Curvature of 3D Graph on Point w/ Directional Vector

[kairama15]

TL;DR Summary

Want to find curvature at a point on a 3d graph if the osculating circle is situated in a certain direction.

I know curvature (k) of a 2 dimensional graph y(x) is equal to y''/(1+(y')^2)^(3/2), were y' is the first derivative of y with respect to x, and y'' is the second derivative of y with respect to x.

Is there a formula for the curvature at a point on a 3 dimensional graph z(x,y)? The curvature will be dependent on which direction the curvature of the fitted osculating circle will face, so assume we care about the curvature going along the graph in the direction of the directional unit vector <a,b,c> where that unit vector is lying flat on the plane tangent to the graph at a point.

 

[anuttarasammyak]

For 2d surface in 3d space we have Gaussian curvature https://en.wikipedia.org/wiki/Gaussian_curvature.
Is that you are looking for ?

 

[kairama15]

Unfortunately no. Looking into this page more, I think what I'm looking for is normal curvature. I found a webpage with a formula:

http://wordpress.discretization.de/...ty-introduction-to-the-curvature-of-surfaces/

I'm not sure how to use it. Suppose I have a graph like z=sqrt(1-2*x^2-y^2) and had a unit vector at point (0,0,1) going in the direction <sqrt(2)/2,sqrt(2)/2,0>. Can I have a hand implementing the formula in the link? The formula is near the beginning of the article.

 

[anuttarasammyak]

From your setting I see
$$\nabla z \cdot t$$
for
$$t=(1/\sqrt{2},1/\sqrt{2},0)$$
at x=0,y=0,z=1
may be a quantity you are looking for. It shows ratio of how high you climb on the surface for a horizontal direction t walk.