Worksheets · AP Calculus AB · Unit 4

AP Calculus AB Unit 4 FRQ Practice: Derivatives in Context

Four exam-style free-response questions on motion, related rates, linear approximation and L'Hospital's rule: one with a graphing calculator, three without. Each is scored out of 9, with a scoring guide and a printable PDF with room to write, free to use in class. Good practice before the Unit 4 Progress Check in AP Classroom. Multiple choice for this unit: Unit 4 worksheet.

The exam rarely builds a whole free-response question on Unit 4, so this set is practice with its skills in the free-response format. The mixed set is the closest to a real exam.

Sigma Prep · sigmaprep.io/worksheets/ap-calculus-ab

Name

AP Calculus AB Unit 4 Free Response: Derivatives in Context

Date Version A

Directions: Show your work in the space under each part: the setup, the values you use and your reasoning. A justification needs a mathematical reason, with the conditions of any theorem or test you use checked. An answer with no supporting work may earn no credit. Give decimal answers correct to three decimal places unless a part says otherwise. Each question is worth 9 points. Unless a question says otherwise, the domain of a function f is every real number x for which f(x) is a real number.

This set is practice with Unit 4 skills in the free-response format: the exam almost never builds a whole question on Unit 4 alone. The mixed free-response set is the closest to a real exam.

Part A: A graphing calculator is required.

Question 1

Set your calculator to radian mode. When you use it to solve an equation, find a derivative at a point or evaluate a definite integral, write down the equation, function or integral you used.

1.
The height of a plant, in centimeters, $t$ weeks after it is planted is modeled by $H (t) = 6 + 24 ⁣ (1 - e^{- t^{2} / 40})$ for $t \geq 0$ .
1. (a)
  Find $H^{'} (7)$ , showing the setup for your calculations, and say what this number tells you about the situation at that moment, using correct units.
2. (b)
  Find how fast $H$ changes on average from $t = 7$ to $t = 9$ . Then find the time in $7 < t < 9$ at which $H^{'} (t)$ equals that average. Show the setup for your calculations.
1. (c)
  At $t = 1$ , is $H^{'} (t)$ , the rate at which the height of the plant changes, itself increasing or decreasing? Give a reason for your answer.
2. (d)
  Find $\lim_{t \to \infty} H (t)$ , and describe, using correct units, what it says about the height of the plant over a long time.

Part B: No calculator is allowed.

Questions 2–4

2.
The point $P = (2, 2)$ lies on the curve $x^{2} + 2 x y + y^{3} = 20$ .
1. (a)
  Show that $\frac{d y}{d x} = - \frac{2 x + 2 y}{2 x + 3 y^{2}}$ .
2. (b)
  Write an equation of the tangent line to the curve at $P$ , and use the line to estimate the $y$ -coordinate of the nearby point of the curve with $x$ -coordinate $1.8$ .
1. (c)
  A bead slides along the curve, with $x$ and $y$ in centimeters and $t$ in seconds. As the bead passes through $P$ , its $y$ -coordinate grows by $6$ centimeters per second. Find how fast the bead's distance from the origin is changing at that moment. Indicate units of measure.
3.
Two particles, $R$ and $S$ , travel along the $y$ -axis. At each time $t \geq 0$ , particle $R$ is at $y_{R} (t) = 2 t^{3} - 12 t^{2} + 18 t - 2$ and particle $S$ is at $y_{S} (t) = 5 + \frac{6}{t + 1}$ .
1. (a)
  Find expressions for the velocities $v_{R} (t)$ and $v_{S} (t)$ of the two particles.
2. (b)
  During $0 < t < 4$ , on which open intervals does particle $R$ travel downward? Give a reason for your answer.
1. (c)
  At the instant $t = 5$ , is particle $R$ speeding up or slowing down? Give a reason for your answer.
2. (d)
  Determine whether the distance between particles $R$ and $S$ is increasing or decreasing at time $t = 6$ . Explain your reasoning.
4.
Let $f$ , $g$ , and $h$ be functions that are twice differentiable for all $x$ . The graphs of $g$ and $h$ both pass through the point $(4, - 4)$ , and at that point both graphs have the tangent line $y = - 4 + (x - 4)$ .
1. (a)
  Find $h^{'} (4)$ .
2. (b)
  Consider $p (x) = \frac{x + 1}{h (x)}$ . Find $p^{'} (x)$ in terms of $h (x)$ and $h^{'} (x)$ , and then compute $p^{'} (4)$ .
1. (c)
  The function $h$ satisfies $h (x) = \frac{e^{x - 4} - 1}{\sqrt{f (x)} - 2}$ for $x \neq 4$ . Explain why $\lim_{x \to 4} \frac{e^{x - 4} - 1}{\sqrt{f (x)} - 2} = - 4$ , and use this to find $f (4)$ . Then use L'Hospital's Rule to find $f^{'} (4)$ , showing the steps you use.
2. (d)
  It is known that $g (x) \leq h (x)$ for $3 < x < 5$ . Let $k$ be a function satisfying $g (x) \leq k (x) \leq h (x)$ for $3 < x < 5$ . Is $k$ continuous at $x = 4$ ? Justify your answer.
  End of Version A

Free AP Calculus AB worksheets at sigmaprep.io/worksheets/ap-calculus-abOriginal questions written by Sigma Prep. AP is a registered trademark of the College Board, which is not affiliated with this sheet.

Scoring guide · Version A

Question 1A rate of change in context · calculator · 9 points

(a) 2 points Derivative value with units interpretation · Topic 4.3

1 point: $H^{'} (7) = 2.468$
1 point: interpretation with the time, the value and units: at time $t = 7$ weeks, the height of the plant is increasing at a rate of $2.468$ centimeters per week

Solution

On the calculator, $H^{'} (7) = 2.468$ ; its units are centimeters per week.
At time $t = 7$ weeks, the height of the plant is increasing at a rate of $2.468$ centimeters per week.

(b) 3 points Instantaneous rate equal to the average rate · Topic 4.3

1 point: average rate of change $\frac{H (9) - H (7)}{9 - 7} = 1.941$
1 point: $H^{'} (t) = 1.941$
1 point: $t = 7.994$

Solution

On the calculator, $H (7) = 22.950$ and $H (9) = 26.832$ , so the average rate of change is $\frac{H (9) - H (7)}{9 - 7} = 1.941$ centimeters per week.
Graph $y = H^{'} (t)$ (the calculator's derivative) with $y = 1.941$ : they meet once for $7 < t < 9$ , at $t = 7.994$ .

(c) 2 points Increasing or decreasing at a time · Topic 4.3

1 point: $H^{''} (1) = 1.112$
1 point: increasing, with the reason ( $H^{''} (1) > 0$ )

Solution

The rate of change of the height of the plant is $H^{'} (t)$ ; it is increasing where its derivative $H^{''} (t)$ is positive and decreasing where it is negative.
On the calculator, $H^{''} (1) = 1.112$ , which is positive, so the rate of change is increasing at $t = 1$ .

(d) 2 points Limit at infinity · Topic 1.15

1 point: $\lim_{t \to \infty} H (t) = 30$
1 point: meaning: as time goes on, the height of the plant approaches $30$ centimeters

Solution

As $t \to \infty$ , $\frac{t^{2}}{40} \to \infty$ , so $e^{- t^{2} / 40} \to 0$ and $H (t) \to 6 + 24 = 30$ .
The limit says that as time goes on, the height of the plant approaches $30$ centimeters.

Question 2Related rates · no calculator · 9 points

(a) 2 points Implicit differentiation · Topic 3.2

1 point: implicit differentiation:
$2 x + 2 y + 2 x \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x} = 0$
1 point: solves for $\frac{d y}{d x}$ (verification)

Solution

Differentiate implicitly, with the product rule on $2 x y$ :
$2 x + 2 y + 2 x \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x} = 0$ .
Collect the $\frac{d y}{d x}$ terms and divide:
$\frac{d y}{d x} = - \frac{2 x + 2 y}{2 x + 3 y^{2}}$ .

(b) 3 points Tangent line or tangent-line approximation · Topic 4.6

1 point: $\frac{d y}{d x} = - \frac{1}{2}$ at $P$
1 point: $y = 2 - \frac{1}{2} (x - 2)$
1 point: $y \approx 2.1$

Solution

At $P$ :
$\frac{d y}{d x} =$ $- \frac{2 (2) + 2 (2)}{2 (2) + 3 {(2)}^{2}} =$ $- \frac{1}{2}$ .
The tangent line is $y = 2 - \frac{1}{2} (x - 2)$ ; at $x = 1.8$ it gives $y \approx 2.1$ .

(c) 4 points Related rates · Topic 4.4

1 point: $\frac{d x}{d t} = - 12$ , from $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ with $\frac{d y}{d x} = - \frac{1}{2}$ at $P$
1 point: $D^{2} = x^{2} + y^{2}$ , so $2 D \frac{d D}{d t} = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}$
Also acceptable: differentiating $D = \sqrt{x^{2} + y^{2}}$ directly:
$\frac{d D}{d t} = \frac{x \frac{d x}{d t} + y \frac{d y}{d t}}{\sqrt{x^{2} + y^{2}}}$
1 point: $\frac{d D}{d t} = - 3 \sqrt{2}$
1 point: units:
$\frac{d D}{d t} = - 3 \sqrt{2}$ centimeters per second

Solution

By the chain rule, $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ , with $\frac{d y}{d x} = - \frac{1}{2}$ at $P$ from part (b):
$6 = - \frac{1}{2} \frac{d x}{d t}$ , so $\frac{d x}{d t} = - 12$ .
$D^{2} = x^{2} + y^{2}$ ; differentiate with respect to $t$ :
$2 D \frac{d D}{d t} = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}$ .
At $P$ , $D = \sqrt{{(2)}^{2} + {(2)}^{2}} = \sqrt{8}$ , so $\frac{d D}{d t} =$ $\frac{2 (2) (- 12) + 2 (2) (6)}{2 \sqrt{8}} =$ $- 3 \sqrt{2}$ centimeters per second; the distance is decreasing.

Question 3Particle motion from a position function · no calculator · 9 points

(a) 2 points Velocity from position · Topic 4.2

1 point: $v_{R} (t) = 6 t^{2} - 24 t + 18$
1 point: $v_{S} (t) = - \frac{6}{{(t + 1)}^{2}}$

Solution

The velocity is the derivative of the position:
$v_{R} (t) = y_{R}^{'} (t) = 6 t^{2} - 24 t + 18$ and $v_{S} (t) = y_{S}^{'} (t) = - \frac{6}{{(t + 1)}^{2}}$ .

(b) 2 points Direction of motion · Topic 4.2

1 point: considers $v_{R} (t) = 0$ : $t = 1$ and $t = 3$
1 point: the interval $1 < t < 3$ , with the reason $v_{R} (t) < 0$ there

Solution

Particle $R$ moves down where $v_{R} (t) < 0$ . In $0 < t < 4$ , $v_{R} (t) = 6 t^{2} - 24 t + 18$ is $0$ only at $t = 1$ and $t = 3$ .
Testing a time in each interval between these zeros: $v_{R} (t) > 0$ for $0 < t < 1$ , $v_{R} (t) < 0$ for $1 < t < 3$ , $v_{R} (t) > 0$ for $3 < t < 4$ .

(c) 3 points Speeding up or slowing down · Topic 4.2

1 point: $a_{R} (t) = v_{R}^{'} (t) = 12 t - 24$
1 point: $v_{R} (5) = 48$ and $a_{R} (5) = 36$
1 point: speeding up, because $v_{R} (5)$ and $a_{R} (5)$ have the same sign

Solution

Differentiate the velocity from part (a): $a_{R} (t) = 12 t - 24$ .
At $t = 5$ : $v_{R} (5) = 48$ , which is positive, and $a_{R} (5) = 36$ , which is positive.
The speed increases when the velocity and the acceleration have the same sign and decreases when they have opposite signs, so particle $R$ is speeding up.

(d) 2 points Two particles relative to each other · Topic 4.2

1 point: positions: $y_{R} (6) = 106$ and $y_{S} (6) = \frac{41}{7}$ , so $R$ is above $S$ :
$D (6) = y_{R} (6) - y_{S} (6) = \frac{701}{7} > 0$
1 point: the distance is increasing (the particles move away from each other), with the reason: $v_{R} (6) = 90$ and $v_{S} (6) = - \frac{6}{49}$ , so $R$ moves up and $S$ moves down
Also acceptable: The derivative of the distance:
$D^{'} (6) = v_{R} (6) - v_{S} (6) = \frac{4416}{49} > 0$ and $D (6) > 0$ have the same sign, so $∣ D (t) ∣$ is increasing at $t = 6$ .

Solution

Positions at $t = 6$ : $y_{R} (6) = 106$ and $y_{S} (6) = \frac{41}{7}$ . $D (6) = y_{R} (6) - y_{S} (6) = \frac{701}{7} > 0$ , so particle $R$ is above particle $S$ .
Velocities, the derivatives of the positions: $v_{R} (6) = 90$ , which is positive, and $v_{S} (6) = - \frac{6}{49}$ , which is negative.
Particle $R$ moves up and particle $S$ moves down, so they are moving away from each other.

Question 4Functions sharing a tangent line: a rule from values, L'Hospital's Rule and the squeeze theorem · no calculator · 9 points

(a) 1 point Tangent line or tangent-line approximation · Topic 2.2

1 point: answer: $h^{'} (4) = 1$

Solution

The tangent line $y = - 4 + (x - 4)$ has slope $1$ , and the slope of the tangent line at $x = 4$ is $h^{'} (4)$ : $h^{'} (4) = 1$ .

(b) 3 points Quotient rule · Topic 2.8

1 point: form of the quotient rule, $\frac{(numerator)^{'} (denominator) - (numerator) (denominator)^{'}}{(denominator)^{2}}$ , applied with the numerator $x + 1$ and the denominator $h (x)$ ; this structure alone earns the point, even with an error in the derivative of the numerator or the denominator
1 point: $p^{'} (x) = \frac{h (x) - (x + 1) h^{'} (x)}{(h (x))^{2}}$
1 point: $p^{'} (4) = - \frac{9}{16}$

Solution

Quotient rule:
$p^{'} (x) = \frac{h (x) - (x + 1) h^{'} (x)}{(h (x))^{2}}$ .
Substitute $x = 4$ , $h (4) = - 4$ , and $h^{'} (4) = 1$ from part (a):
$p^{'} (4) =$ $\frac{- 4 - (4 + 1) (1)}{{(- 4)}^{2}} =$ $- \frac{9}{16}$ .

(c) 4 points L'Hospital's rule · Topic 4.7

1 point: $\lim_{x \to 4} \frac{e^{x - 4} - 1}{\sqrt{f (x)} - 2} = - 4$ , because $h$ is continuous at $x = 4$
1 point: $f (4) = 4$
1 point: L'Hospital's Rule, with the condition that the numerator and denominator both approach $0$ as $x \to 4$ :
$\lim_{x \to 4} \frac{e^{x - 4}}{\frac{f^{'} (x)}{2 \sqrt{f (x)}}} = - 4$
1 point: $f^{'} (4) = - 1$

Solution

$h$ is continuous at $x = 4$ , so $\lim_{x \to 4} \frac{e^{x - 4} - 1}{\sqrt{f (x)} - 2} =$ $h (4) =$ $- 4$ .
The numerator approaches $0$ while the quotient approaches $- 4 \neq 0$ , so the denominator approaches $0$ : $\sqrt{f (4)} = 2,$ so $f (4) = 4$ .
L'Hospital's Rule:
$\lim_{x \to 4} \frac{e^{x - 4}}{\frac{f^{'} (x)}{2 \sqrt{f (x)}}} = - 4$ . At $x = 4$ this is $\frac{1}{\frac{1}{4} \cdot f^{'} (4)} = - 4$ , so $f^{'} (4) = - 1$ .

(d) 1 point Squeeze theorem · Topic 1.8

1 point: continuous, with justification:
$\lim_{x \to 4} k (x) = - 4 = k (4)$ by the squeeze theorem

Solution

$g$ and $h$ are continuous at $x = 4$ (they are differentiable), and both have the value $- 4$ there, so both have limit $- 4$ .
$k$ is trapped between them, so $\lim_{x \to 4} k (x) = - 4$ by the squeeze theorem, and $k (4)$ is trapped between $- 4$ and $- 4$ .
The limit equals the value: $k$ is continuous at $x = 4$ .

What does the AP Calculus AB free response ask on Unit 4?

A lot: Unit 4 is about 19% of the free-response points since 2012, second only to Unit 8.

A particle moving along a line: about 38% of Unit 4 free-response points; 1 on this set.
Tangent-line approximation and L’Hospital’s rule: about 26% of Unit 4 free-response points; not on this set.
Related rates: about 19% of Unit 4 free-response points; 1 on this set.
A rate of change in context: about 13% of Unit 4 free-response points; 1 on this set.
A function given by a formula: about 10% of Unit 4 free-response points; 1 on this set.

This set has 4 questions in that mix, 1 with a calculator (about 36% of Unit 4 free-response points are in the calculator section), each scored out of 9 with a scoring guide.

Counted from the official AP Calculus AB free-response questions, 2012–2026. No official question is reproduced here.

AP Calculus AB Unit 4 free-response settings by share of exam points

Free-response settings on this set and their share of official AP Calculus AB free-response points, 2012–2026
Setting	Questions on this set	Share of Unit 4 free-response points	Share of all free-response points	Last on the exam
A rate of change in contextCalculator	1 of 4	13%	–	2024
Related ratesNo calculator	1 of 4	19%	–	2026
Particle motion from a position functionNo calculator	1 of 4	16%	–	2026
Functions sharing a tangent line: a rule from values, L'Hospital's Rule and the squeeze theoremNo calculator	1 of 4	10%	3%	–

A part counts toward a setting when its primary skill is in Unit 4, in the same section, and the setting asks that skill (in the part’s own setting where there is one), on the main exams 2012–2026, by its points; a part that fits two settings is shared between them.

Questions about this free-response set

Unit 4, Contextual Applications of Differentiation, has 7 topics in the College Board's course description: interpreting a derivative in context, with units; straight-line motion (position, velocity and acceleration, when a particle is at rest or changes direction, and when it speeds up or slows down); rates of change in other applied contexts; related rates, from a given equation and from geometry such as circles, spheres, ladders and cones; approximating a value with the tangent line (linearization); and L'Hospital's rule for 0/0 and ∞/∞ limits. The course description weights the unit at 10–15% of the multiple-choice section.

Each question is worth 9 points, spread over its three or four parts. Points are earned for specific pieces of work: a setup, an antiderivative, a value with its units, a conclusion with its reason. The scoring guide lists every point and what earns it, then a worked solution for each part. On the exam the six free-response questions are half of the score.

Only where the exam allows one. On this set, question 1 is Part A, where a graphing calculator is required, and questions 2–4 are Part B, without one. A calculator question has parts that need one (a value to three decimal places, an equation solved on the calculator); its other parts can be done by hand, as on the exam.

We tagged every free-response part from 2012 to 2026 by the skill it tests and the setting of its question. The Unit 4 settings the exam leans on most get the most questions here, each in its own context with its own parts.

No. Every question here is original, written by Sigma Prep in the style of the exam: the settings, functions, numbers and wording are our own, and no official question or scoring guideline is copied or reworded. Every value and statement in the scoring guide was checked by an independent computer-algebra check before the set was published.

Yes. Each version has its own: open it under the questions, or download the PDF, which prints the questions with room to write and then the scoring guide on its own pages. There is also a PDF without the scoring guide to hand out.

AP is a registered trademark of the College Board, which was not involved in making these questions and does not endorse them. All questions and scoring guides are original to Sigma Prep. Frozen 2026-10-24.

AP Calculus AB Unit 4 FRQ Practice: Derivatives in Context

Part A: A graphing calculator is required.

Part B: No calculator is allowed.

Scoring guide · Version A

Question 1A rate of change in context · calculator · 9 points

Question 2Related rates · no calculator · 9 points

Question 3Particle motion from a position function · no calculator · 9 points

Question 4Functions sharing a tangent line: a rule from values, L'Hospital's Rule and the squeeze theorem · no calculator · 9 points

What does the AP Calculus AB free response ask on Unit 4?

AP Calculus AB Unit 4 free-response settings by share of exam points

More AP Calculus AB free response

Questions about this free-response set

What is on AP Calculus AB Unit 4?+

How is an AP Calculus AB free-response question scored?+

Can students use a calculator?+

How were the questions chosen?+

Are these real AP exam questions?+

Is there a scoring guide?+