Editing Brachistochrone curve (section)

==Johann Bernoulli's solution==

=== Introduction ===

In a letter to L’Hôpital, (21/12/1696), Bernoulli stated that when considering the problem of the curve of quickest descent, after only 2 days he noticed a curious affinity or connection with another no less remarkable problem leading to an ‘indirect method’ of solution. Then shortly afterwards he discovered a ‘direct method’. <ref>{{cite book |last1= Costabel |first1= Pierre| first2= Jeanne|last2= Peiffer| title= Der Briefwechsel von Johann I Bernoulli", Vol. II: "Der Briefwechsel mit Pierre Varignon, Erster Teil: 1692-1702"|date=1988 |publisher= Springer Basel Ag |isbn=978-3-0348-5068-1 |pages=329|edition=Hardback}}</ref>

=== 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 two 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 included only the indirect method in the ''Acta Eruditorum Lipsidae'' of May 1697. He wrote that this was partly because he believed it was sufficient to convince anyone who doubted the conclusion, partly because it also resolved two famous problems in optics that "the late Mr. Huygens" had raised in his treatise on light. In the same letter he criticised Newton for concealing his method.

In addition to his indirect method he also published the five other replies to the problem that he received.

Johann Bernoulli's direct method is historically important as a 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.

In 1718, Bernoulli explained how he solved the brachistochrone problem by his direct method.<ref>Bernoulli, Johann. Mémoires de l'Académie des Sciences (French Academy of Sciences) Vol. 3, 1718, pp. 135–138</ref><ref>''The Early Period of the Calculus of Variations'', by P. Freguglia and M. Giaquinta, pp. 53–57, {{ISBN|978-3-319-38945-5}}.</ref>

He explained that he had not published it in 1697, for reasons that no longer applied in 1718. This paper was largely ignored until 1904 when the depth of the method was first appreciated by [[Constantin 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 ====
[[File:Brachistochrone Bernoulli Direct Method.png|Brachistochrone Bernoulli Direct Method]]

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 line Kne intersects it at n, and they make a small angle CKe at K. Let NK = a, and define a variable point, C on KN extended. 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. To find Mm Bernoulli argues as follows.

Let MN = x. He defines m so that MD = mx, and n so that Mm = nx + na and notes that x is the only variable and that m is finite and n is infinitely small. The small time to travel along arc Mm is <math> \frac{Mm}{MD^{\frac{1}{2}}} = \frac{n(x + a)}{(mx)^{\frac{1}{2}}} </math>, which has to be a minimum (‘un plus petit’). He does not explain that because Mm is so small the speed along it can be assumed to be the speed at M, which is as the square root of MD, the vertical distance of M below the horizontal line AL. Plus MD=mx via Pythagoras theorem.

It follows that, when differentiated this must give 
:<math>
\frac{(x - a)dx}{2x^{\frac{3}{2}}} = 0
</math> so that x = a.

This condition defines the curve that the body slides along in the shortest time possible. For each point, M on the curve, the radius of curvature, MK is cut in 2 equal parts by its axis AL. This property, which Bernoulli says had been known for a long time, is unique to the cycloid.

Finally, he considers the more general case where the speed is an arbitrary function X(x), so the time to be minimised is <math> \frac{(x + a)}{X} </math>.
The minimum condition then becomes 
<math>
X = \frac{(x + a)dX}{dx}
</math>
which he writes as :<math>
X = (x + a)\Delta x
</math>
and which gives MN (=x) as a function of NK (= a). From this the equation of the curve could be obtained from the integral calculus, though he does not demonstrate this.

==== 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.

"The reason for the synthetic demonstration, in the manner of the ancients, is to convince [[Philippe de La Hire|Mr. de la Hire]]. He has little time for our new analysis, describing it as false (He claims he has found 3 ways to prove that the curve is a cubic parabola)" – Letter from Johan Bernoulli to Pierre Varignon dated 27 Jul 1697. <ref>{{cite book |last1= Costabel |first1= Pierre| first2= Jeanne|last2= Peiffer| title= "Der Briefwechsel von Johann I Bernoulli", Vol. II: "Der Briefwechsel mit Pierre Varignon, Erster Teil: 1692-1702"|date=1988 |publisher= Springer Basel Aktiengesellschaft |isbn=978-3-0348-5068-1 |pages=117–118|edition=Hardback}}</ref>

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 <math> \tau </math> and t be the times the body takes to fall along Mm and Ce respectively.

:<math> \tau \propto \frac{Mm}{MD^{\frac{1}{2}}} </math>, <math> t \propto \frac{Ce}{CG^{\frac{1}{2}}} </math>, 
Extend CG to point F where, <math> CF = \frac{CH^2}{MD} </math> and since <math>  \frac{Mm}{Ce}  = \frac{MD}{CH} </math>, it follows that
:<math> \frac {\tau}{t} = \frac{Mm}{Ce}.\left({\frac{CG}{MD}}\right)^{\frac{1}{2}} = \left({\frac{CG}{CF}}\right)^{\frac{1}{2}} </math>
Since MN = NK, for the cycloid:
:<math> GH = \frac{MD.HD}{DK} = \frac{MD.CM}{MK} </math>, <math> CH = \frac{MD.CK}{MK} = \frac{MD.(MK + CM)}{MK} </math>, and <math> CG = CH + GH = \frac{MD.(MK + 2CM)}{MK} </math>

If Ce is closer to K than Mm then 
:<math> CH = \frac{MD.(MK - CM)}{MK} </math> and <math> CG = CH - GH = \frac{MD.(MK - 2CM)}{MK} </math>
In either case, 
:<math> CF = \frac{CH^2}{MD} > CG </math>, and it follows that <math> \tau < t </math>
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 (which obeys [[Snell's law#Derivation from Fermat's principle|Snell's law of refraction]]) 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'').<ref>{{citation |title=The Brachistochrone Problem: Mathematics for a Broad Audience via a Large Context Problem |first1=Jeff |last1=Babb |first2=James |last2=Currie |url=http://www.math.umt.edu/tmme/vol5no2and3/TMME_vol5nos2and3_a1_pp.169_184.pdf |journal=The Montana Mathematics Enthusiast |volume=5 |issue=2&3 |pages=169–184 |date=July 2008 |doi=10.54870/1551-3440.1099 |s2cid=8923709 |url-status=dead |archive-url=https://web.archive.org/web/20110727210743/http://www.math.umt.edu/tmme/vol5no2and3/TMME_vol5nos2and3_a1_pp.169_184.pdf |archive-date=2011-07-27 }}</ref>

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:
:<math>v=\sqrt{2gy}</math>,
The speed of motion of the body along an arbitrary curve does not depend on the horizontal displacement.

Bernoulli noted that Snell's law of refraction gives a constant of the motion for a beam of light in a medium of variable density:
:<math>\frac{\sin{\theta}}{v}=\frac{1}{v}\frac{dx}{ds}=\frac{1}{v_m}</math>,
where ''v<sub>m</sub>'' is the constant and ''<math>\theta</math>'' represents the angle of the trajectory with respect to the vertical.

The equations above lead to two conclusions:
# 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.
# 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'':
:<math>v_m=\sqrt{2gD}</math>.
Rearranging terms in the law of refraction and squaring gives:
:<math>v_m^2 dx^2=v^2 ds^2=v^2 (dx^2+dy^2)</math>
which can be solved for ''dx'' in terms of ''dy'':
:<math>dx=\frac{v\, dy}{\sqrt{v_m^2-v^2}}</math>.
Substituting from the expressions for ''v'' and ''v<sub>m</sub>'' above gives:
:<math>dx=\sqrt{\frac{y}{D-y}}\,dy\,,</math>
which is the [[differential equation]] of an inverted [[cycloid]] generated by a circle of diameter ''D=2r'', whose [[parametric equation]] is:
:<math>\begin{align}
  x &= r(\varphi - \sin \varphi) \\
  y &= r(1 - \cos \varphi).
 \end{align}</math>
where φ is a real [[parameter]], corresponding to the angle through which the rolling circle has rotated. For given φ, the circle's centre lies at {{math|1=(''x'', ''y'') = (''rφ'', ''r'')}}.

In the brachistochrone problem, the motion of the body is given by the time evolution of the parameter:
:<math>\varphi(t)=\omega t\,,\omega=\sqrt{\frac{g}{r}}</math>
where ''t'' is the time since the release of the body from the point (0,0).