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
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.21$: The Cycloid: Problem $2$