Answer
(a) The carrying capacity is $400$
(b) $P'(0) = 17.5$
(c) It takes 4.9 years for the population to reach 50% of the carrying capacity.
Work Step by Step
(a) A logistic equation has this form:
$\frac{dP}{dt} = kP(1-\frac{P}{M})$
We can consider the given logistic equation:
$\frac{dP}{dt} = 0.4P-0.001P^2$
$\frac{dP}{dt} = 0.4P(1-\frac{P}{400})$
The carrying capacity is $400$
(b) We can find $P'(0)$:
$P' = \frac{dP}{dt} = 0.4P-0.001P^2$
$P'(0) = 0.4P(0)-0.001~[P(0)]^2$
$P'(0) = 0.4(50)-0.001(50)^2$
$P'(0) = 17.5$
(c) The solution has this form:
$P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$
We can find $A$:
$A = \frac{M-P_0}{P_0} = \frac{400-50}{50} = 7$
We can write the solution:
$P(t) = \frac{400}{1+7e^{-0.4t}}$
We can find the time $t$ it takes the population to reach 50% of the carrying capacity:
$P(t) = \frac{400}{1+7e^{-0.4t}} = 200$
$1+7e^{-0.4t} = \frac{400}{200}$
$1+7e^{-0.4t} = 2$
$7e^{-0.4t} = 1$
$e^{-0.4t} = \frac{1}{7}$
$-0.4t = ln(\frac{1}{7})$
$t = (\frac{1}{-0.4})~ln(\frac{1}{7})$
$t = 4.9$
It takes 4.9 years for the population to reach 50% of the carrying capacity.
