Difficulty: Advanced | Prerequisites: All derivative and integral techniques, FTC, applications of both
Differential equations and slope fields are the most "new" topic in AP Calculus AB for students coming from pre-calculus. They connect everything: a differential equation tells you the slope at every point, a slope field visualises that, and separation of variables lets you solve for the actual function. The free-response section, meanwhile, tests your ability to communicate mathematical reasoning clearly in writing. Roughly half of a student's total score comes from the six FRQ problems, and the rubric rewards correct setup, supporting justification, and proper notation at least as much as a correct final answer.
A differential equation gives dy/dx as a function of x and y. To solve, separate variables (get all y on one side, all x on the other), integrate both sides, and use an initial condition to find the constant. Slope fields are the visual representation. For free response, show every step, state theorems by name, include units, and write in complete mathematical sentences.
Differential equation
An equation involving a derivative, such as dy/dx = f(x, y). In simple terms, it tells you the rate of change of an unknown function, and your job is to find the function itself.
Slope field (direction field)
A diagram showing short line segments at grid points, each with the slope given by the differential equation. Think of it as a map of the flow of the function: if you drop a solution curve onto the field, it follows the direction of the segments.
Separation of variables
A method for solving differential equations of the form dy/dx = g(x) · h(y). Rearrange to (1/h(y)) dy = g(x) dx, then integrate both sides.
Particular solution
The specific solution to a differential equation that satisfies a given initial condition, such as f(2) = 5. The initial condition pins down the constant of integration.
Proportional to
"dh/dt is proportional to √h" means dh/dt = k√h for some constant k. This phrasing appears frequently in AP problems modelling growth, decay, or flow.
Worked example (from Q4): The rate of change of the height h with respect to time is proportional to the square root of the height.
"Rate of change of h with respect to t" translates to dh/dt. "Proportional to √h" means equal to k√h.
The equation is: dh/dt = k√h
The answer is C) dh/dt = k√h.
Watch the wording: "proportional to the square root of the height" means √h (the dependent variable), not √t (the independent variable). Option A (dh/dt = k√t) is a common wrong choice.
Worked example (from Q9): f'(x) = 4x – 4, and f passes through (2, 3). Find f(x).
This is a straightforward antiderivative problem:
f(x) = ∫(4x – 4) dx = 2x² – 4x + C
Using (2, 3): 3 = 2(4) – 4(2) + C = 8 – 8 + C, so C = 3.
f(x) = 2x² – 4x + 3
The answer is D) 2x² – 4x + 3.
Worked example (from FRQ 5): dy/dx = (2 – y)/x², with f(2) = 5.
Separate: dy/(2 – y) = dx/x²
Integrate both sides: –ln|2 – y| = –1/x + C (or equivalently ln|2 – y| = 1/x + C')
Exponentiate: |2 – y| = e^(1/x + C') = Ke^(1/x)
So: 2 – y = ±Ke^(1/x), giving y = 2 – ke^(–1/x) (absorbing the sign into k).
Apply the initial condition f(2) = 5:
5 = 2 – ke^(–1/2), so –3 = –ke^(–1/2), giving k = 3e^(1/2) = 3√e.
The particular solution is: y = 2 – 3e^(1/2 – 1/x)
Finding the limit: As x → ∞, 1/x → 0, so e^(1/2 – 1/x) → e^(1/2) = √e.
lim(x→∞) y = 2 – 3√e · (1/√e)... let me restate cleanly:
y = 2 – 3e^(1/2) · e^(–1/x) = 2 – 3e^(1/2 – 1/x)
As x → ∞: y → 2 – 3e^(1/2) = 2 – 3√e
To sketch a slope field at a given point (x, y), plug the coordinates into the differential equation to get the slope, then draw a short segment with that slope.
For dy/dx = (2 – y)/x² at the point (1, 1): slope = (2 – 1)/1 = 1 (draw a segment at 45°)
At (1, 3): slope = (2 – 3)/1 = –1 (draw a segment at –45°)
At (2, 2): slope = (2 – 2)/4 = 0 (draw a horizontal segment)
Slopes of zero occur wherever y = 2, regardless of x. This is an equilibrium line.
When a question asks for the "least number of times" a derivative equals zero, use the Intermediate Value Theorem on the derivative (which requires continuity).
From the data: h increases (0 to 3 min), decreases (3 to 5), increases (5 to 9), decreases (9 to 10). Each change of direction requires at least one point where h'(t) = 0. Three direction changes mean at least three zeros of h'(t).
The AP rubric typically awards separate marks for the integral setup, the integrand, the bounds, and the evaluation. Even if you make an arithmetic error, a correct setup earns 2–3 out of 4 marks on a part. Write the integral clearly before evaluating.
When a question says "justify your answer," name the theorem. Write "by the Mean Value Theorem" or "by the Intermediate Value Theorem" rather than just describing the idea. The rubric often requires the name.
If the problem gives units (metres, seconds, litres per minute), your answer must include units. This is typically worth one mark on its own.
Do not skip from the setup to the final answer. Show the antiderivative, show the substitution of bounds, and show the arithmetic. If you use a calculator, write what you entered: "∫₀⁵ 8t · e^(–0.4t) dt = 29.7 litres."
If the question asks for the rate of change, give the rate, not the total change. If it asks for the average height, divide the integral by the interval length. Re-read the question after writing your answer.
Region W is bounded by y₁ = 2cos(πx/6), y₂ = x³ – 4x² + 3x, and the y-axis.
Part a) Area:
A = ∫₀³ (y₁ – y₂) dx ≈ 6.07 (calculator).
The rubric awards marks for: correct integrand (y₁ – y₂), correct bounds (0 to 3), and correct evaluation.
Part c) Volume with square cross-sections:
V = ∫₀³ (y₁ – y₂)² dx ≈ 13.739.
Note: the side length of each square is (y₁ – y₂), so the area of each cross-section is (y₁ – y₂)².
Part d) Volume of the wing:
V = ∫₀³ (y₁ – y₂)(1 – x/4) dx ≈ 3.679.
The thickness function h(x) = 1 – x/4 multiplies the width (y₁ – y₂) to give the cross-sectional area.
Separation of variables: rearrange dy/dx = g(x)h(y) into (1/h(y)) dy = g(x) dx, integrate both sides
Exponential decay/growth: if dy/dt = ky, then y = Ce^(kt)
Slope at a grid point: plug (x, y) into the right-hand side of dy/dx = f(x, y)
Differential equations model nearly every dynamic system in science and engineering: population growth (logistic equation), radioactive decay, Newton's law of cooling (as in Q89), fluid flow, and electrical circuits. The slope field is a qualitative tool that lets you visualise how solutions behave without solving the equation, which is useful when an exact solution does not exist.
Students confuse "proportional to √h" with "proportional to √t". The variable that appears under the square root is determined by the wording: "proportional to the square root of the height" means √h, because h is the height.
When separating variables, students sometimes forget to include the constant of integration. Without it, you cannot apply the initial condition, and the particular solution is lost.
In slope field sketches, segments should be short and centred on the grid point. Students sometimes draw long lines that look like solution curves; the examiner wants segments, not curves.
On free response, students lose marks for writing "by the theorem" or "because of calculus" without naming the specific theorem.
⚠️ Differential equations appear in at least one FRQ every year. Separation of variables is the only solution technique required for AB.
⚠️ Slope fields are tested in both MC and FRQ. Practise sketching them quickly; look for equilibrium values where the slope is zero.
⚠️ FRQ scoring is additive: each correct element earns a mark. A partially correct solution with good setup, proper notation, and stated theorems can earn most of the available marks.
⚠️ Units are worth marks. If the rate is in metres per minute, the answer to "how fast is h changing" is in metres per minute.
True or false: "dh/dt is proportional to √h" means dh/dt = k√t.
Fill in the blank: To solve dy/dx = xy by separation of variables, rearrange to (1/y) dy = ______.
True or false: On a slope field, a horizontal segment means the slope is zero at that point.
Fill in the blank: On a free response problem, you should name theorems because ______.
True or false: An initial condition is needed to find the particular solution from a general solution.
Answers: 1. False; it means dh/dt = k√h. 2. x dx. 3. True. 4. The rubric typically requires the theorem's name for full credit. 5. True.
Q: Grain pours from a silo such that dh/dt is proportional to √h. Write the differential equation.
A: dh/dt = k√h, where k is a constant (negative if h is decreasing).
Q: f'(x) = 4x – 4 and f(2) = 3. Find f(x).
A: f(x) = 2x² – 4x + 3. Integrate, then use the initial condition to find C = 3.
Q: For dy/dx = (2 – y)/x², what is the slope at the point (1, 3)?
A: –1. Substituting: (2 – 3)/1² = –1.
Q: On a free response problem, you compute ∫₀⁵ 8t · e⁻⁰·⁴ᵗ dt = 29.7. How should you present this?
A: Write the integral with bounds, state the calculator result, and include units: "29.7 litres."
Q: The altitude data shows h increasing, then decreasing, then increasing, then decreasing. What is the minimum number of times h'(t) = 0?
A: Three. Each change of direction requires at least one zero of h'.
Differential equations tie together derivatives (the equation itself), integration (the solution method), and applications (modelling real systems). Slope fields connect to graphical analysis of functions. Free-response problems draw on every topic in the course, so the strategies here apply across all six FRQ problems, not just the differential equation question.
differential equation, slope field, direction field, separation of variables, particular solution, initial condition, proportional, exponential growth, exponential decay, Newton's law of cooling, AP Calculus AB differential equations, free response, FRQ strategies, AP exam scoring