- #1
whynothis
- 15
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Homework Statement
Given a parametric curve, \alpha(t) = (x(t),y(t) ), not necessarily arc length parameterized show that the curvature is given by:
k = \frac{x'y'' - y'x''}{|\alpha'|^{3}}
Homework Equations
As I understand this the curvature is defined from the point of view of a arc length parameterization as
k = |\frac{d^{2}\alpha(s)}{ds^{2}}|
So I tried using the chain rule...
The Attempt at a Solution
\frac{d\alpha}{ds} = \frac{d\alpha}{dt}\frac{dt}{ds}where dt/ds = 1/(ds/dt) = 1/|\frac{d\alpha}{dt}(t)|
Then I differentiated again to get...
\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}}
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