Chapter 3.5, Problem 32E

a.

To determine

Check whether the population is increasing or decreasing using value of $k$ .

Answer to Problem 32E

Value of $k$ is positive, then the population is increasing.

Explanation of Solution

Given:

The given model for populations is $p=319.2e^{kt}$ , where $t$ represents the year. The population of Tallahassee in $2006$ was about $347,000$ . $t=5$ for year $2005$ .

Calculation:

  $p=319.2e^{kt}$

Substitute the value of $p=347$ and $t=6$ in the equation.

  $\Rightarrow 347=319.2e^{6k}$

  $\begin{array}{l}\Rightarrow \frac{347}{319.2}=\frac{319.2}{319.2}{e}^{6k}\\ \Rightarrow \frac{347}{319.2}=e^{6k}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{347}{319.2}=\ln e^{6k}\\ \Rightarrow 0.0835=6k\\ \Rightarrow \frac{0.0835}{6}=\frac{6}{6}k\\ \Rightarrow k=0.0139\end{array}$

Hence the value of $k$ is positive, then the population is increasing.

b.

To determine

Find the population of Tallahassee in the given years.

Answer to Problem 32E

The answer is reasonable because the population is increasing.

Explanation of Solution

Given:

The given model for populations is $p=319.2e^{kt}$ , where $t$ represents the year. The population of Tallahassee in $2006$ was about $347,000$ . $t=5$ for year $2005$ .

Calculation:

  $p=319.2e^{kt}$

In $2015$ .

Substitute the value of $t=15$ in the equation.

  $\Rightarrow 347=319.2e^{6k}$

  $\begin{array}{l}\Rightarrow p=319.2e^{6k}\\ \Rightarrow \frac{347}{319.2}=e^{6k}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{347}{319.2}=\ln e^{6k}\\ \Rightarrow 0.0835=6k\\ \Rightarrow \frac{0.0835}{6}=\frac{6}{6}k\\ \Rightarrow k=0.0139\end{array}$

Substitute the value of $t=15$ in the equation.

  $\begin{array}{l}p=319.2e^{kt}\\ \Rightarrow p=319.2e^{0.0139\times 15}\\ \Rightarrow p=319.2\cdot 1.2318\\ \Rightarrow p\approx 393.2\end{array}$

Population in $2015$ was $393.2$ thousands.

Substitute the value of $t=20$ in the equation.

  $\begin{array}{l}p=319.2e^{kt}\\ \Rightarrow p=319.2e^{0.0139\times 20}\\ \Rightarrow p=319.2\cdot 1.32\\ \Rightarrow p\approx 421.5\end{array}$

Population in $2015$ was $421.5$ thousands.

Hence from the above results, the answer is reasonable because the population is increasing.

c.

To determine

Find the year for the given population.

Answer to Problem 32E

The population reaches $410,000$ in $2018$ .

Explanation of Solution

Given:

The given model for populations is $p=319.2e^{kt}$ , where $t$ represents the year. The population of Tallahassee is $410,000$ .

Calculation:

  $p=319.2e^{kt}$

Substitute the value of $p=347$ and $t=6$ in the equation.

  $\Rightarrow 347=319.2e^{6k}$

  $\begin{array}{l}\Rightarrow \frac{347}{319.2}=\frac{319.2}{319.2}{e}^{6k}\\ \Rightarrow \frac{347}{319.2}=e^{6k}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{347}{319.2}=\ln e^{6k}\\ \Rightarrow 0.0835=6k\\ \Rightarrow \frac{0.0835}{6}=\frac{6}{6}k\\ \Rightarrow k=0.0139\end{array}$

Substitute the value of $k=410$ in the equation.

  $\Rightarrow 347=319.2e^{6k}$

  $\begin{array}{l}\Rightarrow \frac{410}{319.2}=\frac{319.2}{319.2}{e}^{6k}\\ \Rightarrow \frac{410}{319.2}=e^{0.0139t}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{410}{319.2}=\ln e^{0.0139t}\\ \Rightarrow 0.25=0.0139t\\ \Rightarrow \frac{0.25}{0.0139}=\frac{0.0139}{0.0139}t\\ \Rightarrow t\approx 18\end{array}$

Hence the population reaches $410,000$ in $2018$ .