Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 9.3 SEPARABLE EQUATIONS ¤ 389
(b) y 0 =1/y ⇒ dy/dx =1/y ⇒
(c)
ydy= dx ⇒ ydy= dx ⇒ 1 2 y2 = x + C ⇒
y 2 =2(x + C) or y = ± 2(x + C).
29. The curves x 2 +2y 2 = k 2 form a family of ellipses with major axis on the x-axis. Differentiating gives
d
dx (x2 +2y 2 )= d
dx (k2 ) ⇒ 2x +4yy 0 =0 ⇒ 4yy 0 = −2x ⇒ y 0 = −x . Thus, the slope of the tangent line
2y
at any point (x, y) on one of the ellipses is y 0 = −x , so the orthogonal trajectories
2y
must satisfy y 0 = 2y x
dy
y =2 dx
x
⇔
dy
dx = 2y x
⇔
dy
y =2= dx x
⇔ ln |y| =2ln|x| + C 1 ⇔ ln |y| =ln|x| 2 + C 1 ⇔
|y| = e ln x2 +C 1
⇔ y = ± x 2 · e C 1
= Cx 2 . This is a family of parabolas.
⇔
31. The curves y = k/x form a family of hyperbolas with asymptotes x =0and y =0. Differentiating gives
d
dx (y) = d k
⇒ y 0 = − k ⇒ y 0 = − xy [since y = k/x ⇒ xy = k] ⇒ y 0 = − y . Thus, the slope
dx x
x 2 x 2 x
of the tangent line at any point (x, y) on one of the hyperbolas is y 0 = −y/x,
so the orthogonal trajectories must satisfy y 0 = x/y ⇔ dy
dx = x y
⇔
ydy= xdx ⇔ ydy= xdx ⇔ 1 2 y2 = 1 2 x2 + C 1 ⇔
y 2 = x 2 + C 2 ⇔ x 2 − y 2 = C. This is a family of hyperbolas with
asymptotes y = ±x.
33. From Exercise 9.2.27, dQ
dt =12− 4Q ⇔
dQ
12 − 4Q =
dt ⇔ − 1 ln|12 − 4Q| = t + C ⇔
4
ln|12 − 4Q| = −4t − 4C ⇔ |12 − 4Q| = e −4t−4C ⇔ 12 − 4Q = Ke −4t [K = ±e −4C ] ⇔
4Q =12− Ke −4t ⇔ Q =3− Ae −4t [A = K/4]. Q(0) = 0 ⇔ 0=3− A ⇔ A =3 ⇔
Q(t) =3− 3e −4t .Ast →∞, Q(t) → 3 − 0=3(the limiting value).
35.
dP
dt = k(M − P ) ⇔
dP
P − M = (−k) dt ⇔ ln|P − M| = −kt + C ⇔ |P − M| = e −kt+C ⇔
P − M = Ae −kt [A = ±e C ] ⇔ P = M + Ae −kt . If we assume that performance is at level 0 when t =0,then
P (0) = 0 ⇔ 0=M + A ⇔ A = −M ⇔ P (t) =M − Me −kt . lim
t→∞
P (t) =M − M · 0=M.