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Vector calc, gradient vector fields

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Is F = (2ye^x)i + x(sin2y)j + 18k a gradient vector field?



    3. The attempt at a solution

    Yeah I just don't know...I started to find some partial derivatives but I really don't know what to do here. Please help!
     
  2. jcsd
  3. Apr 4, 2012 #2

    tiny-tim

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    hi calculusisrad! :smile:

    (try using the X2 button just above the Reply box :wink:)

    learn: div(curl) = 0, curl(grad) = 0

    does that help? :smile:
     
  4. Apr 4, 2012 #3

    HallsofIvy

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    For any g(x,y), [itex]\nabla g= \partial g/\partial x\vec{i}+ \partial g/\partial y\vec{j}+ \partial g/\partial z\vec{k}[/itex].

    So is there a function g such that
    [tex]\frac{\partial g}{\partial x}= 2ye^x[/tex]
    [tex]\frac{\partial g}{\partial z}= 18[/tex]
    and
    [tex]\frac{\partial g}{\partial y}= x sin(2y)[/tex]?


    One way to answer that is to try to find g by finding anti-derivatives. Another is to use the fact that as long as the derivatives are continuous (which is the case here), the mixed second derivatives are equal:
    [tex]\frac{\partial}{\partial x}\left(\frac{\partial g}{\partial y}\right)= \frac{\partial}{\partial y}\left(\frac{\partial g}{\partial x}\right)[/tex]
    Is
    [tex]\frac{\partial x sin(2y)}{\partial x}= \frac{\partial 2ye^x}{\partial y}[/tex]?
    etc.
     
    Last edited: Apr 4, 2012
  5. Apr 4, 2012 #4
    Thank you soo much :)
    I am still confused, though. So if I can prove that dg/dxy = dg/dyx and dg/dxz = dg/dzx and dg/dyz = dg/dzy , I will have proved that F is a gradient vector field, correct???

    I found that dg/dxy = sin2y, and d/dyx = 2e^x. Since they are not equal, that means that F is not a gradient vector field?

    Thanks
     
  6. Apr 5, 2012 #5

    tiny-tim

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    hi calculusisrad! :smile:

    (have a curly d: ∂ and try using the X2 and X2 buttons just above the Reply box :wink:)
    (your notation is terrible :yuck:, but …) yes :smile:

    this is because you're actually proving that curlF = 0,

    and if F is a gradient, then F = ∇φ, and so curlF = (curl∇)φ = 0 :wink:
    yup :biggrin:
     
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