sanctus Posted April 29, 2008 Report Posted April 29, 2008 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? Quote
Pyrotex Posted April 29, 2008 Report Posted April 29, 2008 I am sad to say, you lost me at "...initial curve in R²". :( Quote
sanctus Posted April 30, 2008 Author Report Posted April 30, 2008 Pyro, I know you know that, it is just cartesian 2D space...ok I used R² instead of: [math]\mathcal{R}^2 \text{ or } \mathbb{R}^2[/math] Quote
LaurieAG Posted May 6, 2008 Report Posted May 6, 2008 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? Quote
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