AP Predicted Exam 2026

AP Calculus AB Predicted Exam (May 2026)

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AP® Calculus AB — Predicted Exam (May 2026)

Section II: Free Response

  • Part A: 2 questions, 30 minutes, graphing calculator required

  • Part B: 4 questions, 60 minutes, no calculator

Unless otherwise specified, answers need not be simplified. When numeric results are required with a calculator, report to 3 decimal places—a convention emphasized in recent exams.

AP® Calculus AB

2026 Free-Response Questions

CALCULUS AB

SECTION II, Part A

Time—30 minutes
2 Questions

A GRAPHING CALCULATOR IS REQUIRED FOR THESE QUESTIONS.

Question 1

Water flows into a cylindrical tank through an inlet pipe. The rate at which water flows into the tank at time t minutes is modeled by the function R(t) = 15e−0.1tcos(0.3t) + 20, where R(t) is measured in liters per minute, for 0 ≤ t ≤ 10.

(a)

Find R′(4). Using correct units, interpret the meaning of R′(4) in the context of this problem.

(b)

Is there a time t, for 2 < t < 6, at which R(t) = 25? Justify your answer.

(c)

Use a left Riemann sum with four subintervals of equal length to approximate ∫08 R(t) dt. Show the setup for your calculations.

(d)

Water flows out of the tank at a constant rate of 18 liters per minute. Let V(t) represent the volume of water in the tank at time t minutes. At time t = 0, the tank contains 50 liters of water. Write an expression for V(t) and find the maximum volume of water in the tank over the interval 0 ≤ t ≤ 10. Justify your answer.

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

Question 2

Let f and g be the functions defined by f(x) = 2 + ln(x + 1) and g(x) = x3 − 3x2 + 4. The graphs of f and g intersect at x = 0 and x = B, where B > 0.

(a)

Find the value of B. Show the setup for your calculations.

(b)

Find the area of the region bounded by the graphs of f and g between x = 0 and x = B.

(c)

The region described in part (b) is the base of a solid. For this solid, each cross section perpendicular to the x-axis is an isosceles right triangle with its hypotenuse in the base. Find the volume of the solid.

(d)

A particle moves along the curve y = f(x) so that its position at time t seconds is (x(t), f(x(t))). At time t = 3, the particle is at the point (1, 2.693) and dx/dt = 4. Find dy/dt at time t = 3.

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

END OF PART A

CALCULUS AB

SECTION II, Part B

Time—1 hour
4 Questions

NO CALCULATOR IS ALLOWED FOR THESE QUESTIONS.

Question 3

Consider the differential equation dy/dx = y(3 − y)/4, where y > 0.

(a)

The slope field for this differential equation is shown below. Sketch the solution curve that passes through the point (0, 1).

[Slope field diagram]

(b)

Find the coordinates of all equilibrium solutions. Classify each equilibrium as stable or unstable. Justify your answer.

(c)

Find d2y/dx2 in terms of y only. Use this to determine whether the solution curve through (0, 1) is concave up or concave down at that point.

(d)

Use separation of variables to find y as a function of x for the particular solution with initial condition y(0) = 1.

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

Question 4

Let f be a twice-differentiable function defined on the interval [−3, 5]. The graph of f′, the derivative of f, is shown below. The graph consists of a semicircle centered at (1, 0) and two line segments. The areas of regions A, B, and C are 2, π/2, and 3, respectively.

[Graph of f′ with regions A, B, and C labeled]

Given that f(−3) = 4.

(a)

Find f(1) and f(5).

(b)

On what intervals is the graph of f concave down? Justify your answer.

(c)

Find the absolute minimum value of f on the interval [−3, 5]. Justify your answer.

(d)

Let g(x) = ∫−3x f′(t) dt. Find g″(2).

Question 5

Consider the curve defined by the equation x2y + 2xy2 = 12.

(a)

Find dy/dx in terms of x and y.

(b)

Find the equation of the line tangent to the curve at the point (2, 1).

(c)

Find the coordinates of all points on the curve where the tangent line is horizontal.

(d)

The curve forms a closed loop. A particle travels clockwise around this loop. At the instant when the particle is at the point (2, 1), the x-coordinate of its position is decreasing at a rate of 3 units per second. At what rate is the y-coordinate changing at this instant?

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

Question 6

Two particles, P and Q, move along the x-axis. For t ≥ 0, the position of particle P at time t is given by xP(t) = t3 − 6t2 + 9t + 2, and the position of particle Q at time t is given by xQ(t) = 2t + 5cos(πt/3).

(a)

Find the velocity of particle P at time t = 1.

(b)

At what time(s) in the interval 0 ≤ t ≤ 4 is particle P at rest? Justify your answer.

(c)

Find the total distance traveled by particle Q from time t = 0 to time t = 3.

(d)

At what time t in the interval 0 ≤ t ≤ 4 are the two particles moving in the same direction at the same speed? Justify your answer.

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

STOP

END OF EXAM

Model Answers and Scoring Rubric

Part A

Question 1 Model Answer:

(a) [2 points]

  • R(t) = 15e−0.1tcos(0.3t) + 20
  • R′(t) = 15[−0.1e−0.1tcos(0.3t) + e−0.1t(−0.3sin(0.3t))]
  • R′(t) = −1.5e−0.1tcos(0.3t) − 4.5e−0.1tsin(0.3t)
  • R′(4) = −1.5e−0.4cos(1.2) − 4.5e−0.4sin(1.2)
  • R′(4) ≈ −1.5(0.6703)(0.3624) − 4.5(0.6703)(0.9320) ≈ −3.17 liters per minute per minute

Interpretation: At time t = 4 minutes, the rate at which water flows into the tank is decreasing at approximately 3.17 liters per minute per minute.

(b) [2 points]

  • R(2) = 15e−0.2cos(0.6) + 20 ≈ 15(0.8187)(0.8253) + 20 ≈ 30.14
  • R(6) = 15e−0.6cos(1.8) + 20 ≈ 15(0.5488)(−0.2272) + 20 ≈ 18.13
  • Since R is continuous on [2, 6], R(2) > 25 > R(6), by the Intermediate Value Theorem, there exists at least one time t in (2, 6) where R(t) = 25.

(c) [3 points]

  • Δt = 8/4 = 2
  • Left Riemann sum = 2[R(0) + R(2) + R(4) + R(6)]
  • R(0) = 15e0cos(0) + 20 = 15(1)(1) + 20 = 35
  • R(2) ≈ 30.14 (from part b)
  • R(4) = 15e−0.4cos(1.2) + 20 ≈ 15(0.6703)(0.3624) + 20 ≈ 23.64
  • R(6) ≈ 18.13 (from part b)
  • Left Riemann sum ≈ 2(35 + 30.14 + 23.64 + 18.13) ≈ 213.82 liters

(d) [3 points]

  • Net flow rate = R(t) − 18
  • V(t) = 50 + ∫0t [R(s) − 18] ds
  • To find maximum, solve V′(t) = R(t) − 18 = 0
  • 15e−0.1tcos(0.3t) + 20 − 18 = 0
  • 15e−0.1tcos(0.3t) = −2
  • e−0.1tcos(0.3t) = −2/15
  • Using calculator, t ≈ 2.44
  • Check endpoints and critical point
  • Maximum volume occurs at t ≈ 2.44 minutes

Question 2 Model Answer:

(a) [2 points]

  • Set f(x) = g(x): 2 + ln(x + 1) = x3 − 3x2 + 4
  • ln(x + 1) = x3 − 3x2 + 2
  • Using calculator to solve: B ≈ 2.177

(b) [3 points]

  • Area = ∫0B |f(x) − g(x)| dx
  • Need to determine which function is on top
  • At x = 1: f(1) = 2 + ln(2) ≈ 2.693, g(1) = 1 − 3 + 4 = 2
  • So f(x) > g(x) on the interval
  • Area = ∫02.177 [f(x) − g(x)] dx
  • = ∫02.177 [ln(x + 1) − x3 + 3x2] dx
  • Using calculator: Area ≈ 3.245

(c) [3 points]

  • Height of triangle at position x = |f(x) − g(x)|
  • For isosceles right triangle with hypotenuse h, legs = h/√2
  • Area of triangle = (1/2)(h/√2)(h/√2) = h2/4
  • Volume = ∫0B [f(x) − g(x)]2/4 dx
  • = (1/4)∫02.177 [ln(x + 1) − x3 + 3x2]2 dx
  • Using calculator: Volume ≈ 2.851

(d) [2 points]

  • y = f(x) = 2 + ln(x + 1)
  • dy/dx = 1/(x + 1)
  • At x = 1: dy/dx = 1/2
  • dy/dt = (dy/dx)(dx/dt) = (1/2)(4) = 2

Part B

Question 3 Model Answer:

(a) [2 points]

  • Sketch shows solution curve starting at (0, 1)
  • Curve increases and approaches y = 3 asymptotically
  • Curve is concave up initially, then concave down

(b) [2 points]

  • Equilibrium when dy/dx = 0
  • y(3 − y)/4 = 0
  • y = 0 or y = 3
  • For y = 0: unstable (solutions move away)
  • For y = 3: stable (solutions approach)

(c) [3 points]

  • dy/dx = y(3 − y)/4
  • d2y/dx2 = d/dx[y(3 − y)/4]
  • = (1/4)[dy/dx(3 − y) + y(−dy/dx)]
  • = (1/4)[dy/dx(3 − 2y)]
  • = (1/4)[y(3 − y)/4](3 − 2y)
  • = y(3 − y)(3 − 2y)/16
  • At (0, 1): d2y/dx2 = 1(2)(1)/16 = 1/8 > 0
  • Solution curve is concave up at (0, 1)

(d) [3 points]

  • dy/dx = y(3 − y)/4
  • 4dy/[y(3 − y)] = dx
  • Using partial fractions: 4/[y(3 − y)] = 4/3[1/y + 1/(3 − y)]
  • (4/3)[ln|y| − ln|3 − y|] = x + C
  • (4/3)ln|y/(3 − y)| = x + C
  • When x = 0, y = 1: C = (4/3)ln(1/2)
  • y/(3 − y) = (1/2)e3x/4
  • Solving for y: y = 3e3x/4/[2 + e3x/4]

Question 4 Model Answer:

(a) [3 points]

  • f(1) − f(−3) = ∫−31 f′(x) dx = −Area(A) = −2
  • f(1) = f(−3) − 2 = 4 − 2 = 2
  • f(5) − f(1) = ∫15 f′(x) dx = Area(B) − Area(C) = π/2 − 3
  • f(5) = 2 + π/2 − 3 = π/2 − 1

(b) [2 points]

  • f is concave down where f″(x) < 0
  • This occurs where f′(x) is decreasing
  • From the graph, f′ is decreasing on (−3, −1) and (3, 5)
  • Therefore f is concave down on (−3, −1) ∪ (3, 5)

(c) [3 points]

  • Critical points where f′(x) = 0: x = −1, 1, 3
  • f(−1) = f(−3) + ∫−3−1 f′(x) dx = 4 + (−2) = 2
  • f(3) = f(1) + ∫13 f′(x) dx = 2 + π/2
  • Compare: f(−3) = 4, f(−1) = 2, f(1) = 2, f(3) = 2 + π/2, f(5) = π/2 − 1
  • Since π/2 − 1 ≈ 0.57, absolute minimum is f(5) = π/2 − 1

(d) [2 points]

  • g(x) = ∫−3x f′(t) dt = f(x) − f(−3) = f(x) − 4
  • g′(x) = f′(x)
  • g″(x) = f″(x)
  • From graph, at x = 2, the semicircle has slope 0
  • So f″(2) = 0
  • Therefore g″(2) = 0

Question 5 Model Answer:

(a) [3 points]

  • x2y + 2xy2 = 12
  • Differentiate implicitly: 2xy + x2(dy/dx) + 2y2 + 4xy(dy/dx) = 0
  • (x2 + 4xy)(dy/dx) = −2xy − 2y2
  • dy/dx = −(2xy + 2y2)/(x2 + 4xy) = −2y(x + y)/[x(x + 4y)]

(b) [2 points]

  • At (2, 1): dy/dx = −2(1)(2 + 1)/[2(2 + 4)] = −6/12 = −1/2
  • Tangent line: y − 1 = −1/2(x − 2)
  • y = −x/2 + 2

(c) [3 points]

  • Horizontal tangent when dy/dx = 0
  • −2y(x + y)/[x(x + 4y)] = 0
  • This occurs when y = 0 or x + y = 0
  • If y = 0: x2(0) + 2x(0)2 = 12 gives no solution
  • If x = −y: (−y)2y + 2(−y)y2 = 12
  • y3 − 2y3 = 12
  • −y3 = 12
  • y = −∛12, x = ∛12
  • Point: (∛12, −∛12)

(d) [2 points]

  • Given: dx/dt = −3 at (2, 1)
  • From part (a): dy/dx = −1/2 at (2, 1)
  • dy/dt = (dy/dx)(dx/dt) = (−1/2)(−3) = 3/2
  • The y-coordinate is increasing at 3/2 units per second

Question 6 Model Answer:

(a) [2 points]

  • xP(t) = t3 − 6t2 + 9t + 2
  • vP(t) = 3t2 − 12t + 9
  • vP(1) = 3(1)2 − 12(1) + 9 = 3 − 12 + 9 = 0

(b) [2 points]

  • Particle P at rest when vP(t) = 0
  • 3t2 − 12t + 9 = 0
  • 3(t2 − 4t + 3) = 0
  • 3(t − 1)(t − 3) = 0
  • t = 1 or t = 3
  • Both are in [0, 4], so particle P is at rest at t = 1 and t = 3

(c) [3 points]

  • xQ(t) = 2t + 5cos(πt/3)
  • vQ(t) = 2 − (5π/3)sin(πt/3)
  • Find when vQ(t) = 0: sin(πt/3) = 6/(5π) ≈ 0.382
  • πt/3 = arcsin(0.382) ≈ 0.392 or π − 0.392 ≈ 2.750
  • t ≈ 0.374 or t ≈ 2.626 in [0, 3]
  • xQ(0) = 0 + 5cos(0) = 5
  • xQ(0.374) ≈ 0.748 + 5cos(0.392) ≈ 5.371
  • xQ(2.626) ≈ 5.252 + 5cos(2.750) ≈ 4.348
  • xQ(3) = 6 + 5cos(π) = 6 − 5 = 1
  • Total distance = |5.371 − 5| + |4.348 − 5.371| + |1 − 4.348|
  • = 0.371 + 1.023 + 3.348 = 4.742

(d) [3 points]

  • Same direction and speed when |vP(t)| = |vQ(t)| and signs match
  • vP(t) = 3t2 − 12t + 9 = 3(t − 1)(t − 3)
  • vQ(t) = 2 − (5π/3)sin(πt/3)
  • For 0 < t < 1: vP > 0
  • For 1 < t < 3: vP < 0
  • For 3 < t < 4: vP > 0
  • Need to check where vP(t) = vQ(t) or vP(t) = −vQ(t)
  • Setting equal and solving: 3t2 − 12t + 9 = 2 − (5π/3)sin(πt/3)
  • Using numerical methods or graphing calculator: t ≈ 3.412
  • Check: at t ≈ 3.412, both velocities are positive and approximately equal

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