Tangent plane of level surface

Question

Is the level surface

$N_k = \lbrace (x_1,...,x_n): z(x_1,...,x_n)= k, k\in \mathbb{R}\rbrace$

of the tangent plane $z(x_1,...,x_n)$ of a function $f(x_1,...,x_n)$ at a point $(a_1,..,a_n) \in N_k$

equal to the tangent plane $p(x_1,..,x_n)$ at $(a_1,..,a_n)$

of the level surface $N_k = \lbrace (x_1,...,x_n): f(x_1,...,x_n)=k, k\in\mathbb{R}\rbrace$ of the function?

Answer 1

Yes.

Proof:

1. Equation for tangent plane $x_{n}(x_1,...,x_{n-1})$ of level surface $N_k = \lbrace \textbf{x}: f(\textbf{x}) = k\rbrace$ at $\textbf{a} \in N_k$

$\textbf{n}\cdot (\textbf{x}-\textbf{a}) = 0$

$\nabla f(\textbf{a})\cdot (\textbf{x}-\textbf{a}) = 0$

2. Equation for level surface $N_k = \lbrace \textbf{x}: x_{n+1}(\textbf{x}) = k\rbrace$ of tangent plane:

Tangent plane:

$ x_{n+1} = f(\textbf{a})+\nabla f(\textbf{a})\cdot (\textbf{x}-\textbf{a})$

$ x_{n+1}= k +\nabla f(\textbf{a})\cdot (\textbf{x}-\textbf{a})$

Level surface:

$k = k +\nabla f(\textbf{a})\cdot (\textbf{x}-\textbf{a})$

$\nabla f(\textbf{a})\cdot (\textbf{x}-\textbf{a}) = 0$