# Thread: Line of curvature

1. ## Line of curvature

Show that if a plane touches a surface along a curve, then it is a curvature line and all of its points are parabolic.

Thank you for your help in advance!

2. ## Re: Line of curvature

Originally Posted by doug
Show that if a plane touches a surface along a curve, then it is a curvature line and all of its points are parabolic.
That is not telling us much.
What surface? A parabolic what?
What does 'touch' mean? Cut? Slice how?

3. ## Re: Line of curvature

Originally Posted by Plato
That is not telling us much.
What surface? A parabolic what?
What does 'touch' mean? Cut? Slice how?
This is about 2-dimensional surfaces in $\mathbb{R}^3$, a point is parabolic if $LN-M^2=0$ (where L, N, M are the coefficients of the second fundamental form). I use word 'touch' in the same meaning as a tangent line 'touches' a curve.

Thank you for your help in advance!

4. ## Re: Line of curvature

I think I understood the question:
A plane is tangent to a surface embedding in R^3, at each point of a curve on the surface. Prove the tangent vector to the curve is a principle direction, and the second fundamental form is degenerate( singular, rank 0, or zero determinant) along the curve.

Originally Posted by Plato
That is not telling us much.
What surface? A parabolic what?
What does 'touch' mean? Cut? Slice how?

5. ## Re: Line of curvature

The easiest way to understand the problem is, using the Gauss Weingarten map, N. Normal vectors of surface along the curve is constant, actually they're all identical to the normal vector to the touching plane. So the curve is mapped to a single point by the Gauss map N. So the induced linear map in the tangent space, i.e. the differential dN, maps the tangent vector v of the curve to a 0 vector.
So dN is degenate, with v as a eigenvector, belonging to the eigenvalue 0. And note that dN is nothing but the second fundamental form, we're done.

6. ## Re: Line of curvature

More reference goes here Shape operator