Study Guide and Review - Chapter 9
Choose the correct term from the list to complete each sentence.
absolute value of a complex number
Argand plane argument
cardioid complex plane
imaginary axis lemniscate
limaçon modulus
polar axis polar coordinate system
polar coordinates polar equation
polar form polar graph
pole p th roots of unity
real axis rose
spiral of Archimedes trigonometric form
1. A(n) __________ is the set of all points with
coordinates (r,
equation.
) that satisfy a given polar
SOLUTION: The location of a point in the polar coordinate system
can be identified by polar coordinates of the form (r,
). A polar graph is the set of all points that satisfy
an equation expressed in terms of polar coordinates.
2. A plane that has an axis for the real component and
an axis for the imaginary component is a(n)
__________.
SOLUTION: The horizontal axis is called the real axis and the
vertical axis is called the imaginary axis on the
complex plane.
3. The location of a point in the __________ is
identified using the directed distance from a fixed
point and the angle from a fixed axis.
SOLUTION: A point in the polar coordinate system is identified by
polar coordinates of the form (r, ), where r is the
distance from the center, or the pole, to the given
point and is the measure of the angle formed by the polar axis and a line from the pole through the
point.
4. A special type of limaçon with equation of the form r
= a + b cos where a = b is called a(n)
__________.
SOLUTION: When a = b, the limaçon is a cardiod. When a< b,
the limaçon has an inner loop. When b < a < 2b, a
dimpled limaçon is formed. When a ≥ 2b, a convex
limaçon is formed.
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5. The __________ is the angle
of a complex number written in the form r(cos + i sin ).
polar coordinates of the form (r, ), where r is the
distance from the center, or the pole, to the given
point and is the measure of the angle formed by the polar axis and a line from the pole through the
point.
4. A special type of limaçon with equation of the form r
= a + b cos where a = b is called a(n)
__________.
SOLUTION: When a = b, the limaçon is a cardiod. When a< b,
the limaçon has an inner loop. When b < a < 2b, a
dimpled limaçon is formed. When a ≥ 2b, a convex
limaçon is formed.
5. The __________ is the angle
of a complex number written in the form r(cos + i sin ).
SOLUTION: The argument
−1
of the complex number a + b i is
−1
= tan
or = tan
+ π if a < 0. The
argument is the angle of a complex number written
in the form r(cos + i sin ).
6. The origin of a polar coordinate system is called the
__________.
SOLUTION: The pole is the origin of the polar coordinate system
and is a fixed point that is the initial point of the polar
axis.
7. The absolute value of a complex number is also
called the __________.
SOLUTION: For the complex number written in the form r(cos
+ i sin ), r represents the absolute value, or
modulus, of the complex number and is r =
.
8. The __________ is another name for the complex
plane.
SOLUTION: The complex plane may also be referred to as the
Argand plane and consists of both a real and an
imaginary axis.
9. The graph of a polar equation of the form r2 = a 2
2
2
cos 2 or r = a sin 2
__________.
is called a(n)
SOLUTION: 2
A lemniscate is a classic curve written in the form r
2
2
2
Page 1
= a cos 2 or r = a sin 2 .
10. The __________ is an initial ray from the pole,
SOLUTION: The complex plane may also be referred to as the
Argand
planeand
and consists
of both
a real and9an
Study
Guide
Review
- Chapter
imaginary axis.
9. The graph of a polar equation of the form r2 = a 2
2
2
cos 2 or r = a sin 2
__________.
12. is called a(n)
SOLUTION: SOLUTION: A lemniscate is a classic curve written in the form r
2
2
2
= a cos 2 or r = a sin 2 .
Because
2
= , 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 along the
terminal side of the angle.
10. The __________ is an initial ray from the pole,
usually horizontal and directed toward the right.
SOLUTION: The polar axis is a ray that has an initial point at the
pole. The polar axis can extend in any direction, but
usually extends to the right.
Graph each point on a polar grid.
11. W(–0.5, –210°)
SOLUTION: Because = –210°, locate the terminal side of a –
210° angle with the polar axis as its initial side. –210° + 360° = 150°. Because r = −0.5, plot a point 0.5 units from the pole
in the opposite direction of the terminal side of the
angle.
13. Y(4, –120°)
SOLUTION: Because = –210°, locate the terminal side of a –
210° angle with the polar axis as its initial side. –120° + 360° = 240°. Because r = 4, plot a point 4 units from the pole
along the terminal side of the angle.
12. SOLUTION: Because
= 14. , 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 along the
terminal side of the angle.
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SOLUTION: Because
= , locate the terminal side of a
-
angle with the polar axis as its initial side. Because r
Page 2
= −3, plot a point 3 units from the pole in the opposite
direction of the terminal side of the angle.
Study Guide and Review - Chapter 9
14. 16. r =
SOLUTION: SOLUTION: Because
= , 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 in the opposite
direction of the terminal side of the angle.
Graph each polar equation.
15. = –60°
SOLUTION: The solutions of = −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 = 4.5 are ordered pairs of the form
(4.5, ), where is any real number. The graph consists of all points that are 4.5 units from the pole,
so the graph is a circle centered at the origin with
radius 4.5.
17. r = 7
SOLUTION: The solutions of r = 7 are ordered pairs of the form
(7, ), where is any real number. The graph consists of all points that are 7 units from the pole, so
the graph is a circle centered at the origin with radius
7.
18. =
16. r =
SOLUTION: The solutions of r = 4.5 are ordered pairs of the form
(4.5, ), where is any real number. The graph eSolutions
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consists
of- all
pointsbythat
are 4.5 units from the pole,
so the graph is a circle centered at the origin with
radius 4.5.
SOLUTION: The solutions of
form = are ordered pairs of the , where r is any real number. The
Page 3
graph consists of all points on the line that make an
angle of
with the positive polar axis.
SOLUTION: Use the Polar Distance Formula.
Study Guide and Review - Chapter 9
18. =
22. SOLUTION: SOLUTION: The solutions of
form = are ordered pairs of the Use the Polar Distance Formula.
, 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.
Use symmetry, zeros, and maximum r-values to
graph each function.
23. r = sin 3
SOLUTION: Because the polar equation is a function of the sine
function, it is symmetric with respect to the line = .
Sketch the graph of the rectangular function y = sin
3x on the interval
see that
Find the distance between each pair of points.
. From the graph, you can
= 1 when and y =
0 when
19. SOLUTION: Use the Polar Distance Formula.
20. (–3, 60°), (4, 240°)
SOLUTION: Use the Polar Distance Formula.
21. (–1, –45°), (6, 270°)
Interpreting these results in terms of the polar
equation r = sin 3 , we can say that has a maximum value of 1 when
and r = 0 when
SOLUTION: Use the Polar Distance Formula.
Since the function is symmetric with respect to the
line = , make a table and calculate the values of
22. eSolutions Manual - Powered by Cognero
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r on
SOLUTION: .
and r = 0 when
Study Guide and Review - Chapter 9
Sketch the graph of the rectangular function y = 2
cos x on the interval [0, π]. From the graph, you can
see that
= 2 when x = 0 and π and y = 0 when
Since the function is symmetric with respect to the
line = , make a table and calculate the values of
r on
.
θ
r = sin
3θ
1
0
−0.7
0
−1
0
1
0.7
0
−1
Interpreting these results in terms of the polar
equation r = 2 cos , we can say that has a maximum value of 2 when
and r = 0
when
Since the function is symmetric with respect to the
polar axis, make a table and calculate the values of r
on [0, π].
r=2
θ
cos θ 0
2
Use these and a few additional points to sketch the
graph of the function.
1.7
1.4
1
0
−1
−1.4
π
24. r = 2 cos
−1.7
−2
Use these and a few additional points to sketch the
graph of the function.
SOLUTION: Because the polar equation is a function of the
cosine function, it is symmetric with respect to the
polar axis.
Sketch the graph of the rectangular function y = 2
cos x on the interval [0, π]. From the graph, you can
see that
= 2 when x = 0 and π and y = 0 when
25. r = 5 cos 2
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SOLUTION: Page 5
Because the polar equation is a function of the
cosine function, it is symmetric with respect to the
2.5
5
π
Study Guide and Review - Chapter 9
25. r = 5 cos 2
Use these and a few additional points to sketch the
graph of the function.
SOLUTION: Because the polar equation is a function of the
cosine function, it is symmetric with respect to the
polar axis.
Sketch the graph of the rectangular function y = 5
cos 2x on the interval [0, π]. From the graph, you
can see that
= 5 when x = 0, , and π and y = 0
when x =
and .
26. r = 4 sin 4
SOLUTION: Because the polar equation is a function of the sine
function, it is symmetric with respect to the line = .
Interpreting these results in terms of the polar
equation r = 5 cos 2 , we can say that has a Sketch the graph of the rectangular function y = 4
sin 4x on the interval
maximum value of 5 when
. From the graph, you
and r = 0
= 4 when can see that
when
and y = 0 when
Since the function is symmetric with respect to the
polar axis, make a table and calculate the values of r
on [0, π].
r = 5 cos
θ
2θ
0
5
2.5
0
−2.5
−5
−2.5
0
2.5
5
π
Use these and a few additional points to sketch the
graph of the function.
Interpreting these results in terms of the polar
equation r = 4 sin 4 , we can say that has a maximum value of 4 when
and r = 0 when
Since the function is symmetric with respect to the
line = , make a table and calculate the values of
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Page 6
r on
.
and r = 0 when
Study Guide and Review - Chapter 9
Since the function is symmetric with respect to the
line = , make a table and calculate the values of
r on
polar axis.
Sketch the graph of the rectangular function y = 2 +
2 cos x on the interval [0, π]. From the graph, you
can see that
= 4 when x = 0 and y = 0 when x =
π.
.
θ
r = 4 sin
4θ
0
3.5
0
0
−3.5
0
3.5
0
−3.5
0
Use these and a few additional points to sketch the
graph of the function.
Interpreting these results in terms of the polar
equation r = 2 + 2 cos , we can say that has a maximum value of 4 when
and r = 0 when
Since the function is symmetric with respect to the
polar axis, make a table and calculate the values of r
on [0, π].
r=2+2
θ
cos θ 0
4
3.7
3.4
3
2
1
0.6
0.3
0
π
27. r = 2 + 2 cos
Use these and a few additional points to sketch the
graph of the function.
SOLUTION: Because the polar equation is a function of the
cosine function, it is symmetric with respect to the
polar axis.
Sketch the graph of the rectangular function y = 2 +
2 cos x on the interval [0, π]. From the graph, you
can see that
= 4 when x = 0 and y = 0 when x =
π.
28. r = 1.5 ,
≥ 0
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The equation is of the form r = a + b, so its graph
is a spiral of Archimedes. Replacing (r, ) with (−r,
− ) yields r = 1.5 . Therefore, the function has
Study Guide and Review - Chapter 9
28. r = 1.5 ,
Use symmetry to graph each equation.
29. r = 2 – sin
≥ 0
SOLUTION: The equation is of the form r = a + b, so its graph
is a spiral of Archimedes. Replacing (r, ) with (−r,
− ) yields r = 1.5 . Therefore, the function has
symmetry with respect to the line = . However,
≥ 0, so the symmetry will not be seen in the graph.
Spirals are unbounded. Therefore, the function has
no maximum r-values and only one zero when = 0.
SOLUTION: Because the polar equation is a function of the sine
function, it is symmetric with respect to the line = . Therefore, make a table and calculate the values
of r on
.
θ
r = 2 – sin
3
2.9
Use points on the interval [0, 2π] to sketch the graph
of the function.
2.7
θ
0
r = 1.5θ
0
0
2.5
2
1.2
1.5
1.3
π
2.4
4.7
1
2π
7.1
9.4
1.1
Use these points and symmetry with respect to the
line = to graph the function.
Use symmetry to graph each equation.
29. r = 2 – sin
30. r = 1 + 5 cos
SOLUTION: SOLUTION: Because the polar equation is a function of the sine
function, it is symmetric with respect to the line = . Therefore, make a table and calculate the values
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=1+5
Page 8
θ
cos θ 0
6
of r on
.
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θ
r = 2 – sin
Study Guide and Review - Chapter 9
30. r = 1 + 5 cos
31. r = 3 – 2 cos
SOLUTION: SOLUTION: 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=1+5
θ
cos θ 0
6
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=3−2
θ
cos θ 0
1
5.3
1.3
4.5
1.6
3.5
2
1
3
−1.5
4
−2.5
4.4
−3.3
−4
4.7
5
π
Use these points and polar axis symmetry to graph
the function.
31. r = 3 – 2 cos
π
Use these points and polar axis symmetry to graph
the function.
32. r = 4 + 4 sin
SOLUTION: SOLUTION: 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=3−2
θ
cos θ 0
1
Because the polar equation is a function of the sine
function, it is symmetric with respect to the line = . Therefore, make a table and calculate the values
1.3
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of r on
.
θ
r=4+4
sin θ 1.6
0
2
0.5
Page 9
Study Guide and Review - Chapter 9
32. r = 4 + 4 sin
33. r = −3 sin
SOLUTION: SOLUTION: Because the polar equation is a function of the sine
function, it is symmetric with respect to the line = . Therefore, make a table and calculate the values
Because the polar equation is a function of the sine
function, it is symmetric with respect to the line = . Therefore, make a table and calculate the values
of r on
of r on
.
.
θ
r=4+4
sin θ 0
0
3
0.5
2.6
1.2
2.1
2
4
1.5
0
−1.5
6.8
−2.1
7.5
−2.6
33. r = −3 sin
−3
Use these points and symmetry with respect to the
line = to graph the function.
34. r = −5 + 3 cos
SOLUTION: Because the polar equation is a function of the sine
function, it is symmetric with respect to the line = . Therefore, make a table and calculate the values
.
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θ
0
6
8
Use these points and symmetry with respect to the
line = to graph the function.
of r on
θ
r = −3
sin θ r = −3
sin θ SOLUTION: 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 = −5 +
Page 10
θ
3 cos θ 0
−2
Study Guide and Review - Chapter 9
34. r = −5 + 3 cos
SOLUTION: 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 = −5 +
θ
3 cos θ 0
−2
Find two pairs of polar coordinates for each
point with the given rectangular coordinates if 0
≤ ≤ 2π. Round to the nearest hundredth.
35. (–1, 5)
SOLUTION: For (−1, 5), x = −1 and y = 5.
−1
Since x < 0, use tan
+ π to find
.
−2.4
−2.9
−3.5
−5
−6.5
−7.1
π
−7.6
−8
Use these points and polar axis symmetry to graph
the function.
One set of polar coordinates is (5.10, 1.77).
Another representation that uses a negative r-value
is (−5.10, 1.77 + π) or (−5.10, 4.91).
36. (3, 7)
SOLUTION: For (3, 7), x = 3 and y = 7.
−1
Since x > 0, use tan
to find .
Find two pairs of polar coordinates for each
point with the given rectangular coordinates if 0
≤ ≤ 2π. Round to the nearest hundredth.
35. (–1, 5)
SOLUTION: For (−1, 5), x = −1 and y = 5.
Since x < 0, use tan
−1
+ π to find
.
One set of polar coordinates is (7.62, 1.17).
Another representation that uses a negative r-value
is (−7.62, 1.17+ π) or (−7.62, 4.31).
37. (2a, 0),
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SOLUTION: For (2a, 0), x = 2a and y = 0.
One set of polar coordinates is (7.62, 1.17).
Another
representation
that uses
a negative 9
r-value
Study Guide
and Review
- Chapter
is (−7.62, 1.17+ π) or (−7.62, 4.31).
37. (2a, 0),
38. (4b, –6b),
SOLUTION: SOLUTION: For (2a, 0), x = 2a and y = 0.
Since x > 0, use tan
One set of polar coordinates is (2a, 0). Another
representation that uses a negative r-value is (−2a, 0
+ π) or (−2a, π).
−1
to find .
For (4b, –6b), x = 4b and y = –6b.
−1
Since x > 0, use tan
to find .
One set of polar coordinates is (2a, 0). Another
representation that uses a negative r-value is (−2a, 0
+ π) or (−2a, π).
38. (4b, –6b),
SOLUTION: For (4b, –6b), x = 4b and y = –6b.
Since x > 0, use tan
−1
One set of polar coordinates is (7.21b,
).
Since the domain for is 0 ≤ ≤ 2π, another
representation is (7.21b, −0.98 + 2π) or (7.21b,
5.30). A representation that uses a negative r-value
is (−7.21b,
+ π) or (−7.21b, 2.16).
to find .
Write each equation in rectangular form. Then
identify its graph. Support your answer by
graphing the polar form of the equation.
39. r = 5
SOLUTION: The graph of this equation is a circle with a center at
the origin and radius 5. Evaluate the function for
several -values in its domain and use these points
to graph the function.
One set of polar coordinates is (7.21b,
).
Since the domain for is 0 ≤ ≤ 2π, another
representation is (7.21b, −0.98 + 2π) or (7.21b,
5.30). A representation that uses a negative r-value
is (−7.21b,
+ π) or (−7.21b, 2.16).
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Write each equation in rectangular form. Then
identify its graph. Support your answer by
graphing the polar form of the equation.
Page 12
One set of polar coordinates is (7.21b,
).
Since the domain for is 0 ≤ ≤ 2π, another
representation is (7.21b, −0.98 + 2π) or (7.21b,
5.30).
A representation
that uses
a negative 9
r-value
Study
Guide
and Review
- Chapter
is (−7.21b,
+ π) or (−7.21b, 2.16).
Write each equation in rectangular form. Then
identify its graph. Support your answer by
graphing the polar form of the equation.
39. r = 5
40. r = –4 sin
SOLUTION: SOLUTION: The graph of this equation is a circle with a center at
the origin and radius 5. Evaluate the function for
several -values in its domain and use these points
to graph the function.
The graph of this equation is a circle centered at (0,
−2) with radius 2. Evaluate the function for several
-values in its domain and use these points to graph
the function.
40. r = –4 sin
SOLUTION: 41. r = 6 sec
SOLUTION: The graph of this equation is a circle centered at (0,
−2) with radius 2. Evaluate the function for several
-values in its domain and use these points to graph
the function.
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The graph of this equation is a vertical line through
the x-intercept 6 with an undefined slope. Evaluate
the function for several -values in its domain and
use these points to graph the function.
Page 13
Study Guide and Review - Chapter 9
41. r = 6 sec
42. r =
csc SOLUTION: SOLUTION: The graph of this equation is a vertical line through
the x-intercept 6 with an undefined slope. Evaluate
the function for several -values in its domain and
use these points to graph the function.
The graph of this equation is a horizontal line through
the y-intercept
with slope 0. Evaluate the function
for several -values in its domain and use these
points to graph the function.
42. r =
csc SOLUTION: Determine the eccentricity, type of conic, and
equation of the directrix for each polar
equation.
43. r =
SOLUTION: Write the equation in standard form, r =
The graph of this equation is a horizontal line through
.
the y-intercept
with slope 0. Evaluate the function
for several -values in its domain and use these
points to graph the function.
Since e = 1, the conic is a parabola. For a polar
equation of this form (where sinθ is included), the
equation of the directrix is y = d. From the
numerator, we know that ed = 3.5, so d = 3.5.
Therefore, the equation of the directrix is y = 3.5.
44. r =
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SOLUTION: Write the equation in standard form, r =
Page 14
.
Since e = 1, the conic is a parabola. For a polar
equation of this form (where sinθ is included), the
equation of the directrix is y = d. From the
numerator, we know that ed = 3.5, so d = 3.5.
Study
Guidetheand
Review
Chapter
Therefore,
equation
of the -directrix
is y =93.5.
SOLUTION: Because e = 0.5, the conic is an ellipse. The center
of the ellipse is at (0, 2). This point is above the pole.
Therefore, the directrix will be below the pole at y =
The polar equation of a conic with this directrix
44. r =
is
. Use the value of e and the polar
form of a point on the conic to find the value of d.
The vertex point (0, 6) has polar coordinates
SOLUTION: Write the equation in standard form, r =
.
.
Since e = 0.3, the conic is an ellipse. For a polar
equation of this form (where cosθ is included), the
equation of the directrix is x = d. From the
numerator, we know that ed = 1.2, so d = 4.
Therefore, the equation of the directrix is x = 4.
45. SOLUTION: Write the equation in standard form, r =
.
Therefore, the equation for the ellipse
is
or
. Because d = 6,
the equation of the directrix is y =
Evaluate the function for several -values in its
domain and use these points to graph the function
and its directrix. Since e = 2, the conic is a hyperbola. For a polar
equation of this form (where sinθ is included), the
equation of the directrix is y = d. From the
numerator, we know that ed =14, so d = 7.
Therefore, the equation of the directrix is y = 7.
Substitute and
for θ to determine the location
of the vertices of the ellipse.
46. r =
SOLUTION: Write the equation in standard form, r =
.
Since e = 1, the conic is a parabola. For a polar
equation of this form (where cosθ is included), the
equation of the directrix is x = d. From the
numerator, we know that ed = 6, so d = 6.
Therefore, the equation of the directrix is x = 6.
Write and graph a polar equation and directrix
for the conic with the given characteristics.
47. e = 0.5; vertices at (0, –2) and (0, 6)
SOLUTION: Because e = 0.5, the conic is an ellipse. The center
of the ellipse is at (0, 2). This point is above the pole.
Therefore, the directrix will be below the pole at y =
The polar equation of a conic with this directrix
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is
. Use the value of e and the polar
form of a point on the conic to find the value of d.
Page 15
and its directrix. The vertices are the endpoints of the major axis and
occur when
. Evaluating the function
at
, we find that the vertex lies at polar
coordinate
, which correspond to rectangular
Study Guide and Review - Chapter 9
coordinates
. Evaluating the function at
we find that the vertex lies at polar coordinate
, which correspond to rectangular
,
coordinates
. The graph is a hyperbola with
vertices at (3, 0) and (15, 0)
Substitute 0 and π for θ to determine the location of
the vertices of the hyperbola.
The graph is an ellipse with vertices at (0,
(0, 6).
) and
48. e = 1.5; directrix: x = 5
SOLUTION: Because e = 1.5, the conic is a hyperbola. The
directrix x = 5 is to the right of the pole, so the
equation is of the form r =
and d = 5.
Use the values for e and d to write the equation.
Write each polar equation in rectangular form.
49. r =
Evaluate the function for several -values in its
domain and use these points to graph the function
and its directrix. The vertices are the endpoints of the major axis and
occur when
. Evaluating the function
at
, we find that the vertex lies at polar
coordinate
, which correspond to rectangular
coordinates
. Evaluating the function at
we find that the vertex lies at polar coordinate
, which correspond to rectangular
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,
SOLUTION: Write the equation in standard form.
For this equation, e = 0.2 and d = 1.6 ÷ 0.2 or 8. The
eccentricity and form of the equation determine that
this is an ellipse with directrix y = –8. The general equation of such an ellipse in
Page 16
rectangular form is
.
Write the equation in standard form.
or b=
For Guide
this equation,
= 0.2 and d = 1.6 ÷ 0.2 or 8. The
Study
ande Review
- Chapter 9
eccentricity and form of the equation determine that
this is an ellipse with directrix y = –8. The general equation of such an ellipse in
rectangular form is
.
Substitute the values for h, k, a, and b into the
standard form of an equation for an ellipse.
.
The vertices are the endpoints of the major axis and
. occur when
50. r =
SOLUTION: The equation is in standard form.
The vertices have polar coordinates
and
, which correspond to rectangular
coordinates (0, 2) and (0, –
). The ellipse’s center
For this equation, e = 1 and d = 5. The eccentricity
and form of the equation determine that this is a
parabola that opens horizontally with focus at the
pole and a directrix x = 5. The general equation of such a parabola in
rectangular form is
The vertex lies between the focus F and the directrix
of the parabola, occurring when
. is the midpoint of the segment between the vertices,
so (h, k) = (0,
). The distance a between the
center and each vertex is
. The distance c from
the center to the focus at (0, 0) is . b=
or .
Substitute the values for h, k, a, and b into the
standard form of an equation for an ellipse.
The vertex lies at polar coordinate
correspond to rectangular coordinates
=
, which
. So
. The distance p from the vertex at
to the focus at
is 2.5.
Substitute the values for h, k, and p into the general
equation for rectangular form.
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Page 17
Study Guide and Review - Chapter 9
Graph each number in the complex plane, and
find its absolute value.
51. z = 3 − i
50. r =
SOLUTION: SOLUTION: The equation is in standard form.
For z = 3 − i, (a, b) = (3, –1). Graph the point (3, –1)
in the complex plane.
For this equation, e = 1 and d = 5. The eccentricity
and form of the equation determine that this is a
parabola that opens horizontally with focus at the
pole and a directrix x = 5. The general equation of such a parabola in
rectangular form is
The vertex lies between the focus F and the directrix
of the parabola, occurring when
. Use the absolute value of a complex number
formula.
52. z = 4i
The vertex lies at polar coordinate
correspond to rectangular coordinates
=
SOLUTION: , which
. So
For z = 4i, (a, b) = (0, 4). Graph the point (0, 4) in
the complex plane.
. The distance p from the vertex at
to the focus at
is 2.5.
Substitute the values for h, k, and p into the general
equation for rectangular form.
Use the absolute value of a complex number
formula.
Graph each number in the complex plane, and
find its absolute value.
51. z = 3 − i
SOLUTION: For z = 3 − i, (a, b) = (3, –1). Graph the point (3, –1)
in the complex plane.
53. z = −4 + 2i
SOLUTION: For z = –4 + 2i, (a, b) = (–4, 2). Graph the point (–4,
2) in the complex plane.
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Page 18
Study Guide and Review - Chapter 9
Express each complex number in polar form.
55. 3 +
i
53. z = −4 + 2i
SOLUTION: SOLUTION: For z = –4 + 2i, (a, b) = (–4, 2). Graph the point (–4,
2) in the complex plane.
i
3+
Find the modulus r and argument
.
i is
The polar form of 3 +
3.317(cos 0.441 + i sin 0.441).
Use the absolute value of a complex number
formula.
56. −5 + 8i
SOLUTION: –5 + 8i
Find the modulus r and argument
.
54. z = 6 − 3i
SOLUTION: For z = 6 – 3i, (a, b) = (6, –3). Graph the point (6, –
3) in the complex plane.
The polar form of −5 + 8i is 9.434(cos 2.129 + i sin
2.129).
57. −4 −
i
SOLUTION: i
–4 –
Find the modulus r and argument
.
Use the absolute value of a complex number
formula.
The polar form of −4 –
sin 3.550).
Express each complex number in polar form.
55. 3 +
i
SOLUTION: i
Find the modulus r and argument
58. +
i is 4.359(cos 3.550 + i
i
SOLUTION: i
+
Find the modulus r and argument
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eSolutions
.
.
Page 19
The rectangular form of
i is- 4.359(cos
TheGuide
polar form
of −4
–
Study
and
Review
Chapter3.550
9 +i
is
.
sin 3.550).
58. +
i
60. SOLUTION: i
+
Find the modulus r and argument
SOLUTION: .
The value of r is 5, and the value of
the polar coordinates
is . Plot
.
The polar form of
+ i is
.
Graph each complex number on a polar grid.
Then express it in rectangular form.
59. SOLUTION: The value of r is 3, and the value of
the polar coordinates
is To express the number in rectangular form, evaluate
the trigonometric values and simplify.
. Plot
.
The rectangular form of
is
.
61. SOLUTION: The value of r is
To express the number in rectangular form, evaluate
the trigonometric values and simplify.
The rectangular form of
is
, and the value of
the polar coordinates
is . Plot
.
.
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SOLUTION: The value of r is 5, and the value of
Page 20
is . Plot
To express the number in rectangular form, evaluate
the trigonometric values and simplify.
The rectangular form of
The rectangular form of
Study Guide andisReview - Chapter
9
.
is
61. .
62. SOLUTION: The value of r is
SOLUTION: , and the value of
the polar coordinates
is . Plot
The value of r is 4, and the value of
.
the polar coordinates
is . Plot
.
To express the number in rectangular form, evaluate
the trigonometric values and simplify.
To express the number in rectangular form, evaluate
the trigonometric values and simplify.
The rectangular form of
The rectangular form of
is
.
is
.
Find each product or quotient, and express it in
rectangular form.
62. 63. SOLUTION: The value of r is 4, and the value of
the polar coordinates
is . Plot
.
SOLUTION: Use the Product Formula to find the product in polar
form.
Now find the rectangular form of the product.
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To express the number in rectangular form, evaluate
Page 21
The polar form is
. The
The rectangular form of
The polar form is
Study Guide and Review
- Chapter
9
is
.
Find each product or quotient, and express it in
rectangular form.
63. . The
rectangular form is
.
65. SOLUTION: Use the Quotient Formula to find the quotient in
polar form.
SOLUTION: Use the Product Formula to find the product in polar
form.
Now find the rectangular form of the product.
The polar form is
Now find the rectangular form.
. The
The polar form of the quotient is
rectangular form is
64. 8(cos 225° + i sin 225°) ⋅
.
. The rectangular form of
(cos 120° + i sin 120°)
SOLUTION: the quotient is
.
66. 6(cos 210° + i sin 210°) ÷ 3(cos 150° + i sin 150°)
Use the Product Formula to find the product in polar
form.
SOLUTION: Use the Quotient Formula to find the quotient in
polar form.
Now find the rectangular form of the product.
Now find the rectangular form.
The polar form is
rectangular form is
. The
.
The polar form of the quotient is
. The rectangular form of the
65. SOLUTION: Use the Quotient Formula to find the quotient in
polar form.
quotient is
.
Find each power, and express it in rectangular
form.
67. (4 − i)5
SOLUTION: eSolutions Manual - Powered by Cognero
First, write 4 – i in polar form.
Page 22
The polar form of the quotient is
. The rectangular form of the
Study Guide and Review - Chapter 9
quotient is
.
Find each power, and express it in rectangular
form.
67. (4 − i)5
Therefore,
.
Find all of the distinct p th roots of the complex
number.
69. cube roots of 6 – 4i
SOLUTION: SOLUTION: First, write 4 – i in polar form.
First, write 6 – 4i in polar form.
The polar form of 4 – i is . Now use De
Moivre’s Theorem to find the fifth power.
The polar form of 6 – 4i
is
. Now write an expression for the cube roots.
Therefore,
68. (
+ 3i)
Let n = 0, 1 and 2 successively to find the cube
roots.
Let n = 0.
.
4
SOLUTION: First, write
+ 3i in polar form.
Let n = 1.
+ 3i is
. Now use De Moivre’s
Theorem to find the fourth power.
The polar form of
Let n = 2.
The cube roots of 6 – 4i are approximately 1.895 −
0.376i, −0.622 + 1.829i, −1.273 − 1.453i.
70. fourth roots of 1 + i
Therefore,
.
Find all of the distinct p th roots of the complex
number.
69. cube roots of 6 – 4i
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SOLUTION: First, write 6 – 4i in polar form.
SOLUTION: First, write 1 + i in polar form.
Page 23
0.376 , −0.622 + 1.829 , −1.273 − 1.453 .
70. fourth roots of 1 + i
The fourth roots of 1 + i are approximately 1.07 +
0.21i, –0.21 + 1.07i, –1.07 – 0.21i, 0.21 – 1.07i.
Study
Guide and Review - Chapter 9
SOLUTION: First, write 1 + i in polar form.
71. GAMES An arcade game consists of rolling a ball
up an incline at a target. The region in which the ball
lands determines the number of points earned. The
model shows the point value for each region.
The polar form of 1 + i is
.
Now write an expression for the fourth roots.
a. If, on a turn, a player rolls the ball to the point (3.5,
165°), how many points does he get?
b. Give two possible locations that a player will
receive 50 points.
SOLUTION: Let n = 0, 1, 2 and 3 successively to find the fourth
roots.
Let n = 0.
a. Because = 165°, locate the terminal side of a 165°- angle with the polar axis as its initial side.
Because r = 3.5, plot a point 3.5 units from the pole
along the terminal side of the angle.
Let n = 1.
Let n = 2.
The ball lands in the 20-point region.
b. The 50-point region is the region on the target that
is shaded green. Two locations that a player will
receiver 50 points are (2, 0°) or (2, 180°) as shown.
Let n = 3.
72. LANDSCAPING A landscaping company uses an
The fourth roots of 1 + i are approximately 1.07 +
0.21i, –0.21 + 1.07i, –1.07 – 0.21i, 0.21 – 1.07i.
71. GAMES An arcade game consists of rolling a ball
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up an incline at a target. The region in which the ball
lands determines the number of points earned. The
model shows the point value for each region.
adjustable lawn sprinkler that can rotate 360º and
can cover a circular region with radius of 20 feet.
(Lesson 9-1)
a. Graph the dimensions of the region that the
sprinkler can cover on a polar grid if it is set to rotate
360º.
b. Find the area of the region that the sprinkler Page 24
covers if the rotation is adjusted to
–30º ≤ ≤ 210º.
For the new region, if −30° ≤ ≤ 210° a central angle of 240° is formed. Thus, the new region will Study Guide and Review - Chapter 9
72. LANDSCAPING A landscaping company uses an
adjustable lawn sprinkler that can rotate 360º and
can cover a circular region with radius of 20 feet.
(Lesson 9-1)
a. Graph the dimensions of the region that the
sprinkler can cover on a polar grid if it is set to rotate
360º.
b. Find the area of the region that the sprinkler
covers if the rotation is adjusted to
–30º ≤ ≤ 210º.
SOLUTION: a. Since the radius of the circular region is 20 feet, r
= 20. The graph of the region that the sprinkler can
cover can be represented by the polar equation r =
20. The solutions of r = 20 are ordered pairs of the
form (20, ), where is any real number. The graph consists of all points that are 20 units from the
pole, so the graph is a circle centered at the origin
with radius 20. Since the sprinkler covers the lawn
up to a 20 foot radius, the region from the origin to
the circle represented by r = 20 is shaded.
or about 837.8 square have an area of 400π · feet.
73. BIOLOGY The pattern on the shell of a snail can
be modeled using r =
+ ,
≥ 0. Identify and
graph the classic curve that models this pattern.
SOLUTION: The equation is of the form r = a + b, so its graph
is a spiral of Archimedes.
Make a table of values to find the r-values
corresponding to various values of on the interval [0, 2π]. Round each r-value to the nearest tenth.
r = θ + θ
0
0.5
0.7
0.8
1.0
1.2
π
1.4
1.5
1.7
1.9
2.1
2.2
b. The area of the region found in part a can be
2
found using A = πr and r = 20.
2π
2.4
2.6
Graph the ordered pairs (r,
with a smooth curve.
) and connect them
The area of the entire region is 400π square feet.
For the new region, if −30° ≤ ≤ 210° a central angle of 240° is formed. Thus, the new region will or about 837.8 square have an area of 400π · feet.
73. BIOLOGY The pattern on the shell of a snail can
be modeled using r =
+ ,
≥ 0. Identify and
graph the classic curve that models this pattern.
Page 25
74. RIDES The path of a Ferris wheel can be modeled
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SOLUTION: The equation is of the form r = a
is a spiral of Archimedes.
+ b, so its graph
by r = 50 sin
, where r is given in feet.
axis will also represent the ground. The rider is
located at the rectangular coordinates (12.5, 3.3).
The y-coordinate represents the distance that the
rider is above the ground. Thus, the rider is about 3.3
feet above the ground.
Study Guide and Review - Chapter 9
74. RIDES The path of a Ferris wheel can be modeled
by r = 50 sin
75. ORIENTEERING Orienteering requires
participants to make their way through an area using
a topographic map. One orienteer starts at
Checkpoint A and walks 5000 feet at an angle of 35°
measured clockwise from due east. A second
orienteer starts at Checkpoint A and walks 3000 feet
due west and then 2000 feet due north. How far, to
the nearest foot, are the two orienteers from each
other?
, where r is given in feet.
SOLUTION: a. What are the polar coordinates of a rider located
at = ? Round to the nearest tenth, if
necessary.
b. What are the rectangular coordinates of the
rider’s location? Round to the nearest tenth, if
necessary.
c. What is the rider’s distance above the ground if
the polar axis represents the ground?
Let Checkpoint A represent the origin. The first
orienteer is located at the polar coordinates (5000,
−35°). The second orienteer is located at the rectangular coordinates (−3000, 2000). Convert the
rectangular coordinates (−3000, 2000) to polar
coordinates.
−1
Since x < 0, use tan
SOLUTION: a. Substitute
= into r = 50 sin
and solve for
r.
The polar coordinates of the rider are
.
+ 180° to find .
Polar coordinates for (−3000, 2000) are (3605.55,
146.3°). Use the Polar Distance Formula to find the distance between the first orienteer located at (5000,
−35°) and the second orienteer located at (3605.55, 146.3°).
b. For
, r = 12.9 and
= So, the two orienteers are about 8605 feet from each
other.
.
76. SATELLITE The orbit of a satellite around Earth
The rectangular coordinates of
are (12.5,
3.3).
c. If the polar axis represents the ground, then the xaxis will also represent the ground. The rider is
located at the rectangular coordinates (12.5, 3.3).
The y-coordinate represents the distance that the
rider is above the ground. Thus, the rider is about 3.3
feet above the ground.
75. ORIENTEERING
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Cognero
requires
participants to make their way through an area using
a topographic map. One orienteer starts at
has eccentricity of 0.05, and the distance from a
vertex of the path to the center of Earth is 32,082
miles. Write a polar equation that can be used to
model the path of the satellite if Earth is located at
the focus closest to the given vertex.
SOLUTION: Let the sun be located at the origin. Since e = 0.05,
Page 26
the path of the satellite is the shape of an ellipse. If
the sun is at the focus located closest to the vertex
as shown in the diagram, then the second focus is
77. ELECTRICITY Most circuits in Europe are
So, Guide
the two orienteers
are about
8605 feet from
Study
and Review
- Chapter
9 each
other.
76. SATELLITE The orbit of a satellite around Earth
has eccentricity of 0.05, and the distance from a
vertex of the path to the center of Earth is 32,082
miles. Write a polar equation that can be used to
model the path of the satellite if Earth is located at
the focus closest to the given vertex.
designed to accommodate 220 volts. For parts a and
b, use E = I · Z, where voltage E is measured in
volts, impedance Z is measured in ohms, and current
I is measured in amps. Round to the nearest tenth.
a. If the circuit has a current of 2 + 5j amps, what is
the impedance?
b. If a circuit has an impedance of 1 – 3j ohms, what
is the current?
SOLUTION: a. Express each number is polar form.
For 220, find the modulus r and argument
.
SOLUTION: Let the sun be located at the origin. Since e = 0.05,
the path of the satellite is the shape of an ellipse. If
the sun is at the focus located closest to the vertex
as shown in the diagram, then the second focus is
located to the left of the origin. Thus, the equation
will be of the form r =
. The vertex
occurs at the point (32,082, 0), which has polar
coordinates of (32,082, 0). Substitute this point and
the value of e into the equation to solve for d.
The polar form of 220 is 220(cos 0 + j sin 0).
For 2 + 5j , find the modulus r and argument .
The polar form of 2 + 5j is
(cos 1.19 + j sin
1.19).
Solve for the impedance Z in E = I · Z.
Substitute the values for d and e into the polar
equation.
77. ELECTRICITY Most circuits in Europe are
designed to accommodate 220 volts. For parts a and
b, use E = I · Z, where voltage E is measured in
volts, impedance Z is measured in ohms, and current
I is measured in amps. Round to the nearest tenth.
a. IfManual
the circuit
has by
a current
eSolutions
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Cognero of 2 + 5j amps, what is
the impedance?
b. If a circuit has an impedance of 1 – 3j ohms, what
Now convert the impedance to rectangular form.
The impedance of the circuit is about 15.2 − 37.9j
ohms.
b. Express each number is polar form.
For 220, find the modulus r and argument
.
Page 27
The impedance of the circuit is about 15.2 − 37.9j
ohms.
Study
Guideeach
and
Review
- Chapter
9
b. Express
number
is polar
form.
For 220, find the modulus r and argument
.
square ABCD, as shown below, each of the vertices
can be represented by a complex number in polar
form. Multiplication can then be used to rotate and
dilate the square, producing the square A′B′C′D′. By
what complex number should the programmer
multiply each number to produce this transformation?
The polar form of 220 is 220(cos 0 + j sin 0).
For 1 − 3j , find the modulus r and argument .
SOLUTION: When the complex numbers representing each
vertex of ABCD is multiplied by the complex number
z, the product will result in A′B′C′D′.
Write A(−4, 4) in polar form.
The polar form of 1 − 3j is
[cos (−1.25) + j sin
(−1.25)].
Solve for the current I in E = I · Z.
The polar form of A is
.
Write A′(2, 2) in polar form.
Now convert the current to rectangular form.
The polar form of A′ is
. Use
substitution to solve for z.
The impedance of the circuit is about 21.9 + 66.0j
amps.
78. COMPUTER GRAPHICS Geometric
transformations of figures can be performed using
complex numbers. If a programmer starts with
square ABCD, as shown below, each of the vertices
can be represented by a complex number in polar
form. Multiplication can then be used to rotate and
dilate the square, producing the square A′B′C′D′. By
what complex number should the programmer
multiply each number to produce this transformation?
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To express z in rectangular form, evaluate the
trigonometric values and simplify.
Page 28
The programmer should multiply by
.
Guide and Review - Chapter 9
Study
To express z in rectangular form, evaluate the
trigonometric values and simplify.
The programmer should multiply by
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.
Page 29