Differential equation population growth problem

In summary, a differential equation in the context of population growth is a mathematical equation that relates the rate of change of a quantity to its current value. It is derived by considering the population as a function of time and takes into account factors such as birth and death rates. The carrying capacity in the equation represents the maximum sustainable population size. The equation can be solved using various mathematical techniques and has real-world applications in a variety of fields.
  • #1
paulmdrdo
89
2
A bacterial population B is known to have a rate of growth proportional to B itself. If between noon and 2pm the population triples, at what time no controls being exerted, should B becomes 100 times? what it was at noon?

using this formula $\displaystyle P(t) \;=\;P_oe^{kt}$

please help me get started. thanks!
 
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  • #2
Re: differential equation population growth problem

LATEBLOOMER said:
A bacterial population B is known to have a rate of growth proportional to B itself. If between noon and 2pm the population triples, at what time no controls being exerted, should B becomes 100 times? what it was at noon?

using this formula $\displaystyle P(t) \;=\;P_oe^{kt}$

please help me get started. thanks!

The equation we want to work with is $B(t)=B_0e^{kt}$ (just to remain consistent with the variables used in the original problem statement); I'm sure you know that the starting differential equation is $\dfrac{dB}{dt} = kB$, right?

Anyways, the first bit of information helps you find $k$; if we let $t$ represent the number of hours that have passed since noon (with $t=0$ being noon itself), then we know that at 2pm ($t=2$) that the population has tripled; i.e. $B(2) = 3B_0$, where $B_0$ is the initial population. So what you first want to do is solve $3B_0 = B_0e^{2k}$ for $k$.

Once you have $k$, we can now find the time it takes for the population to be $100B_0$, i.e. you'll need to solve the equation $100B_0 = B_0e^{kt}$ for $t$.

For the last part, are you asking how one would go about finding $B_0$? I don't think that's possible in this case unless more information is provided.

Either way, I hope this is enough to help you make progress with this problem; I hope this all made sense! (Smile)
 
  • #3
Re: differential equation population growth problem

solving for k in $3B_0 = B_0e^{2k}$ I get

$\displaystyle k=\frac{\ln(3)}{2}$

now I'll have

$\displaystyle B(t)=B_0e^{\frac{\ln(3)}{2}t}$

now,

$\displaystyle 100B_0=B_0e^{\frac{\ln(3)}{2}t}$

solving for t i get

$100=3^{\frac{1}{2}t}$

$\ln(100)=\frac{1}{2}t\ln(3)$

$\frac{2\ln(100)}{\ln(3)}=t$

now $t=8.38$hours

so after 8.38 hours the population is 100 times or at 10:22 pm. is it correct?
 
  • #4
kindly check my answer thanks!
 
  • #5
LATEBLOOMER said:
kindly check my answer thanks!

Please don't bump the thread by simply repeating something stated in the previous post. We ask that you be patient and wait for a response.

I get \(\displaystyle t=4\log_3(10)\text{ hr}\approx8.38361309715754\text{ hr}\)

But since this is the number of hours after noon, this would be (to the nearest second):

8:23:01 pm
 
  • #6
Re: differential equation population growth problem

LATEBLOOMER said:
solving for k in $3B_0 = B_0e^{2k}$ I get

$\displaystyle k=\frac{\ln(3)}{2}$

now I'll have

$\displaystyle B(t)=B_0e^{\frac{\ln(3)}{2}t}$

now,

$\displaystyle 100B_0=B_0e^{\frac{\ln(3)}{2}t}$

solving for t i get

$100=3^{\frac{1}{2}t}$

$\ln(100)=\frac{1}{2}t\ln(3)$

$\frac{2\ln(100)}{\ln(3)}=t$

now $t=8.38$hours

so after 8.38 hours the population is 100 times or at 10:22 pm. is it correct?

The math is good, but you want to be careful with your conclusion. The time when the population is 100 times the initial amount is at 8:23pm, not 10:23pm. Other than that, everything else looks great to me. (Smile)
 

1. What is a differential equation in the context of population growth?

A differential equation is a mathematical equation that relates the rate of change of a quantity to its current value. In the context of population growth, it represents the change in population over time based on factors such as birth rate, death rate, and migration.

2. How is the differential equation for population growth derived?

The differential equation for population growth is derived by considering the population as a function of time and taking into account the factors that affect its growth, such as birth and death rates. The equation is typically modeled as a logistic function, which takes into account the carrying capacity of the environment.

3. What is the significance of the carrying capacity in the differential equation for population growth?

The carrying capacity in the differential equation for population growth represents the maximum population size that can be sustained in a given environment. As the population approaches this limit, the growth rate decreases and eventually reaches a steady state.

4. How can the differential equation for population growth be solved?

The differential equation for population growth can be solved using mathematical techniques such as separation of variables, substitution, or using numerical methods. Solutions to the equation can provide insights into the long-term behavior of the population.

5. What are some real-world applications of the differential equation for population growth?

The differential equation for population growth has many real-world applications, including predicting the growth of animal populations, analyzing population dynamics in ecology and epidemiology, and studying the effects of resource availability on human populations. It is also used in economics to model the growth of markets and in engineering to model the spread of diseases.

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