# In the context of curl, what does the difference between two partial derivatives tell me about rotation in a plane?

Let $\mathbf{A}(x,y,z)$ be a vector field. The curl of this vector field is defined as

$$\nabla \times \mathbf{A} = \left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}\right) \mathbf{j} + \left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right) \mathbf{k}$$

Let's consider a simple example from Wikipedia.

$$\mathbf{A}(x,y,z) = y\mathbf{\hat{x}}-x\mathbf{\hat{y}}+0\mathbf{\hat{z}}$$

This corresponds to the following

The curl is

$$\nabla \times \mathbf{A} =0\boldsymbol{\hat{x}}+0\mathbf{\hat{y}}+ \left({\frac{\partial}{\partial x}}(-x) -{\frac{\partial}{\partial y}} y\right)\mathbf{\hat{z}}=-2\mathbf{\hat{z}}$$

But there is where I am confused. I don't get what $\left({\frac{\partial}{\partial x}}(-x) -{\frac{\partial}{\partial y}} y\right)$ has to do with the rotation and size of the vectors in the picture shown above.

# My Question

How do these derivatives tell me anything about the rotation? Any why must they be opposites? Why is it not $\frac{\partial A_x}{\partial x}$? I understand of course why, when you compute the cross product, it comes out that way, but I want to understand the intuition of why that is important for rotation.

• Ah if I had it in front of me there was a really good sort of physical argument about it in a fluids book that made it feel intuitive if not super rigorous. – Triatticus Jan 21 '17 at 3:49