A The Frenet Serret frame

1. Apr 4, 2017

rabbed

What is it useful for? Can it help in creating arc length parameterizations?

2. Apr 4, 2017

WWGD

EDIT As I remember, it is a set of orthonormal moving frames, i.e., they provide a frame at each point with the moving axes : Tangent, Normal and Binormal moving axes are a basis for the derivatives of the curve. Did you have any specific question in mind?.
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Last edited: Apr 4, 2017
3. Apr 5, 2017

rabbed

I'm interested in it's relation to arc length parameterization of a curve.

And let's say I have some curve in explicit form y = f(x) that (when default parameterized by setting x = t, y = f(t)) gives some non-unit-speed velocity along the curve.
If I have the Frenet Serret tangent-vector at any point on the curve, can I calculate the new angle at which I must move in at each point in order to stay on the curve having a speed of 1?

4. Apr 6, 2017

lavinia

What is a Frenet-Serret tangent vector?

5. Apr 6, 2017

6. Apr 6, 2017

7. Apr 6, 2017

rabbed

Are you sure? I have tried to find it for y = x^2, but it doesn't seem possible.

8. Apr 6, 2017

yes.

9. Apr 6, 2017

rabbed

Can you show me how to find it?

10. Apr 6, 2017

Staff: Mentor

What about $L(t) = \int_a^t ds$?

11. Apr 6, 2017

lavinia

The new parameter $s$ is the length of the curve at time $t$ starting at an arbitrary point. This can always be done from an original parameterization by integration if the original parameter never has a derivative of zero.

Last edited: Apr 6, 2017
12. Apr 6, 2017

rabbed

Yep
Yes, I want s = 0 at t = 0.
The derivative of the original parameter t is 1 at all points so it should work, or do you mean the derivative of (t,t^2)?

But doesn't it come down to an integral that can't be expressed in terms of standard mathematical functions, or some expression that cannot be inverted. So in the end it can't be done?

13. Apr 6, 2017

rabbed

What is L(t)?

14. Apr 6, 2017

Staff: Mentor

The length you want to parameterize. It tells you how far you got on your curve, starting at $L(a)=0$. So it's a parameter $s=L(t)$ in accordance to the arc length of the curve. What happens to the formulas if you describe your curve by $s$ depends on the individual cases.

15. Apr 6, 2017

rabbed

But the indefinite integral of 1*ds would always be s, and the definite integral would always be t-a?
That length does not depend on the geometry of the curve from a to t?

16. Apr 6, 2017

WWGD

Why are you computing the integral of 1ds? You need the integral $\int_a^b \sqrt{ 1+ f'(x)^2}dx$ which is the general arc -length formula for a (differentiable, of course) curve given as y=f(x). I think the simplest example is that of the circle C((0,0),1 )given as f(t)=(Cost, Sint), starting at t=0 . At time t, the length of the curve is the perimeter , which is precisely t . EDIT: I think this is what Lavinia and Fresh meant. EDIT 2: in Fresh's post, s is a composite function (the length function) so you need to use the chain rule after defining s.

Last edited: Apr 6, 2017
17. Apr 6, 2017

WWGD

Ultimately it comes down to substituting the arc-length formula where the usual parameter goes. For (t, t^2) find the arc-length formula L(t) and substitute into
(L(t), L(t)^2 ).

18. Apr 7, 2017

rabbed

Since s = L(t), i think you need to substitute with t = L^-1(s) to
(L^-1(s), (L^-1(s))^2 )
In the case of y = x^2 it's possible to find L(t), but i don't think you can find L^-1(s)?
http://planetmath.org/arclengthofparabola

19. Apr 7, 2017

martinbn

I think you expect to find it explicitly. But that is not need (and not possible in general). The inverse function theorem guaranties that it exists. That is all you need to say that you can choose to parametrize your curve by arc length.

20. Apr 7, 2017

WWGD

Well, yes, the issue is that the curve's parametrization is defined in function of its arc-length.