<p>I was totally stumbled on this concept.</p>
<p>dP/dt = (P/5)(1-P/12)</p>
<p>If P(0) = 3, what's P(t) when t approaches infinity?
what value of P is the population growing the fastest?</p>
<p>the form we learned for logistics dy/dt=ky[A-y] then y=A/(1+ce^-kt). In this case, k=1/60 and A=12. y=12/(1+ce^-(1/60)t)</p>
<p>when t=0, y=3.</p>
<p>12/(1+ce^0)=3</p>
<p>12/(1+c)=3</p>
<p>12=3+3c</p>
<p>c=3</p>
<p>you get y=12/(1+3e^-(1/60)t)</p>
<p>as t approaches infinity, 3e^-(1/60)t approaches 0</p>
<p>limit as t approaches infinity = 12/1 = 12</p>
<p>P(t) is 12</p>
<p>For growing the fastest, find the absolute max of dP/dt. Rel. max (which is the ONLY relative max): at t=6. Plug into logistics equation and you get P=3.231</p>
<p>The trick is to know that as t approaches infinity, you’ll will always get the carrying capacity, P (or in the above equation, A).</p>
<p>Yea, all you have to do is make the equation become something like P(A-P), basically, make that second P into a whole number by factoring out whatever you need to. You don’t really know how to derive how to do it (like the guy above me just showed), because all the free response just asked you for the carrying capacity and the only one I’ve seen speficially said “no work is needed” lol.</p>