Show that the cylinder $(x,y,z) \in R^3; x^2+y^2=1 $is a regular surface and find parameterizations whose coordinate neighborhoods cover it.
I'm going to be honest I saw this answer but I don't quite understand it. I am familiar with the propositions that I am given but not sure how its applied.
Proposition 2: If $f: U \subset R^3 \rightarrow R $ is a differentiable function and $a \in f(U)$ is a regular value of $f$, then $f^{-1}(a)$ is a regular surface in $R^3$.
Proof:Define this function $f(x,y,z)= x^2+y^2+z^2-1$. Then the cylinder is the set $f^{-1}(0)$. Computing all partial derivatives, this result is obtained.
$\frac{\partial f}{\partial x}=2x, \frac{\partial f}{\partial y}=2y, \frac{\partial f}{\partial z}=2z.$
It is clear that all partial derivatives are zero if and only if x=y=z=0 or $(0,0,0)$. Further checking shows that $f(0,0,0) \neq 0$, which means that $(0,0,0)$ does not belong to $f^{-1}(0)$. Hence for all $u \in f^{-1}(0)$, not all of partial derivatives at $u$ are zero. By proposition 2, the cylinder hence is regular surface.
Futhermore the cylinder can be parameterized as $g(u,v)=(cos u, sinu,v)$ where $u,v \in \mathbb{R}$
Okay so I'm not sure why we define the function as $f(x,y,z)= x^2+y^2+z^2-1$; is it because we are told it is in $R^3$ so we have to include $z$? I really want to understand all the steps to this. If someone can help I would really appreciate it.
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