Chapter 2.8, Problem 20E

a

To determine

To graph: Graph and find the quadratic model for the data.

Explanation of Solution

Given information:

  

Graph :

Interpretation : A quadratic model finds the equation of the parabola that best fits the set of data and its equation is given by $y=a{x}^2+bx+c$ .

From the graph enter the value of x and y coordinates to find the value of a, b and c

And also the value of r-square to get the best fit value.

From the given formula below we calculate the respective value and get the required equation,

  $\begin{array}{l}{R}^2=1-\frac{SSE}{SST}\\ SSE={\displaystyle \sum _{i=1}^n{(y_i-a{x}_i^2-b{x}_i-c)}^2}\\ SST={\displaystyle \sum _{i=1}^n{(y_i-\overline{y})}^2}\end{array}$ , $\left[\begin{array}{ccc}{\displaystyle \sum _{i=1}^n{X}_i^4}& {\displaystyle \sum _{i=1}^n{X}_i^3}& {\displaystyle \sum _{i=1}^n{X}_i^2}\\ {\displaystyle \sum _{i=1}^n{X}_i^3}& {\displaystyle \sum _{i=1}^n{X}_i^2}& {\displaystyle \sum _{i=1}^n{X}_i}\\ {\displaystyle \sum _{i=1}^n{X}_i^2}& {\displaystyle \sum _{i=1}^n{X}_i}& n\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}{\displaystyle \sum _{i=1}^n{X}^2{}_i{Y}_i}\\ {\displaystyle \sum _{i=1}^n{X}_i{Y}_i}\\ {\displaystyle \sum _{i=1}^n{Y}_i}\\ \end{array}\right]$

  $\begin{array}{l}\Rightarrow a=-15.8445,\\ b=39.5199\\ c=5.04055\end{array}$

Therefore, $\begin{array}{l}y=-15.8445x^2+39.5199x+5.04055\\ R^2=1\end{array}$

b

To determine

To calculate: calculate the time when the javelin hit the ground.

Answer to Problem 20E

  $2.615\text{ sec}$

Explanation of Solution

Given information:

Formula used: A quadratic model finds the equation of the parabola that best fits the set of data and its equation is given by $y=a{x}^2+bx+c$ .

  $\left[\begin{array}{ccc}{\displaystyle \sum _{i=1}^n{X}_i^4}& {\displaystyle \sum _{i=1}^n{X}_i^3}& {\displaystyle \sum _{i=1}^n{X}_i^2}\\ {\displaystyle \sum _{i=1}^n{X}_i^3}& {\displaystyle \sum _{i=1}^n{X}_i^2}& {\displaystyle \sum _{i=1}^n{X}_i}\\ {\displaystyle \sum _{i=1}^n{X}_i^2}& {\displaystyle \sum _{i=1}^n{X}_i}& n\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}{\displaystyle \sum _{i=1}^n{X}^2{}_i{Y}_i}\\ {\displaystyle \sum _{i=1}^n{X}_i{Y}_i}\\ {\displaystyle \sum _{i=1}^n{Y}_i}\\ \end{array}\right]$ and $\begin{array}{l}{R}^2=1-\frac{SSE}{SST}\\ SSE={\displaystyle \sum _{i=1}^n{(y_i-a{x}_i^2-b{x}_i-c)}^2}\\ SST={\displaystyle \sum _{i=1}^n{(y_i-\overline{y})}^2}\end{array}$

Calculation:

From above graph the respective values are $\begin{array}{l}a=-15.8445,\\ b=39.5199\\ c=5.04055\end{array}$ and $R^2=1$ \

Therefore the equation of the quadratic model is given by,

  $y=-15.8445x^2+39.5199x+5.04055$

Now to determine the time when the javelin hit the ground, we consider the value of y be zero and calculate for x.

  $\begin{array}{l}\Rightarrow 0=-15.8445x^2+39.5199x+5.04055\\ 15.8445x^2-39.5199x-5.04055=0\end{array}$

  $x=2.615,-0.1216$

But the negative value can not be considered , therefore the time when the javelin hit the ground is $2.615\text{ sec}$