# Brachistochrone curve

(Redirected from Brachystochrone)
The solution to the brachistochrone problem is not a straight line or some combination thereof but a cycloid.

In mathematics and physics, a brachistochrone curve (from Ancient Greek βράχιστος χρόνος (brákhistos khrónos), meaning 'shortest time'),[1] or curve of fastest descent, is the one lying on 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. Incidentally, 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 only use up to the first half rotation, and always ends at the horizontal.[2]

The problem can be solved with the tools from the calculus of variations and optimal control.[3]

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.[4] If the body is given an initial velocity at A, or if friction is taken into account, then the curve that minimizes time will differ from the one described above.

## History

Johann Bernoulli posed the problem of the brachistochrone to the readers of Acta Eruditorum in June, 1696.[5][6] He said,

I, Johann Bernoulli, address the most brilliant mathematicians in the world. Nothing is more attractive to intelligent people than an honest, challenging problem, whose possible solution will bestow fame and remain as a lasting monument. Following the example set by Pascal, Fermat, etc., I hope to gain the gratitude of the whole scientific community by placing before the finest mathematicians of our time a problem which will test their methods and the strength of their intellect. If someone communicates to me the solution of the proposed problem, I shall publicly declare him worthy of praise

and he wrote the problem statement as

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.

He published his 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.[7][8] 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.[9] On 29 January 1697 the challenge was received by Isaac Newton, who found it in his mail, in a letter from Johann Bernoulli,[10] when he arrived home from the Royal Mint at 4 p.m., and 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.[4][11] Newton also wrote, I do not love to be dunned [pestered] and teased by foreigners about mathematical things ...

At the end, five mathematicians responded with solutions: Newton, Jakob Bernoulli (Johann's brother), 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.[7] 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 (Third Day, Theorem 22, Prop. 36) that the arc of a circle is faster than any number of its chords,[12]

"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 is 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."

We are warned earlier in the Two New Sciences (just after Theorem 6) of possible fallacies and the need for a "higher science." In this dialogue Galileo reviews his own work. The actual solution to Galileo's problem is half a cycloid. Galileo studied the cycloid and gave it its name, but the connection between it and his problem had to wait for advances in mathematics.

## Johann Bernoulli's solution

### Direct Method

In a letter to Henri Basnage, held at the University of Basel Public Library, dated 30th March 1697, Johann Bernoulli stated that he had found 2 methods (always referred to as 'direct' and 'indirect') to show that the Brachistochrone was the ‘common cycloid’, also called the ‘roulette‘. Following advice from Leibniz, he only included the indirect method in the Acta Eruditorum Lipsidae of May 1697. He writes that this is partly because he believed it was sufficient to convince anyone who doubted the conclusion, partly because it also resolved 2 famous problems in optics which ‘the late Mr. Huygens’ had raised in his treatise on Light. In the same letter he criticises Newton for concealing his method.

As well as his indirect method he also published the 5 other replies to the problem that he received.

Johann Bernoulli's direct method is historically important as it was the first proof that the brachistochrone is the cycloid. The method is to determine the curvature of the curve at each point. All the other proofs, including Newton’s (which was not revealed at the time) are based on finding the gradient at each point.

It was only in the 3rd volume of Mémoires de l'Académie des Sciences (French Academy of Sciences), for 1718, pp 135-138 (followed by the Figure above, on the next page), that Bernoulli explained how he solved the brachistochrone problem by his direct method. Details can be found in ‘The Early Period of the Calculus of Variations’, by P. Freguglia and M. Giaquinta pp 53 – 57, ISBN 978-3-319-38945-5.

He explained that he had not published it in 1697, for reasons which no longer applied in 1718. This paper was largely ignored until 1904 when the depth of the method was first appreciated by C. Carathéodory, who stated that it shows that the cycloid is the only possible curve of quickest descent. According to him, the other solutions simply implied that the time of descent is stationary for the cycloid, but not necessarily the minimum possible.

#### Analytic Solution

A body is regarded as sliding along any small circular arc Ce between the radii KC and Ke, with centre K fixed. The first stage of the proof involves finding the particular circular arc, Mm which the body traverses in the minimum time.

The line KNC intersects AL at N, and Kne at n, and they make a small angle at K. Let NK = a, and define a variable point, C on KN extended with NC = x. Of all the possible circular arcs Ce, it is required to find the arc Mm which requires the minimum time to slide between the 2 radii, KM and Km. The speed of the body is assumed to be constant and to be that of a body having reached C by falling from the horizontal line AL under the action of a constant acceleration. Consequently, the speed at C is as the square root of CG, the vertical distance of C below the horizontal line, AL. Ce is proportional to the radius KC and the fixed angle, NKn.

Since

${\displaystyle Ce\propto (x+a)}$ and along the infinitesimal arc Ce the
${\displaystyle speed\propto CG^{\frac {1}{2}}=x^{\frac {1}{2}}}$ so that the time to travel along arc Ce:
${\displaystyle t\propto {\frac {(x+a)}{x^{\frac {1}{2}}}}}$

and the differential,

${\displaystyle dt\propto {\frac {(x-a)dx}{2x^{\frac {3}{2}}}}}$. The stationary condition, dt = 0, occurs when x = a.

The curve AMmB, from point A to B is assumed to be the one that the body slides along in the shortest time possible. By dividing the curve into a large number of circular arcs, all subtending the same infinitesimal angle, NKn at their centre of curvature, from the minimum condition above, the radius joining each arc to its centre of curvature must be bisected by the horizontal line, AL containing the start point, A. This is a well-known property of the cycloid.

#### Synthetic Solution

He then proceeds with what he called his Synthetic Solution, which was a classical, geometrical proof, that there is only a single curve that a body can slide down in the minimum time, and that curve is the cycloid.

Assume AMmB is the part of the cycloid joining A to B, which the body slides down in the minimum time. Let ICcJ be part of a different curve joining A to B which can be closer to AL than AMmB. If the arc Mm subtends the angle MKm at its centre of curvature, K, let the arc on IJ that subtends the same angle be Cc. The circular arc through C with centre K is Ce. Point D on AL is vertically above M. Join K to D and point H is where CG intersects KD, extended if necessary.

Let ${\displaystyle \tau }$ and t be the times the body takes to fall along Mm and Ce respectively.

${\displaystyle \tau \propto {\frac {Mm}{MD^{\frac {1}{2}}}}}$, ${\displaystyle t\propto {\frac {Ce}{CG^{\frac {1}{2}}}}}$,

Extend CG to point F where, ${\displaystyle CF={\frac {CH^{2}}{MD}}}$ and since ${\displaystyle {\frac {Mm}{Ce}}={\frac {MD}{CH}}}$, it follows that

${\displaystyle {\frac {\tau }{t}}={\frac {Mm}{Ce}}.({\frac {CG}{MD}})^{\frac {1}{2}}=({\frac {CG}{CF}})^{\frac {1}{2}}}$

Since MN = NK, for the cycloid:

${\displaystyle GH={\frac {MD.HD}{DK}}={\frac {MD.CM}{MK}}}$, ${\displaystyle CH={\frac {MD.CK}{MK}}={\frac {MD.(MK+CM)}{MK}}}$, and ${\displaystyle CG=CH+GH={\frac {MD.(MK+2CM)}{MK}}}$

If Ce is closer to K than Mm then

${\displaystyle CH={\frac {MD.(MK-CM)}{MK}}}$ and ${\displaystyle CG=CH-GH={\frac {MD.(MK-2CM)}{MK}}}$

In either case,

${\displaystyle CF={\frac {CH^{2}}{MD}}>CG}$, and it follows that ${\displaystyle \tau

If the arc, Cc subtended by the angle infinitesimal angle MKm on IJ is not circular, it must be greater than Ce, since Cec becomes a right-triangle in the limit as angle MKm approaches zero.

Note, Bernoulli proves that CF > CG by a similar but different argument.

From this he concludes that a body traverses the cycloid AMB in less time than any other curve ACB.

### Indirect Method

According to Fermat’s principle, the actual path between two points taken by a beam of light is one that takes the least time. In 1697 Johann Bernoulli used this principle to derive the brachistochrone curve by considering the trajectory of a beam of light in a medium where the speed of light increases following a constant vertical acceleration (that of gravity g).[13]

By the conservation of energy, the instantaneous speed of a body v after falling a height y in a uniform gravitational field is given by:

${\displaystyle v={\sqrt {2gy}}}$,

The speed of motion of the body along an arbitrary curve does not depend on the horizontal displacement.

Bernoulli noted that the law of refraction gives a constant of the motion for a beam of light in a medium of variable density:

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

where vm is the constant and ${\displaystyle \theta }$ represents the angle of the trajectory with respect to the vertical.

The equations above lead to two conclusions:

1. At the onset, the angle must be zero when the particle speed is zero. Hence, the brachistochrone curve is tangent to the vertical at the origin.
2. The speed reaches a maximum value when the trajectory becomes horizontal and the angle θ = 90°.

Assuming for simplicity that the particle (or the beam) with coordinates (x,y) departs from the point (0,0) and reaches maximum speed after falling a vertical distance D:

${\displaystyle v_{m}={\sqrt {2gD}}}$.

Rearranging terms in the law of refraction and squaring gives:

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

which can be solved for dx in terms of dy:

${\displaystyle dx={\frac {v\,dy}{\sqrt {v_{m}^{2}-v^{2}}}}}$.

Substituting from the expressions for v and vm above gives:

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

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

## Jakob Bernoulli's solution

Johann's brother Jakob showed how 2nd differentials can be used to obtain the condition for least time. A modernized version of the proof is as follows. If we make a negligible deviation from the path of least time, then, for the differential triangle formed by the displacement along the path and the horizontal and vertical displacements,

${\displaystyle ds^{2}=dx^{2}+dy^{2}}$.

On differentiation with dy fixed we get,

${\displaystyle 2ds\ d^{2}s=2dx\ d^{2}x}$.

And finally rearranging terms gives,

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

where the last part is the displacement for given change in time for 2nd differentials. Now consider the changes along the two neighboring paths in the figure below for which the horizontal separation between paths along the central line is d2x (the same for both the upper and lower differential triangles). Along the old and new paths, the parts that differ are,

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

For the path of least times these times are equal so for their difference we get,

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

And the condition for least time is,

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

## References

1. ^  Chisholm, Hugh, ed. (1911). "Brachistochrone". Encyclopædia Britannica (11th ed.). Cambridge University Press.
2. ^ Stewart, James. "Section 10.1 - Curves Defined by Parametric Equations." Calculus: Early Transcendentals. 7th ed. Belmont, CA: Thomson Brooks/Cole, 2012. 640. Print.
3. ^ Ross, I. M. The Brachistochrone Paradigm, in A Primer on Pontryagin's Principle in Optimal Control, Collegiate Publishers, 2009. ISBN 978-0-9843571-0-9.
4. ^ a b Hand, Louis N., and Janet D. Finch. "Chapter 2: Variational Calculus and Its Application to Mechanics." Analytical Mechanics. Cambridge: Cambridge UP, 1998. 45, 70. Print.
5. ^ Johann Bernoulli (June 1696) "Problema novum ad cujus solutionem Mathematici invitantur." (A new problem to whose solution mathematicians are invited.), Acta Eruditorum, 18 : 269. From p. 269: "Datis in plano verticali duobus punctis A & B (vid Fig. 5) assignare Mobili M, viam AMB, per quam gravitate sua descendens & moveri incipiens a puncto A, brevissimo tempore perveniat ad alterum punctum B." (Given in a vertical plane two points A and B (see Figure 5), assign to the moving [body] M, the path AMB, by means of which — descending by its own weight and beginning to be moved [by gravity] from point A — it would arrive at the other point B in the shortest time.)
6. ^ Solutions to Johann Bernoulli's problem of 1696:
7. ^ a b Struik, J. D. (1969), A Source Book in Mathematics, 1200-1800, Harvard University Press, ISBN 0-691-02397-2
8. ^ Herman Erlichson (1999), "Johann Bernoulli's brachistochrone solution using Fermat's principle of least time", Eur. J. Phys., 20: 299–304, doi:10.1088/0143-0807/20/5/301
9. ^ Sagan, Carl (2011). Cosmos. Random House Publishing Group. p. 94. ISBN 9780307800985. Retrieved 2 June 2016.
10. ^ Katz, Victor J. (1998), A History of Mathematics / An Introduction (2nd ed.), Addison Wesley Longman, p. 547, ISBN 978-0-321-01618-8
11. ^ D.T.Whiteside, Newton the mathematician, in Bechler, Contemporary Newtonian Research, p. 122.
12. ^ Galileo Galilei (1638), Discourses regarding two new sciences, p. 239 This conclusion had appeared six years earlier in Galileo's Dialogue Concerning the Two Chief World Systems (Day 4).
13. ^ Babb, Jeff; Currie, James (July 2008), "The Brachistochrone Problem: Mathematics for a Broad Audience via a Large Context Problem" (PDF), TMME, 5 (2&3): 169–184