (w) = 4 + 5 cos2 w to get
8
4
i (0) + 4i 4 + 2i 2 + 4i 3
+ 2i () + 4i 5
+ 2i 3
+ 4i 7
+ i (2) E 15=9.
O E V8 = @4
3
4
4
2
4
Now use Simpson’s Rule with q = 8, {w =
60. The length of the major axis is 2d, so d =
1
(1=18
2
× 1010 ) = 5=9 × 109 . The length of the minor axis is 2e, so
e = 12 (1=14 × 1010 ) = 5=7 × 109 . An equation of the ellipse is
{2
|2
+
= 1, or converting into parametric equations,
d2
e2
{ = d cos and | = e sin . So
O=4
U @2 s
U @2 s
(g{@g)2 + (g|@g)2 g = 4 0
d2 sin2 + e2 cos2 g
0
s
, and i () = d2 sin2 + e2 cos2 , we get
Using Simpson’s Rule with q = 10, { = @2103 0 = 20
+ 2i 2
+ · · · + 2i 8
+ 4i 9
+ i 2 E 3=64 × 1010 km
O E 4 · V10 = 4 · 20· 3 i (0) + 4i 20
20
20
20
61.
{2
|2
|2
{2 3 d2
eI 2
3
=
1
i
=
i |=±
{ 3 d2 .
d2
e2
e2
d2
d
s
] f s
f
s
e
d2 39 2e {
D= 2
ln { + {2 3 d2 {2 3 d2 g{ =
{2 3 d2 3
d
d
2
2
d
d
I
e I 2
f f 3 d2 3 d2 ln f + f2 3 d2 + d2 ln |d|
d
I
Since d2 + e2 = f2 > f2 3 d2 = e2 , and f2 3 d2 = e.
e
e
= fe 3 d2 ln(f + e) + d2 ln d = fe + d2 (ln d 3 ln(e + f))
d
d
=
= e2 f@d + de ln[d@(e + f)], where f2 = d2 + e2 .
62. (a)
(b)
{2
|2
|2
d2 3 {2
eI 2
+ 2 =1 i
=
i |=±
d 3 {2 .
2
2
d
e
e
d2
d
2
]
] d s
e
e2 d 2
Y =
d2 3 {2
g{ = 2 2
(d 3 {2 ) g{
d
d 0
3d
ld
2e2 k
4
2e2 2d3
= 2 d2 { 3 13 {3 = 2
= e2 d
d
d
3
3
0
{2
|2
{2
e2 3 | 2
ds 2
+
=
1
i
=
i {=±
e 3 |2 .
d2
e2
d2
e2
e
]
] e s
2
d
d2 e 2
Y =
e2 3 | 2
g| = 2 2
(e 3 | 2 ) g|
e
e 0
3e
le
2d2 k
4
2d2 2e3
= 2 e2 | 3 13 | 3 = 2
= d2 e
e
e
3
3
0
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SECTION 10.5
63. 9{2 + 4| 2 = 36
C
CONIC SECTIONS
¤
73
|2
{2
+
= 1 i d = 3, e = 2. By symmetry, { = 0. By Example 2 in Section 7.3, the area of the
4
9
top half of the ellipse is 12 (de) = 3. Solve 9{2 + 4| 2 = 36 for | to get an equation for the top half of the ellipse:
I
9{2 + 4| 2 = 36 C 4| 2 = 36 3 9{2 C | 2 = 94 (4 3 {2 ) i | = 32 4 3 {2 . Now
2
]
] 2 s
] 2
1 3
1 e1
1
3
[i ({)]2 g{ =
|=
4 3 {2
g{ =
(4 3 {2 ) g{
D d 2
3 32 2 2
8 32
2
] 2
1 3
4
3
3 16
3
2
·2
4{ 3 {
=
(4 3 { ) g{ =
=
=
8
4
3
4
3
0
0
so the centroid is (0> 4@).
64. (a) Consider the ellipse
{2
|2
+
= 1 with d A e, so that the major axis is the {-axis. Let the ellipse be parametrized by
d2
e2
{ = d cos w, | = e sin w, 0 $ w $ 2. Then
2 2
g{
g|
+
= d2 sin2 w + e2 cos2 w = d2 (1 3 cos2 w) + e2 cos2 w = d2 + (e2 3 d2 ) cos2 w = d2 3 f2 cos2 w,
gw
gw
where f2 = d2 3 e2 . Using symmetry and rotating the ellipse about the major axis gives us surface area
] 0s
]
] @2
s
1
x = f cos w
2(e sin w) d2 3 f2 cos2 w gw = 4e
d2 3 x2 3 gx
V = 2| gv = 2
gx = f sin w gw
f
0
f
s
] fs
x f
f l
d2
2e k s 2
4e
30 4e x
d2 3 x2 gx =
d2 3 x2 +
=
=
sin31
f d 3 f2 + d2 sin31
f 0
f
2
2
d 0
f
d
k
l
f
2e
ef + d2 sin31
=
f
d
(b) As in part (a),
2 2
g|
g{
+
= d2 sin2 w + e2 cos2 w = d2 sin2 w + e2 (1 3 sin2 w) = e2 + (d2 3 e2 ) sin2 w = e2 + f2 sin2 w.
gw
gw
Rotating about the minor axis gives us
]
]
] @2
s
2(d cos w) e2 + f2 sin2 w gw = 4d
V = 2{ gv = 2
0
0
I
4d x I 2
e
ln x + e2 + x2
e + x2 +
f 2
2
d + f l
2d k
df + e2 ln
=
f
e
21
=
65. Differentiating implicitly,
line at S is 3
2
{2
|2
+
=1 i
d2
e2
f
0
=
f
s
1
gx
e2 + x2
f
x = f sin w
gx = f cos w gw
I
2d I 2
f e + f2 + e2 ln f + e2 + f2 3 e2 ln e
f
2{ 2||0
e2 {
0
+
=
0
i
|
=
3
d2
e2
d2 |
[| 6= 0]. Thus, the slope of the tangent
e2 {1
|1
|1
and of I2 S is
. By the formula from Problems Plus, we have
. The slope of I1 S is
d2 |1
{1 + f
{1 3 f
|1
e2 {1
+ 2
d2 | 2 + e2 {1 ({1 + f)
d2 e2 + e2 f{1
{1 + f
d |1
tan =
= 2 1
= 2
2
2
d |1 ({1 + f) 3 e {1 |1
f {1 |1 + d2 f|1
e {1 |1
13 2
d |1 ({1 + f)
e2 f{1 + d2
e2
=
=
2
f|1 (f{1 + d )
f|1
using e2 {21 + d2 |12 = d2 e2 ,
and d2 e2 = f2
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¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
and
e2 {1
|1
3
e2 f{1 3 d2
3d2 |12 3 e2 {1 ({1 3 f)
3d2 e2 + e2 f{1
e2
d2 |1
{1 3 f
= 2
=
=
tan =
=
2
d |1 ({1 3 f) 3 e2 {1 |1
f2 {1 |1 3 d2 f|1
f|1 (f{1 3 d2 )
f|1
e {1 |1
13 2
d |1 ({1 3 f)
3
Thus, = .
66. The slopes of the line segments I1 S and I2 S are
2{ 2||0
e2 {
3 2 = 0 i |0 = 2
2
d
e
d |
|1
|1
and
, where S is ({1 > |1 ). Differentiating implicitly,
{1 + f
{1 3 f
i the slope of the tangent at S is
e2 {1
, so by the formula in Problem 19 on text
d2 |1
page 271,
|1
e2 {1
3
e2 {1 ({1 + f) 3 d2 |12
e2 (f{1 + d2 )
d2 |1
{1 + f
= 2
=
tan =
2
d |1 ({1 + f) + e2 {1 |1
f|1 (f{1 + d2 )
e {1 |1
1+ 2
d |1 ({1 + f)
using {21 @d2 |12 @e2 = 1,
2
2
2
and d + e = f
=
e2
f|1
|1
e2 {1
+
e2
3e2 {1 ({1 3 f) + d2 |12
e2 (f{1 3 d2 )
d2 |1
{1 3 f
=
tan =
= 2
=
2
2
2
d |1 ({1 3 f) + e {1 |1
f|1 (f{1 3 d )
f|1
e {1 |1
1+ 2
d |1 ({1 3 f)
3
and
So = .
10.6 Conic Sections in Polar Coordinates
1. The directrix { = 4 is to the right of the focus at the origin, so we use the form with “+ h cos ” in the denominator.
(See Theorem 6 and Figure 2.) An equation is u =
1
·4
4
hg
2
=
=
.
1
1 + h cos 2 + cos 1 + 2 cos 2. The directrix { = 33 is to the left of the focus at the origin, so we use the form with “3 h cos ” in the denominator.
h = 1 for a parabola, so an equation is u =
hg
1·3
3
=
=
.
1 3 h cos 1 3 1 cos 1 3 cos 3. The directrix | = 2 is above the focus at the origin, so we use the form with “+ h sin ” in the denominator. An equation is
u=
1=5(2)
6
hg
=
=
.
1 + h sin 1 + 1=5 sin 2 + 3 sin 4. The directrix { = 3 is to the right of the focus at the origin, so we use the form with “+ h cos ” in the denominator. An
equation is u =
hg
3·3
9
=
=
.
1 + h cos 1 + 3 cos 1 + 3 cos 5. The vertex (4> 3@2) is 4 units below the focus at the origin, so the directrix is 8 units below the focus (g = 8), and we
use the form with “3 h sin ” in the denominator. h = 1 for a parabola, so an equation is
u=
1(8)
8
hg
=
=
.
1 3 h sin 1 3 1 sin 1 3 sin 6. The vertex S (1> @2) is 1 unit above the focus I at the origin, so |S I | = 1 and we use the form with “+ h sin ” in the
denominator. The distance from the focus to the directrix o is g, so
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SECTION 10.6
h=
|S I |
|S o|
i 0=8 =
An equation is u =
1
g31
CONIC SECTIONS IN POLAR COORDINATES
¤
75
i 0=8g 3 0=8 = 1 i 0=8g = 1=8 i g = 2=25.
0=8(2=25)
5
9
hg
=
· =
.
1 + h sin 1 + 0=8 sin 5
5 + 4 sin 7. The directrix u = 4 sec (equivalent to u cos = 4 or { = 4) is to the right of the focus at the origin, so we will use the form
with “+ h cos ” in the denominator. The distance from the focus to the directrix is g = 4, so an equation is
u=
1
(4)
2
4
hg
2
· =
=
.
1 + h cos 2 + cos 1 + 12 cos 2
8. The directrix u = 36 csc (equivalent to u sin = 36 or | = 36) is below the focus at the origin, so we will use the form
with “3 h sin ” in the denominator. The distance from the focus to the directrix is g = 6, so an equation is
u=
9. u =
3(6)
18
hg
=
=
.
1 3 h sin 1 3 3 sin 1 3 3 sin 1@5
4@5
4
, where h =
·
=
5 3 4 sin 1@5
1 3 45 sin (a) Eccentricity = h =
(b) Since h =
4
5
4
5
and hg =
4
5
i g = 1.
4
5
? 1, the conic is an ellipse.
(c) Since “3 h sin ” appears in the denominator, the directrix is below the focus
at the origin, g = |I o| = 1, so an equation of the directrix is | = 31.
.
(d) The vertices are 4> 2 and 49 > 3
2
10. u =
1@3
4
12
, where h =
·
=
3 3 10 cos 1@3
1 3 10
cos 3
(a) Eccentricity = h =
(b) Since h =
10
3
10
3
10
3
3 6
= 5.
and hg = 4 i g = 4 10
A 1, the conic is a hyperbola.
(c) Since “3 h cos ” appears in the denominator, the directrix is to the left of the
focus at the origin. g = |I o| = 65 , so an equation of the directrix is { = 3 65 .
> 0 and 12
> , so the center is midway between them,
(d) The vertices are 3 12
7
13
> .
that is, 120
91
11. u =
1@3
2@3
2
·
=
, where h = 1 and hg =
3 + 3 sin 1@3
1 + 1 sin 2
3
i g = 23 .
(a) Eccentricity = h = 1
(b) Since h = 1, the conic is a parabola.
(c) Since “+ h sin ” appears in the denominator, the directrix is above the focus
at the origin. g = |I o| = 23 , so an equation of the directrix is | = 23 .
(d) The vertex is at 13 > 2 , midway between the focus and directrix.
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
12. u =
3
1@2
3@2
·
=
, where h = 1 and hg =
2 + 2 cos 1@2
1 + 1 cos 3
2
i g = 32 .
(a) Eccentricity = h = 1
(b) Since h = 1, the conic is a parabola.
(c) Since “+h cos ” appears in the denominator, the directrix is to the right of
the focus at the origin. g = |I o| = 32 , so an equation of the directrix is
{ = 32 .
(d) The vertex is at
13. u =
3
4
> 0 , midway between the focus and directrix.
1@6
3@2
9
, where h =
·
=
6 + 2 cos 1@6
1 + 13 cos (a) Eccentricity = h =
(b) Since h =
1
3
1
3
and hg =
3
2
i g = 92 .
1
3
? 1, the conic is an ellipse.
(c) Since “+h cos ” appears in the denominator, the directrix is to the right of
the focus at the origin. g = |I o| = 92 , so an equation of the directrix is
{ = 92 .
(d) The vertices are
9 that is, 16
> .
14. u =
9
8
> 0 and 94 > , so the center is midway between them,
1@4
8
2
, where h =
·
=
4 + 5 sin 1@4
1 + 54 sin (a) Eccentricity = h =
(b) Since h =
5
4
5
4
5
4
and hg = 2 i g = 2 45 = 85 .
A 1, the conic is a hyperbola.
(c) Since “+h sin ” appears in the denominator, the directrix is above the
focus at the origin. g = |I o| = 85 , so an equation of the directrix is | = 85 .
, so the center is midway between them,
(d) The vertices are 89 > 2 and 38> 3
2
40 that is, 9 > 2 .
15. u =
1@4
3@4
3
·
=
, where h = 2 and hg =
4 3 8 cos 1@4
1 3 2 cos 3
4
i g = 38 .
(a) Eccentricity = h = 2
(b) Since h = 2 A 1, the conic is a hyperbola.
(c) Since “3h cos ” appears in the denominator, the directrix is to the left of
the focus at the origin. g = |I o| = 38 , so an equation of the directrix is
{ = 3 38 .
(d) The vertices are 3 34 > 0 and 14 > , so the center is midway between them,
that is, 12 > .
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SECTION 10.6
16. u =
10
1@5
2
, where h =
·
=
6
5 3 6 sin 1@5
1 3 5 sin (a) Eccentricity = h =
(b) Since h =
6
5
6
5
6
5
CONIC SECTIONS IN POLAR COORDINATES
and hg = 2 i g = 2 56 = 53 .
A 1, the conic is a hyperbola.
(c) Since “3h sin ” appears in the denominator, the directrix is below the focus
at the origin. g = |I o| = 53 , so an equation of the directrix is | = 3 53 .
> 3 , so the center is midway between them,
(d) The vertices are 310> 2 and 10
11 2
that is, 60
> 3 .
11 2
17. (a) u =
1
, where h = 2 and hg = 1 i g = 12 . The eccentricity
1 3 2 sin h = 2 A 1, so the conic is a hyperbola. Since “3h sin ” appears in the
denominator, the directrix is below the focus at the origin. g = |I o| = 12 ,
so an equation of the directrix is | = 3 12 . The vertices are 31> 2 and
1 3 , so the center is midway between them, that is, 23 > 3
.
3> 2
2
(b) By the discussion that precedes Example 4, the equation
is u =
18. u =
1
1 3 2 sin 3
3
4
.
4@5
4
, so h =
=
5 + 6 cos 1 + 65 cos An equation of the directrix is { =
2
3
6
5
and hg =
4
5
i u cos =
i g = 23 .
2
3
i u=
2
.
3 cos If the hyperbola is rotated about its focus (the origin) through an angle @3,
its equation is the same as that of the original, with replaced by 3
(see Example 4), so u =
3
4
.
5 + 6 cos 3 3
19. For h ? 1 the curve is an ellipse. It is nearly circular when h is close to 0. As h
increases, the graph is stretched out to the right, and grows larger (that is, its
right-hand focus moves to the right while its left-hand focus remains at the
origin.) At h = 1, the curve becomes a parabola with focus at the origin.
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CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
20. (a) The value of g does not seem to affect the shape of the conic (a parabola) at
all, just its size, position, and orientation (for g ? 0 it opens upward, for
g A 0 it opens downward).
(b) We consider only positive values of h. When 0 ? h ? 1, the conic is an
ellipse. As h < 0+ , the graph approaches perfect roundness and zero size.
As h increases, the ellipse becomes more elongated, until at h = 1 it turns
into a parabola. For h A 1, the conic is a hyperbola, which moves
downward and gets broader as h continues to increase.
h = 0=1
h = 0=5
h = 0=9
h = 1=1
21. |S I | = h |S o|
hg
u=
1 + h sin h = 1=5
h = 10
i u = h[g 3 u cos( 3 )] = h(g + u cos ) i
u(1 3 h cos ) = hg i u =
22. |S I | = h |S o|
h=1
hg
1 3 h cos i u = h[g 3 u sin ] i u(1 + h sin ) = hg i
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SECTION 10.6
23. |S I | = h |S o|
CONIC SECTIONS IN POLAR COORDINATES
¤
79
i u = h[g 3 u sin( 3 )] = h(g + u sin ) i
u(1 3 h sin ) = hg i u =
hg
1 3 h sin 24. The parabolas intersect at the two points where
For the first parabola,
f
g
=
1 + cos 1 3 cos i cos =
f3g
f+g
i u=
f+g
.
2
f sin gu
=
, so
g
(1 + cos )2
g|
(gu@g) sin + u cos f sin2 + f cos (1 + cos )
1 + cos =
=
=
g{
(gu@g) cos 3 u sin f sin cos 3 f sin (1 + cos )
3 sin and similarly for the second,
1 3 cos sin g|
=
=
. Since the product of these slopes is 31, the parabolas intersect
g{
sin 1 + cos at right angles.
25. We are given h = 0=093 and d = 2=28 × 108 . By (7), we have
u=
d(1 3 h2 )
2=28 × 108 [1 3 (0=093)2 ]
2=26 × 108
=
E
1 + h cos 1 + 0=093 cos 1 + 0=093 cos 26. We are given h = 0=048 and 2d = 1=56 × 109
u=
i d = 7=8 × 108 . By (7), we have
7=8 × 108 [1 3 (0=048)2 ]
7=78 × 108
d(1 3 h2 )
=
E
1 + h cos 1 + 0=048 cos 1 + 0=048 cos 27. Here 2d = length of major axis = 36=18 AU
i d = 18=09 AU and h = 0=97. By (7), the equation of the orbit is
2
u=
18=09[1 3 (0=97) ]
1=07
E
. By (8), the maximum distance from the comet to the sun is
1 + 0=97 cos 1 + 0=97 cos 18=09(1 + 0=97) E 35=64 AU or about 3=314 billion miles.
28. Here 2d = length of major axis = 356=5 AU
is u =
i d = 178=25 AU and h = 0=9951. By (7), the equation of the orbit
178=25[1 3 (0=9951)2 ]
1=7426
E
. By (8), the minimum distance from the comet to the sun is
1 + 0=9951 cos 1 + 0=9951 cos 178=25(1 3 0=9951) E 0=8734 AU or about 81 million miles.
29. The minimum distance is at perihelion, where 4=6 × 107 = u = d(1 3 h) = d(1 3 0=206) = d(0=794)
i
d = 4=6 × 107@0=794. So the maximum distance, which is at aphelion, is
u = d(1 + h) = 4=6 × 107@0=794 (1=206) E 7=0 × 107 km.
30. At perihelion, u = d(1 3 h) = 4=43 × 109 , and at aphelion, u = d(1 + h) = 7=37 × 109 . Adding, we get 2d = 11=80 × 109 ,
so d = 5=90 × 109 km. Therefore 1 + h = d(1 + h)@d =
7=37
5=90
E 1=249 and h E 0=249.
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CHAPTER 10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
31. From Exercise 29, we have h = 0=206 and d(1 3 h) = 4=6 × 107 km. Thus, d = 4=6 × 107@0=794. From (7), we can write the
equation of Mercury’s orbit as u = d
1 3 h2
. So since
1 + h cos d(1 3 h2 )h sin gu
=
i
g
(1 + h cos )2
2
d2 (1 3 h2 )2
d2 (1 3 h2 )2 h2 sin2 d2 (1 3 h2 )2
gu
u2 +
=
+
=
(1 + 2h cos + h2 )
2
4
g
(1 + h cos )
(1 + h cos )
(1 + h cos )4
the length of the orbit is
]
O=
2
0
]
s
u2 + (gu@g)2 g = d(1 3 h2 )
0
2
I
1 + h2 + 2h cos g E 3=6 × 108 km
(1 + h cos )2
This seems reasonable, since Mercury’s orbit is nearly circular, and the circumference of a circle of radius d
is 2d E 3=6 × 108 km.
10 Review
1. (a) A parametric curve is a set of points of the form ({> |) = (i (w)> j(w)), where i and j are continuous functions of a
variable w.
(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 i(w) and j(w) for various values of w, either by hand or with a calculator or computer. Sometimes, when
i and j are given by formulas, we can eliminate w from the equations { = i (w) and | = j(w) to get a Cartesian equation
relating { and |. It may be easier to graph that equation than to work with the original formulas for { and | in terms of w.
2. (a) You can find
g|@gw
g|
g|
as a function of w by calculating
=
[if g{@gw 6= 0].
g{
g{
g{@gw
(b) Calculate the area as
Ue
d
| g{ =
U
j(w) i 0 (w)gw [or
U
j(w) i 0 (w)gw if the leftmost point is (i ()> j()) rather
than (i()> j())].
U s
Us
3. (a) O = (g{@gw)2 + (g|@gw)2 gw = [i 0 (w)]2 + [j 0 (w)]2 gw
s
s
U
U
(b) V = 2| (g{@gw)2 + (g|@gw)2 gw = 2j(w) [i 0 (w)]2 + [j0 (w)]2 gw
4. (a) See Figure 5 in Section 10.3.
(b) { = u cos , | = u sin (c) To find a polar representation (u> ) with u D 0 and 0 $ ? 2, first calculate u =
s
{2 + | 2 . Then is specified by
cos = {@u and sin = |@u.
gu
g|
g
g
sin + u cos (|)
(u sin )
g
g|
g
g
g
=
5. (a) Calculate
, where u = i().
=
=
= g{
g
g
g{
gu
({)
(u cos )
cos 3 u sin g
g
g
g
Ue 1
Ue 1 2
2
(b) Calculate D = d 2 u g = d 2 [i()] g
Ues
Ues
Ues
(c) O = d (g{@g)2 + (g|@g)2 g = d u2 + (gu@g)2 g = d [i ()]2 + [i 0 ()]2 g
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CHAPTER 10 REVIEW
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81
6. (a) A parabola is a set of points in a plane whose distances from a fixed point I (the focus) and a fixed line o (the directrix)
are equal.
(b) {2 = 4s|; | 2 = 4s{
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.
(b)
|2
{2
+
= 1.
d2
d2 3 f2
8. (a) A hyperbola 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
3
=1
d2
f2 3 d2
I
f2 3 d2
(c) | = ±
{
d
(b)
9. (a) If a conic section has focus I and corresponding directrix o, then the eccentricity h is the fixed ratio |S I | @ |S o| for points
S of the conic section.
(b) h ? 1 for an ellipse; h A 1 for a hyperbola; h = 1 for a parabola.
(c) { = g: u =
1. False.
hg
hg
hg
hg
. { = 3g: u =
. | = g: u =
. | = 3g: u =
.
1 + h cos 1 3 h cos 1 + h sin 1 3 h sin Consider the curve defined by { = i (w) = (w 3 1)3 and | = j(w) = (w 3 1)2 . Then j0 (w) = 2(w 3 1), so j0 (1) = 0,
but its graph has a vertical tangent when w = 1. Note: The statement is true if i 0 (1) 6= 0 when j0 (1) = 0.
g 2|
g
=
g{2
g{
g|
g{
g
gw
g|
g{
g{
gw
2. False.
If { = i (w) and | = j(w) are twice differentiable, then
3. False.
For example, if i (w) = cos w and j(w) = sin w for 0 $ w $ 4, then the curve is a circle of radius 1, hence its length
U 4 s
U 4
U 4 s
[i 0 (w)]2 + [j 0 (w)]2 gw = 0
(3 sin w)2 + (cos w)2 gw = 0 1 gw = 4, since as w increases
is 2, but 0
=
.
from 0 to 4, the circle is traversed twice.
4. False.
If (u> ) = (1> ), then ({> |) = (31> 0), so tan31 (|@{) = tan31 0 = 0 6= . The statement is true for points in
quadrants I and IV.
5. True.
The curve u = 1 3 sin 2 is unchanged if we rotate it through 180 about R because
1 3 sin 2( + ) = 1 3 sin(2 + 2) = 1 3 sin 2. So it’s unchanged if we replace u by 3u. (See the discussion
after Example 8 in Section 10.3.) In other words, it’s the same curve as u = 3(1 3 sin 2) = sin 2 3 1.
6. True.
The polar equation u = 2, the Cartesian equation {2 + | 2 = 4, and the parametric equations { = 2 sin 3w,
| = 2 cos 3w [0 $ w $ 2] all describe the circle of radius 2 centered at the origin.
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CHAPTER 10
7. False.
PARAMETRIC EQUATIONS AND POLAR COORDINATES
The first pair of equations gives the portion of the parabola | = {2 with { D 0, whereas the second pair of equations
traces out the whole parabola | = {2 .
8. True.
9. True.
| 2 = 2| + 3{ C (| 3 1)2 = 3{ + 1 = 3 { + 13 = 4 34 { + 13 , which is the equation of a parabola with
vertex (3 13 > 1) and focus 3 13 + 34 > 1 , opening to the right.
By rotating and translating the parabola, we can assume it has an equation of the form | = f{2 , where f A 0.
The tangent at the point d> fd2 is the line | 3 fd2 = 2fd({ 3 d); i.e., | = 2fd{ 3 fd2 . This tangent meets
the parabola at the points {> f{2 where f{2 = 2fd{ 3 fd2 . This equation is equivalent to {2 = 2d{ 3 d2
[since f A 0]. But {2 = 2d{ 3 d2 C {2 3 2d{ + d2 = 0 C ({ 3 d)2 = 0 C { = d C
{> f{2 = d> fd2 . This shows that each tangent meets the parabola at exactly one point.
10. True.
hg
, where h A 1.
1 + h cos h cos hg cos =g
6= g.
The directrix is { = g, but along the hyperbola we have { = u cos =
1 + h cos 1 + h cos Consider a hyperbola with focus at the origin, oriented so that its polar equation is u =
1. { = w2 + 4w, | = 2 3 w, 34 $ w $ 1. w = 2 3 |, so
{ = (2 3 |)2 + 4(2 3 |) = 4 3 4| + | 2 + 8 3 4| = |2 3 8| + 12 C
{ + 4 = |2 3 8| + 16 = (| 3 4)2 . This is part of a parabola with vertex
(34> 4), opening to the right.
2. { = 1 + h2w , | = hw .
{ = 1 + h2w = 1 + (hw )2 = 1 + | 2 , | A 0.
3. | = sec =
1
1
= . Since 0 $ $ @2, 0 ? { $ 1 and | D 1.
cos {
This is part of the hyperbola | = 1@{.
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CHAPTER 10 REVIEW
4. { = 2 cos , | = 1 + sin , cos2 + sin2 = 1
{ 2
2
+ (| 3 1)2 = 1 i
¤
83
i
2
{
+ (| 3 1)2 = 1. This is an ellipse,
4
centered at (0> 1), with semimajor axis of length 2 and semiminor axis of
length 1.
5. Three different sets of parametric equations for the curve | =
(i) { = w, | =
I
{ are
I
w
(ii) { = w4 , | = w2
(iii) { = tan2 w, | = tan w, 0 $ w ? @2
There are many other sets of equations that also give this curve.
6. For w ? 31, { A 0 and | ? 0 with { decreasing and | increasing. When
w = 31, ({> |) = (0> 0). When 31 ? w ? 0, we have 31 ? { ? 0 and
0 ? | ? 1@2. When w = 0, ({> |) = (31> 0). When 0 ? w ? 1,
31 ? { ? 0 and 3 12 ? | ? 0. When w = 1, ({> |) = (0> 0) again.
When w A 1, both { and | are positive and increasing.
The Cartesian coordinates are { = 4 cos 2
= 4 3 12 = 32 and
3
I I
I 3
| = 4 sin 2
= 2 3, that is, the point 32> 2 3 .
=
4
3
2
7. (a)
(b) Given { = 33 and | = 3, we have u =
(33> 3) is in the second quadrant, =
I 11 I
3 2> 4 and 33 2> 7
.
4
8. 1 $ u ? 2,
6
$$
s
I
I
|
(33)2 + 32 = 18 = 3 2. Also, tan =
{
3
.
4
i tan =
I
Thus, one set of polar coordinates for (33> 3) is 3 2>
3
4
3
, and since
33
, and two others are
5
6
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CHAPTER 10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
9. u = 1 3 cos . This cardioid is
symmetric about the polar axis.
10. u = sin 4. This is an
eight-leaved rose.
11. u = cos 3. This is a
three-leaved rose. The curve is
traced twice.
12. u = 3 + cos 3. The curve is
symmetric about the horizontal
axis.
13. u = 1 + cos 2. The curve is
symmetric about the pole and
both the horizontal and vertical
axes.
14. u = 2 cos (@2) = The curve is
symmetric about the pole and
both the horizontal and vertical
axes.
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CHAPTER 10 REVIEW
15. u =
3
1 + 2 sin i h = 2 A 1, so the conic is a hyperbola. gh = 3 i
3
2
and the form “+2 sin ” imply that the directrix is above the focus at
.
the origin and has equation | = 32 . The vertices are 1> 2 and 33> 3
2
g=
16. u =
1@2
3@2
3
·
=
2 3 2 cos 1@2
1 3 1 cos parabola. gh =
3
2
i g=
3
2
i h = 1, so the conic is a
and the form “32 cos ” imply that the
directrix is to the left of the focus at the origin and has equation { = 3 32 .
The vertex is 34 > .
17. { + | = 2
C u cos + u sin = 2 C u(cos + sin ) = 2 C u =
18. {2 + | 2 = 2
i u2 = 2 i u =
2
cos + sin I
I
2. [u = 3 2 gives the same curve.]
19. u = (sin )@. As < ±", u < 0.
As < 0, u < 1. In the first figure,
there are an infinite number of
{-intercepts at { = q, q a nonzero
integer. These correspond to pole
points in the second figure.
20. u =
1@2
2
=
4 3 3 cos 1 3 34 cos the directrix is { = 3 23
i h=
3
4
and g = 23 . The equation of
i u = 32@(3 cos ). To obtain the equation
of the rotated ellipse, we replace in the original equation with 3
and get u =
2
4 3 3 cos 3
2
3
21. { = ln w, | = 1 + w2 ; w = 1.
2
,
3
.
g|
g{
1
g|
g|@gw
2w
= 2w and
= , so
=
=
= 2w2 .
gw
gw
w
g{
g{@gw
1@w
When w = 1, ({> |) = (0> 2) and g|@g{ = 2.
22. { = w3 + 6w + 1, | = 2w 3 w2 ;
w = 31.
g|
g|@gw
2 3 2w
g|
4
=
= 2
. When w = 31, ({> |) = (36> 33) and
= .
g{
g{@gw
3w + 6
g{
9
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CHAPTER 10
23. u = h3
PARAMETRIC EQUATIONS AND POLAR COORDINATES
i | = u sin = h3 sin and { = u cos = h3 cos gu
g
gu
g
g|
g|@g
=
=
g{
g{@g
When = ,
i
sin + u cos sin 3 cos 3h3 sin + h3 cos 3h
=
=
·
.
3h3 cos 3 h3 sin 3h
cos + sin cos 3 u sin 0 3 (31)
1
g|
=
=
= 31.
g{
31 + 0
31
24. u = 3 + cos 3
When = @2,
g|@g
g|
=
=
g{
g{@g
i
gu
g
gu
g
sin + u cos 33 sin 3 sin + (3 + cos 3) cos =
.
33 sin 3 cos 3 (3 + cos 3) sin cos 3 u sin (33)(31)(1) + (3 + 0) · 0
3
g|
=
=
= 31.
g{
(33)(31)(0) 3 (3 + 0) · 1
33
25. { = w + sin w, | = w 3 cos w
i
g|
g|@gw
1 + sin w
=
=
g{
g{@gw
1 + cos w
i
g g|
(1 + cos w) cos w 3 (1 + sin w)(3 sin w)
gw g{
1 + cos w + sin w
(1 + cos w)2
g 2|
cos w + cos2 w + sin w + sin2 w
=
=
=
=
2
g{
g{@gw
1 + cos w
(1 + cos w)3
(1 + cos w)3
26. { = 1 + w2 , | = w 3 w3 .
g|
g{
g|
g|@gw
1 3 3w2
= 1 3 3w2 and
= 2w, so
=
=
= 12 w31 3 32 w.
gw
gw
g{
g{@gw
2w
3 12 w32 3
g2 |
g(g|@g{)@gw
=
=
g{2
g{@gw
2w
3
2
= 3 14 w33 3 34 w31 = 3
3w2 + 1
1 2
1
+
3w
=
3
.
4w3
4w3
27. We graph the curve { = w3 3 3w, | = w2 + w + 1 for 32=2 $ w $ 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 w-value at which
g|@gw = 2w + 1 = 0 C w = 3 12 C ({> |) = 11
>3 =
8 4
28. We estimate the coordinates of the point of intersection to be (32> 3). In fact this is exact, since both w = 32 and w = 1 give
the point (32> 3). So the area enclosed by the loop is
U w=1
w=32
| g{ =
=
29. { = 2d cos w 3 d cos 2w
sin w = 0 or cos w =
1
2
2
,
3
or
4
.
3
32
3
i
5
(w2 + w + 1)(3w2 3 3) gw =
1
w5 + 34 w4 3 32 w2 3 3w
32
=
U1
3
5
32
+
(3w4 + 3w3 3 3w 3 3) gw
3
4
3
3
2
3 3 3 3 96
+ 12 3 6 3 (36) =
5
81
20
g{
= 32d sin w + 2d sin 2w = 2d sin w(2 cos w 3 1) = 0 C
gw
i w = 0,
| = 2d sin w 3 d sin 2w i
w = 0,
U1
3,
, or
5
3 .
g|
= 2d cos w 3 2d cos 2w = 2d 1 + cos w 3 2 cos2 w = 2d(1 3 cos w)(1 + 2 cos w) = 0 i
gw
Thus the graph has vertical tangents where w =
,
3
and
5
,
3
and horizontal tangents where w =
2
3
and
4
.
3
To determine
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CHAPTER 10 REVIEW
w<0
w
{
d
3
2
3
3
d
2
1
32d
3
d
2
I
3 3
d
2
3 12 d
3323d
|
0
33d
4
3
5
3
87
g|@gw
= 0, so there is a horizontal tangent there.
g{@gw
what the slope is where w = 0, we use l’Hospital’s Rule to evaluate lim
0
¤
3
d
2
I
0
I
3
I
3
d
2
30. From Exercise 29, { = 2d cos w 3 d cos 2w, | = 2d sin w 3 d sin 2w
i
U
D = 2 (2d sin w 3 d sin 2w)(32d sin w + 2d sin 2w) gw = 4d2 0 (2 sin2 w + sin2 2w 3 3 sin w sin 2w) gw
U
= 4d2 0 (1 3 cos 2w) + 12 (1 3 cos 4w) 3 6 sin2 w cos w gw = 4d2 w 3 12 sin 2w + 12 w 3 18 sin 4w 3 2 sin3 w 0
= 4d2 32 = 6d2
U0
31. The curve u2 = 9 cos 5 has 10 “petals.” For instance, for 3 10
$$
,
10
there are two petals, one with u A 0 and one
with u ? 0.
D = 10
32. u = 1 3 3 sin .
U @10
1 2
u
3@10 2
g = 5
U @10
3@10
9 cos 5 g = 5 · 9 · 2
U @10
0
@10
cos 5 g = 18 sin 5 0
= 18
The inner loop is traced out as goes from = sin31 13 to 3 , so
U @2
U @2 U 3 1 2
u g = (1 3 3 sin )2 g = D= 1 3 6 sin + 92 (1 3 cos 2) g
2
I
@2
= 11
+ 6 cos 3 94 sin 2 = 11
3 11
sin31 13 3 3 2
2
4
2
33. The curves intersect when 4 cos = 2
i cos = 12 i = ± 3
for 3 $ $ . The points of intersection are 2> 3 and 2> 3 3 .
34. The two curves clearly both contain the pole. For other points of intersection, cot = 2 cos( + 2q) or
32 cos( + + 2q), both of which reduce to cot = 2 cos cos = 0 or sin =
1
2
i =
5
6, 2, 6
or
3
2
35. The curves intersect where 2 sin = sin + cos sin = cos i=
,
4
C cos = 2 sin cos C cos (1 3 2 sin ) = 0 i
I I 11 i intersection points are 0> 2 ,
3> 6 , and
3> 6 .
i
and also at the origin (at which =
3
4
on the second curve).
U 3@4
U @4
D = 0 21 (2 sin )2 g + @4 12 (sin + cos )2 g
U 3@4
U @4
= 0 (1 3 cos 2) g + 12 @4 (1 + sin 2) g
@4 3@4
= 3 12 sin 2 0 + 12 3 14 cos 2 @4 = 12 ( 3 1)
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88
¤
CHAPTER 10
36. D = 2
=
U @6
1
3@2 2
PARAMETRIC EQUATIONS AND POLAR COORDINATES
(2 + cos 2)2 3 (2 + sin )2 g
U @6 4 cos 2 + cos2 2 3 4 sin 3 sin2 g
3@2
= 2 sin 2 + 12 +
=
51
16
1
8
I
3
sin 4 + 4 cos 3 12 +
1
4
sin 2
@6
3@2
37. { = 3w2 , | = 2w3 .
I
U2s
U2I
U2I
U2s
(g{@gw)2 + (g|@gw)2 gw = 0 (6w)2 + (6w2 )2 gw = 0 36w2 + 36w4 gw = 0 36w2 1 + w2 gw
0
I
U2
U2 I
U5
x = 1 + w2 , gx = 2w gw
= 0 6 |w| 1 + w2 gw = 6 0 w 1 + w2 gw = 6 1 x1@2 12 gx
k
l5
I
= 6 · 12 · 23 x3@2 = 2(53@2 3 1) = 2 5 5 3 1
O=
1
| = cosh 3w i (g{@gw)2 + (g|@gw)2 = 32 + (3 sinh 3w)2 = 9(1 + sinh2 3w) = 9 cosh2 3w, so
38. { = 2 + 3w,
O=
U1
U1
1
U1I
9 cosh2 3w gw = 0 |3 cosh 3w| gw = 0 3 cosh 3w gw = sinh 3w 0 = sinh 3 3 sinh 0 = sinh 3.
0
U 2 s
U 2 s
39. O = u2 + (gu@g)2 g = (1@)2 + (31@2 )2 g =
]
2
s
2 + 1
g
2
% s
&
I
I
I
2
s
2 + 1
2 + 1
42 + 1
2 + 42 + 1
24
2
I
+ ln + + 1
3
+ ln
= 3
=
2
+ 2 + 1
I
I
I
2 + 42 + 1
2 2 + 1 3 42 + 1
I
+ ln
=
2
+ 2 + 1
40. O =
=
Us
U t 6 1 2 + (gu@g)2 g = u
sin 3 + sin4 13 cos2 13 g
0
0
U
0
41. { = 4
V=
sin2
1 g = 12 3
3
3
2
sin 23 0 = 12 3
3
8
I
3
I
1
w3
+ 2, 1 $ w $ 4 i
w, | =
3
2w
U4
1
= 2
2|
s
U4 t I 2
(g{@gw)2 + (g|@gw)2 gw = 1 2 13 w3 + 12 w32
2@ w + (w2 3 w33 )2 gw
U 4 1
1
3
w3 + 12 w32
s
U4
(w2 + w33 )2 gw = 2 1 13 w5 +
5
6
1 6 5
4
+ 12 w35 gw = 2 18
w + 6 w 3 18 w34 1 =
471,295
1024
| = cosh 3w i (g{@gw)2 + (g|@gw)2 = 32 + (3 sinh 3w)2 = 9(1 + sinh2 3w) = 9 cosh2 3w, so
I
U1
U1
U1
U1
V = 0 2| gv = 0 2 cosh 3w 9 cosh2 3w gw = 0 2 cosh 3w |3 cosh 3w| gw = 0 2 cosh 3w · 3 cosh 3w gw
k
l1
U1
U1
= 6 0 cosh2 3w gw = 6 0 12 (1 + cosh 6w) gw = 3 w + 16 sinh 6w = 3 1 + 16 sinh 6 = 3 + 2 sinh 6
42. { = 2 + 3w,
0
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CHAPTER 10 REVIEW
43. For all f except 31, the curve is asymptotic to the line { = 1. For
f ? 31, the curve bulges to the right near | = 0. As f increases, the
bulge becomes smaller, until at f = 31 the curve is the straight line { = 1.
As f continues to increase, the curve bulges to the left, until at f = 0 there
is a cusp at the origin. For f A 0, there is a loop to the left of the origin,
whose size and roundness increase as f increases. Note that the {-intercept
of the curve is always 3f=
44. For d close to 0, the graph consists of four thin petals. As d increases, the petals get wider, until as d < ", each petal
occupies almost its entire quarter-circle.
45.
d = 0=01
d = 0=1
d=1
d=5
d = 10
d = 25
{2
|2
+
= 1 is an ellipse with center (0> 0).
9
8
I
d = 3, e = 2 2, f = 1 i
foci (±1> 0), vertices (±3> 0).
|2
{2
3
= 1 is a hyperbola
4
16
with center (0> 0), vertices (±2> 0), d = 2, e = 4,
I
I I
f = 16 + 4 = 2 5, foci ±2 5> 0 and
46. 4{2 3 | 2 = 16
C
asymptotes | = ±2{.
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89
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90
¤
CHAPTER 10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
47. 6| 2 + { 3 36| + 55 = 0
48. 25{2 + 4| 2 + 50{ 3 16| = 59
C
2
2
6(| 3 6| + 9) = 3({ + 1) C
2
(| 3 3) =
C
25({ + 1) + 4(| 3 2) = 100 C
3 16 ({
+ 1), a parabola with vertex (31> 3),
1
opening to the left, s = 3 24
i focus 3 25
> 3 and
24
directrix { =
2
3 23
24 .
1
({
4
+ 1)2 +
1
(|
25
3 2)2 = 1 is an ellipse centered at
(31> 2) with foci on the line { = 31, vertices (31> 7)
I
and (31> 33); d = 5, e = 2 i f = 21 i
I foci 31> 2 ± 21 .
49. The ellipse with foci (±4> 0) and vertices (±5> 0) has center (0> 0) and a horizontal major axis, with d = 5 and f = 4,
|2
{2
+
= 1.
25
9
so e2 = d2 3 f2 = 52 3 42 = 9. An equation is
50. The distance from the focus (2> 1) to the directrix { = 34 is 2 3 (34) = 6, so the distance from the focus to the vertex
is 12 (6) = 3 and the vertex is (31> 1). Since the focus is to the right of the vertex, s = 3. An equation is
(| 3 1)2 = 4 · 3[{ 3 (31)], or (| 3 1)2 = 12({ + 1).
51. The center of a hyperbola with foci (0> ±4) is (0> 0), so f = 4 and an equation is
The asymptote | = 3{ has slope 3, so
10e2 = 16 i e2 =
8
5
52. Center is (3> 0), and d =
3
d
=
e
1
and so d2 = 16 3
8
2
an equation of the ellipse is
8
5
i d = 3e and d2 + e2 = f2
=
= 4, f = 2 C e =
72
5 .
Thus, an equation is
|2
{2
3
= 1.
d2
e2
i (3e)2 + e2 = 42
i
{2
5| 2
5{2
|2
3
= 1, or
3
= 1.
72@5
8@5
72
8
I
I
42 3 22 = 12 i
({ 3 3)2
|2
+
= 1.
12
16
53. {2 = 3(| 3 100) has its vertex at (0> 100), so one of the vertices of the ellipse is (0> 100). Another form of the equation of a
parabola is {2 = 4s(| 3 100) so 4s(| 3 100) = 3(| 3 100) i 4s = 31 i s = 3 14 . Therefore the shared focus is
found at 0> 399
3 0 i f = 399
and the center of the ellipse is 0> 399
= 401
and
so 2f = 399
. So d = 100 3 399
4
4
8
8
8
8
2
2
399
399
|3 8
|3 8
4012 3 3992
{2
{2
e2 = d2 3 f2 =
+
=
25.
So
the
equation
of
the
ellipse
is
+
=
1
i
= 1,
401 2
82
e2
d2
25
8
2
2
or
54.
(8| 3 399)
{
+
= 1.
25
160,801
{2
|2
+ 2 =1 i
d2
e
condition with
2{
2| g|
=0 i
+ 2
d2
e g{
g|
e2 {
g|
e2 {
= 3 2 . Therefore
= p C | = 3 2 . Combining this
g{
d |
g{
d p
|2
d2 p
{2
I
+
=
1,
we
find
that
{
=
±
. In other words, the two points on the ellipse where the
d2
e2
d2 p2 + e2
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 10 REVIEW
¤
91
d2 p
e2
. The tangent lines at these points have the equations
tangent has slope p are ± I
>~I
d2 p2 + e2
d2 p2 + e2
I
e2
d2 p2
d2 p
e2
|± I
or | = p{ ~ I
=p {~ I
~I
= p{ ~ d2 p2 + e2 .
2
2
2
2
2
2
2
2
2
2
2
2
d p +e
d p +e
d p +e
d p +e
55. Directrix { = 4
i g = 4, so h =
1
3
i u=
4
hg
=
.
1 + h cos 3 + cos 56. See the end of the proof of Theorem 10.6.1. If h A 1, then 1 3 h2 ? 0 and Equations 10.6.4 become d2 =
h2 g 2
and
(h2 3 1)2
I
h2 g 2
e
e2
e
, so 2 = h2 3 1. The asymptotes | = ± { have slopes ± = ± h2 3 1, so the angles they make with the
2
h 31
d
d
d
I
31
31
2
polar axis are ± tan
h 3 1 = cos (±1@h).
e2 =
57. (a) If (d> e) lies on the curve, then there is some parameter value w1 such that
3w1
3w21
= e. If w1 = 0,
3 = d and
1 + w1
1 + w31
the point is (0> 0), which lies on the line | = {. If w1 6= 0, then the point corresponding to w =
{=
3(1@w1 )2
3w2
3w1
3(1@w1 )
= d. So (e> d) also lies on the curve. [Another way to see
= 3 1 = e, | =
= 3
3
1 + (1@w1 )
w1 + 1
1 + (1@w1 )3
w1 + 1
this is to do part (e) first; the result is immediate.] The curve intersects the line | = { when
w = w2
(b)
1
is given by
w1
i w = 0 or 1, so the points are (0> 0) and
3
2
3w2
3w
=
1 + w3
1 + w3
> 32 .
i
I
g|
(1 + w3 )(6w) 3 3w2 (3w2 )
6w 3 3w4
=
=
= 0 when 6w 3 3w4 = 3w(2 3 w3 ) = 0 i w = 0 or w = 3 2, so there are
3
2
gw
(1 + w )
(1 + w3 )2
I I horizontal tangents at (0> 0) and 3 2> 3 4 . Using the symmetry from part (a), we see that there are vertical tangents at
I I (0> 0) and 3 4> 3 2 .
(c) Notice that as w < 31+ , we have { < 3" and | < ". As w < 313 , we have { < " and | < 3". Also
| 3 (3{ 3 1) = | + { + 1 =
slant asymptote.
(d)
3w + 3w2 + (1 + w3 )
(w + 1)3
(w + 1)2
=
=
< 0 as w < 31. So | = 3{ 3 1 is a
1 + w3
1 + w3
w2 3 w + 1
(1 + w3 )(3) 3 3w(3w2 )
g|@gw
w(2 3 w3 )
3 3 6w3
g|
g|
g{
6w 3 3w4
=
and from part (b) we have
. So
.
=
=
=
=
3
2
3
2
3
2
gw
(1 + w )
(1 + w )
gw
(1 + w )
g{
g{@gw
1 3 2w3
g g|
2
gw g{
g |
2(1 + w3 )4
1
Also 2 =
=
A0 C w? I
.
3
g{
g{@gw
3(1 3 2w3 )3
2
So the curve is concave upward there and has a minimum point at (0> 0)
I I and a maximum point at 3 2> 3 4 . Using this together with the
information from parts (a), (b), and (c), we sketch the curve.
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¤
CHAPTER 10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
3 3
3w
3w2
27w3 + 27w6
27w3 (1 + w3 )
27w3
+
=
=
=
and
3
3
3
3
3
3
1+w
1+w
(1 + w )
(1 + w )
(1 + w3 )2
3w2
27w3
3w
=
3{| = 3
, so {3 + | 3 = 3{|.
3
3
1+w
1+w
(1 + w3 )2
(e) {3 + | 3 =
(f ) We start with the equation from part (e) and substitute { = u cos , | = u sin . Then {3 + | 3 = 3{|
u3 cos3 + u3 sin3 = 3u2 cos sin . For u 6= 0, this gives u =
i
3 cos sin . Dividing numerator and denominator
cos3 + sin3 sin 1
3
cos cos 3 sec tan 3
=
.
by cos , we obtain u =
1 + tan3 sin3 1+
cos3 (g) The loop corresponds to M 0> 2 , so its area is
2
]
]
]
u2
1 @2 3 sec tan 9 @2 sec2 tan2 9 " x2 gx
g =
g
=
g
=
2
2 0
1 + tan3 2 0
(1 + tan3 )2
2 0 (1 + x3 )2
0
e
= lim 92 3 13 (1 + x3 )31 0 = 32
D=
]
@2
[let x = tan ]
e<"
(h) By symmetry, the area between the folium and the line | = 3{ 3 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 12 , and since | = 3{ 3 1 i
1
, the area in the fourth quadrant is
sin + cos %
2 2 &
]
3 sec tan 1
1 3@4
CAS 1
3
3
g = . Therefore, the total area is
2 3@2
sin + cos 1 + tan3 2
u sin = 3u cos 3 1 i u = 3
1
2
+ 2 12 = 32 .
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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PROBLEMS PLUS
1. { =
]
w
1
cos x
gx, | =
x
]
w
1
sin x
g{
cos w
g|
sin w
gx, so by FTC1, we have
=
and
=
. Vertical tangent lines occur when
x
gw
w
gw
w
g{
= 0 C cos w = 0. The parameter value corresponding to ({> |) = (0, 0) is w = 1, so the nearest vertical tangent
gw
occurs when w =
2.
Therefore, the arc length between these points is
v
] @2 u 2
] @2
] @2 2 2
@2
g|
g{
cos w
gw sin2 w
= ln w 1 = ln 2
+
gw =
+
gw
=
O=
2
2
gw
gw
w
w
w
1
1
1
2. (a) The curve {4 + | 4 = {2 + | 2 is symmetric about both axes and about the line | = { (since interchanging {
and | does not change the equation) so we need only consider | D { D 0 to begin with. Implicit differentiation gives
4{3 + 4| 3 | 0 = 2{ + 2|| 0
|4 = |2
i |0 =
{(1 3 2{2 )
|(2| 2 3 1)
i | 0 = 0 when { = 0 and when { = ± I12 . If { = 0, then
i | 2 (| 2 3 1) = 0 i | = 0 or ±1. The point (0> 0) can’t be a highest or lowest point because it is
isolated. [If 31 ? { ? 1 and 31 ? | ? 1, then {4 ? {2 and | 4 ? | 2
If { =
I1 ,
2
I
1+ 2
2
I
I
+ | 4 = 12 + | 2 i 4|4 3 4| 2 3 1 = 0 i | 2 = 4 ± 816+16 = 1 ±2 2 .
t I i | = ± 12 1 + 2 . Near the point (0> 1), the denominator of | 0 is positive and the
then {2 = 12 , {4 = 14 , so
But | 2 A 0, so | 2 =
i {4 + | 4 ? {2 + | 2 , except for (0> 0).]
1
4
numerator changes from negative to positive as { increases through 0, so (0> 1) is a local minimum point. At
t I 1+ 2
I1 >
, | 0 changes from positive to negative, so that point gives a maximum. By symmetry, the highest points
2
2
t I t I on the curve are ± I12 > 1 +2 2 and the lowest points are ± I12 > 3 1 +2 2 .
(b) We use the information from part (a), together with symmetry with respect to the
axes and the lines | = ±{, to sketch the curve.
(c) In polar coordinates, {4 + | 4 = {2 + | 2 becomes u4 cos4 + u4 sin4 = u2 or
1
. By the symmetry shown in part (b), the area enclosed by
cos4 + sin4 ] @4
] @4
1 2
g
CAS I
u g = 4
the curve is D = 8
= 2 .
4
4
2
cos + sin 0
0
u2 =
3. In terms of { and |, we have { = u cos = (1 + f sin ) cos = cos + f sin cos = cos + 12 f sin 2 and
| = u sin = (1 + f sin ) sin = sin + f sin2 . Now 31 $ sin $ 1 i 31 $ sin + f sin2 $ 1 + f $ 2, so
31 $ | $ 2. Furthermore, | = 2 when f = 1 and =
,
2
while | = 31 for f = 0 and =
3
.
2
Therefore, we need a viewing
rectangle with 31 $ | $ 2.
To find the {-values, look at the equation { = cos + 12 f sin 2 and use the fact that sin 2 D 0 for 0 $ $
2
and
sin 2 $ 0 for 3 2 $ $ 0. [Because u = 1 + f sin is symmetric about the |-axis, we only need to consider
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CHAPTER 10 PROBLEMS PLUS
3 2 $ $
So for 3 2 $ $ 0, { has a maximum value when f = 0 and then { = cos has a maximum value
of 1 at = 0. Thus, the maximum value of { must occur on 0> 2 with f = 1. Then { = cos + 12 sin 2 i
g{
g
2 .]
= 3 sin + cos 2 = 3 sin + 1 3 2 sin2 i
g{
g
= 3(2 sin 3 1)(sin + 1) = 0 when sin = 31 or
1
2
[but sin 6= 31 for 0 $ $ 2 ]. If sin = 12 , then = 6 and
I
I
{ = cos 6 + 12 sin 3 = 34 3. Thus, the maximum value of { is 34 3, and,
I
by symmetry, the minimum value is 3 34 3. Therefore, the smallest
viewing rectangle that contains every member of the family of polar curves
I I
u = 1 + f sin , where 0 $ f $ 1, is 3 34 3> 34 3 × [31> 2].
4. (a) Let us find the polar equation of the path of the bug that starts in the upper
right corner of the square. If the polar coordinates of this bug, at a
particular moment, are (u> ), then the polar coordinates of the bug that it is
crawling toward must be u> + 2 . (The next bug must be the same
distance from the origin and the angle between the lines joining the bugs to
the pole must be
.)
2
The Cartesian coordinates of the first bug are
(u cos > u sin ) and for the second bug we have
{ = u cos + 2 = 3u sin , | = u sin + 2 = u cos . So the slope of the line joining the bugs is
sin 3 cos u cos 3 u sin =
. This must be equal to the slope of the tangent line at (u> ), so by
3u sin 3 u cos sin + cos Equation 10.3.3 we have
sin 3 cos gu
(gu@g) sin + u cos =
. Solving for
, we get
(gu@g) cos 3 u sin sin + cos g
gu
gu
gu
gu
sin2 +
sin cos + u sin cos + u cos2 =
sin cos 3
cos2 3 u sin2 + u sin cos i
g
g
g
g
gu 2
gu
sin + cos2 + u cos2 + sin2 = 0 i
= 3u. Solving this differential equation as a separable
g
g
equation (as in Section 9.3), or using Theorem 9.4.2 with n = 31, we get u = Fh3 . To determine F we use the fact that,
at its starting position, =
bug’s path is u =
4
and u =
I1 dh@4 h3
2
or u =
1
I
d,
2
so
I1 d
2
= Fh3@4
i F=
I1 dh@4 .
2
Therefore, a polar equation of the
1
I
dh(@4)3 .
2
(b) The distance traveled by this bug is O =
U" s
gu
d
u2 + (gu@g)2 g, where
= I h@4 (3h3 ) and so
@4
g
2
u2 + (gu@g)2 = 12 d2 h@2 h32 + 12 d2 h@2 h32 = d2 h@2 h32 . Thus
O=
U"
@4
Uw
w<" @4
dh@4 h3 g = dh@4 lim
= dh@4
w
h3 g = dh@4 lim 3h3 @4
w<"
k
l
lim h3@4 3 h3w = dh@4 h3@4 = d
w<"
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CHAPTER 10 PROBLEMS PLUS
5. Without loss of generality, assume the hyperbola has equation
¤
95
{2
|2
3 2 = 1. Use implicit differentiation to get
2
d
e
2{ 2| |0
e2 {
e2 f
3 2 = 0, so |0 = 2 . The tangent line at the point (f> g) on the hyperbola has equation | 3 g = 2 ({ 3 f).
2
d
e
d |
d g
The tangent line intersects the asymptote | =
e
e2 f
e
{ when { 3 g = 2 ({ 3 f) i deg{ 3 d2 g2 = e2 f{ 3 e2 f2
d
d
d g
i
d2 g2 3 e2 f2
dg + ef
e dg + ef
dg + ef
=
and the |-value is
=
.
e(dg 3 ef)
e
d
e
d
ef 3 dg dg 3 ef
e
>
. The midpoint of these intersection points is
Similarly, the tangent line intersects | = 3 { at
d
e
d
1 dg + ef
ef 3 dg
1 dg + ef
dg 3 ef
1 2ef 1 2dg
+
>
+
=
>
= (f> g), the point of tangency.
2
e
e
2
d
d
2 e 2 d
deg{ 3 e2 f{ = d2 g2 3 e2 f2
i {=
Note: If | = 0, then at (±d> 0), the tangent line is { = ±d, and the points of intersection are clearly equidistant from the point
of tangency.
6. (a) Since the smaller circle rolls without slipping around F, the amount of arc
traversed on F (2u in the figure) must equal the amount of arc of the smaller
circle that has been in contact with F. Since the smaller circle has radius u,
it must have turned through an angle of 2u@u = 2. In addition to turning
through an angle 2, the little circle has rolled through an angle against F.
Thus, S has turned through an angle of 3 as shown in the figure. (If the little
circle had turned through an angle of 2 with its center pinned to the {-axis,
then S would have turned only 2 instead of 3. The movement of the little circle around F adds to the angle.) From the
figure, we see that the center of the small circle has coordinates (3u cos > 3u sin ). Thus, S has coordinates ({> |), where
{ = e cos 3 + 3u cos and | = e sin 3 + 3u sin .
(b)
e = 15 u
e = 25 u
e = 35 u
e = 45 u
(c) The diagram gives an alternate description of
point S on the epitrochoid. T moves around
a circle of radius e, and S rotates one-third as
fast with respect to T at a distance of 3u.
Place an equilateral triangle with sides of
I
length 3 3u so that its centroid is at T and
one vertex is at S . (The distance from the centroid to a vertex is
I1
3
times the length of a side of the equilateral triangle.)
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CHAPTER 10 PROBLEMS PLUS
As increases by
2
3 ,
the point T travels once around the circle of radius e, returning to its original position. At the
same time, S (and the rest of the triangle) rotate through an angle of
2
3
about T, so S ’s position is occupied by another
vertex. In this way, we see that the epitrochoid traced out by S is simultaneously traced out by the other two vertices as
well. The whole equilateral triangle sits inside the epitrochoid (touching it only with its vertices) and each vertex traces out
the curve once while the centroid moves around the circle three times.
(d) We view the epitrochoid as being traced out in the same way as in part (c), by a rotor for which the distance from its center
to each vertex is 3u, so it has radius 6u. To show that the rotor fits inside the epitrochoid, it suffices to show that for any
position of the tracing point S , there are no points on the opposite side of the rotor which are outside the epitrochoid. But
I the most likely case of intersection is when S is on the |-axis, so as long as the diameter of the rotor which is 3 3 u is
less than the distance between the |-intercepts, the rotor will fit. The |-intercepts occur when =
2
or =
3
2
i
| = 3e + 3u or | = e 3 3u, so the distance between the intercepts is (3e + 3u) 3 (e 3 3u) = 6u 3 2e, and the rotor will
I I
I
fit if 3 3 u $ 6u 3 2e C 2e $ 6u 3 3 3 u C e $ 32 2 3 3 u.
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11
INFINITE SEQUENCES AND SERIES
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 dq approach 8 as q becomes large. In fact, we can make dq as close to 8 as we like by taking q sufficiently
large.
(c) The terms dq become large as q becomes large. In fact, we can make dq as large as we like by taking q sufficiently large.
2. (a) From Definition 1, a convergent sequence is a sequence for which lim dq exists. Examples: {1@q}, {1@2q }
q<"
(b) A divergent sequence is a sequence for which lim dq does not exist. Examples: {q}, {sin q}
q<"
3. dq =
2q
, so the sequence is
q2 + 1
2
4
6
8
10
4 3 8 5
>
>
>
>
> = = = = 1> > > > > = = = .
1 + 1 4 + 1 9 + 1 16 + 1 25 + 1
5 5 17 13
4. dq =
3q
, so the sequence is
1 + 2q
9
27
81
243
9
81 81
3
>
>
>
>
> = = = = 1> > 3> > > = = = .
1 + 2 1 + 4 1 + 8 1 + 16 1 + 32
5
17 11
5. dq =
(31)q31
, so the sequence is
5q
6. dq = cos
7. dq =
q
, so the sequence is
2
1 31 1 31 1
>
> >
> >===
51 52 53 54 55
=
1 1
1
1
1
>3 >
>3
>
>=== .
5
25 125
625 3125
3
5
cos > cos > cos
> cos 2> cos
> = = = = {0> 31> 0> 1> 0> = = =}.
2
2
2
1
, so the sequence is
(q + 1)!
1 1 1 1 1
> > > > >===
2! 3! 4! 5! 6!
=
1 1 1 1
1
> > >
>
>=== .
2 6 24 120 720
(31)q q
(31)1 1
31
, so d1 =
=
, and the sequence is
q! + 1
1! + 1
2
2
33
4
35
1 2
3 4
5
31
>
>
>
>
>=== = 3 > >3 > >3
>=== .
2 2 + 1 6 + 1 24 + 1 120 + 1
2 3
7 25
121
8. dq =
9. d1 = 1, dq+1 = 5dq 3 3.
Each term is defined in terms of the preceding term. d2 = 5d1 3 3 = 5(1) 3 3 = 2.
d3 = 5d2 3 3 = 5(2) 3 3 = 7. d4 = 5d3 3 3 = 5(7) 3 3 = 32. d5 = 5d4 3 3 = 5(32) 3 3 = 157.
The sequence is {1> 2> 7> 32> 157> = = =}.
10. d1 = 6, dq+1 =
6
6
3
1
dq
d1
d2
d3
d4
. d2 =
= = 6. d3 =
= = 3. d4 =
= = 1. d5 =
= .
q
1
1
2
2
3
3
4
4
The sequence is 6> 6> 3> 1> 14 > = = = .
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CHAPTER 11
11. d1 = 2, dq+1 =
d5 =
INFINITE SEQUENCES AND SERIES
2
2
2
dq
d1
2
d2
2@3
d3
2@5
= . d3 =
= . d4 =
= .
. d2 =
=
=
=
1 + dq
1 + d1
1+2
3
1 + d2
1 + 2@3
5
1 + d3
1 + 2@5
7
2
d4
2@7
= . The sequence is 2> 23 > 25 > 27 > 29 > = = = .
=
1 + d4
1 + 2@7
9
12. d1 = 2, d2 = 1, dq+1 = dq 3 dq31 .
Each term is defined in term of the two preceding terms.
d3 = d2 3 d1 = 1 3 2 = 31. d4 = d3 3 d2 = 31 3 1 = 32. d5 = d4 3 d3 = 32 3 (31) = 31.
d6 = d5 3 d4 = 31 3 (32) = 1. The sequence is {2> 1> 31> 32> 31> 1> = = =}.
13.
14.
15.
1> 13 > 15 > 17 > 19 > = = = . The denominator of the nth term is the nth positive odd integer, so dq =
q31
1
1
> 81
> = = = . Each term is 3 13 times the preceding term, so dq = 3 13
.
1> 3 13 > 19 > 3 27
q31
> = = = . The first term is 33 and each term is 3 23 times the preceding one, so dq = 33 3 23
.
33> 2> 3 43 > 89 > 3 16
27
16. {5> 8> 11> 14> 17> = = =}.
17.
Each term is larger than the preceding term by 3, so dq = d1 + g(q 3 1) = 5 + 3(q 3 1) = 3q + 2.
1
> 3 43 > 94 > 3 16
> 25
>===
2
5
6
we get dq = (31)q+1
. The numerator of the nth term is q2 and its denominator is q + 1. Including the alternating signs,
q2
.
q+1
18. {1> 0> 31> 0> 1> 0> 31> 0> = = =}.
19.
1
.
2q 3 1
q
dq =
Two possibilities are dq = sin
q
(q 3 1)
and dq = cos
.
2
2
3q
1 + 6q
1
0=4286
2
0=4615
3
0=4737
4
0=4800
5
0=4839
6
0=4865
7
0=4884
8
0=4898
9
0=4909
10
0=4918
It appears that lim dq = 0=5.
q<"
lim
q<"
3q
(3q)@q
3
3
1
= lim
= lim
= =
1 + 6q q<" (1 + 6q)@q q<" 1@q + 6
6
2
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SECTION 11.1
20.
q
21.
dq = 2 +
1=0000
2
2=5000
3
1=6667
4
2=2500
5
1=8000
6
2=1667
7
1=8571
8
2=1250
9
1=8889
10
2=1000
It appears that lim dq = 2.
q<"
(31)q
1
(31)q
= lim 2 + lim
= 2 + 0 = 2 since lim
=0
lim 2 +
q<"
q<"
q<"
q<" q
q
q
(31)q
= 0.
q<"
q
and by Theorem 6, lim
1
q
dq = 1 + 3 12
2
1=2500
3
0=8750
4
1=0625
5
0=9688
6
1=0156
7
0=9922
8
1=0039
9
0=9980
10
1=0010
0=5000
22.
q
dq = 1 +
It appears that lim dq = 1.
q<"
q q
= lim 1 + lim 3 12 = 1 + 0 = 1 since
lim 1 + 3 12
q<"
q<"
q<"
q
lim 3 12 = 0 by (9).
q<"
10q
9q
1
2=1111
2
2=2346
3
2=3717
4
2=5242
5
2=6935
6
2=8817
7
3=0908
8
3=3231
9
3=5812
10
3=8680
It appears that the sequence does not have a limit.
q
10
10q
lim
= lim
, which diverges by (9) since
q<" 9q
q<"
9
23. dq = 1 3 (0=2)q , so lim dq = 1 3 0 = 1 by (9) .
q<"
24. dq =
¤
(31)q
q
1
q
SEQUENCES
10
9
A 1.
Converges
q3@q3
q3
1
1
=
= 1 as q < ". Converges
=
, so dq <
q3 + 1
(q3 + 1)@q3
1 + 1@q3
1+0
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¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
25. dq =
3 + 5q2
(3 + 5q2 )@q2
5 + 3@q2
5+0
, so dq <
= 5 as q < ". Converges
=
=
2
2
2
q+q
(q + q )@q
1 + 1@q
1+0
26. dq =
q3@q
q2
q3
=
=
, so dq < " as q < " since lim q2 = " and lim (1 + 1@q2 ) = 1. Diverges
q<"
q<"
q+1
(q + 1)@q
1 + 1@q2
27. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write
lim dq = lim h1@q = hlimq$4 (1@q) = h0 = 1= Converges
q<"
28. dq =
q<"
q
q
3q+2
32 3q
= q = 9 35 , so lim dq = 9 lim 35 = 9 · 0 = 0 by (9) with u = 35 . Converges
q
q<"
q<"
5
5
2q
(2q)@q
2
2
, then lim eq = lim
= lim
=
= . Since tan is continuous at
q<"
q<" (1 + 8q)@q
q<" 1@q + 8
1 + 8q
8
4
2q
2q
Theorem 7, lim tan
= tan lim
= tan = 1. Converges
q<"
q<" 1 + 8q
1 + 8q
4
29. If eq =
,
4
by
30. Using the last limit law for sequences and the continuity of the square root function,
lim dq = lim
q<"
q<"
u
q+1
=
9q + 1
u
q+1
lim
=
q<" 9q + 1
v
1 + 1@q
=
q<" 9 + 1@q
lim
u
1
1
= . Converges
9
3
I
I
I
q2 @ q3
q2
q
I
= I
= s
, so dq < " as q < " since lim q = " and
3
2
3
3
q<"
q + 4q
1 + 4@q
q + 4q@ q
31. dq = I
lim
q<"
s
1 + 4@q2 = 1. Diverges
32. If eq =
(2q)@q
2
2
2q
, then lim eq = lim
= lim
= = 2. Since the natural exponential function is
q<"
q<" (q + 2)@q
q<" 1 + 2@q
q+2
1
continuous at 2, by Theorem 7, lim h2q@(q+2) = hlimq$4 eq = h2 . Converges
q<"
q
(31)
1
1
1
33. lim |dq | = lim I =
= (0) = 0, so lim dq = 0 by (6).
lim
q<"
q<" 2 q
q<"
2 q<" q1@2
2
Converges
(31)q+1 q
q
q@q
1
1
I = lim
I
I =
I has odd-numbered terms
= lim
= 1. Thus, dq =
q<" q +
1+0
q q<" (q + q )@q q<" 1 + 1@ q
q+ q
34. lim
that approach 1 and even-numbered terms that approach 31 as q < ", and hence, the sequence {dq } is divergent.
35. dq = cos( q@2).
This sequence diverges since the terms don’t approach any particular real number as q < ".
The terms take on values between 31 and 1=
36. dq = cos(2@q).
As q < ", 2@q < 0, so cos(2@q) < cos 0 = 1 because cos is continuous. Converges
37. dq =
(2q 3 1)!
(2q 3 1)!
1
=
=
< 0 as q < ". Converges
(2q + 1)!
(2q + 1)(2q)(2q 3 1)!
(2q + 1)(2q)
38. dq =
ln q
ln q
=
=
ln 2q
ln 2 + ln q
1
1
<
= 1 as q < ". Converges
0+1
+1
ln 2
ln q
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SECTION 11.1
SEQUENCES
39. dq =
hq + h3q h3q
1 + h32q
· 3q = q
< 0 as q < " because 1 + h32q < 1 and hq 3 h3q < ". Converges
2q
h 31 h
h 3 h3q
40. dq =
tan31 q
.
q
41. dq = q2 h3q =
lim tan31 q = lim tan31 { =
q<"
{<"
¤
101
by (3), so lim dq = 0. Converges
q<"
2
{2 H
2{ H
2
q2
. Since lim { = lim { = lim { = 0, it follows from Theorem 3 that lim dq = 0. Converges
q
{<" h
{<" h
{<" h
q<"
h
1
q+1
= ln 1 +
< ln (1) = 0 as q < " because ln is continuous. Converges
42. dq = ln(q + 1) 3 ln q = ln
q
q
cos2 q
1
43. 0 $
$ q
2q
2
44. dq =
1
[since 0 $ cos q $ 1], so since lim q = 0,
q<" 2
2
cos2 q
2q
converges to 0 by the Squeeze Theorem.
I
q 1+3q
2
= (21+3q )1@q = (21 23q )1@q = 21@q 23 = 8 · 21@q , so
lim dq = 8 lim 21@q = 8 · 2limq$4 (1@q) = 8 · 20 = 8 by Theorem 7, since the function i ({) = 2{ is continuous at 0.
q<"
q<"
Converges
45. dq = q sin(1@q) =
sin(1@{)
sin w
sin(1@q)
. Since lim
= lim
[where w = 1@{] = 1, it follows from Theorem 3
{<"
1@q
1@{
w
w<0+
that {dq } converges to 1.
46. dq = 23q cos q.
2
, so
i ln | = { ln 1 +
{
1
2
3 2
1 + 2@{
{
ln(1 + 2@{) H
2
lim ln | = lim
= lim
= lim
=2 i
{<"
{<"
{<"
{<" 1 + 2@{
1@{
31@{2
{
q
2
2
= lim hln | = h2 , so by Theorem 3, lim 1 +
= h2 . Converges
lim 1 +
{<"
{<"
q<"
{
q
47. | =
{
2
1+
{
q
cos q 1
1
0$ q $ q =
, so lim |dq | = 0 by (9), and lim dq = 0 by (6)= Converges
q<"
q<"
2
2
2
48. dq =
1
sin 2q
1
1
31
I . |dq | $
I and lim
I = 0, so
I $ dq $
I
q<"
1+ q
1+ q
1+ q
1+ q
1+ q
i
lim dq = 0 by the
q<"
Squeeze Theorem. Converges
49. dq = ln(2q2 + 1) 3 ln(q2 + 1) = ln
50. lim
{<"
2q2 + 1
q2 + 1
= ln
2 + 1@q2
1 + 1@q2
< ln 2 as q < ". Converges
(ln {)2 H
2(ln {)(1@{)
ln { H
1@{
(ln q)2
= 2 lim
= 0, so by Theorem 3, lim
= 0. Converges
= lim
= 2 lim
{<"
{<" {
{<" 1
q<"
{
1
q
51. dq = arctan(ln q). Let i ({) = arctan(ln {). Then lim i ({) =
{<"
Thus, lim dq = lim i (q) =
q<"
q<"
.
2
2
since ln { < " as { < " and arctan is continuous.
Converges
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¤
CHAPTER 11
52. dq = q 3
=
INFINITE SEQUENCES AND SERIES
I
I
I
q3
q + 1 q + 3 = q 3 q2 + 4q + 3 =
I
I
q2 + 4q + 3 q + q2 + 4q + 3
I
·
1
q + q2 + 4q + 3
q2 3 (q2 + 4q + 3)
34q 3 3
(34q 3 3)@q
34 3 3@q
s
I
I
I
,
=
= =
2
2
2
q + q + 4q + 3
q + q + 4q + 3
q + q + 4q + 3 @q
1 + 1 + 4@q + 3@q2
so lim dq =
q<"
34
34 3 0
I
=
= 32. Converges
2
1+ 1+0+0
53. {0> 1> 0> 0> 1> 0> 0> 0> 1> = = =} diverges since the sequence takes on only two values, 0 and 1, and never stays arbitrarily close to
either one (or any other value) for q sufficiently large.
54.
1 1 1 1 1 1 1 1
> > > > > > > >===
1 3 2 4 3 5 4 6
lim d2q31 = lim
q<"
q<"
. d2q31 =
1
1
and d2q =
for all positive integers q. lim dq = 0 since
q<"
q
q+2
1
1
= 0 and lim d2q = lim
= 0. For q sufficiently large, dq can be made as close to 0
q<"
q<" q + 2
q
as we like. Converges
55. dq =
(q 3 1) q
1 q
q!
1 2 3
· D ·
= · · · ··· ·
2q
2 2 2
2
2
2 2
56. 0 ? |dq | =
[for q A 1] =
3 3 3
3
3
3 3 3
3q
= · · · ··· ·
· $ · ·
q!
1 2 3
(q 3 1) q
1 2 q
q
< " as q < ", so {dq } diverges.
4
[for q A 2] =
27
< 0 as q < ", so by the Squeeze
2q
Theorem and Theorem 6, {(33)q@q!} converges to 0.
57.
From the graph, it appears that the sequence converges to 1.
{(32@h)q } converges to 0 by (7), and hence {1 + (32@h)q }
converges to 1 + 0 = 1.
58.
From the graph, it appears that the sequence converges to a number
greater than 3.
I sin @ q
I
I
lim dq = lim q sin I
= lim
·
q<"
q<"
q<"
q
@ q
k
I l
sin {
= lim
·
{ = @ q = 1 · = =
{
{<0+
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SECTION 11.1
59.
¤
SEQUENCES
From the graph, it appears that the sequence converges to 12 .
As q < ",
v
u
3 + 2q2
3@q2 + 2
dq =
=
2
8q + q
8 + 1@q
i
u
0+2
=
8+0
u
1
1
= ,
4
2
so lim dq = 12 .
q<"
60.
From the graph, it appears that the sequence converges to 5=
I
I
I
I
I
q q
q
q
5=
5 $ q 3q + 5q $ q 5q + 5q = 2 5q
k
l
I
q
= 2 · 5 < 5 as q < "
lim 21@q = 20 = 1
q<"
Hence, dq < 5 by the Squeeze Theorem.
Alternate solution: Let | = (3{ + 5{ )1@{ . Then
ln (3{ + 5{ ) H
3{ ln 3 + 5{ ln 5
= lim
= lim
lim ln | = lim
{<"
{<"
{<"
{<"
{
3{ + 5{
I
so lim | = hln 5 = 5, and so q 3q + 5q converges to 5.
3 {
ln 3 + ln 5
5
3 {
= ln 5,
+1
5
{<"
61.
From the graph, it appears that the sequence {dq } =
q2 cos q
1 + q2
is
divergent, since it oscillates between 1 and 31 (approximately). To
prove this, suppose that {dq } converges to O. If eq =
{eq } converges to 1, and lim
q<"
lim
q<"
q2
, then
1 + q2
dq
dq
O
= O. But
=
= cos q, so
eq
1
eq
dq
does not exist. This contradiction shows that {dq } diverges.
eq
62.
From the graphs, it seems that the sequence diverges. dq =
dq D
1 · 3 · 5 · · · · · (2q 3 1)
. We first prove by induction that
q!
q31
3
for all q. This is clearly true for q = 1, so let S (q) be the statement that the above is true for q. We must
2
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104
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
q31
3
2q + 1
3
2q + 1
(induction hypothesis). But
D
·
2
q+1
q+1
2
3 q31 3 3 q
· 2 = 2 which
[since 2 (2q + 1) D 3 (q + 1) C 4q + 2 D 3q + 3 C q D 1], and so we get that dq+1 D 2
q r
3 q31
diverges [by (9)], so does the given sequence {dq }.
is S (q + 1). Thus, we have proved our first assertion, so since
2
show it is then true for q + 1. dq+1 = dq ·
63.
2q + 1
D
q+1
From the graph, it appears that the sequence approaches 0.
0 ? dq =
3
5
2q 3 1
1 · 3 · 5 · · · · · (2q 3 1)
1
·
·
· ··· ·
=
(2q)q
2q 2q 2q
2q
1
1
· (1) · (1) · · · · · (1) =
< 0 as q < "
2q
2q
1 · 3 · 5 · · · · · (2q 3 1)
converges to 0.
So by the Squeeze Theorem,
q
(2q)
$
64. (a) d1 = 1, dq+1 = 4 3 dq for q D 1.
d1 = 1, d2 = 4 3 d1 = 4 3 1 = 3, d3 = 4 3 d2 = 4 3 3 = 1,
d4 = 4 3 d3 = 4 3 1 = 3, d5 = 4 3 d4 = 4 3 3 = 1. Since the terms of the sequence alternate between 1 and 3,
the sequence is divergent.
(b) d1 = 2, d2 = 4 3 d1 = 4 3 2 = 2, d3 = 4 3 d2 = 4 3 2 = 2. Since all of the terms are 2, lim dq = 2 and hence, the
q<"
sequence is convergent.
65. (a) dq = 1000(1=06)q
i d1 = 1060, d2 = 1123=60, d3 = 1191=02, d4 = 1262=48, and d5 = 1338=23.
(b) lim dq = 1000 lim (1=06)q , so the sequence diverges by (9) with u = 1=06 A 1.
q<"
q<"
66. (a) Substitute 1 to 6 for q in Lq = 100
1=0025q 3 1
3 q to get L1 = $0, L2 = $0=25, L3 = $0=75, L4 = $1=50,
0=0025
L5 = $2=51, and L6 = $3=76.
(b) For two years, use 2 · 12 = 24 for q to get $70=28.
67. (a) We are given that the initial population is 5000, so S0 = 5000. The number of catfish increases by 8% per month and is
decreased by 300 per month, so S1 = S0 + 8%S0 3 300 = 1=08S0 3 300, S2 = 1=08S1 3 300, and so on. Thus,
Sq = 1=08Sq31 3 300.
(b) Using the recursive formula with S0 = 5000, we get S1 = 5100, S2 = 5208, S3 = 5325 (rounding any portion of a
catfish), S4 = 5451, S5 = 5587, and S6 = 5734, which is the number of catfish in the pond after six months.
68. dq+1 =
+1
2
dq
if dq is an even number
3dq + 1 if dq is an odd number
When d1 = 11, the first 40 terms are 11, 34, 17, 52, 26, 13, 40, 20, 10, 5,
16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4. When d1 = 25, the first 40 terms are 25, 76, 38,
19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4.
The famous Collatz conjecture is that this sequence always reaches 1, regardless of the starting point d1 .
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SECTION 11.1
¤
SEQUENCES
105
69. If |u| D 1, then {uq } diverges by (9), so {quq } diverges also, since |quq | = q |uq | D |uq |. If |u| ? 1 then
{
lim {u{ = lim
{<" u 3{
{<"
H
1
u{
= 0, so lim quq = 0, and hence {quq } converges
= lim
3{
{<" (3 ln u) u
{<" 3 ln u
q<"
= lim
whenever |u| ? 1.
70. (a) Let lim dq = O. By Definition 2, this means that for every % A 0 there is an integer Q such that |dq 3 O| ? %
q<"
whenever q A Q. Thus, |dq+1 3 O| ? % whenever q + 1 A Q
C q A Q 3 1. It follows that lim dq+1 = O and so
q<"
lim dq = lim dq+1 .
q<"
q<"
(b) If O = lim dq then lim dq+1 = O also, so O must satisfy O = 1@ (1 + O) i O2 + O 3 1 = 0 i O =
q<"
q<"
31 +
2
I
5
(since O has to be nonnegative if it exists).
71. Since {dq } is a decreasing sequence, dq A dq+1 for all q D 1. Because all of its terms lie between 5 and 8, {dq } is a
bounded sequence. By the Monotonic Sequence Theorem, {dq } is convergent; that is, {dq } has a limit O. O must be less than
8 since {dq } is decreasing, so 5 $ O ? 8.
72. The terms of dq = (32)q+1 alternate in sign, so the sequence is not monotonic. The first five terms are 4, 38, 16, 332,
and 64. Since lim |dq | = lim 2q+1 = ", the sequence is not bounded.
q<"
73. dq =
1
1
1
1
is decreasing since dq+1 =
=
?
= dq for each q D 1. The sequence is
2q + 3
2(q + 1) + 3
2q + 5
2q + 3
bounded since 0 ? dq $
74. dq =
q<"
1
5
for all q D 1. Note that d1 = 15 .
2{ 3 3
2q 3 3
defines an increasing sequence since for i ({) =
,
3q + 4
3{ + 4
i 0 ({) =
(3{ + 4)(2) 3 (2{ 3 3)(3)
17
=
A 0. The sequence is bounded since dq D d1 = 3 17 for q D 1,
(3{ + 4)2
(3{ + 4)2
and dq ?
2q
2
2q 3 3
?
= for q D 1.
3q
3q
3
75. The terms of dq = q(31)q alternate in sign, so the sequence is not monotonic. The first five terms are 31, 2, 33, 4, and 35.
Since lim |dq | = lim q = ", the sequence is not bounded.
q<"
q<"
76. dq = qh3q defines a positive decreasing sequence since the function i ({) = {h3{ is decreasing for { A 1.
[i 0 ({) = h3{ 3 {h3{ = h3{ (1 3 {) ? 0 for { A 1.] The sequence is bounded above by d1 =
77. dq =
q2
1
h
and below by 0.
{
q
({2 + 1)(1) 3 {(2{)
1 3 {2
defines a decreasing sequence since for i ({) = 2
, i 0 ({) =
= 2
$0
2
2
+1
{ +1
({ + 1)
({ + 1)2
for { D 1. The sequence is bounded since 0 ? dq $
78. dq = q +
1
2
for all q D 1.
1
1
defines an increasing sequence since the function j({) = { + is increasing for { A 1. [j 0 ({) = 1 3 1@{2 A 0
q
{
for { A 1.] The sequence is unbounded since dq < " as q < ". (It is, however, bounded below by d1 = 2.)
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106
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
t s
I s I
I
q
q
q
79. For
2, 2 2, 2 2 2, = = = , d1 = 21@2 , d2 = 23@4 , d3 = 27@8 , = = =, so dq = 2(2 31)@2 = 213(1@2 ) .
q)
lim dq = lim 213(1@2
q<"
q<"
= 21 = 2.
Alternate solution: Let O = lim dq . (We could show the limit exists by showing that {dq } is bounded and increasing.)
q<"
Then O must satisfy O =
I
2 · O i O2 = 2O i O(O 3 2) = 0. O 6= 0 since the sequence increases, so O = 2.
80. (a) Let Sq be the statement that dq+1 D dq and dq $ 3. S1 is obviously true. We will assume that Sq is true and
then show that as a consequence Sq+1 must also be true. dq+2 D dq+1
2 + dq+1 D 2 + dq
C
I
I
2 + dq+1 D 2 + dq
C dq+1 D dq , which is the induction hypothesis. dq+1 $ 3 C
C
I
2 + dq $ 3 C
2 + dq $ 9 C dq $ 7, which is certainly true because we are assuming that dq $ 3. So Sq is true for all q, and so
d1 $ dq $ 3 (showing that the sequence is bounded), and hence by the Monotonic Sequence Theorem, lim dq exists.
q<"
(b) If O = lim dq , then lim dq+1 = O also, so O =
q<"
q<"
I
2 + O i O2 = 2 + O C O2 3 O 3 2 = 0 C
(O + 1)(O 3 2) = 0 C O = 2 [since O can’t be negative].
81. d1 = 1, dq+1 = 3 3
1
. We show by induction that {dq } is increasing and bounded above by 3. Let Sq be the proposition
dq
that dq+1 A dq and 0 ? dq ? 3. Clearly S1 is true. Assume that Sq is true. Then dq+1 A dq
3
1
1
1
1
A 3 . Now dq+2 = 3 3
A33
= dq+1
dq+1
dq
dq+1
dq
i
1
1
?
dq+1
dq
i
C Sq+1 . This proves that {dq } is increasing and bounded
above by 3, so 1 = d1 ? dq ? 3, that is, {dq } is bounded, and hence convergent by the Monotonic Sequence Theorem.
If O = lim dq , then lim dq+1 = O also, so O must satisfy O = 3 3 1@O i O2 3 3O + 1 = 0 i O =
q<"
q<"
But O A 1, so O =
82. d1 = 2, dq+1 =
I
3+ 5
.
2
1
. We use induction. Let Sq be the statement that 0 ? dq+1 $ dq $ 2. Clearly S1 is true, since
3 3 dq
d2 = 1@(3 3 2) = 1. Now assume that Sq is true. Then dq+1 $ dq
dq+2 =
I
3± 5
.
2
i 3dq+1 D 3dq
i 3 3 dq+1 D 3 3 dq
1
1
$
= dq+1 . Also dq+2 A 0 [since 3 3 dq+1 is positive] and dq+1 $ 2 by the induction
3 3 dq+1
3 3 dq
hypothesis, so Sq+1 is true. To find the limit, we use the fact that lim dq = lim dq+1
q<"
O2 3 3O + 1 = 0 i O =
I
3± 5
.
2
But O $ 2, so we must have O =
q<"
i O=
1
33O
i
I
33 5
.
2
83. (a) Let dq be the number of rabbit pairs in the nth month. Clearly d1 = 1 = d2 . In the nth month, each pair that is
2 or more months old (that is, dq32 pairs) will produce a new pair to add to the dq31 pairs already present. Thus,
dq = dq31 + dq32 , so that {dq } = {iq }, the Fibonacci sequence.
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SECTION 11.1
(b) dq =
iq+1
iq
i dq31 =
¤
107
iq
iq31 + iq32
iq32
1
1
=
=1+
=1+
=1+
. If O = lim dq ,
q<"
iq31
iq31
iq31
iq31 /iq32
dq32
then O = lim dq31 and O = lim dq32 , so O must satisfy O = 1 +
q<"
SEQUENCES
q<"
1
O
i O2 3 O 3 1 = 0 i O =
I
1+ 5
2
[since O must be positive].
84. (a) If i is continuous, then i(O) = i
lim dq = lim i (dq ) = lim dq+1 = lim dq = O by Exercise 70(a).
q<"
q<"
q<"
q<"
(b) By repeatedly pressing the cosine key on the calculator (that is, taking cosine of the previous answer) until the displayed
value stabilizes, we see that O E 0=73909.
85. (a)
From the graph, it appears that the sequence
converges to 0, that is, lim
q<"
q5
q!
q5
= 0.
q!
(b)
From the first graph, it seems that the smallest possible value of Q corresponding to % = 0=1 is 9, since q5 @q! ? 0=1
whenever q D 10, but 95 @9! A 0=1. From the second graph, it seems that for % = 0=001, the smallest possible value for Q
is 11 since q5 @q! ? 0=001 whenever q D 12.
86. Let % A 0 and let Q be any positive integer larger than ln(%)@ ln |u|. If q A Q, then q A ln(%)@ ln |u|
i q ln |u| ? ln %
[since |u| ? 1 i ln |u| ? 0] i ln(|u|q ) ? ln % i |u|q ? % i |uq 3 0| ? %, and so by Definition 2,
lim uq = 0.
q<"
87. Theorem 6: If lim |dq | = 0 then lim 3 |dq | = 0, and since 3 |dq | $ dq $ |dq |, we have that lim dq = 0 by the
q<"
q<"
q<"
Squeeze Theorem.
88. Theorem 7: If lim dq = O and the function i is continuous at O, then lim i(dq ) = i (O).
q<"
q<"
Proof: We must show that, given a number % A 0, there is an integer Q such that |i(dq ) 3 i (O)| ? % whenever q A Q.
Suppose % A 0. Since i is continuous at O, there is a number A 0 such that |i({) 3 i (O)| ? % if |{ 3 O| ? . Since
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108
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
lim dq = O, there is an integer Q such that |dq 3 O| ? if q A Q. Suppose q A Q. Then 0 ? |dq 3 O| ? , so
q<"
|i (dq ) 3 i (O)| ? %.
89. To Prove: If lim dq = 0 and {eq } is bounded, then lim (dq eq ) = 0.
q<"
q<"
Proof: Since {eq } is bounded, there is a positive number P such that |eq | $ P and hence, |dq | |eq | $ |dq | P for
all q D 1. Let % A 0 be given. Since lim dq = 0, there is an integer Q such that |dq 3 0| ?
q<"
|dq eq 3 0| = |dq eq | = |dq | |eq | $ |dq | P = |dq 3 0| P ?
%
if q A Q. Then
P
%
· P = % for all q A Q. Since % was arbitrary,
P
lim (dq eq ) = 0.
q<"
90. (a)
eq+1 3 dq+1
= eq + eq31 d + eq32 d2 + eq33 d3 + · · · + edq31 + dq
e3d
? eq + eq31 e + eq32 e2 + eq33 e3 + · · · + eeq31 + eq = (q + 1)eq
(b) Since e 3 d A 0, we have eq+1 3 dq+1 ? (q + 1)eq (e 3 d) i eq+1 3 (q + 1)eq (e 3 d) ? dq+1
q
q+1
e [(q + 1)d 3 qe] ? d
i
.
(c) With this substitution, (q + 1)d 3 qe = 1, and so eq =
q
q+1
1
1
1+
? dq+1 = 1 +
.
q
q+1
q q
2q
1
1
1
1
?1 i
1+
?2 i
1+
? 4.
(d) With this substitution, we get 1 +
2q
2
2q
2q
(e) dq ? d2q since {dq } is increasing, so dq ? d2q ? 4.
(f ) Since {dq } is increasing and bounded above by 4, d1 $ dq $ 4, and so {dq } is bounded and monotonic, and hence has a
limit by the Monotonic Sequence Theorem.
91. (a) First we show that d A d1 A e1 A e.
d1 3 e1 =
d+e
2
3
I
I
I 2
I
de = 12 d 3 2 de + e = 12
d 3 e A 0 [since d A e] i d1 A e1 . Also
d 3 d1 = d 3 12 (d + e) = 12 (d 3 e) A 0 and e 3 e1 = e 3
I
I I
I de = e e 3 d ? 0, so d A d1 A e1 A e. In the same
way we can show that d1 A d2 A e2 A e1 and so the given assertion is true for q = 1. Suppose it is true for q = n, that is,
dn A dn+1 A en+1 A en . Then
dn+2 3 en+2 = 12 (dn+1 + en+1 ) 3
I
2
s
s
s
dn+1 en+1 = 12 dn+1 3 2 dn+1 en+1 + en+1 = 12
dn+1 3 en+1 A 0,
dn+1 3 dn+2 = dn+1 3 12 (dn+1 + en+1 ) = 12 (dn+1 3 en+1 ) A 0, and
en+1 3 en+2 = en+1 3
s
s
s
I
dn+1 en+1 = en+1
en+1 3 dn+1 ? 0
i
dn+1 A dn+2 A en+2 A en+1 ,
so the assertion is true for q = n + 1. Thus, it is true for all q by mathematical induction.
(b) From part (a) we have d A dq A dq+1 A eq+1 A eq A e, which shows that both sequences, {dq } and {eq }, are
monotonic and bounded. So they are both convergent by the Monotonic Sequence Theorem.
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SECTION 11.1
(c) Let lim dq = and lim eq = . Then lim dq+1 = lim
q<"
q<"
2 = + q<"
q<"
dq + eq
2
i =
+
2
¤
SEQUENCES
109
i
i = .
92. (a) Let % A 0. Since lim d2q = O, there exists Q1 such that |d2q 3 O| ? % for q A Q1 . Since lim d2q+1 = O, there
q<"
q<"
exists Q2 such that |d2q+1 3 O| ? % for q A Q2 . Let Q = max {2Q1 > 2Q2 + 1} and let q A Q. If q is even, then
q = 2p where p A Q1 , so |dq 3 O| = |d2p 3 O| ? %. If q is odd, then q = 2p + 1, where p A Q2 , so
|dq 3 O| = |d2p+1 3 O| ? %. Therefore lim dq = O.
q<"
1
1+1
(b) d1 = 1, d2 = 1 +
d5 = 1 +
1
29@12
d8 = 1 +
1
408@169
=
41
29
=
=
3
2
= 1=5, d3 = 1 +
E 1=413793, d6 = 1 +
577
408
1
5@2
=
1
70@29
=
7
5
= 1=4, d4 = 1 +
99
70
1
12@5
=
E 1=414286, d7 = 1 +
17
12
= 1=416,
1
169@70
=
239
169
E 1=414201,
E 1=414216. Notice that d1 ? d3 ? d5 ? d7 and d2 A d4 A d6 A d8 . It appears that the
odd terms are increasing and the even terms are decreasing. Let’s prove that d2q32 A d2q and d2q31 ? d2q+1 by
mathematical induction. Suppose that d2n32 A d2n . Then 1 + d2n32 A 1 + d2n
1+
1
1
?1+
1 + d2n32
1 + d2n
1
1
A
1 + d2n31
1 + d2n+1
i d2n31 ? d2n+1
i 1+
i
1
1
?
1 + d2n32
1 + d2n
i 1 + d2n31 ? 1 + d2n+1
1
1
A1+
1 + d2n31
1 + d2n+1
i
i
i d2n A d2n+2 . We have thus shown, by
induction, that the odd terms are increasing and the even terms are decreasing. Also all terms lie between 1 and 2, so both
{dq } and {eq } are bounded monotonic sequences and are therefore convergent by the Monotonic Sequence Theorem. Let
lim d2q = O. Then lim d2q+2 = O also. We have
q<"
q<"
dq+2 = 1 +
1
4 + 3dq
1
=1+
=
1 + 1 + 1@(1 + dq )
(3 + 2dq )@(1 + dq )
3 + 2dq
4 + 3d2q
4 + 3O
. Taking limits of both sides, we get O =
i 3O + 2O2 = 4 + 3O i O2 = 2 i
3 + 2d2q
3 + 2O
I
I
I
O = 2 [since O A 0]. Thus, lim d2q = 2. Similarly we find that lim d2q+1 = 2. So, by part (a),
so d2q+2 =
q<"
q<"
I
lim dq = 2.
q<"
93. (a) Suppose {sq } converges to s. Then sq+1 =
s2 + ds = es
(b) sq+1
esq
d + sq
e lim sq
i
lim sq+1 =
q<"
q<"
d + lim sq
q<"
i
s=
es
d+s
i
i
s(s + d 3 e) = 0 i s = 0 or s = e 3 d.
e
sq
d
e
esq
sq
?
sq since 1 +
A 1.
=
=
sq
d + sq
d
d
1+
d
2
3
q
e
e
e
e
e
e
(c) By part (b), s1 ?
s0 , s2 ?
s1 ?
s2 ?
s0 , s3 ?
s0 , etc. In general, sq ?
s0 ,
d
d
d
d
d
d
q
e
e
· s0 = 0 since e ? d. By (7)> lim uq = 0 if 3 1 ? u ? 1. Here u = M (0> 1) .
so lim sq $ lim
q<"
q<"
q<"
d
d
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¤
CHAPTER 11 INFINITE SEQUENCES AND SERIES
(d) Let d ? e. We first show, by induction, that if s0 ? e 3 d, then sq ? e 3 d and sq+1 A sq .
For q = 0, we have s1 3 s0 =
es0
s0 (e 3 d 3 s0 )
3 s0 =
A 0 since s0 ? e 3 d. So s1 A s0 .
d + s0
d + s0
Now we suppose the assertion is true for q = n, that is, sn ? e 3 d and sn+1 A sn . Then
e 3 d 3 sn+1 = e 3 d 3
esn
d(e 3 d) + esn 3 dsn 3 esn
d(e 3 d 3 sn )
=
=
A 0 because sn ? e 3 d. So
d + sn
d + sn
d + sn
sn+1 ? e 3 d. And sn+2 3 sn+1 =
esn+1
sn+1 (e 3 d 3 sn+1 )
3 sn+1 =
A 0 since sn+1 ? e 3 d. Therefore,
d + sn+1
d + sn+1
sn+2 A sn+1 . Thus, the assertion is true for q = n + 1. It is therefore true for all q by mathematical induction.
A similar proof by induction shows that if s0 A e 3 d, then sq A e 3 d and {sq } is decreasing.
In either case the sequence {sq } is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem.
It then follows from part (a) that lim sq = e 3 d.
q<"
LABORATORY PROJECT Logistic Sequences
1. To write such a program in Maple it is best to calculate all the points first and then graph them. One possible sequence of
commands [taking s0 =
1
2
and n = 1=5 for the difference equation] is
t:=’t’;p(0):=1/2;k:=1.5;
for j from 1 to 20
do p(j):=k*p(j-1)*(1-p(j-1)) od;
plot([seq([t,p(t)] t=0..20)],t=0..20,p=0..0.5,style=point);
In Mathematica, we can use the following program:
p[0]=1/2
k=1.5
p[j_]:=k*p[j-1]*(1-p[j-1])
P=Table[p[t],{t,20}]
ListPlot[P]
With s0 =
1
2
q
and n = 1=5:
sq
q
sq
q
sq
0
0=5
7
0=3338465076
14
0=3333373303
1
0=375
8
0=3335895255
15
0=3333353318
2
0=3515625
9
0=3334613309
16
0=3333343326
3
0=3419494629
10
0=3333973076
17
0=3333338329
4
0=3375300416
11
0=3333653143
18
0=3333335831
5
0=3354052689
12
0=3333493223
19
0=3333334582
6
0=3343628617
13
0=3333413274
20
0=3333333958
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LABORATORY PROJECT LOGISTIC SEQUENCES
With s0 =
1
2
q
and n = 2=5:
sq
q
sq
q
sq
0
0=5
7
0=6004164790
14
0=5999967417
1
0=625
8
0=5997913269
15
0=6000016291
2
0=5859375
9
0=6001042277
16
0=5999991854
3
0=6065368651
10
0=5999478590
17
0=6000004073
4
0=5966247409
11
0=6000260637
18
0=5999997964
5
0=6016591486
12
0=5999869664
19
0=6000001018
6
0=5991635437
13
0=6000065164
20
0=5999999491
Both of these sequences seem to converge the first to about 13 , the second to about 0.60 .
With s0 =
7
8
q
and n = 1=5:
sq
q
sq
q
sq
0
0=875
7
0=3239166554
14
0=3332554829
1
0=1640625
8
0=3284919837
15
0=3332943990
2
0=2057189941
9
0=3308775005
16
0=3333138639
3
0=2450980344
10
0=3320963702
17
0=3333235980
4
0=2775374819
11
0=3327125567
18
0=3333284655
5
0=3007656421
12
0=3330223670
19
0=3333308994
6
0=3154585059
13
0=3331777051
20
0=3333321164
q
sq
q
sq
With s0 =
7
8
q
and n = 2=5:
sq
0
0=875
7
0=6016572368
14
0=5999869815
1
0=2734375
8
0=5991645155
15
0=6000065088
2
0=4966735840
9
0=6004159972
16
0=5999967455
3
0=6249723374
10
0=5997915688
17
0=6000016272
4
0=5859547872
11
0=6001041070
18
0=5999991864
5
0=6065294364
12
0=5999479194
19
0=6000004068
6
0=5966286980
13
0=6000260335
20
0=5999997966
The limit of the sequence seems to depend on n, but not on s0 .
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112
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CHAPTER 11 INFINITE SEQUENCES AND SERIES
2. With s0 =
7
8
q
and n = 3=2:
sq
q
sq
q
sq
0
0=875
7
0=5830728495
14
0=7990633827
1
0=35
8
0=7779164854
15
0=5137954979
2
0=728
9
0=5528397669
16
0=7993909896
3
0=6336512
10
0=7910654689
17
0=5131681132
4
0=7428395416
11
0=5288988570
18
0=7994451225
5
0=6112926626
12
0=7973275394
19
0=5130643795
6
0=7603646184
13
0=5171082698
20
0=7994538304
It seems that eventually the terms fluctuate between two values (about 0=5 and 0=8 in this case).
3. With s0 =
7
8
q
and n = 3=42:
sq
q
sq
q
sq
0
0=875
7
0=4523028596
14
0=8442074951
1
0=3740625
8
0=8472194412
15
0=4498025048
2
0=8007579316
9
0=4426802161
16
0=8463823232
3
0=5456427596
10
0=8437633929
17
0=4446659586
4
0=8478752457
11
0=4508474156
18
0=8445284520
5
0=4411212220
12
0=8467373602
19
0=4490464985
6
0=8431438501
13
0=4438243545
20
0=8461207931
q
sq
q
sq
With s0 =
7
8
q
and n = 3=45:
sq
0
0=875
7
0=4670259170
14
0=8403376122
1
0=37734375
8
0=8587488490
15
0=4628875685
2
0=8105962830
9
0=4184824586
16
0=8577482026
3
0=5296783241
10
0=8395743720
17
0=4209559716
4
0=8594612299
11
0=4646778983
18
0=8409445432
5
0=4167173034
12
0=8581956045
19
0=4614610237
6
0=8385707740
13
0=4198508858
20
0=8573758782
From the graphs above, it seems that for n between 3=4 and 3=5, the terms eventually fluctuate between four values. In the
graph below, the pattern followed by the terms is 0=395> 0=832> 0=487> 0=869> 0=395> = = =. Note that even for n = 3=42 (as in the
first graph), there are four distinct “branches”; even after 1000 terms, the first and third terms in the pattern differ by about
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LABORATORY PROJECT LOGISTIC SEQUENCES
2 × 1039 , while the first and fifth terms differ by only 2 × 10310 . With s0 =
7
8
¤
113
and n = 3=48:
4.
s0 = 0=5, n = 3=7
s0 = 0=501, n = 3=7
s0 = 0=75, n = 3=9
s0 = 0=749, n = 3=9
s0 = 0=5, n = 3=999
From the graphs, it seems that if s0 is changed by 0=001, the whole graph changes completely. (Note, however, that this might
be partially due to accumulated round-off error in the CAS. These graphs were generated by Maple with 100-digit accuracy,
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114
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
and different degrees of accuracy give different graphs.) There seem to be some some fleeting patterns in these graphs, but on
the whole they are certainly very chaotic. As n increases, the graph spreads out vertically, with more extreme values close to 0
or 1.
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 sums is a convergent sequence. A series is divergent if it is not convergent.
2.
"
S
dq = 5 means that by adding sufficiently many terms of the series we can get as close as we like to the number 5.
q=1
In other words, it means that limq<" vq = 5, where vq is the qth partial sum, that is,
q
S
dl .
l=1
3.
"
S
q=1
4.
"
S
dq = lim vq = lim [2 3 3(0=8)q ] = lim 2 3 3 lim (0=8)q = 2 3 3(0) = 2
q<"
q<"
q<"
q2 3 1
(q2 3 1)@q2
1 3 1@q2
1
130
=
lim
=
=
lim
=
q<" 4q2 + 1
q<" (4q2 + 1)@q2
q<" 4 + 1@q2
4+0
4
dq = lim vq = lim
q<"
q=1
5. For
q<"
" 1
S
1
1
1
, dq = 3 . v1 = d1 = 3 = 1, v2 = v1 + d2 = 1 + 3 = 1=125, v3 = v2 + d3 r 1=1620,
3
q
1
2
q=1 q
v4 = v3 + d4 r 1=1777, v5 = v4 + d5 r 1=1857, v6 = v5 + d6 r 1=1903, v7 = v6 + d7 r 1=1932, and
v8 = v7 + d8 r 1=1952. It appears that the series is convergent.
6. For
"
S
q=1
1
1
1
1
1
, dq =
. v1 = d1 =
r 1=4427, v2 = v1 + d2 =
+
r 2=3529,
ln(q + 1)
ln(q + 1)
ln(1 + 1)
ln 2
ln 3
v3 = v2 + d3 r 3=0743, v4 = v3 + d4 r 3=6956, v5 = v4 + d5 r 4=2537, v6 = v5 + d6 r 4=7676,
v7 = v6 + d7 r 5=2485, and v8 = v7 + d8 r 5=7036. It appears that the series is divergent.
7. For
"
S
q=1
q
1
2
q
I = 0=5, v2 = v1 + d2 = 0=5 +
I r 1=3284,
I , dq =
I . v1 = d1 =
1+ q
1+ q
1+ 1
1+ 2
v3 = v2 + d3 r 2=4265, v4 = v3 + d4 r 3=7598, v5 = v4 + d5 r 5=3049, v6 = v5 + d6 r 7=0443,
v7 = v6 + d7 r 8=9644, v8 = v7 + d8 r 11=0540. It appears that the series is divergent.
8. For
" (31)q31
S
1
1
1
, dq = (31)q31 . v1 = d1 =
= 1, v2 = v1 + d2 = 1 3
= 0=5,
q!
q!
1!
2!
q=1
v3 = v2 + d3 = 0=5 +
1
r 0=6667, v4 = v3 + d4 = 0=625, v5 = v4 + d5 r 0=6333, v6 = v5 + d6 r 0=6319,
3!
v7 = v6 + d7 r 0=6321, v8 = v7 + d8 r 0=6321. It appears that the series is convergent.
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SECTION 11.2
SERIES
¤
9.
q
vq
1
32=40000
2
31=92000
3
32=01600
4
31=99680
5
32=00064
6
31=99987
7
32=00003
8
31=99999
9
32=00000
10
32=00000
From the graph and the table, it seems that the series converges to 32. In fact, it is a geometric
series with d = 32=4 and u = 3 15 , so its sum is
"
S
q=1
12
32=4
32=4
=
= 32=
=
(35)q
1=2
1 3 3 15
Note that the dot corresponding to q = 1 is part of both {dq } and {vq }.
TI-86 Note: To graph {dq } and {vq }, set your calculator to Param mode and DrawDot mode. (DrawDot is under
GRAPH, MORE, FORMT (F3).) Now under E(t)= make the assignments: xt1=t, yt1=12/(-5)ˆt, xt2=t,
yt2=sum seq(yt1,t,1,t,1). (sum and seq are under LIST, OPS (F5), MORE.) Under WIND use
1,10,1,0,10,1,-3,1,1 to obtain a graph similar to the one above. Then use TRACE (F4) to see the values.
10.
q
vq
1
0=54030
2
0=12416
3
30=86584
4
5
6
7
31=51948
31=23582
30=27565
0=47825
8
0=33275
9
30=57838
10
31=41745
The series
"
S
cos q diverges, since its terms do not approach 0.
q=1
11.
q
vq
1
0=44721
2
1=15432
3
1=98637
4
2=88080
5
3=80927
6
4=75796
7
5=71948
8
6=68962
9
7=66581
10
8=64639
The series
"
S
q=1
q
I
diverges, since its terms do not approach 0.
q2 + 4
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¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
12.
q
vq
1
4=90000
2
8=33000
3
10=73100
4
12=41170
5
13=58819
6
14=41173
7
14=98821
8
15=39175
9
15=67422
10
15=87196
q
vq
1
0=29289
2
0=42265
3
0=50000
4
0=55279
5
0=59175
6
0=62204
7
0=64645
8
0=66667
13.
9
0=68377
10
0=69849
q
vq
2
0=12500
3
0=19167
4
0=23333
5
0=26190
6
0=28274
7
0=29861
8
0=31111
9
0=32121
10
0=32955
14.
11
From the graph and the table, we see that the terms are getting smaller and may approach 0,
and that the series approaches a value near 16. The series is geometric with d1 = 4=9 and
u = 0=7, so its sum is
" 7q+1
S
4=9
4=9
=
= 16=3.
=
q
10
1
3
0=7
0=3
q=1
From the graph and the table, it seems that the series converges.
n
S
1
1
1
1
1
1
1
1
I 3I
I
I
I
I
I
I
3
3
+
3
+ ··· +
=
q
q+1
n+1
1
2
2
3
n
q=1
1
=13 I
,
n+1
"
S
1
1
1
I 3I
= lim 1 3 I
= 1.
so
n<"
q
q+1
n+1
q=1
From the graph and the table, it seems that the series converges.
1
1@2
1@2
1 1
1
=
3
=
3
, so
q(q + 2)
q
q+2
2 q q+2
0=33654
n
S
1
1 1
1
1 1
1
1 1
1
1 1
1
1 1
1
1
1
=
3
+
3
+
3
+ ··· +
3
=
+ 3
3
.
2 2
4
2 3
5
2 4
6
2 n
n+2
2 2
3
n+1
n+2
q=2 q(q + 2)
5
.
As n < ", this sum approaches 12 56 3 0 = 12
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SECTION 11.2
15. (a) lim dq = lim
q<"
q<"
(b) Since lim dq =
q<"
16. (a) Both
q
S
dl and
l=1
(b)
q
S
SERIES
¤
2q
2
= , so the sequence {dq } is convergent by (11.1.1).
3q + 1
3
2
3
q
S
m=1
6= 0, the series
"
S
dq is divergent by the Test for Divergence.
q=1
dm represent the sum of the first q terms of the sequence {dq }, that is, the qth partial sum.
q
S
dm = dm + dm + · · · + dm = qdm , which, in general, is not the same as
dl = d1 + d2 + · · · + dq .

~}
€
l=1
l=1
q terms
17. 3 3 4 +
16
3
18. 4 + 3 +
9
4
3
+
64
9
27
16
+ · · · is a geometric series with ratio u = 3 43 . Since |u| =
+ · · · is a geometric series with ratio 34 . Since |u| =
3
4
4
3
A 1, the series diverges.
? 1, the series converges to
2
19. 10 3 2 + 0=4 3 0=08 + · · · is a geometric series with ratio 3 10
= 3 15 . Since |u| =
1
5
d
4
=
= 16.
13u
1 3 3@4
? 1, the series converges to
d
10
10
50
25
=
=
=
=
.
13u
1 3 (31@5)
6@5
6
3
20. 2 + 0=5 + 0=125 + 0=03125 + · · · is a geometric series with ratio u =
to
21.
0=5
2
= 14 . Since |u| =
1
4
? 1, the series converges
d
2
2
8
=
=
= .
13u
1 3 1@4
3@4
3
"
S
q=1
6(0=9)q31 is a geometric series with first term d = 6 and ratio u = 0=9. Since |u| = 0=9 ? 1, the series converges to
d
6
6
=
=
= 60.
13u
1 3 0=9
0=1
22.
" 10(10)q31
" q31
S
S
10q
3 10
=
= 10
. The latter series is geometric with d = 10 and ratio u = 3 10
9
9 .
q31
q31
q=1 (39)
q=1 (39)
q=1
"
S
Since |u| =
10
9
A 1, the series diverges.
q31
" (33)q31
"
S
1 S
3
23.
=
. The latter series is geometric with d = 1 and ratio u = 3 34 . Since |u| =
3
4q
4 q=1
4
q=1
converges to
24.
"
S
q=0
3
4
? 1, it
1
= 47 . Thus, the given series converges to 14 47 = 17 .
1 3 (33@4)
1
1
1
I q is a geometric series with ratio u = I . Since |u| = I ? 1, the series converges. Its sum is
2
2
2
I
I
I
I
2
2
2 + 1 I I
1
I = I
= I
·I
= 2 2 + 1 = 2 + 2.
1 3 1@ 2
231
231
2+1
25.
"
S
117
q=0
" q
1 S
q
=
is a geometric series with ratio u = . Since |u| A 1, the series diverges.
3q+1
3 q=0 3
3
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118
26.
"
S
q=1
CHAPTER 11
q
h
=3
3q31
INFINITE SEQUENCES AND SERIES
" h q
S
h
is a geometric series with first term 3(h@3) = h and ratio u = . Since |u| ? 1, the series
3
3
q=1
h
3h
=
.
1 3 h@3
33h
converges. Its sum is
27.
" 1
" 1
S
1
1
1
1
1
1 S
+ + +
+
+ ··· =
=
. This is a constant multiple of the divergent harmonic series, so
3
6
9
12
15
3 q=1 q
q=1 3q
it diverges.
28.
1
3
+
2
9
+
1
27
+
2
81
+
1
243
geometric series with sums
29.
30.
2
729
+
+ ··· =
1
3
+
1
27
+
1
243
+ · · · + 29 +
2
81
+
2
729
+ · · · , which are both convergent
1@3
3
2@9
1
=
and
= , so the original series converges and its sum is
1 3 1@9
8
1 3 1@9
4
3
8
+
1
4
= 58 .
" q31
S
q31
1
diverges by the Test for Divergence since lim dq = lim
= 6= 0.
q<"
q<"
3q
3
1
3q
3
1
3
q=1
" n(n + 2)
S
n(n + 2)
1 · (1 + 2@n)
diverges by the Test for Divergence since lim dn = lim
= lim
= 1 6= 0.
2
2
n<"
n<"
n<"
(n
+
3)
(n
+
3)
(1 + 3@n)2
n=1
31. Converges.
q q " 1 + 2q
"
"
S
S
S
1
1
2q
2
=
+ q =
+
q
3q
3
3
3
q=1
q=1 3
q=1
2@3
1
5
1@3
+
= +2=
1 3 1@3
1 3 2@3
2
2
=
32.
" 1 + 3q
"
S
S
=
q
2
q=1
q=1
1
3q
+ q
2q
2
geometric series (|u| =
[sum of two convergent geometric series]
1
2
=
q q q
q
"
"
S
S
3
1
1
3
+
+
. The first series is a convergent
=
2
2
2
2
q=1
q=1
q=1
"
S
? 1), but the second series is a divergent geometric series (|u| =
3
2
D 1), so the original series
is divergent.
33.
" I
I
I
I
S
q
2 = 2 + 2 + 3 2 + 4 2 + · · · diverges by the Test for Divergence since
q=1
lim dq = lim
q<"
34.
q<"
I
q
2 = lim 21@q = 20 = 1 6= 0.
q<"
" "
"
S
S
S
(0=8)q31 3 (0=3)q =
(0=8)q31 3
(0=3)q
q=1
q=1
=
35.
36.
[difference of two convergent geometric series]
q=1
0=3
3
32
1
3
=53 =
1 3 0=8
1 3 0=3
7
7
2
q +1
diverges by the Test for Divergence since
ln
2q2 + 1
q=1
2
q2 + 1
q +1
lim dq = lim ln
=
ln
lim
= ln 12 6= 0.
q<"
q<"
q<" 2q2 + 1
2q2 + 1
"
S
"
S
q=1 1 +
1
2 q diverges by the Test for Divergence since lim
3
q<"
1+
1
2 q =
3
1
= 1 6= 0.
1+0
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SECTION 11.2
37.
" S
n
n=0
38.
"
S
3
is a geometric series with ratio u =
3
(cos 1)n is a geometric series with ratio u = cos 1 E 0=540. It converges because |u| ? 1. Its sum is
"
S
arctan q diverges by the Test for Divergence since lim dq = lim arctan q =
q<"
q=1
40.
"
S
q=1
2
3
+
5q
q
diverges because
q<"
"
S
q=1
3
2
+
5q
q
we have just seen that
41.
" 1
"
S
S
=
q
q=1 h
q=1
3
2
6= 0.
" 1
S
diverges.) If the given series converges, then the
q=1 q
" 3
" 3
" 2
S
S
S
, but
must converge (since
is a convergent geometric series) and equal
q
q
q=1 5
q=1 5
q=1 q
" 2
S
diverges, so the given series must also diverge.
q=1 q
q
1
1
1
1
is a geometric series with first term d = and ratio u = . Since |u| = ? 1, the series converges
h
h
h
h
"
S
1
1@h
h
1
1@h
=
· =
. By Example 7,
= 1. Thus, by Theorem 8(ii),
1 3 1@h
1 3 1@h h
h31
q=1 q(q + 1)
"
" 1
"
S
S
S
1
1
1
1
1
h31
h
+
+
=
=
+1 =
+
=
.
q
q
h
q(q + 1)
h31
h31
h31
h31
q=1
q=1 h
q=1 q(q + 1)
to
42.
" hq
S
hq
h{ H
h{ H
h{
= " 6= 0.
diverges by the Test for Divergence since lim dq = lim 2 = lim 2 = lim
= lim
2
q<"
q<" q
{<" {
{<" 2{
{<" 2
q=1 q
"
S
2
are
q2 3 1
q
q
S
S
2
1
1
=
3
vq =
l31
l+1
l=2 (l 3 1)(l + 1)
l=2
1
1 1
1
1
1
1
1
1
+
3
+
3
+ ··· +
3
+
3
= 13
3
2
4
3
5
q33
q31
q32
q
43. Using partial fractions, the partial sums of the series
q=2
1
1
1
3
3 .
2
q31
q
"
S
2
1
1
1
3
v
=
lim
1
+
Thus,
=
lim
3
3
= .
q
2
q<"
q<"
2
q31
q
2
q=2 q 3 1
This sum is a telescoping series and vq = 1 +
44. For the series
"
S
q=1
ln
119
cos 1
E 1=175.
1 3 cos 1
" 2
" 1
" 1
S
S
S
1
=2
diverges. (If it converged, then · 2
would also converge by
2
q=1 q
q=1 q
q=1 q
Theorem 8(i), but we know from Example 8 that the harmonic series
difference
¤
r 1=047. It diverges because |u| D 1.
n=1
39.
SERIES
q
,
q+1
vq = (ln 1 3 ln 2) + (ln 2 3 ln 3) + (ln 3 3 ln 4) + · · · + [ln q 3 ln(q + 1)] = ln 1 3 ln(q + 1) = 3 ln(q + 1)
[telescoping series]
Thus, lim vq = 3", so the series is divergent.
q<"
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120
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
q
q
S
S
3
3
1
1
, vq =
=
3
[using partial fractions]. The latter sum is
l+3
q=1 q(q + 3)
l=1 l(l + 3)
l=1 l
1
1
1
1
+
+ q1 3
3 q1 + q 132 3 q +
3
1 3 14 + 12 3 15 + 13 3 16 + 14 3 17 + · · · + q33
1
q31
q+2
45. For the series
"
S
=1+
Thus,
3
= lim vq = lim 1 +
q<"
q<"
q=1 q(q + 3)
"
S
1
2
+
1
3
3
1
q +1
1
2
3
+
1
3
3
1
q+2
1
q +1
3
1
q+3
3
1
q+2
3
=1+
1
2
1
q+3
+
1
3
1
q+3
[telescoping series]
=
11
.
6
Converges
1
1
cos 2 3 cos
,
q
(q + 1)2
q=1
q
S
1
1
1
1
1
1
1
cos 2 3 cos
=
cos
1
3
cos
+
cos
3
cos
+
·
·
·
+
cos
3
cos
vq =
l
(l + 1)2
4
4
9
q2
(q + 1)2
l=1
46. For the series
"
S
= cos 1 3 cos
1
(q + 1)2
[telescoping series]
1
1
1
Thus,
cos 2 3 cos
= lim vq = lim cos 1 3 cos
= cos 1 3 cos 0 = cos 1 3 1.
q<"
q<"
q
(q + 1)2
(q + 1)2
q=1
"
S
Converges
47. For the series
" S
h1@q 3 h1@(q+1) ,
q=1
q S
vq =
h1@l 3 h1@(l+1) = (h1 3 h1@2 ) + (h1@2 3 h1@3 ) + · · · + h1@q 3 h1@(q+1) = h 3 h1@(q+1)
l=1
[telescoping series]
" S
Thus,
h1@q 3 h1@(q+1) = lim vq = lim h 3 h1@(q+1) = h 3 h0 = h 3 1. Converges
q=1
q<"
q<"
"
S
1
are
q3 3 q
q
q
q
S
S
1
1
1
1@2
1@2
1 S
2
1
vq =
3 +
=
+
=
3 +
l
l31
l+1
2 l=2 l 3 1
l
l+1
l=2 l(l 3 1)(l + 1)
l=2
1 2
1
1 2
1
1
2
1
1
2
1
1
3 +
+
3 +
+
3 +
+
3 +
+ ···
=
2
1
2
3
2
3
4
3
4
5
4
5
6
1
2
1
1
2
1
1
2
1
+
3
+
+
3
+
+
3 +
q33
q32
q31
q32
q31
q
q31
q q+1
48. Using partial fractions, the partial sums of the series
q=2
Note: In three consecutive expressions in parentheses, the 3rd term in the first expression plus
the 2nd term in the second expression plus the 1st term in the third expression sum to 0.
1 2
1
1
2
1
1
1
1
3 + + 3 +
= 3
+
1
2
2
q q q+1
4
2q
2q + 2
"
S
1
1
1
1
1
=
lim
3
+
= .
v
=
lim
Thus,
q
3
q<"
q<"
4
2q 2q + 2
4
q=2 q 3 q
=
1
2
49. (a) Many people would guess that { ? 1, but note that { consists of an infinite number of 9s.
(b) { = 0=99999 = = = =
u = 0=1. Its sum is
"
S
9
9
9
9
9
+
+
+
+ ··· =
, which is a geometric series with d1 = 0=9 and
q
10
100
1000
10,000
q=1 10
0=9
0=9
=
= 1, that is, { = 1.
1 3 0=1
0=9
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SECTION 11.2
SERIES
¤
121
(c) The number 1 has two decimal representations, 1=00000 = = = and 0=99999 = = = .
(d) Except for 0, all rational numbers that have a terminating decimal representation can be written in more than one way. For
example, 0=5 can be written as 0=49999 = = = as well as 0=50000 = = = .
50. d1 = 1, dq = (5 3 q)dq31
i d2 = (5 3 2)d1 = 3(1) = 3, d3 = (5 3 3)d2 = 2(3) = 6, d4 = (5 3 4)d3 = 1(6) = 6,
d5 = (5 3 5)d4 = 0, and all succeeding terms equal 0. Thus,
51. 0=8 =
"
S
q=1
dq =
4
S
dq = 1 + 3 + 6 + 6 = 16.
q=1
8
1
d
8@10
8
8
8
+ 2 + · · · is a geometric series with d =
and u =
. It converges to
=
= .
10
10
10
10
13u
1 3 1@10
9
52. 0=46 =
46
1
d
46@100
46
46
46
+
and u =
. It converges to
=
=
.
+ · · · is a geometric series with d =
100
1002
100
100
13u
1 3 1@100
99
53. 2=516 = 2 +
516
516
516
516
516
1
+ 6 + · · · . Now 3 + 6 + · · · is a geometric series with d = 3 and u = 3 . It converges to
103
10
10
10
10
10
516@103
516
516
d
516@103
2514
838
=
=
=
. Thus, 2=516 = 2 +
=
=
.
13u
1 3 1@103
999@103
999
999
999
333
54. 10=135 = 10=1 +
to
35
35
35
35
35
1
+ 5 + · · · . Now 3 + 5 + · · · is a geometric series with d = 3 and u = 2 . It converges
103
10
10
10
10
10
35@103
9999 + 35
10,034
5017
35@103
35
35
d
=
. Thus, 10=135 = 10=1 +
=
=
=
.
=
=
2
13u
1 3 1@10
99@102
990
990
990
990
495
55. 1=5342 = 1=53 +
It converges to
42
42
42
42
42
1
+ 6 + · · · . Now 4 + 6 + · · · is a geometric series with d = 4 and u = 2 .
104
10
10
10
10
10
42@104
42@104
42
d
=
=
=
.
2
13u
1 3 1@10
99@102
9900
Thus, 1=5342 = 1=53 +
56. 7=12345 = 7 +
It converges to
12,345 12,345
12,345 12,345
12,345
1
+
+ · · · . Now
+
+ · · · is a geometric series with d =
and u = 5 .
105
1010
105
1010
105
10
12,345@105
12,345
d
12,345@105
=
=
=
.
5
13u
1 3 1@10
99,999@105
99,999
Thus, 7=12345 = 7 +
57.
"
S
(35)q {q =
q=1
42
153
42
15,147
42
15,189
5063
=
+
=
+
=
or
.
9900
100
9900
9900
9900
9900
3300
"
S
699,993
12,345
712,338
237,446
12,345
=
+
=
or
.
99,999
99,999
99,999
99,999
33,333
(35{)q is a geometric series with u = 35{, so the series converges C |u| ? 1 C
q=1
|35{| ? 1 C |{| ? 15 , that is, 3 15 ? { ? 15 . In that case, the sum of the series is
58.
"
S
d
35{
35{
=
=
.
13u
1 3 (35{)
1 + 5{
({ + 2)q is a geometric series with u = { + 2, so the series converges C |u| ? 1 C |{ + 2| ? 1 C
q=1
31 ? { + 2 ? 1
C
33 ? { ? 31. In that case, the sum of the series is
d
{+2
{+2
=
=
.
13u
1 3 ({ + 2)
3{ 3 1
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122
59.
¤
CHAPTER 11
q
" ({ 3 2)q
"
S
S
{32
{32
, so the series converges C |u| ? 1 C
=
is a geometric series with u =
3q
3
3
q=0
q=0
{ 3 2
{32
3 ? 1 C 31 ? 3 ? 1 C 33 ? { 3 2 ? 3 C 31 ? { ? 5. In that case, the sum of the series is
d
=
13u
60.
INFINITE SEQUENCES AND SERIES
"
S
3
1
1
=
=
.
{32
3 3 ({ 3 2)
53{
13
3
3
(34)q ({ 3 5)q =
q=0
"
S
q=0
[34({ 3 5)]q is a geometric series with u = 34({ 3 5), so the series converges C
|u| ? 1 C |34({ 3 5)| ? 1 C |{ 3 5| ?
the series is
62.
63.
C 3 14 ? { 3 5 ?
1
4
C
19
4
?{?
21
.
4
In that case, the sum of
1
1
d
=
=
.
13u
1 3 [34({ 3 5)]
4{ 3 19
" 2q
"
S
S
61.
=
q
{
q=0
q=0
2 ? |{|
1
4
C
q
2
2
is a geometric series with u = , so the series converges C |u| ? 1 C
{
{
{ A 2 or { ? 32. In that case, the sum of the series is
d
1
{
=
=
.
13u
1 3 2@{
{32
2
?1 C
{
q
" sinq {
"
S
S
sin {
sin {
=
is a geometric series with u =
, so the series converges C |u| ? 1 C
q
3
3
3
q=0
q=0
sin { d
1
3
3 ? 1 C |sin {| ? 3, which is true for all {. Thus, the sum of the series is 1 3 u = 1 3 (sin {)@3 = 3 3 sin { .
"
S
hq{ =
q=0
"
S
q=0
(h{ )q is a geometric series with u = h{ , so the series converges C |u| ? 1 C |h{ | ? 1 C
31 ? h{ ? 1 C 0 ? h{ ? 1 C { ? 0. In that case, the sum of the series is
d
1
.
=
13u
1 3 h{
1
1
< 0 and ln is continuous, we have lim ln 1 +
= ln 1 = 0.
q<"
q
q
"
"
"
S
S
S
1
q+1
We now show that the series
ln 1 +
ln
[ln(q + 1) 3 ln q] diverges.
=
=
q
q
q=1
q=1
q=1
64. Because
vq = (ln 2 3 ln 1) + (ln 3 3 ln 2) + · · · + (ln(q + 1) 3 ln q) = ln(q + 1) 3 ln 1 = ln(q + 1)=
As q < ", vq = ln(q + 1) < ", so the series diverges.
65. After defining i, We use convert(f,parfrac); in Maple, Apart in Mathematica, or Expand Rational and
1
1
3q2 + 3q + 1
= 3 3
. So the nth partial sum is
(q2 + q)3
q
(q + 1)3
q
S
1
1
1
1
1
1
1
1
= 13 3 +
=13
3
3 3 + ··· +
3
vq =
3
3
3
3
3
n
(n
+
1)
2
2
3
q
(q
+
1)
(q
+
1)3
n=1
Simplify in Derive to find that the general term is
The series converges to lim vq = 1. This can be confirmed by directly computing the sum using
q<"
sum(f,n=1..infinity); (in Maple), Sum[f,{n,1,Infinity}] (in Mathematica), or Calculus Sum
(from 1 to ") and Simplify (in Derive).
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SECTION 11.2
SERIES
¤
123
66. See Exercise 65 for specific CAS commands.
1
1
1
1
1
1
=
+
3
3
+
. So the qth partial sum is
q5 3 5q3 + 4q
24(q 3 2)
24(q + 2)
6(q 3 1)
6(q + 1)
4q
q
1
4
6
4
1
1 S
3
+ 3
+
vq =
24 n=3 n 3 2
n31
n
n+1
n+2
1
1
4
6
4
1
1
4
6
4
1
=
3 + 3 +
+ ··· +
3
+ 3
+
24
1
2
3
4
5
q32
q31
q
q+1
q+2
The terms with denominator 5 or greater cancel, except for a few terms with q in the denominator. So as q < ",
1 1
3
3
1
1 1
1
vq <
3 + 3
=
=
.
24 1
2
3
4
24 4
96
67. For q = 1, d1 = 0 since v1 = 0. For q A 1,
dq = vq 3 vq31 =
Also,
"
S
dq = lim vq = lim
q=1
68. d1 = v1 = 3 3
q<"
1
2
q<"
q31
(q 3 1) 3 1
(q 3 1)q 3 (q + 1)(q 3 2)
2
3
=
=
q+1
(q 3 1) + 1
(q + 1)q
q(q + 1)
1 3 1@q
= 1.
1 + 1@q
= 52 . For q 6= 1,
l
k
2(q 3 1)
q31 2
q
q32
q
3 q =
dq = vq 3 vq31 = 3 3 q23q 3 3 3 (q 3 1)23(q31) = 3 q + q31 · =
2
2
2
2q
2
2q
"
S
{ H
1
q
Also,
dq = lim vq = lim 3 3 q = 3 because lim { = lim {
= 0.
q<"
q<"
{<"
{<"
2
2
2
ln 2
q=1
69. (a) The quantity of the drug in the body after the first tablet is 150 mg. After the second tablet, there is 150 mg plus 5%
of the first 150- mg tablet, that is, [150 + 150(0=05)] mg. After the third tablet, the quantity is
[150 + 150(0=05) + 150(0=05)2 ] = 157=875 mg. After q tablets, the quantity (in mg) is
150(1 3 0=05q )
3000
=
(1 3 0=05q ).
1 3 0=05
19
(1 3 0=05q ) = 3000
(1 3 0) r 157=895,
(b) The number of milligrams remaining in the body in the long run is lim 3000
19
19
150 + 150(0=05) + · · · + 150(0=05)q31 . We can use Formula 3 to write this as
q<"
only 0=02 mg more than the amount after 3 tablets.
70. (a) The residual concentration just before the second injection is Gh3dW ; before the third, Gh3dW + Gh3d2W ; before the
(q + 1)st, Gh3dW + Gh3d2W + · · · + Gh3dqW . This sum is equal to
Gh3dW 1 3 h3dqW
1 3 h3dW
[Formula 3].
Gh3dW 1 3 h3dqW
Gh3dW (1 3 0) hdW
G
.
=
· dW = dW
3dW
q<"
13h
1 3 h3dW
h
h 31
(b) The limiting pre-injection concentration is lim
(c)
G
DF
31
hdW
i G D F hdW 3 1 , so the minimal dosage is G = F hdW 3 1 .
71. (a) The first step in the chain occurs when the local government spends G dollars. The people who receive it spend a
fraction f of those G dollars, that is, Gf dollars. Those who receive the Gf dollars spend a fraction f of it, that is,
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¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
Gf2 dollars. Continuing in this way, we see that the total spending after q transactions is
Vq = G + Gf + Gf2 + · · · + Gfq–1 =
(b) lim Vq = lim
q<"
q<"
G
=
v
G(1 3 fq )
by (3).
13f
G
G(1 3 fq )
G
=
lim (1 3 fq ) =
13f
1 3 f q<"
13f
[since f + v = 1] = nG
k
since 0 ? f ? 1
i
l
lim fq = 0
q<"
[since n = 1@v]
If f = 0=8, then v = 1 3 f = 0=2 and the multiplier is n = 1@v = 5.
72. (a) Initially, the ball falls a distance K, then rebounds a distance uK, falls uK, rebounds u2 K, falls u2 K, etc. The total
distance it travels is
K + 2uK + 2u2 K + 2u3 K + · · · = K 1 + 2u + 2u2 + 2u3 + · · · = K 1 + 2u 1 + u + u2 + · · ·
1
1+u
= K 1 + 2u
=K
meters
13u
13u
(b) From Example 3 in Section 1.4 [ET Section 2.1], we know that a ball falls 12 jw2 meters in w seconds, where j is the
s
gravitational acceleration. Thus, a ball falls k meters in w = 2k@j seconds. The total travel time in seconds is
u
2K
+2
j
u
2K
u+2
j
u
2K 2
u +2
j
u
2K 3
u + ··· =
j
u
u
l
I 2
I 3
I
2K k
1 + 2 u + 2 u + 2 u + ···
j
l
I k
I
I 2
2K 1 + 2 u 1 + u + u + ···
j
u
u
I
I
2K
2K 1 + u
1
I
I
=
1+2 u
=
j
j 13 u
13 u
=
(c) It will help to make a chart of the time for each descent and each rebound of the ball, together with the velocity just before
s
and just after each bounce. Recall that the time in seconds needed to fall k meters is 2k@j. The ball hits the ground with
s
I
velocity 3j 2k@j = 3 2kj (taking the upward direction to be positive) and rebounds with velocity
s
s
I
nj 2k@j = n 2kj, taking time n 2k@j to reach the top of its bounce, where its velocity is 0. At that point,
its height is n2 k. All these results follow from the formulas for vertical motion with gravitational acceleration 3j:
g2 |
= 3j
gw2
i y=
g|
= y0 3 jw i | = |0 + y0 w 3 12 jw2 .
gw
number of
descent
1
2
3
···
time of
descent
s
2K@j
s
2n2 K@j
s
2n4 K@j
···
speed before
bounce
I
2Kj
s
2n2 Kj
s
2n4 Kj
···
speed after
bounce
I
n 2Kj
s
n 2n2 Kj
s
n 2n4 Kj
···
time of
ascent
s
n 2K@j
s
n 2n2 K@j
s
n 2n4 K@j
···
peak
height
n2 K
n4 K
n6 K
···
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SECTION 11.2
SERIES
¤
125
The total travel time in seconds is
u
2K
+n
j
u
2K
+n
j
u
2K
+ n2
j
u
2K
+ n2
j
u
u
2K 1 + 2n + 2n2 + 2n3 + · · ·
j
u
2K 1 + 2n(1 + n + n2 + · · · )
=
j
u
u
2K
2K 1 + n
1
1 + 2n
=
=
j
13n
j 13n
2K
+ ··· =
j
Another method: We could use part (b). At the top of the bounce, the height is n2 k = uk, so
I
u = n and the result follows
from part (b).
73.
"
S
(1 + f)3q is a geometric series with d = (1 + f)32 and u = (1 + f)31 , so the series converges when
q=2
(1 + f)31 ? 1 C |1 + f| A 1 C 1 + f A 1 or 1 + f ? 31 C f A 0 or f ? 32. We calculate the sum of the
series and set it equal to 2:
(1 + f)32
=2 C
1 3 (1 + f)31
2f2 + 2f 3 1 = 0 C f =
I
So f =
74.
"
S
I
12
=
I
± 331
.
2
1
1+f
2
1
C 1 = 2(1 + f)2 3 2(1 + f) C
=232
1+f
However, the negative root is inadmissible because 32 ?
"
S
hqf =
(hf )q is a geometric series with d = (hf )0 = 1 and u = hf . If hf ? 1, it has sum
q=0
= 1 3 hf
1
i hf =
1
9
10
1
234
q+1
···
=q+1
123
q
1
2
1 + 13 · · · 1 + q1
[h{ A 1 + {]
Thus, hvq A q + 1 and lim hvq = ". Since {vq } is increasing, lim vq = ", implying that the harmonic series is
q<"
q<"
divergent.
76. The area between | = {q31 and | = {q for 0 $ { $ 1 is
]
1
? 0.
1
1
, so
= 10 i
1 3 hf
1 3 hf
9
i f = ln 10
.
75. hvq = h1+ 2 + 3 +···+ q = h1 h1@2 h1@3 · · · h1@q A (1 + 1) 1 +
=
I
3 331
2
331
.
2
q=0
1
10
32 ±
4
q31
({
0
{q
{q+1
3 { ) g{ =
3
q
q+1
q
=
1
0
=
1
1
3
q q+1
(q + 1) 3 q
1
=
q(q + 1)
q(q + 1)
We can see from the diagram that as q < ", the sum of the areas
between the successive curves approaches the area of the unit square,
that is, 1. So
"
S
q=1
1
= 1.
q (q + 1)
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126
CHAPTER 11
INFINITE SEQUENCES AND SERIES
77. Let gq be the diameter of Fq . We draw lines from the centers of the Fl to
the center of G (or F), and using the Pythagorean Theorem, we can write
2 2
C
12 + 1 3 12 g1 = 1 + 12 g1
2 2
1 = 1 + 12 g1 3 1 3 12 g1 = 2g1 [difference of squares] i g1 = 12 .
Similarly,
2 2
1 = 1 + 12 g2 3 1 3 g1 3 12 g2 = 2g2 + 2g1 3 g21 3 g1 g2
= (2 3 g1 )(g1 + g2 )
C
2 2
1
(1 3 g1 )2
3 g1 =
, 1 = 1 + 12 g3 3 1 3 g1 3 g2 3 12 g3
2 3 g1
2 3 g1
g2 =
gq+1 =
C g3 =
[1 3 (g1 + g2 )]2
, and in general,
2 3 (g1 + g2 )
2
S
gl
13 q
1
1
Sl=1
and
. If we actually calculate g2 and g3 from the formulas above, we find that they are =
23 q
g
6
2
·3
l
l=1
1
1
1
=
respectively, so we suspect that in general, gq =
. To prove this, we use induction: Assume that for all
12
3·4
q(q + 1)
n $ q, gn =
q
S
1
1
1
1
q
gl = 1 3
= 3
. Then
=
n(n + 1)
n
n+1
q
+
1
q
+
1
l=1
[telescoping sum]. Substituting this into our
2
1
q
13
q+1
(q + 1)2
1
=
=
, and the induction is complete.
formula for gq+1 , we get gq+1 =
q+2
(q + 1)(q + 2)
q
23
q+1
q+1
S
Now, we observe that the partial sums q
l=1 gl of the diameters of the circles approach 1 as q < "; that is,
"
S
dq =
q=1
"
S
q=1
1
= 1, which is what we wanted to prove.
q(q + 1)
78. |FG| = e sin , |GH| = |FG| sin = e sin2 , |HI | = |GH| sin = e sin3 , = = = . Therefore,
sin sin = e
since this is a geometric series with u = sin |FG| + |GH| + |HI | + |I J| + · · · = e
1 3 sin q=1
and |sin | ? 1
because 0 ? ? 2 .
"
S
q
79. The series 1 3 1 + 1 3 1 + 1 3 1 + · · · diverges (geometric series with u = 31) so we cannot say that
0 = 1 31 + 1 31 + 1 31 + ···.
80. If
"
S
dq is convergent, then lim dq = 0 by Theorem 6, so lim
q<"
q=1
q<"
Divergence.
81.
S"
q=1
82. If
S
fdq = lim
q<"
Sq
l=1
fdl = lim f
fdq were convergent, then
must diverge.
q<"
Sq
l=1
dl = f lim
q<"
Sq
l=1
" 1
S
1
6= 0, and so
is divergent by the Test for
dq
q=1 dq
dl = f
S"
q=1
dq , which exists by hypothesis.
S
S
S
(1@f)(fdq ) =
dq would be also, by Theorem 8(i). But this is not the case, so fdq
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SECTION 11.2
SERIES
¤
127
S
S
S
(dq + eq ) converges. Then (dq + eq ) and dq are convergent series. So by
S
S
S
Theorem 8(iii), [(dq + eq ) 3 dq ] would also be convergent. But [(dq + eq ) 3 dq ] =
eq , a contradiction, since
S
eq is given to be divergent.
83. Suppose on the contrary that
84. No. For example, take
with sum 0.
S
dq =
S
q and
S
eq =
S
S
S
(3q), which both diverge, yet (dq + eq ) =
0, which converges
85. The partial sums {vq } form an increasing sequence, since vq 3 vq31 = dq A 0 for all q. Also, the sequence {vq } is bounded
since vq $ 1000 for all q. So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series
S
dq is convergent.
1
1
iq iq+1 3 iq iq31
iq+1 3 iq31
(iq31 + iq ) 3 iq31
1
3
=
=
=
=
= LHS
iq31 iq
iq iq+1
iq2 iq31 iq+1
iq iq31 iq+1
iq iq31 iq+1
iq31 iq+1
"
"
S
S
1
1
1
[from part (a)]
=
3
(b)
iq iq+1
q=2 iq31 iq+1
q=2 iq31 iq
1
1
1
1
1
1
1
1
= lim
3
3
3
3
+
+
+ ··· +
q<"
i1 i2
i2 i3
i2 i3
i3 i4
i3 i4
i4 i5
iq31 iq
iq iq+1
1
1
1
1
=
= 1 because iq < " as q < ".
3
30 =
= lim
q<" i1 i2
iq iq+1
i1 i2
1·1
86. (a) RHS =
(c)
"
S
iq
iq
iq
[as above]
=
3
iq iq+1
q=2 iq31 iq+1
q=2 iq31 iq
"
S
1
1
3
=
iq+1
q=2 iq31
1
1
1
1
1
1
1
1
1
1
+
+
+
+ ··· +
3
3
3
3
3
= lim
q<"
i1
i3
i2
i4
i3
i5
i4
i6
iq31
iq+1
1
1
1
1
= lim
+
3
3
= 1 + 1 3 0 3 0 = 2 because iq < " as q < ".
q<" i1
i2
iq
iq+1
"
S
> 23 (length 13 ) is removed. At the second step, we remove the intervals 19 > 29 and
3
7 8
2
> , which have a total length of 2 · 13 . At the third step, we remove 22 intervals, each of length 13 . In general,
9 9
q31
q
q
at the nth step we remove 2q31 intervals, each of length 13 , for a length of 2q31 · 13 = 13 23
. Thus, the total
87. (a) At the first step, only the interval
length of all removed intervals is
1
3
"
S
q=1
1
3
2 q31
3
=
1@3
1 3 2@3
= 1 geometric series with d =
1
3
and u =
2
3
. Notice that at
q q , so we never remove 0, and 0 is in the Cantor set. Also,
the qth step, the leftmost interval that is removed is 13 > 23
1 q 2 q
, so 1 is never removed. Some other numbers in the Cantor set
the rightmost interval removed is 1 3 3 > 1 3 3
are 13 , 23 , 19 , 29 , 79 , and 89 .
2
3
(b) The area removed at the first step is 19 ; at the second step, 8 · 19 ; at the third step, (8)2 · 19 . In general, the area
q
q31
removed at the qth step is (8)q31 19 = 19 89
, so the total area of all removed squares is
q31
" 1
S
8
1@9
= 1.
=
9
9
1
3
8@9
q=1
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¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
88. (a)
d1
1
2
4
1
1
1000
d2
2
3
1
4
1000
d3
1=5
2=5
2=5
2=5
500=5
500=5
d4
1=75
2=75
1=75
3=25
750=25
250=75
d5
1=625
2=625
2=125
2=875
625=375
375=625
d6
1=6875
2=6875
1=9375
3=0625
687=813
313=188
d7
1=65625
2=65625
2=03125
2=96875
656=594
344=406
d8
1=67188
2=67188
1=98438
3=01563
672=203
328=797
d9
1=66406
2=66406
2=00781
2=99219
664=398
336=602
d10
1=66797
2=66797
1=99609
3=00391
668=301
332=699
d11
1=66602
2=66602
2=00195
2=99805
666=350
334=650
d12
1=66699
2=66699
1=99902
3=00098
667=325
333=675
1
The limits seem to be 53 , 83 , 2, 3, 667, and 334. Note that the limits appear to be “weighted” more toward d2 . In general, we
guess that the limit is
d1 + 2d2
.
3
(b) dq+1 3 dq = 12 (dq + dq31 ) 3 dq = 3 12 (dq 3 dq31 ) = 3 12 12 (dq31 + dq32 ) 3 dq31
q31
(d2 3 d1 )
= 3 12 3 12 (dq31 3 dq32 ) = · · · = 3 12
Note that we have used the formula dn = 12 (dn31 + dn32 ) a total of q 3 1 times in this calculation, once for each n
between 3 and q + 1. Now we can write
dq = d1 + (d2 3 d1 ) + (d3 3 d2 ) + · · · + (dq31 3 dq32 ) + (dq 3 dq31 )
= d1 +
q31
S
(dn+1 3 dn ) = d1 +
n=1
and so
lim dq = d1 + (d2 3 d1 )
q<"
89. (a) For
q31
S
n=1
" n31
S
3 12
= d1 + (d2 3 d1 )
n=1
3 12
n31
(d2 3 d1 )
1
d1 + 2d2
= d1 + 23 (d2 3 d1 ) =
.
1 3 (31@2)
3
"
S
q
1
2
5
3
23
1
1
5
, v1 =
= , v2 = +
= , v3 = +
=
,
(q
+
1)!
1
·
2
2
2
1
·
2
·
3
6
6
1
·
2
·
3
·
4
24
q=1
v4 =
4
119
23
(q + 1)! 3 1
+
=
. The denominators are (q + 1)!, so a guess would be vq =
.
24
1·2·3·4·5
120
(q + 1)!
(b) For q = 1, v1 =
vn+1 =
=
1
(n + 1)! 3 1
2! 3 1
=
, so the formula holds for q = 1. Assume vn =
. Then
2
2!
(n + 1)!
(n + 1)! 3 1
n+1
(n + 1)! 3 1
n+1
(n + 2)! 3 (n + 2) + n + 1
+
=
+
=
(n + 1)!
(n + 2)!
(n + 1)!
(n + 1)!(n + 2)
(n + 2)!
(n + 2)! 3 1
(n + 2)!
Thus, the formula is true for q = n + 1. So by induction, the guess is correct.
"
S
(q + 1)! 3 1
q
1
(c) lim vq = lim
= lim 1 3
= 1 and so
= 1.
q<"
q<"
q<"
(q + 1)!
(q + 1)!
q=1 (q + 1)!
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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
90.
¤
129
Let u1 = radius of the large circle, u2 = radius of next circle, and so on.
From the figure we have _EDF = 60 and cos 60 = u1 @ |DE|, so
|DE| = 2u1 and |GE| = 2u2 . Therefore, 2u1 = u1 + u2 + 2u2
u1 = 3u2 . In general, we have uq+1 =
1
u ,
3 q
i
so the total area is
1
1
1
D = u12 + 3u22 + 3u32 + · · · = u12 + 3u22 1 + 2 + 4 + 6 + · · ·
3
3
3
1
2
2
2
2
27
= u1 + 3u2 ·
= u1 + 8 u2
1 3 1@9
u1
tan 30
1
1
i u2 = I ,
. Thus, u1 =
= I
1@2
2
2 3
6 3
I
2
2
27
3
1
11
1
I
+
=
+
=
. The area of the triangle is
, so the circles occupy about 83=1%
so D = I
8
12
32
96
4
2 3
6 3
Since the sides of the triangle have length 1, |EF| =
and tan 30 =
1
2
of the area of the triangle.
11.3 The Integral Test and Estimates of Sums
1. The picture shows that d2 =
d3 =
1
31=3
?
]
3
1
{1=3
2
1
21=3
?
]
2
1
g{, and so on, so
1
g{,
{1=3
"
S
q=2
1
?
q1=3
]
1
"
1
g{. The
{1=3
integral converges by (7.8.2) with s = 1=3 A 1, so the series converges.
2. From the first figure, we see that
have
6
S
l=2
dl ?
U6
1
i ({) g{ ?
5
S
U6
1
i({) g{ ?
5
S
dl . From the second figure, we see that
l=1
6
S
l=2
dl ?
U6
1
i ({) g{. Thus, we
dl .
l=1
I
5
3. The function i ({) = 1@ { = {31@5 is continuous, positive, and decreasing on [1> "), so the Integral Test applies.
U"
1
{31@5 g{ = lim
Uw
w<" 1
{31@5 g{ = lim
w<"
k
5 4@5
{
4
lw
1
= lim
w<"
5 4@5
w
4
3
5
4
= ", so
"
S
I
1@ 5 q diverges.
q=1
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130
CHAPTER 11
INFINITE SEQUENCES AND SERIES
4. The function i ({) = 1@{5 is continuous, positive, and decreasing on [1> "), so the Integral Test applies.
]
"
1
1
g{ = lim
w<"
{5
]
w
{35 g{ = lim
w<"
1
{34
34
w
1
1
1
3 4 +
= .
w<"
4w
4
4
= lim
1
Since this improper integral is convergent, the series
1
is continuous, positive, and decreasing on [1> "), so the Integral Test applies.
(2{ + 1)3
5. The function i ({) =
]
"
1
" 1
S
is also convergent by the Integral Test.
5
q=1 q
1
g{ = lim
w<"
(2{ + 1)3
]
w
1
w
1
1
1
1
1
1
g{
=
lim
=
lim
+
3
3
=
.
w<"
(2{ + 1)3
4 (2{ + 1)2 1 w<"
4(2w + 1)2
36
36
Since this improper integral is convergent, the series
"
S
q=1
I
1
is also convergent by the Integral Test.
(2q + 1)3
6. The function i ({) = 1@ { + 4 = ({ + 4)31@2 is continuous, positive, and decreasing on [1> "), so the Integral Test applies.
U"
1
"
S
({ + 4)31@2 g{ = lim
Uw
w<" 1
I
1@ q + 4 diverges.
k
lw
I I
({ + 4)31@2 g{ = lim 2({ + 4)1@2 = lim 2 w + 4 3 2 5 = ", so the series
w<"
1
w<"
q=1
{
is continuous, positive, and decreasing on [1> "), so the Integral Test applies.
{2 + 1
w
] w
{
{
1
1
2
g{ = lim
g{ = lim
ln({ + 1) = lim [ln(w2 + 1) 3 ln 2] = ". Since this improper
w<" 1 {2 + 1
w<" 2
w<"
{2 + 1
2
1
7. The function i ({) =
]
"
1
integral is divergent, the series
"
S
q=1
q
is also divergent by the Integral Test.
q2 + 1
3
8. The function i ({) = {2 h3{ is continuous, positive, and decreasing (B) on [1> "), so the Integral Test applies.
]
"
3
{2 h3{ g{ = lim
w<"
1
]
1
w
w
3
3
3
1
1
1
1
1
03
=
.
{2 h3{ g{ = lim 3 h3{
= 3 lim h3w 3 h31 = 3
w<"
w<"
3
3
3
h
3h
1
Since this improper integral is convergent, the series
"
S
3
q2 h3q is also convergent by the Integral Test.
q=1
3
3
3
(B): i 0 ({) = {2 h3{ (33{2 ) + h3{ (2{) = {h3{ (33{3 + 2) =
9.
10.
{(2 3 3{3 )
? 0 for { A 1
h{3
I
1
I is a s-series with s =
2 A 1, so it converges by (1).
2
q=1 q
"
S
"
S
q=3
q30=9999 =
"
S
q=3
1
is a s-series with s = 0=9999 $ 1, so it diverges by (1). The fact that the series begins with
q0=9999
q = 3 is irrelevant when determining convergence.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
11. 1 +
131
" 1
S
1
1
1
1
+
+
+
+··· =
. This is a s-series with s = 3 A 1, so it converges by (1).
3
8
27
64
125
q=1 q
1
1
1
1
2 2
3 3
4 4
5 5
12. 1 + I + I + I + I + · · · =
13. 1 +
¤
"
S
"
S
1
1
I =
. This is a s-series with s =
3@2
q
q
q
q=1
q=1
3
2
A 1, so it converges by (1).
"
S
1
1
1
1
1
1
+ + + + ··· =
. The function i({) =
is
3
5
7
9
2q
3
1
2{
31
q=1
continuous, positive, and decreasing on [1> "), so the Integral Test applies.
] "
] w
"
w
S
1
1
1
g{ = lim
g{ = lim 12 ln |2{ 3 1| 1 = 12 lim (ln(2w 3 1) 3 0) = ", so the series
w<" 1 2{ 3 1
w<"
w<"
2{ 3 1
q=1 2q 3 1
1
diverges.
14.
"
S
1
1
1
1
1
1
1
+ +
+
+
+ ··· =
. The function i ({) =
is continuous, positive, and decreasing on
5
8
11
14
17
3q
+
2
3{
+2
q=1
[1> "), so the Integral Test applies.
] "
] w
w
1
1
g{ = lim
g{ = lim 13 ln |3{ + 2| 1 =
w<"
w<"
3{
+
2
3{
+
2
1
1
"
S
q=1
15.
1
lim (ln(3w
3 w<"
+ 2) 3 ln 5) = ", so the series
1
diverges.
3q + 2
I
I
"
"
" 4
S
S
S
4
q+4
q
1
=
+
+
.
=
2
2
2
2
3@2
q
q
q
q=1
q=1
q=1 q
q=1 q
"
S
"
S
q=1
1
is a convergent s-series with s =
q3@2
3
2
A 1.
" 4
" 1
S
S
=4
is a constant multiple of a convergent s-series with s = 2 A 1, so it converges. The sum of two
2
2
q
q=1
q=1 q
convergent series is convergent, so the original series is convergent.
16. i ({) =
{2
{(2 3 {3 )
0
is
continuous
and
positive
on
[2>
"),
and
also
decreasing
since
i
({)
=
? 0 for { D 2,
{3 + 1
({3 + 1)2
so we can use the Integral Test [note that i is not decreasing on [1> ")].
] "
"
w
S
{2
q2
g{ = lim 13 ln({3 + 1) 2 = 13 lim ln(w3 + 1) 3 ln 9 = ", so the series
diverges, and so does
3
3
w<"
w<"
{ +1
q=2 q + 1
2
the given series,
q2
.
+1
"
S
q3
q=1
1
is continuous, positive, and decreasing on [1> "), so we can apply the Integral Test.
{2 + 4
w
] w
1
1
1
w
1
31 {
31
31 1
tan
g{
=
lim
g{
=
lim
tan
lim
3
tan
=
w<" 1 {2 + 4
w<" 2
{2 + 4
2 1
2 w<"
2
2
1
1 3 tan31
=
2 2
2
17. The function i ({) =
]
1
"
Therefore, the series
"
S
q=1
1
converges.
q2 + 4
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132
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
18. The function i ({) =
2
1
3{ 3 4
= +
[by partial fractions] is continuous, positive, and decreasing on [3> ") since it
{2 3 2{
{
{32
is the sum of two such functions, so we can apply the Integral Test.
] w
] "
k
lw
2
3{ 3 4
1
g{
=
lim
+
g{ = lim 2 ln { + ln({ 3 2) = lim [2 ln w + ln(w 3 2) 3 2 ln 3] = ".
2
w<" 3
w<"
w<"
{ 3{
{ {32
3
3
The integral is divergent, so the series
19.
" 3q 3 4
S
is divergent.
2
q=3 q 3 q
" ln q
" ln q
S
S
ln 1
ln {
=
since
= 0. The function i ({) = 3 is continuous and positive on [2> ").
3
3
1
{
q=1 q
q=2 q
i 0 ({) =
{3 (1@{) 3 (ln {)(3{2 )
{2 3 3{2 ln {
1 3 3 ln {
=
=
? 0 C 1 3 3 ln { ? 0 C ln { A
3
2
({ )
{6
{4
1
3
C
{ A h1@3 E 1=4, so i is decreasing on [2> "), and the Integral Test applies.
w
] w
] "
" ln q
S
ln {
ln {
(B)
ln {
1
1
1 (BB) 1
3
3
g{
=
lim
g{
=
lim
3
=
lim
(2
ln
w
+
1)
+
= , so the series
3
3
2
2
2
3
w<"
w<"
w<"
{
{
2{
4{
4w
4
4
q=2 q
2
2
1
converges.
(B): x = ln {, gy = {33 g{ i gx = (1@{) g{, y = 3 12 {32 , so
]
]
]
ln {
1 32
1 32
1 32
1
{33 g{ = 3 12 {32 ln { 3 14 {32 + F=
g{
=
3
{
ln
{
3
3
{
(1@{)
g{
=
3
{
ln
{
+
2
2
2
2
{3
2@w
1
2 ln w + 1 H
3
= 3 14 lim 2 = 0.
= 3 lim
w<"
w<" 8w
w<" w
4w2
(BB): lim
20. The function i({) =
Integral Test.
1
1
=
is continuous, positive, and decreasing on [1> "), so we can apply the
{2 + 6{ + 13
({ + 3)2 + 4
]
"
w
1
1
31 { + 3
g{
=
lim
tan
w<" 1 ({ + 3)2 + 4
w<" 2
2
1
w+3
1
1 3 tan31 2 =
3 tan31 2
= lim tan31
2 w<"
2
2 2
i ({) g{ = lim
1
The integral converges, so the series
]
w
"
S
1
converges.
2 + 6q + 13
q
q=1
1
1 + ln {
is continuous and positive on [2> "), and also decreasing since i 0 ({) = 3 2
? 0 for { A 2, so we can
{ ln {
{ (ln {)2
] "
"
S
1
1
g{ = lim [ln(ln {)]w2 = lim [ln(ln w) 3 ln(ln 2)] = ", so the series
diverges.
use the Integral Test.
w<"
w<"
{
ln
{
q
ln
q
q=2
2
21. i ({) =
1
is continuous, positive, and decreasing on [2> "), so the Integral Test applies.
{(ln {)2
w
] w
] "
31
1
1
1
1
3
=
,
i ({) g{ = lim
g{
=
lim
[by
substitution
with
x
=
ln
{]
=
3
lim
w<" 2 {(ln {)2
w<" ln {
w<"
ln
w
ln
2
ln
2
2
2
22. The function i ({) =
so the series
"
S
1
converges.
q(ln
q)2
q=2
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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
¤
133
23. The function i ({) = h1@{@{2 is continuous, positive, and decreasing on [1> "), so the Integral Test applies.
[j({) = h1@{ is decreasing and dividing by {2 doesn’t change that fact.]
] w 1@{
] "
k
lw
" h1@q
S
h
1@{
i ({) g{ = lim
g{
=
lim
= 3 lim (h1@w 3 h) = 3(1 3 h) = h 3 1, so the series
3h
2
2
w<"
w<"
w<"
{
1
q=1 q
1
1
converges.
24. i ({) =
{2
h{
i i 0 ({) =
h{ (2{) 3 {2 h{
{h{ (2 3 {)
{(2 3 {)
=
=
? 0 for { A 2, so i is continuous, positive, and
(h{ )2
(h{ )2
h{
decreasing on [3> ") and so the Integral Test applies.
] "
] w 2
w
(BB) 17
(B)
{
i ({) g{ = lim
g{ = lim 3h3{ ({2 + 2{ + 2) 3 = 3 lim h3w (w2 + 2w + 2) 3 h33 (17) = 3 ,
{
w<"
w<"
w<"
h
h
3
3
so the series
(B):
U
" q2
S
converges.
q
q=3 h
97
{2 h3{ g{ = 3{2 h3{ + 2
U
U
97
{h3{ g{ = 3{2 h3{ + 2 3{h3{ + h3{ g{
[or use integration by parts twice]
= 3{2 h3{ 3 2{h3{ 3 2h3{ + F = 3h3{ ({2 + 2{ + 2) + F.
w2 + 2w + 2 H
2w + 2 H
2
= lim
= lim w = 0.
w<"
w<"
w<" h
hw
hw
(BB): lim
25. The function i ({) =
1
1
1
1
= 2 3 +
[by partial fractions] is continuous, positive and decreasing on [1> "),
{2 + {3
{
{
{+1
so the Integral Test applies.
w
] w
] "
1
1
1
1
+
g{
=
lim
3
ln
{
+
ln({
+
1)
i ({) g{ = lim
3
3
w<" 1
w<"
{2
{
{+1
{
1
1
1
w+1
+ 1 3 ln 2 = 0 + 0 + 1 3 ln 2
= lim 3 + ln
w<"
w
w
The integral converges, so the series
"
S
q=1
26. The function i ({) =
i 0 ({) =
]
1
"
1
converges.
q2 + q3
{
is positive, continuous, and decreasing on [1> "). [Note that
{4 + 1
{4 + 1 3 4{4
1 3 3{4
=
? 0 on [1> ").] Thus, we can apply the Integral Test.
({4 + 1)2
({4 + 1)2
{
g{ = lim
w<"
{4 + 1
so the series
"
S
q=1
]
1
w
1
1
2 (2{)
+ ({2 )2
g{ = lim
w<"
w
1
1
1
tan31 ({2 ) = lim [tan31 (w2 ) 3 tan31 1] =
3
=
w<"
2
2
2 2
4
8
1
q
converges.
q4 + 1
27. The function i ({) =
satisfied for the series
cos {
I
is neither positive nor decreasing on [1> "), so the hypotheses of the Integral Test are not
{
" cos q
S
I .
q
q=1
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134
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
28. The function i ({) =
series
cos2 {
is not decreasing on [1> "), so the hypotheses of the Integral Test are not satisfied for the
1 + {2
" cos2 q
S
.
2
q=1 1 + q
29. We have already shown (in Exercise 21) that when s = 1 the series
i ({) =
"
S
1
diverges, so assume that s 6= 1.
q(ln
q)s
q=2
s + ln {
1
is continuous and positive on [2> "), and i 0 ({) = 3 2
? 0 if { A h3s , so that i is eventually
{(ln {)s
{ (ln {)s+1
decreasing and we can use the Integral Test.
]
2
"
w
1
(ln {)13s
g{
=
lim
w<"
{(ln {)s
13s 2
[for s 6= 1] = lim
w<"
(ln w)13s
(ln 2)13s
3
13s
13s
This limit exists whenever 1 3 s ? 0 C s A 1, so the series converges for s A 1.
30. i ({) =
1
is positive and continuous on [3> "). For s D 0, i clearly decreases on [3> "); and for s ? 0,
{ ln { [ln(ln {)]s
it can be verified that i is ultimately decreasing. Thus, we can apply the Integral Test.
w
] "
] w
g{
[ln(ln {)]3s
[ln(ln {)]3s+1
L=
g{ = lim
= lim
[for s 6= 1]
w<" 3
w<"
{ ln { [ln(ln {)]s
{ ln {
3s + 1
3
3
[ln(ln w)]3s+1
[ln(ln 3)]3s+1
= lim
3
,
w<"
3s + 1
3s + 1
w
which exists whenever 3s + 1 ? 0 C s A 1. If s = 1, then L = lim ln(ln(ln {)) 3 = ". Therefore,
w<"
"
S
q=3
1
converges for s A 1.
q ln q [ln(ln q)]s
31. Clearly the series cannot converge if s D 3 12 , because then lim q(1 + q2 )s 6= 0. So assume s ? 3 12 . Then
q<"
i ({) = {(1 + {2 )s is continuous, positive, and eventually decreasing on [1> "), and we can use the Integral Test.
w
] "
1
1 (1 + {2 )s+1
{(1 + {2 )s g{ = lim
=
·
lim [(1 + w2 )s+1 3 2s+1 ].
w<"
2
s
+
1
2(s
+ 1) w<"
1
1
This limit exists and is finite C s + 1 ? 0 C s ? 31, so the series converges whenever s ? 31.
32. If s $ 0, lim
q<"
ln q
ln {
= " and the series diverges, so assume s A 0. i ({) = s is positive and continuous and i 0 ({) ? 0
qs
{
for { A h1@s , so i is eventually decreasing and we can use the Integral Test. Integration by parts gives
13s
w
] "
k
l
ln {
[(1 3 s) ln { 3 1]
{
1
13s
lim
g{
=
lim
(for
s
=
6
1)
=
w
[(1
3
s)
ln
w
3
1]
+
1
, which exists
2
2
w<"
{s
(1 3 s)
(1 3 s) w<"
1
1
whenever 1 3 s ? 0 C s A 1. Thus,
" ln q
S
converges C s A 1.
s
q=1 q
33. Since this is a s-series with s = {, ({) is defined when { A 1. Unless specified otherwise, the domain of a function i is the
set of real numbers { such that the expression for i({) makes sense and defines a real number. So, in the case of a series, it’s
the set of real numbers { such that the series is convergent.
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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
34. (a)
(b)
(c)
35. (a)
(b)
" 1
" 1
S
S
1
2
31
=
3 2 [subtract d1 ] =
2
2
1
6
q=2 q
q=1 q
"
S
" 1
" 1
S
S
1
=
=
3
2
2
2
q=3 (q + 1)
q=4 q
q=1 q
"
S
1
1
1
+ 2 + 2
12
2
3
"
" 1
S
1
1
1 S
1
=
=
=
2
2
2
(2q)
4q
4
q
4
q=1
q=1
q=1
2
6
=
"
S
n=5
2
49
3
6
36
1
1
+ 4
14
2
135
2
24
4
4
" 81
" 1
S
S
3
94
=
=
=
81
=
81
4
4
90
10
q=1 q
q=1 q
q=1 q
"
S
=
¤
" 1
S
1
1
1
4
1
= 4 + 4 + 4 + ··· =
=
3
4
(n 3 2)4
3
4
5
n
90
n=3
[subtract d1 and d2 ] =
4
17
3
90
16
36. (a) i ({) = 1@{4 is positive and continuous and i 0 ({) = 34@{5 is negative for { A 0, and so the Integral Test applies.
" 1
S
1
1
1
1
E v10 = 4 + 4 + 4 + · · · + 4 E 1=082037.
4
q
1
2
3
10
q=1
U10 $
(b) v10 +
]
"
10
]
"
11
w
1
1
1
1
1
=
, so the error is at most 0=0003.
g{ = lim
= lim 3 3 +
3
w<" 33{3
w<"
{4
3w
3000
3
(10)
10
1
g{ $ v $ v10 +
{4
]
"
10
1
1
1
g{ i v10 +
$ v $ v10 +
{4
3(11)3
3(10)3
i
1=082037 + 0=000250 = 1=082287 $ v $ 1=082037 + 0=000333 = 1=082370, so we get v E 1=08233 with
error $ 0=00005.
(c) The estimate in part (b) is v E 1=08233 with error $ 0=00005. The exact value given in Exercise 35 is 4 @90 E 1=082323.
The difference is less than 0.00001=
(d) Uq $
]
"
1
1
g{ =
. So Uq ? 0=00001 i
{4
3q3
q
1
1
? 5
3q3
10
i 3q3 A 105
i qA
s
3
(10)5 @3 E 32=2,
that is, for q A 32.
1
2
is positive and continuous and i 0 ({) = 3 3 is negative for { A 0, and so the Integral Test applies.
{2
{
37. (a) i ({) =
" 1
S
1
1
1
1
E v10 = 2 + 2 + 2 + · · · + 2 E 1=549768.
2
q
1
2
3
10
q=1
U10 $
(b) v10 +
]
"
10
]
"
11
w
31
1
1
1
1
+
=
, so the error is at most 0=1.
g{
=
lim
=
lim
3
w<"
{2
{ 10 w<"
w
10
10
1
g{ $ v $ v10 +
{2
]
"
10
1
g{ i v10 +
{2
1
11
$ v $ v10 +
1
10
i
1=549768 + 0=090909 = 1=640677 $ v $ 1=549768 + 0=1 = 1=649768, so we get v E 1=64522 (the average of 1=640677
and 1=649768) with error $ 0=005 (the maximum of 1=649768 3 1=64522 and 1=64522 3 1=640677, rounded up).
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136
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
(c) The estimate in part (b) is v E 1=64522 with error $ 0=005. The exact value given in Exercise 34 is 2 @6 E 1=644934.
The difference is less than 0=0003.
(d) Uq $
]
"
q
1
1
1
1
g{ = . So Uq ? 0=001 if ?
{2
q
q
1000
C q A 1000.
38. i ({) = 1@{5 is positive and continuous and i 0 ({) = 35@{6 is negative for { A 0, and so the Integral Test applies. Using (3),
Uq $
]
"
{35 g{ = lim
w<"
q
31
4{4
w
=
q
1
. If we take q = 5, then v5 E 1=036662 and U5 $ 0=0004. So v E v5 E 1=037.
4q4
39. i ({) = 1@(2{ + 1)6 is continuous, positive, and decreasing on [1> "), so the Integral Test applies. Using (3),
Uq $
]
"
(2{ + 1)36 g{ = lim
w<"
q
1
5
$ 6
10(2q + 1)5
10
v4 =
4
S
q=1
31
10(2{ + 1)5
w
=
q
1
. To be correct to five decimal places, we want
10(2q + 1)5
C (2q + 1)5 D 20,000 C q D
1
2
I
5
20,000 3 1 E 3=12, so use q = 4.
1
1
1
1
1
= 6 + 6 + 6 + 6 E 0=001 446 E 0=00145.
(2q + 1)6
3
5
7
9
1
ln { + 2
is positive and continuous and i 0 ({) = 3 2
is negative for { A 1, so the Integral Test applies.
{(ln {)2
{ (ln {)3
40. i ({) =
Using (2), we need 0=01 A
]
"
q
w
1
g{
31
= lim
=
. This is true for q A h100 , so we would have to take this
w<" ln {
{(ln {)2
ln
q
q
many terms, which would be problematic because h100 E 2=7 × 1043 .
41.
"
S
q31=001 =
q=1
Uq $
"
S
q=1
]
"
1
is a convergent s-series with s = 1=001 A 1. Using (2), we get
q1=001
{31=001 g{ = lim
w<"
q
{30=001
30=001
We want Uq ? 0=000 000 005 C
w
q
= 31000 lim
w<"
1000
? 5 × 1039
q0=001
1
{0=001
w
q
= 31000 3
C q0=001 A
1000
5 × 1039
1
q0=001
=
1000
.
q0=001
C
1000
= 21000 × 1011,000 E 1=07 × 10301 × 1011,000 = 1=07 × 1011,301 .
q A 2 × 1011
42. (a) i ({) =
ln {
{
2
is continuous and positive for { A 1, and since i 0 ({) =
the Integral Test. Using a CAS, we get
]
1
"
ln {
{
2
g{ = 2, so the series also converges.
(b) Since the Integral Test applies, the error in v E vq is Uq $
(c) By graphing the functions |1 =
2 ln { (1 3 ln {)
? 0 for { A h, we can apply
{3
]
"
q
ln {
{
2
g{ =
(ln q)2 + 2 ln q + 2
.
q
(ln {)2 + 2 ln { + 2
and |2 = 0=05, we see that |1 ? |2 for q D 1373.
{
(d) Using the CAS to sum the first 1373 terms, we get v1373 E 1=94.
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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
43. (a) From the figure, d2 + d3 + · · · + dq $
i ({) =
1
1
1 1 1
, + + + ··· + $
{ 2
3
4
q
]
Uq
1
q
1
¤
137
i ({) g{, so with
1
g{ = ln q.
{
1
1
1
1
Thus, vq = 1 + + + + · · · + $ 1 + ln q.
2
3
4
q
(b) By part (a), v106 $ 1 + ln 106 E 14=82 ? 15 and
v109 $ 1 + ln 109 E 21=72 ? 22.
44. (a) The sum of the areas of the q rectangles in the graph to the right is
1+
1
1
1
+ + · · · + . Now
2
3
q
]
q+1
1
g{
is less than this sum because
{
the rectangles extend above the curve | = 1@{, so
] q+1
1
1 1
1
g{ = ln(q + 1) ? 1 + + + · · · + , and since
{
2
3
q
1
ln q ? ln(q + 1), 0 ? 1 +
1 1
1
+ + · · · + 3 ln q = wq .
2
3
q
(b) The area under i ({) = 1@{ between { = q and { = q + 1 is
] q+1
g{
= ln(q + 1) 3 ln q, and this is clearly greater than the area of
{
q
1
, so
the inscribed rectangle in the figure to the right which is
q+1
wq 3 wq+1 = [ln(q + 1) 3 ln q] 3
1
A 0, and so wq A wq+1 , so {wq } is a decreasing sequence.
q+1
(c) We have shown that {wq } is decreasing and that wq A 0 for all q. Thus, 0 ? wq $ w1 = 1, so {wq } is a bounded monotonic
sequence, and hence converges by the Monotonic Sequence Theorem.
45. eln q = hln e
ln q
ln e
= hln q
= qln e =
ln e ? 31 C e ? h31
1
. This is a s-series, which converges for all e such that 3 ln e A 1 C
q3 ln e
C e ? 1@h [with e A 0].
f
1
3
,
q q+1
q=1
q
S
f
1
f
1
f
1
f
1
f
1
3
=
3
+
3
+
3
+ ··· +
3
vq =
l
l+1
1
2
2
3
3
4
q
q+1
l=1
f
f31
f31
f31
f31
1
1
1
1
1
1
= +
+
+
+ ··· +
3
= f + (f 3 1)
+ + + ··· +
3
1
2
3
4
q
q+1
2
3
4
q
q+1
46. For the series
Thus,
"
S
q=1
"
S
f
1
3
q q+1
q 1
S
1
f + (f 3 1)
3
. Since a constant multiple of a divergent series
q<"
q+1
l=2 l
= lim vq = lim
q<"
is divergent, the last limit exists only if f 3 1 = 0, so the original series converges only if f = 1.
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138
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
11.4 The Comparison Tests
1. (a) We cannot say anything about
S
dq . If dq A eq for all q and
divergent. (See the note after Example 2.)
(b) If dq ? eq for all q, then
2. (a) If dq A eq for all q, then
S
S
3.
eq is convergent, then
S
dq could be convergent or
dq is convergent. [This is part (i) of the Comparison Test.]
dq is divergent. [This is part (ii) of the Comparison Test.]
(b) We cannot say anything about
divergent.
S
S
dq . If dq ? eq for all q and
S
eq is divergent, then
S
dq could be convergent or
"
" 1
S
S
q
q
q
1
1
?
converges by comparison with
=
? 2 for all q D 1, so
, which converges
3
2
3
2
+1
2q
2q
q
q=1 2q + 1
q=1 q
2q3
because it is a p-series with s = 2 A 1.
4.
"
" 1
S
S
1
q3
q3
q3
A 4 = for all q D 2, so
diverges by comparison with
, which diverges because it is a p-series
4
31
q
q
q=2 q 3 1
q=2 q
q4
with s = 1 $ 1 (the harmonic series).
5.
" q+1
"
S
S
q+1
1
q
1
I A I = I for all q D 1, so
I diverges by comparison with
I , which diverges because it is a
q=1 q
q=1
q q
q q
q
q
q
p-series with s =
6.
1
2
$ 1.
" q31
"
S
S
1
q31
1
q
I ? 2 1@2 = 3@2 for all q D 1, so
I converges by comparison with
, which converges because
3@2
2
2
q q
q
q=1 q
q=1 q
q q
q
it is a p-series with s =
7.
9q
9q
?
=
3 + 10q
10q
3
2
9
10
A 1.
q
for all q D 1.
converges by the Comparison Test.
8.
6q
6q
A q =
q
5 31
5
" S
q=1
9 q
10
is a convergent geometric series |u| =
q
" S
6
6 q
for all q D 1.
is a divergent geometric series |u| =
5
5
q=1
the Comparison Test.
9.
"
S
? 1 , so
q=1
"
S
A 1 , so
q=1
9q
3 + 10q
6q
diverges by
31
5q
" ln n
" 1
S
S
ln n
1
A for all k D 3 [since ln n A 1 for n D 3], so
diverges by comparison with
, which diverges because it
n
n
n
n=3
n=3 n
is a s-series with s = 1 $ 1 (the harmonic series). Thus,
convergence or divergence of a series.
10.
6
5
9
10
" ln n
S
diverges since a finite number of terms doesn’t affect the
n=1 n
" n sin2 n
" 1
S
S
n
n
1
n sin2 n
$
?
=
for
all
n
D
1,
so
converges by comparison with
, which converges
3
2
1 + n3
1 + n3
n3
n2
1
+
n
n=1
n=1 n
because it is a s-series with s = 2 A 1.
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SECTION 11.4
THE COMPARISON TESTS
¤
139
I
I
I
3
3
3
"
"
S
S
n1@3
1
n
n
n
1
I
? I = 3@2 = 7@6 for all n D 1, so
converges by comparison with
11. I
,
7@6
3
3
n
n
n + 4n + 3
n + 4n + 3
n3
n=1
n=1 n
which converges because it is a s-series with s =
12.
7
6
A 1.
" (2n 3 1)(n 2 3 1)
" 1
S
S
(2n 3 1)(n2 3 1)
2n(n2 )
2n3
2
?
= 5 = 2 for all n D 1, so
converges by comparison with 2
,
2
2
2
2
2
2
2
(n + 1)(n + 4)
n(n )
n
n
n=1 (n + 1)(n + 4)
n=1 n
which converges because it is a constant multiple of a s-series with s = 2 A 1.
13.
" arctan q
"
S
@2
S
arctan q
1
? 1=2 for all q D 1, so
converges by comparison with
, which converges because it is a
1=2
1=2
q
q
q
2 q=1 q1=2
q=1
constant times a p-series with s = 1=2 A 1.
14.
I
I
I
"
"
S
S
q
q
q
1
1
I
A
= I , so
diverges by comparison with the divergent (partial) s-series
q31
q
q
q
q=2 q 3 1
q=2
s=
q
q
"
S
4
4
4
=4
is a divergent geometric series |u| =
3
q=1
q=1 3
q
4q+1
4 · 4q
4
A
15. q
=4
for all q D 1.
3 32
3q
3
"
S
4q+1
diverges by the Comparison Test.
q
q=1 3 3 2
"
S
4
3
1
2
$1 .
A 1 , so
"
"
S
S
1
1
1
1
1
1
I
? I
= 4@3 for all q D 1, so
, which
? I
converges by comparison with
3
3
3
4@3
4
4
4
4
q
3q + 1
3q + 1
3q
q
q=1
q=1 q
16. I
3
converges because it is a s-series with s =
4
3
A 1.
1
1
and eq = :
q
q2 + 1
17. Use the Limit Comparison Test with dq = I
lim
q<"
"
S
" 1
S
dq
q
1
= 1 A 0. Since the harmonic series
diverges, so does
= lim I
= lim s
2
2
q<"
q<"
eq
q +1
1 + (1@q )
q=1 q
1
I
.
2 +1
q
q=1
18. Use the Limit Comparison Test with dq =
Since the harmonic series
dq
q
1
1
1
1
and eq = : lim
= lim
= A 0.
= lim
q<" 2q + 3
q<" 2 + (3@q)
2q + 3
q q<" eq
2
" 1
"
S
S
1
diverges, so does
.
q=1 q
q=1 2q + 3
19. Use the Limit Comparison Test with dq =
1 + 4q
4q
and eq = q :
q
1+3
3
1 + 4q
q
dq
1 + 4q 3q
1 + 4q
1
3q
1
lim
= lim 1 +q3 = lim
·
=
lim
·
=
lim
+
1
·
=1A0
1
q<" eq
q<"
q<" 1 + 3q
q<"
q<"
4
4q
4q
1 + 3q
4q
+
1
3q
3q
" 1 + 4q
S 4 q
S
S
diverges, so does
. Alternatively, use the Comparison Test with
Since the geometric series eq =
3
q
q=1 1 + 3
q
1 + 4q
1 4
1 + 4q
4q
=
A q
A
or use the Test for Divergence.
1 + 3q
3 + 3q
2(3q )
2 3
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¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
20. 4q A q for all q D 1 since the function i ({) = 4{ 3 { satisfies i (1) = 3 and i 0 ({) = 4{ ln 4 3 1 A 0 for { D 1, so
" q + 4q
" q
S
S
q + 4q
4q + 4q
2 · 4q
2 q
?
?
= 2 46 , so the series
converges by comparison with 2
, which is a
3
q
q
q
q
q+6
q+6
6
q=1 q + 6
q=1
constant multiple of a convergent geometric series |u| =
Or: Use the Limit Comparison Test with dq =
2
3
?1 .
q
q + 4q
and eq = 23 .
q + 6q
I
q+2
1
and eq = 3@2 :
21. Use the Limit Comparison Test with dq =
2q2 + q + 1
q
s
I
I
I
I
1 + 2@q
dq
q3@2 q + 2
(q3@2 q + 2 )@(q3@2 q )
1
1
= lim
= A 0.
lim
= lim
= lim
=
q<" eq
q<" 2q2 + q + 1
q<"
q<" 2 + 1@q + 1@q2
(2q2 + q + 1)@q2
2
2
Since
"
S
1
q=1
q3@2
is a convergent p-series s =
22. Use the Limit Comparison Test with dq =
3
2
"
S
A 1 , the series
q=1
I
q+2
also converges.
2q2 + q + 1
q+2
1
and eq = 2 :
(q + 1)3
q
" 1
S
1 + q2
q2 (q + 2)
dq
= lim
= lim is a convergent (partial) s-series [s = 2 A 1],
3 = 1 A 0. Since
3
2
1
q<" eq
q<" (q + 1)
q<" 1 +
q=3 q
q
lim
the series
"
S
q+2
also converges.
(q
+ 1)3
q=3
23. Use the Limit Comparison Test with dq =
5 + 2q
1
and eq = 3 :
(1 + q2 )2
q
1@q4
dq
q3 (5 + 2q)
5q3 + 2q4
= lim
= lim
·
= lim 2
2
2
2
q<" eq
q<" (1 + q )
q<" (1 + q )
q<"
1@(q2 )2
lim
s-series [s = 3 A 1], the series
24. If dq =
so
"
S
q=1
5
q
1
q2
"
S
5 + 2q
also converges.
(1
+ q2 )2
q=1
" 1
S
is a convergent
2 = 2 A 0. Since
3
q=1 q
+1
+2
dq
q3 3 5q2
1 3 5@q
q2 3 5q
1
and eq = , then lim
= lim
= lim 3
= 1 A 0,
q<" eq
q<" q + q + 1
q<" 1 + 1@q2 + 1@q3
+q+1
q
q3
" 1
S
q2 3 5q
diverges by the Limit Comparison Test with the divergent harmonic series
.
+q+1
q=1 q
q3
(Note that dq A 0 for q D 6.)
25.
I
I
I
"
S
q4 + 1
q4
q4 + 1
q2
1
=
=
for
all
q
D
1,
so
A
diverges by comparison with
3
2
2
2
3
2
q +q
q (q + 1)
q (q + 1)
q+1
q=1 q + q
"
S
" 1
S
1
=
, which diverges because it is a s-series with s = 1 $ 1.
q
+
1
q=1
q=2 q
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SECTION 11.4
26. If dq =
THE COMPARISON TESTS
¤
141
1
1
I
and eq = 2 , then
2
q
q q 31
"
S
1
dq
q2
q@q
1
1
I
= lim I
= lim s
converges by the
= lim I
= = 1 A 0, so
2
2
2
q<" eq
q<" q q 3 1
q<"
1
q 3 1@q q<" 1 3 1@q
q2 3 1
q=2 q
lim
Limit Comparison Test with the convergent series
"
S
1
2.
q
q=2
2
2
dq
1
1
1+
h3q and eq = h3q : lim
= lim 1 +
= 1 A 0. Since
q<" eq
q<"
q
q
2
" 1
"
S
S
1
1
1+
=
is a convergent geometric series |u| = h ? 1 , the series
h3q also converges.
q
q
q=1 h
q=1
27. Use the Limit Comparison Test with dq =
"
S
h3q
q=1
28.
" h1@q
" 1
S
S
h1@q
1
A for all q D 1, so
diverges by comparison with the harmonic series
.
q
q
q=1 q
q=1 q
29. Clearly q! = q(q 3 1)(q 3 2) · · · (3)(2) D 2 · 2 · 2 · · · · · 2 · 2 = 2q31 , so
series |u| =
30.
1
2
" 1
S
? 1 , so
converges by the Comparison Test.
q=1 q!
"
S
1
1
1
$ q31 .
is a convergent geometric
q31
q!
2
2
q=1
" 2
" q!
S
S
1 · 2 · 3 · · · · · (q 3 1)q
q!
1 2
=
converges [s = 2 A 1],
converges
$ · · 1 · 1 · · · · · 1 for q D 2, so since
q
2
q
q
q · q· q · ··· · q · q
q q
q=1 q
q=1 q
also by the Comparison Test.
S
S
1
1
and eq = . Then dq and eq are series with positive terms and
q
q
31. Use the Limit Comparison Test with dq = sin
"
S
dq
sin(1@q)
sin = lim
eq is the divergent harmonic series,
= lim
= 1 A 0. Since
q<" eq
q<"
<0
1@q
q=1
lim
"
S
sin (1@q) also diverges. [Note that we could also use l’Hospital’s Rule to evaluate the limit:
q=1
lim
{<"
cos(1@{) · 31@{2
sin(1@{) H
1
= lim
= lim cos = cos 0 = 1.]
{<"
{<"
1@{
31@{2
{
1
1
dq
q
1
and eq = . lim
= lim 1+1@q = lim 1@q = 1
q<" q
q<" q
q q<" eq
q1+1@q
l
" 1
"
S
S
1
diverges [harmonic series] i
= 1 by l’Hospital’s Rule , so
diverges.
1+1@q
q=1 q
q=1 q
32. Use the Limit Comparison Test with dq =
k
since lim {1@{
{<"
33.
10
S
1
1
1
1
1
1
1
1
I
E 1=24856. Now I
= I + I + I + ··· + I
? I = 2 , so the error is
q
10,001
q4 + 1
q4 + 1
2
17
82
q4
w
] "
1
1
1
1
1
3
3
+
=
= 0=1.
g{
=
lim
=
lim
U10 $ W10 $
2
w<"
w<"
{
{
w
10
10
10
10
q=1
34.
10 sin2 q
S
sin2 2
sin2 3
sin2 10
sin2 q
sin2 1
1
+
+
+
·
·
·
+
E
0=83253.
Now
=
$ 3 , so the error is
3
3
q
1
8
27
1000
q
q
q=1
w
] "
1
1
1
1
1
g{
=
lim
=
lim
+
U10 $ W10 $
3
3
=
= 0=005.
3
2
2
w<"
w<"
{
2{
2w
200
200
10
10
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142
35.
¤
10
S
CHAPTER 11
cos2 2
cos2 1
cos2 3
cos2 10
cos2 q
1
+
+
+ ··· +
E 0=07393. Now
$ q , so the error is
2
3
10
5
5
5
5
5q
5
3{ w
3w
] w
] "
1
5310
1
5
5
3{
+
= 10
? 6=4 × 1038 .
$
g{
=
lim
5
g{
=
lim
=
lim
3
3
{
w<" 10
w<"
ln 5 10 w<"
ln 5
ln 5
5 ln 5
10 5
53q cos2 q =
q=1
U10 $ W10
36.
INFINITE SEQUENCES AND SERIES
10
S
1
1
1
1
1
1
1
1
= 1
+ 2
+ 3
+ · · · + 10
E 0=19788. Now q
? q
=
, so the
q + 4q
1
2
3
10
q
q
3
3
+
4
3
+
4
3
+
4
3
+
4
3
+
4
3
+
3
2
·
3q
q=1
error is
U10 $ W10 $
]
"
10
1
g{ = lim
w<"
2 · 3{
=
37. Since
]
w
10
w
1 3{
1 33{
1 33w
1 3310
· 3 g{ = lim 3
+
= lim 3
w<"
2
2 ln 3 10 w<"
2 ln 3
2 ln 3
1
? 7=7 × 1036 =
2 · 310 ln 3
"
S
9
gq
9
$ q for each q, and since
is a convergent geometric series |u| =
q
q
10
10
q=1 10
1
10
will always converge by the Comparison Test.
38. Clearly, if s ? 0 then the series diverges, since lim
q<"
qs
" g
S
q
? 1 , 0=g1 g2 g3 = = = =
q
q=1 10
1
= ". If 0 $ s $ 1, then qs ln q $ q ln q i
qs ln q
"
"
S
S
1
1
1
1
D
and
diverges (Exercise 11.3.21), so
diverges. If s A 1, use the Limit Comparison
s
ln q
q ln q
q=2 q ln q
q=2 q ln q
Test with dq =
qs
"
"
S
1
1 S
dq
1
1
eq converges, and lim
= lim
and eq = s .
= 0, so
also converges.
s
q<" eq
q<" ln q
ln q
q q=2
q=2 q ln q
(Or use the Comparison Test, since qs ln q A qs for q A h.) In summary, the series converges if and only if s A 1.
39. Since
S
dq converges, lim dq = 0, so there exists Q such that |dq 3 0| ? 1 for all q A Q
q<"
all q A Q
i 0 $ d2q $ dq . Since
S
dq converges, so does
S
i 0 $ dq ? 1 for
d2q by the Comparison Test.
40. (a) Since lim (dq @eq ) = 0, there is a number Q A 0 such that |dq @eq 3 0| ? 1 for all q A Q, and so dq ? eq since dq
q<"
and eq are positive. Thus, since
(b) (i) If dq =
part (a).
S
eq converges, so does
S
dq by the Comparison Test.
" ln q
S
dq
ln q
ln { H
1@{
ln q
1
= lim
= 0, so
and eq = 2 , then lim
= lim
= lim
converges by
3
3
q<"
q<"
{<"
{<"
q
q
eq
q
{
1
q=1 q
1@{
ln q
1
dq
ln q
ln { H
2
I = lim I = 0. Now
(ii) If dq = I q and eq = q , then lim
= lim I = lim I = lim
q<" eq
q<"
{<"
{<" 1@(2 {)
{<"
h
qh
q
{
{
S
S
eq is a convergent geometric series with ratio u = 1@h [|u| ? 1], so dq converges by part (a).
dq
dq
= ", there is an integer Q such that
A 1 whenever q A Q. (Take P = 1 in Definition 11.1.5.)
eq
eq
S
S
Then dq A eq whenever q A Q and since eq is divergent, dq is also divergent by the Comparison Test.
41. (a) Since lim
q<"
(b) (i) If dq =
dq
q
{ H
1
1
1
and eq = for q D 2, then lim
= lim
= lim { = ",
= lim
= lim
q<" eq
q<" ln q
{<" ln {
{<" 1@{
{<"
ln q
q
so by part (a),
"
S
q=2
1
is divergent.
ln q
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SECTION 11.5
(ii) If dq =
"
S
so
ALTERNATING SERIES
¤
143
"
S
dq
ln q
1
and eq = , then
eq is the divergent harmonic series and lim
= lim ln q = lim ln { = ",
q<"
q<"
{<"
q
q
eq
q=1
dq diverges by part (a).
q=1
42. Let dq =
S
S
dq
1
1
1
= 0, but eq diverges while dq converges.
and eq = . Then lim
= lim
q<" eq
q<" q
q2
q
43. lim qdq = lim
q<"
q<"
dq
1
, so we apply the Limit Comparison Test with eq = . Since lim qdq A 0 we know that either both
q<"
1@q
q
series converge or both series diverge, and we also know that
divergent.
44. First we observe that, by l’Hospital’s Rule, lim
{<0
Theorem 11.2.6. Therefore, lim
q<"
Thus,
S
" 1
S
S
diverges [s-series with s = 1]. Therefore, dq must be
q=1 q
S
ln(1 + {)
1
= lim
= 1. Also, if dq converges, then lim dq = 0 by
{<0 1 + {
q<"
{
S
ln(1 + dq )
ln(1 + {)
= 1 A 0. We are given that dq is convergent and dq A 0.
= lim
{<0
dq
{
ln(1 + dq ) is convergent by the Limit Comparison Test.
45. Yes. Since
S
dq is a convergent series with positive terms, lim dq = 0 by Theorem 11.2.6, and
q<"
series with positive terms (for large enough q). We have lim
q<"
[ET Theorem 3.3.2]. Thus,
46. Yes. Since
S
S
S
eq =
S
sin(dq ) is a
eq
sin(dq )
= lim
= 1 A 0 by Theorem 2.4.2
q<"
dq
dq
eq is also convergent by the Limit Comparison Test.
dq converges, its terms approach 0 as q < ", so for some integer Q, dq $ 1 for all q D Q. But then
SQ 31
S"
SQ 31
S"
q=1 dq eq =
q=1 dq eq +
q=Q dq eq $
q=1 dq eq +
q=Q eq . The first term is a finite sum, and the second term
S
S"
converges since q=1 eq converges. So dq eq converges by the Comparison Test.
S"
11.5 Alternating Series
1. (a) An alternating series is a series whose terms are alternately positive and negative.
(b) An alternating series
"
S
q=1
dq =
"
S
(31)q31 eq , where eq = |dq |, converges if 0 ? eq+1 $ eq for all q and lim eq = 0.
q=1
q<"
(This is the Alternating Series Test.)
(c) The error involved in using the partial sum vq as an approximation to the total sum v is the remainder Uq = v 3 vq and the
size of the error is smaller than eq+1 ; that is, |Uq | $ eq+1 . (This is the Alternating Series Estimation Theorem.)
2.
"
S
2
2
2
2
2
2
2
3 + 3 +
3 ··· =
. Now eq =
A 0, {eq } is decreasing, and lim eq = 0, so the
(31)q+1
q<"
3
5
7
9
11
2q + 1
2q + 1
q=1
series converges by the Alternating Series Test.
3. 3
"
S
2q
2q
2
4
6
8
10
2
2
+ 3 + 3
+ ··· =
. Now lim eq = lim
= lim
= 6= 0. Since
(31)q
q<"
q<" q + 4
q<" 1 + 4@q
5
6
7
8
9
q+4
1
q=1
lim dq 6= 0 (in fact the limit does not exist), the series diverges by the Test for Divergence.
q<"
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144
¤
CHAPTER 11
1
2
1
3
INFINITE SEQUENCES AND SERIES
1
4
1
5
1
6
4. I 3 I + I 3 I + I 3 · · · =
"
S
1
1
. Now eq = I
A 0, {eq } is decreasing, and
(31)q31 I
q+1
q+1
q=1
lim eq = 0, so the series converges by the Alternating Series Test.
q<"
5.
"
S
dq =
q=1
"
S
(31)q31
q=1
"
S
1
1
=
A 0, {eq } is decreasing, and lim eq = 0, so the
(31)q31 eq . Now eq =
q<"
2q + 1 q=1
2q + 1
series converges by the Alternating Series Test.
6.
"
S
dq =
q=1
"
S
(31)q31
q=1
"
S
1
1
(31)q31 eq . Now eq =
=
A 0, {eq } is decreasing, and lim eq = 0, so
q<"
ln(q + 4) q=1
ln(q + 4)
the series converges by the Alternating Series Test.
7.
"
S
dq =
q=1
"
S
(31)q
q=1
"
S
3q 3 1
3 3 1@q
3
(31)q eq . Now lim eq = lim
=
= 6= 0. Since lim dq 6= 0
q<"
q<" 2 + 1@q
q<"
2q + 1 q=1
2
(in fact the limit does not exist), the series diverges by the Test for Divergence.
q
A 0 for q D 1. {eq } is decreasing for q D 2 since
q3 + 2
0
1
({3 + 2)1@2 (1) 3 { · 12 ({3 + 2)31@2 (3{2 )
({3 + 2)31@2 [2({3 + 2) 3 3{3 ]
4 3 {3
{
I
=
= 2
=
? 0 for
I
2
3
1
({ + 2)
2({3 + 2)3@2
{3 + 2
{3 + 2
8. eq = I
{A
"
I
S
q@q
1
q
3
I = lim s
4 E 1=6. Also, lim eq = lim I
= 0. Thus, the series
(31)q I
2
3
2
q<"
q<"
q<"
q3 + 2
q + 2@q
q + 2@ q
q=1
converges by the Alternating Series Test.
9.
"
S
dq =
q=1
"
S
(31)q h3q =
q=1
"
S
(31)q eq . Now eq =
q=1
1
A 0, {eq } is decreasing, and lim eq = 0, so the series converges
q<"
hq
by the Alternating Series Test.
I
q
A 0 for q D 1. {eq } is decreasing for q D 2 since
2q + 3
I 0
1 31@2
(2{ + 3) 12 {31@2 3 {1@2 (2)
{
[(2{ + 3) 3 4{]
{
3 3 2{
=
= 2
= I
? 0 for { A 32 .
2
2{ + 3
(2{ + 3)
(2{ + 3)2
2 { (2{ + 3)2
I I
I
"
S
q@ q
1
q
I = lim I
I = 0. Thus, the series
Also, lim eq = lim
(31)q
converges by the
q<"
q<" (2q + 3)@ q
q<" 2 q + 3@ q
2q
+
3
q=1
10. eq =
Alternating Series Test.
q2
A 0 for q D 1. {eq } is decreasing for q D 2 since
+4
0
{2
({3 + 4)(2{) 3 {2 (3{2 )
{(2{3 + 8 3 3{3 )
{(8 3 {3 )
=
=
= 3
? 0 for { A 2. Also,
3
3
2
3
2
{ +4
({ + 4)
({ + 4)
({ + 4)2
11. eq =
q3
"
S
1@q
q2
= 0. Thus, the series
(31)q+1 3
converges by the Alternating Series Test.
3
q<" 1 + 4@q
q +4
q=1
lim eq = lim
q<"
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.5
12. eq = qh3q =
q<"
{<"
Series Test.
145
"
S
{ H
1
= lim { = 0. Thus, the series
(31)q+1 qh3q converges by the Alternating
{
{<"
h
h
q=1
13. lim eq = lim h2@q = h0 = 1, so lim (31)q31 h2@q does not exist. Thus, the series
q<"
q<"
14. lim eq = lim arctan q =
,
2
q<"
"
S
(31)q31 h2@q diverges by the
q=1
Test for Divergence.
q<"
¤
q
A 0 for q D 1. {eq } is decreasing for q D 1 since ({h3{ )0 = {(3h3{ ) + h3{ = h3{ (1 3 {) ? 0 for
hq
{ A 1. Also, lim eq = 0 since lim
q<"
ALTERNATING SERIES
so lim (31)q31 arctan q does not exist. Thus, the series
q<"
"
S
(31)q31 arctan q diverges
q=1
by the Test for Divergence.
sin q + 12 1
(31)q
I
I . Now eq =
I A 0 for q D 0, {eq } is decreasing, and lim eq = 0, so the series
15. dq =
=
q<"
1+ q
1+ q
1+ q
" sin q + 1 S
I2
converges by the Alternating Series Test.
1+ q
q=0
16. dq =
q cos q
q
= (31)q q = (31)q eq . {eq } is decreasing for q D 2 since
2q
2
({23{ )0 = {(323{ ln 2) + 23{ = 23{ (1 3 { ln 2) ? 0 for { A
lim
{<"
17.
1
[E 1=4]. Also, lim eq = 0 since
q<"
ln 2
" q cos q
S
{ H
1
= lim {
converges by the Alternating Series Test.
= 0. Thus, the series
{
{<" 2 ln 2
2
2q
q=1
. eq = sin
A 0 for q D 2 and sin
D sin
, and lim sin
= sin 0 = 0, so the
(31)q sin
q<"
q
q
q
q+1
q
q=1
"
S
series converges by the Alternating Series Test.
18.
. lim cos
= cos(0) = 1, so lim (31)q cos
does not exist and the series diverges by the Test
(31)q cos
q<"
q q<"
q
q
q=1
"
S
for Divergence.
19.
qq
q· q · ··· · q
=
Dq i
q!
1 · 2 · ··· · q
qq
=" i
q<" q!
lim
by the Test for Divergence.
20. eq =
I
I
I
I
(q + 1) 3 q
1
q+13 q
q+1+ q
·I
I = I
I = I
I A 0 for q D 1. {eq } is decreasing and
1
q+1+ q
q+1+ q
q+1+ q
lim eq = 0, so the series
q<"
"
S
(31)q qq
qq
does not exist. So the series
diverges
(31)q
q<"
q!
q!
q=1
lim
"
S
(31)q
q=1
I
I q + 1 3 q converges by the Alternating Series Test.
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146
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
21.
The graph gives us an estimate for the sum of the series
" (30=8)q
S
of 30=55.
q!
q=1
e8 =
(0=8)q
E 0=000 004, so
8!
" (30=8)q
7 (30=8)q
S
S
E v7 =
q!
q!
q=1
q=1
E 30=8 + 0=32 3 0=0853 + 0=01706 3 0=002 731 + 0=000 364 3 0=000 042 E 30=5507
Adding e8 to v7 does not change the fourth decimal place of v7 , so the sum of the series, correct to four decimal places,
is 30=5507.
22.
The graph gives us an estimate for the sum of the series
"
S
(31)q31
q=1
e6 =
"
S
(31)q31
q=1
q
of 0=1.
8q
6
E 0=000 023, so
86
5
S
q
q
E v5 =
(31)q31 q
q
8
8
q=1
E 0=125 3 0=03125 + 0=005 859 3 0=000 977 + 0=000 153 E 0=0988
Adding e6 to v5 does not change the fourth decimal place of v5 , so the sum of the series, correct to four decimal places,
is 0=0988.
23. The series
" (31)q+1
S
1
1
1
satisfies (i) of the Alternating Series Test because
? 6 and (ii) lim 6 = 0, so the
q<" q
q6
(q + 1)6
q
q=1
series is convergent. Now e5 =
1
1
= 0=000064 A 0=00005 and e6 = 6 E 0=00002 ? 0=00005, so by the Alternating Series
56
6
Estimation Theorem, q = 5. (That is, since the 6th term is less than the desired error, we need to add the first 5 terms to get the
sum to the desired accuracy.)
24. The series
" (31)q
S
1
1
1
satisfies (i) of the Alternating Series Test because
?
and (ii) lim
= 0, so
q
q<" q 5q
(q + 1)5q+1
q 5q
q=1 q 5
the series is convergent. Now e4 =
1
1
= 0=0004 A 0=0001 and e5 =
= 0=000064 ? 0=0001, so by the Alternating
4 · 54
5 · 55
Series Estimation Theorem, q = 4. (That is, since the 5th term is less than the desired error, we need to add the first 4 terms to
get the sum to the desired accuracy.)
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.5
25. The series
ALTERNATING SERIES
¤
147
" (31)q
S
1
1
1
satisfies (i) of the Alternating Series Test because q+1
? q and (ii) lim
= 0,
q q!
q q!
q<"
10
10
(q
+
1)!
10
q!
10
q=0
so the series is convergent. Now e3 =
1
1
E 0=000 167 A 0=000 005 and e4 = 4 = 0=000 004 ? 0=000 005, so by
103 3!
10 4!
the Alternating Series Estimation Theorem, q = 4 (since the series starts with q = 0, not q = 1). (That is, since the 5th term
is less than the desired error, we need to add the first 4 terms to get the sum to the desired accuracy.)
26. The series
"
S
(31)q31 qh3q =
q=1
"
S
(31)q31
q=1
q
satisfies (i) of the Alternating Series Test because
hq
{ 0
q
{ H
1
h{ (1) 3 {h{
h{ (1 3 {)
13{
=
=
=
? 0 for { A 1 and (ii) lim q = lim { = lim { = 0> so the series is
q<" h
{<" h
{<" h
h{
(h{ )2
(h{ )2
h{
convergent. Now e6 = 6@h6 E 0=015 A 0=01 and e7 = 7@h7 E 0=006 ? 0=01, so by the Alternating Series Estimation
Theorem, q = 6. (That is, since the 7th term is less than the desired error, we need to add the first 6 terms to get the sum to the
desired accuracy.)
27. e4 =
1
1
=
E 0=000 025, so
8!
40,320
" (31)q
3 (31)q
S
S
1
1
1
E v3 =
=3 +
3
E 30=459 722
2
24
720
q=1 (2q)!
q=1 (2q)!
Adding e4 to v3 does not change the fourth decimal place of v3 , so by the Alternating Series Estimation Theorem, the sum of
the series, correct to four decimal places, is 30=4597.
28.
" (31)q+1
S
1
1
1
1
1
1
1
1
1
E v9 = 6 3 6 + 6 3 6 + 6 3 6 + 6 3 6 + 6 E 0=985 552. Subtracting e10 = 1@106 from v9
q6
1
2
3
4
5
6
7
8
9
q=1
does not change the fourth decimal place of v9 , so by the Alternating Series Estimation Theorem, the sum of the series, correct
to four decimal places, is 0=9856.
29. e7 =
72
= 0=000 004 9, so
107
" (31)q31 q2
6 (31)q31 q2
S
S
E
v
=
=
6
10q
10q
q=1
q=1
1
10
3
4
100
+
9
1000
3
16
10,000
+
25
100,000
3
36
1,000,000
= 0=067 614
Adding e7 to v6 does not change the fourth decimal place of v6 , so by the Alternating Series Estimation Theorem, the sum of
the series, correct to four decimal places, is 0=0676.
30. e6 =
1
1
=
E 0=000 001 9, so
36 · 6!
524,880
" (31)q
5 (31)q
S
S
E v5 =
= 3 13 +
q
q
q=1 3 q!
q=1 3 q!
1
18
3
1
162
+
1
1944
3
1
29,160
E 30=283 471
Adding e6 to v5 does not change the fourth decimal place of v5 , so by the Alternating Series Estimation Theorem, the sum of
the series, correct to four decimal places, is 30=2835.
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148
31.
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
" (31)q31
S
1
1
1
1
1
1
1
= 1 3 + 3 + ··· +
3
+
3
+ · · · . The 50th partial sum of this series is an
q
2
3
4
49
50
51
52
q=1
" (31)q31
S
1
1
1
1
underestimate, since
= v50 +
3
+
3
+ · · · , and the terms in parentheses are all positive.
q
51
52
53
54
q=1
The result can be seen geometrically in Figure 1.
1
1
1
$ s ({1@qs } is decreasing) and lim s = 0, so the series converges by the Alternating Series Test.
q<" q
(q + 1)s
q
32. If s A 0,
" (31)q31
S
(31)q31
does not exist, so the series diverges by the Test for Divergence. Thus,
s
q<"
q
qs
q=1
If s $ 0, lim
converges C s A 0.
33. Clearly eq =
1
is decreasing and eventually positive and lim eq = 0 for any s. So the series converges (by the
q<"
q+s
Alternating Series Test) for any s for which every eq is defined, that is, q + s 6= 0 for q D 1, or s is not a negative integer.
34. Let i ({) =
(ln {)s
(ln {)s31 (s 3 ln {)
. Then i 0 ({) =
? 0 if { A hs so i is eventually decreasing for every s. Clearly
{
{2
(ln q)s
= 0 if s $ 0, and if s A 0 we can apply l’Hospital’s Rule [[s + 1]] times to get a limit of 0 as well. So the series
q<"
q
lim
converges for all s (by the Alternating Series Test).
35.
S
e2q =
S
1@(2q)2 clearly converges (by comparison with the s-series for s = 2). So suppose that
converges. Then by Theorem 11.2.8(ii), so does
S
(31)q31 eq + eq = 2 1 +
diverges by comparison with the harmonic series, a contradiction. Therefore,
Series Test does not apply since {eq } is not decreasing.
S
1
3
+
1
5
S
S
+··· = 2
(31)q31 eq
1
. But this
2q 3 1
(31)q31 eq must diverge. The Alternating
36. (a) We will prove this by induction. Let S (q) be the proposition that v2q = k2q 3 kq . S (1) is the statement v2 = k2 3 k1 ,
which is true since 1 3
consequence.
1
2
= 1 + 12 3 1. So suppose that S (q) is true. We will show that S (q + 1) must be true as a
1
1
1
1
1
+
3 kq +
= (k2q 3 kq ) +
3
2q + 1 2q + 2
q+1
2q + 1 2q + 2
1
1
= v2q +
3
= v2q+2
2q + 1 2q + 2
k2q+2 3 kq+1 =
k2q +
which is S (q + 1), and proves that v2q = k2q 3 kq for all q.
(b) We know that k2q 3 ln(2q) < and kq 3 ln q < as q < ". So
v2q = k2q 3 kq = [k2q 3 ln(2q)] 3 (kq 3 ln q) + [ln(2q) 3 ln q], and
lim v2q = 3 + lim [ln(2q) 3 ln q] = lim (ln 2 + ln q 3 ln q) = ln 2.
q<"
q<"
q<"
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SECTION 11.6
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
¤
149
11.6 Absolute Convergence and the Ratio and Root Tests
dq+1 = 8 A 1, part (b) of the Ratio Test tells us that the series
1. (a) Since lim q<"
dq S
dq is divergent.
dq+1 = 0=8 ? 1, part (a) of the Ratio Test tells us that the series S dq is absolutely convergent (and
(b) Since lim q<"
dq therefore convergent).
dq+1 = 1, the Ratio Test fails and the series S dq might converge or it might diverge.
(c) Since lim q<"
dq dq+1 (32)q+1
q2 q2 1
= lim
= lim (32)
= 2 lim
2. lim ·
= 2(1) = 2 A 1, so the series
q<"
q<" (1 + 1@q)2
dq q<" (q + 1)2 (32)q q<" (q + 1)2 " (32)q
S
is divergent by the Ratio Test.
q2
q=1
" q
dq+1 q + 1 5q 1 q + 1 1
S
1 + 1@q
1
1
=
= lim ·
lim
= (1) = ? 1, so the series
= lim q+1 ·
3. lim is
q
q<"
q<" 5
q<" 5
dq
q
q
5 q<"
1
5
5
q=1 5
absolutely convergent by the Ratio Test.
4. eq =
q2
"
S
q
q
(31)q31 2
A 0 for q D 1, {eq } is decreasing for q D 2, and lim eq = 0, so
converges by the
q<"
+4
q +4
q=1
1
to get
q
"
S
dq
1@q
q2 + 4
1 + 4@q2
q
= lim
= 1 A 0, so
diverges by the Limit
= lim
= lim
lim
2
q<" eq
q<" q@(q2 + 4)
q<"
q<"
q2
1
q=1 q + 4
Alternating Series Test. To determine absolute convergence, choose dq =
Comparison Test with the harmonic series. Thus, the series
"
S
(31)q31
q=1
5. eq =
q
is conditionally convergent.
q2 + 4
" (31)q
S
1
A 0 for q D 0, {eq } is decreasing for q D 0, and lim eq = 0, so
converges by the Alternating
q<"
5q + 1
q=0 5q + 1
1
to get
q
"
S
dq
1@q
5q + 1
1
lim
= lim
= 5 A 0, so
diverges by the Limit Comparison Test with the
= lim
q<" eq
q<" 1@(5q + 1)
q<"
q
q=1 5q + 1
Series Test. To determine absolute convergence, choose dq =
harmonic series. Thus, the series
" (31)q
S
is conditionally convergent.
q=0 5q + 1
q+1
dq+1 1
1
(2q + 1)! = 3 lim
= lim (33)
= lim (33)
6. lim ·
q<"
q<" [2(q + 1) + 1]!
q<" (2q + 3)(2q + 2)
dq
(33)q q<" (2q + 3)(2q + 2) = 3(0) = 0 ? 1
so the series
(33)q
is absolutely convergent by the Ratio Test.
q=0 (2q + 1)!
"
S
dn+1 = lim
7. lim n<"
dn n<"
%
n+1 &
1
(n + 1) 23
2
n+1 2
1
=
=
lim
1
+
lim
= 23 (1) =
n
n<"
n
3
3 n<"
n
n 23
2
3
? 1, so the series
"
n
S
n 23 is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the
q=1
same as convergence.
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¤
150
CHAPTER 11
INFINITE SEQUENCES AND SERIES
q
"
dq+1 S
q+1
q!
= lim (q + 1)! · 100
8. lim diverges by the Ratio Test.
= lim
= ", so the series
q+1
q
q<"
q<"
q<"
dq
100
q!
100
q=1 100
q+1
dq+1 q4
= lim (1=1)
9. lim ·
q<"
dq q<" (q + 1)4 (1=1)q
= lim
q<"
(1=1)q4
1
1
lim
lim
4 = (1=1) q<"
4 = (1=1) q<"
(1
+
1@q)4
(q + 1)
(q + 1)
4
q
= (1=1)(1) = 1=1 A 1,
"
S
so the series
(31)q
q=1
10.
(1=1)q
diverges by the Ratio Test.
q4
"
S
q
1
q
(31)q I
converges by the Alternating Series Test (see Exercise 11.5.8). Let dq = I with eq = I
. Then
3
3
q +2
q +2
q=1
q
u
I
I
"
S
1
dq
q3 + 2
q3 + 2
q
2
I
I
lim
= lim
= lim I ·
= lim
1 + 3 = 1 A 0, so
diverges by limit
3
3
q<" eq
q<"
q<"
q<"
q
q
q +2
q=1
q
q
comparision with the divergent p-series
11. Since 0 $
"
S
1 I
s=
q=1
q
1
2
"
S
q
$ 1 . Thus,
(31)q I
is conditionally convergent.
q3 + 2
q=1
" 1
" h1@q
S
S
h1@q
h
1
$
=
h
is a convergent s-series [s = 3 A 1],
converges, and so
and
3
3
3
3
3
q
q
q
q=1 q
q=1 q
" (31)q h1@q
S
is absolutely convergent.
q3
q=1
" sin 4q " 1 sin 4q S
S
1
|u| =
12. q $ q , so
4q converges by comparison with the convergent geometric series
q
4
4
q=1
q=1 4
Thus,
" sin 4q
S
is absolutely convergent.
4q
q=1
dq+1 10q+1
(q + 1) 42q+1
= lim
13. lim ·
q<"
dq q<" (q + 2) 42q+3
10q
= lim
q<"
10 q + 1
·
42 q + 2
=
1
4
?1 .
"
S
10q
5
? 1, so the series
2q+1
8
q=1 (q + 1)4
is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the same as
convergence.
10
1 q + 1 10 dq+1 (q + 1)10 (310)q+1 1
1
1
1
= lim = lim lim
(1) =
? 1,
=
14. lim ·
=
1
+
q<" 310
q<"
10 q<"
dq q<" (310)q+2
q10
q
q
10
10
so the series
q10
is absolutely convergent by the Ratio Test.
q+1
q=1 (310)
"
S
" @2
" 1
" (31)q arctan q
(31)q arctan q @2
S
S
S
?
15. , so since
=
converges (s = 2 A 1), the given series
2
2
2
2
2
q
q
q=1
q
2
q=1
q
q
q=1
converges absolutely by the Comparison Test.
16. q2@3 3 2 A 0 for q D 3, so
"
S
3 3 cos q
1
1
1
diverges s =
A 2@3
A 2@3 for q D 3. Since
2@3
q2@3 3 2
q
32
q
q
q=1
" 3 3 cos q
S
by the Comparison Test.
2@3 3 2
q=1 q
2
3
$ 1 , so does
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SECTION 11.6
17.
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
" (31)q
S
1
converges by the Alternating Series Test since lim
= 0 and
q<" ln q
q=2 ln q
1
ln q
¤
151
is decreasing. Now ln q ? q, so
" 1
"
S
S
1
1
1
A , and since
is the divergent (partial) harmonic series,
diverges by the Comparison Test. Thus,
ln q
q
q
ln
q
q=2
q=2
" (31)q
S
is conditionally convergent.
q=2 ln q
q+1
[" q!
dq+1 qq
1
1
= lim (q + 1)!@ (q + 1)
? 1, so the series
18. lim =
lim
q = lim
q =
q<"
q<" (q + 1)
q<" (1 + 1@q)
q=1 qq
dq q<"
q!@qq
h
converges absolutely by the Ratio Test.
19.
" 1
" cos(q@3)
S
S
|cos (q@3)|
1
$
and
converges (use the Ratio Test), so the series
converges absolutely by the
q!
q!
q!
q=1 q!
q=1
Comparison Test.
s
20. lim q |dq | = lim
q<"
21. lim
q<"
q<"
v
q
" (32)q
S
2
q (32) =
0
?
1,
so
the
series
=
lim
is absolutely convergent by the Root Test.
qq q<" q
qq
q=1
"
s
S
q2 + 1
1 + 1@q2
1
q
=
lim
|dq | = lim
= ? 1, so the series
q<" 2q2 + 1
q<" 2 + 1@q2
2
q=1
q2 + 1
2q2 + 1
q
is absolutely convergent by the
Root Test.
y
x
x 32q 5q s
25 q5
1
1
q
q
22. lim
|dq | = lim w
= 32 lim = lim
5 = 32 lim
q<"
q<"
q<"
q<" (1 + 1@q)5
q+1
q<" (q + 1)5
q+1
q
= 32(1) = 32 A 1,
so the series
"
S
q=2
32q
q+1
s
23. lim q |dq | = lim
q<"
q<"
5q
diverges by the Root Test.
v
q2
q
1
1
q
1+
= lim 1 +
= h A 1 [by Equation 7.4.9 (or 7.4*.9) [ ET 3.6.6] ],
q<"
q
q
q2
1
1+
so the series
diverges by the Root Test.
q
q=1
"
S
2
dq+1 = lim [2(q + 1)]! · (q!) = lim (2q + 2)(2q + 1) = lim (2 + 2@q)(2 + 1@q) = 2 · 2 = 4 A 1,
24. lim q<"
q<" (1 + 1@q)(1 + 1@q)
dq q<" [(q + 1)!]2 (2q)! q<" (q + 1)(q + 1)
1·1
so the series
" (2q)!
S
diverges by the Ratio Test.
2
q=1 (q!)
100
q+1
dq+1 (q + 1) 100
= lim 25. lim q<"
dq q<" (q + 1)!
so the series
=0·1=0?1
·
100
100
100
q!
q+1
1
= lim 100
1
+
=
lim
q<" q + 1
q100 100q q<" q + 1
q
q
" q100 100q
S
is absolutely convergent by the Ratio Test.
q!
q=1
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152
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
2
2(q+1)2
dq+1 2q +2q+1
22q+1
q! = lim
· q2 = lim q2
= O. Now
= lim 26. lim q<"
q<" (q + 1)! 2
q<" 2 (q + 1) q<" q + 1
dq
2
lim
{<"
" 2q
S
22{+1 H
22{+1 (2 ln 2)
= lim
= ", so O A 1 and the series
diverges by the Ratio Test.
{<"
{+1
1
q=1 q!
27. Use the Ratio Test with the series
13
"
S
1 · 3 · 5 · · · · · (2q 3 1)
1 · 3 · 5 · · · · · (2q 3 1)
1·3·5
1·3·5·7
1·3
+
3
+ · · · + (31)q31
+ ··· =
.
(31)q31
3!
5!
7!
(2q 3 1)!
(2q 3 1)!
q=1
dq+1 (31)q · 1 · 3 · 5 · · · · · (2q 3 1)[2(q + 1) 3 1]
(2q 3 1)!
= lim lim
·
q<" dq q<"
[2(q + 1) 3 1]!
(31)q31 · 1 · 3 · 5 · · · · · (2q 3 1) (31)(2q + 1)(2q 3 1)! = lim 1 = 0 ? 1,
= lim q<" (2q + 1)(2q)(2q 3 1)! q<" 2q
so the given series is absolutely convergent and therefore convergent.
" 2 · 6 · 10 · 14 · · · · · (4q 3 2)
S
2 2·6
2 · 6 · 10
2 · 6 · 10 · 14
+
+
+
+ ··· =
.
5
5·8
5 · 8 · 11
5 · 8 · 11 · 14
q=1 5 · 8 · 11 · 14 · · · · · (3q + 2)
dq+1 = lim 2 · 6 · 10 · · · · · (4q 3 2)[4(q + 1) 3 2] · 5 · 8 · 11 · · · · · (3q + 2) = lim 4q + 2 = 4 A 1,
lim q<"
q<"
dq
5 · 8 · 11 · · · · · (3q + 2)[3(q + 1) + 2] 2 · 6 · 10 · · · · · (4q 3 2) q<" 3q + 5
3
28. Use the Ratio Test with the series
so the given series is divergent.
29.
" 2 · 4 · 6 · · · · · (2q)
" (2 · 1) · (2 · 2) · (2 · 3) · · · · · (2 · q)
" 2q q!
"
S
S
S
S
=
=
=
2q , which diverges by the Test for
q!
q!
q=1
q=1
q=1 q!
q=1
Divergence since lim 2q = ".
q<"
2q+1 (q + 1)!
5 · 8 · 11 · · · · · (3q + 5) dq+1 = lim 2(q + 1) = 2 ? 1, so the series converges absolutely by the
= lim
30. lim q<" 3q + 5
q<"
2q q!
dq q<" 3
5 · 8 · 11 · · · · · (3q + 2) Ratio Test.
dq+1 = lim 5q + 1 = 5 A 1, so the series diverges by the Ratio Test.
31. By the recursive definition, lim q<"
dq q<" 4q + 3 4
dq+1 2 + cos q = 0 ? 1, so the series converges absolutely by the Ratio Test.
I
= lim
32. By the recursive definition, lim q<"
dq q<" q " eq cos q
"
S
S
eq
1
q
=
(31)q q , where eq A 0 for q D 1 and lim eq = .
q<"
q
q
2
q=1
q=1
q+1 q+1
q
"
dq+1 eq
q
= lim (31)
= lim eq q = 1 (1) = 1 ? 1, so the series S eq cos q is
lim ·
q
q
q<"
q<"
q<"
dq
q+1
(31) eq
q+1
2
2
q
q=1
33. The series
absolutely convergent by the Ratio Test.
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SECTION 11.6
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
¤
153
dq+1 (31)(q + 1)qq (31)q+1 (q + 1)!
qq
qq e1 e2 · · · eq = lim = lim
=
lim
34. lim ·
q+1
q
q+1
q<"
q<"
q<"
q<"
dq
(q + 1)
e1 e2 · · · eq eq+1
(31) q!
eq+1 (q + 1)
eq+1 (q + 1)q
q
q
2
1
1
1
q
1
1
= lim
= lim
= lim
= 1 = ?1
q<" eq+1
q<" eq+1
q<" eq+1 (1 + 1@q)q
q+1
1 + 1@q
h
h
2
so the series
(31)q q!
is absolutely convergent by the Ratio Test.
· · · eq
"
S
q
q=1 q e1 e2 e3
1@(q + 1)3 q3
1
= lim
= lim
= 1. Inconclusive
3
q<" (q + 1)3
q<" (1 + 1@q)3
1@q
35. (a) lim q<"
(q + 1) 2q = lim q + 1 = lim 1 + 1 = 1 . Conclusive (convergent)
(b) lim q+1 ·
q<"
q<" 2q
q<" 2
2
q
2q
2
u
I
u
(33)q
q q
1
(c) lim I
=
3
lim
= 3. Conclusive (divergent)
=
3
lim
·
q31 q<"
q<"
q<"
(33)
q
+
1
1
+
1@q
q+1
%u
&
I
q+1
1@q2 + 1
1 + q2 1
(d) lim · I = lim
1+ ·
= 1. Inconclusive
q<" 1 + (q + 1)2
q<"
q 1@q2 + (1 + 1@q)2
q
36. We use the Ratio Test:
2
dq+1 (q + 1)2
= lim [(q + 1)!] /[n(q + 1)]! = lim lim 2
q<"
q<"
q<" [n(q + 1)] [n(q + 1) 3 1] · · · [nq + 1] dq
(q!) /(nq)!
(q + 1)2 = ", so the series diverges; if n = 2, the limit is
Now if n = 1, then this is equal to lim q<" (q + 1) 1
(q + 1)2
= ? 1, so the series converges, and if n A 2, then the highest power of q in the denominator is
lim q<" (2q + 2)(2q + 1) 4
larger than 2, and so the limit is 0, indicating convergence. So the series converges for n D 2.
q+1
dq+1 {
{ 1
q! ·
= |{| lim
= |{| · 0 = 0 ? 1, so by the Ratio Test the
= lim = lim 37. (a) lim q<"
q<" q + 1
dq q<" (q + 1)! {q q<" q + 1 series
" {q
S
converges for all {.
q=0 q!
(b) Since the series of part (a) always converges, we must have lim
q<"
{q
= 0 by Theorem 11.2.6.
q!
dq+3
dq+4
dq+2
+
+
+ ···
dq+1
dq+1
dq+1
dq+2
+ ···
dq+1
38. (a) Uq = dq+1 + dq+2 + dq+3 + dq+4 + · · · = dq+1 1 +
dq+2
dq+3 dq+2
dq+4 dq+3
= dq+1 1 +
+
+
dq+1
dq+2 dq+1
dq+3 dq+2
= dq+1 (1 + uq+1 + uq+2 uq+1 + uq+3 uq+2 uq+1 + · · · ) (B)
2
3
$ dq+1 1 + uq+1 + uq+1
+ uq+1
+ ···
[since {uq } is decreasing] =
dq+1
1 3 uq+1
(b) Note that since {uq } is increasing and uq < O as q < ", we have uq ? O for all q. So, starting with equation (B),
dq+1
Uq = dq+1 (1 + uq+1 + uq+2 uq+1 + uq+3 uq+2 uq+1 + · · · ) $ dq+1 1 + O + O2 + O3 + · · · =
.
13O
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154
¤
CHAPTER 11
39. (a) v5 =
5
S
q=1
uq =
INFINITE SEQUENCES AND SERIES
1
1
1
1
1
1
661
= + +
+
+
=
E 0=68854. Now the ratios
q2q
2
8
24
64
160
960
dq+1
q2q
q
form an increasing sequence, since
=
=
dq
(q + 1)2q+1
2(q + 1)
q
(q + 1)2 3 q(q + 2)
1
q+1
3
=
=
A 0. So by Exercise 34(b), the error
2(q + 2)
2(q + 1)
2(q + 1)(q + 2)
2(q + 1)(q + 2)
1@ 6 · 26
1
d6
in using v5 is U5 $
=
E 0=00521.
=
1 3 lim uq
1 3 1@2
192
uq+1 3 uq =
q<"
(b) The error in using vq as an approximation to the sum is Uq =
dq+1
2
=
. We want Uq ? 0=00005 C
(q + 1)2q+1
1 3 12
1
? 0=00005 C (q + 1)2q A 20,000. To find such an q we can use trial and error or a graph. We calculate
(q + 1)2q
(11 + 1)211 = 24,576, so v11 =
11
S
q=1
40. v10 =
1
E 0=693109 is within 0=00005 of the actual sum.
q2q
10 q
S
2 3
10
q+1
1
1
1
dq+1
q + 1 2q
+
+
+
·
·
·
+
E
1=988.
The
ratios
u
=
=
1
+
form a
=
=
=
·
q
q
2
4 8
1024
dq
2q+1 q
2q
2
q
q=1 2
decreasing sequence, and u11 =
of the series
11 + 1
12
6
=
=
? 1, so by Exercise 38(a), the error in using v10 to approximate the sum
2(11)
22
11
11
" q
S
d11
121
E 0=0118.
is U10 $
= 20486 =
q
1 3 u11
10,240
1 3 11
q=1 2
41. (i) Following the hint, we get that |dq | ? uq for q D Q, and so since the geometric series
S"
uq converges [0 ? u ? 1],
S
S"
S"
the series "
q=Q |dq | converges as well by the Comparison Test, and hence so does
q=1 |dq |, so
q=1 dq is absolutely
q=1
convergent.
(ii) If lim
q<"
s
s
q
|dq | = O A 1, then there is an integer Q such that q |dq | A 1 for all q D Q, so |dq | A 1 for q D Q. Thus,
lim dq 6= 0, so
q<"
(iii) Consider
S"
q=1
dq diverges by the Test for Divergence.
" 1
" 1
s
S
S
[converges]. For each sum, lim q |dq | = 1, so the Root Test is inconclusive.
[diverges] and
2
q<"
q
q
q=1
q=1
dq+1 [4(q + 1)]! [1103 + 26,390(q + 1)]
(q!)4 3964q
= lim
42. (a) lim ·
q<"
dq q<" (4q)! (1103 + 26,390q) [(q + 1)!]4 3964(q+1)
(4q + 4)(4q + 3)(4q + 2)(4q + 1)(26,390q + 27,493)
44
1
=
= 4 ? 1,
4
4
q<"
(q + 1) 396 (26,390q + 1103)
3964
99
= lim
so by the Ratio Test, the series
(b)
" (4q)! (1103 + 26,390q)
S
converges.
(q!)4 3964q
q=0
I
" (4q)! (1103 + 26,390q)
2 2 S
1
=
9801 q=0
(q!)4 3964q
I
1
2 2 1103
With the first term (q = 0), E
·
9801
1
i E 3=141 592 73, so we get 6 correct decimal places of ,
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SECTION 11.6
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
¤
155
which is 3=141 592 653 589 793 238 to 18 decimal places.
I 1
2 2 1103
4! (1103 + 26,390)
i E 3=141 592 653 589 793 878, so
With the second term (q = 1), E
+
9801
1
3964
we get 15 correct decimal places of .
3
+
3
dq is absolutely convergent, and since d+
q $ |dq | and dq $ |dq | (because dq and dq each equal
S
S 3
dq must be absolutely convergent.
either dq or 0), we conclude by the Comparison Test that both d+
q and
43. (a) Since
S
Or: Use Theorem 11.2.8.
S
S 3
S
(b) We will show by contradiction that both d+
dq must diverge. For suppose that d+
q and
q converged. Then so
S
S + 1 S 1
S + 1 (dq + |dq |) 3 12 dq = 12
|dq |, which
dq 3 2 dq =
would
dq 3 2 dq by Theorem 11.2.8. But
2
S +
S
S
diverges because dq is only conditionally convergent. Hence, dq can’t converge. Similarly, neither can d3
q.
44. Let
S
eq be the rearranged series constructed in the hint. [This series can be constructed by virtue of the result of
Exercise 43(b).] This series will have partial sums vq that oscillate in value back and forth across u. Since lim dq = 0
q<"
(by Theorem 11.2.6), and since the size of the oscillations |vq 3 u| is always less than |dq | because of the way
S
constructed, we have that eq = lim vq = u.
q<"
45. Suppose that
(a)
S
S
S
eq was
dq is conditionally convergent.
q2 dq is divergent: Suppose
S
integer Q A 0 such that q A Q
q2 dq converges. Then lim q2 dq = 0 by Theorem 6 in Section 11.2, so there is an
q<"
i q2 |dq | ? 1. For q A Q, we have |dq | ?
S
1
, so
|dq | converges by
q2
qAQ
S 1
S
. In other words,
dq converges absolutely, contradicting the
2
q
qAQ
S
S
assumption that dq is conditionally convergent. This contradiction shows that q2 dq diverges.
S
Remark: The same argument shows that qs dq diverges for any s A 1.
comparison with the convergent s-series
(b)
" (31)q
S
is conditionally convergent. It converges by the Alternating Series Test, but does not converge absolutely
q=2 q ln q
1
by the Integral Test, since the function i ({) =
is continuous, positive, and decreasing on [2> ") and
{ ln {
] w
] "
k
lw
(31)q
g{
g{
= lim
= lim ln(ln {) = " . Setting dq =
for q D 2, we find that
{ ln { w<" 2 { ln { w<"
q ln q
2
2
"
S
qdq =
q=2
" (31)q
S
converges by the Alternating Series Test.
q=2 ln q
It is easy to find conditionally convergent series
S
dq such that
S
qdq diverges. Two examples are
" (31)q31
S
and
q
q=1
" (31)q31
S
S
I
, both of which converge by the Alternating Series Test and fail to converge absolutely because |dq | is a
q
q=1
S
s-series with s $ 1. In both cases, qdq diverges by the Test for Divergence.
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156
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
11.7 Strategy for Testing Series
1.
1
1
? q =
q + 3q
3
q
1
for all q D 1.
3
converges by the Comparison Test.
s
2. lim q |dq | = lim
q<"
q<"
q
1
is a convergent geometric series |u| =
3
q=1
"
S
1
3
"
S
? 1 , so
q=1
1
q + 3q
v
q
" (2q + 1)q
S
2q + 1
2
1
q (2q + 1) = 0 ? 1, so the series
+
=
lim
=
lim
q2q
q<" q2
2
q<"
q
q
q2q
q=1
converges by the Root Test.
3. lim |dq | = lim
q<"
q<"
"
S
q
q
q
(31)q
= 1, so lim dq = lim (31)q
does not exist. Thus, the series
diverges by
q<"
q<"
q+2
q+2
q
+
2
q=1
the Test for Divergence.
4. eq =
q
A 0 for q D 1. {eq } is decreasing for q D 2 since
q2 + 2
for { D
I
2. Also, lim eq = lim
q<"
q<" q2
Alternating Series Test.
{
{2 + 2
0
=
({2 + 2)(1) 3 {(2{)
2 3 {2
= 2
?0
2
2
({ + 2)
({ + 2)2
"
S
q
1@q
q
= lim
converges by the
= 0. Thus, the series
(31)q 2
2
+ 2 q<" 1 + 2@q
q +2
q=1
2 q
q 2
dq+1 = lim (q + 1) 2 · (35) = lim 2(q + 1) = 2 lim 1 + 1
5. lim q+1
2
q31
2
q<"
q<"
q<"
dq
q 2
5q
5 q<"
q
(35)
2
=
2
2
(1) = ? 1, so the series
5
5
" q2 2q31
S
q converges by the Ratio Test.
q=1 (35)
6. Use the Limit Comparison Test with dq =
Since the harmonic series
1
1
and eq = :
2q + 1
q
lim
q<"
dq
q
1
1
= lim
= lim
= A 0=
q<" 2q + 1
q<" 2 + (1@q)
eq
2
" 1
"
S
S
1
diverges, so does
. [Or: Use the Integral Test.]
q
2q
+1
q=1
q=1
1
I
. Then i is positive, continuous, and decreasing on [2> "), so we can apply the Integral Test.
ln {
]
]
I
1
x = ln {,
I
g{
Since
=
x31@2 gx = 2x1@2 + F = 2 ln { + F, we find
gx
=
g{@{
{ ln {
] "
] w
lw
k I
I
I
g{
g{
I
I
= lim
= lim 2 ln { = lim 2 ln w 3 2 ln 2 = ". Since the integral diverges, the
w<"
2
{ ln { w<" 2 { ln { w<"
2
7. Let i ({) =
given series
{
"
S
q=2
8.
q
1
I
diverges.
ln q
"
S
2n n!
2n
=
. Using the Ratio Test, we get
n=1 (n + 2)!
n=1 (n + 1)(n + 2)
dn+1 2n+1
n+1
(n + 1)(n + 2) = lim 2
·
lim ·
= 2 A 1, so the series diverges.
=
lim
n<"
n<"
dn n<" (n + 2)(n + 3)
2n
n+3
"
S
Or: Use the Test for Divergence.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.7
9.
STRATEGY FOR TESTING SERIES
¤
" n2
S
. Using the Ratio Test, we get
n
n=1
n=1 h
%
&
2
dn+1 (n + 1)2 hn 1
1
n
+
1
1
= lim lim · 2 = lim
·
= 12 · = ? 1, so the series converges.
n<" dn n<" hn+1
n<"
n
n
h
h
h
"
S
157
n2 h3n =
{(2 3 3{3 )
? 0 for { D 1, so i is
h{3
k
l
U"
3
3 w
1
decreasing on [1> ") as well, and we can apply the Integral Test. 1 {2 h3{ g{ = lim 3 13 h3{
= 3h
, so the integral
3
10. Let i({) = {2 h3{ . Then i is continuous and positive on [1> "), and i 0 ({) =
w<"
1
converges, and hence, the series converges.
11.
"
S
q=1
1
1
+ q
q3
3
" 1
"
S
S
+
=
3
q=1 q
q=1
q
1
. The first series converges since it is a s-series with s = 3 A 1 and the second
3
series converges since it is geometric with |u| =
12.
1
3
? 1. The sum of two convergent series is convergent.
"
" 1
S
S
1
1
1
1
I
I
? I = 2 , so
converges by comparison with the convergent s-series
2
2
2
n
n n2 + 1
n
n
+
1
n n
n=1
n=1 n
( s = 2 A 1).
q+1
" 3q q2
dq+1 S
(q + 1)2
3(q + 1)2
q+1
q! = lim 3
=
lim
13. lim =
3
lim
=
0
?
1,
so
the
series
·
q<"
q<" q2
dq q<" (q + 1)!
3q q2 q<" (q + 1)q2
q=1 q!
converges by the Ratio Test.
sin 2q $
14. q
1
1
? q =
1 + 2q
2
1+2
q
"
S
1
with |u| =
2
q=1
2n31 3n+1
nn
15. dn =
q
" sin 2q S
1
converges by comparison with the geometric series
, so the series
q
2
q=1 1 + 2
" sin 2q
S
converges absolutely, implying convergence.
q
q=1 1 + 2
v n
n
6
6
2n 231 3n 31
3 2·3
n
= 0 ? 1, so the series
=
=
.
By
the
Root
Test,
lim
= lim
n<"
n<" n
nn
2
n
n
1
2
? 1. Thus, the series
n
n
" 2n31 3n+1
" 3
S
S
6
6
converges. It follows from Theorem 8(i) in Section 11.2 that the given series,
=
,
n
nn
n
n=1
n=1
n=1 2
"
S
also converges.
16. Using the Limit Comparison Test with dq =
q2 + 1
1
and eq = , we have
q3 + 1
q
2
"
S
dq
q3 + q
1 + 1@q2
q +1 q
= lim
= 1 A 0. Since
eq is the divergent harmonic
·
= lim 3
= lim
3
3
q<" eq
q<" q + 1
q<" q + 1
q<" 1 + 1@q
1
q=1
lim
series,
"
S
dq is also divergent.
q=1
dq+1 = lim 1 · 3 · 5 · · · · · (2q 3 1)(2q + 1) · 2 · 5 · 8 · · · · · (3q 3 1) = lim 2q + 1
17. lim q<"
q<"
dq
2 · 5 · 8 · · · · · (3q 3 1)(3q + 2) 1 · 3 · 5 · · · · · (2q 3 1) q<" 3q + 2
= lim
q<"
so the series
2 + 1@q
2
= ? 1>
3 + 2@q
3
" 1 · 3 · 5 · · · · · (2q 3 1)
S
converges by the Ratio Test.
q=1 2 · 5 · 8 · · · · · (3q 3 1)
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°
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¤
158
CHAPTER 11
INFINITE SEQUENCES AND SERIES
" (31)q31
S
1
I
for q D 2. {eq } is a decreasing sequence of positive numbers and lim eq = 0, so
converges by
q<"
q31
q31
q=2
18. eq = I
the Alternating Series Test.
ln {
19. Let i({) = I . Then i 0 ({) =
{
2 3 ln {
ln q
? 0 when ln { A 2 or { A h2 , so I is decreasing for q A h2 .
2{3@2
q
"
S
ln q
1@q
2
ln q
I = lim I = 0, so the series
By l’Hospital’s Rule, lim I = lim
(31)q I converges by the
q<"
q<"
q<"
q=1
q
q
q
1@ 2 q
Alternating Series Test.
20. dn =
I
I
I
I
3
3
3
3
"
S
n31
n
n
n31
1
n1@3
I
I
? I
? I = 3@2 = 7@6 , so the series
converges by comparison with the
n
n
n( n + 1)
n( n + 1)
n n
n=1 n( n + 1)
convergent s-series
1 s=
7@6
n
n=1
"
S
7
6
A1 .
21. lim |dq | = lim (31)q cos(1@q2 ) = lim cos(1@q2 ) = cos 0 = 1, so the series
q<"
q<"
Test for Divergence.
q<"
1
"
S
(31)q cos(1@q2 ) diverges by the
q=1
1
= lim
, which does not exist (the terms vary between
22. lim |dn | = lim n<"
n<" 2 + sin n n<" 2 + sin n
"
S
n=1
1
3
and 1). Thus, the series
1
diverges by the Test for Divergence.
2 + sin n
1
1
and eq = , we have
q
q
23. Using the Limit Comparison Test with dq = tan
dq
tan(1@q)
tan(1@{) H
sec2 (1@{) · (31@{2 )
= lim
= lim
= lim
= lim sec2 (1@{) = 12 = 1 A 0. Since
q<" eq
q<"
{<"
{<"
{<"
1@q
1@{
31@{2
lim
"
S
eq is the divergent harmonic series,
q=1
"
S
dq is also divergent.
q=1
"
S
sin(1@q)
sin {
1
= lim
= lim
= 1 6= 0, so the series
q sin(1@q) diverges by the
q sin
+
q<"
q<"
q
1@q
{
{<0
q=1
24. lim dq = lim
q<"
Test for Divergence.
2
2
" q!
dq+1 S
(q + 1)q! · hq
hq q+1
= lim (q + 1)!
= lim 2q+1 = 0 ? 1, so
25. Use the Ratio Test. lim = lim
2 ·
2 +2q+1
(q+1)
q
q2
q<"
q<" h
q<" h
dq
q! q<" h
q!
q=1 h
converges.
2
q
dq+1 = lim dq+1 = lim q + 2q + 2 · 5
26. lim q+1
2
q<"
q<" dq
q<"
dq
5
q +1
= lim
q<"
1 + 2@q + 2@q2 1
·
1 + 1@q2
5
converges by the Ratio Test.
27.
]
2
"
w
ln {
ln {
1
3
3
g{
=
lim
w<"
{2
{
{ 1
H
[using integration by parts] = 1. So
=
" q2 + 1
S
1
? 1, so
5
5q
q=1
" ln q
S
converges by the Integral Test, and since
2
q=1 q
"
S
n ln n
ln n
n ln n
n ln n
=
,
the
given
series
converges by the Comparison Test.
3 ?
n3
n2
(n + 1)
(n
+ 1)3
n=1
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.7
STRATEGY FOR TESTING SERIES
¤
159
" h
" h1@q
S
S
1
28. Since
converges (s = 2 A 1), so
is a decreasing sequence, h1@q $ h1@1 = h for all q D 1, and
2
2
q
q=1 q
q=1 q
converges by the Comparison Test. (Or use the Integral Test.)
29.
"
S
dq =
q=1
"
S
(31)q
q=1
"
S
1
1
=
A 0, {eq } is decreasing, and lim eq = 0, so the series
(31)q eq . Now eq =
q<"
cosh q q=1
cosh q
converges by the Alternating Series Test.
Or: Write
" 1
"
S
S
1
2
1
2
? q and
is a convergent geometric series, so
= q
is convergent by the
q
cosh q
h + h3q
h
h
cosh
q
q=1
q=1
"
S
Comparison Test. So
(31)q
q=1
1
is absolutely convergent and therefore convergent.
cosh q
I
{
53{
30. Let i ({) =
? 0 for { A 5, i ({) is
. Then i ({) is continuous and positive on [1> "), and since i 0 ({) = I
{+5
2 { ({ + 5)2
I
q
1
eventually decreasing, so we can use the Alternating Series Test. lim
= lim 1@2
= 0, so the series
q<" q + 5
q<" q
+ 5q31@2
I
"
S
m
converges.
(31) m
m+5
m=1
5n
= [divide by 4n ]
n<" 3n + 4n
31. lim dn = lim
n<"
Thus,
n
n
(5@4)n
3
5
=
"
since
lim
=
0
and
lim
= ".
n<" (3@4)n + 1
n<" 4
n<" 4
lim
5n
diverges by the Test for Divergence.
n
n
n=1 3 + 4
"
S
v
q
q!
q q31 q32 q33
q (q!) ·
·
·
· (q 3 4)!
lim 4 = lim
q4q = q<"
q<"
q<" q
q
q
q
q
2
3
1
13
13
(q 3 4)! = ",
13
= lim
q<"
q
q
q
s
32. lim q |dq | = lim
q<"
so the series
" (q!)q
S
diverges by the Root Test.
4q
q=1 q
s
33. lim q |dq | = lim
q<"
q<"
q
q+1
q2 @q
"
S
1
1
1
= lim
= ? 1, so the series
q =
q<" [(q + 1) @q]
lim (1 + 1@q)q
h
q=1
q<"
q
q+1
q2
converges by the Root Test.
34. 0 $ q cos2 q $ q, so
"
" 1
S
S
1
1
1
1
D
=
. Thus,
diverges by comparison with
, which is
2
2
q + q cos q
q+q
2q
q=1 q + q cos q
q=1 2q
a constant multiple of the (divergent) harmonic series.
35. dq =
1
1
1
=
, so let eq = and use the Limit Comparison Test.
q
q1+1@q
q · q1@q
[see Exercise 3.4.61], so the series
"
S
q=1
lim
q<"
dq
1
= lim 1@q = 1 A 0
q<" q
eq
1
diverges by comparison with the divergent harmonic series.
q1+1@q
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160
¤
CHAPTER 11 INFINITE SEQUENCES AND SERIES
ln q
36. Note that (ln q)
ln q ln q ln ln q
= hln ln q
= h
= qln ln q and ln ln q < " as q < ", so ln ln q A 2 for sufficiently
large q. For these q we have (ln q)ln q A q2 , so
"
S
1
by the Comparison Test.
(ln q)ln q
q=2
37. lim
q<"
" 1
S
1
1
? 2 . Since
converges [s = 2 A 1], so does
ln q
2
q
(ln q)
q=2 q
" I
s
q
S
q
q
|dq | = lim (21@q 3 1) = 1 3 1 = 0 ? 1, so the series
2 3 1 converges by the Root Test.
q<"
q=1
I
38. Use the Limit Comparison Test with dq = q 2 3 1 and eq = 1@q. Then
lim
q<"
21@{ · ln 2 · (31@{2 )
dq
21@q 3 1
21@{ 3 1 H
= lim
= lim
= lim (21@{ · ln 2) = 1 · ln 2 = ln 2 A 0.
= lim
q<"
{<"
{<"
{<"
eq
1@q
1@{
31@{2
So since
"
S
eq diverges (harmonic series), so does
q=1
" I
S
q
231 .
q=1
I
q
231 =
1
1
[rationalize the numerator] D
,
2q
2(q31)@q + 2(q32)@q + 2(q33)@q + · · · + 21@q + 1
" 1
" 1
" I
S
S
1 S
q
=
diverges (harmonic series), so does
2 3 1 by the Comparison Test.
and since
2 q=1 q
q=1 2q
q=1
Alternate solution:
11.8 Power Series
1. A power series is a series of the form
S"
q=0
fq {q = f0 + f1 { + f2 {2 + f3 {3 + · · · , where { is a variable and the fq ’s are
constants called the coefficients of the series.
S
q
2
More generally, a series of the form "
q=0 fq ({ 3 d) = f0 + f1 ({ 3 d) + f2 ({ 3 d) + · · · is called a power series in
({ 3 d) or a power series centered at d or a power series about d, where d is a constant.
S
q
2. (a) Given the power series "
q=0 fq ({ 3 d) , the radius of convergence is:
(i) 0 if the series converges only when { = d
(ii) " if the series converges for all {, or
(iii) a positive number U such that the series converges if |{ 3 d| ? U and diverges if |{ 3 d| A U.
In most cases, U can be found by using the Ratio Test.
(b) The interval of convergence of a power series is the interval that consists of all values of { for which the series converges.
Corresponding to the cases in part (a), the interval of convergence is: (i) the single point {d}, (ii) all real numbers; that is,
the real number line (3"> "), or (iii) an interval with endpoints d 3 U and d + U 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.
3. If dq = (31)q q{q , then
dq+1 (31)q+1 (q + 1){q+1 1
q + 1 1+
{ = lim
|{| = |{|. By the Ratio Test, the
= lim lim (31)
lim
= q<"
q<" dq q<"
q<"
(31)q q{q
q
q
series
"
S
(31)q q{q converges when |{| ? 1, so the radius of convergence U = 1. Now we’ll check the endpoints, that is,
q=1
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SECTION 11.8 POWER SERIES
{ = ±1. Both series
"
S
(31)q q(±1)q =
q=1
"
S
161
(~1)q q diverge by the Test for Divergence since lim |(~1)q q| = ". Thus,
q<"
q=1
the interval of convergence is L = (31> 1).
¤
(31)q {q
I
, then
3
q
u
I
I q+1 q+1
3
dq+1 (31){ 3 q {
q 1
= lim (31)
I
= lim 3
I
|{| = |{|. By the Ratio Test,
=
lim
·
lim 3
3
q {q q<"
q<"
q<"
dq q<" (31)
1
+
1@q
q+1
q+1
4. If dq =
the series
" (31)q {q
" (31)q
S
S
I
I
converges
when
|{|
?
1,
so
U
=
1.
When
{
=
1,
the
series
converges by the Alternating
3
3
q
q
q=1
q=1
Series Test. When { = 31, the series
is (31> 1].
5. If dq =
"
S
q=1
1
I
diverges since it is a s-series s =
3
q
1
3
$ 1 . Thus, the interval of convergence
q+1
dq+1 2 3 1@q
{q
2q 3 1
2q 3 1 = lim {
=
lim
, then lim ·
|{|
=
lim
|{|
= |{|. By
q<"
q<"
2q 3 1
dq q<" 2q + 1
{q q<" 2q + 1
2 + 1@q
"
S
the Ratio Test, the series
q=1
"
S
{q
1
converges when |{| ? 1, so U = 1. When { = 1, the series
diverges by
2q 3 1
2q
31
q=1
" 1
" 1
S
1
1
1 S
since
A
and
diverges since it is a constant multiple of the harmonic series.
comparison with
2q
2q
3
1
2q
2
q=1
q=1 q
When { = 31, the series
" (31)q
S
converges by the Alternating Series Test. Thus, the interval of convergence is [31> 1).
q=1 2q 3 1
(31)q {q
, then
q2
%
2 &
dq+1 (31)q+1 {q+1
(31){q2 q2
q
= lim = lim
= lim
lim
·
|{| = 12 · |{| = |{|.
q<" dq q<"
(q + 1)2
(31)q {q q<" (q + 1)2 q<"
q+1
6. If dq =
When { = 1, the series
" (31)q
" 1
S
S
converges by the Alternating Series Test. When { = 31, the series
converges
2
2
q
q=1
q=1 q
since it is a s-series with s = 2 A 1. Thus, the interval of convergence is [31> 1].
7. If dq =
q+1
dq+1 { 1
q! {q
= |{| lim
= lim {
, then lim ·
= |{| · 0 = 0 ? 1 for all real {.
=
lim
q<"
q<" q + 1
q!
dq q<" (q + 1)! {q q<" q + 1 So, by the Ratio Test, U = " and L = (3"> ").
8. Here the Root Test is easier. If dq = qq {q then lim
q<"
s
q
|dq | = lim q |{| = " if { 6= 0, so U = 0 and L = {0}.
q<"
q2 {q
, then
2q
%
2 &
2 q+1
dq+1 {(q + 1)2 |{| 2
2q = lim (q + 1) {
= lim |{| 1 + 1
lim =
lim
·
=
(1) =
q<"
dq q<" 2q+1
q2 {q q<" 2q2 q<" 2
q
2
9. If dq = (31)q
Ratio Test, the series
"
S
(31)q
q=1
When { = ±2, both series
"
S
q2 {q
converges when
2q
(31)q
q=1
1
2
1
2
|{|. By the
|{| ? 1 C |{| ? 2, so the radius of convergence is U = 2.
"
S
q2 (±2)q
=
(~1)q q2 diverge by the Test for Divergence since
2q
q=1
lim (~1)q q2 = ". Thus, the interval of convergence is L = (32> 2).
q<"
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162
¤
CHAPTER 11 INFINITE SEQUENCES AND SERIES
10. If dq =
10q {q
, then
q3
q+1 q+1
dq+1 10{ q3 {
10 |{|
q3 10 |{|
= lim 10
= lim
=
lim
·
=
= 10 |{|
lim q<"
dq q<" (q + 1)3
10q {q q<" (q + 1)3 q<" (1 + 1@q)3
13
By the Ratio Test, the series
" 10q {q
S
converges when 10 |{| ? 1 C |{| ?
q3
q=1
1
, the series converges by the Alternating Series Test; when { =
When { = 3 10
1 1
with s = 3 A 1. Thus, the interval of convergence is L = 3 10
> 10 .
11. If dq =
1
,
10
1
,
10
so the radius of convergence is U =
1
.
10
the series converges because it is a p-series
(33)q {q
, then
q3@2
3@2 3@2
dq+1 (33)q+1 {q+1
q
q3@2 1
33{
=
3
|{|
lim
= lim =
lim
lim ·
q<"
q<"
q<" q<"
dq
(33)q {q
q+1
1 + 1@q
(q + 1)3@2
= 3 |{| (1) = 3 |{|
By the Ratio Test, the series
" (33)q
S
I {q converges when 3 |{| ? 1 C |{| ? 13 , so U = 13 . When { = 13 , the series
q
q=1 q
" (31)q
"
S
S
1
converges by the Alternating Series Test. When { = 3 13 , the series
is a convergent s-series
3@2
3@2
q
q
q=1
q=1
s = 32 A 1 . Thus, the interval of convergence is 3 13 > 13 .
12. If dq =
{q
, then
q3q
dq+1 {q+1
|{|
|{|
q
1
1
q3q lim
= lim
= |{|
= lim · q = lim
q<" dq q<" (q + 1)3q+1
q<" 3
q<" 3
{
q+1
1 + 1@q
3
By the Ratio Test, the series
" {q
S
converges when
q
q=1 q3
1
3
the divergent harmonic series. When { = 33, the series
by 31). Thus, the interval of convergence is [33> 3).
13. If dq = (31)q
"
S
(31)q
q=2
(31)q
q=2
(31)q
q=1
1
is the convergent alternating harmonic series (multiplied
q
|{|
|{|
. By the Ratio Test, the series converges when
?1
4
4
C
|{| ? 4, so U = 4. When
" [(31)(34)]q
"
" 1
S
S
S
{q
1
1
1
=
=
. Since ln q ? q for q D 2,
A and
is the
q
ln q q=2
4 ln q
ln q
q
q=2 ln q
q=2 q
4q
divergent harmonic series (without the q = 1 term),
"
S
"
S
" 1
S
is
q=1 q
dq+1 {q+1
{q
ln q
4q ln q |{|
|{|
= lim =
,
then
lim
·
lim
=
·1
q<" dq q<" 4q+1 ln(q + 1)
4q ln q
{q 4 q<" ln(q + 1)
4
[by l’Hospital’s Rule] =
{ = 34,
|{| ? 1 C |{| ? 3, so U = 3. When { = 3, the series
"
S
1
is divergent by the Comparison Test. When { = 4,
ln
q
q=2
"
S
{q
1
(31)q
=
, which converges by the Alternating Series Test. Thus, L = (34> 4].
ln q q=2
ln q
4q
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.8 POWER SERIES
¤
163
{2q+1
, then
(2q + 1)!
q+1 2q+3
dq+1 (31){2
{
1
(2q + 1)! = {2 lim
= lim (31)
=
lim
lim ·
q<"
q<" (2q + 3)(2q + 2)
dq q<" (2q + 3)!
(31)q {2q+1 q<" (2q + 3)(2q + 2) 14. If dq = (31)q
= {2 · 0 = 0 ? 1
So, by the Ratio Test, the series
"
S
(31)q
q=0
15. If dq =
{2q+1
converges for all real {. Thus, U = " and L = (3"> ").
(2q + 1)!
q+1
dq+1 q2 + 1
q2 + 1 ({ 3 2)q
= lim ({ 3 2)
,
then
lim
·
= |{ 3 2|. By the
= |{ 3 2| lim
2
2
q
q<"
q<"
q<"
q +1
dq
(q + 1) + 1 ({ 3 2)
(q + 1)2 + 1
" ({ 3 2)q
S
converges when |{ 3 2| ? 1 [U = 1] C 31 ? { 3 2 ? 1 C 1 ? { ? 3. When
2
q=0 q + 1
Ratio Test, the series
"
S
{ = 1, the series
(31)q
q=0
"
S
1
1
converges by the Alternating Series Test; when { = 3, the series
converges by
2
q2 + 1
q=0 q + 1
comparison with the p-series
16. If dq = (31)q
" 1
S
[s = 2 A 1]. Thus, the interval of convergence is L = [1> 3].
2
q=1 q
q+1
dq+1 ({ 3 3)q
2q + 1
2q + 1 = lim ({ 3 3)
, then lim ·
= |{ 3 3|. By the
= |{ 3 3| lim
q<"
q<" 2q + 3
2q + 1
dq q<" 2q + 3
({ 3 3)q "
S
Ratio Test, the series
(31)q
q=0
When { = 2, the series
"
S
q=0
{ = 4, the series
"
S
(31)q
q=0
({ 3 3)q
converges when |{ 3 3| ? 1 [U = 1] C 31 ? { 3 3 ? 1 C 2 ? { ? 4.
2q + 1
1
diverges by limit comparison with the harmonic series (or by the Integral Test); when
2q + 1
1
converges by the Alternating Series Test. Thus, the interval of convergence is L = (2> 4].
2q + 1
I
I
3q+1 ({ + 4)q+1
dq+1 q
q
3q ({ + 4)q
= lim I
I
I
=
3
|{
+
4|
lim
17. If dq =
·
= 3 |{ + 4|.
, then lim q<"
q<"
dq q<" 3q ({ + 4)q q+1
q+1
q
By the Ratio Test, the series
3 13 ? { + 4 ?
1
3
" 3q ({ + 4)q
S
I
converges when 3 |{ + 4| ? 1 C |{ + 4| ?
q=1
q
11
13
C 3 13
3 ? { ? 3 3 . When { = 3 3 , the series
Test; when { = 3 11
, the series
3
18. If dq =
"
S
1
I diverges s =
q=1
q
1
2
"
S
1
3
U = 13
C
1
(31)q I converges by the Alternating Series
q
q=1
.
$ 1 . Thus, the interval of convergence is L = 3 13
> 3 11
3
3
q+1
dq+1 |{ + 1|
q+1
|{ + 1|
q
4q
q
= lim (q + 1)({ + 1)
=
lim
=
.
({
+
1)
,
then
lim
·
q<" dq q<" 4q
4q+1
q({ + 1)q 4 q<" q
4
By the Ratio Test, the series
" q
S
|{ + 1|
({ + 1)q converges when
? 1 C |{ + 1| ? 4 [U = 4] C
q
4
4
q=1
34 ? { + 1 ? 4 C 35 ? { ? 3. When { = 35 or 3, both series
"
S
(~1)q q diverge by the Test for Divergence since
q=1
lim |(~1)q q| = ". Thus, the interval of convergence is L = (35> 3).
q<"
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¤
CHAPTER 11 INFINITE SEQUENCES AND SERIES
19. If dq =
s
|{ 3 2|
({ 3 2)q
, then lim q |dq | = lim
= 0, so the series converges for all { (by the Root Test).
q<"
q<"
qq
q
U = " and L = (3"> ").
(2{ 3 1)q
I
, then
5q q
u
I
u
q+1
dq+1 |2{ 3 1|
q
|2{ 3 1|
1
|2{ 3 1|
5q q = lim (2{ 3I1)
=
lim
=
.
=
lim
·
lim q<"
dq q<" 5q+1 q + 1 (2{ 3 1)q q<"
5
q + 1 q<"
5
1 + 1@q
5
20. If dq =
By the Ratio Test, the series
3 52 ? { 3
1
2
?
5
2
"
S
1
I is a divergent s-series s =
q
q=1
C 32 ? { ? 3, so U = 52 . When { = 3, the series
When { = 32, the series
is L = [32> 3).
" (2{ 3 1)q
S
|2{ 3 1|
I
? 1 C |2{ 3 1| ? 5 C { 3 12 ?
converges when
q
5
5 q
q=1
1
2
" (31)q
S
I
converges by the Alternating Series Test. Thus, the interval of convergence
q
q=1
5
2
C
$1 .
q
({ 3 d)q , where e A 0.
eq
q+1
dq+1 eq
1 |{ 3 d|
|{ 3 d|
= lim (q + 1) |{ 3 d|
·
=
lim
1
+
lim =
.
q
q<"
q<"
dq q<"
eq+1
q |{ 3 d|
q
e
e
21. dq =
By the Ratio Test, the series converges when
|{ 3 d|
? 1 C |{ 3 d| ? e [so U = e] C 3e ? { 3 d ? e C
e
d 3 e ? { ? d + e. When |{ 3 d| = e, lim |dq | = lim q = ", so the series diverges. Thus, L = (d 3 e> d + e).
q<"
q<"
eq
({ 3 d)q , where e A 0.
ln q
q+1
dq+1 ({ 3 d)q+1
ln q
ln q
= lim e
= lim
lim ·
· e |{ 3 d| = e |{ 3 d| since
q
q
q<"
q<"
dq
ln(q + 1)
e ({ 3 d) q<" ln(q + 1)
22. dq =
lim
q<"
ln q
ln {
1@{
{+1 H
1
H
= lim
= lim
= 1. By the Ratio Test, the series
= lim
= lim
q<" ln({ + 1)
{<" 1@({ + 1)
{<"
{<" 1
ln(q + 1)
{
" eq
S
1
({ 3 d)q converges when e |{ 3 d| ? 1 C |{ 3 d| ?
e
q=2 ln q
so U =
C
3
1
1
?{3d?
e
e
C d3
1
1
?{?d+ ,
e
e
"
" 1
S
S
1
1
1
1
1
. When { = d + , the series
diverges by comparison with the divergent s-series
since
A
e
e
ln q
q
q=2 ln q
q=2 q
" (31)q
S
1
for q D 2. When { = d 3 , the series
converges by the Alternating Series Test. Thus, the interval of
e
q=2 ln q
1
1
convergence is L = d 3 > d +
.
e
e
for all { 6= 12 . Since the series diverges for all { 6= 12 , U = 0 and L =
= lim (q + 1) |2{ 3 1| < " as q < "
q<"
q+1 (q + 1)! (2{ 3 1)
dq+1 = lim 23. If dq = q! (2{ 3 1)q , then lim q<"
dq q<" q!(2{ 3 1)q
1
2
.
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SECTION 11.8 POWER SERIES
¤
165
q2 {q
q{q
q2 {q
= q
= q
, so
2 · 4 · 6 · · · · · (2q)
2 q!
2 (q 3 1)!
q+1
dq+1 q + 1 |{|
2q (q 3 1)!
= lim (q + 1) |{|
lim ·
= 0. Thus, by the Ratio Test, the series converges for
= lim
q+1
q<"
q<"
q<" q2
dq
2
q!
q |{|q
2
24. dq =
all real { and we have U = " and L = (3"> ").
(5{ 3 4)q
, then
q3
3
3
dq+1 (5{ 3 4)q+1
q3
q
1
= lim = lim |5{ 3 4|
·
= lim |5{ 3 4|
lim
q<" dq q<"
q<"
(q + 1)3
(5{ 3 4)q q<"
q+1
1 + 1@q
25. If dq =
= |5{ 3 4| · 1 = |5{ 3 4|
By the Ratio Test,
3
5
" (5{ 3 4)q
S
converges when |5{ 3 4| ? 1 C { 3 45 ?
3
q
q=1
? { ? 1, so U = 15 . When { = 1, the series
1
5
C 3 15 ? { 3
4
5
?
1
5
C
" 1
S
is a convergent s-series ( s = 3 A 1). When { = 35 , the series
3
q=1 q
" (31)q
S
converges by the Alternating Series Test. Thus, the interval of convergence is L = 35 > 1 .
q3
q=1
dq+1 {2q+2
{2q
q (ln q)2 2 q (ln q)2
=
lim
=
{
lim
26. If dq =
,
then
lim
·
= {2 .
q<"
q<" (q + 1)[ln(q + 1)]2
q<" (q + 1)[ln(q + 1)]2
q (ln q)2
dq
{2q
{2q
converges when {2 ? 1 C |{| ? 1, so U = 1. When { = ±1, {2q = 1, the
2
q=2 q (ln q)
"
S
By the Ratio Test, the series
series
"
S
q=2
1
converges by the Integral Test (see Exercise 11.3.22). Thus, the interval of convergence is L = [31> 1].
q (ln q)2
{q
, then
1 · 3 · 5 · · · · · (2q 3 1)
dq+1 |{|
1 · 3 · 5 · · · · · (2q 3 1) {q+1
= lim lim
lim ·
= 0 ? 1. Thus, by
= q<"
q<"
dq q<" 1 · 3 · 5 · · · · · (2q 3 1)(2q + 1)
{q
2q + 1
27. If dq =
{q
converges for all real { and we have U = " and L = (3"> ").
q=1 1 · 3 · 5 · · · · · (2q 3 1)
the Ratio Test, the series
"
S
q! {q
, then
1 · 3 · 5 · · · · · (2q 3 1)
dq+1 (q + 1)! {q+1
(q + 1) |{|
1 · 3 · 5 · · · · · (2q 3 1) lim
·
=
= lim lim
= q<"
q<" dq q<" 1 · 3 · 5 · · · · · (2q 3 1)(2q + 1)
q! {q
2q + 1
28. If dq =
By the Ratio Test, the series
"
S
q=1
|dq | =
dq converges when
1
2
1
2
|{|.
|{| ? 1 i |{| ? 2> so U = 2. When { = ±2,
[1 · 2 · 3 · · · · · q] 2q
2 · 4 · 6 · · · · · 2q
q! 2q
=
=
A 1, so both endpoint series
1 · 3 · 5 · · · · · (2q 3 1)
[1 · 3 · 5 · · · · · (2q 3 1)]
1 · 3 · 5 · · · · · (2q 3 1)
diverge by the Test for Divergence. Thus, the interval of convergence is L = (32> 2).
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166
¤
CHAPTER 11 INFINITE SEQUENCES AND SERIES
29. (a) We are given that the power series
S"
q
q=0 fq {
is convergent for { = 4. So by Theorem 3, it must converge for at least
34 ? { $ 4. In particular, it converges when { = 32; that is,
(b) It does not follow that
S"
q
q=0 fq (34)
S"
q=0
fq (32)q is convergent.
is necessarily convergent. [See the comments after Theorem 3 about convergence at
the endpoint of an interval. An example is fq = (31)q @(q4q ).]
30. We are given that the power series
S"
q
q=0 fq {
is convergent for { = 34 and divergent when { = 6. So by Theorem 3 it
converges for at least 34 $ { ? 4 and diverges for at least { D 6 and { ? 36. Therefore:
(a) It converges when { = 1; that is,
(b) It diverges when { = 8; that is,
S
S
(c) It converges when { = 33; that is,
(d) It diverges when { = 39; that is,
31. If dq =
fq is convergent.
fq 8q is divergent.
S
S
fq (33q ) is convergent.
fq (39)q =
S
(31)q fq 9q is divergent.
(q!)n q
{ , then
(nq)!
n
dq+1 (q + 1)n
= lim [(q + 1)!] (nq)! |{| = lim
lim |{|
n
q<"
q<" (q!) [n(q + 1)]!
q<" (nq + n)(nq + n 3 1) · · · (nq + 2)(nq + 1)
dq
(q + 1) (q + 1)
(q + 1)
···
|{|
q<" (nq + 1) (nq + 2)
(nq + n)
q+1
q+1
q+1
lim
· · · lim
|{|
= lim
q<" nq + 1 q<" nq + 2
q<" nq + n
= lim
n
1
=
|{| ? 1
n
C
|{| ? nn for convergence, and the radius of convergence is U = nn =
32. (a) Note that the four intervals in parts (a)–(d) have midpoint p =
know that the power series
"
S
1
2 (s + t)
and radius of convergence u = 12 (t 3 s). We also
{q has interval of convergence (31> 1). To change the radius of convergence to u, we can
q=0
change {q to
{ q
u
. To shift the midpoint of the interval of convergence, we can replace { with { 3 p. Thus, a power
series whose interval of convergence is (s> t) is
(b) Similar to Example 2, we know that
" { 3 p q
S
, where p = 12 (s + t) and u = 12 (t 3 s).
u
q=0
" {q
S
has interval of convergence [31> 1). By introducing the factor (31)q
q=1 q
in dq , the interval of convergence changes to (31> 1]. Now change the midpoint and radius as in part (a) to get
"
S
(31)q
q=1
1 { 3 p q
as a power series whose interval of convergence is (s> t].
q
u
(c) As in part (b),
" 1 { 3 p q
S
is a power series whose interval of convergence is [s> t).
u
q=1 q
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SECTION 11.8 POWER SERIES
¤
167
(d) If we increase the exponent on q (to say, q = 2), in the power series in part (c), then when { = t, the power series
"
S
1 { 3 p q
will converge by comparison to the p-series with s = 2 A 1, and the interval of convergence will
2
u
q=1 q
be [s> t].
33. No. If a power series is centered at d, its interval of convergence is symmetric about d. If a power series has an infinite radius
of convergence, then its interval of convergence must be (3"> "), not [0> ").
34. The partial sums of the series
S"
q=0
{q definitely do not converge
to i({) = 1@(1 3 {) for { D 1, since i is undefined at { = 1 and
negative on (1> "), while all the partial sums are positive on this
interval. The partial sums also fail to converge to i for { $ 31,
since 0 ? i ({) ? 1 on this interval, while the partial sums are
either larger than 1 or less than 0. The partial sums seem to
converge to i on (31> 1). This graphical evidence is consistent
with what we know about geometric series: convergence for
|{| ? 1, divergence for |{| D 1 (see Examples 1 and 5 in Section 11.2).
(31)q {2q+1
, then
q!(q + 1)! 22q+1
dq+1 {2q+3
1
q!(q + 1)! 22q+1 { 2
= lim lim = 0 for all {.
·
lim
= 2 q<"
q<"
dq q<" (q + 1)!(q + 2)! 22q+3
{2q+1
(q + 1)(q + 2)
35. (a) If dq =
So M1 ({) converges for all { and its domain is (3"> ").
(b), (c) The initial terms of M1 ({) up to q = 5 are d0 =
d1 = 3
{
,
2
{3
{5
{7
{9
, d2 =
, d3 = 3
, d4 =
,
16
384
18,432
1,474,560
and d5 = 3
{11
. The partial sums seem to
176,947,200
approximate M1 ({) well near the origin, but as |{| increases,
we need to take a large number of terms to get a good
approximation.
36. (a) D({) = 1 +
"
S
dq , where dq =
q=1
for all {, so the domain is R.
dq+1 1
{3q
= |{|3 lim
, so lim =0
q<"
q<" (3q + 2)(3q + 3)
2 · 3 · 5 · 6 · · · · · (3q 3 1)(3q)
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168
¤
CHAPTER 11 INFINITE SEQUENCES AND SERIES
v0 = 1 has been omitted from the graph. The
(b), (c)
partial sums seem to approximate D({) well
near the origin, but as |{| increases, we need to
take a large number of terms to get a good
approximation.
To plot D, we must first define D({) for the CAS. Note that for q D 1, the denominator of dq is
Tq
(3n 3 2) 3q
(3q)!
(3q)!
T
2 · 3 · 5 · 6 · · · · · (3q 3 1) · 3q =
= q
, so dq = n=1
{ and thus
1 · 4 · 7 · · · · · (3q 3 2)
(3n
3
2)
(3q)!
n=1
Tq
"
S
n=1 (3n 3 2) 3q
D({) = 1 +
{ . Both Maple and Mathematica are able to plot D if we define it this way, and Derive
(3q)!
q=1
is able to produce a similar graph using a suitable partial sum of D({).
Derive, Maple and Mathematica all have two initially known Airy functions, called AI·SERIES(z,m) and
BI·SERIES(z,m) from BESSEL.MTH in Derive and AiryAi and AiryBi in Maple and Mathematica (just Ai and
Bi in older versions of Maple). However, it is very difficult to solve for D in terms of the CAS’s Airy functions, although
I
3 AiryAi({) + AiryBi({)
.
in fact D({) = I
3 AiryAi(0) + AiryBi(0)
37. v2q31 = 1 + 2{ + {2 + 2{3 + {4 + 2{5 + · · · + {2q32 + 2{2q31
= 1(1 + 2{) + {2 (1 + 2{) + {4 (1 + 2{) + · · · + {2q32 (1 + 2{) = (1 + 2{)(1 + {2 + {4 + · · · + {2q32 )
= (1 + 2{)
1 3 {2q
1 + 2{
[by (11.2.3) with u = {2 ] <
as q < " by (11.2.4)], when |{| ? 1.
1 3 {2
1 3 {2
Also v2q = v2q31 + {2q <
approach
1 + 2{
1 + 2{
since {2q < 0 for |{| ? 1. Therefore, vq <
since v2q and v2q31 both
1 3 {2
1 3 {2
1 + 2{
1 + 2{
as q < ". Thus, the interval of convergence is (31> 1) and i ({) =
.
1 3 {2
1 3 {2
38. v4q31 = f0 + f1 { + f2 {2 + f3 {3 + f0 {4 + f1 {5 + f2 {6 + f3 {7 + · · · + f3 {4q31
f0 + f1 { + f2 {2 + f3 {3
= f0 + f1 { + f2 {2 + f3 {3 1 + {4 + {8 + · · · + {4q34 <
as q < "
1 3 {4
[by (11.2.4) with u = {4 ] for {4 ? 1 C |{| ? 1. Also v4q , v4q+1 , v4q+2 have the same limits
(for example, v4q = v4q31 + f0 {4q and {4q < 0 for |{| ? 1). So if at least one of f0 , f1 , f2 , and f3 is nonzero, then the
interval of convergence is (31> 1) and i ({) =
39. We use the Root Test on the series
|{| ? 1@f, so U = 1@f.
S
f0 + f1 { + f2 {2 + f3 {3
.
1 3 {4
fq {q . We need lim
q<"
s
s
q
|fq {q | = |{| lim q |fq | = f |{| ? 1 for convergence, or
q<"
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SECTION 11.9
40. Suppose fq 6= 0. Applying the Ratio Test to the series
S
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
¤
169
fq ({ 3 d)q , we find that
dq+1 fq+1 ({ 3 d)q+1 |{ 3 d|
= lim |{ 3 d| (W) =
(if lim |fq @fq+1 | 6= 0), so the
= lim
O = lim q<"
q<"
dq q<" fq ({ 3 d)q q<" |fq @fq+1 |
lim |fq @fq+1 |
q<"
series converges when
|{ 3 d|
?1
lim |fq @fq+1 |
C
q<"
fq . Thus, U = lim fq . If lim fq = 0
|{ 3 d| ? lim q<" fq+1 q<" fq+1 q<" fq+1 and |{ 3 d| 6= 0, then (W) shows that O = " and so the series diverges, and hence, U = 0. Thus, in all cases,
fq .
U = lim q<" fq+1 41. For 2 ? { ? 3,
S
fq {q diverges and
S
gq {q converges. By Exercise 11.2.69,
converge for |{| ? 2, the radius of convergence of
42. Since
S
fq {q converges whenever |{| ? U,
second series has radius of convergence
I
U.
S
S
(fq + gq ) {q is 2.
fq {2q =
S
S
(fq + gq ) {q diverges. Since both series
I
q
fq {2 converges whenever {2 ? U C |{| ? U, so the
11.9 Representations of Functions as Power Series
1. If i ({) =
"
S
fq {q has radius of convergence 10, then i 0 ({) =
q=0
"
S
qfq {q31 also has radius of convergence 10 by
q=1
Theorem 2.
2. If i ({) =
"
S
eq {q converges on (32> 2), then
q=0
U
i ({) g{ = F +
"
S
q=0
eq
{q+1 has the same radius of convergence
q+1
(by Theorem 2), but may not have the same interval of convergence—it may happen that the integrated series converges at an
endpoint (or both endpoints).
1
, and then use Equation (1) to represent the function as a sum of a power
13u
3. Our goal is to write the function in the form
series. i ({) =
4. i ({) =
|{|2 ?
5. i ({) =
"
"
S
S
1
1
(3{)q =
(31)q {q with |3{| ? 1 C |{| ? 1, so U = 1 and L = (31> 1).
=
=
1+{
1 3 (3{) q=0
q=0
5
=5
1 3 4{2
1
4
1
1 3 4{2
C |{| ? 12 , so U =
2
2
=
33{
3
1
1 3 {@3
1
2
=
=5
"
S
(4{2 )q = 5
q=0
and L = 3 12 > 12 .
"
S
q=0
4q {2q . The series converges when 4{2 ? 1 C
{
" { q
"
S
1
2 S
q
or, equivalently, 2
{
.
The
series
converges
when
? 1,
q+1
3 q=0 3
3
q=0 3
that is, when |{| ? 3, so U = 3 and L = (33> 3).
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170
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
" "
S
1
1
{ q
1
1
1 S
or, equivalently,
(31)q q+1 {q . The series converges
3
=
=
{ + 10
10 1 3 (3{@10)
10 q=0
10
10
q=0
{
when ? 1, that is, when |{| ? 10, so U = 10 and L = (310> 10).
10
6. i ({) =
q
"
"
"
S
1
1
{2q
{2q+1
{ S
{
{
{ S
{
{ 2
=
=
=
(31)q q =
(31)q q+1
=
3
2
2
2
9+{
9 1 + ({@3)
9 1 3 {3({@3) }
9 q=0
3
9 q=0
9
9
q=0
{ 2 q
2
"
{ 2 S
? 1 C { ? 1 C |{|2 ? 9 C |{| ? 3, so
The geometric series
converges when 3
3
3
3
9
q=0
7. i ({) =
U = 3 and L = (33> 3).
"
"
S
S
1
{
=
{
={
(32{2 )q or, equivalently,
(31)q 2q {2q+1 . The series converges when
2{2 + 1
1 3 (32{2 )
q=0
q=0
32{2 ? 1 i {2 ? 1 i |{| ? I1 , so U = I1 and L = 3 I1 > I1 .
2
2
2
2
2
8. i ({) =
9. i ({) =
"
"
"
"
"
"
S
S
S
S
S
S
1+{
1
{q =
{q +
{q+1 = 1 +
{q +
{q = 1 + 2
{q .
= (1 + {)
= (1 + {)
13{
13{
q=0
q=0
q=0
q=1
q=1
q=1
The series converges when |{| ? 1, so U = 1 and L = (31> 1).
A second approach: i ({) =
A third approach:
i({) =
"
"
S
S
3(1 3 {) + 2
1
1+{
=
= 31 + 2
= 31 + 2
{q = 1 + 2
{q .
13{
13{
13{
q=0
q=1
1
1+{
= (1 + {)
= (1 + {)(1 + { + {2 + {3 + · · · )
13{
13{
= (1 + { + {2 + {3 + · · · ) + ({ + {2 + {3 + {4 + · · · ) = 1 + 2{ + 2{2 + 2{3 + · · · = 1 + 2
"
S
{q .
q=1
3 q
"
" {3q+2
S
{
{2
{2
1
{2 S
=
.
=
=
. The series converges when {3 @d3 ? 1
3q+3
d3 3 {3
d3 1 3 {3 @d3
d3 q=0 d3
d
q=0
3 3
{ ? d C |{| ? |d|, so U = |d| and L = (3|d| > |d|).
10. i ({) =
11. i ({) =
3
D
E
3
=
=
+
{2 3 { 3 2
({ 3 2)({ + 1)
{32
{+1
C
i 3 = D({ + 1) + E({ 3 2). Let { = 2 to get D = 1 and
{ = 31 to get E = 31. Thus
" { q
"
S
3
1
1
1
1
1
1 S
=
3
=
3
=3
3
(3{)q
2
{ 3{32
{32
{+1
32 1 3 ({@2)
1 3 (3{)
2 q=0 2
q=0
q
"
"
S
S
1
1 1
q
q
q+1
=
3 1(31) { =
3 q+1 {q
3
(31)
2 2
2
q=0
q=0
We represented i as the sum of two geometric series; the first converges for { M (32> 2) and the second converges for (31> 1).
Thus, the sum converges for { M (31> 1) = L.
12. i ({) =
3 = 3E
{+2
D
E
{+2
=
=
+
2{2 3 { 3 1
(2{ + 1)({ 3 1)
2{ + 1
{31
i E = 1 and { = 3 12 to get
3
2
i { + 2 = D({ 3 1) + E(2{ + 1). Let { = 1 to get
= 3 32 D i D = 31. Thus,
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.9
{+2
31
1
=
+
3{31
2{ + 1
{31
2{2
=3
= 31
" S
(32)q + 1 {q
1
1 3 (32{)
31
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
1
13{
=3
"
S
(32{)q 3
q=0
"
S
¤
171
{q
q=0
q=0
We represented i as the sum of two geometric series; the first converges for { M 3 12 > 12 and the second converges for
(31> 1). Thus, the sum converges for { M 3 12 > 12 = L.
"
1
g
31
g S
(31)q {q
[from Exercise 3]
=3
2 =
g{ 1 + {
g{ q=0
(1 + {)
"
"
S
S
(31)q+1 q{q31 [from Theorem 2(i)] =
(31)q (q + 1){q with U = 1.
=
13. (a) i ({) =
q=1
q=0
In the last step, note that we decreased the initial value of the summation variable q by 1, and then increased each
occurrence of q in the term by 1 [also note that (31)q+2 = (31)q ].
"
1
1 g
1 g S
1
q
q
(b) i ({) =
=
3
(31)
(q
+
1){
=
3
[from part (a)]
2 g{ (1 + {)2
2 g{ q=0
(1 + {)3
= 3 12
"
S
(31)q (q + 1)q{q31 =
q=1
1
2
"
S
(31)q (q + 2)(q + 1){q with U = 1.
q=0
"
{2
1
1 S
= {2 ·
= {2 ·
(31)q (q + 2)(q + 1){q
(1 + {)3
(1 + {)3
2 q=0
"
1 S
=
(31)q (q + 2)(q + 1){q+2
2 q=0
(c) i ({) =
[from part (b)]
To write the power series with {q rather than {q+2 , we will decrease each occurrence of q in the term by 2 and increase
the initial value of the summation variable by 2. This gives us
14. (a)
]
"
1 S
(31)q (q)(q 3 1){q with U = 1.
2 q=2
1
g{ = 3 ln(1 3 {) + F and
13{
]
]
" {q
S
{2
{3
1
2
g{ = (1 + { + { + · · · ) g{ = { +
+
+ ··· + F =
+ F for |{| ? 1.
13{
2
3
q=1 q
So 3 ln(1 3 {) =
" {q
" {q
S
S
+ F and letting { = 0 gives 0 = F. Thus, i ({) = ln(1 3 {) = 3
with U = 1.
q=1 q
q=1 q
(b) i ({) = { ln(1 3 {) = 3{
(c) Letting { =
" {q
" {q+1
S
S
=3
.
q
q=1 q
q=1
"
[
(1@2)q
1
1
gives ln = 3
2
2
q
q=1
15. i ({) = ln(5 3 {) = 3
]
1
g{
=3
53{
5
]
i
ln 1 3 ln 2 = 3
g{
1
=3
1 3 {@5
5
" 1q
S
q
q=1 q2
i
ln 2 =
"
S
1
.
q
q2
q=1
" {
"
"
q
S
S
{q+1
{q
1 S
g{ = F 3
=F3
q
q
5 q=0 5 (q + 1)
q=0 5
q=1 q 5
] Putting { = 0, we get F = ln 5. The series converges for |{@5| ? 1 C |{| ? 5, so U = 5.
16. i ({) = {2 tan31 ({3 ) = {2
"
S
(31)q
q=0
3
{ ? 1
C
|{| ? 1, so U = 1.
"
"
S
S
({3 )2q+1
{6q+3+2
{6q+5
[by Example 7] =
=
for
(31)q
(31)q
2q + 1
2q + 1
2q + 1
q=0
q=0
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172
¤
CHAPTER 11
17. We know that
INFINITE SEQUENCES AND SERIES
"
S
1
1
=
=
(34{)q . Differentiating, we get
1 + 4{
1 3 (34{) q=0
"
"
S
S
34
=
(34)q q{q31 =
(34)q+1 (q + 1){q , so
2
(1 + 4{)
q=1
q=0
i ({) =
"
"
S
34
3{
3{ S
{
·
=
=
(34)q+1 (q + 1){q =
(31)q 4q (q + 1){q+1
2
2
(1 + 4{)
4 (1 + 4{)
4 q=0
q=0
for |34{| ? 1
18.
|{| ? 14 , so U = 14 .
C
"
" { q
"
S
S 1 q
1
1
1 q
1
1 S
g
g
=
=
=
=
{
.
Now
{
i
q+1
23{
2(1 3 {@2)
2 q=0 2
g{ 2 3 {
g{ q=0 2q+1
q=0 2
"
"
S
S 1
1
1
g
1
g
q31
q31
=
q{
and
q{
=
i
q+1
(2 3 {)2
g{ (2 3 {)2
g{ q=1 2q+1
q=1 2
"
" (q + 2)(q + 1)
S
S
2
1
=
q(q 3 1){q32 =
{q .
3
q+1
(2 3 {)
2q+3
q=2 2
q=0
Thus, i({) =
{
23{
3
=
" (q + 2)(q + 1)
" (q + 2)(q + 1)
S
2
{3
{3
{3 S
·
=
=
{q =
{q+3
3
3
q+3
(2 3 {)
2 (2 3 {)
2 q=0
2
2q+4
q=0
{
for ? 1 C |{| ? 2, so U = 2.
2
19. By Example 5,
"
S
1
=
(q + 1){q . Thus,
(1 3 {)2
q=0
i ({) =
"
"
S
S
1
{
1+{
=
+
=
(q + 1){q +
(q + 1){q+1
2
2
2
(1 3 {)
(1 3 {)
(1 3 {)
q=0
q=0
=
"
S
(q + 1){q +
q=0
=1+
"
S
q{q
[make the starting values equal]
q=1
"
S
[(q + 1) + q]{q = 1 +
q=1
"
S
1
=
(q + 1){q , so
(1 3 {)2
q=0
"
S
1
g
g
q
(q
+
1){
=
i
g{ (1 3 {)2
g{ q=0
"
S
(2q + 1){q =
q=1
"
S
(2q + 1){q with U = 1.
q=0
20. By Example 5,
i ({) =
=
=
=
"
S
2
=
(q + 1)q{q31 . Thus,
(1 3 {)3
q=1
{2 + {
{2
{
{2
{
2
2
=
+
=
+ ·
·
(1 3 {)3
(1 3 {)3
(1 3 {)3
2 (1 3 {)3
2 (1 3 {)3
"
"
" (q + 1)q
" (q + 1)q
S
S
{2 S
{ S
{q+1 +
{q
(q + 1)q{q31 +
(q + 1)q{q31 =
2 q=1
2 q=1
2
2
q=1
q=1
" q(q 3 1)
" (q + 1)q
S
S
{q +
{q
2
2
q=2
q=1
" q2 3 q
" q2 + q
S
S
{q + { +
{q
2
2
q=2
q=2
={+
"
S
q=2
q2 {q =
"
S
[make the exponents on { equal by changing an index]
[make the starting values equal]
q2 {q with U = 1=
q=1
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.9
21. i ({) =
{
{
=
{2 + 16
16
1
1 3 (3{2 @16)
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
{
,
The series converges when 3{2@16 ? 1 C {2 ? 16 C |{| ? 4, so U = 4. The partial sums are v1 =
16
v2 = v1 3
173
{3
{5
{7
{9
, v3 = v2 + 3 , v4 = v3 3 4 , v5 = v4 + 5 , = = = . Note that v1 corresponds to the first term of the infinite
2
16
16
16
16
sum, regardless of the value of the summation variable and the value of the exponent.
As q increases, vq ({) approximates i better on the interval of convergence, which is (34> 4).
22. i ({) = ln({2 + 4)
so i ({) =
¤
2 q
"
"
"
S
1
1
{ S
{ S
{
=
=
(31)q q {2q =
(31)q q+1 {2q+1 .
3
16 q=0
16
16 q=0
16
16
q=0
]
"
S
i i 0 ({) =
(31)q
q=0
2{
2{
=
{2 + 4
4
1
1 3 (3{2 @4)
=
2 q
"
"
S
{2q+1
{
{ S
3
=
(31)q 2q+1 ,
2 q=0
4
2
q=0
"
"
S
S
{2q+1
{2q+2
{2q+2
= ln 4 +
g{ = F +
(31)q 2q+1
(31)q
2q+1
2
2
(2q + 2)
(q + 1)22q+2
q=0
q=0
[i (0) = ln 4, so F = ln 4]. The series converges when 3{2@4 ? 1 C {2 ? 4 C |{| ? 2, so U = 2. If
{ = ±2, then i({) = ln 4 +
"
S
(31)q
q=0
are v0 = ln 4 [E 1=39], v1 = v0 +
1
, which converges by the Alternating Series Test. The partial sums
q+1
{2
{4
{6
{8
, v2 = v1 3
, v3 = v2 +
, v4 = v3 3
, ===.
4
6
4
2·2
3·2
4 · 28
As q increases, vq ({) approximates i better on the interval of convergence, which is [32> 2].
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°
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174
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
]
]
]
]
1+{
g{
g{
g{
g{
23. i ({) = ln
= ln(1 + {) 3 ln(1 3 {) =
+
=
+
13{
1+{
13{
1 3 (3{)
13{
]
] "
"
S
S
(31)q {q +
{q g{ = [(1 3 { + {2 3 {3 + {4 3 · · · ) + (1 + { + {2 + {3 + {4 + · · · )] g{
=
q=0
=
]
q=0
]
(2 + 2{2 + 2{4 + · · · ) g{ =
"
S
2{2q g{ = F +
q=0
But i (0) = ln 11 = 0, so F = 0 and we have i ({) =
" 2{2q+1
"
S
S
1
with U = 1. If { = ±1, then i ({) = ±2
,
2q
+
1
q=0
q=0 2q + 1
which both diverge by the Limit Comparison Test with eq =
The partial sums are v1 =
" 2{2q+1
S
q=0 2q + 1
1
.
q
2{
2{3
2{5
, v2 = v1 +
, v3 = v2 +
, ===.
1
3
5
As q increases, vq ({) approximates i better on the interval of
convergence, which is (31> 1).
24. i ({) = tan31 (2{) = 2
=F +2
]
g{
=2
1 + 4{2
]
"
S
q
(31)q 4{2 g{ = 2
q=0
" (31)q 4q {2q+1
" (31)q 22q+1 {2q+1
S
S
=
2q + 1
2q + 1
q=0
q=0
]
"
S
(31)q 4q {2q g{
q=0
[i (0) = tan31 0 = 0, so F = 0]
"
S
The series converges when 4{2 ? 1 C |{| ? 12 , so U = 12 . If { = ± 12 , then i ({) =
(31)q
q=0
i ({) =
"
S
(31)q+1
q=0
v1 =
1
and
2q + 1
1
, respectively. Both series converge by the Alternating Series Test. The partial sums are
2q + 1
2{
23 {3
25 {5
, v2 = v1 3
, v3 = v2 +
, ===.
1
3
5
As q increases, vq ({) approximates i better on the interval of convergence, which is 3 12 > 12 .
25.
]
"
"
" w8q+2
S
S
S
1
w
1
w
8 q
8q+1
=
w
·
=
w
(w
)
=
w
i
gw
=
F
+
converges
. The series for
8
1 3 w8
1 3 w8
1
3
w
8q
+
2
1
3
w8
q=0
q=0
q=0
when w8 ? 1 C |w| ? 1, so U = 1 for that series and also the series for w@(1 3 w8 ). By Theorem 2, the series for
]
w
gw also has U = 1.
1 3 w8
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.9
26.
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
¤
175
]
"
"
"
S
S
S
w
w3q+2
1
w
=w
. The series for
=w·
(3w3 )q =
(31)q w3q+1 i
gw = F +
(31)q
3
3
3
1+w
1 3 (3w )
1+w
3q + 2
q=0
q=0
q=0
1
w
converges when 3w3 ? 1 C |w| ? 1, so U = 1 for that series and also for the series
. By Theorem 2, the
1 + w3
1 + w3
]
w
gw also has U = 1.
series for
1 + w3
"
S
27. From Example 6, ln(1 + {) =
(31)q31
q=1
]
"
S
{q
{q+2
(31)q31
for |{| ? 1, so {2 ln(1 + {) =
and
q
q
q=1
{q+3
. U = 1 for the series for ln(1 + {), so U = 1 for the series representing
q(q + 3)
q=1
]
{2 ln(1 + {) as well. By Theorem 2, the series for {2 ln(1 + {) g{ also has U = 1.
"
S
{2 ln(1 + {) g{ = F +
28. From Example 7, tan31 { =
(31)q
"
S
(31)q
q=0
]
31
"
S
{2q+1
{2q
tan31 {
for |{| ? 1, so
=
and
(31)q
2q + 1
{
2q + 1
q=0
{2q+1
tan {
g{ = F +
(31)q
. U = 1 for the series for tan31 {, so U = 1 for the series representing
{
(2q + 1)2
q=0
]
tan31 {
tan31 {
as well. By Theorem 2, the series for
g{ also has U = 1.
{
{
29.
"
S
" "
q
S
S
1
1
3{5 =
=
=
(31)q {5q i
1 + {5
1 3 (3{5 )
q=0
q=0
]
] "
"
S
S
1
{5q+1
. Thus,
g{
=
(31)q {5q g{ = F +
(31)q
5
1+{
5q + 1
q=0
q=0
0=2
] 0=2
1
{11
(0=2)11
{6
(0=2)6
L=
+
3
·
·
·
+
3 · · · . The series is alternating, so if we use
g{
=
{
3
=
0=2
3
1 + {5
6
11
6
11
0
0
the first two terms, the error is at most (0=2)11@11 E 1=9 × 1039 . So L E 0=2 3 (0=2)6@6 E 0=199 989 to six decimal places.
30. From Example 6, ln(1 + {4 ) =
"
S
(31)q31
q=1
]
ln 1 + {4 g{ =
]
{4q
q
i
"
S
{4q
{4q+1
g{ = F +
. Thus,
(31)q31
q
q(4q + 1)
q=1
q=1
5
0=4
] 0=4
(0=4)5
(0=4)9
(0=4)13
(0=4)17
{
{9
{13
{17
L=
ln 1 + {4 g{ =
=
3
+
3
+ ···
3
+
3
+ ···.
5
18
39
68
5
18
39
68
0
0
"
S
(31)q31
The series is alternating, so if we use the first three terms, the error is at most (0=4)17@68 E 2=5 × 1039 .
So L E (0=4)5@5 3 (0=4)9@18 + (0=4)13@39 E 0=002 034 to six decimal places.
31. We substitute 3{ for { in Example 7, and find that
]
{ arctan(3{) g{ =
So
]
]
0
{
"
S
(31)q
q=0
0=1
(3{)2q+1
g{ =
2q + 1
]
"
S
(31)q
q=0
"
S
32q+1 {2q+2
32q+1 {2q+3
g{ = F +
(31)q
2q + 1
(2q + 1)(2q + 3)
q=0
33 {5
35 {7
37 {9
3{3
3
+
3
+ ···
{ arctan(3{) g{ =
1·3
3·5
5·7
7·9
0=1
0
1
9
243
2187
= 3 3
+
3
+ ···.
10
5 × 105
35 × 107
63 × 109
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°
licated, or posted to a publicly accessible website, in whole or in part.
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[continued]
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¤
176
CHAPTER 11
INFINITE SEQUENCES AND SERIES
The series is alternating, so if we use three terms, the error is at most
]
0=1
{ arctan(3{) g{ E
0
32.
]
0
0=3
{2
g{ =
1 + {4
=
]
0=3
1
9
243
3
+
E 0=000 983 to six decimal places.
103
5 × 105
35 × 107
{2
"
S
(31)q {4q g{ =
q=0
0
3
2187
E 3=5 × 1038 . So
63 × 109
"
S
q=0
7
11
(31)q {4q+3
4q + 3
0=3
0
3
3
3
3
+
3···
3 × 103
7 × 107
11 × 1011
The series is alternating, so if we use only two terms, the error is at most
places,
]
0=3
0
Thus, to five decimal places, arctan 0=2 E 0=2 3
" (31)q {2q
S
(2q)!
q=0
i i 0 ({) =
(0=2)7
E 0=000 002.
7
(0=2)3
(0=2)5
+
E 0=197 40.
3
5
" (31)q 2q{2q31
S
(2q)!
q=1
[the first term disappears], so
" (31)q (2q)(2q 3 1){2q32
" (31)q {2(q31)
" (31)q+1 {2q
S
S
S
=
=
(2q)!
[2(q
3
1)]!
(2q)!
q=1
q=1
q=0
=3
35. (a) M0 ({) =
311
E 0=000 000 16. So, to six decimal
11 × 1011
{5
{7
(0=2)3
(0=2)5
(0=2)7
{3
+
3
+ · · · , so arctan 0=2 = 0=2 3
+
3
+···.
3
5
7
3
5
7
The series is alternating, so if we use three terms, the error is at most
i 00 ({) =
(31)q 34q+3
4q+3
q=0 (4q + 3)10
"
S
33
37
{2
g{ E
3
E 0=008 969.
4
3
1+{
3 × 10
7 × 107
33. By Example 7, arctan { = { 3
34. i ({) =
=
" (31)q {2q
S
= 3i ({)
(2q)!
q=0
i
[substituting q + 1 for q]
i 00 ({) + i ({) = 0.
" (31)q {2q
" (31)q 2q{2q31
" (31)q 2q(2q 3 1){2q32
S
S
S
> M00 ({) =
, and M000 ({) =
, so
2q
2
2q
2
2 (q!)
22q (q!)2
q=0 2 (q!)
q=1
q=1
{2 M000 ({) + {M00 ({) + {2 M0 ({) =
=
" (31)q 2q(2q 3 1){2q
" (31)q 2q{2q
" (31)q {2q+2
S
S
S
+
+
2q
2
2q
2
2 (q!)
2 (q!)
22q (q!)2
q=1
q=1
q=0
" (31)q 2q(2q 3 1){2q
" (31)q 2q{2q
"
S
S
S
(31)q31 {2q
+
+
2q
2
2q
2
2q32
2 (q!)
2 (q!)
[(q 3 1)!]2
q=1
q=1
q=1 2
" (31)q 2q(2q 3 1){2q
" (31)q 2q{2q
" (31)q (31)31 22 q2 {2q
S
S
S
+
+
2q
2
2q
2
2 (q!)
2 (q!)
22q (q!)2
q=1
q=1
q=1
"
S
2q(2q 3 1) + 2q 3 22 q2 2q
=
(31)q
{
22q (q!)2
q=1
2
2
"
S
q 4q 3 2q + 2q 3 4q
{2q = 0
=
(31)
22q (q!)2
q=1
=
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
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SECTION 11.9
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
¤
177
] 1
1
" (31)q {2q
S
{4
{6
{2
+
3
+
·
·
·
g{
(b)
M0 ({) g{ =
g{
=
1
3
2q
2
4
64
2304
q=0 2 (q!)
0
0
0
1
{5
{7
1
1
1
{3
+
3
+ ··· = 1 3
+
3
+ ···
= {3
3·4
5 · 64
7 · 2304
12
320
16,128
0
]
]
1
1
E 0=000062, it follows from The Alternating Series Estimation Theorem that, correct to three decimal places,
Since 16,128
U1
1
1
M ({) g{ E 1 3 12
+ 320
E 0=920.
0 0
36. (a) M1 ({) =
"
S
q=0
" (31)q (2q + 1) {2q
" (31)q (2q + 1) (2q) {2q31
S
S
(31)q {2q+1
0
00
,
M
({)
=
,
and
M
({)
=
.
1
1
2q+1
q! (q + 1)! 22q+1
q! (q + 1)! 22q+1
q=0 q! (q + 1)! 2
q=1
{2 M100 ({) + {M10 ({) + {2 3 1 M1 ({)
=
" (31)q (2q + 1)(2q){2q+1
" (31)q (2q + 1){2q+1
S
S
+
q! (q + 1)! 22q+1
q! (q + 1)! 22q+1
q=1
q=0
+
"
S
"
S
(31)q {2q+3
(31)q {2q+1
3
2q+1
2q+1
q! (q + 1)! 2
q=0 q! (q + 1)! 2
"
S
"
S
(31)q {2q+1
(31)q {2q+1
3
2q31
2q+1
(q 3 1)! q! 2
q=0 q! (q + 1)! 2
q=0
=
" (31)q (2q + 1)(2q){2q+1
" (31)q (2q + 1){2q+1
S
S
+
q! (q + 1)! 22q+1
q! (q + 1)! 22q+1
q=1
q=0
3
q=1
Replace q with q 1
in the third term
2
"
S
{
{
q (2q + 1)(2q) + (2q + 1) 3 (q)(q + 1)2 3 1
= 3 +
{2q+1 = 0
(31)
2
2 q=1
q! (q + 1)! 22q+1
(b) M0 ({) =
M00 ({) =
" (31)q {2q
S
2q (q!)2
q=0 2
37. (a) i ({) =
i
" (31)q (2q){2q31
" (31)q+1 2(q + 1){2q+1
S
S
=
2
22q (q!)
22q+2 [(q + 1)!]2
q=1
q=0
=3
"
S
q=0
" {q
S
q=0 q!
(31)q {2q+1
+ 1)! q!
22q+1 (q
i i 0 ({) =
[Replace q with q + 1]
[cancel 2 and q + 1; take 1 outside sum] = 3M1 ({)
" q{q31
"
" {q
S
S
S
{q31
=
=
= i({)
q!
q=1
q=1 (q 3 1)!
q=0 q!
(b) By Theorem 9.4.2, the only solution to the differential equation gi ({)@g{ = i ({) is i ({) = Nh{ , but i(0) = 1, so
N = 1 and i ({) = h{ .
Or: We could solve the equation gi ({)@g{ = i ({) as a separable differential equation.
38.
" sin q{
S
|sin q{|
1
$ 2 , so
converges by the Comparison Test.
2
q
q
q2
q=1
[n an integer],
"
S
iq0 ({) =
q=1
"
S
q=1
iq00 ({) = 3
"
S
g
g{
sin q{
q2
=
cos q{
, so when { = 2n
q
" cos(2nq)
" 1
S
S
=
, which diverges [harmonic series].
q
q=1
q=1 q
iq00 ({) = 3 sin q{, so
sin q{, which converges only if sin q{ = 0, or { = n [n an integer].
q=1
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178
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
q+1
2
dq+1 {q
q2 q
= lim {
=
|{|
lim
,
then
by
the
Ratio
Test,
lim
·
= |{| ? 1 for
q<" dq q<" (q + 1)2
q<" q + 1
q2
{q " {q "
S
= S 1 which is a convergent s-series (s = 2 A 1), so the interval of
convergence, so U = 1. When { = ±1,
q2 2
q=1
q=1 q
39. If dq =
convergence for i is [31> 1]. By Theorem 2, the radii of convergence of i 0 and i 00 are both 1, so we need only check the
endpoints. i ({) =
" {q
S
2
q=1 q
i i 0 ({) =
" q{q31
"
S
S
{q
=
, and this series diverges for { = 1 (harmonic series)
2
q
q=1
q=0 q + 1
and converges for { = 31 (Alternating Series Test), so the interval of convergence is [31> 1). i 00 ({) =
at both 1 and 31 (Test for Divergence) since lim
q<"
40. (a)
"
S
q{q31 =
q=1
(b) (i)
"
S
q
q{ = {
q=1
"
S
q=2
"
S
q31
q{
q=1
(ii) Put { =
(c) (i)
" g
S
g
{q =
g{
g{
q=0
1
2
in (i):
= {2
1
2
in (i):
q
= 1 6= 0, so its interval of convergence is (31> 1).
q+1
g
1
1
1
=3
{q =
(31) =
, |{| ? 1.
2
g{
1
3
{
(1
3
{)
(1
3
{)2
q=0
"
S
1
={
(1 3 {)2
[from part (a)] =
{
for |{| ? 1.
(1 3 {)2
" q
"
q
S
S
1@2
=
q 12 =
= 2.
q
2
(1 3 1@2)2
q=1
q=1
q(q 3 1){q = {2
(ii) Put { =
" q{q31
S
diverges
q=1 q + 1
"
S
q=2
q(q 3 1){q32 = {2
g
g{
g
1
q{q31 = {2
g{
(1
3
{)2
q=1
"
S
2{2
2
=
for |{| ? 1.
(1 3 {)3
(1 3 {)3
" q2 3 q
"
q
S
S
2(1@2)2
=
q(q 3 1) 12 =
= 4.
q
2
(1 3 1@2)3
q=2
q=2
(iii) From (b)(ii) and (c)(ii), we have
41. By Example 7, tan31 { =
"
S
(31)q
q=0
" q2
" q2 3 q
" q
S
S
S
=
+
= 4 + 2 = 6.
q
q
q
2
2
q=1
q=1
q=1 2
{2q+1
1
for |{| ? 1. In particular, for { = I , we
2q + 1
3
I 2q+1
q
"
"
1@ 3
S
S
1
1
1
1
I
=
= tan31 I
=
, so
(31)q
(31)q
6
2q + 1
3
3
3 2q + 1
q=0
q=0
have
"
"
I S
(31)q
(31)q
6 S
=2 3
.
= I
q
q
3 q=0 (2q + 1)3
q=0 (2q + 1)3
42. (a)
]
0
1@2
s
s
g{
3
3
2
1
1
=
x,
x
=
,
g{
=
gx
{
{
s
2
2
2
2
3
({ 3 1@2)2 + 3@4
0
I
I
] 0
l0
l
3@2 gx
2 3k
2 k
31
I
=
=
tan
= I
x
=
0
3
3
I
I (3@4)(x2 + 1)
3
6
31@
3
3
3 3
31@ 3
g{
=
{2 3 { + 1
]
1@2
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SECTION 11.10
(b)
1
1
=
{3 + 1
({ + 1)({2 3 { + 1)
TAYLOR AND MACLAURIN SERIES
¤
179
i
"
S
1
1
1
=
({
+
1)
= ({ + 1)
(31)q {3q
= ({ + 1)
{2 3 { + 1
1 + {3
1 3 (3{3 )
q=0
=
"
S
(31)q {3q+1 +
q=0
]
]
0
"
S
(31)q {3q
q=0
for |{| ? 1 i
3q+2
3q+1
"
"
S
S
g{
q {
q {
(31)
(31)
=
F
+
+
for |{| ? 1 i
{2 3 { + 1
3q + 2 q=0
3q + 1
q=0
1@2
"
" (31)q
S
g{
1
2
1
1 S
1
q
(31)
=
+
=
+
.
{2 3 { + 1 q=0
4 · 8q (3q + 2)
2 · 8q (3q + 1)
4 q=0 8q
3q + 1
3q + 2
I
" (31)q
2
3 3 S
1
+
.
By part (a), this equals I , so =
4 q=0 8q
3q + 1
3q + 2
3 3
11.10 Taylor and Maclaurin Series
1. Using Theorem 5 with
"
S
q=0
eq ({ 3 5)q , eq =
i (q) (d)
i (8) (5)
, so e8 =
.
q!
8!
2. (a) Using Equation 6, a power series expansion of i at 1 must have the form i (1) + i 0 (1)({ 3 1) + · · · . Comparing to the
given series, 1=6 3 0=8({ 3 1) + · · · , we must have i 0 (1) = 30=8. But from the graph, i 0 (1) is positive. Hence, the given
series is not the Taylor series of i centered at 1.
(b) A power series expansion of i at 2 must have the form i (2) + i 0 (2)({ 3 2) + 12 i 00 (2)({ 3 2)2 + · · · . Comparing to the
given series, 2=8 + 0=5({ 3 2) + 1=5({ 3 2)2 3 0=1({ 3 2)3 + · · · , we must have 12 i 00 (2) = 1=5; that is, i 00 (2) is positive.
But from the graph, i is concave downward near { = 2, so i 00 (2) must be negative. Hence, the given series is not the
Taylor series of i centered at 2.
3. Since i (q) (0) = (q + 1)!, Equation 7 gives the Maclaurin series
" i (q) (0)
" (q + 1)!
"
S
S
S
{q =
{q =
(q + 1){q . Applying the Ratio Test with dq = (q + 1){q gives us
q!
q!
q=0
q=0
q=0
q+1 dq+1 = lim (q + 2){
= |{| lim q + 2 = |{| · 1 = |{|. For convergence, we must have |{| ? 1, so the
lim q<"
q<" q + 1
dq q<" (q + 1){q radius of convergence U = 1.
4. Since i (q) (4) =
(31)q q!
, Equation 6 gives the Taylor series
3q (q + 1)
" i (q) (4)
"
"
S
S
S
(31)q q!
(31)q
({ 3 4)q =
({ 3 4)q =
({ 3 4)q , which is the Taylor series for i
q
q
q!
q=0
q=0 3 (q + 1) q!
q=0 3 (q + 1)
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180
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
centered at 4. Apply the Ratio Test to find the radius of convergence U.
dq+1 (31)q+1 ({ 3 4)q+1
(31)({ 3 4)(q + 1) 3q (q + 1) ·
= lim = lim
lim
q<" dq q<"
3q+1 (q + 2)
(31)q ({ 3 4)q q<" 3(q + 2)
1
q+1
1
|{ 3 4| lim
= |{ 3 4|
q<" q + 2
3
3
=
For convergence,
1
3
|{ 3 4| ? 1 C |{ 3 4| ? 3, so U = 3.
5.
q
i (q) ({)
i (q) (0)
0
(1 3 {)32
1
1
2
3
4
..
.
2(1 3 {)33
2
24(1 3 {)35
24
34
6(1 3 {)
120(1 3 {)36
..
.
120
..
.
q
i (q) ({)
i (q) (0)
0
ln(1 + {)
0
1
(1 + {)31
1
3
4
5
..
.
= 1 + 2{ + 62 {2 +
32
3 (1 + {)
36
24(1 + {)35
..
.
24
..
.
+
120 4
{
24
+ ···
"
S
(q + 1){q
q=0
for convergence, so U = 1.
ln(1 + {) = i(0) + i 0 (0){ +
+
i 00 (0) 2
{
2!
i 000 (0) 3 i (4) (0) 4 i (5) (0) 5
{ +
{ +
{ + ···
3!
4!
5!
= 0 + { 3 12 {2 + 26 {3 3
2
36(1 + {)34
24 3
{
6
q+1 dq+1 = lim (q + 2){
= |{| lim q + 2 = |{| (1) = |{| ? 1
lim q<"
q<" q + 1
dq q<" (q + 1){q 31
2(1 + {)33
i 00 (0) 2 i 000 (0) 3 i (4) (0) 4
{ +
{ +
{ + ···
2!
3!
4!
= 1 + 2{ + 3{2 + 4{3 + 5{4 + · · · =
6
6.
2
(1 3 {)32 = i (0) + i 0 (0){ +
={3
6 4
{
24
+
24 5
{
120
3 ···
" (31)q31
S
{2
{3
{4
{5
+
3
+
3 ··· =
{q
2
3
4
5
q
q=1
q+1
dq+1 |{|
q = lim {
·
= |{| ? 1 for convergence,
= lim
lim q<"
dq q<" q + 1 {q q<" 1 + 1@q
so U = 1.
Notice that the answer agrees with the entry for ln(1 + {) in Table 1, but we obtained it by a different method. (Compare with
Example 6 in Section 11.9.)
7.
q
i (q) ({)
i (q) (0)
0
sin {
0
1
cos {
2
3
4
5
..
.
2
3 sin {
0
sin { = i (0) + i 0 (0){ +
+
33
0
= { 3
5 cos {
..
.
5
..
.
=
4 sin {
i (4) (0) 4 i (5) (0) 5
{ +
{ + ···
4!
5!
= 0 + { + 0 3
33 cos {
"
S
q=0
i 00 (0) 2 i 000 (0) 3
{ +
{
2!
3!
5 5
3 3
{ +0+
{ +···
3!
5!
3 3 5 5 7 7
{ +
{ 3
{ + ···
3!
5!
7!
(31)q
2q+1
{2q+1
(2q + 1)!
2q+3 2q+3
dq+1 {
2 {2
(2q + 1)! · 2q+1 2q+1 = lim
= 0 ? 1 for all {, so U = ".
= lim lim q<"
q<"
q<" (2q + 3)(2q + 2)
dq
(2q + 3)!
{
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.10
h32{ =
8.
q
i (q) ({)
i (q) (0)
0
h32{
1
1
2
3
4
..
.
32{
32h
4h
4
32{
38h
16
..
.
2{ =
9.
q
i (q) ({)
i (q) (0)
2{
1
0
1
{
2 (ln 2)
{
2
2 (ln 2)
(ln 2)2
3
2{ (ln 2)3
(ln 2)3
4
..
.
2{ (ln 2)4
..
.
(ln 2)4
..
.
= lim
q<"
10.
q
i (q) ({)
i (q) (0)
0
{ cos {
0
1
3{ sin { + cos {
1
{ sin { 3 3 cos {
33
3
4
5
6
7
..
.
3{ cos { 3 2 sin {
{ cos { + 4 sin {
= 0 + 1{ + 0 3
={3
=
0
5
{ sin { 3 7 cos {
..
.
37
..
.
0
q
i (q) ({)
i (q) (0)
0
sinh {
0
1
cosh {
1
2
sinh {
0
3
cosh {
1
4
..
.
sinh {
..
.
0
..
.
"
S
i 00 (0) 2 i 000 (0) 3
i (4) (0) 4
{ +
{ +
{ + ···
2!
3!
4!
5
7
3 3
{ + 0 + {5 + 0 3 {7 + · · ·
3!
5!
7!
1
1
1 3
{ + {5 3 {7 + · · ·
2!
4!
6!
(31)q
q=0
1
{2q+1
(2q)!
dq+1 (31)q+1 {2q+3
(2q)!
·
= lim lim
q<" dq q<"
(2q + 2)!
(31)q {2q+1 = lim
q<"
i (q) (0) =
11.
(ln 2) |{|
= 0 ? 1 for all {, so U = ".
q+1
{ cos { = i (0) + i 0 (0){ +
0
3{ sin { + 5 cos {
3{ cos { 3 6 sin {
" i (q) (0)
" (ln 2)q
S
S
{q =
{q .
q!
q!
q=0
q=0
q+1 q+1
dq+1 {
q!
= lim (ln 2)
lim ·
q
q
q<"
q<"
dq
(q + 1)!
(ln 2) { ln 2
2
2
" i (q) (0)
" (32)q
S
S
{q =
{q .
q!
q!
q=0
q=0
= 0 ? 1 for all {, so U = "=
38
16h32{
..
.
¤
q+1 q+1
dq+1 {
q!
= lim (32)
= lim 2 |{|
lim ·
q
q
q<"
q<"
dq
(q + 1)!
(32) { q<" q + 1
32
32{
TAYLOR AND MACLAURIN SERIES
+
0 if q is even
1 if q is odd
{2
= 0 ? 1 for all {, so U = ".
(2q + 2)(2q + 1)
so sinh { =
{2q+1
.
q=0 (2q + 1)!
"
S
{2q+1
, then
(2q + 1)!
2q+3
dq+1 {
(2q + 1)! 1
= lim = {2 · lim
lim
·
q<" dq q<" (2q + 3)!
q<" (2q + 3)(2q + 2)
{2q+1 Use the Ratio Test to find U. If dq =
= 0 ? 1 for all {, so U = ".
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°
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182
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
12.
i
q
i (q) ({)
i (q) (0)
0
cosh {
1
1
sinh {
0
2
cosh {
1
3
..
.
sinh {
..
.
0
..
.
(q)
(0) =
+
1 if q is even
0 if q is odd
so cosh { =
" {2q
S
.
q=0 (2q)!
{2q
, then
(2q)!
2q+2
dq+1 1
(2q)! = lim {
·
= {2 · lim
lim q<"
q<" (2q + 2)(2q + 1)
dq q<" (2q + 2)! {2q Use the Ratio Test to find U. If dq =
= 0 ? 1 for all {, so U = "=
13.
i (q) ({) = 0 for q D 5, so i has a finite series expansion about d = 1.
q
i (q) ({)
i (q) (1)
0
{4 3 3{2 + 1
31
2
6
=
24{
24
4
24
24
5
0
0
24
24
({ 3 1)3 +
({ 3 1)4
3!
4!
= 31 3 2({ 3 1) + 3({ 3 1)2 + 4({ 3 1)3 + ({ 3 1)4
6
..
.
0
..
.
0
..
.
1
4{3 3 6{
2
32
12{ 3 6
3
14.
q
0
1
2
3
4
5
..
.
i
(q)
({)
3
{3{
i
(q)
(32)
6
1 3 3{2
36{
36
311
12
36
0
..
.
0
..
.
0
0
15.
i ({) = {4 3 3{2 + 1 =
31
32
6
({ 3 1)0 +
({ 3 1)1 + ({ 3 1)2
0!
1!
2!
+
A finite series converges for all {, so U = ".
i (q) ({) = 0 for q D 4, so i has a finite series expansion about d = 32.
i ({) = { 3 {3 =
=
i (q) ({)
i (q) (2)
0
ln {
ln 2
1
1@{
2
3
4
5
..
.
1@2
2
31@{
2@{3
4
36@{
24@{5
..
.
3 i (q) (32)
S
({ + 2)q
q!
q=0
6
311
12
36
({ + 2)0 +
({ + 2)1 +
({ + 2)2 +
({ + 2)3
0!
1!
2!
3!
= 6 3 11({ + 2) + 6({ + 2)2 3 ({ + 2)3
A finite series converges for all {, so U = ".
i({) = ln { =
q
=
" i (q) (2)
S
({ 3 2)q
q!
q=0
ln 2
1
31
2
({ 3 2)0 +
({ 3 2)1 +
({ 3 2)2 +
({ 3 2)3
0!
1! 21
2! 22
3! 23
31@22
+
2@23
36@24
24@25
..
.
4 i (q) (1)
S
({ 3 1)q
q!
q=0
= ln 2 +
"
S
36
24
({ 3 2)4 +
({ 3 2)5 + · · ·
4! 24
5! 25
(31)q+1
q=1
= ln 2 +
"
S
(31)q+1
q=1
(q 3 1)!
({ 3 2)q
q! 2q
1
({ 3 2)q
q 2q
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.10
TAYLOR AND MACLAURIN SERIES
¤
q+2
dq+1 |{ 3 2|
q
({ 3 2)q+1
q 2q
= lim (31)
= lim (31)({ 3 2)q = lim
lim ·
q<"
dq q<" (q + 1) 2q+1
(31)q+1 ({ 3 2)q q<" (q + 1)2 q<" q + 1
2
=
|{ 3 2|
? 1 for convergence, so |{ 3 2| ? 2 and U = 2.
2
16.
i ({) =
q
i (q) ({)
i (q) (33)
0
1@{
1
2
3
31@{2
2@{3
36@{4
31@3
4
..
.
24@{5
..
.
=
31@32
32@33
36@34
" i (q) (33)
S
1
=
({ + 3)q
{ q=0
q!
31@3
31@32
32@33
({ + 3)0 +
({ + 3)1 +
({ + 3)2
0!
1!
2!
+
324@35
..
.
=
36@34
324@35
({ + 3)3 +
({ + 3)4 + · · ·
3!
4!
" 3q!@3q+1
" ({ + 3)q
S
S
({ + 3)q = 3
q!
3q+1
q=0
q=0
q+1
dq+1 |{ + 3|
|{ + 3|
3q+1 = lim ({ + 3)
=
? 1 for convergence,
= lim
lim ·
q<"
dq q<" 3q+2
({ + 3)q q<" 3
3
so |{ + 3| ? 3 and U = 3.
i({)= h2{ =
17.
18.
q
i (q) ({)
i (q) (3)
0
h2{
h6
1
2h2{
2h6
2
22 h2{
4h6
3
23 h2{
8h6
4
..
.
24 h2{
..
.
16h6
..
.
" i (q) (3)
S
({ 3 3)q
q!
q=0
h6
2h6
4h6
({ 3 3)0 +
({ 3 3)1 +
({ 3 3)2
0!
1!
2!
=
+
=
8h6
16h6
({ 3 3)3 +
({ 3 3)4 + · · ·
3!
4!
" 2q h6
S
({ 3 3)q
q=0 q!
q+1 6
dq+1 h ({ 3 3)q+1
q!
= lim 2
= lim 2 |{ 3 3| = 0 ? 1 for all {, so U = ".
·
lim q
6
q
q<"
q<"
dq
(q + 1)!
2 h ({ 3 3) q<" q + 1
q
0
1
2
3
4
..
.
i (q) ({)
sin {
cos {
3 sin {
3 cos {
sin {
..
.
i (q) (@2)
1
0
31
0
1
..
.
i({) = sin { =
=13
=
"
S
q=0
" i (n) (@2) S
n
{3
n!
2
n=0
({ 3 @2)4
({ 3 @2)6
({ 3 @2)2
+
3
+ ···
2!
4!
6!
(31)q
({ 3 @2)2q
(2q)!
2q+2
dq+1 |{ 3 @2|2
(2q)!
= lim |{ 3 @2|
lim = lim
·
= 0 ? 1 for all {, so U = ".
2q
q<"
q<"
q<" (2q + 2)(2q + 1)
dq
(2q + 2)!
|{ 3 @2|
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¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
19.
q
i (q) ({)
i (q) ()
0
cos {
1
3 sin {
31
2
3
4
..
.
3 cos {
sin {
i ({) = cos { =
0
= 31 +
1
=
({ 3 )4
({ 3 )6
({ 3 )2
3
+
3 ···
2!
4!
6!
(31)q+1
q=0
0
cos {
..
.
"
S
" i (n) ()
S
({ 3 )n
n!
n=0
({ 3 )2q
(2q)!
2q+2
dq+1 (2q)!
= lim |{ 3 |
lim ·
q<"
dq q<" (2q + 2)!
|{ 3 |2q
31
..
.
|{ 3 |2
= 0 ? 1 for all {, so U = ".
q<" (2q + 2)(2q + 1)
= lim
20.
i({)=
0
i (q) ({)
I
{
1
1 31@2
{
2
1 1
·
2 4
2
3 14 {33@2
1 1
3 · 3
4 4
3
3 35@2
{
8
3 1
·
8 45
4
..
.
3 15
{37@2
16
..
.
q
i (q) (16)
4
3
15 1
·
16 47
..
.
=
" i (q) (16)
I
S
{=
({ 3 16)q
q!
q=0
4
1 1 1
1 1 1
({ 3 16)0 +
· · ({ 3 16)1 3 · 3 · ({ 3 16)2
0!
2 4 1!
4 4 2!
+
3 1 1
15 1 1
·
·
· ({ 3 16)3 3
· ({ 3 16)4 + · · ·
8 45 3!
16 47 4!
"
S
1 · 3 · 5 · · · · · (2q 3 3)
1
({ 3 16)q
(31)q31
= 4 + ({ 3 16) +
8
2q 42q31 q!
q=2
"
S
1 · 3 · 5 · · · · · (2q 3 3)
1
({ 3 16)q
= 4 + ({ 3 16) +
(31)q31
8
25q32 q!
q=2
dq+1 (31)q 1 · 3 · 5 · · · · · (2q 3 1)({ 3 16)q+1
25q32 q!
= lim lim
·
q<" dq q<"
25q+3 (q + 1)!
(31)q31 1 · 3 · 5 · · · · · (2q 3 3)({ 3 16)q = lim
q<"
=
(2q 3 1) |{ 3 16|
2 3 1@q
|{ 3 16|
|{ 3 16|
=
lim
=
·2
25 (q + 1)
32 q<" 1 + 1@q
32
|{ 3 16|
? 1 for convergence, so |{ 3 16| ? 16 and U = 16.
16
21. If i ({) = sin {, then i (q+1) ({) = ±q+1 sin { or ±q+1 cos {. In each case, i (q+1) ({) $ q+1 , so by Formula 9
with d = 0 and P = q+1 , |Uq ({)| $
q+1
|{|q+1
|{|q+1 =
. Thus, |Uq ({)| < 0 as q < " by Equation 10.
(q + 1)!
(q + 1)!
So lim Uq ({) = 0 and, by Theorem 8, the series in Exercise 7 represents sin { for all {.
q<"
22. If i({) = sin {, then i (q+1) ({) = ± sin { or ± cos {. In each case, i (q+1) ({) $ 1, so by Formula 9 with d = 0 and
P = 1, |Uq ({)| $
q+1
1
. Thus, |Uq ({)| < 0 as q < " by Equation 10. So lim Uq ({) = 0 and, by
{ 3 q<"
(q + 1)!
2
Theorem 8, the series in Exercise 18 represents sin { for all {.
23. If i ({) = sinh {, then for all q, i (q+1) ({) = cosh { or sinh {. Since |sinh {| ? |cosh {| = cosh { for all {, we have
(q+1) ({) $ cosh { for all q. If g is any positive number and |{| $ g, then i (q+1) ({) $ cosh { $ cosh g, so by
i
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SECTION 11.10
Formula 9 with d = 0 and P = cosh g, we have |Uq ({)| $
TAYLOR AND MACLAURIN SERIES
¤
185
cosh g
|{|q+1 . It follows that |Uq ({)| < 0 as q < " for
(q + 1)!
|{| $ g (by Equation 10). But g was an arbitrary positive number. So by Theorem 8, the series represents sinh { for all {.
24. If i({) = cosh {, then for all q, i (q+1) ({) = cosh { or sinh {. Since |sinh {| ? |cosh {| = cosh { for all {, we have
(q+1) ({) $ cosh { for all q. If g is any positive number and |{| $ g, then i (q+1) ({) $ cosh { $ cosh g, so by
i
Formula 9 with d = 0 and P = cosh g, we have |Uq ({)| $
cosh g
|{|q+1 . It follows that |Uq ({)| < 0 as q < " for
(q + 1)!
|{| $ g (by Equation 10). But g was an arbitrary positive number. So by Theorem 8, the series represents cosh { for all {.
$
#
1@4
(3{)q = 1 + 14 (3{) +
q
q=0
"
I
S
25. 4 1 3 { = [1 + (3{)]1@4 =
1
4
3
34
(3{)2 +
2!
1
4
3 7
34 34
(3{)3 + · · ·
3!
" (31)q31 (31)q · [3 · 7 · · · · · (4q 3 5)]
S
1
{q
=13 {+
4
4q · q!
q=2
" 3 · 7 · · · · · (4q 3 5)
S
1
{q
=13 {3
4
4q · q!
q=2
and |3{| ? 1
C
|{| ? 1, so U = 1.
#
$
u "
S
{
1@3 { q
{ 1@3
3
8 1+
=2
=2 1+
q
8
8
8
q=0
%
&
2
2 5 1
1
33 33
{ 3
1 { 3 3 3 { 2
3
+
=2 1+
+
+ ···
3 8
2!
8
3!
8
I
26. 3 8 + { =
" (31)q31 · [2 · 5 · · · · · (3q 3 4)]
S
1
q
{
=2 1+ {+
24
3q · 8q · q!
q=2
=2+
{
and ? 1
8
" (31)q31 [2 · 5 · · · · · (3q 3 4)]
S
1
{+2
{q
12
24q · q!
q=2
C
|{| ? 8, so U = 8.
# $
"
1
1
1 S
1
33 { q
{ 33
27.
=
. The binomial coefficient is
1+
3 =
3 =
8
2
8 q=0 q
2
(2 + {)
[2(1 + {@2)]
# $
(33)(34)(35) · · · · · [3(q + 2)]
33
(33)(34)(35) · · · · · (33 3 q + 1)
=
=
q!
q!
q
=
Thus,
(31)q (q + 1)(q + 2)
(31)q · 2 · 3 · 4 · 5 · · · · · (q + 1)(q + 2)
=
2 · q!
2
{
" (31)q (q + 1)(q + 2) {q
" (31)q (q + 1)(q + 2){q
S
1 S
1
=
=
for
? 1 C |{| ? 2, so U = 2.
8 q=0
2
2q
2q+4
2
(2 + {)3
q=0
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186
¤
CHAPTER 11
2@3
28. (1 3 {)
=
"
S
q=0
INFINITE SEQUENCES AND SERIES
# $
2
3
q
q
(3{) = 1 +
2
(3{)
3
= 1 3 23 { +
+
2
3
1
33
(3{)2 +
2!
2
3
1 4 33 33
(3{)3 + · · ·
3!
" (31)q31 (31)q · 2 · [1 · 4 · 7 · · · · · (3q 3 5)]
S
{q
3q · q!
q=2
= 1 3 23 { 3 2
" 1 · 4 · 7 · · · · · (3q 3 5)
S
{q
3q · q!
q=2
and |3{| ? 1 C |{| ? 1, so U = 1.
29. sin { =
"
S
(31)q
q=0
30. cos { =
"
S
(31)q
q=0
31. h{ =
" {q
S
q=0 q!
{2q+1
(2q + 1)!
{2q
(2q)!
i i({) = sin({) =
i i ({) = cos({@2) =
i h2{ =
"
S
(31)q
q=0
"
S
(31)q
q=0
"
S
({)2q+1
2q+1
=
{2q+1 , U = ".
(31)q
(2q + 1)! q=0
(2q + 1)!
"
S
({@2)2q
2q
=
{2q , U = ".
(31)q 2q
(2q)!
2 (2q)!
q=0
" (2{)q
" 2q {q
" 1
" 2q
" 2q + 1
S
S
S
S
S
=
, so i({) = h{ + h2{ =
{q +
{q =
{q ,
q!
q!
q!
q=0
q=0
q=0 q!
q=0 q!
q=0
U = ".
32. h{ =
" {q
S
q=0 q!
33. cos { =
"
S
i 2h3{ = 2
(31)q
q=0
" (3{)q
" (31)q {q
" [1 + 2(31)q ]
S
S
S
=2
, so i({) = h{ + 2h3{ =
{q , U = ".
q!
q!
q!
q=0
q=0
q=0
"
S
(31)q
i cos 12 {2 =
{2q
(2q)!
q=0
"
S
(31)q
i ({) = { cos 12 {2 =
q=0
34. ln(1 + {) =
"
S
(31)q31
q=1
{q
q
1
2q
"
{2
S
{4q
(31)q 2q
=
, so
(2q)!
2 (2q)!
q=0
2
1
{4q+1 , U = ".
22q (2q)!
i
ln(1 + {3 ) =
"
S
(31)q31
q=1
"
S
{3q
{3q+2
(31)q31
, so i({) = {2 ln(1 + {3 ) =
,
q
q
q=1
U = 1.
35. We must write the binomial in the form (1+ expression), so we’ll factor out a 4.
# $ q
31@2
"
{
{
{
{ S
3 12
{
{2
{2
I
= s
=
= s
=
1+
2
2
2
2
4
2 q=0 q
4
4+{
4(1 + { @4)
2 1 + { @4
%
&
1 3 2 2 1 3 5 2 3
1 {2
32 32
32 32 32
{
{
{
1 + 32
+
=
+
+ ···
2
4
2!
4
3!
4
=
=
"
1 · 3 · 5 · · · · · (2q 3 1) 2q
{ { S
(31)q
+
{
2
2 q=1
2q · 4q · q!
"
S
1 · 3 · 5 · · · · · (2q 3 1) 2q+1
{
{2
+
?1 C
(31)q
{
and
2 q=1
q! 23q+1
4
|{|
? 1 C |{| ? 2, so U = 2.
2
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.10
{ 31@2
{2
{2 {2
= I 1+
= s
2
2+{
2
2 (1 + {@2)
36. I
TAYLOR AND MACLAURIN SERIES
¤
# $
"
3 12 { q
{2 S
= I
2
2 q=0 q
%
&
1 3 1 3 5 { 32 32 { 2
32 32 32 { 3
{2
+
= I 1 + 3 12
+
+ ···
2
2!
2
3!
2
2
"
1 · 3 · 5 · · · · · (2q 3 1) q
{2
{2 S
(31)q
{
= I +I
q! 22q
2
2 q=1
37. sin2 { =
U="
38.
{
"
S
1 · 3 · 5 · · · · · (2q 3 1) q+2
{2
= I +
(31)q
{
and ? 1 C |{| ? 2, so U = 2.
2q+1@2
2
q! 2
2 q=1
" (31)q (2{)2q
" (31)q (2{)2q
" (31)q+1 22q31 {2q
S
S
S
1
1
1
(1 3 cos 2{) =
13
=
1313
=
,
2
2
(2q)!
2
(2q)!
(2q)!
q=0
q=1
q=1
"
" (31)q {2q+1
" (31)q {2q+1
S
S
S (31)q+1 {2q+3
1
{ 3 sin {
1
1
=
{
3
{
3
{
3
3
=
=
{3
{3
{3
{3
(2q + 3)!
q=0 (2q + 1)!
q=1 (2q + 1)!
q=0
=
" (31)q {2q+3
" (31)q {2q
S
1 S
=
3
{ q=0 (2q + 3)!
q=0 (2q + 3)!
and this series also gives the required value at { = 0 (namely 1@6); U = ".
"
(16) S
39. cos { =
(31)q
q=0
i({) = cos({2 ) =
= 1 3 12 {4 +
{2q
(2q)!
i
" (31)q ({2 )2q
" (31)q {4q
S
S
=
(2q)!
(2q)!
q=0
q=0
1 8
{
24
3
1
{12
720
+ ···
The series for cos { converges for all {, so the same is true of the series for
i ({), that is, U = ". Notice that, as q increases, Wq ({) becomes a better
approximation to i ({).
"
(11) S
40. h{ =
q=0
" (3{2 )q
"
S
S
2
{q
{2q
, so h3{ =
=
.
(31)q
q!
q!
q!
q=0
q=0
"
(16) S
Also, cos { =
(31)q
q=0
3{2
i({) = h
{2q
, so
(2q)!
+ cos { =
"
S
q
(31)
q=0
1
1
+
{2q
q!
(2q)!
3
13
121 6
{ + ··· =
= 2 3 {2 + {4 3
2
24
720
The series for h{ and cos { converge for all {, so the same is true of the series
for i({); that is, U = ". From the graphs of i and the first few Taylor
polynomials, we see that Wq ({) provides a closer fit to i ({) near 0 as q increases.
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°
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188
¤
CHAPTER 11
"
(11) S
41. h{ =
q=0
INFINITE SEQUENCES AND SERIES
" (3{)q
"
S
S
{q
{q
(31)q
, so h3{ =
=
, so
q!
q!
q!
q=0
q=0
i ({) = {h3{ =
"
S
(31)q
q=0
1 q+1
{
q!
= { 3 {2 + 12 {3 3 16 {4 +
"
S
=
(31)q31
q=1
{
1 5
{
24
3
1
{6
120
+ ···
{q
(q 3 1)!
The series for h converges for all {, so the same is true of the series
for i({); that is, U = ". From the graphs of i and the first few Taylor
polynomials, we see that Wq ({) provides a closer fit to i ({) near 0 as q increases.
"
S
42. From Table 1, tan31 { =
(31)q
q=0
i({) = tan31 ({3 ) =
"
S
(31)q
q=0
{2q+1
, so
2q + 1
"
S
({3 )2q+1
{6q+3
(31)q
=
2q + 1
2q + 1
q=0
= {3 3 13 {9 + 15 {15 3 17 {21 + · · ·
The series for tan31 { has U = 1 and {3 ? 1 C |{| ? 1,
so the series for i ({) also has U = 1. From the graphs of i and
the first few Taylor polynomials, we see that Wq ({) provides a
closer fit to i ({) near 0 as q increases.
43. 5 = 5
cos
"
S
{2q
{2
{4
{6
(31)q
=
radians and cos { =
=13
+
3
+ · · · , so
180
36
(2q)!
2!
4!
6!
q=0
(@36)2
(@36)4
(@36)6
(@36)2
(@36)4
=13
+
3
+ · · · . Now 1 3
E 0=99619 and adding
E 2=4 × 1036
36
2!
4!
6!
2!
4!
does not affect the fifth decimal place, so cos 5 E 0=99619 by the Alternating Series Estimation Theorem.
I
44. 1@ 10 h = h31@10 and h{ =
h31@10 = 1 3
13
" {q
S
{2
{3
= 1+{+
+
+ · · · , so
2!
3!
q=0 q!
1
(1@10)2
(1@10)3
(1@10)4
(1@10)5
+
3
+
3
+ · · · . Now
10
2!
3!
4!
5!
(1@10)2
(1@10)3
(1@10)4
(1@10)5
1
+
3
+
E 0=90484 and subtracting
E 8=3 × 1038 does not affect the fifth
10
2!
3!
4!
5!
decimal place, so h31@10 E 0=90484 by the Alternating Series Estimation Theorem.
1 3 1 3 5 3 2 3 2 2 2
3 2 3 2 3 2 2 3
3{
3{
= 1 + 3 12 3{2 +
+
+ ···
2!
3!
" 1 · 3 · 5 · · · · · (2q 3 1)
S
{2q
= 1+
2q · q!
q=1
I
45. (a) 1@ 1 3 {2 = 1 + 3{2
(b) sin31 { =
]
31@2
" 1 · 3 · 5 · · · · · (2q 3 1)
S
1
I
{2q+1
g{ = F + { +
2
(2q + 1)2q · q!
13{
q=1
={+
" 1 · 3 · 5 · · · · · (2q 3 1)
S
{2q+1
(2q + 1)2q · q!
q=1
since 0 = sin31 0 = F.
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SECTION 11.10
TAYLOR AND MACLAURIN SERIES
¤
189
# $
1 5 1 5 9 "
S
34 34 2
34 34 34 3
3 14
1
q
{ = 13 {+
{ +
{ + ···
=
q
4
2!
3!
q=0
I
46. (a) 1@ 4 1 + { = (1 + {)31@4
"
S
1 · 5 · 9 · · · · · (4q 3 3) q
1
=13 {+
{
(31)q
4
4q · q!
q=2
I
(b) 1@ 4 1 + { = 1 3 14 { +
5 2
32 {
3
15 3
128 {
195 4
2048 {
I
I
3 · · · . 1@ 4 1=1 = 1@ 4 1 + 0=1, so let { = 0=1. The sum of the first four
195
E 0=976. The fifth term is 2048
(0=1)4 E 0=000 009 5, which does not
I
affect the third decimal place of the sum, so we have 1@ 4 1=1 E 0=976. (Note that the third decimal place of the sum of the
terms is then 1 3 14 (0=1) +
5
(0=1)2
32
+
3
15
(0=1)3
128
first three terms is affected by the fourth term, so we need to use more than three terms for the sum.)
"
(16) S
47. cos { =
(31)q
q=0
{ cos({3 ) =
"
S
{2q
(2q)!
(31)q
q=0
(11)
48. h{ =
"
[
{q
q!
q=0
{6q+1
(2q)!
"
S
"
S
({3 )2q
{6q
(31)q
=
(2q)!
(2q)!
q=0
(31)q
q=0
]
i
h{ 3 1 =
i
cos({3 ) =
i
"
[
{q
q!
q=1
{ cos({3 ) g{ = F +
"
S
(31)q
q=0
i
"
[
{q31
h{ 3 1
=
{
q!
q=1
i
{6q+2
, with U = ".
(6q + 2)(2q)!
i
]
"
[
{q
h{ 3 1
g{ = F +
,
{
q · q!
q=1
with U = ".
"
(16) S
49. cos { =
(31)q
q=0
]
{2q
(2q)!
"
S
(31)q
q=0
(31)q
q=1
"
S
i arctan({2 ) =
(31)q
q=0
"
S
(31)q
q=0
]
{2q+1
2q + 1
arctan({2 ) g{ = F +
51. arctan { =
]
"
S
{2q
(2q)!
i
"
S
{2q
cos { 3 1
g{ = F +
, with U = ".
(31)q
{
2q · (2q)!
q=1
50. arctan { =
]
i cos { 3 1 =
"
S
{3 arctan { g{ =
"
S
(31)q
q=0
5
i
{2q+5
. Since
(2q + 1)(2q + 5)
1
2
? 1, we have
(1@2)2q+5
(1@2)5
(1@2)7
(1@2)9
(1@2)11
=
3
+
3
+ · · · . Now
(2q + 1)(2q + 5)
1·5
3·7
5·9
7 · 11
(1@2)
(1@2)9
(1@2)11
(1@2)
3
+
E 0=0059 and subtracting
E 6=3 × 1036 does not affect the fourth decimal place,
1·5
3·7
5·9
7 · 11
U 1@2
so 0 {3 arctan { g{ E 0=0059 by the Alternating Series Estimation Theorem.
52. sin { =
7
q=0
"
S
({2 )2q+1
{4q+2
=
(31)q
2q + 1
2q + 1
q=0
{4q+3
, with U = 1.
(2q + 1)(4q + 3)
(31)q
q=0
0
(31)q
i
"
S
{2q+1
{2q+4
(31)q
for |{| ? 1, so {3 arctan { =
for |{| ? 1 and
2q + 1
2q + 1
q=0
{3 arctan { g{ = F +
1@2
"
S
"
S
{2q31
cos { 3 1
=
(31)q
{
(2q)!
q=1
"
S
(31)q
q=0
]
"
S
{2q+1
{8q+4
(31)q
for all {, so sin({4 ) =
for all { and
(2q + 1)!
(2q + 1)!
q=0
sin({4 ) g{ = F +
"
S
(31)q
q=0
{8q+5
. Thus,
(2q + 1)! (8q + 5)
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¤
190
]
1
CHAPTER 11
sin({4 ) g{ =
INFINITE SEQUENCES AND SERIES
"
S
(31)q
q=0
0
1
1
1
1
1
=
3
+
3
+ · · · . Now
(2q + 1)! (8q + 5)
1! · 5
3! · 13
5! · 21
7! · 29
1
1
1
1
3
+
E 0=1876 and subtracting
E 6=84 × 1036 does not affect the fourth decimal place, so
1! · 5
3! · 13
5! · 21
7! · 29
U1
sin({4 ) g{ E 0=1876 by the Alternating Series Estimation Theorem.
0
$
$
#
#
] s
"
S
1@2
1@2 {4q+1
4 q
4
({ ) , so
and hence, since 0=4 ? 1,
1 + { g{ = F +
q
q 4q + 1
q=0
q=0
"
I
S
53. 1 + {4 = (1 + {4 )1@2 =
we have
$
#
1@2 (0=4)4q+1
q
4q + 1
q=0
0
1
1
3 12 (0=4)9
(0=4)5
(0=4)1
+ 2
+ 2
+
= (1)
0!
1! 5
2!
9
L=
]
0=4
s
"
S
1 + {4 g{ =
= 0=4 +
1
2
1 3 3 2 3 2 (0=4)13
+
3!
13
1
2
1 3 5 3 2 3 2 3 2 (0=4)17
+ ···
4!
17
(0=4)5
(0=4)9
(0=4)13
5(0=4)17
3
+
3
+ ···
10
72
208
2176
(0=4)5
(0=4)9
E 3=6 × 1036 ? 5 × 1036 , so by the Alternating Series Estimation Theorem, L E 0=4 +
E 0=40102
72
10
Now
(correct to five decimal places).
54.
]
0=5
2
{2 h3{ g{ =
0
]
0
with q = 2 is
1@2
"
"
S
S
(31)q {2q+2
(31)q
(31)q {2q+3
g{ =
=
and since the term
2q+3
q!
q!(2q + 3) 0
q=0
q=0
q=0 q!(2q + 3)2
0=5 "
S
1
S
(31)q
1
1
1
? 0=001, we use
3
E 0=0354.
=
2q+3
1792
q!(2q
+
3)2
24
160
q=0
1 3
1 4
1 5
1 2
{ 3 ({ 3 12 {2 + 13 {3 3 14 {4 + 15 {5 3 · · · )
{ 3 ln(1 + {)
2{ 3 3{ + 4{ 3 5{ + ···
=
lim
=
lim
{<0
{<0
{<0
{2
{2
{2
55. lim
= lim ( 12 3 13 { + 14 {2 3 15 {3 + · · · ) =
{<0
1
2
since power series are continuous functions.
56. lim
{<0
1 2
1 4
1 6
1 3 1 3 2!
{ + 4!
{ 3 6!
{ + ···
1 3 cos {
=
lim
{<0 1 + { 3 1 + { + 1 {2 + 1 {3 + 1 {4 + 1 {5 + 1 {6 + · · ·
1 + { 3 h{
2!
3!
4!
5!
6!
= lim
1 2
1 4
1 6
{ 3 4!
{ + 6!
{ 3 ···
2!
1 3
1 4
1 5
1 6
3 3! { 3 4! { 3 5!
{ 3 6!
{
= lim
1
1 2
1 4
3 4!
{ + 6!
{ 3 ···
2!
1
1 2
1 3
1 4
{
3
{
3
{ 3 6!
{
3!
4!
5!
{<0 3 1 {2
2!
{<0 3 1
2!
3
3 ···
3 ···
=
1
30
2
3 12 3 0
= 31
since power series are continuous functions.
57. lim
{<0
1 3
1 5
1 7
{ + 5!
{ 3 7!
{ + · · · 3 { + 16 {3
{ 3 3!
sin { 3 { + 16 {3
=
lim
{<0
{5
{5
1 5
1 7
{ 3 7! { + · · ·
1
{2
{4
1
1
3
+
3
·
·
·
=
=
= lim 5!
=
lim
{<0
{<0 5!
{5
7!
9!
5!
120
since power series are continuous functions.
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SECTION 11.10
58. lim
{<0
tan { 3 {
= lim
{<0
{3
{ + 13 {3 +
2 5
{
15
{3
+ ··· 3 {
= lim
1 3
{
3
{<0
+
2 5
{
15
{3
+ ···
TAYLOR AND MACLAURIN SERIES
1
{<0 3
= lim
since power series are continuous functions.
+
2 2
{
15
+ ··· =
¤
1
3
{4
{6
{2
{4
{2
+
3
+ · · · and we know that cos { = 1 3
+
3 · · · from
1!
2!
3!
2!
4!
2
1 4
{ 3 · · · . Writing only the terms with
Equation 16. Therefore, h3{ cos { = 1 3 {2 + 12 {4 3 · · · 1 3 12 {2 + 24
2
59. From Equation 11, we have h3{ = 1 3
2
degree $ 4, we get h3{ cos { = 1 3 12 {2 +
60. sec { =
1 4
24 {
3 {2 + 12 {4 + 12 {4 + · · · = 1 3 32 {2 +
1 (16)
1
.
=
1 4
cos {
1 3 12 {2 + 24
{ 3 ···
1 + 12 {2 +
1 3 12 {2 +
1 4
{
24
3 ···
1 2
2{
1 2
2{
61.
5 4
{
24
3
3
{3
+
3 ···
1 4
24 {
1 4
4{
+···
5 4
24 {
5 4
24 {
+···
+···
+···
1 3
{
6
1 3
{
6
7
{4
360
7
{4
360
+···
1
{5
120
3···
1
{5
120
1 5
{
36
+···
7
{5
360
7
{5
360
+···
{
{ 3 16 {3 +
{
= 1 + 16 {2 +
From the long division above,
sin {
3···
···
1 + 16 {2 +
1
{5
120
1 4
24 {
+ ···.
{
{ (15)
.
=
1
sin {
{ 3 16 {3 + 120
{5 3 · · ·
1 3
{
6
+···
+ ···.
1
1 3 12 {2 +
From the long division above, sec { = 1 + 12 {2 +
5 4
{
24
25 4
24 {
3
3
+···.
+···
+···
···
{
{2
{3
{2
{3
{4
+
+
+ · · · and that ln(1 + {) = { 3
+
3
+ · · · . Therefore,
1!
2!
3!
2
3
4
{2
{3
{2
{3
{4
{
+
+ ···
{3
+
3
+ · · · . Writing only terms with degree $ 3,
| = h{ ln(1 + {) = 1 + +
1!
2!
3!
2
3
4
62. From Table 1, we have h{ = 1 +
we get h{ ln(1 + {) = { 3 12 {2 + 13 {3 + {2 3 12 {3 + 12 {3 + · · · = { + 12 {2 + 13 {3 + · · · .
63.
"
S
(31)q
q=0
4 q
"
S
3{
4
{4q
=
= h3{ , by (11).
q!
q!
q=0
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192
¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
2q
" (31)q 2q
"
S
S
q 6
64.
(31)
=
= cos 6 =
2q
(2q)!
q=0 6 (2q)!
q=0
65.
"
S
(31)q31
q=1
66.
67.
by (16).
q
"
S
3q
3
8
q31 (3@5)
=
(31)
=
ln
1
+
[from Table 1] = ln
q5q
q
5
5
q=1
" (3@5)q
S
3q
=
= h3@5 , by (11).
q
q!
q=0 5 q!
q=0
"
S
2q+1
" (31)q S
(31)q 2q+1
4
=
= sin 4 =
2q+1 (2q + 1)!
(2q + 1)!
q=0 4
q=0
"
S
68. 1 3 ln 2 +
69. 3 +
70.
I
3
,
2
I1 ,
2
by (15).
" (3 ln 2)q
31
S
(ln 2)3
(ln 2)2
3
+ ··· =
= h3 ln 2 = hln 2
= 231 = 12 , by (11).
2!
3!
q!
q=0
" 3q
" 3q
S
S
9
27
81
31
32
33
34
+
+
+ ··· =
+
+
+
+··· =
=
3 1 = h3 3 1, by (11).
2!
3!
4!
1!
2!
3!
4!
q!
q=1
q=0 q!
2q+1
"
"
S
S
1
1
1
1
1
q
q (1@2)
3
= tan31
+
3
+
·
·
·
=
(31)
=
(31)
1 · 2 3 · 23
5 · 25
7 · 27
(2q + 1)22q+1
2q + 1
q=0
q=0
71. If s is an qth-degree polynomial, then s(l) ({) = 0 for l A q, so its Taylor series at d is s({) =
Put { 3 d = 1, so that { = d + 1. Then s(d + 1) =
q s(l) (d)
S
.
l!
l=0
This is true for any d, so replace d by {: s({ + 1) =
q s(l) ({)
S
l!
l=0
72. The coefficient of {58 in the Maclaurin series of i ({) = (1 + {3 )30 is
3 30
(1 + { )
1
[from Table 1]
2
q s(l) (d)
S
({ 3 d)l .
l!
l=0
i (58) (0)
. But the binomial series for i ({) is
58!
# $
30 3q
{ , so it involves only powers of { that are multiples of 3 and therefore the coefficient of {58 is 0.
=
q
q=0
"
S
So i (58) (0) = 0.
U{
i 000 (w) gw $ d P gw i
U{
U{
i 00 ({) 3 i 00 (d) $ P ({ 3 d) i i 00 ({) $ i 00 (d) + P ({ 3 d). Thus, d i 00 (w) gw $ d [i 00 (d) + P (w 3 d)] gw i
73. Assume that |i 000 ({)| $ P, so i 000 ({) $ P for d $ { $ d + g. Now
U{
d
i 0 ({) 3 i 0 (d) $ i 00 (d)({ 3 d) + 12 P({ 3 d)2 i i 0 ({) $ i 0 (d) + i 00 (d)({ 3 d) + 12 P ({ 3 d)2
U{ 0
U{
i (w) gw $ d i 0 (d) + i 00 (d)(w 3 d) + 12 P(w 3 d)2 gw i
d
i
i ({) 3 i (d) $ i 0 (d)({ 3 d) + 12 i 00 (d)({ 3 d)2 + 16 P({ 3 d)3 . So
i ({) 3 i (d) 3 i 0 (d)({ 3 d) 3 12 i 00 (d)({ 3 d)2 $
1
P({
6
3 d)3 . But
U2 ({) = i ({) 3 W2 ({) = i ({) 3 i (d) 3 i 0 (d)({ 3 d) 3 12 i 00 (d)({ 3 d)2 , so U2 ({) $ 16 P({ 3 d)3 .
A similar argument using i 000 ({) D 3P shows that U2 ({) D 3 16 P({ 3 d)3 . So |U2 ({2 )| $ 16 P |{ 3 d|3 .
Although we have assumed that { A d, a similar calculation shows that this inequality is also true if { ? d.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.10
74. (a) i ({) =
+
2
h31@{
¤
193
2
if { 6= 0
0
TAYLOR AND MACLAURIN SERIES
so i 0 (0) = lim
{<0
if { = 0
i ({) 3 i (0)
h31@{
1@{
{
= lim
= lim 1@{2 = lim 1@{2 = 0
{<0
{<0 h
{<0 2h
{30
{
(using l’Hospital’s Rule and simplifying in the penultimate step). Similarly, we can use the definition of the derivative and
l’Hospital’s Rule to show that i 00 (0) = 0, i (3) (0) = 0, = = =, i (q) (0) = 0, so that the Maclaurin series for i consists
entirely of zero terms. But since i ({) 6= 0 except for { = 0, we see that i cannot equal its Maclaurin series except
at { = 0.
(b)
From the graph, it seems that the function is extremely flat at the origin.
In fact, it could be said to be “infinitely flat” at { = 0, since all of its
derivatives are 0 there.
# $
n q
75. (a) j({) =
{
q
q=0
# $
n
i j ({) =
q{q31 , so
q
q=1
# $
# $
# $
"
"
"
S
S
S
n
n
n
q{q31 =
q{q31 +
q{q
(1 + {)j 0 ({) = (1 + {)
q=1 q
q=1 q
q=1 q
"
S
=
"
S
q=0
=
"
S
0
"
S
$
# $
"
S
n
n
(q + 1){q +
q{q
q+1
q=0 q
#
(q + 1)
q=0
Replace q with q + 1
in the first series
"
S
n(n 3 1)(n 3 2) · · · (n 3 q + 1)(n 3 q) q
n(n 3 1)(n 3 2) · · · (n 3 q + 1) q
{ +
{
(q)
(q + 1)!
q!
q=0
" (q + 1)n(n 3 1)(n 3 2) · · · (n 3 q + 1)
S
[(n 3 q) + q] {q
(q + 1)!
q=0
# $
" n(n 3 1)(n 3 2) · · · (n 3 q + 1)
"
S
S
n q
q
{ = nj({)
{ =n
=n
q!
q=0
q=0 q
=
Thus, j0 ({) =
nj({)
.
1+{
(b) k({) = (1 + {)3n j({) i
k0 ({) = 3n(1 + {)3n31 j({) + (1 + {)3n j0 ({)
= 3n(1 + {)3n31 j({) + (1 + {)3n
nj({)
1+{
[Product Rule]
[from part (a)]
= 3n(1 + {)3n31 j({) + n(1 + {)3n31 j({) = 0
(c) From part (b) we see that k({) must be constant for { M (31> 1), so k({) = k(0) = 1 for { M (31> 1).
Thus, k({) = 1 = (1 + {)3n j({) C j({) = (1 + {)n for { M (31> 1).
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°
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194
¤
CHAPTER 11 INFINITE SEQUENCES AND SERIES
76. Using the binomial series to expand
I
1 + { as a power series as in Example 9, we get
" (31)q31 1 · 3 · 5 · · · · · (2q 3 3){q
I
S
{
, so
1 + { = (1 + {)1@2 = 1 + +
2 q=2
2q · q!
" 1 · 3 · 5 · · · · · (2q 3 3)
1@2
S
1
1 3 {2
{2q and
= 1 3 {2 3
2
2q · q!
q=2
s
" 1 · 3 · 5 · · · · · (2q 3 3)
S
1
1 3 h2 sin2 = 1 3 h2 sin2 3
h2q sin2q . Thus,
2
2q · q!
q=2
] @2 ] @2 s
" 1 · 3 · 5 · · · · · (2q 3 3)
S
1
2q
2q
1 3 h2 sin2 g = 4d
sin
g
1 3 h2 sin2 3
O = 4d
h
2
2q · q!
q=2
0
0
2 q " 1 · 3 · 5 · · · · · (2q 3 3)
S
h2
h
3 V1 3
Vq
= 4d
2
2
q!
2
q=2
where Vq =
]
@2
1 · 3 · 5 · · · · · (2q 3 1) by Exercise 7.1.46.
2 · 4 · 6 · · · · · 2q
2
2 q
" 1 · 3 · 5 · · · · · (2q 3 3)
S
1 · 3 · 5 · · · · · (2q 3 1)
h2 1
h
· 3
2 2 q=2
q!
2
2 · 4 · 6 · · · · · 2q
2q
2
2
2
2
"
S h
1 · 3 · 5 · · · · · (2q 3 3) (2q 3 1)
·
3
q
q! · 2q · q!
q=2 2
&
2
" h2q
S
1 · 3 · · · · · (2q 3 3)
3
(2q 3 1)
q
q!
q=2 4
sin2q g =
0
O = 4d
13
2
h2
= 2d 1 3
4
%
h2
= 2d 1 3
4
h2
3h4
5h6
d
= 2d 1 3
3
3
3 ··· =
(256 3 64h2 3 12h4 3 5h6 3 · · · )
4
64
256
128
LABORATORY PROJECT An Elusive Limit
1. i ({) =
sin(tan {) 3 tan(sin {)
q({)
=
g({)
arcsin(arctan {) 3 arctan(arcsin {)
The table of function values were obtained using Maple with 10 digits of
precision. The results of this project will vary depending on the CAS and
precision level. It appears that as { < 0+ , i ({) <
function, we have i ({) <
10
3
10
.
3
Since i is an even
as { < 0.
{
i ({)
1
1=1838
0=1
0=9821
0=01
2=0000
0=001
3=3333
0=0001
3=3333
2. The graph is inconclusive about the limit of i as { < 0.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 11.11
APPLICATIONS OF TAYLOR POLYNOMIALS
3. The limit has the indeterminate form 00 . Applying l’Hospital’s Rule, we obtain the form
application we obtain lim
{<0
195
six times. Finally, on the seventh
3168
q(7) ({)
=
= 1.
3168
g(7) ({)
3 1 {7 3
q({) CAS
= lim 30
{<0 g({)
{<0 3 1 {7 +
30
1 7
3 { 3
= lim 30
{<0 3 1 {7 +
30
4. lim i ({) = lim
{<0
0
0
¤
29 9
{
756
13 9
756 {
29 9
{
756
13 9
{
756
+ ···
+ ···
+ · · · @{7
31 3
= lim 30
7
{<0 3 1 +
+ · · · @{
30
29 2
{
756
13 2
{
756
31
+ ···
= 30
1 = 1
+ ···
3 30
7!
= 3 5040
Note that q(7) ({) = g(7) ({) = 3 30
30 = 3168, which agrees with the result in Problem 3.
5. The limit command gives the result that lim i ({) = 1.
{<0
6. The strange results (with only 10 digits of precision) must be due to the fact that the terms being subtracted in the numerator
and denominator are very close in value when |{| is small. Thus, the differences are imprecise (have few correct digits).
11.11 Applications of Taylor Polynomials
1. (a)
q
i (q) ({)
0
cos {
1
1
1
3 sin {
0
1
2
3
i (q) (0)
3 cos {
31
sin {
0
4
cos {
1
5
3 sin {
0
6
3 cos {
31
Wq ({)
1 3 12 {2
1 3 12 {2
1 3 12 {2 +
1 3 12 {2 +
1 3 12 {2 +
1 4
{
24
1 4
{
24
1 4
{
24
3
1
{6
720
(b)
{
4
2
W2 = W3
W4 = W5
0=7071
i
W0 = W1
1
0=6916
0=7074
0=7071
0
1
0=0200
31
1
30=2337
30=0009
33=9348
0=1239
W6
31=2114
(c) As q increases, Wq ({) is a good approximation to i ({) on a larger and larger interval.
2. (a)
q
0
1
2
3
i (q) ({)
i (q) (1)
Wq ({)
31
1
32
3{
31
1 3 ({ 3 1) = 2 3 {
36{34
36
1 3 ({ 3 1) + ({ 3 1)2 3 ({ 3 1)3 = 3{3 + 4{2 3 6{ + 4
{
2{33
2
1
1 3 ({ 3 1) + ({ 3 1)2 = {2 3 3{ + 3
[continued]
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CHAPTER 11
INFINITE SEQUENCES AND SERIES
(b)
{
i
W0
W1
W2
W3
0=9
1=1
1
1=1
1=11
1=111
1=3
0=7692
1
0=7
0=79
0=763
(c) As q increases, Wq ({) is a good approximation to i ({) on a larger and larger interval.
3.
q
i (q) ({)
i (q) (2)
0
1@{
1
31@{2
1
2
3 14
1
4
3 38
2@{3
2
36@{4
3
W3 ({) =
=
=
3 i (q) (2)
S
({ 3 2)q
q!
q=0
1
2
0!
1
2
1
4
3
1!
({ 3 2) +
1
4
2!
({ 3 2)2 3
3 14 ({ 3 2) + 18 ({ 3 2)2 3
3
8
3!
1
({
16
({ 3 2)3
3 2)3
4.
q
i (q) ({)
i (q) (0)
0
{ + h3{
1
1
3{
0
13h
3{
2
h
3h3{
3
W3 ({) =
=
1
31
3 i (q) (0)
S
({ 3 0)q
q!
q=0
0
1
1
1
+ { + {2 3 {3 = 1 + 12 {2 3 16 {3
0!
1!
2!
3!
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SECTION 11.11
APPLICATIONS OF TAYLOR POLYNOMIALS
5.
q
i (q) ({)
i (q) (@2)
0
cos {
0
1
3 sin {
31
sin {
1
2
3 cos {
3
0
3 i (q) (@2) q
S
{ 3 2
q!
q=0
3
= 3 { 3 2 + 16 { 3 2
W3 ({) =
6.
q
i (q) ({)
i (q) (0)
0
h3{ sin {
0
3{
1
h
(cos { 3 sin {)
32h3{ cos {
2
2h3{ (cos { + sin {)
3
W3 ({) =
1
32
2
3 i (q) (0)
S
{q = { 3 {2 + 13 {3
q!
q=0
7.
q
i (q) ({)
i (q) (1)
0
ln {
0
1
1@{
1
2
2
31@{
3
3
2@{
W3 ({) =
31
2
3 i (q) (1)
S
({ 3 1)q
q!
q=0
=0+
31
1
2
({ 3 1) +
({ 3 1)2 + ({ 3 1)3
1!
2!
3!
= ({ 3 1) 3 12 ({ 3 1)2 + 13 ({ 3 1)3
8.
q
i (q) ({)
i (q) (0)
0
{ cos {
0
1
3{ sin { + cos {
1
{ sin { 3 3 cos {
33
2
3{ cos { 3 2 sin {
3
0
3 i (q) (0)
S
{q
q!
q=0
33 3
1
1
{ = { 3 {3
=0+ {+0+
1!
3!
2
W3 ({) =
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CHAPTER 11
INFINITE SEQUENCES AND SERIES
9.
q
i (q) ({)
i (q) (0)
0
{h32{
0
1
32{
1
32{
34
(1 3 2{)h
2
4({ 3 1)h
3
32{
4(3 3 2{)h
W3 ({) =
12
3 i (q) (0)
S
{q =
q!
q=0
0
1
· 1 + 11 {1 +
34 2
{
2
+
12 3
{
6
= { 3 2{2 + 2{3
10.
q
i (q) ({)
i (q) (1)
0
tan31 {
4
1
1 + {2
32{
(1 + {2 )2
1
2
1
2
3 12
6{2 3 2
(1 + {2 )3
3
W3 ({) =
=
1
2
3 i (q) (1)
S
1@2
31@2
1@2
({ 3 1)q = +
({ 3 1)1 +
({ 3 1)2 +
({ 3 1)3
q!
4
1
2
6
q=0
4
+ 12 ({ 3 1) 3 14 ({ 3 1)2 +
1
({
12
3 1)3
11. You may be able to simply find the Taylor polynomials for
i ({) = cot { using your CAS. We will list the values of i (q) (@4)
for q = 0 to q = 5.
i
(q)
q
0
1
2
3
4
5
(@4)
1
32
4
316
80
3512
5 i (q) (@4) q
S
{ 3 4
q!
q=0
2
3
= 1 3 2 { 3 4 + 2 { 3 4 3 83 { 3 4 +
W5 ({) =
10
3
4
{ 3 4 3
64
15
5
{ 3 4
q
For q = 2 to q = 5, Wq ({) is the polynomial consisting of all the terms up to and including the { 3 4 term=
12. You may be able to simply find the Taylor polynomials for
i ({) =
I
3
1 + {2 using your CAS. We will list the values of i (q) (0)
for q = 0 to q = 5.
q
i
(q)
0
(0)
W5 ({) =
1
1
2
0
2
3
3
4
5
0
3 83
0
5 i (q) (0)
S
{q = 1 + 13 {2 3 19 {4
q!
q=0
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SECTION 11.11
APPLICATIONS OF TAYLOR POLYNOMIALS
¤
199
For q = 2 to q = 5, Wq ({) is the polynomial consisting of all the terms up to and including the {q term.
Note that W2 = W3 and W4 = W5 .
13.
(a) i ({) =
i (q) (4)
0
i (q) ({)
I
{
1
1 31@2
{
2
1
4
2
3 14 {33@2
1
3 32
q
3
I
1@32
1
{ E W2 ({) = 2 + ({ 3 4) 3
({ 3 4)2
4
2!
= 2 + 14 ({ 3 4) 3
2
3 35@2
{
8
(b) |U2 ({)| $
1
({
64
3 4)2
P
|{ 3 4|3 , where |i 000 ({)| $ P= Now 4 $ { $ 4=2 i
3!
|{ 3 4| $ 0=2 i |{ 3 4|3 $ 0=008. Since i 000 ({) is decreasing
on [4> 4=2], we can take P = |i 000 (4)| = 38 435@2 =
|U2 ({)| $
3
,
256
so
0=008
3@256
(0=008) =
= 0=000 015 625.
6
512
(c)
I
From the graph of |U2 ({)| = | { 3 W2 ({)|, it seems that the
error is less than 1=52 × 1035 on [4> 4=2].
14.
q
i
(q)
({)
i
(q)
(1)
0
{32
1
1
32{33
32
2
3
34
6{
(a) i ({) = {32 E W2 ({) = 132({31)+
(b) |U2 ({)| $
6
({31)2 = 132({31)+3({31)2
2!
P
|{ 3 1|3 , where |i 000 ({)| $ P . Now 0=9 $ { $ 1=1 i
3!
|{ 3 1| $ 0=1 i |{ 3 1|3 $ 0=001. Since i 000 ({) is decreasing on
6
[0=9> 1=1], we can take P = |i 000 (0=9)| =
35
324{
|U2 ({)| $
24
, so
(0=9)5
24@(0=9)5
0=004
(0=001) =
E 0=006 774 04.
6
0=59049
(c)
From the graph of |U2 ({)| = {32 3 W2 ({), it seems that
the error is less than 0=0046 on [0=9> 1=1].
15.
q
i (q) ({)
i (q) (1)
0
{2@3
1
1
2 31@3
3{
2
3
2
3 29 {34@3
3 29
3
8 37@3
{
27
8
27
4
3 56
{310@3
81
(a) i ({) = {2@3 E W3 ({) = 1 + 23 ({ 3 1) 3
2@9
8@27
({ 3 1)2 +
({ 3 1)3
2!
3!
= 1 + 23 ({ 3 1) 3 19 ({ 3 1)2 +
4
({
81
3 1)3
P
|{ 3 1|4 , where i (4) ({) $ P . Now 0=8 $ { $ 1=2 i
4!
|{ 3 1| $ 0=2 i |{ 3 1|4 $ 0=0016. Since i (4) ({) is decreasing
(0=8)310@3 , so
on [0=8> 1=2], we can take P = i (4) (0=8) = 56
81
(b) |U3 ({)| $
|U3 ({)| $
56
(0=8)310@3
81
24
(0=0016) E 0=000 096 97.
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¤
CHAPTER 11
INFINITE SEQUENCES AND SERIES
(c)
From the graph of |U3 ({)| = {2@3 3 W3 ({), it seems that the
error is less than 0=000 053 3 on [0=8> 1=2].
16.
q
i (q) ({)
i (q) (@6)
0
sin {
1
cos {
1@2
I
3@2
2
3 sin {
3
3 cos {
31@2
I
3 3@2
4
sin {
1@2
5
cos {
(a) i ({) = sin { E W4 ({)
I 2
= 12 + 23 { 3 6 3 14 { 3 6 3
I
3
12
3
{ 3 6 +
1
48
4
{ 3 6
5
P { 3 6 , where i (5) ({) $ P . Now 0 $ { $ 3 i
5!
5 5
3 6 $ { 3 6 $ 6 i { 3 6 $ 6 i { 3 6 $ 6 . Since
(5) i ({) is decreasing on 0> 3 , we can take P = i (5) (0) = cos 0 = 1,
(b) |U4 ({)| $
so |U4 ({)| $
1 5
E 0=000 328.
5! 6
(c)
From the graph of |U4 ({)| = |sin { 3 W4 ({)|, it seems that the
error is less than 0=000 297 on 0> 3 .
17.
q
i
(q)
({)
i
(q)
(0)
0
sec {
1
1
sec { tan {
0
2
3
2
sec { (2 sec { 3 1)
(a) i({) = sec { E W2 ({) = 1 + 12 {2
1
2
sec { tan { (6 sec { 3 1)
(b) |U2 ({)| $
P
|{|3 , where i (3) ({) $ P . Now 30=2 $ { $ 0=2 i |{| $ 0=2 i |{|3 $ (0=2)3 .
3!
i (3) ({) is an odd function and it is increasing on [0> 0=2] since sec { and tan { are increasing on [0> 0=2],
(c)
i (3) (0=2)
(0=2)3 E 0=001 447.
so i (3) ({) $ i (3) (0=2) E 1=085 158 892. Thus, |U2 ({)| $
3!
From the graph of |U2 ({)| = |sec { 3 W2 ({)|, it seems that the
error is less than 0=000 339 on [30=2> 0=2].
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SECTION 11.11
18.
q
i (q) ({)
i (q) (1)
0
ln(1 + 2{)
ln 3
1
2@(1 + 2{)
2
3
2
2
34@(1 + 2{)
3 49
3
16@(1 + 2{)3
16
27
4
396@(1 + 2{)4
APPLICATIONS OF TAYLOR POLYNOMIALS
¤
201
(a) i ({) = ln(1 + 2{) E W3 ({)
= ln 3 + 23 ({ 3 1) 3
(b) |U3 ({)| $
4@9
16@27
({ 3 1)2 +
({ 3 1)3
2!
3!
P
|{ 3 1|4 , where i (4) ({) $ P. Now 0=5 $ { $ 1=5 i
4!
30=5 $ { 3 1 $ 0=5 i |{ 3 1| $ 0=5 i |{ 3 1|4 $
letting { = 0=5 gives P = 6, so |U3 ({)| $
1
,
16
and
1
6 1
·
=
= 0=015 625.
4! 16
64
(c)
From the graph of |U3 ({)| = |ln(1 + 2{) 3 W3 ({)|, it seems that the
error is less than 0=005 on [0=5> 1=5].
2
19.
i (q) ({)
q
i (q) (0)
2
0
h{
1
h{ (2{)
1
2
0
2
2
h{ (2 + 4{2 )
3
h{ (12{ + 8{3 )
4
h{ (12 + 48{2 + 16{4 )
2
2
(a) i({) = h{ E W3 ({) = 1 +
(b) |U3 ({)| $
2 2
{ = 1 + {2
2!
P
|{|4 , where i (4) ({) $ P . Now 0 $ { $ 0=1 i
4!
2
{4 $ (0=1)4 , and letting { = 0=1 gives
0
|U3 ({)| $
h0=01 (12 + 0=48 + 0=0016)
(0=1)4 E 0=00006.
24
(c)
2
From the graph of |U3 ({)| = h{ 3 W3 ({), it appears that the
error is less than 0=000 051 on [0> 0=1].
20.
q
i
(q)
({)
i
(q)
(1)
0
{ ln {
0
1
ln { + 1
1
2
1@{
1
3
31@{2
31
4
2@{3
(a) i ({) = { ln { E W3 ({) = ({ 3 1) + 12 ({ 3 1)2 3 16 ({ 3 1)3
P
|{ 3 1|4 , where i (4) ({) $ P . Now 0=5 $ { $ 1=5 i
4!
1
. Since i (4) ({) is decreasing on
|{ 3 1| $ 12 i |{ 3 1|4 $ 16
(b) |U3 ({)| $
[0=5> 1=5], we can take P = i (4) (0=5) = 2@(0=5)3 = 16, so
|U3 ({)| $
16
(1@16)
24
=
1
24
= 0=0416.
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CHAPTER 11
INFINITE SEQUENCES AND SERIES
(c)
From the graph of |U3 ({)| = |{ ln { 3 W3 ({)|, it seems that the error
is less than 0=0076 on [0=5> 1=5].
21.
q
i (q) ({)
i (q) (0)
0
{ sin {
0
1
sin { + { cos {
0
2
2 cos { 3 { sin {
2
3
33 sin { 3 { cos {
0
4
34 cos { + { sin {
34
5
5 sin { + { cos {
(a) i ({) = { sin { E W4 ({) =
2
34
1
({ 3 0)2 +
({ 3 0)4 = {2 3 {4
2!
4!
6
P
|{|5 , where i (5) ({) $ P. Now 31 $ { $ 1 i
5!
|{| $ 1, and a graph of i (5) ({) shows that i (5) ({) $ 5 for 31 $ { $ 1.
(b) |U4 ({)| $
Thus, we can take P = 5 and get |U4 ({)| $
5
1
· 15 =
= 0=0416.
5!
24
(c)
From the graph of |U4 ({)| = |{ sin { 3 W4 ({)|, it seems that the
error is less than 0=0082 on [31> 1].
22.
(q)
({)
i
(q)
q
i
(0)
0
sinh 2{
0
1
2 cosh 2{
2
2
4 sinh 2{
0
3
8 cosh 2{
8
4
16 sinh 2{
0
5
32 cosh 2{
32
6
64 sinh 2{
(a) i ({) = sinh 2{ E W5 ({) = 2{ +
(b) |U5 ({)| $
P
6!
8 3
{
3!
+
32 5
{
5!
= 2{ + 43 {3 +
4 5
{
15
|{|6 , where i (6) ({) $ P. For { in [31> 1], we have
|{| $ 1. Since i (6) ({) is an increasing odd function on [31> 1], we see
that i (6) ({) $ i (6) (1) = 64 sinh 2 = 32(h2 3 h32 ) E 232=119,
so we can take P = 232=12 and get |U5 ({)| $
232=12
720
· 16 E 0=3224.
(c)
From the graph of |U5 ({)| = |sinh 2{ 3 W5 ({)|, it seems that the
error is less than 0=027 on [31> 1].
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SECTION 11.11
APPLICATIONS OF TAYLOR POLYNOMIALS
¤
203
4
3
P { 3 2 + U3 ({), where |U3 ({)| $
{ 3 2 with
4!
(4) = 4
radians, so the error is
i ({) = |cos {| $ P = 1. Now { = 80 = (90 3 10 ) = 2 3 18
9
23. From Exercise 5, cos { = 3 { 3
4 $
U3
9
1
24
4
18
2
+
1
6
E 0=000 039, which means our estimate would not be accurate to five decimal places. However,
$
W3 = W4 , so we can use U4 4
9
1
120
5
E 0=000 001. Therefore, to five decimal places,
18
1 3
+ 6 3 18 E 0=17365.
cos 80 E 3 3 18
24. From Exercise 16, sin { =
1
2
+
I
3
2
2
{ 3 6 3 14 { 3 6 3
I
3
12
3
{ 3 6 +
1
48
4
{ 3 6 + U4 ({), where
5
P radians,
{ 3 6 with i (5) ({) = |cos {| $ P = 1. Now { = 38 = (30 + 8 ) = 6 + 2
45
5!
$ 1 2 5 E 0=000 000 44, which means our estimate will be accurate to five decimal places.
so the error is U4 38
180
120 45
|U4 ({)| $
Therefore, to five decimal places, sin 38 =
25. All derivatives of h{ are h{ , so |Uq ({)| $
Uq (0=1) $
1
2
+
I 3 2
2
45
3
1
4
2 2
45
3
I
3
12
2 3
45
+
1
48
2 4
45
E 0=61566.
h{
|{|q+1 , where 0 ? { ? 0=1. Letting { = 0=1,
(q + 1)!
h0=1
(0=1)q+1 ? 0=00001, and by trial and error we find that q = 3 satisfies this inequality since
(q + 1)!
U3 (0=1) ? 0=0000046. Thus, by adding the four terms of the Maclaurin series for h{ corresponding to q = 0, 1, 2, and 3,
we can estimate h0=1 to within 0=00001. (In fact, this sum is 1=10516 and h0=1 E 1=10517.)
26. From Table 1 in Section 11.10, ln(1 + {) =
"
S
(31)q31
q=1
"
S
{q
(0=4)q
for |{| ? 1. Thus, ln 1=4 = ln(1 + 0=4) =
.
(31)q31
q
q
q=1
Since this is an alternating series, the error is less than the first neglected term by the Alternating Series Estimation Theorem,
and we find that |d6 | = (0=4)6@6 E 0=0007 ? 0=001. So we need the first five (nonzero) terms of the Maclaurin series for the
desired accuracy. (In fact, this sum is approximately 0=33698 and ln 1=4 E 0=33647.)
27. sin { = { 3
1 3
1
{ + {5 3 · · · . By the Alternating Series
3!
5!
Estimation Theorem, the error in the approximation
1 1
sin { = { 3 {3 is less than {5 ? 0=01 C
3!
5!
5
{ ? 120(0=01) C |{| ? (1=2)1@5 E 1=037. The curves
| = { 3 16 {3 and | = sin { 3 0=01 intersect at { E 1=043, so
the graph confirms our estimate. Since both the sine function
and the given approximation are odd functions, we need to check the estimate only for { A 0. Thus, the desired range of
values for { is 31=037 ? { ? 1=037.
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CHAPTER 11
INFINITE SEQUENCES AND SERIES
1 2
1
1
{ + {4 3 {6 + · · · . By the Alternating Series
2!
4!
6!
1 Estimation Theorem, the error is less than 3 {6 ? 0=005 C
6!
28. cos { = 1 3
{6 ? 720(0=005) C |{| ? (3=6)1@6 E 1=238. The curves
| = 1 3 12 {2 +
1 4
{
24
and | = cos { + 0=005 intersect at { E 1=244,
so the graph confirms our estimate. Since both the cosine function
and the given approximation are even functions, we need to check
the estimate only for { A 0. Thus, the desired range of values for { is 31=238 ? { ? 1=238.
{5
{7
{3
+
3
+ · · · . By the Alternating Series
3
5
7
Estimation Theorem, the error is less than 3 17 {7 ? 0=05 C
7
{ ? 0=35 C |{| ? (0=35)1@7 E 0=8607. The curves
29. arctan { = { 3
| = { 3 13 {3 + 15 {5 and | = arctan { + 0=05 intersect at
{ E 0=9245, so the graph confirms our estimate. Since both the
arctangent function and the given approximation are odd functions,
we need to check the estimate only for { A 0. Thus, the desired
range of values for { is 30=86 ? { ? 0=86.
" i (q) (4)
"
"
S
S
S
(31)q q!
(31)q
({ 3 4)q =
({ 3 4)q =
({ 3 4)q . Now
q
q
q!
q=0
q=0 3 (q + 1) q!
q=0 3 (q + 1)
30. i ({) =
i (5) =
"
S
q=0
"
S
(31)q
=
(31)q eq is the sum of an alternating series that satisfies (i) eq+1 $ eq and
+ 1)
q=0
3q (q
(ii) lim eq = 0, so by the Alternating Series Estimation Theorem, |U5 (5)| = |i (5) 3 W5 (5)| $ e6 , and
q<"
e6 =
1
1
=
E 0=000196 ? 0=0002 ; that is, the fifth-degree Taylor polynomial approximates i (5) with error less
36 (7)
5103
than 0=0002.
31. Let v(w) be the position function of the car, and for convenience set v(0) = 0. The velocity of the car is y(w) = v0 (w) and the
acceleration is d(w) = v00 (w), so the second degree Taylor polynomial is W2 (w) = v(0) + y(0)w +
d(0) 2
w = 20w + w2 . We
2
estimate the distance traveled during the next second to be v(1) E W2 (1) = 20 + 1 = 21 m. The function W2 (w) would not be
accurate over a full minute, since the car could not possibly maintain an acceleration of 2 m@s2 for that long (if it did, its final
speed would be 140 m@s E 313 mi@h!).
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SECTION 11.11
32. (a)
q
(q) (w)
(q) (20)
0
20 h(w320)
20
1
20 h(w320)
20
2
2
(w320)
20 h
¤
205
The linear approximation is
W1 (w) = (20) + 0 (20)(w 3 20) = 20 [1 + (w 3 20)]
The quadratic approximation is
2 20
(b)
APPLICATIONS OF TAYLOR POLYNOMIALS
(c)
00 (20)
W2 (w) = (20) + 0 (20)(w 3 20) +
(w 3 20)2
2
= 20 1 + (w 3 20) + 12 2 (w 3 20)2
From the graph, it seems that W1 (w) is within 1% of (w), that
is, 0=99(w) $ W1 (w) $ 1=01(w), for 314 C $ w $ 58 C.
%
32 &
t
t
t
t
t
g
33. H = 2 3
= 2 3 2
= 2 13 1+
.
G
(G + g)2
G
G (1 + g@G)2
G
G
We use the Binomial Series to expand (1 + g@G)32 :
%
#
% $&
&
2
3
2
3
g
g
2·3 g
g
2·3·4 g
t
g
t
+
33
3
+···
= 2 2
+4
3 ···
H= 2 13 132
G
G
2! G
3!
G
G
G
G
G
g
1
t
= 2tg · 3
E 2 ·2
G
G
G
when G is much larger than g; that is, when S is far away from the dipole.
34. (a)
q1
q2
1 q2 vl
q1 vr
+
=
3
[Equation 1] where
cr
cl
U
cl
cr
s
s
cr = U2 + (vr + U)2 3 2U(vr + U) cos ! and cl = U2 + (vl 3 U)2 + 2U(vl 3 U) cos ! (2)
Using cos ! E 1 gives
t
s
s
cr = U2 + (vr + U)2 3 2U(vr + U) = U2 + v2r + 2Uvr + U2 3 2Uvr 3 2U2 = v2r = vr
q1
q2
1
and similarly, cl = vl . Thus, Equation 1 becomes
+
=
vr
vl
U
q2 vl
q1 vr
3
vl
vr
i
q2
q2 3 q1
q1
.
+
=
vr
vl
U
(b) Using cos ! E 1 3 12 !2 in (2) gives us
cr =
=
t
U2 + (vr + U)2 3 2U(vr + U) 1 3 12 !2
s
s
U2 + v2r + 2Uvr + U2 3 2Uvr + Uvr !2 3 2U2 + U2 !2 = v2r + Uvr !2 + U2 !2
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CHAPTER 11
INFINITE SEQUENCES AND SERIES
Anticipating that we will use the binomial series expansion (1 + {)n E 1 + n{, we can write the last expression for cr as
v
v
U
U
U2
U2
2
vr 1 + !
+ 2 and similarly, cl = vl 1 3 !2
3 2 . Thus, from Equation 1,
vr
vr
vl
vl
q2
1
q1
+
=
cr
cl
U
q2 vl
q1 vr
3
cl
cr
31
C q1 c31
=
r + q2 cl
q2 vl
q1 vr
·
·
3
U cl
U cr
C
31@2
31@2
U
U
q1
U2
q2
U2
1 + !2
1 3 !2
+ 2
+
3 2
vr
vr
vr
vl
vl
vl
=
31@2
31@2
U
U
q2
U2
q1
U2
1 3 !2
1 + !2
3 2
3
+ 2
U
vl
vl
U
vr
vr
31
Approximating the expressions for c31
by the first two terms in their binomial series, we get
r and cl
U2
U2
U
q2
U
q1
+ 2
3 2
1 3 12 !2
+
1 + 12 !2
vr
vr
vr
vl
vl
vl
q2
U
q1
U
U2
U2
=
3
C
1 + 12 !2
1 3 12 !2
3 2
+ 2
U
vl
vl
U
vr
vr
q1
q1 !2
3
vr
2vr
U
U2
+ 2
vr
vr
+
q2
q2 !2
+
vl
2vl
U
U2
3 2
vl
vl
=
q2
q2 !2
+
U
2U
U
U2
3 2
vl
vl
3
q1
q1 !2
+
U
2U
U
U2
+ 2
vr
vr
C
q1
q1 !2 U
q2 !2 U
q2 !2 U
q1
q1 !2 U
q2
q2
U2
U2
U2
U2
3
+
+
=
+ 2 +
+ 2 +
3 2 3
3 2
vr
vl
U
U
2vr vr
vr
2U vr
vr
2U vl
vl
2vl vl
vl
1
1
q2 3 q1
q1 !2 U
q2 !2 U
1
U2
1
U2
=
+
+
3
+ 2
+
3 2
U
2
vr
vr
vr
U
2
vl
vl
U
vl
q2 3 q1
1
q2 !2 U2 1
1
q1 !2 U2 1
1
1
1
1
=
+
+
+
+
3
3
U
2vr
U
vr
U
vr
2vl
U
vl
U
vl
%
&
2
2
q1 1
1
1
q2 1
q2 3 q1
+ !2 U2
+
3
+
=
U
2vr U
vr
2vl U
vl
From Figure 8, we see that sin ! = k@U. So if we approximate sin ! with !, we get k = U! and k2 = !2 U2 and hence,
Equation 4, as desired.
35. (a) If the water is deep, then 2g@O is large, and we know that tanh { < 1 as { < ". So we can approximate
tanh(2g@O) E 1, and so y 2 E jO@(2) C y E
(b) From the table, the first term in the Maclaurin series of
s
jO@(2).
tanh { is {, so if the water is shallow, we can approximate
2g
2g
jO 2g
tanh
E
, and so y 2 E
·
O
O
2
O
I
C y E jg.
q
i (q) ({)
i (q) (0)
0
tanh {
0
1
2
3
2
sech {
1
2
32 sech { tanh {
2 sech2 { (3 tanh2 { 3 1)
0
32
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SECTION 11.11
APPLICATIONS OF TAYLOR POLYNOMIALS
¤
(c) Since tanh { is an odd function, its Maclaurin series is alternating, so the error in the approximation
3
3
2g
1 2g
2g
|i 000 (0)| 2g
tanh
=
.
E
is less than the first neglected term, which is
O
O
3!
O
3 O
3
3
1 2g
1
1
3
If O A 10g, then
2 ·
, so the error in the approximation y 2 = jg is less
?
=
3 O
3
10
375
than
jO 3
·
E 0=0132jO.
2 375
36. First note that
%
&
u
I
I
U2
2
2
2
1+ 2 3g
2 g +U 3g =2 g
g
2
I
U 1
E 2 g 1 + 2 · + ··· 3 g
use the binomial series 1 + 12 { + · · · for 1 + {
g
2
U2
U2
=2 g+
+ ··· 3 g E
2g
g
since for large g the other terms are comparatively small. Now Y = 2nh approximation.
37. (a) O is the length of the arc subtended by the angle , so O = U
i
= O@U. Now sec = (U + F)@U i U sec = U + F
i
I
nh U2 by the preceding
g2 + U2 3 g E
g
F = U sec 3 U = U sec(O@U) 3 U.
(b) First we’ll find a Taylor polynomial W4 ({) for i ({) = sec { at { = 0.
q
i (q) ({)
i (q) (0)
0
sec {
1
1
sec { tan {
0
2
2
sec {(2 tan { + 1)
1
3
sec { tan {(6 tan2 { + 5)
0
4
4
2
sec {(24 tan { + 28 tan { + 5)
5
1
5
5 4
Thus, i ({) = sec { E W4 ({) = 1 + 2!
({ 3 0)2 + 4!
({ 3 0)4 = 1 + 12 {2 + 24
{ . By part (a),
%
2
4 &
O2
O4
5O4
1 O
5 O
5
O2
1
+
F EU 1+
+
.
3U =U+ U· 2 + U· 4 3U =
2 U
24 U
2
U
24
U
2U
24U3
(c) Taking O = 100 km and U = 6370 km, the formula in part (a) says that
F = U sec(O@U) 3 U = 6370 sec(100@6370) 3 6370 E 0=785 009 965 44 km.
The formula in part (b) says that F E
O2
5O4
5 · 1004
1002
+
+
=
E 0=785 009 957 36 km.
3
2U
24U
2 · 6370
24 · 63703
The difference between these two results is only 0=000 000 008 08 km, or 0=000 008 08 m!
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208
¤
CHAPTER 11
38. (a) 4
u
O
j
]
@2
0
INFINITE SEQUENCES AND SERIES
u ] @2
31@2
g{
O
s
1 + 3n2 sin2 {
g{
=4
j 0
1 3 n2 sin2 {
u ] @2 1
1
2
3
· 3
· 3 · 5
O
1
1 3 3n2 sin2 { + 2 2 3n2 sin2 { 3 2 2 2 3n2 sin2 { + · · · g{
=4
j 0
2
2!
3!
u ] @2 O
1 2 2
1·3 4 4
1·3·5 6 6
1+
=4
n sin { +
n sin { +
n sin { + · · · g{
j 0
2
2·4
2·4·6
u 1
1 2
1·3 4
1·3·5 6
1·3
1·3·5
O +
·
n +
·
n +
·
n + ···
=4
j 2
2
2 2
2·4
2·4 2
2·4·6
2·4·6 2
= 2
u
[split up the integral and use the result from Exercise 7.1.46]
O
12
12 · 32
12 · 32 · 52
1 + 2 n2 + 2 2 n4 + 2 2 2 n6 + · · ·
j
2
2 ·4
2 ·4 ·6
(b) The first of the two inequalities is true because all of the terms in the series are positive. For the second,
W = 2
$ 2
u
u
O
12
12 · 32
12 · 32 · 52
12 · 32 · 52 · 72
1 + 2 n 2 + 2 2 n 4 + 2 2 2 n6 + 2 2 2 2 n8 + · · ·
j
2
2 ·4
2 ·4 ·6
2 ·4 ·6 ·8
O
1 + 14 n2 + 14 n4 + 14 n6 + 14 n8 + · · ·
j
The terms in brackets (after the first) form a geometric series with d = 14 n2 and u = n2 = sin2
So W $ 2
u
u
O
O 4 3 3n2
n2 @4
.
=
2
1+
j
1 3 n2
j 4 3 4n2
1
2 0
? 1.
(c) We substitute O = 1, j = 9=8, and n = sin(10 @2) E 0=08716, and the inequality from part (b) becomes
2=01090 $ W $ 2=01093, so W E 2=0109. The estimate W E 2
s
O@j E 2=0071 differs by about 0=2%.
If 0 = 42 , then n E 0=35837 and the inequality becomes 2=07153 $ W $ 2=08103, so W E 2=0763.
The one-term estimate is the same, and the discrepancy between the two estimates increases to about 3=4%.
39. Using i ({) = Wq ({) + Uq ({) with q = 1 and { = u, we have i (u) = W1 (u) + U1 (u), where W1 is the first-degree Taylor
polynomial of i at d. Because d = {q , i (u) = i ({q ) + i 0 ({q )(u 3 {q ) + U1 (u). But u is a root of i , so i (u) = 0
and we have 0 = i({q ) + i 0 ({q )(u 3 {q ) + U1 (u). Taking the first two terms to the left side gives us
U1 (u)
i ({q )
= 0
. By the formula for Newton’s
i 0 ({q )
i ({q )
U1 (u) . Taylor’s Inequality gives us
method, the left side of the preceding equation is {q+1 3 u, so |{q+1 3 u| = 0
i ({q ) i 0 ({q )({q 3 u) 3 i({q ) = U1 (u). Dividing by i 0 ({q ), we get {q 3 u 3
|i 00 (u)|
|u 3 {q |2 . Combining this inequality with the facts |i 00 ({)| $ P and |i 0 ({)| D N gives us
2!
P
|{q+1 3 u| $
|{q 3 u|2 .
2N
|U1 (u)| $
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APPLIED PROJECT RADIATION FROM THE STARS
¤
209
APPLIED PROJECT Radiation from the Stars
1. If we write i() =
8kf35
d35
= e@(W )
, then as < 0+ , it is of the form "@", and as < " it is of the form
31
h
31
hkf@(nW )
0@0, so in either case we can use l’Hospital’s Rule. First of all,
H
lim i () = lim
<"
Also,
<"
H
lim i () = 5
<0+
d 3536
2 36
34
dW
dW
lim e@(W ) = 5
lim e@(W ) = 0
=5
eW e@(W )
<"
<"
e
e
h
h
3
h
(W )2
34 H dW
dW
lim e@(W ) = 5
lim
e <0+ h
e <0+
3435
33
dW 2
= 20 2 lim e@(W )
eW e@(W )
e <0+ h
3
h
(W )2
This is still indeterminate, but note that each time we use l’Hospital’s Rule, we gain a factor of in the numerator, as well as a
constant factor, and the denominator is unchanged. So if we use l’Hospital’s Rule three more times, the exponent of in the
numerator will become 0. That is, for some {nl }, all constant,
H
lim i () = n1 lim
<0+
33
<0+ he@(W )
H
= n2 lim
32
<0+ he@(W )
H
= n3 lim
31
<0+ he@(W )
2. We expand the denominator of Planck’s Law using the Taylor series h{ = 1 + { +
H
= n4 lim
<0+
1
=0
he@(W )
{3
kf
{2
+
+ · · · with { =
, and use
2!
3!
nW
the fact that if is large, then all subsequent terms in the Taylor expansion are very small compared to the first one, so we can
approximate using the Taylor polynomial W1 :
i () =
8kf35
8nW
8kf35
8kf35
&
=
E
= %
2
3
31
kf
4
1
kf
kf
1
kf
31
1+
+
+
+ ··· 3 1
1+
nW
nW
2! nW
3! nW
hkf@(nW )
which is the Rayleigh-Jeans Law.
3. To convert to m, we substitute @106 for in both laws. The first figure shows that the two laws are similar for large . The
second figure shows that the two laws are very different for short wavelengths (Planck’s Law gives a maximum at
E 0=51 m; the Rayleigh-Jeans Law gives no minimum or maximum.).
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210
CHAPTER 11 INFINITE SEQUENCES AND SERIES
4. From the graph in Problem 3, i () has a maximum under Planck’s Law at E 0=51 m.
5.
As W gets larger, the total area under the curve increases, as we would expect: the hotter the star, the more energy it emits.
Also, as W increases, the -value of the maximum decreases, so the higher the temperature, the shorter the peak wavelength
(and consequently the average wavelength) of light emitted. This is why Sirius is a blue star and Betelgeuse is a red star: most
of Sirius’s light is of a fairly short wavelength; that is, a higher frequency, toward the blue end of the spectrum, whereas most
of Betelgeuse’s light is of a lower frequency, toward the red end of the spectrum.
11 Review
1. (a) See Definition 11.1.1.
(b) See Definition 11.2.2.
(c) The terms of the sequence {dq } approach 3 as q becomes large.
(d) By adding sufficiently many terms of the series, we can make the partial sums as close to 3 as we like.
2. (a) See the definition on page 721 [ET page 697].
(b) A sequence is monotonic if it is either increasing or decreasing.
(c) By Theorem 11.1.12, every bounded, monotonic sequence is convergent.
3. (a) See (4) in Section 11.2.
(b) The s-series
4. If
S
" 1
S
is convergent if s A 1.
s
q=1 q
dq = 3, then lim dq = 0 and lim vq = 3.
q<"
q<"
5. (a) Test for Divergence: If lim dq does not exist or if lim dq 6= 0, then the series
q<"
q<"
S"
q=1
dq is divergent.
(b) Integral Test: Suppose i is a continuous, positive, decreasing function on [1> ") and let dq = i (q). Then the series
U"
S"
q=1 dq is convergent if and only if the improper integral 1 i({) g{ is convergent. In other words:
U"
S
(i) If 1 i ({) g{ is convergent, then "
q=1 dq is convergent.
U"
S"
(ii) If 1 i ({) g{ is divergent, then q=1 dq is divergent.
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CHAPTER 11 REVIEW
¤
S
S
(c) Comparison Test: Suppose that dq and eq are series with positive terms.
S
S
(i) If eq is convergent and dq $ eq for all q, then dq is also convergent.
S
S
(ii) If eq is divergent and dq D eq for all q, then dq is also divergent.
(d) Limit Comparison Test: Suppose that
S
dq and
S
eq are series with positive terms. If lim (dq @eq ) = f, where f is a
q<"
finite number and f A 0, then either both series converge or both diverge.
(e) Alternating Series Test: If the alternating series
S"
q31
eq
q=1 (31)
= e1 3 e2 + e3 3 e4 + e5 3 e6 + · · · [eq A 0]
satisfies (i) eq+1 $ eq for all q and (ii) lim eq = 0, then the series is convergent.
q<"
(f ) Ratio Test:
"
dq+1 = O ? 1, then the series S dq is absolutely convergent (and therefore convergent).
(i) If lim q<"
dq
q=1
"
dq+1 = O A 1 or lim dq+1 = ", then the series S dq is divergent.
(ii) If lim q<"
q<"
dq
dq
q=1
dq+1 = 1, the Ratio Test is inconclusive; that is, no conclusion can be drawn about the convergence or
(iii) If lim q<"
dq S
divergence of dq.
(g) Root Test:
s
S
q
|dq | = O ? 1, then the series "
q=1 dq is absolutely convergent (and therefore convergent).
q<"
s
s
S
(ii) If lim q |dq | = O A 1 or lim q |dq | = ", then the series "
q=1 dq is divergent.
q<"
q<"
s
(iii) If lim q |dq | = 1, the Root Test is inconclusive.
(i) If lim
q<"
6. (a) A series
S
dq is called absolutely convergent if the series of absolute values
S
|dq | is convergent.
S
(b) If a series dq is absolutely convergent, then it is convergent.
S
(c) A series dq 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.
8. (a)
"
S
q=0
fq ({ 3 d)q
(b) Given the power series
"
S
q=0
fq ({ 3 d)q , the radius of convergence is:
(i) 0 if the series converges only when { = d
(ii) " if the series converges for all {, or
(iii) a positive number U such that the series converges if |{ 3 d| ? U and diverges if |{ 3 d| A U.
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CHAPTER 11 INFINITE SEQUENCES AND SERIES
(c) The interval of convergence of a power series is the interval that consists of all values of { for which the series converges.
Corresponding to the cases in part (b), the interval of convergence is: (i) the single point {d}, (ii) all real numbers, that is,
the real number line (3"> "), or (iii) an interval with endpoints d 3 U and d + U 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) Wq ({) =
(b)
(c)
q i (l) (d)
S
({ 3 d)l
l!
l=0
" i (q) (d)
S
({ 3 d)q
q!
q=0
" i (q) (0)
S
{q
q!
q=0
[d = 0 in part (b)]
(d) See Theorem 11.10.8.
(e) See Taylor’s Inequality (11.10.9).
11. (a)–(f ) See Table 1 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.
1. False.
See Note 2 after Theorem 11.2.6.
2. False.
The series
"
S
q3 sin 1 =
q=1
3. True.
"
S
q=1
1
is a s-series with s = sin 1 E 0=84 $ 1, so the series diverges.
qsin 1
If lim dq = O, then as q < ", 2q + 1 < ", so d2q+1 < O.
q<"
4. True by Theorem 11.8.3.
Or: Use the Comparison Test to show that
5. False.
S
fq (32)q converges absolutely.
For example, take fq = (31)q@(q6q ).
6. True by Theorem 11.8.3.
3
3
3
1@q dq+1 1
q
q 1
= lim = lim = lim
7. False, since lim ·
·
= 1.
q<"
dq q<" (q + 1)3 1 q<" (q + 1)3 1@q3 q<" (1 + 1@q)3
dq+1 1
1
q!
· = lim
= 0 ? 1.
= lim 8. True, since lim q<"
dq q<" (q + 1)! 1 q<" q + 1
9. False.
See the note after Example 2 in Section 11.4.
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CHAPTER 11 REVIEW
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213
q
10. True, since
11. True.
" {q
" (31)
S
S
1
= h31 and h{ =
, so h31 =
.
h
q!
q!
q=0
q=0
See (9) in Section 11.1.
12. True, because if
13. True.
S
|dq | is convergent, then so is
S
dq by Theorem 11.6.3.
By Theorem 11.10.5 the coefficient of {3 is
1
i 000 (0)
=
3!
3
i i 000 (0) = 2.
Or: Use Theorem 11.9.2 to differentiate i three times.
14. False.
Let dq = q and eq = 3q. Then {dq } and {eq } are divergent, but dq + eq = 0, so {dq + eq } is convergent.
15. False.
For example, let dq = eq = (31)q . Then {dq } and {eq } are divergent, but dq eq = 1, so {dq eq } is convergent.
16. True by the Monotonic Sequence Theorem, since {dq } is decreasing and 0 ? dq $ d1 for all q
17. True by Theorem 11.6.3.
18. True.
19. True.
S
[
(31)q dq is absolutely convergent and hence convergent.]
S
dq+1
?1 i
dq converges (Ratio Test) i
q<" dq
lim
i {dq } is bounded.
lim dq = 0 [Theorem 11.2.6].
q<"
0=99999 = = = = 0=9 + 0=9(0=1)1 + 0=9(0=1)2 + 0=9(0=1)3 + · · · =
"
S
(0=9)(0=1)q31 =
q=1
0=9
= 1 by the formula
1 3 0=1
for the sum of a geometric series [V = d1 @(1 3 u)] with ratio u satisfying |u| ? 1.
20. True. Since lim dq = 2, we know that lim dq+3 = 2. Thus, lim (dq+3 3 dq ) = lim dq+3 3 lim dq = 2 3 2 = 0.
q<"
q<"
q<"
q<"
q<"
21. True. A finite number of terms doesn’t affect convergence or divergence of a series.
22. False.
Let dq = (0=1)q and eq = (0=2)q . Then
"
S
q=1
"
S
q=1
1.
2 + q3
1 + 2q3
2. dq =
(0=1)q =
q=1
1
0=1
= = D,
1 3 0=1
9
1
9
·
1
4
=
1
.
36
2@q3 + 1
2 + q3
1
= lim
= .
3
q<" 1 + 2q
q<" 1@q3 + 2
2
converges since lim
9 q
9 q
9q+1
= 9 · 10
, so lim dq = 9 lim 10
= 9 · 0 = 0 by (11.1.9).
q<"
q<"
10q
3. lim dq = lim
q<"
"
S
"
"
S
S
1
1
0=2
0=02
= = E, and
=
, but
eq =
(0=2)q =
dq eq =
(0=02)q =
1
3
0=2
4
1
3
0=02
49
q=1
q=1
q=1
DE =
"
S
dq =
q<"
q3
q
= ", so the sequence diverges.
= lim
q<" 1@q2 + 1
1 + q2
4. dq = cos(q@2), so dq = 0 if q is odd and dq = ±1 if q is even. As q increases, dq keeps cycling through the values
0, 1, 0, 31, so the sequence {dq } is divergent.
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CHAPTER 11 INFINITE SEQUENCES AND SERIES
q sin q q
1
$
? , so |dq | < 0 as q < ". Thus, lim dq = 0. The sequence {dq } is convergent.
q<"
q2 + 1 q2 + 1
q
5. |dq | = ln q
ln { H
1@{
2
I
for { A 0. Then lim i ({) = lim I = lim
= lim I = 0.
{<"
{<"
{<"
{<"
{
{
1@(2 { )
{
ln {
6. dq = I . Let i({) = I
q
Thus, by Theorem 3 in Section 11.1, {dq } converges and lim dq = 0.
q<"
+
4q ,
4{
3
3
7.
. Then
1+
is convergent. Let | = 1 +
q
{
1
3
3 2
1 + 3@{
{
ln(1 + 3@{) H
12
= lim
lim ln | = lim 4{ ln(1 + 3@{) = lim
= lim
= 12, so
{<"
{<"
{<"
{<"
{<" 1 + 3@{
1@(4{)
31@(4{2 )
4q
3
= h12 .
lim | = lim 1 +
{<"
q<"
q
8.
(310)q
q!
lim
q<"
converges, since
q310
10
10q
10 · 10 · 10 · · · · · 10 10 · 10 · · · · · 10
=
·
$ 1010
< 0 as q < ", so
q!
1 · 2 · 3 · · · · · 10
11 · 12 · · · · · q
11
(310)q
= 0 [Squeeze Theorem]. Or: Use (11.10.10).
q!
9. We use induction, hypothesizing that dq31 ? dq ? 2. Note first that 1 ? d2 =
1
3
(1 + 4) =
5
3
? 2, so the hypothesis holds
for q = 2. Now assume that dn31 ? dn ? 2. Then dn = 13 (dn31 + 4) ? 13 (dn + 4) ? 13 (2 + 4) = 2. So dn ? dn+1 ? 2,
and the induction is complete. To find the limit of the sequence, we note that O = lim dq = lim dq+1
q<"
q<"
i
O = 13 (O + 4) i O = 2.
10. lim
{<"
{4 H
4{3 H
12{2 H
24{ H
24
= lim { = lim
= lim { = lim { = 0
{<" h
{<" h{
{<" h
{<" h
h{
Then we conclude from Theorem 11.1.3 that lim q4 h3q = 0.
q<"
From the graph, it seems that 124 h312 A 0=1, but q4 h3q ? 0=1
whenever q A 12. So the smallest value of Q corresponding to
% = 0=1 in the definition of the limit is Q = 12.
11.
"
" 1
S
S
1
q
q
q
[ s = 2 A 1].
? 3 = 2 , so
converges by the Comparison Test with the convergent s-series
3 +1
2
q3 + 1
q
q
q
q=1
q=1 q
12. Let dq =
Since
"
S
q=1
dq
q3 + q
1 + 1@q2
q2 + 1
1
and eq = , so lim
= lim
= lim 3
= 1 A 0.
q<" eq
q<" q + 1
q<" 1 + 1@q3
q3 + 1
q
eq is the divergent harmonic series,
"
S
dq also diverges by the Limit Comparison Test.
q=1
3
" q3
dq+1 S
(q + 1)3 5q
1
1
1
= lim
13. lim · 3 = lim 1 +
· = ? 1, so
converges by the Ratio Test.
q+1
q
q<"
q<"
q<"
dq
5
q
q
5
5
q=1 5
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CHAPTER 11 REVIEW
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215
1
. Then eq is positive for q D 1, the sequence {eq } is decreasing, and lim eq = 0, so the series
q<"
q+1
14. Let eq = I
" (31)q
S
I
converges by the Alternating Series Test.
q+1
q=1
1
I
. Then i is continuous, positive, and decreasing on [2> "), so the Integral Test applies.
ln {
] w
] ln w
] "
k
l
k I lln w
1
1
I
g{ x = ln {, gx = g{ = lim
i ({) g{ = lim
x31@2 gx = lim 2 x
{
w<" 2 {
w<" ln 2
w<"
ln 2
ln {
2
I
I
= lim 2 ln w 3 2 ln 2 = ">
15. Let i ({) =
{
w<"
so the series
"
S
q=2
q
1
I
diverges.
ln q
"
S
q
q
q
1
ln
= , so lim ln
= ln 13 6= 0. Thus, the series
diverges by the Test for
q<" 3q + 1
q<"
3
3q + 1
3q + 1
q=1
16. lim
Divergence.
17. |dq | = q
"
S
cos 3q 5
1
1
$
?
=
, so
|dq | converges by comparison with the convergent geometric
q
q
q
1 + (1=2)
1 + (1=2)
(1=2)
6
q=1
" S
5 q
u=
6
series
q=1
s
18. lim q |dq | = lim
q<"
q<"
Root Test.
5
6
"
S
? 1 . It follows that
dq converges (by Theorem 3 in Section 11.6).
q=1
v
"
S
q2q
q2
1
q2q
1
q = lim
=
?
1,
so
=
lim
converges by the
(1 + 2q2 )q q<" 1 + 2q2
2 q
q<" 1@q2 + 2
2
q=1 (1 + 2q )
dq+1 2q + 1
5q q!
2
= lim 1 · 3 · 5 · · · · · (2q 3 1)(2q + 1) ·
19. lim = lim
= ? 1, so the series
q+1
q<"
q<"
dq
5
(q + 1)!
1 · 3 · 5 · · · · · (2q 3 1) q<" 5(q + 1)
5
converges by the Ratio Test.
20.
" (35)2q
" 1
S
S
=
2
q
2
q=1 q 9
q=1 q
25
9
q
dq+1 q2 · 9q
25
25q+1
25q2
= lim
. Now lim ·
= lim
=
A 1,
2
q
q+1
q<"
q<" (q + 1) · 9
q<" 9(q + 1)2
dq
25
9
so the series diverges by the Ratio Test.
21. eq =
I
I
"
S
q
q
(31)q31
A 0, {eq } is decreasing, and lim eq = 0, so the series
converges by the Alternating
q<"
q+1
q+1
q=1
Series Test.
22. Use the Limit Comparison Test with dq =
eq =
I
I
q+13 q31
2
(rationalizing the numerator) and
I
= I
q
q q+1+ q31
I
"
S
dq
2 q
I
I
= 1, so since
.
lim
=
lim
eq converges s =
3@2
q<"
q<"
eq
q
q+1+ q31
q=1
1
3
2
"
S
A1 ,
dq converges also.
q=1
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216
CHAPTER 11 INFINITE SEQUENCES AND SERIES
23. Consider the series of absolute values:
"
S
q31@3 is a p-series with s =
q=1
1
3
$ 1 and is therefore divergent. But if we apply the
"
S
1
Alternating Series Test, we see that eq = I
(31)q31 q31@3
A 0, {eq } is decreasing, and lim eq = 0, so the series
3
q<"
q
q=1
"
S
converges. Thus,
(31)q31 q31@3 is conditionally convergent.
q=1
24.
" "
"
S
(31)q31 q33 = S q33 is a convergent p-series [s = 3 A 1]. Therefore, S (31)q31 q33 is absolutely convergent.
q=1
q=1
q+2 3
(q + 2)3q+1
1 + (2@q) 3
3
22q+1
=
· =
· < ? 1 as q < ", so by the Ratio
·
2q+3
q
q
2
(31) (q + 1)3 q + 1 4
1 + (1@q) 4
4
q+1
dq+1 (31)
25. =
dq
Test,
q=1
" (31)q (q + 1)3q
S
is absolutely convergent.
22q+1
q=1
I
I
I
I
{ H
1@(2 { )
{
(31)q q
26. lim
= lim
= lim
= ". Therefore, lim
6= 0, so the given series is divergent by the
{<" ln {
{<"
{<" 2
q<"
1@{
ln q
Test for Divergence.
q31
" (33)q31
" (33)q31
" (33)q31
" (33)q31
"
S
S
S
1 S
1 S
1
3
1
27.
=
=
=
=
=
3
3 q
23q
8q
8 q=1 8q31
8 q=1
8
8 1 3 (33@8)
q=1
q=1 (2 )
q=1
=
1 8
1
·
=
8 11
11
"
S
1
1
1
=
3
[partial fractions].
28.
3(q + 3)
q=1 q(q + 3)
q=1 3q
"
S
vq =
q
S
l=1
"
S
q=1
29.
"
S
1
1
1
1
1
1
1
1
3
= + + 3
3
3
(telescoping sum), so
3l
3(l + 3)
3
6
9
3(q + 1)
3(q + 2)
3(q + 3)
1
1
1
11
1
= lim vq = + + =
.
q(q + 3) q<"
3
6
9
18
[tan31 (q + 1) 3 tan31 q] = lim vq
q=1
q<"
= lim [(tan31 2 3 tan31 1) + (tan31 3 3 tan31 2) + · · · + (tan31 (q + 1) 3 tan31 q)]
q<"
= lim [tan31 (q + 1) 3 tan31 1] =
q<"
" (31)q q
"
"
S
S
S
1
1
q
q
30.
(31)
=
(31)q
=
·
·
2q (2q)!
2q
3
(2q)!
3
(2q)!
q=0
q=0
q=0
2
3
4
=
4
I I 2q
"
S
{2q
= cos
(31)q
since cos { =
3
3
(2q)!
q=0
for all {.
31. 1 3 h +
"
" (3h)q
" {q
S
S
S
hq
h3
h4
h2
3
+
3··· =
=
= h3h since h{ =
for all {.
(31)q
2!
3!
4!
q!
q!
q=0
q=0
q=0 q!
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CHAPTER 11 REVIEW
32. 4=17326 = 4=17 +
¤
217
326
416,909
326
326
326@105
417
+
=
+ 8 + · · · = 4=17 +
=
5
10
10
1 3 1@103
100
99,900
99,900
" q
" (3{)q
S
{
1 {
1 S
(h + h3{ ) =
+
2
2 q=0 q! q=0 q!
1
{2
{3
{4
{2
{3
{4
=
1+{+
+
+
+··· + 1 3{ +
3
+
3 ···
2
2!
3!
4!
2!
3!
4!
2
4
2q
" {
S
1
{
{
1
1
=
2+2·
+2·
+ · · · = 1 + {2 +
D 1 + {2 for all {
2
2!
4!
2
(2q)!
2
q=2
33. cosh { =
34.
"
S
q=1
35.
(ln {)q is a geometric series which converges whenever |ln {| ? 1 i 31 ? ln { ? 1 i h31 ? { ? h.
" (31)q+1
S
1
1
1
1
1
1
1
=13
+
3
+
3
+
3
+ ···.
5
q
32
243
1024
3125
7776
16,807
32,768
q=1
Since e8 =
36. (a) v5 =
" (31)q+1
7 (31)q+1
S
S
1
1
=
E
E 0=9721.
? 0=000031,
5
5
8
32,768
q
q5
q=1
q=1
5
" 1
S
S
1
1
1
= 1 + 6 + · · · + 6 E 1=017305. The series
converges by the Integral Test, so we estimate the
6
6
q
2
5
q=1
q=1 q
remainder U5 with (11.3.2): U5 $
(b) In general, Uq $
]
"
q
]
"
5
35 "
{
535
g{
=
3
=
= 0=000064. So the error is at most 0=000064.
6
{
5 5
5
1
1
g{
=
. If we take q = 9, then v9 E 1=01734 and U9 $
E 3=4 × 1036 .
{6
5q5
5 · 95
So to five decimal places,
" 1
9
S
S
1
E
E 1=01734.
5
5
q=1 q
q=1 q
Another method: Use (11.3.3) instead of (11.3.2).
37.
"
S
8
S
1
1
1
1
E
E 0=18976224. To estimate the error, note that
? q , so the remainder term is
q
q
2 + 5q
5
q=1 2 + 5
q=1 2 + 5
U8 =
" 1
S
1
1@59
= 6=4 × 1037 geometric series with d =
?
=
q
q
2
+
5
5
1
3
1@5
q=9
q=9
"
S
1
59
and u =
1
5
.
q
q
1
dq+1 (q + 1)q+1 (2q)! 1
= lim (q + 1) (q + 1) = lim q + 1
·
=
lim
38. (a) lim q<"
q<"
dq q<" [2(q + 1)]!
qq q<" (2q + 2)(2q + 1)qq
q
2(2q + 1)
q
1
1
= lim 1 +
=h·0=0?1
q<"
q
2(2q + 1)
so the series converges by the Ratio Test.
qq
= 0 by Theorem 11.2.6.
q<" (2q)!
(b) The series in part (a) is convergent, so lim dq = lim
q<"
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¤
CHAPTER 11 INFINITE SEQUENCES AND SERIES
q + 1 d q+1
1
q
q
39. Use the Limit Comparison Test. lim = lim 1 +
= 1 A 0.
= lim
q<" q<"
q<" q
dq
q
Since
S
|dq | is convergent, so is
S q + 1
dq , by the Limit Comparison Test.
q
"
dq+1 S
{q+1
1
|{|
{q
|{|
q2 5q = lim =
, so by the Ratio Test,
40. lim · q = lim
(31)q 2 q
2
2
q<"
q<" (q + 1) 5q+1
q<" (1 + 1@q)
dq
{
5
5
q 5
q=1
converges when
|{|
?1
5
C
|{| ? 5, so U = 5. When { = 35, the series becomes the convergent s-series
s = 2 A 1. When { = 5, the series becomes
" 1
S
with
2
q=1 q
" (31)q
S
, which converges by the Alternating Series Test. Thus, L = [35> 5].
q2
q=1
dq+1 |{ + 2|q+1
q |{ + 2|
|{ + 2|
q 4q
=
? 1 C |{ + 2| ? 4, so U = 4.
= lim
41. lim ·
= lim
q<"
q<" q + 1
dq q<" (q + 1) 4q+1 |{ + 2|q
4
4
|{ + 2| ? 4 C 34 ? { + 2 ? 4 C 36 ? { ? 2. If { = 36, then the series
" ({ + 2)q
S
becomes
q 4q
q=1
" (34)q
" (31)q
S
S
, the alternating harmonic series, which converges by the Alternating Series Test. When { = 2, the
=
q
q
q=1 q4
q=1
series becomes the harmonic series
" 1
S
, which diverges. Thus, L = [36> 2).
q=1 q
" 2q ({ 3 2)q
S
({ 3 2)q+1
(q + 2)! 2
· q
|{ 3 2| = 0 ? 1, so the series
q = lim
q<" q + 3
(q + 3)!
2 ({ 3 2)
(q + 2)!
q=1
q+1
2
dq+1 = lim 42. lim q<"
dq q<" converges for all {. U = " and L = (3"> ").
so U = 12 . |{ 3 3| ?
1
2
u
I
({ 3 3)q+1
q + 3 q+3
I
= 2 |{ 3 3| ? 1 C |{ 3 3| ? 12 ,
=
2
|{
3
3|
lim
· q
q<"
2 ({ 3 3)q q+4
q+4
q+1
dq+1 2
= lim 43. lim q<"
dq q<" C 3 12 ? { 3 3 ?
1
2
"
S
1
1
I
=
, which diverges s =
1@2
q
q+3
q=0
q=3
"
S
alternating series, so L =
5
2
> 72 .
q+1
dq+1 (2q + 2)! {
= lim 44. lim q<"
dq q<" [(q + 1)!]2
To converge, we must have 4 |{| ? 1
·
C
1
2
5
2
? { ? 72 . For { = 72 , the series
" 2q ({ 3 3)q
S
I
becomes
q+3
q=1
" (31)q
S
I
, which is a convergent
$ 1 , but for { = 52 , we get
q+3
q=0
(2q + 2)(2q + 1)
(q!)2 |{| = 4 |{|.
= lim
(2q)! {q q<" (q + 1)(q + 1)
C
|{| ? 14 , so U = 14 .
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CHAPTER 11 REVIEW
45.
q
i (q) ({)
0
sin {
1
cos {
2
3
4
..
.
i (q)
6
1
2
I
3
2
3 12
I
3 23
1
2
3 sin {
3 cos {
sin {
..
.
..
.
i 00
2 i (3)
3 i (4)
4
6
6
6
+ i0
{3
+
{3
{3
{3
sin { = i
+
+
+ ···
6
6
6
2!
6
3!
6
4!
6
I 1
1
3
1
2
4
1 3
=
+
3 ··· +
+ ···
13
{3
{3
{3
3
{3
2
2!
6
4!
6
2
6
3!
6
I
"
"
1 S
1 3 S
1
2q
2q+1
=
(31)q
+
(31)q
{3
{3
2 q=0
(2q)!
6
2 q=0
(2q + 1)!
6
46.
q
i (q) ({)
0
cos {
1
3 sin {
2
3
4
..
.
i (q)
3
1
2
I
3 23
3 12
I
3
2
1
2
3 cos {
sin {
cos {
..
.
..
.
00 (3) i 3 2 i
3
cos { = i
+
+i
{3
+
{3
{3
3
3
3
2!
3
3!
I 1
1
3
1
2
4
+
3 ··· +
=
13
{3
{3
3 {3
+
2
2!
3
4!
3
2
3
0
(4) 3 i
4
3
+
+ ···
{3
3
4!
3
1
3
3 ···
{3
3!
3
I
"
"
1 3 S
1
1 S
2q
2q+1
q
=
{3
{3
(31)
+
(31)q+1
2 q=0
(2q)!
3
2 q=0
(2q + 1)!
3
47.
"
"
S
S
1
1
=
=
(3{)q =
(31)q {q for |{| ? 1 i
1+{
1 3 (3{) q=0
q=0
48. tan31 { =
"
S
(31)q
q=0
tan31 ({2 ) =
"
S
(31)q
q=0
Therefore, U = 1.
49.
{2q+1
with interval of convergence [31> 1], so
2q + 1
"
S
{2
=
(31)q {q+2 with U = 1.
1 + { q=0
"
S
({2 )2q+1
{4q+2
=
, which converges when {2 M [31> 1] C { M [31> 1].
(31)q
2q + 1
2q + 1
q=0
]
1
g{ = 3 ln(4 3 {) + F and
43{
]
]
] " q
] " q
"
S {
S {
1
{q+1
1
1
1
1
1 S
g{ =
g{ =
+ F. So
g{ =
g{ =
q
q
43{
4
1 3 {@4
4 q=0 4
4 q=0 4
4 q=0 4 (q + 1)
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220
CHAPTER 11 INFINITE SEQUENCES AND SERIES
ln(4 3 {) = 3
"
"
" {q
S
S
{q+1
{q+1
1 S
+F =3
+F =3
+ F. Putting { = 0, we get F = ln 4.
q
q+1
q
4 q=0 4 (q + 1)
(q + 1)
q=0 4
q=1 q4
Thus, i({) = ln(4 3 {) = ln 4 3
Another solution:
" {q
S
. The series converges for |{@4| ? 1
q
q=1 q4
C
|{| ? 4, so U = 4.
ln(4 3 {) = ln[4(1 3 {@4)] = ln 4 + ln(1 3 {@4) = ln 4 + ln[1 + (3{@4)]
= ln 4 +
"
S
(31)q+1
q=1
50. h{ =
" {q
S
q=0 q!
51. sin { =
i h2{ =
" (31)q {2q+1
S
(2q + 1)!
q=0
"
" {q
S
S
(3{@4)q
{q
(31)2q+1 q = ln 4 3
.
[from Table 1] = ln 4 +
q
q
q4
q=1
q=1 q4
" (2{)q
S
q!
q=0
i {h2{ = {
i sin({4 ) =
convergence is ".
52. h{ =
" {q
S
q=0 q!
i 10{ = h(ln 10){ =
" 2q {q
" 2q {q+1
S
S
=
, U="
q!
q!
q=0
q=0
" (31)q ({4 )2q+1
" (31)q {8q+4
S
S
=
for all {, so the radius of
(2q + 1)!
(2q + 1)!
q=0
q=0
" [(ln 10){]q
" (ln 10)q {q
S
S
=
, U="
q!
q!
q=0
q=0
31@4
1
1
1
1
1
= I
= s
1@4 = 2 1 3 16 {
4
4
1
16 3 {
16(1 3 {@16)
16 1 3 16 {
%
&
1 5 1 5 9
34 34
34 34 34
1 {
{ 2
{ 3
1
1+ 3
3
+
3
3
=
+
+ ···
2
4
16
2!
16
3!
16
53. i ({) = I
4
" 1 · 5 · 9 · · · · · (4q 3 3)
" 1 · 5 · 9 · · · · · (4q 3 3)
S
S
1
1
+
{q
{q = +
2 q=1
2 · 4q · q! · 16q
2 q=1
26q+1 q!
{
for 3 ? 1 C |{| ? 16, so U = 16.
16
# $
"
S
35
(35)(36)
(35)(36)(37)
35
(33{)q = 1 + (35)(33{) +
(33{)2 +
(33{)3 + · · ·
54. (1 3 3{)
=
q
2!
3!
q=0
=
=1+
55. h{ =
]
" 5 · 6 · 7 · · · · · (q + 4) · 3q {q
S
q!
q=1
for |33{| ? 1 C |{| ? 13 , so U = 13 .
" {q
" {q
" {q31
" {q31
" {q31
S
S
S
S
h{
1 S
1
, so
=
=
= {31 +
= +
and
{
{ q=0 q!
{ q=1 q!
q=0 q!
q=0 q!
q=1 q!
"
S
{q
h{
g{ = F + ln |{| +
.
{
q=1 q · q!
56. (1 + {4 )1@2 =
"
S
q=0
# $
1
2
q
({4 )q = 1 +
1 4
{ +
2
1 1 1 1 3 32
32 32
2
({4 )2 + 2
({4 )3 + · · ·
2!
3!
1 12
{ 3 ···
= 1 + 12 {4 3 18 {8 + 16
1
U1
1 5
1 9
1
{ 3 72
{ + 208
{13 3 · · · 0 = 1 +
so 0 (1 + {4 )1@2 g{ = { + 10
1
10
3
1
72
+
1
208
3···.
This is an alternating series, so by the Alternating Series Test, the error in the approximation
U1
1
1
1
(1 + {4 )1@2 g{ E 1 + 10
3 72
E 1=086 is less than 208
, sufficient for the desired accuracy.
0
U1
Thus, correct to two decimal places, 0 (1 + {4 )1@2 g{ E 1=09.
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CHAPTER 11 REVIEW
57. (a)
q
0
1
2
3
4
..
.
i
(q)
({)
i
1@2
{
(q)
(1)
I
1@2
3@8
1@4
{ E W3 ({) = 1 +
({ 3 1) 3
({ 3 1)2 +
({ 3 1)3
1!
2!
3!
= 1 + 12 ({ 3 1) 3 18 ({ 3 1)2 +
1
1 31@2
{
2
1 33@2
34{
3 35@2
{
8
15 37@2
3 16 {
¤
1
({
16
3 1)3
1
2
3 14
3
8
3 15
16
..
.
..
.
(c) |U3 ({)| $
(b)
P
|{ 3 1|4 , where i (4) ({) $ P with
4!
{37@2 . Now 0=9 $ { $ 1=1 i
i (4) ({) = 3 15
16
30=1 $ { 3 1 $ 0=1
i
({ 3 1)4 $ (0=1)4 ,
and letting { = 0=9 gives P =
|U3 ({)| $
15
, so
16(0=9)7@2
15
(0=1)4 E 0=000 005 648
16(0=9)7@2 4!
E 0=000 006 = 6 × 1036
(d)
I
From the graph of |U3 ({)| = | { 3 W3 ({)|, it appears that
the error is less than 5 × 1036 on [0=9> 1=1].
58. (a)
i (q) ({)
q
(b)
0
sec {
1
sec { tan {
sec { E W2 ({) = 1 + 12 {2
i (q) (0)
1
2
0
3
2
sec { tan { + sec {
1
3
..
.
sec { tan3 { + 5 sec3 { tan {
..
.
0
..
.
(c) |U2 ({)| $
P
|{|3 , where i (3) ({) $ P with
3!
i (3) ({) = sec { tan3 { + 5 sec3 { tan {.
3
Now 0 $ { $ 6 i {3 $ 6 , and letting { =
14 3
, so |U2 ({)| $
E 0=111648.
P = 14
3
3·6 6
6
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CHAPTER 11 INFINITE SEQUENCES AND SERIES
(d)
From the graph of |U2 ({)| = |sec { 3 W2 ({)|, it appears that
the error is less than 0=02 on 0> 6 .
{2q+1
{3
{5
{7
{3
{5
{7
= {3
+
3
+ · · · , so sin { 3 { = 3
+
3
+ · · · and
(2q + 1)!
3!
5!
7!
3!
5!
7!
q=0
sin { 3 {
{2
{4
{2
{4
1
1
1
sin { 3 {
+
3
+
·
·
·
.
Thus,
lim
+
3
+
·
·
·
=3 .
=
3
=
lim
3
{<0
{<0
{3
3!
5!
7!
{3
6
120
5040
6
59. sin { =
"
S
(31)q
# $ q
"
S
32
k
pjU2
pj
60. (a) I =
=
=
pj
[binomial series]
q
(U + k)2
(1 + k@U)2
U
q=0
(b) We expand I = pj 1 3 2 (k@U) + 3 (k@U)2 3 · · · .
This is an alternating series, so by the Alternating Series
Estimation Theorem, the error in the approximation I = pj
is less than 2pjk@U, so for accuracy within 1% we want
2pjk@U
2k(U + k)2
? 0=01.
pjU2 @(U + k)2 ? 0=01 C
U3
This inequality would be difficult to solve for k, so we substitute U = 6,400 km and plot both sides of the inequality.
It appears that the approximation is accurate to within 1% for k ? 31 km.
61. i ({) =
"
S
fq {q
q=0
i i (3{) =
"
S
fq (3{)q =
q=0
"
S
(31)q fq {q
q=0
(a) If i is an odd function, then i (3{) = 3i ({) i
"
S
(31)q fq {q =
q=0
q=0
are uniquely determined (by Theorem 11.10.5), so (31)q fq = 3fq .
If q is even, then (31)q = 1, so fq = 3fq
"
S
3fq {q . The coefficients of any power series
i 2fq = 0 i fq = 0. Thus, all even coefficients are 0, that is,
f0 = f2 = f4 = · · · = 0.
(b) If i is even, then i (3{) = i ({) i
"
S
q=0
If q is odd, then (31)q = 31, so 3fq = fq
(31)q fq {q =
"
S
q=0
fq {q
i (31)q fq = fq .
i 2fq = 0 i fq = 0. Thus, all odd coefficients are 0,
that is, f1 = f3 = f5 = · · · = 0.
62. h{ =
" {q
S
q=0 q!
i ({) =
2
i i ({) = h{ =
" ({2 )q
" {2q
" 1
S
S
S
=
=
{2q . By Theorem 11.10.6 with d = 0, we also have
q!
q=0
q=0 q!
q=0 q!
" i (n) (0)
S
1
i (2q) (0)
{n . Comparing coefficients for n = 2q, we have
=
n!
(2q)!
q!
n=0
i i (2q) (0) =
(2q)!
.
q!
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PROBLEMS PLUS
1. It would be far too much work to compute 15 derivatives of i. The key idea is to remember that i (q) (0) occurs in the
coefficient of {q in the Maclaurin series of i . We start with the Maclaurin series for sin: sin { = { 3
Then sin({3 ) = {3 3
i (15) (0) =
{3
{5
+
3 ···.
3!
5!
{15
1
{9
i (15) (0)
+
3 · · · , and so the coefficient of {15 is
= . Therefore,
3!
5!
15!
5!
15!
= 6 · 7 · 8 · 9 · 10 · 11 · 12 · 13 · 14 · 15 = 10,897,286,400.
5!
2. We use the problem-solving strategy of taking cases:
If |{| ? 1, then 0 $ {2 ? 1, so lim {2q = 0 [see Example 11 in Section 11.1]
Case (i):
q<"
2q
and i ({) = lim
q<"
{ 31
031
=
= 31.[]
{2q + 1
0+1
{2q 3 1
131
= lim
= 0.
{2q + 1 q<" 1 + 1
Case (ii):
If |{| = 1, that is, { = ±1, then {2 = 1, so i ({) = lim
Case (iii):
If |{| A 1, then {2 A 1, so lim {2q = " and i ({) = lim
q<"
q<"
;
1
A
A
A
A
A
A
0
A
?
31
Thus, i({) =
A
A
A
A
0
A
A
A
=
1
q<"
{2q 3 1
1 3 (1@{2q )
130
=
lim
=
= 1.
{2q + 1 q<" 1 + (1@{2q )
1+0
if { ? 31
if { = 31
if 31 ? { ? 1
if { = 1
if { A 1
The graph shows that i is continuous everywhere except at { = ±1.
3. (a) From Formula 14a in Appendix D, with { = | = , we get tan 2 =
2 cot 2 =
2 tan 1 3 tan2 , so cot 2 =
2
1 3 tan 2 tan i
1 3 tan2 = cot 3 tan . Replacing by 12 {, we get 2 cot { = cot 12 { 3 tan 12 {, or
tan tan 12 { = cot 12 { 3 2 cot {.
(b) From part (a) with
{
2q31
in place of {, tan
" 1
S
{
{
{
{
= cot q 3 2 cot q31 , so the qth partial sum of
tan q is
q
q
2
2
2
2
q=1 2
tan({@4)
tan({@8)
tan({@2q )
tan({@2)
+
+
+ ··· +
2
4
8
2q
cot({@4)
cot({@8)
cot({@2)
cot({@2)
cot({@4)
3 cot { +
3
+
3
+ ···
=
2
4
2
8
4
cot({@2q31 )
cot({@2q )
cot({@2q )
3
[telescoping sum]
= 3 cot { +
+
q
q31
2
2
2q
vq =
Now
cos({@2q )
{@2q
1
1
cos({@2q )
cot({@2q )
=
·
< · 1 = as q < " since {@2q < 0
= q
q
q
2
2 sin({@2 )
{
sin({@2q )
{
{
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¤
CHAPTER 11 PROBLEMS PLUS
for { 6= 0. Therefore, if { 6= 0 and { 6= n where n is any integer, then
" 1
S
1
1
{
{
3
cot
{
+
= 3 cot { +
tan
=
lim
v
=
lim
cot
q
q
q<"
q<"
2q
2q
2q
{
q=1 2
If { = 0, then all terms in the series are 0, so the sum is 0.
2
2
2
2
4. |DS2 | = 2, |DS3 | = 2 + 22 , |DS4 | = 2 + 22 + 22
2
|DSq |2 = 2 + 22 + 22 + · · · + (2q32 )2
=2+
4(4q32 3 1)
431
So tan _Sq DSq+1 =
Thus, _Sq DSq+1 <
2 2
, |DS5 |2 = 2 + 22 + 22 + 23 , = = = ,
[for q D 3] = 2 + (4 + 42 + 43 + · · · + 4q32 )
[finite geometric sum with d = 4, u = 4] =
6 4q31 3 4
2 4q31
+
= +
3
3
3
3
I
I
|Sq Sq+1 |
4q31
1
2q31
= u
= u
< 3 as q < ".
= u
q31
q31
|DSq |
2
1
2
2
4
4
+
+
+
3 · 4q31
3
3
3
3
3
3
as q < ".
5. (a) At each stage, each side is replaced by four shorter sides, each of length
1
3
v0 = 3
of the side length at the preceding stage. Writing v0 and c0 for the
number of sides and the length of the side of the initial triangle, we
c0 = 1
v1 = 3 · 4
c1 = 1@3
2
c2 = 1@32
v3 = 3 · 43
..
.
c3 = 1@33
..
.
v2 = 3 · 4
generate the table at right. In general, we have vq = 3 · 4q and
q
cq = 13 , so the length of the perimeter at the qth stage of construction
q
q
is sq = vq cq = 3 · 4q · 13 = 3 · 43 .
q31
4q
4
(b) sq = q31 = 4
. Since 43 A 1, sq < " as q < ".
3
3
(c) The area of each of the small triangles added at a given stage is one-ninth of the area of the triangle added at the preceding
stage. Let d be the area of the original triangle. Then the area dq of each of the small triangles added at stage q is
dq = d ·
1
d
= q . Since a small triangle is added to each side at every stage, it follows that the total area Dq added to the
9q
9
figure at the qth stage is Dq = vq31 · dq = 3 · 4q31 ·
curve is D = d + D1 + D2 + D3 + · · · = d + d ·
geometric series with common ratio
d
4q31
= d · 2q31 . Then the total area enclosed by the snowflake
9q
3
1
42
43
4
+ d · 3 + d · 5 + d · 7 + · · · . After the first term, this is a
3
3
3
3
d@3
d 9
4
8d
, so D = d +
· =
. But the area of the original equilateral
4 = d+
9
3 5
5
13 9
I
I
I
3
3
8
2 3
1
. So the area enclosed by the snowflake curve is ·
=
.
triangle with side 1 is d = · 1 · sin =
2
3
4
5 4
5
6. Let the series V = 1 +
1
2
+
1
3
+
1
4
+
1
6
+
1
8
+
1
9
+
1
12
+ · · · . Then every term in V is of the form
1
, p, q D 0, and
2p 3q
furthermore each term occurs only once. So we can write
V=
"
"
S
S
p=0 q=0
"
"
" 1
"
S
S
S
1
1 1
1 S
1
=
=
=
p 3q
p
q
2p 3q
2
2
3
1
3
p=0 q=0
p=0
q=0
1
2
·
1
13
1
3
=2·
3
2
=3
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CHAPTER 11 PROBLEMS PLUS
¤
7. (a) Let d = arctan { and e = arctan |. Then, from Formula 14b in Appendix D,
tan(d 3 e) =
tan(arctan {) 3 tan(arctan |)
{3|
tan d 3 tan e
=
=
1 + tan d tan e
1 + tan(arctan {) tan(arctan |)
1 + {|
Now arctan { 3 arctan | = d 3 e = arctan(tan(d 3 e)) = arctan
{3|
since 32 ? d 3 e ?
1 + {|
.
2
(b) From part (a) we have
1
arctan 120
3 arctan 239
= arctan
119
1
120
1
3 239
119
+ 120
· 1
119 239
= arctan
28,561
28,441
28,561
28,441
(c) Replacing | by 3| in the formula of part (a), we get arctan { + arctan | = arctan
4 arctan
1
5
= 2 arctan 15 + arctan 15 = 2 arctan
= arctan
5
12
+
5
12
13
5
12
·
5
12
1
5
4
{+|
. So
1 3 {|
+ 15
5
5
5
= 2 arctan 12
= arctan 12
+ arctan 12
1 3 15 · 15
= arctan 120
119
1
1
Thus, from part (b), we have 4 arctan 15 3 arctan 239
= arctan 120
119 3 arctan 239 =
(d) From Example 7 in Section 11.9 we have arctan { = { 3
arctan
= arctan 1 =
4.
{5
{7
{9
{11
{3
+
3
+
3
+ · · · , so
3
5
7
9
11
1
1
1
1
1
1
1
= 3
+
3
+
3
+ ···
5
5
3 · 53
5 · 55
7 · 57
9 · 59
11 · 511
This is an alternating series and the size of the terms decreases to 0, so by the Alternating Series Estimation Theorem,
the sum lies between v5 and v6 , that is, 0=197395560 ? arctan 15 ? 0=197395562.
(e) From the series in part (d) we get arctan
1
1
1
1
+
3 · · · . The third term is less than
=
3
239
239
3 · 2393
5 · 2395
2=6 × 10313 , so by the Alternating Series Estimation Theorem, we have, to nine decimal places,
1
1
E v2 E 0=004184076. Thus, 0=004184075 ? arctan 239
? 0=004184077.
arctan 239
1
, so from parts (d) and (e) we have
(f ) From part (c) we have = 16 arctan 15 3 4 arctan 239
16(0=197395560) 3 4(0=004184077) ? ? 16(0=197395562) 3 4(0=004184075) i
3=141592652 ? ? 3=141592692. So, to 7 decimal places, E 3=1415927.
8. (a) Let d = arccot { and e = arccot | where 0 ? d 3 e ? . Then
cot(d 3 e) =
=
1
1
·
+ 1 cot d cot e
1 + tan d tan e
1
·
=
= cot d cot e
1
1
tan(d 3 e)
tan d 3 tan e
cot d cot e
3
cot d
cot e
1 + cot d cot e
1 + cot(arccot {) cot(arccot |)
1 + {|
=
=
cot e 3 cot d
cot(arccot |) 3 cot(arccot {)
|3{
Now arccot { 3 arccot | = d 3 e = arccot(cot(d 3 e)) = arccot
1 + {|
since 0 ? d 3 e ? .
|3{
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CHAPTER 11 PROBLEMS PLUS
(b) Applying the identity in part (a) with { = q and | = q + 1, we have
arccot(q2 + q + 1) = arccot(1 + q(q + 1)) = arccot
1 + q(q + 1)
= arccot q 3 arccot(q + 1)
(q + 1) 3 q
Thus, we have a telescoping series with qth partial sum
vq = [arccot 0 3 arccot 1] + [arccot 1 3 arccot 2] + · · · + [arccot q 3 arccot(q + 1)] = arccot 0 3 arccot(q + 1).
Thus,
"
S
q=0
arccot(q2 + q + 1) = lim vq = lim [arccot 0 3 arccot(q + 1)] =
q<"
"
S
q<"
2
30=
.
2
1
, |{| ? 1, and differentiate:
13{
"
"
"
S q
S
S
1
1
g
{
g
=
=
{
for |{| ? 1 i
q{q = {
q{q31 =
=
2
g{ q=0
g{ 1 3 {
(1 3 {)
(1 3 {)2
q=1
q=1
9. We start with the geometric series
{q =
q=0
"
S
q{q31
q=1
for |{| ? 1. Differentiate again:
"
S
q2 {q31 =
q=1
"
S
q3 {q31 =
q=1
{
g
(1 3 {)2 3 { · 2(1 3 {)(31)
{+1
=
=
2
g{ (1 3 {)
(1 3 {)4
(1 3 {)3
i
"
S
q2 {q =
q=1
g {2 + {
(1 3 {)3 (2{ + 1) 3 ({2 + {)3(1 3 {)2 (31)
{2 + 4{ + 1
=
=
g{ (1 3 {)3
(1 3 {)6
(1 3 {)4
{2 + {
(1 3 {)3
i
i
{3 + 4{2 + {
, |{| ? 1. The radius of convergence is 1 because that is the radius of convergence for the
(1 3 {)4
q=1
S
geometric series we started with. If { = ±1, the series is q3 (±1)q , which diverges by the Test For Divergence, so the
"
S
q3 {q =
interval of convergence is (31> 1).
10. Let’s first try the case n = 1: d0 + d1 = 0
i d1 = 3d0
i
I
I
I
I
I
q+ q+1
I
I
I
I
q3 q+1 I
lim d0 q + d1 q + 1 = lim d0 q 3 d0 q + 1 = d0 lim
q<"
q<"
q<"
q+ q+1
31
I
= d0 lim I
=0
q<"
q+ q+1
In general we have d0 + d1 + · · · + dn = 0 i dn = 3d0 3 d1 3 · · · 3 dn31
I
I
I
I
lim d0 q + d1 q + 1 + d2 q + 2 + · · · + dn q + n
i
q<"
I
I
I
I
I
I
= lim d0 q + d1 q + 1 + · · · + dn31 q + n 3 1 3 d0 q + n 3 d1 q + n 3 · · · 3 dn31 q + n
q<"
I
I
I
I
I
I
= d0 lim
q 3 q + n + d1 lim
q + 1 3 q + n + · · · + dn31 lim
q+n313 q+n
q<"
q<"
q<"
Each of these limits is 0 by the same type of simplification as in the case n = 1. So we have
I
I
I
I
lim d0 q + d1 q + 1 + d2 q + 2 + · · · + dn q + n = d0 (0) + d1 (0) + · · · + dn31 (0) = 0
q<"
2
q 31
1
(q + 1)(q 3 1)
11. ln 1 3 2 = ln
= ln[(q + 1)(q 3 1)] 3 ln q2
= ln
q
q2
q2
= ln(q + 1) + ln(q 3 1) 3 2 ln q = ln(q 3 1) 3 ln q 3 ln q + ln(q + 1)
= ln
q31
q31
q
3 [ln q 3 ln(q + 1)] = ln
3 ln
.
q
q
q+1
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CHAPTER 11 PROBLEMS PLUS
¤
227
n
n
S
S
q31
q
1
ln
3 ln
for n D 2. Then
Let vn =
ln 1 3 2 =
q
q
q+1
q=2
q=2
1
2
2
3
n31
n
1
n
vn = ln 3 ln
+ ln 3 ln
+ · · · + ln
3 ln
= ln 3 ln
, so
2
3
3
4
n
n+1
2
n+1
"
S
n
1
1
1
= ln 3 ln 1 = ln 1 3 ln 2 3 ln 1 = 3 ln 2.
ln 1 3 2 = lim vn = lim ln 3 ln
n<"
n<"
q
2
n+1
2
q=2
12. Place the |-axis as shown and let the length of each book be O. We want to
show that the center of mass of the system of q books lies above the table,
that is, { ? O. The {-coordinates of the centers of mass of the books are
{1 =
O
O
O
O
O
O
, {2 =
+ , {3 =
+
+ , and so on.
2
2(q 3 1)
2
2(q 3 1)
2(q 3 2)
2
Each book has the same mass p, so if there are q books, then
{1 + {2 + · · · + {q
p{1 + p{2 + · · · + p{q
=
pq
q
O
O
O
O
O
1 O
+
+
+
+
+
+ ···
=
q 2
2(q 3 1)
2
2(q 3 1)
2(q 3 2)
2
O
O
O O O
+
+
+ ··· + + +
2(q 3 1)
2(q 3 2)
4
2
2
q32
2 1
q
O
1
q
2q 3 1
O q31
+
+ ··· + + +
=
(q 3 1) +
=
O?O
=
q 2(q 3 1)
2(q 3 2)
4
2
2
q
2
2
2q
{=
This shows that, no matter how many books are added according to the given scheme, the center of mass lies above the table.
S
It remains to observe that the series 12 + 14 + 16 + 18 + · · · = 12
(1@q) is divergent (harmonic series), so we can make the top
book extend as far as we like beyond the edge of the table if we add enough books.
13. (a)
The x-intercepts of the curve occur where sin { = 0 C { = q,
q an integer. So using the formula for disks (and either a CAS or
sin2 { = 12 (1 3 cos 2{) and Formula 99 to evaluate the integral),
the volume of the nth bead is
U q
U q
Yq = (q31) (h3{@10 sin {)2 g{ = (q31) h3{@5 sin2 { g{
=
250
(h3(q31)@5
101
3 h3q@5 )
(b) The total volume is
U"
0
h3{@5 sin2 { g{ =
"
S
Yq =
q=1
250
101
"
S
[h3(q31)@5 3 h3q@5 ] =
q=1
Another method: If the volume in part (a) has been written as Yq =
as a geometric series with d =
250
(1
101
250
101
[telescoping sum].
250 3q@5 @5
h
(h
101
3 1), then we recognize
3 h3@5 ) and u = h3@5 =
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S
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Yq
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CHAPTER 11 PROBLEMS PLUS
14. First notice that both series are absolutely convergent (p-series with s A 1.) Let the given expression be called {. Then
1
1
1
1
1
1
1
1
1 + 2 · s 3 s + s + 2 · s 3 s + ···
+
+
+
·
·
·
2
2
3
4
4
2s
3s
4s
=
{=
1
1
1
1
1
1
1 3 s + s 3 s +···
1 3 s + s 3 s + ···
2
3
4
2
3
4
1
1
1
1
1
1
1 3 s + s 3 s + ··· + 2 · s + 2 · s + 2 · s + ···
2
3
4
2
4
6
=
1
1
1
1 3 s + s 3 s + ···
2
3
4
1
1
1
1
1
1
1
1
1
+
+
+
+
+
·
·
·
+
+
+
·
·
·
2
2s
4s
6s
8s
2s31
2s
3s
4s
=1+
= 1 + 213s {
=1+
1
1
1
1
1
1
1 3 s + s 3 s + ···
1 3 s + s 3 s + ···
2
3
4
2
3
4
1+
Therefore, { = 1 + 213s { C { 3 213s { = 1 C {(1 3 213s ) = 1 C { =
15. If O is the length of a side of the equilateral triangle, then the area is D =
1
2O
·
I
3
2 O
=
1
.
1 3 213s
I
3 2
4 O
and so O2 =
I4 D.
3
Let u be the radius of one of the circles. When there are q rows of circles, the figure shows that
I
I I
O
I
3 u + u + (q 3 2)(2u) + u + 3 u = u 2q 3 2 + 2 3 , so u = .
2 q+ 331
O=
The number of circles is 1 + 2 + · · · + q =
q(q + 1)
, and so the total area of the circles is
2
O2
q(q + 1) 2
q(q + 1)
u =
I
2
2
2
4 q+ 331
I
D
4D@ 3
q(q + 1)
q(q + 1)
=
I
I
2 = 2 I
2
2
3
4 q+ 331
q+ 331
Dq =
i
q(q + 1)
Dq
= I
2 I
D
q+ 331 2 3
1 + 1@q
= I
2 I < I as q < "
2
3
2
3
1+
3 3 1 @q
16. Given d0 = d1 = 1 and dq =
d2 =
(q 3 1)(q 3 2)dq31 3 (q 3 3)dq32
, we calculate the next few terms of the sequence:
q(q 3 1)
1
1
1
1 · 0 · d1 3 (31)d0
2 · 1 · d2 3 0 · d1
3 · 2 · d3 3 1 · d2
1
= , d3 =
= , d4 =
=
. It seems that dq = ,
2·1
2
3·2
6
4·3
24
q!
so we try to prove this by induction. The first step is done, so assume dn =
dn+1
n(n 3 1)dn 3 (n 3 2)dn31
=
=
(n + 1)n
complete. Therefore,
"
S
q=0
dq =
1
1
and dn31 =
. Then
n!
(n 3 1)!
n(n 3 1)
n32
3
n!
(n 3 1)!
(n 3 1) 3 (n 3 2)
1
=
=
and the induction is
(n + 1)n
[(n + 1)(n)](n 3 1)!
(n + 1)!
" 1
S
= h.
q=0 q!
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CHAPTER 11 PROBLEMS PLUS
¤
229
17. As in Section 11.9 we have to integrate the function {{ by integrating series. Writing {{ = (hln { ){ = h{ ln { and using the
" ({ ln {)q
" {q (ln {)q
S
S
=
. As with power series, we can
q!
q!
q=0
q=0
] 1
" 1
S
{q (ln {)q
g{ =
{q (ln {)q g{. We integrate by parts
q!
q=0 q! 0
Maclaurin series for h{ , we have {{ = (hln { ){ = h{ ln { =
integrate this series term-by-term:
]
1
{
{ g{ =
"
S
q=0
0
]
1
0
q31
q(ln {)
{
with x = (ln {)q , gy = {q g{, so gx =
]
1
{q (ln {)q g{ = lim
w<0+
0
]
1
g{ and y =
{q (ln {)q g{ = lim
w<0+
w
q
=03
q+1
]
1
q
q31
{ (ln {)
{q+1
:
q+1
{q+1
(ln {)q
q+1
1
w
3 lim
w<0+
]
1
w
q
{q (ln {)q31 g{
q+1
g{
0
(where l’Hospital’s Rule was used to help evaluate the first limit). Further integration by parts gives
] 1
] 1
n
{q (ln {)n g{ = 3
{q (ln {)n31 g{ and, combining these steps, we get
q+1 0
0
]
1
{q (ln {)q g{ =
0
]
1
{{ g{ =
0
" 1
S
q=0 q!
(31)q q!
(q + 1)q
]
1
]
1
(31)q q!
(q + 1)q+1
{q g{ =
0
i
" 1 (31)q q!
"
" (31)q31
S
S
S
(31)q
=
=
.
q+1
q+1
qq
q=0 q! (q + 1)
q=0 (q + 1)
q=1
{q (ln {)q g{ =
0
18. (a) Since Sq is defined as the midpoint of Sq34 Sq33 , {q =
1
{
2 q
1
({q34
2
+ {q33 ) for q D 5. So we prove by induction that
+ {q+1 + {q+2 + {q+3 = 2. The case q = 1 is immediate, since
1
2
· 0 + 1 + 1 + 0 = 2. Assume that the result
holds for q = n 3 1, that is, 12 {n31 + {n + {n+1 + {n+2 = 2. Then for q = n,
1
{
2 n
+ {n+1 + {n+2 + {n+3 = 12 {n + {n+1 + {n+2 + 12 ({n+334 + {n+333 ) [by above]
= 12 {n31 + {n + {n+1 + {n+2 = 2 [by the induction hypothesis]
Similarly, for q D 5, |q = 12 (|q34 + |q33 ), so the same argument as above holds for |, with 2 replaced by
1
|
2 1
(b) lim
+ |2 + |3 + |4 =
q<"
1
2
1
2
· 1 + 1 + 0 + 0 = 32 . So 12 |q + |q+1 + |q+2 + |q+3 =
1
lim {
2 q<" q
{q + {q+1 + {q+2 + {q+3 =
the limits on the left hand side are the same, we get
7
lim |q
2 q<"
=
3
2
i
lim |q = 37 , so S =
q<"
19. By Table 1 in Section 11.10, tan31 { =
"
S
4
3
7> 7
(31)q
q=0
3
2
for all q.
+ lim {q+1 + lim {q+2 + lim {q+3 = 2. Since all
q<"
7
lim {
2 q<" q
q<"
=2 i
q<"
lim {q = 47 . In the same way,
q<"
.
{2q+1
1
for |{| ? 1. In particular, for { = I , we
2q + 1
3
I 2q+1
q
"
"
S
S
1
1
1
1
31
q 1@ 3
I
I
(31)
(31)q
=
=
, so
have = tan
6
2q
+
1
3
2q
+1
3
3
q=0
q=0
"
"
"
"
I S
I
S
S
(31)q
(31)q
(31)q
(31)q
6 S
=
2
3
=
2
3
1
+
= I 3 1.
i
= I
q
q
q
q
3 q=0 (2q + 1)3
2 3
q=0 (2q + 1)3
q=1 (2q + 1)3
q=1 (2q + 1)3
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¤
CHAPTER 11 PROBLEMS PLUS
d(1 3 uq )
,
13u
l
k
1 1 3 (3{)2q
20. (a) Using vq = d + du + du2 + · · · + duq31 =
1 3 { + {2 3 {3 + · · · + {2q32 3 {2q31 =
(b)
]
0
1
1 3 (3{)
(1 3 { + {2 3 {3 + · · · + {2q32 3 {2q31 ) g{ =
]
1
0
=
1 3 {2q
.
1+{
1 3 {2q
g{ i
1+{
1
] 1 2q
] 1
g{
{
{3
{4
{2q31
{2q
{2
+
3
+ ··· +
3
3
g{ i
=
{3
2
3
4
2q 3 1
2q 0
0 1+{
0 1+{
13
1 1
1
1
1
+ 3 + ··· +
3
=
2
3
4
2q 3 1
2q
(c) Since 1 3
]
0
1
g{
3
1+{
]
0
1
{2q
g{
1+{
1 1 1
1
1
1
1
1
=
, 3 =
>··· >
3
=
, we see from part (b) that
2
1·2 3
4
3·4
2q 3 1
2q
(2q 3 1)(2q)
] 1
] 1 2q
1
g{
{
1
1
+
+ ··· +
3
=3
g{. Thus,
1·2
3·4
(2q 3 1)(2q)
0 1+{
0 1+{
] 1 2q
] 1
] 1
1
g{ {
1
1
=
{2q g{
+
+
·
·
·
+
3
g{
?
1 · 2
3·4
(2q 3 1)(2q)
0 1+{
0 1+{
0
{2q
? {2q for 0 ? { $ 1 .
since
1+{
2q+1 1
] 1
k
l1
{
g{
1
= ln(1 + {) = ln 2 and
. So part (c) becomes
{2q g{ =
=
2q + 1 0
2q + 1
0
0 1+{
0
1
1
1
1
?
+
+
·
·
·
+
3
ln
2
2q + 1 . In other words, the qth partial sum vq of the given series
1 · 2
3·4
(2q 3 1)(2q)
(d) Note that
]
1
satisfies |vq 3 ln 2| ?
1
1
1
1
1
. Thus, lim vq = ln 2, that is,
+
+
+
+ · · · = ln 2.
q<"
2q + 1
1·2
3·4
5·6
7·8
21. Let i({) denote the left-hand side of the equation 1 +
{2
{3
{4
{
+
+
+
+ · · · = 0. If { D 0, then i ({) D 1 and there are
2!
4!
6!
8!
{2
{4
{6
{8
+
3
+
3 · · · = cos {. The solutions of cos { = 0 for
2!
4!
6!
8!
2
{ ? 0 are given by { = 3 n, where n is a positive integer. Thus, the solutions of i ({) = 0 are { = 3
3 n , where
2
2
no solutions of the equation. Note that i (3{2 ) = 1 3
n is a positive integer.
22. Suppose the base of the first right triangle has length d. Then by repeated use of the Pythagorean theorem, we find that the base
of the second right triangle has length
right triangle has base of length
"
S
q=1
q =
I
I
1 + d2 , the base of the third right triangle has length 2 + d2 , and in general, the nth
I
I
I
q 3 1 + d2 and hypotenuse of length q + d2 . Thus, q = tan31 1@ q 3 1 + d2 and
"
S
1
1
tan31 I
tan31 I
=
. We wish to show that this series diverges.
q 3 1 + d2
q + d2
q=1
q=0
"
S
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CHAPTER 11 PROBLEMS PLUS
First notice that the series
"
S
1
2
231
"
S
1
1
I
I
diverges by the Limit Comparison Test with the divergent p-series
2
q
q+d
q=1
q=1
s=
¤
I
u
I
u
1@ q + d2
q
q
1
I
$ 1 since lim
= lim
=
lim
= lim I
= 1 A 0. Thus,
q<"
q<"
q<"
q<"
q + d2
1 + d2 @q
1@ q
q + d2
"
"
S
S
1
1
1
I
I
also diverges. Now
tan31 I
since
diverges by the Limit Comparison Test with
2
2
q+d
q+d
q + d2
q=0
q=0
q=0
"
S
I
I
tan31 1@ q + d2
tan31 1@ { + d2
tan31 (1@|)
I
I
= lim
= lim
lim
2
2
q<"
{<"
|<"
1@|
1@ q + d
1@ { + d
= lim
}<0+
Thus,
"
S
tan31 } } = 1@|
}
H
= lim
}<0+
I
| = { + d2
1@(1 + } 2 )
=1A0
1
q is a divergent series.
q=1
23. Call the series V. We group the terms according to the number of digits in their denominators:
V=
1
1
+
1
2
+ ··· +
~}
j1
1
8
+
1
9
€
+
1
 11
+ ··· +
~}
j2
1
99
€
+
1
 111
+ ··· +
~}
j3
1
999
€
+···
Now in the group jq , since we have 9 choices for each of the q digits in the denominator, there are 9q terms.
9 q31
1
1
[except for the first term in j1 ]. So jq ? 9q · 10q1
= 9 10
.
Furthermore, each term in jq is less than 10q1
Now
"
9 q31
S
9 10
is a geometric series with d = 9 and u =
q=1
V=
"
S
jq ?
q=1
"
9 q31
S
9 10
=
q=1
24. (a) Let i ({) =
9
1 3 9@10
9
10
? 1. Therefore, by the Comparison Test,
= 90.
"
S
{
=
fq {q = f0 + f1 { + f2 {2 + f3 {3 + · · · . Then
2
13{3{
q=0
{ = (1 3 { 3 {2 )(f0 + f1 { + f2 {2 + f3 {3 + · · · )
{ = f0 + f1 { + f2 {2 + f3 {3 + f4 {4 + f5 {5 + · · ·
3 f0 { 3 f1 {2 3 f2 {3 3 f3 {4 3 f4 {5 3 · · ·
3 f0 {2 3 f1 {3 3 f2 {4 3 f3 {5 3 · · ·
{ = f0 + (f1 3 f0 ){ + (f2 3 f1 3 f0 ){2 + (f3 3 f2 3 f1 ){3 + · · ·
Comparing coefficients of powers of { gives us f0 = 0 and
f1 3 f0 = 1
i
f1 = f0 + 1 = 1
f2 3 f1 3 f0 = 0
i
f2 = f1 + f0 = 1 + 0 = 1
f3 3 f2 3 f1 = 0
i
f3 = f2 + f1 = 1 + 1 = 2
In general, we have fq = fq31 + fq32 for q D 3. Each fq is equal to the qth Fibonacci number, that is,
"
S
q=0
fq {q =
"
S
q=1
fq {q =
"
S
iq {q
q=1
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232
¤
CHAPTER 11 PROBLEMS PLUS
(b) Completing the square on {2 + { 3 1 gives us
2
2 I 2
5
1
1
1
1
5
313 = {+
{2 + { +
3 = {+
3
4
4
2
4
2
2
I I I I 5
5
1
1+ 5
13 5
1
{+ 3
= {+
{+
= {+ +
2
2
2
2
2
2
So
3{
{
= = 2
1 3 { 3 {2
{ +{31
{+
so the partial fraction decomposition is
{+
If { =
31 +
2
If { =
31 3
2
I
5
I
5
I
1+
2
3{
5
{+
, then 3 31 +2
, then 3 31 3
2
{
=
1 3 { 3 {2
=
=
=
I 13 5
2
I
5
I
5
=E
3{
I 1+ 5
{
2
=
D
{+
+
I
1+ 5
2
I
5 i E=
I .
13 5
2
The factors in the denominator are linear,
E
+
{+
I
13 5
2
3{=D {+
I 13 5
2
+E {+
I 1+ 5
2
I
1 3I 5
.
2 5
I =D 3 5
i D=
I
1 +I 5
.
32 5
Thus,
I
I
I
I
13 5
1+ 5
2
2
1+ 5
13 5
I
I
I
I
I
I
32 5
2 5
32 5
1+ 5
2 5
13 5
I +
I =
I ·
I ·
+
2
2
1+ 5
13 5
1+ 5
13 5
I
I
{+
{+
{+
{+
1+ 5
13 5
2
2
2
2
I
I
q
q
"
"
1 S
1@ 5
1 S
31@ 5
2
2
I {
I {
+I
+
= 3I
3
3
2
2
5 q=0
1+ 5
5 q=0
13 5
I { 1+
I {
1+
1+ 5
13 5
q q "
32
32
1 S
I
I
I
3
{q
5 q=0 1 3 5
1+ 5
%
I q
I q &
"
(32)q 1 + 5 3 (32)q 1 3 5
1 S
I
I
I
[the q = 0 term is 0]
{q
q q
5 q=1
13 5
1+ 5
5
I q I q 6
(32)q 1 + 5 3 1 3 5
"
1 S
7
8 {q
= I
(1 3 5)q
5 q=1
"
1 S
= I
5 q=1
%
I q I q &
1+ 5 3 13 5
{q
2q
From part (a), this series must equal
"
S
q=1
the nth Fibonacci number.
25. x = 1 +
iq {q , so iq =
[(34)q = (32)q · 2q ]
I q
I q 1+ 5 3 13 5
I
, which is an explicit formula for
2q 5
{6
{9
{4
{7
{10
{2
{5
{8
{3
+
+
+···, y = { +
+
+
+ ···, z =
+
+
+ ···.
3!
6!
9!
4!
7!
10!
2!
5!
8!
Use the Ratio Test to show that the series for x, y, and z have positive radii of convergence (" in each case), so
Theorem 11.9.2 applies, and hence, we may differentiate each of these series:
3{2
6{5
9{8
{2
{5
{8
gx
=
+
+
+ ··· =
+
+
+ ··· = z
g{
3!
6!
9!
2!
5!
8!
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CHAPTER 11 PROBLEMS PLUS
Similarly,
¤
233
{3
{6
{9
gz
{4
{7
{10
gy
=1+
+
+
+ · · · = x, and
={+
+
+
+ · · · = y.
g{
3!
6!
9!
g{
4!
7!
10!
So x0 = z, y 0 = x, and z0 = y. Now differentiate the left-hand side of the desired equation:
g 3
(x + y 3 + z3 3 3xyz) = 3x2 x0 + 3y 2 y 0 + 3z2 z0 3 3(x0 yz + xy 0 z + xyz0 )
g{
= 3x2 z + 3y 2 x + 3z2 y 3 3(yz2 + x2 z + xy 2 ) = 0
i
x3 + y 3 + z3 3 3xyz = F. To find the value of the constant F, we put { = 0 in the last equation and get
13 + 03 + 03 3 3(1 · 0 · 0) = F
i F = 1, so x3 + y 3 + z3 3 3xyz = 1.
26. To prove: If q A 1, then the nth partial sum vq =
q 1
S
of the harmonic series is not an integer.
l=1 l
Proof: Let 2n be the largest power of 2 that is less than or equal to q and let P be the product of all the odd positive integers
that are less than or equal to q. Suppose that vq = p, an integer. Then P 2n vq = P2n p. Since q D 2, we have n D 1, and
hence, P2n p is an even integer. We will show that P 2n vq is an odd integer, contradicting the equality P2n vq = P 2n p
and showing that the supposition that vq is an integer must have been wrong.
P 2n vq = P2n
q P 2n
q 1
S
S
P
=
. If 1 $ l $ q and l is odd, then
is an odd integer since l is one of the odd integers
l
l
l=1 l
l=1
that were multiplied together to form P. Thus,
P 2n
is an even integer in this case. If 1 $ l $ q and l is even, then we can
l
write l = 2u o, where 2u is the largest power of 2 dividing l and o is odd. If u ? n, then
an even integer, the product of the even integer 2n3u and the odd integer
P
2n P
P2n
= u ·
= 2n3u , which is
l
2
o
o
P
. If u = n, then o = 1, since o A 1 = o D 2 i
o
l = 2n o D 2n · 2 = 2n+1 , contrary to the choice of 2n as the largest power of 2 that is less than or equal to q. This shows that
u = n only when l = 2n . In that case,
P 2n
P2n
= P , an odd integer. Since
is an even integer for every l except 2n and
l
l
P 2n
is an odd integer when l = 2n , we see that P 2n vq is an odd integer. This concludes the proof.
l
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12
VECTORS AND THE GEOMETRY OF SPACE
12.1 Three-Dimensional Coordinate Systems
1. We start at the origin, which has coordinates (0> 0> 0). First we move
2.
4 units along the positive {-axis, affecting only the {-coordinate,
bringing us to the point (4> 0> 0). We then move 3 units straight
downward, in the negative }-direction. Thus only the }-coordinate is
affected, and we arrive at (4> 0> 33).
3. The distance from a point to the |}-plane is the absolute value of the {-coordinate of the point. F(2> 4> 6) has the {-coordinate
with the smallest absolute value, so F is the point closest to the |}-plane. D(34> 0> 31) must lie in the {}-plane since the
distance from D to the {}-plane, given by the |-coordinate of D, is 0.
4. The projection of (2> 3> 5) onto the {|-plane is (2> 3> 0);
onto the |}-plane, (0> 3> 5); onto the {}-plane, (2> 0> 5).
The length of the diagonal of the box is the distance between
the origin and (2> 3> 5), given by
s
I
(2 3 0)2 + (3 3 0)2 + (5 3 0)2 = 38 E 6=16
5. The equation { + | = 2 represents the set of all points in
R3 whose {- and |-coordinates have a sum of 2, or
equivalently where | = 2 3 {= This is the set
{({> 2 3 {> }) | { M R> } M R} which is a vertical plane
that intersects the {|-plane in the line | = 2 3 {, } = 0.
6. (a) In R2 , the equation { = 4 represents a line parallel to
the |-axis. In R3 , the equation { = 4 represents the
set {({> |> }) | { = 4}, the set of all points whose
{-coordinate is 4. This is the vertical plane that is
parallel to the |}-plane and 4 units in front of it.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
(b) In R3 , the equation | = 3 represents a vertical plane that is parallel to the {}-plane and 3 units to the right of it. The
equation } = 5 represents a horizontal plane parallel to the {|-plane and 5 units above it. The pair of equations | = 3,
} = 5 represents the set of points that are simultaneously on both planes, or in other words, the line of intersection
of the planes | = 3, } = 5. This line can also be described as
the set {({> 3> 5) | { M R}, which is the set of all points in R3
whose {-coordinate may vary but whose |- and }-coordinates
are fixed at 3 and 5, respectively. Thus the line is parallel to
the {-axis and intersects the |}-plane in the point (0> 3> 5).
7. We can find the lengths of the sides of the triangle by using the distance formula between pairs of vertices:
s
I
(7 3 3)2 + [0 3 (32)]2 + [1 3 (33)]2 = 16 + 4 + 16 = 6
s
I
I
I
|TU| = (1 3 7)2 + (2 3 0)2 + (1 3 1)2 = 36 + 4 + 0 = 40 = 2 10
s
I
|US | = (3 3 1)2 + (32 3 2)2 + (33 3 1)2 = 4 + 16 + 16 = 6
|S T| =
The longest side is TU, but the Pythagorean Theorem is not satisfied: |S T|2 + |US |2 6= |TU|2 . Thus S TU is not a right
triangle. S TU is isosceles, as two sides have the same length.
8. Compute the lengths of the sides of the triangle by using the distance formula between pairs of vertices:
s
I
(4 3 2)2 + [1 3 (31)]2 + (1 3 0)2 = 4 + 4 + 1 = 3
s
I
I
I
|TU| = (4 3 4)2 + (35 3 1)2 + (4 3 1)2 = 0 + 36 + 9 = 45 = 3 5
s
I
|US | = (2 3 4)2 + [31 3 (35)]2 + (0 3 4)2 = 4 + 16 + 16 = 6
|S T| =
Since the Pythagorean Theorem is satisfied by |S T|2 + |US |2 = |TU|2 , S TU is a right triangle. S TU is not isosceles, as
no two sides have the same length.
9. (a) First we find the distances between points:
s
I
(3 3 2)2 + (7 3 4)2 + (32 3 2)2 = 26
s
I
I
|EF| = (1 3 3)2 + (3 3 7)2 + [3 3 (32)]2 = 45 = 3 5
s
I
|DF| = (1 3 2)2 + (3 3 4)2 + (3 3 2)2 = 3
|DE| =
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.
I
I
I
Since 26 + 3 6= 3 5, the three points do not lie on a straight line.
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SECTION 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS
¤
237
(b) First we find the distances between points:
s
I
|GH| = (1 3 0)2 + [32 3 (35)]2 + (4 3 5)2 = 11
s
I
I
|HI | = (3 3 1)2 + [4 3 (32)]2 + (2 3 4)2 = 44 = 2 11
s
I
I
|GI | = (3 3 0)2 + [4 3 (35)]2 + (2 3 5)2 = 99 = 3 11
Since |GH| + |HI | = |GI |, the three points lie on a straight line.
10. (a) The distance from a point to the {|-plane is the absolute value of the }-coordinate of the point. Thus, the distance
is |6| = 6.
(b) Similarly, the distance to the |}-plane is the absolute value of the {-coordinate of the point: |4| = 4.
(c) The distance to the {}-plane is the absolute value of the |-coordinate of the point: |32| = 2.
(d) The point on the {-axis closest to (4> 32> 6) is the point (4> 0> 0). (Approach the {-axis perpendicularly.)
The distance from (4> 32> 6) to the {-axis is the distance between these two points:
s
I
I
(4 3 4)2 + (32 3 0)2 + (6 3 0)2 = 40 = 2 10 E 6=32.
(e) The point on the |-axis closest to (4> 32> 6) is (0> 32> 0). The distance between these points is
s
I
I
(4 3 0)2 + [32 3 (32)]2 + (6 3 0)2 = 52 = 2 13 E 7=21.
(f ) The point on the }-axis closest to (4> 32> 6) is (0> 0> 6). The distance between these points is
s
I
I
(4 3 0)2 + (32 3 0)2 + (6 3 6)2 = 20 = 2 5 E 4=47.
11. An equation of the sphere with center (33> 2> 5) and radius 4 is [{ 3 (33)]2 + (| 3 2)2 + (} 3 5)2 = 42 or
({ + 3)2 + (| 3 2)2 + (} 3 5)2 = 16. The intersection of this sphere with the |}-plane is the set of points on the sphere
whose {-coordinate is 0. Putting { = 0 into the equation, we have 9 + (| 3 2)2 + (} 3 5)2 = 16> { = 0 or
(| 3 2)2 + (} 3 5)2 = 7> { = 0, which represents a circle in the |}-plane with center (0> 2> 5) and radius
I
7.
12. An equation of the sphere with center (2> 36> 4) and radius 5 is ({ 3 2)2 + [| 3 (36)]2 + (} 3 4)2 = 52 or
({ 3 2)2 + (| + 6)2 + (} 3 4)2 = 25. The intersection of this sphere with the {|-plane is the set of points on the sphere
whose }-coordinate is 0. Putting } = 0 into the equation, we have ({ 3 2)2 + (| + 6)2 = 9> } = 0 which represents a circle
in the {|-plane with center (2> 36> 0) and radius 3. To find the intersection with the {}-plane, we set | = 0:
({ 3 2)2 + (} 3 4)2 = 311. Since no points satisfy this equation, the sphere does not intersect the {}-plane. (Also note that
the distance from the center of the sphere to the {}-plane is greater than the radius of the sphere.) To find the intersection with
the |}-plane, we set { = 0: (| + 6)2 + (} 3 4)2 = 21> { = 0, a circle in the |}-plane with center (0> 36> 4) and radius
13. The radius of the sphere is the distance between (4> 3> 31) and (3> 8> 1): u =
Thus, an equation of the sphere is ({ 3 3)2 + (| 3 8)2 + (} 3 1)2 = 30.
I
21.
s
I
(3 3 4)2 + (8 3 3)2 + [1 3 (31)]2 = 30.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
14. If the sphere passes through the origin, the radius of the sphere must be the distance from the origin to the point (1> 2> 3):
u=
s
I
(1 3 0)2 + (2 3 0)2 + (3 3 0)2 = 14. Then an equation of the sphere is ({ 3 1)2 + (| 3 2)2 + (} 3 3)2 = 14.
15. Completing squares in the equation {2 + | 2 + } 2 3 2{ 3 4| + 8} = 15 gives
({2 3 2{ + 1) + (| 2 3 4| + 4) + (} 2 + 8} + 16) = 15 + 1 + 4 + 16 i ({ 3 1)2 + (| 3 2)2 + (} + 4)2 = 36, which we
recognize as an equation of a sphere with center (1> 2> 34) and radius 6.
16. Completing squares in the equation gives ({2 + 8{ + 16) + (| 2 3 6| + 9) + (} 2 + 2} + 1) = 317 + 16 + 9 + 1
i
({ + 4)2 + (| 3 3)2 + (} + 1)2 = 9, which we recognize as an equation of a sphere with center (34> 3> 31) and radius 3.
17. Completing squares in the equation 2{2 3 8{ + 2| 2 + 2} 2 + 24} = 1 gives
2({2 3 4{ + 4) + 2| 2 + 2(} 2 + 12} + 36) = 1 + 8 + 72 i 2({ 3 2)2 + 2| 2 + 2(} + 6)2 = 81 i
({ 3 2)2 + | 2 + (} + 6)2 =
radius
t
81
2
81
2 ,
which we recognize as an equation of a sphere with center (2> 0> 36) and
I
= 9@ 2.
18. Completing squares in the equation 3{2 + 3| 2 3 6| + 3} 2 3 12} = 10 gives
3{2 + 3(| 2 3 2| + 1) + 3(} 2 3 4} + 4) = 10 + 3 + 12 i 3{2 + 3(| 3 1)2 + 3(} 3 2)2 = 25 i
{2 + (| 3 1)2 + (} 3 2)2 =
t
25
3
25
,
3
which we recognize as an equation of a sphere with center (0> 1> 2) and radius
I
= 5@ 3.
19. (a) If the midpoint of the line segment from S1 ({1 > |1 > }1 ) to S2 ({2 > |2 > }2 ) is T =
{ + { | + | } + } 1
2
1
2
1
2
>
>
,
2
2
2
then the distances |S1 T| and |TS2 | are equal, and each is half of |S1 S2 |. We verify that this is the case:
t
|S1 S2 | = ({2 3 {1 )2 + (|2 3 |1 )2 + (}2 3 }1 )2
t
2 2 2
1
|S1 T| =
({1 + {2 ) 3 {1 + 12 (|1 + |2 ) 3 |1 + 12 (}1 + }2 ) 3 }1
2
t
2 2 2
1
=
{ 3 12 {1 + 12 |2 3 12 |1 + 12 }2 3 12 }1
2 2
t
t 1 2
=
({2 3 {1 )2 + (|2 3 |1 )2 + (}2 3 }1 )2 = 12 ({2 3 {1 )2 + (|2 3 |1 )2 + (}2 3 }1 )2
2
=
1
2
|S1 S2 |
t
2 2 2
{2 3 12 ({1 + {2 ) + |2 3 12 (|1 + |2 ) + }2 3 12 (}1 + }2 )
|TS2 | =
t
2 2 2 t 1 2 1
=
{ 3 12 {1 + 12 |2 3 12 |1 + 12 }2 3 12 }1 =
({2 3 {1 )2 + (|2 3 |1 )2 + (}2 3 }1 )2
2 2
2
t
= 12 ({2 3 {1 )2 + (|2 3 |1 )2 + (}2 3 }1 )2 = 12 |S1 S2 |
So T is indeed the midpoint of S1 S2 .
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SECTION 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS
¤
239
(b) By part (a), the midpoints of sides DE, EF and FD are S1 3 12 > 1> 4 , S2 1> 12 > 5 and S3 52 > 32 > 4 . (Recall that a median
of a triangle is a line segment from a vertex to the midpoint of the opposite side.) Then the lengths of the medians are:
|DS2 | =
|ES3 | =
|FS1 | =
t
t
t
2
02 + 12 3 2 + (5 3 3)2 = 94 + 4 = 25
=
4
t
5
2
t
2 2
+ 2 + 32 + (4 3 5)2 = 81
+
4
9
4
+1=
5
2
t
94
4
=
1
2
I
94
t
t
I
2
3 12 3 4 + (1 3 1)2 + (4 3 5)2 = 81
+ 1 = 12 85
4
20. By Exercise 19(a), the midpoint of the diameter (and thus the center of the sphere) is (3> 2> 7). The radius is half the diameter,
so u =
1
2
s
I
I
(4 3 2)2 + (3 3 1)2 + (10 3 4)2 = 12 44 = 11. Therefore an equation of the sphere is
({ 3 3)2 + (| 3 2)2 + (} 3 7)2 = 11.
21. (a) Since the sphere touches the {|-plane, its radius is the distance from its center, (2> 33> 6), to the {|-plane, namely 6.
Therefore u = 6 and an equation of the sphere is ({ 3 2)2 + (| + 3)2 + (} 3 6)2 = 62 = 36.
(b) The radius of this sphere is the distance from its center (2> 33> 6) to the |}-plane, which is 2. Therefore, an equation is
({ 3 2)2 + (| + 3)2 + (} 3 6)2 = 4.
(c) Here the radius is the distance from the center (2> 33> 6) to the {}-plane, which is 3. Therefore, an equation is
({ 3 2)2 + (| + 3)2 + (} 3 6)2 = 9.
22. The largest sphere contained in the first octant must have a radius equal to the minimum distance from the center (5> 4> 9) to
any of the three coordinate planes. The shortest such distance is to the {}-plane, a distance of 4. Thus an equation of the
sphere is ({ 3 5)2 + (| 3 4)2 + (} 3 9)2 = 16.
23. The equation { = 5 represents a plane parallel to the |}-plane and 5 units in front of it.
24. The equation | = 32 represents a plane parallel to the {}-plane and 2 units to the left of it.
25. The inequality | ? 8 represents a half-space consisting of all points to the left of the plane | = 8.
26. The inequality { D 33 represents a half-space consisting of all points on or in front of the plane { = 33.
27. The inequality 0 $ } $ 6 represents all points on or between the horizontal planes } = 0 (the {|-plane) and } = 6.
28. The equation } 2 = 1
C
} = ±1 represents two horizontal planes; } = 1 is parallel to the {|-plane, one unit above it,
and } = 31 is one unit below it.
29. Because } = 31, all points in the region must lie in the horizontal plane } = 31. In addition, {2 + | 2 = 4, so the region
consists of all points that lie on a circle with radius 2 and center on the }-axis that is contained in the plane } = 31.
30. Here | 2 + } 2 = 16 with no restrictions on {, so a point in the region must lie on a circle of radius 4, center on the {-axis, but it
could be in any vertical plane { = n (parallel to the |}-plane). Thus the region consists of all possible circles | 2 + } 2 = 16,
{ = n and is therefore a circular cylinder with radius 4 whose axis is the {-axis.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
s
I
{2 + | 2 + } 2 $ 3, so the region consists of those points whose distance
I
I
from the origin is at most 3. This is the set of all points on or inside the sphere with radius 3 and center (0> 0> 0).
31. The inequality {2 + | 2 + } 2 $ 3 is equivalent to
32. The equation { = } represents a plane perpendicular to the {}-plane and intersecting the {}-plane in the line { = }, | = 0.
33. Here {2 + } 2 $ 9 or equivalently
I
{2 + } 2 $ 3 which describes the set of all points in R3 whose distance from the |-axis is
at most 3. Thus, the inequality represents the region consisting of all points on or inside a circular cylinder of radius 3 with
axis the |-axis.
34. The inequality {2 + | 2 + } 2 A 2}
C {2 + | 2 + (} 3 1)2 A 1 is equivalent to
s
{2 + | 2 + (} 3 1)2 A 1, so the region
consists of those points whose distance from the point (0> 0> 1) is greater than 1. This is the set of all points outside the sphere
with radius 1 and center (0> 0> 1).
35. This describes all points whose {-coordinate is between 0 and 5, that is, 0 ? { ? 5.
36. For any point on or above the disk in the {|-plane with center the origin and radius 2 we have {2 + | 2 $ 4. Also each point
lies on or between the planes } = 0 and } = 8, so the region is described by {2 + | 2 $ 4, 0 $ } $ 8.
37. This describes a region all of whose points have a distance to the origin which is greater than u, but smaller than U. So
inequalities describing the region are u ?
s
{2 + | 2 + } 2 ? U, or u2 ? {2 + | 2 + } 2 ? U2 .
s
{2 + | 2 + } 2 $ 2. Since we want only the upper hemisphere, we restrict the
s
}-coordinate to nonnegative values. Then inequalities describing the region are {2 + | 2 + } 2 $ 2, } D 0, or
38. The solid sphere itself is represented by
{2 + | 2 + } 2 $ 4, } D 0.
39. (a) To find the {- and |-coordinates of the point S , we project it onto O2
and project the resulting point T onto the {- and |-axes. To find the
}-coordinate, we project S onto either the {}-plane or the |}-plane
(using our knowledge of its {- or |-coordinate) and then project the
resulting point onto the }-axis. (Or, we could draw a line parallel to
TR from S to the }-axis.) The coordinates of S are (2> 1> 4).
(b) D is the intersection of O1 and O2 , E is directly below the
|-intercept of O2 , and F is directly above the {-intercept of O2 .
40. Let S = ({> |> }). Then 2 |S E| = |S D|
C 4 |S E|2 = |S D|2
C
4 ({ 3 6)2 + (| 3 2)2 + (} + 2)2 = ({ + 1)2 + (| 3 5)2 + (} 3 3)2
C
4 {2 3 12{ + 36 3 {2 3 2{ + 4 | 2 3 4| + 4 3 | 2 + 10| + 4 } 2 + 4} + 4 3 } 2 + 6} = 35 C
3{2 3 50{ + 3| 2 3 6| + 3} 2 + 22} = 35 3 144 3 16 3 16 C {2 3
50
{
3
+ | 2 3 2| + } 2 +
22
}
3
= 3 141
.
3
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SECTION 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS
By completing the square three times we get { 3
center
25
3
and radius
> 1> 3 11
3
I
332
.
3
25 2
3
+ (| 3 1)2 + } +
41. We need to find a set of points S ({> |> }) |DS | = |ES | .
s
s
({ + 1)2 + (| 3 5)2 + (} 3 3)2 = ({ 3 6)2 + (| 3 2)2 + (} + 2)2
({ + 1)2 + (| 3 5) + (} 3 3)2 = ({ 3 6)2 + (| 3 2)2 + (} + 2)2
11 2
3
=
332
9 ,
¤
241
which is an equation of a sphere with
i
i
{2 + 2{ + 1 + |2 3 10| + 25 + } 2 3 6} + 9 = {2 3 12{ + 36 + |2 3 4| + 4 + } 2 + 4} + 4 i 14{ 3 6| 3 10} = 9.
Thus the set of points is a plane perpendicular to the line segment joining D and E (since this plane must contain the
perpendicular bisector of the line segment DE).
42. Completing the square three times in the first equation gives ({ + 2)2 + (| 3 1)2 + (} + 2)2 = 22 , a sphere with center
(32> 1> 2) and radius 2. The second equation is that of a sphere with center (0> 0> 0) and radius 2. The distance between the
s
I
centers of the spheres is (32 3 0)2 + (1 3 0)2 + (32 3 0)2 = 4 + 1 + 4 = 3. Since the spheres have the same radius,
the volume inside both spheres is symmetrical about the plane containing the circle of intersection of the spheres. The
distance from this plane to the center of the circles is 32 . So the region inside both
spheres consists of two caps of spheres of height k = 2 3
3
2
= 12 . From
Exercise 5.2.49 [ET 6.2.49], the volume of a cap of a sphere is
2 Y = k2 u 3 13 k = 12 2 3 13 · 12 = 11
24 . So the total volume is 2 ·
11
24
=
11
12 .
43. The sphere {2 + | 2 + } 2 = 4 has center (0> 0> 0) and radius 2. Completing squares in {2 3 4{ + | 2 3 4| + } 2 3 4} = 311
gives ({2 3 4{ + 4) + (| 2 3 4| + 4) + (} 2 3 4} + 4) = 311 + 4 + 4 + 4 i ({ 3 2)2 + (| 3 2)2 + (} 3 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 connecting their centers. The distance between (0> 0> 0) and (2> 2> 2) is
s
I
I
(2 3 0)2 + (2 3 0)2 + (2 3 0)2 = 12 = 2 3, and subtracting the radius of each circle, the distance between the
I
I
spheres is 2 3 3 2 3 1 = 2 3 3 3.
44. There are many different solids that fit the given description. However, any possible solid must have a circular horizontal
cross-section at its top or at its base. Here we illustrate a solid with a circular base in the {|-plane. (A circular cross-section at
the top results in an inverted version of the solid described below.) The vertical cross-section through the center of the base
that is parallel to the {}-plane must be a square, and the vertical cross-section parallel
to the |}-plane (perpendicular to the square) through the center of the base must be a
triangle with two vertices on the circle and the third vertex at the center of the top side
of the square. (See the figure.)
[continued]
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
The solid can include any additional points that do not extend beyond these
three "silhouettes" when viewed from directions parallel to the coordinate
axes. One possibility shown here is to draw the circular base and the vertical
square first. Then draw a surface formed by line segments parallel to the
|}-plane that connect the top of the square to the circle.
Problem 8 in the Problems Plus section at the end of the chapter illustrates another possible solid.
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) and direction at any given
location.
(c) If we assume that the initial path is linear, the initial flight path from Houston to Dallas is a vector, because it has both
magnitude (distance) and direction.
(d) The population of the world is a scalar, because it has only magnitude.
2. If the initial point of the vector h4> 7i is placed at the origin, then
h4> 7i is the position vector of the point (4> 7).
3. Vectors are equal when they share the same length and direction (but not necessarily location). Using the symmetry of the
3< 33< 33< 33
< 33< 33<
3< 3
3
<
parallelogram as a guide, we see that DE = GF, GD = FE, GH = HE, and HD = FH.
3
3
<
3<
3<
3
3
<
4. (a) The initial point of EF is positioned at the terminal point of DE, so by the Triangle Law the sum DE + EF is the vector
3<
with initial point D and terminal point F, namely DF.
33< 33<
33
<
(b) By the Triangle Law, FG + GE is the vector with initial point F and terminal point E, namely FE.
33< 3< 33< 3<
3<
3<
(c) First we consider GE 3 DE as GE + 3DE . Then since 3DE has the same length as DE but points in the opposite
3< 3<
33< 3< 33< 3< 33<
direction, we have 3DE = ED and so GE 3 DE = GE + ED = GD.
33< 3< 3< 33< 3< 3< 33< 3< 33<
(d) We use the Triangle Law twice: GF + FD + DE = GF + FD + DE = GD + DE = GE.
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SECTION 12.2 VECTORS
5. (a)
(b)
(c)
(d)
(e)
(f )
6. (a)
(b)
(c)
(d)
(e)
(f )
¤
3<
7. Because the tail of d is the midpoint of TU we have TU = 2d, and by the Triangle Law,
2d = b 3 a i d = 12 (b 3 a) = 12 b 3 12 a. Again by the Triangle Law we have c + d = b so
c = b 3 d = b 3 12 b 3 12 a = 12 a + 12 b.
a + 2d = b
i
8. We are given u + v + w = 0, so w = (3u) + (3v). (See the figure.)
Vectors 3u, 3v, and w form a right triangle, so from the Pythagorean Theorem
we have |3u|2 + |3v|2 = |w|2 . But |3u| = |u| = 1 and |3v| = |v| = 1 so |w| =
9. a = h3 3 (31)> 2 3 1i = h4> 1i
t
I
|3u|2 + |3v|2 = 2.
10. a = h1 3 (34)> 2 3 (31)i = h5> 3i
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¤
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
11. a = h2 3 (31)> 2 3 3i = h3> 31i
12. a = h0 3 2> 6 3 1i = h32> 5i
13. a = h2 3 0> 3 3 3> 31 3 1i = h2> 0> 32i
14. a = h4 3 4> 2 3 0> 1 3 (32)i = h0> 2> 3i
15. h31> 4i + h6> 32i = h31 + 6> 4 + (32)i = h5> 2i
16. h3> 31i + h31> 5i = h3 + (31)> 31 + 5i = h2> 4i
17. h3> 0> 1i + h0> 8> 0i = h3 + 0> 0 + 8> 1 + 0i
18. h1> 3> 32i + h0> 0> 6i = h1 + 0> 3 + 0> 32 + 6i
= h3> 8> 1i
= h1> 3> 4i
19. a + b = h5 + (33) > 312 + (36)i = h2> 318i
2a + 3b = h10> 324i + h39> 318i = h1> 342i
s
I
|a| = 52 + (312)2 = 169 = 13
|a 3 b| = |h5 3 (33)> 312 3 (36)i| = |h8> 36i| =
s
I
82 + (36)2 = 100 = 10
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SECTION 12.2 VECTORS
¤
245
20. a + b = (4 i + j) + (i 3 2 j) = 5 i 3 j
2a + 3b = 2 (4 i + j) + 3 (i 3 2 j) = 8 i + 2 j + 3 i 3 6 j = 11 i 3 4 j
|a| =
I
I
42 + 12 = 17
|a 3 b| = |(4 i + j) 3 (i 3 2 j)| = |3 i + 3 j| =
I
I
I
32 + 32 = 18 = 3 2
21. a + b = (i + 2 j 3 3 k) + (32 i 3 j + 5 k) = 3 i + j + 2k
2a + 3b = 2 (i + 2 j 3 3 k) + 3 (32 i 3 j + 5 k) = 2 i + 4 j 3 6 k 3 6 i 3 3 j + 15 k = 3 4 i + j + 9k
|a| =
s
I
12 + 22 + (33)2 = 14
|a 3 b| = |(i + 2 j 3 3 k) 3 (32 i 3 j + 5 k)| = |3 i + 3 j 3 8 k| =
22. a + b = (2 i 3 4 j + 4 k) + (2 j 3 k) = 2 i 3 2 j + 3 k
s
I
32 + 32 + (38)2 = 82
2a + 3b = 2 (2 i 3 4 j + 4 k) + 3 (2 j 3 k) = 4 i 3 8 j + 8 k + 6 j 3 3 k = 4 i 3 2 j + 5 k
|a| =
s
I
22 + (34)2 + 42 = 36 = 6
|a 3 b| = |(2 i 3 4 j + 4 k) 3 (2 j 3 k)| = |2 i 3 6 j + 5 k| =
23. The vector 33 i + 7 j has length |33 i + 7 j| =
s
I
22 + (36)2 + 52 = 65
s
I
(33)2 + 72 = 58, so by Equation 4 the unit vector with the same
1
3
7
direction is I (33 i + 7 j) = 3 I i + I j.
58
58
58
24. |h34> 2> 4i| =
s
I
(34)2 + 22 + 42 = 36 = 6, so u = 16 h34> 2> 4i = 3 23 > 13 > 23 .
25. The vector 8 i 3 j + 4 k has length |8 i 3 j + 4 k| =
the same direction is 19 (8 i 3 j + 4 k) =
26. |h32> 4> 2i| =
8
9
i3
1
9
j+
s
I
82 + (31)2 + 42 = 81 = 9, so by Equation 4 the unit vector with
4
9
k.
s
I
I
(32)2 + 42 + 22 = 24 = 2 6, so a unit vector in the direction of h32> 4> 2i is u =
A vector in the same direction but with length 6 is 6u = 6 ·
27.
1
I h32> 4> 2i =
2 6
6 12 6
3I > I > I
6
6
6
I
3 I
From the figure, we see that tan =
= 3
1
i
1
I h32> 4> 2i.
2 6
I I I
or 3 6> 2 6> 6 .
= 60 .
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
28.
From the figure we see that tan =
6
8
= 34 , so = tan31
3
4
E 36=9 .
29. From the figure, we see that the {-component of v is
y1 = |v| cos(@3) = 4 ·
1
2
= 2 and the |-component is
I
3
2
y2 = |v| sin(@3) = 4 ·
=2
I v = hy1 > y2 i = 2> 2 3 .
I
3= Thus
30. From the figure, we see that the horizontal component of the
force F is |F| cos 38 = 50 cos 38 E 39=4 N, and the
vertical component is |F| sin 38 = 50 sin 38 E 30=8 N.
31. The velocity vector v makes an angle of 40 with the horizontal and
has magnitude equal to the speed at which the football was thrown.
From the figure, we see that the horizontal component of v is
|v| cos 40 = 60 cos 40 E 45=96 ft/s and the vertical component
is |v| sin 40 = 60 sin 40 E 38=57 ft/s.
32. The given force vectors can be expressed in terms of their horizontal and vertical components as
I
I
I
2 i + 10 2 j and 16 cos 30 i 3 16 sin 30 j = 8 3 i 3 8 j. The resultant force F
I I
I
is the sum of these two vectors: F = 10 2 + 8 3 i + 10 2 3 8 j E 28=00 i + 6=14 j. Then we have
20 cos 45 i + 20 sin 45 j = 10
s
(28=00)2 + (6=14)2 E 28=7 lb and, letting be the angle F makes with the positive {-axis,
I
I
10 2 3 8
10 2 3 8
I
I
I
I
tan =
i = tan31
E 12=4 .
10 2 + 8 3
10 2 + 8 3
|F| E
33. The given force vectors can be expressed in terms of their horizontal and vertical components as 3300 i and
I I
200 cos 60 i + 200 sin 60 j = 200 12 i + 200 23 j = 100 i + 100 3 j. The resultant force F is the sum of
I I
these two vectors: F = (3300 + 100) i + 0 + 100 3 j = 3200 i + 100 3 j. Then we have
|F| E
t
I 2 I
I
(3200)2 + 100 3 = 70,000 = 100 7 E 264=6 N. Let be the angle F makes with the
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SECTION 12.2 VECTORS
¤
247
I
I
100 3
3
positive {-axis. Then tan =
=3
and the terminal point of F lies in the second quadrant, so
3200
2
I 3
+ 180 E 340=9 + 180 = 139=1 .
= tan31 3
2
34. Set up the coordinate axes so that north is the positive |-direction, and east is the positive {-direction. The wind is blowing
at 50 km@h from the direction N45 W, so that its velocity vector is 50 km@h S45 E, which can be written as
vwind = 50(cos 45 i 3 sin 45 j). With respect to the still air, the velocity vector of the plane is 250 km@h N 60 E, or
equivalently vplane = 250(cos 30 i + sin 30 j). The velocity of the plane relative to the ground is
v = vwind + vplane = (50 cos 45 + 250 cos 30 ) i + (350 sin 45 + 250 sin 30 ) j
I I I
= 25 2 + 125 3 i + 125 3 25 2 j E 251=9 i + 89=6 j
s
The ground speed is |v| E (251=9)2 + (89=6)2 E 267 km@h. The angle the velocity vector makes with the {-axis is
89=6 E tan31 251=9
E 20 . Therefore, the true course of the plane is about N(90 3 20) E = N 70 E.
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 |-direction, then
I
v = h0> 22i + h33> 0i = h33> 22i. The woman’s speed is |v| = 9 + 484 E 22=2 mi@h. The vector v makes an angle 22
with the east, where = tan31 33
E 98 . Therefore, the woman’s direction is about N(98 3 90) W = N8 W.
36. Call the two tensile forces T3 and T5 , corresponding to the ropes of length 3 m and 5 m. In terms of vertical and horizontal
components,
T3 = 3 |T3 | cos 52 i + |T3 | sin 52 j (1)
and
T5 = |T5 | cos 40 i + |T5 | sin 40 j (2)
The resultant of these forces, T3 + T5 , counterbalances the force of gravity acting on the decoration [which is
35j j E 35(9=8) j = 349 j]. So T3 + T5 = 49 j. Hence
T3 + T5 = (3 |T3 | cos 52 + |T5 | cos 40 ) i + (|T3 | sin 52 + |T5 | sin 40 ) j = 49 j.
Thus 3 |T3 | cos 52 + |T5 | cos 40 = 0 and |T3 | sin 52 + |T5 | sin 40 = 49.
From the first of these two equations |T3 | = |T5 |
|T5 | =
cos 40
cos 40
. Substituting this into the second equation gives
cos 52
49
cos 40
E 30 N. Therefore, |T3 | = |T5 |
E 38 N. Finally, from (1) and (2),
tan 52 + sin 40
cos 52
T3 E 323 i + 30 j, and T5 E 23 i + 19 j.
37. Let T1 and T2 represent the tension vectors in each side of the
clothesline as shown in the figure. T1 and T2 have equal vertical
components and opposite horizontal components, so we can write
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
T1 = 3d i + e j and T2 = d i + e j [d> e A 0]. By similar triangles,
0=08
e
=
d
4
i d = 50e. The force due to gravity
acting on the shirt has magnitude 0=8j E (0=8)(9=8) = 7=84 N, hence we have w = 37=84 j. The resultant T1 + T2
of the tensile forces counterbalances w, so T1 + T2 = 3w i (3d i + e j) + (d i + e j) = 7=84 j i
(350e i + e j) + (50e i + e j) = 2e j = 7=84 j i e =
7=84
2
= 3=92 and d = 50e = 196. Thus the tensions are
T1 = 3d i + e j = 3196 i + 3=92 j and T2 = d i + e j = 196 i + 3=92 j.
Alternatively, we can find the value of and proceed as in Example 7.
38. We can consider the weight of the chain to be concentrated at its midpoint. The
forces acting on the chain then are the tension vectors T1 , T2 in each end of the
chain and the weight w, as shown in the figure. We know |T1 | = |T2 | = 25 N
so, in terms of vertical and horizontal components, we have
T1 = 325 cos 37 i + 25 sin 37 j
T2 = 25 cos 37 i + 25 sin 37 j
The resultant vector T1 + T2 of the tensions counterbalances the weight w> giving T1 + T2 = 3w= Since w = 3 |w| j,
we have (325 cos 37 i + 25 sin 37 j) + (25 cos 37 i + 25 sin 37 j) = |w| j i 50 sin 37 j = |w| j i
|w| = 50 sin 37 E 30=1. So the weight is 30=1 N, and since z = pj, the mass is
30=1
9=8
E 3=07 kg.
39. (a) Set up coordinate axes so that the boatman is at the origin, the canal is
bordered by the |-axis and the line { = 3, and the current flows in the
negative |-direction. The boatman wants to reach the point (3> 2). Let be
the angle, measured from the positive |-axis, in the direction he should
steer. (See the figure.)
In still water, the boat has velocity ve = h13 sin > 13 cos i and the velocity of the current is vf h0> 33=5i, so the true path
of the boat is determined by the velocity vector v = ve + vf = h13 sin > 13 cos 3 3=5i. Let w be the time (in hours)
after the boat departs; then the position of the boat at time w is given by wv and the boat crosses the canal when
wv = h13 sin > 13 cos 3 3=5i w = h3> 2i. Thus 13(sin )w = 3
Substituting gives (13 cos 3 3=5)
3
=2
13 sin i
i
w=
3
and (13 cos 3 3=5) w = 2.
13 sin 39 cos 3 10=5 = 26 sin (1). Squaring both sides, we have
1521 cos2 3 819 cos + 110=25 = 676 sin2 = 676 1 3 cos2 2197 cos2 3 819 cos 3 565=75 = 0
The quadratic formula gives
cos =
819 ±
s
(3819)2 3 4(2197)(3565=75)
2(2197)
=
819 ±
I
5,642,572
E 0=72699 or 3 0=35421
4394
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SECTION 12.2 VECTORS
249
The acute value for is approximately cos31 (0=72699) E 43=4 . Thus the boatman should steer in the direction that is
43=4 from the bank, toward upstream.
Alternate solution: We could solve (1) graphically by plotting | = 39 cos 3 10=5 and | = 26 sin on a graphing device
and finding the appoximate intersection point (0=757> 17=85). Thus E 0=757 radians or equivalently 43=4 .
(b) From part (a) we know the trip is completed when w =
3
. But E 43=4 , so the time required is approximately
13 sin 3
E 0=336 hours or 20=2 minutes.
13 sin 43=4
40. Let v1 , v2 , and v3 be the force vectors where |v1 | = 25, |v2 | = 12, and |v3 | = 4. Set up coordinate axes so that the object is
at the origin and v1 , v2 lie in the {|-plane. We can position the vectors so that v1 = 25 i, v2 = 12 cos 100 i + 12 sin 100 j,
and v3 = 4 k. The magnitude of a force that counterbalances the three given forces must match the magnitude of the resultant
force. We have v1 + v2 + v3 = (25 + 12 cos 100 ) i + 12 sin 100 j + 4 k, so the counterbalancing force must have
s
magnitude |v1 + v2 + v3 | = (25 + 12 cos 100 )2 + (12 sin 100 )2 + 42 E 26=1 N.
41. The slope of the tangent line to the graph of | = {2 at the point (2> 4) is
g| =
2{
=4
g{ {=2
{=2
and a parallel vector is i + 4 j which has length |i + 4 j| =
I
I
12 + 42 = 17, so unit vectors parallel to the tangent line
are ± I117 (i + 4 j).
42. (a) The slope of the tangent line to the graph of | = 2 sin { at the point (@6> 1) is
I
g| 3 I
=
2
cos
{
=
2
·
= 3
g{ {=@6
2
{=@6
I t
I
I 2 I
3 = 4 = 2, so unit vectors parallel to the
and a parallel vector is i + 3 j which has length i + 3 j = 12 +
I 1
tangent line are ± 2 i + 3 j .
I
(c)
(b) The slope of the tangent line is 3, so the slope of a line
perpendicular to the tangent line is 3 I13 and a vector in this direction
is
I
tI 2
I
3 i 3 j. Since 3 i 3 j =
3 + (31)2 = 2, unit vectors
perpendicular to the tangent line are ± 12
3<
3
3
<
3<
I
3i 3 j .
3<
3
3
<
3<
3<
3<
3<
3<
3<
3<
43. By the Triangle Law, DE + EF = DF. Then DE + EF + FD = DF + FD, but DF + FD = DF + 3DF
3< 3
3
< 3<
So DE + EF + FD = 0.
3<
44. DF =
3<
ED =
3<
1
3 DE
3
2
c3
= 0.
33
<
3<
3< 3<
3<
3<
3
3< 3
3< 3<
3<
and EF = 23 ED. c = RD + DF = a + 13 DE i DE = 3 c 3 3 a. c = RE + EF = RD + 23 ED i
3
2
3<
3<
b. ED = 3DE, so
3
2
c3
3
2
b = 3a 3 3c C c + 2c = 2a + b
C c=
2
3
a+
1
3
b.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
45. (a), (b)
(c) From the sketch, we estimate that v E 1=3 and w E 1=6.
(d) c = v a + w b C 7 = 3v + 2w and 1 = 2v 3 w.
Solving these equations gives v =
9
7
and w =
11
7 .
46. Draw a, b, and c emanating from the origin. Extend a and b to form lines D
and E, and draw lines D0 and E 0 parallel to these two lines through the terminal
point of c. Since a and b are not parallel, D and E 0 must meet (at S ), and D0
3
3
< 3
3
<
and E must also meet (at T). Now we see that RS + RT = c, so if
3
<
3
RS 3
3
<
or its negative, if a points in the direction opposite RS and w =
v=
|a|
then c = va + wb, as required.
3
<
3
RT
|b|
(or its negative, as in the diagram),
Argument using components: Since a, b, and c all lie in the same plane, we can consider them to be vectors in two
dimensions. Let a = hd1 > d2 i, b = he1 > e2 i, and c = hf1 > f2 i. We need vd1 + we1 = f1 and vd2 + we2 = f2 . Multiplying
the first equation by d2 and the second by d1 and subtracting, we get w =
f2 d1 3 f1 d2
e2 f1 3 e1 f2
. Similarly v =
.
e2 d1 3 e1 d2
e2 d1 3 e1 d2
Since a 6= 0 and b 6= 0 and a is not a scalar multiple of b, the denominator is not zero.
47. |r 3 r0 | is the distance between the points ({> |> }) and ({0 > |0 > }0 ), so the set of points is a sphere with radius 1 and
center ({0 > |0 > }0 ).
Alternate method: |r 3 r0 | = 1 C
s
({ 3 {0 )2 + (| 3 |0 )2 + (} 3 }0 )2 = 1 C
({ 3 {0 )2 + (| 3 |0 )2 + (} 3 }0 )2 = 1, which is the equation of a sphere with radius 1 and center ({0 > |0 > }0 ).
48. Let S1 and S2 be the points with position vectors r1 and r2 respectively. Then |r 3 r1 | + |r 3 r2 | is the sum of the distances
from ({> |) to S1 and S2 . Since this sum is constant, the set of points ({> |) represents an ellipse with foci S1 and S2 . The
condition n A |r1 3 r2 | assures us that the ellipse is not degenerate.
49. a + (b + c) = hd1 > d2 i + (he1 > e2 i + hf1 > f2 i) = hd1 > d2 i + he1 + f1 > e2 + f2 i
= hd1 + e1 + f1 > d2 + e2 + f2 i = h(d1 + e1 ) + f1 > (d2 + e2 ) + f2 i
= hd1 + e1 > d2 + e2 i + hf1 > f2 i = (hd1 > d2 i + he1 > e2 i) + hf1 > f2 i
= (a + b) + c
50. Algebraically:
f(a + b) = f (hd1 > d2 > d3 i + he1 > e2 > e3 i) = f hd1 + e1 > d2 + e2 > d3 + e3 i
= hf (d1 + e1 ) > f (d2 + e2 ) > f (d3 + e3 )i = hfd1 + fe1 > fd2 + fe2 > fd3 + fe3 i
= hfd1 > fd2 > fd3 i + hfe1 > fe2 > fe3 i = f a + f b
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SECTION 12.3 THE DOT PRODUCT
¤
251
Geometrically:
3
3
<
3<
According to the Triangle Law, if a = S T and b = TU, then
3<
3<
a + b = S U. Construct triangle S VW as shown so that S V = f a and
3<
VW = f b. (We have drawn the case where f A 1.) By the Triangle Law,
3<
S W = f a + f b. But triangle S TU and triangle S VW are similar triangles
3<
3<
because f b is parallel to b. Therefore, SU and S W are parallel and, in fact,
3<
3<
S W = fSU. Thus, f a + f b = f(a + b).
3<
33
<
3<
51. Consider triangle DEF, where G and H are the midpoints of DE and EF. We know that DE + EF = DF
(1) and
33< 33
< 33<
33<
3<
33<
3
3
<
33<
33<
GE + EH = GH (2). However, GE = 12 DE, and EH = 12 EF. Substituting these expressions for GE and EH into
3<
3
3
< 33<
33<
3<
3<
33<
(2) gives 12 DE + 12 EF = GH. Comparing this with (1) gives GH = 12 DF. Therefore DF and GH are parallel and
33<
GH =
1
2
3<
DF .
52. The question states that the light ray strikes all three mirrors, so it is not parallel to any of them and d1 6= 0, d2 6= 0 and
d3 6= 0. Let b = he1 > e2 > e3 i, as in the diagram. We can let |b| = |a|, since only its direction is important. Then
|d2 |
|e2 |
= sin =
|b|
|a|
i |e2 | = |d2 |.
From the diagram e2 j and d2 j point in opposite directions,
so e2 = 3d2 . |DE| = |EF|, so
|e3 | = sin ! |EF| = sin ! |DE| = |d3 |, and
|e1 | = cos ! |EF| = cos ! |DE| = |d1 |.
e3 k and d3 k have the same direction, as do e1 i and d1 i, so
b = hd1 > 3d2 > d3 i. When the ray hits the other mirrors, similar
arguments show that these reflections will reverse the signs of
the other two coordinates, so the final reflected ray will be
h3d1 > 3d2 > 3d3 i = 3a, which is parallel to a.
12.3 The Dot Product
1. (a) a · b is a scalar, and the dot product is defined only for vectors, so (a · b) · c has no meaning.
(b) (a · b) c is a scalar multiple of a vector, so it does have meaning.
(c) Both |a| and b · c are scalars, so |a| (b · c) is an ordinary product of real numbers, and has meaning.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
(d) Both a and b + c are vectors, so the dot product a · (b + c) has meaning.
(e) a · b is a scalar, but c is a vector, and so the two quantities cannot be added and a · b + c has no meaning.
(f ) |a| is a scalar, and the dot product is defined only for vectors, so |a| · (b + c) has no meaning.
2. a · b = h32> 3i · h0=7> 1=2i = (32)(0=7) + (3)(1=2) = 2=2
3. a · b = 32>
1
3
· h35> 12i = (32)(35) +
1
(12) = 10 + 4 = 14
3
· h6> 33> 38i = (4)(6) + (1)(33) +
4. a · b = h6> 32> 3i · h2> 5> 31i = (6)(2) + (32) (5) + (3)(31) = 12 3 10 3 3 = 31
5. a · b = 4> 1>
1
4
1
4
(38) = 19
6. a · b = hs> 3s> 2si · h2t> t> 3ti = (s)(2t) + (3s)(t) + (2s)(3t) = 2st 3 st 3 2st = 3st
7. a · b = (2 i + j) · (i 3 j + k) = (2)(1) + (1)(31) + (0)(1) = 1
8. a · b = (3 i + 2 j 3 k) · (4 i + 5 k) = (3)(4) + (2)(0) + (31)(5) = 7
9. By Theorem 3, a · b = |a| |b| cos = (6)(5) cos
2
3
= 30 3 12 = 315.
I I I 6 cos 45 = 3 6 22 =
10. By Theorem 3, a · b = |a| |b| cos = (3)
3
2
·2
I
I
3 = 3 3 E 5=20.
11. u> v> and w are all unit vectors, so the triangle is an equilateral triangle. Thus the angle between u and v is 60 and
u · v = |u| |v| cos 60 = (1)(1) 12 = 12 = If w is moved so it has the same initial point as u, we can see that the angle
between them is 120 and we have u · w = |u| |w| cos 120 = (1)(1) 3 12 = 3 12 .
12. u is a unit vector, so w is also a unit vector, and |v| can be determined by examining the right triangle formed by u and v=
Since the angle between u and v is 45 , we have |v| = |u| cos 45 =
I
2
.
2
Since u and w are orthogonal, u · w = 0.
I I
Then u · v = |u| |v| cos 45 = (1) 22 22 = 12 .
13. (a) i · j = h1> 0> 0i · h0> 1> 0i = (1)(0) + (0)(1) + (0)(0) = 0. Similarly, j · k = (0)(0) + (1)(0) + (0)(1) = 0 and
k · i = (0)(1) + (0)(0) + (1)(0) = 0.
Another method: Because i, j, and k are mutually perpendicular, the cosine factor in each dot product (see Theorem 3)
is cos 2 = 0.
(b) By Property 1 of the dot product, i · i = |i|2 = 12 = 1 since i is a unit vector. Similarly, j · j = |j|2 = 1 and
k · k = |k|2 = 1.
14. The dot product A · P is
hd> e> fi · h2> 1=5> 1i = d(2) + e(1=5) + f(1)
= (number of hamburgers sold)(price per hamburger)
+ (number of hot dogs sold)(price per hot dog)
+ (number of soft drinks sold)(price per soft drink)
so it is equal to the vendor’s total revenue for that day.
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SECTION 12.3 THE DOT PRODUCT
¤
253
s
I
I
42 + 32 = 5, |b| = 22 + (31)2 = 5, and a · b = (4)(2) + (3)(31) = 5. From Corollary 6, we have
1
5
a·b
1
I = I . So the angle between a and b is = cos31 I
=
cos =
E 63 .
|a| |b|
5· 5
5
5
15. |a| =
s
I
I
(32)2 + 52 = 29, |b| = 52 + 122 = 13, and a · b = (32) (5) + (5)(12) = 50. Using Corollary 6, we have
a·b
50
50
50
I
cos =
= I and the angle between a and b is = cos31
E 44 .
= I
|a| |b|
29 · 13
13 29
13 29
16. |a| =
17. |a| =
s
s
I
I
32 + (31)2 + 52 = 35, |b| = (32)2 + 42 + 32 = 29, and a · b = (3)(32) + (31)(4) + (5)(3) = 5. Then
cos =
18. |a| =
a·b
5
5
5
I = I
E 81 .
and the angle between a and b is = cos31 I1015
= I
|a| |b|
35 · 29
1015
s
I
I
I
42 + 02 + 22 = 20, |b| = 22 + (31)2 + 02 = 5, and a · b = (4)(2) + (0)(31) + (2)(0) = 8.
Then cos =
8
a·b
4
I = and = cos31 45 E 37 .
= I
|a| |b|
5
20 · 5
s
s
I
I
42 + (33)2 + 12 = 26, |b| = 22 + 02 + (31)2 = 5, and a · b = (4)(2) + (33)(0) + (1)(31) = 7.
a·b
7
7
7
I = I
Then cos =
and = cos31 I
E 52 .
= I
|a| |b|
26 · 5
130
130
19. |a| =
20. |a| =
s
s
I
I
12 + 22 + (32)2 = 9 = 3, |b| = 42 + 02 + (33)2 = 25 = 5, and
a · b = (1)(4) + (2)(0) + (32)(33) = 10. Then cos =
a·b
10
2
=
= and = cos31 23 E 48 .
|a| |b|
3·5
3
21. Let s, t, and u be the angles at vertices S , T, and U respectively.
3
3
<
3<
Then s is the angle between vectors S T and S U, t is the angle
3
3
<
3<
between vectors TS and TU, and u is the angle between vectors
3<
3<
US and UT.
3
3
< 3<
h32> 3i · h1> 4i
10
ST · SU
32 + 12
10
31
s
I
I
I
I
E 48 . Similarly,
=
Thus cos s = 3
=
=
and
s
=
cos
I
< 3<
2 + 32
2 + 42
3
13
17
221
221
(32)
1
S T S U
3
3
< 3<
3
h2> 33i · h3> 1i
TS · TU
633
3
3< = I
I
cos t = 3
= I I = I
so t = cos31 I
E 75 and
3
<
4+9 9+1
13 10
130
130
TS TU
u E 180 3 (48 + 75 ) = 57 .
3<2 3
<2 3<2
3
TU 3 S T 3 S U
3
,
Alternate solution: Apply the Law of Cosines three times as follows: cos s =
< 3<
3
2 S T S U
3<2 3
3
<2 3<2 3<2
<2 3<2
3
3
S U 3 S T 3 TU
S T 3 S U 3 TU
3
3< 3<
cos t =
, and cos u =
.
< 3<
3
2 S T TU
2 S U TU
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
22. Let d, e, and f be the angles at vertices D, E, and F. Then d is the angle
3<
3<
3<
3
3
<
between vectors DE and DF, e is the angle between vectors ED and EF,
3<
3
3
<
and f is the angle between vectors FD and FE.
3< 3<
2
h2> 32> 1i · h0> 3> 4i
2
DE · DF
036+4
E 98 .
=3
and d = cos31 3 15
Thus cos d = 3< 3< = s
=
I
2
2
2
2
2
2
3
·
5
15
2
+
(32)
+
1
0
+
3
+
4
DE DF 3< 3
3
<
h32> 2> 31i · h32> 5> 3i
4 + 10 3 3
11
11
ED · EF
= I
I
I
=
Similarly, cos e = 3< 3
= I so e = cos31 I
E 54 and
3
<
4 + 4 + 1 4 + 25 + 9
3 · 38
3 38
3 38
ED EF f E 180 3 (98 + 54 ) = 28 .
Alternate solution: Apply the Law of Cosines three times as follows:
33
<2 3<2 3<2
EF 3 DE 3 DF 3< 3<
cos d =
2 DE DF 3<2 3<2 3
<2
3
DF 3 DE 3 EF 3< 33
cos e =
<
2 DE EF 3<2 3<2 3
<2
3
DE 3 DF 3 EF 3< 3
cos f =
<
3
2 DF EF 23. (a) a · b = (35)(6) + (3)(38) + (7)(2) = 340 6= 0, so a and b are not orthogonal. Also, since a is not a scalar multiple
of b, a and b are not parallel.
(b) a · b = (4)(33) + (6)(2) = 0, so a and b are orthogonal (and not parallel).
(c) a · b = (31)(3) + (2)(4) + (5)(31) = 0, so a and b are orthogonal (and not parallel).
(d) Because a = 3 23 b, a and b are parallel.
24. (a) Because u = 3 34 v, u and v are parallel vectors (and thus not orthogonal).
(b) u · v =(1)(2) + (31)(31) + (2)(1) = 5 6= 0, so u and v are not orthogonal. Also, u is not a scalar multiple of v,
so u and v are not parallel.
(c) u · v =(d)(3e) + (e)(d) + (f)(0) = 3de + de + 0 = 0, so u and v are orthogonal (and not parallel).
3
3
<
3<
3
3
< 3<
3
3
<
3<
25. TS = h31> 33> 2i, TU = h4> 32> 31i, and TS · TU = 34 + 6 3 2 = 0. Thus TS and TU are orthogonal, so the angle of
the triangle at vertex T is a right angle.
26. By Theorem 3, vectors h2> 1> 31i and h1> {> 0i meet at an angle of 45 when
h2> 1> 31i · h1> {> 0i =
I I
I
I
4 + 1 + 1 1 + {2 + 0 cos 45 or 2 + { 3 0 = 6 1 + {2 ·
I
2
2
C 2+{=
I I
3 1 + {2 .
Squaring both sides gives 4 + 4{ + {2 = 3 + 3{2 C 2{2 3 4{ 3 1 = 0. By the quadratic formula,
s
I
I
I
3(34) ± (34)2 3 4(2)(31)
6
4 ± 24
4±2 6
=
=
=1±
. (You can verify that both values are valid.)
{=
2(2)
4
4
2
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SECTION 12.3 THE DOT PRODUCT
27. Let a = d1 i + d2 j + d3 k be a vector orthogonal to both i + j and i + k. Then a · (i + j) = 0
a · (i + k) = 0
I1
3
i3
I1
3
j3
I1
3
k and a = 3 I13 i +
1
I
3
j+
28. Let u = hd> ei be a unit vector. By Theorem 3 we need u · v = |u| |v| cos 60
5
8
25 2
d
16
255
d1 + d2 = 0 and
d1 + d3 = 0, so d1 = 3d2 = 3d3 . Furthermore a is to be a unit vector, so 1 = d21 + d22 + d23 = 3d21
C
implies d1 = ± I13 . Thus a =
e=
C
¤
I1
3
k are two such unit vectors.
C 3d + 4e = (1)(5) 12
C
I
2
3 34 d. Since u is a unit vector, |u| = d2 + e2 = 1 C d2 + e2 = 1 C d2 + 58 3 34 d = 1 C
= 1 C 100d2 3 60d 3 39 = 0= By the quadratic formula,
s
I
I
I
3(360) ± (360)2 3 4(100)(339)
60 ± 19,200
3±4 3
3+4 3
d=
=
=
. If d =
then
2(100)
200
10
10
I I
I
I I
5
3 3+4 3
433 3
334 3
5
3 334 3
4+3 3
e= 3
=
, and if d =
then e = 3
=
. Thus the two
8
4
10
10
10
8
4
10
10
I
I
I
I
334 3 4+3 3
3+4 3 433 3
>
E h0=9928> 30=1196i and
>
E h30=3928> 0=9196i.
unit vectors are
10
10
10
10
3
15
d
16
+
25
64
29. The line 2{ 3 | = 3
C | = 2{ 3 3 has slope 2, so a vector parallel to the line is a = h1> 2i. The line 3{ + | = 7 C
| = 33{ + 7 has slope 33, so a vector parallel to the line is b = h1> 33i. The angle between the lines is the same as the
I
I
angle between the vectors. Here we have a · b = (1)(1) + (2)(33) = 35, |a| = 12 + 22 = 5, and
I
s
I
2
35
a·b
35
1
= I I = I = 3 I or 3
. Thus = 135 , and the
|b| = 12 + (33)2 = 10, so cos =
|a| |b|
2
5 · 10
5 2
2
acute angle between the lines is 180 3 135 = 45 .
30. The line { + 2| = 7
C | = 3 12 { +
7
2
has slope 3 12 , so a vector parallel to the line is a = h2> 31i. The line
5{ 3 | = 2 C | = 5{ 3 2 has slope 5, so a vector parallel to the line is b = h1> 5i. The lines meet at the same
s
I
angle that the vectors meet at. Here we have a · b = (2)(1) + (31)(5) = 33, |a| = 22 + (31)2 = 5, and
I
I
33
33
a·b
33
E 105=3 . The acute
= I I = I
and = cos31 I
|b| = 12 + 52 = 26, so cos =
|a| |b|
5 · 26
130
130
angle between the lines is approximately 180 3 105=3 = 74=7 .
31. The curves | = {2 and | = {3 meet when {2 = {3
C {3 3 {2 = 0 C {2 ({ 3 1) = 0 C { = 0, { = 1. We have
g 3
g 2
{ = 2{ and
{ = 3{2 , so the tangent lines of both curves have slope 0 at { = 0. Thus the angle between the curves is
g{
g{
g 2 g 3 = 2 and
= 3 so the tangent lines at the point (1> 1) have slopes 2 and
0 at the point (0> 0). For { = 1,
{
{
g{ {=1
g{ {=1
3. Vectors parallel to the tangent lines are h1> 2i and h1> 3i, and the angle between them is given by
cos =
Thus = cos31
7
I
5 2
1+6
7
h1> 2i · h1> 3i
= I I = I
|h1> 2i| |h1> 3i|
5 10
5 2
E 8=1 .
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256
¤
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
32. The curves | = sin { and | = cos { meet when sin { = cos {
C tan { = 1 C { = @4 [0 $ { $ @2]. Thus the
I
I
g
2
= cos {
=
point of intersection is @4> 2@2 . We have
sin {
and
g{
2
{=@4
{=@4
I
I
I
2
2
2
g
cos {
, so the tangent lines at that point have slopes
and 3
. Vectors parallel to
= 3 sin {
=3
g{
2
2
2
{=@4
{=@4
I
2
the tangent lines are 1>
2
I
2
and 1> 3
, and the angle between them is given by
2
I
I
2@2 · 1> 3 2@2
13 1
1
1@2
I
=
cos = I
= t t2 =
1> 2@2 1> 3 2@2 3@2
3
3
3
1>
2
Thus = cos31
1
3
E 70=5 .
2
I
I
4 + 1 + 4 = 9 = 3, using Equations 8 and 9 we have cos = 23 , cos = 13 , and cos = 23 . The
direction angles are given by = cos31 23 E 48 , = cos31 13 E 71 , and = cos31 23 = 48 .
33. Since |h2> 1> 2i| =
I
I
36 + 9 + 4 = 49 = 7, using Equations 8 and 9 we have cos = 67 , cos = 37 , and cos =
The direction angles are given by = cos31 67 E 31 , = cos31 37 E 65 , and = cos31 3 27 = 107 .
34. Since |h6> 3> 32i| =
35. Since | i 3 2 j 3 3 k| =
= cos31
1
36. Since 2
I1
14
I
I
1 + 4 + 9 = 14, Equations 8 and 9 give cos =
I1 ,
14
1
3
32
I
,
14
and cos =
33
I
,
14
while
E 74 , = cos31 3 I214 E 122 , and = cos31 3 I314 E 143 .
t
t
i + j + k = 14 + 1 + 1 = 94 = 32 , Equations 8 and 9 give cos =
= cos31
cos =
32
7 .
E 71 and = = cos31
2
3
1@2
3@2
= 13 , cos = cos =
1
3@2
= 23 , while
E 48 .
I
I
f
1
f2 + f2 + f2 = 3f [since f A 0], so cos = cos = cos = I = I and
3f
3
= = = cos31 I13 E 55 .
37. |hf> f> fi| =
38. Since cos2 + cos2 + cos2 = 1, cos2 = 1 3 cos2 3 cos2 = 1 3 cos2
Thus cos = ± 12 and =
3
or =
2
.
3
4
3 cos2
3
=13
I 2
2
2
3
1 2
2
= 14 .
s
I
35 · 4 + 12 · 6
a·b
=
= 4 and the
(35)2 + 122 = 169 = 13. The scalar projection of b onto a is compa b =
|a|
13
a·b a
1
h35> 12i = 3 20
> 3 48
vector projection of b onto a is proja b =
.
= 4 · 13
13
13
|a|
|a|
39. |a| =
I
I
a·b
14
1·2+4·3
I
12 + 42 = 17. The scalar projection of b onto a is compa b =
= I and the vector
=
|a|
17
17
a·b a
.
= I1417 · I117 h1> 4i = 14
> 56
projection of b onto a is proja b =
17
17
|a|
|a|
40. |a| =
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SECTION 12.3 THE DOT PRODUCT
41. |a| =
I
a·b
=
9 + 36 + 4 = 7 so the scalar projection of b onto a is compa b =
|a|
projection of b onto a is proja b =
42. |a| =
9
7
·
1
7
h3> 6> 32i =
9
49
37 a
= 3 37
·
7
7 |a|
1
7
h32> 3> 36i = 3 37
h32> 3> 36i =
49
.
74
> 3 111
> 222
49
49
49
.
I
I
031+2
a·b
1
= I
4 + 1 + 16 = 21 so the scalar projection of b onto a is compa b =
= I while the vector
|a|
21
21
1
1 a
2i 3 j + 4k
I
= I ·
projection of b onto a is proja b = I
=
21 |a|
21
21
44. |a| =
(3 + 12 3 6) = 97 . The vector
27 54
> > 3 18
49 49
49
h3> 6> 32i =
257
I
a·b
4 + 9 + 36 = 7 so the scalar projection of b onto a is compa b =
, while the
= 17 (310 3 3 3 24) = 3 37
7
|a|
vector projection is proja b = 3
43. |a| =
9 a
=
7 |a|
1
7
¤
1
21 (2 i
3 j + 4 k) =
2
21
i3
1
21
j+
4
21
k.
I
I
a·b
1
131+1
I
1 + 1 + 1 = 3, so the scalar projection of b onto a is compa b =
= I while the vector
=
|a|
3
3
1 a
1 i+j+k
I
projection of b onto a is proja b = I
= 13 (i + j + k).
= I ·
|a|
3
3
3
45. (ortha b) · a = (b 3 proja b) · a = b · a 3 (proja b) · a = b · a 3
a·b
a·b 2
a·a=b·a3
|a| = b · a 3 a · b = 0.
|a|2
|a|2
So they are orthogonal by (7).
46. Using the formula in Exercise 45 and the result of Exercise 40, we have
ortha b = b 3 proja b = h2> 3i 3
14 56
17 > 17
=
20
5
17 > 3 17
.
I
I
a·b
= 2 C a · b = 2 |a| = 2 10. If b = he1 > e2 > e3 i, then we need 3e1 + 0e2 3 1e3 = 2 10.
|a|
I
I One possible solution is obtained by taking e1 = 0, e2 = 0, e3 = 32 10. In general, b = v> w> 3v 3 2 10 , v, w M R.
47. compa b =
48. (a) compa b = compb a
C
b·a
a·b
=
|a|
|b|
C
1
1
=
or a · b = 0 C |b| = |a| or a · b = 0.
|a|
|b|
That is, if a and b are orthogonal or if they have the same length.
a·b
b·a
a=
b C
|a|2
|b|2
(b) proja b = projb a C
But
b
a
2 =
|a|
|b|2
i
|a|
|b|
2 =
|a|
|b|2
So proja b = projb a C
a · b = 0 or
a
b
=
.
|a|2
|b|2
i |a| = |b|. Substituting this into the previous equation gives a = b.
a and b are orthogonal, or they are equal.
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258
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
49. The displacement vector is D = (6 3 0) i + (12 3 10) j + (20 3 8) k = 6 i + 2 j + 12 k so, by Equation 12, the work done is
Z = F · D = (8 i 3 6 j + 9 k) · (6 i + 2 j + 12 k) = 48 3 12 + 108 = 144 joules.
50. Here |D| = 1000 m, |F| = 1500 N, and = 30 . Thus
I I
Z = F · D = |F| |D| cos = (1500)(1000) 23 = 750,000 3 joules.
51. Here |D| = 80 ft, |F| = 30 lb, and = 40 . Thus
Z = F · D = |F| |D| cos = (30)(80) cos 40 = 2400 cos 40 E 1839 ft-lb.
52. Z = F · D = |F| |D| cos = (400)(120) cos 36 E 38,833 ft-lb
53. First note that n = hd> ei is perpendicular to the line, because if T1 = (d1 > e1 ) and T2 = (d2 > e2 ) lie on the line, then
333<
n · T1 T2 = dd2 3 dd1 + ee2 3 ee1 = 0, since dd2 + ee2 = 3f = dd1 + ee1 from the equation of the line.
Let S2 = ({2 > |2 ) lie on the line. Then the distance from S1 to the line is the absolute value of the scalar projection
333< |n · h{ 3 { > | 3 | i|
333<
|d{1 + e|1 + f|
|d{2 3 d{1 + e|2 3 e|1 |
2
1
2
1
I
I
=
=
of S1 S2 onto n. compn S1 S2 =
|n|
d2 + e2
d2 + e2
since d{2 + e|2 = 3f. The required distance is
|(3)(32) + (34)(3) + 5|
13
s
.
=
5
32 + (34)2
54. (r 3 a) · (r 3 b) = 0 implies that the vectors r 3 a and r 3 b are orthogonal.
From the diagram (in which D, E and U are the terminal points of the vectors),
we see that this implies that U lies on a sphere whose diameter is the line from
D to E. The center of this circle is the midpoint of DE, that is,
1
(a + b) = 12 (d1 + e1 ) > 12 (d2 + e2 ) > 12 (d3 + e3 ) , and its radius is
2
1
2
|a 3 b| =
1
2
s
(d1 3 e1 )2 + (d2 3 e2 )2 + (d3 3 e3 )2 .
Or: Expand the given equation, substitute r · r = {2 + | 2 + } 2 and complete the squares.
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 h1> 1> 1i.
The angle between this vector and the vector of the edge which also begins at the origin and runs along the {-axis [that is,
h1> 0> 0i] is given by cos =
1
h1> 1> 1i · h1> 0> 0i
= I
|h1> 1> 1i| |h1> 0> 0i|
3
i = cos31
I1
3
E 55 .
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SECTION 12.3 THE DOT PRODUCT
¤
259
56. Consider a cube with sides of unit length, wholly within the first octant and with edges along each of the three coordinate axes.
i + j + k and i + j are vector representations of a diagonal of the cube and a diagonal of one of its faces. If is the angle
u
t
2
1+1
(i + j + k) · (i + j)
= I I =
i = cos31 23 E 35 .
between these diagonals, then cos =
|i + j + k| |i + j|
3
3 2
57. Consider the H — C — H 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 3 12 > 0 3 12 > 0 3 12 = 12 > 3 12 > 3 12 and
0 3 12 > 1 3 12 > 0 3 12 = 3 12 > 12 > 3 12 . The bond angle, , is therefore given by
1
1
1
· 3 12 > 12 > 3 12
3 14 3 14 + 14
1
2> 32> 32
= t
t
cos = 1
=3
i = cos31 3 13 E 109=5 .
1
1 1 1
1 3
3
3
3
>
3
>
3
>
>
3
2
2
2
2 2
2
4
4
58. Let be the angle between a and c and be the angle between c and b. We need to show that = . Now
cos =
a·c
a · |a| b + a · |b| a
|a| a · b + |a|2 |b|
a · b + |a| |b|
=
=
=
. Similarly,
|a| |c|
|a| |c|
|a| |c|
|c|
cos =
|a| |b| + b · a
b·c
=
. Thus cos = cos . However 0 $ $ 180 and 0 $ $ 180 , so = and
|b| |c|
|c|
c bisects the angle between a and b.
59. Let a = hd1 > d2 > d3 i and = he1 > e2 > e3 i.
Property 2: a · b = hd1 > d2 > d3 i · he1 > e2 > e3 i = d1 e1 + d2 e2 + d3 e3
= e1 d1 + e2 d2 + e3 d3 = he1 > e2 > e3 i · hd1 > d2 > d3 i = b · a
Property 4: (f a) · b = hfd1 > fd2 > fd3 i · he1 > e2 > e3 i = (fd1 )e1 + (fd2 )e2 + (fd3 )e3
= f (d1 e1 + d2 e2 + d3 e3 ) = f (a · b) = d1 (fe1 ) + d2 (fe2 ) + d3 (fe3 )
= hd1 > d2 > d3 i · hfe1 > fe2 > fe3 i = a · (f b)
Property 5: 0 · a = h0> 0> 0i · hd1 > d2 > d3 i = (0)(d1 ) + (0)(d2 ) + (0)(d3 ) = 0
3<
33< 3<
3<
3
3
<
60. Let the figure be called quadrilateral DEFG. The diagonals can be represented by DF and EG. DF = DE + EF and
33< 3
3
< 33< 3
3
< 33< 3
3
< 3<
EG = EF + FG = EF 3 GF = EF 3 DE (Since opposite sides of the object are of the same length and parallel,
3< 33<
DE = GF.) Thus
3< 33< 3< 3
3
< 3
3
< 3< 3< 3
3
< 3< 33
< 3
3
< 3< DF · EG = DE + EF · EF 3 DE = DE · EF 3 DE + EF · EF 3 DE
3<2 3< 3
3
< 3
3
<2 3<2
3< 33
< 3<2 3
= DE · EF 3 DE + EF 3 DE · EF = EF 3 DE 3<2 3
<2
3< 33<
3
But DE = EF because all sides of the quadrilateral are equal in length. Therefore DF · EG = 0, and since both of
these vectors are nonzero this tells us that the diagonals of the quadrilateral are perpendicular.
61. |a · b| = |a| |b| cos = |a| |b| |cos |. Since |cos | $ 1, |a · b| = |a| |b| |cos | $ |a| |b|.
Note: We have equality in the case of cos = ±1, so = 0 or = , thus equality when a and b are parallel.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
62. (a)
The Triangle Inequality states that the length of the longest side of
a triangle is less than or equal to the sum of the lengths of the two
shortest sides.
(b) |a + b|2 = (a + b) · (a + b) = (a · a) + 2(a · b) + (b · b) = |a|2 + 2(a · b) + |b|2
$ |a|2 + 2 |a| |b| + |b|2
[by the Cauchy-Schwartz Inequality]
2
= (|a| + |b|)
Thus, taking the square root of both sides, |a + b| $ |a| + |b|.
63. (a)
The Parallelogram Law states that the sum of the squares of the
lengths of the diagonals of a parallelogram equals the sum of the
squares of its (four) sides.
(b) |a + b|2 = (a + b) · (a + b) = |a|2 + 2(a · b) + |b|2 and |a 3 b|2 = (a 3 b) · (a 3 b) = |a|2 3 2(a · b) + |b|2 .
Adding these two equations gives |a + b|2 + |a 3 b|2 = 2 |a|2 + 2 |b|2 .
64. If the vectors u + v and u 3 v are orthogonal then (u + v) · (u 3 v) = 0. But
(u + v) · (u 3 v) = (u + v) · u 3 (u + v) · v
by Property 3 of the dot product
= u·u+v·u3u·v3v·v
2
2
= |u| + u · v 3 u · v 3 |v|
by Property 3
by Properties 1 and 2
= |u|2 3 |v|2
Thus |u|2 3 |v|2 = 0 i |u|2 = |v|2
i |u| = |v| [since |u|, |v| D 0].
12.4 The Cross Product
i j
k 6 0
6 32 0 32 1. a × b = 6 0 32 = k
j + i 3 0 8
0
8
0
0
0 8
0
= [0 3 (316)] i 3 (0 3 0) j + (48 3 0) k = 16 i + 48 k
Now (a × b) · a = h16> 0> 48i · h6> 0> 32i = 96 + 0 3 96 = 0 and (a × b) · b = h16> 0> 48i · h0> 8> 0i = 0 + 0 + 0 = 0, so
a × b is orthogonal to both a and b.
i j
k 1 1
1 31 1 31 j
+
i
3
2. a × b = 1 1 31 = k
2 4
2
4
6
6
2 4
6
= [6 3 (34)] i 3 [6 3 (32)] j + (4 3 2) k = 10 i 3 8 j + 2 k
Now (a × b) · a = h10> 38> 2i · h1> 1> 31i = 10 3 8 3 2 = 0 and (a × b) · b = h10> 38> 2i · h2> 4> 6i = 20 3 32 + 12 = 0,
so a × b is orthogonal to both a and b.
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SECTION 12.4 THE CROSS PRODUCT
¤
i j
k 1 3
1 32 3 32 1
3
32
3. a × b = k
j + i 3 =
31 0 31
0
5
5
31 0
5
= (15 3 0) i 3 (5 3 2) j + [0 3 (33)] k = 15 i 3 3 j + 3 k
Since (a × b) · a = (15 i 3 3 j + 3 k) · (i + 3 j 3 2 k) = 15 3 9 3 6 = 0, a × b is orthogonal to a.
Since (a × b) · b = (15 i 3 3 j + 3 k) · (3i + 5 k) = 315 + 0 + 15 = 0, a × b is orthogonal to b.
i
j k 0
0 7
1 7
1 1 7=
4. a × b = 0
k
j + i 3 2 31 2 4
31 4 2 31 4 = [4 3 (37)] i 3 (0 3 14) j + (0 3 2) k = 11 i + 14 j 3 2 k
Since (a × b) · a = (11 i + 14 j 3 2 k) · (j + 7 k) = 0 + 14 3 14 = 0, a × b is orthogonal to a.
Since (a × b) · b = (11 i + 14 j 3 2 k) · (2 i 3 j + 4 k) = 22 3 14 3 8 = 0, a × b is orthogonal to b.
j k i
31 31 1 31 1 31 5. a × b = 1 31 31 = i 3 1
j + 1
k
1 1 1
1
2
2
2
2
1
1 1
2
2
= 3 12 3 (31) i 3 12 3 (3 12 ) j + 1 3 (3 12 ) k = 12 i 3 j +
Now (a × b) · a =
1
1
2
i3j+
3
2
k · (i 3 j 3 k) =
(a × b) · b = 2 i 3 j + 32 k · 12 i + j +
1
k =
2
1
4
1
2
+13
31+
3
4
3
2
3
2
k
= 0 and
= 0, so a × b is orthogonal to both a and b.
i
j
k w
w sin w cos w sin w cos w cos w sin w = 6. a × b = w
k
j + i 3 1 3 sin w 1 cos w 3 sin w cos w 1 3 sin w cos w = [cos2 w 3 (3 sin2 w)] i 3 (w cos w 3 sin w) j + (3w sin w 3 cos w) k = i + (sin w 3 w cos w) j + (3w sin w 3 cos w) k
Since
(a × b) · a = [ i + (sin w 3 w cos w) j + (3w sin w 3 cos w) k ] · (w i + cos w j + sin w k)
a × b is orthogonal to a.
= w + sin w cos w 3 w cos2 w 3 w sin2 w 3 sin w cos w
= w 3 w cos2 w + sin2 w = 0
Since
(a × b) · b = [ i + (sin w 3 w cos w) j + (3w sin w 3 cos w) k ] · (i 3 sin w j + cos w k)
a × b is orthogonal to b.
= 1 3 sin2 w + w sin w cos w 3 w sin w cos w 3 cos2 w
= 1 3 sin2 w + cos2 w = 0
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262
¤
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
i j k w 1@w w 1
1 1@w w
1
1@w
7. a × b = i 3 2
j + 2 2 k
= 2
w
w w w
1
1
w2 w2 1 = (1 3 w) i 3 (w 3 w) j + (w3 3 w2 ) k = (1 3 w) i + (w3 3 w2 ) k
Since (a × b) · a = 1 3 w> 0> w3 3 w2 · hw> 1> 1@wi = w 3 w2 + 0 + w2 3 w = 0, a × b is orthogonal to a.
Since (a × b) · b = 1 3 w> 0> w3 3 w2 · w2 > w2 > 1 = w2 3 w3 + 0 + w3 3 w2 = 0, a × b is orthogonal to b.
i j k
8. a × b = 1 0 32 0 1
1
1 0
1 32 0 32 =
k
j + i 3 0 1
0
1
1
1
= 2i 3 j + k
9. According to the discussion preceding Theorem 11, i × j = k, so (i × j) × k = k × k = 0 [by Example 2].
10. k × (i 3 2 j) = k × i + k × (32 j)
by Property 3 of Theorem 11
= k × i + (32) (k × j)
by Property 2 of Theorem 11
= j + (32)(3i) = 2 i + j
by the discussion preceding Theorem 11
11. (j 3 k) × (k 3 i) = (j 3 k) × k + (j 3 k) × (3i)
by Property 3 of Theorem 11
= j × k + (3k) × k + j × (3i) + (3k) × (3i)
by Property 4 of Theorem 11
= (j × k) + (31)(k × k) + (31)(j × i) + (31)2 (k × i)
by Property 2 of Theorem 11
= i + (31) 0 + (31)(3k) + j = i + j + k
by Example 2 and
the discussion preceeding Theorem 11
12. (i + j) × (i 3 j) = (i + j) × i + (i + j) × (3j)
by Property 3 of Theorem 11
= i × i + j × i + i × (3j) + j × (3j)
by Property 4 of Theorem 11
= (i × i) + (j × i) + (31)(i × j) + (31)(j × j)
by Property 2 of Theorem 11
= 0 + (3k) + (31) k + (31) 0 = 32 k
by Example 2 and
the discussion preceeding Theorem 11
13. (a) Since b × c is a vector, the dot product a · (b × c) is meaningful and is a scalar.
(b) b · c is a scalar, so a × (b · c) is meaningless, as the cross product is defined only for two vectors.
(c) Since b × c is a vector, the cross product a × (b × c) is meaningful and results in another vector.
(d) b · c is a scalar, so the dot product a · (b · c) is meaningless, as the dot product is defined only for two vectors.
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SECTION 12.4 THE CROSS PRODUCT
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263
(e) Since (a · b) and (c · d) are both scalars, the cross product (a · b) × (c · d) is meaningless.
(f ) a × b and c × d are both vectors, so the dot product (a × b) · (c × d) is meaningful and is a scalar.
14. Using Theorem 9, we have |u × v| = |u| |v| sin = (4)(5) sin 45 = 20 ·
I
I
2
= 10 2. By the right-hand rule, u × v is
2
directed out of the page.
15. If we sketch u and v starting from the same initial point, we see that the
angle between them is 60 . Using Theorem 9, we have
I
I
3
|u × v| = |u| |v| sin = (12)(16) sin 60 = 192 ·
= 96 3.
2
By the right-hand rule, u × v is directed into the page.
16. (a) |a × b| = |a| |b| sin = 3 · 2 · sin
2
=6
(b) a × b is orthogonal to k, so it lies in the {|-plane, and its }-coordinate is 0.
By the right-hand rule, its |-component is negative and its {-component
is positive.
i
j k 2 31 2 3
31 3 17. a × b = 2 31 3 = k = (31 3 6) i 3 (2 3 12) j + [4 3 (34)] k = 37 i + 10 j + 8 k
j + i 3 4
4 1
2 1
2
4
2 1
i
j k 4
2 1
4 1
2 2 1=
b×a = 4
k = [6 3 (31)] i 3 (12 3 2) j + (34 3 4) k = 7 i 3 10 j 3 8 k
j + i 3 2 31 2 3
31 3 2 31 3 Notice a × b = 3b × a here, as we know is always true by Property 1 of Theorem 11.
i j k 2 1
2 31 1 31 j
+
i
3
18. b × c = 2 1 31 = k = 4 i 3 6 j + 2 k so
0 1
0
1
3
3
0 1
3
i
j k 1
1 1
0 1
0 0 1=
a × (b × c) = 1
k = 6 i + 2 j 3 6 k.
j + i 3 4 36 4 2
36 2 4 36 2 i j k 1 0
1
0
1 1 1 = a×b=1 0
k = 3 i + 3 j + k so
j + i 3 2 1
2 31 1 31 2 1 31 i j k 31 3 31 1 3 1
(a × b) × c = 31 3 1 = k = 8 i + 3 j 3 k.
j + i 3 0 1
0 3
1 3
0 1 3
Thus a × (b × c) 6= (a × b) × c.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
19. By Theorem 8, the cross product of two vectors is orthogonal to both vectors. So we calculate
i
h3> 2> 1i × h31> 1> 0i = 3
31
j
2
1
k 2
1=
1
0
3
1 i 3 31
0
3
1 j + 31
0
2 k = 3i 3 j + 5 k.
1 H
G
h31> 31> 5i
h31> 31> 5i
1
1
5
I
I
I
So two unit vectors orthogonal to both are ± I
=±
>
3
>
, that is, 3 3I
3
3 3 3 3
1 + 1 + 25
3 3
G
H
1
1
5
and 3I
.
> I
> 3 3I
3 3 3
3
20. By Theorem 8, the cross product of two vectors is orthogonal to both vectors. So we calculate
i j k 0 1
0 31 1 31 (j 3 k) × (i + j) = 0 1 31 = k = i 3 j 3 k
j + i 3 1 1
1
1
0
0
1 1
0
Thus two unit vectors orthogonal to both given vectors are ± I13 (i 3 j 3 k), that is,
3 I13 i +
I1
3
j+
I1
3
I1
3
i3
I1
3
j3
1
I
3
k and
k.
21. Let a = hd1 > d2 > d3 i. Then
i j
0×a = 0 0
d d
1
2
i j
a × 0 = d1 d2
0 0
k 0 0 0 0 0 0 0 =
i 3 j + k = 0,
d2 d3 d1 d3 d1 d2 d3 k d1 d3 d1 d2 d2 d3 d3 = i 3 j + k = 0.
0 0 0 0 0 0 0
22. Let a = hd1 > d2 > d3 i and b = he1 > e2 > e3 i.
(a × b) · b =
.
-
d2 d3 d1 d3 d1 d2 d2 d3 d1 d3 d1 d2 >
>
·
he
e
e
>
e
>
e
i
=
3
+
1
2
e3
1 2 3
e2 e3 e1 e3 e1 e2 e2 e3 e1 e3 e1 e2 = (d2 e3 e1 3 d3 e2 e1 ) 3 (d1 e3 e2 3 d3 e1 e2 ) + (d1 e2 e3 3 d2 e1 e3 ) = 0
23. a × b = hd2 e3 3 d3 e2 > d3 e1 3 d1 e3 > d1 e2 3 d2 e1 i
= h(31)(e2 d3 3 e3 d2 ) > (31)(e3 d1 3 e1 d3 ) > (31)(e1 d2 3 e2 d1 )i
= 3 he2 d3 3 e3 d2 > e3 d1 3 e1 d3 > e1 d2 3 e2 d1 i = 3b × a
24. fa = hfd1 > fd2 > fd3 i, so
(f a) × b = hfd2 e3 3 fd3 e2 > fd3 e1 3 fd1 e3 > fd1 e2 3 fd2 e1 i
= fhd2 e3 3 d3 e2 > d3 e1 3 d1 e3 > d1 e2 3 d2 e1 i = f(a × b)
= hfd2 e3 3 fd3 e2 > fd3 e1 3 fd1 e3 > fd1 e2 3 fd2 e1 i
= hd2 (fe3 ) 3 d3 (fe2 ) > d3 (fe1 ) 3 d1 (fe3 ) > d1 (fe2 ) 3 d2 (fe1 )i
= a × fb
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SECTION 12.4 THE CROSS PRODUCT
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265
25. a × (b + c) = a × he1 + f1 > e2 + f2 > e3 + f3 i
= hd2 (e3 + f3 ) 3 d3 (e2 + f2 ) , d3 (e1 + f1 ) 3 d1 (e3 + f3 ) , d1 (e2 + f2 ) 3 d2 (e1 + f1 )i
= hd2 e3 + d2 f3 3 d3 e2 3 d3 f2 , d3 e1 + d3 f1 3 d1 e3 3 d1 f3 , d1 e2 + d1 f2 3 d2 e1 3 d2 f1 i
= h(d2 e3 3 d3 e2 ) + (d2 f3 3 d3 f2 ) , (d3 e1 3 d1 e3 ) + (d3 f1 3 d1 f3 ) , (d1 e2 3 d2 e1 ) + (d1 f2 3 d2 f1 )i
= hd2 e3 3 d3 e2 > d3 e1 3 d1 e3 > d1 e2 3 d2 e1 i + hd2 f3 3 d3 f2 > d3 f1 3 d1 f3 > d1 f2 3 d2 f1 i
= (a × b) + (a × c)
26. (a + b) × c = 3c × (a + b)
by Property 1 of Theorem 11
= 3(c × a + c × b)
by Property 3 of Theorem 11
= 3(3a × c + (3b × c))
by Property 1 of Theorem 11
=a×c+b×c
by Property 2 of Theorem 11
27. By plotting the vertices, we can see that the parallelogram is determined by the
3<
33<
vectors DE = h2> 3i and DG = h4> 32i. We know that the area of the parallelogram
determined by two vectors is equal to the length of the cross product of these vectors.
3<
In order to compute the cross product, we consider the vector DE as the three33<
dimensional vector h2> 3> 0i (and similarly for DG), and then the area of
parallelogram DEFG is
i
j
3< 33< 3
DE × DG = 2
4 32
k 0 = |(0) i 3 (0) j + (34 3 12) k| = |316 k| = 16
0 33<
33<
28. The parallelogram is determined by the vectors NO = h0> 1> 3i and NQ = h2> 5> 0i, so the area of parallelogram NOPQ is
i j k 33
I
< 33< NO × NQ = 0 1 3 = |(315) i 3 (36) j + (32) k| = |315 i + 6 j 3 2 k| = 265 E 16=28
2 5 0 3
3
<
3<
29. (a) Because the plane through S , T, and U contains the vectors S T and S U, a vector orthogonal to both of these vectors
3
3
<
3<
(such as their cross product) is also orthogonal to the plane. Here S T = h33> 1> 2i and S U = h3> 2> 4i, so
3
3
< 3<
S T × S U = h(1)(4) 3 (2)(2)> (2)(3) 3 (33)(4)> (33)(2) 3 (1)(3)i = h0> 18> 39i
Therefore, h0> 18> 39i (or any nonzero scalar multiple thereof, such as h0> 2> 31i) is orthogonal to the plane through S , T,
and U.
(b) Note that the area of the triangle determined by S , T, and U is equal to half of the area of the
parallelogram determined by the three points. From part (a), the area of the parallelogram is
3
< 3< I
I
I
3
S T × S U = |h0> 18> 39i| = 0 + 324 + 81 = 405 = 9 5, so the area of the triangle is
1
2
I
I
· 9 5 = 92 5.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
3
3
<
3<
30. (a) S T = h4> 2> 3i and S U = h3> 3> 4i, so a vector orthogonal to the plane through S , T, and U is
3
3
< 3<
S T × S U = h(2)(4) 3 (3)(3)> (3)(3) 3 (4)(4)> (4)(3) 3 (2)(3)i = h31> 37> 6i (or any nonzero scalar mutiple
thereof).
3
3
<
3< 3
3
< 3<
I
I
(b) The area of the parallelogram determined by S T and S U is S T × S U = |h31> 37> 6i| = 1 + 49 + 36 = 86,
so the area of triangle S TU is
3
3
<
1
2
I
86.
3<
31. (a) S T = h4> 3> 32i and S U = h5> 5> 1i, so a vector orthogonal to the plane through S , T, and U is
3
3
< 3<
S T × S U = h(3)(1) 3 (32)(5)> (32)(5) 3 (4)(1)> (4)(5) 3 (3)(5)i = h13> 314> 5i [or any scalar mutiple thereof ].
3
3
<
3<
(b) The area of the parallelogram determined by S T and S U is
3
s
< 3<
I
I
3
S T × S U = |h13> 314> 5i| = 132 + (314)2 + 52 = 390, so the area of triangle S TU is 12 390.
3
3
<
3<
32. (a) S T = h1> 2> 1i and S U = h5> 0> 32i, so a vector orthogonal to the plane through S , T, and U is
3
3
< 3<
S T × S U = h(2)(32) 3 (1)(0)> (1)(5) 3 (1)(32)> (1) (0) 3 (2)(5)i = h34> 7> 310i [or any scalar multiple thereof ].
3
3
<
3< 3
3
< 3<
I
I
(b) The area of the parallelogram determined by S T and S U is S T × S U = |h34> 7> 310i| = 16 + 49 + 100 = 165,
so the area of triangle S TU is
1
2
I
165.
33. By Equation 14, the volume of the parallelepiped determined by a, b, and c is the magnitude of their scalar triple product,
1 2 3
31 1 31 2 1 2
which is a · (b × c) = 31 1 2 = 1 = 1(4 3 2) 3 2(34 3 4) + 3(31 3 2) = 9.
+ 3
3 2
2 1
2 4
1 4
2 1 4
Thus the volume of the parallelepiped is 9 cubic units.
1 1 0
0 1
0 1
1 1
+
0
3
1
34. a · (b × c) = 0 1 1 = 1 = 0 + 1 + 0 = 1.
1 1
1 1
1 1
1 1 1
So the volume of the parallelepiped determined by a, b, and c is 1 cubic unit.
3
3
<
3<
3<
35. a = S T = h4> 2> 2i, b = S U = h3> 3> 31i, and c = S V = h5> 5> 1i.
4 2
2
3 3
3 31 3 31 a · (b × c) = 3 3 31 = 4 = 32 3 16 + 0 = 16,
+ 2
3 2
5 5
5
5
1
1
5 5
1
so the volume of the parallelepiped is 16 cubic units.
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SECTION 12.4 THE CROSS PRODUCT
3
3
<
3<
¤
267
3<
36. a = S T = h34> 2> 4i, b = S U = h2> 1> 32i and c = S V = h33> 4> 1i.
34
a · (b × c) = 2
33
4
2 1
2 32 1 32 1 32 = 34 = 336 + 8 + 44 = 16, so the volume of the
+ 4
3 2
33 4 33 1 4 1
4 1
2
parallelepiped is 16 cubic units.
1 5 32 3 31 3 0
31 0 37. u · (v × w) = 3 31 0 = 1 = 4 + 60 3 64 = 0, which says that the volume
+ (32) 3 5
5 9
5 34 9 34 5 9 34 of the parallelepiped determined by u, v and w is 0, and thus these three vectors are coplanar.
3<
3<
33<
38. u = DE = h2> 34> 4i, v = DF = h4> 31> 32i and w = DG = h2> 3> 36i.
2 34
4
4 31 4 32 31 32 +
4
3
(34)
u · (v × w) = 4 31 32 = 2 = 24 3 80 + 56 = 0, so the volume of the
2 3
2 36 3 36 2 3 36 parallelepiped determined by u, v and w is 0, which says these vectors lie in the same plane. Therefore, their initial and
terminal points D, E, F and G also lie in the same plane.
39. The magnitude of the torque is | | = |r × F| = |r| |F| sin = (0=18 m)(60 N) sin(70 + 10) = 10=8 sin 80 E 10=6 N·m.
40. |r| =
I
I
42 + 42 = 4 2 ft. A line drawn from the point S to the point of application of the force makes an angle of
180 3 (45 + 30) = 105 with the force vector. Therefore,
I | | = |r × F| = |r| |F| sin = 4 2 (36) sin 105 E 197 ft-lb.
41. Using the notation of the text, r = h0> 0=3> 0i and F has direction h0> 3> 34i. The angle between them can be determined by
cos =
h0> 0=3> 0i · h0> 3> 34i
|h0> 0=3> 0i| |h0> 3> 34i|
100 = 0=3 |F| sin 53=1
i cos =
0=9
(0=3)(5)
i cos = 0=6 i E 53=1 . Then | | = |r| |F| sin i
i |F| E 417 N.
42. Since |u × v| = |u| |v| sin , 0 $ $ , |u × v| achieves its maximum value for sin = 1
i =
,
2
in which case
|u × v| = |u| |v| = 15= The minimum value is zero, which occurs when sin = 0 i = 0 or , so when u, v are
parallel. Thus, when u points in the same direction as v, so u = 3 j, |u × v| = 0. As u rotates counterclockwise, u × v is
directed in the negative }-direction (by the right-hand rule) and the length increases until =
2,
in which case u = 33 i and
|u × v| = 15. As u rotates to the negative |-axis, u × v remains pointed in the negative }-direction and the length of u × v
decreases to 0> after which the direction of u × v reverses to point in the positive }-direction and |u × v| increases. When
u = 3 i (so =
2 ),
|u × v| again reaches its maximum of 15, after which |u × v| decreases to 0 as u rotates to the positive
|-axis.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
43. From Theorem 9 we have |a × b| = |a| |b| sin , where is the angle between a and b, and from Theorem 12.3.3 we have
a · b = |a| |b| cos i
|a| |b| =
a·b
a·b
. Substituting the second equation into the first gives |a × b| =
sin , so
cos cos I
I
|a × b|
3
|a × b|
= tan . Here |a × b| = |h1> 2> 2i| = 1 + 4 + 4 = 3, so tan =
= I = 3
a·b
a·b
3
i
= 60 .
44. (a) Let v = hy1 > y2> y3 i. Then
i j k 2 1 1 1 1 2 h1> 2> 1i × v = 1 2 1 = i 3 j + k = (2y3 3 y2 ) i 3 (y3 3 y1 ) j + (y2 3 2y1 ) k.
y2 y3 y1 y3 y1 y2 y y y 1
2
3
If h1> 2> 1i × v = h3> 1> 35i then h2y3 3 y2 > y1 3 y3 > y2 3 2y1 i = h3> 1> 35i C 2y3 3 y2 = 3 (1), y1 3 y3 = 1 (2),
and y2 3 2y1 = 35 (3). From (3) we have y2 = 2y1 3 5 and from (2) we have y3 = y1 3 1; substitution into (1) gives
2 (y1 3 1) 3 (2y1 3 5) = 3
i
3 = 3, so this is a dependent system. If we let y1 = d then y2 = 2d 3 5 and
y3 = d 3 1, so v is any vector of the form hd> 2d 3 5> d 3 1i.
(b) If h1> 2> 1i × v = h3> 1> 5i then 2y3 3 y2 = 3 (1), y1 3 y3 = 1 (2), and y2 3 2y1 = 5 (3). From (3) we have
y2 = 2y1 + 5 and from (2) we have y3 = y1 3 1; substitution into (1) gives 2 (y1 3 1) 3 (2y1 + 5) = 3
i
37 = 3,
so this is an inconsistent system and has no solution.
Alternatively, if we use matrices to solve the system we could show that the determinant is 0 (and hence the system has no
solution).
45. (a)
The distance between a point and a line is the length of the perpendicular
3<
from the point to the line, here S V = g. But referring to triangle S TV,
3< 3
<
3
3
<
3
g = S V = TS sin = |b| sin . But is the angle between TS = b
3<
|a × b|
and TU = a. Thus by Theorem 9, sin =
|a| |b|
and so g = |b| sin =
|a × b|
|b| |a × b|
=
.
|a| |b|
|a|
3<
3
3
<
(b) a = TU = h31> 32> 31i and b = TS = h1> 35> 37i. Then
a × b = h(32)(37) 3 (31)(35)> (31)(1) 3 (31)(37)> (31)(35) 3 (32)(1)i = h9> 38> 7i.
Thus the distance is g =
|a × b|
=
|a|
I1
6
t
t
I
97
81 + 64 + 49 = 194
=
.
6
3
46. (a) The distance between a point and a plane is the length of the perpendicular from
3<
3<
the point to the plane, here W S = g. But W S is parallel to b × a (because
3<
b × a is perpendicular to b and a) and g = W S = the absolute value of the
scalar projection of c along b × a, which is |c| |cos |. (Notice that this is the same
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SECTION 12.4 THE CROSS PRODUCT
setup as the development of the volume of a parallelepiped with k = |c| |cos |). Thus g = |c| |cos | = k = Y @D
where D = |a × b|, the area of the base. So finally g =
|a · (b × c)|
Y
=
.
D
|a × b|
3<
3<
3
3
<
(b) a = TU = h31> 2> 0i, b = TV = h31> 0> 3i and c = TS = h1> 1> 4i. Then
31 2 0 31 3 0 3
a · (b × c) = 31 0 3 = (31) + 0 = 17
3 2
1 4
1 4
1 1 4
and
Thus g =
i j k 31 2 31 0 2 0
a × b = 31 2 0 = k = 6i + 3j + 2k
j + i 3 31 0 31 3 0 3
31 0 3 |a · (b × c)|
17
17
=
= I
.
|a × b|
7
36 + 9 + 4
47. From Theorem 9 we have |a × b| = |a| |b| sin so
|a × b|2 = |a|2 |b|2 sin2 = |a|2 |b|2 1 3 cos2 = |a|2 |b|2 3 (|a| |b| cos )2 = |a|2 |b|2 3 (a · b)2
by Theorem 12.3.3.
48. If a + b + c = 0 then b = 3 (a + c), so
a × b = a × [3 (a + c)] = 3[a × (a + c)]
by Property 2 of Theorem 11 (with f = 31)
= 3 [(a × a) + (a × c)]
by Property 3 of Theorem 11
= 3 [0 + (a × c)] = 3a × c
by Example 2
=c×a
by Property 1 of Theorem 11
Similarly, a = 3 (b + c) so
c × a = c × [3 (b + c)] = 3[c × (b + c)]
= 3 [(c × b) + (c × c)] = 3 [(c × b) + 0]
= 3c × b = b × c
Thus a × b = b × c = c × a.
49. (a 3 b) × (a + b) = (a 3 b) × a + (a 3 b) × b
by Property 3 of Theorem 11
= a × a + (3b) × a + a × b + (3b) × b
by Property 4 of Theorem 11
= (a × a) 3 (b × a) + (a × b) 3 (b × b)
by Property 2 of Theorem 11 (with f = 31)
= 0 3 (b × a) + (a × b) 3 0
by Example 2
= (a × b) + (a × b)
by Property 1 of Theorem 11
= 2(a × b)
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
50. Let a = hd1 > d2 > d3 i, b = he1 > e2 > e3 i and c = hf1 > f2 > f3 i, so b × c = he2 f3 3 e3 f2 > e3 f1 3 e1 f3 > e1 f2 3 e2 f1 i and
a × (b × c) = hd2 (e1 f2 3 e2 f1 ) 3 d3 (e3 f1 3 e1 f3 ), d3 (e2 f3 3 e3 f2 ) 3 d1 (e1 f2 3 e2 f1 ),
d1 (e3 f1 3 e1 f3 ) 3 d2 (e2 f3 3 e3 f2 )i
= hd2 e1 f2 3 d2 e2 f1 3 d3 e3 f1 + d3 e1 f3 , d3 e2 f3 3 d3 e3 f2 3 d1 e1 f2 + d1 e2 f1 >
d1 e3 f1 3 d1 e1 f3 3 d2 e2 f3 + d2 e3 f2 i
= h(d2 f2 + d3 f3 )e1 3 (d2 e2 + d3 e3 )f1 , (d1 f1 + d3 f3 )e2 3 (d1 e1 + d3 e3 )f2 ,
(d1 f1 + d2 f2 )e3 3 (d1 e1 + d2 e2 )f3 i
(B) = h(d2 f2 + d3 f3 )e1 3 (d2 e2 + d3 e3 )f1 + d1 e1 f1 3 d1 e1 f1 ,
(d1 f1 + d3 f3 )e2 3 (d1 e1 + d3 e3 )f2 + d2 e2 f2 3 d2 e2 f2 ,
(d1 f1 + d2 f2 )e3 3 (d1 e1 + d2 e2 )f3 + d3 e3 f3 3 d3 e3 f3 i
= h(d1 f1 + d2 f2 + d3 f3 )e1 3 (d1 e1 + d2 e2 + d3 e3 )f1 ,
(d1 f1 + d2 f2 + d3 f3 )e2 3 (d1 e1 + d2 e2 + d3 e3 )f2 ,
(d1 f1 + d2 f2 + d3 f3 )e3 3 (d1 e1 + d2 e2 + d3 e3 )f3 i
= (d1 f1 + d2 f2 + d3 f3 ) he1 > e2 > e3 i 3 (d1 e1 + d2 e2 + d3 e3 ) hf1 > f2 > f3 i
= (a · c)b 3 (a · b)c
(B) Here we look ahead to see what terms are still needed to arrive at the desired equation. By adding and subtracting the
same terms, we don’t change the value of the component.
51. a × (b × c) + b × (c × a) + c × (a × b)
= [(a · c)b 3 (a · b)c] + [(b · a)c 3 (b · c)a] + [(c · b)a 3 (c · a)b]
by Exercise 50
= (a · c)b 3 (a · b)c + (a · b)c 3 (b · c)a + (b · c)a 3 (a · c)b = 0
52. Let c × d = v. Then
(a × b) · (c × d) = (a × b) · v = a · (b × v)
by Property 5 of Theorem 11
= a · [b × (c × d)] = a · [(b · d)c 3 (b · c)d]
by Exercise 50
= (b · d)(a · c) 3 (b · c)(a · d)
a· c b · c
=
a · d b · d
by Properties 3 and 4 of the dot product
53. (a) No. If a · b = a · c, then a · (b 3 c) = 0, so a is perpendicular to b 3 c, which can happen if b 6= c. For example,
let a = h1> 1> 1i, b = h1> 0> 0i and c = h0> 1> 0i.
(b) No. If a × b = a × c then a × (b 3 c) = 0, which implies that a is parallel to b 3 c, which of course can happen
if b 6= c.
(c) Yes. Since a · c = a · b, a is perpendicular to b 3 c, by part (a). From part (b), a is also parallel to b 3 c. Thus since
a 6= 0 but is both parallel and perpendicular to b 3 c, we have b 3 c = 0, so b = c.
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DISCOVERY PROJECT THE GEOMETRY OF A TETRAHEDRON
¤
271
54. (a) kl is perpendicular to vl if l 6= m by the definition of kl and Theorem 8.
(b) k1 · v1 =
v1 · (v2 × v3 )
v2 × v3
· v1 =
=1
v1 · (v2 × v3 )
v1 · (v2 × v3 )
k2 · v2 =
v3 × v1
v2 · (v3 × v1 )
(v2 × v3 ) · v1
· v2 =
=
=1
v1 · (v2 × v3 )
v1 · (v2 × v3 )
v1 · (v2 × v3 )
k3 · v3 =
(v1 × v2 ) · v3
v1 · (v2 × v3 )
=
=1
v1 · (v2 × v3 )
v1 · (v2 × v3 )
(c) k1 · (k2 × k3 ) = k1 ·
=
[by Property 5 of Theorem 11]
[by Property 5 of Theorem 11]
v3 × v1
v1 × v2
×
v1 · (v2 × v3 )
v1 · (v2 × v3 )
=
k1
· [(v3 × v1 ) × (v1 × v2 )]
[v1 · (v2 × v3 )]2
k1
· ([(v3 × v1 ) · v2 ] v1 3 [(v3 × v1 ) · v1 ] v2 )
[v1 · (v2 × v3 )]2
[by Exercise 50]
But (v3 × v1 ) · v1 = 0 since v3 × v1 is orthogonal to v1 , and
(v3 × v1 ) · v2 = v2 · (v3 × v1 ) = (v2 × v3 ) · v1 = v1 · (v2 × v3 ). Thus
k1 · (k2 × k3 ) =
1
k1
k1 · v1
=
· [v1 · (v2 × v3 )] v1 =
v1 · (v2 × v3 )
v1 · (v2 × v3 )
[v1 · (v2 × v3 )]2
[by part (b)]
DISCOVERY PROJECT The Geometry of a Tetrahedron
1. Set up a coordinate system so that vertex V is at the origin, U = (0> |1 > 0), T = ({2 > |2 > 0), S = ({3 > |3 > }3 ).
3<
3<
3<
3<
3
3
<
Then VU = h0> |1 > 0i, VT = h{2 > |2 > 0i, VS = h{3 > |3 > }3 i, TU = h3{2 > |1 3 |2 > 0i, and TS = h{3 3 {2 > |3 3 |2 > }3 i.
Let
3< 3
3
<
vV = TU × TS = (|1 }3 3 |2 }3 ) i + {2 }3 j + (3{2 |3 3 {3 |1 + {3 |2 + {2 |1 ) k
Then vV is an outward normal to the face opposite vertex V. Similarly,
3< 3<
3< 3<
vU = VT × VS = |2 }3 i 3 {2 }3 j + ({2 |3 3 {3 |2 ) k, vT = VS × VU = 3|1 }3 i + {3 |1 k, and
3< 3<
vS = VU × VT = 3{2 |1 k i
vV + vU + vT + vS = 0. Now
3<
3
3
<
|vV | = area of the parallelogram determined by TU and TS
= 2 (area of triangle UTS ) = 2|v1 |
So vV = 2v1 , and similarly vU = 2v2 , vT = 2v3 , vS = 2v4 . Thus v1 + v2 + v3 + v4 = 0.
2. (a) Let V = ({0 > |0 > }0 ), U = ({1 > |1 > }1 ), T = ({2 > |2 > }2 ), S = ({3 > |3 > }3 ) be the four vertices. Then
Volume = 13 (distance from V to plane UTS ) × (area of triangle UTS )
3<
3< 3<
N · VU
· 12 UT × US = 13
|N|
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
3< 3<
where N is a vector which is normal to the face UTS . Thus N = UT × US . Therefore
{ 3 { |0 3 |1 }0 3 }1 3< 3< 3< 1 0
Y = 16 UT × US · VU = {2 3 {1 |2 3 |1 }2 3 }1 6 {3 3 {1 |3 3 |1 }3 3 }1 1 3 1
132
1 132
(b) Using the formula from part (a), Y = 1 3 1
6 3 3 1 31 3 2
1 3 3 1
1
2 3 3 = 2(1 3 2) = .
6
3
2 3 3 3. We define a vector v1 to have length equal to the area of the face opposite vertex S , so we can say |v1 | = D, and direction
perpendicular to the face and pointing outward, as in Problem 1. Similarly, we define v2 , v3 , and v4 so that |v2 | = E,
|v3 | = F, and |v4 | = G and with the analogous directions. From Problem 1, we know v1 + v2 + v3 + v4 = 0 i
v4 = 3 (v1 + v2 + v3 ) i |v4 | = |3 (v1 + v2 + v3 )| = |v1 + v2 + v3 | i |v4 |2 = |v1 + v2 + v3 |2
i
v4 · v4 = (v1 + v2 + v3 ) · (v1 + v2 + v3 )
= v1 · v1 + v1 · v2 + v1 · v3 + v2 · v1 + v2 · v2 + v2 · v3 + v3 · v1 + v3 · v2 + v3 · v3
Since the vertex V is trirectangular, we know the three faces meeting at V are mutually perpendicular, so the vectors
v1 , v2 , v3 are also mutually perpendicular. Therefore, vl · vm = 0 for l 6= m and l, m M {1> 2> 3}. Thus we have
v4 · v4 = v1 · v1 + v2 · v2 + v3 · v3
i |v4 |2 = |v1 |2 + |v2 |2 + |v3 |2
i G2 = D2 + E 2 + F 2 .
Another method: We introduce a coordinate system, as shown. Recall that
the area of the parallelogram spanned by two vectors is equal to the length
of their cross product, so since
u × v = h3t> u> 0i × h3t> 0> si = hsu> st> tui, we have
s
|u × v| = (su)2 + (st)2 + (tu)2 , and therefore
2
|u × v| = 14 [(su)2 + (st)2 + (tu)2 ]
2 2 2
= 12 su + 12 st + 12 tu = D2 + E 2 + F 2 .
G2 =
1
2
A third method: We draw a line from V perpendicular to TU, as shown.
Now G = 12 fk, so G2 = 14 f2 k2 . Substituting k2 = s2 + n2 , we get
G2 = 14 f2 s2 + n2 = 14 f2 s2 + 14 f2 n2 . But F = 12 fn, so
G2 = 14 f2 s2 + F 2 . Now substituting f2 = t 2 + u2 gives
G2 = 14 s2 t 2 + 14 t 2 u2 + F 2 = D2 + E 2 + F 2 .
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SECTION 12.5 EQUATIONS OF LINES AND PLANES
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273
12.5 Equations of Lines and Planes
1. (a) True; each of the first two lines has a direction vector parallel to the direction vector of the third line, so these vectors are
each scalar multiples of the third direction vector. Then the first two direction vectors are also scalar multiples of each
other, so these vectors, and hence the two lines, are parallel.
(b) False; for example, the {- and |-axes are both perpendicular to the }-axis, yet the {- and |-axes are not parallel.
(c) True; each of the first two planes has a normal vector parallel to the normal vector of the third plane, so these two normal
vectors are parallel to each other and the planes are parallel.
(d) False; for example, the {|- and |}-planes are not parallel, yet they are both perpendicular to the {}-plane.
(e) False; the {- and |-axes are not parallel, yet they are both parallel to the plane } = 1.
(f ) True; if each line is perpendicular to a plane, then the lines’ direction vectors are both parallel to a normal vector for the
plane. Thus, the direction vectors are parallel to each other and the lines are parallel.
(g) False; the planes | = 1 and } = 1 are not parallel, yet they are both parallel to the {-axis.
(h) True; if each plane is perpendicular to a line, then any normal vector for each plane is parallel to a direction vector for the
line. Thus, the normal vectors are parallel to each other and the planes are parallel.
(i) True; see Figure 9 and the accompanying discussion.
( j) False; they can be skew, as in Example 3.
(k) True. Consider any normal vector for the plane and any direction vector for the line. If the normal vector is perpendicular
to the direction vector, the line and plane are parallel. Otherwise, the vectors meet at an angle , 0 $ ? 90 , and the
line will intersect the plane at an angle 90 3 .
2. For this line, we have r0 = 6 i 3 5 j + 2 k and v = i + 3 j 3
r = r0 + w v = (6 i 3 5 j + 2 k) + w i + 3 j 3
{ = 6 + w, | = 35 + 3w, } = 2 3 23 w.
2
3
k, so a vector equation is
2
k = (6 + w) i + (35 + 3w) j + 2 3 23 w k and parametric equations are
3
3. For this line, we have r0 = 2 i + 2=4 j + 3=5 k and v = 3 i + 2 j 3 k, so a vector equation is
r = r0 + w v = (2 i + 2=4 j + 3=5 k) + w(3 i + 2 j 3 k) = (2 + 3w) i + (2=4 + 2w) j + (3=5 3 w) k and parametric equations are
{ = 2 + 3w, | = 2=4 + 2w, } = 3=5 3 w.
4. This line has the same direction as the given line, v = 2 i 3 3 j + 9 k. Here r0 = 14 j 3 10 k, so a vector equation is
r = (14 j 3 10 k) + w(2 i 3 3 j + 9 k) = 2w i + (14 3 3w) j + (310 + 9w) k and parametric equations are { = 2w,
| = 14 3 3w, } = 310 + 9w.
5. A line perpendicular to the given plane has the same direction as a normal vector to the plane, such as
n = h1> 3> 1i. So r0 = i + 6 k, and we can take v = i + 3 j + k. Then a vector equation is
r = (i + 6 k) + w(i + 3 j + k) = (1 + w) i + 3w j + (6 + w) k, and parametric equations are { = 1 + w, | = 3w, } = 6 + w.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
6. The vector v = h4 3 0> 3 3 0> 31 3 0i = h4> 3> 31i is parallel to the line. Letting S0 = (0> 0> 0), parametric equations are
{ = 0 + 4 · w = 4w, | = 0 + 3 · w = 3w, } = 0 + (31) · w = 3w, while symmetric equations are
{
|
= = 3}.
4
3
{
|
}
= =
or
4
3
31
7. The vector v = 2 3 0> 1 3 12 > 33 3 1 = 2> 12 > 34 is parallel to the line. Letting S0 = (2> 1> 33), parametric equations
are { = 2 + 2w, | = 1 + 12 w, } = 33 3 4w, while symmetric equations are
|31
}+3
{32
=
=
or
2
1@2
34
{32
}+3
= 2| 3 2 =
.
2
34
8. v = h2=6 3 1=0> 1=2 3 2=4> 0=3 3 4=6i = h1=6> 31=2> 34=3i, and letting S0 = (1=0> 2=4> 4=6), parametric equations are
{ = 1=0 + 1=6w, | = 2=4 3 1=2w, } = 4=6 3 4=3w, while symmetric equations are
{ 3 1=0
| 3 2=4
} 3 4=6
=
=
.
1=6
31=2
34=3
9. v = h3 3 (38)> 32 3 1> 4 3 4i = h11> 33> 0i, and letting S0 = (38> 1> 4), parametric equations are { = 38 + 11w,
| = 1 3 3w, } = 4 + 0w = 4, while symmetric equations are
f = 0, so rather than writing
{+8
|31
=
, } = 4. Notice here that the direction number
11
33
}34
in the symmetric equation we must write the equation } = 4 separately.
0
i j k
10. v = (i + j) × ( j + k) = 1 1 0 = i 3 j + k is the direction of the line perpendicular to both i + j and j + k.
0 1 1
With S0 = (2> 1> 0), parametric equations are { = 2 + w, | = 1 3 w, } = w and symmetric equations are { 3 2 =
|31
=}
31
or { 3 2 = 1 3 | = }.
11. The line has direction v = h1> 2> 1i. Letting S0 = (1> 31> 1), parametric equations are { = 1 + w, | = 31 + 2w, } = 1 + w
and symmetric equations are { 3 1 =
|+1
= } 3 1.
2
12. Setting } = 0 we see that (1> 0> 0) satisfies the equations of both planes, so they do in fact have a line of intersection.
The line is perpendicular to the normal vectors of both planes, so a direction vector for the line is
v = n1 × n2 = h1> 2> 3i × h1> 31> 1i = h5> 2> 33i. Taking the point (1> 0> 0) as S0 , parametric equations are { = 1 + 5w,
| = 2w, } = 33w, and symmetric equations are
{31
|
}
= =
.
5
2
33
13. Direction vectors of the lines are v1 = h32 3 (34)> 0 3 (36)> 33 3 1i = h2> 6> 34i and
v2 = h5 3 10> 3 3 18> 14 3 4i = h35> 315> 10i, and since v2 = 3 52 v1 , the direction vectors and thus the lines are parallel.
14. Direction vectors of the lines are v1 = h3> 33> 1i and v2 = h1> 34> 312i. Since v1 · v2 = 3 + 12 3 12 6= 0, the vectors and
thus the lines are not perpendicular.
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SECTION 12.5 EQUATIONS OF LINES AND PLANES
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275
15. (a) The line passes through the point (1> 35> 6) and a direction vector for the line is h31> 2> 33i, so symmetric equations for
the line are
|+5
}36
{31
=
=
.
31
2
33
(b) The line intersects the {|-plane when } = 0, so we need
{31
|+5
036
{31
=
=
or
= 2 i { = 31,
31
2
33
31
|+5
= 2 i | = 31. Thus the point of intersection with the {|-plane is (31> 31> 0). Similarly for the |}-plane,
2
we need { = 0 i
1=
}36
|+5
=
2
33
the {}-plane, we need | = 0 i
at 3 32 > 0> 3 32 .
i | = 33, } = 3. Thus the line intersects the |}-plane at (0> 33> 3). For
5
}36
{31
= =
31
2
33
i
{ = 3 32 , } = 3 32 . So the line intersects the {}-plane
16. (a) A vector normal to the plane { 3 | + 3} = 7 is n = h1> 31> 3i, and since the line is to be perpendicular to the plane, n is
also a direction vector for the line. Thus parametric equations of the line are { = 2 + w, | = 4 3 w, } = 6 + 3w.
(b) On the {|-plane, } = 0. So } = 6 + 3w = 0 i w = 32 in the parametric equations of the line, and therefore { = 0
and | = 6, giving the point of intersection (0> 6> 0). For the |}-plane, { = 0 so we get the same point of interesection:
(0> 6> 0). For the {}-plane, | = 0 which implies w = 4, so { = 6 and } = 18 and the point of intersection is (6> 0> 18).
17. From Equation 4, the line segment from r0 = 2 i 3 j + 4 k to r1 = 4 i + 6 j + k is
r(w) = (1 3 w) r0 + w r1 = (1 3 w)(2 i 3 j + 4 k) + w(4 i + 6 j + k) = (2 i 3 j + 4 k) + w(2 i + 7 j 3 3 k), 0 $ w $ 1.
18. From Equation 4, the line segment from r0 = 10 i + 3 j + k to r1 = 5 i + 6 j 3 3 k is
r(w) = (1 3 w) r0 + w r1 = (1 3 w)(10 i + 3 j + k) + w(5 i + 6 j 3 3 k)
= (10 i + 3 j + k) + w(35 i + 3 j 3 4 k), 0 $ w $ 1.
The corresponding parametric equations are { = 10 3 5w, | = 3 + 3w, } = 1 3 4w, 0 $ w $ 1.
19. Since the direction vectors h2> 31> 3i and h4> 32> 5i are not scalar multiples of each other, the lines aren’t parallel. For the
lines to intersect, we must be able to find one value of w and one value of v that produce the same point from the respective
parametric equations. Thus we need to satisfy the following three equations: 3 + 2w = 1 + 4v, 4 3 w = 3 3 2v,
1 + 3w = 4 + 5v. Solving the last two equations we get w = 1, v = 0 and checking, we see that these values don’t satisfy the
first equation. Thus the lines aren’t parallel and don’t intersect, so they must be skew lines.
20. Since the direction vectors are v1 = h312> 9> 33i and v2 = h8> 36> 2i, we have v1 = 3 32 v2 so the lines are parallel.
21. Since the direction vectors h1> 32> 33i and h1> 3> 37i aren’t scalar multiples of each other, the lines aren’t parallel. Parametric
equations of the lines are O1 : { = 2 + w, | = 3 3 2w, } = 1 3 3w and O2 : { = 3 + v, | = 34 + 3v, } = 2 3 7v. Thus, for the
lines to intersect, the three equations 2 + w = 3 + v, 3 3 2w = 34 + 3v, and 1 3 3w = 2 3 7v must be satisfied simultaneously.
Solving the first two equations gives w = 2, v = 1 and checking, we see that these values do satisfy the third equation, so the
lines intersect when w = 2 and v = 1, that is, at the point (4> 31> 35).
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
22. The direction vectors h1> 31> 3i and h2> 32> 7i are not parallel, so neither are the lines. Parametric equations for the lines are
O1 : { = w, | = 1 3 w, } = 2 + 3w and O2 : { = 2 + 2v, | = 3 3 2v, } = 7v. Thus, for the lines to interesect, the three
equations w = 2 + 2v, 1 3 w = 3 3 2v, and 2 + 3w = 7v must be satisfied simultaneously. Solving the last two equations gives
w = 310, v = 34 and checking, we see that these values don’t satisfy the first equation. Thus the lines aren’t parallel and
don’t intersect, so they must be skew.
23. Since the plane is perpendicular to the vector h1> 32> 5i, we can take h1> 32> 5i as a normal vector to the plane.
(0> 0> 0) is a point on the plane, so setting d = 1, e = 32, f = 5 and {0 = 0, |0 = 0, }0 = 0 in Equation 7 gives
1({ 3 0) + (32)(| 3 0) + 5(} 3 0) = 0 or { 3 2| + 5} = 0 as an equation of the plane.
24. 2 i + j 3 k = h2> 1> 31i is a normal vector to the plane and (5> 3> 5) is a point on the plane, so setting d = 2, e = 1, f = 31>
{0 = 5, |0 = 3, }0 = 5 in Equation 7 gives 2({ 3 5) + 1(| 3 3) + (31)(} 3 5) = 0 or 2{ + | 3 } = 8 as an equation of the
plane.
25. i + 4 j + k = h1> 4> 1i is a normal vector to the plane and 31> 12 > 3 is a point on the plane, so setting d = 1, e = 4, f = 1>
{0 = 31, |0 = 12 , }0 = 3 in Equation 7 gives 1[{ 3 (31)] + 4 | 3 12 + 1(} 3 3) = 0 or { + 4| + } = 4 as an equation of
the plane.
26. Since the line is perpendicular to the plane, its direction vector h3> 31> 4i is a normal vector to the plane. The point (2> 0> 1) is
on the plane, so an equation of the plane is 3({ 3 2) + (31)(| 3 0) + 4(} 3 1) = 0 or 3{ 3 | + 4} = 10.
27. Since the two planes are parallel, they will have the same normal vectors. So we can take n = h5> 31> 31i, and an equation of
the plane is 5({ 3 1) 3 1[| 3 (31)] 3 1[} 3 (31)] = 0 or 5{ 3 | 3 } = 7.
28. Since the two planes are parallel, they will have the same normal vectors. A normal vector for the plane } = { + | or
{ + | 3 } = 0 is n = h1> 1> 31i, and an equation of the desired plane is 1({ 3 2) + 1(| 3 4) 3 1(} 3 6) = 0 or
{ + | 3 } = 0 (the same plane!).
29. Since the two planes are parallel, they will have the same normal vectors. So we can take n = h1> 1> 1i, and an equation of the
plane is 1({ 3 1) + 1 | 3 12 + 1 } 3 13 = 0 or { + | + } =
11
6
or 6{ + 6| + 6} = 11.
30. First, a normal vector for the plane 5{ + 2| + } = 1 is n = h5> 2> 1i. A direction vector for the line is v = h1> 31> 33i, and
since n · v = 0 we know the line is perpendicular to n and hence parallel to the plane. Thus, there is a parallel plane which
contains the line. By putting w = 0, we know that the point (1> 2> 4) is on the line and hence the new plane. We can use the
same normal vector n = h5> 2> 1i, so an equation of the plane is 5({ 3 1) + 2(| 3 2) + 1(} 3 4) = 0 or 5{ + 2| + } = 13.
31. Here the vectors a = h1 3 0> 0 3 1> 1 3 1i = h1> 31> 0i and b = h1 3 0> 1 3 1> 0 3 1i = h1> 0> 31i lie in the plane, so
a × b is a normal vector to the plane. Thus, we can take n = a × b = h1 3 0> 0 + 1> 0 + 1i = h1> 1> 1i. If S0 is the point
(0> 1> 1), an equation of the plane is 1({ 3 0) + 1(| 3 1) + 1(} 3 1) = 0 or { + | + } = 2.
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277
32. Here the vectors a = h2> 34> 6i and b = h5> 1> 3i lie in the plane, so
n = a × b = h312 3 6> 30 3 6> 2 + 20i = h318> 24> 22i is a normal vector to the plane and an equation of the plane is
318({ 3 0) + 24(| 3 0) + 22(} 3 0) = 0 or 318{ + 24| + 22} = 0.
33. Here the vectors a = h8 3 3> 2 3 (31)> 4 3 2i = h5> 3> 2i and b = h31 3 3> 32 3 (31)> 33 3 2i = h34> 31> 35i lie in
the plane, so a normal vector to the plane is n = a × b = h315 + 2> 38 + 25> 35 + 12i = h313> 17> 7i and an equation of
the plane is 313({ 3 3) + 17[| 3 (31)] + 7(} 3 2) = 0 or 313{ + 17| + 7} = 342.
34. If we first find two nonparallel vectors in the plane, their cross product will be a normal vector to the plane. Since the given
line lies in the plane, its direction vector a = h3> 1> 31i is one vector in the plane. We can verify that the given point (1> 2> 3)
does not lie on this line, so to find another nonparallel vector b which lies in the plane, we can pick any point on the line and
find a vector connecting the points. If we put w = 0, we see that (0> 1> 2) is on the line, so
b = h1 3 0> 2 3 1> 3 3 2i = h1> 1> 1i and n = a × b = h1 + 1> 31 3 3> 3 3 1i = h2> 34> 2i. Thus, an equation of the plane
is 2({ 3 1) 3 4(| 3 2) + 2(} 3 3) = 0 or 2{ 3 4| + 2} = 0. (Equivalently, we can write { 3 2| + } = 0.)
35. If we first find two nonparallel vectors in the plane, their cross product will be a normal vector to the plane. Since the given
line lies in the plane, its direction vector a = h32> 5> 4i is one vector in the plane. We can verify that the given point (6> 0> 32)
does not lie on this line, so to find another nonparallel vector b which lies in the plane, we can pick any point on the line and
find a vector connecting the points. If we put w = 0, we see that (4> 3> 7) is on the line, so
b = h6 3 4> 0 3 3> 32 3 7i = h2> 33> 39i and n = a × b = h345 + 12> 8 3 18> 6 3 10i = h333> 310> 34i. Thus, an
equation of the plane is 333({ 3 6) 3 10(| 3 0) 3 4[} 3 (32)] = 0 or 33{ + 10| + 4} = 190.
36. Since the line { = 2| = 3}, or { =
|
}
=
, lies in the plane, its direction vector a = 1> 12 > 13 is parallel to the plane.
1@2
1@3
The point (0> 0> 0) is on the line (put w = 0), and we can verify that the given point (1> 31> 1) in the plane is not on the line.
The vector connecting these two points, b = h1> 31> 1i, is therefore parallel to the plane, but not parallel to h1> 2> 3i. Then
a × b = 12 + 13 > 13 3 1> 31 3 12 = 56 > 3 23 > 3 32 is a normal vector to the plane, and an equation of the plane is
5
({
6
3 0) 3 23 (| 3 0) 3 32 (} 3 0) = 0 or 5{ 3 4| 3 9} = 0.
37. A direction vector for the line of intersection is a = n1 × n2 = h1> 1> 31i × h2> 31> 3i = h2> 35> 33i, and a is parallel to the
desired plane. Another vector parallel to the plane is the vector connecting any point on the line of intersection to the given
point (31> 2> 1) in the plane. Setting { = 0, the equations of the planes reduce to | 3 } = 2 and 3| + 3} = 1 with
simultaneous solution | = 72 and } = 32 . So a point on the line is 0> 72 > 32 and another vector parallel to the plane is
31> 3 32 > 3 12 . Then a normal vector to the plane is n = h2> 35> 33i × 31> 3 32 > 3 12 = h32> 4> 38i and an equation of
the plane is 32({ + 1) + 4(| 3 2) 3 8(} 3 1) = 0 or { 3 2| + 4} = 31.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
38. The points (0> 32> 5) and (31> 3> 1) lie in the desired plane, so the vector v1 = h31> 5> 34i connecting them is parallel to
the plane. The desired plane is perpendicular to the plane 2} = 5{ + 4| or 5{ + 4| 3 2} = 0 and for perpendicular planes,
a normal vector for one plane is parallel to the other plane, so v2 = h5> 4> 32i is also parallel to the desired plane.
A normal vector to the desired plane is n = v1 × v2 = h310 + 16> 320 3 2> 34 3 25i = h6> 322> 329i.
Taking ({0 > |0 > }0 ) = (0> 32> 5), the equation we are looking for is 6({ 3 0) 3 22(| + 2) 3 29(} 3 5) = 0 or
6{ 3 22| 3 29} = 3101.
39. If a plane is perpendicular to two other planes, its normal vector is perpendicular to the normal vectors of the other two planes.
Thus h2> 1> 32i × h1> 0> 3i = h3 3 0> 32 3 6> 0 3 1i = h3> 38> 31i is a normal vector to the desired plane. The point
(1> 5> 1) lies on the plane, so an equation is 3({ 3 1) 3 8(| 3 5) 3 (} 3 1) = 0 or 3{ 3 8| 3 } = 338.
40. n1 = h1> 0> 31i and n2 = h0> 1> 2i. Setting } = 0, it is easy to see that (1> 3> 0) is a point on the line of intersection of
{ 3 } = 1 and | + 2} = 3. The direction of this line is v1 = n1 × n2 = h1> 32> 1i. A second vector parallel to the desired
plane is v2 = h1> 1> 32i, since it is perpendicular to { + | 3 2} = 1. Therefore, a normal of the plane in question is
n = v1 × v2 = h4 3 1> 1 + 2> 1 + 2i = h3> 3> 3i, or we can use h1> 1> 1i. Taking ({0 > |0 > }0 ) = (1> 3> 0), the equation we are
looking for is ({ 3 1) + (| 3 3) + } = 0
C
{ + | + } = 4.
41. To find the {-intercept we set | = } = 0 in the equation 2{ + 5| + } = 10
and obtain 2{ = 10 i { = 5 so the {-intercept is (5> 0> 0). When
{ = } = 0 we get 5| = 10 i | = 2, so the |-intercept is (0> 2> 0).
Setting { = | = 0 gives } = 10, so the }-intercept is (0> 0> 10) and we
graph the portion of the plane that lies in the first octant.
42. To find the {-intercept we set | = } = 0 in the equation 3{ + | + 2} = 6
and obtain 3{ = 6 i { = 2 so the {-intercept is (2> 0> 0). When
{ = } = 0 we get | = 6 so the |-intercept is (0> 6> 0). Setting { = | = 0
gives 2} = 6 i } = 3, so the }-intercept is (0> 0> 3). The figure shows
the portion of the plane that lies in the first octant.
43. Setting | = } = 0 in the equation 6{ 3 3| + 4} = 6 gives 6{ = 6
i
{ = 1, when { = } = 0 we have 33| = 6 i | = 32, and { = | = 0
implies 4} = 6 i } = 32 , so the intercepts are (1> 0> 0), (0> 32> 0), and
(0> 0> 32 ). The figure shows the portion of the plane cut off by the coordinate
planes.
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SECTION 12.5 EQUATIONS OF LINES AND PLANES
44. Setting | = } = 0 in the equation 6{ + 5| 3 3} = 15 gives 6{ = 15
{=
5
,
2
279
i
when { = } = 0 we have 5| = 15 i | = 3, and { = | = 0
implies 33} = 15 i } = 35, so the intercepts are ( 52 > 0> 0), (0> 3> 0),
and (0> 0> 35). The figure shows the portion of the plane cut off by the
coordinate planes.
45. Substitute the parametric equations of the line into the equation of the plane: (3 3 w) 3 (2 + w) + 2(5w) = 9
i
8w = 8 i w = 1. Therefore, the point of intersection of the line and the plane is given by { = 3 3 1 = 2, | = 2 + 1 = 3,
and } = 5(1) = 5> that is, the point (2> 3> 5).
46. Substitute the parametric equations of the line into the equation of the plane: (1 + 2w) + 2(4w) 3 (2 3 3w) + 1 = 0
i
13w = 0 i w = 0. Therefore, the point of intersection of the line and the plane is given by { = 1 + 2(0) = 1,
| = 4(0) = 0, and } = 2 3 3(0) = 2> that is, the point (1> 0> 2).
47. Parametric equations for the line are { = w, | = 1 + w, } =
4(w) 3 (1 + w) + 3 12 w = 8 i
9
2w
1
w
2
and substituting into the equation of the plane gives
= 9 i w = 2. Thus { = 2, | = 1 + 2 = 3, } = 12 (2) = 1 and the point of
intersection is (2> 3> 1).
48. A direction vector for the line through (1> 0> 1) and (4> 32> 2) is v = h3> 32> 1i and, taking S0 = (1> 0> 1), parametric
equations for the line are { = 1 + 3w, | = 32w, } = 1 + w. Substitution of the parametric equations into the equation of the
plane gives 1 + 3w 3 2w + 1 + w = 6 i w = 2. Then { = 1 + 3(2) = 7, | = 32(2) = 34, and } = 1 + 2 = 3 so the point
of intersection is (7> 34> 3).
49. Setting { = 0, we see that (0> 1> 0) satisfies the equations of both planes, so that they do in fact have a line of intersection.
v = n1 × n2 = h1> 1> 1i × h1> 0> 1i = h1> 0> 31i is the direction of this line. Therefore, direction numbers of the intersecting
line are 1, 0, 31.
50. The angle between the two planes is the same as the angle between their normal vectors. The normal vectors of the
two planes are h1> 1> 1i and h1> 2> 3i. The cosine of the angle between these two planes is
u
h1> 1> 1i · h1> 2> 3i
6
6
1+2+3
I
cos =
= I =
= I
.
|h1> 1> 1i| |h1> 2> 3i|
7
1+1+1 1+4+9
42
51. Normal vectors for the planes are n1 = h1> 4> 33i and n2 = h33> 6> 7i, so the normals (and thus the planes) aren’t parallel.
But n1 · n2 = 33 + 24 3 21 = 0, so the normals (and thus the planes) are perpendicular.
52. Normal vectors for the planes are n1 = h31> 4> 32i and n2 = h3> 312> 6i. Since n2 = 33n1 , the normals (and thus the
planes) are parallel.
53. Normal vectors for the planes are n1 = h1> 1> 1i and n2 = h1> 31> 1i. The normals are not parallel, so neither are the planes.
Furthermore, n1 · n2 = 1 3 1 + 1 = 1 6= 0, so the planes aren’t perpendicular. The angle between them is given by
cos =
n1 · n2
1
1
= I I =
|n1 | |n2 |
3
3 3
i = cos31
1
3
E 70=5 .
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
54. The normals are n1 = h2> 33> 4i and n2 = h1> 6> 4i so the planes aren’t parallel. Since n1 · n2 = 2 3 18 + 16 = 0, the
normals (and thus the planes) are perpendicular.
55. The normals are n1 = h1> 34> 2i and n2 = h2> 38> 4i. Since n2 = 2n1 , the normals (and thus the planes) are parallel.
56. The normal vectors are n1 = h1> 2> 2i and n2 = h2> 31> 2i. The normals are not parallel, so neither are the planes.
Furthermore, n1 · n2 = 2 3 2 + 4 = 4 6= 0, so the planes aren’t perpendicular. The angle between them is given by
cos =
n1 · n2
4
4
= I I =
|n1 | |n2 |
9
9 9
i = cos31
4
9
E 63=6 .
57. (a) To find a point on the line of intersection, set one of the variables equal to a constant, say } = 0. (This will fail if the line of
intersection does not cross the {|-plane; in that case, try setting { or | equal to 0.) The equations of the two planes reduce
to { + | = 1 and { + 2| = 1. Solving these two equations gives { = 1, | = 0. Thus a point on the line is (1> 0> 0).
A vector v in the direction of this intersecting line is perpendicular to the normal vectors of both planes, so we can take
v = n1 × n2 = h1> 1> 1i × h1> 2> 2i = h2 3 2> 1 3 2> 2 3 1i = h0> 31> 1i. By Equations 2, parametric equations for the
line are { = 1, | = 3w, } = w.
(b) The angle between the planes satisfies cos =
n1 · n2
5
5
1+2+2
I
E 15=8 .
= I I = I . Therefore = cos31
|n1 | |n2 |
3 9
3 3
3 3
58. (a) If we set } = 0 then the equations of the planes reduce to 3{ 3 2| = 1 and 2{ + | = 3 and solving these two equations
gives { = 1, | = 1. Thus a point on the line of intersection is (1> 1> 0). A vector v in the direction of this intersecting line
is perpendicular to the normal vectors of both planes, so let v = n1 × n2 = h3> 32> 1i × h2> 1> 33i = h5> 11> 7i. By
Equations 2, parametric equations for the line are { = 1 + 5w, | = 1 + 11w, } = 7w.
(b) cos =
63233
n1 · n2
1
= I I =
|n1 | |n2 |
14
14 14
i = cos31
1
14
E 85=9 .
59. Setting } = 0, the equations of the two planes become 5{ 3 2| = 1 and 4{ + | = 6. Solving these two equations gives
{ = 1, | = 2 so a point on the line of intersection is (1> 2> 0). A vector v in the direction of this intersecting line is
perpendicular to the normal vectors of both planes. So we can use v = n1 × n2 = h5> 32> 32i × h4> 1> 1i = h0> 313> 13i or
equivalently we can take v = h0> 31> 1i, and symmetric equations for the line are { = 1,
|32
}
= or { = 1, | 3 2 = 3}.
31
1
60. If we set } = 0 then the equations of the planes reduce to 2{ 3 | 3 5 = 0 and 4{ + 3| 3 5 = 0 and solving these two
equations gives { = 2, | = 31. Thus a point on the line of intersection is (2> 31> 0). A vector v in the
direction of this intersecting line is perpendicular to the normal vectors of both planes, so take
v = n1 × n2 = h2> 31> 31i × h4> 3> 31i = h4> 32> 10i or equivalently we can take v = h2> 31> 5i. Symmetric equations for
the line are
|+1
}
{32
=
= .
2
31
5
61. The distance from a point ({> |> }) to (1> 0> 32) is g1 =
(3> 4> 0) is g2 =
s
({ 3 1)2 + | 2 + (} + 2)2 and the distance from ({> |> }) to
s
({ 3 3)2 + (| 3 4)2 + } 2 . The plane consists of all points ({> |> }) where g1 = g2
i g12 = g22
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SECTION 12.5 EQUATIONS OF LINES AND PLANES
({ 3 1)2 + | 2 + (} + 2)2 = ({ 3 3)2 + (| 3 4)2 + } 2
281
C
{2 3 2{ + | 2 + } 2 + 4} + 5 = {2 3 6{ + | 2 3 8| + } 2 + 25 C 4{ + 8| + 4} = 20 so an equation for the plane is
4{ + 8| + 4} = 20 or equivalently { + 2| + } = 5.
Alternatively, you can argue that the segment joining points (1> 0> 32) and (3> 4> 0) is perpendicular to the plane and the plane
includes the midpoint of the segment.
62. The distance from a point ({> |> }) to (2> 5> 5) is g1 =
to (36> 3> 1) is g2 =
g12 = g22
s
({ 3 2)2 + (| 3 5)2 + (} 3 5)2 and the distance from ({> |> })
s
({ + 6)2 + (| 3 3)2 + (} 3 1)2 . The plane consists of all points ({> |> }) where g1 = g2
C ({ 3 2)2 + (| 3 5)2 + (} 3 5)2 = ({ + 6)2 + (| 3 3)2 + (} 3 1)2
i
C
{2 3 4{ + | 2 3 10| + } 2 3 10} + 54 = {2 + 12{ + | 2 3 6| + } 2 3 2} + 46 C 16{ + 4| + 8} = 8 so an equation
for the plane is 16{ + 4| + 8} = 8 or equivalently 4{ + | + 2} = 2.
63. The plane contains the points (d> 0> 0), (0> e> 0) and (0> 0> f). Thus the vectors a = h3d> e> 0i and b = h3d> 0> fi lie in the
plane, and n = a × b = hef 3 0> 0 + df> 0 + dei = hef> df> dei is a normal vector to the plane. The equation of the plane is
therefore ef{ + df| + de} = def + 0 + 0 or ef{ + df| + de} = def. Notice that if d 6= 0, e 6= 0 and f 6= 0 then we can
rewrite the equation as
}
{ |
+ + = 1. This is a good equation to remember!
d
e
f
64. (a) For the lines to intersect, we must be able to find one value of w and one value of v satisfying the three equations
1 + w = 2 3 v, 1 3 w = v and 2w = 2. From the third we get w = 1, and putting this in the second gives v = 0. These values
of v and w do satisfy the first equation, so the lines intersect at the point S0 = (1 + 1> 1 3 1> 2(1)) = (2> 0> 2).
(b) The direction vectors of the lines are h1> 31> 2i and h31> 1> 0i, so a normal vector for the plane is
h31> 1> 0i × h1> 31> 2i = h2> 2> 0i and it contains the point (2> 0> 2). Then an equation of the plane is
2({ 3 2) + 2(| 3 0) + 0(} 3 2) = 0 C { + | = 2.
65. Two vectors which are perpendicular to the required line are the normal of the given plane, h1> 1> 1i, and a direction vector for
the given line, h1> 31> 2i. So a direction vector for the required line is h1> 1> 1i × h1> 31> 2i = h3> 31> 32i. Thus O is given
by h{> |> }i = h0> 1> 2i + wh3> 31> 32i, or in parametric form, { = 3w, | = 1 3 w, } = 2 3 2w.
66. Let O be the given line. Then (1> 1> 0) is the point on O corresponding to w = 0. O is in the direction of a = h1> 31> 2i
and b = h31> 0> 2i is the vector joining (1> 1> 0) and (0> 1> 2). Then
b 3 proja b = h31> 0> 2i 3
h1> 31> 2i · h31> 0> 2i
h1> 31> 2i = h31> 0> 2i 3 12 h1> 31> 2i = 3 32 > 12 > 1 is a direction vector
12 + (31)2 + 22
for the required line. Thus 2 3 32 > 12 > 1 = h33> 1> 2i is also a direction vector, and the line has parametric equations { = 33w,
| = 1 + w, } = 2 + 2w. (Notice that this is the same line as in Exercise 65.)
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
67. Let Sl have normal vector nl . Then n1 = h3> 6> 33i, n2 = h4> 312> 8i, n3 = h3> 39> 6i, n4 = h1> 2> 31i. Now n1 = 3n4 ,
so n1 and n4 are parallel, and hence S1 and S4 are parallel; similarly S2 and S3 are parallel because n2 = 43 n3 . However, n1
and n2 are not parallel (so not all four planes are parallel). Notice that the point (2> 0> 0) lies on both S1 and S4 , so these two
planes are identical. The point 54 > 0> 0 lies on S2 but not on S3 , so these are different planes.
68. Let Ol have direction vector vl . Rewrite the symmetric equations for O3 as
v2 = h2> 1> 4i, v3 =
{31
|31
}+1
=
=
; then v1 = h6> 33> 12i,
1@2
31@4
1
, and v4 = h4> 2> 8i. v1 = 12v3 , so O1 and O3 are parallel. v4 = 2v2 , so O2 and O4 are
1
> 3 14 > 1
2
parallel. (Note that O1 and O2 are not parallel.) O1 contains the point (1> 1> 5), but this point does not lie on O3 , so they’re not
identical. (3> 1> 5) lies on O4 and also on O2 (for w = 1), so O2 and O4 are the same line.
69. Let T = (1> 3> 4) and U = (2> 1> 1), points on the line corresponding to w = 0 and w = 1. Let
3<
3
3
<
S = (4> 1> 32). Then a = TU = h1> 32> 33i, b = TS = h3> 32> 36i. The distance is
s
u
I
62 + (33)2 + 42
|a × b|
61
61
|h1> 32> 33i × h3> 32> 36i|
|h6> 33> 4i|
= I =
g=
=
=
= s
.
2
2
2
|a|
|h1> 32> 33i|
|h1> 32> 33i|
14
14
1 + (32) + (33)
70. Let T = (0> 6> 3) and U = (2> 4> 4), points on the line corresponding to w = 0 and w = 1. Let
3<
3
3
<
S = (0> 1> 3). Then a = TU = h2> 32> 1i and b = TS = h0> 35> 0i. The distance is
s
I
I
52 + 02 + (310)2
125
|h2> 32> 1i × h0> 35> 0i|
|h5> 0> 310i|
|a × b|
5 5
=
=
= s
.
= I =
g=
|a|
|h2> 32> 1i|
|h2> 32> 1i|
3
9
22 + (32)2 + 12
71. By Equation 9, the distance is G =
|d{1 + e|1 + f}1 + g|
|3(1) + 2(32) + 6(4) 3 5|
|18|
18
I
I
=
= I =
.
7
d2 + e2 + f2
32 + 22 + 62
49
72. By Equation 9, the distance is G =
|1(36) 3 2(3) 3 4(5) 3 8|
|340|
40
s
= I
= I .
21
21
12 + (32)2 + (34)2
73. Put | = } = 0 in the equation of the first plane to get the point (2> 0> 0) on the plane. Because the planes are parallel, the
distance G between them is the distance from (2> 0> 0) to the second plane. By Equation 9,
I
|4(2) 3 6(0) + 2(0) 3 3|
5
5
5 14
G= s
.
= I = I or
28
56
2 14
42 + (36)2 + (2)2
74. Put { = | = 0 in the equation of the first plane to get the point (0> 0> 0) on the plane. Because the planes are parallel the
distance G between them is the distance from (0> 0> 0) to the second plane 3{ 3 6| + 9} 3 1 = 0. By Equation 9,
G=
1
1
|3(0) 3 6(0) + 9(0) 3 1|
s
= I
= I .
2
2
2
126
3 14
3 + (36) + 9
75. The distance between two parallel planes is the same as the distance between a point on one of the planes and the other plane.
Let S0 = ({0 > |0 > }0 ) be a point on the plane given by d{ + e| + f} + g1 = 0. Then d{0 + e|0 + f}0 + g1 = 0 and the
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283
distance between S0 and the plane given by d{ + e| + f} + g2 = 0 is, from Equation 9,
G=
|d{0 + e|0 + f}0 + g2 |
|3g1 + g2 |
|g1 3 g2 |
I
= I
= I
.
2
2
2
2
2
2
d +e +f
d +e +f
d2 + e2 + f2
76. The planes must have parallel normal vectors, so if d{ + e| + f} + g = 0 is such a plane, then for some w 6= 0,
hd> e> fi = wh1> 2> 32i = hw> 2w> 32wi. So this plane is given by the equation { + 2| 3 2} + n = 0, where n = g@w. By
|1 3 n|
Exercise 75, the distance between the planes is 2 = s
12 + 22 + (32)2
C 6 = |1 3 n| C n = 7 or 35. So the
desired planes have equations { + 2| 3 2} = 7 and { + 2| 3 2} = 35.
77. O1 : { = | = }
i { = | (1). O2 : { + 1 = |@2 = }@3 i { + 1 = |@2 (2). The solution of (1) and (2) is
{ = | = 32. However, when { = 32, { = }
i } = 32, but { + 1 = }@3 i } = 33, a contradiction. Hence the
lines do not intersect. For O1 , v1 = h1> 1> 1i, and for O2 , v2 = h1> 2> 3i, so the lines are not parallel. Thus the lines are skew
lines. If two lines are skew, they can be viewed as lying in two parallel planes and so the distance between the skew lines
would be the same as the distance between these parallel planes. The common normal vector to the planes must be
perpendicular to both h1> 1> 1i and h1> 2> 3i, the direction vectors of the two lines. So set
n = h1> 1> 1i × h1> 2> 3i = h3 3 2> 33 + 1> 2 3 1i = h1> 32> 1i. From above, we know that (32> 32> 32) and (32> 32> 33)
are points of O1 and O2 respectively. So in the notation of Equation 8, 1(32) 3 2(32) + 1(32) + g1 = 0
1(32) 3 2(32) + 1(33) + g2 = 0
i
i
g1 = 0 and
g2 = 1.
1
|0 3 1|
= I .
By Exercise 75, the distance between these two skew lines is G = I
1+4+1
6
Alternate solution (without reference to planes): A vector which is perpendicular to both of the lines is
n = h1> 1> 1i × h1> 2> 3i = h1> 32> 1i. Pick any point on each of the lines, say (32> 32> 32) and (32> 32> 33), and form the
vector b = h0> 0> 1i connecting the two points. The distance between the two skew lines is the absolute value of the scalar
projection of b along n, that is, G =
|1 · 0 3 2 · 0 + 1 · 1|
1
|n · b|
I
=
= I .
|n|
1+4+1
6
78. First notice that if two lines are skew, they can be viewed as lying in two parallel planes and so the distance between the skew
lines would be the same as the distance between these parallel planes. The common normal vector to the planes must be
perpendicular to both v1 = h1> 6> 2i and v2 = h2> 15> 6i, the direction vectors of the two lines respectively. Thus set
n = v1 × v2 = h36 3 30> 4 3 6> 15 3 12i = h6> 32> 3i. Setting w = 0 and v = 0 gives the points (1> 1> 0) and (1> 5> 32).
So in the notation of Equation 8, 6 3 2 + 0 + g1 = 0 i g1 = 34 and 6 3 10 3 6 + g2 = 0
i
g2 = 10.
|34 3 10|
14
Then by Exercise 75, the distance between the two skew lines is given by G = I
= 2.
=
7
36 + 4 + 9
Alternate solution (without reference to planes): We already know that the direction vectors of the two lines are
v1 = h1> 6> 2i and v2 = h2> 15> 6i. Then n = v1 × v2 = h6> 32> 3i is perpendicular to both lines. Pick any point on
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
each of the lines, say (1> 1> 0) and (1> 5> 32), and form the vector b = h0> 4> 32i connecting the two points. Then the
distance between the two skew lines is the absolute value of the scalar projection of b along n, that is,
G=
|n · b|
14
1
|0 3 8 3 6| =
= I
= 2.
|n|
7
36 + 4 + 9
79. A direction vector for O1 is v1 = h2> 0> 31i and a direction vector for O2 is v2 = h3> 2> 2i. These vectors are not parallel so
neither are the lines. Parametric equations for the lines are O1 : { = 2w, | = 0, } = 3w, and O2 : { = 1 + 3v, | = 31 + 2v,
} = 1 + 2v. No values of w and v satisfy these equations simultaneously, so the lines don’t intersect and hence are skew. We
can view the lines as lying in two parallel planes; a common normal vector to the planes is n = v1 × v2 = h2> 37> 4i. Line
O1 passes through the origin, so (0> 0> 0) lies on one of the planes, and (1> 31> 1) is a point on O2 and therefore on the other
plane. Equations of the planes then are 2{ 3 7| + 4} = 0 and 2{ 3 7| + 4} 3 13 = 0, and by Exercise 75, the distance
13
|0 3 (313)|
= I .
between the two skew lines is G = I
4 + 49 + 16
69
Alternate solution (without reference to planes): Direction vectors of the two lines are v1 = h2> 0> 31i and v2 = h3> 2> 2i.
Then n = v1 × v2 = h2> 37> 4i is perpendicular to both lines. Pick any point on each of the lines, say (0> 0> 0) and (1> 31> 1),
and form the vector b = h1> 31> 1i connecting the two points. Then the distance between the two skew lines is the absolute
value of the scalar projection of b along n, that is, G =
|2 + 7 + 4|
13
|n · b|
= I
= I .
|n|
4 + 49 + 16
69
80. A direction vector for the line O1 is v1 = h1> 2> 2i. A normal vector for the plane 1 is n1 = h1> 31> 2i. The vector from the
point (0> 0> 1) to (3> 2> 31), h3> 2> 32i, is parallel to the plane 2 , as is the vector from (0> 0> 1) to (1> 2> 1), namely h1> 2> 0i.
Thus a normal vector for 2 is h3> 2> 32i × h1> 2> 0i = h4> 32> 4i, or we can use n2 = h2> 31> 2i, and a direction vector for
the line O2 of intersection of these planes is v2 = n1 × n2 = h1> 31> 2i × h2> 31> 2i = h0> 2> 1i. Notice that the point
(3> 2> 31) lies on both planes, so it also lies on O2 . The lines are skew, so we can view them as lying in two parallel planes; a
common normal vector to the planes is n = v1 × v2 = h32> 31> 2i. Line O1 passes through the point (1> 2> 6), so (1> 2> 6)
lies on one of the planes, and (3> 2> 31) is a point on O2 and therefore on the other plane. Equations of the planes then are
32{ 3 | + 2} 3 8 = 0 and 32{ 3 | + 2} + 10 = 0, and by Exercise 75, the distance between the lines is
18
|38 3 10|
= 6.
=
G= I
3
4+1+4
Alternatively, direction vectors for the lines are v1 = h1> 2> 2i and v2 = h0> 2> 1i, so n = v1 × v2 = h32> 31> 2i is
perpendicular to both lines. Pick any point on each of the lines, say (1> 2> 6) and (3> 2> 31), and form the vector
b = h2> 0> 37i connecting the two points. Then the distance between the two skew lines is the absolute value of the scalar
projection of b along n, that is, G =
|34 + 0 3 14|
18
|n · b|
= I
= 6.
=
|n|
3
4+1+4
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LABORATORY PROJECT PUTTING 3D IN PERSPECTIVE
81. If d 6= 0, then d{ + e| + f} + g = 0
¤
285
i d({ + g@d) + e(| 3 0) + f(} 3 0) = 0 which by (7) is the scalar equation of the
plane through the point (3g@d> 0> 0) with normal vector hd> e> fi. Similarly, if e 6= 0 (or if f 6= 0) the equation of the plane can
be rewritten as d({ 3 0) + e(| + g@e) + f(} 3 0) = 0 [or as d({ 3 0) + e(| 3 0) + f(} + g@f) = 0] which by (7) is the
scalar equation of a plane through the point (0> 3g@e> 0) [or the point (0> 0> 3g@f)] with normal vector hd> e> fi.
82. (a) The planes { + | + } = f have normal vector h1> 1> 1i, so they are all
parallel. Their {-, |-, and }-intercepts are all f. When f A 0 their
intersection with the first octant is an equilateral triangle and when f ? 0
their intersection with the octant diagonally opposite the first is an
equilateral triangle.
(b) The planes { + | + f} = 1 have {-intercept 1, |-intercept 1, and }-intercept 1@f. The plane with f = 0 is parallel to the
}-axis. As f gets larger, the planes get closer to the {|-plane.
(c) The planes | cos + } cos = 1 have normal vectors h0> cos > sin i, which are perpendicular to the {-axis, and so the
planes are parallel to the {-axis. We look at their intersection with the |}-plane. These are lines that are perpendicular to
hcos > sin i and pass through (cos > sin ), since cos2 + sin2 = 1. So these are the tangent lines to the unit circle.
Thus the family consists of all planes tangent to the circular cylinder with radius 1 and axis the {-axis.
LABORATORY PROJECT Putting 3D in Perspective
1. If we view the screen from the camera’s location, the vertical clipping plane on the left passes through the points
(1000> 0> 0), (0> 3400> 0), and (0> 3400> 600). A vector from the first point to the second is v1 = h31000> 3400> 0i
and a vector from the first point to the third is v2 = h31000> 3400> 600i. A normal vector for the clipping plane is
v1 × v2 = 3240,000 i + 600,000 j or 32 i + 5 j, and an equation for the plane is
32({ 3 1000) + 5(| 3 0) + 0(} 3 0) = 0 i 2{ 3 5| = 2000. By symmetry, the vertical clipping plane on the right is
given by 2{ + 5| = 2000. The lower clipping plane is } = 0. The upper clipping plane passes through the points (1000> 0> 0),
(0> 3400> 600), and (0> 400> 600). Vectors from the first point to the second and third points are v1 = h31000> 3400> 600i
and v2 = h31000> 400> 600i, and a normal vector for the plane is v1 × v2 = 3480,000 i 3 800,000 k or 3 i + 5 k. An
equation for the plane is 3({ 3 1000) + 0(| 3 0) + 5(} 3 0) = 0 i 3{ + 5} = 3000.
A direction vector for the line O is v = h630> 390> 162i and taking S0 = (230> 3285> 102), parametric equations
are { = 230 + 630w, | = 3285 + 390w, } = 102 + 162w. O intersects the left clipping plane when
2(230 + 630w) 3 5(3285 + 390w) = 2000 i w = 3 16 . The corresponding point is (125> 3350> 75). O intersects
the right clipping plane when 2(230 + 630w) + 5(3285 + 390w) = 2000 i w =
593
642 .
The corresponding point is
approximately (811=9> 75=2> 251=6), but this point is not contained within the viewing volume. O intersects the upper clipping
plane when 3(230 + 630w) + 5(102 + 162w) = 3000 i w = 23 , corresponding to the point (650> 325> 210), and O
. The corresponding point is
intersects the lower clipping plane when } = 0 i 102 + 162w = 0 i w = 3 17
27
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
approximately (3166=7> 3530=6> 0), which is not contained within the viewing volume. Thus O should be clipped at the
points (125> 3350> 75) and (650> 325> 210).
2. A sight line from the camera at (1000> 0> 0) to the left endpoint (125> 3350> 75) of the clipped line has direction
v = h3875> 3350> 75i. Parametric equations are { = 1000 3 875w, | = 3350w, } = 75w. This line intersects the screen
. Similarly, a sight line from
when { = 0 i 1000 3 875w = 0 i w = 87 , corresponding to the point 0> 3400> 600
7
the camera to the right endpoint (650> 325> 210) of the clipped line has direction h3350> 325> 210i and parametric equations
, corresponding to the point
are { = 1000 3 350w, | = 325w, } = 210w. { = 0 i 1000 3 350w = 0 i w = 20
7
> 600 . Thus the projection of the clipped line is the line segment between the points 0> 3400> 600
and
0> 3 500
7
7
> 600 .
0> 3 500
7
3. From Equation 12.5.4, equations for the four sides of the screen
are r1 (w) = (1 3 w)h0> 3400> 0i + w h0> 3400> 600i,
r2 (w) = (1 3 w)h0> 3400> 600i + w h0> 400> 600i,
r3 (w) = (1 3 w)h0> 400> 0i + w h0> 400> 600i, and
r4 (w) = (1 3 w)h0> 3400> 0i + w h0> 400> 0i. The clipped line
segment connects the points (125> 3350> 75) and
(650> 325> 210), so an equation for the segment is
r5 (w) = (1 3 w)h125> 3350> 75i + w h650> 325> 210i.
The projection of the clipped segment connects the points
0> 3400> 600
and 0> 3 500
+ w 0> 3 500
> 600 , so an equation is r6 (w) = (1 3 w) 0> 3400> 600
> 600 .
7
7
7
7
, so an equation is
The sight line on the left connects the points (1000> 0> 0) and 0> 3400> 600
7
. The other sight line connects (1000> 0> 0) to 0> 3 500
r7 (w) = (1 3 w)h1000> 0> 0i + w 0> 3400> 600
7
7 > 600 , so an equation
is r8 (w) = (1 3 w)h1000> 0> 0i + w 0> 3 500
> 600 .
7
4. The vector from (621> 3147> 206) to (563> 31> 242), v1 = h358> 178> 36i, lies in the plane of the rectangle, as does the
vector from (621> 3147> 206) to (657> 3111> 86), v2 = h36> 36> 3120i. A normal vector for the plane is
v1 × v2 = h31888> 3142> 3708i or h8> 2> 3i, and an equation of the plane is 8{ + 2| + 3} = 5292. The line O intersects
this plane when 8(230 + 630w) + 2(3285 + 390w) + 3(102 + 162w) = 5292 i w =
1858
3153
E 0=589. The corresponding
point is approximately (601=25> 355=18> 197=46). Starting at this point, a portion of the line is hidden behind the rectangle.
The line becomes visible again at the left edge of the rectangle, specifically the edge between the points (621> 3147> 206) and
(657> 3111> 86). (This is most easily determined by graphing the rectangle and the line.) A plane through these two points
and the camera’s location, (1000> 0> 0), will clip the line at the point it becomes visible. Two vectors in this plane are
v1 = h3379> 3147> 206i and v2 = h3343> 3111> 86i. A normal vector for the plane is
v1 × v2 = h10224> 338064> 38352i and an equation of the plane is 213{ 3 793| 3 174} = 213,000. O intersects this plane
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SECTION 12.6
CYLINDERS AND QUADRIC SURFACES
when 213(230 + 630w) 3 793(3285 + 390w) 3 174(102 + 162w) = 213,000 i w =
44,247
203,268
¤
287
E 0=2177. The
corresponding point is approximately (367=14> 3200=11> 137=26). Thus the portion of O that should be removed is the
segment between the points (601=25> 355=18> 197=46) and (367=14> 3200=11> 137=26).
12.6 Cylinders and Quadric Surfaces
1. (a) In R2 , the equation | = {2 represents a parabola.
(b) In R3 , the equation | = {2 doesn’t involve }, so any
horizontal plane with equation } = n intersects the graph
in a curve with equation | = {2 . Thus, the surface is a
parabolic cylinder, made up of infinitely many shifted
copies of the same parabola. The rulings are parallel to
the }-axis.
(c) In R3 , the equation } = | 2 also represents a parabolic
cylinder. Since { doesn’t appear, the graph is formed by
moving the parabola } = | 2 in the direction of the {-axis.
Thus, the rulings of the cylinder are parallel to the {-axis.
2. (a)
(b) Since the equation | = h{ doesn’t
(c) The equation } = h| doesn’t involve {,
involve }, horizontal traces are
so vertical traces in { = n (parallel to the
copies of the curve | = h{ . The
|}-plane) are copies of the curve } = h| .
rulings are parallel to the }-axis.
The rulings are parallel to the {-axis.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
3. Since | is missing from the equation, the vertical traces
2
2
4. Since } is missing from the equation, the horizontal
{ + } = 1, | = n, are copies of the same circle in
traces 4{2 + | 2 = 4, } = n, are copies of the same
the plane | = n. Thus the surface {2 + } 2 = 1 is a
ellipse in the plane } = n. Thus the surface
circular cylinder with rulings parallel to the |-axis.
4{2 + | 2 = 4 is an elliptic cylinder with rulings
parallel to the }-axis.
5. Since { is missing, each vertical trace } = 1 3 | 2 ,
{ = n, is a copy of the same parabola in the plane
2
6. Since { is missing, each vertical trace | = } 2 , { = n,
is a copy of the same parabola in the plane { = n.
{ = n. Thus the surface } = 1 3 | is a parabolic
Thus the surface | = } 2 is a parabolic cylinder with
cylinder with rulings parallel to the {-axis.
rulings parallel to the {-axis.
.
7. Since } is missing, each horizontal trace {| = 1,
} = n, is a copy of the same hyperbola in the plane
} = n. Thus the surface {| = 1 is a hyperbolic
cylinder with rulings parallel to the }-axis.
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SECTION 12.6
CYLINDERS AND QUADRIC SURFACES
¤
289
8. Since { is missing, each vertical trace } = sin |,
{ = n, is a copy of a sine curve in the plane { = n.
Thus the surface } = sin | is a cylindrical surface with
rulings parallel to the {-axis.
9. (a) The traces of {2 + | 2 3 } 2 = 1 in { = n are | 2 3 } 2 = 1 3 n2 , a family of hyperbolas. (Note that the hyperbolas are
oriented differently for 31 ? n ? 1 than for n ? 31 or n A 1.) The traces in | = n are {2 3 } 2 = 1 3 n2 , a similar
family of hyperbolas. The traces in } = n are {2 + | 2 = 1 + n2 , a family of circles. For n = 0, the trace in the
{|-plane, the circle is of radius 1. As |n| increases, so does the radius of the circle. This behavior, combined with the
hyperbolic vertical traces, gives the graph of the hyperboloid of one sheet in Table 1.
(b) The shape of the surface is unchanged, but the hyperboloid is
rotated so that its axis is the |-axis. Traces in | = n are circles,
while traces in { = n and } = n are hyperbolas.
(c) Completing the square in | gives {2 + (| + 1)2 3 } 2 = 1. The
surface is a hyperboloid identical to the one in part (a) but shifted
one unit in the negative |-direction.
10. (a) The traces of 3{2 3 | 2 + } 2 = 1 in { = n are 3| 2 + } 2 = 1 + n2 , a family of hyperbolas, as are the traces in | = n,
3{2 + } 2 = 1 + n2 . The traces in } = n are {2 + | 2 = n2 3 1, a family of circles for |n| A 1. As |n| increases, the radii
of the circles increase; the traces are empty for |n| ? 1. This behavior, combined with the vertical traces, gives the graph of
the hyperboloid of two sheets in Table 1.
(b) The graph has the same shape as the hyperboloid in part (a) but is rotated so
that its axis is the {-axis. Traces in { = n, |n| A 1, are circles, while traces
in | = n and } = n are hyperbolas.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
11. For { = | 2 + 4} 2 , the traces in { = n are | 2 + 4} 2 = n. When n A 0 we
have a family of ellipses. When n = 0 we have just a point at the origin, and
the trace is empty for n ? 0. The traces in | = n are { = 4} 2 + n2 , a
family of parabolas opening in the positive {-direction. Similarly, the traces
in } = n are { = | 2 + 4n2 , a family of parabolas opening in the positive
{-direction. We recognize the graph as an elliptic paraboloid with axis the
{-axis and vertex the origin.
12. 9{2 3 | 2 + } 2 = 0. The traces in { = n are | 2 3 } 2 = 9n2 , a family of
hyperbolas if n 6= 0 and two intersecting lines if n = 0. The traces in | = n
are 9{2 + } 2 = n2 , n D 0, a family of ellipses; the traces in } = n are
| 2 3 9{2 = n2 , a family of hyperbolas for n 6= 0 and two intersecting lines
for n = 0. We recognize the graph as an elliptic cone with axis the |-axis
and vertex the origin.
13. {2 = | 2 + 4} 2 . The traces in { = n are the ellipses | 2 + 4} 2 = n2 . The
traces in | = n are {2 3 4} 2 = n2 , hyperbolas for n 6= 0 and two
intersecting lines if n = 0. Similarly, the traces in } = n are
{2 3 | 2 = 4n2 , hyperbolas for n 6= 0 and two intersecting lines if n = 0.
We recognize the graph as an elliptic cone with axis the {-axis and vertex
the origin.
14. 25{2 + 4| 2 + } 2 = 100. The traces in { = n are 4| 2 + } 2 = 100 3 25n2 ,
a family of ellipses for |n| ? 2. (The traces are a single point for |n| = 2
and are empty for |n| A 2.) Similarly, the traces in | = n are the ellipses
25{2 + } 2 = 100 3 4n2 , |n| ? 5, and the traces in } = n are the ellipses
25{2 + 4|2 = 100 3 n2 , |n| ? 10. The graph is an ellipsoid centered at the
origin with intercepts { = ±2, | = ±5, } = ±10.
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SECTION 12.6
CYLINDERS AND QUADRIC SURFACES
15. 3{2 + 4| 2 3 } 2 = 4. The traces in { = n are the hyperbolas
4| 2 3 } 2 = 4 + n2 . The traces in | = n are {2 + } 2 = 4n2 3 4, a family of
circles for |n| A 1, and the traces in } = n are 4| 2 3 {2 = 4 + n2 , a family
of hyperbolas. Thus the surface is a hyperboloid of two sheets with
axis the |-axis.
16. 4{2 + 9| 2 + } = 0. The traces in { = n are the parabolas } = 39| 2 3 4n2
which open downward. Similarly, the traces in | = n are the parabolas
} = 34{2 3 9n2 , also opening downward, and the traces in } = n are
4{2 + 9| 2 = 3n, n $ 0, a family of ellipses. The graph is an elliptic
paraboloid with axis the }-axis, opening downward, and vertex the origin.
17. 36{2 + | 2 + 36} 2 = 36. The traces in { = n are | 2 + 36} 2 = 36(1 3 n2 ),
a family of ellipses for |n| ? 1. (The traces are a single point for |n| = 1
and are empty for |n| A 1.) The traces in | = n are the circles
36{2 + 36} 2 = 36 3 n2
C
{2 + } 2 = 1 3
1 2
n ,
36
|n| ? 6, and the
traces in } = n are the ellipses 36{2 + | 2 = 36(1 3 n2 ), |n| ? 1. The
graph is an ellipsoid centered at the origin with intercepts { = ±1, | = ±6,
} = ±1.
18. 4{2 3 16| 2 + } 2 = 16. The traces in { = n are } 2 3 16| 2 = 16 3 4n2 , a
family of hyperbolas for |n| 6= 2 and two intersecting lines when |n| = 2.
(Note that the hyperbolas are oriented differently for |n| ? 2 than for
|n| A 2.) The traces in | = n are 4{2 + } 2 = 16(1 + n2 ), a family of
ellipses, and the traces in } = n are 4{2 3 16| 2 = 16 3 n2 , two
intersecting lines when |n| = 4 and a family of hyperbolas when |n| 6= 4
(oriented differently for |n| ? 4 than for |n| A 4). We recognize the graph
as a hyperboloid of one sheet with axis the |-axis.
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¤
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
19. | = } 2 3 {2 . The traces in { = n are the parabolas | = } 2 3 n2 ;
the traces in | = n are n = } 2 3 {2 , which are hyperbolas (note the hyperbolas
are oriented differently for n A 0 than for n ? 0); and the traces in } = n are
the parabolas | = n2 3 {2 . Thus,
}2
|
{2
= 2 3 2 is a hyperbolic paraboloid.
1
1
1
20. { = | 2 3 } 2 . The traces in { = n are | 2 3 } 2 = n, two intersecting lines
when n = 0 and a family of hyperbolas for n 6= 0 (oriented differently for
n A 0 than for n ? 0). The traces in | = n are the parabolas
{ = 3} 2 + n2 , opening in the negative {-direction, and the traces in } = n
are the parabolas { = | 2 3 n2 which open in the positive {-direction. The
graph is a hyperbolic paraboloid with saddle point (0> 0> 0).
21. This is the equation of an ellipsoid: {2 + 4| 2 + 9} 2 = {2 +
|2
}2
= 1, with {-intercepts ±1, |-intercepts ± 12
2 +
(1@2)
(1@3)2
and }-intercepts ± 13 . So the major axis is the {-axis and the only possible graph is VII.
22. This is the equation of an ellipsoid: 9{2 + 4| 2 + } 2 =
{2
|2
+ } 2 = 1, with {-intercepts ± 13 , |-intercepts ± 12
2 +
(1@3)
(1@2)2
and }-intercepts ±1. So the major axis is the }-axis and the only possible graph is IV.
23. This is the equation of a hyperboloid of one sheet, with d = e = f = 1. Since the coefficient of | 2 is negative, the axis of the
hyperboloid is the |-axis, hence the correct graph is II.
24. This is a hyperboloid of two sheets, with d = e = f = 1. This surface does not intersect the {}-plane at all, so the axis of the
hyperboloid is the |-axis and the graph is III.
25. There are no real values of { and } that satisfy this equation for | ? 0, so this surface does not extend to the left of the
{}-plane. The surface intersects the plane | = n A 0 in an ellipse. Notice that | occurs to the first power whereas { and }
occur to the second power. So the surface is an elliptic paraboloid with axis the |-axis. Its graph is VI.
26. This is the equation of a cone with axis the |-axis, so the graph is I.
27. This surface is a cylinder because the variable | is missing from the equation. The intersection of the surface and the {}-plane
is an ellipse. So the graph is VIII.
28. This is the equation of a hyperbolic paraboloid. The trace in the {|-plane is the parabola | = {2 . So the correct graph is V.
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SECTION 12.6
29. | 2 = {2 + 19 } 2 or | 2 = {2 +
}2
represents an elliptic
9
cone with vertex (0> 0> 0) and axis the |-axis.
CYLINDERS AND QUADRIC SURFACES
30. 4{2 3 | + 2} 2 = 0 or | =
¤
293
}2
|
{2
}2
+
or = {2 +
1@4
1@2
4
2
represents an elliptic paraboloid with vertex (0> 0> 0) and
axis the |-axis.
31. {2 + 2| 3 2} 2 = 0 or 2| = 2} 2 3 {2 or | = } 2 3
{2
2
represents a hyperbolic paraboloid with center (0> 0> 0).
32. | 2 = {2 + 4} 2 + 4 or 3{2 + | 2 3 4} 2 = 4 or
3
{2
|2
+
3 } 2 = 1 represents a hyperboloid of two
4
4
sheets with axis the |-axis.
33. Completing squares in | and } gives
4{2 + (| 3 2)2 + 4(} 3 3)2 = 4 or
(| 3 2)2
+ (} 3 3)2 = 1, an ellipsoid with
4
center (0> 2> 3).
{2 +
34. Completing squares in | and } gives
4(| 3 2)2 + (} 3 2)2 3 { = 0 or
{
(} 3 2)2
= (| 3 2)2 +
, an elliptic paraboloid with
4
4
vertex (0> 2> 2) and axis the horizontal line | = 2, } = 2.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
35. Completing squares in all three variables gives
2
2
2
({ 3 2) 3 (| + 1) + (} 3 1) = 0 or
(| + 1)2 = ({ 3 2)2 + (} 3 1)2 , a circular cone with
center (2> 31> 1) and axis the horizontal line { = 2,
} = 1.
36. Completing squares in all three variables gives
({ 3 1)2 3 (| 3 1)2 + (} + 2)2 = 2 or
({ 3 1)2
(| 3 1)2
(} + 2)2
3
+
= 1, a hyperboloid of
2
2
2
one sheet with center (1> 1> 32) and axis the horizontal
line { = 1, } = 32.
s
s
1 + 4{2 + | 2 , so we plot separately } = 1 + 4{2 + | 2 and
37. Solving the equation for } we get } = ±
s
} = 3 1 + 4{2 + | 2 .
To restrict the }-range as in the second graph, we can use the option view = -4..4 in Maple’s plot3d command, or
PlotRange -A {-4,4} in Mathematica’s Plot3D command.
38. We plot the surface } = {2 3 | 2 .
s
s
s
4{2 + | 2 , so we plot separately } = 4{2 + | 2 and } = 3 4{2 + | 2 .
39. Solving the equation for } we get } = ±
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SECTION 12.6
CYLINDERS AND QUADRIC SURFACES
¤
295
40. We plot the surface } = {2 3 6{ + 4| 2 .
41.
42.
43. The surface is a paraboloid of revolution (circular
44. The surface is a right circular cone with vertex at (0, 0, 0)
paraboloid) with vertex at the origin, axis the |-axis and
opens to the right. Thus the trace in the |}-plane is also a
and axis the {-axis. For { = n 6= 0, the trace is a circle
with center (n, 0, 0) and radius u = | =
parabola: | = } 2 , { = 0. The equation is | = {2 + } 2 .
n
{
= . Thus
3
3
the equation is ({@3)2 = | 2 + } 2 or {2 = 9| 2 + 9} 2 .
45. Let S = ({, |, }) be an arbitrary point equidistant from (31, 0, 0) and the plane { = 1. Then the distance from S to
I
s
({ + 1)2 + | 2 + } 2 and the distance from S to the plane { = 1 is |{ 3 1| @ 12 = |{ 3 1|
s
(by Equation 12.5.9). So |{ 3 1| = ({ + 1)2 + | 2 + } 2 C ({ 3 1)2 = ({ + 1)2 + | 2 + } 2 C
(31, 0, 0) is
{2 3 2{ + 1 = {2 + 2{ + 1 + | 2 + } 2
C 34{ = | 2 + } 2 . Thus the collection of all such points S is a circular
paraboloid with vertex at the origin, axis the {-axis, which opens in the negative direction.
46. Let S = ({> |> }) be an arbitrary point whose distance from the {-axis is twice its distance from the |}-plane. The distance
s
s
from S to the {-axis is ({ 3 {)2 + | 2 + } 2 = | 2 + } 2 and the distance from S to the |}-plane ({ = 0) is |{| @1 = |{|.
s
Thus | 2 + } 2 = 2 |{| C | 2 + } 2 = 4{2 C {2 = (| 2@22 ) + (} 2@22 ). So the surface is a right circular cone with
vertex the origin and axis the {-axis.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
47. (a) An equation for an ellipsoid centered at the origin with intercepts { = ±d, | = ±e, and } = ±f is
{2
|2
}2
+ 2 + 2 = 1.
2
d
e
f
Here the poles of the model intersect the }-axis at } = ±6356=523 and the equator intersects the {- and |-axes at
{ = ±6378=137, | = ±6378=137, so an equation is
{2
|2
}2
+
+
=1
2
2
(6378=137)
(6378=137)
(6356=523)2
{2
|2
n2
+
=
1
3
(6378=137)2
(6378=137)2
(6356=523)2
2
6378=137
n2 .
{2 + | 2 = (6378=137)2 3
6356=523
(b) Traces in } = n are the circles
C
(c) To identify the traces in | = p{ we substitute | = p{ into the equation of the ellipsoid:
{2
(p{)2
}2
+
+
=1
2
2
(6378=137)
(6378=137)
(6356=523)2
(1 + p2 ){2
}2
+
=1
2
(6378=137)
(6356=523)2
{2
}2
=1
+
2
2
(6378=137) @(1 + p )
(6356=523)2
As expected, this is a family of ellipses.
48. If we position the hyperboloid on coordinate axes so that it is centered at the origin with axis the }-axis then its equation is
given by
|2
}2
{2
|2
n2
{2
+
3
=
1.
Horizontal
traces
in
}
=
n
are
+
=
1
+
, a family of ellipses, but we know that the
d2
e2
f2
d2
e2
f2
traces are circles so we must have d = e. The trace in } = 0 is
|2
{2
+
= 1 C {2 + | 2 = d2 and since the minimum
d2
d2
radius of 100 m occurs there, we must have d = 100. The base of the tower is the trace in } = 3500 given by
|2
(3500)2
{2
1
+ 2 =1+
but d = 100 so the trace is {2 + | 2 = 1002 + 50,0002 2 . We know the base is a circle of
2
d
d
f2
f
radius 140, so we must have 1002 + 50,0002
tower is
1
= 1402
f2
i
f2 =
50,0002
781,250
=
and an equation for the
1402 3 1002
3
|2
}2
{2
{2
|2
3} 2
+
3
=
1
or
+
3
= 1, 3500 $ } $ 500.
1002
1002
(781,250)@3
10,000
10,000
781,250
49. If (d> e> f) satisfies } = | 2 3 {2 , then f = e2 3 d2 . O1 : { = d + w, | = e + w, } = f + 2(e 3 d)w,
O2 : { = d + w, | = e 3 w, } = f 3 2(e + d)w. Substitute the parametric equations of O1 into the equation
of the hyperbolic paraboloid in order to find the points of intersection: } = |2 3 {2
i
f + 2(e 3 d)w = (e + w)2 3 (d + w)2 = e2 3 d2 + 2(e 3 d)w i f = e2 3 d2 . As this is true for all values of w,
O1 lies on } = | 2 3 {2 . Performing similar operations with O2 gives: } = |2 3 {2
2
2
2
2
2
i
2
f 3 2(e + d)w = (e 3 w) 3 (d + w) = e 3 d 3 2(e + d)w i f = e 3 d . This tells us that all of O2 also lies on
} = | 2 3 {2 .
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CHAPTER 12 REVIEW
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297
50. Any point on the curve of intersection must satisfy both 2{2 + 4| 2 3 2} 2 + 6{ = 2 and 2{2 + 4| 2 3 2} 2 3 5| = 0.
Subtracting, we get 6{ + 5| = 2, which is linear and therefore the equation of a plane. Thus the curve of intersection lies in
this plane.
51.
The curve of intersection looks like a bent ellipse. The projection
of this curve onto the {|-plane is the set of points ({> |> 0) which
satisfy {2 + | 2 = 1 3 | 2
C {2 + 2| 2 = 1 C
|2
{2 + I 2 = 1. This is an equation of an ellipse.
1@ 2
12 Review
1. A scalar is a real number, while a vector is a quantity that has both a real-valued 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 f A 0, f a is a vector with the same direction as a and length f times the length of a. If f ? 0, fa points in the opposite
direction as a and has length |f| times the length of a. (See Figures 7 and 15 in Section 12.2.) Algebraically, to find f a we
multiply each component of a by f.
4. See (1) in Section 12.2.
5. See Theorem 12.3.3 and Definition 12.3.1.
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. Also, the dot product can be used to determine the
work done moving an object given the force and displacement vectors.
7. See the boxed equations as well as Figures 4 and 5 and the accompanying discussion on page 828 [ET 804].
8. 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: |a × b|.
(b) The volume of the parallelepiped determined by a, b, and c is the magnitude of their scalar triple product: |a · (b × c)|.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
11. If an equation of the plane is known, it can be written as d{ + e| + f} + g = 0. A normal vector, which is perpendicular to the
plane, is hd> e> fi (or any scalar multiple of hd> e> fi). If an equation is not known, we can use points on the plane to find two
non-parallel 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.
3
3
<
3<
16. (a) Determine the vectors S T = hd1 > d2 > d3 i and S U = he1 > e2 > e3 i. If there is a scalar w such that
hd1 > d2 > d3 i = w he1 > e2 > e3 i, then the vectors are parallel and the points must all lie on the same line.
3
3
< 3<
3
3
<
3<
Alternatively, if S T × S U = 0, then S T and S U are parallel, so S , T, and U are collinear.
Thirdly, an algebraic method is to determine an equation of the line joining two of the points, and then check whether or
not the third point satisfies this equation.
3
3
<
3<
3<
(b) Find the vectors S T = a, S U = b, S V = c. a × b is normal to the plane formed by S , T and U, and so V lies on this
plane if a × b and c are orthogonal, that is, if (a × b) · c = 0. (Or use the reasoning in Example 5 in Section 12.4.)
Alternatively, find an equation for the plane determined by three of the points and check whether or not the fourth point
satisfies this equation.
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 { = n (parallel to the |}-plane) by setting { = n and determining the curve represented by the resulting
equation. Traces in the planes | = n (parallel to the {}-plane) and } = n (parallel to the {|-plane) are found similarly.
19. See Table 1 in Section 12.6.
1. This is false, as the dot product of two vectors is a scalar, not a vector.
2. False. For example, if u = i and v = 3i then |u + v| = |0| = 0 but |u| + |v| = 1 + 1 = 2.
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CHAPTER 12 REVIEW
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299
3. False. For example, if u = i and v = j then |u · v| = |0| = 0 but |u| |v| = 1 · 1 = 1. In fact, by Theorem 12.3.3,
|u · v| = |u| |v| cos .
4. False. For example, |i × i| = |0| = 0 (see Example 12.4.2) but |i| |i| = 1 · 1 = 1. In fact, by Thereom 12.4.9,
|u × v| = |u| |v| sin .
5. True, by Theorem 12.3.2, property 2.
6. False. Property 1 of Theorem 12.4.11 says that u × v = 3v × u.
7. True. If is the angle between u and v, then by Theorem 12.4.9, |u × v| = |u| |v| sin = |v| |u| sin = |v × u|.
(Or, by Theorem 12.4.11, |u × v| = |3v × u| = |31| |v × u| = |v × u|.)
8. This is true by Theorem 12.3.2, property 4.
9. Theorem 12.4.11, property 2 tells us that this is true.
10. This is true by Theorem 12.4.11, property 4.
11. This is true by Theorem 12.4.11, property 5.
12. In general, this assertion is false; a counterexample is i × (i × j) 6= (i × i) × j. (See the paragraph preceding Theorem
12.4.11.)
13. This is true because u × v is orthogonal to u (see Theorem 12.4.8), and the dot product of two orthogonal vectors is 0.
14. (u + v) × v = u × v + v × v
=u×v+0
[by Theorem 12.4.11, property 4]
[by Example 12.4.2]
= u × v, so this is true.
15. This is false. A normal vector to the plane is n = h6> 32> 4i. Because h3> 31> 2i =
1
n,
2
the vector is parallel to n and hence
perpendicular to the plane.
16. This is false, because according to Equation 12.5.8, d{ + e| + f} + g = 0 is the general equation of a plane.
17. This is false. In R2 , {2 + | 2 = 1 represents a circle, but ({> |> }) | {2 + | 2 = 1 represents a three-dimensional surface,
namely, a circular cylinder with axis the }-axis.
18. This is false. In R3 the graph of | = {2 is a parabolic cylinder (see Example 12.6.1). A paraboloid has an equation such as
} = {2 + | 2 .
19. False. For example, i · j = 0 but i 6= 0 and j 6= 0.
20. This is false. By Corollary 12.4.10, u × v = 0 for any nonzero parallel vectors u, v. For instance, i × i = 0.
21. This is true. If u and v are both nonzero, then by (7) in Section 12.3, u · v = 0 implies that u and v are orthogonal. But
u × 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.
22. This is true. We know u · v = |u| |v| cos where |u| D 0, |v| D 0, and |cos | $ 1, so |u · v| = |u| |v| |cos | $ |u| |v|.
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
1. (a) The radius of the sphere is the distance between the points (31> 2> 1) and (6> 32> 3), namely,
s
I
[6 3 (31)]2 + (32 3 2)2 + (3 3 1)2 = 69. By the formula for an equation of a sphere (see page 813 [ET 789]),
I
an equation of the sphere with center (31> 2> 1) and radius 69 is ({ + 1)2 + (| 3 2)2 + (} 3 1)2 = 69.
(b) The intersection of this sphere with the |}-plane is the set of points on the sphere whose {-coordinate is 0. Putting { = 0
into the equation, we have (| 3 2)2 + (} 3 1)2 = 68> { = 0 which represents a circle in the |}-plane with center (0> 2> 1)
I
and radius 68.
(c) Completing squares gives ({ 3 4)2 + (| + 1)2 + (} + 3)2 = 31 + 16 + 1 + 9 = 25. Thus the sphere is centered at
(4> 31> 33) and has radius 5.
2. (a)
(b)
(c)
(d)
3. u · v = |u| |v| cos 45 = (2)(3)
I
2
2
=3
I
I
I
2. |u × v| = |u| |v| sin 45 = (2)(3) 22 = 3 2.
By the right-hand rule, u × v is directed out of the page.
4. (a) 2 a + 3 b = 2 i + 2 j 3 4 k + 9 i 3 6 j + 3 k = 11 i 3 4 j 3 k
(b) |b| =
I
I
9 + 4 + 1 = 14
(c) a · b = (1)(3) + (1)(32) + (32)(1) = 31
i j k
(d) a × b = 1 1 32 = (1 3 4) i 3 (1 + 6) j + (32 3 3) k = 33 i 3 7 j 3 5 k
3 32 1 i j k
(e) b × c = 3 32 1 = 9 i + 15 j + 3 k,
0 1 35 |b × c| = 3
I
I
9 + 25 + 1 = 3 35
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CHAPTER 12 REVIEW
1 1 32 3 32 3 1
32 1 3
32
1
3
2
3
=
(f ) a · (b × c) = = 9 + 15 3 6 = 18
0 1
0 35 1 35 0 1 35 (g) c × c = 0 for any c.
(h) From part (e),
i j k
a × (b × c) = a × (9 i + 15 j + 3 k) = 1 1 32 9 15 3 = (3 + 30) i 3 (3 + 18) j + (15 3 9) k = 33 i 3 21 j + 6 k
(i) The scalar projection is compa b = |b| cos = a · b@ |a| = 3 I16 .
1
( j) The vector projection is proja b = 3 I
6
(k) cos =
a
|a|
= 3 16 (i + j 3 2 k).
31
31
a·b
31
I
= I I = I and = cos31
E 96 .
|a| |b|
6 14
2 21
2 21
5. For the two vectors to be orthogonal, we need h3> 2> {i · h2{> 4> {i = 0
{2 + 6{ + 8 = 0
C
({ + 2)({ + 4) = 0
C
C
(3)(2{) + (2)(4) + ({)({) = 0
C
{ = 32 or { = 34.
6. We know that the cross product of two vectors is orthogonal to both. So we calculate
( j + 2 k) × (i 3 2 j + 3 k) = [3 3 (34)] i 3 (0 3 2) j + (0 3 1) k = 7 i + 2 j 3 k.
7i + 2j 3 k
1
Then two unit vectors orthogonal to both given vectors are ± s
= ± I (7 i + 2 j 3 k),
3 6
72 + 22 + (31)2
that is,
3
7
I
6
i+
3
2
I
6
j3
3
1
I
6
7
k and 3 3 I
i3
6
3
2
I
6
j+
3
1
I
6
k.
7. (a) (u × v) · w = u · (v × w) = 2
(b) u · (w × v) = u · [3 (v × w)] = 3u · (v × w) = 32
(c) v · (u × w) = (v × u) · w = 3 (u × v) · w = 32
(d) (u × v) · v = u · (v × v) = u · 0 = 0
8. (a × b) · [(b × c) × (c × a)] = (a × b) · ([(b × c) · a] c 3 [(b × c) · c] a)
[by Property 6 of Theorem 12.4.11]
= (a × b) · [(b × c) · a] c = [a · (b × c)] (a × b) · c
= [a · (b × c)] [a · (b × c)] = [a · (b × c)]2
9. For simplicity, consider a unit cube positioned with its back left corner at the origin. Vector representations of the diagonals
joining the points (0> 0> 0) to (1> 1> 1) and (1> 0> 0) to (0> 1> 1) are h1> 1> 1i and h31> 1> 1i. Let be the angle between these
two vectors. h1> 1> 1i · h31> 1> 1i = 31 + 1 + 1 = 1 = |h1> 1> 1i| |h31> 1> 1i| cos = 3 cos = cos31 13 E 71 .
i cos =
1
3
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3<
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
3<
33<
10. DE = h1> 3> 31i, DF = h32> 1> 3i and DG = h31> 3> 1i. By Equation 12.4.13,
1 3 31 32 3 32 1 1 3
3< 3< 33<
DE · DF × DG = 32 1 3 = = 38 3 3 + 5 = 36.
3
3 3
31 1 31 3 3 1
31 3 1 3< 3< 33<
The volume is DE · DF × DG = 6 cubic units.
3<
3<
11. DE = h1> 0> 31i, DF = h0> 4> 3i, so
3< 3<
(a) a vector perpendicular to the plane is DE × DF = h0 + 4> 3(3 + 0)> 4 3 0i = h4> 33> 4i.
3< 3< I
I
(b) 12 DE × DF = 12 16 + 9 + 16 = 241 .
12. D = 4 i + 3 j + 6 k,
Z = F · D = 12 + 15 + 60 = 87 J
13. Let I1 be the magnitude of the force directed 20 away from the direction of shore, and let I2 be the magnitude of the other
force. Separating these forces into components parallel to the direction of the resultant force and perpendicular to it gives
I1 cos 20 + I2 cos 30 = 255 (1), and I1 sin 20 3 I2 sin 30 = 0 i I1 = I2
sin 30
(2). Substituting (2)
sin 20
into (1) gives I2 (sin 30 cot 20 + cos 30 ) = 255 i I2 E 114 N. Substituting this into (2) gives I1 E 166 N.
14. | | = |r| |F| sin = (0=40)(50) sin(90 3 30 ) E 17=3 N·m.
15. The line has direction v = h33> 2> 3i. Letting S0 = (4> 31> 2), parametric equations are
{ = 4 3 3w, | = 31 + 2w, } = 2 + 3w.
16. A direction vector for the line is v = h3> 2> 1i, so parametric equations for the line are { = 1 + 3w, | = 2w, } = 31 + w.
17. A direction vector for the line is a normal vector for the plane, n = h2> 31> 5i, and parametric equations for the line are
{ = 32 + 2w, | = 2 3 w, } = 4 + 5w.
18. Since the two planes are parallel, they will have the same normal vectors. Then we can take n = h1> 4> 33i and an equation of
the plane is 1({ 3 2) + 4(| 3 1) 3 3(} 3 0) = 0 or { + 4| 3 3} = 6.
19. Here the vectors a = h4 3 3> 0 3 (31) > 2 3 1i = h1> 1> 1i and b = h6 3 3> 3 3 (31)> 1 3 1i = h3> 4> 0i lie in the plane,
so n = a × b = h34> 3> 1i is a normal vector to the plane and an equation of the plane is
34({ 3 3) + 3(| 3 (31)) + 1(} 3 1) = 0 or 34{ + 3| + } = 314.
20. If we first find two nonparallel vectors in the plane, their cross product will be a normal vector to the plane. Since the given
line lies in the plane, its direction vector a = h2> 31> 3i is one vector in the plane. We can verify that the given point (1> 2> 32)
does not lie on this line. The point (0> 3> 1) is on the line (obtained by putting w = 0) and hence in the plane, so the vector
b = h0 3 1> 3 3 2> 1 3 (32)i = h31> 1> 3i lies in the plane, and a normal vector is n = a × b = h36> 39> 1i. Thus an
equation of the plane is 36({ 3 1) 3 9(| 3 2) + (} + 2) = 0 or 6{ + 9| 3 } = 26.
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CHAPTER 12 REVIEW
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303
21. Substitution of the parametric equations into the equation of the plane gives 2{ 3 | + } = 2(2 3 w) 3 (1 + 3w) + 4w = 2
i
3w + 3 = 2 i w = 1. When w = 1, the parametric equations give { = 2 3 1 = 1, | = 1 + 3 = 4 and } = 4. Therefore,
the point of intersection is (1> 4> 4).
22. Use the formula proven in Exercise 12.4.45(a). In the notation used in that exercise, a is just the direction of the line; that is,
a = h1> 31> 2i. A point on the line is (1> 2> 31) (setting w = 0), and therefore b = h1 3 0> 2 3 0> 31 3 0i = h1> 2> 31i.
u
|a × b|
|h33> 3> 3i|
|h1> 31> 2i × h1> 2> 31i|
27
3
I
I
Hence g =
=
=
=
= I .
|a|
6
1+1+4
6
2
23. Since the direction vectors h2> 3> 4i and h6> 31> 2i aren’t parallel, neither are the lines. For the lines to intersect, the three
equations 1 + 2w = 31 + 6v, 2 + 3w = 3 3 v, 3 + 4w = 35 + 2v must be satisfied simultaneously. Solving the first two
equations gives w = 15 , v =
2
5
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.
24. (a) The normal vectors are h1> 1> 31i and h2> 33> 4i. Since these vectors aren’t parallel, neither are the planes parallel.
Also h1> 1> 31i · h2> 33> 4i = 2 3 3 3 4 = 35 6= 0 so the normal vectors, and thus the planes, are not perpendicular.
(b) cos =
h1> 1> 31i · h2> 33> 4i
5
I I
= 3 I and = cos31 3 I587 E 122
3 29
87
[or we can say E 58 ].
25. n1 = h1> 0> 31i and n2 = h0> 1> 2i. Setting } = 0, it is easy to see that (1> 3> 0) is a point on the line of intersection of
{ 3 } = 1 and | + 2} = 3. The direction of this line is v1 = n1 × n2 = h1> 32> 1i. A second vector parallel to the desired
plane is v2 = h1> 1> 32i, since it is perpendicular to { + | 3 2} = 1. Therefore, the normal of the plane in question is
n = v1 × v2 = h4 3 1> 1 + 2> 1 + 2i = 3 h1> 1> 1i. Taking ({0 > |0 > }0 ) = (1> 3> 0), the equation we are looking for is
({ 3 1) + (| 3 3) + } = 0 C { + | + } = 4.
3<
3<
26. (a) The vectors DE = h31 3 2> 31 3 1> 10 3 1i = h33> 32> 9i and DF = h1 3 2> 3 3 1> 34 3 1i = h31> 2> 35i lie in the
3< 3<
plane, so n = DE × DF = h33> 32> 9i × h31> 2> 35i = h38> 324> 38i or equivalently h1> 3> 1i is a normal vector to
the plane. The point D(2> 1> 1) lies on the plane so an equation of the plane is 1({ 3 2) + 3(| 3 1) + 1(} 3 1) = 0 or
{ + 3| + } = 6.
(b) The line is perpendicular to the plane so it is parallel to a normal vector for the plane, namely h1> 3> 1i. If the line passes
through E(31> 31> 10) then symmetric equations are
{ 3 (31)
| 3 (31)
} 3 10
|+1
=
=
or { + 1 =
= } 3 10.
1
3
1
3
(c) Normal vectors for the two planes are n1 = h1> 3> 1i and n2 = h2> 34> 33i. The angle between the planes is given by
cos =
h1> 3> 1i · h2> 34> 33i
n1 · n2
2 3 12 3 3
13
s
= I
= I I
= 3I
2
2
2
2
2
2
|n1 | |n2 |
11 29
319
1 +3 +1
2 + (34) + (33)
13
E 137 or 180 3 137 = 43 .
Thus = cos31 3 I
319
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
(d) From part (c), the point (2> 0> 4) lies on the second plane, but notice that the point also satisfies the equation of the first
plane, so the point lies on the line of intersection of the planes. A vector v in the direction of this intersecting line is
perpendicular to the normal vectors of both planes, so take v = n1 × n2 = h1> 3> 1i × h2> 34> 33i = h35> 5> 310i or
equivalently we can take v = h1> 31> 2i. Parametric equations for the line are { = 2 + w, | = 3w, } = 4 + 2w.
|32 3 (324)|
22
= I .
26
32 + 12 + (34)2
27. By Exercise 12.5.75, G = s
28. The equation { = 3 represents a plane parallel to the
|}-plane and 3 units in front of it.
29. The equation { = } represents a plane perpendicular to
the {}-plane and intersecting the {}-plane in the line
{ = }, | = 0.
30. The equation | = } 2 represents a parabolic cylinder
whose trace in the {}-plane is the {-axis and which opens
31. The equation {2 = | 2 + 4} 2 represents a (right elliptical)
cone with vertex at the origin and axis the {-axis.
to the right.
32. 4{ 3 | + 2} = 4 is a plane with intercepts
(1> 0> 0), (0> 34> 0), and (0> 0> 2).
33. An equivalent equation is 3{2 +
|2
3 } 2 = 1, a
4
hyperboloid of two sheets with axis the |-axis. For
||| A 2, traces parallel to the {}-plane are circles.
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CHAPTER 12 REVIEW
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305
34. An equivalent equation is 3{2 + | 2 + } 2 = 1,
a hyperboloid of one sheet with axis the {-axis.
35. Completing the square in | gives
36. Completing the square in | and } gives
}2
4{2 + 4(| 3 1)2 + } 2 = 4 or {2 + (| 3 1)2 +
= 1,
4
an ellipsoid centered at (0> 1> 0).
37. 4{2 + | 2 = 16
C
{ = (| 3 1)2 + (} 3 2)2 , a circular paraboloid with
vertex (0> 1> 2) and axis the horizontal line | = 1, } = 2.
{2
|2
{2
|2
}2
+
= 1. The equation of the ellipsoid is
+
+ 2 = 1, since the horizontal trace in the
4
16
4
16
f
plane } = 0 must be the original ellipse. The traces of the ellipsoid in the |}-plane must be circles since the surface is obtained
by rotation about the {-axis. Therefore, f2 = 16 and the equation of the ellipsoid is
|2
}2
{2
+
+
=1 C
4
16
16
4{2 + | 2 + } 2 = 16.
38. The distance from a point S ({> |> }) to the plane | = 1 is || 3 1|, so the given condition becomes
|| 3 1| = 2
s
({ 3 0)2 + (| + 1)2 + (} 3 0)2
(| 3 1)2 = 4{2 + 4(| + 1)2 + 4} 2
2
= 4{2 + 3 | + 53 + 4} 2
i || 3 1| = 2
s
{2 + (| + 1)2 + } 2
C 33 = 4{2 + (3|2 + 10|) + 4} 2
2
16
9
i 34 {2 + 16
| + 53 + 34 } 2 = 1.
3
This is the equation of an ellipsoid whose center is 0> 3 53 > 0 .
i
C
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PROBLEMS PLUS
1. Since three-dimensional 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 u are contained in a
square of side 1 m so that the circles touch each other and four of the circles touch
two sides of the square, find u.
I
The diagonal of the square is 2. The diagonal is also 4u + 2{. But { is the diagonal of a smaller square of side u. Therefore
I
I
I
I I
2 = 4u + 2{ = 4u + 2 2 u = 4 + 2 2 u i u = 4 + 22I2 .
{ = 2u i
I
I
Let’s use these ideas to solve the original three-dimensional problem. The diagonal of the cube is 12 + 12 + 12 = 3.
The diagonal of the cube is also 4u + 2{ where { is the diagonal of a smaller cube with edge u. Therefore
I
I
I
I
I
I I
3
2 333
2
2
2
I =
.
3 = 4u + 2{ = 4u + 2 3 u = 4 + 2 3 u. Thus u =
{ = u +u + u = 3u i
2
4+2 3
I
3 3 32 m.
The radius of each ball is
2. Try an analogous problem in two dimensions. Consider a rectangle with
length O and width Z and find the area of V in terms of O and Z . Since V
contains E, it has area
D(V) = OZ + the area of two O × 1 rectangles
+ the area of two 1 × Z rectangles
+ the area of four quarter-circles of radius 1
as seen in the diagram. So D(V) = OZ + 2O + 2Z + · 12 .
Now in three dimensions, the volume of V is
OZ K + 2(O × Z × 1) + 2(1 × Z × K) + 2(O × 1 × K)
+ the volume of 4 quarter-cylinders with radius 1 and height Z
+ the volume of 4 quarter-cylinders with radius 1 and height O
+ the volume of 4 quarter-cylinders with radius 1 and height K
+ the volume of 8 eighths of a sphere of radius 1
So
Y (V) = OZ K + 2OZ + 2Z K + 2OK + · 12 · Z + · 12 · O + · 12 · K + 43 · 13
= OZ K + 2(OZ + Z K + OK) + (O + Z + K) + 43 =
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CHAPTER 12 PROBLEMS PLUS
3. (a) We find the line of intersection O as in Example 12.5.7(b). Observe that the point (31> f> f) lies on both planes. Now since
O lies in both planes, it is perpendicular to both of the normal vectors n1 and n2 , and thus parallel to their cross product
i
j k
n1 × n2 = f 1 1 = 2f> 3f2 + 1> 3f2 3 1 . So symmetric equations of O can be written as
1 3f f {+1
|3f
}3f
= 2
= 2
, provided that f 6= 0, ±1.
32f
f 31
f +1
If f = 0, then the two planes are given by | + } = 0 and { = 31, so symmetric equations of O are { = 31, | = 3}. If
f = 31, then the two planes are given by 3{ + | + } = 31 and { + | + } = 31, and they intersect in the line { = 0,
| = 3} 3 1. If f = 1, then the two planes are given by { + | + } = 1 and { 3 | + } = 1, and they intersect in the line
| = 0, { = 1 3 }.
(b) If we set } = w in the symmetric equations and solve for { and | separately, we get { + 1 =
|3f=
(w 3 f)(f2 3 1)
f2 + 1
(w 3 f)(32f)
,
f2 + 1
32fw + (f2 3 1)
(f2 3 1)w + 2f
, |=
. Eliminating f from these equations, we
2
f +1
f2 + 1
i {=
have {2 + | 2 = w2 + 1. So the curve traced out by O in the plane } = w is a circle with center at (0> 0> w) and
I
radius w2 + 1.
(c) The area of a horizontal cross-section of the solid is D(}) = (} 2 + 1), so Y =
U1
0
D(})g} = 1
3
}3 + }
1
0
=
4
.
3
4. (a) We consider velocity vectors for the plane and the wind. Let vl be the initial, intended
velocity for the plane and vj the actual velocity relative to the ground. If w is the velocity
of the wind, vj is the resultant, that is, the vector sum vl + w as shown in the figure. We
know vl = 180 j, and since the plane actually flew 80 km in
1
2
hour, |vj | = 160. Thus
vj = (160 cos 85 ) i + (160 sin 85 ) j E 13=9 i + 159=4 j. Finally, vl + w = vj , so
w = vj 3 vl E 13=9 i 3 20=6 j. Thus, the wind velocity is about 13=9 i 3 20=6 j, and the
wind speed is |w| E
s
(13=9)2 + (320=6)2 E 24=9 km@h.
(b) Let v be the velocity the pilot should have taken. The pilot would need to head a little west of north to compensate
for the wind, so let be the angle v makes with the north direction. Then we can write
v = h180 cos( + 90 )> 180 sin( + 90 )i. With the effect of the wind, the actual velocity (with respect to the ground)
will be v + w, which we want to be due north. Equating horizontal components of the vectors, we must have
180 cos( + 90 ) + 160 cos 85 = 0
i
cos( + 90 ) = 3 160
cos 85 E 30=0775, so
180
E cos31 (30=0775) 3 90 E 4=4 . Thus the pilot should have headed about 4=4 west of north.
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CHAPTER 12 PROBLEMS PLUS
v4 = projv2 v3 =
v2 · 252 v1
v2 · v3
5
52
v2 = 2 2 (v1 · v2 ) v2 = 2 2 v2
2 v2 =
2
2 ·3
2 ·3
|v2 |
|v2 |
v5 = projv3 v4 =
v3 · v4
v3 =
|v3 |2
Thus
309
v1 · v2
5
5
5
v so |v3 | = 2 |v1 | = ,
2 v1 =
2 1
2
2
2
|v1 |
5. v3 = projv1 v2 =
|v5 | =
¤
5
22
2
v1 · 225 32 v2
5 2
2
5
v1
22
=
i |v4 | =
53
52
(v1 · v2 ) v1 = 4 2 v1
2
·3
2 ·3
24
52
52
,
|v
|
=
2
22 · 32
22 · 3
i
q32
53
53
54
55
5q32
|v1 | = 3 2 . Similarly, |v6 | = 4 3 , |v7 | = 5 4 , and in general, |vq | = q32 q33 = 3 56
.
2
·3
2 ·3
2 ·3
2 ·3
2
·3
24
"
S
q=1
|vq | = |v1 | + |v2 | +
=5+
"
S
q=1
"
S
q=3
5 5 q31
2
6
"
q32
q
S
3 56
= 2+3+
3 56
q=1
=5+
5
2
13
5
6
[sum of a geometric series]
= 5 + 15 = 20
6. Completing squares in the inequality {2 + | 2 + } 2 ? 136 + 2({ + 2| + 3})
gives ({ 3 1)2 + (| 3 2)2 + (} 3 3)2 ? 150 which describes the set of all
points ({> |> }) whose distance from the point S (1> 2> 3) is less than
I
I
I
I
150 = 5 6. The distance from S to T (31> 1> 4) is 4 + 1 + 1 = 6,
so the largest possible sphere that passes through T and satisfies the stated
I
conditions extends 5 6 units in the opposite direction, giving a diameter of
I
6 6. (See the figure.)
Thus the radius of the desired sphere is 3
direction is u =
I1
6
I
I
6, and its center is 3 6 units from T in the direction of S . A unit vector in this
h2> 1> 31i, so starting at T(31> 1> 4) and following the vector 3
I
6 u = h6> 3> 33i we arrive at the center
I 2
of the sphere, (5> 4> 1). An equation of the sphere then is ({ 3 5)2 + (| 3 4)2 + (} 3 1)2 = 3 6
or ({ 3 5)2 + (| 3 4)2 + (} 3 1)2 = 54.
7. (a) When = v , the block is not moving, so the sum of the forces on the block
must be 0, thus N + F + W = 0= This relationship is illustrated
geometrically in the figure. Since the vectors form a right triangle, we have
tan(v ) =
q
|F|
= v = v .
|N|
q
(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 F lies along the positive {-axis and the inclined plane
is parallel to the {-axis. (See the following figure.)
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¤
CHAPTER 12 PROBLEMS PLUS
|F| is maximal, so |F| = v q for A v . Then the vectors, in terms of components parallel and perpendicular to the
inclined plane, are
N=qj
F = (v q) i
W = (3pj sin ) i + (3pj cos ) j
H = (kmin cos ) i + (3kmin sin ) j
Equating components, we have
v q 3 pj sin + kmin cos = 0
i
kmin cos + v q = pj sin (1)
q 3 pj cos 3 kmin sin = 0
i
kmin sin + pj cos = q
(2)
(c) Since (2) is solved for q, we substitute into (1):
kmin cos + v (kmin sin + pj cos ) = pj sin i
kmin cos + kmin v sin = pj sin 3 pjv cos kmin = pj
sin 3 v cos cos + v sin From part (a) we know v = tan v , so this becomes kmin
= pj
tan 3 v
1 + v tan tan 3 tan v
= pj
1 + tan v tan i
and using a trigonometric identity,
this is pj tan( 3 v ) as desired.
Note for = v , kmin = pj tan 0 = 0, which makes sense since the block is at rest for v , thus no additional force H
is necessary to prevent it from moving. As increases, the factor tan( 3 v ), and hence the value of kmin , increases
slowly for small values of 3 v but much more rapidly as 3 v becomes significant. This seems reasonable, as the
steeper the inclined plane, the less the horizontal 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 < 90 , corresponding
to the inclined plane being placed vertically, the value of kmin 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 v = 0), we would have
< 90
i kmin < ", and it would be impossible to keep the block from slipping.
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CHAPTER 12 PROBLEMS PLUS
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311
(d) Since kmax is the largest value of k that keeps the block from slipping, 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 = 3(v q) i. (Note that |F| is again maximal.) Following our procedure in parts (b) and (c), we
equate components:
3v q 3 pj sin + kmax cos = 0 i kmax cos 3 v q = pj sin q 3 pj cos 3 kmax sin = 0 i kmax sin + pj cos = q
Then substituting,
kmax cos 3 v (kmax sin + pj cos ) = pj sin i
kmax cos 3 kmax v sin = pj sin + pjv cos kmax = pj
= pj
sin + v cos cos 3 v sin tan + tan v
1 3 tan v tan = pj
tan + v
1 3 v tan i
= pj tan( + v )
We would expect kmax to increase as increases, with similar behavior as we established for kmin , but with kmax values
always larger than kmin . We can see that this is the case if we graph kmax as a function of , as the curve is the graph of
kmin translated 2v to the left, so the equation does seem reasonable. Notice that the equation predicts kmax < " as
< (90 3 v ). In fact, as kmax increases, the normal force increases as well. When (90 3 v ) $ $ 90 , the
horizontal force is completely counteracted by the sum of the normal and frictional forces, so no part of the horizontal
force contributes to moving the block up the plane no matter how large its magnitude.
8. (a) The largest possible solid is achieved by starting with a circular cylinder of diameter 1 with axis the }-axis and with a
height of 1. This is the largest solid that creates a square shadow with side length 1 in the |-direction and a circular disk
shadow in the }-direction. For convenience, we place the base of the
cylinder on the {|-plane so that it intersects the {- and |-axes at ± 12 .
We then remove as little as possible from the solid that leaves an
isosceles triangle shadow in the {-direction. This is achieved by
cutting the solid with planes parallel to the {-axis that intersect the
}-axis at 1 and the |-axis at ± 12 (see the figure).
To compute the volume of this solid, we take vertical slices parallel to the {}-plane. The equation of the base of the solid
is {2 + | 2 = 14 , so a cross-section | units to the right of the origin is a rectangle with base 2
t
1
4
3 | 2 . For 0 $ | $ 12 ,
the solid is cut off on top by the plane } = 1 3 2|, so the height of the rectangular cross-section is 1 3 2| and the
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CHAPTER 12 PROBLEMS PLUS
cross-sectional area is D(|) = 2
]
0
1@2
2
t
1
4
t
1
4
3 | 2 (1 3 2|). The volume of the right half of the solid is
]
3 | 2 (1 3 2|) g| = 2
0
1@2 t
1
4
]
3 | 2 g| 3 4
1@2
0
|
t
1
4
3 | 2 g|
k 3@2 l1@2
= 2 14 area of a circle of radius 12 3 4 3 13 14 3 | 2
0
Thus the volume of the solid is 2 8 3 16 =
k
3@2 l
2 l 4 k
+ 3 0 3 14
=
= 2 14 · 12
4
3
1
3
8
3
1
6
E 0=45.
(b) There is not a smallest volume. We can remove portions of the solid from
part (a) to create smaller and smaller volumes as long as we leave the
“skeleton” shown in the figure intact (which has no volume at all and is not a
solid). Thus we can create solids with arbitrarily small volume.
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13
VECTOR FUNCTIONS
13.1 Vector Functions and Space Curves
1. The component functions
I
4 3 w2 , h33w , and ln(w + 1) are all defined when 4 3 w2 D 0 i 32 $ w $ 2 and
w + 1 A 0 i w A 31, so the domain of r is (31> 2].
2. The component functions
w32
, sin w, and ln(9 3 w2 ) are all defined when w 6= 32 and 9 3 w2 A 0 i 33 ? w ? 3,
w+2
so the domain of r is (33> 32) (32> 3).
3. lim h33w = h0 = 1, lim
w<0
w<0
w2
1
1
1
1
=
= = lim
2 = 2 = 1,
1
sin2 w w<0 sin2 w
sin2 w
sin w
lim 2
lim
w<0
w2
w
w<0
w
and lim cos 2w = cos 0 = 1. Thus
w<0
lim h33w i +
w<0
4. lim
w<1
k
k
l
l
w2
w2
33w
j + lim cos 2w k = i + j + k.
i
+
lim
h
j
+
cos
2w
k
=
lim
2
2
w<0
w<0 sin w
w<0
sin w
I
w2 3 w
w (w 3 1)
sin w
cos w
= lim
= lim w = 1, lim w + 8 = 3, lim
= lim
= 3
w<1
w<1
w<1
w<1 ln w
w<1
w31
w31
1@w
[by l’Hospital’s Rule].
Thus the given limit equals i + 3 j 3 k.
1 + w2
(1@w2 ) + 1
0+1
=
= 31, lim tan31 w =
= lim
2
w<" 1 3 w
w<" (1@w2 ) 3 1
w<"
031
2
32w
1+w
13h
= 31> 2 > 0 .
> tan31 w>
lim
w<"
1 3 w2
w
5. lim
6. lim wh3w = lim
w<"
w<"
and lim w sin
w<"
Thus lim
w<"
, lim
2 w<"
1 3 h32w
1
1
= lim
3 2w = 0 3 0 = 0. Thus
w<" w
w
wh
w
1
w3 + w
1 + (1@w2 )
1+0
1
= lim
=
= ,
= lim w = 0 [by l’Hospital’s Rule], lim
w
3
w<" h
w<" 2w 3 1
w<" 2 3 (1@w3 )
h
230
2
1
1
sin(1@w)
cos(1@w)(31@w2 )
= lim cos = cos 0 = 1 [again by l’Hospital’s Rule].
= lim
= lim
w<"
w<"
w<"
w
1@w
31@w2
w
1
w3 + w
> w sin
wh3w > 3
2w 3 1
w
= 0> 12 > 1 .
7. The corresponding parametric equations for this curve are { = sin w, | = w.
We can make a table of values, or we can eliminate the parameter: w = |
i
{ = sin |, with | M R. By comparing different values of w, we find the direction in
which w increases as indicated in the graph.
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CHAPTER 13 VECTOR FUNCTIONS
8. The corresponding parametric equations for this curve are { = w3 , | = w2 .
We can make a table of values, or we can eliminate the parameter:
I
I 2
{ = w3 i w = 3 { i | = w2 = ( 3 { ) = {2@3 ,
with w M R i { M R. By comparing different values of w, we find the
direction in which w increases as indicated in the graph.
9. The corresponding parametric equations are { = w, | = 2 3 w, } = 2w, which are
parametric equations of a line through the point (0> 2> 0) and with direction vector
h1> 31> 2i.
10. The corresponding parametric equations are { = sin w, | = w, } = cos w.
Note that {2 + } 2 = sin2 w + cos2 w = 1, so the curve lies on the circular
cylinder {2 + } 2 = 1. A point ({> |> }) on the curve lies directly to the left or
right of the point ({> 0> }) which moves clockwise (when viewed from the left)
along the circle {2 + } 2 = 1 in the {}-plane as w increases. Since | = w, the
curve is a helix that spirals toward the right around the cylinder.
11. The corresponding parametric equations are { = 1, | = cos w, } = 2 sin w.
Eliminating the parameter in | and } gives |2 + (}@2)2 = cos2 w + sin2 w = 1
or |2 + } 2 @4 = 1. Since { = 1, the curve is an ellipse centered at (1> 0> 0) in
the plane { = 1.
12. The parametric equations are { = w2 , | = w, } = 2, so we have { = | 2 with } = 2.
Thus the curve is a parabola in the plane } = 2 with vertex (0> 0> 2).
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SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES
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315
13. The parametric equations are { = w2 , | = w4 , } = w6 . These are positive
for w 6= 0 and 0 when w = 0. So the curve lies entirely in the first octant.
The projection of the graph onto the {|-plane is | = {2 , | A 0, a half parabola.
Onto the {}-plane } = {3 , } A 0, a half cubic, and the |}-plane, | 3 = } 2 .
14. If { = cos w, | = 3 cos w, } = sin w, then {2 + } 2 = 1 and | 2 + } 2 = 1,
so the curve is contained in the intersection of circular cylinders along the
{- and |-axes. Furthermore, | = 3{, so the curve is an ellipse in the
plane | = 3{, centered at the origin.
15. The projection of the curve onto the {|-plane is given by r(w) = hw> sin w> 0i
[we use 0 for the }-component] whose graph
is the curve | = sin {, } = 0. Similarly, the projection onto the {}-plane is r(w) = hw> 0> 2 cos wi, whose graph is the cosine
wave } = 2 cos {, | = 0, and the projection onto the |}-plane is r(w) = h0> sin w> 2 cos wi whose graph is the ellipse
| 2 + 14 } 2 = 1, { = 0.
{|-plane
{}-plane
|}-plane
From the projection onto the |}-plane we see that the curve lies on an elliptical
cylinder with axis the {-axis. The other two projections show that the curve
oscillates both vertically and horizontally as we move in the {-direction,
suggesting that the curve is an elliptical helix that spirals along the cylinder.
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CHAPTER 13 VECTOR FUNCTIONS
16. The projection of the curve onto the {|-plane is given by r(w) = hw> w> 0i whose graph is the line | = {, } = 0.
The projection onto the {}-plane is r(w) = w> 0> w2 whose graph is the parabola } = {2 , | = 0.
The projection onto the |}-plane is r(w) = 0> w> w2 whose graph is the parabola } = | 2 , { = 0.
{|-plane
{}-plane
|}-plane
From the projection onto the {|-plane we see that the curve lies on the vertical
plane | = {. The other two projections show that the curve is a parabola contained
in this plane.
17. Taking r0 = h2> 0> 0i and r1 = h6> 2> 32i, we have from Equation 12.5.4
r(w) = (1 3 w) r0 + w r1 = (1 3 w) h2> 0> 0i + w h6> 2> 32i, 0 $ w $ 1 or r(w) = h2 + 4w> 2w> 32wi, 0 $ w $ 1.
Parametric equations are { = 2 + 4w, | = 2w, } = 32w, 0 $ w $ 1.
18. Taking r0 = h31> 2> 32i and r1 = h33> 5> 1i, we have from Equation 12.5.4
r(w) = (1 3 w) r0 + w r1 = (1 3 w) h31> 2> 32i + w h33> 5> 1i, 0 $ w $ 1 or r(w) = h31 3 2w> 2 + 3w> 32 + 3wi, 0 $ w $ 1.
Parametric equations are { = 31 3 2w, | = 2 + 3w, } = 32 + 3w, 0 $ w $ 1.
19. Taking r0 = h0> 31> 1i and r1 =
1 1 1
> >
2 3 4
, we have
r(w) = (1 3 w) r0 + w r1 = (1 3 w) h0> 31> 1i + w
1 1 1
> >
2 3 4
, 0 $ w $ 1 or r(w) =
1
w> 31
2
Parametric equations are { = 12 w, | = 31 + 43 w, } = 1 3 34 w, 0 $ w $ 1.
+ 43 w> 1 3 34 w , 0 $ w $ 1.
20. Taking r0 = hd> e> fi and r1 = hx> y> zi, we have
r(w) = (1 3 w) r0 + w r1 = (1 3 w) hd> e> fi + w hx> y> zi, 0 $ w $ 1 or r(w) = hd + (x 3 d)w> e + (y 3 e)w> f + (z 3 f)wi,
0 $ w $ 1. Parametric equations are { = d + (x 3 d)w, | = e + (y 3 e)w, } = f + (z 3 f)w, 0 $ w $ 1.
21. { = w cos w, | = w, } = w sin w, w D 0.
At any point ({> |> }) on the curve, {2 + } 2 = w2 cos2 w + w2 sin2 w = w2 = | 2 so the
curve lies on the circular cone {2 + } 2 = | 2 with axis the |-axis. Also notice that | D 0; the graph is II.
22. { = cos w, | = sin w, } = 1@(1 + w2 ).
At any point on the curve we have {2 + | 2 = cos2 w + sin2 w = 1, so the curve lies
on a circular cylinder {2 + | 2 = 1 with axis the }-axis. Notice that 0 ? } $ 1 and } = 1 only for w = 0. A point ({> |> }) on
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317
the curve lies directly above the point ({> |> 0), which moves counterclockwise around the unit circle in the {|-plane as w
increases, and } < 0 as w < ±". The graph must be VI.
23. { = w, | = 1@(1 + w2 ), } = w2 .
At any point on the curve we have } = {2 , so the curve lies on a parabolic cylinder parallel
to the |-axis. Notice that 0 ? | $ 1 and } D 0. Also the curve passes through (0> 1> 0) when w = 0 and | < 0, } < " as
w < ±", so the graph must be V.
24. { = cos w, | = sin w, } = cos 2w.
{2 + | 2 = cos2 w + sin2 w = 1, so the curve lies on a circular cylinder with axis the
}-axis. A point ({> |> }) on the curve lies directly above or below ({> |> 0), which moves around the unit circle in the {|-plane
with period 2. At the same time, the }-value of the point ({> |> }) oscillates with a period of . So the curve repeats itself and
the graph is I.
25. { = cos 8w, | = sin 8w, } = h0=8w , w D 0.
{2 + | 2 = cos2 8w + sin2 8w = 1, so the curve lies on a circular cylinder with
axis the }-axis. A point ({> |> }) on the curve lies directly above the point ({> |> 0), which moves counterclockwise around the
unit circle in the {|-plane as w increases. The curve starts at (1> 0> 1), when w = 0, and } < " (at an increasing rate) as
w < ", so the graph is IV.
26. { = cos2 w, | = sin2 w, } = w.
{ + | = cos2 w + sin2 w = 1, so the curve lies in the vertical plane { + | = 1.
{ and | are periodic, both with period , and } increases as w increases, so the graph is III.
27. If { = w cos w, | = w sin w, } = w, then {2 + | 2 = w2 cos2 w + w2 sin2 w = w2 = } 2 ,
so the curve lies on the cone } 2 = {2 + | 2 . Since } = w, the curve is a spiral on
this cone.
28. Here {2 = sin2 w = } and {2 + | 2 = sin2 w + cos2 w = 1, so the
curve is contained in the intersection of the parabolic cylinder
} = {2 with the circular cylinder {2 + | 2 = 1. We get the complete
intersection for 0 $ w $ 2.
29. Parametric equations for the curve are { = w, | = 0, } = 2w 3 w2 . Substituting into the equation of the paraboloid
gives 2w 3 w2 = w2
i 2w = 2w2
i w = 0, 1. Since r(0) = 0 and r(1) = i + k, the points of intersection
are (0> 0> 0) and (1> 0> 1).
30. Parametric equations for the helix are { = sin w, | = cos w, } = w. Substituting into the equation of the sphere gives
sin2 w + cos2 w + w2 = 5 i 1 + w2 = 5 i w = ±2. Since r(2) = hsin 2> cos 2> 2i and
r(32) = hsin(32)> cos(32)> 32i, the points of intersection are (sin 2> cos 2> 2) E (0=909> 30=416> 2) and
(sin(32)> cos(32)> 32) E (30=909> 30=416> 32).
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CHAPTER 13 VECTOR FUNCTIONS
31. r(w) = hcos w sin 2w> sin w sin 2w> cos 2wi.
We include both a regular plot and a plot
showing a tube of radius 0.08 around the
curve.
32. r(w) = w2 > ln w> w
33. r(w) = hw> w sin w> w cos wi
34. r(w) = w> hw > cos w
35. r(w) = hcos 2w> cos 3w> cos 4wi
36. { = sin w, | = sin 2w, } = cos 4w.
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SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES
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319
We graph the projections onto the coordinate planes.
{|-plane
{}-plane
|}-plane
From the projection onto the {|-plane we see that from above the curve appears to be shaped like a “figure eight.”
The curve can be visualized as this shape wrapped around an almost parabolic cylindrical surface, the profile of
which is visible in the projection onto the |}-plane.
37.
{ = (1 + cos 16w) cos w, | = (1 + cos 16w) sin w, } = 1 + cos 16w. At any
point on the graph,
{2 + | 2 = (1 + cos 16w)2 cos2 w + (1 + cos 16w)2 sin2 w
= (1 + cos 16w)2 = } 2 , so the graph lies on the cone {2 + | 2 = } 2 .
From the graph at left, we see that this curve looks like the projection of a
leaved two-dimensional curve onto a cone.
38.
{=
I
I
1 3 0=25 cos2 10w cos w, | = 1 3 0=25 cos2 10w sin w,
} = 0=5 cos 10w. At any point on the graph,
{2 + | 2 + } 2 = (1 3 0=25 cos2 10w) cos2 w
+(1 3 0=25 cos2 10w) sin2 w + 0=25 cos2 w
= 1 3 0=25 cos2 10w + 0=25 cos2 10w = 1,
so the graph lies on the sphere {2 + | 2 + } 2 = 1, and since } = 0=5 cos 10w
the graph resembles a trigonometric curve with ten peaks projected onto the
sphere. We get the complete graph for 0 $ w $ 2.
39. If w = 31, then { = 1, | = 4, } = 0, so the curve passes through the point (1> 4> 0). If w = 3, then { = 9, | = 38, } = 28,
so the curve passes through the point (9> 38> 28). For the point (4> 7> 36) to be on the curve, we require | = 1 3 3w = 7 i
w = 32= But then } = 1 + (32)3 = 37 6= 36, so (4> 7> 36) is not on the curve.
40. The projection of the curve F of intersection onto the {|-plane is the circle {2 + | 2 = 4, } = 0.
Then we can write { = 2 cos w, | = 2 sin w, 0 $ w $ 2. Since F also lies on the surface } = {|, we have
} = {| = (2 cos w)(2 sin w) = 4 cos w sin w, or 2 sin(2w). Then parametric equations for F are { = 2 cos w, | = 2 sin w,
} = 2 sin(2w), 0 $ w $ 2, and the corresponding vector function is r(w) = 2 cos w i + 2 sin w j + 2 sin(2w) k, 0 $ w $ 2.
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CHAPTER 13 VECTOR FUNCTIONS
41. Both equations are solved for }, so we can substitute to eliminate }:
2
{ = 1 + 2|
i |=
1
({2
2
s
{2 + | 2 = 1 + |
i {2 + | 2 = 1 + 2| + | 2
i
3 1). We can form parametric equations for the curve F of intersection by choosing a
parameter { = w, then | = 12 (w2 3 1) and } = 1 + | = 1 + 12 (w2 3 1) = 12 (w2 + 1). Thus a vector function representing F
is r(w) = w i + 12 (w2 3 1) j + 12 (w2 + 1) k.
42. The projection of the curve F of intersection onto the {|-plane is the parabola | = {2 , } = 0. Then we can choose the
parameter { = w i | = w2 . Since F also lies on the surface } = 4{2 + | 2 , we have } = 4{2 + | 2 = 4w2 + (w2 )2 .
Then parametric equations for F are { = w, | = w2 , } = 4w2 + w4 , and the corresponding vector function
is r(w) = w i + w2 j + (4w2 + w4 ) k.
43. The projection of the curve F of intersection onto the {|-plane is the circle {2 + | 2 = 1, } = 0, so we can write { = cos w,
| = sin w, 0 $ w $ 2. Since F also lies on the surface } = {2 3 | 2 , we have } = {2 3 | 2 = cos2 w 3 sin2 w or cos 2w.
Thus parametric equations for F are { = cos w, | = sin w, } = cos 2w, 0 $ w $ 2, and the corresponding vector function
is r(w) = cos w i + sin w j + cos 2w k, 0 $ w $ 2.
44. The projection of the curve F of intersection onto the {}-plane is the circle {2 + } 2 = 1, | = 0, so we can write { = cos w,
} = sin w, 0 $ w $ 2. F also lies on the surface {2 + | 2 + 4} 2 = 4, and since | D 0 we can write
s
I
s
I
I
| = 4 3 {2 3 4} 2 = 4 3 cos2 w 3 4 sin2 w = 4 3 cos2 w 3 4(1 3 cos2 w) = 3 cos2 w = 3 | cos w |
I
Thus parametric equations for F are { = cos w, | = 3 | cos w |, } = sin w, 0 $ w $ 2, and the corresponding vector function
I
is r(w) = cos w i + 3 | cos w | j + sin w k, 0 $ w $ 2.
45.
The projection of the curve F of intersection onto the
{|-plane is the circle {2 + | 2 = 4> } = 0. Then we can write
{ = 2 cos w, | = 2 sin w, 0 $ w $ 2. Since F also lies on
the surface } = {2 , we have } = {2 = (2 cos w)2 = 4 cos2 w.
Then parametric equations for F are { = 2 cos w, | = 2 sin w,
} = 4 cos2 w, 0 $ w $ 2.
46.
{ = w i | = w2
i 4} 2 = 16 3 {2 3 4| 2 = 16 3 w2 3 4w4
i }=
t
2
4 3 12 w 3 w4 .
Note that } is positive because the intersection is with the top half of the ellipsoid. Hence the curve is given
t
by { = w, | = w2 , } = 4 3 14 w2 3 w4 .
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SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES
47. For the particles to collide, we require r1 (w) = r2 (w)
C
321
w2 > 7w 3 12> w2 = 4w 3 3> w2 > 5w 3 6 . Equating components
gives w2 = 4w 3 3, 7w 3 12 = w2 , and w2 = 5w 3 6. From the first equation, w2 3 4w + 3 = 0 C (w 3 3)(w 3 1) = 0 so w = 1
or w = 3. w = 1 does not satisfy the other two equations, but w = 3 does. The particles collide when w = 3, at the
point (9> 9> 9).
48. The particles collide provided r1 (w) = r2 (w)
w> w2 > w3 = h1 + 2w> 1 + 6w> 1 + 14wi. Equating components gives
C
w = 1 + 2w, w2 = 1 + 6w, and w3 = 1 + 14w. The first equation gives w = 31, but this does not satisfy the other equations, so
the particles do not collide. For the paths to intersect, we need to find a value for w and a value for v where r1 (w) = r2 (v) C
w> w2 > w3 = h1 + 2v> 1 + 6v> 1 + 14vi. Equating components, w = 1 + 2v, w2 = 1 + 6v, and w3 = 1 + 14v. Substituting the
first equation into the second gives (1 + 2v)2 = 1 + 6v i 4v2 3 2v = 0 i 2v(2v 3 1) = 0 i v = 0 or v = 12 .
From the first equation, v = 0 i w = 1 and v =
1
2
i w = 2. Checking, we see that both pairs of values satisfy the
third equation. Thus the paths intersect twice, at the point (1> 1> 1) when v = 0 and w = 1, and at (2> 4> 8) when v =
1
2
and w = 2.
49. Let u(w) = hx1 (w)> x2 (w)> x3 (w)i and v(w) = hy1 (w)> y2 (w)> y3 (w)i. In each part of this problem the basic procedure is to use
Equation 1 and then analyze the individual component functions using the limit properties we have already developed for
real-valued functions.
G
H G
H
(a) lim u(w) + lim v(w) = lim x1 (w)> lim x2 (w)> lim x3 (w) + lim y1 (w)> lim y2 (w)> lim y3 (w) and the limits of these
w<d
w<d
w<d
w<d
w<d
w<d
w<d
w<d
component functions must each exist since the vector functions both possess limits as w < d. Then adding the two vectors
and using the addition property of limits for real-valued functions, we have that
G
H
lim u(w) + lim v(w) = lim x1 (w) + lim y1 (w)> lim x2 (w) + lim y2 (w)> lim x3 (w) + lim y3 (w)
w<d
w<d
w<d
w<d
G
w<d
w<d
w<d
w<d
H
= lim [x1 (w) + y1 (w)] > lim [x2 (w) + y2 (w)] > lim [x3 (w) + y3 (w)]
w<d
w<d
w<d
= lim hx1 (w) + y1 (w)> x2 (w) + y2 (w)> x3 (w) + y3 (w)i
w<d
[using (1) backward]
= lim [u(w) + v(w)]
w<d
G
H
(b) lim fu(w) = lim hfx1 (w)> fx2 (w)> fx3 (w)i = lim fx1 (w)> lim fx2 (w)> lim fx3 (w)
w<d
w<d
w<d
w<d
w<d
G
H
G
H
= f lim x1 (w)> f lim x2 (w)> f lim x3 (w) = f lim x1 (w)> lim x2 (w)> lim x3 (w)
w<d
w<d
w<d
w<d
w<d
w<d
= f lim hx1 (w)> x2 (w)> x3 (w)i = f lim u(w)
w<d
w<d
G
H G
H
(c) lim u(w) · lim v(w) = lim x1 (w)> lim x2 (w)> lim x3 (w) · lim y1 (w)> lim y2 (w)> lim y3 (w)
w<d
w<d
w<d
k
w<d
w<d
w<d
w<d
w<d
lk
l k
lk
l k
lk
l
= lim x1 (w) lim y1 (w) + lim x2 (w) lim y2 (w) + lim x3 (w) lim y3 (w)
w<d
w<d
w<d
w<d
w<d
w<d
= lim x1 (w)y1 (w) + lim x2 (w)y2 (w) + lim x3 (w)y3 (w)
w<d
w<d
w<d
= lim [x1 (w)y1 (w) + x2 (w)y2 (w) + x3 (w)y3 (w)] = lim [u(w) · v(w)]
w<d
w<d
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¤
CHAPTER 13 VECTOR FUNCTIONS
G
H G
H
(d) lim u(w) × lim v(w) = lim x1 (w)> lim x2 (w)> lim x3 (w) × lim y1 (w)> lim y2 (w)> lim y3 (w)
w<d
w<d
w<d
=
w<d
w<d
w<d
w<d
w<d
lk
l k
lk
l
lim x2 (w) lim y3 (w) 3 lim x3 (w) lim y2 (w) >
w<d
w<d
w<d
w<d
k
lk
l k
lk
l
lim x3 (w) lim y1 (w) 3 lim x1 (w) lim y3 (w) >
w<d
w<d
w<d
w<d
lk
l k
lk
lH
k
lim x1 (w) lim y2 (w) 3 lim x2 (w) lim y1 (w)
Gk
w<d
G
w<d
w<d
w<d
= lim [x2 (w)y3 (w) 3 x3 (w)y2 (w)] > lim [x3 (w)y1 (w) 3 x1 (w)y3 (w)] >
w<d
w<d
H
lim [x1 (w)y2 (w) 3 x2 (w)y1 (w)]
w<d
= lim hx2 (w)y3 (w) 3 x3 (w)y2 (w)> x3 (w) y1 (w) 3 x1 (w)y3 (w)> x1 (w)y2 (w) 3 x2 (w)y1 (w)i
w<d
= lim [u(w) × v(w)]
w<d
50. The projection of the curve onto the {|-plane is given by the parametric equations { = (2 + cos 1=5w) cos w,
| = (2 + cos 1=5w) sin w. If we convert to polar coordinates, we have
u2 = {2 + | 2 = [(2 + cos 1=5w) cos w]2 + [(2 + cos 1=5w) sin w]2 = (2 + cos 1=5w)2 (cos2 w + sin2 w) = (2 + cos 1=5w)2
u = 2 + cos 1=5w. Also, tan =
i
(2 + cos 1=5w) sin w
|
=
= tan w i = w.
{
(2 + cos 1=5w) cos w
Thus the polar equation of the curve is u = 2 + cos 1=5. At = 0, we have
u = 3, and u decreases to 1 as increases to
2
.
3
For
2
3
$$
4
,
3
u
increases to 3; u decreases to 1 again at = 2, increases to 3 at =
decreases to 1 at =
10
,
3
8
,
3
and completes the closed curve by increasing
to 3 at = 4. We sketch an approximate graph as shown in the figure.
We can determine how the curve passes over itself by investigating the maximum and minimum values of } for 0 $ w $ 4.
Since } = sin 1=5w, } is maximized where sin 1=5w = 1 i 1=5w =
w=
5
, 3,
3
1=5w =
5
, 2,
2
or
9
2
i
or 3. } is minimized where sin 1=5w = 31 i
3 7
, 2,
2
or
11
2
i w = ,
7
,
3
or
11
.
3
Note that these are
precisely the values for which cos 1=5w = 0 i u = 2, and on the graph
of the projection, these six points appear to be at the three self-intersections
we see. Comparing the maximum and minimum values of } at these
intersections, we can determine where the curve passes over itself, as
indicated in the figure.
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SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES
¤
323
We show a computer-drawn graph of the curve from above, as well as views from the front and from the right side.
Top view
Front view
Side view
The top view graph shows a more accurate representation of the projection of the trefoil knot onto the {|-plane (the axes are
rotated 90 ). Notice the indentations the graph exhibits at the points corresponding to u = 1. Finally, we graph several
additional viewpoints of the trefoil knot, along with two plots showing a tube of radius 0=2 around the curve.
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¤
CHAPTER 13 VECTOR FUNCTIONS
51. Let r(w) = hi (w) > j (w) > k (w)i and b = he1 > e2 > e3 i. If lim r(w) = b, then lim r(w) exists, so by (1),
w<d
w<d
H
b = lim r(w) = lim i (w)> lim j(w)> lim k(w) . By the definition of equal vectors we have lim i (w) = e1 , lim j(w) = e2
w<d
G
w<d
w<d
w<d
w<d
w<d
and lim k(w) = e3 . But these are limits of real-valued functions, so by the definition of limits, for every % A 0 there exists
w<d
1 A 0, 2 A 0, 3 A 0 so that if 0 ? |w 3 d| ? 1 then |i (w) 3 e1 | ? %@3, if 0 ? |w 3 d| ? 2 then |j(w) 3 e2 | ? %@3, and
if 0 ? |w 3 d| ? 3 then |k(w) 3 e3 | ? %@3. Letting = minimum of { 1 > 2 > 3 }, then if 0 ? |w 3 d| ? we have
|i (w) 3 e1 | + |j(w) 3 e2 | + |k(w) 3 e3 | ? %@3 + %@3 + %@3 = %. But
s
|r(w) 3 b| = |hi (w) 3 e1 > j(w) 3 e2 > k(w) 3 e3 i| = (i (w) 3 e1 )2 + (j(w) 3 e2 )2 + (k(w) 3 e3 )2
s
s
s
$ [i (w) 3 e1 ]2 + [j(w) 3 e2 ]2 + [k(w) 3 e3 ]2 = |i (w) 3 e1 | + |j(w) 3 e2 | + |k(w) 3 e3 |
Thus for every % A 0 there exists A 0 such that if 0 ? |w 3 d| ? then
|r(w) 3 b| $ |i (w) 3 e1 | + |j(w) 3 e2 | + |k(w) 3 e3 | ? %. Conversely, suppose for every % A 0, there exists A 0 such
that if 0 ? |w 3 d| ? then |r(w) 3 b| ? % C |hi(w) 3 e1 > j(w) 3 e2 > k(w) 3 e3 i| ? % C
s
[i (w) 3 e1 ]2 + [j(w) 3 e2 ]2 + [k(w) 3 e3 ]2 ? % C [i (w) 3 e1 ]2 + [j(w) 3 e2 ]2 + [k(w) 3 e3 ]2 ? %2 . But each term
on the left side of the last inequality is positive, so if 0 ? |w 3 d| ? , then [i (w) 3 e1 ]2 ? %2 , [j(w) 3 e2 ]2 ? %2 and
[k(w) 3 e3 ]2 ? %2 or, taking the square root of both sides in each of the above, |i (w) 3 e1 | ? %, |j(w) 3 e2 | ? % and
|k(w) 3 e3 | ? %. And by definition of limits of real-valued functions we have lim i (w) = e1 , lim j(w) = e2 and
w<d
w<d
H
lim k(w) = e3 . But by (1), lim r(w) = lim i (w)> lim j(w)> lim k(w) , so lim r(w) = he1 > e2 > e3 i = b.
w<d
w<d
G
w<d
w<d
w<d
w<d
13.2 Derivatives and Integrals of Vector Functions
1. (a)
(b)
r(4=5) 3 r(4)
= 2[r(4=5) 3 r(4)], so we draw a vector in the same
0=5
direction but with twice the length of the vector r(4=5) 3 r(4).
r(4=2) 3 r(4)
= 5[r(4=2) 3 r(4)], so we draw a vector in the same
0=2
direction but with 5 times the length of the vector r(4=2) 3 r(4).
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SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
(c) By Definition 1, r0 (4) = lim
k<0
¤
325
r(4 + k) 3 r(4)
r0 (4)
. T(4) = 0
.
k
|r (4)|
(d) T(4) is a unit vector in the same direction as r0 (4), that is, parallel to the
tangent line to the curve at r(4) with length 1.
2. (a) The curve can be represented by the parametric equations { = w2 , | = w, 0 $ w $ 2. Eliminating the parameter, we have
{ = | 2 , 0 $ | $ 2, a portion of which we graph here, along with the vectors r(1), r(1=1), and r(1=1) 3 r(1).
(b) Since r(w) = w2 > w , we differentiate components, giving r0 (w) = h2w> 1i, so r0 (1) = h2> 1i.
h1=21> 1=1i 3 h1> 1i
r(1=1) 3 r(1)
=
= 10 h0=21> 0=1i = h2=1> 1i.
0=1
0=1
As we can see from the graph, these vectors are very close in length and direction. r0 (1) is defined to be
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¤
CHAPTER 13 VECTOR FUNCTIONS
lim
k<0
r(1 + k) 3 r(1)
r(1=1) 3 r(1)
, and we recognize
as the expression after the limit sign with k = 0=1= Since k is
k
0=1
close to 0, we would expect
3. Since ({ + 2)2 = w2 = | 3 1
r(1=1) 3 r(1)
to be a vector close to r0 (1).
0=1
i
(a), (c)
(b) r0 (w) = h1> 2wi,
r0 (31) = h1> 32i
| = ({ + 2)2 + 1, the curve is a
parabola.
4. Since { = w2 = (w3 )2@3 = | 2@3 ,
(a), (c)
r0 (1) = h2> 3i
the curve is the graph of { = | 2@3 .
5. { = sin w, | = 2 cos w so
(a), (c)
(b) r0 (w) = cos w i 3 2 sin w j,
I2
I
0 =
i3 2j
r
4
2
(a), (c)
(b) r0 (w) = hw i 3 h3w j,
{2 + (|@2)2 = 1 and the curve is
an ellipse.
6. Since | = h3w =
1
1
= the
hw
{
r0 (0) = i 3 j
curve is part of the hyperbola
|=
(b) r0 (w) = 2w> 3w2 ,
1
. Note that { A 0, | A 0.
{
7. Since { = h2w = (hw )2 = | 2 , the
(a), (c)
(b) r0 (w) = 2h2w i + hw j,
r0 (0) = 2 i + j
curve is part of a parabola. Note
that here { A 0, | A 0.
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SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
8. { = 1 + cos w, | = 2 + sin w so
curve is a circle.
g 2 g
g
[w sin w] >
w > [w cos 2w]
gw
gw
gw
327
(b) r0 (w) = 3 sin w i + cos w j,
I
3
1
=3 i+
j
r0
6
2
2
(a), (c)
({ 3 1)2 + (| 3 2)2 = 1 and the
9. r0 (w) =
¤
= hw cos w + sin w> 2w> w(3 sin 2w) · 2 + cos 2wi
= hw cos w + sin w> 2w> cos 2w 3 2w sin 2wi
10. r(w) = tan w> sec w> 1@w2
I
r0 (w) =
i r0 (w) = sec2 w> sec w tan w> 32@w3
i r0 (w) = 1 i + 0 j + 2
11. r(w) = w i + j + 2 w k
12. r(w) =
1
w
w2
i+
j+
k i
1+w
1+w
1+w
1 31@2
2w
1
k=i+ I k
w
0 3 1(1)
(1 + w) · 1 3 w(1)
(1 + w) · 2w 3 w2 (1)
1
1
w2 + 2w
i+
j+
k=3
i+
j+
k
2
2
2
2
2
(1 + w)
(1 + w)
(1 + w)
(1 + w)
(1 + w)
(1 + w)2
2
13. r(w) = hw i 3 j + ln(1 + 3w) k
2
i r0 (w) = 2whw i +
3
k
1 + 3w
14. r0 (w) = [dw(33 sin 3w) + d cos 3w] i + e · 3 sin2 w cos w j + f · 3 cos2 w(3 sin w) k
= (d cos 3w 3 3dw sin 3w) i + 3e sin2 w cos w j 3 3f cos2 w sin w k
15. r0 (w) = 0 + b + 2w c = b + 2w c by Formulas 1 and 3 of Theorem 3.
16. To find r0 (w), we first expand r(w) = w a × (b + w c) = w(a × b) + w2 (a × c), so r0 (w) = a × b + 2w(a × c).
17. r0 (w) = 3wh3w + h3w > 2@(1 + w2 )> 2hw
T(0) =
r0 (0)
=
|r0 (0)|
1
3
h1> 2> 2i =
18. r0 (w) = 3w2 + 3> 2w> 3
T(1) =
1 2 2
> >
3 3 3
i r0 (0) = h1> 2> 2i. So |r0 (0)| =
I
I
12 + 22 + 22 = 9 = 3 and
.
i r0 (1) = h6> 2> 3i. Thus
1
r0 (1)
= I
h6> 2> 3i =
|r0 (1)|
62 + 22 + 32
19. r0 (w) = 3 sin w i + 3 j + 4 cos 2w k
T(0) =
1
7
h6> 2> 3i =
6 2 3
7> 7> 7
.
i r0 (0) = 3 j + 4 k. Thus
r0 (0)
1
= I
(3 j + 4 k) = 15 (3 j + 4 k) =
|r0 (0)|
02 + 32 + 42
3
5
j+
4
5
k.
20. r0 (w) = 2 sin w cos w i 3 2 cos w sin w j + 2 tan w sec2 w k
r0 4 = 2 ·
I
2
2
I
I
i
I
I
I
I
· 22 j + 2 · 1 · ( 2)2 k = i 3 j + 4 k and r0 4 = 1 + 1 + 16 = 18 = 3 2. Thus
I
· 22 i 3 2 · 22
r0 1
1
4
1
T 4 = 0 4 = I (i 3 j + 4 k) = I i 3 I j + I k.
r 4
3 2
3 2
3 2
3 2
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¤
CHAPTER 13 VECTOR FUNCTIONS
21. r(w) = w> w2 > w3
T(1) =
r0 (1)
=
|r0 (1)|
I
I
i r0 (w) = 1> 2w> 3w2 . Then r0 (1) = h1> 2> 3i and |r0 (1)| = 12 + 22 + 32 = 14, so
I1
14
22. r(w) = h2w > h32w > wh2w
h1> 2> 3i =
G
I1 > I2 > I3
14
14
14
H
. r00 (w) = h0> 2> 6wi, so
i j k 1 3w2 1 2w 2w 3w2 0
00
2 r (w) × r (w) = 1 2w 3w = k
i 3 j + 0 6w 0 2 2 6w 0 2 6w = (12w2 3 6w2 ) i 3 (6w 3 0) j + (2 3 0) k = 6w2 > 36w> 2
i r0 (w) = 2h2w > 32h32w > (2w + 1)h2w
i r0 (0) = 2h0 > 32h0 > (0 + 1)h0 = h2> 32> 1i
s
r0 (0)
= 13 h2> 32> 1i = 23 > 3 23 > 13 .
22 + (32)2 + 12 = 3. Then T(0) = 0
|r (0)|
i r00 (0) = 4h0 > 4h0 > (0 + 4)h0 = h4> 4> 4i.
r00 (w) = 4h2w > 4h32w > (4w + 4)h2w
and |r0 (0)| =
r0 (w) · r00 (w) = 2h2w > 32h32w > (2w + 1)h2w · 4h2w > 4h32w > (4w + 4)h2w
= (2h2w )(4h2w ) + (32h32w )(4h32w ) + ((2w + 1)h2w )((4w + 4)h2w )
= 8h4w 3 8h34w + (8w2 + 12w + 4)h4w = (8w2 + 12w + 12)h4w 3 8h34w
23. The vector equation for the curve is r(w) = 1 + 2
I 3
I
w> w 3 w> w3 + w , so r0 (w) = 1@ w> 3w2 3 1> 3w2 + 1 . The point
(3> 0> 2) corresponds to w = 1, so the tangent vector there is r0 (1) = h1> 2> 4i. Thus, the tangent line goes through the point
(3> 0> 2) and is parallel to the vector h1> 2> 4i. Parametric equations are { = 3 + w, | = 2w, } = 2 + 4w.
G
24. The vector equation for the curve is r(w) = hw > whw > whw
2
G
H
H
2
2
, so r0 (w) = hw > whw + hw > 2w2 hw + hw . The point (1> 0> 0)
corresponds to w = 0, so the tangent vector there is r0 (0) = h1> 1> 1i. Thus, the tangent line is parallel to the vector h1> 1> 1i
and includes the point (1> 0> 0). Parametric equations are { = 1 + 1 · w = 1 + w, | = 0 + 1 · w = w, } = 0 + 1 · w = w.
25. The vector equation for the curve is r(w) = h3w cos w> h3w sin w> h3w , so
r0 (w) = h3w (3 sin w) + (cos w)(3h3w ), h3w cos w + (sin w)(3h3w ), (3h3w )
= 3h3w (cos w + sin w)> h3w (cos w 3 sin w)> 3h3w
The point (1> 0> 1) corresponds to w = 0, so the tangent vector there is
r0 (0) = 3h0 (cos 0 + sin 0)> h0 (cos 0 3 sin 0)> 3h0 = h31> 1> 31i. Thus, the tangent line is parallel to the vector
h31> 1> 31i and parametric equations are { = 1 + (31)w = 1 3 w, | = 0 + 1 · w = w, } = 1 + (31)w = 1 3 w.
26. The vector equation for the curve is r(w) =
w = 1 and r0 (1) =
I
I
w2 + 3> ln(w2 + 3)> w , so r0 (w) = w@ w2 + 3> 2w@(w2 + 3)> 1 . At (2> ln 4> 1),
. Thus, parametric equations of the tangent line are { = 2 + 12 w, | = ln 4 + 12 w, } = 1 + w.
1 1
> >1
2 2
27. First we parametrize the curve F of intersection. The projection of F onto the {|-plane is contained in the circle
{2 + | 2 = 25, } = 0, so we can write { = 5 cos w, | = 5 sin w. F also lies on the cylinder | 2 + } 2 = 20, and } D 0
s
s
near the point (3> 4> 2), so we can write } = 20 3 | 2 = 20 3 25 sin2 w. A vector equation then for F is
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SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
¤
329
G
H
G
H
s
r(w) = 5 cos w> 5 sin w> 20 3 25 sin2 w
i r0 (w) = 35 sin w> 5 cos w> 12 (20 3 25 sin2 w)31@2 (350 sin w cos w) .
The point (3> 4> 2) corresponds to w = cos31 35 , so the tangent vector there is
2 31@2 r0 cos31 35 = 35 45 > 5 35 > 12 20 3 25 45
350 45 35
= h34> 3> 36i.
The tangent line is parallel to this vector and passes through (3> 4> 2), so a vector equation for the line
is r(w) = (3 3 4w)i + (4 + 3w)j + (2 3 6w)k.
i r0 (w) = 32 sin w> 2 cos w> hw . The tangent line to the curve is parallel to the plane when the
I
curve’s tangent vector is orthogonal to the plane’s normal vector. Thus we require 32 sin w> 2 cos w> hw ·
3> 1> 0 = 0 i
I
32 3 sin w + 2 cos w + 0 = 0 i tan w = I13 i w = 6 [since 0 $ w $ ].
28. r(w) = 2 cos w> 2 sin w> hw
H
I
GI
3> 1> h@6 , so the point is ( 3> 1> h@6 ).
r 6 =
29. r(w) = w> h3w > 2w 3 w2
i r0 (w) = 1> 3h3w > 2 3 2w . At (0> 1> 0),
w = 0 and r0 (0) = h1> 31> 2i. Thus, parametric equations of the tangent
line are { = w, | = 1 3 w, } = 2w.
30. r(w) = h2 cos w> 2 sin w> 4 cos 2wi,
r0 (w) = h32 sin w> 2 cos w> 38 sin 2wi. At
I
3> 1> 2 , w =
6
and
I
I r0 ( 6 ) = 31> 3> 34 3 . Thus, parametric equations of the
tangent line are { =
I
I
I
3 3 w, | = 1 + 3 w, } = 2 3 4 3 w.
31. r(w) = hw cos w> w> w sin wi
i r0 (w) = hcos w 3 w sin w> 1> w cos w + sin wi.
At (3> > 0), w = and r0 () = h31> 1> 3i. Thus, parametric equations
of the tangent line are { = 3 3 w, | = + w, } = 3w.
32. (a) The tangent line at w = 0 is the line through the point with position vector r(0) = hsin 0> 2 sin 0> cos 0i = h0> 0> 1i, and in
the direction of the tangent vector, r0 (0) = h cos 0> 2 cos 0> 3 sin 0i = h> 2> 0i. So an equation of the line is
h{> |> }i = r(0) + x r0 (0) = h0 + x> 0 + 2x> 1i = hx> 2x> 1i.
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¤
330
CHAPTER 13 VECTOR FUNCTIONS
r 12 = sin 2 > 2 sin 2 > cos 2 = h1> 2> 0i ,
r0 12 = cos 2 > 2 cos 2 > 3 sin 2 = h0> 0> 3i .
(b)
So the equation of the second line is
h{> |> }i = h1> 2> 0i + y h0> 0> 3i = h1> 2> 3yi.
The lines intersect where hx> 2x> 1i = h1> 2> 3yi,
so the point of intersection is (1> 2> 1).
33. The angle of intersection of the two curves is the angle between the two tangent vectors to the curves at the point of
intersection. Since r01 (w) = 1> 2w> 3w2 and w = 0 at (0> 0> 0), r01 (0) = h1> 0> 0i is a tangent vector to r1 at (0> 0> 0). Similarly,
r02 (w) = hcos w> 2 cos 2w> 1i and since r2 (0) = h0> 0> 0i, r02 (0) = h1> 2> 1i is a tangent vector to r2 at (0> 0> 0). If is the angle
between these two tangent vectors, then cos = I11I6 h1> 0> 0i · h1> 2> 1i = I16 and = cos31 I16 E 66 .
34. To find the point of intersection, we must find the values of w and v which satisfy the following three equations simultaneously:
w = 3 3 v, 1 3 w = v 3 2, 3 + w2 = v2 . Solving the last two equations gives w = 1, v = 2 (check these in the first equation).
Thus the point of intersection is (1> 0> 4). To find the angle of intersection, we proceed as in Exercise 33. The tangent
vectors to the respective curves at (1> 0> 4) are r01 (1) = h1> 31> 2i and r02 (2) = h31> 1> 4i. So
6
cos = I61I18 (31 3 1 + 8) = 6I
= I13 and = cos31 I13 E 55 .
3
Note: In Exercise 33, the curves intersect when the value of both parameters is zero. However, as seen in this exercise, it is not
necessary for the parameters to be of equal value at the point of intersection.
35.
U2
0
(w i 3 w3 j + 3w5 k) gw =
U
2
0
U
U
2
2
w gw i 3 0 w3 gw j + 0 3w5 gw k
2
2
2
= 12 w2 0 i 3 14 w4 0 j + 12 w6 0 k
= 12 (4 3 0) i 3 14 (16 3 0) j + 12 (64 3 0) k = 2 i 3 4 j + 32 k
36.
]
0
37.
1
U @2
0
1 4
2w
j+
k gw = 4 tan31 w j + ln(1 + w2 ) k 0 = 4 tan31 1 j + ln 2 k 3 4 tan31 0 j + ln 1 k
1 + w2
1 + w2
= 4 4 j + ln 2 k 3 0 j 3 0 k = j + ln 2 k
(3 sin2 w cos w i + 3 sin w cos2 w j + 2 sin w cos w k) gw
U
U
U
@2
@2
@2
= 0 3 sin2 w cos w gw i + 0 3 sin w cos2 w gw j + 0 2 sin w cos w gw k
@2
@2
@2
= sin3 w 0 i + 3 cos3 w 0 j+ sin2 w 0 k = (1 3 0) i + (0 + 1) j + (1 3 0) k = i + j + k
38.
l2 k
2 U 2
U2 2
I
w i + w w 3 1 j + w sin w k gw = 13 w3 i + 25 (w 3 1)5@2 + 23 (w 3 1)3@2 j + 3 1 w cos w 1 + 1
1
1
1
2 3
k=
j
+
3
+
sin
w
= 73 i + 16
2
15
1
7
3
i+
16
15
j3
3
1
k
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SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
39.
U
¤
331
U 2
U 2
sec2 w gw i +
w(w + 1)3 gw j +
w ln w gw k
= tan w i + 18 (w2 + 1)4 j + 13 w3 ln w 3 19 w3 k + C,
(sec2 w i + w(w2 + 1)3 j + w2 ln w k) gw =
U
where C is a vector constant of integration. [For the }-component, integrate by parts with x = ln w, gy = w2 gw.]
40.
] wh2w i +
]
]
]
1
w
1
w
I
wh2w gw i +
j+ I
gw j +
k gw =
gw k
13w
13w
1 3 w2
1 3 w2
]
] U
1
1
I
gw j +
31 +
gw k
= 12 wh2w 3 12 h2w gw i +
13w
1 3 w2
= 12 wh2w 3 14 h2w i + (3w 3 ln | 1 3 w |) j + sin31 w k + C
I
w k i r(w) = w2 i + w3 j + 23 w3@2 k + C, where C is a constant vector.
But i + j = r (1) = i + j + 23 k + C. Thus C = 3 23 k and r(w) = w2 i + w3 j + 23 w3@2 3 23 k.
41. r0 (w) = 2w i + 3w2 j +
i r(w) = 12 w2 i + hw j + whw 3 hw k + C. But i + j + k = r (0) = j 3 k + C.
Thus C = i + 2 k and r(w) = 12 w2 + 1 i + hw j + (whw 3 hw + 2) k.
42. r0 (w) = w i + hw j + whw k
For Exercises 43 – 46, let u(w) = hx1 (w)> x2 (w)> x3 (w)i and v(w) = hy1 (w)> y2 (w)> y3 (w)i. In each of these exercises, the procedure is to apply
Theorem 2 so that the corresponding properties of derivatives of real-valued functions can be used.
43.
g
g
[u(w) + v(w)] =
hx1 (w) + y1 (w)> x2 (w) + y2 (w)> x3 (w) + y3 (w)i
gw
gw
g
g
g
[x1 (w) + y1 (w)] > [x2 (w) + y2 (w)] > [x3 (w) + y3 (w)]
=
gw
gw
gw
= hx01 (w) + y10 (w)> x02 (w) + y20 (w)> x03 (w) + y30 (w)i
= hx01 (w)> x02 (w) > x03 (w)i + hy10 (w)> y20 (w)> y30 (w)i = u0 (w) + v0 (w)
44.
g
g
[i (w) u(w)] =
hi(w)x1 (w)> i (w)x2 (w)> i (w)x3 (w)i
gw
gw
g
g
g
=
[i(w)x1 (w)] > [i(w)x2 (w)] > [i (w)x3 (w)]
gw
gw
gw
= hi 0 (w)x1 (w) + i (w)x01 (w)> i 0 (w)x2 (w) + i (w)x02 (w)> i 0 (w)x3 (w) + i (w)x03 (w)i
= i 0 (w) hx1 (w)> x2 (w)> x3 (w)i + i(w) hx01 (w)> x02 (w)> x03 (w)i = i 0 (w) u(w) + i (w) u0 (w)
45.
g
g
[u(w) × v(w)] =
hx2 (w)y3 (w) 3 x3 (w)y2 (w)> x3 (w)y1 (w) 3 x1 (w)y3 (w)> x1 (w)y2 (w) 3 x2 (w)y1 (w)i
gw
gw
= hx02 y3 (w) + x2 (w)y30 (w) 3 x03 (w)y2 (w) 3 x3 (w)y20 (w)>
x03 (w)y1 (w) + x3 (w)y10 (w) 3 x01 (w)y3 (w) 3 x1 (w)y30 (w)>
x01 (w)y2 (w) + x1 (w)y20 (w) 3 x02 (w)y1 (w) 3 x2 (w)y10 (w)i
= hx02 (w)y3 (w) 3 x03 (w)y2 (w) > x03 (w)y1 (w) 3 x01 (w)y3 (w)> x01 (w)y2 (w) 3 x02 (w)y1 (w)i
+ hx2 (w)y30 (w) 3 x3 (w)y20 (w)> x3 (w)y10 (w) 3 x1 (w)y30 (w)> x1 (w)y20 (w) 3 x2 (w)y10 (w)i
= u0 (w) × v(w) + u(w) × v0 (w)
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332
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CHAPTER 13 VECTOR FUNCTIONS
Alternate solution: Let r(w) = u(w) × v(w). Then
r(w + k) 3 r(w) = [u(w + k) × v(w + k)] 3 [u(w) × v(w)]
= [u(w + k) × v(w + k)] 3 [u(w) × v(w)] + [u(w + k) × v(w)] 3 [u(w + k) × v(w)]
= u(w + k) × [v(w + k) 3 v(w)] + [u(w + k) 3 u(w)] × v(w)
(Be careful of the order of the cross product.) Dividing through by k and taking the limit as k < 0 we have
r0 (w) = lim
k<0
u(w + k) × [v(w + k) 3 v(w)]
[u(w + k) 3 u(w)] × v(w)
+ lim
= u(w) × v0 (w) + u0 (w) × v(w)
k<0
k
k
by Exercise 13.1.49(a) and Definition 1.
46.
g
g
[u(i (w))] =
hx1 (i (w))> x2 (i (w))> x3 (i (w))i =
gw
gw
g
g
g
[x1 (i (w))] > [x2 (i (w))] > [x3 (i (w))]
gw
gw
gw
= hi 0 (w)x01 (i (w))> i 0 (w)x02 (i (w))> i 0 (w)x03 (i (w))i = i 0 (w) u0 (w)
47.
g
[u(w) · v(w)] = u0 (w) · v(w) + u(w) · v0 (w)
gw
[by Formula 4 of Theorem 3]
= hcos w> 3 sin w> 1i · hw> cos w> sin wi + hsin w> cos w> wi · h1> 3 sin w> cos wi
= w cos w 3 cos w sin w + sin w + sin w 3 cos w sin w + w cos w
= 2w cos w + 2 sin w 3 2 cos w sin w
48.
g
[u(w) × v(w)] = u0 (w) × v(w) + u(w) × v0 (w)
gw
[by Formula 5 of Theorem 3]
= hcos w> 3 sin w> 1i × hw> cos w> sin wi + hsin w> cos w> wi × h1> 3 sin w> cos wi
= 3 sin2 w 3 cos w> w 3 cos w sin w> cos2 w + w sin w
+ cos2 w + w sin w> w 3 cos w sin w> 3 sin2 w 3 cos w
= cos2 w 3 sin2 w 3 cos w + w sin w> 2w 3 2 cos w sin w> cos2 w 3 sin2 w 3 cos w + w sin w
49. By Formula 4 of Theorem 3, i 0 (w) = u0 (w) · v(w) + u(w) · v0 (w), and v0 (w) = 1> 2w> 3w2 , so
i 0 (2) = u0 (2) · v(2) + u(2) · v0 (2) = h3> 0> 4i · h2> 4> 8i + h1> 2> 31i · h1> 4> 12i = 6 + 0 + 32 + 1 + 8 3 12 = 35.
50. By Formula 5 of Theorem 3, r0 (w) = u0 (w) × v(w) + u(w) × v0 (w), so
r0 (2) = u0 (2) × v(2) + u(2) × v0 (2) = h3> 0> 4i × h2> 4> 8i + h1> 2> 31i × h1> 4> 12i
= h316> 316> 12i + h28> 313> 2i = h12> 329> 14i
51.
g
[r(w) × r0 (w)] = r0 (w) × r0 (w) + r(w) × r00 (w) by Formula 5 of Theorem 3. But r0 (w) × r0 (w) = 0 (by Example 2 in
gw
Section 12.4). Thus,
52.
g
[r(w) × r0 (w)] = r(w) × r00 (w).
gw
g
g
(u(w) · [v (w) × w(w)])= u0 (w) · [v(w) × w(w)] + u(w) ·
[v(w) × w (w)]
gw
gw
= u0 (w) · [v(w) × w(w)] + u(w) · [v0 (w) × w(w) + v(w) × w0 (w)]
= u0 (w) · [v(w) × w(w)] + u(w) · [v0 (w) × w(w)] + u(w) · [v(w) × w0 (w)]
= u0 (w) · [v(w) × w(w)] 3 v0 (w) · [u(w) × w(w)] + w0 (w) · [u(w) × v(w)]
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SECTION 13.3 ARC LENGTH AND CURVATURE
53.
¤
333
g
g
1
|r(w)| =
[r(w) · r(w)]1@2 = 12 [r(w) · r(w)]31@2 [2r(w) · r0 (w)] =
r(w) · r0 (w)
gw
gw
|r(w)|
54. Since r(w) · r0 (w) = 0, we have 0 = 2r(w) · r0 (w) =
g
g
[r(w) · r(w)] =
|r(w)|2 . Thus |r(w)|2 , and so |r(w)|, is a constant,
gw
gw
and hence the curve lies on a sphere with center the origin.
55. Since u(w) = r(w) · [r0 (w) × r00 (w)],
u0 (w) = r0 (w) · [r0 (w) × r00 (w)] + r(w) ·
g 0
[r (w) × r00 (w)]
gw
= 0 + r(w) · [r00 (w) × r00 (w) + r0 (w) × r000 (w)]
[since r0 (w) z r0 (w) × r00 (w)]
= r(w) · [r0 (w) × r000 (w)]
[since r00 (w) × r00 (w) = 0]
56. The tangent vector r0 (w) is defined as lim
k<0
r(w + k) 3 r(w)
. Here we assume that this limit exists and r0 (w) 6= 0; then we know
k
that this vector lies on the tangent line to the curve. As in Figure 1, let points S and T have position vectors r(w) and r(w + k).
3
3
<
The vector r(w + k) 3 r(w) points from S to T, so r(w + k) 3 r(w) = S T. If k A 0 then w ? w + k, so T lies “ahead”
3
3
<
of S on the curve. If k is sufficiently small (we can take k to be as small as we like since k < 0) then S T approximates
the curve from S to T and hence points approximately in the direction of the curve as w increases. Since k is positive,
3
< r(w + k) 3 r(w)
13
ST =
points in the same direction. If k ? 0, then w A w + k so T lies “behind” S on the curve. For k
k
k
3
3
<
sufficiently small, S T approximates the curve but points in the direction of decreasing w. However, k is negative, so
3
< r(w + k) 3 r(w)
13
ST =
points in the opposite direction, that is, in the direction of increasing w. In both cases, the difference
k
k
quotient
r(w + k) 3 r(w)
points in the direction of increasing w. The tangent vector r0 (w) is the limit of this difference quotient,
k
so it must also point in the direction of increasing w.
13.3 Arc Length and Curvature
i r0 (w) = h1> 33 sin w> 3 cos wi i
s
s
I
|r0 (w)| = 12 + (33 sin w)2 + (3 cos w)2 = 1 + 9(sin2 w + cos2 w) = 10.
I
I
U5
U5 I
5
Then using Formula 3, we have O = 35 |r0 (w)| gw = 35 10 gw = 10 w 35 = 10 10.
1. r(w) = hw> 3 cos w> 3 sin wi
i r0 (w) = 2> 2w> w2
i
s
s
I
|r0 (w)| = 22 + (2w)2 + (w2 )2 = 4 + 4w2 + w4 = (2 + w2 )2 = 2 + w2 for 0 $ w $ 1. Then using Formula 3, we have
1
U1
U1
O = 0 |r0 (w)| gw = 0 (2 + w2 ) gw = 2w + 13 w3 0 = 73 .
2. r(w) = 2w> w2 > 13 w3
I
I
2 w i + hw j + h3w k i r0 (w) = 2 i + hw j 3 h3w k i
tI s
I
2
2 + (hw )2 + (3h3w )2 = 2 + h2w + h32w = (hw + h3w )2 = hw + h3w [since hw + h3w A 0].
|r0 (w)| =
3. r(w) =
Then O =
U1
0
|r0 (w)| gw =
U1
0
1
(hw + h3w ) gw = hw 3 h3w 0 = h 3 h31 .
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¤
CHAPTER 13 VECTOR FUNCTIONS
4. r(w) = cos w i + sin w j + ln cos w k
|r0 (w)| =
i r0 (w) = 3 sin w i + cos w j +
3 sin w
k = 3 sin w i + cos w j 3 tan w k,
cos w
I
s
I
(3 sin w)2 + cos2 w + (3 tan w)2 = 1 + tan2 w = sec2 w = |sec w|. Since sec w A 0 for 0 $ w $ @4, here we
can say |r0 (w)| = sec w. Then
]
k
l@4
sec w gw = ln |sec w + tan w|
= ln sec + tan 3 ln |sec 0 + tan 0|
4
4
0
0
I
I
= ln 2 + 1 3 ln |1 + 0| = ln( 2 + 1)=
O=
5. r(w) = i + w2 j + w3 k
Then O =
U1
0
@4
i r0 (w) = 2w j + 3w2 k i |r0 (w)| =
|r0 (w)| gw =
U1 I
w 4 + 9w2 gw =
0
1
18
· 23 (4 + 9w2 )3@2
I
I
4w2 + 9w4 = w 4 + 9w2
l1
=
0
1
(133@2
27
3 43@2 ) =
[since w D 0].
1
(133@2
27
3 8).
I
i r0 (w) = 12 i + 12 w j + 6w k i
s
I
|r0 (w)| = 144 + 144w + 36w2 = 36(w + 2)2 = 6 |w + 2| = 6(w + 2) for 0 $ w $ 1. Then
6. r(w) = 12w i + 8w3@2 j + 3w2 k
O=
U1
0
|r0 (w)| gw =
7. r(w) = w2 > w3 > w4
O=
U2
0
|r0 (w)| gw =
8. r(w) = w> h3w > wh3w
|r0 (w)| =
O=
U3
1
U1
0
k
l1
6(w + 2) gw = 3w2 + 12w = 15.
0
i r0 (w) = 2w> 3w2 > 4w3
i |r0 (w)| =
U2I
4w2 + 9w4 + 16w6 gw E 18=6833.
0
i r0 (w) = 1> 3h3w > (1 3 w)h3w
t
I
(2w)2 + (3w2 )2 + (4w3 )2 = 4w2 + 9w4 + 16w6 , so
i
s
s
s
12 + (3h3w )2 + [(1 3 w)h3w ]2 = 1 + h32w + (1 3 w)2 h32w = 1 + (2 3 2w + w2 )h32w , so
|r0 (w)| gw =
U3s
1 + (2 + 2w + w2 )h32w gw E 2=0454.
1
9. r(w) = hsin w> cos w> tan wi
|r0 (w)| =
i r0 (w) = cos w> 3 sin w> sec2 w i
s
I
U @4 0
U @4 I
cos2 w + (3 sin w)2 + (sec2 w)2 = 1 + sec4 w and O = 0
|r (w)| gw = 0
1 + sec4 w gw E 1=2780.
10. We plot two different views of the curve with parametric equations { = sin w, | = sin 2w, } = sin 3w. To help visualize the
curve, we also include a plot showing a tube of radius 0=07 around the curve.
The complete curve is given by the parameter interval [0> 2] and we have r0 (w) = hcos w> 2 cos 2w> 3 cos 3wi i
I
U 2
U 2 I
|r0 (w)| = cos2 w + 4 cos2 2w + 9 cos2 3w, so O = 0 |r0 (w)| gw = 0
cos2 w + 4 cos2 2w + 9 cos2 3w gw E 16=0264.
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SECTION 13.3 ARC LENGTH AND CURVATURE
11. The projection of the curve F onto the {|-plane is the curve {2 = 2| or | =
1 2
2{ ,
¤
335
} = 0. Then we can choose the parameter
Since F also lies on the surface 3} = {|, we have } = 13 {| = 13 (w)( 12 w2 ) = 16 w3 . Then parametric
equations for F are { = w, | = 12 w2 , } = 16 w3 and the corresponding vector equation is r(w) = w> 12 w2 > 16 w3 . The origin
{=w i |=
1 2
w .
2
corresponds to w = 0 and the point (6> 18> 36) corresponds to w = 6, so
U6 2
U6t
U6t
U6
O = 0 |r0 (w)| gw = 0 1> w> 12 w2 gw = 0 12 + w2 + 12 w2 gw = 0 1 + w2 + 14 w4 gw
=
6
U6
U6t
(1 + 12 w2 )2 gw = 0 (1 + 12 w2 ) gw = w + 16 w3 0 = 6 + 36 = 42
0
12. Let F be the curve of intersection. The projection of F onto the {|-plane is the ellipse 4{2 + | 2 = 4 or {2 + | 2 @4 = 1,
} = 0. Then we can write { = cos w, | = 2 sin w, 0 $ w $ 2. Since F also lies on the plane { + | + } = 2, we have
} = 2 3 { 3 | = 2 3 cos w 3 2 sin w. Then parametric equations for F are { = cos w, | = 2 sin w, } = 2 3 cos w 3 2 sin w,
0 $ w $ 2, and the corresponding vector equation is r(w) = hcos w> 2 sin w> 2 3 cos w 3 2 sin wi. Differentiating gives
r0 (w) = h3 sin w> 2 cos w> sin w 3 2 cos wi i
s
s
|r0 (w)| = (3 sin w)2 + (2 cos w)2 + (sin w 3 2 cos w)2 = 2 sin2 w + 8 cos2 w 3 4 sin w cos w. The length of F is
U 2
U 2 s
O = 0 |r0 (w)| gw = 0
2 sin2 w + 8 cos2 w 3 4 sin w cos w gw E 13=5191.
I
I
i r0 (w) = 2 i 3 3 j + 4 k and gv
= |r0 (w)| = 4 + 9 + 16 = 29. Then
gw
I
Uw I
Uw
v = v(w) = 0 |r0 (x)| gx = 0 29 gx = 29 w. Therefore, w = I129 v, and substituting for w in the original equation, we
have r(w(v)) = I229 v i + 1 3 I329 v j + 5 + I429 v k.
13. r(w) = 2w i + (1 3 3w) j + (5 + 4w) k
i r0 (w) = 2h2w (cos 2w 3 sin 2w) i + 2h2w (cos 2w + sin 2w) k,
s
s
I
= |r0 (w)| = 2h2w (cos 2w 3 sin 2w)2 + (cos 2w + sin 2w)2 = 2h2w 2 cos2 2w + 2 sin2 2w = 2 2 h2w .
14. r(w) = h2w cos 2w i + 2 j + h2w sin 2w k
gv
gw
v = v(w) =
Uw
0
|r0 (x)| gx =
Substituting, we have
Uw
0
2
I 2x
I
I
w
2 h gx = 2 h2x 0 = 2 (h2w 3 1) i
Iv
2
+ 1 = h2w
i w=
1
2
ln Iv2 + 1 .
2 1 ln sv +1
2 1 ln sv +1
2
2
cos 2 12 ln Iv2 + 1 i + 2 j + h 2
sin 2 12 ln Iv2 + 1 k
r(w(v)) = h 2
= Iv2 + 1 cos ln Iv2 + 1 i + 2 j + Iv2 + 1 sin ln Iv2 + 1 k
15. Here r(w) = h3 sin w> 4w> 3 cos wi, so r0 (w) = h3 cos w> 4> 33 sin wi and |r0 (w)| =
s
I
9 cos2 w + 16 + 9 sin2 w = 25 = 5.
The point (0> 0> 3) corresponds to w = 0, so the arc length function beginning at (0> 0> 3) and measuring in the positive
Uw
Uw
direction is given by v(w) = 0 |r0 (x)| gx = 0 5 gx = 5w. v(w) = 5 i 5w = 5 i w = 1, thus your location after
moving 5 units along the curve is (3 sin 1> 4> 3 cos 1).
2
2w
34w
32w2 + 2
3
1
i+ 2
j i r0 (w) = 2
i+ 2
j,
2
2
w +1
w +1
(w + 1)
(w + 1)2
v
v
2 2 v 4
u
4(w2 + 1)2
32w2 + 2
2
4w + 8w2 + 4
4
gv
34w
0
+
=
=
=
= 2
= |r (w)| =
.
gw
(w2 + 1)2
(w2 + 1)2
(w2 + 1)4
(w2 + 1)4
(w2 + 1)2
w +1
16. r(w) =
Since the initial point (1> 0) corresponds to w = 0, the arc length function
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CHAPTER 13 VECTOR FUNCTIONS
]
w
0 r (x) gx =
]
w
2
gx = 2 arctan w. Then arctan w = 12 v i w = tan 12 v. Substituting, we have
x2 + 1
&
%
2 tan 12 v
1 3 tan2 12 v
2 tan 12 v
2
j
3
1
i
+
j
=
r(w(v)) =
i
+
tan2 12 v + 1
tan2 12 v + 1
1 + tan2 12 v
sec2 12 v
1 3 tan2 12 v
i + 2 tan 12 v cos2 12 v j = cos2 12 v 3 sin2 12 v i + 2 sin 12 v cos 12 v j = cos v i + sin v j
=
sec2 12 v
v(w) =
0
0
With this parametrization, we recognize the function as representing the unit circle. Note here that the curve approaches, but
does not include, the point (31> 0), since cos v = 31 for v = + 2n (n an integer) but then w = tan 12 v is undefined.
s
I
i r0 (w) = h1> 33 sin w> 3 cos wi i |r0 (w)| = 1 + 9 sin2 w + 9 cos2 w = 10.
G
H
r0 (w)
Then T(w) = 0
= I110 h1> 33 sin w> 3 cos wi or I110 > 3 I310 sin w> I310 cos w .
|r (w)|
s
T0 (w) = I110 h0> 33 cos w> 33 sin wi i |T0 (w)| = I110 0 + 9 cos2 w + 9 sin2 w = I310 . Thus
I
T0 (w)
1@ 10
I h0> 33 cos w> 33 sin wi = h0> 3 cos w> 3 sin wi.
N(w) =
=
|T0 (w)|
3@ 10
17. (a) r(w) = hw> 3 cos w> 3 sin wi
(b) (w) =
I
|T0 (w)|
3
3@ 10
I
=
=
|r0 (w)|
10
10
18. (a) r(w) = w2 > sin w 3 w cos w> cos w + w sin w
i
r0 (w) = h2w> cos w + w sin w 3 cos w, 3sin w + w cos w + sin wi = h2w> w sin w> w cos wi i
s
s
I
I
|r0 (w)| = 4w2 + w2 sin2 w + w2 cos2 w = 4w2 + w2 (cos2 w + sin2 w) = 5w2 = 5 w [since w A 0]. Then
T(w) =
r0 (w)
1
= I h2w> w sin w> w cos wi =
|r0 (w)|
5w
|T0 (w)| =
1
I
5
s
0 + cos2 w + sin2 w =
I1 .
5
I1
5
h2> sin w> cos wi. T0 (w) =
Thus N(w) =
1
I
5
h0> cos w> 3 sin wi i
I
1@ 5
T0 (w)
I h0> cos w> 3 sin wi = h0> cos w> 3 sin wi.
=
|T0 (w)|
1@ 5
I
1@ 5
|T0 (w)|
1
= I
(b) (w) = 0
=
|r (w)|
5w
5w
19. (a) r(w) =
Then
T(w) =
i
|r0 (w)| =
s
I
2 + h2w + h32w = (hw + h3w )2 = hw + h3w .
I w
I w 2w
1
1
r0 (w)
= w
2> h > 3h3w = 2w
2 h > h > 31
|r0 (w)|
h + h3w
h +1
hw
and
after multiplying by w
h
I w 2w
I w 2w 1
2h2w
2 h > 2h > 0 3
2 h > h > 31
2
2w
+1
(h + 1)
I w 2w
I w
I
1
1
(h2w + 1) 2 hw > 2h2w > 0 3 2h2w
2 h > h > 31 = 2w
2 h 1 3 h2w > 2h2w > 2h2w
= 2w
2
2
(h + 1)
(h + 1)
T0 (w) =
Then
I
I w
i r0 (w) =
2 w> hw > h3w
2> h > 3h3w
h2w
s
s
1
1
2h2w (1 3 2h2w + h4w ) + 4h4w + 4h4w =
2h2w (1 + 2h2w + h4w )
2
2w
(h2w + 1)2
(h + 1)
I w
I w
t
1
2 h (1 + h2w )
2h
2w (1 + h2w )2 =
= 2w
2h
=
(h + 1)2
(h2w + 1)2
h2w + 1
|T0 (w)| =
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SECTION 13.3 ARC LENGTH AND CURVATURE
¤
Therefore
I w
T0 (w)
1
h2w + 1
2 h (1 3 h2w )> 2h2w > 2h2w
= I
0
2w
2
w
|T (w)|
2 h (h + 1)
I w
I
I
1
1
1 3 h2w > 2 hw > 2 hw
2 h (1 3 h2w )> 2h2w > 2h2w = 2w
= I
w
2w
h
+
1
2 h (h + 1)
N(w) =
I w
I 2w
I w
I 2w
2h
2h
2h
2h
1
|T0 (w)|
=
·
=
= 3w
= 4w
(b) (w) = 0
|r (w)|
h2w + 1 hw + h3w
h + 2hw + h3w
h + 2h2w + 1
(h2w + 1)2
20. (a) r(w) = w> 12 w2 > w2
T(w) =
i r0 (w) = h1> w> 2wi
i
|r0 (w)| =
I
I
1 + w2 + 4w2 = 1 + 5w2 . Then
1
r0 (w)
= I
h1> w> 2wi.
|r0 (w)|
1 + 5w2
35w
1
h1> w> 2wi + I
h0> 1> 2i
[by Formula 3 of Theorem 13.2.3]
(1 + 5w2 )3@2
1 + 5w2
1
1
35w> 35w2 > 310w2 + 0> 1 + 5w2 > 2 + 10w2 =
=
h35w> 1> 2i
(1 + 5w2 )3@2
(1 + 5w2 )3@2
I I
I
I
I
5 5w2 + 1
5
1
1
2 +1+4 =
2 +5 =
|T0 (w)| =
25w
25w
=
1 + 5w2
(1 + 5w2 )3@2
(1 + 5w2 )3@2
(1 + 5w2 )3@2
T0 (w) =
T0 (w)
1
1
1 + 5w2
I
h35w> 1> 2i = I
h35w> 1> 2i.
·
=
|T0 (w)|
(1 + 5w2 )3@2
5 + 25w2
5
I
I
5@(1 + 5w2 )
5
|T0 (w)|
= I
=
(b) (w) = 0
|r (w)|
(1 + 5w2 )3@2
1 + 5w2
Thus N(w) =
21. r(w) = w3 j + w2 k
i r0 (w) = 3w2 j + 2w k, r00 (w) = 6w j + 2 k, |r0 (w)| =
r0 (w) × r00 (w) = 36w2 i, |r0 (w) × r00 (w)| = 6w2 . Then (w) =
s
I
02 + (3w2 )2 + (2w)2 = 9w4 + 4w2 ,
|r0 (w) × r00 (w)|
6w2
6w2
= I
=
.
3
3
4
(9w + 4w2 )3@2
|r0 (w)|
9w4 + 4w2
i r0 (w) = i + 2w j + hw k, r00 (w) = 2 j + hw k,
s
I
|r0 (w)| = 12 + (2w)2 + (hw )2 = 1 + 4w2 + h2w , r0 (w) × r00 (w) = (2w 3 2)hw i 3 hw j + 2 k,
s
s
s
|r0 (w) × r00 (w)| = [(2w 3 2)hw ]2 + (3hw )2 + 22 = (2w 3 2)2 h2w + h2w + 4 = (4w2 3 8w + 5)h2w + 4.
s
s
(4w2 3 8w + 5)h2w + 4
(4w2 3 8w + 5)h2w + 4
|r0 (w) × r00 (w)|
Then (w) =
=
=
.
I
3
3
0
|r (w)|
(1 + 4w2 + h2w )3@2
1 + 4w2 + h2w
22. r(w) = w i + w2 j + hw k
i r0 (w) = 3 i + 4 cos w j 3 4 sin w k, r00 (w) = 34 sin w j 3 4 cos w k,
s
I
|r0 (w)| = 9 + 16 cos2 w + 16 sin2 w = 9 + 16 = 5, r0 (w) × r00 (w) = 316 i + 12 cos w j 3 12 sin w k,
23. r(w) = 3w i + 4 sin w j + 4 cos w k
|r0 (w) × r00 (w)| =
s
I
|r0 (w) × r00 (w)|
20
4
.
256 + 144 cos2 w + 144 sin2 w = 400 = 20. Then (w) =
= 3 =
5
25
|r0 (w)|3
i r0 (w) = h2w> 1@w> 1 + ln wi, r00 (w) = 2> 31@w2 > 1@w . The point (1> 0> 0) corresponds
I
I
to w = 1, and r0 (1) = h2> 1> 1i, |r0 (1)| = 22 + 12 + 12 = 6, r00 (1) = h2> 31> 1i, r0 (1) × r00 (1) = h2> 0> 34i,
I
I
I
s
I
I
|r0 (1) × r00 (1)|
2 5
2 5
30
I
=
=
|r0 (1) × r00 (1)| = 22 + 02 + (34)2 = 20 = 2 5. Then (1) =
or
.
I 3
18
|r0 (1)|3
6
6
6
24. r(w) = w2 > ln w> w ln w
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CHAPTER 13 VECTOR FUNCTIONS
i r0 (w) = 1> 2w> 3w2 . The point (1> 1> 1) corresponds to w = 1, and r0 (1) = h1> 2> 3i i
I
I
|r0 (1)| = 1 + 4 + 9 = 14. r00 (w) = h0> 2> 6wi i r00 (1) = h0> 2> 6i. r0 (1) × r00 (1) = h6> 36> 2i, so
u
I
I
I
|r0 (1) × r00 (1)|
1 19
76
0
00
= I 3 =
|r (1) × r (1)| = 36 + 36 + 4 = 76. Then (1) =
.
7 14
|r0 (1)|3
14
25. r(w) = w> w2 > w3
26.
Note that we get the complete curve for 0 $ w ? 2.
r(w) = hcos w> sin w> sin 5wi i r0 (w) = h3 sin w> cos w> 5 cos 5wi,
r00 (w) = h3 cos w> 3 sin w> 325 sin 5wi. The point (1> 0> 0)
corresponds to w = 0, and r0 (0) = h0> 1> 5i i
I
I
|r0 (0)| = 02 + 12 + 52 = 26, r00 (0) = h31> 0> 0i,
r0 (0) × r00 (0) = h0> 35> 1i i
s
I
|r0 (0) × r00 (0)| = 02 + (35)2 + 12 = 26. The curvature at
I
|r0 (0) × r00 (0)|
1
26
the point (1> 0> 0) is (0) =
= I 3 =
.
26
|r0 (0)|3
26
i 0 ({) = 4{3 , i 00 ({) = 12{2 , ({) =
27. i ({) = {4 ,
12{2 |i 00 ({)|
12{2
=
=
0
2
3@2
3
2
3@2
[1 + (i ({)) ]
[1 + (4{ ) ]
(1 + 16{6 )3@2
i 0 ({) = sec2 {, i 00 ({) = 2 sec { · sec { tan { = 2 sec2 { tan {,
2 sec2 { tan {
2 sec2 { |tan {|
|i 00 ({)|
=
=
({) =
0
2
3@2
2
2
3@2
[1 + (i ({)) ]
[1 + (sec {) ]
(1 + sec4 {)3@2
28. i ({) = tan {,
29. i ({) = {h{ ,
({) =
30. | 0 =
i 0 ({) = {h{ + h{ , i 00 ({) = {h{ + 2h{ ,
|{h{ + 2h{ |
|{ + 2| h{
|i 00 ({)|
=
=
0
2
3@2
{
{
2
3@2
[1 + (i ({)) ]
[1 + ({h + h ) ]
[1 + ({h{ + h{ )2 ]3@2
1
1
, |00 = 3 2 ,
{
{
31 1
|| 00 ({)|
1 ({2 )3@2
|{|
{
({) = =
= 2 2
= 2
= 2
3@2
2
2
3@2
3@2
3@2
2
{
{
(1
+
1@{
)
({
+
1)
({
+
1)
({
+
1)3@2
0
1 + (| ({))
[since { A 0].
To find the maximum curvature, we first find the critical numbers of ({):
0
({) =
({2 + 1)3@2 3 { 32 ({2 + 1)1@2 (2{)
2
[({2 + 1)3@2 ]
=
({2 + 1)1@2 [({2 + 1) 3 3{2 ]
1 3 2{2
=
;
({2 + 1)3
({2 + 1)5@2
0 ({) = 0 i 1 3 2{2 = 0, so the only critical number in the domain is { =
and 0 ({) ? 0 for { A
Since lim
{<"
I1 ,
2
({) attains its maximum at { =
1
I
.
2
I1 .
2
Since 0 ({) A 0 for 0 ? { ?
Thus, the maximum curvature occurs at
{
= 0, ({) approaches 0 as { < ".
({2 + 1)3@2
I1 > ln I1
2
2
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SECTION 13.3 ARC LENGTH AND CURVATURE
31. Since | 0 = | 00 = h{ , the curvature is ({) =
[1 +
|| 00 ({)|
(| 0 ({))2 ]3@2
=
¤
339
h{
= h{ (1 + h2{ )33@2 .
(1 + h2{ )3@2
To find the maximum curvature, we first find the critical numbers of ({):
2{
1 + h2{ 3 3h2{
{ 1 3 2h
0 ({) = h{ (1 + h2{ )33@2 + h{ 3 32 (1 + h2{ )35@2 (2h2{ ) = h{
=
h
.
(1 + h2{ )5@2
(1 + h2{ )5@2
0 ({) = 0 when 1 3 2h2{ = 0, so h2{ =
1
2
or { = 3 12 ln 2. And since 1 3 2h2{ A 0 for { ? 3 12 ln 2 and 1 3 2h2{ ? 0
for { A 3 12 ln 2, the maximum curvature is attained at the point 3 12 ln 2> h(3 ln 2)@2 = 3 12 ln 2> I12 .
Since lim h{ (1 + h2{ )33@2 = 0> ({) approaches 0 as { < ".
{<"
32. We can take the parabola as having its vertex at the origin and opening upward, so the equation is i ({) = d{2 > d A 0. Then by
Equation 11, ({) =
|i 00 ({)|
|2d|
2d
=
=
, thus (0) = 2d. We want (0) = 4, so
[1 + (i 0 ({))2 ]3@2
[1 + (2d{)2 ]3@2
(1 + 4d2 {2 )3@2
d = 2 and the equation is | = 2{2 .
33. (a) F appears to be changing direction more quickly at S than T, so we would expect the curvature to be greater at S .
(b) First we sketch approximate osculating circles at S and T. Using the
axes scale as a guide, we measure the radius of the osculating circle
at S to be approximately 0=8 units, thus =
=
1
i
1
1
E
E 1=3. Similarly, we estimate the radius of the
0=8
osculating circle at T to be 1=4 units, so =
1
1
E
E 0=7.
1=4
i | 0 = 4{3 3 4{, | 00 = 12{2 3 4, and
12{2 3 4
|| 00 |
({) = =
3@2
3@2 . The graph of the
1 + (|0 )2
1 + (4{3 3 4{)2
34. | = {4 3 2{2
curvature here is what we would expect. The graph of | = {4 3 2{2
appears to be bending most sharply at the origin and near { = ±1.
35. | = {32
i | 0 = 32{33 , | 00 = 6{34 , and
34 6{ || 00 |
6
({) = = =
.
3@2
2
2 3@2
4 (1 + 4{36 )3@2
0
33
{
1 + (| )
1 + (32{ )
The appearance of the two humps in this graph is perhaps a little surprising, but it is
explained by the fact that | = {32 increases asymptotically at the origin from both
directions, and so its graph has very little bend there. [Note that (0) is undefined.]
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CHAPTER 13 VECTOR FUNCTIONS
i r0 (w) = h1 3 cos w> sin w> 32 sin(w@2)i, r00 (w) = hsin w> cos w> 3 cos(w@2)i.
I
Using a CAS, r0 (w) × r00 (w) = 32 sin3 (w@2)> 3 sin(w@2) sin w> cos w 3 1 , |r0 (w) × r00 (w)| = 3 3 4 cos w + cos 2w or
36. r(w) = hw 3 sin w> 1 3 cos w> 4 cos(w@2)i
I
I
I
2 2 sin2 (w@2), and |r0 (w)| = 2 1 3 cos w or 2 2 |sin(w@2)|. (To compute cross products in Maple, use the
VectorCalculus or LinearAlgebra package and the CrossProduct(a,b) command. Here loading
the RealDomain package will give simpler results. In Mathematica, use Cross[a,b].)
I
3 3 4 cos w + cos 2w
1
1
|r0 (w) × r00 (w)|
. We plot the space curve and its
or
=
or I
Then (w) =
8 |sin(w@2)|
4 2 3 2 cos w
|r0 (w)|3
8 (1 3 cos w)3@2
curvature function for 0 $ w $ 8 below.
The asymptotes in the graph of (w) correspond to the sharp cusps we see in the graph of r(w). The space curve bends most
sharply as it approaches these cusps (mostly in the {-direction) and bends most gradually between these, near its intersections
with the {|-plane, where w = + 2q (q an integer). (The bending we see in the }-direction on the curve near these points is
deceiving; most of the curvature occurs in the {-direction.) The curvature graph has local minima at these values of w.
37. r(w) = whw > h3w >
I I 2w i r0 (w) = (w + 1)hw > 3h3w > 2 , r00 (w) = (w + 2)hw > h3w > 0 . Then
s
I
I
r0 (w) × r00 (w) = 3 2h3w > 2(w + 2)hw > 2w + 3 , |r0 (w) × r00 (w)| = 2h32w + 2(w + 2)2 h2w + (2w + 3)2 ,
|r0 (w)| =
s
|r0 (w) × r00 (w)|
(w + 1)2 h2w + h32w + 2, and (w) =
=
|r0 (w)|3
s
2h32w + 2(w + 2)2 h2w + (2w + 3)2
[(w + 1)2 h2w + h32w + 2]3@2
.
We plot the space curve and its curvature function for 35 $ w $ 5 below.
From the graph of (w) we see that curvature is maximized for w = 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 (w) becomes nearly 0 as |w| increases.
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SECTION 13.3 ARC LENGTH AND CURVATURE
¤
341
38. Notice that the curve d is highest for the same {-values at which curve e is turning more sharply, and d is 0 or near 0 where e is
nearly straight. So, d must be the graph of | = ({), and e is the graph of | = i ({).
39. Notice that the curve e has two inflection points at which the graph appears almost straight. We would expect the curvature to
be 0 or nearly 0 at these values, but the curve d isn’t near 0 there. Thus, d must be the graph of | = i ({) rather than the graph
of curvature, and e is the graph of | = ({).
40. (a) The complete curve is given by 0 $ w $ 2. Curvature
appears to have a local (or absolute) maximum at 6
points. (Look at points where the curve appears to turn
more sharply.)
(b) Using a CAS, we find (after simplifying)
I s
3 2 (5 sin w + sin 5w)2
. (To compute cross
(w) =
(9 cos 6w + 2 cos 4w + 11)3@2
products in Maple, use the VectorCalculus or
LinearAlgebra package and the
CrossProduct(a,b) command; in Mathematica, use
Cross[a,b].) The graph shows 6 local (or absolute)
maximum points for 0 $ w $ 2, as observed in part (a).
41. Using a CAS, we find (after simplifying)
(w) =
6
I
4 cos2 w 3 12 cos w + 13
. (To compute cross
(17 3 12 cos w)3@2
products in Maple, use the VectorCalculus or
LinearAlgebra package and the CrossProduct(a,b)
command; in Mathematica, use Cross[a,b].) Curvature is
largest at integer multiples of 2.
r0 (w) = hi 0 (w)> j0 (w)i, r00 (w) = hi 00 (w)> j 00 (w)i,
l3
ks
3
(i 0 (w))2 + (j0 (w))2 = [(i 0 (w))2 + (j 0 (w))2 ]3@2 = ({ 2 + | 2 )3@2 , and
|r0 (w)| =
42. Here r(w) = hi(w)> j (w)i,
1@2
|{¨
| 3 |¨
{|
|r0 (w) × r00 (w)| = |h0> 0> i 0 (w) j 00 (w) 3 i 00 (w) j 0 (w)i| = ({¨
| 3 {
¨|)
2
= |{¨
| 3 |¨
{|. Thus (w) = 2
.
[{ + | 2 ]3@2
43. { = w2
i { = 2w i {
¨ = 2, | = w3 i | = 3w2 i |¨ = 6w.
2
(2w)(6w) 3 (3w2 )(2)
12w 3 6w2 |{¨
| 3 |¨
{|
6w2
Then (w) = 2
=
=
=
.
[{ + | 2 ]3@2
[(2w)2 + (3w2 )2 ]3@2
(4w2 + 9w4 )3@2
(4w2 + 9w4 )3@2
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CHAPTER 13 VECTOR FUNCTIONS
44. { = d cos $w
i { = 3d$ sin $w i {
¨ = 3d$2 cos $w,
| = e sin $w i | = e$ cos $w i |¨ = 3e$2 sin $w. Then
(3d$ sin $w)(3e$2 sin $w) 3 (e$ cos $w)(3d$2 cos $w)
|{¨
| 3 |¨
{|
=
(w) = 2
[{ + | 2 ]3@2
[(3d$ sin $w)2 + (e$ cos $w)2 ]3@2
de$ 3 sin2 $w + de$3 cos2 $w
de$3 = 2 2 2
=
(d $ sin $w + e2 $2 cos2 $w)3@2
(d2 $2 sin2 $w + e2 $2 cos2 $w)3@2
45. { = hw cos w
i { = hw (cos w 3 sin w) i {
¨ = hw (3 sin w 3 cos w) + hw (cos w 3 sin w) = 32hw sin w,
| = hw sin w i | = hw (cos w + sin w) i |¨ = hw (3 sin w + cos w) + hw (cos w + sin w) = 2hw cos w. Then
w
h (cos w 3 sin w)(2hw cos w) 3 hw (cos w + sin w)(32hw sin w)
|{¨
| 3 |¨
{|
=
(w) = 2
[{ + | 2 ]3@2
([hw (cos w 3 sin w)]2 + [hw (cos w + sin w)]2 )3@2
2w
2w 2h (cos2 w 3 sin w cos w + sin w cos w + sin2 w)
2h (1)
2h2w
1
= =
= 3w 3@2 = I
3@2
3@2
2
2
2w
h (2)
2 hw
[h (1 + 1)]
h2w (cos2 w 3 2 cos w sin w + sin w + cos2 w + 2 cos w sin w + sin w)
i 0 ({) = fhf{ , i 00 ({) = f2 hf{ . Using Formula 11 we have
2 f{ f h |i 00 ({)|
f2 hf{
({) =
=
=
so the curvature at { = 0 is
0
2
3@2
f{
2
3@2
[1 + (i ({)) ]
[1 + (fh ) ]
(1 + f2 h2f{ )3@2
46. i ({) = hf{ ,
f2
f2
.
To
determine
the
maximum
value
for
(0),
let
i
(f)
=
. Then
(1 + f2 )3@2
(1 + f2 )3@2
(1 + f2 )1@2 2f(1 + f2 ) 3 3f3
2f 3 f3
2f · (1 + f2 )3@2 3 f2 · 32 (1 + f2 )1@2 (2f)
0
=
=
. We have a critical
i (f) =
(1 + f2 )3
[(1 + f2 )3@2 ]2
(1 + f2 )5@2
I
I
I
number when 2f 3 f3 = 0 i f(2 3 f2 ) = 0 i f = 0 or f = ± 2. i 0 (f) is positive for f ? 3 2, 0 ? f ? 2
(0) =
and negative elsewhere, so i achieves its maximum value when f =
of the family with the largest value of (0) are i ({) = h
I
2{
I
I
2 or 3 2. In either case, (0) =
I
and i ({) = h3
2{
2
, so the members
33@2
.
2 2w> 2w2 > 1
2w> 2w2 > 1
r0 (w)
I
=
, so T(1) = 23 > 23 > 13 .
47. 1> 3 > 1 corresponds to w = 1. T(w) = 0
=
|r (w)|
2w2 + 1
4w2 + 4w4 + 1
[by Formula 3 of Theorem 13.2.3]
T0 (w) = 34w(2w2 + 1)32 2w> 2w2 > 1 + (2w2 + 1)31 h2> 4w> 0i
2
32
2
2
3
3
38w + 4w + 2> 38w + 8w + 4w> 34w = 2(2w2 + 1)32 1 3 2w2 > 2w> 32w
= (2w + 1)
2(2w2 + 1)32 1 3 2w2 > 2w> 32w
1 3 2w2 > 2w> 32w
1 3 2w2 > 2w> 32w
T0 (w)
s
I
=
=
=
|T0 (w)|
1 + 2w2
1 3 4w2 + 4w4 + 8w2
2(2w2 + 1)32 (1 3 2w2 )2 + (2w)2 + (32w)2
N(1) = 3 13 > 23 > 3 23 and B(1) = T(1) × N(1) = 3 49 3 29 > 3 3 49 + 19 > 49 + 29 = 3 23 > 13 > 23 .
N(w) =
48. (1> 0> 0) corresponds to w = 0.
r(w) = hcos w> sin w> ln cos wi, and in Exercise 4 we found that r0 (w) = h3 sin w> cos w> 3 tan wi
and |r0 (w)| = |sec w|. Here we can assume 3 2 ? w ?
T(w) =
2
and then sec w A 0
i
|r0 (w)| = sec w.
h3 sin w> cos w> 3 tan wi
r0 (w)
=
= 3 sin w cos w> cos2 w> 3 sin w and T(0) = h0> 1> 0i.
0
|r (w)|
sec w
T0 (w) = h3[(sin w)(3 sin w) + (cos w)(cos w)]> 2(cos w)(3 sin w)> 3 cos wi = sin2 w 3 cos2 w> 32 sin w cos w> 3 cos w , so
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SECTION 13.3 ARC LENGTH AND CURVATURE
¤
H
G
T0 (0)
h31> 0> 31i
1
= I h31> 0> 31i = 3 I12 > 0> 3 I12 .
= I
0
|T (0)|
1+0+1
2
G
H G
H
Finally, B(0) = T(0) × N(0) = h0> 1> 0i × 3 I12 > 0> 3 I12 = 3 I12 > 0> I12 .
N(0) =
49. (0> > 32) corresponds to w = .
T(w) =
T() =
r(w) = h2 sin 3w> w> 2 cos 3wi i
r0 (w)
h6 cos 3w> 1> 36 sin 3wi
1
= I h6 cos 3w> 1> 36 sin 3wi.
= s
|r0 (w)|
37
36 cos2 3w + 1 + 36 sin2 3w
I1
37
h36> 1> 0i is a normal vector for the normal plane, and so h36> 1> 0i is also normal. Thus an equation for the
plane is 36 ({ 3 0) + 1(| 3 ) + 0(} + 2) = 0 or | 3 6{ = .
s
182 sin2 3w + 182 cos2 3w
18
I
= I
T0 (w) = I137 h318 sin 3w> 0> 318 cos 3wi i |T0 (w)| =
37
37
T0 (w)
= h3 sin 3w> 0> 3 cos 3wi. So N() = h0> 0> 1i and B() =
|T0 (w)|
N(w) =
I1
37
i
h36> 1> 0i × h0> 0> 1i =
I1
37
h1> 6> 0i.
Since B() is a normal to the osculating plane, so is h1> 6> 0i.
An equation for the plane is 1({ 3 0) + 6(| 3 ) + 0(} + 2) = 0 or { + 6| = 6.
50. w = 1 at (1> 1> 1).
r0 (w) = 1> 2w> 3w2 . r0 (1) = h1> 2> 3i is normal to the normal plane, so an equation for this plane
is 1({ 3 1) + 2(| 3 1) + 3(} 3 1) = 0, or { + 2| + 3} = 6.
r0 (w)
1
1> 2w> 3w2 . Using the product rule on each term of T(w) gives
= I
0
2
4
|r (w)|
1 + 4w + 9w
T(w) =
T0 (w) =
1
3 12 (8w + 36w3 )> 2(1 + 4w2 + 9w4 ) 3 12 (8w + 36w3 )2w>
(1 + 4w2 + 9w4 )3@2
6w(1 + 4w2 + 9w4 ) 3 12 (8w + 36w3 )3w2
=
32
1
34w 3 18w3 > 2 3 18w4 > 6w + 12w3 =
h11> 8> 39i when w = 1.
(1 + 4w2 + 9w4 )3@2
(14)3@2
N(1) k T0 (1) k h11> 8> 39i and T(1) k r0 (1) = h1> 2> 3i i a normal vector to the osculating plane is
h11> 8> 39i × h1> 2> 3i = h42> 342> 14i or equivalently h3> 33> 1i.
An equation for the plane is 3({ 3 1) 3 3(| 3 1) + (} 3 1) = 0 or 3{ 3 3| + } = 1.
51. The ellipse is given by the parametric equations { = 2 cos w, | = 3 sin w, so using the result from Exercise 42,
(w) =
|{¨
| 3 {
¨||
|(32 sin w)(33 sin w) 3 (3 cos w)(32 cos w)|
6
=
=
.
[{ 2 + | 2 ]3@2
(4 sin2 w + 9 cos2 w)3@2
(4 sin2 w + 9 cos2 w)3@2
At (2> 0), w = 0. Now (0) =
6
27
= 29 , so the radius of the osculating circle is
2
and its center is 3 52 > 0 . Its equation is therefore { + 52 + | 2 = 81
.
4
At (0> 3), w = 2 , and 2 = 68 = 34 . So the radius of the osculating circle is 43 and
1@(0) =
9
2
2
its center is 0> 53 . Hence its equation is {2 + | 3 53 =
16
.
9
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¤
CHAPTER 13 VECTOR FUNCTIONS
1
. So the curvature at (0> 0) is (0) = 1 and
(1 + {2 )3@2
the osculating circle has radius 1 and center (0> 1), and hence equation {2 + (| 3 1)2 = 1. The curvature at 1> 12
52. | =
i | 0 = { and | 00 = 1, so Formula 11 gives ({) =
1 2
2{
is (1) =
1
1
= I . The tangent line to the parabola at 1> 12
(1 + 12 )3@2
2 2
has slope 1, so the normal line has slope 31. Thus the center of the
H
G
osculating circle lies in the direction of the unit vector 3 I12 > I12 .
I
The circle has radius 2 2, so its center has position vector
H
I G
1> 12 + 2 2 3 I12 > I12 = 31> 52 . So the equation of the circle
2
is ({ + 1)2 + | 3 52 = 8.
53. The tangent vector is normal to the normal plane, and the vector h6> 6> 38i is normal to the given plane.
But T(w) k r0 (w) and h6> 6> 38i k h3> 3> 34i, so we need to find w such that r0 (w) k h3> 3> 34i.
i r0 (w) = 3w2 > 3> 4w3 k h3> 3> 34i when w = 31. So the planes are parallel at the point (31> 33> 1).
r(w) = w3 > 3w> w4
54. To find the osculating plane, we first calculate the unit tangent and normal vectors.
In Maple, we use the VectorCalculus package and set r:= ?tˆ3,3*t,tˆ4A;. After differentiating, the
Normalize command converts the tangent vector to the unit tangent vector: T:=Normalize(diff(r,t));. After
3w2 > 3> 4w3
simplifying, we find that T(w) = I
. We use a similar procedure to compute the unit normal vector,
16w6 + 9w4 + 9
3w(8w6 3 9)> 33w3 (3 + 8w2 )> 6w2 (w4 + 3)
N:=Normalize(diff(T,t));. After simplifying, we have N(w) = s
. Then
w2 (4w6 + 36w2 + 9)(16w6 + 9w4 + 9)
6w2 > 32w4 > 33w
.
we use the command B:=CrossProduct(T,N);. After simplification, we find that B(w) = s
w2 (4w6 + 36w2 + 9)
In Mathematica, we define the vector function r={tˆ3,3*t,tˆ4} and use the command Dt to differentiate. We find
T(w) by dividing the result by its magnitude, computed using the Norm command. (You may wish to include the option
Element[t,Reals] to obtain simpler expressions.) N(w) is found similarly, and we use Cross[T,N] to find B(w).
Now B(w) is parallel to 6w2 > 32w4 > 33w , so if B(w) is parallel to h1> 1> 1i for some w 6= 0 [since B(0) = 0], then
6w2 > 32w4 > 33w = n h1> 1> 1i for some value of n. But then 6w2 = 32w4 = 33w which has no solution for w 6= 0. So there is
no such osculating plane.
55. First we parametrize the curve of intersection. We can choose | = w; then { = | 2 = w2 and } = {2 = w4 , and the curve is
given by r(w) = w2 > w> w4 . r0 (w) = 2w> 1> 4w3 and the point (1> 1> 1) corresponds to w = 1, so r0 (1) = h2> 1> 4i is a normal
vector for the normal plane. Thus an equation of the normal plane is
2({ 3 1) + 1(| 3 1) + 4(} 3 1) = 0 or 2{ + | + 4} = 7. T(w) =
r0 (w)
1
2w> 1> 4w3 and
= I
|r0 (w)|
4w2 + 1 + 16w6
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SECTION 13.3 ARC LENGTH AND CURVATURE
¤
345
T0 (w) = 3 12 (4w2 + 1 + 16w6 )33@2 (8w + 96w5 ) 2w> 1> 4w3 + (4w2 + 1 + 16w6 )31@2 2> 0> 12w2 . A normal vector for
the osculating plane is B(1) = T(1) × N(1), but r0 (1) = h2> 1> 4i is parallel to T(1) and
T0 (1) = 3 12 (21)33@2 (104)h2> 1> 4i + (21)31@2 h2> 0> 12i =
2
I
21 21
h331> 326> 22i is parallel to N(1) as is h331> 326> 22i,
so h2> 1> 4i × h331> 326> 22i = h126> 3168> 321i is normal to the osculating plane. Thus an equation for the osculating
plane is 126({ 3 1) 3 168(| 3 1) 3 21(} 3 1) = 0 or 6{ 3 8| 3 } = 33.
56. r(w) = w + 2> 1 3 w> 12 w2
i r0 (w) = h1> 31> wi, T(w) =
1
r0 (w)
= I
h1> 31> wi,
|r0 (w)|
2 + w2
T0 (w)= 3 12 (2 + w2 )33@2 (2w)h1> 31> wi + (2 + w2 )31@2 h0> 0> 1i
= 3(2 + w2 )33@2 wh1> 31> wi 3 (2 + w2 )h0> 0> 1i = (2+w31
2 )3@2 hw> 3w> 32i
A normal vector for the osculating plane is B(w) = T(w) × N(w), but r0 (w) = h1> 31> wi is parallel to T(w) and hw> 3w> 32i
is parallel to T0 (w) and hence parallel to N(w), so h1> 31> wi × hw> 3w> 32i = w2 + 2> w2 + 2> 0 is normal to the
osculating plane for any w. All such vectors are parallel to h1> 1> 0i, so at any point w + 2> 1 3 w> 12 w2 on the curve, an
equation for the osculating plane is 1[{ 3 (w + 2)] + 1[| 3 (1 3 w)] + 0 } 3 12 w2 = 0 or { + | = 3. Because the osculating
plane at every point on the curve is the same, we can conclude that the curve itself lies in that same plane. In fact, we can
easily verify that the parametric equations of the curve satisfy { + | = 3.
gT gT@gw |gT@gw|
gT@gw
=
=
57. = and N =
, so N =
gv gv@gw gv@gw
|gT@gw|
gT gT
gw gw
gT@gw
gT
gT gv = gv@gw = gv by the Chain Rule.
gw gw
58. For a plane curve, T = |T| cos ! i + |T| sin ! j = cos ! i + sin ! j. Then
gT
=
gv
gT
g!
g!
gv
gT g! g! g!
= (3 sin ! i + cos ! j)
and = |3 sin ! i + cos ! j| = . Hence for a plane
gv
gv gv
gv
curve, the curvature is = |g!@gv|.
59. (a) |B| = 1
i B·B=1 i
g
gB
(B · B) = 0 i 2
·B=0 i
gv
gv
gB
zB
gv
(b) B = T × N i
g
g
1
g
1
1
gB
=
(T × N) =
(T × N)
=
(T × N) 0
= [(T0 × N) + (T × N0 )] 0
gv
gv
gw
gv@gw
gw
|r (w)|
|r (w)|
=
T0
1
T × N0
T0 × 0 + (T × N0 )
=
|T |
|r0 (w)|
|r0 (w)|
(c) B = T × N i
i
gB
zT
gv
T z N, B z T and B z N. So B, T and N form an orthogonal set of vectors in the three-
dimensional space R3 . From parts (a) and (b), gB@gv is perpendicular to both B and T, so gB@gv is parallel to N.
Therefore, gB@gv = 3 (v)N, where (v) is a scalar.
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CHAPTER 13 VECTOR FUNCTIONS
(d) Since B = T × N, T z 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 gB@gv = 0, but gB@gv = 3 (v)N and N 6= 0, so (v) = 0.
60. N = B × T
i
gN
g
gB
gT
=
(B × T) =
×T+B×
gv
gv
gv
gv
[by Formula 5 of Theorem 13.2.3]
= 3 N × T + B × N
[by Formulas 3 and 1]
= 3 (N × T) + (B × N)
[by Property 2 of Theorem 12.4.11 ]
But B × N = B × (B × T) = (B · T) B 3 (B · B) T [by Property 6 of Theorem 12.4.11 ] = 3T i
gN@gv = (T × N) 3 T = 3 T + B.
61. (a) r0 = v0 T
i r00 = v00 T + v0 T0 = v00 T + v0
gT 0
v = v00 T + (v0 )2 N by the first Serret-Frenet formula.
gv
(b) Using part (a), we have
r0 × r00 = (v0 T) × [v00 T + (v0 )2 N]
= [(v0 T) × (v00 T)] + (v0 T) × ((v0 )2 N)
[by Property 3 of Theorem 12.4.11 ]
= (v0 v00 )(T × T) + (v0 )3 (T × N) = 0 + (v0 )3 B = (v0 )3 B
(c) Using part (a), we have
r000 = [v00 T + (v0 )2 N]0 = v000 T + v00 T0 + 0 (v0 )2 N + 2v0 v00 N + (v0 )2 N0
= v000 T + v00
gT 0
gN 0
v + 0 (v0 )2 N + 2v0 v00 N + (v0 )2
v
gv
gv
= v000 T + v00 v0 N + 0 (v0 )2 N + 2v0 v00 N + (v0 )3 (3 T + B)
[by the second formula]
= [v000 3 2 (v0 )3 ] T + [3v0 v00 + 0 (v0 )2 ] N + (v0 )3 B
(d) Using parts (b) and (c) and the facts that B · T = 0, B · N = 0, and B · B = 1, we get
(v0 )3 B · [v000 3 2 (v0 )3 ] T + [3v0 v00 + 0 (v0 )2 ] N + (v0 )3 B
(r0 × r00 ) · r000
=
2
|r0 × r00 |
|(v0 )3 B|2
=
(v0 )3 (v0 )3
= .
[(v0 )3 ]2
62. First we find the quantities required to compute :
r0 (w) = h3d sin w> d cos w> ei
r00 (w) = h3d cos w> 3d sin w> 0i i r000 (w) = hd sin w> 3d cos w> 0i
s
I
|r0 (w)| = (3d sin w)2 + (d cos w)2 + e2 = d2 + e2
i
j
k r0 (w) × r00 (w) = 3d sin w d cos w e = de sin w i 3 de cos w j + d2 k
3d cos w 3d sin w 0 i
|r0 (w) × r00 (w)| =
s
I
(de sin w)2 + (3de cos w)2 + (d2 )2 = d2 e2 + d4
(r0 (w) × r00 (w)) · r000 (w) = (de sin w)(d sin w) + (3de cos w)(3d cos w) + (d2 )(0) = d2 e
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SECTION 13.3 ARC LENGTH AND CURVATURE
¤
347
I
I
d2 e2 + d4
|r0 (w) × r00 (w)|
d d2 + e2
d
Then by Theorem 10, (w) =
=
=
which is a constant.
I
3
I
3 = 2
d + e2
|r0 (w)|3
d2 + e2
d2 + e2
From Exercise 61(d), the torsion is given by =
i r0 = 1> w> w2 , r00 = h0> 1> 2wi, r000 = h0> 0> 2i i r0 × r00 = w2 > 32w> 1 i
w2 > 32w> 1 · h0> 0> 2i
2
(r0 × r00 ) · r000
= 4
=
=
2
4 + 4w2 + 1
2 +1
0
00
w
w
+
4w
|r × r |
63. r = w> 12 w2 > 13 w3
(r0 × r00 ) · r000
d2 e
e
=
which is also a constant.
I
2 = 2
2
0
00
2
2
4
d
+
e2
|r × r |
d e +d
i r0 = hcosh w> sinh w> 1i, r00 = hsinh w> cosh w> 0i, r000 = hcosh w> sinh w> 0i i
r0 × r00 = 3 cosh w> sinh w> cosh2 w 3 sinh2 w = h3 cosh w> sinh w> 1i i
s
cosh2 w + sinh2 w + 1
1
|r0 × r00 |
|h3 cosh w> sinh w> 1i|
1
=
,
=
=
3@2 =
3
3 = 2
2
0
2
2
cosh w + sinh w + 1
2 cosh2 w
|r |
|hcosh w> sinh w> 1i|
cosh w + sinh w + 1
64. r = hsinh w> cosh w> wi
=
(r0 × r00 ) · r000
h3 cosh w> sinh w> 1i · hcosh w> sinh w> 0i
3 cosh2 w + sinh2 w
31
=
=
=
2
2
2
cosh w + sinh w + 1
2 cosh2 w
2 cosh2 w
|r0 × r00 |
So at the point (0> 1> 0), w = 0, and =
1
2
and = 3 12 .
65. For one helix, the vector equation is r(w) = h10 cos w> 10 sin w> 34w@(2)i (measuring in angstroms), because the radius of each
helix is 10 angstroms, and } increases by 34 angstroms for each increase of 2 in w. Using the arc length formula, letting w go
from 0 to 2=9 × 108 × 2, we find the approximate length of each helix to be
O=
U 2=9×108 ×2
0
|r0 (w)| gw =
8
t
34 2 2=9×10 ×2
U 2=9×108 ×2 t
2 + (10 cos w)2 + 34 2 gw =
(310
sin
w)
100
+
w
2
2
0
0
t
34 2
= 2=9 × 108 × 2 100 + 2
E 2=07 × 1010 Å — more than two meters!
;
0
if { ? 0
A
?
S
({)
if
0?{?1
66. (a) For the function I ({) =
A
=
1
if { D 1
to be continuous, we must have S (0) = 0 and S (1) = 1.
For I 0 to be continuous, we must have S 0 (0) = S 0 (1) = 0. The curvature of the curve | = I ({) at the point ({> I ({))
|I 00 ({)|
00
00
is ({) = 3@2 . For ({) to be continuous, we must have S (0) = S (1) = 0.
1 + [I 0 ({)]2
Write S ({) = d{5 + e{4 + f{3 + g{2 + h{ + i . Then S 0 ({) = 5d{4 + 4e{3 + 3f{2 + 2g{ + h and
S 00 ({) = 20d{3 + 12e{2 + 6f{ + 2g. Our six conditions are:
S (0) = 0
0
i
i =0
(1)
S (1) = 1
0
i
d+e+f+g+h+i = 1
(2)
S (0) = 0
i
h=0
(3)
S (1) = 0
i
5d + 4e + 3f + 2g + h = 0
(4)
S 00 (0) = 0
i
g=0
(5)
S 00 (1) = 0
i
20d + 12e + 6f + 2g = 0
(6)
From (1), (3), and (5), we have g = h = i = 0. Thus (2), (4) and (6) become (7) d + e + f = 1, (8) 5d + 4e + 3f = 0,
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CHAPTER 13 VECTOR FUNCTIONS
and (9) 10d + 6e + 3f = 0. Subtracting (8) from (9) gives (10) 5d + 2e = 0. Multiplying (7) by 3 and subtracting from
(8) gives (11) 2d + e = 33. Multiplying (11) by 2 and subtracting from (10) gives d = 6. By (10), e = 315.
By (7), f = 10. Thus, S ({) = 6{5 3 15{4 + 10{3 .
(b)
13.4 Motion in Space: Velocity and Acceleration
1. (a) If r(w) = {(w) i + | (w) j + }(w) k is the position vector of the particle at time t, then the average velocity over the time
interval [0> 1] is
vave =
(4=5 i + 6=0 j + 3=0 k) 3 (2=7 i + 9=8 j + 3=7 k)
r(1) 3 r(0)
=
= 1=8 i 3 3=8 j 3 0=7 k. Similarly, over the other
130
1
intervals we have
[0=5> 1] :
vave =
r(1) 3 r(0=5)
(4=5 i + 6=0 j + 3=0 k) 3 (3=5 i + 7=2 j + 3=3 k)
=
= 2=0 i 3 2=4 j 3 0=6 k
1 3 0=5
0=5
[1> 2] :
vave =
(7=3 i + 7=8 j + 2=7 k) 3 (4=5 i + 6=0 j + 3=0 k)
r(2) 3 r(1)
=
= 2=8 i + 1=8 j 3 0=3 k
231
1
[1> 1=5] :
vave =
(5=9 i + 6=4 j + 2=8 k) 3 (4=5 i + 6=0 j + 3=0 k)
r(1=5) 3 r(1)
=
= 2=8 i + 0=8 j 3 0=4 k
1=5 3 1
0=5
(b) We can estimate the velocity at w = 1 by averaging the average velocities over the time intervals [0=5> 1] and [1> 1=5]:
v(1) E 12 [(2 i 3 2=4 j 3 0=6 k) + (2=8 i + 0=8 j 3 0=4 k)] = 2=4 i 3 0=8 j 3 0=5 k. Then the speed is
|v(1)| E
s
(2=4)2 + (30=8)2 + (30=5)2 E 2=58.
2. (a) The average velocity over 2 $ w $ 2=4 is
r(2=4) 3 r(2)
= 2=5 [r(2=4) 3 r(2)], so we sketch a vector in the same
2=4 3 2
direction but 2=5 times the length of [r(2=4) 3 r(2)] .
(b) The average velocity over 1=5 $ w $ 2 is
r(2) 3 r(1=5)
= 2[r(2) 3 r(1=5)], so we sketch a vector in the
2 3 1=5
same direction but twice the length of [r(2) 3 r(1=5)].
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SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
(c) Using Equation 2 we have v(2) = lim
k<0
r(2 + k) 3 r(2)
.
k
(d) v(2) is tangent to the curve at r(2) and points in the direction of
increasing w. Its length is the speed of the particle at w = 2. We can
estimate the speed by averaging the lengths of the vectors found in
parts (a) and (b) which represent the average speed over 2 $ w $ 2=4 and
1=5 $ w $ 2 respectively. Using the axes scale as a guide, we estimate the
vectors to have lengths 2=8 and 2=7. Thus, we estimate the speed at w = 2
to be |v(2)| E 12 (2=8 + 2=7) = 2=75 and we draw the velocity vector v(2)
with this length.
3. r(w) = 3 12 w2 > w
i
At w = 2:
v(w) = r0 (w) = h3w> 1i
v(2) = h32> 1i
a(w) = r00 (w) = h31> 0i
a(2) = h31> 0i
|v(w)| =
I
w2 + 1
I w
i
I v(w) = r0 (w) = 31> 2@ w
H
G
a(w) = r00 (w) = 0> 31@w3@2
4. r(w) = 2 3 w> 4
|v(w)| =
v(1) = h31> 2i
a(1) = h0> 31i
s
1 + 4@w
5. r(w) = 3 cos w i + 2 sin w j
v(w) = 33 sin w i + 2 cos w j
a(w) = 33 cos w i 3 2 sin w j
|v(w)| =
At w = 1:
i
At w = @3:
I
v 3 = 3 3 2 3 i + j
I
a 3 = 3 32 i 3 3 j
s
s
9 sin2 w + 4 cos2 w = 4 + 5 sin2 w
Notice that {2 @9 + | 2@4 = sin2 w + cos2 w = 1, so the path is an ellipse.
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CHAPTER 13 VECTOR FUNCTIONS
6. r(w) = hw i + h2w j
i
At w = 0:
v(w) = hw i + 2h2w j
v(0) = i + 2 j
a(w) = hw i + 4h2w j
a(0) = i + 4 j
I
I
h2w + 4h4w = hw 1 + 4h2w
2
Notice that | = h2w = hw = {2 , so the particle travels along a parabola,
|v(w)| =
but { = hw , so { A 0.
7. r(w) = w i + w2 j + 2 k
i
At w = 1:
v(w) = i + 2w j
v(1) = i + 2 j
a(w) = 2 j
a(1) = 2 j
|v(w)| =
I
1 + 4w2
Here { = w, | = w2
i | = {2 and } = 2, so the path of the particle is a
parabola in the plane } = 2.
8. r(w) = w i + 2 cos w j + sin w k
i
At w = 0:
v(w) = i 3 2 sin w j + cos w k
v(0) = i + k
a(w) = 32 cos w j3 sin w k
a(0) = 32 j
|v(w)| =
s
s
1 + 4 sin2 w + cos2 w = 2 + 3 sin2 w
Since |2 @4 + } 2 = 1, { = w, the path of the particle is an elliptical helix
about the {-axis.
9. r(w) = w2 + w> w2 3 w> w3
|v(w)| =
s
I
(2w + 1)2 + (2w 3 1)2 + (3w2 )2 = 9w4 + 8w2 + 2.
10. r(w) = h2 cos w> 3w> 2 sin wi
|v(w)| =
i v(w) = r0 (w) = 2w + 1> 2w 3 1> 3w2 , a(w) = v0 (w) = h2> 2> 6wi,
i v(w) = r0 (w) = h32 sin w> 3> 2 cos wi, a(w) = v0 (w) = h32 cos w> 0> 32 sin wi,
s
I
4 sin2 w + 9 + 4 cos2 w = 13.
I
I
2 w i + hw j + h3w k i v(w) = r0 (w) = 2 i + hw j 3 h3w k, a(w) = v0 (w) = hw j + h3w k,
s
I
|v(w)| = 2 + h2w + h32w = (hw + h3w )2 = hw + h3w .
11. r(w) =
i v(w) = r0 (w) = 2w i + 2 j + (1@w) k, a(w) = v0 (w) = 2 i 3 (1@w2 ) k,
s
s
|v(w)| = 4w2 + 4 + (1@w2 ) = [2w + (1@w)]2 = |2w + (1@w)|.
12. r(w) = w2 i + 2w j + ln w k
13. r(w) = hw hcos w> sin w> wi
i
v(w) = r0 (w) = hw hcos w> sin w> wi + hw h3 sin w> cos w> 1i = hw hcos w 3 sin w> sin w + cos w> w + 1i
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SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
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351
a(w) = v0 (w) = hw hcos w 3 sin w 3 sin w 3 cos w> sin w + cos w + cos w 3 sin w> w + 1 + 1i
= hw h32 sin w> 2 cos w> w + 2i
s
cos2 w + sin2 w 3 2 cos w sin w + sin2 w + cos2 w + 2 sin w cos w + w2 + 2w + 1
I
= hw w2 + 2w + 3
|v(w)| = hw
14. r(w) = w2 > sin w 3 w cos w> cos w + w sin w
i
v(w) = r0 (w) = h2w> cos w 3 (3w sin w + cos w)> 3 sin w + w cos w + sin wi = h2w> w sin w> w cos wi,
a(w) = v0 (w) = h2> w cos w + sin w> 3w sin w + cos wi,
s
I
I
I
|v(w)| = 4w2 + w2 sin2 w + w2 cos2 w = 4w2 + w2 = 5w2 = 5 w [since w D 0].
U
a(w) gw = (i + 2 j) gw = w i + 2w j + C and k = v (0) = C,
U
U
so C = k and v(w) = w i + 2w j + k. r(w) = v(w) gw = (w i + 2w j + k) gw = 12 w2 i + w2 j + w k + D.
But i = r (0) = D, so D = i and r(w) = 12 w2 + 1 i + w2 j + w k.
15. a(w) = i + 2 j
i
v(w) =
U
U
i v(w) = (2 i + 6w j + 12w2 k) gw = 2w i + 3w2 j + 4w3 k + C, and i = v(0) = C,
U
so C = i and v(w) = (2w + 1)i + 3w2 j + 4w3 k. r(w) =
(2w + 1)i + 3w2 j + 4w3 k gw = (w2 + w) i + w3 j + w4 k + D.
16. a(w) = 2 i + 6w j + 12w2 k
But j 3 k = r(0) = D, so D = j 3 k and r(w) = (w2 + w) i + (w3 + 1) j + (w4 3 1) k.
17. (a) a(w) = 2w i + sin w j + cos 2w k
v(w) =
U
i
(b)
2
(2w i + sin w j + cos 2w k) gw = w i 3 cos w j +
and i = v (0) = 3j + C, so C = i + j
and v(w) = w2 + 1 i + (1 3 cos w) j +
1
2
1
4
sin 2w k + C
sin 2w k.
U r(w) = [ w2 + 1 i + (1 3 cos w) j + 12 sin 2w k]gw
= 13 w3 + w i + (w 3 sin w) j 3 14 cos 2w k + D
But j = r (0) = 3 14 k + D, so D = j +
1
2
k and r(w) =
18. (a) a(w) = w i + hw j + h3w k
1
3
w3 + w i + (w 3 sin w + 1) j + 14 3
i
U
w i + hw j + h3w k gw = 12 w2 i + hw j 3 h3w k + C
v (w) =
(b)
1
4
cos 2w k.
and k = v (0) = j 3 k + C, so C = 3j + 2 k
and v(w) = 12 w2 i + hw 3 1 j + 2 3 h3w k.
r(w) =
U 1
2
2w
i + (hw 3 1) j + (2 3 h3w ) k gw
= 16 w3 i + (hw 3 w) j + (h3w + 2w) k + D
But j + k = r(0) = j + k + D, so D = 0 and r(w) = 16 w3 i + (hw 3 w) j + (h3w + 2w) k=
19. r(w) = w2 > 5w> w2 3 16w
and
i v(w) = h2w> 5> 2w 3 16i, |v(w)| =
I
I
4w2 + 25 + 4w2 3 64w + 256 = 8w2 3 64w + 281
g
|v(w)| = 12 (8w2 3 64w + 281)31@2 (16w 3 64). This is zero if and only if the numerator is zero, that is,
gw
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CHAPTER 13 VECTOR FUNCTIONS
16w 3 64 = 0 or w = 4. Since
I
g
g
|v(w)| ? 0 for w ? 4 and
|v(w)| A 0 for w A 4, the minimum speed of 153 is attained
gw
gw
at w = 4 units of time.
20. Since r(w) = w3 i + w2 j + w3 k, a(w) = r00 (w) = 6w i + 2 j + 6w k. By Newton’s Second Law,
F(w) = p a(w) = 6pw i + 2p j + 6pw k is the required force.
21. |F(w)| = 20 N in the direction of the positive }-axis, so F(w) = 20 k. Also p = 4 kg, r(0) = 0 and v(0) = i 3 j.
Since 20k = F(w) = 4 a(w), a(w) = 5 k. Then v(w) = 5w k + c1 where c1 = i 3 j so v(w) = i 3 j + 5w k and the
I
I
speed is |v(w)| = 1 + 1 + 25w2 = 25w2 + 2. Also r(w) = w i 3 w j + 52 w2 k + c2 and 0 = r(0), so c2 = 0
and r(w) = w i 3 w j + 52 w2 k.
22. The argument here is the same as that in Example 13.2.4 with r(w) replaced by v(w) and r0 (w) replaced by a(w).
23. |v(0)| = 200 m@s and, since the angle of elevation is 60 , a unit vector in the direction of the velocity is
(cos 60 )i + (sin 60 )j = 12 i +
I
3
2
j. Thus v(0) = 200 12 i +
I
3
2
I
j = 100 i + 100 3 j and if we set up the axes so that the
projectile starts at the origin, then r(0) = 0. Ignoring air resistance, the only force is that due to gravity, so
F(w) = pa(w) = 3pj j where j E 9=8 m@s2 . Thus a(w) = 39=8 j and, integrating, we have v(w) = 39=8w j + C. But
I
I
100 i + 100 3 j = v(0) = C, so v(w) = 100 i + 100 3 3 9=8w j and then (integrating again)
I
r(w) = 100 w i + 100 3 w 3 4=9w2 j + D where 0 = r(0) = D. Thus the position function of the projectile is
I
r(w) = 100 w i + 100 3 w 3 4=9w2 j.
(a) Parametric equations for the projectile are {(w) = 100w, |(w) = 100
I
3 w 3 4=9w2 . The projectile reaches the ground when
I
I
|(w) = 0 (and w A 0) i 100 3 w 3 4=9w2 = w 100 3 3 4=9w = 0 i w =
I I 3
3
{ 100
= 100 100
E 3535 m.
4=9
4=9
I
100 3
4=9
E 35=3 s. So the range is
(b) The maximum height is reached when |(w) has a critical number (or equivalently, when the vertical component
of velocity is 0): | 0 (w) = 0 i 100
|
I 100 3
9=8
= 100
I
3 3 9=8w = 0 i w =
I 2
I 100I3 3
3 4=9 100
3
E 1531 m.
9=8
9=8
I
100 3
9=8
E 17=7 s. Thus the maximum height is
I
3
s. Thus, the velocity at impact is
(c) From part (a), impact occurs at w = 100
4=9
I l
k
I I
I
3
3
= 100 i + 100 3 3 9=8 100
j = 100 i 3 100 3 j and the speed is
v 100
4=9
4=9
I I
3 v 100
= 10,000 + 30,000 = 200 m@s.
4=9
24. As in Exercise 23, v(w) = 100 i + 100
I
I
3 3 9=8w j and r(w) = 100 w i + 100 3 w 3 4=9w2 j + D.
I
But r(0) = 100 j, so D = 100 j and r(w) = 100 w i + 100 + 100 3 w 3 4=9w2 j.
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SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
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353
I
I
(a) | = 0 i 100 + 100 3 w 3 4=9w2 = 0 or 4=9w2 3 100 3 w 3 100 = 0. From the quadratic formula we have
t
I
I
I
I
100 3 ± (3100 3 )2 3 4(4=9)(3100)
100 3 ± 31,960
=
. Taking the positive w-value gives
w=
2(4=9)
9=8
w=
I
I
100 3 + 31,960
9=8
E 35=9 s. Thus the range is { = 100 ·
I
I
100 3 + 31,960
9=8
E 3592 m.
I
g|
= 0 i 100 3 3 9=8w = 0 i w =
gw
I 2
I I3 3
3 4=9 100
E 1631 m.
maximum height is 100 + 100 3 100
9=8
9=8
(b) The maximum height is attained when
I
100 3
9=8
E 17=7 s and the
Alternate solution: Because the projectile is fired in the same direction and with the same velocity as in Exercise 23,
but from a point 100 m higher, the maximum height reached is 100 m higher than that found in Exercise 23, that is,
1531 m + 100 m = 1631 m.
I
I
31,960
s. Thus the velocity at impact is
(c) From part (a), impact occurs at w = 100 3 +
9=8
k
l
I I
I
I
I
I
31,960
31,960
= 100 i + 100 3 3 9=8 100 3 +
j = 100 i 3 31,960 j and the speed is
v 100 3 +
9=8
9=8
|v| =
I
I
10,000 + 31,960 = 41,960 E 205 m@s.
I
I
y0 2 w i + y0 2 w 3 jw2 j . The ball lands when
I
I
I y0 2
y0 2
s. Now since it lands 90 m away, 90 = { = 12 y0 2
or y02 = 90j and the initial
w=
j
j
25. As in Example 5, r(w) = (y0 cos 45 )w i + (y0 sin 45 )w 3 12 jw2 j =
| = 0 (and w A 0) i
velocity is y0 =
I
90j E 30 m@s.
26. As in Example 5, r(w) = (y0 cos 30 )w i + (y0 sin 30 )w 3 12 jw2 j =
v(w) = r0 (w) =
1
2
1
2
1
2
I
y0 3 w i + (y0 w 3 jw2 ) j and then
I
y0 3 i + (y0 3 2jw) j . The shell reaches its maximum height when the vertical component of velocity
is zero, so 12 (y0 3 2jw) = 0 i w =
y0
. The vertical height of the shell at that time is 500 m,
2j
% 2 &
y0
y0
1
y0
3j
so
= 500 i
2
2j
2j
s
I
y02
= 500 i y0 = 4000j = 4000(9=8) E 198 m@s.
8j
27. Let be the angle of elevation. Then y0 = 150 m@s and from Example 5, the horizontal distance traveled by the projectile is
g=
y02 sin 2
1502 sin 2
800j
E 0=3484 i 2 E 20=4 or 180 3 20=4 = 159=6 .
. Thus
= 800 i sin 2 =
j
j
1502
Two angles of elevation then are E 10=2 and E 79=8 .
28. Here y0 = 115 ft@s, the angle of elevation is = 50 , and if we place the origin at home plate, then r(0) = 3 j.
As in Example 5, we have r (w) = 3 12 jw2 j + w v0 + D where D = r(0) = 3 j and v0 = y0 cos i + y0 sin j,
so r(w) = (y0 cos )w i + (y0 sin )w 3 12 jw2 + 3 j. Thus, parametric equations for the trajectory of the ball are
{ = (y0 cos )w, | = (y0 sin )w 3 12 jw2 + 3. The ball reaches the fence when { = 400 i
(y0 cos )w = 400 i w =
400
400
=
E 5=41 s. At this time, the height of the ball is
y0 cos 115 cos 50
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CHAPTER 13 VECTOR FUNCTIONS
| = (y0 sin )w 3 12 jw2 + 3 E (115 sin 50 )(5=41) 3 12 (32)(5=41)2 + 3 E 11=2 ft. Since the fence is 10 ft high, the ball
clears the fence.
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 y0 = 80 m@s and let be the angle the catapult is set at. As in Example 5, the trajectory of
the catapulted rock is given by r (w) = (80 cos )w i + (80 sin )w 3 4=9w2 j. The top of the near city wall is at (100> 15),
which the rock will hit when (80 cos ) w = 100 i w =
80 sin ·
5
and (80 sin )w 3 4=9w2 = 15 i
4 cos 2
5
5
3 4=9
= 15 i 100 tan 3 7=65625 sec2 = 15. Replacing sec2 with tan2 + 1 gives
4 cos 4 cos 7=65625 tan2 3 100 tan + 22=65625 = 0. Using the quadratic formula, we have tan E 0=230635, 12=8306 i
E 13=0 , 85=5 . So for 13=0 ? ? 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 )w = 600 i w =
80 sin ·
15
and (80 sin )w 3 4=9w2 = 0 i
2 cos 2
15
15
3 4=9
= 0 i 600 tan 3 275=625 sec2 = 0 i
2 cos 2 cos 275=625 tan2 3 600 tan + 275=625 = 0. Solutions are tan E 0=658678, 1=51819 i E 33=4 , 56=6 . Thus the
rock lands beyond the enclosed city ground for 33=4 ? ? 56=6 , and the angles that allow the rock to land on city ground
are 13=0 ? ? 33=4 , 56=6 ? ? 85=5 . If you consider that the rock can hit the far wall and bounce back into the city, we
calculate the angles that cause the rock to hit the top of the wall at (600> 15): (80 cos )w = 600 i w =
15
and
2 cos (80 sin )w 3 4=9w2 = 15 i 600 tan 3 275=625 sec2 = 15 i 275=625 tan2 3 600 tan + 290=625 = 0.
Solutions are tan E 0=727506, 1=44936 i E 36=0 , 55=4 , so the catapult should be set with angle where
13=0 ? ? 36=0 , 55=4 ? ? 85=5 .
30. If we place the projectile at the origin then, as in Example 5, r(w) = (y0 cos )w i + (y0 sin )w 3 12 jw2 j and
v(w) = (y0 cos ) i + [(y0 sin ) 3 jw] j. The maximum height is reached when the vertical component of velocity is zero, so
(y0 sin ) 3 jw = 0 i w =
y0 sin , and the corresponding height is the vertical component of the position function:
j
(y0 sin )w 3 12 jw2 = (y0 sin )
Half that time is w =
y0 sin j
3 12 j
y0 sin j
2
1 2 2
y sin 2j 0
=
y0 sin , when the height of the projectile is
2j
(y0 sin )w 3 12 jw2 = (y0 sin )
y0 sin 2j
3 12 j
y0 sin 2j
2
1 2 2
1 2 2
3 2 2
3
y sin 3
y sin =
y sin =
=
2j 0
8j 0
8j 0
4
1 2 2
y sin 2j 0
or three-quarters of the maximum height.
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¤
355
31. Here a(w) = 34 j 3 32 k so v(w) = 34w j 3 32w k + v0 = 34w j 3 32w k + 50 i + 80 k = 50 i 3 4w j + (80 3 32w) k and
r(w) = 50w i 3 2w2 j + (80w 3 16w2 ) k (note that r0 = 0). The ball lands when the }-component of r(w) is zero
and w A 0: 80w 3 16w2 = 16w(5 3 w) = 0
i
w = 5. The position of the ball then is
r(5) = 50(5) i 3 2(5)2 j + [80(5) 3 16(5)2 ] k = 250 i 3 50 j or equivalently the point (250> 350> 0). This is a distance of
s
50 I
E 11=3 from the eastern direction
2502 + (350)2 + 02 = 65,000 E 255 ft from the origin at an angle of tan31 250
toward the south. The speed of the ball is |v(5)| = |50 i 3 20 j 3 80 k| =
s
I
502 + (320)2 + (380)2 = 9300 E 96=4 ft/s.
32. Place the ball at the origin and consider j to be pointing in the northward direction with i pointing east and k pointing
upward. Force = mass × acceleration i acceleration = force@mass, so the wind applies a constant acceleration of
4 N@0=8 kg = 5 m@s2 in the easterly direction. Combined with the acceleration due to gravity, the acceleration acting
U
on the ball is a(w) = 5 i 3 9=8 k. Then v(w) = a(w) gw = 5w i 3 9=8w k + C where C is a constant vector.
I
I
We know v(0) = C = 330 cos 30 j + 30 sin 30 k = 315 3 j + 15 k i C = 315 3 j + 15 k and
I
I
U
v(w) = 5w i 3 15 3 j + (15 3 9=8w) k. r(w) = v(w) gw = 2=5w2 i 3 15 3 w j + 15w 3 4=9w2 k + D but r(0) = D = 0
so r(w) = 2=5w2 i 3 15
I
3 w j + 15w 3 4=9w2 k. The ball lands when 15w 3 4=9w2 = 0 i w = 0, w = 15@4=9 E 3=0612 s,
so the ball lands at approximately r(3=0612) E 23=43 i 3 79=53 j which is 82.9 m away in the direction S 16.4 E. Its speed is
I
approximately |v(3=0612)| E 15=306 i 3 15 3 j 3 15 k E 33=68 m@s.
33. (a) After w seconds, the boat will be 5w meters west of point D. The velocity
of the water at that location is
3
(5w)(40
400
3 5w) j. The velocity of the
boat in still water is 5 i> so the resultant velocity of the boat is
3
3 2
v(w) = 5 i + 400
(5w)(40 3 5w) j = 5i + 32 w 3 16
w j. Integrating, we obtain
1 3
r(w) = 5w i + 34 w2 3 16
w j + C. If we place the origin at D (and consider j
1 3
to coincide with the northern direction) then r(0) = 0 i C = 0 and we have r(w) = 5w i + 34 w2 3 16
w j. The boat
1
reaches the east bank after 8 s, and it is located at r(8) = 5(8)i + 34 (8)2 3 16
(8)3 j = 40 i + 16 j. Thus the boat is 16 m
downstream.
(b) Let be the angle north of east that the boat heads. Then the velocity of the boat in still water is given by
5(cos ) i + 5(sin ) j. At w seconds, the boat is 5(cos )w meters from the west bank, at which point the velocity
3
[5(cos )w][40
400
3 5(cos )w] j. The resultant velocity of the boat is given by
2
3
3
3 2
400 (5w cos )(40 3 5w cos ) j = (5 cos ) i + 5 sin + 2 w cos 3 16 w cos j.
1 3
w cos2 j (where we have again placed
Integrating, r(w) = (5w cos ) i + 5w sin + 34 w2 cos 3 16
of the water is
v(w) = 5(cos ) i + 5 sin +
the origin at D). The boat will reach the east bank when 5w cos = 40 i w =
In order to land at point E(40> 0) we need 5w sin + 34 w2 cos 3
1 3
w
16
8
40
=
.
5 cos cos cos2 = 0 i
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356
¤
CHAPTER 13 VECTOR FUNCTIONS
2
8
8
5
cos 3
sin + 34
cos cos 1
16
8
cos 3
cos2 = 0 i
1
(40 sin + 48 3 32) = 0 i
cos 40 sin + 16 = 0 i sin = 3 25 . Thus = sin31 3 25 E 323=6 , so the boat should head 23=6 south of
east (upstream). The path does seem realistic. The boat initially heads
upstream to counteract the effect of the current. Near the center of the river,
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 E.
34. As in Exercise 33(b), let be the angle north of east that the boat heads, so the velocity of the boat in still water is given
by 5(cos ) i + 5(sin ) j. At w seconds, the boat is 5(cos )w meters from the west bank, at which point the velocity
of the water is 3 sin({@40) j = 3 sin[ · 5(cos )w@40] j = 3 sin 8 w cos j. The resultant velocity of the boat
then is given by v(w) = 5(cos ) i + 5 sin + 3 sin 8 w cos j. Integrating,
24
r (w) = (5w cos ) i + 5w sin 3
cos 8 w cos j + C.
cos 24
24
j+C =0 i C=
j and
cos cos 24
24
cos 8 w cos +
j. The boat will reach the east bank when
r(w) = (5w cos ) i + 5w sin 3
cos cos If we place the origin at D then r(0) = 0 i 3
5w cos = 40 i w =
8
. In order to land at point E(40> 0) we need
cos 24
24
cos 8 w cos +
=0 i
cos cos 8
8
24
24
1
24
24
5
sin 3
cos
cos +
=0 i
40 sin 3
cos +
=0 i
cos cos 8 cos cos cos 6
6
48
= 0 i sin = 3 . Thus = sin31 3
E 322=5 , so the boat should head 22=5 south of east.
40 sin +
5
5
5w sin 3
35. If r0 (w) = c × r(w) then r0 (w) is perpendicular to both c and r(w). Remember that r0 (w) points in the direction of motion, so if
r0 (w) is always perpendicular to c, the path of the particle must lie in a plane perpendicular to c. But r0 (w) is also perpendicular
to the position vector r(w) 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.
36. (a) From Equation 7 we have a = y 0 T + y 2 N. If a particle moves along a straight line, then = 0 [see Section 13.3], so the
acceleration vector becomes a = y 0 T. Because the acceleration vector is a scalar multiple of the unit tangent vector, it is
parallel to the tangent vector.
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SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
¤
357
(b) If the speed of the particle is constant, then y 0 = 0 and Equation 7 gives a = y 2 N. Thus the acceleration vector is
parallel to the unit normal vector (which is perpendicular to the tangent vector and points in the direction that the curve is
turning).
i r0 (w) = (3 3 3w2 ) i + 6w j,
s
s
I
|r0 (w)| = (3 3 3w2 )2 + (6w)2 = 9 + 18w2 + 9w4 = (3 3 3w2 )2 = 3 + 3w2 ,
37. r(w) = (3w 3 w3 ) i + 3w2 j
r00 (w) = 36w i + 6 j, r0 (w) × r00 (w) = (18 + 18w2 ) k. Then Equation 9 gives
k
(3 3 3w2 )(36w) + (6w)(6)
r0 (w) · r00 (w)
18w + 18w3
18w(1 + w2 )
=
= 6w or by Equation 8,
=
=
0
2
2
2
|r (w)|
3 + 3w
3 + 3w
3(1 + w )
18 + 18w2
g |r0 (w) × r00 (w)|
18(1 + w2 )
dW = y 0 =
and Equation 10 gives dQ =
3 + 3w2 = 6w
=
= 6.
=
0
2
gw
|r (w)|
3 + 3w
3(1 + w2 )
dW =
i r0 (w) = i + (2w 3 2) j, |r0 (w)| =
38. r(w) = (1 + w) i + (w2 3 2w) j
r00 (w) = 2 j, r0 (w) × r00 (w) = 2 k. Then Equation 9 gives dW =
gives dQ =
s
I
12 + (2w 3 2)2 = 4w2 3 8w + 5,
r0 (w) · r00 (w)
2(2w 3 2)
and Equation 10
= I
|r0 (w)|
4w2 3 8w + 5
2
|r0 (w) × r00 (w)|
= I
.
|r0 (w)|
4w2 3 8w + 5
39. r(w) = cos w i + sin w j + w k
i r0 (w) = 3 sin w i + cos w j + k, |r0 (w)| =
r00 (w) = 3 cos w i 3 sin w j, r0 (w) × r00 (w) = sin w i 3 cos w j + k.
s
I
sin2 w + cos2 w + 1 = 2,
sin w cos w 3 sin w cos w
r0 (w) · r00 (w)
|r0 (w) × r00 (w)|
I
=
=
Then dW =
= 0 and dQ =
0
|r (w)|
|r0 (w)|
2
40. r(w) = w i + w2 j + 3w k
i r0 (w) = i + 2w j + 3 k, |r0 (w)| =
r00 (w) = 2j, r0 (w) × r00 (w) = 36i + 2k.
s
I
sin2 w + cos2 w + 1
2
I
= I = 1.
2
2
s
I
12 + (2w)2 + 32 = 4w2 + 10,
I
r0 (w) · r00 (w)
|r0 (w) × r00 (w)|
4w
2 10
Then dW =
and dQ =
.
= I
= I
|r0 (w)|
|r0 (w)|
4w2 + 10
4w2 + 10
41. r(w) = hw i +
I
2 w j + h3w k
i
s
I
I
2 j 3 h3w k, |r(w)| = h2w + 2 + h32w = (hw + h3w )2 = hw + h3w ,
h2w 3 h32w
(hw + h3w )(hw 3 h3w )
=
= hw 3 h3w = 2 sinh w
w
3w
h +h
hw + h3w
I 3w
I
s
2h i 3 2 j 3 2hw k
I
2(h32w + 2 + h2w ) I hw + h3w
=
=
= 2 w
= 2.
hw + h3w
hw + h3w
h + h3w
r00 (w) = hw i + h3w k. Then dW =
and dQ
r0 (w) = hw i +
42. r(w) = w i + cos2 w j + sin2 w k
i r0 (w) = i 3 2 cos w sin w j + 2 sin w cos w k = i 3 sin 2w j + sin 2w k,
s
1 + 2 sin2 2w, r00 (w) = 2(sin2 w 3 cos2 w) j + 2(cos2 w 3 sin2 w) k = 32 cos 2w j + 2 cos 2w k. So
I
2 sin 2w cos 2w + 2 sin 2w cos 2w
|32 cos 2w j 3 2 cos 2w k|
4 sin 2w cos 2w
2 2 |cos 2w|
s
s
dW =
= s
and dQ =
= s
.
1 + 2 sin2 2w
1 + 2 sin2 2w
1 + 2 sin2 2w
1 + 2 sin2 w
|r0 (w)| =
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CHAPTER 13 VECTOR FUNCTIONS
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 length to be 9=0. Thus dW E 4=5 cm@s2 and
dQ E 9=0 cm@s2 .
44. L(w) = p r(w) × v(w)
i
L0 (w) = p[r0 (w) × v(w) + r (w) × v0 (w)]
[by Formula 5 of Theorem 13.2.3]
= p[v(w) × v(w) + r(w) × v0 (w)] = p[0 + r(w) × a (w)] = (w)
So if the torque is always 0, then L0 (w) = 0 for all w, and so L(w) is constant.
45. If the engines are turned off at time w, then the spacecraft will continue to travel in the direction of v(w), so we need a w such
that for some scalar v A 0, r(w) + v v(w) = h6> 4> 9i. v(w) = r0 (w) = i +
r(w) + v v(w) =
v
4
8vw
3 + w + v> 2 + ln w + > 7 3 2
+
w
w + 1 (w2 + 1)2
8(3 3 w)w
4
+
=9 C
w2 + 1 (w2 + 1)2
so 7 3
]
0
gv
gp
=
vh
gw
gw
w
gv
gx = vh
gx
i 3 + w + v = 6 i v = 3 3 w,
24w 3 12w2 3 4
= 2 C w4 + 8w2 3 12w + 3 = 0.
(w2 + 1)2
It is easily seen that w = 1 is a root of this polynomial. Also 2 + ln 1 +
46. (a) p
8w
1
j+ 2
k i
w
(w + 1)2
331
= 4, so w = 1 is the desired solution.
1
gv
1 gp
=
vh . Integrating both sides of this equation with respect to w gives
gw
p gw
] v(w)
] p(w)
1 gp
gp
gx i
[Substitution Rule] i
gv = vh
p gx
v(0)
p(0) p
C
]
0
w
p(w)
v(w) 3 v(0) = ln
vh
p(0)
p(0)
i v(w) = v(0) 3 ln
vh .
p(w)
p(0)
vh i
(b) |v(w)| = 2 |vh |, and |v(0)| = 0. Therefore, by part (a), 2 |vh | = 3 ln
p(w)
p(0)
p(0)
|vh |. Note: p(0) A p(w) so that ln
A0
i p(w) = h32 p(0).
2 |vh | = ln
p (w)
p(w)
Thus
p(0) 3 h32 p(0)
= 1 3 h32 is the fraction of the initial mass that is burned as fuel.
p(0)
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APPLIED PROJECT KEPLER’S LAWS
¤
359
APPLIED PROJECT Kepler's Laws
1. With r = (u cos ) i + (u sin ) j and h = k where A 0,
g
g
i + u0 sin + u cos j
u0 cos 3 u sin gw
gw
g
g
g
= uu0 cos sin + u2 cos2 3 uu0 cos sin + u2 sin2 k = u2
k
gw
gw
gw
(a) h = r × r0 = [(u cos ) i + (u sin ) j] ×
(b) Since h = k, A 0, = |h|. But by part (a), = |h| = u2 (g@gw).
(c) D(w) =
1
2
U
0
|r|2 g =
1
2
u2 g
gD
=
.
gw
2 gw
(d)
Uw
w0
u2 (g@gw) gw in polar coordinates. Thus, by the Fundamental Theorem of Calculus,
gD
u2 g
k
=
= = constant since h is a constant vector and k = |h|.
gw
2 gw
2
2. (a) Since gD@gw =
1
k,
2
a constant, D(w) = 12 kw + f1 . But D(0) = 0, so D(w) = 12 kw. But D(W ) = area of the ellipse = de
and D(W ) = 12 kW , so W = 2de@k.
(b) k2@(JP ) = hg where h is the eccentricity of the ellipse. But d = hg@(1 3 h2 ) or hg = d(1 3 h2 ) and 1 3 h2 = e2@d2 .
Hence k2@(JP ) = hg = e2@d.
(c) W 2 =
4d2 e2
42 3
d
= 42 d2 e2
=
d .
2
2
k
JPe
JP
3. From Problem 2, W 2 =
d3 =
seconds
4 2 3
d . W E 365=25 days × 24 · 602
E 3=1558 × 107 seconds. Therefore
JP
day
(6=67 × 10311 )(1=99 × 1030 )(3=1558 × 107 )2
JPW 2
E
E 3=348 × 1033 m3
2
4
42
i d E 1=496 × 1011 m. Thus, the
length of the major axis of the earth’s orbit (that is, 2d) is approximately 2=99 × 1011 m = 2=99 × 108 km.
2
4. We can adapt the equation W =
42 3
d from Problem 2(c) with the earth at the center of the system, so W is the period of the
JP
satellite’s orbit about the earth, P is the mass of the earth, and d is the length of the semimajor axis of the satellite’s orbit
(measured from the earth’s center). Since we want the satellite to remain fixed above a particular point on the earth’s equator,
W must coincide with the period of the earth’s own rotation, so W = 24 h = 86,400 s. The mass of the earth is
P = 5=98 × 1024 kg, so d =
W 2 JP
42
1@3
E
(86,400)2 (6=67 × 10311 )(5=98 × 1024 )
42
1@3
E 4=23 × 107 m. If we
assume a circular orbit, the radius of the orbit is d, and since the radius of the earth is 6=37 × 106 m, the required altitude
above the earth’s surface for the satellite is 4=23 × 107 3 6=37 × 106 E 3=59 × 107 m, or 35,900 km.
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¤
CHAPTER 13 VECTOR FUNCTIONS
13 Review
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 function.
2. The tip of the moving vector r(w) of a continuous vector function traces out a space curve.
3. The tangent vector to a smooth curve at a point S with position vector r(w) is the vector r0 (w). The tangent line at S is the line
through S parallel to the tangent vector r0 (w). The unit tangent vector is T(w) =
r0 (w)
.
|r0 (w)|
4. (a) (a) – (f ) See Theorem 13.2.3.
5. Use Formula 13.3.2, or equivalently, 13.3.3.
gT where T is the unit tangent vector.
gv 6. (a) The curvature of a curve is = 0 T (w) (b) (w) = 0
r (w) 7. (a) The unit normal vector: N(w) =
(c) (w) =
|r0 (w) × r00 (w)|
|r0 (w)|3
(d) ({) =
|i 00 ({)|
[1 + (i 0 ({))2 ]3@2
T0 (w)
. The binormal vector: B(w) = T(w) × N(w).
|T0 (w)|
(b) See the discussion preceding Example 7 in Section 13.3.
8. (a) If r(w) is the position vector of the particle on the space curve, the velocity v(w) = r0 (w), the speed is given by |v(w)|,
and the acceleration a(w) = v0 (w) = r00 (w).
(b) a = dW T + dQ N where dW = y0 and dQ = y 2 .
9. See the statement of Kepler’s Laws on page 892 [ET 868].
1. True. If we reparametrize the curve by replacing x = w3 , we have r(x) = x i + 2x j + 3x k, which is a line through the origin
with direction vector i + 2 j + 3 k.
2. True. Parametric equations for the curve are { = 0, | = w2 , } = 4w, and since w = }@4 we have | = w2 = (}@4)2 or
|=
1 2
16 } ,
{ = 0. This is an equation of a parabola in the |}-plane.
3. False. The vector function represents a line, but the line does not pass through the origin; the {-component is 0 only for w = 0
which corresponds to the point (0> 3> 0) not (0> 0> 0).
4. True. See Theorem 13.2.2.
5. False. By Formula 5 of Theorem 13.2.3,
g
[u(w) × v(w)] = u0 (w) × v(w) + u(w) × v0 (w).
gw
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CHAPTER 13 REVIEW
6. False. For example, let r(w) = hcos w> sin wi. Then |r(w)| =
|r0 (w)| = |h3 sin w> cos wi| =
s
(3 sin w)2 + cos2 w = 1.
s
cos2 w + sin2 w = 1 i
¤
361
g
|r(w)| = 0, but
gw
7. False. is the magnitude of the rate of change of the unit tangent vector T with respect to arc length v, not with respect to w.
8. False. The binormal vector, by the definition given in Section 13.3, is B(w) = T(w) × N(w) = 3 [N(w) × T(w)].
9. True. At an inflection point where i is twice continuously differentiable we must have i 00 ({) = 0, and by Equation 13.3.11,
the curvature is 0 there.
10. True. From Equation 13.3.9 , (w) = 0
C |T0 (w)| = 0 C T0 (w) = 0 for all w. But then T(w) = C, a constant vector,
which is true only for a straight line.
11. False. If r(w) is the position of a moving particle at time w and |r(w)| = 1 then the particle lies on the unit circle or the unit
sphere, but this does not mean that the speed |r0 (w)| must be constant. As a counterexample, let r(w) = w>
I
1 3 w2 , then
s
I
I
I
r0 (w) = 1> 3w@ 1 3 w2 and |r(w)| = w2 + 1 3 w2 = 1 but |r0 (w)| = 1 + w2 @(1 3 w2 ) = 1@ 1 3 w2 which is not
constant.
12. True. See Example 4 in Section 13.2 .
13. True. See the discussion preceding Example 7 in Section 13.3.
14. False. For example, r1 (w) = hw> wi and r2 (w) = h2w> 2wi both represent the same plane curve (the line | = {), but the tangent
vector r01 (w) = h1> 1i for all w, while r02 (w) = h2> 2i. In fact, different parametrizations give parallel tangent vectors at a point,
but their magnitudes may differ.
1. (a) The corresponding parametric equations for the curve are { = w,
| = cos w, } = sin w. Since |2 + } 2 = 1, the curve is contained in a
circular cylinder with axis the {-axis. Since { = w, the curve is a helix.
(b) r(w) = w i + cos w j + sin w k i
r0 (w) = i 3 sin w j + cos w k i
r00 (w) = 32 cos w j 3 2 sin w k
2. (a) The expressions
I
2 3 w, (hw 3 1)@w, and ln(w + 1) are all defined when 2 3 w D 0 i w $ 2, w 6= 0,
and w + 1 A 0 i w A 31. Thus the domain of r is (31> 0) (0> 2].
I
I
hw 3 1
hw
(b) lim r(w) = lim 2 3 w> lim
> lim ln(w + 1) =
2 3 0> lim > ln(0 + 1)
w<0
w<0
w<0
w<0
w<0 1
w
I
2> 1> 0
[using l’Hospital’s Rule in the |-component]
=
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¤
CHAPTER 13 VECTOR FUNCTIONS
(c) r0 (w) =
g I
g hw 3 1 g
> ln(w + 1)
2 3 w>
gw
gw w
gw
=
1
whw 3 hw + 1
1
3 I
>
>
w2
w+1
2 23w
3. The projection of the curve F of intersection onto the {|-plane is the circle {2 + | 2 = 16> } = 0. So we can write
{ = 4 cos w, | = 4 sin w, 0 $ w $ 2. From the equation of the plane, we have } = 5 3 { = 5 3 4 cos w, so parametric
equations for F are { = 4 cos w, | = 4 sin w, } = 5 3 4 cos w, 0 $ w $ 2, and the corresponding vector function is
r(w) = 4 cos w i + 4 sin w j + (5 3 4 cos w) k, 0 $ w $ 2.
4. The curve is given by r(w) = h2 sin w> 2 sin 2w> 2 sin 3wi, so
I r0 (w) = h2 cos w> 4 cos 2w> 6 cos 3wi. The point 1> 3> 2 corresponds to w = 6
I
(or 6 + 2n, n an integer), so the tangent vector there is r0 ( 6 ) =
3> 2> 0 .
I
3> 2> 0 and includes the point
Then the tangent line has direction vector
I
I
I 1> 3> 2 , so parametric equations are { = 1 + 3 w, | = 3 + 2w, } = 2.
5.
U1
0
U
U
1
1
w2 gw i + 0 w cos w gw j + 0 sin w gw k
1 U 1
1
1
= 13 w3 0 i + w sin w 0 3 0 1 sin w gw j + 3 1 cos w 0 k
(w2 i + w cos w j + sin w k) gw =
=
U
1
0
1
3
i+
1
2
1
cos w 0 j +
2
k=
1
3
i3
2
2
j+
2
k
where we integrated by parts in the |-component.
6. (a) F intersects the {}-plane where | = 0
3
is 2 3 12 > 0> ln 12 = 15
8 > 0> 3 ln 2 .
i 2w 3 1 = 0 i w = 12 , so the point
(b) The curve is given by r(w) = 2 3 w3 > 2w 3 1> ln w , so r0 (w) = 33w2 > 2> 1@w . The point (1> 1> 0) corresponds to w = 1, so
the tangent vector there is r0 (1) = h33> 2> 1i. Then the tangent line has direction vector h33> 2> 1i and includes the point
(1> 1> 0), so parametric equations are { = 1 3 3w, | = 1 + 2w, } = w.
(c) The normal plane has normal vector r0 (1) = h33> 2> 1i and equation 33({ 3 1) + 2(| 3 1) + } = 0 or 3{ 3 2| 3 } = 1.
I
i r0 (w) = 2w> 3w2 > 4w3
i |r0 (w)| = 4w2 + 9w4 + 16w6 and
I
U3I
U3
O = 0 |r0 (w)| gw = 0 4w2 + 9w4 + 16w6 gw. Using Simpson’s Rule with i (w) = 4w2 + 9w4 + 16w6 and q = 6 we
7. r(w) = w2 > w3 > w4
have {w =
330
6
=
1
2
and
i(0) + 4i 12 + 2i (1) + 4i 32 + 2i (2) + 4i 52 + i (3)
t s
4
6
I
2
0 + 0 + 0 + 4 · 4 12 + 9 12 + 16 12 + 2 · 4(1)2 + 9(1)4 + 16(1)6
= 16
OE
{w
3
t s
4
6
2
4 32 + 9 32 + 16 32 + 2 · 4(2)2 + 9(2)4 + 16(2)6
t 4
6 s
2
+ 4 · 4 52 + 9 52 + 16 52 + 4(3)2 + 9(3)4 + 16(3)6
+4·
E 86=631
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CHAPTER 13 REVIEW
¤
363
G
H
s
I
8. r0 (w) = 3w1@2 > 32 sin 2w> 2 cos 2w , |r0 (w)| = 9w + 4(sin2 2w + cos2 2w) = 9w + 4.
Thus O =
U1I
U 13
9w + 4 gw = 4 19 x1@2 gx =
0
· 23 x3@2
1
9
l13
=
4
2
(133@2
27
3 8).
9. The angle of intersection of the two curves, , is the angle between their respective tangents at the point of intersection.
For both curves the point (1> 0> 0) occurs when w = 0.
r01 (w) = 3 sin w i + cos w j + k i r01 (0) = j + k and r02 (w) = i + 2w j + 3w2 k i r02 (0) = i.
r01 (0) · r02 (0) = (j + k) · i = 0. Therefore, the curves intersect in a right angle, that is, =
.
2
10. The parametric value corresponding to the point (1> 0> 1) is w = 0.
r0 (w) = hw i + hw (cos w + sin w) j + hw (cos w 3 sin w) k i |r0 (w)| = hw
I
Uw I
and v(w) = 0 hx 3 gx = 3(hw 3 1) i w = ln 1 + I13 v .
Therefore, r(w(v)) = 1 +
I1 v
3
i+ 1+
I1 v
3
sin ln 1 +
w2 > w> 1
w2 > w> 1
r0 (w)
I
=
=
|r0 (w)|
|hw2 > w> 1i|
w4 + w2 + 1
11. (a) T(w) =
1
I
v
3
s
I
1 + (cos w + sin w)2 + (cos w 3 sin w)2 = 3 hw
j+ 1+
I1 v
3
cos ln 1 +
1
I
v
3
k.
(b) T0 (w) = 3 12 (w4 + w2 + 1)33@2 (4w3 + 2w) w2 > w> 1 + (w4 + w2 + 1)31@2 h2w> 1> 0i
1
32w3 3 w
w2 > w> 1 + 4
h2w> 1> 0i
(w4 + w2 + 1)3@2
(w + w2 + 1)1@2
w3 + 2w> 3w4 + 1> 32w3 3 w
32w5 3 w3 > 32w4 3 w2 > 32w3 3 w + 2w5 + 2w3 + 2w> w4 + w2 + 1> 0
=
=
(w4 + w2 + 1)3@2
(w4 + w2 + 1)3@2
=
I
I
w6 + 4w4 + 4w2 + w8 3 2w4 + 1 + 4w6 + 4w4 + w2
w8 + 5w6 + 6w4 + 5w2 + 1
=
4
2
3@2
(w + w + 1)
(w4 + w2 + 1)3@2
w3 + 2w> 1 3 w4 > 32w3 3 w
.
N(w) = I
w8 + 5w6 + 6w4 + 5w2 + 1
|T0 (w)| =
(c) (w) =
|T0 (w)|
=
|r0 (w)|
I
w8 + 5w6 + 6w4 + 5w2 + 1
(w4 + w2 + 1)2
or
I
w4 + 4w2 + 1
4
(w + w2 + 1)3@2
12. Using Exercise 13.3.42, we have r0 (w) = h33 sin w> 4 cos wi,
3
|r0 (w)| =
s
3
9 sin2 w + 4 cos2 w and then
and
r00 (w) = h33 cos w> 34 sin wi,
12
|(33 sin w)(34 sin w) 3 (4 cos w)(33 cos w)|
=
.
(9 sin2 w + 16 cos2 w)3@2
(9 sin2 w + 16 cos2 w)3@2
3
At (3> 0), w = 0 and (0) = 12@(16)3@2 = 12
= 16
. At (0> 4), w = 2 and 2 = 12@93@2 =
64
(w) =
12
27
= 49 .
12{2 || 00 |
12
13. | = 4{ , | = 12{ and ({) =
=
, so (1) = 3@2 .
[1 + (|0 )2 ]3@2
(1 + 16{6 )3@2
17
0
3
00
2
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CHAPTER 13 VECTOR FUNCTIONS
12{2 3 2
[1 + (4{3 3 2{)2 ]3@2
14. ({) =
i (0) = 2.
So the osculating circle has radius
1
2
and center 0> 3 12 .
2
Thus its equation is {2 + | + 12 = 14 .
15. r(w) = hsin 2w> w> cos 2wi
T0 (w) =
I1
5
i r0 (w) = h2 cos 2w> 1> 32 sin 2wi i T(w) =
I1
5
h2 cos 2w> 1> 32 sin 2wi i
h34 sin 2w> 0> 34 cos 2wi i N(w) = h3 sin 2w> 0> 3 cos 2wi. So N = N() = h0> 0> 31i and
B = T×N =
I1
5
h31> 2> 0i. So a normal to the osculating plane is h31> 2> 0i and an equation is
31({ 3 0) + 2(| 3 ) + 0(} 3 1) = 0 or { 3 2| + 2 = 0.
16. (a) The average velocity over [3> 3=2] is given by
r(3=2) 3 r(3)
= 5[r(3=2) 3 r(3)], so we draw a
3=2 3 3
vector with the same direction but 5 times the length
of the vector [r(3=2) 3 r(3)].
(b) v(3) = r0 (3) = lim
k<0
(c) T(3) =
r(3 + k) 3 r(3)
k
r0 (3)
, a unit vector in the same direction as
|r0 (3)|
r0 (3), that is, parallel to the tangent line to the curve at
r(3), pointing in the direction corresponding to
increasing w, and with length 1.
17. r(w) = w ln w i + w j + h3w k,
v(w) = r0 (w) = (1 + ln w) i + j 3 h3w k,
s
s
|v (w)| = (1 + ln w)2 + 12 + (3h3w )2 = 2 + 2 ln w + (ln w)2 + h32w , a(w) = v0 (w) =
18. v(w) =
U
1
w
i + h3w k
U
a(w) gw = (6w i + 12w2 j 3 6w k) gw = 3w2 i + 4w3 j 3 3w2 k + C, but i 3 j + 3 k = v(0) = 0 + C,
so C = i 3 j + 3 k and v(w) = (3w2 + 1) i + (4w3 3 1) j + (3 3 3w2 ) k=
U
r(w) = v(w) gw = (w3 + w) i + (w4 3 w) j + (3w 3 w3 ) k + D.
But r(0) = 0> so D = 0 and r(w) = (w3 + w) i + (w4 3 w) j + (3w 3 w3 ) k.
19. We set up the axes so that the shot leaves the athlete’s hand 7 ft above the origin. Then we are given r(0) = 7j,
|v(0)| = 43 ft@s, and v(0) has direction given by a 45 angle of elevation. Then a unit vector in the direction of v(0) is
I1 (i
2
+ j) i v(0) =
43
I
(i
2
+ j). Assuming air resistance is negligible, the only external force is due to gravity, so as in
Example 13.4.5 we have a = 3j j where here j E 32 ft@s2 . Since v0 (w) = a(w), we integrate, giving v(w) = 3jw j + C
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CHAPTER 13 REVIEW
where C = v(0) =
r(w) =
43
I
wi
2
+
43
I
(i
2
43
I
w
2
+ j) i v (w) =
43
I
2
i+
43
I
2
3 12 jw2 j + D. But D = r(0) = 7 j i r(w) =
(a) At 2 seconds, the shot is at r(2) =
43
I
(2) i
2
+
43
I
(2)
2
¤
365
3 jw j. Since r0 (w) = v(w) we integrate again, so
43
I
wi
2
+
43
I
w
2
3 12 jw2 + 7 j.
3 12 j(2)2 + 7 j E 60=8 i + 3=8 j, 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:
43
I
2
3 jw = 0 i
43
w= I
E 0=95 s. Then r(0=95) E 28=9 i + 21=4 j, so the maximum height is approximately 21=4 ft.
2j
(c) The shot hits the ground when the vertical component of r(w) is 0, so
316w2 +
43
I
w
2
43
I
w
2
3 12 jw2 + 7 = 0 i
+ 7 = 0 i w E 2=11 s. r(2=11) E 64=2 i 3 0=08 j, thus the shot lands approximately 64=2 ft from the
athlete.
20. r0 (w) = i + 2 j + 2w k,
Then dW =
r00 (w) = 2 k, |r0 (w)| =
I
I
1 + 4 + 4w2 = 4w2 + 5.
I
2 5
r0 (w) · r00 (w)
|r0 (w) × r00 (w)|
4w
|4 i 3 2 j|
I
I
I
and
d
=
.
=
=
=
Q
|r0 (w)|
|r0 (w)|
4w2 + 5
4w2 + 5
4w2 + 5
21. (a) Instead of proceeding directly, we use Formula 3 of Theorem 13.2.3: r(w) = w R(w)
i
v = r0 (w) = R(w) + w R0 (w) = cos $w i + sin $w j + w vg .
(b) Using the same method as in part (a) and starting with v = R(w) + w R0 (w), we have
a = v0 = R0 (w) + R0 (w) + w R00 (w) = 2 R0 (w) + w R00 (w) = 2 vg + w ag .
(c) Here we have r(w) = h3w cos $w i + h3w sin $w j = h3w R(w). So, as in parts (a) and (b),
v = r0 (w) = h3w R0 (w) 3 h3w R(w) = h3w [R0 (w) 3 R(w)] i
a = v0 = h3w [R00 (w) 3 R0 (w)] 3 h3w [R0 (w) 3 R(w)] = h3w [R00 (w) 3 2 R0 (w) + R(w)]
= h3w ag 3 2h3w vg + h3w R
Thus, the Coriolis acceleration (the sum of the “extra” terms not involving ag ) is 32h3w vg + h3w R.
;
1
A
A
A
A
?I1 3 {2
22. (a) I ({) = I
A
A
A
23{
A
=
if { $ 0
if 0 ? { ?
if { D
;
0
A
A
?
2 3@2
00
I ({) = 31@(1 3 { )
A
A
=
0
I1
2
i
I1
2
if { ? 0
if 0 ? { ?
if { A
;
0
A
A
?
I
I 0 ({) = 3{@ 1 3 {2
A
A
=
31
if { ? 0
if 0 ? { ?
if { A
I1
2
I1
2
i
I1
2
I1
2
g
[3{(1 3 {2 )31@2 ] = 3(1 3 {2 )31@2 3 {2 (1 3 {2 )33@2 = 3(1 3 {2 )33@2 .
g{
I
I
lim
1 3 {2 = I12 = I I12 , so I is continuous. Also, since
Now lim 1 3 {2 = 1 = I (0) and
I
{<0+
{<(1@ 2)
since
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CHAPTER 13 VECTOR FUNCTIONS
lim I 0 ({) = 0 = lim I 0 ({) and
{<0+
{<0
lim
I {<(1@ 2)
I 0 ({) = 31 =
lim
I +
{<(1@ 2)
I 0 ({), I 0 is continuous. But
lim I 00 ({) = 31 6= 0 = lim I 00 ({), so I 00 is not continuous at { = 0. (The same is true at { =
{<0+
{<0
I1 .)
2
So I does not have continuous curvature.
(b) Set S ({) = d{5 + e{4 + f{3 + g{2 + h{ + i . The continuity conditions on S are S (0) = 0, S (1) = 1, S 0 (0) = 0 and
S 0 (1) = 1. Also the curvature must be continuous. For { $ 0 and { D 1, ({) = 0; elsewhere
({) =
|S 00 ({)|
, so we need S 00 (0) = 0 and S 00 (1) = 0.
(1 + [S 0 ({)]2 )3@2
The conditions S (0) = S 0 (0) = S 00 (0) = 0 imply that g = h = i = 0.
The other conditions imply that d + e + f = 1, 5d + 4e + 3f = 1, and
10d + 6e + 3f = 0. From these, we find that d = 3, e = 38, and f = 6.
Therefore S ({) = 3{5 3 8{4 + 6{3 . Since there was no solution with
d = 0, this could not have been done with a polynomial of degree 4.
23. (a) r(w) = U cos $w i + U sin $w j
i v = r0 (w) = 3$U sin $w i + $U cos $w j, so r = U(cos $w i + sin $w j) and
v = $U(3 sin $w i + cos $w j). v · r = $U2 (3 cos $w sin $w + sin $w cos $w) = 0, so v z r. Since r points along a
radius of the circle, and v z r, v is 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 $U u, where u is the unit vector 3 sin $w i + cos $w j. Clearly |v| = $U |u| = $U. At
speed $U, the particle completes one revolution, a distance 2U, in time W =
(c) a =
2U
2
=
.
$U
$
gv
= 3$2 U cos $w i 3 $2 U sin $w j = 3$ 2 U(cos $w i + sin $w j), so a = 3$2 r. This shows that a is proportional
gw
to r and points in the opposite direction (toward the origin). Also, |a| = $2 |r| = $2 U.
(d) By Newton’s Second Law (see Section 13.4), F = pa, so |F| = p |a| = pU$2 =
24. (a) Dividing the equation |F| sin =
(b) U = 400 ft and = 12 , so yU =
p |v|2
p ($U)2
=
.
U
U
2
pyU
y2
2
= Uj tan .
by the equation |F| cos = pj, we obtain tan = U , so yU
U
Uj
I
I
Uj tan E 400 · 32 · tan 12 E 52=16 ft@s E 36 mi@h.
(c) We want to choose a new radius U1 for which the new rated speed is
3
2
of the old one:
I
I
U1 j tan 12 = 32 Uj tan 12 .
Squaring, we get U1 j tan 12 = 94 Uj tan 12 , so U1 = 94 U = 94 (400) = 900 ft.
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PROBLEMS PLUS
g|
g
=
[(y0 sin )w 3 12 jw2 ] = y0 sin 3 jw; that is, when
gw
gw
2
y0 sin 1
y0 sin y0 sin y 2 sin2 w=
and | = (y0 sin )
3 j
. This is the maximum height attained when
= 0
j
j
2
j
2j
1. (a) The projectile reaches maximum height when 0 =
the projectile is fired with an angle of elevation . This maximum height is largest when =
and the maximum height is
.
2
In that case, sin = 1
y02
.
2j
1 2 U
{ + .
(b) Let U = y02 j. We are asked to consider the parabola {2 + 2U| 3 U2 = 0 which can be rewritten as | = 3
2U
2
The points on or inside this parabola are those for which 3U $ { $ U and 0 $ | $
31 2 U
{ + . When the projectile is
2U
2
fired at angle of elevation , the points ({> |) along its path satisfy the relations { = (y0 cos ) w and
| = (y0 sin )w 3 12 jw2 , where 0 $ w $ (2y0 sin )@j (as in Example 13.4.5). Thus
2
y0 2y0 sin y02
$ = |U|. This shows that 3U $ { $ U.
=
sin
2
|{| $ y0 cos j
j
j
2y0 sin 3 w D 0 and
For w in the specified range, we also have | = w y0 sin 3 12 jw = 12 jw
j
| = (y0 sin )
|3
j
{
3
y0 cos 2
31 2 U
{ +
2U
2
{
y0 cos 2
= (tan ) { 3
j
1
{2 = 3
{2 + (tan ) {. Thus
2y02 cos2 2U cos2 31
1 2
U
{2 +
{ + (tan ) { 3
2U cos2 2U
2
1
U
{2 (1 3 sec2 ) + 2U (tan ) { 3 U2
{2
13
+ (tan ) { 3
=
=
2
2U
cos 2
2U
=
=
3(tan2 ) {2 + 2U (tan ) { 3 U2
3 [(tan ) { 3 U]2
=
$0
2U
2U
We have shown that every target that can be hit by the projectile lies on or inside the parabola | = 3
Now let (d> e) be any point on or inside the parabola | = 3
1 2 U
{ + .
2U
2
1 2 U
1 2 U
{ + . Then 3U $ d $ U and 0 $ e $ 3
d + .
2U
2
2U
2
We seek an angle such that (d> e) lies in the path of the projectile; that is, we wish to find an angle such that
31
1
d2 + (tan ) d or equivalently e =
(tan2 + 1)d2 + (tan ) d. Rearranging this equation we get
2U cos2 2U
2
d
d2
2
tan 3 d tan +
+ e = 0 or d2 (tan )2 3 2dU(tan ) + (d2 + 2eU) = 0 (B) . This quadratic equation
2U
2U
e=3
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CHAPTER 13 PROBLEMS PLUS
for tan has real solutions exactly when the discriminant is nonnegative. Now E 2 3 4DF D 0 C
(32dU)2 3 4d2 (d2 + 2eU) D 0 C 4d2 (U2 3 d2 3 2eU) D 0 C 3d2 3 2eU + U2 D 0 C
e$
31 2 U
1
(U2 3 d2 ) C e $
d + . This condition is satisfied since (d> e) is on or inside the parabola
2U
2U
2
1 2 U
{ + . It follows that (d> e) lies in the path of the projectile when tan satisfies (B), that is, when
2U
2
s
I
2dU ± 4d2 (U2 3 d2 3 2eU)
U ± U2 3 2eU 3 d2
tan =
.
=
2d2
d
|=3
If the gun is pointed at a target with height k at a distance G downrange, then
(c)
tan = k@G. When the projectile reaches a distance G downrange (remember
we are assuming that it doesn’t hit the ground first), we have G = { = (y0 cos )w,
so w =
jG2
G
and | = (y0 sin )w 3 12 jw2 = G tan 3 2
.
y0 cos 2y0 cos2 Meanwhile, the target, whose {-coordinate is also G, has fallen from height k to height
k 3 12 jw2 = G tan 3
jG2
. Thus the projectile hits the target.
cos2 2y02
2. (a) As in Problem 1, r(w) = (y0 cos )w i + (y0 sin )w 3 12 jw2 j, so { = (y0 cos )w and | = (y0 sin )w 3 12 jw2 . The
difference here is that the projectile travels until it reaches a point where { A 0 and | = 3(tan ){. (Here 0 $ $
From the parametric equations, we obtain w =
(y0 sin ){
{
j{2
j{2
and | =
3 2
= (tan ){ 3 2
.
2
y0 cos y0 cos 2y0 cos 2y0 cos2 Thus the projectile hits the inclined plane at the point where (tan ){ 3
j{2
= 3(tan ){. Since
cos2 2y02
j{
j{2
= (tan + tan ){ and { A 0, we must have 2
= tan + tan . It follows that
cos2 2y0 cos2 2y02
{
2y0 cos 2y02 cos2 (tan + tan ) and w =
=
(tan + tan ). This means that the parametric
j
y0 cos j
2y0 cos (tan + tan ) .
equations are defined for w in the interval 0>
j
{=
(b) The downhill range (that is, the distance to the projectile’s landing point as
measured along the inclined plane) is U() = { sec , where { is the
coordinate of the landing point calculated in part (a). Thus
U() =
=
.)
2
2y02 cos2 2y 2
(tan + tan ) sec = 0
j
j
sin cos cos2 sin +
cos cos2 2y 2 cos sin( + )
2y02 cos (sin cos + cos sin ) = 0
2
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369
U() is maximized when
0 = U0 () =
2y02
[3 sin sin( + ) + cos cos( + )]
j cos2 2y2 cos(2 + )
2y02
cos[( + ) + ] = 0
2
j cos j cos2 This condition implies that cos(2 + ) = 0 i 2 + = 2 i = 12 2 3 .
=
(c) The solution is similar to the solutions to parts (a) and (b). This time the projectile travels until it reaches a point where
{ A 0 and | = (tan ){. Since tan = 3 tan(3), we obtain the solution from the previous one by replacing with 3.
The desired angle is = 12 2 + .
(d) As observed in part (c), firing the projectile up an inclined plane with angle of inclination involves the same equations as
in parts (a) and (b) but with replaced by 3. So if U is the distance up an inclined plane, we know from part (b) that
U=
2y02 cos sin( 3 )
j cos2 (3)
i y02 =
Uj cos2 . y 2 is minimized (and hence y0 is minimized) with
2 cos sin( 3 ) 0
respect to when
0=
=
Since ? ?
,
2
Uj cos2 3(cos cos ( 3 ) 3 sin sin ( 3 ))
g
(y02 ) =
·
g
2
[cos sin( 3 )]2
cos(2 3 )
3Uj cos2 cos[ + ( 3 )]
3Uj cos2 ·
·
=
2
2
[cos sin( 3 )]
2
[cos sin( 3 )]2
this implies cos(2 3 ) = 0 C 2 3 =
hence the energy required, is minimized for =
3. (a) a = 3j j
1
2
2
+ .
2
i =
1
2
2
+ . Thus the initial speed, and
i v = v0 3 jw j = 2 i 3 jw j i s = s0 + 2w i 3 12 jw2 j = 3=5 j + 2w i 3 12 jw2 j i
s
s
s = 2w i + 3=5 3 12 jw2 j. Therefore | = 0 when w = 7@j seconds. At that instant, the ball is 2 7@j E 0=94 ft to the
right of the table top. Its coordinates (relative to an origin on the floor directly under the table’s edge) are (0=94> 0). At
I
I
impact, the velocity is v = 2 i 3 7j j, so the speed is |v| = 4 + 7j E 15 ft@s.
(b) The slope of the curve when w =
and E 7=6 .
(c) From (a), |v| =
u
s
I
I
3j 7@j
7 g|
7j
g|@gw
3jw
3 7j
is
=
=
=
=
. Thus cot =
j g{
g{@gw
2
2
2
2
I
I
4 + 7j. So the ball rebounds with speed 0=8 4 + 7j E 12=08 ft@s at angle of inclination
90 3 E 82=3886 . By Example 13.4.5, the horizontal distance traveled between bounces is g =
y02 sin 2
, where
j
y0 E 12=08 ft@s and E 82=3886 . Therefore, g E 1=197 ft. So the ball strikes the floor at about
s
2 7@j + 1=197 E 2=13 ft to the right of the table’s edge.
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¤
CHAPTER 13 PROBLEMS PLUS
1
4. By the Fundamental Theorem of Calculus, r0 (w) = sin
2
2 w
> cos 12 w2 , |r0 (w)| = 1 and so T(w) = r0 (w).
s
Thus T0 (w) = w cos 12 w2 > 3 sin 12 w2 and the curvature is = |T0 (w)| = (w)2 (1) = |w|.
5. The trajectory of the projectile is given by r(w) = (y cos )w i + (y sin )w 3 12 jw2 j, so
v(w) = r0 (w) = y cos i + (y sin 3 jw) j and
s
s
|v(w)| = (y cos )2 + (y sin 3 jw)2 = y 2 3 (2yj sin ) w + j 2 w2 =
v 2y
y2
2
2
j w 3
(sin ) w + 2
j
j
v
v
2
2
y
y
y2
y2
y2
2
w 3 sin + 2 3 2 sin = j
w 3 sin + 2 cos2 =j
j
j
j
j
j
The projectile hits the ground when (y sin )w 3 12 jw2 = 0 i w =
O() =
]
(2y@j) sin |v(w)| gw =
0
5
]
(2y@j) sin 0
2y
j
sin , so the distance traveled by the projectile is
v
2
y2
y
j
w 3 sin + 2 cos2 gw
j
j
v
2 2
y
y
w 3 (y@j) sin 7
w 3 sin +
cos =j
2
j
j
46(2y@j) sin 3
v
2 2
[(y@j) cos ]2 C
y
y
y
+
w 3 sin +
ln w 3 sin +
cos D8
2
j
j
j
0
5
j 7y
=
2 j
[using Formula 21 in the Table of Integrals]
3
4
v
v
2 2 2
2 2
y
y
y
y
y
y
sin sin +
cos +
cos lnC sin +
sin +
cos D
j
j
j
j
j
j
3
46
v
v
2 2 2
2 2
y
y
y
y
y
y
y
+ sin sin +
cos 3
cos lnC3 sin +
sin +
cos D8
j
j
j
j
j
j
j
=
=
y
y2
y
y
y
y2
y
j y
y
y
sin · + 2 cos2 ln
sin +
+ sin · 3 2 cos2 ln 3 sin +
2 j
j
j
j
j
j
j
j
j
j
y2
y2
y2
(y@j) sin + y@j
1 + sin y2
sin +
cos2 ln
=
sin +
cos2 ln
j
2j
3 (y@j) sin + y@j
j
2j
1 3 sin We want to maximize O() for 0 $ $ @2.
y2
1 3 sin 1 + sin y2
2 cos 2
O () =
3 2 cos sin ln
cos +
cos ·
·
j
2j
1 + sin (1 3 sin )2
1 3 sin y2
1 + sin y2
2
cos +
cos2 ·
3 2 cos sin ln
=
j
2j
cos 1 3 sin 2
2
y
1 + sin y2
1 + sin y
cos +
cos 1 3 sin ln
=
cos 2 3 sin ln
=
j
j
1 3 sin j
1 3 sin 0
1 + sin = 0 [since cos 6= 0].
O() has critical points for 0 ? ? @2 when O0 () = 0 i 2 3 sin ln
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CHAPTER 13 PROBLEMS PLUS
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371
Solving by graphing (or using a CAS) gives E 0=9855. Compare values at the critical point and the endpoints:
O(0) = 0, O(@2) = y 2 @j, and O(0=9855) E 1=20y 2 @j. Thus the distance traveled by the projectile is maximized
for E 0=9855 or E 56 .
6. As the cable is wrapped around the spool, think of the top or bottom of the
cable forming a helix of radius U + u. Let k be the vertical distance
between coils. Then, from similar triangles,
2u
2(u + U)
I
=
k
k2 3 4u2
i k2 u2 = 2 (u + U)2 (k2 3 4u2 ) i
2u(u + U)
.
k= s
2 (u + U)2 3 u2
If we parametrize the helix by {(w) = (U + u) cos w, |(w) = (U + u) sin w, then we must have }(w) = [k@(2)]w.
The length of one complete cycle is
c=
]
2
0
= 2
]
s
[{0 (w)]2 + [| 0 (w)]2 + [} 0 (w)]2 gw =
0
v
(U +
u)2
2
v
u2 (U + u)2
+ 2
= 2(U + u)
(U + u)2 3 u2
(U +
v
u)2
1+
+
k
2
2
gw = 2
v
(U + u)2 +
k
2
2
u2
22 (U + u)2
s
=
2 (U + u)2 3 u2
2 (U + u)2 3 u2
The number of complete cycles is [[O@c]], and so the shortest length along the spool is
%% s
&&
O 2 (U + u)2 3 u2
2u(U + u)
O
k
= s
c
22 (U + u)2
2 (U + u)2 3 u2
7. We can write the vector equation as r(w) = aw2 + bw + c where a = hd1 > d2 > d3 i, b = he1 > e2 > e3 i, and c = hf1 > f2 > f3 i.
Then r0 (w) = 2w 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 b are nonparallel. Otherwise the tangent vectors are all parallel and the curve lies along a single line.] A normal
vector for the plane is a × b = hd2 e3 3 d3 e2 > d3 e1 3 d1 e3 > d1 e2 3 d2 e1 i. The point (f1, f2 , f3 ) lies on the plane (when
w = 0), so an equation of the plane is
(d2 e3 3 d3 e2 )({ 3 f1 ) + (d3 e1 3 d1 e3 )(| 3 f2 ) + (d1 e2 3 d2 e1 )(} 3 f3 ) = 0
or
(d2 e3 3 d3 e2 ){ + (d3 e1 3 d1 e3 )| + (d1 e2 3 d2 e1 )} = d2 e3 f1 3 d3 e2 f1 + d3 e1 f2 3 d1 e3 f2 + d1 e2 f3 3 d2 e1 f3
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14
PARTIAL DERIVATIVES
14.1 Functions of Several Variables
1. (a) From Table 1, i(315> 40) = 327, which means that if the temperature is 315 C and the wind speed is 40 km@h, then the
air would feel equivalent to approximately 327 C without wind.
(b) The question is asking: when the temperature is 320 C, what wind speed gives a wind-chill index of 330 C? From
Table 1, the speed is 20 km@h.
(c) The question is asking: when the wind speed is 20 km@h, what temperature gives a wind-chill index of 349 C? From
Table 1, the temperature is 335 C.
(d) The function Z = i (35> y) means that we fix W at 35 and allow y to vary, resulting in a function of one variable. In
other words, the function gives wind-chill index values for different wind speeds when the temperature is 35 C. From
Table 1 (look at the row corresponding to W = 35), the function decreases and appears to approach a constant value as y
increases.
(e) The function Z = i (W> 50) means that we fix y at 50 and allow W to vary, again giving a function of one variable. In
other words, the function gives wind-chill index values for different temperatures when the wind speed is 50 km@h . From
Table 1 (look at the column corresponding to y = 50), the function increases almost linearly as W increases.
2. (a) From Table 3, i (95> 70) = 124, which means that when the actual temperature is 95 F and the relative humidity is 70%,
the perceived air temperature is approximately 124 F.
(b) Looking at the row corresponding to W = 90, we see that i(90> k) = 100 when k = 60.
(c) Looking at the column corresponding to k = 50, we see that i (W> 50) = 88 when W = 85.
(d) L = i (80> k) means that W is fixed at 80 and k is allowed to vary, resulting in a function of k that gives the humidex values
for different relative humidities when the actual temperature is 80 F. Similarly, L = i (100> k) is a function of one
variable that gives the humidex values for different relative humidities when the actual temperature is 100 F. Looking at
the rows of the table corresponding to W = 80 and W = 100> we see that i (80> k) increases at a relatively constant rate of
approximately 1 F per 10% relative humidity, while i(100> k) increases more quickly (at first with an average rate of
change of 5 F per 10% relative humidity) and at an increasing rate (approximately 12 F per 10% relative humidity for
larger values of k).
3. S (120> 20) = 1=47(120)0=65 (20)0=35 E 94=2, so when the manufacturer invests $20 million in capital and 120,000 hours of
labor are completed yearly, the monetary value of the production is about $94.2 million.
4. If the amounts of labor and capital are both doubled, we replace O> N in the function with 2O> 2N, giving
S (2O> 2N) = 1=01(2O)0=75 (2N)0=25 = 1=01(20=75 )(20=25 )O0=75 N 0=25 = (21 )1=01O0=75 N 0=25 = 2S (O> N)
Thus, the production is doubled. It is also true for the general case S (O> N) = eO N 1 3 :
S (2O> 2N) = e(2O) (2N)13 = e(2 )(213 )O N 13 = (2+13 )eO N 13 = 2S (O> N).
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¤
CHAPTER 14
PARTIAL DERIVATIVES
5. (a) i (160> 70) = 0=1091(160)0=425 (70)0=725 E 20=5, which means that the surface area of a person 70 inches (5 feet 10
inches) tall who weighs 160 pounds is approximately 20.5 square feet.
(b) Answers will vary depending on the height and weight of the reader.
6. We compare the values for the wind-chill index given by Table 1 with those given by the model function:
Modeled Wind-Chill Index Values Z (W> y)
The values given by the function appear to be fairly close (within 0=5) to the values in Table 1.
7. (a) According to Table 4, i(40> 15) = 25, which means that if a 40-knot wind has been blowing in the open sea for 15 hours,
it will create waves with estimated heights of 25 feet.
(b) k = i (30> w) means we fix y at 30 and allow w to vary, resulting in a function of one variable. Thus here, k = i (30> w)
gives the wave heights produced by 30-knot winds blowing for w hours. From the table (look at the row corresponding to
y = 30), the function increases but at a declining rate as w increases. In fact, the function values appear to be approaching a
limiting value of approximately 19, which suggests that 30-knot winds cannot produce waves higher than about 19 feet.
(c) k = i (y> 30) means we fix w at 30, again giving a function of one variable. So, k = i (y> 30) gives the wave heights
produced by winds of speed y blowing for 30 hours. From the table (look at the column corresponding to w = 30), the
function appears to increase at an increasing rate, with no apparent limiting value. This suggests that faster winds (lasting
30 hours) always create higher waves.
8. (a) The cost of making { small boxes, | medium boxes, and } large boxes is F = i({> |> }) = 8000 + 2=5{ + 4| + 4=5}
dollars.
(b) i (3000> 5000> 4000) = 8000 + 2=5(3000) + 4(5000) + 4=5(4000) = 53,500 which means that it costs $53,500 to make
3000 small boxes, 5000 medium boxes, and 4000 large boxes.
(c) Because no partial boxes will be produced, each of {, |, and } must be a positive integer or zero.
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SECTION 14.1
FUNCTIONS OF SEVERAL VARIABLES
¤
375
9. (a) j(2> 31) = cos(2 + 2(31)) = cos(0) = 1
(b) { + 2| is defined for all choices of values for { and | and the cosine function is defined for all input values, so the domain
of j is R2 .
(c) The range of the cosine function is [31> 1] and { + 2| generates all possible input values for the cosine function, so the
range of cos({ + 2|) is [31> 1].
10. (a) I (3> 1) = 1 +
(b)
I
I
4 3 12 = 1 + 3
s
4 3 | 2 is defined only when 4 3 |2 D 0, or | 2 $ 4 C
32 $ | $ 2. So the domain of I is {({> |) | 32 $ | $ 2 }.
(c) We know 0 $
11. (a) i (1> 1> 1) =
(b)
s
s
4 3 | 2 $ 2 so 1 $ 1 + 4 3 | 2 $ 3. Thus the range of I is [1> 3].
I
I
I
1 + 1 + 1 + ln(4 3 12 3 12 3 12 ) = 3 + ln 1 = 3
I I I
{, |, } are defined only when { D 0, | D 0, } D 0, and ln(4 3 {2 3 | 2 3 } 2 ) is defined when
4 3 {2 3 | 2 3 } 2 A 0 C {2 + | 2 + } 2 ? 4, thus the domain is
({> |> }) | {2 + | 2 + } 2 ? 4> { D 0> | D 0> } D 0 , the portion of the interior of a sphere of radius 2, centered at the
origin, that is in the first octant.
12. (a) j(1> 2> 3) = 13 · 22 · 3
I
I
10 3 1 3 2 3 3 = 12 4 = 24
(b) j is defined only when 10 3 { 3 | 3 } D 0 C } $ 10 3 { 3 |, so the domain is {({> |> }) | } $ 10 3 { 3 |}, the
points on or below the plane { + | + } = 10.
13.
I
2{ 3 | is defined only when 2{ 3 | D 0, or | $ 2{.
So the domain of i is {({> |) | | $ 2{}.
14. We need {| D 0, so G = {({> |) | {| D 0}, the first and
third quadrants.
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CHAPTER 14
PARTIAL DERIVATIVES
15. ln(9 3 {2 3 9| 2 ) is defined only when
2
2
1 2
9{
16.
2
2
+ | ? 1. So the domain of i
9 3 { 3 9| A 0, or
1 2
is ({> |) 9 { + | 2 ? 1 , the interior of an ellipse.
17.
I
1 3 {2 is defined only when 1 3 {2 D 0, or
s
{2 $ 1 C 31 $ { $ 1, and 1 3 |2 is defined
s
{2 3 | 2 is defined only when {2 3 | 2 D 0
2
| ${
C
||| $ |{|
C
C
3 |{| $ | $ |{|. So
the domain of i is {({> |) | 3 |{| $ | $ |{|}.
18.
s
I
| + 25 3 {2 3 | 2 is defined only when | D 0 and
25 3 {2 3 | 2 D 0 C {2 + | 2 $ 25. So the domain
only when 1 3 | 2 D 0, or | 2 $ 1 C 31 $ | $ 1.
of i is ({> |) | {2 + | 2 $ 25> | D 0 , a half disk of
Thus the domain of i is
radius 5.
{({> |) | 31 $ { $ 1> 3 1 $ | $ 1}.
19.
s
| 3 {2 is defined only when | 3 {2 D 0, or | D {2 .
20. arcsin({2 + | 2 3 2) is defined only when
In addition, i is not defined if 1 3 {2 = 0 C
31 $ {2 + | 2 3 2 $ 1 C 1 $ {2 + | 2 $ 3. Thus
{ = ±1. Thus the domain of i is
the domain of i is ({> |) | 1 $ {2 + | 2 $ 3 .
({> |) | | D {2 > { 6= ±1 .
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SECTION 14.1
21. We need 1 3 {2 3 | 2 3 } 2 D 0 or {2 + | 2 + } 2 $ 1,
2
2
2
so G = ({> |> }) | { + | + } $ 1 (the points inside
or on the sphere of radius 1, center the origin).
FUNCTIONS OF SEVERAL VARIABLES
¤
22. i is defined only when 16 3 4{2 3 4| 2 3 } 2 A 0
2
2
377
i
2
{
|
}
+
+
? 1. Thus,
4
4
16
2
{
|2
}2
+
+
? 1 , that is, the points
G = ({> |> }) 4
4
16
inside the ellipsoid
|2
}2
{2
+
+
= 1.
4
4
16
23. } = 1 + |, a plane which intersects the |}-plane in the
24. } = 2 3 {, a plane which intersects the {}-plane in the
line } = 1 + |, { = 0. The portion of this plane for
line } = 2 3 {, | = 0. The portion of this plane for
{ D 0, } D 0 is shown.
| D 0, } D 0 is shown.
25. } = 10 3 4{ 3 5| or 4{ + 5| + } = 10, a plane with
26. } = h3| , a cylinder.
intercepts 2=5, 2, and 10.
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CHAPTER 14
PARTIAL DERIVATIVES
27. } = | 2 + 1, a parabolic cylinder
28. } = 1 + 2{2 + 2| 2 , a circular paraboloid with vertex at
(0> 0> 1).
29. } = 9 3 {2 3 9| 2 , an elliptic paraboloid opening
downward with vertex at (0> 0> 9).
31. } =
30. } =
s
4{2 + | 2 so 4{2 + | 2 = } 2 and } D 0, the top
half of an elliptic cone.
s
4 3 4{2 3 | 2 so 4{2 + | 2 + } 2 = 4 or
{2 +
}2
|2
+
= 1 and } D 0, the top half of an
4
4
ellipsoid.
32. All six graphs have different traces in the planes { = 0 and | = 0, so we investigate these for each function.
(a) i ({> |) = |{| + |||. The trace in { = 0 is } = |||, and in | = 0 is } = |{|, so it must be graph VI.
(b) i ({> |) = |{||. The trace in { = 0 is } = 0> and in | = 0 is } = 0, so it must be graph V.
(c) i ({> |) =
1
1
1
. The trace in { = 0 is } =
, and in | = 0 is } =
. In addition, we can see that i is
1 + {2 + | 2
1 + |2
1 + {2
close to 0 for large values of { and |, so this is graph I.
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SECTION 14.1
FUNCTIONS OF SEVERAL VARIABLES
¤
379
(d) i ({> |) = ({2 3 | 2 )2 . The trace in { = 0 is } = | 4 , and in | = 0 is } = {4 . Both graph II and graph IV seem plausible;
notice the trace in } = 0 is 0 = ({2 3 | 2 )2
i | = ±{, so it must be graph IV.
(e) i ({> |) = ({ 3 |)2 . The trace in { = 0 is } = | 2 , and in | = 0 is } = {2 . Both graph II and graph IV seem plausible;
notice the trace in } = 0 is 0 = ({ 3 |)2
i | = {, so it must be graph II.
(f ) i ({> |) = sin (|{| + |||). The trace in { = 0 is } = sin |||, and in | = 0 is } = sin |{|. In addition, notice that the
oscillating nature of the graph is characteristic of trigonometric functions. So this is graph III.
33. The point (33> 3) lies between the level curves with }-values 50 and 60. Since the point is a little closer to the level curve with
} = 60, we estimate that i (33> 3) E 56. The point (3> 32) appears to be just about halfway between the level curves with
}-values 30 and 40, so we estimate i (3> 32) E 35. The graph rises as we approach the origin, gradually from above, steeply
from below.
34. (a) F (Chicago) lies between level curves with pressures 1012 and 1016 mb, and since F appears to be located about
one-fourth the distance from the 1012 mb isobar to the 1016 mb isobar, we estimate the pressure at Chicago to be about
1013 mb. Q lies very close to a level curve with pressure 1012 mb so we estimate the pressure at Nashville to be
approximately 1012 mb. V appears to be just about halfway between level curves with pressures 1008 and 1012 mb, so we
estimate the pressure at San Francisco to be about 1010 mb. Y lies close to a level curve with pressure 1016 mb but we
can’t see a level curve to its left so it is more difficult to make an accurate estimate. There are lower pressures to the right
of Y and Y is a short distance to the left of the level curve with pressure 1016 mb, so we might estimate that the pressure at
Vancouver is about 1017 mb.
(b) Winds are stronger where the isobars are closer together (see Figure 13), and the level curves are closer near V than at the
other locations, so the winds were strongest at San Francisco.
35. The point (160> 10), corresponding to day 160 and a depth of 10 m, lies between the isothermals with temperature values
of 8 and 12 C. Since the point appears to be located about three-fourths the distance from the 8 C isothermal to the 12 C
isothermal, we estimate the temperature at that point to be approximately 11 C. The point (180> 5) lies between the 16 and
20 C isothermals, very close to the 20 C level curve, so we estimate the temperature there to be about 19=5 C.
36. If we start at the origin and move along the {-axis, for example, the }-values of a cone centered at the origin increase at a
constant rate, so we would expect its level curves to be equally spaced. A paraboloid with vertex the origin, on the other hand,
has }-values which change slowly near the origin and more quickly as we move farther away. Thus, we would expect its level
curves near the origin to be spaced more widely apart than those farther from the origin. Therefore contour map I must
correspond to the paraboloid, and contour map II the cone.
37. Near D, the level curves are very close together, indicating that the terrain is quite steep. At E, the level curves are much
farther apart, so we would expect the terrain to be much less steep than near D, perhaps almost flat.
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CHAPTER 14
PARTIAL DERIVATIVES
38.
39.
40.
41.
42.
43. The level curves are (| 3 2{)2 = n or | = 2{ ±
n D 0, a family of pairs of parallel lines.
I
n,
44. The level curves are {3 3 | = n or | = {3 3 n, a family
of cubic curves.
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SECTION 14.1
45. The level curves are
I
I
{ + | = n or | = 3 { + n, a
family of vertical translations of the graph of the root
I
function | = 3 {.
47. The level curves are |h{ = n or | = nh3{ , a family of
FUNCTIONS OF SEVERAL VARIABLES
¤
381
46. The level curves are ln({2 + 4| 2 ) = n or {2 + 4| 2 = hn ,
a family of ellipses.
48. n = | sec { or | = n cos {, { 6=
2
+ q [q an integer].
exponential curves.
49. The level curves are
s
| 2 3 {2 = n or | 2 3 {2 = n2 ,
n D 0. When n = 0 the level curve is the pair of lines
| = ±{. For n A 0, the level curves are hyperbolas
with axis the |-axis.
50. For n 6= 0 and ({> |) 6= (0> 0), n =
|
{2 + | 2
C
|
1
1 2
= 2 , a family
= 0 C {2 + | 3 2n
n
4n
1
1
of circles with center 0> 2n
and radius 2n
(without the
{2 + | 2 3
origin). If n = 0, the level curve is the {-axis.
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CHAPTER 14
PARTIAL DERIVATIVES
51. The contour map consists of the level curves n = {2 + 9| 2 , a family of
ellipses with major axis the {-axis. (Or, if n = 0, the origin.)
The graph of i ({> |) is the surface } = {2 + 9| 2 , an elliptic paraboloid.
If we visualize lifting each ellipse n = {2 + 9|2 of the contour map to the plane
} = n, we have horizontal traces that indicate the shape of the graph of i .
52.
The contour map consists of the level curves n =
s
36 3 9{2 3 4| 2
9{2 + 4| 2 = 36 3 n2 , n D 0, a family of ellipses with major axis the
|-axis. (Or, if n = 6, the origin.)
s
36 3 9{2 3 4| 2 , or equivalently the upper half of the ellipsoid
s
9{2 + 4| 2 + } 2 = 36. If we visualize lifting each ellipse n = 36 3 9{2 3 4| 2 of the contour map to the plane } = n,
The graph of i ({> |) is the surface } =
we have horizontal traces that indicate the shape of the graph of i .
53. The isothermals are given by n = 100@(1 + {2 + 2| 2 ) or
{2 + 2| 2 = (100 3 n)@n [0 ? n $ 100], a family of ellipses.
f
or
u2 3 {2 3 | 2
54. The equipotential curves are n = s
{2 + | 2 = u2 3
f 2
n
, a family of circles (n D f@u).
Note: As n < ", the radius of the circle approaches u.
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SECTION 14.1
FUNCTIONS OF SEVERAL VARIABLES
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383
55. i ({> |) = {| 2 3 {3
The traces parallel to the |}-plane (such as the left-front trace in the graph above) are parabolas; those parallel to the {}-plane
(such as the right-front trace) are cubic curves. The surface is called a monkey saddle because a monkey sitting on the surface
near the origin has places for both legs and tail to rest.
56. i ({> |) = {| 3 3 |{3
The traces parallel to either the
|}-plane or the {}-plane are cubic
curves.
2 +| 2 )@3
57. i ({> |) = h3({
sin({2 ) + cos(| 2 )
58. i ({> |) = cos { cos |
The traces parallel to either the
|}- or {}-plane are cosine curves
with amplitudes that vary
from 0 to 1.
59. } = sin({|)
(a) C
(b) II
Reasons: This function is periodic in both { and |, and the function is the same when { is interchanged with |, so its graph is
symmetric about the plane | = {. In addition, the function is 0 along the {- and |-axes. These conditions are satisfied only by
C and II.
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CHAPTER 14
60. } = h{ cos |
PARTIAL DERIVATIVES
(a) A
(b) IV
Reasons: This function is periodic in | but not {, a condition satisfied only by A and IV. Also, note that traces in { = n are
cosine curves with amplitude that increases as { increases.
61. } = sin({ 3 |)
(a) F
(b) I
Reasons: This function is periodic in both { and | but is constant along the lines | = { + n, a condition satisfied only
by F and I.
62. } = sin { 3 sin |
(a) E
(b) III
Reasons: This function is periodic in both { and |, but unlike the function in Exercise 61, it is not constant along lines such as
| = { + , so the contour map is III. Also notice that traces in | = n are vertically shifted copies of the sine wave } = sin {,
so the graph must be E.
63. } = (1 3 {2 )(1 3 | 2 )
(a) B
(b) VI
Reasons: This function is 0 along the lines { = ±1 and | = ±1. The only contour map in which this could occur is VI. Also
note that the trace in the {}-plane is the parabola } = 1 3 {2 and the trace in the |}-plane is the parabola } = 1 3 |2 , so the
graph is B.
64. } =
{3|
1 + {2 + | 2
(a) D
(b) V
Reasons: This function is not periodic, ruling out the graphs in A, C, E, and F. Also, the values of } approach 0 as we use
points farther from the origin. The only graph that shows this behavior is D, which corresponds to V.
65. n = { + 3| + 5} is a family of parallel planes with normal vector h1> 3> 5i.
66. n = {2 + 3| 2 + 5} 2 is a family of ellipsoids for n A 0 and the origin for n = 0.
67. Equations for the level surfaces are n = | 2 + } 2 . For n A 0, we have a family of circular cylinders with axis the {-axis and
radius
I
n. When n = 0 the level surface is the {-axis. (There are no level surfaces for n ? 0.)
68. Equations for the level surfaces are {2 3 | 2 3 } 2 = n. For n = 0, the equation becomes | 2 + } 2 = {2 and the surface is a
right circular cone with vertex the origin and axis the {-axis. For n A 0, we have a family of hyperboloids of two sheets with
axis the {-axis, and for n ? 0, we have a family of hyperboloids of one sheet with axis the {-axis.
69. (a) The graph of j is the graph of i shifted upward 2 units.
(b) The graph of j is the graph of i stretched vertically by a factor of 2.
(c) The graph of j is the graph of i reflected about the {|-plane.
(d) The graph of j({> |) = 3i ({> |) + 2 is the graph of i reflected about the {|-plane and then shifted upward 2 units.
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SECTION 14.1
FUNCTIONS OF SEVERAL VARIABLES
¤
385
70. (a) The graph of j is the graph of i shifted 2 units in the positive {-direction.
(b) The graph of j is the graph of i shifted 2 units in the negative |-direction.
(c) The graph of j is the graph of i shifted 3 units in the negative {-direction and 4 units in the positive |-direction.
71. i ({> |) = 3{ 3 {4 3 4| 2 3 10{|
Three-dimensional view
Front view
It does appear that the function has a maximum value, at the higher of the two “hilltops.” From the front view graph, the
maximum value appears to be approximately 15. Both hilltops could be considered local maximum points, as the values of i
there are larger than at the neighboring points. There does not appear to be any local minimum point; although the valley shape
between the two peaks looks like a minimum of some kind, some neighboring points have lower function values.
2 3| 2
72. i ({> |) = {|h3{
Three-dimensional view
Front view
The function does have a maximum value, which it appears to achieve at two different points (the two “hilltops”). From the
front view graph, we can estimate the maximum value to be approximately 0=18. These same two points can also be
considered local maximum points. The two “valley bottoms” visible in the graph can be considered local minimum points, as
all the neighboring points give greater values of i .
73.
i ({> |) =
{+|
. As both { and | become large, the function values
{2 + | 2
appear to approach 0, regardless of which direction is considered. As
({> |) approaches the origin, the graph exhibits asymptotic behavior.
From some directions, i ({> |) < ", while in others i ({> |) < 3".
(These are the vertical spikes visible in the graph.) If the graph is
examined carefully, however, one can see that i({> |) approaches 0
along the line | = 3{.
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CHAPTER 14
74.
PARTIAL DERIVATIVES
i({> |) =
{|
. The graph exhibits different limiting values as { and |
{2 + | 2
become large or as ({> |) approaches the origin, depending on the direction
being examined. For example, although i is undefined at the origin, the
function values appear to be
1
2
along the line | = {, regardless of the distance
from the origin. Along the line | = 3{, the value is always 3 12 . Along the
axes, i({> |) = 0 for all values of ({> |) except the origin. Other directions,
heading toward the origin or away from the origin, give various limiting
values between 3 12 and 12 .
2 +| 2
75. i ({> |) = hf{
. First, if f = 0, the graph is the cylindrical surface
2
} = h| (whose level curves are parallel lines). When f A 0, the vertical trace
above the |-axis remains fixed while the sides of the surface in the {-direction
“curl” upward, giving the graph a shape resembling an elliptic paraboloid. The
level curves of the surface are ellipses centered at the origin.
f=0
For 0 ? f ? 1, the ellipses have major axis the {-axis and the eccentricity increases as f < 0.
f = 0=5 (level curves in increments of 1)
For f = 1 the level curves are circles centered at the origin.
f = 1 (level curves in increments of 1)
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SECTION 14.1
FUNCTIONS OF SEVERAL VARIABLES
¤
387
When f A 1, the level curves are ellipses with major axis the |-axis, and the eccentricity increases as f increases.
f = 2 (level curves in increments of 4)
For values of f ? 0, the sides of the surface in the {-direction curl downward and approach the {|-plane (while the vertical
trace { = 0 remains fixed), giving a saddle-shaped appearance to the graph near the point (0> 0> 1). The level curves consist of
a family of hyperbolas. As f decreases, the surface becomes flatter in the {-direction and the surface’s approach to the curve in
the trace { = 0 becomes steeper, as the graphs demonstrate.
f = 30=5 (level curves in increments of 0=25)
f = 32 (level curves in increments of 0=25)
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CHAPTER 14
PARTIAL DERIVATIVES
2
76. } = (d{2 + e| 2 )h3{
3| 2
. There are only three basic shapes which can be obtained (the fourth and fifth graphs are the
reflections of the first and second ones in the {|-plane). Interchanging d and e rotates the graph by 90 about the }-axis.
d = 1, e = 1
d = 2, e = 1
d = 1, e = 31
d = 31, e = 31
d = 32, e = 31
If d and e are both positive (d 6= e), we see that the graph has two maximum points whose height increases as d and e increase.
If d and e have opposite signs, the graph has two maximum points and two minimum points, and if d and e are both negative,
the graph has one maximum point and two minimum points.
77. } = {2 + | 2 + f{|. When f ? 32, the surface intersects the plane } = n 6= 0 in a hyperbola. (See the following graph.)
It intersects the plane { = | in the parabola } = (2 + f){2 , and the plane { = 3| in the parabola } = (2 3 f){2 . These
parabolas open in opposite directions, so the surface is a hyperbolic paraboloid.
When f = 32 the surface is } = {2 + | 2 3 2{| = ({ 3 |)2 . So the surface is constant along each line { 3 | = n. That
is, the surface is a cylinder with axis { 3 | = 0, } = 0. The shape of the cylinder is determined by its intersection with the
plane { + | = 0, where } = 4{2 , and hence the cylinder is parabolic with minima of 0 on the line | = {.
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SECTION 14.1
f = 35, } = 2
FUNCTIONS OF SEVERAL VARIABLES
f = 310
¤
389
f = 32
When 32 ? f $ 0, } D 0 for all { and |. If { and | have the same sign, then
{2 + | 2 + f{| D {2 + | 2 3 2{| = ({ 3 |)2 D 0. If they have opposite signs, then f{| D 0. The intersection with the
surface and the plane } = n A 0 is an ellipse (see graph below). The intersection with the surface and the planes { = 0 and
| = 0 are parabolas } = | 2 and } = {2 respectively, so the surface is an elliptic paraboloid.
When f A 0 the graphs have the same shape, but are reflected in the plane { = 0, because
{2 + | 2 + f{| = (3{)2 + | 2 + (3f)(3{)|. That is, the value of } is the same for f at ({> |) as it is for 3f at (3{> |).
f = 31, } = 2
f=0
f = 10
So the surface is an elliptic paraboloid for 0 ? f ? 2, a parabolic cylinder for f = 2, and a hyperbolic paraboloid for f A 2.
78. First, we graph i ({> |) =
s
{2 + | 2 .
i ({> |) =
s
{2 + | 2
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CHAPTER 14
PARTIAL DERIVATIVES
Graphs of the other four functions follow.
I
2
2
i ({> |) = h { + |
i ({> |) = ln
s
i ({> |) = sin
{2 + | 2
Notice that each graph i ({> |) = j
1
i ({> |) = s
{2 + | 2
s
{2 + | 2 exhibits radial symmetry about the }-axis and the trace in the {}-plane for
{ D 0 is the graph of } = j({), { D 0. This suggests that the graph of i ({> |) = j
of j by graphing } = j({) in the {}-plane and rotating the curve about the }-axis.
S
= eO N 3
N
O
S
= ln e + ln
ln
N
N
79. (a) S = eO N 13
s
{2 + | 2
i
i
O
S
=e
N
N
i ln
s
{2 + | 2 is obtained from the graph
O
S
= ln e
i
N
N
(b) We list the values for ln(O@N) and ln(S@N) for the years 1899 –1922. (Historically, these values were rounded to
2 decimal places.)
Year
{ = ln(O@N)
| = ln(S@N)
Year
{ = ln(O@N)
| = ln(S@N)
1899
1900
1901
1902
1903
1904
1905
1906
1907
1908
1909
1910
0
30=02
30=04
30=04
30=07
30=13
30=18
30=20
30=23
30=41
30=33
30=35
0
30=06
30=02
0
30=05
30=12
30=04
30=07
30=15
30=38
30=24
30=27
1911
1912
1913
1914
1915
1916
1917
1918
1919
1920
1921
1922
30=38
30=38
30=41
30=47
30=53
30=49
30=53
30=60
30=68
30=74
31=05
30=98
30=34
30=24
30=25
30=37
30=34
30=28
30=39
30=50
30=57
30=57
30=85
30=59
INSTRUCTOR
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SECTION 14.2
LIMITS AND CONTINUITY
¤
391
After entering the ({> |) pairs into a calculator or CAS, the resulting least squares regression line through the points is
approximately | = 0=75136{ + 0=01053, which we round to | = 0=75{ + 0=01.
(c) Comparing the regression line from part (b) to the equation | = ln e + { with { = ln(O@N) and | = ln(S@N), we have
= 0=75 and ln e = 0=01 i e = h0=01 E 1=01. Thus, the Cobb-Douglas production function is
S = eO N 13 = 1=01O0=75 N 0=25 .
14.2 Limits and Continuity
1. In general, we can’t say anything about i (3> 1)!
lim
({>|)<(3>1)
i ({> |) = 6 means that the values of i ({> |) approach 6 as
({> |) approaches, but is not equal to, (3> 1). If i is continuous, we know that
lim
({>|)<(3>1)
lim
({>|)<(d>e)
i ({> |) = i(d> e), so
i ({> |) = i (3> 1) = 6.
2. (a) The outdoor temperature as a function of longitude, latitude, and time is continuous. Small changes in longitude, latitude,
or time can produce only small changes in temperature, as the temperature doesn’t jump abruptly from one value to
another.
(b) Elevation is not necessarily continuous. If we think of a cliff with a sudden drop-off, a very small change in longitude or
latitude can produce a comparatively large change in elevation, without all the intermediate values being attained.
Elevation can jump from one value to another.
(c) The cost of a taxi ride is usually discontinuous. The cost normally increases in jumps, so small changes in distance traveled
or time can produce a jump in cost. A graph of the function would show breaks in the surface.
3. We make a table of values of
i ({> |) =
{2 | 3 + {3 | 2 3 5
for a set
2 3 {|
of ({> |) points near the origin.
As the table shows, the values of i ({> |) seem to approach 32=5 as ({> |) approaches the origin from a variety of different
directions. This suggests that
lim
({>|)<(0>0)
i ({> |) = 32=5. Since i is a rational function, it is continuous on its domain. i is
defined at (0> 0), so we can use direct substitution to establish that
lim
({>|)<(0>0)
i ({> |) =
our guess.
02 03 + 03 02 3 5
5
= 3 , verifying
230·0
2
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¤
CHAPTER 14
PARTIAL DERIVATIVES
4. We make a table of values of
i ({> |) =
2{|
for a set of ({> |)
{2 + 2| 2
points near the origin.
It appears from the table that the values of i ({> |) are not approaching a single value as ({> |) approaches the origin. For
verification, if we first approach (0> 0) along the {-axis, we have i ({> 0) = 0, so i ({> |) < 0. But if we approach (0> 0) along
the line | = {, i ({> {) =
2{2
2
= ({ 6= 0), so i ({> |) < 23 . Since i approaches different values along different paths
{2 + 2{2
3
to the origin, this limit does not exist.
5. i ({> |) = 5{3 3 {2 | 2 is a polynomial, and hence continuous, so
lim
({>|)<(1>2)
i ({> |) = i (1> 2) = 5(1)3 3 (1)2 (2)2 = 1.
6. 3{| is a polynomial and therefore continuous. Since hw is a continuous function, the composition h3{| is also continuous.
Similarly, { + | is a polynomial and cos w is a continuous function, so the composition cos({ + |) is continuous.
The product of continuous functions is continuous, so i ({> |) = h3{| cos({ + |) is a continuous function and
lim
({>|)<(1>31)
7. i ({> |) =
i ({> |) = i (1> 31) = h3(1)(31) cos(1 + (31)) = h1 cos 0 = h.
4 3 {|
is a rational function and hence continuous on its domain.
{2 + 3| 2
(2> 1) is in the domain of i , so i is continuous there and
8.
lim
({>|)<(2>1)
i ({> |) = i (2> 1) =
4 3 (2)(1)
2
= .
(2)2 + 3(1)2
7
1 + |2
is a rational function and hence continuous on its domain, which includes (1> 0). ln w is a continuous function for
{2 + {|
1 + |2
1 + |2
w A 0, so the composition i({> |) = ln
is continuous wherever 2
A 0. In particular, i is continuous at
2
{ + {|
{ + {|
1 + 02
1
(1> 0) and so
lim
= ln = 0.
i ({> |) = i (1> 0) = ln
({>|)<(1>0)
12 + 1 · 0
1
9. i ({> |) = ({4 3 4| 2 )@({2 + 2| 2 ).
First approach (0> 0) along the {-axis. Then i ({> 0) = {4 @{2 = {2 for { 6= 0, so
i ({> |) < 0. Now approach (0> 0) along the |-axis. For | 6= 0, i(0> |) = 34| 2 @2| 2 = 32, so i ({> |) < 32. Since i has
two different limits along two different lines, the limit does not exist.
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SECTION 14.2
10. i ({> |) = (5| 4 cos2 {)@({4 + | 4 ).
LIMITS AND CONTINUITY
¤
393
First approach (0> 0) along the {-axis. Then i ({> 0) = 0@{4 = 0 for { 6= 0, so
i ({> |) < 0. Next approach (0> 0) along the |-axis. For | 6= 0, i (0> |) = 5|4 @| 4 = 5, so i ({> |) < 5. Since i has two
different limits along two different lines, the limit does not exist.
11. i ({> |) = (| 2 sin2 {)@({4 + | 4 ).
On the {-axis, i ({> 0) = 0 for { 6= 0, so i ({> |) < 0 as ({> |) < (0> 0) along the
{-axis. Approaching (0> 0) along the line | = {, i ({> {) =
lim
{<0
{2 sin2 {
sin2 {
1
=
=
4
4
{ +{
2{2
2
sin {
{
2
for { 6= 0 and
sin {
= 1, so i ({> |) < 12 . Since i has two different limits along two different lines, the limit does not exist.
{
12. i ({> |) =
{| 3 |
. On the {-axis, i ({> 0) = 0@({ 3 1)2 = 0 for { 6= 1, so i({> |) < 0 as ({> |) < (1> 0) along
({ 3 1)2 + | 2
the {-axis. Approaching (1> 0) along the line | = { 3 1, i ({> { 3 1) =
1
2
so i ({> |) <
({ 3 1)2
1
{({ 3 1) 3 ({ 3 1)
=
= for { 6= 1,
2
2
({ 3 1) + ({ 3 1)
2({ 3 1)2
2
along this line. Thus the limit does not exist.
{|
. We can see that the limit along any line through (0> 0) is 0, as well as along other paths through
{2 + | 2
13. i ({> |) = s
(0> 0) such as { = | 2 and | = {2 . So we suspect that the limit exists and equals 0; we use the Squeeze Theorem to prove our
s
{|
assertion. 0 $ s
lim
i ({> |) = 0.
$ |{| since ||| $ {2 + | 2 , and |{| < 0 as ({> |) < (0> 0). So
{2 + | 2 ({>|)<(0>0)
14. i ({> |) =
{4 3 | 4
({2 + | 2 )({2 3 | 2 )
=
= {2 3 | 2 for ({> |) 6= (0> 0). Thus the limit as ({> |) < (0> 0) is 0.
{2 + | 2
{2 + | 2
15. Let i ({> |) =
{2 |h|
. Then i ({> 0) = 0 for { 6= 0, so i ({> |) < 0 as ({> |) < (0> 0) along the {-axis. Approaching
+ 4| 2
{4
(0> 0) along the |-axis or the line | = { also gives a limit of 0. But i {> {2 =
i ({> |) < h0 @5 =
1
5
17.
2
2
as ({> |) < (0> 0) along the parabola | = {2 . Thus the limit doesn’t exist.
16. We can use the Squeeze Theorem to show that
0$
2
{2 {2 h{
{4 h{
h{
for { 6= 0, so
=
=
{4 + 4({2 )2
5{4
5
{2 sin2 |
= 0:
({>|)<(0>0) {2 + 2| 2
lim
{2 sin2 |
{2
{2 sin2 |
2
2
$
sin
|
since
$
1,
and
sin
|
<
0
as
({>
|)
<
(0>
0),
so
lim
= 0.
({>|)<(0>0) {2 + 2| 2
{2 + 2| 2
{2 + 2| 2
lim
({>|)<(0>0)
s
{2 + | 2 + 1 + 1
{2 + | 2
{2 + | 2
s
s
=
lim
·s
2
2
2
2
({>|)<(0>0)
{ +| +131
{ +| +131
{2 + | 2 + 1 + 1
s
2
s
{ + |2
{2 + | 2 + 1 + 1
=
lim
=
lim
{2 + | 2 + 1 + 1 = 2
2
2
({>|)<(0>0)
({>|)<(0>0)
{ +|
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CHAPTER 14
PARTIAL DERIVATIVES
18. i ({> |) = {| 4 @({2 + | 8 ).
On the {-axis, i({> 0) = 0 for { 6= 0, so i({> |) < 0 as ({> |) < (0> 0) along the {-axis.
Approaching (0> 0) along the curve { = | 4 gives i (| 4 > |) = | 8 @2| 8 =
1
2
for | 6= 0, so along this path i ({> |) <
1
2
as
({> |) < (0> 0). Thus the limit does not exist.
2
19. h| is a composition of continuous functions and hence continuous. {} is a continuous function and tan w is continuous for
w 6=
2
+ q (q an integer), so the composition tan({}) is continuous for {} 6=
2
i ({> |> }) = h| tan({}) is a continuous function for {} 6=
lim
+ q. Thus the product
+ q. If { = and } =
2
({>|>})<(>0>1@3)
20. i ({> |> }) =
2
2
i ({> |> }) = i (> 0> 1@3) = h0 tan( · 1@3) = 1 · tan(@3) =
1
3
then {} 6=
2
+ q, so
I
3.
{| + |}
. Then i ({> 0> 0) = 0@{2 = 0 for { 6= 0, so as ({> |> }) < (0> 0> 0) along the {-axis,
{2 + | 2 + } 2
i ({> |> }) < 0. But i({> {> 0) = {2 @(2{2 ) =
1
2
for { 6= 0, so as ({> |> }) < (0> 0> 0) along the line | = {, } = 0,
i ({> |> }) < 12 . Thus the limit doesn’t exist.
21. i ({> |> }) =
{| + |} 2 + {} 2
. Then i({> 0> 0) = 0@{2 = 0 for { 6= 0, so as ({> |> }) < (0> 0> 0) along the {-axis,
{2 + | 2 + } 4
i ({> |> }) < 0. But i({> {> 0) = {2 @(2{2 ) =
1
2
for { 6= 0, so as ({> |> }) < (0> 0> 0) along the line | = {, } = 0,
i ({> |> }) < 12 . Thus the limit doesn’t exist.
22. i ({> |> }) =
|}
. Then i ({> 0> 0) = 0 for { 6= 0, so as ({> |> }) < (0> 0> 0) along the {-axis, i ({> |> }) < 0.
{2 + 4| 2 + 9} 2
But i (0> |> |) = | 2 @(13| 2 ) =
1
13
for | 6= 0, so as ({> |> }) < (0> 0> 0) along the line } = |, { = 0, i({> |> }) <
1
.
13
Thus the limit doesn’t exist.
23.
From the ridges on the graph, we see that as ({> |) < (0> 0) along the
lines under the two ridges, i ({> |) approaches different values. So the
limit does not exist.
24.
From the graph, it appears that as we approach the origin along the lines
{ = 0 or | = 0, the function is everywhere 0, whereas if we approach the
origin along a certain curve it has a constant value of about 12 . [In fact,
i (| 3 > |) = | 6@(2| 6 ) =
1
2
for | 6= 0, so i({> |) <
1
2
as ({> |) < (0> 0)
along the curve { = | 3 .] Since the function approaches different values
depending on the path of approach, the limit does not exist.
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SECTION 14.2
2
25. k({> |) = j(i ({> |)) = (2{ + 3| 3 6) +
LIMITS AND CONTINUITY
¤
395
I
2{ + 3| 3 6. Since i is a polynomial, it is continuous on R2 and j is
continuous on its domain {w | w D 0}. Thus k is continuous on its domain.
G = {({> |) | 2{ + 3| 3 6 D 0} = ({> |) | | D 3 23 { + 2 , which consists of all points on or above the line | = 3 23 { + 2.
26. k({> |) = j(i ({> |)) =
1 3 {|
1 3 {|
+
ln
. i is a rational function, so it is continuous on its domain. Because
1 + {2 | 2
1 + {2 | 2
1 + {2 | 2 A 0, the domain of i is R2 , so i is continuous everywhere. j is continuous on its domain {w | w A 0}. Thus k is
1 3 {|
continuous on its domain ({> |) A 0 = {({> |) | {| ? 1} which consists of all points between (but not on)
1 + {2 | 2
the two branches of the hyperbola | = 1@{.
27.
From the graph, it appears that i is discontinuous along the line | = {.
If we consider i({> |) = h1@({3|) as a composition of functions,
j({> |) = 1@({ 3 |) is a rational function and therefore continuous except
where { 3 | = 0 i | = {. Since the function k(w) = hw is continuous
everywhere, the composition k(j({> |)) = h1@({3|) = i ({> |) is
continuous except along the line | = {, as we suspected.
28.
We can see a circular break in the graph, corresponding approximately to
the unit circle, where i is discontinuous. Sincei({> |) =
1
is
1 3 {2 3 | 2
a rational function, it is continuous except where 1 3 {2 3 | 2 = 0 i
{2 + | 2 = 1, confirming our observation that i is discontinuous on the
circle {2 + | 2 = 1.
29. The functions {| and 1 + h{3| are continuous everywhere, and 1 + h{3| is never zero, so I ({> |) =
{|
is continuous
1 + h{3|
on its domain R2 .
30. I ({> |) = cos
I
I
1 + { 3 | = j(i ({> |)) where i ({> |) = 1 + { 3 |, continuous on its domain
{({> |) | 1 + { 3 | D 0} = {({> |) | | $ { + 1}, and j(w) = cos w is continuous everywhere. Thus I is continuous on its
domain {({> |) | | $ { + 1}.
31. I ({> |) =
1 + {2 + | 2
is a rational function and thus is continuous on its domain
1 3 {2 3 | 2
({> |) | 1 3 {2 3 | 2 6= 0 = ({> |) | {2 + | 2 6= 1 .
32. The functions h{ + h| and h{| 3 1 are continuous everywhere, so K({> |) =
h{| 3 1 = 0
i
{| = 0
i
h{ + h|
is continuous except where
h{| 3 1
{ = 0 or | = 0. Thus K is continuous on its domain {({> |) | { 6= 0> | 6= 0}.
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CHAPTER 14
PARTIAL DERIVATIVES
33. J({> |) = ln({2 + | 2 3 4) = j(i ({> |)) where i ({> |) = {2 + | 2 3 4, continuous on R2 , and j(w) = ln w, continuous on its
domain {w | w A 0}. Thus J is continuous on its domain ({> |) | {2 + | 2 3 4 A 0 = ({> |) | {2 + | 2 A 4 , the exterior
of the circle {2 + | 2 = 4.
34. J({> |) = j(i({> |)) where i ({> |) = ({ + |)32 , a rational function that is continuous on R2 except where { + | = 0, and
j(w) = tan31 w, continuous everywhere. Thus J is continuous on its domain {({> |) | { + | 6= 0} = {({> |) | | 6= 3{}.
35. i ({> |> }) = k(j({> |> })) where j({> |> }) = {2 + | 2 + } 2 , a polynomial that is continuous
everywhere, and k(w) = arcsin w, continuous on [31> 1]. Thus i is continuous on its domain
({> |> }) | 31 $ {2 + | 2 + } 2 $ 1 = ({> |> }) | {2 + | 2 + } 2 $ 1 , so i is continuous on the unit ball.
36.
s
| 3 {2 is continuous on its domain ({> |) | | 3 {2 D 0 = ({> |) | | D {2 and ln } is continuous on its domain
s
{} | } A 0}, so the product i ({> |> }) = | 3 {2 ln } is continuous for | D {2 and } A 0, that is,
({> |> }) | | D {2 , } A 0 .
;
A
?
37. i ({> |) =
{2 | 3
2{2 + | 2
A
=1
if ({> |) 6= (0> 0)
The first piece of i is a rational function defined everywhere except at the
if ({> |) = (0> 0)
origin, so i is continuous on R2 except possibly at the origin. Since {2 $ 2{2 + | 2 , we have {2 | 3@(2{2 + | 2 ) $ | 3 . We
know that | 3 < 0 as ({> |) < (0> 0). So, by the Squeeze Theorem,
lim
({>|)<(0>0)
i ({> |) =
{2 | 3
= 0.
({>|)<(0>0) 2{2 + | 2
lim
But i (0> 0) = 1, so i is discontinuous at (0> 0). Therefore, i is continuous on the set {({> |) | ({> |) 6= (0> 0)}.
;
?
38. i ({> |) =
{|
{2 + {| + |2
if ({> |) 6= (0> 0)
=
0
The first piece of i is a rational function defined everywhere except
if ({> |) = (0> 0)
at the origin, so i is continuous on R2 except possibly at the origin. i({> 0) = 0@{2 = 0 for { 6= 0, so i ({> |) < 0 as
({> |) < (0> 0) along the {-axis. But i ({> {) = {2 @(3{2 ) =
line | = {. Thus
lim
({>|)<(0>0)
1
3
for { 6= 0, so i ({> |) <
1
3
as ({> |) < (0> 0) along the
i ({> |) doesn’t exist, so i is not continuous at (0> 0) and the largest set on which i is continuous
is {({> |) | ({> |) 6= (0> 0)}.
39.
40.
lim
({>|)<(0>0)
lim
(u cos )3 + (u sin )3
{3 + | 3
= lim
= lim (u cos3 + u sin3 ) = 0
2
2
+
{ +|
u2
u<0
u<0+
({2 + | 2 ) ln({2 + | 2 ) = lim u2 ln u2 = lim
u<0+
({>|)<(0>0)
u<0+
ln u2
(1@u2 )(2u)
=
lim
1@u2
32@u3
u<0+
[using l’Hospital’s Rule]
= lim (3u2 ) = 0
u<0+
2
41.
lim
({>|)<(0>0)
2
2
2
h3{ 3| 3 1
h3u 3 1
h3u (32u)
= lim
= lim
2
2
2
+
+
{ +|
u
2u
u<0
u<0
[using l’Hospital’s Rule]
2
= lim 3h3u = 3h0 = 31
u<0+
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SECTION 14.2
42.
lim
({>|)<(0>0)
LIMITS AND CONTINUITY
¤
397
sin({2 + | 2 )
sin(u2 )
= lim
, which is an
2
2
+
{ +|
u2
u<0
indeterminate form of type 0@0. Using l’Hospital’s Rule, we get
lim
u<0+
sin(u2 ) H
2u cos(u2 )
= lim cos(u2 ) = 1.
= lim
2
u
2u
u<0+
u<0+
Or: Use the fact that lim
<0
;
? sin({|)
{|
43. i ({> |) =
=
1
sin = 1.
if ({> |) 6= (0> 0)
if ({> |) = (0> 0)
From the graph, it appears that i is continuous everywhere. We know
{| is continuous on R2 and sin w is continuous everywhere, so
sin({|) is continuous on R2 and
sin({|)
is continuous on R2
{|
except possibly where {| = 0. To show that i is continuous at those points, consider any point (d> e) in R2 where de = 0.
Because {| is continuous, {| < de = 0 as ({> |) < (d> e). If we let w = {|, then w < 0 as ({> |) < (d> e) and
lim
({>|)<(d>e)
sin({|)
sin(w)
= lim
= 1 by Equation 2.4.2 [ET 3.3.2]. Thus
lim
i ({> |) = i (d> e) and i is continuous
w<0
({>|)<(d>e)
{|
w
on R2 .
44. (a) i ({> |) =
+
0
1
if | $ 0 or | D {4
if 0 ? | ? {4
Consider the path | = p{d , 0 ? d ? 4. [The path does not pass through
(0> 0) if d $ 0 except for the trivial case where p = 0.] If p{d $ 0 then i ({> p{d ) = 0. If p{d A 0 then
p{d = |p{d | = |p| |{d | and p{d D {4
defined. Then p{d D {4
C |p| |{d | D {4
C
{4
$ |p| C |{|43d $ |p| whenever {d is
|{d |
C |{| $ |p|1@(43d) so i ({> p{d ) = 0 for |{| $ |p|1@(43d) and i ({> |) < 0 as
({> |) < (0> 0) along this path.
(b) If we approach (0> 0) along the path | = {5 , { A 0 then we have i ({> {5 ) = 1 for 0 ? { ? 1 because 0 ? {5 ? {4 there.
Thus i({> |) < 1 as ({> |) < (0> 0) along this path, but in part (a) we found a limit of 0 along other paths, so
lim
({>|)<(0>0)
i ({> |) doesn’t exist and i is discontinuous at (0> 0).
(c) First we show that i is discontinuous at any point (d> 0) on the {-axis. If we approach (d> 0) along the path { = d, | A 0
then i (d> |) = 1 for 0 ? | ? d4 , so i ({> |) < 1 as ({> |) < (d> 0) along this path. If we approach (d> 0) along the path
{ = d, | ? 0 then i (d> |) = 0 since | ? 0 and i ({> |) < 0 as ({> |) < (d> 0). Thus the limit does not exist and i is
discontinuous on the line | = 0. i is also discontinuous on the curve | = {4 : For any point (d> d4 ) on this curve,
approaching the point along the path { = d, | A d4 gives i (d> |) = 0 since | A d4 , so i ({> |) < 0 as ({> |) < (d> d4 ).
But approaching the point along the path { = d, | ? d4 gives i (d> |) = 1 for | A 0, so i ({> |) < 1 as ({> |) < (d> d4 )
and the limit does not exist there.
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CHAPTER 14
PARTIAL DERIVATIVES
45. Since |x 3 a| = |x| + |a| 3 2 |x| |a| cos D |x| + |a| 3 2 |x| |a| = (|x| 3 |a|) , we have |x| 3 |a| $ |x 3 a|. Let
2
2
2
2
2
2
A 0 be given and set = . Then if 0 ? |x 3 a| ? , |x| 3 |a| $ |x 3 a| ? = . Hence limx<a |x| = |a| and
i (x) = |x| is continuous on Rq .
46. Let A 0 be given. We need to find A 0 such that if 0 ? |x 3 a| ? then |i (x) 3 i (a)| = |c · x 3 c · a| ? .
But |c · x 3 c · a| = |c · (x 3 a)| and |c · (x 3 a)| $ |c| |x 3 a| by Exercise 12.3.61 (the Cauchy-Schwartz Inequality). Set
= @ |c|. Then if 0 ? |x 3 a| ? , |i (x) 3 i (a)| = |c · x 3 c · a| $ |c| |x 3 a| ? |c| = |c| (@ |c|) = . So i is
continuous on Rq .
14.3 Partial Derivatives
1. (a) CW @C{ represents the rate of change of W when we fix | and w and consider W as a function of the single variable {, which
describes how quickly the temperature changes when longitude changes but latitude and time are constant. CW @C|
represents the rate of change of W when we fix { and w and consider W as a function of |, which describes how quickly the
temperature changes when latitude changes but longitude and time are constant. CW @Cw represents the rate of change of W
when we fix { and | and consider W as a function of w, which describes how quickly the temperature changes over time for
a constant longitude and latitude.
(b) i{ (158> 21> 9) represents the rate of change of temperature at longitude 158 W, latitude 21 N at 9:00 AM when only
longitude varies. Since the air is warmer to the west than to the east, increasing longitude results in an increased air
temperature, so we would expect i{ (158> 21> 9) to be positive. i| (158> 21> 9) represents the rate of change of temperature
at the same time and location when only latitude varies. Since the air is warmer to the south and cooler to the north,
increasing latitude results in a decreased air temperature, so we would expect i| (158> 21> 9) to be negative. iw (158> 21> 9)
represents the rate of change of temperature at the same time and location when only time varies. Since typically air
temperature increases from the morning to the afternoon as the sun warms it, we would expect iw (158> 21> 9) to be
positive.
2. By Definition 4, iW (92> 60) = lim
k<0
i (92 + k> 60) 3 i (92> 60)
, which we can approximate by considering k = 2 and
k
k = 32 and using the values given in Table 1: iW (92> 60) E
iW (92> 60) E
111 3 105
i (94> 60) 3 i (92> 60)
=
= 3,
2
2
i (90> 60) 3 i (92> 60)
100 3 105
=
= 2=5. Averaging these values, we estimate iW (92> 60) to be
32
32
approximately 2=75. Thus, when the actual temperature is 92 F and the relative humidity is 60%, the apparent temperature
rises by about 2=75 F for every degree that the actual temperature rises.
Similarly, iK (92> 60) = lim
k<0
iK (92> 60) E
i (92> 60 + k) 3 i (92> 60)
which we can approximate by considering k = 5 and k = 35:
k
108 3 105
103 3 105
i(92> 65) 3 i (92> 60)
i (92> 55) 3 i(92> 60)
=
= 0=6, iK (92> 60) E
=
= 0=4.
5
5
35
35
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SECTION 14.3
PARTIAL DERIVATIVES
¤
399
Averaging these values, we estimate iK (92> 60) to be approximately 0=5. Thus, when the actual temperature is 92 F and the
relative humidity is 60%, the apparent temperature rises by about 0=5 F for every percent that the relative humidity increases.
3. (a) By Definition 4, iW (315> 30) = lim
k<0
i (315 + k> 30) 3 i (315> 30)
, which we can approximate by considering k = 5
k
and k = 35 and using the values given in the table:
iW (315> 30) E
320 3 (326)
6
i (310> 30) 3 i (315> 30)
=
= = 1=2,
5
5
5
iW (315> 30) E
333 3 (326)
37
i (320> 30) 3 i (315> 30)
=
=
= 1=4. Averaging these values, we estimate
35
35
35
iW (315> 30) to be approximately 1=3. Thus, when the actual temperature is 315 C and the wind speed is 30 km@h, the
apparent temperature rises by about 1=3 C for every degree that the actual temperature rises.
Similarly, iy (315> 30) = lim
k<0
and k = 310: iy (315> 30) E
iy (315> 30) E
i (315> 30 + k) 3 i (315> 30)
which we can approximate by considering k = 10
k
327 3 (326)
31
i(315> 40) 3 i (315> 30)
=
=
= 30=1,
10
10
10
324 3 (326)
2
i (315> 20) 3 i (315> 30)
=
=
= 30=2. Averaging these values, we estimate
310
310
310
iy (315> 30) to be approximately 30=15. Thus, when the actual temperature is 315 C and the wind speed is 30 km@h, the
apparent temperature decreases by about 0=15 C for every km@h that the wind speed increases.
(b) For a fixed wind speed y, the values of the wind-chill index Z increase as temperature W increases (look at a column of
the table), so
CZ
is positive. For a fixed temperature W , the values of Z decrease (or remain constant) as y increases
CW
(look at a row of the table), so
CZ
is negative (or perhaps 0).
Cy
(c) For fixed values of W , the function values i (W> y) appear to become constant (or nearly constant) as y increases, so the
corresponding rate of change is 0 or near 0 as y increases. This suggests that lim (CZ@Cy) = 0.
y<"
4. (a) Ck@Cy represents the rate of change of k when we fix w and consider k as a function of y, which describes how quickly the
wave heights change when the wind speed changes for a fixed time duration. Ck@Cw represents the rate of change of k
when we fix y and consider k as a function of w, which describes how quickly the wave heights change when the duration
of time changes, but the wind speed is constant.
(b) By Definition 4, iy (40> 15) = lim
k<0
i (40 + k> 15) 3 i (40> 15)
which we can approximate by considering
k
k = 10 and k = 310 and using the values given in the table: iy (40> 15) E
iy (40> 15) E
36 3 25
i (50> 15) 3 i(40> 15)
=
= 1=1,
10
10
16 3 25
i (30> 15) 3 i (40> 15)
=
= 0=9. Averaging these values, we have iy (40> 15) E 1=0. Thus, when
310
310
a 40-knot wind has been blowing for 15 hours, the wave heights should increase by about 1 foot for every knot that the
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¤
CHAPTER 14
PARTIAL DERIVATIVES
wind speed increases (with the same time duration). Similarly, iw (40> 15) = lim
k<0
can approximate by considering k = 5 and k = 35: iw (40> 15) E
iw (40> 15) E
i (40> 15 + k) 3 i (40> 15)
which we
k
28 3 25
i(40> 20) 3 i (40> 15)
=
= 0=6,
5
5
21 3 25
i (40> 10) 3 i (40> 15)
=
= 0=8. Averaging these values, we have iw (40> 15) E 0=7. Thus, when a
35
35
40-knot wind has been blowing for 15 hours, the wave heights increase by about 0=7 feet for every additional hour that the
wind blows.
(c) For fixed values of y, the function values i(y> w) appear to increase in smaller and smaller increments, becoming nearly
constant as w increases. Thus, the corresponding rate of change is nearly 0 as w increases, suggesting that
lim (Ck@Cw) = 0.
w<"
5. (a) If we start at (1> 2) and move in the positive {-direction, the graph of i increases. Thus i{ (1> 2) is positive.
(b) If we start at (1> 2) and move in the positive |-direction, the graph of i decreases. Thus i| (1> 2) is negative.
6. (a) The graph of i decreases if we start at (31> 2) and move in the positive {-direction, so i{ (31> 2) is negative.
(b) The graph of i decreases if we start at (31> 2) and move in the positive |-direction, so i| (31> 2) is negative.
7. (a) i{{ =
C
(i{ ),
C{
so i{{ is the rate of change of i{ in the {-direction. i{ is negative at (31> 2) and if we move in the
positive {-direction, the surface becomes less steep. Thus the values of i{ are increasing and i{{ (31> 2) is positive.
(b) i|| is the rate of change of i| in the |-direction. i| is negative at (31> 2) and if we move in the positive |-direction, the
surface becomes steeper. Thus the values of i| are decreasing, and i|| (31> 2) is negative.
8. (a) i{| =
C
(i{ ),
C|
so i{| is the rate of change of i{ in the |-direction. i{ is positive at (1> 2) and if we move in the positive
|-direction, the surface becomes steeper, looking in the positive {-direction. Thus the values of i{ are increasing and
i{| (1> 2) is positive.
(b) i{ is negative at (31> 2) and if we move in the positive |-direction, the surface gets steeper (with negative slope), looking
in the positive {-direction. This means that the values of i{ are decreasing as | increases, so i{| (31> 2) is negative.
9. First of all, if we start at the point (3> 33) and move in the positive |-direction, we see that both e and f decrease, while d
increases. Both e and f have a low point at about (3> 31=5), while d is 0 at this point. So d is definitely the graph of i| , and
one of e and f is the graph of i . To see which is which, we start at the point (33> 31=5) and move in the positive {-direction.
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SECTION 14.3
PARTIAL DERIVATIVES
¤
401
e traces out a line with negative slope, while f traces out a parabola opening downward. This tells us that e is the {-derivative
of f. So f is the graph of i, e is the graph of i{ , and d is the graph of i| .
10. i{ (2> 1) is the rate of change of i at (2> 1) in the {-direction. If we start at (2> 1), where i (2> 1) = 10, and move in the
positive {-direction, we reach the next contour line [where i ({> |) = 12] after approximately 0=6 units. This represents an
average rate of change of about
2
.
0=6
If we approach the point (2> 1) from the left (moving in the positive {-direction) the
output values increase from 8 to 10 with an increase in { of approximately 0=9 units, corresponding to an average rate of
change of
2
0=9 .
A good estimate for i{ (2> 1) would be the average of these two, so i{ (2> 1) E 2=8. Similarly, i| (2> 1) is the
rate of change of i at (2> 1) in the |-direction. If we approach (2> 1) from below, the output values decrease from 12 to 10 with
a change in | of approximately 1 unit, corresponding to an average rate of change of 32. If we start at (2> 1) and move in the
positive |-direction, the output values decrease from 10 to 8 after approximately 0.9 units, a rate of change of
32
.
0=9
Averaging
these two results, we estimate i| (2> 1) E 32=1.
11. i ({> |) = 16 3 4{2 3 | 2
i i{ ({> |) = 38{ and i| ({> |) = 32|
i i{ (1> 2) = 38 and i| (1> 2) = 34. The graph
of i is the paraboloid } = 16 3 4{2 3 | 2 and the vertical plane | = 2 intersects it in the parabola } = 12 3 4{2 , | = 2
(the curve F1 in the first figure). The slope of the tangent line
to this parabola at (1> 2> 8) is i{ (1> 2) = 38. Similarly the
plane { = 1 intersects the paraboloid in the parabola
} = 12 3 | 2 , { = 1 (the curve F2 in the second figure) and
the slope of the tangent line at (1> 2> 8) is i| (1> 2) = 34.
12. i ({> |) = (4 3 {2 3 4| 2 )1@2
i i{ ({> |) = 3{(4 3 {2 3 4| 2 )31@2 and i| ({> |) = 34|(4 3 {2 3 4| 2 )31@2
i
i{ (1> 0) = 3 I13 , i| (1> 0) = 0. The graph of i is the upper half of the ellipsoid } 2 + {2 + 4| 2 = 4 and the plane | = 0
intersects the graph in the semicircle {2 + } 2 = 4, } D 0 and the slope of the tangent line W1 to this semicircle
I at 1> 0> 3 is i{ (1> 0) = 3 I13 . Similarly the plane { = 1
intersects the graph in the semi-ellipse } 2 + 4| 2 = 3, } D 0
and the slope of the tangent line W2 to this semi-ellipse at
I 1> 0> 3 is i| (1> 0) = 0.
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¤
CHAPTER 14
13. i ({> |) = {2 | 3
PARTIAL DERIVATIVES
i i{ = 2{| 3 , i| = 3{2 | 2
Note that traces of i in planes parallel to the {}-plane are parabolas which open downward for | ? 0 and upward for | A 0,
and the traces of i{ in these planes are straight lines, which have negative slopes for | ? 0 and positive slopes for | A 0. The
traces of i in planes parallel to the |}-plane are cubic curves, and the traces of i| in these planes are parabolas.
14. i ({> |) =
i| =
|
1 + {2 | 2
i i{ =
(1 + {2 | 2 )(0) 3 |(2{| 2 )
2{| 3
=3
,
2
2
2
(1 + { | )
(1 + {2 | 2 )2
(1 + {2 | 2 )(1) 3 |(2{2 |)
1 3 {2 | 2
=
2
2
2
(1 + { | )
(1 + {2 | 2 )2
Note that traces of i in planes parallel to the {}-plane have only one extreme value (a minimum for | ? 0, a maximum for
| A 0), and the traces of i{ in these planes have only one zero (going from negative to positive if | ? 0 and from positive to
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SECTION 14.3
PARTIAL DERIVATIVES
¤
negative if | A 0). The traces of i in planes parallel to the |}-plane have two extreme values, and the traces of i| in these
planes have two zeros.
15. i ({> |) = | 5 3 3{|
i i{ ({> |) = 0 3 3| = 33|, i| ({> |) = 5| 4 3 3{
16. i ({> |) = {4 | 3 + 8{2 |
i
i{ ({> |) = 4{3 · | 3 + 8 · 2{ · | = 4{3 | 3 + 16{|, i| ({> |) = {4 · 3| 2 + 8{2 · 1 = 3{4 | 2 + 8{2
17. i ({> w) = h3w cos {
18. i ({> w) =
I
{ ln w
i i{ ({> w) = h3w (3 sin {) () = 3h3w sin {, iw ({> w) = h3w (31) cos { = 3h3w cos {
i
I
I 1 I
i{ ({> w) = 12 {31@2 ln w = (ln w)@(2 {), iw ({> w) = { · = {@w
w
C}
C}
= 10(2{ + 3|)9 · 2 = 20(2{ + 3|)9 ,
= 10(2{ + 3|)9 · 3 = 30(2{ + 3|)9
C{
C|
19. } = (2{ + 3|)10
i
20. } = tan {|
C}
C}
= (sec2 {|)(|) = | sec2 {|,
= (sec2 {|)({) = { sec2 {|
C{
C|
i
21. i ({> |) = {@| = {| 31
22. i ({> |) =
i| ({> |) =
23. i ({> |) =
i| ({> |) =
24. z =
{
({ + |)2
i i{ ({> |) = | 31 = 1@|, i| ({> |) = 3{| 32 = 3{@| 2
i i{ ({> |) =
({ + |)2 (1) 3 ({)(2)({ + |)
{ + | 3 2{
|3{
=
=
,
[({ + |)2 ]2
({ + |)3
({ + |)3
({ + |)2 (0) 3 ({)(2)({ + |)
2{
=3
[({ + |)2 ]2
({ + |)3
d{ + e|
f{ + g|
i i{ ({> |) =
(f{ + g|)(d) 3 (d{ + e|)(f)
(dg 3 ef)|
=
,
(f{ + g|)2
(f{ + g|)2
(f{ + g|)(e) 3 (d{ + e|)(g)
(ef 3 dg){
=
(f{ + g|)2
(f{ + g|)2
hy
x + y2
i
0(x + y 2 ) 3 hy (1)
hy (x + y 2 ) 3 hy (2y)
Cz
hy
Cz
hy (x + y 2 3 2y)
=
=
=3
,
=
2
2
2
2
2
2
Cx
(x + y )
(x + y ) Cy
(x + y )
(x + y 2 )2
25. j(x> y) = (x2 y 3 y 3 )5
i jx (x> y) = 5(x2 y 3 y 3 )4 · 2xy = 10xy(x2 y 3 y 3 )4 ,
jy (x> y) = 5(x2 y 3 y 3 )4 (x2 3 3y 2 ) = 5(x2 3 3y 2 )(x2 y 3 y 3 )4
26. x(u> ) = sin(u cos )
i xu (u> ) = cos(u cos ) · cos = cos cos(u cos ),
x (u> ) = cos(u cos )(3u sin ) = 3u sin cos(u cos )
27. U(s> t) = tan31 (st 2 )
28. i ({> |) = {|
i Us (s> t) =
1
t2
1
2st
· t2 =
, Ut (s> t) =
· 2st =
2
2
1 + (st )
1 + s2 t 4
1 + (st 2 )2
1 + s2 t 4
i i{ ({> |) = |{|31 , i| ({> |) = {| ln {
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404
¤
CHAPTER 14
29. I ({> |) =
]
{
|
I| ({> |) =
30. I (> ) =
cos(hw ) gw i I{ ({> |) =
C
C|
]
s
w3 + 1 gw i
{
|
C
C
]
]
C
C{
|
{
cos hw gw = cos(h{ ) by the Fundamental Theorem of Calculus, Part 1;
] |
] |
C
C
cos hw gw =
cos hw gw = 3
cos hw gw = 3 cos(h| ).
3
C|
C|
{
{
]
I (> ) =
PARTIAL DERIVATIVES
] s
] s
s
s
C
C
3
w3 + 1 gw =
w3 + 1 gw = 3
w3 + 1 gw = 3 3 + 1 by the Fundamental
C
C
Theorem of Calculus, Part 1; I (> ) =
31. i ({> |> }) = {} 3 5{2 | 3 } 4
C
C
]
t
s
w3 + 1 gw = 3 + 1.
i i{ ({> |> }) = } 3 10{| 3 } 4 , i| ({> |> }) = 315{2 | 2 } 4 , i} ({> |> }) = { 3 20{2 | 3 } 3
32. i ({> |> }) = { sin(| 3 })
i i{ ({> |> }) = sin(| 3 }), i| ({> |> }) = { cos(| 3 }),
i} ({> |> }) = { cos(| 3 })(31) = 3{ cos(| 3 })
33. z = ln({ + 2| + 3})
34. z = }h{|}
1
Cz
2
Cz
3
Cz
=
,
=
,
=
C{
{ + 2| + 3} C|
{ + 2| + 3} C}
{ + 2| + 3}
i
i
Cz
Cz
Cz
= }h{|} · |} = |} 2 h{|} ,
= }h{|} · {} = {} 2 h{|} ,
= }h{|} · {| + h{|} · 1 = ({|} + 1)h{|}
C{
C|
C}
35. x = {| sin31 (|})
i
Cx
Cx
{|}
1
(}) + sin31 (|}) · { = s
+ { sin31 (|}),
= | sin31 (|}),
= {| · s
C{
C|
1 3 (|})2
1 3 |2 }2
{| 2
Cx
1
(|) = s
= {| · s
C}
1 3 (|})2
1 3 |2 }2
36. x = {|@}
i x{ =
{|@}
| (|@})31
1
3|
|{|@}
{
ln {, x} = {|@} ln { · 2 = 3 2 ln {
, x| = {|@} ln { · =
}
}
}
}
}
37. k({> |> }> w) = {2 | cos(}@w)
i k{ ({> |> }> w) = 2{| cos(}@w), k| ({> |> }> w) = {2 cos(}@w),
k} ({> |> }> w) = 3{2 | sin(}@w)(1@w) = (3{2 |@w) sin(}@w), kw ({> |> }> w) = 3{2 | sin(}@w)(3}w32 ) = ({2 |}@w2 ) sin(}@w)
38. !({> |> }> w) =
{ + | 2
} + w2
i !{ ({> |> }> w) =
1
() =
,
} + w2
} + w2
!| ({> |> }> w) =
1
2|
(} + w2 )(0) 3 ({ + |2 )()
3({ + | 2 )
(2|)
=
,
!
({>
|>
}>
w)
=
=
,
}
} + w2
} + w2
(} + w2 )2
(} + w2 )2
!w ({> |> }> w) =
(} + w2 )(0) 3 ({ + | 2 )(2w)
2w({ + | 2 )
=3
2
2
(} + w )
(} + w2 )2
39. x =
s
31@2
{l
{21 + {22 + · · · + {2q . For each l = 1, = = =, q, x{l = 12 {21 + {22 + · · · + {2q
(2{l ) = s 2
.
{1 + {22 + · · · + {2q
40. x = sin({1 + 2{2 + · · · + q{q ).
For each l = 1, = = =, q, x{l = l cos({1 + 2{2 + · · · + q{q ).
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SECTION 14.3
s
i
41. i ({> |) = ln { + {2 + | 2
k
l
1
1
s
s
1 + 12 ({2 + | 2 )31@2 (2{) =
{ + {2 + | 2
{ + {2 + | 2
3
1
I
1+ I
= 18 1 + 35 = 15 .
so i{ (3> 4) =
3 + 32 + 42
32 + 42
i{ ({> |) =
42. i ({> |) = arctan(|@{)
so i{ (2> 3) = 3
so i| (2> 1> 31) =
1
2
405
#
$
{
1+ s
,
{2 + | 2
1
3|
|
(3|{32 ) = 2
,
=3 2
1 + (|@{)2
{ (1 + | 2 @{2 )
{ + |2
i i| ({> |> }) =
1({ + | + }) 3 |(1)
{+}
=
,
({ + | + })2
({ + | + })2
2 + (31)
1
= .
(2 + 1 + (31))2
4
s
sin2 { + sin2 | + sin2 }
i} ({> |> }) =
¤
3
3
=3 .
22 + 32
13
|
{+|+}
43. i ({> |> }) =
44. i ({> |> }) =
i i{ ({> |) =
PARTIAL DERIVATIVES
i
2
31@2
sin } cos }
sin { + sin2 | + sin2 }
(0 + 0 + 2 sin } · cos }) = s 2
,
sin { + sin2 | + sin2 }
sin 4 cos 4
so i} 0> 0> 4 = t
sin2 0 + sin2 0 + sin2
45. i ({> |) = {| 2 3 {3 |
I
4
= u
2
2
·
I
0+0+
2
2
I
2
2
2 =
1
2
I
2
2
I
1
2
= I or
.
2
2
i
i ({ + k> |) 3 i ({> |)
({ + k)| 2 3 ({ + k)3 | 3 ({| 2 3 {3 |)
= lim
k<0
k<0
k
k
i{ ({> |) = lim
k(| 2 3 3{2 | 3 3{|k 3 |k2 )
= lim (| 2 3 3{2 | 3 3{|k 3 |k2 ) = | 2 3 3{2 |
k<0
k<0
k
= lim
i ({> | + k) 3 i ({> |)
{(| + k)2 3 {3 (| + k) 3 ({| 2 3 {3 |)
k(2{| + {k 3 {3 )
= lim
= lim
k<0
k<0
k<0
k
k
k
i| ({> |) = lim
= lim (2{| + {k 3 {3 ) = 2{| 3 {3
k<0
46. i ({> |) =
{
{ + |2
i
i ({ + k> |) 3 i ({> |)
= lim
k<0
k<0
k
i{ ({> |) = lim
2
{+k
{+k+|2
3
{
{+| 2
k
·
({ + k + | 2 )({ + | 2 )
({ + k + | 2 )({ + | 2 )
2
({ + k)({ + | ) 3 {({ + k + | )
|2 k
= lim
2
2
k<0
k<0 k({ + k + | 2 )({ + | 2 )
k({ + k + | )({ + | )
= lim
= lim
k<0
|2
|2
=
2
2
({ + k + | )({ + | )
({ + | 2 )2
[continued]
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406
¤
CHAPTER 14
PARTIAL DERIVATIVES
{
{
3 {+|
2
{ + (| + k)2 { + | 2
i ({> | + k) 3 i ({> |)
{+(|+k)2
= lim
· k<0
k<0
k
k
{ + (| + k)2 ({ + | 2 )
{({ + | 2 ) 3 { { + (| + k)2
k(32{| 3 {k)
= lim
= lim
k<0
k<0 k[{ + (| + k)2 ]({ + | 2 )
k[{ + (| + k)2 ]({ + | 2 )
i| ({> |) = lim
= lim
k<0
47. {2 + 2| 2 + 3} 2 = 1
32{| 3 {k
32{|
=
2
2
({
+ | 2 )2
[{ + (| + k) ]({ + | )
C
C
({2 + 2| 2 + 3} 2 ) =
(1)
C{
C{
i
C
C}
32{
{
C
=
= 3 , and
({2 + 2| 2 + 3} 2 ) =
(1)
C{
6}
3}
C|
C|
i
i
2{ + 0 + 6}
0 + 4| + 6}
C}
C}
= 0 i 6}
= 32{ i
C{
C{
C}
C}
= 0 i 6}
= 34|
C|
C|
i
34|
2|
C}
=
=3 .
C|
6}
3}
48. {2 3 | 2 + } 2 3 2} = 4
i (2} 3 2)
C
C
({2 3 | 2 + } 2 3 2}) =
(4)
C{
C{
i
i
2{ 3 0 + 2}
C}
C}
32
=0
C{
C{
C}
32{
{
=
=
, and
C{
2} 3 2
13}
C}
= 32{ i
C{
C
C
({2 3 | 2 + } 2 3 2}) =
(4)
C|
C|
i
0 3 2| + 2}
C}
C}
C}
32
= 0 i (2} 3 2)
= 2|
C|
C|
C|
i
C}
2|
|
=
=
.
C|
2} 3 2
}31
}
49. h = {|}
(h} 3 {|)
C
C
(h} ) =
({|})
C{
C{
i
i
C}
C}
C}
C}
h
=| {
+}·1
i h}
3 {|
= |}
C{
C{
C{
C{
}
i
C}
C}
|}
= |}, so
= }
.
C{
C{
h 3 {|
C }
C
(h ) =
({|})
C|
C|
h}
i
C}
C}
C}
C}
={ |
+}·1
i h}
3 {|
= {}
C|
C{
C|
C|
i (h} 3 {|)
C}
= {}, so
C|
{}
C}
= }
.
C|
h 3 {|
50. |} + { ln | = } 2
ln | = (2} 3 |)
i
C
C
(|} + { ln |) =
(} 2 )
C{
C{
|
C}
C}
+ ln | = 2}
C{
C{
i ln | = 2}
C}
C}
3|
C{
C{
C}
ln |
C}
, so
=
.
C{
C{
2} 3 |
C
C
(|} + { ln |) =
(} 2 )
C|
C|
}+
i
i
|
1
C}
C}
+ } · 1 + { · = 2}
C|
|
C|
i }+
{
C}
C}
= 2}
3|
|
C|
C|
i
{
C}
C}
} + ({@|)
{ + |}
= (2} 3 |)
, so
=
=
.
|
C|
C|
2} 3 |
|(2} 3 |)
51. (a) } = i ({) + j(|)
i
C}
= i 0 ({),
C{
C}
= j 0 (|)
C|
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SECTION 14.3
(b) } = i ({ + |). Let x = { + |. Then
PARTIAL DERIVATIVES
¤
407
gi Cx
gi
C}
=
=
(1) = i 0 (x) = i 0 ({ + |),
C{
gx C{
gx
C}
gi Cx
gi
=
=
(1) = i 0 (x) = i 0 ({ + |).
C|
gx C|
gx
52. (a) } = i ({)j(|)
i
C}
= i 0 ({)j(|),
C{
(b) } = i ({|). Let x = {|. Then
C}
= i({)j0 (|)
C|
Cx
C}
gi Cx
gi
Cx
= | and
= {. Hence
=
=
· | = |i 0 (x) = |i 0 ({|)
C{
C|
C{
gx C{
gx
C}
gi Cx
gi
=
=
· { = {i 0 (x) = {i 0 ({|).
C|
gx C|
gx
i 0 ({@|)
C}
1
{
{
Cx
1
Cx
{
gi Cx
(c) } = i
. Let x = . Then
= and
= 3 2 . Hence
=
= i 0 (x) =
|
|
C{
|
C|
|
C{
gx C{
|
|
0
{i ({@|)
gi Cx
{
C}
=
= i 0 (x) 3 2 = 3
.
and
C|
gx C|
|
|2
and
53. i ({> |) = {3 | 5 + 2{4 |
i
i{ ({> |) = 3{2 | 5 + 8{3 |, i| ({> |) = 5{3 | 4 + 2{4 . Then i{{ ({> |) = 6{| 5 + 24{2 |,
i{| ({> |) = 15{2 | 4 + 8{3 , i|{ ({> |) = 15{2 | 4 + 8{3 , and i|| ({> |) = 20{3 | 3 .
54. i ({> |) = sin2 (p{ + q|)
i
i{ ({> |) = 2 sin(p{ + q|) cos(p{ + q|) · p = p sin(2p{ + 2q|) [using the
identity sin 2 = 2 sin cos ], i| ({> |) = 2 sin(p{ + q|) cos(p{ + q|) · q = q sin(2p{ + 2q|).
Then i{{ ({> |) = p cos(2p{ + 2q|) · 2p = 2p2 cos(2p{ + 2q|),
i{| ({> |) = p cos(2p{ + 2q|) · 2q = 2pq cos(2p{ + 2q|),
i|{ ({> |) = q cos(2p{ + 2q|) · 2p = 2pq cos(2p{ + 2q|), and
i|| ({> |) = q cos(2p{ + 2q|) · 2q = 2q2 cos(2p{ + 2q|).
55. z =
zxx
I
x2 + y 2
x
y
zx = 12 (x2 + y 2 )31@2 · 2x = I
, zy = 12 (x2 + y 2 )31@2 · 2y = I
. Then
x2 + y 2
x2 + y 2
I
I
I
1 · x2 + y 2 3 x · 12 (x2 + y 2 )31@2 (2x)
x2 + y 2 3 x2 @ x2 + y 2
x2 + y 2 3 x2
y2
=
=
=
= 2
,
I
2
2
2
2
2
3@2
x +y
(x + y )
(x + y 2 )3@2
x2 + y 2
i
33@2
(2y) = 3
zxy = x 3 12 x2 + y 2
zyy
33@2
xy
xy
, zyx = y 3 12 x2 + y 2
(2x) = 3 2
,
2
3@2
+y )
(x + y 2 )3@2
I
I
I
1 · x2 + y 2 3 y · 12 (x2 + y 2 )31@2 (2y)
x2 + y 2 3 y 2 @ x2 + y 2
x2 + y 2 3 y 2
x2
=
=
=
=
.
I
2
x2 + y 2
(x2 + y 2 )3@2
(x2 + y 2 )3@2
x2 + y 2
56. y =
{|
{3|
y| =
y{ =
|({ 3 |) 3 {|(1)
|2
=3
,
2
({ 3 |)
({ 3 |)2
{({ 3 |) 3 {|(31)
{2
2|2
=
. Then y{{ = 3| 2 (32)({ 3 |)33 (1) =
,
2
2
({ 3 |)
({ 3 |)
({ 3 |)3
y{| = 3
y|{ =
i
(x2
2|({ 3 |)2 3 | 2 · 2({ 3 |)(31)
2|({ 3 |) + 2| 2
2{|
=3
=3
,
2
({ 3 |)3
({ 3 |)3
[({ 3 |)2 ]
2{({ 3 |)2 3 {2 · 2({ 3 |)(1)
2{({ 3 |) 3 2{2
2{|
2{2
2
33
=
=
3
,
y
=
{
(32)({
3
|)
(31)
=
.
||
({ 3 |)3
({ 3 |)3
({ 3 |)3
[({ 3 |)2 ]2
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408
¤
CHAPTER 14
{+|
1 3 {|
57. } = arctan
}{ =
1+
=
PARTIAL DERIVATIVES
1
{+|
13{|
i
2 ·
(1)(1 3 {|) 3 ({ + |)(3|)
1 + |2
1 + |2
=
=
2
2
2
2
(1 3 {|)
(1 3 {|) + ({ + |)
1 + { + | 2 + {2 | 2
1 + |2
1
>
=
(1 + {2 )(1 + | 2 )
1 + {2
}| =
1+
1
{+|
13{|
2 ·
(1)(1 3 {|) 3 ({ + |)(3{)
1 + {2
1 + {2
1
=
=
.
=
2
2
2
(1 3 {|)
(1 3 {|) + ({ + |)
(1 + {2 )(1 + | 2 )
1 + |2
Then }{{ = 3(1 + {2 )32 · 2{ = 3
58. y = h{h
|
2{
2|
, }{| = 0, }|{ = 0, }|| = 3(1 + | 2 )32 · 2| = 3
.
(1 + {2 )2
(1 + | 2 )2
|
|
|
|
|
|
i y{ = h{h · h| = h|+{h , y| = h{h · {h| = {h|+{h . Then y{{ = h|+{h · h| = h2|+{h ,
|
|
|
|
y{| = h|+{h (1 + {h| ), y|{ = {h|+{h (h| ) + h|+{h (1) = h|+{h (1 + {h| ),
|
|
y|| = {h|+{h (1 + {h| ) = h|+{h ({ + {2 h| ).
59. x = {4 | 3 3 | 4
i x{ = 4{3 | 3 , x{| = 12{3 | 2 and x| = 3{4 | 2 3 4| 3 , x|{ = 12{3 | 2 .
Thus x{| = x|{ .
60. x = h{| sin |
i x{ = |h{| sin |, x{| = |h{| cos | + (sin |)(| · {h{| + h{| · 1) = h{| (| cos | + {| sin | + sin |),
x| = h{| cos | + (sin |)({h{| ) = h{| (cos | + { sin |),
x|{ = h{| · sin | + (cos | + { sin |) · |h{| = h{| (sin | + | cos | + {| sin |). Thus x{| = x|{ .
61. x = cos({2 |)
i x{ = 3 sin({2 |) · 2{| = 32{| sin({2 |),
x{| = 32{| · cos({2 |) · {2 + sin({2 |) · (32{) = 32{3 | cos({2 |) 3 2{ sin({2 |) and
x| = 3 sin({2 |) · {2 = 3{2 sin({2 |), x|{ = 3{2 · cos({2 |) · 2{| + sin({2 |) · (32{) = 32{3 | cos({2 |) 3 2{ sin({2 |).
Thus x{| = x|{ .
62. x = ln({ + 2|)
x| =
i x{ =
1
2
= ({ + 2|)31 , x{| = (31)({ + 2|)32 (2) = 3
and
{ + 2|
({ + 2|)2
1
2
. Thus x{| = x|{ .
· 2 = 2({ + 2|)31 , x|{ = (32)({ + 2|)32 = 3
{ + 2|
({ + 2|)2
63. i ({> |) = {4 | 2 3 {3 |
i i{ = 4{3 | 2 3 3{2 |, i{{ = 12{2 | 2 3 6{|, i{{{ = 24{| 2 3 6| and
i{| = 8{3 | 3 3{2 , i{|{ = 24{2 | 3 6{.
64. i ({> |) = sin(2{ + 5|)
i i| = cos(2{ + 5|) · 5 = 5 cos(2{ + 5|), i|{ = 35 sin(2{ + 5|) · 2 = 310 sin(2{ + 5|),
i|{| = 310 cos(2{ + 5|) · 5 = 350 cos(2{ + 5|)
65. i ({> |> }) = h{|}
2
2
2
2
2
2
i i{ = h{|} · |} 2 = |} 2 h{|} , i{| = |} 2 · h{|} ({} 2 ) + h{|} · } 2 = ({|} 4 + } 2 )h{|} ,
2
2
2
i{|} = ({|} 4 + } 2 ) · h{|} (2{|}) + h{|} · (4{|} 3 + 2}) = (2{2 | 2 } 5 + 6{|} 3 + 2})h{|} .
66. j(u> v> w) = hu sin(vw)
i ju = hu sin(vw), juv = hu cos(vw) · w = whu cos(vw),
juvw = whu (3 sin(vw) · v) + cos(vw) · hu = hu [cos(vw) 3 vw sin(vw)].
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SECTION 14.3
67. x = hu sin i
PARTIAL DERIVATIVES
¤
409
Cx
= hu cos + sin · hu (u) = hu (cos + u sin ),
C
C2x
= hu (sin ) + (cos + u sin ) hu () = hu (sin + cos + u sin ),
Cu C
C3x
= hu ( sin ) + (sin + cos + u sin ) · hu () = hu (2 sin + cos + u sin ).
Cu2 C
68. } = x
I
y 3 z = x(y 3 z)1@2
k
l
C}
= x 12 (y 3 z)31@2 (31) = 3 12 x(y 3 z)31@2 ,
Cz
i
C2}
C3}
= 3 12 x 3 12 (y 3 z)33@2 (1) = 14 x(y 3 z)33@2 ,
= 14 (y 3 z)33@2 .
Cy Cz
Cx Cy Cz
69. z =
{
= {(| + 2})31
| + 2}
i
Cz
C2z
= (| + 2})31 ,
= 3(| + 2})32 (1) = 3(| + 2})32 ,
C{
C| C{
C3z
4
Cz
and
= 3(32)(| + 2})33 (2) = 4(| + 2})33 =
= {(31)(| + 2})32 (1) = 3{(| + 2})32 ,
C} C| C{
(| + 2})3
C|
C3z
C2z
= 3(| + 2})32 ,
= 0.
C{ C|
C{2 C|
70. x = {d | e } f .
If d = 0, or if e = 0 or 1, or if f = 0, 1, or 2, then
C6x
Cx
= f{d | e } f31 ,
= 0. Otherwise
C{ C| 2 C} 3
C}
C2x
C3x
C4x
= f(f 3 1){d | e } f32 ,
= f(f 3 1)(f 3 2){d | e } f33 ,
= ef(f 3 1)(f 3 2){d | e31 } f33 ,
C} 2
C} 3
C| C} 3
C5x
C6x
= e(e 3 1)f(f 3 1)(f 3 2){d | e32 } f33 , and
= de(e 3 1)f(f 3 1)(f 3 2){d31 | e32 } f33 .
2
3
C| C}
C{ C|2 C} 3
71. Assuming that the third partial derivatives of i are continuous (easily verified), we can write i{}| = i|{} . Then
I i ({> |> }) = {| 2 } 3 + arcsin { }
i i| = 2{|} 3 + 0, i|{ = 2|} 3 , and i|{} = 6|} 2 = i{}| .
72. Let i({> |> }) =
I
I
1 + {} and k({> |> }) = 1 3 {| so that j = i + k. Then i| = 0 = i|{ = i|{} and
k} = 0 = k}{ = k}{| . But (since the partial derivatives are continous on their domains) i{|} = i|{} and k{|} = k}{| , so
j{|} = i{|} + k{|} = 0 + 0 = 0.
73. By Definition 4, i{ (3> 2) = lim
k<0
i{ (3> 2) E
i(3 + k> 2) 3 i (3> 2)
which we can approximate by considering k = 0=5 and k = 30=5:
k
22=4 3 17=5
10=2 3 17=5
i (3=5> 2) 3 i(3> 2)
i (2=5> 2) 3 i (3> 2)
=
= 9=8, i{ (3> 2) E
=
= 14=6. Averaging
0=5
0=5
30=5
30=5
these values, we estimate i{ (3> 2) to be approximately 12=2. Similarly, i{ (3> 2=2) = lim
k<0
we can approximate by considering k = 0=5 and k = 30=5: i{ (3> 2=2) E
i{ (3> 2=2) E
i(3 + k> 2=2) 3 i (3> 2=2)
which
k
26=1 3 15=9
i (3=5> 2=2) 3 i (3> 2=2)
=
= 20=4,
0=5
0=5
9=3 3 15=9
i (2=5> 2=2) 3 i(3> 2=2)
=
= 13=2. Averaging these values, we have i{ (3> 2=2) E 16=8.
30=5
30=5
To estimate i{| (3> 2), we first need an estimate for i{ (3> 1=8):
i{ (3> 1=8) E
i (3=5> 1=8) 3 i(3> 1=8)
i (2=5> 1=8) 3 i (3> 1=8)
20=0 3 18=1
12=5 3 18=1
=
= 3=8, i{ (3> 1=8) E
=
= 11=2.
0=5
0=5
30=5
30=5
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410
¤
CHAPTER 14
PARTIAL DERIVATIVES
C
[i{ ({> |)] and i{ ({> |) is itself a function of two
C|
Averaging these values, we get i{ (3> 1=8) E 7=5. Now i{| ({> |) =
variables, so Definition 4 says that i{| ({> |) =
i{| (3> 2) = lim
k<0
i{| (3> 2) E
i{ ({> | + k) 3 i{ ({> |)
C
[i{ ({> |)] = lim
k<0
C|
k
i
i{ (3> 2 + k) 3 i{ (3> 2)
. We can estimate this value using our previous work with k = 0=2 and k = 30=2:
k
16=8 3 12=2
7=5 3 12=2
i{ (3> 2=2) 3 i{ (3> 2)
i{ (3> 1=8) 3 i{ (3> 2)
=
= 23, i{| (3> 2) E
=
= 23=5.
0=2
0=2
30=2
30=2
Averaging these values, we estimate i{| (3> 2) to be approximately 23=25.
74. (a) If we fix | and allow { to vary, the level curves indicate that the value of i decreases as we move through S in the positive
{-direction, so i{ is negative at S .
(b) If we fix { and allow | to vary, the level curves indicate that the value of i increases as we move through S in the positive
|-direction, so i| is positive at S .
(c) i{{ =
C
(i{ ), so if we fix | and allow { to vary, i{{ is the rate of change of i{ as { increases. Note that at points to the
C{
right of S the level curves are spaced farther apart (in the {-direction) than at points to the left of S , demonstrating that i
decreases less quickly with respect to { to the right of S= So as we move through S in the positive {-direction the
(negative) value of i{ increases, hence
(d) i{| =
C
(i{ ) = i{{ is positive at S .
C{
C
(i{ ) > so if we fix { and allow | to vary, i{| is the rate of change of i{ as | increases. The level curves are
C|
closer together (in the {-direction) at points above S than at those below S , demonstrating that i decreases more quickly
with respect to { for |-values above S . So as we move through S in the positive |-direction, the (negative) value of i{
decreases, hence i{| is negative.
(e) i|| =
C
(i| ) > so if we fix { and allow | to vary, i|| is the rate of change of i| as | increases. The level curves are
C|
closer together (in the |-direction) at points above S than at those below S , demonstrating that i increases more quickly
with respect to | above S . So as we move through S in the positive |-direction the (positive) value of i| increases, hence
C
(i| ) = i|| is positive at S .
C|
2 n2 w
75. x = h3
sin n{ i x{ = nh3
2 n2 w
2 n2 w
cos n{, x{{ = 3n2 h3
2 n2 w
sin n{, and xw = 32 n2 h3
sin n{.
Thus 2 x{{ = xw .
76. (a) x = {2 + | 2
i x{ = 2{, x{{ = 2; x| = 2|, x|| = 2. Thus x{{ + x|| 6= 0 and x = {2 + | 2 does not satisfy
Laplace’s Equation.
(b) x = {2 3 | 2 is a solution: x{{ = 2, x|| = 32 so x{{ + x|| = 0.
(c) x = {3 + 3{| 2 is not a solution: x{ = 3{2 + 3| 2 , x{{ = 6{; x| = 6{|, x|| = 6{.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 14.3
PARTIAL DERIVATIVES
¤
411
s
1
{
1
(d) x = ln {2 + | 2 is a solution: x{ = s
,
({2 + | 2 )31@2 (2{) = 2
{ + |2
{2 + | 2 2
x{{ =
({2 + | 2 ) 3 {(2{)
| 2 3 {2
{2 3 | 2
= 2
. By symmetry, x|| = 2
, so x{{ + x|| = 0.
2
2
2
2
2
({ + | )
({ + | )
({ + | 2 )2
(e) x = sin { cosh | + cos { sinh | is a solution: x{ = cos { cosh | 3 sin { sinh |> x{{ = 3 sin { cosh | 3 cos { sinh |,
and x| = sin { sinh | + cos { cosh |, x|| = sin { cosh | + cos { sinh |.
(f) x = h3{ cos | 3 h3| cos { is a solution: x{ = 3h3{ cos | + h3| sin {, x{{ = h3{ cos | + h3| cos {, and
x| = 3h3{ sin | + h3| cos {, x|| = 3h3{ cos | 3 h3| cos {.
77. x = s
{2
i x{ = 3 12 ({2 + | 2 + } 2 )33@2 (2{) = 3{({2 + | 2 + } 2 )33@2 and
1
+ |2 + }2
x{{ = 3({2 + | 2 + } 2 )33@2 3 { 3 32 ({2 + | 2 + } 2 )35@2 (2{) =
By symmetry, x|| =
2{2 3 | 2 3 } 2
.
({2 + | 2 + } 2 )5@2
2| 2 3 {2 3 } 2
2} 2 3 {2 3 | 2
and
x
=
.
}}
({2 + | 2 + } 2 )5@2
({2 + | 2 + } 2 )5@2
Thus x{{ + x|| + x}} =
78. (a) x = sin(n{) sin(dnw)
2{2 3 | 2 3 } 2 + 2| 2 3 {2 3 } 2 + 2} 2 3 {2 3 | 2
= 0.
({2 + | 2 + } 2 )5@2
i xw = dn sin(n{) cos(dnw), xww = 3d2 n2 sin(n{) sin(dnw), x{ = n cos(n{) sin(dnw),
x{{ = 3n2 sin(n{) sin(dnw). Thus xww = d2 x{{ .
(b) x =
d2 w2
xww =
w
3 {2
i xw =
(d2 w2 3 {2 ) 3 w(2d2 w)
d2 w2 + {2
=3 2 2
,
2
2
2
2
(d w 3 { )
(d w 3 {2 )2
32d2 w(d2 w2 3 {2 )2 + (d2 w2 3 {2 )(2)(d2 w2 3 {2 )(2d2 w)
2d4 w3 + 6d2 w{2
=
,
(d2 w2 3 {2 )4
(d2 w2 3 {2 )3
x{ = w(31)(d2 w2 3 {2 )32 (2{) =
x{{ =
2w{
,
(d2 w2 3 {2 )2
2w(d2 w2 3 {2 )2 3 2w{ (2)(d2 w2 3 {2 )(32{)
2d2 w3 3 2w{2 + 8w{2
2d2 w3 + 6w{2
=
= 2 2
.
2
2
2
4
2
2
2
3
(d w 3 { )
(d w 3 { )
(d w 3 {2 )3
Thus xww = d2 x{{ .
(c) x = ({ 3 dw)6 + ({ + dw)6
i xw = 36d({ 3 dw)5 + 6d({ + dw)5 , xww = 30d2 ({ 3 dw)4 + 30d2 ({ + dw)4 ,
x{ = 6({ 3 dw)5 + 6({ + dw)5 , x{{ = 30({ 3 dw)4 + 30({ + dw)4 . Thus xww = d2 x{{ .
(d) x = sin({ 3 dw) + ln({ + dw) i xw = 3d cos({ 3 dw) +
x{ = cos({ 3 dw) +
1
1
, x{{ = 3 sin({ 3 dw) 3
. Thus xww = d2 x{{ .
{ + dw
({ + dw)2
79. Let y = { + dw, z = { 3 dw.
xww =
d
d2
,
, xww = 3d2 sin({ 3 dw) 3
{ + dw
({ + dw)2
Then xw =
gi (y) Cy
gj(z) Cz
C[i (y) + j(z)]
=
+
= di 0 (y) 3 dj0 (z) and
Cw
gy Cw
gz Cw
C[di 0 (y) 3 dj0 (z)]
= d[di 00 (y) + dj 00 (z)] = d2 [i 00 (y) + j 00 (z)]. Similarly, by using the Chain Rule we have
Cw
x{ = i 0 (y) + j0 (z) and x{{ = i 00 (y) + j00 (z). Thus xww = d2 x{{ .
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
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412
¤
CHAPTER 14
PARTIAL DERIVATIVES
80. For each l, l = 1> = = = > q, Cx@C{l = dl hd1 {1 +d2 {2 +···+dq {q and C 2 x@C{2l = d2l hd1 {1 +d2 {2 +···+dq {q .
Then
C2x
C2x
C2x
= d21 + d22 + · · · + d2q hd1 {1 +d2 {2 +···+dq {q = hd1 {1 +d2 {2 +···+dq {q = x
2 +
2 + ··· +
2
C{1
C{2
C{q
since d21 + d22 + · · · + d2q = 1.
81. } = ln(h{ + h| )
i
h{
h|
C}
h{
C}
C}
C}
h|
h{ + h|
= {
=
+
=
and
,
so
+
=
= 1.
C{
h + h|
C|
h{ + h|
C{
C|
h{ + h|
h{ + h|
h{ + h|
C2}
0 3 h| (h{ )
h{ (h{ + h| ) 3 h{ (h{ )
h{+|
C2}
h{+|
=
=
=
,
=
3
, and
C{2
(h{ + h| )2
(h{ + h| )2 C{ C|
(h{ + h| )2
(h{ + h| )2
C2}
h| (h{ + h| ) 3 h| (h| )
h{+|
=
= {
. Thus
2
{
|
2
C|
(h + h )
(h + h| )2
C2} C2}
3
C{2 C| 2
C2}
C{ C|
2
=
2
(h{+| )2
(h{+| )2
h{+|
h{+|
h{+|
·
3
3
= {
3 {
=0
{
|
2
{
|
2
{
|
2
|
4
(h + h ) (h + h )
(h + h )
(h + h )
(h + h| )4
82. (a) CW @C{ = 360(2{)@(1 + {2 + | 2 )2 , so at (2> 1), W{ = 3240@(1 + 4 + 1)2 = 3 20
.
3
. Thus from the point (2> 1) the temperature is
(b) CW @C| = 360(2|)@(1 + {2 + | 2 )2 , so at (2> 1), W| = 3120@36 = 3 10
3
decreasing at a rate of
20 C@m
3
in the {-direction and is decreasing at a rate of
10 C@m
3
in the |-direction.
83. By the Chain Rule, taking the partial derivative of both sides with respect to U1 gives
CU
CU31 CU
C [(1@U1 ) + (1@U2 ) + (1@U3 )]
CU
U2
=
or 3U32
= 3U132 . Thus
= 2.
CU CU1
CU1
CU1
CU1
U1
84. S = eO N , so
O
CS
CS
= eO31 N and
= eO N 31 . Then
CO
CN
CS
CS
+N
= O(eO31 N ) + N(eO N 31 ) = eO1+31 N + eO N 1+31 = ( + )eO N = ( + )S
CO
CN
85. If we fix N = N0 > S (O> N0 ) is a function of a single variable O, and
gS
gO
=
S
O
i
]
gS
=
S
]
gO
O
S
gS
= is a separable differential equation. Then
gO
O
i ln |S | = ln |O| + F (N0 ), where F(N0 ) can depend on N0 . Then
|S | = h ln|O| + F(N0 ) , and since S A 0 and O A 0, we have S = h ln O hF(N0 ) = hF(N0 ) hln O = F1 (N0 )O where
F1 (N0 ) = hF(N0 ) .
86. (a) S (O> N) = 1=01O0=75 N 0=25
i
SO (O> N) = 1=01(0=75O30=25 )N 0=25 = 0=7575O30=25 N 0=25 and
SN (O> N) = 1=01O0=75 (0=25N 30=75 ) = 0=2525O0=75 N 30=75 .
(b) The marginal productivity of labor in 1920 is SO (194> 407) = 0=7575(194)30=25 (407)0=25 E 0=912. Recall that S , O, and
N are expressed as percentages of the respective amounts in 1899, so this means that in 1920, if the amount of labor is
increased, production increases at a rate of about 0.912 percentage points per percentage point increase in labor. The
marginal productivity of capital in 1920 is SN (194> 407) = 0=2525(194)0=75 (407)30=75 E 0=145, so an increase in capital
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SECTION 14.3
PARTIAL DERIVATIVES
¤
413
investment would cause production to increase at a rate of about 0.145 percentage points per percentage point increase in
capital.
(c) The value of SO (194> 407) is greater than the value of SN (194> 407), suggesting that increasing labor in 1920 would have
increased production more than increasing capital.
q2 d
87. S + 2 (Y 3 qe) = qUW
Y
We can also write S +
1
CW
q2 d
1
Y 3 qe
i W =
S + 2 (Y 3 qe), so
=
(1)(Y 3 qe) =
.
qU
Y
CS
qU
qU
q2 d
qUW
=
Y2
Y 3 qe
i S =
qUW
q2 d
3 2 = qUW (Y 3 qe)31 3 q2 dY 32 , so
Y 3 qe
Y
CS
2q2 d
qUW
= 3qUW (Y 3 qe)32 (1) + 2q2 dY 33 =
3
.
CY
Y3
(Y 3 qe)2
88. S =
Thus
CS
3pUW
CY
pU
SY
CW
Y
pUW
pUW
so
=
, so
=
; W =
, so
=
.
; Y =
Y
CY
Y2
S
CW
S
pU
CS
pU
CS CY CW
3pUW pU Y
3pUW
=
=
= 31, since S Y = pUW .
CY CW CS
Y2
S pU
SY
89. By Exercise 88, S Y = pUW
Since W =
90.
i S =
pUW
CS
pU
, so
=
. Also, S Y = pUW
Y
CW
Y
i Y =
pUW
CY
pU
and
=
.
S
CW
S
SY
CS CY
S Y pU pU
, we have W
=
·
·
= pU.
pU
CW CW
pU Y
S
CZ
CZ
= 0=6215 + 0=3965y 0=16 . When W = 315 C and y = 30 km@h,
= 0=6215 + 0=3965(30)0=16 E 1=3048, so we
CW
CW
would expect the apparent temperature to drop by approximately 1=3 C if the actual temperature decreases by 1 C.
CZ
= 311=37(0=16)y 30=84 + 0=3965W (0=16)y 30=84 and when W = 315 C and y = 30 km@h,
Cy
CZ
= 311=37(0=16)(30)30=84 + 0=3965(315)(0=16)(30)30=84 E 30=1592, so we would expect the apparent temperature
Cy
to drop by approximately 0=16 C if the wind speed increases by 1 km@h.
91.
CN
CN C 2 N
CN
C2 N
= p. Thus
= 12 y 2 p = N.
= 12 y 2 ,
= py,
·
2
Cp
Cy
Cy
Cp Cy 2
92. The Law of Cosines says that d2 = e2 + f2 3 2ef cos D. Thus
2d = 32ef (3 sin D)
CD
d
CD
, implying that
=
. Taking the partial derivative of both sides with respect to e gives
Cd
Cd
ef sin D
0 = 2e 3 2f(cos D) 3 2ef (3 sin D)
93. i{ ({> |) = { + 4|
C(d2 )
C(e2 + f2 3 2de cos D)
=
or
Cd
Cd
CD
CD
f cos D 3 e
CD
e cos D 3 f
. Thus
=
. By symmetry,
=
.
Ce
Ce
ef sin D
Cf
ef sin D
i i{| ({> |) = 4 and i| ({> |) = 3{ 3 |
i i|{ ({> |) = 3. Since i{| and i|{ are continuous
everywhere but i{| ({> |) 6= i|{ ({> |), Clairaut’s Theorem implies that such a function i({> |) does not exist.
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¤
CHAPTER 14
PARTIAL DERIVATIVES
94. Setting { = 1, the equation of the parabola of intersection is
} = 6 3 1 3 1 3 2|2 = 4 3 2| 2 . The slope of the tangent is
C}@C| = 34|, so at (1> 2> 34) the slope is 38. Parametric
equations for the line are therefore { = 1, | = 2 + w,
} = 34 3 8w.
95. By the geometry of partial derivatives, the slope of the tangent line is i{ (1> 2). By implicit differentiation of
4{2 + 2| 2 + } 2 = 16, we get 8{ + 2} (C}@C{) = 0 i C}@C{ = 34{@}, so when { = 1 and } = 2 we have
C}@C{ = 32. So the slope is i{ (1> 2) = 32. Thus the tangent line is given by } 3 2 = 32({ 3 1), | = 2. Taking the
parameter to be w = { 3 1, we can write parametric equations for this line: { = 1 + w, | = 2, } = 2 3 2w.
96. W ({> w) = W0 + W1 h3{ sin($w 3 {)
(a) CW @C{ = W1 h3{ [cos($w 3 {)(3)] + W1 (3h3{ ) sin($w 3 {) = 3W1 h3{ [sin($w 3 {) + cos($w 3 {)].
This quantity represents the rate of change of temperature with respect to depth below the surface, at a given time w.
(b) CW @Cw = W1 h3{ [cos($w 3 {)($)] = $W1 h3{ cos($w 3 {). This quantity represents the rate of change of
temperature with respect to time at a fixed depth {.
(c) W{{ =
C
C{
CW
C{
= 3W1 h3{ [cos($w 3 {)(3) 3 sin($w 3 {)(3)] + h3{ (3) [sin($w 3 {) + cos($w 3 {)]
= 22 W1 h3{ cos($w 3 {)
But from part (b), Ww = $W1 h3{ cos($w 3 {) =
(d)
$
$
W{{ . So with n =
, the function W satisfies the heat equation.
22
22
Note that near the surface (that is, for small {) the
temperature varies greatly as w changes, but deeper
(for large {) the temperature is more stable.
(e) The term 3{ is a phase shift: it represents the fact that since heat diffuses slowly through soil, it takes time for changes
in the surface temperature to affect the temperature at deeper points. As { increases, the phase shift also increases. For
example, when = 0=2, the highest temperature at the surface is reached when w E 91, whereas at a depth of 5 feet the
peak temperature is attained at w E 149, and at a depth of 10 feet, at w E 207.
97. By Clairaut’s Theorem, i{|| = (i{| )| = (i|{ )| = i|{| = (i| ){| = (i| )|{ = i||{ .
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SECTION 14.3
PARTIAL DERIVATIVES
¤
415
98. (a) Since we are differentiating q times, with two choices of variable at each differentiation, there are 2q qth-order partial
derivatives.
(b) If these partial derivatives are all continuous, then the order in which the partials are taken doesn’t affect the value of the
result, that is, all qth-order partial derivatives with s partials with respect to { and q 3 s partials with respect to | are
equal. Since the number of partials taken with respect to { for an qth-order partial derivative can range from 0 to q, a
function of two variables has q + 1 distinct partial derivatives of order q if these partial derivatives are all continuous.
(c) Since q differentiations are to be performed with three choices of variable at each differentiation, there are 3q qth-order
partial derivatives of a function of three variables.
99. Let j({) = i ({> 0) = {({2 )33@2 h0 = { |{|33 . But we are using the point (1> 0), so near (1> 0), j({) = {32 . Then
j 0 ({) = 32{33 and j0 (1) = 32, so using (1) we have i{ (1> 0) = j 0 (1) = 32.
i (0 + k> 0) 3 i (0> 0)
(k3 + 0)1@3 3 0
k
= lim
= lim = 1.
k<0
k<0
k<0 k
k
k
I
Or: Let j({) = i ({> 0) = 3 {3 + 0 = {. Then j0 ({) = 1 and j0 (0) = 1 so, by (1), i{ (0> 0) = j 0 (0) = 1.
100. i{ (0> 0) = lim
101. (a)
(b) For ({> |) 6= (0> 0),
i{ ({> |) =
=
(3{2 | 3 | 3 )({2 + | 2 ) 3 ({3 | 3 {| 3 )(2{)
({2 + | 2 )2
{4 | + 4{2 | 3 3 | 5
({2 + | 2 )2
and by symmetry i| ({> |) =
{5 3 4{3 | 2 3 {| 4
.
({2 + | 2 )2
i (k> 0) 3 i (0> 0)
(0@k2 ) 3 0
i (0> k) 3 i (0> 0)
= lim
= 0 and i| (0> 0) = lim
= 0.
k<0
k<0
k<0
k
k
k
(c) i{ (0> 0) = lim
(d) By (3), i{| (0> 0) =
i|{ (0> 0) =
Ci{
i{ (0> k) 3 i{ (0> 0)
(3k5 3 0)@k4
= lim
= lim
= 31 while by (2),
k<0
k<0
C|
k
k
i| (k> 0) 3 i| (0> 0)
k5@k4
Ci|
= lim
= lim
= 1.
k<0
k<0
C{
k
k
(e) For ({> |) 6= (0> 0), we use a CAS to compute
i{| ({> |) =
{6 + 9{4 | 2 3 9{2 | 4 3 | 6
({2 + | 2 )3
Now as ({> |) < (0> 0) along the {-axis, i{| ({> |) < 1 while as
({> |) < (0> 0) along the |-axis, i{| ({> |) < 31. Thus i{| isn’t
continuous at (0> 0) and Clairaut’s Theorem doesn’t apply, so there is
no contradiction. The graphs of i{| and i|{ are identical except at the
origin, where we observe the discontinuity.
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¤
CHAPTER 14
PARTIAL DERIVATIVES
14.4 Tangent Planes and Linear Approximations
1. } = i ({> |) = 3| 2 3 2{2 + {
i i{ ({> |) = 34{ + 1, i| ({> |) = 6|, so i{ (2> 31) = 37, i| (2> 31) = 36.
By Equation 2, an equation of the tangent plane is } 3 (33) = i{ (2> 31)({ 3 2) + i| (2> 31)[| 3 (31)] i
} + 3 = 37({ 3 2) 3 6(| + 1) or } = 37{ 3 6| + 5.
2. } = i ({> |) = 3({ 3 1)2 + 2(| + 3)2 + 7
i i{ ({> |) = 6({ 3 1), i| ({> |) = 4(| + 3), so i{ (2> 32) = 6 and
i| (2> 32) = 4. By Equation 2, an equation of the tangent plane is } 3 12 = i{ (2> 32)({ 3 2) + i| (2> 32) [| 3 (32)] i
} 3 12 = 6({ 3 2) + 4(| + 2) or } = 6{ + 4| + 8.
3. } = i ({> |) =
s
{|
i i{ ({> |) = 12 ({|)31@2 · | =
1
2
s
s
|@{, i| ({> |) = 12 ({|)31@2 · { = 12 {@|, so i{ (1> 1) =
1
2
and i| (1> 1) = 12 . Thus an equation of the tangent plane is } 3 1 = i{ (1> 1)({ 3 1) + i| (1> 1)(| 3 1) i
} 3 1 = 12 ({ 3 1) + 12 (| 3 1) or { + | 3 2} = 0.
4. } = i ({> |) = {h{|
i i{ ({> |) = {|h{| + h{| , i| ({> |) = {2 h{| , so i{ (2> 0) = 1, i| (2> 0) = 4, and an equation of
the tangent plane is } 3 2 = i{ (2> 0)({ 3 2) + i| (2> 0)(| 3 0) i } 3 2 = 1({ 3 2) + 4(| 3 0) or } = { + 4|.
5. } = i ({> |) = { sin({ + |)
i i{ ({> |) = { · cos({ + |) + sin({ + |) · 1 = { cos({ + |) + sin({ + |),
i| ({> |) = { cos({ + |), so i{ (31> 1) = (31) cos 0 + sin 0 = 31, i| (31> 1) = (31) cos 0 = 31 and an equation of the
tangent plane is } 3 0 = (31)({ + 1) + (31)(| 3 1) or { + | + } = 0.
6. } = i ({> |) = ln({ 3 2|)
i i{ ({> |) = 1@({ 3 2|), i| ({> |) = 32@({ 3 2|), so i{ (3> 1) = 1, i| (3> 1) = 32, and
an equation of the tangent plane is } 3 0 = i{ (3> 1)({ 3 3) + i| (3> 1)(| 3 1) i } = 1({ 3 3) + (32)(| 3 1) or
} = { 3 2| 3 1.
7. } = i ({> |) = {2 + {| + 3| 2 , so i{ ({> |) = 2{ + |
i i{ (1> 1) = 3, i| ({> |) = { + 6|
i i| (1> 1) = 7 and an
equation of the tangent plane is } 3 5 = 3({ 3 1) + 7(| 3 1) or } = 3{ + 7| 3 5. After zooming in, the surface and the
tangent plane become almost indistinguishable. (Here, the tangent plane is below the surface.) If we zoom in farther, the
surface and the tangent plane will appear to coincide.
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SECTION 14.4
8. } = i({> |) = arctan({| 2 )
i{ (1> 1) =
1
1+1
} = 12 { + | 3
3
2
i i{ =
= 12 , i| (1> 1) =
2
1+1
TANGENT PLANES AND LINEAR APPROXIMATIONS
¤
417
1
|2
1
2{|
(| 2 ) =
, i| =
(2{|) =
,
2
2
1 + ({| )
1 + {2 | 4
1 + ({| 2 )2
1 + {2 | 4
= 1, so an equation of the tangent plane is } 3
4
= 12 ({ 3 1) + 1(| 3 1) or
+ 4 . After zooming in, the surface and the tangent plane become almost indistinguishable. (Here the
tangent plane is above the surface.) If we zoom in farther, the surface and the tangent plane will appear to coincide.
9. i ({> |) =
i| ({> |) =
{| sin ({ 3 |)
| sin ({ 3 |) + {| cos ({ 3 |)
2{2 | sin ({ 3 |)
. A CAS gives i{ ({> |) =
3
and
2
2
2
2
1+{ +|
1+{ +|
(1 + {2 + | 2 )2
{ sin ({ 3 |) 3 {| cos ({ 3 |)
2{| 2 sin ({ 3 |)
3
. We use the CAS to evaluate these at (1> 1), and then
2
2
1+{ +|
(1 + {2 + | 2 )2
substitute the results into Equation 2 to compute an equation of the tangent plane: } = 13 { 3 13 |. The surface and tangent
plane are shown in the first graph below. After zooming in, the surface and the tangent plane become almost indistinguishable,
as shown in the second graph. (Here, the tangent plane is shown with fewer traces than the surface.) If we zoom in farther, the
surface and the tangent plane will appear to coincide.
10. i ({> |) = h3{|@10
I
s
s { + | + {| . A CAS gives
1
|h3{|@10
i{ ({> |) = 3 10
1
{h3{|@10
i| ({> |) = 3 10
I
s
s 1
{ + | + {| + h3{|@10 2 I
+ I|
{
2
{|
I
s
s 1
I
{ + | + {| + h3{|@10
+ I{
2
|
2
{|
and
. We use the CAS to evaluate these at (1> 1),
and then substitute the results into Equation 2 to get an equation of the tangent plane: } = 0=7h30=1 { + 0=7h30=1 | + 1=6h30=1 .
The surface and tangent plane are shown in the first graph below. After zooming in, the surface and the tangent plane become
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CHAPTER 14
PARTIAL DERIVATIVES
almost indistinguishable, as shown in the second graph. (Here, the tangent plane is above the surface.) If we zoom in farther,
the surface and the tangent plane will appear to coincide.
11. i ({> |) = 1 + { ln({| 3 5).
and i| ({> |) = { ·
The partial derivatives are i{ ({> |) = { ·
{|
1
(|) + ln({| 3 5) · 1 =
+ ln({| 3 5)
{| 3 5
{| 3 5
1
{2
({) =
, so i{ (2> 3) = 6 and i| (2> 3) = 4. Both i{ and i| are continuous functions for
{| 3 5
{| 3 5
{| A 5, so by Theorem 8, i is differentiable at (2> 3). By Equation 3, the linearization of i at (2> 3) is given by
O({> |) = i (2> 3) + i{ (2> 3)({ 3 2) + i| (2> 3)(| 3 3) = 1 + 6({ 3 2) + 4(| 3 3) = 6{ + 4| 3 23.
12. i ({> |) = {3 | 4 .
The partial derivatives are i{ ({> |) = 3{2 | 4 and i| ({> |) = 4{3 | 3 , so i{ (1> 1) = 3 and i| (1> 1) = 4.
Both i{ and i| are continuous functions, so i is differentiable at (1> 1) by Theorem 8. The linearization of i at (1> 1) is given
by O({> |) = i (1> 1) + i{ (1> 1)({ 3 1) + i| (1> 1)(| 3 1) = 1 + 3({ 3 1) + 4(| 3 1) = 3{ + 4| 3 6.
13. i ({> |) =
1({ + |) 3 {(1)
{
= |@({ + |)2 and
. The partial derivatives are i{ ({> |) =
{+|
({ + |)2
i| ({> |) = {(31)({ + |)32 · 1 = 3{@({ + |)2 , so i{ (2> 1) =
1
9
and i| (2> 1) = 3 29 . Both i{ and i| are continuous
functions for | 6= 3{, so i is differentiable at (2> 1) by Theorem 8. The linearization of i at (2> 1) is given by
O ({> |) = i (2> 1) + i{ (2> 1)({ 3 2) + i| (2> 1)(| 3 1) =
14. i ({> |) =
2
3
+ 19 ({ 3 2) 3 29 (| 3 1) = 19 { 3 29 | + 23 .
I
{ + h4| = ({ + h4| )1@2 . The partial derivatives are i{ ({> |) = 12 ({ + h4| )31@2 and
i| ({> |) = 12 ({ + h4| )31@2 (4h4| ) = 2h4| ({ + h4| )31@2 , so i{ (3> 0) = 12 (3 + h0 )31@2 =
1
4
and
i| (3> 0) = 2h0 (3 + h0 )31@2 = 1. Both i{ and i| are continuous functions near (3> 0), so i is
differentiable at (3> 0) by Theorem 8. The linearization of i at (3> 0) is
O({> |) = i (3> 0) + i{ (3> 0)({ 3 3) + i| (3> 0)(| 3 0) = 2 + 14 ({ 3 3) + 1(| 3 0) = 14 { + | + 54 .
15. i ({> |) = h3{| cos |.
The partial derivatives are i{ ({> |) = h3{| (3|) cos | = 3|h3{| cos | and
i| ({> |) = h3{| (3 sin |) + (cos |)h3{| (3{) = 3h3{| (sin | + { cos |), so i{ (> 0) = 0 and i| (> 0) = 3.
Both i{ and i| are continuous functions, so i is differentiable at (> 0), and the linearization of i at (> 0) is
O({> |) = i (> 0) + i{ (> 0)({ 3 ) + i| (> 0)(| 3 0) = 1 + 0({ 3 ) 3 (| 3 0) = 1 3 |.
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SECTION 14.4
16. i ({> |) = | + sin({@|).
i{ (0> 3) =
1
3
TANGENT PLANES AND LINEAR APPROXIMATIONS
¤
419
The partial derivatives are i{ ({> |) = (1@|) cos({@|) and i| ({> |) = 1 + (3{@| 2 ) cos({@|), so
and i| (0> 3) = 1. Both i{ and i| are continuous functions for | 6= 0, so i is differentiable at (0> 3), and the
linearization of i at (0> 3) is
O({> |) = i (0> 3) + i{ (0> 3)({ 3 0) + i| (0> 3)(| 3 3) = 3 + 13 ({ 3 0) + 1(| 3 3) = 13 { + |.
17. Let i ({> |) =
2{ + 3
2
38{ 3 12
. Then i{ ({> |) =
and i| ({> |) = (2{ + 3)(31)(4| + 1)32 (4) =
. Both i{ and i|
4| + 1
4| + 1
(4| + 1)2
are continuous functions for | 6= 3 14 , so by Theorem 8, i is differentiable at (0> 0). We have i{ (0> 0) = 2, i| (0> 0) = 312
and the linear approximation of i at (0> 0) is i ({> |) E i (0> 0) + i{ (0> 0)({ 3 0) + i| (0> 0)(| 3 0) = 3 + 2{ 3 12|.
18. Let i ({> |) =
s
s
| + cos2 {. Then i{ ({> |) = 12 (| + cos2 {)31@2 (2 cos {)(3 sin {) = 3 cos { sin {@ | + cos2 { and
s
i| ({> |) = 12 (| + cos2 {)31@2 (1) = 1@ 2 | + cos2 { . Both i{ and i| are continuous functions for | A 3 cos2 {, so i
is differentiable at (0> 0) by Theorem 8. We have i{ (0> 0) = 0 and i| (0> 0) = 12 , so the linear approximation of i at (0> 0) is
i ({> |) E i (0> 0) + i{ (0> 0)({ 3 0) + i| (0> 0)(| 3 0) = 1 + 0{ + 12 | = 1 + 12 |.
19. We can estimate i(2=2> 4=9) using a linear approximation of i at (2> 5), given by
i ({> |) E i(2> 5) + i{ (2> 5)({ 3 2) + i| (2> 5)(| 3 5) = 6 + 1({ 3 2) + (31)(| 3 5) = { 3 | + 9. Thus
i (2=2> 4=9) E 2=2 3 4=9 + 9 = 6=3.
20. i ({> |) = 1 3 {| cos |
i i{ ({> |) = 3| cos | and
i| ({> |) = 3{[|(3 sin |) + (cos |)(1)] = {| sin | 3 { cos |, so i{ (1> 1) = 1, i| (1> 1) = 1. Then the linear
approximation of i at (1> 1) is given by
i({> |) E i (1> 1) + i{ (1> 1)({ 3 1) + i| (1> 1)(| 3 1)
= 2 + (1)({ 3 1) + (1)(| 3 1) = { + |
Thus i (1=02> 0=97) E 1=02 + 0=97 = 1=99. We graph i and its
tangent plane near the point (1> 1> 2) below. Notice near | = 1 the
surfaces are almost identical.
21. i ({> |> }) =
s
{2 + | 2 + } 2
{
|
i i{ ({> |> }) = s
, i| ({> |> }) = s
, and
2
2
2
2
{ +| +}
{ + |2 + }2
}
, so i{ (3> 2> 6) = 37 , i| (3> 2> 6) = 27 , i} (3> 2> 6) = 67 . Then the linear approximation of i
i} ({> |> }) = s
{2 + | 2 + } 2
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420
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CHAPTER 14
PARTIAL DERIVATIVES
at (3> 2> 6) is given by
i ({> |> }) E i (3> 2> 6) + i{ (3> 2> 6)({ 3 3) + i| (3> 2> 6)(| 3 2) + i} (3> 2> 6)(} 3 6)
= 7 + 37 ({ 3 3) + 27 (| 3 2) + 67 (} 3 6) = 37 { + 27 | + 67 }
Thus
s
(3=02)2 + (1=97)2 + (5=99)2 = i (3=02> 1=97> 5=99) E 37 (3=02) + 27 (1=97) + 67 (5=99) E 6=9914.
22. From the table, i (40> 20) = 28. To estimate iy (40> 20) and iw (40> 20) we follow the procedure used in Exercise 14.3.4. Since
i (40 + k> 20) 3 i (40> 20)
, we approximate this quantity with k = ±10 and use the values given in the
k
iy (40> 20) = lim
k<0
table:
iy (40> 20) E
40 3 28
i (50> 20) 3 i (40> 20)
=
= 1=2,
10
10
iy (40> 20) E
Averaging these values gives iy (40> 20) E 1=15. Similarly, iw (40> 20) = lim
k<0
17 3 28
i (30> 20) 3 i (40> 20)
=
= 1=1
310
310
i (40> 20 + k) 3 i (40> 20)
, so we use k = 10
k
and k = 35:
iw (40> 20) E
31 3 28
i (40> 30) 3 i(40> 20)
=
= 0=3,
10
10
iw (40> 20) E
25 3 28
i (40> 15) 3 i(40> 20)
=
= 0=6
35
35
Averaging these values gives iw (40> 15) E 0=45. The linear approximation, then, is
i (y> w) E i (40> 20) + iy (40> 20)(y 3 40) + iw (40> 20)(w 3 20) E 28 + 1=15(y 3 40) + 0=45(w 3 20)
When y = 43 and w = 24, we estimate i (43> 24) E 28 + 1=15(43 3 40) + 0=45(24 3 20) = 33=25, so we would expect the
wave heights to be approximately 33=25 ft.
23. From the table, i (94> 80) = 127. To estimate iW (94> 80) and iK (94> 80) we follow the procedure used in Section 14.3. Since
iW (94> 80) = lim
k<0
i(94 + k> 80) 3 i (94> 80)
, we approximate this quantity with k = ±2 and use the values given in the
k
table:
iW (94> 80) E
135 3 127
i (96> 80) 3 i(94> 80)
=
= 4,
2
2
iW (94> 80) E
Averaging these values gives iW (94> 80) E 4. Similarly, iK (94> 80) = lim
k<0
iK (94> 80) E
132 3 127
i (94> 85) 3 i (94> 80)
=
= 1,
5
5
119 3 127
i(92> 80) 3 i (94> 80)
=
=4
32
32
i (94> 80 + k) 3 i (94> 80)
, so we use k = ±5:
k
iK (94> 80) E
122 3 127
i(94> 75) 3 i (94> 80)
=
=1
35
35
Averaging these values gives iK (94> 80) E 1. The linear approximation, then, is
i (W> K) E i (94> 80) + iW (94> 80)(W 3 94) + iK (94> 80)(K 3 80)
E 127 + 4(W 3 94) + 1(K 3 80)
[or 4W + K 3 329]
Thus when W = 95 and K = 78, i (95> 78) E 127 + 4(95 3 94) + 1(78 3 80) = 129, so we estimate the heat index to be
approximately 129 F.
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SECTION 14.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
¤
421
24. From the table, i(315> 50) = 329. To estimate iW (315> 50) and iy (315> 50) we follow the procedure used in Section 14.3.
Since iW (315> 50) = lim
k<0
i (315 + k> 50) 3 i (315> 50)
, we approximate this quantity with k = ±5 and use the values
k
given in the table:
iW (315> 50) E
i (310> 50) 3 i(315> 50)
322 3 (329)
=
= 1=4
5
5
iW (315> 50) E
335 3 (329)
i (320> 50) 3 i (315> 50)
=
= 1=2
35
35
Averaging these values gives iW (315> 50) E 1=3. Similarly iy (315> 50) = lim
k<0
i(315> 50 + k) 3 i(315> 50)
,
k
so we use k = ±10:
iy (315> 50) E
330 3 (329)
i (315> 60) 3 i (315> 50)
=
= 30=1
10
10
iy (315> 50) E
i (315> 40) 3 i (315> 50)
327 3 (329)
=
= 30=2
310
310
Averaging these values gives iy (315> 50) E 30=15. The linear approximation to the wind-chill index function, then, is
i (W> y) E i (315> 50) + iW (315> 50)(W 3 (315)) + iy (315> 50)(y 3 50) E 329 + (1=3)(W + 15) 3 (0=15)(y 3 50).
Thus when W = 317 C and y = 55 km@h, i (317> 55) E 329 + (1=3)(317 + 15) 3 (0=15)(55 3 50) = 332=35, so we
estimate the wind-chill index to be approximately 332=35 C.
25. } = h32{ cos 2w
C}
C}
g{ +
gw = h32{ (32) cos 2w g{ + h32{ (3 sin 2w)(2) gw = 32h32{ cos 2w g{ 3 2h32{ sin 2w gw
C{
Cw
g} =
26. x =
s
{2 + 3| 2 = ({2 + 3| 2 )1@2
gx =
i
Cx
Cx
{
3|
g{ +
g| = 12 ({2 + 3| 2 )31@2 (2{) g{ + 12 ({2 + 3| 2 )31@2 (6|) g| = s
g{ + s
g|
C{
C|
{2 + 3| 2
{2 + 3| 2
27. p = s5 t 3
28. W =
i
i gp =
y
1 + xyz
Cp
Cp
gs +
gt = 5s4 t 3 gs + 3s5 t 2 gt
Cs
Ct
i
gW =
CW
CW
CW
gx +
gy +
gz
Cx
Cy
Cz
= y(31)(1 + xyz)32 (yz) gx +
29. U = 2 cos 1(1 + xyz) 3 y(xz)
gy + y(31)(1 + xyz)32 (xy) gz
(1 + xyz)2
=3
1
xy 2
y2 z
gx +
gy 3
gz
2
2
(1 + xyz)
(1 + xyz)
(1 + xyz)2
i
gU =
CU
CU
CU
g +
g +
g = 2 cos g + 2 cos g 3 2 sin g
C
C
C
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¤
CHAPTER 14
30. O = {}h3|
gO =
2 3} 2
PARTIAL DERIVATIVES
i
2
2
2
2
2
2
2
2
CO
CO
CO
g{ +
g| +
g} = }h3| 3} g{ + {}h3| 3} (32|) g| + {[} · h3| 3} (32}) + h3| 3} · 1] g}
C{
C|
C}
= }h3|
2
3} 2
g{ 3 2{|}h3|
2
3} 2
g| + {(1 3 2} 2 )h3|
2
3} 2
g}
31. g{ = {{ = 0=05, g| = {| = 0=1, } = 5{2 + | 2 , }{ = 10{, }| = 2|. Thus when { = 1 and | = 2,
g} = }{ (1> 2) g{ + }| (1> 2) g| = (10)(0=05) + (4)(0=1) = 0=9 while
{} = i (1=05> 2=1) 3 i (1> 2) = 5(1=05)2 + (2=1)2 3 5 3 4 = 0=9225.
32. g{ = {{ = 30=04, g| = {| = 0=05, } = {2 3 {| + 3| 2 , }{ = 2{ 3 |, }| = 6| 3 {. Thus when { = 3 and | = 31,
g} = (7)(30=04) + (39)(0=05) = 30=73 while {} = (2=96)2 3 (2=96)(30=95) + 3(30=95)2 3 (9 + 3 + 3) = 30=7189.
33. gD =
CD
CD
g{ +
g| = | g{ + { g| and |{{| $ 0=1, |{|| $ 0=1. We use g{ = 0=1, g| = 0=1 with { = 30, | = 24; then
C{
C|
the maximum error in the area is about gD = 24(0=1) + 30(0=1) = 5=4 cm2 .
34. Let Y be the volume. Then Y = u2 k and {Y E gY = 2uk gu + u2 gk is an estimate of the amount of metal. With
gu = 0=05 and gk = 0=2 we get gY = 2(2)(10)(0=05) + (2)2 (0=2) = 2=80 E 8=8 cm3 .
35. The volume of a can is Y = u2 k and {Y E gY is an estimate of the amount of tin. Here gY = 2uk gu + u2 gk, so put
gu = 0=04, gk = 0=08 (0=04 on top, 0=04 on bottom) and then {Y E gY = 2(48)(0=04) + (16)(0=08) E 16=08 cm3 .
Thus the amount of tin is about 16 cm3 .
36. Z = 13=12 + 0=6215W 3 11=37y 0=16 + 0=3965W y 0=16 , so the differential of Z is
CZ
CZ
gW +
gy = (0=6215 + 0=3965y 0=16 ) gW + 311=37(0=16)y 30=84 + 0=3965W (0=16)y 30=84 gy
CW
Cy
= (0=6215 + 0=3965y 0=16 ) gW + (31=8192 + 0=06344W )y 30=84 gy
gZ =
Here we have |{W | $ 1, |{y| $ 2, so we take gW = 1, gy = 2 with W = 311, y = 26. The maximum error in the calculated
value of Z is about gZ = (0=6215 + 0=3965(26)0=16 )(1) + (31=8192 + 0=06344(311))(26)30=84 (2) E 0=96.
37. W =
pjU
, so the differential of W is
2u2 + U2
gW =
=
CW
(2u2 + U2 )(pj) 3 pjU(2U)
CW
(2u2 + U2 )(0) 3 pjU(4u)
gU +
gu =
gU +
gu
2
2
2
CU
Cu
(2u + U )
(2u2 + U2 )2
pj(2u2 3 U2 )
4pjUu
gU 3
gu
(2u2 + U2 )2
(2u2 + U2 )2
Here we have {U = 0=1 and {u = 0=1, so we take gU = 0=1, gu = 0=1 with U = 3, u = 0=7. Then the change in the
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SECTION 14.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
¤
423
tension W is approximately
gW =
pj[2(0=7)2 3 (3)2 ]
4pj(3)(0=7)
(0=1) 3
(0=1)
[2(0=7)2 + (3)2 ]2
[2(0=7)2 + (3)2 ]2
=3
0=84pj
1=642
0=802pj
pj E 30=0165pj
3
=3
(9=98)2
(9=98)2
99=6004
Because the change is negative, tension decreases.
38. Here gY = {Y = 0=3, gW = {W = 35, S = 8=31
gS =
8=31
Y
39. First we find
C
CU1
1
U
gW 3
W
, so
Y
5
8=31 · W
310 3
gY
=
8=31
3
3
·
E 38=83. Thus the pressure will drop by about 8=83 kPa.
Y2
12
144 10
CU
implicitly by taking partial derivatives of both sides with respect to U1 :
CU1
=
C [(1@U1 ) + (1@U2 ) + (1@U3 )]
CU1
i 3U32
CU
= 3U132
CU1
i
CU
17
U2 CU
U2
1
=
= 2,
= 2 . When U1 = 25, U2 = 40 and U3 = 50,
CU2
U2 CU3
U3
U
200
CU
U2
= 2 . Then by symmetry,
CU1
U1
C U=
200
17
l. Since the possible error
for each Ul is 0=5%, the maximum error of U is attained by setting {Ul = 0=005Ul . So
CU
CU
CU
1
1
1
{U E gU =
{U1 +
{U2 +
{U3 = (0=005)U2
+
+
= (0=005)U =
CU1
CU2
CU3
U1
U2
U3
1
17
E 0=059 l.
40. Let {> |> } and z be the four numbers with s({> |> }> z) = {|}z. Since the largest error due to rounding
for each number is 0=05, the maximum error in the calculated product is approximated by
gs = (|}z)(0=05) + ({}z)(0=05) + ({|z)(0=05) + ({|})(0=05). Furthermore, each of the numbers is positive but less than
50, so the product of any three is between 0 and (50)3 . Thus gs $ 4(50)3 (0=05) = 25,000.
{z $ 0=02 and {k $ 0=02. The relative error in the calculated surface
z
k 41. The errors in measurement are at most 2%, so area is
gV 0=1091(0=425z0=42531 )k0=725 gz + 0=1091z0=425 (0=725k0=72531 ) gk
gk
gz
{V
E
=
+ 0=725
= 0=425
V
V
0=1091z0=425 k0=725
z
k
{z gz
= 0=02 and gk = {k = 0=02 i
= To estimate the maximum relative error, we use
k z
z k
gV
= 0=425 (0=02) + 0=725 (0=02) = 0=023. Thus the maximum percentage error is approximately 2=3%.
V
42. r1 (w) = 2 + 3w> 1 3 w2 > 3 3 4w + w2
r01 (w) = h3> 32w> 34 + 2wi, r2 (x) = 1 + x2 > 2x3 3 1> 2x + 1
i
r02 (x) = 2x> 6x2 > 2 . Both curves pass through S since r1 (0) = r2 (1) = h2> 1> 3i, so the tangent vectors r01 (0) = h3> 0> 34i
i
and r02 (1) = h2> 6> 2i are both parallel to the tangent plane to V at S . A normal vector for the tangent plane is
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¤
CHAPTER 14
PARTIAL DERIVATIVES
r01 (0) × r02 (1) = h3> 0> 34i × h2> 6> 2i = h24> 314> 18i, so an equation of the tangent plane is
24({ 3 2) 3 14(| 3 1) + 18(} 3 3) = 0 or 12{ 3 7| + 9} = 44.
43. {} = i (d + {{> e + {|) 3 i (d> e) = (d + {{)2 + (e + {|)2 3 (d2 + e2 )
= d2 + 2d {{ + ({{)2 + e2 + 2e {| + ({|)2 3 d2 3 e2 = 2d {{ + ({{)2 + 2e {| + ({|)2
But i{ (d> e) = 2d and i| (d> e) = 2e and so {} = i{ (d> e) {{ + i| (d> e) {| + {{ {{ + {| {|, which is Definition 7
with %1 = {{ and %2 = {|. Hence i is differentiable.
44. {} = i (d + {{> e + {|) 3 i (d> e) = (d + {{)(e + {|) 3 5(e + {|)2 3 (de 3 5e2 )
= de + d {| + e {{ + {{ {| 3 5e2 3 10e {| 3 5({|)2 3 de + 5e2
= (d 3 10e) {| + e {{ + {{ {| 3 5 {| {|,
but i{ (d> e) = e and i| (d> e) = d 3 10e and so {} = i{ (d> e) {{ + i| (d> e) {| + {{ {| 3 5{| {|, which is Definition 7
with %1 = {| and %2 = 35 {|. Hence i is differentiable.
45. To show that i is continuous at (d> e) we need to show that
equivalently
lim
({{>{|)<(0>0)
lim
({>|)<(d>e)
i ({> |) = i (d> e) or
i (d + {{> e + {|) = i (d> e). Since i is differentiable at (d> e),
i (d + {{> e + {|) 3 i (d> e) = {} = i{ (d> e) {{ + i| (d> e) {| + %1 {{ + %2 {|, where 1 and 2 < 0 as
({{> {|) < (0> 0). Thus i (d + {{> e + {|) = i(d> e) + i{ (d> e) {{ + i| (d> e) {| + %1 {{ + %2 {|. Taking the limit of
both sides as ({{> {|) < (0> 0) gives
46. (a) lim
k<0
lim
({{>{|)<(0>0)
i (d + {{> e + {|) = i (d> e). Thus i is continuous at (d> e).
i (k> 0) 3 i (0> 0)
i (0> k) 3 i (0> 0)
030
030
= lim
= 0 and lim
= lim
= 0. Thus i{ (0> 0) = i| (0> 0) = 0.
k<0
k<0
k<0
k
k
k
k
To show that i isn’t differentiable at (0> 0) we need only show that i is not continuous at (0> 0) and apply Exercise 45. As
({> |) < (0> 0) along the {-axis i ({> |) = 0@{2 = 0 for { 6= 0 so i ({> |) < 0 as ({> |) < (0> 0) along the {-axis. But
as ({> |) < (0> 0) along the line | = {, i ({> {) = {2 @ 2{2 =
line. Thus
lim
({>|)<(0>0)
1
2
for { 6= 0 so i ({> |) <
1
2
as ({> |) < (0> 0) along this
i ({> |) doesn’t exist, so i is discontinuous at (0> 0) and thus not differentiable there.
(b) For ({> |) 6= (0> 0), i{ ({> |) =
i{ ({> |) = i{ (0> |) =
({2 + | 2 )| 3 {|(2{)
|(| 2 3 {2 )
= 2
. If we approach (0> 0) along the |-axis, then
({2 + | 2 )2
({ + | 2 )2
|3
1
= , so i{ ({> |) < ±" as ({> |) < (0> 0). Thus
lim
i{ ({> |) does not exist and
({>|)<(0>0)
|4
|
i{ ({> |) is not continuous at (0> 0)= Similarly, i| ({> |) =
({2 + | 2 ){ 3 {|(2|)
{({2 3 | 2 )
= 2
for ({> |) 6= (0> 0), and
2
2
2
({ + | )
({ + | 2 )2
if we approach (0> 0) along the {-axis, then i| ({> |) = i{ ({> 0) =
{3
1
lim
= . Thus
i| ({> |) does not exist and
({>|)<(0>0)
{4
{
i| ({> |) is not continuous at (0> 0)=
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SECTION 14.5
THE CHAIN RULE
¤
425
14.5 The Chain Rule
1. } = {2 + | 2 + {|, { = sin w, | = hw
2. } = cos({ + 4|), { = 5w4 , | = 1@w
i
g}
C} g{
C} g|
=
+
= (2{ + |) cos w + (2| + {)hw
gw
C{ gw
C| gw
i
g}
C} g{
C} g|
=
+
= 3 sin({ + 4|)(1)(20w3 ) + [3 sin({ + 4|)(4)](3w32 )
gw
C{ gw
C| gw
4
4
3
3
20w
= 320w3 sin({ + 4|) + 2 sin({ + 4|) =
sin({ + 4|)
w
w2
3. } =
s
1 + {2 + | 2 , { = ln w, | = cos w i
{
1
1
C} g{ C} g|
g}
=
+
= 12 (1 + {2 + | 2 )31@2 (2{) · + 12 (1 + {2 + | 2 )31@2 (2|)(3 sin w) = s
3 | sin w
gw
C{ gw C| gw
w
1 + {2 + | 2 w
4. } = tan31 (|@{), { = hw , | = 1 3 h3w
i
1
g}
C} g{
C} g|
1
(3|{32 ) · hw +
(1@{) · (3h3w )(31)
=
+
=
gw
C{ gw
C| gw
1 + (|@{)2
1 + (|@{)2
=3
{2
1
{h3w 3 |hw
|
· h3w =
· hw +
2
2
+|
{ + | @{
{2 + | 2
5. z = {h|@} , { = w2 , | = 1 3 w, } = 1 + 2w
i
|
gz
1
{ 2{|
Cz g{ Cz g|
Cz g}
=
+
+
= h|@} · 2w + {h|@}
· (31) + {h|@} 3 2 · 2 = h|@} 2w 3 3 2
gw
C{ gw
C| gw
C} gw
}
}
}
}
6. z = ln
s
{2 + | 2 + } 2 =
1
2
ln({2 + | 2 + } 2 ), { = sin w, | = cos w, } = tan w i
gz
Cz g{ Cz g|
Cz g}
1
2{
2|
2}
1
1
=
+
+
= · 2
· cos w + · 2
· (3 sin w) + · 2
· sec2 w
gw
C{ gw
C| gw
C} gw
2 { + |2 + }2
2 { + |2 + }2
2 { + |2 + }2
=
{ cos w 3 | sin w + } sec2 w
{2 + | 2 + } 2
7. } = {2 | 3 , { = v cos w, | = v sin w
i
C} C{
C} C|
C}
=
+
= 2{|3 cos w + 3{2 | 2 sin w
Cv
C{ Cv
C| Cv
C} C{
C} C|
C}
=
+
= (2{| 3 )(3v sin w) + (3{2 | 2 )(v cos w) = 32v{| 3 sin w + 3v{2 | 2 cos w
Cw
C{ Cw
C| Cw
8. } = arcsin({ 3 |), { = v2 + w2 , | = 1 3 2vw
i
C}
C} C{
C} C|
1
1
2v + 2w
=
+
= s
(1) · 2v + s
(31) · (32w) = s
Cv
C{ Cv
C| Cv
1 3 ({ 3 |)2
1 3 ({ 3 |)2
1 3 ({ 3 |)2
1
2v + 2w
C}
C} C{
C} C|
1
(1) · 2w + s
(31) · (32v) = s
=
+
= s
Cw
C{ Cw
C| Cw
1 3 ({ 3 |)2
1 3 ({ 3 |)2
1 3 ({ 3 |)2
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¤
CHAPTER 14
PARTIAL DERIVATIVES
9. } = sin cos !, = vw2 , ! = v2 w
i
C}
C} C
C} C!
=
+
= (cos cos !)(w2 ) + (3 sin sin !)(2vw) = w2 cos cos ! 3 2vw sin sin !
Cv
C Cv
C! Cv
C} C
C} C!
C}
=
+
= (cos cos !)(2vw) + (3 sin sin !)(v2 ) = 2vw cos cos ! 3 v2 sin sin !
Cw
C Cw
C! Cw
10. } = h{+2| , { = v@w, | = w@v
i
C} C{
C} C|
C}
=
+
= (h{+2| )(1@w) + (2h{+2| )(3wv32 ) = h{+2|
Cv
C{ Cv
C| Cv
C} C{
C} C|
C}
=
+
= (h{+2| )(3vw32 ) + (2h{+2| )(1@v) = h{+2|
Cw
C{ Cw
C| Cw
11. } = hu cos , u = vw, =
I
v2 + w2
1
2w
3 2
w
v
2
v
3 2
v
w
i
v
C}
C} Cu
C} C
=
+
= hu cos · w + hu (3 sin ) · 12 (v2 + w2 )31@2 (2v) = whu cos 3 hu sin · I
Cv
Cu Cv
C Cv
v2 + w2
v
= hu w cos 3 I
sin v2 + w2
w
C}
C} Cu
C} C
=
+
= hu cos · v + hu (3 sin ) · 12 (v2 + w2 )31@2 (2w) = vhu cos 3 hu sin · I
Cw
Cu Cw
C Cw
v2 + w2
w
= hu v cos 3 I
sin v2 + w2
12. } = tan(x@y), x = 2v + 3w, y = 3v 3 2w
i
C} Cx
C} Cy
C}
=
+
= sec2 (x@y)(1@y) · 2 + sec2 (x@y)(3xy 32 ) · 3
Cv
Cx Cv
Cy Cv
x 3x
x 2y 3 3x
2
2 x
sec
= sec2
3 2 sec2
=
y
y
y
y
y2
y
C}
C} Cx
C} Cy
=
+
= sec2 (x@y)(1@y) · 3 + sec2 (x@y)(3xy 32 ) · (32)
Cw
Cx Cw
Cy Cw
x 2x
x 2x + 3y
x
3
= sec2
+ 2 sec2
=
sec2
2
y
y
y
y
y
y
13. When w = 3, { = j(3) = 2 and | = k(3) = 7. By the Chain Rule (2),
g}
Ci g{
Ci g|
=
+
= i{ (2> 7)j 0 (3) + i| (2> 7) k0 (3) = (6)(5) + (38)(34) = 62.
gw
C{ gw
C| gw
14. By the Chain Rule (3),
CZ Cx
CZ Cy
CZ
=
+
. Then
Cv
Cx Cv
Cy Cv
Zv (1> 0) = Ix (x(1> 0)> y(1> 0)) xv (1> 0) + Iy (x(1> 0)> y(1> 0)) yv (1> 0) = Ix (2> 3)xv (1> 0) + Iy (2> 3)yv (1> 0)
= (31)(32) + (10)(5) = 52
Similarly,
CZ
CZ Cx
CZ Cy
=
+
Cw
Cx Cw
Cy Cw
i
Zw (1> 0) = Ix (x(1> 0)> y(1> 0)) xw (1> 0) + Iy (x(1> 0)> y(1> 0)) yw (1> 0) = Ix (2> 3)xw (1> 0) + Iy (2> 3)yw (1> 0)
= (31)(6) + (10)(4) = 34
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SECTION 14.5
15. j(x> y) = i({(x> y)> |(x> y)) where { = hx + sin y, | = hx + cos y
THE CHAIN RULE
¤
427
i
C{
C|
Cj
Ci C{
Ci C|
C{
C|
= hx ,
= cos y,
= hx ,
= 3 sin y. By the Chain Rule (3),
=
+
. Then
Cx
Cy
Cx
Cy
Cx
C{ Cx
C| Cx
jx (0> 0) = i{ ({(0> 0)> |(0> 0)) {x (0> 0) + i| ({(0> 0)> |(0> 0)) |x (0> 0) = i{ (1> 2)(h0 ) + i| (1> 2)(h0 ) = 2(1) + 5(1) = 7.
Similarly,
Ci C{
Ci C|
Cj
=
+
. Then
Cy
C{ Cy
C| Cy
jy (0> 0) = i{ ({(0> 0)> |(0> 0)) {y (0> 0) + i| ({(0> 0)> |(0> 0)) |y (0> 0) = i{ (1> 2)(cos 0) + i| (1> 2)(3 sin 0)
= 2(1) + 5(0) = 2
16. j(u> v) = i ({(u> v)> |(u> v)) where { = 2u 3 v, | = v2 3 4u
By the Chain Rule (3)
i
C{
C{
C|
C|
= 2,
= 31,
= 34,
= 2v.
Cu
Cv
Cu
Cv
Ci C{ Ci C|
Cj
=
+
. Then
Cu
C{ Cu
C| Cu
ju (1> 2) = i{ ({(1> 2)> |(1> 2)) {u (1> 2) + i| ({(1> 2)> |(1> 2)) |u (1> 2) = i{ (0> 0)(2) + i| (0> 0)(34)
= 4(2) + 8(34) = 324
Similarly,
Ci C{ Ci C|
Cj
=
+
. Then
Cv
C{ Cv
C| Cv
jv (1> 2) = i{ ({(1> 2)> |(1> 2)) {v (1> 2) + i| ({(1> 2)> |(1> 2)) |v (1> 2) = i{ (0> 0)(31) + i| (0> 0)(4)
= 4(31) + 8(4) = 28
17.
x = i({> |), { = {(u> v> w), | = |(u> v> w) i
Cx C{
Cx C|
Cx
=
+
,
Cu
C{ Cu
C| Cu
Cx
Cx C{
Cx C|
=
+
,
Cv
C{ Cv
C| Cv
Cx
Cx C{
Cx C|
=
+
Cw
C{ Cw
C| Cw
18.
U = i({> |> }> w), { = {(x> y> z), | = |(x> y> z), } = }(x> y> z),
w = w(x> y> z) i
CU
CU C{
CU C|
CU C}
CU Cw
=
+
+
+
,
Cx
C{ Cx
C| Cx
C} Cx
Cw Cx
CU C{ CU C|
CU C}
CU Cw
CU
=
+
+
+
,
Cy
C{ Cy
C| Cy
C} Cy
Cw Cy
CU C{
CU C|
CU C}
CU Cw
CU
=
+
+
+
Cz
C{ Cz
C| Cz
C} Cz
Cw Cz
19.
z = i (u> v> w), u = u({> |), v = v({> |), w = w({> |) i
Cz
Cz Cu
Cz Cv
Cz Cw Cz
Cz Cu
Cz Cv
Cz Cw
=
+
+
,
=
+
+
C{
Cu C{
Cv C{
Cw C{ C|
Cu C|
Cv C|
Cw C|
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
°
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428
¤
CHAPTER 14
PARTIAL DERIVATIVES
20.
w = i (x> y> z), x = x(s> t> u> v), y = y(s> t> u> v), z = z(s> t> u> v) i
Cw
Cw Cx
Cw Cy
Cw Cz Cw
Cw Cx
Cw Cy
Cw Cz
=
+
+
,
=
+
+
,
Cs
Cx Cs
Cy Cs
Cz Cs Ct
Cx Ct
Cy Ct
Cz Ct
Cw
Cw Cx
Cw Cy
Cw Cz Cw
Cw Cx
Cw Cy
Cw Cz
=
+
+
,
=
+
+
Cu
Cx Cu
Cy Cu
Cz Cu Cv
Cx Cv
Cy Cv
Cz Cv
21. } = {4 + {2 |, { = v + 2w 3 x, | = vwx2
i
C}
C} C{
C} C|
=
+
= (4{3 + 2{|)(1) + ({2 )(wx2 ),
Cv
C{ Cv
C| Cv
C}
C} C{
C} C|
=
+
= (4{3 + 2{|)(2) + ({2 )(vx2 ),
Cw
C{ Cw
C| Cw
C} C{
C} C|
C}
=
+
= (4{3 + 2{|)(31) + ({2 )(2vwx).
Cx
C{ Cx
C| Cx
When v = 4, w = 2, and x = 1 we have { = 7 and | = 8,
so
C}
C}
C}
= (1484)(1) + (49)(2) = 1582,
= (1484) (2) + (49)(4) = 3164,
= (1484)(31) + (49)(16) = 3700.
Cv
Cw
Cx
I
I
22. W = y@(2x + y) = y(2x + y)31 , x = st u, y = s t u
i
I
CW
CW Cx
CW Cy
(2x + y)(1) 3 y(1) I
=
+
= [3y(2x + y)32 (2)](t u) +
( t u)
Cs
Cx Cs
Cy Cs
(2x + y)2
=
I
32y
2x
I
(t u) +
( t u),
(2x + y)2
(2x + y)2
I
su
CW Cx
CW Cy
32y
2x
CW
=
+
=
(s u) +
I ,
Ct
Cx Ct
Cy Ct
(2x + y)2
(2x + y)2 2 t
CW
st
2x
CW Cx
CW Cy
32y
I
I +
(s t).
=
+
=
Cu
Cx Cu
Cy Cu
(2x + y)2 2 u
(2x + y)2
When s = 2, t = 1, and u = 4 we have x = 4 and y = 8,
so
1
1
1
1
1 1 1 CW
CW
CW
(2) + 32
(4) = 0,
(4) + 32
(4) = 3 18 ,
+ 32 (2) =
= 3 16
= 3 16
= 3 16
2
Cs
Ct
Cu
23. z = {| + |} + }{, { = u cos , | = u sin , } = u
1
32 .
i
Cz
Cz C{
Cz C|
Cz C}
=
+
+
= (| + })(cos ) + ({ + })(sin ) + (| + {)(),
Cu
C{ Cu
C| Cu
C} Cu
Cz
Cz C{
Cz C|
Cz C}
=
+
+
= (| + })(3u sin ) + ({ + })(u cos ) + (| + {)(u).
C
C{ C
C| C
C} C
When u = 2 and = @2 we have { = 0, | = 2, and } = , so
Cz
= (2 + )(0) + (0 + )(1) + (2 + 0)(@2) = 2 and
Cu
Cz
= (2 + )(32) + (0 + )(0) + (2 + 0)(2) = 32.
C
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SECTION 14.5
24. S =
I
x2 + y2 + z2 = (x2 + y 2 + z2 )1@2 , x = {h| , y = |h{ , z = h{|
THE CHAIN RULE
¤
429
i
CS Cx
CS Cy
CS Cz
CS
=
+
+
C{
Cx C{
Cy C{
Cz C{
= 12 (x2 + y 2 + z2 )31@2 (2x)(h| ) + 12 (x2 + y 2 + z2 )31@2 (2y)(|h{ ) + 12 (x2 + y 2 + z2 )31@2 (2z)(|h{| )
=
xh| + y|h{ + z|h{|
I
,
x2 + y 2 + z2
y
z
CS
CS Cx
CS Cy
CS Cz
x
({h| ) + I
(h{ ) + I
({h{| )
=
+
+
= I
C|
Cx C|
Cy C|
Cz C|
x2 + y 2 + z2
x2 + y 2 + z2
x2 + y 2 + z2
=
x{h| + yh{ + z{h{|
I
.
x2 + y 2 + z2
When { = 0 and | = 2 we have x = 0, y = 2, and z = 1, so
25. Q =
s+t
, s = x + yz, t = y + xz, u = z + xy
s+u
6
CS
2
CS
0+4+2
0+2+0
I
I
= I and
= I .
=
=
C{
C|
5
5
5
5
i
CQ Cs
CQ Ct
CQ Cu
CQ
=
+
+
Cx
Cs Cx
Ct Cx
Cu Cx
=
(s + u)(1) 3 (s + t)(0)
(s + u)(0) 3 (s + t)(1)
(s + u)(1) 3 (s + t)(1)
(1) +
(z) +
(y)
(s + u)2
(s + u)2
(s + u)2
=
(u 3 t) + (s + u)z 3 (s + t)y
,
(s + u)2
CQ
s+u
3(s + t)
(u 3 t)z + (s + u) 3 (s + t)x
CQ Cs
CQ Ct
CQ Cu
u3t
(z) +
(1) +
(x) =
,
=
+
+
=
Cy
Cs Cy
Ct Cy
Cu Cy
(s + u)2
(s + u)2
(s + u)2
(s + u)2
s+u
3(s + t)
(u 3 t)y + (s + u)x 3 (s + t)
CQ
CQ Cs
CQ Ct
CQ Cu
u3t
(y) +
(x) +
(1) =
.
=
+
+
=
Cz
Cs Cz
Ct Cz
Cu Cz
(s + u)2
(s + u)2
(s + u)2
(s + u)2
When x = 2, y = 3, and z = 4 we have s = 14, t = 11, and u = 10, so
31 + (24)(4) 3 (25)(3)
20
CQ
5
=
=
=
,
Cx
(24)2
576
144
(31)(4) + 24 3 (25)(2)
(31)(3) + (24)(2) 3 25
330
20
5
CQ
5
CQ
=
=
=
= 3 , and
=
=
.
Cy
(24)2
576
96
Cz
(24)2
576
144
26. x = {hw| , { = 2 , | = 2 , w = 2 i
Cx
Cx C{
Cx C|
Cx Cw
=
+
+
= hw| (2) + {whw| (0) + {|hw| ( 2 ) = hw| (2 + {| 2 ),
C
C{ C
C| C
Cw C
Cx
Cx C{
Cx C|
Cx Cw
=
+
+
= hw| (2 ) + {whw| (2) + {|hw| (0) = hw| (2 + 2{w),
C
C{ C
C| C
Cw C
Cx
Cx C{
Cx C|
Cx Cw
=
+
+
= hw| (0) + {whw| ( 2 ) + {|hw| (2) = hw| ({w 2 + 2{|).
C
C{ C
C| C
Cw C
When = 31, = 2, and = 1 we have { = 2, | = 4, and w = 31, so
Cx
= h34 (34 + 8) = 4h34 ,
C
Cx
Cx
= h34 (1 3 8) = 37h34 , and
= h34 (38 3 16) = 324h34 .
C
C
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°
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430
¤
CHAPTER 14
PARTIAL DERIVATIVES
27. | cos { = {2 + | 2 , so let I ({> |) = | cos { 3 {2 3 | 2 = 0. Then by Equation 6
I{
2{ + | sin {
3| sin { 3 2{
g|
=3
=
.
=3
g{
I|
cos { 3 2|
cos { 3 2|
28. cos({|) = 1 + sin |, so let I ({> |) = cos({|) 3 1 3 sin | = 0. Then by Equation 6
g|
I{
| sin({|)
3 sin({|)(|)
=3
=3
.
=3
g{
I|
3 sin({|)({) 3 cos |
cos | + { sin({|)
29. tan31 ({2 |) = { + {| 2 , so let I ({> |) = tan31 ({2 |) 3 { 3 {| 2 = 0. Then
and
I{ ({> |) =
1
2{|
2{| 3 (1 + |2 )(1 + {4 | 2 )
2
2
(2{|)
3
1
3
|
=
3
1
3
|
=
,
1 + ({2 |)2
1 + {4 | 2
1 + {4 | 2
I| ({> |) =
1
{2
{2 3 2{|(1 + {4 | 2 )
({2 ) 3 2{| =
3 2{| =
2
2
4
2
1 + ({ |)
1+{ |
1 + {4 | 2
I{
(1 + | 2 )(1 + {4 | 2 ) 3 2{|
g|
[2{| 3 (1 + | 2 )(1 + {4 | 2 )]@(1 + {4 | 2 )
=3
=
=3
g{
I|
[{2 3 2{|(1 + {4 | 2 )]@(1 + {4 | 2 )
{2 3 2{|(1 + {4 | 2 )
=
1 + {4 | 2 + | 2 + {4 | 4 3 2{|
{2 3 2{| 3 2{5 | 3
30. h| sin { = { + {|, so let I ({> |) = h| sin { 3 { 3 {| = 0. Then
I{
1 + | 3 h| cos {
g|
h| cos { 3 1 3 |
=3
=
.
=3
g{
I|
h| sin { 3 {
h| sin { 3 {
31. {2 + 2| 2 + 3} 2 = 1, so let I ({> |> }) = {2 + 2| 2 + 3} 2 3 1 = 0. Then by Equations 7
2{
C}
I{
{
=3
=3
=3
C{
I}
6}
3}
and
4|
C}
I|
2|
=3
=3
=3 .
C|
I}
6}
3}
32. {2 3 | 2 + } 2 3 2} = 4, so let I ({> |> }) = {2 3 | 2 + } 2 3 2} 3 4 = 0. Then by Equations 7
2{
C}
I{
{
=3
=3
=
C{
I}
2} 3 2
13}
and
32|
C}
I|
|
=3
=3
=
.
C|
I}
2} 3 2
}31
33. h} = {|}, so let I ({> |> }) = h} 3 {|} = 0. Then
I{
|}
C}
3|}
=3
= }
=3 }
C{
I}
h 3 {|
h 3 {|
and
C}
I|
{}
3{}
=3
= }
.
=3 }
C|
I}
h 3 {|
h 3 {|
34. |} + { ln | = } 2 , so let I ({> |> }) = |} + { ln | 3 } 2 = 0. Then
I{
ln |
C}
ln |
=3
=
=3
C{
I}
| 3 2}
2} 3 |
and
C}
I|
{ + |}
} + ({@|)
=3
=
=3
.
C|
I}
| 3 2}
2|} 3 | 2
35. Since { and | are each functions of w, W ({> |) is a function of w, so by the Chain Rule,
3 seconds, { =
Then
I
I
1 + w = 1 + 3 = 2, | = 2 +
1
3
w = 2 + 13 (3) = 3,
CW g{
CW g|
gW
=
+
. After
gw
C{ gw
C| gw
1
1
g{
1
g|
1
= I
= , and
= I
= .
gw
4
gw
3
2 1+w
2 1+3
g{
g|
gW
= W{ (2> 3)
+ W| (2> 3)
= 4 14 + 3 13 = 2. Thus the temperature is rising at a rate of 2 C@s.
gw
gw
gw
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 14.5
THE CHAIN RULE
¤
431
36. (a) Since CZ@CW is negative, a rise in average temperature (while annual rainfall remains constant) causes a decrease in
wheat production at the current production levels. Since CZ@CU is positive, an increase in annual rainfall (while the
average temperature remains constant) causes an increase in wheat production.
(b) Since the average temperature is rising at a rate of 0=15 C@year, we know that gW @gw = 0=15. Since rainfall is
decreasing at a rate of 0=1 cm@year, we know gU@gw = 30=1. Then, by the Chain Rule,
gZ
CZ gW
CZ gU
=
+
= (32)(0=15) + (8)(30=1) = 31=1. Thus we estimate that wheat production will decrease
gw
CW gw
CU gw
at a rate of 1=1 units@year.
37. F = 1449=2 + 4=6W 3 0=055W 2 + 0=00029W 3 + 0=016G, so
CF
CF
= 4=6 3 0=11W + 0=00087W 2 and
= 0=016.
CW
CG
According to the graph, the diver is experiencing a temperature of approximately 12=5 C at w = 20 minutes, so
CF
= 4=6 3 0=11(12=5) + 0=00087(12=5)2 E 3=36. By sketching tangent lines at w = 20 to the graphs given, we estimate
CW
1
1
gW
1
gF
CF gW
CF gG
gG
+ (0=016) 12 E 30=33.
E and
E 3 . Then, by the Chain Rule,
=
+
E (3=36) 3 10
gw
2
gw
10
gw
CW gw
CG gw
Thus the speed of sound experienced by the diver is decreasing at a rate of approximately 0=33 m@s per minute.
38. Y = u2 k@3, so
CY gu
CY gk
2uk
u2
gY
=
+
=
1=8 +
(32=5) = 20,160 3 12,000 = 8160 in3@s.
gw
Cu gw
Ck gw
3
3
39. (a) Y = czk, so by the Chain Rule,
gY
CY gc
CY gz
CY gk
gc
gz
gk
=
+
+
= zk
+ ck
+ cz
= 2 · 2 · 2 + 1 · 2 · 2 + 1 · 2 · (33) = 6 m3@s.
gw
Cc gw
Cz gw
Ck gw
gw
gw
gw
(b) V = 2(cz + ck + zk), so by the Chain Rule,
gV
CV gc
CV gz
CV gk
gc
gz
gk
=
+
+
= 2(z + k)
+ 2(c + k)
+ 2(c + z)
gw
Cc gw
Cz gw
Ck gw
gw
gw
gw
= 2(2 + 2)2 + 2(1 + 2)2 + 2(1 + 2)(33) = 10 m2@s
(c) O2 = c2 + z2 + k2
i 2O
gO
gc
gz
gk
= 2c
+ 2z
+ 2k
= 2(1)(2) + 2(2)(2) + 2(2)(33) = 0 i
gw
gw
gw
gw
gO@gw = 0 m@s.
40. L =
Y
U
i
CL gY
CL gU
1 gY
Y gU
1 gY
L gU
1
0=08
gL
=
+
=
3 2
=
3
=
(30=01) 3
(0=03) = 30=000031 A@s
gw
CY gw
CU gw
U gw
U gw
U gw
U gw
400
400
41.
gS
gW
W
gY
8=31 gW
W gS
= 0=05,
= 0=15, Y = 8=31 and
=
3 8=31 2
. Thus when S = 20 and W = 320,
gw
gw
S
gw
S gw
S gw
0=15
(0=05)(320)
gY
= 8=31
3
E 30=27 L@s.
gw
20
400
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432
¤
CHAPTER 14
PARTIAL DERIVATIVES
42. S = 1=47O0=65 N 0=35 and considering S , O, and N as functions of time w we have
gO
gN
CS gO
CS gN
gS
=
+
= 1=47(0=65)O30=35 N 0=35
+ 1=47(0=35)O0=65 N 30=65
. We are given
gw
CO gw
CN gw
gw
gw
that
gN
gS
gO
= 32 and
= 0=5, so when O = 30 and N = 8, the rate of change of production
is
gw
gw
gw
1=47(0=65)(30)30=35 (8)0=35 (32) + 1=47(0=35)(30)0=65 (8)30=65 (0=5) E 30=596. Thus production at that time
is decreasing at a rate of about $596,000 per year.
43. Let { be the length of the first side of the triangle and | the length of the second side. The area D of the triangle is given by
D = 12 {| sin where is the angle between the two sides. Thus D is a function of {, |, and , and {, |, and are each in turn
functions of time w. We are given that
g|
gD
g{
= 3,
= 32, and because D is constant,
= 0. By the Chain Rule,
gw
gw
gw
CD g{
CD g|
CD g
gD
=
+
+
gw
C{ gw
C| gw
C gw
i
gD
g{
g|
g
= 12 | sin ·
+ 12 { sin ·
+ 12 {| cos ·
. When { = 20, | = 30,
gw
gw
gw
gw
and = @6 we have
g
0 = 12 (30) sin 6 (3) + 12 (20) sin 6 (32) + 12 (20)(30) cos 6
gw
I
I g
3 g
·
= 25
+ 150 3
= 45 · 12 3 20 · 12 + 300 ·
2
2 gw
gw
Solving for
g
g
325@2
1
I = 3 I , so the angle between the sides is decreasing at a rate of
gives
=
gw
gw
150 3
12 3
I 1@ 12 3 E 0=048 rad@s.
44. ir =
f + yr
f 3 yv
iv =
45. (a) By the Chain Rule,
(b)
C}
Cu
2
C}
C{
2
460 E 576=6 Hz. yr and yv are functions of time w, so
1
gir
Cir gyr
Cir gyv
f + yr
gyr
gyv
=
+
=
+
iv ·
2 iv ·
gw
Cyr gw
Cyv gw
f 3 yv
gw
gw
(f 3 yv )
1
332+34
(460) (1=2) + (332340)
= 332340
2 (460) (1=4) E 4=65 Hz@s
C}
C}
C}
C}
C}
C}
=
cos +
sin ,
=
(3u sin ) +
u cos .
Cu
C{
C|
C
C{
C|
cos2 + 2
C} C}
cos sin +
C{ C|
C}
C|
2
sin2 ,
2
2
C}
C}
C} C} 2
u cos sin +
u2 sin2 3 2
u2 cos2 . Thus
C{
C{ C|
C|
2 2
2 % 2 2 &
2
1 C}
C}
C}
C}
C}
C}
+ 2
=
+
+
.
(cos2 + sin2 ) =
Cu
u
C
C{
C|
C{
C|
C}
C
2
=
332+34
332340
=
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SECTION 14.5
46. By the Chain Rule,
Cx
Cv
2
=
Cx
C{
2
h2v cos2 w + 2
Cx Cx 2v
h cos w sin w +
C{ C|
Cx
C|
2
h2v sin2 w and
Cx
Cw
2
=
Cx
C{
2
h2v sin2 w 3 2
Cx Cx 2v
h cos w sin w +
C{ C|
Cx
C|
2
h2v sin2 w. Thus
Cx
Cv
2
+
Cx
Cw
2 &
2 2
Cx
Cx
h32v =
+
.
C{
C|
47. Let x = { 3 |. Then
48.
¤
433
Cx
Cx v
Cx
Cx v
Cx v
Cx
=
h cos w +
h sin w,
=
(3hv sin w) +
h cos w. Then
Cv
C{
C|
Cw
C{
C|
%
THE CHAIN RULE
g} Cx
g}
C}
g}
C}
C}
C}
=
=
and
=
(31). Thus
+
= 0.
C{
gx C{
gx
C|
gx
C{
C|
C}
C}
C}
C}
C}
C}
C} C}
=
+
and
=
3
. Thus
=
Cv
C{
C|
Cw
C{
C|
Cv Cw
C}
C{
2
3
C}
C|
2
.
49. Let x = { + dw, y = { 3 dw. Then } = i(x) + j(y), so C}@Cx = i 0 (x) and C}@Cy = j 0 (y).
C} Cx
C} Cy
C}
=
+
= di 0 (x) 3 dj 0 (y) and
Cw
Cx Cw
Cy Cw
0
C2}
gj 0 (y) Cy
C 0
gi (x) Cx
0
[i
3
= d2 i 00 (x) + d2 j00 (y).
=d
(x) 3 j (y)] = d
Cw2
Cw
gx Cw
gy Cw
Thus
Similarly
C2}
C2}
C2}
C}
= i 00 (x) + j00 (y). Thus 2 = d2
.
= i 0 (x) + j 0 (y) and
2
C{
C{
Cw
C{2
50. By the Chain Rule,
Then
Cx
Cx
Cx
Cx
Cx
Cx
= hv cos w
+ hv sin w
and
= 3hv sin w
+ hv cos w
.
Cv
C{
C|
Cw
C{
C|
Cx
C
C2x
= hv cos w
+ hv cos w
Cv2
C{
Cv
Cx
C{
+ hv sin w
Cx
C
+ hv sin w
C|
Cv
C
Cv
Cx
C{
=
C 2 x C{
C 2 x C|
C2x
C2x
+
= hv cos w 2 + hv sin w
and
2
C{ Cv
C| C{ Cv
C{
C| C{
C
Cv
Cx
C|
=
C2x
C2 x
C 2 x C|
C 2 x C{
+
= hv sin w 2 + hv cos w
.
2
C| Cv
C{ C| Cv
C|
C{ C|
Also, by continuity of the partials,
Cx
. But
C|
C2x
C2 x
=
. Thus
C{ C|
C| C{
C2x
Cx
C2x
C2x
Cx
C2x
C2x
v
v
v
v
v
v
v
v
+
h
+
h
+
h
=
h
cos
w
cos
w
h
cos
w
+
h
sin
w
sin
w
sin
w
h
sin
w
+
h
cos
w
Cv2
C{
C{2
C{ C|
C|
C| 2
C{ C|
= hv cos w
Cx
Cx
C2x
C2x
C2x
+ hv sin w
+ h2v cos2 w 2 + 2h2v cos w sin w
+ h2v sin2 w 2
C{
C|
C{
C{ C|
C|
[continued]
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434
¤
CHAPTER 14
PARTIAL DERIVATIVES
Similarly
Cx
C Cx
Cx
C Cx
C2x
v
v
v
v
3 h sin w
3 h sin w
+ h cos w
= 3h cos w
Cw2
C{
Cw C{
C|
Cw C|
2
2
Cx
C x
C x
3 hv sin w 3hv sin w 2 + hv cos w
= 3hv cos w
C{
C{
C{ C|
Cx
C2x
C2x
3hv sin w
+ hv cos w hv cos w 2 3 hv sin w
C|
C|
C{ C|
= 3hv cos w
Thus h32v
51.
C2x
C2x
+ 2
2
Cv
Cw
Cx
Cx
C2x
C2x
C2x
3 hv sin w
+ h2v sin2 w 2 3 2h2v cos w sin w
+ h2v cos2 w 2
C{
C|
C{
C{ C|
C|
2
C2x
C x
C2x
C2x
= (cos2 w + sin2 w)
=
+
+
, as desired.
2
2
2
C{
C|
C{
C| 2
C}
C}
C}
=
2v +
2u. Then
Cv
C{
C|
C}
C C}
2v +
2u
C{
Cu C|
C 2 } C{
C C} C|
C} C
C 2 } C|
C C} C{
C}
=
2v
+
2v
+
2v
+
2u
+
2u +
2
C{2 Cu
C| C{ Cu
C{ Cu
C| 2 Cu
C{ C| Cu
C|
C
C2}
=
Cu Cv
Cu
= 4uv
C2}
C2}
C2 }
C}
C2}
2
4v
4u2 + 2
+
+
0
+
4uv
+
C{2
C| C{
C| 2
C{ C|
C|
By the continuity of the partials,
C2}
C2}
C2}
C2}
C}
= 4uv 2 + 4uv 2 + (4u2 + 4v2 )
+2 .
CuCv
C{
C|
C{ C|
C|
52. By the Chain Rule,
(a)
C}
C}
C}
=
cos +
sin Cu
C{
C|
(c)
C2}
C2}
C
=
=
Cu C
C Cu
C
= 3 sin C}
C{
= 3 sin C}
C{
(b)
C}
C}
C}
=3
u sin +
u cos C
C{
C|
C}
C}
C}
C C}
C}
C C}
cos +
sin = 3 sin + cos + cos + sin C{
C|
C{
C C{
C|
C C|
2
C } C{
C 2 } C|
C}
C 2 } C|
C 2 } C{
+ cos +
+ cos + sin 2
+
2
C{ C
C| C{ C
C|
C| C
C{ C| C
C2}
C}
C2}
C2}
C2}
+ cos 3u sin 2 + u cos + cos + sin u cos 2 3 u sin C{
C| C{
C|
C|
C{ C|
C}
C2}
C}
C2}
C2}
C2}
3 u cos sin 2 + u cos2 + cos + u cos sin 2 3 u sin2 C{
C{
C| C{
C|
C|
C| C{
2
C }
C}
C2}
C2}
C}
3 sin + u cos sin 3
= cos + u(cos2 3 sin2 )
2
2
C|
C{
C|
C{
C| C{
= 3 sin c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 14.5
53.
THE CHAIN RULE
¤
C}
C}
C}
C}
C}
C}
=
cos +
sin and
=3
u sin +
u cos . Then
Cu
C{
C|
C
C{
C|
2
2
C2}
C }
C }
C2}
C2}
sin
+
sin
cos
=
cos
cos
+
sin
+
Cu2
C{2
C| C{
C| 2
C{ C|
= cos2 C2}
C2}
C2}
2
+
sin
+
2
cos
sin
C{2
C{ C|
C| 2
and
C2}
C2}
u
cos
(3u
sin
)
+
C{2
C| C{
2
C}
C2}
C }
3u sin u
cos
+
+ u cos (3u
sin
)
C|
C| 2
C{ C|
C2}
C}
+ (3u sin )
= 3u cos C{
C2
= 3u cos C}
C2}
C2}
C2}
C}
3 u sin + u2 sin2 2 3 2u2 cos sin + u2 cos2 2
C{
C|
C{
C{ C|
C|
Thus
C2}
C2}
1 C2}
1 C}
C2} = (cos2 + sin2 ) 2 + sin2 + cos2 + 2 2 +
2
Cu
u C
u Cu
C{
C| 2
C}
1
C}
1
C}
C}
1
3 sin +
cos + sin 3 cos u
C{
u
C|
u
C{
C|
=
54. (a)
(b)
C2}
C2}
+ 2 as desired.
2
C{
C|
C} C{
C} C|
C}
=
+
. Then
Cw
C{ Cw
C| Cw
C2}
C C} C|
C C} C{
C 2 { C}
C C} C|
C 2 | C}
C C} C{
+
=
+
+
+
=
Cw2
Cw C{ Cw
Cw C| Cw
Cw C{ Cw
Cw2 C{ Cw C| Cw
Cw2 C|
2
2
C 2 } C{
C 2 } C{ C|
C 2 } C| C{ C 2 | C}
C 2 { C}
C 2 } C|
=
+
+
+ 2
+ 2
+ 2
2
C{
Cw
C| C{ Cw Cw
Cw C{
C|
Cw
C{ C| Cw Cw
Cw C|
2
2
C 2 } C|
C 2 | C}
C 2 } C{
C 2 } C{ C|
C 2 { C}
+ 2
+ 2
=
+2
+ 2
2
C{
Cw
C{ C| Cw Cw
C|
Cw
Cw C{
Cw C|
C C} C{
C} C|
C2}
=
+
Cv Cw
Cv C{ Cw
C| Cw
2
2
C } C{
C } C|
C 2 } C| C{
C} C 2 {
C 2 } C{ C|
C} C 2 |
=
+
+
+
+
+
C{2 Cv
C| C{ Cv Cw
C{ Cv Cw
C| 2 Cv
C{ C| Cv Cw
C| Cv Cw
C 2 } C{ C{
C2}
C| C{ C| C{
C} C 2 {
C} C 2 |
C 2 } C| C|
=
+
+
+
+
+ 2
2
C{ Cv Cw
C{ C| Cv Cw
Cw Cv
C{ Cv Cw
C| Cv Cw
C| Cv Cw
55. (a) Since i is a polynomial, it has continuous second-order partial derivatives, and
i (w{> w|) = (w{)2 (w|) + 2(w{)(w|)2 + 5(w|)3 = w3 {2 | + 2w3 {| 2 + 5w3 | 3 = w3 ({2 | + 2{| 2 + 5| 3 ) = w3 i ({> |).
Thus, i is homogeneous of degree 3.
(b) Differentiating both sides of i (w{> w|) = wq i ({> |) with respect to w using the Chain Rule, we get
C
C q
i (w{> w|) =
[w i ({> |)] C
Cw
Cw
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436
¤
CHAPTER 14
PARTIAL DERIVATIVES
C
C(w{)
C
C(w|)
C
C
i (w{> w|) ·
+
i(w{> w|) ·
={
i (w{> w|) + |
i (w{> w|) = qwq31 i({> |).
C(w{)
Cw
C(w|)
Cw
C(w{)
C(w|)
Setting w = 1: {
C
C
i ({> |) + |
i ({> |) = qi ({> |).
C{
C|
56. Differentiating both sides of i (w{> w|) = wq i({> |) with respect to w using the Chain Rule, we get
C(w{)
C
C(w|)
C
C
C
i (w{> w|) ·
+
i (w{> w|) ·
={
i (w{> w|) + |
i(w{> w|) = qwq31 i({> |) and
C(w{)
Cw
C(w|)
Cw
C(w{)
C(w|)
differentiating again with respect to w gives
C2
C (w{)
C2
C (w|)
{
i
(w{>
w|)
·
+
i
(w{>
w|)
·
Cw
C (w|) C (w{)
Cw
C (w{)2
C2
C (w{)
C2
C (w|)
i(w{> w|) ·
+
= q(q 3 1)wq31 i ({> |).
+|
i
(w{>
w|)
·
C (w{) C (w|)
Cw
Cw
C (w|)2
Setting w = 1 and using the fact that i|{ = i{| , we have {2 i{{ + 2{|i{| + | 2 i|| = q(q 3 1)i ({> |).
57. Differentiating both sides of i (w{> w|) = wq i({> |) with respect to { using the Chain Rule, we get
C q
C
i (w{> w|) =
[w i ({> |)] C
C{
C{
C
C
C (w{)
C
C (w|)
i (w{> w|) ·
+
i (w{> w|) ·
= wq
i ({> |) C wi{ (w{> w|) = wq i{ ({> |).
C (w{)
C{
C (w|)
C{
C{
Thus i{ (w{> w|) = wq31 i{ ({> |).
58. I ({> |> }) = 0 is assumed to define } as a function of { and |, that is, } = i ({> |). So by (7),
I{
C}
=3
since I} 6= 0.
C{
I}
Similarly, it is assumed that I ({> |> }) = 0 defines { as a function of | and }, that is { = k({> }). Then I (k(|> })> |> }) = 0
C{
C|
C}
C{
C}
C|
+ I|
+ I}
= 0. But
= 0 and
= 1, so I{
+ I| = 0 i
C|
C|
C|
C|
C|
C|
I}
I{
C} C{ C|
C|
I|
I}
= 3 . Thus
= 3
A similar calculation shows that
3
3
= 31.
C}
I|
C{ C| C}
I}
I{
I|
and by the Chain Rule, I{
59. Given a function defined implicitly by I ({> |) = 0, where I is differentiable and I| 6= 0, we know that
I|
C{
=3 .
C|
I{
g|
I{
= 3 . Let
g{
I|
I{
g|
= J({> |). Differentiating both sides with respect to { and using the Chain Rule gives
so
I|
g{
CJ g{
I| I{{ 3 I{ I|{ CJ
I| I{| 3 I{ I||
g2 |
CJ g|
CJ
C
I{
C
I{
=
,
.
=
3
=3
+
where
=
3
=
3
g{2
C{ g{
C| g{
C{
C{
I|
I|2
C|
C|
I|
I|2
J({> |) = 3
Thus
g2 |
=
g{2
I| I{{ 3 I{ I|{
I| I{| 3 I{ I||
I{
3
(1)
+
3
3
I|2
I|2
I|
=3
I{{ I|2 3 I|{ I{ I| 3 I{| I| I{ + I|| I{2
I|3
But I has continuous second derivatives, so by Clauraut’s Theorem, I|{ = I{| and we have
I{{ I|2 3 2I{| I{ I| + I|| I{2
g2 |
=3
as desired.
2
g{
I|3
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 14.6
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
¤
437
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 of S, 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 the distance from K to S is 300 km). Then the rate of change of
pressure in the direction given is approximately
996 3 1000
50
= 30=08 millibar@km.
2. First we draw a line passing through Dubbo and Sydney. We approximate the directional derivative at Dubbo in the direction
of Sydney by the average rate of change of temperature between the points where the line intersects the contour lines closest to
Dubbo. In the direction of Sydney, the temperature changes from 30 C to 27 C. We estimate the distance between these two
points to be approximately 120 km, so the rate of change of maximum temperature in the direction given is approximately
27 3 30
120
= 30=025 C@km.
3. Gu i (320> 30) = Qi (320> 30) · u = iW (320> 30)
iW (320> 30) = lim
k<0
+ iy (320> 30) I12 .
i (315> 30) 3 i (320> 30)
326 3 (333)
=
= 1=4,
5
5
i (325> 30) 3 i (320> 30)
339 3 (333)
=
= 1=2. Averaging these values gives iW (320> 30) E 1=3.
35
35
Similarly, iy (320> 30) = lim
k<0
iy (320> 30) E
i(320 + k> 30) 3 i(320> 30)
, so we can approximate iW (320> 30) by considering k = ±5 and
k
using the values given in the table: iW (320> 30) E
iW (320> 30) E
I1
2
i (320> 30 + k) 3 i (320> 30)
, so we can approximate iy (320> 30) with k = ±10:
k
i (320> 40) 3 i(320> 30)
334 3 (333)
=
= 30=1,
10
10
330 3 (333)
i (320> 20) 3 i (320> 30)
=
= 30=3. Averaging these values gives iy (320> 30) E 30=2.
310
310
Then Gu i (320> 30) E 1=3 I12 + (30=2) I12 E 0=778.
iy (320> 30) E
4. i ({> |) = {3 | 4 + {4 | 3
direction of =
5. i ({> |) = |h3{
,
6
i i{ ({> |) = 3{2 | 4 + 4{3 | 3 and i| ({> |) = 4{3 | 3 + 3{4 | 2 . If u is a unit vector in the
then from Equation 6, Gu i(1> 1) = i{ (1> 1) cos 6 + i| (1> 1) sin 6 = 7 ·
I
3
2
+7·
1
2
=
I
7+7 3
.
2
i i{ ({> |) = 3|h3{ and i| ({> |) = h3{ . If u is a unit vector in the direction of = 2@3, then
from Equation 6, Gu i (0> 4) = i{ (0> 4) cos 2
+ i| (0> 4) sin 2
= 34 · 3 12 + 1 ·
3
3
6. i ({> |) = h{ cos |
I
3
2
= 2+
I
3
.
2
i i{ ({> |) = h{ cos | and i| ({> |) = 3h{ sin |. If u is a unit vector in the direction of =
from Equation 6, Gu i (0> 0) = i{ (0> 0) cos 4 + i| (0> 0) sin 4 = 1 ·
I
2
2
+0 =
I
2
2 .
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438
¤
CHAPTER 14
PARTIAL DERIVATIVES
7. i ({> |) = sin(2{ + 3|)
(a) Qi ({> |) =
Ci
Ci
i+
j = [cos(2{ + 3|) · 2] i + [cos(2{ + 3|) · 3] j = 2 cos (2{ + 3|) i + 3 cos (2{ + 3|) j
C{
C|
(b) Qi (36> 4) = (2 cos 0) i + (3 cos 0) j = 2 i + 3 j
1
2
(c) By Equation 9, Gu i(36> 4) = Qi (36> 4) · u = (2 i + 3 j) ·
8. i ({> |) = | 2 @{
(a) Qi ({> |) =
I
3i 3 j =
1
2
I
I
2 3 3 3 = 3 3 32 .
Ci
|2
2|
Ci
i+
j = | 2 (3{32 )i + (2|@{) j = 3 2 i +
j
C{
C|
{
{
(b) Qi (1> 2) = 34 i + 4 j
(c) By Equation 9, Gu i(1> 2) = Qi (1> 2) · u = (34 i + 4 j) ·
1
3
9. i ({> |> }) = {2 |} 3 {|} 3
I 2i + 5j =
1
3
I 38 + 4 5 =
(a) Qi ({> |> }) = hi{ ({> |> })> i| ({> |> })> i} ({> |> })i = 2{|} 3 |} 3 > {2 } 3 {} 3 > {2 | 3 3{|} 2
(b) Qi (2> 31> 1) = h34 + 1> 4 3 2> 34 + 6i = h33> 2> 2i
(c) By Equation 14, Gu i(2> 31> 1) = Qi (2> 31> 1) · u = h33> 2> 2i · 0> 45 > 3 35 = 0 +
8
5
3
6
5
4
3
I
532 .
= 25 .
10. i ({> |> }) = | 2 h{|}
(a) Qi ({> |> }) = hi{ ({> |> })> i| ({> |> })> i} ({> |> })i = | 2 h{|} (|})> | 2 · h{|} ({}) + h{|} · 2|> | 2 h{|} ({|)
= | 3 }h{|} > ({| 2 } + 2|)h{|} > {| 3 h{|}
(b) Qi (0> 1> 31) = h31> 2> 0i
(c) Gu i (0> 1> 31) = Qi (0> 1> 31) · u = h31> 2> 0i ·
11. i ({> |) = h{ sin |
1
(36)2 +82
Gu i (0> @3) = Qi (0> @3) · u =
{
12. i ({> |) = 2
{ + |2
3
4
> 3 25
25
GI
3 1
2 > 2
i Qi ({> |) =
3
= 3 13
+
H
h36> 8i =
1
10
8
13
3
4
> 3 25
25
3 1
>2
2
h36> 8i = 3 35 >
I
· 3 35 > 45 = 3 3103 +
4
10
=
9
I
25 34
3
20
I
25 34
5
13
H
, and a
I
433 3
10 .
({2 + | 2 )(1) 3 {(2{) 0 3 {(2|)
> 2
({2 + | 2 )2
({ + | 2 )2
H
G 3
· I34 > I534 =
+0=
GI
, and a unit vector in the direction of v = h3> 5i is u =
Gu i (1> 2) = Qi (1> 2) · u =
13. j(s> t) = s4 3 s2 t 3
i Qi ({> |) = hh{ sin |> h{ cos |i, Qi (0> @3) =
unit vector in the direction of v is u = I
Qi (1> 2) =
3
4 12
13 > 13 > 13
=
4
5
, so
I 1
9+25
|2 3 {2
2{|
,
>3 2
({2 + | 2 )2
({ + | 2 )2
h3> 5i =
G
I3 > I5
34
34
H
, so
I .
= 3 2511
34
i Qj(s> t) = 4s3 3 2st 3 i + 33s2 t 2 j, Qj(2> 1) = 28 i 3 12 j, and a unit
vector in the direction of v is u = I
1
12 +32
(i + 3 j) =
Gu j(2> 1) = Qj(2> 1) · u = (28 i 3 12 j) ·
I1 (i
10
I1 (i
10
+ 3 j) =
+ 3 j), so
I1
10
I
(28 3 36) = 3 I810 or 3 4
10
.
5
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SECTION 14.6
14. j(u> v) = tan31 (uv)
Qj(1> 2) =
2
5
i+
1
5
i Qj(u> v) =
1
1 + (uv)2
15. i ({> |> }) = {h| + |h} + }h{
j) · ( I15 i +
I
1
h5> 1> 32i
25+1+4
I1
30
Gu i (0> 0> 0) = Qi (0> 0> 0) · u = h1> 1> 1i ·
Qi ({> |> }) =
Qi (3> 2> 6) =
G
G
j) =
2
I
5 5
+
2
I
5 5
(5 i + 10 j) =
=
=
I1
30
· |}>
1
({|})31@2
2
12
18
I
> I
> I6
36 2 36 2 36
direction of v is u =
4
I
5 5
or
1
I
(5 i
5 5
+ 10 j) =
I1
5
i+
I2
5
j,
I
4 5
.
25
h5> 1> 32i =
h5> 1> 32i, so
I4 .
30
i
1
({|})31@2
2
2
1
i Qi ({> |> }) = hh| + }h{ > {h| + h} > |h} + h{ i, Qi (0> 0> 0) = h1> 1> 1i, and a unit
vector in the direction of v is u =
s
{|}
I2
5
439
1
v
u
·u j =
i+
j,
1 + (uv)2
1 + u2 v2
1 + u2 v2
52 +102
1
5
¤
j, and a unit vector in the direction of v is u = I
so Gu j(1> 2) = Qj(1> 2) · u = ( 25 i +
16. i ({> |> }) =
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
·v i+
H
· {}>
= 1> 32 >
1
2
1
({|})31@2
2
· {| =
, and a unit vector in the
= 3 13 > 3 23 > 23 , so
I 1
h31> 32> 2i
1+4+4
-
H
Gu i (3> 2> 6) = Qi (3> 2> 6) · u = 1> 32 > 12 · 3 13 > 3 23 > 23 = 3 13 3 1 +
1
3
.
|}
{}
{|
s
> s
> s
,
2 {|} 2 {|} 2 {|}
= 31.
17. k(u> v> w) = ln(3u + 6v + 9w)
i Qk(u> v> w) = h3@(3u + 6v + 9w)> 6@(3u + 6v + 9w)> 9@(3u + 6v + 9w)i,
Qk(1> 1> 1) = 16 > 13 > 12 , and a unit vector in the direction of v = 4 i + 12 j + 6 k
is u =
I
1
16+144+36
(4 i + 12 j + 6 k) =
2
7
1 1 1
> >
6 3 2
Gu k(1> 1> 1) = Qk(1> 1> 1) · u =
6
7
i+
j+
2 6 3
> >
7 7 7
·
3
7
k, so
=
1
21
+
2
7
+
3
14
23
.
42
=
18. Gu i(2> 2) = Qi (2> 2) · u, the scalar projection of Qi (2> 2) onto u, so we draw a
perpendicular from the tip of Qi (2> 2) to the line containing u. We can use the
point (2> 2) to determine the scale of the axes, and we estimate the length of the
projection to be approximately 3.0 units. Since the angle between Qi (2> 2) and u
is greater than 90 , the scalar projection is negative. Thus Gu i (2> 2) E 33.
19. i ({> |) =
s
{|
i Qi ({> |) =
G
H
1
({|)31@2 (|)> 12 ({|)31@2 ({)
2
=
-
3
3
<
The unit vector in the direction of S T = h5 3 2> 4 3 8i = h3> 34i is u =
Gu i (2> 8) = Qi (2> 8) · u = 1>
20. i ({> |> }) = {| + |} + }{
1
4
·
3
> 3 45
5
.
|
{
s > s
, so Qi (2> 8) = 1> 14 .
2 {| 2 {|
3
> 3 45
5
= 25 .
, so
i Qi ({> |> }) = h| + }> { + }> | + {i, so Qi (1> 31> 3) = h2> 4> 0i. The unit vector in the
3
3
<
direction of S T = h1> 5> 2i is u =
I1 h1> 5> 2i,
30
so Gu i (1> 31> 3) = Qi(1> 31> 3) · u = h2> 4> 0i ·
I1 h1> 5> 2i
30
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440
¤
CHAPTER 14
PARTIAL DERIVATIVES
G
I H
I
I
i Qi ({> |) = 4| · 12 {31@2 > 4 { = h2|@ {> 4 { i.
I
21. i ({> |) = 4| {
Qi (4> 1) = h1> 8i is the direction of maximum rate of change, and the maximum rate is |Qi (4> 1)| =
i Qi (v> w) = whvw (w)> whvw (v) + hvw (1) = w2 hvw > (vw + 1)hvw .
22. i (v> w) = whvw
Qi (0> 2) = h4> 1i is the direction of maximum rate of change, and the maximum rate is |Qi (0> 2)| =
23. i ({> |) = sin({|)
I
I
1 + 64 = 65.
I
I
16 + 1 = 17.
i Qi ({> |) = h| cos({|)> { cos({|)i, Qi (1> 0) = h0> 1i. Thus the maximum rate of change is
|Qi (1> 0)| = 1 in the direction h0> 1i.
24. i ({> |> }) =
{+|
}
i
Qi ({> |> }) =
change is |Qi (1> 1> 31)| =
25. i ({> |> }) =
Qi({> |> }) =
=
2
1
2 ({
-
Qi (3> 6> 32) =
i
+ | 2 + } 2 )31@2 · 2{> 12 ({2 + | 2 + } 2 )31@2 · 2|> 12 ({2 + | 2 + } 2 )31@2 · 2}
|
}
{
s
>s
>s
{2 + | 2 + } 2
{2 + | 2 + } 2
{2 + | 2 + } 2
G
|Qi (3> 6> 32)| =
1 1
{+|
, Qi(1> 1> 31) = h31> 31> 32i. Thus the maximum rate of
> >3 2
} }
}
I
I
1 + 1 + 4 = 6 in the direction h31> 31> 32i.
s
{2 + | 2 + } 2
G
32
I3 > I6 > I
49
49
49
t 3 2
7
26. i (s> t> u) = arctan(stu)
+
H
=
3 6
2
7> 7> 37
.
>
. Thus the maximum rate of change is
6 2
2 t 36 + 4
+ 3 27 = 9 + 49
= 1 in the direction
i
Qi(s> t> u) =
7
the maximum rate of change is |Qi(1> 2> 1)| =
t
H
3 6
> > 3 27
7 7
or equivalently h3> 6> 32i.
tu
su
st
, Qi(1> 2> 1) =
>
>
1 + (stu)2 1 + (stu)2 1 + (stu)2
4
25
+
1
25
+
4
25
=
t
9
25
=
3
5
in the direction
2 1 2
> >
5 5 5
h2> 1> 2i.
2 1 2
5> 5> 5
. Thus
or equivalently
27. (a) As in the proof of Theorem 15, Gu i = |Qi | cos . Since the minimum value of cos is 31 occurring when = , the
minimum value of Gu i is 3 |Qi | occurring when = , that is when u is in the opposite direction of Qi
(assuming Qi 6= 0).
(b) i ({> |) = {4 | 3 {2 | 3
i Qi ({> |) = 4{3 | 3 2{|3 > {4 3 3{2 | 2 , so i decreases fastest at the point (2> 33) in the
direction 3Qi (2> 33) = 3 h12> 392i = h312> 92i.
28. i ({> |) = |h3{|
i i{ ({> |) = |h3{| (3|) = 3| 2 h3{| , i| ({> |) = |h3{| (3{) + h3{| = (1 3 {|)h3{| and
i{ (0> 2) = 34h0 = 34, i| (0> 2) = (1 3 0)h0 = 1. If u is a unit vector which makes an angle with the positive {-axis,
then Gu i (0> 2) = i{ (0> 2) cos + i| (0> 2) sin = 34 cos + sin . We want Gu i(0> 2) = 1, so 34 cos + sin = 1 i
sin = 1 + 4 cos i sin2 = (1 + 4 cos )2
i 1 3 cos2 = 1 + 8 cos + 16 cos2 i
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SECTION 14.6
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
¤
8
17 cos2 + 8 cos = 0 i cos (17 cos + 8) = 0 i cos = 0 or cos = 3 17
. If cos = 0 then = 2 or =
8
8
8
or = 2 3 cos31 3 17
but
but 3
does not satisfy the original equation. If cos = 3 17
then = cos31 3 17
2
8
= cos31 3 17
is not a solution of the original equation. Thus the directions are = 2 or
8
= 2 3 cos31 3 17
E 4=22 rad.
441
3
2
29. The direction of fastest change is Qi ({> |) = (2{ 3 2) i + (2| 3 4) j, so we need to find all points ({> |) where Qi ({> |) is
parallel to i + j
C
(2{ 3 2) i + (2| 3 4) j = n (i + j) C n = 2{ 3 2 and n = 2| 3 4. Then 2{ 3 2 = 2| 3 4 i
| = { + 1> so the direction of fastest change is i + j at all points on the line | = { + 1.
= 3 45 > 3 35 ,
and if the depth of the lake is given by i ({> |) = 200 + 0=02{2 3 0=001| 3 , then Qi({> |) = 0=04{> 30=003| 2 .
Gu i (80> 60) = Qi (80> 60) · u = h3=2> 310=8i · 3 45 > 3 35 = 3=92. Since Gu i (80> 60) is positive, the depth of the lake is
30. The fisherman is traveling in the direction h380> 360i. A unit vector in this direction is u =
1
h380> 360i
100
increasing near (80> 60) in the direction toward the buoy.
31. W = s
{2
(a) u =
n
n
and 120 = W (1> 2> 2) = so n = 360.
2
2
3
+| +}
h1> 31> 1i
I
,
3
k
l
33@2
h{> |> }i
Gu W (1> 2> 2) = QW (1> 2> 2) · u = 3360 {2 + | 2 + } 2
(1>2>2)
40
I
I1
· u = 3 40
3 h1> 2> 2i · 3 h1> 31> 1i = 3 3 3
33@2
(b) From (a), QW = 3360 {2 + | 2 + } 2
h{> |> }i, and since h{> |> }i is the position vector of the point ({> |> }), the
vector 3 h{> |> }i, and thus QW , always points toward the origin.
2 33| 2 39} 2
32. QW = 3400h3{
(a) u =
I1 h1> 32> 1i,
6
Gu W (2> 31> 2) =
h{> 3|> 9}i
QW (2> 31> 2) = 3400h343 h2> 33> 18i and
I
400h343
5200 6
3 I
(26) = 3
3h43
6
C@m.
(b) QW (2> 31> 2) = 400h343 h32> 3> 318i or equivalently h32> 3> 318i.
(c) |QW | = 400h3{
s
{2 + 9| 2 + 81} 2 C@m is the maximum rate of increase. At (2> 31> 2) the maximum rate
2 3 3| 2 3 9} 2
I
of increase is 400h343 337 C@m.
33. QY ({> |> }) = h10{ 3 3| + |}> {} 3 3{> {|i, QY (3> 4> 5) = h38> 6> 12i
(a) Gu Y (3> 4> 5) = h38> 6> 12i ·
I1 h1> 1> 31i
3
=
32
I
3
(b) QY (3> 4> 5) = h38> 6> 12i, or equivalently, h19> 3> 6i.
(c) |QY (3> 4> 5)| =
I
I
I
382 + 62 + 122 = 1624 = 2 406
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442
¤
CHAPTER 14
PARTIAL DERIVATIVES
34. } = i ({> |) = 1000 3 0=005{2 3 0=01| 2
i Qi({> |) = h30=01{> 30=02|i and Qi (60> 40) = h30=6> 30=8i.
(a) Due south is in the direction of the unit vector u = 3j and
Gu i (60> 40) = Qi (60> 40) · h0> 31i = h30=6> 30=8i · h0> 31i = 0=8. Thus, if you walk due south from (60> 40> 966)
you will ascend at a rate of 0=8 vertical meters per horizontal meter.
(b) Northwest is in the direction of the unit vector u = I12 h31> 1i and
Gu i (60> 40) = Qi (60> 40) ·
I1
2
h31> 1i = h30=6> 30=8i ·
I1
2
I E 30=14. Thus, if you walk northwest
h31> 1i = 3 0=2
2
from (60> 40> 966) you will descend at a rate of approximately 0=14 vertical meters per horizontal meter.
(c) Qi (60> 40) = h30=6> 30=8i is the direction of largest slope with a rate of ascent given by
s
|Qi (60> 40)| = (30=6)2 + (30=8)2 = 1. The angle above the horizontal in which the path begins is given by
tan = 1 i = 45 .
3<
3<
35. A unit vector in the direction of DE is i and a unit vector in the direction of DF is j. Thus G33< i (1> 3) = i{ (1> 3) = 3 and
DE
G33< i(1> 3) = i| (1> 3) = 26. Therefore Qi (1> 3) = hi{ (1> 3)> i| (1> 3)i = h3> 26i, and by definition,
DF
33<
G33< i (1> 3) = Qi · u where u is a unit vector in the direction of DG, which is
DG
G33< i (1> 3) = h3> 26i ·
DG
5 12
>
13 13
=3·
5
13
+ 26 ·
12
13
=
5 12
13 > 13
327
.
13
. Therefore,
36. The curves of steepest ascent or descent are perpendicular to all of the contour lines (see Figure 12) so we sketch curves
beginning at A and B that head toward lower elevations, crossing each contour line at a right angle.
37. (a) Q(dx + ey) =
C(dx + ey) C(dx + ey)
>
C{
C|
=
Cx
Cy Cx
Cy
d
+e
>d
+e
C{
C{ C|
C|
=d
Cx Cx
>
C{ C|
+e
Cy Cy
>
C{ C|
= d Qx + e Qy
(b) Q(xy) =
Cx
Cy Cx
Cy
y
+x
>y
+x
C{
C{ C|
C|
=y
Cx Cx
>
C{ C|
+x
Cy Cy
>
C{ C|
= y Qx + x Qy
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SECTION 14.6
x
=
(c) Q
y
(d) Qxq =
Cy y Cx 3 x Cy . y
- Cx
3x
y
C|
C{
C{ > C|
=
y2
y2
C(xq ) C(xq )
>
C{
C|
=
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
Cx Cx
>
C{ C|
Cx
Cx
qxq31
> qxq31
C{
C|
3x
y2
Cy Cy
>
C{ C|
=
¤
443
y Qx 3 x Qy
y2
= qxq31 Qx
38. If we place the initial point of the gradient vector Qi (4> 6) at (4> 6), the vector is perpendicular to the level curve of i that
includes (4> 6), so we sketch a portion of the level curve through (4> 6) (using the nearby level curves as a guideline)
and draw a line perpendicular to the curve at (4> 6). The gradient vector is
parallel to this line, pointing in the direction of increasing function values, and
with length equal to the maximum value of the directional derivative of i at
(4> 6). We can estimate this length by finding the average rate of change in the
direction of the gradient. The line intersects the contour lines corresponding to
32 and 33 with an estimated distance of 0=5 units. Thus the rate of change is
approximately
32 3 (33)
= 2, and we sketch the gradient vector with
0=5
length 2.
39. i ({> |) = {3 + 5{2 | + | 3
i
Gu i({> |) = Qi ({> |) · u = 3{2 + 10{|> 5{2 + 3| 2 ·
= 95 {2 + 6{| + 4{2 +
Gu2 i({> |) = Gu [Gu i ({> |)] = Q [Gu i ({> |)] · u = 58
{ + 6|> 6{ + 24
| · 35 > 45
5
5
=
174
{
25
and Gu2 i (2> 1) =
+
18
|
5
294
(2)
25
+
+
24
{
5
+
186
(1)
25
96
|
25
=
=
294
{
25
+
3 4
>
5 5
12 2
|
5
=
29 2
{
5
+ 6{| +
12 2
| .
5
Then
186
|
25
774
.
25
40. (a) From Equation 9 we have Gu i = Qi · u = hi{ > i| i · hd> ei = i{ d + i| e and from Exercise 39 we have
Gu2 i = Gu [Gu i] = Q [Gu i] · u = hi{{ d + i|{ e> i{| d + i|| ei · hd> ei = i{{ d2 + i|{ de + i{| de + i|| e2 .
But i|{ = i{| by Clairaut’s Theorem, so Gu2 i = i{{ d2 + 2i{| de + i|| e2 .
(b) i ({> |) = {h2|
i{ = h2| , i| = 2{h2| , i{{ = 0, i{| = 2h2| , i|| = 4{h2| and a
H
G
unit vector in the direction of v is u = I 21 2 h4> 6i = I213 > I313 = hd> ei. Then
i
4 +6
Gu2 i = i{{ d2 + 2i{| de + i|| e2 = 0 ·
I2
13
2
+ 2 · 2h2|
I2
13
I3
13
+ 4{h2|
I3
13
2
=
24 2|
13 h
+
2|
36
13 {h .
41. Let I ({> |> }) = 2({ 3 2)2 + (| 3 1)2 + (} 3 3)2 . Then 2({ 3 2)2 + (| 3 1)2 + (} 3 3)2 = 10 is a level surface of I .
I{ ({> |> }) = 4({ 3 2) i I{ (3> 3> 5) = 4, I| ({> |> }) = 2(| 3 1) i I| (3> 3> 5) = 4, and
I} ({> |> }) = 2(} 3 3) i I} (3> 3> 5) = 4.
(a) Equation 19 gives an equation of the tangent plane at (3> 3> 5) as 4({ 3 3) + 4(| 3 3) + 4(} 3 5) = 0 C
4{ + 4| + 4} = 44 or equivalently { + | + } = 11.
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CHAPTER 14
PARTIAL DERIVATIVES
(b) By Equation 20, the normal line has symmetric equations
|33
}35
{33
=
=
or equivalently
4
4
4
{ 3 3 = | 3 3 = } 3 5. Corresponding parametric equations are { = 3 + w, | = 3 + w, } = 5 + w.
42. Let I ({> |> }) = {2 3 } 2 3 |. Then | = {2 3 } 2
C {2 3 } 2 3 | = 0 is a level surface of I . I{ ({> |> }) = 2{ i
I{ (4> 7> 3) = 8, I| ({> |> }) = 31 i I| (4> 7> 3) = 31, and I} ({> |> }) = 32}
i I} (4> 7> 3) = 36.
(a) An equation of the tangent plane at (4> 7> 3) is 8({ 3 4) 3 1(| 3 7) 3 6(} 3 3) = 0 or 8{ 3 | 3 6} = 7.
(b) The normal line has symmetric equations
{34
|37
}33
=
=
and parametric equations { = 4 + 8w, | = 7 3 w,
8
31
36
} = 3 3 6w.
43. Let I ({> |> }) = {|} 2 . Then {|} 2 = 6 is a level surface of I and QI ({> |> }) = |} 2 > {} 2 > 2{|} .
(a) QI (3> 2> 1) = h2> 3> 12i is a normal vector for the tangent plane at (3> 2> 1), so an equation of the tangent plane
is 2({ 3 3) + 3(| 3 2) + 12(} 3 1) = 0 or 2{ + 3| + 12} = 24.
(b) The normal line has direction h2> 3> 12i, so parametric equations are { = 3 + 2w, | = 2 + 3w, } = 1 + 12w, and
symmetric equations are
{33
|32
}31
=
=
.
2
3
12
44. Let I ({> |> }) = {| + |} + }{. Then {| + |} + }{ = 5 is a level surface of I and QI ({> |> }) = h| + }> { + }> { + |i.
(a) QI (1> 2> 1) = h3> 2> 3i is a normal vector for the tangent plane at (1> 2> 1), so an equation of the tangent plane
is 3({ 3 1) + 2(| 3 2) + 3(} 3 1) = 0 or 3{ + 2| + 3} = 10.
(b) The normal line has direction h3> 2> 3i, so parametric equations are { = 1 + 3w, | = 2 + 2w, } = 1 + 3w, and symmetric
equations are
|32
}31
{31
=
=
.
2
1
3
45. Let I ({> |> }) = { + | + } 3 h{|} . Then { + | + } = h{|} is the level surface I ({> |> }) = 0,
and QI ({> |> }) = h1 3 |}h{|} > 1 3 {}h{|} > 1 3 {|h{|} i.
(a) QI (0> 0> 1) = h1> 1> 1i is a normal vector for the tangent plane at (0> 0> 1), so an equation of the tangent plane
is 1({ 3 0) + 1(| 3 0) + 1(} 3 1) = 0 or { + | + } = 1.
(b) The normal line has direction h1> 1> 1i, so parametric equations are { = w, | = w, } = 1 + w, and symmetric equations are
{ = | = } 3 1.
46. Let I ({> |> }) = {4 + | 4 + } 4 3 3{2 | 2 } 2 . Then {4 + | 4 + } 4 = 3{2 | 2 } 2 is the level surface I ({> |> }) = 0,
and QI ({> |> }) = 4{3 3 6{| 2 } 2 > 4| 3 3 6{2 |} 2 > 4} 3 3 6{2 | 2 } .
(a) QI (1> 1> 1) = h32> 32> 32i or equivalently h1> 1> 1i is a normal vector for the tangent plane at (1> 1> 1), so an equation
of the tangent plane is 1({ 3 1) + 1(| 3 1) + 1(} 3 1) = 0 or { + | + } = 3.
(b) The normal line has direction h1> 1> 1i, so parametric equations are { = 1 + w, | = 1 + w, } = 1 + w, and symmetric
equations are { 3 1 = | 3 1 = } 3 1 or equivalently { = | = }.
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SECTION 14.6
47. I ({> |> }) = {| + |} + }{,
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
¤
445
48. I ({> |> }) = {|}, QI ({> |> }) = h|}> {}> |{i,
QI ({> |> }) = h| + }> { + }> | + {i,
QI (1> 2> 3) = h6> 3> 2i, so an equation of the tangent
QI (1> 1> 1) = h2> 2> 2i, so an equation of the tangent
plane is 6{ + 3| + 2} = 18, and the normal line is given
plane is 2{ + 2| + 2} = 6 or { + | + } = 3, and the
normal line is given by { 3 1 = | 3 1 = } 3 1 or
{ = | = }. To graph the surface we solve for }:
}=
by
{31
|32
}33
=
=
or { = 1 + 6w, | = 2 + 3w,
6
3
2
} = 3 + 2w. To graph the surface we solve for }: } =
3 3 {|
.
{+|
49. i ({> |) = {|
i Qi ({> |) = h|> {i, Qi (3> 2) = h2> 3i. Qi (3> 2)
is perpendicular to the tangent line, so the tangent line has equation
Qi (3> 2) · h{ 3 3> | 3 2i = 0 i h2> 3i · h{ 3 3> { 3 2i = 0 i
2({ 3 3) + 3(| 3 2) = 0 or 2{ + 3| = 12.
50. j({> |) = {2 + | 2 3 4{
i Qj({> |) = h2{ 3 4> 2|i,
Qj(1> 2) = h32> 4i. Qj(1> 2) is perpendicular to the tangent line, so
the tangent line has equation Qj(1> 2) · h{ 3 1> | 3 2i = 0 i
h32> 4i · h{ 3 1> | 3 2i = 0 i 32({ 3 1) + 4(| 3 2) = 0 C
32{ + 4| = 6 or equivalently 3{ + 2| = 3.
2{0 2|0 2}0
>
>
. Thus an equation of the tangent plane at ({0 > |0 > }0 ) is
d2 e2 f2
2
2|0
2}0
{0
|02
}02
2{0
{ + 2 | + 2 } = 2 2 + 2 + 2 = 2(1) = 2 since ({0 > |0 > }0 ) is a point on the ellipsoid. Hence
d2
e
f
d
e
f
51. QI ({0 > |0 > }0 ) =
{0
|0
}0
{ + 2 | + 2 } = 1 is an equation of the tangent plane.
d2
e
f
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.
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446
¤
CHAPTER 14
PARTIAL DERIVATIVES
2{0 2|0 32}0
>
> 2
, so an equation of the tangent plane at ({0 > |0 > }0 ) is
d2 e2
f
2
{0
2{0
2|0
2}0
|02
}02
|0
}0
{0
{
+
|
3
}
=
2
+
3
= 2 or 2 { + 2 | 3 2 } = 1.
d2
e2
f2
d2
e2
f2
d
e
f
52. QI ({0 > |0 > }0 ) =
2{0
2{20
2{0 2|0 31
2|0
1
2|02
}0
,
so
an
equation
of
the
tangent
plane
is
}
=
>
>
{
+
|
3
+
3
d2 e2 f
d2
e2
f
d2
e2
f
2
2|0
}
|2
|2
2{0
}0
{
}0
{2
or 2 { + 2 | = + 2 20 + 20 3 . But
= 20 + 20 , so the equation can be written as
d
e
f
d
e
f
f
d
e
53. QI ({0 > |0 > }0 ) =
2|0
} + }0
2{0
.
{+ 2 | =
d2
e
f
54. Let I ({> |> }) = {2 + } 2 3 |; then the paraboloid | = {2 + } 2 is a level surface of I . QI ({> |> }) = h2{> 31> 2}i is a
normal vector to the surface at ({> |> }) and so it is a normal vector for the tangent plane there. The tangent plane is parallel
to the plane { + 2| + 3} = 1 when the normal vectors of the planes are parallel, so we need a point ({0 > |0 > }0 ) on the
paraboloid where h2{0 > 31> 2}0 i = n h1> 2> 3i. Comparing |-components we have n = 3 12 , so
h2{0 > 31> 2}0 i = 3 12 > 31> 3 32 and 2{0 = 3 12 i {0 = 3 14 , 2}0 = 3 32 i }0 = 3 34 . Then
2 2
|0 = {20 + }02 = 3 14 + 3 34 =
5
8
and the point is 3 14 > 58 > 3 34 .
55. The hyperboloid {2 3 | 2 3 } 2 = 1 is a level surface of I ({> |> }) = {2 3 | 2 3 } 2 and QI ({> |> }) = h2{> 32|> 32}i is a
normal vector to the surface and hence a normal vector for the tangent plane at ({> |> }). The tangent plane is parallel to the
plane } = { + | or { + | 3 } = 0 if and only if the corresponding normal vectors are parallel, so we need a point ({0 > |0 > }0 )
on the hyperboloid where h2{0 > 32|0 > 32}0 i = f h1> 1> 31i or equivalently h{0 > 3|0 > 3}0 i = n h1> 1> 31i for some n 6= 0.
Then we must have {0 = n, |0 = 3n, }0 = n and substituting into the equation of the hyperboloid gives
n2 3 (3n)2 3 n2 = 1 C 3n2 = 1, an impossibility. Thus there is no such point on the hyperboloid.
56. First note that the point (1> 1> 2) is on both surfaces. The ellipsoid is a level surface of I ({> |> }) = 3{2 + 2| 2 + } 2 and
QI ({> |> }) = h6{> 4|> 2}i. A normal vector to the surface at (1> 1> 2) is QI (1> 1> 2) = h6> 4> 4i and an equation of the
tangent plane there is 6({ 3 1) + 4(| 3 1) + 4(} 3 2) = 0 or 6{ + 4| + 4} = 18 or 3{ + 2| + 2} = 9. The sphere is a level
surface of J({> |> }) = {2 + | 2 + } 2 3 8{ 3 6| 3 8} + 24 and QJ({> |> }) = h2{ 3 8> 2| 3 6> 2} 3 8i. A normal vector to
the sphere at (1> 1> 2) is QJ(1> 1> 2) = h36> 34> 34i and the tangent plane there is 36({ 3 1) 3 4(| 3 1) 3 4(} 3 2) = 0 or
3{ + 2| + 2} = 9. Since these tangent planes are identical, the surfaces are tangent to each other at the point (1> 1> 2).
57. Let ({0 > |0 > }0 ) be a point on the cone [other than (0> 0> 0)]. The cone is a level surface of I ({> |> }) = {2 + | 2 3 } 2 and
QI ({> |> }) = h2{> 2|> 32}i, so QI ({0 > |0 > }0 ) = h2{0 > 2|0 > 32}0 i is a normal vector to the cone at this point and an
equation of the tangent plane there is 2{0 ({ 3 {0 ) + 2|0 (| 3 |0 ) 3 2}0 (} 3 }0 ) = 0 or {0 { + |0 | 3 }0 } = {20 + |02 3 }02 .
But {20 + |02 = }02 so the tangent plane is given by {0 { + |0 | 3 }0 } = 0, a plane which always contains the origin.
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SECTION 14.6
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
58. Let ({0 > |0 > }0 ) be a point on the sphere. Then the normal line is given by
(0> 0> 0) to be on the line, we need 3
¤
447
{ 3 {0
| 3 |0
} 3 }0
=
=
. For the center
2{0
2|0
2}0
{0
|0
}0
=3
=3
or equivalently 1 = 1 = 1, which is true.
2{0
2|0
2}0
59. Let I ({> |> }) = {2 + | 2 3 }. Then the paraboloid is the level surface I ({> |> }) = 0 and QI ({> |> }) = h2{> 2|> 31i, so
QI (1> 1> 2) = h2> 2> 31i is a normal vector to the surface. Thus the normal line at (1> 1> 2) is given by { = 1 + 2w,
| = 1 + 2w, } = 2 3 w. Substitution into the equation of the paraboloid } = {2 + | 2 gives 2 3 w = (1 + 2w)2 + (1 + 2w)2
2 3 w = 2 + 8w + 8w2
C 8w2 + 9w = 0 C
C
w(8w + 9) = 0. Thus the line intersects the paraboloid when w = 0,
corresponding to the given point (1> 1> 2), or when w = 3 98 , corresponding to the point 3 54 > 3 54 > 25
.
8
60. The ellipsoid is a level surface of I ({> |> }) = 4{2 + | 2 + 4} 2 and QI ({> |> }) = h8{> 2|> 8}i, so QI (1> 2> 1) = h8> 4> 8i
or equivalently h2> 1> 2i is a normal vector to the surface. Thus the normal line to the ellipsoid at (1> 2> 1) is given
by { = 1 + 2w, | = 2 + w, } = 1 + 2w. Substitution into the equation of the sphere gives
(1 + 2w)2 + (2 +w)2 + (1 + 2w)2 = 102
C
6 + 12w+ 9w2 = 102
C
9w2 + 12w3 96 = 0
C
3(w+ 4)(3w 3 8) = 0.
Thus the line intersects the sphere when w = 34, corresponding to the point (37> 32> 37), and when w = 83 , corresponding to
the point
19
3
.
> 14
> 19
3
3
61. Let ({0 > |0 > }0 ) be a point on the surface. Then an equation of the tangent plane at the point is
|
}
{
I + I + I =
2 {0
2 |0
2 }0
I
I
I
{0 + |0 + }0
I
I
I
I
. But {0 + |0 + }0 = f, so the equation is
2
I
I
I
{
|
}
I
I + I + I = f. The {-, |-, and }-intercepts are f{0 , f|0 and f}0 respectively. (The {-intercept is found by
{0
|0
}0
setting | = } = 0 and solving the resulting equation for {, and the |- and }-intercepts are found similarly.) So the sum of the
I I
I I
intercepts is f {0 + |0 + }0 = f, a constant.
62. The surface {|} = 1 is a level surface of I ({> |> }) = {|} and QI ({> |> }) = h|}> {}> {|i is normal to the surface, so a
normal vector for the tangent plane to the surface at ({0 > |0 > }0 ) is h|0 }0 > {0 }0 > {0 |0 i. An equation for the tangent plane there
is |0 }0 ({ 3 {0 ) + {0 }0 (| 3 |0 ) + {0 |0 (} 3 }0 ) = 0 i |0 }0 { + {0 }0 | + {0 |0 } = 3{0 |0 }0 or
{
|
}
+
+
= 3.
{0
|0
}0
If ({0 > |0 > }0 ) is in the first octant, then the tangent plane cuts off a pyramid in the first octant with vertices (0> 0> 0),
(3{0 > 0> 0), (0> 3|0 > 0), (0> 0> 3}0 ). The base in the {|-plane is a triangle with area
1
2
(3{0 ) (3|0 ) and the height (along the
}-axis) of the pyramid is 3}0 . The volume of the pyramid for any point ({0 > |0 > }0 ) on the surface {|} = 1 in the first octant is
1
3
(base) (height) =
1
3
·
1
2
(3{0 ) (3|0 ) · 3}0 = 92 {0 |0 }0 =
9
2
since {0 |0 }0 = 1.
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¤
CHAPTER 14
PARTIAL DERIVATIVES
63. If i ({> |> }) = } 3 {2 3 | 2 and j({> |> }) = 4{2 + | 2 + } 2 , then the tangent line is perpendicular to both Qi and Qj
at (31> 1> 2). The vector v = Qi × Qj will therefore be parallel to the tangent line.
We have Qi ({> |> }) = h32{> 32|> 1i i Qi (31> 1> 2) = h2> 32> 1i, and Qj({> |> }) = h8{> 2|> 2}i i
i j k
Qj(31> 1> 2) = h38> 2> 4i. Hence v = Qi × Qj = 2 32 1 = 310 i 3 16 j 3 12 k.
38 2 4 Parametric equations are: { = 31 3 10w, | = 1 3 16w, } = 2 3 12w.
64. (a) Let i ({> |> }) = | + } and j({> |> }) = {2 + | 2 . Then the required tangent
(b)
line is perpendicular to both Qi and Qj at (1> 2> 1) and the vector
v = Qi × Qj is parallel to the tangent line. We have
Qi ({> |> }) = h0> 1> 1i i Qi(1> 2> 1) = h0> 1> 1i, and
Qj({> |> }) = h2{> 2|> 0i i Qj(1> 2> 1) = h2> 4> 0i. Hence
i j k
v = Qi × Qj = 0 1 1 = 34 i + 2 j 3 2 k. So parametric equations
2 4 0
of the desired tangent line are { = 1 3 4w, | = 2 + 2w, } = 1 3 2w.
65. (a) The direction of the normal line of I is given by QI , and that of J by QJ. Assuming that
QI 6= 0 6= QJ, the two normal lines are perpendicular at S if QI · QJ = 0 at S
hCI@C{> CI@C|> CI@C}i · hCJ@C{> CJ@C|> CJ@C}i = 0 at S
C
C I{ J{ + I| J| + I} J} = 0 at S .
(b) Here I = {2 + | 2 3 } 2 and J = {2 + | 2 + } 2 3 u2 , so
QI · QJ = h2{> 2|> 32}i · h2{> 2|> 2}i = 4{2 + 4| 2 3 4} 2 = 4I = 0, since the point ({> |> }) lies on the graph of
I = 0. To see that this is true without using calculus, note that J = 0 is the equation of a sphere centered at the origin and
I = 0 is the equation of a right circular cone with vertex at the origin (which is generated by lines through the origin). At
any point of intersection, the sphere’s normal line (which passes through the origin) lies on the cone, and thus is
perpendicular to the cone’s normal line. So the surfaces with equations I = 0 and J = 0 are everywhere orthogonal.
66. (a) The function i({> |) = ({|)1@3 is continuous on R2 since it is a composition of a polynomial and the cube root function,
both of which are continuous. (See the text just after Example 14.2.8.)
i (0 + k> 0) 3 i (0> 0)
(k · 0)1@3 3 0
= lim
= 0,
k<0
k<0
k
k
i{ (0> 0) = lim
i (0> 0 + k) 3 i (0> 0)
(0 · k)1@3 3 0
= lim
= 0.
k<0
k<0
k
k
i| (0> 0) = lim
Therefore, i{ (0> 0) and i| (0> 0) do exist and are equal to 0. Now let u be any unit vector other than i and j (these
correspond to i{ and i| respectively.) Then u = d i + e j where d 6= 0 and e 6= 0. Thus
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SECTION 14.7
MAXIMUM AND MINIMUM VALUES
¤
449
s
I
3
3
(kd)(ke)
i(0 + kd> 0 + ke) 3 i (0> 0)
de
= lim
= lim 1@3 and this limit does not exist, so
Gu i (0> 0) = lim
k<0
k<0
k<0 k
k
k
Gu i (0> 0) does not exist.
Notice that if we start at the origin and proceed in the direction of
(b)
the {- or |-axis, then the graph is flat. But if we proceed in any
other direction, then the graph is extremely steep.
67. Let u = hd> ei and v = hf> gi. Then we know that at the given point, Gu i = Qi · u = di{ + ei| and
Gv i = Qi · v = fi{ + gi| . But these are just two linear equations in the two unknowns i{ and i| , and since u and v are
not parallel, we can solve the equations to find Qi = hi{ > i| i at the given point. In fact,
g Gu i 3 e Gv i d Gv i 3 f Gu i
>
.
Qi =
dg 3 ef
dg 3 ef
68. Since } = i ({> |) is differentiable at x0 = ({0 > |0 ), by Definition 14.4.7 we have
{} = i{ ({0 > |0 ) {{ + i| ({0 > |0 ) {| + %1 {{ + %2 {| where %1 > %2 < 0 as ({{> {|) < (0> 0). Now
{} = i (x) 3 i (x0 ), h{{> {|i = x 3 x0 so ({{> {|) < (0> 0) is equivalent to x < x0 and
hi{ ({0 > |0 ) > i| ({0 > |0 )i = Qi(x0 ). Substituting into 14.4.7 gives i (x) 3 i(x0 ) = Qi (x0 ) · (x 3 x0 ) + h%1 > %2 i · h{{> {|i
or h%1 > %2 i · (x 3 x0 ) = i (x) 3 i(x0 ) 3 Qi (x0 ) · (x 3 x0 ),
and so
lim
x<x0
h%1 > %2 i · (x 3 x0 )
x 3 x0
i (x) 3 i(x0 ) 3 Qi (x0 ) · (x 3 x0 )
=
. But
is a unit vector so
|x 3 x0 |
|x 3 x0 |
|x 3 x0 |
h%1 > %2 i · (x 3 x0 )
i (x) 3 i (x0 ) 3 Qi (x0 ) · (x 3 x0 )
= 0 since %1 > %2 < 0 as x < x0 . Hence lim
= 0.
x<x0
|x 3 x0 |
|x 3 x0 |
14.7 Maximum and Minimum Values
1. (a) First we compute G(1> 1) = i{{ (1> 1) i|| (1> 1) 3 [i{| (1> 1)]2 = (4)(2) 3 (1)2 = 7. Since G(1> 1) A 0 and
i{{ (1> 1) A 0, i has a local minimum at (1> 1) by the Second Derivatives Test.
(b) G(1> 1) = i{{ (1> 1) i|| (1> 1) 3 [i{| (1> 1)]2 = (4)(2) 3 (3)2 = 31. Since G(1> 1) ? 0, i has a saddle point at (1> 1) by
the Second Derivatives Test.
2. (a) G = j{{ (0> 2) j|| (0> 2) 3 [j{| (0> 2)]2 = (31)(1) 3 (6)2 = 337. Since G ? 0, j has a saddle point at (0> 2) by the
Second Derivatives Test.
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¤
CHAPTER 14
PARTIAL DERIVATIVES
(b) G = j{{ (0> 2) j|| (0> 2) 3 [j{| (0> 2)]2 = (31)(38) 3 (2)2 = 4. Since G A 0 and j{{ (0> 2) ? 0, j has a local
maximum at (0> 2) by the Second Derivatives Test.
(c) G = j{{ (0> 2) j|| (0> 2) 3 [j{| (0> 2)]2 = (4)(9) 3 (6)2 = 0. In this case the Second Derivatives Test gives no
information about j at the point (0> 2).
3. In the figure, a point at approximately (1> 1) is enclosed by level curves which are oval in shape and indicate that as we move
away from the point in any direction the values of i 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 of i increase in some
directions and decrease in others, so we would expect to find a saddle point there.
To verify our predictions, we have i({> |) = 4 + {3 + | 3 3 3{|
i i{ ({> |) = 3{2 3 3|, i| ({> |) = 3| 2 3 3{. We
have critical points where these partial derivatives are equal to 0: 3{2 3 3| = 0, 3|2 3 3{ = 0. Substituting | = {2 from the
first equation into the second equation gives 3({2 )2 3 3{ = 0 i 3{({3 3 1) = 0 i { = 0 or { = 1. Then we have
two critical points, (0> 0) and (1> 1). The second partial derivatives are i{{ ({> |) = 6{, i{| ({> |) = 33, and i|| ({> |) = 6|,
so G({> |) = i{{ ({> |) i|| ({> |) 3 [i{| ({> |)]2 = (6{)(6|) 3 (33)2 = 36{| 3 9. Then G(0> 0) = 36(0)(0) 3 9 = 39,
and G(1> 1) = 36(1)(1) 3 9 = 27. Since G(0> 0) ? 0, i has a saddle point at (0> 0) by the Second Derivatives Test. Since
G(1> 1) A 0 and i{{ (1> 1) A 0, i has a local minimum at (1> 1).
4. In the figure, points at approximately (31> 1) and (31> 31) are enclosed by oval-shaped level curves which indicate that as we
move away from either point in any direction, the values of i are increasing. Hence we would expect local minima at or near
(31> ±1). Similarly, the point (1> 0) appears to be enclosed by oval-shaped level curves which indicate that as we move away
from the point in any direction the values of i are decreasing, so we should have a local maximum there. We also show
hyperbola-shaped level curves near the points (31> 0), (1> 1), and (1> 31). The values of i increase along some paths leaving
these points and decrease in others, so we should have a saddle point at each of these points.
To confirm our predictions, we have i ({> |) = 3{ 3 {3 3 2| 2 + | 4
i i{ ({> |) = 3 3 3{2 , i| ({> |) = 34| + 4| 3 .
Setting these partial derivatives equal to 0, we have 3 3 3{2 = 0 i { = ±1 and 34| + 4|3 = 0 i
| | 2 3 1 = 0 i | = 0> ±1. So our critical points are (±1> 0), (±1> ±1).
The second partial derivatives are i{{ ({> |) = 36{, i{| ({> |) = 0, and i|| ({> |) = 12| 2 3 4, so
G({> |) = i{{ ({> |) i|| ({> |) 3 [i{| ({> |)]2 = (36{)(12| 2 3 4) 3 (0)2 = 372{| 2 + 24{.
We use the Second Derivatives Test to classify the 6 critical points:
Critical Point
G
(1> 0)
24
(1> 1)
i{{
36
Conclusion
G A 0, i{{ ? 0
i i has a local maximum at (1> 0)
348
G?0
i i has a saddle point at (1> 1)
(1> 31)
348
G?0
i i has a saddle point at (1> 31)
(31> 0)
324
G?0
i i has a saddle point at (31> 0)
(31> 1)
48
6
G A 0, i{{ A 0
i i has a local minimum at (31> 1)
(31> 31)
48
6
G A 0, i{{ A 0
i i has a local minimum at (31> 31)
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SECTION 14.7
5. i ({> |) = {2 + {| + | 2 + |
¤
MAXIMUM AND MINIMUM VALUES
451
i i{ = 2{ + |, i| = { + 2| + 1, i{{ = 2, i{| = 1, i|| = 2. Then i{ = 0 implies
| = 32{, and substitution into i| = { + 2| + 1 = 0 gives { + 2 (32{) + 1 = 0 i 33{ = 31 i { = 13 .
Then | = 3 23 and the only critical point is
1
3
> 3 23 .
G({> |) = i{{ i|| 3 (i{| )2 = (2)(2) 3 (1)2 = 3, and since
G 13 > 3 23 = 3 A 0 and i{{ 13 > 3 23 = 2 A 0, i 13 > 3 23 = 3 13 is a local
minimum by the Second Derivatives Test.
6. i ({> |) = {| 3 2{ 3 2| 3 {2 3 | 2
i i{ = | 3 2 3 2{,
i| = { 3 2 3 2|, i{{ = 32, i{| = 1, i|| = 32. Then i{ = 0 implies
| = 2{ + 2, and substitution into i| = 0 gives { 3 2 3 2(2{ + 2) = 0 i
33{ = 6
i
{ = 32. Then | = 32 and the only critical point is
(32> 32). G({> |) = i{{ i|| 3 (i{| )2 = (32)(32) 3 12 = 3, and since
G(32> 32) = 3 A 0 and i{{ (32> 32) = 32 ? 0, i (32> 32) = 4 is a
local maximum by the Second Derivatives Test.
7. i ({> |) = ({ 3 |)(1 3 {|) = { 3 | 3 {2 | + {| 2
i i{ = 1 3 2{| + | 2 , i| = 31 3 {2 + 2{|, i{{ = 32|,
i{| = 32{ + 2|, i|| = 2{. Then i{ = 0 implies 1 3 2{| + | 2 = 0 and i| = 0 implies 31 3 {2 + 2{| = 0. Adding the
two equations gives 1 + | 2 3 1 3 {2 = 0
1 + 2{2 + {2 = 0
i
| 2 = {2
i
| = ±{, but if | = 3{ then i{ = 0 implies
3{2 = 31 which has no real solution. If | = {
i
then substitution into i{ = 0 gives 1 3 2{2 + {2 = 0 i {2 = 1 i
{ = ±1, so the critical points are (1> 1) and (31> 31). Now
G(1> 1) = (32)(2) 3 02 = 34 ? 0 and
G(31> 31) = (2)(32) 3 02 = 34 ? 0, so (1> 1) and (31> 31) are
saddle points.
2 32| 2
8. i ({> |) = {h32{
i i{ = (1 3 4{2 )h32{
2 32| 2
i{| = (16{2 3 4)|h32{
2 32| 2
, i| = 34{|h32{
2 32| 2
, i|| = (16| 2 3 4){h32{
2 32| 2
2 32| 2
, i{{ = (16{2 3 12){h32{
,
. Then i{ = 0 implies 1 3 4{2 = 0 i { = ± 12 , and
substitution into i| = 0 i 34{| = 0 gives | = 0, so the critical points are ± 12 > 0 . Now
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¤
G
CHAPTER 14
1
i{{
2
PARTIAL DERIVATIVES
> 0 = (34h31@2 )(32h31@2 ) 3 02 = 8h31 A 0 and
1
2>0
= 34h31@2 ? 0, so i
1
= 12 h31@2 is a local maximum.
2>0
G 3 12 > 0 = (4h31@2 )(2h31@2 ) 3 02 = 8h31 A 0 and
i{{ 3 12 > 0 = 4h31@2 A 0, so i 3 12 > 0 = 3 12 h31@2
is a local minimum.
9. i ({> |) = | 3 + 3{2 | 3 6{2 3 6| 2 + 2
i i{ = 6{| 3 12{, i| = 3| 2 + 3{2 3 12|, i{{ = 6| 3 12, i{| = 6{,
i|| = 6| 3 12. Then i{ = 0 implies 6{(| 3 2) = 0, so { = 0 or | = 2. If { = 0 then substitution into i| = 0 gives
3| 2 3 12| = 0
i
3|(| 3 4) = 0
i
| = 0 or | = 4, so we have critical points (0> 0) and (0> 4). If | = 2,
substitution into i| = 0 gives 12 + 3{2 3 24 = 0 i {2 = 4 i
{ = ±2, so we have critical points (±2> 2).
G(0> 0) = (312)(312) 3 02 = 144 A 0 and i{{ (0> 0) = 312 ? 0, so
i (0> 0) = 2 is a local maximum. G(0> 4) = (12)(12) 3 02 = 144 A 0
and i{{ (0> 4) = 12 A 0, so i (0> 4) = 330 is a local minimum.
G(±2> 2) = (0)(0) 3 (±12)2 = 3144 ? 0, so (±2> 2) are saddle points.
10. i ({> |) = {|(1 3 { 3 |) = {| 3 {2 | 3 {| 2
i i{ = | 3 2{| 3 | 2 , i| = { 3 {2 3 2{|, i{{ = 32|,
i{| = 1 3 2{ 3 2|, i|| = 32{. Then i{ = 0 implies |(1 3 2{ 3 |) = 0, so | = 0 or | = 1 3 2{. If | = 0 then
substitution into i| = 0 gives { 3 {2 = 0 i {(1 3 {) = 0 i { = 0 or { = 1, so we have critical points (0> 0) and
(1> 0). If | = 1 3 2{, substitution into i| = 0 gives { 3 {2 3 2{(1 3 2{) = 0 i 3{2 3 { = 0 i {(3{ 3 1) = 0
i { = 0 or { = 13 . If { = 0 then | = 1, and if { = 13 then | = 13 , so (0> 1) and 13 > 13 are critical points.
G(0> 0) = (0)(0) 3 12 = 31 ? 0,
G(1> 0) = (0)(32) 3 (31)2 = 31 ? 0, and
G(0> 1) = (32)(0) 3 (31)2 = 31 ? 0, so (0> 0), (1> 0), and (0> 1) are
saddle points. G 13 > 13 = (3 23 )(3 23 ) 3 (3 13 )2 = 13 A 0 and
1
is a local maximum.
i{{ 13 > 13 = 3 23 ? 0, so i 13 > 13 = 27
11. i ({> |) = {3 3 12{| + 8| 3
i i{ = 3{2 3 12|, i| = 312{ + 24| 2 , i{{ = 6{, i{| = 312, i|| = 48|. Then i{ = 0
implies {2 = 4| and i| = 0 implies { = 2|2 . Substituting the second equation into the first gives (2| 2 )2 = 4|
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SECTION 14.7
4| 4 = 4|
MAXIMUM AND MINIMUM VALUES
i 4|(| 3 3 1) = 0 i | = 0 or | = 1. If | = 0 then
{ = 0 and if | = 1 then { = 2, so the critical points are (0> 0) and (2> 1).
G(0> 0) = (0)(0) 3 (312)2 = 3144 ? 0, so (0> 0) is a saddle point.
G(2> 1) = (12)(48) 3 (312)2 = 432 A 0 and i{{ (2> 1) = 12 A 0 so
i (2> 1) = 38 is a local minimum.
12. i ({> |) = {| +
i{| = 1, i|| =
1
1
+
{
|
i i{ = | 3
1
1
2
, i| = { 3 2 , i{{ = 3 ,
{2
|
{
2
1
. Then i{ = 0 implies | = 2 and i| = 0 implies
|3
{
{=
1
. Substituting the first equation into the second gives
|2
{=
1
(1@{2 )2
i { = {4
i {({3 3 1) = 0 i { = 0 or { = 1.
i is not defined when { = 0, and when { = 1 we have | = 1, so the only critical point is (1> 1).
G(1> 1) = (2)(2) 3 12 = 3 A 0 and i{{ (1> 1) = 2 A 0, so i (1> 1) = 3 is a local minimum.
13. i ({> |) = h{ cos |
i i{ = h{ cos |, i| = 3h{ sin |.
Now i{ = 0 implies cos | = 0 or | = 2 + q for q an integer.
But sin 2 + q 6= 0, so there are no critical points.
14. i ({> |) = | cos {
i i{ = 3| sin {, i| = cos {, i{{ = 3| cos {,
i{| = 3 sin {, i|| = 0. Then i| = 0 if and only if { = 2 + q for q an
integer. But sin 2 + q 6= 0, so i{ = 0 i | = 0 and the critical
points are 2 + q> 0 , q an integer.
G 2 + q> 0 = (0)(0) 3 (±1)2 = 31 ? 0, so each critical point is
a saddle point.
15. i ({> |) = ({2 + | 2 )h|
2
3{2
i
i{ = ({2 + | 2 )h|
2 3{2
(32{) + 2{h|
i| = ({2 + | 2 )h|
2 3{2
(2|) + 2|h|
2 3{2
2 3{2
= 2{h|
= 2|h|
2 3{2
2 3{2
(1 3 {2 3 | 2 ),
(1 + {2 + | 2 ),
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454
¤
CHAPTER 14
i{{ = 2{h|
2 3{2
2
2
2
2
2
2
(32{) + (1 3 {2 3 | 2 ) 2{ 32{h| 3{ + 2h| 3{ = 2h| 3{ ((1 3 {2 3 | 2 )(1 3 2{2 ) 3 2{2 ),
2 3{2
2
2
(1 3 {2 3 | 2 ) = 34{|h| 3{ ({2 + | 2 ),
2
2
2
2
2
2
2
2
= 2|h| 3{ (2|) + (1 + {2 + | 2 ) 2| 2|h| 3{ + 2h| 3{ = 2h| 3{ ((1 + {2 + | 2 )(1 + 2| 2 ) + 2| 2 ).
i{| = 2{h|
i||
2 3{2
PARTIAL DERIVATIVES
(32|) + 2{(2|)h|
i| = 0 implies | = 0, and substituting into i{ = 0 gives
2
2{h3{ (1 3 {2 ) = 0 i { = 0 or { = ±1. Thus the critical points are
(0> 0) and (±1> 0). Now G(0> 0) = (2)(2) 3 0 A 0 and i{{ (0> 0) = 2 A 0,
so i (0> 0) = 0 is a local minimum. G(±1> 0) = (34h31 )(4h31 ) 3 0 ? 0
so (±1> 0) are saddle points.
16. i ({> |) = h| (| 2 3 {2 )
i i{ = 32{h| , i| = (2| + |2 3 {2 )h| ,
i{{ = 32h| , i{| = 32{h| , i|| = (2 + 4| + |2 3 {2 )h| . Then i{ = 0
implies { = 0 and substituting into i| = 0 gives (2| + | 2 )h| = 0 i
|(2 + |) = 0 i | = 0 or | = 32, so the critical points are (0> 0) and
(0> 32). G(0> 0) = (32)(2) 3 (0)2 = 34 ? 0 so (0> 0) is a saddle point.
G(0> 32) = (32h32 )(32h32 ) 3 (0)2 = 4h34 A 0 and i{{ (0> 32) = 32h32 ? 0, so i (0> 32) = 4h32 is a local
maximum.
17. i ({> |) = | 2 3 2| cos {
i i{ = 2| sin {, i| = 2| 3 2 cos {,
i{{ = 2| cos {, i{| = 2 sin {, i|| = 2. Then i{ = 0 implies | = 0 or
sin { = 0 i { = 0, , or 2 for 31 $ { $ 7. Substituting | = 0 into
i| = 0 gives cos { = 0 i { =
2
or
3
2 ,
substituting { = 0 or { = 2
into i| = 0 gives | = 1, and substituting { = into i| = 0 gives | = 31.
> 0 , and (2> 1).
Thus the critical points are (0> 1), 2 > 0 , (> 31), 3
2
G
2
> 0 = G 3
> 0 = 34 ? 0 so 2 > 0 and 3
> 0 are saddle points. G(0> 1) = G(> 31) = G(2> 1) = 4 A 0 and
2
2
i{{ (0> 1) = i{{ (> 31) = i{{ (2> 1) = 2 A 0, so i(0> 1) = i (> 31) = i (2> 1) = 31 are local minima.
18. i ({> |) = sin { sin |
i i{ = cos { sin |, i| = sin { cos |, i{{ = 3 sin { sin |, i{| = cos { cos |,
i|| = 3 sin { sin |. Here we have 3 ? { ? and 3 ? | ? , so i{ = 0 implies cos { = 0 or sin | = 0. If cos { = 0
then { = 3 2 or
,
2
and if sin | = 0 then | = 0. Substituting { = ± 2 into i| = 0 gives cos | = 0 i | = 3 2 or
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SECTION 14.7
MAXIMUM AND MINIMUM VALUES
¤
455
substituting | = 0 into i| = 0 gives sin { = 0 i { = 0. Thus the critical points are 3 2 > ± 2 , 2 > ± 2 , and (0> 0).
G(0> 0) = 31 ? 0 so (0> 0) is a saddle point.
G 3 2 > ± 2 = G 2 > ± 2 = 1 A 0 and
i{{ 3 2 > 3 2 = i{{ 2 > 2 = 31 ? 0 while
i{{ 3 2 > 2 = i{{ 2 > 3 2 = 1 A 0, so i 3 2 > 3 2 = i 2 > 2 = 1
are local maxima and i 3 2 > 2 = i 2 > 3 2 = 1 are local minima.
19. i ({> |) = {2 + 4| 2 3 4{| + 2
i i{ = 2{ 3 4|, i| = 8| 3 4{, i{{ = 2, i{| = 34, i|| = 8. Then i{ = 0
and i| = 0 each implies | = 12 {, so all points of the form {0 > 12 {0 are critical points and for each of these we have
G {0 > 12 {0 = (2)(8) 3 (34)2 = 0. The Second Derivatives Test gives no information, but
i ({> |) = {2 + 4| 2 3 4{| + 2 = ({ 3 2|)2 + 2 D 2 with equality if and only if | = 12 {. Thus i {0 > 12 {0 = 2 are all local
(and absolute) minima.
20. i ({> |) = {2 |h3{
2
3| 2
i
2 3| 2
(32{) + 2{|h3{
2 3| 2
(32|) + {2 h3{
i{ = {2 |h3{
i| = {2 |h3{
2 3| 2
2 3| 2
2 3| 2
i{{ = 2|(2{4 3 5{2 + 1)h3{
i{| = 2{(1 3 {2 )(1 3 2| 2 )h3{
2 3| 2
= 2{|(1 3 {2 )h3{
= {2 (1 3 2| 2 )h3{
2 3| 2
,
,
,
2 3| 2
2 3| 2
, i|| = 2{2 |(2| 2 3 3)h3{
.
i{ = 0 implies { = 0, | = 0, or { = ±1. If { = 0 then i| = 0 for any |-value, so all points of the form (0> |) are critical
2
points. If | = 0 then i| = 0 i {2 h3{ = 0 i { = 0, so (0> 0) (already included above) is a critical point. If { = ±1
2
then (1 3 2|2 )h313| = 0 i | = ± I12 , so ±1> I12 and ±1> 3 I12 are critical points. Now
I
G ±1> I12 = 8h33 A 0, i{{ ±1> I12 = 32 2 h33@2 ? 0 and G ±1> 3 I12 = 8h33 A 0,
I
i{{ ±1> 3 I12 = 2 2 h33@2 A 0, so i ±1> I12 =
I1 h33@2
2
are local maximum points while
i ±1> 3 I12 = 3 I12 h33@2 are local minimum points. At all critical points (0> |) we have G(0> |) = 0, so the Second
2 3| 2
Derivatives Test gives no information. However, if | A 0 then {2 |h3{
D 0 with equality only when { = 0, so we have
local minimum values i (0> |) = 0, | A 0. Similarly, if | ? 0 then {2 |h3{
2
3|2
$ 0 with equality when { = 0 so
i (0> |) = 0, | ? 0 are local maximum values, and (0> 0) is a saddle point.
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456
CHAPTER 14
PARTIAL DERIVATIVES
21. i ({> |) = {2 + | 2 + {32 | 32
From the graphs, there appear to be local minima of about i (1> ±1) = i (31> ±1) E 3 (and no local maxima or saddle
points). i{ = 2{ 3 2{33 | 32 , i| = 2| 3 2{32 | 33 , i{{ = 2 + 6{34 | 32 , i{| = 4{33 | 33 , i|| = 2 + 6{32 | 34 . Then
i{ = 0 implies 2{4 | 2 3 2 = 0 or {4 | 2 = 1 or | 2 = {34 . Note that neither { nor |