Theorem: Equation of Catenary/Cartesian/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 $\tuple {0, \dfrac 1 a}$.

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 mass 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.
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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 Solution to Separable Differential Equation:
:$(2): \quad \ds \int \frac {\d p} {\sqrt {1 + p^2} } = \int a \rd x$
The {{LHS}} 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 {{RHS}} 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:
{{begin-eqn}}
{{eqn | l = a x
      | r = \map \ln {\sqrt {1 + p^2} + p}
      | c = 
}}
{{eqn | ll = \leadsto
      | l = e^{a x}
      | r = \sqrt {1 + p^2} + p
      | c = Exponential of Natural Logarithm
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = -a x
      | r = -\map \ln {\sqrt {1 + p^2} + p}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = e^{- a x}
      | r = \dfrac 1 {\sqrt {1 + p^2} + p}
      | c = Exponential of Natural Logarithm and Logarithm of Reciprocal
}}
{{eqn | r = \dfrac 1 {\sqrt {1 + p^2} + p} \paren {\dfrac {\sqrt {1 + p^2} - p} {\sqrt {1 + p^2} - p} }
      | c = multiplying by $1$
}}
{{eqn | r = \dfrac {\sqrt {1 + p^2} - p} {1 + p^2 - p^2}
      | c = 
}}
{{eqn | r = \sqrt {1 + p^2} - p
      | c = 
}}
{{end-eqn}}
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:
:$y = \dfrac {e^{a x} + e^{-a x} } {2 a}$
{{qed}}