CHAPTER 26
Exponential Growth and Decay
26.1
26.2
26.3
26.4
26.5
26.6
26.7
26.8
26.9
215
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 = y 0 e
Kt ,
where y a is the value of y at time t = 0.
Hence, yle
Kl is a constant C, y= Ce
Kl . When
( = 0, y 0 =Ce° = C. Thus, y = y 0 e
Kl .
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 = y 0 e
Kl . By the given information, 2y 0 = y 0 e*
K , 2 = e
}K ,
In 2 = In e
3
* = 3K, K = (ln2)/3. When f = 9, y = y 0 e
9K = y 0 e^"
2 = y 0 (e
la2 )
3 = y a -2
3 = 8y 0 . Thus, the
initial number has been multiplied by 8.
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 = y n e
Kl . The given information tells us that
50 = 200e
K ,
3.125 grams.
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 = y 0 e
Kl . Let A be a fixed time interval, and let t be any time. Then
y(t + A)/y(0 = y a e
K( '
+ ^/y 0 e
K> = e*
A
, which does not depend on t.
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
, = 50, y = 4.5e°
04
n 2 )
2 = 4.5(2)
2 = 18 billion people.
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.
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
(usine a table for the exponential function). Hence,
over 10 years, there will be an increase of about 21.9 percent.
years, y = y 0 e" = y 0 e'
u
""""
u = y 0 e" "° ~ (1.219)j> 0
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, y 0 (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 y 0 (l + r/100n)*".
An amount of money y a 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?
When
r = 3,
y = 200e
3
* = 200(e
A:
)
3
y = y n e
K> . When /=!, y = (1 + r/100)y n . Hence, (1 + r/100)y n = y n e
K , l + r/100 = e*
So
A" = In (! + /•/100) and r= 100(e" - 1).
Exponential Growth and Decay
26.1
26.2
26.3
26.4
26.5
26.6
26.7
26.8
26.9
215
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 = y 0 e
Kt ,
where y a is the value of y at time t = 0.
Hence, yle
Kl is a constant C, y= Ce
Kl . When
( = 0, y 0 =Ce° = C. Thus, y = y 0 e
Kl .
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 = y 0 e
Kl . By the given information, 2y 0 = y 0 e*
K , 2 = e
}K ,
In 2 = In e
3
* = 3K, K = (ln2)/3. When f = 9, y = y 0 e
9K = y 0 e^"
2 = y 0 (e
la2 )
3 = y a -2
3 = 8y 0 . Thus, the
initial number has been multiplied by 8.
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 = y n e
Kl . The given information tells us that
50 = 200e
K ,
3.125 grams.
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 = y 0 e
Kl . Let A be a fixed time interval, and let t be any time. Then
y(t + A)/y(0 = y a e
K( '
+ ^/y 0 e
K> = e*
A
, which does not depend on t.
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
, = 50, y = 4.5e°
04
n 2 )
2 = 4.5(2)
2 = 18 billion people.
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.
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
(usine a table for the exponential function). Hence,
over 10 years, there will be an increase of about 21.9 percent.
years, y = y 0 e" = y 0 e'
u
""""
u = y 0 e" "° ~ (1.219)j> 0
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, y 0 (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 y 0 (l + r/100n)*".
An amount of money y a 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?
When
r = 3,
y = 200e
3
* = 200(e
A:
)
3
y = y n e
K> . When /=!, y = (1 + r/100)y n . Hence, (1 + r/100)y n = y n e
K , l + r/100 = e*
So
A" = In (! + /•/100) and r= 100(e" - 1).