Editing Implicit function theorem (section)

=== Example: polar coordinates ===
As a simple application of the above, consider the plane, parametrised by [[polar coordinates]] {{math|(''R'', ''θ'')}}. We can go to a new coordinate system ([[cartesian coordinates]]) by defining functions {{math|1=''x''(''R'', ''θ'') = ''R'' cos(''θ'')}} and {{math|1=''y''(''R'', ''θ'') = ''R'' sin(''θ'')}}. This makes it possible given any point {{math|(''R'', ''θ'')}} to find corresponding Cartesian coordinates {{math|(''x'', ''y'')}}. When can we go back and convert Cartesian into polar coordinates? By the previous example, it is sufficient to have {{math|1=det ''J'' ≠ 0}}, with
<math display="block">J =\begin{bmatrix}
 \frac{\partial x(R,\theta)}{\partial R} & \frac{\partial x(R,\theta)}{\partial \theta} \\
 \frac{\partial y(R,\theta)}{\partial R} & \frac{\partial y(R,\theta)}{\partial \theta} \\
\end{bmatrix}=
 \begin{bmatrix}
 \cos \theta & -R \sin \theta \\
 \sin \theta & R \cos \theta
\end{bmatrix}.</math>
Since {{math|1=det ''J'' = ''R''}}, conversion back to polar coordinates is possible if {{math|1=''R'' ≠ 0}}. So it remains to check the case {{math|1=''R'' = 0}}. It is easy to see that in case {{math|1=''R'' = 0}}, our coordinate transformation is not invertible: at the origin, the value of θ is not well-defined.