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Question Number 186077 by Mastermind last updated on 31/Jan/23

Prove that

▽ ∙ (\emptyset \bar{A}) = (▽ \emptyset) ∙ A + \emptyset (▽ ∙ \bar{A}) .

Help!

Answered by aleks041103 last updated on 30/Apr/23

I n t e r e s t i n g c h o i c e o f n o t a t i o n \dots

w h a t y o u w a n t i s

d i v (a \vec{v}) = g r a d (a) . \vec{v} + a d i v (\vec{v})

F i r s t w a y :

d i v (a v) = \frac{\partial ({a v}_{x})}{\partial x} + \frac{\partial ({a v}_{y})}{\partial y} + \frac{\partial ({a v}_{z})}{\partial z} =

= v_{x} \frac{\partial a}{\partial x} + v_{y} \frac{\partial a}{\partial y} + v_{z} \frac{\partial a}{\partial z} + a (\frac{\partial v_{x}}{\partial x} + \frac{\partial v_{y}}{\partial y} + \frac{\partial v_{z}}{\partial z}) =

= (\begin{matrix} v_{x} \\ v_{y} \\ v_{z} \end{matrix}) . (\begin{matrix} \partial a / \partial x \\ \partial a / \partial y \\ \partial a / \partial z \end{matrix}) + a d i v (v) =

= v . g r a d (a) + a d i v (v) = d i v (a v)

S e c o n d w a y : u s i n g e i n s t e i n n o t a t i o n

d i v (a v) = \partial_{i} ({a v}^{i}) = v^{i} (\partial_{i} a) + a (\partial_{i} v^{i}) =

= v . g r a d (a) + a d i v (v) = d i v (a v)

N o t e :

I d r o p p e d t h e \vec{} o n t o p o f t h e v f o r e a s y

t y p e s e t t i n g .