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HallsofIvy · #4 July 4th, 2009, 06:04 AM

And what is the independent variable? In other words, to find the inverse function, what should we solve for? r? $\theta$ ?

If the problem is to solve $y=\frac{P_1 A r^{-\alpha}}{N+P_2 A(r^2-2Rcos\theta+R^2)^{-\frac{\alpha}{2}}}$ for $\omega$ , this is relatively straight forward. Multiply both sides by the denominator: $y(N+P_2 A(r^2- 2Rcos\theta+ R^2)^\frac{\alpha}{2}= P_1 A r^{-\alpha}$ .

Take the $2/\alpha$ power of both sides: $y^{\frac{2}{\alpha}}(N+ P_2 A(r^2- 2Rcos(\theta)+ R^2))= P_1^{\frac{2}{\alpha}}A^{\frac{2}{\alpha}}r^-2$

Multiply both sides by $y^{\frac{\alpha}{2}}$ : $N+ P_2 A(r^2- 2Rcos(\theta)+ R^2)= y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^{\frac{2}{\alpha}}r^-2$

Subtract N from both sides: $P_2 A(r^2- 2Rcos(\theta)+ R^2)= y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^{\frac{2}{\alpha}}r^-2- N$

Divide both sides by $P_2 A$ : $r^2- 2Rcos(\theta)+ R^2= \frac{y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^{\frac{2}{\alpha}}r^-2- N}{P_2 A}$

Subtract $r^2+ R^2$ from both sides: $- 2Rcos(\theta) = \frac{y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^{\frac{2}{\alpha}}r^-2- N}{P_2 A}- r^2- R^2$

Finally, divide both sides by -2R: $cos(\theta) = -\frac{\frac{y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^{\frac{2}{\alpha}}r^-2- N}{P_2 A}- r^2- R^2}{2R}$
and take the arccosine of both sides:
: $\theta = arccos\left(-\frac{\frac{y^{\frac{\alpha}{2}}P_1^{\frac{2}{\alpha}}A^{\frac{2}{\alpha}}r^-2- N}{P_2 A}- r^2- R^2}{2R}\right)$

If you want to solve for r, having both $r^2$ and $r^{-\alpha}$ makes it much harder.