Calculus 2 Quiz: Logistic Models

20 questions · exam conditions

0:00

Logistic ModelsQuestion 1 of 20

A rumor spreads through a school of 1200 students according to the logistic model $\frac{dR}{dt} = 0.002R(1200 - R)$ , where $R(t)$ is the number of students who have heard the rumor at time $t$ days. If 20 students initially know the rumor, approximately how long will it take for 90% of the school to hear it?

2.1 days

2.8 days

3.4 days

4.1 days

← Back to quizzes

Calculus 2 Quiz

Calculus 2 Quiz: Logistic Models

Practice Logistic Models in Calculus 2 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, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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

2.1 days
2.8 days
3.4 days (correct answer)
4.1 days

Explanation: The logistic solution is

R(t) = \frac{1200}{1 + Ae^{-kt}}

where

k = 0.002 \cdot 1200 = 2.4

. From

R(0) = 20

20 = \frac{1200}{1+A}

, so

A = 59

. We want

R(t) = 0.9 \times 1200 = 1080

. Setting up:

1080 = \frac{1200}{1 + 59e^{-2.4t}}

. Solving:

1 + 59e^{-2.4t} = \frac{1200}{1080} = \frac{10}{9}

, so

59e^{-2.4t} = \frac{1}{9}

, giving

e^{-2.4t} = \frac{1}{531}

. Therefore

t = \frac{\ln(531)}{2.4} \approx \frac{6.27}{2.4} \approx 3.4

days. Choice A uses 95% instead of 90%. Choice B results from computational error in the exponential. Choice D confuses the calculation with 95% coverage.

Question 2

In a logistic growth model, the relative growth rate is defined as $\frac{1}{P}\frac{dP}{dt}$ . For the equation $\frac{dP}{dt} = 0.02P\left(1 - \frac{P}{800}\right)$ , at what population level is the relative growth rate equal to 75% of its maximum value?

$P = 100$
$P = 150$
$P = 200$ (correct answer)
$P = 250$

Explanation: The relative growth rate is

\frac{1}{P}\frac{dP}{dt} = \frac{1}{P} \cdot 0.02P\left(1 - \frac{P}{800}\right) = 0.02\left(1 - \frac{P}{800}\right)

. The maximum relative growth rate occurs when

P = 0

, giving a maximum of 0.02. We want 75% of this maximum:

0.02\left(1 - \frac{P}{800}\right) = 0.75 \times 0.02 = 0.015

. Solving:

1 - \frac{P}{800} = 0.75

, so

\frac{P}{800} = 0.25

, which gives

P = 200

. Choice A represents 50% of maximum relative growth rate. Choice B results from incorrect percentage calculation. Choice D corresponds to 62.5% of maximum relative growth rate.

Question 3

A population $P(t)$ is modeled by the logistic differential equation $\frac{dP}{dt} = 0.05P(1 - \frac{P}{1000})$ . At what population size is the population growing most rapidly, and what is this maximum rate?

The population grows most rapidly at a size of 500, with a rate of 12.5 individuals per unit time. (correct answer)
The population grows most rapidly at a size of 1000, with a rate of 0 individuals per unit time.
The population grows most rapidly at a size of 500, with a rate of 25 individuals per unit time.
The population grows most rapidly at a size of 1000, with a rate of 50 individuals per unit time.

Explanation: The logistic equation is

\frac{dP}{dt} = kP(1 - \frac{P}{L})

, where

L

is the carrying capacity. Here,

L=1000

. The population growth rate is maximized when the population is at half the carrying capacity, so

P = L/2 = 1000/2 = 500

. To find the maximum rate, substitute

P=500

into the differential equation:

\frac{dP}{dt} = 0.05(500)(1 - \frac{500}{1000}) = 25(1 - 0.5) = 25(0.5) = 12.5

Question 4

The rate of change of a population $y$ is modeled by the differential equation $\frac{dy}{dt} = 3y - 0.01y^2$ . What is the carrying capacity of the population, and at what population size is the growth rate maximized?

The carrying capacity is 300, and the growth rate is maximized when the population is 150. (correct answer)
The carrying capacity is 3, and the growth rate is maximized when the population is 1.5.
The carrying capacity is 150, and the growth rate is maximized when the population is 75.
The carrying capacity is 300, and the growth rate is maximized when the population is 300.

Explanation: To find the carrying capacity, we need to write the equation in the standard form

\frac{dy}{dt} = ky(1 - \frac{y}{L})

\frac{dy}{dt} = ky(L-y)

. Factoring the given equation:

\frac{dy}{dt} = y(3 - 0.01y) = 0.01y(300 - y)

. Comparing this to the form

\frac{dy}{dt} = k'y(L-y)

, we can identify the carrying capacity

L = 300

. The growth rate is maximized at half the carrying capacity, which is

y = L/2 = 300/2 = 150

Question 5

The population of a species in a protected habitat is given by the logistic model $P(t) = \frac{1200}{1 + 29e^{-0.2t}}$ . For what value of $t$ is the population growing fastest?

$t = 5\ln(29)$ (correct answer)
$t = \ln(29)$
$t = 5\ln(2)$
$t = 0$

Explanation: The population grows fastest at the point of inflection, which occurs when the population

P(t)

is at half its carrying capacity. From the model

P(t) = \frac{L}{1 + Ae^{-kt}}

, the carrying capacity is

L = 1200

. The population is growing fastest when

P(t) = L/2 = 600

. We set the population function equal to 600 and solve for

t

600 = \frac{1200}{1 + 29e^{-0.2t}} \implies 1 + 29e^{-0.2t} = 2 \implies 29e^{-0.2t} = 1 \implies e^{-0.2t} = \frac{1}{29} \implies -0.2t = \ln(\frac{1}{29}) = -\ln(29) \implies t = \frac{\ln(29)}{0.2} = 5\ln(29)

Question 6

The spread of a flu virus in a community with a carrying capacity of 2000 people is modeled by the logistic equation $\frac{dP}{dt} = kP(2000-P)$ , where $P(t)$ is the number of infected people. If a group of 2500 infected people moves into the community at $t=0$ , what is the long-term behavior of $P(t)$ ?

$P(t)$ will decrease and approach 2000. (correct answer)
$P(t)$ will increase and approach 2500.
$P(t)$ will decrease and approach 0.
$P(t)$ will increase indefinitely.

Explanation: The carrying capacity is

L = 2000

. The initial population is

P(0) = 2500

. Since the initial population is greater than the carrying capacity (

P_0 > L

), the term

(2000-P)

will be negative. This makes

\frac{dP}{dt}

negative, so the population will decrease over time. The population will approach the stable equilibrium, which is the carrying capacity

L=2000

, from above.

Question 7

The growth of a yeast culture is modeled by the function $P(t) = \frac{600}{1+19e^{-0.5t}}$ . Let $t_{1/2}$ be the time it takes for the population to reach half of its carrying capacity, and let $t_{3/4}$ be the time it takes to reach three-quarters of its carrying capacity. What is the value of $t_{3/4} - t_{1/2}$ ?

$2\ln(3)$ (correct answer)
$2\ln(2)$
$2\ln(19/3)$
$\ln(3)$

Explanation: The carrying capacity is

L=600

. Half capacity is

P=300

, and three-quarters capacity is

P=450

. First, find

t_{1/2}

by setting

P(t_{1/2}) = 300

300 = \frac{600}{1+19e^{-0.5t_{1/2}}} \implies 1+19e^{-0.5t_{1/2}} = 2 \implies e^{-0.5t_{1/2}} = 1/19 \implies t_{1/2} = 2\ln(19)

. Next, find

t_{3/4}

by setting

P(t_{3/4}) = 450

450 = \frac{600}{1+19e^{-0.5t_{3/4}}} \implies 1+19e^{-0.5t_{3/4}} = 600/450 = 4/3 \implies 19e^{-0.5t_{3/4}} = 1/3 \implies e^{-0.5t_{3/4}} = 1/57 \implies t_{3/4} = 2\ln(57)

. The difference is

t_{3/4} - t_{1/2} = 2\ln(57) - 2\ln(19) = 2(\ln(57) - \ln(19)) = 2\ln(57/19) = 2\ln(3)

Question 8

The number of students at a university who have heard a rumor is modeled by $N(t) = \frac{5000}{1+499e^{-0.8t}}$ . How many students had heard the rumor initially, and how many had heard it when the rumor was spreading the fastest?

Initially 10 students; 5000 students when spreading fastest.
Initially 1 student; 2500 students when spreading fastest.
Initially 499 students; 5000 students when spreading fastest.
Initially 10 students; 2500 students when spreading fastest. (correct answer)

Explanation: This question involves a logistic growth model, which commonly appears when studying population dynamics or diffusion processes. When you see this type of function, you need to find two key points: the initial value and the inflection point where growth is fastest. To find the initial number of students, substitute

t = 0

into

N(t) = \frac{5000}{1+499e^{-0.8t}}

N(0) = \frac{5000}{1+499e^{0}} = \frac{5000}{1+499(1)} = \frac{5000}{500} = 10

For logistic models, the rumor spreads fastest at the inflection point, which occurs when

N(t)

equals half the carrying capacity (the horizontal asymptote). As

t \to \infty

e^{-0.8t} \to 0

, so

N(t) \to \frac{5000}{1+0} = 5000

. Therefore, the rumor spreads fastest when

N(t) = \frac{5000}{2} = 2500

students have heard it. Looking at the wrong answers: Choice A correctly identifies 10 initial students but incorrectly claims maximum spread occurs at 5000 students—this is actually the final equilibrium, not the fastest growth point. Choice B gets the inflection point right (2500) but wrongly calculates the initial value as 1 student. Choice C incorrectly uses 499 (the coefficient from the denominator) as the initial value and mistakes the carrying capacity for the inflection point. Study tip: For logistic models

\frac{L}{1+ae^{-kt}}

, remember that initial value comes from substituting

t=0

, the carrying capacity is

L

, and fastest growth occurs at

L/2

Question 9

For the logistic model $\frac{dP}{dt} = kP(L-P)$ with $k>0$ and $L>0$ , there are two equilibrium points, $P=0$ and $P=L$ . Considering initial populations $P_0 > 0$ , which statement correctly describes the stability of these equilibria?

Both $P=0$ and $P=L$ are unstable equilibria.
$P=0$ is a stable equilibrium and $P=L$ is an unstable equilibrium.
Both $P=0$ and $P=L$ are stable equilibria.
$P=0$ is an unstable equilibrium and $P=L$ is a stable equilibrium. (correct answer)

Explanation: When analyzing equilibrium stability in differential equations, you need to examine what happens to solutions when they start near each equilibrium point. For the logistic equation

\frac{dP}{dt} = kP(L-P)

, think about the sign of the derivative on either side of each equilibrium. At

P = 0

: If you start with a small positive population

P_0 > 0

, then

P > 0

and

(L-P) > 0

since

P < L

. This makes

\frac{dP}{dt} = kP(L-P) > 0

, so the population grows away from zero. Since solutions move away from

P = 0

, this equilibrium is unstable. At

P = L

: If you start slightly below

L

, then

P > 0

and

(L-P) > 0

, so

\frac{dP}{dt} > 0

and the population increases toward

L

. If you start slightly above

L

, then

P > 0

but

(L-P) < 0

, making

\frac{dP}{dt} < 0

, so the population decreases toward

L

. Since solutions from both sides converge to

P = L

, this equilibrium is stable. Choice A incorrectly claims both equilibria are unstable. Choice B reverses the stability—it suggests populations would decay to zero, contradicting the growth away from

P = 0

. Choice C incorrectly claims both are stable, missing that

P = 0

repels nearby solutions. Choice D correctly identifies the behavior:

P = 0

is unstable (populations grow away) and

P = L

is stable (the carrying capacity that populations approach). Study tip: For logistic models, remember that zero population is always unstable with positive growth rates, while the carrying capacity acts as a stable attractor.

Question 10

A population $P(t)$ follows a logistic model $\frac{dP}{dt} = kP(1-\frac{P}{1000})$ . The carrying capacity is 1000. At $t=0$ , the population is 100. At $t=5$ , the population is 250. Which expression represents the growth constant $k$ ?

$5\ln(3)$
$\frac{1}{5}\ln(9/7)$
$\frac{1}{5}\ln(7/3)$
$\frac{1}{5}\ln(3)$ (correct answer)

Explanation: When you encounter a logistic growth problem, you're working with a differential equation that models population growth with environmental limits. The key is solving for the unknown parameter using the given data points. The logistic equation

\frac{dP}{dt} = kP(1-\frac{P}{1000})

has the general solution

P(t) = \frac{1000}{1 + Ae^{-kt}}

, where

A

is determined by initial conditions. With

P(0) = 100

, we get

100 = \frac{1000}{1 + A}

, so

A = 9

. This gives us

P(t) = \frac{1000}{1 + 9e^{-kt}}

. Using the second condition

P(5) = 250

250 = \frac{1000}{1 + 9e^{-5k}}

. Solving:

1 + 9e^{-5k} = 4

, so

9e^{-5k} = 3

, which means

e^{-5k} = \frac{1}{3}

. Taking the natural logarithm:

-5k = \ln(\frac{1}{3}) = -\ln(3)

. Therefore,

k = \frac{1}{5}\ln(3)

. Choice A gives

5\ln(3)

, which would result from incorrectly multiplying by 5 instead of dividing. Choice B represents

\frac{1}{5}\ln(9/7)

, which might come from algebraic errors when manipulating the fraction

\frac{1000}{250}

. Choice C gives

\frac{1}{5}\ln(7/3)

, possibly from confusing the relationship between the exponential terms. The correct answer is D:

\frac{1}{5}\ln(3)

. Remember: with logistic growth problems, always set up the general solution first, then use both given conditions systematically. Don't rush the algebra—small errors in fraction manipulation are common traps in these problems.

Question 11

A rumor spreads according to a logistic model in a school of 1600 students. The rumor starts with 10 students. After 2 days, 40 students have heard the rumor. At what time $t$ (in days) is the rumor spreading the fastest?

$2\frac{\ln(800)}{\ln(40)}$
$\frac{\ln(159)}{\ln(39)}$
$\frac{2\ln(159)}{\ln(159) - \ln(39)}$ (correct answer)
$2\frac{\ln(159)}{\ln(13/53)}$

Explanation: When you encounter a logistic growth problem asking when something is "spreading fastest," you're looking for the maximum rate of change, which occurs at the inflection point of the logistic curve. The logistic model is

P(t) = \frac{L}{1 + Ae^{-kt}}

where L is the carrying capacity. Here, L = 1600 students. Using the initial conditions: P(0) = 10 gives us

10 = \frac{1600}{1 + A}

, so A = 159. With P(2) = 40, we get

40 = \frac{1600}{1 + 159e^{-2k}}

, which simplifies to

1 + 159e^{-2k} = 40

, giving us

e^{-2k} = \frac{39}{159}

and

k = \frac{\ln(159) - \ln(39)}{2}

. For logistic growth, the maximum rate occurs when the population reaches half the carrying capacity (800 students). Setting

800 = \frac{1600}{1 + 159e^{-kt}}

gives us

1 + 159e^{-kt} = 2

, so

e^{-kt} = \frac{1}{159}

. Therefore

kt = \ln(159)

, and

t = \frac{\ln(159)}{k} = \frac{2\ln(159)}{\ln(159) - \ln(39)}

. Choice A incorrectly uses ln(800) and ln(40). Choice B uses the wrong denominator and is missing the factor of 2. Choice D has an incorrect denominator with ln(13/53), which doesn't relate to our problem parameters. Study tip: In logistic growth problems, maximum rate always occurs at half the carrying capacity. Set up your equations systematically using the given data points, and remember that the inflection point is your target.

Question 12

The number of people infected by a disease follows a logistic curve. It takes 10 days for the number of infected people to grow from 10% to 50% of the carrying capacity. Due to the symmetry of the logistic curve, approximately how many more days will it take for the number of infected people to grow from 50% to 90% of the carrying capacity?

12.5 days
8 days
10 days (correct answer)
16 days

Explanation: When you encounter logistic growth problems, remember that the logistic curve has a distinctive S-shape with perfect symmetry around its inflection point, which occurs at exactly 50% of the carrying capacity. The key insight here is understanding what "symmetry" means for a logistic curve. The curve is symmetric about the point where the population reaches 50% of carrying capacity. This means the time it takes to go from any percentage below 50% to 50% equals the time to go from 50% to the corresponding percentage above 50%. Since 10% and 90% are equidistant from 50% (both are 40 percentage points away), the symmetry property tells us that the time from 10% to 50% must equal the time from 50% to 90%. Given that it takes 10 days to grow from 10% to 50%, it will take exactly 10 more days to grow from 50% to 90%. Looking at the wrong answers: (A) 12.5 days might tempt you if you think the growth slows down significantly in the upper portion, but this ignores the symmetry property. (B) 8 days could result from incorrectly thinking the growth accelerates as it approaches carrying capacity. (D) 16 days represents a common error of assuming exponential-like behavior where growth becomes increasingly difficult. Study tip: For logistic growth questions, always identify the inflection point (50% of carrying capacity) and use the symmetry property. Equal distances from 50% correspond to equal time intervals—this is the most powerful tool for solving these problems quickly.

Question 13

The logistic equation $\frac{dP}{dt} = kP(1-\frac{P}{L})$ is a nonlinear differential equation. However, it can be transformed into a linear differential equation using the substitution $y = 1/P$ . What is the resulting linear differential equation for $y$ ?

$\frac{dy}{dt} + \frac{k}{L}y = k$
$\frac{dy}{dt} - ky = -\frac{k}{L}$
$\frac{dy}{dt} + ky = \frac{k}{L}$ (correct answer)
$\frac{dy}{dt} - ky = \frac{k}{L}$

Explanation: When you encounter a nonlinear differential equation like the logistic equation, substitution methods can often transform it into a more manageable linear form. The key is carefully applying the chain rule when making the substitution. Starting with

\frac{dP}{dt} = kP(1-\frac{P}{L})

and the substitution

y = \frac{1}{P}

, we need to find

\frac{dy}{dt}

. Since

y = P^{-1}

, using the chain rule:

\frac{dy}{dt} = -P^{-2}\frac{dP}{dt} = -\frac{1}{P^2}\frac{dP}{dt}

From our substitution,

P = \frac{1}{y}

, so

P^2 = \frac{1}{y^2}

. Therefore:

\frac{dy}{dt} = -y^2\frac{dP}{dt}

Substituting the original logistic equation:

\frac{dy}{dt} = -y^2 \cdot kP(1-\frac{P}{L}) = -y^2 \cdot k \cdot \frac{1}{y}(1-\frac{1}{yL}) = -ky(1-\frac{1}{yL})

Expanding:

\frac{dy}{dt} = -ky + \frac{k}{L}

Rearranging:

\frac{dy}{dt} + ky = \frac{k}{L}

This confirms answer C is correct. Answer A has the wrong sign on the

ky

term and wrong constant term. Answer B has both terms with incorrect signs compared to our result. Answer D has the correct constant term but the wrong sign on the

ky

term. Remember: when using substitutions in differential equations, always apply the chain rule carefully and track your signs. The goal is transforming a nonlinear equation into the standard linear form

\frac{dy}{dt} + p(t)y = q(t)

Question 14

Consider two distinct populations, A and B, both governed by logistic growth. Population A is modeled by $\frac{dP_A}{dt} = 0.1P_A(1-\frac{P_A}{100})$ and Population B by $\frac{dP_B}{dt} = 0.2P_B(1-\frac{P_B}{100})$ . If both populations start with the same initial value $P_0=10$ , which statement accurately compares their behavior?

Population A has a higher maximum growth rate than Population B.
Population B reaches its inflection point at an earlier time than Population A. (correct answer)
Both populations approach their carrying capacity at the same rate.
Population A will have a larger carrying capacity than Population B.

Explanation: When you encounter logistic growth problems, focus on identifying the key parameters and understanding how they affect population dynamics over time. Both populations follow the standard logistic model

\frac{dP}{dt} = rP(1-\frac{P}{K})

where

r

is the intrinsic growth rate and

K

is the carrying capacity. For both populations,

K = 100

, but Population A has

r = 0.1

while Population B has

r = 0.2

. The inflection point occurs when the population reaches half its carrying capacity (

P = K/2 = 50

). At this point, growth rate is maximized. Since Population B has twice the intrinsic growth rate of Population A, it will reach any given population level faster, including the inflection point at

P = 50

. This makes answer B correct. Let's examine why the other options are wrong. Answer A is backwards: the maximum growth rate occurs at the inflection point and equals

\frac{rK}{4}

. For Population A:

\frac{0.1 \times 100}{4} = 2.5

. For Population B:

\frac{0.2 \times 100}{4} = 5.0

. Population B has the higher maximum growth rate. Answer C is incorrect because the rate of approach to carrying capacity depends on the intrinsic growth rate

r

– Population B approaches faster. Answer D is wrong because both populations have identical carrying capacities of 100. Remember: in logistic growth comparisons, a higher intrinsic growth rate means faster progression through all phases of growth, including reaching the inflection point earlier, while the carrying capacity depends only on the environmental limits, not the growth rate.

Question 15

Euler's method with a step size of $\Delta t = 1$ is used to approximate the solution to the logistic differential equation $\frac{dP}{dt} = 0.5P(1-\frac{P}{100})$ . If the initial population is $P_0 = 20$ , what is the approximate population $P_2$ after two steps?

36.00
38.08 (correct answer)
28.00
39.20

Explanation: When you encounter Euler's method problems, you're using a numerical technique to approximate solutions to differential equations by taking small linear steps. For the logistic equation

\frac{dP}{dt} = 0.5P(1-\frac{P}{100})

, you'll apply the formula

P_{n+1} = P_n + \Delta t \cdot f(P_n)

where

f(P) = 0.5P(1-\frac{P}{100})

. Starting with

P_0 = 20

and

\Delta t = 1

: Step 1: Calculate

f(20) = 0.5(20)(1-\frac{20}{100}) = 0.5(20)(0.8) = 8

P_1 = 20 + 1(8) = 28

Step 2: Calculate

f(28) = 0.5(28)(1-\frac{28}{100}) = 0.5(28)(0.72) = 10.08

P_2 = 28 + 1(10.08) = 38.08

Answer A (36.00) likely comes from computational errors in the second step or rounding mistakes. Answer C (28.00) represents stopping after just one step instead of completing both required iterations. Answer D (39.20) suggests an arithmetic error, possibly miscalculating the growth rate in the second step or confusing decimal operations. The correct answer is B (38.08). Study tip: In Euler's method problems, organize your work step-by-step and double-check each calculation. The logistic model's growth rate decreases as population approaches carrying capacity, so verify that your growth increments are getting smaller with each step. Always count your iterations carefully to ensure you've completed the requested number of steps.

Question 16

For the logistic differential equation $\frac{dP}{dt} = kP(L-P)$ , the second derivative is $\frac{d^2P}{dt^2} = k(L-2P)\frac{dP}{dt}$ . If a population is at $P = L/4$ , which statement accurately describes the population's growth at that moment?

The growth rate is positive and decreasing.
The growth rate is positive and increasing. (correct answer)
The growth rate is at its maximum value.
The growth rate is negative and decreasing.

Explanation: When analyzing logistic differential equations, you need to understand both the growth rate (first derivative) and how that rate is changing (second derivative). The logistic model describes population growth that starts slowly, accelerates, then slows as it approaches a carrying capacity. At

P = L/4

, let's examine what's happening. First, substitute into the original equation:

\frac{dP}{dt} = kP(L-P) = k \cdot \frac{L}{4} \cdot (L - \frac{L}{4}) = k \cdot \frac{L}{4} \cdot \frac{3L}{4} = \frac{3kL^2}{16}

. Since

k > 0

and

L > 0

, this is positive, so the population is growing. Now check the second derivative:

\frac{d^2P}{dt^2} = k(L-2P)\frac{dP}{dt} = k(L - 2 \cdot \frac{L}{4})\frac{dP}{dt} = k(L - \frac{L}{2})\frac{dP}{dt} = k \cdot \frac{L}{2} \cdot \frac{dP}{dt}

. Since both terms are positive, the second derivative is positive, meaning the growth rate itself is increasing. Choice A is wrong because while the growth rate is positive, it's increasing, not decreasing. Choice C is incorrect because maximum growth occurs at

P = L/2

(where the second derivative equals zero), not at

P = L/4

. Choice D is wrong because the growth rate is positive, not negative. The answer is B: the growth rate is positive and increasing. Study tip: In logistic growth, the inflection point at

P = L/2

divides two phases: accelerating growth below

L/2

and decelerating growth above

L/2

. Always check both derivatives to understand the complete behavior.

Question 17

The harvesting of a fish population is modeled by $\frac{dP}{dt} = 0.08P(1-\frac{P}{1000}) - H$ , where $H$ is the constant harvest rate. The maximum sustainable yield (MSY) is the largest value of $H$ for which there is a stable equilibrium population. What is this maximum sustainable yield?

40
20 (correct answer)
1000
80

Explanation: When you encounter a maximum sustainable yield problem, you're looking for the harvest rate that allows the largest possible catch while maintaining a stable fish population. This requires finding where the population growth rate equals the harvest rate at equilibrium. At equilibrium,

\frac{dP}{dt} = 0

, so we have

0.08P(1-\frac{P}{1000}) = H

. This means

H = 0.08P - 0.08\frac{P^2}{1000}

. To find the maximum sustainable yield, you need to maximize this function with respect to

P

. Taking the derivative:

\frac{dH}{dP} = 0.08 - \frac{0.16P}{1000} = 0.08 - 0.00016P

. Setting this equal to zero:

0.08 = 0.00016P

, which gives

P = 500

. Substituting back:

H = 0.08(500)(1-\frac{500}{1000}) = 40(0.5) = 20

. Choice A (40) represents the value you'd get if you forgot to multiply by the factor

(1-\frac{P}{1000})

when

P = 500

. Choice C (1000) is the carrying capacity of the population, not the harvest rate. Choice D (80) would result from incorrectly using the full carrying capacity in your calculation, giving

0.08(1000) = 80

. For sustainable yield problems, remember that the maximum occurs at exactly half the carrying capacity. The population that produces maximum growth rate is always at the midpoint of the logistic curve, making calculations more predictable once you recognize this pattern.

Question 18

The solution to a logistic equation is given by $P(t) = \frac{L}{1+Ae^{-kt}}$ . Consider the ratio of the population to the 'remaining capacity,' defined as $R(t) = \frac{P(t)}{L-P(t)}$ . Which statement correctly describes $R(t)$ ?

$R(t)$ grows logistically with time.
$R(t)$ approaches a constant value as $t \to \infty$ .
$R(t)$ grows exponentially with time. (correct answer)
$R(t)$ decreases exponentially with time.

Explanation: When you encounter logistic growth problems, pay close attention to how transformations of the basic logistic function behave—they often reveal exponential patterns hidden within the S-curve. Let's work with the given ratio

R(t) = \frac{P(t)}{L-P(t)}

. Substituting the logistic function

P(t) = \frac{L}{1+Ae^{-kt}}

: First, find

L - P(t) = L - \frac{L}{1+Ae^{-kt}} = \frac{L \cdot Ae^{-kt}}{1+Ae^{-kt}}

Now compute the ratio:

R(t) = \frac{\frac{L}{1+Ae^{-kt}}}{\frac{L \cdot Ae^{-kt}}{1+Ae^{-kt}}} = \frac{L}{L \cdot Ae^{-kt}} = \frac{1}{Ae^{-kt}} = \frac{e^{kt}}{A}

This shows

R(t) = \frac{1}{A}e^{kt}

, which is pure exponential growth. Answer C is correct because

R(t)

grows exponentially with rate constant

k

. Answer A is wrong because logistic growth has an S-shaped curve that levels off, while

R(t)

increases without bound. Answer B is incorrect because

R(t)

doesn't approach a constant—it grows indefinitely as

t \to \infty

. Answer D is wrong because

R(t)

increases rather than decreases, and the exponential has a positive exponent

kt

. Study tip: When analyzing ratios or transformations of logistic functions, substitute the full expression and simplify algebraically. The logistic function often contains hidden exponential behavior that becomes apparent through mathematical manipulation.

Question 19

The solution to the logistic differential equation $\frac{dP}{dt} = kP(L-P)$ is of the form $P(t) = \frac{L}{1+Ae^{-Lkt}}$ . For a fish population with carrying capacity $L=800$ and initial population $P(0)=100$ , which equation must be solved to find the time $t$ when the population reaches 600?

$600 = \frac{800}{1+7e^{-800kt}}$ (correct answer)
$600 = \frac{800}{1+7e^{-kt}}$
$600 = 100e^{800kt}$
$400 = \frac{800}{1+7e^{-800kt}}$

Explanation: First, we determine the constant

A

using the initial condition

P(0)=100

and carrying capacity

L=800

. The formula for

A

A = \frac{L-P_0}{P_0}

. So,

A = \frac{800-100}{100} = \frac{700}{100} = 7

. This gives the specific solution

P(t) = \frac{800}{1+7e^{-800kt}}

. To find the time when the population reaches 600, we set

P(t) = 600

, which results in the equation

600 = \frac{800}{1+7e^{-800kt}}

Question 20

The growth rate of a population, $\frac{dP}{dt}$ , is plotted against the population size, $P$ . The resulting graph for a logistic model is a parabola. If the parabola has its vertex at the point $(500, 20)$ , which differential equation describes this model?

$\frac{dP}{dt} = 0.08P(1-\frac{P}{500})$
$\frac{dP}{dt} = 0.04P(1-\frac{P}{1000})$
$\frac{dP}{dt} = 0.08P(1-\frac{P}{1000})$ (correct answer)
$\frac{dP}{dt} = 0.04P(1-\frac{P}{500})$

Explanation: When you encounter a logistic model question involving a parabolic graph of growth rate versus population, you need to understand how the standard logistic equation creates this relationship and what the vertex tells you about the model's parameters. The logistic differential equation

\frac{dP}{dt} = rP(1-\frac{P}{K})

can be rewritten as

\frac{dP}{dt} = rP - \frac{r}{K}P^2

. This is a quadratic function in P, which explains why plotting

\frac{dP}{dt}

versus P gives you a parabola opening downward. The vertex at (500, 20) provides two crucial pieces of information. The x-coordinate (500) tells you where maximum growth occurs, which happens at

P = \frac{K}{2}

for any logistic model. Therefore,

\frac{K}{2} = 500

, so

K = 1000

. The y-coordinate (20) gives you the maximum growth rate:

\frac{rK}{4} = 20

. Substituting K = 1000:

\frac{r(1000)}{4} = 20

, which gives you

r = 0.08

. This confirms that the correct equation is

\frac{dP}{dt} = 0.08P(1-\frac{P}{1000})

, which is option C. Option A has the wrong carrying capacity (K = 500 instead of 1000). Option B has the wrong growth rate (r = 0.04 instead of 0.08). Option D combines both errors, using both the wrong carrying capacity and wrong growth rate. Remember: in logistic models, maximum growth always occurs at half the carrying capacity, and you can find both parameters directly from the vertex coordinates of the parabolic growth rate graph.