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Math Help - Hypocycloid

  1. #1
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    Hypocycloid

    Hello, I have to prove that a unit length ladder is tangent to the hypocycloid x^(2/3) + y^(2/3) = 1 at all times in the first quadrant as it falls to the ground. The simple image of the rotating circle creating the hypocycloid confirms this. I'm having a little trouble proving this, however, as it seems obvious that the slopes should just equal each other, however I'm having trouble determining a formula for the ladder. Can I use a^2 + b^2 = 1 or is there an easier way/better forumla? Thanks for all help!
    Last edited by Dudealadude; November 28th 2010 at 07:13 PM.
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  2. #2
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    Okay, here's what I have so far:

    The slope of the ladder would be rise over run, making the slope -L/sqrt(1-L^2).
    The derivative of the hypocyloid would start out to be:
    (2/3)x^(-1/3) x' + (2/3)y^(-1/3)y'= 0
    but how can I take the derivative of two things at once? I would need the speed in the x or speed in the y wouldn't I? To prove they these two things were tangent, I'd have to make the slopes equal, right?
    Nevermind, got it!
    Last edited by Dudealadude; November 30th 2010 at 09:11 AM.
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