1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook