Editing Tautochrone curve (section)

== Lagrangian solution ==
For a [[Simple harmonic motion#Dynamics|simple harmonic oscillator]] released from rest, regardless of its initial displacement, the time it takes to reach the lowest potential energy point is always a quarter of its period, which is independent of its amplitude. Therefore, the Lagrangian of a simple harmonic oscillator is [[Isochronous timing|isochronous]].

In the tautochrone problem, if the particle's position is parametrized by the [[arclength]] {{math|''s''(''t'')}} from the lowest point, the kinetic energy is then proportional to <math>\dot{s}^2</math>, and the potential energy is proportional to the height {{math|''h''(''s'')}}. One way the curve in the tautochrone problem can be an isochrone is if the Lagrangian is mathematically equivalent to a simple harmonic oscillator; that is, the height of the curve must be proportional to the arclength squared:

{{block indent|1=<math> h(s) = s^2/(8r), </math>}}

where the constant of proportionality is <math>1/(8r)</math>. Compared to the simple harmonic oscillator's [[Lagrangian mechanics#Lagrangian|Lagrangian]], the equivalent spring constant is <math>k=mg/(4r)</math>, and the time of descent is <math>T/4=\frac{\pi}{2} \sqrt{\frac{m}{k}}=\pi \sqrt{\frac{r}{g}}.</math> However, the physical meaning of the constant <math>r</math> is not clear until we determine the exact analytical equation of the curve.

To solve for the analytical equation of the curve, note that the differential form of the above relation is

{{block indent|1=<math>\begin{align}
dh &= s \,ds/(4r), \\
dh^2 &= s^2 \,ds^2/(16r^2) = h \left(dx^2 + dh^2\right)/(2r),\\
\left(\frac{dx}{dh}\right)^2&=\frac{2r}{h}-1
\end{align}</math>}}

which eliminates {{math|''s''}}, and leaves a differential equation for {{math|''dx''}} and {{math|''dh''}}. This is the differential equation for a [[Cycloid#Equations|cycloid]] when the vertical coordinate {{math|''h''}} is counted from its vertex (the point with a horizontal tangent) instead of the [[Cusp (singularity)|cusp]].

To find the solution, integrate for {{math|''x''}} in terms of {{math|''h''}}:

{{block indent|1=<math>\begin{align}
\frac{dx}{dh} &= -\frac{\sqrt{2r-h}}{\sqrt{h}}, \\
x &= -4r\int \sqrt{1-u^2} \, du,
\end{align}</math>}}

where <math>u = \sqrt{h/(2r)}</math>, and the height decreases as the particle moves forward <math>dx/dh<0</math>. This integral is the area under a circle, which can be done with another substitution <math>u=\cos (t/2)</math> and yield:

{{block indent|1=<math>\begin{align}
x &=r(t - \sin t), \\
h &=r(1 + \cos t).
\end{align}</math>}}

This is the standard parameterization of a [[cycloid]] with <math>h=2r-y</math>. It's interesting to note that the arc length squared is equal to the height difference multiplied by the full arch length <math>8r</math>.