Chapter 2
Models of Growth: Rates of Change
2.4 Exponential Functions
2.4.1 Interest Rates and Exponential Growth
When we take up population growth in Section 2.5, we will study functions \(P\) of time \(t\) whose rate of change is proportional to \(P\) of time \(t\) itself, i.e., functions such that
for some \(k\), where \(k\) is a proportionality constant. In this section we investigate what sort of functions have this property. We begin by considering another context in which the rate of growth is proportional to the amount present.
Your savings account grows — because interest is added — at a rate proportional to the amount present in your account.
Activity 1 Use
our calculator or yours to carry out the following exploration.
Suppose you start with \(\$300\) in an account that pays interest once a year (i.e., interest is “compounded annually”) at a rate of \(1.1\)%, and you do not make any additional deposits or withdraw any of the money.
- Find the balance at \(1, 2, 3, 4,\) and \(5\) years.
- Find the rate of growth for each of those
years.
- For each of those years, find the ratio of
the rate of growth to the balance at the start of the year.
We now repeat the calculation from Activity 1 in a different way. At the end of the first year, your new balance will be the starting amount plus interest. That is, your balance will be
\(\$ 300 + \$ 300 \times 0.011 = \$ 300 \times 1.011 = \$ 303.30\).
At the end of the second year, you will have
\(\$ 303.30 + \$ 303.30 \times 0.011 = \$ 303.30 \times 1.011 = \$ 300 \times {1.011} ^2 \).
(Check that this last expression gives \(\$306.64\), as you found in Exploration Activity 1 by adding.) At the end of the third year, you will have
\(\$ 300 \times {1.011}^3 \)
or \(\$310.01\). In general, after \(n\) years you will have
\(\$ 300 \times {1.011}^n \).
Notice how the repeated addition of interest also can be calculated
by repeated multiplication
by (\(1 +\) the interest rate).
If you start with \(A\) dollars instead of \(\$300\), then the amount you would have after \(n\) years is
dollars. If the percentage interest rate is \(r\)% rather than 1.1%, then the decimal interest rate is
and the amount you have after n years is
dollars. In this formula, \(A\) is the initial amount invested, \(k\) is the interest rate as a decimal fraction, and \(n\) is the number of years of the investment. Notice that time (represented by \(n\)) appears in the exponent in the expression \(A(1 + k)^n\).
