Thread: Differential geometry question

Let p be a point on a curve γ on a surface S, and let Π be the tangent plane to S at p. Let μ be the curve obtained by projecting γ orthogonally onto Π . Show that the curvature of the plane curve μ at p is equal up to sign,to the geodesic curvature of γ at p.

I have no idea how to prove this. All i know than is that the normal curvature is given by

Let T be the tangent vector( velocity) of γ, then dT/ds is the curvature vector( acceleration), and the norm of its projection onto Π is the geodesic curvature. Then the statement is a consequence of the following statement:
Projection of the acceleration of a curve equals to the acceleration of its projection.
This is easy to see if your choose a coordinate system so that the projection is (x,y,z) |->(x,y), so the acceleration is (x'', y'', z''), its projection is (x'',y''). And the projection of the curve is (x,y), its acceleration is (x'',y''). DONE.
The deep idea is, the projection onto Π is an isometry in a infinisimal neighborhood of p.