Total curveture of simple closed curve.

**talisman** · November 1st 2012, 01:43 PM

I curently have started reading differential geometry from these notes http://www.matematik.lu.se/matematik...igma/Gauss.pdf and I am trying to solve the exercise 2,7 which says:
Let the positively oriented $\gamma : \mathbb{R} \rightarrow \mathbb{R}^2$ parametrize a simple closed curve by arclength. Show that if the period of $\gamma$ is $L>0$ then the total curvature satisfies $\int\limits_0^L k(s)ds = 2 \pi$ .
Any idea?

**chiro** · November 1st 2012, 06:20 PM

Hey talisman.

Can you use contour integration and the magnitude of said result by using the fact that the contour is a closed contour (so it will involve a 2*pi*i)?

**talisman** · November 1st 2012, 09:21 PM

I haven't yet learn to use contour integration.

**chiro** · November 1st 2012, 10:59 PM

I apologize for asking (I should have asked in the first response), but is k(s) the curvature at a parametrized s value?

**talisman** · November 2nd 2012, 05:10 AM

Yes.

**chiro** · November 3rd 2012, 03:31 PM

Try looking at definition 2.11 and using the composition of taking a circle and deforming it while still keeping the curve a closed curve.

**talisman** · November 5th 2012, 02:57 AM

I found it like this: becuase $\gamma$ is parametrize by arclength then $\parallel T(s) \parallel =1, \forall s$ , that means that $T(s)$ can be written as $T(s)=(cos \theta (s) , sin \theta (s))$ , where $\theta (s)$ is the angle beetwen $T(s)$ and $e_{1} = (1,0)$ , then because $\parallel T(s) \parallel =1$ we have that $N(s)= \frac{ \dot{T(s)} }{\parallel \dot{T(s)} \parallel}$ . we get $k(s)=\langle \dot{T(s)} ,N(s) \rangle = \theta '(s)$ and we get the result we want.

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