Why does the arc-length formula have form $\int_a^b\left|\left|\frac{d\vec{f}(t)}{dt}\right|\right|_2dt$ for C1 curves?

240 Views Asked by At

This discussion focuses on $\mathcal{C}^1$ curve on $\mathbb{R}^n$. But feel free to talk about the case where we only have a continuous curve or the scenario with a manifold with a metric in general.


For any curve $f\in\mathcal{C}^1([a,b],\mathbb{R}^n)$ we have

$$ \text{arc length}=\sup_{\text{all partitions }\mathcal{P}}\sum_{i=1}^N||f(x_i)-f(x_{i-1})||=:L\quad...(1) $$ where $\mathcal{P}=\{ a=x_0\leq\cdots\leq x_N=b\}$.

How does the following arc length formula arise from (1)?

$$ L=\int_a^b\left|\left|\frac{d\vec{f}(t)}{dt}\right|\right|_2dt=\int_a^b\sqrt{\left(\frac{df^1}{dt}\right)^2+\cdots+\left(\frac{df^n}{dt}\right)^2}dt \quad...(2) $$

Please focus on formulating on a rigorous proof along with whatever necessary conditions one may need for (2) to work. It is nice to have pictorial explanations on the side. But a step-by-step proof is the focus here.

1

There are 1 best solutions below

2
On BEST ANSWER

Some smoothness is required.

Read baby Rudin (3rd edition), pp. 136-137 .