How can I show that arc length $L(\gamma)$ of a curve is unchanged after reparametrization? Show that the arc length $L(\gamma)$ of a curve $\gamma$ is unchanged if $\gamma$ is reparametrized
Can you help me please?
 A: Let $\gamma = \gamma(t)$ 
Then $$L(\gamma) = \int_0^t |\frac{d\gamma}{dt}|dt$$
If $\gamma$ is reparametrized so $\gamma = \gamma(s(t))$ then $\displaystyle\gamma' = \frac{d\gamma}{ds}\frac{ds}{dt}$ by the chain rule so $$L(\gamma) = \int_0^s \left|\frac{d\gamma}{ds}\frac{ds}{dt}\right|ds$$
Can you show that $L(\gamma)$ is the same computed either way?
Hopefully this helps, this is my first post so go easy on me!
Edit: Be careful on the limits of integration, it will depend on the domain of the curve.
A: The way I learned it was to let the curve $\gamma$ be described by two functions, $X$ and $Y$ such that $Y(t)=X[u(t)]$ for $c\le t\le d$. (Here $u$ defines the change of parameter.) Also assume that $u'(t)>0$ for $c\le t \le d$. Now prove that
$$\int_{u(c)}^{u(d)} \! ||X'(h)||\,\mathrm d h=\int_c^d \! || Y'(t)|| \, \mathrm dt\,.$$
(If this is the part you're stuck on, let me know and I'll try to think of another hint.)
A: Here is another way to look at the problem:
In order to define the length of a curve 
$$\gamma:\ t\mapsto{\bf x}(t)\quad(a\leq t\leq b)$$
one  considers partitions
$$T:\quad a=t_0<t_1<\ldots< t_N=b$$
of the parameter interval $[a,b]$. Such a partition $T$ induces a "piecewise linear approximation" $\gamma_T$ of $\gamma$ whose length is given by 
$$L(\gamma_T):=\sum_{k=1}^N\bigl|{\bf x}(t_k)-{\bf x}(t_{k-1})\bigr|\ .$$
One then defines
$$L(\gamma):=\sup_{T\in{\cal T}} L(\gamma_T)\ ,$$
where ${\cal T}$ denotes the set of all partitions of $[a,b]$. Given the geometric interpretation of this definition it is obvious that $L(\gamma)$ stays the same when $\gamma$ is reparametrized.
(It is a theorem that the above $L(\gamma)$ equals $\int_a^b\bigl|\dot{\bf x}(t)\bigr|\ dt$ when $\gamma$ is continuously differentiable.)
