Tangent to Cycloid

Theorem

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

$x = a \paren {\theta - \sin \theta}$

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

Then the tangent to $C$ at a point $\tuple {x, y}$ on $C$ is given by the equation:

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

Proof

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

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

This tangent to $C$ also passes through the point $\tuple {a \paren {\theta - \sin \theta}, a \paren {1 - \cos \theta} }$.

We have:

\(\ds \tan \dfrac \theta 2\)

\(=\)

\(\ds \dfrac {1 - \cos \theta} {\sin \theta}\)

Half Angle Formula for Tangent: Corollary $2$

\(\ds \leadsto \ \ \)

\(\ds \cot \dfrac \theta 2\)

\(=\)

\(\ds \dfrac {\sin \theta} {1 - \cos \theta}\)

Cotangent is Reciprocal of Tangent

By Equation of Straight Line in Plane we have:

$y - y_0 = m \paren {x - x_0}$

where $m$ is the slope and $\tuple {x_0, y_0}$ a given point on the straight line.

Substituting for the slope $m$ we obtain:

$y - y_0 = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - x_0}$

Substituting for the given point $\tuple {a \paren {\theta - \sin \theta}, a \paren {1 - \cos \theta} }$

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

Hence the result.

$\blacksquare$

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.21$: The Cycloid: Example $3$