Chapter 3.5, Problem 42E

a.

To determine

Find the population of the city in the given years.

Answer to Problem 42E

The populations of city in $2000$ , $2005$ and $2010$ are $2430.29$ thousands, $2378.5$ thousands and $2315.27$ thousands respectively.

Explanation of Solution

Given:

The populations of city from $2000$ through $2010$ are modeled by $p=\frac{2632}{1+0.083e^{0.050t}}$ .

Calculation:

  $p=\frac{2632}{1+0.083e^{0.050t}}$

The number of population in $2000$ .

  $t=2000-2000=0$

Substitute the value of $t=0$ in the equation.

  $\begin{array}{l}\Rightarrow p=\frac{2632}{1+0.083e^{0.050t}}\\ =\frac{2632}{1+0.083e^{0.050\cdot 0}}\\ =\frac{2632}{1+0.083}\\ =\frac{2632}{1.083}\\ =2430.286\\ \approx 2430.29\end{array}$

The number of population in $2005$ .

  $t=2005-2000=5$

Substitute the value of $t=5$ in the equation.

  $\begin{array}{l}\Rightarrow p=\frac{2632}{1+0.083e^{0.050t}}\\ =\frac{2632}{1+0.083e^{0.050\cdot 5}}\\ =\frac{2632}{1+0.10657}\\ =\frac{2632}{1.10657}\\ =2378.5\end{array}$

The number of population in $2010$ .

  $t=2010-2000=10$

Substitute the value of $t=10$ in the equation.

  $\begin{array}{l}\Rightarrow p=\frac{2632}{1+0.083e^{0.050t}}\\ =\frac{2632}{1+0.083e^{0.050\cdot 10}}\\ =\frac{2632}{1+0.1368}\\ =\frac{2632}{1.1368}\\ =2315.27\end{array}$

Hence the populations of city in $2000$ , $2005$ and $2010$ are $2430.29$ thousands, $2378.5$ thousands and $2315.27$ thousands respectively.

b.

To determine

Draw a graph of the given function using graphing utility.

Explanation of Solution

Given:

The populations of city from $2000$ through $2010$ are modeled by $p=\frac{2632}{1+0.083e^{0.050t}}$ .

.Calculation:

The given function is $p=\frac{2632}{1+0.083e^{0.050t}}$ .

By the use of graphing utility, the graph of the function is given below.

  

c.

To determine

Find the year in which the populations of city will reachedat the given number.

Answer to Problem 42E

The populations of city will reach $2.2$ million in

Explanation of Solution

Given:

The populations of city from $2000$ through $2010$ are modeled by $p=\frac{2632}{1+0.083e^{0.050t}}$ .

.Calculation:

The given function is $p=\frac{2632}{1+0.083e^{0.050t}}$ .

By the use of graphing utility, the graph of the function is given below.

  

From the graph, the populations of city will reach $2.2$ million in $17.223\approx 17$ years.

  $2000+17=2017$

Hence the populations of city will reach $2.2$ million in the year $2017$ .

d.

To determine

Find the year in which the populations of city will reached at the given number.

Answer to Problem 42E

The populations of city will reach $2.2$ million in the year $2017$ .

Explanation of Solution

Given:

The populations of city from $2000$ through $2010$ are modeled by $p=\frac{2632}{1+0.083e^{0.050t}}$ and the given number of populations are $2.2$ millions.

.Calculation:

The given function is $p=\frac{2632}{1+0.083e^{0.050t}}$ .

Substitute the value of $p=2200$ in the function.

  $2200=\frac{2632}{1+0.083e^{0.050t}}$

Multiply both sides by $1+0.083e^{0.050t}$ .

  $\begin{array}{l}\Rightarrow 2200\left(1+0.083e^{0.050t}\right)=\frac{2632}{\left(1+0.083e^{0.050t}\right)}\cdot \left(1+0.083e^{0.050t}\right)\\ \Rightarrow 2200\left(1+0.083e^{0.050t}\right)=2632\end{array}$

Substitute both sides by $2200$ .

  $\begin{array}{l}\Rightarrow \frac{2200}{2200}\left(1+0.083e^{0.050t}\right)=\frac{2632}{2200}\\ \Rightarrow 1+0.083e^{0.050t}=\frac{2632}{2200}\end{array}$

Subtract $1$ from both sides.

  $\begin{array}{l}\Rightarrow 1+0.083e^{0.050t}-1=\frac{2632}{2200}-1\\ \Rightarrow 0.083e^{0.050t}=\frac{2632-2200}{2200}\\ \Rightarrow 0.083e^{0.050t}=\frac{432}{2200}\end{array}$

Divide both sides by $0.083$ .

  $\begin{array}{l}\Rightarrow \frac{0.083}{0.083}{e}^{0.050t}=\frac{432}{2200\cdot 0.083}\\ \Rightarrow e^{0.050t}=\frac{432}{182.6}\end{array}$

Taking natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln e^{0.050t}=\ln \frac{432}{182.6}\\ \Rightarrow 0.050t=0.8611276\end{array}$

Divided both sides by $0.050$ .

  $\begin{array}{l}\Rightarrow \frac{0.050}{0.050}t=\frac{0.8611276}{0.050}\\ \Rightarrow t=\frac{0.8611276}{0.050}\\ \Rightarrow t=17.22\\ \Rightarrow t\approx 17\end{array}$

Now,

  $2000+17=2017$

Hence the populations of city will reach $2.2$ million in the year $2017$ .