Chapter 10.2, Problem 73AYU

Parabolic Arch Bridge A bridge is built in the shape of a parabolic arch. The bridge has a span of 120 feet and a maximum height of 25 feet. See the illustration. Choose a suitable rectangular coordinate system and find the height of the arch at distances of 10,30, and 50 feet from the center.

To determine

To find: Suitable rectangular coordinate system and the height of the arch at a distances of 10, 30, 50 feet from the center if the bridge has a span of 120 feet and a maximum height of 25 feet.

Answer to Problem 73AYU

$24.31$ feet, $18.75$ feet, $7.64$ feet.

Explanation of Solution

Given:

A bridge is built in the shape of a parabolic arch. The bridge has a span of 120 feet and a maximum height of 25 feet. Choose a suitable rectangular coordinate system and find the height of the arch at distances of 10, 30, and 50 feet from the center.

Formula used:

Vertex	Focus	Directrix	Equation	Description
$\left(0,0\right)$	$\left(0,a\right)$	$y=a$	$x²=-4ay$	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 120 feet across and 25 feet deep.

The parabolic representation is given below.

Hence $\left(60,-25\right)$ and $\left(-60,- 25\right)$ lies on the parabola.

To solve for $a$, we substitute $\left(60,-25\right)$ in $x^2=-4ay$.

$60²=-4\ast a\ast -25$

$a=\frac{3600}{100}=36$feet

Hence, the equation of the parabola is given by $x^2=-4\ast 36\ast y=-144y$.

The equation of the parabola is $x²=-144y$.

To find the height of the bridge 10 feet from the centre, the point $\left(10,y\right)$ is a point on the parabola.

Substituting $\left(10,y\right)$ in $x^2=-144y$.

$100=-144\ast y$

$y=\frac{100}{-144}=0.6944~.69$feet.

$h=25-.69=24.31$

Height of the arch at distance of 10 feet from the centre is $24.31$ feet.

To find the height of the bridge 30 feet from the centre, the point $\left(30,y\right)$ is a point on the parabola.

Substituting $\left(30,y\right)$ in $x^2=-144y$.

$30²=-144\ast y$

$y=\frac{900}{-144}=-6.25$feet.

$h=25-6.25=18.75$

Height of the arch at distance of 30 feet from the centre is $18.75$ feet.

To find the height of the bridge 50 feet from the centre, the point (50,y) is a point on the parabola.

Substituting $\left(50,y\right)$ in $x^2=-144y$.

$50²=-144\ast y$

$y=\frac{2500}{-144}=-17.36111~-17.361$feet

$h=25-17.36=7.64$

Height of the arch at distance of 50 feet from the centre is $7.64$ feet.