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Old November 5th, 2008, 03:58 PM
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Default parabolic or planar points

Show that if a surface is tangent to a plane along a curve, then the points of this curve are either parabolic or planar.

Defition: A point of a surface is parabolic if det(dN_p) = 0 with dN_p =/= 0. A point of a surface is planar if dN_p = 0.
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Old June 8th, 2009, 09:52 PM
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Let the surface be parametrized by F(u,v), \ (u,v)\in D and the curve by x(t)=F(u(t),v(t)), \ t\in I.
Since x is also a plane curve, we have that the binormal vector b(t) satisfies b(t)=\pm N(u(t),v(t)) and that b'=0 (as there is no torsion). So for points p along the curve 0=b'=N_uu'+N_vv'={\rm d}N_p(u',v')=0. Now this gives\{N_u,N_v\} are linearly dependent at p, so if they are not zero (and the point p planar) then {\rm det}[{\rm d}N_p]=0 (and so the point is parabolic).
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