Chapter 6 Applications of Trigonometry

6.1 VECTORS IN THE PLANE

Slide 6- 3

Quick Review

-1

1. Find the values of and .

32. Solve for in degrees. sin

11

3. A naval ship leaves Port Northfolk and averages 43 knots (nautical

mph) traveling for 3 hr on a bearing of 35 and then 4 h

x y

r on a course

of 120 . What is the boat's bearing and distance from Port Norfolk

after 7 hr.

Slide 6- 4

Quick Review

The point is on the terminal side of the angle . Find the

measure of if 0 360 .

4. (5,7)

5. (-5,7)

P

P

P

Slide 6- 5

Quick Review Solutions

-1

1. Find the values of and .

32. Solv

7.5, 7.5 3

64.8e for in degrees. sin 11

3. A naval ship leaves Port Northfolk and averages 43

x y

x y

knots (nautical

mph) traveling for 3 hr on a bearing of 35 and then 4 hr on a course

of 120 . What is the boat's bearing and distance from Port N

distance=

orfolk

a 224.2; fter 7 bearin ghr. =84.9

Slide 6- 6

Quick Review Solutions

The point is on the terminal side of the angle . Find the

measure of if 0 360 .

4. (5 54.5

125.

,7)

5. (-5,7 5)

P

P

P

Slide 6- 7

Directed Line Segment

Slide 6- 8

Two-Dimensional Vector

A is an ordered pair of real numbers,

denoted in as , . The numbers and are

the of the vector . The

of the vector ,

a b a b

a b

two - dimensional vector

component form

components standard representation

v

v

is the arrow from the origin to the point ( , ).

The of is the length of the arrow and the

of is the direction in which the arrow is pointing. The vector

= 0,0 , called the

a b

magnitude direction

0 ze

v

v

, has zero length and no direction.ro vector

Slide 6- 9

Initial Point, Terminal Point, Equivalent

Find the vector for both

• RS= <3,4>• QP= <3,4>

Slide 6- 11

Magnitude

1 1 2 2

2 2

2 1 2 1

2 2

If is represented by the arrow from , to , , then

.

If , , then .

x y x y

x x y y

a b v a b

v

v

v

Slide 6- 12

Example Finding Magnitude of a Vector

Find the magnitude of represented by , where (3, 4) and

(5,2).

PQ P

Q

v

Slide 6- 13

Example Finding Magnitude of a Vector

Find the magnitude of represented by , where (3, 4) and

(5,2).

PQ P

Q

v

2 2

2 1 2 1

2 2

5 3 2 ( 4)

2 10

v x x y y

Find Vector

• Find the magnitude of v represented by , where S=(2, -8) and T= (-3, 7)

Slide 6- 15

Vector Addition and Scalar Multiplication

1 2 1 2

1 1 2 2

Let , and , be vectors and let be a real number

(scalar). The (or ) is the vector

, .

The and the vector is

u u v v k

u v u v

k k u

sum resultant of the vectors and

product of the scalar

u v

u v

u v

k u

u1 2 1 2, , .u ku ku

Slide 6- 16

Example Performing Vector Operations

Let 2, 1 and 5,3 . Find 3 . u v u v

Slide 6- 17

Example Performing Vector Operations

Let 2, 1 and 5,3 . Find 3 . u v u v

3 3 2 , 3 1 = 6, 3

3 = 6, 3 5,3 6 5, 3 3 11,0

u

u v

Group Work

• Let u=<-1,3> and v=<5, -6>

• Find• A) u+v

• B) 3u

• C) 2u+(-1)v

Slide 6- 19

Unit Vectors

A vector with | | 1 is a . If is not the zero vector

10,0 , then the vector is a

| | | |

.

unit vector

unit vector in the direction

of

u u v

vu v

v v

v

Slide 6- 20

Example Finding a Unit Vector

Find a unit vector in the direction of 2, 3 . v

Slide 6- 21

Example Finding a Unit Vector

Find a unit vector in the direction of 2, 3 . v

222, 3 2 3 13, so

1 2 32, 3 ,

13 13 13

v

v

v

Find the unit vector

• P=<3,9>

• Q=<1, 6>

Slide 6- 23

Standard Unit Vectors

The two vectors 1,0 and 0,1 are the standard

unit vectors. Any vector can be written as an expression

in terms of the standard unit vector:

,

,0 0,

1,0 0,1

a b

a b

a b

a b

i j

v

v

i j

Slide 6- 24

Resolving the Vector

If has direction angle , the components of can be computed

using the formula = cos , sin .

From the formula above, it follows that the unit vector in the

direction of is cos ,sin .

v v

v v v

vv u

v

Slide 6- 25

Example Finding the Components of a Vector

Find the components of the vector with direction angle 120 and

magnitude 8.

v

, 8cos120 ,8sin120

1 3 8 ,8

2 2

4, 4 3

So 4 and 4 3.

a b

a b

v

Slide 6- 26

Example Finding the Components of a Vector

Find the components of the vector with direction angle 120 and

magnitude 8.

v

, 8cos120 ,8sin120

1 3 8 ,8

2 2

4, 4 3

So 4 and 4 3.

a b

a b

v

Slide 6- 27

Example Finding the Direction Angle of a Vector

Find the magnitude and direction angle of 2,3 .u

Slide 6- 28

Example Finding the Direction Angle of a Vector

Find the magnitude and direction angle of 2,3 .u

2 2| | 2 3 13

Let be the direction angle of , then

2,3 13 cos , 13 sin

2 13 cos

56.3

u

u

u

Slide 6- 29

Velocity and SpeedThe velocity of a moving object is a vector because velocity has both magnitude and direction. The magnitude of velocity is speed.

Word Problem

• The pilot pilots the plane from San Franciso due east. There is a 65 mph wind with the bearing 60 degrees (from the y-axis). Find the compass heading the pilot should follow, and determine what the airplane’s ground speed will be (assuming its speed with no wind is 450 mph).

Answer

• Bearing should be approx 94.14 degrees

• Speed is 505.12 mph

Word Problem

• A jet is flying on a bearing of 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.

Answer

• <453.15,211.31>

Homework Practice

• Pg 511 #1-45 eoo

POLAR COORDINATES

Slide 6- 36

Quick Review

1. Determine the quadrants containing the terminal side of the angle:

4 / 3

2. Find a positive and negative angle coterminal with the given angle:

/ 3

3. Write a standard form equation for the circle wit

h center at ( 6,0)

and a radius of 4.

Slide 6- 37

Quick ReviewUse the Law of Cosines to find the measure of the third side of the given triangle.

4.

40º8

10

5.

35º6

11

Slide 6- 38

Quick Review Solutions

1. Determine the quadrants containing the terminal side of the angle:

4 / 3

2. Find a positive and negative angle coterminal with the given angle:

/ 3

3. Write a s

II

5 /3,

tandard form e

7 /3

qu

2 2

ation for the circle with center at ( 6,0)

and a radius ( 6) 1f 6o 4. x y

Slide 6- 39

Quick Review Solutions

Use the Law of Cosines to find the measure of the third side of the given triangle.

4.

40º8

10

5.

35º6

11

6.47

Slide 6- 40

The Polar Coordinate System

Slide 6- 41

Example Plotting Points in the Polar Coordinate System

Plot the points with the given polar coordinates.

(a) (2, / 3) (b) ( 1, 3 / 4) (c) (3, 45 )P Q R

Slide 6- 42

Example Plotting Points in the Polar Coordinate System

Plot the points with the given polar coordinates.

(a) (2, / 3) (b) ( 1, 3 / 4) (c) (3, 45 )P Q R

Slide 6- 43

Finding all Polar Coordinates of a Point

Let the point have polar coordinates ( , ). Any other polar coordinate of

must be of the form ( , 2 ) or ( , (2 1) ) where is any

integer. In particular, the pole has polar coordinates (0,

P r

P r n r n n

), where is

any angle.

Slide 6- 44

Coordinate Conversion Equations

2 2 2

Let the point have polar coordinates ( , ) and rectangular

coordinates ( , ). Then

cos , sin , , tan .

P r

x y

yx r y r x y r

x

Slide 6- 45

Example Converting from Polar to Rectangular Coordinates

Find the rectangular coordinate of the point with the polar

coordinate (2, 7 / 6).

Slide 6- 46

Example Converting from Polar to Rectangular Coordinates

Find the rectangular coordinate of the point with the polar

coordinate (2, 7 / 6).

For point (2,7 / 6), 2 and 7 / 6.

cos sin

2cos 7 / 6 2sin 7 / 6

3 12 2

2 2

3 1

The recangular coordinate is 3, 1 .

r

x r y r

x y

x y

x y

Slide 6- 47

Example Converting from Rectangular to Polar Coordinates

Find two polar coordinate pairs for the point with the rectangular

coordinate (1, 1).

Slide 6- 48

Example Converting from Rectangular to Polar Coordinates

Find two polar coordinate pairs for the point with the rectangular

coordinate (1, 1).

2 2 2

2 2 2

2

For the point (1, 1), 1 and 1.

tan

11 ( 1) tan

1

2 tan 1

2 4

x y

yr x y

x

r

r

r n

3Two polar coordinates pairs are 2, and 2, .

4 4

Slide 6- 49

Example Converting from Polar Form to Rectangular Form

Convert 2sec to rectangular form.r

Slide 6- 50

Example Converting from Polar Form to Rectangular Form

Convert 2sec to rectangular form.r

2sec

2sec

cos 2

2

r

r

r

x

Slide 6- 51

Example Converting from Polar Form to Rectangular Form

2 2

Convert - 2 3 13 to polar form.x y

Slide 6- 52

Example Converting from Polar Form to Rectangular Form

2 2

Convert - 2 3 13 to polar form.x y

2 2

2 2

2 2

2 2 2

2

2 3 13

4 4 6 9 13

4 6 0

Substitute , cos , and sin .

4 cos 6 sin 0

4cos 6sin 0

0 or 4cos 6sin

0 is a single point that is also on the other graph

x y

x x y y

x y x y

r x y x r y r

r r r

r r

r r

r

.

Thus the equation is 4cos 6sin .r

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Transparency 6 - 53

6.4Polar Coordinates

Page 537

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Transparency 6 - 54

6.4Polar Coordinates (cont’d)

Page 537

Homework Practice

• Pg 539 #1-50 eoe

LIMITS AND MOTION: THE TANGENT PROBLEM

Slide 10- 57

Quick Review

1. Find the slope of the line determined by the points (2, 3) and (4,4).

Write an equation for the specified line.

2. Through (3,4) with slope 4.

3. Through (3,4) and parallel to 2 3.

4. Through (3,4)

y x

3

and perpendicular to 2 3.

5. Simplify the expression assuming 0

1 1

y x

h

h

h

Slide 10- 58

Quick Review Solutions

1. Find the slope of the line determined by the points (2, 3) and (4,4).

Write an equation for the specified line.

2. Through (3,4) with slope 4.

3. Through (3,4) and

7/2

4 4( 3

parallel to 2

)

y

y x

x

2

3

3.

4. Through (3,4) and perpendicular to 2 3.

5. Simplify the expre

4 2( 3)

14 (

ssion as

3)2

suming

13

3

0

1

y x

y x

h

y

h

h

x

h

h

What is Tangent?

Slide 10- 60

Average VelocityAverage velocity is the change in position divided by the change in time.

Slide 10- 61

Limits at a (Informal)

When we write "lim ( ) ," we mean that ( ) gets arbitrarily

close to as gets arbitrarily close (but not equal) to .x af x L f x

L x a

Slide 10- 62

Example Finding the Slope of a Tangent Line

3

Use limits to find the slope of the tangent line to the graph of

at the point (1,1).s t

Slide 10- 63

Example Finding the Slope of a Tangent Line

3

Use limits to find the slope of the tangent line to the graph of

at the point (1,1).s t

3 3

1 1

2

1

2

1

1lim lim

11 1

lim1

lim 1

1 1 1

3

t t

t

t

s t

t tt t t

tt t

Example:

• A ball rolls down a ramp so that its distance s from the top of the ramp after t seconds is exactly feet. What is its instantaneous velocity after 3 second?

2t

Slide 10- 65

Average Rate of Change

If ( ), then the of with respect to

( ) ( )on the interval [ , ] is .

Geometrically, this is the slope of the secant line through ( , ( )) and

( , ( )).

y f x y x

y f b f aa b

x b aa f a

b f b

average rate of change

Slide 10- 66

Derivative at a Point

The , denoted by '( ) and read

( ) ( )" prime of " is '( ) lim , provided the limit exists.

Geometrically, this is the slope of the tangent line through

x a

f a

f x f af a f a

x a

derivative of the function at = f x a

( , ( )).a f a

Slide 10- 67

Derivative at a Point (easier for computing)

0

The , denoted by '( ) and read

( ) ( )" prime of " is '( ) lim , provided the limit exists.

h

f a

f a h f af a f a

h

derivative of the function at = f x a

Slide 10- 68

Example Finding a Derivative at a Point

Find '(3) if ( ) 2 4.f f x x

Slide 10- 69

Example Finding a Derivative at a Point

Find '(3) if ( ) 2 4.f f x x

0

0

0

(3 ) (3)'(3) lim

2(3 ) 4 2(3) 4 lim

lim 2

2

h

h

h

f h ff

hh

h

Slide 10- 70

Derivative

0

If ( ), then

, is the function ' whose value at is

( ) ( )'( ) lim for all values of where the

limit exists.

h

y f x

f x

f x h f xf x x

h

derivative of the function with respect to

at =

f

x x a

Slide 10- 71

Example Finding the Derivative of a Function

2Find '( ) if ( ) 2 .f x f x x

Slide 10- 72

Example Finding the Derivative of a Function

2Find '( ) if ( ) 2 .f x f x x

0

2 2

0

2 2 2

0

0

( ) ( )'( ) lim

2( ) 2 lim

2 4 2 2 lim

lim 4 2

4

h

h

h

h

f x h f xf x

hx h x

hx xh h x

hx h

x

Example:

• Find if )(' xf 2)( xxf

Example:

• Find if dx

dy

xy

1

Homework Practice

• P 801 #1-32 eoe

INTEGRAL: THE AREA PROBLEM

Slide 10- 77

Quick Review

2

5

1

52

1

1. List the elements of the sequence for 1, 2,3, 4.

Find the sum.

2. ( 1)

3.

4. A car travels at an average speed of 56 mph for 3 hours.

How far does it travel?

5. A pump working at 4 g

k

k

k

a k k

k

k

al/min pumps for 3 hours.

How many gallons are pumped?

Slide 10- 78

Quick Review Solutions

2

5

1

52

1

1,4,9,16

2

1. List the elements of the sequence for 1,2,3,4.

Find the sum.

2. ( 1)

3.

4. A car travels at an average speed of 56 mph for 3 hours.

How far does it travel?

0

55

1 68 m

k

k

k

a k k

k

k

5. A pump working at 4 gal/min pumps for 3 hours.

How many gallons are pumped?

iles

720 gal lons

Slide 10- 79

Example Computing Distance Traveled

A car travels at an average rate of 56 miles per hour for 3 hours and 30 minutes. How far does the car travel?

Slide 10- 80

Example Computing Distance Traveled

A car travels at an average rate of 56 miles per hour for 3 hours and 30 minutes. How far does the car travel?

The distance traveled is , the time interval has length ,

and is the average velocity. Therefore,

56 mph 3.5 hours 196 miles.

s t

s

ts

s tt

Slide 10- 81

Limits at Infinity (Informal)

When we write "lim ( ) ," we mean that ( ) gets arbitrarily close

to as gets arbitrarily large.xf x L f x

L x

Slide 10- 82

1( )

a

iif x x

1 2

Let ( ) be a continuous function over an interval [ , ].

Divide [ , ] into subintervals of length ( ) / .

Choose any value in the first subinterval, in the second, and so on.

Compute

y f x a b

a b n x b a n

x x

1 2 3

1

( ), ( ), ( ),... ( ) multiply each value by and sum

up the products. In sigma notation, the sum of the products is ( )

n

n

ii

f x f x f x f x x

f x x

Slide 10- 83

Definite Integral

1

1

Let be a function on [ , ] and let ( ) be defined as above.

The definite integral of over [ , ], denoted ( ) , is given by

( ) lim ( ) , provided the limit exists.

If t

n

ii

b

a

b a

in ia

f a b f x x

f a b f x dx

f x dx f x x

he limit exists, we say is integrable on [ , ].f a b