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How do I undo a gradient?

  1. Jan 6, 2016 #1
    1. The problem statement, all variables and given/known data
    ##\nabla U = 2 r^4 \vec r## Find U.

    2. Relevant equations
    ##\vec r = x \hat i + y \hat j + z \hat j##
    ##r = \sqrt (x^2 + y^2 + z^2)##

    3. The attempt at a solution
    ##\nabla U = 2 (x^2 + y^2 + z^2)^2 (x \hat i + y \hat j + z \hat j)##

    I multiplied everything out,

    ##\nabla U = (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)x\hat i + (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)y\hat j + (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)z\hat k##

    Since ##\nabla U = \frac{\partial U}{\partial x} \hat i + \frac{\partial U}{\partial y} \hat j + \frac{\partial U}{\partial z} \hat k##, we really only need one of the partial derivatives.

    ##\frac{\partial U}{\partial x} = (2 x^5 + 4 x^3 y^2 + 4 x^3 z^2 + 4 x y^2 z^2 + 2 x y^4 + 2 x z^4)##

    Then I took the integral of only the x terms,

    ##\frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (\frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4)##

    Therefore,

    ##U = \frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4##

    Am I correct so far? If so, where do I go from here?
     
    Last edited: Jan 6, 2016
  2. jcsd
  3. Jan 6, 2016 #2

    SteamKing

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    If ##r = \sqrt{x^2 + y^2 + z^2}##, then what is r2?
     
  4. Jan 6, 2016 #3
    Sorry about that, I screwed up the question. I'll edit right now. It should be r^4.
     
  5. Jan 6, 2016 #4

    SammyS

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    You need a "constant" of integration. Of, course it's only constant w.r.t. ##x## . It's potentially a function of ##y## and ##z## .
     
  6. Jan 6, 2016 #5
    So ##\frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (\frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + constant)##?
     
  7. Jan 6, 2016 #6

    SammyS

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    I would put it more like:
    ##\displaystyle U = \frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + f(y,z) ##​

    Now, take the partial derivative w.r.t. y or z.
     
  8. Jan 6, 2016 #7
    But f(y,z) is a function, not a constant? And why do I need to take the PD w.r.t. y or z? Just trying to understand.
     
  9. Jan 6, 2016 #8

    SammyS

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    ##\displaystyle \ \frac{\partial}{\partial x}f(y\,,z)=0\,,\ ## right ?

    So ƒ(y, z) plays the same role here as the constant of integration plays for a single variable integral .
     
  10. Jan 6, 2016 #9
    Yes.

    So taking the PD w.r.t. y yields,

    ##U = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3 + f(y,z)##
     
  11. Jan 6, 2016 #10

    fresh_42

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    Yes (but), if I made no mistake by scrolling up and down. However, can the constant here be a function of y?
     
    Last edited: Jan 6, 2016
  12. Jan 6, 2016 #11

    vela

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    Why didn't you take the partial derivative of f(y,z)? And that's supposed to be ##\partial U/\partial y## on the LHS, right?
     
  13. Jan 7, 2016 #12

    Ray Vickson

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    Because you have not yet matched ##\partial U/ \partial y## and ##\partial U/ \partial z## to the given forms that you started with. Notice that if you did not have the ##f(y,z)## term, the ##y## and ##z## partials would come out wrong. Try it and see! Don't agonize about it; just do it.
     
  14. Jan 7, 2016 #13
    ##{\partial U}/{\partial y} = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3##

    ##{\partial U}/{\partial z} = 4 x^2 y^2 z##
     
    Last edited: Jan 7, 2016
  15. Jan 7, 2016 #14
    Are you not allowed to solve this in spherical coordinates? dU/dr = 2 r5?
     
  16. Jan 7, 2016 #15
    I don't see why not. The book doesn't say I can't. Although, I am trying to get used to partial derivatives and the del operator.
     
  17. Jan 7, 2016 #16
    So.....
     
  18. Jan 7, 2016 #17

    fresh_42

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    Thanks a lot. I just did it by hand and at the end I came out with: "There must be a way to see it immediately!" Thank you.
     
  19. Jan 7, 2016 #18

    vela

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    You're missing a bunch of terms in ##{\partial U}/{\partial z}##.

    Note that the result you got for ##{\partial U}/{\partial y}## by neglecting f(y,z) only has the first three terms of ##\hat{j}## component of ##\nabla U##. The missing terms only depend on y and z. That's why the f(y,z) has to be there.
     
  20. Jan 7, 2016 #19
    I see, and f(y,z) contains only y and z terms, correct?
    So ##U = x^6 / 3 + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + f(y,z)##
    ##f(y,z) = \int {4 y^3 z^2 + 2 y^5 +2 y z^4} dy##
    Something doesn't seem right with that above equation though. I have not learned multiple integrals yet. That's after this section.
     
  21. Jan 7, 2016 #20

    SammyS

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    What you have for ## \displaystyle {\partial U}/{\partial y}\ ## is missing ## \displaystyle \frac{\partial }{\partial y}f(y,\,z)\ ##.
     
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