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AmTran_Bus

  • one year ago

Operator help

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  1. AmTran_Bus
    • one year ago
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  2. beginnersmind
    • one year ago
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    So what's the question? Calculate A*f(x,y,x) ?

  3. AmTran_Bus
    • one year ago
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    Exactly.

  4. beginnersmind
    • one year ago
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    Ok, so |dw:1441654151789:dw| That's what this operator is. It takes partial derivatives of a function and adds them together. Do you understand how to take partial derivatives?

  5. AmTran_Bus
    • one year ago
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    Thanks so much. I am really not all that sure. I know you start with the x and then go with the y and then the z. Like, the x only has the far left to deal with, right?

  6. beginnersmind
    • one year ago
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    I mean, if you are given a function f(x,y,z) = x*y + 2z, can you calculate its partial derivative with respect to x? If not, you need to review that first.

  7. AmTran_Bus
    • one year ago
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    Yea, so never had that class. In P chem, thrown at us blindly. So I know the answer, but I would really just app. if you could show me the steps.

  8. beginnersmind
    • one year ago
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    Can you take a quick guess first?

  9. AmTran_Bus
    • one year ago
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    Sure. 6xy^2z^4 + 2x^3z^4+12x^3y^2z^2

  10. beginnersmind
    • one year ago
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    One sec

  11. Jhannybean
    • one year ago
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    begin by writing your first partials \[f' =\langle 3x^2y^2z^4~,~ 2x^3yz^4~,~ 4x^3y^2z^3\rangle \]

  12. Jhannybean
    • one year ago
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    Then simply take their derivative in component form :)

  13. beginnersmind
    • one year ago
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    Ok, so that's actually the correct answer. I guess the question is which point do you have trouble with?

  14. Jhannybean
    • one year ago
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    I wrote it in component form so you could see their derivatives, it's asking for the additive operative, so just add them instead.

  15. Jhannybean
    • one year ago
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    (just how I would approach it!)

  16. AmTran_Bus
    • one year ago
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    Ohh ok. Thanks so much @Jhannybean

  17. Jhannybean
    • one year ago
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    but yeah, along with what @beginnersmind stated, where were ya having difficulty?

  18. AmTran_Bus
    • one year ago
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    Ive never seen this stuff before and its on a pchem homework problem

  19. AmTran_Bus
    • one year ago
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    like, i understand taking the derivative, but where does the 6, 2, and 12 come from @Jhannybean

  20. AmTran_Bus
    • one year ago
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    or @beginnersmind

  21. beginnersmind
    • one year ago
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    Well, you tell me, it's you who calculated it.

  22. beginnersmind
    • one year ago
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    (it is correct though)

  23. AmTran_Bus
    • one year ago
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    No no, it was the soln man that did!!!!

  24. beginnersmind
    • one year ago
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    Oh. So the question remains, do you know what a partial derivative is and how you calculate it? If you do please calculate |dw:1441655336477:dw|

  25. beginnersmind
    • one year ago
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    If you don't know what it means, that's ok too, but I need to know where to start my explanation.

  26. Jhannybean
    • one year ago
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    hmm... \[f' =\langle 3x^2y^2z^4~,~ 2x^3yz^4~,~ 4x^3y^2z^3\rangle\]\[f''=\langle \color{red}{6x}y^2z^4~,~ \color{red}{2}x^3z^4~,~ \color{red}{12z^2}x^3y^2\rangle \]

  27. Jhannybean
    • one year ago
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    The 6, 2 and 12 come from deriving each component based on what variable you are taking the partial derivative of.

  28. Jhannybean
    • one year ago
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    do you remember basic derivatives? Like.... taking the second derivative of \(f(x) = 2x^4\) would be \(f'(x) = (2 \cdot 4)x^3 = 8x^3 \rightarrow f''(x) = (8\cdot 3)x^2 = 16x^2 \) But now you're doing it with each separate component, \(x\) , \(y\) AND \(z\), hence the gradient.

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