Question

Given a vector function F = (3x)i + (4y)j -(7z)k, how can I find all vectors G such that F = curl(G)?

Answer 1

curl(F)=0
curl(curl(G))=0
laplacian(G)-gard(Div(G))=0

Answer 2

I know...but what's your point?

Answer 3

actually i'm lost here...it seems to have many solutions..i tried to solve it using the help of Helmholt'z equation..

Answer 4

yeah its a little difficult and bugging me. I'll keep working at it, thanks though

Answer 5

\[\frac{d}{dy}G_z-\frac{d}{dz}G_y=3x\] and so on..you can also see that this type of problems generally have infinite solution..

Answer 6

yeah i have those bunch of equations written down too! I'll keep at it.

Answer 7

one thing i can suggest that take G(x,y,z)=X(x)Y(y)Z(z). this may help you

Answer 8

yup ive done that and came up with those eqns. thanks tho.

Answer 9

this problem seems to look like that you have given the mag. vector potential and you need to find the mag. field

Answer 10

haha its exactly that if you apply the physics application.

Answer 11

i gotta go...so bye..and sorry that i couldn't help you.

Answer 12

its all good

Answer 13

G(x,y,z) can not be equal to X(x)Y(y)Z(z) becuase G is a vector field

Answer 14

he meant write it in the form of of G = Pi + Qj + Rk

Answer 15

It can be
G=4yzi-3xzj+3zk

Answer 16

i need a general solution,

Answer 17

G=4yzi-3xzj+Z(z)k also satisfying

Answer 18

its easy to come up with a single example...i need for all vectors G

Answer 19

OK wait..!!

Answer 20

ok

Answer 21

so?

Answer 22

it's a lengthy sol

Answer 23

just describe in short what u did, ill understand

Answer 24

Gx=X1(x)Y1(y)Z1(z)
Gy=X2(x)Y2(y)Z2(z)
Gz=X3(x)Y3(y)Z3(z)

Answer 25

that means every component of G is a fcn of x,y,z... got that already :) proceed

Answer 26

X3*Y3'*Z3-Z2'*X2*Y2=3x
Z1' *Y1*X1-X3'*Y3*Z3=4y
X2'*Y2*Z2-Y1'*X1*Z1=-7z

Answer 27

now can you solve this system by your self??

Answer 28

Here's what I have, cuz youre notation is a little strange....

Let G = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k

So if curl(G) = F,

3x = (dR/dy)-(dQ/dz)
4y = (dP/dz)-(dR/dx)
-7z = (dQ/dx)-(dP/dy)

Answer 29

these are partial derivatives not exact. put Gx,Gy,Gz into that equation you wrote

Answer 30

i know they are partials.
ex dR/dy is my partial deriv of R with respect to y

Answer 31

put
Gx=X1(x)Y1(y)Z1(z)
Gy=X2(x)Y2(y)Z2(z)
Gz=X3(x)Y3(y)Z3(z)
into
3x = (dR/dy)-(dQ/dz)
4y = (dP/dz)-(dR/dx)
-7z = (dQ/dx)-(dP/dy)

Answer 32

By Gx, do you mean the x-component of G?

If so, then why do you write it as a product X1*Y1*Z1

Answer 33

x component can be function of x,y,z, , you are dealing with vector fields!!

Answer 34

i know that

Answer 35

i dont understand your notation of X3*Y3*Z3 for example....

Answer 36

you can write whatever you want I just wanted to write X1(x) is different to X3(x)

Answer 37

ok fine i agree.

but it doesn't necessarily have to be a product of xyz for Gx,Gy and Gz...

Answer 38

for example??

Answer 39

a sum

Answer 40

(x1+x2)(y1+y2)=x1y1+x1y2+x2y1+x2y2

Answer 41

your point?

Answer 42

my point is nearly every expression can be factored so we can show in product, so in product nearly every possiblity is described

Answer 43

hmm ok.

Answer 44

even still, how would i solve that system you gave me before

Answer 45

if you look at the first equation it's function of x so y3' z3=z2' y2=c
so y3'/y2=z2'/z3=c
so y3=c[integral]y2 dy+C1
and z2=c[integrall] z3 dz+C2
so first eqution becomes
c(x3-x2)=3x
I think you can proceed further by yourself :)

Answer 46

hmm ill see. thanks tho

Answer 47

I'm thinking about one more approach

Answer 48

let G=(G1,G2,G3)
G1=G1(x,y,z) so
dG1=G1x dx+G1y dy+G1z dz
from equations
4y = (dP/dz)-(dR/dx)
-7z = (dQ/dx)-(dP/dy) we'll have G1=7zy+4yz+[integral] (partial derivativ w.r.t xG1)).dr similarly you can do for G2 and G3

Answer 49

one more approach
3x = (dR/dy)-(dQ/dz)
4y = (dP/dz)-(dR/dx)
-7z = (dQ/dx)-(dP/dy)
let R=constant
Q=-3xz+C(x,y)
P=4yz+C(x,y).....