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I'm going through past exam questions and I came across part of a question I wasn't sure how to tackle. I don't think I've ever seen something like this:

For $(x,y) \neq (0,0)$, define a vector field F$(x,y)$ and a scalar field $\theta(x,y)$ as follows: $$\mathbf{F}(x,y) = \left(\frac{-y}{x^2 + y^2}\right)\mathbf{i} +\left(\frac{x}{x^2 + y^2}\right)\mathbf{j} $$

$\theta(x,y)$ = the polar angle of $\theta$ of $(x,y)$ such that $0 \le \pi \le 2\pi$

Therefore $x = r\cos\theta(x,y)$ and $y = r\sin\theta(x,y)$, where $r^2 = x^2 + y^2$.

So, verify that $\nabla\theta(x,y) = \mathbf{F}(x,y)$ for all $(x,y) \neq (0,0)$ such that $0 \le \pi \le 2\pi$

I've done some light research about the topic but I think I'd understand it more if it was explained/solved. Any help is appreciated.

Thanks

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Observe that in fact $\;\theta(x,y)=\arctan\frac yx\;$ , so:

$$\theta(x,y):=\arctan\frac yx\implies\begin{cases}& \theta'_x=-\cfrac y{x^2}\cfrac1{1+\frac{y^2}{x^2}}=\cfrac{-y}{x^2+y^2}\\{}\\&\theta'_y=\cfrac1x\cfrac1{1+\frac{y^2}{x^2}}=\cfrac x{x^2+y^2}\end{cases}$$

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  • $\begingroup$ So since we know what F(x,y) is and because we want the grad of θ(x,y), we take the polar form of x and y, combining them through the tanx = sinx/cosx identity, and ultimately taking the derivatives with respect to x and y. That's what I managed to ascertain from what you've shown me. Am I correct? $\endgroup$ – Nathan Lowe Mar 28 '17 at 11:56
  • $\begingroup$ @NathanLowe If I understood you correctly I think you're mostly correct...but observe we don't need to know what $\;F\;$ is, we only need the definition of $\;\theta\;$ . With only that we deduce that $\;\theta=\arctan\frac yx\;$ and etc. $\endgroup$ – DonAntonio Mar 28 '17 at 12:00
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You’re being asked to verify that $F(x,y)={\partial\theta\over\partial x}\mathbf i+{\partial\theta\over\partial y}\mathbf j$. You can do this without an explicit formula for $\theta(x,y)$ by using the Inverse Function theorem. Start by computing the Jacobian matrix of $\phi:(r,\theta)\mapsto(x,y)=(r\cos\theta,r\sin\theta)$: $$\pmatrix{{\partial x\over\partial r} & {\partial y\over\partial r} \\ {\partial x\over\partial\theta} & {\partial y\over\partial\theta}}=\pmatrix{\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta}.$$ In a neighborhood of a point where this matrix is nonsingular, $\phi^{-1}$ exists and its Jacobian matrix is simply the inverse of this matrix. This inverse is quite easy to compute. The determinant of the matrix is $r$ (therefore the Jacobian is nonsingular throughout the domain of $F$), and so $$\pmatrix{{\partial r\over\partial x} & {\partial\theta\over\partial x} \\ {\partial r\over\partial y} & {\partial\theta\over\partial y}}=\pmatrix{\cos\theta & -\frac1r\sin\theta \\ \sin\theta & \frac1r\cos\theta}$$ from which $${\partial\theta\over\partial x}=-{\sin\theta\over r}=-{y\over r^2}=-{y\over x^2+y^2}.$$ A similar calculation gives ${\partial\theta\over\partial y}={x\over x^2+y^2}$. QED.

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  • $\begingroup$ Note that, unlike in single-variable calculus, $\partial x/\partial u$ and $\partial u/\partial x$ are not in general reciprocals, so you need to go through this sort of matrix inversion to calculate one from the other. $\endgroup$ – amd Mar 28 '17 at 20:28

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