Intuitive idea tangent plane to a surface

Question

Given a function $f:\mathbb{R}^2\to\mathbb{R}$ I'm trying to reach, intuitively, to the definition of tangent plane to the surface $G(f)$ defined by $z=f(x,y)$ at the point $p=(a,b,f(a,b))$. In the book I'm using it is said that, to get a satisfactory definition, we require that the set $\{(v,D_vf(a,b)):v\in \mathbb{R}^2\}$ is a vector subspace and $\{(v,D_vf(a,b)):v\in \mathbb{R}^2\} = T_p(G(f))$. My first question is: what does this set $\{(v,D_vf(a,b)):v\in \mathbb{R}^2\}$ represents intuitively?

Answer 1

The vectors $v_x(1,0,f'_x(a,b) )$ and $v_y(0,1,f'_y(a,b) )$ are tangent to G . Then their cross product is normal to G at point $(a,b) $

Answer 2

What do you seek beyond the familiar graphical representation?

Answer 3

The set $S=\{(v,D_vf(a,b)):v\in \mathbb{R}^2\}$ can be mapped to the tangent plane $T_p$ at point $p = (a,b,f(a,b))$ with the mapping:

$S \rightarrow T_P: (v_x, v_y, D_vf) \rightarrow (a+v_x, b+v_y, f(a,b)+D_vf)$

$S$ will be a vector subspace of $\mathbb{R}^3$ provided:

1) $0 \in S$ : this is fine as long as we define $D_0 f(a,b)=0 \space \forall (a,b) \in \mathbb{R}^2$

2) $s \in S \Rightarrow \lambda s \in S \space \forall \lambda \in \mathbb{R}$ : this requires $D_{\lambda v}f=\lambda D_vf$

3) $s_1,s_2 \in S \Rightarrow s_1+s_2 \in S$ : this requires $D_{v_1+v_2}f=D_{v_1}f+D_{v_2}f$

Properties (2) and (3) are simply saying that $D_vf$ must be linear in its argumentv $v$.