2021 AP Calc AB FRQ Answers & Solutions

Question 1

Calculator OK

Bacteria Density in a Petri Dish

Rate & Data from Tables

Hard

The density of a bacteria population in a circular petri dish at a distance

r

centimeters from the center of the dish is given by an increasing, differentiable function

f

, where

f(r)

is measured in milligrams per square centimeter. Values of

f(r)

for selected values of

r

are given in the table shown.

Values of f(r)

r (centimeters)	0	1	2	2.5	4
f(r) (mg per cm²)	1	2	6	10	18

Part AEasy2 points

Use the data in the table to estimate

f'(2.25)

. Using correct units, interpret the meaning of your answer in the context of this problem.

Answer

f'(2.25)\approx8

mg/cm² per cm

Full Solution & Work

Set up the difference quotient

Since

2.25

is the midpoint of

2

and

2.5

:

f'(2.25) \approx \frac{f(2.5)-f(2)}{2.5-2} = \frac{10-6}{0.5}=8

Interpret the answer

At a distance

r=2.25

cm from the center, the density of bacteria is **increasing at a rate of about 8 mg/cm² per cm**.

AP Scoring — 2 Points

**P1**: The difference quotient, correctly set up and evaluated.
**P2**: Correct interpretation with units, tied to

r=2.25

.

Part BMedium3 points

The total mass, in milligrams, of bacteria in the petri dish is given by the integral expression

2\pi\displaystyle\int_0^4 rf(r)\,dr

. Approximate the value of

2\pi\displaystyle\int_0^4 rf(r)\,dr

using a right Riemann sum with the four subintervals indicated by the data in the table.

Answer

\approx 269\pi \approx 845.088

mg

Full Solution & Work

Set up the right Riemann sum for the inner integral

\int_0^4 rf(r)\,dr \approx (1)\big(1\cdot f(1)\big)+(1)\big(2\cdot f(2)\big)+(0.5)\big(2.5\cdot f(2.5)\big)+(1.5)\big(4\cdot f(4)\big)

Substitute values and compute

= (1)(2)+(1)(12)+(0.5)(25)+(1.5)(72) = 2+12+12.5+108=134.5

Multiply by 2π

2\pi\int_0^4 rf(r)\,dr \approx 2\pi(134.5)=269\pi\approx845.088 \text{ mg}

AP Scoring — 3 Points

**P1**: Correct form of the right Riemann sum, using

r\cdot f(r)

at each right endpoint with correct widths.
**P2**: Correctly computes the inner sum, 134.5.
**P3**: Correct final answer including the factor of

2\pi

:

269\pi\approx845.088

mg.

Part CMedium1 point

Is the approximation found in part (b) an overestimate or an underestimate of the total mass of bacteria in the petri dish? Explain your reasoning.

Answer

Overestimate.

Full Solution & Work

Reason about the integrand

f

is given to be an **increasing** function, and

r

is also increasing (it's the variable of integration), so the product

r\cdot f(r)

is increasing on

[0,4]

.

Apply the right-sum rule for increasing integrands

For an increasing integrand, a **right** Riemann sum uses the largest value on each subinterval, so it **overestimates** the true value of the integral.

AP Scoring — 1 Point

**P1**: "Overestimate," with reasoning that

f

(and hence

r f(r)

) is increasing, so the right Riemann sum overestimates.

Part DHard3 points

The density of bacteria in the petri dish, for

1\leq r\leq4

, is modeled by the function

g

defined by

g(r)=2-16\big(\cos(1.57\sqrt r)\big)^3

. For what value of

k

,

1<k<4

, is

g(k)

equal to the average value of

g(r)

on the interval

1\leq r\leq4

?

Answer

k\approx2.497

Full Solution & Work

Set up the average value

\text{Average value} = \frac{1}{4-1}\int_1^4 g(r)\,dr

Evaluate the average value

\approx 9.876

Solve g(k) = average value for k

g(k) = 9.876 \implies k \approx 2.497

AP Scoring — 3 Points

**P1**: Correct average-value setup

\frac13\int_1^4 g(r)\,dr

.
**P2**: Correctly evaluates the average value,

\approx9.876

.
**P3**: Correctly solves

g(k)=9.876

for

k\approx2.497

within

(1,4)

.

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Question 2

Calculator OK

Two Particles Moving on the x-axis

Particle Motion

Open this question on its own page

Hard

A particle,

P

, is moving along the

x

-axis. The velocity of particle

P

at time

t

is given by

v_P(t)=\sin(t^{1.5})

for

0\leq t\leq\pi

. At time

t=0

, particle

P

is at position

x=5

. A second particle,

Q

, also moves along the

x

-axis. The velocity of particle

Q

at time

t

is given by

v_Q(t)=(t-1.8)\cdot1.25^t

for

0\leq t\leq\pi

. At time

t=0

,

Q

is at position

x=10

.

Part AMedium2 points

Find the positions of particles

P

and

Q

at time

t=1

.

Answer

x_P(1)\approx5.371

;

x_Q(1)\approx8.564

Full Solution & Work

Set up each position as initial position plus accumulated change

x_P(1) = 5+\int_0^1 v_P(t)\,dt, \qquad x_Q(1)=10+\int_0^1 v_Q(t)\,dt

Evaluate with a calculator

x_P(1) \approx 5+0.371 = 5.371, \qquad x_Q(1) \approx 10-1.436 = 8.564

AP Scoring — 2 Points

**P1**: Correct value,

x_P(1)\approx5.371

.
**P2**: Correct value,

x_Q(1)\approx8.564

.

Part BMedium2 points

Are particles

P

and

Q

moving toward each other or away from each other at time

t=1

? Explain your reasoning.

Answer

Moving toward each other.

Full Solution & Work

Evaluate velocities at t = 1

v_P(1)=\sin(1)\approx0.841>0, \qquad v_Q(1)=(1-1.8)(1.25)\approx-1<0

Interpret using relative positions

From part (a),

Q

is ahead of

P

(

x_Q(1)\approx8.564 > x_P(1)\approx5.371

). Since

v_P(1)>0

,

P

is moving right (toward

Q

); since

v_Q(1)<0

,

Q

is moving left (toward

P

).

Conclude

Since

P

moves toward

Q

and

Q

moves toward

P

, the two particles are **moving toward each other** at

t=1

.

AP Scoring — 2 Points

**P1**: Correctly evaluates the signs of

v_P(1)

and

v_Q(1)

.
**P2**: Correct conclusion ("toward each other"), tied to the relative positions of

P

and

Q

.

Part CMedium2 points

Find the acceleration of particle

Q

at time

t=1

. Is the speed of particle

Q

increasing or decreasing at time

t=1

? Give a reason for your answer.

Answer

a_Q(1)\approx1.027

; speed is decreasing.

Full Solution & Work

Differentiate v_Q(t) with the Product Rule

a_Q(t) = v_Q'(t) = 1.25^t + (t-1.8)\cdot1.25^t\ln(1.25) = 1.25^t\big[1+(t-1.8)\ln1.25\big]

Evaluate at t = 1

a_Q(1) \approx 1.027

Compare signs of v_Q and a_Q

v_Q(1)\approx-1<0

and

a_Q(1)\approx1.027>0

— opposite signs, so the speed of particle

Q

is **decreasing** at

t=1

.

AP Scoring — 2 Points

**P1**: Correct value,

a_Q(1)\approx1.027

, via the Product Rule.
**P2**: Correct conclusion ("decreasing") tied to the opposite signs of

v_Q(1)

and

a_Q(1)

.

Part DHard3 points

Find the total distance traveled by particle

P

over the time interval

0\leq t\leq\pi

.

Answer

≈ 1.931

Full Solution & Work

Set up the total-distance integral

\text{Distance} = \int_0^\pi |v_P(t)|\,dt = \int_0^\pi \big|\sin(t^{1.5})\big|\,dt

Note the sign change

\sin(t^{1.5})=0

when

t^{1.5}=\pi

, i.e.

t=\pi^{2/3}\approx2.145

.

v_P(t)>0

on

(0,2.145)

and

v_P(t)<0

on

(2.145,\pi)

.

Evaluate

\int_0^\pi |v_P(t)|\,dt \approx 1.931

AP Scoring — 3 Points

**P1**: Correct setup,

\int_0^\pi |v_P(t)|\,dt

, with reference to the sign change at

t=\pi^{2/3}\approx2.145

.
**P2**: Splits the integral correctly at the sign change (or uses a calculator's absolute-value integration directly).
**P3**: Correct final answer, 1.931.

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Question 3

No Calculator

Spinning Toy — Volume of Revolution

Area & Volume

Open this question on its own page

Hard

A company designs spinning toys using the family of functions

y=cx\sqrt{4-x^2}

, where

c

is a positive constant. The region in the first quadrant bounded by the

x

-axis and the graph of

y=cx\sqrt{4-x^2}

, for some

c

, is revolved about the

x

-axis to form each spinning toy (a solid of revolution). Both

x

and

y

are measured in inches. It is known that

\dfrac{dy}{dx}=\dfrac{c(4-2x^2)}{\sqrt{4-x^2}}

.

Part AMedium2 points

Find the area of the region in the first quadrant bounded by the

x

-axis and the graph of

y=cx\sqrt{4-x^2}

for

c=6

.

Answer

16

in²

Full Solution & Work

Set up the area integral

\text{Area} = \int_0^2 6x\sqrt{4-x^2}\,dx

Substitute u = 4 − x²

du=-2x\,dx

. When

x=0,u=4

; when

x=2,u=0

.

= 6\cdot\left(-\frac12\right)\int_4^0 \sqrt u\,du = 3\int_0^4 u^{1/2}\,du = 3\left[\frac23 u^{3/2}\right]_0^4

Evaluate

= 2\big(4^{3/2}\big) = 2(8)=16 \text{ in}^2

AP Scoring — 2 Points

**P1**: Correct integral setup with the substitution

u=4-x^2

.
**P2**: Correct answer, 16.

Part BMedium2 points

For a particular spinning toy, the radius of the largest cross-sectional circular slice is 1.2 inches. What is the value of

c

for this spinning toy?

Answer

c=0.6

Full Solution & Work

Find where y is maximized

\frac{dy}{dx}=0 \implies 4-2x^2=0 \implies x=\sqrt2

(the only critical point in

[0,2]

).

Find the maximum radius in terms of c

y\big(\sqrt2\big) = c\sqrt2\cdot\sqrt{4-2} = c\sqrt2\cdot\sqrt2=2c

Solve for c

2c = 1.2 \implies c=0.6

AP Scoring — 2 Points

**P1**: Correctly finds the maximum occurs at

x=\sqrt2

, with the largest radius equal to

2c

.
**P2**: Correct answer,

c=0.6

.

Part CHard3 points

For another spinning toy, the volume is

2\pi

cubic inches. What is the value of

c

for this spinning toy?

Answer

c=\sqrt{\dfrac{15}{32}}=\dfrac{\sqrt{30}}{8}\approx0.685

Full Solution & Work

Set up the volume of revolution

V = \pi\int_0^2 \big(cx\sqrt{4-x^2}\big)^2\,dx = \pi c^2\int_0^2 x^2(4-x^2)\,dx

Evaluate the integral

\int_0^2 (4x^2-x^4)\,dx = \left[\frac{4x^3}{3}-\frac{x^5}{5}\right]_0^2 = \frac{32}{3}-\frac{32}{5} = \frac{64}{15}

Solve for c

\pi c^2\cdot\frac{64}{15} = 2\pi \implies c^2 = \frac{30}{64}=\frac{15}{32} \implies c=\sqrt{\frac{15}{32}}=\frac{\sqrt{30}}{8}\approx0.685

AP Scoring — 3 Points

**P1**: Correct volume-of-revolution setup,

\pi\int_0^2(cx\sqrt{4-x^2})^2\,dx

.
**P2**: Correctly evaluates

\int_0^2 x^2(4-x^2)\,dx=\frac{64}{15}

.
**P3**: Correct final answer,

c=\frac{\sqrt{30}}{8}\approx0.685

.

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Question 4

No Calculator

Graph Analysis with an Accumulation Function

Graph Analysis & Curve Sketching

Open this question on its own page

Hard

Let

f

be a continuous function defined on the closed interval

-4\leq x\leq6

. The graph of

f

, consisting of four line segments connecting

(-4,-2)

,

(-2,6)

,

(0,4)

,

(2,-4)

, and

(6,2)

, is shown. Let

G

be the function defined by

G(x)=\displaystyle\int_0^x f(t)\,dt

.

2021 AP Calculus AB FRQ Question 4 Graph of f — Graph of $f$ on $[-4,6]$.

Part AMedium2 points

On what open intervals is the graph of

G

concave up? Give a reason for your answer.

Answer

(-4,-2)

and

(2,6)

Full Solution & Work

Relate concavity of G to the slope of f

G'(x)=f(x)

, so

G''(x)=f'(x)

equals the (constant) slope of

f

on each line segment.

G

is concave up where

f'(x)>0

, i.e. where

f

has positive slope.

Compute each segment's slope

(-4,-2)\to(-2,6)

: slope

=\frac{8}{2}=4>0

.

(-2,6)\to(0,4)

: slope

=\frac{-2}{2}=-1<0

.

(0,4)\to(2,-4)

: slope

=\frac{-8}{2}=-4<0

.

(2,-4)\to(6,2)

: slope

=\frac{6}{4}=1.5>0

.

Conclude

G

is concave up where the slope of

f

is positive: on **

(-4,-2)

and

(2,6)

**.

AP Scoring — 2 Points

**P1**: Correctly relates

G''=f'

to the slopes of the line segments.
**P2**: Correct intervals,

(-4,-2)

and

(2,6)

.

Part BHard3 points

Let

P

be the function defined by

P(x)=G(x)\cdot f(x)

. Find

P'(3)

.

Answer

P'(3)=1.375

Full Solution & Work

Apply the Product Rule

P'(x) = G'(x)f(x)+G(x)f'(x) = \big(f(x)\big)^2+G(x)f'(x)

Find f(3) and f'(3)

On

[2,6]

,

f(x)=-4+1.5(x-2)

. So

f(3)=-4+1.5=-2.5

, and

f'(3)=1.5

(the segment's constant slope).

Find G(3)

G(3)=\int_0^3 f(t)\,dt = \int_0^2 f(t)\,dt+\int_2^3 f(t)\,dt = \frac{4+(-4)}{2}(2)+\frac{-4+(-2.5)}{2}(1) = 0-3.25=-3.25

Combine

P'(3) = \big(f(3)\big)^2+G(3)f'(3) = (-2.5)^2+(-3.25)(1.5) = 6.25-4.875=1.375

AP Scoring — 3 Points

**P1**: Correct Product Rule setup,

P'(x)=(f(x))^2+G(x)f'(x)

.
**P2**: Correctly finds

f(3)=-2.5

,

f'(3)=1.5

, and

G(3)=-3.25

.
**P3**: Correct final answer, 1.375.

Part CHard2 points

Find

\lim\limits_{x\to2} \dfrac{G(x)}{x^2-2x}

.

Answer

-2

Full Solution & Work

Confirm the indeterminate form

G(2)=\int_0^2 f(t)\,dt=0

(computed in part (b)), and

x^2-2x=x(x-2)\to0

as

x\to2

. This is

\frac00

, so L'Hôpital's Rule applies.

Apply L'Hôpital's Rule

\lim_{x\to2}\frac{G(x)}{x^2-2x} = \lim_{x\to2}\frac{G'(x)}{2x-2} = \lim_{x\to2}\frac{f(x)}{2x-2}

Evaluate

f(2)=-4

(continuous at the shared endpoint of the two segments), and

2(2)-2=2

:

= \frac{-4}{2}=-2

AP Scoring — 2 Points

**P1**: Confirms the

\frac00

form and applies L'Hôpital's Rule (or an equivalent factoring approach).
**P2**: Correct final answer,

-2

.

Part DMedium2 points

Find the average rate of change of

G

on the interval

[-4,2]

. Does the Mean Value Theorem guarantee a value

c

,

-4<c<2

, for which

G'(c)

is equal to this average rate of change? Justify your answer.

Answer

Average rate of change

=\dfrac{7}{3}

; yes, the MVT applies.

Full Solution & Work

Find G(-4) and G(2)

G(2)=0

(from part b).

G(-4)=\int_0^{-4}f(t)\,dt = -\int_{-4}^0 f(t)\,dt

, and

\int_{-4}^0 f(t)\,dt = \frac{-2+6}{2}(2)+\frac{6+4}{2}(2) = 4+10=14

, so

G(-4)=-14

.

Compute the average rate of change

\frac{G(2)-G(-4)}{2-(-4)} = \frac{0-(-14)}{6} = \frac{14}{6}=\frac73

Check the hypotheses of the MVT

G(x)=\int_0^x f(t)\,dt

with

f

continuous everywhere on

[-4,6]

, so

G

is differentiable (with

G'=f

) — and hence continuous — on all of

[-4,2]

.

Conclude

Since

G

is differentiable on

(-4,2)

and continuous on

[-4,2]

, the **Mean Value Theorem** guarantees a

c\in(-4,2)

with

G'(c)=\frac73

.

AP Scoring — 2 Points

**P1**: Correctly computes the average rate of change,

\frac73

.
**P2**: Correct "yes" with justification citing differentiability of

G

(via continuity of

f

).

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Question 5

No Calculator

Implicit Curve — Tangent Lines & Extrema

Implicit Differentiation

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Hard

Consider the function

y=f(x)

whose curve is given by the equation

2y^2-6=y\sin x

, for

y>0

.

Part AMedium2 points

Show that

\dfrac{dy}{dx}=\dfrac{y\cos x}{4y-\sin x}

.

Answer

\dfrac{dy}{dx}=\dfrac{y\cos x}{4y-\sin x}

(shown by implicit differentiation)

Full Solution & Work

Differentiate both sides implicitly

\frac{d}{dx}(2y^2-6) = \frac{d}{dx}(y\sin x) \implies 4y\frac{dy}{dx} = \frac{dy}{dx}\sin x+y\cos x

Solve for dy/dx

\frac{dy}{dx}(4y-\sin x) = y\cos x \implies \frac{dy}{dx}=\frac{y\cos x}{4y-\sin x}

AP Scoring — 2 Points

**P1**: Correct implicit differentiation, including the Product Rule on

y\sin x

.
**P2**: Requires P1. Correctly isolates

\frac{dy}{dx}

to reach the given expression.

Part BMedium2 points

Write an equation for the line tangent to the curve at the point

(0,\sqrt3)

.

Answer

y=\dfrac14 x+\sqrt3

Full Solution & Work

Find the slope at (0, √3)

\left.\frac{dy}{dx}\right|_{(0,\sqrt3)} = \frac{\sqrt3\cos(0)}{4\sqrt3-\sin(0)} = \frac{\sqrt3}{4\sqrt3}=\frac14

Write the tangent line

y-\sqrt3 = \frac14(x-0) \implies y=\frac14 x+\sqrt3

AP Scoring — 2 Points

**P1**: Correct slope,

\frac14

.
**P2**: Correct tangent-line equation.

Part CMedium2 points

For

0\leq x\leq\pi

and

y>0

, find the coordinates of the point where the line tangent to the curve is horizontal.

Answer

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

Full Solution & Work

Set dy/dx = 0

Since

y>0

,

\frac{dy}{dx}=0

requires

\cos x=0

, so

x=\frac{\pi}{2}

within

[0,\pi]

.

Solve for y at x = π/2

Substitute into the curve equation:

2y^2-6=y\sin(\pi/2)=y

.

2y^2-y-6=0 \implies y=\frac{1\pm7}{4} \implies y=2 \text{ or } y=-1.5

Since

y>0

,

y=2

.

State the point

The tangent line is horizontal at

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

.

AP Scoring — 2 Points

**P1**: Correctly sets

\cos x=0

to find

x=\pi/2

.
**P2**: Correctly solves the resulting quadratic for

y

and selects the positive root,

y=2

.

Part DHard2 points

Determine whether

f

has a relative minimum, a relative maximum, or neither at the point found in part (c). Justify your answer.

Answer

Relative maximum.

Full Solution & Work

Analyze the sign of dy/dx near x = π/2

Near

x=\frac{\pi}{2}

,

y\approx2>0

and

4y-\sin x\approx8-1=7>0

stay positive, so the sign of

\frac{dy}{dx}=\frac{y\cos x}{4y-\sin x}

is governed by

\cos x

.

Determine the sign change

\cos x>0

for

x

slightly less than

\frac{\pi}{2}

, and

\cos x<0

for

x

slightly greater than

\frac\pi2

. So

\frac{dy}{dx}

changes from **positive to negative** as

x

increases through

\frac{\pi}{2}

.

Conclude

Since

y'

changes from positive to negative at

x=\frac\pi2

,

f

has a **relative maximum** there.

AP Scoring — 2 Points

**P1**: Correct sign analysis of

\frac{dy}{dx}

on either side of

x=\pi/2

(via the sign of

\cos x

, with

y

and

4y-\sin x

remaining positive).
**P2**: Correct conclusion, "relative maximum," tied to that sign change.

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Question 6

No Calculator

Medication in the Bloodstream — Differential Equations

Differential Equations

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Hard

A medication is administered to a patient. The amount, in milligrams, of the medication in the patient at time

t

hours is modeled by a function

y=A(t)

that satisfies the differential equation

\dfrac{dy}{dt}=\dfrac{12-y}{3}

. At time

t=0

hours, there are 0 milligrams of the medication in the patient.

Part AEasy1 point

A portion of the slope field for the differential equation

\dfrac{dy}{dt}=\dfrac{12-y}{3}

is given. Sketch the solution curve through the point

(0,0)

.

Answer

An increasing, concave-down curve from

(0,0)

approaching

y=12

.

Full Solution & Work

Determine the shape from the differential equation

Since

y(0)=0<12

,

\frac{dy}{dt}=\frac{12-0}{3}=4>0

initially, so the curve rises steeply from the origin. As

y

increases toward 12,

(12-y)

shrinks, so the slope flattens — the curve is concave down throughout and levels off, asymptotically approaching

y=12

(matching part (b)'s limit) without ever crossing it.

AP Scoring — 1 Point

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

(0,0)

, stay below

y=12

, rise steeply at first and then level off, matching the given slope field.

Part BEasy1 point

Using correct units, interpret the statement

\lim\limits_{t\to\infty} A(t)=12

in the context of the problem.

Answer

As time increases without bound, the amount of medication in the patient approaches 12 milligrams.

Full Solution & Work

State the interpretation

As time

t

increases without bound, the amount of medication in the patient's body **approaches (stabilizes at) 12 milligrams**.

AP Scoring — 1 Point

**P1**: Interpretation must reference both the long-run behavior (as

t\to\infty

) and the units (milligrams).

Part CHard4 points

Use separation of variables to find

y=A(t)

, the particular solution to the differential equation

\dfrac{dy}{dt}=\dfrac{12-y}{3}

with initial condition

A(0)=0

.

Answer

A(t)=12-12e^{-t/3}

Full Solution & Work

Separate variables

\frac{dy}{12-y} = \frac13\,dt

Integrate both sides

-\ln|12-y| = \frac{t}{3}+C

Apply the initial condition A(0) = 0

-\ln|12-0|=0+C \implies C=-\ln12

Since

y(0)=0<12

and

y

increases toward 12,

12-y>0

, so

|12-y|=12-y

.

-\ln(12-y) = \frac t3-\ln12 \implies \ln(12-y)=-\frac t3+\ln12

Solve for y

12-y = 12e^{-t/3} \implies y = 12-12e^{-t/3}

AP Scoring — 4 Points

**P1**: Correct separation of variables.
**P2**: Correct antiderivatives on both sides.
**P3**: Correctly includes the constant of integration and applies the initial condition

A(0)=0

.
**P4**: Correct final answer,

A(t)=12-12e^{-t/3}

.

Part DHard3 points

A different procedure is used to administer the medication to a second patient. The amount, in milligrams, of the medication in the second patient at time

t

hours is modeled by a function

y=B(t)

that satisfies the differential equation

\dfrac{dy}{dt}=3-\dfrac{y}{t+2}

. At time

t=1

hour, there are 2.5 milligrams of the medication in the second patient. Is the rate of change of the amount of medication in the second patient increasing or decreasing at time

t=1

? Give a reason for your answer.

Answer

Decreasing, since

\dfrac{d^2y}{dt^2}\Big|_{t=1}=-\dfrac{4}{9}<0

.

Full Solution & Work

Differentiate the differential equation with respect to t

Applying the Quotient Rule to

\frac{y}{t+2}

:

\frac{d^2y}{dt^2} = -\frac{y'(t+2)-y}{(t+2)^2}

Find B'(1) first

B'(1) = 3-\frac{2.5}{1+2} = 3-\frac{5}{6}=\frac{13}{6}

Evaluate the second derivative at t = 1

With

y=2.5=\frac52

,

y'=\frac{13}{6}

,

t+2=3

:

\frac{d^2y}{dt^2}\bigg|_{t=1} = -\frac{\frac{13}{6}(3)-\frac52}{3^2} = -\frac{\frac{13}{2}-\frac52}{9} = -\frac{4}{9}

Conclude

Since

\dfrac{d^2y}{dt^2}\Big|_{t=1}=-\frac49<0

, the rate of change

\frac{dy}{dt}

(i.e.

B'(t)

) is **decreasing** at

t=1

.

AP Scoring — 3 Points

**P1**: Correctly finds

B'(1)=\frac{13}{6}

from the given differential equation.
**P2**: Correctly sets up

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

using the Quotient Rule (or equivalent) on the right side of the differential equation.
**P3**: Correct final value,

-\frac49

, and correct conclusion ("decreasing").

Common mistake: This part is asking about the second derivative — whether the *rate itself* is speeding up or slowing down — not the sign of $B'(1)$ alone. Confusing "is $y$ increasing" with "is $\frac{dy}{dt}$ increasing" is the most common error on this type of question.

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