Derivative of Arc Length

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Theorem

Let $C$ be a curve in the cartesian coordinate 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:

DSbyDX.png

By Pythagoras's Theorem, we have:

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


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

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

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:

\(\displaystyle s\) \(=\) \(\displaystyle \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} }^2} \rd u\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d s} {\d x}\) \(=\) \(\displaystyle \frac {\d} {\d x} \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} } ^2} \rd u\) differentiating both sides WRT $x$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {1 + \paren {\frac {\d y} {\d x} }^2}\) Fundamental Theorem of Calculus/First Part

$\blacksquare$