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: First Order ODE Growth and Decay Modelling

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A culture of bacteria have a growth rate (as a percent) given by [itex]kb[/itex] per year, constant [itex]k>0[/itex] and [itex]b[/itex] is the number of bacteria. A virus removes bacteria at a rate of [itex]m[/itex] bacteria per year. I am trying to model this information using an ODE, but might be making a mistake.

    2. Relevant equations

    3. The attempt at a solution
    If the percent growth rate is kb, [itex](dy/y)/dt = kb[/itex] so [itex]dy/dt = kb^2[/itex].
    Add this to the removal rate, and [itex]dy/dt = kb^2 - m[/itex].
    I am not sure the model is as simple as this, it would mean the units of k are [itex]1/(bacteria*year)[/itex], which sounds odd.
    Last edited: Sep 9, 2012
  2. jcsd
  3. Sep 9, 2012 #2
    I think that the real confusion comes about when we try to determine what the question means by a "growth rate (as a percent)". To be honest with you, I have no idea what that means. The rate (in percent, per year) is kb, which depends on the number of bacteria. This suggests that the "actual" rate is [itex]kb^2[/itex], as you inferred. However, bacteria don't reproduce proportional to the square of their population; they reproduce by binary fission. The rate of reproduction should be proportional to [itex]b[/itex], not [itex]b^2[/itex].

    If you're concerned, ask your professor about this problem, and how you should interpret "as a percent".

    EDIT: Welcome to the forum, by the way.
  4. Sep 9, 2012 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think that growth rate as a percent refers to fractional growth rate i.e. growth rate as a fraction of the existing population. So if the fractional growth rate is 5%, it means that if the population starts out with 'b' individuals at the beginning of the year, then by the end of a year, the number of additional individuals will be 0.05b. So k=0.05 in this example. So if dy/dt is the fractional growth rate, then I think that what the problem means is that dy/dt = k. But the fractional growth rate is the ratio of the growth rate to the population size. Hence $$\frac{dy}{dt} = \frac{1}{b}\frac{db}{dt} = k$$ OR $$\frac{db}{dt} = kb$$ which gives you the standard exponential growth for populations of this type. I think that this is what the problem is saying, but I could be wrong.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook