Derivative of Arc Length

Theorem

Let $C$ be a curve in the cartesian coordinate plane described by the equation $y = f \left({x}\right)$.

Let $s$ be the length along the arc of the curve from some reference point $P$.

Then the derivative of $s$ with respect to $x$ is given by:

$\displaystyle \frac{\mathrm d s}{\mathrm d x} = \sqrt{1 + \left({\frac{\mathrm d y}{\mathrm d x}}\right)^2}$

Proof 1

Consider a length $\mathrm{d}{s}$ of $C$, short enough for it to be approximated to a straight line segment:

By Pythagoras's Theorem, we have:

$\mathrm d s^2 = \mathrm d x^2 + \mathrm d y^2$

Dividing by $\mathrm d x^2$ we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\frac {\mathrm d s}{\mathrm d x} }\right)^2\)	\(=\)	\(\displaystyle \left({\frac {\mathrm d x}{\mathrm d x} }\right)^2 + \left({\frac {\mathrm d y}{\mathrm d x} }\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \left({\frac {\mathrm d y}{\mathrm d x} }\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Hence the result, by taking the square root of both sides.

$\blacksquare$

Proof 2

From Continuously Differentiable Curve Has Finite Arc Length, $s$ exists and is given by:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle s\)	\(=\)	\(\displaystyle \int_P^x \sqrt{1 + \left({\frac {\mathrm dy}{\mathrm du} }\right)^2}\ \mathrm d u\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm ds}{\mathrm dx}\)	\(=\)	\(\displaystyle \frac {\mathrm d}{\mathrm dx} \int_P^x \sqrt{1 + \left({\frac {\mathrm dy}{\mathrm du} }\right)^2}\ \mathrm d u\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	differentiate both sides WRT $x$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt{1 + \left({\frac{\mathrm d y}{\mathrm d x} }\right)^2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Fundamental Theorem of Calculus/First Part

$\blacksquare$