2025 AP Calc AB FRQ Answers & Solutions

Question 1

Calculator OK

Invasive Species Spread Rate

Rate, Average Value & Optimization

Hard

An invasive species of plant appears in a fruit grove at time

t = 0

and begins to spread. The function

C

defined by

C(t) = 7.6\arctan(0.2t)

models the number of acres in the fruit grove affected by the species

t

weeks after the species appears. It can be shown that

C'(t) = \dfrac{38}{25 + t^2}

. (Calculator should be in radian mode.)

Part AEasy2 points

Find the average number of acres affected by the invasive species from time

t = 0

to time

t = 4

weeks. Show the setup for your calculations.

Answer

\dfrac{1}{4-0}\int_0^4 C(t)\,dt = \dfrac{1}{4}(11.112896) = 2.778224 \approx 2.778

acres

Full Solution & Work

Apply the average value formula

The average value of

C

on

[0,4]

is:

\frac{1}{4-0}\int_0^4 C(t)\,dt

Evaluate with a calculator

\int_0^4 C(t)\,dt = 11.112896

\frac{1}{4}(11.112896) = 2.778224

From

t=0

to

t=4

weeks, the average number of acres affected was **2.778 acres**.

AP Scoring — 2 Points

**P1**: Correct average-value setup

\frac{1}{4}\int_0^4 C(t)\,dt

(division by 4 can be shown implicitly via the correct answer).
**P2**: Correct answer 2.778, accurate to three decimal places.

Part BMedium2 points

Find the time

t

when the instantaneous rate of change of

C

equals the average rate of change of

C

over the time interval

0 \leq t \leq 4

. Show the setup for your calculations.

Answer

t = 2.154298 \approx 2.154

weeks

Full Solution & Work

Find the average rate of change of C on [0,4]

\frac{C(4)-C(0)}{4-0} = 1.282008

(since

C(0)=0

and

C(4) = 7.6\arctan(0.8) = 5.128031

)

Set the instantaneous rate equal to it and solve

C'(t) = \frac{38}{25+t^2} = 1.282008

Solving with a calculator:

t = 2.154298

.

AP Scoring — 2 Points

**P1**: Uses the correct average rate of change

\frac{C(4)-C(0)}{4-0}

(or an equivalent expression like

1.282

).
**P2**: Correct answer

t = 2.154

, supported by the equation

C'(t) = 1.282

.

Common mistake: Presenting only "$t = 1.282$" without the supporting equation earns neither point — the average rate of change and the equation $C'(t) = \text{average rate}$ must both appear.

Part CEasy2 points

Assume that the invasive species continues to spread according to the given model for all times

t > 0

. Write a limit expression that describes the end behavior of the rate of change in the number of acres affected by the species. Evaluate this limit expression.

Answer

\lim_{t \to \infty} C'(t) = \lim_{t \to \infty} \dfrac{38}{25+t^2} = 0

Full Solution & Work

Write the limit of the rate function

The rate of change is

C'(t)

, so the end behavior is described by:

\lim_{t \to \infty} C'(t) = \lim_{t \to \infty} \frac{38}{25+t^2}

Evaluate

As

t \to \infty

, the denominator

25+t^2 \to \infty

while the numerator stays fixed at 38, so:

\lim_{t \to \infty} \frac{38}{25+t^2} = 0

AP Scoring — 2 Points

**P1**: Correct limit expression — either

\lim_{t\to\infty} C'(t)

or

\lim_{t\to\infty} \frac{38}{25+t^2}

. Writing

\lim_{t\to\infty} C(t)

instead does **not** earn this point (that describes the acres, not the rate).
**P2**: Correct value 0.

Common mistake: A very common slip: writing $\lim_{t\to\infty} C(t)$ instead of $C'(t)$. The question asks about the *rate of change*, so it must be the derivative inside the limit.

Part DHard3 points

At time

t = 4

weeks after the invasive species appears in the fruit grove, measures are taken to counter the spread of the species. The function

A

, defined by

A(t) = C(t) - \int_4^t 0.1\ln(x)\,dx

, models the number of acres affected by the species over the time interval

4 \leq t \leq 36

. At what time

t

, for

4 \leq t \leq 36

, does

A

attain its maximum value? Justify your answer.

Answer

t = 11.442

(or

11.441

) weeks

Full Solution & Work

Differentiate A using FTC

By the Fundamental Theorem of Calculus:

A'(t) = C'(t) - 0.1\ln t

Set A'(t) = 0 and solve

A maximum on

[4,36]

occurs where

A'(t)=0

or at an endpoint.

A'(t) = C'(t) - 0.1\ln t = 0 \implies C'(t) = 0.1\ln t

Solving numerically:

t = 11.441700

.

Candidates test

\begin{array}{c|c} t & A(t) \\\hline 4 & 5.128031 \\ 11.4417 & 7.316978 \\ 36 & 1.743056 \end{array}

The largest value of

A(t)

occurs at

t = 11.442

.

AP Scoring — 3 Points

**P1**: Considers

A'(t) = 0

, i.e. sets

C'(t) - 0.1\ln t = 0

(equivalently

C'(t)=0.1\ln t

). Just presenting

t=11.4417

without this equation does **not** earn the point.
**P2**: A complete justification — either a candidates-test table evaluating

A(4)

,

A(11.442)

,

A(36)

, or a sign analysis of

A'(t)

around

t=11.442

.
**P3**: Correct final answer

t = 11.442

(or 11.441), supported by the work above.

Common mistake: A first- or second-derivative test alone (a *local* argument) does not earn the justification point on a closed-interval max/min question — AP requires a *global* argument: either compare $A$ at all candidates (endpoints + critical points), or show $A'$ changes from $+$ to $-$ specifically at $t=11.442$ and note it's the only critical point in the interval.

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Question 2

Calculator OK

Area and Volume Between f and g

Area & Volume

Open this question on its own page

Hard

The shaded region

R

is bounded by the graphs of the functions

f

and

g

, where

f(x) = x^2 - 2x

and

g(x) = x + \sin(\pi x)

, as shown in the figure. The two graphs intersect at

(0,0)

and

(3,3)

. (Calculator should be in radian mode.)

2025 AP Calculus AB FRQ Question 2 Graph — region R bounded by f and g — Region $R$, bounded by $y=f(x)$ and $y=g(x)$ on $[0,3]$.

Part AMedium2 points

Find the area of

R

. Show the setup for your calculations.

Answer

\text{Area} = \int_0^3 (g(x)-f(x))\,dx = 5.136620 \approx 5.137

Full Solution & Work

Set up the integral

On

[0,3]

,

g(x) \geq f(x)

, so:

\text{Area} = \int_0^3 \big(g(x)-f(x)\big)\,dx

Evaluate with a calculator

\int_0^3 (g(x)-f(x))\,dx = 5.136620

The area of

R

is **5.137** (or 5.136).

AP Scoring — 2 Points

**P1**: Correct integrand

g(x)-f(x)

(or

|g(x)-f(x)|

,

f(x)-g(x)

,

|f(x)-g(x)|

) in a definite integral over

[0,3]

.
**P2**: Correct answer 5.137 (or 5.136).

Part BMedium2 points

Region

R

is the base of a solid. For this solid, at each

x

the cross section perpendicular to the

x

-axis is a rectangle with height

x

and base in region

R

. Find the volume of the solid. Show the setup for your calculations.

Answer

V = \int_0^3 x\big(g(x)-f(x)\big)\,dx = 7.704930 \approx 7.705

Full Solution & Work

Identify the cross-sectional area

At each

x

, the rectangle's base (in region

R

) has length

g(x)-f(x)

, and its height is

x

. So the cross-sectional area is:

A(x) = x\big(g(x)-f(x)\big)

Integrate over [0,3] and evaluate

V = \int_0^3 x\big(g(x)-f(x)\big)\,dx = 7.704930

The volume of the solid is **7.705** (or 7.704).

AP Scoring — 2 Points

**P3**: A definite integral with integrand presented as a product of two nonconstant factors, one of which is

x

,

g(x)-f(x)

, or

f(x)-g(x)

.
**P4**: Correct answer 7.705 (or 7.704). The presence of

dx

is not required for scoring.

Part CHard3 points

Write, but do not evaluate, an integral expression for the volume of the solid generated when the region

R

is rotated about the horizontal line

y = -2

.

Answer

V = \pi \int_0^3 \Big[\big(g(x)+2\big)^2 - \big(f(x)+2\big)^2\Big]\,dx

Full Solution & Work

Identify outer and inner radii

Rotating about

y=-2

, the outer radius (farther curve,

g

) is

g(x)-(-2) = g(x)+2

, and the inner radius (closer curve,

f

) is

f(x)-(-2) = f(x)+2

.

Apply the washer method

V = \pi\int_0^3 \Big[\big(g(x)+2\big)^2 - \big(f(x)+2\big)^2\Big]\,dx

AP Scoring — 3 Points

**P5**: Definite integral with integrand of the form

R^2-r^2

, where one of

\{R,r\}

is correct or a difference between

g

and a nonzero constant, and the other similarly for

f

.
**P6**: The correct integrand

\big(g(x)+2\big)^2-\big(f(x)+2\big)^2

(or equivalent).
**P7**: Correct limits

[0,3]

, the constant

\pi

, and the differential — all present together. (A response missing

\pi

can still earn P5–P6 but not P7.)

Part DMedium2 points

It can be shown that

g'(x) = 1 + \pi\cos(\pi x)

. Find the value of

x

, for

0 < x < 1

, at which the line tangent to the graph of

f

is parallel to the line tangent to the graph of

g

.

Answer

x = 0.675819 \approx 0.676

Full Solution & Work

Set the two derivatives equal

Tangent lines are parallel when the slopes match:

f'(x)=g'(x)

.

f'(x) = 2x-2, \qquad 2x - 2 = 1 + \pi\cos(\pi x)

Solve numerically

Using a calculator:

x = 0.675819

.

AP Scoring — 2 Points

**P8**: A general statement

f'(x)=g'(x)

, or the correct equation formed by substituting

2x-2

for

f'(x)

,

1+\pi\cos(\pi x)

for

g'(x)

, or both.
**P9**: Correct answer

x = 0.676

(or 0.675).

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Question 3

No Calculator

Reading Rates — Table Data

Rate & Data from Tables

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Hard

A student starts reading a book at time

t = 0

minutes and continues reading for the next 10 minutes. The rate at which the student reads is modeled by the differentiable function

R

, where

R(t)

is measured in words per minute. Selected values of

R(t)

are given in the table shown.

Values of R(t)

t (minutes)	0	2	8	10
R(t) (words per minute)	90	100	150	162

Part AEasy2 points

Approximate

R'(1)

using the average rate of change of

R

over the interval

0 \leq t \leq 2

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

Answer

R'(1) \approx \dfrac{100-90}{2-0} = 5

words per minute per minute

Full Solution & Work

Set up the difference quotient

R'(1) \approx \frac{R(2)-R(0)}{2-0} = \frac{100-90}{2} = \frac{10}{2} = 5

State the units

The units are

\dfrac{\text{words/minute}}{\text{minute}} = \text{words per minute per minute}

(or words/min²).

AP Scoring — 2 Points

**P1**: The answer, supported by the difference quotient

\frac{100-90}{2-0}

using values from the table. Presenting the quotient with no numerical follow-through does not earn this point on its own.
**P2**: Correct units — "words per minute per minute" (or words/minute²), whether or not attached to a numerical value.

Part BMedium2 points

Must there be a value

c

, for

0 < c < 10

, such that

R(c) = 155

? Justify your answer.

Answer

Yes. Since

R

is differentiable (hence continuous) and

R(0)=90 < 155 < 162=R(10)

, by the Intermediate Value Theorem there is a

c \in (0,10)

with

R(c)=155

.

Full Solution & Work

Establish continuity

R

is given to be differentiable on

[0,10]

, so

R

is continuous on

[0,10]

.

Compare 155 to the endpoint values

R(0) = 90 < 155 < 162 = R(10)

Apply the Intermediate Value Theorem

Since

R

is continuous on

[0,10]

and 155 lies between

R(0)

and

R(10)

, the **Intermediate Value Theorem** guarantees a value

c

, with

0<c<10

, such that

R(c)=155

.

AP Scoring — 2 Points

**P3**: States that

R

is continuous *because*

R

is differentiable — simply asserting "

R

is continuous" with no justification does not earn this point.
**P4**: Indicates

R(0)<155

(or equivalently uses

R(2)<155

or

R(8)<155

) and

R(10)>155

, and answers "yes." (P4 does not require P3 to be earned.)

Common mistake: You don't have to name the Intermediate Value Theorem to earn credit, but if you do name a theorem, it must be the correct one — citing the Mean Value Theorem here is a common and costly mix-up.

Part CMedium2 points

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

\int_0^{10} R(t)\,dt

. Show the work that leads to your answer.

Answer

\int_0^{10} R(t)\,dt \approx \dfrac{90+100}{2}(2) + \dfrac{100+150}{2}(6) + \dfrac{150+162}{2}(2) = 190+750+312 = 1252

Full Solution & Work

Set up the trapezoidal sum

\int_0^{10} R(t)\,dt \approx \frac{R(0)+R(2)}{2}(2-0) + \frac{R(2)+R(8)}{2}(8-2) + \frac{R(8)+R(10)}{2}(10-8)

Substitute values and compute

= \frac{90+100}{2}(2) + \frac{100+150}{2}(6) + \frac{150+162}{2}(2)

= 190 + 750 + 312 = 1252

AP Scoring — 2 Points

**P5**: The form of a trapezoidal sum — three terms, each a product of two factors where one factor is a

\frac{1}{2}

(sum of endpoints)

\times

width. At least five of the six numeric factors must be correct.
**P6**: Correct answer 1252, with supporting work. To be eligible for P6, P5 must be earned. A fully correct left or right Riemann sum (1080 or 1424) earns P5 but **not** P6 — it isn't a trapezoidal sum.

Part DMedium3 points

A teacher also starts reading at time

t = 0

minutes and continues reading for the next 10 minutes. The rate at which the teacher reads is modeled by the function

W

defined by

W(t) = -\dfrac{3}{10}t^2 + 8t + 100

, where

W(t)

is measured in words per minute. Based on the model, how many words has the teacher read by the end of the 10 minutes? Show the work that leads to your answer.

Answer

\int_0^{10} W(t)\,dt = 1300

words

Full Solution & Work

Set up the integral

\int_0^{10} W(t)\,dt = \int_0^{10}\left(-\frac{3}{10}t^2+8t+100\right)dt

Find the antiderivative

= \left[-\frac{1}{10}t^3 + 4t^2 + 100t\right]_0^{10}

Evaluate

= \left(-\frac{1}{10}(1000) + 4(100) + 100(10)\right) - 0 = (-100+400+1000) = 1300

The teacher has read **1300 words** by the end of the 10 minutes.

AP Scoring — 3 Points

**P7**: Correct integral (definite or indefinite) with integrand

W(t)

.
**P8**: Correct antiderivative

-\frac{1}{10}t^3+4t^2+100t

.
**P9**: Correct final answer 1300, with P8 required to be eligible.

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Question 4

No Calculator

Accumulation Function from a Graph

Graph Analysis & Curve Sketching

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Hard

The continuous function

f

is defined on the closed interval

-6 \leq x \leq 12

. The graph of

f

, consisting of two semicircles (radius 3) and one line segment, is shown in the figure: a lower semicircle from

(-6,0)

down to

(-3,-3)

and back up to

(0,0)

, an upper semicircle from

(0,0)

up to

(3,3)

and back down to

(6,0)

, and a line segment from

(6,0)

to

(12,3)

. Let

g

be the function defined by

g(x) = \int_6^x f(t)\,dt

.

2025 AP Calculus AB FRQ Question 4 Graph of f — two semicircles and a line segment — Graph of $f$ on $[-6, 12]$.

Part AEasy2 points

Find

g'(8)

. Give a reason for your answer.

Answer

g'(8) = f(8) = 1

Full Solution & Work

Apply the Fundamental Theorem of Calculus

Since

g(x)=\int_6^x f(t)\,dt

, by FTC Part 1:

g'(x) = f(x)

.

Evaluate f(8) from the line segment

For

6 \leq x \leq 12

,

f

is the line through

(6,0)

and

(12,3)

, i.e.

f(x) = \frac{1}{2}(x-6)

.

f(8) = \frac{1}{2}(8-6) = 1 \implies g'(8) = 1

AP Scoring — 2 Points

**P1**: Considers

g'(x)=f(x)

.
**P2**: Correct answer

g'(8)=1

. (An implied use of FTC, like writing

g'(8)=1

directly by reading the graph, can also earn P2 without P1.)

Part BHard2 points

Find all values of

x

in the open interval

-6 < x < 12

at which the graph of

g

has a point of inflection. Give a reason for your answer.

Answer

x = -3

,

x = 3

, and

x = 6

— the graph of

g

has points of inflection at all three.

Full Solution & Work

Relate inflection points of g to f

g''(x) = f'(x)

, so

g

has an inflection point wherever

f'

changes sign — equivalently, wherever

f

changes from decreasing to increasing or increasing to decreasing (a local extremum of

f

).

Read local extrema off the graph of f

At

x=-3

:

f

decreases (left semicircle going down) then increases (coming back up) — a local minimum of

f

.
At

x=3

:

f

increases (right semicircle going up) then decreases (coming back down) — a local maximum of

f

.
At

x=6

:

f

is still decreasing just before

x=6

(end of the upper semicircle) and increasing just after (start of the line segment) — another local minimum of

f

, even though the graph has a corner there.

Conclusion

f

changes from decreasing to increasing at

x=-3

and

x=6

, and from increasing to decreasing at

x=3

. So the graph of

g

has points of inflection at

x=-3

,

x=3

, and

x=6

.

AP Scoring — 2 Points

**P3**: Correct answer — earned *only* for exactly

x=-3, 3, 6

. Naming any additional or different value forfeits both points.
**P4**: A reason tied to the given graph of

f

(e.g. "

f

changes from increasing to decreasing / decreasing to increasing there", or "

f

has a local extremum there"). A reason based on an ambiguous "the function" or "the graph," without specifying

f

, does not earn P4.

Common mistake: The corner at $x=6$ is easy to miss — students often only report $x=-3$ and $x=3$ since those look like the "obvious" smooth extrema, but $f$ still switches from decreasing to increasing there, so $g$ still has an inflection point.

Part CMedium2 points

Find

g(12)

and

g(0)

. Label your answers.

Answer

g(12) = 9

and

g(0) = -\dfrac{9\pi}{2}

Full Solution & Work

Compute g(12) as a signed area

g(12) = \int_6^{12} f(t)\,dt = \frac{1}{2}(6)(3) = 9

(area of the triangle under the line segment from

x=6

to

x=12

)

Compute g(0) as a signed area

g(0) = \int_6^0 f(x)\,dx = -\int_0^6 f(x)\,dx = -\frac{1}{2}\pi(3)^2 = -\frac{9\pi}{2}

(the upper semicircle from

x=0

to

x=6

has area

\frac{1}{2}\pi r^2

; the direction of integration flips the sign)

AP Scoring — 2 Points

**P5**:

g(12)=9

, with or without supporting work.
**P6**:

g(0)=-\frac{9\pi}{2}

, with or without supporting work. Both values must be clearly labeled — unlabeled numbers do not earn credit.

Part DHard3 points

Find the value of

x

at which

g

attains an absolute minimum on the closed interval

-6 \leq x \leq 12

. Justify your answer.

Answer

x = 0

Full Solution & Work

Find critical points of g

g

attains an absolute minimum either at a critical point (

g'(x)=f(x)=0

) or an endpoint. From the graph,

f(x)=0

at

x=0

and

x=6

.

Evaluate g at all candidates

\begin{array}{c|c} x & g(x) \\\hline -6 & 0 \\ 0 & -9\pi/2 \\ 6 & 0 \\ 12 & 9 \end{array}

(using

g(-6)=\int_6^{-6}f(t)\,dt = -\left(-\frac{9\pi}{2}+\frac{9\pi}{2}\right) = 0

, and the values of

g(0)

,

g(12)

from part C)

Conclusion

The smallest value in the table is

-\frac{9\pi}{2}

at

x=0

, so

g

attains its absolute minimum on

[-6,12]

at

x = 0

.

AP Scoring — 3 Points

**P7**: Considers

g'(x)=0

or

f(x)=0

(not earned by simply stating

x=0,6

with no equation).
**P8**: A complete justification — a candidates-test table evaluating

g

at

-6

,

0

,

6

,

12

(values may be imported from part C), or an equivalent global sign argument.
**P9**: Correct answer

x=0

.

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Question 5

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Two Particles in Motion

Particle Motion

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Hard

Two particles,

H

and

J

, are moving along the

x

-axis. For

0 \leq t \leq 5

, the position of particle

H

at time

t

is given by

x_H(t) = e^{t^2-4t}

and the velocity of particle

J

at time

t

is given by

v_J(t) = 2t(t^2-1)^3

.

Part AEasy2 points

Find the velocity of particle

H

at time

t = 1

. Show the work that leads to your answer.

Answer

v_H(1) = x_H'(1) = -2e^{-3}

Full Solution & Work

Differentiate x_H(t) with the Chain Rule

x_H'(t) = v_H(t) = (2t-4)e^{t^2-4t}

Evaluate at t = 1

v_H(1) = (2-4)e^{1-4} = -2e^{-3}

AP Scoring — 2 Points

**P1**: Considers

x_H'(t)

— presenting

x_H'(t)

,

x_H'(1)

, or

(2t-4)e^{t^2-4t}

earns this.
**P2**: Correct answer

-2e^{-3}

. An unsupported answer of

-2e^{-3}

alone earns P2 but not P1.

Part BHard3 points

During what open intervals of time

t

, for

0 < t < 5

, are particles

H

and

J

moving in opposite directions? Give a reason for your answer.

Answer

1 < t < 2

Full Solution & Work

Analyze the sign of v_H(t)

v_H(t) = (2t-4)e^{t^2-4t} = 0 \implies t=2

(the exponential factor is never 0).

v_H(t) < 0

for

0<t<2

(moving left);

v_H(t) > 0

for

2<t<5

(moving right).

Analyze the sign of v_J(t)

v_J(t) = 2t(t^2-1)^3 = 0 \implies t=0

or

t=1

on

(0,5)

.
For

0<t<1

:

(t^2-1)<0

, so

(t^2-1)^3<0

, and

2t>0

, giving

v_J(t)<0

(moving left).
For

1<t<5

:

(t^2-1)>0

, so

(t^2-1)^3>0

, giving

v_J(t)>0

(moving right).

Compare directions

H

: left on

(0,2)

, right on

(2,5)

.

J

: left on

(0,1)

, right on

(1,5)

.
The signs disagree only on

(1,2)

:

H

is still moving left there while

J

has already turned right.

AP Scoring — 3 Points

**P3**: Sets

v_H(t)=0

or

v_J(t)=0

(or identifies

t=2

for

H

and no other value, or

t=1

for

J

and no other value, or identifies the interval

1<t<2

outright).
**P4**: A correct sign/direction analysis for at least one of the two particles on

(0,5)

.
**P5**: Correct analyses for **both** particles, leading to the answer

1<t<2

.

Part CMedium1 point

It can be shown that

v_J'(2) > 0

. Is the speed of particle

J

increasing, decreasing, or neither at time

t = 2

? Give a reason for your answer.

Answer

Increasing, because

v_J(2) > 0

and

v_J'(2) > 0

have the same sign.

Full Solution & Work

Find the sign of v_J(2)

v_J(2) = 2(2)\big(2^2-1\big)^3 = 4(27) = 108 > 0

Compare signs of velocity and acceleration

Since

v_J(2)=108>0

and

v_J'(2)>0

(given), velocity and acceleration have the **same sign**, so the speed of particle

J

is **increasing** at

t=2

.

AP Scoring — 1 Point

**P6**: "Increasing," with the reason that

v_J(2)

and

v_J'(2)

have the same sign. (An explicit value for

v_J(2)

is not required, but if given it must be correct —

v_J(2)=108

.)

Part DMedium3 points

Particle

J

is at position

x = 7

at time

t = 0

. Find the position of particle

J

at time

t = 2

. Show the work that leads to your answer.

Answer

x_J(2) = 7 + \dfrac{1}{4}\big[(3)^4-(-1)^4\big] = 27

Full Solution & Work

Set up position as initial position + accumulated change

x_J(2) = x_J(0) + \int_0^2 v_J(t)\,dt = 7 + \int_0^2 2t(t^2-1)^3\,dt

Substitute u = t² − 1

Let

u=t^2-1

,

du=2t\,dt

. When

t=0

,

u=-1

; when

t=2

,

u=3

.

\int_0^2 2t(t^2-1)^3\,dt = \int_{-1}^3 u^3\,du = \left[\frac{u^4}{4}\right]_{-1}^3 = \frac{81-1}{4}=20

Combine

x_J(2) = 7+20 = 27

AP Scoring — 3 Points

**P7**: An indefinite or definite integral with integrand

v_J(t)

or

2t(t^2-1)^3

.
**P8**: A correct antiderivative of the form

k(t^2-1)^4

(equivalently) with

k=\frac{1}{4}

— a different constant

k

forfeits eligibility for P9.
**P9**: Correct final answer

x_J(2)=27

, requiring P8.

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Question 6

No Calculator

Implicit Curve — Tangent Lines & Related Rates

Implicit Differentiation

Open this question on its own page

Hard

Consider the curve

G

defined by the equation

y^3 - y^2 - y + \dfrac{1}{4}x^2 = 0

.

Part AMedium2 points

Show that

\dfrac{dy}{dx} = \dfrac{-x}{2(3y^2-2y-1)}

.

Answer

\dfrac{dy}{dx} = \dfrac{-x}{2(3y^2-2y-1)}

(shown by implicit differentiation)

Full Solution & Work

Differentiate both sides implicitly

\frac{d}{dx}\left(y^3-y^2-y+\frac{1}{4}x^2\right) = \frac{d}{dx}(0)

3y^2\frac{dy}{dx} - 2y\frac{dy}{dx} - \frac{dy}{dx} + \frac{x}{2} = 0

Solve for dy/dx

(3y^2-2y-1)\frac{dy}{dx} = -\frac{x}{2} \implies \frac{dy}{dx} = \frac{-x}{2(3y^2-2y-1)}

AP Scoring — 2 Points

**P1**: Correct implicit differentiation of

y^3-y^2-y+\frac14 x^2=0

(alternate notations for

\frac{dy}{dx}

, such as

y'

, are accepted).
**P2**: Algebraic verification that isolates

\frac{dy}{dx}

to reach the given expression — earned only if P1 was earned. Presenting

(3y^2-2y-1)\frac{dy}{dx}=-\frac{x}{2}

is sufficient, provided no later error appears.

Part BMedium2 points

There is a point

P

on the curve

G

near

(2,-1)

with

x

-coordinate 1.6. Use the line tangent to the curve at

(2,-1)

to approximate the

y

-coordinate of point

P

.

Answer

y \approx -1 - \dfrac{1}{4}(1.6-2) = -0.9

Full Solution & Work

Find the slope at (2, −1)

\left.\frac{dy}{dx}\right|_{(2,-1)} = \frac{-2}{2(3(1)-2(-1)-1)} = \frac{-2}{2(3+2-1)} = \frac{-2}{8} = -\frac{1}{4}

Apply the tangent line approximation

y \approx -1 + \left(-\frac{1}{4}\right)(1.6-2) = -1 - \frac{1}{4}(-0.4) = -1+0.1 = -0.9

AP Scoring — 2 Points

**P3**: Correct slope of the tangent line,

-\frac14

, at

(2,-1)

— a response that declares any other nonzero slope cannot earn this point (though it may still earn P4 for consistent use of that value).
**P4**: Correct tangent-line approximation, clearly evaluated at

x=1.6

:

y\approx -0.9

.

Part CMedium2 points

For

x > 0

and

y > 0

, there is a point

S

on the curve

G

at which the line tangent to the curve at that point is vertical. Find the

y

-coordinate of point

S

. Show the work that leads to your answer.

Answer

y = 1

Full Solution & Work

A vertical tangent occurs where dy/dx is undefined

From part A,

\frac{dy}{dx}=\frac{-x}{2(3y^2-2y-1)}

is undefined (vertical tangent, for

x\neq0

) when the denominator is 0:

2(3y^2-2y-1) = 0

Solve for y

3y^2-2y-1 = 0 \implies (3y+1)(y-1) = 0 \implies y = -\frac{1}{3} \text{ or } y = 1

Apply the y > 0 restriction

Since the problem requires

y>0

, only

y=1

is valid. So the

y

-coordinate of point

S

is

\boxed{1}

.

AP Scoring — 2 Points

**P5**: Sets the denominator equal to 0 — any of

2(3y^2-2y-1)=0

,

3y^2-2y-1=0

, or its factored form.
**P6**: Correct answer

y=1

, clearly identified as the only solution satisfying

y>0

(earned only if P5 was earned, and only if the algebra to isolate

y

is correct).

Part DHard3 points

A particle moves along the curve

H

defined by the equation

2xy + \ln y = 8

. At the instant when the particle is at the point

(4,1)

,

\dfrac{dx}{dt} = 3

. Find

\dfrac{dy}{dt}

at that instant. Show the work that leads to your answer.

Answer

\dfrac{dy}{dt} = -\dfrac{2}{3}

Full Solution & Work

Differentiate implicitly with respect to t

\frac{d}{dt}\big(2xy+\ln y\big) = \frac{d}{dt}(8)

2\frac{dx}{dt}y + 2x\frac{dy}{dt} + \frac{1}{y}\frac{dy}{dt} = 0

Substitute the known values

At

(x,y)=(4,1)

with

\frac{dx}{dt}=3

:

2(3)(1) + 2(4)\frac{dy}{dt} + \frac{1}{1}\frac{dy}{dt} = 0

6 + 9\frac{dy}{dt} = 0

Solve for dy/dt

\frac{dy}{dt} = -\frac{6}{9} = -\frac{2}{3}

AP Scoring — 3 Points

**P7**: Attempts implicit differentiation of

2xy+\ln y=8

with respect to

t

, with at most one error.
**P8**: A correct equation equivalent to

2\frac{dx}{dt}y+2x\frac{dy}{dt}+\frac{1}{y}\frac{dy}{dt}=0

— requires P7 with no errors.
**P9**: Correct final value

\frac{dy}{dt}=-\frac{2}{3}

, requiring both P7 and P8.

Common mistake: An equally valid alternate route: solve $\frac{dy}{dx}$ implicitly with respect to $x$ first, evaluate it at $(4,1)$ to get $-\frac{2}{9}$, then multiply by $\frac{dx}{dt}=3$ using the chain rule $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$.

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2025 AP Calculus AB Free-Response Questions — Full Solutions

Invasive Species Spread Rate

Apply the average value formula

Evaluate with a calculator

AP Scoring — 2 Points

Find the average rate of change of C on [0,4]

Set the instantaneous rate equal to it and solve

AP Scoring — 2 Points

Write the limit of the rate function

Evaluate

AP Scoring — 2 Points

Differentiate A using FTC

Set A'(t) = 0 and solve

Candidates test

AP Scoring — 3 Points

Area and Volume Between f and g

Set up the integral

Evaluate with a calculator

AP Scoring — 2 Points

Identify the cross-sectional area

Integrate over [0,3] and evaluate

AP Scoring — 2 Points

Identify outer and inner radii

Apply the washer method

AP Scoring — 3 Points

Set the two derivatives equal

Solve numerically

AP Scoring — 2 Points

Reading Rates — Table Data

Set up the difference quotient

State the units

AP Scoring — 2 Points

Establish continuity

Compare 155 to the endpoint values

Apply the Intermediate Value Theorem

AP Scoring — 2 Points

Set up the trapezoidal sum

Substitute values and compute

AP Scoring — 2 Points

Set up the integral

Find the antiderivative

Evaluate

AP Scoring — 3 Points

Accumulation Function from a Graph

Apply the Fundamental Theorem of Calculus

Evaluate f(8) from the line segment

AP Scoring — 2 Points

Relate inflection points of g to f

Read local extrema off the graph of f

Conclusion

AP Scoring — 2 Points

Compute g(12) as a signed area

Compute g(0) as a signed area

AP Scoring — 2 Points

Find critical points of g

Evaluate g at all candidates

Conclusion

AP Scoring — 3 Points

Two Particles in Motion

Differentiate x_H(t) with the Chain Rule

Evaluate at t = 1

AP Scoring — 2 Points

Analyze the sign of v_H(t)

Analyze the sign of v_J(t)

Compare directions

AP Scoring — 3 Points

Find the sign of v_J(2)

Compare signs of velocity and acceleration

AP Scoring — 1 Point

Set up position as initial position + accumulated change

Substitute u = t² − 1

Combine

AP Scoring — 3 Points

Implicit Curve — Tangent Lines & Related Rates

Differentiate both sides implicitly

Solve for dy/dx

AP Scoring — 2 Points

Find the slope at (2, −1)

Apply the tangent line approximation

AP Scoring — 2 Points