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Recovering function from its gradient

  1. Dec 10, 2012 #1
    Hi all,
    I've to recover a function ∅ such that F=∇∅,F=3x^2y i +(x^3+2yz)j+y^2k.
    So, ∂∅/∂x=3x^2y, Integrating w.r.t. x
    ∅=x^3y+f(y,z) ,Assuming there may be function of y and z.----------1
    ∴∂∅/∂y=x^3+f'(y,z)
    now, ∂∅/∂y=(x^3+2yz)=x^3+f'(y,z)
    ∴f'(y,z)=2yz
    To find ∅ from 1 I've to find f(y,z).
    My problem is how to get f(y,z) from f'(y,z),should I integrate it w.r.t. y or z?
    Thanks for your help.
     
    Last edited: Dec 10, 2012
  2. jcsd
  3. Dec 10, 2012 #2

    LCKurtz

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    OK so far


    You shouldn't use ' for a partial derivative. What you have is ##\phi_y = x^3 +f_y(y,z)## which must equal ##x^3+2yz##.
    And that illustrates perfectly why you should use the partial symbol instead of the '. So what you actually have to solve at this step is ##f_y(y,z) = 2yz##. Can you finish it From there?
     
    Last edited: Dec 10, 2012
  4. Dec 10, 2012 #3
    Thanks a lot for your help.
    Now I got it, I've to integrate it w.r.t. y to get ##f_y(y,z)## to get f(y,z),Right?
    Btw, I've got a new problem,F=i(x3z-2xyz)+j(xy-3x2yz)+k(yz2-zx) is a solenoidal vector field.I've to find a V such that F=∇XV
    So,now it comes to find a vector field from its curl.How can I find it? In my book only from gradient was explained but not from curl :cry: So please help me out.
     
  5. Dec 10, 2012 #4

    haruspex

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    Don't know if there's a smarter way, but the obvious is to write out three PDEs, one for each component of F.
     
  6. Dec 10, 2012 #5
    Sorry,I exactly didn't get you.
    I think PDE=partial differential equations.If so with respect to which variable x,y,or z or all of them?
    Please say something more,
    Thanks a lot for your help.
     
  7. Dec 11, 2012 #6

    haruspex

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    Yes. You know the definition of and of cross-product, so given the vector that results, F, you can write down three PDEs, one for each component of ∇XV. That gives you three equations involving ∂V/∂x etc.
     
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