Differential Equations Quiz: Growth Model Parameters

What this quiz covers

This quiz focuses on Growth Model Parameters, 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. In the exponential model, $r$ is the constant relative growth rate, while in the logistic model, $r$ is the maximum possible relative growth rate, approached only as $P \to 0$. (correct answer)
2. In both models, $r$ represents the constant relative growth rate, $(dP/dt)/P$, for any population size $P$.
3. In the exponential model, $r$ is the constant growth rate, while in the logistic model, $r$ is the carrying capacity.
4. In the exponential model, $r$ is the initial population size, while in the logistic model, $r$ represents the maximum rate of change of the population.

Explanation: When analyzing population models, you need to understand how parameters change meaning as models become more sophisticated. The key insight is recognizing what "relative growth rate" means in each context. In the exponential model $\frac{dP}{dt} = rP$, the relative growth rate is $\frac{1}{P}\frac{dP}{dt} = r$, which remains constant regardless of population size. Here, $r$ represents a fixed percentage growth rate that never changes. In the logistic model $\frac{dP}{dt} = rP(1 - \frac{P}{K})$, the relative growth rate becomes $\frac{1}{P}\frac{dP}{dt} = r(1 - \frac{P}{K})$. Notice that $r$ now represents the maximum possible relative growth rate, achieved only when $P$ approaches zero (making the term $(1 - \frac{P}{K})$ approach 1). As population grows toward the carrying capacity $K$, the actual relative growth rate decreases. This makes answer A correct. Answer B is wrong because the relative growth rate is not constant in the logistic model—it decreases as $P$ increases. Answer C incorrectly states that $r$ is the carrying capacity in the logistic model, when that's actually $K$. Answer D completely misidentifies what $r$ represents in both models, confusing it with initial population size and maximum rate of change. Study tip: Remember that in logistic models, the original exponential parameters become "maximum" values that are reduced by limiting factors. Always check whether a parameter maintains the same meaning across different model complexities.

Question 2

1. The maximum possible population size is 50.
2. An initial population between 50 and 400 will cause the population to grow and stabilize at 400. (correct answer)
3. The population's growth rate is always positive as long as the initial population is greater than 50.
4. The intrinsic growth rate of the population is 0.02.

Explanation: The equilibrium points of the model are found by setting $dP/dt = 0$, which gives $P=0$, $P=50$, and $P=400$. We analyze the sign of $dP/dt$ in the intervals between these points. For $0 < P < 50$, $(\frac{P}{50} - 1)$ is negative and $(1 - \frac{P}{400})$ is positive, so $dP/dt < 0$ (population dies out). For $50 < P < 400$, $(\frac{P}{50} - 1)$ is positive and $(1 - \frac{P}{400})$ is positive, so $dP/dt > 0$ (population grows). For $P > 400$, $(\frac{P}{50} - 1)$ is positive and $(1 - \frac{P}{400})$ is negative, so $dP/dt < 0$ (population declines). Therefore, $P=50$ is an unstable equilibrium (a population threshold), and $P=400$ is a stable equilibrium (the carrying capacity). An initial population between 50 and 400 will grow towards 400.

Question 3

1. It increases to a maximum value at $N=K/e$ and then decreases.
2. It decreases monotonically as $N$ approaches $K$. (correct answer)
3. It remains constant for all values of $N < K$.
4. It increases monotonically as $N$ approaches $K$.

Explanation: The relative growth rate is given by the expression $R(N) = \frac{1}{N}\frac{dN}{dt} = -\alpha \ln(\frac{N}{K})$. This can be rewritten as $R(N) = \alpha \ln(K) - \alpha \ln(N)$. To see how this changes with $N$, we can take its derivative with respect to $N$: $\frac{dR}{dN} = -\frac{\alpha}{N}$. Since $\alpha$ and $N$ are both positive, $\frac{dR}{dN}$ is always negative for $N>0$. This indicates that the relative growth rate $R(N)$ is a strictly decreasing function of $N$. As $N$ increases towards $K$, the relative growth rate decreases. The absolute growth rate $\frac{dN}{dt}$ is what peaks at $N=K/e$, a common point of confusion.

Question 4

1. $k = 0.04$ and $M = 50,000$
2. $k = 0.08$ and $M = 50,000$
3. $k = 0.08$ and $M = 100,000$
4. $k = 0.04$ and $M = 100,000$ (correct answer)

Explanation: When you encounter logistic growth problems, you need to extract parameter values from two key behavioral clues: the relative growth rate for small populations and the population size at maximum growth rate. The logistic equation $\frac{dP}{dt} = kP(1 - \frac{P}{M})$ has two parameters to determine. First, for very small populations where $P \ll M$, the term $(1 - \frac{P}{M}) \approx 1$, so the equation simplifies to $\frac{dP}{dt} \approx kP$. The relative growth rate is $\frac{1}{P}\frac{dP}{dt} = k$, so $k = 0.04$. Second, the growth rate $\frac{dP}{dt}$ reaches its maximum when its derivative equals zero. Taking the derivative: $\frac{d}{dP}\left[kP(1 - \frac{P}{M})\right] = k(1 - \frac{2P}{M}) = 0$. This gives $P = \frac{M}{2}$. Since maximum growth occurs at $P = 50,000$, we have $\frac{M}{2} = 50,000$, so $M = 100,000$. Choice A incorrectly sets $M = 50,000$, confusing the maximum growth population with the carrying capacity. Choice B makes this same error while also incorrectly doubling $k$. Choice C correctly identifies $M = 100,000$ but wrongly doubles $k$ to 0.08, possibly misinterpreting the relative growth rate. The correct answer is D: $k = 0.04$ and $M = 100,000$. Remember: in logistic growth, maximum growth rate occurs at half the carrying capacity, and the intrinsic growth rate $k$ equals the relative growth rate for small populations.

Question 5

1. Population A will initially grow faster because its carrying capacity $M$ is larger, providing more room for growth.
2. Population B will initially grow faster because its intrinsic growth rate $k$ is larger, which is the dominant factor for small populations. (correct answer)
3. Both populations will initially grow at the same rate because their initial population sizes are identical.
4. Population A has a higher initial growth rate, but Population B will reach its carrying capacity in less time.

Explanation: The initial growth rate is found by evaluating $dP/dt$ at $t=0$ (i.e., $P=50$). For Population A: $\frac{dP_A}{dt} = 0.1(50)(1 - \frac{50}{1000}) = 5(1 - 0.05) = 4.75$. For Population B: $\frac{dP_B}{dt} = 0.2(50)(1 - \frac{50}{800}) = 10(1 - 0.0625) = 9.375$. Since $9.375 > 4.75$, Population B has a higher initial growth rate. This is because for populations that are small relative to their carrying capacity ($P \ll M$), the term $(1 - P/M)$ is close to 1, so the growth is approximately exponential: $dP/dt \approx kP$. In this regime, the larger intrinsic growth rate $k$ leads to faster growth.

Question 6

1. 300
2. 600
3. 9 (correct answer)
4. 18

Explanation: The model is a logistic equation, which can be written in the standard form $\frac{dP}{dt} = kP(1 - \frac{P}{M})$. Factoring the given equation, we get $\frac{dP}{dt} = 0.06P(1 - \frac{0.0001}{0.06}P) = 0.06P(1 - \frac{P}{600})$. From this, we identify the intrinsic growth rate $k=0.06$ and the carrying capacity $M=600$. The maximum growth rate for a logistic model occurs when the population is at half its carrying capacity, $P = M/2 = 300$. To find this maximum rate, we substitute $P=300$ into the original differential equation: $\frac{dP}{dt} |_{P=300} = 0.06(300) - 0.0001(300)^2 = 18 - 0.0001(90000) = 18 - 9 = 9$ The maximum growth rate is 9 individuals per hour. Alternatively, the maximum growth rate is given by the formula $kM/4 = (0.06)(600)/4 = 36/4 = 9$.

Question 7

1. It is the maximum sustainable yield, the highest harvesting rate for which a stable, non-zero population can be maintained. (correct answer)
2. It is the harvesting rate that maintains the fish population at its original carrying capacity of 1,000 thousand.
3. It is the natural growth rate of the population when it has been reduced to half its original carrying capacity.
4. It is the harvesting rate that causes the population to stabilize at exactly 25 thousand fish.

Explanation: The natural growth of the fish population is governed by the logistic term $G(P) = 0.1P(1 - \frac{P}{1000})$. For the population to be sustainable, the harvesting rate $H$ must not exceed the population's natural growth rate. The maximum growth rate, which corresponds to the maximum sustainable yield (MSY), occurs at the vertex of the parabola $G(P)$, which is at $P = M/2 = 1000/2 = 500$. The value of this maximum rate is $G(500) = 0.1(500)(1 - \frac{500}{1000}) = 50(0.5) = 25$. If the harvesting rate $H$ exceeds this value, $H > 25$, then $\frac{dP}{dt}$ will be negative for all $P$, leading to extinction. Thus, $H=25$ is the maximum sustainable yield.

Question 8

1. 0.1
2. 0.2
3. 0.4 (correct answer)
4. 10000

Explanation: The differential equation can be rewritten by factoring out $P$: $\frac{dP}{dt} = P \left[ 0.4(1 - \frac{P}{10000}) - h \right] = P \left[ (0.4 - h) - \frac{0.4}{10000}P \right]$ This is a logistic-type equation with an effective intrinsic growth rate of $k' = (0.4 - h)$. For the population to have any capacity for growth when its size is small, this effective growth rate must be positive. If $k' \le 0$, then $\frac{dP}{dt}$ will be negative for all $P > 0$, leading to extinction. The critical point is when $k' = 0$, which is $0.4 - h = 0$, or $h = 0.4$. For any value of $h > 0.4$, the term $(0.4-h)$ is negative, and since the $-P^2$ term is also negative, $\frac{dP}{dt}$ will always be negative, causing the population to decline to zero.

Question 9

1. $M$ represents a threshold; if the initial population is below $M$, it dies out.
2. $M$ no longer represents a stable carrying capacity, as populations above $M$ grow infinitely.
3. $M$ is a stable carrying capacity, and the population grows fastest at $M/2$, same as the standard model.
4. $M$ is a stable carrying capacity, but the population grows fastest at a level greater than $M/2$. (correct answer)

Explanation: When analyzing modified differential equation models, you need to examine equilibrium points and their stability, then determine where growth rates are maximized. For this modified logistic model $\frac{dP}{dt} = kP(1 - (\frac{P}{M})^2)$, let's find the equilibrium points by setting the derivative equal to zero. This occurs when $P = 0$ or when $1 - (\frac{P}{M})^2 = 0$, giving us $P = M$ as the non-trivial equilibrium. Since the growth rate is positive when $P < M$ and negative when $P > M$, populations are attracted to $M$, confirming it's a stable carrying capacity. To find the maximum growth rate, we take the derivative: $\frac{d^2P}{dt^2} = k(1 - (\frac{P}{M})^2) - k P \cdot \frac{2P}{M^2} = k(1 - 3(\frac{P}{M})^2)$. Setting this equal to zero gives $P = \frac{M}{\sqrt{3}} \approx 0.577M$, which is greater than $M/2 = 0.5M$. Answer A is wrong because populations below $M$ actually grow toward $M$, not die out. Answer B incorrectly claims $M$ isn't stable—we showed populations above $M$ decrease back toward it. Answer C correctly identifies $M$ as stable but wrongly assumes the inflection point occurs at $M/2$ like in the standard logistic model. The correct answer is D: $M$ remains a stable carrying capacity, but the squared term in the model shifts the maximum growth rate to occur above $M/2$. Study tip: When analyzing modified population models, always check equilibrium stability and recalculate inflection points—don't assume they match standard models.

Question 10

1. $26.5°C$ because the larger temperature difference increases the cooling rate proportionally
2. $27.8°C$ because the cooling follows the same exponential pattern with adjusted parameters
3. $28.6°C$ because the cooling constant remains the same but the driving force changes (correct answer)
4. $29.4°C$ because the temperature difference affects only the initial rate, not the overall pattern

Explanation: First, find $k$ from the original scenario: $T(t) = T_a + (T_0 - T_a)e^{-kt} = 20 + 17e^{-kt}$. At $t = 1$: $32 = 20 + 17e^{-k}$, so $e^{-k} = 12/17$, giving $k = \ln(17/12) \approx 0.347$. For $T_a = 15°C$: $T(t) = 15 + 22e^{-kt}$. At $t = 1$: $T(1) = 15 + 22 \times (12/17) = 15 + 264/17 ≈ 28.6°C$. Choice A incorrectly assumes linear proportionality. Choice B uses an incorrect adjustment method. Choice D wrongly assumes the driving force only affects initial conditions.

Question 11

1. The equilibrium is unstable, and perturbations grow exponentially with rate $0.1$ while oscillating with period $\frac{2\pi}{0.3}$ (correct answer)
2. The system exhibits neutral stability with oscillations of increasing amplitude at frequency $0.3$ radians per unit time
3. The real part indicates the average growth rate of the system's energy function over one complete oscillatory cycle
4. The equilibrium is a stable focus with perturbations decaying at rate $0.1$ while maintaining oscillatory motion

Explanation: For complex eigenvalues $\alpha ± \beta i$, the real part $\alpha$ determines stability: $\alpha > 0$ means unstable (growing), $\alpha < 0$ means stable (decaying), $\alpha = 0$ means neutrally stable. Here $\alpha = 0.1 > 0$, so the equilibrium is unstable. Perturbations grow like $e^{0.1t}$ while oscillating with frequency $0.3$, giving period $T = 2\pi/0.3$. Choice B incorrectly suggests neutral stability (would need $\alpha = 0$). Choice C misinterprets the meaning of the real part. Choice D incorrectly calls it stable (would need $\alpha < 0$).

Question 12

1. The ratio is $2^{1.5} \approx 2.83$ because the second substance decays more slowly
2. The ratio is $2^{0.75} \approx 1.68$ because half-life relationships are exponential
3. The ratio is $e^{1.0395} \approx 2.83$ because the decay constants differ by a factor of 2 (correct answer)
4. The ratio is $4$ because after 3 years, twice the half-life difference compounds quadratically

Explanation: For the first substance: $\lambda_1 = 0.693$, so $t_{1/2} = \ln(2)/0.693 = 1$ year. The second substance has $t_{1/2} = 2$ years, so $\lambda_2 = \ln(2)/2 = 0.3465$. After 3 years: $N_1(3) = N_0 e^{-0.693 \times 3} = N_0 e^{-2.079}$ and $N_2(3) = N_0 e^{-0.3465 \times 3} = N_0 e^{-1.0395}$. The ratio is $\frac{N_2(3)}{N_1(3)} = e^{-1.0395 + 2.079} = e^{1.0395} \approx 2.83$. Choice A gets the right answer but uses incorrect reasoning about powers of 2. Choice B miscalculates the time ratios. Choice D assumes a quadratic relationship that doesn't exist.

Question 13

1. Since $R_{eff} = 0.5 < 1$, the infection will decline exponentially with rate $\gamma - \beta S = 0.05 \text{ day}^{-1}$
2. Since $R_{eff} = 0.5 < 1$, the infection will decline exponentially with rate $\gamma(1 - R_{eff}) = 0.05 \text{ day}^{-1}$
3. Since $R_{eff} = 0.5 < 1$, each infected person will infect $0.5$ others on average before recovering
4. Since $R_{eff} = 5 > 1$, the infection will grow exponentially with rate $\beta S - \gamma = 0.4 \text{ day}^{-1}$ (correct answer)

Explanation: $R_{eff} = \frac{\beta S}{\gamma} = \frac{0.0001 \times 5000}{0.1} = \frac{0.5}{0.1} = 5$. Since $R_{eff} = 5 > 1$, the epidemic grows. The growth rate is $\frac{dI}{dt}/I = \beta S - \gamma = 0.5 - 0.1 = 0.4 \text{ day}^{-1}$. Choice A incorrectly calculates $R_{eff} = 0.5$. Choice B has the same calculation error but correct formula structure. Choice C has the wrong $R_{eff}$ value and misinterprets its meaning ($R_{eff}$ is the effective reproduction number under current conditions, not the basic reproduction number).