Derivative of Arc Length/Proof 1
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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
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.
$\blacksquare$