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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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By the sound of it, you think that $x^2-y^2=1$ after plugging in $x=r\cos(\theta)$ and $y=r\sin(\theta)$. This is false. –  Alex R. Nov 1 '12 at 2:56
    
whoops- sorry - I mistyped –  scuba Nov 1 '12 at 2:57
    
Can you show your working for $u=2r\cos\theta$? –  wj32 Nov 1 '12 at 3:03
1  
Unfortunately, $r^2 \cos^2 \theta + r^2 \sin^2 \theta = r^2$, not 1. So you should get $u=2r^{-1}\cos\theta$. –  wj32 Nov 1 '12 at 3:06

3 Answers 3

up vote 1 down vote accepted

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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wait up, isn't your v/u the other way around if it's to equal tan? –  scuba Nov 1 '12 at 3:11
    
@scuba: No, why? –  wj32 Nov 1 '12 at 3:19

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? –  scuba Nov 1 '12 at 3:14

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