Results 1 to 3 of 3

Math Help - Suspension Bridge

  1. #1
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, magentarita!

    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.



    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    wow...

    Quote Originally Posted by Soroban View Post
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Suspension homology
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: December 17th 2011, 11:22 AM
  2. Catenary differential equation (suspension bridge)
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: July 13th 2011, 10:42 PM
  3. Suspension Bridge Parabola Problem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 2nd 2011, 10:17 AM
  4. Parabola word problem about a suspension bridge
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 19th 2010, 04:35 PM
  5. Chain maps inducing the suspension isomorphism
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 22nd 2010, 05:26 AM

Search Tags


/mathhelpforum @mathhelpforum