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Here's the full problem: Assume that the osculating plane of a line of curvature $C \subset S$, which is nowhere tangent to an asymptotic direction, makes a constant angle with the tangent plane of $S$ along $C$. Prove that $C$ is a plane curve.

My attempt:

Let $\alpha : I \to S$ be some regular parametrization of $C$. Then the binormal vector $b(s)$ determines the osculating plane of $\alpha$, and the tangent plane to $S$ is determined by the normal vector $N$. Thus the stipulation is that

$$b(s) \cdot N(s) = const$$ or

$$ b'(s) \cdot N(s) + b(s) \cdot N'(s) = 0$$

Since $C$ is a line of curvature we have $N'(s) = \lambda(s) \alpha'(s)$ and so $b(s) \cdot N'(s) = 0$. Then we are left with

$$\tau(s) n(s) \cdot N(s) = 0 $$

If I can show that $n(s) \cdot N(s) \neq 0$ then I have my result, since this means $\tau(s) = 0$. But I don't think this should be true in general, which means I've messed up somewhere.

Any help would be greatly appreciated!

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If you had $n(s)\cdot N(s)=0$ for some $s$, the normal curvature of the curve would be $0$ at $\alpha(s)$, contradicting the hypothesis that $\alpha$ is never tangent to an asymptotic direction.

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  • $\begingroup$ Ted, thank you very much for the response! I missed that condition when I was working through this last night - maybe too tired... $\endgroup$ – Nitin Mar 18 '15 at 17:24

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