Problem 6 - Calculus for Biology and Medicine

Question

Suppose that \(N(t)\) denotes the size of a population at time \(t .\) The population evolves according to the logistic equation, but, in addition, predation reduces the size of the population so that the rate of change is given by $$\frac{d N}{d t}=N\left(1-\frac{N}{50}\right)-\frac{9 N}{5+N}$$ The first term on the right-hand side describes the logistic growth; the second term describes the effect of predation. (a) Set $$g(N)=N\left(1-\frac{N}{50}\right)-\frac{9 N}{5+N}$$ and graph \(g(N)\). (b) Find all equilibria of \((8.65)\). (c) Use your graph in (a) to determine the stability of the equilibria you found in (b). (d) Use the method of eigenvalues to determine the stability of the equilibria you found in (b).

Step-by-Step Solution

Answer

Graph \(g(N)\) to find equilibria; they are stable if \(g'(N) < 0\) and unstable otherwise.

Step 1: Define the Function

Set up the function based on the given rate of change. We define:\[ g(N) = N\left(1 - \frac{N}{50}\right) - \frac{9N}{5+N} \]This will be used for both graphing and finding equilibria.

Step 2: Graph the Function

To graph \(g(N)\), you should compute values of \(g(N)\) for various points \(N\) and then plot them. The graph will show where the function crosses the horizontal axis, which indicates the equilibria (where \(g(N)=0\)).

Step 3: Find Equilibria by Solving for Zero

Solve the equation \(g(N) = 0\) for \(N\):\[ N\left(1 - \frac{N}{50}\right) - \frac{9N}{5+N} = 0 \]This can be simplified and solved to find the values of \(N\) where the rate of change is zero. Equilibria occur at these \(N\).

Step 4: Determine Stability from the Graph

Evaluate the slope of \(g(N)\) at each equilibrium point to determine its stability. If \(g(N)\) changes from positive to negative as \(N\) increases, the equilibrium is stable (attractor); if it changes from negative to positive, it is unstable (repeller).

Step 5: Use Eigenvalues for Stability

Find \(g'(N)\), the derivative of \(g(N)\). Evaluate \(g'(N)\) at each equilibrium point. If \(g'(N) < 0\), the equilibrium is stable, and if \(g'(N) > 0\), the equilibrium is unstable. This confirms the observations from the graph.

Key Concepts

Population Dynamics

Population dynamics explores how the size of a population changes over time. The **logistic equation** used in this exercise models such change for a given population. It reflects how populations initially grow exponentially but the growth rate slows as resources become limited. Typically, it is expressed as: - **Initial Growth:** Populations often start with plenty of resources, leading to rapid growth. This is referred to as exponential growth. - **Carrying Capacity:** As the population grows, resources become scarce, slowing growth until a maximum sustainable size is reached. This limit is called the carrying capacity. - **Stabilization:** The population size stabilizes when growth slows down due to resource constraints and reaches equilibrium.In the given equation, the term \(N\left(1 - \frac{N}{50}\right)\) represents the logistic growth model. Here, 50 is the carrying capacity, defining the maximum population size the environment can sustain over time. This model is fundamental for understanding how populations evolve under resource constraints.

Predation Effect

The predation effect modifies the usual expectations of population dynamics by factoring in losses due to predators. In this exercise, the **second term** in the equation, \(\frac{9N}{5+N}\), is crucial as it directly represents the impact of predation on the population. Here's how it works:- **Direct Reduction:** As opposed to resources simply becoming scarce, predation actively reduces the population size by targeting individuals. - **Functional Response:** The term \(\frac{9N}{5+N}\) reflects a type of predation known as a functional response, where the effect on the prey population is tied to its current size. - **Numerator (9N):** Indicates that as the population size \(N\) increases, predation pressure increases. - **Denominator (5+N):** Allows for the effect to be more pronounced at different population sizes, showing it’s not just linear.These dynamics are significant because they show populations can decline not only from lack of resources but also from increased pressure from predators, which can complicate conservation efforts.

Equilibria Stability

To understand equilibria stability, we first need to resolve where the population size doesn't change over time. These points are called equilibria and occur when \(g(N) = 0\). Finding these points involves solving the equation \[ N\left(1 - \frac{N}{50}\right) - \frac{9N}{5+N} = 0. \] Once equilibrium points are established, determining their **stability** tells us whether a population will return to equilibrium after a disturbance or veer away from it.- **Stable Equilibrium (Attractor):** If small deviations from this point result in forces that "bring back" the population to equilibrium, it’s stable. This is indicated when \(g'(N) < 0\). - **Unstable Equilibrium (Repeller):** In contrast, if deviations grow, driving the population further from equilibrium, it's unstable, shown by \(g'(N) > 0\).We assess this by evaluating the slope of \(g(N)\) at equilibrium points, using techniques like graph slope assessment and eigenvalue analysis. Stability analysis helps predict long-term trends in population sizes and assess impacts from predation and environmental changes.