computing the normal unit vector Let $U = (0,∞) × \mathbb{R}$ and consider the function $f : U → \mathbb{R}^{3}$ defined by
$f(u,v) = (\sinh(u)\cos(v), \sinh(u)\sin(v), \cosh(u))$.
(a) Compute the matrices representing the first and second fundamental forms of the
surface $S = f(U)$. Compute the Gauss curvature of $S$, showing it to be everywhere
positive
(b) Show $f$ is a surface patch.
So I've got the first fundamental form $I=\begin{pmatrix} \cosh(2t)& \  0 \\ 0& \ \sinh^2(u) \end{pmatrix}$ but I'm so unsure on how to calculate the normal unit vector $\hat{N}$ and I badly need it to get the second fundamental form and finish part (a). I'm sure I can finish (a) if anyone could help me work out $\hat{N}$. I have $f_{u}= (\cosh(u)\cos(v), \cosh(u)\sin(v), \sinh(u))$ and $f_{v}= (-
\sinh(u)\sin(v), \sinh(u)\cos(v), 0)$.
Then for (b), I've looked at so many definitions for a surface patch and I'm just not seeing how I should be applying it to $f$
Edit: Okay so I've tried an attempt at the unit normal vector formula but I'm just not sure about my answer:
$\hat{N}=\frac{f_{u} \times f_{v}}{|f_{u} \times f_{v}|}$ = $\frac{(-sinh^{2}(u)cos(v))i +(-sinh^{2}(u)sin(v))j + (2cosh(u)sinh(u))k}{\sqrt{sinh^{4}(u)cos^{2}(v) + sinh^{4}(u)sin^{2}(v) + 4cosh^{2}(u)sinh^{2}(u)}} = \frac{(-sinh^{2}(u)cos(v))i +(-sinh^{2}(u)sin(v))j + (2cosh(u)sinh(u))k}{\sqrt{sinh^{4}(u)+ 4cosh^{2}(u)sinh^{2}(u)}} $
 A: I add it as an answer then:
If you compute $\hat{N} = \frac{f_u \times f_v}{\| f_u \times f_v\|}$ then you should be getting the following
$$ \hat{N} = \frac{1}{\sqrt{\cosh(2u)}}\left( \begin {array}{c} 
-{{\sinh \left( u \right) \cos \left( v\right)}}\\ 
-{{\sinh \left( u \right) \sin \left( v\right)}}\\ 
\cosh(u)
\end {array} \right).
$$
which most probably can be obtained from your computation following a trigonometric simplification.
Now, you can compute the second fundamental form:
$$
\mathbf{II} = 
\left(\begin{array}{cc}
\langle f_{uu}, N\rangle & \langle f_{uv}, N\rangle\\
\langle f_{uv}, N\rangle & \langle f_{vv}, N\rangle
\end{array}\right) =
\frac{1}{\sqrt{2\cosh^2(u) - 1}}
\left(\begin{array}{cc}
1 & 0\\
0 & \sinh^2(u)
\end{array}\right) 
$$
Since the Gaussian curvature is $K = \frac{\det{\mathbf{II}}}{\det\mathbf{I}}$ we get
$$
K = \frac{1}{\cosh^2(2u)}
$$
which is clearly everywhere positive.
Regarding the surface patch, I know this definition:
Definition: Let $M \subset \mathbb{R}^3$ be a smooth surface. A surface patch or a local parametrization of $M$ is an injective immersion whose image is contained in $M$.
You already have your map $f(u,v)$ is an immersion (note that the z-coordinate of $\hat{N}$ never vanishes). Therefore, it just remains to prove that $f$ is injective. Which this part is also straight forward as you are dealing with a surface of revolution with the profile curve $u\mapsto(\sinh(u),0,\cosh(u))$ which does not have a self-intersection.
