Brachistochrone curve

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In mathematics and physics, a brachistochrone curve (from Ancient Greek ????????? ?????? (brákhistos khrónos), meaning 'shortest time'), 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.

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.

The problem can be solved with the 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 will differ from the one described above.

Video Brachistochrone curve

History

Johann Bernoulli posed the problem of the brachistochrone to the readers of Acta Eruditorum in June, 1696. 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. However, his proof is marred by his use of a single constant instead of the three constants, v_m, 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. On 29 January 1697 the challenge was received by Isaac Newton, who found it in his mail, in a letter from Johann Bernoulli, 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. Newton also wrote, I do not love to be dunned [pestered] and teased by foreigners about mathematical things ... and Newton had already solved the Newton's minimal resistance problem, which is considered as the first of the kind in calculus of variations.

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. 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,

"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.

Maps Brachistochrone curve

Johann Bernoulli's solution

Direct Method

In a letter to Henri Basnage, held at the University of Basel Public Library, dated 30 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 below, 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

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

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 ${\displaystyle \tau <t}$

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).

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 v_m 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 ? = 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 v_m 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.

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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,

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

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Newton's Solution

Introduction

In June 1696, Johann Bernoulli had proposed a mathematical challenge in the Acta Eruditorum Lipsidae to find the form of the curve joining 2 fixed points that a mass will slide down in the minimum time. The solution was originally to be submitted within 6 months. At the suggestion of Leibniz he extended the challenge till Easter 1697, by means of a printed text, called 'Programma', published in Groeningen, the Netherlands.

The Programma is dated 'New Year's Day', 1 January 1697, in the Gregorian Calendar. This was 22 December 1696 in the Julian Calendar, in use in Britain. Newton, claimed he had been unaware of the challenge until he first saw it at 4 pm on 29 January, some 5 weeks after its publication.

He further claims that he solved it by 4 am the following morning, and his solution is dated 30 January. It seems highly suspicious that it would take so long for a communication from Groeningen to arrive in London. Newton's account should be viewed with scepticism as he was not always truthful about his accomplishments, especially when it comes to his dealings with Leibniz and his associates.

Newton's solution was that the curve was the cycloid, though he never publicly demonstrated his method of proof. Bernoulli, writing to Henri Basnage in March 1697, recognised that although the author, 'by an excess of modesty' had not revealed his name, yet even from the scant details supplied he knew that it was from Mr Newton, 'as the lion by its claw' [ex ungue Leonem].

It is even possible that Newton had previous knowledge of the challenge. His teacher, Wallis in Oxford, who was 80 had been made aware of it in September 1696, by Bernoulli's youngest brother, Hieronymus, and had spent 3 months attempting a solution before passing it to David Gregory in December, who also failed to solve it. After Newton had submitted his solution, Gregory asked him for the details and made notes from their conversation. These can be found in the University of Edinburgh Library, manuscript A ${\displaystyle 78^{1}}$, dated 7 March 1697.

Either Gregory did not understand Newton's argument, or Newton's explanation was very brief. However, it is possible, with a high degree of confidence, to construct Newton's proof from Gregory's notes, by analogy with his method to determine the solid of minimum resistance (Principia, Book 2, Proposition 34, Scholium 2). A detailed description of his solution of this latter problem is included in the draft of a letter in 1694, also to David Gregory.

Note, a lot of the information here was taken from, No. 737 Bulletin De L'Union Des Physiciens, 'Chute d'une bille le long d'une gouttière cycloïdale; Tautochrone et brachistochrone; Propriétés et historique.', Pages 1251 - 1289, Vol 85, Oct 1991 - Jacques Dubois, 37000 Tours, France

In addition to the minimum time curve problem there was a second problem which Newton also solved at the same time. Both solutions appeared anonymously in Philosophical Transactions of the Royal Society, for January 1697.

The Brachistochrone Problem

Fig. 1, shows Gregory's diagram (except the additional line IF is absent from it, and Z, the start point has been added). The curve ZVA is a cycloid and CHV is its generating circle. Since it appears that the body is moving upwards from e to E, it must be assumed that a small body is released from Z and slides along the curve to A, without friction, under the action of gravity.

Consider a small arc eE which the body is ascending. Assume that it traverses the straight line eL to point L, horizontally displaced from E by a small distance, o, instead of the arc eE. Note, that eL is not the tangent at e, and that o will be negative when L is between B and E. Draw the line through E parallel to CH, cutting eL at n. From a property of the cycloid, En is the normal to the tangent at E, and similarly the tangent at E is parallel to VH.

Since the displacement, EL is small it differs little in direction from the tangent at E so that the angle EnL is close to a right-angle. In the limit as the arc eE approaches zero, eL becomes parallel to VH, provided o is small compared to eE making the triangles EnL and CHV similar.

Also en approaches the length of chord eE, and the increase in length, ${\displaystyle eL-eE=nL={\frac {o.CH}{CV}}}$, ignoring terms in ${\displaystyle o^{2}}$ and higher, which represent the error due to the approximation that eL and VH are parallel

The speed along eE or eL can be taken as that at E, proportional to ${\displaystyle {\sqrt {CB}}}$ which is as CH, since ${\displaystyle CH={\sqrt {CB.CV}}}$

This appears to be all that Gregory's note contains.

Let t be the additional time to reach L,

${\displaystyle t\propto {\frac {nL}{\sqrt {CB}}}={\frac {o.CH}{CV.{\sqrt {CB}}}}={\frac {o}{\sqrt {CV}}}}$

Therefore, the increase in time to traverse a small arc displaced at one endpoint depends only on the displacement at the endpoint and is independent of the position of the arc. However, by Newton's method, this is just the condition required for the curve to be traversed in the minimum time possible. Therefore, he concludes that the minimum curve must be the cycloid.

He argues as follows.

Assuming now that Fig. 1 is the minimum curve not yet determined, with vertical axis CV, and the circle CHV removed, and Fig. 2 shows part of the curve between the infinitesimal arc eE and a further infinitesimal arc Ff a finite distance along the curve. The extra time, t, to traverse eL (rather than eE) is nL divided by the speed at E (proportional to ?CB), ignoring terms in ${\displaystyle o^{2}}$ and higher:

${\displaystyle t\propto {\frac {o.DE}{eE.{\sqrt {CB}}}}}$,

At L the particle continues along a path LM, parallel to the original EF, to some arbitrary point M. As it has the same speed at L as at E, the time to traverse LM is the same as it would have been along the original curve EF. At M it returns to the original path at point f. By the same reasoning, the reduction in time, T, to reach f from M rather than from F is

${\displaystyle T\propto {\frac {o.FG}{Ff.{\sqrt {CI}}}}}$

The difference (t - T) is the extra time it takes along the path eLMf compared to the original eEFf :

${\displaystyle (t-T)\propto \left({\frac {DE}{eE{\sqrt {CB}}}}-{\frac {FG}{Ff{\sqrt {CI}}}}\right).o}$ plus terms in ${\displaystyle o^{2}}$ and higher (1)

Because eEFf is the minimum curve, (t - T) is must be greater than zero, whether o is positive or negative. It follows that the coefficient of o in (1) must be zero:

${\displaystyle {\frac {DE}{eE{\sqrt {CB}}}}={\frac {FG}{Ff{\sqrt {CI}}}}}$ (2) in the limit as eE and fF approach zero. Note since eEFf is the minimum curve it has to be assumed that the coefficient of ${\displaystyle o^{2}}$ is greater than zero.

Clearly there has to be 2 equal and opposite displacements, or the body would not return to the endpoint, A, of the curve.

If e is fixed, and if f is considered a variable point higher up the curve, then for all such points, f, ${\displaystyle {\frac {FG}{Ff{\sqrt {CI}}}}}$ is constant (equal to ${\displaystyle {\frac {DE}{eE{\sqrt {CB}}}}}$). By keeping f fixed and making e variable it is clear that ${\displaystyle {\frac {DE}{eE{\sqrt {CB}}}}}$ is also constant.

But, since points, e and f are arbitrary, equation (2) can only be true if ${\displaystyle {\frac {DE}{eE{\sqrt {CB}}}}=constant}$, everywhere, and this condition characterises the curve that is sought. This is the same technique he uses to find the form of the Solid of Least Resistance.

For the cycloid, ${\displaystyle {\frac {DE}{eE}}={\frac {BH}{VH}}={\frac {CH}{CV}}}$ , so that ${\displaystyle {\frac {DE}{eE{\sqrt {CB}}}}={\frac {CH}{CV.{\sqrt {CB}}}}}$ which was shown above to be constant, and the Brachistochrone is the cycloid.

Newton gives no indication of how he discovered that the cycloid satisfied this last relation. It may have been by trial and error, or he may have recognised immediately that it implied the curve was the cycloid.

The Other Problem

The second problem proposed by Johann Bernoulli is: 'Given a fixed point P, a curve is sought, such that for each straight line PKL cutting it in 2 points K and L, the sum of the distances PK and PL, raised to a given power n, is a constant'. Although its only connection with the Brachistochrone problem is that it was posed at the same time, Newton's solution is elegant and worth including, since it is not well known.

He defines an infinite line ADB not containing P, with a fixed point A and variable point D, having AD = x. The required curve is obtained from the quadratic expression in y: ${\displaystyle y^{2}+Q(x).y+R(x)=0}$, where Q(x) and R(x) are continuous functions of x. Q(x) and R(x) are not independent, their particular dependence is determined by the value of n.

PDK is set equal to the smaller root, and PDL to the larger, whenever the 2 roots are real.

Since ${\displaystyle -Q(x)=PK+PL,R(x)=PK.PL}$, the sum of PK and PL both raised to any power, n can be obtained from polynomials in Q and R, which can be set to a constant.

For example, ${\displaystyle PK^{3}+PL^{3}=-(Q^{3}-3QR)}$, so if the latter is made a constant, then the condition is satisfied for n = 3.

Similarly, for n = 4, ${\displaystyle PK^{4}+PL^{4}=Q^{4}-4Q^{2}R+2R^{2}=constant}$ and so on for any value of n.

Having solved the problem he goes on to consider the case where a curve is cut in 3 points, so that the sum of the distances raised to any power from a fixed point, P to each of them is constant.

His curve is now ${\displaystyle y^{3}+Q(x).y^{2}+R(x).y+S(x)=0}$, where Q(x), R(x) and S(x) are continuous functions of x. Setting PDK, PDL and PDM to the 3 roots,

${\displaystyle PK^{2}+PL^{2}+PM^{2}=Q^{2}-2R=constant}$, and in addition, ${\displaystyle PK.PL.PM=S}$, when n = 2

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See also

• Beltrami identity
• Calculus of variations
• Catenary
• Cycloid
• Tautochrone curve
• Newton's minimal resistance problem
• Uniformly accelerated motion

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References

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External links

• Hazewinkel, Michiel, ed. (2001) [1994], "Brachistochrone", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4 
• Weisstein, Eric W. "Brachistochrone Problem". MathWorld. 
• Brachistrochrone ( at MathCurve, with excellent animated examples)
• The Brachistochrone, Whistler Alley Mathematics.
• Table IV from Bernoulli's article in Acta Eruditorum 1697
• Brachistochrones by Michael Trott and Brachistochrone Problem by Okay Arik, Wolfram Demonstrations Project.
• The Brachistochrone problem at MacTutor
• Geodesics Revisited -- Introduction to geodesics including two ways of derivation of the equation of geodesic with brachistochrone as a special case of a geodesic.
• Optimal control solution to the Brachistochrone problem in Python.
• The straight line, the catenary, the brachistochrone, the circle, and Fermat Unified approach to some geodesics.

Source of the article : Wikipedia