Therefore f must increase more than 4 units on 0 4 c

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Unformatted text preview: ve downward on 1, 5 . Points of inflection at x 1, 5. 67. a t vt v0 st s0 st 6 60 32 ft sec2 32 dt C1 32t C2 16t2 60t 6 Position function 60 dt 16t 2 60t C2 32t C1 −6 69. From Exercise 68, we have: st 16t2 v0t 0 when t v0 32 v0 32 time to reach st 32t v0 maximum height. s v02 64 v0 32 v02 32 v02 v0 16 550 35,200 v0 32 2 v0 550 The ball reaches its maximim height when vt 32t 60 32t t s 15 18 16 9.8 9.8 dt v0 9.8t C1 9.8t 4.9t 2 0 60 15 seconds 8 2 187.617 ft sec 15 8 60 15 8 6 62.26 feet 73. From Exercise 71, f t vt 9.8t v0 t v0t v0t C2 s0 f 10 9.8 9.8t 10 10 9.8 7.1 m 10 4.9t2 10t 2. 0.) 71. a t vt v0 ft f0 s0 a vt st s 20 0 (Maximum height when v C1 ⇒ v t 9.8t v0 dt C2 ⇒ f t 1.6 1.6 dt 1.6t dt 0⇒ 1.6t 4.9t 2 75. v0 s0 s0 s0 1.6t, since the stone was dropped, v0 0. 0.8t2 0.8 20 2 0 320 Thus, the height of the cliff is 320 meters. vt v 20 1.6t 32 m sec S ection 4.1 Antiderivatives and Indefinite Integration 181 77. x t (a) v t t3 6t2 xt 3t 2 9t 3t 4t 6t 2 2 12t 3 12 0≤t≤5 9 3t 6t 1t 2 3 79. v t xt x1 xt at 1 t t 12 t>0 2 v t dt 4 2t1 2 vt 21 2t1 C 2 at vt C⇒C (b) v t > 0 when 0 < t < 1 or 3 < t < 5. (c) a t v2 81. (a) v 0 v 13 at vt v0 v 13 6t 31 2 1 0 when t 3 25 80 1000 3600 1000 3600 250 m sec 36 800 m sec 36 2. 2 position function 1 t 2 32 1 acceleration 2t 3 2 25 km hr 80 km hr 83. Truck: v t st 30 30t Let s 0 6 6t Let v 0 3t2 Let s 0 0. 0. 0. Automobile: a t vt st a constant acceleration at C at 250 36 250 36 250 ⇒ vt 36 800 36 550 36 a 13a 13a 550 468 250 t 36 2 At the point where the automobile overtakes the truck: 30t 0 0 (a) s 10 3t2 3t2 3t t 3 10 6 10 30t 10 when t 2 10 sec. 300 ft 60 ft sec 41 mph 275 234 s0 250 13 36 22 ft sec 15 (b) v 10 1.175 m sec2 0 189.58 m (b) st s 13 a t2 2 275 13 234 2 85. 1 mi hr 5280 ft mi 3600 sec hr (a) t V1 ft sec V2 ft sec (c) S1 t S2 t 0 0 0 5 10 10.27 55.73 15 23.47 74.8 0.0416 2 t 2 6.7991t2 2 20 42.53 88 25 66 93.87 30 95.33 95.33 (b) V1 t V2 t 0.1068t2 0.1208t2 0.0416t 6.7991t 0.3679 0.0707 3.67 30.8 V1 t dt V2 t dt 0.1068 3 t 3 0.1208t3 3 0.3679t 0.0707t S2 0 0 In both cases, the constant of integration is 0 because S1 0 S1 30 S2 30 953.5 feet 1970.3 feet The second car was going faster than the first until the end. 182 Chapter 4 Integration 87. a t vt st k kt k2 t since v 0 2 s0 0. 160 and k 2 t2 0.7. Since k 2 t2 0.7, At the time of lift-off, kt t v 1.4 k 1.4k k 1.4 k 1.4 k 160 1602 1.4 1602 ⇒ k 18,285.714 mi hr2 7.45 ft sec2. 89. True x3 3 91. True x2 2 x2 2 93. False. For example, x x dx x dx x dx because C C1 C2 95. f x 1, 0 ≤ x < 2 3x, 2 ≤ x ≤ 5 x C1, 0≤x<2 3x2 C2, 2 ≤ x ≤ 5 2 3⇒1 C1 3 ⇒ C1 2 2: fx f1 f is continuous: Values must agree at x 4 fx 6 C2 ⇒ C2 2 0≤x<2 x 2, 3x 2 2, 2 ≤ x ≤ 5 2 2 do not agree. Hence f is not differentiable at x 2. The left and right hand derivatives at x Section 4.2 5 5 Area 5 1. i 4 1 2i 1 0 9 1 2 i i 1 i 1 1 21 1 10 9. j 1 2 1 17 8 3 158 85 4 5 5 35 4 3. k k2 1 1 1 2 1 5 5. k 1 c 2n ni 1 20 c 2i n 1 2 3 c c 2i n 19 c 4c 3n 21 ni 1 3i n 2 7. i 1 3i 1 20 5 j 8 3 11. 13. 20 15. i 1 2i 2 i i 1 2 20 21 2 420 17. i 1 i i2 i 1 19 20 39 6 2470 S ection 4.2 Area 183 15 15 15 15 19. i 1 ii 1 i 1 2 i 1 i 1 20 > 2 i3 152 16 4 14,400 12,040 2 i2 i 15 16 2 21. sum seq x i2 i 1 2 3 3, x, 1, 20, 1 20 20 2930 (TI-82) 1 3 20 15 16 31 2 6 2,480 120 1 2 20 6 60 20 21 41 6 2930 23. S s 3 1 4 3 9 2 51 9 2 33 2 25 2 16.5 12.5 25. S s 31 44 1 4 1 1 2 8 1 5 1 6 1 6 1 7 81 4 2 8 3 3 2 3 3 2 2 51 31 11 7 4 1 27. S 4 s4 0 1 4 1 5 11...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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