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Problem solving equation with negative exponent

  1. Feb 22, 2010 #1
    I'm trying to calculate maximum power. Once I computed power as a function of time I took its derivative and set it equal to zero. Now it need to solve for time and I can't seem to get it done. What I have is...

    [tex]-1000e^{-500t}-2000e^{-1000t}=0[/tex]

    My first inclination is to use ln, but ln(0) is undefined. Any tips?
     
  2. jcsd
  3. Feb 22, 2010 #2
    Are you sure that equation is right? Looks to me like you have -(something which is always positive) - (something else that is always positive) = 0, but the LHS will always be negative, and never zero, unless t is infinite?
     
  4. Feb 22, 2010 #3

    CRGreathouse

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    That would seem to indicate that the power has no maximum; perhaps it increases without bound.
     
  5. Feb 22, 2010 #4
    As it is now, there's no solution. You probably made a sign error while computing the derivative.
    If that is indeed the case you can try substituting [itex] e^{-500t} = u [/itex]
     
  6. Feb 22, 2010 #5
    Thanks to all who responded.

    I believe I was doing my computations incorrectly. I am given voltage:

    [tex]100e^{-500t}V[/tex]

    And current:

    [tex]20-20e^{-500t}mA[/tex]

    Power being the product:

    [tex]2e^{-1000t}(e^{500t}-1)W[/tex]

    And derivative:

    [tex]-1000e^{-1000t}(e^{500t}-2)[/tex]

    To find out when power is maximized:

    [tex]-1000e^{-1000t}(e^{500t}-2)=0[/tex]

    [tex]2000e^{-1000t}=1000e^{-500t}[/tex]

    [tex]ln2=500t[/tex]

    Putting t at about .0013863 s.
     
  7. Feb 22, 2010 #6

    Redbelly98

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    I can't find any error so far.

    Just out of curiousity, what is the purpose of the circuit?
     
  8. Feb 22, 2010 #7
    It isn't real. This was one part of a practice problem I was solving as to lead me up to DEs, which I need to reacquaint myself with.
     
  9. Feb 22, 2010 #8

    Redbelly98

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    Oh. Questions like that should get posted in the Homework & Coursework Questions area, even if it's for personal study and not an actual homework assignment.

    I have moved this thread.
     
  10. Feb 23, 2010 #9
    Cool. Thanks for the info.
     
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