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anonymous

  • one year ago

Please, help me

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  1. anonymous
    • one year ago
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    Proof \[\frac{ \textbf{x }- \textbf{x}_i }{ \left| \textbf{x} - {\textbf{x}_i} \right|^3 } = \bigtriangledown \left( \frac{ 1 }{ x-x' } \right)\]

  2. TheCatMan
    • one year ago
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    i say to find the area of the triangle using other info to the question (if there is none i have no clue) distribute to the right side of the equation and cross eliminate to solve

  3. UsukiDoll
    • one year ago
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    WHAT THE! SOlving has nothing to do with writing a complete proof. And I don't know which subject this came from. It could be from some grad school Math course :S

  4. UsukiDoll
    • one year ago
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    I know that for proofs like this you either start at the left apply some theorems...definitions lalalalalla and achieve the right or start at the right apply some theorems...definitions lalalla and achieve the left.

  5. anonymous
    • one year ago
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    mathematical physics

  6. UsukiDoll
    • one year ago
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    that could be in the later part of Mathematical Physics...... I only know the first half and have not done a proof like this

  7. TheCatMan
    • one year ago
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    i wish i could help but im dumber than a fart in a crowded room.

  8. alekos
    • one year ago
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    it really means nothing without some sort of context. tel us more about the problem

  9. Michele_Laino
    • one year ago
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    here are some steps: we can write: \[{\mathbf{x}} - {{\mathbf{x}}_i} = {\mathbf{r}}\] where: \[{\mathbf{r}} \cdot {\mathbf{r}} = {x^2} + {y^2} + {z^2}\]

  10. Michele_Laino
    • one year ago
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    so we have: \[\large \begin{gathered} \frac{\partial }{{\partial x}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial x}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{x}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial y}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial y}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{y}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial z}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial z}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{z}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \end{gathered} \]

  11. Michele_Laino
    • one year ago
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    \[\begin{gathered} \frac{\partial }{{\partial x}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial x}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{x}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial y}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial y}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{y}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial z}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial z}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{z}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \end{gathered} \]

  12. Michele_Laino
    • one year ago
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    and finally: \[\Large \nabla \left( {\frac{1}{r}} \right) = - \frac{{\mathbf{r}}}{{{r^3}}}\]

  13. anonymous
    • one year ago
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    awesome @Michele_Laino , thank you

  14. anonymous
    • one year ago
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    i also thank to @TheCatMan , @UsukiDoll , @alekos

  15. UsukiDoll
    • one year ago
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    what @Michele_Laino latexed is similar to an example from a book I had to use almost a year ago. There was partial derivatives, product rule, and chain rule involved.

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