I'm asked if given a map $F:\mathbb{R}^{2}\rightarrow\mathbb{R}^{3}$, $F(u,v)=(11 \cos(u),11 \sin(u),v)$ is a parametrization of a surface. What I have thought is, well:
1) It's differentiable (each component is differentiable).
2) It's differential is: \begin{equation} \begin{pmatrix} -11 \sin(u) & 0\\ 11 \cos(u)& 0\\ 0 & 1 \end{pmatrix} \end{equation}.
It's determinant minors are: (a) $11 \cos(u)$, (b) $-11 \sin(u)$, (c) $0$. Respectively are zero if $u=\frac{\pi}{2}+k\pi, v=l\pi, k,l\in\mathbb{Z}$. Fixed (u_{0},v_{0}), DF(u_{0},v_{0}) has rank two always (because there exist some minor determinant that is not zero, is clear that it is no posible that a) and b) be zero at the same time). So, DF is inyective.
3) $F$ is continuous (is clear). To prove that it is an homeomophism I have built it's inverse: $z=v$, $\frac{y}{x}= \tan(u)$, $u= \arctan(\frac{y}{x})$ (whenever $x\neq 0$). So $F^{-1}(x,y,z)=(\arctan(\frac{y}{x}),z)$ is continuous.
Is $F$ a parametrization if $|u|<1$? If $|v<|1$? What I thought is that in both cases yes, because they don't include points where the condition (2) is false.
Is $F$ a parametrization if we restrict to $100u < v < 100u +1$? In this case, it is not (because it includes points where the condition (2) is not true).
I have to determine if it is a local isometry. What I have done is calculate $T_{p}S=\{(x,y,z)\in\mathbb{R}^{3}:<(x,y,z),(\cos(u), \sin(u),0)>=0\}$. But to prove $<DF(u),DF(v)>=<u,v>$ for all $u,v\in T_{p}S$.
Any hint to continue is appreciated!
I'm going to write this as a quick answer to point out a few things. As I stated in my comments, you have enough to show that you have a surface. Now we need to check for local isometry. I think you may have used the same notation for a few concepts (such as $u,v$ as coordinates on $\mathbb{R}^2$ but also as tangent vectors to $F(\mathbb{R}^2) = S$) so I'm going to straighten it here.
Let $p = (u,v) \in \mathbb{R}^2$. Then $T_p\mathbb{R}^2 \cong \mathbb{R}^2$. Let us denote two tangent vectors here by $x, y \in T_p\mathbb{R}^2$. Note that $DF_p$ is a linear map from $T_p(\mathbb{R}^2) \to T_{F(p)}S$ (represented by the matrix you have, evaluated at $p$). Now, the condition for a local isometry is that $$\langle x, y \rangle = \langle DF_p(x), DF_p(y) \rangle$$
for all $x, y \in T_p\mathbb{R}^2$ (and not $T_{F(p)} S$ as you wrote).
Instead of checking on every $x_1,x_2$, it suffices to check on every basis combination, because the inner product can be extended by bilinearity. If you're not sure what this means or why it is, let me know. So let us check for the four basis combinations $\langle e_1, e_1 \rangle, \langle e_1, e_2 \rangle, \langle e_2, e_1 \rangle, \langle e_2, e_2 \rangle$. The calculations are below
$$\langle e_1, e_1 \rangle = 1, \langle DF_p(e_1), DF_p(e_1) \rangle = \langle (-11 \sin u, 11 \cos u, 0), (-11 \sin u, 11 \cos u, 0)\rangle = 121 \neq 1$$
$$\langle e_1, e_2 \rangle = 0, \langle DF_p(e_1), DF_p(e_2) \rangle = \langle (-11 \sin u, 11 \cos u, 0), (0, 0, 1)\rangle = 0 $$
$$\langle e_2, e_1 \rangle = 0, \langle DF_p(e_2), DF_p(e_1) \rangle = \langle (0, 0, 1),(-11 \sin u, 11 \cos u, 0)\rangle = 0 $$
$$\langle e_2, e_2 \rangle = 1, \langle DF_p(e_2), DF_p(e_2) \rangle = \langle (0,0,1), (0,0,1)\rangle = 1$$
So here you see it fails for the $e_1, e_1$ pair, so this is not an isometry.
One last thing I would point out to quickly show it's a surface. Note that $S$ is also described by the equation $x^2 + y^2 = 11$. Then note that its a regular value of the function $G: \mathbb{R}^3 \to \mathbb{R}$ such that $G(x,y,z) = x^2 + y^2 - 11$, and $S = G^{-1}(0)$. Since it's a regular value, it's a surface.