Solution do Livro de Calculo do Anton, Exercícios de Cálculo

Baixe Solution do Livro de Calculo do Anton e outras Exercícios em PDF para Cálculo, somente na Docsity!

Physics

ACT

http://PHYSICSACT.wordpress.com

1

CHAPTER 1

Functions

EXERCISE SET 1.

1. (a) around 1943 (b) 1960; (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, (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

80

20 80

(d) Lmin ≈ 89 .44 ft

10. (a) V = lwh = (6 − 2 x)(6 − 2 x)x (b) From the figure it is clear that 0 < x < 3.

(c) 20

0

0 3

(d) Vmax ≈ 16 in^3

Exercise Set 1.2 3

4. (a) x = − (^75) (b) x − 3 x^2 must be nonnegative; y = x − 3 x^2 is a parabola that crosses the x-axis at x = 0, (^13) and opens downward, thus 0 ≤ x ≤ (^13)

(c)

x^2 − 4 x − 4

0, so x^2 − 4 > 0 and x − 4 > 0, thus x > 4;or x^2 − 4 < 0 and x − 4 < 0, thus − 2 < x < 2 (d) x = − 1 (e) cos x ≤ 1 < 2, 2 − cos x > 0, all x

5. (a) x ≤ 3 (b) − 2 ≤ x ≤ 2 (c) x ≥ 0 (d) all x (e) all x
6. (a) x ≥ 23 (b) − 32 ≤ x ≤ 32 (c) x ≥ 0 (d) x = 0 (e) x ≥ 0
7. (a) yes (b) yes (c) no (vertical line test fails) (d) no (vertical line test fails)
8. The sine of θ/2 is (L/2)/10 (side opposite over hypotenuse), so that L = 20 sin(θ/2).
9. The cosine of θ is (L − h)/L (side adjacent over hypotenuse), so h = L(1 − cos θ).

10. T

t

t

h (^) 12.

5 10 15

t

w

13. (a) If x < 0, then |x| = −x so f (x) = −x + 3x + 1 = 2x + 1. If x ≥ 0, then |x| = x so f (x) = x + 3x + 1 = 4x + 1;

f (x) =

2 x + 1, x < 0 4 x + 1, x ≥ 0

(b) If x < 0, then |x| = −x and |x − 1 | = 1 − x so g(x) = −x + 1 − x = 1 − 2 x. If 0 ≤ x < 1, then |x| = x and |x − 1 | = 1 − x so g(x) = x + 1 − x = 1. If x ≥ 1, then |x| = x and |x − 1 | = x − 1 so g(x) = x + x − 1 = 2x − 1;

g(x) =

1 − 2 x, x < 0 1 , 0 ≤ x < 1 2 x − 1 , x ≥ 1

14. (a) If x < 5 /2, then | 2 x − 5 | = 5 − 2 x so f (x) = 3 + (5 − 2 x) = 8 − 2 x. If x ≥ 5 /2, then | 2 x − 5 | = 2x − 5 so f (x) = 3 + (2x − 5) = 2x − 2;

f (x) =

8 − 2 x, x < 5 / 2 2 x − 2 , x ≥ 5 / 2

(b) If x < −1, then |x − 2 | = 2 − x and |x + 1| = −x − 1 so g(x) = 3(2 − x) − (−x − 1) = 7 − 2 x. If − 1 ≤ x < 2, then |x − 2 | = 2 − x and |x + 1| = x + 1 so g(x) = 3(2 − x) − (x + 1) = 5 − 4 x. If x ≥ 2, then |x − 2 | = x − 2 and |x + 1| = x + 1 so g(x) = 3(x − 2) − (x + 1) = 2x − 7;

g(x) =

7 − 2 x, x < − 1 5 − 4 x, − 1 ≤ x < 2 2 x − 7 , x ≥ 2

4 Chapter 1

15. (a) V = (8 − 2 x)(15 − 2 x)x (b) −∞ < x < +∞, −∞ < V < +∞ (c) 0 < x < 4

(d) minimum value at x = 0 or at x = 4;maximum value somewhere in between (can be approximated by zooming with graphing calculator)

16. (a) x = 3000 tan θ (b) θ = nπ + π/2 for n an integer, −∞ < n < ∞

(c) 0 ≤ θ < π/2, 0 ≤ x < +∞ (d) 3000 ft 6000

0

0 6

17. (i) x = 1, −2 causes division by zero (ii) g(x) = x + 1, all x
18. (i) x = 0 causes division by zero (ii) g(x) =

x + 1 for x ≥ 0

19. (a) 25 ◦F (b) 2 ◦F (c) − 15 ◦F
20. If v = 48 then −60 = WCI = 1. 6 T − 55;thus T = (−60 + 55)/ 1. 6 ≈ − 3 ◦F.
21. If v = 8 then −10 = WCI = 91.4 + (91. 4 − T )(0.0203(8) − 0. 304

8 − 0 .474);thus T = 91.4 + (10 + 91.4)/(0.0203(8) − 0. 304

8 − 0 .474) and T = 5◦F

22. The WCI is given by three formulae, but the first and third don’t work with the data. Hence −15 = WCI = 91.4 + (91. 4 − 20)(0. 0203 v − 0. 304

v − 0 .474);set x =

v so that v = x^2 and obtain 0. 0203 x^2 − 0. 304 x − 0 .474 + (15 + 91.4)/(91. 4 − 20) = 0. Use the quadratic formula to find the two roots. Square them to get v and discard the spurious solution, leaving v ≈ 25.

23. Let t denote time in minutes after 9:23 AM. Then D(t) = 1000 − 20 t ft.

EXERCISE SET 1.

1. (e) seems best, though only (a) is bad.

-0.

y

-1 1

x

2. (e) seems best, though only (a) is bad and (b) is not good.

-0.

y

-1 1

x

3. (b) and (c) are good; (a) is very bad.

12

13

14

y

-1 1

x

6 Chapter 1

14. [− 3 , 20] × [− 3500 , 3000]

1000

5 10 15

x

y

15. [− 2 , 2] × [− 20 , 20]

10

20

y

x -2 -1 1 2

16. [1. 6 , 2] × [0, 2]

1

2

y

1.6 1.7 1.8 2

x

17. depends on graphing utility

2

4

6

y

-4 2 4

x

18. depends on graphing utility

2

4

6

y

-4 -2 2 4

x

19. (a) f (x) =

16 − x^2

1

3

-2 2 4

x

y

(b) f (x) = −

16 − x^2

y x -4 -2 2 4

(c)

2 x

y

-2 2

(d)

1

4

1 4

y

x

(e) No;the vertical line test fails.

20. (a) y = ± 3

1 − x^2 / 4

3

-2 2

x

y

(b) y = ±

x^2 + 1

2

4

-4 -2 2 4

x

y

Exercise Set 1.3 7

21. (a)

-1 1

x

y

1

(b)

1

x

1

2

y

(c)

-1 (^1)

x

y (^) (d)

c 2 c

1

y

x

(e)

c

1

y

x C

(f )

1

x

1

y

1

x

y

-1 1

23. The portions of the graph of y = f (x) which lie below the x-axis are reflected over the x-axis to give the graph of y = |f (x)|.
24. Erase the portion of the graph of y = f (x) which lies in the left-half plane and replace it with the reflection over the y-axis of the portion in the right-half plane (symmetry over the y-axis) and you obtain the graph of y = f (|x|).
25. (a) for example, let a = 1. 1

a

y

x

(b)

1

2

3

y

1 2 3

x

26. They are identical.

x

y

5

10

15

y

-1 1 2 3

x

Exercise Set 1.4 9

32. (a) y

2

1 2

x

(b) x-intercepts at x = 0, a, b. Assume a < b and let a approach b. The two branches of the curve come together. If a moves past b then a and b switch roles.

1

3

1 2 3

x

a = 0. b = 1.

y

1

3

1 2 3

x

a = 1 b = 1.

y

1

3

2 3

x

a = 1. b = 1.

y

33. The curve oscillates between the lines y = x and y = −x with increasing rapidity as |x| increases.

10

30

y

-30 -10 10 30

x

34. The curve oscillates between the lines y = +1 and y = −1, infinitely many times in any neighborhood of x = 0. y

–2 4

x

EXERCISE SET 1.

1. (a)

0

1

y

-1 1 2

x

(b) (^2)

1

y

1 2 3

x

(c) 1^ y

-1 1 2

x

(d) 2

y

-4 -2 2

x

10 Chapter 1

2. (a) y x -3 1

(b)

x 1

y

1

(c)

x 1

y

1 (d)

x 1

y 1

3. (a)

1

y

x -2 -1 1 2

(b)

1

y

1

x

(c)

1 y

-1 1 2 3

x

(d)

1

y

-1 1 2 3

x

4. 1 y

2

x

5. Translate right 2 units, and up one unit.

10

-2 2 6

x

y

6. Translate left 1 unit, reflect over x-axis, and translate up 2 units. y

x

1

7. Translate left 1 unit, stretch vertically by a factor of 2, reflect over x-axis, translate down 3 units. - -

-6 -2 2 6

y x

12 Chapter 1

20. y = 1 − 1 /x; reflect over x-axis, translate up 1 unit. y

x

5

2

21. Translate left 2 units and down 2 units.

-4 -

x

y

22. Translate right 3 units, reflect over x-axis, translate up 1 unit. y

x

1

5

23. Stretch vertically by a factor of 2, translate right 1 unit and up 1 unit. y

x

2

4

2

24. y = |x − 2 |;translate right 2 units.

y

x

1

2

2 4

25. Stretch vertically by a factor of 2, reflect over x-axis, translate up 1 unit.

1

3

-2 2

x

y

26. Translate right 2 units and down 3 units. y (^) x

2

27. Translate left 1 unit and up 2 units.

1

3

y

-3 -1 1

x

28. Translate right 2 units, reflect over x-axis. y

x

1

4

29. (a) 2 y

-1 1

x

(b) y =

0 if x ≤ 0 2 x if 0 < x

Exercise Set 1.4 13

30. y

x

2

31. (f + g)(x) = x^2 + 2x + 1, all x; (f − g)(x) = 2x − x^2 − 1, all x; (f g)(x) = 2x^3 + 2x, all x; (f /g)(x) = 2x/(x^2 + 1), all x
32. (f + g)(x) = 3x − 2 + |x|, all x; (f − g)(x) = 3x − 2 − |x|, all x; (f g)(x) = 3x|x| − 2 |x|, all x; (f /g)(x) = (3x − 2)/|x|, all x = 0
33. (f + g)(x) = 3

x − 1, x ≥ 1;( f − g)(x) =

x − 1, x ≥ 1;( f g)(x) = 2x − 2, x ≥ 1; (f /g)(x) = 2, x > 1

34. (f + g)(x) = (2x^2 + 1)/[x(x^2 + 1)], all x = 0; (f − g)(x) = − 1 /[x(x^2 + 1)], all x = 0; (f g)(x) = 1 /(x^2 + 1), all x = 0; (f /g)(x) = x^2 /(x^2 + 1), all x = 0
35. (a) 3 (b) 9 (c) 2 (d) 2
36. (a) π − 1 (b) 0 (c) −π^2 + 3π − 1 (d) 1
37. (a) t^4 + 1 (b) t^2 + 4t + 5 (c) x^2 + 4x + 5 (d)

x^2

(e) x^2 + 2xh + h^2 + 1 (f ) x^2 + 1 (g) x + 1 (h) 9 x^2 + 1

38. (a)

5 s + 2 (b)

x + 2 (c) 3

5 x (d) 1 /

x (e) 4

x (f ) 0 (g) 1 / 4

x (h) |x − 1 |

39. (f ◦ g)(x) = 2x^2 − 2 x + 1, all x; (g ◦ f )(x) = 4x^2 + 2x, all x
40. (f ◦ g)(x) = 2 − x^6 , all x; (g ◦ f )(x) = −x^6 + 6x^4 − 12 x^2 + 8, all x
41. (f ◦ g)(x) = 1 − x, x ≤ 1;( g ◦ f )(x) =

1 − x^2 , |x| ≤ 1

42. (f ◦ g)(x) =

x^2 + 3 − 3, |x| ≥

6;( g ◦ f )(x) =

x, x ≥ 3

43. (f ◦ g)(x) =

1 − 2 x , x =

, 1;( g ◦ f )(x) = −

2 x

, x = 0, 1

44. (f ◦ g)(x) =

x x^2 + 1

, x = 0; (g ◦ f )(x) =

x

• x, x = 0

45. x−^6 + 1 46. x x + 1
46. (a) g(x) =

x, h(x) = x + 2 (b) g(x) = |x|, h(x) = x^2 − 3 x + 5

48. (a) g(x) = x + 1, h(x) = x^2 (b) g(x) = 1/x, h(x) = x − 3
49. (a) g(x) = x^2 , h(x) = sin x (b) g(x) = 3/x, h(x) = 5 + cos x

Exercise Set 1.4 15

64. (a)

x

y (^) (b)

x

y

(c) x

y

65. (a) (^) x − 3 − 2 − 1 0 1 2 3

f (x) 1 − 5 − 1 0 − 1 − 5 1

(b) (^) x − 3 − 2 − 1 0 1 2 3 f (x) 1 5 − 1 0 1 − 5 − 1

66. (a)

x

y (^) (b)

x

y

67. (a) even (b) odd (c) odd (d) neither
68. neither;odd;even
69. (a) f (−x) = (−x)^2 = x^2 = f (x), even (b) f (−x) = (−x)^3 = −x^3 = −f (x), odd

(c) f (−x) = | − x| = |x| = f (x), even (d) f (−x) = −x + 1, neither

(e) f (−x) =

(−x)^3 − (−x) 1 + (−x)^2

x^3 + x 1 + x^2

= −f (x), odd

(f ) f (−x) = 2 = f (x), even

70. (a) x-axis, because x = 5(−y)^2 + 9 gives x = 5y^2 + 9

(b) x-axis, y-axis, and origin, because x^2 − 2(−y)^2 = 3, (−x)^2 − 2 y^2 = 3, and (−x)^2 − 2(−y)^2 = 3 all give x^2 − 2 y^2 = 3

(c) origin, because (−x)(−y) = 5 gives xy = 5

71. (a) y-axis, because (−x)^4 = 2y^3 + y gives x^4 = 2y^3 + y

(b) origin, because (−y) =

(−x) 3 + (−x)^2

gives y =

x 3 + x^2

(c) x-axis, y-axis, and origin because (−y)^2 = |x| − 5, y^2 = | − x| − 5 , and (−y)^2 = | − x| − 5 all give y^2 = |x| − 5

16 Chapter 1

-4 4

-3 3

74. (a) Whether we replace x with −x, y with −y, or both, we obtain the same equation, so by Theorem 1.4.3 the graph is symmetric about the x-axis, the y-axis and the origin. (b) y = (1 − x^2 /^3 )^3 /^2 (c) For quadrant II, the same;for III and IV use y = −(1 − x^2 /^3 )^3 /^2. (For graphing it may be helpful to use the tricks that precede Exercise 29 in Section 1.3.)

2

5

y

1 2

x

76. y

x

2

2

77. (a) 1

y

C

x O c^ o

(b) 2

y

O

x C c^ o

78. (a)

x 1

y

1

(b)

x 1

y

1

• ‚ 2 -1 ‚ (^2) ‚ 3

(c)

x 1

y

3

(d)

x

y 1

C^ c/

79. Yes, e.g. f (x) = xk^ and g(x) = xn^ where k and n are integers.
80. If x ≥ 0 then |x| = x and f (x) = g(x). If x < 0 then f (x) = |x|p/q^ if p is even and f (x) = −|x|p/q if p is odd;in both cases f (x) agrees with g(x).

18 Chapter 1

17. Parallel means the lines have equal slopes, so y = 4x + 7. 12

0

-1 1

18. The slope of both lines is − 3 /2, so y − 2 = (− 3 /2)(x − (−1)), or y = − 32 x + (^12) 4

-2 1

19. The negative reciprocal of 5 is − 1 /5, so y = − 15 x + 6. 12

0

-9 9

20. The slope of x − 4 y = 7 is 1/ 4 whose negative reciprocal is −4, so y − (−4) = −4(x − 3) or y = − 4 x + 8. 9

0 18

21. m =

so y − (−7) = 11(x − 1), or y = 11x − 18 7

0 4

22. m =

= −5, so

y − 6 = −5(x − (−3)), or y = − 5 x − 9 15

–3 0

23. (a) m 1 = m 2 = 4, parallel (b) m 1 = 2 = − 1 /m 2 , perpendicular

(c) m 1 = m 2 = 5/3, parallel (d) If A = 0 and B = 0 then m 1 = −A/B = − 1 /m 2 , perpendicular;if A = 0 or B = 0 (not both) then one line is horizontal, the other vertical, so perpendicular. (e) neither

24. (a) m 1 = m 2 = −5, parallel (b) m 1 = 2 = − 1 /m 2 , perpendicular (c) m 1 = − 4 /5 = − 1 /m 2 , perpendicular (d) If B = 0, m 1 = m 2 = −A/B; (e) neither if^ B^ = 0 both are vertical, so parallel
25. (a) m = (0 − (−3))/(2 − 0)) = 3/2 so y = 3x/ 2 − 3

(b) m = (− 3 − 0)/(4 − 0) = − 3 /4 so y = − 3 x/ 4

26. (a) m = (0 − 2)/(2 − 0)) = −1 so y = −x + 2

(b) m = (2 − 0)/(3 − 0) = 2/3 so y = 2x/ 3

Exercise Set 1.5 19

27. (a) The velocity is the slope, which is

= 9/10 ft/s.

(b) x = − 4 (c) The line has slope 9/10 and passes through (0, −4), so has equation x = 9t/ 10 − 4;at t = 2, x = − 2 .2. (d) t = 80/ 9

28. (a) v =

= 2 m/s (b) x − 1 = 2(t − 2) or x = 2t − 3 (c) x = − 3

29. (a) The acceleration is the slope of the velocity, so a =

ft/s^2.

(b) v − 3 = − 43 (t − 1), or v = − 43 t + 133 (c) v = 133 ft/s

30. (a) The acceleration is the slope of the velocity, so a =

ft/s^2.

(b) v = 5 ft/s (c) v = 4 ft/s (d) t = 4 s

31. (a) It moves (to the left) 6 units with velocity v = −3 cm/s, then remains motionless for 5 s, then moves 3 units to the left with velocity v = −1 cm/s.

(b) vave =

cm/s

(c) Since the motion is in one direction only, the speed is the negative of the velocity, so save = 109 cm/s.

32. It moves right with constant velocity v = 5 km/h;then accelerates;then moves with constant, though increased, velocity again;then slows down.
33. (a) If x 1 denotes the final position and x 0 the initial position, then v = (x 1 − x 0 )/(t 1 − t 0 ) = 0 mi/h, since x 1 = x 0. (b) If the distance traveled in one direction is d, then the outward journey took t = d/40 h. Thus save =

total dist total time

2 d t + (2/3)t

80 t t + (2/3)t

= 48 mi/h.

(c) t + (2/3)t = 5, so t = 3 and 2d = 80t = 240 mi round trip

34. (a) down, since v < 0 (b) v = 0 at t = 2 (c) It’s constant at 32 ft/s^2.
35. (a) v

t

20

60

20 80

(b) v =

10 t if 0 ≤ t ≤ 10 100 if 10 ≤ t ≤ 100 600 − 5 t if 100 ≤ t ≤ 120

36. x

t