-1
$\begingroup$

I am trying to make $\theta$ disappear from the function $f(r,\theta)=r^2 + a^2 \cos^2 \theta$ by multiplying it by another real function $g(r,\theta)$ then use some identity such as $r^{2}+a^2\cos^{2}\theta+a^2\sin^{2}\theta=r^2+a^2$ so that $\theta$ disappears from the resulting function $h(r,\theta)=f(r,\theta)g(r,\theta)$.

This is why I am looking for this

I have problem that involves the following

$$ \frac{\text{some function of}\,\,r\,\,\text{and}\,\,\theta}{r^2 + a^2 \cos^2 \theta} $$

I am trying to multiply the top and the bottom by a function so that $\theta$ disappears from the bottom and only appears in the top.

Note: both $r$ and $\theta$ are real.

I hope you now what I mean.

Any help is appreciated.

  • 0
    Is there a reason that you expect this is possible? Why do you need this particular property? – 2017-02-15
  • 2
    $g(r,\theta) = 1/(r^2+a^2 \cos^2 \theta)$ – 2017-02-15
  • 0
    @mercio, sorry I forgot to mention that $g(r,\theta)=1/(r^2+a^{2}\cos^2\theta)$ is not allowed. – 2017-02-15
  • 0
    First, make a list of what's allowed and what's not. – 2017-02-15
  • 0
    @SchrodingersCat, please see my edited question. – 2017-02-15

1 Answers 1

2

If I understand correctly you want a function $g(r,\theta)$ such that $$ \frac{\partial}{\partial\theta}\bigl(g(r,\theta)\,f(r,\theta)\bigr)=0. $$ This leads to $$ \frac{1}{g}\frac{\partial g}{\partial\theta}=-\frac{1}{f}\frac{\partial f}{\partial\theta}\implies\frac{\partial}{\partial\theta}\,\log g=-\frac{\partial}{\partial\theta}\,\log f. $$ It follows that $g=C(r)/f$ for some function $C(r)$, which I guess is not allowed.