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Math Help - need help on path and vector field problem

  1. #1
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    need help on path and vector field problem

    how do I find a scalar function on problem 4?


    Last edited by yzc717; December 19th 2008 at 08:02 AM.
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  2. #2
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    Sorry the answer I gave you was wrong. Don't use it!
    Last edited by JD-Styles; December 19th 2008 at 04:31 AM. Reason: wrong answer
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  3. #3
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    vector field problem with volumn

    do i have to draw the vector field first, then i can compute the volume?

    how do i do it?
    Last edited by mr fantastic; December 19th 2008 at 12:04 AM.
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  4. #4
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    Quote Originally Posted by JD-Styles View Post
    \nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})

    So, if we're working in the reverse order:

    f=\int \frac{\partial f}{\partial x}dx+ \int \frac{\partial f}{\partial y}dy+\int \frac{\partial f}{\partial z}dz+C

    So basically, integrate 2z^2 wrt x, (e^zcosy+x^3) wrt y, and (e^zsiny-5z) wrt z.
    I'm not sure the OP will get the right idea from this.

    IF a scalar function f exists such that F = \nabla f then a solution to the following simultaneous partial differential equations will exist:

    \frac{\partial f}{\partial x} = 2z^2 .... (1)

    \frac{\partial f}{\partial y} = e^z \cos y + x^3 .... (2)

    \frac{\partial f}{\partial z} = e^z \sin y - 5z .... (3)
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  5. #5
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    Quote Originally Posted by yzc717 View Post
    how do I find a scalar function on problem 4?




     \nabla f = F

    Now... nabla f is given by:

     \frac{d f}{dx}\vec{i}+\frac{d f}{dy}\vec{j}+ \frac{d f}{dz}\vec{k}

    And you know that  F = 2z^2\vec{i} + (e^z cos(y) + x^3)\vec{j} + (e^z sin(y) -5z)\vec{k}

    So, in essence, you have to find a scalar function f that satisfies these equations:

    \frac{d f}{dx}=2z^2

    \frac{d f}{dy}=(e^z cos(y) + x^3)

     \frac{d f}{dz}=(e^z sin(y) -5z)

    A little integration is needed...
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  6. #6
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    I'll try to redeem myself here. I'd argue that no such function exists because:

    Suppose such a function did exist. Look at the partial derivative wrt y:

    \frac{\partial f}{\partial y}=e^zcosy+x^3

    f(x,y,z)=e^zsiny+x^3y+h(x,z) (2)

    Here, h(x,z) is a function that depends only on x and z (which if derived wrt y becomes zero). Now if we take the partial derivative wrt x we should get 2z^2:

    \frac{\partial f}{\partial x}=3x^2y+\frac{\partial h(x,z)}{\partial x}=2z^2

    \frac{\partial h(x,z)}{\partial x}=2z^2-3x^2y

    h(x,z)=2xz^2-x^3y+g(z)

    But this is a contradiction because we have a term with y in a function that can only contain x and z. Thus no such function exists.


    Another, maybe simpler, way to look at it is to argue that if such a function existed then:

    \frac{\partial ^2f}{\partial y \partial x}=\frac{\partial ^2f}{\partial x \partial y}

    \frac{\partial }{\partial y}(2z^2)=\frac{\partial }{\partial x}(e^zcosy+x^3)

    0=3x^2 which is clearly a contradiction.
    Last edited by JD-Styles; December 19th 2008 at 05:05 AM.
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  7. #7
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    how to find the volume,

    and i got the first one, since curl does not equal to zero, f is not a gradient vector field, it not conservative.
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