Brachistochrone Problem

Find the shape of the curve down which a bead sliding from rest and accelerated by gravity will slip (without friction) from one point to another in the least time. The term derives from the Greek

(brachistos) "the shortest" and (chronos) "time, delay."

The brachistochrone problem was one of the earliest problems posed in the calculus of variations. Newton was challenged to solve the problem in 1696, and did so the very next day (Boyer and Merzbach 1991, p. 405). In fact, the solution, which is a segment of a cycloid, was found by Leibniz, L'Hospital, Newton, and the two Bernoullis. Johann Bernoulli solved the problem using the analogous one of considering the path of light refracted by transparent layers of varying density (Mach 1893, Gardner 1984, Courant and Robbins 1996). Actually, Johann Bernoulli had originally found an incorrect proof that the curve is a cycloid, and challenged his brother Jakob to find the required curve. When Jakob correctly did so, Johann tried to substitute the proof for his own (Boyer and Merzbach 1991, p. 417).

In the solution, the bead may actually travel uphill along the cycloid for a distance, but the path is nonetheless faster than a straight line (or any other line).

The time to travel from a point

to another point is given by the integral

where

is the arc length and is the speed. The speed at any point is given by a simple application of conservation of energy equating kinetic energy to gravitational potential energy,

giving

Plugging this into (◇) together with the identity

then gives

			(5)
			(6)

The function to be varied is thus

To proceed, one would normally have to apply the full-blown Euler-Lagrangedifferential equation

However, the function

is particularly nice since does not appear explicitly. Therefore, , and we can immediately use the Beltrami identity

Computing

subtracting

from , and simplifying then gives

Squaring both sides and rearranging slightly results in

			(12)
			(13)

where the square of the old constant

has been expressed in terms of a new (positive) constant . This equation is solved by the parametric equations

			(14)
			(15)

which are--lo and behold--the equations of a cycloid.

If kinetic friction is included, the problem can also be solved analytically, although the solution is significantly messier. In that case, terms corresponding to the normal component of weight and the normal component of the acceleration (present because of path curvature) must be included. Including both terms requires a constrained variational technique (Ashby et al. 1975), but including the normal component of weight only gives an approximate solution. The tangent and normal vectors are

			(16)
			(17)

gravity and friction are then

			(18)
			(19)
			(20)

and the components along the curve are

			(21)
			(22)

so Newton's Second Law gives

But

			(24)
			(25)

so

Using the Euler-Lagrange differentialequation gives

This can be reduced to

Now letting

the solution is

			(32)
			(33)