# 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:

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

From Slope of Tangent to Cycloid:

\(\displaystyle y'\) | \(=\) | \(\displaystyle \cot \frac \theta 2\) | $\quad$ Slope of Tangent to Cycloid | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\mathrm d y'} {\mathrm d x}\) | \(=\) | \(\displaystyle \frac {\mathrm d} {\mathrm d \theta} \cot \frac \theta 2 \frac {\mathrm d \theta} {\mathrm d x}\) | $\quad$ Chain Rule | $\quad$ | ||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\frac 1 2 \cot \frac \theta 2 \csc \frac \theta 2 / \frac {\mathrm d x} {\mathrm d \theta}\) | $\quad$ Derivative of Cotangent Function | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\frac 1 2 \cot \frac \theta 2 \csc \frac \theta 2 \left({\frac 1 {a \left({1 - \cos \theta}\right)} }\right)\) | $\quad$ Derivative of Sine Function | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\frac 1 2 y' \sqrt{\left({1 + y'^2}\right)} \left({\frac 1 {a \left({1 - \cos \theta}\right)} }\right)\) | $\quad$ Difference of Squares of Cosecant and Cotangent | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\frac 1 2 y' \sqrt{\left({1 + y'^2}\right)} \frac 1 y\) | $\quad$ | $\quad$ |

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.21$: The Cycloid: Problem $2$