Chapter 4, Problem 38REAE

Solution

To Draw: a scatter plot of given points and also find the power model of the given point.

  $y=28\cdot {\left(x\right)}^{0.7480}$ .

Given: points are given in $\left(x,y\right)$ as $\left(1,28\right)$ , $\left(2,47\right)$ , $\left(3,64\right)$ , $\left(4,79\right)$ , $\left(5,94\right)$ and $\left(6,107\right)$ .

Concept used:

(1) If a function $y=f\left(x\right)$ passes through a point $\left(x_0,y_0\right)$ , then the point $\left(x_0,y_0\right)$ satisfies the equation of the function, that is, $y_0=f\left(x_0\right)$ .

(2) A power model or function is defined as:

  $y=a\cdot {(x)}^b$ ; where $a\ne 0$ and $b\ge 0$ .

(3) Equation of a line passing through the given two points as like $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

Then, $\left(y-y_1\right)=m\left(x-x_1\right)$ .

Here, $m$ is the slope. And $m$ is as follows:

  $m=\frac{y_2-y_1}{x_2-x_1}$ .

Calculation:

The table of data pairs $\left(x,y\right)$ written as;

  $\begin{array}{ccccccc}x& 1& 2& 3& 4& 5& 6\\ y& 28& 47& 64& 79& 94& 107\end{array}$

The Scatter graph of $\left(x,y\right)$ is given as;

  

Now, the general power function form is $y=a\cdot {\left(x\right)}^b$ ;

Here, choose two points $\left(1,28\right)$ and $\left(6,107\right)$ then the equation of the line;

  $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Now, substitute the value;

  $\begin{array}{c}y-28=\frac{107-28}{6-1}\left(x-1\right)\\ y-28=\frac{79}{5}\left(x-1\right)\\ y-28=15.8\left(x-1\right)\\ y=15.8x-15.8+28\end{array}$

Thus,

  $y=15.8x+12.2$

This is a required equation between given two points of a line.

Now, the general power function of the model form is $y=a\cdot {\left(x\right)}^b$   ....... (1)

The point $\left(1,28\right)$ passes through the line. Then, the above equation (1) is as follows:

  $28=a{\left(1\right)}^b$

  $a=\frac{28}{{\left(1\right)}^b}$  ....... (2)

Again similarly the point $\left(6,107\right)$ passes through the line. Then, the above equation (1) is as follows:

  $107=a{\left(6\right)}^b$  ....... (3)

Now substitute the value of $a$ from equation (2) in equation (3) becomes;

  $\begin{array}{c}107=\frac{28}{{(1)}^b}\cdot {\left(6\right)}^b\\ \frac{107}{28}={\left(\frac{6}{1}\right)}^b\\ 3.82={\left(6\right)}^b\end{array}$

Take on both sides logarithm with base $\left(6\right)$ ;

  $\begin{array}{c}{\log }_{6}3.82={\log }_6{\left(6\right)}^b\\ 0.7480=b{\log }_{6}6\left[{\log }_a{m}^n=n{\log }_{a}m\right]\\ b=0.7480\left[{\log }_{a}a=1\right]\\ \end{array}$

Thus,

  $\begin{array}{c}{\log }_{6}3.82={\log }_6{\left(6\right)}^b\\ 0.7480=b{\log }_{6}6\left[{\log }_a{m}^n=n{\log }_{a}m\right]\\ b=0.7480\left[{\log }_{a}a=1\right]\\ \end{array}$

Substitute the value of $b=0.7480$ in the above equation (2);

  $\begin{array}{c}a=\frac{28}{{\left(1\right)}^{0.7480}}\\ a=\frac{28}{1}\left[{\left(1\right)}^n=1;n\ge 1\right]\end{array}$

Thus,

  $a=28$

Therefore, substitute the value of $a$ and $b$ in equation (1);

  $y=28\cdot {\left(x\right)}^{0.7480}$

This is a power function of the model of given data pairs.

Conclusion:

Hence, the required power function of the model that satisfies the given points is $y=28\cdot {\left(x\right)}^{0.7480}$ .