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2021 AP Calculus AB FRQ Question 6: Medication in the Bloodstream — Differential Equations

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2021.

Question 6

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Medication in the Bloodstream — Differential Equations

Differential Equations

Hard
A medication is administered to a patient. The amount, in milligrams, of the medication in the patient at time tt hours is modeled by a function y=A(t)y=A(t) that satisfies the differential equation dydt=12−y3\dfrac{dy}{dt}=\dfrac{12-y}{3}. At time t=0t=0 hours, there are 0 milligrams of the medication in the patient.
Part AEasy1 point
A portion of the slope field for the differential equation dydt=12−y3\dfrac{dy}{dt}=\dfrac{12-y}{3} is given. Sketch the solution curve through the point (0,0)(0,0).

Answer

An increasing, concave-down curve from (0,0)(0,0) approaching y=12y=12.
Full Solution & Work

Determine the shape from the differential equation

Since y(0)=0<12y(0)=0<12, dydt=12−03=4>0\frac{dy}{dt}=\frac{12-0}{3}=4>0 initially, so the curve rises steeply from the origin. As yy increases toward 12, (12−y)(12-y) shrinks, so the slope flattens — the curve is concave down throughout and levels off, asymptotically approaching y=12y=12 (matching part (b)'s limit) without ever crossing it.

AP Scoring — 1 Point

**P1**: The solution curve must pass through (0,0)(0,0), stay below y=12y=12, rise steeply at first and then level off, matching the given slope field.
Part BEasy1 point
Using correct units, interpret the statement lim⁡t→∞A(t)=12\lim\limits_{t\to\infty} A(t)=12 in the context of the problem.

Answer

As time increases without bound, the amount of medication in the patient approaches 12 milligrams.
Full Solution & Work

State the interpretation

As time tt increases without bound, the amount of medication in the patient's body **approaches (stabilizes at) 12 milligrams**.

AP Scoring — 1 Point

**P1**: Interpretation must reference both the long-run behavior (as t→∞t\to\infty) and the units (milligrams).
Part CHard4 points
Use separation of variables to find y=A(t)y=A(t), the particular solution to the differential equation dydt=12−y3\dfrac{dy}{dt}=\dfrac{12-y}{3} with initial condition A(0)=0A(0)=0.

Answer

A(t)=12−12e−t/3A(t)=12-12e^{-t/3}
Full Solution & Work

Separate variables

dy12−y=13 dt\frac{dy}{12-y} = \frac13\,dt

Integrate both sides

−ln⁡∣12−y∣=t3+C-\ln|12-y| = \frac{t}{3}+C

Apply the initial condition A(0) = 0

−ln⁡∣12−0∣=0+C  ⟹  C=−ln⁡12-\ln|12-0|=0+C \implies C=-\ln12

Since y(0)=0<12y(0)=0<12 and yy increases toward 12, 12−y>012-y>0, so ∣12−y∣=12−y|12-y|=12-y.
−ln⁡(12−y)=t3−ln⁡12  ⟹  ln⁡(12−y)=−t3+ln⁡12-\ln(12-y) = \frac t3-\ln12 \implies \ln(12-y)=-\frac t3+\ln12

Solve for y

12−y=12e−t/3  ⟹  y=12−12e−t/312-y = 12e^{-t/3} \implies y = 12-12e^{-t/3}

AP Scoring — 4 Points

**P1**: Correct separation of variables.
**P2**: Correct antiderivatives on both sides.
**P3**: Correctly includes the constant of integration and applies the initial condition A(0)=0A(0)=0.
**P4**: Correct final answer, A(t)=12−12e−t/3A(t)=12-12e^{-t/3}.
Part DHard3 points
A different procedure is used to administer the medication to a second patient. The amount, in milligrams, of the medication in the second patient at time tt hours is modeled by a function y=B(t)y=B(t) that satisfies the differential equation dydt=3−yt+2\dfrac{dy}{dt}=3-\dfrac{y}{t+2}. At time t=1t=1 hour, there are 2.5 milligrams of the medication in the second patient. Is the rate of change of the amount of medication in the second patient increasing or decreasing at time t=1t=1? Give a reason for your answer.

Answer

Decreasing, since d2ydt2∣t=1=−49<0\dfrac{d^2y}{dt^2}\Big|_{t=1}=-\dfrac{4}{9}<0.
Full Solution & Work

Differentiate the differential equation with respect to t

Applying the Quotient Rule to yt+2\frac{y}{t+2}:
d2ydt2=−y′(t+2)−y(t+2)2\frac{d^2y}{dt^2} = -\frac{y'(t+2)-y}{(t+2)^2}

Find B'(1) first

B′(1)=3−2.51+2=3−56=136B'(1) = 3-\frac{2.5}{1+2} = 3-\frac{5}{6}=\frac{13}{6}

Evaluate the second derivative at t = 1

With y=2.5=52y=2.5=\frac52, y′=136y'=\frac{13}{6}, t+2=3t+2=3:
d2ydt2∣t=1=−136(3)−5232=−132−529=−49\frac{d^2y}{dt^2}\bigg|_{t=1} = -\frac{\frac{13}{6}(3)-\frac52}{3^2} = -\frac{\frac{13}{2}-\frac52}{9} = -\frac{4}{9}

Conclude

Since d2ydt2∣t=1=−49<0\dfrac{d^2y}{dt^2}\Big|_{t=1}=-\frac49<0, the rate of change dydt\frac{dy}{dt} (i.e. B′(t)B'(t)) is **decreasing** at t=1t=1.

AP Scoring — 3 Points

**P1**: Correctly finds B′(1)=136B'(1)=\frac{13}{6} from the given differential equation.
**P2**: Correctly sets up d2ydt2\dfrac{d^2y}{dt^2} using the Quotient Rule (or equivalent) on the right side of the differential equation.
**P3**: Correct final value, −49-\frac49, and correct conclusion ("decreasing").

Common mistake: This part is asking about the second derivative — whether the *rate itself* is speeding up or slowing down — not the sign of $B'(1)$ alone. Confusing "is $y$ increasing" with "is $\frac{dy}{dt}$ increasing" is the most common error on this type of question.

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