# How to find inverse of general curvilinear coordinates

Lets say I have a curvilinear coordinate system

$$A=A(x,y,z) = \frac{x^2+y^2+z^2}{2z}$$, $$B=B(x,y,z)= \frac{x^2+y^2+z^2}{2\sqrt{x^2+y^2}}$$, $$C=C(x,y,z)=\tan^{-1}(y/x)$$

How do I find the inverse of those i.e

$$x=x(A,B,C)$$, $$y=y(A,B,C)$$, $$z=z(A,B,C)$$

I know that in cylindrical or spherical coordinates I could do it based on the geometry, but I dont have the geometry, I just have the equations for A, B, and C in terms of x, y, and z

• So what are the equations? Not sure there is a general answer. It is like asking "how do I solve $f(x)=0$?" – Andrei Oct 11 '18 at 2:22
• I have added the equations. – Stone Preston Oct 11 '18 at 2:44

## 1 Answer

I will assume that $$x,y,z>0$$. You will need to check if it make sense to have them negative.

The first equation we are going to use is $$\tan C=\frac{y}{x}$$ Since everything else we have squares, I will write this as $$y^2=x^2\tan^2C\tag{1}$$ The second equation can be written from the ratio $$A/B$$ or in fact we can use the square of that: $$\frac{A^2}{B^2}=\frac{x^2+y^2}{z^2}$$ This yields $$z^2=\frac{B^2}{A^2}(x^2+y^2)\tag{2}$$ The third equation is $$\frac{1}{A^2}+\frac{1}{B^2}=\frac{4(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2}=\frac{4}{x^2+y^2+z^2}$$ I will rewrite this as $$x^2+y^2+z^2=\frac{4}{\frac{1}{A^2}+\frac{1}{B^2}}\tag{3}$$ Now it should be trivial to find $$x^2$$, $$y^2$$, and $$z^2$$ from $$(1)$$, $$(2)$$, and $$(3)$$.