Chapter 7.4, Problem 6E

a.

To determine

To predict : the population in $2001$ by using the exponential model for $1951$ and $1961$

Answer to Problem 6E

The $y\left(50\right)=960.05$

Explanation of Solution

Given information : the exponential table is

Year	Population
1951	361
1961	439

Calculation :

  $\begin{array}{l}\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(10\right)=361\times e^{k\times 10}\\ \frac{439}{361}=e^{k\times 10}\\ \ln \left(\frac{439}{361}\right)=k\times 10\\ \frac{1}{10}\times \ln \left(\frac{439}{361}\right)=k\end{array}$

Let $y\left(0\right)$

  $=361$ or the population in $1951$

Let $y\left(10\right)=$

  $439$ or the population of $1961$

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(t\right)=361\times e^{\frac{1}{10}\times \ln \left(\frac{439}{361}\right)\times t}\\ y\left(t\right)=361\times e^{\frac{t}{10}\times \ln \left(\frac{439}{361}\right)}\\ y\left(t\right)=361\times {\left(\frac{439}{361}\right)}^{\frac{t}{10}}\end{array}$

Rule 1: $y\times \ln \left(x\right)=\ln \left(x^y\right)$

Rule2: $e^{\ln \left(x\right)}=x$

  $\begin{array}{l}y\left(50\right)=361\times {\left(\frac{439}{361}\right)}^{\frac{50}{10}}\\ y\left(50\right)=960.05\\ \text{At 2001},t=50\text{ because }2001-1951=50\\ y\left(50\right)=960.05\text{ v/s actual population }1029\end{array}$

b.

To determine

To predict : the population in $2010\&2020$ by using the exponential model for $1961$ and $1981$

Answer to Problem 6E

The $y\left(40\right)=1062.62$

Explanation of Solution

Given information : the exponential table is

Year	Population
1961	439
1981	683

Calculation :

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(20\right)=439\times e^{k\times 20}\\ \frac{683}{439}=e^{k\times 20}\\ \ln \left(\frac{683}{439}\right)=k\times 20\\ \frac{1}{20}\times \ln \left(\frac{683}{439}\right)=k\end{array}$

Let $y\left(0\right)$

  $=439$ or the population in $1961$

Let $y\left(20\right)=$

  $683$ or the population of $1981$

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(t\right)=439\times e^{\frac{1}{20}\times \ln \left(\frac{683}{439}\right)\times t}\\ y\left(t\right)=439\times e^{\frac{t}{20}\times \ln \left(\frac{683}{439}\right)}\\ y\left(t\right)=439\times {\left(\frac{683}{439}\right)}^{\frac{t}{20}}\end{array}$

Rule 1: $y\times \ln \left(x\right)=\ln \left(x^y\right)$

Rule2: $e^{\ln \left(x\right)}=x$

  $\begin{array}{l}y\left(40\right)=439\times {\left(\frac{683}{439}\right)}^{\frac{40}{20}}\\ y\left(40\right)=1062.62\\ \text{At 2001},t=40\text{ because }2001-1961=50\\ y\left(40\right)=1062.62\text{ v/s actual population }1029\end{array}$

c.

To determine

To graph : both the exponential function of part(a) and part(b)

Explanation of Solution

Given information : the equation is $y=439\times {\left(\frac{683}{439}\right)}^{\frac{x}{20}}$ & $y=361\times {\left(\frac{431}{361}\right)}^{\frac{x}{10}}$

Graph : Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $439\times {\left(\frac{683}{439}\right)}^{\frac{x}{20}}$ & $361\times {\left(\frac{431}{361}\right)}^{\frac{x}{10}}$ which is required to graph.

Step 4: Press $\boxed{\text{GRAPH}}$ button to graph the function.

The graph is obtained as:

  

Interpretation :

The pink graph is $y=439\times {\left(\frac{683}{439}\right)}^{\frac{x}{20}}$

The green graph is $y=361\times {\left(\frac{431}{361}\right)}^{\frac{x}{10}}$

The black dots represent the actual population