Slope of Tangent to Cycloid/Proof 2

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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

Consider a polygon $ABCD$ being rolled along a straight line in the same way as the generating circle of $C$.

Let $A', B', C', D'$ be the points around which the $ABCD$ rotates while rolling.


TangentToCycloid-construction.png


The point $A$ traces out in succession several arcs of circles with centers $B', C', D'$.

The tangent to each of these arcs is perpendicular to the line joining the point of tangency to the corresponding point of rotation.

Consider a circle as being a polygon with an infinite number of sides.

It follows that in the limit the tangent to $C$ at any given point is perpendicular to the line joining the point of tangency to the corresponding point where the generating circle touches the straight line along which the generating circle rolls.


TangentToCycloid.png


Let $O$ be the center of the generating circle.

Let $P$ be the point of tangency.

Let $Q$ be the point where the generating circle touches the straight line along which the generating circle rolls.

Let $S$ be the point where the tangent to $C$ meets the straight line along which the generating circle rolls.


As $OP$ and $OQ$ are both radii of the generating circle:

$OP = OQ = a$

and so $\triangle POQ$ is isosceles.

Let $R$ be dropped perpendicular to $PQ$ at $R$.

As $\triangle POQ$ is isosceles, $\triangle OPR = \triangle OQR$ and so $PR = RQ$.

Thus $\angle POR = \angle QOR = \dfrac \theta 2$.

As $OR \perp PR$ and $PR \perp PS$, we have that $\triangle OPR$ is similar to $\triangle QPS$.

Thus $\angle PQS = \angle POR$ and so $\angle PSQ$ is $\dfrac \pi 2 - \angle POR = \dfrac \pi 2 - \dfrac \theta 2$.

By definition, $\tan \angle PSQ$ is the slope of the tangent $\dfrac {\d y} {\d x}$ to $C$.

Thus:

\(\ds \dfrac {\d y} {\d x}\) \(=\) \(\ds \tan \angle PSQ\)
\(\ds \) \(=\) \(\ds \map \tan {\dfrac \pi 2 - \dfrac \theta 2}\)
\(\ds \) \(=\) \(\ds \cot \frac \theta 2\) Tangent of Complement equals Cotangent

$\blacksquare$


Historical Note

Some sources suggest that the first to solve the problem of constructing a tangent to the cycloid was Galileo Galilei, along with his pupil at the time Vincenzo Viviani.

Other sources cite the geometric proof of René Descartes, which dates to $1638$, although yet other sources suggest it was Gilles Personne de Roberval in the same year.

It has been documented that Pierre de Fermat solved the problem immediately once it had been proposed to him.


Sources

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