Worksheets · AP Calculus AB · Unit 7

AP Calculus AB Unit 7 FRQ Practice: Differential Equations

Four exam-style free-response questions on separation of variables, particular solutions, slope fields and exponential growth and decay, all without a calculator. 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 7 Progress Check in AP Classroom. Multiple choice for this unit: Unit 7 worksheet.

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

Name

AP Calculus AB Unit 7 Free Response: Differential Equations

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.

Part B: No calculator is allowed.

Questions 1–4

1.
A rectangular box 4 feet long and 2 feet wide holds water, which drains out through a layer of gravel in its base. (The volume of water in the box is its length times its width times the depth of the water.) The depth of the water at time $t$ minutes is $h$ feet, and $h$ satisfies the differential equation $\frac{d h}{d t} = - \sqrt{h}$ while water remains in the box.
1. (a)
  Find the rate at which the volume of water in the box is changing when the depth of the water is 1 foot. Indicate units of measure.
2. (b)
  When the depth of the water is 4 feet, is $\frac{d h}{d t}$ increasing or decreasing? Explain your reasoning.
1. (c)
  Given that $h (0) = 4$ , separate the variables to find $h$ as a function of $t$ .
2.
Let $P (t)$ be the number of people, in thousands, in a city who have seen a news report, where $t$ is measured in hours. The function $P$ is the solution of $\frac{d P}{d t} = \frac{t}{4} (6 - P)$ with $P (0) = 1$ . A slope field for this differential equation is shown.
1. (a)
  Sketch, on the given slope field, the particular solution whose graph contains the point $(0, 1)$ .
2. (b)
  It can be shown that $\lim_{t \to \infty} P (t) = 6$ . Explain the meaning of this limit in the context of the problem, using correct units.
1. (c)
  Given $P (0) = 1$ , use separation of variables to find $P (t)$ .
2. (d)
  In a second city, the number of people, in thousands, who have seen the report at time $t$ hours is $J (t)$ , where $\frac{d J}{d t} = \frac{t + 1}{J^{2} + 2}$ and $J (3) = 1$ . Is the rate of change of $J$ increasing or decreasing at $t = 3$ ? Give a reason for your answer.
3.
The graph of $y = f (x)$ passes through $(1, 0)$ , and $f$ solves $\frac{d y}{d x} = x^{2} e^{- y}$ . A slope field for this equation is shown.
1. (a)
  The point $(1, 0)$ is marked on the slope field. Sketch the solution curve through it.
2. (b)
  Use the line tangent to the graph of $f$ at $x = 1$ to approximate $f (0.8)$ , and give an equation of that tangent line.
1. (c)
  On the interval $0.8 \leq x \leq 1$ , the second derivative of $f$ is positive. Is the approximation from part (b) an overestimate or an underestimate of $f (0.8)$ ? Give a reason for your answer.
2. (d)
  Solve the differential equation by separating the variables: find $f (x)$ if $f (1) = 0$ .
4.
Let $G (t)$ model the amount of grain, in hundreds of tons, stored in a silo at time $t$ weeks, for $0 \leq t \leq 3$ . It is known that $G (0) = 2$ , that $\frac{d G}{d t} = \frac{2 t (2 - t)}{G}$ , and that $G (t) > 0$ on this interval. The slope field of the differential equation is shown.
1. (a)
  The point $(0, 2)$ is marked on the slope field. Sketch the solution curve through it.
2. (b)
  Find the value of $t$ , for $0 < t < 3$ , at which $G$ has a critical point. Is it a relative minimum, a relative maximum, or neither? Justify your answer.
1. (c)
  Solve the differential equation by separating the variables: find $G (t)$ if $G (0) = 2$ .
  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 draining tank: related rates, the second derivative from the equation, separation · no calculator · 9 points

(a) 2 points Related rates · Topic 4.5

1 point: $\frac{d V}{d t}$ in terms of $\frac{d h}{d t}$ :
$\frac{d V}{d t} = 8 \frac{d h}{d t}$
1 point: answer with units:
$\frac{d V}{d t} = - 8$ cubic feet per minute

Solution

The volume of water when the depth is $h$ feet is $V = 4 (2) h = 8 h$ cubic feet, so $\frac{d V}{d t} = 8 \frac{d h}{d t}$ .
When $h = 1$ , $\frac{d h}{d t} = - 1$ , so $\frac{d V}{d t} = - 8$ cubic feet per minute: the volume is decreasing at $8$ cubic feet per minute.

(b) 3 points Second derivative at a point · Topic 3.6

1 point: $\frac{d}{d h} ⁣ (- \sqrt{h}) = - \frac{1}{2 \sqrt{h}}$
Also acceptable: the argument from the equation below, in place of this point
1 point: $\frac{d^{2} h}{d t^{2}} = - \frac{1}{2 \sqrt{h}} \cdot \frac{d h}{d t}$
Also acceptable: the argument from the equation below, in place of this point
1 point: increasing, with an explanation from the sign of $\frac{d^{2} h}{d t^{2}}$
Also acceptable: An argument from the equation: $h$ is decreasing because $\frac{d h}{d t} < 0$ , and as $h$ decreases, $- \sqrt{h}$ increases toward 0, so $\frac{d h}{d t}$ is increasing. A complete argument of this kind earns all 3 points.

Solution

By the chain rule, $\frac{d^{2} h}{d t^{2}} =$ $\frac{d}{d h} ⁣ (- \sqrt{h}) \cdot \frac{d h}{d t} =$ $- \frac{1}{2 \sqrt{h}} \cdot \frac{d h}{d t}$ .
Substituting $\frac{d h}{d t} = - \sqrt{h}$ :
$\frac{d^{2} h}{d t^{2}} =$ $(- \frac{1}{2 \sqrt{h}}) (- \sqrt{h}) =$ $\frac{1}{2}$ .
When $h = 4$ , $\frac{d^{2} h}{d t^{2}} = \frac{1}{2}$ , which is positive, so the rate of change of the depth is increasing when the depth of the water is 4 feet.

(c) 4 points Separable differential equation · Topic 7.7

1 point: separation of variables (without it, 0 of the 4 points)
1 point: both antiderivatives
1 point: constant of integration and uses the initial condition (without a constant of integration, at most the first 2 points)
1 point: solves for the dependent variable (eligible only with the first 3 points)

Solution

Separate the variables:
$\frac{1}{\sqrt{h}} d h = - d t$ .
Integrate: $2 \sqrt{h} = - t + C$ .
With $h = 4$ at $t = 0$ , $C = 4$ .
So $\sqrt{h} = 2 - \frac{1}{2} t$ , and $h = {(2 - \frac{1}{2} t)}^{2}$ while water remains in the box.

Question 2A model with a limiting value, and a second differential equation · no calculator · 9 points

(a) 1 point Solution curve on a slope field · Topic 7.3

1 point: solution curve through $(0, 1)$ that follows the slope field

Solution

At $(0, 1)$ , $\frac{d P}{d t} = 0$ , so the graph of $P$ has a horizontal tangent there.
Starting at that point, follow the segments to the right: the curve is tangent to the segment at every point it passes and continues to the edge of the slope field (see the figure).

(b) 1 point Limit at infinity · Topic 1.15

1 point: interpretation with units: the number of people who have seen the report approaches 6 thousand people

Solution

As time goes on, the number of people who have seen the report gets closer and closer to $6$ thousand people: in the long run it approaches 6 thousand people.

(c) 4 points Separable differential equation · Topic 7.7

1 point: separation of variables (without it, 0 of the 4 points)
1 point: both antiderivatives
1 point: constant of integration and uses the initial condition (without a constant of integration, at most the first 2 points)
1 point: solves for the dependent variable (eligible only with the first 3 points)

Solution

Separate the variables:
$\frac{1}{6 - P} d P = \frac{t}{4} d t$ .
Integrate:
$- \ln ∣ 6 - P ∣ = \frac{t^{2}}{8} + C$ .
With $P (0) = 1$ : $- \ln 5 = C$ , so $C = - ⁣ \ln 5$ .
Since $P (0) = 1 < 6$ , $6 - P > 0$ and $∣ 6 - P ∣ = 6 - P$ . Then $6 - P = 5 e^{- \frac{t^{2}}{8}}$ , so $P (t) = 6 - 5 e^{- \frac{t^{2}}{8}}$ .

(d) 3 points Second derivative at a point · Topic 3.6

1 point: $\frac{d^{2} J}{d t^{2}}$ by the quotient rule
Also acceptable: Implicit differentiation of $(J^{2} + 2) \frac{d J}{d t} = t + 1$ with the product rule, solved for $\frac{d^{2} J}{d t^{2}}$ .
1 point: $\frac{d^{2} J}{d t^{2}} = - \frac{23}{27}$ at $t = 3$
1 point: decreasing, with a reason from the sign of $\frac{d^{2} J}{d t^{2}}$

Solution

By the quotient rule, $\frac{d^{2} J}{d t^{2}} = \frac{(J^{2} + 2) (1) - (t + 1) (2 J) \frac{d J}{d t}}{{(J^{2} + 2)}^{2}}$ .
At $t = 3$ , $J = 1$ and $\frac{d J}{d t} = \frac{4}{3}$ , so $\frac{d^{2} J}{d t^{2}} =$ $\frac{3 - 4 \cdot 2 \cdot \frac{4}{3}}{9} =$ $- \frac{23}{27}$ .
Since $\frac{d^{2} J}{d t^{2}} < 0$ at $t = 3$ , the rate of change of $J$ is decreasing at $t = 3$ .

Question 3Slope field, tangent-line approximation, concavity and separation of variables · no calculator · 9 points

(a) 1 point Solution curve on a slope field · Topic 7.3

1 point: solution curve through $(1, 0)$ that follows the slope field

Solution

At $(1, 0)$ , $\frac{d y}{d x} = 1$ , so the graph of $f$ is increasing there.
Starting at that point, follow the segments to the left and to the right: the curve is tangent to the segment at every point it passes and continues to the edge of the slope field (see the figure).

(b) 2 points Tangent line or tangent-line approximation · Topic 4.6

1 point: tangent line
1 point: approximation

Solution

At $(1, 0)$ , $\frac{d y}{d x} = 1$ , so an equation of the tangent line is $y = x - 1$ .
$f (0.8) \approx 0.8 - 1 = - 0.2$ .

(c) 1 point Over- or underestimate from concavity · Topic 4.6

1 point: underestimate, with a reason based on concavity

Solution

Because $f^{''} (x) > 0$ for $0.8 \leq x \leq 1$ , the graph of $f$ is concave up on that interval.
A tangent line to a graph that is concave up lies below the graph, so the approximation $f (0.8) \approx - 0.2$ is an underestimate.

(d) 5 points Separable differential equation · Topic 7.7

1 point: separation of variables (without it, 0 of the 5 points)
1 point: antiderivative of the side in the dependent variable
1 point: antiderivative of the side in the independent variable
1 point: constant of integration and uses the initial condition (without a constant of integration, at most the first 3 points)
1 point: solves for the dependent variable (eligible only with the first 4 points)

Solution

Separate the variables: $e^{y} d y = x^{2} d x$ .
Integrate both sides:
$e^{y} = \frac{1}{3} x^{3} + C$ .
Since $f (1) = 0$ :
$1 = \frac{1}{3} + C$ , so $C = \frac{2}{3}$ .
So $e^{y} = \frac{1}{3} x^{3} + \frac{2}{3}$ , and $f (x) = \ln ⁣ (\frac{1}{3} x^{3} + \frac{2}{3})$ .

Question 4A differential equation in context: slope field, relative extremum, separation · no calculator · 9 points

(a) 1 point Solution curve on a slope field · Topic 7.3

1 point: solution curve through $(0, 2)$ that follows the slope field

Solution

At $(0, 2)$ , $\frac{d G}{d t} = 0$ , so the graph of $G$ has a horizontal tangent there.
Starting at that point, follow the segments to the right: the curve is tangent to the segment at every point it passes and continues to the edge of the slope field (see the figure).

(b) 3 points Relative min/max classification · Topic 5.4

1 point: sets $\frac{d G}{d t} = 0$
1 point: identifies $t = 2$
1 point: relative maximum, with a justification (the sign change of $\frac{d G}{d t}$ )
Also acceptable: The second derivative test: since the numerator of $\frac{d G}{d t}$ is 0 at $t = 2$ , $\frac{d^{2} G}{d t^{2}} = - 4 \cdot \frac{1}{G (2)}$ there, which is negative because $G (2) > 0$ , so $G$ has a relative maximum at $t = 2$ .

Solution

Because $G (t) > 0$ , $\frac{d G}{d t} = \frac{2 t (2 - t)}{G}$ equals 0 only where its numerator is 0: $t (2 - t) = 0$ gives $t = 2$ (the value $t = 0$ is not inside the interval).
For $0 < t < 2$ , $\frac{d G}{d t} > 0$ , and for $2 < t < 3$ , $\frac{d G}{d t} < 0$ , since the numerator is positive before $t = 2$ and negative after it while the denominator stays positive.
So $\frac{d G}{d t}$ changes sign from positive to negative at $t = 2$ , and $G$ has a relative maximum at $t = 2$ .

(c) 5 points Separable differential equation · Topic 7.7

1 point: separation of variables (without it, 0 of the 5 points)
1 point: antiderivative of the side in the dependent variable
1 point: antiderivative of the side in the independent variable
1 point: constant of integration and uses the initial condition (without a constant of integration, at most the first 3 points)
1 point: solves for the dependent variable (eligible only with the first 4 points)

Solution

Separate the variables:
$G d G = (4 t - 2 t^{2}) d t$ .
Integrate:
$\frac{G^{2}}{2} = 2 t^{2} - \frac{2}{3} t^{3} + C$ .
With $G (0) = 2$ :
$\frac{4}{2} = C$ , so $C = 2$ and $G^{2} = 4 + 4 t^{2} - \frac{4}{3} t^{3}$ .
Because $G (t) > 0$ , take the positive square root:
$G (t) = \sqrt{4 + 4 t^{2} - \frac{4}{3} t^{3}}$ .

What does the AP Calculus AB free response ask on Unit 7?

A fair share: Unit 7 is about 10% of the free-response points since 2012.

A differential equation: every Unit 7 free-response question since 2012; all 4 on this set.
Separation of variables with an initial condition: about 79% of Unit 7 free-response points.
A slope field to sketch: about 16% of Unit 7 free-response points.
Writing or checking a differential equation: about 3% of Unit 7 free-response points.

This set has 4 questions in that mix, all without a calculator (all Unit 7 free-response points are in the no-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 7 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 7 free-response points	Share of all free-response points	Last on the exam
A differential equation in context: slope field, relative extremum, separationNo calculator	1 of 4	20%	15%	–
A model with a limiting value, and a second differential equationNo calculator	1 of 4	20%	15%	–
A draining tank: related rates, the second derivative from the equation, separationNo calculator	1 of 4	20%	15%	–
Slope field, tangent-line approximation, concavity and separation of variablesNo calculator	1 of 4	20%	15%	–

A part counts toward a setting when its primary skill is in Unit 7, 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 7, Differential Equations, has 7 AB topics in the College Board's course description: modeling situations with differential equations; verifying solutions; sketching slope fields; reasoning using slope fields; finding general solutions by separation of variables; finding particular solutions from an initial condition; and exponential models, dy/dt = ky. Topics 7.5 and 7.9 (Euler's method and logistic models) are BC only. The course description weights the unit at 5–10% of the multiple-choice section, and a separable differential equation was on the free-response section of 12 of the 15 exams from 2012 to 2026.

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, all four are in Part B, without a calculator, as almost every Unit 7 free-response part on the exam has been.

We tagged every free-response part from 2012 to 2026 by the skill it tests and the setting of its question. The Unit 7 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 7 FRQ Practice: Differential Equations

Part B: No calculator is allowed.

Scoring guide · Version A

Question 1A draining tank: related rates, the second derivative from the equation, separation · no calculator · 9 points

Question 2A model with a limiting value, and a second differential equation · no calculator · 9 points

Question 3Slope field, tangent-line approximation, concavity and separation of variables · no calculator · 9 points

Question 4A differential equation in context: slope field, relative extremum, separation · no calculator · 9 points

What does the AP Calculus AB free response ask on Unit 7?

AP Calculus AB Unit 7 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 7?+

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?+