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I'm confused about the concept of equivalent parametric curves. Based on my understanding, two parametric curves, $\phi$ and $\psi$, are equivalent, if there is a strictly monotonically increasing function $g$ such that $\psi(g(t)) = \phi(t)$. Intuitively speaking, the two curves must have the same direction and the same image, and they "travel" through their image for the same number of times.

Three questions:

(1) Why does $g$ have to be monotonically increasing? Why can't it be decreasing?

(2) They are allowed to have different speeds, correct?

(3) Are these two equivalent: $\phi(\theta) = (cos(\theta),sin(\theta))$ and $\psi(\theta) = (cos(2\theta),sin(2\theta))$, where the domain of $\phi$ is $[0,2\pi]$, and the domain of $\psi$ is $[0,\pi]$?

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You're right, there is definitely a distinction.

The two curves have the same trace, but are not considered the same parametrisation. As you have correctly pointed out, they have distinct velocities, which you can easily see when you calculate tangent vectors at all points, $\psi'(\theta)=2\cdot\phi'(\theta)$, so $\psi(\theta)$ travels twice as fast as $\phi(\theta)$. Note that this means that you cannot distinguish parametrised differentiable curve simply by looking at their image sets.

$g$ will have to be monotonically increasing rather than decreasing, to preserve the orientation / direction of the curve.

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  • $\begingroup$ Are you then saying the speed matters and for two curves to be equivalent, they must have the same speed? $\endgroup$ – user1691278 Apr 19 '18 at 4:22
  • $\begingroup$ Yes, it depends precisely on how you define this equivalence class of curves, but if you are simply considering curves that have the same trace, then speed would not matter in this case because you can reparametrise your curve with a different parameter, for example, with arclength. $\endgroup$ – T J. Kim Apr 19 '18 at 4:23
  • $\begingroup$ But in hindsight your definition of equivalence of parametrised curves seems to allow for reparametrisations, so answers to your query would be: (2) yes, (3) yes. $\endgroup$ – T J. Kim Apr 19 '18 at 4:32

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