The Brachistochrone

8/3/2019 The Brachistochrone

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A bit of optics

We begin with an apparently unrelated problem from optics.

A ray of light travels from point A to point P with velocity 1v and then,

entering a denser medium, travels from P to B with a smaller velocity 2v .

The total time T required for the journey is given by

( )

2

22

1

22

v

xcb

v

xaT

++

+= .

If we assume that this ray of light is able to select its path from A to B byway of P in such a way as to minimize T, then 0=dx

dT and by some

elementary calculus we find that

( ) 222

22

1 xcbv

xc

xav

x

+

=

+

or2

2

1

1sinsin

vv

=

This is Snells law of refraction. The assumption that light travels from one

point to another along the path requiring the shortest time is called

Fermats principle of least time. This principle not only provides a rational

basis for Snells law, but can also be applied to find the path of a ray of light

through a medium of variable density, where in general light will travel along

curves instead of straight lines.


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In the individual layers the velocity of light is constant, but the velocity

decreases from each layer to the one below it. As the descending ray of

light passes from layer to layer, it is refracted more and more toward the

vertical, and when Snells law is applied to the boundaries between the

layers, we obtain

4

4

3

3

2

2

1

1sinsinsinsin

vvvv

=== .

If we next allow these layers to grow thinner and more numerous, then in

the limit the velocity of light decreases continuously as the ray descends,

and we conclude that

=v

sina constant.


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A return to Bernoullis problem

Imagine the following coordinate system and that the bead (like the ray of

light) is capable of selecting the path down which it will slide from A to B in

the shortest possible time.

The argument given above yields

=v

sina constant. (1)

If the bead has mass m, so that mgis the downward force that gravity

exerts on it, then the principle of conservation of energy, the fact that the

work done by gravity in pulling the bead down the wire equals the increase in

kinetic energy of the bead tells us that 22

1m vm g y= . This gives

gyv 2= . (2)

From the geometry of the situation we also have

( ) 22 1

1

tan1

1

sec

1cossin

y+=

+

===

. (3)

On combining equations (1), (2), and (3) obtained from optics, mechanics,

and calculus we get

( ) cyy =+ 21

(4)


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as the differential equation of the brachistochrone.

Solving the brachistochrone

When y is replaced bydx

dyand the variables are separated, (4) becomes

dyyc

ydx

21

= . (5)

At this point we introduce a new variable by putting

tan

21

=

ycy

, (6)

so that 2sincy = , dcdy cossin2= , and

dydx tan=

dc 2sin2=

( ) dc 2cos1= .

Integration now yields

( )1

2sin22

cc

x += .

Our curve is to pass through the origin, so by (6) we have x= y= 0 when0= , and consequently 01 =c . Thus

( ) 2sin22

=c

x (7)

and

( ) 2cos12

sin 2 == ccy . (8)

if we now put2

ca = and 2= , then (7) and (8) become


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( ) sin=ax and ( )cos1=ay .

(9)

The Cycloid

The cycloidis the curve traced out by a point as a circle rolls along a line

(the x-axis).

We will use parametric equations to get the equation for the cycloid.

Since the distance the circle rolls is equal the amount of the circumference

that has been traced out, to get the x-value of our parametric equation we

need only to subtract the horizontal distance P is from C (i.e. PQ) from the

size of the segment of the circumference that has been traced out.

Since QP = sina , we get

( ) sinsin == aaax . (10)

For the y-value we simply subtract CQ from the radius, giving us

( ) cos1cos == aaay . (11)


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But (10) and (11) are the equations for the brachistochrone!

The Brachistochrone

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