Tangent of an injective regular curve independent of its parametrization

Question

I am trying to show that the tangent of a regular curve is independent of its parametrization.

Let $c, \tilde c: \mathbb R \to \mathbb R^2$ be injective $C^1$-curves and $c'(t) \not=0, \tilde c'(s) \not= 0$. Now assume that $c$ and $c'$ define the same curve as point set, i.e. $c(\mathbb R)=\tilde c (\mathbb R)$. I want to show that $c'(t_0)$ and $\tilde c'(s_0)$ differ just by a $\mathbb R$-multiple, whenever $c(t_0)=c(s_0)$. We may assume $t_0=s_0=0,c(s_0)=t(s_0)=0$.

• I tried to parametrize by arc length, but how can I know that then the parametrizations must be the same up to orientation?
• I tried to look at the parameter change between $c$ and $\tilde c$, but how can I know that $\tilde c ^{-1} \circ c$ is differentiable?

Answer 1

Let $c: I \to \mathbb{R}^2$ and suppose $|c'(t)| \not = 0$ for all $t \in I$. Suppose now that $\gamma: J \to \mathbb{R}^2$ is just a reparametrization of $c$ i.e there exists diffeomorphism $g: J \to I$ such that $c \circ g = \gamma$. By the chain-rule we have:

$$\frac{d}{dt} \gamma = \frac{d}{dt} (c \circ g) =\nabla c\Bigr|_{g(t)} \cdot g'(t)$$

Suppose $c(t_0) = \gamma(s_0) = (c \circ g)(s_0)$. If we take the above evaluation at $t = s_0$ then we have that $\gamma'(s_0) = \nabla c(g(s_0)) \cdot g'(s_0)$. If you can suppose as well that $c$ is injective, then:

$$c(t_0) = c(g(s_0)) \Rightarrow t_0 = g(s_0) \Rightarrow \frac{d}{dt}\Bigr|_{t = s_0} \gamma=\nabla c(t_0) \cdot g'(t_0)$$

The place where I see regularity helping is when we consider $c'(t_0)$. As it stands, we have no alternate form for $c'(t)$. If we can assume $|c'(t)| \not = 0$ then we can parametrize by arc-length parameter $t = h(s)$. Then we have:

$$\frac{d}{dt} c(t) = \frac{d}{ds} c(h(s)) = \nabla c(h(s)) \cdot h'(s) = \nabla c(t) \cdot \frac{dt}{ds} = \frac{\nabla c(t)}{\frac{ds}{dt}} = \nabla c(t) \cdot \frac{1}{\|c'(t)\|} = \nabla c(t)$$

i.e the two become multiples once we take the above evaluation at $t = t_0$ since $g'(t_0) \in \mathbb{R}$.

Answer 2

We can describe the reparametrization explicitly by parametrizing by arc length. Let $s_c(t) = \int_{t_0}^t \lvert c'(u) \rvert du$ and $s_\tilde{c}(t) = \int_{s_0}^t \lvert \tilde{c}'(u) \rvert du$. Then $s_c$ and $s_\tilde{c}$ are both differentiable and injective, and so have differentiable inverses. Moreover, their derivatives are non-zero everywhere, which therefore also holds true for the differentiable map $h = s_\tilde{c}^{-1} \circ s_c$.

Now, note that $c \circ s_c^{-1} (0) = c(t_0)$ and $\tilde{c} \circ s_\tilde{c}^{-1}(0) = \tilde{c}(s_0)$ and $\lvert (c \circ s_c^{-1})'(t) \rvert = \lvert (\tilde{c} \circ s_\tilde{c}^{-1})'(t) \rvert = 1$ for all $t$, so it follows that (up to an overall change of orientation at least) $c \circ s_c^{-1} = \tilde{c} \circ s_{\tilde{c}}^{-1}$, or in other words, $c = \tilde{c} \circ h$. At this point, the claim follows from the chain rule, since

$$c'(t_0) = \tilde{c}'(h(t_0))h'(t_0) = \tilde{c}'(s_0)h'(t_0).$$