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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers

Worked solutions and scoring guides follow. Award separation and integration points independently, and require the constant of integration. Answers without an initial-condition step cannot earn full credit on particular-solution questions.
Standards: AP Calculus AB Unit 7 · Level: Grade 12 AP Calculus AB
Question 1 Solution with an initial condition (2 points)
Write dydx = 3(y − 2). Separation gives y − 2 = Ae3x. With y(0) = 7, A = 5. Answer (A).
Scoring: 2 points for the correct choice; 1 point for a correct separated form.
Question 2 Matching a slope field (2 points)
The segments are horizontal along y = x, so the slope involves x − y. At (0, 2) the segment slopes steeply downward: x − y = −2. This rules out (C), whose slope there would be +2. Answer (B).
Scoring: 2 points for the correct choice; 1 point for checking a point.
Question 3 Separable equation (3 points)
(a) Separate: y dy = 6x2 dx. Integrate: y2/2 = 2x3 + C.
(b) At (1, 3): 9/2 = 2 + C, so C = 5/2. Then y2 = 4x3 + 5. Since y(1) > 0, y = √(4x3 + 5).
Scoring: Part (a): 1 point for separating, 1 point for integrating both sides with a constant. Part (b): 1 point.
Question 4 Radioactive decay (3 points)
(a) m(t) = 250e−0.04t.
(b) Solve e−0.04t = 1/2: t = ln 2/0.04 ≈ 17.3 years.
(c) Solve 40 = 250e−0.04t: t = ln(6.25)/0.04 ≈ 45.8 years.
Scoring: 1 point per part.
Question 5 Verifying a solution (3 points)
(a) y′ = 2e2x + 1. Then y′ − 2y = 2e2x + 1 − 2e2x − 2x = 1 − 2x. The equation holds.
(b) y(0) = e0 + 0 = 1.
Scoring: Part (a): 1 point for the derivative, 1 point for the substitution. Part (b): 1 point.
Question 6 Slope field reasoning (3 points)
(a) The slope is zero for y = 0 and y = 4.
(b) The solution starts at 1 with positive slope and cannot cross the equilibrium y = 4, so the limit is 4.
(c) d2ydx2 = (4 − 2y)dydx. For 0 < y < 2, both factors are positive, so the solution is concave up.
Scoring: 1 point per part.
Question 7 Cooling model (2 points)
(a) Solving gives T = 20 + 60e−0.1t. So T(5) = 20 + 60e−0.5 ≈ 56.4°F.
(b) As t → ∞, e−0.1t → 0, so T approaches 20°F.
Scoring: 1 point per part.
Question 8 Tangent line approximation (2 points)
(a) The slope is x2 − y = 1 − 2 = −1 at (1, 2). The tangent line is y = 2 − (x − 1).
(b) At x = 1.1: 2 − 0.1 = 1.9.
Scoring: 1 point per part.
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