## Calculus: Early Transcendentals 8th Edition

(a) The carrying capacity is $50$ $k = 0.02$ (b) $P(t) = \frac{50}{1+0.25e^{-0.02t}}$ (c) 41.5
(a) A logistic equation has this form: $\frac{dP}{dt} = kP(1-\frac{P}{M})$ We can consider the given logistic equation: $\frac{dP}{dt} = 0.02P-0.0004P^2$ $\frac{dP}{dt} = 0.02P(1-\frac{P}{50})$ The carrying capacity is $50$ $k = 0.02$ (b) The solution has this form: $P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$ We can find $A$: $A = \frac{M-P_0}{P_0} = \frac{50-40}{40} = 0.25$ We can write the solution: $P(t) = \frac{50}{1+0.25e^{-0.02t}}$ (c) We can find the population after 10 weeks: $P(t) = \frac{50}{1+0.25e^{(-0.02)(10)}} = 41.5$