Chapter 10, Problem 78RE

To determine

The height of the arch at distances of $5,$ $10$ and $20$ feet from the centre.

Answer to Problem 78RE

At $x=5$ , $\boxed{y=19.45feet}$

At $x=10$ , $\boxed{y=17.78feet}$

At $x=20$ , $\boxed{y=11.12feet}$

Explanation of Solution

Given information:

A bridge is built in the shape of a parabolic arch. The bridge has a span of $60$ feet and a maximum height of $20$ feet. Find the height of the arch at distances of $5,$

  $10$ and $20$ feet from the centre.

Calculation:

Let the vertex of the parabolic arch bridge be the point at the maximum height which is given by coordinates $\left(0,20\right)$ since the origin of the plane in which the parabola resides is the center point of the parabola. The equation of such a parabola is given below.

  $x^2=-4a\left(y-20\right)$

The value of $a$ can be determined substituting coordinates of any point that lies on the parabola. One such point that lies on the parabola is $\left(30,0\right)$ that are the coordinates of one of the end-point of the span of parabola.

  ${\left(30\right)}^2=-4a\left(0-20\right)$

  $a=\frac{900}{80}$

  $a=\frac{45}{4}$

Hence, the equation of the parabolic bridge is $x^2=-45\left(y-20\right)$ .

The height of the bridge when the distance from the centre of the origin is $x=5$ feet can be determined as follows:

  ${\left(5\right)}^2=-45\left(y-20\right)$

  $y=\frac{25-45\times 20}{-45}$

  $y=\frac{875}{45}$

  $\boxed{y=19.45feet}$

The height of the bridge when the distance from the centre of origin is $x=10$ feet can be determined as follows:

  ${\left(10\right)}^2=-45\left(y-20\right)$

  $y=\frac{10-45\times 20}{-45}$

  $y=\frac{800}{45}$

  $\boxed{y=17.78feet}$

The height of the bridge when the distance from the centre or the origin is $x=20$ feet can be determined as follows:

  ${\left(20\right)}^2=-45\left(y-20\right)$

  $y=\frac{400-45\times 20}{-45}$

  $y=\frac{500}{45}$

  $\boxed{y=11.12feet}$ .