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\begin{center}
\textbf{Vector Calculus}

\textit{\textbf{Grad, Div and Curl Identities}}
\end{center}

\textbf{Question}

It is given that $\phi$ and $\psi$ are scalar fields and $\un{F}$ and
$\un{G}$ are vector fields. They are all assumed to be smooth
functions. Prove the following identity

$$\nabla(\un{f}\bullet\un{G}) = \un{F}\times(\nabla\times\un{G}) +
\un{G}\times(\nabla \times\un{F}) + (\un{F}\bullet\nabla) \un{G} +
(\un{G} \bullet \nabla ) \un{F}$$


\textbf{Answer}

The first component of $\nabla (\un{F}\bullet\un{G})$ is
$$\frac{\pa F_1}{\pa x}G_1 + F_1\frac{\pa G_1}{\pa x} + \frac{\pa
F_2}{\pa x}G_2 + F_2\frac{\pa G_2}{\pa x} + \frac{\pa F_3}{\pa x}G_3 +
F_3\frac{\pa G_3}{\pa x}.$$

We now need the calculate the first components of the terms on the
right hand side of the equation.

For $\un{F}\times(\nabla\times\un{G})$, the first component is
$$F_2 \left ( \frac{\pa G_2}{\pa x} - \frac{\pa G_1}{\pa y} \right ) -
F_3 \left ( \frac{\pa G_1}{\pa z} - \frac{\pa G_3}{\pa x} \right ).$$
For $\un{G}\times(\nabla\times\un{F})$, the first component is
$$G_2 \left ( \frac{\pa F_2}{\pa x} - \frac{\pa F_1}{\pa y} \right ) -
G_3 \left ( \frac{\pa F_1}{\pa z} - \frac{\pa F_3}{\pa x} \right ).$$ 
For $(\un{F}\bullet\nabla)\un{G}$, the first component is
$$F_1 \frac{\pa G_1}{\pa x} + F_2\frac{\pa G_1}{\pa y} + F_3\frac{\pa
G_1}{\pa z}.$$
For $(\un{G}\bullet\nabla)\un{F}$, the first component is
$$G_1 \frac{\pa F_1}{\pa x} + G_2\frac{\pa F_1}{\pa y} + G_3\frac{\pa
F_1}{\pa z}.$$

By adding these first components together it can be seen that all of
the terms cancel out except those in the first component of
$\nabla(\un{F}\bullet\un{G}$. Similar calculations for the other
components yield similar results, hence 
$$\nabla(\un{f}\bullet\un{G})
= \un{F}\times(\nabla\times\un{G}) +
\un{G}\times(\nabla \times\un{F}) + (\un{F}\bullet\nabla) \un{G} +
(\un{G} \bullet \nabla ) \un{F}$$

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