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

$\blacksquare$