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Population Growth with Logistic Model

Created by @nathanedwards
 at November 1st 2023, 7:39:24 am.
AP_Calculus_AB-Questions
#differential-equations
#logistic-growth-model
#initial-conditions

AP Calculus AB Exam Question

Consider a population of bacteria that grows according to logistic growth model. Let P(t)P(t) represent the population of bacteria at time tt, where tt is measured in days. The differential equation that models the growth of population is given by:

dPdt=k⋅P(t)⋅(1−P(t)M)\frac{dP}{dt} = k \cdot P(t) \cdot (1 - \frac{P(t)}{M})

where kk is the growth rate constant, and MM is the maximum population size that the environment can support.

Suppose the initial population is P(0)=100P(0) = 100 bacteria and the growth rate constant is k=0.05k = 0.05. The maximum population size MM is unknown.

(a) Find the particular solution P(t)P(t) that satisfies the initial condition.

(b) Find the value of MM if the population doubles in size after 5 days.

Solution

(a) To find the particular solution P(t)P(t), we need to solve the differential equation with the initial condition P(0)=100P(0) = 100.

The differential equation is:

dPdt=k⋅P(t)⋅(1−P(t)M)\frac{dP}{dt} = k \cdot P(t) \cdot (1 - \frac{P(t)}{M})

Substituting the given values k=0.05k = 0.05 and P(0)=100P(0) = 100, we have:

dPdt=0.05⋅P(t)⋅(1−P(t)M)\frac{dP}{dt} = 0.05 \cdot P(t) \cdot (1 - \frac{P(t)}{M})

To solve this separable differential equation, we can rewrite it as:

dPP(t)⋅(1−P(t)M)=0.05⋅dt\frac{dP}{P(t) \cdot (1 - \frac{P(t)}{M})} = 0.05 \cdot dt

We can use partial fraction decomposition to simplify the left-hand side of the equation:

dPP(t)⋅(1−P(t)M)=AP(t)+B1−P(t)M\frac{dP}{P(t) \cdot (1 - \frac{P(t)}{M})} = \frac{A}{P(t)} + \frac{B}{1 - \frac{P(t)}{M}}

where AA and BB are constants.

Multiplying both sides of the equation by P(t)⋅(1−P(t)M)P(t) \cdot (1 - \frac{P(t)}{M}), we get:

1=A⋅(1−P(t)M)+B⋅P(t)1 = A \cdot (1 - \frac{P(t)}{M}) + B \cdot P(t)

To solve for AA and BB, we can substitute two convenient values of P(t)P(t).

Let's choose P(t)=0P(t) = 0 and P(t)=MP(t) = M.

Substituting P(t)=0P(t) = 0, we have:

1=A⋅(1−0M)+B⋅01 = A \cdot (1 - \frac{0}{M}) + B \cdot 0
1=A1 = A

Substituting P(t)=MP(t) = M, we have:

1=A⋅(1−MM)+B⋅M1 = A \cdot (1 - \frac{M}{M}) + B \cdot M
1=B⋅M1 = B \cdot M

Since we have A=1A = 1, we can rewrite the equation as:

1=1⋅(1−P(t)M)+B⋅P(t)1 = 1 \cdot (1 - \frac{P(t)}{M}) + B \cdot P(t)

Simplifying further, we get:

1=1−P(t)M+B⋅P(t)1 = 1 - \frac{P(t)}{M} + B \cdot P(t)

Now, substituting P(t)=0P(t) = 0, we can solve for BB:

1=1−0M+B⋅01 = 1 - \frac{0}{M} + B \cdot 0
1=11 = 1

Since 1=11 = 1, the equation is satisfied, and we can conclude that BB can be any value.

Returning to the original differential equation and using the partial fraction decomposition, we have:

∫dPP(t)⋅(1−P(t)M)=∫0.05⋅dt\int\frac{dP}{P(t) \cdot (1 - \frac{P(t)}{M})} = \int 0.05 \cdot dt
∫dPP(t)+∫B⋅dP(1−P(t)M)=∫0.05⋅dt\int\frac{dP}{P(t)} + \int\frac{B \cdot dP}{(1 - \frac{P(t)}{M})} = \int 0.05 \cdot dt
ln⁡(∣P(t)∣)−ln⁡(∣1−P(t)M∣)=0.05⋅t+C\ln(|P(t)|) - \ln(|1 - \frac{P(t)}{M}|) = 0.05 \cdot t + C
ln⁡(∣P(t)1−P(t)M∣)=0.05⋅t+C\ln(|\frac{P(t)}{1 - \frac{P(t)}{M}}|) = 0.05 \cdot t + C

We can simplify the expression inside the logarithm using the property of logarithms:

ln⁡(∣P(t)⋅MM−P(t)∣)=0.05⋅t+C\ln(|\frac{P(t) \cdot M}{M - P(t)}|) = 0.05 \cdot t + C

Exponentiating both sides of the equation, we get:

∣P(t)⋅MM−P(t)∣=e0.05⋅t+C|\frac{P(t) \cdot M}{M - P(t)}| = e^{0.05 \cdot t + C}

Since eCe^C is just another constant, let's rewrite the equation as:

P(t)⋅MM−P(t)=±⋅e0.05⋅t\frac{P(t) \cdot M}{M - P(t)} = \pm \cdot e^{0.05 \cdot t}

where ±\pm represents the positive or negative sign. Solving this equation for P(t)P(t), we get:

P(t)⋅M=±⋅e0.05⋅t⋅(M−P(t))P(t) \cdot M = \pm \cdot e^{0.05 \cdot t} \cdot (M - P(t))
P(t)⋅M=±⋅e0.05⋅t⋅M∓⋅P(t)⋅e0.05⋅tP(t) \cdot M = \pm \cdot e^{0.05 \cdot t} \cdot M \mp \cdot P(t) \cdot e^{0.05 \cdot t}
P(t)⋅M±⋅P(t)⋅e0.05⋅t=±⋅e0.05⋅t⋅MP(t) \cdot M \pm \cdot P(t) \cdot e^{0.05 \cdot t} = \pm \cdot e^{0.05 \cdot t} \cdot M
P(t)⋅(M±⋅e0.05⋅t)=±⋅e0.05⋅t⋅MP(t) \cdot (M \pm \cdot e^{0.05 \cdot t}) = \pm \cdot e^{0.05 \cdot t} \cdot M
P(t)=±⋅e0.05⋅t⋅MM±⋅e0.05⋅tP(t) = \frac{\pm \cdot e^{0.05 \cdot t} \cdot M}{M \pm \cdot e^{0.05 \cdot t}}

Now, we can substitute the given initial condition P(0)=100P(0) = 100 to find the particular solution. Substituting t=0t = 0, we have:

P(0)=±⋅e0.05⋅0⋅MM±⋅e0.05⋅0P(0) = \frac{\pm \cdot e^{0.05 \cdot 0} \cdot M}{M \pm \cdot e^{0.05 \cdot 0}}
100=±⋅MM±100 = \frac{\pm \cdot M}{M \pm}

To satisfy the initial condition, we can choose the positive sign:

100=MM+1100 = \frac{M}{M + 1}
100⋅(M+1)=M100 \cdot (M + 1) = M
100⋅M+100=M100 \cdot M + 100 = M
99⋅M+100=099 \cdot M + 100 = 0

Since this equation has no real solutions, we can conclude that there is no particular solution that satisfies the initial condition.

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