I have the following problem largely figured out, and just want some pointers with the details that actually justify what is being done working fast and loose with differentials.
Suppose $\gamma$ is a curve in $\mathbb{R}^n$ and let $\tilde\gamma$ be a reparametrization of $\gamma$ with map $\phi$ (so that $\tilde\gamma(\tilde t) = \gamma(\phi(\tilde t))).$ Let $\tilde t_0$ be a fixed value of $\tilde t$ and let $t_0=\phi(\tilde t_0)$. Let $s$ and $\tilde s$ be the arc-lengths of $\gamma $ and $\tilde\gamma$ starting at the point $\gamma(t_0) = \tilde \gamma(\tilde t_0).$ Prove that $s=\tilde s$ if $d\phi/d\tilde t>0$ for all $\tilde t,$ and $s=-\tilde s$ if $d\phi/d\tilde t<0$ for all $\tilde t$.
The proof is very short. Recall the arc-length $s$ of $\gamma$ is given by $$s(t)=\int_{t_0}^t\bigg|\frac{d}{du}\gamma(u)\bigg|\space du.$$
Then make the $u$-substitution $u=\phi(\tilde v)\implies du = d\phi\space d\tilde v$ so that
\begin{equation} \begin{split} s(t)& =\int_{\tilde t_0}^{\tilde t}\bigg| \frac{d}{d\phi \space d\tilde v}\gamma(\phi(\tilde v))\bigg|d\phi \space d\tilde v \\ & =\int_{\tilde t_0}^{\tilde t}\bigg| \frac{d}{d\tilde v}\tilde\gamma(\tilde v)\bigg|\frac{d\phi}{|d\phi|} \space d\tilde v\\ & = \int_{\tilde t_0}^{\tilde t}\bigg| \frac{d}{d\tilde v}\tilde\gamma(\tilde v)\bigg|\frac{d\phi/d\tilde t}{|d\phi|/d\tilde t} \space d\tilde v \end{split} \end{equation}
Now here is where I'm unsure. Already the manipulation of differentials as such here makes me nervous, but how does one consider the sign of $|d\phi|/d\tilde t$? Naively, one just says what the question suggests: when $d\phi/d\tilde t>0$ we have $|d\phi|/d\tilde t = d\phi/d\tilde t >0,$ and when $d\phi/d\tilde t<0$ we have $|d\phi|/d\tilde t= -d\phi/d\tilde t>0$, but these manipulations appear to assume $d\tilde t>0$. Can we assume $d\tilde t>0$ is valid, and true for any denominator in such a differential? Furthermore, are the manipulations of the differentials above justified?