Curl operator for time-varying vector

[Malamala]

Hello! I have the following vector $\mathbf{A} = R(-\sin(\omega t)\mathbf{x}+\cos(\omega t)\mathbf{y})$, where $\mathbf{x}$ and $\mathbf{y}$ are orthogonal unit vectors (aslo orthogonal to $\mathbf{z}$). I want to calculate $\nabla \times \mathbf{A}$, but I am a bit confused. The curl operator involves partial derivatives with respect to x, y and z, so in the form above I would have $\nabla \times \mathbf{A} = 0$ . However, I know that $\sin(\omega t) = \frac{y(t)}{R}$ and $\cos(\omega t) = \frac{x(t)}{R}$ so I can write $\mathbf{A} = R(-\frac{y}{R}\mathbf{x}+\frac{x}{R}\mathbf{y})$, in which case $\nabla \times \mathbf{A} = 2 \mathbf{z}$. Which is the right way to do this and why? Thank you!

 

[andrewkirk]

You need to decide whether you are defining a vector or a vector field. A vector field has a curl. A single vector does not.
The definition you've given above says it is a vector.

Further, a vector field must be defined at every point in the space we are considering, so there is no role for specifying "location at time t" which is what your use of y(t) and x(t) appears to do. Location of what?

To calculate a curl you need a vector field, which means you need a formula that specifies the vector at each point in the space. Although your definition above says it is for a vector, not a vector field, we can interpret it as saying that the field is uniform across space, having that magnitude and direction everywhere. With that interpretation, the curl is zero.