Editing Brachistochrone curve (section)

== Newton's solution ==
=== Introduction ===

In June 1696, Johann Bernoulli had used the pages of the ''Acta Eruditorum Lipsidae'' to pose a challenge to the international mathematical community: to find the form of the curve joining two fixed points so that a mass will slide down along it, under the influence of gravity alone, in the minimum amount of time. The solution was originally to be submitted within six months. At the suggestion of Leibniz, Bernoulli extended the challenge until Easter 1697, by means of a printed text called "Programma", published in [[Groningen]], in the Netherlands.

The ''Programma'' is dated 1 January 1697, in the Gregorian Calendar. This was 22 December 1696 in the Julian Calendar, in use in Britain.
According to Newton's niece, Catherine Conduitt, Newton learned of the challenge at 4 pm on 29 January and had solved it by 4 am the following morning.  His solution, communicated to the Royal Society, is dated 30 January.  This solution, later published anonymously in the ''Philosophical Transactions'', is correct but does not indicate the method by which Newton arrived at his conclusion.  Bernoulli, writing to Henri Basnage in March 1697, indicated that even though its author, "by an excess of modesty", had not revealed his name, yet even from the scant details supplied it could be recognised as Newton's work, "as the lion by its claw" (in Latin, ''ex ungue Leonem'').

D. T. Whiteside notes that the letter in French has ''ex ungue Leonem'' preceded by the French word ''comme''. The much quoted version ''tanquam ex ungue Leonem'' is due to David Brewster's 1855 book on the life and works of Newton. Bernoulli's intention was, Whiteside argues, simply to indicate he could tell the anonymous solution was Newton's, just as it was possible to tell that an animal was a lion given its claw; it was not meant to suggest that Bernoulli considered Newton to be the lion among mathematicians, as it has since come to be interpreted.<ref>{{cite book |last1=Whiteside |first1=Derek Thomas |title=The Mathematical Papers of Isaac Newton Vol. 8 |date=2008 |publisher=Cambridge University Press |isbn=978-0-521-20103-2 |pages=9–10, notes (21) and (22) |edition=Paperback}}</ref>

[[John Wallis]], who was 80 years old at the time, had learned of the problem in September 1696 from Johann Bernoulli's youngest brother Hieronymus, and had spent three months attempting a solution before passing it in December to [[David Gregory (mathematician)|David Gregory]], 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 <math>78^1</math>, 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.<ref name="Dubois">{{cite journal |last=Dubois |first=Jacques |date=1991 |title=Chute d'une bille le long d'une gouttière cycloïdale; Tautochrone et brachistochrone; Propriétés et historique |url=http://sciences-physiques-moodle.ac-orleans-tours.fr/moodle/pluginfile.php/2162/mod_resource/content/0/gouin/cab_gouin12novembre2006/cycloid_fichiers/cyclo_rectif.pdf |journal=Bulletin de l'Union des Physiciens |volume=85 |issue=737 |pages=1251–1289 }}</ref>  In addition to the minimum time curve problem, there was a second problem that Newton also solved at the same time. Both solutions appeared anonymously in Philosophical Transactions of the Royal Society, for January 1697.

=== The Brachistochrone problem ===
[[File:Bernoulli Challenge to Newton 1.png|Bernoulli Challenge to Newton 1]]

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 upward 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 is 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, <math>eL - eE = nL= \frac{o.CH}{CV}</math>, ignoring terms in <math> o^2 </math> 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 <math>\sqrt{CB}</math>, which is as CH, since <math>CH=\sqrt{CB.CV}</math>

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

Let t be the additional time to reach L,

<math>t \propto \frac{nL}{\sqrt{CB}} =\frac{o.CH}{CV.\sqrt{CB}} = \frac{o}{\sqrt{CV}}</math>

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 <math>\sqrt{CB}</math>), ignoring terms in <math> o^2 </math> and higher:

<math>t \propto \frac{o.DE}{eE.\sqrt{CB}}</math>,

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

<math>T \propto \frac{o.FG}{Ff.\sqrt{CI}}</math>

The difference (t – T) is the extra time it takes along the path {{not a typo|eLMf}} compared to the original {{not a typo|eEFf}} :

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

Because {{not a typo|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:

<math> \frac{DE}{eE\sqrt{CB}} = \frac{FG}{Ff\sqrt{CI}} </math>     (2) in the limit as eE and fF approach zero. Note since {{not a typo|eEFf}} is the minimum curve it has to be assumed  that the coefficient of <math> o^2 </math> 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, <math> \frac{FG}{Ff\sqrt{CI}} </math> is constant (equal to <math> \frac{DE}{eE\sqrt{CB}} </math>). By keeping f fixed and making e variable it is clear that <math> \frac{DE}{eE\sqrt{CB}} </math> is also constant.

But, since points, e and f are arbitrary, equation (2) can be true only if <math> \frac{DE}{eE\sqrt{CB}} = \text{constant} </math>,   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, <math> \frac{DE}{eE} = \frac{BH}{VH} = \frac{CH}{CV} </math>  , so that <math>  \frac{DE}{eE\sqrt{CB}} = \frac{CH}{CV.\sqrt{CB}} </math>, 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.