Chapter 3.2, Problem 111E

a.

To determine

To complete:

The given table and interpret your results.

Answer to Problem 111E

The complete table would be:

K	1	2	4	6	8	10	12
t	0	19.80	39.61	51.19	59.41	65.79	71.00

Explanation of Solution

Given:

A principalP , invested at $3\frac{1}{2}\%$ and compounded continuously, increases to an amount K times the original principal after t years, where $T=\frac{\ln (K)}{0.035}$.

K	1	2	4	6	8	10	12
t

Calculation:

To complete the given table, we will substitute the values of K in our given function as shown below:

  $T=\frac{\ln (K)}{0.035}=\frac{\ln (1)}{0.035}=\frac{0}{0.035}=0$

  $T=\frac{\ln (K)}{0.035}=\frac{\ln (2)}{0.035}=\frac{\text{0}\text{.6931471805599}}{0.035}=\text{19}\text{.8042051588}\approx 19.80$

It will take approximately 19.80 years for the investment to double.

  $T=\frac{\ln (K)}{0.035}=\frac{\ln (4)}{0.035}=\frac{1.38629436112}{0.035}=39.6084103177\approx 39.61$

It will take approximately 39.61 years for the investment to be 4 times original amount.

  $T=\frac{\ln (K)}{0.035}=\frac{\ln (6)}{0.035}=\frac{1.79175946923}{0.035}=51.1931276922\approx 51.19$

It will take approximately 51.19 years for the investment to be 6 times original amount.

  $T=\frac{\ln (K)}{0.035}=\frac{\ln (8)}{0.035}=\frac{2.07944154168}{0.035}=59.4126154766\approx 59.41$

It will take approximately 59.41 years for the investment to be 8 times original amount.

  $T=\frac{\ln (K)}{0.035}=\frac{\ln (10)}{0.035}=\frac{2.30258509299}{0.035}=65.7881455141\approx 65.79$

It will take approximately 65.79 years for the investment to be 10 times original amount.

  $T=\frac{\ln (K)}{0.035}=\frac{\ln (12)}{0.035}=\frac{2.48490664979}{0.035}=70.9973328511\approx 71.00$

It will take approximately 71.00 years for the investment to be 12 times original amount.

Therefore, the complete table would be:

K	1	2	4	6	8	10	12
t	0	19.80	39.61	51.19	59.41	65.79	71.00

b.

To determine

To graph:

The given function using graphing utility.

Answer to Problem 111E

The graph of our given function would be:

  

Explanation of Solution

Given:

A principal P , invested at $3\frac{1}{2}\%$ and compounded continuously, increases to an amount K times the original principal after t years, where $T=\frac{\ln (K)}{0.035}$.

Calculation:

Upon graphing our given function, we will get our required graph as shown below: