Calculus of Variations:
Functionals
Principle of Least Action
The shortest path
The brachistochrone curve

2. Equations of the brachistochrone curve

We will first express the travel time between A and B, then look for the curve, that is the shape of the slide, that minimizes this time.

Let s the curvilinear abscissa. An element ds of the curve is expressed in cartesian coordinates as

ds = [1 + (dy/dx)²]^1/2 dx = [1 + y'²]^1/2 dx

ds/dx = [1 + (dy/dx)²]^1/2 dx = [1 + y'²]^1/2

The speed of the object is given by v = ds/dt.

The conservation of the mechanical energy at a point (x,y) is written as:

(1/2)mv² + mgy = mgh

That gives:

v = [2g(h - y)]^1/2

v = ds/dt = [2g(h - y)]^1/2

Let express the travel time T[y]. It is a functional that depends on the function y(x) of the bead. We have

dx/ds = (dx/dt) (dt/ds).

Therefore

dt/dx = (dt/ds)/(dx/ds) = (1/v)/(dx/ds).

Then

dt = (1/v) (ds/dx) dx = [1 + y'²]^1/2/[2g(h - y)]^1/2 dx
= [(1 + y'²)/2g(h - y)]^1/2

dt = [(1 + y'²)/2g(h - y)]^1/2 dx

The travel time T[y] is then written :

T[y] = ∫₀^d f(y, y')dx

Where f(y,y') = [(1 + y'²)/2g(h - y)]^1/2.
f(y,y') does not depend on x.


We search them te function y(x) that minimizes the fonctional T[y].

Since f does not depend on x, we can then use the Euler-Lagrange simplified (Beltrami formula):

f - y'(∂f/∂y') = constant.

Let constant = C1. First, let's calculate the partial derivative ∂f/∂y':

∂f/∂y' = (1/2) (1 + y'²)^-1/2 2y'/[2g(h - y)]^1/2 =
y'/[2g(h - y) (1 + y'²)]^1/2

∂f/∂y' = y'/[2g(h - y) (1 + y'²)]^1/2

Then:

f - y'(∂f/∂y') = (1 + y'²)/2g(h - y)^1/2 - y'²/[2g(h - y) (1 + y'²)]^1/2 =
(1 + y'²)^1/2 - y'²/(1 + y'²)^1/2 = C1 . [2g(h - y)]^1/2

Hence:

1/(1 + y'²)^1/2 = C1. [2g(h - y)]^1/2

Squaring, the two members of the equality, we obtain:

1 + y'² = C2 /2g(h - y) = C2/(h - y) , with C2 = C1²

Let the constant R = C2/2g, then

1 + y'² = R/(h - y)

Let y' = tan(θ/2). So 1 + y'² = 1 + tan² (θ/2) = 1/cos²(θ/2) = R /(h - y)

Therefore

y = h - R cos² (θ/2) = h - (R/2)(1 + cos θ)

with dy/dθ = (R/2) sin θ

In addition, we have:

y' = tan (θ/2) = dy/dx = (dy/dθ)(dθ/dx) = (R/2) sin θ . (dθ/dx)

Then :

tan (θ/2) = (R/2) sin θ . (dθ/dx)

sin (θ/2) /cos (θ/2) = 2 (R/2) sin(θ/2) . cos(θ/2) . (dθ/dx)

1 /cos (θ/2) = R cos(θ/2) . (dθ/dx)

Therefore:

dx = 1 /cos (θ/2) = R cos²(θ/2) . (dθ) = (R/2) (1 + cos θ) dθ

dx = (R/2) (1 + cos θ) dθ .

Its integral is x = (R/2) (θ + sin θ) + sct

We have then:

x = (R/2) (θ + sin θ) + cst
y = h - (R/2)(1 + cos θ)

These are the equations of a cycloid.

With the change of the variable θ as θ --> θ - π
and the condition x(0) = 0 , these equations become:

x = (R/2) (θ - sin θ)
y = h - (R/2)(1 - cos θ)

We can make other conditions, by requiring that the curve pass through the point B(XB, yB), to fix the constant R.

We can also take h = 0 and direct the y axis to - y axis, and end with the two following simple equations:

x = a (θ - sin θ)
y = a (1 - cos θ)

ds² = dx² + dy² = a² [(1 - cos θ)² + sin²θ] (dθ)² =

2 a² (1 - cos θ) (dθ)² = 4 a² sin²(θ/2) (dθ)²

So ds = 2a sin(θ/2) (dθ)

ds = 2a sin(θ/2) (dθ)

Let the displacement s mesured, along the cycloid, from the bottom at &theta = π. The integration gives:

s = ∫_π^θ2a sin(θ/2) (dθ) = - 4 a cos(θ/2)

s = - 4 a cos(θ/2)

The displacement s ranges from s(0) = - 4a to s(2π) = 4 a.

The kinetic energy of the bead of mass m which slides along the cycloid is :

T = (1/2) m (ds/dt)²

T = (m/2) (ds/dt)²

The gravitational potential energy of the bead , mesured from the bottom of the cycloid is:

V = mg(2a - y) = mg( 2a - a (1 - cos θ)) = mga (1 + cos θ) =
2 m g a cos² (θ/2) = (1/2) m (g/4a) s²

V = (m/2) (g/4a) s²

The lagrangian is :

L = T - V = (m/2) (ds/dt)² - (m/2)(g/4a) s²

L = (m/2) [(ds/dt)² - (g/4a) s²]

The lagrage's equation, with v = ds/dt, is written as:

d[∂L/∂v]/dt = ∂L/∂s .

We have: :

∂L/∂v = (m/2) 2 (ds/dt). So
d[∂L/∂v] = m d²s/dt², and
∂L/∂s = (m/2)2 (- (g/4a) s) = - m (g/4a) s

Thus

d²s/dt² + (g/4a) s = 0

d²s/dt² + (g/4a) s = 0

The time to fall from the point A to the point B on the cycloid is independent of the starting point A.

We the following two parametric equations of the cycloid:

x = a(θ - sin θ)
y = a(1 - cos θ)

The correspondant derivative are:

dx = a(1 - cosθ) dθ
dy = a sin θ dθ

so

ds² = dx² + dy² = 2a² (1 - cos θ) (dθ)²

• The origine of potential energy at y = 0:

At any point M(x, y) on the curve, using the concervation of energy law , we obtain

(1/2) m v² = m g y

Since v = ds/dt,
dt = ds/v = a (√(2(1 - cos θ)) dθ)/(√(2gy)) =
a (√(2(1 - cos θ)) dθ)/(√(2ga(1 - cosθ))) = √(a/g) dθ

dt = √(a/g) dθ

Therefore, the time to travel to the point A(θ = 0) to the point B(θ = π) is

T = ∫₀^π dt = π √(a/g)

T = π √(a/g)

• The origine of potential energy at y = yo:

At any other point, say M(x, Y = y - yo), we will have :

y = a(1 - cos θ)
yo = a(1 - cos θo)

y - yo = cos θo - cos θ

(1/2) m v² = m g Y = m g (y - yo) = mga (cos θo - cos θ)

dt = ds/v = a (√(2(1 - cos θ)) dθ)/(√(2gY)) =
a (√(2(1 - cos θ)) dθ)/(√(2ga(cos θo - cos θ))) =
√(a/g) √ [(1 - cos θ)/(cos θo - cos θ)] dθ

dt = √(a/g) √ [(1 - cos θ)/(cos θo - cos θ)] dθ

Therefore, the time to travel to the point C(θ = θo)
to the point B(θ = π) is

T = ∫₀^π dt = √(a/g) ∫_θo^π √ [(1 - cos θ)/(cos θo - cos θ)] dθ

Substituting the following trigonometric formulas:

cos (θ/2) = √[(1 + cos θ)/2],
sin (θ/2) = √[(1 - cos θ)/2],

The integral becomes:

T = ∫₀^π dt = √(a/g) ∫_θo^π √ [(sin (θ/2))/(cos (θo/2) - cos (θ/2)] dθ

Using the following change of variables :

u = cos(θ/2)/cos(θo/2) ,so
du = - (1/2) dθ sin(θ/2)/cos(θo/2)

Therefore:

T = 2 √(a/g) ∫₀¹ du/√(1 - u)² = 2 √(a/g) [Arcsin (u)]₀¹ = π √(a/g).

T = π √(a/g)

Conclusion:

The time is the same from any starting point in the cycloid.