# Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises - Page 238: 40

(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.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.