Chapter 4.6, Problem 9E

a.

To determine

Find an exponential model for the population.

Answer to Problem 9E

  $\boxed{n\left(t\right)={112000.2}^{t/18}}$

Explanation of Solution

Given information:

The population of a certain city was $112,000$ in $2006$ , and the observed doubling time for the population is $18$ years.

Find an exponential model $n\left(t\right)=n_0{.2}^{t/a}$ for the population $t$ years after $2006$ .

Calculation:

Consider the population of a certain city was $112,000$ in $2006$ ,

Doubling time = $18$ years.

Exponentially growth model,

  $n\left(t\right)=n_0{2}^{t/a}$

Initial population size, $n_0=112000$

  $r=18$

Solve for population in $t$ years,

  $\begin{array}{l}n\left(t\right)=n_0{.2}^{t/a}\\ \boxed{n\left(t\right)={112000.2}^{t/18}}\end{array}$

The projected population is $\boxed{n\left(t\right)={112000.2}^{t/18}}$

Hence, the solution is, $\boxed{n\left(t\right)={112000.2}^{t/18}}$

b.

To determine

Find an exponential model for the population.

Answer to Problem 9E

  $n\left(t\right)=\boxed{112000e^{0.0385t}}$

Explanation of Solution

Given information:

The population of a certain city was $112,000$ in $2006$ , and the observed doubling time for the population is $18$ years.

Find an exponential model $n\left(t\right)=n_0.e^{rt}$ for the population $t$ years after $2006$ .

Calculation:

Consider the population of a certain city was $112,000$ in $2006$ ,

Doubling time = $18$ years.

Exponentially growth model,

  $\begin{array}{l}n\left(t\right)=n_0.e^{rt}\\ 224000=112000e^{18r}\\ 2=e^{18r}\end{array}$

  $\begin{array}{l}\ln 2=\ln e^{18r}\\ 0.69=18r\\ r=.0385\\ n\left(t\right)=n_0.e^{rt}\\ =112000e^{0.0385t}\end{array}$

Hence, the solution is, $n\left(t\right)=\boxed{112000e^{0.0385t}}$

c.

To determine

Find an exponential model for the population.

Answer to Problem 9E

  

Explanation of Solution

Given information:

The population of a certain city was $112,000$ in $2006$ , and the observed doubling time for the population is $18$ years.

Sketch a graph of the population at time $t$ .

Calculation:

The graph is as shown,

  

Hence, the solution is a graph.

d.

To determine

Find an exponential model for the population.

Answer to Problem 9E

  $\boxed{t=38.85t}$

Explanation of Solution

Given information:

The population of a certain city was $112,000$ in $2006$ , and the observed doubling time for the population is $18$ years.

Estimate when the population will reach $500,000$ .

Calculation:

Consider the population will reach $500,000$ .

Put, $n\left(t\right)=500,000$ and solve for $t$ .

  $\begin{array}{l}500,000=112000e^{0.0385t}\\ 4.46=e^{0.0385t}\\ \ln 4.46=\ln e^{0.0385t}\\ 1.5=0.0385t\\ \boxed{t=38.85t}\end{array}$

Hence, the solution is, $\boxed{t=38.85t}$