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Math Help - Differential Model

  1. #1
    Junior Member
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    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
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    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

    dx=\dfrac{dy}{\sqrt{k/y-1}}=dy\sqrt{\dfrac{y}{k-y}}.

    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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