Solução_Calculo_Stewart_6e

F.
408 ¤ CHAPTER 9 PROBLEMS PLUS
TX.10
4e 20k − 7e 10k +3=0. This is a quadratic equation in e 10k . 4e 10k − 3 e 10k − 1 =0 ⇒ e 10k = 3 4 or 1 ⇒
10k =ln 3 4
or ln 1 ⇒ k =
1
10 ln 3 4 since k is a nonzero constant of proportionality. Substituting 3 4 for e10k in (3) gives us
−20 = M · 3
4 − M ⇒ −20 = − 1 4 M ⇒ M =80.NowR =100− M so R =20◦ C.
9. (a) While running from (L, 0) to (x, y), the dog travels a distance
s = L
1+(dy/dx)2 dx = − x
1+(dy/dx)2 dx,so
x
L
ds
dx = − 1+(dy/dx) 2 . The dog and rabbit run at the same speed, so the
rabbit’s position when the dog has traveled a distance s is (0,s). Since the
dog runs straight for the rabbit, dy
dx = s − y (see the figure).
0 − x
Thus, s = y − x dy
dx ⇒ ds
dx = dy
dx −
x d 2 y
dx 2 +1dy dx
gives us x d 2 y
dx 2 =
1+
(b) Letting z = dy
dx
ln x =
dz
√
1+z
2
2 dy
, as claimed.
dx
, we obtain the differential equation x
dz
dx = √ 1+z 2 ,or
= −x d2 y
ds
. Equating the two expressions for
dx2 dx
dz
√
1+z
2 = dx x .Integrating:
25
=ln
z +
1+z 2 + C. Whenx = L, z = dy/dx =0,soln L =ln1+C. Therefore,
C =lnL, soln x =ln √ 1+z 2 + z +lnL =ln L √ 1+z 2 + z ⇒ x = L √ 1+z 2 + z ⇒
√
1+z2 = x x
2
L − z ⇒ 2xz
x
2 x 1+z2 = −
L L + z2 ⇒ − 2z − 1=0 ⇒
L L
z = (x/L)2 − 1
2(x/L)
= x2 − L 2
2Lx
= x
2L − L 1
2 x
dy
[for x>0]. Since z =
dx , y = x2
4L − L ln x + C1.
2
Since y =0when x = L, 0= L 4 − L 2 ln L + C1 ⇒ C1 = L 2 ln L − L 4 .Thus,
y = x2
4L − L 2 ln x + L 2 ln L − L 4 = x2 − L 2
− L x
4L 2 ln .
L
(c) As x → 0 + , y →∞, so the dog never catches the rabbit.
11. (a) We are given that V = 1 3 πr2 h, dV/dt =60,000π ft 3 /h, and r =1.5h = 3 2 h.SoV = 1 3 π 3
2 h2 h = 3 4 πh3 ⇒
dV
dt = 3 π · dh
4 3h2 dt = 9 dh
dh
4 πh2 . Therefore,
dt dt = 4(dV/dt) = 240,000π = 80,000 () ⇒
9πh 2 9πh 2 3h 2
3h 2 dh = 80,000 dt ⇒ h 3 =80,000t + C. Whent =0, h =60.Thus,C =60 3 = 216,000, so
h 3 =80,000t + 216,000. Leth =100.Then100 3 =1,000,000 = 80,000t +216,000
⇒
80,000t =784,000 ⇒ t =9.8, so the time required is 9.8 hours.