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Differentials and Implicit Differentiation

  1. Dec 31, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm reviewing physics using Feynman's Lectures, and I'm finding that he frequently uses implicit differentiation in his lessons. This is unfortunate for me because I never got the hang of it beyond the simplest cases. I'm currently going through the proof that the gravitational potential inside a hollow sphere is zero. I almost got through it, but at one point (outlined below in red) he uses an implicit derivative and loses me. Could someone please explain how this works?

    Implicit_zps7c650b1b.png

    I'm used to two-variable implicit differentiation, where one variable (y) is dependent upon the independent variable (x). Thus, if you have an implicitly defined function such as

    x^2 + y(x)^2 = 25

    it can be solved by deriving with respect to this independent variable:

    d(x^2)/dx + d(y^2)/dx = d(25)/dx

    2x + 2y*dy/dx = 0

    dy/dx = -x / y


    Is this an incorrect way of thinking about it? I can't relate it to what Feynman is doing in the image above. First, he doesn't seem to be deriving with respect to anything; he's just taking the differential. What are the rules for this operation? Second, it's as if he's deriving with respect to r on the left side (r^2 ---> 2r dr) and with respect to x on the right side (a^2 + R^2 -2Rx ---> -2R dx). How is it "mathematically legal" to arbitrarily pick different variables to differentiate with respect to on the two sides of the equation?

    On a hopefully more simple note, in the blue box within the red box, where did he hide the negative sign which appears in front of 2R dx?


    Thanks to anyone who can help clear this up. I never really got implicit differentiation down in high school calculus and somehow I got through multivariate and diff eq without really understanding the concept.



    2. Relevant equations

    d/dx (F(y(x)) = dF/dy * dy/dx

    3. The attempt at a solution

    Stated above
     
  2. jcsd
  3. Dec 31, 2012 #2

    lurflurf

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    It is just the differential
    d(r^2)=2r dr
    d(a^2+R^2-2Rx)=-2R dx

    [tex]\text{d}F=\sum_{k=1}^n \dfrac{\partial F}{\partial x_k} \text{d}x_k[/tex]

    for however many variable there are, in this case only one variable on each side.
     
  4. Dec 31, 2012 #3

    joshmccraney

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    as an aside, whenever you differentiate over functions the d(whatever) is multiplied by the chain rule, quick example:

    y=x^2
    dy=2xdx
    dy/dx=2x

    a natural extension for multi variables.
     
  5. Dec 31, 2012 #4
    Thank you! I didn't notice that R is also held constant by the physical situation. So if R was also a variable in this problem, the differential would be 2r dr = 2R dR - 2xdR - 2Rdx?

    That makes a lot of sense. Thanks for that explanation. So the differential is the total infinitesimal distance moved across all available degrees of freedom? Is there notation that specifies the differential across a subset of available dimensions? For example, for a 3D function F(x,y), how would you denote the differential only along x?

    Finally, in the blue box from my image above, did Feynman make a mistake in dropping the negative sign, or am I missing something?

    Thanks again for all the help!
     
  6. Dec 31, 2012 #5

    joshmccraney

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    yes, but only if we assume a^2 is a constant (is this true? it seems x^2+y^2=a^2, and if this is the case we need to differentiate over that a^2. without more knowledge of this problem i cant say.


    not necessarily over all degrees of freedom. consider partial derivatives. lurflurf outlined general notation for the F(x,y). specifically, let n=2 recognizing x_2 is equivalent to y (this is natural at higher dimensions). it seems with your hypothetical R you understand, so not sure where the confusion is.

    it seems so, though sometimes physicists are permitted to make assumptions mathematicians are not. nonetheless, what about the x that's in the a (assuming x^2+y^2=a^2)?
     
  7. Dec 31, 2012 #6

    lurflurf

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    Thank you! I didn't notice that R is also held constant by the physical situation. So if R was also a variable in this problem, the differential would be 2r dr = 2R dR - 2xdR - 2Rdx?
    Yes that is right.


    That makes a lot of sense. Thanks for that explanation. So the differential is the total infinitesimal distance moved across all available degrees of freedom? Is there notation that specifies the differential across a subset of available dimensions? For example, for a 3D function F(x,y), how would you denote the differential only along x?
    I prefer to thing of it as the closest linear function. Often it is helpful to approximate differences of a function by small but not infinitesimal differentials. Also dealing with infinitesimals carelessly leads to logical errors. There is such a thing as a partial differential
    [tex]d_{x_k}F=\dfrac{\partial F}{\partial x_k} \text{d}x_k[/tex]
    so
    [tex]\text{d}F=\sum_{k=1}^n d_{x_k}F=\sum_{k=1}^n \dfrac{\partial F}{\partial x_k} \text{d}x_k[/tex]

    The total differential is invariant, meaning it is the same in any coordinate system. Partial differentials do not share this advantage, limiting their usefulness.


    Finally, in the blue box from my image above, did Feynman make a mistake in dropping the negative sign, or am I missing something?
    Yes a minus disappeared.

    Some of these confusions become clear when we realize each (partial or total) derivative (or differential) we see depends upon how the variables are related. It is convenient to omit such information when obvious. It is a minor abuse to denote a function f(x) as y because F(x,y)=F(x,f(x)) for example.
     
  8. Jan 1, 2013 #7
    I think there was no reason to omit the minus sign..
     
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