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Logistic Growth of Rabbit Population

Created by @nathanedwards
 at November 1st 2023, 12:09:41 am.
AP_Calculus_AB-Questions
#population-growth
#differential-equations
#logistic-growth-model

Question:

The population PP of rabbits on a certain island follows a logistic growth model, given by the differential equation:

dPdt=kP(1−PM)\frac{dP}{dt} = kP\left(1-\frac{P}{M}\right)

where kk is the growth rate constant and MM is the carrying capacity of the island.

  1. Determine the general solution to the differential equation.

  2. Given that the initial population P0=100P_0 = 100 rabbits and the carrying capacity M=500M = 500 rabbits, find the particular solution to the differential equation.

  3. Use the particular solution obtained in part (2) to determine the maximum rate of population growth on the island.

Answer:

  1. To solve the given differential equation dPdt=kP(1−PM)\frac{dP}{dt} = kP\left(1-\frac{P}{M}\right), we can rearrange it as a separable differential equation:
1P(1−PM)dP=kdt\frac{1}{P\left(1-\frac{P}{M}\right)} dP = k dt

Now, we can integrate both sides of the equation:

∫1P(1−PM)dP=∫kdt\int \frac{1}{P\left(1-\frac{P}{M}\right)} dP = \int k dt

Let's solve each integral separately.

Integrating the left side:

∫1P(1−PM)dP=∫(1P+1M−P)dP\int \frac{1}{P\left(1-\frac{P}{M}\right)} dP = \int \left(\frac{1}{P} + \frac{1}{M-P}\right) dP

Using the property of logarithms, we can rewrite the integral as:

∫1P(1−PM)dP=∫(1P−1M−P)dP\int \frac{1}{P\left(1-\frac{P}{M}\right)} dP = \int \left(\frac{1}{P} - \frac{1}{M-P}\right) dP

Integrating each term:

=ln⁡∣P∣−ln⁡∣M−P∣+C1= \ln |P| - \ln |M-P| + C_1

where C1C_1 is the constant of integration.

Integrating the right side:

∫kdt=kt+C2\int k dt = kt + C_2

where C2C_2 is another constant of integration.

Putting it all together, we have the general solution:

ln⁡∣P∣−ln⁡∣M−P∣=kt+C\ln |P| - \ln |M-P| = kt + C

where C=C2−C1C = C_2 - C_1 is the combined constant of integration.

  1. Now, we need to find the particular solution to the given differential equation, given that the initial population P0=100P_0 = 100 rabbits and the carrying capacity M=500M = 500 rabbits.

Using the initial condition P0=100P_0 = 100, we can substitute t=0t = 0 and P=100P = 100 into the general solution:

ln⁡∣100∣−ln⁡∣500−100∣=k(0)+C\ln |100| - \ln |500-100| = k(0) + C
ln⁡∣100∣−ln⁡∣400∣=C\ln |100| - \ln |400| = C
ln⁡100−ln⁡400=C\ln 100 - \ln 400 = C
ln⁡100−ln⁡42=C\ln 100 - \ln 4^2 = C
2ln⁡10−2ln⁡2=C2\ln 10 - 2\ln 2 = C
2(ln⁡10−ln⁡2)=C2(\ln 10 - \ln 2) = C
C=2(ln⁡5)C = 2(\ln 5)

Therefore, the particular solution is:

ln⁡∣P∣−ln⁡∣M−P∣=kt+2(ln⁡5)\ln |P| - \ln |M-P| = kt + 2(\ln 5)
  1. To determine the maximum rate of population growth on the island, we need to find the point where the derivative dPdt\frac{dP}{dt} is equal to zero.

Taking the derivative of the particular solution, we have:

ddt(ln⁡∣P∣−ln⁡∣M−P∣)=ddt(kt+2(ln⁡5))\frac{d}{dt} (\ln |P| - \ln |M-P|) = \frac{d}{dt} (kt + 2(\ln 5))
1P⋅dPdt−1M−P⋅dPdt=k\frac{1}{P} \cdot \frac{dP}{dt} - \frac{1}{M-P} \cdot \frac{dP}{dt} = k

Substituting the logistic growth differential equation dPdt=kP(1−PM)\frac{dP}{dt} = kP\left(1-\frac{P}{M}\right):

1P⋅kP(1−PM)−1M−P⋅kP(1−PM)=k\frac{1}{P} \cdot kP\left(1-\frac{P}{M}\right) - \frac{1}{M-P} \cdot kP\left(1-\frac{P}{M}\right) = k

Simplifying:

k(1P−1M−P)(1−PM)=kk\left(\frac{1}{P} - \frac{1}{M-P}\right)\left(1-\frac{P}{M}\right) = k

Since the last factor, (1−PM)\left(1-\frac{P}{M}\right), is always positive, we can divide both sides of the equation by it:

1P−1M−P=1\frac{1}{P} - \frac{1}{M-P} = 1

Multiplying both sides of the equation by P(M−P)P(M-P):

P(M−P)−P=(M−P)P(M-P) - P = (M-P)

Expanding and simplifying:

PM−P2−P=M−PPM - P^2 - P = M - P
P2−PM+P=0P^2 - PM + P = 0

Factoring out a PP:

P(P−M+1)=0P(P-M+1) = 0

Since population (PP) cannot be negative, the only solution that satisfies the condition is P=M−1P = M-1. Therefore, the maximum rate of population growth happens when P=M−1=500−1=499P = M-1 = 500-1 = 499 rabbits.

Conclusion:

  1. The general solution to the given logistic growth model is ln⁡∣P∣−ln⁡∣M−P∣=kt+C\ln |P| - \ln |M-P| = kt + C, where CC is the constant of integration.
  2. The particular solution to the given logistic growth model, with an initial population P0=100P_0 = 100 and carrying capacity M=500M = 500, is ln⁡∣P∣−ln⁡∣M−P∣=kt+2(ln⁡5)\ln |P| - \ln |M-P| = kt + 2(\ln 5).
  3. The maximum rate of population growth occurs when the population PP is equal to the carrying capacity M−1M-1, which in this case is 499 rabbits.

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