Chapter 3.5, Problem 30E

a.

To determine

Find the exponential growth or decay model and the population of each country.

Answer to Problem 30E

Bulgaria

  $y=7.64e^{-0.0073t}$

Population in $2030$ is $6.137$ million.

Canada

  $y=31.4e^{0.0074t}$

Population in $2030$ is $39.2$ million.

China

  $y=1278.9e^{0.004t}$

Population in $2030$ is $1441.96$ million.

United Kingdom

  $y=58.9e^{0.0055t}$

Population in $2030$ is $69.47$ million.

United States

  $y=282e^{0.0096t}$

Population in $2030$ is $376.12$ million.

Explanation of Solution

Given:

The table shows the mid-year populations of five countries inthe given years.

  

Calculation:

Write the equations for Bulgaria.

  $7.1=a{e}^{10b}---\left(i\right)$

  $6.6=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{6.6}{7.1}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{6.6}{7.1}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{6.6}{7.1}=\ln e^{10b}\\ \Rightarrow -0.073=10b\\ \Rightarrow \frac{-0.073}{10}=\frac{10}{10}b\\ \Rightarrow b=-0.0073\end{array}$

Substitute the value of $b=-0.0073$ in equation $\left(i\right)$ .

  $\begin{array}{l}7.1=a{e}^{10b}\\ \Rightarrow 7.1=a{e}^{10\cdot \left(-0.0073\right)}\\ \Rightarrow 7.1=a\cdot 0.9296\\ \Rightarrow \frac{7.1}{0.9296}=a\cdot \frac{0.9296}{0.9296}\\ \Rightarrow 7.637=a\end{array}$

Now the model is $y=7.64e^{-0.0073t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=7.64e^{-0.0073\cdot 30}\\ \Rightarrow y=7.64\cdot 0.8\\ \Rightarrow y=6.137\end{array}$

Hence the model for Bulgaria is $y=7.64e^{-0.0073t}$ and the population in $2030$ is $6.137$ million.

Write the equations for Canada.

  $33.8=a{e}^{10b}---\left(i\right)$

  $36.4=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{36.4}{33.8}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{36.4}{33.8}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{36.4}{33.8}=\ln e^{10b}\\ \Rightarrow 0.0741=10b\\ \Rightarrow \frac{0.0741}{10}=\frac{10}{10}b\\ \Rightarrow b=0.00741\end{array}$

Substitute the value of $b=0.0074$ in equation $\left(i\right)$ .

  $\begin{array}{l}33.8=a{e}^{10b}\\ \Rightarrow 33.8=a{e}^{10\cdot \left(0.0074\right)}\\ \Rightarrow 33.8=a\cdot 1.0768\\ \Rightarrow \frac{33.8}{1.0768}=a\cdot \frac{1.0768}{1.0768}\\ \Rightarrow 31.389=a\end{array}$

Now the model is $y=31.4e^{0.0074t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=31.4e^{0.0074\cdot 30}\\ \Rightarrow y=31.4\cdot 1.24857\\ \Rightarrow y=39.2\end{array}$

Hence the model for Canada is $y=31.4e^{0.0074t}$ and the population in $2030$ is $39.2$ million.

Write the equations for China.

  $1330.1=a{e}^{10b}---\left(i\right)$

  $1384.5=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{1384.5}{1330.1}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{1384.5}{1330.1}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{1384.5}{1330.1}=\ln e^{10b}\\ \Rightarrow 0.04=10b\\ \Rightarrow \frac{0.04}{10}=\frac{10}{10}b\\ \Rightarrow b=0.004\end{array}$

Substitute the value of $b=0.004$ in equation $\left(i\right)$ .

  $\begin{array}{l}1330.1=a{e}^{10b}\\ \Rightarrow 1330.1=a{e}^{10\cdot \left(0.004\right)}\\ \Rightarrow 1330.1=a\cdot 1.04\\ \Rightarrow \frac{1330.1}{1.04}=a\cdot \frac{1.04}{1.04}\\ \Rightarrow 1278.9=a\end{array}$

Now the model is $y=1278.9e^{0.004t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=1278.9e^{0.004\cdot 30}\\ \Rightarrow y=1278.9\cdot 1.127\\ \Rightarrow y\approx 1441.96\end{array}$

Hence the model for china is $y=1278.9e^{0.004t}$ and the population in $2030$ is $1441.96$ million.

Write the equations for United Kingdom.

  $62.3=a{e}^{10b}---\left(i\right)$

  $65.8=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{65.8}{62.3}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{65.8}{62.3}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{65.8}{62.3}=\ln e^{10b}\\ \Rightarrow 0.055=10b\\ \Rightarrow \frac{0.055}{10}=\frac{10}{10}b\\ \Rightarrow b=0.0055\end{array}$

Substitute the value of $b=0.0055$ in equation $\left(i\right)$ .

  $\begin{array}{l}62.3=a{e}^{10b}\\ \Rightarrow 62.3=a{e}^{10\cdot \left(0.0055\right)}\\ \Rightarrow 62.3=a\cdot 1.057\\ \Rightarrow \frac{62.3}{1.057}=a\cdot \frac{1.057}{1.057}\\ \Rightarrow 58.9=a\end{array}$

Now the model is $y=58.9e^{0.0055t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=58.9e^{0.0055\cdot 30}\\ \Rightarrow y=58.9\cdot 1.179\\ \Rightarrow y\approx 69.47\end{array}$

Hence the model for United Kingdom is $y=58.9e^{0.0055t}$ and the population in $2030$ is $69.47$ million.

Write the equations for United States.

  $310.2=a{e}^{10b}---\left(i\right)$

  $341.4=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{341.4}{310.2}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{341.4}{310.2}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{341.4}{310.2}=\ln e^{10b}\\ \Rightarrow 0.0958=10b\\ \Rightarrow \frac{0.0958}{10}=\frac{10}{10}b\\ \Rightarrow b\approx 0.0096\end{array}$

Substitute the value of $b\approx 0.0096$ in equation $\left(i\right)$ .

  $\begin{array}{l}310.2=a{e}^{10b}\\ \Rightarrow 310.2=a{e}^{10\cdot \left(0.0096\right)}\\ \Rightarrow 310.2=a\cdot 1.1\\ \Rightarrow \frac{310.2}{1.1}=a\cdot \frac{1.1}{1.1}\\ \Rightarrow 282=a\end{array}$

Now the model is $y=282e^{0.0096t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=282e^{0.0096\cdot 30}\\ \Rightarrow y=282\cdot 1.334\\ \Rightarrow y\approx 376.12\end{array}$

Hence the model for United States is $y=282e^{0.0096t}$ and the population in $2030$ is $376.12$ million.

b.

To determine

Which constant defines the growth rate in the equation.

Answer to Problem 30E

The value of ‘ $b$ ’ defines the growth rate.

Explanation of Solution

Given:

The table shows the mid-year populations of five countries in the given years.

  

Calculation:

Write the equations for United Kingdom.

  $62.3=a{e}^{10b}---\left(i\right)$

  $65.8=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{65.8}{62.3}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{65.8}{62.3}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{65.8}{62.3}=\ln e^{10b}\\ \Rightarrow 0.055=10b\\ \Rightarrow \frac{0.055}{10}=\frac{10}{10}b\\ \Rightarrow b=0.0055\end{array}$

Substitute the value of $b=0.0055$ in equation $\left(i\right)$ .

  $\begin{array}{l}62.3=a{e}^{10b}\\ \Rightarrow 62.3=a{e}^{10\cdot \left(0.0055\right)}\\ \Rightarrow 62.3=a\cdot 1.057\\ \Rightarrow \frac{62.3}{1.057}=a\cdot \frac{1.057}{1.057}\\ \Rightarrow 58.9=a\end{array}$

Now the model is $y=58.9e^{0.0055t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=58.9e^{0.0055\cdot 30}\\ \Rightarrow y=58.9\cdot 1.179\\ \Rightarrow y\approx 69.47\end{array}$

Write the equations for United States.

  $310.2=a{e}^{10b}---\left(i\right)$

  $341.4=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{341.4}{310.2}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{341.4}{310.2}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{341.4}{310.2}=\ln e^{10b}\\ \Rightarrow 0.0958=10b\\ \Rightarrow \frac{0.0958}{10}=\frac{10}{10}b\\ \Rightarrow b\approx 0.0096\end{array}$

Substitute the value of $b\approx 0.0096$ in equation $\left(i\right)$ .

  $\begin{array}{l}310.2=a{e}^{10b}\\ \Rightarrow 310.2=a{e}^{10\cdot \left(0.0096\right)}\\ \Rightarrow 310.2=a\cdot 1.1\\ \Rightarrow \frac{310.2}{1.1}=a\cdot \frac{1.1}{1.1}\\ \Rightarrow 282=a\end{array}$

Now the model is $y=282e^{0.0096t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=282e^{0.0096\cdot 30}\\ \Rightarrow y=282\cdot 1.334\\ \Rightarrow y\approx 376.12\end{array}$

The constant ‘ $b$ ’ defines the growth rate in the equation. Large value of $b$ means higher growth rate and lower value of $b$ means lower growth rate.

Hence the value of ‘ $b$ ’ defines the growth rate.

c.

To determine

Which constant defines the increasingor decreasing process in the equation.

Answer to Problem 30E

The value of ‘ $b$ ’ defines the increasing or decreasing process.

Explanation of Solution

Given:

The table shows the mid-year populations of five countries in the given years.

  

Calculation:

Write the equations for Bulgaria.

  $7.1=a{e}^{10b}---\left(i\right)$

  $6.6=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{6.6}{7.1}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{6.6}{7.1}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{6.6}{7.1}=\ln e^{10b}\\ \Rightarrow -0.073=10b\\ \Rightarrow \frac{-0.073}{10}=\frac{10}{10}b\\ \Rightarrow b=-0.0073\end{array}$

Substitute the value of $b=-0.0073$ in equation $\left(i\right)$ .

  $\begin{array}{l}7.1=a{e}^{10b}\\ \Rightarrow 7.1=a{e}^{10\cdot \left(-0.0073\right)}\\ \Rightarrow 7.1=a\cdot 0.9296\\ \Rightarrow \frac{7.1}{0.9296}=a\cdot \frac{0.9296}{0.9296}\\ \Rightarrow 7.637=a\end{array}$

Now the model is $y=7.64e^{-0.0073t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=7.64e^{-0.0073\cdot 30}\\ \Rightarrow y=7.64\cdot 0.8\\ \Rightarrow y=6.137\end{array}$

Write the equations for China.

  $1330.1=a{e}^{10b}---\left(i\right)$

  $1384.5=a{e}^{20b}---\left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{1384.5}{1330.1}=\frac{a{e}^{20b}}{a{e}^{10b}}\\ \Rightarrow \frac{1384.5}{1330.1}=e^{10b}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{1384.5}{1330.1}=\ln e^{10b}\\ \Rightarrow 0.04=10b\\ \Rightarrow \frac{0.04}{10}=\frac{10}{10}b\\ \Rightarrow b=0.004\end{array}$

Substitute the value of $b=0.004$ in equation $\left(i\right)$ .

  $\begin{array}{l}1330.1=a{e}^{10b}\\ \Rightarrow 1330.1=a{e}^{10\cdot \left(0.004\right)}\\ \Rightarrow 1330.1=a\cdot 1.04\\ \Rightarrow \frac{1330.1}{1.04}=a\cdot \frac{1.04}{1.04}\\ \Rightarrow 1278.9=a\end{array}$

Now the model is $y=1278.9e^{0.004t}$ .

Population in $2030$ is

  $t=30$

  $\begin{array}{l}y=1278.9e^{0.004\cdot 30}\\ \Rightarrow y=1278.9\cdot 1.127\\ \Rightarrow y\approx 1441.96\end{array}$

The constant ‘ $b$ ’ defines the increasing or decreasing process in the equation. A Large value of $b$ means that thepopulation is increasing and lower value means that the population is decreasing.

Hence the value of ‘ $b$ ’ defines the increasing or decreasing process.