Problem 40 - Calculus for Biology and Medicine

Question

Suppose that the size of a population, denoted by \(N(t)\), satisfies $$\frac{d N}{d t}=0.7 N\left(1-\frac{N}{35}\right)$$ (a) Determine all equilibria by solving \(d N / d t=0\). (b) Solve \((8.46)\) for (i) \(N(0)=10\), (ii) \(N(0)=35\), (iii) \(N(0)=50\), and (iv) \(N(0)=0\). Find \(\lim _{t \rightarrow \infty} N(t)\) for each of the four initial conditions. (c) Compare your answer in (a) with the limiting values you found in (b).

Step-by-Step Solution

Answer

Equilibria are at 0 and 35. For all initial conditions, \( N(t) \to 35 \) except \( N(0) = 0 \), which remains 0.

Step 1: Determine Equilibria

To find the equilibria, set the equation for population change to zero: \( \frac{dN}{dt} = 0.7N\left(1 - \frac{N}{35}\right) = 0 \). Factoring, we get \( N(1 - \frac{N}{35}) = 0 \). Solving for \( N \), the equilibria are \( N = 0 \) and \( N = 35 \).

Step 2: Solve the Equation for Different Initial Conditions

Given the logistic differential equation \( \frac{dN}{dt} = 0.7N(1 - \frac{N}{35}) \), we use the equilibrium points and analyze limitations. The general solution has a form respecting those equilibria. Let's consider the analysis for stability at and around equilibrium points specifically for the initial values: \( N(0) = 10, 35, 50, 0 \).(i) For \( N(0) = 10 \), since \( 0 < N(0) < 35 \), the solution will approach \( N = 35 \) as \( t \to \infty \).(ii) For \( N(0) = 35 \), starting on an equilibrium point, \( N(t) \) remains at 35.(iii) For \( N(0) = 50 \), since \( N(0) > 35 \), \( N(t) \) will decrease toward \( N = 35 \) as \( t \to \infty \).(iv) For \( N(0) = 0 \), \( N(t) \) remains at 0 as \( t \to \infty \).

Step 3: Compare Equilibria and Limit Values

When compared to the limiting values calculated:- For (i) \( N(0) = 10 \), the limit \( \lim_{t \rightarrow \infty} N(t) = 35 \) matches the stable equilibrium.- For (ii) \( N(0) = 35 \), the population remains constant at the equilibrium \( 35 \).- For (iii) \( N(0) = 50 \), the population decreases to the stable equilibrium \( 35 \).- For (iv) \( N(0) = 0 \), the population stays at the unstable equilibrium \( 0 \).Thus, every initial condition aligns with one of the equilibria.

Key Concepts

Population Dynamics

Population dynamics involves understanding how and why the size of populations changes over time. It studies patterns like growth, decline, and stability in populations under various conditions.
In the given exercise, the change in population size is modeled using a differential equation. It describes how the population size, denoted by \( N(t) \), evolves based on the current population and a given rate of growth or decay.
For the equation \( \frac{dN}{dt} = 0.7N(1 - \frac{N}{35}) \), each part has a significance:

• \( N(t) \): The population size at time \( t \).
• \( 0.7N \): A term indicating that the growth rate is proportional to the population size. The factor 0.7 is the intrinsic growth rate.
• \( \left(1 - \frac{N}{35}\right) \): Introduces the concept of carrying capacity, the maximum population size the environment can sustain. As \( N \) approaches 35, this term approaches 0, slowing growth.

As such, population dynamics help predict how a species might change over time and the factors that influence these changes.

Equilibrium Analysis

Equilibrium analysis involves finding points at which the system, such as a population, does not change. We identify these points by setting the right-hand side of the differential equation to zero.
In the exercise, the equilibrium points are found by solving \( \frac{dN}{dt} = 0 \), leading to conclusions about population stability:

• \( N = 0 \): An unstable equilibrium. Any small population introduced will grow, showing that \( N = 0 \) cannot sustain itself if perturbed.
• \( N = 35 \): A stable equilibrium. Here, the population remains constant since \( \frac{dN}{dt} = 0 \) when \( N = 35 \). For any initial populations between 0 and 35, or above 35, the population will tend toward this stable point over time.

This demonstrates how equilibrium analysis can indicate conditions under which populations stabilize, grow, or decline.

Logistic Growth

Logistic growth models describe how populations grow in environments with limited resources. Unlike exponential growth, which assumes unlimited resources, logistic growth assumes growth rates slow as populations near a carrying capacity.
The given equation \( \frac{dN}{dt} = 0.7N(1 - \frac{N}{35}) \) is a classic example of logistic growth. Here's what happens:

• When populations are small, growth is approximately exponential because resources are abundant.
• As the population size approaches the carrying capacity of 35, the growth rate diminishes, stabilizing the population.
• Populations starting above the carrying capacity decrease toward it, as resources become insufficient to sustain such levels.

Understanding logistic growth is crucial for predicting how populations interact with their environments and what factors may limit their growth.