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Help With Partial Differentiation & Integration

  1. Aug 27, 2008 #1
    I know I should know this... it looks so ridiculously easy. In the course of getting d'Alembert's wave equation solution, we get the following equation:


    The primes are derivatives wrt t. Then we re-order the equation and "integrate the relation" to get an expression for p:


    I have to be missing something very, very simple. How can I differentiate [tex]p\left(x\right)[/tex] wrt [tex]t[/tex] then integrate wrt something (x, I suppose, in this case) and recover the original function p? Or am I NOT recovering it... just something new named [tex]p\left(\xi\right)[/tex]? Thanks for the help!

  2. jcsd
  3. Aug 27, 2008 #2


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    Hi Tom! :smile:

    [tex]\int^{\xi}_{0}f'(x)dx\ =\ \left[f(x)\right]^{\xi}_{0}\ =\ f(\xi)\ -\ f(0)[/tex] :smile:
  4. Aug 27, 2008 #3
    Alright, cool. Why is that true, though? If I have f(x) = 3x^2, differentiate wrt t and integrate wrt x from 0 to xi I don't think I will get 3(xi)^2?

  5. Aug 27, 2008 #4


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    f'(x) = 6x.

    [tex]\int^{\xi}_{0} 6x dx\ =\ [3x^2]^{\xi}_{0}\ =\ 3\xi^2[/tex] :smile:

    Integration is the opposite of differentiation …

    that's how it works!​
  6. Aug 27, 2008 #5
    The fancy name of what tiny-tim show you is "Fundamental Theorem of Calculus". You may wish to read up more about it :)
  7. Aug 27, 2008 #6
    Yes... when it was covered waaaay back in Calc I, it was something more like:

    I was just concerned with differentiating wrt t AND integrating wrt x:

    I felt like I had variables flying everywhere, hehe. Thanks!

    Last edited: Aug 27, 2008
  8. Aug 27, 2008 #7
    Your first formula is incorrect I believe. You should only be left with f(x) and I believe you want [tex] \frac{d}{dx} [/tex] instead of [tex] \frac{d}{dt} [/tex]. The "general" version of this


    [tex] F(x) = \int_{g(x)}^{h(x)} f(t) dt [/tex]


    [tex] F'(x) = f(h(x)) h'(x) - f(g(x)) g'(x) [/tex]
  9. Aug 27, 2008 #8
    alright... thanks.

    now, here's the thing... this all makes sense, except in my problem.

    I have [tex]f(x) = 3x^{2}[/tex] and [tex]\frac{df}{dt}=f'\left(x\right) = 6x\frac{dx}{dt}[/tex]

    [tex]\int^{\xi}_{0}6x\frac{dx}{dt}dx[/tex] = ??

    or am I waaay confused?
    Last edited: Aug 27, 2008
  10. Aug 27, 2008 #9
    If you are doing implicit differentiation you should get [tex] 6x \frac{dx}{dt}[/tex] (minor error).
  11. Aug 27, 2008 #10
    My bad, just in a hurry. Still... [tex]\frac{dx}{dt}dx[/tex]??
  12. Aug 27, 2008 #11
    Ok, I am an idiot... these aren't t-derivatives, I guess.


    [tex]\frac{\partial u}{\partial t}\right|_{t=0}=\frac{\partial u}{\partial \xi}\right|_{t=0}\frac{\partial \xi}{\partial t}\right|_{t=0} + \frac{\partial u}{\partial \eta}\right|_{t=0}\frac{\partial \eta}{\partial t}\right|_{t=0}[/tex]

    but [tex]\xi=\left(x+ct\right)\right|_{t=0}=x[/tex], same for [tex]\eta[/tex]. So...

    [tex]\frac{\partial u}{\partial t}\right|_{t=0}=\frac{\partial u}{\partial x}\frac{\partial \xi}{\partial t}\right|_{t=0} + \frac{\partial u}{\partial x}\frac{\partial \eta}{\partial t}\right|_{t=0}=\frac{\partial u}{\partial x}\cdot c + \frac{\partial u}{\partial x}\cdot \left(-c\right)[/tex]

    So [tex]p'\left(x\right)=\frac{dp}{dx}[/tex], not [tex]\frac{dp\left(x\right)}{dt}[/tex].

    Does that seem correct?
  13. Aug 27, 2008 #12


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    Surely that's not what you meant! If you differentiate f(x) wrt t, you get 0. Integrating that with respect to x still gives 0.
  14. Aug 27, 2008 #13
    Exactly... yet my professor was adamant. I'm sure now that he didn't understand my question.
  15. Aug 27, 2008 #14


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    Your second statement is incorrect. It should be
    [tex]\int^{t_i}_0 6x \frac{dx}{dt}dt= \int^{x_i}_0 6x dx= 3x_i^2[/itex]
    Where [itex]x_i= x(t_i)[/itex]
  16. Aug 27, 2008 #15
    right... this is why i was so confused. i thought my prof was saying to differentiate wrt t, then integrate wrt x, not t.
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