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

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- Thread starter rabbed
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- #1

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

- #2

WWGD

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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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- #3

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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

lavinia

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What is a Frenet-Serret tangent vector?

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?

- #5

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The vector T here:What is a Frenet-Serret tangent vector?

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..?

- #6

lavinia

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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

- #7

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Are you sure? I have tried to find it for y = x^2, but it doesn't seem possible.You can always parameterize a curve by arc length

- #8

lavinia

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

yes.

- #9

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Can you show me how to find it?

- #10

fresh_42

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What about ##L(t) = \int_a^t ds##?Are you sure? I have tried to find it for y = x^2, but it doesn't seem possible.

- #11

lavinia

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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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- #12

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

Yes, I want s = 0 at t = 0.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.

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

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

- #14

fresh_42

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- #15

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That length does not depend on the geometry of the curve from a to t?

- #16

WWGD

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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.

That length does not depend on the geometry of the curve from a to t?

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- #17

WWGD

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(L(t), L(t)^2 ).

- #18

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Since s = L(t), i think you need to substitute with t = L^-1(s) toFor (t, t^2) find the arc-length formula L(t) and substitute into

(L(t), L(t)^2 ).

(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

martinbn

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- #20

WWGD

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

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