# Logistic Equation Calculus Question

## 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 dont 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 dont really know wat to do next or whether or not I am understanding this question correctly. Any help would be greatly appreciated

Related Calculus and Beyond Homework Help News on Phys.org
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
Staff Emeritus
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?

Last edited:
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