Section 8.3
Applications of Exponential Functions
789
Version: Fall 2007
8.3
Applications of Exponential Functions
In the preceding section, we examined a population growth problem in which the popu-
lation grew at a fixed percentage each year. In that case, we found that the population
can be described by an exponential function.
A similar analysis will show that any
process in which a quantity grows by a fixed percentage each year (or each day, hour,
etc.) can be modeled by an exponential function. Compound interest is a good example
of such a process.
Discrete Compound Interest
If you put money in a savings account, then the bank will pay you interest (a percentage
of your account balance) at the end of each time period, typically one month or one
day.
For example, if the time period is one month, this process is called
monthly
compounding
. The term compounding refers to the fact that interest is added to your
account each month and then in subsequent months you earn interest on the interest.
If the time period is one day, it’s called
daily compounding
.
Let’s look at monthly compounding in more detail. Suppose that you deposit $100
in your account, and the bank pays interest at an annual rate of 5%. Let the function
P
(
t
)
represent the amount of money that you have in your account at time
t
, where
we measure
t
in years. We will start time at
t
= 0
when the initial amount, called the
principal
, is $100. In other words,
P
(0) = 100
.
In the discussion that follows, we will compute the account balance at the end of
each month. Since one month is 1/12 of a year,
P
(
1
/
12
)
represents the balance at the
end of the first month,
P
(
2
/
12
)
represents the balance at the end of the second month,
etc.
At the end of the first month, interest is added to the account balance. Since the
annual interest rate 5%, the monthly interest rate is 5%/12, or .05/12 in decimal form.
Although we could approximate .05/12 by a decimal, it will be more useful, as well as
more accurate, to leave it in this form. Therefore, at the end of the first month, the
interest earned will be
100(
.
05
/
12)
, so the total amount will be
P
(
1
/
12
) = 100 + 100
.
05
12
= 100
1 +
.
05
12
.
(1)
Now at the end of the second month, you will have the amount that you started that
month with, namely
P
(
1
/
12
)
, plus another month’s worth of interest on that amount.
Therefore, the total amount will be
P
(
2
/
12
) =
P
(
1
/
12
) +
P
(
1
/
12
)
.
05
12
=
P
(
1
/
12
)
1 +
.
05
12
.
(2)
If we replace
P
(
1
/
12
)
in
equation (2)
with the result found in
equation (1)
, then
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