91Ó°ÊÓ

given,

$\frac{dP}{dt}=kP\left(1−\frac{P}{100}\right)$

$\frac{dP}{dt}=kP\left(1−\frac{P}{L}\right)$

It is given that the population $P\left(t\right)$is growing. This means that the rate of growth of the population is positive. Mathematically, it implies that $\frac{dP}{dt}$is positive or $\frac{dP}{dt}>0$. So, the left-hand side of equation (1) is positive. Now, on the right-hand side $P$is positive, and P is less than $500$, the factor $1-\frac{P}{500}$is positive and less than $1$. That is $0<1−\frac{P}{500}<1$Hence, for the right-hand side to be positive the constant $k$must be positive. Therefore, in the growing population model (1), the constant $k$is positive.

The quality $\frac{P}{500}$will be negatively small. This implies the differential equation can be approximated by

$\begin{array}{r}\frac{dP}{dt}≈kP(1−0)\\ ≈kP\end{array}$

Therefore, for a small value of $P,\begin{array}{r}\frac{dP}{dt}≈kP\\ \end{array}$

In such case

$\begin{array}{r}\frac{dP}{dt}≈kPâ‹\dots 0\\ ≈0\end{array}$

That is the growth rate is zero when the population approaches carrying capacity.