Homework Help: Conservative force fields

1. Apr 21, 2005

Hoppa

hello im having a few troubles on these two force field problems, determining whether that are conservative or not.

F = (x, y, z) / (x^2 + y^2 + z^2)^3/2

and

F = (x, y, z) / (x^3 + y^3 + z^3)

i know that when the force is independent of the path then the force is said to be conservative, but how to i show that.
also i know i got to use the equation
W = Integral (t1 - t0) F mulitplied by v dt
but how do i implement this?

2. Apr 21, 2005

James R

Are you familiar with the curl of a vector field?

F is conservative if $\nabla \times F = 0$.

3. Apr 21, 2005

Theelectricchild

you can also check your work--- that is, when a vector field is curl-free, it will do no work on, for example, a particle along a closed path.

4. Apr 22, 2005

Hoppa

no i not familiar with the curl of a vector field? should i be to be able to do these questions?

5. Apr 22, 2005

HallsofIvy

A force field is conservative if and only if there exist some "potential" from which the force can be derived. That is, mathematically, if there exist some scalar function $\phi$ such that $F= \nabla \phi$.
That is, the x coordinate of F is the derivative of $\phi[/tex] with respect to x, the y coordinate the derivative with respect to y, the z coordinate the derivative with respect to z. It isn't necessary to FIND [itex]phi$ to determine if F is conservative- use the fact that the mixed second derivatives are independent of order (as long as they are continuous- which is true here). That, if F(x,y,z) is given by the vector f(x,y,z)i+ g(x,y,z)j+ h(x,y,z)k then F is conservative if and only if
$f_y= g_x, f_z= h_x, g_z= h_y$ (subscripts indicate partial derivatives).

Last edited by a moderator: Apr 22, 2005
6. Apr 25, 2005

Hoppa

ok what does that mean?

7. Apr 25, 2005

whozum

It means for vector field f(x,y,z), it is conservative if and only if the partial derivatives mentioned in his post are equal. If you have f(x,y,z) you can take these partial dreivatives and compare.

8. Apr 25, 2005

Hoppa

ok how do i show whether it is conservative or not? like what is the working that i need?

Last edited: Apr 25, 2005
9. Apr 25, 2005

whozum

Can you break your vector field into its components?

Is it F(x,y,z) = (x^2i + y^2j + z^2k)^(3/2)?

If so, I dont think it can get simpler than what HallsofIvy said:

10. Apr 25, 2005

Theelectricchild

Understanding the basics of divergence and curl are essential to know the meanings of the math that you are doing--- if your instructor has not explained the curl of a vector field to you, it baffles me as to why you are required to solve the aforementioned problem.

11. Apr 25, 2005

whozum

We learned curl and divergence way later than what he's doing here. curl and divergence were the last section in vector calculus here.

12. Apr 25, 2005

Hippo

Really? That makes no sense whatsoever. :yuck:

(Make no mistake, though; I believe you.)

13. Apr 25, 2005

whozum

Surely youve done this kind of problem in class, without using curl of divergence. Do you have ANY kind of idea on what to do?

14. Apr 25, 2005

James R

Of course, you CAN show that a force is NOT conservative without using curl. All you need to do is to choose two different paths in space and show that the work done by the force along each path is different.

15. Apr 25, 2005

Hippo

ANY kind of idea? Well, yeah.

I would evaluate $$\nabla \times \vec{F}$$.

That's a simple and straightforward way.

Look, I understand the definition of a conservative field, and, as it demands a "potential" function, it implies that the curl of a conservative field will always be zero.

I think one or two have suggested evaluating $$\oint_C \vec{F} \bullet ds$$ which is fine and dandy.

However, line integration over a closed curve is really a silly way of doing it; the Curl Theorem implies that such a thing will vanish so long as the curl(F) is zero, and the integration itself is guarunteed to be more difficult than evaluating $$\nabla \times \vec{F}$$.

I was just commenting on how strange the subject order you mentioned seemed to me, nothing more.

Can you clarify that, James? Or explain further?

Last edited by a moderator: Apr 25, 2005
16. Apr 25, 2005

jdavel

HallsofIvy,

But fy = gx, fz = hx, gz = hy are exactly the conditions that the z, y and x components (respectively) of curlF are all zero. Right?

Hippo, calculating a closed line integral can disprove that F is conservative if the integral isn't zero. But if it is zero, you haven't proven that F is conservative. That requires that EVERY closed loop line integral be zero. You either need to learn HallsofIvy's conditions or learn what the curlF is.

Last edited: Apr 25, 2005
17. Apr 25, 2005

whozum

He's telling you to evaluate the line integral for two different lines.
If you know how to solve this, then whats the problem here?

edit: confused your name with the OP.

18. Apr 26, 2005

James R

Hippo:

Are you the same person as Hoppa?

I'm confused.

19. Apr 26, 2005

Theelectricchild

how could he be the same person?

20. Apr 26, 2005

James R

I assume it is possible for one person to register here under 2 different names. Isn't it?

21. Apr 26, 2005

whozum

Both people talk in the first person about the same problem, as if they have to solve it, yet one of them defines the curl function as a solution and the other seemingly has never heard of curl.

22. Apr 26, 2005

Hippo

No, I'm not Hoppa.

Whozum, you wrote that.
I assumed the above was a response to my comment and felt rather disgruntled at the way it was phrased.
That's why I gave my explanation of why I'd use curl(F) for a solution, in preference to any other method.

I'm sorry for the misunderstanding, whozum.

"Two different lines"?

I mean to say, that's a very vague way to put it.

If he was referring to this condition for a conservative field, $$\vec{F}$$,

$$\int_{C_1} \vec{F} \bullet ds = \int_{C_2} \vec{F} \bullet ds$$

where the curves $${C_1}$$ and $${C_2}$$ have the same endpoints, then I understand completely.

However, I couldn't tell if he meant that.

Last edited by a moderator: Apr 26, 2005
23. Apr 26, 2005

Theelectricchild

24. Apr 26, 2005

whozum

It wasnt in response to you, I'm sorry, and your names are so similar I thought the OP was talking.

The "two different lines" was in reference to

"All you need to do is to choose two different paths in space and show that the work done by the force along each path is different."
by James R

taking 'paths' as 'lines'. It lost some of its meaning in the translation I guess.