Show that the arc length $L(\gamma)$ of a curve $\gamma$ is unchanged if $\gamma$ is reparametrized
Can you help me please?
Show that the arc length $L(\gamma)$ of a curve $\gamma$ is unchanged if $\gamma$ is reparametrized
Can you help me please?
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.
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.)
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
(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.)