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: Differential equation for exponential growth

  1. Dec 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Viruses are reproducing exponentially, while the body eliminates the viruses.
    The elimination rate is constant, 50000 per hour. I decided to take down on the minute level, so it would be 50000/60.
    Pinitial is 10^6
    k is ln(1,6)/240, since the growth rate is 160% in 4 hours.
    Then the exponential and differential equation would be:
    P(t) = 10^6 * (ln(1,6) * t / 240)

    2. Relevant equations
    Then the exponential and differential equation would be:
    P(t) = 10^6 * (ln(1,6) * t / 240)

    3. The attempt at a solution
    dP/dt = P*k ...., right? How can I add up the elimination rate to the differential equation.
    And I should integrate this to find the specific moment when the population reaches 10^12? I am somehow not feeling good about this solution. I would appreciate if you help.
  2. jcsd
  3. Dec 17, 2008 #2


    User Avatar
    Science Advisor

    Look at the units of dP/dt. P is number of virus, t is time, in seconds. So dP/dt has units of "viruses per second". The right hand side of the equation dP/dt= ... must also have units of "viruses per second" and tells how the number of viruses changes. There are two reasons why that would change:
    1) The viruses reproduce at a rate proportional to their number.

    2) Some viruses are killed- "50000 per hour" so -5000 viruses per hour has precisely the correct units. Just subtract 5000 from the right side.
  4. Dec 17, 2008 #3
    Sorry, I posted wrong again. The function is P(t) = 10^6 * e^(ln(1,6) * t / 240)
    You mean 50000/60 = 833,33? and
    Should I write:
    dP/dt = P*k -833,33 and than integrate? Bu that would be the same as
    P(t) = 10^6 * e^(ln(1,6) * t / 240) - 833,33t
    I don't think that would work. Or should I take the limit of
    P(t) = 10^6 * e^(ln(1,6) * (t/n) / 240) - 833,33(t/n) when n goes to infinity?
  5. Dec 17, 2008 #4


    User Avatar
    Homework Helper

    You can use 50k directly. As Halls said there is no need to do it per minute.

    What is n supposed to be here? And what did you work k out to be?
  6. Dec 18, 2008 #5
    Wouldn't the operation be
    ((( * e^(kt))-50000)* e^(kt))-50000)-50000.....

    and e^(kt) must be 1.6 since it grows into %160 of its initial population in 4 hours. Therefore t is 4 or 240 depending on the time interval.

    I solved e^(k * 240) = 1,6 to find k.
  7. Dec 18, 2008 #6


    User Avatar
    Science Advisor

    No, it would not. You are not integrating the equation correctly.

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