Day 27 · Free-Response Practice Exam

FRQ Practice Exam 1

A full free-response paper in the style of the AP exam: 6 questions, 9 points each. Give yourself about 15 minutes per question, show every setup, then open each solution to grade yourself.

Each question is worth 9 points. A part typically gives 1 point for the setup (the integral, derivative, or equation), 1 for the value, and — where asked — 1 for units and 1 for a complete justification. Do not simplify: an unsimplified correct answer earns full credit.

Questions

1 2 3 4 5 6

Part A — Graphing Calculator Required

2 questions · 30 minutes · a graphing calculator is required

Question 1

Rate in / rate out

Hard

Water flows into a tank at a rate of

R(t) = 12 + 5\sin\!\left(\dfrac{t}{2}\right)

gallons per hour and is pumped out at a rate of

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

gallons per hour, for

0 \le t \le 8

hours. The tank contains 20 gallons of water at time

t = 0

APart A

Medium

t = 3

, is the amount of water in the tank increasing or decreasing? Justify your answer.

Need a Hint?

The net rate is

R(t) - D(t)

. Increasing means the net rate is positive.

Show Solution

Solution · 2 points

R(3) - D(3) = \big(12 + 5\sin(1.5)\big) - 4.5 \approx 12.487 > 0

, so the amount of water is increasing at

t = 3

BPart B

Easy

Write an expression involving an integral for the amount of water in the tank at time

t = 8

Need a Hint?

Amount = initial amount + net accumulation.

Show Solution

Solution · 1 point

Amount

(8) = 20 + \displaystyle\int_0^8 \big[R(t) - D(t)\big]\,dt

CPart C

Medium

Find the amount of water in the tank at

t = 8

Need a Hint?

Evaluate the integral from Part B on your calculator.

Show Solution

Solution · 2 points

\displaystyle\int_0^8 [R(t) - D(t)]\,dt \approx 27.203

, so Amount

(8) \approx 47.203

gallons.

DPart D

Hard

For

0 \le t \le 8

, at what time

t

is the amount of water in the tank a maximum? Justify your answer.

Need a Hint?

The maximum occurs where the net rate

R(t) - D(t)

changes from positive to negative.

Show Solution

Solution · 4 points

Solving

12 + 5\sin(t/2) = \tfrac{1}{2}t^2

gives

t \approx 5.340

. The net rate is positive for

t

just less than

5.340

and negative for

t

just greater, so the amount of water is a maximum at

t \approx 5.340

hours.

Practice more of this type — Accumulation / Rate In & Out FRQs, with a full walkthrough and a timed set

Question 2

Particle motion (calculator)

Hard

A particle moves along the

x

-axis with velocity

v(t) = e^{-t/2}\sin t

for

0 \le t \le 6

. At time

t = 0

the particle is at position

x(0) = 1

APart A

Medium

Find the acceleration of the particle at

t = 2

Need a Hint?

a(t) = v'(t)

— use the numerical derivative.

Show Solution

Solution · 2 points

a(2) = v'(2) \approx -0.320

BPart B

Medium

Find the position of the particle at

t = 4

Need a Hint?

Position

= x(0) + \int_0^4 v(t)\,dt

Show Solution

Solution · 2 points

x(4) = 1 + \displaystyle\int_0^4 e^{-t/2}\sin t\,dt \approx 1 + 0.912 = 1.912

CPart C

Hard

Find the total distance traveled by the particle over

0 \le t \le 6

Need a Hint?

Total distance

= \int_0^6 |v(t)|\,dt

. Split where

v(t) = 0

Show Solution

Solution · 3 points

v(t) = 0

t = \pi

;

v > 0

(0, \pi)

and

v < 0

(\pi, 6)

. Total distance

= \displaystyle\int_0^{\pi} v\,dt - \int_{\pi}^{6} v\,dt \approx 0.966 + 0.199 = 1.165

DPart D

Easy

For

0 < t < 6

, find all times at which the particle is at rest.

Need a Hint?

At rest means

v(t) = 0

Show Solution

Solution · 2 points

e^{-t/2}\sin t = 0 \Rightarrow \sin t = 0

. On

(0, 6)

the only solution is

t = \pi \approx 3.142

Practice more of this type — Particle Motion FRQs, with a full walkthrough and a timed set

Part B — No Calculator

4 questions · 60 minutes · no calculator

Question 3

Analysis of f from the graph of f'

Medium

The function

f

is continuous on

[-2, 6]

. Its derivative

f'

is piecewise linear, passing through the points

(-2, 3)

(0, 0)

(2, -2)

(4, 0)

(6, 4)

, connected by line segments. Also

f(0) = 5

APart A

Medium

On what open intervals is

f

increasing? Justify your answer.

Need a Hint?

f

increases where

f' > 0

Show Solution

Solution · 2 points

f' > 0

(-2, 0)

(values fall from

3

0

) and on

(4, 6)

(values rise from

0

4

);

f' < 0

(0, 4)

. So

f

is increasing on

(-2, 0)

and

(4, 6)

because

f' > 0

there.

BPart B

Medium

Find the

x

-coordinate of each local extremum of

f

(-2, 6)

. Justify your answer.

Need a Hint?

Local extrema occur where

f'

changes sign.

Show Solution

Solution · 2 points

f'

changes from

+

-

x = 0

: local maximum.

f'

changes from

-

+

x = 4

: local minimum. At

x = 2

f'

has a minimum but does not change sign, so

f

has no extremum there.

CPart C

Medium

On what open interval(s) is the graph of

f

concave up? Justify your answer.

Need a Hint?

f''

is the slope of

f'

Show Solution

Solution · 2 points

The slope of

f'

is negative on

(-2, 2)

and positive on

(2, 6)

, so

f'' > 0

(2, 6)

and the graph of

f

is concave up on

(2, 6)

DPart D

Medium

Find

f(4)

Need a Hint?

f(4) = f(0) + \int_0^4 f'(t)\,dt

; read the integral as signed area.

Show Solution

Solution · 3 points

\displaystyle\int_0^4 f'(t)\,dt

is two triangles: on

[0, 2]

\tfrac{1}{2}(2)(-2) = -2

; on

[2, 4]

\tfrac{1}{2}(2)(-2) = -2

. So

\int_0^4 f'\,dt = -4

and

f(4) = 5 + (-4) = 1

Practice more of this type — Graph & Function Analysis FRQs, with a full walkthrough and a timed set

Question 4

Particle motion (analytic)

Medium

A particle moves along the

x

-axis with velocity

v(t) = t^2 - 6t + 8

for

0 \le t \le 5

. At time

t = 0

the particle is at position

x(0) = 2

APart A

Easy

Find the acceleration of the particle at

t = 1

Need a Hint?

a(t) = v'(t)

Show Solution

Solution · 2 points

a(t) = 2t - 6

, so

a(1) = -4

BPart B

Medium

Find the displacement of the particle over

0 \le t \le 5

Need a Hint?

Displacement

= \int_0^5 v(t)\,dt

(no absolute value).

Show Solution

Solution · 2 points

\displaystyle\int_0^5 (t^2 - 6t + 8)\,dt = \left[\frac{t^3}{3} - 3t^2 + 8t\right]_0^5 = \frac{125}{3} - 75 + 40 = \frac{20}{3}

CPart C

Hard

Find the total distance traveled over

0 \le t \le 5

Need a Hint?

v(t) = (t-2)(t-4)

. Split at

t = 2

and

t = 4

Show Solution

Solution · 3 points

\int_0^2 v = \frac{20}{3}

\int_2^4 v = -\frac{4}{3}

\int_4^5 v = \frac{4}{3}

. Total distance

= \frac{20}{3} + \frac{4}{3} + \frac{4}{3} = \frac{28}{3}

DPart D

Medium

Is the particle speeding up or slowing down at

t = 3

? Justify your answer.

Need a Hint?

Compare the signs of

v(3)

and

a(3)

Show Solution

Solution · 2 points

v(3) = -1 < 0

and

a(3) = 2(3) - 6 = 0

. Since

a(3) = 0

, the particle is neither speeding up nor slowing down at that instant.

Practice more of this type — Particle Motion FRQs, with a full walkthrough and a timed set

Question 5

Area and volume

Medium

Let

R

be the region in the first quadrant bounded by

y = \sqrt{x}

and

y = \dfrac{x}{2}

. The curves meet where

\sqrt{x} = \dfrac{x}{2}

, at

x = 0

and

x = 4

; on

(0, 4)

\sqrt{x} > \dfrac{x}{2}

APart A

Medium

Find the area of

R

Need a Hint?

Integrate top minus bottom.

Show Solution

Solution · 2 points

\displaystyle\int_0^4 \left(\sqrt{x} - \frac{x}{2}\right) dx = \left[\frac{2}{3}x^{3/2} - \frac{x^2}{4}\right]_0^4 = \frac{16}{3} - 4 = \frac{4}{3}

BPart B

Medium

R

is revolved about the

x

-axis. Find the volume of the resulting solid.

Need a Hint?

Washers: outer radius

\sqrt{x}

, inner radius

\dfrac{x}{2}

. Square each radius first.

Show Solution

Solution · 4 points

V = \pi\displaystyle\int_0^4 \left[x - \frac{x^2}{4}\right] dx = \pi\left[\frac{x^2}{2} - \frac{x^3}{12}\right]_0^4 = \pi\left(8 - \frac{16}{3}\right) = \frac{8\pi}{3}

CPart C

Hard

R

is the base of a solid whose cross sections perpendicular to the

x

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

Need a Hint?

The side of each square is

\sqrt{x} - \dfrac{x}{2}

; there is no

\pi

Show Solution

Solution · 3 points

V = \displaystyle\int_0^4 \left(\sqrt{x} - \frac{x}{2}\right)^2 dx = \int_0^4 \left(x - x^{3/2} + \frac{x^2}{4}\right) dx = \left[\frac{x^2}{2} - \frac{2}{5}x^{5/2} + \frac{x^3}{12}\right]_0^4 = \frac{8}{15}

Practice more of this type — Area & Volume FRQs, with a full walkthrough and a timed set

Question 6

Function defined by an integral

Medium

Let

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

, where

f(t) = 3t^2 - 12

APart A

Medium

Find

g(2)

and

g'(2)

Need a Hint?

Evaluate the integral for

g(2)

; use the FTC for

g'

Show Solution

Solution · 2 points

g(2) = \displaystyle\int_0^2 (3t^2 - 12)\,dt = [t^3 - 12t]_0^2 = -16

g'(x) = f(x) = 3x^2 - 12

, so

g'(2) = 0

BPart B

Medium

On what interval(s) is

g

decreasing? Justify your answer.

Need a Hint?

g

decreases where

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

Show Solution

Solution · 2 points

g'(x) = 3x^2 - 12 < 0

when

-2 < x < 2

, so

g

is decreasing on

(-2, 2)

CPart C

Medium

Find the

x

-coordinate of each point of inflection of the graph of

g

(0, \infty)

. Justify your answer.

Need a Hint?

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

; look for a sign change.

Show Solution

Solution · 3 points

g''(x) = f'(x) = 6x > 0

for all

x

(0, \infty)

, so

g''

never changes sign there — there is no point of inflection on

(0, \infty)

DPart D

Medium

Find the absolute minimum value of

g

[0, 3]

Need a Hint?

Compare

g

at the critical number and at the endpoints.

Show Solution

Solution · 2 points

g'(x) = 0

x = 2

g(0) = 0

g(2) = -16

g(3) = 27 - 36 = -9

. The absolute minimum value is

-16

, at

x = 2

Practice more of this type — Graph & Function Analysis FRQs, with a full walkthrough and a timed set

Drill the question types on this paper

Accumulation / Rate In & Out Particle Motion Graph & Function Analysis Area & Volume