The Tautochrone Problem

Submitted by:

Candidate Number:

Session: May 2015

Subject Area: Mathematics

Word Count: 2926

Abstract

The research statement of this Mathematics Extended Essay is the following:

“On a frictionless wire, there are two beads in a uniform gravitational field; show that

the shape the wire must take on, such that the two beads will reach the end of the wire

at the same time when they are simultaneously released from rest, from any point on

the wire, is half of an inverted cycloid cut at the non-inverted apex.” This is known as

the Tautochrone Problem. In investigating this problem I used an energy approach,

which allowed me to express the speed of the bead as a function of the height it

dropped from during its motion. I then used the definition of instantaneous speed to

express a fraction of the time of decent as a function of the speed of the bead at a

certain height, and the infinitesimally small distance it covers during that time. The

distance the bead covers in the infinitesimally small time interval was approximated

to be a straight line. A method for determining the time of decent down any strictly

decreasing and differentiable curve was then devised, which was ultimately used to

prove that the curve that solves the Tautochrone Problem is a cycloid. A by-product of

the proof states that the time of decent for a bead sliding down a cycloid is equal to

where is the radius of the circle that defines the cycloid and is the

strength of the gravitational field. Finally it was also argued that rolling bodies on a

cycloid also exhibit tautochronous behaviour, but this is only true for bodies of the

same moment of inertia.

Word Count: 274

1

Table of Contents

Heading Page

Introduction 3

Pre-examination of the Problem 5

Defining the Curve 6

Expressing Time 8

An Example of Determining the Time of Descent 14

Proving The Tautochronous Property of a Cycloid 17

The Case with Rolling Bodies 23

Further Investigations 25

Bibliography 26

2

Introduction:

From the Greek tauto, for “the same,” and chronos, for “time,” the

Tautochrone Problem describes a frictionless ramp that causes a body to travel to the

bottom in a set amount of time, regardless of the starting position of the body on the

ramp.1 This means that two bodies released at the same time, but starting at different

positions on the Tautochrone, will arrive at the bottom at the same time.

This problem was first solved by the Dutch mathematician and physicist,

Christiaan Huygens, and was published in his book Horologium oscillatoriumi in

1673.2 It should be noted that this was before Leibniz and Newton published their

discoveries of calculus.3 Huygens attempted to use his discovery of the Tautochrone

to make a more accurate pendulum clock that forced the pendulum to follow a

tautochronous path, causing the period of the oscillating pendulum to be independent

of the amplitude. This later proved to be impractical, due to the friction between the

wire and the Tautochrone.4

1 Darling, David. The Universal Book of Mathematics: From Abracadbra to Zeno's

Paradoxes. New York: John Wiley & Sons, 2004. Print.

2 Pickover, Clifford A. The Math Book: From Pythagoras to the 57th Dimension, 250

Milestones in the History of Mathematics. New York, NY: Sterling Pub.,

2009. Print.

3 Clifford 152

4 Clifford 156

3

This Extended Essay will focus on proving that the shape of the Tautochrone

curve is a cycloid. The research statement of the essay will be as follows:

On a frictionless wire, there are two beads in a uniform

gravitational field; show that the shape the wire must take on, such that

the two beads will reach the end of the wire at the same time when they

are simultaneously released from rest, from any point on the wire, is half

of an inverted cycloid cut at the non-inverted apex.

The reason why beads on a wire are chosen, as opposed to balls on ramps, is

because for this extended essay, the rotational kinetic energies, of the bodies will be

neglected. After the analysis, this point will be discussed in further detail.

My primary reason for deciding to investigate the Tautochrone is because of

the way the curve seems unreal at first inspection, and its special property in the field

of mechanics. I am personally fascinated by mechanics problems in general and I

knew about this problem for quite some time, which is why I decided to investigate it

further. The special property of the curve is so appealing that it is even mentioned in

the novel Moby Dick by Herman Melville, where a soapstone slides inside a “try-pot”5

shaped like a Tautochrone.6

It should be noted that for the sake of simplicity, this paper will focus on a

calculus based proof and not the proof of Huygens, which relied on Greek geometry

and Galileo’s laws of motion. Huygens’s proof makes use of limited mathematical

tools, but despite that fact, it was remarkably still able to prove that the curve that

5 A “try-pot” is a pot, which is used to process whale blubber.

6 Clifford 156.

4

solves the Tautochrone Problem is a cycloid.7 By examining Huygens’s proof I

personally believe he took a similar approach to me, in that he did initially speculate

that the Tautochrone is a cycloid. He most likely came to that result through physical

experimentation, and then proved his claim mathematically.

Pre-examination of The Problem:

In order to begin, one has to consider what the curve that solves the

Tautochrone Problem would roughly have to look like. Consider a scenario where one

bead starts further away from the end of the wire, than the other. The bead that starts

further away from the end of the wire will have to be moving quicker at the end of its

motion, than the other. Therefore, the bead further away has to experience a greater

acceleration. This would imply the segment of the wire further away from the end of

the wire would have to be steeper. Thus, the shape of the curve the wire will mimic

will have to strictly decrease at a decreasing rate. By this logic, the Tautochrone curve

would have to look something like figure 1 on the next page, where two beads,

marked 1 and 2, are placed at arbitrarily points on the wire.

7 Huygens, Christian. "Christian Huygens : Horologium Oscillatorium." Christian

Huygens : Horologium Oscillatorium. Trans. Ian Bruce. 17 Century Maths,

Aug. 2013. Web. 10 Aug. 2014.

<http://www.17centurymaths.com/contents/huygenscontents.html>.

5

Figure 1

The arrow in figure 1 represents the direction of the uniform gravitational

field, . Notice that bead 1 initially accelerates more than bead 2, due to the wire

being steeper at that point.

Defining the Curve:

Now that an idea for what a Tautochrone should look like has been

established, a cycloid can be formally defined. This coming definition will be used in

a later proof to illustrate the cycloid’s tautochronous quality.

A cycloid is a curve defined by a point on the edge of a circle, which is rolling

along a defined axis.8 When working with Cartesian coordinates, it is difficult to

explicitly state the equation of a cycloid, and therefore describe the vertical position

of a bead, , in terms of its horizontal displacement, . It is much simpler to work

with parametric equations, which describe the position of a bead in terms of a

parameter , the angle through witch the cycloid-defining circle has rolled.

8 Clifford 156.

6

Figure 2

If the cycloid-defining circle is rolling along the x-axis and the cycloid

defining point on the edge of the circle started from the point , then the centre of

the circle can be described as , were is the radius of the circle and is

measured in radians. The coordinate of the centre of the circle (the length of the

line segment from 0 to A in the figure) comes from the fact that the arc length PA in

figure 2 is equal to the coordinate of the centre of the circle, because the circle is

rolling.

7

Now, determining the coordinate of point P is relatively simple. The x-

coordinate is given by and the y-coordinate is given by , which

can be determined by analysing figure 2 carefully.9

The equations above will be used later in the paper, but for now a method for

determining the time of decent of a bead needs to be devised.

Expressing Time:

In order to determine the time of descent for a bead that starts at an arbitrary

point on the wire, the way time passes has to be mathematically expressed. There are

multiple ways of mathematically expressing how time passes; the method used in this

paper will make use of the definition of instantaneous speed, which states that the

instantiations speed of an object, is equal to the distance, , it covers over a time

interval, , which is allowed to become infinitesimally small. Symbolically this

would imply the following.

9 Explanation for parametric equations based on:

Gilbert, Gayle, and Greg Schmidt. "Parametric Equations for a Cycloid." The

University of Georgia Mathematics Education. The University of Georgia,

n.d. Web. 20 Apr. 2014.

<http://jwilson.coe.uga.edu/EMAT6680Fa07/Gilbert/Assignment%2010/Gayl

e&Greg-10.htm>.

8

The speed of the bead at any given instant can be described using an energy

approach. As the bead moves from a higher position to a lower position, the bead

looses gravitational potential energy (PE) and gains kinetic energy (KE) and thus

speed. (The red vector with the next to it stands for velocity, which describes a

particle’s speed and direction.)

Figure 3

There are only two forces acting on the bead at any time: the force due to

gravity, w for “weight,” and the normal force, N , the wire exerts on the bead.

Because the normal force is always perpendicular to the instantaneous displacement

of the bead, only gravity does work on the bead. This means, gravity is the only force

responsible for the bead gaining energy as it descends.10

10 Henderson, Tom. "Energy Conservation on an Incline." Energy Conservation on an

Incline. N.p., n.d. Web. 18 Apr. 2014.

<http://www.physicsclassroom.com/mmedia/energy/ie.cfm>.

9

Figure 4

It should be noted that w is equal to mg , where is the mass of the bead and

g is the gravitational field strength.

By the conservation of energy, it can be stated that after the bead looses

gravitational potential energy, after falling through a height , the bead gains kinetic

energy equal to the amount of gravitational potential energy lost.11 From the

definitions of potential energy, and kinetic energy, this mathematically implies,

(Potential Energy) (Kinetic Energy)

and therefore,

where is the speed of the instantaneous speed of the bead and where is the

strength of the gravitational field. Notice that the speed of a bead is independent of its

mass, and that is a scalar.

11 Henderson.

10

With the knowledge of instantaneous speeds, one is able to derive an

expression that will give a means of determining the time of descent for a bead. Recall

the definition of instantaneous speed is the following.

Now the following can be written.

Which implies,

where is a very small distance that the bead covers and is the very short time

interval, in which the bead covers the distance .

In order to simplify the analysis, will be expressed in another way. If a very

small portion of the curve is examined, one could argue that the small distance the

bead has to travel for that small portion, is close to a straight line. Therefore the

following can be written:

11

Figure 5

Where and are infinitesimally small distances covered by the bead in

the horizontal and vertical direction respectively. Using the just derived equation, the

amount of time required to slide down a fraction of the curve can be expressed as:

.

The equation above can now be integrated with respect to either or . It is

also important to note in the equation above, only magnitudes of variables will be

considered. The only problem with this approach is that in the picture that is usually

used to solve a time of decent problem, is pointing downwards and is thus negative.

In order to avoid problems involving negative values in square roots, the picture that

is used to solve problems of time of descents will be inverted, such that it looks like

the bead is sliding toward the top of the page.

12

Figure 6

It should be noted that time of decent problems can also be solved using a non-

inverted picture, but one would also have to make negative, which can complicate

the analysis, due to the square root.

By inspecting the right hand portion of the figure above, it can be seen that

can be expressed as .

Now, in order to determine the total time required for a bead released from

rest to travel between , the initial height, and , the final height, the equation

found earlier will be integrated with respect to , from , to . This leads to the

equation,

which is only valid for strictly decreasing and differentiable curves.

It should be noted that the equation above is not truly valid as the right most

function being integrated does not exist when y= y0 . This is an example of an

improper integral. There are also a few situations where the end of the curve is

13

horizontal, which leads to not existing, or becoming infinitely large. For this

reason, the equation above should actually be altered to the following.

But it is cumbersome to repeat the limits in proofs, so they will be omitted but

implied in this paper.

Below is a summary on how to solve time of decent problems, down strictly

decreasing and differentiable curves of interest12:

Invert the curve. The curve can also be shifted parallel to to make the

analysis simpler.

Apply

In some situations, recall that in the equation above, the integrated function on

the right accepts values in the open interval .

An Example of Determining the Time of Descent:

12 The idea to approach time of decent problems like this is partially based on:

Weisstein, Eric W. "Tautochrone Problem." Wolfram MathWorld.

Wolfram, n.d. Web. 20 Apr. 2014.

<http://mathworld.wolfram.com/TautochroneProblem.html>.

14

In order to demonstrate the equation derived in the previous section, a simple

situation where the answer is already known will be considered. Assume a bead slides

down a straight wire, with a 45° incline from to as illustrated in the

figure on the next page.

Figure 7

According to Galileo Galilei’s work concerning the motion of objects under

the influence of gravity, the time of decent for an object sliding down an incline is

equal to

where is the distance the object has to slide down, is the height of the

incline, and is the magnitude of the gravitational field strength13.

13 Sanchis, Gabriela R. "Historical Activities for Calculus - Module 3: Optimization –

Galileo and the Brachistochrone Problem." MAA. Mathematical Association of

America, July 2014. Web. 08 Oct. 2014.

<http://www.maa.org/publications/periodicals/convergence/historical-

activities-for-calculus-module-3-optimization-galileo-and-the-

brachistochrone-problem>.

15

In the case of a 45° incline, the time of decent should be:

In order to try and achieve the expected result using the derived equation

from the last section, we first need the equation of the curve being considered in

this example. It is clear that the equation that describes the curve is simply:

.

When the curve is inverted and shifted so that , the inverted curve

has the equation .

Figure 8

It should also be noted that .

Since this is a strictly decreasing function, the time of decent from to

can now be determined.

First the equation derived from the previous section will be stated again.

When making the appropriate substitution for , it can be seen that,

16

When and

And since,

Which is the same result devised by Galileo Galilei.

Proving The Tautochronous Property of a Cycloid:

The tautochronous property of a cycloid can now be proven using what was

derived in the previous sections of this paper. First of all, it should be noted that in the

section where the equation of a cycloid was determined, the cycloid was already

effectively “inverted” so that it would look like the bead would slide towards the top

of the page, so there is no need to alter the equations of the cycloid derived earlier.

Recall that the parametric equations of a cycloid are,

17

And thus

In order to simplify the analysis, it will first be assumed that the bead starts

from the top of the cycloid as shown in the edited picture from a previous section

below.

Figure 9

First, in order to make the analysis simpler, recall that

When the four previously mentioned cycloid-defining equations are utilized in the

equation defining , the following can be written.

Let and therefore

18

Recall the Pythagorean Trigonometric Identity:

19

When the bead starts at the top of the inverted cycloid, and

Note that in the open interval ,

When the above equation is integrated from 0 to π, the time of decent can be

determined.

Which gives the total time of decent from the top to the bottom of the cycloid.

Now, if a cycloid is in fact the curve that solves the Tautochrone Problem, then the

time of decent should remain the same, regardless of where the bead starts its motion.

Recall that from before.

When the equation is integrated from to

20

Now it has to be demonstrated that the right most integral above is equal to π.

Substituting the half angle identities: and

.

Let and thus since is a constant.

21

Note, . The bead cannot begin its motion at the end of the wire.

Note that in the open interval , .

on the interval

Which proves that the curve that solves the Tautochrone Problem is half of a

cycloid cut at the non-inverted apex.14

14 Proof based on:

Weisstein.

22

On a final note, the reason that q0 cannot exactly equal 0 or p can also be

understood on a physical level. First of all, it is obvious that the bead cannot begin its

motion at the end of the wire for the Tautochrone problem, since it takes no time for it

to arrive at the end of the wire, as the bead is already there. On the other hand, the

bead cannot start at exactly q0=0 , because when the wire is allowed to continue to

take on the shape of the graph the parametric equations of the cycloid, for negative

values of q as well, then it can be seen that placing the bead at the exact top of the

cycloid can be problematic, as the bead can slide down either side of the cycloid, or

not slide down at all.

Figure 10

This is the primary reason for why the limits were mentioned earlier in this paper.

The Case with Rolling Bodies:

Earlier in the essay it was explicitly stated that beads would be sliding down

the wire as opposed to balls on a ramp. It turns out it does not matter which type of

object is moving down the cycloid, as long as the object uniformly rolls or slides

down the cycloid. This means, the object cannot roll and then slide as it descends

down the Tautochrone.

23

If a solid sphere were to roll down the cycloid, then a different energy

approach would have to be taken. As the sphere looses potential gravitational energy,

it gains translational and rotational kinetic energy. This would mathematically imply:

where is the moment of inertia of the sphere about the diameter and is the

angular velocity of the ball. The angular velocity of the ball is a function of how many

revolutions per second the ball undergoes at a specific time. It is known that for a

solid sphere the moment of inertia is given by:

where is the mass and is the radius of the sphere. Generally speaking, the

moment of inertia of a body rotating about an axis is a function of the mass

distribution about the axis of rotation. The translational speed of the ball can also be

expressed as15:

Through multiple algebraic manipulations and substitutions it can be shown that:

.

Now this equation for instantaneous speed has the same form as with the beads, but

the coefficient before the changed from 2 to . It can be argued that the cycloid is

still tautochronous for rolling objects of the same moment of inertia, but the time of

decent is different for different objects.

15 Rotational physics from:

Tipler, Paul Allen, and Gene Mosca. "Chapter 9: Rotation." Physics for Scientists

and Engineers. New York: W.H. Freeman, 2008. N. pag. Print.

24

From a separate calculation, which uses the same techniques described earlier,

one is able to determine that the time of decent is equal to the equation below, which

is independent of the starting height of the bead.

Further investigations:

Even though the Tautochrone Problem has been thoroughly solved already,

one cannot help but wonder whether there might be a Tautochrone curve that takes

friction into account by introducing a frictional coefficient in a set of equations, or

where the gravitational field varies as the ball descends, like a Tautochrone that

hypothetically goes up all the way to space. One might need to neglect the atmosphere

when doing such an analysis, but one cannot help but wonder whether a Tautochrone

like that could even exist. Even though Huygens might have been unsuccessful at

utilizing the Tautochrone’s special property, there could be a chance that the

Tautochrone will be a key component in a future technology or theory, where the

mathematics of the Tautochrone are already discovered.

25

Bibliography

Darling, David. The Universal Book of Mathematics: From Abracadbra to Zeno's

Paradoxes. New York: John Wiley & Sons, 2004. Print.

Gilbert, Gayle, and Greg Schmidt. "Parametric Equations for a Cycloid." The

University of Georgia Mathematics Education. The University of Georgia, n.d.

Web. 20 Apr. 2014.

<http://jwilson.coe.uga.edu/EMAT6680Fa07/Gilbert/Assignment%2010/Gayl

e&Greg-10.htm>.

Henderson, Tom. "Energy Conservation on an Incline." Energy Conservation on an

Incline. N.p., n.d. Web. 18 Apr. 2014.

<http://www.physicsclassroom.com/mmedia/energy/ie.cfm>.

Huygens, Christian. "Christian Huygens : Horologium Oscillatorium." Christian

Huygens : Horologium Oscillatorium. Trans. Ian Bruce. 17 Century Maths,

Aug. 2013. Web. 10 Aug. 2014.

<http://www.17centurymaths.com/contents/huygenscontents.html>.

Pickover, Clifford A. The Math Book: From Pythagoras to the 57th Dimension, 250

Milestones in the History of Mathematics. New York, NY: Sterling Pub.,

2009. Print.

26

Sanchis, Gabriela R. "Historical Activities for Calculus - Module 3: Optimization –

Galileo and the Brachistochrone Problem." MAA. Mathematical Association of

America, July 2014. Web. 08 Oct. 2014.

<http://www.maa.org/publications/periodicals/convergence/historical-

activities-for-calculus-module-3-optimization-galileo-and-the-

brachistochrone-problem>.

Tipler, Paul Allen, and Gene Mosca. "Chapter 9: Rotation." Physics for Scientists and

Engineers. New York: W.H. Freeman, 2008. N. pag. Print.

Weisstein, Eric W. "Tautochrone Problem." Wolfram MathWorld. Wolfram, n.d.

Web. 20 Apr. 2014.

<http://mathworld.wolfram.com/TautochroneProblem.html>.

27