Chapter 3.5, Problem 20E

To determine

Tocomplete:

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

Answer to Problem 20E

The completed table is:

  $\begin{array}{cccc}\begin{array}{l}\text{Initial }\\ \text{Investment}\end{array}& \begin{array}{l}\text{Annual }\\ \text{\% Rate}\end{array}& \begin{array}{l}\text{Time to }\\ \text{Double}\end{array}& \begin{array}{l}\text{Amount After}\\ \text{10 Years}\end{array}\\ 1761.72& 3.5\%& 19.80& \text{\$2500}\end{array}$

Explanation of Solution

Given:

Rate of interest in a continuously compounded account is 3.5% and the amount after 10 years is $2500.

Formula used:

Amount in a continuously compounded account is given by:

  $A=P{e}^{rt}$

Calculation:

Upon substituting the given values in the formula for continuous compound interest, we get:

  $\begin{array}{l}2500=P{e}^{\left(0.035\times 10\right)}\\ \Rightarrow 2500=P{e}^{\left(0.35\right)}\\ \Rightarrow P=\frac{2500}{e^{0.35}}\\ \Rightarrow P=1761.72\end{array}$

We can rewrite the formula as: $A=1761.72e^{0.035t}$

In order to find the doubling time, we will replace A by 3523.44 and solve for t as shown below:

  $\begin{array}{l}\text{3523}\text{.44}=\text{1761}\text{.72}{e}^{0.035t}\\ \Rightarrow e^{0.035t}=2\\ \Rightarrow 0.035t=\ln 2\\ \Rightarrow t=\frac{\ln 2}{0.035}\approx 19.80\end{array}$

Therefore, the complete table is:

  $\begin{array}{cccc}\begin{array}{l}\text{Initial }\\ \text{Investment}\end{array}& \begin{array}{l}\text{Annual }\\ \text{\% Rate}\end{array}& \begin{array}{l}\text{Time to }\\ \text{Double}\end{array}& \begin{array}{l}\text{Amount After}\\ \text{10 Years}\end{array}\\ 1761.72& 3.5\%& 19.80& \text{\$2500}\end{array}$