Chapter 4.1, Problem 39PS

Solution

a.

To write: A model giving the population $P$ (in thousands) of Austin $t$ years 1990.

  $P=494.29{\left(1.03\right)}^t$ .

The population in 2000 is 664.284 thousands.

Given:

The population of Austin in 1990 is 494,290.

The population is increased by 3% per year for the next 10 years.

Calculation:

Here, the initial population is

  $a=494.29$ thousands.

The growth rate is,

  $\begin{array}{c}r=\frac{3}{100}\\ =0.03\end{array}$

Now the model is,

  $\begin{array}{c}P=a{\left(1+r\right)}^t\\ P=494.29{\left(1+0.03\right)}^t\\ P=494.29{\left(1.03\right)}^t\end{array}$

Now since 10 years after 1990 is 2000, therefore to find the population in 2000, put $t=10$ in the above model and simplify. That is,

  $\begin{array}{c}P=494.29{\left(1.03\right)}^{10}\\ =\text{664}\text{.284}\end{array}$

Hence, the population in 2000 is 664.284 thousands.

b.

To find: The given model and to estimate when the value of the car will be $10,000.

  

The domain is $\left[0,10\right]$ , and the range is $\left[494.29,664.284\right]$ .

Given:

  $P=494.29{\left(1.03\right)}^t$

Calculation:

The given function passes through the points $\left(0,494.29\right),\left(1,\text{509}\text{.1187}\right)$ .

Now the graph is,

  

From the graph, it can be seen that the time is ranging from $t=0$ (1990) to $t=10$ (2000).

So, the domain is $\left[0,10\right]$ .

The population is ranging from 494.29 thousands to 664.284 thousands.

So, the range is $\left[494.29,664.284\right]$ .

c.

To find: The year in which the population is 590,000.

The population is 590,000 in the year 1996.

Given:

  $P=494.29{\left(1.03\right)}^t$

Calculation:

By substituting $P=590$ in $P=494.29{\left(1.03\right)}^t$ , it follows that

  $\begin{array}{c}590=494.29{\left(1.03\right)}^t\\ {\left(1.03\right)}^t=\text{1}\text{.1936}\\ t={\log }_{1.03}\left(1.1936\right)\\ t=5.988\\ t\approx 6\end{array}$

Therefore, the population is 590,000 in about 6 years after 1990.

Hence, the population is 590,000 in the year 1996.