Solution Manual for Single Variable Calculus: Concepts and Contexts, 4th Edition

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COMPLETE SOLUTIONS MANUAL
for Stewart’s
SINGLE VARIABLE CALCULUS
CONCEPTS AND CONTEXTS
FOURTH EDITION
JEFFERY A. COLE
Anoka Ramsey Community College
TIMOTHY J. FLAHERTY
Carnegie Mellon University

ABBREVIATIONS AND SYMBOLS
CD concave downward
CU concave upward
D the domain of
FDT First Derivative Test
HA horizontal asymptote(s)
I interval of convergence
IP inection 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.

= indicates the use of Formula in the Table of Integrals in the back endpapers.
s
= indicates the use of the substitution { = sin = cos }.
c
= indicates the use of the substitution { = cos = sin }.
I/D Increasing/Decreasing Test

vii
CONTENTS
■ DIAGNOSTIC TESTS 1
1 ■ FUNCTIONS AND MODELS 9
1.1 Four Ways to Represent a Function 9
1.2 Mathematical Models: A Catalog of Essential Functions 20
1.3 New Functions from Old Functions 26
1.4 Graphing Calculators and Computers 38
1.5 Exponential Functions 45
1.6 Inverse Functions and Logarithms 51
1.7 Parametric Curves 59
Laboratory Project ■
Running Circles Around Circles 71
Review 73
Principles of Problem Solving 83
2 ■ LIMITS AND DERIVATIVES 87
2.1 The Tangent and Velocity Problems 87
2.2 The Limit of a Function 90
2.3 Calculating Limits Using the Limit Laws 96
2.4 Continuity 104
2.5 Limits Involving Infinity 113
2.6 Derivatives and Rates of Change 123
2.7 The Derivative as a Function 134
2.8 What Does Say about ? 146
Review 152
Focus on Problem Solving 163
3 ■ DIFFERENTIATION RULES 167
3.1 Derivatives of Polynomials and Exponential Functions 167
Applied Project ■
Building a Better Roller Coaster 177
3.2 The Product and Quotient Rules 179
ff
QQ

viii ■ CONTENTS
3.3 Derivatives of Trigonometric Functions 188
3.4 The Chain Rule 194
Laboratory Project ■
Bézier Curves 207
Applied Project ■
Where Should a Pilot Start Descent? 208
3.5 Implicit Differentiation 209
3.6 Inverse Trigonometric Functions and Their Derivatives 219
3.7 Derivatives of Logarithmic Functions 225
Discovery Project ■
Hyperbolic Functions 230
3.8 Rates of Change in the Natural and Social Sciences 232
3.9 Linear Approximations and Differentials 241
Laboratory Project ■
Taylor Polynomials 247
Review 249
Focus on Problem Solving 261
4 ■ APPLICATIONS OF DIFFERENTIATION 271
4.1 Related Rates 271
4.2 Maximum and Minimum Values 278
Applied Project ■
The Calculus of Rainbows 288
4.3 Derivatives and the Shapes of Curves 289
4.4 Graphing with Calculus and Calculators 310
4.5 Indeterminate Forms and l’Hospital’s Rule 328
4.6 Optimization Problems 340
Applied Project ■
The Shape of a Can 358
4.7 Newton’s Method 359
4.8 Antiderivatives 368
Review 375
Focus on Problem Solving 393
5 ■ INTEGRALS 403
5.1 Areas and Distances 403
5.2 The Definite Integral 412
5.3 Evaluating Definite Integrals 421
Discovery Project ■
Area Functions 428
5.4 The Fundamental Theorem of Calculus 430
5.5 The Substitution Rule 436
5.6 Integration by Parts 4445.65.6 Integration by Parts 444Integration by Parts 444

CONTENTS ■ ix
5.7 Additional Techniques of Integration 452
5.8 Integration Using Tables and Computer Algebra Systems 459
Discovery Project ■
Patterns in Integrals 465
5.9 Approximate Integration 467
5.10 Improper Integrals 479
Review 491
Focus on Problem Solving 505
6 ■ APPLICATIONS OF INTEGRATION 513
6.1 More about Areas 513
6.2 Volumes 523
Discovery Project ■
Rotating on a Slant 537
6.3 Volumes by Cylindrical Shells 538
6.4 Arc Length 546
Discovery Project ■
Arc Length Contest 554
6.5 Average Value of a Function 554
Applied Project ■
Where To Sit at the Movies 557
6.6 Applications to Physics and Engineering 557
Discovery Project ■
Complementary Coffee Cups 568
6.7 Applications to Economics and Biology 569
6.8 Probability 572
Review 575
Focus on Problem Solving 585
7 ■ DIFFERENTIAL EQUATIONS 593
7.1 Modeling with Differential Equations 593
7.2 Direction Fields and Euler’s Method 596
7.3 Separable Equations 604
Applied Project ■
How Fast Does a Tank Drain? 616
Applied Project ■
Which Is Faster, Going Up or Coming Down? 618
7.4 Exponential Growth and Decay 619
Applied Project ■
Calculus and Baseball 624
7.5 The Logistic Equation 625
7.6 Predator-Prey Systems 635
Review 640
Focus on Problem Solving 647Focus on Problem Solving 647Focus on Problem Solving 6

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x ■ CONTENTS
8 ■ INFINITE SEQUENCES AND SERIES 653
8.1 Sequences 653
Laboratory Project ■
Logistic Sequences 661
8.2 Series 665
8.3 The Integral and Comparison Tests; Estimating Sums 678
8.4 Other Convergence Tests 684
8.5 Power Series 691
8.6 Representations of Functions as Power Series 698
8.7 Taylor and Maclaurin Series 707
Laboratory Project ■
An Elusive Limit 721
8.8 Applications of Taylor Polynomials 722
Applied Project ■
Radiation from the Stars 733
Review 735
Focus on Problem Solving 747
■ APPENDIXES 753
A Intervals, Inequalities, and Absolute Values 753
B Coordinate Geometry 756
C Trigonometry 762
D Precise Definitions of Limits 766
F Sigma Notation 771
G Integration of Rational Functions by Partial Fractions 774
H Polar Coordinates 785
Discovery Project ■
Conic Sections in Polar Coordinates 808
I Complex Numbers 809

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DIAGNOSTIC TESTS
Test A Algebra
1. (a) (3)4 = (3)(3)(3)(3) = 81 (b) 34 = (3)(3)(3)(3) = 81
(c) 34 = 1
34 = 1
81 (d) 523
521 = 52321 = 52 = 25
(e) 2
3
2 = 3
2
2 = 9
4 (f ) 1634 = 1
1634 = 1
4
16 3 = 1
23 = 1
8
2. (a) Note that 200 = 100 · 2 = 10 2 and 32 = 16 · 2 = 4 2. Thus 200 32 = 10 2 4 2 = 6 2.
(b) (33 3)(4 2)2 = 33 3162 4 = 485 7
(c)
332 3
2 12
2
=
2 12
332 3
2
= (2 12)2
(332 3)2 = 4 1
93 6 = 4
93 6 =
9 7
3. (a) 3( + 6) + 4(2 5) = 3 + 18 + 8 20 = 11 2
(b) ( + 3)(4 5) = 42 5 + 12 15 = 42 + 7 15
(c)
+
=

2
+
2
=
Or: Use the formula for the difference of two squares to see that
+
=

2

2
= .
(d) (2 + 3)2 = (2 + 3)(2 + 3) = 42 + 6 + 6 + 9 = 42 + 12 + 9.
Note: A quicker way to expand this binomial is to use the formula ( + )2 = 2 + 2 + 2 with = 2 and = 3:
(2 + 3)2 = (2)2 + 2(2)(3) + 32 = 42 + 12 + 9
(e) See Reference Page 1 for the binomial formula ( + )3 = 3 + 32 + 3 2 + 3. Using it, we get
( + 2)3 = 3

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TEST B ANALYTIC GEOMETRY ¤ 3
(c) The inequality ( 1)( + 2) 0 has critical values of 2 0 and 1. The corresponding possible intervals of solution
are ( 2), (2 0), (0 1) and (1 ). By choosing a single test value from each interval, we see that both intervals
(2 0) and (1 ) satisfy the inequality. Thus, the solution is the union of these two intervals: (2 0) (1 ).
(d) | 4| 3 3 4 3 1 7. In interval notation, the answer is (1 7).
(e) 2 3
+ 1 1 2 3
+ 1 1 0 2 3
+ 1 + 1
+ 1 0 2 3 1
+ 1 0 4
+ 1 0.
Now, the expression 4
+ 1 may change signs at the critical values = 1 and = 4, so the possible intervals of solution
are ( 1), (1 4], and [4 ). By choosing a single test value from each interval, we see that (1 4] is the only
interval that satises the inequality.
10. (a) False. In order for the statement to be true, it must hold for all real numbers, so, to show that the statement is false, pick
= 1 and = 2 and observe that (1 + 2)2 6 = 12 + 22. In general, ( + )2 = 2 + 2 + 2.
(b) True as long as and are nonnegative real numbers. To see this, think in terms of the laws of exponents:
= ( )12 = 12 12 = .
(c) False. To see this, let = 1 and = 2, then 12 + 22 6 = 1 + 2.
(d) False. To see this, let = 1 and = 2, then 1 + 1(2)
2 6 = 1 + 1.
(e) False. To see this, let = 2 and = 3, then 1
2 3 6 = 1
2 1
3 .
(f ) True since 1
·
= 1
, as long as 6 = 0 and 6 = 0.
Test B Analytic Geometry
1. (a) Using the point (2 5) and = 3 in the point-slope equation of a line, 1 = ( 1), we get
(5) = 3( 2) + 5 = 3 + 6 = 3 + 1.
(b) A line parallel to the -axis must be horizontal and thus have a slope of 0. Since the line passes through the point (2 5),
the -coordinate of every point on the line is 5, so the equation is = 5.
(c) A line parallel to the -axis is vertical with undened slope. So the -coordinate of every point on the line is 2 and so the
equation is = 2.
(d) Note that 2 4 = 3 4 = 2 + 3 = 1
2 3
4 . Thus the slope of the given line is = 1
2 . Hence, the
slope of the line we’re looking for is also 1
2 (since the line we’re looking for is required to be parallel to the given line).
So the equation of the line is (5) = 1
2 ( 2) + 5 = 1
2 1 = 1
2 6.
2. First we’ll nd the distance between the two given points in order to obtain the radius, , of the circle:

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4 ¤ DIAGNOGSTIC TESTS
3. We must rewrite the equation in standard form in order to identify the center and radius. Note that
2 + 2 6 + 10 + 9 = 0 2 6 + 9 + 2 + 10 = 0. For the left-hand side of the latter equation, we
factor the rst three terms and complete the square on the last two terms as follows: 2 6 + 9 + 2 + 10 = 0
( 3)2 + 2 + 10 + 25 = 25 ( 3)2 + ( + 5)2 = 25. Thus, the center of the circle is (3 5) and the radius is 5.
4. (a) (7 4) and (5 12) = 12 4
5 (7) = 16
12 = 4
3
(b) 4 = 4
3 [ (7)] 4 = 4
3 28
3 3 12 = 4 28 4 + 3 + 16 = 0. Putting = 0,
we get 4 + 16 = 0, so the -intercept is 4, and substituting 0 for results in a -intercept of 16
3 .
(c) The midpoint is obtained by averaging the corresponding coordinates of both points:
7+5
2 4+(12)
2

= (1 4).
(d) = [5 (7)]2 + (12 4)2 = 122 + (16)2 = 144 + 256 = 400 = 20
(e) The perpendicular bisector is the line that intersects the line segment at a right angle through its midpoint. Thus the
perpendicular bisector passes through (1 4) and has slope 3
4 [the slope is obtained by taking the negative reciprocal of
the answer from part (a)]. So the perpendicular bisector is given by + 4 = 3
4 [ (1)] or 3 4 = 13.
(f ) The center of the required circle is the midpoint of , and the radius is half the length of , which is 10. Thus, the
equation is ( + 1)2 + ( + 4)2 = 100.
5. (a) Graph the corresponding horizontal lines (given by the equations = 1 and
= 3) as solid lines. The inequality 1 describes the points ( ) that lie
on or above the line = 1. The inequality 3 describes the points ( )
that lie on or below the line = 3. So the pair of inequalities 1 3
describes the points that lie on or between the lines = 1 and = 3.
(b) Note that the given inequalities can be written as 4 4 and 2 2,
respectively. So the region lies between the vertical lines = 4 and = 4 and
between the horizontal lines = 2 and = 2. As shown in the graph, the
region common to both graphs is a rectangle (minus its edges) centered at the
origin.
(c) We rst graph = 1 1
2 as a dotted line. Since 1 1
2 , the points in the
region lie below this line.

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TEST C FUNCTIONS ¤ 5
(d) We rst graph the parabola = 2 1 using a solid curve. Since 2 1,
the points in the region lie on or above the parabola.
(e) We graph the circle 2 + 2 = 4 using a dotted curve. Since2 + 2 2, the
region consists of points whose distance from the origin is less than 2, that is,
the points that lie inside the circle.
(f ) The equation 92 + 16 2 = 144 is an ellipse centered at (0 0). We put it in
standard form by dividing by 144 and get 2
16 +
2
9 = 1. The -intercepts are
located at a distance of 16 = 4 from the center while the -intercepts are a
distance of 9 = 3 from the center (see the graph).
Test C Functions
1. (a) Locate 1 on the -axis and then go down to the point on the graph with an -coordinate of 1. The corresponding
-coordinate is the value of the function at = 1, which is 2. So, (1) = 2.
(b) Using the same technique as in part (a), we get (2) 28.
(c) Locate 2 on the -axis and then go left and right to nd all points on the graph with a -coordinate of 2. The corresponding
-coordinates are the -values we are searching for. So = 3 and = 1.
(d) Using the same technique as in part (c), we get 25 and 03.
(e) The domain is all the -values for which the graph exists, and the range is all the -values for which the graph exists.
Thus, the domain is [3 3], and the range is [2 3].
2. Note that (2 + ) = (2 + )3 and (2) = 23 = 8. So the difference quotient becomes
(2 + ) (2) = (2 + )3 8 = 8 + 12 + 6 2 + 3 8 = 12 + 6 2 + 3
= (12 + 6 + 2) = 12 + 6 + 2.
3. (a) Set the denominator equal to 0 and solve to nd restrictions on the domain: 2 + 2 = 0
( 1)( + 2) = 0 = 1 or = 2. Thus, the domain is all real numbers except 1 or 2 or, in interval
notation, ( 2) (2 1) (1 ).
(b) Note that the denominator is always greater than or equal to 1, and the numerator is dened for all real numbers. Thus, the
domain is ( ).
(c) Note that the function is the sum of two root functions. So is dened on the intersection of the domains of these two
root functions. The domain of a square root function is found by setting its radicand greater than or equal to 0. Now,

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6 ¤ DIAGNOGSTIC TESTS
4 0 4 and 2 1 0 ( 1)( + 1) 0 1 or 1. Thus, the domain of
is ( 1] [1 4].
4. (a) Reect the graph of about the -axis.
(b) Stretch the graph of vertically by a factor of 2, then shift 1 unit downward.
(c) Shift the graph of right 3 units, then up 2 units.
5. (a) Make a table and then connect the points with a smooth curve:
2 1 0 1 2
8 1 0 1 8
(b) Shift the graph from part (a) left 1 unit.
(c) Shift the graph from part (a) right 2 units and up 3 units.
(d) First plot = 2. Next, to get the graph of () = 4 2,
reect about the x-axis and then shift it upward 4 units.
(e) Make a table and then connect the points with a smooth curve:
0 1 4 9
0 1 2 3
(f ) Stretch the graph from part (e) vertically by a factor of two.

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TEST D TRIGONOMETRY ¤ 7
(g) First plot = 2. Next, get the graph of = 2 by reecting the graph of
= 2 about the x-axis.
(h) Note that = 1 + 1 = 1 + 1. So rst plot = 1 and then shift it
upward 1 unit.
6. (a) (2) = 1 (2)2 = 3 and (1) = 2(1) + 1 = 3
(b) For 0 plot () = 1 2 and, on the same plane, for 0 plot the graph
of () = 2 + 1.
7. (a) ( )() = (()) = (2 3) = (2 3)2 + 2(2 3) 1 = 42 12 + 9 + 4 6 1 = 42 8 + 2
(b) ( )() = ( ()) = (2 + 2 1) = 2(2 + 2 1) 3 = 22 + 4 2 3 = 22 + 4 5
(c) ( )() = ((())) = ((2 3)) = (2(2 3) 3) = (4 9) = 2(4 9) 3
= 8 18 3 = 8 21
Test D Trigonometry
1. (a) 300 = 300

180

= 300
180 = 5
3 (b) 18 = 18

180

= 18
180 =
10
2. (a) 5
6 = 5
6
180

= 150 (b) 2 = 2
180

= 360
1146
3. We will use the arc length formula, = , where is arc length, is the radius of the circle, and is the measure of the
central angle in radians. First, note that 30 = 30

180

=
6 . So = (12)

6

= 2 cm.
4. (a) tan(3) = 3 You can read the value from a right triangle with sides 1, 2, and 3.
(b) Note that 76 can be thought of as an angle in the third quadrant with reference angle 6. Thus, sin(76) = 1
2 ,
since the sine function is negative in the third quadrant.
(c) Note that 53 can be thought of as an angle in the fourth quadrant with reference angle 3. Thus,
sec(53) = 1
cos(53) = 1
12 = 2, since the cosine function is positive in the fourth quadrant.
5. sin = 24 = 24 sin and cos = 24 = 24 cos 5.5. sinsin == 2424 = 24 sin= 24 sin andand coscos == 2424 = 24 cos= 24 co

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8 ¤ DIAGNOGSTIC TESTS
6. sin = 1
3 and sin2 + cos2 = 1 cos = 1 1
9 = 2 2
3 . Also, cos = 4
5 sin = 1 16
25 = 3
5 .
So, using the sum identity for the sine, we have
sin( + ) = sin cos + cos sin = 1
3 · 4
5 + 2 2
3 · 3
5 = 4 + 6 2
15 = 1
15
4 + 6 2
7. (a) tan sin + cos = sin
cos sin + cos = sin2
cos + cos2
cos = 1
cos = sec
(b) 2 tan
1 + tan2 = 2 sin (cos )
sec2 = 2 sin
cos cos2 = 2 sin cos = sin 2
8. sin 2 = sin 2 sin cos = sin 2 sin cos sin = 0 sin (2 cos 1) = 0
sin = 0 or cos = 1
2 = 0,
3 , , 5
3 , 2.
9. We rst graph = sin 2 (by compressing the graph of sin
by a factor of 2) and then shift it upward 1 unit.

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1 FUNCTIONS AND MODELS
1.1 Four Ways To Represent a Function
In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here.
1. (a) The point (1 3) is on the graph of , so (1) = 3.
(b) When = 1, is about 02, so (1) 02.
(c) () = 1 is equivalent to = 1 When = 1, we have = 0 and = 3.
(d) A reasonable estimate for when = 0 is = 08.
(e) The domain of consists of all -values on the graph of . For this function, the domain is 2 4, or [2 4].
The range of consists of all -values on the graph of . For this function, the range is 1 3, or [1 3].
(f ) As increases from 2 to 1, increases from 1 to 3. Thus, is increasing on the interval [2 1].
2. (a) The point (4 2) is on the graph of , so (4) = 2. The point (3 4) is on the graph of , so (3) = 4.
(b) We are looking for the values of for which the -values are equal. The -values for and are equal at the points
(2 1) and (2 2), so the desired values of are 2 and 2.
(c) () = 1 is equivalent to = 1. When = 1, we have = 3 and = 4.
(d) As increases from 0 to 4, decreases from 3 to 1. Thus, is decreasing on the interval [0 4].
(e) The domain of consists of all -values on the graph of . For this function, the domain is 4 4, or [4 4].
The range of consists of all -values on the graph of . For this function, the range is 2 3, or [2 3].
(f ) The domain of is [4 3] and the range is [05 4].
3. From Figure 1 in the text, the lowest point occurs at about ( ) = (12 85). The highest point occurs at about (17 115).
Thus, the range of the vertical ground acceleration is 85 115. Written in interval notation, we get [85 115].
4. Example 1: A car is driven at 60 mih for 2 hours. The distance
traveled by the car is a function of the time . The domain of the
function is { | 0 2}, where is measured in hours. The range
of the function is { | 0 120}, where is measured in miles.
Example 2: At a certain university, the number of students on
campus at any time on a particular day is a function of the time after
midnight. The domain of the function is { | 0 24}, where is
measured in hours. The range of the function is { | 0 },
where is an integer and is the largest number of students on
campus at once.

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