info prev up next book cdrom email home

Tautochrone Problem

Find the curve down which a bead placed anywhere will fall to the bottom in the same amount of time. The solution is a Cycloid, a fact first discovered and published by Huygens in Horologium oscillatorium (1673). Huygens also constructed the first pendulum clock with a device to ensure that the pendulum was isochronous by forcing the pendulum to swing in an arc of a Cycloid.

The parametric equations of the Cycloid are

$\displaystyle x$ $\textstyle =$ $\displaystyle a(\theta-\sin\theta)$ (1)
$\displaystyle y$ $\textstyle =$ $\displaystyle a(1-\cos\theta).$ (2)

To see that the Cycloid satisfies the tautochrone property, consider the derivatives
$\displaystyle x'$ $\textstyle =$ $\displaystyle a(1-\cos\theta)$ (3)
$\displaystyle y'$ $\textstyle =$ $\displaystyle a\sin\theta,$ (4)

$\displaystyle x'^2+y'^2$ $\textstyle =$ $\displaystyle a^2[(1-2\cos\theta+\cos^2\theta)+\sin^2\theta]$  
  $\textstyle =$ $\displaystyle 2a^2(1-\cos\theta).$ (5)

{\textstyle{1\over 2}}mv^2=mgy
\end{displaymath} (6)

v={ds\over dt}=\sqrt{2gy}
\end{displaymath} (7)

$\displaystyle dt$ $\textstyle =$ $\displaystyle {ds\over \sqrt{2gy}} = {\sqrt{dx^2+dy^2}\over \sqrt{2gy}}$  
  $\textstyle =$ $\displaystyle {a\sqrt{2(1-\cos\theta)}\,d\theta\over\sqrt{2ga(1-\cos\theta)}} = \sqrt{a\over g}\,d\theta,$ (8)

so the time required to travel from the top of the Cycloid to the bottom is
T=\int_0^\pi dt=\sqrt{a\over g} \,\pi.
\end{displaymath} (9)

However, from an intermediate point $\theta_0$,
v={ds\over dt}=\sqrt{2g(y-y_0)},
\end{displaymath} (10)

$\displaystyle T$ $\textstyle =$ $\displaystyle \int_{\theta_0}^\pi {\sqrt{2a^2(1-\cos\theta)}\over 2ag(\cos\theta_0-\cos\theta)}\,d\theta$  
  $\textstyle =$ $\displaystyle \sqrt{a\over g} \int_{\theta_0}^\pi\sqrt{1-\cos\theta\over\cos\theta_0-\cos\theta}\,d\theta$  
  $\textstyle =$ $\displaystyle \sqrt{a\over g} \int_{\theta_0}^\pi {\sin({\textstyle{1\over 2}}\...
...{\cos^2({\textstyle{1\over 2}}\theta_0)-\cos^2({\textstyle{1\over 2}}\theta)}}.$ (11)

Now let
$\displaystyle u$ $\textstyle =$ $\displaystyle {\cos({\textstyle{1\over 2}}\theta)\over\cos({\textstyle{1\over 2}}\theta_0)}$ (12)
$\displaystyle du$ $\textstyle =$ $\displaystyle -{\sin({\textstyle{1\over 2}}\theta)d\theta\over 2\cos(\theta_0)},$ (13)

T=-2\sqrt{a\over g}\int_1^0 {du\over\sqrt{1-u^2}} = 2\sqrt{a\over g}\, [\sin^{-1}u]^1_0 = \pi\sqrt{a\over g}\,,
\end{displaymath} (14)

and the amount of time is the same from any point!

See also Brachistochrone Problem, Cycloid


Muterspaugh, J.; Driver, T.; and Dick, J. E. ``The Cycloid and Tautochronism.''

Muterspaugh, J.; Driver, T.; and Dick, J. E. ``P221 Tautochrone Problem.''

Wagon, S. Mathematica in Action. New York: W. H. Freeman, pp. 54-60 and 384-385, 1991.

info prev up next book cdrom email home

© 1996-9 Eric W. Weisstein