(1) Construct the di
ff
erential equation for
Q
(
t
).
(2) Solve the di
ff
erential equation.
(3) Predict the long term quantity of the salt in the reservoir.
Solution: (1)
Rate of change of total quantity of salt = Rate In  Rate Out;
Rate of change of total quantity of salt =
dQ
dt
;
Rate In = salt IN per day = salt water volume IN per day
×
concentration of salt water
IN;
Rate Out = salt OUT per day = salt water volume OUT per day
×
concentration of
salt water OUT.
We have
dQ
dt
= 0
.
1
×
50

1
×
Q
100
,
⇒
dQ
dt
=
500

Q
100
,
Q
(0) = 0
.
(2)
Q
(
t
) = 500(1

e

0
.
01
t
)
.
(3)
lim
t
→∞
Q
(
t
) = 500
.
23
7.4 Exponential Growth and Decay
If the population growth
dP
dt
is proportional to the population size
P
(
t
), then
dP
dt
=
kP.
The solution is
P
(
t
) =
P
(0)
e
kt
.
k >
0
→
growth,
k <
0
→
decay.
•
Absolute growth rate = rate of change of population =
dP/dt
,
•
Relative growth rate = percent change per unit time = (
dP/dt
)
/P,
•
Doubling time (
D
) of exponential growth = the time required for it to double:
P
(
t
) =
P
(0)2
t/D
.
•
Half life (
H
) of exponential growth = the time required for it to be half:
P
(
t
) =
P
(0)
1
2
t/H
.
•
Newtons Law of Cooling states that the rate of change of the temperature of a cooling
body is proportional to the di
ff
erence between its temperature T and the temperature
of its surrounding medium
T
s
. The model is:
dT
dt
=

k
(
T

T
s
)
,
where
k
is a constant.
Example.
A bacteria culture growth at a rate proportional to its size. After 2 hours there
are 40 bacteria and after 4 hours the count is 120. Find an expression for the population
after t hours.
Solution: We measure the time
t
in hours. Let
P
(
t
) be the population at
t
hours, then
we have
dP
dt
=
kP.
The solution of the equation is
P
(
t
) =
P
(0)
e
kt
.
24
Note that
P
(2) = 40 and
P
(4) = 120, we obtain
40 =
P
(0)
e
2
k
,
120 =
P
(0)
e
4
k
.
These imply that
P
(0) =
40
3
and
e
2
k
= 3
,
or
k
= ln 3
/
2
.
We thus have
P
(
t
) =
40
3
3
t/
2
=
40
3
√
3
t
=
40
3
e
(ln 3
/
2)
t
.
Example.
The halflife of Sodium24 is 15 hours. Suppose you have 100 grams of Sodium
24. How many grams remaining after 27 minutes (keep three decimals)?
Solution: Assume
m
(
t
) be the amount after
t
hours. Then
m
(
t
) =
m
(0)(
1
2
)
t/H
,
where
m
(0) = 100,
H
= 15 hours. Note that 27 minutes = 27/60=0.45 hours. Thus
m
(0
.
45) = 100(
1
2
)
0
.
45
/
15
= 100(
1
2
)
0
.
03
= 97
.
942
g.
Example.
Use Newton’s Law of Cooling to determine the time of death of a healthy man.
•
He died in his room some time before noon;
•
At noon, his body temperature was found to be 70 degrees;
•
His body cooled another 5 degrees in 1 hour after noon;
•
The room temperature was a constant 60 degrees;
•
Normal temperature of people’s body is 98.6 degree.
Solution: Let
T
(
t
) be the temperature of the body at time
t
. Taking noon as
t
= 0, we
have
T
(0) = 70. Note that
T
s
= 60, we have
dT
dt
=

k
(
T

60)
,
this equation is separable. We obtain
dT
T

60
=

kdt,
⇒
T
(
t
) = 60 +
Ce

kt
.
25
From
T
(0) = 70 we get 70 = 60 +
C
,
C
= 10 and
T
(
t
) = 60 + 10
e

kt
.
At 1:00pm, his body temperature is 70

5 = 65. Hence
T
(1) = 65.
65 = 60 + 10
e

k
(1)
,
⇒
5 = 10
e

k
,
⇒
k
= ln 2
,
⇒
T
(
t
) = 60 + 10
e

t
ln 2
.
Thus
98
.
6 = 60 + 10
e

t
ln 2
,
⇒
t
=

ln 3
.
86
/
ln 2 =

1
.
95
,
which means 1 hour and 0.95(60) minutes before noon, or 1 hour and 57 minutes before
noon. Or the time of death is 10:03AM.