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Partial Differentiation

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Hello all,
    New to partial derivatives. I was wondering if someone could look over my work and determine if my final step is as far as I can take the proble (ie. that will be my solution). Thanks in advance.

    Let the temperature of a 2D domain in polar coordinates [itex](r, \varphi)[/itex] be given by [itex]T=f(r, \varphi, t)[/itex], where t is time. Suppose a probe follows the straight path, given in Cartesian coordinates by [itex]x=X(t), y=Y_0=constant[/itex]. Using the fact that [itex]r^2=x^2+y^2, \varphi=arctan(y/x)[/itex], find [itex]dT/dt[/itex].


    3. The attempt at a solution

    [itex]\frac{dr}{dt}=\frac{\delta r}{\delta x}\frac{dx}{dt}+\frac{\delta r}{\delta y}\frac{dy}{dt}=\frac{\delta r}{\delta x}\frac{dx}{dt}[/itex]

    and

    [itex]\frac{d\varphi}{dt}=\frac{\delta \varphi}{\delta x}\frac{dx}{dt}+\frac{\delta\varphi}{\delta y}\frac{dy}{dt}=\frac{\delta\varphi}{\delta x}\frac{dx}{dt}[/itex]

    thus,

    [itex]\frac{dT}{dt}=\frac{\delta T}{\delta r}\frac{dr}{dt}+\frac{\delta T}{\delta\varphi}\frac{d\varphi}{dt}+\frac{\delta T}{\delta t}[/itex]

    Next, we have,

    [itex]\frac{dT}{dt}=\frac{\delta T}{\delta r}\frac{\delta r}{\delta x}\frac{dx}{dt}+\frac{\delta T}{\delta\varphi}\frac{\delta\varphi}{\delta x}\frac{dx}{dt}+\frac{\delta T}{\delta t}[/itex]
    [itex]=\frac{\delta T}{\delta r}(x)\frac{d[X(t)]}{dt}-\frac{\delta T}{\delta\varphi}\frac{1}{1+(y/x)^2}\frac{y}{x^2}\frac{d[X(t)]}{dt}+\frac{\delta T}{\delta t}[/itex]

    Is this as far as I can take it with the information given?
     
  2. jcsd
  3. Oct 17, 2011 #2
    also wanted to inform everyone that the lowercase deltas were supposed to be partials
     
  4. Oct 18, 2011 #3
    No one?
     
  5. Oct 18, 2011 #4

    cepheid

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    You should try \partial

    Your application of the chain rule seems okay.

    I think that partial deriv of r wrt x should be 2x, not x, right?
     
  6. Oct 18, 2011 #5

    cepheid

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    Actually nvmind my third statement. I forgot about the square root.
     
  7. Oct 18, 2011 #6
    Thanks. My real reason for asking was because the answer seems kind of odd. It just feels like I could go furthur with it, but I dont see how.
     
  8. Oct 18, 2011 #7
    Now that I have had a second look at the problem I'm going to say that it is 100 percent wrong....

    [itex]\frac{dT}{dt}=\frac{\partial T}{\partial r}\frac{dr}{dt}+\frac{\partial T}{\partial \varphi}\frac{d\varphi}{dt}+\frac{\partial T}{\partial t}[/itex]

    Hence that couldnt be right since
    [itex]\frac{dr}{dr}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial t}[/itex]

    and a similar expression for [itex]\varphi[/itex] can be found...
     
    Last edited: Oct 18, 2011
  9. Oct 18, 2011 #8

    cepheid

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    Okay, well, this is definitely true:

    That's just how the chain rule works for multivariable functions.

    Doesn't the second term above assume that r has some sort of explicit dependence on t that is outside of the implicit dependence on t that comes via its dependence on x? In this case, I don't think that's true. Since there is no explicit dependence on t that is not via some other variable, I think that [itex]\partial r / \partial t [/itex] would be 0.

    EDIT: Mathematical way of saying this: [itex] r(x,t) = r(x(t)) [/itex] (no explicit dependence on t).

    Another way to look at it: if you posit that r depends on x and on t, then the way to compute [itex]\partial r / \partial t [/itex] is to hold x constant. But for x to be constant, t must be constant as well.

    By the way, the itex tags are for LaTeX equations that are inline with regular text. For equations that are not inline, you can just use tex tags instead of itex like so:

    [tex]\frac{dT}{dt}=\frac{\partial T}{\partial r}\frac{dr}{dt}+\frac{\partial T}{\partial \varphi}\frac{d\varphi}{dt}+\frac{\partial T}{\partial t}[/tex]
     
  10. Oct 18, 2011 #9
    If I understand what you are saying correctly, since dr/dt=0, then essentially dT/dT will simply be the partial derivative of T with respect to the partial derivative of time.
     
  11. Oct 19, 2011 #10

    cepheid

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    NO, you did not understand me correctly. dr/dt is NOT 0. I was arguing that ∂r/∂t = 0. There's an important difference there. With the latter, the derivative dr/dt reduces to:

    [tex] \frac{dr}{dt} = \frac{\partial r}{\partial x} \frac{dx}{dt} [/tex]

    So it's just the normal chain rule for composite functions. In other words, we return to what you had in your original post.
     
    Last edited: Oct 19, 2011
  12. Oct 19, 2011 #11
    Oh wow, thanks. Mind went blank. Thanks a lot!
     
  13. Oct 19, 2011 #12

    cepheid

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    By the way, I also don't think you computed ∂r/∂x correctly. We have:

    [tex] \frac{\partial r}{\partial x} = \frac{\partial }{\partial x}(x^2 + y^2)^{1/2} [/tex]

    How would you go about differentiating this?
     
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