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Partial Derivatives. Did I make a mistake or my professor

  1. Feb 1, 2016 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    the equation is E= k((xy)x[hat] +(2yz)y[hat] +(3xz)z[hat])

    2. Relevant equations
    partial of x with respect to y on the x component
    partial of y with respect to x on the y component

    3. The attempt at a solution
    my professor said during class that the partial of x with respect to y is y and the partial of y with respect to x is 0
    shouldn't the partial of x with respect to y be x?
     
  2. jcsd
  3. Feb 1, 2016 #2

    Mark44

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    Moved from Precalc section. Questions about derivatives should be posted in the Calculus section.
    In LaTeX, the equation is ##E = k(xy\hat{x} + 2yz\hat{y} + 3xz\hat{z})##
    I'm not sure these make sense. Do you mean, partial of E with respect to x and partial of E with respect to y? In symbols, ##\frac{\partial E}{\partial x}## and ##\frac{\partial E}{\partial y}##.

    If x, y, and z are independent variables, then ##\frac{\partial x}{\partial y}## = 0 and ##\frac{\partial y}{\partial x}## = 0. The partial of any of the three variables with respect to a different variable is zero.
     
  4. Feb 1, 2016 #3

    grandpa2390

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    I don't know. the only explanation I can offer is that each component is an equation to itself. ##\hat{x}=xy## and ##\hat{y}=2yz##

    Physics professors do often joke about there being a difference between Physics math and Mathematician math... Not being a mathematician, I can't tell you how it is supposed to be. Perhaps I should email this question to my professor?
     
  5. Feb 1, 2016 #4

    Mark44

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    I don't think so, at least based on your first post.
    In this equation, ##E = k(xy\hat{x} + 2yz\hat{y} + 3xz\hat{z})##, E is apparently a vector with components ##\hat{x}, \hat{y},## and ##\hat{z}##. The coefficients of these vectors (which I believe are unit vectors, given the "hat" notation), respectively, are kxy, 2kyz, and 3kxz.

    It makes no sense, as I read this problem, to say that ##\hat{x} = xy## or the other two equations you show here.
     
  6. Feb 1, 2016 #5

    grandpa2390

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    ok. I'm going to go visit my professor. If you would like, I'll post his response. Just to make sure I understand what you are saying. it should be the partial of E with respect to each component. otherwise, if each variable is independent, then what I wrote should be 0.

    I don't understand what you mean though that the coefficient of the vectors. they are variables. and the coefficients given the partial derivatives would be (kx)y and (2kyz). therefore kx and 0. that's what it seems like to me. though, then again, I don't know this stuff as well as you do, so I am not arguing with you.
     
  7. Feb 1, 2016 #6

    SammyS

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    Definitely not that.

    Perhaps he/she meant that ##\displaystyle \ \frac{\partial E_x}{\partial y} = (k)\cdot x \ ## and ##\displaystyle \ \frac{\partial E_y}{\partial x} = 0 \ .##
     
  8. Feb 1, 2016 #7

    grandpa2390

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    Actually, that sounds right. He has used that notation before. and I think when he was saying it, he was just saying it quickly assuming we understood that. But still, the first partial would be kx right? he left the k in front because it was irrelevant to what he was doing. He was just proving that the the first partial and the second weren't equal. But otherwise your interpretation makes sense, and does the first partial equal kx then? rather than y?
     
  9. Feb 1, 2016 #8

    Ray Vickson

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    I think you mean ##\vec{E} = E_x \hat{x} + E_y \hat{y} + E_z \hat{z}##, where ##E_x = kxy, E_y=2kyz, E_z=3kxz## are the ##x,y,z##-components of ##\vec{E}## and ##\hat{x}, \hat{y}, \hat{z}## are the unit vectors along the ##x,y,z##-axes. It makes sense to speak of quantities such as ##\partial E_x/\partial x## or ##\partial E_x/\partial y##, for example, but makes no sense at all to speak of "partial of x with respect to y on the x component". Are you stating the question exactly and completely?
     
  10. Feb 1, 2016 #9

    grandpa2390

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    It's not a question. it is just something he said in class. and I think @SammyS got it.
    looks like you got it too
     
  11. Feb 1, 2016 #10

    Mark44

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    Good idea.
    Yes, please do.
    Yes, that's what I was saying.
    Yes. The partial of any variable with respect to some other variable would be zero.
    By coefficients, I mean the things that multiply each of ##\hat{x}, \hat{y},## and ##\hat{z}##.
    It's not clear to me what you're saying here.
    If ##E = k(xy\hat{x} + 2yz\hat{y} + 3xz\hat{z})##, then ##\frac{\partial E}{\partial x}## would be ##ky\hat{x} + 0k\hat{y} + 3kz\hat{z}##, which could also be written as ##k(y\hat{x} + 3z\hat{z} ).##

    The only thing that bothers me about this, is not knowing what ##\hat{x}, \hat{y},## and ##\hat{z}## are. I believe they are unit vectors, based on the "hat" notation, but if they are not all mutually orthogonal, I'm not sure what complexities that adds.
     
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