Re: Population growth rate

Regarding the population, what does $\displaystyle \frac{dP}{dt}>0$ imply ? What does $\displaystyle \frac{dP}{dt}<0$ imply ?

Now look at the right-hand side. If $\displaystyle P>k$ is it positive or negative ? If $\displaystyle P<k$ is it positive or negative ?

Just put the two conclusions together.

Re: Population growth rate

You should start (as they suggest) by determining the sign of dP/dt. It's reasonable to assume P is positive, and we are given that r is positive. So it all depends on 1-P/k.

If P_0 is less than k, will P ever be greater than k?

- Hollywood

Re: Population growth rate

Quote:

Originally Posted by

**BobP** Regarding the population, what does $\displaystyle \frac{dP}{dt}>0$ imply ? What does $\displaystyle \frac{dP}{dt}<0$ imply ?

Now look at the right-hand side. If $\displaystyle P>k$ is it positive or negative ? If $\displaystyle P<k$ is it positive or negative ?

Just put the two conclusions together.

dP/dt > 0 implies that the population is increasing, dP/dt < 0 implies that the population is decreasing. If P > k , it is negative, and if P < k, it is positive. Is that so? Then it is safe for me to assume that the P_0(initial population) as P ?

Re: Population growth rate

Quote:

Originally Posted by

**hollywood** You should start (as they suggest) by determining the sign of dP/dt. It's reasonable to assume P is positive, and we are given that r is positive. So it all depends on 1-P/k.

If P_0 is less than k, will P ever be greater than k?

- Hollywood

If P_0 is less than k, then I will get dP/dt as a negative value, but how was I suppose to find P so that I can compare with k?

Re: Population growth rate

P is a function of t, and P_0 is just the value of the function at t=0.

If P is less than k at any time, then dP/dt is positive so P increases as t increases. But since dP/dt would be zero if P were equal to k, P can never be greater than k.

This is the type of thing you do to analyze differential equations qualitatively.