Brachistochrone Curve Problem

A Brachistochrone Curve is the curve that would carry a bead from rest along the curve, without friction, under constant gravity, to an end point in the shortest amount of time.

Source: storyofmathematics

The Brachistochrone problem was one of the earliest problems posed in calculus of variations. The solution, which is a segment of a cycloid, was found individually by Leibniz, L’Hospital, Newton and both the Bernoulli’s.

The following solutions are taken from wikipedia.

Johann Bernoulli’s Solution

Johann Bernoulli used Fermat’s principle that “the actual path between two points taken by a beam of light is the one which is traversed in the least time” in order to derive the brachistochrone curve. He did this by considering the path that a beam of light would take in a medium where the speed of light increases due to a constant vertical acceleration equal to g.

Due to the conservation of energy, v={\sqrt  {2gy}}, where y is the vertical distance. Furthermore, the law of refraction gives us a constant (vm) of the motion for a beam of light in a medium of variable density:

{\frac  {\sin {\theta }}{v}}={\frac  {1}{v}}{\frac  {dx}{ds}}={\frac  {1}{v_{m}}}

Rearranging this gives us

v_{m}^{2}dx^{2}=v^{2}ds^{2}=v^{2}(dx^{2}+dy^{2})

which can be manipulated to give

dx=\frac{v\, dy}{\sqrt{v_m^2-v^2}}

If we assume that the beam, with coordinates (x,y) departs from the origin and reaches a maximum speed after falling a vertical distance D:

v_{m}={\sqrt  {2gD}}

we can rearrange the equation to give us the following:

dx={\sqrt  {{\frac  {y}{D-y}}}}dy

which is the differential equation of an inverted cycloid generated by a circle of diameter D, as required.

Jakob Bernoulli’s Solution

Jakob Bernoulli’s approach was to use second order differentials to find the condition for the least time.  The differential triangle formed by the displacement along the path, the horizontal displacement and the vertical displacement is a right-handed triangle, therefore:

ds^2=dx^2+dy^2

Differentiating this gives

2ds\ d^{2}s=2dx\ d^{2}x

{\frac  {dx}{ds}}d^{2}x=d^{2}s=v\ d^{2}t

Consider the follow diagram:

Path function 2.PNG

The horizontal separation between paths along the central line is d2x.

The diagram gives us two separate equations:

d^{2}t_{1}={\frac  {1}{v_{1}}}{\frac  {dx_{1}}{ds_{1}}}d^{2}x

d^{2}t_{2}={\frac  {1}{v_{2}}}{\frac  {dx_{2}}{ds_{2}}}d^{2}x

For the path of the least time, these times are equal hence their difference is equal to zero.

d^{2}t_{2}-d^{2}t_{1}=0={\bigg (}{\frac  {1}{v_{2}}}{\frac  {dx_{2}}{ds_{2}}}-{\frac  {1}{v_{1}}}{\frac  {dx_{1}}{ds_{1}}}{\bigg )}d^{2}x

Consequently, the condition for the least time is

{\frac  {1}{v_{2}}}{\frac  {dx_{2}}{ds_{2}}}={\frac  {1}{v_{1}}}{\frac  {dx_{1}}{ds_{1}}}

as required.

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Hope you enjoy, M x

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