Suppose we have a plane (or space, I don't think it matters here) curve $\boldsymbol{\alpha}(u) : I \rightarrow \mathbb{R}^2$ with any parametrisation $u$. Is the unit tangent vector $$ \boldsymbol{t}(u) = \frac{\boldsymbol{\alpha'(u)}}{||\boldsymbol{\alpha'(u)}||} $$ (where $||\boldsymbol{\alpha'(u)}|| \neq 0$) with the parametrisation $u$ the same as $\boldsymbol{t}(s)=\boldsymbol{\alpha'(s)}$ where $s$ is the arc-length parameter? Certainly, $||\boldsymbol{t}(u)||=||\boldsymbol{t}(s)||=1$, but I don't know if this is sufficient.
If $l: I \rightarrow \mathbb{R}$ is the arc length of $\boldsymbol{\alpha}$ and $J=l(I)$, then $\boldsymbol{\beta} = \boldsymbol{\alpha} \circ l^{-1} : J \rightarrow \mathbb{R}^2$ is an arc length parametrisation for $\boldsymbol{\alpha}$. I was thinking of perhaps taking the derivative of $\boldsymbol{\beta}(s)$ and showing equality/non-equality to $\boldsymbol{t}(u)$ but I'm not quite sure if this leads anywhere useful.
Many thanks for any answers.
Indeed, taking the derivative of $\boldsymbol{\beta}$ we have
\begin{align} \boldsymbol{t}(s) = \frac{d}{ds} \boldsymbol{\beta}(s) &= \frac{d}{ds} (\boldsymbol{\alpha} ( l^{-1}(s) ) \\ &= \frac{d}{d u} (\boldsymbol{\alpha} ( l^{-1}(s) ) \cdot \frac{d l^{-1}(s)}{ds} \\ &= \frac{d}{d u} (\boldsymbol{\alpha} ( u )) \cdot \frac{1}{l'(u)} \\ &= \frac{\boldsymbol{\alpha'(u)}}{||\boldsymbol{\alpha'(u)}||} \\ &= \boldsymbol{t}(u) \end{align} so $\boldsymbol{t}(s)=\boldsymbol{t}(u)$ and the equality is true.