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Math Help - Total curveture of simple closed curve.

  1. #1
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    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 \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?
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  2. #2
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    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)?
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  3. #3
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    Re: Total curveture of simple closed curve.

    I haven't yet learn to use contour integration.
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  4. #4
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    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?
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  5. #5
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    Re: Total curveture of simple closed curve.

    Yes.
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  6. #6
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    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.
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  7. #7
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    Re: Total curveture of simple closed curve.

    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.
    Thanks from xxp9
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