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Just a quick question about partial derivatives

  1. Apr 29, 2010 #1
    Not a homework question, but It will help me none the less,

    In my book it says

    [tex] \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx [/tex]

    is equivalent to

    [tex] \int_{-\infty}^{\infty} \frac{\partial}{\partial t}|\Psi(x,t)|^2 dx [/tex]

    I understand how It becomes a partial derivative, since I'm differentiating psi with respect to x and t

    But what I don't understand Is how you can just move [tex] \frac{d}{dt} [/tex]
    into an integral, why can you do that? I don't understand the rules behind it,

    for example
    (i'm just making this up)
    [tex] \frac{d}{dt} \int_0^2 2xt dx [/tex]
    I'm first going to integrate the inside function then differentiate it with respect to t
    before I try the other way of moving dt inside the integral

    [tex] \int_0^2 2xt dx = \left \frac{2x^2t}{2} \right|_0^2 [/tex]
    the equation becomes
    [tex] \left x^2t\right|_0^2 [/tex]
    putting in 2, I get 2^2 *t
    = 4t
    differentiating that,
    [tex] \frac{d}{dt} 4t = 4 [/tex]

    now i'll try the other way by moving in the partial derivative

    [tex] \int_{-\infty}^{\infty} \frac{\partial}{\partial t} 2xt dx [/tex]
    I've only done one math paper before, and 5 physics papers,
    so I don't know how to solve partial derivatives in the conventional sense,
    but what I have picked up is, that a physicists first line of attack to any partial derivative, is separation of variables,
    so uh, i'm not too sure how i do it in this context, since I can't separate x,t the way I can separate [tex] \Psi , \textrm{ which I can rewrite as } \psi(x)\phi(x) [/tex]
     
  2. jcsd
  3. Apr 29, 2010 #2
    [tex] \int_{-\infty}^{\infty} \frac{\partial}{\partial t} 2xt dx, = 2x [/tex]

    that's from differentiating 2xt, treating x as a constant, d/dt 2t = 2, so the t disappears

    [tex] \int_0^2 2x dx = \int_0^2 \frac{2x^2}{2} dx = \left x^2\right|_0^2 [/tex]

    so 2^2 = 4

    so i've just proved for the equation 2xy

    [tex] \frac{d}{dy} \int_a^b F(x,y) dx \textrm{ is equivalent to } \int_a^b \frac{\partial}{\partial y} F(x,y) dx [/tex]

    meh

    Is this true for all equations?
     
  4. Apr 29, 2010 #3

    D H

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    Staff Emeritus
    Science Advisor

    This is an application of the general concept of "differentiation under the integral sign" or the Leibiz Integral Rule. http://mathworld.wolfram.com/LeibnizIntegralRule.html.
     
  5. Apr 29, 2010 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    If a and b are constants, yes. This is essentially the fundamental theorem of calculus.

    If either a or b is a function of y (or if both are a function of y) then the above is not true. You need to use the more generic form of the Liebniz Integral Rule to compute derivative.
     
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