Question

Exercises 1. Verify equation (3) 2. Use the techniques of Section 13.7 and the fact that P(0) = 10 to solve equation (5). 3. The carrying capacity of Atlantic harp seals has been estimated to be C = 10 million seals. Let 1 = 0 correspond to the year 1980 when this seal population was estimated to be about 2 mil- lion. (Data from: Fisheries and Oceans Canada.) (a) Use a logistic growth model = kP(C - P) with k =.007 to find a population function similar to the one given in equation (6) (b) According to this model, what is the seal population in the year 2018? (C) If the model remains accurate, when will the scal population reach 8 million? 4. The logistic growth model can be used in other contexts where growth is bounded. For example, the percentage of women par- ticipating in the Olympic Games was about 20% in 1976 and grew rapidly for a period of time, but the rate of growth slowed as the percentage of women participants drew closer to 50%. (Data from: www.olympic.org.) (a) Let P represent the percentage of women in the Olympic Games in year 1, with 7 = 0 corresponding to the year 1976. Use a logistic growth model = kP(C - P), with k = .065 and a carrying capacity of C = 50, to find a function for P similar to the one given in equation (6). (b) According to this model, what percentage of Olympians were women (rounded to the nearest percent) in the year 2000 (when I = 24)? What percentage were women in 2016? What is the projected percentage of women athletes for the 2020 games? ,dp ri- of em

Answer 1

(1) Varification of equation (3)

Equation (3) is given by:

PC-P) PC-P

RHS = PC-P

            

* CIC - P)

             $= \frac{C - P}{CP(C - P)} + \frac{P}{CP(C - P)}$

             $= \frac{C - P + P}{CP(C - P)}$

             $= \frac{C}{CP(C - P)}$

            

PC-P)

                 Equation (3) varified.

(2) We have equation (5) is:

$\frac{1}{200}\ln(P) - \frac{1}{200}\ln(200 - P) = 0.005t + K, ~~ P(0) = 10.$

Substituting 0 for t and 10 for P we get:

$\frac{1}{200}\ln(10) - \frac{1}{200}\ln(200 - 10) = 0.005(0) + K$

$\therefore \frac{1}{200}(1) - \frac{1}{200}\ln(190) = K$

$K \approx -0.021$ Substituting in equation 5 we get:

$\frac{1}{200}\ln(P) - \frac{1}{200}\ln(200 - P) = 0.005t - 0.021$

$\frac{\ln(P) - \ln(200 - P)}{200} = 0.005t - 0.021$

$\ln(\frac{P}{200 - P}) = 200(0.005t - 0.021)$

$\ln(\frac{P}{200 - P}) = t - 4.2$

$(\frac{P}{200 - P}) = e^{t - 4.2}$

$P = \frac{200e^{t - 4.2}}{1 + e^{t - 4.2}}$

$P(t) = \frac{3e^{t}}{1 + 0.015e^{t}}$