Chapter 11.1, Problem 34E

Solution

a)

To find: The average velocity of the ball.

The average speed from one point $\left(0.8,21.24\right)$ to another $\left(1.0,15.76\right)$ is $-27.4$ ft/sec.

Given information:

Ball is dropped from a two-story building. The distance in feet above the ground of the falling is given.

Formula used:

The average speed from one point $\left(a,f\left(a\right)\right)$ to another point $\left(b,f\left(b\right)\right)$ is given as

  $\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

The average velocity from one point $\left(0.8,21.24\right)$ to another $\left(1.0,15.76\right)$ is

  $\begin{array}{c}\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{15.76-21.24}{1.0-0.8}\\ =-\frac{5.48}{0.2}\\ =-27.4\end{array}$

Thus, the average speed from one point $\left(0.8,21.24\right)$ to another $\left(1.0,15.76\right)$ is

  $-27.4$ ft/sec.

b)

To find:the quadratic regression model for the given data

The quadratic regression model for the given data is $s\left(t\right)=-16.01t^2+1.43t+30.35$ ,

  $t=$ time in seconds

Given information:

Given the distance data of the Lead Ball.

Formula used:

Calculator for quadratic regression model used.

Calculation:

The scatter plot of the given is shown in Figure (1).

  

Figure (1)

Quadratic regression model fit is shown below.

  

Figure (2)

Using the calculator for quadratic regression model based given data is $s\left(t\right)=-16.01t^2+1.43t+30.35$ , $t=$ time in seconds

c)

To find: the derivative of the regression equation.

The derivative of the function $s\left(t\right)=-16.01t^2+1.43t+30.35$ is $-32.02t+1.43$ .

The velocity of the ball at $t=1$ is $-30.6$ ft/sec.

Given information:

From part (b), quadratic regression model for given data is $s\left(t\right)=-16.01t^2+1.43t+30.35$ , $t=$ time in seconds

Formula used:

  $\frac{ds}{dt}=\underset{h\to 0}{\lim }\frac{s\left(t+h\right)-s\left(t\right)}{h}$

Calculation:

Derivative of the function$s\left(t\right)$:

  $\begin{array}{c}\frac{ds}{dt}=\underset{h\to 0}{\lim }\frac{s\left(t+h\right)-s\left(t\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(-16.01{\left(t+h\right)}^2+1.43\left(t+h\right)+30.35\right)-\left(-16.01t^2+1.43t+30.35\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(-16.01\left(t^2+h^2+2th\right)+1.43\left(t+h\right)+30.35\right)-\left(-16.01t^2+1.43t+30.35\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(-16.01\left(h+2t\right)+1.43\right)}{h}\end{array}$

Further, simplify.

  $\begin{array}{c}\frac{ds}{dt}=\underset{h\to 0}{\lim }\left(-16.01\left(h+2t\right)+1.43\right)\\ =-16.01\left(2t\right)+1.43\\ =-32.02t+1.43\end{array}$

Thus, the derivative of the function $s\left(t\right)=-16.01t^2+1.43t+30.35$ is $-32.02t+1.43$ .

Estimate the velocity of the ball at $t=1$:

Substitute $t=1$ in the equation $s^{\prime }\left(t\right)=-32.02t+1.43$ :

  $\begin{array}{c}{s}^{\prime }\left(t\right)=-32.02t+1.43\\ =-32.02\left(1\right)+1.43\\ =-30.6\end{array}$

The velocity of the ball at $t=1$ is $-30.6$ ft/sec.