Curvature of an orthogonal projection

[Catria]

Homework Statement

Let \vec{X(t)}: I \rightarrow ℝ³ be a parametrized curve, and let I \ni t be a fixed point where k(t) \neq 0. Define π: ℝ³ \rightarrow ℝ³ as the orthogonal projection of ℝ³ onto the osculating plane to \vec{X(t)} at t. Define γ=π\circ\vec{X(t)} as the orthogonal projection of the space curve \vec{X(t)} onto the opsculating plane. Prove that the curvature k(t) is equal to the curvature of the plane curve \vec{γ}.

Homework Equations

k=\frac{\left\|\vec{X'(t)}\times\vec{X''(t)}\right\|}{\left\|\vec{X'(t)}\right\|^{3}} = Curvature

The Attempt at a Solution

I don't even know how to formulate the equation for the orthogonal projection of X onto the osculating plane, so I can't even begin to understand how to solve the problem in question.

 

[haruspex]

If v and B are vectors, the projection of v in the direction of B can be obtained by subtracting out the component of v in the B direction: v - (B.v)B/(B.B).
The tangent vector to the curve should be something like T = X'(t), and the normal N is like T'. The normal to the osculating plane is then B = TxN.
Does that help?

 

[Catria]

T, N and B form an orthonormal base of R^3 as well... but beyond that is it possible that the curvature of a curve may not be the same as that of its orthogonal projection for all s?

 

[haruspex]

That the curvature is the same in the projection is intuitively obvious, but proving it is the hard part. The use of variables in the question is a bit confusing, using t for a variable and t for a particular value of it, so I'm going to use s for the variable.
Based on what I wrote before, define Y(s) as the projection of X(s). That gives you an equation for Y(s) in terms of X(s) and B, where B is defined in terms of X and its derivatives at the point s = t. You should then be able to write down an expression for the curvature of Y at t. With luck, you can reduce it to k(t).

 

[Catria]

I realized that the question was perhaps improperly formulated and it should have said that the curvature of the curve C is the same as the curvature of its orthogonal projection at s0 (and not for all s)

 

[haruspex]

Yes, that's how I interpreted it.