Chapter 3.5, Problem 32E

a .

To determine

To find: the value of k and also is the population increasing or decreasing.

Answer to Problem 32E

  $k=0.0155$

Population is increasing.

Explanation of Solution

Given:

The population P (in thousands) from 2000 through 2014 can be approximated by the model

  $P=150.9e^{kt}$.........(1)

Where t represents the year, with $t=0$ corresponding to 2000.

Also in the year 2005, population was 163,075.

Formula used:

  $\ln e^x=x$

The year 2005 corresponds to $t=5$ .

The given population model gives the population in thousands Substitute $P=163.075$ and $t=5$ in equation (1), to find value of k.

  $\begin{array}{l}163.075=150.9e^{k\left(5\right)}\\ \frac{163.075}{150.9}=e^{5k}\end{array}$

Take natural logarithms on both sides,

  $\begin{array}{c}\ln e^{5k}=\ln \frac{163.075}{150.9}\\ 5k=\ln \frac{163.075}{150.9}\\ k=\frac{1}{5}\ln \frac{163.075}{150.9}\\ k\approx 0.0155\end{array}$

Substituting $k=0.0155$ in equation (1), the population P is given by,

  $P=150.9e^{0.0155t}$.........(2)

The graph of the equation $P=150.9e^{0.0155t}$ is given below:

By looking at the graph it is clear that the population is increasing.

b.

To determine

To find: the populations in 2020 and 2025.

Answer to Problem 32E

Population in 2020 is nearly 205,741.

Population in 2025 is nearly 222.320

Explanation of Solution

Given:

The population P (in millions) from 2000 through 2014 can be approximated by the model

  $P=150.9e^{0.0155t}$.........(2)

Where t represents the year, with $t=0$ corresponding to 2000.

The population in 2020 is found by substituting $t=20$ in equation (2),

  $\begin{array}{c}P\left(20\right)=150.9e^{0.0155\left(20\right)}\\ =205.741\end{array}$

Thus, population in 2020 is nearly 205,741.

The population in 2025 is found by substituting $t=25$ in equation (2),

  $\begin{array}{c}P\left(25\right)=150.9e^{0.0155\left(25\right)}\\ =222.320\end{array}$

Thus, population in 2025 is nearly 222320.

This answer is reasonable as population as expected is increasing.

c.

To determine

To find: the year when the population will reach 200,000

Answer to Problem 32E

2018

Explanation of Solution

Given:

The population P (in millions) from 2000 through 2014 can be approximated by the model

  $P=150.9e^{0.0155t}$.........(2)

Where t represents the year, with $t=0$ corresponding to 2000.

Formula used:

  $\ln e^x=x$

Calculation:

The given population model gives the population in thousands Substitute $P=200$ in equation (2),

  $\begin{array}{c}200=150.9e^{0.0155t}\\ e^{0.0155t}=\frac{200}{150.9}\end{array}$

Take natural logarithm on both sides,

  $\begin{array}{c}\ln e^{0.0155t}=\ln \left(\frac{200}{150.9}\right)\\ 0.0155t=\ln \left(\frac{200}{150.9}\right)\\ t=\frac{1}{0.0155}\ln \left(\frac{200}{150.9}\right)\\ t\approx 18.17\end{array}$

Now, $t=0$ corresponding to 2000, so it can be said that during 18^th year population will be 200,000.

That is, in the year 2018 population will be 200,000.