# The Frenet Serret frame

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

WWGD
Gold Member
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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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
Gold Member
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?

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

yes.

Can you show me how to find it?

fresh_42
Mentor
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
Gold Member
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.

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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?

What is L(t)?

fresh_42
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.

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
Gold Member
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.

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WWGD
Gold Member
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 ).

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
Gold Member
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.