Problem 34 - Calculus: Graphical, Numerical, Algebraic

Question

Gorilla Population A certain wild animal preserve can support no more than 250 lowland gorillas. Twenty-eight gorillas were known to be in the preserve in \(1970 .\) Assume that the rate of growth of the population is \(\frac{d P}{d t}=0.0004 P(250-P)\) where time \(t\) is in years. (a) Find a formula for the gorilla population in terms of \(t\) . (b) How long will it take for the gorilla population to reach the carrying capacity of the preserve?

Step-by-Step Solution

Answer

The exact functions will depend on the outcome of the integrations and the constant of integration, but the end result should yield the gorilla population as a function of time, \(P(t)\), and the time \(t\) for the population to reach the preserve's carrying capacity.

Step 1: Separation of Variables

To solve the differential equation, we need to separate variables. Firstly, rewrite the differential equation into the form where all terms of \(P\) are on one side and those with \(t\) on the other side: \(\frac{dP}{P(250-P)} = 0.0004 dt\)

Step 2: Integrate both sides

Next, integrate both sides of the equation to solve for \(P\):\(\int{\frac{1}{P(250-P)}} dP = \int{0.0004} dt\)

Step 3: Use Partial Fraction Decomposition

The left hand side requires partial fraction decomposition. We rewrite \(\frac{1}{P(250-P)}\) as \(\frac{A}{P} + \frac{B}{250-P}\). Solve for A and B to get \(\frac{1}{P(250-P)} = \frac{-1}{P} + \frac{1}{250-P}\)

Step 4: Integrate using the partial fractions

Integrate both sides using these new fractions:\(-\int{\frac{1}{P}} dP + \int{\frac{1}{250-P}} dP = 0.0004\int{dt}\)

Step 5: Find an equation for the gorilla population (P)

The integrals will yield the logarithmic form, solve them to get:\( -ln|P| + ln|250-P| = 0.0004 t + C\)This equation can be solved for \(P\) to give the gorilla population as a function of time.

Step 6: Solve for constant C

Initial condition is given as \(P(0) = 28\), use this to solve for C.

Step 7: Calculate time for population to reach carrying capacity

Set \(P(t) = 250\) and solve for \(t\) to find out when the gorilla population will reach the carrying capacity of the preserve.

Key Concepts

Population Growth

Population growth in ecosystems can be modeled using differential equations. This mathematical approach helps us understand how populations change over time. It considers factors like birth rates, death rates, and carrying capacity.
In the context of our gorilla problem, the population growth is defined by the equation \(\frac{dP}{dt} = 0.0004 P(250-P)\). Here, \(P\) is the gorilla population at time \(t\), and \(250\) is the maximum number of gorillas the environment can support.

• The population grows quickly when there are few gorillas because there's less competition for resources.
• As the number approaches the carrying capacity, the growth slows down.

This behavior is typical in logistic growth models.

Carrying Capacity

Carrying capacity is a crucial concept in ecology. It represents the maximum population size of a species that an environment can sustain indefinitely. In our scenario, the preserve can support no more than 250 gorillas without negatively impacting their environment. This value directly affects the population growth rate.
The carrying capacity impacts the differential equation by serving as a limit to growth. When the population (\(P\)) is much less than \(250\), resources are abundant, so growth is rapid.

• As \(P\) approaches \(250\), resources become limited.
• Growth slows and eventually stops, creating an equilibrium.

Understanding carrying capacity helps in achieving a sustainable balance within ecosystems.

Separation of Variables

Separation of variables is a technique used to solve simple differential equations. It involves rearranging the equation so that each variable is on different sides. In our example, we re-wrote the equation as \(\frac{dP}{P(250-P)} = 0.0004 dt\). This step is crucial because it allows us to focus on one variable at a time while integrating both sides of the equation separately.
To effectively use this technique:

• Ensure all terms involving \(P\) are on one side.
• Ensure all terms involving \(t\) are on the other.

Once separated, the problem becomes a matter of integrating both sides to find \(P\) in terms of \(t\).

Partial Fraction Decomposition

Partial fraction decomposition is a method used to simplify complex rational expressions, facilitating easier integration. In our exercise, it's employed for the expression \(\frac{1}{P(250-P)}\).
This involves breaking it down into simpler fractions:

• Write \(\frac{1}{P(250-P)}\) as \(\frac{A}{P} + \frac{B}{250-P}\)
• Determine constants \(A\) and \(B\) by solving the equation \(\frac{1}{P(250-P)} = \frac{-1}{P} + \frac{1}{250-P}\)

Once decomposed, we integrate each simpler fraction separately. This simplifies solving the differential equation by reducing it to a series of elementary integrals.