Answer to Question #275781 in Calculus for Nawrin A

Question

 If A

 and B

 are vector fields, prove the following:

∇(A⋅B)=(B⋅∇)A+(A⋅∇)B+B×(∇×A)+A×(∇×B).

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Answer 1

Consider,

$\nabla(\vec{A}\cdot\vec{B})=\sum\vec{i}\frac{d}{dx}(\vec{A}\cdot\vec{B})=\sum\vec{i}(\frac{d\vec{A}}{dx}\cdot\vec{B}+\vec{A}\cdot\frac{d\vec{B}}{dx})$

$=\sum\vec{i}(\frac{d\vec{A}}{dx}\cdot\vec{B})+\sum(\vec{A}\cdot\frac{d\vec{B}}{dx})$ ..............(i)

Now $\vec{B}(\vec{i}*\frac{d\vec{A}}{dx})=(\vec{B}\cdot\frac{d\vec{A}}{dx})\vec{i}-(\vec{B}\cdot\vec{i})\frac{d\vec{A}}{dx}$

$=(\vec{B}\cdot\frac{d\vec{A}}{dx})\vec{i}=\vec{B}*(\vec{i}*\frac{d\vec{A}}{dx})+(\vec{B}\cdot\vec{i})\frac{d\vec{A}}{dx}$

$\therefore \sum(\vec{B}\cdot\frac{d\vec{A}}{dx})\vec{i}=\vec{B}*(\vec{i}*\frac{d\vec{A}}{dx})+(\vec{B}\cdot\vec{i})\frac{d\vec{A}}{dx}$

$=\vec{B}*\sum(\vec{i}*\frac{d\vec{A}}{dx})+(\vec{B}\cdot\sum\vec{i}\frac{d}{dx})\vec{A}$

$=\vec{B}*(\nabla*\vec{A})+(\vec{B}\cdot\nabla)\vec{A}$

$\therefore\sum\vec{i}(\frac{d\vec{A}}{dx}\cdot\vec{B})=\vec{B}*(\nabla*\vec{A})+(\vec{B}\cdot\nabla)\vec{A}............(ii)$

Similarly, if we interchange the role of $\vec{A}$ and $\vec{B}$ we can prove;

$\sum\vec{i}(\vec{A}\cdot\frac{d\vec{B}}{dx})=\vec{A}*(\nabla*\vec{B})+(\vec{A}\cdot\nabla)\vec{B}............(iii)$

Substituting (ii) and (iii) in (i), we get:

$\nabla(\vec{A}\cdot\vec{B})=(\vec{B}\cdot\nabla)\vec{A}+(\vec{A}\cdot\nabla)\vec{B}+\vec{B}*(\nabla*\vec{A})+\vec{A}*(\nabla*\vec{A})$