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Homework Help: Evolute question

  1. Aug 22, 2007 #1
    Prove that the evolute of the rectangular hyperbola [tex] x=a\cosh\theta [/tex], [tex] y=a\sinh\theta [/tex] is [tex] x^{\frac{2}{3}} - y^{\frac{2}{3}}= (2a)^{\frac{2}{3}}\\ [/tex].
    Let (X,Y) be a pair of coordinates on the center of curvature (the evolute) of the hyperbola. [tex] X =x -\frac{y^{'}(1 + (y^')^2)}{y^{''}} \\ [/tex] and [tex] Y = y-\frac{(1 + (y^')^2)}{y^{''}} \\ [/tex], where [tex] y^{'}=\frac{dy}{dx} \\[/tex].
    Now [tex] \frac{dx}{d\theta} = a sinh\theta [/tex], [tex] \frac{dy}{d\theta}=a cosh\theta \\ [/tex]; therefore [tex] \frac{dy}{dx} = \coth\theta \\ [/tex];
    hence [tex] \frac{d^2y}{dx^2} = \frac{d \coth\theta}{d\theta}\frac{d\theta}{dx} =-\cosech^2\theta \frac{d\theta}{dx}\\ [/tex].
    Therefore [tex] y^{''} =-\frac{cosech^2 \theta}{a\sinh\theta}\\ [/tex].
    Therefore [tex] X = a\cosh\theta + a\cosh\theta(\sinh^2\theta + \cosh^2\theta)\\[/tex] and [tex] Y = a\sinh\theta + a\sinh\theta(\sinh^2\theta + \cosh^2\theta) \\[/tex].
    Thest are the parametric equations of the evolute. How do I get the cartesian form of the equations for the evolute. I mean how do I eliminate [tex] \theta [/tex]. Thanks for the help.
     
    Last edited: Aug 23, 2007
  2. jcsd
  3. Aug 31, 2007 #2
    Therefore I get [tex] x^{\frac{1}{3}} - y^{\frac{1}{3}} = (2a)^{\frac{1}{3}}((\cosh^3\theta)^{\frac{1}{3}} \\ - ( \sinh^{\frac{1}{3}}\theta + \sinh^3\theta)^{\frac{1}{3}}) \\[/tex] which gives [tex]x^{\frac{2}{3}} - y^{\frac{2}{3}} = (2a)^{\frac{2}{3}}( 1 - \sinh^{\frac{2}{3}}\theta)\\ [/tex] Can anyone see where I went wrong? Thanks for the help.
     
  4. Sep 3, 2007 #3
    John

    There is a typo in the formula for Y above. It should be Y=y + etc. Then you get the required result.

    Regards
     
  5. Sep 3, 2007 #4
    Thanks for the reply.
     
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