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2023 AP Calculus AB FRQ Question 1: Gas Station Flow Rate — Table Data

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

Question 1

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Gas Station Flow Rate — Table Data

Rate & Data from Tables

Hard
A customer at a gas station is pumping gasoline into a gas tank. The rate of flow of gasoline is modeled by a differentiable function ff, where f(t)f(t) is measured in gallons per second and tt is measured in seconds since pumping began. Selected values of f(t)f(t) are given in the table shown.

Values of f(t)

t (seconds)06090120135150
f(t) (gallons per second)00.10.150.10.050
Part AMedium3 points
Using correct units, interpret the meaning of ∫60135f(t) dt\int_{60}^{135} f(t)\,dt in the context of the problem. Use a right Riemann sum with the three subintervals [60,90][60,90], [90,120][90,120], and [120,135][120,135] to approximate the value of ∫60135f(t) dt\int_{60}^{135} f(t)\,dt.

Answer

≈ 8.25 gallons
Full Solution & Work

Interpret the integral

∫60135f(t) dt\int_{60}^{135} f(t)\,dt represents the **total number of gallons of gasoline pumped into the gas tank** from time t=60t=60 seconds to time t=135t=135 seconds.

Set up the right Riemann sum

∫60135f(t) dt≈f(90)(90−60)+f(120)(120−90)+f(135)(135−120)\int_{60}^{135} f(t)\,dt \approx f(90)(90-60)+f(120)(120-90)+f(135)(135-120)

Substitute and compute

=(0.15)(30)+(0.1)(30)+(0.05)(15)=4.5+3+0.75=8.25= (0.15)(30)+(0.1)(30)+(0.05)(15) = 4.5+3+0.75=8.25

AP Scoring — 3 Points

**P1**: The interpretation must reference gallons of gasoline added/pumped and the time interval t=60t=60 to t=135t=135.
**P2**: At least five of the six numerical factors in the Riemann sum must be correct.
**P3**: The final answer 8.25 — an error anywhere in the Riemann sum forfeits this point.

Common mistake: A fully correct **left** Riemann sum with supporting work (e.g. $f(60)(30)+f(90)(30)+f(120)(15)=9$) earns 1 of the last 2 points — it demonstrates a valid Riemann-sum process but isn't the right sum the question asked for.

Part BMedium2 points
Must there exist a value of cc, for 60<c<12060<c<120, such that f′(c)=0f'(c)=0? Justify your answer.

Answer

Yes, by the Mean Value Theorem.
Full Solution & Work

Compute the average rate of change on [60,120]

ff is differentiable, so ff is continuous on [60,120][60,120].
f(120)−f(60)120−60=0.1−0.160=0\frac{f(120)-f(60)}{120-60} = \frac{0.1-0.1}{60}=0

Apply the Mean Value Theorem

By the **Mean Value Theorem**, since ff is differentiable on (60,120)(60,120) and continuous on [60,120][60,120], there must exist a cc, for 60<c<12060<c<120, such that f′(c)f'(c) equals the average rate of change over [60,120][60,120], which is 0.

AP Scoring — 2 Points

**P1**: Presenting either f(120)−f(60)=0f(120)-f(60)=0, 0.1−0.1=00.1-0.1=0, or f(60)=f(120)f(60)=f(120).
**P2**: Requires P1. The response must also state ff is continuous because ff is differentiable, and answer "yes." Citing the Intermediate Value Theorem here (instead of the Mean Value Theorem) does **not** earn this point.
Part CMedium2 points
The rate of flow of gasoline, in gallons per second, can also be modeled by g(t)=(t500)cos⁡ ⁣((t120)2)g(t) = \left(\dfrac{t}{500}\right)\cos\!\left(\left(\dfrac{t}{120}\right)^2\right) for 0≤t≤1500\leq t\leq 150. Using this model, find the average rate of flow of gasoline over the time interval 0≤t≤1500\leq t\leq 150. Show the setup for your calculations.

Answer

≈ 0.096 (or 0.095) gallons per second
Full Solution & Work

Set up the average value formula

1150−0∫0150g(t) dt\frac{1}{150-0}\int_0^{150} g(t)\,dt

Evaluate with a calculator

=0.0959967= 0.0959967

The average rate of flow of gasoline is **0.096** (or 0.095) gallons per second.

AP Scoring — 2 Points

**P1**: The correct average-value formula, whether presented in one or multiple steps.
**P2**: Correct answer 0.096 (or 0.095). Degree-mode calculators give the wrong value (0.150 or 0.003) and forfeit this point.
Part DMedium2 points
Using the model gg defined in part (c), find the value of g′(140)g'(140). Interpret the meaning of your answer in the context of the problem.

Answer

g′(140)≈−0.005g'(140)\approx -0.005 (or −0.004-0.004)
Full Solution & Work

Differentiate g and evaluate at t = 140

g′(140)≈−0.004908≈−0.005 (or −0.004)g'(140) \approx -0.004908 \approx -0.005 \text{ (or } -0.004\text{)}

Interpret the meaning

The rate at which gasoline is flowing into the tank is **decreasing** at a rate of **0.005** (or 0.004) gallon per second per second at time t=140t=140.

AP Scoring — 2 Points

**P1**: A numerical value for g′(140)g'(140) — this value may only appear inside the interpretation.
**P2**: The interpretation must include "the rate of flow of gasoline is changing at a rate of [the declared g′(140)g'(140)]" and "at t=140t=140" — saying only "decreasing at a rate of −0.005-0.005" (a double-negative) does not earn this point.

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