Chapter 3.5, Problem 21E

To determine

Tocomplete:

The table for time t, such that the investment P triples when the interest is compounded continuously at rate r. Create a scatter plot.

Answer to Problem 21E

The completed table is:

  $\begin{array}{ccccccc}r& 2\%& 4\%& 6\%& 8\%& 10\%& 12\%\\ t& 54.93& 27.47& 18.31& 13.73& 10.97& 9.16\end{array}$

The scatter plot for this table is shown below:

  

Explanation of Solution

Given:

An incomplete table:

  $\begin{array}{ccccccc}r& 2\%& 4\%& 6\%& 8\%& 10\%& 12\%\\ t& & & & & & \end{array}$

Formula used:

Amount of a continuously compounded account is given by:

  $A=P{e}^{rt}$

Calculation:

We can set up the times for the different interest rates given in the table as shown below:

  $\begin{array}{l}r=0.02\Rightarrow 3P=P{e}^{0.02t}\Rightarrow 0.02t=\ln 3\Rightarrow t=\frac{\ln 3}{0.02}\approx 54.93\\ r=0.04\Rightarrow 3P=P{e}^{0.04t}\Rightarrow 0.04t=\ln 3\Rightarrow t=\frac{\ln 3}{0.04}\approx 27.47\\ r=0.06\Rightarrow 3P=P{e}^{0.06t}\Rightarrow 0.06t=\ln 3\Rightarrow t=\frac{\ln 3}{0.06}\approx 18.31\\ r=0.08\Rightarrow 3P=P{e}^{0.08t}\Rightarrow 0.08t=\ln 3\Rightarrow t=\frac{\ln 3}{0.08}\approx 13.73\\ r=0.10\Rightarrow 3P=P{e}^{0.10t}\Rightarrow 0.10t=\ln 3\Rightarrow t=\frac{\ln 3}{0.10}\approx 10.97\\ r=0.12\Rightarrow 3P=P{e}^{0.12t}\Rightarrow 0.12t=\ln 3\Rightarrow t=\frac{\ln 3}{0.12}\approx 9.16\end{array}$

Therefore, we can write the completed table as shown below:

  $\begin{array}{ccccccc}r& 2\%& 4\%& 6\%& 8\%& 10\%& 12\%\\ t& 54.93& 27.47& 18.31& 13.73& 10.97& 9.16\end{array}$

The scatter plot for this table is shown below: