Second Derivative of Locus of Cycloid

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Theorem

Consider a circle of radius $a$ rolling without slipping along the x-axis of a cartesian 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 $\tuple {x, y}$ 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:

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


From Slope of Tangent to Cycloid:

\(\ds y'\) \(=\) \(\ds \cot \dfrac \theta 2\) Slope of Tangent to Cycloid
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d y'} {\d x}\) \(=\) \(\ds \dfrac {\d} {\d \theta} \cot \dfrac \theta 2 \frac {\d \theta} {\d x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds -\dfrac 1 2 \csc^2 \dfrac \theta 2 / \dfrac {\d x} {\d \theta}\) Derivative of Cotangent Function
\(\ds \) \(=\) \(\ds -\dfrac 1 2 \csc^2 \dfrac \theta 2 \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }\) Derivative of Sine Function
\(\ds \) \(=\) \(\ds -\dfrac 1 {2 \sin^2 \dfrac \theta 2} \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }\) Definition of Cosecant
\(\ds \) \(=\) \(\ds -\dfrac 1 {1 - \cos \theta} \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }\) Double Angle Formula for Cosine: Corollary $5$
\(\ds \) \(=\) \(\ds -\dfrac a {y^2}\) from $y = a \paren {1 - \cos \theta}$

$\blacksquare$


Sources