Mid-Chapter Quiz: Lessons 9-1 through 9-3
Graph each point on a polar grid.
1. A(–2, 45°)
3. SOLUTION: SOLUTION: Because θ = 45°, locate the terminal side of a 45° angle with the polar axis as its initial side. Because r
= −2, plot a point 2 units from the pole in the opposite
direction of the terminal side of the angle.
Because θ =
, locate the terminal side of a
- angle with the polar axis as its initial side. Because r = −1.5, plot a point 1.5 units from the pole
in the opposite direction of the terminal side of the
angle.
2. D(1, 315°)
SOLUTION: Because θ = 315°, locate the terminal side of a 315° - angle with the polar axis as its initial side. Because
r = 1, plot a point 1 unit from the pole along the
terminal side of the angle.
4. SOLUTION: Because θ =
, locate the terminal side of a
- angle with the polar axis as its initial side. Because r = 3, plot a point 3 units from the pole
along the terminal side of the angle.
3. SOLUTION: Because θ =
, locate the terminal side of a
- angle with the polar axis as its initial side. Because
−1.5, plot
a point 1.5 units from the pole
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in the opposite direction of the terminal side of the
angle.
Page 1
Mid-Chapter Quiz: Lessons 9-1 through 9-3
4. 6. SOLUTION: SOLUTION: Because θ =
, locate the terminal side of a
- angle with the polar axis as its initial side. The solutions of θ =
are ordered pairs of the
form
, where r is any real number. The
graph consists of all points on the line that make an
angle of
with the positive polar axis.
Because r = 3, plot a point 3 units from the pole
along the terminal side of the angle.
7. θ= 60°
SOLUTION: Graph each polar equation.
5. r = 3
SOLUTION: The solutions of θ = 60° are ordered pairs of the form (r, 60°), where r is any real number. The graph
consists of all points on the line that make an angle
of 60° with the positive polar axis.
The solutions of r = 3 are ordered pairs of the form
(3, θ), where θ is any real number. The graph
consists of all points that are 3 units from the pole, so
the graph is a circle centered at the origin with radius
3.
8. r = −1.5
SOLUTION: The solutions of r = −1.5 are ordered pairs of the
form (−1.5, θ), where θ is any real number. The
graph consists of all points that are 1.5 units from the
pole, so the graph is a circle centered at the origin
with radius 1.5.
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SOLUTION: Page 2
Mid-Chapter Quiz: Lessons 9-1 through 9-3
8. r = −1.5
SOLUTION: The solutions of r = −1.5 are ordered pairs of the
form (−1.5, θ), where θ is any real number. The
graph consists of all points that are 1.5 units from the
pole, so the graph is a circle centered at the origin
with radius 1.5.
Since the blades of the rotor create 5 angles, the
angle between each pair of adjacent blades is 360° ÷
5 or 72°. The length of each blade is 11.5 feet. So, r
= 11.5 for each blade. The angle blade A makes with the polar axis is 3°, therefore the tip of blade A can be represented by
the polar coordinates (11.5, 3°). The angle that blade
B makes with the polar axis is 3° + 72° or 75°. Thus,
the tip of blade B can be represented by the polar
coordinates (11.5, 75°). By adding 72° to 75°, the polar coordinates for blade C can be found as (11.5, 147°). By adding 72° to 147°, the polar coordinates for blade D can be found
as (11.5, 219°). By adding 72° to 219°, the polar coordinates for blade E can be found as (11.5, 291°).
9. HELICOPTERS A helicopter rotor consists of five
equally spaced blades. Each blade is 11.5 feet long.
b. Blade B has the polar coordinates (11.5, 75°) and blade C has the polar coordinates (11.5, 147°). Use the Polar Distance Formula to find the distance
between them.
The distance between the tips of the two blades is
approximately 13.5 feet.
Graph each equation.
10. r =
a. If the angle blade A makes with the polar axis is
3°, write an ordered pair to represent the tip of each blade on a polar grid. Assume that the rotor is
centered at the pole.
b. What is the distance d between the tips of the
helicopter blades to the nearest tenth of a foot?
sec θ
SOLUTION: Make a table of values to find the r-values
corresponding to various values of θ on the interval
[0, 2π]. Round each r-value to the nearest tenth.
SOLUTION: a. Sample answer: Sketch a diagram of the situation.
r=
θ
0
sec
θ
0.3
0.3
0.5
−
−0.5
Since the blades of the rotor create 5 angles, the
angle between each pair of adjacent blades is 360° ÷
5 or 72°. The length of each blade is 11.5 feet. So, r
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= 11.5 for each blade. The angle blade A makes with the polar axis is 3°, π
−0.3
−0.3
−0.3
Page 3
−0.5
−
The distance between
the tips of the
blades is 9-3
Mid-Chapter
Quiz: Lessons
9-1twothrough
approximately 13.5 feet.
Graph each equation.
10. r =
11. r =
cos θ
sec θ
SOLUTION: SOLUTION: Make a table of values to find the r-values
corresponding to various values of θ on the interval
[0, 2π]. Round each r-value to the nearest tenth.
r=
θ
0
π
sec
Because the polar equation is a function of the
cosine function, it is symmetric with respect to the
polar axis. Therefore, make a table and calculate the
values of r on [0, π].
r = cos
θ
0
θ
0.3
θ
0.3
0.3
0.3
0.5
0.2
−
0.2
−0.5
0
−0.3
−0.3
−0.2
−0.2
−0.3
−0.5
−
0.5
2π
−0.3
−0.3
π
Use these points and polar axis symmetry to graph
the function.
0.3
0.3
Graph the ordered pairs (r, θ) and connect them with
a line.
12. r = 3 csc θ
SOLUTION: Make a table of values to find the r-values
corresponding to various values of θ on the interval
[0, 2π]. Round each r-value to the nearest tenth.
11. r =
cos θ
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SOLUTION: Because the polar equation is a function of the
cosine function, it is symmetric with respect to the
θ
0
r = 3 csc
θ
−
6
Page 4
Mid-Chapter Quiz: Lessons 9-1 through 9-3
13. r = 4 sin θ
12. r = 3 csc θ
SOLUTION: SOLUTION: Make a table of values to find the r-values
corresponding to various values of θ on the interval
[0, 2π]. Round each r-value to the nearest tenth.
Because the polar equation is a function of the sine
function, it is symmetric with respect to the line θ =
. Therefore, make a table and calculate the values
θ
0
r = 3 csc
θ
−
6
of r on
.
θ
r = 4 sin
θ
3.5
−4
3
−3.5
3.5
−2.8
6
−
−2
0
π
0
−6
2
−3.5
2.8
3.5
−3
−3.5
−6
−
4
Use these points and symmetry with respect to the
line θ = to graph the function.
Graph the ordered pairs (r, θ) and connect them with
a line.
14. STAINED GLASS A rose window is a circular
13. r = 4 sin θ
SOLUTION: Because the polar equation is a function of the sine
function, it is symmetric with respect to the line θ =
. Therefore, make a table and calculate the values
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of r on
.
window seen in gothic architecture. The pattern of
the window radiates from the center. The window
shown can be approximated by the equation r = 3 sin
6θ. Use symmetry, zeros, and maximum r-values of
the function to graph the function.
Refer to the image on Page 560.
SOLUTION: Because the polar equation is a function of the sine
Page 5
function, it is symmetric with respect to the line θ =
.
shown can be approximated by the equation r = 3 sin
6θ. Use symmetry, zeros, and maximum r-values of
the function to graph the function.
3
Refer to the image
onLessons
Page 560. 9-1 through 9-3
Mid-Chapter
Quiz:
−2.1
0
0
SOLUTION: 2.1
Because the polar equation is a function of the sine
function, it is symmetric with respect to the line θ =
.
Sketch the graph of the rectangular function y = 3
sin 6x on the interval
. From the graph, you
−3
2.1
Use these and a few additional points to sketch the
graph of the function.
= 3 when can see that
and y = 0 when
Identify and graph each classic curve.
15. r =
sin θ
SOLUTION: Interpreting these results in terms of the polar
equation r = 3 sin 6θ, we can say that has a maximum value of 3 when
and r = 0 when
The equation is of the form r = a sin θ, so its graph is
a circle. Because the polar equation is a function of
the sine function, it is symmetric with respect to the
line θ = . Therefore, make a table and calculate
the values of r on
.
r=
θ
Since the function is symmetric with respect to the
line θ = , make a table and calculate the values of
r on
sin
θ
−0.5
−0.4
.
−0.4
θ
r = 3 sin
6θ
0
0.3
−2.1
0.4
3
0
−2.1
0
2.1
−3
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2.1
−0.3
0
0.4
0.5
Use these points and symmetry with respect to the
line θ = to graph the function.
Page 6
Mid-Chapter Quiz: Lessons 9-1 through 9-3
Identify and graph each classic curve.
15. r =
16. r =
θ + 3, θ ≥ 0
sin θ
SOLUTION: SOLUTION: The equation is of the form r = a sin θ, so its graph is
a circle. Because the polar equation is a function of
the sine function, it is symmetric with respect to the
line θ = . Therefore, make a table and calculate
the values of r on
The equation is of the form r = aθ + b, so its graph is
a spiral of Archimedes.
Use points on the interval [0, 2π] to sketch the graph
of the function.
r= θ+
.
θ
0
3
3
3.3
r=
sin
θ
θ
π
3.5
4.0
2π
4.6
5.1
−0.5
−0.4
−0.4
0
−0.3
0
0.3
0.4
0.4
0.5
Use these points and symmetry with respect to the
line θ = to graph the function.
17. r = 1 + 2 cos θ
SOLUTION: 16. r =
θ + 3, θ ≥ 0
The equation is of the form r = a + b cos θ, so its
graph is a limacon. Since a < b, the graph with have
an inner loop. Because this polar equation is a
function of the cosine function, it is symmetric with
respect to the polar axis.
Therefore, make a table and calculate the values of r
on
.
r=1+2
cos θ
θ
0
3
SOLUTION: 2.7
The equation is of the form r = aθ + b, so its graph is
a spiral of Archimedes.
Use points on the interval [0, 2π] to sketch the graph
of the function.
2.4
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2
1
Page 7
Mid-Chapter Quiz: Lessons 9-1 through 9-3
17. r = 1 + 2 cos θ
18. r = 5 sin 3θ
SOLUTION: SOLUTION: The equation is of the form r = a + b cos θ, so its
graph is a limacon. Since a < b, the graph with have
an inner loop. Because this polar equation is a
function of the cosine function, it is symmetric with
respect to the polar axis.
Therefore, make a table and calculate the values of r
on
.
r=1+2
cos θ
θ
0
3
The equation is of the form r = a sin nθ, so its graph
is a rose. Because this polar equation is a function of
the sine function, it is symmetric with respect to the
line θ = . Therefore, make a table and calculate
the values of r on
.
r = 5 sin
3θ
θ
5
2.7
0
2.4
−3.5
2
−5
0
0
1
5
0
3.5
−0.4
0
−0.7
−1
π
Use these points and polar axis symmetry to graph
the function.
−5
Use these points and symmetry with respect to the
line θ = to graph the function.
18. r = 5 sin 3θ
19. MULTIPLE CHOICE Identify the polar graph of
SOLUTION: The equation is of the form r = a sin nθ, so its graph
is a rose. Because this polar equation is a function of
the sine function, it is symmetric with respect to the
line θ = . Therefore, make a table and calculate
the values of r on
.
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r = 5 sin
2
y =
x.
Page 8
corresponds to graph B. The The point
Mid-Chapter Quiz: Lessons 9-1 through 9-3
correct answer is B.
Find the rectangular coordinates for each point
with the given polar coordinates. 19. MULTIPLE CHOICE Identify the polar graph of
2
y =
x.
20. SOLUTION: , r = 4 and θ =
For
.
are The rectangular coordinates of
SOLUTION: 2
Write the rectangular equation y =
.
x in polar
form.
21. SOLUTION: , r = −2 and θ =
For
.
Graph r =
2
cos θ csc θ using a graphing
calculator. Let θ =
The rectangular coordinates of
and solve for r.
are .
22. (–1, 210°)
SOLUTION: For (–1, 210°), r = −1 and θ = 210°.
The point
corresponds to graph B. The correct answer is B.
Find the rectangular coordinates for each point
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The rectangular coordinates of (–1, 210°) are The rectangular coordinates of
are The rectangular coordinates of (3, 30°) are .
Mid-Chapter
. Quiz: Lessons 9-1 through 9-3
22. (–1, 210°)
SOLUTION: For (–1, 210°), r = −1 and θ = 210°.
Find two pairs of polar coordinates for each
point with the given rectangular coordinates if 0
≤ θ ≤ 2π. Round to the nearest hundredth.
24. (−3, 5)
SOLUTION: For (−3, 5), x = −3 and y = 5.
to find θ.
Since x < 0, use
The rectangular coordinates of (–1, 210°) are .
23. (3, 30°)
SOLUTION: For (3, 30°), r = 3 and θ = 30°.
One set of polar coordinates is (5.83, 2.11). Another
representation that uses a negative r-value is (−5.83,
2.11 + π) or (−5.83, 5.25).
25. (8, 1)
The rectangular coordinates of (3, 30°) are SOLUTION: For (8, 1), x = 8 and y = 1.
.
Since x > 0, use
Find two pairs of polar coordinates for each
point with the given rectangular coordinates if 0
≤ θ ≤ 2π. Round to the nearest hundredth.
24. (−3, 5)
to find θ.
SOLUTION: For (−3, 5), x = −3 and y = 5.
Since x < 0, use
to find θ.
One set of polar coordinates is (8.06, 0.12). Another
representation that uses a negative r-value is (−8.06,
0.12 + π) or (−8.06, 3.27).
26. (7, −6)
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SOLUTION: For (7, −6), x = 7 and y = −6.
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One set of polar coordinates is (10.77, 4.33).
Another representation that uses a negative r-value
is (−10.77, 4.33 − π) or (−10.77, 1.19).
One set of polar coordinates is (8.06, 0.12). Another
representationQuiz:
that uses
a negative9-1
r-value
is (−8.06,
Mid-Chapter
Lessons
through
9-3
0.12 + π) or (−8.06, 3.27).
Write a rectangular equation for each graph.
26. (7, −6)
SOLUTION: For (7, −6), x = 7 and y = −6.
to find θ.
Since x > 0, use
28. SOLUTION: One set of polar coordinates is (9.22, −0.71). Since
this set is not in the required domain, two more sets
have to be found. A representation that uses a
positive r-value is (9.22, −0.71 + 2π) or (9.22, 5.57).
A representation that uses a negative r-value is
(−9.22, −0.71 + π) or (−9.22, 2.43).
27. (−4, −10)
SOLUTION: 29. For (−4, −10), x = −4 and y = −10.
Since x < 0, use tan
−1
SOLUTION: + π to find θ.
One set of polar coordinates is (10.77, 4.33).
Another representation that uses a negative r-value
is (−10.77, 4.33 − π) or (−10.77, 1.19).
Write a rectangular equation for each graph.
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