Differential equation population growth problem

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  • #1
paulmdrdo
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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
Chris L T521
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Re: differential equation population growth problem

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
paulmdrdo
89
2
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
paulmdrdo
89
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kindly check my answer thanks!
 
  • #5
MarkFL
Gold Member
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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
Chris L T521
Gold Member
MHB
915
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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?

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)
 

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