Page 186 - How To Solve Word Problems In Calculus

1 −1.5t
= e
100
1
−1.5t = ln
100
1.5t = ln 100

ln 100
t = ≈ 3.07
1.5
The rate of infection starts to decline 3.07 weeks after the
outbreak begins.
10
(d) lim Q(t) = lim −1.5t = 10
t→∞ t→∞ 1 + 100e
10,000 people would contract the disease if no medical
treatment were received.
11. Let u(t) represent the temperature of the drink t minutes after it is
taken out of the refrigerator. Since the temperature changes at a
rate proportional to the difference in temperature between the
du
object and the outside medium, = k(u − 30).
dt
1 du
= k
u − 30 dt

d The antiderivative of 1
ln(u − 30) = k u − 30
dt is ln|u − 30|. Since u < 30,
this reduces to ln(30 − u).
ln(30 − u) = kt + C

The initial temperature, u(0) = 5soln25 = C .

ln(30 − u) = kt + ln 25
ln(30 − u) − ln 25 = kt

30 − u
ln = kt
25
30 − u
= e kt
25

30 − u = 25e kt

u(t) = 30 − 25e kt
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