Editing Tautochrone curve (section)

== "Virtual gravity" solution ==
The simplest solution to the tautochrone problem is to note a direct relation between the angle of an incline and the gravity felt by a particle on the incline.  A particle on a 90° vertical incline undergoes full gravitational acceleration <math>g</math>, while a particle on a horizontal plane undergoes zero gravitational acceleration.  At intermediate angles, the acceleration due to "virtual gravity" by the particle is <math>g\sin\theta</math>. Note that <math>\theta</math> is measured between the tangent to the curve and the horizontal, with angles above the horizontal being treated as positive angles. Thus, <math>\theta</math> varies from <math>-\pi/2</math> to <math>\pi/2</math>.

The position of a mass measured along a tautochrone curve, <math>s(t)</math>, must obey the following differential equation:

{{block indent|1=<math>\frac{d^2s}{{dt}^2} = - \omega^2s</math>}}

which, along with the initial conditions <math>s(0)=s_0</math> and <math>s'(0)=0</math>, has solution:

{{block indent|1=<math>s(t) = s_0 \cos \omega t </math>}}

It can be easily verified both that this solution solves the differential equation and that a particle will reach <math>s=0</math> at time <math>\pi/2\omega</math> from any starting position <math>s_0</math>.  The problem is now to construct a curve that will cause the mass to obey the above motion. [[Newton's second law]] shows that the force of gravity and the acceleration of the mass are related by:

{{block indent|1=<math>
\begin{align}
-g \sin \theta & = \frac{d^2s}{{dt}^2} \\
& = - \omega^2 s \,
\end{align}
</math>}}

The explicit appearance of the distance, <math>s</math>, is troublesome, but we can [[derivative|differentiate]] to obtain a more manageable form:

{{block indent|1=<math>\begin{align}
g \cos \theta \,d\theta &= \omega^2 \,ds \\
\Longrightarrow ds &= \frac{g}{\omega^2} \cos \theta \,d\theta
\end{align}</math>}}

This equation relates the change in the curve's angle to the change in the distance along the curve.  We now use [[trigonometry]] to relate the angle <math>\theta</math> to the differential lengths <math>dx</math>, <math>dy</math> and <math>ds</math>:

{{block indent|1=<math>
\begin{align}
ds = \frac{dx}{\cos \theta} \\
ds = \frac{dy}{\sin \theta}
\end{align}
</math>}}

Replacing <math>ds</math> with <math>dx</math> in the above equation lets us solve for <math>x</math> in terms of <math>\theta</math>:

{{block indent|1=<math>
\begin{align}
ds & = \frac{g}{\omega^2} \cos \theta \,d\theta \\
 \frac{dx}{\cos\theta}  & = \frac{g}{\omega^2} \cos \theta\, d\theta \\
dx & = \frac{g}{\omega^2} \cos^2 \theta \,d\theta  \\
 & = \frac{g}{2 \omega^2} \left ( \cos 2 \theta + 1 \right ) d\theta  \\
x & = \frac{g}{4 \omega^2} \left ( \sin 2 \theta + 2 \theta \right )  + C_x
\end{align}
</math>}}

Likewise, we can also express <math>ds</math> in terms of <math>dy</math> and solve for <math>y</math> in terms of <math>\theta</math>:

{{block indent|1=<math>
\begin{align}
ds & = \frac{g}{\omega^2} \cos \theta \,d\theta \\
 \frac{dy}{\sin\theta}  & = \frac{g}{\omega^2} \cos \theta\, d\theta \\
dy & = \frac{g}{\omega^2} \sin \theta \cos \theta \,d\theta \\
   & = \frac{g}{2\omega^2} \sin 2 \theta \,d\theta  \\
 y & = -\frac{g}{4\omega^2} \cos 2 \theta + C_y
\end{align}
</math>}}

Substituting <math>\phi = 2\theta</math> and <math display="inline">r = \frac{g}{4\omega^2}\,</math>, we see that these [[parametric equations]] for <math>x</math> and <math>y</math> are those of a point on a circle of radius <math>r</math> rolling along a horizontal line (a [[cycloid]]), with the circle center at the coordinates <math>(C_x + r\phi, C_y)</math>:

{{block indent|1=<math>
\begin{align}
x & = r \left( \sin \phi + \phi \right) + C_x \\
y & = -r \cos \phi + C_y
\end{align}
</math>}}

Note that <math>\phi</math> ranges from <math>-\pi \le \phi \le \pi</math>. It is typical to set <math>C_x = 0</math> and <math>C_y = r</math> so that the lowest point on the curve coincides with the origin. Therefore:

{{block indent|1=<math>
\begin{align}
x & = r \left( \phi + \sin \phi\right)\\
y & = r \left( 1 - \cos \phi\right)\\
\end{align}
</math>}}

Solving for <math>\omega</math> and remembering that <math>T = \frac{\pi}{2\omega}</math> is the time required for descent, being a quarter of a whole cycle, we find the descent time in terms of the radius <math>r</math>:

{{block indent|1=<math>
\begin{align}
r & = \frac{g}{4\omega^2} \\
\omega & = \frac{1}{2} \sqrt{\frac{g}{r}} \\
T & = \pi \sqrt{\frac{r}{g}}\\
\end{align}
</math>}}

(Based loosely on ''Proctor'', pp.&nbsp;135–139)