So the problem says to compute the gauss map and it's derivative for the cone parametrized by $$\mathbf x (u,v)=(v\cos u, v\sin u, v).$$
From what I understand, a Gauss map takes the unit normal vectors of this cone and maps them into the unit sphere in order to give geometric meaning to the equation by examining the shape traced on the sphere by the normal vectors of the cone. What I don't understand is what this should look like. It seems to me that the Gauss map should just be the equation for the unit normal vector, which I found like this:
$$\begin{split} \mathbf x_u&= (-v\sin u, v\cos u, 0)\\ \mathbf x_v&= (\cos u, \sin u, 1)\\ U &= \frac{\mathbf x_u \times \mathbf x_v}{|\mathbf x_u \times \mathbf x_v|}\\ &= \frac{1}{\sqrt2}(\cos u,\sin u,-1). \end{split}$$
But I'm not sure how I should state my answer.
Should it be $f(u,v)= \frac{1}{\sqrt2}(\cos u,\sin u,-1)$?
And for finding the derivative of the Gauss map, also called the second fundamental form, I know this is equal to the negative of the shape operator. This I am also a bit confused about. Here is what I know about the shape operator:
I think this should look like a matrix $S$ so that if:
$$\begin{split} S(\mathbf x_u)&= a\mathbf x_u + b\mathbf x_v\\ S(\mathbf x_v)&=c\mathbf x_u + d\mathbf x_v. \end{split}$$
then
$$S= \left( \begin{matrix} a & c \\ b & d \\ \end{matrix}\right) $$
And using
$$E=\mathbf x_u \cdot\mathbf x_u = v^2, \ \ F= \mathbf x_u \cdot\mathbf x_v = 0, \ \ \ G= \mathbf x_v\cdot\mathbf x_v = 2.$$
$$l= \mathbf x_{uu}\cdot U = -\frac{v}{\sqrt 2},\ \ m= \mathbf x_{uv}\cdot U = \mathbf x_{vu}\cdot U = 0,\ \ n= \mathbf x_{vv}\cdot U = 0.$$
I can find $$ \begin{split} a&=-\frac{ Fm - Gl }{ EG - F^2 } = -\frac{1}{v\sqrt2}\\ b&=-\frac{ Fl + Em }{ EG - F^2 } = 0 \\ c&= -\frac{ -mG + Fn }{ EG - F^2 } = 0\\ d&= \frac{ En - Fm }{ EG - F^2 } = 0. \end{split}$$
so the derivative of the gauss map is
$$\left( \begin{matrix} -\frac{1}{v\sqrt2} & 0 \\ 0 & 0 \\ \end{matrix}\right) $$
Am I on the right path? Any insight would be appreciated.