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Maximum charge differential equation

  1. Dec 10, 2011 #1
    Hi, I am having a small problem with this equation

    Charge Q=t^3+5t^2+10t+130

    What is the maxiumum charge?
    If I differentiate with repect to time, at t=6, for example then I have value of current flowing at that time.
    What is the maximum charge then? I dont understand the question. Do I need to differentaite it again? Thanks
     
  2. jcsd
  3. Dec 10, 2011 #2
    At maximum Q, dQ/dt = 0.

    But, so long as there is no mistake in the equation given, there is no value of t for which dQ/dt = 0.
     
  4. Dec 10, 2011 #3
    The equation has some value changed by me, but if dQ/dt=0 then the current is 0. Is that right?. About the equation, if I differentiate Q, then I can have two roots t. If I substitute them in the Q equation I have two real, different values. Is that right? Do I choose the higher value obtained?
     
  5. Dec 10, 2011 #4

    HallsofIvy

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    A quadratic equation can have 2, 1, or no real roots. Read what grzz said again. Are there limits on the values of t?
     
  6. Dec 10, 2011 #5
    No, there are no limits.I know that quadratic equation can have 2,1, or no ral roots. But in this case it has two different real ones.
     
  7. Dec 10, 2011 #6

    NascentOxygen

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    Each term on the right hand side increases with time, and these are all summed. So Q continuously increases with time. It will reach a maximum only at the end of time.
    I think it's worth taking another look at the question. Something might be amiss.
     
  8. Dec 10, 2011 #7

    SammyS

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    It asks for maximum charge, so just differentiate once. Yes, it so happens that when dQ/dt = 0 , in addition to having maximum (or minimum) charge, the current is also zero, because current = dQ/dt .
     
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