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Exponential Growth and Decay solved questions
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CHAPTER 26Exponential Growth and Decay
26.1 A quantity y is said to grow or decay exponentially in time if D,y = Ky for some constant K. (K is called the
growth constant or decay constant, depending on whether it is positive or negative.) Show that y = y0eKt,
where ya is the value of y at time t = 0.
Hence, yleKl is a constant C, y= CeKl. When
Kl
( = 0, y0=Ce° = C. Thus, y = y0e .
26.2 A bacteria culture grows exponentially so that the initial number has doubled in 3 hours. How many times the
initial number will be present after 9 hours?
Let y be the number of bacteria. Then y = y0eKl. By the given information, 2y0 = y0e*K, 2 = e}K,
In 2 = In e3* = 3K, K = (ln2)/3. When f = 9, y = y0e9K = y0e^"2 = y0(ela2)3 = ya -2 3 = 8y0. Thus, the
initial number has been multiplied by 8.
26.3 A certain chemical decomposes exponentially. Assume that 200 grams becomes 50 grams in 1 hour. How much
will remain after 3 hours?
Let y be the number of grams present at time t. Then y = yneKl. The given information tells us that
50 = 200eK, When r = 3, y = 200e3* = 200(eA:)3
3.125 grams.
26.4 Show that, when a quantity grows or decays exponentially, the rate of increase over a fixed time interval is
constant (that is, it depends only on the time interval, not on the time at which the interval begins).
The formula for the quantity is y = y0eKl. Let A be a fixed time interval, and let t be any time. Then
y(t + A)/y(0 = yaeK('+^/y0eK> = e*A, which does not depend on t.
26.5 If the world population in 1980 was 4.5 billion and if it is growing exponentially with a growth constant
K = 0.04 In 2, find the population in the year 2030.
Let y be the population in billions in year t, with t = 0 in 1980. Then y = 4.5e° 04<ln 2)'. In 2030, when
, = 50, y = 4.5e° 04<ln 2)'50 = 4.5(e'n 2 ) 2 = 4.5(2)2 = 18 billion people.
26.6 If a quantity y grows exponentially with a growth constant K and if during each unit of time there is an increase in y
of r percent, find the relationship between K and r.
y = yneK>. When /=!, y = (1 + r/100)y n . Hence, (1 + r/100)yn = yneK, l + r/100 = e* So
A" = In (! + /•/100) and r = 100(e" - 1).
26.7 If a population is increasing exponentially at the rate of 2 percent per year, what will be the percentage increase
over a period of 10 years?
In the notation of Problem 26.6, r = 2, K- In (1.02) = 0.0198 (by a table of logarithms). Hence, after 10
years, y = y0e" = y0e'u""""u = y0e" "° ~ (1.219)j>0 (usine a table for the exponential function). Hence,
over 10 years, there will be an increase of about 21.9 percent.
26.8 If an amount of money v 0 is invested at a rate of r percent per year, compounded n times per year, what is the
amount of money that will be available after k years?
After the first period of interest (sth of a year), the amount will be y 0 (l + r/lOOn); after the second
period, y0(l + r/lOOn)2, etc. The interval of k years contains kn periods of interest, and, therefore, the
amount present after k years will be y0(l + r/100n)*".
26.9 An amount of money ya earning r percent per year is compounded continuously (that is, assume that it is
compounded n times per year, and then let n approach infinity). How much is available after k years?
215
, 216 CHAPTER 26
By Problem 26.8, y = y 0 (l+ r/100n)*" if the money is compounded n times per year. If we
let n approach infinity, we get
*,(«"IOO)* = */-0"*. (Here we have used Problem 24.74.) Thus, the money grows exponentially, with growth
constant K = 0.0lr.
26.10 If an amount of money earning 8 percent per year is compounded quarterly, what is the equivalent yearly rate of
return?
By Problem 26.8, the amount present after 1 year will be Thus, the
equivalent yearly rate is 8.24 percent.
26.11 If an amount of money earning 8 percent per year is compounded 10 times per year, what is the equivalent yearly
rate of return?
By Problem 26.8, the amount after 1 year will be .y 0 (H-Tggg) 10 = 3>0(1.008)IO = 1.0829>>. Thus, the
equivalent yearly rate is 8.29 percent.
26.12 If an amount of money receiving interest of 8 percent per year is compounded continuously, what is the equivalent
yearly rate of return?
By Problem 26.9, the amount after 1 year will be yae° °8, which, by a table of values of e*, is approximately
l.OS33y0. Hence, the equivalent yearly rate is about 8.33 percent.
26.13 A sum of money, compounded continuously, is multiplied by 5 in 8 years. If it amounts to $10,000 after 24 years,
what was the initial sum of money?
By the formula of Problem 26.9, y = y0e0'airk, where r is the rate of interest and k is the number of
years. Hence, 5y0 = y0e°'"*', 5 = e°08'. After 24 years, 10,000 = y 0 e° 024r = ^0(e°-08r)3 = >-0(5)3 = 125>-0.
Hence, the initial quantity y0 was 80 dollars.
26.14 If a quantity of money, earning interest compounded continuously, is worth 55 times the initial amount after 100
years, what was the yearly rate of interest?
By Problem 26.9, is the percentage rate of interest, ya is the initial amount, and k is the
number of years. Then, Hence, and, by a table of logarithms,
In Thus, the rate is approximately 4 percent per year.
26.15 Assume that a quantity y decays exponentially, with a decay constant K. The half-life Tis defined to be the time
interval after which half of the original quantity remains. Find the relationship between K and T.
y = y0eK>. By definition, So
26.16 The half-life of radium is 1690 years. If 10 percent of an original quantity of radium remains, how long ago was
the radium created?
Let y be the number of grams of radium t years after the radium was created. Then y = y0eKl, where
1690K = -In 2, by Problem 26.15. If at the present time then
Hence, -(In2/1690)f = -In 10, t= 1690 In 10/ln 2 = 5614.477. Thus, the radium was
created about 5614 years ago.
26.17 If radioactive carbon-14 has a half-life of 5750 years, what will remain of 1 gram after 3000 years?
The amount of carbon is We know that Since and
(from a table for e *). Thus, about 0.7 gram will remain.
26.18 If 20 percent of a radioactive element disappears in 1 year, compute its half-life.
Let y n be the original amount, and let T be the half-life. Then 0.8y0 remains when t = 1. Thus,
0.&y0 = y0e, 0.8 = eK, K = In 0.8 « -0.2231 (from a table of logarithms). But KT = -In 2= -0.6931.
So -0.2231 T= -0.6931, 7=3.1067 years.
26.19 Fruit flies are being bred in an enclosure that can hold a maximum of 640 flies. If the flies multiply exponentially,
with a growth constant K = 0.05 and where time is measured in days, how long will it take an initial population
of 20 to fill the enclosure?
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