Surface integrals and conditions for surfaces I have researched online for the conditions that are required for a surface to be called smooth. One of which is that the derivative matrix of the map $F$, mapping a region from $\mathbb{R}^2$ to $\mathbb{R}^3$ should have rank 2 for all points $(u,v)\in \mathbb{R}^2$. I understand that the rank cannot be three as this would contradict the rank nullity theorem, but why can't the rank be 1? Would this imply that you have a degenerate line? But surely if the rank is 1 it means that for all points on the surface the derivatives are multiples of one vector and so lie on a plane? 
Thank you
 A: Every point of a smooth surface must have a tangent plane. Let $$\vec r(u,v)=x(u,v)\hat i+y(u,v)\hat j+z(u,v)\hat k$$ be the parametrization of a surface. The derivative of the map $\vec r:\mathbb{R}^2\to\mathbb{R^3}$ (assuming it exists) is given by 
$$F=\begin{bmatrix}
{\partial x\over\partial u}&{\partial x\over\partial v}\\
{\partial y\over\partial u}&{\partial y\over\partial v}\\
{\partial z\over\partial u}&{\partial z\over\partial v}
\end{bmatrix}=\begin{bmatrix}{\partial\vec r\over \partial u}&{\partial\vec r\over \partial v}\end{bmatrix}$$ 
Suppose $u_0,v_0$ are constants in $\mathbb{R}$. Then $$\vec r(u,v_0)=x(u,v_0)\hat i+y(u,v_0)\hat j+z(u,v_0)\hat k$$ is a curve that lies on our surface and ${\partial\vec r\over \partial u}\Big{|}_{(u,v_0)}$ is the tangent vector along the curve. This vector lies in the tangent plane of our surface at $(u,v_0)$. Similarly ${\partial\vec r\over \partial v}\Big{|}_{(u_0,v)}$ lies in the tangent plane at $(u_0,v)$. So the tangent plane at $(u_0,v_0)$ contains the vectors ${\partial\vec r\over \partial u}\Big{|}_{(u_0,v_0)},\;{\partial\vec r\over \partial v}\Big{|}_{(u_0,v_0)}$. 
In case $F$ has rank $2$, the two columns of $F$ are independent. So the vectors ${\partial\vec r\over \partial u}\Big{|}_{(u_0,v_0)},\;{\partial\vec r\over \partial v}\Big{|}_{(u_0,v_0)}$ generate a tangent plane at $(u_0,v_0)$. If $F$ has rank $1$ or $1$ linearly independent column, ${\partial\vec r\over \partial u}\Big{|}_{(u_0,v_0)}||\;{\partial\vec r\over \partial v}\Big{|}_{(u_0,v_0)}$ and consequently they can only generate a straight line and not a plane.
Of course ${\partial\vec r\over \partial u}\Big{|}_{(u_0,v_0)}$ lies on infinitely many planes in this case but we can't decide which one of these planes to call the tangent plane. For example if our surface is the cone
$$
x=(1-u)\cos v,\quad y=(1-u)\sin v,\quad z=u\quad(u\in[0,1],v\in[0,2\pi)),\\
F=\begin{bmatrix}-\cos v & -(1-u)\sin v\\
-\sin v & (1-u)\cos v\\
1 & 0\\
\end{bmatrix}
$$ 
At the vertex $(u,v)=(1,0),\,F$ becomes
$$
\begin{bmatrix}-1 & 0\\
0 & 0\\
1 & 0\\
\end{bmatrix}
$$ 
Now any plane $a(x+z)+by=a$ at the vertex $(0,0,1)$ contains the derivative vectors ${\partial\vec r\over \partial u}\Big{|}_{(1,0)},{\partial\vec r\over \partial v}\Big{|}_{(1,0)}$. To see this note that the normal to the plane is $[a\,b\,a]^\top$ and $[a\,b\,a]^\top\cdot[-1\,0\,1]^\top=0=[a\,b\,a]^\top\cdot[0\,0\,0]^\top$ so $[-1\,0\,1]^\top={\partial\vec r\over \partial u}\Big{|}_{(1,0)},\vec 0={\partial\vec r\over \partial v}\Big{|}_{(1,0)}$ lie on the plane. So we're unable to define the tangent plane and the cone is not smooth at the vertex. However at $(u,v)=(0,\pi),\,F$ becomes
$$
\begin{bmatrix}1 & 0\\
0 & -1\\
1 & 0\\
\end{bmatrix}
$$ 
And in this case $x-z+1=0$ the only plane containing $[1\,0\,1]^\top,[0\,-1\,0]^\top$ and passing through $(-1,0,0)$ and hence is the tangent plane.
A: Very simple example: what is the tangent plane of $(u,v)\to(u+v,u+v,u+v)$? (the image is the straight line $x=y=z$)
