WEBVTT

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Which slide gets a bead
from the same high point

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to the same low point in the least time?

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The straight line is the shortest
route, so it is the obvious favorite.

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Let us race it
before trusting that intuition.

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Every candidate begins at A and ends at B.

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The blue track is the straight ramp.

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The green track is a circular arc.

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The red track is a cycloid, a curve traced
by a point on the rim of a rolling circle.

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The three beads start from rest.

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Gravity is the same for all
of them, the track is frictionless,

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and only the shape of the route changes.

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Ready.

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Go.

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The red bead dives sharply, gathers
speed early, and reaches the bottom first.

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The green bead follows close behind.

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The blue bead
on the direct ramp arrives last.

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The measured times put the cycloid first
at about one point one four seconds.

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The circular arc takes
about one point one seven.

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The straight ramp takes
about one point three zero.

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So shortest does not mean fastest.

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The straight ramp saves distance,

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but the curved tracks spend more
of their journey moving quickly.

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A steep initial drop can repay
the extra distance later.

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The circular arc already exploits
that bargain, but the cycloid balances

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early acceleration against
later distance even better.

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Our task is to turn that visual bargain

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into a condition that selects one curve
from every possible curve joining A to B.

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Let y measure vertical depth below
the release point, down toward the bottom.

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At depth y, energy conservation says
the lost gravitational potential energy

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has become kinetic energy.

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The mass cancels, and a bead released from
rest has speed square root of two g y.

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Deeper parts of the track
are therefore traversed faster.

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Now inspect one tiny piece of path.

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Its horizontal change is d x,

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its vertical change is d y,
and Pythagoras gives the arc length d s.

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Time is distance divided by speed,
so this small piece costs d s over v.

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Replace d s by square root of one
plus y prime squared, times d x.

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Adding the costs from the start
to the finish gives this integral.

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It depends on the height y and the slope
y prime at every x along the route.

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This notation T of y matters.

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The input is not one number.

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The input is the entire function y
of x, and the output is one travel time.

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Ordinary calculus
might minimize f of alpha.

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We move one number left or right
and compare the resulting values.

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Here the choice is a curve. Moving
one part can leave another part alone,

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and there are infinitely many
places where the route can be nudged.

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A single slider can display one family

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of curves, but it cannot
represent every possible deformation.

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The mathematical variation
needs an arbitrary function

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to describe all those independent nudges.

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Write a nearby path as y
plus epsilon times eta.

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Epsilon controls
the overall size of the change.

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Eta of x controls its shape
from place to place.

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The endpoints are prescribed,
so the variation vanishes at both ends.

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Between them, eta is otherwise arbitrary.

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That freedom is what will turn one
integral statement into a differential

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equation for the fastest path.

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Call the integrand F of y and y prime.

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A fastest path must have zero
first-order change in travel time

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for every allowed deformation eta.

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Differentiate under the integral sign.

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Changing y contributes F sub y times eta.

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Changing the slope contributes
F sub y prime times eta prime.

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The eta-prime term is inconvenient because
the arbitrary function is differentiated.

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Integrate that term by parts,
moving the derivative onto F sub y prime.

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The boundary term vanishes because
eta is zero at both fixed endpoints.

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What remains is one integral
of eta times a coefficient.

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Now use the crucial freedom.

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Eta can be a small smooth
bump wherever we choose.

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If the coefficient were positive

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or negative anywhere, a bump placed
there would make the integral nonzero.

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Therefore the coefficient
itself must vanish.

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This is the Euler-Lagrange equation,

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the differential condition
satisfied by every stationary path.

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It is the analogue of setting
an ordinary derivative equal to zero,

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but now the derivative
tests every possible direction

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in a space of functions rather than
the two directions on a number line.

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For this particular problem
there is a useful shortcut.

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The integrand depends on y and y
prime, but it has no explicit x in it.

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Differentiate F minus y
prime times F sub y prime.

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The chain rule

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and the Euler-Lagrange equation
reduce that derivative to F sub x.

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But F sub x is zero,
so the expression is constant.

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This conserved quantity is often
called the Beltrami identity.

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Substitute our square-root integrand.

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After the y-prime terms combine,

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F minus y prime F sub y prime becomes
the reciprocal of these two square roots.

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Set that expression equal to C,
square both sides, and rearrange.

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The fastest curve must satisfy
y times one plus y prime squared

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equals a positive constant.

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We name the constant two a because that
choice will make the geometry transparent.

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This equation is already the answer to the
variational problem in differential form.

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We now solve y times one plus
y prime squared equals two a.

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The constant is positive, and y is the
downward depth below the release point.

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The combination two

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a minus y suggests the substitution y
equals a times one minus cosine theta.

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It automatically starts at y
equals zero when theta equals zero.

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Solve the condition for y prime
squared, then insert the substitution.

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The ratio becomes one plus
cosine over one minus cosine,

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which is cotangent
squared of theta over two.

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Differentiate y with respect to theta.

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That gives a sine theta.

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The slope equation also says d x
over d y is tangent of theta over two.

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Multiply d y by d x over d y.

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The result is a sine theta times tangent

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of theta over two, which simplifies
to a times one minus cosine theta.

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Integrate with respect to theta
and choose x equals zero at the cusp.

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We obtain x equals a times
theta minus sine theta.

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Together with the substitution for y,

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these are the parametric
equations of a cycloid.

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Let theta grow.

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The point traces the curve from its cusp,

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initially almost straight downward,
then gradually turns toward the finish.

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The two endpoint equations
determine the two remaining constants.

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For the race, a is about

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one point five three and the finishing
parameter is about two point eight eight.

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Near the release point,

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the vertical depth y is tiny,
so the curve begins extremely steep.

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Farther down, the bead already has speed,

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and the curve spends that speed
moving toward the destination.

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The straight line
minimizes distance alone.

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The cycloid minimizes the integral
of distance divided by speed,

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so it deliberately buys
early speed with extra length.

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That is why it wins the race.

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The fastest curve
has another remarkable property.

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Put three beads at different
heights on the same cycloid,

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and release all three from rest.

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The red bead
has the longest arc to travel.

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The yellow bead starts lower.

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The green bead begins lower
still, close to the bottom.

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Release them together.

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Their speeds and remaining distances
are different throughout the descent,

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yet all three reach
the bottom at the same instant.

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This is the tautochrone property.

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Tauto means the same,
and chronos means time.

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On an ideal cycloid,

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the travel time to the bottom
is independent of the release point.

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Let theta measure position from the cusp.

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The downward depth is a times
one minus cosine theta.

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The bottom occurs at theta
equals pi and depth two a.

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Let h be vertical height above the bottom.

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Subtracting the depth from two
a gives a times one plus cosine theta,

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or two a cosine squared of theta over two.

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Now let s be arc length
measured upward from the bottom.

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The cycloid's arc element is two
a sine of theta over two, d theta.

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Integrating back to pi gives
four a cosine of theta over two.

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Eliminate the cosine
between the two formulas.

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The height above the bottom
is exactly s squared divided by eight a.

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That quadratic relation is the secret.

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Gravitational potential energy is m g h.

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Because h is quadratic in s, the potential
energy is a constant times s squared.

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Differentiate the potential
with respect to s and reverse the sign

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to get the force.

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Newton's law gives
an acceleration proportional to minus s.

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After cancelling the mass, this is
the equation of simple harmonic motion.

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Its angular frequency
is square root of g over four a.

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A bead released from rest begins
at one extreme of that oscillation.

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The bottom is the equilibrium point,
one quarter of a period later.

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One quarter period is pi over two omega,

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which simplifies to pi
times square root of a over g.

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Notice what is missing:
the starting arc length s zero.

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The higher bead travels farther,

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but it also begins
with a larger restoring acceleration.

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The lower bead has less distance
but receives a weaker pull.

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The cycloid balances
those effects exactly.

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Return the beads
to their three starting places.

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Their different amplitudes
are now visible again.

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Release them once more.

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One shared harmonic clock
carries every amplitude

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to zero at the same quarter period, so the
three beads meet at the bottom together.

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The brachistochrone and tautochrone
are therefore two faces of the same curve.

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Between fixed endpoints, the cycloid
minimizes gravitational travel time.

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Along one cycloidal bowl, every
release height shares one arrival time.