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Math Help - Parabolic arches/canals

  1. #1
    Newbie hrose21's Avatar
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    Parabolic arches/canals

    I have two questions I need help with. They are for a pre-calc class....

    1) A building has an entry with the shape of a parabolic arch 4ft high and 26ft wide at the base. Find an equation for the parabola if the vertex is at the origin of the coordinate system.

    2) A cross-section of an irrigation canal is a parabola. If the surface of the water is 33ft wide and the canal is 14ft deep in the center, how deep is it 1ft from the edge?

    If anyone can help it would be much appreciated. Thanks!
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  2. #2
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    y=a(x-h)^2+k Vertex is at (h,k). Vertex at (0,0)

    y=a(x-0)^2+0=y=ax^2

    Another point is (13,-4). The arch is 26 ft so 13 feet from the origin it should be 4 ft below the top of the arch.

    -4=a(-13)^2 . Therefore a=-\frac{4}{169}

    Equation is then
    y=-\frac{4}{169}(x^2)
    Last edited by jabroni1212; November 29th 2007 at 07:57 PM.
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  3. #3
    Newbie hrose21's Avatar
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    Thanks

    Thanks so much! For the second problem I would just follow the same formula and then plug in 15.5 for x, correct?
    Last edited by hrose21; November 29th 2007 at 07:54 PM. Reason: oops
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  4. #4
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    Exactly yes.
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  5. #5
    Newbie hrose21's Avatar
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    Thanks so much!
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