Logistic Models with Differential Equations - AP Calculus BC

A lab culture satisfies $\frac{dy}{dt}=ky\left(1-\frac{y}{400}\right)$ with $k>0$ . If $y=400$ , what is $\frac{dy}{dt}$ ?
Answer: $0$ This question tests logistic model reasoning by evaluating the growth rate at the carrying capacity. The carrying capacity K=400 is where dy/dt=0, as the term (1 - y/K) becomes zero, halting population change. The inflection point at y=200 would show maximum growth, but at y=400, resources are fully utilized, leading to equilibrium. In logistic equations, substituting y=K directly yields dy/dt=0, confirming stability. A tempting distractor like choice A, 400k, fails because it ignores the (1 - y/K)=0 factor, mistakenly applying only the r y part. A transferable strategy for logistic models is to set dy/dt=0 and solve for y to find equilibria, then evaluate stability based on the sign of dy/dt around those points.
A yeast population satisfies $\frac{dy}{dt}=0.5y\left(1-\frac{y}{600}\right)$ . For which $y$ does the population increase?
Answer: $0<y<600$ This question analyzes when logistic populations increase. For dy/dt = 0.5y(1 - y/600), the sign of dy/dt determines whether the population increases (dy/dt > 0) or decreases (dy/dt < 0). Since 0.5 and y are positive for living populations, the sign depends on (1 - y/600). When 0 < y < 600, this factor is positive, making dy/dt > 0 (population increases). When y = 600, dy/dt = 0 (equilibrium). When y > 600, the factor is negative, making dy/dt < 0 (population decreases). Choice A incorrectly identifies the decreasing region as the increasing region. The key principle is that logistic populations increase when below carrying capacity and decrease when above it.
An invasive plant area $y$ (m $^2$ ) satisfies $\frac{dy}{dt}=0.1y\left(1-\frac{y}{300}\right)$ . When $y>300$ , which statement is true?
Answer: $\frac{dy}{dt}<0$ so $y$ decreases toward $300$ This question tests logistic model reasoning by analyzing behavior when the population exceeds the carrying capacity. The carrying capacity K = 300 is the stable equilibrium where dy/dt = 0, and for y > K, the term (1 - y/300) < 0, making dy/dt < 0 so y decreases toward K. This reflects overcrowding reducing growth. The inflection point at y = 150 is below K, but for y > K, the trajectory is decreasing without passing through acceleration. A tempting distractor is dy/dt > 0 leading to unbounded growth, but this ignores the negative sign for y > K in logistic models. A transferable logistic-model strategy is to evaluate the sign of dy/dt in regions divided by equilibria (y=0 and y=K) to predict qualitative behavior.
A deer herd is modeled by $\frac{dy}{dt}=0.2y\left(1-\frac{y}{1200}\right)$ . For which initial value $y(0)$ is the herd increasing at $t=0$ ?
Answer: $y(0)=900$ This question tests logistic model reasoning by evaluating the initial growth direction based on the sign of $dy/dt$ at t=0. The carrying capacity $K$ = 1200 represents the long-term population limit where $dy/dt$ = 0, and populations between 0 and $K$ grow positively toward it. For y(0) = 900, which is between 0 and 1200, $dy/dt$ = 0.2*900*(1-900/1200) > 0, indicating initial increase. The inflection point at y = $K$ /2 = 600 is where growth is fastest, but the sign of $dy/dt$ determines increase or decrease regardless. A tempting distractor is y(0)=1300, but for y > $K$ , $dy/dt$ < 0, so the population decreases initially. A transferable logistic-model strategy is to plug initial conditions into $dy/dt$ and check its sign to determine if the population is increasing, decreasing, or stable.
A fish population $y$ satisfies $\frac{dy}{dt}=0.4y\left(1-\frac{y}{800}\right)$ . Which value is the carrying capacity?
Answer: $800$ This question tests logistic model reasoning by identifying the carrying capacity in the differential equation. The carrying capacity, denoted as K, is the maximum population size that the environment can sustain indefinitely, where the growth rate becomes zero when y equals K. In the logistic equation dy/dt = r y (1 - y/K), K appears in the term (1 - y/K), limiting growth as y approaches K. The inflection point occurs at y = K/2, where the population growth rate is maximized before slowing down due to resource constraints. A tempting distractor like choice A, 0.4, represents the intrinsic growth rate r, not the carrying capacity. A transferable strategy for logistic models is to always identify K as the value in the denominator of the (1 - y/K) term, which sets the upper limit for population growth.
An invasive plant satisfies $\frac{dy}{dt}=0.6y\left(1-\frac{y}{300}\right)$ with $y(0)=50$ . Which is true about $y(t)$ ?
Answer: $y(t)$ increases and approaches $300$ as $t$ increases. This question examines logistic model behavior with initial conditions below carrying capacity. Given dy/dt = 0.6y(1 - y/300) with y(0) = 50, we analyze the solution's trajectory. Since y(0) = 50 < 300 (the carrying capacity), the factor (1 - y/300) is positive, making dy/dt > 0, so the population increases. As y approaches 300, the factor (1 - y/300) approaches zero, causing dy/dt to approach zero, which means y(t) levels off at 300. The population cannot exceed 300 because dy/dt would become negative above this value. Choice C incorrectly assumes exponential growth by ignoring the limiting factor. For logistic models starting below carrying capacity, populations always increase toward and asymptotically approach the carrying capacity.
A population follows $\frac{dy}{dt}=ky\left(1-\frac{y}{L}\right)$ with $k>0$ . Which statement correctly describes equilibria of the model?
Answer: The equilibria are $y=0$ and $y=L$ , with $y=L$ stable for $y>0$ This question tests logistic model reasoning by identifying equilibria and their stability. The carrying capacity L is the stable equilibrium where $dy/dt = 0$ for $y = L$ , attracting populations from above or below for $y > 0$ . The other equilibrium $y = 0$ is unstable, as small positive perturbations grow away from it. The inflection point at $y = L/2$ separates regions of accelerating and decelerating growth toward L. A tempting distractor is that $y = L$ is unstable, but stability analysis shows $dy/dt > 0$ for $0 < y < L$ and $< 0$ for $y > L$ , confirming attraction to L. A transferable logistic-model strategy is to set $dy/dt = 0$ to find equilibria, then test stability by checking the sign of $dy/dt$ around them.
A lab population is modeled by $\frac{dy}{dt}=0.8y\left(1-\frac{y}{1500}\right)$ . Which value of $y$ makes $\frac{dy}{dt}=0$ ?
Answer: $y=0$ or $y=1500$ This question tests logistic model reasoning by finding values where the growth rate is zero. The carrying capacity $K = 1500$ is one equilibrium solution where $\frac{dy}{dt} = 0$ , representing the stable long-term population. Setting $\frac{dy}{dt} = 0$ gives $y(1 - \frac{y}{1500}) = 0$ , so $y = 0$ or $y = 1500$ . The inflection point at $y = 750$ is not an equilibrium but where growth maximizes. A tempting distractor is $y=1500$ only, but $y=0$ is also an equilibrium, though unstable. A transferable logistic-model strategy is to factor the right-hand side of $\frac{dy}{dt}$ to solve for equilibria, then confirm by substitution.
A deer herd satisfies $\frac{dy}{dt}=0.18y\left(1-\frac{y}{1200}\right)$ . For which $y$ is the growth rate greatest?
Answer: $y=600$ This question tests logistic model reasoning by finding where the population growth rate is greatest. The carrying capacity $K=1200$ is the upper limit where $\frac{dy}{dt}=0$ , stabilizing the population at that level. The inflection point, where the growth rate $\frac{dy}{dt}$ is maximized, occurs at $y = K/2$ , as this is halfway to the carrying capacity when competition is minimal. At $y = K/2$ , the product $y(1 - \frac{y}{K})$ reaches its maximum value, leading to the peak growth rate. A tempting distractor like choice A, $y=1200$ , fails because at the carrying capacity, $\frac{dy}{dt}=0$ , not the maximum. A transferable strategy for logistic models is to calculate the inflection point as $K/2$ to locate where growth accelerates most rapidly before environmental limits take effect.
A population satisfies $\frac{dy}{dt}=0.6y\left(1-\frac{y}{200}\right)$ . At what population size is the growth rate $\frac{dy}{dt}$ maximized?
Answer: $y=100$ This question tests logistic model reasoning by finding the population size where the growth rate dy/dt is maximized. The carrying capacity K = 200 is the upper limit where dy/dt = 0, and the growth rate peaks at the inflection point y = K/2 due to the quadratic nature of the logistic equation. To find the maximum, take d/dt(dy/dt) = 0, which yields y = 100, confirming the inflection point. At y = 100, the model shifts from concave up to concave down, marking the transition in growth acceleration. A tempting distractor is y=200, but at K, dy/dt=0, which is the minimum growth rate, not the maximum. A transferable logistic-model strategy is to remember that maximum growth occurs at half the carrying capacity in standard logistic models, verifiable by differentiating dy/dt.
A lab culture follows $\frac{dy}{dt}=1.5y\left(1-\frac{y}{200}\right)$ . Which value of $y$ is a stable equilibrium?
Answer: $y=200$ This question tests identifying stable equilibria in logistic models. For $dy/dt = 1.5y(1 - \frac{y}{200})$ , equilibrium points occur where $dy/dt = 0$ , which happens when $y = 0$ or $y = 200$ . To determine stability, we analyze the sign of $dy/dt$ near each equilibrium: for $0 < y < 200$ , $dy/dt > 0$ (solutions move away from $y = 0$ toward $y = 200$ ), and for $y > 200$ , $dy/dt < 0$ (solutions move back toward $y = 200$ ). This makes $y = 200$ a stable equilibrium (attractor) and $y = 0$ an unstable equilibrium (repeller). Choice C incorrectly identifies the midpoint rather than the carrying capacity as the stable equilibrium. In logistic models, the carrying capacity $K$ is always the stable equilibrium that non-zero populations approach.
A city's number of electric scooters follows $\frac{dy}{dt}=0.3y\left(1-\frac{y}{10{,}000}\right)$ . What does $10{,}000$ represent in this model?
Answer: The carrying capacity (long‑term limiting number of scooters) This question tests logistic model reasoning by interpreting the parameters in the differential equation. The carrying capacity K = 10,000 is the long-term limiting value of y, where dy/dt approaches zero as y nears K due to resource constraints. This parameter models environmental limits, contrasting with unbounded exponential growth. The inflection point at y = K/2 = 5,000 is where the growth rate is highest and concavity changes. A tempting distractor is the growth constant 0.3, but this is r, not K, and affects speed, not the limit. A transferable logistic-model strategy is to identify K as the value that makes the (1 - y/K) term zero in the equation, representing the sustainable maximum.
A population satisfies $\frac{dy}{dt}=0.3y\left(1-\frac{y}{900}\right)$ . Which statement about concavity is correct for $0<y<900$ ?
Answer: The solution changes from concave up to concave down at $y=450$ . This question requires analyzing concavity changes in logistic solutions. For dy/dt = 0.3y(1 - y/900), concavity is determined by the second derivative d²y/dt². Using the chain rule: d²y/dt² = (d/dy)[0.3y(1 - y/900)] × dy/dt = 0.3(1 - 2y/900) × dy/dt. Since dy/dt > 0 for 0 < y < 900, the sign of d²y/dt² depends on (1 - 2y/900). When y < 450, this factor is positive (concave up); when y > 450, it's negative (concave down). The inflection point occurs at y = 450, exactly half the carrying capacity. Choice A incorrectly claims uniform concavity throughout the interval. The key insight is that logistic curves always have their inflection point at y = K/2, where growth transitions from accelerating to decelerating.
Bacteria follow $\frac{dy}{dt}=1.2y\left(1-\frac{y}{5000}\right)$ with $y(0)=6000$ . What happens to $y(t)$ as $t$ increases?
Answer: $y(t)$ decreases toward $5000$ This question tests logistic model reasoning by analyzing long-term behavior given an initial condition above the carrying capacity. The carrying capacity K=5000 represents the stable equilibrium where dy/dt=0, and populations above K decrease toward it due to resource limitations. The inflection point at y=K/2=2500 marks where growth would be fastest if starting below, but here y starts above K, so the population declines. In logistic models, if y(0) > K, the term (1 - y/K) is negative, making dy/dt negative and causing y to approach K from above. A tempting distractor like choice D fails because y=6000 is above K, so it decreases rather than staying constant. A transferable strategy for logistic models is to compare the initial y to K to predict whether the population will increase or decrease toward the carrying capacity.
A fish population satisfies $\frac{dy}{dt}=0.4y\left(1-\frac{y}{800}\right)$ . Which value represents the carrying capacity of the lake?
Answer: $800$ This question tests logistic model reasoning by identifying the carrying capacity from the differential equation. The carrying capacity, denoted as K, is the maximum population the environment can sustain long-term, where the growth rate becomes zero when $y = K$ . In the equation $\frac{dy}{dt} = 0.4 y \left(1 - \frac{y}{800}\right)$ , K = 800, as setting the term inside parentheses to zero gives $y = 800$ . The inflection point occurs at $y = \frac{K}{2} = 400$ , where the population growth transitions from accelerating to decelerating, marking the point of maximum growth rate. A tempting distractor is 0.4, but this represents the intrinsic growth rate r, not the carrying capacity. A transferable logistic-model strategy is to rewrite the equation in standard form $\frac{dy}{dt} = r y \left(1 - \frac{y}{K}\right)$ to directly extract the parameters r and K.
A population satisfies $\frac{dy}{dt}=0.8y\left(1-\frac{y}{1000}\right)$ . Which value is the carrying capacity $L$ ?
Answer: $L=1000$ This question tests identifying the carrying capacity from a logistic equation. The standard logistic form is $\frac{dy}{dt} = r y \left(1 - \frac{y}{L}\right)$ , where L is the carrying capacity. Comparing $\frac{dy}{dt} = 0.8 y \left(1 - \frac{y}{1000}\right)$ to this standard form, we see that $r = 0.8$ and $L = 1000$ . The carrying capacity represents the maximum sustainable population where $\frac{dy}{dt} = 0$ (when $y = L$ ). Choice A incorrectly identifies the growth rate coefficient as the carrying capacity. The strategy for finding carrying capacity is to rewrite the equation in standard form and identify the denominator in the fraction $\frac{y}{L}$ .
A population satisfies $\frac{dy}{dt}=0.5y\left(1-\frac{y}{1000}\right)$ . If $y(0)=100$ , which describes $y(t)$ initially?
Answer: Increasing because $0<y<1000$ This question tests logistic model reasoning by determining initial behavior based on position relative to the carrying capacity. The carrying capacity K=1000 sets the threshold where dy/dt=0, with populations below K growing toward it. The inflection point at y=500 marks the fastest growth, but since y(0)=100 <500, it's in the initial accelerating phase. For 0<y<K, dy/dt>0 because both y and (1 - y/K) are positive, indicating increase. A tempting distractor like choice A fails because y is below K, so it grows rather than decreases. A transferable strategy for logistic models is to analyze the sign of dy/dt in intervals defined by equilibria (0 and K) to predict increasing or decreasing trends.
A population satisfies $\frac{dy}{dt}=0.08y\left(1-\frac{y}{1500}\right)$ . At what population is the growth rate $0$ ?
Answer: At $y=0$ and $y=1500$ This question tests logistic model reasoning by identifying equilibrium points where growth rate is zero. The carrying capacity K=1500 is a stable equilibrium where dy/dt=0, and y=0 is an unstable equilibrium. The inflection point at y=750 is not an equilibrium but where concavity changes. Setting dy/dt=0 gives y=0 or y=K, the points of no population change. A tempting distractor like choice D fails because y=750 is the maximum growth point, not zero growth. A transferable strategy for logistic models is to factor the differential equation and set each factor to zero to find all equilibria, then classify them as stable or unstable.
A population follows $\frac{dy}{dt}=0.25y\left(1-\frac{y}{200}\right)$ . Which value of $y$ makes $\frac{dy}{dt}<0$ ?
Answer: $y=250$ This question tests logistic model reasoning by finding where the growth rate is negative. The carrying capacity $K=200$ divides regions: $\frac{dy}{dt} > 0$ for $0 < y < 200$ (growth) and $\frac{dy}{dt} < 0$ for $y > 200$ (decline toward K). The inflection point at $y=100$ is within the growth region, not affecting the sign. For $y=250 > 200$ , $(1 - \frac{y}{K}) < 0$ , so $\frac{dy}{dt} < 0$ . A tempting distractor like choice A, $y=150$ , fails because $150 < 200$ , making $\frac{dy}{dt} > 0$ , not negative. A transferable strategy for logistic models is to test values in intervals ( $0$ to $K$ and above $K$ ) to determine where $\frac{dy}{dt}$ is positive or negative, predicting population trends.
A city's population $y$ satisfies $\frac{dy}{dt}=0.03y\left(1-\frac{y}{2,000,000}\right)$ . Which statement is correct?
Answer: The population approaches $2,000,000$ as $t\to\infty$ This question tests logistic model reasoning by interpreting statements about the population's long-term behavior. The carrying capacity $K=2,000,000$ is the asymptotic limit that the population approaches as t goes to infinity, regardless of starting value above zero. The inflection point at $y=K/2=1,000,000$ is where the growth curve changes from concave up to concave down, marking the transition to slower growth. In logistic models, the term (1 - y/K) ensures bounded growth, preventing unlimited increase. A tempting distractor like choice C fails because growth is not linear; it's sigmoidal, slowing as y nears K. A transferable strategy for logistic models is to recognize that populations always approach K asymptotically for positive initial conditions, providing a reliable long-term prediction.

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