I'm interested in finding geodesics of particular revolution surfaces,
I've not studied "Clairaut theorem" or "Clairaut's relation" in course ,
However, a relation is given on Wikipedia :
I have to say that for the moment I don't really understand how to easily extract Clairaut's relation from a particular parametrization of a surface of revolution, so let's choose the example of the cathenoid, with a choosen parametrization
$$(r,\theta)\rightarrow (\cosh(r)\cos(\theta),\cosh(r)\sin(\theta),r).$$
On a question I link, the author says the translation from the Clairaut's relation would be
$$\cosh(r)^2 \dot \theta \equiv c$$
However I don't really understand how to find it trivially.
Best Answer
Not giving here how Clairaut's Law is found. Consult any text book on differential geometry for it.
You are not supposed to find general laws from particular situations.
A way to find geodesics on a catenoid is given here by application of this law.
Angle $ \psi$ is between meridian and geodesic, $\phi$ is slope; Using cylindrical coordinates for catenoid $ (r,\theta,z);$ In cartesian coordinates $z$ is from $r= c \cosh \frac{z}{c},$ and $ (x,y)= ( r \cos \theta, r \sin \theta) $.
Clairaut's Law: At $ r=r_{min}$ geodesics are tangential.
$$ r^2 \frac{d \theta}{ds}=r \sin \psi = r _{min} \tag1 $$
$$ \cot \psi= \sqrt{\frac{r^2}{r_{min}^2}-1} \tag2$$
Draw a differential triangle in tangent plane
$$ \frac{dr}{\sin \phi }= r \;d \theta \;cot \psi \tag 3 $$
Meridian of minimum catenoid radius $r=c$.
$$ r =c \cosh {\frac{z}{c}}; \; \tan \phi =\frac{dr}{dz} =\sqrt{\frac{r^2}{c^2}-1}\tag 4 $$
Eliminate $(\phi,\psi)$ combining all these and integrate to get polar projection/ view of the geodesic:
$$ \frac{dr}{d \theta}= \frac{\sqrt{(r^2-r_{min}^2) (r^2-c^2)}}{r_{min}}. \tag 5$$
$r_{min}$ can be greater,less or equal to $c$. There would be two or one cover of geodesics.