2023 AP Calc AB FRQ Answers & Solutions

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

to

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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Question 2

Calculator OK

Stephen's Swim — Particle Motion

Particle Motion

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Hard

Stephen swims back and forth along a straight path in a 50-meter-long pool for 90 seconds. Stephen's velocity is modeled by

v(t) = 2.38e^{-0.02t}\sin\!\left(\dfrac{\pi}{56}t\right)

, where

t

is measured in seconds and

v(t)

is measured in meters per second.

Part AEasy2 points

Find all times

t

in the interval

0<t<90

at which Stephen changes direction. Give a reason for your answer.

Answer

t=56

seconds

Full Solution & Work

Solve v(t) = 0 on (0, 90)

v(t)=0 \implies t=56

Interpret the sign change

Stephen changes direction when his velocity changes sign. This occurs at

t=56

seconds.

AP Scoring — 2 Points

**P1**: Considers

v(t)=0

.
**P2**: The answer,

t=56

, with reason — a response of just "

t=56

" with no supporting work does not earn this point.

Part BMedium3 points

Find Stephen's acceleration at time

t=60

seconds. Show the setup for your calculations, and indicate units of measure. Is Stephen speeding up or slowing down at time

t=60

seconds? Give a reason for your answer.

Answer

a(60)\approx -0.036

m/sec²; Stephen is speeding up.

Full Solution & Work

Relate acceleration to v'(t)

v'(60) = a(60) = -0.0360162

Stephen's acceleration at

t=60

seconds is

-0.036

meter per second per second.

Compare signs of v and a

v(60) = -0.1595124 < 0

. Because Stephen's velocity and acceleration are both negative at

t=60

, Stephen is **speeding up** at that time.

AP Scoring — 3 Points

**P1**: The minimum work needed is

v'(60)=-0.036

— evaluating only

a(60)=-0.0360162

without explicitly connecting it to

v'(t)

is not sufficient.
**P2**: A declared value for

a(60)

(units).
**P3**: Requires a presented value of

v(60)

correct to at least 1 decimal place, and consistent conclusion — "speeding up because

a(60)

and

v(60)

have the same sign" is sufficient without an explicit numeric

v(60)

.

Part CMedium2 points

Find the distance between Stephen's position at time

t=20

seconds and his position at time

t=80

seconds. Show the setup for your calculations.

Answer

≈ 23.384 (or 23.383) meters

Full Solution & Work

Set up the definite integral

\int_{20}^{80} v(t)\,dt

Evaluate

= 23.383997

The distance between Stephen's positions at

t=20

and

t=80

seconds is **23.384** (or 23.383) meters.

AP Scoring — 2 Points

**P1**: Only for

\int_{20}^{80} v(t)\,dt

(with or without the differential).
**P2**: Only for the answer 23.384 (or 23.383), regardless of whether the first point was earned.

Part DMedium2 points

Find the total distance Stephen swims over the time interval

0 \leq t \leq 90

seconds. Show the setup for your calculations.

Answer

≈ 62.164 meters

Full Solution & Work

Set up the total-distance integral

\int_0^{90} |v(t)|\,dt

Evaluate

= 62.164216

The total distance Stephen swims over

0\leq t\leq 90

seconds is **62.164** meters.

AP Scoring — 2 Points

**P1**: Only for

\int_0^{90}|v(t)|\,dt

, or the equivalent

\int_0^{56}v(t)\,dt - \int_{56}^{90}v(t)\,dt

, with or without differentials.
**P2**: Only for the answer 62.164, regardless of whether the first point was earned.

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Question 3

No Calculator

Warming Milk — Differential Equation

Differential Equations

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Hard

A bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed. The increasing function

M

models the temperature of the milk at time

t

, where

M(t)

is measured in degrees Celsius and

t

is the number of minutes since the bottle was placed in the pan.

M

satisfies the differential equation

\dfrac{dM}{dt} = \dfrac{1}{4}(40-M)

. At time

t=0

, the temperature of the milk is 5°C. It can be shown that

M(t) < 40

for all values of

t

.

Part AEasy1 point

A slope field for the differential equation

\dfrac{dM}{dt}=\dfrac{1}{4}(40-M)

is shown. Sketch the solution curve through the point

(0,5)

.

Answer

The curve rises from

(0,5)

, concave down, approaching the asymptote

M=40

.

Full Solution & Work

Follow the slope field from (0, 5)

Starting at

(0,5)

, follow the slope-field arrows: since

M<40

throughout,

\frac{dM}{dt}=\frac14(40-M)>0

always, so the curve rises steadily. As

M

climbs toward 40,

(40-M)

shrinks, so the slope decreases toward 0 — the curve is concave down and levels off, approaching the horizontal asymptote

M=40

without crossing it.

AP Scoring — 1 Point

**P1**: The solution curve must pass through

(0,5)

, extend reasonably close to the left and right edges of the given rectangle, lie entirely below

M=40

, and have no obvious conflicts with the given slope lines.

Part BEasy2 points

Use the line tangent to the graph of

M

at

t=0

to approximate

M(2)

, the temperature of the milk at time

t=2

minutes.

Answer

M(2)\approx 22.5°C

Full Solution & Work

Find the slope at t = 0

\left.\frac{dM}{dt}\right|_{t=0} = \frac{1}{4}(40-5) = \frac{35}{4}

Apply the tangent-line approximation

y = 5+\frac{35}{4}(t-0) \implies M(2)\approx 5+\frac{35}{4}\cdot 2 = 22.5

The temperature of the milk at time

t=2

minutes is approximately **22.5°C**.

AP Scoring — 2 Points

**P1**:

\frac{1}{4}(40-5)

— a subsequent simplification error forfeits the second point.
**P2**: An approximation using a tangent line through

(0,5)

with slope

\frac{35}{4}

(or a declared value of

\frac{dM}{dt}

), evaluated at

t=2

. An unsupported approximation does not earn this point.

Part CMedium2 points

Write an expression for

\dfrac{d^2M}{dt^2}

in terms of

M

. Use

\dfrac{d^2M}{dt^2}

to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of

M(2)

. Give a reason for your answer.

Answer

\dfrac{d^2M}{dt^2}=-\dfrac{1}{16}(40-M)

; the tangent line approximation is an overestimate.

Full Solution & Work

Differentiate dM/dt with respect to t

\frac{d^2M}{dt^2} = \frac{1}{4}\left(-\frac{dM}{dt}\right) = -\frac{1}{4}\left(\frac{1}{4}(40-M)\right)=-\frac{1}{16}(40-M)

Determine the sign and conclude

Because

M(t)<40

for all

t

,

\frac{d^2M}{dt^2}<0

, so the graph of

M

is **concave down**. Therefore, the tangent line approximation of

M(2)

is an **overestimate**.

AP Scoring — 2 Points

**P1**: Either

\frac{d^2M}{dt^2}=-\frac{1}{4}\left(\frac{1}{4}(40-M)\right)

or the simplified

-\frac{1}{16}(40-M)

— a subsequent simplification error forfeits this point.
**P2**: Requires stating

\frac{d^2M}{dt^2}<0

(or "

\frac{dM}{dt}

is decreasing" or "

M

is concave down") **and** concluding "overestimate." An argument based on concavity at a single point does not earn this point.

Part DHard4 points

Use separation of variables to find an expression for

M(t)

, the particular solution to the differential equation

\dfrac{dM}{dt}=\dfrac{1}{4}(40-M)

with initial condition

M(0)=5

.

Answer

M(t) = 40-35e^{-t/4}

Full Solution & Work

Separate variables

\frac{dM}{40-M} = \frac{1}{4}\,dt

Integrate both sides

-\ln|40-M| = \frac{1}{4}t+C

Apply the initial condition

-\ln|40-5|=0+C \implies C=-\ln 35

Because

M(t)<40

for all

t

,

40-M>0

, so

|40-M|=40-M

.

-\ln(40-M) = \frac{1}{4}t-\ln 35 \implies \ln(40-M)=-\frac{1}{4}t+\ln 35

Solve for M

40-M = 35e^{-t/4} \implies M = 40-35e^{-t/4}

AP Scoring — 4 Points

**P1**: Correct separation of variables — no separation earns 0 of 4 points.
**P2**: Finds antiderivatives on both sides (presenting

-\ln(40-M)

without absolute value symbols is still eligible for all 4 points).
**P3**: Correctly includes the constant of integration and uses the initial condition. Requires the first 2 points.
**P4**: Earned only for the final answer

M=40-35e^{-t/4}

(or equivalent) — requires the first 3 points.

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Question 4

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Analyzing a Function from its Derivative Graph

Graph Analysis & Curve Sketching

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Hard

The function

f

is defined on the closed interval

[-2,8]

and satisfies

f(2)=1

. The graph of

f'

, the derivative of

f

, consists of two line segments and a semicircle, as shown in the figure: a line segment from

(-2,2)

down to

(0,-2)

, a line segment from

(0,-2)

up to

(4,2)

, and a lower semicircle from

(4,2)

down and back up to

(8,2)

(dipping to a minimum of

0

at

x=6

).

2023 AP Calculus AB FRQ Question 4 Graph of f' — Graph of $f'$ on $[-2,8]$.

Part AMedium1 point

Does

f

have a relative minimum, a relative maximum, or neither at

x=6

? Give a reason for your answer.

Answer

Neither —

f'(x)

does not change sign at

x=6

.

Full Solution & Work

Check the sign of f' around x = 6

f'(x)>0

on

(2,6)

and

f'(x)>0

on

(6,8)

(the semicircle dips to touch 0 at

x=6

but never goes negative).

Conclude

f'(x)

does not change sign at

x=6

, so there is **neither** a relative maximum nor a relative minimum there.

AP Scoring — 1 Point

**P1**: Declaring

f'(x)

does not change sign at

x=6

, so neither, is sufficient — a response is not required to give specific sign intervals, but any given must be correct. Declaring

f'(x)>0

before **and** after

x=6

does not, by itself, earn this point without the "does not change sign" conclusion.

Part BMedium2 points

On what open intervals, if any, is the graph of

f

concave down? Give a reason for your answer.

Answer

(-2,0)

and

(4,6)

Full Solution & Work

Relate concavity of f to the behavior of f'

f

is concave down where

f'

is **decreasing**.

Read decreasing intervals off the graph of f'

f'

decreases on

(-2,0)

(the first line segment, negative slope) and on

(4,6)

(the left half of the semicircle, coming down from 2 to 0). So the graph of

f

is concave down on **

(-2,0)

and

(4,6)

**.

AP Scoring — 2 Points

**P1**: Earned only for the answer

(-2,0)

and

(4,6)

(endpoints may or may not be included).
**P2**: Requires P1. Must correctly discuss the behavior of

f'

(or the slopes of

f'

) as the reason.

Common mistake: A response with exactly one of the two correct intervals, with a correct reason, still earns 1 of the 2 points as a special case.

Part CHard3 points

Find the value of

\lim\limits_{x\to2} \dfrac{6f(x)-3x}{x^2-5x+6}

, or show that it does not exist. Justify your answer.

Answer

3

Full Solution & Work

Confirm the indeterminate form

Because

f

is differentiable at

x=2

,

f

is continuous there, so

\lim_{x\to2}f(x)=f(2)=1

.

\lim_{x\to2}\big(6f(x)-3x\big) = 6(1)-3(2)=0, \qquad \lim_{x\to2}(x^2-5x+6)=0

The limit is of indeterminate form

\frac{0}{0}

, so L'Hôpital's Rule can be applied.

Apply L'Hôpital's Rule

\lim_{x\to2}\frac{6f(x)-3x}{x^2-5x+6} = \lim_{x\to2}\frac{6f'(x)-3}{2x-5} = \frac{6\cdot f'(2)-3}{2(2)-5}

Read f'(2) off the graph and finish

From the graph,

f'(2)=0

(the middle line segment crosses the

x

-axis at

x=2

).

= \frac{6(0)-3}{4-5} = \frac{-3}{-1}=3

AP Scoring — 3 Points

**P1**: Presenting the two separate limits for the numerator and denominator (a response that presents a limit explicitly equal to

\frac{0}{0}

does not earn this point).
**P2**: Applying L'Hôpital's Rule — presenting at least one correct derivative in the limit of a ratio of derivatives.
**P3**: The correct answer, 3, with supporting work.

Part DHard3 points

Find the absolute minimum value of

f

on the closed interval

[-2,8]

. Justify your answer.

Answer

f(2)=1

Full Solution & Work

Find critical points

f'(x)=0 \implies x=-1,\ x=2,\ x=6

Evaluate f at critical points and endpoints

f

is continuous on

[-2,8]

, so the candidates for the absolute minimum are

x=-2,-1,2,6,8

.

\begin{array}{c|c} x & f(x) \\\hline -2 & 3 \\ -1 & 4 \\ 2 & 1 \\ 6 & 7-\pi \\ 8 & 11-2\pi\end{array}

(these are found by adding/subtracting the signed areas between consecutive

x

-values, using

f(2)=1

as the anchor value)

Conclude

The smallest value in the table is

f(2)=1

. The absolute minimum value of

f

on

[-2,8]

is

\boxed{1}

.

AP Scoring — 3 Points

**P1**: Stating

f'(x)=0

(or equivalent) — merely listing the zeros of

f'

is not sufficient.
**P2**: A justification:

f

is continuous on

[-2,8]

and correctly evaluated at all 5 candidates. Any error in evaluating

f

at a candidate point forfeits this point. (A response need not evaluate

f(-1)

if it argues

x=-1

is a local max via other reasoning, and need not evaluate

f(8)

if it argues

f'(x)\geq0

for

x>2

, hence

f(8)>f(2)

.)
**P3**: Earned only for indicating the minimum value is 1 — noting only that the minimum *occurs at*

x=2

is not sufficient.

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Question 5

No Calculator

Tabular Data, Chain Rule & FTC

Limits, Chain Rule & FTC

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Hard

The functions

f

and

g

are twice differentiable. The table shown gives values of the functions and their first derivatives at selected values of

x

.

Values of f, f', g, and g'

x	0	2	4	7
f(x)	10	7	4	5
f′(x)	3/2	−8	3	6
g(x)	1	2	−3	0
g′(x)	5	4	2	8

Part AMedium2 points

Let

h

be the function defined by

h(x)=f(g(x))

. Find

h'(7)

. Show the work that leads to your answer.

Answer

h'(7)=12

Full Solution & Work

Apply the Chain Rule

h'(x) = f'(g(x))\cdot g'(x)

Evaluate at x = 7

h'(7) = f'(g(7))\cdot g'(7) = f'(0)\cdot 8 = \frac{3}{2}\cdot 8 = 12

AP Scoring — 2 Points

**P1**: Either

h'(x)=f'(g(x))\cdot g'(x)

or

h'(7)=f'(g(7))\cdot g'(7)

.
**P2**: Requires P1 — only for the answer 12 (if P1 is not earned, this point can still be earned only for the response

f'(0)\cdot 8=12

or

\frac{3}{2}\cdot 8

).

Part BHard3 points

Let

k

be a differentiable function such that

k'(x)=(f(x))^2\cdot g(x)

. Is the graph of

k

concave up or concave down at the point where

x=4

? Give a reason for your answer.

Answer

Concave down, since

k''(4)=-40<0

.

Full Solution & Work

Differentiate k'(x) using Product and Chain Rules

k''(x) = 2f(x)\cdot f'(x)\cdot g(x) + (f(x))^2 \cdot g'(x)

Evaluate at x = 4

k''(4) = 2(4)(3)(-3) + (4)^2(2) = -72+32=-40

Conclude

The graph of

k

is **concave down** at

x=4

because

k''(4)<0

(and

k''

is continuous, since

f

,

g

are twice differentiable).

AP Scoring — 3 Points

**P1**: Either

k''(x)=2f(x)\cdot f'(x)\cdot g(x)+(f(x))^2\cdot g'(x)

or

k''(4)=2f(4)\cdot f'(4)\cdot g(4)+(f(4))^2\cdot g'(4)

— several partially-correct expressions with a single product/chain-rule error still earn this point.
**P2**: Requires correctly computing

k''(4)=-40

with supporting work.
**P3**: A correct answer and reason consistent with the declared value of

k''(4)

.

Part CMedium1 point

Let

m

be the function defined by

m(x) = 5x^3+\displaystyle\int_0^x f'(t)\,dt

. Find

m(2)

. Show the work that leads to your answer.

Answer

m(2)=37

Full Solution & Work

Apply the Fundamental Theorem of Calculus

m(2) = 5(2)^3+\int_0^2 f'(t)\,dt = 40+\big(f(2)-f(0)\big) = 40+(7-10)=37

AP Scoring — 1 Point

**P1**: Earned only for the answer 37 (or equivalent) with supporting work equivalent to

5\cdot 8+(f(2)-f(0))

,

40+(7-10)

, or similar. An unsupported answer of just "37" does not earn this point.

Part DMedium3 points

Is the function

m

defined in part (c) increasing, decreasing, or neither at

x=2

? Justify your answer.

Answer

Increasing, because

m'(2)=52>0

.

Full Solution & Work

Differentiate m(x)

m'(x) = 15x^2+f'(x)

Evaluate at x = 2

m'(2) = 15(4)+f'(2)=60+(-8)=52

Conclude

The graph of

m

is **increasing** at

x=2

because

m'(2)=52>0

.

AP Scoring — 3 Points

**P1**: Considering

m'(x)

,

m'(2)

, or

m'

— this consideration may appear inside a justification statement.
**P2**: Correctly evaluates

m'(2)=15\cdot2^2+f'(2)

,

=60+f'(2)

, or

=60-8

— an unsupported response of

m'(2)=52

alone does not earn this point.
**P3**: An answer and justification consistent with the declared value of

m'(2)

.

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Question 6

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Implicit Curve — Tangent Lines & Related Rates

Implicit Differentiation

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Hard

Consider the curve given by the equation

6xy = 2+y^3

.

Part AMedium2 points

Show that

\dfrac{dy}{dx}=\dfrac{2y}{y^2-2x}

.

Answer

\dfrac{dy}{dx}=\dfrac{2y}{y^2-2x}

(shown by implicit differentiation)

Full Solution & Work

Differentiate both sides implicitly

\frac{d}{dx}(6xy) = \frac{d}{dx}(2+y^3) \implies 6y+6x\frac{dy}{dx} = 3y^2\frac{dy}{dx}

Solve for dy/dx

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

AP Scoring — 2 Points

**P1**: Correct implicit differentiation of

6xy=2+y^3

(alternate notations for

\frac{dy}{dx}

, such as

y'

, are accepted).
**P2**: Requires P1. It is sufficient to present

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

, provided there are no subsequent errors.

Part BMedium2 points

Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.

Answer

No such point exists.

Full Solution & Work

Set dy/dx = 0

For the tangent line to be horizontal, it is necessary that

2y=0

(so

y=0

) and that

y^2-2x\neq0

.

Test y = 0 in the original curve equation

Substituting

y=0

into

6xy=2+y^3

yields

6x\cdot0=2

, i.e.

0=2

, which has **no solution**.

Conclude

Therefore, there is **no point** on the curve at which the tangent line is horizontal.

AP Scoring — 2 Points

**P1**: Earned with any of

2y=0

,

y=0

,

\frac{dy}{dx}=0

,

dy=0

,

y'=0

, or

\frac{2y}{y^2-2x}=0

.
**P2**: The answer with reason — a response does not need to also state that

y^2-2x\neq0

at that point.

Part CHard3 points

Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.

Answer

\left(\dfrac{1}{2},1\right)

Full Solution & Work

Set the denominator equal to 0

For a vertical tangent it is necessary that

2y\neq0

and that

y^2-2x=0

(so

x=\frac{y^2}{2}

).

Substitute into the curve equation

Substituting

x=\frac{y^2}{2}

into

6xy=2+y^3

:

3y^2\cdot y = 2+y^3 \implies 3y^3=2+y^3 \implies 2y^3=2 \implies y=1

Find x

Substituting

y=1

into

6xy=2+y^3

:

6x=2+1=3 \implies x=\frac{1}{2}

.
The tangent line to the curve is vertical at the point

\left(\frac12,1\right)

.

AP Scoring — 3 Points

**P1**: Presenting

y^2=2x

or

y=\sqrt{2x}

.
**P2**: Substituting

y=\sqrt{2x}

(or

x=\frac{y^2}{2}

from

y^2-2x=0

) into the original curve equation.
**P3**: Both coordinates of the point

\left(\frac12,1\right)

, labeled — a response that identifies the point without verifying it satisfies the curve equation does not earn P2 or P3.

Part DHard2 points

A particle is moving along the curve. At the instant when the particle is at the point

\left(\dfrac{1}{2},-2\right)

, its horizontal position is increasing at a rate of

\dfrac{dx}{dt}=\dfrac{2}{3}

unit per second. What is the value of

\dfrac{dy}{dt}

, the rate of change of the particle's vertical position, at that instant?

Answer

\dfrac{dy}{dt}=-\dfrac{8}{9}

unit per second

Full Solution & Work

Differentiate 6xy = 2+y³ implicitly with respect to t

6y\frac{dx}{dt}+6x\frac{dy}{dt} = 3y^2\frac{dy}{dt}

Substitute the known values

At

(x,y)=\left(\frac12,-2\right)

with

\frac{dx}{dt}=\frac23

:

6(-2)\left(\frac23\right)+6\left(\frac12\right)\frac{dy}{dt} = 3(-2)^2\frac{dy}{dt}

-8+3\frac{dy}{dt}=12\frac{dy}{dt}

Solve for dy/dt

-8 = 9\frac{dy}{dt} \implies \frac{dy}{dt} = -\frac{8}{9} \text{ unit per second}

AP Scoring — 2 Points

**P1**: Presenting one or more of the terms

6y\frac{dx}{dt}

,

6x\frac{dy}{dt}

, or

3y^2\frac{dy}{dt}

.
**P2**: The correct answer

-\frac89

— an unsupported response of

-\frac89

alone earns no points.

Common mistake: Alternate route: use $\frac{dy}{dx}\big|_{(1/2,-2)}=\frac{2(-2)}{(-2)^2-2(1/2)}=-\frac43$ from part (a)'s formula, then $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}=-\frac43\cdot\frac23=-\frac89$.

Study the concept

Learn the theory behind this question type with worked examples and strategy tips.

Practice More

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