Chapter 13, Problem 31CCSR

Solution

a.

To find: The rate of interest.

The rate of interest is 4%.

Given:

Years	0	1	2	3
Value	$750.00	$780.61	$812.47	$845.62

The investment is compounded continuously.

Concept used:

The amount accumulated in $t$ years at an interest rate of $r\%$ compounded continuously is $A=P{e}^{rt}$ , where $P$ is the principal amount.

Calculation:

Here, the principal amount is $750.00.

Let $r$ be the rate of interest.

Now, the amount after 1 year is

  $\begin{array}{c}780.61=750e^{r\cdot 1}\\ e^r=\frac{780.61}{750}\\ r=\ln \left(\frac{780.61}{750}\right)\\ r=0.04\\ r=4\%\end{array}$

b.

To find: An equation that models the growth of an investment.

  $A=750e^{0.04t}$ .

Given:

Years	0	1	2	3
Value	$750.00	$780.61	$812.47	$845.62

The investment is compounded continuously.

Concept used:

The amount accumulated in $t$ years at an interest rate of $r\%$ compounded continuously is $A=P{e}^{rt}$ , where $P$ is the principal amount.

Calculation:

Here,

  $P=750,r=0.04$

Now, the equation is

  $\begin{array}{l}A=P{e}^{rt}\\ A=750e^{0.04t}\end{array}$

c.

To find: The year in which the investment is more than $1000.

The investment worth more than $1000 in 7 years.

Given:

Years	0	1	2	3
Value	$750.00	$780.61	$812.47	$845.62

The investment is compounded continuously.

Concept used:

The amount accumulated in $t$ years at an interest rate of $r\%$ compounded continuously is $A=P{e}^{rt}$ , where $P$ is the principal amount.

Calculation:

From subpart (b),

  $A=750e^{0.04t}$

Now since the investment must be worth more than $1000, therefore

  $\begin{array}{c}750e^{0.04t}>1000\\ e^{0.04t}>1.3333\\ 0.04t>\ln \left(1.3333\right)\\ t>\frac{\ln \left(1.3333\right)}{0.04}\\ t>7.19\end{array}$

So, the investment worth more than $1000 in 7 years.