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Calc III Chain rule - which vars to put in?

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data

    z = cos(x^2 + 3y^2)
    x = ucosv
    y=usinv

    find dz/dv

    2. Relevant equations



    3. The attempt at a solution


    I think I can do these fairly well, but I'm a little unsure of the "protocol" for which variables to put back in. Sometimes (in this case) I can't really put everything as v. So I choose to put everything as u and v since they are on the same "level."

    for example, I get that

    (dv/dv) = (dy/dv)(dz/dy) + (dx/dv)(dz/dx)

    When taking those partials and multiplying out ( leaving variables as they are.. )

    = - ucos(v)sin(x^2 + 3y^2)6y + usin(v)sin(x^2 + 3y^2)2x

    So what is the technique? Replace all x and y with ucosv and usinv respectively, and leave u in the partial derivative?
     
  2. jcsd
  3. Feb 29, 2012 #2

    BruceW

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    Homework Helper

    This is not quite right. The partial derivatives you have defined are keeping x constant (to get dy/dv) and y constant to get dx/dv. So u is not necessarily kept constant, but you have taken the partial derivative as if it were constant.

    EDIT: So sorry! I got myself confused! you have done it correctly! Ignore Everything I was saying please!
     
  4. Feb 29, 2012 #3
    I have no idea what you mean.

    My partials are
    dy/dv = ucos(v)
    dz/dy = -sin(x^2 + 3y^2)6y
    dx/dv = -usinv
    dz/dx = -sin(x^2 + 3y^2)2x

    Then I multiplied them and added as such:

    (dy/dv)(dz/dy) + (dx/dv)(dz/dx)

    What part of this is wrong?



    EDIT:

    Saw your edit, no apologies needed. Is it correct to use u and v in the final partial and phase out x and y?
     
  5. Feb 29, 2012 #4

    BruceW

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    OK, so now I've looked at it again, I see what you are saying:
    [tex]\frac{dz}{dv} = \frac{\partial z}{\partial x} \LARGE | \small _y \normalsize \ \ \frac{dx}{dv} + \frac{\partial z}{\partial y} \LARGE | \small _x \normalsize \ \ \frac{dy}{dx} [/tex]
    Right? (I hope I did the latex correctly). Anyway, your evaluation of the partial derivatives are correct, but you derivatives (i.e. dx/dv and dy/dv) are not correct, because they are full derivatives, yet you have treated u as constant.

    Edit: my latex skills aren't amazing, but the line and small x (or y) show what is being held constant when the partial derivative is taken.
     
    Last edited: Feb 29, 2012
  6. Feb 29, 2012 #5
    I'm sorry, it's my fault for not being clear (latex stopped working on my browser) but, EVERY derivative I've written here is intended a partial, including dz/dv. My goal is the partial derivative of function z with respect to variable v. That's why I hold u constant.
     
  7. Feb 29, 2012 #6

    BruceW

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    Oh, I get you. So the goal is the partial derivative of z with respect to v, while holding u constant. Yes, I think you've got the right answer then. So in the final answer you can have u, v, x and y in there, it doesn't matter, you can rearrange it whichever way you like really.
     
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