Question

Vector calc

Answer 1

@rational Let \[\textbf{r} = x\textbf{i}+y\textbf{j}+z \textbf{k}~~~~\text{and}~~~r=|\textbf{r}|\]
If \[\textbf{F} = \frac{ \textbf{r} }{ r^p }\], find div F. Is there a value of p for which div F = 0.

Answer 2

Oh so silly, I kept thinking |r| was absolute value, as I'm used to ||r||...haha. No wonder it didn't make sense.

Answer 3

. . . . . . . . . .

Answer 4

Ok it's not that bad I guess, just have to find the partials using product rule, but now the tricky part must be where it equals = 0.

Answer 5

dang...im so confused....u must be in high school...XD

Answer 6

Nvm, it's simple.

Answer 7

High school, pfft :P

Answer 8

yep i knew it...

Answer 9

th pfft is because he isn't in high school silly

Answer 10

oh...

Answer 11

Thanks for coming rational but I figured it out haha :P

Answer 12

. . . . . . . . .

Answer 13

whats the answer

Answer 14

jesus crist....

Answer 15

\[F = \frac{ x i + yj + z k }{ \left( \sqrt{x^2+y^2+z^2} \right)^p}\] set up and getting all the partials and adding..\[ = \frac{ 3-p }{ \sqrt{x^2+y^2+z^2} }\] so obviously it's 0 when p = 3

Answer 16

woah O_O

Answer 17

Bella please dont take God's name in vein

Answer 18

vain

Answer 19

lol

Answer 20

does that mean in 2D the divergence becomes 0 when p=2 ?

Answer 21

Trick question

Answer 22

It looks like it

Answer 23

it looks very interesting
for a field pointing along position vector, the divergence becomes 0 after scaling it down.. hmm weird too

Answer 24

also that exponent matches exactly with the dimension of space.. worth thinking..

Answer 25

Fascinating, I did not notice that