Chapter ISG, Problem 9.11STP

(a)

To determine

To find: The quadratic model of the shape of the mound.

Answer to Problem 9.11STP

  $h=-0.0327x^2+0.5827x+0.0462$

Explanation of Solution

Given: A cross section of the mound is shown below.

  

According to shape of mound. It is parabolic shape of mound.

Passing point of parabola, $\left(3,1.5\right),\left(11,2.5\right),\left(16,1\right)$

Let the quadratic model be $h=a{x}^2+bx+c$

Where, h is height of mound and x is horizontal distance.

  $\begin{array}{c}1.5=9a+3b+c\mathrm{...}\left(i\right)\\ 2.5=121a+11b+c\mathrm{...}\left(ii\right)\\ 1=256a+16b+c\mathrm{...}\left(iii\right)\end{array}$

Solve above three equation find out the coefficients a, b and c.

  $\begin{array}{l}a\approx -0.0327\\ b\approx 0.5827\\ c\approx 0.0462\end{array}$

Therefore, quadratic model is $h=-0.0327x^2+0.5827x+0.0462$

(b)

To determine

To find: The width of the bound.

Answer to Problem 9.11STP

  $18\text{ m}$

Explanation of Solution

Given: A cross section of the mound is shown below.

  

The quadratic model is $h=-0.0327x^2+0.5827x+0.0462$ .

Width of the bound is difference of x -intercept of the model.

  $\begin{array}{l}0=-0.0327x^2+0.5827x+0.0462\\ x=-0.079,17.899\end{array}$

Width of the mound $=17.899-\left(-0.079\right)=17.978\approx 18\text{ m}$

Hence, the width of the mound is 18 m

(c)

To determine

To find: The depth of the mound.

Answer to Problem 9.11STP

  $2.6\text{ m}$

Explanation of Solution

Given: A cross section of the mound is shown below.

  

The quadratic model is $h=-0.0327x^2+0.5827x+0.0462$ .

Depth of the mound is the distance of base to vertex.

So, y -coordinate of vertex is depth at its deepest point.

  $x=-\frac{b}{2a}=-\frac{0.5827}{2\left(-0.0327\right)}\approx 8.91$

Now find the value of h at $x=8.91$

  $h=2.6$

Hence, the depth of the mound at deepest point is 2.6 m