Mathematical Functions - Fall 2012
Quiz 9 November 25, 2012
This part consists of 8 multiple choice problems. Nothing more than the answer is required; consequently no partial credit will be awarded.
1. Express 36◦ in radians.
A
π
4
B
π
5
C
D
E
π
6
π
7
π
8
Solution: We have
π 18 · 2π
2π
π
36π
rad =
rad =
rad = rad
rad =
36 = 36
180
180
18 · 10
10
5
◦
π
rad in degrees.
540
2. Express
◦
A
0.1
B
0.2
C
0.3
D
0.4
E
0.5
◦
◦
◦
◦
Solution: We have
π π
rad =
540
540
180
π
=
1
18
1
π · 180
◦
=
= = 0.3
540 · π
54
3
3. Find an angle that is coterminal with the angle θ = −719◦ in standard position.
A
−3◦
B
−2◦
C
−1◦
D
1◦
E
2◦
Solution: Note that
−719◦ = −720◦ + 1◦ = −2 · 360◦ + 1◦
therefore 1◦ is coterminal with θ = −719◦ .
4. Find an angle that is coterminal with the angle θ = −
A
π
5
B
−π
5
C
2π
5
D
−2π
5
E
3π
5
F
−3π
5
19π
in standard position.
5
Solution: Note that
20π − π
19π
=−
=−
−
5
5
therefore
20π π
−
5
5
π
19π
is coterminal with θ = −
.
5
5
2
=−
π
20π π
+ = −4π +
5
5
5
5. Find an angle with measure between 0◦ and 360◦ that is coterminal with the angle of measure
−997◦ in standard position.
A
53◦
B
63◦
C
73◦
D
E
83◦
93◦
Solution: Note that
−997◦ + 3 · 360◦ = 83◦
therefore 83◦ is coterminal with θ = −997◦ .
6. Find the length of an arc of a circle with radius 10 m that subtends a central angle of 135◦ .
A
15π
2
B
15π
C
7π
2
D
7π
E
5π
2
Solution: We first note that 135◦ =
3π
rad. So the length of the arc is
4
s = rθ = (10)
3π
15π
=
m
4
2
3
7. A central angle θ in a circle of radius
the measure of θ in radians.
A
1
2
B
2
C
1
4
D
4
E
None of the above
√
5 m is subtended by an arc of length
√
20 m. Find
Solution: By the formula θ = s/r, we have
r
√
s
20
20 √
θ= = √ =
= 4 = 2 rad
r
5
5
8. Find the area of a sector of a circle with central angle 210◦ if the radius of the circle is
m.
A
π
6
B
5π
6
C
7π
6
D
E
√
2
11π
6
π
3
Solution: To use the formula for the area of a circular sector, we must find the central angle of
the sector in radians:
π 21π
7π
210π
210◦ = 210
rad =
rad =
rad
rad =
180
180
18
6
Thus, the area of the sector is
1
1 √
A = r2 θ = ( 2)2
2
2
7π
6
4
1
= (2)
2
7π
6
=
7π 2
m
6
In the following problems you are required to show all your work and provide the necessary explanations everywhere to get full credit.
1. A DVD is approximately 12 centimeters in diameter. The drive motor of the DVD player is
controlled to rotate precisely between 200 and 500 revolutions per minute, depending on what
track is being read.
(a) Find intervals for the angular and linear speed of a disc as it rotates.
(b) Find the linear speed of a point on the outermost track as the disc rotates.
Solution:
(a) We have
200 ≤
revolutions
≤ 500
minute
2π(200) ≤angular speed ≤ 2π(500)
400π rad/minute ≤angular speed ≤ 1000π rad/minute
Since v = rw, it follows that
400π · 6 cm/minute ≤linear speed ≤ 1000π · 6 cm/minute
2400π cm/minute ≤linear speed ≤ 6000π cm/minute
(b) It immediately follows from (a) that the linear speed of a point on the outermost track is
6000π cm/minute.
5
2. In traveling across flat land you notice a mountain directly in front of you. Its angle of
elevation (to the peak) is 3.5◦ . After you drive 13 miles closer to the mountain, the angle of
elevation is 9◦ (see the figure below). Approximate the height of the mountain.
h
3.5°
13
Solution: We have
c
13 + c
and cot 3.5◦ =
h
h
(see the Figure above on the right). Subtracting, we get
cot 9◦ =
13 + c c
−
h
h
c
13 c
+ −
cot 3.5◦ − cot 9◦ =
h
h h
13
cot 3.5◦ − cot 9◦ =
h
−1
13
(cot 3.5◦ − cot 9◦ )−1 =
h
cot 3.5◦ − cot 9◦ =
cot 3.5◦
1
h
=
◦
− cot 9
13
13
=h
cot 3.5◦ − cot 9◦
which is ≈ 1.3 miles.
6
9°
c