Slope of Tangent to Cycloid/Proof 1

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:

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

By 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 $

$=$

$\ds \frac 1 {\map \tan {\theta / 2} }$

$\ds $

$=$

$\ds \cot \frac \theta 2$

$\blacksquare$

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