Differential Equations Quiz: Predator Prey Models

What this quiz covers

This quiz focuses on Predator Prey Models, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The phase lag indicates that the system parameters satisfy the relationship $b/a = d/c$ for symmetric coupling
2. The system exhibits resonance behavior where predator oscillations are driven by external forcing at prey frequency
3. The quarter-period lag confirms the model prediction that predator growth responds to prey abundance with inherent delay (correct answer)
4. The observed lag suggests significant deviation from Lotka-Volterra assumptions, possibly due to age structure or handling time

Explanation: When analyzing predator-prey dynamics, understanding the phase relationship between population oscillations reveals crucial insights about the underlying biological mechanisms. The Lotka-Volterra model predicts specific timing patterns that reflect how predator populations respond to changes in prey abundance. The quarter-period lag observed in real systems directly confirms a fundamental prediction of the Lotka-Volterra model. In these equations, predator growth ($\frac{dy}{dt} = -cy + dxy$) depends on the interaction term $dxy$, meaning predators can only increase when prey are abundant. This creates an inherent delay: prey must first grow, reach peak abundance, then decline before predators—having capitalized on the high prey density—reach their own peak. The quarter-period lag mathematically corresponds to this biological reality where predator responses trail prey availability. Choice A incorrectly suggests the lag indicates symmetric coupling through parameter ratios, but $b/a = d/c$ relates to equilibrium stability, not phase relationships. Choice B misinterprets the system as driven oscillations with external forcing, when Lotka-Volterra describes autonomous oscillations arising from internal dynamics alone. Choice D argues the lag indicates model failure, but this phase relationship is actually a key prediction that validates the model's core assumptions about predator-prey interactions. Remember that in predator-prey models, phase lags are expected features, not model failures. When you see questions about population timing, focus on the biological mechanism: predators need time to respond to prey changes, creating the characteristic delay pattern that confirms rather than contradicts basic model predictions.

Question 2

1. The prey population is initially increasing, and the predator population is initially decreasing.
2. The prey population is initially decreasing, and the predator population is initially increasing.
3. Both the prey and predator populations are initially decreasing.
4. Both the prey and predator populations are initially increasing. (correct answer)

Explanation: To determine the initial behavior, we evaluate the derivatives at the initial point $(x(0), y(0)) = (4, 1)$. For the prey population: $\frac{dx}{dt}|_{(4,1)} = 4(2 - 1) = 4(1) = 4$. Since $\frac{dx}{dt} > 0$, the prey population is initially increasing. For the predator population: $\frac{dy}{dt}|_{(4,1)} = 1(4 - 3) = 1(1) = 1$. Since $\frac{dy}{dt} > 0$, the predator population is also initially increasing. The non-trivial equilibrium point is at $(3,2)$, and the point $(4,1)$ is in the fourth quadrant relative to this center, where the flow is up and to the right.

Question 3

1. The average prey population remains the same, while the average predator population increases. (correct answer)
2. Both the average prey and average predator populations increase.
3. The average prey population increases, while the average predator population remains the same.
4. The average prey population increases, and the average predator population decreases.

Explanation: The coexistence equilibrium point for a Lotka-Volterra system $\frac{dx}{dt} = ax - bxy$ and $\frac{dy}{dt} = -cy + dxy$ is $(x_E, y_E) = (c/d, a/b)$. The average populations over a cycle correspond to these equilibrium values. Initially, $a=0.5, b=0.01, c=0.2, d=0.002$. The equilibrium is $(0.2/0.002, 0.5/0.01) = (100, 50)$. After the change, the parameter $a$ becomes $a' = 0.6$. The new equilibrium is $(0.2/0.002, 0.6/0.01) = (100, 60)$. Thus, the average prey population remains at 100, while the average predator population increases from 50 to 60.

Question 4

1. The equilibrium predator population doubles, while the equilibrium prey population remains unchanged.
2. The equilibrium prey population is halved, while the equilibrium predator population remains unchanged. (correct answer)
3. Both the prey and predator equilibrium populations are halved.
4. The equilibrium prey population is doubled, while the equilibrium predator population is halved.

Explanation: The coexistence equilibrium point $(x_E, y_E)$ is found by setting both derivatives to zero (for $x,y > 0$): $a - by = 0$ and $-c + dx = 0$. This gives the equilibrium point $(x_E, y_E) = (c/d, a/b)$. If the parameter $d$ is doubled to become $d' = 2d$, the new equilibrium point $(x_E', y_E')$ will be $(c/d', a/b) = (c/(2d), a/b)$. Comparing this to the original, the new equilibrium prey population is $x_E' = x_E/2$, and the new equilibrium predator population is $y_E' = y_E$. The prey population is halved, and the predator population is unchanged.

Question 5

1. $H(x, y) = ax - b \ln(x) + cy - d \ln(y)$
2. $H(x, y) = dx - c \ln(x) + by - a \ln(y)$ (correct answer)
3. $H(x, y) = (x-\frac{c}{d})^2 + (y-\frac{a}{b})^2$
4. $H(x, y) = c d x^2 - a b y^2$

Explanation: A conserved quantity $H(x,y)$ must satisfy $\frac{dH}{dt} = 0$ along trajectories. This means $\frac{\partial H}{\partial x} \frac{dx}{dt} + \frac{\partial H}{\partial y} \frac{dy}{dt} = 0$. Let's test choice B: $H(x, y) = dx - c \ln(x) + by - a \ln(y)$. We have $\frac{\partial H}{\partial x} = d - \frac{c}{x}$ and $\frac{\partial H}{\partial y} = b - \frac{a}{y}$. Then $\frac{dH}{dt} = (d - \frac{c}{x})(x(a-by)) + (b - \frac{a}{y})(y(-c+dx)) = (\frac{dx-c}{x})(x(a-by)) + (\frac{by-a}{y})(y(-c+dx)) = (dx-c)(a-by) - (a-by)(-c+dx) = 0$. Thus, $H(x, y)$ is conserved. The other options do not satisfy this condition.

Question 6

1. Both populations will remain at their initial values, as they are on a nullcline.
2. The prey population will remain constant, while the predator population begins to decrease.
3. The predator population is at its maximum for its cycle, and the prey population will immediately begin to decrease. (correct answer)
4. The predator population is at its maximum for its cycle, and the prey population will immediately begin to increase.

Explanation: When analyzing Lotka-Volterra predator-prey dynamics, you need to understand how populations change based on their positions relative to the nullclines—curves where population growth rates are zero. In this scenario, the prey population starts exactly at its equilibrium value $x(0) = c/d$, which places it on the prey nullcline where $\frac{dx}{dt} = 0$. However, the predator population exceeds equilibrium with $y(0) > a/b$, positioning it above the predator nullcline. At this specific point, the prey experiences maximum predation pressure because the predator population is at its peak for this cycle. Since we're on the prey nullcline with maximum predators present, any slight perturbation will cause the prey population to decline immediately. The predator population, being above its nullcline, will also begin decreasing as it has exceeded the carrying capacity supported by the current prey level. Option A is incorrect because being on only one nullcline (not both) doesn't create equilibrium—populations will still change. Option B wrongly assumes the prey stays constant; while it starts on its nullcline, the excessive predator population will immediately drive prey numbers down. Option D makes the correct observation about predator population being at maximum but incorrectly predicts prey will increase—with maximum predation pressure, prey must decrease. Remember: in Lotka-Volterra systems, when one population is at equilibrium but the other isn't, focus on which nullcline you're crossing and the direction of flow around the phase portrait to predict immediate population changes.

Question 7

1. $\begin{cases} \frac{dx}{dt} = ax(1 - x/K) - bxy \\ \frac{dy}{dt} = -cy + dxy \end{cases}$
2. $\begin{cases} \frac{dx}{dt} = ax - bxy \\ \frac{dy}{dt} = -cy + dxy \end{cases}$
3. $\begin{cases} \frac{dx}{dt} = -cx + dxy \\ \frac{dy}{dt} = ay(1 - y/K) - bxy \end{cases}$ (correct answer)
4. $\begin{cases} \frac{dx}{dt} = -cx + dxy \\ \frac{dy}{dt} = ay - bxy \end{cases}$

Explanation: When modeling predator-prey interactions, you need to identify which species plays each role and understand how their populations naturally behave. Here, herbivores are the "predators" consuming plants as "prey." For any species, consider its growth pattern without the other species present. Plants have limited growth due to space and resources, following logistic growth: $ay(1-y/K)$ where $K$ is the carrying capacity. Herbivores depend entirely on plants for survival, so without plants they would decline exponentially: $-cx$ (negative because the population decreases). The interaction term $xy$ represents encounters between species. When herbivores consume plants, the plant population decreases ($-bxy$ in the plant equation) while the herbivore population benefits and increases ($+dxy$ in the herbivore equation). Option C correctly captures this: $\frac{dx}{dt} = -cx + dxy$ shows herbivores declining without food but growing through consumption, while $\frac{dy}{dt} = ay(1-y/K) - bxy$ shows plants following logistic growth but being reduced by herbivore consumption. Option A incorrectly gives plants unlimited exponential growth and herbivores logistic growth with self-limitation. Option B has plants growing exponentially without resource limits. Option D swaps the roles, making plants the predators and herbivores the prey, which contradicts the biological setup. Study tip: Always identify which species has intrinsic growth limitations (usually prey due to resources) versus which depends on the other for survival (usually predators), then match the mathematical terms accordingly.

Question 8

1. Both average populations decrease.
2. The average prey population decreases, while the average predator population increases.
3. The average prey population increases, while the average predator population decreases. (correct answer)
4. Both average populations increase.

Explanation: The average populations correspond to the non-trivial equilibrium point. For the original system, $(x_E, y_E) = (c/d, a/b)$. For the harvested system, the parameters $a$ and $c$ are effectively replaced by $a' = a-k$ and $c' = c+k$. The new equilibrium is $(x_H, y_H) = (c'/d, a'/b) = ((c+k)/d, (a-k)/b)$. Comparing the new and old equilibrium values: $x_H = (c+k)/d > c/d = x_E$, so the average prey population increases. $y_H = (a-k)/b < a/b = y_E$, so the average predator population decreases. This counter-intuitive result is Volterra's principle.

Question 9

1. It represents a constant number of prey, $A$, that are harvested each time unit.
2. It signifies that predators will only hunt prey if the prey population exceeds a threshold $A$.
3. It indicates that the prey's carrying capacity is reduced by a value $A$ due to environmental factors.
4. It introduces a population threshold $A$, below which the prey population will decline to extinction even without predators. (correct answer)

Explanation: In the absence of predators ($y=0$), the prey growth equation is $\frac{dx}{dt} = rx(1 - \frac{x}{K})(x-A)$. The sign of the growth rate depends on the term $(x-A)$. If the prey population $x$ is between 0 and $A$, the term $(x-A)$ is negative, making $\frac{dx}{dt}$ negative (since all other terms are positive in this range). A negative growth rate means the population will decline, leading to extinction. This value $A$ is known as the Allee threshold, a critical minimum population size for the species to be viable.

Question 10

1. The period will decrease because the predator death rate $c$ increases.
2. The period will increase because the product $ac$ will decrease.
3. The period will increase because the prey birth rate $a$ decreases and the predator death rate $c$ increases.
4. The effect on the period cannot be determined because the changes to $a$ and $c$ have opposing effects on the product $ac$. (correct answer)

Explanation: The period of oscillation is approximately $T \approx 2\pi / \sqrt{ac}$. The virus causes the prey's intrinsic birth rate, $a$, to decrease. It also causes the predator's intrinsic death rate, $c$, to increase. The period depends on the product $ac$. Since $a$ decreases and $c$ increases, the change in the product $ac$ is ambiguous. For example, if $a$ goes from 2 to 1 and $c$ goes from 1 to 2, the product is unchanged. If $a$ goes from 2 to 1 and $c$ goes from 1 to 3, the product increases. If $a$ goes from 4 to 1 and $c$ goes from 1 to 2, the product decreases. Without knowing the relative magnitudes of these changes, the effect on the period cannot be determined.

Question 11

1. In the absence of predators, the prey population exhibits logistic growth with a carrying capacity of 100. (correct answer)
2. The prey population's growth is limited by a fixed harvesting rate proportional to its carrying capacity.
3. The interaction between prey and predators leads to a decrease in the prey's carrying capacity to 100.
4. In the absence of predators, the prey population grows exponentially with a rate of $0.4$ but is limited by a factor of 100.

Explanation: The term $0.4x(1 - \frac{x}{100})$ is the standard mathematical representation of logistic growth. In this model, if the predator population $y$ were zero, the prey equation would be $\frac{dx}{dt} = 0.4x(1 - \frac{x}{100})$. This equation describes a population with an intrinsic growth rate of $r=0.4$ and a carrying capacity of $K=100$. This means the prey population's growth slows as it approaches 100 due to internal competition or resource limitation, independent of predation.

Question 12

1. A saddle point. (correct answer)
2. A stable node.
3. An unstable node.
4. A center.

Explanation: To classify the equilibrium point at the origin, we compute the Jacobian matrix of the system: $J(x,y) = \begin{pmatrix} a-by & -bx \\ dy & -c+dx \end{pmatrix}$. Evaluating at $(0,0)$ gives $J(0,0) = \begin{pmatrix} a & 0 \\ 0 & -c \end{pmatrix}$. The eigenvalues of this diagonal matrix are its diagonal entries, $\lambda_1 = a$ and $\lambda_2 = -c$. Since all parameters are positive, we have one positive eigenvalue and one negative eigenvalue. An equilibrium point with real eigenvalues of opposite signs is a saddle point.

Question 13

1. Both populations will oscillate periodically with constant amplitude, maintaining their initial energy level indefinitely (correct answer)
2. Both populations will spiral inward toward the non-trivial equilibrium point, eventually reaching stable coexistence
3. The prey population will grow exponentially while the predator population approaches zero asymptotically
4. Both populations will spiral outward from any initial condition, leading to unbounded growth over time

Explanation: In the Lotka-Volterra model, trajectories in the first quadrant are closed orbits around the equilibrium point (c/d, a/b). The system conserves energy (has a conserved quantity H(x,y) = dx - cln(x) + by - aln(y)), so populations oscillate periodically with constant amplitude determined by initial conditions. Choice B is wrong because trajectories don't spiral inward (no damping). Choice C is wrong because it describes behavior near (0,0) only, not typical first-quadrant behavior. Choice D is wrong because trajectories are bounded closed curves, not outward spirals.

Question 14

1. These terms represent intraspecific competition within each species, causing logistic growth limitations for both populations (correct answer)
2. These terms model immigration rates that increase proportionally with current population sizes in each species
3. These terms represent cooperative hunting behavior that enhances predation efficiency at higher population densities
4. These terms model seasonal migration patterns that remove individuals from the system at predictable rates

Explanation: The terms -0.1x² and -0.05y² (from factoring) represent intraspecific competition - members of the same species competing for limited resources. This creates logistic-type growth limitations, making the equilibrium point stable unlike the classical Lotka-Volterra neutral center. Choice B is wrong because negative terms represent removal, not immigration. Choice C is wrong because these terms affect each species independently, not predator-prey interaction. Choice D is wrong because the terms are density-dependent, not time-dependent seasonal effects.

Question 15

1. The period would become undefined because the system would no longer exhibit periodic behavior with these initial conditions
2. The period would decrease because higher predator density leads to faster prey consumption and more rapid population changes
3. The period would remain exactly the same because it depends only on the system parameters, not initial conditions
4. The period would increase because the trajectory moves to a larger closed orbit farther from the equilibrium point (correct answer)

Explanation: When analyzing Lotka-Volterra systems, remember that these predator-prey models create closed orbital trajectories around an equilibrium point, and the characteristics of these orbits depend crucially on initial conditions. The Lotka-Volterra system has a conserved quantity (first integral) that remains constant along each trajectory. This means different initial conditions create different closed orbits of varying sizes around the equilibrium point. The equilibrium occurs at $(\frac{c}{d}, \frac{a}{b}) = (3, 2)$. When you increase the initial predator population from 4 to 6 while keeping prey at 10, you move to a trajectory that's farther from this equilibrium point, creating a larger orbital path. Larger orbits in Lotka-Volterra systems correspond to longer periods because the populations must traverse greater distances in phase space during each cycle. The trajectory extends to higher maximum values and lower minimum values for both species, requiring more time to complete one full oscillation. Answer D correctly identifies that the period increases due to the larger closed orbit farther from equilibrium. Answer A is wrong because Lotka-Volterra systems maintain periodic behavior for any positive initial conditions. Answer B incorrectly assumes that higher predator density always leads to faster dynamics—while it does initially accelerate prey decline, the overall cycle becomes longer. Answer C reflects a common misconception; unlike some dynamical systems, Lotka-Volterra periods definitely depend on initial conditions, not just parameters. Study tip: For Lotka-Volterra problems, always consider how initial conditions affect orbit size and remember that larger orbits mean longer periods.

Question 16

1. The prey carrying capacity is 3 times larger than the predator carrying capacity, indicating asymmetric resource limitations
2. The ratio of prey growth rate to predation efficiency equals 1.5, and the ratio of predator death rate to conversion efficiency equals 3 (correct answer)
3. The predation rate parameter is exactly twice the prey intrinsic growth rate, ensuring balanced population dynamics
4. The system exhibits neutral stability because the predator conversion efficiency equals the prey vulnerability coefficient

Explanation: When analyzing Lotka-Volterra systems, the key insight is that equilibrium points and their coordinates directly encode information about the underlying parameters. The standard predator-prey system has the form $\frac{dx}{dt} = ax - bxy$ and $\frac{dy}{dt} = -cy + dxy$, where $x$ is prey, $y$ is predator, and $a, b, c, d$ are positive parameters. The equilibrium point occurs where both derivatives equal zero. Setting $ax - bxy = 0$ and $-cy + dxy = 0$ (excluding the trivial solution), we get $x^* = \frac{c}{d}$ and $y^* = \frac{a}{b}$. Given the equilibrium at $(3, 1.5)$, we have $\frac{c}{d} = 3$ and $\frac{a}{b} = 1.5$. This means the ratio of predator death rate to conversion efficiency equals 3, and the ratio of prey growth rate to predation efficiency equals 1.5. Choice A incorrectly interprets the equilibrium coordinates as carrying capacities, but Lotka-Volterra systems don't have carrying capacities—they exhibit neutral oscillations. Choice C makes an unfounded claim about specific parameter relationships ($b = 2a$) that cannot be deduced from equilibrium coordinates alone. Choice D uses non-standard terminology and incorrectly suggests that equal parameters create neutral stability, when neutral stability actually results from the system's conservative nature. Remember: In Lotka-Volterra systems, equilibrium coordinates directly give you parameter ratios. The prey equilibrium equals the ratio of predator parameters, and the predator equilibrium equals the ratio of prey parameters.

Question 17

1. The equilibrium point shifts to lower prey and higher predator densities, with trajectories remaining as closed orbits around the new equilibrium
2. The equilibrium point shifts to higher prey and lower predator densities, while trajectories change from closed orbits to spiral sinks (correct answer)
3. Harvesting creates a new boundary equilibrium where prey population reaches zero, causing predator extinction in finite time
4. The system maintains the same equilibrium point, but harvesting introduces damping that causes trajectories to spiral inward slowly

Explanation: When analyzing modified predator-prey models, you need to examine how additional terms affect both the equilibrium points and the system's stability properties. The classic Lotka-Volterra system has a center equilibrium with closed orbital trajectories, but modifications typically change this behavior. To find the new equilibrium, set both derivatives to zero. From the second equation: $x^* = \frac{c}{d}$. Substituting into the first equation: $a \cdot \frac{c}{d} - b \cdot \frac{c}{d} \cdot y^* - h = 0$, which gives $y^* = \frac{ac - hd}{bcd}$. Since $h < \frac{ac}{b}$, we have $y^* > 0$. Compared to the unharvested equilibrium $(\frac{c}{d}, \frac{a}{b})$, the prey density stays the same but predator density decreases because $\frac{ac - hd}{bcd} < \frac{a}{b}$. More importantly, harvesting breaks the conservative structure of the original system. The Jacobian analysis reveals that the harvested system no longer has purely imaginary eigenvalues, introducing a spiral sink behavior where trajectories converge to the equilibrium. Answer A is incorrect because predator density decreases, not increases, and orbits are no longer closed. Answer C is wrong because with moderate harvesting ($h < \frac{ac}{b}$), a positive interior equilibrium exists—prey don't go extinct. Answer D fails because the equilibrium point definitely shifts due to the harvesting term. Study tip: Remember that adding non-conservative terms (like constant harvesting) to Lotka-Volterra systems typically breaks the neutral stability, converting centers to spiral points and changing population dynamics fundamentally.

Question 18

1. System 1 has a larger amplitude of oscillation because its equilibrium point is farther from the initial condition
2. Both systems follow identical trajectories in phase space and have the same oscillation period due to parameter scaling
3. System 2 oscillates with exactly twice the frequency of System 1, but both have identical trajectory shapes in phase space (correct answer)
4. System 2 exhibits faster approach to equilibrium due to higher parameter values, while System 1 maintains larger oscillations

Explanation: When analyzing Lotka-Volterra systems, you need to understand how parameter scaling affects trajectory shape and oscillation frequency. The general form is $\frac{dx}{dt} = ax - bxy$ and $\frac{dy}{dt} = -cy + dxy$. Let's examine the parameter relationships between these systems. Notice that System 2's parameters are exactly double System 1's: $a_2 = 2a_1$, $b_2 = 2b_1$, $c_2 = 2c_1$, and $d_2 = 2d_1$. This uniform scaling has specific effects on the system behavior. When all parameters are scaled by the same factor, the equilibrium points remain identical: both systems have equilibrium at $(\frac{c}{d}, \frac{a}{b}) = (2, 2)$. The trajectory shapes in phase space are also preserved because the ratios between parameters stay constant. However, the time scaling changes dramatically - System 2 evolves twice as fast as System 1, meaning it completes oscillations in half the time, giving exactly twice the frequency. Option A is incorrect because both systems have the same equilibrium point $(2, 2)$ and identical distances from the initial condition. Option B misses the frequency difference - while trajectory shapes are identical, the periods differ by a factor of 2. Option D is wrong because Lotka-Volterra systems don't approach equilibrium; they oscillate perpetually, and the "faster" behavior refers to oscillation frequency, not convergence. Remember: uniform parameter scaling in Lotka-Volterra systems preserves trajectory shapes and equilibrium points but changes the time scale proportionally. Look for this scaling pattern to quickly identify frequency relationships.