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

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

 

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

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)

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.

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, 201015 yr by triclino
correction

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.

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.

[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?

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.

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.

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

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 .

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

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

I get to learn so much more now!