# Derivative of Arc Length

## Theorem

Let $C$ be a curve in the cartesian plane described by the equation $y = \map f x$.

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:

$\dfrac {\d s} {\d x} = \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$

## Proof 1

Consider a length $\d s$ of $C$, short enough for it to be approximated to a straight line segment: By Pythagoras's Theorem, we have:

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

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

 $\ds \paren {\frac {\d s} {\d x} }^2$ $=$ $\ds \paren {\frac {\d x} {\d x} }^2 + \paren {\frac {\d y} {\d x} }^2$ $\ds$ $=$ $\ds 1 + \paren {\frac {\d y} {\d x} }^2$

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

$\blacksquare$

## Proof 2

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

 $\ds s$ $=$ $\ds \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} }^2} \rd u$ $\ds \leadsto \ \$ $\ds \frac {\d s} {\d x}$ $=$ $\ds \frac {\d} {\d x} \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} } ^2} \rd u$ differentiating both sides with respect to $x$ $\ds$ $=$ $\ds \sqrt {1 + \paren {\frac {\d y} {\d x} }^2}$ Fundamental Theorem of Calculus: First Part

$\blacksquare$