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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?

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Well, actually you can also use the vector $(0,1)$, since the tangent vectors to the coordinate lines $x$ and $\theta$ are orthogonal at each point. Then your curve measures a constant angle $\alpha_0$ with the tangent $x-$vector if and only if it also measures a constant angle $\pi/2 - \alpha_0$ with the tangent $\theta-$vector, so you can go this way too. But anyway, the result is going to be virtually the same. You end up with the dot product expression (i.e. the dot product formula $(a \cdot b) = |a||b|\cos{\alpha_0} \,\,$): \begin{align} (1,0) \,\, G_I(x) \, \, (x',\theta')^T &= \sqrt{\Big((1,0) \,\, G_I(x) \,\, (1,0)^T\Big)} \sqrt{\Big((x',\theta') \,\, G_I(x) \,\, (x',\theta')^T\Big)} \, \cos{(\alpha_0)} \\ \left(1+ \frac{1}{x^4}\right) x' &= \cos{(\alpha_0)}\sqrt{\left(1+ \frac{1}{x^4}\right)} \sqrt{\left(1+ \frac{1}{x^4}\right)(x')^2 + x^2 (\theta')^2} \end{align} Notice that all expressions, apart from the derivative $\theta'$, depend only on the variable $x$! This is not a coincidence as the surface is a surface of revolution and is therefore rotatiationally invariant with respect to rotations around the $z$ axis, which in the $x,\theta$ coordinates manifest themselves as translations of the $\theta$ coordinate, i.e. $\theta \mapsto \theta + c$. Consequently, when you have one curve on the surface of constant angle $\alpha_0$ with respect to the $x-$coordinate curves, if you rotate it around $z$, you get again a curve on the surface of the same constant angle $\alpha_0$ with respect to the $x-$coordinate curves. Therefore, your equations for the curves of angle $\alpha_0$ should be invariant with respect to the symmetries $\theta \mapsto \theta + c$, which means that the equations do not depend explicitly on $\theta$. Thus, we can simply write the curve as $\theta = \theta(x)$, i.e. it is parametrized as $x \mapsto (x, \theta(x))$, so $x' = 1$ and $\theta' = \frac{d\theta}{dx}$. So the equation becomes \begin{align} \left(1+ \frac{1}{x^4}\right) &= \cos{(\alpha_0)}\sqrt{\left(1+ \frac{1}{x^4}\right)} \sqrt{\left(1+ \frac{1}{x^4}\right) + x^2 \Big(\frac{d\theta}{dx}\Big)^2};\\ \sqrt{1+ \frac{1}{x^4}} &= \cos{(\alpha_0)} \sqrt{\left(1+ \frac{1}{x^4}\right) + x^2 \Big(\frac{d\theta}{dx}\Big)^2};\\ {1+ \frac{1}{x^4}} &= \cos^2{(\alpha_0)}\left( 1+ \frac{1}{x^4}+ x^2 \Big(\frac{d\theta}{dx}\Big)^2\right);\\ \frac{1}{\cos^2{(\alpha_0)}}\left({1+ \frac{1}{x^4}}\right) &= 1+ \frac{1}{x^4}+ x^2 \Big(\frac{d\theta}{dx}\Big)^2;\\ x^2 \Big(\frac{d\theta}{dx}\Big)^2 &= \left(\frac{1}{\cos^2{(\alpha_0)}} - 1\right)\left(1+ \frac{1}{x^4}\right) \\ \Big(\frac{d\theta}{dx}\Big)^2 &= \tan^2{(\alpha_0)}\,\frac{1}{x^2}\left(1+ \frac{1}{x^4}\right) = \tan^2{(\alpha_0)}\,\left(\frac{x^4+1}{x^6}\right);\\ \frac{d\theta}{dx} &= \pm \tan{(\alpha_0)}\,\left(\frac{\sqrt{x^4+1}}{x^3}\right); \end{align} So as you can see, there is almost no actual differential equation, the problem is solved by simply explicitly integrating a derivative \begin{align} \theta(x) &= \theta_0 \pm k_0 \,\int \frac{\sqrt{x^4+1}}{x^3}dx, \end{align} where $k_0 = \tan{\alpha_0}$.

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  • $\begingroup$ Great answer, thank you very much. $\endgroup$
    – Lonidard
    Aug 19, 2016 at 13:54
  • $\begingroup$ You are welcome. I just want to add for future reference that if you have a surface of revolution and you want to find a family of curves defined by a condition that is invariant with respect to rotations around the $z$ axis, then the family of curves will be also invariant and thus the corresponding differential equation will be invariant with respect to these rotations and so it would not depend on $\theta$ explicitly, i.e. it will be of the form $F\Big(x, \frac{d\theta}{dx}\Big) = 0$. After you solve for $ \frac{d\theta}{dx}$ you get $ \frac{d\theta}{dx} = f(x)$. $\endgroup$ Aug 19, 2016 at 17:30
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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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