# The intuitive behind the equivalence of two parametrized curves

I'm learning differential geometry and I found this definition about the equivalence of two parametrized curves.

I want to understand the intuition behind those conditions. What I have been understanding so far is the condition $$\gamma_2(\varphi(t))=\gamma_1(t), \forall t \in I_1$$ means the two parametrized curves have the same trace. What about the first condition $$\varphi'(t) \neq 0, \forall t \in I_1$$? What is the intuition behind it?

UPDATE: Consider the two parametrized curves $$(C_1): r_3(t)=(t^3, 0), t \in \mathbb{R}$$ and $$(C_1): r_4(t)=(t, 0), t \in \mathbb{R}$$. They are not equivalent even they have the same direction and the same trace. The only thing here is the kind of velocity they travel is different. Like the third one has the constant acceleration but the fourth is not. How the kind of velocity affect the equivalence?

Thank you for reading my question. Any help would be appreciated.

If $$\gamma_2\bigl(\varphi(t)\bigr)=\gamma_1(t)$$, then $$\gamma_1\bigl(\varphi^{-1}(t)\bigr)=\gamma_2(t)$$. However, without the assumption that you always have $$\varphi'(t)\ne0$$, $$\varphi^{-1}$$ would not be differentiable, and therefore two parametrizations of the same curve being equivalent would not be an equivalence relation. But it is natural to expect that it is.
• Concerning your assertion that “the first condition is to make the inverse map differentiable”, that is exactly what I wrote in my answer. And this has nothing to do with direction. The curves $\gamma_1,\gamma_2\colon[0,2\pi]\longrightarrow\Bbb R^2$ defined by $\gamma_1(t)=(\cos(t),\sin(t))$ and by $\gamma_2(t)=(\cos(t),\sin(-t))$ are equivalent; just take $\varphi(t)=2\pi-t$. Commented Jun 27, 2021 at 14:26
• What do you expect to say other than “Yes, $r_1$ and $r_2$ are indeed not equivalent”? The only thing that I can add is that the speed of $r_3$ is $0$ at some point, whereas with $r_4$ that does not occur. Commented Jun 27, 2021 at 16:18