# Slope of Tangent to Cycloid/Proof 1

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## Theorem

Let $C$ be a cycloid generated by the equations:

- $x = a \paren {\theta - \sin \theta}$
- $y = a \paren {1 - \cos \theta}$

The slope of the tangent to $C$ at the point $\tuple {x, y}$ is given by:

- $\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$

## Proof

By definition of Derivative of Curve at Point, the tangent to $C$ at the point $\tuple {x, y}$ is the derivative of its equation at that point.

Thus:

\(\ds \frac {\d x} {\d \theta}\) | \(=\) | \(\ds a \paren {1 - \cos \theta}\) | ||||||||||||

\(\ds \frac {\d y} {\d \theta}\) | \(=\) | \(\ds a \sin \theta\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac {\d y} {\d x}\) | \(=\) | \(\ds \frac {\d y} {\d \theta} / \frac {\d x} {\d \theta}\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {a \sin \theta} {a \paren {1 - \cos \theta} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\sin \theta} {1 - \cos \theta}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {\map \tan {\theta / 2} }\) | Half Angle Formula for Tangent: Corollary 2 | |||||||||||

\(\ds \) | \(=\) | \(\ds \cot \frac \theta 2\) |

$\blacksquare$

## Sources

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