Equation of Catenary

Curve
Consider a flexible chain of uniform linear density hanging from two points under its own weight.

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

Then the shape of the chain describes a curve given by the equation:
 * $y = \dfrac {e^{ax} + e^{-ax}} {2 a}$

where $a$ is a constant.

The lowest point of the chain is at $\left({0, \dfrac 1 a}\right)$.

This curve is called a catenary.

Alternative formulation
Putting $\dfrac 1 a$ for $a$, we have:


 * $y = \dfrac a 2 \left({e^{\frac x a} + e^{- \frac x a}}\right) = a \cosh \dfrac x a$

where the lowest point of the chain is at $\left({0, a}\right)$.

Proof
Let $\left({x, y}\right)$ be an arbitrary point on the chain.

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

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 $\left({x, y}\right)$ 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 $\left({x, y}\right)$
 * 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.


 * [[File:Catenary.png]]

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 $a = w_0 / T_0$:
 * $\tan \theta = a s = \dfrac{\mathrm d y}{\mathrm d x}$

Differentiating with respect to $x$:


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

From Derivative of Arc Length, we have:
 * $\dfrac{\mathrm d s}{\mathrm d x} = \sqrt{1 + \left({\dfrac{\mathrm d y}{\mathrm d x}}\right)^2}$

So we have this differential equation to solve:


 * $(1): \quad \dfrac{\mathrm d^2 y}{\mathrm d x^2} = a \sqrt{1 + \left({\dfrac{\mathrm d y}{\mathrm d x}}\right)^2}$

Let us make the substitution $p = \dfrac{\mathrm d y}{\mathrm d x}$.

This transforms $(1)$ into:
 * $\dfrac{\mathrm d p}{\mathrm d x} = a \sqrt{1 + p^2}$

This can be solved by Separation of Variables:
 * $(2): \quad \displaystyle \int \frac {\mathrm d p}{\sqrt{1 + p^2}} = \int a \mathrm d x$

The LHS is worked by using Integral of Reciprocal of Root a x + b:
 * $\displaystyle \int \frac {\mathrm d p}{\sqrt{1 + p^2}} = \ln \left({\sqrt{1 + p^2} + p}\right) + c_1$

The RHS is worked by using Integral of Constant:
 * $\displaystyle \int a \mathrm d x = a x + c_2$

So $(2)$ becomes:


 * $\ln \left({\sqrt{1 + p^2} + p}\right) = a x + c_3$

When $x = 0$ we have that $\theta = \dfrac{\mathrm d y}{\mathrm d x} = p = 0$ and so $c_3 = 0$, so:
 * $\ln \left({\sqrt{1 + p^2} + p}\right) = a x$

After some algebra, this gives us:
 * $p = \dfrac{\mathrm d y}{\mathrm d x} = \dfrac {e^{a x} - e^{-a x}} 2$

By Derivative of Exponential Function, we get:
 * $y = \dfrac {e^{a x} + e^{-a x}} {2 a} + c_4$

All we need to do now is place the coordinate axes at just the right height so that $y = 1/a$ when $x = 0$ and we make $c_4 = 0$.

We end up with:
 * $y = \ddfrac {e^{a x} + e^{-a x}} {2 a}$

which is what we wanted to show.

We can put this in a form which uses the hyperbolic cosine:
 * $y = \dfrac {\cosh ax} a$

Replacing $a$ with $\dfrac 1 a$ gives the alternative form:


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

Historical Note
The problem of determining the shape of the catenary was posed in 1690 by as a challenge.

It had been thought by to be a parabola.

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

In 1691,, and  all independently published solutions.

Lingustic Note
The word catenary comes from the Latin word catena meaning chain.