AP Calculus Question

Question

Let P(t) represent the number of grizzly bears in a population at time t years, when t>=0. The population P(t) is increasing at a rate directly proportional to 600-P(t), where the constant of proportionality is k. If P(0)=400, find P(t) in terms of t and k.

Answer 1

Let P(t) be the population in t years.

dP/dt = k(600 - P)

[1 / (600 - P)] dP = kdt

∫ [1 / (600 - P)]dP = ∫kdt

-ln(600 - P) = kt + c

ln(600 - P) = -kt - c

600 - P = e^{-kt - c} = e^-ce^-kt

P(t) = 600 - Ce^-kt (e^-c was relabeled as C)

Since P(0) = 400, we have 400 = 600 - C. So, C = 200.

P(t) = 600 - 200e^-kt

Answer 2

The rate of change of the population is P'(t), so the given information implies that:

P'(t) = k(600 - P(t))

or, equivalently,

(*) P'(t) = 600 k - kP(t)

Solutions to such an equation are of the from

P(t) = a e^-kt + c

for some constants a and c. Since P'(t) = - ake^-kt and 600k - kP(t) = 600k - ake^-kt - kc, we need that c=600 in order for equation (*) to hold. The initial condition P(0)=400 implies that a = -200. So the final answer is:

P(t) = -200 e^-kt + 600