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## SectionCompound Growth

###### Supplemental Videos

The main topics of this section are also presented in the following videos:

### 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 $$2\text{.}$ 1. Write a function that gives the price of a pound of butter $t$ years after inflation began. 2. How much did a pound of butter cost after $3$ years? After $15$ months? Solution 1. Looking at our compound interest formula \begin{equation*} A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt}, \end{equation*} we know that we are compounding twice a year, so $n=2\text{,}$ and the nominal rate is $24$%, so $r=.24\text{.}$ The problem also tells us the original price of a pound of butter was$$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*}
2. 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.

1. Write a formula for $C(t)\text{,}$ the cost of one year of college $t$ years after $1998\text{.}$

2. Complete the table and sketch a graph of $C(t)\text{.}$

 $t$ $0$ $5$ $10$ $15$ $20$ $25$ $C(t)$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$

3. If the growth rate remained steady, how much did a year of college cost in $2005\text{?}$

4. If the growth rate continues to remain steady, how much will a year of college cost in $2020\text{?}$

Solution
1. $C(t) = 10,069(1.06)^t$

2.  $t$ $0$ $5$ $10$ $15$ $20$ $25$ $C(t)$ $10,069$ $13,475$ $18,032$ $24,131$ $32,293$ $43,215$
3. 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

4. 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