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Thread: Differential Model

  1. #1
    Junior Member
    Aug 2009

    Differential Model

    A classical problem in the calculus of variations is to find the shape of a curve such that a bead, under the influence of gravity, will slide from point A(0,0) to point B (x1,y1) in the least time. It can be shown that a nonlinear differential for the shape y(x) of the path is y[1 + (dy/dx)^2] = k, where k is a constant. First solve for dx in terms of y and dy and then use the substitution y = ksin^2(theta) to obtain a parametric form of the solution. The curve turns out to be a cycloid.

    Not entirely sure what to do with this problem, mainly b/c I don't really understand what the problem is asking, which includes what a cycloid is...or how you find the parametric solution for one.

    Not sure if this is right, but it's what I've done so far:
    (dy/dx)^2 = k/y - 1
    dy/dx = (k/y - 1)^1^/^2
    dx = dy/[(k/y - 1)^1^/^2]

    Thanks in advance.
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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    Looking good so far. I'm assuming that the problem defines y positive downward, so that you want the positive square root for the derivative. If y is positive up, then you'd need to take the negative square root, since the bead obviously has to be going down under the influence of gravity. So you have


    The suggested substitution is y=k\sin^{2}(\theta), with

    dy=2k\sin(\theta)\cos(\theta)\,d\theta. We obtain the new DE

    dx=2k\sin(\theta)\cos(\theta)\,d\theta\sqrt{\dfrac  {k\sin^{2}(\theta)}{k-k\sin^{2}(\theta)}}.

    Can you continue from here?

    Here is a link to information about cycloids.
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