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Homework Help: Two part question vectors

  1. Jan 5, 2009 #1
    consider the vector field q=yzcos(xy)i+xzcos(xy)j+(sin(xy)+2z)k

    verify that x q=0 and concequently determine a function f(x,y,z) such that q=f

    now i have completed the first part by using the determinant and multiplying out, it is quite obvious that the whole thing equals zero, however i dont know how to do the second part "concequently determine a function f(x,y,z) such that q=f"

    my first question is does f simply mean the dot product when there is no dot there? i assume it does

    and my second question is how do i approach this question? thanks
     
  2. jcsd
  3. Jan 5, 2009 #2

    tiny-tim

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    Hi terryfields! :smile:

    No, f is the gradient …

    for an ordinary function f(x) of a vector x, f is the gradient vector (∂f/∂x1, … ,∂f/∂xn) :smile:

    And if x q = 0, then there must be an f with q = f.

    (and basically you find f by integrating each component of q, and using a little common-sense for the "constants"! :wink:)
     
  4. Jan 5, 2009 #3
    ah ok, so do i just need to integrate each term in respect to x or in respect to x,y and z?
     
  5. Jan 5, 2009 #4

    tiny-tim

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    The i term with respect to x, the j term with respect to y, and the k term with respect to z. :wink:
     
  6. Jan 5, 2009 #5
    ok, so by my calculations i would get (zsin(xy))i+(zsin(xy))j+(zsin(xy)+z2)k i assume this is how it's writen? as i now assume i need to do ∇ x f = 0 to find my constants?
     
  7. Jan 5, 2009 #6

    HallsofIvy

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    And, as tiny tim said, "using a little common-sense for the "constants"!" The reason he put "constants" in quotes is because the partial derivative with respect to one variable treats the other variables as constants- and you have to consider that in taking an "ant-derivative".

    For example, if the v were the vector [itex]2xy\vec{i}+ (x^2+ z)\vec{j}+ (1+ y)\vec{k}[/itex], then you must have [itex]\partial f/\partial x= 2xy[/itex], [itex]\partial f/\partial y= x^2+ z[/itex], and [itex]\partial f/\partial z= 1+ y[/itex]. Integrating [itex]\partial f/\partial x= 2xy[/itex] gives [itex]f= x^2y+ \phi(y,z)[/itex] where [itex]\phi[/itex] can be any function of y and z since its derivative, with respect to x, is 0. If we differentiate that with respect to y, we get [itex]\partial f/\partial y= x^2+ \partial \phi/\partial y= x^2+ z[/itex] or [itex]\partial \phi/\partial y= z[/itex]. (Notice how the "[itex]x^2[/itex]" cancelled- that HAS to happen because [itex]\phi[/itex] is a function of y and z only. That's why not every function is a "gradient".)

    Integrating [itex]\partial\ph/\partial y= z[/itex] gives us [itex]\phi(y,z)= yz+ \psi(z)[/itex]. Again, the "constant" of integration may be a function of z. That means that we now have [itex]f(x,y,z)= x^2y+ yz+ \psi(z)[/itex] and differentiating that with respect to z, [itex]\partial f/\partial z= y+ d\psi/dz= 1+ y[/itex] and the "y" term cancels leaving [itex]d\psi/dz= 1[/itex]. (That is an ordinary derivative because [itex]\psi[/itex] is a function of z only.) Integrating that, [itex]\psi(z)= z+ C[/itex] where now the "constant" of integration is now really a constant.

    That gives [itex]f(x,y,z)= x^2y+ yz+ z+ C[/itex] and you can check that [itex]\nabla (x^2y+ yz+ z)= 2xy\vec{i}+ (x^2+ z)\vec{j}+ (y+1)\vec{k} for any constant C.
     
  8. Jan 5, 2009 #7
    i think i understand

    so when we get zsin(xy)i +zsin(xy)j+z2 from integrating we have to check what we would get from differentiating this?

    in this case the i term and j term dont need constants because they would differentiate right back to what they were integrated from? however for the k term we do because if we differentiate z2 we just get 2z not 2z+ sin(xy)
    so we need to add on a term that will differentiate with respects to z to get sin(xy) namely zsin(xy)?

    giving our full answer to be zsin(xy)i+zsin(xy)j+z2+zsin(xy)
     
  9. Jan 5, 2009 #8

    tiny-tim

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    Nooo … f is an ordinary function, not a vector …

    when you integrate, you're aiming to get the same function f, each of the three times …

    that's where the "constants" come in …

    how can you combine zsin(xy) and z2 to get a single f that fits? :smile:
     
  10. Jan 5, 2009 #9
    f=z2 +zsin(xy)?? is there a method to do this by or is it just inspection to find it?
     
  11. Jan 5, 2009 #10

    tiny-tim

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    That's right! :smile:

    (and the method is what you just did …

    didn't you like it? :wink:)
     
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