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Hard maths question can you help? watch

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    km
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    (Original post by opi453)
    1.Calculate the divergence of the vector field a=(x,y^2,z^3) at the point P(-2,4,5).

    2.Calculate the divergence of the vector field a=(z^2,x^2,y^2) at the point P(1,2,3).
    There is nothing hard about this question ...

    Look at the definition of the divergence in your notes and it is 1 line each.

    (They are so simple I cannot even give you a hint)
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    (Original post by TeeEm)
    There is nothing hard about this question ...

    Look at the definition of the divergence in your notes and it is 1 line each.

    (They are so simple I cannot even give you a hint)
    tyu
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    (Original post by opi453)
    Please im really struggling
    What, exactly, have you done so far?
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    (Original post by opi453)
    Please im really struggling
    \displaystyle \text{div} {F} = \nabla \cdot F

    So calculate the div for both fields and then sub in values.

    As said above 1 line working.
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    (Original post by poorform)
    \displaystyle \text{div} {F} = \nabla \cdot F

    So calculate the div for both fields and then sub in values.

    As said above 1 line working.
    lio
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    (Original post by opi453)
    Still struggling cant you just give me the answer?
    No. This is not a "do my homework for me" service.

    Please read the guide to posting stickied at the top of the forum.
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    (Original post by DFranklin)
    No. This is not a "do my homework for me" service.

    Please read the guide to posting stickied at the top of the forum.
    This isent my homework:mad:
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    (Original post by opi453)
    Please im really struggling
    divergence of a vector field F is defined as ∇. F where ∇ operator is defined as [∂/∂x, ∂/∂y, ∂/∂z]

    Example

    divergence of [x2-3y, y2-z2-y, x+2y] at (1,1,1)

    [∂/∂x, ∂/∂y, ∂/∂z] . [x2-3y, y2-z2-y, x+2y] =

    ∂/∂x(x2-3y) + ∂/∂y(y2-z2-y) + ∂/∂z(x+2y) =

    2x +2y-1+0=

    2x+2y-1

    at the required point 3
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    (Original post by opi453)
    This isent my homework:mad:
    Then why do you need the answer so badly?
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    (Original post by DFranklin)
    Then why do you need the answer so badly?
    lo
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    (Original post by opi453)
    Need the answer so I can understand the method not that's its any of you ****ing business
    Wow that's how you're going to treat people trying to help you.
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    (Original post by opi453)
    Thanks understand it now
    no worries
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    (Original post by opi453)
    Need the answer so I can understand the method not that's its any of you ****ing business
    Well, the answer isn't going to help you learn the method, is it? I mean, if I say the divergence is 6, how is that going to help you understand anything?

    Spoiler:
    Show
    It isn't 6. (Edit, actually maybe it is, who knows?).


    Anyhow, thanks for removing the last sliver of doubt about whether it's worth wasting any time trying to help you.
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    (Original post by DFranklin)
    Well, the answer isn't going to help you learn the method, is it? I mean, if I say the divergence is 6, how is that going to help you understand anything?

    Spoiler:
    Show
    It isn't 6. (Edit, actually maybe it is, who knows?).


    Anyhow, thanks for removing the last sliver of doubt about whether it's worth wasting any time trying to help you.
    sorry I did not see his comment as I was busy typing an example for him, so I missed the whole conversation
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    (Original post by TeeEm)
    sorry I did not see his comment as I was busy typing an example for him, so I missed the whole conversation
    Not a problem. If you managed to help him that's great.
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    (Original post by opi453)
    Need the answer so I can understand the method not that's its any of you ****ing business
    Understanding a method will teach you nothing about the mathematics.

    You can spot method-people a mile off. Give them an integral like \displaystyle\int_{-1}^{1} \dfrac{1}{x^2}\ \textrm{d}x and when they come back with -2, you know they don't know what they're really doing.
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    (Original post by FireGarden)
    Understanding a method will teach you nothing about the mathematics.

    You can spot method-people a mile off. Give them an integral like \displaystyle\int_{-1}^{1} \dfrac{1}{x^2}\ \textrm{d}x and when they come back with -2, you know they don't know what they're really doing.
    Is it 0 or does it just not converge? I thought the former but wolfram outputs the latter.
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    (Original post by FireGarden)
    Understanding a method will teach you nothing about the mathematics.

    You can spot method-people a mile off. Give them an integral like \displaystyle\int_{-1}^{1} \dfrac{1}{x^2}\ \textrm{d}x and when they come back with -2, you know they don't know what they're really doing.
    You can't integrate over the discontinuity at x=0, right?
 
 
 
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