Review Eercises for Chapter 367 Review Eercises for Chapter. f 1 1 f f f lim lim 1 1 1 1 lim 1 1 1 1 lim 1 1 lim lim 1 1 1 1 1 1 1 1 1 4. 8. f f f f lim lim lim lim lim f 4, 1 4, if < if (a) Nonremovable discontinuit at. (b) Not differentiable at because the function is discontinuous there. 6. f is differentiable for all 3. 1. Using the limit defintion, ou obtain h 3 4. 8 At, h 3 8 4 67 8. 4 1 4 1 1 1 1. (a) Using the limit definition, f 1. (b) 4 (, ) At, f. The tangent line is 6 6 4
368 Chapter Differentiation 14. f lim f f 16. 1 1 1 3 lim lim 3 1 13 π 1 1 π f f 1 lim 13 1 9 18. 1. g 1. g 1 11 f t 8t ft 4t 4 4. 3. 34. gs 4s 4 s 6. f 1 1 8. gs 16s 3 1s f 1 1 1 3 1 3 g 4 cos 63. g 4 sin s 16t s First ball: 16t 1 g sin 3 g cos 3 h 9 h 4 9 3 4 9 3 t 1 16 Second ball: 1 4 16t 7. seconds to hit ground t 7 3.16 seconds to hit ground 16 4 Since the second ball was released one second after the first ball, the first ball will hit the ground first. The second ball will hit the ground 3.16..66 second later. 36. st 16t 14,4 16t 14,4 t 3 sec Since 6 mph 1 mi/sec, in 3 seconds the bomb will move horizontall 1 6 63 miles.
Review Eercises for Chapter 369 38. v 64 v 18 v v (, 64 18 ) v (, 3 ) (a) (c) 3 v 1 3 v if or v 3. Projectile strikes the ground when v 3. Projectile reaches its maimum height at (one-half the distance) 3 v 1 3 v when and 3. Therefore, the range is v 3. When the initial velocit is doubled the range is v 3 4v 3 or four times the initial range. From part (a), the maimum height occurs when v 64. The maimum height is v 64 v 3 64 v v 64 v 64 v 18 v 18. If the initial velocit is doubled, the maimum height is v 64 v 18 4 v 18 or four times the original maimum height. v 64. (b) (d) 1 64 When v 64, v 7 ftsec Range: Maimum height: v 1 64 v 7 13.1 ft 3 3 16 v v. 64 v 7 38.8 ft 18 18 4. (a).14 4.43 8.4 (b) 3 6 (c) 1 (d) If 6, 36 feet. 6 (e) As the speed increases, the stopping distance increases at an increasing rate.
37 Chapter Differentiation 4. g 3 3 44. g 3 31 3 3 3 3 3 3 6 3 6 4 3 6 6 6 ft t 3 cos t ft t 3 sin t cos t3t t 3 sin t 3t cos t 46. f 1 1 48. f 6 1 f 11 11 1 f 16 6 1 1 3 3 1. 4. f 93 1. f 93 6 181 3 3 tan 6. sec tan sin cos sin 1 sin 1 sin 4 cos 1 sin cos sin 3 1 sin cos 1 sin cos 1 sin 8. vt 36 t, t 6 6. at vt t v4 36 16 msec a4 8 msec f 1 14 f 3 34 f 9 4 74 9 4 74 6. 66. ht 4 sin t cos t 64. ht 4 cos t sin t ht 4 sin t cos t f 1 13 68. cos 1 sin 1 cos sin sin sin f 1 f 1 3 1 3 3 1 3 f 1 4 1
Review Eercises for Chapter 371 7. 7. h 1 3 1 h 3 31 1 1 6 1 1 1 1 6 1 4 3 1 cos cos sin 4 cos sin sin cos 4 sin cos 74. csc 3 cot 3 76. 3 csc 3 cot 3 3 csc 3 3 csc 3cot 3 csc 3 sec7 7 sec sec 6 sec tan sec 4 sec tan sec tan sec 1 sec tan 3 78. f 3 1 1 3 1 1 1 f 1 3 1 3 1 3 3 1 3 1 3 8. cos 1 1 1 sin 1 cos 11 1 1 1 1 sin 1 cos 1 8. f 4 8 f 4 3 3 6 8 4 1 4 The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 1 f f 7 6 84. g 1 1 86. g 1 1 g does not equal zero for an value of. The graph of g has no horizontal tangent lines. g 3 3 3 7 3 does not equal zero for an in the domain. The graph has no horizontal tangent lines. 7 6 6 g 3 3 3
37 Chapter Differentiation 88. csc 3 1 3 csc3 cot The zero of corresponds to the point on the graph of where the tangent line is horizontal. 1 8 4 9. 1 tan 9. sin 94. sec sin cos sin 3 sec sec tan cos 3 sec tan g 6 1 g 3 3 1 g 63 1 18 1 3 96. h 1 98. h 1 1 h 3 1 3 v gh 3h 8h dv dh 4 h dv (a) When h 9, dh 4 3 ftsec. dv (b) When h 4, ftsec. dh 1. 9 4 3 1. 18 4 3 36 1 4 4 36 1 3 3 3 4 4 3 3 4 14. cos 16. 16 1 1 sin 1 sin 1 sin 1 sin sin csc 1 1 At, 3: 3 1 1 1 Tangent line: 3 3 3 16 Normal line: 3 3 3 3 18. Surface area A 6, length of edge. d dt da d 1 dt dt 14. 7 cm sec
Problem Solving for Chapter 373 11. tan d sec d d dt dt When 1, dt 3 radmin d dt tan 16 6 1 1 θ d dt 6 1 4 1 1 kmmin 4 kmhr. Problem Solving for Chapter. Let a, a and b, b b be the points of tangenc. 1 8 6 4 8 6 4 4 6 8 1 4 6 For, and for,. Thus, a b a b 1, or a 1 b. Furthermore, the slope of the common tangent line is a b b a b 1 b b b b 1 b b 4b 6 4b 6b b b 4 b b b b 1 b, 1 1 b b b 1 b b b b For b, a 1 b 1and the points of tangenc are 1, 1,,. The tangent line has slope : 1 1 1 For b 1, a 1 b and the points of tangenc are, 4 and 1, 8. The tangent line has slope 4: 4 4 4 4. 4. (a),. Slope 4 at, 4. Tangent line: 4 4 (b) Slope of normal line: 1 4. Normal line: 4 1 4 1 4 9 4 18 4 9, 9 4. Second intersection point: (c) Tangent line: Normal line: CONTINUED 4 4 1 4 9 9 4, 81 16
374 Chapter Differentiation 4. CONTINUED (d) Let a, a, a, be a point on the parabola. Tangent line at a, a is a a a. Normal line at a, a is 1 To find points of intersection, solve a a a. 1 a a a 1 a a 1 1 a 1 16a a 1 1 16a 1 4a a 1 4a 1 4a ± a 1 4a 1 1 a a Point of tangenc 4a 4a 1 4a a 4a 1 1 1 a a a a The normal line intersects a second time at a 1. a 6. f a b cos c f bc sin c At, 1: a b 1 Equation 1 At 4, 3 : a b cos c 4 3 bc sin c 4 1 Equation Equation 3 From Equation 1, a 1 b. Equation becomes From Equation 3, b 1 cos c 4 1 c sin c 4 Graphing the equation 1 c sin c 4. One answer: c, b 1, a 3 Thus gc 1 c sin c 4 cos c 4 1 c sin c 4 1 f 3 1 cos 1 b b cos c 4 3 c b b cos 4 1 c sin c 4 cos c 4 1 1, ou see that man values of c will work.
Problem Solving for Chapter 37 8. (a) b 3 a ; a, b > (b) a determines the -intercept on the right: a,. 3 a b Graph 1 3 a and b (c) Differentiating implicitl. 3 a b b affects the height. b 3 a 3 3a 4 3 3a 4 3 b 3a 4 3 3a 4 3a 4. b 3a 4 3 3a 7a3 a 4 64 1 4 a 7a4 ±33a 6b 16b Two points: 3a 4, 33a 16b, 3a 4, 33a 16b 1. (a) (b) (c) 13 d dt 1 d 3 3 dt 1 1 d 83 3 dt d 1 cmsec dt D dd dt 1 d d dt dt tan sec d dt d d dt dt 81 1 64 4 d dt d dt 98 49 68 17 cmsec. θ 8 68 From the triangle, sec Hence d 81 1 68 8. dt 64 68 64 16 68 4 17 radsec
376 Chapter Differentiation 1. E E E lim EE E lim lim E E 1 E 1 E lim E E E 1 But, E lim lim 1. Thus, E EE E eists for all. For eample: E e. 14. (a) (b) vt 7 t 7 ftsec at 7 ftsec vt 7 t 7 7 t 7 t seconds S 7 1 7 6 73. feet (c) The acceleration due to gravit on Earth is greater in magnitude than that on the moon.