# Curvature of an orthogonal projection

1. Oct 25, 2012

### Catria

1. The problem statement, all variables and given/known data

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 $\vec{γ}$.

2. Relevant equations

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

3. 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.

2. Oct 26, 2012

### 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?

3. Oct 26, 2012

### 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?

4. Oct 26, 2012

### 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).

5. Oct 26, 2012

### 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)

6. Oct 26, 2012

### haruspex

Yes, that's how I interpreted it.