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!