Results 1 to 3 of 3

Math Help - Parabolic arch problem..

  1. #1
    NKS
    NKS is offline
    Newbie
    Joined
    Dec 2009
    Posts
    14

    Parabolic arch problem..

    I'm having some trouble understanding if a parabolic arch problem can be set up the same way as a semiellipse.. for example this problem has me utterly confused:

    An experimental model for a suspension bridge is built in the shape of a parabolic arch. In one section, cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top of the second tower. The towers stand 80 inches apart. At a point between the towers and 28 inches along the road from the base of one tower, the cable is 1.44 inches above the roadway. Find the height of the towers.




    now this problem, I think I get

    A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 196 feet and a max height of 35 feet. Find the height of the arch at 25 feet from its center.

    the answer I got is about 33.8ft. I set it up as x^2/a^2 + y^2/b^2 = 1. then plugged in the values and solved for y.
    is this correct??

    I appreciate any and all help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by NKS View Post
    I'm having some trouble understanding if a parabolic arch problem can be set up the same way as a semiellipse.. for example this problem has me utterly confused:

    An experimental model for a suspension bridge is built in the shape of a parabolic arch. In one section, cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top of the second tower. The towers stand 80 inches apart. At a point between the towers and 28 inches along the road from the base of one tower, the cable is 1.44 inches above the roadway. Find the height of the towers.




    now this problem, I think I get

    A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 196 feet and a max height of 35 feet. Find the height of the arch at 25 feet from its center.

    the answer I got is about 33.8ft. I set it up as x^2/a^2 + y^2/b^2 = 1. then plugged in the values and solved for y.
    is this correct?? <<<< No, you used the equation of an allipse

    I appreciate any and all help
    To [I]:

    1. Draw a sketch.
    2. Determine a coordinate system: The road is placed on the x-axis, the y-axis is the perpendicular bisector of the distance between the towers.

    3. The equation of the parabola would be:

    p: y = ax^2

    Plug in the coordinates of the given point (28, 1.44):
    1.44 = a \cdot 28^2~\implies~a = \dfrac9{4900}

    4. Now you have the equation of the parabola. The foot of the right tower is at x = 40. Thus the height of the tower is:

    h = \dfrac9{4900} \cdot 40^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    NKS
    NKS is offline
    Newbie
    Joined
    Dec 2009
    Posts
    14
    thanks a bunch. that is all pretty clear to me now. DUH i should have been using a parabolic equation..

    the only thing I noticed is that the given point is not (28, 1.44). 28 is the distance from the base of one of the towers.. but 40-28 is 12. so the distance from the origin to that same point is 12. thus the x coordinate would be 12.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parabolic Arch Bridge
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 16th 2008, 06:14 PM
  2. Parabolic Arch Bridge
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: July 31st 2008, 01:22 PM
  3. Parabolic Arch
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 17th 2007, 11:11 AM
  4. Parabolic arch
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 18th 2007, 02:11 AM
  5. parabolic arch
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 7th 2007, 04:52 AM

Search Tags


/mathhelpforum @mathhelpforum