# Total curveture of simple closed curve.

#### talisman

I curently have started reading differential geometry from these notes http://www.matematik.lu.se/matematiklu/personal/sigma/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?

#### chiro

MHF Helper
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

I haven't yet learn to use contour integration.

#### chiro

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

Yes.

#### chiro

MHF Helper
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

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.

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