Logistic models with differential equations (BC only) — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The standard logistic differential equation form, equilibrium solutions, separation of variables for general/particular solutions, carrying capacity and intrinsic growth rate identification, growth rate analysis, concavity, inflection points, and contextual interpretation of logistic models.

You should already know: Separation of variables for first-order differential equations. Exponential growth modeling. Logarithm rules and partial fraction decomposition.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Logistic models with differential equations (BC only)?

Logistic differential equation models describe bounded growth of a quantity limited by finite resources, correcting the unrealistic unlimited growth of exponential growth models. This topic is exclusive to AP Calculus BC, appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and makes up 2-3% of total exam score per the official AP Calculus CED, within Unit 7: Differential Equations. Unlike exponential growth, which assumes a constant per-capita growth rate regardless of population size, logistic growth accounts for resource scarcity: as the population approaches a maximum sustainable size called the carrying capacity, the growth rate slows to zero. Logistic models are used to model population growth, spread of disease, adoption of new technology, and drug concentration in the bloodstream, among other real-world quantities that approach an upper limit over time. Most exam questions ask you to identify carrying capacity from the differential equation, solve for the explicit solution, analyze growth rate and concavity, or interpret results in context.

2. The Logistic Differential Equation and Equilibrium Solutions

The standard form of the logistic differential equation is: $$\frac{dP}{dt}=kP\left(1-\frac{P}{K}\right)$$ where $P(t)$ is the size of the quantity (e.g., population) at time $t$, $k>0$ is the intrinsic growth rate (the per-capita growth rate when the population is very small), and $K>0$ is the carrying capacity (the maximum sustainable size of $P$). To build intuition for this form: when $P$ is very small, $\frac{P}{K}\approx 0$, so $\frac{dP}{dt}\approx kP$, which matches exponential growth, as expected for a small population with abundant resources. When $P=K$, $\frac{dP}{dt}=0$, so the population stops growing. When $P>K$, $\frac{dP}{dt}<0$, so the population decreases toward $K$.

Equilibrium solutions are constant solutions where the growth rate $\frac{dP}{dt}=0$. For the logistic equation, the equilibria are $P=0$ (the trivial solution with no population) and $P=K$. $P=0$ is unstable: any small positive population will grow away from 0. $P=K$ is stable: any non-zero population will approach $K$ as $t\to \infty$. The growth rate $\frac{dP}{dt}$ is a quadratic function of $P$, so its maximum occurs at $P=\frac{K}{2}$, meaning the population grows fastest when it reaches half the carrying capacity.

Worked Example

Given the differential equation $\frac{dP}{dt}=0.3P\left(1-\frac{P}{1800}\right)$, (a) identify the carrying capacity and intrinsic growth rate, (b) find all equilibrium solutions and classify their stability, (c) find the value of $P$ that maximizes the growth rate.

1. Compare the given equation directly to standard form $\frac{dP}{dt}=kP\left(1-\frac{P}{K}\right)$. We immediately get $k=0.3$ (intrinsic growth rate) and $K=1800$ (carrying capacity).
2. Set $\frac{dP}{dt}=0$: $0.3P\left(1-\frac{P}{1800}\right)=0$, so solutions are $P=0$ and $P=1800$. For $0<P<1800$, $\frac{dP}{dt}$ is positive, so $P$ moves away from $0$ toward $1800$. For $P>1800$, $\frac{dP}{dt}$ is negative, so $P$ moves toward $1800$. Thus, $P=0$ is unstable, and $P=1800$ is stable.
3. The growth rate $\frac{dP}{dt}$ is a quadratic function of $P$ opening downward, with roots at $P=0$ and $P=1800$. The maximum of a quadratic occurs at the midpoint of its roots, so maximum growth rate occurs at $P=\frac{0+1800}{2}=900$.

Exam tip: Always rewrite the logistic equation to match the standard form where the leading term inside the parentheses is 1 before identifying K. Do not guess K from non-standard factored forms—this is the most common MCQ mistake.

3. Solving the Logistic Differential Equation

The logistic differential equation is separable, so we can use separation of variables and partial fraction decomposition to find an explicit solution for $P(t)$. Start with the standard equation and separate variables: $$\frac{dP}{P\left(1-\frac{P}{K}\right)}=kdt$$ Using partial fractions, the left-hand side simplifies to $\left(\frac{1}{P}+\frac{1}{K-P}\right)dP$. Integrate both sides: $$\ln |P|-\ln |K-P|=kt+C$$ Combine logarithms and exponentiate to get: $$\frac{P}{K-P}=A{e}^{kt}$$ where $A=e^C$ is a non-zero constant. Rearrange to solve for $P(t)$ to get the general solution: $$P(t)=\frac{KA{e}^{kt}}{1+A{e}^{kt}}$$ If we have an initial condition $P(0)=P_0$ (the population at time $t=0$), substitute $t=0$ to solve for $A=\frac{P_0}{K-P_0}$. Substitute back to get the standard particular solution: $$P(t)=\frac{K{P}_0}{P_0+(K-P_0)e^{-kt}}$$ As $t\to \infty$, the exponential term $e^{-kt}\to 0$, so $P(t)\to K$, which confirms the stable equilibrium at $P=K$.

Worked Example

A population grows according to $\frac{dP}{dt}=0.2P\left(1-\frac{P}{600}\right)$, with initial population $P(0)=150$. Find the particular solution $P(t)$.

1. Identify the parameters: $k=0.2$, $K=600$, $P_0=150$.
2. Substitute directly into the particular solution formula: $$P(t)=\frac{K{P}_0}{P_0+(K-P_0)e^{-kt}}=\frac{600\cdot 150}{150+(600-150)e^{-0.2t}}$$
3. Simplify the expression: $$P(t)=\frac{90000}{150+450e^{-0.2t}}=\frac{600}{1+3e^{-0.2t}}$$
4. Verify the initial condition: $P(0)=\frac{600}{1+3}=150$, which matches the given initial value, confirming no algebra errors.

Exam tip: Always check your particular solution by plugging in $t=0$ to confirm it gives $P_0$. This catches 90% of common algebra errors from rearranging terms to solve for $P(t)$.

4. Concavity and Inflection Points of Logistic Solutions

To analyze the shape of the logistic solution curve $P(t)$, we need the second derivative $\frac{d^{2}P}{d{t}^2}$ to find concavity and inflection points. Use the chain rule to differentiate $\frac{dP}{dt}$ with respect to $t$: $$\frac{d^{2}P}{d{t}^2}=\frac{d}{dt}\left(\frac{dP}{dt}\right)=\frac{d}{dP}\left(kP\left(1-\frac{P}{K}\right)\right)\cdot \frac{dP}{dt}=k\left(1-\frac{2P}{K}\right)\frac{dP}{dt}$$ For $0<P<K$, $\frac{dP}{dt}$ is positive, so the sign of $\frac{d^{2}P}{d{t}^2}$ matches the sign of $\left(1-\frac{2P}{K}\right)$. When $P<\frac{K}{2}$, $\frac{d^{2}P}{d{t}^2}>0$, so $P(t)$ is concave up. When $P>\frac{K}{2}$, $\frac{d^{2}P}{d{t}^2}<0$, so $P(t)$ is concave down. This means the logistic curve has an inflection point (where concavity changes) exactly at $P=\frac{K}{2}$, which matches our earlier result that the growth rate is maximized at $P=\frac{K}{2}$. If the initial population $P_0>K$, $P(t)$ is always between $K$ and $P_0$ as it decreases to $K$, so it never crosses $P=\frac{K}{2}$ and has no inflection point for $t>0$.

Worked Example

For the logistic model $\frac{dP}{dt}=0.3P\left(1-\frac{P}{900}\right)$ with $P(0)=100$, find the $t$-coordinate of the inflection point and state the intervals of concavity.

1. We know inflection occurs at $P=\frac{K}{2}=\frac{900}{2}=450$.
2. Write the particular solution for $P(t)$: $P(t)=\frac{900\cdot 100}{100+800e^{-0.3t}}=\frac{900}{1+8e^{-0.3t}}$.
3. Set $P(t)=450$ and solve for $t$: $$450=\frac{900}{1+8e^{-0.3t}}⟹1+8e^{-0.3t}=2⟹e^{-0.3t}=\frac{1}{8}⟹t=\frac{\ln 8}{0.3}=10\ln 2\approx 6.93$$
4. For $0<t<10\ln 2$, $P<450$ so $\frac{d^{2}P}{d{t}^2}>0$, so $P(t)$ is concave up on $(0,10\ln 2)$. For $t>10\ln 2$, $P>450$ so $\frac{d^{2}P}{d{t}^2}<0$, so $P(t)$ is concave down on $(10\ln 2,\infty )$.

Exam tip: Remember we find concavity of $P(t)$ as a function of $t$, not of $\frac{dP}{dt}$ as a function of $P$. Always use the chain rule to get the second derivative with respect to $t$, do not stop at the derivative of $\frac{dP}{dt}$ with respect to $P$.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For $\frac{dP}{dt}=0.2P(1000-2P)$, calculating carrying capacity as $K=1000$ instead of $K=500$. Why: Students assume the constant term in the factored binomial is always $K$, without rewriting the equation to standard form. Correct move: Always factor out the coefficient of $P$ from the binomial to get a leading constant term, so the denominator under $P$ is $K$. For this example, $0.2P(1000-2P)=0.4P(500-P)=200P\left(1-\frac{P}{500}\right)$, so $K=500$.
• Wrong move: After integrating, incorrectly combining $\ln |P|-\ln |K-P|$ as $\ln |P(K-P)|$. Why: Students misremember logarithm rules when rushing through the partial fraction step. Correct move: Write the logarithm rule explicitly $\ln A-\ln B=\ln \left(\frac{A}{B}\right)$ before exponentiating to confirm.
• Wrong move: Solving for the constant $A$ and getting $A=\frac{K-P_0}{P_0}$ instead of $A=\frac{P_0}{K-P_0}$. Why: Simple fraction flip when rearranging the initial condition equation. Correct move: After finding $P(t)$, plug in $t=0$ to confirm you get $P_0$, which will catch this error immediately.
• Wrong move: Claiming an inflection point at $P=K/2$ for a solution starting at $P_0>K$. Why: Students memorize that inflection is always at $K/2$, but forget that if $P$ starts above $K$, it decreases toward $K$ and never crosses $K/2$. Correct move: Always check the range of $P(t)$ for your initial condition before stating there is an inflection point.
• Wrong move: Answering a question asking for maximum growth rate by reporting the carrying capacity $K$. Why: Students confuse "maximum population size" with "maximum growth rate of the population". Correct move: If asked for maximum growth rate, calculate $\frac{kK}{4}$, the value of $\frac{dP}{dt}$ at $P=K/2$, not $K$.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

What is the carrying capacity of a population growing according to the differential equation $\frac{dP}{dt}=0.04P(800-P)$? A) 32 B) 200 C) 800 D) 3200

Worked Solution: To find carrying capacity, rewrite the equation in standard form $\frac{dP}{dt}=kP\left(1-\frac{P}{K}\right)$. Factor the constant term 800 out of the binomial: $\frac{dP}{dt}=0.04P\cdot 800\left(1-\frac{P}{800}\right)=32P\left(1-\frac{P}{800}\right)$. Matching to standard form, $K=800$. The common error is multiplying 0.04 and 800 to get 32, which is the intrinsic growth rate $k$, not $K$. The correct answer is C.

Question 2 (Free Response)

Consider the logistic differential equation $\frac{dP}{dt}=0.4P\left(1-\frac{P}{200}\right)$ with initial condition $P(0)=40$. (a) Find all equilibrium solutions and classify each as stable or unstable. (b) Find the particular solution $P(t)$. (c) Find the $t$-coordinate of the inflection point and state the intervals of concavity for $t>0$.

Worked Solution: (a) Set $\frac{dP}{dt}=0$: $0.4P\left(1-\frac{P}{200}\right)=0⟹P=0$ or $P=200$. For $0<P<200$, $\frac{dP}{dt}>0$, so $P$ grows away from 0 toward 200. For $P>200$, $\frac{dP}{dt}<0$, so $P$ decreases toward 200. Thus, $P=0$ is unstable, $P=200$ is stable. (b) Use the particular solution formula: $P(t)=\frac{K{P}_0}{P_0+(K-P_0)e^{-kt}}=\frac{200\cdot 40}{40+160e^{-0.4t}}=\frac{8000}{40+160e^{-0.4t}}=\frac{200}{1+4e^{-0.4t}}$. Verify $P(0)=\frac{200}{1+4}=40$, which matches the initial condition. (c) Inflection occurs at $P=K/2=100$. Set $P(t)=100$: $100=\frac{200}{1+4e^{-0.4t}}⟹1+4e^{-0.4t}=2⟹e^{-0.4t}=1/4⟹t=\frac{\ln 4}{0.4}=5\ln 2\approx 3.47$. $P(t)$ is concave up on $(0,5\ln 2)$ and concave down on $(5\ln 2,\infty )$.

Question 3 (Application / Real-World Style)

A coffee shop is introducing a new seasonal drink. The total number of regular customers that can try the drink is 800. The rate of new customers trying the drink follows a logistic model, where $C(t)$ is the total number of customers that have tried the drink after $t$ days, with $\frac{dC}{dt}=0.5C\left(1-\frac{C}{800}\right)$, and 20 customers tried the drink on opening day (so $C(0)=20$). How many customers have tried the drink after 5 days? Round to the nearest whole number, and interpret the result.

Worked Solution: Write the particular solution using the logistic formula: $$C(t)=\frac{K{P}_0}{P_0+(K-P_0)e^{-kt}}=\frac{800\cdot 20}{20+780e^{-0.5t}}=\frac{16000}{20+780e^{-0.5t}}$$ Substitute $t=5$: $e^{-0.5(5)}=e^{-2.5}\approx 0.082085$. Calculate the denominator: $20+780(0.082085)\approx 20+64.026=84.026$. Then $C(5)\approx 16000/84.026\approx 190$. After 5 days, approximately 190 of the 800 total regular customers have tried the new seasonal drink.

7. Quick Reference Cheatsheet

8. What's Next

Logistic models build on your knowledge of separation of variables and differential equations, and they are the primary example of bounded growth used for applied modeling on the AP Calculus BC exam. Immediately after this topic, you will explore more general autonomous differential equations, their equilibrium solutions, and other solution methods like Euler’s method for differential equations that cannot be solved algebraically. Mastery of logistic differential equations is critical for cross-topic FRQ questions that combine differential equations with real-world modeling, a common exam scenario. This topic also lays the groundwork for analyzing stability of equilibrium solutions, a key skill for more advanced calculus courses and extension questions on the BC exam.

Separation of Variables Autonomous Differential Equations Euler's Method Modeling with Differential Equations