Chapter 3.5, Problem 29E

a .

To determine

To complete: the table given below:

Year	Population
1980
1990
2000
2010

Answer to Problem 29E

Year	Population
1980	104,752
1990	143,251
2000	195,899
2010	267,896

Explanation of Solution

Given:

The population P (in thousands) is given by the model:

  $P=76.6e^{0.0313t}$

Where t represents the year, with $t=1$ corresponding to 1971.

Calculation:

Consider the given population model:

  $P=76.6e^{0.0313t}$ ….. (1)

Now, as $t=1$ corresponding to 1971, so $t=10$ corresponds to 1980, $t=20$ corresponding to 1990, so $t=30$ corresponds to 2000, $t=40$ corresponds to 2010.

The population in year 1980:

Substitute $t=10$ in equation (1),

  $\begin{array}{c}P=76.6e^{0.0313\times 10}\\ \approx 104.752\end{array}$

The population in year 1990:

Substitute $t=20$ in equation (1),

  $\begin{array}{c}P=76.6e^{0.0313\times 20}\\ \approx 143.251\end{array}$

The population in year 2000:

Substitute $t=30$ in equation (1),

  $\begin{array}{c}P=76.6e^{0.0313\times 30}\\ \approx 195.899\end{array}$

The population in year 2010:

Substitute $t=40$ in equation (1),

  $\begin{array}{c}P=76.6e^{0.0313\times 20}\\ \approx 267.896\end{array}$

Since, the given population model gives the population in thousands, so the actual population is obtained by multiplying the value obtained for that year by 1000. Thus, the complete table is:

Year	Population
1980	104,752
1990	143,251
2000	195,899
2010	267,896

b.

To determine

To find: the year when the population will reach 360,000

Answer to Problem 29E

2019

Explanation of Solution

Given:

The population P (in thousands) is given by the model:

  $P=76.6e^{0.0313t}$ ….. (1)

Where t represents the year, with $t=1$ corresponding to 1971.

Formula used:

  $\ln e^x=x$

Calculation:

The given population model gives the population in thousands Substitute $P=360$ in equation (1),

  $\begin{array}{c}360=76.6e^{0.0313t}\\ e^{0.0313t}=\frac{360}{76.6}\end{array}$

Take natural logarithm on both sides,

  $\begin{array}{c}\ln e^{0.0313t}=\ln \left(\frac{360}{76.6}\right)\\ 0.0313t=\ln \left(\frac{360}{76.6}\right)\\ t=\frac{1}{0.0313}\ln \left(\frac{360}{76.6}\right)\\ t\approx 49.44\end{array}$

Now, $t=1$ corresponding to 1971, so it can be said that during 49^th year population will be 360,000.

That is, in the year 2019 population will be 360,000.

c.

To determine

To explain: is the model valid for long-term predictions of the population

Answer to Problem 29E

No

Explanation of Solution

Observe the table obtained in part (a),

Year	Population
1980	104,752
1990	143,251
2000	195,899
2010	267,896

It can be said that according to this population model, as the time t increases, the population is increasing very rapidy without any bound.