Question

Biology A conservation organization releases 100 animals of an endangered species into a game preserve. During the first 2 years, the population increases to 134 animals. The organization believes that the preserve has a capacity of 1000 animals and that the herd will grow according to a logistic growth model. That is, the size \(y\) of the herd will follow the equation \(\int \frac{1}{y(1-y / 1000)} d y=\int k d t\) where \(t\) is measured in years. Find this logistic curve. (To solve for the constant of integration \(C\) and the proportionality constant \(k\), assume \(y=100\) when \(t=0\) and \(y=134\) when \(t=2 .\) Use a graphing utility to graph your solution.

Short answer

The derived logistic growth function is: \( y(t) = 1000 / (1 + 9e^{-0.288t}) \).

Step-by-step solution

Identify given variables

There are a few given variables in the problem: Initial population \(y(0) = 100\), and after 2 years, the population increased to \(y(2) = 134\). The carrying capacity is \(1000\), which corresponds to the maximum sustainable population. The task is to find the unkown \(k\) which represents the growth rate and \(C\) is the constant of integration.

Solve the integral

Take the given integral equation \(\int \frac{1}{y(1-y / 1000)} d y=\int k d t\) and solve each side separately. The left side gives \( \ln |y(1 - \frac{y}{1000})| = kt + C \).

Simplify equation

Exponentiate to get rid of natural log: \(y(1 - \frac{y}{1000}) = e^{kt+C}, \) where \(e^{C}\) is just some constant A.

Use initial conditions to solve for constants

Substitute the initial conditions \(y(0) = 100\) and \(y(2) = 134\) into the equation to solve for \(A\) and \(k\) respectively.

Finalize equation

Consolidate the equation by placing solved constants back into the equation to get the final logistic function that models population growth in that preserve.

Key concepts

Carrying Capacity

In population dynamics, the carrying capacity represents the maximum number of individuals that an environment can sustainably support. In the context of our exercise, the game preserve has a carrying capacity of 1000 animals. This number is pivotal because it sets an upper limit for the population size. Without such a limit, a population might grow indefinitely, although in realistic settings, resource limitations always apply.

Carrying capacity (\( K \)) in the logistic growth model ensures that as the population size approaches this limit, the growth rate decreases. This concept reflects the impact of limiting factors such as food availability, space, and other environmental conditions. It's observed in the logistic growth curve, where the growth rate decreases as the population nears the carrying capacity.

Initial Conditions

The initial conditions in a mathematical model provide the starting values of the variables involved. They are critical to finding a specific solution to a differential equation. In our problem, we have two such conditions: when time (\( t=0 \)), the population size (\( y \)) is 100, and after 2 years (\( t=2 \)), the size is 134.

These initial conditions help us determine unknown constants in our equation, such as the integration constant (\( C \)) and the growth rate coefficient (\( k \)). Without initial conditions, the solution to a differential equation would include arbitrary constants, making it impossible to pin down a precise model for predictions.

Population Dynamics

Population dynamics refers to how populations change over time, considering factors like birth rates, death rates, and migration. The logistic growth model, used in our exercise, is a fundamental concept that describes how populations grow rapidly when numbers are low but slow down as they approach carrying capacity.

The equation's logistic nature accounts for the self-limiting factors that prevent exponential growth, which is not sustainable in real-world situations. Population dynamics in the logistic model show a characteristic S-shaped curve, allowing us to understand trends and predict future population sizes based on current data and growth rates.

Integration of Differential Equations

Integration of differential equations is a mathematical process used to solve equations that describe changing quantities. In this exercise, the integration of the logistic differential equation \( \int \frac{1}{y(1-y/1000)} \, dy = \int k \, dt \) helps us find the explicit equation that models population growth over time.

By integrating both sides of the equation separately, we derive the relationship \( \ln |y(1 - \frac{y}{1000})| = kt + C \). This result allows us to express the population size (\( y \)) explicitly as a function of time (\( t \)). After solving for this function, initial conditions are used to find specific values for constants, ensuring the equation correctly models the real-world scenario discussed.