Supplemental Videos
The main topics of this section are also presented in the following videos:
SectionCompound Growth
SubsectionCompounding Once a Year
If the interest on a savings account is compounded annually, the amount of money in the account grows exponentially.
Consider a principal of $100 invested at 5% interest compounded annually. At the end of \(1\) year, the amount is \begin{align*} \text{Amount} \amp = \text{Principal} + \text{Interest} \\ A \amp = P + Pr\\ \amp = 100 + 100(0.05) = 105. \end{align*} It will be more useful to write the formula for the amount after \(1\) year in factored form. \begin{align*} A \amp = P + Pr \amp\amp \text{Factor out P.} \\ \amp = P(1 + r) \end{align*} With this version of the formula, the calculation for the amount at the end of \(1\) year looks like this: \begin{align*} A \amp = P(1 + r ) \\ \amp = 100(1 + 0.05) \\ \amp = 100(1.05) = \alert{105}. \end{align*}
The amount, $105, becomes the new principal for the second year. To find the amount at the end of the second year, we apply the formula again, with \(P = 105\text{.}\) \begin{align*} A \amp = P(1 + r ) \\ \amp = 105(1 + 0.05) \\ \amp = \alert{105}(1.05) = 110.25. \end{align*} Observe that to find the amount at the end of each year, we multiply the principal by a factor of \(1 + r = 1.05\text{.}\) Thus, we can express the amount at the end of the second year as \begin{align*} A \amp = [100(1.05)](1.05) \\ \amp = 100(1.05)^2 \end{align*} and at the end of the third year as \begin{align*} A \amp = \left[100(1.05)^2\right](1.05) \\ \amp = 100(1.05)^3. \end{align*}
At the end of each year, we multiply the old balance by another factor of \(1.05\) to get the new amount. We organize our results into Table196, where \(A(t)\) represents the amount of money in the account after \(t\) years. For this example, a formula for the amount after \(t\) years is
\begin{equation*}
A(t) = 100(1.05)^t.
\end{equation*}
| \(t\) | \(P(1 + r)^t\) | \(A(t)\) |
| \(0\) | \(100\) | \(100\) |
| \(1\) | \(100(1.05)\) | \(105\) |
| \(2\) | \(100(1.05)^2\) | \(110.25\) |
| \(3\) | \(100(1.05)^3\) | \(115.76\) |
In general, for an initial investment of \(P\) dollars at an interest rate \(r\) compounded annually, we have the following formula for the amount accumulated after \(t\) years.
Annual Interest
The amount, \(A(t)\text{,}\) accumulated (principal plus interest) in an account bearing interest compounded annually is
\begin{equation*}
A(t) = P(1 + r)^t
\end{equation*}
where \begin{align*} \amp P \amp\amp \text{is the principal invested,} \\ \amp r \amp\amp \text{is the effective annual interest rate (as a decimal),} \\ \amp t \amp\amp \text{is the time period, in years,}\\ \amp b=1+r \amp\amp \text{is the growth factor.} \end{align*}
This function describes exponential growth with an initial value of \(P\) and a growth factor of \(b = 1 + r\text{.}\) The interest rate \(r\text{,}\) which indicates the percent increase in the account each year, corresponds to a growth factor of \(1 + r\text{.}\) The notion of percent increase is often used to describe the growth factor for quantities that grow exponentially.
SubsectionCompounding \(n\)-Times a Year
If we have a principal of \(P\) and an interest rate \(r\text{,}\) then by compounding this interest rate \(n\)-times a year we mean the following:
Rather than apply the interest rate \(r\) one time over the course of a year, we instead apply an interest of \(\dfrac{r}{n}\) a total of \(n\) times over the course of the year.
Compound Interest
The amount \(A(t)\) accumulated (principal plus interest) in an account bearing interest compounded \(n\)-times a year is
\begin{equation*}
A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt}
\end{equation*}
where \begin{align*} \amp P \amp\amp \text{is the principal invested,} \\ \amp r \amp\amp \text{is the nominal interest rate,} \\ \amp n \amp\amp \text{is the number of times the interest is compounded,} \\ \amp t \amp\amp \text{is the time period, in years.} \end{align*}
Caution197
The growth factor of
\begin{equation*}
A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt}
\end{equation*}
is not \(1+\dfrac{r}{n}\text{.}\) Recall that when a function is written in the form \(P(b)^t \) the growth factor is \(b\text{.}\) So, we need to rewrite our function in this form. We do this as follows:
\begin{equation*}
A(t)=P\left(1 + \dfrac{r}{n}\right)^{nt} = P\left[\left(1+\dfrac{r}{n}\right)^n\right]^t.
\end{equation*}
If we set
\begin{equation*}
b=\left(1+\dfrac{r}{n}\right)^n,
\end{equation*}
then we have rewritten \(A(t)\) in the form \(P(b)^t\text{,}\) and thus the growth factor is
\begin{equation*}
\left(1+\dfrac{r}{n}\right)^n.
\end{equation*}
If you have a sharp eye, then you've realized that the Annual Interest formula can come from the Compound Inteterest formula! Since annual means once a year, we let \(n=1\) in the Compound Interest formula to get:
\begin{equation*}
A(t) = P\left(1 + \dfrac{r}{1}\right)^{1t} = P\left(1 + r\right)^{t}.
\end{equation*}
So, the nominal interest rate is the same as the annual interest rate if your \(n\) is equal to \(1\text{.}\) In general, these are NOT the same. You will be asked to translate between nominal and annual. The NOTE after Example198 shows how to go from nominal to annual.
Example198
During a period of rapid inflation, prices of a pound of butter rose by \(24\)% compounded twice a year. At the beginning of the inflationary period, a pound of butter cost $\(2\text{.}\)
Write a function that gives the price of a pound of butter \(t\) years after inflation began.
How much did a pound of butter cost after \(3\) years? After \(15\) months?
Solution
-
Looking at our compound interest formula
\begin{equation*} A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt}, \end{equation*}we know that we are compounding twice a year, so \(n=2\text{,}\) and the nominal rate is \(24\)%, so \(r=.24\text{.}\) The problem also tells us the original price of a pound of butter was $\(2\text{,}\) so \(P=2\text{.}\) Using this information, we get the formula
\begin{equation*} A(t) = 2\left(1 + \dfrac{.24}{2}\right)^{2t}= 2(1.12)^{2t}. \end{equation*} To find the price of butter at any time after inflation began, we evaluate the function at the appropriate value of \(t\text{.}\) \begin{align*} P(\alert{3}) \amp = 2(1.12)^{2(\alert{3})} \\ \amp = 2(1.12)^6 \approx 3.95 \end{align*} After \(3\) years, the price was $\(3.95\text{.}\) Fifteen months is \(1.25\) years, so we evaluate \(P(1.25)\text{.}\) \begin{align*} P(\alert{1.25}) \amp = 2(1.12)^{2(\alert{1.25})} \\ \amp = 2(1.12)^{2.5} \approx 2.66 \end{align*} After \(15\) months, the price of butter was $\(2.66\text{.}\)
NOTE: In Example198, we can rewrite the formula for \(P(t)\) as follows: \begin{align*} P(t)\amp = 2(1.12)^{2t} \\ \amp = 2\left[(1.12)^2\right]^t = 2(1.2544)^t. \end{align*} Thus, the annual growth factor for the price of butter is \(1.2544\text{,}\) and the annual percent growth rate, or effective annual rate, is \(25.44\)%.
Example199
In \(1998\text{,}\) the average annual cost of attending a public college was \(\$10,069\text{,}\) and costs were climbing by \(6\)% per year.
Write a formula for \(C(t)\text{,}\) the cost of one year of college \(t\) years after \(1998\text{.}\)
-
Complete the table and sketch a graph of \(C(t)\text{.}\)
\(t\) \(0\) \(5\) \(10\) \(15\) \(20\) \(25\) \(C(t)\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) If the growth rate remained steady, how much did a year of college cost in \(2005\text{?}\)
If the growth rate continues to remain steady, how much will a year of college cost in \(2020\text{?}\)
Solution
\(C(t) = 10,069(1.06)^t\)
-
\(t\) \(0\) \(5\) \(10\) \(15\) \(20\) \(25\) \(C(t)\) \(10,069\) \(13,475\) \(18,032\) \(24,131\) \(32,293\) \(43,215\) 
-
Since \(2005\) is \(7\) years after \(1998\text{,}\)
\begin{equation*} C(7)=10,069(1.06)^7\approx15,140 \end{equation*}So in \(2005\text{,}\) a year of college will cost about \(\$15,140\) per year
-
Since \(2020\) is \(22\)years after \(1998\text{,}\)
\begin{equation*} C(22)=10,069(1.06)^22\approx36,284 \end{equation*}So in \(2020\text{,}\) a year of college will cost about \(\$36284\) per year