Population problem

L

legacyofpiracy

Junior Member
Joined
Oct 20, 2005
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I'm having a bit of trouble with this population problem:

Code: 
Let P(T) represent a population of bunnies at time t years. The population P(t) is increasing at a rate directly proportional to 800-P(t), where the constant of proportionality is k

if P(0)=500, find P(t) in terms of t and k

find the limit of P(t) as t approaches infinity

I'm having trouble getting started in general without a clear equation to follow
 
"The population P(t) is increasing at a rate directly proportional to 800-P(t)" means we have

\(\displaystyle \mbox{ \frac{dP}{dt} = k\left(800 - P\right)}\)


Oddly enough Joey had a similar question about wolves earlier today.
 
Oh wow that strange, the problem is almost identical...must be the same book. Thanks I'll just look at that thread
 
There should be a negative in front of the k.

Separate variables:

\(\displaystyle \mbox{ \frac{1}{800 - P} \frac{dP}{dt} = k}\)

and integrate both sides wrt t.
 
stapel said:
Please review the discussion regarding this exercise posted in another thread today.

Thank you.

Eliz.
Click to expand...
Don't flood the thread. Legacy just asked a question about it two posts earlier.

Take up any issues you have via PM.

"Thank you."
 
\(\displaystyle \mbox{ \int \frac{1}{800 - P} dP = \int k dt}\)

\(\displaystyle \mbox{ -\ln{|800 - P|} = kt + C}\)

Now continue.
 
Have you never seen this before?

\(\displaystyle \mbox{ \ln{|800 - P|} = -kt - C}\)

\(\displaystyle \mbox{ |800 - P| = e^{\left(-kt - C\right)} = e^{\left(-kt\right)}\cdot e^{\left(-C\right)}}\)

\(\displaystyle \mbox{ 800 - P = Ae^{-kt} }\) where \(\displaystyle \mbox{A = \pm e^{\left(-C\right)}}\)

Then solve for P, substitute initial conditions, etc.
 
i swear i have...its all coming back now. I understand it perfectly when it is shown to me..im trying to work on realizing it myself
 
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