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Source: Brachistochrone curve

In physics and mathematics, a brachistochrone curve (from Ancient Greek βράχιστος χρόνος (brákhistos khrónos) 'shortest time'), 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

= Galileo's problem

brachistochrone

= Introduction of the problem

brachistochrone

curve

brachistochrone

Johann Bernoulli's solution

= Introduction

curve

= Direct method

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 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).
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:

v
=

2
g
y

{\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 Snell's law of refraction gives a constant of the motion for a beam of light in a medium of variable density:

sin
⁡

θ

v

=

1
v

d
x

d
s

=

1

v

m

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

,
where vm is the constant and

θ

{\displaystyle \theta }

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:

v

m

=

2
g
D

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

.
Rearranging terms in the law of refraction and squaring gives:

v

m

2

d

x

2

=

v

2

d

s

2

=

v

2

(
d

x

2

+
d

y

2

)

{\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:

d
x
=

v

d
y

v

m

2

−

v

2

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

.
Substituting from the expressions for v and vm above gives:

d
x
=

y

D
−
y

d
y

,

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

which is the differential equation of an inverted cycloid generated by a circle of diameter D=2r, whose parametric equation is:

x

=
r
(
φ
−
sin
⁡
φ
)

y

=
r
(
1
−
cos
⁡
φ
)
.

{\displaystyle {\begin{aligned}x&=r(\varphi -\sin \varphi )\\y&=r(1-\cos \varphi ).\end{aligned}}}

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 (x, y) = (rφ, r).
In the brachistochrone problem, the motion of the body is given by the time evolution of the parameter:

φ
(
t
)
=
ω
t

,
ω
=

g
r

{\displaystyle \varphi (t)=\omega t\,,\omega ={\sqrt {\frac {g}{r}}}}

where t is the time since the release of the body from the point (0,0).

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,

d

s

2

=
d

x

2

+
d

y

2

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

.
On differentiation with dy fixed we get,

2
d
s

d

2

s
=
2
d
x

d

2

x

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

.
And finally rearranging terms gives,

d
x

d
s

d

2

x
=

d

2

s
=
v

d

2

t

{\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,

d

2

t

1

=

1

v

1

d

x

1

d

s

1

d

2

x

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

d

2

t

2

=

1

v

2

d

x

2

d

s

2

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,

d

2

t

2

−

d

2

t

1

=
0
=

(

1

v

2

d

x

2

d

s

2

−

1

v

1

d

x

1

d

s

1

)

d

2

x

{\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,

1

v

2

d

x

2

d

s

2

=

1

v

1

d

x

1

d

s

1

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

which agrees with Johann's assumption based on the law of refraction.

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.
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, 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

78

1

{\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. 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

=

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,

e
L
−
e
E
=
n
L
=

o
.
C
H

C
V

{\displaystyle eL-eE=nL={\frac {o.CH}{CV}}}

, ignoring terms in

o

2

{\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

C
B

{\displaystyle {\sqrt {CB}}}

, which is as CH, since

C
H
=

C
B
.
C
V

{\displaystyle CH={\sqrt {CB.CV}}}

This appears to be all that Gregory’s note contains.
Let t be the additional time to reach L,

t
∝

n
L

C
B

=

o
.
C
H

C
V
.

C
B

=

o

C
V

{\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

C
B

{\displaystyle {\sqrt {CB}}}

), ignoring terms in

o

2

{\displaystyle o^{2}}

and higher:

t
∝

o
.
D
E

e
E
.

C
B

{\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

T
∝

o
.
F
G

F
f
.

C
I

{\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 :

(
t
−
T
)
∝

(

D
E

e
E

C
B

−

F
G

F
f

C
I

)

.
o

{\displaystyle (t-T)\propto \left({\frac {DE}{eE{\sqrt {CB}}}}-{\frac {FG}{Ff{\sqrt {CI}}}}\right).o}

plus terms in

o

2

{\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:

D
E

e
E

C
B

=

F
G

F
f

C
I

{\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

o

2

{\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,

F
G

F
f

C
I

{\displaystyle {\frac {FG}{Ff{\sqrt {CI}}}}}

is constant (equal to

D
E

e
E

C
B

{\displaystyle {\frac {DE}{eE{\sqrt {CB}}}}}

). By keeping f fixed and making e variable it is clear that

D
E

e
E

C
B

{\displaystyle {\frac {DE}{eE{\sqrt {CB}}}}}

is also constant.
But, since points, e and f are arbitrary, equation (2) can be true only if

D
E

e
E

C
B

=

constant

{\displaystyle {\frac {DE}{eE{\sqrt {CB}}}}={\text{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,

D
E

e
E

=

B
H

V
H

=

C
H

C
V

{\displaystyle {\frac {DE}{eE}}={\frac {BH}{VH}}={\frac {CH}{CV}}}

, so that

D
E

e
E

C
B

=

C
H

C
V
.

C
B

{\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.

References

External links

"Brachistochrone", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
Weisstein, Eric W. "Brachistochrone Problem". MathWorld.
Brachistochrone ( 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.