Given a curve defined as any differentiable function, e.g. $f(x,y)=ax^2+bxy+cy^2+d $, how can I parameterize it into a vector-valued function $c(t) = (x(t), y(t))$? I appreciate any suggestion.

  • $\begingroup$ Depends on how complicated your function $f$ is. If it is quadratic like here, then your curve is a conic section, for which parametric equations are already known. In general, this is not an easy thing to do. See this discussion: math.stackexchange.com/questions/3329/… $\endgroup$ – user53153 Mar 7 '13 at 3:08
  • $\begingroup$ Question is not clear. is it $z=f(z,y)$, or is it $f(x,y)=0$ $\endgroup$ – Maesumi Mar 7 '13 at 3:16
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    $\begingroup$ if it is $f(x,y)=0$ then the standard approach is to use "polar equation for conic sections" $\endgroup$ – Maesumi Mar 7 '13 at 3:20

For curves described by implicit equations of degree 2, it's easy, as others have pointed out. The curves are conic sections, and there are many well-known ways to parameterize them. If you still need help with this case, let us know.

For curves described by implicit equations of degree 3 and higher, the situation is more complex, and sometimes rational parameterizations (i.e. parameterizations that use rational functions) don't exist. You can get a bit further if you're willing to use square roots, in addition to rational functions.

A series of papers by Abhayankar and Bajaj studied this issue in the 1980's.

A more up-to-date reference is this one.

If your original implicit equation is something more complicated than a polynomial, then I don't know of any general results. However, I'm not an expert in this field, so it may be that results exist.

In CAD and computer graphics, parametric equations are usually defined using some bounded interval, like $[0,1]$, as their domain. So, if the equations are continuous functions, this means that the curve itself must form a connected set. On the other hand, curves described by implicit equations are not necessarily connected. For example, the standard hyperbola $xy = 1$ has two disconnected branches. For this reason (and for other reasons) you may find that your parameterisations sometimes do not "cover" the entire curve, only some portion of it. But, depending on your application, this might be OK -- sometimes you just want to parameterize some portion of a curve, rather than the whole thing.


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