How do I tell if a Vector Field is a Gradient Vector Field?
I have the F=-2xi-3yj+5zk

Does it have something to do with it being conservative or not?

2. Originally Posted by Jen
How do I tell if a Vector Field is a Gradient Vector Field?
I have the F=-2xi-3yj+5zk

Does it have something to do with it being conservative or not?
not exactly. conservative vector fields are a special kind of vector field. this problem is just asking you whether F is a vector field or not.

if $\bold{F}$ is a vector field, then there is some scalar function $f$ (particularly here, one in three variables) such that $\nabla f = \bold{F}$. you are required to find such a function. can you do it?

3. Del f=F

So showing that there is a function such that delf=F says that it is a gradient vector field. I think that I can find f. I just didn't know that this showed it being a Gradient Vector field.

Thank you!

4. Originally Posted by Jen
So showing that there is a function such that delf=F says that it is a gradient vector field. I think that I can find f. I just didn't know that this showed it being a Gradient Vector field.

Thank you!
well, now you know. it just so happens that such a function f exists if F is a conservative vector field. but there are ways to prove F is conservative without going through the trouble of finding f

just remember that $\nabla f$ is the gradient vector of a function $f$. it is only appropriate that if the components of $\nabla f$ are functions, that it be called a gradient vector field, since that's what vector fields are

5. Other ways

What are the other ways of determining if it's conservative?
Can't you show that it is conservative by showing that partial P/dx, and partial Q/dy are equal? by Clairaut's Theorem (I don't think I spelled that right)? Are there other way's?

6. Originally Posted by Jhevon
well, now you know. it just so happens that such a function f exists if F is a conservative vector field. but there are ways to prove F is conservative without going through the trouble of finding f

just remember that $\nabla f$ is the gradient vector of a function $f$. it is only appropriate that if the components of $\nabla f$ are functions, that it be called a gradient vector field, since that's what vector fields are

That makes sense. I can't believe that I didn't think of that!!

Thanks

7. If $F = I(x,y,z)i + J(x,y,z)j + K(x,y,z)k$ then F is conservative if $\left\{ {\begin{array}{c} {K_y = J_z } \\ {K_x = I_z } \\ {J_x = I_y } \\\end{array}} \right.$.

8. Originally Posted by Jen
What are the other ways of determining if it's conservative?
Can't you show that it is conservative by showing that partial Pdx, and partial Qdy are equal? by Clairot's (I don't think I spelled that right)? Are there other way's?
you mean $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$. and there are other things to consider as well. $\bold{F}$ must be defined on an open simply connected region, call it $D$, and $P$ and $Q$ must have continuous first partial derivatives throughout $D$. (of course, here, $\bold {F} = P \bold{i} + Q \bold {j}$).

another way to tell if $F$ is conservative is the following:
again, if $\bold{F}$ is on an open simply-connected region, $D$: if $\oint_C \bold{F} \cdot d \bold{r}$ is independent of path in $D$ (iff $\oint_C \bold{F} \cdot d \bold{r} = 0$ for every closed curve $C$ in $D$) then $\bold{F}$ is conservative.

those are the only two ways i remember.

EDIT: Oh, Plato gave another way

9. Originally Posted by Plato
If $F = I(x,y,z)i + J(x,y,z)j + K(x,y,z)k$ then F is conservative if $\left\{ {\begin{array}{c} {K_y = J_z } \\ {K_x = I_z } \\ {J_x = I_y } \\\end{array}} \right.$.
do you have a mnemonic for that set of equations, Plato?

10. Originally Posted by Jhevon
do you have a mnemonic for that set of equations, Plato?
Yes I do. I have used the notation for F with classes for years.
The capital I, J, and K are function fields that go naturally with the vector notation i, j, and k.
Conservative means that the curl(F)=0 so the determinate is zero.
Note that in each case the corresponding variable is missing in the partials.
Dose that help?

11. So the curl of F being zero also means that it is conservative? I thought that just meant that it didn't tend to rotate about that point. O.k. I just found it in my book. That makes sense.

12. Originally Posted by Plato
Yes I do. I have used the notation for F with classes for years.
The capital I, J, and K are function fields that go naturally with the vector notation i, j, and k.
Conservative means that the curl(F)=0 so the determinate is zero.
Note that in each case the corresponding variable is missing in the partials.
Dose that help?
Yes, indeed! that does help! thanks. i never realized that conservative means $\mbox{curl} \bold{F} = \bold{0}$.

13. Originally Posted by Jhevon
Yes, indeed! that does help! thanks. i never realized that conservative means $\mbox{curl} \bold{F} = \bold{0}$.
You have never been in one of my classes then.

14. O.k. so I got $f=-x^2-\frac{3}{2}y^2+\frac{5}{2}z^2$ because del of this would be the F I wanted. Right?

Also, if curlF=0 means that the vector field is conservative, does Del dot F =0 mean anything.

Also, how do I do the integral, Del, and Dot, symbols in this math type?

15. Originally Posted by Plato
You have never been in one of my classes then.
nope. i never took a class in vector calculus before. we touched on some in advanced calc 2, but we never did things like conservative vector fields. what i know about that comes from personal study

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