Chapter 3.5, Problem 17E

To determine

Tocomplete:

The given table for a savings account which is compounded continuously.

Answer to Problem 17E

The completed table is:

  $\begin{array}{cccc}\begin{array}{l}\text{Initial }\\ \text{Investment}\end{array}& \begin{array}{l}\text{Annual }\\ \text{\% Rate}\end{array}& \begin{array}{l}\text{Time to }\\ \text{Double}\end{array}& \begin{array}{l}\text{Amount After}\\ \text{10 Years}\end{array}\\ \$5000& 1.25\%& 55.45& \text{\$5665}\text{.74}\end{array}$

Explanation of Solution

Given:

Initial investment in a continuously compounded account is $5000 and after 10 years it becomes $5665.74.

Formula used:

Amount in a continuously compounded account is given by:

  $A=P{e}^{rt}$

Calculation:

We have been given the initial investment as $5000 and that it becomes $5665.74.Upon substituting these values in the formula for continuously compounded interest, we get:

  $\begin{array}{l}5665.74=5000e^{r\left(10\right)}\\ \Rightarrow e^{10r}=\frac{5665.74}{5000}=\text{1}\text{.133148}\\ \Rightarrow 10r=\ln \left(\text{1}\text{.133148}\right)\\ \Rightarrow r=\frac{\text{0}\text{.1249996}}{10}=0.01249994\\ \Rightarrow r=1.25\%\end{array}$

We can rewrite the formula as: $A=5000e^{0.0125t}$

In order to find the doubling time, we will replace A by 10000 and solve for t as shown below:

  $\begin{array}{l}A=5000e^{0.0125t}\\ \Rightarrow 10000=5000e^{0.0125t}\\ \Rightarrow e^{0.0125t}=\frac{10000}{5000}=2\\ \Rightarrow 0.0125t=\ln 2\\ \Rightarrow t=\frac{\ln 2}{0.0125}\approx 55.45\end{array}$

Therefore, the complete table is:

  $\begin{array}{cccc}\begin{array}{l}\text{Initial }\\ \text{Investment}\end{array}& \begin{array}{l}\text{Annual }\\ \text{\% Rate}\end{array}& \begin{array}{l}\text{Time to }\\ \text{Double}\end{array}& \begin{array}{l}\text{Amount After}\\ \text{10 Years}\end{array}\\ \$5000& 1.25\%& 55.45& \text{\$5665}\text{.74}\end{array}$