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\leftline{MATH 392 -- Geometry Through History -- Solutions for Quiz 2 -- April 15, 2016}
%\vskip 5pt
%\leftline{Your Name: $\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$}
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\noindent
Let $S$ be the surface parametrized by
$$\vec{x}(u,v) = (u,\cosh(u)\cos(v),\cosh(u)\sin(v)),\quad (u,v) \in (-\infty,\infty) \times (-\infty,\infty)$$
where the function $\cosh(u) = \frac{e^u + e^{-u}}{2}$
is the hypberbolic cosine.  
\begin{enumerate}
\item[(A)] (5)  Compute the coefficients $E,F,G$ of the first fundamental form.
\item[] {\it Solution:} We have 
\begin{eqnarray*}
\vec{x}_u &=& (1,\sinh(u)\cos(v),\sinh(u)\sin(v))\\
\vec{x}_v &=& (0,-\cosh(u)\sin(v),\cosh(u)\cos(v))\\
E &=& \langle \vec{x}_u,\vec{x}_u\rangle = 1 + \sinh^2(u)(\cos^2(v) + \sin^2(v)) = \cosh^2(u)\\
F &=& \langle \vec{x}_u,\vec{x}_v\rangle = 0\\
G &=& \langle \vec{x}_v,\vec{x}_v\rangle = \cosh^2(u)(\sin^2(v) +\cos^2(v)) = \cosh^2(u).
\end{eqnarray*}
\item[(B)] (10) Compute the coefficients $e,f,g$ of the second fundamental form.
\item[] {\it Solution:}
From part (A), the unit normal is computed like this:
$$ \vec{x}_u \times \vec{x}_v = (\sinh(u)\cosh(u),-\cosh(u)\cos(v),-\cosh(u)\sin(v))$$
and 
$$\|\vec{x}_u \times \vec{x}_v\| = \sqrt{\cosh^2(u)(1 + \sinh^2(u))} = \cosh^2(u).$$
So 
$$N = \left(\frac{\sinh(u)}{\cosh(u)},-\frac{\cos(v)}{\cosh(u)},-\frac{\sin(v)}{\cosh(u)}\right).$$
We have 
\begin{eqnarray*}
\vec{x}_{uu} &=& (0,\cosh(u)\cos(v),\cosh(u)\sin(v))\\
\vec{x}_{uv} &=& (0,-\sinh(u)\sin(v),\sinh(u)\cos(v))\\
\vec{x}_{vv} &=& (0,-\cosh(u)\cos(v),-\cosh(u)\sin(v))\\
e &=& \langle \vec{x}_{uu},N\rangle = -1\\
f &=& \langle \vec{x}_{uv},N\rangle = 0\\
g &=& \langle \vec{x}_{vv},N\rangle = 1.
\end{eqnarray*}
\item[(C)] (5) Find the Gaussian curvature of $S$.
\item[] {\it Solution:}  The Gaussian curvature is 
$$K = \frac{eg-f^2}{EG-F^2} = \frac{-1}{\cosh^4(u)}.$$
\end{enumerate}
\end{document}
