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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

    HallsofIvy

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    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

    Defennder

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    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

    HallsofIvy

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

     
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