Chapter 3.5, Problem 13E

To determine

Tocomplete:

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

Answer to Problem 13E

The completed table is:

  $\begin{array}{cccc}\begin{array}{l}\text{Initial }\\ \text{Investment}\end{array}& \begin{array}{l}\text{Annual }\\ \text{\% Rate}\end{array}& \begin{array}{l}\text{Time to }\\ \text{Double}\end{array}& \begin{array}{l}\text{Amount After}\\ \text{10 Years}\end{array}\\ \$10,000& 4\%& 17.33& \$14,918.25\end{array}$

Explanation of Solution

Given:

Initial investment in a continuously compounded account is $10,000 and rate of interest is 4%.

Formula used:

Amount in a continuously compounded account is given by:

  $A=P{e}^{rt}$

Calculation:

We have been given the initial investment as $10000 and rate of interest is 4%. That is, we have been given the values: $P=10000,r=0.04$

First of all, we will substitute these values in the formula for compound interest and write a general rule as shown below:

  $A=10000e^{0.04t}$

In order to find the time to double, we will substitute $A=20000$ and solve for t as shown below:

  $\begin{array}{l}20000=10000e^{0.04t}\\ 2=e^{0.04t}\\ 0.04t=\ln 2\\ t=\frac{\ln 2}{0.04}\approx 17.33\end{array}$

In order to find the amount after 10 years, we will substitute $t=10$ and solve for A as shown below:

  $\begin{array}{l}A=10000e^{0.04\left(10\right)}\\ \Rightarrow A=10000e^{0.4}\\ \Rightarrow A=10000\left(\text{1}\text{.4918246}\right)\\ \Rightarrow A=14918.25\end{array}$

Therefore, the complete table is:

  $\begin{array}{cccc}\begin{array}{l}\text{Initial }\\ \text{Investment}\end{array}& \begin{array}{l}\text{Annual }\\ \text{\% Rate}\end{array}& \begin{array}{l}\text{Time to }\\ \text{Double}\end{array}& \begin{array}{l}\text{Amount After}\\ \text{10 Years}\end{array}\\ \$10,000& 4\%& 17.33& \$14,918.25\end{array}$