Equation of Catenary/Formulation 1/Proof

Proof
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.


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

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

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

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

The is worked by using Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$:
 * $\displaystyle \int \frac {\d p} {\sqrt {1 + p^2} } = \map \ln {\sqrt {1 + p^2} + p} + c_1$

The is worked by using Primitive of Constant:
 * $\displaystyle \int a \rd x = a x + c_2$

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

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

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

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.