Question

In each case, carry out the following steps. a. Identify which equation corresponds to the predator and which corresponds to the prey. b. Find the lines along which \(x^{\prime}(t)=0 .\) Find the lines along which \(y^{\prime}(t)=0\) c. Find the equilibrium points for the system. d. Identify the four regions in the first quadrant of the xy-plane in which \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. e. Sketch a representative solution curve in the xy-plane and indicate the direction in which the solution evolves. $$x^{\prime}(t)=-3 x+6 x y, y^{\prime}(t)=y-4 x y$$

Short answer

Based on the predator-prey model presented in the solution, describe the predator and prey's population dynamics, and explain how these dynamics are represented in each region of the phase plane.

Step-by-step solution

Identify predator and prey equations

Predators are those whose population increases as the prey population increases, and prey are those whose population decreases as the predator's population increases. In this case, we can see that: - \(x'(t)=-3x+6xy\): The prey equation, since \(x\) population decreases if there is no \(y\) around (predator absent). - \(y'(t)=y-4xy\): The predator equation, since \(y\) population increases if there is no \(x\) around (prey absent).

Lines when \(x'(t)=0\) and \(y'(t)=0\)

We will now find when the first derivatives are zero: a. \(x'(t)=0\): \(-3x+6xy = 0 \Rightarrow x(-3+6y)=0\). There are two lines: \(x=0\) and \(y=\frac{1}{2}\). b. \(y'(t)=0\): \(y-4xy=0 \Rightarrow y(1-4x)=0\). There are two lines: \(y=0\) and \(x=\frac{1}{4}\).

Equilibrium points

Equilibrium points occur when \(x'(t)=0\) and \(y'(t)=0\) simultaneously. This means finding the intersection points of the lines determined in the previous step. We have: 1. \((0,0)\): Intersection of \(x=0\) and \(y=0\). 2. \(\left(\frac{1}{4},\frac{1}{2}\right)\): Intersection of \(x=\frac{1}{4}\) and \(y=\frac{1}{2}\).

Four regions in the first quadrant

We will now analyze the signs of \(x'(t)\) and \(y'(t)\) in the four regions formed by the lines \(x=\frac{1}{4}\) and \(y=\frac{1}{2}\) in the first quadrant. Region 1 (\(00\) : Prey population is increasing - \(y'(t)<0\) : Predator population is decreasing Region 2 (\(\frac{1}{4}0\) : Prey population is increasing - \(y'(t)>0\) : Predator population is increasing Region 4 (\(\frac{1}{4}0\) : Predator population is increasing

Sketch a representative solution curve

To sketch a representative solution curve in the xy-plane, follow these steps: 1. Draw the lines \(x=0\), \(y=0\), \(x=\frac{1}{4}\), and \(y=\frac{1}{2}\). 2. Mark the equilibrium points \((0,0)\) and \(\left(\frac{1}{4},\frac{1}{2}\right)\). 3. Based on the signs of \(x'(t)\) and \(y'(t)\) in each region, draw arrows indicating the direction of the solution. 4. Draw a representative solution curve that follows the arrows, typically a closed curve around the equilibrium point \(\left(\frac{1}{4},\frac{1}{2}\right)\). The resulting sketch will show that the predator-prey relationship leads to periodic oscillations in their populations, with the prey population peaking prior to the predator population.

Key concepts

Equilibrium Points

In the predator-prey model, equilibrium points are crucial as they represent the states where populations of both species remain constant over time. To determine these points, we look for conditions where both the predator and prey rates of change are zero simultaneously. Mathematically, this means solving for when both equations \(x'(t)\) and \(y'(t)\) equal zero.

For the system given:

• \(x'(t) = -3x + 6xy\) leads to zero when either \(x = 0\) or \(y = \frac{1}{2}\).
• \(y'(t) = y - 4xy\) leads to zero when either \(y = 0\) or \(x = \frac{1}{4}\).

By intersecting these conditions, we find critical points at \((0,0)\) and \(\left(\frac{1}{4}, \frac{1}{2}\right)\). These points where both populations are unchanged are called equilibrium points. The point \((0,0)\) represents extinction for both species, while \(\left(\frac{1}{4}, \frac{1}{2}\right)\) indicates a possible cycle in their populations.

Solution Curve Sketch

Drawing a solution curve for a predator-prey model helps visualize how the populations evolve over time under the given conditions. The critical part of sketching is to recognize the equilibrium points and how the populations behave in these different regions.

To sketch:

• Begin by sketching lines at \(x = 0\), \(y = 0\), \(x = \frac{1}{4}\), and \(y = \frac{1}{2}\), marking the equilibrium points \((0,0)\) and \(\left(\frac{1}{4}, \frac{1}{2}\right)\).
• Next, analyze the direction of populations in the four regions created by these lines. Draw arrows indicating population growth or decline based on the positive or negative value of the derivatives \(x'(t)\) and \(y'(t)\).
• Finally, sketch curves around these points to show the cyclical nature of the predator-prey interaction.

The resulting design often shows periodic oscillations, illustrating the dynamic balance between predator and prey populations over time.

First Derivative Zero Lines

The lines where the first derivatives become zero reveal key insights about population stability and dynamics. Finding these lines involves setting each derivative to zero and solving for the variables.

For \(x'(t) = -3x + 6xy = 0\):

• The zero lines are \(x = 0\), indicating no prey, or \(y = \frac{1}{2}\), where the prey's growth rate matches its depletion by predators.

For \(y'(t) = y - 4xy = 0\):

• The zero lines are \(y = 0\), indicating no predators, or \(x = \frac{1}{4}\), where predator growth perfectly balances prey consumption.

These lines partition the xy-plane into regions where predator and prey populations grow at different rates. Understanding these lines helps predict how the system evolves: towards equilibrium points or oscillations around them.