SOLUTIONS MANUAL
Prepared by
Neil Wigley
University of Windsor
Albert Herr
To Accompany
CALCULUS
EARLY TRANSCENDENTALS
Seventh Edition
CONTENTS
Chapter 1. Functions .................................................................................................1
Chapter 2. Limits and Continuity ...........................................................................44
Chapter 3. The Derivative.......................................................................................70
Chapter 4. Exponential, Logarithmic, and Inverse Trigonometric Functions ........122
Chapter 5. The Derivative in Graphing and Applications .....................................150
Chapter 6. Integration ...........................................................................................223
Chapter 7. Applications of the Definite Integral in Geometry,
Science, and Engineering ....................................................................278
Chapter 8. Principles of Integral Evaluation .........................................................317
Chapter 9. Mathematical Modeling with Differential Equations ..........................372
Chapter 10. Infinite Series ......................................................................................397
Chapter 11. Analytic Geometry in Calculus ...........................................................447
Chapter 12. Three-Dimensional Space; Vectors .....................................................492
Chapter 13. Vector-Valued Functions ......................................................................534
Chapter 14. Partial Derivatives ...............................................................................569
, Chapter 15. Multiple Integrals ................................................................................620
Chapter 16. Topics in Vector Calculus ....................................................................657
Appendix A. Real Numbers, Intervals, and Inequalities ...........................................690
Appendix B. Absolute Value....................................................................................697
Appendix C. Coordinate Planes and Lines ...............................................................700
Appendix D. Distance, Circles, and Quadratic Equations ........................................709
Appendix E. Trigonometry Review .........................................................................719
Appendix F. Solving Polynomial Equations ............................................................725
CHAPTER 1
Functions
EXERCISE SET 1.1
1. (a) around 1943 (b) 1960; 4200
(c) no; you need the year‘s population (d) war; marketing techniques
(e) news of health risk; social pressure, antismoking campaigns, increased taxation
2. (a) 1989; $35,600 (b) 1975, 1983; $32,000
(c) the first two years; the curve is steeper (downhill)
3. (a) −2.9, −2.0, 2.35, 2.9 (b) none (c) y = 0
(d) −1.75 ≤ x ≤ 2.15 (e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2
4. (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
5. (a) x = 2, 4 (b) none (c) x ≤ 2; 4 ≤ x (d) ymin = −1; no maximum value
6. (a) x=9 (b) none (c) x ≥ 25 (d) ymin = 1; no maximum value
7. (a) Breaks could be caused by war, pestilence, flood, earthquakes, for example.
(b) C decreases for eight hours, takes a jump upwards, and then repeats.
8. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperature
change.
(b) No; the number is always an integer, so the changes are in movements (jumps) of at least
one unit.
, 9. (a) The side adjacent to the building has length x, so L = x + 2y. Since A = xy =
1000, L = x + 2000/x.
(b) x > 0 and x must be smaller than the width of the building, which was not given.
(c) 120 (d) Lmin ≈ 89.44 ft
20 80
80
10. (a) V = lwh = (6 − 2x)(6 − 2x)x (b) From the figure it is clear that 0 < x < 3.
(c) 20 (d) Vmax ≈ 16 in3
0 3
0
, 500
11. (a) V = 500 = πr2h so h = . Then 7
πr2
500
C = (0.02)(2)πr2 + (0.01)2πrh = 0.04πr2 + 0.02πr
πr2
10
= 0.04πr2 + ; Cmin ≈ 4.39 at r ≈ 3.4, h ≈ 13.8.
1.5 6
4
(b) C = (0.02)(2)(2r)2 + (0.01)2πrh = 0.16r2 + 10 . Since 7
r
0.04π < 0.16, the top and bottom now get more weight.
Since they cost more, we diminish their sizes in the
solution, and the cans become taller.
1.5 5.5
4
(c) r ≈ 3.1 cm, h ≈ 16.0 cm, C ≈ 4.76
cents
12. (a) The length of a track with straightaways of length L and semicircles of radius r is
P = (2)L + (2)(πr) ft. Let L = 360 and r = 80 to get P = 720 + 160π = 1222.65
ft. Since this is less than 1320 ft (a quarter-mile), a solution is possible.
(b) P = 2L + 2πr = 1320 and 2r = 2x + 160, 450
so
L = 1 (1320 − 2πr) = 1 (1320 − 2π(80 + x))
2 2
= 660 − 80π − πx.
0 100
0
(c) The shortest straightaway is L = 360, so x = 15.49 ft.
(d) The longest straightaway occurs when x = 0, so L = 660 − 80π = 408.67 ft.
EXERCISE SET 1.2
1. (a) f (0) = 3(0)2 −2 = −2; f (2) = 3(2)2 −2 = 10; f (−2) = 3(−2)2 −2 = 10; f (3) = 3(3)2 −2 = 25;
f ( 2) = 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
2. (a) g(3) = = 2; g(−1) = = 0; g(π) = −1.1) = = = ;
; g(
3−1 −1 − π−1 −1.1 − 1 −2.1 21
2 2
1
t − 1+1 t
g(t2 − 1) = =
t2 − 1 − 1 t2 − 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.
√ √
3. (a) x /= 3 (b) x ≤ − 3 or x ≥ 3
(c) x2 − 2x + 5 = 0 has no real solutions so x2 − 2x + 5 is always positive or always negative. If
x = 0, then x2 − 2x + 5 = 5 > 0; domain: (−∞, +∞).
(d) x /= 0 (e) sin x /= 1, so x /= (2n + 1 )π, n = 0, ±1, ±2,...
Prepared by
Neil Wigley
University of Windsor
Albert Herr
To Accompany
CALCULUS
EARLY TRANSCENDENTALS
Seventh Edition
CONTENTS
Chapter 1. Functions .................................................................................................1
Chapter 2. Limits and Continuity ...........................................................................44
Chapter 3. The Derivative.......................................................................................70
Chapter 4. Exponential, Logarithmic, and Inverse Trigonometric Functions ........122
Chapter 5. The Derivative in Graphing and Applications .....................................150
Chapter 6. Integration ...........................................................................................223
Chapter 7. Applications of the Definite Integral in Geometry,
Science, and Engineering ....................................................................278
Chapter 8. Principles of Integral Evaluation .........................................................317
Chapter 9. Mathematical Modeling with Differential Equations ..........................372
Chapter 10. Infinite Series ......................................................................................397
Chapter 11. Analytic Geometry in Calculus ...........................................................447
Chapter 12. Three-Dimensional Space; Vectors .....................................................492
Chapter 13. Vector-Valued Functions ......................................................................534
Chapter 14. Partial Derivatives ...............................................................................569
, Chapter 15. Multiple Integrals ................................................................................620
Chapter 16. Topics in Vector Calculus ....................................................................657
Appendix A. Real Numbers, Intervals, and Inequalities ...........................................690
Appendix B. Absolute Value....................................................................................697
Appendix C. Coordinate Planes and Lines ...............................................................700
Appendix D. Distance, Circles, and Quadratic Equations ........................................709
Appendix E. Trigonometry Review .........................................................................719
Appendix F. Solving Polynomial Equations ............................................................725
CHAPTER 1
Functions
EXERCISE SET 1.1
1. (a) around 1943 (b) 1960; 4200
(c) no; you need the year‘s population (d) war; marketing techniques
(e) news of health risk; social pressure, antismoking campaigns, increased taxation
2. (a) 1989; $35,600 (b) 1975, 1983; $32,000
(c) the first two years; the curve is steeper (downhill)
3. (a) −2.9, −2.0, 2.35, 2.9 (b) none (c) y = 0
(d) −1.75 ≤ x ≤ 2.15 (e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2
4. (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
5. (a) x = 2, 4 (b) none (c) x ≤ 2; 4 ≤ x (d) ymin = −1; no maximum value
6. (a) x=9 (b) none (c) x ≥ 25 (d) ymin = 1; no maximum value
7. (a) Breaks could be caused by war, pestilence, flood, earthquakes, for example.
(b) C decreases for eight hours, takes a jump upwards, and then repeats.
8. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperature
change.
(b) No; the number is always an integer, so the changes are in movements (jumps) of at least
one unit.
, 9. (a) The side adjacent to the building has length x, so L = x + 2y. Since A = xy =
1000, L = x + 2000/x.
(b) x > 0 and x must be smaller than the width of the building, which was not given.
(c) 120 (d) Lmin ≈ 89.44 ft
20 80
80
10. (a) V = lwh = (6 − 2x)(6 − 2x)x (b) From the figure it is clear that 0 < x < 3.
(c) 20 (d) Vmax ≈ 16 in3
0 3
0
, 500
11. (a) V = 500 = πr2h so h = . Then 7
πr2
500
C = (0.02)(2)πr2 + (0.01)2πrh = 0.04πr2 + 0.02πr
πr2
10
= 0.04πr2 + ; Cmin ≈ 4.39 at r ≈ 3.4, h ≈ 13.8.
1.5 6
4
(b) C = (0.02)(2)(2r)2 + (0.01)2πrh = 0.16r2 + 10 . Since 7
r
0.04π < 0.16, the top and bottom now get more weight.
Since they cost more, we diminish their sizes in the
solution, and the cans become taller.
1.5 5.5
4
(c) r ≈ 3.1 cm, h ≈ 16.0 cm, C ≈ 4.76
cents
12. (a) The length of a track with straightaways of length L and semicircles of radius r is
P = (2)L + (2)(πr) ft. Let L = 360 and r = 80 to get P = 720 + 160π = 1222.65
ft. Since this is less than 1320 ft (a quarter-mile), a solution is possible.
(b) P = 2L + 2πr = 1320 and 2r = 2x + 160, 450
so
L = 1 (1320 − 2πr) = 1 (1320 − 2π(80 + x))
2 2
= 660 − 80π − πx.
0 100
0
(c) The shortest straightaway is L = 360, so x = 15.49 ft.
(d) The longest straightaway occurs when x = 0, so L = 660 − 80π = 408.67 ft.
EXERCISE SET 1.2
1. (a) f (0) = 3(0)2 −2 = −2; f (2) = 3(2)2 −2 = 10; f (−2) = 3(−2)2 −2 = 10; f (3) = 3(3)2 −2 = 25;
f ( 2) = 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
2. (a) g(3) = = 2; g(−1) = = 0; g(π) = −1.1) = = = ;
; g(
3−1 −1 − π−1 −1.1 − 1 −2.1 21
2 2
1
t − 1+1 t
g(t2 − 1) = =
t2 − 1 − 1 t2 − 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.
√ √
3. (a) x /= 3 (b) x ≤ − 3 or x ≥ 3
(c) x2 − 2x + 5 = 0 has no real solutions so x2 − 2x + 5 is always positive or always negative. If
x = 0, then x2 − 2x + 5 = 5 > 0; domain: (−∞, +∞).
(d) x /= 0 (e) sin x /= 1, so x /= (2n + 1 )π, n = 0, ±1, ±2,...
