Differential equation for exponential growth

In summary, the exponential and differential equation would be:P(t) = 10^6 * (ln(1,6) * t / 240)The Attempt at a SolutiondP/dt = P*k -833,33And I should integrate this to find the specific moment when the population reaches 10^12? I am somehow not feeling good about this solution.
  • #1
nrslmz
15
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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.
 
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  • #2
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
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
nrslmz said:
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
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
nrslmz said:
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?
 

FAQ: Differential equation for exponential growth

1. What is a differential equation for exponential growth?

A differential equation for exponential growth is a mathematical equation that describes the rate of change of a quantity over time in terms of its current value. It is commonly used to model natural phenomena such as population growth, bacterial growth, and compound interest.

2. How is the differential equation for exponential growth expressed?

The differential equation for exponential growth is typically expressed as dP/dt = kP, where P represents the quantity being modeled, t represents time, and k is a constant that determines the rate of change. This equation is also known as the "growth equation".

3. What does the constant "k" represent in the differential equation for exponential growth?

The constant "k" in the differential equation for exponential growth is known as the growth rate or the relative growth rate. It determines the rate at which the quantity being modeled changes over time. A larger value of k indicates a faster growth rate, while a smaller value of k indicates a slower growth rate.

4. How is the differential equation for exponential growth used in real-life applications?

The differential equation for exponential growth is used in various fields such as biology, economics, and finance to model the growth of populations, economies, and investments. It can also be used to predict future values of a quantity based on its current growth rate.

5. What are some limitations of the differential equation for exponential growth?

One limitation of the differential equation for exponential growth is that it assumes a constant growth rate, which may not always be the case in real-life situations. It also does not take into account external factors that may affect the growth of a quantity. Additionally, the model may break down when the quantity being modeled reaches its carrying capacity or limit.

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