Editing Cycloid (section)

== Equations ==
The cycloid through the origin, generated by a circle of radius {{mvar|r}} rolling over the ''{{mvar|x}}-''axis on the positive side ({{math|''y'' ≥ 0}}), consists of the points {{math|(''x'', ''y'')}}, with
<math display="block">\begin{align}
  x &= r(t - \sin t) \\
  y &= r(1 - \cos t),
 \end{align}</math>
where {{mvar|t}} is a real [[parameter]] corresponding to the angle through which the rolling circle has rotated. For given {{mvar|t}}, the circle's centre lies at {{math|1=(''x'', ''y'') = (''rt'', ''r'')}}.

The [[Cartesian coordinate system|Cartesian equation]] is obtained by solving the ''{{mvar|y}}''-equation for {{mvar|t}} and substituting into the ''{{mvar|x}}-''equation:<math display="block">x = r \cos^{-1} \left(1 - \frac{y}{r}\right) - \sqrt{y(2r - y)},</math>or, eliminating the multiple-valued inverse cosine:<blockquote><math>r \cos\!\left(\frac{x+\sqrt{y(2r-y)}}{r}\right) + y = r.</math></blockquote>When {{mvar|y}} is viewed as a function of {{mvar|x}}, the cycloid is [[Differentiable function|differentiable]] everywhere except at the [[Cusp (singularity)|cusps]] on the {{mvar|x}}-axis, with the derivative tending toward <math>\infty</math> or <math>-\infty</math> near a cusp (where {{mvar|1=y=0}}). The map from {{mvar|t}} to {{math|(''x'', ''y'')}} is differentiable, in fact of class {{mvar|C}}<sup>∞</sup>, with derivative 0 at the cusps.

The slope of the [[tangent]] to the cycloid at the point <math>(x,y)</math> is given by <math display="inline">\frac{dy}{dx} = \cot(\frac{t}{2})</math>.

A cycloid segment from one cusp to the next is called an arch of the cycloid, for example the points with <math>0 \le t \le 2 \pi</math> and <math>0 \leq x \leq 2\pi</math>.

Considering the cycloid as the graph of a function <math>y = f(x)</math>, it satisfies the [[ordinary differential equation|differential equation]]:<ref>{{cite book |title=Elementary Differential Equations: Applications, Models, and Computing |edition=2nd illustrated |first1=Charles |last1=Roberts |publisher=CRC Press |year=2018 |isbn=978-1-4987-7609-7 |page=141 |url=https://books.google.com/books?id=touADwAAQBAJ}} [https://books.google.com/books?id=touADwAAQBAJ&pg=PA141 Extract of page 141, equation (f) with their ''K''=2''r'']</ref>

:<math>\left(\frac{dy}{dx}\right)^2 = \frac{2r}{y} - 1.</math>
If we define <math>h=2r-y= r(1 + \cos t)</math> as the height difference from the cycloid's vertex (the point with a horizontal tangent and <math>\cos t=-1</math>), then we have:
:<math>\left(\frac{dx}{dh }\right)^2 = \frac{2r}{h} - 1.</math>