Conservative Vector Fields and Associated Potential

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Homework Help Overview

The discussion revolves around finding the potential function for a conservative vector field defined by the vector \(\vec{F}(x, y, z)=[(2xy-z^2), (2yz+x^2), (y^2-2zx)]\). Participants are exploring the steps involved in determining the potential function and identifying where mistakes may have occurred in the process.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integration of the vector field components and the formulation of the scalar potential function. There are questions about the correct application of partial derivatives and the handling of terms in the equations, particularly regarding the placement of the \(-2zx\) term.

Discussion Status

Some participants have provided guidance on taking partial derivatives and equating them with the components of the vector field. There is an acknowledgment of a potential oversight in the original poster's reasoning, but no consensus has been reached on the specific error.

Contextual Notes

Participants are working within the constraints of homework guidelines, indicating that they are not seeking complete solutions but rather clarification on their reasoning and approach.

sandy.bridge
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Homework Statement


Having issues determining what I am doing wrong, so perhaps one of you can pin point it. I have the solution, and I am extremely close to the same result, however, I am nonetheless wrong.

Find the conservative vector fields potential.

[tex]\vec{F}(x, y, z)=[(2xy-z^2), 2yz+x^2), y^2-2zx)][/tex]



The Attempt at a Solution


[tex]\vartheta=\int(2xy-z^2)dx=x^2y-z^2x+C(y, z)[/tex]
then we have
[tex]2yz+x^2=x^2+∂C(y, z)/∂y[/tex]
therefore,
[tex]C(y, z)=zy^2+C(z)[/tex]
It's at the following step that I mess something up.
[tex]y^2-2zx=y^2+∂C(z)/∂z[/tex]
In the solutions however, they have,
[tex]y^2-2zx=y^2-2zx+∂C(z)/∂z[/tex]
However, wouldn't the "-2zx" term go with ∂C(z)/∂z?
 
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Your scalar function ##\phi = x^2y - z^2x + c(y,z)## Then take the partial of this with respect to ##z## and equate with the ##z## component of ##\underline{F}##. Did you mistakenly take the partial with respect to z of something else? You can write $$\frac{\partial c(z)}{\partial z} = c'(z).$$
 
I'm betting that you're going to kick yourself for this !

sandy.bridge said:

Homework Statement


Having issues determining what I am doing wrong, so perhaps one of you can pin point it. I have the solution, and I am extremely close to the same result, however, I am nonetheless wrong.

Find the conservative vector fields potential.

[tex]\vec{F}(x, y, z)=[(2xy-z^2), 2yz+x^2), y^2-2zx)][/tex]

The Attempt at a Solution


[tex]\vartheta=\int(2xy-z^2)dx=x^2y-z^2x+C(y, z)[/tex]
then we have
[tex]2yz+x^2=x^2+∂C(y, z)/∂y[/tex]
therefore,
[tex]C(y, z)=zy^2+C(z)[/tex]
It's at the following step that I mess something up.
So, at this point you have [itex]\displaystyle \ \ \vartheta=x^2y-z^2x+C(y, z)=x^2y-z^2x+zy^2+C(z)\ .[/itex]

Now look at [itex]\displaystyle \ \ ∂\vartheta/∂z\ .[/itex]
[tex]y^2-2zx=y^2+∂C(z)/∂z[/tex]
In the solutions however, they have,
[tex]y^2-2zx=y^2-2zx+∂C(z)/∂z[/tex]
However, wouldn't the "-2zx" term go with ∂C(z)/∂z?
 
Perfect! Not entirely sure how I didn't see that. Figured I should ask rather than stare at it for long. Thanks!
 

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