Gaussian Curvature of Pseudosphere I have the parametrization
$x(u,v)=(\cos u \sin v, \sin u \sin v , \cos v+\log (\tan {v/2}))$
with $0<v<\pi $ , $0<u<2\pi$.
From this parametrization, how can I compute (optimally) the Gaussian curvature?
I know for example that the pseudosphere is a revolution surface, then there should exist a more easy way to calculate its curvature.
Thanks!
 A: Take $\partial_1=\frac{\partial x}{\partial u}$ and $\partial_2=\frac{\partial x}{\partial v}$ as the tangent frame. Then the normal will be $N=\frac{\partial_1\times\partial_2}{||\partial_1\times\partial_2||}$.
Now the derivatives $D_{\partial_1}N=\frac{\partial N}{\partial u}$ and $D_{\partial_2}N=\frac{\partial N}{\partial v}$ are going to be tangent too, so you will get a base change
$$D_{\partial_1}N=A\partial_1+B\partial_2$$
$$D_{\partial_2}N=C\partial_1+D\partial_2$$
for some scalars $A,B,C,D$. Then the determinant $AD-BC$ is the Gaussian curvature.
A: A simplified procedure is through its meridian (of its surface of revolution). We set $u=0$ in its parametrization ( what was there  before rotation).
$$ y= \sin v, z= \cos v+\log (\tan {v/2}) $$
I prefer to linearly dimension the co-ordinates:
$$ y= a \sin v, z= a( \cos v+\log (\tan {v/2})) \tag1 $$
$$\dfrac{dy}{dz}=\dfrac{dy/dv}{dz/dv}= \dfrac 
{a \cos v }{a(-\sin v + \csc v )}=-\tan v = \tan \phi $$
Thus slope of meridian is $ \; v = -\phi$
Principal curvature are $$ \quad k_2= \dfrac{\cos \phi }{y}= \dfrac{\cot v}{a} $$
$$k_1=\dfrac{d\phi}{ds} =  \dfrac{-dv}{dy/\sin \phi}=\dfrac{ -\sin v \;d(\sin^{-1}y)}{dy}=\dfrac{-\sin v}{a \cos v} =\dfrac{-\tan v}{a}$$
Gauss curvature product is a negative constant
$$k_1 k_2= -1/a^2$$
where $a$ is the maximum torsion radius of pseudosphere.
From (1)  $ \dfrac{y}{\sin \phi} =a $ represents constant tangent length upto axis, a property of the Tractrix.
