# Unit Speed Curves: Showing Frenet Frames Agree at s

• Lee33
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?Since they sum to 3 they are parallel hence their dot product equals 1.

## Homework Statement

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##.

Frenet frames

## 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?

Lee33 said:

## Homework Statement

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##.

Frenet frames

## 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?

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?

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Dick -

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

Lee33 said:
Dick -

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

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?

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

Is it really that simple or am I missing something?

Lee33 said:
Since they sum to 3 they are parallel hence their dot product equals 1.

Is it really that simple or am I missing something?

No, I think it's really that simple.

Thanks Dick!

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

Lee33 said:
Thanks Dick!

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

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?

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