Calculus-Master
27

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

123456

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+5sin⁡ ⁣(t2)R(t) = 12 + 5\sin\!\left(\dfrac{t}{2}\right) gallons per hour and is pumped out at a rate of D(t)=12t2D(t) = \dfrac{1}{2}t^2 gallons per hour, for 0≤t≤80 \le t \le 8 hours. The tank contains 20 gallons of water at time t=0t = 0.
APart A
Medium
At t=3t = 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)R(t) - D(t). Increasing means the net rate is positive.
Show Solution

Solution · 2 points

R(3)−D(3)=(12+5sin⁡(1.5))−4.5≈12.487>0R(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=3t = 3.
BPart B
Easy
Write an expression involving an integral for the amount of water in the tank at time t=8t = 8.
Need a Hint?
Amount = initial amount + net accumulation.
Show Solution

Solution · 1 point

Amount(8)=20+∫08[R(t)−D(t)] dt(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=8t = 8.
Need a Hint?
Evaluate the integral from Part B on your calculator.
Show Solution

Solution · 2 points

∫08[R(t)−D(t)] dt≈27.203\displaystyle\int_0^8 [R(t) - D(t)]\,dt \approx 27.203, so Amount(8)≈47.203(8) \approx 47.203 gallons.
DPart D
Hard
For 0≤t≤80 \le t \le 8, at what time tt 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)R(t) - D(t) changes from positive to negative.
Show Solution

Solution · 4 points

Solving 12+5sin⁡(t/2)=12t212 + 5\sin(t/2) = \tfrac{1}{2}t^2 gives t≈5.340t \approx 5.340. The net rate is positive for tt just less than 5.3405.340 and negative for tt just greater, so the amount of water is a maximum at t≈5.340t \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 xx-axis with velocity v(t)=e−t/2sin⁡tv(t) = e^{-t/2}\sin t for 0≤t≤60 \le t \le 6. At time t=0t = 0 the particle is at position x(0)=1x(0) = 1.
APart A
Medium
Find the acceleration of the particle at t=2t = 2.
Need a Hint?
a(t)=v′(t)a(t) = v'(t) — use the numerical derivative.
Show Solution

Solution · 2 points

a(2)=v′(2)≈−0.320a(2) = v'(2) \approx -0.320.
BPart B
Medium
Find the position of the particle at t=4t = 4.
Need a Hint?
Position =x(0)+∫04v(t) dt= x(0) + \int_0^4 v(t)\,dt.
Show Solution

Solution · 2 points

x(4)=1+∫04e−t/2sin⁡t dt≈1+0.912=1.912x(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≤t≤60 \le t \le 6.
Need a Hint?
Total distance =∫06∣v(t)∣ dt= \int_0^6 |v(t)|\,dt. Split where v(t)=0v(t) = 0.
Show Solution

Solution · 3 points

v(t)=0v(t) = 0 at t=πt = \pi; v>0v > 0 on (0,π)(0, \pi) and v<0v < 0 on (π,6)(\pi, 6). Total distance =∫0πv dt−∫π6v dt≈0.966+0.199=1.165= \displaystyle\int_0^{\pi} v\,dt - \int_{\pi}^{6} v\,dt \approx 0.966 + 0.199 = 1.165.
DPart D
Easy
For 0<t<60 < t < 6, find all times at which the particle is at rest.
Need a Hint?
At rest means v(t)=0v(t) = 0.
Show Solution

Solution · 2 points

e−t/2sin⁡t=0⇒sin⁡t=0e^{-t/2}\sin t = 0 \Rightarrow \sin t = 0. On (0,6)(0, 6) the only solution is t=π≈3.142t = \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 ff is continuous on [−2,6][-2, 6]. Its derivative f′f' is piecewise linear, passing through the points (−2,3)(-2, 3), (0,0)(0, 0), (2,−2)(2, -2), (4,0)(4, 0), (6,4)(6, 4), connected by line segments. Also f(0)=5f(0) = 5.
APart A
Medium
On what open intervals is ff increasing? Justify your answer.
Need a Hint?
ff increases where f′>0f' > 0.
Show Solution

Solution · 2 points

f′>0f' > 0 on (−2,0)(-2, 0) (values fall from 33 to 00) and on (4,6)(4, 6) (values rise from 00 to 44); f′<0f' < 0 on (0,4)(0, 4). So ff is increasing on (−2,0)(-2, 0) and (4,6)(4, 6) because f′>0f' > 0 there.
BPart B
Medium
Find the xx-coordinate of each local extremum of ff on (−2,6)(-2, 6). Justify your answer.
Need a Hint?
Local extrema occur where f′f' changes sign.
Show Solution

Solution · 2 points

f′f' changes from ++ to −- at x=0x = 0: local maximum. f′f' changes from −- to ++ at x=4x = 4: local minimum. At x=2x = 2, f′f' has a minimum but does not change sign, so ff has no extremum there.
CPart C
Medium
On what open interval(s) is the graph of ff concave up? Justify your answer.
Need a Hint?
f′′f'' is the slope of f′f'.
Show Solution

Solution · 2 points

The slope of f′f' is negative on (−2,2)(-2, 2) and positive on (2,6)(2, 6), so f′′>0f'' > 0 on (2,6)(2, 6) and the graph of ff is concave up on (2,6)(2, 6).
DPart D
Medium
Find f(4)f(4).
Need a Hint?
f(4)=f(0)+∫04f′(t) dtf(4) = f(0) + \int_0^4 f'(t)\,dt; read the integral as signed area.
Show Solution

Solution · 3 points

∫04f′(t) dt\displaystyle\int_0^4 f'(t)\,dt is two triangles: on [0,2][0, 2], 12(2)(−2)=−2\tfrac{1}{2}(2)(-2) = -2; on [2,4][2, 4], 12(2)(−2)=−2\tfrac{1}{2}(2)(-2) = -2. So ∫04f′ dt=−4\int_0^4 f'\,dt = -4 and f(4)=5+(−4)=1f(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 xx-axis with velocity v(t)=t2−6t+8v(t) = t^2 - 6t + 8 for 0≤t≤50 \le t \le 5. At time t=0t = 0 the particle is at position x(0)=2x(0) = 2.
APart A
Easy
Find the acceleration of the particle at t=1t = 1.
Need a Hint?
a(t)=v′(t)a(t) = v'(t).
Show Solution

Solution · 2 points

a(t)=2t−6a(t) = 2t - 6, so a(1)=−4a(1) = -4.
BPart B
Medium
Find the displacement of the particle over 0≤t≤50 \le t \le 5.
Need a Hint?
Displacement =∫05v(t) dt= \int_0^5 v(t)\,dt (no absolute value).
Show Solution

Solution · 2 points

∫05(t2−6t+8) dt=[t33−3t2+8t]05=1253−75+40=203\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≤t≤50 \le t \le 5.
Need a Hint?
v(t)=(t−2)(t−4)v(t) = (t-2)(t-4). Split at t=2t = 2 and t=4t = 4.
Show Solution

Solution · 3 points

∫02v=203\int_0^2 v = \frac{20}{3}, ∫24v=−43\int_2^4 v = -\frac{4}{3}, ∫45v=43\int_4^5 v = \frac{4}{3}. Total distance =203+43+43=283= \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=3t = 3? Justify your answer.
Need a Hint?
Compare the signs of v(3)v(3) and a(3)a(3).
Show Solution

Solution · 2 points

v(3)=−1<0v(3) = -1 < 0 and a(3)=2(3)−6=0a(3) = 2(3) - 6 = 0. Since a(3)=0a(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 RR be the region in the first quadrant bounded by y=xy = \sqrt{x} and y=x2y = \dfrac{x}{2}. The curves meet where x=x2\sqrt{x} = \dfrac{x}{2}, at x=0x = 0 and x=4x = 4; on (0,4)(0, 4), x>x2\sqrt{x} > \dfrac{x}{2}.
APart A
Medium
Find the area of RR.
Need a Hint?
Integrate top minus bottom.
Show Solution

Solution · 2 points

∫04(x−x2)dx=[23x3/2−x24]04=163−4=43\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
RR is revolved about the xx-axis. Find the volume of the resulting solid.
Need a Hint?
Washers: outer radius x\sqrt{x}, inner radius x2\dfrac{x}{2}. Square each radius first.
Show Solution

Solution · 4 points

V=π∫04[x−x24]dx=π[x22−x312]04=π(8−163)=8π3V = \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
RR is the base of a solid whose cross sections perpendicular to the xx-axis are squares. Find the volume of that solid.
Need a Hint?
The side of each square is x−x2\sqrt{x} - \dfrac{x}{2}; there is no π\pi.
Show Solution

Solution · 3 points

V=∫04(x−x2)2dx=∫04(x−x3/2+x24)dx=[x22−25x5/2+x312]04=815V = \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)=∫0xf(t) dtg(x) = \displaystyle\int_0^x f(t)\,dt, where f(t)=3t2−12f(t) = 3t^2 - 12.
APart A
Medium
Find g(2)g(2) and g′(2)g'(2).
Need a Hint?
Evaluate the integral for g(2)g(2); use the FTC for g′g'.
Show Solution

Solution · 2 points

g(2)=∫02(3t2−12) dt=[t3−12t]02=−16g(2) = \displaystyle\int_0^2 (3t^2 - 12)\,dt = [t^3 - 12t]_0^2 = -16. g′(x)=f(x)=3x2−12g'(x) = f(x) = 3x^2 - 12, so g′(2)=0g'(2) = 0.
BPart B
Medium
On what interval(s) is gg decreasing? Justify your answer.
Need a Hint?
gg decreases where g′(x)=f(x)<0g'(x) = f(x) < 0.
Show Solution

Solution · 2 points

g′(x)=3x2−12<0g'(x) = 3x^2 - 12 < 0 when −2<x<2-2 < x < 2, so gg is decreasing on (−2,2)(-2, 2).
CPart C
Medium
Find the xx-coordinate of each point of inflection of the graph of gg on (0,∞)(0, \infty). Justify your answer.
Need a Hint?
g′′(x)=f′(x)g''(x) = f'(x); look for a sign change.
Show Solution

Solution · 3 points

g′′(x)=f′(x)=6x>0g''(x) = f'(x) = 6x > 0 for all xx in (0,∞)(0, \infty), so g′′g'' never changes sign there — there is no point of inflection on (0,∞)(0, \infty).
DPart D
Medium
Find the absolute minimum value of gg on [0,3][0, 3].
Need a Hint?
Compare gg at the critical number and at the endpoints.
Show Solution

Solution · 2 points

g′(x)=0g'(x) = 0 at x=2x = 2. g(0)=0g(0) = 0, g(2)=−16g(2) = -16, g(3)=27−36=−9g(3) = 27 - 36 = -9. The absolute minimum value is −16-16, at x=2x = 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 & OutParticle MotionGraph & Function AnalysisArea & Volume