Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 9.3 SEPARABLE EQUATIONS ¤ 391
43. Let y(t) be the amount of alcohol in the vat after t minutes. Then y(0) = 0.04(500) = 20 gal. The amount of beer in the vat
is 500 gallons at all times, so the percentage at time t (in minutes) is y(t)/500 × 100, and the change in the amount of alcohol
with respect to time t is dy
dt = rate in − rate out =0.06 5 gal
− y(t)
5 gal
=0.3 − y
min 500 min 100 = 30 − y gal
100 min .
Hence,
dy dt
30 − y = 1
and − ln |30 − y| =
100
t + C. Because y(0) = 20, wehave− ln 10 = C, so
100
− ln |30 − y| = 1
100 t − ln 10 ⇒ ln |30 − y| = −t/100 + ln 10 ⇒ ln |30 − y| =lne−t/100 +ln10 ⇒
ln |30 − y| =ln(10e −t/100 ) ⇒ |30 − y| =10e −t/100 .Sincey is continuous, y(0) = 20, and the right-hand side is
never zero, we deduce that 30 − y is always positive. Thus, 30 − y =10e −t/100 ⇒ y =30− 10e −t/100 .The
percentage of alcohol is p(t) =y(t)/500 × 100 = y(t)/5 =6− 2e −t/100 . The percentage of alcohol after one hour is
p(60) = 6 − 2e −60/100 ≈ 4.9.
45. Assume that the raindrop begins at rest, so that v(0) = 0. dm/dt = km and (mv) 0 = gm ⇒ mv 0 + vm 0 = gm ⇒
mv 0 + v(km) =gm ⇒ v 0 + vk = g ⇒ dv
dt = g − kv ⇒ dv
g − kv = dt ⇒
− (1/k)ln|g − kv| = t + C ⇒ ln |g − kv| = −kt − kC ⇒ g − kv = Ae −kt . v(0) = 0 ⇒ A = g.
So kv = g − ge −kt ⇒ v =(g/k)(1 − e −kt ).Sincek>0,ast →∞, e −kt → 0 and therefore, lim
t→∞
v(t) =g/k.
47. (a) The rate of growth of the area is jointly proportional to A(t) and M − A(t); that is, the rate is proportional to the
product of those two quantities. So for some constant k, dA/dt = k √ A (M − A). We are interested in the maximum of
the function dA/dt (when the tissue grows the fastest), so we differentiate, using the Chain Rule and then substituting for
dA/dt from the differential equation:
d dA √A dA
= k (−1) 1 dA
+(M − A) ·
2
dt dt
dt A−1/2 = 1 dA
2
dt
kA−1/2 [−2A +(M − A)]
dt
=
k √
1
2 kA−1/2 A(M − A) [M − 3A] = 1 2 k2 (M − A)(M − 3A)
This is 0 when M − A =0[this situation never actually occurs, since the graph of A(t) is asymptotic to the line y = M,
as in the logistic model] and when M − 3A =0 ⇔ A(t) =M/3. This represents a maximum by the First Derivative
Test, since d dA
goes from positive to negative when A(t) =M/3.
dt dt
Ce √ 2 Mkt − 1
(b) From the CAS, we get A(t) =M
Ce √ .TogetC in terms of the initial area A 0 and the maximum area M,
Mkt
+1
we substitute t =0and A = A 0 = A(0): A 0 = M
2 C − 1
⇔ (C +1) √ A 0 =(C − 1) √ M ⇔
C +1
C √ A 0 + √ A 0 = C √ M − √ M ⇔ √ M + √ A 0 = C √ M − C √ A 0 ⇔
√ √ √M √
√ √ M + A0
M + A0 = C
− A0 ⇔ C = √ √ . [Notice that if A 0 =0,thenC =1.]
M − A0