EXPONENTIAL GROWTH AND DECAY
219
The world population at the beginning of 1970 was 3.6 billion. The weight of the earth is 6.586 x 10
21 tons. If
the population continues to increase exponentially, with a growth constant K = 0.02 and with time measured in
years, in what year will the weight of all people equal the weight of the earth, if we assume that the average person
weighs 120 pounds?
26.38
26.39
26.40
26.41
26.42
26.43
A radioactive substance decays exponentially. What is the average quantity present over the first half-life?
But by Problem 26.15, AT =-In 2. Hence,
If money is invested at 5 percent, compounded continuously, in how many years will it double in value?
Show that if interest is compounded continuously at an annual percentage r the effective annual percentage rate of
interest is 100(e°
Olr - 1).
Let s be the effective annual percentage rate. Then y 0 e°'
olr = y 0 (l + 0.01s). So e°'
olr = 1 + 0.01s,
0.01s = e
oolr -l, s = 100(e
oolr -l).
2y 0 = y 0 e°05 ', by Problem 26.9. So 2 = e
005 ', 0.05/ = ln2, t = 20 In 2 = 20(0.6931) = 13.862. Thus,
the money will double in a little less than 13 years and 315 days.
An object cools from 120 to 95°F in half an hour when surrounded by air whose temperature is 70°F. Use
Newton's law of cooling (Problem 26.22) to find its temperature at the end of another half an hour.
The earth weighs 6.586 x 10
21 x 2000 pounds. When this is equal to the weight of y billion people,
120 x 10"y = 6.586 x 10
21 x 2000, or y = l.lxl0
14 . Thus we must solve 1.1 x 10
14 = 3.6e
002 ' for (.
Taking logarithms, In 1.1 + 14 In 10 = In 3.6 + 0.02f, 0.02f = 31.05, t*= 1552.5 years. The date would be
1970 + 1552 = 3522.
Let y be the difference in temperature between the object and the air. By Newton's law, y = y 0 e
Kt .
When t = \,
Since y 0 = 120 - 70 = 50, y = 50e . At
y = 50e* = 50
Hence, the temperature of the object is 70 + 12.5 = 82.5°F.
What is the present value of a sum of money which if invested at 5 percent interest, continuously compounded, will
become $1000 in 10 years?
Let y be the value of the money at time t, and let y 0 be its present value. Then, by Problem 26.9,
y = y 0 e
005 '. In 10 years, 1000 = y 0 e°
05(10) = y 0 e
05 . So y 0 = 1000/e
05 = 1000/1.64872 = 606.53. Hence,
the present value is about $606.53.
y = y«eK'- We are told that y = 2ya when t-l. Hence, 2y0 = y0e*, 2 = eK. If we start with
1000, we obtain a billion when 10
9 = 10V, 10
6 = (e
K )' = 2', ln(10
6 ) = f In2, 6In 10 = fin2, t =
(6 In 10)/ln 2 = 6(2.3026)/0.6931«19.9 hours.
so
219
The world population at the beginning of 1970 was 3.6 billion. The weight of the earth is 6.586 x 10
21 tons. If
the population continues to increase exponentially, with a growth constant K = 0.02 and with time measured in
years, in what year will the weight of all people equal the weight of the earth, if we assume that the average person
weighs 120 pounds?
26.38
26.39
26.40
26.41
26.42
26.43
A radioactive substance decays exponentially. What is the average quantity present over the first half-life?
But by Problem 26.15, AT =-In 2. Hence,
If money is invested at 5 percent, compounded continuously, in how many years will it double in value?
Show that if interest is compounded continuously at an annual percentage r the effective annual percentage rate of
interest is 100(e°
Olr - 1).
Let s be the effective annual percentage rate. Then y 0 e°'
olr = y 0 (l + 0.01s). So e°'
olr = 1 + 0.01s,
0.01s = e
oolr -l, s = 100(e
oolr -l).
2y 0 = y 0 e°05 ', by Problem 26.9. So 2 = e
005 ', 0.05/ = ln2, t = 20 In 2 = 20(0.6931) = 13.862. Thus,
the money will double in a little less than 13 years and 315 days.
An object cools from 120 to 95°F in half an hour when surrounded by air whose temperature is 70°F. Use
Newton's law of cooling (Problem 26.22) to find its temperature at the end of another half an hour.
The earth weighs 6.586 x 10
21 x 2000 pounds. When this is equal to the weight of y billion people,
120 x 10"y = 6.586 x 10
21 x 2000, or y = l.lxl0
14 . Thus we must solve 1.1 x 10
14 = 3.6e
002 ' for (.
Taking logarithms, In 1.1 + 14 In 10 = In 3.6 + 0.02f, 0.02f = 31.05, t*= 1552.5 years. The date would be
1970 + 1552 = 3522.
Let y be the difference in temperature between the object and the air. By Newton's law, y = y 0 e
Kt .
When t = \,
Since y 0 = 120 - 70 = 50, y = 50e . At
y = 50e* = 50
Hence, the temperature of the object is 70 + 12.5 = 82.5°F.
What is the present value of a sum of money which if invested at 5 percent interest, continuously compounded, will
become $1000 in 10 years?
Let y be the value of the money at time t, and let y 0 be its present value. Then, by Problem 26.9,
y = y 0 e
005 '. In 10 years, 1000 = y 0 e°
05(10) = y 0 e
05 . So y 0 = 1000/e
05 = 1000/1.64872 = 606.53. Hence,
the present value is about $606.53.
y = y«eK'- We are told that y = 2ya when t-l. Hence, 2y0 = y0e*, 2 = eK. If we start with
1000, we obtain a billion when 10
9 = 10V, 10
6 = (e
K )' = 2', ln(10
6 ) = f In2, 6In 10 = fin2, t =
(6 In 10)/ln 2 = 6(2.3026)/0.6931«19.9 hours.
so