2022 AP Calc AB FRQ Answers & Solutions

Question 1

Calculator OK

Toll Plaza Arrival Rate

Rate, Average Value & Optimization

Hard

From 5 A.M. to 10 A.M., the rate at which vehicles arrive at a certain toll plaza is given by

A(t) = 450\sqrt{\sin(0.62t)}

, where

t

is the number of hours after 5 A.M. and

A(t)

is measured in vehicles per hour. Traffic is flowing smoothly at 5 A.M. with no vehicles waiting in line.

Part AEasy1 point

Write, but do not evaluate, an integral expression that gives the total number of vehicles that arrive at the toll plaza from 6 A.M. (

t=1

) to 10 A.M. (

t=5

).

Answer

\displaystyle\int_1^5 A(t)\,dt = \int_1^5 450\sqrt{\sin(0.62t)}\,dt

Full Solution & Work

Set up the accumulation integral

Total vehicles = accumulated arrival rate over the interval:

\int_1^5 A(t)\,dt = \int_1^5 450\sqrt{\sin(0.62t)}\,dt

AP Scoring — 1 Point

**P1**: Correct integrand

A(t)

with correct limits

t=1

to

t=5

.

Part BMedium2 points

Find the average value of the rate, in vehicles per hour, at which vehicles arrive at the toll plaza from 6 A.M. (

t=1

) to 10 A.M. (

t=5

).

Answer

≈ 375.537 vehicles per hour

Full Solution & Work

Apply the average value formula

\frac{1}{5-1}\int_1^5 A(t)\,dt

Evaluate with a calculator

\approx 375.537 \text{ vehicles per hour}

AP Scoring — 2 Points

**P1**: Correct average-value setup

\frac14\int_1^5 A(t)\,dt

.
**P2**: Correct answer, 375.537 (accept 375.536–375.538).

Part CMedium2 points

Is the rate at which vehicles arrive at the toll plaza at 6 A.M. (

t=1

) increasing or decreasing? Give a reason for your answer.

Answer

Increasing, because

A'(1)\approx148.947>0

.

Full Solution & Work

Differentiate A(t)

A'(t) = 450\cdot\frac12\big(\sin(0.62t)\big)^{-1/2}\cdot0.62\cos(0.62t) = \frac{139.5\cos(0.62t)}{\sqrt{\sin(0.62t)}}

Evaluate at t = 1

A'(1) \approx 148.947 > 0

Conclude

Since

A'(1)>0

, the rate at which vehicles arrive is **increasing** at

t=1

.

AP Scoring — 2 Points

**P1**: A correct expression or numerical value for

A'(1)

.
**P2**: Correct conclusion ("increasing") tied to the sign of

A'(1)

.

Part DHard4 points

A line forms whenever

A(t)\geq400

. The number of vehicles in line at time

t

, for

a\leq t\leq4

, is given by

N(t)=\displaystyle\int_a^t (A(x)-400)\,dx

, where

a

is the time when a line first begins to form. To the nearest whole number, find the greatest number of vehicles in line at the toll plaza in the time interval

a\leq t\leq4

. Justify your answer.

Answer

≈ 71 vehicles

Full Solution & Work

Find a, the time the line first forms

a

is the first positive solution to

A(t)=400

:

450\sqrt{\sin(0.62t)}=400 \implies t = a \approx 1.469

Find where N could be maximized

N'(t)=A(t)-400

. Since

A

rises above 400 at

t=a

and later falls back below 400 at a second crossing,

t\approx3.598

(within

[a,4]

),

N'(t)>0

on

(a,3.598)

and

N'(t)<0

on

(3.598,4)

. So

N

is maximized at

t\approx3.598

.

Evaluate N at the maximizing time

N(3.598) = \int_a^{3.598}\big(A(x)-400\big)\,dx \approx 71.254

(compare to

N(4)\approx62.338

, confirming the interior critical point gives the larger value)

Conclude

To the nearest whole number, the greatest number of vehicles in line is **71**.

AP Scoring — 4 Points

**P1**: Correctly finds

a\approx1.469

, the first time

A(t)=400

.
**P2**: Considers

N'(t)=A(t)-400=0

, i.e. the second crossing

t\approx3.598

, as the critical point of interest.
**P3**: A justification comparing

N

at the critical point to

N(4)

(global argument), or a sign analysis of

N'

around

t\approx3.598

.
**P4**: Correct final answer, 71 vehicles, with supporting work.

Common mistake: A common shortcut error: forgetting that the domain is restricted to $[a,4]$ and instead searching for a maximum over the full $[0,5]$ range covered by $A(t)$ — the question only asks about the line's length up to $t=4$.

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Question 2

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Area and Volume Between f and g

Area & Volume

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Hard

Let

f

and

g

be the functions defined by

f(x)=\ln(x+3)

and

g(x)=x^4+2x^3

. The graphs of

f

and

g

intersect at

x=-2

and

x=B

, where

B>0

.

Part AMedium2 points

Find the area of the region enclosed by the graphs of

f

and

g

.

Answer

≈ 3.604

Full Solution & Work

Find B

Solving

f(x)=g(x)

numerically for the positive root:

B\approx0.782

.

Set up and evaluate the area integral

On

[-2,B]

,

f(x)\geq g(x)

(confirmed by checking

f-g>0

at interior points):

\text{Area}=\int_{-2}^{B}\big(f(x)-g(x)\big)\,dx \approx 3.604

AP Scoring — 2 Points

**P1**: Correct integrand

f(x)-g(x)

(or equivalent) with correct limits

-2

to

B

.
**P2**: Correct answer, 3.604 (accept 3.603–3.605).

Part BMedium2 points

For

-2\leq x\leq B

, let

h(x)

be the vertical distance between the graphs of

f

and

g

. Is

h

increasing or decreasing at

x=-0.5

? Give a reason for your answer.

Answer

Decreasing, since

h'(-0.5)=-0.6<0

.

Full Solution & Work

Write h(x) and differentiate

Since

f\geq g

on this interval,

h(x)=f(x)-g(x)=\ln(x+3)-x^4-2x^3

.

h'(x) = \frac{1}{x+3}-4x^3-6x^2

Evaluate at x = -0.5

h'(-0.5) = \frac{1}{2.5}-4(-0.125)-6(0.25) = 0.4+0.5-1.5 = -0.6

Conclude

Since

h'(-0.5)=-0.6<0

,

h

is **decreasing** at

x=-0.5

.

AP Scoring — 2 Points

**P1**: Correct expression for

h'(x)

or a correctly-evaluated

h'(-0.5)

.
**P2**: Correct conclusion ("decreasing") tied to the sign of

h'(-0.5)

.

Part CHard3 points

The region enclosed by the graphs of

f

and

g

is the base of a solid. Cross sections of the solid taken perpendicular to the

x

-axis are squares. Find the volume of the solid.

Answer

≈ 5.340

Full Solution & Work

Set up the volume integral

Each square cross section has side length

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

, so area

\big(h(x)\big)^2

:

V = \int_{-2}^{B} \big(f(x)-g(x)\big)^2\,dx

Evaluate with a calculator

V \approx 5.340

AP Scoring — 3 Points

**P1**: Integrand of the form

(f(x)-g(x))^2

.
**P2**: Correct limits

-2

to

B

.
**P3**: Correct answer, 5.340 (accept 5.339–5.341).

Part DHard2 points

A vertical line in the

xy

-plane travels from left to right along the base of the solid described in part (c). The vertical line is moving at a constant rate of 7 units per second. Find the rate of change of the area of the cross section above the vertical line with respect to time when the vertical line is at position

x=-0.5

.

Answer

≈ −9.272 (square units per second)

Full Solution & Work

Write the cross-sectional area as a function of x, then apply the chain rule

A_{\text{cross}}(x) = \big(h(x)\big)^2 \implies \frac{dA_{\text{cross}}}{dt} = 2h(x)\cdot h'(x)\cdot\frac{dx}{dt}

Substitute values at x = -0.5

From parts (b):

h(-0.5)\approx1.104

,

h'(-0.5)=-0.6

. Given

\frac{dx}{dt}=7

:

\frac{dA_{\text{cross}}}{dt} = 2(1.104)(-0.6)(7) \approx -9.272

AP Scoring — 2 Points

**P1**: Correct related-rates setup

\frac{dA_{\text{cross}}}{dt}=2h(x)h'(x)\frac{dx}{dt}

.
**P2**: Correct answer,

-9.272

(accept

-9.270

to

-9.274

).

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Question 3

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Analyzing a Function from its Derivative Graph

Graph Analysis & Curve Sketching

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Hard

Let

f

be a differentiable function with

f(4)=3

. On the interval

0\leq x\leq7

, the graph of

f'

, the derivative of

f

, consists of a semicircle (radius 2, centered at

(2,0)

, dipping below the axis) and two line segments: from

(4,0)

to

(6,2)

, and from

(6,2)

to

(7,1)

.

2022 AP Calculus AB FRQ Question 3 Graph of f' — Graph of $f'$ on $[0,7]$.

Part AMedium2 points

Find

f(0)

and

f(5)

.

Answer

f(0) = 3+2\pi

;

f(5)=3.5

Full Solution & Work

Find f(0) using the semicircle's signed area

f(0) = f(4) - \int_0^4 f'(x)\,dx = 3 - \left(-\frac12\pi(2)^2\right) = 3+2\pi

Find f(5) using the triangle's area

f(5) = f(4)+\int_4^5 f'(x)\,dx = 3+\frac12(1)(1) = 3.5

AP Scoring — 2 Points

**P1**:

f(0)=3+2\pi

.
**P2**:

f(5)=3.5

.

Part BHard2 points

Find the

x

-coordinates of all points of inflection of the graph of

f

for

0<x<7

. Justify your answer.

Answer

x=2

and

x=6

Full Solution & Work

Relate inflection points to local extrema of f'

f

has a point of inflection where

f''=(f')'

changes sign — i.e., where

f'

has a local min or max.

Locate local extrema of f'

f'

has a local minimum at

x=2

(bottom of the semicircle) and a local maximum at

x=6

(peak of the "tent" formed by the two line segments).

Conclude

So

f

has points of inflection at

x=2

and

x=6

(at

x=4

,

f'

keeps increasing through the corner, so there's no sign change and no inflection point there).

AP Scoring — 2 Points

**P1**: Correct values

x=2

and

x=6

(no additional/incorrect values).
**P2**: Justification tied to

f'

having a local extremum (or

f''

changing sign) at each location.

Part CHard3 points

Let

g

be the function defined by

g(x)=f(x)-x

. On what intervals, if any, is

g

decreasing for

0\leq x\leq7

? Show the analysis that leads to your answer.

Answer

(0,5)

Full Solution & Work

Set up g'(x) = f'(x) - 1

g

is decreasing where

g'(x)=f'(x)-1<0

, i.e. where

f'(x)<1

.

Compare f'(x) to 1 on each piece

On

[0,4]

(the semicircle):

f'(x)\leq0<1

throughout.
On

[4,6]

(line, slope 1, from

(4,0)

to

(6,2)

):

f'(x)=x-4<1

for

x<5

.
On

[6,7]

(line from

(6,2)

to

(7,1)

):

f'(x)=8-x\geq1

throughout, so

f'\geq1

here.

Conclude

f'(x)<1

(so

g

is decreasing) precisely on

(0,5)

.

AP Scoring — 3 Points

**P1**: Setting up

g'(x)=f'(x)-1

.
**P2**: Correctly comparing

f'(x)

to 1 on each piece of the domain.
**P3**: Correct interval,

(0,5)

, with supporting analysis.

Part DHard2 points

For the function

g

defined in part (c), find the absolute minimum value on the interval

0\leq x\leq7

. Justify your answer.

Answer

-1.5

, at

x=5

Full Solution & Work

Identify the critical point

From part (c),

g

decreases on

(0,5)

and increases on

(5,7)

(since

f'(x)\geq1

throughout

(5,7)

), so

x=5

is the location of the global minimum on

[0,7]

.

Evaluate g(5)

g(5) = f(5)-5 = 3.5-5=-1.5

Confirm against the endpoints

g(0)=f(0)-0=3+2\pi\approx9.283

and

g(7)=f(7)-7

, where

f(7)=f(6)+\int_6^7f'(x)\,dx=5+1.5=6.5

, so

g(7)=-0.5

. Both exceed

-1.5

, confirming the minimum is at

x=5

.

AP Scoring — 2 Points

**P1**: Correctly identifies

x=5

as the global minimum location, with justification (sign change of

g'

from negative to positive).
**P2**: Correct value,

g(5)=-1.5

.

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Question 4

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Melting Ice Sculpture — Table Data

Rate & Data from Tables

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Hard

An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function

r

, where

r(t)

is measured in centimeters and

t

is measured in days. The table above gives selected values of

r'(t)

, the rate of change of the radius, over the time interval

0\leq t\leq12

.

Values of r′(t)

t (days)	0	3	7	10	12
r′(t) (centimeters per day)	−6.1	−5.0	−4.4	−3.8	−3.5

Part AEasy2 points

Approximate

r''(8.5)

using the average rate of change of

r'

over the interval

7\leq t\leq10

. Show the computations that lead to your answer, and indicate units of measure.

Answer

r''(8.5)\approx0.2

cm/day²

Full Solution & Work

Set up the difference quotient

r''(8.5) \approx \frac{r'(10)-r'(7)}{10-7} = \frac{-3.8-(-4.4)}{3} = \frac{0.6}{3}=0.2

State the units

The units are centimeters per day per day (cm/day²).

AP Scoring — 2 Points

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

Part BMedium2 points

Is there a time

t

,

0\leq t\leq3

, for which

r'(t)=-6

? Justify your answer.

Answer

Yes, by the Intermediate Value Theorem.

Full Solution & Work

Set up the IVT argument

r

is twice-differentiable, so

r'

is differentiable, hence continuous, on

[0,3]

.

Compare -6 to the endpoint values

r'(0)=-6.1<-6<-5.0=r'(3)

Conclude

By the **Intermediate Value Theorem**, since

r'

is continuous on

[0,3]

and

-6

lies between

r'(0)

and

r'(3)

, there must exist a

t\in(0,3)

with

r'(t)=-6

.

AP Scoring — 2 Points

**P1**: Establishes continuity of

r'

on

[0,3]

.
**P2**: Correct comparison and conclusion citing the IVT.

Part CMedium2 points

Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of

\displaystyle\int_0^{12} r'(t)\,dt

.

Answer

-51

Full Solution & Work

Set up the right Riemann sum

\int_0^{12} r'(t)\,dt \approx 3\cdot r'(3)+4\cdot r'(7)+3\cdot r'(10)+2\cdot r'(12)

Substitute and compute

= 3(-5.0)+4(-4.4)+3(-3.8)+2(-3.5) = -15-17.6-11.4-7=-51

AP Scoring — 2 Points

**P1**: Correct form of the right Riemann sum with correct widths.
**P2**: Correct answer,

-51

.

Part DHard3 points

The height of the cone decreases at a rate of 2 centimeters per day. At time

t=3

days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time

t=3

days. (The volume

V

of a cone with radius

r

and height

h

is

V=\dfrac13\pi r^2 h

.)

Answer

-\dfrac{70000\pi}{3}\approx -73303.83

cm³/day

Full Solution & Work

Differentiate V with respect to t

\frac{dV}{dt} = \frac{1}{3}\pi\left(2r\frac{dr}{dt}h+r^2\frac{dh}{dt}\right)

Substitute values at t = 3

r=100

,

\frac{dr}{dt}=r'(3)=-5

(from the table),

h=50

,

\frac{dh}{dt}=-2

:

\frac{dV}{dt} = \frac{\pi}{3}\Big[2(100)(-5)(50)+(100)^2(-2)\Big] = \frac{\pi}{3}\big[-50000-20000\big]

Simplify

= \frac{\pi}{3}(-70000) = -\frac{70000\pi}{3} \approx -73303.83 \text{ cm}^3/\text{day}

AP Scoring — 3 Points

**P1**: Correct related-rates differentiation of

V=\frac13\pi r^2 h

using the Product Rule.
**P2**: Correct substitution, including using the table value

r'(3)=-5

for

\frac{dr}{dt}

.
**P3**: Correct final answer,

-\frac{70000\pi}{3}

(or its decimal equivalent).

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Question 5

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A Differential Equation with a Square-Root Factor

Differential Equations

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Hard

Consider the differential equation

\dfrac{dy}{dx}=\dfrac{1}{2}\sin\!\left(\dfrac{\pi}{2}x\right)\sqrt{y+7}

. Let

y=f(x)

be the particular solution to the differential equation with the initial condition

f(1)=2

. The function

f

is defined for all real numbers.

Part AEasy1 point

A portion of the slope field for the differential equation is given. Sketch the solution curve through the point

(1,2)

.

Answer

The curve rises through

(1,2)

, following the slope field.

Full Solution & Work

Follow the slope field from (1, 2)

At

(1,2)

:

\frac{dy}{dx}=\frac12\sin(\pi/2)\sqrt9=\frac12(1)(3)=1.5

, a fairly steep positive slope. Following the field, the curve continues rising to the right and falls off to the left of

x=1

, consistent with the shown slope field.

AP Scoring — 1 Point

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

(1,2)

and have no obvious conflicts with the given slope lines.

Part BMedium2 points

Write an equation for the line tangent to the solution curve in part (a) at the point

(1,2)

. Use the equation to approximate

f(0.8)

.

Answer

y=2+1.5(x-1)

;

f(0.8)\approx1.7

Full Solution & Work

Find the slope at (1,2)

\left.\frac{dy}{dx}\right|_{(1,2)} = \frac12\sin\!\left(\frac{\pi}{2}\right)\sqrt{2+7} = \frac12(1)(3)=1.5

Write the tangent line and approximate

y=2+1.5(x-1)

f(0.8) \approx 2+1.5(0.8-1)=2-0.3=1.7

AP Scoring — 2 Points

**P1**: Correct slope, 1.5, at

(1,2)

.
**P2**: Correct approximation,

f(0.8)\approx1.7

.

Part CMedium1 point

It is known that

f''(x)>0

for

-1\leq x\leq1

. Is the approximation found in part (b) an overestimate or an underestimate for

f(0.8)

? Give a reason for your answer.

Answer

Underestimate.

Full Solution & Work

Relate concavity to tangent-line approximation

Since

f''(x)>0

on

[-1,1]

, the graph of

f

is **concave up** there, so the tangent line lies **below** the curve on this interval.

Conclude

Since

0.8\in[-1,1]

, the tangent-line approximation

f(0.8)\approx1.7

is an **underestimate** of the actual value.

AP Scoring — 1 Point

**P1**: "Underestimate," with reasoning tied to

f''(x)>0

(concave up).

Part DHard3 points

Use separation of variables to find

y=f(x)

, the particular solution to the differential equation

\dfrac{dy}{dx}=\dfrac12\sin\!\left(\dfrac{\pi}{2}x\right)\sqrt{y+7}

with the initial condition

f(1)=2

.

Answer

y = \left[3-\dfrac{1}{2\pi}\cos\!\left(\dfrac{\pi}{2}x\right)\right]^2-7

Full Solution & Work

Separate variables

\frac{dy}{\sqrt{y+7}} = \frac12\sin\!\left(\frac{\pi}{2}x\right)dx

Integrate both sides

2\sqrt{y+7} = -\frac{1}{\pi}\cos\!\left(\frac{\pi}{2}x\right)+C

Apply the initial condition f(1) = 2

At

x=1,y=2

:

2\sqrt9=6=-\frac1\pi\cos(\pi/2)+C=0+C \implies C=6

.

2\sqrt{y+7} = 6-\frac{1}{\pi}\cos\!\left(\frac\pi2 x\right)

Solve for y

\sqrt{y+7} = 3-\frac{1}{2\pi}\cos\!\left(\frac\pi2 x\right)

y = \left[3-\frac{1}{2\pi}\cos\!\left(\frac\pi2 x\right)\right]^2-7

AP Scoring — 3 Points

**P1**: Correct separation of variables.
**P2**: Correct antiderivatives on both sides, with constant of integration correctly evaluated using

f(1)=2

.
**P3**: Correctly solves for

y

to reach the final closed form.

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Question 6

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Two Particles in Motion

Particle Motion

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Hard

Particle

P

moves along the

x

-axis such that, for time

t>0

, its position is given by

x_P(t)=6-4e^{-t}

. Particle

Q

moves along the

y

-axis such that, for time

t>0

, its velocity is given by

v_Q(t)=\dfrac{1}{t^2}

. At time

t=1

, the position of particle

Q

is

y_Q(1)=2

.

Part AEasy1 point

Find

v_P(t)

, the velocity of particle

P

at time

t

.

Answer

v_P(t)=4e^{-t}

Full Solution & Work

Differentiate x_P(t)

v_P(t) = x_P'(t) = -4(-e^{-t}) = 4e^{-t}

AP Scoring — 1 Point

**P1**: Correct answer,

v_P(t)=4e^{-t}

.

Part BHard3 points

Find

a_Q(t)

, the acceleration of particle

Q

at time

t

. Find all times

t

, for

t>0

, when the speed of particle

Q

is decreasing. Justify your answer.

Answer

a_Q(t)=-\dfrac{2}{t^3}

; speed is decreasing for all

t>0

.

Full Solution & Work

Differentiate v_Q(t)

a_Q(t) = v_Q'(t) = -\frac{2}{t^3}

Compare signs of v_Q and a_Q for t > 0

v_Q(t)=\frac{1}{t^2}>0

for all

t>0

.

a_Q(t)=-\frac{2}{t^3}<0

for all

t>0

. So

v_Q

and

a_Q

have **opposite signs** for every

t>0

.

Conclude

The speed of particle

Q

is **decreasing for all

t>0

** — the particle continually slows as it moves.

AP Scoring — 3 Points

**P1**: Correct

a_Q(t)=-\frac{2}{t^3}

.
**P2**: Correct sign analysis showing

v_Q(t)

and

a_Q(t)

have opposite signs.
**P3**: Correct conclusion — for **all**

t>0

, with justification.

Part CMedium2 points

Find

y_Q(t)

, the position of particle

Q

at time

t

.

Answer

y_Q(t)=3-\dfrac{1}{t}

Full Solution & Work

Set up position as initial position plus accumulated change

y_Q(t) = y_Q(1)+\int_1^t v_Q(s)\,ds = 2+\int_1^t s^{-2}\,ds

Evaluate the integral

= 2+\left[-s^{-1}\right]_1^t = 2+\left(-\frac1t+1\right) = 3-\frac1t

AP Scoring — 2 Points

**P1**: Correct setup

y_Q(1)+\int_1^t v_Q(s)\,ds

.
**P2**: Correct final answer,

y_Q(t)=3-\frac1t

.

Part DMedium2 points

As

t\to\infty

, which particle will eventually be farther from the origin? Give a reason for your answer.

Answer

Particle P.

Full Solution & Work

Find the limiting positions

\lim_{t\to\infty} x_P(t) = \lim_{t\to\infty}\big(6-4e^{-t}\big) = 6, \qquad \lim_{t\to\infty} y_Q(t) = \lim_{t\to\infty}\left(3-\frac1t\right)=3

Compare

Both

x_P(t)

and

y_Q(t)

increase monotonically toward their respective limits, staying below them for all finite

t

. Since

\lim x_P = 6 > \lim y_Q=3

, particle

P

's distance from the origin approaches a strictly larger value.

Conclude

As

t\to\infty

, particle **

P

** will eventually be farther from the origin.

AP Scoring — 2 Points

**P1**: Correctly computes both limits, 6 and 3.
**P2**: Correct conclusion (particle

P

) with a reason comparing the two limits.

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