91Ó°ÊÓ

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 $$ \frac{d P}{d t}=0.0004(250-P) P $$ where time \(t\) is in years. $$ \begin{array}{llllll}{t(\text { sec })} & {s(\mathbf{m})} & {t(\text { sec })} & {s(\mathbf{m})} & {t(\text { sec })} & {s(\mathbf{m})} \\ \hline 0 & {0} & {1.5} & {0.89} & {3.1} & {1.30} \\ {0.1} & {0.07} & {1.7} & {0.97} & {3.3} & {1.31} \\ {0.3} & {0.22} & {1.9} & {1.05} & {3.5} & {1.32} \\ {0.5} & {0.36} & {2.1} & {1.11} & {3.7} & {1.32} \\ {0.7} & {0.49} & {2.3} & {1.17} & {3.9} & {1.32} \\ {0.9} & {0.60} & {2.5} & {1.22} & {4.1} & {1.32} \\ {1.1} & {0.71} & {2.7} & {1.25} & {4.3} & {1.32} \\ {1.3} & {0.81} & {2.9} & {1.28} & {4.5} & {1.32}\end{array} $$ 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

01

Recognize the type of differential equation

The given differential equation, \(\frac{dP}{dt} = 0.0004(250-P)P\), is a separable differential equation often used in population dynamics, called the logistic growth model. This type of equation models populations that have a carrying capacity, denoted by \(M = 250\) in this problem.

02

Separate variables

To solve the differential equation, separate the variables \(P\) and \(t\) by dividing both sides by \((250-P)P\) and multiplying both sides by \(dt\): \[\frac{1}{(250-P)P}dP = 0.0004\,dt\].

03

Integrate both sides

Integrate the left-hand side with respect to \(P\) and the right-hand side with respect to \(t\). This results in the equation:\[\int \frac{1}{(250-P)P} dP = \int 0.0004 \, dt\].The left-hand side decomposes using partial fraction decomposition and integrates:\[\frac{1}{250} \ln\left| \frac{P}{250-P} \right| = 0.0004t + C\]where \(C\) is the constant of integration.

04

Solve for the constant using initial conditions

Use the initial condition \(P(0) = 28\) to find \(C\). Substitute \(t = 0\) and \(P = 28\) into the integrated equation:\[\frac{1}{250} \ln\left| \frac{28}{250-28} \right| = C\].Calculate to find \( C \).

05

Simplify and solve for P(t)

Simplify the equation from the previous step. Solve for \(P(t)\) explicitly by exponentiating both sides and rearranging:\[P(t) = \frac{250}{1 + Ae^{-0.1t}}\]where the constant \(A\) is determined from step 4.

06

Determine time to reach carrying capacity

The carrying capacity is reached as \(t\) approaches infinity. Practically, calculate \(t\) when \(P(t)\) is a number very close to 250, such as 249. Use the equation:\[249 = \frac{250}{1 + Ae^{-0.1t}}\].Solve for \(t\) to find the approximate time when \(P(t)\) is effectively equal to 250.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. They are used to describe how a certain quantity, such as population, changes over time. In this exercise, we encounter a specific type of differential equation known as a separable differential equation, which is solved by separating the variables and integrating both sides.
This can be represented in the form \( \frac{dP}{dt} = f(P,t) \), where \( P \) is the population as a function of time \( t \). The solution involves manipulating the equation to express \( dP \) and \( dt \) on opposite sides, and then performing integration.

• Firstly, divide to isolate terms: \( \frac{1}{(250-P)P}dP = 0.0004dt \)
• Next, integrate both sides: \( \int \frac{1}{(250-P)P}dP = \int 0.0004dt \)
• The integration requires techniques like partial fraction decomposition, resulting in a logarithmic expression.

Simplifying these equations provides a model for population over time, which forms the foundation for understanding the population dynamics of the gorillas in the preserve.

Population Dynamics

Population dynamics is the study of how and why populations change over time. This field utilizes mathematical models to predict population growth trends. In the context of this exercise, we're examining a gorilla population in a wildlife preserve.
The logistic growth model, as applied here, suggests that the population will grow rapidly when it is small and slow down as it approaches a maximum sustainable number of individuals, known as the carrying capacity.
The core idea of population dynamics in this situation includes:

• Initial population setup: Starting with 28 gorillas in 1970.
• Growth rate represented by the differential equation \( \frac{dP}{dt} = 0.0004(250-P)P \), indicating growth limited by the difference between the current population \( P \) and the carrying capacity 250.
• The balance between growth factors (births, available resources) and environmental limits (space, food supply), which shapes the logistic curve.

This approach helps predict how soon the population stabilizes as resources become a limiting factor.

Carrying Capacity

Carrying capacity is a fundamental concept within ecology and population dynamics, representing the maximum number of individuals an environment can sustainably support. In this exercise, it is set at 250 gorillas, meaning the preserve can support this number without depleting resources.
As populations near this limit, growth slows due to limited resources. This concept is crucial in the logistic growth model, where the equation \( (250-P) \) represents the difference between the current population and the carrying capacity.
To determine how long the population will take to reach this capacity, the process involved:

• Solving the equation for \( P(t) = \frac{250}{1 + Ae^{-0.1t}} \), which describes the population at any given time \( t \).
• Calculating \( t \) when \( P(t) \) approaches a value close to 250, like 249, to estimate when the carrying capacity is reached.

This helps in predicting the point of demographic equilibrium, which is vital for effective wildlife management and conservation plans. In summary, carrying capacity helps manage natural resources efficiently while maintaining ecological balance.