# Homework Help: Find the curvature of the curve

1. Sep 8, 2008

### whynothis

1. The problem statement, all variables and given/known data

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

$$k = \frac{x'y'' - y'x''}{|\alpha'|^{3}}$$

2. Relevant 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...

3. 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

2. Sep 8, 2008

### Dick

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.

3. Sep 8, 2008

### whynothis

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?

4. Sep 8, 2008

### Dick

Re: curvature

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

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