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Finding a potential function

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-05-30at23129AM.png

    When they say therefore g is a function of z alone, I don't understand why.

    Also when they integrate the second to the last equation with respect to y, I just want to make sure that the y's in bold cancel, that they're really there in the actual answer

    1/y * xzy + e^x cos y
     
    Last edited: May 30, 2012
  2. jcsd
  3. May 30, 2012 #2

    SammyS

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    It says:
    "We write the constant of integration as a function of y and z because its value may change if y and z change."​
    Then they take ∂f/∂y and find that ∂g/∂y =0 . That means that in this case, g does not change if y changes.

    Therefore, g is not a function of y.

    Taking ∂f/∂z will give you g as a function of z, if indeed it is a function of z. So I suppose that technically, until you do this step, you can't say that g is actually a function of z. Added in Edit: Well, at this point they've changed the name of g(y,z) to h(z) .
     
    Last edited: May 30, 2012
  4. May 30, 2012 #3
    They never integrated with respect to y.
     
  5. May 30, 2012 #4
    As far as I can tell I don't even see what dg/dy refers to, so I can't tell if it is zero or something else. How do you know dg/dy = 0

    I can type greek letters with this computer, even when I click on the quick symbols.
     
  6. May 30, 2012 #5

    Why not? Were they supposed to?
     
  7. May 30, 2012 #6

    SammyS

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    The equation
    [itex]\displaystyle -e^x\sin(y)+xz+\frac{\partial g}{\partial y}=xz-e^x\sin(y)[/itex]​
    should tell you that ∂g/∂y =0 .
     
  8. May 30, 2012 #7

    Do you agree that [itex]\frac{\partial f}{\partial y}=xz-e^x\sin y[/itex]?

    Do you agree that [itex]f(x,y,z)=e^x\cos y+xyz+g(y,z)[/itex]?

    Please take the partial of [itex]f[/itex] with respect to [itex]y[/itex] using the last equation.

    Relate the two equations to conclude that [itex]\frac{\partial}{\partial y}g(y,z)=0[/itex].

    Let us know if this does not clear things up.

    (Is it necessary to integrate in [itex]y[/itex] to solve the problem?)
     
  9. May 30, 2012 #8
    As far as I can tell dg/dy will always be zero because there is no g anywhere in the problem because the problem starts out with x y z. I don't get what's going on with dg
     
  10. May 30, 2012 #9
    I will try to suggest where [itex]g(y,z)[/itex] came from.

    Do you agree that [itex]\frac{\partial f}{\partial x}=e^x\cos y+yz[/itex]?

    If so, use integration to find [itex]f(x,y,z)[/itex].
     
  11. May 30, 2012 #10
     
  12. May 30, 2012 #11
    I'm not sure why you think the variables x,y,z imply that g has some properties.
     
    Last edited: May 30, 2012
  13. May 30, 2012 #12
    Well, why g? And is there any circumstance when the partial of g with respect to y would not be zero?
     
  14. May 30, 2012 #13
    We have g because something is missing, and we call it g. Yes, it's partial is not always zero.
     
  15. May 30, 2012 #14
    Have you found f yet? Also, try ingtegrating N and P in y and z respectively. That may shed some light on what's going on.
     
  16. May 30, 2012 #15

    SammyS

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    OK, Now I think we see where you are having difficulty is understanding.

    The textbook solution points-out why there is a g(y.z) --- It's the "constant" of integration, but it's only constant in regards to x, not y or z, because your integration is W.R.T. x, treating y and z as if they were constants.
     
  17. May 30, 2012 #16
    I've given up on this problem. The book only had about 7 problems on these, and I understood about 5 of them, so that's enough to move on.
     
  18. May 30, 2012 #17

    SammyS

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    Let's try the method you used in your more recent thread:
    finding a potential function pt 2


    You had the partials of f(x,y,z) W.R.T. the variables, from the vector field components.

    [itex]\displaystyle \frac{\partial f}{\partial x}=e^x\cos(y)+yz
    \quad\to\quad
    f(x,\,y,\,z)=e^x\cos(y)+xyz+\text{Some term not containing }x\text{ but including the constant, }C[/itex]

    [itex]\displaystyle \frac{\partial f}{\partial y}=xz-e^x\sin(y)
    \quad\to\quad
    f(x,\,y,\,z)=xyz+e^x\cos(y)+\text{Some term not containing }y\text{ but including the constant, }C[/itex]

    [itex]\displaystyle \frac{\partial f}{\partial z}=xy+z
    \quad\to\quad
    f(x,\,y,\,z)=xyz+\frac{z^2}{2}+\text{Some term not containing }z\text{ but including the constant, }C[/itex]

    So by inspection we have that [itex]\displaystyle f(x,\,y,\,z)=e^x\cos(y)+xyz+\frac{z^2}{2}+C\ .[/itex]

    I hope that makes more sense to you!
     
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