A Complete Detailed Solution to the Brachistochrone Problem

A Complete Detailed Solution to the Brachistochrone Problem N.H Nguyen Eastern Oregon University June 3, 2014 Abstract This paper consists of some detailed analysis of the classic mathematical problem known as the brachistochrone, which gave rise to the field calculus of variations. A brachistochrone curve is also known as the curve of quickest descent. Despite the fact that this problem has been around for more than 300 years, there are still ongoing studies focusing on this problem. The aim for this paper is to give a thorough and detailed approach to the problem, which involves presenting a new proof showing why the cycloid is the fastest curve in the family of solutions arrived from solving the Euler-Lagrange equation. The proof of this fact has been done before by showing that extremals must belong in the class of C 2 functions, but it is often left out of many papers. This proof is very involved and requires a good background in functional analysis. However, the proof that I will present 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 quickest route to get to your destination is what we do in our daily life. Most of the time each of us has a device called GPS to help us find the quickest route to our destination. Now suppose that we are in a two dimensional space and you want to travel from point A to point B in the minimal amount of time. The only transportation system available is gravitational force, but you do not need to worry because B is at the same level or lower than A. Furthermore, the weather condition for this trip is perfect since there is no friction or air resistance. You may think to yourself that the straight line is the fastest route since it is the shortest, 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 quickest route, it does not always gives you the shortest route, since the shortest route may have a slower speed limit. Often the fastest route given by a GPS will take you 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. So back to the problem of you trying to find the quickest path of descents from A to B. What path should you take, knowing the only transportation system available is gravitational force?

Well, this problem turned out to be a very famous problem in the late 1600s. It had many mathematicians scratching their heads. The problem was published in the Acta Eruditorum in 1696 by Johann Bernoulli. He challenged all mathematicians 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 problem, whose possible solution will be 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 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 Eruditorum: 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 starts at A and reaches B in the shortest time.[3] The solution to this problem is called ‘Brachistochrone’, which means Shortest path in Greek. Even though this is a classic problem in mathematics and has been around for more than 300 years, it is hard to find a paper that gives a detailed step by step solution to the problem. Therefore, the main goal of this paper is focusing on giving a detail step by step solution, which includes giving a brief history section on the problem itself. It is true that this problem could be solved by several different methods as you will find out in the history lesson but in this paper, the main tool I will use is calculus of variation. As you will see later in the history section, there were five different solutions with five different techniques used to find the solution when the brachistochrone problem was published in 1696. There were many other methods developed afterward, but none is more powerful than the calculus of variation and that is why I wanted to use it as my main tool. Also later in the paper, I will briefly mention a modified brachistochrone problem, where we take the force of friction into consideration. 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 time of Galileo Galilei in 1638. He was an Italian physicist, mathematician, engineer, astronomer and philosopher. One of his most famous works was the discovery of the four largest moons of Jupiter. Galileo’s version of the problem was finding the straight line from a point A to a point B in the vertical line in which it would reach the quickest. He then calculated the time taken for the point to move from A to B in a straight line. He then proved that the point would reach B 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 straight line, but Galileo unfortunately made an untrue claim when he stated that the path of quickest descent from A to B must be an arc of a circle. This turned out to be incorrect, but it is not too far off from the actual solution. Given that Galileo attempted this problem before the invention of calculus and classical mechanics 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 across Europe. Johann claimed that the solution to the problem is a curve that is well known to geometers. He published the problem in Acta Eruditorium on June, 1696, after he had already solved the problem himself. He originally gave the time limit of 6 months, but then later extended it to a year at Leibniz’s suggestion because they wanted foreign mathematicians to have a chance to work the problem as well. We must remember that this was in 1696 so there was no electronic delivery. By the time the deadline was reached, there were a total of five solutions submitted. The five solutions were from Johann Bernoulli, his competitive brother Jacob Bernoulli, his mentor Gottfried Leibniz, his good friend L’Hospital and the powerful Isaac Newton. When the problem was published, Johann and Leibniz tempted Newton with the problem by saying “...There are fewer who are likely to solve our excellent problems, aye, fewer even among the very mathematicians who boast that [they]... have wonderfully extended its bounds by means of the golden theorems which (they thought) were known to no one, but which in fact had long previously been published by others.”[3] According 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 up all night and solved the problem before four in the morning.[3] There are also claims that Newton could have solved this problem in a few hours in his younger days. Newton then sent his solution anonymously to Johann Bernoulli in the next post. Upon reading Newton’s solution, Johann Bernoulli said ‘ Ahh... I recognize the lion by his paw’.[4].

It is the nature of this problem that led to the new field of mathematics called Calculus of Variations. The name Calculus of Variations was given by its original father Leonhard Euler, who later became a preeminent mathematicians 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 this problem. Quite often people jump to the conclusion that the quickest path for a particle to go from point A to point B is a straight line. It is correct that the straight line is the shortest path but it’s not the quickest path in time. Let us see why a straight line can’t be the quickest path. Supposed we have 2 points A and 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 horizontal line.

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

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 law of conservation of energy the particle should be able to travel from point A to point B. But which one will be faster? The answer is that both of these paths will require the same amount of time to get from point A to point B, and that time is infinite. To follow any of these two paths, the particle must undergo an instantaneous change in direction, which is physically impossible since the

4

particle has mass. This change in direction would take an infinite amount of energy. Therefore, we now arrive at the conclusion that the quickest path cannot have angled points in the interval from A to B. Which implies any paths that have a potential to be a brachistochrone must be differentiable on the interval between A and B. Remember what we are trying to achieve here: we want to find a path that generates the quickest time from point A to point B with the parameters I mentioned in the beginning of this paper. It is logical for us to derive the equation to calculate the time it would take a particle to travel from A to B along any path. In physics we remember that the change in position with respective to time is velocity. Hence:

ds =v dta,b

dta,b =

Zb

ds v

which

implies

ta,b = a

Conservation of energy stated: 1 mv 2 = mgy 2

Hence

v=

p 2gy

Furthermore, let remember what ds is:

Pythagorean theorem implies:

ds =

p

dx2 + dy 2 =

p

1 + y 02 dx

Putting it all together gives:

Zb s ta,b = a

5

1 + y 02 dx 2gy

ds v

We now have the equation that enables us to calculate the time traveled by a particle along any path. Also we must not forget the fundamental assumption we are making to get this equation. We assumed that there is no friction or any other force acting on the particle other than gravity because only with this are we able to apply conservation of energy. Now that we have this equation, let me 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 is located at (1, −1). Let us calculate the time the particle would take if it travels p along 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 = 21 .

Circular arc:

Z1 t0,1 =

1 dx = 2.622058 (2x − x2 )3/4

0

Parabolic arc: t0,1 =

1 2

Z1 √ 1 + 4x √ 4 dx = 2.587229 x3 0

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

1 + y 02 dx 2gy

a

q 1+y 02 Which then implies that our answer varies by 2gy . Observe that it is not just a variable that varies but the function itself varies. So how can we deal with something like this? It turns out that this is called a “functional” and it belongs in the field of a mathematics called “The Calculus of Variation.” In the next section, I will briefly introduce the calculus of variation and how can we use it to solve the brachistochrone problem.

4

Calculus of Variation

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

6

functionals instead of functions, which make the problem non-treatable by elementary calculus. A functional is a mapping from a set of functions to real numbers. Thus the candidates for an extremum (stationary) here would be functions as compared to vectors in Rn .[8] This is exactly what the brachistochrone came down to once we got the equation to find the time traveled by the particle. q The functional that we are working with in the brachistochrone is L(x, y, y 0 ) =

1+y 02 y

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

is the dependent variable, and y 0 as usual is the derivative

dy dx .

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:

Zx1

L(x, y, y 0 )dx

x0

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

4.1

Euler-Lagrange equation

Theorem 4.1. Let

S=

1

y ∈ C [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 the form Zb J = L(x, y, y 0 )dx a

If y ∈ S is an extremal for J, then

d ∂L ∂L − =0 ∂y dx ∂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-Lagrange equation. 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 differential equation, but in some special cases, it can be reduced to a first order differential equation 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 must be true. Therefore, below will be a full derivation of the one dimensional EulerLagrange equation.

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

L(x, y(x), y 0 (x))dx

a

Assume L(x,y(x),y’(x)) has continuous first partial derivatives. (This assumption help to make the problem easier, however, a weaker assumption can be made.)

From the definition of extremize, the functional J would either increase or decrease if there is any slight perturbation that preserves the boundary condition occurring in the function y. Now let g (x) = y(x) + Ρ(x) be the result of such perturbation Ρ(x) of y, where is very small and Ρ(x) must be differentiable function that also satisfies the boundary condition Ρ(a) = Ρ(b) = 0.

Now let us define Zb J =

L(x, g (x), g 0 (x))dx

a

8

Let L = L(x, g (x), g 0 (x)), hence J =

Rb

L dx. Now let us consider the deriva-

a

tive of J with respect to : d dJ = d d

Zb

Zb L dx =

dL dx d

a

a

Let us also keep in mind that dx d = 0 since x does not depend on , 0 0

= f (x) + Ρ (x) .

Hence, and dg dx

dg d

= Ρ(x),

dL dx ∂L dg ∂L dg 0 ∂L = + + 0 d d ∂x d ∂g d ∂g dL dg ∂L dg 0 ∂L = + 0 d d ∂g d ∂g ∂L dL ∂L = Ρ(x) + Ρ 0 (x) 0 d ∂g ∂g Hence, dJ = d

Zb h ∂L ∂L i Ρ(x) + Ρ 0 (x) 0 dx ∂g ∂g a

J has an extremum value when = 0, which implies g (x) = y(x) and L = L(x, y(x), y 0 (x)). Thus, let us then evaluate the change in J with respect to when evaluate at = 0.

Zb h ∂L ∂L i dJ

0 = Ρ(x) + Ρ (x) dx = 0 d =0 ∂y ∂y 0 a

Zb Zb dJ

∂L ∂L = Ρ(x) dx + Ρ 0 (x) 0 dx = 0

d =0 ∂y ∂y a

a

∂L For the integral on the right, using integration by part with u = ∂y 0 and dv = d ∂L Ρ 0 (x) then du = dx ∂y0 and v = Ρ(x). The equation then becomes,

Zb a

b Zb ∂L ∂L

d ∂L Ρ(x) dx + Ρ(x) 0 − Ρ(x) dx ∂y ∂y a dx ∂y 0 a

Zb a

b

∂L d ∂L ∂L

Ρ(x) − dx + Ρ(x) ∂y dx ∂y 0 ∂y 0 a

9

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

∂L d ∂L η(x) dx = 0 − ∂y dx ∂y 0

(2)

a

We know that η(x) can be any function as long as it is differentiable and satisfies the boundary conditions so this clearly implies that ∂L d ∂L =0 − ∂y dx ∂y 0 must be true to satisfy the equation (2). This should complete the derivation for the one dimensional Euler-Lagrange equation. If you are skeptical about that final step as a reader, then just notice that it is coming straight from the fundamental lemma of calculus of variations. Thus, functions that satisfy the Euler-Lagrange equation on some interval are called extremals or stationary functions or critical curves. However, we must remember that such functions may or may not be extrema, or even local extrema. This is the same concept that 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 remind ourselves the functional we arrived at for our problem in the earlier section. The q 02

dy functional that we need to vary is L(x, y, y 0 ) = 1+y where as before y 0 = dx . y We can go ahead and apply the complete Euler-Lagrange equation but a sharp q

reader would notice that

1+y 02 y

does not explicitly depends on x. In another

∂L ∂x

word, = 0. This is nice because as you notice, the Euler-Lagrange equation is a second order (generally nonlinear) differential equation that is often too difficult to solve. However, not depends on x reduces the equation to L(y, y 0 ) − y 0

∂L =c ∂y 0

where

c = constant.

This is also known as the Beltrani identity.[8] This is much simpler than the complete Euler-Lagrange equation. I want to stay focused on our task of finding the brachistochrone curve here so I will not go into the derivation of this property, but for any curious reader, you can find a full derivation of this property in [16] and [10] . Applying the Beltrani identity to our functional gives: L(y, y 0 ) − y 0 s

∂L =c ∂y 0

1 + y 02 y0 − y0 p =c y y(1 + y 02 )

10

1 + y 02 p

y(1 +

y 02 )

−p

y 02 y(1 +

y 02 )

y(1 + y 02 ) = k

=p

f or

Hence, this yields the differential equation: 2 dy k−y = dx y

1 y(1 + y 02 )

=c

1 c2

k=

(3)

This is a fairly complicated differential equation. However at this point, without doing any work we can already see one solution that satisfies this differential equation. This obvious solution is a constant function y(x) = k. The other solutions, can be simply obtained by the separation of variables method. Doing this gives: Z Z dy q (4) dx = k−y y

Or equivalently: Z r

Z dx =

y dy k−y

(5)

There are several ways of approaching the integral (4) or (5). The most obvious way is just to start doing regular substitutions but there is a more elegant way of doing it by using trigonometric substitution. For the purpose of this paper, I will only show the trigonometric substitution method explicitly; however, I will include the result I arrived using regular substitution as well. The trigonometric substitution method I will show could be found in [9], [10] and many other calculus texts Trigonometric substitution: y sin2 (φ) = tan(φ) = k−y cos2 (φ) The function φ is a new function of x. Now isolating y gives y = ksin2 (φ) Differentiating in φ, we get: dy = ksin(2φ) = 2ksin(φ)cos(φ) dφ Hence, dx = dφ

r

y dy = 2ksin2 (φ) = k(1 − cos(2φ)) k − y dφ 11

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

Integrating gives:

Z 1 − cosφ k x=k φ= 2φ − sin(2φ) + C d 2 The boundaries conditions that we have on the problem make it quite clear that the integration constant C must be 0. Now let 2φ = θ then we have the parametric 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 that cycloid is a stationary function as well as the constant function since both satisfies the differential equation. As promised, below is the solution I obtained when I solved the integral (4) using the regular substitution method: s s

k−y k−y x(y) = −y − k · arctan y y This is also the equation for a cycloid. Which of course agrees with the parametric 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 determine the time the particle travels along the cycloid. Another property is that given two distinct points A and B, there exist only one unique cycloid through A and B.[10] A suitable radius will give this cycloid. Notice, we have not yet arrived at the optimal curve (path) yet. More work needs to be done to unveil the brachistochrone curve.

5

The Brachistochrone Unveiled

Congratulations! You have reached the most exciting part of the paper. Here the brachistochrone will be unveiled. Let me remind you that the cycloid and 12

the constant function both satisfy the brachistochrone but I have shown earlier that the constant by itself cannot be a brachistochrone. This leads us to a question: Since the constant function by itself cannot be a brachistochrone, does this mean the cycloid is the only curve left which implies that it must be the brachistochrone?

I have read through many papers and books about the brachistochrone and one way or another, they all claim that the cycloid must be the brachistochrone, since it satisfies the differential equation (3). R.Coleman addressed this same issue in [6]. It is true that the constant function can be eliminated, but let me remind you that the problem does not prevent piecewise continuous functions.

As long as my piecewise function is differentiable everywhere then it is fine. Let me 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, we always 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 = − π2 x then the cycloid is the only solution. Otherwise there will be infinitely many solutions, hence infinitely many critical curves. Hence, the question arrives, What is so special about this line y = − π2 x? notice the negative is there only because of the way I look at the problem. This sign would be positive if I just imagine rolling the ball from right to left, so do not get confused by the sign here 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 maximum point if you look at a regular cycloid, is evaluated at θ = π. This is obvious from 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 dy/dt rsin(θ) = = =0 dx dx/dt r(1 − cos(θ)) 13

Thus, the cycloid is at its maximum point when θ = π. This is a fascinating result because what this means is if your starting point is the same, no matter what cycloid you choose, the optimal point of your cycloid will always be touched by the line y = − π2 x. The picture below expresses that fact.

This is why when your destination point is below the line y = − π2 x you will get only one solution, which is a cycloid. Otherwise, you can have one solution be a cycloid, a constant function and infinitely many piecewise definition functions.

Notice that these piecewise functions do not violate the problem at all. We know that the brachistochrone do not have a discontinuous derivative, which means 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 it slopes 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 prove that the cycloid satisfies the differential equation, and therefore it must be the brachistochrone. Other than [6], which also showed why the cycloid must be the fastest in the family and discussed about the facts that many papers and articles out there never address this same issue. The proof [6] presented is quite complicated and required a good amount of preliminary knowledge. It showed that the extremals (Critical curves) must be C 2 class function. All the papers I came across other than [6] seem to assume that the extremals belong in the class of C 2 but that’s not necessary and can be proved. It may not be obvious, but one can see that if C 2 was the requirement for extremals, then there is only one extremal function, no matter where you chose your end point, which is a cycloid. Let us take a look and see why this is true. Recall that for the cycloid, sin(θ) dy dy dy dx = (1−cos(θ)) . But dx is a function with parameter θ, hence let dx = g(θ).

Which implies

d2 y d dy d = = g(θ) (1) dx2 dx dx dx 14

But

dg dθ

=

dg dx

·

dx dθ

which shows

d2 y = dx2

d dθ

dy dx

dx dθ

dg dx

d dθ

=

=

dg/dθ dx/dθ .

But

sin(θ) (1−cosθ)

dg dθ

=

d dθ

dy dx

Hence,

r(1 − cos(θ))

=

cos(θ)−1 (1−cos(θ))2

r(1 − cos(θ))

Therefore, 1 d2 y =− 2 dx r(1 − cos(θ))2 Thus the second derivative of a cycloid is never zero. However, for the constant function, its second derivative is zero, which shows that the piecewise solutions are in class C 1 functions. This implies that if the requirement of class C 2 is enforced on the extremals then the solution to the brachistochrone is indeed the cyloid. However, what I am going to do next is prove this same fact by showing the cycloid is in fact the fastest curve in the family using some basic mathematics.

Theorem 5.1. Given two distinct points A and B, where point B is some distance d away from A and at the same level as A. The cycloid is the fastest curve 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 is a 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 full arc length (L) of a cycloid is 8r since s Z2π 2 2 Z2πp dx dy L= + dθ = r2 (1 − cos(θ))2 + r2 sin2 (θ)dθ = 8r dθ dθ 0

0

The last step may not be obvious but I encourage you as reader to clarify this on your 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, given a cycloidal pendulum with radius r then the period of that pendulum is: s r L 4r T = 2π = 2π g = gravitational constant g g q This indicates that the period for any cycloid must be T = 2π 4r g . However, a period is a round trip, in other words, the time the pendulum would take to swing from one end to another and back. For the purposeqof the brachistochrone, we are only interested in half a period or simply T = π 4r g .

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 many critical curves connecting point A and B; however there is only one cycloid, and the others are piecewise functions.

Suppose curve 1, denotes C1 is a cycloid with radius r2 that connects A and B. Note the minimum point of this cycloid is (r2 π, 2r2 ) where 2r2 = y2 = − π2 x2 .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 = − π2 x1 , and a constant function from (x1 , y1 ) to (x3 , y3 ) where x1 < x3 and similarly y3 = − π2 x3 , then followed by another cycloid from (x3 , y3 ) to point B. Note, x1 3 = x2 . and x3 are picked arbitrarily and x1 +x 2 Now let T1 denoted the time for the particle take on path 1. Similarly, let T2 denoted the time the particle would take on path 2. r 4r1 x3 − x1 T1 = π + √ g π 2y1 16

Let g = π 2 gives: T1 =

p

2y1 +

x3 − x1 √ π 2y1

Using the facts x3 − x1 = 2(x2 − x1 ) r √ √ p 2(x2 − x1 ) 4 2(x2 − x1 ) q T1 = 2y1 + = x1 + √ π y1 π π π2 x Notice, the negative sign of the equation y = − π2 x was left out and the reason for that is the negative sign was arbitrary. It was there because the way we picked our starting point and thus if we rolled the ball from left to right then it would be positive. Furthermore, this will apply consistently through the proof and so the negative sign can be ignored. r r r x2 x2 x1 x1 x1 +√ = +√ − T1 = 2 π πx1 π π πx1

x1 2 x2 x2 T12 = + x2 + π π x1 π We know x1 < x2 , hence

x2 x1

= (1 + k) for k > 0. Therefore, T12 =

x1 k 3 x2 + + x2 π π π

Now let consider T2 . Similarly to before we have r p 4 T2 = 2y2 = x2 π Hence

4 3 x2 x2 = x2 + π π π Notice the difference between T1 and T2 are just those few terms. However, observe x1 k x1 k x1 k k2 + x2 = + · (1 + k)x1 = + x1 + x1 π π π π π π π And x2 (1 + k)x1 x1 k = = + x1 π π π π T22 =

2

Thus, this shows that with the given conditions T1 > T2 by a factor of kπ x1 .

Therefore, the cycloid is the fastest path in the family and hence the brachistochrone curve.

A sharp reader, like my advisor Dr. Steve Tanner, would see that there is an alternative once we arrived at r r x1 x2 4 T1 = +√ T2 = x2 π πx1 π 17

We could take the ratio between T1 and T2 , so p x1 √x2 x1 + x2 T1 π + πx1 q = √ = T2 2 x1 x2 4 x π

2

But notice,

√

x2 −

√

2 x1

>0

⇒

x2 − 2x1 x2 + x1 > 0

Hence,

x2 + x1 >1 2x1 x2 Thus T1 > T2 , which also implies that the cycloid is in fact the brachistochrone curve. x2 + x1 > 2x1 x2

⇒

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

2 u u s sinθ 2π 2π r Z u 1 + 1−cosθ Z t 1 + y 02 4r 1 dx = r(1 − cosθ)dθ = π T = 2gy 2g r(1 − cosθ) g 0

0

Again, the last step is a very simple integral after some algebra manipulations so I encourage you as the reader to confirm the result for yourself. So it is completely legitimate to use the Christian Huygens pendulum clock to describe the movement of the particle along the cycloid curve.

Therefore, the cycloid is in fact the fastest curve for a particle to take to go from point A to point B where B is not higher than A under gravitational force alone. One thing I want to point out is that the proof I gave required you to pick your end point (B) to be on the same level as A. Although, I did not explicitly show the case where we picked point B such that it is above the line y = − π2 x but lower than A, it is easy to see that the cycloid still must be the fastest 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 is gravity. So given the same problem, what is the curve that would satisfy the property of being a brachistochrone? Well this is a hard question and often left out of the discussion by many since it is fairly involved and the result is not that much more astonishing than the original brachistochrone problem. I will however briefly present the results to the problem when we take friction into 18

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

Now that we applied friction into the problem, what will change is the velocity, which of course makes p sense. If you remember from physics, when friction force is applied then v = 2g(y − µx) where µ is the friction coefficient. Hence, our new equation for the time period is: Z s 1 + y 02 dx t= 2g(y − µx) We can now apply the Euler Lagrange equation which gives [1 + y 02 ](1 + µy 0 ) + 2(y − µx)y” = 0 This is a highly nonlinear differential equation and note we cannot use the Beltrami identity. This process points us to the fact that, although, the EulerLagrange equation is great, often times, it can lead us to a differential equation that is very complicated to solve. I’ll not go into detail of how to solve this equation, 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 when µ is zero then we have a cycloid again. There is no unique solution in this case because each solution depends on µ. Since no solution exist such that it would not depends on µ make it less interesting to mathematicians and often left out of discussions. However, if you are trying to build the world fastest slide then the solution with friction involved would be more appropriate.

7

Conclusion

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

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8

Bibliography

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ISSN:1887-1997 [2] Jeff Babb, James Currie, The brachistochrone broblem: mathematics for a broad audience via a large context problem. TMME, vol5, nos.23,p.169 [3] J.J Oâ&#x20AC;&#x2122;Connor, E.F Robertsn, History Topic: The Brachistochrone Problem, MacTutor History of Mathematics [4] John Martin, The Helen of Geometry, the college mathematics journal Vol.41, No.1, January 2010.

[5] William Dunham, Journey through genius: The great theorems of mathematics. WIley 1990, page 199-202 [6] Coleman, Rodney. A detailed analysis of the brachistocrhone problem. arXiv preprint arXiv: 1001.2181(2010).

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[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 brachistochrone problem. arXiv preprint math-ph/061052 (2006).

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

Web 13 May 2014.

[14] Weisstein, Eric W. Brachistochrone Problem. Mathworld-A Wolfram web resource. http://mathworld.wolfram.com/BrachistochroneProblem.html [15] Nishiyama, Yutaka. The Brachistochrone Curve: The Problem of Quickest Descent. 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, Huygenâ&#x20AC;&#x2122;s clocks. Proceeding:Mathematics, Physicial and Engineering Science (2002): 553-579.

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