Editing Brachistochrone curve (section)

=== Indirect method ===
According to [[Fermat’s principle]], the actual path between two points taken by a beam of light (which obeys [[Snell's law#Derivation from Fermat's principle|Snell's law of refraction]]) is one that takes the least time. In 1697 [[Johann Bernoulli]] used this principle to derive the brachistochrone curve by considering the trajectory of a beam of light in a medium where the speed of light increases following a constant vertical acceleration (that of gravity ''g'').<ref>{{citation |title=The Brachistochrone Problem: Mathematics for a Broad Audience via a Large Context Problem |first1=Jeff |last1=Babb |first2=James |last2=Currie |url=http://www.math.umt.edu/tmme/vol5no2and3/TMME_vol5nos2and3_a1_pp.169_184.pdf |journal=The Montana Mathematics Enthusiast |volume=5 |issue=2&3 |pages=169–184 |date=July 2008 |doi=10.54870/1551-3440.1099 |s2cid=8923709 |url-status=dead |archive-url=https://web.archive.org/web/20110727210743/http://www.math.umt.edu/tmme/vol5no2and3/TMME_vol5nos2and3_a1_pp.169_184.pdf |archive-date=2011-07-27 }}</ref>

By the [[conservation of energy]], the instantaneous speed of a body ''v'' after falling a height ''y'' in a uniform gravitational field is given by:
:<math>v=\sqrt{2gy}</math>,
The speed of motion of the body along an arbitrary curve does not depend on the horizontal displacement.

Bernoulli noted that Snell's law of refraction gives a constant of the motion for a beam of light in a medium of variable density:
:<math>\frac{\sin{\theta}}{v}=\frac{1}{v}\frac{dx}{ds}=\frac{1}{v_m}</math>,
where ''v<sub>m</sub>'' is the constant and ''<math>\theta</math>'' represents the angle of the trajectory with respect to the vertical.

The equations above lead to two conclusions:
# At the onset, the angle must be zero when the particle speed is zero. Hence, the brachistochrone curve is [[tangent]] to the vertical at the origin.
# The speed reaches a maximum value when the trajectory becomes horizontal and the angle θ = 90°.

Assuming for simplicity that the particle (or the beam) with coordinates (x,y) departs from the point (0,0) and reaches maximum speed after falling a vertical distance ''D'':
:<math>v_m=\sqrt{2gD}</math>.
Rearranging terms in the law of refraction and squaring gives:
:<math>v_m^2 dx^2=v^2 ds^2=v^2 (dx^2+dy^2)</math>
which can be solved for ''dx'' in terms of ''dy'':
:<math>dx=\frac{v\, dy}{\sqrt{v_m^2-v^2}}</math>.
Substituting from the expressions for ''v'' and ''v<sub>m</sub>'' above gives:
:<math>dx=\sqrt{\frac{y}{D-y}}\,dy\,,</math>
which is the [[differential equation]] of an inverted [[cycloid]] generated by a circle of diameter ''D=2r'', whose [[parametric equation]] is:
:<math>\begin{align}
  x &= r(\varphi - \sin \varphi) \\
  y &= r(1 - \cos \varphi).
 \end{align}</math>
where φ is a real [[parameter]], corresponding to the angle through which the rolling circle has rotated. For given φ, the circle's centre lies at {{math|1=(''x'', ''y'') = (''rφ'', ''r'')}}.

In the brachistochrone problem, the motion of the body is given by the time evolution of the parameter:
:<math>\varphi(t)=\omega t\,,\omega=\sqrt{\frac{g}{r}}</math>
where ''t'' is the time since the release of the body from the point (0,0).