Conservative force field

[triclino]

In how many different ways can we show that the following force field is conservative??

 

 

F= [math](2xy+z^3[/math]) i + [math]x^2 [/math]j + [math]3xz^2[/math] k

[Cap'n Refsmmat]

http://math.scienceforums.net/Vector%20calculus/Conservative%20fields

 

Quite a few, as I wrote in the above link.

[triclino]

http://math.scienceforums.net/Vector%20calculus/Conservative%20fields

 

Quite a few, as I wrote in the above link.

 

Thanks ,how do we find the function Φ(x,y,z) ,such that :

 

 

F = gradΦ(x,y,z)

[timo]

In this case the best way would be to know what the integral over and the derivative of a polynomial function looks like. Not sure if this is homework so I'll not give a potential. It should be straightforward to find one by playing around with possible choices a bit.

[triclino]

Since F = gradΦ(x,y,z) =>

 

F(x) = [math]\frac{\partial\Phi(x,y,z)}{\partial x}[/math]

 

F(y) = [math]\frac{\partial\Phi(x,y,z)}{\partial y}[/math]

 

F(z) = [math]\frac{\partial\Phi(x,y,z)}{\partial z}[/math] and

 

 

[math]\Phi(x,y,z) = \int(3xz^3)dz + C[/math]

 

 

 

AND now what??

Edited April 7, 2010 by triclino
correction

[Cap'n Refsmmat]

Thanks ,how do we find the function Φ(x,y,z) ,such that :

 

 

F = gradΦ(x,y,z)

 

I usually don't -- I just compare the mixed partial derivatives to be sure they're equal, as given in the last section of my link.

[swaha]

Since F = gradΦ(x,y,z) =>

 

F(x) = [math]\frac{\partial\Phi(x,y,z)}{\partial x}[/math]

 

F(y) = [math]\frac{\partial\Phi(x,y,z)}{\partial y}[/math]

 

F(z) = [math]\frac{\partial\Phi(x,y,z)}{\partial z}[/math] and

 

 

[math]\Phi(x,y,z) = \int(2xy+z^3)dz + C[/math]

 

 

 

AND now what??

 

just compare the results. when we take the partial derivative wrt x we cancel out the chance that a function of y & z may remain & similar is the case of y & z partial derivatives. hence the requirement of comparison. the function of (x,y,z) will remain in all three & will be taken once in the phi.

[timo]

[math]\Phi(x,y,z) = \int(2xy+z^3)dz + C[/math]

AND now what??

1) I'd start with [math]\Phi = \int 2xy + 3z^2 dx \ + C[/math] for a start, i.e. you integrate [math]\frac{\partial \Phi}{\partial x}[/math] over x to get [math]\Phi[/math], not over z.

2) Your constant can still depend on x and y, i.e. [math]\Phi = \int 2xy + 3z^2 dx \ + C(x,y)[/math].

3) What happens when you do the same for y and z?

[D H]

Since F = gradΦ(x,y,z) =>

 

F(x) = [math]\frac{\partial\Phi(x,y,z)}{\partial x}[/math]

F(y) = [math]\frac{\partial\Phi(x,y,z)}{\partial y}[/math]

F(z) = [math]\frac{\partial\Phi(x,y,z)}{\partial z}[/math] and

 

[math]\Phi(x,y,z) = \int(2xy+z^3)dz + C[/math]

Where did you get this last equation? It isn't right, so let's drop it.

 

Next, regarding the first three equations: let's get the nomenclature right. The first is better written as [math]F_x[/math] rather than [math]F(x)[/math]. The former indicates the x-component of the force vector; the latter implies that F is a function of x -- and x only. Secondly, you are missing a minus sign. The convention is [math]\vec F = - \nabla \phi[/math].

 

That said, given that [math]F_x = 2xy+z^3[/math] suggests that the potential is of the form

 

[math]\phi(x,y,z) = -x(xy + z^3) + f(y,z)[/math].

 

Continuing with the y-component of the force vector,

 

[math]-\,\frac{\partial \phi(x,y,z)}{\partial y} = x^2 - \frac{\partial f(y,z)}{\partial y}[/math]

 

Since [math]F_y = x^2[/math], this says that f(y,z)[/math] has no dependence on y. So let's call it f(z). Doing the same with the z-component, we find that that f(z) can have no dependence on z, either. All that is left is a constant, so

 

[math]\phi(x,y,z) = -x(xy + z^3) + C[/math]

 

If you can construct a potential function you know that the force is a conservative force. The contrapositive is not necessarily true. Failure to construct a potential function might just mean that you couldn't construct a potential function. You would have to prove that no such function could possibly exist to prove that a force is not conservative.

 

One way to prove this is to find a pair of states [math](x_0, y_0, z_0)[/math] and [math](x_1, y_1, z_1)[/math] and two distinct paths between these points. If the change in mechanical energy depends on the path taken the force is necessarily not conservative.

[timo]

Failure to construct a potential function might just mean that you couldn't construct a potential function.

Considering that the derivatives are all polynomials I would expect that the construction of a potential function by straightforward integration (also polynomials) either leads to a potential function or a contradiction at some point. Neither do I have a sketch for a proof of that statement (I'd expect that you can actually construct an algorithm for this case) on my notebook not am I completely sober at the moment, though.

[D H]

In this simple case the force components are all polynomials. In the general case that is not true.

[triclino]

Where did you get this last equation? It isn't right, so let's drop it.

 

Next, regarding the first three equations: let's get the nomenclature right. The first is better written as [math]F_x[/math] rather than [math]F(x)[/math]. The former indicates the x-component of the force vector; the latter implies that F is a function of x -- and x only. Secondly, you are missing a minus sign. The convention is [math]\vec F = - \nabla \phi[/math].

 

That said, given that [math]F_x = 2xy+z^3[/math] suggests that the potential is of the form

 

[math]\phi(x,y,z) = -x(xy + z^3) + f(y,z)[/math].

 

Continuing with the y-component of the force vector,

 

[math]-\,\frac{\partial \phi(x,y,z)}{\partial y} = x^2 - \frac{\partial f(y,z)}{\partial y}[/math]

 

Since [math]F_y = x^2[/math], this says that f(y,z)[/math] has no dependence on y. So let's call it f(z). Doing the same with the z-component, we find that that f(z) can have no dependence on z, either. All that is left is a constant, so

 

[math]\phi(x,y,z) = -x(xy + z^3) + C[/math]

 

If you can construct a potential function you know that the force is a conservative force. The contrapositive is not necessarily true. Failure to construct a potential function might just mean that you couldn't construct a potential function. You would have to prove that no such function could possibly exist to prove that a force is not conservative.

 

One way to prove this is to find a pair of states [math](x_0, y_0, z_0)[/math] and [math](x_1, y_1, z_1)[/math] and two distinct paths between these points. If the change in mechanical energy depends on the path taken the force is necessarily not conservative.

 

Thank you all .

D.H ,how would you calculate the Φ(x,y,z) of the force field:

 

F = (x+2y+4z) i + (2x-3y-z) j + (4x-y +2z) k.

 

You do not have to use the minus sign .

 

I just want to see the procedure .

[D H]

This is starting to look too much like homework, triclino, so for now, I pass.

[PaulS1950]

I sure am glad the internet wasn't around when I was in school.

I get to learn so much more now!