In physics and mathematics, a brachistochrone curve (), or curve of fastest descent, is the one lying on the plane between a point A and a lower point B, where B is not directly below A, on which a bead slides frictionlessly under the influence of a uniform gravitational field to a given end point in the shortest time. The problem was posed by Johann Bernoulli in 1696.
The brachistochrone curve is the same shape as the tautochrone curve; both are cycloids. However, the portion of the cycloid used for each of the two varies. More specifically, the brachistochrone can use up to a complete rotation of the cycloid (at the limit when A and B are at the same level), but always starts at a cusp. In contrast, the tautochrone problem can use only up to the first half rotation, and always ends at the horizontal. The problem can be solved using tools from the calculus of variations and optimal control.
The curve is independent of both the mass of the test body and the local strength of gravity. Only a parameter is chosen so that the curve fits the starting point A and the ending point B. If the body is given an initial velocity at A, or if friction is taken into account, then the curve that minimizes time differs from the tautochrone curve.
History
Johann Bernoulli posed the problem of the brachistochrone to the readers of Acta Eruditorum in June, 1696. He said:
Bernoulli wrote the problem statement as:
{ { Quote
|text=Given two points A and B in a vertical plane, what is the curve traced out by a point acted on only by gravity, which starts at A and reaches B in the shortest time.}}
Johann and his brother Jakob Bernoulli derived the same solution, but Johann's derivation was incorrect, and he tried to pass off Jakob's solution as his own. Johann published the solution in the journal in May of the following year, and noted that the solution is the same curve as Huygens's tautochrone curve. After deriving the differential equation for the curve by the method given below, he went on to show that it does yield a cycloid. However, his proof is marred by his use of a single constant instead of the three constants, vm, 2g and D, below.
Bernoulli allowed six months for the solutions but none were received during this period. At the request of Leibniz, the time was publicly extended for a year and a half. At 4 p.m. on 29 January 1697 when he arrived home from the Royal Mint, Isaac Newton found the challenge in a letter from Johann Bernoulli. Newton stayed up all night to solve it and mailed the solution anonymously by the next post. Upon reading the solution, Bernoulli immediately recognized its author, exclaiming that he "recognizes a lion from his claw mark". This story gives some idea of Newton's power, since Johann Bernoulli took two weeks to solve it.D.T. Whiteside, Newton the Mathematician, in Bechler, Contemporary Newtonian Research, p. 122. Newton also wrote, "I do not love to be dunned [pestered] and teased by foreigners about mathematical things...", and Newton had already solved Newton's minimal resistance problem, which is considered the first of the kind in calculus of variations.
In the end, five mathematicians responded with solutions: Newton, Jakob Bernoulli, Gottfried Leibniz, Ehrenfried Walther von Tschirnhaus and Guillaume de l'Hôpital. Four of the solutions (excluding l'Hôpital's) were published in the same edition of the journal as Johann Bernoulli's. In his paper, Jakob Bernoulli gave a proof of the condition for least time similar to that below before showing that its solution is a cycloid. According to Newtonian scholar Tom Whiteside, in an attempt to outdo his brother, Jakob Bernoulli created a harder version of the brachistochrone problem. In solving it, he developed new methods that were refined by Leonhard Euler into what the latter called (in 1766) the calculus of variations. Joseph-Louis Lagrange did further work that resulted in modern infinitesimal calculus.
Earlier, in 1638, Galileo had tried to solve a similar problem for the path of the fastest descent from a point to a wall in his Two New Sciences. He draws the conclusion that the arc of a circle is faster than any number of its chords, From the preceding it is possible to infer that the quickest path of all [lationem omnium velocissimam], from one point to another, is not the shortest path, namely, a straight line, but the arc of a circle. ... Consequently the nearer the inscribed polygon approaches a circle the shorter the time required for descent from A to C. What has been proven for the quadrant holds true also for smaller arcs; the reasoning is the same.