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Homework Help: Showing a vector field is conservative.

  1. Jun 8, 2009 #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.

    1. The problem statement, all variables and given/known data
    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.
    Last edited: Jun 8, 2009
  2. jcsd
  3. Jun 8, 2009 #2


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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?
    \nabla \times \nabla \phi=?

    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.
    \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]?
    Last edited: Jun 8, 2009
  4. Jun 8, 2009 #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.

  5. Jun 8, 2009 #4


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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.
  6. Jun 8, 2009 #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?
  7. Jun 8, 2009 #6


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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.

    2x+y & =2x+y \\
    x &=x+\partial_y h(y,z)\\
    2z & =\partial_z h(y,z)

    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].
  8. Jun 8, 2009 #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.
  9. Jun 8, 2009 #8


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    If the derivative of a function is 0 then what do you know about that function?
  10. Jun 8, 2009 #9
    It is a constant.
  11. Jun 8, 2009 #10


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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).
  12. Jun 8, 2009 #11
    oh okay yes, that's what I was trying to clarify.

    Thank you very much!
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