precalculus with limits a graphing approach · · 2012-04-06the jobs of a civil engineer. to...

efore a bridge, road, tunnel, or other large structure can be designed andbuilt, the land on which it will be built needs to be surveyed. This is one ofthe jobs of a civil engineer.

To survey a site, civil engineers rely heavily on¡¡r trigonometry. One of the primary skills a surveyor must

master is that of indirect measurement. For instance, thedistances, a and c labeled in the diagram below, can befound using the Law of Sines. (See Exercises 27 and 28on page 514.)

sin 69° sin 55°

Using a graphing utility in degree mode, you find

17.779 meters.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

460 6 Additional Topics in Trigonometry

Low of Sines

Think About the Proof

To prove the Law of Sines, let Abe the altitude of either triangleshown in the figure at the right.Then you have

hsin A = or h = b sin A

b

and

sin B = - or = α sin B.α

By equating the two values of A,you can establish part of the Lawof Sines. Can you see how toestablish the other part? Thedetails of the proof are given inthe appendix.

Introduction The Ambiguous Case (SSA)

Area of an Oblique Triangle Application

Introduction

In Chapter 4 you looked at techniques for solving right triangles. In this section

and the next, you will solve oblique triangles—triangles that have no right

angles. As standard notation, the angles of a triangle are labeled A, B, and C,

and their opposite sides are labeled α, b, and c, as shown in Figure 6.1.

To solve an oblique triangle, you need to know the measure of at least one side

and any two other parts of the triangle—two sides, two angles, or one angle and

one side. This breaks down into the following four cases.

1. Two angles and any side (AAS or ASA)

2. Two sides and an angle opposite one of them (SSA)

3. Three sides (SSS)

4. Two sides and their included angle (SAS)

The first two cases can be solved using the Law of Sines, whereas the last two

cases require the Law of Cosines (Section 6.2).

Law of Sines

If ABC is a triangle with sides α, b, and c, then

_b c

s'mB

α

sin A sin C

Oblique TrianglesA is obtuse.

C

Note The Law of Sines can also be written in the reciprocal form

sin A _ sin 5 ^ sin C

α b c


6.7 I Law of Sines 461

Figure 6.2

b = 21A ft

Sttcdy

When you are solving triangles,a careful sketch is useful as aquick test for the feasibility of ananswer. Remember that thelongest side lies opposite thelargest angle, and the shortest sidelies opposite the smallest angle.

Figure 6.3

EXAMPLE 1 Given Two Angles and One Side—AAS

For the triangle in Figure 6.2, C = 102.3°, B = 28.7°, and b = 27 A feet. Findthe remaining angle and sides.

Solution

The third angle of the triangle is

A = 180° - B - C = 180° - 28.7° - 102.3° = 49.0°.

By the Law of Sines, you have

a b c

sin 49° ¯ sin 28.7° ¯ sin 102.3°'

Using b = 27.4 produces

27.4a = : - - (sin 49°)

sin 28.743.06 feet

and

c = —g r (sin 102.3°) 55.75 feet.

EXAMPLE 2 Given Two Angles and One Side—ASA

¯ c = 22 ft

A pole tilts toward the sun at an 8° angle from the vertical, and it casts a 22-footshadow. The angle of elevation from the tip of the shadow to the top of the poleis 43°. How tall is the pole?

Solution

From Figure 6.3, note that A = 43° and B = 90° + 8° = 98°. Thus, the thirdangle is

C = 180° - A - B = 180° - 43° - 98° = 39°.


a c

sin 43° " sin 39°'

Because c = 22 feet, the length of the pole is

22 0 (sin 43°) 23.84 feet. ^

sin 39

Note F°r practice, try reworking Example 2 for a pole that tilts away from thesun under the same conditions.



The Ambiguous Case (SSA)

In Examples 1 and 2 you saw that two angles and one side determine a uniquetriangle. However, if two sides and one opposite angle are given, three possiblesituations can occur: (1) no such triangle exists, (2) one such triangle exists, or(3) two distinct triangles may satisfy the conditions.

The Ambiguous Case (SSA)

Consider a triangle in which you are given α, b, and A. (h = b sin A)

Sketch

A is acute. A is acute. A is acute. A is obtuse. A is obtuse.

Necessarycondition

Possibletriangles

a<h

None

a =

One

a > b

One

h<a<b

Two None

a>b

One

Figure 6.4

One solution: a > ¡

C

=l2in./ `¯\j ϊ = 22in.

Note Notice in Example 3 that A isacute and a>b, which results in onepossible triangle.

EXAMPLE 3 Single-Solution Case— SSA

For the triangle in Figure 6.4, a = 22 inches, b = 12 inches, and A = 42°. Findthe remaining side and angles.

SolutionBy the Law of Sines, you have

_22_

sin 42°

_I2_

si β

sinβ

¯ i a ίsin 42C

\ 22

B 21.41°.

0.3649803

β is acute.

Now you can determine that C \%Q° - 42° ~ 2\A\° = 116.59°, and theremaining side is given by

sin 116.59° sin 42°

c = sin 116.59,/_J2_

I sin 42'29.40 inches.



Figure 6.5

No solution: a < h

b = 25

a= 15

The Interactive CD-ROM shows everyexample with its solution; clicking onthe Try It! button brings up similarproblems. Guided Examples andIntegrated Examples show step-by-stepsolutions to additional examples.Integrated Examples are related toseveral concepts in the section.

EXAMPLE 4 ^ No-Solution Case—SSA

Show that there is no triangle for which α = 15, b = 25, and A = 85°.

Solution

Begin by making the sketch shown in Figure 6.5. From this figure it appearsthat no triangle is formed. You can verify this by using the Law of Sines.

sin A

15

sin 85°

sin B

25

sinβ

/sin 85'sin B = 251

\ 151.660 > 1

This contradicts the fact that ¡sinβ < 1. Hence, no triangle can be formedhaving sides α — 15 and b = 25 and an angle of A = 85°. ί^/

Note Because h = b sin A =31 (sin 20.5°) 10.86 meters you canconclude that there are two possibletriangles (because h < α < b).

EXAMPLE 5 ^ Two-Solution Case—SSA

Find two triangles for which α = 12 meters, b = 31 meters, and A = 20.5°.

Solution


Figure 6.6

b = 31 m--""'/ \ .,J^"^ ' \α = 12 rn

O.5° /64.8°λ

α = 12 m

sin A sin B

^s inA\ sin 20.5'

120.9047.

There are two angles B 64.8° and B2 115.2° between 0° and 180° whosesine is 0.9047. For B¦ 64.8°, you obtain

C 180° - 20.5° - 64.8° = 94.7°

α 12c = - —(sin C) = - —(sin 94.7°) 34.15 meters,

sin A sin 20.5

ForZ?2 H5.2°, you obtain

C 180° - 20.5° - 115.2° = 44.3°

α 12c = -.—:(sin C) = -(sin 44.3°) 23.93 meters.

sin A sin 2O.5°

The resulting triangles are shown in Figure 6.6.


464 6 I Additional Topics in Trigonometry

Area of an Oblique Triangle

The procedure used to prove the Law of Sines leads to a simple formula for thearea of an oblique triangle. Referring to Figure 6.7, note that each triangle hasa height of

h = b sin A.

Consequently, the area of each triangle is given by

Area = (base)(height) = r¯(c) ( sin A) = be sin A.

By similar arguments, you can develop the formulas

Area = ~αb sin C = αc sin B.

Note Note that if angle A is 90°, theformula gives the area for a righttriangle as

Area = 2bc = ase)(height).

Similar results are obtained for anglesC and B equal to 90°.

Figure 6.7

A is acute.

Area of an Oblique Triangle

The area of any triangle is given by one-half the product of the lengthsof two sides times the sine of their included angle. That is,

1 1 1Area = —be sin A = ~αb sin C = —αc sin B.

Figure 6.8

> = 52 m

C α = 90 m

EXAMPLE 6 Finding the Area of an Oblique Triangle*¿,¿

Find the area of a triangular lot having two sides of lengths 90 meters and52 meters and an included angle of 102°.

SolutionConsider α = 90 m, b ~ 52 m, and angle C = 102°, as shown in Figure 6.8.Then the area of the triangle is

Area = ~αb sin C = (9O)(52)(sin 102°) 2289 square meters. ^


6.1 Law of Sines 465

Figure (

f·

w-<—`1

β<

>.9

{ ^JX2XI ·̂'

-~E ,s

`` S' 52°

»'\ t

`4O°\

\

Λ1

t

8 1an

D

Figure 6.10

C

= 8 km

Application

EXAMPLE 7 An Application of the Law of SinesI5e«ί¿ί¿e

The course for a boat race starts at point A and proceeds in the directionS 52° W to point B, then in the direction S 40° E to point C, and finally back toA, as shown in Figure 6.9. The point C lies 8 kilometers directly south of pointA. Approximate the total distance of the race course.

Because lines BD and AC are parallel, it follows that ΔBCA = ΔDBC.Consequently, triangle ABC has the measures shown in Figure 6.10. For angleB, you have B = 180° - 52° - 40° = 88°. Using the Law of Sines

sin 52° sin 88° sin 40°

you can let b = 8 and obtain the following.

- (sin 52°) 6.308sin 88 sin

The total length of the course is approximately

Length 8 + 6.308 + 5.145 = 19.453 kilometers.

;(sin 40°) 5.145

Group Mm\\\ Using the Law of Sines

In this section, you have been using the Law of Sines to solve obliquetriangles. Can the Law of Sines also be used to solve a right triangle?If so, write a short paragraph explaining how to use the Law of Sinesto solve the following two triangles. Is there an easier way to solvethese triangles?

b. (SSA)

B

α = ! 5

C

50° c = 2O



6.1 /// EXERCISES

The Interactive CD-ROM contains step-by-step solutions to all odd-numbered Section and Review Exercises. It also provides TutorialExercises, which link to Guided Examples for additional help.

In Exercises 1-14, use the given information to solvethe triangle.

1. A = 30°, α = 20, B = 45°

2. C = 120°, c = 15, B = 45°

3. A = 10°, α = 7.5, B = 60°

4. C = 135°, c = 45, B = 10°

5. A = 36°, α = 8, = 5

6. A = 60°, α = 9, c = 10

7. A = 150°, C = 20°, α = 200

8. A = 24.3°, C = 54.6°, c = 2.68

9. A = 83° 20', C = 54.6°, c= 1 8 . 1

10. A = 5° 40', β = 8°!5', b = 4.8

11. 5 = 15° 30', α = 4.5, b = 6.8

12. C = 85° 20', α = 35, c = 50

13. A = 1 1 0 ° 15', α = 48, b = 16

14. β = 2° 45', b = 6.2, c = 5.8

In Exercises 15-18, use the given information to solvethe triangle. If two solutions exist, find both.

15. A = 58°, α = 4.5, b= 12.8

16. A = 58°, α = 11.4, b = 12.8

17. A = 58°, α = 4.5, fe = 5

18. A = 110°, α = 125, b = 100

In Exercises 19 and 20, find a value for such that thetriangle has (a) one, (b) two, and (c) no solution(s).

19. A = 36°, α = 5 20. A = 60°, α = 10

21. Height A flagpole is located on a slope that makesan angle of 12° with the horizontal. The pole casts a16-meter shadow up the slope when the angle ofelevation of the sun is 20°.

(a) Draw a triangle that represents the problem.

(b) Write an equation involving the unknown.

(c) Find the height of the flagpole.

22. Height Because of prevailing winds, a tree grew sothat it was leaning 6° from the vertical. At a point 30meters from the tree, the angle of elevation to the topof the tree is 22° 50' (see figure). Find the height h ofthe tree.

• 30 m

23. Angle of Elevation A 10-meter telephone pole castsa 17-meter shadow directly down a slope when theangle of elevation of the sun is 42° (sec figure). Findθ, the angle of elevation of the ground.

48

24. Flight Path A plane flies 500 kilometers with abearing of N 44° W from β to C (see figure). Theplane then flies 720 kilometers from C to A. Find thebearing of the flight from C to A.

N C

720 km

N

500 km

\ 44°

46°Λ



25. Bridge Design A bridge is to be built across a smalllake from B to C (see figure). The bearing from B toC is S 41° W. From a point A, 100 meters from β, thebearings to B and C are S 74° E and S 28° E, respec-tively. Find the distance from B to C.

26. Railroad Track Design The circular arc of a railroadcurve has a chord of length 3000 feet and acentral angle of 40°.

(a) Draw a figure that gives a visual representation ofthe problem. Show the known quantities on thefigure and use variables r and s to represent theradius of the arc and the length of the arc, respec-tively.

(b) Find the radius r of the circular arc.

(c) Find the length Λ` of the circular arc.

27. Glide Path A pilot has just started on the glide pathfor landing at an airport where the length of the run-way is 9000 feet. The angles of depression from theplane to the ends of the runway are 17.5° and 18.8°.

(a) Draw a figure that gives a visual representation ofthe problem.

(b) Find the air distance the plane must travel unti ltouching down on the near end of the runway.

(c) Find the ground distance the plane must traveluntil touching down.

(d) Find the altitude of the plane when the pilotbegins the descent.

28. Altitude The angles of elevation to an airplane fromtwo points A and B on level ground are 51° and 68°,respectively. The points A and B are 2.5 miles apart,and the airplane is east of both points in the samevertical plane. Find the altitude of the plane.

29. Locating a Fire Two fire towers A and B are 30kilometers apart. The bearing from A to B is N 65° E.A fire is spotted by a ranger in each tower, and itsbearings from A and B are N 28° E and N 16.5° W,respectively (see figure). Find the distance of the firefrom each tower.

N

30. Distance A boat is sailing due east parallel to theshoreline at a speed of 10 miles per hour. At a giventime the bearing to the lighthouse is S 70° E, and 15minutes later the bearing is S 63° E (see figure). Findthe distance from the boat to the shoreline if the light-house is at the shoreline.

m N

τ «*SS · c.J-63° W

31. Distance A family is traveling due west on a roadthat passes a famous landmark. At a given time thebearing to the landmark is N 62° W, and after thefamily travels 5 miles farther the bearing is N 38° W.What is the closest the family will come to the land-mark while on the road?



32. Engine Design The connecting rod in an engine is6 inches in length and the radius of the crankshaft is

2 inches (see figure). Let dbe the distance the pistonis from the top of its stroke for the angle θ.

(a) Use a graphing utility to complete the table.

θ

d

0° 45° 90° 135° 180°

(b) The spark plug fires at θ = 5° before top deadcenter. How far is the piston from the top of itsstroke at this time?

Piston

1 ' ¯I- in.

33. Graphical and Numerical Analysis In the figure, aand β are positive angles.

(a) Write a as a function of β.

(b) Use a graphing u t i l i t y to graph the function.Determine its domain and range.

(c) Write c as a function of β.

(d) Use a graphing utility to graph the function inpart (c). Determine its domain and range.

(e) Use a graphing utility to complete the table. Whatcan you infer?

βa

c

0 0.4 0.8 1. 2 1.6 2.0 2.4 2.8

34. Distance The angles of elevation θ and φ, to an air-plane are being continuously monitored at two obser-vation points A and B, which are 2 miles apart (seefigure). Write an equation giving the distance dbetween the plane and point B in terms of θ and φ.

In Exercises 35-40, find the area of the triangle havingthe indicated sides and angle.

35. C = 120°, α = 4, b = 6

36. B = 72° 30', a = 105, c = 64

37. A = 43° 45', b = 57, c = 85

38. A = 5° 15', b = 4.5, c = 22

39. B = 130°, a = 62, c = 20

40. C = 84° 30', a =16, = 20

41. Graphical Analysis

(a) Write the area A of the shaded region in thefigure as a function of ft

(b) Use a graphing utility to graph the area function.

(c) Determine the domain of the area function.Explain how the area of the region and thedomain of the function would change if the 8-centimeter line segment were decreased in length.

20cm

30 em


6.2 I Law of Cosines

Lain of CosinesIntroduction Heron's Formula

Introduction

Two cases remain in the list of conditions needed to solve an oblique triangle—SSS and SAS. The Law of Sines does not work in either of these cases. To seetwhy, consider the three ratios given in the Law of Sines.

_ α b

sin A sin B

c

s inC

To use the Law of Sines, you must know at leasangle. If you are given three sides (SSS), or two s(SAS), none of the above ratios would be complethe Law of Cosines.

me side and its oppositeand their included anglesuch cases you can use

Standard Form

a1 = b2 + c2 - 2¿>ccosA cos A =

Law of CosinesAlternative Form

7 9 ι 9 9b2 + c2 - α2

b2 = a2 + c2 — 2ac cos B cos B =

2bc7 . 9 7 7

a2 + c2 - b¿

2ac

c2 = a2 + b2 - 2ab cos C cos C = -b2 - c2

lab

B = (c, 0)

Proof /// Consider a triangle that has three acute angles, as shown in Figure6.11. In the figure, note that vertex B has coordinates (c, 0). Furthermore, C hascoordinates ( x , y ) , where x = fecosA and y = sinA. Because α is thedistance from vertex C to vertex B, it follows that

a = J(x - c)2+(y- O)2

a2 = ( cosA - c)2 + ( s inA) 2

a2 = b2 cos2 A — 2bc cos A + c2 + b2 sin2 A

a2 = ¿>2(sin2A + cos2 A) + c2 — 2 ccosA

a2 = b2 + c2 - 2bc cos A. s in 2 A + cos2 A = 1

A similar argument can be used for a triangle having an obtuse angle. ///

WATERFORD KETTERING HIGH SCHOOL

2800 KETTERING DR.WATERFORD, Ml 48329Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.


Figure 6.12

ί·= 1 4 f t

b= 19f t

Note that if A = 90° in Figure 6 .11, then cos A = 0 and the first form of theLaw of Cosines becomes the Pythagorean Theorem.

Thus, the Pythagorean Theorem is actually just a special case of the more

general Law of Cosines.

EXAMPLE 1 Three Sides of a Triangle—SSS

Find the three angles of the triangle whose sides have lengths α = 8 feet,

b = 19 feet, and c = 14 feet.

Solution

It is a good idea first to find the angle opposite the longest side—side b in thiscase (see Figure 6.12). Using the Law of Cosines, you find that

cos B =α2 + c2 - b2 82 + 142 - 192

2αc 2(8)(I4)-0.45089.

Because cos β is negative, you know that B is an obtuse angle given by B

116.80°. At this point you could use the Law of Cosines to find cos A and cos C.However, knowing that B 116.80°, it is simpler to use the Law of Sines to

obtain the following.

_b

sinS

sin A = α

α

sin A

si β

b

sin 116.80'

190.37582

Because B is obtuse, you know that A must be acute, because a triangle can

have at most one obtuse angle. Thus, A 22.08° and

C 180° - 22.08° -

= 41.12°.

16.80°

Do you see why it was wise to find the largest angle first in Example 1?Knowing the cosine of an angle, you can determine whether the angle is acute

or obtuse. That is,

c o s f l > O for O°<Θ<9O°

cosθ<O for 9 O ° < Θ < I 8 O ° .

Acute

Obtuse

So, in Example 1, once you found that angle B was obtuse, you subsequentlyknew that angles A and C were both acute. If the largest angle is acute, theremaining two angles will be acute also.


6.2 I Law of Cosines 471

Figure 6.13

The pitcher's mound is not halfwaybetween home plate and second base.

EXAMPLE 2 <β Two Sides and the IncludedAngle—SAS

The pitcher's mound on a softball field is 46 feet from home plate and the dis-tance between the bases in 60 feet, as shown in Figure 6.13. How far is thepitcher's mound from first base?

SolutionIn triangle HPF, H = 45° (line HP bisects the right angle at H),f= 46, andp = 60. Using the Law of Cosines for this SAS case, you have

2 = f2 + p2 - 2fpCOSl·l

= 462 + 6O2 - 2(46)(6O) cos 45°

1812.8.

Therefore, the approximate distance from the pitcher's mound to first base is

h 7l8I2.8 42.58 feet. mi

EXAMPLE 3 IS Two Sides and the IncludedAngle—SAS

`Kea¿ ¿¢{e.

A ship travels 60 miles due east, then adjusts its course 15° northward, as shownin Figure 6.14. After traveling 80 miles in the new direction, how far is the shipfrom its point of departure?

SolutionYou have c = 60, B = 180° - 15° = 165°, and α = 80. Consequently, by theLaw of Cosines, it follows that

h2 = a2 + c2 - 2accosB

= 8O2 + 6O2 - 2(8O)(6O) cos 165°

19,273.

Therefore, the distance b is

b 7l9,273 138.8 miles.



Heron's work provided a scientificfoundation for land surveying andengineering. He invented a simpleform of the steam engine knownas Heron's Fountain.

Heron's Formula

The Law of Cosines can be used to establish the following formula for thearea of a triangle. This formula is credited to the Greek mathematician Heron(c. 100 B.C.).

Given any triangle with sides of lengths a, b, and c, the area of thetriangle is

Area = v ¯ ¯ α)(s - b)(s — c)

where s = (α + b + c)/2.

Proof /// From the preceding section, you know that

1Area = - be sin A

b2c2 sirrA

= A - ¿ 2(1 - cos2A)4

bc(\ + cosA) l l '- bc(\ - cos A) |.¿ Δ.

Using the Law of Cosines, you can show that

bc(\ + cos A) =

and

+ c - α +

1 α — b + c α + b — c~bc(l - cos A) = - — - ¯

(See Exercises 41 and 42.) Letting s = (α + b + c)/2, these two equations canbe rewritten as

- bc(\ + cos A) = s(s — α)

and

1bc(\ - cos A) = (s - b)(s - c).

Thus, you can conclude that

Area = -Js(s - α)(s - b)(s — c). Ill



EXAMPLE 4 Using Heron's Area Formula

Find the area of the triangular region having sides of lengths a = 43 meters,b = 53 meters, and c = 72 meters.

SolutionBecause

s = - (a + b + c`) = = 84,

Heron's Formula yields

Area = v/s(ίΓ1 a)(s - b)(s - c)

= >/84(4l)(3l)(l¯2 )

1131.89 square meters.

Group Rc The Area of a Triangle

You have now discussed three different formulas for the area of a triangle.

Standard Formula: Area — ¿ bh

Oblique Triangle: Area = 2 be sin A — j ¤b sin C = , αc sin S

Heron's Formula: Area = Vs(5 — α)(s — fe)(5 — c)

Use the most appropriate formula to find the area of each triangle.Show your work and give your reasons for choosing each formula.

a.2 f t

b.

2 f t

4ft

3 f t

4 f t

c. d.

3f t

4 f t



6.2 /// EXERCISESIn Exercises 1-10, use the Law of Cosines to solve thetriangle.

1. α = 6, b = 8, c = 12

2. α = 8, b = 3, c = 9

3. A = 30°, ft = 15, c = 30

4. C = 105°, α = 10, b = 4.5

5. α = 9, ft = 12, c = 15

6. α = 55, ft = 25, c = 72

7. α = 75.4, ft = 52, c = 52

8. α = 1.42, = 0.75, c = 1.25

9. β = 8° 45', α = 25, c = 15

10. B = 75° 20', α = 6.2, c = 9.5

In Exercises 11-16, solve the parallelogram. (Thelengths of the diagonals are given by c and d.)

11.

12.

13.

14.

15.

16.

α

4

25

10

40

10

b

6

35

14

60

25

20

18

50

d θ φ

30°

120°

80

12

35

18. Navigation A boat race runs along a triangularcourse marked by buoys A, B, and C. The race startswith the boats headed west for 2500 meters. Theother two sides of the course lie to the north of thefirst side, and their lengths are 1100 meters and 2000

meters. Draw a figure that gives a visual representa-tion of the problem and find the bearings for the lasttwo legs of the race.

Surveying To approximate the length of a marsh, asurveyor walks 300 meters from point A to point B,then turns 80° and walks 250 meters to point C (seefigure). Approximate the length AC of the marsh.

19.

20. Streetlight Design Determine the angle θ in thedesign of the streetlight shown in the figure.

17. Navigation A plane flies 810 miles from A to B with

a bearing of N 75° E. Then it flies 648 miles from Bto C with a bearing of N 32° E. Draw a figure thatgives a visual representation of the problem and find

the straight-line distance and bearing from C to A.

21. Distance Two ships leave a port at 9 A.M. One

travels at a bearing of N 53° W at 12 miles per hourand the other travels at a bearing of S 67° W at 16miles per hour. Approximate how far apart they are atnoon that day.



22. Distance A 100-foot vertical tower is to be erectedon the side of a hill that makes a 6° angle with thehorizontal (see figure). Find the length of each of thetwo guy wires that will be anchored 75 feet uphill anddownhill from the base of the tower.

23. Navigation On a map, Orlando is 178 millimetersdue south of Niagara Falls, Denver is 273 millimetersfrom Orlando, and Denver is 235 millimeters fromNiagara Falls (see figure).

(a) Find the bearing of Denver from Orlando.

(b) Find the bearing of Denver from Niagara Falls.

235 mm'

273 mm"

ι Niagara Falls

i78 mm

s Orlando

24. Navigation On a map, Minneapolis is 165 millime-ters due west of Albany, Phoenix is 216 millimetersfrom Minneapolis, and Phoenix is 368 millimetersfrom Albany.

(a) Find the bearing of Minneapolis from Phoenix.

(b) Find the bearing of Albany from Phoenix.

25. Baseball On a baseball diamond with 90-foot sides,the pitcher's mound is 60.5 feet from home plate.How far is it from the pitcher's mound to third base?

26. Baseball The baseball player in center field is play-ing approximately 330 feet from the televisioncamera that is behind home plate. A batter hits a f]yball that goes to the wall 420 feet from the camera(see figure). Approximate the number of feet that thecenter fielder has to run to make the catch if thecamera turns 8° to follow the play.

27. Engineering If Q is the midpoint of the line seg-ment PR in the truss rafter shown in the figure, findthe lengths of the line segments PQ, QS, and RS.

28. Aircraft Tracking To determine the distancebetween two aircraft, a tracking station continuouslydetermines the distance to each aircraft and the angleA between them. Determine the distance a betweenthe planes when A = 42°, b = 35 miles, and c = 20miles.



29. Engine Design An engine has a 7-inch connectingrod fastened to a crank (see figure).

(a) Use the Law of Cosines to write an equationgiving the relationship between x and θ.

(b) Write x as a function of θ. (Select the sign thatyields positive values of A'.)

(c) Use a graphing ut i l i ty to graph the function in

part (b).

(d) Use the graph in part (c) to determine the maxi-mum distance the piston moves in one cycle.

30. Paper Manufacturing In a certain process with con-tinuous paper, the paper passes across three rollers ofradii 3 inches, 4 inches, and 6 inches (see figure). Thecenters of the 3-inch and 6-inch rollers are d inchesapart, and the length of the arc in contact with thepaper on the 4-inch roller is ,v inches. Complete thefollowing table.

d (inches)

θ (degrees)

ί (inches)

9 10 12 13 ,4 15 16

Figure for 30

3 in.

Figure for 31

31. Awning Design A retractable awning lowers at anangle of 50° from the top of a patio door that is 7 feethigh (see figure). Find the length * of the awning if nodirect sunlight is to enter the door when the angle ofelevation of the sun is greater than 70°.

32. Circumscribed and Inscribed Circles Let R and rbe the radii of the circumscribed and inscribed circlesof a triangle ABC, respectively, and let s =(a + b + c)/2. Prove the following.

a b c(a) 2R= , 7 =

(b) r =

sin A sin B sin C

^- a)(s - b)(s - c)

Circumscribed and Inscribed Circles In Exercises 33and 34, use the results of Exercise 32.

33. Given a triangle with a = 25, b = 55, and c = 72,find the area of (a) the triangle, (b) the circumscribedcircle, and (c) the inscribed circle.

34. Find the length of the largest circular track that can bebuil t on a triangular piece of property whose sides are200 feet, 250 feet, and 325 feet.

In Exercises 35-40, use Heron's Area Formula to findthe area of the triangle.

35. a = 5, b = 1, c = 10

36. a = 2.5, b = 10.2, c = 9

37. a = 12, b = 15, c = 9

38. a = 75.4, h = 52, c = 52

39. a = 20, b = 20, c = 10

40. a = 4.25, b = 1.55, c = 3.00

41. Use the Law of Cosines to prove that

- bc(\ + cos A) = - - · -2 2 2

42. Use the Law of Cosines to prove that

1 a — b + c a + b~c-be (I ~ cos A) =- —— .


6.3 I Vectors in the Plane

Vectors in the PlaneIntroduction Component Form of a Vector Vector Operations SUnit Vectors Direction Angles Applications of Vectors

Figure 6.15

Introduction

Many quantities in geometry and physics, such as area, time, and temperature,can be represented by a single real number. Other quantities, such as forceand velocity, involve both magnitude and direction and cannot be completelycharacterized by a single real number. To represent such a quantity, you can usea directed line segment, as shown in Figure 6.15. The directed line segmentPQ has initial point P and terminal point Q. Its length is denoted by |

Two directed line segments that have the same length (or magnitude) and direc-tion are called equivalent. For example, the directed line segments in Figure6. 16 are all equivalent. The set of all directed line segments that are equivalentto a given directed line segment PQ is a vector v in the plane, written v = PQ.Vectors are denoted by lowercase, boldface letters such as u, v, and w.

Be sure you see that a vector in the plane can be represented by manydifferent directed line segments.

EXAMPLE 1 Vector Representation by Directed Line Segments

Let u be represented by the directed line segment from P = (0, 0) to Q =(3, 2), and let v be represented by the directed line segment from R = (1, 2) to5¯ = (4, 4), as shown in Figure 6.17. Show that u = v.

SolutionFrom the distance formula, it follows that ¯PQ and 7?5 have the same length.

\\PQ\\ - - O)2 =

= v/(4 - I)2 + (4 - 2)2 = 713

Moreover, both line segments have the same direction, because they are bothn ^

directed toward the upper right on lines having a slope of 3. Thus, PQ and RShave the same length and direction, and it follows that u = v. ^β>



Note Two vectors u = {u¡, u2) andv = (v j , iλ,} are equal if and only ifu¡ = v, and ιι2 = v2. For instance, inExample 1, the vector u from P =(0, 0) to Q = (3, 2) is

u = P¯Q = (3 - 0, 2 - 0) = (3, 2}

and the vector v from R = (1, 2) toS = (4, 4) is

v = R¯S = (4 - 1,4 - 2) = (3,2).

Component Form of a Vector

The directed line segment whose initial point is the origin is often the mostconvenient representative of a set of equivalent directed line segments. Thisrepresentative of the vector v is in standard position.

A vector whose initial point is at the origin (0, 0) can be uniquely representedby the coordinates of its terminal point (v,, v2). This is the component form ofa vector v, written

v = {v,, v2).

The coordinates v, and v2 are the components of v. If both the initial point andthe terminal point lie at the origin, v is the zero vector and is denoted by0 = (0, 0).

Component Form of a Vector

The component form of the vector with initial point P = (p\, p2) andterminal point Q = ¡.¾) 's

PQ = (<7i - Pi, ¾ ¯ ¯ P2> = \ vι, V2> = v.

The length (or magnitude) of v is given by

=If || v ¡| = ] , v is a unit vector. Moreover, ¡v¡ = 0 if and only if v is thezero vector 0.

EXAMPLE 2 Finding the Component Form of a VectorFigure 6.18

Find the component form and length of the vector v that has initial point(4, —7) and terminal point ( — 1 , 5).

Solution

Let P = (4, -7) = (pt,p2) and β = (-1,5) = ,, ). Then, the compo-nents of v = {v¡, v2) are given by

vι = <?i ¯ P i = "I ¯ 4 = ¯ 5

v2 = ¾ - p2 = 5 - (-7) = 12.

Thus, v = (—5, 12} and the length of v is

||v|| = vT-5)2 + W = vΊ69 = 13,

as shown in Figure 6.18. L7


6.3 I Vectors in the Plane 479

Figure 6.19

A computer animation of this conceptappears in the Interactive CD-ROM.

Figure 6.21

Vector Operations

The two basic vector operations are scalar multiplication and vector addition.Geometrically, the product of a vector v and a scalar k is the vector that is | |times as long as v. If k is positive, k\ has the same direction as v, and if k isnegative, k\ has the opposite direction of v, as shown in Figure 6.19.

To add two vectors geometrically, position them (without changing length ordirection) so that the initial point of one coincides with the terminal point of theother. The sum u + v is formed by joining the initial point of the second vec-tor v with the terminal point of the first vector u, as shown in Figure 6.20. Thistechnique is called the parallelogram law for vector addition because thevector u + v, often called the resultant of vector addition, is the diagonal of aparallelogram having u and v as its adjacent sides.

Figure 6.20

Definition of Vector Addition and Scalar Multiplication

Let u = {M|, M,) and v = {v,, v2) be vectors and let k be a scalar(a real number). Then the sum of u and v is the vector

u + v = {u¦ + v,, u2 + v2) Sum

and the scalar multiple of k times u is the vector

έu = k(u , M2) = (ku¡, ku2). Scalar multiple

The negative of v = (v , , v2) is

-v = ( l)v = (-v,, -v2) Negative

and the difference of u and v is

u - v = u + (-v) = (M, - v,, «2 - V2). Difference

To represent u — v graphically, you can use directed line segments with thesame initial points. The difference u — v is the vector from the terminal pointof v to the terminal point of u, as shown in Figure 6.21.



The component definitions of vector addition and scalar multiplication are

illustrated in Example 3. In this example, notice that each of the vector opera-

tions can be interpreted geometrically.

Note Figure 6.22(b) shows the vectordifference w — v as the sum w + (—v).

EXAMPLE 3 β1 Vector Operations

Let v = ( — 2, 5) and w = (3, 4), and find each of the following vectors.

a. 2v b. w - v c. v + 2w

Solution

a. Because v = ( — 2, 5), you have

2v = <2(-2), 2(5)}

= ( , 10).

A sketch of 2v is shown in Figure 6.22(a).

b. The difference of w and v is given by

w - v = (3 - (-2), 4 - 5)

= <5,-1).

A sketch of w — v i s shown in Figure 6.22(b).

c. Because 2w = (6, 8), it follows that

v + 2w = (-2, 5} + (6, 8}

= (-2 + 6, 5 + 8)

= (4, 13).

A sketch of v + 2w is shown in Figure 6.22(c).

(b)

(3,4)

(c)

(4, 13)



Note Property 9 can be stated asfollows: The length of the vector c\ isthe absolute value of c times thelength of v.

Vector addition and scalar multiplication share many of the properties ofordinary arithmetic.

Properties of Vector Addition and Scalar Multiplication

Lei u, v, and w be vectors and let c and d be scalars. Then thefollowing properties are true.

1. u + v = v + u2. (u + v) + w = u + (v + w)3. u + 0 = u4. u + (-u) = 05. c(du) = (cd)u

6. (c + d)u = cu + du7. c(u + v) = cu + cv8. l(u) = u, O(u) = 0

Some of the earliest work with

vectors was done by the Irish

mathematician William RowanHamilton (1805-1865). Hamilton

spent many years developing asystem of vector-like quantities

called quaternions. Although

Hamilton was convinced of the

benefits of quaternions, the

operations he defined did not

produce good models for physicalphenomena. It wasn't until the

latter half of the nineteenth

century that the Scottish physicist

James Maxwell (1831-1879)

restructured Hamilton's quater-

nions in a form useful for

representing physical quantities

such as force, velocity, andacceleration.

Unit Vectors

In many applications of vectors it is useful to find a unit vector that has thesame direction as a given nonzero vector v. To do this, you can divide v by itslength to obtain

u = unit vector =1

Note that u is a scalar multiple of v. The vector u has length 1 and the samedirection as v. The vector u is called a unit vector in the direction of v.

EXAMPLE 4 Finding a Unit Vector

Find a unit vector in the direction of v = ( — 2, 5} and verify that the result haslength 1.

Solution

The unit vector in the direction of v is

v (-2,5)V(-2)2 + (5)2

¯ 729 λ \729' 729

This vector has length 1 because

ίί^l·W^=V\729/ l,729/



Figure 6.23 The unit vectors (1,0) and (0, 1} are called the standard unit vectors and aredenoted by

i = < l , O ) and J = (O,l>

as shown in Figure 6.23. (Note that the lowercase letter i is written in boldfaceto distinguish it from the imaginary number i = J— \.) These vectors can beused to represent any vector v = {v¡, v2) as follows.

v = (Vl,v2)

= v , < l , O > + v2{0, 1}

= v,i + v2j

The scalars v, and v2 are called the horizontal and vertical components of v,respectively. The vector sum v,i + v2j is called a linear combination of thevectors i and j. Any vector in the plane can be expressed as a linear combina-tion of the standard unit vectors i and j.

Figure 6.24 EXAMPLES Writing a Linear Combination of Unit Vectors

Let u be the vector with initial point (2, —5) and terminal point (— 1 , 3). Writeu as a linear combination of the standard unit vectors i and j.

SolutionBegin by writing the component form of the vector u.

u = (-1 -2 ,3 + 5)

= <-3,8>= -3i + 8j

This result is shown graphically in Figure 6.24. ^f

EXAMPLE 6 d7 Vector Operations

Let u = —3i + 8j and v = 2i j. Find 2u — 3v.

SolutionYou could solve this problem by converting u and v to component form. This,however, is not necessary. It is just as easy to perform the operations in unitvector form.

2u - 3v = 2(-3i + 8j) - 3(2i - j)

= -6i + I6j - 6i + 3j

= -I2i + I9j ββ



Figure 6.25

-I

(cos θ, sin θ)

Direction Angles

If u is a unit vector such that θ is the angle (measured counterclockwise) fromthe positive i¯-axi s to u, the terminal point of u lies on the unit circle andyou have

u = (cos θ, sin θ) = (cos θ)i + (sin 0)j

as shown in Figure 6.25. The angle θ is the direction angle of the vector u.

Suppose that u is a unit vector with direction angle θ. If v is any vector thatmakes an angle θ with the positive jc-axis, then it has the same direction as uand you can write

v = | |v | | (cos θ, sin θ)

= IMl(cos0)i + IHl(sin0)j.

Because v = α\ + ¿>j = ||v|| (cos 0) i + [|v|| (sin 0) j, it follows that thedirection angle 0 for v is determined from

tan 0 =sin θ

cos 0

v|] sin 0

Ml cos 0

b

α

Figure 6.26

(3,3)

EXAMPLE 7 ^P Finding Direction Angles of Vectors

Find the direction angle of each vector.

a. u = 3i + 3j b. v = 3i 4j

Solution

a. The direction angle is given by

tan 0 = - = = I.α 3

Therefore, 0 = 45°, as shown in Figure 6.26.

b. The direction angle is given by

b -4tan 0 = = —.

α 3

Moreover, because v = 3i — 4j lies in Quadrant IV, 0 lies in Quadrant IVand its reference angle is

4arctan — -53.I3°I = 53.13°.

Therefore, it follows that θ 360° - 53.13° = 306.87°, as shown inFigure 6.27. ^S



Applications of Vectors

Figure 6.28

EXAMPLE 8 IS Finding the Component Form of a Vector

Find the component form of the vector that represents the velocity of anairplane descending at a speed of 100 miles per hour at an angle 30° below thehorizontal, as shown in Figure 6.28.

Solution

The velocity vector v has a magnitude of 100 and a direction angle of 0 = 210°.

v = ||v|[(cos 0)i + ||v|| (sin 0)j

= !OO(cos2lO°)i + lOO(s 2lO°)j

+• loof-

= -5Os/3i - 5Oj

= {-5073, -50)

You should check to see that l l v l l = 100.

Figure 7.69

Note In Figure 6.29, note that AC isparallel to the ramp.

EXAMPLE 9 An Application

A force of 600 pounds is required to pull a boat and trailer up a ramp inclinedat 15° from the horizontal. Find the combined weight of the boat and trailer.

Solution

Based on Figure 6.29, you can make the following observations.

\\BA\\ = force of gravity = combined weight of boat and trailer

\\BC\\ = force against ramp

||AC || = force required to move boat up ramp = 600 pounds

By construction, triangles BWD and ABC are similar. Hence, angle ABC is 15°.Therefore, in triangle ABC you have

sin 15° =

\BA\ =

l]ACJj 600

\\BA\\ ¯

6002318.

sin 15°

Consequently, the combined weight is approximately 2318 pounds.



Figure 6.30

112.6°

EXAMPLE 10 An Application¦ξ ¿,¿¢

An airplane is traveling at a fixed altitude with a negligible wind factor. The air-plane is headed N 30° W at a speed of 500 miles per hour, as shown in Figure6.30. As the airplane reaches a certain point, it encounters a wind with a veloc-ity of 70 miles per hour in the direction N 45° E. What are the resultant speedand direction of the airplane?

Solution

Using Figure 6.30, the velocity of the airplane (alone) is given by

v, = 5OO{cos 120°, sin 120°) = (-250, 25073)

and the velocity of the wind is given by

v2 = 7O{eos 45°, sin 45°) = (35^/2,35^2).

Thus, the velocity of the airplane (in the wind) is given by

v = v, + v2 = (-250 + 35χ/2, 25O>/3¯ + 35v¾ (-200.5,482.5)

and the speed of the airplane is

||v|| = 7(^2C>O.5)2 +¯(¯482.5) 2¯ 522.5 miles per hour.

Finally, if θ is the direction angle of the flight path, you have

482.5tan θ = -2.4065

-200.5

which implies that

θ 180° + arctan(-2.4065) 180° - 67.4° = 112.6°.

Group flc Verifying the Associativity of VectorAddition

On page 481, you learned that vector addition is associative, that is, forvectors u, v, and w, (α + v) + w = u + (v + w). Use graph paperand the information in the table to demonstrate geometrically that theresultant vector is the same regardless of whether you add u and v or vand w first.

Initial Point

Terminal Point

u

(-3,2)

(5, -3)

v

( 1 , 6 )

(4, -3)

w

(2, -2)

(-4,3)



6.3 /// EXERCISES ¡In Exercises 1-10, find the component form and the In Exercises 11-16, use the figure to sketch a graph ofmagnitude of the vector v.

1. y

(4 ,3)3 -<- J»

jfV.S

JΓ

\ - jrL/_

1 2 3 4

3.

the specified vector.

2 >`* y

`\— ¡— 4 — ¦ ¦ — \ - x\ ^̀ ^N. 2 3 4 i

-i 4- v̂ ^ I\̂. u

:!̂ JL·-3 |

V

&. x

11. v 12. 3v

, 4 )5-t 6J ( 3 , 5 ) 13· u + v 14. u + 2v

' SsL 4J 15. u - v 16. v - ¦u

24Γ`¯N ,! . (2, 2)

|

-3 -2 -1 i ί 2 3i

5. >`i

4 ( 3 , 3 )3 4- f

ίί ,1! T~¯ ¦ · · ' ί ¯* *

- 2 - 1 - 1 - 1 2 4 5-2 4- <, ί _2`\— j -1- y,,j¦i ^7

2Ϊ71 In Exercises 17-20, find (a)

-2 *4- 1 2 3 4 5 2u - 3v.

(-I,-1)4-17. u = (1,2), v = (3, 1)

18. u = <2, 3), v = (4, 0)6. >

19. u = i + j, v = 2i - 3j4-j_ j' J

^JL.. ^`-*-^-^·`'~ ~ ' · 1

u + v, (b) u — v, and (c)

r 4 n-2 · L v "n Exercises 21-28, find a unit vector in the direction-3 4- of the given vector.—4 ~¦"^

21. u = (5,0) 22. u = (0, -3)

23. v = (-2,2) 24. v = (5, -12)7 V

. J

2-1 / 5 ,

i j > ]

¿/"^

0 v

1 25. v = 4i - 3j 26. v = i + j

5 [°' 5'8) 27. w = 2j 28. w = i - 2j| \

4·J-\

¯ 2^ — 4V ' 9 4- \ In Exercises 29-32, find the vector v with the given

( 7 3 ) ΐ— ¿*<* — ^ ¯ ι` 2 / i

ι -f \(3.4. 0) magnitude and the same direction as u.

T -i 1 1 2 3 4 5 Magnitude Direction

9. Initial point: (-3, -5)Terminal point: (5, 1)

10. Initial point: (-3, 1 1 )Terminal noint: CQ. 40)

29. | |v | | = 5 u = (3

30. | | v |[ = 3 u = (4

, 3 )

, -4)

31. |v| = 7 u = (-3,4)

32. | |v| | = 1 0 u = (-10,0)



In Exercises 33-38, find the component form of v andsketch the specified vector operations geometrically,where u = 2i — j and w = i + 2j.

33. v = ¿u

35. v = u + 2w

37. v = (3u + w)

34. v = u + w

36. v = -u + w

38. v = u - 2w

In Exercises 39-42, find the magnitude and directionangle of the vector v.

39. v = 5(cos 3O°i + sin 3O°j)

40. v = 8(cos I35°i + sin I35°j)

41. v = 6ί 6j

42. v = -2i + 5j

In Exercises 43-48, find the component form of vgiven its magnitude and the angle it makes with thepositive *-axis. Sketch v.

43.

44.

45.

46.

Magnitude

HI = 3

= 372

= 947. ||v|| = 2

48. l l v l l = 3

Angle

θ = 0°

θ = 45°

θ = 150°

θ= 90°

v in the direction i + 3j

v in the direction 3i + 4j

In Exercises 49-52, find the component form of thesum of u and v with direction angles θu and 0V.

49.

50.

51.

52.

Magnitude

H u l l = 5,!MI = s,HI = 2,IMI = 2,HI = 20,IMI = 50,||u|| = 35,

HI = so,

Angle

βr

fl.

¾έ'v

θ.

= 0°

= 90°

= 30°

= 90°

= 45°

= 180'

= 25°

= 120'

In Exercises 53-56, use the Law of Cosines to find theangle α between the given vectors. (Assume 0° < α <180°.)

53. v = i + j, w = 2(i - j)

54. v = 3i + j, w ~ 2i — j

55. v = i + j, w = 3i - j

56. v = i + 2j, w ~ 2i — j

In Exercises 57 and 58, find the angle between theforces given the magnitude of their resultant. (Hint:Write force one as a vector in the direction of thepositive x-axis and force two as a vector at an angle θwith the positive Λ`-axis.)

Force 1

57. 45 pounds

58. 3000 pounds

Force 2

60 pounds

1000 pounds

Resultant Force

90 pounds

3750 pounds

59. Think About It Consider two forces of equal magni-tude acting on a point.

(a) If the magnitude of the resultant is the sum of themagnitudes of the two forces, make a conjectureabout the angle between the forces.

(b) If the resultant of the forces is 0, make a conjec-ture about the angle between the forces.

(c) Can the magnitude of the resultant be greater thanthe sum of the magnitudes of the two forces?Explain.

60. Graphical Reasoning Consider two forces F, —(10, 0}andF 2 = 5(cos θ, sin θ).

(a) Find | |F, + F2||.

(b) Determine the magnitude of the resultant as afunction of θ. Use a graphing utility to graph thefunction for 0 < θ < 2ττ.

(c) Use the graph in part (b) to determine the rangeof the function. What is its maximum and forwhat value of θ does it occur? What is its mini-mum and for what value of θ does it occur?

(d) Explain why the magnitude of the resultant isnever 0.



61. Numerical and Graphical Analysis Forces withmagnitudes of 150 newtons and 220 newtons act on ahook (see figure).

(a) If θ = 30°, find the direction and magnitude ofthe resultant of these forces.

(b) Express the magnitude M of the resultant and thedirection a of the resultant as functions of θ,where 0° < θ < 180°.

(c) Use a graphing util ity to complete the table.

θ

M

a

0° 30° 60° 90° 120° 150° 180°

(d) Use a graphing utility to graph the two functions.

(c) Explain why one function decreases for increas-ing θ, whereas the other doesn't.

150 newtons

62. Resultant Force Forces with magnitudes of 2000newtons and 900 newtons act on a machine part atangles of 30° and —45°, respectively, with the ;c-axis(see figure). Find the direction and magnitude of theresultant of these forces.

2000 newlons

63. Resultant Force Three forces with magnitudes of75 pounds, 100 pounds, and 125 pounds act on anobject at angles of 30°, 45°, 120°, respectively, withthe positive jc-axis. Find the direction and magnitudeof the resultant of these forces.

64. Resultant Force Three forces with magnitudes of70 pounds, 40 pounds, and 60 pounds act on an objectat angles of -30°, 45°, and 135°, respectively, withthe positive jc-axis. Find the direction and magnitudeof the resultant of these forces.

In Exercises 65 and 66, use a graphing utility to graphthe vectors and the resultant of the vectors. Find themagnitude and direction of the resultant.

66.

67. Horizontal and Vertical Components of Velocity Aball is thrown with an initial velocity of 80 feet persecond, at an angle of 40° with the horizontal (see fig-ure). Find the vertical and horizontal components ofthe velocity.

68. Horizontal and Vertical Components of Velocity Agun with a muzzle velocity of 1200 feet per second isfired at an angle of 6° with the horizontal. Find thevertical and horizontal components of the velocity.

900 newtons



Cable Tension In Exercises 69 and 70, use the figureto determine the tension in each cable supporting thegiven load.

69.

2000 lb

70. l() in. 20 in.

71. Numerical and Graphical Analysis A loaded bargeis being towed by two tugboats, and the magnitude ofthe resultant is 6000 pounds directed along the axis ofthe barge (see figure). Each tow line makes an angleof θ degrees with the axis of the barge.

(a) Find the tension in the tow lines if θ = 20°.

(b) Write the tension T of each line as a function ofθ. Determine the domain of the function.

(c) Use a graphing u t i l i t y to complete the table.

θ

T

10° 20° 30° 40° 50° 60°

(d) Use a graphing utility to graph the tension func-tion.

(e) Explain why the tension increases as θ increases.

72. Numerical and Graphical Analysis To carry a 100-

pound cylindrical weight, two people lift on the endsof short ropes that are tied to an eyelet on the top cen-

ter of the cylinder. Each rope makes an angle of θdegrees with the vertical.

(a) Find the tension in the ropes if θ = 30°.

(b) Write the tension T of each rope as a function of

θ. Determine the domain of the function.

(c) Use a graphing util ity to complete the table.

θ

T

10° 20° 30° 40° 50° 60°

(d) Use a graphing uti l i ty to graph the tension func-tion.

(e) Explain why the tension increases as θ increases.

73. Navigation An airplane is flying in the directionS 32° E, with an airspeed of 875 kilometers per hour.Because of the wind, its groundspeed and directionare 800 kilometers per hour and S 40° E, respectively(see figure). Find the direction and speed of the wind.

N

W ¯

800 kilometers per hour

875 kilometers per hour



74.

75.

Navigation An airplane's velocity with respect tothe air is 580 miles per hour, and it is headingN 60° W. The wind, at the altitude of the plane, isfrom the southwest and has a velocity of 60 miles perhour. Draw a figure that gives a visual representationof the problem. What is the true direction of the plane,and what is its speed with respect to the ground?

Work A heavy implement is pulled 20 feet across afloor, using a force of 85 pounds. Find the work doneif the direction of the force is 60° above thehorizontal (see figure). (Use the formula for work,W — FD, where F is the component of the force inthe direction of motion and D is the distance.)

85 Ib

Figure for 76

- 20 ft -

76. Numerical and Graphical Analysis A tetherballweighing 1 pound is pulled outward from the pole bya horizontal force u until the rope makes an angle ofθ degrees with the pole (see figure).

(a) Determine the resulting tension in the rope andthe magnitude of u when θ = 30°.

(b) Write the tension T in the rope and the magnitudeof u as functions of θ. Determine the domains ofthe functions.

(c) Use a graphing utility to complete the table.

θ

r

u

0° 10° 20° 30° 40° 50° 60°

Tension

1 l b -

True or False? In Exercises 77-80, decide whetherthe statement is true or false. If it is false, explain whyor give an example that shows it is false.

77. If u and v have the same magnitude and direction,then u = v.

78. If u is a unit vector in the direction of v, thenv = | j v | | u.

79. If v = a\ + bj = 0, then a = -b.

80. If u = «i + bj is a unit vector, then a2 + b2 = 1.

81. Prove that (cos θ)\ + (sin θ)j is a unit vector for anyvalue of θ.

82. Technology Write a program for your graphingutility that graphs two vectors and their differencegiven the vectors in component form.

In Exercises 83 and 84, use the program in Exercise 82to find the difference of the vectors shown in the figure.

83. 84.ί

100 + (80, 80)(-20, 70) . 4-

(-100,0)-50 -

(d) Use a graphing utility to graph the two functionsfor 0° < θ < 60°.

(e) Compare 7`and | ju| | as θ increases.


6.4 I Vectors and Dot Products 491

Vectors and Oof Products

V Think About the Proof

To prove the second, third, andfifth properties of the dot product,consider the component formsof vectors u, v, and w. You areasked to prove these properties inExercise 60.

The Dot Product of Two Vectors I The Angle Between Two Vectors IFinding Vector Components I Work

The Dot Product of Two Vectors

So far you have studied two vector operations—vector addition and multipli-cation by a scalar—each of which yields another vector. In this section you willstudy a third vector operation, the dot product. This product yields a scalar,rather than a vector.

Definition of Dot Product

The dot product of u = (u¡, u2) and v = (v , , v2) is

U · V — M | V , + W2V2.

Properties of the Dot ProducLet u, v, and w be vectors in the plane or in space and let c be a scalar.

1. u · v — v · u2. 0 · v = 03. u · ( v + w) = u·v + u · w4. v · v = ||v||2

5. c(u · v) = cu · v = u · cv

Proof /// To prove the first property, let u = (u¦, «,) and v = (v, v2). Then

U · V = + M2V2

= v · u.

For the fourth property, let v = (v,, v2). Then

V · V = V , 2 + V22



Note In Example 1, be sure you seethat the dot product of two vectors is ascalar (a real number), not a vector.Moreover, notice that the dot productcan be positive, zero, or negative.

!··A graphing utility can be used to

find the angle between two vec-

tors. The following Tί-82/Tί-83

program sketches two vectors

u = (α, b) and v = (c, d) in

standard position and finds the

measure of the angle between

them. Use the program to verify

Example 4. (Before running the

program, set an appropriate

viewing rectangle.)

:`V`ECANGL:ClrHome:Disp "EBTER(A,B)":lnput "ENTER A", A:lrtp¯u t "ENTER B",B:ClrHome:Disp "ENTER¢C,D)":lnput "ENTER C",G:lnput "ENTER D",D:Line(O,O,A,B):Line(O,O,C,D):Pause:AC+BD E

:cos >(E/(UV)) 0: ClrDraw: ClrHome:Disp "0=",θ:Stop

EXAMPLE 1 IS Finding Dot Products

Find each dot product.

a. (4, 5) · (2, 3)

b. (2,-1) · (1,2)

c. <O, 3) · (4, -2)

Solution

a. (4, 5} · (2, 3) = 4(2) + 5(3) = 8 + 15 = 23

b. (2, -1) · (1,2) = 2(1) + (-l)(2) = 2 - 2 = 0

c. (0, 3} · (4, -2) = 0(4) + 3(-2) = 0 - 6 = -6

EXAMPLE 2 SS Using Properties of Dot Products

Letu = (-1, 3), v = (2, -4), and w = (1, -2). Find each dot product.

a. (u · v)w b. u · 2v

Solution

Begin by finding the dot product of u and v.

u · v = (-1,3) · <2, -4}

= (-l)(2) + 3(-4)

= -14

a. (u · v)w = -14(1, -2} = (-14,28)

b. u · 2v = 2(u · v) = 2(-l4) = -28

Notice that the first product is a vector, whereas the second is a scalar. Can you

see why?

EXAMPLE 3 fiB7 Dot Product and Length

The dot product of u with itself is 5. What is the length of u?

SolutionBecause | |u||2 = u · u = 5, it follows that

u = y · u



The Angle Between Two Vectors

The angle between two nonzero vectors is the angle θ,Q<θ π, between itsrespective standard position vectors, as shown in Figure 6.31. This angle can befound using the dot product. (Note that the angle between the zero vector andanother vector is not defined.)

H yie between iwo veciors

If θ is the angle between two nonzero vectors u and v, then

cos θ =u

l u l lV

II vl

Figure 6.31

Origin

Proof /// Consider the triangle determined by vectors u, v, and v — u, asshown in Figure 6.31. By the Law of Cosines, you can write

(v — u) · (v — u) = || u

(v — u) · v — (v — u) · u = || u

v — u · v — v

l lv l ¡2 _ 9,

u + u · u

_92U v = -2||u|| |U ' V

[|V||¿ - 2||u|| ||v|| cos θ

||v||2 - 2| |u | | |v | |cos0

||v| |2 — 2||u|| | |v| | cos θ

\\v\\2 — 2||u|| ||v|| cos θ

\\\\\2 — 2\\ u|| | |v | | cos θ

v[)cos θ

cos t> =u v

• u = (4, 3}

2 3 4 5 6

EXAMPLE 4 ^P Finding the Angle Between Two Vectors

Find the angle between u = (4, 3) and v = (3, 5).

Solution

u^^ < _3>_· (3±5)_ 2T^m " || u|| || v || H ( 4 , 3 ) l l l l { 3 , 5 } l l 5734

This implies that the angle between the two vectors is

θ = arccos —, - 22.2°,5 734

as shown in Figure 6.32.



Figure 6.33

θ= π

cos θ = -1

Opposite Direction

Rewriting the expression for the angle between two vectors in the form

u · v = || u|| ||v|| cos θ Alternative form of dot product

produces an alternative way to calculate the dot product. From this form, you

can see that because ||u|| and | |v | | are always positive, u · v and cos θ will

always have the same sign. Figure 6.33 shows the five possible orientations of

two vectors.

7Γ f)2 < θ< π

-1 <cos Θ<O

Obtuse Angle

cos θ = 0

90° Angle

O < ( 9 < f0 < cos θ < 1

Acute Angle

Ue ΠHIOΠ OF U nOyOΠdl VeiΛOΓ5

θ = 0

cos θ = 1

Same Direction

The vectors u and v are orthogonal if u · v = 0.

The terms "orthogonal" and "perpendicular" mean essentially the same

thing—meeting at right angles. By definition, however, the zero vector is

orthogonal to every vector u, because 0 · u = 0.

ι 6.34

• u = (2,-3)

EXAMPLE 5 ^ Determining Orthogonal Vectors

Are the vectors u = (2, -3) and v = (6, 4) orthogonal?

Solution

Begin by finding the dot product of the two vectors.

u · v = <2, -3) · (6,4)

= 2(6) + (¯3)(4 )

= 0

Because the dot product is 0, the two vectors are orthogonal, as shown in Figure6.34. CZ7



Figure 6.35

Finding Vector Components

You have already seen applications in which two vectors are added to produce aresultant vector. Many applications in physics and engineering pose the reverseproblem—decomposing a given vector into the sum of two vector components.

Consider a boat on an inclined ramp, as shown in Figure 6.35. The force F dueto gravity pulls the boat down the ramp and against the ramp. These twoorthogonal forces, w, and w2, are vector components of Γ. That is,

F = w, + w2. Vector components of F

The negative of component w, represents the force needed to keep the boatfrom rolling down the ramp, whereas w2 represents the force that the tires mustwithstand against the ramp. A procedure for finding w, and w2 is shown below.

Definition of Vector Components

Let u and v be nonzero vectors such that

U = W , + W2

where w, and w2 are orthogonal and w, is parallel to v, as shown inFigure 6.36. The vectors w, and w2 are called vector components of u.The vector w, is the projection of u onto v and is denoted by

w, = projvu.

The vector w2 is given by w, = u — w,.

Figure 6.36

PrrFrom the definition of vectorcomponents, you can see that it iseasy to find the component W2

once you have found the projec-tion of u onto v. To find theprojection, you can use the dotproduct.

' is acute

Projection of u onto v

Let u and v be nonzero vectors. The projection of u onto v is

P roJvuu · v



Figure 6.37

u = <3, -5)

EXAMPLE 6 £¯J ' Decomposing a Vector into Components

Find the projection of u = (3, -5} onto v = (6, 2). Then write u as the sumof two orthogonal vectors, one of which is projvu.

SolutionThe projection of u onto v is

u · v\ / 8 V /6 2= l Γ 2) = \5· 5

as shown in Figure 6.37. The other component, w2, is

/6 2\ /9 27w2 = u - w, = (3, -5) -

5' 5 5 5

Thus, u = w, + w7 = (-,6 2

5 5

9 27\-,-—} <3,-5>.5 5 '

Figure 6.38

30

EXAMPLE 7 ^ Finding a Force

A 600-pound boat sits on a ramp inclined at 30°, as shown in Figure 6.38. Whatforce is required to keep the boat from rolling down the ramp?

SolutionBecause the force due to gravity is vertical and downward, you can representthe gravitational force by the vector

F = -6OOj. Force due to gravity

To find the force required to keep the boat from rolling down the ramp, projectF onto a unit vector v in the direction of the ramp, as follows.

v = (cos 3O°)i + (sin 3O°)j = γ¡ + ^j

Therefore, the projection of F onto v is given by

w, = projvF

F · v

Unit vector along ramp

( 1 \ I Γ\

- v = -3OOR-i + ;2/ \ 2 , _

The magnitude of mis force is 300, and therefore a force of 300 pounds isrequired to keep the boat from rolling down the ramp. ^β



Figure 6.39

(a)

F

(b)

β

Work

The work W done by a constant force F acting along the line of motion of anobject is given by

W = (magnitude of force)(distance)

as shown in Figure 6.39(a). If the constant force F is not directed along the lineof motion, you can see from Figure 6.39(b) that the work W done by the forceis

= F·PQ.

This notion of work is summarized in the following definition.

Definition of Work

The work W done by a constant force F as its point of application movesalong the vector PQ is given by either of the following.

1. W= ||pro F|||[7 > | l Projection form

2. W = F · ~PQ Dot product form

Figure 6.40

EXAMPLE 8 Finding Work

To close a sliding door, a person pulls on a rope with a constant force of 50pounds at a constant angle of 60°, as shown in Figure 6.40. Find the work donein moving the door 12 feet to its closed position.

Solution

Using a projection, you can calculate the work as follows.

= (cos60°)||F||||Pβ||

= 300 ft-lb

Thus, the work done is 300 foot-pounds.



Group flctivitii The Sign of the Dot Product

On page 494, you were given the alternative form of the dot product oftwo vectors.

U « V = U V COS θ Alternative form of dot product

Use this form to determine the sign of the dot product of u and v for thevectors shown below. Explain your reasoning.

115°

b. c.

6.4 /// EXERCISESIn Exercises 1-4, find the dot product of u and v.

1. u = (3,4) 2. u = (5, 12}

v = {2,-3} v = {-3,2)

3. u = 4i - 2j

v = i

4. u

v = 9i - 3j

In Exercises 5-8, use the vectors u = ^2, 2) andv = {—3, 4^> to find the indicated quantity. Statewhether the result is a vector or a scalar.

5. u · u

7. (u · v)v

6. ||u|| - 2

8. u · 2v

In Exercises 9-12, use the dot product to find ||u||.

9. u = (-5, 12} 10. u = (2, -4}

11. u = 2Oi + 25j 12. u = 6j

13. Revenue The vector u = (1245,2600) gives thenumber of units of two products produced by a con>pany. The vector v = (12.20, 8.50} gives the price(in dollars) of each unit, respectively. Find the dotproduct u · v, and explain what information it gives.

14. Revenue Repeat Exercise 13 after increasing theprices by 5%. Identify the vector operation used toincrease the prices by 5%.

In Exercises 15-20, find the angle θ between the vectors.

15. u = {1,0} 16. u = {4,4}

v = {O,-2} v = {2, 0}

17. u = 3i + 4j 18. u = 2i - 3j

v = -2j v = i - 2j



19. u

v

M= cos[-Ji 4

/3τή.= cos - - i

1 4 I

l·¡r\

- s in ( 3 J j

. (3iff sin

\ 4 ,

lΓ/7Γ\20. u = c o s j i

M· · ·v = cos^Ji + s (~lj

In Exercises 21-24, use a graphing utility to sketch thevectors and find the degree measure of the anglebetween the vectors.

21. u = 3i + 4j

v = -7i + 5j

23. u = 5i + 5j

v = -6i + 6j

22. u = -6i - 3j

v = -8i + 4j

24. u = 2i - 3j

v = 4i + 3j

In Exercises 25 and 26, use vectors to find the interiorangles of the triangle with the given vertices.

25. (1,2), (3,4), (2, 5)

26. (-3, 0), (2, 2), (0, 6)

In Exercises 27 and 28, find u · v, where θ is the anglebetween u and v.

277-27. | |u|| = 4, | |v | | = 10, θ =

28. | |u | | = 100, | | v | | = 250, θ =7Γ

In Exercises 29-34, determine whether u and v areorthogonal, parallel, or neither.

29. u = (-12, 30} 30. u ={15,45}

v = < L - 4 > v=<-5,12)

31. u = i(3i - j) 32. u = j

v = 5i + 6j v = i - 2j

33. u = 2i 2j 34. u = (cos θ, sin θ)

v = —i — j v = (sin θ, —cos θ)

In Exercises 35-38, find the projection of u onto v, andthe vector component of u orthogonal to v.

35. u = (3, 4} 36. u = (4, 2}

v = <8,2} v = {l,-2}

37. u = {0,3} 38. u = {-5, -1}

v = {2, 15} v = (-1, 1}

In Exercises 39-42, use the figure to mentally deter-mine the projection of u onto v. (The coordinates of theterminal points of the vectors in standard position aregiven.) Use the formula to verify your result.

39.(6,4)

41.

(2,-3)

In Exercises 43-46, find two vectors in opposite direc-tions that are orthogonal to the vector u. (The answersare not unique.)

43. u = {3,5} 44. u = {-8,3}

45. u = |i - |j 46. u = -|i 3j



47. Braking Load A truck with a gross weight of36,000 pounds is parked on a 10° slope (see figure).Assume the only force to overcome is that due togravity.

(a) Find the force required to keep the truck fromrolling down the hill.

(b) Find the force perpendicular to the hill.

Weight = 36,000 Ib

48. Braking Load Rework Exercise 47 for a truck thatis parked on a 12° slope.

49. Think About It What is known about θ, theangle between two nonzero vectors u and v, if thefollowing are true?

(a) u * v = 0 (b) u · v > 0 (c) u · v < 0

50. Think About It What can be said about the vectors uand v if the following are true?

(a) The projection of u onto v equals u.

(b) The projection of u onto v equals 0.

51. Work A 25-kilogram (245-newton) bag of sugar islifted 3 meters. Determine the work done.

52. Work Determine the work done by a crane lifting a2400-pound car 5 feet.

53. Work A force of 45 pounds in the direction of 30°above the horizontal is required to slide an implementacross a floor (see figure). Find the work done if theimplement is dragged 20 feet.

45 Ib

54. Work A tractor pulls a log 800 meters and the ten-sion in the cable connecting the tractor and log isapproximately 1600 kilograms (15,691 newtons).Approximate the work done if the direction of theforce is 35° above the horizontal.

Work In Exercises 55 and 56, find the work done inmoving a particle from P to Q if the magnitude anddirection of the force are given by v.

55. P = ( O , 0), Q = (4,7), v = <l,4)

56. P = (l,3), Q = (-3,5), v = -2i + 3j

57. For nonzero vectors u and v, prove thatv\

pr¤JvU =

58. Use vectors to prove that the diagonals of a rhombusare perpendicular.

59. Prove the following.

u - v r = u + l l v l l 2 - 2u

60. Prove the following properties of the dot product.

(a) 0 · v = 0

(b) u · (v + w) = u · v + u · w

(c) c(u · v) = cu · v = u · cv

61. Prove that if u is orthogonal to v and w, then u isorthogonal to cv + <¿w for any scalars c and d.

Review Solve Exercises 62-65 as a review of the skillsand problem-solving techniques you learned in previ-ous sections. Perform the additions or subtractionsand, if possible, simplify the results.

y262. (a) — - x

x

(b) cotjccoΐx

64. (a) - -

63. (a) 1 -

(b) 1 —sec^ x

1 + v65. (a)

1 + z

tan x sec x(b)

sec.r 1 + lίmx

sinjc 1 + cos x(b) - - +

1 + cosx smx

20ft


6.5 I DeMoivre's Theorem

Figure 6.41

Imaginary

(-1,3)or-1 + 3f

f ------ H —-2 -1 !

•(-2, -1) Ij

or-2-7

(3,2)or3 + 2/

Real

DeMoivre's TheoremThe Complex Plane Trigonometric Form of a Complex NumberMultiplication and Division of Complex NumbersPowers of Complex Numbers Roots of Complex Numbers

The Complex Plane

Recall from Section 2.4 that you can represent a complex number

z = α + hi

as the point (α, b) in a coordinate plane (the complex plane). The horizontalaxis is called the real axis and the vertical axis is called the imaginary axis, asshown in Figure 6.41.

The absolute value of the complex number α + bi is defined as the distancebetween the origin (0, 0) and the point (α, b).

Definition of the Absolute Value of a Complex Number

The absolute value of the complex number z — α + bus given by

α + bl\ = 7α2 + b2.

Note If the complex number α + bi is a real number (that is, if b = 0), thisdefinition agrees with that given for the absolute value of a real number

+ Oz¯ = Jα2 + O2 =

Figure 6.42

Imaginaryaxis

(-2, 5)

(0, -3)

EXAMPLE 1 ^ F i n d i n g the Absolute Value of a Complex Number

Plot each complex number and find its absolute value.

a. z = -3i b. z = ~2 + Si-

SolutionThe points are shown in Figure 6.42.

a. The complex number z = 0 + ( —3)¿ has an absolute value of

z = Vθ2 + (^ψ

= 3.

b. The complex number z = —2 + 5ί has an absolute value of

= V(-2)2 + 52

= 729. L



Figure 6.43

Imaginaryax i s

I

Realaxis

Trigonometric Form of a Complex Number

In Section 2.4 you learned how to add, subtract, multiply, and divide complexnumbers. To work effectively with powers and roots of complex numbers, it ishelpful to write complex numbers in trigonometric form. In Figure 6.43,consider the nonzero complex number α + hi. By letting θ be the angle fromthe positive _r-axis (measured counterclockwise) to the line segment connectingthe origin and the point (α, b), you can write

α = r cos θ and b = r sin θ

where r = Jα2 + b2. Consequently, you have

α + hi = (r cos θ) + (r sin θ)i

from which you can obtain the trigonometric form of a complex number.

Note The trigonometric form of acomplex number is also called thepolar form. Because there are infinite-ly many choices for θ, the trigonomet-ric form of a complex number is notunique. Normally, θ is restricted to theinterval 0 < θ < 2π, although onoccasion it is convenient to use 0 < 0.

i yo omei c rorrπ oτ a complex iMuπmcr

The trigonometric form of the complex number z = α + hi is

z — r(cos θ + i sin θ)

where α = r cos θ, b r sin θ, r = -Jα~ + b2, and tan θ = b/α. Thenumber r is the modulus of z, and θ is called an argument of z.

Figure 6.44

z = -2-2v -4-1'

EXAMPLE 2 ββ Writing a Complex Number in Trigonometric Form

Write the complex number z = ¯- 2 — 2v3¡` in trigonometric form.

Solution

The absolute value of z is

= 4-2 - 273; = V(-2)2 + (-273)2 =

and the angle θ is given by

b -2·v/3tan θ = = - = v/3.

a -2

Because tan(ττ/3) = v 3 and z = —2 — 2 v3/` lies in Quadrant III, you chooseθ to be 6> = π + π/3 = 4ττ/3. Thus, the trigonometric form is

4ττ 4τrz = r(cos θ + i sin θ) = 4 cos — + i sin ¯

See Figure 6.44.


6.5 DeMoivre's Theorem 503

mmmA graphing utility can be used toconvert a complex number inpolar form to rectangular form,and vice versa. For instance, toillustrate Example 3 on a 77-52or 77-83, use the followingsteps.

1. Press | ANGLE | and choosePD>Rx(.

2. Enter the values for r and θas (r, θ). Press | ENTER] toobtain the ;c-coordmate.

3. Press | ANGLE | and chooseP>Ry(.

4. Enter the values for r and θas (r, θ). Press | ENTER] toobtain the y-coordinate.

Write the result in rectangularform. Convert the complexnumber —14- ,/3z to trigono-metric form using ROPr ( andR>Pfl(.

The Interactive CD-ROM offers graphingutility emulators of the TI-82 and 77-83,which can be used with the Examples,Explorations, Technology notes, andExercises.

EXAMPLE 3 ί~l Writing a Complex Number in Standard Form

Write the complex number in standard form α + hi.

, , , TΓ\ I 7Γ= v/8lcos l --l + isinl--

Solution

Because cos(—τr/3) = 1/2 and sin(—ττ/3) = — %/3/2, you can write

3

V3= , / x ι ι

; sin —7Γ

Multiplication and Division of Complex Numbers

The trigonometric form adapts nicely to multiplication and division of complexnumbers. Suppose you are given two complex numbers

z, = r,(cos 0, + ¡ sin θ,) and = r2(cos Θ2 + Ί sin Θ2).

The product of z and z2 is

z,z2 = Γ|Γ2(cos 0, + i sin 0,)(cos 02 + ί sin 02)

- ^[(cos 0, cos ¾ - sin0, s 02) + /(sin θ¡ cos02 + cos 0, s 02)].

Using the sum and difference formulas for cosine and sine, you can rewrite thisequation as

z,z2 = r,r2[cos(0, + 02) + /s in(0, + 02)].

This establishes the first part of the following rule. The second part is left to you(see Exercise 65).

Product and Quotient of Two Complex NumbersLet z, = r,(cos 0, + i sin 0,) and = r2(cos 02 + i sin 02) be complexnumbers.

z,z2 = r,r2[cos(0, + 02) + /sin(0, + 02)] Product

9, - 02) + sin(0, - 02)], z2 0 Quotient



Note that this rule says that to multiply two complex numbers you multiplymoduli and add arguments, whereas to divide two complex numbers you dividemoduli and subtract arguments.

EXAMPLE 4 E^ Multiplying Complex Numbers inTrigonometric Form

Find the product of the following complex numbers.

/ 2 < 7 Γ 2-7Γ\ 11 7Γ l l 7 Γ `z¡ = 2 cos — + z sin - -\ z2 = 8 cos - — + i sin —

\ 3 3 \ 6 6

Solution

mmmmm 2π 2π\ j ¶ n¶·\z,¾ = 2 cos — + ¿ sm — · 8 cos — - + i sin-— ¦

..•• V 3 3 \ 6 6i /2ττ l l 7 Γ \ (2¶ \\ΊT\

Some graphing utilities, such as = 16 cosl — ·ι- — 1 + i sml —- ·¦ — Ithe Tl-92, can multiply and divide ' ' ' `' / ;

complex numbers in tr onomet- .J 5ττ 5τr. f ,f , =16 cos - + ¡ sinπc rorm. It you have access to V 2

such a graphing utility, use it tofind zλz2 and z,/¿2 in Examples 4 = I6( cos — + i sin -and 5. V 2 .2

= I6[O + i(l)] = I6¿

Check this result by first converting to the standard forms z, = — 1 + V3i` and22 = 4v — 4i and then multiplying algebraically, as in Section 2.4. C^

EXAMPLE 5 CZ7 Dividing Complex Numbers in Trigonometric Form

Find z, / 2` f°r me following complex numbers.

z, = 24(cos 300° + i sin 300°) z2 = 8(cos 75° + I sin 75°)

Solution

z, _ 24<cos3Op^jf J_sin_3OO°)

Z2 ~¯ 8(cos 75° + i sin 75°)

94= - -[cos(3OO° - 75°) + i sin(3OO° - 75°)]

8

= 3(cos 225° + i sin 225°)Γ To" \ /of -f

__ 3_v/2 _ 3V2

"̄ ¯ ¯ 2 2~'


6.5 I DeMoivre's Theorem 505

Powers of Complex Numbers

To raise a complex number to a power, consider repeated use of the multipli-cation rule.

z = r(cos θ + i sin 0)

z2 = r(cos θ + i sin 0)r(cos θ + ί sin θ) = r2(cos 2Θ + i sin 2Θ)

z3 = r2(cos 2Θ + i sin 20)r(cos θ + i sin θ) = r3(cos 3Θ + i sin 30)

z4 = r4(cos4θ+ / s i n 40)

z5 = r5(cos50 + ¿ sin 50)

E X P L O R A T I O N

Plot the numbers ¿, z¯ 2, ι¯ 3, ί*, andi5 in the complex plane. Writeeach number in trigonometricform and describe what happensto the angle θ as you form higherpowers of z¯" .

This pattern leads to the following important theorem, which is named afterthe French mathematician Abraham DeMoivrc (1667-1754).

If Z = r(cos θ + i sin θ) is a complex number and n is a positive integer,then

:" = [r(cos θ + i sin 0)]" = r"(cos nθ + i sin«0).

EXAMPLE 6 ¿¯¯Z 7 Finding Powers of a Complex Number

Use DcMoivre's Theorem to find ( — 1 + ,/3¡j ·

Solution

First convert to trigonometric form.

2ΊT 1¶

I + v3z = 2( cos + z̀ sin —

Then, by DeMoivre's Theorem, you have

(-1 + /3z¯)' 2 =2 77

ίSin

= 21 2Γcos[l2 · —\ + z¯sin[l 2 ·

= 4O96(cos 8τr + ί sin 877)

= 4096(1 + 0)

= 4096.

Are you surprised to see a real number as the answer?



Roots of Complex Numbers

Recall that a consequence of the Fundamental Theorem of Algebra is that apolynomial equation of degree n has n solutions in the complex number system.Hence, an equation such as xb = 1 has six solutions, and in this particular caseyou can find the six solutions by factoring and using the Quadratic Formula.

jc6 - 1 = (x"` - l)(jc3 + 1)

= (x - ])(x2 + x + l)(;c + l)¢c2 - x+ 1) = 0

Consequently, the solutions are

x = ±1,/Λ .'3ι

and1 ± J3i

Each of these numbers is a sixth root of 1. In general, the nth root of acomplex number is defined as follows.

Definition of nth Root of a Complex NumberThe complex number u = α + bi is an nth root of the complexnumber z if

z = H" = (α + bi)".

•Γ E X P L O R A T I O N

The nth roots of a complex num-ber are useful for solving somepolynomial equations. Forinstance, explain how you can useDeMoivre's Theorem to solve thepolynomial equation

x4 + 16 = 0.

[Hint: Write —16 as

I6(cos TΓ -t¯ i sin τr).|

To find a formula for an nth root of a complex number, let u be an nth root ofz, where

u = s(cos β + i sin β) and z = r(cos θ + ί sin θ).

By DeMoivre's Theorem and the fact that u" = z, you have

a" (cos nβ + ί sin nβ) = r(cos θ + i sin θ).

Taking the absolute value of both sides of this equation, it follows that s" = r.Substituting back into the previous equation and dividing by r, you get

cos nβ + ί sin nβ = cos θ + ί sin θ.

Thus, it follows that

cos nβ = cos θ and sin nβ = sin θ.

Because both sine and cosine have a period oΐ2π, these last two equations havesolutions if and only if the angles differ by a multiple of 2ττ. Consequently,there must exist an integer k such that

nβ = θ + 2πk

_ θ +_^πk

n

By substituting this value for β into the trigonometric form of u, you get theresult stated in the theorem on the following page.


6.5 I DeMo/vre's Theorem 507

Note When k exceeds n — 1, theroots begin to repeat. For instance, ifk = n, the angle

θ+2¶n θ- = + 2ττ

n n

is coterminal with B/n, which is alsoobtained when k = 0.

iπ Koois oτ a v_ompiex i

For a positive integer n, the complex number z — r(cos θ + i sin θ) hasexactly n distinct nth roots given by

,,rί θ+2πk θ + 2πkV π cos— - + z sin

\ n n

where k — 0,1,2,. . . ,n — 1.

Figure 6.45

Imaginary

Figure 6.46

Imaginary

Realaxis

Realaxis

This formula for the nth roots of a complex number z has a nice geometrical inter-

pretation, as shown in Figure 6.45. Note that because the nth roots of z all havethe same magnitude "¦/r, they all lie on a circle of radius "vr with center at theorigin. Furthermore, because successive nth roots have arguments that differ by2π/n, the n roots arc equally spaced along the circle.

You have already found the sixth roots of 1 by factoring and by using theQuadratic Formula. Example 7 shows how you can solve the same problem

with the formula for nth roots.

EXAMPLE 7 ^ Finding the nth Roots of a Real Number

Find all the sixth roots of 1.

SolutionFirst write 1 in the trigonometric form 1 = l(cos 0 + z sin 0). Then, by the nthroot formula, with n = 6 and r = 1, the roots have the form

0 + 2πk 0 + 2πkcos i sin

or simply cos(ττ /3) + i sin(τ7 /3). Thus, for k = 0, 1, 2, 3, 4, and 5, the sixthroots are as follows. (See Figure 6.46.)

cos 0 + ¡ sin 0 = 1

π 77 1cos 3 + ''sin- = -

2ττ 2τrcos — + i sin — = - +

9

cos π + i sin τι = —

4ττcos i sin

4ττ

3

5ττ 5ττ 1 s/3cos 3¯ + i sin - == 2 - — i



E X P L O R A T I O N

Use a graphing utility, set inparametric and radian modes, todisplay the graph given by

Xιτ = cos T

and

Y ι τ = sinT.

Set the viewing rectangle so that-L5 X < L 5 a n d -l< Y 1.Then, using 0 T < 2τr, set the"Tstep" to 2¶/n for variousvalues of n. Explain how thegraphing utility can be used toobtain the «th roots of unity.

In Figure 6.46, notice that the roots obtained in Example 7 all have amagnitude of 1 and are equally spaced around this unit circle. Also notice thatthe complex roots occur in conjugate pairs, as discussed in Section 2.5. The ndistinct nth roots of 1 are called the nth roots of unity.

EXAMPLE 8 S> Finding the nth Roots of a Complex Number

Find the three cube roots of z = -2 + 2¡.

SolutionBecause ?. lies in Quadrant II, the trigonometric form for z is

z = -2 + 2/ = vT(cos 135° + i sin 135°).

By the formula for «th roots, the cube roots have the form

135°ξ/8

+ 36O°fc 135° + 36O°fe .cos - - -— - + i sin - — —

Finally, for k = 0, 1, and 2, you obtain the roots

v/2(cos 45° + i sin 45°) = 1 + i

v/2(cos 165° + / s i n 165°) -1.3660 + O.366Oi

V¾cos 285° + i sin 285°) 0.3660 - l.366O¿.

Group Rcίivifij A Famous Mathematical Formula

The famous formula

e" + bi = e°(cos b + i sin ft)

is called Euler's Formula, after the German mathematician LeonhardEuler (1707-1783). Although the interpretation of this formula isbeyond the scope of this text, we decided to include it because it givesrise to one of the most wonderful equations in mathematics.

e™ + 1 = 0

This elegant equation relates the five most famous numbers in mathe-matics—0, 1, 7Γ, e. and —in a single equation. Show how Euler'sFormula can be used to derive this equation.


6.5 DeMoivre's Theorem 509

G.5 /// EXERCISESIn Exercises 1-6, plot the complex number and find its In Exercises 27-30, use a graphing utility to representabsolute value. the complex number in trigonometric form.

1. -5¿ 2. -5

3. -4 + 4¿ 4. 5 - I2¿

5. 6 - 7¿ 6. -8 + 3¿

27. 5 + 2¿ 28. -3 + ¿

29. 372 - 7¿ 30. -8 - 5 73¿

In Exercises 31-40, represent the complexIn Exercises 7-10, write in trigonometric form. graphically, and find the standard form of the

7. Imaginary 8. Imaginaryaxis axis

4-- 4-[

2--Z ' 2t ¿ = 4

j-~

1 | | j-_ .

-2-1 1 2 axls -4t

9. Imaginary 10. Imagaxis ax

μ L*. Real 3 -

-3 -2 -1/J axis Z = -l+^ί.

• -2-fz = -2-2ΐ \ -i_i__4_J

t -3 -2 -1

31. 2(cos 150° + ¿sin 150°)

32. 5(cos 135° + ¿sin 135°)

33. \ (cos 300° + ¿sin 300°)

Real 34. 1 (cos 3 15° + ¿sin 3 15°)

"axis 3ττ 3ττ\35. 3.75 cos l· ¿sin- ¡

\ 4 4

36. sLs^ + i s in^)nary

37. 4 cos l· ¿ sin\ 2 2 ]

38. 7 (cos 0 + ¿ sin 0)

Rea| 39. 3[cos(18°45') + ¿ sin(!8° 45')]

40. 6[cos(23O° 30') + ¿ sin(23O° 30')]

numbernumber.

In Exercises 11-26, represent the complex numbergraphically, and find the trigonometric form of thenumber.

11. 3 - 3¿

13. 73 + ¿

15. -2(l + 73¿)

17. 6/

19. -7 + 4¿

21. 7

23. 1 + 6¿

25. -3 - ¿

In Exercises 41^44, use a graphing utility to representthe complex number in standard form.

41. 5 cos 1- i sin — ' 42. 9(cos 58° + ¿ sin 58°)12.

14.

16.

18.

20.

22.

24.

26.

2

i'4

3

+ 2¿

- 1 + 73¿

(73 - ¿)

— i

-2/

272 - ¿

1 + 3¿

43.

44.

\ 9

3ττ12 cos + ¿

4(cos 216.5° +

9

3 7Γ\sm yj

¿sin 2 16.5°)

In Exercises 45 and 46, represent the powers z, z2, z3,and z4 graphically. Describe the pattern.

45. = J2( , Λ 46. 7 = ίl + ,/3/)



In Exercises 47-58, perform the operation and leavethe result in trigonometric form.

47.7Γ 7Γ

3( cos + z sm -7Γ 7Γ

4| cos - + z sin —6 6

48. | _-[ cos —|- ¡ sm2\¯

6| cos -4

TΓz sm

49. 3(005 140° + z sin 140°) ¾ (cos 60° + i sin 6O°)j

50. |2(coslOO°+zsinlOO°)| (cos300°+ zsin3OO°)

51. [I (cos 310° + z sin 3lO°)j i(cos 200° + z sin 200°;

52. (cos 5° + z sin 5°)(cos 20° + z sin 20°)

cos 40° + z sin 40°

55.

cos 10° + zsin 10°

2jcos 120° +J sin 120°)

4(cos 40° + z sin 40°)

. cos(5ττ/3) + zsin(5τr/3)56.

cos 7Γ + z sin 7Γ

_!2(cos_52° + z sin 52°)

3(cos 110° + z s i n 110°)

9(cos_2OM- z_sin 20°)

5(cos 75° + i sin 75°)

5(cos 4.3 + z sin 4.3)

¯ 4(cos 2.1 + z sin 2.1)

57.

In Exercises 59-64, (a) give the trigonometric form ofthe complex numbers, (b) perform the indicated oper-ation using the trigonometric form, and (c) performthe indicated operation using the standard form, andcheck your result with that of part (b).

59. (2 + 2/)(1 ¯ 0 60. (73 + z ) ( l + 0

61. -2z(l + 0 62.-^-¯ .

_5__2 + 3/

64.4z

-4 + 2i

65. Given two complex numbers z | = r,(cos 0, + zs in 0¡)and z2 ¯ r2(cos Θ2 + i sin 02), z, 0, prove that

— = - [cos(0, - 02) + z sin(6>, - ¾)].

66. Show that z = r[cos(— θ) + i sin(—θ)] is the com-plex conjugate of z = r(cos θ + i sin θ).

67. Use the trigonometric forms of z and z in Exercise 66to find (a) zz and (b) z/z, z 0.

68. Show that the negative of z = r(cos θ + ί sin θ) is-z = λ¯[cos( 0 + 7r) + z sin(0 + 7r)].

In Exercises 69 and 70, sketch the graph of all complexnumbers z satisfying the given condition.

69. \z\ =2 70. θ = τr/6

In Exercises 71-82, use DeMoivre's Theorem to findthe indicated power of the complex number. Expressthe result in standard form.

71. (1 + z)5

73. (-1 + z)'°

75. 2(73 + i)7

77. [5(cos 20° + z sin 2O°)]3

78. [3(cos 150° + / s i n I5O°)]4

72. (2 + 2z)fi

74. (1 - z)12

76. 4(l - 73zf

79.5ττ . 5τr\ l ( )

cos — + ί sm —4 4

80. 2l cos - + i sin -

81. [5(cos 3.2 + z sin 3.2)]4

82. (cos 0 + z sin 0)2ϋ

In Exercises 83-86, use a graphing utility andDeMoivre's Theorem to find the indicated power ofthe complex number. Express the result in standardform.

83. (3 - 2z)5

85. [3(cos 15° + z s i n I5°)]4

84. (75 - 4z)3

86. 2 cos7Γ

10z sin

7Γ

10

87. Show that —|(l + 73z`) is a sixth root of 1.

88. Show that 2¯ ' / 4(l - z) is a fourth root of -2.


6.5 I DeMoivre's Theorem 511

Graphical Reasoning In Exercises 89 and 90, use thegraph of the roots of a complex number, (a) Write eachof the roots in trigonometric form, (b) Identify thecomplex number whose roots are given, (c) Use agraphing utility to verify the results of part (b).

89. Imaginaryaxis

t

90. Imaginaryaxis

In Exercises 91-102, (a) use the theorem on page 507to find the indicated roots of the complex number, (b)represent each of the roots graphically, and (c) expresseach of the roots in standard form.

91. Square roots of 5(cos 120° + / s i n 120°)

92. Square roots of I6(cos 60° + i sin 60°)

93. Fourth roots of 16 cos4ττ

3i sin

4-π¯

3

94. Fifth roots of 32 cos

95. Square roots of — 25 i

96. Fourth roots of 625i

5 77 5ττ\i*rn-\

97. Cube roots of ¯ 'f (l + >/3ϊ)

98. Cube roots of-4y/2( I - /)

99. Cube roots of 8

100. Fourth roots of i

101. Fifth roots of 1

102. Cube roots of 1000

In Exercises 103-106, (a) use the theorem on page 507and a graphing utility to find the indicated roots of thecomplex number, (b) represent each of the rootsgraphically, and (c) express each of the roots instandard form.

103. Cube roots of - 125

104. Fourth roots of —4

105. Fifth roots of I28(-I + z)

106. Sixth roots of 64z`

In Exercises 107-114, use the theorem on page 507 tofind all the solutions of the equation and represent thesolutions graphically.

107. x4 - i = 0

109. x5 + 243 = 0

111. A-3 + 64i = 0

108. XΛ + 1 = 0

110. A-4 - 81 = 0

112. x6 - 64i = 0

113. jr1 - (1 - i) = 0 114. x4 + (] + i) = 0

Review Solve Exercises 115 and 116 as a review of theskills and problem-solving techniques you learned inprevious sections. Find the altitude of the triangle.

115. 20 ^~~7· 116.

18

-28.1


¢*t @¢*tcefttά>In this chapter, you studied the methods for solving oblique triangles and vectors in theplane. You can use the following questions to check your understanding of several of thesebasic concepts. The answers to these questions are given in the back of the book.

1. State the Law of Sines from memory.

2. State the Law of Cosines from memory.

3. True or False? The Law of Sines is true if one of

the angles in the triangle is a right angle.

4. If one of the angles in the triangle is a right angle,

the Law of Cosines simplifies to what famous theo-rem?

5. True or False ? When the Law of Sines is used, the

solution is always unique. Explain.

6. What characterizes a vector in the plane?

7. Which vectors in the figure appear to be equivalent?

8. The vectors u and v have the same magnitudes in

the two figures. In which figure will the magnitude

of the resultant be greater? Give a reason for youranswer.

(a) (b)

9. Give a geometric description of the scalar multiple

fcu of the vector u.

10. Give a geometric description of the sum of thevectors u and v.

11. Which of the two figures shows the difference

u — v? Give a geometric description of the differ-

ence and state how you determine its direction.

(a) y (b)

12. The figure shows z, and z2. Describe ZjZ 2 and z¡/z

13. One of the fourth roots of a complex number z isshown in the figure.

(a) How many roots are not shown?

(b) Describe the other roots.


I Review Exercises 513

6 /// REVIEW EXERCISES

In Exercises 1-16, use the given information to solvethe triangle (if possible). Tf two solutions exist, listboth.

10

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

a = 5, b = 8, c = 10

a = 6, = 9, C = 45°

A = 12°, B = 58°, α = 5

B = 110°, C = 30°, c = 10.5

£ = 110°, a = 4, c = 4

α = 80, = 60, c = 100

A = 75°, α = 2.5, b= 16.5

A = 130°, a = 50, = 30

B = 115°, a = 7, fc = 14.5

C = 50°, α = 25, c = 22

A = 15°, a = 5, & = 10

S = 150°, α = 64, b = 10

S = 150°, a = 10, c = 20

α = 2.5, b = 15.0, c = 4.5

B = 25°, α = 6.2, b = 4

B = 90°, a = 5, c = 12

In Exercises 17-20, find the area of the triangle.

17. a = 4, b = 5, c = Ί

18. a = 15, b = 8, c = 10

19. A = 27°, b = 5, c = 8

20. B = 80°, α = 4, c = 8

21. Height From a certain distance, the angle of eleva-tion to the top of a building is 17°. At a point 50meters closer to the building, the angle of elevation is31°. Approximate the height of the building.

22. Geometry The lengths of the diagonals of a paral-lelogram are 10 feet and 16 feet. Find the lengths ofthe sides of the parallelogram if the diagonals inter-sect at an angle of 28°.

23. Height of a Tree Find the height of a tree that standson a hillside of slope 28° (from the horizontal) if,from a point 75 feet down the hill, the angle of eleva-tion to the top of the tree is 45° (see figure).

24. Surveying To approximate the length of a marsh, asurveyor walks 425 meters from point A to point B.Then the surveyor turns 65° and walks 300 metersto point C. Approximate the length AC of the marsh(see figure).

65

300m 425 m



25. Navigation Two planes leave an airport at approxi-mately the same time. One is flying at 425 miles perhour at a bearing of N 5° W, and the other is flyingat 530 miles per hour at a bearing of N 67° E (secfigure). Determine the distance between the planesafter flying for 2 hours.

N

W < I > E

67°

26. River Width Determine the width of a river thatflows due east, if a surveyor finds that a tree on theopposite bank has a bearing of N 22° 30' E from acertain point and a bearing of N 15° W from a point400 feet downstream.

27. Chapter Opener Approximate the height of thebridge deck from the water level in the figure on page531.

28. Chapter Opener Approximate a in the figure onpage 531 if the distance between A and C is 27meters.

In Exercises 29-34, find the component form of thevector v satisfying the given conditions.

29. 30.

" (2,-l)

-(0, 1)

2 4 6

31. Initial point: (0, 10), Terminal point: (7, 3)

32. Initial point: (1, 5), Terminal point: (15, 9)

33. | |v| | = 8, θ = 120°

34. l l v l l = , θ= 225°

In Exercises 35 and 36, write the vector v in the form| |v | |( isin0 + jcosθ) .

35. v = -10i + lϋj 36. v = 4i - j

In Exercises 37-40, find the component form of thespecified vector and sketch its graph given thatu = 6i — 5j and v = lOi + 3j.

37. 7 Γ , , un

39. 4u 5v

38. 3v

40. v

In Exercises 41 and 42, use a graphing utility to graphthe vectors and the resultant of the vectors. Find themagnitude and direction of the resultant.

41. 42.

43. Resultant Force Find the direction and magnitudeof the resultant of the three forces shown in the f urc.

Figure for 43

l ( ) O l h

Figure for 45

30°v

250 Ib

fΠ8Olb

200 ib

tan β = | tan a = ~-

44. Resultant Force Forces of 85 pounds and 50pounds act on a single point. The angle between theforces is 15°. Describe the resultant force.

45. Rope Tension A 180-pound weight is supported bytwo ropes, as shown in the figure. Find the tension ineach rope.


I Review Exercises 515

46. Cable Tension In a manufacturing process, an elec-

tric hoist lifts 200-pound ingots (see figure). Find thetension in the supporting cables.

4in.-

47.

48.

49.

Braking Force A 500-pound motorcycle is headed

up a hill inclined at 12°. What force is required to

keep the motorcycle from roll ing back down the hil l

when stopped at a red light?

Navigation An airplane has an airspeed of 430

miles per hour at a bearing of S 45° E. If the wind

velocity is 35 miles per hour in the direction N 30° E,find the groundspeed and the direction of the plane.

Navigation An airplane has an airspeed of 724 kilo-

meters per hour at a bearing of N 30° E. If the wind

velocity is 32 kilometers per hour from the west, find

the groundspeed and the direction of the plane.

50. Angle Between Forces Forces of 60 pounds and 100pounds have a resultant force of 125 pounds. Find theangle between the two forces.

In Exercises 51 and 52, find a unit vector in the direc-tion of PQ.

51. P(l, -4), β(-3, 2)

52. />((), 3), β(5, -8)

In Exercises 53 and 54, decide whether the vectors are

orthogonal, parallel, or neither.

53. u = (39, -12}

v = (-26,8)

54. u = (8, 5)

v = (-2,4>

In Exercises 55-58, find the angle between u and v.

7ττ 7ττ55. u = cos — i + sin j,

4 4

5ττ 5ττv = cos - i + sin — j

6 6

56. u = (-6, -3), v = (4,2)

57. u = <2v/2, -4), v = {-72, 1}

58. u = (3, 1), v = {4,5}

In Exercises 59-62, use a graphing utility to sketch thevectors and find the degree measure of the anglebetween the vectors.

59. u = 4i + j

v = i - 4j

61. u = 7i - 5j

v = lOi + 3j

60. u = 6i + 2j

v = -3i - j

62. u = -5.3i + 2.8j

v = -8.li - 4j

In Exercises 63-66, find projvu.

63. u = {-4,3}, v = {-8, -2}

64. u = {5,6}, v = {10,0}

65. u = {2,7}, v = {1, -1}

66. u = {-3,5}, v = {-5,2}

In Exercises 67-70, find the trigonometric form of the

complex number.

67. Imaginaryax i s

68. Imaginary

Real

69. Imaginaryaxis

Real

70. Imaginaryaxis

I

Real



In Exercises 71-74, find the trigonometric form of thecomplex number.

71. 5 - 5i

73. 5 + I2¿72. -3V3 + 3/74. -7

In Exercises 75-78, write the complex number in stan-dard form.

75. !OO(cos24O° + sin 240°)

76. 24(cos 330° + ί sin 330°)

77. !3(cosO + ί s i n O )

5ττ 5π78. 8( cos + i sin -

6 6

In Exercises 79 and 80, (a) express the two complexnumbers in trigonometric form, and (b) use thetrigonometric form to find z,z2

a d zjz2·

79. z, = 273 ~ 2/, z, = -lO¿

80. z, = -3(1 + ;), = 2U/3 + i

In Exercises 81-84, use the theorem on page 507 tofind the indicated power of the complex number.Express the result in standard form.

81. I s ίcos-12

82.4π 4τr

cos I- ;' sin15 15

83. (2 + 3i)6

84. (1 - /)8

Graphical Reasoning In Exercises 85-88, use thegraph of the roots of a complex number, (a) Write eachof the roots in trigonometric form, (b) Identify thecomplex number whose roots are given, (c) Use agraphing utility to verify the results of part (b).

85. Imaginary 86.

87. Imaginaryaxi>>

In Exercises 89 and 90, use the theorem on page 507 tofind the roots of the complex number.

89. Sixth roots of -729ϊ` 90. Fourth roots of 256

In Exercises 91-94, find all solutions of the equationand represent the solutions graphically.

91. A·4 + 8 1 = 0

93. + 8/ = 0

92. .r̀ ` - 32 = 0

94. (x3 - l)tϊ2 + 1) = 0


I Chapter Project 517

C H A P T E R P R O J E C T Adding Vectors Graphically

-2

The program below is written for a 77-82 or TI~83 graphing calculator. The

program sketches two vectors u = αi + j and v = c'\ + dj in standardposition. Then, using the parallelogram law for vector addition, the programalso sketches the vector sum u + v. Before running the program, you shouldset values that produce an appropriate viewing rectangle.

TI-82 or TI~83 Program

PROGRAM:ADDVECT:Input "ENTER A",A:Input "ENTER B",B: Input "ENTER C",C:lnput "ENTER D",D

:Line(O,O,A,B):Line(O,O,C,D):A+C E:B+D F:Line(O,O,E,F)

:Line(A,B,E,F):Line(C,D,E,F):Pause:ClrDraw:Stop

(a) Use the program listed above to sketch the sum of the vectorsu = 5i + 2j and v = —4i + 3j. Set your viewing window as indicatedat the left. Identify the vectors u, v, and u + v in the graph.

(b) An airplane is headed N 60° W at a speed of 400 miles per hour. Theairplane encounters wind of velocity 75 miles per hour in the directionN 40° E. Use the program listed above to find the resultant speed and

direction of the airplane.

Questions for Further Exploration

In Questions l-4, use the program listed above (ora comparable program on some other graphing utility)to sketch the sum of the vectors. Use the result toestimate graphically the components of the sum. Then

check your result analytically.

1. u = 3i + 4j, v = -5i + j

2. u = 5i — 4j, v = 3i + 2j

3. u = —4i + 4j, v = —2i — 6j

4. u = 7i + 3j, v = -2i 6j

5. After encountering the wind, is the airplane in

Exercise (b) traveling at a higher speed or a lower

speed? Explain.

6. Consider the airplane described in Exercise (b),headed N 60° W at a speed of 400 miles per hour.What wind velocity, in the direction of N 40° E,will produce a resultant direction of N 50° W?

Explain how to use the program listed above toobtain the answer experimentally. Then explainhow to obtain the answer analytically.

7. Consider the airplane described in Exercise (b),headed N 60° W at a speed of 400 miles per hour.What wind direction, at a speed of 75 miles perhour, will produce a resultant direction of N 50°

W? Explain how to use the program listed above toobtain the answer experimentally. Then explainhow to obtain the answer analytically.


518 I Cumulative Test for Chapters 4-6

4-6 /// CUMOL1TIΪETEST

Take this test as you would take a test in class. After you are done, check yourwork against the answers given in the back of the book.

1. Consider the angle θ = -120°.

(a) Sketch the angle in standard position.

(b) Determine a coterminal angle in the interval [0°, 360°).

(c) Convert the angle to radian measure.

(d) Find its reference angle θ'.

(e) Find the exact values of the six trigonometric functions of θ.

2. Convert the angle of magnitude 2.35 radians to degrees. Round the answer to onedecimal place.

3. Find cos θ if tan θ = — 3 and sin θ < 0.

4. Use a graphing uti l i ty to sketch the graphs of (a) f(x) = 3 — 2 sin πx and (b) g(x) =7Γ`

The Interactive CD-ROMprovides answers to theChapter Tests and CumulativeTests. It also offers ChapterPre-Tests (that test key skillsand concepts covered in pre-vious chapters) and ChapterPost-Tests, both of which haverandomly generated exerciseswith diagnostic capabilities.

5. Find a, b, and c such that the graph of the function h(x) = a co$(bx + c) matches thegraph in the figure.

6. Write an algebraic expression equivalent to sin(arccos 2x).

sin θ — 1 cos θ7. Subtract and simpliίy: - —.

cos θ sin θ - 1

8. Prove the identities.

(a) cot2 o(sec2 a — 1) = 1 (b) sin(jc + v) sin(.x — v) = sin 2 x — sin2}'

9. Find all solutions of the equations in the interval [0, 2π).

(a) 2 cos2 β - cos β = 0 (b) 3 tan θ - cot θ = 0

10. Use a graphing uti l i ty to approximate the remaining side and angles of the triangleshown in the figure. Verify analytically.

(a) A = 30°, α = 9, b = 8 (b) A = 30°, b = 8, c = 10

11. Find the trigonometric form of the complex number —2 + 2i.

12. Find the product of [4(cos 30° + i sin 3O°)][6(cos 120° + i sin 120°)]. Write theanswer in standard form.

13. Use DeMoivre's Theorem to find the three cube roots of 1.

14. From a point 200 feet from a flagpole, the angles of elevation to the bottom and top ofthe flag are 16° 45' and 18°, respectively. Approximate the height of the flag to thenearest foot.

15. An airplane's velocity with respect to the air is 500 kilometers per hour, with a bearingof N 30° E. The wind at the altitude of the plane has a velocity of 50 kilometers perhour with a bearing of N 60° E. What is the true direction of the plane, and what is itsspeed relative to the ground?

Figure for 5

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precalculus with limits a graphing approach · · 2012-04-06the jobs of a civil engineer. to...

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