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

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

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 $(3)$.


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