I need to prove that, if there is a coordinate neighborhood $V$ of a regular surface $S$ that can have a parameterization of form
$$X(u,v)=\alpha_1(u)+\alpha_2(v)$$
where $\alpha_1,\alpha_2$ are parametric regular curvers, then the tangent planes through one coordinate curve of this neighborhood are all paralell to a line.
Well, I think that for some $q=(u_0,v_0)$ s.t. $X(q)\in V$, I have
$$dX_q(x,y)=\begin{bmatrix}\alpha'_1(u_0)&0\\ 0&\alpha_2(v_0)\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}\alpha_1'(u_0)x\\ \alpha_2'(v_0)y\end{bmatrix}.$$
But I do not know how I can study this through a coordinate curve. For instance, a curve that fix $u$. So should I put $\alpha_1'(u_0)=0$?
Many thanks!
Best Answer
Do it as if it were a Calc III exercise: fix $u_0$ and let $v$ vary, and consider $x=(u_0,v)$.Then, $T_xS$ is the vector space spanned by $\partial_uX(u_0,v)$ and $\partial_vX(u_0,v)$, and these are, respectively, $\alpha_1'(u_0)$ and $\alpha_2'(v)$.
But, any line into $V$ has the form $\ell:x(t)=u+ta$, so $every$ tangent plane in $V$ will be parallel to $\ell$ as soon as $a=\alpha_1'(u_0)$.