Modeling situations with differential equations — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Translating verbal contextual descriptions to differential equations, modeling exponential growth/decay, Newton’s Law of Cooling/Heating, logistic population growth, mixing problems, and motion scenarios for AP exam questions.

You should already know: Derivative as a rate of change, basic algebra for equation setup, chain rule for composite functions.

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 Modeling situations with differential equations?

A differential equation (DE) is any equation that contains one or more derivatives of an unknown function. Modeling with DEs means translating real-world or abstract scenarios about changing quantities into this mathematical form, and it is the mandatory first step before solving or analyzing any DE problem. According to the AP Calculus BC Course and Exam Description (CED), this topic is a foundational skill for 6-12% of the entire differential equations unit, so it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, almost always as the opening step of a longer problem. Unlike solving DEs, this topic tests your ability to interpret rate information rather than compute integrals or draw slope fields. Many students rush through this setup step, which leads to losing points on entire FRQs even if their later solving steps are correct. Synonyms for this process include "setting up a differential equation from context" or "translating rate descriptions to DE form." It is the gateway to all subsequent work with differential equations on the AP exam.

2. Translating Verbal Rate Descriptions to Differential Equations

The core idea of this skill is that any statement about how a quantity changes can be written as $\frac{dy}{dt}$ (or whatever variable pair you are using) equal to the net rate of change of the quantity. The key rule of thumb: if a problem says "the rate of change of $y$ is proportional to $y$", this directly translates to $\frac{dy}{dt}=ky$, where $k$ is the constant of proportionality. If it says "the rate of change of $y$ is proportional to the difference between $y$ and a fixed value $C$", this becomes $\frac{dy}{dt}=k(y-C)$. You must always check the sign of $k$ to match the context: if $y$ is decreasing towards $C$, $k$ will be negative. For more complex descriptions with multiple rates, you simply add all increasing rates and subtract all decreasing rates to get the net rate on the right-hand side. The step-by-step process for any translation is: 1) name all unknown functions and their independent variables, 2) write the derivative of the unknown function of interest, 3) translate all proportionality relationships by adding constants, 4) combine all inflow/outflow rates to get the final DE.

Worked Example

Problem: The number of customers $C(t)$ at a coffee shop increases at a rate proportional to the difference between the current number of customers and the shop's maximum capacity of 120 customers. Write the differential equation that models $C(t)$.

Solution:

1. Identify variables: $C(t)$ = number of customers at time $t$, maximum capacity = 120, so the difference described is $120-C$.
2. The rate of change of $C$ is $\frac{dC}{dt}$, which is proportional to $120-C$.
3. Since the number of customers is increasing toward capacity, $\frac{dC}{dt}$ is positive when $C<120$, so the constant of proportionality $k>0$.
4. Final DE: $\frac{dC}{dt}=k(120-C)$ for some positive constant $k$.

Exam tip: Always explicitly name the constant of proportionality in your DE for FRQs; AP exam graders require you to include the constant, even if you do not calculate its value, to earn full setup points.

3. Standard Contextual Models

The AP exam frequently tests a set of standard contextual models that you should recognize immediately, though you should always be able to derive them from the problem description instead of relying solely on memorization. The most common standard models are:

1. Exponential growth/decay: If a quantity changes at a rate proportional to the size of the quantity, $\frac{dy}{dt}=ky$, with $k>0$ for growth (population, continuous compounding) and $k<0$ for decay (radioactive decay).
2. Newton's Law of Cooling/Heating: The rate of change of an object's temperature is proportional to the difference between the object's temperature $T(t)$ and the constant ambient temperature $T_s$, so $\frac{dT}{dt}=k(T-T_s)$, with $k<0$ for a cooling object.
3. Logistic population growth: For a population with carrying capacity $K$, the growth rate is proportional to both the population size and the remaining capacity, giving $\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right)$ with $r>0$.
4. Rectilinear motion: Acceleration is the rate of change of velocity, so if acceleration is proportional to position or velocity, you get a DE relating these quantities. BC also tests second-order DEs for simple harmonic motion here.

Worked Example

Problem: A population of deer in a forest has a carrying capacity of 800 deer, and the initial growth rate constant is 0.3 per year. Write the differential equation for the deer population $P(t)$ at time $t$ years.

Solution:

1. We use the logistic growth model, which matches the description of a population with a fixed carrying capacity.
2. The standard logistic DE is $\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right)$, where $r$ is the growth rate constant and $K$ is carrying capacity.
3. Substitute $r=0.3$ and $K=800$: $$\frac{dP}{dt}=0.3P\left(1-\frac{P}{800}\right)$$
4. Confirm the sign: when $0<P<800$, $\frac{dP}{dt}$ is positive, which matches the expected growth of the population, so this is correct.

Exam tip: Always confirm that the sign of $\frac{dP}{dt}$ matches your expectation for the context: for logistic growth, $\frac{dP}{dt}$ must be positive when the population is between 0 and the carrying capacity.

4. Net Rate Models for Inflow/Outflow Systems

Many AP problems model mixing problems, bank accounts, or pollution systems where a quantity is both added and removed at different rates. The core rule for these problems is: $$\frac{dQ}{dt}=(\text{rate of }Q\text{ entering the system})-(\text{rate of }Q\text{ leaving the system})$$ where $Q(t)$ is the amount of the quantity at time $t$. For mixing problems (the most common inflow/outflow problem), the inflow rate of solute (e.g. salt, plastic) is equal to the concentration of solute in the inflow multiplied by the flow rate of the inflow. The outflow rate of solute is equal to the concentration of solute in the well-mixed system at time $t$ multiplied by the outflow rate. The concentration at time $t$ is $\frac{Q(t)}{V(t)}$, where $V(t)$ is the total volume of liquid in the system at time $t$. If inflow and outflow rates are equal, $V(t)$ is constant; if not, $V(t)=V_0+(r_{in}-r_{out})t$, where $V_0$ is the initial volume.

Worked Example

Problem: A 50-liter tank initially contains 10 liters of water with 2 kg of sugar dissolved in it. Water containing 0.5 kg of sugar per liter flows into the tank at 2 liters per minute, and well-mixed sugar water flows out at 1 liter per minute. Write the differential equation for the mass of sugar $Q(t)$ in kg at time $t$ minutes.

Solution:

1. Calculate the volume at time $t$: initial volume $V_0=10$ L, net inflow = $2-1=1$ L/min, so $V(t)=10+t$.
2. Calculate inflow rate of sugar: $0.5\text{ kg/L}\times 2\text{ L/min}=1$ kg/min.
3. Calculate outflow rate of sugar: concentration = $\frac{Q(t)}{10+t}$ kg/L, so outflow rate = $1\times \frac{Q}{10+t}=\frac{Q}{10+t}$ kg/min.
4. Net rate gives the DE: $$\frac{dQ}{dt}=1-\frac{Q}{10+t}$$

Exam tip: Always calculate $V(t)$ first in mixing problems; do not assume the volume is equal to the maximum tank capacity unless the problem says the tank is always full.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing $\frac{dT}{dt}=k(T-T_s)$ with $k>0$ for a cooling object and claiming the DE is wrong. Why: Students memorize a specific sign convention for Newton's Law and mark any other form incorrect, or get confused about what sign they need. Correct move: Check that $\frac{dT}{dt}$ is negative when $T>T_s$ (cooling) and positive when $T<T_s$ (heating); any DE that satisfies this sign condition is acceptable.
• Wrong move: Omitting the constant of proportionality when the problem says "proportional", e.g. writing $\frac{dP}{dt}=(K-P)$ instead of $\frac{dP}{dt}=k(K-P)$. Why: Students think the "proportional" description is just context and forget that proportional means multiplied by an unknown constant. Correct move: Every time you see the word "proportional" in a problem, add a constant of proportionality to your DE before moving on.
• Wrong move: In mixing problems with different inflow/outflow rates, using the maximum tank volume instead of the current volume $V(t)$ to calculate concentration. Why: Students assume the tank starts full, even when the problem says otherwise. Correct move: Always calculate $V(t)=V_0+(r_{in}-r_{out})t$ explicitly before setting up the outflow rate.
• Wrong move: Writing the logistic DE as $\frac{dP}{dt}=rP\left(\frac{P}{K}-1\right)$ with $r>0$ for a growing population. Why: Students swap the order of the terms, leading to a negative growth rate when $P<K$. Correct move: Always confirm that $\frac{dP}{dt}$ is positive for $0<P<K$; if it is negative, flip the order of the terms or change the sign of $r$.
• Wrong move: Writing acceleration as $\frac{dx}{dt}$ when the problem describes the rate of change of velocity. Why: Students confuse position, velocity, and acceleration derivatives. Correct move: Remember that velocity is $\frac{dx}{dt}$ and acceleration is $\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$, so match the left-hand side of your DE to what is changing.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

The amount of a radioactive substance $A(t)$ decays at a rate proportional to the amount of substance remaining at time $t$. Which of the following differential equations correctly models this situation? A) $\frac{dA}{dt}=kA$, $k>0$ B) $\frac{dA}{dt}=-kA$, $k>0$ C) $\frac{dA}{dt}=kA$, $k=0$ D) $\frac{dA}{dt}=k(A-A_0)$, $k>0$

Worked Solution: The problem states the rate of change of $A$ is proportional to $A$, so the DE has the form $\frac{dA}{dt}=kA$. Since the substance is decaying, the amount $A$ is decreasing over time, so $\frac{dA}{dt}$ must be negative for positive $A$. If we write this as $\frac{dA}{dt}=-kA$ with $k>0$, the derivative is negative, which matches the decay context. Option A has positive $k$, leading to growth, so it is incorrect. Option C has $k=0$, which means no change, so it is incorrect. Option D models rate proportional to the difference between $A$ and $A_0$, which is not the description given. The correct answer is B.

Question 2 (Free Response)

A coffee shop opens at $t=0$ with no customers inside. Customers enter at a constant rate of 5 customers per minute, and customers leave at a rate proportional to the number of customers currently inside. Let $C(t)$ be the number of customers inside at time $t$ minutes. (a) Write the differential equation that models $C(t)$. (b) If the constant of proportionality for customers leaving is 0.2, write the differential equation for this case. (c) What is the equilibrium number of customers for the DE from part (b), and what does this mean in context?

Worked Solution: (a) The rate of change of customers $\frac{dC}{dt}$ equals entry rate minus exit rate. Entry rate is a constant 5 customers per minute. Exit rate is proportional to $C$, so exit rate is $kC$ for $k>0$. The differential equation is: $$\frac{dC}{dt}=5-kC,k>0$$ (b) Substitute $k=0.2$: $\frac{dC}{dt}=5-0.2C$. (c) Equilibrium occurs when $\frac{dC}{dt}=0$, so set $0=5-0.2C$, solving gives $C=25$. In context, this means the number of customers in the coffee shop will approach 25 over time, and will stabilize at that number once enough time has passed.

Question 3 (Application / Real-World Style)

A 500 million liter reservoir initially has no nitrogen pollution. Agricultural runoff carries water containing 0.2 mg of nitrogen per liter into the reservoir at a rate of 10 million liters per year. A outflow river carries well-mixed water out of the reservoir at the same 10 million liters per year rate. Write the differential equation for the total mass of nitrogen $N(t)$ in mg at time $t$ years, and find the equilibrium mass of nitrogen that $N(t)$ approaches as $t\to \infty$.

Worked Solution: Inflow and outflow rates are equal, so the volume of the reservoir is constant at $V=500\times 10^6=5\times 10^8$ liters. Rate of nitrogen entering is $(0.2\text{ mg/L})\times (10\times 10^6\text{ L/year})=2\times 10^6$ mg/year. Rate of nitrogen leaving is $\left(\frac{N(t)}{5\times 10^8}\text{ mg/L}\right)\times (10\times 10^6\text{ L/year})=0.02N(t)$ mg/year. The differential equation is: $$\frac{dN}{dt}=2\times 10^6-0.02N$$ At equilibrium, $\frac{dN}{dt}=0$, so $0=2\times 10^6-0.02N$, solving gives $N=10^8$ mg. In context, this means the total mass of nitrogen in the reservoir will approach 100 grams (100 million milligrams) over the long term.

7. Quick Reference Cheatsheet

8. What's Next

Modeling with differential equations is the non-negotiable prerequisite for every subsequent topic in Unit 7: Differential Equations. Next, you will learn to draw slope fields to visualize DE solutions, use separation of variables to solve first-order DEs, and find particular solutions to initial value problems. Without correctly setting up the differential equation from context, you will not be able to get the correct final solution on any modeling FRQ, which makes up the majority of the differential equations FRQ on the AP exam. Beyond this unit, these modeling skills are foundational for solving parametric motion problems and second-order differential equation problems exclusive to BC. The ability to translate verbal context to mathematical equations is also useful for all applied problems across the AP exam.

• Slope fields and solution curves
• Separation of variables
• Exponential and logistic models
• Second-order differential equations