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Rebesques · #2 June 8th, 2009, 09:52 PM

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).