1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

Homework Help: Another potential function

  1. Apr 22, 2007 #1
    How about the potential function of the vector field (x+y,x-z,z-y)?

    Then we have that Df/Dx=x+y, Df/Dy=x-z and Df/Dz=z-y

    I integrate the first of these equations with respect to x:

    f(x,y,z)= (1/2)x^2+yx+C(x,y)

    Then I derivate it with respect to y:

    Df/Dy=x + DC(x,z)/Dy which means that DC(x,z)/Dy = -z and C(x,z)=-zy

    Then I derivate f(x,y,z) with respect to z:

    Df/Dz=DC/Dz which means that DC/Dz=z-y and C=(1/2)z^2-yz



    The correct answer, however, is: f(x,y,z)=(1/2)(x^2+z^2)+yx-yz

    I can't find my mistake!
  2. jcsd
  3. Apr 22, 2007 #2
    DC/Dz= z not z - y

    the -y on both sides cancel out.
    Last edited: Apr 22, 2007
  4. Apr 23, 2007 #3
    I don't understand that.

    From before I've found Df/Dz=z-y

    Then I calculate Df/Dz again, and get DC/Dz. That should mean that DC/Dz=z-y...
  5. Apr 23, 2007 #4
    because C(y,z)= -yz , DC/Dz= - y
  6. Apr 23, 2007 #5
    So in general, when you have a vector field F(Df/Dx, Df/Dy, Df/Dz), you should start by integrating Df/Dx, and then derivate this with respect to both y and z?
  7. Apr 23, 2007 #6


    User Avatar

    http://folk.ntnu.no/bronner/temp/temp1177357693.6875.png [Broken]

    Might not be the best method but in this case it's quite easy to see (by inspection)
    Last edited by a moderator: May 2, 2017
  8. Apr 23, 2007 #7
    I understand everything here except how you find C1, C2 and C3. How do you calculate those?
  9. Apr 24, 2007 #8


    User Avatar
    Science Advisor

    ?? Finding C1, C2, and C3 are the whole exercize!

    You want to find some function F(x,y,z) such that Fx= x+y, Fy= x-z, and Fz= z-y.

    Integrating Fx= x+ y with respect to x gives F(x,y,z)= (1/2)x2+ xy+ C1(y,z). The "constant of integration" may be a function both y and z since we were integrating "with respect to x".

    Differentiating with respect to y, Fy[/sup]= x+ C1y(y,z)= x-z. The 'x' terms cancel- that has to happen because C1 is a function of y and z only, not x. From C1y(y,z)= -z we get C1(y,z)= -zy+ C2(z). Again, the "constant of integration" may be a function of z because C1 was a function of y and z and we integrated only with respect to y.
    Now we have F(x,y,z)= (1/2)x2+ xy+ C1(y,z)= (1/2)x2+ xy- zy+ C2(z). Differentiating with respect to z, Fz= -y+ C2'(z)= z-y. The "-y" terms cancel- again, that HAD to happen. C2 is a function of z only and its derivative cannot contain a "y". From C2'= z we get C2= (1/2)z2+ C3 where C3 now really is a constant.

    We have F(x,y,z)= (1/2)x2+ xy- zy+ C2(z)= (1/2)x2+ xy- zy+ (1/2)z2+ C.
  10. Apr 24, 2007 #9
    Very well explained, HallsofIvy!

    In my book they have omitted the C in the final equation.
  11. Apr 27, 2007 #10
    Why is it that we can ommit C?
  12. Apr 27, 2007 #11
    Double post ...
  13. Apr 28, 2007 #12


    User Avatar
    Science Advisor

    If you are determining the "anti-derivative" (your "potential function". As I have pointed out before that is "physics terminology" that I dislike using.) in order to integrate from one point to another, then the constant will cancel anyway. Of course, if you were specifically asked to find "the potential function" you should have the "C". If you were asked to find "a" potential function, then you don't need the "C" but should be aware that there are an infinite number of other functions that would work as well.

    That is, by the way, why the physics "potential energy" at a point is always "relative" to another point- the other point determines the value of "C".
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook