Question

Caclulus III - Determine whether or not the vector field is conservative ?
F(x,y,z)=2xy,(x^2+2yz),y^z)

this page is quite useful
http://tutorial.math.lamar.edu/Classes/CalcIII/ConservativeVectorField.aspx#LineInt_Conserv_Ex1a
bt only deals with 2 instead of 3 sets of equations so abit confused

Answer 1

nope I was wrong

Answer 2

Let F(x,y,z)=<P(x,y,z),Q(x,y,z),R(x,y,z)>, then F is conservative if:
\[1) P_y=Q_x\]
\[2) P_z=R_x\]
\[3)Q_z=R_y\]

Answer 3

that's what I found :)

Answer 4

That's good! :)

Answer 5

thanks for the replies
the link talks about finding P and Q, what are my P and Q ?

Answer 6

In the given function, P=2xy, Q=x^2+2yz, R=y^z. I think something is wrong with the R though.

Answer 7

It's probably y^2, right?

Answer 8

F(x,y,z)=(2xy,(x^2+2yz),y^2)

P= 2xy Dp/dy=2x

Answer 9

sorry yes it is, y^2, my mistake

Answer 10

Q=x^2+2yz ...Dq/Dx = 2x ?

Answer 11

Yep!

Answer 12

R=y^2 Dr/dz= 0 ?
do P,Q,R have to be the same for it to be conservative or just P and Q

Answer 13

My text book doesnt refer to i and j, so I just assume they are present

Answer 14

The notation <P,Q,R> is the same as Pi+Qj+Rk.

Answer 15

Thanks,
im not sure which order to differentiate it
e.g
should be 2xy be differentiated with respect to x,y or z ? the link uses y , but doenst give reason for why the first equation is done with respect to y

Answer 16

Following Mr.Math's formula we get\[P_y=Q_x\to2x=2x\]\[P_z=R_x\to0=0\]\[Q_z=R_y\to2y=2y\]So it appears to be conservative.
I don't normally give answers like this, but this was news to me as well, so I wanted to try it.

Answer 17

By definition, a vector field V in \( \mathbb{R}^3 \) is conservative if there is a scalar field \( \phi \) such
\[ V = \nabla \phi \]
Now, that implies
\[ \nabla \times V = \nabla \times \nabla \phi =0 \]
It turns out that if we're in \( \mathbb{R}^3 \), the opposite is also true:
if \( \nabla \times V = 0 \) then there exists a scalar field \( \phi \) such that \( V = \nabla \phi \).

Hence to show if a vector field V is conservative, it is sufficient to show that
\[ \nabla \times V = 0 \]

Writing V = (P,Q,R), that last condition translates into the equations Mr.M wrote above. But they are not the definition of a vector field being conservative; rather they are the consequence of it.

This concept is really useful in Physics, because a vector field of forces being conservative means we can write a function which expresses the potential energy of that vector force field. For example, gravity, or electric force.

Answer 18

grait explanations