Equation of Catenary/Formulation 1/Proof

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

The is worked by using Primitive 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:
 * $y = \dfrac {e^{a x} + e^{-a x}} {2 a}$

which is what we wanted to show.