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    I dont really understand how to do this question. I get that the divergence of the field would be 3, But id have thought the divergence of the unit vector would just be the divergence of the vector itself divided by the magnitude, but it appears that this isnt the case?
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    (Original post by Eremor)
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    I dont really understand how to do this question. I get that the divergence of the field would be 3, But id have thought the divergence of the unit vector would just be the divergence of the vector itself divided by the magnitude, but it appears that this isnt the case?
    You can write the unit vector \hat{v} = \frac{v}{|v|} = \frac{v}{\sqrt{v^2}} now use the product/quotient rule on this to find the divergence.

    If you do it this way - you should end up with \nabla \cdot \hat{v} = \frac{\nabla \cdot v - 1}{|r|}.
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    Unit vector W (which is V hat in the question) = (\frac{x}{\sqrt{x^2+y^2+z^2}}, \frac{y}{\sqrt{x^2+y^2+z^2}}, \frac{z}{\sqrt{x^2+y^2+z^2}})

    I will find d/dx for you. The rest is similar.

    (Using quotient rule)

    \frac{d}{dx}(\frac{x}{\sqrt{x^2+  y^2+z^2}}) = \frac{(\sqrt{x^2+y^2+z^2})(1)-(x)(2x*\frac{1}{2}*(x^2+y^2+z^2)  ^{-\frac{1}{2}})}{x^2+y^2+z^2} = \frac{y^2+z^2}{(x^2+y^2+z^2)^{ \frac{3}{2}}}

    You will get 2(x^2+y^2+z^2) at the top after you find the other derivatives and add them (definition of divergence). Cancel some stuff, you get the answer.
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    (Original post by Eremor)
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    I dont really understand how to do this question. I get that the divergence of the field would be 3, But id have thought the divergence of the unit vector would just be the divergence of the vector itself divided by the magnitude, but it appears that this isnt the case?
    From a mathematical POV, this is much easier if you convert the field to spherical polars, then find the divergence, since then we get:

    \bold{V} = r\bold{\hat{r}}

    with no component in the direction of \bold{\hat{\theta}}, \bold{\hat{\phi}}
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    (Original post by atsruser)
    From a mathematical POV, this is much easier if you convert the field to spherical polars, then find the divergence, since then we get:

    \bold{V} = r\bold{\hat{r}}

    with no component in the direction of \bold{\hat{\theta}}, \bold{\hat{\phi}}
    The divergence in SPC is disgusting
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    (Original post by M14B)
    The divergence in SPC is disgusting
    Also I think this is an exam question and I imagine they are looking to do it in cartesian since there is information about what divergence in spherical is. Of course OP might have a good memory!!
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    (Original post by rayquaza17)
    Also I think this is an exam question and I imagine they are looking to do it in cartesian since there is information about what divergence in spherical is. Of course OP might have a good memory!!
    Exactly!

    Btw is to possible to sticky again
    http://www.thestudentroom.co.uk/show....php?t=3330283
    as I am finding increasingly difficult to find it down the pages
    Thanks
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    (Original post by M14B)
    Exactly!

    Btw is to possible to sticky again
    http://www.thestudentroom.co.uk/show....php?t=3330283
    as I am finding increasingly difficult to find it down the pages
    Thanks
    Done!
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    (Original post by rayquaza17)
    Done!
    Great
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    (Original post by M14B)
    Great
    Hey just in case you're looking for it, I've merged the undergrad resources and the a level resources and they're now in a new sticky in maths called "maths resources".
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    (Original post by rayquaza17)
    Hey just in case you're looking for it, I've merged the undergrad resources and the a level resources and they're now in a new sticky in maths called "maths resources".
    No problem
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    (Original post by M14B)
    The divergence in SPC is disgusting
    The general formula is more complex, if that's what you mean, but we only need the radial term here, which makes the divergence calculation trivial.
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    (Original post by atsruser)
    The general formula is more complex, if that's what you mean, but we only need the radial term here, which makes the divergence calculation trivial.
    Agreed it will be trivial for the OP.
 
 
 
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