Showing a vector field is conservative.

  • #1
Ok so I'm new to vector analysis, just started about a week or 2 ago. I'm using Paul C. Matthews' book, "Vector Calculus". This is an example problem from it which I have difficulty understanding because of integration with partial derivatives. The problem is solved, I just have trouble understanding the solution.

Homework Statement


Show that the vector field [tex]F = (2x+y, x, 2z)[/tex] is conservative.

So if [tex]F[/tex] is conservative, it can be written as the gradient of some scalar field [tex]\phi[/tex]. This gives the three equations:

[tex]\frac{\partial \phi}{\partial x}[/tex], [tex]\frac{\partial \phi}{\partial y}[/tex] and [tex]\frac{\partial \phi}{\partial z}[/tex]

After this, they integrate first of the equations with respect to x and this gives [tex]\phi = x^2+xy+h(x,y)[/tex]. This is still alright, because h is analogous to the constant of integration. But after this it says:

"The second equation forces the partial derivative of h with respect to y to be zero so that h only depends on z. The third equation yields [tex]\frac{dh}{dz}=2z[/tex] so [tex]h(z) = z^2+c[/tex] where c is any constant. Therefore, all three equations are satisfied by the potential function [tex]\phi = x^2+xy+z^2[/tex] and F is a conservative vector field"

I didn't really understand much of what is in the quotes except this part:

[tex]\frac{\partial \phi}{\partial y} = x[/tex] and [tex]\frac{\partial (x^2+xy+h(x,y))}{\partial y} = x+\frac{\partial h(x,y)}{\partial y}[/tex] and so h(x,y) is constant but why does this mean that h only depends on z?

Thanks if you can help.
 
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Answers and Replies

  • #2
Cyosis
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You're complicating things. If F is conservative then it can indeed be written as the gradient of a scalar field. What do you know about the curl of the gradient of a function?
[tex]
\nabla \times \nabla \phi=?
[/tex]

Edit:The question is whether F is conservative or not? In the attempt you're trying to determine the scalar field itself. Is this part of your question or?

Either way [itex]\phi = x^2+xy+h(x,y)[/itex] should be [itex]\phi = x^2+xy+h(y,z)[/itex]. Now lets take the gradient of this scalar field.
[tex]
\nabla \phi=(2x+y,x+\partial_y h(y,z),\partial_z h(y,z)[/tex]

Do you see now why [itex]\partial_y h(y,z)=0[/itex]?
 
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  • #3
Yes, the solution to the question is to determine the scalar field. At this point, curl has not been covered, so we're supposed to use the idea that a conservative vector field F is the gradient of a scalar function.

Ty
 
  • #4
Cyosis
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Ah I edited my first post to explain the method you listed. The nice part of the curl equation is that the curl of a gradient is always 0. So if F is conservative then F can be written as grad f and therefore [itex]\nabla \times F=\nabla \times \nabla f=0[/itex]. Which is a very quick way to check if a field is conservative.
 
  • #5
Oh, so is h dependent only on z because it's partial with respect to y is 0 so the function doesn't change for any change in y? And partial phi with respect to z confirms this?

Is that right?
 
  • #6
Cyosis
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I am not sure what you mean with your last sentence, but we know that [itex]\nabla phi=F[/itex].

We know that [itex]F= (2x+y,x,2z)=\nabla \phi=(2x+y,x+\partial_y h(y,z),\partial_z h(y,z))[/itex]. So lets compare components.

[tex]
\begin{align}
2x+y & =2x+y \\
x &=x+\partial_y h(y,z)\\
2z & =\partial_z h(y,z)
\end{align}[/tex]

Equation (2) seems to be the troublesome one for you, but when you look at it like this there is only one way to satisfy the equation and that is [itex]\partial_y h(y,z)=0[/itex].
 
  • #7
Yes I understand that part, but I'm not clear with why it means that h only depends on z now.
Thanks for your help.
 
  • #8
Cyosis
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If the derivative of a function is 0 then what do you know about that function?
 
  • #10
Cyosis
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Yes or slightly more accurate, the function is constant with respect to the variable you're differentiating to. Because we know that [itex]\partial_y h(y,z)=0[/itex] we know that this function has to be constant with respect to y. In other words there cannot be a y variable in it, because then the derivative would not be zero. Therefore the only variable that is left is z, h(y,z)=h(z).
 
  • #11
oh okay yes, that's what I was trying to clarify.

Thank you very much!
 

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