Chapter 11, Problem 18CT

(a).

To determine

To calculate: Calculate a quadratic model of y

Answer to Problem 18CT

The required quadratic model of y, $y=8.786x^2-6.243x-0.428$

Explanation of Solution

Given information: $y=a{x}^2+bx+c$ and data

Time, x ( In second)	Height, y (In feet)
0	0
1	1
2	23
3	60
4	115
5	188

Calculation:

Quadratic regression: $y=a{x}^2+bx+c$

Function	Value
Mean of x	2.5
Mean of y	64.5
Correlation coefficient r	0.99962917
a	-0.4285714
b	-6.2428571
c	8.7857143

Hence, the required quadratic model of y, $y=8.786x^2-6.243x-0.428$

  

(b).

To determine

To calculate: Calculate the velocity of the shuttle at after 5 second.

Answer to Problem 18CT

Explanation of Solution

Given information: quadratic model of y, $y=8.786x^2-6.243x-0.428$

Let, time after 5 second is 6 second, since x = 6 sec

Calculation:

Differentiate the quadratic model of y with respect to ‘ x’ .

Now,

  $\begin{array}{l}\therefore \frac{dy}{dx}=\frac{d}{dx}\left(8.786x^2-6.243x-0.428\right)\\ =8.786\times 2\times x-6.243\\ =17.572x-6.243\\ \therefore \text{Velocity }\left(\frac{dy}{dx}\right)=17.572x-6.243\mathrm{...}\left(1\right)\end{array}$

Put, $x=6\text{ sec}$ in equation (1), we get

  $\begin{array}{l}\therefore \text{Velocity }\left(\frac{dy}{dx}\right)=17.572\times 6-6.243\\ =105.432-6.243\\ =99.189\text{feet per second}\end{array}$

Hence, the required velocity of the shuttle is 99.189 per second.