Second Derivative of Locus of Cycloid

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Theorem

Consider a circle of radius $a$ rolling without slipping along the x-axis of a cartesian coordinate plane.

Consider the point $P$ on the circumference of this circle which is at the origin when its center is on the y-axis.

Consider the cycloid traced out by the point $P$.

Let $\left({x, y}\right)$ be the coordinates of $P$ as it travels over the plane.


The second derivative of the locus of $P$ is given by:

$y'' = - \dfrac a {y^2}$


Proof

From Equation of Cycloid:

$x = a \left({\theta - \sin \theta}\right)$
$y = a \left({1 - \cos \theta}\right)$


From Slope of Tangent to Cycloid:

\(\displaystyle y'\) \(=\) \(\displaystyle \cot \frac \theta 2\) Slope of Tangent to Cycloid
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\mathrm d y'} {\mathrm d x}\) \(=\) \(\displaystyle \frac {\mathrm d} {\mathrm d \theta} \cot \frac \theta 2 \frac {\mathrm d \theta} {\mathrm d x}\) Chain Rule
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 2 \cot \frac \theta 2 \csc \frac \theta 2 / \frac {\mathrm d x} {\mathrm d \theta}\) Derivative of Cotangent Function
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 2 \cot \frac \theta 2 \csc \frac \theta 2 \left({\frac 1 {a \left({1 - \cos \theta}\right)} }\right)\) Derivative of Sine Function
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 2 y' \sqrt{\left({1 + y'^2}\right)} \left({\frac 1 {a \left({1 - \cos \theta}\right)} }\right)\) Difference of Squares of Cosecant and Cotangent
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 2 y' \sqrt{\left({1 + y'^2}\right)} \frac 1 y\)



Sources