Gaussian curvature of the pseudosphere

Question

I am asked to show that the pseudosphere has Gaussian Curvature $-1$ at all points. So I parametrized the tractrix as: \begin{equation} \alpha(t) = (\sin{t},\cos{t}+\log{\tan{\frac{t}{2}}}) \end{equation} So the surface of revolution would be: \begin{equation} x(\theta,t)=(\sin{t}\cos{\theta},\sin{t}\sin{\theta},\cos{t}+\log{\tan{\frac{t}{2}}}) \end{equation} By calculating the first and second fundamental forms we have that $F=f=0$ and $E=(\sin{t})^2,G=(\cot{t})^2,e=(\cos{t})^2$ and $g=-(\cot{t})^2$. So the Gaussian curvature is: \begin{equation} K = \frac{eg}{EG} = -(\cot{t})^2 \end{equation} What am I doing wrong?

Answer 1

We have \begin{align} x_\theta &= (-\sin t\sin\theta, \sin t \cos \theta, 0) = \sin t(-\sin\theta, \cos\theta, 0) \\ x_t &= (\cos t \cos\theta, \cos t \cos\theta, -\sin t + \frac{\sec^2(t/2)}{2\tan(t/2)}) = \cos t(\cos\theta, \sin\theta, \cot t), \end{align} so \begin{align} E&= \langle x_\theta, x_\theta \rangle = \sin^2 t\\ F&= 0 \quad\text{[surface of revolution]}\\ G&= \langle x_t, x_t \rangle = \cos^2 t (1+\cot^2t) = \cot^2t. \end{align} So far so good. Now $x_\theta \times x_t$ is parallel to $$ (-\sin\theta, \cos\theta, 0) \times(\cos\theta,\sin\theta,\cot t) = (\cos\theta\cot t, \sin\theta\cot t, -1) $$ so we can (after normalizing to make it a unit vector, i.e. multiplying by $\sin t$) choose the Gauss map to be $$ N = (\cos\theta\cos t, \sin\theta\cos t, -\sin t). $$ This gives \begin{align} N_\theta &= (-\sin\theta\cos t, \cos\theta\cos t, 0)\\ N_t &= (-\cos\theta\sin t, -\sin\theta\sin t, -\cos t), \end{align} so that \begin{align} e &= - \langle N_\theta, x_\theta \rangle = -\sin t \cos t\\ f &= 0 \quad\text{[surface of revolution]}\\ g &= - \langle N_t, x_t \rangle = -\cos t (-\sin t - \cos t \cot t) = \cot t, \end{align} which gives $$ K = \frac{eg-f^2}{EG-F^2} = \frac{-\cos^2t}{\sin^2t\cot^2t} = -1. $$