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rabbed
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What is it useful for? Can it help in creating arc length parameterizations?
What is a Frenet-Serret tangent vector?rabbed said: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?
The vector T here:lavinia said:What is a Frenet-Serret tangent vector?
rabbed said: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..?
Are you sure? I have tried to find it for y = x^2, but it doesn't seem possible.lavinia said:You can always parameterize a curve by arc length
rabbed said: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##?rabbed said:Are you sure? I have tried to find it for y = x^2, but it doesn't seem possible.
lavinia said:I assume it is the curve ##c(t) = (t,t^2)##?
Yeplavinia said:I 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.lavinia said: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.
What is L(t)?fresh_42 said:What about L(t)=∫tadsL(t)=∫atdsL(t) = \int_a^t ds?
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.rabbed said: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?
Since s = L(t), i think you need to substitute with t = L^-1(s) toWWGD said:For (t, t^2) find the arc-length formula L(t) and substitute into
(L(t), L(t)^2 ).
Well, yes, the issue is that the curve's parametrization is defined in function of its arc-length.rabbed said: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
The Frenet Serret frame is a mathematical concept used in differential geometry to describe the local behavior of a curve in three-dimensional space. It consists of three mutually perpendicular unit vectors that are tangent, normal, and binormal to the curve at any given point.
The Frenet Serret frame is calculated using the derivative of the position vector of the curve. This derivative is then used to find the unit tangent vector, which is then used to find the normal vector and binormal vector.
The Frenet Serret frame is significant because it provides a way to describe the curvature and torsion of a curve at any point. This information is important in many fields, including physics, engineering, and computer graphics.
The Frenet Serret frame is used in many real-world applications, such as describing the motion of objects in space, analyzing the shape of DNA molecules, and creating smooth animations in computer graphics.
One limitation of the Frenet Serret frame is that it can only be used for smooth curves, meaning curves that have a continuous first derivative. Additionally, it is only applicable in three-dimensional space and cannot be used for curves in higher dimensions.