Download Name: Growth, Decay, and Others Solve each of the problems

Name:
Growth, Decay, and Others
Solve each of the problems showing all work and figures necessary. Label answers
with appropriate units.
(1) (a) How long will it take an investment to double in value if the interest rate
is 6% compounded continuously?
ln 2
≈ 11.55 years
.06
2P0 = P0 e.06t ⇒ ln 2 = .06t ⇒ t =
(b) Suppose the same amount is invested at an annual interest rate, r, for
the same amount of time as part (a) and it doubles. Calculate r.
2P0 = P0 (1 + r)t ⇒ 2 = (1 + r)t ⇒ 21/t = 1 + r ⇒ r = 21/t − 1
ln 2
.06
Using t =
then
1
t
=
.06
ln 2
.06
and r = 2 ln 2 − 1 ≈ 6.18%
(2) David Caruso is investigating a murder. After dramatically removing his
sunglasses, the medical investigator informs him that the temperature of the
corpse was 32.5◦ C at 1:30PM and 30.3◦ C an hour later. Normal body temperature is 37.0◦ C and the temperature of the surroundings was 20.0◦ C. When
did the murder take place?
We have T (t) = the temperature of the body at time T . Then let y(t) =
T (t) − Ts where Ts is the temperature of the surroundings, 20.0◦ C. Then,
y(t) = y0 ekt and, by substitution, T (t) − Ts = (T (0) − Ts ) ekt . Thus, T (t) =
17ekt + 20.
32.5 = 17ekt1 + 20 and 30.3 = 17ek(t1 )+1) + 20 where t1 is the time elapsed
from the murder to 1:30PM.
ln 12.5
= kt1 and ln 10.3
= k(t1 + 1) = kt1 + k. By substitution,
17
17
ln
t1 =
10.3
12.5
10.3
12.5
= ln
+ k ⇒ ln
− ln
=k
17
17
17
17
ln 10.3
17
ln 12.5
ln 12.5
ln 12.5 − ln 17
17
17
=
= 10.3
≈ 1.588
ln
10.3
−
ln
12.5
− ln 12.5
=
k
ln
17
12.5
Since 1.588 hours is 1 hour and 35.3 minutes, the murder occurred just before
11:55AM.
(3) 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 heard the rumor.
(a) Write a differential equation that is satisfied by y.
dy
= ky(1 − y)
dt
(b) Solve the differential equation:
dy
y(1−y)
= k dt. By partial fractions,
2
A
B
1
= +
⇒ 1 = A(1 − y) + By = A + (B − A)y
y(1 − y)
y
1−y
⇒ A = 1, B = 1
Thus,
Z
Z
Z
dy
dy
+
= k dt ⇒ ln |y| − ln |1 − y] = kt + C
y
1−y
y = kt + C ⇒ y = Aekt
⇒ ln 1 − y
1−y
1−y
1
= Ae−kt ⇒ − 1 = Ae−kt
y
y
1
1
⇒ = Ae−kt + 1 ⇒ y =
−kt
y
Ae + 1
(c) A small town has 1000 inhabitants. At 8AM, 80 people have heard a
rumor that the local Cookout will be giving away free milkshakes. By
noon half the town has heard it. At what time will 90% of the population
have heard the rumor?
2
1
25
23
80
=
=
⇒A=
−1=
y(0) =
1000
25
A+1
2
2
1
⇒ y(t) =
23/2e−kt + 1
1
1
y(4) = =
⇒ 2 = 23/2e−4k + 1
−4k
2
23/2e
+1
2
1 = 23/2e−4k ⇒ ln
= −4k
23
ln 2
k = 23 ≈ 0.61
−4
⇒
.9 =
1
23/2e−kt
1/9 = 23/2e
t=
+1
−kt
⇒ 10/9 = 23/2e−kt + 1
⇒ ln (2/207) = −kt
ln (2/207)
ln (2/207)
ln 2 − ln 207
=
=4
≈ 7.60 hours
2
ln
−k
ln 2 − ln 23
23
4
So, by 3:36PM, 90% of the population will have heard the rumor.