91Ó°ÊÓ

Let \(y(t)\) be the number of thousands of mice that live on a farm; assume time \(t\) is measured in years. \({ }^{1}\) a. The population of the mice grows at a yearly rate that is twenty times the number of mice. Express this as a differential equation. b. At some point, the farmer brings \(C\) cats to the farm. The number of mice that the cats can eat in a year is $$ M(y)=C \frac{y}{2+y} $$ thousand mice per year. Explain how this modifies the differential equation that you found in part a). c. Sketch a graph of the function \(M(y)\) for a single cat \(C=1\) and explain its features by looking, for instance, at the behavior of \(M(y)\) when \(y\) is small and when \(y\) is large. d. Suppose that \(C=1\). Find the equilibrium solutions and determine whether they are stable or unstable. Use this to explain the long-term behavior of the mice population depending on the initial population of the mice. e. Suppose that \(C=60 .\) Find the equilibrium solutions and determine whether they are stable or unstable. Use this to explain the long-term behavior of the mice population depending on the initial population of the mice. f. What is the smallest number of cats you would need to keep the mice population from growing arbitrarily large?

Step by step solution

01

- Express the Population Growth Rate

To express the population growth rate of the mice population, let’s denote the population of the mice as a function of time, which is given by \( y(t) \). The problem states that the population grows at a yearly rate that is twenty times the number of mice. Therefore, the differential equation representing this growth is given by: \[ \frac{dy}{dt} = 20y \]

02

- Modify the Differential Equation with Cats Introduced

When \( C \) cats are introduced, the number of mice they eat per year is given by: \[ M(y) = C \frac{y}{2+y} \] This modifies the differential equation found in step 1, as the loss in population due to the cats has to be subtracted from the growth rate. Thus, the new differential equation becomes: \[ \frac{dy}{dt} = 20y - C \frac{y}{2 + y} \]

03

- Graph the Function for a Single Cat

To sketch the graph of \( M(y) \) when \( C = 1 \), plot the function \[ M(y) = \frac{y}{2 + y} \] Key features: For small \( y \), \( M(y) \approx \frac{y}{2} \) (near-linear growth). For large \( y \), \( M(y) \approx 1 \) (saturation point).

04

- Determine Equilibrium Solutions for C=1

To find equilibrium solutions where \( \frac{dy}{dt} = 0 \) for \( C = 1 \), solve the equation: \[ 20y - \frac{y}{2 + y} = 0 \] Simplify to find values of \( y \): \[ \frac{20y(2 + y) - y}{2 + y} = 0 \] \[ 20y(2 + y) - y = 0 \] \[ 40y + 20y^2 - y = 0 \] \[ y(40 + 20y - 1) = 0 \] \[ y(20y + 39) = 0 \] The equilibrium solutions are \( y = 0 \) and \( y = -\frac{39}{20} \). Since the population cannot be negative, the only feasible equilibrium is \( y = 0 \). Stability: \( y = 0 \) is unstable because if the initial population is slightly greater than 0, \( dy/dt \) remains positive, causing the population to increase.

05

- Determine Equilibrium Solutions for C=60

For \( C = 60 \), solve the equation: \[ 20y - 60 \frac{y}{2 + y} = 0 \] Simplify to find values of \( y \): \[ \frac{20y(2 + y) - 60y}{2 + y} = 0 \] \[ 20y(2 + y) - 60y = 0 \] \[ 40y + 20y^2 - 60y = 0 \] \[ 20y^2 - 20y = 0 \] \[ 20y(y - 1) = 0 \] The equilibrium solutions are \( y = 0 \) and \( y = 1 \). Stability: \( y = 0 \) is unstable, and \( y = 1 \) is stable because if the initial population is slightly less or more than 1, the tendency is to move back to \( y = 1 \).

06

- Find the Minimum Number of Cats

To keep the mice population from growing arbitrarily large, we need to find the smallest number of cats \(C\) such that the population stabilizes. Set \[ 20y - C \frac{y}{2 + y} = 0 \] Solve for \( C \): \[ 20y(2 + y) = Cy \] For large y, to prevent growth, \[ 20(2 + y) \approx C \] Simplify to find \( C \): \[ C \geq 40 \]

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Key Concepts

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

population growth rate

The population growth rate describes how quickly the number of individuals in a population increases or decreases over time. It is especially important in understanding population dynamics.
In the given problem, the population of mice grows at a yearly rate that is twenty times the number of mice. This can be expressed by the differential equation: \( \frac{dy}{dt} = 20y \).
Here, \( y(t) \) represents the population of mice at time \( t \), measured in years. The term \( 20y \) indicates that the growth rate is proportional to the current population.

modification by external factors

External factors, such as predators or environmental changes, can significantly alter population dynamics. In this example, the introduction of cats affects the mice population.
When \( C \) cats are introduced, they eat mice, leading to a modified differential equation: \( \frac{dy}{dt} = 20y - C \frac{y}{2 + y} \).
The term \( C \frac{y}{2+y} \) models the mice eaten by the cats per year. As \( y \) (the mice population) increases, the number of mice eaten approaches a maximum determined by the number of cats (\( C \)). This factor modifies the straightforward exponential growth, accounting for population control.

equilibrium solutions

Equilibrium solutions occur where the population remains constant over time, meaning \( \frac{dy}{dt} = 0 \).
For \( C = 1 \), solving \( 20y - \frac{y}{2 + y} = 0 \) gives the equilibrium \( y = 0 \). However, this is unstable, meaning any small deviation will lead to population growth.
For \( C = 60 \), solving \( 20y - 60 \frac{y}{2 + y} = 0 \) gives equilibria \( y = 0 \) and \( y = 1 \). Here, \( y = 1 \) is stable, meaning small deviations will return to this equilibrium, while \( y = 0 \) is still unstable.

stability of equilibrium

Stability determines whether a population remains at equilibrium following a small perturbation.
For \( C = 1 \), the equilibrium \( y = 0 \) is unstable. If the population is slightly above zero, it will continue to grow.
For \( C = 60 \), the equilibrium \( y = 0 \) remains unstable while \( y = 1 \) is stable. Small deviations from \( y = 1 \) will result in the population returning to this stable point, thus preventing unchecked growth.

predator-prey model

The predator-prey model describes interactions between species where one is the predator and the other is the prey. In this problem, mice (prey) are hunted by cats (predator).
The modification to the differential equation introduces a term \( C \frac{y}{2 + y} \) that models predation. For small mice populations, this term grows linearly, while for large populations, it reaches a maximum.
The simplest predator-prey relationships can be complex, leading to various dynamic behaviors such as steady states and oscillations. Here, we focus on how many predators (cats) are needed to keep the mice population under control, demonstrating the importance of predators in regulating prey populations.