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C H A P T E R
6
6.1
Vectors in the Plane
6.2
Dot Product of
Vectors
6.3
Parametric
Equations and
Motion
6.4
Polar Coordinates
6.5
Graphs of Polar
Equations
6.6
De Moivre’s
Theorem and
n
th
Roots
501
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502
CHAPTER 6
Applications of Trigonometry
JAMES BERNOULLI (1654–1705)
The first member of the Bernoulli family
(driven out of Holland by the Spanish persecutions and settled in Switzerland) to achieve mathematical fame, James defined the numbers now known as
Bernoulli numbers. He determined the form (the elastica) taken by an elastic rod acted on at one end by a given force and fixed at the other end.
We introduce vectors in the plane, perform vector operations, and use vectors to represent quantities such as force and velocity. Vector methods are used extensively in physics, engineering, and applied mathematics. Vectors are used to plan airplane flight paths. The trigonometric form of a complex number is used to obtain De Moivre’s theorem and find the
n
th roots of a complex number.
Parametric equations are studied and used to simulate motion. One of the principal applications of parametric equations is the analysis of motion in space. Polar coordinates another of Newton’s inventions, although James Bernoulli usually gets the credit because he published first are used to represent points in the coordinate plane. Planetary motion is best described with polar coordinates. We convert rectangular coordinates to polar coordinates, polar coordinates to rectangular coordinates, and study graphs of polar equations.
6.1
What you’ll learn about
■ TwoDimensional Vectors
■ Vector Operations
■ Unit Vectors
■ Direction Angles
■ Applications of Vectors
. . . and why
These topics are important in many realworld applications, such as calculating the effect of the wind on an airplane’s path.
OBJECTIVE
Students will be able to apply the arithmetic of vectors and use vectors to solve realworld problems.
MOTIVATE
Discuss the difference between the statements: “Jose lives 3 miles away from
Mary” and “Jose lives 3 miles west of
Mary.”
LESSON GUIDE
Day 1: TwoDimensional Vectors; Vector
Operations
Day 2: Unit Vectors, Direction Angles;
Applications of Vectors
Some quantities, like temperature, distance, height, area, and volume, can be represented by a single real number that indicates
magnitude
or
size
. Other quantities, such as force, velocity, and acceleration, have
magnitude
and
direction
. Since the number of possible directions for an object moving in a plane is infinite, you might be surprised to learn that two numbers are all that we need to represent both the magnitude of an object’s velocity and its direction of motion. We simply look at ordered pairs of real numbers in a new way. While the pair (
a
,
b
) determines a point in the plane, it also determines a
directed line segment
(or
arrow
) with its tail at the origin and its head at (
a
,
b
) (Figure 6.1). The length of this arrow represents magnitude, while the direction in which it points represents direction. Because in this context the ordered pair (
a
,
b
) represents a mathematical object with both magnitude and direction, we call it the
position vector of (
a
,
b
)
, and denote it as
a
,
b
to distinguish it from the point (
a
,
b
).
y
(
a
,
b
)
y a, b
(
a
,
b
)
x x
O O
(a) (b)
FIGURE 6.1
The point represents the ordered pair (
a
,
b
). The arrow (directed line segment) represents the vector
a
,
b
.
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SECTION 6.1
Vectors in the Plane
503
IS AN ARROW A VECTOR?
While an arrow
represents
a vector, it is not a vector itself, since each vector can be represented by an infinite number of equivalent arrows. Still, it is hard to avoid referring to “the vector
PQ
” in practice, and we will often do that ourselves.
When we say “the vector
u
PQ
,” we really mean “the vector
u
represented by
PQ
.”
S
(–1, 6)
y
P
(3, 4)
R
(–4, 2)
x
O
(0, 0)
FIGURE 6.2
The arrows
RS
and
OP
both represent the vector
3, 4
, as would any arrow with the same length pointing in the same direction. Such arrows are called
equivalent
.
DEFINITION
TwoDimensional Vector
A
twodimensional vector v component form
as
a
,
b
is an ordered pair of real numbers, denoted in
. The numbers
a
and
b
are the
components
of the
standard representation
the origin to the point (
a
,
b
). The of the vector
magnitude
of
v
a
,
b
is the arrow from is the length of the arrow, and the tor
0 direction
of
v
is the direction in which the arrow is pointing. The vec
0, 0 , called the
zero vector ,
has zero length and no direction.
It is often convenient in applications to represent vectors with arrows that begin at points other than the origin. The important thing to remember is that
any two arrows with the same length and pointing in the same direction represent the same vector
.
In Figure 6.2, for example, the vector 3, 4 is shown represented by
RS
, an arrow with
initial point
R
and
terminal point
S
, as well as by its standard representation
OP
.
Two arrows that represent the same vector are called
equivalent
.
The quick way to associate arrows with the vectors they represent is to use the following rule.
Head Minus Tail (HMT) Rule
If an arrow has initial point vector
x
2
x
1
,
y
2
y
1
.
x
1
,
y
1 and terminal point
x
2
,
y
2
, it represents the
EXAMPLE 1 Showing Arrows are Equivalent
Show that the arrow from
R
P
(2, 1) to
Q
( 4, 2) to
(5, 3) (Figure 6.3).
S
( 1, 6) is equivalent to the arrow from
y
S
( – 1, 6)
Q
(5, 3)
R
( – 4, 2)
x
O
P
(2, – 1)
FIGURE 6.3
The arrows
RS
and
PQ
appear to have the same magnitude and direction. The
Head Minus Tail Rule proves that they represent the same vector (Example 1).
SOLUTION
Applying the HMT rule, we see that
RS
represents the vector
3, 4 , while
PQ
represents the vector 5 2, 3 ( 1) 3, 4
1 ( 4), 6 2
. Although they have different positions in the plane, these arrows represent the same vector and are therefore equivalent.
Now try Exercise 1.
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CHAPTER 6
Applications of Trigonometry y
Q
(
x
2
,
y
2
)
EXPLORATION 1
Vector Archery
See how well you can direct arrows in the plane using vector information and the HMT Rule.
1.
An arrow has initial point (2, 3) and terminal point (7, 5). What vector does it represent?
5, 2
2.
An arrow has initial point (3, 5) and represents the vector is the terminal point?
0, 11
3, 6 . What
3.
4.
If
P
is the point (4,
If
Q
is the point (4,
3) and
PQ
represents 2,
3) and
PQ
represents 2,
4 , find
Q
.
6,
4 , find
P
.
2, 1
7
P
(
x
1
,
y
1
)
x
FIGURE 6.4
The magnitude of
v
is the length of the arrow
PQ
, which is found using the distance formula:
x
2
x
1
2
y
2
y
1
2
.
v
WHAT ABOUT DIRECTION?
You might expect a quick computational rule for
direction
to accompany the rule for magnitude, but direction is less easily quantified. We will deal with vector direction later in the section.
y
P
(–3, 4)
Q
(–5, 2)
v
O
(0, 0)
(–2, –2)
x
FIGURE 6.5
The vector
v
of Example 2.
If you handled Exploration 1 with relative ease, you have a good understanding of how vectors are represented geometrically by arrows. This will help you understand the algebra of vectors, beginning with the concept of magnitude.
The magnitude of a vector
v
denoted by
v
is also called the
. (You might see
v
absolute value of
v
, so it is usually in some textbooks.) Note that it is a nonnegative real number, not a vector. The following computational rule follows directly from the distance formula in the plane (Figure 6.4).
Magnitude
If
v
is represented by the arrow from
x
1
,
y
1
v
x
2
x
1
2
y
2
y
1
2
.
If
v
a
,
b
, then
v
a
2
b
2
.
EXAMPLE 2
SOLUTION
to
x
2
,
y
2
, then
Finding Magnitude of a Vector
Find the magnitude of the vector
v
represented by
PQ
, where
P
Q
( 5, 2).
Working directly with the arrow,
HMT Rule shows that
v
2,
(See Figure 6.5.)
v
(
2 , so
v
5
(
( 3))
2
2)
2
(
(2
2 )
2
4)
2
2
2
(
2 .
3, 4) and
2 . Or, the
Now try Exercise 5.
The algebra of vectors sometimes involves working with vectors and numbers at the same time. In this context we refer to the numbers as
scalars
. The two most basic algebraic operations involving vectors are
vector addition
(adding a vector to a vector) and
scalar multiplication
(multiplying a vector by a number). Both operations are easily represented geometrically, and both have immediate applications to many realworld problems.
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SECTION 6.1
Vectors in the Plane
505
WHAT ABOUT VECTOR
MULTIPLICATION?
There is a useful way to define the multiplication of two vectors—in fact, there are two useful ways, but neither one of them follows the simple pattern of vector addition. (You may recall that matrix multiplication did not follow the simple pattern of matrix addition either, and for similar reasons.) We will look at the
dot product
in Section 6.2. The
cross product
requires a third dimension, so we will not deal with it in this course.
u
2
u
(1/2)
u
–2
u
FIGURE 6.7
Representations of
u
and several scalar multiples of
u
.
DEFINITION
Vector Addition and Scalar Multiplication
Let
u
(or
u
1
,
u
2
resultant
) and
v
v
1
,
v
2 be vectors and let
of the vectors u and v
k
be a real number (scalar). The is the vector
sum u v
u
1
v
1
,
u
2
v
2
.
k
u
k u
1
,
u
2
ku
1
,
ku
2
.
The sum of the vectors
u
ways.
and
v
can be represented geometrically by arrows in two
In the
tailtohead
origin to
u
1
,
u
2 representation, the standard representation of
u
. The arrow from
u
1
,
u
2 to
u
1
v
1
,
u
2 verify by the HMT Rule). The arrow from the origin to
u
1
v
2 represents
v
(as you can
v
1
,
u
2 points from the
v
2 then represents
u v
(Figure 6.6a).
In the
parallelogram
representation, the standard representations of
u
and
v
determine a parallelogram, the diagonal of which is the standard representation of
u v
(Figure 6.6b).
y y
v u u + v u v u + v
x x
(a) (b)
FIGURE 6.6
Two ways to represent vector addition geometrically: (a) tailtohead, and (b) parallelogram.
The product
k
u
of the scalar
k
and the vector
u
can be represented by a stretch (or shrink) of
u
by a factor of
k
. If
k
> 0, then
k
u
points in the same direction as
u
; if
k
< 0, then
k
u
points in the opposite direction (Figure 6.7).
EXAMPLE 3
Let
u
(a) u v
1, 3 and
v
Performing Vector Operations
4, 7 . Find the component form of the following vectors:
(b)
3
u
(c)
2
u
( 1)
v
SOLUTION
(a) u v
Using the vector operations as defined, we have:
1, 3 4, 7 1 4, 3 7 3, 10
(b)
3
u
3 1, 3 3, 9
(c)
2
u
( 1)
v
2 1, 3
Geometric representations of
u
( 1) 4, 7 2, 6 4, 7 6, 1
v
and 3
u
are shown in Figure 6.8 on the next page.
continued
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CHAPTER 6
Applications of Trigonometry y
(3, 10)
(–1, 3)
u v u
+
v
x
3
u
= –3, 9
y
u
= –1, 3
x
A WORD ABOUT VECTOR NOTATION
Both notations,
a
,
b
and
a
i
b
j
, are designed to convey the idea that a single vector
v
has two separate components. This is what makes a twodimensional vector twodimensional.
You will see both
a
,
b
,
c
and
a
i
b
j
c
k
used for threedimensional vectors, but scientists stick to the notation for dimensions higher than three.
(a)
(b)
FIGURE 6.8
Given that
u
1, 3 and
v
4, 7
, we can (a) represent
u
tailtohead method, and (b) represent 3
u
as a stretch of
u
by a factor of 3.
v
by the
Now try Exercise 13.
A vector
u
0, 0 with length
, then the vector
u
1 is a
unit vector
. If
v
is not the zero vector
u v v
1
v v
is a
unit vector in the direction of v
. Unit vectors provide a way to represent the direction of any nonzero vector. Any vector in the direction of
v
, or the opposite direction, is a scalar multiple of this unit vector
u
.
EXAMPLE 4
Finding a Unit Vector
Find a unit vector in the direction of
v
3, 2 , and verify that it has length 1.
SOLUTION
3
2
2
2
1 3 , so
v
3, 2
v v
1
1 3
3, 2
3
1 3
,
2
1 3
The magnitude of this vector is
3
1 3
,
2
1 3
(
9
1 3
3
1 3
2
)
4
1 3
(
2
1 3
2
)
1 3
1 3
1
Thus, the magnitude of
v
tive scalar multiple of
v
.
v
is 1. Its direction is the same as
v
because it is a posi
Now try Exercise 21.
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SECTION 6.1
Vectors in the Plane
507
y
v
=
a
,
b b
j
x a
i
FIGURE 6.9
a
i
b
j
.
The vector
v
is equal to
y

v
 sin
θ
v
θ

v
 cos
θ
x
FIGURE 6.10
The horizontal and vertical components of
v.
The two unit vectors
i
1, 0 and
j
0, 1 are the
standard unit vectors
tor
v
can be written as an expression in terms of the standard unit vectors:
. Any vec
v
a
,
b a
, 0
a
1, 0
a
i
b
j
0,
b b
0, 1
Here the vector
v
tors
i
and
j
a
,
. The scalars
b a
is expressed as the and
b
are the
linear combination horizontal
and
vertical
a
i
b
j
of the vec
components
, respectively, of the vector
v
. See Figure 6.9.
You may recall from our applications in Section 4.8 that direction is measured in different ways in different contexts, especially in navigation. A simple but precise way to specify the direction of a vector
v
is to state its
direction angle
, the angle that
v
makes with the positive
x
axis, just as we did in Section 4.3. Using trigonometry
(Figure 6.10), we see that the horizontal component of
v
component is
v
is
v
sin . Solving for these components is called cos and the vertical
resolving the vector
.
Resolving the Vector
If
v
has direction angle , the components of
v
can be computed using the formula
v v
cos ,
v
sin .
v
=
a
,
b y
6
O
115
°
x
From the formula above, it follows that the unit vector in the direction of
v
is
u v v
cos , sin .
EXAMPLE 5 Finding the Components of a Vector
Find the components of the vector
v
Figure 6.11
.
with direction angle 115 and magnitude 6
SOLUTION v
, then
If
a
and
b
are the horizontal and vertical components, respectively, of
v
a
,
b
6 cos 115 , 6 sin 115 .
So,
a
6 cos 115 2.54 and
b
6 sin 115 5.44.
Now try Exercise 29.
FIGURE 6.11
The direction angle of
v
is 115°. (Example 5)
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CHAPTER 6
Applications of Trigonometry y
β
u
α
u
= 3, 2
x
v v
= –2, –5
FIGURE 6.12
Example 6.
The two vectors of
y
65
°
25
°
500 mph
v
x
FIGURE 6.13
ing) in Example 7.
The airplane’s path (bear
TEACHING NOTE
Encourage students to draw pictures to analyze the geometry of various situations.
EXAMPLE 6 Finding the Direction Angle of a Vector
Find the magnitude and direction angle of each vector:
(a) u
3, 2
(b) v
2, 5
SOLUTION
(a) u u
cos ,
See Figure 6.12.
3
u
2
2 sin
2
.
1 3 . If
3
3
u
3 cos
2
2
2 cos
3 1 3 cos cos
(
1
3
1 3
) is the direction angle of
u
, then
u
33.69
Horizontal component of
u u
3 2 is acute.
2 2
(b)
2,
v
5
2
2
2
3, 2
v v
2
2 cos , cos
v
5
2 sin .
2 9 . If is the direction angle of
v
, then
v
Horizontal component of
v
2
2
5
2 cos
v
( 2)
2
( 5 )
2
2 9 cos
360 cos
(
1
2
2 9
)
248.2
180° 270°
Now try Exercise 33.
The
velocity
of a moving object is a vector because velocity has both magnitude and direction. The magnitude of velocity is
speed
.
EXAMPLE 7 Writing Velocity as a Vector
A DC10 jet aircraft is flying on a bearing of 65 at 500 mph. Find the component form of the velocity of the airplane. Recall that the bearing is the angle that the line of travel makes with due north, measured clockwise see Section 4.1, Figure 4.2
.
SOLUTION
Let
v
be the velocity of the airplane. A bearing of 65 a direction angle of 25 is equivalent to
. The plane’s speed, 500 mph, is the magnitude of vector
v
; that is,
v
500. See Figure 6.13.
The horizontal component of
500 sin 25 , so
v
is 500 cos 25 and the vertical component is
v
500 cos 25
i
500 sin 25
j
500 cos 25 , 500 sin 25 453.15, 211.31
The components of the velocity give the eastward and northward speeds. That is, the airplane travels about 453.15 mph eastward and about 211.31 mph northward as it travels at 500 mph on a bearing of 65 .
Now try Exercise 41.
5144_Demana_Ch06pp501566 01/11/06 9:32 PM Page 509
y
A
60
°
C
65 mph
θ
450 mph
v
B
D x
FIGURE 6.14
The
x
axis represents the flight path of the plane in Example 8.
FOLLOWUP
Have students discuss why it does not make sense to add a scalar to a vector.
ASSIGNMENT GUIDE
Day 1: Ex. 3–27, multiples of 3, 39, 40
Day 2: Ex. 29, 32, 34, 37, 42, 43, 45, 46, 49
COOPERATIVE LEARNING
Group Activity: Ex. 53–54
NOTES ON EXERCISES
Ex. 43–50 are problems that students would typically encounter in a physics course.
Ex. 55–60 provide practice with standardized tests.
Ex. 62 and 64 demonstrate connections between vectors and geometry.
ONGOING ASSESSMENT
SelfAssessment: Ex. 1, 5, 13, 21, 29, 33,
41, 43, 47
Embedded Assessment: Ex. 45, 46, 62
SECTION 6.1
Vectors in the Plane
509
A typical problem for a navigator involves calculating the effect of wind on the direction and speed of the airplane, as illustrated in Example 8.
EXAMPLE 8 Calculating the Effect of Wind Velocity
Pilot Megan McCarty’s flight plan has her leaving San Francisco International
Airport and flying a Boeing 727 due east. There is a 65mph wind with the bearing
60 . Find the compass heading McCarty should follow, and determine what the airplane’s ground speed will be assuming that its speed with no wind is 450 mph .
SOLUTION
plane alone,
See Figure 6.14. Vector
AB
represents the velocity produced by the air
AC
represents the velocity of the wind, and is the angle
DAB
. Vector
v
AD
represents the resulting velocity, so
v
AD AC AB
.
We must find the bearing of
AB
and
v
.
Resolving the vectors, we obtain
AC
AB
65 cos 30 , 65 sin 30
450 cos , 450 sin
AD AC AB
65 cos 30 450 cos , 65 sin 30 450 sin
Because the plane is traveling due east, the second component of
AD
must be zero.
65 sin 30 450 sin 0 sin
(
1
65 s i n 30
4 5 0
)
4.14
0
Thus, the compass heading McCarty should follow is
90 94.14
.
The ground speed of the airplane is
v
0
2
AD
6 5 c o s 3 0 °
65 cos 30 450 cos
4 5 0 c o s
2
505.12
Bearing 90°
Using the unrounded value of .
McCarty should use a bearing of approximately 94.14
east at approximately 505.12 mph.
. The airplane will travel due
Now try Exercise 43.
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CHAPTER 6
Applications of Trigonometry
A
D
20
°
w
B
20
°
C
FIGURE 6.15
The force of gravity
AB
has a component AC that holds the box against the surface of the ramp, and a component
AD
CB
that tends to push the box down the ramp. (Example 9)
EXAMPLE 9 Finding the Effect of Gravity
A force of 30 pounds just keeps the box in Figure 6.15 from sliding down the ramp inclined at 20 . Find the weight of the box.
w
; then
SOLUTION
We are given that
AD
30. Let
AB
sin 20
C w
B
3
w
0
.
Thus,
w
sin
30
20°
87.71.
The weight of the box is about 87.71 pounds.
Now try Exercise 47.
PROBLEM:
During one part of its migration, a salmon is swimming at 6 mph, and the current is flowing downstream at 3 mph at an angle of 7 degrees. How fast is the salmon moving upstream?
SOLUTION:
of the water.
Assume the salmon is swimming in a plane parallel to the surface
A
θ current salmon swimming in still water
B
salmon net velocity
C
In the figure, vector
AB
is 7 degrees, the vector represents the current of 3 mph,
CA
is the angle
CAB
, which represents the velocity of the salmon of 6 mph, and the vector
CB
is the net velocity at which the fish is moving upstream.
So we have
AB
3 cos 83 , 3 sin 83 0.37, 2.98
CA
0, 6
Thus
CB CA AB
0.37, 3.02
3 cos 83
The speed of the salmon is then
CB
, 3 sin
0.37
2
83
+ 3.02
2
6
3.04 mph upstream.
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SECTION 6.1
Vectors in the Plane
511
(For help, go to Sections 4.3 and 4.7.)
In Exercises 1– 4, find the values of
x
and
y
.
1.
y
9
2
3
; 4.5
2.
3.
y x
7
9
30
°
x
(
x
,
y
)
y x y
5.36;
220
°
4.50
In
x
4.
(
x
,
y
)
y x
15
y y
120
°
7.5;
15
2
3
x
3.86; 4.60
6
x
–50
°
y x
Exercises 5 and 6, solve for
5.
sin
1
(
3
2 9 in degrees.
)
33.85°
6.
cos
(
1
1
)
104.96°
1 5
In Exercises 7–9, the point
P
is on the terminal side of the angle .
Find the measure of if 0 360 .
7.
P
8.
P
5, 9
60.95°
5, 7
305.54°
9.
P
2, 5
180 tan
–1
(5 2) 248.20
10.
A naval ship leaves Port Norfolk and averages 42 knots nautical mph traveling for 3 hr on a bearing of 40 and then 5 hr on a course of 125 . What is the boat’s bearing and distance from Port Norfolk after 8 hr?
Distance: 254.14 naut mi.; Bearing: 95.40°
(
x
,
y
)
(
x
,
y
)
In Exercises 1– 4, prove that
RS
and
PQ
are equivalent by showing that they represent the same vector.
1.
R
2.
R
3.
R
4.
R
4, 7 ,
S
7,
2,
3
2, 1 ,
S
,
S
1 ,
S
1, 5
2, 4 ,
,
O
0,
4, 5 ,
O
1 ,
O
O
0, 0 , and
P
0, 0 , and
P
1, 4 , and
P
3, 1 , and
P
3,
3,
1, 2
2
2
1, 4
In Exercises 5 –12, let
P
S
2, 8
2, 2 ,
Q
3, 4 ,
R
2, 5 , and
. Find the component form and magnitude of the vector.
5.
PQ
7.
QR
9.
2
QS
11.
3
QR
5, 2 ;
5, 1 ;
2,
PS
2 9
2 6
24 ; 2
11,
1 4 5
7 ; 170
6.
RS
8.
PS
10.
4,
4,
2
PR
13 ;
10 ; 2
0, 3
1 8 5
2 9
2 ; 3
12.
PS
3
PQ
11, 16 ;
2
377
In Exercises 13 – 20, let
u
1, 3 ,
v
Find the component form of the vector.
13.
15.
u u v w
1, 7
3, 8
17.
2
u
19.
2
u
3
w
3
v
4, 9
4, 18
2, 4 , and
w
2,
14.
u
16.
3
v
18.
2
u
20.
u v
1
6, 12
4
v v
10,
1,
3,
7
1
10
5 .
In Exercises 21– 24, find a unit vector in the direction of the given vector.
21.
u
23.
w i
2, 4
2
j
22.
24.
v w
1,
5
i
1
5
j
In Exercises 25– 28, find the unit vector in the direction of the given vector. Write your answer in
(a)
component form and
(b)
as a linear combination of the standard unit vectors
i
and
j
.
25.
u
27.
u
2, 1
4, 5
26.
u
28.
u
3,
3, 2
4
In Exercises 29– 32, find the component form of the vector
v
.
29.
y
16.31, 7.61
30.
y
8.03, 11.47
18
25
°
v
x
14
v
55
°
x
21.
23.
0.45
i
0.89
j
0.45
i
0.89
j
22.
24.
0.71
i
0.71
j
0.71
i
0.71
j
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512
CHAPTER 6
Applications of Trigonometry
31.
v
47
y
14.52, 44.70
108
°
x
32.
v
33
y
23.74, 22.92
136
°
x
In Exercises 33–38, find the magnitude and direction angle of the vector.
33.
3, 4
35.
37.
3
7
i
5;
4
j
5; cos 135
i
53.13°
306.87° sin 135
34.
1, 2
36.
j
7; 135°
38.
3
i
5
2 cos 60
j i
5 ; 116.57°
3 4 ; sin 60
239.04°
j
2; 60°
In Exercises 39 and 40, find the vector
v
with the given magnitude and the same direction as
u
.
39.
v
2,
u
3, 3
40.
v
5,
u
5, 7
41.
Navigation
An airplane is flying on a bearing of 335 at
530 mph. Find the component form of the velocity of the airplane.
223.99, 480.34
42.
Navigation
An airplane is flying on a bearing of 170 at
460 mph. Find the component form of the velocity of the airplane.
79.88, 453.01
43.
Flight Engineering
ing bearing of 340
An airplane is flying on a compass headat 325 mph. A wind is blowing with the bearing 320 at 40 mph.
(a)
Find the component form of the velocity of the airplane.
(b)
Find the actual ground speed and direction of the plane.
44.
Flight Engineering
ing bearing of 170
An airplane is flying on a compass headat 460 mph. A wind is blowing with the bearing 200 at 80 mph.
(a)
Find the component form of the velocity of the airplane.
(b)
Find the actual ground speed and direction of the airplane.
45.
Shooting a Basketball
A basketball is shot at a 70 the horizontal direction with an initial speed of 10 m sec.
angle with
(a)
Find the component form of the initial velocity.
(b) Writing to Learn
Give an interpretation of the horizontal and vertical components of the velocity.
46.
Moving a Heavy Object
pushed up a 15
In a warehouse a box is being inclined plane with a force of 2.5 lb, as shown in the figure.
v
2.5 lb
15
°
(a)
Find the component form of the force.
2.41, 0.65
(b) Writing to Learn
Give an interpretation of the horizontal and vertical components of the force.
47.
Moving a Heavy Object
Suppose the box described in
Exercise 46 is being towed up the inclined plane, as shown in the figure below. Find the force
w
needed in order for the component of the force parallel to the inclined plane to be 2.5 lb. Give the answer in component form.
2.20, 1.43
w
33
°
15
°
48.
Combining Forces
Juana and Diego Gonzales, ages six and four respectively, own a strong and stubborn puppy named
Corporal. It is so hard to take Corporal for a walk that they devise a scheme to use two leashes. If Juana and Diego pull with forces of 23 lb and 27 lb at the angles shown in the figure, how hard is Corporal pulling if the puppy holds the children at a standstill?
about 47.95 lb
23 lb
18
°
15
°
27 lb
In Exercises 49 and 50, find the direction and magnitude of the resultant force.
49.
Combining Forces
A force of 50 lb acts on an object at an angle of 45 of 30 .
F
. A second force of 75 lb acts on the object at an angle
100.33 lb and 1.22
50.
Combining Forces
80 lb, act on an object at angles of 50 tively.
F
Three forces with magnitudes 100, 50, and
, and 20 , respec
113.81 lb and 35.66
, 160
51.
Navigation
A ship is heading due north at 12 mph. The current is flowing southwest at 4 mph. Find the actual bearing and speed of the ship.
342.86
; 9.6 mph
52.
Navigation
site shore.
A motor boat capable of 20 mph keeps the bow of the boat pointed straight across a milewide river. The current is flowing left to right at 8 mph. Find where the boat meets the oppo
0.4 mi downstream
53.
Group Activity
A ship heads due south with the current flowing northwest. Two hours later the ship is 20 miles in the direction 30 west of south from the original starting point.
Find the speed with no current of the ship and the rate of the current.
13.66 mph; 7.07 mph
54.
Group Activity
Express each vector in component form and prove the following properties of vectors.
(a)
(b)
(c) u u u v
0 v v u
,
w u u
where
v
0 w
0, 0
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SECTION 6.1
Vectors in the Plane
513
(d)
(e) u
a
u v u
a
0 u
, where
a
v
(g)
(i)
1
ab
u u u
,
a b
u
1
u u
a
,
b
(f)
(h)
a
,
a a
0
(j)
a
u
b b
u
a
u
0
, 0
u
a
u
0
b
u
55.
True or False
If
u
is a unit vector, then
Justify your answer.
u
is also a unit vector.
56.
True or False
If
u
is a unit vector, then 1
u
is also a unit vector.
Justify your answer.
False. 1/
u
is not a vector.
In Exercises 57–60, you may use a graphing calculator to solve the problem.
57.
Multiple Choice
the vector
(A)
(D)
1
5
2, 1 ?
Which of the following is the magnitude of
D
(B)
3
(C)
5
5
(E)
5
58.
Multiple Choice
Let
u
the following is equal to
u v
?
2, 3
E and
v
4, 1 . Which of
(A)
6,
(D)
4
6, 2
(B)
(E)
2, 2
6, 4
(C)
2, 2
59.
Multiple Choice
Which of the following represents the vector
v
shown in the figure below?
A
y
Explorations
61.
Dividing a Line Segment in a Given Ratio
two points in the plane, as shown in the figure.
B
Let
A
and
B
be
(a)
Prove that
BA
origin.
OA OB
, where
O
is the
C
A
(b)
Let
C
be a point on the line segment
BA
which divides the segment in the ratio
x
:
y
where
x y
1. That is,
B
C
C
A x y
.
O
Show that
OC xOA yOB
.
62.
Medians of a Triangle
Perform the following steps to use vectors to prove that the medians of a triangle meet at a point
O
which divides each median in the ratio 1 : 2.
M
1
,
M
2
, and
M
midpoints of the sides of the triangle shown in the figure.
3 are
C
M
3
O
M
2
O
30
°
3
v
x
(A)
(C)
3 cos 30
3 cos 60
, 3 sin 30
, 3 sin 60
(B)
3 sin 30 , 3 cos 30
(D)
3 cos 30 , 3 sin 30
(E)
60.
Multiple Choice
direction of
v i
Which of the following is a unit vector in the
3
j
?
C
(A)
1
1 0
i
1
3
0
j (B)
1
1 0
i
1
3
0
j (C)
1
10
i
3
10
j
(D)
3 cos 30 ,
1
10
i
3
10
j
3 sin 30
(E)
1
8
i
3
8
j
A
M
1
B
(a)
Use Exercise 61 to prove that
OM
1
1
2
OA
1
2
OB
OM
2
1
2
OC
1
2
OB
OM
3
1
2
OA
1
2
OC
(b)
Prove that each of 2
OM
1 equal to
OA OB OC
.
OC
, 2
OM
2
OA
, 2
OM
3
OB
(c) Writing to Learn
desired result.
Explain why part b establishes the is
63.
Vector Equation of a Line
points
A
and
B
. Prove that
C
OC
origin.
t OA
1
t OB
, where
Let
t x
,
y
L
be the line through the two is on the line
L
if and only if is a real number and
O
is the
64.
Connecting Vectors and Geometry
Prove that the lines which join one vertex of a parallelogram to the midpoints of the opposite sides trisect the diagonal.
55.
True.
u
and length of
u
have the same length but opposite directions. Thus, the
u
is also 1.
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514
CHAPTER 6
Applications of Trigonometry
6.2
What you’ll learn about
■ The Dot Product
■ Angle Between Vectors
■ Projecting One Vector onto
Another
■ Work
. . . and why
Vectors are used extensively in mathematics and science applications such as determining the net effect of several forces acting on an object and computing the work done by a force acting on an object.
Vectors can be multiplied in two different ways, both of which are derived from their usefulness for solving problems in vector applications. The
cross product
(or
vector product
or
outer product
) results in a vector perpendicular to the plane of the two vectors being multiplied, which takes us into a third dimension and outside the scope of this chapter. The
dot product
(or
scalar product
or
inner product
) results in a scalar. In other words, the dot product of two vectors is not a vector but a real number! It is the important information conveyed by that number that makes the dot product so worthwhile, as you will see.
Now that you have some experience with vectors and arrows, we hope we won’t confuse you if we occasionally resort to the common convention of using arrows to name the vectors they represent. For example, we might write “
u
hand for “
u
is the vector represented by
PQ
PQ
” as a short
.” This greatly simplifies the discussion of concepts like vector projection. Also, we will continue to use both vector notations,
a
,
b
and
a
i
b
j
, so you will get some practice with each.
DEFINITION
Dot Product u
•
v
of
v
u
1
v
1
u
2
v
2
.
v
1
,
v
2 is
DOT PRODUCT AND STANDARD
UNIT VECTORS
(
u
1
i
u
2
j
) • (
v
1
i
v
2
j
)
u
1
v
1
u
2
v
2
Dot products have many important properties that we make use of in this section. We prove the first two and leave the rest for the Exercises.
OBJECTIVE
Students will be able to calculate dot products and projections of vectors.
MOTIVATE
Ask students to guess the meaning of a projection of one vector onto another.
LESSON GUIDE
Day 1: The Dot Product; Angle Between
Vectors
Day 2: Projecting One Vector Onto
Another; Work
Properties of the Dot Product
Let
u
,
v
, and
w
be vectors and let
c
be a scalar.
1.
u
•
v
4.
u
•
v w
2.
3.
u
•
u
0
•
u v
•
u u
2
0
5.
u
•
v u
•
w u
c
u
•
v
•
w v u
•
u
•
w
c
v v
•
w
c
u
•
v
Proof
Let
u
u
1
,
u
2 and
v
Property 1 u
•
v
v
1
,
v
2
.
u
1
v
1
v
1
u
1
v
•
u
u
2
v
2
v
2
u
2
Property 2 u
•
u
Definition of
u
•
v
Commutative property of real numbers
Definition of
u
•
v
u
2
1
u
2
u
2
1
u
2
2
u
2
2
2
Definition of
u
•
u
Definition of
u
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SECTION 6.2
Dot Product of Vectors
515
TEACHING NOTE
If you do not plan to cover Chapter 8 and you want to cover vectors in threedimensional space, you can cover the relevant parts of Section 8.6 after you finish Section 6.2.
EXAMPLE 1
Find each dot product.
Finding Dot Products
(a)
3, 4
•
5, 2
(b)
(c)
1,
2
i
2
•
j
•
3
i
4, 3
5
j
SOLUTION
(a)
3, 4
•
5, 2
(b)
1,
(c)
2
i
2
j
•
3
•
3
i
4, 3
5
j
5
1
2 3
4
4
2 23
1
2 3 10
5 11
Now try Exercise 3.
Property 2 of the dot product gives us another way to find the length of a vector, as illustrated in Example 2.
DOT PRODUCTS ON
CALCULATORS
It is really a waste of time to compute a simple dot product of twodimensional vectors using a calculator, but it can be done. Some calculators do vector operations outright, and others can do vector operations via matrices. If you have learned about matrix multiplication already, you will know why the matrix product [
u
1
,
u
2
]
•
v
1
v
2
[ ]
v
1
,
v
2 yields the dot as a 1by1 product
u
1
,
u
2
• matrix. (The same trick works with vectors of higher dimensions.) This book will cover matrix multiplication in
Chapter 7.
v
–
u v u
FIGURE 6.16
The angle between nonzero vectors
u
and
v
.
EXAMPLE 2 Using Dot Product to Find Length
Use the dot product to find the length of the vector
u
4, 3 .
SOLUTION
It follows from Property 2 that
u
4, 3 4 , 3
•
4 , 3 4 4
u
•
u
. Thus,
3 3 2 5 5.
Now try Exercise 9.
Let
u
and
v
be two nonzero vectors in standard position as shown in Figure 6.16. The
angle between u and v
is the angle , 0 or 0 180 . The angle between any two nonzero vectors is the corresponding angle between their respective standard position representatives.
We can use the dot product to find the angle between nonzero vectors, as we prove in the next theorem.
THEOREM
Angle Between Two Vectors
If is the angle between the nonzero vectors
u
and
v
, then cos and
u u
•
v v
cos
(
1
u
•
v u v
)
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516
CHAPTER 6
Applications of Trigonometry
v
= –2, 5
y
θ
u
= 2, 3
x
(a)
y
θ
u
= 2, 1
x
v
= –1, –3
(b)
FIGURE 6.17
The vectors in
(a) Example 3a and (b) Example 3b.
Proof
We apply the Law of Cosines to the triangle determined by
u
,
v
, and
v
and use the properties of the dot product.
v
•
v v
•
u v
2
v u
•
v u
2
v u u
2
u
2
u
2
v
2
v
2
v
2
2
u
2
u u
•
v u
•
u
2
u
2
u
•
v u
2
u
2
v
2
2
u
•
v
2
u
cos cos
u u
•
v v
(
1
v u
•
v u v
cos
)
2
u v v v v
cos cos cos cos
u
in Figure 6.16,
EXAMPLE 3 Finding the Angle Between Vectors
Find the angle between the vectors
u
and
v
.
(a) u
2, 3 ,
v
2, 5
(b) u
2, 1 ,
v
1, 3
SOLUTION
(a)
See Figure 6.17a. Using the Angle Between Two Vectors Theorem, we have cos
u u
•
v v
2 , 3
2 , 3
•
2 ,
2 ,
5
5
1
11
3 2 9
.
So, cos cos
1
1
1
2
11
1 3 2 9
135 .
55.5
.
(b)
See Figure 6.17b. Again using the Angle Between Two Vectors Theorem, we have cos
u u
•
v v
2 , 1
2 , 1
•
1
1
,
,
3
3
5
5
1 0
1
2
.
So,
Now try Exercise 13.
If vectors
u
and
v
are perpendicular, that is, if the angle between them is 90 , then
u
•
v u v
cos 90 0 because cos 90 0.
DEFINITION
Orthogonal Vectors
The vectors
u
and
v
are
orthogonal
if and only if
u
•
v
0.
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SECTION 6.2
Dot Product of Vectors
517
EXPLORATION EXTENSIONS
Now suppose
B
(
x
,
y
) is a point that is not on the given circle. If
x
2
y
2
a
2
, what can you say about
u
•
v
? If
x
2
y
2
a
2
, what can you say about
u
•
v
?
The terms “perpendicular” and “orthogonal” almost mean the same thing. The zero vector has no direction angle, so technically speaking, the zero vector is not perpendicular to any vector. However, the zero vector is orthogonal to every vector. Except for this special case, orthogonal and perpendicular are the same.
EXAMPLE 4
Prove that the vectors
u
Proving Vectors are Orthogonal
2, 3 and
v
6, 4 are orthogonal.
SOLUTION
We must prove that their dot product is zero.
u
•
v
2, 3
•
6, 4 12 12
The two vectors are orthogonal.
0
Now try Exercise 23.
y
A
(–
a
, 0)
θ
B
(
x
,
y
)
C
(
a
, 0)
x
FIGURE 6.18
The angle in the upper half of the circle
x
2
ABC
inscribed
y
2
a
2
.
(Exploration 1)
Q
u v
S
R
P
FIGURE 6.19
v
PS
,
The vectors
u
PQ
, and the vector projection of
u
onto
v
,
PR
proj
v u
.
FOLLOWUP
Ask students to name a pair of vectors that are orthogonal but not perpendicular.
ASSIGNMENT GUIDE
Day 1: Ex. 1–21, multiples of 3, 30–42, multiples of 3
Day 2: Ex. 27–51, multiples of 3, 61–66
COOPERATIVE LEARNING
Group Activity: Ex. 58, 59
EXPLORATION 1
Angles Inscribed in Semicircles
Figure 6.18 shows
x
2
y
2
a
2
.
ABC
inscribed in the upper half of the circle
1.
2.
3.
For
a
v
BC
.
2, find the component form of the vectors
u
2
x
,
y
, 2
x
,
y
BA
and
Find
u
• vectors?
v
. What can you conclude about the angle between these two
90
Repeat parts 1 and 2 for arbitrary a.
Answers will vary
vector projection
mined by dropping a perpendicular from
Q
v
to the line
PS
PS
is the vector
PR
deter
(Figure 6.19). We have resolved
u
into components
PR
and
RQ
u
PR RQ
with
PR
and
RQ
perpendicular.
The standard notation for this notation,
RQ
u
PR
, the vector projection of
u
onto
v
, is
PR
proj
v u
. We ask you to establish the following formula in the
Exercises (see Exercise 58).
proj
v u
. With
Projection of u onto v
If
u
and
v
are nonzero vectors, the projection of
u
onto
v
is proj
v u u
•
v
2
v v
.
5144_Demana_Ch06pp501566 01/11/06 9:32 PM Page 518
518
CHAPTER 6
Applications of Trigonometry y
–1
–1
–2
–3
–4
–5
–6
3
2
1
u
1
u
= 6, 2
u
2
1 2 3 4 5 6 7
v
= 5, –5
x
FIGURE 6.20
v
5, 5
(Example 5)
,
u
1
The vectors
u
proj
v u
, and
u
2
6, 2 ,
u u
1
.
u
proj
v
u v
FIGURE 6.21
If we pull on a box with force
u
, the effective force in the direction of
v
is proj
v u
, the vector projection of
u
onto
v.
F
1
F
45
°
FIGURE 6.22
The sled in Example 6.
EXAMPLE 5 Decomposing a Vector into
Perpendicular Components
Find the vector projection of
u
6, 2 onto
v
two orthogonal vectors, one of which is proj
v u
.
5, 5 . Then write
u
as the sum of
SOLUTION
We write
u u u
1
(Figure 6.20).
u
1 proj
v u u
1
u
2 where
u
1
u
•
v
2
v v
2 0
5 0
5, proj
v u
and
u
2
5 2, 2
Thus,
u
1
u
2
u
2
u
2,
u
1
2
6, 2
4, 4
2,
6, 2
2
u
.
4, 4
Now try Exercise 25.
If
u
is a force, then proj
v u
6.21).
represents the effective force in the direction of
v
(Figure
We can use vector projections to determine the amount of force required in problem situations like Example 6.
EXAMPLE 6 Finding a Force
Juan is sitting on a sled on the side of a hill inclined at 45 . The combined weight of
Juan and the sled is 140 pounds. What force is required for Rafaela to keep the sled from sliding down the hill? (See Figure 6.22.)
SOLUTION
We can represent the force due to gravity as
F
140
j
because gravity acts vertically downward. We can represent the side of the hill with the vector
v
cos 45
i
sin 45
j
2
2
i
2
2
j
.
The force required to keep the sled from sliding down the hill is
F
1 proj
v
F
(
F v
•
2
v
)
v F
•
v v
because
v
1. So,
F
1
F
•
v v
140
(
2
2
)
v
70
i j
.
The magnitude of the force that Rafaela must exert to keep the sled from sliding down the hill is 70 2 99 pounds.
Now try Exercise 45.
If
F work
is a constant force whose direction is the same as the direction of
AB
, then the
W
done by
F
in moving an object from
A
to
B
is
W
F
AB
.
5144_Demana_Ch06pp501566 01/11/06 9:32 PM Page 519
SECTION 6.2
Dot Product of Vectors
519
NOTES ON EXERCISES
Ex. 19–20 can be completed by using dot products or by using common sense.
Encourage students to try both methods.
Ex. 51–56 involve work done by a force that is not parallel to the direction of motion.
Ex. 61–66 provide practice with standardized tests.
ONGOING ASSESSMENT
SelfAssessment: Ex. 3, 9, 13, 23, 25,
45, 53
Embedded Assessment: Ex. 67, 68
UNITS FOR WORK
Work is usually measured in footpounds or Newtonmeters. One
Newtonmeter is commonly referred to as one Joule.
If
F
is a constant force in any direction, then the object from
A
to
B
is
work
W
done by
F
in moving an
W
F
•
AB
F
AB
cos where is the angle between
F
and
AB
. Except for the sign, the work is the magnitude of the effective force in the direction of
AB
times
AB
.
EXAMPLE 7 Finding Work
Find the work done by a 10 pound force acting in the direction object 3 feet from 0, 0 to 3, 0 .
1, 2 in moving an
SOLUTION
The force
F
has magnitude 10 and acts in the direction 1, 2 , so
F
The direction of motion is from
A
work done by the force is
10
1 ,
1 ,
2
2
0, 0 to
B
10
5
1, 2 .
3, 0 , so
AB
F
•
AB
10
5
1, 2
•
3, 0
30
5
3, 0 . Thus, the
13.42 footpounds.
Now try Exercise 53.
(For help, go to Section 6.1
.
)
In Exercises 1– 4, find
u
.
1.
u
2, 3
1 3
2.
u
3
i
4
3.
u
cos 35
i
sin 35
j
1
4.
u
2 cos 75
i
sin 75
j
2
In Exercises 5 – 8, the points
A
and
B
lie on the circle
x
2
Find the component form of the vector
AB
.
5.
A
2, 0 ,
B
1, 3
6.
A
2, 0 ,
B
j
5
y
2
1,
3, 3 1, 3
4.
3
7.
A
8.
A
2, 0 ,
B
2, 0 ,
B
1,
1,
3
3
1,
3, 3
3
In Exercises 9 and 10, find a vector
u
with the given magnitude in the direction of
v
.
9.
u
2,
4
1 3
,
v
6
1 3
2, 3
10.
u
3,
v
4
i
1 2
,
5
9
5
3
j
In Exercises 1– 8, find the dot product of
u
and
v
.
1.
u
2.
u
5, 3 ,
v
5, 2 ,
v
12, 4
8, 13
72
14
3.
u
4, 5 ,
v
3, 7
47
4.
u
2, 7
5.
u
4
i
,
9
j v
,
5,
v
3
i
8
46
2
j
30
6.
7.
8.
u u u
2
4
i i
7
i
,
4
j
,
v
8
i v
2
i
11
j
,
5
j v
3
j
7
14
33
j
44
In Exercises 9 –12, use the dot product to find
u
.
9.
11.
u
5,
u
4
i
12
4
13
10.
12.
u u
3
j
8, 15
3
17
5144_Demana_Ch06pp501566 01/11/06 9:32 PM Page 520
520
CHAPTER 6
Applications of Trigonometry
In Exercises 13 – 22, find the angle between the vectors.
13.
u
4, 3 ,
v
1, 5
115.6
14.
u
2, 2 ,
v
3, 3
90
16.
165
u
5, 2
15.
u
17.
u
2, 3 ,
3
i v
3
j
,
3, 5
64.65°
v
2
i
2 3
j
90
19.
u
20.
u v
18.
u
2
i
,
(
(
2 cos
4 cos
3
)
i
)
i
5
j
( sin
(
2 sin
4
)
j
,
3
v
)
j
,
v
( cos
3
2
)
i
(
3 cos
5
6
)
i
,
v
6, 1
167.66°
( sin
3
2
)
j
135°
(
3 sin
5
6
)
j
90°
21.
y
94.86
22.
(–3, 4)
v
6
5
4
3
2
1
–4 –3 –2 –1
–1
u
(8, 5)
1 2 3 4 5 6 7 8 9
y
153.10
(–3, 8)
(–1, –9)
–9
–10
9
8
7
6
5
4
3
2
–4 –3 –2 –1
–1
–2
1
x x
In Exercises 23–24, prove that the vectors
u
and
v
are orthogonal.
23.
u
24.
u
2, 3 ,
v
4, 1 ,
v
3 2,
1,
1
4
In Exercises 25–28, find the vector projection of
u
onto
v
. Then write
u
as a sum of two orthogonal vectors, one of which is proj
v u
.
25.
u
8, 3 ,
v
6, 2
26.
u
3, 7 ,
v
2, 6
27.
u
8, 5 ,
v
4, 5 , 1, 10 ,
9,
3, 1
2
28.
u
4, 1 ,
2, 8 ,
v
1, 6 ,
9,
5,
3
In Exercises 29 and 30, find the interior angles of the triangle with given vertices.
29.
30.
1
In Exercises 31 and 32, find
u
where is the angle between
u
•
v
satisfying the given conditions and
v
.
31.
150 ,
u
3,
v
8
32.
3
,
u
12,
v
40
In Exercises 33– 38, determine whether the vectors
u
and
v
are parallel, orthogonal, or neither.
33.
u
34.
u
35.
u
36.
u
37.
u
38.
u
5, 3 ,
v
1 0
,
4
3
2
Parallel
2, 5 ,
v
15, 12 ,
v
1 0
,
3
4
3
4, 5
Neither
Neither
5, 6 ,
v
3, 4 ,
v
2, 7 ,
v
12, 10
Orthogonal
20, 15
Orthogonal
4, 14
Parallel
In Exercises 39– 42, find
(a)
the
x
intercept
A
and
y
intercept
B
of the line.
(b)
the coordinates of the point
P
so that
AP
is perpendicular to the line and
AP
1. (There are two answers.)
39.
41.
3
3
x x
4
y
7
y
12
21
40.
42.
x
2
x
2
y
5
y
6
10
In Exercises 43 and 44, find the vector(s)
v
satisfying the given conditions.
43.
44.
u u
2, 3 ,
u
•
v
10,
v
2
2, 5 ,
u
•
v
11,
v
17
2
10
45.
Sliding Down a Hill
Ojemba is sitting on a sled on the side of a hill inclined at 60 . The combined weight of Ojemba and the sled is 160 pounds. What is the magnitude of the force required for Mandisa to keep the sled from sliding down the hill?
46.
Revisiting Example 6
Suppose Juan and Rafaela switch positions. The combined weight of Rafaela and the sled is 125 pounds.
What is the magnitude of the force required for Juan to keep the sled from sliding down the hill?
88.39 pounds
47.
Braking Force
A 2000 pound car is parked on a street that makes an angle of 12 with the horizontal (see figure).
12
°
(a)
Find the magnitude of the force required to keep the car from rolling down the hill.
415.82 pounds
(b)
Find the force perpendicular to the street.
1956.30 pounds
5144_Demana_Ch06pp501566 01/11/06 9:32 PM Page 521
SECTION 6.2
Dot Product of Vectors
521
48.
Effective Force
A 60 pound force
F
that makes an angle of 25 with an inclined plane is pulling a box up the plane.The inclined plane makes an 18 angle with the horizontal (see figure). What is the magnitude of the effective force pulling the box up the plane?
54.38 pounds
25
°
18
°
49.
Work
Find the work done lifting a
2600 pound car 5.5 feet.
14,300 footpounds
50.
Work
3 feet.
Find the work done lifting a 100 pound bag of potatoes
300 footpounds
51.
Work
Find the work done by a force
F
of 12 pounds acting in the direction 1, 2 in moving an object 4 feet from 0, 0 to 4, 0 .
21.47 footpounds
52.
Work
Find the work done by a force
F
of 24 pounds acting in the direction 4, 5 in moving an object 5 feet from 0, 0 to 5, 0 .
74.96 footpounds
53.
Work
Find the work done by a force
F
of 30 pounds acting in the direction 2, 2 in moving an object 3 feet from 0, 0 to a point in the first quadrant along the line
y
1 2
x
.
54.
Work
Find the work done by a force
F
of 50 pounds acting in the direction 2, 3 in moving an object 5 feet from point in the first quadrant along the line
y x
.
0, 0 to a
55.
Work
AB
2
i
The angle between a 200 pound force
F
and
3 object from
A
j
is 30 to
B
.
. Find the work done by
F
in moving an
100 3 9 624.5 footpounds
56.
Work
The angle between a 75 pound force
F
and
AB
is 60 where
A
1, 1 and
B
moving an object from
A
4, 3 . Find the work done by
F
to
B
.
201.94 footpounds
, in
57.
Properties of the Dot Product
and let
c
Let
u
,
v
, and
w
be vectors be a scalar. Use the component form of vectors to prove the following properties.
(a) 0
•
u
(b)
(c) u u
•
v v
0
•
w w u
•
v u
•
w u v
•
•
w w
(d)
c
u
•
v u
•
c
v
c
u
•
v
58.
Group Activity Projection of a Vector
nonzero vectors. Prove that
(a)
proj
v u
(
u
•
v v
2
v
)
u
proj
v u
proj
v u
(b)
•
0
Let
u
and
v
be
59.
Group Activity Connecting Geometry and Vectors
Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
60.
If
u
is any vector, prove that we can write
u
as
u u
•
i i u
•
j j
.
61.
True or False
If
u
•
Justify your answer.
v
0, then
u
and
v
are perpendicular.
62.
True or False
If
u
is a unit vector, then
u
answer.
True.
u
•
u u
2
(1)
2
1
•
u
1. Justify your
In Exercises 63–66, you may use a graphing calculator to solve the problem.
63.
Multiple Choice
Let
u
1, 1 and
v
following is the angle between
u
and
v
?
D
(A)
0
(B)
45
(C)
60
(D)
90
1, 1 . Which of the
64.
Multiple Choice
Let
u
the following is equal to
u
•
v
?
4,
C
5 and
v
(A)
23
(B)
7
(C)
7
(D)
23
(E)
135
2, 3 . Which of
(E)
65.
Multiple Choice
Let
u
3 the following is equal to proj
v u
?
2,
A
3 2
(A)
3 2, 0
(B)
3, 0 and
v
(C)
7
2, 0 . Which of
3 2, 0
(D)
3 2, 3 2
(E)
3 2, 3 2
66.
Multiple Choice
Which of the following vectors describes a 5 lb force acting in the direction of
u
1, 1 ?
B
5
(A)
5 1, 1
(B)
1, 1
(C)
5 1, 1
2
(D)
5
2
1, 1
(E)
5
2
1, 1
67.
Distance from a Point to a Line
equation 2
x
5
y
10 and the point
P
Consider the line
L
with
3, 7 .
(a)
Verify that
A x
intercepts of
L
.
0, 2 and
B
5, 0 are the
y
 and
(b)
Find
w
1 proj
AB
AP
and
w
2
AP
(c) Writing to Learn
Explain why to
L
. What is this distance?
w
2 proj
AB
AP
.
is the distance from
P
(d)
(e)
Find a formula for the distance of any point line
ax by c
.
P
Find a formula for the distance of any point
P x
0
,
y
0
x
0
,
y
0 to
L
.
to the
68.
Writing to Learn
are not parallel.
Let
w
cos
t
u
sin
t
v
where
u
and
v
(a)
Can the vector
w
be parallel to the vector
u
? Explain.
(b)
Can the vector
w
be parallel to the vector
v
? Explain.
(c)
Can the vector
w
be parallel to the vector
u v
? Explain.
69.
If the vectors
u
and
v
are not parallel, prove that
a
u
b
v
c
u
d
v
⇒
a c
,
b d
.
5144_Demana_Ch06pp501566 01/11/06 9:32 PM Page 522
522
CHAPTER 6
Applications of Trigonometry
6.3
What you’ll learn about
■ Parametric Equations
■ Parametric Curves
■ Eliminating the Parameter
■ Lines and Line Segments
■ Simulating Motion with a
Grapher
. . . and why
These topics can be used to model the path of an object such as a baseball or a golf ball.
Imagine that a rock is dropped from a 420ft tower. The rock’s height
y
in feet above the ground
t
seconds later (ignoring air resistance) is modeled by
y
16
t
2
420 as we saw in Section 2.1. Figure 6.23 shows a coordinate system imposed on the scene so that the line of the rock’s fall is on the vertical line
x
2.5.
The rock’s original position and its position after each of the first 5 seconds are the points
2.5, 420 , 2.5, 404 , 2.5, 356 , 2.5, 276 , 2.5, 164 , 2.5, 20 , which are described by the pair of equations
x
2.5,
y
16
t
2
420, when
t
0, 1, 2, 3, 4, 5. These two equations are an example of
parametric equations
with
parameter t
. As is often the case, the parameter
t
represents time.
t
= 0,
y
= 420
t t
= 1,
y
= 2,
y
= 404
= 356
t
= 3,
y
= 276
In this section we study the graphs of
parametric equations
of objects that can be modeled with parametric equations.
and investigate motion
t
= 4,
y
= 164
t
= 5,
y
= 20
[0, 5] by [–10, 500]
FIGURE 6.23
The position of the rock at
0, 1, 2, 3, 4, and 5 seconds.
OBJECTIVE
Students will be able to define parametric equations, graph curves parametrically, and solve application problems using parametric equations.
MOTIVATE
Have students use a grapher to graph the parametric equations
x
for 5
t t
and
y t
5. Have them write the
2 equation for this graph in the form
y f
(
x
).
(
y x
2
)
LESSON GUIDE
Day 1: Parametric Equations; Parametric
Curves; Eliminating the Parameter; Lines and Line Segments
Day 2: Simulating Motion with a Grapher
DEFINITION
Parametric Curve, Parametric Equations
The graph of the ordered pairs
x
,
y
where
x f t
,
y g t
are functions defined on an interval
I
of
t
values is a
parametric curve
. The equations are
parametric equations parameter interval
.
for the curve, the variable
t
is a
parameter
, and
I
is the
When we give parametric equations and a parameter interval for a curve, we have
parametrized
the curve. A
parametrization
of a curve consists of the parametric equations and the interval of
t
values. Sometimes parametric equations are used by companies in their design plans. It is then easier for the company to make larger and smaller objects efficiently by just changing the parameter
t
.
Graphs of parametric equations can be obtained using parametric mode on a grapher.
EXAMPLE 1 Graphing Parametric Equations
For the given parameter interval, graph the parametric equations
(a)
3
t
1
(b)
x t
2
2
t
3
2,
y
(c)
3
t
.
3
t
3
continued
5144_Demana_Ch06pp501566 01/11/06 9:32 PM Page 523
NOTES ON EXAMPLES
Example 1 is important because it shows how a parametric graph is affected by the chosen range of
t
values.
SECTION 6.3
Parametric Equations and Motion
523
SOLUTION
In each case, set Tmin equal to the left endpoint of the interval and
Tmax equal to the right endpoint of the interval. Figure 6.24 shows a graph of the parametric equations for each parameter interval. The corresponding relations are different because the parameter intervals are different.
Now try Exercise 7.
[–10, 10] by [–10, 10]
(a)
[–10, 10] by [–10, 10]
(b)
FIGURE 6.24
Three different relations defined parametrically. (Example 1)
TEACHING NOTE
If students are not familiar with parametric graphing, it might be helpful to show them the graph of the linear function
f
(
x
) 3
x
2 and compare it to one defined parametrically as
x y
3
t t
and
2, using a trace key to show how
t
,
x
, and
y
are related.
[–10, 5] by [–5, 5]
FIGURE 6.25
(Example 2)
The graph of
y
0.5
x
1.5.
[–10, 10] by [–10, 10]
(c)
When a curve is defined parametrically it is sometimes possible to
eliminate the parameter
and obtain a rectangular equation in
x
and
y
that represents the curve. This often helps us identify the graph of the parametric curve as illustrated in Example 2.
EXAMPLE 2
Eliminating the Parameter
Eliminate the parameter and identify the graph of the parametric curve
x
1 2
t
,
y
2
t
,
∞
t
∞
.
SOLUTION
We solve the first equation for
t
:
x
1 2
t
2
t
1
x t
1
2
1
x
Then we substitute this expression for
t
into the second equation:
y y y
2
t
2
1
2
0.5
x
1
1.5
x
The graph of the equation
y
1.5 Figure 6.25
.
0.5
x
1.5 is a line with slope 0.5 and
y
intercept
Now try Exercise 11.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 524
524
CHAPTER 6
Applications of Trigonometry
EXPLORATION EXTENSIONS
Determine the smallest possible range of
t
values that produces the graph shown in
Figure 6.25, using the given parametric equations.
ALERT
Many students will confuse range values of
t
with range values on the function grapher. Point out that while the scale factor does not affect the way a graph is drawn, the Tstep does affect the way the graph is displayed.
PARABOLAS
The inverse of a parabola that opens up or down is a parabola that opens left or right. We will investigate these curves in more detail in Chapter 8.
EXPLORATION 1
Graphing the Curve of Example 2
Parametrically
1.
Use the parametric mode of your grapher to reproduce the graph in
Figure 6.25. Use 2 for Tmin and 5.5 for Tmax.
1.5. Find the
2.
Prove that the point 17, 10 corresponding value of
t
is on the graph of
y
that produces this point.
0.5
x t
8
3.
Repeat part 2 for the point 23, 10 .
t
12
4.
Assume that
a
,
b
is on the graph of
y
value of
t
that produces this point.
t
1 2
0.5
x a
2
1.5. Find the corresponding
2
b
5.
How do you have to choose Tmin and Tmax so that the graph in
Figure 6.25 fills the window?
Tmin 2 and Tmax 5.5
If we do not specify a parameter interval for the parametric equations
x f t
,
y g t
it is understood that the parameter
t
can take on all values which produce real num
, bers for
x
and
y
. We use this agreement in Example 3.
EXAMPLE 3 Eliminating the Parameter
Eliminate the parameter and identify the graph of the parametric curve
x t
2
2,
y
3
t.
SOLUTION
obtaining
t y
Here
t
can be any real number. We solve the second equation for
3 and substitute this value for
y
into the first equation.
t x t
2
2
x
( )
y
3
2
2
y x
2
y
2
9
2
9
x
2
Figure 6.24c shows what the graph of these parametric equations looks like. In
Chapter 8 we will call this a parabola that opens to the right. Interchanging we can identify this graph as the inverse of the graph of the parabola
x
2
9
x y
and
y
2 .
Now try Exercise 15.
[–4.7, 4.7] by [–3.1, 3.1]
FIGURE 6.26
The graph of the circle of Example 4.
EXAMPLE 4 Eliminating the Parameter
Eliminate the parameter and identify the graph of the parametric curve
x
2 cos
t
,
y
2 sin
t
, 0
t
2 .
SOLUTION
The graph of the parametric equations in the square viewing window of Figure 6.26 suggests that the graph is a circle of radius 2 centered at the origin.
We confirm this result algebraically.
continued
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 525
SECTION 6.3
Parametric Equations and Motion
525
A
(–2, 3)
y
B
(3, 6)
P
(
x
,
y
)
x
O
1
FIGURE 6.27
Example 5 uses vectors to construct a parametrization of the line through
A
and
B
.
TEACHING NOTE
The parametrization in Example 5 is not unique. You may want to have your students find alternate parametrizations.
x
2
y
2
4 cos
2
t
4 sin
2
t
4 cos
2
t
sin
2
t
4 1 cos
2
t
sin
2
t
1
4
The graph of
x
2
y
2 length of the interval 0
4 is a circle of radius 2 centered at the origin. Increasing the
t
2 will cause the grapher to trace all or part of the circle more than once. Decreasing the length of the interval will cause the grapher to only draw a portion of the complete circle. Try it!
Now try Exercise 23.
In Exercise 65, you will find parametric equations for any circle in the plane.
We can use vectors to help us find parametric equations for a line as illustrated in
Example 5.
EXAMPLE 5 Finding Parametric Equations for a Line
Find a parametrization of the line through the points
A
2, 3 and
B
3, 6 .
SOLUTION
Let
P x
,
y
be an arbitrary point on the line through can see from Figure 6.27, the vector
OP
A
and
B
. As you is the tailtohead vector sum of
OA
and
AP
.
You can also see that
AP
is a scalar multiple of
AB
.
If we let the scalar be
t
, we have
OP OA AP
OP x
,
y x
,
y
OA t
•
AB
2, 3
2, 3
t t
3
5, 3
( 2), 6 3
x
,
y
2 5
t
, 3 3
t
This vector equation is equivalent to the parametric equations
y
3 3
t
. Together with the parameter interval ( ,
x
2 5
t
and
), these equations define the line.
We can confirm our work numerically as follows: If
t
which gives the point
A
. Similarly, if
t
1, then
x
0, then
x
3 and
y
2 and
y
3,
6, which gives the point
B
.
Now try Exercise 27.
The fact that
t
0 yields point
A
and
t
1 yields point
B
in Example 5 is no accident, as a little reflection on Figure 6.27 and the vector equation
OP OA t
•
AB
should suggest. We use this fact in Example 6.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 526
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CHAPTER 6
Applications of Trigonometry
T=0
X=8.5
Y=5
Start,
t
= 0
(a)
EXAMPLE 6 Finding Parametric Equations for a Line Segment
Find a parametrization of the line segment with endpoints
A
2, 3 and
B
3, 6 .
SOLUTION
and
B
:
In Example 5 we found parametric equations for the line through
A x
2 5
t
,
y
3 3
t.
We also saw in Example 5 that
t
0 produces the point
A
point
B
. A parametrization of the line segment is given by and
t x
2 5
t
,
y
3 3
t
, 0
t
1.
1 produces the
As
t
varies between 0 and 1 we pick up every point on the line segment between
A
and
B
.
Now try Exercise 29.
T=5
X=–9 Y=5
5 sec later,
t
= 5
(b)
T=8
X=–2.7
Y=5
3 sec after that,
t
= 8
(c)
[
FIGURE 6.28
Three views of the graph
C
1
:
x
1
y
1
0.1(
t
3
5, 0
t
20
t
2
12 in the [
110
t
85),
12, 12] by
10, 10] viewing window. (Example 7)
GRAPHER NOTE
The equation
y
2
t
is typically used in the parametric equations for the graph
C
2 in Figure 6.29. We have chosen
y
2
t
to get two curves in Figure 6.29 that do not overlap. Also notice that the
y
coordinates of
C
1 that the
y
coordinates of
C
time
t
(
y
2 are constant (
y
1
5), and vary with
t
).
2
Example 7 illustrates several ways to simulate motion along a horizontal line using parametric equations. We use the variable t for the parameter to represent time.
EXAMPLE 7 Simulating Horizontal Motion
Gary walks along a horizontal line think of it as a number line of his position in meters given by with the coordinate
s
0.1
t
3
20
t
2
110
t
85 where 0
t
12. Use parametric equations and a grapher to simulate his motion.
Estimate the times when Gary changes direction.
SOLUTION
The graph
C
1
We arbitrarily choose the horizontal line
y
of the parametric equations,
C
1
:
x
1
0.1
t
3
20
t
2
110
t
85 ,
y
1
5 to display this motion.
5, 0
t
12, simulates the motion. His position at any time
t x
1
t
, 5 .
is given by the point
Using trace in Figure 6.28 we see that when
t y
axis at the point 8.5, 5
0, Gary is 8.5 m to the right of the
, and that he initially moves left. Five seconds later he is 9 m to the left of the
y
axis at the point 9, 5 . And after 8 seconds he is only 2.7 m to the left of the
y
axis. Gary must have changed direction during the walk. The motion of the trace cursor simulates Gary’s motion.
A variation in
y t
,
C
2
:
x
2
0.1
t
3
20
t
2
110
t
85 ,
y
2
t
, 0
t
12, can be used to help visualize where Gary changes direction. The graph
C
2 shown in
Figure 6.29 suggests that Gary reverses his direction at 3.9 seconds and again at
9.5 seconds after beginning his walk.
Now try Exercise 37.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 527
SECTION 6.3
Parametric Equations and Motion
527
C
1
C
1
T=1
X=5.5
Y=135
[0, 6] by [0, 200]
(a)
C
2
T=3.9
X=–9.9119
Y=–3.9
C
2
T=9.5
X=–1.2375
Y=–9.5
[–12, 12] by [–15, 15]
(a)
[–12, 12] by [–15, 15]
(b)
[
t
FIGURE 6.29
Two views of the graph
C
1
12 and the graph
C
2
:
x
2
0.1(
t
3
20
t
2
:
x
1
12, 12] by [
0.1(
t
3
110
t
85),
15, 15] viewing window. (Example 7)
y
20
t
2
2
110
t t
, 0
t
85),
y
1
5, 0
12 in the
T=2
X=5.5
Y=163
[0, 6] by [0, 200]
(b)
T=4
X=5.5
Y=123
[0, 6] by [0, 200]
(c)
T=5
X=5.5
Y=55
[0, 6] by [0, 200]
(d)
FIGURE 6.30
x
1
t
,
y
1 time) and
Simultaneous graphing of
16
t
2
76
t
75 (height against
x
2
5.5,
y
2
16
t
2
76
t
75
(the actual path of the flare). (Example 8)
Example 8 solves a projectilemotion problem. Parametric equations are used in two ways: to find a graph of the modeling equation and to simulate the motion of the projectile.
EXAMPLE 8 Simulating Projectile Motion
A distress flare is shot straight up from a ship’s bridge 75 ft above the water with an initial velocity of 76 ft sec. Graph the flare’s height against time, give the height of the flare above water at each time, and simulate the flare’s motion for each length of time.
(a)
1 sec
(b)
2 sec
(c)
4 sec
(d)
5 sec
SOLUTION
after launch is
An equation that models the flare’s height above the water
t
seconds
y
16
t
2
76
t
75.
A graph of the flare’s height against time can be found using the parametric equations
x
1
t
,
y
1
16
t
2
76
t
75.
To simulate the flare’s flight straight up and its fall to the water, use the parametric equations
x
2
5.5,
y
2
16
t
2
76
t
75.
We chose
x
2
5.5 so that the two graphs would not intersect.
Figure 6.30 shows the two graphs in simultaneous graphing mode for b 0
t
2, c 0
t
4, and d 0
t
a 0
t
1,
5. We can read that the height of the flare above the water after 1 sec is 135 ft, after 2 sec is 163 ft, after 4 sec is 123 ft, and after
5 sec is 55 ft.
Now try Exercise 39.
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CHAPTER 6
Applications of Trigonometry y v
0
y
0
v
0 cos
v
0 sin
x
FIGURE 6.31
Throwing a baseball.
[0, 450] by [0, 80]
FIGURE 6.32
The fence and path of the baseball in Example 9. See Exploration 2 for ways to draw the wall.
In Example 8 we modeled the motion of a projectile that was launched straight up.
Now we investigate the motion of objects, ignoring air friction, that are launched at angles other than 90° with the horizontal.
Suppose that a baseball is thrown from a point
y
0 tial speed of
v
0 ft sec at an angle feet above ground level with an iniwith the horizontal Figure 6.31
. The initial velocity can be represented by the vector
v
v
0 cos ,
v
0 sin .
The path of the object is modeled by the parametric equations
x v
0 cos
t
,
The
x
component is simply
y
16
t
2
v
0 sin
t y
0
.
distance
x
component of initial velocity time.
The
y
component is the familiar vertical projectilemotion equation using the
y
component of initial velocity.
EXAMPLE 9 Hitting a Baseball
Kevin hits a baseball at 3 ft above the ground with an initial speed of 150 ft an angle of 18 sec at with the horizontal. Will the ball clear a 20ft wall that is 400 ft away?
SOLUTION
The path of the ball is modeled by the parametric equations
x
150 cos 18
t
,
y
16
t
2
150 sin 18
t
3.
A little experimentation will show that the ball will reach the fence in less than 3 sec.
Figure 6.32 shows a graph of the path of the ball using the parameter interval 0
t
3 and the 20ft wall. The ball does not clear the wall.
Now try Exercise 43.
EXPLORATION EXTENSIONS
Using trial and error, determine the minimum angle, to the nearest 0.05°, such that the ball clears the fence.
30 ft
10 ft
FIGURE 6.33
Example 10.
The Ferris wheel of
A
EXPLORATION 2
Extending Example 9
1.
If your grapher has a line segment feature, draw the fence in Example 9.
2.
Describe the graph of the parametric equations
x
400,
y
20
t
3 , 0
t
3.
3.
Repeat Example 9 for the angles 19 , 20 , 21 , and 22 .
In Example 10 we see how to write parametric equations for position on a moving
Ferris wheel using time
t
as the parameter.
EXAMPLE 10 Riding on a Ferris Wheel
Jane is riding on a Ferris wheel with a radius of 30 ft. As we view it in Figure 6.33, the wheel is turning counterclockwise at the rate of one revolution every 10 sec.
Assume the lowest point of the Ferris wheel that Jane is at the point marked
A
3 o’clock
6 o’clock at time
t
is 10 ft above the ground, and
0. Find parametric equations to model Jane’s path and use them to find Jane’s position 22 sec into the ride.
continued
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 529
FOLLOWUP
Have students explain how the parametric equations in Example 10 were determined.
ASSIGNMENT GUIDE
Day 1: Ex. 1–4, 6–30, multiples of 3,
33–36
Day 2: Ex. 39–51, multiples of 3, 59–64
COOPERATIVE LEARNING
Group Activity: Ex. 55–58, 66
NOTES ON EXERCISES
Ex. 37–51 and 67–70 include a variety of interesting applications.
Ex. 53–54 relate to cycloids and hypocycloids. A Spirograph can be used to help illustrate these curves.
Ex. 59–64 provide practice with standardized tests.
ONGOING ASSESSMENT
SelfAssessment: Ex. 7, 11, 15, 23, 27,
29, 37, 39, 43, 51
Embedded Assessment: Ex. 57, 58, 65, 66
y
40
30
θ
P
A
SECTION 6.3
Parametric Equations and Motion
529
x
FIGURE 6.34
A model for the Ferris wheel of Example 10.
SOLUTION
Figure 6.34 shows a circle with center 0, 40 and radius 30 that models the Ferris wheel. The parametric equations for this circle in terms of the parameter , the central angle of the circle determined by the arc
AP
, are
x
30 cos ,
y
40 30 sin , 0 2 .
To take into account the rate at which the wheel is turning we must describe function of time sec, or 2 10
t
in seconds. The wheel is turning at the rate of 2
5 rad sec. So, 5
t
as a radians every 10
. Thus, parametric equations that model
Jane’s path are given by
x
30 cos
(
5
)
,
t y
40 30 sin
(
5
)
,
t t
0.
We substitute
t x
22 into the parametric equations to find Jane’s position at that time:
30 cos
(
5
•
22
)
y
40 30 sin
(
5
•
22
)
9.27
y
68.53
x
After riding for 22 sec, Jane is approximately 68.5 ft above the ground and approximately 9.3 ft to the right of the
y
axis using the coordinate system of Figure 6.34.
Now try Exercise 51.
(For help, go to Sections P.2, P.4, 1.3, 4.1, and 6.1.)
In Exercises 1 and 2, find the component form of the vectors
(a)
OA
,
(b)
OB
, and
(c)
AB
where
O
is the origin.
1.
A
3, 2 ,
B
4, 6
2.
A
1, 3 ,
B
4, 3
In Exercises 3 and 4, write an equation in pointslope form for the line through the two points.
3.
3, 2 , 4, 6
4.
1, 3 , 4, 3
In Exercises 5 and 6, find and graph the two functions defined implicitly by each given relation.
5.
y
2
8
x
6.
y
2
5
x
In Exercises 7 and 8, write an equation for the circle with given center and radius.
7.
0, 0 , 2
x
2
y
2
4
8.
2, 5 , 3
In Exercises 9 and 10, a wheel with radius
r
spins at the given rate.
Find the angular velocity in radians per second.
9.
r
13 in., 600 rpm
10.
r
12 in., 700 rpm
3.
4.
9.
y y
2
3
8
7
(
x
6
5
(
x
20 rad/sec
3) or
10.
y
1) or
y
7 0 rad/sec
3
6
8
7
(
x
3
6
5
4)
(
x
4)
8.
(
x
2)
2
(
y
5)
2
9
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CHAPTER 6
Applications of Trigonometry
In Exercises 1–4, match the parametric equations with their graph.
Identify the viewing window that seems to have been used.
(a) (b)
(c) (d)
1.
3.
4.
x x x
4 cos
3
t
,
y
2 sin
3
t
2.
x
3 cos
t
,
y
2 cos
t
2 cos
2 sin
t t
cos
t
,
y t
,
y
cos
2 sin
t t
sin 2
t t
sin
t
sin 2
t
In Exercises 5 and 6,
(a)
tions and
(b)
complete the table for the parametric equaplot the corresponding points.
11.
x
13.
x
14.
x
15.
x
16.
x
17.
x
18.
x
5.
x t t x y
6.
x t x y
cos
t
,
y
0
1
0
2,
y
2
0
1/2
1
2 sin
t
/2
0
1
1
1
3
t
0
2 und.
0
1
1
3
4
3
0
/2
1
2
4
5/2
2
1
0
In Exercises 7–10, graph the parametric equations
x
3
t
2
,
y
2
t
, in the specified parameter interval. Use the standard viewing window.
7.
9.
0
t
3
t
10
3
8.
10.
10
2
t t
4
0
In Exercises 11–26, eliminate the parameter and identify the graph of the parametric curve.
1
t
2
,
t
5
t
2
,
y y t
,
y
2
t
2
t
,
y
3,
3
t
,
y t t
2
3
3
2
t
3
1,
t y y t
9
2
1 [
t
[
12.
x t
4
t
, 3
,
t
1
t
5
3
Hint:
Hint:
Eliminate
t
Eliminate
t
2 3
t
,
y
and solve for
x
in terms of
y
.] and solve for
x
5
t
in terms of
y
.]
19.
x
20.
x
21.
x
22.
x
23.
x
25.
x
26.
x t t
4
0.5
t
,
t
2
,
y
3,
2,
y y y t
[
Hint:
Eliminate
t
and solve for
x
in terms of
y
.]
2
t
2
3
t
,
4
t
,
t
3,
5
2
t t
5
2
2
5 cos
2 sin
3 cos
t t
,
t
,
,
y y y
5 sin
t
24.
x
2 cos
t
, 0
3 sin
t
, 0
t
3
t
2
4 cos
t
,
y
4 sin
t
In Exercises 27– 32 find a parametrization for the curve.
27.
The line through the points
28.
The line through the points
2, 5
3, 3 and 4, 2 .
and 5, 1 .
29.
The line segment with endpoints 3, 4 and 6,
30 .
The line segment with endpoints 5, 2 and 2,
3 .
4 .
31.
The circle with center 5, 2 and radius 3.
32.
The circle with center 2, 4 and radius 2.
Exercises 33–36 refer to the graph of the parametric equations
x
2
t
,
y t
0.5, 3
t
3 given below. Find the values of the parameter
t
that produces the graph in the indicated quadrant.
[–5, 5] by [–5, 5]
33.
Quadrant I
35.
Quadrant III
0.5
t
2
3
t
2
34.
36.
Quadrant II
Quadrant IV
2
2
t
3
t
0.5
37.
Simulating a Foot Race
Ben can sprint at the rate of 24 ft sec.
Jerry sprints at 20 ft sec. Ben gives Jerry a 10ft head start. The parametric equations can be used to model a race.
x
1
x
2
20
24
t t
,
10,
y y
1
2
3
5
(a)
Find a viewing window to simulate a 100yd dash. Graph simulaneously with
t
starting at
t
0 and Tstep 0.05.
(b)
Who is ahead after 3 sec and by how much?
Ben is ahead by 2 ft.
38.
Capture the Flag
Two opposing players in “Capture the Flag” are 100 ft apart. On a signal, they run to capture a flag that is on the ground midway between them. The faster runner, however, hesitates for 0.1 sec. The following parametric equations model the race to the flag:
x
1
x
2
10
100
t
0.1
9
t
,
,
y
1
y
2
3
3
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SECTION 6.3
Parametric Equations and Motion
531
(a)
(b)
Simulate the game in a 0, 100 by 1, 10 with
t
starting at 0. Graph simultaneously.
viewing window
Who captures the flag and by how many feet?
50 ft 50 ft
39.
Famine Relief Air Drop
A relief agency drops food containers from an airplane on a wartorn famine area. The drop was made from an altitude of 1000 ft above ground level.
(a)
Use an equation to model the height of the containers (during free fall) as a function of time
t
.
y
16
t
2
1000
(b)
Use parametric mode to simulate the drop during the first 6 sec.
(c)
After 4 sec of free fall, parachutes open. How many feet above the ground are the food containers when the parachutes open?
744 ft
40.
Height of a Popup
A baseball is hit straight up from a height of 5 ft with an initial velocity of 80 ft sec.
(a)
Write an equation that models the height of the ball as a function of time
t
.
y
16
t
2
80
t
5
(b)
Use parametric mode to simulate the popup.
(c)
Use parametric mode to graph height against time. [
Hint:
Let
x t t
.]
(d)
How high is the ball after 4 sec?
69 ft
(e)
What is the maximum height of the ball? How many seconds does it take to reach its maximum height?
41.
The complete graph of the parametric equations
x
2 cos
t
,
y
sin
t
is the circle of radius 2 centered at the origin. Find an interval of values for
t
so that the graph is the given portion of the circle.
2
(a)
(b)
The portion in the first quadrant
The portion above the
x
axis
0
0
t t
2
2
t
(c)
The portion to the left of the
y
axis
3 2
42.
Writing to Learn
tions
x
0
t
3 cos
t
,
2 .
y
Consider the two pairs of parametric equa
3 sin
t
and
x
3 sin
t
,
y
3 cos
t
for
(a)
Give a convincing argument that the graphs of the pairs of parametric equations are the same.
(b)
Explain how the parametrizations are different.
43.
Hitting a Baseball
Consider Kevin’s hit discussed in Example 9.
(a)
Approximately how many seconds after the ball is hit does it hit the wall?
about 2.80 sec
(b)
How high up the wall does the ball hit?
7.18 ft
(c) Writing to Learn
Explain why Kevin’s hit might be caught by an outfielder. Then explain why his hit would likely not be caught by an outfielder if the ball had been hit at a 20° angle with the horizontal.
44.
Hitting a Baseball
Kirby hits a ball when it is 4 ft above the ground with an initial velocity of 120 ft sec. The ball leaves the bat at a 30° angle with the horizontal and heads toward a 30ft fence 350 ft from home plate.
(a)
Does the ball clear the fence?
no
(b)
If so, by how much does it clear the fence? If not, could the ball be caught?
not catchable
45.
Hitting a Baseball
Suppose that the moment Kirby hits the ball in Exercise 44 there is a 5ft sec splitsecond wind gust. Assume the wind acts in the horizontal direction out with the ball.
(a)
Does the ball clear the fence?
yes
(b)
If so, by how much does it clear the fence? If not, could the ball be caught?
1.59 ft
46.
TwoSoftball Toss
Chris and Linda warm up in the outfield by tossing softballs to each other. Suppose both tossed a ball at the same time from the same height, as illustrated in the figure. Find the minimum distance between the two balls and when this minimum distance occurs.
6.60 ft; 1.21 sec
45 ft/sec
44
°
Linda
5 ft
78 ft
39
°
41 ft/sec
Chris
47.
Yard Darts
Tony and Sue are launching yard darts 20 ft from the front edge of a circular target of radius 18 in. on the ground. If
Tony throws the dart directly at the target, and releases it 3 ft above the ground with an initial velocity of 30 ft angle, will the dart hit the target?
no sec at a 70°
48.
Yard Darts
In the game of darts described in Exercise 47, Sue releases the dart 4 ft above the ground with an initial velocity of
25 ft sec at a 55° angle. Will the dart hit the target?
yes
49.
Hitting a Baseball
Orlando hits a ball when it is 4 ft above ground level with an initial velocity of 160 ft sec. The ball leaves the bat at a 20° angle with the horizontal and heads toward a 30ft fence 400 ft from home plate. How strong must a splitsecond wind gust be (in feet per second) that acts directly with or against the ball in order for the ball to hit within a few inches of the top of the wall? Estimate the answer graphically and solve algebraically.
50.
Hitting Golf Balls
Nancy hits golf balls off the practice tee with an initial velocity of 180 ft sec with four different clubs.
How far down the fairway does the ball hit the ground if it comes off the club making the specified angle with the horizontal?
(a)
15°
506.25 ft
(b)
20°
650.82 ft
(c)
25°
775.62 ft
(d)
30°
876.85 ft
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 532
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CHAPTER 6
Applications of Trigonometry
51.
Analysis of a Ferris Wheel
Ron is on a Ferris wheel of radius
35 ft that turns counterclockwise at the rate of one revolution every
12 sec. The lowest point of the Ferris wheel (6 o’clock) is 15 ft above ground level at the point 0, 15 on a rectangular coordinate system. Find parametric equations for the position of Ron as a function of time at the point
t
(in seconds) if the Ferris wheel starts
35, 50 .
t
0 with Ron
52.
Revisiting Example 5
Eliminate the parameter
t
from the parametric equations of Example 5 to find an equation in
x
and
y
for the line. Verify that the line passes through the points
A
and
B
of the example.
y
(3 5)
x
21 5
53.
Cycloid
y
1
The graph of the parametric equations
x
cos
t
is a
cycloid
.
t
sin
t
,
[–2, 16] by [–1, 10]
(a)
What is the maximum value of
y
value related to the graph?
1 cos
t
? How is that
(b)
What is the distance between neighboring
x
intercepts?
54.
Hypocycloid
x
2 cos
t
The graph of the parametric equations cos 2
t
,
y
2 sin
t
sin 2
t
is a
hypocycloid
. The graph is the path of a point
P
on a circle of radius 1 rolling along the inside of a circle of radius 3, as illustrated in the figure.
y
3
t
C
1
P x
–3 3
–3
(a)
Graph simultaneously this hypocycloid and the circle of radius 3.
(b)
Suppose the large circle had a radius of 4. Experiment!
How do you think the equations in part (a) should be changed to obtain defining equations? What do you think the hypocycloid would look like in this case? Check your guesses.
All 2’s should be changed to 3’s.
Group Activity
In Exercises 55– 58, a particle moves along a horizontal line so that its position at any time
t
is given by description of the motion. [
Hint:
See Example 7.]
s t
. Write a
55.
56.
57.
58.
s s s s t t t t t
2
t
2
t
0.5
3
t
3
5
t
3
t
,
4
t
,
2
7
t
2
4
t
,
2
t
1
t
4
5
2
t
,
1
t
1
5
t
7
59.
True or False
y
1
3
t
1 and
The two sets of parametric equations
x
1
x
2
2 3
t
4 3,
y
2 rectangular equation. Justify your answer.
2
t t
1, correspond to the same
60.
True or False
The graph of the parametric equations
x y
2, 5
2
t
1, 1
t
3 is a line segment with endpoints
. Justify your answer.
0, 1
t
1, and
In Exercises 61–64, solve the problem without using a calculator.
61.
t
Multiple Choice
Which of the following points corresponds to
1 in the parametrization
x t
2
4,
y t
1
t
?
A
(A)
3, 2
(B)
3, 0
(C)
5, 2
(D)
5, 0
(E)
3, 2
62.
Multiple Choice
the same point as
t y
2 sin
t
?
A
Which of the following values of
2 3 in the parametrization
x t
produces
2 cos
t
,
(A)
(D)
t t
4
4
3
3
(B)
(E)
t t
2
7
3
3
(C)
t
3
63.
Multiple Choice
A rock is thrown straight up from level ground with its position above ground at any time
t x
5,
y
16
t
2 above ground?
D
80
t
0 given by
7. At what time will the rock be 91 ft
(A)
1.5 sec
(B)
2.5 sec
(C)
3.5 sec
(D)
1.5 sec and 3.5 sec
(E)
The rock never goes that high.
64.
Multiple Choice
Which of the following describes the graph of the parametric equations
x
1
t
,
y
3
t
2,
t
0?
C
(A)
a straight line
(B)
a line segment
(C)
a ray
(D)
a parabola
(E)
a circle
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 533
SECTION 6.3
Parametric Equations and Motion
533
65.
Parametrizing Circles
(a)
x a
cos
t
,
Graph the parametric equations for
a
square viewing window.
Consider the parametric equations
y a
sin
t
, 0
t
2 .
1, 2, 3, 4 in the same
(b)
Eliminate the parameter
t
in the parametric equations to verify that they are all circles. What is the radius?
Now consider the parametric equations
(c)
x h a
cos
t
,
Graph the equations for ues for
h
and
k: y a k a
sin
t
, 0
t
2 .
1 using the following pairs of val
h k
2
3
2
3
4
2
3
3
(d)
Eliminate the parameter
t
in the parametric equations and identify the graph.
(e)
Write a parametrization for the circle with center radius 3.
x
3 cos
t
1;
y
3 sin
t
4
1, 4 and
66.
Group Activity
parametrization
Parametrization of Lines
Consider the
x at b
,
y ct d
, where
a
and
c
are not both zero.
(a)
Graph the curve for
a
2,
b
3,
c
1, and
d
4,
c
1, and
d
3.
2.
(b)
Graph the curve for
a
3,
b
(c) Writing to Learn
Eliminate the parameter
t
and write an equation in
x
and
y
for the curve. Explain why its graph is a line.
(d) Writing to Learn
Find the slope,
y
intercept, and
x
intercept of the line if they exist? If not, explain why not.
(e)
Under what conditions will the line be horizontal?
Vertical?
c
0;
a
0
67.
Throwing a Ball at a Ferris Wheel
A 20ft Ferris wheel turns counterclockwise one revolution every 12 sec (see figure).
Eric stands at point
D
, 75 ft from the base of the wheel. At the instant Jane is at point
A
, Eric throws a ball at the Ferris wheel, releasing it from the same height as the bottom of the wheel. If the ball’s initial speed is 60 ft sec and it is released at an angle of
120° with the horizontal, does Jane have a chance to catch the ball? Follow the steps below to obtain the answer.
(a)
Assign a coordinate system so that the bottom car of the Ferris wheel is at 0, 0 and the center of the wheel is at 0, 20 . Then
Eric releases the ball at the point 75, 0 metric equations for Jane’s path are:
x
1
20 cos
6
t
,
y
1
20 20 sin
. Explain why para
6
t
,
t
0.
(b)
Explain why parametric equations for the path of the ball are:
x
2
30
t
75,
y
2
16
t
2
30 3
t
,
t
0.
(c)
Graph the two paths simultaneously and determine if Jane and the ball arrive at the point of intersection of the two paths at the same time.
(d)
Find a formula for the distance
d
at any time
t
.
t
between Jane and the ball
(e) Writing to Learn
Use the graph of the parametric equations
x
3
t
,
y
3
d t
, to estimate the minimum distance between
Jane and the ball and when it occurs. Do you think Jane has a chance to catch the ball?
A
20 ft
75 ft
D
68.
Throwing a Ball at a Ferris Wheel
A 71ftradius Ferris wheel turns counterclockwise one revolution every 20 sec. Tony stands at a point 90 ft to the right of the base of the wheel. At the instant Matthew is at point A (3 o’clock), Tony throws a ball toward the Ferris wheel with an initial velocity of 88 ft sec at an angle with the horizontal of 100°. Find the minimum distance between the ball and Matthew.
about 3.47 ft
69.
Two Ferris Wheels Problem
center 0, 20
Chang is on a Ferris wheel of and radius 20 ft turning counterclockwise at the rate of one revolution every 12 sec. Kuan is on a Ferris wheel of center
15, 15 and radius 15 turning counterclockwise at the rate of one revolution every 8 sec. Find the minimum distance between Chang and Kuan if both start out
t
0 at 3 o’clock.
about 4.11 ft
70.
Two Ferris Wheels Problem
Chang and Kuan are riding the
Ferris wheels described in Exercise 69. Find the minimum distance between Chang and Kuan if Chang starts out
t
0 at 3 o’clock and Kuan at 6 o’clock. about 10.48 ft
Exercises 71– 73 refer to the graph
C
of the parametric equations
x tc
where
P
1
a
,
b
and
P
2
c
,
d
1
t a
,
y td
1 are two fixed points.
t b
71.
Using Parametric Equations in Geometry
point
P x
,
y
on
C
is equal to
Show that the
72.
(a)
P
1
a
,
b
when
t
0.
(b)
P
2
c
,
d
when
t
1.
Using Parametric Equations in Geometry
Show that if
t
0.5, the corresponding point line segment with endpoints
a
,
b x
,
y
and on
C
is the midpoint of the
c
,
d
.
73.
What values of
t
will find two points that divide the line segment
P
1 into three equal pieces? Four equal pieces?
t
1
3
,
2
3
;
t
1
4
,
1
2
,
3
4
P
2
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 534
534
CHAPTER 6
Applications of Trigonometry
6.4
What you’ll learn about
■ Polar Coordinate System
■ Coordinate Conversion
■ Equation Conversion
■ Finding Distance Using Polar
Coordinates
. . . and why
Use of polar coordinates sometimes simplifies complicated rectangular equations and they are useful in calculus.
A
polar coordinate system
the
polar axis
, as shown in Figure 6.35. Each point
polar coordinates
follows:
r
P
is the
directed distance
in the plane is assigned as from
O
to
P
, and is the
directed angle
whose initial side is on the polar axis and whose terminal side is on the line
OP
.
is a plane with a point
O
, the
pole
, and a ray from
O
,
As in trigonometry, we measure as positive when moving counterclockwise and negative when moving clockwise. If
r
then
P
angle is on the terminal side of as illustrated in Example 1.
0, then
P
is on the terminal side of . If
r
0,
. We can use radian or degree measure for the
P
(
r
, )
Pole
O
θ
Polar axis
FIGURE 6.35
The polar coordinate system.
EXAMPLE 1 Plotting Points in the Polar Coordinate
System
Plot the points with the given polar coordinates.
(a)
P
2, 3
(b)
Q
1, 3 4
(c)
R
3, 45
SOLUTION
Figure 6.36 shows the three points.
Now try Exercise 7.
OBJECTIVE
Students will be able to convert points and equations from polar to rectangular coordinates and vice versa.
MOTIVATE
Ask students to suggest other methods
(besides Cartesian coordinates) of describing the location of a point on a plane.
LESSON GUIDE
Day 1: Polar Coordinate System;
Coordinate Conversion; Equation
Conversion (Polar to Rectangular)
Day 2: Equation Conversion (Rectangular to Polar); Finding Distance Using Polar
Coordinates
P
a
2,
π
3 b
3
π
4
O
2
π
3
O
1
Q
a
–1,
3
π
4 b
(b) (a)
FIGURE 6.36
The three points in Example 1.
O
–45
°
3
(c)
R
(3, –45
°
)
Each polar coordinate pair determines a unique point. However, the polar coordinates of a point
P
in the plane are not unique.
ALERT
Because of their extensive use of the
Cartesian coordinate system, many students will be surprised that the polar coordinates of a point are not unique. Emphasize the fact that neither
r
nor is uniquely defined.
EXAMPLE 2 Finding all Polar Coordinates for a Point
If the point
P
has polar coordinates 3, 3 , find all polar coordinates for
P
.
SOLUTION
nates for
P
are
Point
P
is shown in Figure 6.37. Two additional pairs of polar coordi
3,
(
3
2
)
3,
(
7
3
) and
(
3,
3
) (
3,
4
3
)
.
continued
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 535
SECTION 6.4
Polar Coordinates
535
P
a
3,
π
3 b
4
π
3
O
3
π
3
FIGURE 6.37
The point
P
in Example 2.
We can use these two pairs of polar coordinates for
P
to write the rest of the possibilities:
(
3,
3
(
3,
Where
n
is any integer.
3
2
n
1
2
n
)
) (
(
3,
3,
6
n
6
n
3
1
3
4
)
) or
Now try Exercise 23.
The coordinates
r
, ,
r
, 2 , and eral, the point with polar coordinates
r
,
r
, all name the same point. In genalso has the following polar coordinates:
Finding all Polar Coordinates of a Point
Let
P
have polar coordinates form
r
, . Any other polar coordinate of
P
must be of the
r
, 2
n
or
r
, 2
n
1 where
n
is any integer. In particular, the pole has polar coordinates is any angle.
0, , where
y
Pole
O
(0, 0)
r y
θ
x
Polar axis
P
(
r
, )
P
(
x
,
y
)
x
FIGURE 6.38
coordinates for
P
.
Polar and rectangular
When we use both polar coordinates and Cartesian coordinates, the pole is the origin and the polar axis is the positive
x
axis as shown in Figure 6.38. By applying trigonometry we can find equations that relate the polar coordinates
r
, and the rectangular coordinates
x
,
y
of a point
P
.
Coordinate Conversion Equations
Let the point
P x
,
y
. Then have polar coordinates
r
,
x y r
cos ,
r
sin ,
r
2 and rectangular coordinates
x
2 tan
y x
.
y
2
,
These relationships allow us to convert from one coordinate system to the other.
EXAMPLE 3 Converting from Polar to Rectangular Coordinates
Find the rectangular coordinates of the points with the given polar coordinates.
(a)
P
3, 5 6
(b)
Q
2, 200
continued
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 536
536
CHAPTER 6
Applications of Trigonometry y
P
a
3,
5
π
6 b
3
5
π
6
(a)
y
Q
(2, –200
°
)
2
x
–200
°
(b)
FIGURE 6.39
Example 3.
The points
P
and
Q
in
NOTES ON EXAMPLES
Example 4 provides an opportunity to monitor students’ use of the inverse keys on their graphers. You may need to assist some students in the correct choice of the quadrant for this example.
y x
SOLUTION
(a)
For
P
3, 5 6 ,
r x x r
cos
3 cos
5
6
x
(
3
2
)
3
3 and
2.60
5 6: and
y y y r
sin
3 sin
5
6
3
2
(
1
)
1.5
The rectangular coordinates for
P
are 3 3 2, 1.5
2.60, 1.5
Figure 6.39a
.
(b)
For
Q
2, 200 ,
r
2 and 200 :
x r
cos
200
y r
sin
x
2 cos 1.88
and
y
The rectangular coordinates for
Q
are approximately
2 sin 200 0.68
1.88, 0.68
Figure 6.39b
.
Now try Exercise 15.
When converting rectangular coordinates to polar coordinates, we must remember that there are infinitely many possible polar coordinate pairs. In Example 4 we report two of the possibilities.
EXAMPLE 4
Converting from Rectangular to Polar
Coordinates
Find two polar coordinate pairs for the points with given rectangular coordinates.
(a)
P
1, 1
(b)
Q
3, 0
SOLUTION
(a)
For
P
1, 1 ,
x
1 and
y r
2
x
2
y
2
1: tan
y x r r
2
1
2
2
1
2 and tan
1
1
1 tan
1
1
n
4
n
P
(–1, 1)
2
π
+ tan
–1
(–1) =
3
π
4 tan
–1
(–1) = –
x
π
4
FIGURE 6.40
The point
P
in Example 4a.
2 ,
3
4
)
.
(b)
For
Q
3, 0 ,
x
3 and
y
0. Thus,
r
3 and and . So two polar coordinates pairs for point
Q
are
3, 0 and 3, .
n
. We use the angles 0
Now try Exercise 27.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 537
SECTION 6.4
Polar Coordinates
537
EXPLORATION EXTENSIONS
Describe how your grapher chooses what values to give when converting rectangular coordinates to polar coordinates. For example, according to your grapher, what are the possible values for
r
and for ?
FOLLOWUP
Ask whether it is possible for two polar equations that are not algebraically equivalent to have identical graphs.
(Yes)
[–4.7, 4.7] by [–3.1, 3.1]
FIGURE 6.41
equation
r
The graph of the polar
4 cos in 0 2 .
[–2, 8] by [–10, 10]
FIGURE 6.42
line
r
4 sec (
x
The graph of the vertical
4). (Example 5)
EXPLORATION 1
Using a Grapher to Convert Coordinates
Most graphers have the capability to convert polar coordinates to rectangular coordinates and vice versa. Usually they give just one possible polar coordinate pair for a given rectangular coordinate pair.
1.
Use your grapher to check the conversions in Examples 3 and 4.
2.
Use your grapher to convert the polar coordinate pairs
1, 2 , 2, , 5, 3 2 , 3, 2
2, 3 ,
, to rectangular coordinate pairs.
(1, 3 ), (0, 1), ( 2, 0), (0, 5), (3, 0)
3.
Use your grapher to convert the rectangular coordinate pairs
1, pairs.
(
3
2,
, 0, 2 , 3, 0 ,
3), (2, 2), (3, 0), (1,
1, 0 , 0,
), (4, 3 2)
4 to polar coordinate
We can use the Coordinate Conversion Equations to convert polar form to rectangular form and vice versa. For example, the polar equation
r
4 cos can be converted to rectangular form as follows:
r r
2
4 cos
4
r
cos
x
2
y
2
4
x r
2
x
2
y
2
,
r
cos
x x
2
4
x
4
y
2
4
Subtract 4
x
and add 4.
x
2
2
y
2
4
Factor.
Thus the graph of
r
4 cos is all or part of the circle with center 2, 0 and radius 2.
Figure 6.41 shows the graph of
r
4 cos for 0 graphing mode of our grapher. So, the graph of
r
2
4 cos obtained using the polar is the entire circle.
Just as with parametric equations, the domain of a polar equation in
r
stood to be all values of for which the corresponding values of
r
and is underare real numbers.
You must also select a value for min and max to graph in polar mode.
You may be surprised by the polar form for a vertical line in Example 5.
EXAMPLE 5
Converting from Polar Form to Rectangular Form
Convert
r
4 sec to rectangular form and identify the graph. Support your answer with a polar graphing utility.
SOLUTION
r
4 sec se
r
c
4
r
cos 4
Divide by sec .
cos se
1 c
.
x
4
r
cos
x
The graph is the vertical line
x
4 Figure 6.42
.
Now try Exercise 35.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 538
538
CHAPTER 6
Applications of Trigonometry
[–5, 10] by [–2, 8]
r
FIGURE 6.43
6 cos
The graph of the circle
4 sin . (Example 6)
y
(8, 110
°
)
(5, 15
°
)
x
FIGURE 6.44
The distance and direction of two airplanes from a radar source. (Example 7)
ASSIGNMENT GUIDE
Day 1: Ex. 1–30, multiples of 3
Day 2: Ex. 33–54, multiples of 3, 55–60
COOPERATIVE LEARNING
Group Activity: Ex. 68–71
NOTES ON EXERCISES
Ex. 23–30 emphasize the fact that the polar coordinates of a point are not unique.
EXAMPLE 6
Convert
x
3
2
y
Converting from Rectangular Form to Polar Form
2
2
13 to polar form.
SOLUTION
x
2
6
x x
9
3
2
y
2
y
4
y
2
2
4
13
13
x
2
y
2
6
x
4
y
0
Substituting
r
2 for
x
2
r
0 or
r y
2
,
r
cos for
x
, and
r
sin for
y
gives the following:
2
6
r
cos 4
r
sin 0
r r r
6 cos
6 cos
4 sin
4 sin
0
0
The graph of of
r
6 cos
r
0 consists of a single point, the origin, which is also on the graph
4 sin 0. Thus, the polar form is
r
6 cos 4 sin .
The graph of
r
6 cos 4 sin appears to be a circle with center 3, 2 for 0 and radius
2 is shown in Figure 6.43 and
1 3 , as expected.
Now try Exercise 43.
A radar tracking system sends out highfrequency radio waves and receives their reflection from an object. The distance and direction of the object from the radar is often given in polar coordinates.
EXAMPLE 7
Using a Radar Tracking System
Radar detects two airplanes at the same altitude. Their polar coordinates are 8 mi, 110 and 5 mi, 15 . See Figure 6.44.
How far apart are the airplanes?
SOLUTION
By the Law of Cosines Section 5.6
,
d
2
8
2
5
2
2
•
8
•
5 cos 110 15
d
8
2
5
2
2
•
8
•
5 c o s 9 5
d
9.80
The airplanes are about 9.80 mi apart.
Now try Exercise 51.
We can also use the Law of Cosines to derive a formula for the distance between points in the polar coordinate system. See Exercise 61.
Ex. 55–60 provide practice with standardized tests.
Ex. 67–71 show the connection between polar equations and parametric equations.
ONGOING ASSESSMENT
SelfAssessment: Ex. 7, 15, 23, 27, 35,
43, 51
Embedded Assessment: Ex. 61
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 539
SECTION 6.4
Polar Coordinates
539
(For help, go to Sections P.2, 4.3, and 5.6.)
In Exercises 1 and 2, determine the quadrants containing the terminal side of the angles.
1. (a)
5 6
2. (a)
300
II
I
(b)
(b)
3 4
210 III
III
In Exercises 3 – 6, find a positive and a negative angle coterminal with the given angle.
3.
5.
4 7 4,
160 520 ,
9 4
200
4.
6.
3
120
7 3,
240 ,
5 3
480
In Exercises 7 and 8, write a standard form equation for the circle.
7.
Center 3, 0 and radius 2
8.
Center 0, 4 and radius 3
In Exercises 9 and 10, use The Law of Cosines to find the measure of the third side of the given triangle.
9.
11.14
10.
5.85
12
9
40
°
6
60
°
10
7.
(
x
3)
2
y
2
4
8.
x
2
(
y
4)
2
9
In Exercises 1– 4, the polar coordinates of a point are given. Find its rectangular coordinates.
1.
y
3
2
,
3
2
3
2.
y
a
3,
2
π
3 b
(2 2 , 2 a
–4,
5
π
4 b
2 )
x x
3.
(–2, 60
°
)
y
( 1, 3 )
x
4.
(–1, 315
°
)
y
2
2
,
2
2
x
In Exercises 5 and 6,
(b)
(a)
complete the table for the polar equation and plot the corresponding points.
5.
r
3 sin
4
r
3 2 2
2 5 6
3 3 2 0
4 3
3
2
3 2 0
6.
r
2 csc
4
r
2 2
2
2
5 6
4 und.
4 3
4
2
3 3 und.
In Exercises 7–14, plot the point with the given polar coordinates.
7.
10.
13.
3, 4 3
3, 17 10
2, 120
8.
11.
14.
2, 5
2, 30
6
3, 135
9.
12.
1, 2 5
3, 210
In Exercises 15 – 22, find the rectangular coordinates of the point with given polar coordinates.
15.
17.
19.
21.
1.5, 7
3,
3
29
2,
7
(2, 0)
2, 270
(0, 2)
16.
18.
20.
22.
2.5, 17
2,
4
14
1, 2
5
(0, 1)
(1.62, 1.18)
3, 360
( 3, 0)
In Exercises 23 – 26, polar coordinates of point
P
are given. Find all of its polar coordinates.
23.
25.
P
P
2,
1.5,
6
20
24.
26.
P
P
1, 4
2.5, 50
In Exercises 27– 30, rectangular coordinates of point
P
are given. Find all polar coordinates of
P
that satisfy
(a)
0 2
27.
P
1, 1
29.
P
2, 5
(b) (c)
0
28.
30.
P
P
1, 3
1,
4
2
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540
CHAPTER 6
Applications of Trigonometry
In Exercises 31– 34, use your grapher to match the polar equation with its graph.
(a) (b)
31.
33.
r r
(c)
5 csc
4 cos 3
(b)
(c)
32.
34.
r r
(d)
4 sin
4 sin 3
(d)
(a)
In Exercises 35 – 42, convert the polar equation to rectangular form and identify the graph. Support your answer by graphing the polar equation.
35.
r
38.
41.
r r
3 sec
4 cos
36.
39.
2 sin 4 cos
r
2 csc
r
csc 1
42.
r
37.
40.
r r
3 sin sec 3
4 cos 4 sin
In Exercises 43 – 50, convert the rectangular equation to polar form.
Graph the polar equation.
43.
x
45.
47.
49.
2
x x x
2
3
3
3
y
2
2
5
y
2
y
9
3
2
18
44.
46.
48.
50.
x
3
x x x
2
5
4
1
y y
2
1
2
2
y
1
4
2
17
51.
Tracking Airplanes
2 3 3.46 mi
The location, given in polar coordinates, of two planes approaching the Vicksburg airport are
4 mi, 12 and 2 mi, 72 . Find the distance between the airplanes.
52.
Tracking Ships
The location of two ships from Mays Landing
Lighthouse, given in polar coordinates, are 3 mi, 170 and 5 mi,
150 . Find the distance between the ships.
2.41 mi
53.
Using Polar Coordinates in Geometry
A square with sides of length
a
and center at the origin has two sides parallel to the
x
axis. Find polar coordinates of the vertices.
54.
Using Polar Coordinates in Geometry
A regular pentagon whose center is at the origin has one vertex on the positive
x
axis at a distance
a
from the center. Find polar coordinates of the vertices.
55.
True or False
Every point in the plane has exactly two polar coordinates. Justify your answer.
56.
True or False
are not 0, and if represent the same point in the plane, then
r
1 answer.
If
r
1 and
r
2
r
1
,
r
and
2
r
2
,
. Justify your
In Exercises 57–60, solve the problem without using a calculator.
57.
Multiple Choice
If
r
0, which of the following polar coordinate pairs represents the same point as the point with polar coordinates
r
, ?
C
(A)
r
,
(B)
r
,
(C)
r
, 3
(D)
r
,
(E)
r
, 3
2
58.
Multiple Choice
Which of the following are the rectangular coordinates of the point with polar coordinate 2, 3 ?
C
(A)
3
, 1
(B)
1,
3
(C)
1,
3
(D)
1,
3
(E)
1,
3
59.
Multiple Choice
Which of the following polar coordinate pairs represent the same point as the point with polar coordinates 2, 110 ?
A
(A)
2, 70
(B)
2, 110
(C)
2, 250
(D)
60.
Multiple Choice
Which of the following polar coordinate pairs does
not
represent the point with rectangular coordinates 2, 2 ?
E
(A)
(D)
2,
2
70
2
2 , 135
2 , 45
(E)
(B)
(E)
2
2, 290
2
2 , 225
2 , 135
(C)
2 2 , 315
61.
Polar Distance Formula
r
1
,
1 and
r
2
,
2
Let
P
1
, respectively.
(a)
and
P
2 have polar coordinates
If
1 2 between
P
1 is a multiple of , write a formula for the distance and
P
2
.
(b)
Use the Law of Cosines to prove that the distance between
P
1 and
P
2 is given by
d r
2
1
r
2
2
2
r
1
r
2 cos (
1 2
)
(c) Writing to Learn
the formula s
Does the formula in part you found in part a ? Explain.
b agree with
62.
Watching Your Step
Consider the polar curve
Describe the graph for each of the following.
r
(a)
0
(c)
0 3
2
2
(b)
(d)
0
0
3
4
4
4 sin .
In Exercises 63– 66, use the results of Exercise 61 to find the distance between the points with given polar coordinates.
63.
65.
2, 10
3, 25
,
,
5, 130
5, 160
6.24
7.43
64.
66.
4, 20
6, 35
, 6, 65
, 8, 65
4.25
4.11
67.
Graphing Polar Equations Parametrically
ric equations for the polar curve
r f
.
Find paramet
Group Activity
In Exercises 68– 71, use what you learned in
Exercise 67 to write parametric equations for the given polar equation.
Support your answers graphically.
68.
70.
r r
2 cos
2 sec
69.
71.
r r
5 sin
4 csc
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 541
SECTION 6.5
Graphs of Polar Equations
541
6.5
What you’ll learn about
■ Polar Curves and Parametric
Curves
■ Symmetry
■ Analyzing Polar Curves
■ Rose Curves
■ Limaçon Curves
■ Other Polar Curves
. . . and why
Graphs that have circular or cylindrical symmetry often have simple polar equations, which is very useful in calculus.
Polar curves are actually just special cases of parametric curves. Keep in mind that polar curves are graphed in the (
x
, of
r y
) plane, despite the fact that they are given in terms and . That is why the polar graph of
r
4 cos is a circle (see Figure 6.41 in
Section 6.4) rather than a cosine curve.
In function mode, points are determined by a vertical coordinate that changes as the horizontal coordinate moves left to right. In polar mode, points are determined by a directed distance from the pole that changes as the angle sweeps around the pole. The connection is provided by the Coordinate Conversion Equations from Section 6.4, which show that the graph of
r f
( ) is really just the graph of the parametric equations
x f
( ) cos
y f
( ) sin for all values of in some parameter interval that suffices to produce a complete graph.
(In many of our examples, 0 2 will do.)
Since modern graphing calculators produce these graphs so easily in polar mode, we are frankly going to assume that you do not have to sketch them by hand. Instead we will concentrate on analyzing the properties of the curves. In later courses you can discover further properties of the curves using the tools of calculus.
OBJECTIVE
Students will be able to graph polar equations and determine the maximum
r
value and the symmetry of a graph.
MOTIVATE
Ask students to use a grapher to compare the graphs of the polar equations
r
tan and
r
tan for 0
(The graphs are identical.)
2 .
LESSON GUIDE
Day 1: Polar Curves and Parametric
Curves; Symmetry; Analyzing Polar
Graphs; Rose Curves
Day 2: Limaçon Curves; Other Polar
Curves
You learned algebraic tests for symmetry for equations in rectangular form in Section
1.2. Algebraic tests also exist for polar form.
Figure 6.45 on the next page shows a rectangular coordinate system superimposed on a polar coordinate system, with the origin and the pole coinciding and the positive
x
axis and the polar axis coinciding.
The three types of symmetry figures to be considered will have are:
1.
2.
3.
The
x
axis (polar axis) as a line of symmetry (Figure 6.45a).
The
y
axis (the line 2) as a line of symmetry (Figure 6.45b).
The origin (the pole) as a point of symmetry (Figure 6.45c).
All three algebraic tests for symmetry in polar forms require replacing the pair
r
, which satisfies the polar equation, with another coordinate pair and determining
, whether it also satisfies the polar equation.
TEACHING NOTE
When determining symmetry, sometimes only one of the replacements will appear to produce an equivalent polar equation. For example, the graph of
r
is symmetric about the polar axis, but at first glance this equation does not appear to be equivalent to
r
. Although the pairs (
r
, ) that solve each equation are different, the graphs of these two equations are the same.
Symmetry Tests for Polar Graphs
The graph of a polar equation has the indicated symmetry if either replacement produces an equivalent polar equation.
To Test for Symmetry Replace By
1.
2.
3.
about the about the
x y
axis,
axis, about the origin,
r
,
r
,
r
,
r
,
r
,
r
, or or or
r r
,
r
,
,
.
.
.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 542
542
CHAPTER 6
Applications of Trigonometry y y y
(
r
, )
(
r
,
π
–
θ
) = (–
r
, – ) (
r
, )
(
r
, )
θ
–
θ
x
θ
x
θ
x
(
r
, – ) = (–
r
,
π
–
θ
)
(–
r
, ) = (
r
,
θ
+
π
)
(a)
(b)
FIGURE 6.45
Symmetry with respect to (a) the
x
axis (polar axis), (b) the
y
axis (the line
(c)
2), and (c) the origin (the pole).
[–6,6] by [–4, 4]
FIGURE 6.46
The graph of
r
4 sin 3 is symmetric about the
y
axis. (Example 1)
TEACHING NOTE
It is customary to draw graphs of polar curves in radian mode.
EXAMPLE 1
Testing for Symmetry
Use the symmetry tests to prove that the graph of
r y
axis.
4 sin 3 is symmetric about the
SOLUTION
Figure 6.46 suggests that the graph of
r
the
y
axis and not symmetric about the
x
axis or origin.
4 sin 3 is symmetric about
r
4 sin 3
r r
4 sin 3
4 sin
r
4 sin 3
r
4 sin 3
3
Because the equations
r
symmetry about the
y
axis.
Replace
r
, by
(Same as original.)
4 sin 3 and
r r
, .
sin is an odd function of .
4 sin 3 are equivalent, there is
Now try Exercise 13.
ALERT
Students are used to specifying different viewing windows to control what they see on the graphing screen. On polar graphs, students should be careful to note their selection of the range values for
x
,
y
, and
. Since the environment into which the polar coordinates are patched is the rectangular system, students need to be concerned about controlling both the viewing window and the polar environment.
Encourage students to experiment with different input values for both the polar function
r
and the input variable and to observe the effects on the graphs.
Analyzing Polar Graphs
We analyze graphs of polar equations in much the same way that we analyze the graphs of rectangular equations. For example, the function
r
function of . Also
r
0 when 0 and when of Example 1 is a continuous is any integer multiple of 3. The domain of this function is the set of all real numbers.
Trace can be used to help determine the range of this polar function (Figure 6.47). It can be shown that 4
r
4.
Usually, we are more interested in the maximum value of in polar equations. In this case,
r r
rather than the range of
4 so we can conclude that the graph is bounded.
r
A maximum value for
r
is a
maximum
r
value
for a polar equation. A maximum
r
value occurs at a point on the curve that is the maximum distance from the pole. In
Figure 6.47, a maximum maximum
r r
value at every
value occurs at
r
,
4, 6 and 4, 2 . In fact, we get a which represents the tip of one of the three petals.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 543
1
SECTION 6.5
Graphs of Polar Equations
543
1
r
= 2 + 2 cos
θ
[–4.7, 4.7] by [–3.1, 3.1]
Polar coordinates
(a)
y
= 2 + 2 cos
x
[0, 2
π
] by [–4, 4]
Rectangular coordinates
(b)
FIGURE 6.48
With
x
, the
y
values in
(b) are the same as the directed distance from the pole to (
r
, ) in (a).
R=4
θ
=.52359878
[–6,6] by [–5, 3]
(a)
FIGURE 6.47
The values of
r
in
r
R= 4
θ
=1.5707963
[–6,6] by [–5, 3]
(b)
4 sin 3 vary from (a) 4 to (b) 4.
To find maximum
r
values we must find maximum values of
r
as opposed to the directed distance
r
. Example 2 shows one way to find maximum
r
values graphically.
EXAMPLE 2
Finding Maximum
r
Values
Find the maximum
r
value of
r
2 2 cos .
SOLUTION
Figure 6.48a shows the graph of
r
2 2 cos for 0 2 .
Because we are only interested in the values of
r
, we use the graph of the rectangular equation
y
2 2 cos
x
in function graphing mode (Figure 6.48b). From this graph we can see that the maximum value of
r
, or
y
, is 4. It occurs when is any multiple of 2 .
Now try Exercise 21.
EXAMPLE 3
Finding Maximum
r
Values
Identify the points on the graph of
r r
values.
3 cos 2 for 0 2 that give maximum
SOLUTION
the graph of
r
Using trace in Figure 6.49 we can show that there are four points on
3 cos 2 in 0 2 at maximum distance of 3 from the pole:
3, 0 , 3, 2 , 3, , and 3, 3 2 .
Figure 6.50a shows the directed distances
Figure 6.50b shows the distances
r
as the
y r
as the
y
values of
y
1
values of
y
2
3 cos 2
x
maximum values of
y
1
(i.e.,
r
) in part (a).
y
2
(i.e.,
r
3 cos 2
x
, and
. There are four
) in part (b) corresponding to the four extreme values of
Now try Exercise 23.
a
–3,
3
π
2 b
(3,
π
)
y
Maximum
r
values
Maximum
r
values
(3, 0)
x
a
–3,
π
2 b
FIGURE 6.49
(Example 3)
The graph of
r
3 cos 2 .
[0, 2
π
] by [–5, 5]
(a)
FIGURE 6.50
The graph of (a)
y
1 mode. (Example 3)
[0, 2
π
] by [–5, 5]
(b)
3 cos 2
x
and (b)
y
2
3 cos 2
x
in function graphing
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 544
544
CHAPTER 6
Applications of Trigonometry
ALERT
Some students will find the maximum value of
r
instead of the maximum value of
r
. Emphasize that we are looking for the maximum distance from the pole, and
r
may be either positive or negative at this point.
[–4.7, 4.7] by [–3.1, 3.1]
FIGURE 6.51
curve
r
The graph of 8petal rose
3 sin 4 . (Example 4)
TEACHING NOTE
Polar graphs are an invitation for students to explore mathematics. Students will be able to produce elaborate graphs using polar graphing techniques. Encourage students to graph different modifications of
r a b
cos to see the effects of their choices.
The curve in Example 1 is a 3petal rose curve and the curve in Example 3 is a 4petal rose curve. The graphs of the polar equations
r a
cos
n
and
r a
sin
n
, where
n
is an integer greater than 1, are
rose curves
. If
n
is odd there are
n
petals, and if
n
is even there are 2
n
petals.
EXAMPLE 4
Analyzing a Rose Curve
Analyze the graph of the rose curve
r
3 sin 4 .
SOLUTION
Figure 6.51 shows the graph of the 8petal rose curve
r
3 sin 4 . The maximum
r
value is 3. The graph appears to be symmetric about the
x
axis,
y
axis, and the origin. For example, to prove that the graph is symmetric about the
x
axis we replace
r
, by
r
, :
r
3 sin 4
r r r
3 sin 4
3 sin 4 4
3 sin 4 cos 4
r r
3 0 cos 4
r
3 sin 4
3 sin 4
1 cos 4 sin 4 sin 4
Sine difference identity sin 4 0, cos 4 1
Because the new polar equation is the same as the original equation, the graph is symmetric about the
x
axis. In a similar way, you can prove that the graph is symmetric about the
y
axis and the origin. (See Exercise 58.)
Domain: All reals.
Range: 3, 3
Continuous
Symmetric about the
x
axis, the
y
axis, and the origin.
Bounded
Maximum
r
value: 3
No asymptotes.
Now try Exercise 29.
Here are the general characteristics of rose curves. You will investigate these curves in more detail in Exercises 67 and 68.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 545
SECTION 6.5
Graphs of Polar Equations
545
A ROSE IS A ROSE…
Budding botanists like to point out that the rose curve doesn’t look much like a rose.
However, consider the beautiful stainedglass window shown here, which is a feature of many great cathedrals and is called a “rose window.”
Graphs of Rose Curves
The graphs of
r a
cos following characteristics:
n
and
r a
sin
n
, where
n
Domain: All reals
Range:
a
,
a
Continuous
Symmetry:
n
even, symmetric about
x
,
n
odd,
r y
axis, origin
a
cos
n
symmetric about
x
axis
n
odd,
r a
sin
n
symmetric about
y
axis
Bounded
Maximum
r
value:
a
No asymptotes
Number of petals:
n
, if
n
is odd
2
n
, if
n
is even
1 is an integer, have the
The
limaçon curves
r
are graphs of polar equations of the form
a b
sin and
r a b
cos , where
a
0 and
b
0.
Limaçon
, pronounced “LEEmasohn,” is Old French for “snail.” There are four different shapes of limaçons, as illustrated in Figure 6.52.
Limaçon with an inner loop:
a b
< 1
(a)
Cardioid:
a b
= 1
(b)
FIGURE 6.52
The four types of limaçons.
1
R=6
θ
=4.712389
[–7, 7] by [–8, 2]
FIGURE 6.53
of Example 5.
The graph of the cardioid
Dimpled limaçon: 1 <
a b
< 2
(c)
Convex limaçon:
a b
≥
2
(d)
EXAMPLE 5
Analyzing a Limaçon Curve
Analyze the graph of
r
3 3 sin .
SOLUTION
We can see from Figure 6.53 that the curve is a cardioid with maximum
r
value 6. The graph is symmetric only about the
y
axis.
Domain: All reals.
Range: 0, 6
Continuous
Symmetric about the
y
axis.
Bounded
Maximum
r
value: 6
No asymptotes.
Now try Exercise 33.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 546
546
CHAPTER 6
Applications of Trigonometry
1
R=5
θ
=0
[–3, 8] by [–4, 4]
FIGURE 6.54
The graph of a limaçon with an inner loop. (Example 6)
EXAMPLE 6
Analyzing a Limaçon Curve
Analyze the graph of
r
2 3 cos .
SOLUTION
We can see from Figure 6.54 that the curve is a limaçon with an inner loop and maximum
r
value 5. The graph is symmetric only about the
x
axis.
Domain: All reals.
Range: 1, 5
Continuous
Symmetric about the
x
axis.
Bounded
Maximum
r
value: 5
No asymptotes.
Now try Exercise 39.
Graphs of Limaçon Curves
The graphs of
r a b
sin and
r
the following characteristics:
a b
cos , where
Domain: All reals
Range:
a b
,
a b
Continuous
Symmetry:
r r
Bounded
Maximum
r
value:
a b
sin , symmetric about
y
axis
a b
cos , symmetric about
x
axis
a b
No asymptotes
a
0 and
b
0, have
[–30, 30] by [–20, 20]
(a)
[–30, 30] by [–20, 20]
(b)
FIGURE 6.55
(a) step max
0 (set min
0, step
The graph of
0, max
r
45,
0.1). (Example 7) for
0.1) and (b) 0 (set min 45,
EXPLORATION 1
Limaçon Curves
Try several values for
a
and
b
to convince yourself of the characteristics of limaçon curves listed above.
All the polar curves we have graphed so far have been bounded. The spiral in Example 7 is unbounded.
EXAMPLE 7
Analyzing the Spiral of Archimedes
Analyze the graph of
r
.
SOLUTION
We can see from Figure 6.55 that the curve has no maximum
r
value and is symmetric about the
y
axis.
Domain: All reals.
Range: All reals.
Continuous
Symmetric about the
y
axis.
Unbounded
No maximum
r
value.
No asymptotes.
Now try Exercise 41.
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 547
SECTION 6.5
Graphs of Polar Equations
547
[–4.7, 4.7] by [–3.1, 3.1]
FIGURE 6.56
r
2
The graph of the lemniscate
4 cos 2 . (Example 8)
FOLLOWUP
Have students try to confirm the
x
axis symmetry of Example 4 by using the
( replacement (
r
,
r
, ) instead of using
). Discuss the results.
ASSIGNMENT GUIDE
Day 1: Ex. 1–8 all, 15–30, multiples of 3
Day 2: Ex. 9–12, 33–48, multiples of 3, 58,
61–65
COOPERATIVE LEARNING
Group Activity: Ex. 57
NOTES ON EXERCISES
Ex. 45–48 require students to find the distance from the pole to the furthest point on the petal, not the arc length.
Ex. 58–60 provide additional discussion of examples in the text.
Ex. 61–66 provide practice with standardized tests.
ONGOING ASSESSMENT
SelfAssessment: Ex. 13, 21, 23, 29, 33,
39, 41, 43
Embedded Assessment: Ex. 59, 60, 73
The
lemniscate curves
are graphs of polar equations of the form
r
2
a
2 sin 2 and
r
2
a
2 cos 2 .
EXAMPLE 8
Analyzing a Lemniscate Curve
Analyze the graph of
r
2
4 cos 2 for 0, 2 .
SOLUTION
It turns out that you can get the complete graph using
r
You also need to choose a very small
2 cos 2 step to produce the graph in Figure 6.56.
.
Domain:
Range:
0,
2, 2
4 3 4, 5 4 7 4, 2
Symmetric about the
x
axis, the
y
axis, and the origin.
Continuous (on its domain)
Bounded
Maximum
r
value: 2
No asymptotes.
Now try Exercise 43.
EXPLORATION 2
Revisiting Example 8
1.
2.
3.
4.
5.
Prove that values in the intervals in the domain of the polar equation
r
Explain why
r
in the interval
2
0, 2 .
2
4, 3 4 and
4 cos 2 .
5 4, 7 co s 2 produces the same graph as
r
Use the symmetry tests to show that the graph of metric about the
x
axis.
r
2
4
2 are not co s
4 cos 2 is sym
2
Use the symmetry tests to show that the graph of metric about the
y
axis.
r
2
4 cos 2 is sym
Use the symmetry tests to show that the graph of metric about the origin.
r
2
4 cos 2 is sym
EXPLORATION EXTENSIONS
Graph
r
2
4 sin 2 . How is this graph related to the graph of
r
2
4 cos 2 ?
(For help, go to Sections 1.2 and 5.3
.
)
In Exercises 1 – 4, find the absolute maximum value and absolute minimum value in 0, 2 and where they occur.
1.
3.
y y
3 cos 2
x
2 co s 2
x
2.
4.
y y
2
3
3 cos
3 sin
x x
In Exercises 5 and 6, determine if the graph of the function is symmetric about the
(a)
x
axis,
(b)
y
axis, and
(c)
origin.
5.
y
sin 2
x
no; no; yes
6.
y
cos 4
x
no; yes; no
In Exercises 7–10, use trig identities to simplify the expression.
7.
sin
9.
cos 2 sin cos
2 sin
2
8.
cos
10.
sin 2 cos
2 sin cos
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 548
548
CHAPTER 6
Applications of Trigonometry
In Exercises 1 and 2,
(a)
complete the table for the polar equation, and
(b)
plot the corresponding points.
1.
r r
3 cos 2
0 4
3 0
2
3
3
0
4
3
5
0
4 3
3
2 7
0
4
2.
r r
2 sin 3
0
0
6
2 0
3 2
2
2
0
3 5
2
6
0
In Exercises 3 – 6, draw a graph of the rose curve. State the smallest
interval 0
k
that will produce a complete graph.
3.
5.
r r
3 sin 3
3 cos 2
4.
6.
r r
3 cos 2
3 sin 5
Exercises 7 and 8 refer to the curves in the given figure.
[–4.7, 4.7] by [–3.1, 3.1]
(a)
[–4.7, 4.7] by [–3.1, 3.1]
(b)
7.
The graphs of which equations are shown?
r
3
r
1
3 cos 6
r
2
3 sin 8
r
3
3 is graph (b).
cos 3
8.
Use trigonometric identities to explain which of these curves is the graph of
r
6 cos 2 sin 2 .
(a)
In Exercises 9 – 12, match the equation with its graph without using your graphing calculator.
[–4.7, 4.7] by [–4.1, 2.1]
(a)
[–4.7, 4.7] by [–3.1, 3.1]
(b)
9.
10.
11.
Does the graph of the figure? Explain.
Does the graph of
r r
the figure? Explain.
2 2 sin or
r
2 2 cos appear in
Graph (b) is
r
2 2 cos .
2 3 cos or
r
Graph (c) is
r
2
Is the graph in (a) the graph of
Explain.
Graph (a) is
r
2
r
2 sin .
2
2
3 cos .
3 cos appear in
2 sin or
r
2 2 cos ?
12.
Is the graph in (d) the graph of
r
sin ? Explain.
Graph (d) is
r
2
2 1.5 cos
1.5 sin .
or
r
2 1.5
In Exercises 13– 20, use the polar symmetry tests to determine if the graph is symmetric about the
x
axis, the
y
axis, or the origin.
13.
15.
17.
r r r
3
4
3 sin
3 cos
5 cos 2
19.
r
1
3 sin
14.
16.
18.
20.
r r r r
1
1
7 sin 3
1
2 cos
3 sin
2 cos
In Exercises 21– 24, identify the points for 0 2 where maximum
r
values occur on the graph of the polar equation.
21.
23.
r r
2 3 cos
3 cos 3
22.
24.
r
3
r
2 sin
4 sin 2
25.
r
27.
29.
r
31.
r
33.
r
35.
r
37.
r
39.
r
41.
r
43.
r
2
In Exercises 25 – 44, analyze the graph of the polar curve.
3
3
2 sin 3
5
4
4 sin
4 cos
5
2
1
2 cos
5 cos cos
2 sin 2 , 0 2
26.
28.
30.
32.
34.
36.
38.
40.
42.
44.
r
2
4
r
3 cos 4
r
6 5 cos
r r r r r r
2
5
3
3
2
5 sin sin
4 sin sin
4
9 cos 2 , 0 2
In Exercises 45–48, find the length of each petal of the polar curve.
45.
47.
r r
2
1
4 sin 2
4 cos 5
46.
48.
r r
3
3
5 cos 2
4 sin 5
In Exercises 49–52, select the two equations whose graphs are the same curve. Then, even though the graphs of the equations are identical, describe how the two paths are different as increases from 0 to 2 .
49.
50.
51.
52.
r r r r
1
1
1
1
1
1
1
2
3 sin ,
2 cos ,
2 cos ,
2 sin ,
r
2
r
2
r
2
1
1
1
r
2
2
3 sin ,
2 cos ,
2 cos ,
2 sin ,
r
3
r
3
r
3
1
1
r
3
1
2
3 sin
2 cos
2 cos
2 sin
[–3.7, 5.7] by [–3.1, 3.1]
(c)
[–4.7, 4.7] by [–4.1, 2.1]
(d)
5144_Demana_Ch06pp501566 01/11/06 9:33 PM Page 549
SECTION 6.5
Graphs of Polar Equations
549
In Exercises 53 – 56,
(b)
(a)
describe the graph of the polar equation, state any symmetry that the graph possesses, and
(c)
state its maximum
r
value if it exists.
53.
55.
r r
2 sin
2
2
1 3 cos 3 sin 2
54.
56.
r r
3 cos 2
1 3 sin 3 sin 3
57.
Group Activity
r a
cos
n
and
r
Analyze the graphs of the polar equations
a
sin
n
when
n
is an even integer.
58.
Revisiting Example 4
that the graph of the curve
r
Use the polar symmetry tests to prove
3 sin 4 is symmetric about the
y
axis and the origin.
59.
Writing to Learn Revisiting Example 5
Confirm the range stated for the polar function graphing
y
3 3 sin
x r
for 0
3 3 sin of Example 5 by
x
2 . Explain why this works.
60.
Writing to Learn Revisiting Example 6
stated for the polar function graphing
y
2 3 cos
x r
for 0
2
Confirm the range
x
3 cos of Example 6 by
2 . Explain why this works.
Standardized Test Questions
61.
62.
True or False
answer.
A polar curve is always bounded. Justify your
False. The spiral
r
is unbounded.
True or False
The graph of
r
2 cos is symmetric about the
x
axis. Justify your answer.
In Exercises 63–66, solve the problem without using a calculator.
63.
Multiple Choice
Which of the following gives the number of petals of the rose curve
r
3 cos 2 ?
D
(A)
1
(B)
2
(C)
3
(D)
4
(E)
6
64.
Multiple Choice
Which of the following describes the symmetry of the rose graph of
r
3 cos 2 ?
D
(A)
only the
x
axis
(B)
only the
y
axis
(C)
only the origin
(D)
the
x
axis, the
y
axis, the origin
(E)
Not symmetric about the
x
axis, the
y
axis, or the origin
65.
Multiple Choice
r
value for
r
2
(A)
6
(B)
5
Which of the following is a maximum
3 cos
?
B
(C)
3
(D)
2
(E)
1
66.
Multiple Choice
Which of the following is the number of petals of the rose curve
r
5 sin 3
?
B
(A)
1
(B)
3
(C)
6
(D)
10
(E)
15
Explorations
67.
Analyzing Rose Curves
r a
cos
n
for
n,
Consider the polar equation an odd integer.
(a)
Prove that the graph is symmetric about the
x
axis.
(b)
Prove that the graph is not symmetric about the
y
axis.
(c)
Prove that the graph is not symmetric about the origin.
(d)
Prove that the maximum
r
value is
a
.
(e)
Analyze the graph of this curve.
68.
Analyzing Rose Curves
r
Consider the polar equation
a
sin
n
for
n
an odd integer.
(a)
Prove that the graph is symmetric about the
y
axis.
(b)
Prove that the graph is not symmetric about the
x
axis.
(c)
Prove that the graph is not symmetric about the origin.
(d)
Prove that the maximum
r
value is
a
.
(e)
Analyze the graph of this curve.
69.
Extended Rose Curves
r
2
3 sin 7 2
The graphs of
r
1 may be called rose curves.
(a)
3 sin 5 2 and
Determine the smallest interval that will produce a complete graph of
r
1
; of
r
2
.
(b)
How many petals does each graph have?
Extending the Ideas
In Exercises 70– 72, graph each polar equation. Describe how they are related to each other.
70. (a)
r
1
3 sin 3
(b)
r
2
3 sin 3
(
1 2
)
(c)
r
3
3 sin 3
(
4
)
71. (a)
r
1
2 sec
(b)
r
2
2 sec
(
4
)
(c)
r
3
2 sec
(
3
)
72. (a)
73.
r
1
2 2 cos
(b)
r
2
r
1
(
4
)
(c)
r
3
r
1
(
3
)
Writing to Learn
r f
, and
r
Describe how the graphs of
f r f
, are related. Explain why you think this generalization is true.
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 550
550
CHAPTER 6
Applications of Trigonometry
6.6
n
What you’ll learn about
■ The Complex Plane
■ Trigonometric Form of
Complex Numbers
■ Multiplication and Division of
Complex Numbers
■ Powers of Complex Numbers
■ Roots of Complex Numbers
. . . and why
This material extends your equationsolving technique to include equations of the form
z n c
,
n
an integer and
c
a complex number.
You might be curious as to why we reviewed complex numbers in Section P.6, then proceeded to ignore them for the next six chapters. (Indeed, after this section we will pretty much ignore them again.) The reason is simply because the key to understanding calculus is the graphing of functions in the Cartesian plane, which consists of two perpendicular real (not complex) lines.
We are not saying that complex numbers are impossible to graph. Just as every real number is associated with a point of the real number line, every complex number can be associated with a point of the
complex plane
. This idea evolved through the work of Caspar Wessel (1745–1818), JeanRobert Argand (1768–1822) and Carl Friedrich
Gauss (1777–1855). Real numbers are placed along the horizontal axis (the
real axis
) and imaginary numbers along the vertical axis (the
imaginary axis
), thus associating the complex number
2 3
i
as an example.
a bi
with the point (
a
,
b
). In Figure 6.57 we show the graph of
Imaginary axis
a
+
bi bi a
(a)
Imaginary axis
2 + 3
i
3
i
Real axis
EXAMPLE 1
Plotting Complex Numbers
Plot
u
1 3
i
,
v
2
i
, and
u v
in the complex plane. These three points and the origin determine a quadrilateral. Is it a parallelogram?
SOLUTION
u
,
v
, and
u
First notice that
v u v
(1 3
i
) (2
i
) 3 2
i
. The numbers are plotted in Figure 6.58a. The quadrilateral is a parallelogram because the arithmetic is exactly the same as in vector addition (Figure 6.58b).
Now try Exercise 1.
y
Imaginary axis
u
= 1 + 3
i u
= 1, 3
u
+
v
= 3 + 2
i u
+
v
= 3, 2
2
Real axis
(b)
FIGURE 6.57
complex plane.
Plotting points in the
IS THERE A CALCULUS OF COMPLEX
FUNCTIONS?
There is a calculus of complex functions.
If you study it someday, it should only be after acquiring a pretty firm algebraic and geometric understanding of the calculus of real functions.
Real axis
O x
O v
= 2 –
i v
= 2, –1
(a) (b)
FIGURE 6.58
(a) Two numbers and their sum are plotted in the complex plane. (b) The arithmetic is the same as in vector addition. (Example 1)
Example 1 shows how the complex plane representation of complex number addition is virtually the same as the Cartesian plane representation of vector addition. Another similarity between complex numbers and twodimensional vectors is the definition of absolute value.
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 551
SECTION 6.6
De Moivre’s Theorem and nth Roots
551
OBJECTIVE
Students will be able to represent complex numbers in the complex plane and write them in trigonometric form. They will be able to use trigonometric form to simplify some algebraic operations with complex numbers.
MOTIVATE
Have students find all solutions of the equation
z
4 number. (
z
1, where
z
1,
z i
is a complex
)
POLAR FORM
What’s in a cis?
Trigonometric (or polar) form appears frequently enough in scientific texts to have an abbreviated form. The expression “cos
i
sin ” is often shortened to “cis ” (pronounced “kiss ”). Thus
z r
cis .
DEFINITION
Absolute Value (Modulus) of a Complex Number
The
absolute value
or
modulus
of a complex number
z z a bi a
2
b
2
.
In the complex plane,
a bi
is the distance of
a a bi
is
bi
from the origin.
Figure 6.59 shows the graph of
z a bi
in the complex plane. The distance
r
from the origin is the modulus of
z
. If we define a direction angle did with vectors, we see that
a r
cos and
b r
sin for
z
just as we
. Substituting these expressions for
a
and
b
gives us the
trigonometric form
(or
polar form
) of the complex number
z
.
Imaginary axis
z = a
+
bi
LESSON GUIDE
Day 1: The Complex Plane;
Trigonometric Form of Complex
Numbers; Multiplication and Division of
Complex Numbers
Day 2: Powers of Complex Numbers;
Roots of Complex Numbers
r
θ
a = r
cos u
b = r
sin u
Real axis
FIGURE 6.59
If
r
angle shown, then
z
is the distance of
r
(cos
z a bi
from the origin and is the directional
i
sin ), which is the trigonometric form of
z
.
DEFINITION
Trigonometric Form of a Complex Number
The
trigonometric form
of the complex number
z a bi
is
z r
cos
i
sin where
a r
cos ,
b r
sin ,
r a
2
absolute value
or
modulus
of
z
, and is an
b
2
, and tan
argument
of
z
.
b a
. The number
r
is the
An angle for the trigonometric form of
z
can always be chosen so that 0 2 , although any angle coterminal with could be used. Consequently, the
angle
and
argument
of a complex number
z
are not unique. It follows that the trigonometric form of a complex number
z
is not unique.
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 552
552
CHAPTER 6
Applications of Trigonometry
TEACHING NOTE
It may be useful to review complex numbers as introduced in Section P.6.
Imaginary axis
θ
θ ′
1 – 3
i
Real axis
FIGURE 6.60
Example 2a.
The complex number for
Imaginary axis
θ
θ ′
Real axis
–3 – 4
i
FIGURE 6.61
The complex number for
Example 2b.
EXAMPLE 2
Finding Trigonometric Forms
Find the trigonometric form with 0 2 for the complex number.
(a)
1 3
i
(b)
3 4
i
SOLUTION
(a)
For 1 3
i
,
r
1 3
i
1
2
3
2
Because the reference angle
2 for is
(
3 (Figure 6.60),
)
3
5
3
.
2.
Thus,
1 3
i
2 cos
5
3
2
i
sin
5
.
3
(b)
For 3 4
i
,
3 4
i
3
2
4
2
5.
The reference angle for (Figure 6.61) satisfies the equation tan
4
3
, so tan
1
4
3
0.927
. . . .
Because the terminal side of is in the third quadrant, we conclude that
4.07.
Therefore,
3 4
i
5 cos 4.07
i
sin 4.07
.
Now try Exercise 5.
The trigonometric form for complex numbers is particularly convenient for multiplying and dividing complex numbers. The product involves the product of the moduli and the sum of the arguments. (
Moduli
is the plural of
modulus
.) The quotient involves the quotient of the moduli and the difference of the arguments.
Product and Quotient of Complex Numbers
Let
1.
z
1
z
1
•
z
2
r
1 cos
r
1
r
2
1 cos
i
sin
1
1 2 and
z i
2 sin
r
1
2 cos
2
2
.
i
sin
2
. Then
2.
z
1
z
2
r r
1
2 cos
1 2
i
sin
1 2
,
r
2
0.
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 553
SECTION 6.6
De Moivre’s Theorem and nth Roots
553
TEACHING NOTE
The proofs of the product and quotient formulas are good applications of the sum and difference identities studied in
Section 5.3
Proof of the Product Formula
z
1
•
z
2
r
1
r r
1
1
r r
2 cos
2 cos
1
1
i
sin
1 cos
1 cos
2
2
•
r
2 cos
2 sin
1 sin
2
i
sin
1 2
i
sin
2
i
sin
1 cos
2
You will be asked to prove the quotient formula in Exercise 63.
cos
1 sin
2
TEACHING NOTE
Many of the calculations discussed in this section can be performed using a grapher’s builtin functions for converting between rectangular and polar coordinates.
EXAMPLE 3
SOLUTION
Multiplying Complex Numbers
Express the product of
z
1 and
z
2 in standard form:
z
1
25 2
( cos
4
i
sin
4
)
,
z
2
14
( cos
3
i
sin
3
)
.
z
1
•
z
2
25 2
( cos
4
i
sin
25
•
14
350
2 cos
(
4 3
)
2
( cos
1 2
i
sin
1 2
)
4
)
•
14
( cos
3
i
sin
3
)
i
sin
(
4 3
)
478.11
128.11
i
Now try Exercise 19.
EXAMPLE 4
Dividing Complex Numbers
Express the quotient
z
1
z
2 in standard form:
z
1
SOLUTION
2 2 cos 135°
i
sin 135° ,
z
2
6 cos 300°
z
1
z
2
2 2 cos 135°
6 cos 300°
i
sin 135°
i
sin 300°
3
2 cos 135° 300°
i
sin 135° 300°
3
2 cos
0.46
165°
0.12
i i
sin 165°
i
sin 300° .
Now try Exercise 23.
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 554
554
CHAPTER 6
Applications of Trigonometry
Imaginary axis
z
2
r
2
2
θ
r z
θ
FIGURE 6.62
of
z
2
.
A geometric interpretation
TEACHING NOTE
This section provides a nice opportunity to bring geometry and algebra together.
Providing geometric motivations to numerical work helps students connect different mathematical ideas.
Imaginary axis
1 +
i
3
2
1
3
Real axis
Real axis
FIGURE 6.63
Example 5.
The complex number in
We can use the product formula to raise a complex number to a power. For example, let
z r
cos
i
sin . Then
z
2
z
•
z r
cos
r r
2
2 cos cos 2
i
sin
•
r
cos
i
sin
i
sin 2
i
sin
Figure 6.62 gives a geometric interpretation of squaring a complex number: its argument is doubled and its distance from the origin is multiplied by a factor of
r
, increased if
r
1 or decreased if
r
1.
We can find
z
3 by multiplying
z
by
z
2
:
z
3
z
•
z
2
r r
3 cos cos
r
3 cos 3
i
sin
2
•
i
sin 3
r
2 cos 2
i
sin 2
i
sin 2
Similarly,
z
4
z
5
.
.
.
r r
4
5 cos 4 cos 5
i i
sin 4 sin 5
This pattern can be generalized to the following theorem, named after the mathematician Abraham De Moivre (1667–1754), who also made major contributions to the field of probability.
De Moivre’s Theorem
Let
z r
cos
z n i
sin and let
r
cos
i n
be a positive integer. Then sin
n r n
cos
n i
sin
n
.
EXAMPLE 5
Find 1
i
Using De Moivre’s theorem
3
3 using De Moivre’s theorem.
SOLUTION
Solve Algebraically
and its modulus is
z
1
i
See Figure 6.63. The argument of
z
3 1 3 2. Therefore,
2
( cos
3
i
sin
3
)
z
3
2
3 cos
(
3
•
3
)
i
sin
(
3
•
)
3
8 cos
8 1
i
sin
0
i
8
1
i
3 is 3,
continued
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 555
NOTES ON EXAMPLES
Problems like Example 6 are frequently found on tests in math contests. They are easy if a student knows De Moivre’s theorem.
TEACHING NOTE
It is worth pointing out that “unity” simply means “one.”
SECTION 6.6
De Moivre’s Theorem and nth Roots
555
Support Numerically
Figure 6.64a sets the graphing calculator we use in complex number mode. Figure 6.64b supports the result obtained algebraically.
Now try Exercise 31.
Normal Sci Eng
Float 0123456789
Radian Degree
Func Par Pol Seq
Connected Dot
Sequential Simul
Real a+bi re^
θ i
Full Horiz G–T
(a)
(1+i (3)) 3
–8
(b)
FIGURE 6.64
(1
i
(a) Setting a graphing calculator in complex number mode. (b) Computing
3 )
3 with a graphing calculator.
EXAMPLE 6
Find 2 2
i
Using De Moivre’s Theorem
2 2
8 using De Moivre’s theorem.
SOLUTION
modulus is
The argument of
z
2 2
i
2 2 is
2
2
i
2
2 1
2
1
2
1.
Therefore,
z z
8
3 4, and its
1 cos
3
4
i
sin
3
4 cos
(
8
•
3
4
) cos 6
i i
sin 6 sin
(
8
•
3
4
)
i
•
0 1
Now try Exercise 35.
The complex number 1 and the complex number
The complex number 1 eighth root of 1.
i
3
2
i
2
3
i
in Example 5 is a solution of
2 2
z
3 in Example 6 is a solution of
z
is a third root of 8 and 2 2
i
2 2
8,
8
1.
is an
n
th Root of a Complex Number
A complex number
v a bi
is an
n
th root of
z
v n z
.
if
If
z
1, then
v
is an
n
th root of unity
.
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 556
556
CHAPTER 6
Applications of Trigonometry
FOLLOWUP
Ask . . .
The complex number cos 3
i
sin 3 an 8th root of unity. Why is this number is not listed in the solution to Example 9 (or is it)?
(It is the same as cos
i
sin 1 0
i
.)
ASSIGNMENT GUIDE
Day 1: Ex. 3–30, multiples of 3
Day 2: Ex. 33–45, multiples of 3, 59,
65–70
COOPERATIVE LEARNING
Group Activity: Ex. 64
NOTES ON EXERCISES
Ex. 19–30 provide an opportunity to show how the product and quotient formulas can simplify calculations when applicable.
Ex. 39–56 involve
n
th roots of complex numbers. Encourage students to think about the radius of the circle in which the roots fall and the angular spacing between the roots.
Ex. 65–70 provide practice with standardized tests.
Ex. 75–76 give a graphical interpretation of the product of two complex numbers.
ONGOING ASSESSMENT
SelfAssessment: Ex. 1, 7, 19, 23, 31, 35,
45, 57, 59
Embedded Assessment: Ex. 71, 72, 78
z
We use De Moivre’s theorem to develop a general formula for finding the
n
th roots of a nonzero complex number. Suppose that
r
cos
i
sin . Then
v n z v s
cos
i
sin is an
n
th root of
s
[
n s
cos cos
n i i
sin sin
n
]
n r r
cos cos
i
sin
i
sin (1)
Next, we take the absolute value of both sides:
s n
cos
n s
2
n
co s
2
n i
sin s in
2
na n s
2
n r
cos
r
2 c o s
2
i
sin s in
2
r
2
s n s r n r s
0,
r
0
Substituting
s n r
into Equation (1), we obtain cos
n i
sin
n
cos
i
sin .
Therefore,
n
can be any angle coterminal with . Consequently, for any integer
k
,
v n
an
n
th root of
z
if
s r
and is
n
2
k n
2
k
.
0, 1, …,
n
1, and the values The expression for
v
start to repeat for
k
takes on
n
,
n n
different values for
k
1, ….
We summarize this result.
Finding
n
th Roots of a Complex Number
If
z r
cos
i
sin , then the
n
distinct complex numbers
n r
( cos
n
2
k i
sin
n
2
k
)
, where
k
0, 1, 2, . . . ,
n
1, are the
n
th roots of the complex number
z
.
EXAMPLE 7
Finding Fourth Roots
Find the fourth roots of
z
5 cos 3
i
sin 3 .
SOLUTION
The fourth roots of
z
are the complex numbers
4
5
( cos
3
4
2
k i
sin
3
4
2
k
) for
k
0, 1, 2, 3.
continued
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 557
SECTION 6.6
De Moivre’s Theorem and nth Roots
557
Taking into account that
z
1
4
5 cos
(
3 2
k
1 2
0
2
)
4
4
5 cos
1 2
i
sin
1 2
i
12 sin
(
k
2, the list becomes
1 2
0
2
)
z
2
4
5 cos
(
1 2 2
4
5 cos
7
1 2
)
i
sin
7
1 2
i
sin
(
1 2 2
)
z
3
z
4
4
5 cos
(
1 2
2
2
)
4
5 cos
1 3
1 2
i
sin
1 3
1 2
4
5 cos
(
1 2
3
2
)
i
sin
(
1 2
2
2
)
i
sin
(
1 2
3
2
)
4
5 cos
1 9
1 2
i
sin
1 9
1 2
Now try Exercise 45.
TEACHING NOTE
Example 8 can also be solved by writing the equation
z
3
1 0, factoring, and using the quadratic formula. It is useful for students to see that this method gives the same answer.
z
1
z
2
z
3
[–2.4, 2.4] by [–1.6, 1.6]
FIGURE 6.65
The three cube roots
z
1
,
z
2
, and
z
3 of 1 displayed on the unit circle
(dashed). (Example 8)
EXAMPLE 8
Finding Cube Roots
Find the cube roots of 1 and plot them.
SOLUTION
First we write the complex number
z
1 in trigonometric form
The third roots of
z
1
z
1 cos cos
3
2
0
i i
sin .
i
sin are the complex numbers
k
cos
i
sin
3
2
k
, for
k
0, 1, 2. The three complex numbers are
z z
1
2 cos
3
i
sin
3 cos
3
2
i
sin
1
2 2
3
i
,
3
2
1 0
i
,
z
3 cos
3
4
i
sin
3
4 1
2 2
3
i
.
Figure 6.65 shows the graph of the three cube roots
z
1 spaced (with distance of 2
,
z
2
, and
3 radians) around the unit circle.
z
3
. They are evenly
Now try Exercise 57.
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 558
558
CHAPTER 6
Applications of Trigonometry
Imaginary axis
z
4
z
3
z
2
z
5
z
6
z
7
z
8
z
1
Real axis
FIGURE 6.66
The eight eighth roots of unity are evenly spaced on a unit circle.
(Example 9)
EXAMPLE 9
Finding Roots of Unity
Find the eight eighth roots of unity.
SOLUTION
First we write the complex number
z
1 in trigonometric form
The eighth roots of
z z
1 0
i
cos 0
i
sin 0.
1 0
i
cos
0 cos 0
i
sin 0 are the complex numbers
8
2
k i
sin
0
8
2
k
, for
k
0, 1, 2, . . . , 7.
z z z z
1
2
3
4 cos 0
i
sin 0 1 0
i
cos
4
i
sin
4 2
2
2
2
i
cos
2
i
sin
2
0
i
cos
3
4
i
sin
3
4 2
2
2
2
i z z z
5
6
7 cos
i
sin 1 0
i
cos
5
4
i
sin
5
4 2
2
2
2
i
cos
3
2
i
sin
3
2
0
i z
8 cos
7
4
i
sin
7
4 2
2
2
2
i
Figure 6.66 shows the eight points. They are spaced 2 8 4 radians apart.
Now try Exercise 59.
(For help, go to Sections P.5, P.6, and 4.3.)
In Exercises 1 and 2, write the roots of the equation in
a
1.
2.
x
2
5(
x
2
13 4
x
2 3
i
, 2 3
i
1) 6
x
0.6 0.8
i
, 0.6 0.8
i bi
form.
In Exercises 3 and 4, write the complex number in standard form
a bi
.
3.
1
i
5
4 4
i
4.
1
i
4
4 0
i
In Exercises 5 – 8, find an angle in 0 equations.
5.
sin
1
2 and cos
2
3
2
5
6 which satisfies both
6.
sin
2
2 and cos
2
2
7
4
7.
sin
2
3 and cos
1
2
4
3
8.
sin
2
2 and cos
2
2
5
4
In Exercises 9 and 10, find all real solutions.
9.
x
3
1 0
1
10.
x
4
1 0
1
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 559
SECTION 6.6
De Moivre’s Theorem and nth Roots
559
In Exercises 1 and 2, plot all four points in the same complex plan.
1.
1
2.
2
2
i
, 3
3
i
, 1
i, i,
3,
2
2
2
i
,
i i
,
In Exercises 3–12, find the trigonometric form of the complex number where the argument satisfies 0 2 .
3.
5.
7.
3
2
i
2
2
i
2
i
3
4.
6.
8.
3
2
i
3
3
i i
9.
3
11.
2
i y
10.
4
12.
7
i y
3
30
°
z x x
45
°
4
z
In Exercises 13–18, write the complex number in standard form
a i
sin 210°
14.
8 cos 210°
13.
3 cos 30°
i
sin 30°
15.
5 cos
i
sin
5 2
60° 60°
(5 2)
16.
5
17.
18.
cos
7
(
( cos
4
i
2
( cos
7
6
1 2 sin
4
)
5
2
2
5
2
2
i i
sin
i
sin
1 2
7
6
)
2
6
2
2
i
)
2.56
0.68
i
3
i
In Exercises 19 – 22, find the product of
z
1 trigonometric form.
and
z
2
. Leave the answer in
19.
z
1
z
2
20.
z
1
z
2
21.
z
1
22.
z
1
7 cos 25°
i
sin 25°
14 (cos 155°
i
sin 155°)
2 cos 130°
0.5
2 cos 118° cos 19°
i
sin 130°
i
sin 118°
i
sin 19°
2
2
(cos 99°
i
sin 99°)
5
( cos
4
i
3
( cos
3
4 sin
4
i
sin
)
3
4
)
z
2
z
2
3
( cos
5
3
1
3
i
sin
5
3
)
( cos
6
i
sin
6
)
In Exercises 23 – 26, find the trigonometric form of the quotient.
bi
.
23.
2 cos 30°
3 cos 60°
25.
6
3 cos 5 cos 2
i
sin 30°
i
sin 60°
i
sin 5
i
sin 2
24.
5 cos 220°
2 cos 115°
26.
cos cos
2
4
i
sin 220°
i
sin 115°
i
sin
i
sin
2
4
In Exercises 27– 30, find the product
z
1 in two ways,
(a)
•
z
2 and quotient using the trigonometric form for
z
1 and
z z
1
2
z
2
.and
(b)
using the standard form for
z
1 and
z
2
.
27.
28.
29.
30.
z z z z
1
1
1
1
3
1
3
2
i
2
i
and
z
2 and
z
2
i
3
i
and
z
2 and
z
2
5
1
1
3
3
i i i
3
i
In Exercises 31– 38, use De Moivre’s theorem to find the indicated power of the complex number. Write your answer in standard form
a bi
.
31.
33.
2
( cos
4
i
( cos
3
4 sin
4
i
sin
3
)
3
4
)
i
5
35.
1
4 4
i
3
37.
1 3
i
3
8
32.
34.
36.
38.
3
6
(
( cos
3 cos
2
i
sin
5
6
i
sin
3
2
)
5
6
)
5
4
243
i
3
(
1
2
i
4
i
20
2
3
3
)
5
20
(0.95
0.30
i
)
1
In Exercises 39 – 44, find the cube roots of the complex number.
39.
2 cos 2
41.
43.
3
3
( cos
4
3
4
i i
sin 2
i
sin
4
3
)
40.
42.
2
( cos
4
i
sin
4
27
( cos
11
6
i
sin
)
11
6
)
44.
2 2
i
In Exercises 45 – 50, find the fifth roots of the complex number.
45.
47.
i
cos
2
( cos
6
i
sin sin
6
)
46.
48.
2
32
( cos
( cos
4
2
i
sin
i
sin
2
4
)
)
49.
2
i
3
i
50.
1
In Exercises 51– 56, find the
n
th roots of the complex number for the specified value of
n
.
51.
1
53.
2
55.
2
i
,
i
,
2
i
,
n n n
6
4
3
52.
1
54.
2
i
,
n
2
i
,
56.
32,
n
5
n
6
4
In Exercises 57– 60, express the roots of unity in standard form
a
Graph each root in the complex plane.
bi
.
57.
Cube roots of unity
59.
Sixth roots of unity
58.
60.
Fourth roots of unity
Square roots of unity
61.
Determine
z
1 3
i
.
and the three cube roots of
8; 2 and 1 3
i z
if one cube root of
z
is
62.
Determine
z
and the four fourth roots of
z
if one fourth root of
z
is
2 2
i
.
64; 2 2
i
and 2 2
i
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 560
560
CHAPTER 6
Applications of Trigonometry
63.
Quotient Formula
z
2
r
2 cos
2
Let
z
1
i
sin
2
,
r
2
z
1
z
2
r
1
r
2 cos
1 2
i r
1 cos
1
i
0. Verify that sin
1 sin
1 2
.
and
64.
Group Activity
complex number
r
n
th Roots
cos on a circle with radius
n r
.
i
Show that the
n
th roots of the sin are spaced 2
n
radians apart
65.
True or False
The trigonometric form of a complex number is unique. Justify your answer.
66.
True or False
The complex number
i
Justify your answer.
True.
i
3
i
, so
i
is a cube root of is a cube root of
i
.
i
.
In Exercises 67–70, you may use a graphing calculator to solve the problem.
67.
Multiple Choice
Which of the following is a trigonometric form of the complex number 1 3
i
?
B
(A)
2 cos
3
i
sin
3
(B)
2 cos
2
3
i
sin
2
3
(C)
2 cos
4
3
i
sin
4
3
(D)
2 cos
5
3
i
sin
5
3
(E)
2 cos
7
3
i
sin
7
3
68.
Multiple Choice
Which of the following is the number of distinct complex number solutions of
z
5
1
i
?
E
(A)
0
(B)
1
69.
Multiple Choice
(D)
4
(E)
5
Which of the following is the standard form for the product of 2 cos
4
A
i
sin
4
(C)
3 and 2 cos
7
4
i
sin
7
4
?
(A)
2
(B)
2
(C)
2
i
(D)
1
i
(E)
1
i
70.
Multiple Choice
Which of the following is not a fourth root of 1?
E
(A)
i
2
(B)
i
2
(C)
1
(D)
1
(E)
i
71.
Complex Conjugates
z a bi
. Let
z r
The complex conjugate of
z
cos
i
sin .
(a)
Prove that
z r
cos
i
sin .
(b)
Use the trigonometric form to find
z
•
z
.
r
2
72.
(c)
Use the trigonometric form to find
z z
, if
z
(d)
Prove that
z r
cos
Modulus of Complex Numbers
i
sin
Let
z
0.
r
.
cos
(a)
Prove that
z r
.
i a
sin
bi
.
is
(b)
Use the trigonometric form for the complex numbers
z
1 to prove that
z
1
•
z
2
z
1
•
z
2
.
and
z
2
73.
Using Polar Form on a Graphing Calculator
The complex number
r
cos
i
sin some graphing calculators as
re i
can be entered in polar form on
.
(a)
Support the result of Example 3 by entering the complex numbers
z
1 and
z
2 in polar form on your graphing calculator and computing the product with your graphing calculator.
(b)
Support the result of Example 4 by entering the complex numbers
z
1 and
z
2 in polar form on your graphing calculator and computing the quotient with your graphing calculator.
(c)
Support the result of Example 5 by entering the complex number in polar form on your graphing calculator and computing the power with your graphing calculator.
74.
Visualizing Roots of Unity
Set your graphing calculator in parametric mode with 0
T
Xmax
8, Tstep
2.4, Ymin 1.6, and Ymax
1, Xmin
1.6.
2.4,
(a)
Let
x
cos 2 8
t
and
y
sin 2 8
t
. Use trace to visualize the eight eighth roots of unity. We say that 2 8
generates
the eighth roots of unity. (Try both dot mode and connected mode.)
(b)
Replace 2 8 in part (a) by the arguments of other eighth roots of unity. Do any others
generate
the eighth roots of unity?
Yes. 6 8, 10 8, 14 8
(c)
Repeat parts (a) and (b) for the fifth, sixth, and seventh roots of unity, using appropriate functions for
x
and
y
.
(d)
What would you conjecture about an
n
th root of unity that generates all the
n
th roots of unity in the sense of part (a)?
75.
Parametric Graphing
represent representing
2
2
i
Write parametric equations that
n
for
n i n
for
n t
. Draw and label an
accurate
0, 1, 2, 3, 4.
spiral
76.
Parametric Graphing
represent representing
1
1
i
Write parametric equations that
n
for
i n n
for
n t
. Draw and label an
accurate
0, 1, 2,
y z
1
z
2 spiral
3, 4.
77.
Explain why the triangles formed by 0,
1, and
z
1 and by 0,
z
2 and
z
1
z
2 shown in the figure are similar triangles.
z
2
78.
Compass and Straightedge
Construction
Using only a compass and straightedge, construct the location of
z
1
z
2 of 0, 1,
z
1
, and
z
2
.
given the location 0 1
z
1
x
In Exercises 79–84, find all solutions of the equation (real and complex).
79.
81.
83.
x
3
1 0
x
3
1 0
x
5
1 0
80.
82.
84.
x
4
x
4
x
5
1 0
1 0
1 0
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 561
CHAPTER 6
Key Ideas
561
CHAPTER 6
PROPERTIES, THEOREMS, AND FORMULAS
Component Form of a Vector 503
The Magnitude or Length of a Vector 504
Vector Addition and Scalar Multiplication 505
Unit Vector in the Direction of the Vector
Dot Product of Two Vectors 514
v
506
Properties of the Dot Product 514
Theorem Angle Between Two Vectors 515
Projection of the Vector
u
onto the Vector
v
517
PROCEDURES
Head Minus Tail Rule for Vectors
Resolving a Vector 507
503
GALLERY OF FUNCTIONS
Rose Curves:
r a
cos
n
and
r a
sin
n
Work 518
Coordinate Conversion Equations
Symmetry Tests for Polar Graphs
The Complex Plane 551
535
541
Modulus or Absolute Value of a Complex
Number 551
Trigonometric Form of a Complex Number
De Moivre’s Theorem 554
551
Product and Quotient of Complex Numbers
n
th Root of a Complex Number 556
551
Limaçon Curves:
r
[–6,6] by [–4, 4]
a r
4 sin 3
b
sin and
r a b
cos with
a
[–4.7, 4.7] by [–3.1, 3.1]
r
3 sin 4
0 and
b
0
Limaçon with an inner loop:
a b
1 Cardioid:
a b
1
Dimpled limaçon: 1
a b
2
Spiral of Archimedes: Lemniscate Curves:
r
2
a
2
a
Convex limaçon:
b
sin 2
2 and
r
2
a
2 cos 2
[–30, 30] by [–20, 20]
r
, 0 45
[–4.7, 4.7] by [–3.1, 3.1]
r
2
4 cos 2
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 562
562
CHAPTER 6
Applications of Trigonometry
CHAPTER 6
The collection of exercises marked in red could be used as a chapter test.
In Exercises 1 – 6, let
u
2, 1 ,
v
vectors. Find the indicated expression.
4, 2 , and
w
1, 3 be
1.
u
3.
u v v
5.
u
•
v
6
2,
3 7
3
2.
4.
2
u w
3
w
2
u
6.
u
•
w
5
1, 7
1 0
In Exercises 7–10, let
A
1, 5
2, 1 ,
B
3, 1 ,
C
4, 2 , and
. Find the component form and magnitude of the vector.
D
7.
3
AB
9.
AC
3, 6 ; 3
BD
8,
5
3 ; 7 3
8.
AB
10.
CD
CD
AB
6,
4,
5 ;
9 ;
6 1
9 7
In Exercises 11 and 12, find
(a)
a unit vector in the direction of
AB
and
(b)
a vector of magnitude 3 in the opposite direction.
11.
12.
A
A
4, 0
3, 1 ,
,
B
B
2, 1
5, 1
12. (a)
1, 0
(b)
3, 0
In Exercises 13 and 14, find
(a)
the direction angles of
u
and
v
and
(b)
the angle between
u
and
v
.
13.
u
4, 3 ,
v
2, 5
14.
u
2, 4 ,
v
6, 4
In Exercises 15 – 18, convert the polar coordinates to rectangular coordinates.
15.
17.
2,
2.5, 25°
(
4
( 2 ,
2.27,
2 )
1.06)
16.
18.
3.1, 135°
(1.55
3.6, 3 4
( 1.8
2 , 1.55
2 , 1.8
2 )
2 )
In Exercises 19 and 20, polar coordinates of point
P
are given. Find all of its polar coordinates.
19.
P
1, 2 3
20.
P
2, 5 6
In Exercises 21 – 24, rectangular coordinates of point
P
are given.
Find polar coordinates of
P
that satisfy these conditions:
(a)
21.
0
P
2
2, 3
(b)
22.
P
(c)
0 4
10, 0
23.
P
5, 0
24.
P
0, 2
In Exercises 25 – 30, eliminate the parameter
t
and identify the graph.
25.
x
26.
x
27.
x
29.
x
3
4
2
t
2
e
2
t
5
t
,
y
3,
y
4
t
,
y
8
t
1,
y e t
3
t
5
t
,
1
3
t
28.
30.
5
x x
3 cos
t
,
y t
3
,
y
ln
t
,
t
3 sin
0
t
In Exercises 31 and 32, find a parametrization for the curve.
31.
The line through the points 1,
32.
The line segment with endpoints
2 and 3, 4 .
2, 3 and 5, 1 .
Exercises 33 and 34 refer to the complex number
z
1 figure.
shown in the
Imaginary axis
z
1
4
–3
Real axis
33.
If
z
1
a bi
, find
a
,
b
, and
z
1
.
34.
Find the trigonometric form of
z
1
.
a
3,
b
4,
z
1
5
In Exercises 35 – 38, write the complex number in standard form.
35.
6 cos 30°
37.
2.5
( cos
4
3
i
sin 30°
i
sin
4
3
)
36.
3 cos 150°
38.
4 cos 2.5
i
sin 150°
i
sin 2.5
In Exercises 39 – 42, write the complex number in trigonometric form where 0 2 . Then write three other possible trigonometric forms for the number.
39.
3
41.
3
3
i
5
i
40.
42.
1
2
i
2
i
2
In Exercises 43 and 44, write the complex numbers
z
1 trigonometric form.
•
z
2 and
z
1
z
2 in
43.
z
1
44.
z
1
3 cos 30°
5 cos 20°
i
sin 30°
i
sin 20° and
z
2
4 cos 60° and
z
2
2 cos 45°
i
sin 60°
i
sin 45°
In Exercises 45 – 48, use De Moivre’s theorem to find the indicated power of the complex number. Write your answer in
(a)
trigonometric form and
(b)
standard form.
45.
47.
3
( cos
4
i
5
( cos
5
3 sin
)
4
i
sin
5
3
)
5
3
46.
48.
2
( cos
1 2
i
7
( cos
2 4
i
sin
1 2
) sin
2 4
)
8
6
In Exercises 49 – 52, find and graph the
n
th roots of the complex number for the specified value of
n
.
49.
3
51.
1,
n
3
i
,
5
n
4
50.
8,
52.
1,
n n
3
6
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 563
CHAPTER 6
Review Exercises
563
In Exercises 53 – 60, decide whether the graph of the given polar equation appears among the four graphs shown.
(a)
(c)
(b)
(d)
53.
55.
57.
59.
r r r r
3 sin 4
2
(b)
2 sin
(a)
2 2 sin not shown
3 cos 5
(c)
54.
56.
58.
60.
r r r r
2
3
1
3 sin sin 3 not shown not shown
2 cos
2 tan
(d) not shown
In Exercises 61– 64, convert the polar equation to rectangular form and identify the graph.
61.
r
63.
r
2
3 cos 2 sin
62.
64.
r r
2 sin
3 sec
In Exercises 65 – 68, convert the rectangular equation to polar form.
Graph the polar equation.
65.
y
4
67.
x
3
2
y
1
2
10
66.
68.
x
2
x
5
3
y
4
In Exercises 69 –72, analyze the graph of the polar curve.
69.
71.
r r
2 5 sin
2 sin 3
(c)
Let
y mx b
. Prove that
70.
72.
r r
2
4 4 cos
2 sin 2 , 0 2
73.
Graphing Lines Using Polar Equations
(a)
(b)
Explain why
Explain why
r r a
sec
b
csc is a polar form for the line
x
is a polar form for the line
y a
.
b
.
r
sin
b m
cos is a polar form for the line. What is the domain of
r
?
(d)
Illustrate the result in part (c) by graphing the line
y
using the polar form from part (c).
2
x
74. Flight Engineering
An airplane is flying on a bearing of 80° at 540 mph. A wind is blowing with the bearing 100° at 55 mph.
3
(a)
Find the component form of the velocity of the airplane.
(b)
Find the actual speed and direction of the airplane.
75.
Flight Engineering
An airplane is flying on a bearing of 285° at 480 mph. A wind is blowing with the bearing 265° at 30 mph.
(a)
Find the component form of the velocity of the airplane.
(b)
Find the actual speed and direction of the airplane.
76.
Combining Forces
angle of 20°. A second force of 300 lb acts on the object at an angle of 5°. Find the direction and magnitude of the resultant force.
411.89 lb;
A force of 120 lb acts on an object at an
2.07°
77. Braking Force
A 3000 pound car is parked on a street that makes an angle of 16° with the horizontal (see figure).
(a)
Find the force required to keep the car from rolling down the hill.
826.91 pounds
(b)
Find the component of the force perpendicular to the street.
2883.79 pounds
16
°
78.
Work
Find the work done by a force
F
of 36 pounds acting in the direction given by the vector 3, 5 in moving an object 10 feet from 0, 0 to 10, 0 .
185.22 footpounds
79.
Height of an Arrow
Stewart shoots an arrow straight up from the top of a building with initial velocity of 245 ft sec. The arrow leaves from a point 200 ft above level ground.
(a)
Write an equation that models the height of the arrow as a function of time
t
.
h
16
t
2
245
t
200
(b)
Use parametric equations to simulate the height of the arrow.
(c)
Use parametric equations to graph height against time.
(d)
How high is the arrow after 4 sec?
924 ft
(e)
What is the maximum height of the arrow? When does it reach its maximum height?
1138 ft;
t
7.66
(f)
How long will it be before the arrow hits the ground?
80.
Ferris Wheel Problem
Lucinda is on a Ferris wheel of radius
35 ft that turns at the rate of one revolution every 20 sec. The lowest point of the Ferris wheel (6 o’clock) is 15 ft above ground level at the point 0, 15 of a rectangular coordinate system. Find parametric equations for the position of Lucinda as a function of time
t
in seconds if Lucinda starts
t
0 at the point 35, 50 .
81.
Ferris Wheel Problem
The lowest point of a Ferris wheel
(6 o’clock) of radius 40 ft is 10 ft above the ground, and the center is on the
y
axis. Find parametric equations for
Henry’s position as a function of time
t
in seconds if his starting position
t
0 is the point 0, 10 and the wheel turns at the rate of one revolution every 15 sec.
x
40 sin
2
t
1 5
,
y
50 40 cos
2
t
1 5
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 564
564
CHAPTER 6
Applications of Trigonometry
82.
Ferris Wheel Problem
Sarah rides the Ferris wheel described in Exercise 81. Find parametric equations for Sarah’s position as a function of time
t
in seconds if her starting position
t
0 is the point 0, 90 every 18 sec.
and the wheel turns at the rate of one revolution
83.
Epicycloid
The graph of the parametric equations
x
4 cos
t
cos 4
t
,
y
4 sin
t
sin 4
t
is an
epicycloid
. The graph is the path of a point
P
on a circle of radius 1 rolling along the outside of a circle of radius 3, as suggested in the figure.
(a)
Graph simultaneously this epicycloid and the circle of radius 3.
(b)
Suppose the large circle has a radius of 4. Experiment! How do you think the equations in part (a) should be changed to obtain defining equations? What do you think the epicycloid would look like in this case? Check your guesses.
y
All 4’s should be changed to 5’s.
84.
85.
–3
3
–3
t
C
1
P
3
Throwing a Baseball
the ground with an initial velocity of 66 ft with the horizontal. How many seconds after the ball is thrown will it hit the ground? How far from Sharon will the ball be when it hits the ground? it hits the ground?
t t
0.71 sec,
x
Throwing a Baseball
1.06 sec,
x x
Sharon releases a baseball 4 ft above
46.75 ft the ground with an initial velocity of 66 ft
68.65 ft sec at an angle of 5°
Diego releases a baseball 3.5 ft above sec at an angle of 12° with the horizontal. How many seconds after the ball is thrown will it hit the ground? How far from Diego will the ball be when
86.
Field Goal Kicking
Spencer practices kicking field goals
40 yd from a goal post with a crossbar 10 ft high. If he kicks the ball with an initial velocity of 70 ft sec at a 45° angle with the horizontal (see figure), will Spencer make the field goal if the kick sails “true”?
It clears the crossbar.
70 ft/sec
45
°
40 yd
87.
Hang Time
An NFL placekicker kicks a football downfield with an initial velocity of 85 ft sec. The ball leaves his foot at the
15 yard line at an angle of 56° with the horizontal. Determine the following:
(a)
The ball’s maximum height above the field.
77.59 ft
(b)
The “hang time” (the total time the football is in the air).
88. Baseball Hitting
Brian hits a baseball straight toward a
15fthigh fence that is 400 ft from home plate. The ball is hit when it is 2.5 ft above the ground and leaves the bat at an angle of
30° with the horizontal. Find the initial velocity needed for the ball to clear the fence.
just over 125 ft/sec
89.
Throwing a Ball at a Ferris Wheel
A 60ftradius Ferris wheel turns counterclockwise one revolution every 12 sec. Sam stands at a point 80 ft to the left of the bottom (6 o’clock) of the wheel. At the instant Kathy is at 3 o’clock, Sam throws a ball with an initial velocity of 100 ft sec and an angle with the horizontal of
70°. He releases the ball from the same height as the bottom of the
Ferris wheel. Find the minimum distance between the ball and
Kathy.
17.65 ft
90.
Yard Darts
Gretta and Lois are launching yard darts 20 ft from the front edge of a circular target of radius 18 in. If Gretta releases the dart 5 ft above the ground with an initial velocity of 20 ft sec and at a 50° angle with the horizontal, will the dart hit the target?
no
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CHAPTER 6
Project
565
CHAPTER 6
Parametrizing Ellipses
As you discovered in the Chapter 4 Data Project, it is possible to model the displacement of a swinging pendulum using a sinusoidal equation of the form
x a
sin
b t c d
where
x
represents the pendulum’s distance from a fixed point and
t
represents total elapsed time. In fact, a pendulum’s velocity behaves sinusoidally as well:
y ab
cos
b t c
, where
y
represents the pendulum’s velocity and
a
,
b
, and
c
are constants common to both the displacement and velocity equations.
Use a motion detection device to collect distance, velocity, and time data for a pendulum, then determine how a resulting plot of velocity versus displacement (called a phasespace plot) can be modeled using parametric equations.
COLLECTING THE DATA
Construct a simple pendulum by fastening about 1 meter of string to the end of a ball. Collect time, distance, and velocity readings for between 2 and 4 seconds (enough time to capture at least one complete swing of the pendulum). Start the pendulum swinging in front of the detector, then activate the system.
The data table below shows a sample set of data collected as a pendulum swung back and forth in front of a CBR where
t
is total elapsed time in seconds,
d
distance from the CBR in meters,
v
velocity in meters second.
t
0
0.1
0.2
0.3
0.4
0.5
0.6
d
1.021
1.038
1.023
0.977
0.903
0.815
0.715
v
0.325
0.013
–0.309
–0.598
–0.819
–0.996
–0.979
t
0.7
0.8
0.9
1.0
1.1
1.2
1.3
d
0.621
0.544
0.493
0.473
0.484
0.526
0.596
v
–0.869
–0.654
–0.359
–0.044
0.263
0.573
0.822
t
1.4
1.5
1.6
1.7
1.8
1.9
2.0
d
0.687
0.785
0.880
0.954
1.008
1.030
1.020
v
0.966
1.013
0.826
0.678
0.378
0.049
–0.260
EXPLORATIONS
1.
Create a scatter plot for the data you collected or the data above.
2.
With your calculator computer in function mode, find values for
a
,
b
,
c
, and
d
so that the equation
y a
sin
b x c d
(where
y
is distance and time data plot.
y x
is time) fits the distance versus
0.28 sin(3.46(
x
1.47)) 0.75
3.
Make a scatter plot of velocity versus time. Using the same
a
,
b
, and
c
values you found in 2 , verify that the equation
y ab
cos
b x c
(where
y
is velocity and fits the velocity versus time data plot.
x
is time)
4.
What do you think a plot of velocity versus distance
(with velocity on the vertical axis and distance on the horizontal axis) would look like? Make a rough sketch of your prediction, then create a scatter plot of velocity versus distance. How well did your predicted graph match the actual data plot?
5.
With your calculator computer in parametric mode, graph the parametric curve
x ab
cos
b t c
, 0
t a
sin
2 where
x b t c d
,
y
represents distance,
y
represents velocity, and
t
is the time parameter. How well does this curve match the scatter plot of velocity versus time?
5144_Demana_Ch06pp501566 01/11/06 9:34 PM Page 566
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