Math 21
HW7 solutions
Spring 2010
Here are the solutions to the problems on the seventh homework. Please do not print unless necessary. 7.4# 15, 16,
18a, 22
7.5# 4, 9, 11, 19a
1. (7.4#15) When a cold drink is taken from a fridge its temperature is 5 degrees C. After 25 minutes in a 20
degree C room, its temp increased to 10 degrees C. what the temperature of the drink after 50 minutes. When
will the temperature be 15 degrees C?
Solution:
The problem involves Newton’s law of cooling.
dT
= k(T − 20), T (0) = 5,
dt
where t is time in minutes and T is temp of the drink in degrees celsuis. We know from class that the solution
is given by
T (t) = 20 − 15ekt .
Let us find k. Noting that T (25) = 10 we find
1
2
ln .
25 3
10 = 20 − 15e25k → k =
Now
T (t) = 20 − 15e−f ract25 ln(2/3) .
We may compute T (50) = 13.3 degrees C, and the time t where T (t) = 15 is t = 67.74 minutes.
2. (7.4 #16) A freshly brewed cup of coffee has temperature 95 degrees celsuis in a 20 degree celsuis room. When
its temperature is 70 degrees, it is cooling at a rate of 1 degree per minute. When does this happen?
Solution:
This is a problem involving Newton’s law of cooling:
dT
= k(T − 20), T (0) = 95.
dt
We know from class that the solution is given by
T (t) = 20 + 75ekt .
In order to find k, we return to the differential equation as we know the value of dT /dt:
−1 = k(70 − 20),
1
where we have chosen a negative sign as the coffee is cooling. So k = − 50
. This means the model is
t
T (t) = 20 + 75e− 50 .
Now we solve for the time when T = 70:
t
70 = 20 + 75e− 50 ,
So t = −50 ln(2/3) = 20.27 minutes.
3. (7.4 #18a) If $1000 is borrowed at 8% interest, find the amounts due at the end of 3 years if the interest if
compounded annually, quarterly, monthly, weekly, daily, hourly and continuously.
Solution:
The main idea is to use the formula
r
A(t) = 1000(1 + )mt ,
m
where t is the time in years, r is the annual interest rate and m is the number of compound periods per year.
For this problem t = 3, r = 0.08 and:
m = 1 → A(3) = 1259.71.
m = 4 → A(3) = 1268.24.
m = 12 → A(3) = 1270.24.
m = 52 → A(3) = 1271.01.
m = 365 → A(3) = 1271.22.
m = 365 · 24 → A(3) = 1271.25.
For continuous compounding we find
A(3) = 1000e0.08·3 = 1271.25.
Math 21
HW7 solutions
Spring 2010
4. (7.4 #22) Consider the doomsday equation
dy
= ky 1+c ,
dt
where k, c are positive constants.
Solution:
First we solve the equation. Separating variables we find
dy
= kdt
y 1+c
Z
Z
−1−c
y
dy = kdt
y −c
= kt + D
−c
y −c = −ckt + D = D − ckt
At this point, lets solve for D using y(0) = y0 . We find
y0−c = D,
so that
y −c = y0−c − ckt
1
y= q
c
−c
y0 − ckt
= q
c
1
1−y0c ckt
y0c
y0
= p
c
1 − y0c ckt
Observe that this function has a vertical asymptote when the denominator is zero. This is the doomsdays time.
This occurs when
1
1 − y0c kt = 0 → t = c ,
y0 ck
i.e.
lim1
t→ yc ck
p
c
0
y0
= ∞.
1 − y0c ckt
Finally for part (c), we have from the text that c = 0.01, y0 = 2 and y(3) = 16. We may solve for k.
16 =
√
0.01
2
1 − 20.01 0.01 · k · 3
Once we have this, plut into the doomsday formula
1
y0c ck
to obtain a doomsday of 145.77 months.
5. (7.5 #4) Suppose a population P (t) satisfies:
dP
= 0.4P − 0.001P 2 , P (0) = 50.
dt
where t is measured in years.
(a) What is the carrying capacity?
Solution:
Here let’s rewrite the equation to match the algebraic form of the logistic model:
dP
0.001
P
2
= 0.4P − 0.001P = 0.4P 1 −
P = 0.4 1 − 0.4
.
dt
0.4
0.001
So M = 0.4/0.001 = 400.
(b) What is P 0 (0)?
Solution:
P 0 (0) =
dP
(0) = 0.4P (0) − 0.001P (0)2 = 0.4(50) − 0.001(50)2 = 17.5.
dt
Math 21
HW7 solutions
Spring 2010
(c) When will the population reach 50% of the carrying capacity?
Solution:
Noting that k = 0.4, we may use the solution for the logistic model derived in class:
P (t) =
We solve for t where
400
.
350/50e−0.4t + 1
400
400
= −0.4t
2
7e
+1
So
2 = 7e−0.4t + 1
and t = 4.86 years
6. (7.5 # 9) One model for the spread of a rumor is that the rate of spread is proportional to the product of the
fraction y of the population who have heard the rumor and the fraction who have not.
(a) Write a DE modeling this.
Solution:
Let t be time and y(t) be the fraction of the population that have heard the rumor.
Note that if 25% of the population have heard the rumor, then 75% have not heard it. So 1 − y represents
the fraction who have not heard the rumor. This means dy/dt is proportional to y(1 − y), or
dy
= ky(1 − y),
dt
for some constant k.
(b) Solve the DE.
Solution:
We recognize that the DE is just the logistic equation with M = 1. So from the notes,
y(t) =
1
1−y0 −kt
y0 e
+1
,
where y(0) = y0 . There are other correct ways to write this equation, and it is ok if you solved for this
equation instead of citing the result.
(c) A small town has 1000 inhabitants. Suppose at 8 am, 80 people hear a rumor. By noon half the town has
heard it. At what time will 90% of the population have heard it?
Solution:
Let us take t = 0 to be 8 am. We are told y(0) = 80/1000 = 0.08. The fact that
y(4) = 0.5 means that
0.5 = y(4) =
1
1−0.08 −4k
0.08 e
+1
1
11.5e−4k + 1
5.75e−4k + 0.5 = 1
0.5 =
e−4k =0.086956522
1
k = − ln 0.086956522
4
k = −0.612
Now then,
y=
1
11.5e−0.612t
+1
We now solve for t where y(t) = 0.9. So
0.9 = y(t) =
1
11.5e−0.612t + 1
t = 7.58
So 90% of the town hears the rumor 7.58 hours after 8 am. This is about 3:35 pm.
Math 21
HW7 solutions
Spring 2010
7. (7.5 #11) Show that if P satifies the logistic growth equation then
d2 P
P
2P
2
=
k
P
1
−
1
−
,
dt2
M
M
and deduce that a population grows fastest when it reaches half its carrying capacity.
Solution:
Begin with the logistic equation
P
dP
= kP 1 −
.
dt
M
Before differentiating we write this as
dP
k 2
= kP −
P ,
dt
M
but this step is not strictly necessary. We now differentiate in t. This means we have to remember to use
implicit differentiation, since P = P (t):
d2 P
dP
k
dP
=k
−
2P
.
2
dt
dt
M
dt
The idea now is we have an equation for dP/dt! So using the logistic equation,
d2 P
P
k
P
=
k
kP
1
−
−
2P
kP
1
−
.
dt2
M
M
M
We now simplify. Factor out common terms:
d2 P
= k2 P
dt2
1−
P
M
2P
1−
,
M
as we needed. Now the population is growing fastest when dP/dt is at a max. But to detect this we take
2
its derivative, ddtP2 and set it equal to zero to find critical numbers. The above calculation shows the critical
numbers are when P = 0, M and M
2 . We can reject the first two possibilities as the logistic model is constant
when P = 0 and P = M . So the population grows fastest when P = M/2.
8. (7.5 #19a) Find a solution to the seasonal growth model
dP
= kP cos(rt − φ), P (0) = P0 ,
dt
where k, r, φ are positive constants.
Solution:
We separate the variables, as always.
Z
Z
dP
= k cos(rt − φ)dt
P
k
ln |P | = sin(rt − φ) + C, (use u = rt − φ)
r
k
P = Ce r sin(rt−φ) .
We solve for C:
k
k
P0 = Ce r sin(0−φ) = Ce r sin(φ)
So
C=
We conclude
P0
e
k
r
k
sin(φ)
k
= P0 e− r sin(φ)
k
P = P0 e− r sin(φ) e r sin(rt−φ)
and we simplify a bit:
k
P = P0 e r (sin(rt−φ)−sin(φ))