I would like to know if there is a way to determine mathematically (given some parameter) the equivalence of two curves given by different parameterizations.

For example, the two curves defined respectively by

$x_1 = \sin(t),\ y_1 = \cos(t)$ for $t\in[0,2\pi]$

$x_2 = \cos(2t),\ y_1 = \sin(2t)$ for $t\in[0,\pi]$

trace the same circle and are both solutions to the equation of the circle: $x^2+y^2 = 1$. However, I am wondering if there is a on operation that can be done on both curves to determine whether they trace the same implicit curve or not?

  • 1
    $\begingroup$ This might be strictly technical, but did you mean $t \in [0,\pi]$ for the second curve? Because strictly speaking the two curves are different. First one is unit circle traversed only once and second one is unit circle traversed 4 times. $\endgroup$
    – jgsmath
    Nov 21, 2017 at 18:59
  • $\begingroup$ Actually you are right. But this mistakes serves my point. The traced curve is still equivalent. So to generalize my point even further: is there a way to determine the equivalence of a curve with respect to a part of another curve? However, for simplicity, I will edit the second curve, thanks for pointing that out. $\endgroup$
    – Snifkes
    Nov 21, 2017 at 19:10
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    $\begingroup$ I don't think there's an exact algorithm, but if two curves $\alpha: I_{1} \to \mathbb{R}^n$ and $\beta: I_{2} \to \mathbb{R}^n$ are equivalent, that means that there's a diffeomorphism $\phi: I_{1} \to I_{2}$ such that $\beta \circ \phi = \alpha$. You should be able to determine whether this diffeomorphism exists for most of the basic examples. An algorithm for determining equivalence seems very unlikely, considering there's uncountably many ways to parametrize a circle. I'm sorry if you knew all this already :P . $\endgroup$ Nov 21, 2017 at 19:18
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    $\begingroup$ lets add $x = \frac {1-t^2}{1+t^2}, y = \frac {2t}{1+t^2}, t\in (-\infty, \infty)$ to the mix. No, I don't have any "mechanical" tests to show that two parmeterization are equivalent. (note that this one is not quite equivalent as there is no $t$ mapping to the point $(-1,0)$) $\endgroup$
    – Doug M
    Nov 21, 2017 at 19:23
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    $\begingroup$ From a pragmatic point of view the answer is that (most) parameterizations will be different. As humans, if we go to plot these functions we'll use a finite number of points. With each parameterization we'll have points that are spaced differently and so, on a microscale, at least, the curves will be different. And this only becomes exacerbated as we take derivatives or look at more complex shapes, such as cusps. I've developed a family of curves for which I have both Cartesian and polar parameterizations. For various applications, one or other of these can provide a superior representation. $\endgroup$ Nov 21, 2017 at 22:32

1 Answer 1


$\def\RR{\mathbb{R}} \def\vr{\mathbf{r}} \def\vv{\mathbf{v}} \def\vnull{\mathbf{0}} \def\ss{\subseteq} \def\To{\rightarrow} \def\p{\pi}$Consider a curve in $\RR^n$ given by $\vr(t) = (x_1(t),\ldots,x_n(t))$ for $t\in I\ss\RR$. Suppose $\vr$ is differentiable and non-self intersecting. (The restriction of non-self intersecting can be relaxed, but introduces technicalities that we wish to avoid.) Let $\vv(t) = \vr'(t)$ and further suppose that $\vv(t)\ne \vnull$ for any $t$. The arclength along this curve will be given by \begin{equation}s=\int_{t_0}^t |\vv(t)|dt,\tag{1}\end{equation} where $t_0\in I$ is some reference value for the parameter $t$ corresponding to an arclength of $s=0$. (Note that $t<t_0$ corresponds to a negative arclength. This sign agrees with the orientation of the curve.) By assumption, $|\vv|>0$. Thus, $s$ is a strictly monotone increasing function of $t$ and so will have an inverse function, $t=t(s)$. Thus, $\vr(s)\equiv\vr(t(s))$, where $s\in I'\ss\RR$, will be this same curve parametrized by arclength. The interval $I'$ can be determined from (1).

After this procedure has been done, it should be clear that two parametrizations of the same curve, $\vr_1(s)$ and $\vr_2(s)$, can differ in their representation only through the following transformation: $s\To ks+s_0$, where $k=\pm 1$. That is, the curves can only differ by a shift in $s$ or by having opposite orientations. (It is also possible that after this transformation further transformations on $I'$ may be allowed and necessary, depending on the periodicity of the components of $\vr(s)$.)

A quick way that one can distinguish different curves is by examining the intervals $I'$. For example, if the intervals are finite subsets of $\RR$ and are different in length, then the curves are not the same. (They do not have the same length.)


Let \begin{align*} \vr_1(t) &= (\sin t,\cos t), & t\in[0,2\p) \\ \vr_2(t) &= (\cos 2t,\sin 2t), & t\in[0,\p) \\ \vr_3(t) &= \left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right), & t\in\RR \end{align*} (The last example is @Doug M's from the comments.) It is a straightforward exercise to show that \begin{align*} \vr_1(s) &= (\sin s,\cos s), & s\in[0,2\p) \\ \vr_2(s) &= (\cos s,\sin s), & s\in[0,2\p) \\ \vr_3(s) &= \left(\cos s,\sin s\right), & s\in(-\p,\p) \end{align*} Here we have let $t_0=0$ for each parametrized curve for simplicity. Note that $\vr_1(-s+\p/2) = \vr_2(s)$. Under this transformation, $I_1=[0,2\p)\To[-3\p/2,\pi/2)$. Using periodicity, $I_1\To[0,2\p)$. Thus, the first and second parametrized curves correspond to the same graph. Using periodicity we find $I_3=(-\p,\p)\To[0,2\p)\setminus\{\p\}$. Thus, the third parametrized curve corresponds to a different graph.

As an exercise, using this procedure one should be able to determine that the following curves correspond to the same graph. \begin{align*} &(\sin t,\cos t), & t\in[0,\p] \\ &(\cos 2t,\sin 2t), & t\in[-\p/4,\p/4] \\ &\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right), & t\in[-1,1] \end{align*}


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