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Day 28 · Free-Response Practice Exam

FRQ Practice Exam 2

A second full free-response paper: 6 questions, 9 points each. About 15 minutes per question, show every setup, then grade yourself against the solution.

Each question is worth 9 points. Setup, value, units, and justification are scored separately. For a differential equation, separate the variables in writing and keep the +C+C until the initial condition. 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
People enter a fair at a rate of E(t)=120+80cos⁡ ⁣(t2)E(t) = 120 + 80\cos\!\left(\dfrac{t}{2}\right) people per hour and leave at a rate of L(t)=25tL(t) = 25t people per hour, for 0≤t≤60 \le t \le 6 hours. At time t=0t = 0 there are 200 people at the fair.
APart A
Medium
How many people entered the fair during the first 3 hours?
Need a Hint?
"Entered" is ∫E\int E only — not the net change.
Show Solution

Solution · 2 points

∫03E(t) dt=[120t+160sin⁡(t/2)]03=360+160sin⁡(1.5)≈520\displaystyle\int_0^3 E(t)\,dt = \big[120t + 160\sin(t/2)\big]_0^3 = 360 + 160\sin(1.5) \approx 520 people.
BPart B
Medium
Is the number of people at the fair increasing or decreasing at t=4t = 4? Justify your answer.
Need a Hint?
Compare E(4)E(4) and L(4)L(4).
Show Solution

Solution · 2 points

E(4)−L(4)=(120+80cos⁡(2))−100≈−13.3<0E(4) - L(4) = \big(120 + 80\cos(2)\big) - 100 \approx -13.3 < 0, so the number of people is decreasing at t=4t = 4.
CPart C
Medium
Write an expression involving an integral for the number of people at the fair at t=6t = 6, and evaluate it.
Need a Hint?
N(6)=200+∫06[E(t)−L(t)] dtN(6) = 200 + \int_0^6 [E(t) - L(t)]\,dt.
Show Solution

Solution · 3 points

N(6)=200+∫06[E(t)−L(t)] dt≈200+292.6≈493N(6) = 200 + \displaystyle\int_0^6 \big[E(t) - L(t)\big]\,dt \approx 200 + 292.6 \approx 493 people.
DPart D
Hard
At what time tt, 0≤t≤60 \le t \le 6, is the number of people at the fair a maximum? Justify your answer.
Need a Hint?
Where E(t)−L(t)E(t) - L(t) changes from positive to negative.
Show Solution

Solution · 2 points

Solving 120+80cos⁡(t/2)=25t120 + 80\cos(t/2) = 25t gives t≈3.784t \approx 3.784. The net rate is positive just before and negative just after, so the number of people is a maximum at t≈3.784t \approx 3.784 hours.
Practice more of this type — Accumulation / Rate In & Out FRQs, with a full walkthrough and a timed set

Question 2

Table analysis

Medium
A cup of coffee cools over time. The table gives its temperature h(t)h(t) in degrees Celsius at selected times tt in minutes. Assume hh is differentiable and decreasing.

Temperature h(t) in °C at time t in minutes

t (min)0481216
h(t) (°C)9075666056
APart A
Easy
Approximate h′(10)h'(10) using data from the table. Show the computation and give units.
Need a Hint?
Use the interval [8,12][8, 12] that straddles t=10t = 10.
Show Solution

Solution · 2 points

h′(10)≈h(12)−h(8)12−8=60−664=−1.5 ∘C/minh'(10) \approx \dfrac{h(12) - h(8)}{12 - 8} = \dfrac{60 - 66}{4} = -1.5\ ^\circ\mathrm{C}/\text{min}.
BPart B
Medium
Use a left Riemann sum with the four subintervals in the table to approximate ∫016h(t) dt\int_0^{16} h(t)\,dt. Using units, explain the meaning of 116∫016h(t) dt\dfrac{1}{16}\int_0^{16} h(t)\,dt.
Need a Hint?
A left sum weights each left value by its subinterval width (here 44).
Show Solution

Solution · 3 points

Left sum =4(90+75+66+60)=1164= 4(90 + 75 + 66 + 60) = 1164. The quantity 116∫016h(t) dt\dfrac{1}{16}\int_0^{16} h(t)\,dt is the average temperature of the coffee, in ∘C^\circ\mathrm{C}, over the 16-minute interval.
CPart C
Medium
Using the Mean Value Theorem, explain why there must be a time tt in (0,16)(0, 16) with h′(t)=−2.125h'(t) = -2.125.
Need a Hint?
Compute the average rate of change of hh over [0,16][0, 16].
Show Solution

Solution · 2 points

The average rate of change on [0,16][0, 16] is 56−9016=−2.125\dfrac{56 - 90}{16} = -2.125. Since hh is differentiable (hence continuous) on [0,16][0, 16], by the MVT there is a tt in (0,16)(0, 16) with h′(t)=−2.125 ∘C/minh'(t) = -2.125\ ^\circ\mathrm{C}/\text{min}.
DPart D
Easy
Find the average rate of change of hh over 8≤t≤168 \le t \le 16.
Need a Hint?
h(16)−h(8)16−8\dfrac{h(16) - h(8)}{16 - 8}.
Show Solution

Solution · 2 points

56−668=−1.25 ∘C/min\dfrac{56 - 66}{8} = -1.25\ ^\circ\mathrm{C}/\text{min}.
Practice more of this type — Rate & Data from Tables FRQs, with a full walkthrough and a timed set

Part B — No Calculator

4 questions · 60 minutes · no calculator

Question 3

Differential equation

Medium
Consider the differential equation dydx=2xy\dfrac{dy}{dx} = \dfrac{2x}{y}.
APart A
Easy
Find the slope of a solution curve at the point (2,3)(2, 3) and at the point (1,−2)(1, -2).
Need a Hint?
Substitute each point into 2xy\dfrac{2x}{y}.
Show Solution

Solution · 2 points

At (2,3)(2, 3): 2(2)3=43\dfrac{2(2)}{3} = \dfrac{4}{3}. At (1,−2)(1, -2): 2(1)−2=−1\dfrac{2(1)}{-2} = -1.
BPart B
Medium
Find the particular solution y=f(x)y = f(x) with f(0)=1f(0) = 1.
Need a Hint?
Separable: y dy=2x dxy\,dy = 2x\,dx. Apply f(0)=1f(0) = 1, then choose the sign of the root.
Show Solution

Solution · 5 points

y dy=2x dx⇒y22=x2+Cy\,dy = 2x\,dx \Rightarrow \dfrac{y^2}{2} = x^2 + C. f(0)=1f(0) = 1: 12=C\dfrac{1}{2} = C. So y2=2x2+1y^2 = 2x^2 + 1, and since f(0)=1>0f(0) = 1 > 0, f(x)=2x2+1f(x) = \sqrt{2x^2 + 1}.
CPart C
Medium
For that particular solution, evaluate lim⁡x→∞f(x)x\displaystyle\lim_{x \to \infty} \dfrac{f(x)}{x}.
Need a Hint?
Divide inside the root by x2x^2.
Show Solution

Solution · 2 points

f(x)x=2x2+1x2=2+1x2→2\dfrac{f(x)}{x} = \sqrt{\dfrac{2x^2 + 1}{x^2}} = \sqrt{2 + \dfrac{1}{x^2}} \to \sqrt{2}.
Practice more of this type — Differential Equations FRQs, with a full walkthrough and a timed set

Question 4

Area and volume

Hard
Let RR be the region bounded by y=4−x2y = 4 - x^2 and the xx-axis. The curve meets the xx-axis at x=−2x = -2 and x=2x = 2.
APart A
Medium
Find the area of RR.
Need a Hint?
∫−22(4−x2) dx\int_{-2}^{2} (4 - x^2)\,dx.
Show Solution

Solution · 2 points

∫−22(4−x2) dx=[4x−x33]−22=16−163=323\displaystyle\int_{-2}^{2} (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = 16 - \frac{16}{3} = \frac{32}{3}.
BPart B
Medium
RR is revolved about the xx-axis. Find the volume of the solid.
Need a Hint?
Disks of radius 4−x24 - x^2.
Show Solution

Solution · 4 points

V=π∫−22(16−8x2+x4) dx=π[16x−8x33+x55]−22=512π15V = \pi\displaystyle\int_{-2}^{2} (16 - 8x^2 + x^4)\,dx = \pi\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{-2}^{2} = \frac{512\pi}{15}.
CPart C
Hard
RR is the base of a solid whose cross sections perpendicular to the xx-axis are squares. Find the volume.
Need a Hint?
Side =4−x2= 4 - x^2; same integrand as Part B without π\pi.
Show Solution

Solution · 3 points

V=∫−22(4−x2)2 dx=[16x−8x33+x55]−22=51215V = \displaystyle\int_{-2}^{2} (4 - x^2)^2\,dx = \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{-2}^{2} = \frac{512}{15}.
Practice more of this type — Area & Volume FRQs, with a full walkthrough and a timed set

Question 5

Accumulation from a graph

Medium
The function ff is piecewise linear on [0,6][0, 6], passing through (0,0)(0, 0), (2,2)(2, 2), (4,−2)(4, -2), (6,0)(6, 0), connected by line segments. Let g(x)=∫0xf(t) dtg(x) = \displaystyle\int_0^x f(t)\,dt.
APart A
Medium
Find g(2)g(2) and g(6)g(6).
Need a Hint?
gg is signed area under ff. On [2,4][2, 4], ff crosses zero at x=3x = 3.
Show Solution

Solution · 3 points

g(2)=12(2)(2)=2g(2) = \tfrac{1}{2}(2)(2) = 2. On [2,4][2, 4]: ff contributes +1+1 then −1-1; on [4,6][4, 6]: −2-2. So g(6)=2+1−1−2=0g(6) = 2 + 1 - 1 - 2 = 0.
BPart B
Medium
On what open interval(s) is gg decreasing? Justify your answer.
Need a Hint?
g′(x)=f(x)g'(x) = f(x).
Show Solution

Solution · 2 points

f<0f < 0 on (3,6)(3, 6), so gg is decreasing on (3,6)(3, 6).
CPart C
Medium
Find the xx-coordinate of the absolute maximum of gg on [0,6][0, 6]. Justify your answer.
Need a Hint?
Where does g′=fg' = f change from ++ to −-?
Show Solution

Solution · 2 points

g′(x)=f(x)g'(x) = f(x) changes from positive to negative at x=3x = 3, and gg is increasing on (0,3)(0, 3) and decreasing on (3,6)(3, 6), so the absolute maximum of gg on [0,6][0, 6] is at x=3x = 3.
DPart D
Easy
Find g′(5)g'(5).
Need a Hint?
g′(5)=f(5)g'(5) = f(5); use the line from (4,−2)(4, -2) to (6,0)(6, 0).
Show Solution

Solution · 2 points

g′(5)=f(5)=−1g'(5) = f(5) = -1.
Practice more of this type — Graph & Function Analysis FRQs, with a full walkthrough and a timed set

Question 6

Related rates

Hard
A tank has the shape of a cone with its vertex pointing down. The cone is 12 ft tall and has radius 6 ft at the top. Water drains so that the volume decreases at a constant rate of 2 ft3^3/min. Let hh be the depth of the water.
APart A
Medium
Write the volume VV of water as a function of hh alone.
Need a Hint?
By similar triangles rh=612\dfrac{r}{h} = \dfrac{6}{12}.
Show Solution

Solution · 2 points

r=h2r = \dfrac{h}{2}, so V=13π(h2)2h=πh312V = \dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2 h = \dfrac{\pi h^3}{12}.
BPart B
Hard
Find the rate at which the water depth is changing when h=4h = 4 ft.
Need a Hint?
Differentiate VV with respect to tt, then substitute dVdt=−2\dfrac{dV}{dt} = -2 and h=4h = 4.
Show Solution

Solution · 4 points

dVdt=πh24dhdt\dfrac{dV}{dt} = \dfrac{\pi h^2}{4}\dfrac{dh}{dt}. With dVdt=−2\dfrac{dV}{dt} = -2, h=4h = 4: −2=4πdhdt-2 = 4\pi\dfrac{dh}{dt}, so dhdt=−12π\dfrac{dh}{dt} = -\dfrac{1}{2\pi} ft/min.
CPart C
Medium
Find the rate at which the radius of the water's surface is changing when h=4h = 4 ft.
Need a Hint?
r=h2r = \dfrac{h}{2}.
Show Solution

Solution · 3 points

drdt=12dhdt=−14π\dfrac{dr}{dt} = \dfrac{1}{2}\dfrac{dh}{dt} = -\dfrac{1}{4\pi} ft/min.

Drill the question types on this paper

Accumulation / Rate In & OutRate & Data from TablesDifferential EquationsArea & VolumeGraph & Function Analysis