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!

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

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Start with:
    [tex]y(t)=y_{0}e^{kt}[/tex]
    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:
    [tex]2y_{0}=y(20)=y_{0}e^{20k}[/tex],
    which means:
    [tex]e^{20k}=2\to{k}=\frac{\ln(2)}{20}[/tex]
    Thus, we get:
    [tex]y(t)=63e^{\frac{t\ln(2)}{20}}=63*2^{\frac{t}{20}}[/tex]
    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.

    -Hover
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook