2023 AP Calculus AB Free-Response Questions — Full Solutions

Complete worked solutions with scoring breakdowns for all 6 FRQs

6 Questions24 Parts TotalFull Solutions Included

Questions 1–2 (Part A, 30 min) permit a graphing calculator. Questions 3–6 (Part B, 60 min) do not.

Question 1

Calculator OK

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

f

, where

f(t)

is measured in gallons per second and

t

is measured in seconds since pumping began. Selected values of

f(t)

are given in the table shown.

Values of f(t)

t (seconds)	0	60	90	120	135	150
f(t) (gallons per second)	0	0.1	0.15	0.1	0.05	0

Part AMedium3 points

Using correct units, interpret the meaning of

\int_{60}^{135} f(t)\,dt

in the context of the problem. Use a right Riemann sum with the three subintervals

[60,90]

[90,120]

, and

[120,135]

to approximate the value of

\int_{60}^{135} f(t)\,dt

Answer

≈ 8.25 gallons

Full Solution & Work

Interpret the integral

\int_{60}^{135} f(t)\,dt

represents the **total number of gallons of gasoline pumped into the gas tank** from time

t=60

seconds to time

t=135

seconds.

Set up the right Riemann sum

\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

AP Scoring — 3 Points

**P1**: The interpretation must reference gallons of gasoline added/pumped and the time interval

t=60

t=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

c

, for

60<c<120

, such that

f'(c)=0

? Justify your answer.

Answer

Yes, by the Mean Value Theorem.

Full Solution & Work

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

f

is differentiable, so

f

is continuous on

[60,120]

\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

f

is differentiable on

(60,120)

and continuous on

[60,120]

, there must exist a

c

, for

60<c<120

, such that

f'(c)

equals the average rate of change over

[60,120]

, which is 0.

AP Scoring — 2 Points

**P1**: Presenting either

f(120)-f(60)=0

0.1-0.1=0

, or

f(60)=f(120)

.
**P2**: Requires P1. The response must also state

f

is continuous because

f

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) = \left(\dfrac{t}{500}\right)\cos\!\left(\left(\dfrac{t}{120}\right)^2\right)

for

0\leq t\leq 150

. Using this model, find the average rate of flow of gasoline over the time interval

0\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

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

Evaluate with a calculator

= 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

g

defined in part (c), find the value of

g'(140)

. Interpret the meaning of your answer in the context of the problem.

Answer

g'(140)\approx -0.005

(or

-0.004

)

Full Solution & Work

Differentiate g and evaluate at t = 140

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=140

AP Scoring — 2 Points

**P1**: A numerical value for

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)

]" and "at

t=140

" — saying only "decreasing at a rate of

-0.005

" (a double-negative) does not earn this point.

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