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2026 AP Calculus AB FRQ Question 3: Cooling Pie — Differential Equations

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2026.

Question 3

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Cooling Pie — Differential Equations

Differential Equations

Hard
A pie is taken from a hot oven and put on a table. The internal temperature of the pie at time tt minutes can be modeled by the function HH that satisfies the differential equation dHdt=−115(H−20)\dfrac{dH}{dt} = -\dfrac{1}{15}(H - 20), where H(t)H(t) is measured in degrees Celsius and H(0)=75H(0) = 75. For t>0t > 0, it is known that 20<H(t)<7520 < H(t) < 75.
2026 AP Calculus AB FRQ Question 3 Graph
Part AMedium2 points
Explain why the following could not be a slope field for the differential equation dHdt=−115(H−20)\dfrac{dH}{dt} = -\dfrac{1}{15}(H-20).

Answer

The slope field shows positive slopes for H>20H > 20, but the differential equation gives dHdt<0\frac{dH}{dt} < 0 whenever H>20H > 20. Therefore the slope field is incorrect.
Full Solution & Work

Analyze the sign of dH/dt

When H>20H > 20: (H−20)>0(H - 20) > 0, so dHdt=−115(H−20)<0\dfrac{dH}{dt} = -\dfrac{1}{15}(H-20) < 0. Slopes must be **negative** for all H>20H > 20.

Identify the contradiction

The shown slope field displays positive (upward) slopes for H>20H > 20, which contradicts the differential equation. This is why the slope field could not be correct.

AP Scoring — 2 Points

**P1**: States dHdt<0\frac{dH}{dt} < 0 when H>20H > 20.
**P2**: Connects this to the slope field showing positive slopes, hence the contradiction.
Part BEasy2 points
Find the slope of the line tangent to the graph of HH at time t=0t = 0. Show the work that leads to your answer.

Answer

dHdt∣t=0=−115(75−20)=−5515=−113\dfrac{dH}{dt}\Big|_{t=0} = -\dfrac{1}{15}(75 - 20) = -\dfrac{55}{15} = -\dfrac{11}{3}
Full Solution & Work

Substitute H(0) = 75 into the differential equation

dHdt∣t=0=−115(H(0)−20)=−115(75−20)=−5515=−113\frac{dH}{dt}\bigg|_{t=0} = -\frac{1}{15}(H(0) - 20) = -\frac{1}{15}(75 - 20) = -\frac{55}{15} = -\frac{11}{3}

AP Scoring — 2 Points

**P1**: Substitutes H=75H = 75 correctly into the ODE.
**P2**: Correct slope −113-\frac{11}{3} (or equivalent decimal ≈−3.667\approx -3.667).
Part CMedium2 points
It can be shown that d2Hdt2=1225(H−20)\dfrac{d^2H}{dt^2} = \dfrac{1}{225}(H-20). The line tangent to the graph of HH at time t=0t = 0 is used to approximate H(5)H(5), the internal temperature of the pie at time t=5t = 5. Is this approximation an overestimate or an underestimate for the actual value of H(5)H(5)? Give a reason for your answer.

Answer

Underestimate. Since d2Hdt2>0\dfrac{d^2H}{dt^2} > 0 for H>20H > 20, the graph of HH is concave up, so the tangent line lies below the curve.
Full Solution & Work

Determine the sign of the second derivative

For 20<H<7520 < H < 75: (H−20)>0(H - 20) > 0, so d2Hdt2=(H−20)225>0\dfrac{d^2H}{dt^2} = \dfrac{(H-20)}{225} > 0. The graph of HH is **concave up**.

Conclude overestimate or underestimate

When a function is concave up, the tangent line lies **below** the curve. Therefore the tangent line approximation of H(5)H(5) is an **underestimate**.

AP Scoring — 2 Points

**P1**: Identifies d2Hdt2>0\frac{d^2H}{dt^2} > 0 (concave up).
**P2**: States 'underestimate' with concavity reasoning.
Part DHard6 points
Use separation of variables to find an expression for H(t)H(t), the particular solution to the given differential equation with initial condition H(0)=75H(0) = 75.

Answer

H(t)=20+55e−t/15H(t) = 20 + 55e^{-t/15}
Full Solution & Work

Separate variables

dHH−20=−115 dt\frac{dH}{H - 20} = -\frac{1}{15}\,dt

Integrate both sides

∫dHH−20=∫−115 dt\int \frac{dH}{H-20} = \int -\frac{1}{15}\,dt

ln⁡∣H−20∣=−t15+C\ln|H - 20| = -\frac{t}{15} + C

Exponentiate and solve for H

∣H−20∣=eC⋅e−t/15|H - 20| = e^C \cdot e^{-t/15}

H−20=Ae−t/15(where A=±eC)H - 20 = Ae^{-t/15} \quad \text{(where } A = \pm e^C\text{)}

H(t)=20+Ae−t/15H(t) = 20 + Ae^{-t/15}

Apply the initial condition H(0) = 75

75=20+Ae0  ⟹  A=5575 = 20 + Ae^{0} \implies A = 55

Write the particular solution

H(t)=20+55e−t/15\boxed{H(t) = 20 + 55e^{-t/15}}

Verification: as t→∞t \to \infty, H→20H \to 20 ✓ (the pie cools to room temperature).

AP Scoring — 6 Points

**P1**: Correct separation of variables.
**P2**: Correct antiderivatives on both sides.
**P3**: ln⁡∣H−20∣\ln|H-20| on left side.
**P4**: Correct general solution form H=20+Ae−t/15H = 20 + Ae^{-t/15}.
**P5**: Uses H(0)=75H(0) = 75 to find A=55A = 55.
**P6**: Correct particular solution H(t)=20+55e−t/15H(t) = 20 + 55e^{-t/15}.

Common mistake: Forgetting the absolute value in $\ln|H-20|$ or dropping the constant of integration are the two most common errors. Both cost points.

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