Arc length intuition

The result should be easy to prove now:

\begin{align*} l = \int_a^b \sqrt{1 + (f'(x))^2} dx \end{align*}In the context of parametric functions, this can be written as:

This is true whether $dx/dt$ is positive or negative. The positive case is trivial. When it's negative, $\alpha > \beta$, we need to switch order, then the sigan becomes positive again.

Denis Aurous has a nice interpretation of this formula. $(dx/dt, \; dy/dt)$

, or $d\vec{r}/dt$ where $\vec{r}$ stands for the position vector $(x, y)$,

is the velocity vector, and
$\sqrt{(dx/dt)^2 + (dy/dt)^2}$ is just the length of
the vector, which is speed $dl/dt$. i.e.
\begin{align*}
\frac{dl}{dt} = |\frac{dr}{dt}|
\end{align*}
He has another interesting observation: \begin{align*} \frac{dr}{dt} &= \frac{dr}{dl} \frac{dl}{dt} \\ &= \frac{dr}{dl} |\frac{dr}{dt}| \\ \end{align*}

As you can see, $\frac{dr}{dl}$ is the direction vector, or tangent vector (figure below).

He also kindly reminds us of the difference between $dr$ and $d|r|$:

integral, integration, arc, calculus

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