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Differential equation population growth problem

LATEBLOOMER

New member
Aug 7, 2013
21
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!
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
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)
 

LATEBLOOMER

New member
Aug 7, 2013
21
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?
 

LATEBLOOMER

New member
Aug 7, 2013
21
kindly check my answer thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
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)