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Partial Derivatives and Polar Coordinates

  1. Mar 2, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Write the chain rule for the following composition using a tree diagram:
    z =g(x,y)
    where x=x(r,theta) and y=y(r,theta). Write formulas for the partial derivatives dz/dr and dz/dtheta. Use them to answer: Find first partial derivatives of the function z=e^x+yx^2, in polar coordinates, that is find dz/dr and dz/dtheta as a function of polar coordinates (r, theta).

    2. Relevant equations


    3. The attempt at a solution
    Tree diagram was relatively simple
    g extends to x and y. x and y both extend to r and theta.

    **all "d's" symbolize partial derivatives**
    dz/dr = dg/dx*dx/dr+ dg/dy*dy/dr
    dz/dtheta = dg/dx*dx/dtheta+dg/dy*dy/dtheta

    Ok.
    So
    partial x = e^x+2yx
    partial y = x^2

    Now the problem I'm having trouble with is taking the partial derivative of the polar coordinate functions.
    If I am right, then:
    x=rcos(theta)
    y=rsin(theta)
    Taking the partial derivatives dx/dr and dy/dr the answers would be cos(theta) for x and sin(theta) for y, as we treat these as 'constants'.
    From here it is merely a plug-n-chug.

    Is this correct? Do you have any opinions? Thank you.
     
    Last edited: Mar 3, 2015
  2. jcsd
  3. Mar 3, 2015 #2

    Svein

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    Why is [itex] \frac{\partial g}{\partial x}\frac{\partial x}{\partial r}[/itex] missing?
     
  4. Mar 3, 2015 #3

    RJLiberator

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    Ah, typo, it should be dx/dr instead of dx/dtheta. My apologies.
     
  5. Mar 3, 2015 #4

    Svein

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    OK. So we have [itex]\frac{\partial g}{\partial x}=\frac{\partial g}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial g}{\partial \theta}\frac{\partial \theta}{\partial x} [/itex] and [itex]\frac{\partial g}{\partial y}=\frac{\partial g}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial g}{\partial \theta}\frac{\partial \theta}{\partial y} [/itex]...
     
  6. Mar 3, 2015 #5

    RJLiberator

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    I'm not quite sure why you are using that.

    I believe this would be correct, no?:
    dz/dr = dg/dx*dx/dr+ dg/dy*dy/dr
    dz/dtheta = dg/dx*dx/dtheta+dg/dy*dy/dtheta

    So if x=x(r,theta) and y=y(r,theta) then x=rcos(theta) and y=rcos(theta) ??
     
  7. Mar 4, 2015 #6

    Svein

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    Not quite...
    OK.
     
  8. Mar 4, 2015 #7

    RJLiberator

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    How could I make it correct? I feel like the x=rcos(theta) is a very important part of the problem. Why is it wrong?
     
  9. Mar 4, 2015 #8

    Svein

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    It isn't wrong. Look at your entire statement.
     
  10. Mar 4, 2015 #9

    RJLiberator

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    Ugh, another typo. Apologies. y=rsin(theta).
    So, if these are right, then their respective derivatives are
    cos(theta) with respects to dx/dr
    sin(theta) dy/dr
    and -rsin(theta) dx/dtheta
    rcos(theta) dy/dtheta

    And then it's just plugging in to the chain rule equations with the partial derivatives of the g function.
     
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