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Bacteria growth rate

  1. Nov 3, 2007 #1
    If all I have given is that
    1. Bacteria grows at a rate proportional to it's size.
    2. It doubles in 2 days.
    3. At 10 days, population is 1000.

    I'm not given the initial bacteria count, I need help setting up the equation.

    I did:
    dy/dt = ky => dy/y = kdt => lny= kt + c => y=e^(kt) + c

    y(10)= 1000 = e^(10k) + c

    But I'm lost how to complete this. Any idea?
     
  2. jcsd
  3. Nov 3, 2007 #2
     
  4. Nov 3, 2007 #3
    so it should be y=ce^(kt)

    y(o) = c e^0 = c
    then y(2) = c e^(2k) = 2 * y (0)
    => 2k = ln(2) => k = ln(2) / 2

    now y(10) = 1000 = c e^(10k)

    => c = 1000 / (e^(5ln2))

    correct?
     
  5. Nov 3, 2007 #4
    Yes! great job, keep up the good work! very good.
     
  6. Nov 4, 2007 #5

    HallsofIvy

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    Of course, it would have been much simpler to argue that, since the population doubles every 2 days, we must have P(t)= C2t/2 where t is in days. Then P(10)= C 25= 1000 so C= 1000/32= 100/16= 10/8= 5/2.

    P(t)= (5/2) 2t/2.

    (But Antineutron is right- great job!)
     
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