Total curveture of simple closed curve.

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 $\displaystyle \gamma : \mathbb{R} \rightarrow \mathbb{R}^2$ parametrize a simple closed curve by arclength. Show that if the period of $\displaystyle \gamma$ is $\displaystyle L>0$ then the total curvature satisfies $\displaystyle \int\limits_0^L k(s)ds = 2 \pi$.

Any idea?

Re: Total curveture of simple closed curve.

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)?

Re: Total curveture of simple closed curve.

I haven't yet learn to use contour integration.

Re: Total curveture of simple closed curve.

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

Re: Total curveture of simple closed curve.

Re: Total curveture of simple closed curve.

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.

Re: Total curveture of simple closed curve.

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