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Math Help - can anyone plz help me out with this problem?? maybe this is a piece of cake 4 u

  1. #1
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    can anyone plz help me out with this problem?? maybe this is a piece of cake 4 u

    A culture of bacteria is growing at a rate proportional to the number,
    N, of bacteria present, so that
    dN/dt= kN
    where t is the time elapsed in hours and k is a constant. Initially, the culture contains 3000 bacteria, and after three hours of uninterrupted growth, it contains 9000 bacteria.
    (i) Find k.

    Thereafter (that is, for t > 3) the culture is carefully washed with a solution
    which is harmful to the bacteria, and the bacteria are killed o by the solution
    at a constant rate of 75 each minute. [Thus, the rate of change in the number
    of bacteria is as described above, but diminished by 4500 per hour.]
    (ii) Find the number of bacteria in the culture for any time t > 3.
    (iii) After how much washing time (in hours and minutes) do the bacteria die out?
    --------------------------------------------------------------

    For (i)

    dN/dt=kN

    1/N*dN/dt=k

    int(1/N)*(dN/dt)*dt=int k dt
    int(1/N)*dN=kt+C

    ln ㅣNㅣ=kt+C
    ㅣNㅣ=e^(kt+c)
    ㅣNㅣ=e^c*e^kt

    and N=ħe^c*e^kt
    =Ae^kt, where A=ħe^c

    ∴N=Ae^kt

    when N(0)=3000,

    3000=A

    so, N(t)=3000e^kt,

    when N(3)=9000

    9000=3000e^kt
    3=e^3k
    3k ln e= ln 3
    3k= ln 3

    k=(1/3) ln (3), which is 0.3662 in 4 decimal places....
    --------------------------------------------------------
    First, can u plz check whether this answer is right or wrong??

    and..i have no idea how to approach to question (ii) and (iii)

    Thank you for your kindness
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  2. #2
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    your value of k is correct ...

    k = \frac{\ln{3}}{3}


    see new post below ...
    Last edited by skeeter; May 24th 2009 at 10:12 AM. Reason: error
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  3. #3
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    let me try this again ... the washing will affect the rate of growth, so a new DE is required.

    \frac{dN}{dt} = kN - 4500

    \frac{dN}{kN-4500} = dt

    \frac{k}{kN-4500} \, dN = k \, dt

    \ln|kN-4500| = kt + C

    kN - 4500 = Ae^{kt}

    at t = 3 , N = 9000 , k = \frac{\ln{3}}{3}

    3000\ln(3) - 4500 = 3A

    A = 1000\ln(3) - 1500

    kN = 4500 - (1500 - 1000\ln{3})e^{kt}

    N = \frac{3}{\ln{3}}[4500 - (1500 - 1000\ln{3})e^{kt}]
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