Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1. We use the Fundamental Theorem of Calculus to differentiate the given equation:
[f(x)] 2 =100+
x
[f(t)] 2 +[f 0 (t)] 2 dt ⇒ 2f(x)f 0 (x) =[f(x)] 2 +[f 0 (x)] 2 ⇒
0
[f(x)] 2 +[f 0 (x)] 2 − 2f(x)f 0 (x) =0 ⇒ [f(x) − f 0 (x)] 2 =0 ⇔ f(x) =f 0 (x).Wecansolvethisasaseparable
equation, or else use Theorem 9.4.2 with k =1, which says that the solutions are f(x) =Ce x .Now[f(0)] 2 =100,so
f(0) = C = ±10, and hence f(x) =±10e x are the only functions satisfying the given equation.
3. f 0 f(x + h) − f(x) f(x)[f(h) − 1]
(x) =lim
=lim
[since f(x + h) =f(x)f(h)]
h→0 h
h→0 h
f(h) − 1
f(h) − f(0)
= f(x) lim = f(x) lim
= f(x)f 0 (0) = f(x)
h→0 h
h→0 h − 0
Therefore, f 0 (x) =f(x) for all x and from Theorem 9.4.2 we get f(x) =Ae x .Nowf(0) = 1 ⇒ A =1 ⇒
f(x) =e x .
5. “The area under the graph of f from 0 to x is proportional to the (n +1)st power of f(x)” translates to
x
0 f(t) dt = k[f(x)]n+1 for some constant k. ByFTC1,
d
dx
x
0
f(t) dt = d
k[f(x)]
n+1
dx
f(x) =k(n +1)[f(x)] n f 0 (x) ⇒ 1=k(n +1)[f(x)] n−1 f 0 (x) ⇒ 1=k(n +1)y n−1 dy
dx
k(n +1)y n−1 dy = dx ⇒ k(n +1)y n−1 dy = dx ⇒ k(n +1) 1 n yn = x + C.
Now f(0) = 0 ⇒ 0=0+C ⇒ C =0and then f(1) = 1 ⇒ k(n +1) 1 n =1 ⇒ k = n
n +1 ,
so y n = x and y = f(x) =x 1/n .
⇒
⇒
7. Let y(t) denote the temperature of the peach pie t minutes after 5:00 PM and R the temperature of the room. Newton’s Law of
Cooling gives us dy/dt = k(y − R). Solving for y we get
dy
= kdt ⇒ ln|y − R| = kt + C ⇒
y − R
|y − R| = e kt+C ⇒ y − R = ±e kt · e C ⇒ y = Me kt + R, whereM is a nonzero constant. We are given
temperatures at three times.
Substituting 100 − M for R in (1) and (2) gives us
y(0) = 100 ⇒ 100 = M + R ⇒ R =100− M
y(10) = 80 ⇒ 80 = Me 10k + R (1)
y(20) = 65 ⇒ 65 = Me 20k + R (2)
Dividing (3) by (4) gives us −20
−35 = M e 10k − 1
M(e 20k − 1)
−20 = Me 10k − M (3) and −35 = Me 20k − M (4)
⇒ 4 7 = e10k − 1
e 20k − 1
⇒ 4e 20k − 4=7e 10k − 7 ⇒
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