# Suspension Bridge

• Oct 14th 2008, 06:03 AM
magentarita
Suspension Bridge
The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are 400 feet apart and 100 feet high. If the cables are at a height of 10 feet midway between the towers, what is the height of the cable at a point 50 feet from the center of the bridge?
• Oct 14th 2008, 09:25 AM
Soroban
Hello, magentarita!

Quote:

The cables of a suspension bridge are in the shape of a parabola.
The towers supporting the cable are 400 feet apart and 100 feet high.
If the cable is at a height of 10 feet midway between the towers,
what is the height of the cable at a point 50 feet from the center of the bridge?

Place the parabola on a graph.
Code:

                  | (-200,100)*      |      *(200,100)           :      |      :           :*      |      *:           : *    |    * :           :  *  |  *  :           :      *      :           :      |(0,10) :           :      |      :       - - + - - - + - - - + --         -200      |      200

The parabola opens upward and is symmetric to the y-axis.
. . Its general form is: . $y \;=\;ax^2 + c$

Its y-intercept is (0, 10) . . . Hence: . $y \;=\;ax^2 + 10$

It passes through (200, 100).
. . We have: . $100 \:=\:a\cdot200^2 + 10 \quad\Rightarrow\quad a \:=\:\frac{9}{4000}$
Hence: . $y \;=\;\tfrac{9}{4000}x^2 + 10$

When $x = \pm50,\;\;y \:=\:\tfrac{9}{4000}(50^2) + 10 \:=\:\frac{125}{8}$

Therefore, 50 feet from the center, the cable is $15\tfrac{5}{8}$ feet high.

• Oct 14th 2008, 08:56 PM
magentarita
wow...
Quote:

Originally Posted by Soroban
Hello, magentarita!

Place the parabola on a graph.
Code:

                  | (-200,100)*      |      *(200,100)           :      |      :           :*      |      *:           : *    |    * :           :  *  |  *  :           :      *      :           :      |(0,10) :           :      |      :       - - + - - - + - - - + --         -200      |      200
The parabola opens upward and is symmetric to the y-axis.
. . Its general form is: . $y \;=\;ax^2 + c$

Its y-intercept is (0, 10) . . . Hence: . $y \;=\;ax^2 + 10$

It passes through (200, 100).
. . We have: . $100 \:=\:a\cdot200^2 + 10 \quad\Rightarrow\quad a \:=\:\frac{9}{4000}$
Hence: . $y \;=\;\tfrac{9}{4000}x^2 + 10$

When $x = \pm50,\;\;y \:=\:\tfrac{9}{4000}(50^2) + 10 \:=\:\frac{125}{8}$

Therefore, 50 feet from the center, the cable is $15\tfrac{5}{8}$ feet high.

What a marvelous explanation. How do you make your diagrams?