# Show a curve is not a regular surface

A set $S\subseteq\mathbb{R}^3$ is said to be a regular surface, if for every $p\in S$ there exists an open ball $B_p(r)$ around $p$, an open set $A\subseteq\mathbb{R}^2$ and a function $f:A\to\mathbb{R}^3$ such that $f(A) = B_p(r)\cap S$, $f$ is continuously differentiable and $rank(D_f) = 2$.

I need to show that a straight line $\gamma(t) = \gamma(0) + tv$ where $v\in\mathbb{R}^3$ is not a regular surface. I don't think the definition has fully sinked in, and this seems really intuitive but I just can't think how to start.

Let's write out the condition ${\rm rank}(D_f) = 2$. If $(u,v)$ are coordinates on $A \subset \mathbb R^2$, this is saying that $$D_f = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} \\ \end{bmatrix}$$ has rank two for any choice of $(u,v) \in A$.
A more intuitive way of saying this is as follows. Pick a point $p = f(u_0, v_0)$ on the surface, and consider the curves $$t \mapsto f(u_0 + t, v_0), \ \ \ \ \ \ \ \ \ t \mapsto f(u_0, v_0 + t)$$ within the surface. The tangent vectors at $p$ to these curves are the partial derivatives $\frac{\partial \vec r}{\partial u}$ and $\frac{\partial \vec r}{\partial v}$ evaluated at $(u_0, v_0)$. The condition ${\rm rank}(D_f) = 2$ is saying that the two tangent vectors $\frac{\partial \vec r}{\partial u}$ and $\frac{\partial \vec r}{\partial v}$ are linearly independent. In other words, there are two independent directions in which you can move within the surface, and it is in this sense that the surface is two-dimensional.