1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $(3)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.