Given $u = \frac{2x}{x^2 + y^2}$ and $v = \frac{-2y}{x^2 + y^2}$, write $x$ in terms of $u$ and $v$.

Question

I have these two equations

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

And I need to write $x$ in terms of $u$ and $v$. If I take polar coordinates and plug them in, I get (in the case of $u$): because $(r \cos \theta )^2 + (r \sin \theta )^2 = 1$,$$u = 2r\cos \theta .$$

Can I simply change that back to$$u = 2x ?$$

Answer 1

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 $v$.

Answer 2

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.

Answer 3

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...