2

The normal curvature of a curve $C$ on a surface $S$ - please correct - can be estimated as the dot product of the second derivative with respect to arc length with the normal (orthogonal) to the plane at $p:$

$$\kappa_n=\langle C_{ss},\vec N\rangle$$

Now, this $C_{ss}$ is orthogonal to the curve but not necessarily to the surface (?).

I thought after listening to this video on Khan Academy, as well as in Henrik Schlichtkrull's Curves and Surfaces, that the curvature was

$$k(t) =\frac{\Vert S'(t)\times S''(t)\Vert}{\Vert S'(t)\Vert^3 }$$

  1. I am sure there is no contradiction, barring misunderstandings, but how can these be reconciled?

And

  1. Why is the second derivative orthogonal to the curve?
Antoni Parellada
  • 8,889
  • 1
    No, you have misunderstood the formula in the Khan Academy. And the correct formula (magnitude of cross product in the numerator) is for the curvature of a space curve, nothing to do with normal curvature on a surface. The answer to your second question is that $C_s$ is everywhere a unit vector and therefore its derivative is orthogonal to it. You might find my differential geometry text, linked in my profile, helpful as you continue. – Ted Shifrin Jan 05 '20 at 23:32

0 Answers0