(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Homework Help: Find the curvature of the curve

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