# Second Derivative of Locus of Cycloid

## 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:

$x = a \left({\theta - \sin \theta}\right)$
$y = a \left({1 - \cos \theta}\right)$
 $\displaystyle y'$ $=$ $\displaystyle \cot \frac \theta 2$ 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 for Derivatives $\displaystyle$ $=$ $\displaystyle -\frac 1 2 \cot \frac \theta 2 \csc \frac \theta 2 / \frac {\mathrm d x} {\mathrm d \theta}$ 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 $\displaystyle$ $=$ $\displaystyle -\frac 1 2 y' \sqrt{\left({1 + y'^2}\right)} \frac 1 y$