Curvature of an orthogonal projection

Catria
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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.
 
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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?
 
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?
 
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).
 
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)
 
Yes, that's how I interpreted it.
 
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