. The brachistochrone problem asks what shape the wire should be in order to minimize the bead’s time of descent from P to Q.

Question

The figure below shows a bead sliding down a frictionless wire from point P to point Q. The brachistochrone problem asks what shape the wire should be in order to minimize the bead’s time of descent from P to Q. In June of 1696, John Bernoulli proposed this problem as a public challenge, with a 6-month deadline (later extended to Easter 1697 at George Leibniz’s request). Isaac Newton, then retired from academic life and serving as Warden of the Mint in London, received Bernoulli’s challenge on January 29, 1697. The very next day he communicated his own solution - the curve of minimal descent time is an arc of an inverted cycloid - to the Royal Society of London. For a modern derivation of this result, suppose the bead starts from rest at the origin P and let y=y(x) be the equation of the desired curve in a coordinate system with the y-axis pointing downward. Then a mechanical analogue of Snell’s law in optics implies that

where c denotes the angle of deflection (from the vertical) of the tan gent line to the curve - so cotα = y’(x) (why?) - and  u=√(2gh) is the bead’s velocity when it has descended a distance y vertically (from KE =1/2 mv² = mgy = -PE). 

 Substitute y =2asin²t, dy=4a sin t cos t dt in the above differential equation to derive the solution

x = a(2t−sin2t),   y = a(1−cos2t) for which t = y = 0 when x = 0. Finally, the substitution of θ = 2a in the equations for x and y yields the standard parametric equations x = a(θ−sinθ),y = a(1−cosθ) of the cycloid that is generated by a point on the rim of a circular wheel of radius a as it rolls along the x-axis.

Answer 1

 The equation

becomes

From this we get:

If we use the trigonometric identity sin² t = (1−cos 2t) / 2 this equation becomes:

Integrating this we get:

x(t) = a(2t−sin(2t)) + C.

If x(0) = 0 this implies C = 0, so

x(t) = a(2t−sin(st)).

On the other hand, we have 

So, combining what we’ve derived we get:

x(t) = a(2t−sin(2t))

y(t) = a(1−cos(2t)).