1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential function for conservative vector fields

  1. Apr 21, 2007 #1

    Find a potential function for the conservative vector field (y,x,1)

    My work:

    (Df/Dx, Df/Dy and Df/Dz are the partial derivatives)

    Df/Dx=1, Df/Dy=x, Df/Dz=1

    I take the first eq. and integrate, so that I get

    f(x,y,z) = yx + C

    I then derivate with respect to y:

    Df/Dy= x + C '

    I compare this to the other eq. Df/Dy=x, hence C ' is 0, and consequently C is also 0.

    My answer is then


    Which is wrong! Correct answer is yx+z. Where do I fail?
  2. jcsd
  3. Apr 22, 2007 #2

    I'm confused what you're doing. If you're looking for the potential function, you're looking for f such that it's gradient is the vector field you gave. Note also, that when integrating a function g(x,y,z) with respect to one variable say x, you won't just have a constant of integration, but a function dependent on y and z.
  4. Apr 22, 2007 #3
    Since the gradient of f is the vector field, we have that the partials are

    Df/Dx=y, Df/Dy=x and Df/Dz=1, right?

    I integrate Df/Dx=y with respect to x. This gives: f=yx+C(y,z). Then I derivate this with respect to y and get: Df/Dy=x + C ' (x,y). From before we have that Df/Dy=x, which means that C ' (x,y)= 0 and C(x,y)=0.

    Southe potential function is


    See what I've done?
  5. Apr 22, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Here, C should be a function of y and z. Also, the prime notation is ambiguous, since C is a function of two variables. You should write Df/Dy=x+DC/Dy, which tells us that C is not a function of y (on comparing this with your second equation above). You must then differentiate f with respect to z, which gives Df/Dz=DC/Dz, but from the third equation, this is equal to 1, which says that C=z; hence f=xy+z
    Last edited: Apr 22, 2007
  6. Apr 22, 2007 #5
    Its not a normal integral youre doing, its like a 'partial integral.' I remember this confusing me too.
  7. Apr 22, 2007 #6
    OK, I figured it our.
    Last edited: Apr 22, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook