Solução_Calculo_Stewart_6e

F.
136 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
93. (a) y(t) =y(0)e kt =200e kt ⇒ y(0.5) = 200e 0.5k =360 ⇒ e 0.5k =1.8 ⇒ 0.5k =ln1.8 ⇒
k =2ln1.8 =ln(1.8) 2 =ln3.24 ⇒ y(t) = 200e (ln 3.24)t = 200(3.24) t
(b) y(4) = 200(3.24) 4 ≈ 22,040 bacteria
(c) y 0 (t) = 200(3.24) t · ln 3.24,soy 0 (4) = 200(3.24) 4 · ln 3.24 ≈ 25,910 bacteria per hour
(d) 200(3.24) t =10,000 ⇒ (3.24) t =50 ⇒ t ln 3.24 = ln 50 ⇒ t =ln50/ ln 3.24 ≈ 3.33 hours
95. (a) C 0 (t) =−kC(t) ⇒ C(t) =C(0)e −kt by Theorem 9.4.2. But C(0) = C 0 ,soC(t) =C 0 e −kt .
(b) C(30) = 1 2 C0 since the concentration is reduced by half. Thus, 1 2 C0 = C0e−30k ⇒ ln 1 2 = −30k ⇒
k = − 1 ln 1 = 1 ln 2. Since10% of the original concentration remains if 90% is eliminated, we want the value of t
30 2 30
such that C(t) = 1 10 C 0. Therefore,
1
C 10 0 = C 0 e −t(ln 2)/30 ⇒ ln 0.1 =−t(ln 2)/30 ⇒ t = − 30 ln 0.1 ≈ 100 h.
ln 2
97. If x = edge length, then V = x 3 ⇒ dV/dt =3x 2 dx/dt =10 ⇒ dx/dt =10/(3x 2 ) and S =6x 2 ⇒
dS/dt =(12x) dx/dt =12x[10/(3x 2 )] = 40/x. Whenx =30, dS/dt = 40
30 = 4 3 cm2 /min.
99. Given dh/dt =5and dx/dt =15, find dz/dt. z 2 = x 2 + h 2 ⇒
2z dz dx dh
=2x +2h
dt dt dt
⇒
dz
dt = 1 (15x +5h). Whent =3,
z
h =45+3(5)=60and x = 15(3) = 45 ⇒ z = √ 45 2 +60 2 =75,
so dz
dt = 1 [15(45) + 5(60)] = 13 ft/s.
75
101. We are given dθ/dt = −0.25 rad/h. tan θ =400/x ⇒
x =400cotθ ⇒ dx
dt = −400 csc2 θ dθ
dt .Whenθ = π , 6
dx
dt = −400(2)2 (−0.25) = 400 ft/h.
103. (a) f(x) = 3√ 1+3x =(1+3x) 1/3 ⇒ f 0 (x) =(1+3x) −2/3 , so the linearization of f at a =0is
L(x) =f(0) + f 0 (0)(x − 0) = 1 1/3 +1 −2/3 x =1+x. Thus,
3√
1.03 =
3 1+3(0.01) ≈ 1+(0.01) = 1.01.
3√ 1+3x ≈ 1+x
⇒
(b) The linear approximation is 3√ 1+3x ≈ 1+x,sofortherequiredaccuracy
we want 3√ 1+3x − 0.1 < 1+x< 3√ 1+3x +0.1. From the graph,
itappearsthatthisistruewhen−0.23