Calculating curl in cylindrical and cartesian coordinates

So I have this vector function $$\mathbf{R}=\mathbf{i}r\cos\omega t + \mathbf{j} r\sin\omega t$$, where $$x=r\cos\omega t$$ and $$y=r\sin\omega t$$.

I want to find the curl of it's time derivative, $$\frac{\mathrm{d}R}{\mathrm{d}t}=\mathbf{v}=\omega (-\mathbf{i}y+\mathbf{j}x)$$. I want to do this in both cartesian and cylindrical coordinates. Since the curl is along the $$z$$ axis, I expected both answers to be the same. However, I think I made some mistake.

Here is what I did:

Cartesian:

$$\nabla \times \mathbf{v}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -\omega y & \omega x & 0 \\ \end{vmatrix}=\mathbf{i}\cdot 0+ \mathbf{j}\cdot0+\mathbf{k}\cdot (\omega-[-\omega])=2\mathbf{k}\omega$$

Cylindrical:

Coordinates: $$x^2 + y^2 = r^2 \Rightarrow r=\sqrt{x^2+y^2}=\sqrt{r^2(\sin^2\omega t + \cos^2\omega t)}=r$$ and $$\frac{y}{x}=\tan{\theta} \Rightarrow \theta=\arctan\frac{y}{x}$$. But in the case of $$\mathbf{v}$$, the x-coordinate is $$-y$$ and the y-coordinate is $$x$$. So for $$\mathbf{v}$$, we get: $$\theta=\arctan(\frac{r \cos \omega t}{-r\sin\omega t})=\arctan\left(\frac{1}{\tan(\omega t)}\right)$$. This looks somewhat nasty.

Ok, so if we do take $$\mathbf{R}$$ in cylindical coordinates instead, we get $$r=r$$ and $$\theta=\arctan(\tan(\omega t))=\omega t$$, which is much easier to work with.

Taking $$\frac{\mathrm{d}R}{\mathrm{d}t}$$ we note that $$r$$ is not a function of time, and thus we get $$\frac{\mathrm{d}R}{\mathrm{d}t}=\mathbf{r}r+\mathbf{\theta}\omega$$. Alternativley, we can do $$\nabla \times \frac{\mathrm{d}\mathbf{R}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left( \nabla\times \mathbf{R}\right)$$.

The curl in cylindrical coordinates is:

$$(\nabla \times \mathbf{F})_r=\frac{1}{r}\frac{\partial F_z}{\partial \theta}-\frac{\partial F_\theta}{\partial z}$$

$$(\nabla \times \mathbf{F})_\theta=\frac{\partial F_r}{\partial z}-\frac{\partial F_z}{\partial r}$$

$$(\nabla \times \mathbf{F})_z=\frac{1}{r}\frac{\partial}{\partial r}(rF_\theta)-\frac{1}{r}\frac{\partial F_r}{\partial \theta}$$

In this case, $$v_z=0$$ and $$r$$ and $$\theta$$ doesn't change in the $$z$$ direction, so all terms with $$F_z$$ or $$\frac{\partial}{\partial z}$$ vanish.

$$\Rightarrow \left( \nabla\times \mathbf{R}\right)=\left( \nabla\times \mathbf{R}\right)_z=\frac{1}{r}\frac{\partial}{\partial r}(r\cdot \omega t)-\frac{1}{r}\frac{\partial r}{\partial \theta}=\frac{\omega t}{r}-0$$

I used that $$\frac{\partial r}{\partial \theta}=0$$ because the length $$r$$ doesn't change even though the unit vector $$\mathbf{r}$$ changes direction. (When varying the angle, the value of the radius stays constant).

If we take the time derivative we get:

$$\frac{\mathrm{d}}{\mathrm{d}t}\left( \nabla\times \mathbf{R}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\omega t}{r}\mathbf{k} \right)=\mathbf{k}\frac{\omega}{r}\neq 2\mathbf{k}\omega$$

Where did I go wrong? Was my assumption that $$\nabla \times \frac{\mathrm{d}\mathbf{R}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left( \nabla\times \mathbf{R}\right)$$ wrong? I feel like that one should hold true since $$x,y$$ and $$z$$ are continous, differentiable functions in both $$t$$ and coordinates. Or can't I convert $$\mathbf{R}$$ to cylindrical and then take the derivative? Or did I make some other math mistake?

I feel like the answers should be the same?

As to avoid problems, I'd define $$\mathbf R=r\cos\theta\,\mathbf i+r\sin\theta\, \mathbf j=r\mathbf{\hat r}\ \ \text{ with }\ \ r\neq r(t)\text{ and }\theta=\theta(t)$$ and after having differentiated wrt $$t$$ and taken the curl or viceversa, I'd do $$\theta=\omega t$$.
Let's first do the case $$\dfrac{\mathrm d}{\mathrm dt}(\nabla\times\mathbf R)$$; we'll do the other one later to corroborate that they're not the same. Let's start by computing the curl: $$\nabla\times\mathbf R=\frac{1}{r} \begin{vmatrix} \mathbf{\hat r} & r\boldsymbol{\hat \theta} & \mathbf{\hat z}\\ \partial_r & \partial_\theta & \partial_z\\ r & 0 & 0 \end{vmatrix}=0$$ Notice how it yields $$0$$ because $$\mathbf R$$ is a central field and thus a conservative field (their curl is $$0$$). The temporal derivative will still be $$0$$ which insinuates $$\dfrac{\mathrm d}{\mathrm dt}(\nabla\times\mathbf R)\neq \nabla\times\dfrac{\mathrm d \mathbf R}{\mathrm dt}$$ As for the other case, let's start by computing the temporal derivative: $$\dfrac{\mathrm d \mathbf R}{\mathrm dt}=r\dfrac{\mathrm d \mathbf{\hat r}}{\mathrm dt}=r\dot\theta\boldsymbol{\hat \theta}$$ Now we can apply the curl: $$\nabla\times\dfrac{\mathrm d \mathbf R}{\mathrm dt}=\frac{1}{r} \begin{vmatrix} \mathbf{\hat r} & r\boldsymbol{\hat \theta} & \mathbf{\hat z}\\ \partial_r & \partial_\theta & \partial_z\\ 0 & r\cdot r\dot\theta & 0 \end{vmatrix}=\frac{1}{r}\frac{\partial(r^2\dot\theta)}{\partial r}\mathbf{\hat z}=2\dot\theta\mathbf{\hat z}$$ Substituting $$\theta=\omega t$$, we get $$\nabla\times\dfrac{\mathrm d \mathbf R}{\mathrm dt}=2\omega\mathbf{\hat z}$$ The main problem here was that the time derivative of a unit vector with constant modulus changes its direction perpendicular to the original unit vector which means the curl isn't the same for each of them.
• Well, if you factor $r$ you'll get $\cos\theta \mathbf i+\sin\theta\mathbf j$ which is by definition the radial unit vector. Commented Apr 30 at 19:47