AP Calculus BC Quiz: Logistic Models With Differential Equations

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Logistic Models With Differential EquationsQuestion 1 of 20

A bacteria culture follows $\frac{dy}{dt}=ky\left(1-\frac{y}{5000}\right)$ with $k>0$ . If $y(0)=6000$ , what happens to $y$ as $t\to\infty$ ?

y\to 0

y\to 5000

y\to 6000

y\to\infty

y

oscillates around

5000

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AP Calculus BC Quiz

AP Calculus BC Quiz: Logistic Models With Differential Equations

Practice Logistic Models With Differential Equations in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Logistic Models With Differential Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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.

All questions

Question 1

A bacteria culture follows $\frac{dy}{dt}=ky\left(1-\frac{y}{5000}\right)$ with $k>0$ . If $y(0)=6000$ , what happens to $y$ as $t\to\infty$ ?

$y\to 0$
$y\to 5000$ (correct answer)
$y\to 6000$
$y\to\infty$
$y$ oscillates around $5000$

Explanation: This question tests logistic model reasoning by determining the long-term behavior of the population given an initial condition above the carrying capacity. The carrying capacity

K = 5000

is the stable equilibrium that the population approaches as

t

increases, regardless of whether it starts below or above

K

for

y > 0

. When

y(0) = 6000 > K

, the growth rate

\frac{dy}{dt}

is negative because

1 - \frac{y}{5000} < 0

, causing

y

to decrease toward

K

. The inflection point at

y = \frac{K}{2} = 2500

indicates where the concavity changes, but since

y

starts above

K

, it will pass through this point while decreasing. A tempting distractor is

y \to 6000

, but this ignores that

\frac{dy}{dt} < 0

for

y > K

, so it cannot stabilize at the initial value. A transferable logistic-model strategy is to analyze the sign of

\frac{dy}{dt}

relative to the equilibria

y = 0

and

y = K

to predict asymptotic behavior.

Question 2

A lab culture satisfies $\frac{dy}{dt}=ky\left(1-\frac{y}{400}\right)$ with $k>0$ . If $y=400$ , what is $\frac{dy}{dt}$ ?

$400k$
$k$
$0$ (correct answer)
$-400k$
$-k$

Explanation: 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.

Question 3

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?

$y(t)$ increases without bound
$y(t)$ decreases toward $0$
$y(t)$ decreases toward $5000$ (correct answer)
$y(t)$ stays constant at $6000$
$y(t)$ oscillates around $5000$

Explanation: 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.

Question 4

A fish population satisfies $\frac{dy}{dt}=0.4y\left(1-\frac{y}{800}\right)$ . Which statement best interprets the carrying capacity?

The population approaches $800$ as $t$ increases, and growth slows near $800$ . (correct answer)
The population approaches $0$ as $t$ increases, and growth slows near $0$ .
The population approaches $0.4$ as $t$ increases, and growth slows near $0.4$ .
The population increases without bound as $t$ increases, since the rate is proportional to $y$ .
The population approaches $320$ as $t$ increases, and growth stops at $320$ .

Explanation: This question tests understanding of logistic model parameters and carrying capacity interpretation. In the logistic differential equation dy/dt = ry(1 - y/K), the parameter K represents the carrying capacity—the maximum sustainable population. Here, we have dy/dt = 0.4y(1 - y/800), so the carrying capacity is 800. As t increases, the population approaches this carrying capacity of 800, and the growth rate slows as y gets closer to 800 because the factor (1 - y/800) approaches zero. Choice D incorrectly assumes exponential growth by ignoring the limiting factor (1 - y/800). The key strategy for logistic models is to identify K in the standard form dy/dt = ry(1 - y/K) as the long-term population limit.

Question 5

A population follows $\frac{dy}{dt}=0.25y\left(1-\frac{y}{200}\right)$ . Which value of $y$ makes $\frac{dy}{dt}<0$ ?

$y=150$
$y=200$
$y=0$
$y=250$ (correct answer)
$y=100$

Explanation: 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

K

and above

K

) to determine where

\frac{dy}{dt}

is positive or negative, predicting population trends.

Question 6

A deer population satisfies $\frac{dy}{dt}=0.2y\left(1-\frac{y}{1200}\right)$ . At what population is $\frac{dy}{dt}$ maximized?

$y=0$
$y=240$
$y=600$ (correct answer)
$y=1200$
$y=1400$

Explanation: This question tests finding the inflection point of logistic growth. For

dy/dt = 0.2y(1 - y/1200)

, we need to maximize the growth rate function

f(y) = 0.2y(1 - y/1200) = 0.2y - 0.2 y^2 / 1200

. Taking the derivative:

f'(y) = 0.2 - 0.4y/1200 = 0.2 - y/3000

. Setting

f'(y) = 0

gives

y = 600

, which is exactly half the carrying capacity of 1200. This inflection point at

y = K/2

is where the population transitions from accelerating growth (concave up) to decelerating growth (concave down). Choice D incorrectly suggests the maximum occurs at the carrying capacity, where growth actually equals zero. The universal rule for logistic models is that maximum growth rate always occurs at half the carrying capacity.

Question 7

A city's population $y$ satisfies $\frac{dy}{dt}=0.03y\left(1-\frac{y}{2,000,000}\right)$ . Which statement is correct?

The population approaches $2,000,000$ as $t\to\infty$ (correct answer)
The population approaches $0.03$ as $t\to\infty$
The population grows linearly at a constant rate
The population must decrease whenever $y<2,000,000$
The population reaches a maximum at $y=0.03$

Explanation: 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.

Question 8

A population satisfies $\frac{dy}{dt}=0.3y\left(1-\frac{y}{900}\right)$ . Which statement about concavity is correct for $0<y<900$ ?

The solution is concave down for all $0<y<900$ .
The solution is concave up for all $0<y<900$ .
The solution changes from concave up to concave down at $y=450$ . (correct answer)
The solution changes from concave down to concave up at $y=900$ .
Concavity cannot be determined from the differential equation.

Explanation: 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.

Question 9

An invasive plant satisfies $\frac{dy}{dt}=0.6y\left(1-\frac{y}{300}\right)$ with $y(0)=50$ . Which is true about $y(t)$ ?

$y(t)$ decreases to $0$ because the factor $1-\frac{y}{300}$ is negative initially.
$y(t)$ increases and approaches $300$ as $t$ increases. (correct answer)
$y(t)$ increases without bound because $\frac{dy}{dt}$ is proportional to $y$ .
$y(t)$ approaches $0.6$ as $t$ increases because $k=0.6$ .
$y(t)$ oscillates about $300$ because the differential equation is nonlinear.

Explanation: 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.

Question 10

Bacteria follow $\frac{dy}{dt}=ky\left(1-\frac{y}{5000}\right)$ with $k>0$ . For which $y$ is the growth rate negative?

$0<y<5000$
$y=2500$
$y>5000$ (correct answer)
$y=5000$
$y<0$

Explanation: This question requires analyzing when logistic growth becomes negative. In the logistic equation dy/dt = ky(1 - y/5000) with k > 0, the growth rate's sign depends on the factor (1 - y/5000). When y < 5000, this factor is positive, making dy/dt > 0 (growth). When y = 5000, the factor equals zero, making dy/dt = 0 (equilibrium). When y > 5000, the factor becomes negative, making dy/dt < 0 (decline). Choice A incorrectly identifies the region where growth is positive, not negative. For logistic models, populations above the carrying capacity always experience negative growth as they return to equilibrium.

Question 11

A yeast population satisfies $\frac{dy}{dt}=0.5y\left(1-\frac{y}{600}\right)$ . For which $y$ does the population increase?

$y>600$
$0<y<600$ (correct answer)
$y=600$
$y<0$
$y=0$

Explanation: 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.

Question 12

A rabbit population follows $\frac{dy}{dt}=ky\left(1-\frac{y}{L}\right)$ with $k>0$ . If $y(0)>L$ , what happens?

$y(t)$ increases without bound because $y(0)$ is large.
$y(t)$ remains constant at $y(0)$ for all $t$ .
$y(t)$ decreases toward $L$ as $t$ increases. (correct answer)
$y(t)$ decreases toward $0$ as $t$ increases.
$y(t)$ oscillates and crosses $L$ repeatedly.

Explanation: This question examines logistic behavior when starting above carrying capacity. With dy/dt = ky(1 - y/L) where k > 0 and y(0) > L, we analyze the dynamics. Since y > L, the factor (1 - y/L) is negative, making dy/dt = ky(1 - y/L) < 0 because k > 0 and y > 0. This negative growth rate causes the population to decrease. As y decreases toward L, the factor (1 - y/L) approaches zero from below, causing dy/dt to approach zero, so y(t) asymptotically approaches L from above. Choice D incorrectly suggests the population approaches zero rather than the carrying capacity. For logistic models, populations always converge to the carrying capacity L, whether starting above or below it.

Question 13

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?

$\frac{dy}{dt}>0$ so $y$ increases without bound
$\frac{dy}{dt}<0$ so $y$ decreases toward $300$ (correct answer)
$\frac{dy}{dt}=0.1$ so $y$ increases linearly
$\frac{dy}{dt}=0$ so $y$ stays constant at its current value
$\frac{dy}{dt}$ is undefined so the model no longer applies

Explanation: 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.

Question 14

A fish population satisfies $\frac{dy}{dt}=0.4y\left(1-\frac{y}{800}\right)$ . Which value represents the carrying capacity of the lake?

$0.4$
$800$ (correct answer)
$\frac{1}{800}$
$320$
$\infty$

Explanation: 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.

Question 15

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?

$y=200$
$y=100$ (correct answer)
$y=0.6$
$y=400$
$y=0$

Explanation: 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.

Question 16

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 $t=0$ ?

$y(0)=1300$
$y(0)=1200$
$y(0)=0$
$y(0)=900$ (correct answer)
$y(0)=-50$

Explanation: 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

= 0.2900(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 >

,

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.

Question 17

A rabbit population satisfies $\frac{dy}{dt}=0.5y\left(1-\frac{y}{1000}\right)$ . If $y(0)=200$ , what is the sign of $\frac{d^2y}{dt^2}$ initially?

Positive, because growth is accelerating when $y<500$ (correct answer)
Negative, because growth is always decelerating in logistic models
Zero, because $y(0)\ne 0$ implies constant growth
Positive, because $y(0)>1000$ implies increasing rate
Negative, because $y(0)<0$ implies decreasing rate

Explanation: This question tests logistic model reasoning by determining the sign of the second derivative to assess concavity initially. The carrying capacity

K = 1000

sets the limit, with the inflection point at

y = K/2 = 500

dividing the growth phases. Below

y = 500

\frac{d^2 y}{dt^2} > 0

, meaning accelerating growth (concave up), as seen by differentiating

\frac{dy}{dt}

. For

y(0) = 200 < 500

, this holds initially. A tempting distractor is negative, assuming always decelerating, but logistic models accelerate below the inflection point. A transferable logistic-model strategy is to compute

\frac{d^2 y}{dt^2} = r (1 - y/K) - r y/K \cdot \frac{dy}{dt}

or recognize concavity changes at

K/2

for qualitative analysis.

Question 18

A population follows $\frac{dy}{dt}=ky\left(1-\frac{y}{L}\right)$ with $k>0$ . Which statement correctly describes equilibria of the model?

The only equilibrium is $y=L$ , and it is unstable
The equilibria are $y=0$ and $y=L$ , with $y=L$ stable for $y>0$ (correct answer)
The equilibria are $y=0$ and $y=k$ , with $y=k$ stable
The equilibria are $y=0$ and $y=1/L$ , both stable
There are no equilibria because $\frac{dy}{dt}$ depends on $t$

Explanation: 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.

Question 19

A deer herd satisfies $\frac{dy}{dt}=0.18y\left(1-\frac{y}{1200}\right)$ . For which $y$ is the growth rate greatest?

$y=1200$
$y=0$
$y=600$ (correct answer)
$y=300$
$y=900$

Explanation: 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.

Question 20

A population satisfies $\frac{dy}{dt}=0.5y\left(1-\frac{y}{1000}\right)$ . If $y(0)=100$ , which describes $y(t)$ initially?

Decreasing because $y<1000$
Increasing because $0<y<1000$ (correct answer)
Constant because $y$ is far from $1000$
Increasing without bound because $0.5>0$
Oscillating because the model is nonlinear

Explanation: 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.