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Finding the intergral function (dQ/dt) = kQ

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the integral (I think I'm finding the integral) of:
    [itex]\frac{dQ}{dt}[/itex] =kQ

    2. Relevant equations



    3. The attempt at a solution
    I just got feedback from my teacher, and he told me I make a mistake somewhere in this question. But I don't know where I've gone wrong?

    [itex]\frac{dQ}{dt}[/itex]=kQ
    [itex]\frac{dQ}{Q}[/itex]=kdt
    ∫[itex]\frac{dQ}{Q}[/itex]=∫kdt
    lnQ=kt+c
    Q=ekt+ec
    Where ec is a constant, so let ec=A
    Q=Aekt

    ∴ Q=Aekt

    Can anyone see where I've gone wrong?
     
    Last edited: Apr 27, 2012
  2. jcsd
  3. Apr 27, 2012 #2
    If you substitute your solution back into the differential equation, the LHS equals the RHS, so the solution is fine. However, you wrote in an intermediate calculation that Q = ekt + ec. It can't be an addition. Try and think why it can't be an addition.

    Hint: What are rules for algebraic operations with indices?

    You made the same mistake in your next post.
     
  4. Apr 27, 2012 #3
    oh... Kt + c will have to go to

    ekt+c
    ektx ec

    right?
     
  5. Apr 27, 2012 #4
    Yup, I can't see any other mistakes. This answer should score full marks now.
     
  6. Apr 27, 2012 #5
    awesome!

    Thanks man, I owe you one
     
  7. Apr 27, 2012 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Technically, there is one other error. The integral, [itex]\int (1/Q)dQ= ln|Q|[/itex], not ln(Q).
     
  8. Apr 28, 2012 #7
    ahhh... ok, thanks for that
     
  9. Apr 28, 2012 #8

    Mark44

    Staff: Mentor

    Minor point: there is no such word as "intergral" in the English language. Seeing as you have started two threads with this in the title, I thought I should point it out.
     
  10. Apr 28, 2012 #9
    Sorry, I have terrible spelling :frown:
     
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