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Math Help - Parabolic Arch Bridge

  1. #1
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    Parabolic Arch Bridge

    A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch, a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.
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    Quote Originally Posted by magentarita View Post
    A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch, a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.
    Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.

    The points (-50, 0) and (50, 0) lie on the parabola on the x-axis since the span is 100.

    The points (-40, 10) and (40, 10) also lie on the parabola.

    We use (x-h)^2=4p(y-k) for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.

    Substituting point (50, 0) into this equation, we get:

    (50-0)^2=4p(0-k)

    \boxed{2500=0p-4pk}

    Substituting point (40, 10) into this equation, we get:

    (40-0)^2=4p(10-k)

    \boxed{1600=40p-4pk}

    Use the two boxed equations to solve for p.

    2500= \ \ 0p-4pk
    1500=40p-4pk

    Subtract the two equations to get:

    900=-40p

    p=-\frac{45}{2}

    Now, to find k, we substitute p back into one of our boxed equations.

    2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k

    2500=90k

    k=\frac{350}{9} \approx 27.8 feet.
    Attached Thumbnails Attached Thumbnails Parabolic Arch Bridge-parabola.bmp  
    Last edited by masters; October 15th 2008 at 03:33 PM.
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  3. #3
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    ok....

    Quote Originally Posted by masters View Post
    Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.

    The points (-50, 0) and (50, 0) line on the parabola on the x-axis since the span is 100.

    The points (-40, 10) and (40, 10) also lie on the parabola.

    We use (x-h)^2=4p(y-k) for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.

    Substituting point (50, 0) into this equation, we get:

    (50-0)^2=4p(0-k)

    \boxed{2500=0p-4pk}

    Substituting point (40, 10) into this equation, we get:

    (40-0)^2=4p(10-k)

    \boxed{1600=40p-4pk}

    Use the two boxed equations to solve for p.

    2500= \ \ 0p-4pk
    1500=40p-4pk

    Subtract the two equations to get:

    900=-40p

    p=-\frac{45}{2}

    Now, to find k, we substitute p back into one of our boxed equations.

    2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k

    2500=90k

    k=\frac{350}{9} \approx 27.8 feet.
    What a great reply.
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    A riddle wrapped in an enigma
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    Quote Originally Posted by magentarita View Post
    What a great reply.
    You are toooooo kind! Blush Blush
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    ok.....

    Quote Originally Posted by masters View Post
    Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.

    The points (-50, 0) and (50, 0) lie on the parabola on the x-axis since the span is 100.

    The points (-40, 10) and (40, 10) also lie on the parabola.

    We use (x-h)^2=4p(y-k) for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.

    Substituting point (50, 0) into this equation, we get:

    (50-0)^2=4p(0-k)

    \boxed{2500=0p-4pk}

    Substituting point (40, 10) into this equation, we get:

    (40-0)^2=4p(10-k)

    \boxed{1600=40p-4pk}

    Use the two boxed equations to solve for p.

    2500= \ \ 0p-4pk
    1500=40p-4pk

    Subtract the two equations to get:

    900=-40p

    p=-\frac{45}{2}

    Now, to find k, we substitute p back into one of our boxed equations.

    2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k

    2500=90k

    k=\frac{350}{9} \approx 27.8 feet.
    Another fabulous reply!
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