Chapter 6.4, Problem 64E

(a)

To determine

Area under the parabolic arch.

Answer to Problem 64E

Area under the parabolic arch,

  $\text{Area}=\frac{125}{6}$

Explanation of Solution

Given information:

  

  $y=6-x-x^2$

Such that

  $-3\le x\le 2$

For

  $y=6-x-x^2$

With

  $-3\le x\le 2$

Area under the parabolic arch,

  $\begin{array}{l}\text{Area}={\displaystyle {\int }_{-3}^2\left(6-x-x^2\right)dx}\\ ={6x-\frac{x^2}{2}-\frac{x^3}{3}|}_{-3}^2\\ =6\left(2\right)-\frac{{\left(2\right)}^2}{2}-\frac{{\left(2\right)}^3}{3}-\left[6\left(-3\right)-\frac{{\left(-3\right)}^2}{2}-\frac{{\left(-3\right)}^3}{3}\right]\\ =\frac{22}{3}-\left(-\frac{27}{2}\right)\\ =\frac{125}{6}\end{array}$

(b)

To determine

Height of the arch.

Answer to Problem 64E

Height of the arch is $\frac{25}{4}$ .

Explanation of Solution

Given information:

  

  $y=6-x-x^2$

The height of the arch is the y − coordinate of the parabola’s vertex.

The vertex of a parabola of the form

  $y=a{x}^2+bx+c$

Is at

  $x=-\frac{b}{2a}$

For

  $y=6-x-x^2$

With

  $a=b=-1$

Thus,

The vertex is at:

  $x=\frac{-\left(-1\right)}{2\left(-1\right)}=-\frac{1}{2}$

Now,

To find the y − coordinate of the vertex:

Substitute

  $x=-\frac{1}{2}$

Into

  $y=6-x-x^2$

Thus,

  $y=6-\left(-\frac{1}{2}\right)-{\left(-\frac{1}{2}\right)}^2=\frac{25}{4}$

(c)

To determine

Area is two − thirds the base times the height.

Answer to Problem 64E

Area will be same as area under the parabolic arch.

Explanation of Solution

Given information:

  

  $y=6-x-x^2$

Base,

  $2-\left(-3\right)=5$

From Part (b) result,

The height is $\frac{25}{4}$ .

Now,

  $\begin{array}{l}\text{Area}=\frac{2}{3}\times \text{Base}\times \text{Height}\\ \text{=}\frac{2}{3}\times 5\times \frac{25}{4}\\ =\frac{125}{6}\end{array}$

This is same as value for area under parabolic arch obtained in Part (a).