Chapter 10.2, Problem 74AYU

Parabolic Arch Bridge A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch a distance of 40 feet from the center is to be 10 feet. Find the height of the arch at its center.

To determine

To find: Height of the arch at its center, if arch has a span of 100 feet and its height is 10 feet at a distance 40 feet from the center.

Answer to Problem 74AYU

Height/depth of the arch at its centre is about $27.78$ feet.

Explanation of Solution

Given:

A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch at a distance of 40 feet from the center is to be 10 feet.

Formula used:

Vertex	Focus	Directrixc	Equation	Description
$\left(0,0\right)$	$(0,-a)$	$y=a$	$x²=-4ay$	Parabola, axis of symmetry is $y\text{-axis}$, opens down

Calculation:

Let the vertex of the parabola be $\left(0,0\right)$ and it opens down.

The equation of the parabola is given by $x²=-4ay$ where a is the distance from vertex to focus.

The parabola is 100 feet across.

Hence $\left(50,-h\right)$ and $\left(40,-h+10\right)$ lies on the parabola.

To solve for $a$ and $\text{h}$, we substitute the points in $x^2=-4ay$.

${50}^2=-4\ast a\ast -\text{h}$

$2500=4ah$

Similarly,

${40}^2=-4a\left(-h+10\right)$

$1600=4ah+40a$

$40a=-2500+1600$

$4a=-900/10=-90$

By substitution,

$2500=4ah=-90h$

$h=\frac{2500}{-90}$

$h\approx -27.78$feet

Height/depth of the arch at its centre is about $27.78$ feet.