Applications of Trigonometry

Applications of Trigonometry

5144_Demana_Ch06pp501-566 01/11/06 9:31 PM Page 501

C H A P T E R

6

Applications of Trigonometry

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

Young salmon migrate from the fresh water they are born in to salt water and live in the ocean for several years. When it’s time to spawn, the salmon return from the ocean to the river’s mouth, where they follow the organic odors of their homestream to guide them upstream. Researchers believe the fish use currents, salinity, temperature, and the magnetic field of the Earth to guide them. Some fish swim as far as

3500 miles upstream for spawning. See a related problem on page 510.

501

5144_Demana_Ch06pp501-566 01/11/06 9:31 PM Page 502

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.

Chapter 6 Overview

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

Vectors in the Plane

What you’ll learn about

■ Two-Dimensional Vectors

■ Vector Operations

■ Unit Vectors

■ Direction Angles

■ Applications of Vectors

. . . and why

These topics are important in many real-world 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 real-world 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: Two-Dimensional Vectors; Vector

Operations

Day 2: Unit Vectors, Direction Angles;

Applications of Vectors

Two-Dimensional 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

.

5144_Demana_Ch06pp501-566 01/11/06 9:31 PM Page 503

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

Two-Dimensional Vector

A

two-dimensional 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.

5144_Demana_Ch06pp501-566 01/11/06 9:31 PM Page 504

504

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.

Vector Operations

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 real-world problems.

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 505

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

tail-to-head

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) tail-to-head, 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

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 506

506

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 two-dimensional.

You will see both

a

,

b

,

c

and

a

i

b

j

c

k

used for three-dimensional 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

tail-to-head method, and (b) represent 3

u

as a stretch of

u

by a factor of 3.

v

by the

Now try Exercise 13.

Unit Vectors

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.

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 507

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.

Direction Angles

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)

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 508

508

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.

Applications of Vectors

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 DC-10 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_Ch06pp501-566 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.

FOLLOW-UP

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

Self-Assessment: 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 65-mph 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.

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 510

510

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.

CHAPTER OPENER PROBLEM

(from page 501)

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.

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 511

SECTION 6.1

Vectors in the Plane

511

QUICK REVIEW 6.1

(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

)

SECTION 6.1 EXERCISES

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

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 512

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 mile-wide 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

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 513

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

Standardized Test Questions

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

Extending the Ideas

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.

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 514

514

CHAPTER 6

Applications of Trigonometry

6.2

Dot Product of Vectors

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.

The Dot Product

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

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 515

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 two-dimensional 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 1-by-1 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.

Angle Between Vectors

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

)

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 516

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.

5144_Demana_Ch06pp501-566 01/11/06 9:32 PM Page 517

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

.

FOLLOW-UP

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

Projecting One Vector onto Another

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_Ch06pp501-566 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.

Work

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_Ch06pp501-566 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

Self-Assessment: Ex. 3, 9, 13, 23, 25,

45, 53

Embedded Assessment: Ex. 67, 68

UNITS FOR WORK

Work is usually measured in footpounds or Newton-meters. One

Newton-meter 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 foot-pounds.

Now try Exercise 53.

QUICK REVIEW 6.2

(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

SECTION 6.2 EXERCISES

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_Ch06pp501-566 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_Ch06pp501-566 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 foot-pounds

50.

Work

3 feet.

Find the work done lifting a 100 pound bag of potatoes

300 foot-pounds

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 foot-pounds

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 foot-pounds

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 foot-pounds

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 foot-pounds

, 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

.

Standardized Test Questions

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

Explorations

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

Extending the Ideas

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_Ch06pp501-566 01/11/06 9:32 PM Page 522

522

CHAPTER 6

Applications of Trigonometry

6.3

Parametric Equations and Motion

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.

Parametric Equations

Imagine that a rock is dropped from a 420-ft 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

Parametric Curves

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_Ch06pp501-566 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)

Eliminating the Parameter

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_Ch06pp501-566 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_Ch06pp501-566 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.

Lines and Line Segments

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 tail-to-head 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_Ch06pp501-566 01/11/06 9:33 PM Page 526

526

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

Simulating Motion with a Grapher

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_Ch06pp501-566 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 projectile-motion 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.

5144_Demana_Ch06pp501-566 01/11/06 9:33 PM Page 528

528

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 projectile-motion 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 20-ft 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 20-ft 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_Ch06pp501-566 01/11/06 9:33 PM Page 529

FOLLOW-UP

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

Self-Assessment: 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.

Quick Review 6.3

(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 point-slope 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

5144_Demana_Ch06pp501-566 01/11/06 9:33 PM Page 530

530

CHAPTER 6

Applications of Trigonometry

SECTION 6.3 EXERCISES

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 10-ft 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 100-yd 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

5144_Demana_Ch06pp501-566 01/11/06 9:33 PM Page 531

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 war-torn 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 Pop-up

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 pop-up.

(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 30-ft 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 5-ft sec split-second 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.

Two-Softball 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 30-ft fence 400 ft from home plate. How strong must a split-second 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_Ch06pp501-566 01/11/06 9:33 PM Page 532

532

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

Standardized Test Questions

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_Ch06pp501-566 01/11/06 9:33 PM Page 533

SECTION 6.3

Parametric Equations and Motion

533

Explorations

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 20-ft 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 71-ft-radius 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

Extending the Ideas

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_Ch06pp501-566 01/11/06 9:33 PM Page 534

534

CHAPTER 6

Applications of Trigonometry

6.4

Polar Coordinates

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.

Polar Coordinate System

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_Ch06pp501-566 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

Coordinate Conversion

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_Ch06pp501-566 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_Ch06pp501-566 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 ?

FOLLOW-UP

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

Equation Conversion

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_Ch06pp501-566 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.

Finding Distance Using Polar Coordinates

A radar tracking system sends out high-frequency 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

Self-Assessment: Ex. 7, 15, 23, 27, 35,

43, 51

Embedded Assessment: Ex. 61

5144_Demana_Ch06pp501-566 01/11/06 9:33 PM Page 539

SECTION 6.4

Polar Coordinates

539

QUICK REVIEW 6.4

(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

SECTION 6.4 EXERCISES

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

5144_Demana_Ch06pp501-566 01/11/06 9:33 PM Page 540

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.

Standardized Test Questions

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

Explorations

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

Extending the Ideas

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_Ch06pp501-566 01/11/06 9:33 PM Page 541

SECTION 6.5

Graphs of Polar Equations

541

6.5

Graphs of Polar Equations

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 and Parametric Curves

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

Symmetry

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_Ch06pp501-566 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_Ch06pp501-566 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_Ch06pp501-566 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 8-petal 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.

Rose Curves

The curve in Example 1 is a 3-petal rose curve and the curve in Example 3 is a 4-petal 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 8-petal 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_Ch06pp501-566 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

Limaçon Curves

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 “LEE-ma-sohn,” 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_Ch06pp501-566 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.

Other Polar Curves

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_Ch06pp501-566 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)

FOLLOW-UP

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

Self-Assessment: 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 ?

QUICK REVIEW 6.5

(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_Ch06pp501-566 01/11/06 9:33 PM Page 548

548

CHAPTER 6

Applications of Trigonometry

SECTION 6.5 EXERCISES

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_Ch06pp501-566 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_Ch06pp501-566 01/11/06 9:34 PM Page 550

550

CHAPTER 6

Applications of Trigonometry

6.6

De Moivre’s Theorem and

n

th Roots

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 equation-solving technique to include equations of the form

z n c

,

n

an integer and

c

a complex number.

The Complex Plane

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), Jean-Robert 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 two-dimensional vectors is the definition of absolute value.

5144_Demana_Ch06pp501-566 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.

Trigonometric Form of Complex Numbers

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_Ch06pp501-566 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.

Multiplication and Division of Complex Numbers

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_Ch06pp501-566 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 built-in 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_Ch06pp501-566 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

Powers of Complex Numbers

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_Ch06pp501-566 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.

Roots of Complex Numbers

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_Ch06pp501-566 01/11/06 9:34 PM Page 556

556

CHAPTER 6

Applications of Trigonometry

FOLLOW-UP

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

Self-Assessment: 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_Ch06pp501-566 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_Ch06pp501-566 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.

QUICK REVIEW 6.6

(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_Ch06pp501-566 01/11/06 9:34 PM Page 559

SECTION 6.6

De Moivre’s Theorem and nth Roots

559

SECTION 6.6 EXERCISES

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_Ch06pp501-566 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

Standardized Test Questions

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

Explorations

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

Extending the Ideas

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_Ch06pp501-566 01/11/06 9:34 PM Page 561

CHAPTER 6

Key Ideas

561

CHAPTER 6

Key Ideas

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_Ch06pp501-566 01/11/06 9:34 PM Page 562

562

CHAPTER 6

Applications of Trigonometry

CHAPTER 6

Review Exercises

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_Ch06pp501-566 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 foot-pounds

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_Ch06pp501-566 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 place-kicker 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

15-ft-high 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 60-ft-radius 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

5144_Demana_Ch06pp501-566 01/11/06 9:34 PM Page 565

CHAPTER 6

Project

565

CHAPTER 6

Project

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 phase-space 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_Ch06pp501-566 01/11/06 9:34 PM Page 566

Was this manual useful for you? yes no
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Download PDF

advertisement