Answer
(a) $\frac{dP}{dt} = 0~~$ corresponds to a stable population.
(b) The stable population level is 2000 fish.
(c) The population would decrease to 0.
Work Step by Step
(a) A stable population means the size of the population does not change.
$\frac{dP}{dt} = 0~~$ corresponds to a stable population.
(b) $\frac{dP}{dt}= r(1 - \frac{P(t)}{P_c})P(t)- \beta~P(t) = 0$
$rP(t)- \frac{r[P(t)]^2}{Pc}- \beta~P(t) = 0$
$\frac{r[P(t)]^2}{Pc}= rP(t) - \beta~P(t)$
$[P(t)]^2= \frac{P_c~[rP(t) - \beta~P(t)]}{r}$
$P(t)= \frac{P_c~(r - \beta)}{r}$
$P(t)= \frac{(10,000)~(0.05 - 0.04)}{0.05}$
$P(t) = 2000$
The stable population level is 2000 fish.
(c) Suppose that $\beta$ is raised to 5%
Then:
$P(t)= \frac{P_c~(r - \beta)}{r}$
$P(t)= \frac{(10,000)~(0.05 - 0.05)}{0.05}$
$P(t) = 0$
The population would decrease to 0.
