Differential equation for exponential growth

  • Thread starter nrslmz
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  • #1
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Homework Statement


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)




Homework Equations


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



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.
 

Answers and Replies

  • #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.
 
  • #3
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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?
 
  • #4
Defennder
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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?
You can use 50k directly. As Halls said there is no need to do it per minute.

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?
What is n supposed to be here? And what did you work k out to be?
 
  • #5
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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.
 
  • #6
HallsofIvy
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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
No, it would not. You are not integrating the equation correctly.

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?
 

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