Chapter 3.2, Problem 32E

Solution

(a).

To calculate: The population in $1930$ and $1945$ .

The population in $1930$ is $6554$ and in $1945$ is $9151$ .

Given information:

The population of River City in the year $1910$ was $4200$ . Assume the population increased at a rate of $2.25\%$ per year.

Calculation:

The model formula:

  $P\left(t\right)=a{b}^t$ , where $a$ is the initial population and $b$ is the rate of growth.

Let $t$ represent the number of years since $1910.$

Here, $a=4200$ and $b=1+2.25\%=1.0225$

Therefore, the equation is $P\left(t\right)=4200\cdot {1.0225}^t$

Now, find the populations in $1930$ and $1945$ by letting $t=20$ and $t=35$ respectively.

  $\begin{array}{l}P\left(20\right)=4200\cdot {1.0225}^{20}\\ \approx 6554\\ P\left(35\right)=4200\cdot {1.0225}^{35}\\ \approx 9151\end{array}$

Hence, the population in $1930$ is $6554$ and in $1945$ is $9151$ .

(b).

To predict: When the population reached $20,000.$ .

The population reached $50,000$ when $t\approx 70.1$ or in early $1980.$

Given information:

The population of River City in the year $1910$ was $4200$ . Assume the population increased at a rate of $2.25\%$ per year.

Calculation:

The model formula:

  $P\left(t\right)=a{b}^t$ , where $a$ is the initial population and $b$ is the rate of growth.

Let $t$ represent the number of years since $1910.$

Here, $a=4200$ and $b=1+2.25\%=1.0225$

Therefore, the equation is $P\left(t\right)=4200\cdot {1.0225}^t$

If the population is $20,000$ , then $P\left(t\right)=20,000$ .

Substitute the values and solve for $t:$

  $\begin{array}{l}20,000=4200\cdot {1.0225}^t\\ {1.0225}^t=\frac{20,000}{4200}\\ {1.0225}^t=4.761905\end{array}$

Taking natural log to both sides:

  $\begin{array}{l}\ln {1.0225}^t=\ln 4.761905\\ t\ln 1.0225=4.761905\\ t=\frac{4.761905}{\ln 1.0225}\\ t\approx 70.1\end{array}$

Hence, the population reached $50,000$ when $t\approx 70.1$ or in early $1980.$