Determine the maximum value of dP/dt and interpret the meaning of this maximum value.

Question

The population of Aedes mosquitoes which carry the Dengue virus can be modeled by a differential equation which describes the rate of growth of the population. The population growth rate dP/dt is given by dP/dt = rP(1-P/k) , where r is a positive constant and k is the carrying capacity.

Answer 1

The maximum value of dP/dt is found by differentiating the differential with respect to t, but to do this we need to find dP/dt in terms of t, which requires integrating the dP/dt.

integral(dP/(rP(1-P/k)))=t+c, where c is a constant. Let A/rP+B/(1-P/k)=1/(rP(1-P/k)), where A and B are constants. Then, A(1-P/k)+BrP=1; A-AP/k+BrP=1. Therefore, A=1 and -A/k+Br=0, from which B=1/rk. So integral becomes:

integral(dP/rP)+integral(dP/(r(k-P)))=t+c; (1/r)ln(P)-(1/r)ln|k-P|=(1/r)ln|P/(k-P)|=t+c.

ln|P/(k-P)|=rt+rc; P/(k-P)=e^r(t+c). At t=0 population is P0, some initial value, so P0/(k-P0)=e^rc, c=(1/r)ln|P0/(k-P0)|.

P/(k-P)=e^rt*e^ln|P0/(k-P0)|=(P0/(k-P0))e^rt.

dP/dt=rP(1-P/k)=rkP/(k-P)=rk(P0/(k-P0))e^rt;

P"=kr^2(P0/(k-P0))e^rt. This expression can never be zero because e^rt cannot be zero, so dP/dt has no maximum value. However, when dP/dt=0 this is a turning point for P, and dP/dt=0 when P=k, the carrying capacity. The population becomes static at this point. If the initial population P0 is less than the carrying capacity the population increases, and if it's greater than the carrying capacity the population decreases. When it has decreased sufficiently to reach the carrying capacity, it remains static. When it has increased sufficiently to reach the carrying capacity it becomes static.