Logistic Equation Calculus Question

[saralsaigh]

Homework Statement

Suppose that a population develops according to the logistic equation
dp/dt = 0.03 p - 0.006 p^2
where t is measured in weeks.
-What is the carrying capacity and the value of k?

Homework Equations

dp/dt = kP ( 1 - (p/K)) where K is the carrying capacity

The Attempt at a Solution

Well i thought that in order to solve this i need to get the differenctial expression give in the question to resemble that of the logistic equation so i can get the values. so far my attempts have failed...so i don't know if THAT is wat i am actually supposed to do.
I rearranged the equation and i got:
dp/dt = P^2 ( (o.o3/P) - 0.oo6)
= 0.006 P^2 ( (5/P) - 1)
= -0.006 P^2 ( 1- (5/P))
that is the farthest I went so far i don't really know wat to do next or whether or not I am understanding this question correctly. Any help would be greatly appreciated

 

[gammamcc]

You have the general form, but then you factored out a p^2. Now you are inventing another form...

To paraphrase "Only Euclid has seen beauty bare" ... but not here.

 

[cristo]

So, you want to get your equation in the form $$\frac{dp}{dt}=\frac{rp(K-p)}{K}=rp(1-\frac{p}{K})$$, where here r is the Malthusian parameter (your k) and K is the carrying capacity.

Your mistake was factoring out p², since this gives us a term which looks like 1/p instead of p inside the brackets. You should do this: $$\frac{dp}{dt}=0.006p(5-p)=0.03p(1-\frac{p}{5})$$. Can you solve now?

 

[saralsaigh]

So, you want to get your equation in the form $$\frac{dp}{dt}=\frac{rp(K-p)}{K}=rp(1-\frac{p}{K})$$, where here r is the Malthusian parameter (your k) and K is the carrying capacity.

Your mistake was factoring out p², since this gives us a term which looks like 1/p instead of p inside the brackets. You should do this: $$\frac{dp}{dt}=0.006p(5-p)=0.03p(1-\frac{p}{5})$$. Can you solve now?

thanks for the help i got the right answer