Vector Field Verification

Question

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

Answer 1

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

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

Answer 2

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.