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Homework Help: Find the curvature of the curve

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Given a parametric curve, [tex] \alpha(t) = (x(t),y(t) )[/tex], not necessarily arc length parameterized show that the curvature is given by:

    [tex] k = \frac{x'y'' - y'x''}{|\alpha'|^{3}} [/tex]

    2. Relevant equations

    As I understand this the curvature is defined from the point of view of a arc length parameterization as
    [tex] k = |\frac{d^{2}\alpha(s)}{ds^{2}}| [/tex]

    So I tried using the chain rule...

    3. The attempt at a solution

    [tex] \frac{d\alpha}{ds} = \frac{d\alpha}{dt}\frac{dt}{ds} [/tex]where dt/ds = 1/(ds/dt) = [tex]1/|\frac{d\alpha}{dt}(t)| [/tex]
    Then I differentiated again to get...
    [tex]\frac{d}{ds} \frac{d\alpha}{dt}\frac{1}{|\frac{d\alpha}{dt}(t)|} = \frac{d}{dt}\frac{dt}{ds} \frac{d\alpha}{dt}\frac{1}{|\frac{d\alpha}{dt}(t)|}=\frac{d^{2}\alpha}{dt^{2}}\frac{1}{|\frac{d\alpha}{dt}(t)|^{2}}}-\frac{d\alpha}{dt}\frac{\alpha'\cdot\alpha''}{|\frac{d\alpha}{dt}(t)|^{4}}[/tex]

    This isn't what I should get. It seems that I am clearly missing something since all the problems I try and do along these lines I am stuck on. Does anyone have any insight on this?

    Thanks in advanced
  2. jcsd
  3. Sep 8, 2008 #2


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    Re: curvature

    Don't you think you ought to start by representing the curve in parametric form (x(t),y(t))? Now start differentiating that. The unit tangent is (x',y')/sqrt(x'^2+y'^2). The curvature is the norm of the derivative of that.
  4. Sep 8, 2008 #3
    Re: curvature

    Thanks for the advice. The expression turned out to be right I just needed to do a little foot work to get it in the same form : )

    I am curious to know, however, if you had a more elegant way in mind?
  5. Sep 8, 2008 #4


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    Re: curvature

    Nah. Just differentiate, cancel some terms and collect stuff.
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