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AP Calc

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When a chain hangs under its own weight, its shape is catenary. The equation of a catenary with a vertex on the y-axis is y=e^x+e^-x. Find the length of the chain from x=-2 to x=2. Then use the same technique from Problem #8(Previous AP calc problem I posted) to find the length of the parabola with the same vertex and endpoints.

I go the first part that asks to find the length to be 12.225, but I don't understand what the next part is asking or how to do it!

  • AP Calc - ,

    when x = 0, middle, y = e^0 + e^-0 = 2

    when x = 2 or -2 (end) y = e^2 + 1/e^2
    = 7.524

    so parabola has vertex at (0, 2)
    and goes through (-2, 7.524 ) and (-2,7.524)

    we might as well move this down weo so the vertex is at the origin
    vertex at (0,0)
    (2, 5.524) and (-2, 5.524)
    so form is
    y = a x^2 + b x
    b is 0 because symmetric about y axis
    (even function)
    y = a x^2
    when x = 2, y = 5.524
    5.524 = a (4)
    a = 1.381
    so our parabola is
    y = 1.381 x^2
    now use your technique (I do not know which you are using) to find the length of y = 1.381 x^2 from 0 to 2 and double that for -2 to + 2

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