I've come across something in my notes that jumps from:
$${d\rho \over dz} = \sqrt{\left({\rho \over C}\right)^2 - 1}$$
to:
$$\rho(z) = C \cosh\frac{z-z_0}{C}$$
I know that separation of variables is used so what I'm really asking is how to integrate this part:
$$\int{1 \over \left(\left(\dfrac{\rho}{C}\right)^2 - 1\right)^{1/2}} \, d\rho ,$$
where $\rho$ is a function of $z$ and $C$ is a constant.
Help with this one would be greatly appreciated, thanks!
\begin{align} \frac \rho C & = \sec\theta \\[10pt] \sqrt{\left(\frac\rho C\right)^2 -1} & = \sqrt{\sec^2\theta-1}= \tan\theta \\[10pt] \frac{d\rho}{C} & = \sec^2\theta\,d\theta \end{align}
\begin{align} \int{1 \over \sqrt{\left(\dfrac{\rho}{C}\right)^2 - 1}} \, d\rho & = \int\frac{1}{\tan\theta} \Big( C\sec^2\theta\,d\theta\Big) = C\int \frac{d\theta}{\sin\theta\cos\theta} \\[15pt] & = C\int\frac{2\,d\theta}{2\sin\theta\cos\theta} = C\int\frac{2\,d\theta}{\sin(2\theta)} = C\int\frac{d\eta}{\sin\eta} \\[15pt] & = -C\log(\csc\theta+\cot\theta)+\text{constant}. \end{align}
\begin{align} \csc\operatorname{arcsec}\frac\rho C & = \frac{\rho}{\sqrt{\rho^2-C^2}} \\[15pt] \cot\operatorname{arcsec}\frac\rho C & = \frac{C}{\sqrt{\rho^2-C^2}} \end{align}
So you get $$ -C\log\frac{\rho+C}{\sqrt{\rho^2-C^2}} = z + \text{constant} = z - z_0. $$
Consequently $$ \frac{\rho+C}{\sqrt{\rho^2-C^2}} = \exp\left(- \frac{z-z_0}{C} \right) = A $$ $$ \rho+C = A \sqrt{\rho^2-C^2} $$ Now if you square both sides you get an equation that is quadratic in $\rho$. $$ \rho^2+2\rho C+C^2 = A^2\rho^2 - A^2 C^2 $$ $$ \rho^2(1-A^2)+2\rho C + C^2(1+A^2) = 0 $$ Now apply the "quadratic formula": $$ \rho = \frac{-2C \pm \sqrt{4C^2 - 4(1-A^4)}}{2(1-A^2)} = \frac{-C\pm\sqrt{C^2+A^4-1}}{1-A^2}. $$
Notice that \begin{align} 1-A^2 & = 1 - \exp\left( -2\frac{z-z_0}{C} \right) \\[10pt] & = 2\exp\left( -\frac{z-z_0}{C} \right)\left( \frac{\exp\left( +\frac{z-z_0}{C}\right) - \exp\left( -\frac{z-z_0}{C} \right)}{2} \right) \\[10pt] & = 2\exp\left( -\frac{z-z_0}{C} \right) \cosh\left(\frac{z-z_0}{C}\right). \end{align}
(Maybe I'll be back later; I may want to check details . . . )
[ . . . ]
