# Equation of Catenary

## Curve

Consider a **catenary**.

Let a cartesian plane be arranged so that the y-axis passes through the lowest point of the catenary.

### Formulation 1

The **catenary** is described by the equation:

- $y = \dfrac {e^{a x} + e^{-a x} } {2 a} = \dfrac {\cosh a x} a$

where $a$ is a constant.

The lowest point of the catenary is at $\tuple {0, \dfrac 1 a}$.

### Formulation 2

The **catenary** is described by the equation:

- $y = \dfrac a 2 \paren {e^{x / a} + e^{-x / a} } = a \cosh \dfrac x a$

where $a$ is a constant.

The lowest point of the chain is at $\tuple {0, a}$.

This curve is called a **catenary**.

## Proof of Formulation 1

Let $\tuple {x, y}$ be an arbitrary point on the chain.

Let $s$ be the length along the arc of the chain from the lowest point to $\tuple {x, y}$.

Let $w_0$ be the linear density of the chain, that is, its weight per unit length.

The section of chain between the lowest point and $\tuple {x, y}$ is in static equilibrium under the influence of three forces, as follows:

- The tension $T_0$ at the lowest point
- The tension $T$ at the point $\tuple {x, y}$
- The weight $w_0 s$ of the chain between these two points.

As the chain is (ideally) flexible, the tension $T$ is along the line of the chain, and therefore along a tangent to the chain.

We can resolve this system of forces to obtain the horizontal and vertical components:

- $T_0 = T \cos \theta$
- $w_0 s = T \sin \theta$

We divide one by the other to eliminate $T$ and set $a = w_0 / T_0$:

- $\tan \theta = a s = \dfrac {\d y} {\d x}$

Differentiating with respect to $x$:

- $\dfrac {\d^2 y} {\d x^2} = a \dfrac {\d s} {\d x}$

From Derivative of Arc Length, we have:

- $\dfrac {\d s} {\d x} = \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$

So we have this differential equation to solve:

- $(1): \quad \dfrac {\d^2 y} {\d x^2} = a \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$

Let $p = \dfrac {\d y} {\d x}$.

This transforms $(1)$ into:

- $\dfrac {\d p} {\d x} = a \sqrt {1 + p^2}$

This can be solved by Separation of Variables:

- $(2): \quad \ds \int \frac {\d p} {\sqrt {1 + p^2} } = \int a \rd x$

The left hand side is worked by using Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$:

- $\ds \int \frac {\d p} {\sqrt {1 + p^2} } = \map \ln {\sqrt {1 + p^2} + p} + c_1$

The right hand side is worked by using Primitive of Constant:

- $\ds \int a \rd x = a x + c_2$

So $(2)$ becomes:

- $\map \ln {\sqrt {1 + p^2} + p} = a x + c_3$

When $x = 0$ we have that $\theta = \dfrac {\d y} {\d x} = p = 0$ and so $c_3 = 0$, so:

- $\map \ln {\sqrt {1 + p^2} + p} = a x$

Therefore:

\(\ds a x\) | \(=\) | \(\ds \map \ln {\sqrt {1 + p^2} + p}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds e^{a x}\) | \(=\) | \(\ds \sqrt {1 + p^2} + p\) | Exponential of Natural Logarithm |

and:

\(\ds -a x\) | \(=\) | \(\ds -\map \ln {\sqrt {1 + p^2} + p}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds e^{- a x}\) | \(=\) | \(\ds \dfrac 1 {\sqrt {1 + p^2} + p}\) | Exponential of Natural Logarithm and Logarithm of Reciprocal | ||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {1 + p^2} + p} \paren {\dfrac {\sqrt {1 + p^2} - p} {\sqrt {1 + p^2} - p} }\) | multiplying by $1$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {\sqrt {1 + p^2} - p} {1 + p^2 - p^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \sqrt {1 + p^2} - p\) |

which gives us:

- $p = \dfrac {\d y} {\d x} = \dfrac {e^{a x} - e^{-a x} } 2$

By Derivative of Exponential Function, we get:

- $y = \dfrac {e^{a x} + e^{-a x} } {2 a} + c_4$

We now need to position the coordinate axes so as to make $c_4 = 0$.

So, setting $x = 0$ and $c_4 = 0$:

- $y = \dfrac 1 a$

and the lowest point of the catenary is seen to be at $\tuple {0, \dfrac 1 a}$.

Hence the result:

- $y = \dfrac {e^{a x} + e^{-a x} } {2 a}$

$\blacksquare$

## Proof of Formulation 2

Take the equation of the catenary according to Formulation 1:

- $y = \dfrac {e^{ax} + e^{-ax}} {2 a}$

Put this in a form which uses the hyperbolic cosine:

- $y = \dfrac {\cosh a x} a$

Replace $a$ with $\dfrac 1 a$:

- $y = a \cosh \dfrac x a$

Hence the result.

$\blacksquare$

## Historical Note

The problem of determining the shape of the catenary was posed in $1690$ by Jacob Bernoulli as a challenge.

It had been thought by Galileo to be a parabola.

Huygens showed in $1646$ by physical considerations that it could not be so, but he failed to establish its exact nature.

In $1691$, Leibniz, Huygens and Johann Bernoulli all independently published solutions.

It was Leibniz who gave it the name **catenary**.

From a letter that Johann Bernoulli wrote in $1718$:

*The efforts of my brother were without success. For my part, I was more fortunate, for I found the skill (I say it without boasting; why should I conceal the truth?) to solve it in full ... It is true that it cost me study that robbed me of rest for an entire night. It was a great achievement for those days and for the slight age and experience I then had. The next morning, filled with joy, I ran to my brother, who was still struggling miserably with this Gordian knot without getting anywhere, always thinking like Galileo that the catenary was a parabola. Stop! Stop! I say to him, don't torture yourself any more trying to prove the identity of the catenary with the parabola, since it is entirely false.*

However, Jacob Bernoulli was first to demonstrate that of all possible shapes, the catenary has the lowest center of gravity, and hence the smallest potential energy.

This discovery was significant.

## Linguistic Note

The word **catenary** comes from the Latin word **catena** meaning **chain**.

## Sources

- 2008: Ian Stewart:
*Taming the Infinite*... (previous) ... (next): Chapter $8$: The System of the World: The English get left behind