Showing if two regular parametrized curves are equivalent How do I know if the following two regular parametrized curves, $c_1$ and $c_2$, are equivalent? 
(1) $c_1$ : [0, 2$\pi$] $\to$ $\mathbb{R}^2$ : $t$ $\mapsto$ (cos $t$, sin $t$)
(2) $c_2$ : [0, 2$\pi$] $\to$ $\mathbb{R}^2$ : $t$ $\mapsto$ (cos 2$t$, sin 2$t$)
I know that two curves are equivalent if one is a reparametrization of the other. What exactly does that mean and how do I show that one is a reparametrization of the other? 
 A: Let $\tilde{\gamma}:\tilde{I} \to \mathbb{R}^n$ and  $\gamma:I \to \mathbb{R}^n$ be smooth maps. We say that $\tilde{\gamma}$ is a reparametrization of $\gamma$ if there exists a diffeomorphism $\phi: \tilde{I} \to I$ such that $(\gamma \circ \phi)(\tilde{t}) = \tilde{\gamma}(\tilde{t})$.
A: We say that two regular parametric curves, $\phi$ and $\psi$, are equivalent, if there is a strictly monotonically increasing function $g$ such that $\psi(g(t)) = \phi(t)$. Reparametrisation means that you have parametrised $\psi$ by some new parameter $g(t)$ that is a function of $t$, instead of the original parameter, $t$. For example, you can reparametrise a curve with its arclength, given as a function of $t$.
Now for your question, if we let $\\c_1(t) = (cos(t),sin(t))$ and $c_2(t) = (cos(2t),sin(2t))$, where the domain of $c_1$ is $[0,2\pi]$, and the domain of $c_2$ is $[0,2\pi]$, then you can reparametrise $c_2$ with a new parameter $s(t)=\frac{t}{2}$. Now you have $c_1(t) = c_2(s(t)) = (cos(t), sin(t))$, but the domain of $c_2$ is now $[0, \pi]$, which is different from $c_1$. Hence the two parametric curves are not equivalent.
A: Intuitively, $c_1$ goes around the circle once, but $c_2$ goes around the circle twice. They have different 'paths', so they are not equivalent.
Rigorously, length of a curve is preserved under reparametrization. Note that $c_1$ and $c_2$ have different lengths.
