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Involute of a circle

  1. Aug 23, 2007 #1
    Show that for the involute of a circle ([tex] x=a(\cos\theta + \theta\sin\theta) \\ [/tex], [tex] y=a(\sin\theta - \theta\cos\theta) [/tex] for 0 <=[tex]\theta [/tex]<=[tex]\pi \\ [/tex]) radius a, the radius of curvature, is [tex] \sqrt{2as} [/tex], where s is the arc length. The radius of curvature is [tex]\rho = \frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}\\ [\tex];
    My attempt: [tex]\frac{dx}{d\theta} = a \theta\cos\theta \\[/tex] and [tex]\frac{dy}{d\theta} = a\theta\sin\theta \\[/tex],
    that implies that [tex] \frac{dy}{dx} = \tan\theta \\[/tex], which implies that [tex]\frac{d^2y}{dx^2} = \sec^{2}\theta[/tex], therefore [tex] \rho = \sec\theta\\[/tex]. Which is not the answer they suggest. Please help, thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 23, 2007 #2

    dextercioby

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    Check again the second derivative of y wrt x. Be careful about using the chain rule. Remember that

    [tex]\frac{d^2 y}{dx^2}=\frac{d}{d\Huge{\mathbf{x}}}\frac{dy}{dx} [/tex]
     
    Last edited: Aug 23, 2007
  4. Aug 24, 2007 #3
    I checked [tex] \frac{d^2y}{dx^2} [/tex] and got the following [tex] \frac{d^2y}{dx^2} = \frac{d\tan\theta}{d\theta} \cdot \frac{d\theta}{dx} = \sec^{2}\theta \cdot \frac{d\theta}{dx} = \frac{\sec^{2}\theta}{\frac{dx}{d\theta}} \\[/tex]. This gives [tex] \frac{d^2y}{dx^2} = \frac{\sec^{2}\theta}{a\theta \cos\theta} \\[/tex], which yields [tex] \rho =a\theta[/tex]. Which is not the correct answer either, but the integral of [tex] a\theta [/tex] w.r.t, [tex] \theta [/tex] gives s. Thanks for the reply.
     
  5. Aug 28, 2007 #4
    Is anyone interest as to why I can't get the correct answer. I get [tex] \rho = a\theta \mbox{ not the } \sqrt{2sa}[/tex]? Thanks.
     
    Last edited: Aug 29, 2007
  6. Aug 28, 2007 #5

    dextercioby

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    What is "s" equal to ? In terms of [itex] \rho,\theta [/itex], of course ?
     
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