Results 1 to 3 of 3

Math Help - Finding a Potential for a field

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    36

    Finding a Potential for a field

    So im trying to find the potential for the following gradient

    F = <2xy,+3, x^2-3z, -3y>

    I've gotten a bunch of different answers, but none of them have been right

    Ive tried the method of integrating one component with its respective variable, (as in intergrating 2xy+3 with respect to x) and then taking the partial with respect to another variable and setting it equal to another term in the gradient, eventually solving for the new functions g(y,z) then h(z).

    my answer has been

     \phi = x^2y-3yz

    which then gives you back your gradient

    the online software is telling me this is wrong, and im not sure if im just punching it in wrong or if im doing it wrong

    thanks guys
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by TheUnfocusedOne View Post
    F = <2xy,+3, x^2-3z, -3y>


    thanks guys
    Is the field F = <2xy +3, x^2-3z, -3y> or

    F = <2xy, x^2-3z, -3y> ??

    If the  +3 vanishes , i think it may go to your answer

    From  \frac{\partial \phi}{\partial x} = 2xy

      \phi = x^2 y + h(y,z)

     \frac{\partial \phi}{\partial y } = x^2 + h_y(y,z) = x^2 - 3z

     -3z = h_y(y,z)

    Also

     \frac{\partial \phi}{\partial z} = h_z(x,y)= -3y

    Finally , i obtain  h(y,z) = -3yz + C

    If you omit the constant  \phi = x^2y - 3yz

    But if  +3 is present ,

     \phi = x^2y - 3yz + 3x not sure but you still can take a look !



    ps : Is it a 4-D field !?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,237
    Thanks
    1795
    Quote Originally Posted by TheUnfocusedOne View Post
    So im trying to find the potential for the following gradient

    F = <2xy,+3, x^2-3z, -3y>

    I've gotten a bunch of different answers, but none of them have been right

    Ive tried the method of integrating one component with its respective variable, (as in intergrating 2xy+3 with respect to x) and then taking the partial with respect to another variable and setting it equal to another term in the gradient, eventually solving for the new functions g(y,z) then h(z).

    my answer has been

     \phi = x^2y-3yz

    which then gives you back your gradient

    the online software is telling me this is wrong, and im not sure if im just punching it in wrong or if im doing it wrong

    thanks guys
    That is correct except for the missing constant: \phi(x,y,z)= x^2- 3yz+ C for any constant C. Perhaps your online software is expecting that.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Electric Field and Potential (conceptual help)
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 27th 2011, 06:44 AM
  2. Find the potential function f for the field F
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 6th 2010, 05:34 AM
  3. [SOLVED] General potential of a field
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 21st 2010, 04:22 PM
  4. Potential Function of a Vector Field
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 13th 2009, 10:05 PM
  5. find the potential and electric field
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: January 27th 2009, 04:58 AM

Search Tags


/mathhelpforum @mathhelpforum