Differential Geometry and Its Application - Exercise 1.2.2

Problem:

Recall that the arclength of a curve $\alpha : [a, b] \to \mathbf{R}^3$ is given by $L(\alpha) = \int | \alpha'(t)| dt$. Let $\beta(r) : [c, d] \to \mathbf{R}^3$ be a reparametrization of $\alpha$ defined by taking a map $h : [c, d] \to [a, b]$ with $h[c] = a, h[d] = b$ and $h'(r) \ge 0$ for all $r \in [c, d]$. Show that the arclength does not change under this type of reparametrization.

Solution:

Intuitively, arclength of a curve should not change because we parametrize it differently. To show that, let's compute the arclength of $\beta$.

First, we notice $\beta(r) = \alpha(h(r))$. By chain rule, we know $\frac{d\beta}{dr} = \frac{d\alpha}{dh}\frac{dh}{dr}$

$\begin{eqnarray*} & & \int\limits_{c}^{d}{|\beta'(r)|dr} \\ &=& \int\limits_{c}^{d}{|\frac{d\alpha}{dh}\frac{dh}{dr}|dr} \\ &=& \int\limits_{c}^{d}{|\frac{d\alpha}{dh}|\frac{dh}{dr}dr} & & (\text{We are using } h'(r) \ge 0 \text{ here.}) \\ &=& \int\limits_{a}^{b}{|\frac{d\alpha}{dh}|dh} \end{eqnarray*}$

So the arclength in unchanged after reparametrization!