Three ways to find the normal of a hyperboloid? In the hyperboloid $x^2 + y^2 − z^2 = 4$, where $z \ge 0$, I have found three ways to get the normal vector, but my problem is they do not seem equivalent.
The first is as I have been taught that you can find the gradient of the function, and that will be the normal:
$$\langle 2x,2y,-2z \rangle$$ 
The unit normal should be 
$$\langle {2x\over \sqrt{4x^2+4y^2+4z^2} },{2y\over \sqrt{4x^2+4y^2+4z^2}},{-2z\over \sqrt{4x^2+4y^2+4z^2}} \rangle$$
$$=\langle {2x\over \sqrt{4x^2+4y^2+4(x^2+y^2-4)} },{2y\over \sqrt{4x^2+4y^2+4(x^2+y^2-4)}},{-2\sqrt{x^2+y^2-4}\over \sqrt{4x^2+4y^2+4(x^2+y^2-4)}} \rangle$$
$$=\langle {2x\over 2\sqrt{2x^2+2y^2-4} },{2y\over 2\sqrt{2x^2+2y^2-4}},{-2\sqrt{x^2+y^2-4}\over 2\sqrt{2x^2+2y^2-4}} \rangle$$
A second way would be to rewrite the hyperboloid as $z=\sqrt{x^2+y^2-4}$
and then use the identity that the normal is $\langle f_x,f_y,-1 \rangle$
Then the normal is $$\langle {x\over \sqrt{x^2+y^2-4} },{y\over \sqrt{x^2+y^2-4}},-1 \rangle$$
The unit normal is then
$$\langle {x \sqrt{x^2+y^2-4} \over 2x^2+2y^2-4 },{y  \sqrt{x^2+y^2-4} \over 2x^2+2y^2-4},-\big( {x^2+y^2-4\over 2x^2+2y^2-4} \big) \rangle$$
The third is to use the parametrization $x=r \cos t, y=r\sin t ,z= r^2$. Then the normal is given by  $$r_r \times r_t= \langle -2r^2\cos t, -2r^2\sin t, r \rangle$$
The unit normal should be 
$$\langle {-2r\cos t\over \sqrt{4r^2+1}}, {-2r\sin t \over \sqrt{4r^2+1}}, {1\over \sqrt{4r^2+1}} \rangle$$
However, it does not seem that these are equivalent, so which ones are incorrect and why?
 A: Ignoring normalization ...


*

*Your first vector, divided-through by $2$, is $(x,y,−z)$.

*Your second vector, multiplied-through by $\sqrt{x^2+y^2−4}$, is $(x,y,−\sqrt{x^2+y^2−4})=(x,y,−z)$
So, those vectors are equivalent. 
Your third vector is problematic, which isn't surprising, since your parameterization doesn't satisfy the surface's equation:
$$(r\cos t)^2 + ( r\sin t)^2 - ( r^2 )^2 = 4 \quad\to\quad r^2\cos^2t+r^2\sin^2t-r^4=4\quad\to\quad r^2-r^4=4$$ 
which, for arbitrary $r$, generally isn't true.
One correct parameterization would be
$$x = r \cos t \qquad y = r \sin t \qquad z = \sqrt{r^2-4}$$
(Verifying that this satisfies the surface's equation is left as an exercise to the reader.)
If you don't care for the square root, you can make the substitution $r=2\sec s$, so that the parameterization becomes
$$x = 2\sec s \cos t \qquad y = 2 \sec s \sin t \qquad z = \sqrt{4\sec^2s-4}=\sqrt{4\tan^2s}=2\tan s$$
In the latter case, 
$$\begin{align}
\frac{\partial}{\partial s}(x,y,z) \times \frac{\partial}{\partial t}(x,y,z) &= ( 2 \sec s \tan s \cos t, 2 \sec s \tan s \sin t, 2 \sec^2 s ) \tag{1}\\
&\quad\times (-2 \sec s \sin t, 2 \sec s \cos t, 0) \\[4pt]
&\propto ( \sin s \cos t, \sin s \sin t, 1 ) \times (-\sin t, \cos t, 0) \tag{2}\\[4pt]
&= (-\cos t, -\sin t, \sin s) \tag{3}\\
&\propto (2\sec s\cos t, 2\sec s\sin t, -2\tan s) \tag{4}\\
&=(x,y,-z) \tag{5}
\end{align}$$
In $(3)$, I scaled the vectors by dividing-through by $2\sec^2s$ and $2\sec s$, respectively, to simplify the calculation of their cross-product. In $(4)$, I multiplied-through by $-2\sec s$ to make the equivalence with $(x,y,-z)$ apparent.
