Can the Frenet Serret Frame Aid in Arc Length Parameterizations?

[rabbed]

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

 

[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?.
:

 

[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?

 

[lavinia]

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?

What is a Frenet-Serret tangent vector?

 

[rabbed]

What is a Frenet-Serret tangent vector?

The vector T here:
https://en.wikipedia.org/wiki/Frenet–Serret_formulas

I have no idea if you can always find this vector, since it seems related to arc length..?

 

[lavinia]

The vector T here:
https://en.wikipedia.org/wiki/Frenet–Serret_formulas

I have no idea if you can always find this vector, since it seems related to arc length..?

You can always parameterize a curve by arc length

 

[rabbed]

You can always parameterize a curve by arc length

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

 

[lavinia]

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

yes.

 

[rabbed]

Can you show me how to find it?

 

[fresh_42]

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

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

 

[lavinia]

I assume it is the curve $c(t) = (t,t^2)$?

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.

 

[rabbed]

I assume it is the curve c(t)=(t,t2)c(t)=(t,t2)c(t) = (t,t^2)?

Yep

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

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?

 

[rabbed]

What about L(t)=∫tadsL(t)=∫atdsL(t) = \int_a^t ds?

What is L(t)?

 

[fresh_42]

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.

 

[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?

 

[WWGD]

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?

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.

 

[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 ).

 

[rabbed]

For (t, t^2) find the arc-length formula L(t) and substitute into
(L(t), L(t)^2 ).

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

 

[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.

 

[WWGD]

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

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