Chapter 3.5, Problem 68E

Solution

a.

To graph: Create a scatter plot of the $\left(t,P\right)$ pairs.

Given information:

The number of years after $1900$ is $t$ .

The U.S. population (in millions) is $P$ .

Year	Population
$1900$	$76.2$
$1910$	$92.2$
$1920$	$106.0$
$1930$	$123.2$
$1940$	$132.2$
$1950$	$151.3$
$1960$	$199.3$
$1970$	$203.3$
$1980$	$226.5$
$1990$	$248.7$
$2000$	$281.4$
$2010$	$308.7$
$2016$	$323.1$

Graph:

Let $P\left(t\right)$ be the population in millions of the United States $t$ years after $\text{19}00$ . Using exponential regression,

  $P\left(t\right)=\left(\text{81}.\text{2263}\right){\left(\text{1}.0\text{1263}\right)}^t$

By given data table:

  

Interpretation:

Use of the data from the specified table, obtained the scatter plot of $\left(t,P\right)$ pairs.

b.

To graph: Plot a scatter of the $\left(P,t\right)$ pairs with the scatter plot of the $\left(t,P\right)$ pairs.

Given information:

The number of years after $1900$ is $t$ .

The U.S. population (in millions) is $P$ .

Year	Population
$1900$	$76.2$
$1910$	$92.2$
$1920$	$106.0$
$1930$	$123.2$
$1940$	$132.2$
$1950$	$151.3$
$1960$	$199.3$
$1970$	$203.3$
$1980$	$226.5$
$1990$	$248.7$
$2000$	$281.4$
$2010$	$308.7$
$2016$	$323.1$

Graph:

  

Interpretation:

Use of the data from the specified given table, Obtained the scatter plot of $\left(P,t\right)$ pairs with scatter plot of $\left(t,P\right)$ .

c.

To graph: Show that the two plots are reflections of each other about the line $y=x$ .

Given information:

The number of years after $1900$ is $t$ .

The U.S. population (in millions) is $P$ .

Graph:

  

Interpretation:

The reflection of the point $\left(x,y\right)$ across the line $y=x$ is the point $\left(y,x\right)$ .

d.

To calculate: Find an exponential model $\left(y_1\right)$ for the $\left(t,P\right)$ pairs and a logarithmic model $\left(y_2\right)$ for the $\left(P,t\right)$ pairs.

Exponential regression:

  $f\left(x\right)=76.2{\left(1.013\right)}^x$

Logarithm regression:

  $g\left(x\right)=\left(-340.43\right)+78.62\ln x$

Given information:

The number of years after $1900$ is $t$ .

The U.S. population (in millions) is $P$ .

Exponential regression: $y_1=f\left(x\right)$

Logarithm regression: $y_2=g\left(x\right)$

Concept used:

Exponential regression: $y=a{b}^x$

Logarithm regression: $y=a+b\ln x$

Calculation:

To find logarithmic and exponential regression use formulas:

Exponential regression: $y=a{b}^x$

Logarithm regression: $y=a+b\ln x$

Find $a$ and $b$ ;

For Exponential regression: $y=a{b}^x$

Take two points $\left(0,\text{76}.\text{2}\right)$ and $\left(\text{111},\text{311}.\text{6}\right)$ ;

  $\begin{array}{l}{x}_1=0;\\ y_1=76.2;\\ x_2=111;\\ y_2=311.6;\end{array}$

Substitute the values:

  $\begin{array}{l}y=a{b}^x\\ 76.2=a{b}^0\\ a{b}^{111}=311.6\\ a=76.2\end{array}$

And;

  $\begin{array}{l}311.6=76.2b^{111}\\ b^{111}=4.09\\ b=1.013\end{array}$

Thus obtained Exponential regression:

  $f\left(x\right)=76.2{\left(1.013\right)}^x$

Find $a$ and $b$ ;

For Logarithm regression: $y=a+b\ln x$

Take two points $\left(\text{76}.\text{2,0}\right)$ and $\left(\text{311}.\text{6,111}\right)$ ;

  $\begin{array}{l}{x}_1=76.2;\\ y_1=0;\\ x_2=311.6;\\ y_2=111;\end{array}$

Substitute the values:

  $\begin{array}{l}0=a+b\ln \left(76.2\right)\\ a=-b\ln \left(76.2\right)\\ a=-b\left(4.33\right)\end{array}$

Now,

  $a+b\ln \left(311.6\right)=111$

Substitute the $a=-b\left(4.33\right)$ ;

  $\begin{array}{l}a+b\ln \left(311.6\right)=111\\ -b\left(4.33\right)+b\ln \left(311.6\right)=111\\ b\left(1.412\right)=111\\ b=78.62\end{array}$

And;

  $\begin{array}{l}a=-b\left(4.33\right)\\ a=-340.43\end{array}$

Thus obtained Logarithm regression:

  $g\left(x\right)=\left(-340.43\right)+78.62\ln x$

Conclusion:

  $\begin{array}{l}{y}_1=76.2{\left(1.013\right)}^x\\ y_2=\left(-340.43\right)+78.62\ln x\end{array}$

e.

To graph: Graph the exponential regression model.

Given information:

The number of years after $1900$ is $t$ .

The U.S. population (in millions) is $P$ .

Exponential regression: $y_1=f\left(x\right)$

Logarithm regression: $y_2=g\left(x\right)$

Graph:

Exponential regression:

  $f\left(x\right)=76.2{\left(1.013\right)}^x$

Logarithm regression:

  $g\left(x\right)=\left(-340.43\right)+78.62\ln x$

  

Interpretation:

Obtained the graph of functions $f\left(x\right)$ and $g\left(x\right)$ .

f.

To calculate: Analytically show how to compute the logarithmic regression value $g\left(300\right)$ using only the exponential regression equation. Interpret your result.

  $g\left(311.6\right)=111.30604$

Given information:

The number of years after $1900$ is $t$ .

The U.S. population (in millions) is $P$ .

Exponential regression: $y_1=f\left(x\right)$

Logarithm regression: $y_2=g\left(x\right)$

Formula used:

Exponential regression:

  $f\left(x\right)=76.2{\left(1.013\right)}^x$

Logarithm regression:

  $g\left(x\right)=\left(-340.43\right)+78.62\ln x$

Calculation:

  $g\left(x\right)=\left(-340.43\right)+78.62\ln x$

For $x=311.6$ ;

  $\begin{array}{l}g\left(311.6\right)=\left(-340.43\right)+78.62\ln \left(311.6\right)\\ =\left(-340.43\right)+78.62\left(5.742\right)\\ =-340.43+451.44\\ =111.30604\end{array}$

Conclusion:

  $g\left(311.6\right)=111.30604$