<s>[INST] Loxodromics (curves with fixed angle with meridians) of a revolution surface I am trying to find the curves with a fixed angle $\phi$ with meridians for the revolution surface given by the revolution of the graph of $z=\frac{1}{x}$ around the $z$ axis. The surface is easily parametrized as
$$(x,\theta)\mapsto(x \cos \theta, x\sin \theta,\frac{1}{x}) \qquad x\in \mathbb R_{>0},\theta \in (0,2\pi)$$
Now, some tedious computations (I hope they are correct) show that the first and the second fundamental form for this parametrization are
$$G_{I}=\begin{pmatrix}1+\frac{1}{x^4} &0 \\ 0 & x^2 \end{pmatrix} \qquad G_{II}=\frac{1}{\sqrt{x^4+1}}\begin{pmatrix}\frac{1}{x} & 0 \\ 0 & -x \end{pmatrix}$$
Recall that the angle $\phi$ between two $C^1$ curves $\gamma_1,\gamma_2$ is given by
$$\cos \phi=\frac{\gamma_1'^TG_I \gamma_2'}{\sqrt{\gamma_1'^T G_I \gamma_1'}\sqrt{\gamma_2'^TG_I\gamma_2'}}$$
Since we are looking for curves $(x(t),\theta(t))^T$ with constant angle with meridians, we want that
$$\frac{\begin{pmatrix}1 & 0 \end{pmatrix}\begin{pmatrix} 1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t)\end{pmatrix}\begin{pmatrix} x'(t) \\ \theta'(t) \end{pmatrix}}{\sqrt{\begin{pmatrix} x(t) & \theta(t)\end{pmatrix}\begin{pmatrix}1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t) \end{pmatrix}\begin{pmatrix} x'(t)\\ \theta'(t) \end{pmatrix}}\sqrt{\begin{pmatrix} 1 & 0\end{pmatrix}\begin{pmatrix}1+\frac{1}{x^4(t)} &0 \\ 0 & x^2(t) \end{pmatrix}\begin{pmatrix} 1 \\ 0\end{pmatrix}}}=K$$
which, dropping the dependency on $t$, yields the differential equation
$$x'\left (1+\frac{1}{x^4}\right )=K\sqrt{1+\frac{1}{x^4}}\sqrt{x'^2\left ( 1+\frac{1}{x^4}\right)+x^2\theta'^2}$$
Not looking very good. Through a few manipulations, I brought this to the form
$$x'(t)=K\sqrt{x'^2+\frac{x^6\theta'^2}{x^4+1}}$$
Since this is an exercise from an exam, I would expect the solution to have a nice closed form, but maybe I am mistaken. 
Is there any elementary way of solving such differential equation or of finding an explicit equation for the loxodromics?
 [/INST] After Futurologist's answer, you just need $$I=\int \frac{\sqrt{x^4+1}}{x^3}dx=\int \frac{\sqrt{x^4+1}}{x^4}x\,dx$$ So, define $$x^2=y\implies x=\sqrt y\implies dx=\frac{dy}{2 \sqrt{y}}$$ which make $$I=\frac 12\int \frac{\sqrt{y^2+1}}{ y^2}\,dy=\frac{1}{2} \left(\sinh ^{-1}(y)-\frac{\sqrt{y^2+1}}{y}\right)$$
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