Math 116 — Practice for Exam 2
Generated October 21, 2015
Name:
SOLUTIONS
Instructor:
Section Number:
1. This exam has 7 questions. Note that the problems are not of equal difficulty, so you may want to skip
over and return to a problem on which you are stuck.
2. Do not separate the pages of the exam. If any pages do become separated, write your name on them
and point them out to your instructor when you hand in the exam.
3. Please read the instructions for each individual exercise carefully. One of the skills being tested on
this exam is your ability to interpret questions, so instructors will not answer questions about exam
problems during the exam.
4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that the
graders can see not only the answer but also how you obtained it. Include units in your answers where
appropriate.
5. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad).
However, you must show work for any calculation which we have learned how to do in this course. You
are also allowed two sides of a 3′′ × 5′′ note card.
6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of the
graph, and to write out the entries of the table that you use.
7. You must use the methods learned in this course to solve all problems.
Semester
Exam
Problem
Name
Points
Winter 2010
2
8
Fall 2010
2
2
Fall 2010
3
5
Winter 2014
2
9
pond rock
Winter 2015
2
2
lilypad
6
Fall 2013
2
4
volcano
12
Winter 2014
3
7
alpaca pool
14
cardioid
14
9
Total
Recommended time (based on points): 71 minutes
10
8
73
Score
Math 116 / Exam 2 (March 23, 2010)
page 9
8. [14 points] Consider the area contained above the line y = 0.8, and below the curve r =
2 sin(3θ). You may find the following figure helpful.
y
r = 2sin(31)
y = 0.8
x
a. [4 points] Find the (x, y) coordinates for the two points where y = 0.8 and r = 2 sin(3θ)
intersect as shown in the figure above. Show enough work to support your answer.
0.8
0.8
Solution: We convert y = 0.8 into r = sin
θ , then solve for sin θ = 2 sin(3θ), which gives
θ ≈ 0.4296, 0.8623. The corresponding coordinates are (1.7464, 0.8) and (0.6854, 0.8).
b. [4 points] Write an expression for the area that is specified. You do not need to evaluate
your expression.
Solution:
Area =
1
2
Z
.8623
4 sin2 (3θ)dθ −
.4296
1
2
Z
.8623
.4296
0.64
dθ
sin2 (θ)
c. [6 points] Calculate the perimeter that surrounds the specified area. You may round your
final answer to two decimal places.
Solution: The distance along the line y = 0.8, is 1.7464 − 0.6854 = 1.0610. Along the
curve r = 2 sin(3θ), x = 2 sin(3θ) cos(θ) and y = 2 sin(3θ) sin(θ). We use this to find
dx
dy
= −2 sin(3θ) sin(θ) + 6 cos(3θ) cos(θ) and
= 2 sin(3θ) cos(θ) + 6 cos(3θ) sin(θ).
dθ
dθ
arc length =
Z
.8623
.4296
s
dx
dθ
2
+
dy
dθ
2
≈ 1.3737
The total perimeter is approximately 1.3737 + 1.0610 = 2.4347 ≈ 2.43.
University of Michigan Department of Mathematics
Winter, 2010 Math 116 Exam 2 Problem 8 Solution
Math 116 / Exam 2 (November 17, 2010)
page 3
2. [14 points] The graph of the circle r = 4 and and the cardioid r = 2 sin θ − 2 are shown below.
4
2
0
-2
-4
-4
-2
0
2
4
a. [3 points] Write a formula for the area inside the circle and outside the cardioid in the
first quadrant.
Solution: Area of the quarter of a circle= 4π
R 3π
Area of cardioid= π 2 21 (2 sin θ − 2)2 dθ
R 3π
Area=4π − π 2 21 (2 sin θ − 2)2 dθ
b. [7 points] At what angles 0 ≤ θ < 2π is the minimum value of the y coordinate on
the cardioid attained? No credit will be given for answers without proper mathematical
justification.
Solution:
y(θ) = (2 sin θ − 2) sin θ
y ′ (θ) = 2 cos θ sin θ + (2 sin θ − 2) cos θ = 4 cos θ sin θ − 2 cos θ
Critical points
(4 sin θ − 2) cos θ = 0
3π
then θ = π6 , π2 , 5π
cos θ = 0
or
sin θ = 21
6 , 2 .
Minimum y coordinate at θ = π6 , 5π
6 .
c. [4 points] Write an integral that computes the value of the length of the piece of the
cardioid lying below the x-axis.
Solution:
x(θ) = (2 sin θ − 2) cos θ
L=
Rπp
0
x′ (θ) = 2 cos2 θ − (2 sin θ − 2) sin θ
(2 cos2 θ − (2 sin θ − 2) sin θ)2 + (4 cos θ sin θ − 2 cos θ)2 dθ
University of Michigan Department of Mathematics
Fall, 2010 Math 116 Exam 2 Problem 2 (cardioid) Solution
Math 116 / Final (December 17, 2010)
page 7
5. [9 points]
a. [3 points] In the picture below, the graphs of r = 2 and r = 2 − sin(5θ) are shown.
Write a definite integral that computes the shaded area.
2
1
-2
1
-1
2
-1
-2
-3
Solution: Endpoints satisfy: 2 = 2 − sin(5θ), hence sin(5θ) = 0. Then the
= .628, 1.25 radians.
endpoints are θ = π5 , 2π
5
1
A=
2
Z
2π
5
(2 − sin(5θ))2 − 4dθ
π
5
b. [6 points] Find parametric equations of the tangent line to the limacon r = sin(θ)− 12
at θ = π4 .
Solution:
1
x(θ) = (sin θ − ) cos θ
2
1
x′ (θ) = (sin θ − )(− sin θ) + cos2 (θ)
2
1
y(θ) = (sin θ − ) sin θ
2
1
y ′ (θ) = (sin θ − ) cos θ + sin θ cos θ
2
π
x( ) = .146
4
π
x′ ( ) = .353
4
π
y( ) = .146
4
π
y ′ ( ) = .646
4
xtan (t) = .146 + .353 t
ytan (t) = .146 + .646 t
University of Michigan Department of Mathematics
Fall, 2010 Math 116 Exam 3 Problem 5 Solution
Math 116 / Exam 2 (March 24th, 2014)
page 8
9. [10 points] Linda is designing a pond with a flat rock at one end. The rock plus the pond are
in the shape of a cardioid. Plans for her pond design are depicted below. The cardioid has
equation r = 20 + 40 sin θ where r is in feet and θ is in radians. The inner loop of the cardioid
forms the shape of the rock and the outer loop forms the boundary of the pond.
60
40
20
−40
20
−20
40
−20
a. [2 points] Find all values of θ between 0 and 2π for which r = 0.
Solution: First we set r = 20 + 40 sin θ = 0. Therefore rearranging sin θ = − 21 . The
solutions between 0 and 2π are θ = 7π/6, 11π/6.
b. [4 points] Write an integral or sum of integrals which give(s) the perimeter of the boundary
of the pond. Note this is the perimeter of the part of the cardioid drawn with a solid line.
Solution:
We will need to use the polar arc length formula so we need to calculate r′ = 40 cos θ.
R 7π/6 p
The arc length can be written as a single integral −π/6 (40 cos θ)2 + (20 + 40 sin θ)2 dθ.
R 7π/6 p
Writing the arc length as two integrals we get 0
(40 cos θ)2 + (20 + 40 sin θ)2 dθ +
R 2π p
(40 cos θ)2 + (20 + 40 sin θ)2 dθ
11π/6
c. [4 points] Write an integral or sum of integrals which give(s) the area of the top of the
rock. Note this is the area enclosed by the dashed part of the cardioid.
Solution:
R 11π/6
Now we need to use the polar area formula 7π/6 12 (20 + 40 sin θ)2 dθ.
University of Michigan Department of Mathematics
Winter, 2014 Math 116 Exam 2 Problem 9 (pond rock) Solution
Math 116 / Exam 2 (March 23, 2015)
page 4
2. [6 points] Your friend the goliath frog is going to decorate the boundary of his lily pad with
a string of tiny flowers. The boundary of the lily pad is given by a portion of the curve
r = 13 + 26 cos(θ) where r is measured in inches and θ is measured in radians. The part of the
curve that traces out the lily pad is shown below in the xy-plane.
y
20
10
-10
10
20
30
40
x
-10
-20
If the goliath frog is going to decorate only the part of the boundary of the lily pad for which
x ≤ 0, write an expression involving integrals for the length of the string of flowers required.
Do not evaluate your integral.
Solution: The length of the string of flowers is given by
Z
− π2
− 2π
3
q
2
2
(13 + 26 cos(θ)) + (26 sin(θ)) dθ +
Z
2π
3
π
2
q
(13 + 26 cos(θ))2 + (26 sin(θ))2 dθ.
3. [4 points] We can approximate the value of ln(1.5) by using the fact that y = ln(x) solves the
differential equation
1
dy
=
dx
x
Approximate ln(1.5) by using Euler’s method for the differential equation above with initial
condition y(1) = 0 and with ∆x = 0.25. Fill in the table with the y-values obtained at each
step.
Solution: We are given that y(1) = 0. Using Euler’s method with ∆x = 0.25 we compute
y(1.25) ≈ y(1) + y ′ (1)∆x = 0 + (1)(0.25) = 0.25,
y(1.50) ≈ y(1.25) + y ′ (1.25)∆x ≈ 0.25 + (0.8)(0.25) = 0.45.
x
1.00
1.25
1.50
Thus, ln(1.5) ≈
University of Michigan Department of Mathematics
y
0
0.25
0.45
0.45
Winter, 2015 Math 116 Exam 2 Problem 2 (lilypad) Solution
Math 116 / Exam 2 (November 13, 2013)
page 6
4. [12 points] A volcano has erupted, covering its surroundings with ash. The area covered with
2
ash is bounded by the polar curve r =
, where the volcano is located at the
2 + sin(θ + π4 )
origin and r is measured in km (see the shaded region in Figure 1).
Figure 1. Affected area. Figure 2. Area to be cleaned.
a. [3 points] Local authorities want to estimate the area of the region covered with ash.
Write an integral that computes the total area covered with ash. You do not need to
compute the value of the integral.
Solution:
A=
Z
2π
0
1
2
2
2 + sin(θ + π4 )
2
dθ.
b. [7 points] Days after the eruption, the government designs a plan to remove the ashes
outside a circle of radius one km around the volcano (see the shaded region in Figure 2).
Find an expression containing definite integrals that computes the cost of cleaning this
region, assuming it costs 3 million dollars to clean the ashes in one km square of land.
You do not need to compute the value of the integral.
Solution: The curves intersect when θ satisfies 1 =
2
. This equation leads
2 + sin(θ + π4 )
to sin(θ + π4 ) = 0. Then θ = 34 π and θ = 47 π.
Cost = $3 million ·
Z
7
π
4
3
π
4
or
Cost = $3 million ·
Z
1
2
7
π
4
3
π
4
2
2 + sin(θ + π4 )
1
2
2
dθ −
2
2 + sin(θ + π4 )
2
Z
7
π
4
3
π
4
dθ −
1
dθ .
2
π
.
2
c. [2 points] How far from the volcano is the farthest point covered with ash? Justify your
answer.
2
and −1 ≤ sin(θ + π4 ) ≤ 1, then
2 + sin(θ + π4 )
2
2
2
2
=
≤
= 2. Hence the the farthest point covered with ash
π ≤
3
2+1
2 + sin(θ + 4 )
2−1
is 2 km away from the volcano.
Solution: Since r =
University of Michigan Department of Mathematics
Fall, 2013 Math 116 Exam 2 Problem 4 (volcano) Solution
Math 116 / Final (April 28, 2014)
page 8
7. [8 points] Roy the alpaca is designing a pool and a deck for his family. The pool has the shape
of a cardioid whose equation is given by r = 4 − 4 sin(θ) where r is in meters and θ is a number
between 0 and 2π. The deck will be built in the region that lies inside the circle x2 + y 2 = 4
and outside the cardioid. The deck is depicted in the figure as the region enclosed by the solid
lines
2
−6
−4
2
−2
4
6
−2
−4
−6
−8
a. [1 point] Write the equation for the circle x2 + y 2 = 4 in polar coordinates.
Solution: r2 = 4 so r = 2
b. [2 points] Find the values of θ between 0 and 2π where the cardioid and the circle intersect.
Solution: Setting the two equations equal to each other we have 2 = 4 − 4 sin(θ) thus
sin(θ) = 12 . Therefore θ = π/6, 5π/6.
c. [5 points] Write an expression involving integrals that gives the area of the region where
the deck will be built. Do not evaluate your expression.
R 5π/6
R 5π/6
R 5π/6
Solution: π/6 2 dθ − π/6 12 (4 − 4 sin(θ)2 dθ = 4π/3 − π/6 12 (4 − 4 sin(θ)2 dθ.
University of Michigan Department of Mathematics
Winter, 2014 Math 116 Exam 3 Problem 7 (alpaca pool) Solution