Differential Equations Quiz: Logistic Growth And Carrying Capacity

What this quiz covers

This quiz focuses on Logistic Growth And Carrying Capacity, 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. $N = 2.5 \times 10^7$ bacteria, where the population is 25% of the carrying capacity
2. $N = 5.0 \times 10^7$ bacteria, where the population is exactly half of the carrying capacity
3. $N = 1.5 \times 10^8$ bacteria, where the population exceeds the carrying capacity and growth becomes negative
4. $N \approx 1.46 \times 10^7$ or $N \approx 8.54 \times 10^7$ bacteria, representing two symmetric points around the inflection point (correct answer)

Explanation: Maximum growth rate occurs at N = K/2 = 5×10^7, giving dN/dt = 1.5×10^7. Half of this is 7.5×10^6. Setting 0.6N(1-N/10^8) = 7.5×10^6 and solving the quadratic: N = 1.46×10^7 or 8.54×10^7. Choice A gives 25% of max rate, not 50%. Choice B is where maximum rate occurs. Choice C involves negative growth and misses that there are two solutions where growth rate equals half the maximum.

Question 2

1. $h=500$ fish per year
2. $h=1000$ fish per year (correct answer)
3. $h=2000$ fish per year
4. $h=4000$ fish per year

Explanation: A steady-state population exists if $dP/dt = 0$ has real, positive solutions for $P$. The equation is $0.4P(1 - P/10000) - h = 0$. The term $G(P) = 0.4P(1 - P/10000)$ represents the natural growth rate of the population. To sustain a harvest, the harvest rate $h$ cannot exceed the maximum possible growth rate. The maximum of $G(P)$ occurs when $G'(P) = 0$. $G'(P) = 0.4 - 0.00008P = 0$, which gives $P = 5000$. This is half the carrying capacity, as expected. The maximum sustainable yield (MSY) is the growth rate at this population level: $h_{max} = G(5000) = 0.4(5000)(1 - 5000/10000) = 0.4(5000)(0.5) = 1000$. If $h > 1000$, $dP/dt$ will always be negative, leading to extinction.

Question 3

1. $t = 18$ years
2. $t = 20$ years (correct answer)
3. $t = 22.5$ years
4. $t = 25$ years

Explanation: The logistic growth curve is symmetric about its inflection point, which occurs at $P=K/2$. Here, $K=500$, so the inflection point occurs at $P=250$, which the problem states is reached at $t=10$ years. The initial population is $P(0)=50$. The target population is $P(t)=450$. Note that $450 = 500 - 50 = K - P(0)$. Due to the symmetry of the logistic curve around the inflection point $(t_{infl}, K/2)$, the time required to grow from $P_0$ to $K/2$ is the same as the time required to grow from $K/2$ to $K-P_0$. The time to grow from $P(0)=50$ to $P(10)=250$ is 10 years. Therefore, the time to grow from $P=250$ to $P=450$ is also 10 years. The total time is $10 + 10 = 20$ years.

Question 4

1. $r=0.1$
2. $r=0.2$
3. $r=0.4$ (correct answer)
4. $r=0.5$

Explanation: The logistic equation is $dP/dt = rP(1 - P/K)$. The population grows most rapidly at the inflection point, which occurs at $P = K/2$. The problem states this occurs at $P=400$, consistent with $K=800$. At this point, we are given that $dP/dt = 80$. We can substitute these values into the logistic equation to solve for $r$: $80 = r(400)(1 - 400/800)$. This simplifies to $80 = r(400)(1 - 0.5) = r(400)(0.5) = 200r$. Solving for $r$, we get $r = 80/200 = 0.4$.

Question 5

1. $dy/dt = ry(1-y)$ (correct answer)
2. $dy/dt = r(1-y)$
3. $dy/dt = (r/K)y(1-y)$
4. $dy/dt = ry(K-y)$

Explanation: We are given the substitution $y = P/K$. This implies $P = Ky$. We differentiate this with respect to $t$ to find $dP/dt$: $dP/dt = K(dy/dt)$. Now we substitute $P = Ky$ and $dP/dt = K(dy/dt)$ into the original logistic equation: $K(dy/dt) = r(Ky)(1 - Ky/K)$. Simplifying the term in the parenthesis gives: $K(dy/dt) = rKy(1 - y)$. Dividing both sides by the constant $K$ (assuming $K ≠ 0$) yields the dimensionless form of the logistic equation: $dy/dt = ry(1-y)$.

Question 6

1. $T_1 < T_2$
2. $T_1 > T_2$
3. $T_1 = T_2$ (correct answer)
4. The relationship depends on the value of the intrinsic growth rate $r$.

Explanation: The logistic growth curve $P(t)$ is symmetric about its inflection point, which occurs at time $t_{infl}$ when $P(t_{infl}) = K/2$. This symmetry means that for any time interval $\Delta t$, the population at $t_{infl} - \Delta t$ and $t_{infl} + \Delta t$ are symmetric with respect to the carrying capacity. Specifically, if $P(t_{infl} - \Delta t) = P_1$, then $P(t_{infl} + \Delta t) = K - P_1$. In this problem, the first interval ends at $K/2$ and starts at $P_1 = K/4$. The second interval starts at $K/2$ and ends at $P_2 = 3K/4$. Notice that $P_2 = 3K/4 = K - K/4 = K - P_1$. Because of the symmetry, the time taken to go from $P_1$ to $K/2$ must be the same as the time taken to go from $K/2$ to $K-P_1$. Therefore, $T_1 = T_2$.

Question 7

1. $h = 0.125$
2. $h = 0.25$ (correct answer)
3. $h = 0.375$
4. $h = 0.5$

Explanation: To find the equilibrium populations, we set $dP/dt = 0$. We can factor out $P$: $P[0.5(1-P/2000) - h] = 0$. This gives two equilibria: $P=0$ and the solution to $0.5(1-P/2000) - h = 0$. We are interested in the non-zero equilibrium. Solving for $P$: $0.5 - P/4000 - h = 0$, which gives $P/4000 = 0.5 - h$, so the new equilibrium is $P_{eq} = 4000(0.5 - h) = 2000 - 4000h$. We want this new equilibrium to be half of the original carrying capacity, so we set $P_{eq} = K/2 = 2000/2 = 1000$. $1000 = 2000 - 4000h$. Solving for $h$: $-1000 = -4000h$, so $h = 1000/4000 = 0.25$.

Question 8

1. $0$ (correct answer)
2. $0.4$
3. $0.8$
4. $4000$

Explanation: The differential equation is $dN/dt = N(0.8 - 0.0002N)$. The relative growth rate is $R(N) = (1/N)(dN/dt) = 0.8 - 0.0002N$. To find the carrying capacity $K$, we set $dN/dt=0$, which gives $N=0$ or $0.8 - 0.0002N = 0$, so $K = 0.8/0.0002 = 4000$. For a positive initial population, the population will approach the carrying capacity as $t \to \infty$, i.e., $\lim_{t\to\infty} N(t) = K = 4000$. We need to find the limit of the relative growth rate: $\lim_{t\to\infty} R(N(t)) = R(\lim_{t\to\infty} N(t)) = R(4000)$. Plugging $N=4000$ into the expression for the relative growth rate gives $R(4000) = 0.8 - 0.0002(4000) = 0.8 - 0.8 = 0$.

Question 9

1. The initial population is 19.
2. The population reaches its carrying capacity when $t=10$.
3. The intrinsic growth rate is 19.
4. The population grows fastest when $P=500$. (correct answer)

Explanation: When you encounter a logistic growth function, you need to identify its key parameters and understand what they represent. The standard form is $P(t) = \frac{K}{1 + Ae^{-rt}}$, where $K$ is the carrying capacity, $r$ is the intrinsic growth rate, and $A$ determines the initial condition. From the given function $P(t) = \frac{1000}{1 + 19e^{-0.1t}}$, you can identify: $K = 1000$ (carrying capacity), $r = 0.1$ (intrinsic growth rate), and $A = 19$. The population grows fastest at the inflection point, which occurs at exactly half the carrying capacity: $P = K/2 = 500$. This is where the curve transitions from accelerating to decelerating growth. Let's examine why each answer is wrong: A) The initial population occurs when $t = 0$: $P(0) = \frac{1000}{1 + 19} = 50$, not 19. The value 19 is just the coefficient $A$ in the exponential term. B) The population never truly reaches carrying capacity—it approaches 1000 asymptotically. At $t = 10$, $P(10) = \frac{1000}{1 + 19e^{-1}} \approx 130$, still far from the carrying capacity. C) The intrinsic growth rate is the coefficient in the exponent: $r = 0.1$, not 19. D) is correct because logistic populations always grow fastest at half their carrying capacity—this is a fundamental property of the logistic model. Study tip: For logistic growth problems, immediately identify $K$, $r$, and remember that maximum growth rate occurs at $P = K/2$. Don't confuse the coefficient $A$ with meaningful population parameters.

Question 10

1. $P = 80$ (correct answer)
2. $P = 100$
3. $P = 150$
4. $P = 210$

Explanation: First, identify the carrying capacity $K$. The equation can be written as $dP/dt = 0.02 \cdot 200 \cdot P(1 - P/200) = 4P(1-P/200)$. So, $K=200$. The population is 'growing' when $dP/dt > 0$, which occurs for $0 < P < 200$. The population grows 'at an increasing rate' when the solution curve $P(t)$ is concave up, meaning $d^2P/dt^2 > 0$. The inflection point, where the rate of growth is maximal and the concavity changes from up to down, occurs at $P = K/2 = 200/2 = 100$. The growth rate is increasing for $0 < P < 100$. Both conditions (growing and at an increasing rate) are met only when $0 < P < 100$. Among the choices, only $P=80$ satisfies this condition.

Question 11

1. $dP/dt = (\ln(4)/2) P(1 - P/100)$
2. $dP/dt = 9 P(1 - P/100)$
3. $dP/dt = (\ln(6)/2) P(1 - P/100)$ (correct answer)
4. $dP/dt = (\ln(6)/2) P(1 - P/40)$

Explanation: From the problem statement, the initial population is $P_0=10$ and the carrying capacity is $K=100$. The solution to the logistic equation is $P(t) = K / (1 + Ae^{-rt})$, where $A = (K-P_0)/P_0$. First, we find $A = (100 - 10)/10 = 9$. So, $P(t) = 100 / (1 + 9e^{-rt})$. We are given that $P(2) = 40$. We use this to find $r$: $40 = 100 / (1 + 9e^{-2r})$. Rearranging gives $1 + 9e^{-2r} = 100/40 = 2.5$, so $9e^{-2r} = 1.5$, and $e^{-2r} = 1.5/9 = 1/6$. Taking the natural logarithm of both sides, $-2r = \ln(1/6) = -\ln(6)$, so $r = \ln(6)/2$. The differential equation is $dP/dt = rP(1 - P/K)$, which is $dP/dt = (\ln(6)/2) P(1 - P/100)$.

Question 12

1. The growth rate will be -140 fish per year, indicating rapid population decline toward the new carrying capacity
2. The growth rate will be -35 fish per year, showing gradual adjustment to the reduced environmental capacity (correct answer)
3. The growth rate will be +85 fish per year, as the population is still below the original carrying capacity
4. The growth rate will be 0 fish per year, since the population must stabilize before adjusting to new conditions

Explanation: First find r: 120 = r(8000)(1-8000/10000) = r(8000)(0.2), so r = 0.0075. With new K = 6000 and P = 7000: dP/dt = 0.0075(7000)(1-7000/6000) = 52.5(-1/6) = -35. Choice A uses incorrect calculation methods. Choice C ignores that the new carrying capacity is what matters. Choice D incorrectly assumes growth stops during transitions.

Question 13

1. The growth rate is decreasing too rapidly, suggesting environmental factors are limiting growth more severely than predicted by logistic theory
2. The growth rate is not decreasing fast enough as the population approaches the carrying capacity, indicating possible model parameter errors
3. The time intervals are too short to establish a reliable logistic growth pattern, requiring longer-term data collection
4. The population is growing at a nearly constant rate rather than showing the characteristic S-curve deceleration expected in logistic growth (correct answer)

Explanation: The population increases by 25 individuals (2020-2022) then 20 individuals (2022-2024), showing minimal deceleration. For logistic growth with K=200, starting at P=50, we'd expect more significant deceleration by P=95. The growth appears nearly linear rather than logistic. Choice A suggests faster deceleration than observed. Choice B incorrectly states deceleration is too slow when it's actually too minimal. Choice C focuses on data collection rather than the pattern mismatch.

Question 14

1. $h = 20$ individuals per time unit, achieved when the population stabilizes at exactly 1000 individuals (correct answer)
2. $h = 40$ individuals per time unit, corresponding to the maximum intrinsic growth rate of the unharvested population
3. $h = 80$ individuals per time unit, representing the theoretical maximum before population collapse becomes inevitable
4. $h = 10$ individuals per time unit, providing a safety margin below the theoretical maximum sustainable yield

Explanation: For sustainable harvesting, dP/dt = 0, so h = 0.04P(1-P/2000). To find maximum h, differentiate: dh/dP = 0.04(1-P/1000) = 0, giving P = 1000. At P = 1000: h = 0.04(1000)(1-0.5) = 20. Choice B confuses maximum with carrying capacity relationship. Choice C incorrectly calculates using K instead of K/2. Choice D arbitrarily reduces the true maximum without justification.

Question 15

1. $P_1 \approx 400$ and $P_2 \approx 500$, with Species 1 dominating due to its higher intrinsic growth rate
2. $P_1 \approx 600$ and $P_2 \approx 320$, representing a stable coexistence where competition effects are balanced (correct answer)
3. $P_1 \approx 200$ and $P_2 \approx 650$, with Species 2 dominating despite its lower intrinsic growth rate
4. No stable coexistence equilibrium exists; one species will eventually exclude the other through competitive dominance

Explanation: For coexistence equilibrium, both dP₁/dt = 0 and dP₂/dt = 0: (P₁ + 0.8P₂)/1000 = 1 and (0.6P₁ + P₂)/800 = 1. This gives P₁ + 0.8P₂ = 1000 and 0.6P₁ + P₂ = 800. Solving simultaneously: P₁ = 600, P₂ = 320. Choice A uses incorrect calculation. Choice C reverses the dominance relationship. Choice D incorrectly concludes no equilibrium exists when the system actually has a stable coexistence point.

Question 16

1. The carrying capacity is exactly 1200, confirming the model parameters are correct
2. The carrying capacity must be greater than 1200 since the actual growth rate exceeds the predicted rate
3. The carrying capacity is approximately 960, indicating the model needs recalibration (correct answer)
4. The carrying capacity cannot be determined without additional population measurements over time

Explanation: Using the given model with P = 400: dP/dt = 0.08(400)(1 - 400/1200) = 32(2/3) ≈ 21.33. Since the actual growth rate is 20, not 21.33, the model parameters don't match reality. Working backwards: 20 = 0.08(400)(1 - 400/K), solving gives K ≈ 960. Choice A assumes the model is correct. Choice B incorrectly thinks we need a larger K when actually we need smaller. Choice D ignores that we can solve for K using the current data point.

Question 17

1. The modified model has a lower effective carrying capacity of 4800 instead of 5000 due to the Allee effect constraint
2. The modified model requires a minimum viable population of 200 individuals; populations below this threshold will decline to extinction (correct answer)
3. The modified model exhibits faster growth rates at intermediate population sizes due to enhanced cooperative breeding effects
4. The modified model shows oscillatory behavior around the carrying capacity rather than smooth exponential approach to equilibrium

Explanation: The Allee effect introduces a critical threshold at P = 200. For P < 200, (P/200 - 1) < 0, making dP/dt < 0, so the population declines. For P > 200, growth is positive until approaching K = 5000. This creates a minimum viable population threshold absent in standard logistic growth. Choice A incorrectly calculates carrying capacity. Choice C misunderstands that Allee effects generally reduce growth at low densities. Choice D describes dynamics not present in this deterministic model.