Conservative Vector Fields and Associated Potential

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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!