# Implicit Function Theorem and Change of Coordinates(?)

Suppose $\rho:\mathbb{R^n} \to \mathbb{R}$ is $C^k$ for some $k \geq 1$ and consider the level set $\rho \equiv 0$. Suppose that $\rho(x_0) = 0$ and $\frac{\partial \rho}{\partial x_n}(x_0) \neq 0$. By the implicit function theorem, there is a $C^k$ function $R$ so that in a neighborhood of the first $n-1$ coordinates of $x_0$, $R(x_1, ..., x_{n-1}) = x_n$ and $\rho(x_1, ..., x_{n-1}, R(x_1, ..., x_{n-1})) = 0$.

I have seen multiple times in textbooks that we can assume without loss of generality that $\frac{\partial R}{\partial x_j}(x_0) = 0$ for $j = 1, 2, ... n-1$. Why is that? I believe this is some sort of change of variables/rotation of the level set, but I have never seen the details provided. Can someone explain this?

$\mathbf{Update:}$

Here is an example of where I have seen this construction

"Represent $\partial \Omega$ near 0 in the form $\partial \Omega =\{x=(x_1,...,x_4)|\text{ } x_4=R(x′) \text{ for } x′=(x_1,x_2,x_3) \in V′\}$ with a $C^\infty$ function $R$ on a neighborhood $V′$ of $0 \in \mathbb{R^3}$ such that $R(0)=0, R_i(0)=0, i=1,2,3$."

I have been told that when we do this construction that we can make $\nabla \rho(0) = (0, 0, 0, 1)$ where $\rho$ is a $C^\infty$ function with its zero set equal to $\partial \Omega$ and $\frac{\partial \rho}{\partial x_n} \neq 0$. Is this true?

To keep notation nice and simple, let's assume we're working around the origin. Let $\nabla_i \rho ( 0)$ be the gradient vector of $\rho$ at the origin.
Rotate the coordinate axes to obtain a new orthonormal set of coordinates $\{ y_i \}$ such that the $y_n$-axis is aligned with the gradient vector $\nabla_i \rho (0)$. (So the $y_1$, $y_2, \dots , y_{n-1}$-axes are orthogonal to the gradient vector $\nabla_i \rho (0)$.) Then the partial derivatives w.r.t. the $y_i$-coordinates are $$\frac{\partial \rho}{\partial y_i}(0) = \begin{cases} 0 & i = 1, 2, \dots, n-1 \\ |\nabla_i \rho (0) | & i = n\end{cases}$$
By the implicit function theorem, we can find a function $R(y_1, \dots, y_{n-1})$, defined on a neighbourhood of $0$, such that $\rho (y_1, \dots, y_{n-1}, R(y_1, \dots, y_{n-1})) = 0$, and $$\frac {\partial R}{ \partial y_i} (0)= - \left( \frac{\partial \rho}{\partial y_n} (0)\right)^{-1} \frac { \partial \rho}{\partial y_i} (0)= 0$$ for all $i \in \{ 1, 2, \dots, n-1 \}$.