Finding a basis of $T_pS$ where $S = \{(u,v,f(u,v)): u,v\in \Bbb R\}$ is a surface

Question

I want to prove the following result about tangent spaces to surfaces in $\Bbb R^3$. $T_pS$ denotes the set of all tangent vectors at $p\in S$.

Suppose $S = \{(u,v,f(u,v)): u,v\in \Bbb R\}$ is a surface in $\Bbb R^3$ where $f:\Bbb R^2\to\Bbb R$ is a differentiable map. Prove that $(1,0, f_u)$ and $(0,1,f_v)$ form a basis of $T_pS$, the tangent space at $p\in S$. Here, $f_u = \frac{\partial f}{\partial u}$ and $f_v = \frac{\partial f}{\partial v}$.

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Some useful definitions:

Definition $1$. $S\subset \Bbb R^3$ is called a regular surface if for any $p\in S$, there exists an open set $U\subset \Bbb R^3$ containing $p$, an open set $V\subset \Bbb R^2$, and a differentiable$\color{red}{^1}$ map $\varphi: V\to U$ such that:

1. The restriction map $\varphi:V\to U\cap S$ is a homeomorphism.
2. For all $(x,y)\in V$, $D\varphi_{(x,y)}: \Bbb R^2\to\Bbb R^3$ (the derivative map of $\varphi$) is injective.

Definition $2$. $v\in T_p(S)$ iff there exists $\epsilon > 0$, and a differentiable map $c:(-\epsilon,\epsilon)\to\Bbb R^3$ such that $c(t)\in S$ for all $t\in (-\epsilon,\epsilon)$, $c(0) = p$ and $c'(0) = v$.

Let $p(u,v) = (u,v, f(u,v))$. Then, I noticed that $\frac{\partial p}{\partial u} = (1,0,f_u)$ and $\frac{\partial p}{\partial v} = (0,1,f_v)$. The result above claims that these are basis vectors of $T_pS$, which is a subspace of dimension $2$ in $\Bbb R^3$. However, I am not sure why partial differentiation yields the basis vectors in this case. It would be helpful if I could get some hints on how to go ahead. Thanks!

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Footnotes:
$\color{red}{1.}$ $\varphi$ is differentiable in the sense that if $\varphi(u,v) = (\mathbf x(u,v), \mathbf y(u,v), \mathbf z(u,v))$ then $\mathbf x, \mathbf y, \mathbf z: \Bbb R^2\to\Bbb R$ have partial derivatives of all orders.

Answer 1

This is a compilation of comments. The surface is a level set of $z-f(x,y)$, its equation is $z-f(x,y)=0$. The gradient of this function is $(-f_x,-f_y,1)$. It is perpendicular to vectors $\vec p=(1,0, f_x)$ and $\vec q= (0,1,f_y)$. Hence vectors $\vec p$ and $\vec q$ are parallel to the tangent plane. These vectors are linearly independent, so they form a basic of the tangent space.

Answer 2

Let $p = (u_0, v_0, f(u_0, v_0))$.

The two vectors lie in $T_pS$:

Consider the curve $\gamma(t) = (u_0 + t, v_0, f(u_0 + t, v_0))$. Clearly this lies on the surface, and $\gamma'(t) = (1, 0, \frac{\partial f}{\partial u}(u_0, v_0))$. Hence $(1, 0, \frac{\partial f}{\partial u}(u_0, v_0)) \in T_pS$.

Similarly, the other vector $(0, 1, \frac{\partial f}{\partial v})$ lies in $T_pS$ as well.

The two vectors span $T_pS$:

Consider an arbitrary curve on $S$: $\gamma(t) = (u(t), v(t), f(u(t), v(t)))$, with $(u(0), v(0)) = (u_0, v_0)$. Chain rule gives

$$\gamma'(t) = (u'(t), v'(t), u'(t) \frac{\partial f}{\partial u} + v'(t) \frac{\partial f}{\partial v}) = u'(t) (1, 0, \frac{\partial f}{\partial u}) + v'(t) (0, 1, \frac{\partial f}{\partial v})$$

So these two vectors span $T_pS$ as well, which also shows that the two vectors form a basis of $T_pS$.