
#1
Feb1110, 02:09 PM

P: 625

Hello,
if one wants to represent a parametric curve in a Ndimensional vector space, the straightforward way is to express it as: [tex]\mathbf{v} = f_1(t)\mathbf{e}_1 + \ldots + f_n(t)\mathbf{e}_n[/tex] However there are infinite representations! I could make any substitution [tex] t = g(t)[/tex] as long as g is invertible and I would get the same curve in the space. How should a curve (or surface) be expressed in a general way, independent of the type of parametrization? 



#2
Feb1110, 06:15 PM

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PF Gold
P: 9,017

A curve C in a vector space V is by definition a function [itex]C:I\rightarrow V[/itex], where I is some interval. Even if I=[0,1] and [itex]f:I\rightarrow I[/itex] is an increasing smooth bijection such that f(0)=0 and f(1)=1, [itex]B=C\circ f[/itex] is a different curve according to this definition (which is pretty standard). What you call a curve is what I would call the "graph" or the "image" of the curve. It's the set of "values" of the curve. You can e.g. write that set as [itex]C(I)=\{C(t)t\in I\}[/itex].




#3
Feb1110, 07:14 PM

P: 211

In the study of the geometric properties of a curve (i.e. those properties that are invariant under reparametrization, like curvature), a standard assumption is that curves are "parametrized by arclength." That is, one can look at maps [itex]f:[0,L] \rightarrow \mathbb{R}^n[/itex] with [itex]f'(t)=1[/itex] for all t. This representation is unique modulo orientation and the extra assumption simplifies many formulas.




#4
Feb1310, 02:22 AM

P: 625

how to represent a parametric curve?
Uhm ... does tensor calculus have anything to say in this type of problem?
the reparametrization t=g(t), with g invertible, is after all a coordinate change. Isn't tensor calculus supposed to provide tools to study geometric properties invariant to coordinate changes? If my question makes sense, is it possible to generalize and express any hypersurface in such a manner that is invariant to reparametrizations? 



#5
Feb1410, 09:57 AM

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P: 9,017

I'm not sure what that would mean. Suppose that we define a 2dimensional hypersurface in [itex]\mathbb R^3[/itex] by a function [itex]\phi:U\rightarrow\mathbb R^3[/itex], where U is some open subset of [itex]\mathbb R^2[/itex]. The range of this function can be written as [itex]\phi(U)[/itex]. This is the set of points that belongs to the submanifold we're defining.
By a reparametrization, I assume that you're talking about a nice enough function g that takes U to itself. Now [itex]f\circ g[/itex] has the same range as [itex]f[/itex], so I guess that you can say that the submanifold is invariant to reparametrizations, if this is how we define a reparametrization and what it means to be invariant under a reparametrization. But this is getting a little weird. Statements about something being independent of parametrizations or coordinates don't mean much without an accompanying description of how the things we're talking about are related to parameters or coordinates in the first place. For example if V is a tangent vector at a point p in a manifold M, and [itex]E_i[/itex] are the basis vectors for the tangent space at p constructed from a specific coordinate system, then we can write [itex]V=V^iE_i[/itex]. Here V is coordinate independent, and [itex]V^i[/itex] isn't, because [itex]E_i[/itex] isn't, by definition of [itex]E_i[/itex]. Not sure if this helps. 



#6
Feb1410, 10:27 AM

P: 625

[tex] \mathbf{p} = f_1(t)\mathbf{e}_1 + f_2(t)\mathbf{e}_2 + f_3(t)\mathbf{e}_3 [/tex] This expresses, for example the positions of a particle in space respect to the time t. You will probably agree that after applying the substition [tex]t\rightarrow t^3[/tex], the trajectory is really the same: it is just traversed at different "velocities". So is there a way to get rid of these infinitely different representation of the same trajectory? I am looking for a way to give a single general representation for such a trajectory. Basically if two parametric curves "pass" exactly through the same set of positions in space, then I define them to be equal. Wikipedia seem to have something, but I dont know if it can be generalized to surfaces: http://en.wikipedia.org/wiki/Differe...lence_relation 



#7
Feb1410, 10:33 AM

Math
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Thanks
PF Gold
P: 38,900

The standard method is to use arclength of the curve itself as the parameter. You still have an ambiguity in where you start measuring and, of course, your unit length. But you must understand that applying numbers to a purely geometric object necessarily requires choices that can be made in infinitely many ways.




#8
Feb1410, 10:35 AM

P: 625

Thanks for the hint!
...what about surfaces in general? Can that be generalized? And by the way, does it make sense to represent a general surface in the following way: [tex] \mathbf{p}=f_1(\phi_1(u_1,\ldots,u_k),\ldots,\phi_k(u_1,\ldots,u_k))\ma thbf{e}_1 + \ldots + f_n(\phi_1(u_1,\ldots,u_k),\ldots,\phi_k(u_1,\ldots,u_k))\mathbf{e}_n [/tex] where the arbitrary functions [itex]\phi_i(u_1,\ldots,u_k)[/itex] define an admissible coordinate change (e.g.: Jacobian [itex]\neq[/itex] 0), so that there is a onetoone correspondence between the ktuples [itex](u_1,\ldots,u_k)[/itex] and [itex](\phi_1,\ldots,\phi_k)[/itex]. If that is correct, one would need to replace the parameter with quantities which are invariant to coordinate changes. That's why I said that tensor calculus here "rings a bell". 



#9
Feb1510, 04:48 AM

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P: 9,017

You can do what HallsofIvy suggested in #7, which is to always use the "preferred" parameter (i.e. arc length), or you can do what I suggested in #2, which is to specify the graph of the curve instead of the curve itself. What you're talking about here is a curve [itex]x:\mathbb R\rightarrow\mathbb R^3[/itex], and another curve [itex]y:\mathbb R\rightarrow\mathbb R^3[/itex], defined by [itex]y(t)=x(s(t))[/itex], where [itex]s:\mathbb R\rightarrow\mathbb R[/itex] is a smooth strictly increasing function. ([itex]s(t)=t^3[/itex] is a poor choice because it's not strictly increasing at 0. We would have [itex]y'(0)=x'(s(0))s'(0)=0[/itex], so y wouldn't have a tangent vector at 0). Both curves have the same graph: [itex]x(\mathbb R)=y(\mathbb R)[/itex]. HallsofIvy's suggestion doesn't work on all curves in all spaces, because arc length isn't always defined. It works fine on almost all interesting spaces though. 


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