Total curveture of simple closed curve.

Nov 2012
4
1
Greece
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
Sep 2012
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Australia
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)?
 
Nov 2012
4
1
Greece
I haven't yet learn to use contour integration.
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
I apologize for asking (I should have asked in the first response), but is k(s) the curvature at a parametrized s value?
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
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.
 
Nov 2012
4
1
Greece
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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