Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 4.9 ANTIDERIVATIVES ¤ 213 (c) The velocity with which the stone strikes the ground is v(t 1)=−9.8 450/4.9 ≈−93.9 m/s. (d) This is just reworking parts (a) and (b) with v(0) = −5. Usingv(t) =−9.8t + C, v(0) = −5 ⇒ 0+C = −5 ⇒ v(t) =−9.8t − 5. Sos(t) =−4.9t 2 − 5t + D and s(0) = 450 ⇒ D =450 ⇒ s(t) =−4.9t 2 − 5t + 450. Solving s(t) =0by using the quadratic formula gives us t = 5 ± √ 8845 (−9.8) ⇒ t 1 ≈ 9.09 s. 65. By Exercise 64 with a = −9.8, s(t) =−4.9t 2 + v 0 t + s 0 and v(t) =s 0 (t) =−9.8t + v 0 .So [v(t)] 2 =(−9.8t + v 0 ) 2 =(9.8) 2 t 2 − 19.6v 0 t + v0 2 = v0 2 +96.04t 2 − 19.6v 0 t = v0 2 − 19.6 −4.9t 2 + v 0 t . But −4.9t 2 + v 0t is just s(t) without the s 0 term;thatis,s(t) − s 0.Thus,[v(t)] 2 = v0 2 − 19.6[s(t) − s 0]. 67. Using Exercise 64 with a = −32, v 0 =0,ands 0 = h (the height of the cliff ), we know that the height at time t is s(t) =−16t 2 + h. v(t) =s 0 (t) =−32t and v(t) =−120 ⇒ −32t = −120 ⇒ t =3.75, so 0=s(3.75) = −16(3.75) 2 + h ⇒ h = 16(3.75) 2 = 225 ft. 69. Marginal cost =1.92 − 0.002x = C 0 (x) ⇒ C(x) =1.92x − 0.001x 2 + K. ButC(1) = 1.92 − 0.001 + K =562 ⇒ K =560.081. Therefore, C(x) =1.92x − 0.001x 2 +560.081 ⇒ C(100) = 742.081, so the cost of producing 100 items is $742.08. 71. Taking the upward direction to be positive we have that for 0 ≤ t ≤ 10 (using the subscript 1 to refer to 0 ≤ t ≤ 10), a 1 (t) =− (9 − 0.9t) =v1(t) 0 ⇒ v 1 (t) =−9t +0.45t 2 + v 0 ,butv 1 (0) = v 0 = −10 ⇒ v 1(t) =−9t +0.45t 2 − 10 = s 0 1(t) ⇒ s 1(t) =− 9 2 t2 +0.15t 3 − 10t + s 0.Buts 1(0) = 500 = s 0 ⇒ s 1 (t) =− 9 2 t2 +0.15t 3 − 10t + 500. s 1 (10) = −450 + 150 − 100 + 500 = 100,soittakes more than 10 seconds for the raindrop to fall. Now for t>10, a(t) =0=v 0 (t) ⇒ v(t) =constant = v 1 (10) = −9(10) + 0.45(10) 2 − 10 = −55 ⇒ v(t) =−55. At 55 m/s, it will take 100/55 ≈ 1.8 stofallthelast100 m. Hence, the total time is 10 + 100 = 130 ≈ 11.8 s. 55 11 73. a(t) =k, the initial velocity is 30 mi/h =30· 5280 3600 =44ft/s, and the final velocity (after 5 seconds) is 50 mi/h =50· 5280 = 220 ft/s. So v(t) =kt + C and v(0) = 44 ⇒ C =44.Thus,v(t) =kt +44 ⇒ 3600 3 v(5) = 5k +44.Butv(5) = 220 220 ,so5k +44= ⇒ 5k = 88 ⇒ k = 88 ≈ 5.87 3 3 3 15 ft/s2 . 75. Let the acceleration be a(t) =k km/h 2 .Wehavev(0) = 100 km/h and we can take the initial position s(0) to be 0. We want the time t f for which v(t) =0to satisfy s(t) < 0.08 km. In general, v 0 (t) =a(t) =k, sov(t) =kt + C,where C = v(0) = 100. Nows 0 (t) =v(t) =kt + 100, sos(t) = 1 2 kt2 + 100t + D, whereD = s(0) = 0. Thus, s(t) = 1 2 kt2 +100t. Sincev(t f )=0,wehavekt f + 100 = 0 or t f = −100/k, so s(t f )= 1 2 k − 100 2 +100 − 100 k k − 5,000 k 1 =10,000 2k − 1 k = − 5,000 k . The condition s(t f ) must satisfy is < 0.08 ⇒ − 5,000 0.08 >k [k is negative] ⇒ k
F. 214 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION TX.10 77. (a) First note that 90 mi/h =90× 5280 3600 ft/s = 132 ft/s. Then a(t) =4ft/s2 ⇒ v(t) =4t + C,butv(0) = 0 ⇒ C =0.Now4t =132when t = 132 4 =33s, so it takes 33 s to reach 132 ft/s. Therefore, taking s(0) = 0,wehave s(t) =2t 2 , 0 ≤ t ≤ 33. Sos(33) = 2178 ft. 15 minutes = 15(60) = 900 s, so for 33
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