Curl

Curl Derivation

$$Curl\vec{F}=⟨P,Q,R⟩=⟨P,Q,R⟩=⟨\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z},\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x},\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}⟩$$

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Recall that the gradient of $f$ is represented by $\nabla f=⟨\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}⟩$ and that scalar multiplication is $k\vec{V}$.

Therefore, with these two properties, factoring out $f$ will give:

$$f⟨\frac{\partial }{\partial x}\frac{\partial }{\partial y},\frac{\partial }{\partial z}⟩=f\nabla$$

Using $\nabla$, we can take the cross product with $\vec{F}$.

$$\nabla \times \vec{F}=\left|\begin{array}{ccccc}\mathbf{i}& & \mathbf{j}& & \mathbf{k}\\ \frac{\partial }{\partial x}& & \frac{\partial }{\partial y}& & \frac{\partial }{\partial z}\\ P& & Q& & R\end{array}\right|=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\mathbf{i}-\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)\mathbf{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathbf{k}$$

Simplifying,

$$⟨\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z},\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x},\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}⟩=\nabla \times \vec{F}=Curl\vec{F}$$