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My question comes from a paper I'm reading and I know it is quite basic stuff, but atm I'm lost so better asking...

 

Suppose we have a smooth, simple and closed initial curve [imath]\gamma(0)[/imath] in R². Supposte that [imath]\gamma(t)\quad t\in[0,\infty)[/imath] is a one parameter family of curves generated by moving the initial curve along the normal vector field with speed F.

 

Furthermore let X(s,t)=(x(s,t),y(s,t)) be the position vector which parametrizes [imath]\gamma(t)[/imath] by s, [imath]0\leq s \leq S \quad X(0,t)=X(S,t)[/imath].

 

Also the curve is parametrized so that the interior is on the left in the direction of increasing s.

 

Eventually the questions which are mainly things I forgot because for a few years I didn't use them anymore...

  • the unit normal is given by
    [math]\frac{1}{\vert \partial_sX\vert}(\partial_s y, -\partial_s x)[/math]
     
    Graphically it is easy to see that this is normal but how do you show it? I mean usually you would take the gradient, but how would it be defined here? I mean you derive only with respect to the parametrization parameter not x and y...
     
  • the curvature which is usually the laplacian of the curve is said to be:
    [math]\frac{\partial^2_sy\partial_sx-\partial^2_sx\partial_s y}{(\partial_sx)^2+(\partial_sy)^2}[/math]
     
    How is this found?

 

Thanks for telling me how to do second (or first?) year stuff, but I guess everybody can get confused sometimes, no?

Posted
Graphically it is easy to see that this is normal but how do you show it? I mean usually you would take the gradient, but how would it be defined here? I mean you derive only with respect to the parametrization parameter not x and y...

 

While you might not derive with x and y you would have to regard the gradient as the line between the start and end positions of each curve, for each curve, wouldn't you?

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