Find the curvature of the curve

[whynothis]

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

 

[Dick]

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.

 

[whynothis]

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.

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?

 

[Dick]

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?

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