A Complete Detailed Solution to the

Brachistochrone Problem

N.H NguyenEastern Oregon University

June 3, 2014

Abstract

This paper consists of some detailed analysis of the classic mathematicalproblem known as the brachistochrone, which gave rise to the field calcu-lus of variations. A brachistochrone curve is also known as the curve ofquickest descent. Despite the fact that this problem has been around formore than 300 years, there are still ongoing studies focusing on this prob-lem. The aim for this paper is to give a thorough and detailed approachto the problem, which involves presenting a new proof showing why thecycloid is the fastest curve in the family of solutions arrived from solvingthe Euler-Lagrange equation. The proof of this fact has been done beforeby showing that extremals must belong in the class of C2 functions, but itis often left out of many papers. This proof is very involved and requiresa good background in functional analysis. However, the proof that I willpresent in this paper is elementary.

1 Introduction

Have you ever tried to find the quickest path to go from one point to another?We may not think of it in deep mathematical concepts but finding the quickestroute to get to your destination is what we do in our daily life. Most of thetime each of us has a device called GPS to help us find the quickest route to ourdestination. Now suppose that we are in a two dimensional space and you wantto travel from point A to point B in the minimal amount of time. The onlytransportation system available is gravitational force, but you do not need toworry because B is at the same level or lower than A. Furthermore, the weathercondition for this trip is perfect since there is no friction or air resistance. Youmay think to yourself that the straight line is the fastest route since it is theshortest, but be aware that you want to get to B quickest in terms of time,not distance. If you think about it, when you search your GPS for the quickestroute, it does not always gives you the shortest route, since the shortest routemay have a slower speed limit. Often the fastest route given by a GPS will takeyou to a highway, since you can drive at a higher speed there. So the trade off

1

in speed will make up for the longer distance that one may have to travel. Soback to the problem of you trying to find the quickest path of descents fromA to B. What path should you take, knowing the only transportation systemavailable is gravitational force?

Well, this problem turned out to be a very famous problem in the late 1600s. Ithad many mathematicians scratching their heads. The problem was publishedin the Acta Eruditorum in 1696 by Johann Bernoulli. He challenged all math-ematicians across Europe, by stated the following[5] :

“I, Johann Bernoulli, address the most brilliant mathematicians in the world.Nothing is more attractive to intelligent people than an honest, challenging prob-lem, whose possible solution will be bestow fame and remain as a lasting mon-ument. Following the example set by Pascal, Fermat,etc., I hope to gain thegratitude of the whole scientific community by placing before the finest mathe-maticians of our time a problem which will test their methods and the strength oftheir intellect. If someone communicate to me the solution of proposed problem,I shall publicly declare him worthy of praise ” [3]

Here is the exact statement of the problem that he published in the Acta Eru-ditorum:

Given two points A and B in the vertical plane, with A not lower than B,what is the curve traced out by a point ac ted on only by gravity, which startsat A and reaches B in the shortest time.[3]

The solution to this problem is called ‘Brachistochrone’, which means Shortestpath in Greek. Even though this is a classic problem in mathematics and hasbeen around for more than 300 years, it is hard to find a paper that gives adetailed step by step solution to the problem. Therefore, the main goal of thispaper is focusing on giving a detail step by step solution, which includes givinga brief history section on the problem itself. It is true that this problem couldbe solved by several different methods as you will find out in the history lessonbut in this paper, the main tool I will use is calculus of variation. As youwill see later in the history section, there were five different solutions with fivedifferent techniques used to find the solution when the brachistochrone problemwas published in 1696. There were many other methods developed afterward,but none is more powerful than the calculus of variation and that is why Iwanted to use it as my main tool. Also later in the paper, I will briefly mentiona modified brachistochrone problem, where we take the force of friction intoconsideration. However,no detailed analysis will be done for this part.

2

2 History

The problem of finding the quickest path has been studied as early as the timeof Galileo Galilei in 1638. He was an Italian physicist, mathematician, engineer,astronomer and philosopher. One of his most famous works was the discovery ofthe four largest moons of Jupiter. Galileo’s version of the problem was findingthe straight line from a point A to a point B in the vertical line in which itwould reach the quickest. He then calculated the time taken for the point tomove from A to B in a straight line. He then proved that the point would reachB quicker if it first travels to C then to B where C is a point on a circle [1].

It is indeed correct that the circular arc generates a faster time than a straightline, but Galileo unfortunately made an untrue claim when he stated that thepath of quickest descent from A to B must be an arc of a circle. This turnedout to be incorrect, but it is not too far off from the actual solution. Given thatGalileo attempted this problem before the invention of calculus and classicalmechanics and arrived at such a solution, is fascinating on its own.

Some 58 years after, Johann Bernoulli posted the brachistochrone problem,which somewhat is the same problem and challenged all mathematicians acrossEurope. Johann claimed that the solution to the problem is a curve that is wellknown to geometers. He published the problem in Acta Eruditorium on June,1696, after he had already solved the problem himself. He originally gave thetime limit of 6 months, but then later extended it to a year at Leibniz’s sugges-tion because they wanted foreign mathematicians to have a chance to work theproblem as well. We must remember that this was in 1696 so there was no elec-tronic delivery. By the time the deadline was reached, there were a total of fivesolutions submitted. The five solutions were from Johann Bernoulli, his com-petitive brother Jacob Bernoulli, his mentor Gottfried Leibniz, his good friendL’Hospital and the powerful Isaac Newton. When the problem was published,Johann and Leibniz tempted Newton with the problem by saying “...Thereare fewer who are likely to solve our excellent problems, aye, fewer even amongthe very mathematicians who boast that [they]... have wonderfully extended itsbounds by means of the golden theorems which (they thought) were known tono one, but which in fact had long previously been published by others.”[3] Ac-cording to Conduitt, who is a biographer for Newton, Newton saw the problem

3

after returning from the Royal Mint at four in the afternoon. He then stayed upall night and solved the problem before four in the morning.[3] There are alsoclaims that Newton could have solved this problem in a few hours in his youngerdays. Newton then sent his solution anonymously to Johann Bernoulli in thenext post. Upon reading Newton’s solution, Johann Bernoulli said ‘ Ahh... Irecognize the lion by his paw’.[4].

It is the nature of this problem that led to the new field of mathematicscalled Calculus of Variations. The name Calculus of Variations was given byits original father Leonhard Euler, who later became a preeminent mathemati-cians in the 18th century and one of the greatest mathematician to ever live.Ironically, Leonhard Euler was a student of Johann Bernoulli.

3 Brainstorming

Prior to diving into all the equations, let us build some intuitions about thisproblem. Quite often people jump to the conclusion that the quickest path fora particle to go from point A to point B is a straight line. It is correct that thestraight line is the shortest path but it’s not the quickest path in time. Let ussee why a straight line can’t be the quickest path. Supposed we have 2 points Aand B, where point A and point B are at the same height (on a level ground).Now draw the straight line from A to B which of course generates a horizontalline.

It is obvious that the ball will take an infinite amount of time to get from A topoint B just under the force of gravity itself. This arrives straight from the lawof conservation of energy. The particle needs to drop below its starting point toget some momentum. Therefore, the straight line cannot be a solution to theproblem.

Then what about a rectangular path or a triangular path one may wonder?

In both of these cases, the particle first travels down then back up. By the lawof conservation of energy the particle should be able to travel from point A topoint B. But which one will be faster? The answer is that both of these pathswill require the same amount of time to get from point A to point B, and thattime is infinite. To follow any of these two paths, the particle must undergoan instantaneous change in direction, which is physically impossible since the

4

particle has mass. This change in direction would take an infinite amount of en-ergy. Therefore, we now arrive at the conclusion that the quickest path cannothave angled points in the interval from A to B. Which implies any paths thathave a potential to be a brachistochrone must be differentiable on the intervalbetween A and B.Remember what we are trying to achieve here: we want to find a path thatgenerates the quickest time from point A to point B with the parameters I men-tioned in the beginning of this paper. It is logical for us to derive the equationto calculate the time it would take a particle to travel from A to B along anypath. In physics we remember that the change in position with respective totime is velocity. Hence:

ds

dta,b= v

dta,b =ds

vwhich implies ta,b =

b∫a

ds

v

Conservation of energy stated:

1

2mv2 = mgy Hence v =

√2gy

Furthermore, let remember what ds is:

Pythagorean theorem implies:

ds =√dx2 + dy2 =

√1 + y′2dx

Putting it all together gives:

ta,b =

b∫a

√1 + y′2

2gydx

5

We now have the equation that enables us to calculate the time traveled by aparticle along any path. Also we must not forget the fundamental assumptionwe are making to get this equation. We assumed that there is no friction or anyother force acting on the particle other than gravity because only with this arewe able to apply conservation of energy. Now that we have this equation, letme show why a circular arc is not the quickest path like Galileo thought it was.Let us consider two points A and B such that A is at the origin (0, 0) and B islocated at (1,−1). Let us calculate the time the particle would take if it travelsalong the circular arc y = −

√1− (x− 1)2 versus if it travels along a parabolic

arc y = −√x

To make the calculations easier let us set g = 12 .

Circular arc:

t0,1 =

1∫0

1

(2x− x2)3/4dx = 2.622058

Parabolic arc:

t0,1 =1

2

1∫0

√1 + 4x(√x3)4 dx = 2.587229

Thus the circular arc is not the fastest curve and the parabolic arc is faster butthis does not answer our question. There are infinitely many paths one coulddraw from point A to point B, so brute force is not a strategy here. We wantto minimize the time, which means we must minimize

b∫a

√1 + y′2

2gydx

Which then implies that our answer varies by√

1+y′2

2gy . Observe that it is not

just a variable that varies but the function itself varies. So how can we dealwith something like this? It turns out that this is called a “functional” and itbelongs in the field of a mathematics called “The Calculus of Variation.” In thenext section, I will briefly introduce the calculus of variation and how can weuse it to solve the brachistochrone problem.

4 Calculus of Variation

The field of calculus was developed mainly by Leonhard Euler and Joseph La-grange in the motivation of solving problems such as the brachistochrone prob-lem. It is mainly about finding paths such that some integral along the pathis extremized. Calculus of variation deals with finding extrema (stationary) for

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functionals instead of functions, which make the problem non-treatable by el-ementary calculus. A functional is a mapping from a set of functions to realnumbers. Thus the candidates for an extremum (stationary) here would befunctions as compared to vectors in Rn.[8] This is exactly what the brachis-tochrone came down to once we got the equation to find the time traveled bythe particle. The functional that we are working with in the brachistochrone is

L(x, y, y′) =√

1+y′2

y when we let g = 12 . Here x is the independent variable, y

is the dependent variable, and y′ as usual is the derivative dydx .

The Problem: We want to find the path y(x) connecting two points (x0, y0)and (x1, y1) such that the integral given below is extremized:

x1∫x0

L(x, y, y′)dx

This type of integral condition is known as the variation principle and the func-tion L(x,y,y’) is usually known as the Lagrangian.[7] Fortunately for us, Eu-ler and Lagrange solved this problem and it turned out remarkably that anyextrema (stationary) to a functional must satisfy what is known as the Euler-Lagrange equation.

4.1 Euler-Lagrange equation

Theorem 4.1. Let

S =

{y ∈ C1[a, b] : y(a) = ya and y(b) = yb

},

where ya and yb are given real numbers. Now let J : S → R be a function of theform

J =

b∫a

L(x, y, y′)dx

If y ∈ S is an extremal for J , then

∂L

∂y− d

dx

(∂L

∂y′

)= 0 (1)

for all x ∈ [a, b].

The differential equation (1) is second-order that is mandatory for any extremal(stationary) y must satisfy. This equation (1) is called the Euler-Lagrangeequation. As mentioned in the theorem above, the boundary values associated

7

with this equation for a fixed endpoint problem are y(a) = ya and y(b) = yb.The Euler-Lagrange equation is generally a nonlinear second order differentialequation, but in some special cases, it can be reduced to a first order differentialequation or where its solution can be obtained entirely by evaluating integrals.It is logical for one to be curious about why the Euler-Lagrange equation mustbe true. Therefore, below will be a full derivation of the one dimensional Euler-Lagrange equation.

Derivation for one-dimensional Euler-Lagrange equation: We wish tofind a function y(x) which satisfy the boundary condition y(a) = ya, y(b) = yb,and it optimized the functional

J =

b∫a

L(x, y(x), y′(x))dx

Assume L(x,y(x),y’(x)) has continuous first partial derivatives. (This assump-tion help to make the problem easier, however, a weaker assumption can bemade.)From the definition of extremize, the functional J would either increase or de-crease if there is any slight perturbation that preserves the boundary conditionoccurring in the function y. Now let gε(x) = y(x) + εη(x) be the result of suchperturbation εη(x) of y, where ε is very small and η(x) must be differentiablefunction that also satisfies the boundary condition η(a) = η(b) = 0.

Now let us define

Jε =

b∫a

L(x, gε(x), g′ε(x))dx

8

Let Lε = L(x, gε(x), g′ε(x)), hence Jε =b∫a

Lεdx. Now let us consider the deriva-

tive of Jε with respect to ε:

dJε

dε=

d

dε

b∫a

Lεdx =

b∫a

dLεdε

dx

Let us also keep in mind that dxdε = 0 since x does not depend on ε, dgεdε = η(x),

and dgεdx = f ′(x) + η′(x)ε. Hence,

dLεdε

=dx

dε

∂Lε∂x

+dgεdε

∂Lε∂gε

+dg′εdε

∂Lε∂g′ε

dLεdε

=dgεdε

∂Lε∂gε

+dg′εdε

∂Lε∂g′ε

dLεdε

= η(x)∂Lε∂gε

+ η′(x)∂Lε∂g′ε

Hence,

dJεdε

=

b∫a

[η(x)

∂Lε∂gε

+ η′(x)∂Lε∂g′ε

]dx

Jε has an extremum value when ε = 0, which implies gε(x) = y(x) and Lε =L(x, y(x), y′(x)). Thus, let us then evaluate the change in Jε with respect to εwhen evaluate at ε = 0.

dJεdε

∣∣∣∣ε=0

=

b∫a

[η(x)

∂L

∂y+ η′(x)

∂L

∂y′

]dx = 0

dJεdε

∣∣∣∣ε=0

=

b∫a

η(x)∂L

∂ydx+

b∫a

η′(x)∂L

∂y′dx = 0

For the integral on the right, using integration by part with u = ∂L∂y′ and dv =

η′(x) then du = ddx

∂L∂y′ and v = η(x). The equation then becomes,

b∫a

η(x)∂L

∂ydx+ η(x)

∂L

∂y′

∣∣∣∣ba

−b∫a

η(x)d

dx

∂L

∂y′dx

b∫a

η(x)

[∂L

∂y− d

dx

∂L

∂y′

]dx+ η(x)

∂L

∂y′

∣∣∣∣ba

9

From the boundary conditions η(a) = η(b) = 0 the equation become,

b∫a

η(x)

[∂L

∂y− d

dx

∂L

∂y′

]dx = 0 (2)

We know that η(x) can be any function as long as it is differentiable and satisfiesthe boundary conditions so this clearly implies that

∂L

∂y− d

dx

∂L

∂y′= 0

must be true to satisfy the equation (2). This should complete the derivationfor the one dimensional Euler-Lagrange equation. If you are skeptical aboutthat final step as a reader, then just notice that it is coming straight from thefundamental lemma of calculus of variations. Thus, functions that satisfy theEuler-Lagrange equation on some interval are called extremals or stationaryfunctions or critical curves. However, we must remember that such functionsmay or may not be extrema, or even local extrema. This is the same conceptthat not all critical points are maxima or minima.

4.2 Approaching the brachistochrone problem

After all this, we finally have the tools to find the brachistochrone. Let us remindourselves the functional we arrived at for our problem in the earlier section. The

functional that we need to vary is L(x, y, y′) =√

1+y′2

y where as before y′ = dydx .

We can go ahead and apply the complete Euler-Lagrange equation but a sharp

reader would notice that√

1+y′2

y does not explicitly depends on x. In another

word, ∂L∂x = 0. This is nice because as you notice, the Euler-Lagrange equation

is a second order (generally nonlinear) differential equation that is often toodifficult to solve. However, not depends on x reduces the equation to

L(y, y′)− y′ ∂L∂y′

= c where c = constant.

This is also known as the Beltrani identity.[8] This is much simpler than thecomplete Euler-Lagrange equation. I want to stay focused on our task of findingthe brachistochrone curve here so I will not go into the derivation of this prop-erty, but for any curious reader, you can find a full derivation of this propertyin [16] and [10] . Applying the Beltrani identity to our functional gives:

L(y, y′)− y′ ∂L∂y′

= c

√1 + y′2

y− y′ y′√

y(1 + y′2)= c

10

1 + y′2√y(1 + y′2)

− y′2√y(1 + y′2)

=1√

y(1 + y′2)= c

y(1 + y′2) = k for k =1

c2

Hence, this yields the differential equation:(dy

dx

)2

=k − yy

(3)

This is a fairly complicated differential equation. However at this point, withoutdoing any work we can already see one solution that satisfies this differentialequation. This obvious solution is a constant function y(x) = k. The othersolutions, can be simply obtained by the separation of variables method. Doingthis gives: ∫

dx =

∫dy√k−yy

(4)

Or equivalently: ∫dx =

∫ √y

k − ydy (5)

There are several ways of approaching the integral (4) or (5). The most obviousway is just to start doing regular substitutions but there is a more elegant wayof doing it by using trigonometric substitution. For the purpose of this paper, Iwill only show the trigonometric substitution method explicitly; however, I willinclude the result I arrived using regular substitution as well. The trigonomet-ric substitution method I will show could be found in [9], [10] and many othercalculus texts

Trigonometric substitution:

y

k − y= tan(φ) =

sin2(φ)

cos2(φ)

The function φ is a new function of x. Now isolating y gives

y = ksin2(φ)

Differentiating in φ, we get:

dy

dφ= ksin(2φ) = 2ksin(φ)cos(φ)

Hence,dx

dφ=

√y

k − ydy

dφ= 2ksin2(φ) = k(1− cos(2φ))

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This indicates the relationship between the two infinitesimal values dx and dφ.Integrating gives:

x = k

∫1− cosφ

dφ =

k

2

(2φ− sin(2φ)

)+ C

The boundaries conditions that we have on the problem make it quite clearthat the integration constant C must be 0. Now let 2φ = θ then we have theparametric equation below: {

x = r(θ − sin(θ)),

y = r(1− cos(θ))

where r = k2 . One may immediately realize that this is the parametric equation

for the cycloid with radius r. So the cycloid curve is also the solution to equation(3) which we arrived at using the Euler-Lagrange equation. This implies thatcycloid is a stationary function as well as the constant function since bothsatisfies the differential equation. As promised, below is the solution I obtainedwhen I solved the integral (4) using the regular substitution method:

x(y) = −y

√k − yy− k · arctan

(√k − yy

)This is also the equation for a cycloid. Which of course agrees with the para-metric equation. Below is a picture demonstrating how a cycloid can be made:

One of the properties of the cycloid is it cannot express it as a function y(x).Later we will see how this can create some complications when we want to de-termine the time the particle travels along the cycloid. Another property is thatgiven two distinct points A and B, there exist only one unique cycloid throughA and B.[10] A suitable radius will give this cycloid. Notice, we have not yetarrived at the optimal curve (path) yet. More work needs to be done to unveilthe brachistochrone curve.

5 The Brachistochrone Unveiled

Congratulations! You have reached the most exciting part of the paper. Herethe brachistochrone will be unveiled. Let me remind you that the cycloid and

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the constant function both satisfy the brachistochrone but I have shown earlierthat the constant by itself cannot be a brachistochrone. This leads us to a ques-tion:Since the constant function by itself cannot be a brachistochrone, does thismean the cycloid is the only curve left which implies that it must be the brachis-tochrone?I have read through many papers and books about the brachistochrone and oneway or another, they all claim that the cycloid must be the brachistochrone,since it satisfies the differential equation (3). R.Coleman addressed this sameissue in [6]. It is true that the constant function can be eliminated, but let meremind you that the problem does not prevent piecewise continuous functions.As long as my piecewise function is differentiable everywhere then it is fine. Letme give you an idea what this family of solutions looks like by the picture below:

In the picture above, I used the cycloid of radius 1 and since I start at the origin,the cycloid must finish at the point (2π, 0). For the purpose of this problem, wealways think about the point we initially released the particle at is the origin.I claim that if you pick an end point A such that A is below the line y = − 2

πxthen the cycloid is the only solution. Otherwise there will be infinitely manysolutions, hence infinitely many critical curves. Hence, the question arrives,What is so special about this line y = − 2

πx? notice the negative is there onlybecause of the way I look at the problem. This sign would be positive if I justimagine rolling the ball from right to left, so do not get confused by the signhere but rather focus more about what this line represents.Let us take a closer look at the cycloid equation:

x = r(θ − sin(θ)) y = r(1− cos(θ))

As you can see, the minimum point if you look at an inverted cycloid or maxi-mum point if you look at a regular cycloid, is evaluated at θ = π. This is obviousfrom the fact that the largest value y can take on is 2r, and y = r(1− cos(π)) =2r. If you are a calculus person then this can be arrived at as follow:

dy

dx=dy/dt

dx/dt=

rsin(θ)

r(1− cos(θ))= 0

13

Thus, the cycloid is at its maximum point when θ = π. This is a fascinatingresult because what this means is if your starting point is the same, no mat-ter what cycloid you choose, the optimal point of your cycloid will always betouched by the line y = − 2

πx. The picture below expresses that fact.

This is why when your destination point is below the line y = − 2πx you will get

only one solution, which is a cycloid. Otherwise, you can have one solution be acycloid, a constant function and infinitely many piecewise definition functions.Notice that these piecewise functions do not violate the problem at all. Weknow that the brachistochrone do not have a discontinuous derivative, whichmeans it cannot slope upward until after the first derivative first reached zero.But that does not mean that it can continue horizontally for a while before itslopes upward. That path would not conflict with the differential equation (3)we arrived at using the Euler-Lagrange equation.

As mentioned before, many published papers, somehow or another always provethat the cycloid satisfies the differential equation, and therefore it must be thebrachistochrone. Other than [6], which also showed why the cycloid must bethe fastest in the family and discussed about the facts that many papers andarticles out there never address this same issue. The proof [6] presented is quitecomplicated and required a good amount of preliminary knowledge. It showedthat the extremals (Critical curves) must be C2 class function. All the papersI came across other than [6] seem to assume that the extremals belong in theclass of C2 but that’s not necessary and can be proved. It may not be obvious,but one can see that if C2 was the requirement for extremals, then there is onlyone extremal function, no matter where you chose your end point, which is acycloid. Let us take a look and see why this is true. Recall that for the cycloid,dydx = sin(θ)

(1−cos(θ)) . But dydx is a function with parameter θ, hence let dy

dx = g(θ).

Which impliesd2y

dx2=

d

dx

(dy

dx

)=

d

dxg(θ) (1)

14

But dgdθ = dg

dx ·dxdθ which shows dg

dx = dg/dθdx/dθ . But dg

dθ = ddθ

(dydx

)Hence,

d2y

dx2=

ddθ

(dydx

)dxdθ

=

ddθ

(sin(θ)

(1−cosθ)

)r(1− cos(θ))

=

cos(θ)−1(1−cos(θ))2

r(1− cos(θ))

Therefore,d2y

dx2= − 1

r(1− cos(θ))2

Thus the second derivative of a cycloid is never zero. However, for the constantfunction, its second derivative is zero, which shows that the piecewise solutionsare in class C1 functions. This implies that if the requirement of class C2 isenforced on the extremals then the solution to the brachistochrone is indeedthe cyloid. However, what I am going to do next is prove this same fact byshowing the cycloid is in fact the fastest curve in the family using some basicmathematics.

Theorem 5.1. Given two distinct points A and B, where point B is somedistance d away from A and at the same level as A. The cycloid is the fastestcurve in term of time in the family of solutions, hence a brachistochrone.

Notice that a cycloid can be made from an involute cycloid, which simply isa shifted copy of the original cycloid. Thus the cycloid is its own involute.[14]Inverted involuted cycloids are known as the Christian Huygens pendulums [10].

Christian Huygens Pendulum

The length of this pendulum is half the arc length (l) of a cycloid where the fullarc length (L) of a cycloid is 8r since

L =

2π∫0

√(dx

dθ

)2

+

(dy

dθ

)2

dθ =

2π∫0

√r2(1− cos(θ))2 + r2sin2(θ)dθ = 8r

The last step may not be obvious but I encourage you as reader to clarify this onyour own. Thus, the length of this pendulum is 4r. The interesting fact here is

15

the time period for this pendulum can be calculated exactly. Specifically, givena cycloidal pendulum with radius r then the period of that pendulum is:

T = 2π

√L

g= 2π

√4r

gg = gravitational constant

This indicates that the period for any cycloid must be T = 2π√

4rg . However,

a period is a round trip, in other words, the time the pendulum would take toswing from one end to another and back. For the purpose of the brachistochrone,

we are only interested in half a period or simply T = π√

4rg .

Figure 1

Proof. Consider two points A and B located at (xa, 0), (xb, 0) respectively,where xa 6= xb. By the Euler-Lagrange equations, there exist infinitely manycritical curves connecting point A and B; however there is only one cycloid, andthe others are piecewise functions.Suppose curve 1, denotes C1 is a cycloid with radius r2 that connects A andB. Note the minimum point of this cycloid is (r2π, 2r2) where 2r2 = y2 =− 2πx2.Curve 2, denotes C2 is made up by a piecewise function that consists of a

cycloid with radius r1 from point A to (x1, y1) where x0 < x1 and y1 = − 2πx1,

and a constant function from (x1, y1) to (x3, y3) where x1 < x3 and similarlyy3 = − 2

πx3, then followed by another cycloid from (x3, y3) to point B. Note, x1and x3 are picked arbitrarily and x1+x3

2 = x2.

Now let T1 denoted the time for the particle take on path 1. Similarly, let T2denoted the time the particle would take on path 2.

T1 = π

√4r1g

+x3 − x1π√

2y1

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Let g = π2 gives:

T1 =√

2y1 +x3 − x1π√

2y1

Using the facts x3 − x1 = 2(x2 − x1)

T1 =√

2y1 +

√2(x2 − x1)

π√y1

=

√4

πx1 +

√2(x2 − x1)

π√

2πx

Notice, the negative sign of the equation y = − 2πx was left out and the reason

for that is the negative sign was arbitrary. It was there because the way wepicked our starting point and thus if we rolled the ball from left to right then itwould be positive. Furthermore, this will apply consistently through the proofand so the negative sign can be ignored.

T1 = 2

√x1π

+x2√πx1−√x1π

=

√x1π

+x2√πx1

T 21 =

x1π

+2

πx2 +

(x2x1

)(x2π

)We know x1 < x2, hence x2

x1= (1 + k) for k > 0. Therefore,

T 21 =

3

πx2 +

x1π

+k

πx2

Now let consider T2. Similarly to before we have

T2 =√

2y2 =

√4

πx2

Hence

T 22 =

4

πx2 =

3

πx2 +

x2π

Notice the difference between T1 and T2 are just those few terms. However,observe

x1π

+k

πx2 =

x1π

+k

π· (1 + k)x1 =

x1π

+k

πx1 +

k2

πx1

Andx2π

=(1 + k)x1

π=x1π

+k

πx1

Thus, this shows that with the given conditions T1 > T2 by a factor of k2

π x1 .Therefore, the cycloid is the fastest path in the family and hence the brachis-tochrone curve.

A sharp reader, like my advisor Dr. Steve Tanner, would see that there is analternative once we arrived at

T1 =

√x1π

+x2√πx1

T2 =

√4

πx2

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We could take the ratio between T1 and T2, so

T1T2

=

√x1

π + x2√πx1√

4πx2

=x1 + x22√x1x2

But notice, (√x2 −

√x1

)2

> 0 ⇒ x2 − 2x1x2 + x1 > 0

Hence,

x2 + x1 > 2x1x2 ⇒ x2 + x12x1x2

> 1

Thus T1 > T2, which also implies that the cycloid is in fact the brachistochronecurve.

One may question the legitimacy of using this time as the time for the brachis-tochrone problem since we are not dealing with the case where the particle ishanging on a string. However, notice that if we solve for the time the particlewould take to travel along the cycloid path, we would get

T =

2π∫0

√1 + y′2

2gydx =

1

2g

2π∫0

√√√√√1 +

(sinθ

1−cosθ

)2

r(1− cosθ)r(1− cosθ)dθ = π

√4r

g

Again, the last step is a very simple integral after some algebra manipulationsso I encourage you as the reader to confirm the result for yourself. So it iscompletely legitimate to use the Christian Huygens pendulum clock to describethe movement of the particle along the cycloid curve.Therefore, the cycloid is in fact the fastest curve for a particle to take to gofrom point A to point B where B is not higher than A under gravitational forcealone. One thing I want to point out is that the proof I gave required you topick your end point (B) to be on the same level as A. Although, I did notexplicitly show the case where we picked point B such that it is above the liney = − 2

πx but lower than A, it is easy to see that the cycloid still must be thefastest path physically if it is the fastest when B is on the same level as A.

6 Brachistochrone in an ideal world

In an ideal world, you will never find an environment where the only force isgravity. So given the same problem, what is the curve that would satisfy theproperty of being a brachistochrone? Well this is a hard question and often leftout of the discussion by many since it is fairly involved and the result is notthat much more astonishing than the original brachistochrone problem. I willhowever briefly present the results to the problem when we take friction into

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consideration. A more detailed and rigorous analysis of this problem can befound at [13].

Now that we applied friction into the problem, what will change is the velocity,which of course makes sense. If you remember from physics, when friction forceis applied then v =

√2g(y − µx) where µ is the friction coefficient. Hence, our

new equation for the time period is:

t =

∫ √1 + y′2

2g(y − µx)dx

We can now apply the Euler Lagrange equation which gives

[1 + y′2](1 + µy′) + 2(y − µx)y” = 0

This is a highly nonlinear differential equation and note we cannot use theBeltrami identity. This process points us to the fact that, although, the Euler-Lagrange equation is great, often times, it can lead us to a differential equationthat is very complicated to solve. I’ll not go into detail of how to solve thisequation, but the solution to this turned out to be:

x = r[(θ − sinθ) + µ(1− cosθ)]

y = r[(1− cosθ) + µ(θ + sinθ)]

Note that the brachistochrone here is dependent on the coefficient of friction(µ) and therefore it is not the same curve every time. It is obvious that whenµ is zero then we have a cycloid again. There is no unique solution in this casebecause each solution depends on µ. Since no solution exist such that it wouldnot depends on µ make it less interesting to mathematicians and often left outof discussions. However, if you are trying to build the world fastest slide thenthe solution with friction involved would be more appropriate.

7 Conclusion

Throughout this paper I have established and shown that the cycloid is in factthe brachistochrone in an ideal environment. I have also shown why the cycloidis the optimal extremal curve out of the family of extremal solutions using afairly elementary approach. Although the brachistochrone problem has beenknown for more than 300 years, it continues to be the problem to introducethe field of calculus of variation and many are still trying to find new methodsto solve it. Before I end this paper, I also want to mention the fact that thebrachistochrone in an ideal environment is also a tautochrome or isochrone. Iwill leave it to the readers to prove this fact.

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8 Bibliography

References

[1] Douglas S. Shafer, The Brachistochrone: historical gateway to the calcu-lus of variations. MATerials MATematics, Volum 2007, Treball no.5,14 pp.ISSN:1887-1997

[2] Jeff Babb, James Currie, The brachistochrone broblem: mathematics for abroad audience via a large context problem. TMME, vol5, nos.23,p.169

[3] J.J O’Connor, E.F Robertsn, History Topic: The Brachistochrone Problem,MacTutor History of Mathematics

[4] John Martin, The Helen of Geometry, the college mathematics journalVol.41, No.1, January 2010.

[5] William Dunham, Journey through genius: The great theorems of mathe-matics. WIley 1990, page 199-202

[6] Coleman, Rodney. A detailed analysis of the brachistocrhone problem. arXivpreprint arXiv: 1001.2181(2010).

[7] Bruce van Brunt, The caclculus of variations, Springer, 2004.

[8] Tai L. Chow, Mathematical methods for physicist: A concise introduction,New York, 2000.

[9] V. Arnold, Mathematical Methods of Classical Mechanics, Sringer-Verlag,1978.

[10] Byron Jr, Fredick W., and Robert W.Fuller. Mathematics of classical andquantum physics. Courier Dover Publication, 2012.

[11] Weisstein, Eric W. Cycloid involute. Mathworld-A Wolfram web resource.http://mathworld.wolfram.com/Cycloidinvolute.html

[12] Gemmer, John, R. Umble, and M. Nolan. Generalization of the brachis-tochrone problem. arXiv preprint math-ph/061052 (2006).

[13] Lipp, Stephen C. Brachistochrone with Coulomb Friction. SIAM JournalOn Control and Optimization 35.2(1997): 562. Academic Search Preimier.Web 13 May 2014.

[14] Weisstein, Eric W. Brachistochrone Problem. Mathworld-A Wolfram webresource. http://mathworld.wolfram.com/BrachistochroneProblem.html

[15] Nishiyama, Yutaka. The Brachistochrone Curve: The Problem of QuickestDescent. Osaka Keidai Ronshu, Vol.61. No.6 March 2011

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[16] Wisstein, Eric W. Beltrami identity. Mathworld- A wolfram web resource(2009).

[17] Bennet Matthew, Huygen’s clocks. Proceeding:Mathematics, Physicial andEngineering Science (2002): 553-579.

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