2024 AP Calculus AB Free-Response Questions — Full Solutions

Complete worked solutions with scoring breakdowns for all 6 FRQs

6 Questions21 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

Coffee Temperature — Table Data

Rate & Data from Tables

Hard

The temperature of coffee in a cup at time

t

minutes is modeled by a decreasing differentiable function

C

, where

C(t)

is measured in degrees Celsius. For

0 \leq t \leq 12

, selected values of

C(t)

are given in the table shown.

Values of C(t)

t (minutes)	0	3	7	12
C(t) (degrees Celsius)	100	85	69	55

Part AEasy2 points

Approximate

C'(5)

using the average rate of change of

C

over the interval

3 \leq t \leq 7

. Show the work that leads to your answer and include units of measure.

Answer

C'(5) \approx \dfrac{C(7)-C(3)}{7-3} = \dfrac{69-85}{4} = -4

degrees Celsius per minute

Full Solution & Work

Set up the difference quotient

C'(5) \approx \frac{C(7)-C(3)}{7-3} = \frac{69-85}{4} = \frac{-16}{4} = -4

State the units

The units are degrees Celsius per minute.

AP Scoring — 2 Points

**P1**: A response must include a difference and a quotient as supporting work —

\frac{-16}{4}

\frac{69-85}{4}

, or equivalent is sufficient.
**P2**: Units of "degrees Celsius per minute" attached to a numerical value. Units alone with no numerical approximation do not earn this point.

Part BMedium3 points

Use a left Riemann sum with the three subintervals indicated by the data in the table to approximate the value of

\int_0^{12} C(t)\,dt

. Interpret the meaning of

\dfrac{1}{12}\int_0^{12} C(t)\,dt

in the context of the problem.

Answer

\int_0^{12} C(t)\,dt \approx 3\cdot C(0) + 4\cdot C(3) + 5\cdot C(7) = 300+340+345=985

Full Solution & Work

Set up the left Riemann sum

\int_0^{12} C(t)\,dt \approx (3-0)\cdot C(0) + (7-3)\cdot C(3) + (12-7)\cdot C(7)

Substitute values and compute

= 3\cdot 100 + 4\cdot 85 + 5\cdot 69 = 300+340+345=985

Interpret the average value expression

\dfrac{1}{12}\int_0^{12} C(t)\,dt

is the **average temperature of the coffee** (in degrees Celsius) **over the time interval from

t=0

t=12

**.

AP Scoring — 3 Points

**P1**: Correct form of the left Riemann sum — at least five of the six numerical factors must be correct.
**P2**: Correct estimate, 985 — requires P1 (a fully correct right Riemann sum, e.g.

4\cdot 85+4\cdot 69+5\cdot 55

, earns P1 but not P2).
**P3**: Interpretation must include both "average temperature" and the time interval; units are not required, but incorrect units forfeit this point.

Part CMedium3 points

For

12 \leq t \leq 20

, the rate of change of the temperature of the coffee is modeled by

C'(t) = \dfrac{-24.55e^{0.01t}}{t}

, where

C'(t)

is measured in degrees Celsius per minute. Find the temperature of the coffee at time

t=20

. Show the setup for your calculations.

Answer

C(20) = C(12) + \displaystyle\int_{12}^{20} C'(t)\,dt = 55-14.670812=40.329

Full Solution & Work

Set up C(20) using C(12) plus accumulated change

C(20) = C(12) + \int_{12}^{20} C'(t)\,dt

Substitute the known value and evaluate

= 55 + \int_{12}^{20} \frac{-24.55e^{0.01t}}{t}\,dt = 55 - 14.670812 = 40.329188

State the answer

The temperature of the coffee at time

t=20

is **40.329** degrees Celsius.

AP Scoring — 3 Points

**P1**: A definite integral with integrand

C'(t)

. If the limits of integration are incorrect, this response is not eligible for the third point.
**P2**: Adding

C(12)

(or 55) to a definite integral with lower limit 12.
**P3**: Correct answer

55-14.671

(with supporting work) — an answer of just 40.329 with no supporting work earns no points at all.

Common mistake: A "linkage error" like writing $C(20)=\int_{12}^{20}C'(t)\,dt = 55-14.670812$ (treating the whole right side as equal to the integral alone) still earns the first two points but not the third — keep the equation chain unambiguous.

Part DMedium1 point

For the model defined in part (c), it can be shown that

C''(t) = \dfrac{0.2455e^{0.01t}(100-t)}{t^2}

. For

12 < t < 20

, determine whether the temperature of the coffee is changing at a decreasing rate or at an increasing rate. Give a reason for your answer.

Answer

Increasing rate, because

C''(t) > 0

12 < t < 20

Full Solution & Work

Determine the sign of C''(t)

For

12<t<20

0.01t < 1

e^{0.01t}>0

;

100-t>0

(since

t<20<100

);

t^2>0

. So

C''(t)>0

throughout this interval.

Interpret the sign

Because

C''(t)>0

12<t<20

, the rate of change

C'(t)

is **increasing** on this interval — that is, the temperature is changing at an **increasing rate**.

AP Scoring — 1 Point

**P1**: Earned only for a correct answer with a correct reason that references the sign of the **second derivative** of

C

. A reason based on evaluating

C''(t)

at only a single point does not earn this point, and neither does an ambiguous pronoun like "it is positive, so increasing."

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