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2024 AP Calculus AB FRQ Question 3: Seawater Depth — Differential Equation

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2024.

Question 3

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Seawater Depth — Differential Equation

Differential Equations

Hard
The depth of seawater at a location can be modeled by the function HH that satisfies the differential equation dHdt=12(H−1)cos⁡ ⁣(t2)\dfrac{dH}{dt} = \dfrac{1}{2}(H-1)\cos\!\left(\dfrac{t}{2}\right), where H(t)H(t) is measured in feet and tt is measured in hours after noon (t=0t=0). It is known that H(0)=4H(0)=4.
Part AEasy1 point
A portion of the slope field for the differential equation is provided. Sketch the solution curve, y=H(t)y=H(t), through the point (0,4)(0,4).

Answer

The curve passes through (0,4)(0,4), rises to a local max around t=πt=\pi, then decreases.
Full Solution & Work

Follow the slope field from (0, 4)

Starting at (0,4)(0,4), follow the slope-field arrows: since H>1H>1 and cos⁡(t/2)>0\cos(t/2)>0 for small t>0t>0, dHdt>0\frac{dH}{dt}>0 initially, so the curve rises. Slopes flatten and turn negative once tt passes π\pi (where cos⁡(t/2)=0\cos(t/2)=0), so the curve peaks near t=πt=\pi and then decreases through the rest of the shown domain (matches the official figure, which peaks around H≈9H\approx 9 near t≈3t\approx 3).

AP Scoring — 1 Point

**P1**: The solution curve must pass through (0,4)(0,4), extend to at least t=4.5t=4.5, and have no obvious conflicts with the given slope lines.
Part BHard3 points
For 0<t<50<t<5, it can be shown that H(t)>1H(t)>1. Find the value of tt, for 0<t<50<t<5, at which HH has a critical point. Determine whether the critical point corresponds to a relative minimum, a relative maximum, or neither a relative minimum nor a relative maximum of the depth of seawater at the location. Justify your answer.

Answer

t=πt=\pi is a relative maximum.
Full Solution & Work

Set dH/dt = 0

Because H(t)>1H(t)>1, dHdt=0\frac{dH}{dt}=0 implies cos⁡ ⁣(t2)=0\cos\!\left(\frac{t}{2}\right)=0, which gives t=πt=\pi as a critical point in 0<t<50<t<5.

Sign-analyze dH/dt around t = π

For 0<t<π0<t<\pi: dHdt>0\frac{dH}{dt}>0. For π<t<5\pi<t<5: dHdt<0\frac{dH}{dt}<0. Therefore, t=πt=\pi is the location of a **relative maximum** value of HH.

AP Scoring — 3 Points

**P1**: Considers dHdt=0\frac{dH}{dt}=0, dHdt>0\frac{dH}{dt}>0, dHdt<0\frac{dH}{dt}<0, cos⁡(t/2)=0\cos(t/2)=0, or cos⁡(t/2)>0\cos(t/2)>0/<0<0.
**P2**: Identifies t=πt=\pi, with or without supporting work.
**P3**: Cannot be earned without P1. Earned only for a correct justification and the correct answer "relative maximum." The justification can be a sign analysis of dHdt\frac{dH}{dt} on either side of π\pi, or a Second Derivative Test showing d2Hdt2∣t=π<0\frac{d^2H}{dt^2}\big|_{t=\pi}<0.
Part CHard5 points
Use separation of variables to find y=H(t)y=H(t), the particular solution to the differential equation dHdt=12(H−1)cos⁡ ⁣(t2)\dfrac{dH}{dt}=\dfrac{1}{2}(H-1)\cos\!\left(\dfrac{t}{2}\right) with initial condition H(0)=4H(0)=4.

Answer

H(t)=1+3esin⁡(t/2)H(t) = 1+3e^{\sin(t/2)}
Full Solution & Work

Separate variables

dHH−1=12cos⁡ ⁣(t2)dt\frac{dH}{H-1} = \frac{1}{2}\cos\!\left(\frac{t}{2}\right)dt

Integrate both sides

∫dHH−1=∫12cos⁡ ⁣(t2)dt\int \frac{dH}{H-1} = \int \frac{1}{2}\cos\!\left(\frac{t}{2}\right)dt

ln⁡∣H−1∣=sin⁡ ⁣(t2)+C\ln|H-1| = \sin\!\left(\frac{t}{2}\right)+C

Apply the initial condition

ln⁡∣4−1∣=sin⁡(0)+C  ⟹  C=ln⁡3\ln|4-1| = \sin(0)+C \implies C=\ln 3

Because H(0)=4H(0)=4, H>1H>1, so ∣H−1∣=H−1|H-1|=H-1.
ln⁡(H−1)=sin⁡ ⁣(t2)+ln⁡3\ln(H-1) = \sin\!\left(\frac{t}{2}\right)+\ln 3

Solve for H

H−1=esin⁡(t/2)+ln⁡3=3esin⁡(t/2)H-1 = e^{\sin(t/2)+\ln 3} = 3e^{\sin(t/2)}

H(t)=1+3esin⁡(t/2)H(t) = 1+3e^{\sin(t/2)}

AP Scoring — 5 Points

**P1**: Correct separation of variables — a response with no separation of variables earns 0 of 5 points.
**P2**: One correct antiderivative (either side).
**P3**: Both correct antiderivatives — presenting ln⁡(H−1)\ln(H-1) without absolute value symbols still earns this point.
**P4**: Correctly includes the constant of integration and uses the initial condition H(0)=4H(0)=4. Requires P1 and at least one of P2/P3.
**P5**: Earned only for the final answer H(t)=1+3esin⁡(t/2)H(t)=1+3e^{\sin(t/2)} (or an equivalent form) — requires all 4 previous points.

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