SOLUTION MANUAL
, Contents
Chapter 0. Before Calculus ………..…………………………………………………………………………..……. 1
Chapter 1. Limits and Continuity ……………………………………………………………………………….. 39
Chapter 2. The Derivative ……………………………………………………………………………………..…….71
Chapter 3. Topics in Differentiation ……………………………..………………………………………..…….109
Chapter 4. The Derivative in Graphing and Applications ……………………………………..………. 153
Chapter 5. Integration …………………………………………………………………………………………..…… 243
Chapter 6. Applications of the Definite Integral in Geometry, Science, and Engineering… 305
Chapter 7. Principals of Integral Evaluation ……………………………………………………………….. 363
Chapter 8. Mathematical Modeling with Differential Equations …………………………………… 413
Chapter 9. Infinite Series ……………………………………………………………………………………..…….. 437
Chapter 10. Parametric and Polar Curves; Conic Sections ……………………………………….…….. 485
Chapter 11. Three-Dimensional Space; Vectors ………………………………………………….…………. 545
Chapter 12. Vector-Valued Functions ………………………………………………………………….……….. 589
Chapter 13. Partial Derivatives ………………………………………………………………………………….… 627
Chapter 14. Multiple Integrals ……………………………………………………………………………………… 675
Chapter 15. Topics in Vector Calculus ……………………………………………………………………….….. 713
Appendix A. Graphing Functions Using Calculators and Computer Algebra Systems .………. 745
Appendix B. Trigonometry Review ……………………………………………………………………………….. 753
Appendix C. Solving Polynomial Equations …………………………………………………………………… 759
,Before Calculus
Exercise Set 0.1
1. (a) −2.9, −2.0, 2.35, 2.9 (b) None (c) y = 0 (d) −1.75 ≤ x ≤ 2.15, x = −3, x = 3
(e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2
2. (a) x = −1, 4 (b) None (c) y = −1 (d) x = 0, 3, 5
(e) ymax = 9 at x = 6; ymin = −2 at x = 0
3. (a) Yes (b) Yes (c) No (vertical line test fails) (d) No (vertical line test fails)
4. (a) The natural domain of f is x 6= −1, and for g it is the set of all x. f (x) = g(x) on the intersection of their
domains.
(b) The domain of f is the set of all x ≥ 0; the domain of g is the same, and f (x) = g(x).
5. (a) 1999, $47,700 (b) 1993, $41,600
(c) The slope between 2000 and 2001 is steeper than the slope between 2001 and 2002, so the median income was
declining more rapidly during the first year of the 2-year period.
47.7 − 41.6 6.1
6. (a) In thousands, approximately = per yr, or $1017/yr.
6 6
(b) From 1993 to 1996 the median income increased from $41.6K to $44K (K for ‘kilodollars’; all figures approx-
imate); the average rate of increase during this time was (44 − 41.6)/3 K/yr = 2.4/3 K/yr = $800/year. From
1996 to 1999 the average rate of increase was (47.7 − 44)/3 K/yr = 3.7/3 K/yr ≈ $1233/year. The increase was
larger during the last 3 years of the period.
(c) 1994 and 2005.
2 2 2 2
√
√ f2(0) = 3(0) − 2 = −2; f2 (2) = 3(2)2 − 2 = 10; f (−2) = 3(−2) − 2 = 10; f (3) = 3(3) − 2 = 25; f ( 2) =
7. (a)
3( 2) − 2 = 4; f (3t) = 3(3t) − 2 = 27t − 2.
√ √
(b) f (0) = 2(0) = 0; f (2) = 2(2) = 4; f (−2) = 2(−2) = −4; f (3) = 2(3) = 6; f ( 2) = 2 2; f (3t) = 1/(3t) for
t > 1 and f (3t) = 6t for t ≤ 1.
3+1 −1 + 1 π+1 −1.1 + 1 −0.1 1
8. (a) g(3) = = 2; g(−1) = = 0; g(π) = ; g(−1.1) = = = ; g(t2 − 1) =
3−1 −1 − 1 π−1 −1.1 − 1 −2.1 21
t2 − 1 + 1 t2
2
= 2 .
t −1−1 t −2
√ √
(b)
√ g(3) = 3 + 1 = 2; g(−1) = 3; g(π) = π + 1; g(−1.1) = 3; g(t2 − 1) = 3 if t2 < 2 and g(t2 − 1) =
t2 − 1 + 1 = |t| if t2 ≥ 2.
1
, 2 Chapter 0
9. (a) Natural domain: x 6= 3. Range: y 6= 0. (b) Natural domain: x 6= 0. Range: {1, −1}.
√ √
(c) Natural domain: x ≤ − 3 or x ≥ 3. Range: y ≥ 0.
√
(d) x2 − 2x + 5 = (x − 1)2 + 4 ≥ 4. So G(x) is defined for all x, and is ≥ 4 = 2. Natural domain: all x. Range:
y ≥ 2.
(e) Natural domain: sin x 6= 1, so x 6= (2n+ 12 )π, n = 0, ±1, ±2, . . .. For such x, −1 ≤ sin x < 1, so 0 < 1−sin x ≤ 2,
1 1 1
and 1−sin x ≥ 2 . Range: y ≥ 2 .
2
−4
(f ) Division by 0 occurs for x = 2. For all other x, xx−2 = x + 2, which is nonnegative for x ≥ −2. Natural
√ √
domain: [−2, 2) ∪ (2, +∞). The range of x + 2 is [0, +∞). But we must exclude x = 2, for which x + 2 = 2.
Range: [0, 2) ∪ (2, +∞).
10. (a) Natural domain: x ≤ 3. Range: y ≥ 0. (b) Natural domain: −2 ≤ x ≤ 2. Range: 0 ≤ y ≤ 2.
(c) Natural domain: x ≥ 0. Range: y ≥ 3. (d) Natural domain: all x. Range: all y.
(e) Natural domain: all x. Range: −3 ≤ y ≤ 3.
√ √
√ ) For x to exist, we must have x ≥ 0. For H(x) to exist, we2 must also have sin x 6= 0, which is equivalent
(f √ to
x 6= πn√for n = 0, 1, 2, . . .. Natural domain: x > 0, x 6= (πn) for n = 1, 2, . . .. For such x, 0 < | sin x| ≤ 1, so
0 < (sin x)2 ≤ 1 and H(x) ≥ 1. Range: y ≥ 1.
11. (a) The curve is broken whenever someone is born or someone dies.
(b) C decreases for eight hours, increases rapidly (but continuously), and then repeats.
12. (a) Yes. The temperature may change quickly under some conditions, but not instantaneously.
(b) No; the number is always an integer, so the changes are in movements (jumps) of at least one unit.
h
t
13.
T
t
14.
√
15. Yes. y = 25 − x2 .
√
16. Yes. y = − 25 − x2 .
√
2
17. Yes. y = √25 − x , −5 ≤ x ≤ 0
− 25 − x2 , 0<x≤5
, Contents
Chapter 0. Before Calculus ………..…………………………………………………………………………..……. 1
Chapter 1. Limits and Continuity ……………………………………………………………………………….. 39
Chapter 2. The Derivative ……………………………………………………………………………………..…….71
Chapter 3. Topics in Differentiation ……………………………..………………………………………..…….109
Chapter 4. The Derivative in Graphing and Applications ……………………………………..………. 153
Chapter 5. Integration …………………………………………………………………………………………..…… 243
Chapter 6. Applications of the Definite Integral in Geometry, Science, and Engineering… 305
Chapter 7. Principals of Integral Evaluation ……………………………………………………………….. 363
Chapter 8. Mathematical Modeling with Differential Equations …………………………………… 413
Chapter 9. Infinite Series ……………………………………………………………………………………..…….. 437
Chapter 10. Parametric and Polar Curves; Conic Sections ……………………………………….…….. 485
Chapter 11. Three-Dimensional Space; Vectors ………………………………………………….…………. 545
Chapter 12. Vector-Valued Functions ………………………………………………………………….……….. 589
Chapter 13. Partial Derivatives ………………………………………………………………………………….… 627
Chapter 14. Multiple Integrals ……………………………………………………………………………………… 675
Chapter 15. Topics in Vector Calculus ……………………………………………………………………….….. 713
Appendix A. Graphing Functions Using Calculators and Computer Algebra Systems .………. 745
Appendix B. Trigonometry Review ……………………………………………………………………………….. 753
Appendix C. Solving Polynomial Equations …………………………………………………………………… 759
,Before Calculus
Exercise Set 0.1
1. (a) −2.9, −2.0, 2.35, 2.9 (b) None (c) y = 0 (d) −1.75 ≤ x ≤ 2.15, x = −3, x = 3
(e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2
2. (a) x = −1, 4 (b) None (c) y = −1 (d) x = 0, 3, 5
(e) ymax = 9 at x = 6; ymin = −2 at x = 0
3. (a) Yes (b) Yes (c) No (vertical line test fails) (d) No (vertical line test fails)
4. (a) The natural domain of f is x 6= −1, and for g it is the set of all x. f (x) = g(x) on the intersection of their
domains.
(b) The domain of f is the set of all x ≥ 0; the domain of g is the same, and f (x) = g(x).
5. (a) 1999, $47,700 (b) 1993, $41,600
(c) The slope between 2000 and 2001 is steeper than the slope between 2001 and 2002, so the median income was
declining more rapidly during the first year of the 2-year period.
47.7 − 41.6 6.1
6. (a) In thousands, approximately = per yr, or $1017/yr.
6 6
(b) From 1993 to 1996 the median income increased from $41.6K to $44K (K for ‘kilodollars’; all figures approx-
imate); the average rate of increase during this time was (44 − 41.6)/3 K/yr = 2.4/3 K/yr = $800/year. From
1996 to 1999 the average rate of increase was (47.7 − 44)/3 K/yr = 3.7/3 K/yr ≈ $1233/year. The increase was
larger during the last 3 years of the period.
(c) 1994 and 2005.
2 2 2 2
√
√ f2(0) = 3(0) − 2 = −2; f2 (2) = 3(2)2 − 2 = 10; f (−2) = 3(−2) − 2 = 10; f (3) = 3(3) − 2 = 25; f ( 2) =
7. (a)
3( 2) − 2 = 4; f (3t) = 3(3t) − 2 = 27t − 2.
√ √
(b) f (0) = 2(0) = 0; f (2) = 2(2) = 4; f (−2) = 2(−2) = −4; f (3) = 2(3) = 6; f ( 2) = 2 2; f (3t) = 1/(3t) for
t > 1 and f (3t) = 6t for t ≤ 1.
3+1 −1 + 1 π+1 −1.1 + 1 −0.1 1
8. (a) g(3) = = 2; g(−1) = = 0; g(π) = ; g(−1.1) = = = ; g(t2 − 1) =
3−1 −1 − 1 π−1 −1.1 − 1 −2.1 21
t2 − 1 + 1 t2
2
= 2 .
t −1−1 t −2
√ √
(b)
√ g(3) = 3 + 1 = 2; g(−1) = 3; g(π) = π + 1; g(−1.1) = 3; g(t2 − 1) = 3 if t2 < 2 and g(t2 − 1) =
t2 − 1 + 1 = |t| if t2 ≥ 2.
1
, 2 Chapter 0
9. (a) Natural domain: x 6= 3. Range: y 6= 0. (b) Natural domain: x 6= 0. Range: {1, −1}.
√ √
(c) Natural domain: x ≤ − 3 or x ≥ 3. Range: y ≥ 0.
√
(d) x2 − 2x + 5 = (x − 1)2 + 4 ≥ 4. So G(x) is defined for all x, and is ≥ 4 = 2. Natural domain: all x. Range:
y ≥ 2.
(e) Natural domain: sin x 6= 1, so x 6= (2n+ 12 )π, n = 0, ±1, ±2, . . .. For such x, −1 ≤ sin x < 1, so 0 < 1−sin x ≤ 2,
1 1 1
and 1−sin x ≥ 2 . Range: y ≥ 2 .
2
−4
(f ) Division by 0 occurs for x = 2. For all other x, xx−2 = x + 2, which is nonnegative for x ≥ −2. Natural
√ √
domain: [−2, 2) ∪ (2, +∞). The range of x + 2 is [0, +∞). But we must exclude x = 2, for which x + 2 = 2.
Range: [0, 2) ∪ (2, +∞).
10. (a) Natural domain: x ≤ 3. Range: y ≥ 0. (b) Natural domain: −2 ≤ x ≤ 2. Range: 0 ≤ y ≤ 2.
(c) Natural domain: x ≥ 0. Range: y ≥ 3. (d) Natural domain: all x. Range: all y.
(e) Natural domain: all x. Range: −3 ≤ y ≤ 3.
√ √
√ ) For x to exist, we must have x ≥ 0. For H(x) to exist, we2 must also have sin x 6= 0, which is equivalent
(f √ to
x 6= πn√for n = 0, 1, 2, . . .. Natural domain: x > 0, x 6= (πn) for n = 1, 2, . . .. For such x, 0 < | sin x| ≤ 1, so
0 < (sin x)2 ≤ 1 and H(x) ≥ 1. Range: y ≥ 1.
11. (a) The curve is broken whenever someone is born or someone dies.
(b) C decreases for eight hours, increases rapidly (but continuously), and then repeats.
12. (a) Yes. The temperature may change quickly under some conditions, but not instantaneously.
(b) No; the number is always an integer, so the changes are in movements (jumps) of at least one unit.
h
t
13.
T
t
14.
√
15. Yes. y = 25 − x2 .
√
16. Yes. y = − 25 − x2 .
√
2
17. Yes. y = √25 − x , −5 ≤ x ≤ 0
− 25 − x2 , 0<x≤5
