# Inverse function theorem and parametric surfaces as graphs

Let $\psi$ be a regular surface at the point $(u_{0}, v_{0})$ ($\psi \in C^{1}, T_{u} \times T_{v} \neq 0$ at $u_{0}, v_{0}$). Use the implicit function theorem to show the image of $\psi$ near $(u_{0}, v_{0})$ can be a graph of a $C^{1}$ function of two variables. Use this result to show that the tangent plane at $\psi(u_{0}, v_{0})$ defined by $T_{u}$ and $T_{v}$ coincides with the tangent plane of a graph $z = f(x,y)$.

Until now, we know that we can write the parametric surface as $\psi(u,v) = (x(u,v), y(u,v), z(u,v))$. I assumed without loss of generality that the $z$ component of $T_{u} \times T_{v}$ at $u_{0}, v_{0}$ is not $0$ (as this vector itself will be nonzero). Then, this implies that the following Jacobian is nonzero:

$\displaystyle\frac{\partial(x,y)}{\partial(u,v)} \neq 0$ at the point $(u_{0}, v_{0})$

By using the inverse function theorem, we can then write $u = u(x,y)$, $v=v(x,y)$ in a neighborhood of $(u_{0}, v_{0})$. If we use result again in our equation:

$\psi(u(x,y),v(x,y)) = (x(u(x,y),v(x,y)), y(u(x,y),v(x,y)), z(u(x,y),v(x,y)))$

$\psi(u(x,y),v(x,y)) = (x, y, f(x,y))$

If we let $z(u(x,y),v(x,y)) = f(x,y)$. This will be the graph of the function. However, when I try to find the tangent plane of the surface, I would have that I'm only finding the normal vector spanned by $T_{x}, T_{y}$ (which reduces to the well-known formula of the tangent plane of a graph). However, if I try to find $T_{u}, T_{v}$, I don't know how to proceed. It would make more sense if we let $x=u$, $y=v$ (hence by finding both tangent vectors we would get that expression). If $x=u, v=y$, we have $T_{u} \times T_{v} = (-\displaystyle\frac{\partial f}{\partial x}, -\displaystyle\frac{\partial f}{\partial y}, 1)$. Else, we have:

$T_{u} = (\displaystyle\frac{\partial x}{\partial u}, \displaystyle\frac{\partial y}{\partial u}, \displaystyle\frac{\partial z}{\partial u})$

$T_{v} = (\displaystyle\frac{\partial x}{\partial v}, \displaystyle\frac{\partial y}{\partial v}, \displaystyle\frac{\partial z}{\partial v})$

What would be the way to solve this issue? I'm sure I'm missing a point in my argument for the second part of the question. Or when the question asks for $T_{u}, T_{v}$, does it refer to $T_{x}, T_{y}$? Thank you for your time.