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Intermediate variable chain rule question.

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose that w=f(x,y), x=r*cos(θ), y=r*sin(θ). Show that:
    [tex](\frac{\partial w}{\partial x})^2 + (\frac{\partial w}{\partial y})^2 =(\frac{\partial w}{\partial r})^2 + \frac{1}{r^2} (\frac{\partial w}{\partial \theta})^2 [/tex]


    2. Relevant equations
    the multivariable chain rule


    3. The attempt at a solution
    we just were taught this yesterday, but my prof didn't exactly do a good job, and i'm doing a good job at figuring it out and understanding it, but all i'm able to do is use the chain rule with respect to the independent variables (r and θ in this case), and i can't figure out how to use it for the intermediate variables x and y and my book doesn't have any examples of this. I'm sure it's something really obvious that i'm missing, but i just haven't had that lightbulb moment...
     
  2. jcsd
  3. Sep 18, 2009 #2
    w depends on x and y

    x depends on r and theta; y depends on r and theta

    If r changes, then x and y are influenced, and each of them influences w, so the formula for [tex]\frac{\partial w}{\partial r}[/tex] will have to include x and y.

    This helps you remember that the formula is
    [tex]\frac{\partial w}{\partial r}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial r}[/tex]

    The intuition and formula for [tex]\frac{\partial w}{\partial \theta} [/tex] are similar.

    --


    Now turning to your problem, on the left hand side, there is nothing to do. [tex]\frac{\partial w}{\partial x} [/tex] and [tex]\frac{\partial w}{\partial y} [/tex] do not simplify.

    On the right hand side, use
    [tex]\frac{\partial w}{\partial r}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial r}[/tex]
    and the analogous formula for [tex]\frac{\partial w}{\partial \theta} [/tex], find the partials with respect to r and theta, and simplify.
     
  4. Sep 18, 2009 #3

    LCKurtz

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    Part of the confusion in this type of problem stems from the abuse of notation whereby the same notation, in this case w is used for the function whether it is expressed in terms of x and y or r and theta. To more properly phrase the question it should be given as follows:

    If [tex] w(x,y) = W(r,\theta)[/tex], where [tex]x = r\cos(\theta)\, y = r \sin(\theta)[/tex], show that

    [tex]w_{x}^2 + w_{y}^2 = W_r^2+\frac 1 {r^2}W_{\theta}^2[/tex]

    Now start with the right side and use the chain rule in this form:

    [tex]W_r = w_r = w_x x_r + w_y y_r[/tex]

    [tex]W_{\theta} = w_{\theta} = w_x x_{\theta} + w_y y_{\theta}[/tex]

    The partials [tex] x_r,\ y_r,\ x_{\theta},\ y_{\theta}[/tex] are easy to calculate from your equations. Manipulate that a bit and you should get the required equation.

    Once you have done this a couple of times, it likely won't confuse you to use the lower case w for W. Although they aren't the same function, many texts do this.
     
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