Question

\[A=(2yz)i-(x ^{2}y)j+(xz ^{2})k,B=(x ^{2})i+(yz)j-(xy)k\]what is\[(B.\nabla)A\]

Answer 1

\[(\vec B\cdot\nabla)\vec A\]?

Answer 2

...as in the scalar multiple of \(\text{div}\vec B\) with \(\vec A\) ???

Answer 3

should be, lol

Answer 4

i cant obtain the answer provided by my tutor

Answer 5

do you know how to get the divergence of \(\vec B\)\[\text{div}\vec B=?\]

Answer 6

yea. i get \[(2x+z+0)\]

Answer 7

or i get it wrongly
LOL

Answer 8

nope that's right

Answer 9

then the scalar multiplication, i get\[(2yz(2x+z), -x ^{2}y(2x+z), 0)\]but the answer provided is \[(2y(z ^{2}-xy), -x ^{2y}(2x+z), x ^{2}z(z-2y))\]

Answer 10

well that's not \[(\nabla\cdot\vec B)\vec A\]

Answer 11

let's see if it'swell that's not \[(\nabla\cdot\vec A)\vec B\]real quick maybe?

Answer 12

u mean the answer's has problem?

Answer 13

yeah, if the question is\[(\text{div}\vec B)\vec A\]then the book made a mistake or something
...unless I've lost my mind, which happens from time to time at 3:30am

Answer 14

if u dont mind, 1more question?

Answer 15

sure, try me

Answer 16

given\[\phi =2x ^{2}yz ^{3}\]
then what is \[(A \times \nabla)\phi\]

Answer 17

the cross product i get\[(z ^{2}-2y)j+(2z+2xy)k\], correct?

Answer 18

is that just\[\vec A\times\nabla\]up there?
'cause that's right as far as I can see for that, but I don't see \(\phi\)

Answer 19

yes its vector of A crossed with nabla, and outside the phi is given as\[2x ^{2}yz ^{3}\]

Answer 20

T_T the test is 2mr and his tutorial aint helping anytg lol

Answer 21

lol
just multiply be \(\phi\) as a scalar now and I guess that's it
kinda dumb problem if you ask me

Answer 22

it want's the negative curl and a simple scalar multiplication
just an exercise I guess

Answer 23

i get \[2x ^{2}yz ^{3}(z ^{2}-2y)j+2x ^{2}yz ^{3}(2z+2xy)k\]
the answer given is \[-2x ^{3}z ^{2}(3xy ^{2}+z ^{3})i+4x ^{2}yz ^{2}(z ^{2}-3y)j+4x ^{2}yz ^{3}(z+xy)k\]._.

Answer 24

the i popped out from nowhere

Answer 25

we are not understanding what they want it seems

Answer 26

now the problem is to figure out what they are asking lol
because it ain't\[(\vec A\times\nabla)\phi\]with phi as a scalar multiple

Answer 27

like you said, the \(\hat i\) component is clearly zero after \(\nabla\times A\)

Answer 28

i think i'll just leave those aside.. lol. mayb my tutor is blur afterall

Answer 29

I feel bad I can't help, but something is amiss here
better luck next time, promise
as for me, it's goodnight :)

Answer 30

kk nitez