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Population growth

  1. Mar 18, 2005 #1
    A bacteria culture starts with 300 bacteria and grows at a rate proportional to its size. After 3 hours, there are 9000 bacteria

    A.) Find an expression for the number of bacteria after t hours. for this part, i got [tex]300*e^{1.133732*t}[/tex]

    B.) Find the number of bacteria after 4 hours. well using the expression from above and subbing in 4, i get 27965.04104

    C.) Find the growth rate after 4 hours. this just means that i need to solve for k right?
    p(t) = 300e^{kt}
    p(3) = 300e^{3k} = 9000
    solved for k and got 1.133% right?

    D.) After how many hours will the population reach 30000
    well setting the equation from part A equal to 30000 and solved for t and got 4.319991


    i know that at least one of these are wrong, but i cant figure out which one. can someone tell me what im doing wrong?
     
  2. jcsd
  3. Mar 18, 2005 #2
    I think Part C is wrong. Isn't the growth rate = dp/dt = kp? (k is constant, by the way, and is equal to 1.133732). So use your answer from part B to get the answer to part C. (Simple as multiplying by k).
     
  4. Mar 18, 2005 #3
    awesome, thanks for the help
     
  5. Mar 18, 2005 #4

    xanthym

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    Science Advisor

    From problem statement:
    {(dB/dt) = k*B} ⇒ B(t) = B0*exp(k*t)
    {B(t=0) = 300} ⇒ B0 = 300
    {B(t=3) = 9000} ⇒ 9000 = 300*exp{k*(3)} ⇒ k = (1/3)*Loge{9000/300} = (1.1337325)

    ITEM #A:
    B(t) = 300*exp{(1.1337325)*t}

    ITEM #B:
    B(4) = 300*exp{(1.1337325)*(4)} = (27,965 bacteria)

    ITEM #C:
    (dB/dt)t=4 = k*B(4) = (1.1337325)*(27965) = (31,705 bacteria/hr)

    ITEM #D:
    (30000) = 300*exp{(1.1337325)*t} ⇒ t = (1.1337325)(-1)*Loge{30000/300}
    t = (4.06195 hr)


    ~~
     
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