# Unit speed curves

1. Oct 1, 2013

### Lee33

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

Let $\alpha(s)$ and $\beta(s)$ be two unit speed curves and assume that $\kappa_{\alpha}(s)=\kappa_{\beta}(s)$ and $\tau_{\alpha}(s)=\tau_{\beta}(s)$, where $\kappa$ and $\tau$ are respectively the curvature and torsion. Let $J(s) = T_{\alpha}(s)\dot\ T_{\beta}(s)+N_{\alpha}(s) \dot\ N_{\beta}(s) +B_{\alpha}(s) \dot\ B_{\beta}(s).$

Show that:

$J(0)=3$ and $J(s)=3$ implies that the Frenet frames of $\alpha$ and $\beta$ agree at $s$

$J'(s) = 0$ and $\alpha(s) = \beta(s)$ for all $s$.

2. Relevant equations

Frenet frames

3. The attempt at a solution

For the first question, I know that the Frenet frame vectors $T,B,N$ are unit vectors but how can I formally prove the given statement?

I know for the second statement that $J'(s) = 0$ everywhere implies that the Frenet frames agree and since $\alpha$ and $\beta$ are unit speed, they are equal to the integral of their tangent vector. Also, how can I formally prove this second part?

2. Oct 1, 2013

### Dick

For any two vectors v.w=|v||w|cos(θ). If v and w are unit vectors then that's less than one unless the vectors are parallel. So?

Last edited: Oct 1, 2013
3. Oct 1, 2013

### Lee33

Dick -

If they are unit vectors and are parallel then there dot product is 1.

4. Oct 1, 2013

### Dick

The point is the dot product is ONLY 1 if they are parallel, otherwise it's less. You have three of them summing to 3. So?

5. Oct 1, 2013

### Lee33

Since they sum to 3 they are parallel hence their dot product equals 1.

Is it really that simple or am I missing something?

6. Oct 1, 2013

### Dick

No, I think it's really that simple.

7. Oct 2, 2013

### Lee33

Thanks Dick!

For D(s), is it essentially the same thing?

8. Oct 2, 2013

### Dick

What's D(s)? Look, if J(s)=3 then all three Frenet frame vectors must be equal in $\alpha$ and $\beta$, yes? Am I missing something?

Last edited: Oct 2, 2013
9. Oct 2, 2013

### Lee33

Opps, I meant J(s), sorry about that. But yes they will be equal, thanks again!