Solução_Calculo_Stewart_6e

F.
224 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
63. f(t) =cost + t − t 2 ⇒ f 0 (t) =− sin t +1− 2t. f 0 (t) exists for all
t,sotofind the maximum of f, we can examine the zeros of f 0 .
From the graph of f 0 ,weseethatagoodchoicefort 1 is t 1 =0.3.
Use g(t) =− sin t +1− 2t and g 0 (t) =− cos t − 2 to obtain
t 2 ≈ 0.33535293, t 3 ≈ 0.33541803 ≈ t 4 .Sincef 00 (t) =− cos t − 2 < 0
for all t, f(0.33541803) ≈ 1.16718557 is the absolute maximum.
65. f 0 (x) =cosx − (1 − x 2 ) −1/2 =cosx −
1
√
1 − x
2
⇒
f(x) =sinx − sin −1 x + C
67. f 0 (x) = √ x 3 + 3√ x 2 = x 3/2 + x 2/3 ⇒ f(x) = x5/2
5/2 + x5/3
5/3 + C = 2 5 x5/2 + 3 5 x5/3 + C
69. f 0 (t) =2t − 3sint ⇒ f(t) =t 2 +3cost + C.
f(0) = 3 + C and f(0) = 5 ⇒ C =2,sof(t) =t 2 +3cost +2.
71. f 00 (x) =1− 6x +48x 2 ⇒ f 0 (x) =x − 3x 2 +16x 3 + C. f 0 (0) = C and f 0 (0) = 2 ⇒ C =2,so
f 0 (x) =x − 3x 2 +16x 3 +2and hence, f(x) = 1 2 x2 − x 3 +4x 4 +2x + D.
f(0) = D and f(0) = 1 ⇒ D =1,sof(x) = 1 2 x2 − x 3 +4x 4 +2x +1.
73. v(t) =s 0 (t) =2t − 1
1+t 2 ⇒ s(t) =t 2 − tan −1 t + C.
s(0) = 0 − 0+C = C and s(0) = 1 ⇒ C =1,sos(t) =t 2 − tan −1 t +1.
75. (a) Since f is 0 just to the left of the y-axis, we must have a minimum of F at the same place since we are increasing through
(0, 0) on F . There must be a local maximum to the left of x = −3,sincef changes from positive to negative there.
(b) f(x) =0.1e x +sinx ⇒
F (x) =0.1e x − cos x + C. F (0) = 0
0.1 − 1+C =0 ⇒ C =0.9,so
F (x) =0.1e x − cos x +0.9.
⇒
(c)
77. Choosing the positive direction to be upward, we have a(t) =−9.8 ⇒ v(t) =−9.8t + v 0 ,butv(0) = 0 = v 0 ⇒
v(t) =−9.8t = s 0 (t) ⇒ s(t) =−4.9t 2 + s 0,buts(0) = s 0 =500 ⇒ s(t) =−4.9t 2 +500.Whens =0,
−4.9t 2 500
+500=0 ⇒ t 1 = ≈ 10.1 ⇒ v(t 500
4.9 1)=−9.8 ≈−98.995 m/s. Since the canister has been
4.9
designed to withstand an impact velocity of 100 m/s, the canister will not burst.