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I have these two equations

$u = \frac{2x}{x^2 + y^2} \\ v = \frac{-2y}{x^2 + y^2}$

And I need to put $x$ in terms of $u$ and $v$. If I take polar co-ordinates and plug them in I get(in the case of $u$), because

(rcos(theta))^2 + (rsin(theta))^2 = 1

$u = 2r\cos(\theta)$

Can I simply change that back to

$u = 2x$

?

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    Unfortunately, $r^2 \cos^2 \theta + r^2 \sin^2 \theta = r^2$, not 1. So you should get $u=2r^{-1}\cos\theta$.2012-11-01

3 Answers 3

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Let $x = r \cos \theta$ and $y = r \sin \theta$. Then $ u = \frac{2x}{x^2 + y^2} = \frac{2 \cos \theta}{r} $ $ v = \frac{-2y}{x^2 + y^2} = \frac{- 2 \sin \theta}{r} $ Dividing these gives $ \tan \theta = - \frac{v}{u} \Rightarrow \theta = \tan^{-1} \left(-\frac{v}{u}\right), $ and squaring and then adding gives $ u^2 + v^2 = \frac{4}{r^2} \Rightarrow r = \frac{2}{\sqrt{u^2+v^2}}. $ You now have $r$ and $\theta$, and thus $x$ and $y$, in terms of $u$ and $v$.

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    @scuba: No, why?2012-11-01
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In polar/Cartesian coordinates, using standard variables, $x^2+y^2 = r^2$, so your first equation could be written $u=\frac{2x}{r^2}$ But then also, $x=r\cos\theta$, so $u=\frac{2\cos(\theta)}{r}$ So there is your mistake.

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Using polar coordinates:

$u=\frac{2x}{x^2+y^2}\longrightarrow\frac{2r\cos\theta}{r^2}=\frac{2\cos\theta}{r}$

$v=-\frac{2y}{x^2+y^2}\longrightarrow -\frac{2r\sin\theta}{r^2}=-\frac{2\sin\theta}{r}$

which is different from what you wrote...

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    mhmm ok, so after I get that, how do I put x in terms of u and v?2012-11-01