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The population P(t) of a species satisfies the logistic differential equation dP/dt= P(2(P/5000)) where the initial population 3,000 and t is the time in years. What is the lim t>infinity P(t)?
 one year ago
 one year ago
The population P(t) of a species satisfies the logistic differential equation dP/dt= P(2(P/5000)) where the initial population 3,000 and t is the time in years. What is the lim t>infinity P(t)?
 one year ago
 one year ago

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zepdrixBest ResponseYou've already chosen the best response.1
What part are you stuck on Ms Jocelyn? :) Were you able to find P(t)?
 one year ago

jocelynevsqBest ResponseYou've already chosen the best response.0
i don't know how to start it
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
There is some function P(t) which satisfies the given differential equation. So we'll start with our differential equation, move some stuff around, and integrate to solve for P(t).
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large \frac{dP}{dt}=P\left(2\frac{P}{5000}\right)\]This is a separable differential equation, so we'll move all of the P's to one side, and move the dt to the other. Hmm it's going to take some work though D:
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
We'll start by getting a common denominator in the brackets, giving us this, \[\large \frac{dP}{dt}=P\left(\frac{10000P}{5000}\right)\]Let's go ahead and move that extra P on top as well.\[\large \frac{dP}{dt}=\frac{P(10000P)}{5000}\]We'll multiply both sides by the `reciprocal` oh the giant term on the right.\[\large \frac{5000}{P(10000P)}\frac{dP}{dt}=1\]Then we'll "multiply" the dt to the other side,\[\large \frac{5000}{P(10000P)}dP=dt\]We've successfully separated the P's and t's! :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
We'll have to apply `partial fraction decomposition` to the fraction so we can integrate it.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large \frac{5000}{P(10000P)} \quad = \quad \frac{A}{P}+\frac{B}{10000P}\]Multiplying through by the denominator gives us,\[\large 5000=A(10000P)+BP\] Hopefully I did my math correctly, I came up with \(A=1/2 \quad B=1/2\). So we can write the left side like this,\[\large \frac{\frac{1}{2}}{P}+\frac{\frac{1}{2}}{10000P}dP=dt\]We'll factor out the 1/2, and take the integral.\[\large \frac{1}{2}\int\limits \frac{1}{P}+\frac{1}{10000P}dP=dt\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Integrating will give you a couple of natural logs,\[\large \frac{1}{2}\left[\ln P\ln\left(10000P\right)\right]=t+c\]Multiplying both sides by 2 gives us,\[\large \left[\ln P\ln\left(10000P\right)\right]=2t+c\](The 2 was distributed but the other one got absorbed into the c). We'll combine our logs using a rule of logs,\[\large \ln\left(\frac{P}{10000P}\right)=2t+c\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
We'll exponentiate both sides, rewrite them with a base of e,\[\huge e^{\ln\left(\frac{P}{10000P}\right)}=e^{2t+c}\] On the left, the exponential e, and the natural log are inverse operations of one another, so they're "cancel out".\[\large \frac{P}{10000P}=e^{2t+c}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Recalling a rule of exponents allows us to write the right side like this,\[\large \frac{P}{10000P}=e^{2t}\cdot e^c\]e^c is just another constant, let's call it such,\[\large \frac{P}{10000P}=Ce^{2t}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Following any of this? :) Or this is way too confusing?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
It's a ton of steps to get through this one... very easy to get lost :(
 one year ago

jocelynevsqBest ResponseYou've already chosen the best response.0
i get it..it's really helpful!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
We'll multiply both sides by the denominator on the left.\[\large P=(10000P)Ce^{2t}\]Multiply out the right side,\[\large P=10000Ce^{2t}PCe^{2t}\]Move the clunky P term to the left side by adding it,\[\large P+PCe^{2t}=10000Ce^{2t}\]Factor out a P on the left,\[\large P\left(1+Ce^{2t}\right)=10000Ce^{2t}\]Then do some division to solve for P, \[\large P=\frac{10000Ce^{2t}}{1+Ce^{2t}}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
I checked my work a few minutes ago on Wolfram, and they came up with something `very slightly` different. They don't have the constant C attached to the 10,000 on the top. So I hope I didn't make a silly mistake somewhere in there.. hmm
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
They told us also that the `Initial population` is 3000. We can write it like this,\[P(0)=3000 \qquad \rightarrow \qquad \large 3000=\frac{10000Ce^{0}}{1+Ce^{0}}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
And ummm hopefully that will get us the C value...
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
I came up with C=3/7. I dunno this problem is sooooo long, I have a feeling I may have made an error somewhere along the way lol. I hope that at least gives you an idea of how to progress through this of problem D:
 one year ago

jocelynevsqBest ResponseYou've already chosen the best response.0
i got 3/7 for C also
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
So if we take the limit, \[\huge \lim_{t \rightarrow \infty} \frac{10000\cdot\frac{3}{7}e^{2t}}{1+\frac{3}{7}e^{2t}}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Ok so the exponentials are increasing at the same rate. The +1 on the bottom will become insignificant as t>infty. So this is approaching a value of \(1\) times the coefficients on the exponentials.\[\large \frac{10000\cdot\frac{3}{7}}{\frac{3}{7}}\] There are some little algebra tricks you can do to convince yourself of this, but I didn't feel like being that thorough XD heh.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
As times progresses on and on and on, the population of the given species will approach but not exceed 10,000. I would word it something like that. And again, I would strongly suggest you go through this a bit more and make sure I didn't make any mistakes :D
 one year ago

jocelynevsqBest ResponseYou've already chosen the best response.0
10,000 is the correct answer :D thank you so much this was very helpful!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
err maybe the question was just looking for a number, so you don't have to say it all fancy like that :) lol
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
It was correct? yay team \c:/
 one year ago

jocelynevsqBest ResponseYou've already chosen the best response.0
yay for you actually lol xD
 one year ago
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