Chapter 4, Problem 29T

Solution

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

(A) $y=0.2236\cdot x^{0.4999}$

(B) Speed of the particle is $y=2.4482\text{meter/second}$ .

Given: points are given in $\left(x,y\right)$ as $\left(0.2,0.10\right)$ , $\left(5,0.50\right)$ , $\left(11,0.75\right)$ , $\left(20,1.00\right)$ and $\left(45,1.50\right)$ .

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:

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

  $\begin{array}{cccccc}\ln x& -1.609& 1.609& 2.397& 2.995& 3.806\\ \ln y& -2.302& -0.693& -0.287& 0& 0.405\end{array}$

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

  

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

Here, choose two points $\left(-1.609,-2.302\right)$ and $\left(3.806,0.405\right)$ then the equation of the line;

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

Now, substitute the value;

  $\begin{array}{c}\ln y-(-2.302)=\frac{0.405-(-2.302)}{3.806-(-1.609)}\left[\ln x-(-1.609)\right]\\ \ln y+2.302=\frac{0.405+2.302}{3.806+1.609}\left(\ln x+1.609\right)\\ \ln y+2.302=\frac{2.707}{5.415}\left(\ln x+1.609\right)\\ \ln y+2.302=0.4999\left(\ln x+1.609\right)\end{array}$

Thus,

  $\begin{array}{c}\ln y+2.302=0.4999\ln x+0.4999\cdot 1.609\\ \ln y=\ln x^{0.4999}+0.8043-2.302\left[a\ln b=\ln b^a\right]\\ \ln y-\ln x^{0.4999}=-1.4977\\ \ln \left(\frac{y}{x^{0.4999}}\right)=-1.4977\left[\ln a-\ln b=\ln \left(\frac{a}{b}\right)\right]\end{array}$

Now, take the exponentiation of each side by e;

  $\begin{array}{c}{e}^{\ln \left(\frac{y}{x^{0.4999}}\right)}=e^{-1.4977}\left[e^{\ln a}=a\right]\\ \frac{y}{x^{{}^{0.4999}}}=e^{{}^{-1.4977}}\\ y\approx 0.2236\cdot x^{0.4999}\left[\begin{array}{c}e\approx 2.7182\\ e^{-1.4977}\approx 0.2236\end{array}\right]\end{array}$

Therefore, the power function of the model is $y=0.2236\cdot x^{0.4999}$ .

(B) Now take the power function of the model with the help of data pairs is given as follows:

  $y=0.2236\cdot x^{0.4999}$   ....... (1)

And also given as $x=120$ and substitute this valsue in above equation (1);

  $\begin{array}{c}y=0.2236\cdot {\left(120\right)}^{0.4999}\\ y=0.2236\cdot 10.9492\\ y=2.482\text{meter/second}\end{array}$

Conclusion:

Hence, the required power function of the model that satisfies the given points is $y=0.2236\cdot x^{0.4999}$ and speed of the particle is $y=2.4482\text{meter/second}$ .