Why the length of a curve involves derivatives

Question

Suppose we have a parametrised curve with position $A$ at $\underline{r}(t_0)$ and position $B$ at $\underline{r}(t_1)$ given by $$\underline{r}(t) = x(t)\underline{i}+y(y)\underline{j}+z(t)\underline{k}$$ In all the textbooks I could find, they give out the formula giving the length of the curve, but in none of them it is explained why that's the correct formula. Here it is $$L = \int_{t_0}^{t_1}|\dot{\underline{r}}(t)|dt = \int_{t_0}^{t_1}\sqrt{\dot{\underline{x}}(t)^2+\dot{\underline{y}}(t)^2+\dot{\underline{z}}(t)^2}dt$$ Can someone please explain to me in a detailed way, why this is true? I understand why $|\dot{\underline{r}}(t)| = \sqrt{\dot{\underline{x}}(t)^2+\dot{\underline{y}}(t)^2+\dot{\underline{z}}(t)^2}$ but not why we put this (and not the position vector or something else) under the integral. Why are we summing up the speeds?

Answer 1

As always for intiution, its a good idea to think of approximations to the integral. Choose, given $\eta >0 $ points $t_0 = \tau_0 < \ldots < \tau_n = t_1$ with $\tau_{i+1}-\tau_i < \eta$ for each $i$. Then the length of the curve we want to calculate is approximated by the length of the curve consisting of the segments from $r(\tau_i)$ to $r(\tau_{i+1})$. The length of this segment is $$ \def\abs#1{\left|#1\right|} \abs{r(\tau_{i+1}) - r(\tau_i)} = \abs{r'(\theta_i)}\cdot (\tau_{i+1} - \tau_i) $$ for some $\theta_i \in(\tau_i, \tau_{i+1})$ by the mean value theorem. Hence the length is approximated by $$ \sum_{i=0}^{n-1} \abs{r'(\theta_i)} (\tau_{i+1} -\tau_i) $$ This is a Riemann sum for the integral $\int_{t_0}^{t_1} \abs{r'(\tau)}\,d\tau$, hence for $\eta \to 0$ this converges to $$ \int_{t_0}^{t_1} \abs{r'(\tau)}\, d\tau. $$

Answer 2

SKETCH OF PROOF: It is an application of sums. Consider a parametrised curve in $\mathbb{R^2}$ and $t \in [t_1,t_2]. $ then you take a partition of $[t_1,t_2]$ . Then break your curve to straight parts. That is, take a partition of your curve. Then the length of this curve is the sum of the length of those straight parts. Use the mean value theorem and you get the result for a curve in $\mathbb{R^2}$ . Generalize this for higher dimensions.