Brachistochrone Problem

## **Problem**

The **brachistochrone problem** (Ancient Greek: βράχιστος + χρόνος *(brákhistos + khrónos),* "shortest time"), is a classic mechanics problem posed by Johann Bernoulli (1667 – 1748) in 1696.

Suppose a wire in a vertical plane connects two points, point A and point B. A bead with mass is placed at point A and slides without friction down the wire to point B under the influence of gravity.

**Part 1:** Show that the shape of the wire so that the bead travels from point A to point B in the least time satisfies the differential equation

d
y

d
x

=

2
a
−
y

y

.

{\displaystyle \frac{dy}{dx} = \sqrt{\frac{2a-y}{y}}. }

.
**Part 2:** Show that the inverted cycloid

{

x
(
θ
# )
a
(
1
−
sin
⁡
θ
)

y
(
θ
# )
a
(
θ
−
cos
⁡
θ
)

{\displaystyle \begin{cases}
x(\theta) = a(1 - \sin \theta) \\[5 pt]
y(\theta) = a(\theta - \cos \theta)
\end{cases}}

satisfies the differential equation in Part 1.

## **Solution**

**Part 1**

We assume that at any given time

t

{\displaystyle t}

 the bead is a point

(
x
,
y
)

{\displaystyle (x,y)}

 on an arc

s

{\displaystyle s}

. For convenience, let point A be

A
(
0
,
0
)

{\displaystyle A(0,0)}

 and

B
(

x
2

,

y
2

)

{\displaystyle B(x\_2,y\_2) }

. By the conservation of energy, while the bead slides down the wire, we get

K

E
1

P

E
1

=
K

E
2

P

E
2

0
0

=

1
2

m

(

d
s

d
t

)

2

−
m
g
y
.

{\displaystyle \begin{align}
KE\_1 + PE\_1 &= KE\_2 + PE\_2\\[5 pt]
0 + 0 &= \frac{1}{2}m{\left(\frac{ds}{dt}\right)}^{2} - mgy.
\end{align}}

Simplify and isolate for the velocity

d
s

d
t

=

2
g
y

.

{\displaystyle \frac{ds}{dt} = \sqrt{2gy}. }

Thus, the time for the bead to slide from point A to point B is

∫

0

T

d
# t

∫

0

x
2

1

2
g
y

d
s
.

{\displaystyle \int\_{0}^{T} dt = \int\_{0}^{x\_2} \frac{1}{\sqrt{2gy}}\, ds. }

Writing the arclength element as

d
# s

1

y
′

2

d
x

{\displaystyle ds = \sqrt{1+{y'}^{2}}\,dx }

, we get

∫

0

T

d
# t

1

2
g

∫

0

x
2

1

y
′

2

y

d
x
.

{\displaystyle \int\_{0}^{T} dt = \frac{1}{\sqrt{2g}}\int\_{0}^{x\_2} \frac{\sqrt{1+{y'}^{2}}}{\sqrt{y}}\, dx. }

The minimum time is determined when the integral is a minimum; hence, we can use the Beltrami identity for this optimization problem. Letting

# L

1

y
′

2

y

{\displaystyle L = \frac{\sqrt{1+{y'}^{2}}}{\sqrt{y}} }

 and using the Beltrami identity

L
−

y
˙

∂
L

∂

y
˙

=
C

{\displaystyle L - \dot{y} \frac{\partial L}{\partial \dot{y}} = C }

we find that

1

y
′

2

y

−

y
′

y
′

1

y
′

2

y

=
C

{\displaystyle \frac{\sqrt{1+{y'}^{2}}}{\sqrt{y}} - y'\frac{y'}{\sqrt{1+{y'}^{2}}\sqrt{y}} = C }

which simplifies to

y

1

y
′

2

=

1
C

.

{\displaystyle \sqrt{y} \sqrt{1+{y'}^{2}} = \frac{1}{C}.}

Let

2
# a

1

c
2

{\displaystyle 2a = \frac{1}{c^2} }

, then we can rewrite the above expression as

y
′

=

2
a
−
y

y

[
5
p
t
]

d
y

d
x

=

2
a
−
y

y

.

{\displaystyle \begin{align}
y' &= \sqrt{\frac{2a-y}{y}} \\ [5 pt]
\frac{dy}{dx} &= \sqrt{\frac{2a-y}{y}}.
\end{align} }

**Part 2**

To verify that the parametric function of the cycloid satisfies the differential equation in Part 1, we need to check if the left and right side of the differential equation agree. For parametric functions, the derivative on the right is calculated using the formula

d
y

d
x

=

d
y

/

d
θ

d
x

/

d
θ

{\displaystyle \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}}

.

The parametric equations of cycloid is:

{

x
(
θ
# )
a
(
1
−
sin
⁡
θ
)

y
(
θ
# )
a
(
θ
−
cos
⁡
θ
)
.

{\displaystyle \begin{cases}
x(\theta) = a(1 - \sin \theta) \\[5 pt]
y(\theta) = a(\theta - \cos \theta).
\end{cases}}

Hence the derivatives with respect to

θ

{\displaystyle \theta}

 are

d
x

d
θ

=
a
(
1
−
cos
⁡
θ
)

d
y

d
θ

=
a
sin
⁡
θ
.

{\displaystyle \begin{align}
\frac{dx}{d\theta} &= a(1 - \cos \theta) \\[5 pt]
\frac{dy}{d\theta} &= a\sin \theta.
\end{align}}

Consequently, we get

d
y

d
x

=

a
sin
⁡
θ

a
(
1
−
cos
⁡
θ
)

=

sin
⁡
θ

1
−
cos
⁡
θ

.

{\displaystyle \begin{align}
\frac{dy}{dx} &= \frac{a\sin \theta}{a(1-\cos \theta)} \\[5 pt]
&= \frac{\sin \theta}{1-\cos \theta}.
\end{align}}

Direct substitution to the right hand side of the differential equation yields:

2
a
−
y

y

=

2
a
−
a
(
1
−
cos
⁡
θ
)

a
(
1
−
cos
⁡
θ
)

[
5
p
t
]

=

1
cos
⁡
θ

1
−
cos
⁡
θ

.

{\displaystyle \begin{align}
\sqrt{\frac{2a-y}{y}} &= \sqrt{\frac{2a - a(1-\cos \theta)}{a(1-\cos \theta)}} \\ [5 pt]
&= \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}.
\end{align} }

For now, both sides of the differential equation appears completely different. Nevertheless, if one uses the trigonometric identities

sin
⁡
# x
2
sin
⁡

x
2

cos
⁡

x
2

{\displaystyle \sin x = 2\sin \frac{x}{2} \cos \frac{x}{2} }

,

1
−
cos
⁡
# x
2

sin
2

⁡

x
2

{\displaystyle 1 - \cos x = 2 \sin^2 \frac{x}{2} }

, and

1
cos
⁡
# x
2

cos
2

⁡

x
2

{\displaystyle 1 + \cos x = 2 \cos^2 \frac{x}{2} }

, then both sides can be simplified to

cot
⁡

x
2

{\displaystyle \cot \frac{x}{2} }

!

Behold, the inverted cycloid is the brachistochrone curve!