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Application of differentiation

  1. Apr 6, 2005 #1
    Given a function f(t) = (2t-1) / (t^2-t+0.5)
    max supported species = 120, and t is the time.
    if 75 animals are introduced into te refuge the population changes at a rate of f(t)
    What needs to be done to obtain
    a ) Find when the population will be a minimum
    and to find
    b) When the rate of change of the population will be a maximum
     
    Last edited: Apr 7, 2005
  2. jcsd
  3. Apr 6, 2005 #2

    dextercioby

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    Theoretically

    a)f'(x)=0,f''(x)>0
    b)f''(x)=0,f'''(x)<0...

    Daniel.
     
  4. Apr 7, 2005 #3
    i still dont get it
     
  5. Apr 7, 2005 #4
    Are you sure you don't mean [itex]f(t)[/itex]?

    I'll assume that you do. In that case, what conditions are necessary for a differentiable function to have a minimum at a point? Look at your notes if you don't remember.
     
  6. Apr 7, 2005 #5

    dextercioby

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    No need to look at the notes,i wrote them,before he's edited his post & spell out the function...

    Daniel.
     
  7. Apr 7, 2005 #6
    yup is f(t)
     
  8. Apr 7, 2005 #7

    dextercioby

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    Then solve it...Set those derivatives to zero and verify the nature of the critical points.

    Daniel.
     
  9. Apr 7, 2005 #8
    i found the minimum at t= 1/2
    how do i go about part b?
    for part b i got
    f''(t)= 2(2t-1)
    2(2t-1)=120 -> t =30.5
    is that the time of the max population?
     
    Last edited: Apr 7, 2005
  10. Apr 7, 2005 #9

    HallsofIvy

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    Okay we have a function.

    ?? What happened to f? what does f have to do with the population?
    Have you left something out- like f(t) is the population at time t? But I notice that f(0)= -2, not 75 and the maximum of f is 2, not 120. What does f have to do with the population?
     
  11. Apr 7, 2005 #10
    if 75 animals are introduced into te refuge the population changes at a rate of f(t)
     
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