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Exponential growth and decay

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data
    A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 63 cells.
    (a) Find the relative growth rate.

    2. Relevant equations
    Well this has to do with exponential growth and decay. So equations that could apply are

    [tex] \frac{dy}{dt}=ky [/tex]
    where y is some function, k is a constant and dy/dt is a change in that function

    [tex] y(t)=y_0e^k^t[/tex]

    where y of t is a function, y of 0 is an initial value, k is a constant and t is time

    [tex] y^-^1* \frac{dy}{dt}=k[/tex]

    3. The attempt at a solution

    (a) Find the relative growth rate in cells per hour.

    well I thought since I know the change in the change of the number of bacteria per minute and the starting number of bacteria, I thought i could do this-

    [tex] \frac{1}{63}*\frac{2}{20}[/tex]

    [tex] \frac{1}{63}*\frac{1}{10}[/tex]

    [tex] \frac{1}{630} [/tex] cells per minute

    [tex] \frac{60}{630} [/tex]cells per hour

    [tex] .095 [/tex]cells per hour

    that answer is wrong. I don't know how to go about this as you can clearly tell. can someone give me a kick start?

    thanks :)
  2. jcsd
  3. Oct 31, 2009 #2
    Plug in the values to:

    [tex]y(t) = y_0e^{kt}[/tex]

    After 20 min you have twice as many...

    [tex]126 = 63 e^{k(20)}[/tex]

    Now solve for k. Then you will have the equation for how many there are after t min.
  4. Oct 31, 2009 #3


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    Start with:
    Since we have:
    [tex]63=y(0)=y_{0}[/tex], we ave determined ONE of the two constants, y_0.

    The relative growt rate is, indeed, k.
    We know that after t=20 minutes, the population has doubled.
    Thus, we have:
    which means:
    Thus, we get:
    if you want to make the doubling time explicit, t being understood to be measured in minutes.
  5. Oct 31, 2009 #4
    Thanks guys :). I knew I kind of overlooked this question. I should have known better. k=2.07 cells per hour. I got part a right and once I got that, all the other parts of the question (a-e) fell right out. I thank all of you.

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