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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.