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2023 AP Calculus AB FRQ Question 3: Warming Milk — Differential Equation

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2023.

Question 3

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Warming Milk — Differential Equation

Differential Equations

Hard
A bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed. The increasing function MM models the temperature of the milk at time tt, where M(t)M(t) is measured in degrees Celsius and tt is the number of minutes since the bottle was placed in the pan. MM satisfies the differential equation dMdt=14(40−M)\dfrac{dM}{dt} = \dfrac{1}{4}(40-M). At time t=0t=0, the temperature of the milk is 5°C. It can be shown that M(t)<40M(t) < 40 for all values of tt.
Part AEasy1 point
A slope field for the differential equation dMdt=14(40−M)\dfrac{dM}{dt}=\dfrac{1}{4}(40-M) is shown. Sketch the solution curve through the point (0,5)(0,5).

Answer

The curve rises from (0,5)(0,5), concave down, approaching the asymptote M=40M=40.
Full Solution & Work

Follow the slope field from (0, 5)

Starting at (0,5)(0,5), follow the slope-field arrows: since M<40M<40 throughout, dMdt=14(40−M)>0\frac{dM}{dt}=\frac14(40-M)>0 always, so the curve rises steadily. As MM climbs toward 40, (40−M)(40-M) shrinks, so the slope decreases toward 0 — the curve is concave down and levels off, approaching the horizontal asymptote M=40M=40 without crossing it.

AP Scoring — 1 Point

**P1**: The solution curve must pass through (0,5)(0,5), extend reasonably close to the left and right edges of the given rectangle, lie entirely below M=40M=40, and have no obvious conflicts with the given slope lines.
Part BEasy2 points
Use the line tangent to the graph of MM at t=0t=0 to approximate M(2)M(2), the temperature of the milk at time t=2t=2 minutes.

Answer

M(2)≈22.5°CM(2)\approx 22.5°C
Full Solution & Work

Find the slope at t = 0

dMdt∣t=0=14(40−5)=354\left.\frac{dM}{dt}\right|_{t=0} = \frac{1}{4}(40-5) = \frac{35}{4}

Apply the tangent-line approximation

y=5+354(t−0)  ⟹  M(2)≈5+354⋅2=22.5y = 5+\frac{35}{4}(t-0) \implies M(2)\approx 5+\frac{35}{4}\cdot 2 = 22.5

The temperature of the milk at time t=2t=2 minutes is approximately **22.5°C**.

AP Scoring — 2 Points

**P1**: 14(40−5)\frac{1}{4}(40-5) — a subsequent simplification error forfeits the second point.
**P2**: An approximation using a tangent line through (0,5)(0,5) with slope 354\frac{35}{4} (or a declared value of dMdt\frac{dM}{dt}), evaluated at t=2t=2. An unsupported approximation does not earn this point.
Part CMedium2 points
Write an expression for d2Mdt2\dfrac{d^2M}{dt^2} in terms of MM. Use d2Mdt2\dfrac{d^2M}{dt^2} to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of M(2)M(2). Give a reason for your answer.

Answer

d2Mdt2=−116(40−M)\dfrac{d^2M}{dt^2}=-\dfrac{1}{16}(40-M); the tangent line approximation is an overestimate.
Full Solution & Work

Differentiate dM/dt with respect to t

d2Mdt2=14(−dMdt)=−14(14(40−M))=−116(40−M)\frac{d^2M}{dt^2} = \frac{1}{4}\left(-\frac{dM}{dt}\right) = -\frac{1}{4}\left(\frac{1}{4}(40-M)\right)=-\frac{1}{16}(40-M)

Determine the sign and conclude

Because M(t)<40M(t)<40 for all tt, d2Mdt2<0\frac{d^2M}{dt^2}<0, so the graph of MM is **concave down**. Therefore, the tangent line approximation of M(2)M(2) is an **overestimate**.

AP Scoring — 2 Points

**P1**: Either d2Mdt2=−14(14(40−M))\frac{d^2M}{dt^2}=-\frac{1}{4}\left(\frac{1}{4}(40-M)\right) or the simplified −116(40−M)-\frac{1}{16}(40-M) — a subsequent simplification error forfeits this point.
**P2**: Requires stating d2Mdt2<0\frac{d^2M}{dt^2}<0 (or "dMdt\frac{dM}{dt} is decreasing" or "MM is concave down") **and** concluding "overestimate." An argument based on concavity at a single point does not earn this point.
Part DHard4 points
Use separation of variables to find an expression for M(t)M(t), the particular solution to the differential equation dMdt=14(40−M)\dfrac{dM}{dt}=\dfrac{1}{4}(40-M) with initial condition M(0)=5M(0)=5.

Answer

M(t)=40−35e−t/4M(t) = 40-35e^{-t/4}
Full Solution & Work

Separate variables

dM40−M=14 dt\frac{dM}{40-M} = \frac{1}{4}\,dt

Integrate both sides

−ln⁡∣40−M∣=14t+C-\ln|40-M| = \frac{1}{4}t+C

Apply the initial condition

−ln⁡∣40−5∣=0+C  ⟹  C=−ln⁡35-\ln|40-5|=0+C \implies C=-\ln 35

Because M(t)<40M(t)<40 for all tt, 40−M>040-M>0, so ∣40−M∣=40−M|40-M|=40-M.
−ln⁡(40−M)=14t−ln⁡35  ⟹  ln⁡(40−M)=−14t+ln⁡35-\ln(40-M) = \frac{1}{4}t-\ln 35 \implies \ln(40-M)=-\frac{1}{4}t+\ln 35

Solve for M

40−M=35e−t/4  ⟹  M=40−35e−t/440-M = 35e^{-t/4} \implies M = 40-35e^{-t/4}

AP Scoring — 4 Points

**P1**: Correct separation of variables — no separation earns 0 of 4 points.
**P2**: Finds antiderivatives on both sides (presenting −ln⁡(40−M)-\ln(40-M) without absolute value symbols is still eligible for all 4 points).
**P3**: Correctly includes the constant of integration and uses the initial condition. Requires the first 2 points.
**P4**: Earned only for the final answer M=40−35e−t/4M=40-35e^{-t/4} (or equivalent) — requires the first 3 points.

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